Organic Chemistry
Organic Chemistry Structure and Function PETER VOLLHARDT University of California at Berkeley
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Organic Chemistry
Organic Chemistry Structure and Function PETER VOLLHARDT University of California at Berkeley
NEIL SCHORE University of California at Davis
W. H. Freeman and Company New York
SIXTH EDITION
Publisher: Clancy Marshall Acquisitions Editor: Jessica Fiorillo Developmental Editors: Randi Rossignol and David Chelton Media and Supplements Editor: Dave Quinn Director of Marketing: John Britch Editorial Assistant: Brittany Murphy Photo Editor: Bianca Moscatelli Photo Researcher: Christina Micek Cover Designer: Blake Logan Text Designer: Mike Suh Project Editor: Aptara®, Inc. Illustrations: Network Graphics, Precision Graphics Senior Illustration Coordinator: Bill Page Production Coordinator: Susan Wein Composition: Aptara®, Inc. Printing and Binding: World Color (USA) Corp Library of Congress Cataloging-in-Publication Data Vollhardt, K. Peter C. Organic chemistry : structure and function / K. Peter C. Vollhardt, Neil E. Schore.—6th ed. p. cm. ISBN-13: 978-1-4292-0494-1 ISBN-10: 1-4292-0494-X 1. Chemistry, Organic—Textbooks. I. Schore, Neil Eric, 1948– II. Title. QD251.3.V65 2009 547—dc22 2009040318 © 1999, 2003, 2007, and 2011 W. H. Freeman and Company All rights reserved Printed in the United States of America First printing W. H. Freeman and Company 41 Madison Avenue New York, NY 10010 Houndmills, Basingstoke RG21 6XS, England www.whfreeman.com
B R I E F
C O N T E N T S
PREFACE: A User’s Guide to Organic Chemistry: Structure and Function
xxiii
1
STRUCTURE AND BONDING IN ORGANIC MOLECULES
2
STRUCTURE AND REACTIVITY
49
3
REACTIONS OF ALKANES
95
4
CYCLOALKANES
131
5
STEREOISOMERS
169
6
PROPERTIES AND REACTIONS OF HALOALKANES
215
7
FURTHER REACTIONS OF HALOALKANES
251
8
HYDROXY FUNCTIONAL GROUP: Alcohols
287
9
FURTHER REACTIONS OF ALCOHOLS AND THE CHEMISTRY OF ETHERS
333
USING NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY TO DEDUCE STRUCTURE
387
ALKENES; INFRARED SPECTROSCOPY AND MASS SPECTROMETRY
445
10 11
INTERLUDE: Solving Problems in Organic Chemistry
1
499
12
REACTIONS OF ALKENES
507
13
ALKYNES
567
14
DELOCALIZED PI SYSTEMS
609
INTERLUDE: A Summary of Organic Reaction Mechanisms
15
BENZENE AND AROMATICITY
668 673 v
vi
Brief Contents
16
ELECTROPHILIC ATTACK ON DERIVATIVES OF BENZENE
731
17
ALDEHYDES AND KETONES
775
18
ENOLS, ENOLATES, AND THE ALDOL CONDENSATION
827
19
CARBOXYLIC ACIDS
871
20
CARBOXYLIC ACID DERIVATIVES
925
21
AMINES AND THEIR DERIVATIVES
971
22
CHEMISTRY OF BENZENE SUBSTITUENTS
1019
23
ESTER ENOLATES AND THE CLAISEN CONDENSATION
1081
24
CARBOHYDRATES: Polyfunctional Compounds in Nature
1117
HETEROCYCLES: Heteroatoms in Cyclic Organic Compounds
1165
AMINO ACIDS, PEPTIDES, PROTEINS, AND NUCLEIC ACIDS: Nitrogen-Containing Polymers in Nature
1211
25 26
MCAT® Questions Answers to Exercises Photograph Credits Index
MCAT-1 A-1 C-1 I-1
C O N T E N T S
PREFACE: A User’s Guide to Organic Chemistry: Structure and Function
1
STRUCTURE AND BONDING IN ORGANIC MOLECULES
1-1 The Scope of Organic Chemistry: An Overview Chemical Highlight 1-1 Urea: From Urine to Wöhler’s
Synthesis to Industrial Fertilizer 1-2 Coulomb Forces: A Simplified View of Bonding 1-3 Ionic and Covalent Bonds: The Octet Rule 1-4 Electron-Dot Model of Bonding: Lewis Structures 1-5 Resonance Forms 1-6 Atomic Orbitals: A Quantum Mechanical Description of Electrons Around the Nucleus 1-7 Molecular Orbitals and Covalent Bonding 1-8 Hybrid Orbitals: Bonding in Complex Molecules 1-9 Structures and Formulas of Organic Molecules Chapter Integration Problems Important Concepts Problems
2
STRUCTURE AND REACTIVITY
xxiii
1 2 4 5 7 13 18 23 28 31 37 40 43 44
49
2-1
Kinetics and Thermodynamics of Simple Chemical Processes 2-2 Acids and Bases; Electrophiles and Nucleophiles; Using Curved “Electron-Pushing” Arrows Chemical Highlight 2-1 Stomach Acid and Food Digestion 2-3 Functional Groups: Centers of Reactivity 2-4 Straight-Chain and Branched Alkanes 2-5 Naming the Alkanes 2-6 Structural and Physical Properties of Alkanes Chemical Highlight 2-2 “Sexual Swindle” by Means of Chemical Mimicry 2-7 Rotation about Single Bonds: Conformations
50 56 59 67 70 71 76 79 79 vii
viii
Contents
2-8
3
Rotation in Substituted Ethanes Chapter Integration Problems Important Concepts Problems
82 86 88 89
REACTIONS OF ALKANES
95
3-1 Strength of Alkane Bonds: Radicals 3-2 Structure of Alkyl Radicals: Hyperconjugation 3-3 Conversion of Petroleum: Pyrolysis Chemical Highlight 3-1 Sustainability and the Needs of the
96 99 100
21st Century: “Green” Chemistry 3-4 Chlorination of Methane: The Radical Chain Mechanism 3-5 Other Radical Halogenations of Methane 3-6 Chlorination of Higher Alkanes: Relative Reactivity and Selectivity 3-7 Selectivity in Radical Halogenation with Fluorine and Bromine 3-8 Synthetic Radical Halogenation Chemical Highlight 3-2 Chlorination, Chloral, and DDT 3-9 Synthetic Chlorine Compounds and the Stratospheric Ozone Layer 3-10 Combustion and the Relative Stabilities of Alkanes Chapter Integration Problems Important Concepts Problems
103 104 109
4
CYCLOALKANES
111 115 116 118 119 122 124 126 127
131
4-1 Names and Physical Properties of Cycloalkanes 4-2 Ring Strain and the Structure of Cycloalkanes 4-3 Cyclohexane: A Strain-Free Cycloalkane 4-4 Substituted Cyclohexanes 4-5 Larger Cycloalkanes 4-6 Polycyclic Alkanes Chemical Highlight 4-1 Cyclohexane, Adamantane, and Diamandoids:
132 135 140 144 149 149
Diamond “Molecules” 4-7 Carbocyclic Products in Nature Chemical Highlight 4-2 Cholesterol: How Is It Bad and How Bad Is It? Chemical Highlight 4-3 Controlling Fertility: From “the Pill” to RU-486 Chapter Integration Problems Important Concepts Problems
151 152 155 156 158 161 162
Contents
5
STEREOISOMERS
169
5-1 Chiral Molecules Chemical Highlight 5-1 Chiral Substances in Nature 5-2 Optical Activity 5-3 Absolute Configuration: R – S Sequence Rules 5-4 Fischer Projections Chemical Highlight 5-2 Absolute Configuration:
171 173 174 177 182
A Historical Note 5-5 Molecules Incorporating Several Stereocenters: Diastereomers Chemical Highlight 5-3 Stereoisomers of Tartaric Acid 5-6 Meso Compounds 5-7 Stereochemistry in Chemical Reactions Chemical Highlight 5-4 Chiral Drugs: Racemic or Enantiomerically Pure? Chemical Highlight 5-5 Why Is Nature “Handed”? 5-8 Resolution: Separation of Enantiomers Chapter Integration Problems Important Concepts Problems
183
6
PROPERTIES AND REACTIONS OF HALOALKANES
187 190 191 193 196 199 202 205 207 208
215
6-1 Physical Properties of Haloalkanes Chemical Highlight 6-1 Halogenated Steroids as Anti-
215
Inflammatory and Anti-Asthmatic Agents 6-2 Nucleophilic Substitution 6-3 Reaction Mechanisms Involving Polar Functional Groups: Using “Electron-Pushing” Arrows 6-4 A Closer Look at the Nucleophilic Substitution Mechanism: Kinetics 6-5 Frontside or Backside Attack? Stereochemistry of the SN2 Reaction 6-6 Consequences of Inversion in SN2 Reactions 6-7 Structure and SN2 Reactivity: The Leaving Group 6-8 Structure and SN2 Reactivity: The Nucleophile 6-9 Structure and SN2 Reactivity: The Substrate Chemical Highlight 6-2 The Dilemma of Bromomethane: Highly Useful but Also Highly Toxic Chapter Integration Problems Important Concepts Problems
217 218 221 223 226 228 231 233 240 241 244 245 246
ix
x
Contents
7
FURTHER REACTIONS OF HALOALKANES
7-1 7-2 7-3 7-4
Solvolysis of Tertiary and Secondary Haloalkanes Unimolecular Nucleophilic Substitution Stereochemical Consequences of SN1 Reactions Effects of Solvent, Leaving Group, and Nucleophile on Unimolecular Substitution 7-5 Effect of the Alkyl Group on the SN1 Reaction: Carbocation Stability Chemical Highlight 7-1 Unusually Stereoselective SN1 Displacement in Anticancer Drug Synthesis 7-6 Unimolecular Elimination: E1 7-7 Bimolecular Elimination: E2 7-8 Competition Between Substitution and Elimination: Structure Determines Function 7-9 Summary of Reactivity of Haloalkanes Chapter Integration Problems New Reactions Important Concepts Problems
8 8-1 8-2 8-3 8-4
HYDROXY FUNCTIONAL GROUP: Alcohols
Naming the Alcohols Structural and Physical Properties of Alcohols Alcohols as Acids and Bases Industrial Sources of Alcohols: Carbon Monoxide and Ethene 8-5 Synthesis of Alcohols by Nucleophilic Substitution 8-6 Synthesis of Alcohols: Oxidation - Reduction Relation Between Alcohols and Carbonyl Compounds Chemical Highlight 8-1 Biological Oxidation and Reduction Chemical Highlight 8-2 The Breath Analyzer Test 8-7 Organometallic Reagents: Sources of Nucleophilic Carbon for Alcohol Synthesis 8-8 Organometallic Reagents in the Synthesis of Alcohols 8-9 Complex Alcohols: An Introduction to Synthetic Strategy Chemical Highlight 8-3 Transition Metal-Catalyzed Cross-Coupling Reactions Chapter Integration Problems New Reactions Important Concepts Problems
251 251 252 256 258 260 263 264 267 270 273 275 277 277 278
287 288 289 292 295 295 297 298 302 304 307 309 310 319 322 326 326
Contents
9
FURTHER REACTIONS OF ALCOHOLS AND THE CHEMISTRY OF ETHERS
9-1 9-2
Reactions of Alcohols with Base: Preparation of Alkoxides Reactions of Alcohols with Strong Acids: Alkyloxonium Ions in Substitution and Elimination Reactions of Alcohols 9-3 Carbocation Rearrangements 9-4 Esters from Alcohols and Haloalkane Synthesis 9-5 Names and Physical Properties of Ethers 9-6 Williamson Ether Synthesis Chemical Highlight 9-1 Chemiluminescence of 1,2-Dioxacyclobutanes 9-7 Synthesis of Ethers: Alcohols and Mineral Acids 9-8 Reactions of Ethers Chemical Highlight 9-2 Protecting Groups in Synthesis 9-9 Reactions of Oxacyclopropanes Chemical Highlight 9-3 Hydrolytic Kinetic Resolution of Oxacyclopropanes 9-10 Sulfur Analogs of Alcohols and Ethers 9-11 Physiological Properties and Uses of Alcohols and Ethers Chemical Highlight 9-4 Garlic and Sulfur Chapter Integration Problems New Reactions Important Concepts Problems
10
USING NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY TO DEDUCE STRUCTURE
10-1 Physical and Chemical Tests 10-2 Defining Spectroscopy 10-3 Hydrogen Nuclear Magnetic Resonance Chemical Highlight 10-1 Recording an NMR Spectrum 10-4 Using NMR Spectra to Analyze Molecular Structure:
The Proton Chemical Shift 10-5 Tests for Chemical Equivalence Chemical Highlight 10-2 Magnetic Resonance Imaging
in Medicine 10-6 Integration 10-7 Spin – Spin Splitting: The Effect of Nonequivalent Neighboring Hydrogens 10-8 Spin – Spin Splitting: Some Complications Chemical Highlight 10-3 The Nonequivalence of Diastereotopic Hydrogens
333 334 335 338 344 347 350 351 355 357 359 360 362 365 367 371 372 374 376 377
387 388 388 390 393 395 400 404 405 407 414 418
xi
xii
Contents
10-9 Carbon-13 Nuclear Magnetic Resonance Chemical Highlight 10-4 Correlated NMR Spectra: COSY and
422
HETCOR
428
Chemical Highlight 10-5 Structural Characterization of Natural Products:
Antioxidants from Grape Seeds Chapter Integration Problems Important Concepts Problems
11
ALKENES; INFRARED SPECTROSCOPY AND MASS SPECTROMETRY
430 432 435 436
445
11-1 Naming the Alkenes 11-2 Structure and Bonding in Ethene: The Pi Bond 11-3 Physical Properties of Alkenes 11-4 Nuclear Magnetic Resonance of Alkenes Chemical Highlight 11-1 Prostaglandins 11-5 Catalytic Hydrogenation of Alkenes: Relative Stability of
446 449 452 453 459
Double Bonds 11-6 Preparation of Alkenes from Haloalkanes and Alkyl Sulfonates: Bimolecular Elimination Revisited 11-7 Preparation of Alkenes by Dehydration of Alcohols 11-8 Infrared Spectroscopy 11-9 Measuring the Molecular Mass of Organic Compounds: Mass Spectrometry Chemical Highlight 11-2 Security in the 21st Century: Applications of IR and MS Chemical Highlight 11-3 Detecting Performance-Enhancing Drugs Using Mass Spectrometry 11-10 Fragmentation Patterns of Organic Molecules 11-11 Degree of Unsaturation: Another Aid to Identifying Molecular Structure Chapter Integration Problems New Reactions Important Concepts Problems INTERLUDE: Solving Problems in Organic Chemistry
459
12 12-1 12-2 12-3
REACTIONS OF ALKENES Why Addition Reactions Proceed: Thermodynamic Feasibility Catalytic Hydrogenation Nucleophilic Character of the Pi Bond: Electrophilic Addition of Hydrogen Halides
462 466 468 473 475 476 478 482 484 486 487 488 499
507 507 509 512
Contents
12-4
Alcohol Synthesis by Electrophilic Hydration: Thermodynamic Control 12-5 Electrophilic Addition of Halogens to Alkenes 12-6 The Generality of Electrophilic Addition 12-7 Oxymercuration – Demercuration: A Special Electrophilic Addition Chemical Highlight 12-1 Juvenile Hormone Analogs in the Battle against Insect-Borne Diseases 12-8 Hydroboration – Oxidation: A Stereospecific Anti-Markovnikov Hydration 12-9 Diazomethane, Carbenes, and Cyclopropane Synthesis 12-10 Oxacyclopropane (Epoxide) Synthesis: Epoxidation by Peroxycarboxylic Acids 12-11 Vicinal Syn Dihydroxylation with Osmium Tetroxide Chemical Highlight 12-2 Synthesis of Antitumor Drugs: Sharpless Enantioselective Oxacyclopropanation and Dihydroxylation 12-12 Oxidative Cleavage: Ozonolysis 12-13 Radical Additions: Anti-Markovnikov Product Formation 12-14 Dimerization, Oligomerization, and Polymerization of Alkenes 12-15 Synthesis of Polymers Chemical Highlight 12-3 Polymers in the Clean-up of Oil Spills 12-16 Ethene: An Important Industrial Feedstock 12-17 Alkenes in Nature: Insect Pheromones Chemical Highlight 12-4 Metal-Catalyzed Alkene Metathesis for Constructing Medium and Large Rings Chapter Integration Problems New Reactions Important Concepts Problems
13 13-1 13-2 13-3 13-4 13-5 13-6
ALKYNES
Naming the Alkynes Properties and Bonding in the Alkynes Spectroscopy of the Alkynes Preparation of Alkynes by Double Elimination Preparation of Alkynes from Alkynyl Anions Reduction of Alkynes: The Relative Reactivity of the Two Pi Bonds 13-7 Electrophilic Addition Reactions of Alkynes 13-8 Anti-Markovnikov Additions to Triple Bonds 13-9 Chemistry of Alkenyl Halides Chemical Highlight 13-1 Metal-Catalyzed Stille, Suzuki, and Sonogashira Coupling Reactions
516 518 521 525 526 528 531 532 535 536 538 540 542 543 545 547 548 548 550 553 555 556
567 568 568 571 576 577 579 582 585 587 588
xiii
xiv
Contents
13-10 Ethyne as an Industrial Starting Material 13-11 Naturally Occurring and Physiologically Active Alkynes
14
Chapter Integration Problems New Reactions Important Concepts Problems
590 592 594 595 599 599
DELOCALIZED PI SYSTEMS
609
14-1
Overlap of Three Adjacent p Orbitals: Electron Delocalization in the 2-Propenyl (Allyl) System 14-2 Radical Allylic Halogenation 14-3 Nucleophilic Substitution of Allylic Halides: SN1 and SN2 14-4 Allylic Organometallic Reagents: Useful Three-Carbon Nucleophiles 14-5 Two Neighboring Double Bonds: Conjugated Dienes 14-6 Electrophilic Attack on Conjugated Dienes: Kinetic and Thermodynamic Control Chemical Highlight 14-1 “Face-to-Face” Interaction of Two Double Bonds 14-7 Delocalization Among More than Two Pi Bonds: Extended Conjugation and Benzene 14-8 A Special Transformation of Conjugated Dienes: Diels-Alder Cycloaddition Chemical Highlight 14-2 Conducting Polyenes: Novel Functional Materials Chemical Highlight 14-3 The Diels-Alder Reaction is “Green” 14-9 Electrocyclic Reactions Chemical Highlight 14-4 An Electrocyclization Cascade in Nature: Immunosuppressants from Streptomyces Cultures 14-10 Polymerization of Conjugated Dienes: Rubber 14-11 Electronic Spectra: Ultraviolet and Visible Spectroscopy Chemical Highlight 14-5 The Contributions of IR, MS, and UV to the Characterization of Viniferone Chapter Integration Problems New Reactions Important Concepts Problems INTERLUDE: A Summary of Organic Reaction Mechanisms
15 15-1 15-2
610 612 614 616 617 621 622 626 628 630 637 640 644 647 650 655 656 659 660 661 668
BENZENE AND AROMATICITY
673
Naming the Benzenes Structure and Resonance Energy of Benzene: A First Look at Aromaticity
674 677
Contents
15-3 Pi Molecular Orbitals of Benzene 15-4 Spectral Characteristics of the Benzene Ring 15-5 Polycyclic Aromatic Hydrocarbons Chemical Highlight 15-1 The Allotropes of Carbon: Graphite,
679 682 687
Diamond, and Fullerenes 15-6 Other Cyclic Polyenes: Hückel’s Rule Chemical Highlight 15-2 Juxtaposing Aromatic and Antiaromatic Rings in Fused Hydrocarbons 15-7 Hückel’s Rule and Charged Molecules 15-8 Synthesis of Benzene Derivatives: Electrophilic Aromatic Substitution 15-9 Halogenation of Benzene: The Need for a Catalyst 15-10 Nitration and Sulfonation of Benzene 15-11 Friedel-Crafts Alkylation 15-12 Limitations of Friedel-Crafts Alkylations 15-13 Friedel-Crafts Acylation (Alkanoylation) Chapter Integration Problems New Reactions Important Concepts Problems
688 693
16
ELECTROPHILIC ATTACK ON DERIVATIVES OF BENZENE
694 699 701 704 705 708 712 714 717 720 722 722
731
16-1
Activation or Deactivation by Substituents on a Benzene Ring 16-2 Directing Inductive Effects of Alkyl Groups 16-3 Directing Effects of Substituents in Conjugation with the Benzene Ring Chemical Highlight 16-1 Explosive Nitroarenes: TNT and Picric Acid 16-4 Electrophilic Attack on Disubstituted Benzenes 16-5 Synthetic Strategies Toward Substituted Benzenes 16-6 Reactivity of Polycyclic Benzenoid Hydrocarbons 16-7 Polycyclic Aromatic Hydrocarbons and Cancer Chapter Integration Problems New Reactions Important Concepts Problems
17 17-1 17-2 17-3 17-4
732 734 738 741 745 749 754 758 760 764 765 766
ALDEHYDES AND KETONES
775
Naming the Aldehydes and Ketones Structure of the Carbonyl Group Spectroscopic Properties of Aldehydes and Ketones Preparation of Aldehydes and Ketones
776 778 779 785
xv
xvi
Contents
17-5 Reactivity of the Carbonyl Group: Mechanisms of Addition 17-6 Addition of Water to Form Hydrates 17-7 Addition of Alcohols to Form Hemiacetals and Acetals 17-8 Acetals as Protecting Groups 17-9 Nucleophilic Addition of Ammonia and Its Derivatives Chemical Highlight 17-1 Imines in Biology 17-10 Deoxygenation of the Carbonyl Group 17-11 Addition of Hydrogen Cyanide to Give Cyanohydrins 17-12 Addition of Phosphorus Ylides: The Wittig Reaction 17-13 Oxidation by Peroxycarboxylic Acids: The Baeyer-
787 789 791 793 797 799 802 804 804
Villiger Oxidation 17-14 Oxidative Chemical Tests for Aldehydes Chapter Integration Problems New Reactions Important Concepts Problems
808 809 810 812 815 815
18 18-1 18-2 18-3 18-4 18-5
ENOLS, ENOLATES, AND THE ALDOL CONDENSATION
Acidity of Aldehydes and Ketones: Enolate Ions Keto – Enol Equilibria Halogenation of Aldehydes and Ketones Alkylation of Aldehydes and Ketones Attack by Enolates on the Carbonyl Function: Aldol Condensation 18-6 Crossed Aldol Condensation Chemical Highlight 18-1 Enzyme-Catalyzed Stereoselective Aldol Condensations in Nature Chemical Highlight 18-2 Enzymes in Synthesis: Stereoselective Crossed Aldol Condensations 18-7 Intramolecular Aldol Condensation Chemical Highlight 18-3 Reactions of Unsaturated Aldehydes in Nature: The Chemistry of Vision 18-8 Properties of a,b-Unsaturated Aldehydes and Ketones 18-9 Conjugate Additions to a,b-Unsaturated Aldehydes and Ketones 18-10 1,2- and 1,4-Additions of Organometallic Reagents 18-11 Conjugate Additions of Enolate Ions: Michael Addition and Robinson Annulation Chemical Highlight 18-4 Alexander Borodin: Composer, Chemist, and Pioneering Teacher Chapter Integration Problems New Reactions Important Concepts Problems
827 828 829 832 834 837 840 841 842 843 844 846 848 850 852 853 856 858 861 861
Contents
19
CARBOXYLIC ACIDS
19-1 19-2 19-3
Naming the Carboxylic Acids Structural and Physical Properties of Carboxylic Acids Spectroscopy and Mass Spectrometry of Carboxylic Acids 19-4 Acidic and Basic Character of Carboxylic Acids 19-5 Carboxylic Acid Synthesis in Industry 19-6 Methods for Introducing the Carboxy Functional Group 19-7 Substitution at the Carboxy Carbon: The Addition – Elimination Mechanism 19-8 Carboxylic Acid Derivatives: Acyl Halides and Anhydrides 19-9 Carboxylic Acid Derivatives: Esters 19-10 Carboxylic Acid Derivatives: Amides 19-11 Reduction of Carboxylic Acids by Lithium Aluminum Hydride 19-12 Bromination Next to the Carboxy Group: The HellVolhard-Zelinsky Reaction 19-13 Biological Activity of Carboxylic Acids Chemical Highlight 19-1 Soaps from Long-Chain Carboxylates Chemical Highlight 19-2 Trans Fatty Acids and Your Health Chemical Highlight 19-3 Plastics, Fibers, and Energy from Biomass-Derived Hydroxyesters Chapter Integration Problems New Reactions Important Concepts Problems
20
CARBOXYLIC ACID DERIVATIVES
871 872 874 875 879 882 883 886 889 892 896 897 898 899 900 903 905 907 910 913 913
925
20-1
Relative Reactivities, Structures, and Spectra of Carboxylic Acid Derivatives 20-2 Chemistry of Acyl Halides 20-3 Chemistry of Carboxylic Anhydrides 20-4 Chemistry of Esters 20-5 Esters in Nature: Waxes, Fats, Oils, and Lipids Chemical Highlight 20-1 Greener Alternatives to Petroleum: Fuels from Vegetable Oil 20-6 Amides: The Least Reactive Carboxylic Acid Derivatives Chemical Highlight 20-2 Battling the Bugs: Antibiotic Wars 20-7 Amidates and Their Halogenation: The Hofmann Rearrangement Chemical Highlight 20-3 Methyl Isocyanate, Carbamate-Based Insecticides, and Safety in the Chemical Industry
926 930 934 936 942 944 944 946 950 953
xvii
xviii
Contents
20-8
21
Alkanenitriles: A Special Class of Carboxylic Acid Derivatives Chapter Integration Problems New Reactions Important Concepts Problems
953 957 960 963 964
AMINES AND THEIR DERIVATIVES
971
21-1 Naming the Amines 21-2 Structural and Physical Properties of Amines Chemical Highlight 21-1 Physiologically Active Amines and
972 973
Weight Control
974 977 981
21-3 Spectroscopy of the Amine Group 21-4 Acidity and Basicity of Amines Chemical Highlight 21-2 Separation of Amines from Other
Organic Compounds by Aqueous Extraction Techniques
984
21-5 Synthesis of Amines by Alkylation 986 21-6 Synthesis of Amines by Reductive Amination 989 21-7 Synthesis of Amines from Carboxylic Amides 992 21-8 Reactions of Quaternary Ammonium Salts: Hofmann Elimination 992 21-9 Mannich Reaction: Alkylation of Enols by Iminium Ions 994 21-10 Nitrosation of Amines 996 Chemical Highlight 21-3 N-Nitrosodialkanamines and Cancer 998 Chemical Highlight 21-4 Amines in Industry: Nylon 1000
Chapter Integration Problems New Reactions Important Concepts Problems
22
CHEMISTRY OF BENZENE SUBSTITUENTS
1003 1006 1011 1011
1019
22-1
Reactivity at the Phenylmethyl (Benzyl) Carbon: Benzylic Resonance Stabilization 22-2 Benzylic Oxidations and Reductions 22-3 Names and Properties of Phenols Chemical Highlight 22-1 Two Phenols in the News: Bisphenol A and Resveratrol 22-4 Preparation of Phenols: Nucleophilic Aromatic Substitution 22-5 Alcohol Chemistry of Phenols Chemical Highlight 22-2 Aspirin: A Phenyl Alkanoate Drug 22-6 Electrophilic Substitution of Phenols 22-7 An Electrocyclic Reaction of the Benzene Ring: The Claisen Rearrangement
1020 1024 1026 1030 1030 1041 1043 1044 1048
Contents
22-8 Oxidation of Phenols: Benzoquinones Chemical Highlight 22-3 Chemical Warfare in Nature:
1051
The Bombardier Beetle 22-9 Oxidation-Reduction Processes in Nature 22-10 Arenediazonium Salts 22-11 Electrophilic Substitution with Arenediazonium Salts: Diazo Coupling Chemical Highlight 22-4 William Perkin and the Origins of Industrial and Medicinal Chemistry Chapter Integration Problems New Reactions Important Concepts Problems
1053 1053 1058
23
ESTER ENOLATES AND THE CLAISEN CONDENSATION
1061 1062 1064 1066 1071 1071
1081
23-1 b-Dicarbonyl Compounds: Claisen Condensations Chemical Highlight 23-1 Claisen Condensations in
1082
Biochemistry 23-2 b-Dicarbonyl Compounds as Synthetic Intermediates 23-3 b-Dicarbonyl Anion Chemistry: Michael Additions 23-4 Acyl Anion Equivalents: Preparation of a-Hydroxyketones Chemical Highlight 23-2 Thiamine: A Natural, Metabolically Active Thiazolium Salt Chapter Integration Problems New Reactions Important Concepts Problems
1086 1090 1095
24 24-1 24-2 24-3 24-4
CARBOHYDRATES: Polyfunctional Compounds in Nature
Names and Structures of Carbohydrates Conformations and Cyclic Forms of Sugars Anomers of Simple Sugars: Mutarotation of Glucose Polyfunctional Chemistry of Sugars: Oxidation to Carboxylic Acids 24-5 Oxidative Cleavage of Sugars 24-6 Reduction of Monosaccharides to Alditols 24-7 Carbonyl Condensations with Amine Derivatives 24-8 Ester and Ether Formation: Glycosides Chemical Highlight 24-1 Protecting Groups in Vitamin C Synthesis 24-9 Step-by-Step Buildup and Degradation of Sugars Chemical Highlight 24-2 Sugar Biochemistry
1098 1100 1104 1107 1109 1109
1117 1117 1122 1127 1128 1130 1131 1132 1133 1135 1136 1138
xix
xx
Contents
24-10 Relative Configurations of the Aldoses: An Exercise in
Structure Determination 24-11 Complex Sugars in Nature: Disaccharides Chemical Highlight 24-3 Carbohydrate-Derived Sugar
Substitutes 24-12 Polysaccharides and Other Sugars in Nature Chemical Highlight 24-4 Sialic Acid, “Bird Flu”, and Rational Drug Design Chapter Integration Problem New Reactions Important Concepts Problems
25
HETEROCYCLES: Heteroatoms in Cyclic Organic Compounds
1139 1142 1144 1146 1152 1153 1156 1158 1159
1165
25-1 Naming the Heterocycles 25-2 Nonaromatic Heterocycles Chemical Highlight 25-1 Smoking, Nicotine, Cancer, and
1167 1168
Medicinal Chemistry 25-3 Structure and Properties of Aromatic Heterocyclopentadienes 25-4 Reactions of the Aromatic Heterocyclopentadienes 25-5 Structure and Preparation of Pyridine: An Azabenzene 25-6 Reactions of Pyridine Chemical Highlight 25-2 Pyridinium Salts in Nature: Nicotinamide Adenine Dinucleotide, Dihydropyridines, and Organocatalysis 25-7 Quinoline and Isoquinoline: The Benzopyridines Chemical Highlight 25-3 Folic Acid, Vitamin D, Cholesterol, and the Color of Your Skin 25-8 Alkaloids: Physiologically Potent Nitrogen Heterocycles in Nature Chemical Highlight 25-4 Nature Is Not Always Green: Natural Pesticides Chapter Integration Problems New Reactions Important Concepts Problems
1170
26
AMINO ACIDS, PEPTIDES, PROTEINS, AND NUCLEIC ACIDS: Nitrogen-Containing Polymers in Nature
1172 1175 1179 1184
1186 1188 1188 1191 1192 1195 1198 1200 1200
1211
26-1 Structure and Properties of Amino Acids Chemical Highlight 26-1 Arginine and Nitric Oxide in
1212
Biochemistry and Medicine
1216
Contents
26-2
Synthesis of Amino Acids: A Combination of Amine and Carboxylic Acid Chemistry 26-3 Synthesis of Enantiomerically Pure Amino Acids Chemical Highlight 26-2 Enantioselective Synthesis of Optically Pure Amino Acids: Phase-Transfer Catalysis 26-4 Peptides and Proteins: Amino Acid Oligomers and Polymers 26-5 Determination of Primary Structure: Amino Acid Sequencing 26-6 Synthesis of Polypeptides: A Challenge in the Application of Protecting Groups 26-7 Merrifield Solid-Phase Peptide Synthesis 26-8 Polypeptides in Nature: Oxygen Transport by the Proteins Myoglobin and Hemoglobin 26-9 Biosynthesis of Proteins: Nucleic Acids Chemical Highlight 26-3 Synthetic Nucleic Acid Bases and Nucleosides in Medicine 26-10 Protein Synthesis Through RNA 26-11 DNA Sequencing and Synthesis: Cornerstones of Gene Technology Chemical Highlight 26-4 DNA Fingerprinting Chapter Integration Problems New Reactions Important Concepts Problems MCAT® Questions Answers to Exercises Photograph Credits Index
1217 1220 1222 1222 1230 1234 1238 1239 1241 1243 1246 1248 1256 1260 1262 1264 1264 MCAT-1 A-1 C-1 I-1
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P R E F A C E
A User’s Guide to ORGANIC CHEMISTRY: Structure and Function
I
n this edition of Organic Chemistry: Structure and Function, we maintain our goal of helping students organize all the information presented in the course and fit it into a logical framework for understanding contemporary organic chemistry. This framework emphasizes that the structure of an organic molecule determines how that molecule functions in a chemical reaction. In the sixth edition, we have strengthened the themes of understanding reactivity, mechanisms, and synthetic analysis to apply chemical concepts to realistic situations. We have incorporated new applications of organic chemistry in the life sciences, industrial practices, green chemistry, and environmental monitoring and clean-up. This edition includes more than 100 new or substantially revised problems, including new problems on synthesis and green chemistry, and new “challenging” problems. Organic Chemistry: Structure and Function is offered in an online version to give students cost-effective access to all content from the text plus all student media resources. For more information, please visit our Web site at http://ebooks.bfwpub.com.
CONNECTING STRUCTURE AND FUNCTION This textbook emphasizes that the structure of an organic molecule determines how that molecule functions in a chemical reaction. By understanding the connection between structure and function, we can learn to solve practical problems in organic chemistry. Chapters 1 through 5 lay the foundation for making this connection. In particular, Chapter 1 shows how electronegativity is the basis for polar bond formation, setting the stage for an understanding of polar reactivity. Chapter 2 makes an early connection between acidity and electrophilicity, as well as their respective counterparts, basicitynucleophilicity. Chapter 3 relates the structure of radicals to their relative stability and reactivity. Chapter 4 illustrates how ring size affects the properties of cyclic systems, and Chapter 5 provides an early introduction to stereochemistry. The structures of haloalkanes and how they determine haloalkane behavior in nucleophilic substitution and elimination reactions are the main topics of Chapters 6 and 7. Subsequent chapters present material on functional-group compounds according to the same scheme introduced for haloalkanes: nomenclature, structure, spectroscopy, preparations, reactions, and biological and other applications. The emphasis on structure and function allows us to discuss the mechanisms of all new important reactions concurrently, rather than scattered throughout the text. We believe this unified presentation of mechanisms benefits students by teaching them how to approach understanding reactions rather than memorizing them.
N O O
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O
most of the energy ized society. We s lack functional does combustion that alkanes are undergo several cesses, of which nvolve acid-base adical reactions. dical reactions in significant roles isease processes), he Earth’s ozone synthetic fabrics e breaking of a bond, or bond dissociation. We examine iscuss the conditions under which it occurs. The majority enation, a radical reaction in which a hydrogen atom in The importance of halogenation lies in the fact that it up, turning the alkane into a haloalkane, which is suitable ach of these processes, we shall discuss the mechanism the reaction occurs. We shall see that different alkanes, same alkane molecule, may react at different rates, and C A carbon radical
er of mechanisms are needed to describe the very large mistry. Mechanisms enable us to understand how and why are likely to form in them. In this chapter we apply mechcts of halogen-containing chemicals on the stratospheric rief discussion of alkane combustion and show how that thermodynamic information about organic molecules.
NASA’s X-43A hypersonic research aircraft being dropped from under the wing of a B-52B Stratofortress on June 2, 2001. Most supersonic aircraft produce exhaust gases containing molecules such as nitric oxide (NO), whose radical reactions are destructive to the Earth’s stratospheric ozone (O3) layer. In the 1970s the United States abandoned plans to build a fleet of supersonic aircraft (SSTs, or supersonic transports) for just this reason. In contrast, the X-43A is hydrogen fueled, posing no risk to stratospheric ozone, and may represent the first step toward the development of environmentally acceptable high-speed flight. In 2008, Boeing flew successfully the first manned hydrogen-fuelcell-powered aircraft, another aviation milestone.
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UNDERSTANDING AND VISUALIZING REACTION MECHANISMS The emphasis on structure and function in the early chapters primes the students for building a true understanding of reaction mechanisms, encouraging understanding over memorization. Because visualizing chemical reactivity can be challenging for many students, we use many different visual cues and models to help students visualize reactions and how they proceed mechanistically. • NEW. Improved and expanded coverage of electronpushing arrows in Section 2-2. The use of electronpushing arrows, introduced in Section 2-2, is reinforced in Section 6-3 and applied extensively in all subsequent chapters.
Curved arrows show how starting materials convert to products Bonds consist of electrons. Chemical change is defined as a process in which bonds are broken and/or formed. Therefore, when chemistry takes place, electrons move. Following the basic principles of electrostatics, electrons, being negatively charged, are attracted to sites of electron deficiency, or positive charge. Either highly electronegative or electron-deficient (and, therefore, electron-attracting) atoms, positively charged ions, or the d1 atom in a polar covalent bond can be the destination for electron movement. The vast majority of chemical processes we will introduce in this text will involve the movement of one or more pairs of electrons. A curved arrow ([) will show the flow of an electron pair from its point of origin, either a lone pair or a covalent bond, to its destination. The electron-pair movement that interconverts resonance forms (Section 1-5) follows these same principles. However, we know that resonance forms do not represent distinct entities. When we use curved arrows to depict the electron movement associated with a chemical reaction, we are describing an actual structural change, from the Lewis structures of the starting materials to those of the products. The examples below illustrate the various ways in which curved arrows are employed in this movement. In each case a red color denotes an electron pair that moves.
• Interlude: A Summary of Organic Reaction Mechanisms, following Chapter 1. Dissociation of a polar covalent bond into ions (B more electronegative than A) 14, summarizes the relaA⫹ ⫹ ðB⫺ AO B tively few types of reaction Movement of an electron pair converts the A – B covalent bond into a lone pair on atom B mechanisms that drive the majority of organic reactions, thereby encouraging understanding over memorization.
• Computer-generated ball-and-stick and space-filling models help students visualize steric factors in many kinds of reactions. Icons in the page margins indicate where model building by students will be especially helpful for visualizing three-dimensional structures and dynamics. • Electrostatic potential maps of many species help students see how electron distributions affect the behavior of species in various interactions.
MECHANISM
• Mechanism icons in the page margins highlight the locations of important mechanisms.
• Model-building icons in the page margins indicate all of the structures MODEL BUILDING that are found on the book’s Web site as 3-D and rotatable structures. • Reaction Summary Road Maps, found at the ends of some chapters, summarize the principal reactions for preparation and applications of each major functional group. The Preparation maps indicate the possible origins of a functionality—that is, the precursor functional groups. The Reaction maps show what each functional group does. In both maps, reaction arrows are labeled with particular reagents and start from or end at specific reactants or products. The reaction arrows are also labeled with new section numbers indicating where the transformation is discussed in the new edition. • Approximately 50 animated mechanisms are offered on the W. H. Freeman Web site. All mechanisms are indicated by Media Link icons in the page margins.
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STRONGER PEDAGOGY FOR SOLVING PROBLEMS
Exercise 7-4
Improved Problem-Solving Approaches
Working with the Concepts: Stereochemical Consequences of SN1 Displacement
• NEW. In this edition we have revised all of the in-chapter exercises, called Working with the Concepts. Each exercise now begins with a Strategy section that emphasizes the reasoning students need to apply in attacking problems. The Solution arranges the steps logically and carefully, modeling good problem-solving skills.
Gentle warming of (2R,4R)-2-iodo-4-methylhexane in methanol gives two stereoisomeric methyl ethers. How are they related to each other? Explain mechanistically. Strategy The substrate is secondary; therefore, substitution can proceed by either the SN1 or the SN2 mechanism. Let’s consider the reaction conditions to see which is more likely and what its consequences will be. Solution • The reaction takes place in methanol, CH3OH, a poor nucleophile (disfavoring SN2) but a very polar, protic solvent, well suited for dissociation of secondary and tertiary haloalkanes into ions (favoring SN1). • Dissociation of the excellent leaving group I2 from C2 gives a trigonal planar carbocation. Methanol may attack from either face (compare mechanism steps 1 and 2 in Section 7-2), giving two stereoisomeric oxonium ions. The positively charged oxygen makes the attached hydrogen very acidic; after proton loss, two stereoisomeric ethers result (mechanism step 3 in Section 7-2; see also Figure 7-3). We have here another example of solvolysis (specifically, methanolysis), because the nucleophile is the solvent (methanol). b
ðš Ið H $C R H3C R ( @ H 3C H
H
$C
⫹
H3C ⫺ðš Ið⫺ š
a
š HOCH 3 R
a (@ CH3š OH H3C H b
H
H 3C
H
H Ok ⫹ CH3 $C R S
(@ H3C H ⫺H⫹
š 3 H ðOCH $C R
H 3C S ( @ H3C H
• NEW. Try It Yourself Exercises. Each in-chapter worked exercise is paired with a Try It Yourself problem that follows up on the concept being taught.
⫹
ýO H $C R H 3C R ( @ H 3C H
H 3C
⫺H⫹
šð H H3CO $C R H 3C R ( @ H 3C H
• NEW. Caution statements appear in many of the exercises, alerting students to potential pitfalls and how to avoid them. The exercises chosen for solution are typical homework or test questions, enabling students to acquire a feel for solving complex problems, rather than artificially simplified ones. In response to strong positive feedback on this feature, we have increased substantially the number of these solutions.
[
(Caution! When describing the SN1 mechanism, avoid the following two very common errors: (1) Do not dissociate CH3OH to give methoxide (CH3O2) and a proton before bonding with the cationic carbon. Methanol is a weak acid whose dissociation is thermodynamically not favored. (2) Do not dissociate CH3OH to give a methyl cation and hydroxide ion. Although the presence of the OH functional group in alcohols may remind you of the formulas of inorganic hydroxides, alcohols are not sources of hydroxide ion.) • The two stereoisomeric ether products are diastereomers; 2S,3R and 2R,3R. At the reaction site, C2, both R and S configurations result from the two possible pathways of methanol attack, a and b. At C3 a stereocenter where no reaction occurs, the original R configuration remains unchanged.
Exercise 7-5
Try It Yourself Hydrolysis of molecule A (shown in the margin) gives two alcohols. Explain.
• NEW. Interlude: Solving Problems in Organic Chemistry, following Chapter 11, examines the different types of problems common in organic chemistry and teaches students how to approach and proceed through each type of problem.
]
Interlude
Solving Problems in Organic Chemistry
With approximately one-third of the year’s course in organic chemistry behind you, let’s take stock of your ability to solve problems, and look to possible remedies for the difficulties you may have encountered. This interlude has the following structure: Understanding the Question Types of Problems in Organic Chemistry A General Approach to Problem Solving: The “WHIP” Strategy Solving Problems That Ask “What” Nomenclature Acidity Energy Stability Spectroscopy Solving Problems That Ask “How” and/or “Why” What is the product of a reaction? How does the product form? What reagent(s) do you need to convert one molecule into a specific other?
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Problem I-7. Which reaction below proceeds faster, (a) or (b)? What is the product? Explain. Cl
%
(a)
Cl Excess NaOH, CH3CH2OH
´
(b)
Excess NaOH, CH3CH2OH
What does the question ask? “Explain!” There’s your cue: This is a mechanism problem. How to start? As in the preceding example, characterize the substrates: Here they are secondary chloroalkanes, with additional steric hindrance on the adjacent carbon. Again, the reagent is a strong nucleophile and base. Information: Displacement by SN2 is unlikely; as in Problem I-6, E2 should be favored (Table 7-4). What next? Proceed, logically: Use the question as a clue to how to proceed. Why should these two reactions proceed at different rates? How do the substrates differ? Answer: stereochemically. So, redraw both starting materials in the more instructive cyclohexane chair forms: H3C (a)
Cl
CH3 H H
• NEW. “WHIP” strategy. Beginning in Chapter 3 we introduce a novel and powerful approach to problem solving, the “WHIP” approach. We teach students how to recognize the fundamental types of questions they are likely to encounter, and explain the solution strategy in full detail in a new Interlude section that follows Chapter 11. The “WHIP” strategy encourages students to ask the following questions: WHAT does the problem ask? HOW to begin? INFORMATION needed? PROCEED logically; do not skip steps! Solutions of several problem types illustrate the strategy.
H 3C Axial Cl is trans and anti to hydrogen atoms on the adjacent carbons
Cl CH3
(b)
Equatorial Cl is cis a gauche to hydrogen on the adjacent carb
H H
• Many more exercises. We have increased the number of end-of-chapter problems to give students more practice solving problems.
A Wide Variety of Problem Types Users and reviewers of past editions have often cited the end-of-chapter problems as a major strength of the book, both for the range of difficulty levels and the variety of practical applications. We highlight those end-of-chapter problems that are more difficult with a special icon. • The Chapter Integration Problems include worked-out, step-by-step solutions to problems involving several concepts from within chapters and from among several chapters. These solutions place particular emphasis on problem analysis, deductive reasoning, and logical conclusions. • Team Problems encourage discussion and collaborative learning among students. They can be assigned as regular homework or as projects for groups of students to work on. • For students planning careers in medicine or related fields, Preprofessional Problems offer a multiple-choice format typical of problems on the MCAT®, GRE, and DAT. In addition, a selection of actual test passages and questions from past MCAT® exams appears in an appendix.
REAL CHEMISTRY BY PRACTICING CHEMISTS: An Emphasis on Practical Applications Every chapter of this text features discussions of biological, medical, and industrial applications of organic chemistry, many of them new to this edition. Some of these applications are found in the text discussion, others in the exercises and problems, and still others in the Chemical Highlight boxes. Topics range from the chemistry behind the effects on human health of “compounds in the news” (cholesterol, trans fatty acids, grape seed extracts, green tea), to advances in the development of “green,” environmentally friendly methods in the chemical industry, new chemically based methods of disease diagnosis and treatment, and uses of transition metals and enzymes to catalyze reactions in pharmaceutical and medicinal chemistry. A major application of organic chemistry, stressed throughout the text, is the synthesis of new products and materials. We emphasize the development of good synthetic strategies and the avoidance of pitfalls, illustrating these ideas with many Working with the Concepts and Integration Problems. Many chapters contain specific syntheses of biological and medicinal importance.
Preface
New Applications Include:
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CHEM I CAL HI GHL I GHT 11-2
Detecting Performance Enhancing Drugs Security in the 21st Century: Applications of IR and MS Using Mass Spectrometry (Ch. 11, Spectroscopic methods are revolutionizing our ability to unambiguous identification of the original molecule by detect dangerous environmental substances in real time. comparison with a standard database. In 10 s or less, these p. 476), Portable infrared imaging detectors (high-tech cameras) devices can detect both positive and negative ions deriving that can identify and pinpoint the location, extent, and from minute amounts (less than 10 g) of a variety of Synthesis of Immunosuppressants (Ch. 14, explosive substances, toxic industrial chemicals, illegal movement of clouds of toxic gases are commercially availnarcotics, and chemical warfare agents. able. These devices monitor the IR spectrum of the image p. 644), of each pixel in the camera’s field of view. The detectors are programmed to alert a user to the presence of a variety A Green Approach to Ibuprofen (Ch. 16, of agents by matching the fingerprint regions of observed IR spectra with a customized database loaded into the p. 736), device’s memory. Technological advances in the identification of chemical Synthetic Differentiation of Carbonyl Groups substances by their molecular masses have led to the development of the “puffer” machine that you may have seen (Ch. 17, p. 796), at an airport security checkpoint. A passenger stands under the device’s archway while a puff of air is released, which Plastics Fibers and Energy from Biomasspasses his or her body and is wafted into a detector. The detector uses many of the basic features of the mass specDerived Hydroxyesters (Ch. 19, p. 905), trometer: It ionizes molecules in the air stream and detects their masses. It differs in that masses are distinguished not Synthesis of the Natural Product Buflavine by the amount their paths are curved by electric fields in a vacuum, but by how fast they drift through a series of (Ch. 21, p. 991), charged rings at normal atmospheric pressure — thus the name ion-mobility spectrometry (IMS). Ion mobility, a The Explosive Detection Trace Portal in operation at the Industrial Production of Phenol from San Francisco International Airport. function of the mass, shape, and size of a particle, allows Cumene (Ch. 22, p. 1041), Synthesis of the Antihypertensive Phentolamine (Ch. 22, p. 1047), Synthesis of the Natural Product Resveratrol (Ch. 22, p. 1073), Drug Design and “Bird Flu” (Ch. 24, p. 1152), Synthesis of the Drug Varenicline (Chantix) (Ch. 25, p. 1171), A “Super Green” Hantzsch Pyridine Synthesis (Ch. 25, p. 1182), Organocatalytic Reductions (Ch. 25, p. 1186), Enantioselective Phase Transfer Catalysis (Ch. 26, p. 1222). 29
NEW AND UPDATED TOPICS As with all new editions, each chapter has Applications and Hazards of Haloalkanes: “Greener” Alternatives been carefully reviewed and revised. The properties of haloalkanes have made this class of compounds a rich source of comUpdates and improvements, many of which mercially useful substances. For example, fully halogenated liquid bromomethanes, such as CBrF3 and CBrClF2 (“Halons”), are extremely effective fire retardants. Heat-induced cleavinvolve “green” chemistry, include: age of the weak C–Br bond releases bromine atoms, which suppress combustion by inhibiting the free-radical chain reactions occurring in flames (see Chapter 3, Problem 40). Like New section: Curved arrows show how Freon refrigerants, however, bromoalkanes are ozone depleting (Section 3-9) and have been starting materials convert to products banned for all uses except fire-suppression systems in aircraft engines. Phosphorus tribromide, PBr3, a non-ozone-depleting liquid with a high weight percent of bromine, is a promis(Ch. 2, p. 57) ing replacement. In 2006, a PBr3-based fire-suppression cartridge system (under the trade name PhostrEx™) was approved by both the U.S. Environmental Protection Agency (EPA) New section: Sustainability and Green and the U.S. Federal Aviation Administration (FAA). It is now in commercial use in the Chemistry (Ch 3, p. 103) Eclipse 500 jet aircraft. The polarity of the carbon – halogen bond makes haloalkanes useful for applications such as New section: Applications and Hazards of dry cleaning of clothing and degreasing of mechanical and electronic components. Alternatives Haloalkanes: “Greener” Alternatives (Ch. 6, p. 217) Expanded coverage of solvation effects in SN2 reactions (Ch. 6, p. 235) New section: Which is “Greener”: SN1 or SN2? (Ch. 7, p. 263) Expanded coverage of green uses of ethanol (Ch. 9, p. 368) Expanded coverage of resonance in NMR (Ch. 10, p. 390) New section: Sharpless Oxidations and the Synthesis of Antitumor Drugs (Ch. 12, p. 536) Expanded coverage of the Diels-Alder reaction (Ch.14, p. 628) and Electrocyclic Reactions (Ch.14, p. 641) Expanded coverage of annulenes and aromaticity (Ch. 15, p. 697) Expanded coverage of the stereochemistry of the Wittig reaction (Ch. 17, p. 806) Expanded coverage of intramolecular aldol condensations (Ch. 18, p. 843)
The Eclipse 500 jet over San Francisco.
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New section: Substitution at the Carboxy Carbon Occurs by Addition–Elimination (Ch. 19, p. 886) Expanded coverage of relative reactivity of carboxylic acid derivatives (Ch. 20, p. 926) Expanded coverage of the mechanism of the base-mediated ester hydrolysis (Ch. 20, p. 938) Expanded coverage of IR spectroscopy of amines (Ch. 21, p. 977) Expanded coverage of phenol syntheses from haloarenes, including a new section on Pd catalysis (Ch. 22, p. 1038) Expanded coverage of cell-surface carbohydrate recognition (Ch. 24, p. 1150)
SUPPLEMENTAL MATERIAL FOR STUDENTS AND INSTRUCTORS We believe a student needs to interact with a concept several times in a variety of scenarios to obtain a practical understanding. With that in mind, W. H. Freeman has developed the most comprehensive student learning package available.
FOR INSTRUCTORS Instructors Resource CD-ROM ISBN: 1-4292-3149-1 To help instructors create lecture presentations, Web sites, and other resources, this CD-ROM allows instructors to export resources directly to their computers. Some of these resources include: • All textbook images, in both JPEG and PPT format • Animated Mechanisms • Full solutions to all problems within the text • Lecture PowerPoints • Clicker Questions • Text Test Bank, which includes hundred of multiple-choice questions Electronic Resources Instructors can access valuable teaching tools at www.whfreeman.com/organic6e. These password-protected resources are designed to enhance lecture presentations, and include Textbook Images (available in JPEG and PowerPoint format), the Online Quiz Gradebook, Clicker Questions, Video Lectures, MCAT Solutions, and more. WebAssign Premium For instructors interested in online homework management, W. H. Freeman and WebAssign have partnered to deliver WebAssign Premium—a comprehensive and flexible suite of resources for your course. Combining the most widely used online homework platform with a wealth of visualization and tutorial resources, WebAssign Premium extends and enhances the classroom experience for instructors and students by combining algorithmically generated versions of selected end-of-chapter questions with a fully interactive eBook at an affordable price. See page xxx for more details, or visit www.webassign.net to sign up for a faculty demo account. • NEW. Molecular Modeling Problems With this edition we now offer new molecular modeling problems for almost every chapter, which can be found on the text’s companion Web site. The problems were written to be performed using the popular Spartan Student EditionTM software from Wavefunction, Inc. With purchase of this text, students can buy Spartan Student Edition software at a significant discount from www.wavefun.com/cart/spartaned.html using the code WHFOCHEM.
Preface
While the problems are written to be performed using Spartan Student Edition software, they can be completed using any electronic structure program that allows Hartree-Fock, density functional, and MP2 calculations.
FOR STUDENTS Study Guide and Solutions Manual by Neil Schore, University of California, Davis ISBN: 1-4292-3136-X Written by Organic Chemistry coauthor Neil Schore, this invaluable manual includes chapter introductions that highlight new materials, chapter outlines, detailed comments for each chapter section, a glossary, and solutions to the end-of-chapter problems, presented in a way that shows students how to reason their way to the answer. Workbook for Organic Chemistry: Supplemental Problems and Solutions by Jerry Jenkins, Otterbein College ISBN: 1-4292-4758-4 Jerry Jenkins’ extensive workbook provides approximately 80 problems per topic with full worked-out solutions. The perfect aid for students in need of more problem-solving, the Workbook for Organic Chemistry can be paired with any organic chemistry text on the market. For instructors interested in online homework, W.H. Freeman has also placed these problems in WebAssign. Molecular Model Set ISBN: 0-7167-4822-3 A modeling set offers a simple, practical way to see, manipulate, and investigate molecular behavior. Polyhedra mimic atoms, pegs serve as bonds, oval discs become orbitals. Freeman is proud to offer this inexpensive, best-of-its-kind kit containing everything you need to represent double and triple bonds, radicals, and long pairs of electrons—including more carbon pieces than are offered in other sets. Spartan Student Discount With purchase of this text, student can buy Spartan Student at a significant discount at www.wavefun.com/cart/spartaned.html using the code WHFOCHEM.
PREMIUM MULTIMEDIA RESOURCES The Organic Chemistry Book Companion Web site, which can be accessed at www.whfreeman.com/organic6e, contains a wealth of Premium Student Resources. Students can unlock these resources with the click of a button, putting extensive concept and problemsolving support right at their fingertips. Some of the resources available are listed below. NEW. ChemCasts replicate the face-to-face experience of watching an instructor work a problem. Using a virtual whiteboard, the Organic ChemCast tutors show students the steps involved in solving key worked examples, while explaining the concepts along the way. The worked examples featured in the ChemCasts were chosen with the input of organic chemistry students. NEW. Organic Flashcards contain over 200 terms and images designed to test students’ organic chemistry knowledge and reinforce
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key concepts along the way. Selected from the key terms and concepts in the text, the Organic Flashcards also include nomenclature exercise information and reaction summary road map content. These dynamic Web-based cards also allow students to create their own cards, start review quizzes, and print for on-the-go studying. NEW. ChemNews from Scientific American provides an up-to-the-minute streaming feed of organic chemistry-related new stories direct from Scientific American magazine. Stay on top of the latest happenings in chemistry, all in one, easy place. NEW. The multimedia-enhanced eBook contains the complete text with a wealth of helpful functions. All student multimedia, including the ChemCasts, Organic Flashcards, and ChemNews from Scientific American, are linked directly to the eBook pages. Students are thus able to access supporting resources when they need them—taking advantage of the “teachable moment” as students read. Customization functions include instructor and student notes, document linking, and editing capabilities. The eBook also includes 3D molecule images for all model-building exercises, courtesy of ChemSpider (www.chemspider.com). Online Learning Environments The above resources are available in two platforms. WebAssign Premium offers the most effective and widely used online homework system in the sciences, and is designed specifically for those instructors seeking graded homework management. The Student Companion Web site provides the student-oriented support materials independent of any homework system. WebAssign Premium For instructors interested in online homework management, WebAssign Premium features a time-tested, secure online environment used by millions of students worldwide. Featuring algorithmic problem generation and supported by a wealth of chemistry-specific learning tools, WebAssign Premium for Organic Chemistry presents instructors with a powerful assignment manager and study environment. WebAssign Premium provides the following resources: • • • •
Algorithmically generated problems, where applicable: Students receive homework problems containing unique values for computation, encouraging them to work out the problems on their own. Complete access to the interactive eBook, from a live table of contents, as well as from relevant problem statements. Links to ChemCasts, Organic Flashcards, ChemNews from Scientific American, and other interactive tools are provided as hints and feedback to ensure a clearer understanding of the problems and the concepts they reinforce. Graded molecular drawing problems using the popular MarvinSketch application allow instructors to evaluate student understanding of molecular structure. The system evaluates virtually “drawn” molecular structures, returning a grade as well as helpful feedback for common errors.
Student Companion Web site The Organic Chemistry Book Companion Web site, (www.whfreeman.com/organic6e), provides a range of tools for problem solving and chemical explorations. They include, among others: • • • •
Student self-quizzes An interactive periodic table of the elements Author Lecture Videos Reaction and nomenclature exercises, which are drag-and-drop exercises designed for memorization and rote memorization
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• •
Animated Mechanisms for reference and quizzing To access additional support, including the ChemCasts, Organic Flashcards, and ChemNews from Scientific American, students can upgrade their access through a direct subscription to the Premium component of the Web site.
ACKNOWLEDGMENTS We are grateful to the following professors who reviewed the manuscript for the sixth edition: Michael Barbush, Baker University Debbie J. Beard, Mississippi State University Robert Boikess, Rutgers University Cindy C. Browder, Northern Arizona University Kevin M. Bucholtz, Mercer University Kevin C. Cannon, Penn State Abington J. Michael Chong, University of Waterloo Jason Cross, Temple University Alison Flynn, Ottawa University Roberto R. Gil, Carnegie Mellon University Sukwon Hong, University of Florida Jeffrey Hugdahl, Mercer University Colleen Kelley, Pima Community College
Vanessa McCaffrey, Albion College Keith T. Mead, Mississippi State University James A. Miranda, Sacramento State University David A. Modarelli, University of Akron Thomas W. Ott, Oakland University Hasan Palandoken, Western Kentucky University Gloria Silva, Carnegie Mellon University Barry B. Snider, Brandeis University David A. Spiegel, Yale University Paul G. Williard, Brown University Shmuel Zbaida, Rutgers University Eugene Zubarev, Rice University
We are also grateful to the following professors who reviewed the manuscript for the fifth edition: Donald H. Aue, University of California, Santa Barbara Robert C. Badger, University of Wisconsin-Stevens Point Masimo D. Bezoari, Huntingdon College Michael Burke, North Dakota State College of Science Allen Clabo, Francis Marion University A. Gilbert Cook, Valparaiso University Loretta T. Dorn, Fort Hays State University Graham W. L. Ellis, Bellarmine University Kevin L. Evans, Glenville State College John D. Fields, Methodist College Douglas Flournoy, Indian Hills Community College Larry G. French, St. Lawrence University Allan A. Gahr, Gordon College Gamini U. Gunawardena, Utah Valley State College Sapna Gupta, Park University Ronald L. Halterman, University of Oklahoma, Norman
Gene Hiegel, California State University, Fullerton D. Koholic-Hehemann, Cuyahoga Community College Joseph W. Lauher, SUNY Stony Brook David C. Lever, Ohio Wesleyan University Charles A. Lovelette, Columbus State University Alan P. Marchand, University of North Texas Daniel M. McInnes, East Central University S. Shaun Murphree, Allegheny College Raj Pandian, University of New Orleans P. J. Persichini III, Allegheny College Venkatesh Shanbhag, Nova Southeastern University Douglass F. Taber, University of Delaware Dasan M Thamattoor, Colby College Leon J. Tilley, Stonehill College Nanette M. Wachter, Hofstra University
Peter Vollhardt thanks his synthetic and physical colleagues at UC Berkeley, in particular Professors Bob Bergman, Carolyn Bertozzi, Ron Cohen, Jean Frechet, Steve Pedersen, Rich Saykally, Andrew Streitwieser, Dirk Trauner, Dave Wernner, and Evan Williams, for general and very specific suggestions. He would also like to thank his administrative assistant, Bonnie Kirk, for helping with the logistics of producing and handling manuscript and galleys, and his graduate student Robin Padilla for general help. Neil Schore thanks his organic colleagues, especially Dr. Melekeh Nasiri, who was constantly on the lookout for inconsistencies and errors in the textbook problems and study guide solutions.
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Our thanks go to the many people who helped with this edition. Jessica Fiorillo, acquisitions editor, and Randi Rossignol, development editor, at W. H. Freeman and Company, guided this edition from concept to completion. David Chelton used persistence and humor to help us stick with our plan. Dave Quinn, media editor, managed the media and supplements with great skill, and Brittany Murphy, editorial assistant, coordinated our efforts. Also many thanks to Blake Logan, our designer; and Susan Wein, production coordinator, for their fine work and attention to the smallest detail. Thanks also to Dennis Free at Aptara, for his unlimited patience.
C H A P T E R
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ONE
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Structure and Bonding in Organic Molecules
ow do chemicals regulate your body? Why did your muscles ache this morning after last night’s long jog? What is in the pill you took to get rid of that headache you got after studying all night? What happens to the gasoline you pour into the gas tank of your car? What is the molecular composition of the things you wear? What is the difference between a cotton shirt and one made of silk? What is the origin of the odor of garlic? You will find the answers to these questions, and many others that you may have asked yourself, in this book on organic chemistry. Chemistry is the study of the structure of molecules and the rules that govern their interactions. As such, it interfaces closely with the fields of biology, physics, and mathematics. What, then, is organic chemistry? What distinguishes it from other chemical disciplines, such as physical, inorganic, or nuclear chemistry? A common definition provides a partial answer: Organic chemistry is the chemistry of carbon and its compounds. These compounds are called organic molecules. Organic molecules constitute the chemical building blocks of life. Fats, sugars, proteins, and the nucleic acids are compounds in which the principal component is carbon. So are countless substances that we take for granted in everyday use. Virtually all the clothes that we wear are made of organic molecules—some of natural fibers, such as cotton and silk; others artificial, such as polyester. Toothbrushes, toothpaste, soaps, shampoos, deodorants, perfumes—all contain organic compounds, as do furniture, carpets, the plastic in light fixtures and cooking utensils, paintings, food, and countless other items. Consequently, organic chemical industries are among the largest in the world, including petroleum refining and processing, agrochemicals, plastics, pharmaceuticals, paints and coatings, and the food conglomerates. Organic substances such as gasoline, medicines, pesticides, and polymers have improved the quality of our lives. Yet the uncontrolled disposal of organic chemicals has polluted the environment, causing deterioration of animal and plant life as well as injury and disease to humans. If we are to create useful molecules — and learn to control their effects — we need a knowledge of their properties and an understanding of their behavior. We must be able to apply the principles of organic chemistry.
H
Tetrahedral carbon, the essence of organic chemistry, exists as a lattice of six-membered rings in diamonds. In 2003, a family of molecules called diamandoids was isolated from petroleum. Diamandoids are subunits of diamond in which the excised pieces are capped off with hydrogen atoms. An example is the beautifully crystalline pentamantane (molecular model on top right and picture on the left; © 2004 Chevron U.S.A. Inc. Courtesy of MolecularDiamond Technologies, ChevronTexaco Technology Ventures LLC), which
consists of five “cages” of the diamond lattice. The top right of the picture shows the carbon frame of pentamantane stripped off its hydrogens and its superposition on the lattice of diamond.
2
Chapter 1
Structure and Bonding in Organic Molecules
This chapter explains how the basic ideas of chemical structure and bonding apply to organic molecules. Most of it is a review of topics that you covered in your general chemistry courses, including molecular bonds, Lewis structures and resonance, atomic and molecular orbitals, and the geometry around bonded atoms.
1-1 The Scope of Organic Chemistry: An Overview Almost everything you see in this picture is made of organic chemicals.
R
Function
Carbon frame provides structure
Functional group imparts reactivity
H3COCH3 Ethane
A goal of organic chemistry is to relate the structure of a molecule to the reactions that it can undergo. We can then study the steps by which each type of reaction takes place, and we can learn to create new molecules by applying those processes. Thus, it makes sense to classify organic molecules according to the subunits and bonds that determine their chemical reactivity: These determinants are groups of atoms called functional groups. The study of the various functional groups and their respective reactions provides the structure of this book.
Functional groups determine the reactivity of organic molecules We begin with the alkanes, composed of only carbon and hydrogen atoms (“hydrocarbons”) connected by single bonds. They lack any functional groups and as such constitute the basic scaffold of organic molecules. As with each class of compounds, we present the systematic rules for naming alkanes, describe their structures, and examine their physical properties (Chapter 2). An example of an alkane is ethane. Its structural mobility is the starting point for a review of thermodynamics and kinetics. This review is then followed by a discussion of the strength of alkane bonds, which can be broken by heat, light, or chemical reagents. We illustrate these processes with the chlorination of alkanes (Chapter 3).
H2 C H2C H2C
A Chlorination Reaction CH2
C H2
CH2
Cyclohexane
1
CH4
Cl2
H3C
C
CH3
Acetone (A ketone)
H3C
CH3OCl
1
O
O
NH2
Methyl amine (An amine)
CH3OCl
1
HCl
A Substitution Reaction
Acetylene (An alkyne)
H2C
uy
Next we look at cyclic alkanes (Chapter 4), which contain carbon atoms in a ring. This arrangement can lead to new properties and changes in reactivity. The recognition of a new type of isomerism in cycloalkanes bearing two or more substituents — either on the same side or on opposite sides of the ring plane — sets the stage for a general discussion of stereoisomerism. Stereoisomerism is exhibited by compounds with the same connectivity but differing in the relative positioning of their component atoms in space (Chapter 5). We shall then study the haloalkanes, our first example of compounds containing a functional group — the carbon – halogen bond. The haloalkanes participate in two types of organic reactions: substitution and elimination (Chapters 6 and 7). In a substitution reaction, one halogen atom may be replaced by another; in an elimination process, adjacent atoms may be removed from a molecule to generate a double bond.
HCqCH
Formaldehyde (An aldehyde)
Energy
K1I2
uy
CH3OI
1
K1Cl2
An Elimination Reaction ⫹⫺
CH2
CH2 ⫹ K
H
I
OH
H2C
CH2 ⫹ HOH ⫹ K⫹ I⫺
Like the haloalkanes, each of the major classes of organic compounds is characterized by a particular functional group. For example, the carbon – carbon triple bond is the functional group of alkynes (Chapter 13); the smallest alkyne, acetylene, is the chemical burned in a welder’s torch. A carbon – oxygen double bond is characteristic of aldehydes and ketones (Chapter 17); formaldehyde and acetone are major industrial commodities. The amines (Chapter 21), which include drugs such as nasal decongestants and amphetamines, contain
1-1
The Scope of Organic Chemistry: An Overview
Chapter 1
nitrogen in their functional group; methyl amine is a starting material in many syntheses of medicinal compounds. We shall study the tools for identifying these molecular subunits, especially the various forms of spectroscopy (Chapters 10, 11, and 14). Organic chemists rely on an array of spectroscopic methods to elucidate the structures of unknown compounds. All of these methods depend on the absorption of electromagnetic radiation at specific wavelengths and the correlation of this information with structural features. Subsequently, we shall encounter organic molecules that are especially crucial in biology and industry. Many of these, such as the carbohydrates (Chapter 24) and amino acids (Chapter 26), contain multiple functional groups. However, in every class of organic compounds, the principle remains the same: The structure of the molecule determines the reactions that it can undergo.
Synthesis is the making of new molecules Carbon compounds are called “organic” because it was originally thought that they could be produced only from living organisms. In 1828, Friedrich Wöhler* proved this idea to be false when he converted the inorganic salt lead cyanate into urea, an organic product of protein metabolism in mammals (Chemical Highlight 1-1). Wöhler’s Synthesis of Urea
Pb(OCN)2 ⫹ 2 H2O ⫹ 2 NH3 Lead cyanate
Water
Ammonia
O B 2 H2NCNH C 2 ⫹ Pb(OH)2 Urea
Lead hydroxide
Synthesis, or the making of molecules, is a very important part of organic chemistry (Chapter 8). Since Wöhler’s time, many millions of organic substances have been synthesized from simpler materials, both organic and inorganic.† These substances include many that also occur in nature, such as the penicillin antibiotics, as well as entirely new compounds. Some, such as cubane, have given chemists the opportunity to study special kinds of bonding and reactivity. Others, such as the artificial sweetener saccharin, have become a part of everyday life. Typically, the goal of synthesis is to construct complex organic chemicals from simpler, more readily available ones. To be able to convert one molecule into another, chemists must know organic reactions. They must also know the physical factors that govern such processes, such as temperature, pressure, solvent, and molecular structure. This knowledge is equally valuable in analyzing reactions in living systems. As we study the chemistry of each functional group, we shall develop the tools both for planning effective syntheses and for predicting the processes that take place in nature. But how? The answer lies in looking at reactions step by step.
H
H
C C
C C H
C H
C C H
H
O C
H
H
H H H ´] S C H C C H C N C C O ( H H O C H H H N ´ [ C
H
H C
H C
C
H
An organic molecular architect at work.
HC
C
C
H
O
C
C H
C
C
C
C
C
H
NH H
C
H
H
C H
O Benzylpenicillin
O S
Cubane
*Professor Friedrich Wöhler (1800 –1882), University of Göttingen, Germany. In this and subsequent biographical notes, only the scientist’s last known location of activity will be mentioned, even though much of his or her career may have been spent elsewhere. † As of September 2009, the Chemical Abstracts Service had registered over 50 million chemical substances and more than 61 million genetic sequences.
Saccharin
C O
3
4
Chapter 1
Structure and Bonding in Organic Molecules
C H E M I C A L H IGH LIG H T 1-1 Urea: From Urine to Wöhler’s Synthesis to Industrial Fertilizer Urination is the main process by which we excrete nitrogen from our bodies. Urine is produced by the kidneys and then stored in the bladder, which begins to contract when its volume exceeds about 200 mL. The average human excretes about 1.5 L of urine daily, and a major component is urea, about 20 g per liter. In an attempt to probe the origins of kidney stones, early (al)chemists, in the 18th century, attempted to isolate the components of urine by crystallization, but they were stymied by the cocrystallization with the also present sodium chloride. William Prout,* an English chemist and physician, is credited with the preparation of pure urea in 1817 and the determination of its accurate elemental analysis as CH4N2O. Prout was an avid proponent of the then revolutionary thinking that disease has a molecular basis and could be understood as such. This view clashed with that of the so-called vitalists, who believed that the functions of a living organism are controlled by a “vital principle” and cannot be explained by chemistry (or physics). Into this argument entered Wöhler, an inorganic chemist, who attempted to make ammonium cyanate, NH41OCN2 (also CH4N2O), from lead cyanate and ammonia in 1828, but who obtained the same compound that Prout had characterized as urea. To one of his mentors, Wöhler wrote, “I can make urea without a kidney, or even a living creature.” In his landmark paper, “On the Artificial Formation of Urea,” he commented on his synthesis as a “remarkable fact, as it is an example of the artificial generation of an organic material from inorganic materials.” He also alluded to the significance of the finding that a compound with an identical elemental composition as ammonium cyanate can have such completely different chemical properties, a forerunner to the recognition of isomeric compounds. Wöhler’s synthesis of *Dr. William Prout (1785–1850), Royal College of Physicians, London.
urea forced his contemporary vitalists to accept the notion that simple organic compounds could be made in the laboratory. As you shall see in this book, over the ensuing decades, synthesis has yielded much more complex molecules than urea, some of them endowed with self-replicating and other “lifelike” properties, such that the boundaries between what is lifeless and what is alive are dwindling. Apart from its function in the body, urea’s high nitrogen content makes it an ideal fertilizer. It is also a raw material in the manufacture of plastics and glues, an ingredient of some toiletry products and fire extinguishers, and an alternative to rock salt for deicing roads. It is produced industrially from ammonia and carbon dioxide to the tune of 100 million tons per year.
The effect of nitrogen fertilizer on wheat growth: treated on the left; untreated on the right.
Reactions are the vocabulary and mechanisms are the grammar of organic chemistry When we introduce a chemical reaction, we will first show just the starting compounds, or reactants (also called substrates), and the products. In the chlorination process mentioned earlier, the substrates — methane, CH4, and chlorine, Cl2 — may undergo a reaction to give chloromethane, CH3Cl, and hydrogen chloride, HCl. We described the overall transformation as CH4 1 Cl2 n CH3Cl 1 HCl. However, even a simple reaction such as this one may proceed through a complex sequence of steps. The reactants could have first formed one or more unobserved substances — call these X — that rapidly changed into the observed products. These underlying details of the reaction constitute the reaction mechanism. In our example, the mechanism consists of two major parts: CH4 1 Cl2 n X followed by X n CH3Cl 1 HCl. Each part is crucial in determining whether the overall reaction will proceed. Substances X in our chlorination reaction are examples of reaction intermediates, species formed on the pathway between reactants and products. We shall learn the mechanism of this chlorination process and the nature of the reaction intermediates in Chapter 3.
1-2 Coulomb Forces: A Simplified View of Bonding
How can we determine reaction mechanisms? The strict answer to this question is, we cannot. All we can do is amass circumstantial evidence that is consistent with (or points to) a certain sequence of molecular events that connect starting materials and products (“the postulated mechanism”). To do so, we exploit the fact that organic molecules are no more than collections of bonded atoms. We can, therefore, study how, when, and how fast bonds break and form, in which way they do so in three dimensions, and how changes in substrate structure affect the outcome of reactions. Thus, although we cannot strictly prove a mechanism, we can certainly rule out many (or even all) reasonable alternatives and propose a most likely pathway. In a way, the “learning” and “using” of organic chemistry is much like learning and using a language. You need the vocabulary (i.e., the reactions) to be able to use the right words, but you also need the grammar (i.e., the mechanisms) to be able to converse intelligently. Neither one on its own gives complete knowledge and understanding, but together they form a powerful means of communication, rationalization, and predictive analysis. To highlight the interplay between reaction and mechanism, icons are displayed in the margin at appropriate places throughout the text. Before we begin our study of the principles of organic chemistry, let us review some of the elementary principles of bonding. We shall find these concepts useful in understanding and predicting the chemical reactivity and the physical properties of organic molecules.
5
Chapter 1
REACTION
MECHANISM
1-2 Coulomb Forces: A Simplified View of Bonding The bonds between atoms hold a molecule together. But what causes bonding? Two atoms form a bond only if their interaction is energetically favorable, that is, if energy — heat, for example — is released when the bond is formed. Conversely, breaking that bond requires the input of the same amount of energy. The two main causes of the energy release associated with bonding are based on Coulomb’s law of electric charge: 1. Opposite charges attract each other (electrons are attracted to protons). 2. Like charges repel each other (electrons spread out in space).
Bonds are made by simultaneous coulombic attraction and electron exchange Each atom consists of a nucleus, containing electrically neutral particles, or neutrons, and positively charged protons. Surrounding the nucleus are negatively charged electrons, equal in number to the protons so that the net charge is zero. As two atoms approach each other, the positively charged nucleus of the first atom attracts the electrons of the second atom; similarly, the nucleus of the second atom attracts the electrons of the first atom. As a result, the nuclei are held together by the electrons located between them. This sort of bonding is described by Coulomb’s* law: Opposite charges attract each other with a force inversely proportional to the square of the distance between the centers of the charges.
Charge separation is rectified by Coulomb’s law, appropriately in the heart of Paris.
The Atom Neutron
Electron
−
Coulomb’s Law Attracting force 5 constant 3
(1) charge 3 (2) charge distance2
− −
This attractive force causes energy to be released as the neutral atoms are brought together. This energy is called the bond strength. *Lieutenant-Colonel Charles Augustin de Coulomb (1736–1806), Inspecteur Général of the University of Paris, France.
− + + ++ + ++
−
Nucleus (Protons and Neutrons)
−
−
Proton
6
Chapter 1
Structure and Bonding in Organic Molecules
Absorbed
Figure 1-1 The changes in energy, E, that result when two atoms are brought into close proximity. At the separation defined as bond length, maximum bonding is achieved.
0 Released
E
Bond length Interatomic distance
When the atoms reach a certain closeness, no more energy is released. The distance between the two nuclei at this point is called the bond length (Figure 1-1). Bringing the atoms closer together than this distance results in a sharp increase in energy. Why? As stated above, just as opposite charges attract, like charges repel. If the atoms are too close, the electron – electron and nuclear – nuclear repulsions become stronger than the attractive forces. When the nuclei are the appropriate bond length apart, the electrons are spread out around both nuclei, and attractive and repulsive forces balance for maximum bonding. The energy content of the two-atom system is then at a minimum, the most stable situation (Figure 1-2). An alternative to this type of bonding results from the complete transfer of an electron from one atom to the other. The result is two charged ions: one positively charged, a cation, and one negatively charged, an anion (Figure 1-3). Again, the bonding is based on coulombic attraction, this time between two ions. The coulombic bonding models of attracting and repelling charges shown in Figures 1-2 and 1-3 are highly simplified views of the interactions that take place in the bonding of atoms. Nevertheless, even these simple models explain many of the properties of organic molecules. In the sections to come, we shall examine increasingly more sophisticated views of bonding.
Figure 1-2 Covalent bonding. Attractive (solid-line) and repulsive (dashed-line) forces in the bonding between two atoms. The large spheres represent areas in space in which the electrons are found around the nucleus. The small circled plus sign denotes the nucleus.
Attraction e−
e−
e−
+
+
e−
+
+
+
Repulsion Atom 1
Atom 2
Covalently Bonded Molecule
e transfer
Figure 1-3 Ionic bonding. An alternative mode of bonding results from the complete transfer of an electron from atom 1 to atom 2, thereby generating two ions whose opposite charges attract each other.
e−
+
Atom 1
e−
+
+
Atom 2
+
e− + e−
Anion Cation Ionically Bonded Molecule
1-3 Ionic and Covalent Bonds: The Octet Rule
Chapter 1
7
1-3 Ionic and Covalent Bonds: The Octet Rule We have seen that attraction between negatively and positively charged particles is a basis for bonding. How does this concept work in real molecules? Two extreme types of bonding explain the interactions between atoms in organic molecules: 1. A covalent bond is formed by the sharing of electrons (as shown in Figure 1-2). 2. An ionic bond is based on the electrostatic attraction of two ions with opposite charges (as shown in Figure 1-3). We shall see that many atoms bind to carbon in a way that is intermediate between these extremes: Some ionic bonds have covalent character and some covalent bonds are partly ionic (polarized). What are the factors that account for the two types of bonds? To answer this question, let us return to the atoms and their compositions. We start by looking at the periodic table and at how the electronic makeup of the elements changes as the atomic number increases.
The periodic table underlies the octet rule The partial periodic table depicted in Table 1-1 includes those elements most widely found in organic molecules: carbon (C), hydrogen (H), oxygen (O), nitrogen (N), sulfur (S), chlorine (Cl), bromine (Br), and iodine (I). Certain reagents, indispensable for synthesis and commonly used, contain elements such as lithium (Li), magnesium (Mg), boron (B), and phosphorus (P). (If you are not familiar with these elements, refer to Table 1-1 or the periodic table on the inside cover.) The elements in the periodic table are listed according to nuclear charge (number of protons), which equals the number of electrons. The nuclear charge increases by one with each element listed. The electrons occupy energy levels, or “shells,” each with a fixed capacity. For example, the first shell has room for two electrons; the second, eight; and the third, 18. Helium, with two electrons in its shell, and the other noble gases, with eight electrons (called octets) in their outermost shells, are especially stable. These elements show very little chemical reactivity. All other elements (including carbon, see margin) lack octets in their outermost electron shells. Atoms tend to form molecules in such a way as to reach an octet in the outer electron shell and attain a noble-gas configuration. In the next two sections, we describe two extreme ways in which this goal may be accomplished: by the formation of pure ionic or pure covalent bonds.
Carbon Atom Filled 1st shell
C 2, 4 Unfilled 2nd shell: four valence electrons
Exercise 1-1 (a) Redraw Figure 1-1 for a weaker bond than the one depicted. (b) Write the elements in Table 1-1 from memory.
Table 1-1
Partial Periodic Table
Period First Second Third Fourth Fifth
Halogens H1 Li2,1 Na2,8,1 K2,8,8,1
2,2
Be Mg2,8,2
2,3
B Al2,8,3
2,4
C Si2,8,4
2,5
N P2,8,5
Note: The superscripts indicate the number of electrons in each principal shell of the atom.
2,6
O S2,8,6
2,7
F Cl2,8,7 Br2,8,18,7 I2,8,18,18,7
Noble gases He2 Ne2,8 Ar2,8,8 Kr2,8,18,8 Xe2,8,18,18,8
8
Chapter 1
Structure and Bonding in Organic Molecules
In pure ionic bonds, electron octets are formed by transfer of electrons Sodium (Na), a reactive metal, interacts with chlorine, a reactive gas, in a violent manner to produce a stable substance: sodium chloride. Similarly, sodium reacts with fluorine (F), bromine, or iodine to give the respective salts. Other alkali metals, such as lithium and potassium (K), undergo the same reactions. These transformations succeed because both reaction partners attain noble-gas character by the transfer of outer-shell electrons, called valence electrons, from the alkali metals on the left side of the periodic table to the halogens on the right. Let us see how this works for the ionic bond in sodium chloride. Why is the interaction energetically favorable? First, it takes energy to remove an electron from an atom. This energy is the ionization potential (IP) of the atom. For sodium gas, the ionization energy amounts to 119 kcal mol21.* Conversely, energy may be released when an electron attaches itself to an atom. For chlorine, this energy, called its electron affinity (EA), is 283 kcal mol21. These two processes result in the transfer of an electron from sodium to chlorine. Together, they require a net energy input of 119 2 83 5 36 kcal mol21. 21 e
[Na2,8,1 uy [Na2,8]1 Sodium cation
IP 5 119 kcal mol21 (498 kJ mol21) Energy input required
(Neon configuration) 11 e
Cl2,8,7 uy [Cl2,8,8]2 Chloride anion
EA 5 283 kcal mol21 (2347 kJ mol21) Energy released
(Argon configuration)
Na 1 Cl uy Na1 1 Cl2
Na
Sodium atom
Cl
+
Chlorine atom
Total 5 119 2 83 5 36 kcal mol21 (151 kJ mol21)
Why, then, do the atoms readily form NaCl? The reason is their electrostatic attraction, which pulls them together in an ionic bond. At the most favorable interatomic distance [about 2.8 Å (angstroms) in the gas phase], this attraction releases (see Figure 1-1) about 120 kcal mol21 (502 kJ mol21). This energy release is enough to make the reaction of sodium with chlorine energetically highly favorable [136 2 120 5 284 kcal mol21 (2351 kJ mol21)]. Formation of Ionic Bonds by Electron Transfer Na2,8,1 1 Cl2,8,7 ¡ [Na2,8 ] 1 [Cl2,8,8 ] 2 , or NaCl (284 kcal mol21 )
Na+
Cl –
Sodium chloride
More than one electron may be donated (or accepted) to achieve noble-gas electronic configurations. Magnesium, for example, has two valence electrons. Donation to an appropriate acceptor produces the corresponding doubly charged cation with the electronic structure of neon. In this way, the ionic bonds of typical salts are formed. A representation of how charge (re)distributes itself in molecules is given by electrostatic potential maps. These computer-generated maps not only show a form of the molecule’s “electron cloud” but also use color to depict deviations from charge neutrality. Excess electron density — for example, a negative charge — is shown in colors shaded toward red; conversely, diminishing electron density — ultimately, a positive charge — is shown in colors shaded toward blue. Charge-neutral regions are indicated by green. The reaction of a sodium atom with a chlorine atom to produce Na1Cl2 is pictured this way in the margin. In the product, Na1 is blue, Cl2 is red. *This book will cite energy values in the traditional units of kcal mol21, in which mol is the abbreviation for mole and a kilocalorie (kcal) is the energy required to raise the temperature of 1 kg (kilogram) of water by 18C. In SI units, energy is expressed in joules (kg m2 s22, or kilogram-meter2 per second2). The conversion factor is 1 kcal 5 4184 J 5 4.184 kJ (kilojoule), and we will list these values in parentheses in key places.
1-3 Ionic and Covalent Bonds: The Octet Rule
A more convenient way of depicting valence electrons is by means of dots around the symbol for the element. In this case, the letters represent the nucleus including all the electrons in the inner shells, together called the core configuration. Valence Electrons as Electron Dots Li
Be
B
C
N
O
F
Na
Mg
Al
Si
P
S
Cl
Electron-Dot Picture of Salts š Naj ⫹ jCl ð
1 e transfer
⫺ š Na⫹ðCl ð
Š ⫹ 2jCl š jMg ð
2 e transfer
⫺ š Mg2⫹ [ðCl ð] 2
The hydrogen atom is unique because it may either lose an electron to become a bare nucleus, the proton, or accept an electron to form the hydride ion, [H, i.e., H;]2, which possesses the helium configuration. Indeed, the hydrides of lithium, sodium, and potassium (Li1H2, Na1H2, and K1H2) are commonly used reagents. Hj
⫺1 e
[H]⫹
Bare nucleus
IP ⫽ 314 kcal mol⫺1 (1314 kJ mol⫺1)
Helium configuration
EA ⫽ ⫺18 kcal mol⫺1 (⫺75 kJ mol⫺1)
Proton
Hj
⫹1 e
[Hð]⫺
Hydride ion
Exercise 1-2 Draw electron-dot pictures for ionic LiBr, Na2O, BeF2, AlCl3, and MgS.
In covalent bonds, electrons are shared to achieve octet configurations Formation of ionic bonds between two identical elements is difficult because the electron transfer is usually very unfavorable. For example, in H2, formation of H1H2 would require an energy input of nearly 300 kcal mol21 (1255 kJ mol21). For the same reason, none of the halogens, F2, Cl2, Br2, and I2, has an ionic bond. The high IP of hydrogen also prevents the bonds in the hydrogen halides from being ionic. For elements nearer the center of the periodic table, the formation of ionic bonds is unfeasible, because it becomes more and more difficult to donate or accept enough electrons to attain the noble-gas configuration. Such is the case for carbon, which would have to shed four electrons to reach the helium electronic structure or add four electrons for a neonlike arrangement. The large amount of charge that would develop makes these processes very energetically unfavorable. C 4⫹ Helium configuration
⫺4 e
jŠ Cj
⫹4 e
Very difficult
ðš Cð4⫺ Neon configuration
Instead, covalent bonding takes place: The elements share electrons so that each atom attains a noble-gas configuration. Typical products of such sharing are H2 and HCl. In HCl, the chlorine atom assumes an octet structure by sharing one of its valence electrons with that of hydrogen. Similarly, the chlorine molecule, Cl2, is diatomic because both component atoms gain octets by sharing two electrons. Such bonds are called covalent single bonds.
Chapter 1
9
10
Chapter 1
Structure and Bonding in Organic Molecules
Electron-Dot Picture of Covalent Single Bonds Hj ⫹ jH
HðH
š Hj ⫹ jClð š šj ⫹ jC š ðCl lð
š HðClð š š Cl š ðClð ð
Shared electron pairs
Because carbon has four valence electrons, it must acquire a share of four electrons to gain the neon configuration, as in methane. Nitrogen has five valence electrons and needs three to share, as in ammonia; and oxygen, with six valence electrons, requires only two to share, as in water. H š HðC ðH H
Hðš NðH H
š HðO ðH
Methane
Ammonia
Water
It is possible for one atom to supply both of the electrons required for covalent bonding. This occurs upon addition of a proton to ammonia, thereby forming NH41, or to water, thereby forming H3O1. H Hðš Nð ⫹ H⫹ H
H Hðš N ðH H
⫹
Hðš Oð ⫹ H⫹ H
Hðš OðH H
⫹
Hydronium ion
Ammonium ion
Besides two-electron (single) bonds, atoms may form four-electron (double) and sixelectron (triple) bonds to gain noble-gas configurations. Atoms that share more than one electron pair are found in ethene and ethyne. H H
ýCððCý ½ ½
H
HðCðððCðH
H
Ethene (Ethylene)*
Ethyne (Acetylene)*
The drawings above, with pairs of electron dots representing bonds, are also called Lewis† structures. We shall develop the general rules for formulating such structures in Section 1-4. Exercise 1-3 Draw electron-dot structures for F2, CF4, CH2Cl2, PH3, BrI, HO2, H2N2, and H3C2. (Where applicable, the first element is at the center of the molecule.) Make sure that all atoms have noble-gas electron configurations.
In most organic bonds, the electrons are not shared equally: polar covalent bonds The preceding two sections presented two extreme ways in which atoms attain noble-gas configurations by entering into bonding: pure ionic and pure covalent. In reality, most bonds are of a nature that lies between these two extremes: polar covalent. Thus, the ionic bonds in most salts have some covalent character; conversely, the covalent bonds to carbon have some ionic or polar character. Recall (Section 1-2) that both sharing of electrons and *In labels of molecules, systematic names (introduced in Section 2-5) will be given first, followed in parentheses by so-called common names that are still in frequent usage. † Professor Gilbert N. Lewis (1875–1946), University of California, Berkeley.
1-3 Ionic and Covalent Bonds: The Octet Rule
Table 1-2
Electronegativities of Selected Elements
H 2.2 Li 1.0 Na 0.9 K 0.8
Be 1.6 Mg 1.3
B 2.0 Al 1.6
C 2.6 Si 1.9
N 3.0 P 2.2
O 3.4 S 2.6
F 4.0 Cl 3.2 Br 3.0 I 2.7
Increasing electronegativity
Increasing electronegativity
Note: Values established by L. Pauling and updated by A. L. Allred (see Journal of Inorganic and Nuclear Chemistry, 1961, 17, 215).
coulombic attraction contribute to the stability of a bond. How polar are polar covalent bonds and what is the direction of the polarity? We can answer these questions by considering the periodic table and keeping in mind that the positive nuclear charge of the elements increases from left to right. Thus, the elements on the left of the periodic table are often called electropositive, electron donating, or “electron pushing,” because their electrons are held by the nucleus less tightly than are those of elements to the right. These elements at the right of the periodic table are described as electronegative, electron accepting, or “electron pulling.” Table 1-2 lists the relative electronegativities of some elements. On this scale, fluorine, the most electronegative of them all, is assigned the value 4. Consideration of Table 1-2 readily explains why the most ionic (least covalent) bonds occur between elements at the two extremes (e.g., the alkali metal salts, such as sodium chloride). On the other hand, the purest covalent bonds are formed between atoms of equal electronegativity (i.e., identical elements, as in H2, N2, O2, F2, and so on) or in carbon –carbon bonds. However, most covalent bonds are between atoms of differing electronegativity, resulting in their polarization. The polarization of a bond is the consequence of a shift of the center of electron density in the bond toward the more electronegative atom. It is indicated in a very qualitative manner (using the Greek letter delta d) by designating a partial positive charge, d1, and partial negative charge, d2, to the respective less or more electronegative atom. The larger the difference in electronegativity, the bigger is the charge separation. As a rule of thumb, electronegativity differences of 0.3 to 2.0 units indicate polar covalent bonds; lesser values are typical of essentially “pure” covalent bonds, larger values of “pure” ionic ones. The separation of opposite charges is called an electric dipole, symbolized by an arrow crossed at its tail and pointing from positive to negative. A polarized bond can impart polarity to a molecule as a whole, as in HF, ICl, and CH3F. Polarization results in a dipole
Polar Bonds ␦⫹
⫹ ⫺ A␦ ðB ␦
Less electronegative
A dipole
⫺ ␦⫹Hðš Fð ␦
⫺ ␦⫹ðš šlð ␦ IðC
Hydrogen fluoride
Iodine monochloride
More electronegative
␦⫹
H ⫺ Cðš Fð ␦ Hðš H
H ␦⫺ ⫹ Cðš OðH␦ Hðš H
Fluoromethane
Methanol
In symmetrical structures, the polarizations of the individual bonds may cancel, thus leading to molecules with no net polarization, such as CO2 and CCl4 (margin). To know whether a molecule is polar, we have to know its shape, because the net polarity is the vector sum of the bond dipoles. The electrostatic potential maps in the margin clearly
Chapter 1
11
12
Chapter 1
Molecules Can Have Polar Bonds but No Net Polarization
Dipoles cancel δ −O
2δ +
C
Oδ
−
Structure and Bonding in Organic Molecules
illustrate the polarization in CO2 and CCl4, showing the respective carbon atoms shaded relatively blue, the attached, more electronegative atoms relatively red. Moreover, you can recognize how the shape of each molecule renders it nonpolar as a whole. There are two cautions in viewing electrostatic potential maps: (1) The scale on which the color differentials are rendered may vary. For example, a much more sensitive scale is used for the molecules in the margin, in which the charges are only partial, than for NaCl on p. 8, in which the atoms assume full charges. Thus, it may be misleading to compare the electrostatic potential maps of one set of molecules with those of another, electronically very different group. Most organic structures shown in this book will be on a comparative scale, unless mentioned otherwise. (2) Because of the way in which the potential at each point is calculated, it will contain contributions from all nuclei and electrons in the vicinity. As a consequence, the color of the spatial regions around individual nuclei is not uniform.
Carbon dioxide
Valence electron repulsion controls the shapes of molecules δ− δ−
4δ +
Cl Cl C Cl δ − Cl δ−
Molecules adopt shapes in which electron repulsion (including both bonding and nonbonding electrons) is minimized. In diatomic species such as H2 or LiH, there is only one bonding electron pair and one possible arrangement of the two atoms. However, beryllium fluoride, BeF2, is a triatomic species. Will it be bent or linear? Electron repulsion is at a minimum in a linear structure, because the bonding and nonbonding electrons are placed as far from each other as possible, at 1808.* Linearity is also expected for other derivatives of beryllium, as well as of other elements in the same column of the periodic table. BeF2 Is Linear ðš FðBeðš Fð not 180º
Electrons are farthest apart
Tetrachloromethane
BCl3 Is Trigonal
Beðš Fð ý Fý ½ ½ Electrons are closer
šð ðCl š š not ðCl BðCl ð ðš
ýClk ýClk ½ ½ B 120º šð ðCl š
In boron trichloride, the three valence electrons of boron allow it to form covalent bonds with three chlorine atoms. Electron repulsion enforces a regular trigonal arrangement — that is, the three halogens are at the corners of an equilateral triangle, the center of which is occupied by boron, and the bonding (and nonbonding) electron pairs of the respective chlorine atoms are at maximum distance from each other, that is, 1208. Other derivatives of boron, and the analogous compounds with other elements in the same column of the periodic table, are again expected to adopt trigonal structures. Applying this principle to carbon, we can see that methane, CH4, has to be tetrahedral. Placing the four hydrogens at the vertices of a tetrahedron minimizes the electron repulsion of the corresponding bonding electron pairs. H
H
H
H
109.5°
or
C H
H H
but not H
H
H
C H
H
This method for determining molecular shape by minimizing electron repulsion is called the valence-shell electron-pair repulsion (VSEPR) method. Note that we often draw molecules such as BCl3 and CH4 as if they were flat and had 908 angles. This depiction is for *This is true only in the gas phase. At room temperature, BeF2 is a solid (it is used in nuclear reactors) that exists as a complex network of linked Be and F atoms, not as a distinct linear triatomic structure.
1-4 Electron-Dot Model of Bonding: Lewis Structures
Chapter 1
ease of drawing only. Do not confuse such two-dimensional drawings with the true threedimensional molecular shapes (trigonal for BCl3 and tetrahedral for CH4). Exercise 1-4 Show the bond polarization in H2O, SCO, SO, IBr, CH4, CHCl3, CH2Cl2, and CH3Cl by using dipole arrows to indicate separation of charge. (In the last four examples, place the carbon in the center of the molecule.)
Exercise 1-5 š Ammonia, :NH3, is not trigonal but pyramidal, with bond angles of 107.38. Water, H2 O , is not linear but bent (104.58). Why? (Hint: Consider the effect of the nonbonding electron pairs.)
In Summary There are two extreme types of bonding, ionic and covalent. Both derive favorable energetics from Coulomb forces and the attainment of noble-gas electronic structures. Most bonds are better described as something between the two types: the polar covalent (or covalent ionic) bonds. Polarity in bonds may give rise to polar molecules. The outcome depends on the shape of the molecule, which is determined in a simple manner by arrangement of its bonds and nonbonding electrons to minimize electron repulsion.
1-4 Electron-Dot Model of Bonding: Lewis Structures Lewis structures are important for predicting geometry and polarity (hence reactivity) of organic compounds, and we shall use them for that purpose throughout this book. In this section, we provide rules for writing such structures correctly and for keeping track of valence electrons.
Lewis structures are drawn by following simple rules The procedure for drawing correct electron-dot structures is straightforward, as long as the following rules are observed. Rule 1. Draw the (given or desired) molecular skeleton. As an example, consider methane. The molecule has four hydrogen atoms bonded to one central carbon atom. H H C H H
H H C H H
Correct
Incorrect
Rule 2. Count the number of available valence electrons. Add up all the valence electrons of the component atoms. Special care has to be taken with charged structures (anions or cations), in which case the appropriate number of electrons has to be added or subtracted to account for extra charges. CH4
4H 1C
4 3 1 electron 5 1 3 4 electrons 5 Total
H3O1
3H 1O Charge
3 3 1 electron 5 3 electrons 1 3 6 electrons 5 6 electrons 11 5 21 electron Total 8 electrons
4 electrons 4 electrons 8 electrons
HBr
1H 1 Br
1 3 1 electron 5 1 3 7 electrons 5 Total
NH22
2H 1N Charge
2 3 1 electron 5 2 electrons 1 3 5 electrons 5 5 electrons 21 5 11 electron Total 8 electrons
1 electron 7 electrons 8 electrons
13
14
Chapter 1
Structure and Bonding in Organic Molecules
Rule 3. (The octet rule) Depict all covalent bonds by two shared electrons, giving as many atoms as possible a surrounding electron octet, except for H, which requires a duet. Make sure that the number of electrons used is exactly the number counted according to rule 2. Elements at the right in the periodic table may contain pairs of valence electrons not used for bonding, called lone electron pairs or just lone pairs. Nonbonding or lone electron pairs
šð Hj j Br Duet
Octet Octet
Correct Lewis Structure
Incorrect Lewis Structures
š HðBr ð
š jHðBr ð
šð HððBr
šð HðBr
j j
H
Consider, for example, hydrogen bromide. The shared electron pair supplies the hydrogen atom with a duet, the bromine with an octet, because the bromine carries three lone electron pairs. Conversely, in methane, the four C – H bonds satisfy the requirement of the hydrogens and, at the same time, furnish the octet for carbon. Examples of correct and incorrect Lewis structures for HBr are shown below.
j j
Hj jC H j j
3 electrons around H
No octet
H
Duets
4 electrons around H
Wrong number of electrons
Frequently, the number of valence electrons is not sufficient to satisfy the octet rule only with single bonds. In this event, double bonds (two shared electron pairs) and even triple bonds (three shared pairs) are necessary to obtain octets. An example is the nitrogen molecule, N2, which has ten valence electrons. An N – N single bond would leave both atoms with electron sextets, and a double bond provides only one nitrogen atom with an octet. It is the molecule with a triple bond that satisfies both atoms. You may find a simple procedure useful that gives you the total number of bonds needed in a molecule to give every atom an octet (or duet). Thus, after you have counted the supply of available electrons (rule 2), add up the total “electron demand,” that is, two electrons for each hydrogen atom and eight for each other element atom. Then subtract “demand” from supply and divide by 2. For N2, demand is 16 electrons, supply is 10, and hence the number of bonds is 3. Sextets
Octets Sextet Octet
jš ðš Nj Nð
ðš Nj jðNð
ðNðj jðNð
Single bond
Double bond
Triple bond
Further examples of molecules with double and triple bonds are shown below. Correct Lewis Structures H H
ýCððCý ½ ½
H
ý Ok HðCðððCðH
H
Ethene (Ethylene)
H Ethyne (Acetylene)
C½ ½š
H
Formaldehyde
In practice, another simple sequence may help. First, connect all mutually bonded atoms in your structure by single bonds (i.e., shared electron pairs); second, if there are any electrons left, distribute them as lone electron pairs to maximize the number of octets; and finally, if some of the atoms lack octet structures, change as many lone electron pairs into shared electron pairs as required to complete the octet shells (see also the Chapter Integration Problems 1-23 and 1-24).
1-4 Electron-Dot Model of Bonding: Lewis Structures
Chapter 1
Exercise 1-6 Draw Lewis structures for the following molecules: HI, CH3CH2CH3, CH3OH, HSSH, SiO2 (OSiO), O2, CS2 (SCS).
Rule 4. Assign (formal) charges to atoms in the molecule. Each lone pair contributes two electrons to the valence electron count of an atom in a molecule, and each bonding (shared) pair contributes one. An atom is charged if this total is different from the outer-shell electron count in the free, nonbonded atom. Thus we have the formula
number of outer-shell number of unshared 1 number of bonding Formal charge 5 ° electrons on the ¢ 2 ° electrons on the atom ¢ 2 ° electrons surrounding the ¢ 2 atom in the molecule in the molecule free, neutral atom or simply Formal charge 5 number of valence electrons 2 number of lone pair electrons 1 2 number of bonding electrons 2
The reason for the term formal is that, in molecules, charge is not localized on one atom but is distributed to varying degrees over its surroundings. As an example, which atom bears the positive charge in the hydronium ion? Each hydrogen has a valence electron count of 1 from the shared pair in its bond to oxygen. Because this value is the same as the electron count in the free atom, the (formal) charge on each hydrogen is zero. The electron count on the oxygen in the hydronium ion is 2 (the lone pair) 1 3 (half of 6 bonding electrons) 5 5. This value is one short of the number of outer-shell electrons in the free atom, thus giving the oxygen a charge of 11. Hence the positive charge is assigned to oxygen. Another example is the nitrosyl cation, NO1. The molecule bears a lone pair on nitrogen, in addition to the triple bond connecting the nitrogen to the oxygen atom. This gives nitrogen five valence electrons, a value that matches the count in the free atom; therefore the nitrogen atom has no charge. The same number of valence electrons (5) is found on oxygen. Because the free oxygen atom requires six valence electrons to be neutral, the oxygen in NO1 possesses the 11 charge. Other examples are shown below.
H H
⫹
ýCððCðH ½
Ethenyl (vinyl) cation
Hðš CðH H
H ⫹ Hðš NðH H
H Hðš Cðš Sð⫺ H
Methyl anion
Ammonium ion
Methanethiolate ion
⫺
⫹ý Ok
H
C½ ½š
H
H
Protonated formaldehyde
Sometimes the octet rule leads to charges on atoms even in neutral molecules. The Lewis structure is then said to be charge separated. An example is carbon monoxide, CO. Some compounds containing nitrogen – oxygen bonds, such as nitric acid, HNO3, also exhibit this property.
The octet rule does not always hold The octet rule strictly holds only for the elements of the second row and then only if there is a sufficient number of valence electrons to satisfy it. Thus, there are three exceptions to be considered.
⫹
Hðš O ðH H Hydronium ion
⫹
ðNðððOð Nitrosyl cation
Charge-separated Lewis structures ⫺
⫹
ðCqOð Carbon monoxide
š Oð k k Hðš O ð N ⫹ ⫺ ýO ½ Nitric acid
15
16
Chapter 1
Structure and Bonding in Organic Molecules
Exception 1. You will have noticed that all our examples of “correct” Lewis structures contain an even number of electrons; that is, all are distributed as bonding or lone pairs. This distribution is not possible in species having an odd number of electrons, such as nitrogen oxide (NO) and neutral methyl (methyl radical, ?CH3; see Section 3-1). Odd electron
Electron deficient
š ŠððO ðN
Š Hð CðH H
HðBeðH
Nitrogen oxide
Methyl radical
Beryllium hydride
H
ý Bk H
H
Borane
Exception 2. Some compounds of the early second-row elements, such as BeH2 and BH3, have a deficiency of valence electrons. Because compounds falling under exceptions 1 and 2 do not have octet configurations, they are unusually reactive and transform readily in reactions that lead to octet structures. For example, two molecules of ?CH3 react with each other spontaneously to give ethane, CH3 – CH3, and BH3 reacts with hydride, H2, to give borohydride, BH42. H H š Hðš C j ⫹ j CðH H H
H H Hðš Cðš CðH H H Ethane
H ½ H B ⫹ ðH⫺ H
H ⫺ Hðš BðH H Borohydride
Exception 3. Beyond the second row, the simple Lewis model is not strictly applied, and elements may be surrounded by more than eight valence electrons, a feature referred to as valence-shell expansion. For example, phosphorus and sulfur (as relatives of nitrogen and oxygen) are trivalent and divalent, respectively, and we can readily formulate Lewis octet structures for their derivatives. But they also form stable compounds of higher valency, among them the familiar phosphoric and sulfuric acids. Some examples of octet and expanded-octet molecules containing these elements are shown below.
š š ðCl PðCl ðš ð ðCl ð Phosphorous trichloride
ý Ok
Octet
Hðš Oðš Pðš O ðH Oð ð 10 electrons H Phosphoric acid
ð Oð š Hðš Oð Sðš OðH
Hðš SðH Octet
Hydrogen sulfide
ðš Oð
12 electrons
Sulfuric acid
An explanation for this apparent violation of the octet rule is found in a more sophisticated description of atomic structure by quantum mechanics (Section 1-6). However, you will notice that, even in these cases, you can construct dipolar forms in which the Lewis octet rule is preserved (see Section 1-5). 2⫹ ⫺
ðš Oð ⫹ š š Pðš Hð Oð OðH ð Oð H
⫺
šð ðO š Sðš HðO O ðš ðH ⫺ ðO ð
Covalent bonds can be depicted as straight lines Electron-dot structures can be cumbersome, particularly for larger molecules. It is simpler to represent covalent single bonds by single straight lines; double bonds are represented by
1-4 Electron-Dot Model of Bonding: Lewis Structures
two lines and triple bonds by three. Lone electron pairs can either be shown as dots or simply omitted. The use of such notation was first suggested by the German chemist August Kekulé,* long before electrons were discovered; structures of this type are often called Kekulé structures. Straight-Line Notation for the Covalent Bond H
H HO COH
ðNqNð
H Methane
Diatomic nitrogen
H H G D CPC D G H H Ethene
⫹D
HO½O G
⫹D
H
ýO
H
C D G H H
Hydronium ion
Protonated formaldehyde
Exercise 1-7
Working with the Concepts: Drawing Lewis Structures Draw the Lewis structure of HClO2 (HOClO), including the assignment of any charges to atoms. Strategy To solve such a problem, it is best to follow one by one the rules given in this section for drawing Lewis structures. Solution • Rule 1: The molecular skeleton is given as unbranched, as shown. • Rule 2: Count the number of valence electrons: H 5 1, 2 O 5 12, Cl 5 7, total 5 20 • Rule 3: How many bonds (shared electron pairs) do we need? The supply of electrons is 20; the electron requirement is 2 for H and 3 3 8 5 24 electrons for the other three atoms, for a total of 26 electrons. Thus we need (26 2 20)/2 5 3 bonds. To distribute all valence electrons according to the octet rule, we first connect all atoms by two-electron bonds, H: O: Cl: O, using up 6 electrons. Second, we distribute the remaining 14 electrons to provide octets for all nonhydrogen atoms, (arbitrarily) starting at the left oxygen. This process requires in turn 4, 4, and 6 electrons, resulting in octet structures without needing additional electron sharing: š š OðCl Hð Oð ðš • Rule 4: We determine any formal charges by noting any discrepancies between the “effective” valence electron count around each atom in the molecule we have found and its outer-shell count when isolated. For H in HOClO, the valence electron count is 1, which is the same as in the H atom, so it is neutral in the molecule. For the neighboring oxygen, the two values are again the same, 6. For Cl, the effective electron count is 6, but the neutral atom requires 7. Therefore, Cl bears a positive charge. For the terminal O, the electron counts are 7 (in the molecule) and 6 (neutral atom), giving it a negative charge. The final result is ⫹
⫺
š š OðCl Hð Oð ðš
Exercise 1-8
Try It Yourself Draw Lewis structures of the following molecules, including the assignment of any charges to atoms (the order in which the atoms are attached is given in parentheses, when it may not be obvious from the formula as it is commonly written): SO, F2O (FOF), BF3NH3 (F3BNH3), CH3OH21 (H3COH21), Cl2C “ O, CN2, C222. *Professor F. August Kekulé von Stradonitz (1829–1896), University of Bonn, Germany.
Chapter 1
17
18
Chapter 1
Structure and Bonding in Organic Molecules
In Summary Lewis structures describe bonding by the use of electron dots or straight lines. Whenever possible, they are drawn so as to give hydrogen an electron duet and other atoms an electron octet. Charges are assigned to each atom by evaluating its electron count.
1-5 Resonance Forms In organic chemistry, we also encounter molecules for which there are several correct Lewis structures.
The carbonate ion has several correct Lewis structures Let us consider the carbonate ion, CO322. Following our rules, we can easily draw a Lewis structure (A) in which every atom is surrounded by an octet. The two negative charges are located on the bottom two oxygen atoms; the third oxygen is neutral, connected to the central carbon by a double bond and bearing two lone pairs. But why choose the bottom two oxygen atoms as the charge carriers? There is no reason at all — it is a completely arbitrary choice. We could equally well have drawn structure B or C to describe the carbonate ion. The three correct Lewis pictures are called resonance forms. Resonance Forms of the Carbonate Ion šð O ⫺
ýO
C A
⫺ Ok
šð ðO
⫺
⫺
ýO
C O B
šð ðO
⫺
ýO
C
Red arrows denote the movement of electron pairs: “electron pushing” k⫺
O C
The individual resonance forms are connected by double-headed arrows and are placed within one set of square brackets. They have the characteristic property of being interconvertible by electron-pair movement only, indicated by red arrows, the nuclear positions in the molecule remaining unchanged. Note that, to turn A into B and then into C, we have to shift two electron pairs in each case. Such movement of electrons can be depicted by curved arrows, a procedure informally called “electron pushing.” The use of curved arrows to depict electron-pair movement is a useful technique that will prevent us from making the common mistake of changing the total number of electrons when we draw resonance forms. It is also advantageous in keeping track of electrons when formulating mechanisms (Sections 2-2 and 6-3).
But what is its true structure?
Carbonate ion
Does the carbonate ion have one uncharged oxygen atom bound to carbon through a double bond and two other oxygen atoms bound through a single bond each, both bearing a negative charge, as suggested by the Lewis structures? Or, to put it differently, are A, B, and C equilibrating isomers? The answer is no. If that were true, the carbon – oxygen bonds would be of different lengths, because double bonds are normally shorter than single bonds. But the carbonate ion is perfectly symmetrical and contains a trigonal central carbon, all C – O bonds being of equal length — between the length of a double and that of a single bond. The negative charge is evenly distributed over all three oxygens: It is said to be delocalized, in accord with the tendency of electrons to “spread out in space” (Section 1-2). In other words, none of the individual Lewis representations of this
1-5 Resonance Forms
molecule is correct on its own. Rather, the true structure is a composite of A, B, and C. The resulting picture is called a resonance hybrid. Because A, B, and C are equivalent (i.e., each is composed of the same number of atoms, bonds, and electron pairs), they contribute equally to the true structure of the molecule, but none of them by itself accurately represents it. Because it minimizes coulombic repulsion, delocalization by resonance has a stabilizing effect: The carbonate ion is considerably more stable than would be expected for a doubly negatively charged organic molecule. The word resonance may imply to you that the molecule vibrates or equilibrates from one form to another. This inference is incorrect. The molecule never looks like any of the individual resonance forms; it has only one structure, the resonance hybrid. Unlike substances in ordinary chemical equilibria, resonance forms are not real, although each makes a partial contribution to reality. The trigonal symmetry of carbonate is clearly evident in its electrostatic potential map shown on page 18. An alternative convention used to describe resonance hybrids such as carbonate is to represent the bonds as a combination of solid and dotted lines. The 232 sign here indicates that a partial charge ( 23 of a negative charge) resides on each oxygen atom (see margin). The equivalence of all three carbon – oxygen bonds and all three oxygens is clearly indicated by this convention. Other examples of resonance hybrids of octet Lewis structures are the acetate ion and the 2-propenyl (allyl) anion. ýOð
H HO CO C H
H
⫺
HO CO C
G ⫺ k ½O
ýO k D ½Oð
H
19
Chapter 1
Dotted-Line Notation of Carbonate as a Resonance Hybrid OV⫺ V⫺
O
C D G V⫺ O
H
U O−⫺ D or HO CO C G −⫺ OU H
Acetate ion
Acetate ion
H G C D H
D C G
H ⫺
šOH C
H H G D COC D⫺ H š COH
H
H
H H U G −⫺ D or C OC U G −⫺ D H C OH H 2-Propenyl anion
2-Propenyl (allyl) anion
Resonance is also possible for nonoctet molecules. For example, the 2-propenyl (allyl) cation is stabilized by resonance. H G C D H
H D C G⫹ C OH
H H G⫹ D COC D H COH
H
H
H H U D G −⫹ or C OC U G −⫹ D H C OH H
2-Propenyl cation
2-Propenyl (allyl) cation
When drawing resonance forms, keep in mind that (1) pushing one electron pair toward one atom and away from another results in a movement of charge; (2) the relative positions of all the atoms stay unchanged — only electrons are moved; (3) equivalent resonance forms contribute equally to the resonance hybrid; and (4) the arrows connecting resonance forms are double headed (4). The recognition and formulation of resonance forms is important in predicting reactivity. For example, reaction of carbonate with acid can occur at any two of the three oxygens to give carbonic acid, H2CO3 (which is actually in equilibrium with CO2 and H2O). Similarly, acetate ion is protonated at either oxygen to form acetic acid (see margin). Analogously, the 2-propenyl anion is protonated at either terminus to furnish propene, and
ðOð H
½O ½
C
½O ½
H
Carbonic acid
ðOð C H3C
½O ½
Acetic acid
H
20
Structure and Bonding in Organic Molecules
Chapter 1
the corresponding cation reacts with hydroxide at either end to give the corresponding alcohol (see below).
H G C D H
H D C ⫺ G šOH C
H H ⫺ D G š COC ⫹ H⫹ D COH H
H G C D H
H
H
H H H G D D OC C is the DC G D C OH COH same as H H DD H H H Propene
H G C D H
H D C G⫹ C OH
H H D G⫹ ⫹ HO⫺ COC D COH H
H G C D H
H
H
H HO H D G D C is the OC DC G D H same as C OH COOH H DD H H H 2-Propen-1-ol
Exercise 1-9 (a) Draw two resonance forms for nitrite ion, NO22. What can you say about the geometry of this molecule (linear or bent)? (Hint: Consider the effect of electron repulsion exerted by the lone pair on nitrogen.) (b) The possibility of valence-shell expansion increases the number of feasible resonance forms, and it is often difficult to decide on one that is “best.” One criterion that is used is whether the Lewis structure predicts bond lengths and bond angles with reasonable accuracy. Draw Lewis octet and valence-shell-expanded resonance forms for SO2 (OSO). Considering the Lewis structure for SO (Exercise 1-8), its experimental bond length of 1.48 Å, and the measured S–O distance in SO2 of 1.43 Å, which one of the various structures would you consider “best”?
Not all resonance forms are equivalent The molecules described above all have equivalent resonance forms. However, many molecules are described by resonance forms that are not equivalent. An example is the enolate ion. The two resonance forms differ in the locations of both the double bond and the charge. The Two Nonequivalent Resonance Forms of the Enolate Ion H H
Enolate ion
G C D
C
H D G⫺ šO C C D H ½Ok
H D G ⫺ ½Ok
H
Although both forms are contributors to the true structure of the ion, we shall see that one contributes more than the other. The question is, which one? If we extend our consideration of nonequivalent resonance forms to include those containing atoms without electron octets, the question becomes more general. [Octet mn Nonoctet] Resonance Forms
C⫹
C H
šð⫺ ðO
šð⫺ ðO
ðOð H
H
Formaldehyde
H
š HO
2⫹
S
ðOð š OH
š HO
⫺
ð Oð
S ðOð
Sulfuric acid
š OH
1-5 Resonance Forms
21
Chapter 1
Such an extension requires that we relax our definitions of “correct” and “incorrect” Lewis structures and broadly regard all resonance forms as potential contributors to the true picture of a molecule. The task is then to recognize which resonance form is the most important one. In other words, which one is the major resonance contributor? Here are some guidelines. Guideline 1. Structures with a maximum of octets are most important. In the enolate ion, all component atoms in either structure are surrounded by octets. Consider, however, the nitrosyl cation, NO1: One resonance form has a positive charge on oxygen with electron octets around both atoms; the other form places the positive charge on nitrogen, thereby resulting in an electron sextet on this atom. Because of the octet rule, the second structure contributes less to the hybrid. Thus, the N – O linkage is closer to being a triple than a double bond, and more of the positive charge is on oxygen than on nitrogen. Similarly, the dipolar resonance form for formaldehyde (p. 20) generates an electron sextet around carbon, rendering it a minor contributor. The possibility of valence shell expansion for third-row elements (Section 1-4) makes the non-charge-separated picture of sulfuric acid with 12 electrons around sulfur a feasible resonance form, but the dipolar octet structure is better.
⫹
ðN
Oð
Major resonance contributor
⫹
ðN
š O
Minor resonance contributor
Nitrosyl cation
Guideline 2. Charges should be preferentially located on atoms with compatible electronegativity. Consider again the enolate ion. Which is the major contributing resonance form? Guideline 2 requires it to be the first, in which the negative charge resides on the more electronegative oxygen atom. Indeed, the electrostatic potential map shown in the margin on p. 20 confirms this expectation. Looking again at NO1, you might find guideline 2 confusing. The major resonance contributor to NO1 has the positive charge on the more electronegative oxygen. In cases such as this, the octet rule overrides the electronegativity criterion; that is, guideline 1 takes precedence over guideline 2.
H H
G C D
H D C G ⫺ ½Ok
Major resonance contributor
Enolate ion
Guideline 3. Structures with less separation of opposite charges are more important resonance contributors than those with more charge separation. This rule is a simple consequence of Coulomb’s law: Separating opposite charges requires energy; hence neutral structures are better than dipolar ones. An example is formic acid, shown below. However, the influence of the minor dipolar resonance form is evident in the electrostatic potential map in the margin: The electron density at the carbonyl oxygen is greater than that at its hydroxy counterpart. šð⫺ ðO
ðOð C H
š O
H
C H
⫹
H
O
Major
Formic acid
Minor
Formic acid
In some cases, to draw octet Lewis structures, charge separation is necessary; that is, guideline 1 takes precedence over guideline 3. An example is carbon monoxide. Other examples are phosphoric and sulfuric acids, although valence-shell expansion allows the formulation of expanded octet structures (see also Section 1-4 and guideline 1). When there are several charge-separated resonance forms that comply with the octet rule, the most favorable is the one in which the charge distribution best accommodates the relative electronegativities of the component atoms (guideline 2). In diazomethane, for example, nitrogen is more electronegative than carbon, thus allowing a clear choice between the two resonance contributors (see also the electrostatic potential map in the margin). H C
⫹
N
H
⫺
H
⫺
š C
Nð
⫹
N
ðC
Ok
Minor
⫺
ðC
Major
Carbon monoxide
Nð
H Major
Minor
Diazomethane
⫹
Oð
Diazomethane
22
Chapter 1
Structure and Bonding in Organic Molecules
Exercise 1-10
Working with the Concepts: Drawing Resonance Forms Draw two all-octet resonance forms for nitrosyl chloride, ONCl. Which one is better? Strategy To formulate any octet structure, we follow the rules given in Section 1-4 for drawing Lewis structures. Once this task is completed, we can apply the procedures and guidelines from the present section to derive resonance forms and evaluate their relative contributions. Solution • Rule 1: The molecular skeleton is given as shown. • Rule 2: Count the number of valence electrons: N 5 5, O 5 6, Cl 5 7, total 5 18 • Rule 3: How many bonds (shared electron pairs) do we need? The supply of electrons is 18; and the electron requirement is 3 3 8 5 24 electrons for the three atoms. Thus we need (24 2 18)/2 5 3 bonds. Because there are only three atoms, there has to be a double bond. To distribute all valence electrons according to the octet rule, we first connect all atoms by two-electron bonds, O:N:Cl, using up 4 electrons. Second, we use 2 electrons for a double bond, arbitrarily added to the left portion to render O ::N:Cl. Third, we distribute the remaining 12 electrons to provide octets for all atoms, (again arbitrarily) starting at the left with oxygen. This .. .. .. process requires in turn 4, 2, and 6 electrons, resulting in the octet structure O . . :: N: Cl . . : , which we shall label A. • Rule 4: We determine any formal charges by noting any discrepancies between the “effective” valence electron count around each atom in A and its outer-shell count when it is isolated. For O, the two lone pairs and the double bond give a valence electron count of 6, as in the O atom; for N it is 5 and for Cl it is 7, again as found in the respective neutrals. Thus, there is no formal charge in A. • We are now ready to formulate resonance forms of A by moving electron pairs. You should try to do so for all such electrons. You will soon find that only one type of electron movement furnishes an all-octet structure, namely, that shown on the left below, leading to B:
šððš šð O NðCl š š
H H ⫹ šðš ðO NððClð Compare: CððCðš C⫺ š š H H
⫺
A
H
H ⫺š
CðCððC
H
H
B
Nitrosyl chloride
2-Propenyl (allyl) anion
This movement is similar to that in the 2-propenyl (allyl) anion (as shown on the right above) and related allylic resonance systems described in this section. Because we started with a chargeneutral formula, the electron movement to its resonance form generates charges: a positive one at the origin of the electron movement, a negative one at its terminus. • Which one of the two resonance forms of ONCl is better? Inspection of the three guidelines provided in this section helps us to the answer: Guideline 3 tells us that less charge separation is better than more. Thus, charge-neutral A is a better descriptor of nitrosyl chloride than chargeseparated B.
Exercise 1-11
Try It Yourself Draw resonance forms for the following two molecules. Indicate the more favorable resonance contributor in each case. (a) CNO2; (b) NO2.
1-6 Atomic Orbitals: A Quantum Mechanical Description
Chapter 1
23
To what extent can the assignment of a major resonance form help us predict reactivity? The answer is not simple, as it depends on the reagent, product stability, and other factors. For example, we shall see that the enolate ion can react at either oxygen or carbon with positively charged (or polarized) species (Section 18-1), even though, as shown earlier, more of the negative charge is at the oxygen. Another important example is the case of carbonyl compounds: Although the non-charge-separated form is dominant, the minor dipolar contributor, as shown earlier for formaldehyde, is the origin of the reactivity of the carbon– oxygen double bond, electron-rich species attacking the carbon, electron-poor ones the oxygen (Chapter 17).
In Summary Some molecules cannot be described accurately by one Lewis structure but exist as hybrids of several resonance forms. To find the most important resonance contributor, consider the octet rule, make sure that there is a minimum of charge separation, and place on the relatively more electronegative atoms as much negative and as little positive charge as possible.
1-6 Atomic Orbitals: A Quantum Mechanical Description of Electrons Around the Nucleus So far, we have considered bonds in terms of electron pairs arranged around the component atoms in such a way as to maximize noble-gas configurations (e.g., Lewis octets) and minimize electron repulsion. This approach is useful as a descriptive and predictive tool with regard to the number and location of electrons in molecules. However, it does not answer some simple questions that you may have asked yourself while dealing with this material. For example, why are some Lewis structures “incorrect” or, ultimately, why are noble gases relatively stable? Why are some bonds stronger than others, and how can we tell? What is so good about the two-electron bond, and what do multiple bonds look like? To get some answers, we start by learning more about how electrons are distributed around the nucleus, both spatially and energetically. The simplified treatment presented here has as its basis the theory of quantum mechanics developed independently in the 1920s by Heisenberg, Schrödinger, and Dirac.* In this theory, the movement of an electron around a nucleus is expressed in the form of equations that are very similar to those characteristic of waves. The solutions to these equations, called atomic orbitals, allow us to describe the probability of finding the electron in a certain region in space. The shapes of these regions depend on the energy of the electron.
The electron is described by wave equations The classical description of the atom (Bohr† theory) assumed that electrons move on trajectories around the nucleus. The energy of each electron was thought to relate to the electron’s distance from the nucleus. This view is intuitively appealing because it coincides with our physical understanding of classical mechanics. Yet it is incorrect for several reasons. First, an electron moving in an orbit would give rise to the emission of electromagnetic radiation, which is characteristic of any moving charge. The resulting energy loss from the system would cause the electron to spiral toward the nucleus, a prediction that is completely at odds with reality. Second, Bohr theory violates the Heisenberg uncertainty principle, because it defines the precise position and momentum of an electron at the same time.
*Professor Werner Heisenberg (1901–1976), University of Munich, Germany, Nobel Prize 1932 (physics); Professor Erwin Schrödinger (1887–1961), University of Dublin, Ireland, Nobel Prize 1933 (physics); Professor Paul Dirac (1902–1984), Florida State University, Tallahassee, Nobel Prize 1933 (physics). † Professor Niels Bohr (1885–1962), University of Copenhagen, Denmark, Nobel Prize 1922 (physics).
Nucleus
Neutron Proton Electron
Classical atom: electrons in “orbit” around the nucleus
24
Chapter 1
Structure and Bonding in Organic Molecules
A better model is afforded by considering the wave nature of moving particles. According to de Broglie’s* relation, a particle of mass m that moves with velocity v has a wavelength l. de Broglie Wavelength l5
h mv
in which h is Planck’s† constant. As a result, an orbiting electron can be described by equations that are the same as those used in classical mechanics to describe waves (Figure 1-4). These “matter waves” have amplitudes with alternating positive and negative signs. Points at which the sign changes are called nodes. Waves that interact in phase reinforce each other, as shown in Figure 1-4B. Those out of phase interfere with each other to make smaller waves (and possibly even cancel each other), as shown in Figure 1-4C. This theory of electron motion is called quantum mechanics. The equations developed in this theory, the wave equations, have a series of solutions called wave functions, usually described by the Greek letter psi, c. Their values around the nucleus are not directly identifiable with any observable property of the atom. However, the squares (c2) of their values at each point in space describe the probability of finding an electron at that point. The physical realities of the atom make solutions attainable only for certain specific energies. The system is said to be quantized. Exercise 1-12 Draw a picture similar to Figure 1-4 of two waves overlapping such that their amplitudes cancel each other.
*Prince Louis-Victor de Broglie (1892 –1987), Nobel Prize 1929 (physics). † Professor Max K. E. L. Planck (1858–1947), University of Berlin, Germany, Nobel Prize 1918 (physics).
Positive amplitude
+
Amplitude 0
+ −
Node
A
−
Negative amplitude
Note: The 1 and 2 signs in
Figure 1-4 refer to signs of the mathematical functions describing the wave amplitudes and have nothing to do with electrical charges.
Amplitude 0
+
−
B Figure 1-4 (A) A wave. The signs of the amplitude are assigned arbitrarily. At points of zero amplitude, called nodes, the wave changes sign. (B) Waves with amplitudes of like sign (in phase) reinforce each other to make a larger wave. (C) Waves that are out of phase subtract from each other to make a smaller wave.
Amplitude 0
+
+ −
−
In phase
+
Larger wave amplitude
+
−
+ −
−
−
+
−
Smaller wave amplitude C
Out of phase
1-6 Atomic Orbitals: A Quantum Mechanical Description
Chapter 1
25
Atomic orbitals have characteristic shapes Plots of wave functions in three dimensions typically have the appearance of spheres or dumbbells with flattened or teardrop-shaped lobes. For simplicity, we may regard artistic renditions of atomic orbitals as indicating the regions in space in which the electron is likely to be found. Nodes separate portions of the wave function with opposite mathematical signs. The value of the wave function at a node is zero; therefore, the probability of finding electron density there is zero. Higher-energy wave functions have more nodes than do those of low energy. Let us consider the shapes of the atomic orbitals for the simplest case, that of the hydrogen atom, consisting of a proton surrounded by an electron. The single lowest-energy solution of the wave equation is called the 1s orbital, the number 1 referring to the first (lowest) energy level. An orbital label also denotes the shape and number of nodes of the orbital. The 1s orbital is spherically symmetric (Figure 1-5) and has no nodes. This orbital can be represented pictorially as a sphere (Figure 1-5A) or simply as a circle (Figure 1-5B). The next higher-energy wave function, the 2s orbital, also is unique and, again, spherical. The 2s orbital is larger than the 1s orbital; the higher-energy 2s electron is on the average farther from the positive nucleus. In addition, the 2s orbital has one node, a spherical surface of zero electron density separating regions of the wave function of opposite sign (Figure 1-6). As in the case of classical waves, the sign of the wave function on either side of the node is arbitrary, as long as it changes at the node. Remember that the sign of the wave function is not related to “where the electron is.” As mentioned earlier, the probability of electron occupancy at any point of the orbital is given by the square of the value of the wave function. Moreover, the node does not constitute a barrier of any sort to the electron, which, in this description, is regarded not as a particle but as a wave. After the 2s orbital, the wave equations for the electron around a hydrogen atom have three energetically equivalent solutions, the 2px, 2py, and 2pz orbitals. Solutions of equal energy of this type are called degenerate (degenus, Latin, without genus or kind). As shown in Figure 1-7, p orbitals consist of two lobes that resemble a solid figure eight or dumbbell. A p orbital is characterized by its directionality in space. The orbital axis can be aligned with any one of the x, y, and z axes, hence the labels px, py, and pz. The two lobes of opposite sign of each orbital are separated by a nodal plane through the atom’s nucleus and perpendicular to the orbital axis. y
y
Hydrogen nucleus
+
x
z
x
Sign of wave function (not a charge) B
A y
ANIMATION: Fig. 1-5, 1s orbital
Figure 1-5 Representations of a 1s orbital. (A) The orbital is spherically symmetric in three dimensions. (B) A simplified twodimensional view. The plus sign denotes the mathematical sign of the wave function and is not a charge.
y Node Node
+ −
x
− +
x
z A
MATION AN I
B
Figure 1-6 Representations of a 2s orbital. Notice that it is larger than the 1s orbital and that a node is present. The 1 and 2 denote the signs of the wave function. (A) The orbital in three dimensions, with a section removed to allow visualization of the node. (B) The more conventional two-dimensional representation of the orbital.
26
Structure and Bonding in Organic Molecules
Chapter 1
y
y
2px
y
2py
2pz
+ +
−
x
z
−
x z
−
+
x
z
A y
y
y
+ +
−
z
− x
x z
−
+
x
z
B Figure 1-7 Representations of 2p orbitals (A) in three dimensions and (B) in two dimensions. Remember that the 1 and 2 signs refer to the wave functions and not to electrical charges. Lobes of opposite sign are separated by a nodal plane that is perpendicular to the axis of the orbital. For example, the px orbital is divided by a node in the yz plane. MATION AN I
ANIMATIONS: Fig. 1-7, px orbital; py orbital; pz orbital
The third set of solutions furnishes the 3s and 3p atomic orbitals. They are similar in shape to, but more diffuse than, their lower-energy counterparts and have two nodes. Still higher-energy orbitals (3d, 4s, 4p, etc.) are characterized by an increasing number of nodes and a variety of shapes. They are of much less importance in organic chemistry than are the lower orbitals. To a first approximation, the shapes and nodal properties of the atomic orbitals of other elements are very similar to those of hydrogen. Therefore, we may use s and p orbitals in a description of the electronic configurations of helium, lithium, and so forth.
The Aufbau principle assigns electrons to orbitals Approximate relative energies of the atomic orbitals up to the 5s level are shown in Figure 1-8. With its help, we can give an electronic configuration to every atom in the periodic table. To do so, we follow three rules for assigning electrons to atomic orbitals: That the description of electrons as waves is not simply a mathematical construct but is “visibly real” was demonstrated by researchers at IBM in 1993. Using a device called a scanning tunneling microscope, which allows pictures to be taken at the atomic level, they generated this computerenhanced view of a circle of iron atoms deposited on a copper surface. The image, which they called a “quantum corral,” reveals the electrons moving in waves over the surface, the maxima defining the “corral” hovering over the individual iron atoms. [Photograph courtesy of Dr. Donald Eigler, IBM, San Jose, California.]
1. Lower-energy orbitals are filled before those with higher energy. 2. No orbital may be occupied by more than two electrons, according to the Pauli* exclusion principle. Furthermore, these two electrons must differ in the orientation of their intrinsic angular momentum, their spin. There are two possible directions of the electron spin, usually depicted by vertical arrows pointing in opposite directions. An orbital is filled when it is occupied by two electrons of opposing spin, frequently referred to as paired electrons. 3. Degenerate orbitals, such as the p orbitals, are first occupied by one electron each, all of these electrons having the same spin. Subsequently, three more, each of opposite spin, are added to the first set. This assignment is based on Hund’s† rule.
*Professor Wolfgang Pauli (1900 – 1958), Swiss Federal Institute of Technology (ETH) Zurich, Switzerland, Nobel Prize 1945 (physics). † Professor Friedrich Hund (1896 – 1997), University of Göttingen, Germany.
1-6 Atomic Orbitals: A Quantum Mechanical Description
5s
4py
27
Figure 1-8 Approximate relative energies of atomic orbitals, corresponding roughly to the order in which they are filled in atoms. Orbitals of lowest energy are filled first; degenerate orbitals are filled according to Hund’s rule.
High
4px
Chapter 1
4pz 3d
4s Energy E
3px
3py
3pz
2px
2py
2pz
3s
2s 1s Low
With these rules in hand, the determination of electronic configuration becomes simple. Helium has two electrons in the 1s orbital and its electronic structure is abbreviated (1s)2. Lithium [(1s)2(2s)1] has one and beryllium [(1s)2(2s)2] has two additional electrons in the 2s orbital. In boron [(1s)2(2s)2(2p)1], we begin the filling of the three degenerate 2p orbitals. This pattern continues with carbon and nitrogen, and then the addition of electrons of opposite spin for oxygen, fluorine, and neon fills all p levels. The electronic configurations of four of the elements are depicted in Figure 1-9. Atoms with completely filled sets of atomic orbitals are said to have a closed-shell configuration. For example, Hund: 1 e each
2px E
2s 1s
2py
2pz
2px
1s
C
E
N
2py
2pz
2py
2px
2s
2s
1s
1s O
2pz
2s
Pauli: 2 e maximum occupancy
2px
2py
F
2pz
Figure 1-9 The most stable electronic configurations of carbon, (1s)2(2s)2(2p)2; nitrogen, (1s)2(2s)2(2p)3; oxygen, (1s)2(2s)2(2p)4; and fluorine (1s)2(2s)2(2p)5. Notice that the unpaired electron spins in the p orbitals are in accord with Hund’s rule, and the paired electron spins in the filled 1s and 2s orbitals are in accord with the Pauli principle and Hund’s rule. The order of filling the p orbitals has been arbitrarily chosen as px, py , and then pz. Any other order would have been equally good.
28
Chapter 1
Structure and Bonding in Organic Molecules
Figure 1-10 Closed-shell configurations of the noble gases helium, neon, and argon.
3p
3s
2p
E
1s He
2p
2s
2s
1s
1s Ne
Ar
helium, neon, and argon have this attribute (Figure 1-10). Carbon, in contrast, has an openshell configuration. The process of adding electrons one by one to the orbital sequence shown in Figure 1-8 is called the Aufbau principle (Aufbau, German, build up). It is easy to see that the Aufbau principle affords a rationale for the stability of the electron octet and duet. These numbers are required for closed-shell configurations. For helium, the closed-shell configuration is a 1s orbital filled with two electrons of opposite spin. In neon, the 2s and 2p orbitals are occupied by an additional eight electrons; in argon, the 3s and 3p levels accommodate eight more (Figure 1-10). The availability of 3d orbitals for the third-row elements provides an explanation for the phenomenon of valence-shell expansion (Section 1-4) and the loosening up of the strict application of the octet rule beyond neon. Experimental verification and quantification of the relative ordering of the orbital energy levels depicted in Figures 1-8 through 1-10 can be obtained by measuring the ionization potentials of the corresponding electrons, that is, the energies required to remove these electrons from their respective orbitals. It takes more energy to do so from a 1s orbital than from a 2s orbital; similarly, ejecting an electron from a 2s level is more difficult than from its 2p counterpart, and so on. This makes intuitive sense: As we go from lower- to higherlying orbitals, they become more diffuse and their associated electrons are located (on average) at increasing distances from the positively charged nucleus. Coulomb’s law tells us that such electrons become increasingly less “held” by the nucleus. Exercise 1-13 Using Figure 1-8, draw the electronic configurations of sulfur and phosphorus.
In Summary The motion of electrons around the nucleus is described by wave equations. Their solutions, atomic orbitals, can be symbolically represented as regions in space, with each point given a positive, negative, or zero (at the node) numerical value, the square of which represents the probability of finding the electron there. The Aufbau principle allows us to assign electronic configurations to all atoms.
1-7 Molecular Orbitals and Covalent Bonding We shall now see how covalent bonds are constructed from the overlap of atomic orbitals.
The bond in the hydrogen molecule is formed by the overlap of 1s atomic orbitals Let us begin by looking at the simplest case: the bond between the two hydrogen atoms in H2. In a Lewis structure of the hydrogen molecule, we would write the bond as an electron pair shared by both atoms, giving each a helium configuration. How do we construct H2 by
1-7 Molecular Orbitals and Covalent Bonding
High electron density
In phase
+ 1s
1s
or Bonding molecular orbital
Node
+
No electron density
Out of phase
Antibonding molecular orbital
using atomic orbitals? An answer to this question was developed by Pauling*: Bonds are made by the in-phase overlap of atomic orbitals. What does this mean? Recall that atomic orbitals are solutions of wave equations. Like waves, they may interact in a reinforcing way (Figure 1-4B) if the overlap is between areas of the wave function of the same sign, or in phase. They may also interact in a destructive way if the overlap is between areas of opposite sign, or out of phase (Figure 1-4C). The in-phase overlap of the two 1s orbitals results in a new orbital of lower energy called a bonding molecular orbital (Figure 1-11). In the bonding combination, the wave function in the space between the nuclei is strongly reinforced. Thus, the probability of finding the electrons occupying this molecular orbital in that region is very high: a condition for bonding between the two atoms. This picture is strongly reminiscent of that shown in Figure 1-2. The use of two wave functions with positive signs for representing the in-phase combination of the two 1s orbitals in Figure 1-11 is arbitrary. Overlap between two negative orbitals would give identical results. In other words, it is overlap between like lobes that makes a bond, regardless of the sign of the wave function. On the other hand, out-of-phase overlap between the same two atomic orbitals results in a destabilizing interaction and formation of an antibonding molecular orbital. In the antibonding molecular orbital, the amplitude of the wave function is canceled in the space between the two atoms, thereby giving rise to a node (Figure 1-11.) Thus, the net result of the interaction of the two 1s atomic orbitals of hydrogen is the generation of two molecular orbitals. One is bonding and lower in energy; the other is antibonding and higher in energy. Because the total number of electrons available to the system is only two, they are placed in the lower-energy molecular orbital: the two-electron bond. The result is a decrease in total energy, thereby making H2 more stable than two free hydrogen atoms. This difference in energy levels corresponds to the strength of the H – H bond. The interaction can be depicted schematically in an energy diagram (Figure 1-12A). It is now readily understandable why hydrogen exists as H2, whereas helium is monatomic. The overlap of two filled atomic orbitals, as in He2, with a total of four electrons, leads to bonding and antibonding orbitals, both of which are filled (Figure 1-12B). Therefore, making a He – He bond does not decrease the total energy.
The overlap of atomic orbitals gives rise to sigma and pi bonds The formation of molecular orbitals by the overlap of atomic orbitals applies not only to the 1s orbitals of hydrogen but also to other atomic orbitals. In general, overlap of any n atomic orbitals gives rise to n molecular orbitals. For a simple two-electron bond, n 5 2, and the two molecular orbitals are bonding and antibonding, respectively. The amount of energy by which the bonding level drops and the antibonding level is raised is called the *Professor Linus Pauling (1901–1994), Stanford University, Nobel Prizes 1954 (chemistry) and 1963 (peace).
Chapter 1
29
Figure 1-11 In-phase (bonding) and out-of-phase (antibonding) combinations of 1s atomic orbitals. The 1 and 2 signs denote the sign of the wave function, not charges. Electrons in bonding molecular orbitals have a high probability of occupying the space between the atomic nuclei, as required for good bonding (compare Figure 1-2). The antibonding molecular orbital has a nodal plane where the probability of finding electrons is zero. Electrons in antibonding molecular orbitals are most likely to be found outside the space between the nuclei and therefore do not contribute to bonding.
30
Chapter 1
Structure and Bonding in Organic Molecules
Figure 1-12 Schematic representation of the interaction of (A) two singly occupied atomic orbitals (as in H2) and (B) two doubly occupied atomic orbitals (as in He2) to give two molecular orbitals (MO). (Not drawn to scale.) Formation of an H–H bond is favorable because it stabilizes two electrons. Formation of an He–He bond stabilizes two electrons (in the bonding MO) but destabilizes two others (in the antibonding MO). Bonding between He and He thus results in no net stabilization. Therefore, helium is monatomic.
No energy absorbed
Unfilled antibonding molecular orbital
⫹ΔE E
H
H ⫺ΔE
Net energy decrease
Energy released
Filled bonding molecular orbital
A Filled antibonding molecular orbital
Energy absorbed
⫹ΔE He
E
He ⫺ΔE
No net change
Energy released
Filled bonding molecular orbital
B
energy splitting. It indicates the strength of the bond being made and depends on a variety of factors. For example, overlap is best between orbitals of similar size and energy. Therefore, two 1s orbitals will interact with each other more effectively than a 1s and a 3s. Geometric factors also affect the degree of overlap. This consideration is important for orbitals with directionality in space, such as p orbitals. Such orbitals give rise to two types of bonds: one in which the atomic orbitals are aligned along the internuclear axis (parts A, B, C, and D in Figure 1-13) and the other in which they are perpendicular to it (part E). The first type is called a sigma (S) bond, the second a pi (P) bond. All carbon – carbon single bonds are of the s type; however, we shall find that double and triple bonds also have p components (Section 1-8).
Figure 1-13 Bonding between atomic orbitals. (A) 1s and 1s (e.g., H2), (B) 1s and 2p (e.g., HF), (C) 2p and 2p (e.g., F2), (D) 2p and 3p (e.g., FCl) aligned along internuclear axes, s bonds; (E) 2p and 2p perpendicular to internuclear axis (e.g. H2CPCH2), a p bond. Note the arbitrary use of 1 and 2 signs to indicate in-phase interactions of the wave functions. In (D), also note the “figure 8 within a figure 8” dumbbell shape and more diffuse appearance of the 3p orbital when compared to its 2p counterpart.
1s 1s bond, as in H – H A
2 2p bond, as in H – F B
2 2p
Relatively more diffuse: weaker overlap
Parallel orientation: weaker overlap
1s
2 2p
3 3p bond, as in F – Cl D
2 2p bond, as in F – F C
22p 2p 2 bond, – CH2 as in H2C– E
1-8 Hybrid Orbitals: Bonding in Complex Molecules
Exercise 1-14
Working with the Concepts: Orbital Splitting Diagrams Construct a molecular-orbital and energy-splitting diagram of the bonding in He21. Is it favorable? Strategy To derive the molecular orbitals of the helium–helium bond, we first need to pick the appropriate atomic orbitals for overlap. The periodic table (Table 1-1) and the Aufbau principle (Figure 1-10) tell us that the orbital to use is 1s. Therefore, a bond between two He atoms is made in the same manner as the bond between two H atoms (Figure 1-11)—by the overlap of two 1s atomic orbitals. Solution • In-phase interaction results in the lower-energy (relative to the starting 1s orbital) bonding molecular orbital. Out-of-phase interaction produces the higher-energy antibonding molecular orbital. The resulting energy diagram is essentially identical to those shown in Figure 1-12A and B, except that He21 contains only three electrons. • The Aufbau principle instructs us to “fill from the bottom up.” Therefore two (bonding) electrons go into the lower energy level and one (antibonding) electron goes into the higher energy level.
He⫹
He
He2⫹
• The net effect is a favorable interaction (in contrast to neutral He2, Figure 1-12B). Indeed, He21 can be made in an electrical discharge of He1 with He, showing that bond formation is favorable.
Exercise 1-15
Try It Yourself Construct a molecular-orbital and energy-splitting diagram of the bonding in LiH. Is it favorable? (Caution: The overlapping orbital energies are not the same in this case. Hint: Consult Section 1-6, specifically the part describing the Aufbau principle. What are the electronic configurations of Li and H? The energy splitting between orbitals of unequal energies occurs such as to push the higher-lying level up, the lower-lying one down.)
In Summary We have come a long way in our description of bonding. First, we thought of bonds in terms of Coulomb forces, then in terms of covalency and shared electron pairs, and now we have a quantum mechanical picture. Bonds are a result of the overlap of atomic orbitals. The two bonding electrons are placed in the bonding molecular orbital. Because it is stabilized relative to the two initial atomic orbitals, energy is given off during bond formation. This decrease in energy represents the bond strength.
1-8 Hybrid Orbitals: Bonding in Complex Molecules Let us now use quantum mechanics to construct bonding schemes for more complex molecules. How can we use atomic orbitals to build linear (as in BeH2), trigonal (as in BH3), and tetrahedral molecules (as in CH4)?
Chapter 1
31
32
Chapter 1
Structure and Bonding in Organic Molecules
Figure 1-14 Promotion of an electron in beryllium to allow the use of both valence electrons in bonding.
2p
2p
2s
2s Energy
E 1s
1s
Be[(1s)2 (2s)2] No unpaired electrons
Be[(1s)2 (2s)1 (2p)1] Two unpaired electrons
2s
Figure 1-15 Possible but incorrect bonding in BeH2 by separate use of a 2s and a 2p orbital on beryllium. The node in the 2s orbital is not shown. Moreover, the other two empty p orbitals and the lower-energy filled 1s orbital are omitted for clarity. The dots indicate the valence electrons.
Be
+
2
2 2p
Electron
Be
1s
Electron
Incorrect structure
Consider the molecule beryllium hydride, BeH2. Beryllium has two electrons in the 1s orbital and two electrons in the 2s orbital. Without unpaired electrons, this arrangement does not appear to allow for bonding. However, it takes a relatively small amount of energy to promote one electron from the 2s orbital to one of the 2p levels (Figure 1-14) — energy that is readily compensated by eventual bond formation. Thus, in the 1s22s12p1 configuration, there are now two singly filled atomic orbitals available for bonding overlap. One could propose bond formation in BeH2 by overlap of the Be 2s orbital with the 1s orbital of one H and the Be 2p orbital with the second H (Figure 1-15). This scheme predicts two different bonds of unequal length, probably at an angle. However, if electron repulsion predicts that compounds such as BeH2 should have linear structures (Section 1-3). Experiments on related compounds confirm this prediction and also show that the bonds to beryllium are of equal length.*
sp Hybrids produce linear structures How can we explain this geometry in orbital terms? To answer this question, we use a quantum mechanical approach called orbital hybridization. Just as the mixing of atomic orbitals on different atoms forms molecular orbitals, the mixing of atomic orbitals on the same atom forms new hybrid orbitals. When we mix the 2s and one of the 2p wave functions on beryllium, we obtain two new hybrids, called sp orbitals, made up of 50% s and 50% p character. This treatment rearranges the orbital lobes in space, as shown in Figure 1-16. The major parts of the orbitals, also called front lobes, point away from each other at an angle of 1808. There are two additional minor back lobes (one for each sp hybrid) with opposite sign. The remaining two p orbitals are left unchanged. Overlap of the sp front lobes with two hydrogen 1s orbitals yields the bonds in BeH2. The 1808 angle that results from this hybridization scheme minimizes electron repulsion. The oversized front lobes of the hybrid orbitals also overlap better than do lobes of unhybridized orbitals; the result is energy reduction due to improved bonding. *These predictions cannot be tested for BeH2 itself, which exists as a complex network of Be and H atoms. However, both BeF2 and Be(CH3)2 exist as individual molecules in the gas phase and possess the predicted structures.
1-8 Hybrid Orbitals: Bonding in Complex Molecules
2s
Chapter 1
Front lobe
2 2p
Back lobe
2 Hybridization
Be
Be
Be
costs energy
Bonding releases energy
180°
A
sp
Beryllium hydride: linear
sp
2 2p
2 2p sp hybrid
Hybridization
2s
1s
1s
B Figure 1-16 Hybridization in beryllium to create two sp hybrids. (A) The resulting bonding gives BeH2 a linear structure. Again, both remaining p orbitals and the 1s orbital have been omitted for clarity. The sign of the wave function for the large sp lobes is opposite that for the small lobes. (B) The energy changes occurring on hybridization. The 2s orbital and one 2p orbital combine into two sp hybrids of intermediate energy. The 1s and remaining 2p energies remain the same.
MATION AN I
ANIMATION: Fig. 1-16, sp hybrids
Note that hybridization does not change the overall number of orbitals available for bonding. Hybridization of the four orbitals in beryllium gives a new set of four: two sp hybrids and two essentially unchanged 2p orbitals. We shall see shortly that carbon uses sp hybrids when it forms triple bonds.
sp 2 Hybrids create trigonal structures Now let us consider an element in the periodic table with three valence electrons. What bonding scheme can be derived for borane, BH3? Promotion of a 2s electron in boron to one of the 2p levels gives the three singly filled atomic orbitals (one 2s, two 2p) needed for forming three bonds. Mixing these atomic orbitals creates three new hybrid orbitals, which are designated sp2 to indicate the component atomic orbitals: 67% p and 33% s (Figure 1-17). The third p orbital is left unchanged, so the total number of orbitals stays the same—namely, four. Front lobe 2
sp
2py
H
Back lobe
+ Hybridization
B
120° costs energy 2px
− B +
3
−
+
−
120°
2s
H+ B
sp2
H
Bonding releases energy
H sp2
Boron hydride: trigonal
Figure 1-17 Hybridization in boron to create three sp2 hybrids. The resulting bonding gives BH3 a trigonal planar structure. There are three front lobes of one sign and three back lobes of opposite sign. The remaining p orbital (pz) is perpendicular to the molecular plane (the plane of the page; one pz lobe is above, the other below that plane) and has been omitted. In analogy to Figure 1-16B, the energy diagram for the hybridized boron features three singly occupied, equal-energy sp2 levels and one remaining empty 2p level, in addition to the filled 1s orbital.
MATION AN I
ANIMATION: Fig. 1-17, sp 2 hybrids
33
34
Structure and Bonding in Organic Molecules
Chapter 1
The front lobes of the three sp2 orbitals of boron overlap the respective 1s orbitals of the hydrogen atoms to give trigonal planar BH3. Again, hybridization minimizes electron repulsion and improves overlap, conditions giving stronger bonds. The remaining unchanged p orbital is perpendicular to the plane incorporating the sp2 hybrids. It is empty and does not enter significantly into bonding. The molecule BH3 is isoelectronic with the methyl cation, CH31; that is, they have the same number of electrons. Bonding in CH31 requires three sp2 hybrid orbitals, and we shall see shortly that carbon uses sp2 hybrids in double-bond formation.
sp3 Hybridization explains the shape of tetrahedral carbon compounds Consider the element whose bonding is of most interest to us: carbon. Its electronic configuration is (1s)2(2s)2(2p)2, with two unpaired electrons residing in two 2p orbitals. Promotion of one electron from 2s to 2p results in four singly filled orbitals for bonding. We have learned that the arrangement in space of the four C – H bonds of methane that would minimize electron repulsion is tetrahedral (Section 1-3). To be able to achieve this geometry, the 2s orbital on carbon is hybridized with all three 2p orbitals to make four equivalent sp3 orbitals with tetrahedral symmetry, each of 75% p and 25% s character and occupied by one electron. Overlap with four hydrogen 1s orbitals furnishes methane with four equal C – H bonds. The HCH bond angles are typical of a tetrahedron: 109.58 (Figure 1-18). H
sp3
2py
+
109.5° Hybridization
C
+
costs energy sp3
2s
H+
4
C
+
Bonding releases energy
+
2px
2pz
− −− C −
ANIMATIONS: Fig. 1-18, sp 3 hybrids and methane orbitals
H H
sp3 3
sp MATION AN I
H
109.5°
Methane: tetrahedral
Figure 1-18 Hybridization in carbon to create four sp hybrids. The resulting bonding gives CH4 and other carbon compounds tetrahedral structures. The sp3 hybrids contain small back lobes of sign opposite that of the front lobes. In analogy to Figure 1-16B, the energy diagram of sp3-hybridized carbon contains four singly occupied, equal-energy sp3 levels, in addition to the filled 1s orbital. 3
Any combination of atomic and hybrid orbitals may overlap to form bonds. For example, the four sp3 orbitals of carbon can combine with four chlorine 3p orbitals, resulting in tetrachloromethane, CCl4. Carbon – carbon bonds are generated by overlap of hybrid orbitals. In ethane, CH3 – CH3 (Figure 1-19), this bond consists of two sp3 hybrids, one from each of two CH3 units. Any hydrogen atom in methane and ethane may be replaced by CH3 or other groups to give new combinations. In all of these molecules, and countless more, carbon is approximately tetrahedral. It is this ability of carbon to form chains of atoms bearing a variety of additional substituents that gives rise to the extraordinary diversity of organic chemistry.
MATION AN I
ANIMATION: Fig. 1-19, ethane orbitals
Figure 1-19 Overlap of two sp3 orbitals to form the carbon–carbon bond in ethane.
2
C
C
C
Ethane
1-8 Hybrid Orbitals: Bonding in Complex Molecules
Chapter 1
35
Hybrid orbitals may contain lone electron pairs: ammonia and water What sort of orbitals describe the bonding in ammonia and water (see Exercise 1-5)? Let us begin with ammonia. The electronic configuration of nitrogen, (1s)2(2s)2(2p)3, explains why nitrogen is trivalent, three covalent bonds being needed for octet formation. We could use p orbitals for overlap, leaving the nonbonding electron pair in the 2s level. However, this arrangement does not minimize electron repulsion. The best solution is again sp3 hybridization. Three of the sp3 orbitals are used to bond to the hydrogen atoms, and the fourth contains the lone electron pair. The HNH bond angles (107.38) in ammonia are almost tetrahedral (Figure 1-20). The effect of the lone electron pair is to reduce the bond angles in NH3 from the ideal tetrahedral value of 109.58. Because it is not shared, the lone pair is relatively close to the nitrogen. As a result, it exerts increased repulsion on the electrons in the bonds to hydrogen, thereby leading to the observed bond-angle compression. In principle, the bonding in water could also be described by formal sp3 hybridization on oxygen. However, the energetic cost is now too large (see Exercise 1-17). Nevertheless, for the sake of simplicity, we can extend the picture for ammonia to water as in Figure 1-20, with an HOH bond angle of 104.58. Lone electron pair in sp3 orbital
N H H
Lone electron pairs
O
Electron repulsion
H
Figure 1-20 Bonding and electron repulsion in ammonia and water. The arcs indicate increased electron repulsion by the lone pairs located close to the central nucleus.
H H
107.3°
Ammonia
104.5°
Water
Pi bonds are present in ethene (ethylene) and ethyne (acetylene) The double bond in alkenes, such as ethene (ethylene), and the triple bond in alkynes, such as ethyne (acetylene), are the result of the ability of the atomic orbitals of carbon to adopt sp2 and sp hybridization, respectively. Thus, the s bonds in ethene are derived entirely from carbon-based sp2 hybrid orbitals: Csp2 – Csp2 for the C – C bond, and Csp2 – H1s for holding the four hydrogens (Figure 1-21). In contrast to BH3, with an empty p orbital on boron, the leftover unhybridized p orbitals on the ethene carbons are occupied by one electron each, overlapping to form a p bond (recall Figure 1-13E). In ethyne, the s frame is made up of bonds consisting of Csp hybrid orbitals. The arrangement leaves two singly occupied p orbitals on each carbon and allows the formation of two p bonds (Figure 1-21). π bond π bond σ bond
H C
H C
120°
H
C
C
H
H
H
2p orbital
180°
sp2 orbital
Ethene
π bond
σ bond
Ethyne
Figure 1-21 The double bond in ethene (ethylene) and the triple bond in ethyne (acetylene).
MATION AN I
ANIMATIONS: Fig. 1-21, ethene and ethyne orbitals
36
Structure and Bonding in Organic Molecules
Chapter 1
Exercise 1-16 Draw a scheme for the hybridization and bonding in methyl cation, CH31, and methyl anion, CH32.
Exercise 1-17
Working with the Concepts: Orbital Overlap in Water
2 y 2p
O 2 x 2p 2 z 2p
2s (lone pair)
Water
Although it was convenient to describe water as containing an sp3-hybridized oxygen atom, hybridization is unfavorable when compared to C in methane and N in ammonia. The reason is that the energy difference between the 2s and p orbitals in O is now so large that the energetic cost of hybridization can no longer be matched by the smaller number of bonds to H atoms (two, instead of four or three). Instead, the oxygen uses (essentially) unhybridized orbitals. Why is there a larger energy separation between the 2s and p orbitals in O? (Hint: As you proceed horizontally from C to F in the periodic table, the positive nuclear charge increases steadily. To review what effect this has on the energies of orbitals, consult the end of Section 1-6.) Why should this phenomenon affect the process of hybridization adversely? Ponder these questions and then draw a picture of the orbitals in unhybridized O and their bonding in water. Use electron-repulsion arguments to explain the HOH bond angle of 104.58. Strategy The intraatomic overlap of orbitals underlying hybridization is affected by the same features governing the quality of a bond made by interatomic overlap of orbitals: Overlap is best between orbitals of similar size and energy. Also, bonding is subject to the geometric constraints imposed by the molecule containing the hybridized atom (see also Problem 48). Proceeding along the series methane, ammonia, water, and hydrogen fluoride, the spherically symmetric 2s orbital is lowered more in energy than the corresponding p orbitals are by the increasing nuclear charge. One way to understand this trend is to picture the electrons in their respective orbitals: Those relatively closer to the nucleus (2s) are held increasingly more tightly than those farther away (2p). To the right of N in the periodic table, this discrepancy in energy makes hybridization of the orbitals difficult, as this process effectively moves the 2s electrons farther away from the nucleus. Thus the reduction in electron repulsion by hybridization is offset by the cost in Coulomb energy. However, we can still draw a reasonable overlap picture. Solution • We use the two singly occupied p orbitals on oxygen to overlap with the two respective hydrogen 1s orbitals (margin). In this picture, the two lone electron pairs reside in a p and the 2s orbital, respectively. • Why, then, is the bond angle in water not 908? Well, Coulomb’s law (electron repulsion) still operates, whether we hybridize or not. Thus the two pairs of bonding electrons increase their distance by distorting bond angles to the observed value.
Exercise 1-18
Try It Yourself Extrapolate the picture for water in the preceding exercise to the bonding in HF, which also uses unhybridized orbitals.
In Summary To minimize electron repulsion and maximize bonding in triatomic and larger molecules, we apply the concept of atomic-orbital hybridization to construct orbitals of appropriate shape. Combinations of s and p atomic orbitals create hybrids. Thus, a 2s and a 2p orbital mix to furnish two linear sp hybrids, the remaining two p orbitals being unchanged. Combination of the 2s with two p orbitals gives three sp2 hybrids used in trigonal molecules. Finally, mixing the 2s with all three p orbitals results in the four sp3 hybrids that produce the geometry around tetrahedral carbon.
1-9 Structures and Formulas of Organic Molecules
37
Chapter 1
1-9 Structures and Formulas of Organic Molecules A good understanding of the nature of bonding allows us to learn how chemists determine the identity of organic molecules and depict their structures. Do not underestimate the importance of drawing structures. Sloppiness in drawing molecules has been the source of many errors in the literature and, perhaps of more immediate concern, in organic chemistry examinations.
Ethanol and Methoxymethane: Two Isomers
To establish the identity of a molecule, we determine its structure Organic chemists have many diverse techniques at their disposal with which to determine molecular structure. Elemental analysis reveals the empirical formula, which summarizes the kinds and ratios of the elements present. However, other procedures are usually needed to determine the molecular formula and to distinguish between structural alternatives. For example, the molecular formula C2H6O corresponds to two known substances: ethanol and methoxymethane (dimethyl ether). We can tell them apart on the basis of their physical properties — for example, their melting points, boiling points (b.p.’s), refractive indices, specific gravities, and so forth. Thus ethanol is a liquid (b.p. 78.58C) commonly used as a laboratory and industrial solvent and present in alcoholic beverages. In contrast, methoxymethane is a gas (b.p. 2238C) used as a refrigerant. Their other physical and chemical properties differ as well. Molecules such as these, which have the same molecular formula but differ in the sequence (connectivity) in which the atoms are held together, are called constitutional or structural isomers (see also Chemical Highlight 1-1).
H H A A H O CO CO O OH A A H H Ethanol (b.p. 78.5°C)
H H A A H O C O O O C OH A A H H Methoxymethane (Dimethyl ether) (b.p. ⫺23°C)
Exercise 1-19 Draw the two constitutional isomers with the molecular formula C4H10, showing all atoms and their bonds.
COOH O
Two naturally occurring substances illustrate the biological consequences of such structural differences. Prostacyclin I2 prevents blood inside the circulatory system from clotting. Thromboxane A2, which is released when bleeding occurs, induces platelet aggregation, causing clots to form over wounds. Incredibly, these compounds are constitutional isomers (both have the molecular formula C20H32O5) with only relatively minor connectivity differences. Indeed, they are so closely related that they are synthesized in the body from a common starting material (see Section 19-13 for details). When a compound is isolated in nature or from a reaction, a chemist may attempt to identify it by matching its properties with those of known materials. Suppose, however, that the chemical under investigation is new. In this case, structural elucidation requires the use of other methods, most of which are various forms of spectroscopy. These methods will be dealt with and applied often in later chapters. The most complete methods for structure determination are X-ray diffraction of single crystals and electron diffraction or microwave spectroscopy of gases. These techniques reveal the exact position of every atom, as if viewed under very powerful magnification. The structural details that emerge in this way for the two isomers ethanol and methoxymethane are depicted in the form of ball-and-stick models in Figure 1-22A and B.
107.2°
109.5°
´
≥ HO
108.7°
O
A
B
∞ O -
∑O
O
Figure 1-22 Three-dimensional representations of (A) ethanol and (B) methoxymethane, depicted by ball-and-stick molecular models. Bond lengths are given in angstrom units, bond angles in degrees. (C) Space-filling rendition of methoxymethane, taking into account the effective size of the electron “clouds” around the component nuclei.
COOH ≥ OH
Thromboxane A2
1.41 Å
C
≥ OH
´
1.10 Å
110.8°
-
Prostacyclin I2
111.7°
O
C
38
Chapter 1
MODEL BUILDING
Structure and Bonding in Organic Molecules
Note the tetrahedral bonding around the carbon atoms and the bent arrangement of the bonds to oxygen, which is similar to that in water. A more accurate picture of the relative size of the component atoms in methoxymethane is given in Figure 1-22C, a spacefilling model. The visualization of organic molecules in three dimensions is essential for understanding their structures and, frequently, their reactivities. You may find it difficult to visualize the spatial arrangements of the atoms in even very simple systems. A good aid is a molecular model kit. You should acquire one and practice the assembly of organic structures. To encourage you in this practice and to indicate particularly good examples where building a molecular model can help you, the icon displayed in the margin will appear at the appropriate places in the text. Exercise 1-20 Use your molecular model kit to construct the two isomers with the molecular formula C4H10.
Several types of drawings are used to represent molecular structures The representation of molecular structures was first addressed in Section 1-4, which outlined rules for drawing Lewis structures. We learned that bonding and nonbonding electrons are depicted as dots. A simplification is the straight-line notation (Kekulé structure), with lone pairs (if present) added again as dots. To simplify even further, chemists use condensed formulas in which most single bonds and lone pairs are omitted. The main carbon chain is written horizontally, the attached hydrogens usually to the right of the associated carbon atom. Other groups (the substituents on the main stem) are added through connecting vertical lines. The most economical notation of all is the bond-line formula. It portrays the carbon frame by zigzag straight lines, omitting all hydrogen atoms. Each unsubstituted terminus represents a methyl group, each apex a carbon atom, and all unspecified valences are assumed to be satisfied by single bonds to hydrogen. Kekulé
H
H
H
H
H
C
C
C
H
H
H
Condensed
CH3CH2CH3
H
šð H H ðBr
H
C
C
C
C
H
H
H
H
C
C
Br
Br š Brð Br š
H ýO k H H
H C
Br
CH3CHCH2CH2Br
O
C
Bond-Line Formulas
CH3CCH
O CH2
H H H H
C qC
C H
š OH š
HCq CCH2OH
OH
The Big Picture
´C%
´C%
HH HH ´C% H
A
H B
H H H ∑ i [ C OC i H & H H
H H H ∑ i [ C OC i ½ H & H ½OH
C
D
∑
0
´C%
Δ C
∑
´% C
H
Chapter 1
ýO k H ´C% ´C% H H H H
Figure 1-23 Hashed (red) and wedged (blue) line notation for (A) a carbon chain; (B) methane; (C) ethane; (D) ethanol; and (E) methoxymethane. Atoms attached by ordinary straight lines lie in the plane of the page. Groups at the ends of hashed lines lie below that plane; groups at the ends of wedges lie above it.
Exercise 1-21 Draw condensed and bond-line formulas for each C4H10 isomer.
Figure 1-22 calls attention to a problem: How can we draw the three-dimensional structures of organic molecules accurately, efficiently, and in accord with generally accepted conventions? For tetrahedral carbon, this problem is solved by the hashed-wedged/solid-wedged line notation. It uses a zigzag convention to depict the main carbon chain, now defined to lie in the plane of the page. Each apex (carbon atom) is then connected to two additional lines, one hashed wedged and one solid wedged, both pointing away from the chain. These represent the remaining two bonds to carbon; the hashed-wedged line corresponds to the bond that lies below the plane of the page, and the solid-wedged line to that lying above that plane (Figure 1-23). Substituents are placed at the appropriate termini. This convention is applied to molecules of all sizes, even methane (see Figure 1-23B – E). For simplicity, we will refer to the two types of bonds as “hashed” (instead of “hashed wedged”) and “wedged” (instead of “solid wedged”). Exercise 1-22 Draw hashed-wedged line formulas for each C4H10 isomer.
In Summary Determination of organic structures relies on the use of several experimental techniques, including elemental analysis and various forms of spectroscopy. Molecular models are useful aids for the visualization of the spatial arrangements of the atoms in structures. Condensed and bond-line notations are useful shorthand approaches to drawing two-dimensional representations of molecules, whereas hashed-wedged line formulas provide a means of depicting the atoms and bonds in three dimensions.
THE BIG PICTURE What have we learned and where do we go from here? Much of the material covered in this introductory chapter is probably familiar to you from introductory chemistry or even high school, perhaps in a different context. Here, the purpose was to recapitulate this knowledge as it pertains to the structure and reactivity of organic molecules. The fundamental take-home lessons for organic chemistry are these: 1. The importance of Coulomb’s law (Section 1-2), as evidenced, for example, by atomic
attraction (Section 1-3), relative electronegativity (Table 1-2), the electron repulsion model for the shapes of molecules (Section 1-3), and the choice of dominant resonance contributors (Section 1-5). 2. The tendency of electrons to spread out (delocalize), as manifested in resonance forms
(Section 1-5) or bonding overlap (Section 1-7). 3. The correlation of the valence electron count (Sections 1-3 and 1-4) with the Aufbau
principle (Section 1-6), and the associated stability of the elements in noble-gas – octet– closed-shell configurations obtained by bond formation (Sections 1-3, 1-4, and 1-7).
E
39
40
Chapter 1
Structure and Bonding in Organic Molecules
4. The characteristic shapes of atomic and molecular orbitals (Section 1-6), which provide
a feeling for the location of the “reacting” electrons around the nuclei. 5. The overlap model for bonding (Section 1-7), which allows us to make judgments with
respect to the energetics, directions, and overall feasibility of reactions. With all this information in our grasp, we now have the tools to examine the structural and dynamic diversity of organic molecules, as well as their sites of reactivity.
CHAPTER INTEGRATION PROBLEMS 1-23. Sodium borohydride, Na12BH4, is a reagent used in the conversion of aldehydes and ketones to alcohols (Section 8-6). It can be made by treating BH3 with Na1H2: BH3 ⫹ Na⫹H⫺
H Na⫹ H B H H
⫺
a. Draw the Lewis structure of 2BH4. SOLUTION We follow the rules of Section 1-4. Step 1. The molecular skeleton (Rule 1) is indicated in the bracketed part of the above equation: a boron atom surrounded by four hydrogen atoms. Step 2. What is the number of valence electrons? Answer (Rule 2): 4H 5 4 3 1 electron 5 4 electrons 1B 5 3 electrons Charge 5 21 5 1 electron Total
8 electrons
Step 3. What is the Lewis structure? Answer (Rule 3): This one is simple; just placing two electrons each between boron and its four bonded hydrogens consumes all the valence electrons at our disposal: H š BðH Hð H Step 4. Where is the charge? Answer (Rule 4): Because each hydrogen has a valence electron count of one, namely, that of the neutral atom, the charge has to reside on boron. This conclusion can be verified by counting the valence electrons around boron: four, one more than the number associated with the atom. The correct Lewis structure is therefore H ⫺ š BðH Hð H
H b. What is the shape of the borohydride ion? B− H
H H Borohydride ion
SOLUTION For the answer, we have to remember that electron repulsion controls the shape of simple molecules (Section 1-3). The best arrangement for 2BH4 is therefore the same as that for CH4: tetrahedral and, hence, sp3 hybridized (margin). c. Draw an orbital picture of the attack of H:2 on BH3. Which are the orbitals involved in overlap for bond formation?
Chapter Integration Problems
SOLUTION To approach the answer to this question, let us draw the orbital pictures of the species involved (Sections 1-6 and 1-8). For H:2, it is a doubly occupied 1s orbital; for BH3 it is an sp2-hybridized, trigonal boron with its characteristic three bonds to hydrogen and a remaining empty 2p orbital perpendicular to the molecular plane (Figure 1-17). sp2
Attack
H 1s
H−
H
sp2
B
H
Hydride ion 2
sp
2p (empty orbital, ready
to accept electrons from hydride ion) Boron hydride (borane)
Which part of BH3 is the hydride ion most likely to attack? Hybridization provides the answer: the empty p orbital. You can readily imagine how the boron atom undergoes rehybridization from sp2 to sp3 during the course of the reaction. The initial overlap of the hydrogen 1s orbital with the boron 2p orbital changes to the final overlap of the hydrogen 1s with the boron sp3 hybrid orbital. 1-24. Propyne can be deprotonated twice with very strong base (i.e., the base removes the two protons labeled by arrows) to give a dianion. H H
CqC
C
H
strong base (⫺2H⫹)
CCCH2
2⫺
H Propyne
Propyne dianion
Two resonance forms of the anion can be constructed in which all three carbons have Lewis octets. a. Draw both structures and indicate which is the more important one. SOLUTION Let us analyze the problem one step at a time: Step 1. What structural information is embedded in the picture given for propyne dianion? Answer (Section 1-4, Rule 1): The picture shows the connectivity of the atoms: a chain of three carbons, one of the terminal atoms bearing two hydrogens. Step 2. How many valence electrons are available? Answer (Section 1-4, Rule 2): 2H 5 2 3 1 electron 5 2 electrons 3C 5 3 3 4 electrons 5 12 electrons Charge 5 22 5 2 electrons Total
16 electrons
Step 3. How do we get a Lewis octet structure for this ion? Answer (Section 1-4, Rule 3): Using the connectivity given in the structure for propyne dianion, we can immediately dispose of eight of the available electrons: H CðCðš CðH Now, let us use the remaining eight electrons in the form of lone electron pairs to surround as many carbons as possible with octets. A good place to start is at the right, because that carbon requires
Chapter 1
41
42
Chapter 1
Structure and Bonding in Organic Molecules
only two electrons for this purpose. The center carbon needs two lone pairs and the carbon at the left has to make do (for the time being) with one additional pair of electrons: H š ðCð Cðš CðH This structure leaves the carbon at the left with only four electrons. Thus, we have to change the two lone pairs at the center into two shared pairs, furnishing the following dot structure: H ðCðððCðš CðH Step 4. Now every atom has its duet or octet satisfied, but we still have to concern ourselves with charges. What are the charges on each atom? Answer (Section 1-4, Rule 4): Starting again at the right, we can quickly ascertain that the hydrogens are charge neutral. Each is attached to carbon through a shared electron pair, giving it an effective electron count of one, the same as in a free, neutral hydrogen atom. On the other hand, the carbon atom bears three shared electron pairs and one lone pair, thus having an effective electron count of five, one more electron than the number associated with the neutral nucleus. Hence one of the negative charges is located on the carbon at the right. The central carbon nucleus is surrounded by four shared electron pairs and is therefore neutral. Finally, the carbon at the left is attached to its neighbor by three shared pairs and has, in addition, two unbound electrons, giving it the other negative charge. The result is H ⫺ ðCðððCðš CðH ⫺
Step 5. We can now address the question of resonance forms. Is it possible to move pairs of electrons in such a way as to generate another Lewis octet form? Answer (Section 1-5): Yes, by shifting the lone pair on the carbon at the right into a sharing position and at the same time moving one of the three shared pairs to the left into an unshared location: H ⫺ ðCðððCðš CðH ⫺
H 2⫺ CððCððš CðH ð
The consequence of this movement is the transfer of the negative charge from the carbon at the right to the carbon at the left, which therefore becomes doubly negative. Step 6. Which one of the two resonance pictures is more important? Answer (Section 1-5): Electron repulsion makes the structure at the left a more important resonance contributor. A final point: You could have derived the first resonance structure much more quickly by considering the information given in the reaction scheme leading to the dianion. Thus, the bond-line formula of propyne represents its Lewis structure and the process of removing a proton each from the respective terminal carbons leaves these carbons with two lone electron pairs and the associated charges: H A HO Cq CO COH A H
⫺2H⫹
H A C OH ðCqCO ⫺
⫺
The important lesson to be learned from this final point is that, whenever you are confronted with a problem, you should take the time to complete an inventory (write it down) of all the information given explicitly or implicitly in stating the problem. b. Propyne dianion adopts the following hybridization: [CspCspCsp2H2]22, in which the CH2 terminus is sp2 hybridized, unlike methyl anion, in which the carbon is sp3 hybridized (Exercise 1-16). Provide an orbital drawing of the dianion to explain this preference in hybridization.
Important Concepts
SOLUTION You can construct an orbital picture simply by attaching one half (the CH2 group) of the rendition of ethene in Figure 1-21 to that of ethyne without its hydrogens.
− H
− C
C
C H
You can clearly see how the doubly filled p orbital of the CH2 group enters into overlap with one of the p bonds of the alkyne unit, allowing the charge to delocalize as represented by the two resonance structures.
Important Concepts 1. Organic chemistry is the chemistry of carbon and its compounds. 2. Coulomb’s law relates the attractive force between particles of opposite electrical charge to the distance between them. 3. Ionic bonds result from coulombic attraction of oppositely charged ions. These ions are formed by the complete transfer of electrons from one atom to another, typically to achieve a noble-gas configuration. 4. Covalent bonds result from electron sharing between two atoms. Electrons are shared to allow the atoms to attain noble-gas configurations. 5. Bond length is the average distance between two covalently bonded atoms. Bond formation releases energy; bond breaking requires energy. 6. Polar bonds are formed between atoms of differing electronegativity (a measure of an atom’s ability to attract electrons). 7. The shape of molecules is strongly influenced by electron repulsion. 8. Lewis structures describe bonding by the use of valence electron dots. They are drawn so as to give hydrogen an electron duet and the other atoms electron octets (octet rule). Formal charge separation should be minimized but may be enforced by the octet rule. 9. When two or more Lewis structures differing only in the positions of the electrons are needed to describe a molecule, they are called resonance forms. None correctly describes the molecule, its true representation being an average (hybrid) of all its Lewis structures. If the resonance forms of a molecule are unequal, those which best satisfy the rules for writing Lewis structures and the electronegativity requirements of the atoms are more important. 10. The motion of electrons around the nucleus can be described by wave equations. The solutions to these equations are atomic orbitals, which roughly delineate regions in space in which there is a high probability of finding electrons. 11. An s orbital is spherical; a p orbital looks like two touching teardrops or a “spherical figure eight.” The mathematical sign of the orbital at any point can be positive, negative, or zero (node). With increasing energy, the number of nodes increases. Each orbital can be occupied by a maximum of two electrons of opposite spin (Pauli exclusion principle, Hund’s rule). 12. The process of adding electrons one by one to the atomic orbitals, starting with those of lowest energy, is called the Aufbau principle. 13. A molecular orbital is formed when two atomic orbitals overlap to generate a bond. Atomic orbitals of the same sign overlap to give a bonding molecular orbital of lower energy. Atomic orbitals of opposite sign give rise to an antibonding molecular orbital of higher energy and containing a node. The number of molecular orbitals equals the number of atomic orbitals from which they derive. 14. Bonds made by overlap along the internuclear axis are called S bonds; those made by overlap of p orbitals perpendicular to the internuclear axis are called P bonds.
Chapter 1
43
44
Chapter 1
Structure and Bonding in Organic Molecules
15. The mixing of orbitals on the same atom results in new hybrid orbitals of different shape. One s and one p orbital mix to give two linear sp hybrids, used, for example, in the bonding of BeH2. One s and two p orbitals result in three trigonal sp2 hybrids, used, for example, in BH3. One s and three p orbitals furnish four tetrahedral sp3 hybrids, used, for example, in CH4. The orbitals that are not hybridized stay unchanged. Hybrid orbitals may overlap with each other. Overlapping sp3 hybrid orbitals on different carbon atoms form the carbon – carbon bonds in ethane and other organic molecules. Hybrid orbitals may also be occupied by lone electron pairs, as in NH3. 16. The composition (i.e., ratios of types of atoms) of organic molecules is revealed by elemental analysis. The molecular formula gives the number of atoms of each kind. 17. Molecules that have the same molecular formula but different connectivity order of their atoms are called constitutional or structural isomers. They have different properties. 18. Condensed and bond-line formulas are abbreviated representations of molecules. Hashedwedged line drawings illustrate molecular structures in three dimensions.
Problems 25. Draw a Lewis structure for each of the following molecules and assign charges where appropriate. The order in which the atoms are connected is given in parentheses. (a) ClF (d) CH3NH2 (g) CH2CO
O (c) SOCl2(ClSCl) (f) N2H2 (HNNH) (i) N2O (NNO)
(b) BrCN (e) CH3OCH3 (h) HN3 (HNNN)
26. Using electronegativity values from Table 1-2 (in Section 1-3), identify polar covalent bonds in several of the structures in Problem 25 and label the atoms d1 and d2, as appropriate. 27. Draw a Lewis structure for each of the following species. Again, assign charges where appropriate. (b) CH32 (e) CH3NH31 (h) HC22 (HCC)
(a) H2 (d) CH3 (g) CH2
(c) CH31 (f) CH3O2 (i) H2O2 (HOOH)
28. For each of the following species, add charges wherever required to give a complete, correct Lewis structure. All bonds and nonbonded valence electrons are shown.
(a) H
H
H
O
C
H H
(b)
H N
š O
O
(c) H
H
H
H
(d) H
C
H
H
H
H
H
š O
H
š O
(e)
š C
B
š O
(f) H
š O
š N
š O
H
29. (a) The structure of the bicarbonate (hydrogen carbonate) ion, HCO32, is best described as a hybrid of several contributing resonance forms, two of which are shown here. š ⫺ ðOð
ðOð š HO
⫺ š O ð
š HO ⫹
⫺ š O ð
(i) Draw at least one additional resonance form. (ii) Using curved “electron-pushing” arrows, show how these Lewis structures may be interconverted by movement of electron pairs. (iii) Determine which form or forms will be the major contributor(s) to the real structure of bicarbonate, explaining your answer on the basis of the criteria in Section 1-5. (b) Draw two resonance forms for formaldehyde oxime, H2CNOH. As in parts (ii) and (iii) of (a), use curved arrows to interconvert the resonance forms and determine which form is the major contributor. (c) Repeat the exercises in (b) for formaldehyde oximate ion, [H2CNO]2.
Problems
Chapter 1
30. Several of the compounds in Problems 25 and 28 can have resonance forms. Identify these molecules and write an additional resonance Lewis structure for each. Use electron-pushing arrows to illustrate how the resonance forms for each species are derived from one another, and in each case indicate the major contributor to the resonance hybrid. 31. Draw two or three resonance forms for each of the following species. Indicate the major contributor or contributors to the hybrid in each case. (a) OCN2 (d) O3 (OOO) (g) HOCHNH21
O (c) HCONH2 (HCNH2) (f) ClO22 (OClO)
(b) CH2CHNH2 (e) CH2CHCH22 (h) CH3CNO
32. Compare and contrast the Lewis structures of nitromethane, CH3NO2, and methyl nitrite, CH3ONO. Write at least two resonance forms for each molecule. Based on your examination of the resonance forms, what can you say about the polarity and bond order of the two NO bonds in each substance? 33. Write a Lewis structure for each substance. Within each group, compare (i) number of electrons, (ii) charges on atoms, if any, (iii) nature of all bonds, and (iv) geometry. (a) (b) (c) (d) (e)
chlorine atom, Cl, and chloride ion, Cl2 borane, BH3, and phosphine, PH3 CF4 and BrF42 (C and Br are in the middle) nitrogen dioxide, NO2, and nitrite ion, NO22 (nitrogen is in the middle) NO2, SO2, and ClO2 (N, S, and Cl are in the middle)
34. Use a molecular-orbital analysis to predict which species in each of the following pairs has the stronger bonding between atoms. (Hint: Refer to Figure 1-12.) (a) H2 or H21
(b) He2 or He21
(c) O2 or O21
(d) N2 or N21
35. For each molecule below, predict the approximate geometry about each indicated atom. Give the hybridization that explains each geometry. O
Br Br
(a) H2C
CH2
(b) H3C
(d) H3C
NH2
(e) HC q CO CH2 O OH
C
CH3
(c) H3C
O
(f) H2C
NH2
CH
CH2
⫹
36. For each molecule in Problem 35, describe the orbitals that are used to form every bond to each of the indicated atoms (atomic s, p, hybrid sp, sp2, or sp3). 37. Draw and show the overlap of the orbitals involved in the bonds discussed in Problem 36. 38. Describe the hybridization of each carbon atom in each of the following structures. Base your answer on the geometry about the carbon atom. (a) CH3Cl
(b) CH3OH
(d) CH2“CH2 (trigonal carbons)
(c) CH3CH2CH3
(e) HC‚CH (linear structure)
(f)
(g)
C H3C
O⫺
O
O ⫺
H2C
H
C
C H
39. Depict the following condensed formulas in Kekulé (straight-line) notation. (See also Problem 42.) H2N O (a) CH3CN
(b) (CH3)2CHCHCOH
(c) CH3CHCH2CH3 OH
O (d) CH2BrCHBr2
O
(e) CH3CCH2COCH3
(f) HOCH2CH2OCH2CH2OH
H2C
H
45
46
Structure and Bonding in Organic Molecules
Chapter 1
40. Convert the following bond-line formulas into Kekulé (straight-line) structures. O
Br
OH (b)
(a)
(c)
N
Br (d)
(e)
O
S
(f)
CN
41. Convert the following hashed-wedged line formulas into condensed formulas. H H H [Δ i C O C~ H i & H i (b) C O½O H } H & C x N
H H C H i [ O (a) }C Ci H2š N & š NH2 H
Br i (c) HO C~ & Br Br
42. Depict the following Kekulé (straight-line) formulas in their condensed forms.
(a) H
H
H
H
C
N
C
H
(d) F
H
F
H
C
C
F
(b) H
H
H
O
H
H
H
C
C
N
C
C
H
H
H
H š O
H
(e)
H
H
H
C C
H
H
(c) H
C H
H
š S
C
C
C
C
H
H
H
H
H
H C
C
H
H
H D š H ðO H
H H C
C
(f) H
H C H
C
H
H
O
H
43. Redraw the structures depicted in Problems 39 and 42 using bond-line formulas. 44. Convert the following condensed formulas into hashed-wedged line structures. SH
CN (a) CH3CHOCH3
(b) CHCl3
(c) (CH3)2NH
(d) CH3CHCH2CH3
45. Construct as many constitutional isomers of each molecular formula as you can for (a) C5H12; (b) C3H8O. Draw both condensed and bond-line formulas for each isomer. 46. Draw condensed formulas showing the multiple bonds, charges, and lone electron pairs (if any) for each molecule in the following pairs of constitutional isomers. (Hint: First make sure that you can draw a proper Lewis structure for each molecule.) Do any of these pairs consist of resonance forms? O (a) HCCCH3 and H2CCCH2 (b) CH3CN and CH3NC (c) CH3CH and H2CCHOH 47.
Two resonance forms can be written for a bond between trivalent boron and an atom with a lone pair of electrons. (a) Formulate them for (i) (CH3)2BN(CH3)2; (ii) (CH3)2BOCH3; (iii) (CH3)2BF. (b) Using the guidelines in Section 1-5, determine which form in each pair of resonance forms is more important. (c) How do the electronegativity differences between N, O, and F affect the relative importance of the resonance forms in each case? (d) Predict the hybridization of N in (i) and O in (ii).
48.
The unusual molecule [2.2.2]propellane is pictured in the margin. On the basis of the given structural parameters, what hybridization scheme best describes the carbons marked by asterisks? (Make a model to help you visualize its shape.) What types of orbitals are used in the bond between them? Would you expect this bond to be stronger or weaker than an ordinary carbon – carbon single bond (which is usually 1.54 Å long)?
49.
(a) On the basis of the information in Problem 38, give the likely hybridization of the orbital that contains the unshared pair of electrons (responsible for the negative charge) in each of the following species: CH3CH22; CH2“CH2; HC‚C2. (b) Electrons in sp, sp2, and
CH2 H2C H2C H2C
1.60Å
C* CH2
C* 120ⴗ
CH2
[2.2.2]Propellane
Problems
sp3 orbitals do not have identical energies. Because the 2s orbital is lower in energy than aΔ 2p, the more s character a hybrid orbital has, the lower its energy will be. Therefore the sp3 (14 s and 3 1 1 4 p in character) is highest in energy, and the sp (2 s, 2 p) lowest. Use this information to determine the relative abilities of the three anions in (a) to accommodate the negative charge. (c) The strength of an acid HA is related to the ability of its conjugate base A2 to accommodate negative charge. In other words, the ionization HA Δ H1 1 A2 is favored for a more stable A2. Although CH3CH3, CH2“CH2, and HC‚CH are all weak acids, they are not equally so. On the basis of your answer to (b), rank them in order of acid strength. 50. A number of substances containing positively polarized carbon atoms have been labeled as “cancer suspect agents” (i.e., suspected carcinogens or cancer-inducing compounds). It has been suggested that the presence of such carbon atoms is responsible for the carcinogenic properties of these molecules. Assuming that the extent of polarization is proportional to carcinogenic potential, how would you rank the following compounds with regard to cancercausing potency? (a) CH3Cl (d) CH3OCH2Cl
(b) (CH3)4Si (e) (CH3)3C1
(c) ClCH2OCH2Cl
(Note: Polarization is only one of the many factors known to be related to carcinogenicity. Moreover, none of them shows the type of straightforward correlation implied in this question.) 51. Certain compounds, such as the one pictured below, show strong biological activity against cell types characteristic of prostate cancer. In this structure, locate an example of each of the following types of atoms or bonds: (a) a highly polarized covalent single bond; (b) a highly polarized covalent double bond; (c) a nearly nonpolar covalent bond; (d) an sp-hybridized carbon atom; (e) an sp2-hybridized carbon atom; (f ) an sp3-hybridized carbon atom; (g) a bond between atoms of different hybridization; (h) the longest bond in the molecule; (i) the shortest bond in the molecule (excluding bonds to hydrogen). H H3C
Cl
O
CH H3
O
N CH3
Team Problems Team problems are meant to encourage discussion and collaborative learning. Try to solve team problems with a partner or small study group. Notice that the problems are divided into parts. Rather than tackling each part individually, discuss each section of the problem together. Try out the vocabulary that you learned in the chapter to question and convince yourselves that you are on the right track before you move to the next part. In general, the more you use the terms and apply the concepts presented in the text, the better you will become at correlating molecular structure and reactivity and thus visualizing bond breaking and bond making. You will begin to see the elegant patterns of organic chemistry and will not be a slave to memorization. The collaborative process used in partner or group study will force you to articulate your ideas. Talking out a solution with an “audience” instead of to yourself builds in checks and balances. Your teammates will not let you get away with, “Well, you know what I mean,” because they probably do not. You become responsible to others as well as yourself. By learning from and teaching others, you solidify your own understanding. 52. Consider the following reaction: O CH3CH2CH2CCH3 ⫹ HCN A
H O CH3CH2CH2CCH3 C N B
Chapter 1
47
48
Chapter 1
Structure and Bonding in Organic Molecules
(a) Draw these condensed formulas as Lewis dot structures. Label the geometry and hybridization of the bold carbons in compounds A and B. Did the hybridization change in the course of the reaction? (b) Draw the condensed formulas as bond-line structures. (c) Examine the components of the reaction in light of bond polarity. Using the notation for partial charge separation, d1 and d2, indicate, on the bond-line structures, any polar bonds. (d) This reaction is actually a two-step process: cyanide attack followed by protonation. Depict these processes by using the same “electron pair pushing” technique that we employed for resonance structures in Section 1-5, but now show the flow of electrons for the two steps. Clearly position the beginning (an electron pair) and the end (a positively polarized or charged nucleus) of your arrows.
Preprofessional Problems Preprofessional problems are included to give you practice solving the type of problems found on exams required for entry into professional schools, such as the MCAT, DAT, chemistry GRE, and ACS exams, as well as on many undergraduate tests. Do these multiple-choice questions as you go through this course and then return to them before you take a professional school exam. These questions are to be answered “closed book” — that is, no periodic table, calculator, or the like. 53. A certain organic compound was found on combustion analysis to contain 84% carbon and 16% hydrogen (C 5 12.0, H 5 1.00). A molecular formula for the compound could be (a) CH4O
(b) C6H14O2
(c) C7H16
(d) C6H10
Br CH3 A A 54. The compound Br O Al ONO CH2CH3 has a formal charge of A A Br CH3 (a) 21 on N (d) 11 on Br CH2 P C
D G
CH3 H
(b) 12 on N (e) none of the above
(c) 21 on Al
55. The arrow in the structure in the margin points to a bond that is formed by (a) (b) (c) (d)
overlap of an s orbital on H and an sp2 orbital on C overlap of an s orbital on H and an sp orbital on C overlap of an s orbital on H and an sp3 orbital on C none of the above
56. Which compound has bond angles nearest to 1208? (a) O“C“S (d) HOC‚COH
(b) CHI3 (e) CH4
(c) H2C“O
57. The pair of structures that are resonance hybrids is ⫹
⫹
š (a) HO P CHCH3 O CHCH3 and HO
(b)
CH and A CH
H ðOð B C
(c) CH3
and H
ðš O A C K
CH2
H
⫹
KCH2 N CH2
⫹
(d) CH3CH2 and CH2CH3
(e) C14H22
C H A P T E R
[
TWO
]
Structure and Reactivity Acids and Bases, Polar and Nonpolar Molecules
n Chapter 1, we saw organic molecules containing several different types of bonds between various elements. Can we predict, on the basis of these structures, what kinds of chemical reactivity these substances will display? This chapter will begin to answer this question by showing how certain structural combinations of atoms in organic molecules, called functional groups, display characteristic and predictable behavior. We will see how the chemistry of acids and bases serves as a simple model for understanding the reactions of many functional groups, especially those containing polar bonds. We will pursue this analogy throughout the course, through the concepts of electrophiles and nucleophiles. Most organic molecules contain a structural skeleton that carries the functional groups. This skeleton is a relatively nonpolar assembly consisting of carbon and hydrogen atoms connected by single bonds. The simplest class of organic compounds, the alkanes, lacks functional groups and is constructed entirely of singly bonded carbon and hydrogen. Therefore, alkanes serve as excellent models for the backbones of functionalized organic molecules. They are also useful compounds in their own right, as illustrated by the structure shown at the top of this page, which is called 2,2,4-trimethylpentane and is an alkane in gasoline. By studying the alkanes we can prepare ourselves to better understand the properties of molecules containing functional groups. Therefore, in this chapter we will explore the names, physical properties, and structural characteristics of the members of the alkane family.
I
∑ [
∑ i [ C OC i
Alkane single bond
The branched alkane 2,2,4-trimethylpentane is an important component of gasoline and the standard on which the “octane rating” system for fuel efficiency is based. The car engine shown above requires high-octane fuel to achieve the performance for which it is famous.
50
Chapter 2
Structure and Reactivity
2-1 Kinetics and Thermodynamics of Simple Chemical Processes The simplest chemical reactions may be described as equilibration between two distinct species. Such processes are governed by two fundamental considerations: 1. Chemical thermodynamics, which deals with the changes in energy that take place when processes such as chemical reactions occur. Thermodynamics controls the extent to which a reaction goes to completion. 2. Chemical kinetics, which concerns the velocity or rate at which the concentrations of reactants and products change. In other words, kinetics describes the speed at which a reaction goes to completion. These two principles are frequently related, but not necessarily so. Reactions that are thermodynamically very favorable often proceed faster than do less favorable ones. Conversely, some reactions are faster than others even though they result in a comparatively less stable product. A transformation that yields the most stable products is said to be under thermodynamic control. Its outcome is determined by the net favorable change in energy in going from starting materials to products. A reaction in which the product obtained is the one formed fastest is defined as being under kinetic control. Such a product may not be the thermodynamically most stable one. Let us put these statements on a more quantitative footing.
Equilibria are governed by the thermodynamics of chemical change All chemical reactions are reversible, and reactants and products interconvert to various degrees. When the concentrations of reactants and products no longer change, the reaction is in a state of equilibrium. In many cases, equilibrium lies extensively (say, more than 99.9%) on the side of the products. When this occurs, the reaction is said to go to completion. (In such cases, the arrow indicating the reverse reaction is usually omitted and, for practical purposes, the reaction is considered to be irreversible.) Equilibria are described by equilibrium constants, K. To find an equilibrium constant, divide the arithmetic product of the concentrations of the components on the right side of the reaction by that of the components on the left, all given in units of moles per liter (mol L21). A large value for K indicates that a reaction goes to completion; it is said to have a large driving force. Reaction K
A Δ B
Equilibrium Constant K5
[B] [A]
K5
[C][D] [A][B]
K
A1B Δ C1D
If a reaction has gone to completion, a certain amount of energy has been released. The equilibrium constant can be related directly to the thermodynamic function called the Gibbs* standard free energy change, DG 8.† At equilibrium, ¢G° 5 2RT ln K 5 22.303 RT log K (in kcal mol21 or kJ mol21 )
*Professor Josiah Willard Gibbs (1839–1903), Yale University, Connecticut. † The symbol DG8 refers to the free energy of a reaction with the molecules in their standard states (e.g., ideal molar solutions) after the reaction has reached equilibrium.
2-1 Kinetics and Thermodynamics of Simple Chemical Processes
Table 2-1
Equilibria and Free Energy for A Δ B: K 5 [B]/[A] DG8
K
B
A
(kcal mol21 at 258C)
(kJ mol21 at 258C)
0.01 0.1 0.33 1 2 3 4 5 10 100 1,000 10,000
0.99 9.1 25 50 67 75 80 83 90.9 99.0 99.9 99.99
99.0 90.9 75 50 33 25 20 17 9.1 0.99 0.1 0.01
12.73 11.36 10.65 0 20.41 20.65 20.82 20.95 21.36 22.73 24.09 25.46
111.42 15.69 12.72 0 21.72 22.72 23.43 23.97 25.69 211.42 217.11 222.84
More negative DG8
Increasing equilibrium constant
Percentage
in which R is the gas constant (1.986 cal K21 mol21 or 8.315 J K21 mol21) and T is the absolute temperature in kelvin* (K). A negative DG 8 signifies a release of energy. The equation shows that a large value for K indicates a large favorable free energy change. At room temperature (25 8C, 298 K), the preceding equation becomes ¢G° 5 21.36 log K (in kcal mol21 ) This expression tells us that an equilibrium constant of 10 would have a DG 8 of 21.36 kcal mol21 and, conversely, a K of 0.1 would have a DG 8 5 11.36 kcal mol21. Because the relation is logarithmic, changing the DG 8 value affects the K value exponentially. When K 5 1, starting materials and products are present in equal concentrations and DG 8 is zero (Table 2-1).
The free energy change is related to changes in bond strengths and the degree of energy dispersal in the system The Gibbs standard free energy change is related to two other thermodynamic quantities: the change in enthalpy, DH 8, and the change in entropy, DS 8. Gibbs Standard Free Energy Change DG 8 5 DH 8 2 TDS 8 In this equation, T is again in kelvin and DH 8 in kcal mol21 or kJ mol21, whereas DS 8 is in cal K21 mol21, also called entropy units (e.u.), or J K21 mol21. The enthalpy change of a reaction, DH 8, is the heat absorbed or released at constant pressure during the course of the reaction. Enthalpy changes in chemical reactions relate mainly to differences between the strengths of the bonds in the products compared with those in the starting materials. Bond strengths are described quantitatively by bonddissociation energies, DH 8. The value of DH 8 for a reaction may be estimated by subtracting the sum of the DH 8 values of the bonds formed from those of the bonds broken. Chapter 3 explores in detail bond-dissociation energies and their value in understanding chemical reactions.
*Temperature intervals in kelvin and degrees Celsius are identical. Temperature units are named after Lord Kelvin, Sir William Thomson (1824–1907), University of Glasgow, Scotland, and Anders Celsius (1701–1744), University of Uppsala, Sweden.
Chapter 2
51
52
Chapter 2
Structure and Reactivity
Enthalpy Change in a Reaction a
Sum of DH° sum of DH° b2a b 5 ¢H° of bonds broken of bonds formed
If the bonds formed are stronger than those broken, the value of DH 8 is negative and the reaction is defined as exothermic (releasing heat). In contrast, a positive DH 8 is characteristic of an endothermic (heat-absorbing) process. An example of an exothermic reaction is the combustion of methane, the main component of natural gas, to carbon dioxide and liquid water. CH4 1 2 O2 uy CO2 1 2 H2Oliq
DH 8 5 2213 kcal mol21 (2891 kJ mol21) Exothermic
DH 8 5 (sum of DH 8 values of all the bonds in CH4 1 2 O2) 2 (sum of DH 8 values of all the bonds in CO2 1 2 H2O) The exothermic nature of this reaction is due to the very strong bonds formed in the products. Many hydrocarbons release a lot of energy on combustion and are therefore valuable fuels. If the enthalpy of a reaction depends strongly on changes in bond strength, what is the significance of DS 8, the entropy change? You may be familiar with the concept that entropy is related to the order of a system: Increasing disorder correlates with an increase in the value of S 8. However, the concept of “disorder” is not readily quantifiable and cannot be applied in a precise way to scientific situations. Instead, for chemical purposes, DS 8 is used to describe changes in energy dispersal. Thus, the value of S 8 increases with increasing dispersal of energy content among the constituents of a system. Because of the negative sign in front of the TDS 8 term in the equation for DG 8, a positive value for DS 8 makes a negative contribution to the free energy of the system. In other words, going from lesser to greater energy dispersal is thermodynamically favorable. What is meant by energy dispersal in a chemical reaction? Consider a transformation in which the number of reacting molecules differs from the number of product molecules formed. For example, upon strong heating, 1-pentene undergoes cleavage into ethene and propene. This process is endothermic, primarily because a C – C bond is lost. It would not occur, were it not for entropy. Thus, two molecules are made from one, and this is associated with a relatively large positive DS 8. After bond cleavage, the energy content of the system is distributed over a greater number of particles. At high temperatures, the 2TDS 8 term in our expression for DG 8 overrides the unfavorable enthalpy, making this a feasible reaction. CH3CH2CH2CH“CH2 uy CH2“CH2 1 CH3CH“CH2 1-Pentene
Ethene (Ethylene)
Propene
DH 8 5 122.4 kcal mol21 (193.7 kJ mol21) Endothermic
DS 8 5 133.3 cal K21 mol21 or e.u. (1139.3 J K21 mol21)
Exercise 2-1 Calculate the DG8 at 258C for the preceding reaction. Is it thermodynamically feasible at 258C? What is the effect of increasing T on DG8? What is the temperature at which the reaction becomes favorable? (Caution: DS8 is in the units of cal K21 mol21, whereas DH 8 is in kcal mol21. Don’t forget that factor of 1000!)
In contrast, energy dispersal and entropy decrease when the number of product molecules is less than the number of molecules of starting materials. For example, the reaction of
2-1 Kinetics and Thermodynamics of Simple Chemical Processes
Chapter 2
53
ethene (ethylene) with hydrogen chloride to give chloroethane is exothermic by 215.5 kcal mol21, but the entropy makes an unfavorable contribution to the DG 8; DS 8 5 231.3 e.u. CH2“CH2 1 HCl uy CH3CH2Cl
DH 8 5 215.5 kcal mol21 (264.9 kJ mol21) DS 8 5 231.3 e.u. (2131.0 J K21 mol21)
Exercise 2-2 Calculate the DG8 at 258C for the preceding reaction. In your own words, explain why a reaction that combines two molecules into one should have a large negative entropy change.
In many organic reactions the change in entropy is small, and it often suffices to consider only the changes in bonding energy to estimate whether they are likely to occur or not. In those cases we will equate approximately DG 8 to DH 8. Exceptions are transformations in which the number of molecules on each side of the chemical equation differ (as shown in the examples above) or in which energy dispersal is greatly affected by profound structural changes, such as, for example, ring closures and ring openings.
The rate of a chemical reaction depends on the activation energy How fast is equilibrium established? The thermodynamic features of chemical reactions do not by themselves tell us anything about their rates. Consider the combustion of methane, mentioned earlier. This process releases 213 kcal mol21 (2891 kJ mol21), a huge amount of energy, but we know that methane does not spontaneously ignite in air at room temperature. Why is this highly favorable combustion process so slow? The answer is that during the course of this reaction the potential energy of the system changes in the manner shown in Figure 2-1. This figure, which is an example of a potential-energy diagram, plots energy as a function of reaction progress. We measure reaction progress by the reaction coordinate, which describes the combined processes of bond breaking and bond formation that constitute the overall change from the structures of the starting compounds into those of the products. The energy first rises to a maximum, a point called the transition state (TS), before decreasing to the final value, which is the energy content of the product molecules. The energy of the transition state may be viewed as a barrier to be overcome in order for the reaction to take place. The energy input required to raise the energy of the starting compounds to that of the transition state is called the activation energy, Ea, of the Transition state: maximum energy
Ea large: reaction is slow
E
CH4 + 2 O2 Starting materials
ΔH ° = −213 kcal mol −1: reaction is very exothermic
CO2 + 2 H2O Products Reaction coordinate B
Figure 2-1 A (highly oversimplified) potential-energy diagram for the combustion reaction of methane. Despite its thermodynamic favorability, as shown by the large negative DH8, the process is very slow because it has a high-energy transition state and a large activation energy. (In actuality, the process has many individual bondbreaking and bond-forming steps, and the applicable potentialenergy diagram therefore has multiple maxima and minima.)
54
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reaction. The higher its value, the slower the process. The transition state for methane combustion is very high in energy, corresponding to a high Ea and a very low rate. How can exothermic reactions have such high activation energies? As atoms move from their initial positions in the starting molecules, energy input is required for bond breaking to begin. At the transition state, where both partially broken old bonds and incompletely formed new ones are present, the overall loss of bonding reaches its greatest extent, and the energy content of the system is at its maximum. Beyond this point, continued strengthening of the new bonds releases energy until the atoms reach their final, fully bonded positions in the products.
Collisions supply the energy to get past the activation-energy barrier
Figure 2-2 Boltzmann curves at two temperatures. At the higher temperature (green curve), there are more molecules of kinetic energy E than at the lower temperature (blue curve). Molecules with higher kinetic energy can more easily overcome the activation-energy barrier.
Number of molecules
Where do molecules get the energy to overcome the barrier to reaction? Molecules have kinetic energy as a result of their motion, but at room temperature the average kinetic energy is only about 0.6 kcal mol21 (2.5 kJ mol21), far below many activation-energy barriers. To pick up enough energy, molecules must collide with each other or with the walls of the container. Each collision transfers energy between molecules. A graph called a Boltzmann* distribution curve depicts the distribution of kinetic energy. Figure 2-2 shows that, although most molecules have only average speed at any given temperature, some molecules have kinetic energies that are much higher.
Higher temperature: more molecules with energy E Lower temperature: fewer molecules with energy E
E
Kinetic energy (speed) of molecules
The shape of the Boltzmann curve depends on the temperature. At higher temperatures, as the average kinetic energy increases, the curve flattens and shifts toward higher energies. More molecules now have energy higher than is required by the transition state, so the speed of reaction increases. Conversely, at lower temperatures, the reaction rate decreases.
The concentration of reactants can affect reaction rates Consider the addition of reagent A to reagent B to give product C: A 1 B uy C In many transformations of this type, increasing the concentration of either reactant increases the rate of the reaction. In such cases, the transition-state structure incorporates both molecules A and B. The experimentally observed rate is expressed by Rate 5 k[A][B] in units of mol L21 s21 in which the proportionality constant, k, is also called the rate constant of the reaction. The rate constant equals the rate of the reaction at 1 molar concentrations of the two reactants, *Professor Ludwig Boltzmann (1844–1906), University of Vienna, Austria.
2-1 Kinetics and Thermodynamics of Simple Chemical Processes
A and B. A reaction for which the rate depends on the concentrations of two molecules in this way is said to be a second-order reaction. In some processes, the rate depends on the concentration of only one reactant, such as the hypothetical reaction A uy B Rate 5 k[A] in units of mol L21 s21 A reaction of this type is said to be of first order.
Exercise 2-3
Working with the Concepts: Using Rate Equations What would be the reduction in rate for a reaction following the first-order rate law, rate 5 k[A], when half of A is consumed (i.e., after 50% conversion of the starting material)? Answer the same question for a second-order reaction for which the rate 5 k[A][B]. Strategy Let us denote the initial concentration of A as [A0]. The first question asks us to compare the initial rate, which equals k[A0], with the new rate when [A] has dropped to 0.5[A0]. We write out the first-order rate law, substituting the new concentration, and compare the two rates. Solution • The rate constant k does not change, so new rate 5 k[Anew] 5 k(0.5[A0]) 5 0.5k[A0] 5 0.5(initial rate) The rate is decreased to one-half its initial value. • In the second-order process, we follow the same approach. Assuming that the initial concentrations of A and B are equal, after 50% conversion, [Anew] 5 0.5[A0] and [Bnew] 5 0.5[B0]. Replacing [A] and [B] in the rate law with 0.5[A0] and 0.5[B0], respectively, we obtain new rate 5 k[Anew][Bnew] 5 k(0.5[A0])(0.5[B0]) 5 0.25k[A0][B0] 5 0.25(initial rate) The rate drops to one-quarter its initial value.
Exercise 2-4
Try It Yourself The reaction described by the equation CH3Cl 1 NaOH uy CH3OH 1 NaCl follows the second-order rate law, rate 5 k[CH3Cl][NaOH]. When this reaction is carried out with starting concentrations [CH3Cl] 5 0.2 M and [NaOH] 5 1.0 M, the measured rate is 1 3 1024 mol L21 s21. What is the rate after one-half of the CH3Cl has been consumed? (Caution: The initial concentrations of the starting materials are not identical in this experiment. Hint: Determine how much of the NaOH has been consumed at this point and what its new concentration is, compared with its initial concentration.)
The Arrhenius equation describes how temperature affects reaction rates The kinetic energy of molecules increases when they are heated, which means that a larger fraction of them have sufficient energy to overcome the activation barrier Ea (Figure 2-2). A useful rule of thumb applies to many reactions: Raising the reaction temperature by 10 degrees (Celsius) causes the rate to increase by a factor of 2 to 3. The Swedish chemist
Chapter 2
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Chapter 2
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Arrhenius* noticed the dependence of reaction rate k on temperature T. He found that his measured data conformed to the equation Arrhenius Equation k 5 Ae2Ea yRT 5 Aa
1 eEayRT
b
The Arrhenius equation describes how rates of reactions with different activation energies vary with temperature. In this equation R is again the gas constant and A is a factor with a value characteristic of a specific reaction. You can see readily that the larger the activation energy Ea, the slower the reaction. Conversely, the higher the temperature T, the faster the reaction. The A term can be imagined as the maximum rate constant that the reaction would have if every molecule had sufficient collisional energy to overcome the activation barrier. This will occur at very high temperature, when Ea /RT will be close to zero and e2Ea /RT approaches 1, thus rendering k nearly equal to A.
Exercise 2-5 (a) Calculate DG 8 at 25 8C for the reaction CH3CH2Cl S CH2“CH2 1 HCl (the reverse of the reaction in Exercise 2-2). (b) Calculate DG 8 at 500 8C for the same reaction. (Hint: Apply DG 8 5 DH 8 2 TDS 8 and do not forget to first convert degrees Celsius into kelvin.)
Exercise 2-6 For the reaction in Exercise 2-5, A 5 1014 and Ea 5 58.4 kcal mol21. Using the Arrhenius equation, calculate k at 500 8C for this reaction. R 5 1.986 cal K21 mol21. (Caution: Activation ener-
gies are given in the units of kcal mol21, while R is in cal K21 mol21. Don’t forget that factor of 1000!)
In Summary All reactions are described by equilibrating the concentrations of starting materials and products. On which side the equilibrium lies depends on the size of the equilibrium constant, which in turn is related to the Gibbs free energy changes, DG 8. An increase in the equilibrium constant by a factor of 10 is associated with a change in DG 8 of about 21.36 kcal mol21 (25.69 kJ mol21) at 258C. The free energy change of a reaction is composed of changes in enthalpy, DH 8, and entropy, DS 8. Contributions to enthalpy changes stem mainly from variations in bond strengths; contributions to entropy changes arise from the relative dispersal of energy in starting materials and products. Whereas these terms define the position of an equilibrium, the rate at which it is established depends on the concentrations of starting materials, the activation barrier separating reactants and products, and the temperature. The relation among rate, Ea, and T is expressed by the Arrhenius equation.
2-2 Acids and Bases; Electrophiles and Nucleophiles; Using Curved “Electron-Pushing” Arrows Let us turn to a fundamental application of thermodynamics: the chemistry of acids and bases. This section reviews the way in which acids and bases interact and how we quantify this process. Acid-base reactions provide a model for the reactivity of organic molecules that possess polar covalent bonds. We first introduce the use of curved arrows to indicate how the electrons move when a chemical reaction takes place. *Professor Svante Arrhenius (1859–1927), Technical Institute of Stockholm, Sweden, Nobel Prize 1903 (chemistry), director of the Nobel Institute from 1905 until shortly before his death.
2-2 Acids and Bases; Electrophiles and Nucleophiles
Curved arrows show how starting materials convert to products Bonds consist of electrons. Chemical change is defined as a process in which bonds are broken and/or formed. Therefore, when chemistry takes place, electrons move. Following the basic principles of electrostatics, electrons, being negatively charged, are attracted to sites of electron deficiency, or positive charge. Either highly electronegative or electron-deficient (and, therefore, electron-attracting) atoms, positively charged ions, or the d1 atom in a polar covalent bond can be the destination for electron movement. The vast majority of chemical processes we will introduce in this text will involve the movement of one or more pairs of electrons. A curved arrow ([) will show the flow of an electron pair from its point of origin, either a lone pair or a covalent bond, to its destination. The electron-pair movement that interconverts resonance forms (Section 1-5) follows these same principles. However, we know that resonance forms do not represent distinct entities. When we use curved arrows to depict the electron movement associated with a chemical reaction, we are describing an actual structural change, from the Lewis structures of the starting materials to those of the products. The examples below illustrate the various ways in which curved arrows are employed in this movement. In each case a red color denotes an electron pair that moves. 1. Dissociation of a polar covalent bond into ions (B more electronegative than A) A⫹ ⫹ ðB⫺
AO B
Movement of an electron pair converts the A – B covalent bond into a lone pair on atom B
2. Formation of a polar covalent bond from ions A⫹ ⫹ ðB⫺
AO B
The reverse of the previous process: A lone pair on B moves toward A, becoming a new covalent bond between them
As these examples show, both the formation and the cleavage of a covalent bond by electron-pair movement changes the formal charges on the atoms involved. Notice that the arrow usage in Example 2 does not imply that the electron pair originally on atom B leaves that atom entirely. It simply converts from a lone pair to a shared pair. 3. Simultaneous formation and dissociation of two covalent bonds (A – B polarized as shown) Xð⫺ ⫹
␦⫹
⫺
A O B␦
XO A ⫹ ðB⫺
Movement of two electron pairs results in substitution of one bond for another
4. Addition of (a) an electron-pair donor or (b) an electron-pair acceptor to a multiple bond (a) Xð⫺ ⫹ (b) A
␦⫹
A
B ⫹ Y⫹
⫺
B␦
XO A O Bð⫺ ⫹
AO B OY
As in Example 3, but A remains bonded to B in each product, because only one of the two original bonds between them is broken
These are among the most common types of transformations that occur in organic chemistry. Notice that proper use of curved arrows automatically generates the correct Lewis structure of the product of the reaction. As you read further, try to associate each new reaction with one of the electron-movement patterns shown above. Exercise 2-7 From the categories above, identify the one to which each of the following reactions belongs. Draw appropriate curved arrows to show electron movement, and give the structure of the product. (Hint: First complete all Lewis structures by adding any missing lone pairs.) (a) HO2 1 H1; (b) H1 1 H2CPCH2; (c) (CH3)2N2 1 HCl; (d) CH3O2 1 H2CPO.
Chapter 2
57
58
Structure and Reactivity
Chapter 2
Acid and base strengths are measured by equilibrium constants
Notice that the arrow does not point to show the H moving from the acid to the base. Instead, it shows the electron pair of the base “grabbing” the H away from the acid.
D O G
H H
Brønsted and Lowry* have given us a simple definition of acids and bases: An acid is a proton donor and a base is a proton acceptor. Acidity and basicity are commonly measured in water. An acid donates protons to water, forming hydronium ions, whereas a base removes protons from water, forming hydroxide ions. Examples are the acid hydrogen chloride and the base ammonia. In these equations we employ curved arrows to indicate the shift of a pair of electrons, much as we did in Section 1-5. In each case, a lone pair from the base creates a new bond with an acidic proton, and the electron pair originally linking the proton to the remainder of the acid shifts to become a lone pair on the departing conjugate base. The electron flow in the case of the reaction of water with hydrogen chloride is indicated in the electrostatic potential maps below the equation. The red oxygen of water is protonated by the blue hydrogen in the acid to furnish as products the blue hydronium ion and red chloride ion.
⫹
Water (Base)
š HOCl ð
H D⫹ HOOð G H
Hydrogen chloride (Acid)
Hydronium ion (Conjugate acid of water)
⫺ š ðCl ð
⫹
⫹
Chloride ion (Conjugate base of HCl)
⫹
H ⫹
š š NH3 ⫹ H O OH Ammonia (Base)
Water (Acid)
NH3
⫹
Ammonium ion (Conjugate acid of NH3)
š ðOH
⫺
Hydroxide ion (Conjugate base of water)
Water itself is neutral. It forms an equal number of hydronium and hydroxide ions by self-dissociation. The process is described by the equilibrium constant Kw, the self-ionization constant of water. At 25 8C, Kw
H2O 1 H2O Δ H3O 1 1 OH 2
Kw 5 [H3O 1 ][OH 2 ] 5 10214 mol2 L22
From the value for Kw, it follows that the concentration of H3O1 in pure water is 1027 mol L21. The pH is defined as the negative logarithm of the value for [H3O1]. pH 5 2log [H3O 1 ] Thus, for pure water, the pH is 17. An aqueous solution with a pH lower than 7 is acidic; one with a pH higher than 7 is basic.
*Professor Johannes Nicolaus Brønsted (1879 –1947), University of Copenhagen, Denmark. Professor Thomas Martin Lowry (1874 –1936), University of Cambridge, England.
2-2 Acids and Bases; Electrophiles and Nucleophiles
Chapter 2
C H E M I C A L H IGH LIG H T 2-1 Stomach Acid and Food Digestion The human stomach produces, on average, 2 L of 0.02 M hydrochloric acid each day. The pH of stomach juice falls in the range between 1.0 and 2.5, dropping as HCl production rises in response to the stimuli of tasting, smelling, or even looking at food. Stomach acid disrupts the natural folded shapes of protein molecules in food, exposing them to attack and breakdown by a variety of digestive enzymes. You may wonder how the stomach protects itself from such strongly acidic conditions—after all, stomach tissue itself is constructed of protein molecules. The interior lining of the stomach is coated with a layer of cells called gastric mucosa, whose mucous secretions insulate the stomach wall from acidic gastric juices. When certain cells just beneath the gastric mucosa are activated by the stimuli described above, they release signaling molecules called histamine that cause parietal cells located in pits in the lining to secrete HCl into the stomach. Cimetidine, famotidine, and ranitidine, active ingredients in so-called acid-reducer medications, block histamine from reaching the parietal cells, interrupting the signal that would cause production of stomach acid. These products are useful in treating conditions such as hyperacidity, the secretion of unnecessarily large amounts of acid, and peptic ulcers, sores that result from bacterial infections that weaken the mucosa, exposing the lining to acid attack.
The parietal cells in the gastric pits of the stomach secrete hydrochloric acid upon activation by histamine.
The acidity of a general acid, HA, is expressed by the following general equation, together with its associated equilibrium constant. K
HA 1 H2O Δ H3O1 1 A2 K 5
[H3O1 ][A2 ] [HA][H2O]
In dilute aqueous solution, [H2O] is constant at 55 mol L21, so this number may be incorporated into a new constant, the acid dissociation constant, Ka. Ka 5 K[H2O] 5
[H3O1 ][A2 ] mol L21 [HA]
Like the concentration of H3O1 and its relation to pH, this measurement may be put on a logarithmic scale by the corresponding definition of pKa. pKa 5 2log Ka*
*Ka carries the units of molarity, or mol L21, because it is the product of a dimensionless equilibrium constant K and the concentration [H2O], which equals 55 mol L21. However, the logarithm function can operate only on dimensionless numbers. Therefore, pKa is properly defined as the negative log of the numerical value of Ka, which is Ka divided by the units of concentration. (For purposes of simplicity, we will omit the units of Ka in exercises and problems.)
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Table 2-2
Relative Acidities of Common Compounds (258C) Ka
Acid ,1.0 ,1.0 ,1.0 ,1.0
Hydrogen iodide, HI (strongest acid) Hydrogen bromide, HBr Hydrogen chloride, HCl Sulfuric acid, H2SO4 Hydronium ion, H3O2 Nitric acid, HNO3 Methanesulfonic acid, CH3SO3H Hydrogen fluoride, HF Acetic acid, CH3COOH Hydrogen cyanide, HCN Ammonium ion, NH41 Methanethiol, CH3SH Methanol, CH3OH Water, H2O Ammonia, NH3 Methane, CH4 (weakest acid)
6.3 2.0 6.3 5.7 1.0 3.2 2.0 1.0 ,1.0
3 1010 3 109 3 108 3 103 50 25 16 3 1024 3 1025 3 10210 3 10210 3 10210 3 10216 3 10216 3 10235 3 10250
pKa
Increasing acidity
60
210.0 29.0 28.0 23.0a 21.7 21.4 21.2 3.2 4.7 9.2 9.3 10.0 15.5 15.7 35 ,50
Note: Ka 5 [H3O1][A2]/[HA] mol L21. a First dissociation equilibrium
The pKa is the pH at which the acid is 50% dissociated. An acid with a pKa lower than 1 is defined as strong, one with a pKa higher than 4 as weak. The acidities of several common acids are compiled in Table 2-2 and compared with those of compounds with higher pKa values. Sulfuric acid and, with the exception of HF, the hydrogen halides, are very strong acids. Hydrogen cyanide, water, methanol, ammonia, and methane are decreasingly acidic, the last two being exceedingly weak. The species A2 derived from acid HA is frequently referred to as its conjugate base (conjugatus, Latin, joined). Conversely, HA is the conjugate acid of base A2. The strengths of two substances that are related as a conjugate acid-base pair are inversely related: The conjugate bases of strong acids are weak, as are the conjugate acids of strong bases. For example, HCl is a strong acid, because the equilibrium for its dissociation into H1 and Cl2 is very favorable. The reverse process, reaction of Cl2 to combine with H1, is unfavorable, therefore identifying Cl2 as a weak base.
HO š Clð Strong acid
⫺ š H⫹ ⫹ ðCl ð
Equilibrium lies on the right
Conjugate base is weak
In contrast, dissociation of CH3OH to produce CH3O2 and H1 is unfavorable; CH3OH is a weak acid. The reverse, combination of CH3O2 with H1 is favorable; we therefore consider CH3O2 to be a strong base.
CH3š OOH Weak acid
H⫹ ⫹ CH3š Oð⫺ Conjugate base is strong
Equilibrium lies on the left
2-2 Acids and Bases; Electrophiles and Nucleophiles
In these examples we have used curved arrows to show the electron-pair movement in both the forward and reverse directions. For the sake of simplicity, we have also used the symbol H1 to depict the dissociating proton. In reality, the free proton does not exist in solution, but is always associated with the electron pair of another species present, typically the solvent. As we have seen, in water, H1 is represented by the hydronium ion, H3O1. In methanol, it would be methoxonium ion CH3OH21, in methoxymethane (Figure 1-22B) it would be (CH3)2OH1, and so on. We shall see later that in many organic reactions carried out under acidic conditions, there are several potential recipients of a dissociating proton, and it becomes cumbersome to show them all. In these cases, we will stick with the short notation H1.
Exercise 2-8 Write the formula for the conjugate base of each of the following acids. (a) Sulfurous acid, H2SO3; (b) chloric acid, HClO3; (c) hydrogen sulfide, H2S; (d) dimethyloxonium, (CH3)2OH1; (e) hydrogen sulfate, HSO42.
Exercise 2-9 Write the formula for the conjugate acid of each of the following bases. (a) Dimethylamide, (CH3)2N2; (b) sulfide, S22; (c) ammonia, NH3; (d) acetone, (CH3)2C“O; (e) 2,2,2-trifluoroethoxide, CF3CH2O2.
Exercise 2-10 Which is the stronger acid, nitrous (HNO2, pKa 5 3.3) or phosphorous acid (H3PO3, pKa 5 1.3)? Calculate Ka for each.
We can estimate relative acid and base strengths from a molecule’s structure The relationship between structure and function is very evident in acid-base chemistry. Indeed, three structural features allow us to estimate, at least qualitatively, the relative strength of an acid HA (and hence the weakness of its conjugate base): 1. The increasing electronegativity of A as we proceed from left to right across a row in the periodic table. The more electronegative the atom to which the acidic proton is attached, the more polar the bond, and the more acidic the proton will be. For example, the increasing order of acidity in the series H4C , H3N , H2O , HF parallels the increasing electronegativity of A (Table 1-2).
Increasing electronegativity of A
H4C
H3N
H2O HF
Increasing acidity
2. The increasing size of A as we proceed down a column in the periodic table. The acid strengths of the hydrogen halides increase in the order HF , HCl , HBr , HI. Dissociation to give H1 and A2 is favored for larger A, because the overlap of its larger outer-shell orbital with the 1s hydrogen orbital is poor, weakening the H – A bond. In
Chapter 2
61
62
Chapter 2
Structure and Reactivity
addition, the larger outer-shell orbitals permit the electrons to occupy a larger volume of space, thus reducing electron – electron repulsion in the resulting anion.*
Increasing size of A
HF
HCl
HBr
HI
Increasing acidity
3. The resonance in A2 that allows delocalization of charge over several atoms. This effect is frequently enhanced by the presence of additional electronegative atoms in A2. For example, acetic acid is more acidic than methanol. In both cases, an O – H bond dissociates into ions. Methoxide, the conjugate base of methanol, possesses a localized negative charge on oxygen. In contrast, the acetate ion has two resonance forms and is able to delocalize its charge onto a second oxygen atom. Thus in the acetate ion the negative charge is better accommodated (Section 1-5), stabilizing acetate and making it the weaker base. Acetic Acid Is Stronger than Methanol Because Acetate Is Stabilized by Resonance CH3š OOH ⫹ H2š O
š⫹ CH3 O š Oð⫺ ⫹ H3O
Weaker acid
Stronger base
ðOð B CH3COš OOH ⫹ H2š O
ðš Oð⫺ A š CH3C PO O⫹ ⫹ H3š
ðOð B CH3C Oš Oð⫺
Stronger acid
Weaker base
The effect of resonance is even more pronounced in sulfuric acid. The availability of d orbitals on sulfur enables us to write valence-shell-expanded Lewis structures containing as many as 12 electrons (Sections 1-4 and 1-5). Alternatively, charge-separated structures with one or two positive charges on sulfur can be used. Both representations indicate that the pKa of H2SO4 should be low. ðš Oð⫺ š HO
S
2⫹
ð Oð⫺
ðOð š Oð⫺
š HO
S
⫹
ð Oð⫺
ðš Oð⫺
ðOð š Oð⫺
š HO
S ðOð
š Oð⫺
š HO
S
š O
etc.
ðOð
Hydrogen sulfate ion
As a rule, the acidity of HA increases to the right and down in the periodic table. Therefore, the basicity of A2 decreases in the same fashion.
*The bond-strength argument is often the only reason given for the acidity order in the hydrogen halides: HF possesses the strongest bond, HI the weakest. However, this correlation fails for the series H4C, H3N, H2O, HF, in which the weakest acid, CH4, also has the weakest H–A bond. As we will see in Chapter 3, bond strengths are only indirectly applicable to the process of dissociation of a bond into ions.
2-2 Acids and Bases; Electrophiles and Nucleophiles
63
Chapter 2
The same molecule may act as an acid under one set of conditions and as a base under another. Water is the most familiar example of this behavior, but many other substances possess this capability as well. For instance, nitric acid acts as an acid in the presence of water but behaves as a base toward the more powerfully acidic H2SO4: Nitric Acid Acting as an Acid š O O2NO OH ⫹ H2š
⫺ š O2NO O⫹ ð ⫹ H3š
Nitric Acid Acting as a Base š š HO3SO 2 OH ⫹ HONO
⫹
⫺ š HO3SO ð ⫹ H2ONO 2
Similarly, acetic acid protonates water, as shown earlier in this section, but is protonated by stronger acids such as HBr: HOð⫹
ðOð š Br OH ⫹ CH3COH ðš
š ðš Brð⫺ ⫹ CH3COH
Exercise 2-11 Explain the site of protonation of acetic acid in the preceding equation. (Hint: Try placing the proton first on one, and then on the other of the two oxygen atoms in the molecule, and consider which of the two resulting structures is better stabilized by resonance.)
Exercise 2-12
Working with the Concepts: Determining pKa
2
C6H5CO2H Δ C6H5CO2 1 H
1
The thermodynamic parameters for this process are DH 8 5 267 cal mol21 and DS 8 5 219.44 cal K21 mol21. Calculate the acid dissociation constant, Ka, as well as the pKa for benzoic acid. How does benzoic acid compare in strength with acetic acid, whose pKa 5 4.7? Strategy Acid dissociation constants describe equilibria and are related directly to DG 8 values. Therefore, to calculate the acid dissociation constant we must first determine DG 8 for the dissociation reaction at 25 8C (298 K). Then we need to convert Ka to pKa; the lower pKa will indicate the stronger acid. Solution • We have DG 8 5 DH 8 2 TDS 8 5 2(67 cal mol21) 2 [(298 K) 3 (219.44 cal K21 mol21)] 5 2(67 cal mol21) 1 (5793 cal mol21) 5 15726 cal mol21 5 15.726 kcal mol21 • Using DG8 5 21.36 log K, we obtain log Ka 5 24.2, corresponding to a pKa of 4.2, and a Ka of 6.4 3 1025. • The pKa that we have determined for benzoic acid is lower than the value for acetic acid (4.7). Benzoic acid is stronger.
O
P
Benzoic acid, C6H5CO2H (margin), undergoes dissociation in water at 258C according to the following dissociation equation:
C OH Benzoic acid
64
Structure and Reactivity
Chapter 2
Exercise 2-13
Try It Yourself How much stronger an acid is acetic acid (pKa 5 4.7) compared to water (pKa 5 15.7)?
Lewis acids and bases interact by sharing an electron pair A more generalized description of acid-base interaction in terms of electron sharing was introduced by Lewis. A Lewis acid is a species that contains an atom that is at least two electrons short of a closed outer shell. A Lewis base contains at least one lone pair of electrons. The symbol X denotes any halogen, while R represents an organic group (Section 2-3). Lewis Acids have unfilled valence shells (X)H ⫹
H
(R)H B
⫹
H(X)
(X)H
C
MgX2 , AlX3 , many transition metal halide salts
H(R)
(R)H
Lewis Bases have available electron pairs (R)H ⫺
O
H(R)
O
H(R)
(R)H
S
H(R)
(R)H N
(R)H
H(R)
P
(R)H
H(R)
X
⫺
(R)H
A Lewis base shares its lone pair with a Lewis acid to form a new covalent bond. A Lewis base–Lewis acid interaction may therefore be pictured by means of an arrow pointing in the direction that the electron pair moves — from the base to the acid. The Brønsted acid-base reaction between hydroxide ion and a proton is an example of a Lewis acid-base process as well. Lewis Acid-Base Reactions H⫹ ⫹ ⫺ðš O Cl Cl
Al ⫹ ðN
CH2CH3
H
Cl
CH3
Cl
⫺
Al
N⫹ CH3
Cl
CH3 F ⫺
⫹D
CH2CH3
CH2CH3
FO BO O G CH2CH3 F
⫹
D O G
F
CH3
⫹
B F
CH3
š O
H
CH3
Cl F
H
2-2 Acids and Bases; Electrophiles and Nucleophiles
The reaction between boron trifluoride and ethoxyethane (diethyl ether) to give the Lewis acid-base reaction product is shown on the previous page, and also in the form of electrostatic potential maps. As electron density is transferred, the oxygen becomes more positive (blue) and the boron more negative (red). The dissociation of a Brønsted acid HA is just the reverse of the combination of the Lewis acid H1 and the Lewis base A2. We write it as follows: Dissociation of a Brønsted Acid { HO A ¡ H1 1 : A2
Electrophiles and nucleophiles also interact through movement of an electron pair Many processes in organic chemistry exhibit characteristics of acid-base reactions. For example, heating an aqueous mixture of sodium hydroxide and chloromethane, CH3Cl, produces methanol and sodium chloride. This process involves the same kind of movement of two pairs of electrons as does the acid-base reaction between hydroxide ion and HCl, and may be described by the same curved “electron-pushing” arrows: Reaction of Sodium Hydroxide and Chloromethane . . H2O, ¢ .. .. .. CH3Cl .. : 1 NaO .. H ----¡ CH3O .. H 1 NaCl .. : Methanol
Flow of Electrons Using Curved-Arrow Representation
Compare the Brønsted acid-base reaction:
⫺ š š HO ð ⫹ H O Clð
ð
⫺ š š HO ð ⫹ CH3O Clð
⫺ š HO O CH3 ⫹ ððClð
⫺ š š HO O H ⫹ ðClð
Like the hydrogen atom in HCl, the carbon atom in chloromethane is at the positive end of a polarized bond (Section 1-3). It therefore exhibits a tendency to react with species possessing unshared pairs of electrons. The carbon is said to be electrophilic (literally, “electron friendly”; philos, Greek, friend). In turn, atoms bearing lone pairs, such as the oxygen in hydroxide ion, are described as nucleophilic (“nucleus friendly”). This reaction is an example of a nucleophilic substitution, because a nucleophile substitutes itself for an atom or group in the original starting material. By definition, the terms nucleophile and Lewis base are synonymous. All nucleophiles are Lewis bases. We commonly use the term nucleophile to refer to a Lewis base that is attacking a nonhydrogen electrophilic atom, typically carbon. Nucleophiles, often denoted by the abbreviation Nu, may be negatively charged, such as hydroxide, or neutral, such as water, but every nucleophile contains at least one unshared pair of electrons. All Lewis acids are electrophiles, as shown earlier in the examples of Lewis acid-base reactions. However, species such as HCl and CH3Cl, whose atoms have closed outer shells and therefore are not Lewis acids, nonetheless may behave as electrophiles because they possess polar bonds.
Chapter 2
65
66
Chapter 2
Structure and Reactivity
Nucleophilic substitution is a general reaction of haloalkanes, organic compounds possessing carbon – halogen bonds. The two following equations are additional examples: H
H
CH3CCH2CH3 ⫹ ðš Ið⫺
Brð⫺ CH3CCH2CH3 ⫹ ðš
ð Brð
Ið ð H ⫹
CH3CH2š Ið ⫹ ðNH3
CH3CH2NH ⫹ ðš Ið⫺ H
Exercise 2-14
Working with the Concepts: Using Curved Arrows Using earlier examples in this section as models, add curved arrows to the first of the two reactions immediately above. Strategy In the organic substrate, identify the likely reactive bond and its polarization. Classify the other reacting species, and look for reactions in the text between similar types of substances. Solution • The C – Br bond is the site of reactivity in the substrate and is polarized (d1)C – Br(d2). Iodide is a Lewis base and therefore a potential nucleophile (electron-pair donor). Thus the situation resembles that found in the reaction between hydroxide and CH3Cl just above and Example 3 at the beginning of the section. • Follow the earlier patterns in order to add the appropriate arrows. CH2CH3 ␦⫹
␦⫺
Ið⫺ ⫹ H O C Oš Brð ðš CH3
CH2CH3 ⫺ š I O C O H ⫹ ðBrð
CH3
Exercise 2-15
Try It Yourself Add curved arrows to the second of the reactions immediately above Exercise 2-14 in the text.
Although the haloalkanes in these examples contain different halogens and varied numbers and arrangements of carbon and hydrogen atoms, they behave very similarly toward nucleophiles. We conclude that it is the presence of the carbon – halogen bond that governs the chemical behavior of haloalkanes: The C – X bond is the structural feature that directs the chemical reactivity — structure determines function. The C – X bond constitutes the functional group, or the center of chemical reactivity, of the haloalkanes. In the next section, we shall introduce the major classes of organic compounds, identify their functional groups, and briefly preview their reactivity.
In Summary In Brønsted-Lowry terms, acids are proton donors and bases are proton acceptors. Acid-base interactions are governed by equilibria, which are quantitatively described by an acid dissociation constant Ka. Removal of a proton from an acid generates its conjugate base; attachment of a proton to a base forms its conjugate acid. Lewis bases donate
2-3 Functional Groups: Centers of Reactivity
67
Chapter 2
an electron pair to form a covalent bond with Lewis acids, a process depicted by a curved arrow pointing from the lone pair of the base toward the acid. Electrophiles and nucleophiles are species in organic chemistry that interact very much like acids and bases. The carbon – halogen bond in the haloalkane is its functional group. It contains an electrophilic carbon atom, which reacts with nucleophiles in a process called nucleophilic substitution.
2-3 Functional Groups: Centers of Reactivity Many organic molecules consist predominantly of a backbone of carbons linked by single bonds, with only hydrogen atoms attached. However, they may also contain doubly or triply bonded carbons, as well as other elements. These atoms or groups of atoms tend to be sites of comparatively high chemical reactivity and are referred to as functional groups or functionalities. Such groups have characteristic properties, and they control the reactivity of the molecule as a whole.
R
Function
Carbon frame provides structure
Functional group imparts reactivity
Hydrocarbons are molecules that contain only hydrogen and carbon We begin our study with hydrocarbons, which have the general empirical formula CxHy. Those containing only single bonds, such as methane, ethane, and propane, are called alkanes. Molecules such as cyclohexane, whose carbons form a ring, are called cycloalkanes. Alkanes lack functional groups; as a result, they are relatively nonpolar and unreactive. The properties and chemistry of the alkanes are described in this chapter and in Chapters 3 and 4. Cycloalkanes
Alkanes CH4
CH3 OCH3
CH3O CH2 OCH3
CH3OCH2 OCH2 OCH3
Methane
Ethane
Propane
Butane
H2 C H2C
Double and triple bonds are the functional groups of alkenes and alkynes, respectively. Their properties and chemistry are the topics of Chapters 11 through 13.
H2C
CH2 CH2
Cyclopentane
Alkenes and Alkynes H G CPCH2 D CH3
CH2 PCH2 Ethene (Ethylene)
Propene
HCq CH
CH3 OCq CH
Ethyne (Acetylene)
Propyne
A special hydrocarbon is benzene, C6H6, in which three double bonds are incorporated into a six-membered ring. Benzene and its derivatives are traditionally called aromatic, because some substituted benzenes do have a strong fragrance. Aromatic compounds, also called arenes, are discussed in Chapters 15, 16, 22, and 25. Aromatic Compounds (Arenes) H H
C C
H
H
C
C H
CH3
C C
C C C
H
H
H C C
C
H
H
Benzene
Methylbenzene (Toluene)
H
H2 C H2C
CH2
H2C
CH2 C H2
Cyclohexane
68
Chapter 2
Table 2-3
Structure and Reactivity
Common Functional Groups
Compound class
General structurea
Alkanes (Chapters 3, 4)
RO H
Haloalkanes (Chapters 6, 7)
ROš Xð(X ⫽ F, Cl, Br, I)
None
Example
CH3CH2CH2CH3 Butane
š CH3CH2 OBr ð
š OX ð
Bromoethane
Alcohols (Chapters 8, 9)
RO š OH
Ethers (Chapter 9)
RO š O O R⬘
Thiols (Chapter 9)
ROš SH
H A (CH3)2C O š OH
Oš OH
2-Propanol (Isopropyl alcohol)
Oš OO
O O CH3 CH3CH2 O š Methoxyethane (Ethyl methyl ether)
Oš SH
SH CH3CH2 Oš Ethanethiol
(H)R Alkenes (Chapters 11, 12)
Functional group
(H)R
G D CPC D G
R(H)
CH3
G D CPC D G
R(H)
CH3
G C P CH2 D
2-Methylpropene
Alkynes (Chapter 13)
(H)RO C q C O R(H)
OCq CO
CH3C q CCH3 2-Butyne
CH3
R(H) (H)R Aromatic compounds (Chapters 15, 16, 22)
R(H)
C C C
(H)R
C
C C
C
C
C R(H)
C
C C
HC
C
HC
R(H)
Aldehydes (Chapters 17, 18)
Ketones (Chapters 17, 18)
Carboxylic acids (Chapters 19, 20)
Anhydrides (Chapters 19, 20)
Esters (Chapters 19, 20, 23) a
ðOð B RO C O H
CH CH C H
Methylbenzene (Toluene)
ðOð B O C OH
ðOð B CH3CH2CH Propanal
ðOð B RO C O R⬘
ðOð B OCO
ðOð B CH3CH2CCH2CH2CH3 3-Hexanone
ðOð B RO C O š O OH
ðOð B O C Oš OH
ðOð B š CH3CH2COH Propanoic acid
ðOð ðOð B B RO C O š O O C O R⬘(H)
ðOð ðOð B B O C Oš OOCO
ðOððOð B B š CH3CH2COCCH 2CH3 Propanoic anhydride
O ðOð B (H)RO C O š O O R⬘
ðOð B OCOš OO
The letter R denotes an alkyl group (see text). Different alkyl groups can be distinguished by adding primes to the letter R: R9, R0, and so forth.
ðOð B š CH3CH2COCH 3 Methyl propanoate (Methyl propionate)
2-3 Functional Groups: Centers of Reactivity
Chapter 2
(continued)
Table 2-3
Compound class
General structure
Functional group
Example
ðOð B RO C Oš N O R⬘(H) A R⬙(H)
ðOð B D O C Oš N G
ðOð B š 2 CH3CH2CH2CNH
Nitriles (Chapter 20)
RO C q Nð
O C q Nð
CH3C qNð
Amines (Chapter 21)
ROš N O R⬘(H) A R⬙(H)
Amides (Chapters 19, 20, 26)
Butanamide
Ethanenitrile (Acetonitrile)
D Oš N G
(CH3)3Nð N,N-Dimethylmethanamine (Trimethylamine)
Many functional groups contain polar bonds Polar bonds determine the behavior of many classes of molecules. Recall that polarity is due to a difference in the electronegativity of two atoms bound to each other (Section 1-3). We have already introduced the haloalkanes, which contain polar carbon – halogen bonds as their functional groups. In Chapters 6 and 7 we shall explore their chemistry in depth. Another functionality is the hydroxy group, – O – H, characteristic of alcohols. The characteristic functional unit of ethers is an oxygen bonded to two carbon atoms ƒ ƒ (OCOOOC O). The functional group in alcohols and in some ethers can be converted into ƒ ƒ a large variety of other functionalities and are therefore important in synthetic transformations. This chemistry is the subject of Chapters 8 and 9. Haloalkanes .. CH3Cl .. :
.. CH3CH2 Cl .. :
Chloromethane (Methyl chloride)
Chloroethane (Ethyl chloride)
(Topical anesthetics)
Alcohols .. .. CH3 O CH3CH2 O ..H ..H
Ethers .. CH3 O . . CH3
.. CH3CH2 O . . CH2CH3
Methanol
Ethanol
Methoxymethane (Dimethyl ether)
Ethoxyethane (Diethyl ether)
(Wood alcohol)
(Grain alcohol)
(A refrigerant)
(An inhalation anesthetic)
The carbonyl functional group, CPO, is found in aldehydes, in ketones, and, in conjunction with an attached 2OH, in the carboxylic acids. Aldehydes and ketones are discussed in Chapters 17 and 18, the carboxylic acids and their derivatives in Chapters 19 and 20. Aldehydes
Ketones
ðOð B HCH
ðOð B CH3CH or CH3CHO
ðOð B CH3CCH3
ðOð B CH3CH2CCH3
Formaldehyde
Acetaldehyde
Acetone
Butanone (Methyl ethyl ketone)
(A disinfectant)
(A hypnotic)
(Common solvents)
Carboxylic Acids ðOð B š HCOH or HCOOH or HCO2H Formic acid (Strong irritant)
ðOð B š CH3COH or CH3COOH or CH3CO2H Acetic acid (In vinegar)
69
70
Chapter 2
Structure and Reactivity
Other elements give rise to further characteristic functional groups. For example, alkyl nitrogen compounds are amines. The replacement of oxygen in alcohols by sulfur furnishes thiols. Amines
A Thiol
š 2 CH3NH
H A CH3 NCH3 or (CH3)2š NH
CH3š SH
Methanamine (Methylamine)
N-Methylmethanamine (Dimethylamine)
Methanethiol
(Used in tanning leather)
(Excreted after we eat asparagus)
R represents a part of an alkane molecule Table 2-3 depicts a selection of common functional groups, the class of compounds to which they give rise, a general structure, and an example. In the general structures, we commonly use the symbol R (for radical or residue) to represent an alkyl group, a molecular fragment derived by removal of one hydrogen atom from an alkane (Section 2-5). Therefore, a general formula for a haloalkane is R – X, in which R stands for any alkyl group and X for any halogen. Alcohols are similarly represented as R – O – H. In structures that contain multiple alkyl groups, we add a prime (9) or double prime (0) to R to distinguish groups that differ in structure from one another. Thus a general formula for an ether in which both alkyl groups are the same (a symmetrical ether) is R – O – R, whereas an ether with two dissimilar groups (an unsymmetrical ether) is represented by R – O – R9.
2-4 Straight-Chain and Branched Alkanes The functional groups in organic molecules are typically attached to a hydrocarbon scaffold constructed only with single bonds. Substances consisting entirely of single-bonded carbon and hydrogen atoms and lacking functional groups are called alkanes. They are classified into several types according to structure: the linear straight-chain alkanes; the branched alkanes, in which the carbon chain contains one or several branching points; and the cyclic alkanes, or cycloalkanes, which we shall cover in Chapter 4. A Straight-Chain Alkane
A Branched Alkane
A Cycloalkane
CH3O CH2 O CH2 O CH3
CH3 A CH3O COH A CH3
CH2 O CH2 A A CH2 O CH2
Butane, C4 H10
2-Methylpropane, C4H10 (Isobutane)
MODEL BUILDING
Cyclobutane, C4 H8
Straight-chain alkanes form a homologous series In the straight-chain alkanes, each carbon is bound to its two neighbors and to two hydrogen atoms. Exceptions are the two terminal carbon nuclei, which are bound to only one carbon atom and three hydrogen atoms. The straight-chain alkane series may be described by the general formula H – (CH2)n – H. Each member of this series differs from the next lower one by the addition of a methylene group, – CH2 – . Molecules that are related in this way are homologs of each other (homos, Greek, same as), and the series is a homologous series. Methane (n 5 1) is the first member of the homologous series of the alkanes, ethane (n 5 2) the second, and so forth.
2-5 Naming the Alkanes
Branched alkanes are constitutional isomers of straight-chain alkanes Branched alkanes are derived from the straight-chain systems by removal of a hydrogen from a methylene (CH2) group and replacement with an alkyl group. Both branched and straight-chain alkanes have the same general formula, CnH2n12. The smallest branched alkane is 2-methylpropane. It has the same molecular formula as that of butane (C4H10) but different connectivity; the two compounds therefore form a pair of constitutional isomers (Section 1-9). For the higher alkane homologs (n . 4), more than two isomers are possible. There are three pentanes, C5H12, as shown below. There are five hexanes, C6H14; nine heptanes, C7H16; and eighteen octanes, C8H18.
The Isomeric Pentanes
CH3O CH2 O CH2 OCH2 OCH3 Pentane
CH3 A CH3 O CH2 O CH A CH3
CH3 A CH3O CO CH3 A CH3
2-Methylbutane (Isopentane)
2,2-Dimethylpropane (Neopentane)
The number of possibilities in connecting n carbon atoms to each other and to 2n 1 2 surrounding hydrogen atoms increases dramatically with the size of n (Table 2-4).
Exercise 2-16 (a) Draw the structures of the five isomeric hexanes. (b) Draw the structures of all the possible next higher and lower homologs of 2-methylbutane.
2-5 Naming the Alkanes The multiple ways of assembling carbon atoms and attaching various substituents accounts for the existence of the very large number of organic molecules. This diversity poses a problem: How can we systematically differentiate all these compounds by name? Is it possible, for example, to name all the C6H14 isomers so that information on any of them (such as boiling points, melting points, reactions) might easily be found in the index of a handbook? And is there a way to name a compound that we have never seen in such a way as to be able to draw its structure? This problem of naming organic molecules has been with organic chemistry from its very beginning, but the initial method was far from systematic. Compounds have been named after their discoverers (“Nenitzescu’s hydrocarbon”), after localities (“sydnones”), after their shapes (“cubane,” “basketane”), and after their natural sources (“vanillin”). Many of these common or trivial names are still widely used. However, there now exists a precise system for naming the alkanes. Systematic nomenclature, in which the name of a compound describes its structure, was first introduced by a chemical congress in Geneva, Switzerland, in 1892. It has continually been revised since then, mostly by the International Union of Pure and Applied Chemistry (IUPAC). Table 2-5 gives the systematic names of the first 20 straight-chain alkanes. Their stems, mainly of Greek origin, reveal the number of carbon atoms in the chain. For example, the name heptadecane is composed of the Greek words hepta, seven, and deka, ten. The first four alkanes have special names that have been accepted as part of the systematic
Chapter 2
71
Number of Table 2-4 Possible Isomeric Alkanes, Cn H2n 12 n
Isomers
1 2 3 4 5 6 7 8 9 10 15 20
1 1 1 2 3 5 9 18 35 75 4,347 366,319
MODEL BUILDING
72
Chapter 2
Structure and Reactivity
Table 2-5
Propane, stored under pressure in liquefied form in canisters such as these, is a common fuel for torches, lanterns, and outdoor cooking stoves.
CH3 A CH3 O CO (CH2 )n O CH3 A H An isoalkane (e.g., n ⴝ 1, isopentane)
CH3 A CH3 O CO (CH2 )n O H A CH3 A neoalkane (e.g., n ⴝ 2, neohexane)
Names and Physical Properties of Straight-Chain Alkanes, CnH2n12
n
Name
Formula
Boiling point (8C)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Methane Ethane Propane Butane Pentane Hexane Heptane Octane Nonane Decane Undecane Dodecane Tridecane Tetradecane Pentadecane Hexadecane Heptadecane Octadecane Nonadecane Icosane
CH4 CH3CH3 CH3CH2CH3 CH3CH2CH2CH3 CH3(CH2)3CH3 CH3(CH2)4CH3 CH3(CH2)5CH3 CH3(CH2)6CH3 CH3(CH2)7CH3 CH3(CH2)8CH3 CH3(CH2)9CH3 CH3(CH2)10CH3 CH3(CH2)11CH3 CH3(CH2)12CH3 CH3(CH2)13CH3 CH3(CH2)14CH3 CH3(CH2)15CH3 CH3(CH2)16CH3 CH3(CH2)17CH3 CH3(CH2)18CH3
2161.7 288.6 242.1 20.5 36.1 68.7 98.4 125.7 150.8 174.0 195.8 216.3 235.4 253.7 270.6 287 301.8 316.1 329.7 343
Melting point (8C)
Density at 208C (g mL21)
2182.5 2183.3 2187.7 2138.3 2129.8 295.3 290.6 256.8 253.5 229.7 225.6 29.6 25.5 5.9 10 18.2 22 28.2 32.1 36.8
0.466 (at 21648C) 0.572 (at 21008C) 0.5853 (at 2458C) 0.5787 0.6262 0.6603 0.6837 0.7026 0.7177 0.7299 0.7402 0.7487 0.7564 0.7628 0.7685 0.7733 0.7780 0.7768 0.7855 0.7886
nomenclature but also all end in -ane. It is important to know these names, because they serve as the basis for naming a large fraction of all organic molecules. A few smaller branched alkanes have common names that still have widespread use. They make use of the prefixes iso- and neo- (margin), as in isobutane, isopentane, and neohexane.
Exercise 2-17 Draw the structures of isohexane and neopentane.
Alkyl groups CH3 — Methyl
CH3 — CH2 — Ethyl
CH3 — CH2 — CH2 — Propyl
As mentioned in Section 2-4, an alkyl group is formed by the removal of a hydrogen from an alkane. It is named by replacing the ending -ane in the corresponding alkane by -yl, as in methyl, ethyl, and propyl. Table 2-6 shows a few branched alkyl groups having common names. Note that some have the prefixes sec- (or s-), which stands for secondary, and tert- (or t-), for tertiary. These prefixes are used to classify sp3-hybridized (tetrahedral) carbon atoms in organic molecules. A primary carbon is one attached directly to only one other carbon atom. For example, all carbon atoms at the ends of alkane chains are primary. The hydrogens attached to such carbons are designated primary hydrogens, and an alkyl group created by removing a primary hydrogen also is called primary. A secondary carbon is attached directly to two other carbon atoms, and a tertiary carbon to three others. Their hydrogens are labeled similarly. As shown in Table 2-6, removal of a secondary hydrogen results in a secondary alkyl group, and removal of a tertiary hydrogen in a tertiary alkyl group. Finally, a carbon bearing four alkyl groups is called quaternary.
2-5 Naming the Alkanes
Chapter 2
73
Branched Alkyl Groups
Table 2-6 Structure
Common name
CH3 A CH3 O C O A H CH3 A CH3 O C O CH2 O A H CH3 A CH3 O CH2 O C O A H CH3 A CH3 O C O A CH3 CH3 A CH3 O C O CH2 O A CH3
Example of common name in use
Systematic name
Type of group
Isopropyl
CH3 A CH3 O C O Cl (Isopropyl chloride) A H
1-Methylethyl
Secondary
Isobutyl
CH3 A CH3 O C O CH3 (Isobutane) A H
2-Methylpropyl
Primary
sec-Butyl
CH3 A CH3 O CH2 O C O NH2 (sec-Butyl amine) A H
1-Methylpropyl
Secondary
tert-Butyl
CH3 A CH3 O C O Br (tert-Butyl bromide) A CH3
1,1-Dimethylethyl
Tertiary
Neopentyl
CH3 A CH3 O C O CH2 O OH (Neopentyl alcohol) A CH3
2,2-Dimethylpropyl
Primary
Primary, Secondary, and Tertiary Carbons and Hydrogens Primary C Secondary C
Primary H
CH3 Tertiary C A CH3CH2CCH2CH3 A H Secondary H Tertiary H
3-Methylpentane
Exercise 2-18 Label the primary, secondary, and tertiary hydrogens in 2-methylpentane (isohexane).
The information in Table 2-5 enables us to name the first 20 straight-chain alkanes. How do we go about naming branched systems? A set of IUPAC rules makes this a relatively simple task, as long as they are followed carefully and in sequence. IUPAC Rule 1. Find the longest chain in the molecule and name it. This task is not as easy as it seems. The problem is that, in the condensed formula, complex alkanes may be written in ways that mask the identity of the longest chain. Do not assume that it is always depicted horizontally! In the following examples, the longest chain, or stem chain, is clearly marked; the alkane stem gives the molecule its name. Groups other than hydrogen attached to the stem chain are called substituents. Methyl
CH3 A CH3CHCH2CH3
CH3CH2 CH2CH2CH2CH3 A A Ethyl CH3CHCH2CH2CHCH2CH3
A methyl-substituted butane (A methylbutane)
An ethyl- and methyl-substituted decane (An ethylmethyldecane)
The stem chain is shown in black in the examples in this section.
74
Structure and Reactivity
Chapter 2
If a molecule has two or more chains of equal length, the chain with the largest number of substituents is the base stem chain. CH3 CH3 A A CH3CHCHCHCHCH2CH3 A A CH3 CH2 A CH2 A CH3
CH3 CH3 A A CH3CHCHCHCHCH2CH3 A A CH3 CH2 A CH2 A CH3
not
4 substituents
3 substituents
A heptane Correct stem chain
A heptane Incorrect stem chain
Here are two more examples, drawn with the use of bond-line notation: Ethyl Methyl
A methylbutane
An ethylmethyldecane
IUPAC Rule 2. Name all groups attached to the longest chain as alkyl substituents. For straight-chain substituents, Table 2-5 can be used to derive the alkyl name. However, what if the substituent chain is branched? In this case, the same IUPAC rules apply to such complex substituents: First, find the longest chain in the substituent; next, name all its substituents. IUPAC Rule 3. Number the carbons of the longest chain beginning with the end that is closest to a substituent. CH3
7
CH3CHCH2CH3 1
2
3
4
not 4
3
2
1
3
8
not 1
5
4
4
5
2
1
6 2
3
6
7
8
If there are two substituents at equal distance from the two ends of the chain, use the alphabet to decide how to number. The substituent to come first in alphabetical order is attached to the carbon with the lower number.
CH3CH2
CH3
16
12
14
10
8
6
4
2
CH3CH2CHCH2CH2CHCH2CH3 1
2
3
4
5
6
7
1
8
17
15
11
13
Ethyl before methyl
9
7
5
3
Butyl before propyl
What if there are three or more substituents? Then number the chain in the direction that gives the lower number at the first difference between the two possible numbering schemes. This procedure follows the first point of difference principle. CH3
CH3
CH3
CH3CH2CHCH2CH2CH2CH2CHCH2CHCH2CH3 1
2
3
4
5
6
7
8
9
10
11
12
12
11
10
9
8
7
6
5
4
3
2
1
Numbers for substituted carbons: 3, 8, and 10 (incorrect) 3, 5, and 10 (correct; 5 lower than 8)
3,5,10-Trimethyldodecane
Substituent groups are numbered outward from the main chain, with C1 of the group being the carbon attached to the main stem.
2-5 Naming the Alkanes
IUPAC Rule 4. Write the name of the alkane by first arranging all the substituents in alphabetical order (each preceded by the carbon number to which it is attached and a hyphen) and then adding the name of the stem. Should a molecule contain more than one of a particular substituent, its name is preceded by the prefix di, tri, tetra, penta, and so forth. The positions of attachment to the stem are given collectively before the substituent name and are separated by commas. These prefixes, as well as sec- and tert-, are not considered in the alphabetical ordering, except when they are part of a complex substituent name. CH3
CH3
CH3
CH3CHCH2CH3
CH3CHCHCH3
2-Methylbutane
2,3-Dimethylbutane
75
Chapter 2
3 2
5
5-Ethyl-2,2-dimethyloctane (‘‘di’’ not counted in alphabetical ordering) but
CH3
CH3CHCH2CH2CHCH2CCH3
CH3
CH3CH2
CH3
4-Ethyl-2,2,7-trimethyloctane
5
CH2CH3 CH3CH2CHCHCH3 CH3 3-Ethyl-2-methylpentane
The five common group names in Table 2-6 are permitted by IUPAC: isopropyl, isobutyl, sec-butyl, tert-butyl, and neopentyl. These five are used universally in the course of normal communication between scientists, and it is necessary to know the structures to which they refer. Nonetheless, it is preferable to use systematic names, especially when searching for information about a chemical compound. The online databases containing such information are constructed to recognize the systematic names; therefore, use of a common name as input may not result in retrieval of a complete set of the information being sought. The systematic name of a complex substituent should be enclosed in parentheses to avoid possible ambiguities. If a particular complex substituent is present more than once, a special set of prefixes is placed in front of the parenthesis: bis, tris, tetrakis, pentakis, and so on, for 2, 3, 4, 5, etc. In the chain of a complex substituent, the carbon numbered one (C1) is always the carbon atom directly attached to the stem chain. CH3
3
CH2
2 1 4
2
CH2
First substituent at position 2 determines numbering
Complex alkyl group has carbon 1 attached to 3 the base stem
8
6
1 5
7
9
Longest chain chosen has highest number of substituents
H
4-(1-Ethylpropyl)-2,3,5-trimethylnonane
H CH3 CH3CH CH3CH2CH2CHCH2CH2CH3 4-(1-Methylethyl)heptane (4-Isopropylheptane)
CH2
CH3
C
C
CH2
H
CH2
CH3
C
C
CH2
H
CH3
CH3
CH2 CH2 CH3 5,8-Bis(1-methylethyl)dodecane
1
3 4
2
5-(1,1-Dimethylethyl)-3ethyloctane (‘‘di’’ counted: part of substituent name)
4,5-Diethyl-3,6-dimethyldecane
1
4
76
Chapter 2
Structure and Reactivity
Exercise 2-19 Write down the names of the preceding eight branched alkanes, close the book, and reconstruct their structures from those names.
To name haloalkanes, we treat the halogen as a substituent to the alkane framework. As usual, the longest (stem) chain is numbered so that the first substituent from either end receives the lowest number. Substituents are ordered alphabetically, and complex appendages are named according to the rules used for complex alkyl groups.
CH3I
CH3 A CH3COBr A CH3
CH3 A FCH2CCH3 A H
CH3 A CH3CCH3 A ClCCH3 A CH2 A CH3CH2CH2CH2CH2CHCH2CH2CH2CH2CH3
Iodomethane
2-Bromo-2-methylpropane
1-Fluoro-2-methylpropane
6-(2-Chloro-2,3,3-trimethylbutyl)undecane
Common names are based on the older term alkyl halide. For example, the first three structures above have the common names methyl iodide, tert-butyl bromide, and isobutyl fluoride, respectively. Some chlorinated solvents have common names: for example, carbon tetrachloride, CCl4; chloroform, CHCl3; and methylene chloride, CH2Cl2. Exercise 2-20 Draw the structure of 5-butyl-3-chloro-2,2,3-trimethyldecane.
Further instructions on nomenclature will be presented when new classes of compounds, such as the cycloalkanes, are introduced.
In Summary Four rules should be applied in sequence when naming a branched alkane: (1) Find the longest chain; (2) find the names of all the alkyl groups attached to the stem; (3) number the chain; (4) name the alkane, with substituent names in alphabetical order and preceded by numbers to indicate their locations. Haloalkanes are named in accord with the rules that apply to naming the alkanes, the halo substituent being treated the same as alkyl groups.
2-6 Structural and Physical Properties of Alkanes The common structural feature of all alkanes is the carbon chain. This chain influences the physical properties of not only alkanes but also any organic molecules possessing such a backbone. This section will address the properties and physical appearance of such structures.
Alkanes exhibit regular molecular structures and properties The structural features of the alkanes are remarkably regular. The carbon atoms are tetrahedral, with bond angles close to 1098 and with regular C –H (< 1.10 Å) and C – C (< 1.54 Å) bond lengths. Alkane chains often adopt the zigzag patterns used in bond-line notation (Figure 2-3). To depict three-dimensional structures, we shall make use of the hashed-wedged line notation (see Figure 1-23). The main chain and a hydrogen at each end are drawn in the plane of the page (Figure 2-4). Exercise 2-21 Draw zigzag hashed-wedged line structures for 2-methylbutane and 2,3-dimethylbutane.
Chapter 2
Figure 2-3 Ball-and-stick (top) and space-filling molecular models of hexane, showing the zigzag pattern of the carbon chain typical of the alkanes. [Model sets courtesy of Maruzen Co., Ltd., Tokyo.]
77
H 109.5⬚ H C 1.10 Å ( H H H
109.5⬚
C H ( H
C
H H &
2-6 Structural and Physical Properties of Alkanes
1.095 Å
H
1.54 Å
The regularity in alkane structures suggests that their physical constants would follow predictable trends. Indeed, inspection of the data presented in Table 2-5 reveals regular incremental increases along the homologous series. For example, at room temperature (25 8C), the lower homologs of the alkanes are gases or colorless liquids, the higher homologs are waxy solids. From pentane to pentadecane, each additional CH2 group causes a 20 – 30 8C increase in boiling point (Figure 2-5). 300
Temperature (°C)
0.75
Density 100
Boiling point 0.65
0
Melting point − 100
− 200
1
2
3
4
5
6
7
8
9
10
11
Solid-state density at 90 K (g cm−3)
200
0.55 12
Number of carbons (n) Figure 2-5 The physical constants of straight-chain alkanes. Their values increase with increasing size because London forces increase. Note that even-numbered systems have somewhat higher melting points than expected; these systems are more tightly packed in the solid state (notice their higher densities), thus allowing for stronger attractions between molecules.
Attractive forces between molecules govern the physical properties of alkanes Why are the physical properties of alkanes predictable? Such trends exist because of intermolecular or van der Waals* forces. Molecules exert several types of attractive forces on each other, causing them to aggregate into organized arrangements as solids and liquids. Most solid substances exist as highly ordered crystals. Ionic compounds, such as salts, are rigidly held in a crystal lattice, mainly by strong Coulomb forces. Nonionic but polar molecules, such as chloromethane (CH3Cl), are attracted by weaker dipole – dipole interactions, also of coulombic origin (Sections 1-2 and 6-1). Finally, the nonpolar alkanes attract each other by London† forces, which are due to electron correlation. When one alkane molecule *Professor Johannes D. van der Waals (1837–1923), University of Amsterdam, The Netherlands, Nobel Prize 1910 (physics). † Professor Fritz London (1900–1954), Duke University, North Carolina. Note: In older references the term “van der Waals forces’’ referred exclusively to what we now call London forces; van der Waals forces now refers collectively to all intermolecular attractions.
H
H H * C
H C C & ¥ & ¥ H H H H
H H H H * * C C H C H C & ¥ & ¥ H H H H H H H H H H * * * C C C C H H C & ¥ & ¥ H H H H Figure 2-4 Hashed-wedged line structures of methane through pentane. Note the zigzag arrangement of the principal chain and two terminal hydrogens.
78
Structure and Reactivity
Chapter 2
London Forces: Unsymmetrical Charge Distribution
Ion-ion Interactions: Full Charges
Dipole-dipole Interactions: Partial Charges
Na+
δ− δ+
δ− δ+
O C ⴚ
A
δ− δ+
CH3
O
B
C
Figure 2-6 (A) Coulombic attraction in an ionic compound: crystalline sodium acetate, the sodium salt of acetic acid. (B) Dipole–dipole interactions in solid chloromethane. The polar molecules arrange to allow for favorable coulombic attraction. (C) London forces in crystalline pentane. In this simplified picture, the electron clouds as a whole mutually interact to produce very small partial charges of opposite sign. The charge distributions in the two molecules change continually as the electrons continue to correlate their movements.
This surfboard is being waxed with paraffin to improve performance.
Octane
2,2,3,3-Tetramethylbutane
approaches another, repulsion of the electrons in one molecule by those in the other results in correlation of their movement. Electron motion causes temporary bond polarization in one molecule; correlated electron motion in the bonds of the other induces polarization in the opposite direction, resulting in attraction between the molecules. Figure 2-6 is a simple picture comparing ionic, dipolar, and London attractions. London forces are very weak. In contrast to Coulomb forces, which change with the square of the distance between charges, London forces fall off as the sixth power of the distance between molecules. There is also a limit to how close these forces can bring molecules together. At small distances, nucleus – nucleus and electron – electron repulsions outweigh these attractions. How do these forces account for the physical constants of elements and compounds? The answer is that it takes energy, usually in the form of heat, to melt solids and boil liquids. For example, to cause melting, the attractive forces responsible for the crystalline state must be overcome. In an ionic compound, such as sodium acetate (Figure 2-6A), the strong interionic forces require a rather high temperature (3248C) for the compound to melt. In alkanes, melting points rise with increasing molecular size: Molecules with relatively large surface areas are subject to greater London attractions. However, these forces are still relatively weak, and even high-molecular-weight alkanes have rather low melting points. For example, a mixture of straightchain alkanes from C20H42 to C40H82 constitutes paraffin wax, which melts below 648C. For a molecule to escape these same attractive forces in the liquid state and enter the gas phase, more heat has to be applied. When the vapor pressure of a liquid equals atmospheric pressure, boiling occurs. Boiling points of compounds are also relatively high if the intermolecular forces are relatively large. These effects lead to the smooth increase in boiling points seen in Figure 2-5. Branched alkanes have smaller surface areas than do their straight-chain isomers. As a result, they are generally subject to smaller London attractions and are unable to pack as well in the crystalline state. The weaker attractions result in lower melting and boiling points. Branched molecules with highly compact shapes are exceptions. For example, 2,2,3,3-tetramethylbutane melts at 1101 8C because of highly favorable crystal packing (compare octane, m.p. 257 8C). On the other hand, the greater surface area of octane compared with that of the more spherical 2,2,3,3-tetramethylbutane is clearly demonstrated in their boiling points (126 8C and 106 8C, respectively). Crystal packing differences also
2-7 Rotation about Single Bonds: Conformations
Chapter 2
79
C H E M I C A L H IGH LIG H T 2-2 “Sexual Swindle” by Means of Chemical Mimicry Bees pollinate flowers. We have all watched nature programs and been told by extremely authoritative-sounding narrators that “Instinct tells the bees which flower to pollinate . . . ,” etc., etc. Instinct, schminstinct. Sex tells the bee which flower to pollinate. Female bees of the species Andrena nigroaenea produce a complex mixture of at least 14 alkanes and alkenes containing 21 to 29 carbon atoms. The fragrance of this mixture attracts males of the same species. Such sex attractants, or pheromones (see Section 12-17), are ubiquitous in the animal kingdom and are typically quite species specific. The orchid Ophrys sphegodes relies on the male Andrena bee for pollination. Remarkably, in the orchid, the leaf wax has a composition almost identical to that of the Andrena pheromone mixture: The three major compounds in both the pheromone and in the wax are the straight-chain alkanes tricosane (C23H48), pentacosane (C25H52), and heptacosane (C27H56) in a 3:3:1 ratio. This is an example of what is termed “chemical mimicry,” the use by one species of a chemical substance to elicit a desired, but not necessarily normal response from another species.
The orchid is even more innovative than most plants, because its flower, whose shape and color already resemble those of the insect, also produces the pheromonelike mixture in high concentration. Thus the male bee is hopelessly attracted to this specific orchid by what is termed by the discoverers of this phenomenon a case of “sexual swindle.”
account for the slightly lower than expected melting points of odd-membered straight-chain alkanes relative to those of even-membered systems (Figure 2-5).
In Summary Straight-chain alkanes have regular structures. Their melting points, boiling points, and densities increase with molecular size and surface area because of increasing attraction between molecules.
2-7 Rotation about Single Bonds: Conformations We have considered how intermolecular forces can affect the physical properties of molecules. These forces act between molecules. In this section, we shall examine how the forces present within molecules (i.e., intramolecular forces) make some geometric arrangements of the atoms energetically more favorable than others. Later chapters will show how molecular geometry affects chemical reactivity.
Rotation interconverts the conformations of ethane If we build a molecular model of ethane, we can see that the two methyl groups are readily rotated with respect to each other. The energy required to move the hydrogen atoms past each other, the barrier to rotation, is only 2.9 kcal mol21 (12.1 kJ mol21). This value turns out to be so low that chemists speak of “free rotation” of the methyl groups. In general, there is free rotation about all single bonds at room temperature. Figure 2-7 depicts the rotational movement in ethane by the use of hashed-wedged line structures (Section 1-9). There are two extreme ways of drawing ethane: the staggered conformation and the eclipsed one. If the staggered conformation is viewed along the C – C axis, each hydrogen atom on the first carbon is seen to be positioned perfectly between two hydrogen atoms on the second. The second extreme is derived from the first by a 60 8 turn of one of the methyl groups about the C – C bond. Now, if this eclipsed conformation is viewed along the C – C axis, all hydrogen atoms on the first carbon are directly opposite those on the second — that is, those on the first eclipse those on the second. A further 60 8 turn converts the eclipsed form into a new but equivalent staggered arrangement. Between these two extremes, rotation of the methyl group results in numerous additional positions, referred to collectively as skew conformations.
MODEL BUILDING
80
Structure and Reactivity
Chapter 2
C
60°
C
H H
C
H Staggered
60°
C
H H
H H
H
H
H
H H
H
C
C
H H
H H Eclipsed
H Staggered
60°
60°
A
B
C
Figure 2-7 Rotation in ethane: (A and C) staggered conformations; (B) eclipsed. There is virtually “free rotation” between conformers.
The many forms of ethane (and, as we shall see, substituted analogs) created by such rotations are called conformations (also called conformers). All of them rapidly interconvert at room temperature. The study of their thermodynamic and kinetic behavior is conformational analysis.
Newman projections depict the conformations of ethane A simple alternative to the hashed-wedged line structures for illustrating the conformers of ethane is the Newman* projection. We can arrive at a Newman projection from the hashedwedged line picture by turning the molecule out of the plane of the page toward us and viewing it along the C – C axis (Figure 2-8A and B). In this notation, the front carbon H View
C H H
H
H H
H
H
H
H
Back carbon
H Front carbon
C C
C H
H
H
H
H
H H C
A
B
Figure 2-8 Representations of ethane. (A) Side-on views of the molecule. (B) End-on views of ethane, showing the carbon atoms directly in front of each other and the staggered positions of the hydrogens. (C) Newman projection of ethane derived from the view shown in (B). The “front” carbon is represented by the intersection of the bonds to its three attached hydrogens. The bonds from the remaining three hydrogens connect to the large circle, which represents the “back” carbon. *Professor Melvin S. Newman (1908–1993), Ohio State University.
2-7 Rotation about Single Bonds: Conformations
H H
HH
H
H 60°
H
H
H
H
H
H
60°
H H
H H
H
60°
Eclipsed
Staggered
obscures the back carbon, but the bonds emerging from both are clearly seen. The front carbon is depicted as the point of juncture of the three bonds attached to it, one of them usually drawn vertically and pointing up. The back carbon is represented by a circle (Figure 2-8C). The bonds to this carbon project from the outer edge of the circle. The extreme conformational shapes of ethane are readily drawn in this way (Figure 2-9). To make the three rear hydrogen atoms more visible in eclipsed conformations, they are drawn somewhat rotated out of the perfectly eclipsing position.
The conformers of ethane have different potential energies The conformers of ethane do not all have the same energy content. The staggered conformer is the most stable and lowest energy state of the molecule. As rotation about the C – C bond axis occurs, the potential energy rises as the structure moves away from the staggered geometry, through skewed shapes, finally reaching the eclipsed conformation. At the point of eclipsing, the molecule has its highest energy content, about 2.9 kcal mol21 above the staggered conformation. The change in energy resulting from bond rotation from the staggered to the eclipsed conformation is called rotational or torsional energy, or torsional strain. The origin of the torsional strain in ethane remains controversial. As rotation to the eclipsed geometry brings pairs of C–H bonds on the two carbons closer to each other, repulsion between the electrons in these bonds increases. Rotation also causes subtle changes in molecular orbital interactions, weakening the C – C bond in the eclipsed conformation. The relative importance of these effects has been a matter of debate for decades, with the most recent published theoretical research (2007) favoring electron repulsion as the major contributor to the rotational energy. A potential-energy diagram (Section 2-1) can be used to picture the energy changes associated with bond rotation. In the diagram for rotation of ethane (Figure 2-10), the x axis denotes degrees of rotation, usually called torsional angle. Figure 2-10 sets 08 at the energy minimum of a staggered conformation, the most stable geometry of the ethane molecule.* Notice that the eclipsed conformer occurs at an energy maximum: Its lifetime is extremely short (less than 10212 s), and it is in fact only a transition state between rapidly equilibrating staggered
*Strictly speaking, the torsional (also called dihedral) angle in a chain of atoms A–B–C–D is defined as the angle between the planes containing A,B,C and B,C,D, respectively. Thus, in Figure 2-10 and subsequent figures in the next section, a torsional angle of 08 would correspond to one of the eclipsed conformations.
81
Figure 2-9 Newman projections and ball-and-stick models of staggered and eclipsed conformers of ethane. In these representations, the back carbon is rotated clockwise in increments of 608.
H
60°
Staggered
Chapter 2
Staggered: More stable, because facing C–H bonds are at maximum distance (dashed lines)
Eclipsed: Less stable, because facing C–H bonds are at minimum distance (dashed lines)
82
Structure and Reactivity
Chapter 2
HH
HH
HH
H H H H Transition state
H H H H Transition state
All eclipsed: energy maxima
H H H H Transition state
Barrier to rotation
E Ea = 2.9 kcal mol −1
H
H H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
Rotation
H
H
H
H 0
60
120
H 180
240
All staggered: energy minima
H 300
360
Torsional angle Figure 2-10 Potential-energy diagram of the rotational isomerism in ethane. Because the eclipsed conformations have the highest energy, they correspond to peaks in the diagram. These maxima may be viewed as transition states between the more stable staggered conformers. The activation energy (Ea ) corresponds to the barrier to rotation.
arrangements. The 2.9 kcal mol21 energy difference between the staggered and eclipsed conformations therefore corresponds to the activation energy for the rotational process. All organic molecules with alkanelike backbones exhibit such rotational behavior. The sections that follow will illustrate these principles in more complex alkanes. Later chapters will show how the chemical reactivity of functionalized molecules can depend on their conformational characteristics.
In Summary Intramolecular forces control the arrangement of substituents on neighboring and bonded carbon atoms. In ethane, the relatively stable staggered conformations are interconverted by rotation through higher-energy transition states in which the hydrogen atoms are eclipsed. Because the energy barrier to this motion is so small, rotation is extremely rapid at ordinary temperatures. A potential-energy diagram conveniently depicts the energetics of rotation about the C – C bond.
2-8 Rotation in Substituted Ethanes How does the potential-energy diagram change when a substituent is added to ethane? Consider, for example, propane, whose structure is similar to that of ethane, except that a methyl group replaces one of ethane’s hydrogen atoms.
Steric hindrance raises the energy barrier to rotation MODEL BUILDING
A potential-energy diagram for the rotation about a C – C bond in propane is shown in Figure 2-11. The Newman projections of propane differ from those of ethane only by the substituted methyl group. Again, the extreme conformations are staggered and eclipsed. However, the activation barrier separating the two is 3.2 kcal mol21 (13.4 kJ mol21), slightly higher than that for ethane. This energy difference is due to unfavorable interference between the methyl substituent and the eclipsing hydrogen in the transition state, a phenomenon called steric hindrance. This effect arises from the fact that two atoms or groups of atoms cannot occupy the same region in space.
2-8 Rotation in Substituted Ethanes
H CH3
H CH3
H H
H H
H H
Eclipsed
Chapter 2
H CH3
H H
H H
Eclipsed
Steric Hindrance
H H Eclipsed
E Ea = 3.2 kcal mol −1
CH3
CH3
CH3
CH3
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H 0
60
120
H 180
240
H 300
360
Torsional angle Figure 2-11 Potential-energy diagram of rotation about either C – C bond in propane. Steric hindrance increases the relative energy of the eclipsed form.
Steric hindrance in propane is actually worse than the Ea value for rotation indicates. Methyl substitution raises the energy not only of the eclipsed conformation, but also of the staggered (lowest-energy, or ground state) conformation, the staggered to a lesser extent because of less hindrance. However, because the activation energy is equal to the difference in energy between ground and transition states, the net result is only a small increase in Ea.
There can be more than one staggered and one eclipsed conformation: conformational analysis of butane If we build a model and look at the rotation about the central C – C bond of butane, we find that there are more conformations than one staggered and one eclipsed (Figure 2-12). Consider the staggered conformer in which the two methyl groups are as far away from each other as possible. This arrangement, called anti (i.e., opposed), is the most stable because steric hindrance is minimized. Rotation of the rear carbon in the Newman projections in either direction (in Figure 2-12, the direction is clockwise) produces an eclipsed conformation with two CH3 – H interactions. This conformer is 3.6 kcal mol21 (15.1 kJ mol21) higher in energy than the anti precursor. Further rotation furnishes a new staggered structure in which the two methyl groups are closer than they are in the anti conformation. To distinguish this conformer from the others, it is named gauche (gauche, French, in the sense of awkward, clumsy). As a consequence of steric hindrance, the gauche conformer is higher in energy than the anti conformer by about 0.9 kcal mol21 (3.8 kJ mol21). Further rotation (Figure 2-12) results in a new eclipsed arrangement in which the two methyl groups are superposed. Because the two bulkiest substituents eclipse in this conformer, it is energetically highest, 4.9 kcal mol21 (20.5 kJ mol21) higher than the most stable anti structure. Further rotation produces another gauche conformer. The activation energy for gauche 3 4 gauche interconversion is 4.0 kcal mol21 (16.7 kJ mol21). A potential-energy diagram summarizes the energetics of the rotation (Figure 2-13). The most stable anti conformer is the most abundant in solution (about 72% at 25 8C). Its less stable gauche counterpart is present in lower concentration (28%).
MODEL BUILDING
83
84
Structure and Reactivity
Chapter 2
Steric Hindrance CH3
H CH3
H
CH3 H3C
H
H3C CH3
CH3
H
CH3
H
etc.
H
H H
H CH3
H CH3
H
H
H H
H H H
H
H
H H
etc.
Anti Most stable
Gauche Less stable than anti
Gauche Less stable than anti
Figure 2-12 Clockwise rotation of the rear carbon along the C2 – C3 bond in a Newman projection (top) and a ball-and-stick model (bottom) of butane.
H3C CH3
H CH3
H H
H CH3
H CH3
H H
H H
Ea1 = 3.6 kcal mol−1
CH3 H
H H
Ea2 = 4.0 kcal mol−1
E
Steric Hindrance
Ea3 = 2.7 kcal mol−1
0.9 kcal mol−1
CH3
CH3
CH3
CH3
H
H
H3C
H
H
CH3
H
H
H
H
H
H
H
H
H
H
CH3 anti 0
60
H
H
gauche
gauche
120
180
240
CH3 anti 300
360
Torsional angle Figure 2-13 Potential-energy diagram of the rotation about the C2 – C3 bond in butane. There are three processes: anti n gauche conversion with Ea1 5 3.6 kcal mol21; gauche n gauche rotation with Ea2 5 4.0 kcal mol21; and gauche n anti transformation with Ea3 5 2.7 kcal mol21.
2-8 Rotation in Substituted Ethanes
We can see from Figure 2-13 that knowing the difference in thermodynamic stability of two conformers (e.g., 0.9 kcal mol21 between the anti and gauche isomers) and the activation energy for proceeding from the first to the second (e.g., 3.6 kcal mol21, 15.1 kJ mol21) allows us to estimate the activation barrier of the reverse reaction. In this case, Ea for the gaucheto-anti conversion is 3.6 2 0.9 5 2.7 kcal mol21 (11.3 kJ mol21). Exercise 2-22
Working with the Concepts: Conformations Draw a qualitative potential-energy diagram for rotation about the C3– C4 bond in 2-methylpentane. Show Newman projections for all conformations located at the maximum and minimum points on your graph. Describe similarities and differences with other molecules discussed in this section. Strategy First identify the structure of the molecule in question and the bond to be examined: CH3 CH3
C3– C4 bond
CH
CH2
CH2
CH3
We have been asked to examine rotation about a bond between two CH2 groups, each of which contains one alkyl substituent. This situation therefore resembles rotation about the C2 – C3 bond in butane (see Figures 2-12 and 2-13), and the Newman projections and potential-energy diagram will be similar. Solution • Using Figures 2-12 and 2-13 as guides, we can produce the following Newman projections and potential-energy diagram: H CH(CH3)2 H
CH3
H3C CH(CH3)2
H H
H
H
H CH(CH3)2
H H
H
H
CH3 H
E
H
CH(CH3)2 H
H
H3C H
H CH3
0°
CH(CH3)2 H H H
60°
120°
H
CH(CH3)2 CH3
H
H H
H H
180° 240° Torsional angle
CH(CH3)2 H H CH3
300°
360°
• The single difference is that one of the alkyl groups is a 1-methylethyl (isopropyl) substituent instead of CH3. Because of its greater size, the energies of all its steric interactions will be larger, especially in conformations that bring the two alkyl groups close together. Therefore, the energy differences between the anti and all other conformations will increase, with the greatest increase at 1808.
Exercise 2-23
Try It Yourself Draw the expected potential-energy diagram for the rotation about the C2 – C3 bond in 2,3dimethylbutane. Include the Newman projections of each staggered and eclipsed conformation.
Chapter 2
85
86
Chapter 2
Structure and Reactivity
THE BIG PICTURE The familiar chemistry of acids and bases provides a framework for understanding many of the most important reactions between organic molecules. Much of the chemistry we explore in the upcoming chapters expands on the concept that electrophiles and nucleophiles are mutually attractive species, analogous to acids and bases. By identifying polar sites in molecules, we can develop the ability to understand, and even to predict, what kinds of reactions these molecules will undergo. The sites of reactivity in organic molecules are called functional groups. Functional groups serve to characterize the main classes of organic compounds. Beginning in Chapter 6 and continuing through to the end of the book, we will discuss these compound classes one by one, examining how their properties and reactivity arise from their structural characteristics. Most organic molecules consist of hydrocarbon skeletons with functional groups attached. We began a study of these skeletons with a discussion of the names and structures of alkanes, hydrocarbons that lack functional groups. We presented concepts of molecular motion and their associated energy changes. These ideas will underlie the behavior of molecules of all types, and we will return to them frequently.
CHAPTER INTEGRATION PROBLEMS 2-24. Consider the alkane shown in the margin. a. Name this molecule according to the IUPAC system. SOLUTION Step 1. Locate the main, or stem, chain, the longest one in the molecule (shown in black below). Do not be misled: The drawing of the stem chain can have almost any shape. The stem has eight carbons, so the base name is octane. Step 2. Identify and name all substituents (shown in color): two methyl groups, an ethyl group, and a fourth, branched substituent. The branched substituent is named by first giving the number 1 (italicized in the illustration below) to the carbon that connects it to the main stem. By numbering away from the stem, we reach the number 2; therefore, the substituent is a derivative of the ethyl group (in green), onto which is attached a methyl group (red) at carbon 1. Thus, this substituent is called a 1-methylethyl group. Step 3. Number the stem chain, starting at the end closest to a carbon bearing a substituent. The numbering shown gives a methyl-substituted carbon the number 3. Numbering the opposite way would have C4 as the lowest numbered substituted carbon. Step 4. Arrange the names of substituents alphabetically in the final name: ethyl comes first; then methyl comes before methylethyl (the “di” in dimethyl, denoting two methyl groups, is not considered in alphabetization because it is a multiplier of a substituent name and is therefore not considered part of the name). For more practice naming, see Problem 35. 2 3 4
1 5
6
7
8
1 2
4-Ethyl-3,4-dimethyl-5-(1-methylethyl)octane
b. Draw structures to represent rotation about the C6 – C7 bond. Correlate the structures that you draw with a qualitative potential energy diagram. SOLUTION Step 1. Identify the bond in question. Notice that much of the molecule can be treated simply as a large, complicated substituent on C6, the specific structure of which is unimportant. For the purpose
Chapter Integration Problems
87
Chapter 2
of this question, this large substituent may be replaced by R. The “action” in this problem takes place between C6 and C7:
6 7
⫽R
6 7
Step 2. Recognize that step 1 simplified the problem: Rotation about the C6 – C7 bond will give results very similar to rotation about the C2 – C3 bond in butane. The only difference is that a large R group has replaced one of the smaller methyl groups of butane. Step 3. Draw conformations modeled after those of butane (Section 2-8) and superimpose them on an energy diagram similar to that in Figure 2-13. The only difference between this diagram and that for butane is that we do not know the exact heights of the energy maxima relative to the energy minima. However, we can expect them to be higher, qualitatively, because our R group is larger than a methyl group and thus can be expected to cause greater steric hindrance. c. Two alcohols derived from this alkane are illustrated in the margin. Alcohols are categorized on the basis of the type of carbon atom that contains the – OH group (primary, secondary, or tertiary). Characterize the alcohols shown in the margin. SOLUTION In alcohol 1, the – OH group is located on a carbon atom that is directly attached to one other carbon, a primary carbon. Therefore, alcohol 1 is a primary alcohol. Similarly, the – OH group in alcohol 2 resides on a tertiary carbon (one attached to three other carbon atoms). It is a tertiary alcohol.
OH
Alcohol 1
d. The – O – H bond in an alcohol is acidic to a similar degree to that in water. Primary alcohols have Ka < 10216; tertiary alcohols Ka < 10218. What are the approximate pKa values for alcohols 1 and 2? Which is the stronger acid? SOLUTION The pKa for alcohol 1 is approximately 16 (2log Ka); that for alcohol 2 is about 18. Alcohol 1, with the lower pKa value, is the stronger acid. e. In which direction does the following equilibrium lie? Calculate K, the equilibrium constant, and DG 8, the free energy change, associated with the reaction as written in the left-to-right direction. O⫺
OH
⫹
⫹
O⫺
OH
SOLUTION The stronger acid (alcohol 1) is on the left; the weaker (alcohol 2) is on the right. Recall the relation between conjugate acids and bases: Stronger acids have weaker conjugate bases, and vice versa. Relatively speaking, therefore, we have
Alcohol 1 (Stronger acid)
1
Conjugate base of alcohol 2 (Stronger base)
z Oy
Conjugate base of alcohol 1 (Weaker base)
1
Alcohol 2 (Weaker acid)
The equilibrium lies to the right, on the side of the weaker acid-base pair. Recall that K . 1 and DG 8 , 0 for a reaction that is thermodynamically favorable as written from left to right; use this information to be sure to get the magnitude of K and the sign of DG 8 correct. The equilibrium constant, K, for the process is the ratio of the Ka values, (10216y10218) 5 102 (not 1022). With
OH Alcohol 2
88
Chapter 2
Structure and Reactivity
reference to Table 2-1, a K value of 100 corresponds to a DG 8 of 22.73 kcal mol21 (not 12.73). If the reaction were written in the opposite direction, with the equilibrium lying to the left, the correct values would be those in parentheses. For more practice with acids and bases, see Problem 27. 2-25. a. Calculate the equilibrium concentrations of gauche and anti butane at 25 8C using the data from Figure 2-13. SOLUTION We are given the relevant equations in Section 2-1. In particular, at 25 8C the relationship between the Gibbs free energy and the equilibrium constant simplifies to DG 8 5 21.36 log K. The energy difference between the conformations is 0.9 kcal mol21. Substitution of this value into the equation gives K 5 0.219 5 [gauche]y[anti]. Conversion to percentages may be achieved by recognizing that % gauche 5 100% 3 [gauche]y([anti] 1 [gauche]). Thus we find % gauche 5 100% 3 (0.219)y (1.0 1 0.219) 5 18%, and therefore % anti 5 82%. The problem is, we are given the answer on page 83: At 25 8C, butane consists of 28% gauche and 72% anti. What’s gone wrong? The error derives from the fact that the energy values given in Sections 2-7 and 2-8 are enthalpies, not free energies, and we have failed to include an entropy contribution to the free energy equation DG 8 5 DH 8 2 TDS 8. How do we correct this problem? We may look for equations to calculate DS 8, but let us use another, more intuitive approach. Examine Figure 2-13 again. Note that over a 360 8 rotation the butane molecule passes from its anti conformation through two distinct gauche conformations before returning to its original anti geometry. The entropy term arises from the availability of two gauche conformers equilibrating with a single anti. Thus we really have three species in equilibrium, not two. Is there a way to solve this problem without going through calculations to determine DS 8 and DG 8? The answer is yes, and it is not very difficult. Returning to Figure 2-13, let us label the two gauche forms A and B to distinguish them from each other. When we calculated K in our original solution, what we actually determined was the value associated with equilibration between the anti and just one of the two gauche conformers, say gaucheA. Of course, the value for gaucheB is identical, because the two gauche conformations are equal in energy. So, K 5 [gaucheA]y[anti] 5 0.219, and K 5 [gaucheB]y[anti] 5 0.219. Finally, recognizing that total [gauche] 5 [gaucheA] 1 [gaucheB], we therefore have % gauche 5 100% 3 ([gaucheA] 1 [gaucheB])y([anti] 1 [gaucheA] 1 [gaucheB]), giving us % gauche 5 100% 3 (0.219 1 0.219)y(1.0 1 0.219 1 0.219) 5 30%, and therefore % anti 5 70%, in much better agreement with the values given for the equilibrium percentages in the text. b. Calculate the equilibrium concentrations of gauche and anti butane at 100 8C. SOLUTION As above, we determine K from the enthalpy difference in Figure 2-13, using this value as to input for DG 8 in the more general equation DG 8 5 22.303 RT log K. We follow by correcting for the presence of two gauche conformations in the overall equilibrium. Don’t forget that we must use degrees Kelvin, 373 K for T. Solving, we obtain K 5 0.297 5 [gaucheA]y[anti] 5 [gaucheB]y[anti]. Therefore, % gauche 5 100% 3 (0.297 1 0.297)y(1.0 1 0.297 1 0.297) 5 37%, and therefore % anti 5 63%. The less stable conformations are more prevalent at higher temperatures, a straightforward consequence of the Boltzmann distribution, which states that more molecules have higher energies at higher temperatures (Section 2-1). For more practice with conformations, see Problems 42, 44, and 50.
Important Concepts 1. Chemical reactions can be described as equilibria controlled by thermodynamic and kinetic parameters. The change in the Gibbs free energy, DG 8, is related to the equilibrium constant by DG 8 5 2RT ln K 5 21.36 log K (at 25 8C). The free energy has contributions from changes in enthalpy, DH 8, and entropy, DS 8: DG 8 5 DH 8 2 TDS 8. Changes in enthalpy are due mainly to differences between the strengths of the bonds made and those of the bonds broken. A reaction is exothermic when the former is larger than the latter. It is endothermic when there is a net loss in combined bond strengths. Changes in entropy are controlled by the relative degree of energy dispersal in starting materials compared with that in products. The greater the increase in energy dispersal, the larger a positive DS 8.
Problems
89
Chapter 2
2. The rate of a chemical reaction depends mainly on the concentrations of starting material(s), the activation energy, and temperature. These correlations are expressed in the Arrhenius equation: rate constant k 5 Ae2Ea /RT . 3. If the rate depends on the concentration of only one starting material, the reaction is said to be of first order. If the rate depends on the concentrations of two reagents, the reaction is of second order. 4. Brønsted acids are proton donors; bases are proton acceptors. Acid strength is measured by the acid dissociation constant Ka; pKa 5 2log Ka. Acids and their deprotonated forms have a conjugate relation. Lewis acids and bases are electron pair acceptors and donors, respectively. 5. Electron-deficient atoms attack electron rich atoms and are called electrophiles. Conversely, electron-rich atoms attack electron-poor atoms and are called nucleophiles. When a nucleophile, which may be either negatively charged or neutral, attacks an electrophile, it donates a lone electron pair to form a new bond with the electrophile. 6. An organic molecule may be viewed as being composed of a carbon skeleton with attached functional groups. 7. Hydrocarbons are made up of carbon and hydrogen only. Hydrocarbons possessing only single bonds are also called alkanes. They do not contain functional groups. An alkane may exist as a single continuous chain or it may be branched or cyclic. The empirical formula for the straightchain and branched alkanes is CnH2n12. 8. Molecules that differ only in the number of methylene groups, CH2, in the chain are called homologs and are said to belong to a homologous series. 9. An sp3 carbon attached directly to only one other carbon is labeled primary. A secondary carbon is attached to two and a tertiary to three other carbon atoms. The hydrogen atoms bound to such carbon atoms are likewise designated primary, secondary, or tertiary. 10. The IUPAC rules for naming saturated hydrocarbons are (a) find the longest continuous chain in the molecule and name it; (b) name all groups attached to the longest chain as alkyl substituents; (c) number the carbon atoms of the longest chain; (d) write the name of the alkane, citing all substituents as prefixes arranged in alphabetical order and preceded by numbers designating their positions. 11. Alkanes attract each other through weak London forces, polar molecules through stronger dipole – dipole interactions, and salts mainly through very strong ionic interactions. 12. Rotation about carbon–carbon single bonds is relatively easy and gives rise to conformations (conformers). Substituents on adjacent carbon atoms may be staggered or eclipsed. The eclipsed conformation is a transition state between staggered conformers. The energy required to reach the eclipsed state is called the activation energy for rotation. When both carbons bear alkyl or other groups, there may be additional conformers: Those in which the groups are in close proximity (608) are gauche; those in which the groups are directly opposite (1808) each other are anti. Molecules tend to adopt conformations in which steric hindrance, as in gauche conformations, is minimized.
Problems 26. The hydrocarbon propene (CH3OCHPCH2) can react in two different ways with bromine (Chapters 12 and 14).
(i) CH3 OCHPCH2 ⫹ Br2 (ii) CH3 OCHPCH2 ⫹ Br2
Br Br A A CH3 O CHO CH2 CH2 O CHP CH2 ⫹ HBr A Br
(a) Using the bond strengths (kcal mol21) given in the margin, calculate DH 8 for each of these reactions. (b) DS 8 < 0 cal K21 mol21 for one of these reactions and 235 cal K21 mol21 for the other. Which reaction has which DS 8? Briefly explain your answer. (c) Calculate DG 8 for each reaction at 25 8C and at 600 8C. Are both of these reactions thermodynamically favorable at 25 8C? At 600 8C?
Bond COC CPC COH BrOBr HOBr COBr
Average strength (kcal mol21) 83 146 99 46 87 68
90
Chapter 2
Structure and Reactivity
27. (i) Determine whether each species in the following equations is acting as a Brønsted acid or base, and label it. (ii) Indicate whether the equilibrium lies to the left or to the right. (iii) Estimate K for each equation if possible. (Hint: Use the data in Table 2-2.) (a) (b) (c) (d) (e) (f)
H2O 1 HCN 34 H3O1 1 CN2 CH3O2 1 NH3 34 CH3OH 1 NH22 HF 1 CH3COO2 34 F2 1 CH3COOH CH32 1 NH3 34 CH4 1 NH22 H3O1 1 Cl2 34 H2O 1 HCl CH3COOH 1 CH3S2 34 CH3COO2 1 CH3SH
28. Use curved arrows to show electron movement in each acid-base reaction in Problem 27. 29. Identify each of the following species as either a Lewis acid or a Lewis base, and write an equation illustrating a Lewis acid-base reaction for each one. Use curved arrows to depict electron-pair movement. Be sure that the product of each reaction is depicted by a complete, correct Lewis structure. (a) CN2 (d) MgBr2
(c) (CH3)2CH1 (f) CH3S2
(b) CH3OH (e) CH3BH2
30. For each example in Table 2-3, identify all polarized covalent bonds and label the appropriate atoms with partial positive or negative charges. (Do not consider carbon – hydrogen bonds.) 31. Characterize each of the following atoms as being either nucleophilic or electrophilic. (a) (b) (c) (d) (e) (f )
Iodide ion, I2 Hydrogen ion, H1 Carbon in methyl cation, 1CH3 Sulfur in hydrogen sulfide, H2S Aluminum in aluminum trichloride, AlCl3 Magnesium in magnesium oxide, MgO
32. Circle and identify by name each functional group in the compounds pictured. O
OH
(b)
(a)
(c)
(d)
I
O
O H
(e)
(f )
(g)
O O
O
(h)
H
O
O
(i)
N
O
( j) OH
HO
O
33. On the basis of electrostatics (Coulomb attraction), predict which atom in each of the following organic molecules is likely to react with the indicated reagent. Write “no reaction’’ if none seems likely. (See Table 2-3 for the structures of the organic molecules.) (a) Bromoethane, with the oxygen of HO2; (b) propanal, with the nitrogen of NH3; (c) methoxyethane, with H1; (d) 3-hexanone, with the carbon of CH32; (e) ethanenitrile (acetonitrile), with the carbon of CH31; (f ) butane, with HO2. 34. Use curved arrows to show the electron movement in each reaction in Problem 33. 35. Name the following molecules according to the IUPAC system of nomenclature. CH3CH2CHCH3
(a)
CH3CHCH2CH3
(b) CH3CHCH2CH2C CH2CH2CH2CH3
CH H 3C
CH3
CH3
CH3CHCH3
Problems
Chapter 2
CH3 CH2
H
CH3
CH3
C
C
C
CH2
CH2
CH2
CH2
CH3
CH3
CH3
CH
(c) CH3CH2C CH2 CH3
(d) CH3
CH2CH2CH2CH2CH3
CH3
CH3
CH3CH2 ƒ (f) CH2CH2CH2CH3
(e) CH3CH(CH3)CH(CH3)CH(CH3)CH(CH3)2
(g)
(h)
(i)
(j)
36. Convert the following names into the corresponding molecular structures. After doing so, check to see if the name of each molecule as given here is in accord with the IUPAC system of nomenclature. If not, name the molecule correctly. (a) 2-methyl-3-propylpentane; (b) 5-(1,1-dimethylpropyl)nonane; (c) 2,3,4-trimethyl-4-butylheptane; (d) 4-tert-butyl-5isopropylhexane; (e) 4-(2-ethylbutyl)decane; (f) 2,4,4-trimethylpentane; (g) 4-sec-butylheptane; (h) isoheptane; (i) neoheptane. 37. Draw the structures that correspond to the following names. Correct any names that are not in accord with the rules of systematic nomenclature. (a) (b) (c) (d)
4-Chloro-5-methylhexane 3-Methyl-3-propylpentane 1,1,1-Trifluoro-2-methylpropane 4-(3-Bromobutyl)nonane
38. Draw and name all possible isomers of C7H16 (isomeric heptanes). 39. Identify the primary, secondary, and tertiary carbon atoms and the hydrogen atoms in each of the following molecules: (a) ethane; (b) pentane; (c) 2-methylbutane; (d) 3-ethyl-2,2,3,4tetramethylpentane. 40. Identify each of the following alkyl groups as being primary, secondary, or tertiary, and give it a systematic IUPAC name. CH3
CH3
(a)
CH2
(d) CH3
CH
CH2
CH3
CH2
CH2
CH
CH3
CH2
(b) CH3
(e) CH3
CH
CH2
CH3
CH
CH2
CH
CH2
CH3
(c) CH3
(f) CH3
41. Rank the following molecules in order of increasing boiling point (without looking up the real values!): (a) 3-methylheptane; (b) octane; (c) 2,4-dimethylhexane; (d) 2,2,4trimethylpentane. 42. Using Newman projections, draw each of the following molecules in its most stable conformation with respect to the bond indicated: (a) 2-methylbutane, C2 – C3 bond; (b) 2,2-dimethylbutane, C2 – C3 bond; (c) 2,2-dimethylpentane, C3 – C4 bond; (d) 2,2,4-trimethylpentane, C3 – C4 bond. 43. Based on the energy differences for the various conformations of ethane, propane, and butane in Figures 2-10, 2-11, and 2-13, determine the following: (a) (b) (c) (d)
The energy associated with a single hydrogen – hydrogen eclipsing interaction The energy associated with a single methyl – hydrogen eclipsing interaction The energy associated with a single methyl – methyl eclipsing interaction The energy associated with a single methyl – methyl gauche interaction
CH3
CH3
CH
CH
CH3
CH2
CH2
C
CH3
91
92
Structure and Reactivity
Chapter 2
44. At room temperature, 2-methylbutane exists primarily as two alternating conformations of rotation about the C2 – C3 bond. About 90% of the molecules exist in the more favorable conformation and 10% in the less favorable one. (a) Calculate the free energy change (DG 8, more favorable conformation 2 less favorable conformation) between these conformations. (b) Draw a potential-energy diagram for rotation about the C2 – C3 bond in 2-methylbutane. To the best of your ability, assign relative energy values to all the conformations on your diagram. (c) Draw Newman projections for all staggered and eclipsed conformers in (b) and indicate the two most favorable ones. 45. For each of the following naturally occurring compounds, identify the compound class(es) to which it belongs, and circle all functional groups. NH2 H
O
HSCH2CHCOH B O
C C HC
O B HCCHCH2OH
CH3
O B CH3CHCH2CH2OCCH3
Cysteine
CH
HC
(In proteins)
CH C H
OH
CH3CHP CHCqCCq CCHP CHCH2OH
3-Methylbutyl acetate
2,3-Dihydroxypropanal
Benzaldehyde
Matricarianol
(In banana oil)
(The simplest sugar)
(In fruit pits)
(From chamomile)
O CH 3 H3CE C ECH3 C H2C CH2 CH H2C CH2
CH3 A C H2C CH H2 C
CH2
CH A C D M CH2 CH3
H2C H2C H2C
H C N
CH3 f C CH
H3C
CH3 A CH
OP C
CH2
C H
CH2
C
CH3 f
CH CH2
Cineole
Limonene
Heliotridane
Chrysanthenone
(From eucalyptus)
(In lemons)
(An alkaloid)
(In chrysanthemums)
46. Give IUPAC names for all alkyl groups marked by dashed lines in each of the following biologically important compounds. Identify each group as a primary, secondary, or tertiary alkyl substituent. H3C
CH2 CH
H3C CH2 CH
H3C
CH2
CH3 Cholesterol (A steroid)
HO
HO
CH3 O
H3C Vitamin D4
CH3 CH
CH CH3
HO
CH2
CH H3 CH3
CH2
CH2 CH
CH3
CH3
CH3
CH3
CH2CH2CH2CH CH2CH2CH2CH CH2CH2CH2CH CH3 Vitamin E
H3C H3C
CH3 CH
NH2
CH
CH3 CH
CH3 H3C
CH2 CO2H
NH2
CH
CH2 CH
CO2H
NH2
CH
CO2H
Valine
Leucine
Isoleucine
(An amino acid)
(Another amino acid)
(Still another amino acid)
Problems
47.
Using the Arrhenius equation, calculate the effect on k of increases in temperature of 10, 30, and 50 degrees (Celsius) for the following activation energies. Use 300 K (approximately room temperature) as your initial T value, and assume that A is a constant. (a) Ea 5 15 kcal mol21; (b) Ea 5 30 kcal mol21; (c) Ea 5 45 kcal mol21.
48.
The Arrhenius equation can be reformulated in a way that permits the experimental determination of activation energies. For this purpose, we take the natural logarithm of both sides and convert into the base 10 logarithm. ln k 5 ln (Ae2EayRT ) 5 ln A 2 EayRT becomes log k 5 log A 2
Ea 2.3RT
The rate constant k is measured at several temperatures T and a plot of log k versus 1yT is prepared, a straight line. What is the slope of this line? What is its intercept (i.e., the value of log k at 1yT 5 0)? How is Ea calculated? 49. Reexamine your answers to Problem 33. Rewrite each one in the form of a complete equation describing a Lewis acid-base process, showing the product and using curved arrows to depict electron-pair movement. [Hint: For (b) and (d), start with a Lewis structure that represents a second resonance form of the starting organic molecule.] 50.
The equation relating DG8 to K contains a temperature term. Refer to your answer to Problem 44(a) to calculate the answers to the questions that follow. You will need to know that DS 8 for the formation of the more stable conformer of 2-methylbutane from the next most stable conformer is 11.4 cal K21 mol21. (a) Calculate the enthalpy difference (DH 8) between these two conformers from the equation DG 8 5 DH 8 2 TDS 8. How well does this agree with the DH 8 calculated from the number of gauche interactions in each conformer? (b) Assuming that DH 8 and DS 8 do not change with temperature, calculate DG 8 between these two conformations at the following three temperatures: 2250 8C; 2100 8C; 1500 8C. (c) Calculate K for these two conformations at the same three temperatures.
Team Problem 51. Consider the difference in the rate between the following two second-order substitution reactions. Reaction 1: The reaction of bromoethane and iodide ion to produce iodoethane and bromide ion is second order; that is, the rate of the reaction depends on the concentrations of both bromoethane and iodide ion: Rate 5 k[CH3CH2Br][I2] mol L21 s21 Reaction 2: The reaction of 1-bromo-2,2-dimethylpropane (neopentyl bromide) with iodide ion to produce neopentyl iodide and bromide ion is more than 10,000 times slower than the reaction of bromoethane with iodide ion. Rate 5 k[neopentyl bromide][I2] mol L21 s21 (a) Formulate each reaction by using bond-line structural drawings in your reaction scheme. (b) Identify the reactive site of the starting haloalkane as primary, secondary, or tertiary. (c) Discuss how the reaction might take place; that is, how the species would have to interact in order for the reaction to proceed. Remember that, because the reaction is second order, both reagents must be present in the transition state. Use your model kit to help you visualize the trajectory of approach of the iodide ion toward the bromoalkane that enables the simultaneous iodide bond making and bromide bond breaking required by the second-order kinetics of these two reactions. Of all the possibilities, which one best explains the experimentally determined difference in rate between the reactions? (d) Use hashed-wedged line structures to make a three-dimensional drawing of the trajectory on which you agree.
Chapter 2
93
94
Chapter 2
Structure and Reactivity
Preprofessional Problems 52. The compound 2-methylbutane has (a) no secondary H’s (b) no tertiary H’s (d) twice as many secondary H’s as tertiary H’s (e) twice as many primary H’s as secondary H’s
(c) no primary H’s
Potential energy
53.
A+B
C Progress of reaction This energy profile diagram represents (a) an endothermic reaction (c) a fast reaction
CH3 H
H
H
54. In 4-(1-methylethyl)heptane, any H – C – C angle has the value
H CH3 OK
OCH3
H %
HO
0 H 0 OH Genipin
(b) an exothermic reaction (d) a termolecular reaction
O
(a) 120 8 (d) 90 8
(b) 109.5 8 (e) 360 8
(c) 180 8
55. The structural representation shown in the margin is a Newman projection of the conformer of butane that is (a) gauche eclipsed
(b) anti gauche
(c) anti staggered
(d) anti eclipsed
56. Genipin (margin) is a Chinese herbal remedy that is effective against diabetes. To which of the compound classes below does genipin not belong? (a) alcohol
(b) alkene
(c) ester
(d) ether
(e) ketone
C H A P T E R
[
THREE
]
Reactions of Alkanes
Bond-Dissociation Energies, Radical Halogenation, and Relative Reactivity
N O O ombustion of alkanes releases most of the energy that powers modern industrialized society. We saw in Chapter 2 that alkanes lack functional groups; that being the case, how does combustion occur? We will see in this chapter that alkanes are not very reactive, but that they do undergo several types of transformations. These processes, of which combustion is one example, do not involve acid-base chemistry. Instead, they are called radical reactions. Although we shall not explore radical reactions in great depth in this course, they play significant roles in biochemistry (such as aging and disease processes), the environment (destruction of the Earth’s ozone layer), and industry (manufacture of synthetic fabrics and plastics). Radical reactions begin with the breaking of a bond, or bond dissociation. We examine the energetics of this process and discuss the conditions under which it occurs. The majority of the chapter will deal with halogenation, a radical reaction in which a hydrogen atom in an alkane is replaced by halogen. The importance of halogenation lies in the fact that it introduces a reactive functional group, turning the alkane into a haloalkane, which is suitable for further chemical change. For each of these processes, we shall discuss the mechanism involved to explain in detail how the reaction occurs. We shall see that different alkanes, and indeed different bonds in the same alkane molecule, may react at different rates, and we shall see why this is so. C
O
C
A carbon radical
Only a relatively limited number of mechanisms are needed to describe the very large number of reactions in organic chemistry. Mechanisms enable us to understand how and why reactions occur, and what products are likely to form in them. In this chapter we apply mechanistic concepts to explain the effects of halogen-containing chemicals on the stratospheric ozone layer. We conclude with a brief discussion of alkane combustion and show how that process serves as a useful source of thermodynamic information about organic molecules.
NASA’s X-43A hypersonic research aircraft being dropped from under the wing of a B-52B Stratofortress on June 2, 2001. Most supersonic aircraft produce exhaust gases containing molecules such as nitric oxide (NO), whose radical reactions are destructive to the Earth’s stratospheric ozone (O3) layer. In the 1970s the United States abandoned plans to build a fleet of supersonic aircraft (SSTs, or supersonic transports) for just this reason. In contrast, the X-43A is hydrogen fueled, posing no risk to stratospheric ozone, and may represent the first step toward the development of environmentally acceptable high-speed flight. In 2008, Boeing flew successfully the first manned hydrogen-fuelcell-powered aircraft, another aviation milestone.
O
96
Chapter 3
Reactions of Alkanes
3-1 Strength of Alkane Bonds: Radicals In Section 1-2 we explained how bonds are formed and that energy is released on bond formation. For example, bringing two hydrogen atoms into bonding distance produces 104 kcal mol21 (435 kJ mol21) of heat (refer to Figures 1-1 and 1-12). Bond making
DH 8 5 2104 kcal mol21 (2435 kJ mol21)
H? 1 H? uuuuy HOH
Released heat: exothermic
Consequently, breaking such a bond requires heat — in fact, the same amount of heat that was released when the bond was made. This energy is called bond-dissociation energy, DH8, and is a quantitative measure of the bond strength. Bond breaking
HOH uuuuy H? 1 H?
DH 8 5 DH 8 5 104 kcal mol21 (435 kJ mol21) Consumed heat: endothermic
Radicals are formed by homolytic cleavage In our example, the bond breaks in such a way that the two bonding electrons divide equally between the two participating atoms or fragments. This process is called homolytic cleavage or bond homolysis. The separation of the two bonding electrons is denoted by two singlebarbed or “fishhook” arrows that point from the bond to each of the atoms. A single-barbed arrow shows the movement of a single electron.
Homolytic Cleavage: Bonding Electrons Separate AOB
AjjB Radicals
šj ðCl Chlorine atom
H H
C
H
Methyl radical
H H3C
Cj H
Ethyl radical
The fragments that form have an unpaired electron, for example, H ?, Cl ?, CH3 ?, and CH3CH2 ?. When these species are composed of more than one atom, they are called radicals. Because of the unpaired electron, radicals and free atoms are very reactive and usually cannot be isolated. However, radicals and free atoms are present in low concentration as unobserved intermediates in many reactions, such as the production of polymers (Chapter 12) and the oxidation of fats that leads to the spoilage of perishable foods (Chapter 22). In Section 2-2 we introduced an alternative way of breaking a bond, in which the entire bonding electron pair is donated to one of the atoms. This process is heterolytic cleavage and results in the formation of ions. Heterolytic Cleavage: Bonding Electrons Move as Pair AOB
A ðB Ions
A normal, double-barbed curved arrow shows the movement of a pair of electrons.
Homolytic cleavage may be observed in nonpolar solvents or even in the gas phase. In contrast, heterolytic cleavage normally occurs in polar solvents, which are capable of stabilizing ions. Heterolytic cleavage is also restricted to situations where the electronegativies of atoms A and B and the groups attached to them stabilize positive and negative charges, respectively. Dissociation energies, DH 8, refer only to homolytic cleavages. They have characteristic values for the various bonds that can be formed between the elements. Table 3-1 lists dissociation energies of some common bonds. The larger the value for DH 8, the stronger the corresponding bond. Note the relatively strong bonds to hydrogen, as in H – F and H – OH. However, even though these bonds have high DH 8 values, they readily undergo heterolytic cleavage in water to H1 and F2 or HO2; do not confuse homolytic with heterolytic processes.
3-1 Strength of Alkane Bonds: Radicals
Table 3-1
97
Chapter 3
Bond-Dissociation Energies of Various A – B Bonds in the Gas Phase [DH 8 in kcal mol21 (kJ mol21)] B in A – B –H
A in A – B HO CH3O CH3CH2O CH3CH2CH2O (CH3)2CHO (CH3)3CO
104 (435) 105 (439) 101 (423) 101 (423) 98.5 (412) 96.5 (404)
–F 136 110 111 110 111 110
(569) (460) (464) (460) (464) (460)
– Cl 103 85 84 85 84 85
(431) (356) (352) (356) (352) (356)
– Br 87 70 70 70 71 71
(364) (293) (293) (293) (297) (297)
–I 71 57 56 56 56 55
(297) (238) (234) (234) (234) (230)
– OH 119 93 94 92 96 96
(498) (389) (393) (385) (402) (402)
– NH2 108 84 85 84 86 85
Note: (a) DH 8 5 DH 8 for the process A – B n A . 1 . B. (b) These numbers are being revised continually because of improved methods for their measurement. (c) The trends observed for A – H bonds are significantly altered for polar A – B bonds because of dipolar contributions to DH 8.
Bonds are strongest when made by overlapping orbitals that are closely matched in energy and size (Section 1-7). For example, the strength of the bonds between hydrogen and the halogens decreases in the order F . Cl . Br . I, because the p orbital of the halogen contributing to the bonding becomes larger and more diffuse along the series. Thus, the efficiency of its overlap with the relatively small 1s orbital on hydrogen diminishes. A similar trend holds for bonding between the halogens and carbon. Increasing size of the halogen
CH3 – F DH8 5 110
CH3 – Cl 85
CH3 – Br 70
CH3 – l 57 kcal mol21
Decreasing bond strength
Exercise 3-1
Working with the Concepts: Understanding Bond Strengths Compare the bond-dissociation energies of CH3 – F, CH3 – OH, and CH3 – NH2. Why do the bonds get weaker along this series even though the orbitals participating in overlap become better matched in size and energy? Strategy What factors contribute to the strength of a bond? As mentioned above, the sizes and energies of the orbitals are very important. However, coulombic contributions may enhance covalent bond strength. Let’s look at each factor separately to see if one outweighs the other in these three bonds. Solution • The better the match in size and energy of orbitals between two atoms in a bond, the better the bonding overlap will be (note Figure 1-2). However, the decrease in orbital size in the series C, N, O, F is slight, and therefore the overlap changes by only a very small amount going from C – F to C – O to C – N. • In the progression from N to O to F, nuclear charge increases, giving rise to stronger attraction between the nucleus and the electrons. The increasing electronegativity in this series confirms this effect (Table 1-2). As the electronegativity of the element attached to carbon increases, so does its attraction for the shared electron pair in the covalent bond. Thus the polarity and the charge separation in the bond both increase, giving rise to a partial positive charge (d1) on carbon and a partial negative charge (d2) on the more electronegative atom. • Coulombic attraction between these opposite charges supplements the bonding resulting from covalent overlap and makes the bond stronger. In this series of bonds, increasing coulombic attraction outweighs decreasing overlap going from N to O to F, giving the observed result.
(452) (352) (356) (352) (360) (356)
Chapter 3
Bond-Dissociation Energies for Some Alkanes
CH3O CH3 C2H5O CH3 C2H5O C2H5 (CH3)2CHO CH3 (CH3)3CO CH3 (CH3)2CHO CH(CH3)2 (CH3)3CO C(CH3)3
, , ,
,
90 (377) 89 (372) 88 (368) 88 (368) 87 (364) 85.5 (358) 78.5 (328)
,
105 (439) 101 (423) 101 (423) 101 (423) 98.5 (412) 96.5 (404)
,
,
,
,
, , ,
CH3O H C2H5O H C3H7O H (CH3)2CHCH2O H (CH3)2CHO H (CH3)3CO H
DH8 [kcal mol21 (kJ mol21)]
Compound
Decreasing DH8
DH8 [kcal mol21 (kJ mol21)]
Compound
,
Table 3-2
Reactions of Alkanes
Decreasing DH8
98
Note: See footnote for Table 3-1.
Exercise 3-2
Try It Yourself In the series C – C (in ethane, H3C – CH3), N – N (in hydrazine, H2N – NH2), O – O (in hydrogen peroxide, HO – OH), the bonds decrease in strength from 90 to 60 to 50 kcal mol21. Explain. (Hint: Lone pairs on adjacent atoms repel each other.)
The stability of radicals determines the C – H bond strengths How strong are the C – H and C – C bonds in alkanes? The bond-dissociation energies of various alkane bonds are given in Table 3-2. Note that bond energies generally decrease with the progression from methane to primary, secondary, and tertiary carbon. For example, the C – H bond in methane is strong and has a high DH 8 value of 105 kcal mol21. In ethane, this bond energy is less: DH 8 5 101 kcal mol21. The latter number is typical for primary C – H bonds, as can be seen for the bond in propane. The secondary C – H bond is even weaker, with a DH 8 of 98.5 kcal mol21, and a tertiary carbon atom bound to hydrogen has a DH 8 of only 96.5 kcal mol21. C – H Bond Strength in Alkanes DH8 5
CH3 – H 105
.
Rprimary – H 101
.
Rsecondary – H 98.5
. Rtertiary – H 96.5 kcal mol21
Decreasing bond strength
A similar trend is seen for C – C bonds (Table 3-2). The extremes are the bonds in H3C – CH3 (DH 8 5 90 kcal mol21) and (CH3)3C – CH3 (DH 8 5 87 kcal mol21). Why do all of these dissociations exhibit different DH 8 values? The radicals formed have different energies. For reasons we will address in the next section, radical stability increases along the series from primary to secondary to tertiary; consequently, the energy required to create them decreases. Radical Stabilities Increasing stability
?CH3
,
?Rprimary
, ?Rsecondary ,
Decreasing DH8 of alkane R – H
?Rtertiary
3-2 Structure of Alkyl Radicals: Hyperconjugation
2.5 kcal mol1 (10.5 kJ mol1)
Hj CH3 CHCHR 2 o Secondary radical
E
99
Figure 3-1 The different energies needed to form radicals from an alkane CH3CH2CHR2. Radical stability increases from primary to secondary to tertiary.
Hj CH o 2 CH2CHR2 Primary radical
2.0 kcal mol1 (8.4 kJ mol1)
101 kcal mol1
Chapter 3
Hj CH3 CH2 CR o 2 Tertiary radical
98.5 kcal mol1 96.5 kcal mol1
CH2 CH2 CHR2
CH3 CHCHR2
CH 3CH2 CR2
H Secondary C H bond
H Tertiary C H bond
H
Primary C H bond
Figure 3-1 illustrates this finding in an energy diagram. We start (at the bottom) with an alkane containing primary, secondary, and tertiary C – H bonds. Primary bond dissociation is endothermic by DH 8 5 101 kcal mol21, that is, we are going up in energy by this amount to reach the primary radical. Secondary radical formation costs less, 98.5 kcal mol21. Thus, the secondary radical is more stable than the primary by 2.5 kcal mol21. To form the tertiary radical requires even less, 96.5 kcal mol21, and this radical is more stable than its secondary analog by 2.0 kcal mol21 (or its primary analog by 4.5 kcal mol21). Exercise 3-3 Which C – C bond would break first, the bond in ethane or that in 2,2-dimethylpropane?
In Summary Bond homolysis in alkanes yields radicals and free atoms. The heat required to do so is called the bond-dissociation energy, DH 8. Its value is characteristic only for the bond between the two participating elements. Bond breaking that results in tertiary radicals demands less energy than that furnishing secondary radicals; in turn, secondary radicals are formed more readily than primary radicals. The methyl radical is the most difficult to obtain in this way.
3-2 Structure of Alkyl Radicals: Hyperconjugation What is the reason for the ordering in stability of alkyl radicals? To answer this question, we need to inspect the alkyl radical structure more closely. Consider the structure of the methyl radical, formed by removal of a hydrogen atom from methane. Spectroscopic measurements have shown that the methyl radical, and probably other alkyl radicals, adopts a nearly planar configuration, best described by sp2 hybridization (Figure 3-2). The unpaired electron occupies the remaining p orbital perpendicular to the molecular plane. Csp 3 —H1s
p
−
H
Csp 3 —H1s
H
C H H
Csp2 —H1s
C H
H Nearly planar
Figure 3-2 The hybridization change upon formation of a methyl radical from methane. The nearly planar arrangement is reminiscent of the hybridization in BH3 (Figure 1-17).
100
Reactions of Alkanes
Chapter 3
H H
H
H H
H H
H2C
H C
C
C H
H
H
H2C
H
CH2CH3
C
C
C
H
H2C H
H
CH3
CH
CH3
CH3
C
CH3
CH3 Ethyl radical A Figure 3-3 Hyperconjugation (green dashed lines) resulting from the donation of electrons in filled sp3 hybrids to the partly filled p orbital in (A) ethyl and (B) 1-methylethyl and 1,1-dimethylethyl radicals. The resulting delocalization of electron density has a net stabilizing effect. MODEL BUILDING
1, 1-Dimethylethyl radical (tert-Butyl)
1-Methylethyl radical (Isopropyl) B
Let us see how the planar structures of alkyl radicals help explain their relative stabilities. Figure 3-3A shows that there is a conformer in the ethyl radical in which a C – H bond of the CH3 group is aligned with and overlaps one of the lobes of the singly occupied p orbital on the radical center. This arrangement allows the bonding pair of electrons in the s orbital to delocalize into the partly empty p lobe, a phenomenon called hyperconjugation. The interaction between a filled orbital and a singly occupied orbital has a net stabilizing effect (recall Exercise 1-14). Both hyperconjugation and resonance (Section 1-5) are forms of electron delocalization. They are distinguished by type of orbital: Resonance normally refers to p-type overlap of p orbitals, whereas hyperconjugation incorporates overlap with the orbitals of s bonds. Radicals are stabilized by hyperconjugation. What if we replace the remaining hydrogens on the radical carbon with alkyl groups? Each additional alkyl group increases the hyperconjugation interactions further (Figure 3-3B). The order of stability of the radicals is a consequence of this effect. Notice in Figure 3-1 that the degree of stabilization arising from each hyperconjugative interaction is relatively small [2.0 – 2.5 kcal mol21 (8.4 –10.5 kJ mol21)]: We shall see later that stabilization of radicals by resonance is considerably greater (Chapter 14). Another contribution to the relative stability of secondary and tertiary radicals is the greater relief of steric crowding between the substituent groups as the geometry changes from tetrahedral in the alkane to planar in the radical. Even a cursory glance at the bond-dissociation energies between carbon and the more electronegative atoms in Table 3-1 suggests that hyperconjugation and radical stabilities alone do not provide a complete picture. For example, bonds between carbon and any of the halogens all display essentially identical DH 8 regardless of the type of carbon. Several interpretations have been proposed to explain these observations. Polar effects are likely involved (as mentioned in the table footnote). In addition, the longer bonds between carbon and large atoms reduce the steric repulsion between atoms around that carbon, diminishing its influence on bond-dissociation energies.
3-3 Conversion of Petroleum: Pyrolysis Alkanes are produced naturally by the slow decomposition of animal and vegetable matter in the presence of water but in the absence of oxygen, a process lasting millions of years. The smaller alkanes — methane, ethane, propane, and butane — are present in natural gas, methane being by far its major component. Many liquid and solid alkanes are obtained from crude petroleum, but distillation alone does not meet the enormous demand for the lowermolecular-weight hydrocarbons needed for gasoline, kerosene, and other hydrocarbon-based fuels. Additional heating is required to break up longer-chain petroleum components into smaller molecules. How does this occur? Let us first look into the effect of strong heating on simple alkanes and then move on to petroleum.
3-3 Conversion of Petroleum: Pyrolysis
High temperatures cause bond homolysis When alkanes are heated to a high temperature, both C – H bonds and C – C bonds rupture, a process called pyrolysis. In the absence of oxygen, the resulting radicals can combine to form new higher- or lower-molecular-weight alkanes. Radicals can also remove hydrogen atoms from the carbon atom adjacent to another radical center to give alkenes, a process called hydrogen abstraction. Indeed, very complicated mixtures of alkanes and alkenes form in pyrolyses. Under special conditions, however, these transformations can be controlled to obtain a large proportion of hydrocarbons of a defined chain length. Pyrolysis of Hexane Examples of cleavage into radicals:
1
2
3
4
CH3CH2CH2CH2CH2CH3 Hexane
C1–C2 cleavage
CH3j jCH2CH2CH2CH2CH3
C2 –C3 cleavage
CH3CH2j jCH2CH2CH2CH3
C3 –C4 cleavage
CH3CH2CH2j jCH2CH2CH3
Examples of radical combination reactions: CH3j
jCH2CH3
CH3CH2CH3 Propane
CH3CH2CH2CH2CH2j
jCH2CH2CH3
CH3CH2CH2CH2CH2CH2CH2CH3 Octane
Examples of hydrogen abstraction reactions: CH3CH2j
H A CH3CHO CH2j
H A CH3CH2 CH3CHPCH2 Ethane
CH3CH2CH2j
H A CH2O CH2j
Propene
H A CH3CH2CH2 CH2 P CH2 Propane
Ethene
Note how we used single-barbed (fishhook) arrows in these examples to show the formation of a new covalent bond by the combination of two electrons. In the hydrogen abstraction reactions, electrons from a bond being broken combine with unshared electrons to form new bonds. Control of such processes frequently requires the use of special catalysts, such as crystalline sodium aluminosilicates, also called zeolites. For example, zeolite-catalyzed pyrolysis of dodecane yields a mixture in which hydrocarbons containing from three to six carbons predominate. Zeolite, 482 8C, 2 min
Dodecane uuuuuuuy C3 1 C4 1 C5 1 C6 1 other products 17% 31% 23% 18% 11%
The Function of a Catalyst What is the function of the zeolite catalyst? It speeds up the pyrolysis, so that the process occurs at a lower temperature than otherwise would be the case. The catalyst also causes certain products to form preferentially. Such enhanced reaction selectivity is a feature frequently observed in catalyzed reactions. How does this happen? In general, catalysts are additives that accelerate reactions. They enable new pathways through which reactants and products are interconverted, pathways that have lower activation energies, Ecat, than those available in the absence of the catalyst, Ea. In Figure 3-4, both uncatalyzed and catalyzed processes are shown in a simplified manner to consist of only a single step with one activation barrier. We shall see that most reactions involve more than one step. However, regardless of the number of steps, the catalyzed version of a reaction
Chapter 3
101
102
Reactions of Alkanes
Chapter 3
Figure 3-4 Potential-energy diagram comparing catalyzed and uncatalyzed processes. Although each is shown as consisting of a single step, catalyzed reactions in particular typically proceed via multistep pathways.
Lower barrier: faster Ea
E Ecat
Reaction coordinate
always has greatly reduced activation energies. While a catalyst is not consumed during the reaction it facilitates, it actively participates in it through the formation of intermediate reactive species from which it is ultimately regenerated. Therefore, only a small amount of a catalyst is needed to effect the conversion of a large amount of reactants. Catalysts modify the kinetics of reactions; that is, they change the rate at which equilibrium is established. However, catalysts do not affect the position of the equilibrium. The overall DH 8, DS 8, and hence DG 8 values for catalyzed and uncatalyzed processes are identical: A catalyst does not affect the overall thermodynamics of a reaction. Many organic reactions occur at a useful rate only because of the presence of a catalyst. The catalyst may be an acid (a proton), a base (hydroxide), a metal surface or metal compound, or a complex organic molecule. In nature, enzymes usually fulfill this function. The degree of reaction acceleration induced by a catalyst can amount to many orders of magnitude. Enzyme-catalyzed reactions are known to take place 1019 times faster than the uncatalyzed processes. The use of catalysts allows many transformations to take place at lower temperatures and under much milder reaction conditions than would otherwise be possible.
Petroleum is an important source of alkanes Breaking an alkane down into smaller fragments is also known as cracking. Such processes are important in the oil-refining industry for the production of gasoline and other liquid fuels from petroleum. As mentioned in the introduction to this chapter, petroleum, or crude oil, is believed to be the product of microbial degradation of living organic matter that existed several hundred million years ago. Crude oil, a dark viscous liquid, is primarily a mixture of several hundred different hydrocarbons, particularly straight-chain alkanes, some branched alkanes, and varying quantities of aromatic hydrocarbons. Distillation yields several fractions with a typical product distribution, as shown in Table 3-3. The composition of petroleum varies widely, depending on the origin of the oil.
Table 3-3
Product Distribution in a Typical Distillation of Crude Petroleum
Amount (% of volume)
a
Boiling point (8C)
Carbon atoms
1–2
,30
C1 – C4
15 – 30
30 – 200
C4 – C12
5 – 20 10 – 40 8 – 69
200 – 300 300 – 400 .400 (Nonvolatiles)
C12 – C15 C15 – C25 .C25
This refers to gasoline straight from petroleum, without having been treated in any way.
Products Natural gas, methane, propane, butane, liquefied petroleum gas (LPG) Petroleum ether (C5,6), ligroin (C7), naphtha, straight-run gasolinea Kerosene, heater oil Gas oil, diesel fuel, lubricating oil, waxes, asphalt Residual oil, paraffin waxes, asphalt (tar)
3-3 Conversion of Petroleum: Pyrolysis
Chapter 3
103
To increase the quantity of the much-needed gasoline fraction, the oils with higher boiling points are cracked by pyrolysis. Cracking the residual oil from crude petroleum distillation gives approximately 30% gas, 50% gasoline, and 20% higher-molecular-weight oils and a residue called coke.
C H E M I C A L H IG H LIG H T 3-1 Sustainability and the Needs of the 21st Century: “Green” Chemistry Oil and natural gas supply much of the energy requirement of both the United States and most of the rest of the industrialized nations of the world. In 2007, U.S. energy sources included gas (23%), oil (40%), coal (22%), nuclear (8%), hydroelectric and hydrothermal (3%), and other renewable (4%). Significantly, these percentages have not changed appreciably in over a decade. Much of this requirement is met by imported petroleum products, the rapidly rising cost of which has had a significant impact on the domestic economy. These substances also constitute the raw materials of much of U.S. chemical industry, via their conversion into simpler hydrocarbons such as alkenes, the starting points for countless manufacturing processes. However, this petroleumbased economy is plagued by significant problems: It is energy intensive, suffers from the frequent necessity for toxic discharge, and generates waste in the form of by-products, solvents, and inorganic salts. It is also not sustainable in the future, because the earth’s supply of petroleum is limited. Chemists in academia and industry have responded by actively exploring alternative sources of raw materials. Less exploited fossil fuels such as methane are under investigation, as are renewable starting materials, typically derived from agricultural sources. The latter, consisting of wood, grain, plant parts and plant oils, and carbohydrates, are by far the most abundant. Plant growth consumes CO2 via photosynthesis, a desirable feature in a time of concern over the increasing concentration of CO2 in the atmosphere and its long-term effect on global climate. Conversion of these raw materials into useful products presents a significant challenge, however. Ideally, processes developed for these conversions should be both efficient and environmentally acceptable. What does this mean? Over the past decade the term green chemistry has been used increasingly to describe processes that meet a number of environmental requirements. The term was coined in 1994 by Dr. Paul T. Anastas of the U.S. Environmental Protection Agency (EPA) to denote chemical activities that strive to achieve the goals of environmental protection and sustainable development. Specifically, this means pollution prevention by reducing or eliminating the use or generation of hazardous materials in design, manufacture, and application of chemical H2 C H3C
C H2
CH3
The Alyeska Pipeline Marine Terminal, Valdez, Alaska. Alaska is second only to Texas in oil production in the United States.
products, and switching from oil-based chemicals to those generated by nature. Some of the principles of green chemistry are (in abbreviated form): 1. It is better to prevent waste than to have to clean it up. 2. Synthetic methods should maximize the incorporation of
all starting materials into the final products (“atom economy”). 3. Reactions should use and generate substances that possess
efficacy of function but little or no toxicity. 4. Energy requirements should be minimized by conducting
reactions at ambient temperature and pressure. 5. Feedstocks should be renewable. 6. Catalytic processes are superior to stoichiometric ones.
A case of a green approach to the cracking of petroleum is the recent discovery of a catalytic process that converts linear alkanes into their higher and lower homologs with good selectivity. For example, when butane is passed over a Ta catalyst deposited on silica at 150 8C, it undergoes metathesis (metatithenai, Greek, to transpose) to mainly propane and pentane: H2 C
Ta on SiO2, 150C
H3 C
Butane
This process has no waste, complete atom economy, is nontoxic, occurs at a much lower temperature than conventional cracking, and is catalytic, fulfilling all the requirements
CH3
Propane
H2 C H3C
H2 C C H2
CH3
Pentane
of a green reaction. Methods such as these are forming a new paradigm for the practice of chemistry in the 21st century.
104
Reactions of Alkanes
Chapter 3
3-4 Chlorination of Methane: The Radical Chain Mechanism We have seen that alkanes undergo chemical transformations when subjected to pyrolysis, and that these processes include the formation of radical intermediates. Do alkanes participate in other reactions? In this section, we consider the effect of exposing an alkane (methane) to a halogen (chlorine). A chlorination reaction, in which radicals again play a key role, takes place, producing chloromethane and hydrogen chloride. We shall analyze each step in this transformation to establish the mechanism of the reaction.
Chlorine converts methane into chloromethane When methane and chlorine gas are mixed in the dark at room temperature, no reaction occurs. The mixture must be heated to a temperature above 300 8C (denoted by D) or irradiated with ultraviolet light (denoted by hn) before a reaction takes place. One of the two initial products is chloromethane, derived from methane in which a hydrogen atom is removed and replaced by chlorine. The other product of this transformation is hydrogen chloride. Further substitution leads to dichloromethane (methylene chloride), CH2Cl2; trichloromethane (chloroform), CHCl3; and tetrachloromethane (carbon tetrachloride), CCl4. Why should this reaction proceed? Consider its DH 8. Note that a C – H bond in methane (DH 8 5 105 kcal mol21) and a Cl – Cl bond (DH 8 5 58 kcal mol21) are broken, whereas the C – Cl bond of chloromethane (DH 8 5 85 kcal mol21) and an H – Cl linkage (DH 8 5 103 kcal mol21) are formed. The net result is the release of 25 kcal mol21 in forming stronger bonds: The reaction is exothermic (heat releasing).
REACTION
CH3
š H ðCl 105
58
DH ⴗ (kcal molⴚ1)
š Clð
or h
CH3
š Cl ð H
85
Chloromethane
103
š Cl ð
H energy input energy output DH (bonds broken) DH (bonds formed) (105 58) (85 103) 25 kcal mol1 (105 kJ mol1)
Why, then, does the thermal chlorination of methane not occur at room temperature? The fact that a reaction is exothermic does not necessarily mean that it proceeds rapidly and spontaneously. Remember (Section 2-1) that the rate of a chemical transformation depends on its activation energy, which in this case is evidently high. Why is this so? What is the function of irradiation when the reaction does proceed at room temperature? Answering these questions requires an investigation of the mechanism of the reaction.
The mechanism explains the experimental conditions required for reaction MECHANISM
MATION AN I
ANIMATED MECHANISM: Chlorination of methane
A mechanism is a detailed, step-by-step description of all the changes in bonding that occur in a reaction (Section 1-1). Even simple reactions may consist of several separate steps. The mechanism shows the sequence in which bonds are broken and formed, as well as the energy changes associated with each step. This information is of great value in both analyzing possible transformations of complex molecules and understanding the experimental conditions required for reactions to occur. The mechanism for the chlorination of methane, in common with the mechanisms of most radical reactions, consists of three stages: initiation, propagation, and termination. Let us look at these stages and the experimental evidence for each of them in more detail.
The chlorination of methane can be studied step by step Experimental observation. Chlorination occurs when a mixture of CH4 and Cl2 is either heated to 300 8C or irradiated with light, as mentioned earlier. Under such conditions, methane by itself is completely stable, but Cl2 undergoes homolysis to two atoms of chlorine. Interpretation. The first step in the mechanism of chlorination of methane is the heat- or light-induced homolytic cleavage of the Cl – Cl bond (which happens to be the weakest bond
3-4 Chlorination of Methane: The Radical Chain Mechanism
105
Chapter 3
in the starting mixture, with DH 8 5 58 kcal mol21). This event is required to start the chlorination process and is therefore called the initiation step. As implied by its name, the initiation step generates reactive species (in this case, chlorine atoms) that permit the subsequent steps in the overall reaction to take place. Initiation: Homolytic cleavage of the Cl – Cl bond š ðCl
š Cl ð
or h
š 2 ðCl j
Note: In this scheme and in
H DH(Cl2) 58 kcal mol1 (243 kJ mol1)
Chlorine atom
those that follow, all radicals and free atoms are in green.
Experimental observation. Only a relatively small number of initiation events (e.g., photons of light) are necessary to enable a great many methane and chlorine molecules to undergo conversion into products. Interpretation. After initiation has taken place, the subsequent steps in the mechanism are self-sustaining, or self-propagating; that is, they can occur many times without the addition of further chlorine atoms from the homolysis of Cl2. Two propagation steps fulfill this requirement. In the first step, a chlorine atom attacks methane by abstracting a hydrogen atom. The products are hydrogen chloride and a methyl radical. .. . Propagation step 1: Abstraction of an H atom by : Cl .. H š ðCl j H
H
C
H
š ðCl
H
jC
H DH(CH3–H) DH(H– Cl) 2 kcal mol1 (8 kJ mol1)
H
H
H 105
103
DH ⴗ (kcal molⴚ1)
Methyl radical
The DH 8 for this transformation is positive; it is endothermic (heat absorbing), and its equilibrium is slightly unfavorable. What is its activation energy, Ea? Is there enough heat to overcome this barrier? In this case, the answer is yes. An orbital description of the transition state (Section 2-1) of hydrogen removal from methane (Figure 3-5) reveals the details Bond breaking
H
H C
H
H
Ea = 4 kcal mol −1
Methane
1s
‡
Cl
C
Cl sp 3
Bond making
H
H
3 3p
Growing back lobe of the sp 3 hybrid orbital as it rehybridizes to p
Chlorine atom
H
sp 2
Transition state
H C
2 2p
H Methyl radical
+
Cl 1s
3 3p
Hydrogen chloride
Figure 3-5 Approximate orbital description of the abstraction of a hydrogen atom by a chlorine atom to give a methyl radical and hydrogen chloride. Notice the rehybridization at carbon in the planar methyl radical. The additional three nonbonded electron pairs on chlorine have been omitted. The orbitals are not drawn to scale. The symbol ‡ identifies the transition state.
106
Reactions of Alkanes
Chapter 3
Figure 3-6 Potential-energy diagram of the reaction of methane with a chlorine atom. Partial bonds in the transition state are depicted by dotted lines. This process, propagation step 1 in the radical chain chlorination of methane, is slightly endothermic.
[H3C
H
Cl]‡
Transition state Ea = 4 kcal mol −1 (17 kJ mol1)
CH3 + HCl Products
E
ΔH° = +2 kcal mol −1 (8 kJ mol1) Endothermic
CH4 + Cl Starting molecules
Reaction coordinate
of the process. The reacting hydrogen is positioned between the carbon and the chlorine, partly bound to both: H – Cl bond formation has occurred to about the same extent as C – H bond-breaking. The transition state, which is labeled by the symbol ‡, is located only about 4 kcal mol21 above the starting materials. A potential-energy diagram describing this step is shown in Figure 3-6. Propagation step 1 gives one of the products of the chlorination reaction: HCl. What about the organic product, CH3Cl? Chloromethane is formed in the second propagation step. Here the methyl radical abstracts a chlorine atom from one of the starting Cl2 molecules, thereby furnishing chloromethane and a new chlorine atom. The latter reenters propagation step 1 to react with a new molecule of methane. Thus, one propagation cycle is closed, and a new one begins, without the need for another initiation step to take place. Note how exothermic propagation step 2 is, 227 kcal mol21. It supplies the overall driving force for the reaction of methane with chlorine. Propagation step 2: Abstraction of a Cl atom by ?CH3 H H
H
š Cj ðCl
š Cl ð
H
H
C
š Cl jCl ð
H 58
H DH(Cl2 ) DH (CH3 Cl) 27 kcal mol1 (113 kJ mol1)
85
DHⴗ (kcal molⴚ1)
Because propagation step 2 is exothermic, the unfavorable equilibrium in the first propagation step is pushed toward the product side by the rapid depletion of its methyl radical product in the subsequent reaction. Cl2
z CH3? 1 HCl 88z CH4 1 Cl? y88 y CH3Cl 1 Cl? 1 HCl Slightly unfavorable
Very favorable; “drives” first equilibrium
The potential-energy diagram in Figure 3-7 illustrates this point by continuing the progress of the reaction begun in Figure 3-6. Propagation step 1 has the higher activation energy and is therefore slower than step 2. The diagram also shows that the overall DH 8 of the reaction is made up of the DH 8 values of the propagation steps: 12 2 27 5 225 kcal mol21. You can see that this is so by adding the equations for the two propagation steps. š ðCl j CH4
š CH3j HCl ð
H [kcal mol1 (kJ mol1)] 2 (8)
CH3j Cl2
š š CH3C lð ðCl j
27 (113)
CH 4 Cl2
š š CH3 Cl ð HCl ð
25 (105)
3-4 Chlorination of Methane: The Radical Chain Mechanism
PROPAGATION STEP 1
PROPAGATION STEP 2
[H3C
H
Cl]‡
[H3C
Transition state step 1
Cl
Cl]‡
Transition state step 2
Ea = 4 kcal mol −1 faster Ea < 1 kcal mol −1 slower
CH3 + Cl2 (+ HCl ) CH4 + Cl E
(+ Cl2)
ΔH° of step 1 = + 2 kcal mol −1 ΔH° of overall reaction = − 25 kcal mol −1 ΔH° of step 2 = − 27 kcal mol −1
CH3Cl + HCl + Cl Reaction coordinate
Experimental observation. Small amounts of ethane are identified among the products of chlorination of methane. Interpretation. Radicals and free atoms are capable of undergoing direct covalent bonding with one another. In the methane chlorination process, three such combination processes are possible, one of which — the reaction of two methyl groups — furnishes ethane. The concentrations of radicals and free atoms in the reaction mixture are very low, however, and hence the chance of one radical or free atom finding another is small. Such combinations are therefore relatively infrequent. When such an event does take place, the propagation of the chains giving rise to the radicals or atoms is terminated. We thus describe these combination processes as termination steps. Chain termination: Radical – radical combination š š ðClj jCl ð š ðClj jCH3
Cl CH3
š C lð
CH3j jCH3
CH3
CH3
Cl
The mechanism for the chlorination of methane is an example of a radical chain mechanism. A Radical Chain Mechanism Initiation
X2
Propagation steps
š 2ðX j
Consumed
š ð Xj RH X2 Rj
Chain termination
š Rj HX ð š R Xð ðš Xj Regenerated
š š ðX j ðXj š Xj Rj ð
RX
Rj Rj
R2
X2
Chapter 3
107
Figure 3-7 Complete potentialenergy diagram for the formation of CH3Cl from methane and chlorine. Propagation step 1 has the higher transition-state energy and is therefore slower. The DH 8 of the overall reaction CH4 1 Cl2 y CH3Cl 1 HCl amounts to 225 kcal mol21 (2105 kJ mol21), the sum of the DH 8 values of the two propagation steps.
108
Chapter 3
Reactions of Alkanes
Only a few halogen atoms are. .necessary for initiating the reaction, because the propagation steps are self-sufficient in : X .. ?. The first propagation step consumes a halogen atom, the second produces one. The newly generated halogen atom then reenters the propagation cycle in the first propagation step. In this way, a radical chain is set in motion that can drive the reaction for many thousands of cycles.
Exercise 3-4
Working with the Concepts: Radical Chain Mechanisms Write a detailed mechanism for the light-initiated monochlorination of ethane, which furnishes chloroethane. Calculate DH 8 for each step. Strategy Begin by writing the overall equation for the reaction and calculating its DH 8, using data in Tables 3-1 and 3-2 and the formula DH 8 5 SDH 8 (bonds broken) 2 SDH 8 (bonds formed). Then write the initiation, propagation, and termination steps for a typical radical chain mechanism. Solution • The reaction equation is CH3CH2
š H ðC l
š C lð
CH3CH2
š C lð H
š C lð
¢H° 5 101 1 58 2 84 2 103 5 228 kcal mol21 The reaction is more exothermic than chlorination of methane, mainly because breaking the C – H bond in ethane requires less energy than breaking the C – H bond in methane. • The initiation step in the mechanism is light-induced dissociation of Cl2: Initiation
š ðC l
š C lð
h
š 2ðC l Ý
DH 8 5 158 kcal mol21
• Using the propagation steps for chlorination of methane as a model, write analogous steps for ethane. In step 1, a chlorine atom abstracts a hydrogen: Propagation 1
CH3CH2
š H ÝC lð
CH3CH2 Ý H
š C lð
¢H° 5 101 2 103 5 22 kcal mol21 • In step 2, the ethyl radical formed in step 1 abstracts a chlorine atom from Cl2: Propagation 2
š CH3CH2 Ý ðC l
š C lð
CH3CH2
š š C lð lð Ý C
¢H° 5 58 2 84 5 226 kcal mol21 Note that the sum of the DH 8 values for the two propagation steps gives us DH 8 for the overall reaction. This is because summing the species present in the two propagation steps cancels both ethyl radicals and chlorine atoms, leaving just the overall stoichiometry. • Finally, we formulate the termination steps as follows: Termination
š š ðC l Ý Ý C lð š CH3CH2 Ý Ý C lð
CH3CH2 Ý ÝCH2CH3
Cl2
DH 8 5 258 kcal mol21
š CH3CH2C lð CH3CH2CH2CH3
DH 8 5 284 kcal mol21 DH 8 5 288 kcal mol21
One of the practical problems in chlorinating methane is the control of product selectivity. As mentioned earlier, the reaction does not stop at the formation of chloromethane but continues to form di-, tri-, and tetrachloromethane by further substitution. A practical solution to this problem is the use of a large excess of methane in the reaction. Under
3-5 Other Radical Halogenations of Methane
109
Chapter 3
such conditions, the reactive intermediate chlorine atom is at any given moment surrounded by many more methane molecules than product CH3Cl. Thus, the chance of Cl? finding CH3Cl to eventually make CH2Cl2 is greatly diminished, and product selectivity is achieved.
Exercise 3-5
Try It Yourself Write out the overall equation and the propagation steps of the mechanism for the chlorination of chloromethane to furnish dichloromethane, CH2Cl2. (Caution: Write out each step of the mechanism separately and completely. Be sure to include full Lewis structures of all species and all appropriate arrows to show electron movement.)
In Summary Chlorine transforms methane into chloromethane. The reaction proceeds through a mechanism in which heat or light causes a small number of Cl2 molecules to undergo homolysis to chlorine atoms (initiation). The chlorine atoms induce and maintain a radical chain sequence consisting of two (propagation) steps: (1) hydrogen abstraction to generate the methyl radical and HCl and (2) conversion of CH3? by Cl2 into CH3Cl and regenerated Cl?. The chain is terminated by various combinations of radicals and free atoms. The heats of the individual steps are calculated by comparing the strengths of the bonds that are being broken with those of the bonds being formed.
3-5 Other Radical Halogenations of Methane Fluorine and bromine, but not iodine, also react with methane by similar radical mechanisms to furnish the corresponding halomethanes. The dissociation energies of X2 (X 5 F, Br, I) are lower than that of Cl2, thus ensuring ready initiation of the radical chain (Table 3-4).
DH 8 Values for Table 3-4 the Elemental Halogens
Fluorine is most reactive, iodine least reactive Let us compare the enthalpies of the first propagation step in the different halogenations of methane (Table 3-5). For fluorine, this step is exothermic by 231 kcal mol21. We have already seen that, for chlorine, the same step is slightly endothermic; for bromine, it is substantially so, and for iodine even more so. This trend has its origin in the decreasing bond strengths of the hydrogen halides in the progression from fluorine to iodine (Table 3-1). The strong hydrogen – fluorine bond is the cause of the high reactivity of fluorine atoms in hydrogen abstraction reactions. Fluorine is more reactive than chlorine, chlorine is more reactive than bromine, and the least reactive halogen atom is iodine. The contrast between fluorine and iodine is illustrated by comparing potential-energy diagrams for their respective hydrogen abstractions from methane (Figure 3-8). The highly exothermic reaction of fluorine has a negligible activation barrier. Moreover, in its transition state, the fluorine atom is relatively far from the hydrogen that is being
Table 3-5
Halogen F2 Cl2 Br2 I2
DH 8 [kcal mol21 (kJ mol21)] 38 58 46 36
(159) (243) (192) (151)
Enthalpies of the Propagation Steps in the Halogenation of Methane [kcal mol21 (kJ mol21)]
Reaction
F
Cl
Br
I
ðš Xj CH4
šð jCH3 HX
231 (2130)
12 (18)
118 (175)
134 (1142)
jCH3 X2
Xj CH3 š Xððš
272 (2301)
227 (2113)
224 (2100)
221 (288)
CH4 X2
š Xð Xð CH3 š H
2103 (2431)
225 (2105)
26 (225)
113 (154)
110
Chapter 3
Reactions of Alkanes
Early transition state
Late transition state
Transfer of H not very advanced
[H3C
Transfer of H far advanced
H I]‡ CH3 + HI
E
[H3C H
F]‡
CH4 + F H° 31 kcal mol1 Exothermic
CH4 + I
H° 34 kcal mol1 Endothermic
CH3 + HF Reaction coordinate: extent of H shift
Figure 3-8 Potential-energy diagrams: (left) the reaction of a fluorine atom with CH4, an exothermic process with an early transition state; and (right) the reaction of an iodine atom with CH4, an endothermic transformation with a late transition state. Both are thus in accord with the Hammond postulate.
Relative Reactivities of X? in Hydrogen Abstractions F? . Cl? . Br ? . I? Decreasing reactivity
transferred, and the H – CH3 distance is only slightly greater than that in CH4 itself. Why should this be so? The H – CH3 bond is about 30 kcal mol21 (125 kJ mol21) weaker than that of H – F (Table 3-1). Only a small shift of the H toward the F? is necessary for bonding between the two to overcome the bonding between hydrogen and carbon. If we view the reaction coordinate as a measure of the degree of hydrogen shift from C to F, the transition state is reached early and is much closer in appearance to the starting materials than to the products. Early transition states are frequently characteristic of fast, exothermic processes. On the other hand, reaction of I? with CH4 has a very high Ea [at least as large as its endothermicity, 134 kcal mol21 (1142 kJ mol21); Table 3-5]. Thus, the transition state is not reached until the H – C bond is nearly completely broken and the H – I bond is almost fully formed. The transition state is said to occur late in the reaction pathway. It is substantially further along the reaction coordinate and is much closer in structure to the products of this process, CH3? and HI. Late transition states are frequently typical of relatively slow, endothermic reactions. Together these rules concerning early and late transition states are known as the Hammond* postulate.
The second propagation step is exothermic Let us now consider the second propagation step for each halogenation in Table 3-5. This process is exothermic for all the halogens. Again, the reaction is fastest and most exothermic for fluorine. The combined enthalpies of the two steps for the fluorination of methane result in a DH 8 of 2103 kcal mol21 (2431 kJ mol21). The formation of chloromethane is less exothermic and that of bromomethane even less so. In the latter case, the appreciably endothermic nature of the first step [DH 8 5 118 kcal mol21 (175 kJ mol21)] is barely overcome by the enthalpy of the second [DH 8 5 224 kcal mol21 (2100 kJ mol21)], resulting in an energy change of only 26 kcal mol21 (225 kJ mol21) for the overall substitution. Finally, iodine does not react with methane to furnish methyl iodide and hydrogen iodide: The first step costs so much energy that the second step, although exothermic, cannot drive the reaction.
*Professor George S. Hammond (1921 – 2005), Georgetown University, Washington, D.C.
3-6 Chlorination of Higher Alkanes: Relative Reactivity
Chapter 3
Exercise 3-6 When methane is allowed to react with an equimolar mixture of chlorine and bromine, only hydrogen abstraction by chlorine atoms is observed. Explain.
In Summary Fluorine, chlorine, and bromine react with methane to give halomethanes. All three reactions follow the radical chain mechanism described for chlorination. In these processes, the first propagation step is always the slower of the two. It becomes more exothermic and its activation energy decreases in the progression from bromine to chlorine to fluorine. This trend explains the relative reactivity of the halogens, fluorine being the most reactive. Iodination of methane is endothermic and does not occur. Strongly exothermic reaction steps are often characterized by early transition states. Conversely, endothermic steps typically have late transition states.
3-6 Chlorination of Higher Alkanes: Relative Reactivity and Selectivity What happens in the radical halogenation of other alkanes? Will the different types of R – H bonds — namely, primary, secondary, and tertiary — react in the same way as those in methane? As we saw in Exercise 3-4, the monochlorination of ethane gives chloroethane as the product. REACTION
Chlorination of Ethane D or hv
CH3CH3 1 Cl2 uuy CH3CH2Cl 1 HCl Chloroethane
21
DH 8 5 228 kcal mol (2117 kJ mol21)
This reaction proceeds by a radical chain mechanism analogous to the one observed for methane. As in methane, all of the hydrogen atoms in ethane are indistinguishable from one another. Therefore, we observe only one monochlorination product, chloroethane, regardless of which hydrogen is initially abstracted by chlorine in the first propagation step. Propagation Steps in the Mechanism of the Chlorination of Ethane š CH3CH3 ðClj CH3CH2j Cl2
š CH3CH2j HCl ð
MECHANISM
1
H 2 kcal mol (8 kJ mol1) š š CH3CH2Cl H 26 kcal mol1 ð ðClj (109 kJ mol1)
What can be expected for the next homolog, propane?
Secondary C – H bonds are more reactive than primary ones The eight hydrogen atoms in propane fall into two groups: six primary and two secondary hydrogens. If chlorine atoms were to abstract and replace primary and secondary hydrogens at equal rates, we should expect to find a product mixture containing three times as much 1-chloropropane as 2-chloropropane. We call this outcome a statistical product ratio, because it derives from the statistical fact that in propane there are three times as many primary sites for reaction as secondary sites. In other words, it is three times as likely for a chlorine atom to collide with a primary hydrogen in propane, of which there are six, as with a secondary hydrogen atom, of which there are only two. However, secondary C – H bonds are weaker than primary ones (DH 8 5 98.5 versus 101 kcal mo121). Abstraction of a secondary hydrogen is therefore more exothermic and
CH3CH2CH3 Propane Six primary hydrogens (blue) Two secondary hydrogens (red) Ratio 3:1
111
112
Reactions of Alkanes
Chapter 3
Ea = 1 kcal mol −1 (4 kJ mol−1): slower Ea = 0.5 kcal mol −1 (2 kJ mol−1): faster
CH3CH2CH3 + Cl
E
ΔH ° = −2 kcal mol −1 (−8 kJ mol−1)
Removal of primary hydrogen: less exothermic
CH3CH2CH2 + HCl Removal of secondary hydrogen: more exothermic
ΔH ° = − 4.5 kcal mol −1 (− 19 kJ mol−1)
CH3CHCH3 + HCl
Reaction coordinate Figure 3-9 Hydrogen abstraction by a chlorine atom from the secondary carbon in propane is more exothermic and faster than that from the primary carbon.
proceeds with a smaller activation barrier (Figure 3-9). We might thus expect secondary hydrogens to react faster, leading to more 2-chloro- than 1-chloropropane. What is actually observed? At 25 8C, the experimental ratio of 1-chloropropane;2-chloropropane is found to be 43;57. This result indicates that both statistical and bond-energy factors determine the product formed. Chlorination of Propane
Cl2 CH3CH2CH3
hv
Cl A CH3CH2CH2Cl CH3CHCH3 HCl 1-Chloropropane
Expected statistical ratio Expected C – H bond reactivity ratio Experimental ratio (258C)
3 Less 43
2-Chloropropane : : :
1 More 57
It would be good to know the relative reactivity of secondary and primary hydrogens in the chlorination of propane and, by extrapolation, the alkanes in general. For this purpose, we need to factor out the statistical contribution of the hydrogens in propane to the product ratio. In other words, what we want to know is: How much does one primary hydrogen contribute to the “43” in the product ratio, and how much does one secondary hydrogen do so to the “57” part? To find out, we simply divide these values by the respective numbers of hydrogens giving rise to them: 43 is divided by 6 (¯7) and 57 by 2 (¯28). Reactivity of Hsec yield of 2-chloropropane/number of Hsec 57/2 28 5 5 < 54 Reactivity of Hprim yield of 1-chloropropane/number of Hprim 43/6 7 Thus, the relative reactivity of a secondary versus primary H in propane is about 28/7 5 4. We say that chlorine exhibits a selectivity of 4:1 in the removal of secondary versus primary hydrogens at 25 8C. We could predict from this analysis that all secondary hydrogens are four times as reactive as all primary hydrogens in all radical chain reactions. Is this prediction true? Unfortunately, no. Although secondary C – H bonds generally undergo dissociation faster than do their primary counterparts, their relative reactivity very much depends on the nature of the attacking species, X?, the strength of the resulting H – X bond, and the temperature. For example, at 600 8C, the chlorination of propane results in the statistical distribution of products: roughly three times as much 1-chloropropane as 2-chloropropane. At such a high
3-6 Chlorination of Higher Alkanes: Relative Reactivity
Chapter 3
temperature, virtually every collision between a chlorine atom and any hydrogen in the propane molecule takes place with sufficient energy to lead to successful reaction. Chlorination is said to be unselective at this temperature, and the product ratio is governed by statistical factors.
Exercise 3-7 What do you expect the products of monochlorination of butane to be? In what ratio will they be formed at 25 8C?
Tertiary C – H bonds are more reactive than secondary ones Let us now find the relative reactivity of a tertiary hydrogen in the chlorination of alkanes. For this purpose, we expose 2-methylpropane, a molecule containing one tertiary and nine primary hydrogens, to chlorination conditions at 25 8C. The resulting two products, 2-chloro2-methylpropane (tert-butyl chloride) and 1-chloro-2-methylpropane (isobutyl chloride), are formed in the relative yields of 36 and 64%, respectively. Chlorination of 2-Methylpropane CH3 Cl2 CH3
C CH3
CH3 H
hv
Cl CH2
C
CH3
H
CH3
CH3
C
Cl
CH3
64%
36%
1-Chloro-2-methylpropane (Isobutyl chloride)
2-Chloro-2-methylpropane (tert-Butyl chloride)
Expected statistical ratio Expected C – H bond reactivity ratio Experimental ratio (258C)
HCl
9 Less 64
: 1 : More : 36
We determine the reactivity of tertiary relative to primary hydrogens in the same manner as that employed above for secondary relative to primary hydrogens. We combine the experimental result of 64% primary chlorination and 36% tertiary chlorination in 2-methylpropane with the statistical presence of nine primary hydrogens and one tertiary hydrogen atom in the starting alkane. Dividing the proportionate amount observed of each product by the number of hydrogens that contribute to its formation gives a measure of reactivity per hydrogen atom: Reactivity of Htert yield of 2-chloro-2-methylpropane/number of Htert 36/1 36 5 5 < J
First-order
Expansion (CH3)4Si
4
B
3
2
1
0
ppm (δ )
independent of field (Section 10-7). Thus, at increasingly higher field, the individual multiplets become more and more separated (resolved), and the non-first-order effect of a close-lying absorption gradually disappears. The effect is similar to that of a magnifying glass in the observation of ordinary objects. The eye has a relatively low resolving power and cannot discern the fine structure of a sample that becomes evident only on appropriate magnification. Greater field strength has a dramatic effect on the spectrum of 2-chloro-1-(2chloroethoxy)ethane (Figure 10-24). In this compound, the deshielding effect of the oxygen is about equal to that of the chlorine substituent. As a consequence, the two sets of methylene hydrogens give rise to very close-lying peak patterns. At 90 MHz, the resulting absorption has a symmetric shape but is very complicated, exhibiting more than 32 peaks of various intensities. However, recording the NMR spectrum with a 500-MHz spectrometer (Figure 10-24B) produces a first-order pattern.
10-8 Spin–Spin Splitting: Some Complications
1H NMR
CH3 c
Hb
Ha
C
C
Cl
Cl
3H 1 H neighbor: doublet
Cl
1H
8
1.8 1.7 ppm 4.4 4.3 ppm
5.9 5.8 ppm
9
Hb
1 H neighbor: doublet Jab = 3.6 Hz
Jab = 3.6 Hz
Figure 10-25 300-MHz 1H NMR spectrum of 1,1,2-trichloropropane. Nucleus Hb gives rise to a quartet of doublets at d 5 4.35 ppm: eight peaks.
Jbc = 6.8 Hz
3 H neighbors: quartet Jbc = 6.8 Hz
1 H neighbor: doublet
417
Chapter 10
(CH3)4Si
1H
7
6
5
4
3
2
1
0
ppm (δ )
Coupling to nonequivalent neighbors may modify the simple N 1 1 rule When hydrogens are coupled to two sets of nonequivalent neighbors, complicated splitting patterns may result. The spectrum of 1,1,2-trichloropropane illustrates this point (Figure 10-25). In this compound, the hydrogen at C2 is located between a methyl and a CHCl2 group, and it is coupled to the hydrogens of each group independently and with different coupling constants. Let us analyze the spectrum in detail. We first notice two doublets, one at low field (d 5 5.86 ppm, J 5 3.6 Hz, 1 H) and one at high field (d 5 1.69 ppm, J 5 6.8 Hz, 3 H). The low-field absorption is assignable to the hydrogen at C1 (Ha), adjacent to two deshielding halogens; and the methyl hydrogens (Hc) resonate as expected at highest field. In accord with the N 1 1 rule, each signal is split into a doublet because of coupling with the hydrogen at C2 (Hb). The resonance of Hb, however, is quite different in appearance from what we expect. The nucleus giving rise to this absorption has a total of four hydrogens as its neighbors: Ha and three Hc. Application of the N 1 1 rule suggests that a quintet should be observed. However, the signal for Hb at d 5 4.35 ppm consists of eight lines, with relative intensities that do not conform to those expected for ordinary splitting patterns (see Tables 10-4 and 10-5). What is the cause of this complexity? The N 1 1 rule strictly applies only to splitting by equivalent neighbors. In this molecule, we have two sets of different adjacent nuclei that couple to Hb with different coupling constants. The effect of these couplings can be understood, however, if we apply the N 1 1 rule sequentially. The methyl group causes a splitting of the Hb resonance into a quartet, with the relatively larger Jbc 5 6.8 Hz. Then, coupling with Ha further splits each peak in this quartet into a doublet, with the smaller Jab 5 3.6 Hz, both splits resulting in the observed eight-line pattern (Figure 10-26). The hydrogen at C2 is said to be split into a quartet of doublets. The hydrogens on C2 of 1-bromopropane also couple to two nonequivalent sets of neighbors. In this case, however, the resulting splitting pattern appears to conform with the N 1 1 rule, and a (slightly distorted) sextet is observed (Figure 10-27). The reason is that the coupling constants to the two different groups are very similar, about 6– 7 Hz. Although
Hb
Jbc
Jab
Jbc
Jab
Jbc
Jab
Jab
Figure 10-26 The splitting pattern for Hb in 1,1,2-trichloropropane follows the sequential N 1 1 rule. Each of the four lines arising from coupling to the methyl group is further split into a doublet by the hydrogen on C1.
418
Chapter 10
Using NMR Spectroscopy to Deduce Structure
CHE M I C AL H I G H L I G H T 1 0 - 3 The Nonequivalence of Diastereotopic Hydrogens H
You have probably assumed that all methylene (CH2) groups contain equivalent hydrogens, which give rise to only one signal in the NMR spectrum. Such is indeed the case when there is a symmetry element present that renders them equivalent, such as a mirror plane or axis of rotation. For example, the methylene hydrogens in butane or cyclohexane have this property. However, this symmetry is readily removed by substitution (Chapters 3 and 5), a change that has profound effects not only with respect to stereoisomerism (Chapter 5), but also to NMR spectroscopy. Consider, for example, turning cyclohexane into bromocyclohexane, such as by radical bromination. Not only does this alteration make C1, C2, C3, and C4 different, but also all of the hydrogens cis to the bromine substituent have become different from those trans. In other words, the CH2 groups exhibit nonequivalent geminal hydrogens. These hydrogens have separate chemical shifts and show mutual geminal (as well as vicinal) coupling. The resulting 300-MHz spectrum is quite complex, as shown on the right. Only the C1 hydrogen at d 5 4.17 ppm, deshielded by the neighboring bromine, is readily assigned. Methylene hydrogens that are not equivalent are called diastereotopic. This word is based on the stereochemical outcome of replacing either of these hydrogens by a substituent: diastereomers. For example, in the case of bromocyclohexane, substituting the wedged (red) hydrogen at C2 results in a cis product. The same alteration involving the hashed (blue) hydrogen generates the trans isomer. You can verify the same outcome of this procedure when applied to C3 and C4, respectively.
Br
H
Br2 hv Br
H
H
1 4
H H
H
Diastereotopic
H
2 3
H H
H
Diastereotopic
Diastereotopic
Bromocyclohexane
1H NMR
Br
4
3
2
1
ppm (δ )
1H NMR
CH3CH2CH2Br
3H2H5H neighbors: sextet
2 H neighbors: triplet
2 H neighbors: triplet
1.1 1.0 ppm 1.9 1.8 ppm
(CH3)4Si
3.5 3.4 ppm
3H
2H
9
8
7
6
5
4
3
2H 2
ppm (d )
Figure 10-27 300-MHz 1H NMR spectrum of 1-bromopropane.
1
0
10-8 Spin–Spin Splitting: Some Complications
You might think that it is the rigidity of the cyclic frame that imparts this property to diastereotopic hydrogens. However, this is not true, as you can realize by reviewing Section 5-5. The chlorination of 2-bromobutane at C3 resulted in two diastereomers; therefore, the methylene group in 2-bromobutane also contains diastereotopic protons. We can recognize and generalize the origin of this effect: The presence of the stereocenter precludes a mirror plane through the CH2 carbon, and rotation is ineffective in making the two hydrogens equivalent. To illustrate the presence of diastereotopic hydrogens in such chiral, acyclic molecules, we return to one of the three monochlorination products of 1-chloropropane, discussed in Section 10-6, namely, 1,2-dichloropropane. The 300-MHz 1H NMR spectrum shown reveals four signals (instead of the normally expected three), two of which are due to the diastereotopic hydrogens at C1. Specifically, these hydrogens appear as doublets of doublets at d 5 3.58 and 3.76 ppm, because they coupled to each other with Jgeminal 5 10.8 Hz, and then with individual coupling constants to the hydrogen at C2 (Jvicinal 5 4.7 and 9.1 Hz, respectively). Frequently, diastereotopic hydrogens have such close chemical shifts that their nonequivalence is not visible in an NMR spectrum. For example, in the chiral 2-bromohexane, all three methylene groups contain diastereotopic hydrogens. In this compound, only the two methylenes closest to the stereocenter reveal their presence. The third one is too far removed for the asymmetry of the stereocenter to have a measurable effect.
1H NMR
Chapter 10
419
Diastereotopic H
H
H
*
3
H 1
2
H Cl
1.7 1.6 ppm
Cl H
1,2-Dichloropropane
3.8 3.7 3.6 3.5 ppm 4.2 4.1 ppm
6
5
4
3
2
1
ppm (δ )
an analysis similar to that given earlier for 1,1,2-trichloropropane would lead us to predict as many as 12 lines in this signal (a quartet of triplets), the nearly equal coupling constants cause many of the lines to overlap, thus simplifying the pattern (Figure 10-28). The hydrogens in many simple alkyl derivatives display similar coupling constants and, therefore, spectra that are in accord with the N 1 1 rule. Hb
CH3 a
CH2 b
CH2
Jbc
X
Jbc
c
Jab
Jab
Jab
Jab
Jab
Figure 10-28 Splitting pattern expected for Hb in a propyl derivative when Jab < Jbc. Several of the peaks coincide, giving rise to a deceptively simple spectrum: a sextet.
Chapter 10
Using NMR Spectroscopy to Deduce Structure
Exercise 10-15
Working with the Concepts: Applying the N 1 1 Rule Predict the coupling patterns for the boldface hydrogen in the structure shown, first according to the simple N 1 1 rule and then according to the sequential N 1 1 rule. CH3 A H3C O C O CH2 O OH A H Strategy In general, to predict splitting patterns of specific hydrogens, you need to be clear about the identity of their “neighborhood.” It is therefore often useful to build a model and to draw the structure out fully, showing all hydrogens. In this way, you can recognize symmetry and establish the different types of hydrogen neighbors and their relative abundance. Solution • Applying the simple N 1 1 rule is easy: All we have to do is count the number of neighbors to the bold hydrogen to determine N. In this case, we have two CH3 units and one CH2 unit, adding up to 8 H. Therefore, the bold hydrogen should appear as a nonet or nine lines, their relative intensities following Pascal’s triangle (Table 10-4): 1 : 8 : 28 : 56 : 70 : 56 : 28 : 8 : 1. In practice, you don’t have to figure out these exact numbers, and it suffices to look for a symmetrical pattern around the most intense center line, ascending from the left and descending to the right. You would need to amplify the spectrum to make sure that you don’t miss the outermost lines.
However, a symmetrical nonet is what is expected only if the coupling constants to the methyl and methylenes hydrogens, respectively, are the same. • The sequential N 1 1 rule applies if the J values to unequal neighbors differ. In that case, we first determine the splitting pattern expected if there was only one type of neighbor present. For example, if we pick the CH3 groups, it would be 6 H, giving rise to a septet. We then apply the additional splitting caused by another neighbor, in our case the CH2 group, and split each line of the original septet into a triplet. The result is a septet of triplets, 21 lines. This sequence is arbitrary and could have been reversed, resulting in a triplet of septets, the same 21 lines. In practice, the coupling pattern with the larger J value is quoted first, followed by the others, in order of decreasing size. You may be curious what the actual experimental splitting pattern looks like: It is a nonet, with J 5 6.5 Hz.
Exercise 10-16
Try It Yourself Predict the coupling patterns for the boldface hydrogens in the structures below, first according to the simple N 1 1 rule and then according to the sequential N 1 1 rule. (a) BrCH2CH2CH2Cl
(b) CH3CHCHCl2 OCH3
(d)
O
S
H3C CH3 H &@ H H3C H &@
(c) Cl2CHCHCHCH3 A A CH3S Br
@&
420
Exercise 10-17 In Section 10-6 you learned how to distinguish among the three monochlorination products of 1-chloropropane—1,1-, 1,2-, and 1,3-dichloropropane—by using just chemical-shift data and chemical integration. Would you also expect to tell them apart on the basis of their coupling patterns?
10-8 Spin–Spin Splitting: Some Complications
Chapter 10
421
Fast proton exchange decouples hydroxy hydrogens With our knowledge of vicinal coupling, let us now return to the NMR of alcohols. We note in the NMR spectrum of 2,2-dimethyl-1-propanol (Figure 10-11) that the OH absorption appears as a single peak, devoid of any splitting. This is curious, because the hydrogen is adjacent to two others, which should cause its appearance as a triplet. The CH2 hydrogens that show up as a singlet should in turn appear as a doublet with the same coupling constant. Why, then, do we not observe spin – spin splitting? The reason is that the weakly acidic OH hydrogens are rapidly transferring, both between alcohol molecules and to traces of water, on the NMR time scale at room temperature. As a consequence of this process, the NMR spectrometer sees only an average signal for the OH hydrogen. No coupling is visible, because the binding time of the proton to the oxygen is too short (about 1025 s). It follows that the CH2 nuclei are similarly uncoupled, a condition resulting in the observed singlet. Rapid Proton Exchange Between Alcohols with Various CH2–␣, Spin Combinations Averages ␦OH H␣ A RO CO OH A H
H␣ A RO CO OH A H␣
H A RO CO OH A H
Observed average ␦OH
Rapid ␣-Proton with -Proton Exchange Averages ␦CH2 RO CH2 O OH
RO CH2 O OH␣ Observed average ␦CH2
Absorptions of this type are said to be decoupled by fast proton exchange. The exchange may be slowed by removal of traces of water or acid or by cooling. In these cases, the OH bond retains its integrity long enough (more than 1 s) for coupling to be observed on the NMR time scale. An example is shown in Figure 10-29 for methanol. At 378C, two singlets are observed, corresponding to the two types of hydrogens, both devoid of spin–spin splitting. However, at 2658C, the expected coupling pattern is detectable: a quartet and a doublet.
CH3
CH3
+37°C
−65°C OH
OH
Figure 10-29 Temperature dependence of spin – spin splitting in methanol. The singlets at 378C illustrate the effect of fast proton exchange in alcohols. (After H. Günther, NMR-Spektroskopie, Georg Thieme Verlag, Stuttgart, 1973.)
Rapid magnetic exchange “self-decouples” chlorine, bromine, and iodine nuclei All halogen nuclei are magnetic. Therefore, the 1H NMR spectra of haloalkanes would be expected to exhibit spin – spin splitting owing to the presence of these nuclei (in addition to the normal H – H coupling). In practice, only fluorine exhibits this effect, in much the same way as the proton does but with much larger J values. Thus, for example, the 1H NMR spectrum of CH3F exhibits a doublet with J 5 81 Hz. Because fluoroorganic compounds are a relatively specialised area, we shall not deal with their NMR spectroscopy any further.
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Turning to the other halides, inspection of the spectra of the haloalkanes depicted in Figures 10-16, 10-21, 10-22, 10-24, 10-25, and 10-27 (fortunately) reveals the absence of any visible spin – spin splitting by these nuclei. The reason for this observation lies in their relatively fast internal magnetic equilibration on the NMR time scale, precluding their recognition by the adjacent hydrogens as having differing alignments with respect to the external magnetic field H0. They “self-decouple,” in contrast with the “exchange decoupling” exhibited by the hydroxy protons.
In Summary The peak patterns in many NMR spectra are not first order, because the differences between the chemical shifts of nonequivalent hydrogens are close to the values of the corresponding coupling constants. Use of higher-field NMR instruments may improve the appearance of such spectra. Coupling to nonequivalent hydrogen neighbors occurs separately, with different coupling constants. In some cases, they are sufficiently dissimilar to allow for an analysis of the multiplets. In many simple alkyl derivatives, they are sufficiently similar (J 5 6 – 7 Hz) that the spectra observed are simplified to those predicted in accordance with the N 1 1 rule. Vicinal coupling through the oxygen in alcohols is frequently not observed because of decoupling by fast proton exchange.
10-9 Carbon-13 Nuclear Magnetic Resonance Proton nuclear magnetic resonance is a powerful method for determining organic structures because most organic compounds contain hydrogens. Of even greater potential utility is NMR spectroscopy of carbon. After all, by definition, all organic compounds contain this element. In combination with 1H NMR, it has become the most important analytical tool in the hands of the organic chemist. We shall see in this section that 13C NMR spectra are much simpler than 1H NMR spectra, because we can avoid the complications of spin – spin splitting.
Carbon NMR utilizes an isotope in low natural abundance: 13C Carbon NMR is possible. However, there is a complication: The most abundant isotope of carbon, carbon-12, is not active in NMR. Fortunately, another isotope, carbon-13, is present in nature at a level of about 1.11%. Its behavior in the presence of a magnetic field is the same as that of hydrogen. One might therefore expect it to give spectra very similar to those observed in 1H NMR spectroscopy. This expectation turns out to be only partly correct, because of a couple of important (and very useful) differences between the two types of NMR techniques. Carbon-13 NMR (13C NMR) spectra used to be much more difficult to record than hydrogen spectra, not only because of the low natural abundance of the nucleus under observation, but also because of the much weaker magnetic resonance of 13C. Thus, under comparable conditions, 13C signals are about 1y6000 as strong as those for hydrogen. FT NMR (Sections 10-3 and 10-4) is particularly valuable here, because multiple radio-frequency pulsing allows the accumulation of much stronger signals than would normally be possible. One advantage of the low abundance of 13C is the absence of carbon – carbon coupling. Just like hydrogens, two adjacent carbons, if magnetically nonequivalent (as they are in, e.g., bromoethane), split each other. In practice, however, such splitting is not observed. Why? Because coupling can occur only if two 13C isotopes come to lie next to each other. With the abundance of 13C in the molecule at 1.11%, this event has a very low probability (roughly 1% of 1%; i.e., 1 in 10,000). Most 13C nuclei are surrounded by only 12C nuclei, which, having no spin, do not give rise to spin–spin splitting. This feature simplifies 13C NMR spectra appreciably, reducing the problem of their analysis to a determination of the coupling patterns resulting from any attached hydrogens. Figure 10-30 depicts the 13C NMR spectrum of bromoethane (for the 1H NMR spectrum, see Figure 10-21). The chemical shift d is defined as in 1H NMR and is determined relative to an internal standard, normally the carbon absorption in (CH3)4Si. The chemical-shift range of carbon is much larger than that of hydrogen. For most organic compounds, it covers a distance of about 200 ppm, in contrast with the relatively narrow spectral “window” (10 ppm) of hydrogen. Figure 10-30 reveals the relative complexity of the absorptions caused by extensive 13C – H spin – spin splittings. Not surprisingly, directly bound hydrogens are most strongly coupled (,125 – 200 Hz). Coupling tapers off, however, with increasing
10-9 Carbon-13 Nuclear Magnetic Resonance
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Chapter 10
13C NMR
C H
H
H
C
C
H
H
H
C Br
H
C
C
H
H
Br
C is coupled to 3 attached H: quartet with a large J
C is coupled to 2 attached H: triplet with a large J
CH3CH2Br
A second, smaller coupling to the neighboring 2 Hs splits each line of the quartet into a triplet
A second, smaller coupling to the neighboring 3 Hs splits each line of the triplet into a quartet
32
H
30
28
26
24
22
20
18
16
14
12
(CH3)4Si 3 attached Hs: quartet
10
ppm (δ ) Expansion
80
70
60
50
40 ppm (δ )
30
Figure 10-30 13C NMR spectrum of bromoethane, showing the complexity of 13C – H coupling. There is an upfield quartet (d 5 18.3 ppm, J 5 126 Hz) and a downfield triplet (d 5 26.6 ppm, J 5 151 Hz) resonance for the two carbon atoms. Note the large chemical-shift range. Tetramethylsilane, defined to be located at d 5 0 ppm as in 1H NMR, absorbs as a quartet (J 5 118 Hz) because of coupling of each carbon to three equivalent hydrogens. The inset shows a part of the spectrum expanded horizontally to reveal the fine splitting of each of the main peaks that is due to coupling of each 13C to protons on the neighboring carbon. Thus, each line of the upfield quartet (red) is split again into a triplet with J 5 3 Hz, and each line of the downfield triplet (blue) is split again into a quartet with J 5 5 Hz.
distance from the 13C nucleus under observation, such that the two-bond coupling constant J13C– C– H is in the range of only 0.7 – 6.0 Hz. You may wonder why it is that we observe hydrogen couplings in 13C NMR spectroscopy, yet we do not notice the converse, namely, carbon couplings, in 1H NMR. The answer lies in the low natural abundance of the NMR-active 13C isotope and the high natural abundance of the 1H nucleus. Thus, we could not detect 13C coupling in our proton spectra, because 99% of the attached carbons were 12C. On the other hand, the corresponding carbon
20
10
0
424
Using NMR Spectroscopy to Deduce Structure
Chapter 10
spectra reveal 1H coupling, because 99.9% of our sample contains this isotope of hydrogen (Table 10-1).
13C NMR
Exercise 10-18 Predict the 13C NMR spectral pattern of 1-bromopropane (for the 1H NMR spectrum, see Figure 10-27). (Hint: Use the sequential N 1 1 rule.) CH3CH2Br
(CH3)4Si
30
20
10
0
ppm (δ )
Figure 10-31 This 62.8-MHz 13 C NMR spectrum of bromoethane was recorded with broad-band decoupling at 250 MHz. All lines simplify to singlets, including the absorption for (CH3)4Si.
Hydrogen decoupling gives single lines A technique that completely removes 13C – H coupling is called broad-band hydrogen (or proton) decoupling. This method employs a strong, broad radio-frequency signal that covers the resonance frequencies of all the hydrogens and is applied at the same time as the 13 C spectrum is recorded. For example, in a magnetic field of 7.05 T, carbon-13 resonates at 75.3 MHz, hydrogen at 300 MHz (Figure 10-7). To obtain a proton-decoupled carbon spectrum at this field strength, we irradiate the sample at both frequencies. The first radiofrequency signal produces carbon magnetic resonance. Simultaneous exposure to the second signal causes all the hydrogens to undergo rapid a 3 4 b spin flips, fast enough to average their local magnetic field contributions. The net result is the absence of coupling. Use of this technique simplifies the 13C NMR spectrum of bromoethane to two single lines, as shown in Figure 10-31. The power of proton decoupling becomes evident when spectra of relatively complex molecules are recorded. Every magnetically distinct carbon gives only one single peak in the 13C NMR spectrum. Consider, for example, a hydrocarbon such as methylcyclohexane. Analysis by 1H NMR is made very difficult by the close-lying chemical shifts of the eight different types of hydrogens. However, a proton-decoupled 13C spectrum shows only five peaks, clearly depicting the presence of the five different types of carbons and revealing the twofold symmetry in the structure (Figure 10-32). These spectra also exhibit a limitation in 13C NMR spectroscopy: Integration is not usually straightforward. As a consequence of the broad-band decoupling, peak intensities no longer correspond to numbers of nuclei.
13C NMR
CH3
Figure 10-32 13C NMR spectrum of methylcyclohexane with hydrogen decoupling. Each of the five magnetically different types of carbon in this compound gives rise to a distinct peak: d 5 23.1, 26.7, 26.8, 33.1, and 35.8 ppm.
(CH3)4Si 70
60
50
40
30 ppm (δ )
20
10
0
10-9 Carbon-13 Nuclear Magnetic Resonance
Chapter 10
Table 10-6 Typical 13C NMR Chemical Shifts Type of carbon
Chemical shift D (ppm)
Primary alkyl, RCH3 Secondary alkyl, RCH2R9 Tertiary alkyl, R3CH Quaternary alkyl, R4C Allylic, R2C P CCH2R9 A R0 Chloroalkane, RCH2Cl Bromoalkane, RCH2Br Ether or alcohol, RCH2OR9 or RCH2OH Carboxylic acids, RCOOH O O B B Aldehyde or ketone, RCH or RCR9 Alkene, aromatic, R2C P CR2 Alkyne, RC q CR
5 – 20 20 – 30 30 – 50 30 – 45 20 – 40 25 – 50 20 – 40 50 – 90 170 – 180 190 – 210 100 – 160 65 – 95
RCH2Br RCOOH
RCH2Cl R2C P CCH2R⬘
RCH2OR⬘ or RCH2OH
A
R⬙
R2C P CR2
O O B B RCH or RCR⬘
R4C R3CH RCH2R⬘
P CR RC O
220 210 200 190 180 170 160 150 140 130 120 110 100
90
80
70
RCH3 60
50
40
ppm (␦)
Table 10-6 shows that carbon, like hydrogen (Table 10-2), has characteristic chemical shifts depending on its structural environment. As in 1H NMR, electron-withdrawing groups cause deshielding, and the chemical shifts go up in the order primary , secondary , tertiary carbon. Apart from the diagnostic usefulness of such d values, knowing the number of different carbon atoms in the molecule can aid structural identification. Consider, for example, the difference between methylcyclohexane and its isomers with the molecular formula C7H14. Many of them differ in the number of nonequivalent carbons and therefore give distinctly different carbon spectra. Notice how much the (lack of ) symmetry in a molecule affects the complexity of the carbon spectrum.
Four peaks
CH3 H3 C H3 C
Four peaks
/∑
? CH3
H3 C ?
C Peaks in Some C7H14 Isomers
∑/
CH3 %
13
@
Number of
CH3 CH3
Three peaks
One peak
30
20
10
0
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Using NMR Spectroscopy to Deduce Structure
Exercise 10-19 13
How many peaks would you expect in the proton-decoupled compounds? (Hint: Look for symmetry.)
C NMR spectra of the following
H %
(a) 2,2-Dimethyl-1-propanol
(b)
H H % ł H
(d)
ł H
^
(c)
`
%
Chapter 10
%
426
H
Exercise 10-20
Working with the Concepts: Differentiating Isomers by 13C NMR In Exercise 2-16(a) you formulated the structures of the five possible isomers of hexane, C6H14. One of them shows three peaks in the 13C NMR spectrum at d 5 13.7, 22.7, and 31.7 ppm. Deduce its structure. Strategy We need to write down all possible isomers and see to what extent symmetry (or the lack thereof ) influences the number of peaks expected for each (using the labels a, b, c, and d.) Solution a
a
a
b a
a
d c a
b
d e a
c
b
a
c
b
b
a a
Hexane
2-Methylpentane
b
a
2,2-Dimethylbutane 2,3-Dimethylbutane
c
d 3-Methylpentane
• Only one isomer has only three different carbons: hexane.
Exercise 10-21
Try It Yourself A researcher finds two unlabeled bottles in the stock room. She knows that one of them contains the sugar d-ribose, the other d-arabinose (both shown as Fischer projections below), but she does not know which bottle contains which sugar. Armed with a supply of Na12BH4 and equipped with an NMR spectrometer, how could she tell?
CHO
CHO H
OH
HO
H
OH
H
OH
H
OH
H
OH
CH2OH D-Ribose
H
CH2OH D-Arabinose
10-9 Carbon-13 Nuclear Magnetic Resonance
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427
Advances in FT NMR greatly aid structure elucidation: DEPT 13C and 2D-NMR The FT technique for the measurement of NMR spectra is extremely versatile, allowing data to be collected and presented in a variety of ways, each providing information about the structure of molecules. Most recent advances are due to the development of sophisticated time-dependent pulse sequences, including the application of two-dimensional NMR, or 2D NMR (Chemical Highlight 10-4). With these methods, it is now possible to establish coupling (and therefore bonding) between close-lying hydrogens (homonuclear correlation) or connected carbon and hydrogen atoms (heteronuclear correlation). Thus, 1H and 13C NMR allow the determination of molecular connectivity by measuring the magnetic effect of neighboring atoms on one another along a carbon chain. An example of such a pulse sequence, now routine in the research laboratory, is the DEPT 13C NMR spectrum (distortionless enhanced polarization transfer), which tells you what type of carbon gives rise to a specific signal in the normal 13C spectrum: CH3, CH2, CH, or Cquaternary. It avoids the complications arising from a proton-coupled 13C NMR spectrum (see Figure 10-30), particularly overlapping multiplets of close-lying carbon signals. The DEPT method consists of a combination of spectra run with differing pulse sequences: the normal broad-band decoupled spectrum and a set of spectra that reveals signals only of carbons bound to three (CH3), two (CH2), and one hydrogen (CH), respectively. Figure 10-33 depicts a series of such spectra for limonene (see also Problem 5-29). The first is the normal proton-decoupled spectrum (Figure 10-33A), which exhibits the expected number of lines (10) and groups them into the six alkyl and four alkenyl carbon signals at high and at low field, respectively. The remaining spectra specifically identify the three types of possible carbons bearing hydrogens, CH3 (red; Figure 10-33B),
13C NMR 7 1 6
2
5
3 4 8
10
9
Limonene
A
Normal 13C NMR spectrum
B
Spectrum showing only CH3 peaks
C
Spectrum showing only CH2 peaks
D
Spectrum showing only CH peaks
180
160
140
120
100
80 ppm (δ )
60
40
20
0
Figure 10-33 The DEPT 13C NMR method applied to limonene: (A) broad-band decoupled spectrum revealing the six alkyl carbon signals at high field (20 – 40 ppm) and four alkenyl carbon signals at low field (108 – 150 ppm; see Table 10-6); (B) spectrum showing only the two CH3 signals for C7 and C10 (red); (C) spectrum showing only the four CH2 signals for C3, C5, C6, and C9 (blue); (D) spectrum showing only the two CH signals for C2 and C4 (green). The additional lines in (A) are assigned to the quaternary carbons C1 and C8 (black).
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CHE M I C AL H I G H L I G H T 1 0 - 4 Correlated NMR Spectra: COSY and HETCOR We have learned that NMR is of great help in proving the structure of molecules. It tells us what types of nuclei are present and how many of them are in a compound. It can also reveal connectivity by exploiting spin coupling through bonds. This last phenomenon can be exploited in a special type of plot, which is obtained by the application of sophisticated pulse techniques and the associated computer analysis: two-dimensional (2D) NMR. In this method, two spectra of the molecule are plotted on the horizontal
and vertical axes. Mutually coupled signals are indicated as “blotches,” or correlation peaks, on the x – y graph. Because this plot correlates nuclei that are in close proximity (and therefore exhibit spin – spin splitting), this type of spectroscopy is also called correlation spectroscopy or COSY. Let us pick an example we already know: 1-bromopropane (see Figure 10-27). The COSY spectrum is shown below.
1H NMR
0.5
CH3CH2CH2Br
1.0
2.0
ppm (δ )
1.5
2.5
3.0
3.5
3.5
3.0
2.5
2.0
ppm (δ )
The two spectra along the x and y axes are identical to that in Figure 10-27. The marks on the diagonal line correlate the same peaks in each spectrum (indicated by the colored “blotches”) and can be ignored. It is the off-diagonal correlations (indicated by the orange lines) that prove the connectivity of the hydrogens in the molecule. Thus, the (red) methyl triplet at d 5 1.05 ppm has a cross peak (indicated
1.5
1.0
0.5
by the black “blotches”) with the (green) methylene sextet at d 5 1.88 ppm, establishing their relation as neighbors. Similarly, the (blue) peaks for the methylene hydrogens, deshielded by the bromine to appear at d 5 3.40 ppm, also correlate with the center signal at d 5 1.88 ppm. Finally, the center absorption shows cross peaks with the signals of both of its neighbors, thus proving the structure.
10-9 Carbon-13 Nuclear Magnetic Resonance
Of course, the example of 1-bromopropane is overly simple and was chosen only to demonstrate the workings of this technique. Think of 1-bromohexane instead. Here, the rapid assignment of the individual methylene groups is made possible by COSY spectroscopy. Starting from either of the two signals that are readily recognized — namely, those due to the methyl (a unique high-field triplet) or the CH2Br protons (a unique low-field triplet) — we can “walk along” the correlated signals from one methylene neighbor
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429
to the next until the complete connectivity of the structure is evident. Connectivity is disclosed even more powerfully by heteronuclear correlation spectroscopy (HETCOR). Here, an 1 H NMR spectrum is juxtaposed with a 13C spectrum. This allows the mapping of a molecule along its entire C – H framework by revealing the identity of C–H bonded fragments. The principle is illustrated again with 1-bromopropane in the spectrum shown below.
05 CH3CH2CH2Br 10
15
25
ppm (δ )
20
30
35
40
45
3.5
3.0
2.5
2.0
ppm (δ )
The 13C NMR spectrum is plotted on the y axis and exhibits three peaks. On the basis of what we know about carbon NMR, we can readily assign the signal at d 5 13.0 ppm to the methyl carbon, that at d 5 26.2 ppm to the methylene, and that at d 5 36.0 ppm to the bromine-bearing carbon. But even without this knowledge, these assignments are immediately evident from the cross-correlations of the
1.5
1.0
0.5
respective hydrogen signals (which we have already assigned) with those of their bound carbon counterparts. Thus, the methyl hydrogen triplet correlates with the carbon signal at d 5 13.0 ppm, the sextet for the methylenes has a cross peak with the carbon signal at d 5 26.2 ppm, and the downfield proton triplet is connected to the carbon with a signal at d 5 36.0 ppm.
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Using NMR Spectroscopy to Deduce Structure
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CHE M I C AL H I G H L I G H T 1 0 - 5 Structural Characterization of Natural Products: Antioxidants from Grape Seeds The world of plants is a rich source of useful pharmacological, therapeutic, and chemoprotective substances. The O
H H H O %
HO
O
≥ H
4 ( & ¥ OH O OH H H
H
H ?H `
%
2 3 `H
OH
O O O
O
O O O
O Viniferone
Three Signals CH3CH2CHCl2 10.1
34.9
Three Signals CH3CHClCH2Cl 55.8
49.5 ppm
Moderately deshielded
Two Signals ClCH2CH2CH2Cl 42.2
Moderately deshielded
CH2 (blue; Figure 10-33C), and CH (green; Figure 10-33D). The quaternary carbon signals (black) don’t show up in the last three experiments and are located by subtracting all of the lines in Figure 10-33B – D from the complete spectrum in Figure 10-33A. For the remainder of the text, whenever the depiction or description of a 13C NMR spectrum includes spectral assignments to CH3 CH2, CH, or Cquaternary carbons, they are based on a DEPT experiment.
73.2 ppm
Strongly deshielded
22.4
15-carbon compound viniferone (a sesquiterpene; see Section 4-7) was isolated from grape seeds in 2004. It is one of a variety of substances called grape seed proanthocyanidins, which are highly active against radicals (Chapter 3) and oxidative stress (see Section 22-9). Only 40 mg of the compound was obtained from 10.5 kg of grape seeds; therefore, elemental analysis, chemical tests, and any other procedures that would destroy the tiny amount of available material could not be used in structural elucidation. Instead, a combination of spectroscopic techniques (NMR, IR, MS, and UV) was employed, leading to the structure shown here, which was later confirmed by X-ray crystallography. The characterization of viniferone relied in part on 1 H and 13C NMR data. Proton chemical shifts provided evidence for key pieces of the structure. Thus, there were three signals in the range reserved for internal alkene and aromatic hydrogens (shown in orange in the structural drawings) between d 5 5.9 and 6.2 (consult Table 10-2 for typical proton chemical shifts). Another three signals were observed at d . 3.8 ppm, reflecting the presence of three protons (red) attached to oxygen-bearing carbons. Finally, there were four absorptions with relatively low d values between 2.5 and 3.1 ppm, assignable to the two pairs of (green) diastereotopic hydrogens (see Chemical Highlight 10-3). 13 C NMR was equally helpful (consult Table 10-6 for
35.6 ppm
We can apply 13C NMR spectroscopy to the problem of the monochlorination of 1-chloropropane In Section 10-6 we learned how we could use 1H NMR chemical shifts and integration to distinguish among the three isomers of dichloropropane arising from the chlorination of 1-chloropropane. Exercise 10-12 addressed the use of spin – spin splitting patterns as a complementary means of solving this problem. How does 13C NMR fare in this task? Our prediction is straightforward: Both 1,1- and 1,2-dichloropropane should exhibit three carbon signals each, but spaced significantly differently because the 1,1-isomer has the two electronwithdrawing chlorine atoms located on the same carbon (hence no chlorines on the remaining two), whereas 1,2-dichloropropane bears one chlorine each on C1 and C2. In contrast, 1,3-dichloropropane would be clearly distinct from the other two isomers because of its symmetry: Only two lines should be observed. The experimental data are shown in the margin and confirm our expectations. The specific assignments of the two deshielded chlorine-bearing carbons in 1,2-dichloropropane (as indicated in the margin) can be made on the basis of the DEPT technique: The signal at 49.5 ppm appears as a CH2, and that at 55.8 ppm as a CH in the DEPT-90 experiment.
10-9 Carbon-13 Nuclear Magnetic Resonance
Chapter 10
431
typical carbon chemical shifts), as it clearly revealed the two C P O carbons at d 5 171.0 and 173.4 ppm and indicated the presence of eight alkene and benzene carbons (d . 95 ppm). The three tetrahedral carbons attached to oxygen showed up as peaks between d 5 67 and 81 ppm, and the two remaining tetrahedral carbon signals occurred at d 5 28.9 and 37.4 ppm. All of these were identified further by establishing the number of attached hydrogens by techniques equivalent to DEPT NMR. Spin – spin splitting and correlated (COSY) proton spectra (Chemical Highlight 10-4) confirmed the structural assignment. For example, focusing on the six-membered ether ring, the proton on C2 (d 5 4.61 ppm) showed a doublet pattern due to coupling to its neighbor on C3 (d 5 3.90 ppm). Similarly, the two diastereotopic protons at C4 (d 5 2.53 and 3.02 ppm) gave rise to a doublet of doublets each, because of mutual coupling and independent coupling of each to the proton on C3. Not surprisingly, the proton on C3 produced an unresolved multiplet. Additional aspects of the structural determination of viniferone will be presented in Chapter 14 (Chemical Highlight 14-5).
The extracts of grape seeds are being touted in the prevention of a range of maladies, including heart disease, cancer, and psoriasis.
You can see from this example how 1H NMR and 13C NMR spectroscopy complement each other. 1H NMR spectra provide an estimate of the electronic environment (i.e., electron rich versus electron poor) of a hydrogen nucleus under observation (d), a measure of its relative abundance (integration), and an indication of how many neighbors (and their number of types) it has (spin – spin splitting). Proton-decoupled 13C NMR provides the total number of chemically distinct carbons, their electronic environment (d), and, in the DEPT mode, even the quantity of their attached hydrogens. Application of both techniques to the solution of a structural problem is not unlike the methods used to solve a crossword puzzle. The horizontal entries (such as the data provided by 1H NMR spectroscopy) have to fit the vertical ones (i.e., the corresponding 13C NMR information) to provide the correct answer. Exercise 10-22 Are bicyclic compounds A and B shown below readily distinguished by their proton-decoupled 13 C NMR spectra? Would DEPT spectra be of use in solving this problem? H3C
H3C
/ ;
H %
CH3 ?
H %
0 H
' CH3
0 H
A
B
In Summary 13C NMR requires FT techniques because of the low natural abundance of the carbon-13 isotope and its intrinsically lower sensitivity in this experiment. 13C–13C coupling is
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Using NMR Spectroscopy to Deduce Structure
Chapter 10
not observed, because the scarcity of the isotope in the sample renders the likelihood of neighboring 13C nuclei negligible. 13C–H coupling can be measured but is usually removed by broadband proton decoupling, providing single lines for each distinct carbon atom in the molecule under investigation. The 13C NMR chemical-shift range is large, about 200 ppm for organic structures. 13C NMR spectra cannot usually be integrated, but the DEPT experiment allows the identification of each signal as arising from CH3, CH2, CH, or Cquaternary units, respectively.
THE BIG PICTURE In your study of organic chemistry so far, there may have been moments during which you wondered, how do chemists know this? How do they establish the veracity of a structure? How do they follow the kinetics of disappearance of a molecule? How is an equilibrium constant determined? When do they know that a reaction is over? The introduction of NMR spectroscopy provides a first glimpse of the variety of practical tools available to answer these questions. Other forms of spectroscopy can yield other important information about molecules. We shall introduce them in conjunction with the functional groups for which they are particularly diagnostic. Spectroscopy, especially NMR spectroscopy, is the key to identifying the different classes of organic compounds and is an important part of each subsequent chapter in the book. You may find it helpful to review the material in this chapter as you examine new spectra later on.
CHAPTER INTEGRATION PROBLEMS 10-23. A researcher executed the following reaction sequence in a preparation of (S)-2-chlorobutane: H H3C ?
SOCl2
/
H H3C
HO H H3C
CH3
CH3
μ
/∑
CH3Li
∑
O
⫺SO2,⫺HCl
(R)-2-Methyloxacyclopropane
Cl (S)-2-Chlorobutane
Careful preparative gas chromatography of the reaction product (b.p. 68.28C) allowed the separation of a very small amount of another compound, C4H9Cl (b.p. 68.58C), which was optically inactive and exhibited the NMR spectra depicted below. What is the structure of this compound and how could it have been formed? 1
13
H NMR
CH3
C NMR
6H
CH2
2H 2.05
2.00
1.95
CH
1.90
1H
3.6
3.4
3.2
3.0
2.8
2.6
2.4
2.2
2.0
(CH3)4 Si
1.8
1.6
300-MHz 1H NMR spectrum ppm (δ) δ
1.4
1.2
1.0
0.8
50
25
13
0
C NMR spectrum ppm (δ) δ
Chapter Integration Problems
SOLUTION As we did in Chapter Integration Problems 3-14 and 6-30, let us apply the WHIP approach to break down the process of solving this problem. Again, this strategy will be described in detail in the Interlude that follows Chapter 11. What is the problem asking? We are first to deduce the structure of a minor product of the given reaction sequence. Subsequently, we are asked to explain “how” this compound forms. As mentioned earlier, “how” and “why” questions normally require some mechanistic insight. We’ll take things in order. How to begin? The given molecular composition of the unknown compound, C4H9Cl, tells us that it is an isomer of the observed major product. We have both its 1H and 13C NMR spectra available for analysis. We also know the reagents that convert the starting material to the major product. Which pieces of information are the most useful ones to help us get started? We could try to come up with alternative results of the two reactions in the sequence without examining the spectroscopic data at all, but, realistically, sooner or later it will be necessary to use the spectra to confirm any hypothetical structure we propose. It makes much more sense to start with them and at least find out whether they give us an unambiguous answer. Information needed? Sections 10-4 through 10-7 reveal the information contained in an 1H NMR spectrum; Section 10-9 covers 13C NMR. The carbon spectrum is simpler. As a general rule of strategy, try to extract the most information you can from the sources that are the simplest to analyze. Start with the 13C NMR spectrum; then proceed to the 1H NMR spectrum. Proceed. The formula of the unknown is C4H9Cl. It has four carbon atoms, but the 13C NMR spectrum (assignments by DEPT) displays only three lines. Therefore, one of these three lines must arise from two equivalent carbon atoms. The line farthest downfield, at d 5 51 (CH2) ppm, is the most deshielded and therefore most likely to be the unique carbon bearing the chlorine atom (Table 10-6). This reasoning suggests the presence of substructure – CH2Cl. The other two lines are in the alkyl region, at about d 5 20 (CH3) and 31 (CH) ppm. One of these two must be due to two equivalent carbon nuclei. We can use the molecular formula and our knowledge of substructure – CH2Cl to identify which one. Thus, if the signal (for CH) at d 5 31 ppm were due to two identical carbons, then we would arrive at a molecular formula of CH3 1 2 CH 1 CH2Cl 5 C4H7Cl, not a match for that given. On the other hand, if the signal at d 5 20 (CH3) ppm corresponded to two carbons, then we have 2 CH 3 1 CH 1 CH 2Cl 5 C 4H 9Cl, the correct formula. Therefore, the unknown contains two equivalent CH3 groups. We could connect these fragments to give us an immediate likely solution, but let us exercise a little patience and see what the 1H NMR spectrum has to tell us. This spectrum reveals three types of hydrogens at about d 5 1.0, 1.9, and 3.4 ppm. Following the same logic as that applied in the above assignment of the most deshielded carbon atom, the most deshielded hydrogens must be those attached to the chlorine-bearing carbon (Section 10-4). The integrated values for the three hydrogen signals are 6, 1, and 2, respectively, totaling to the nine hydrogens present (Section 10-6). Finally, there is spin-spin splitting. Both the highest and lowest field signals are doublets, with almost identical J values. This means that each of these sets of hydrogens (6 1 2, or 8 in all) has a single neighbor (the ninth H). That hydrogen shows up in the middle as a nine-line pattern, as expected by the N 1 1 rule (N 5 8; Section 10-7). Now let us combine this information in a structural assignment. As with most puzzles, one can arrive at the answer in several ways. In NMR spectral problems, it is often best to start with the formulation of partial structures, as dictated by the 1H NMR spectrum, and use the other information as corroborating evidence. Thus, the high-field doublet integrating for 6 H indicates a (CH3)2CH – substructure. Similarly, the low-field counterpart points to – CH2CH – . Combining the two provides – CH2CH(CH3)2 and, adding the Cl atom, the solution: the achiral (hence optically inactive) ClCH2CH(CH3)2. This assignment is confirmed by the 13C NMR spectrum, the highest field line being due to the presence of the two equivalent methyl carbons. The center line is due to the tertiary, the most deshielded absorption due to the chlorine-bearing carbon (Table 10-6). You can confirm your solution in another manner, taking advantage of the relatively small molecular formula; thus, there are only four possible chlorobutane isomers: CH3CH2CH2CH2Cl, CH3CHCH2CH3 (our major product), ƒ Cl (CH3)2CHCH2Cl (the minor product), and (CH3)3CCl. They differ drastically in their 1H and 13 C NMR spectra with respect to number of signals, chemical shifts, integration, and multiplicities. (Verify.)
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Using NMR Spectroscopy to Deduce Structure
The second aspect of this problem is mechanistic. How do we get 1-chloro-2-methylpropane from the preceding reaction sequence? The answer presents itself on retrosynthetic analysis, using the reagents given in our initial scheme: CH3 A H3C O C O CH2Cl A H
SOCl2
CH3 A H3C O C O CH2OH A H
O
CH3Li
/∑
434
H H3C
The only way to obtain a product with two CH3 groups attached to the same carbon is to have the methyl group of CH3Li add to the carbon of the oxacyclopropane that already contains one CH3. Therefore, the observed minor product is the result of nucleophilic ring opening of the starting compound by attack at the more hindered position, usually neglected because it is less favored. 10-24. a. A graduate student took the 1H and 13C NMR spectra of optically pure (1R, 2R)-trans1-bromo-2-methylcyclohexane (A) in deuterated nitromethane (CD3NO2) as a solvent (Table 6-5) and recorded the following values: 1H NMR spectrum: d 5 1.06 (d, 3 H), 1.42 (m, 6 H), 1.90 (m, 2 H), 2.02 (m, 1 H), 3.37 (m, 1 H); 13C NMR (DEPT): d 5 16.0 (CH3), 23.6 (CH2), 23.9 (CH2), 30.2 (CH2), 33.1 (CH2), 35.0 (CH), 43.2 (CH) ppm. Assign these spectra as best you can with the help of Tables 10-2 and 10-6, respectively. SOLUTION H NMR spectrum. All of the signals, with the exception of the highest field doublet, appear as complex multiplets. This is not surprising, considering that each hydrogen (except for the three CH3 nuclei) is magnetically distinct, that there are multiple vicinal and geminal couplings all around the ring, and that the d values (except for those in the immediate vicinity of the bromine substituent) are close. The exception is the methyl group appearing at highest field, as expected, and as a doublet due to coupling to the neighboring tertiary hydrogen. To complete our assignments, we have to rely more than usual on chemical shifts and integration. What is the expected effect of bromine on its vicinity? Answer: Br is deshielding (Section 10-4), mostly its direct neighbors, the effect tapering off with distance. Indeed, the d values divide into two groups. One set has higher (d 5 3.37, 2.02, 1.90 ppm), the other lower values (d 1.42, 1.06 ppm). The most deshielded single hydrogen at d 5 3.37 ppm is readily assignable to that at C1 next to bromine. The next most deshielded positions are its neighbors at C2 and C6. Since the signal at d 5 2.02 ppm integrates for only 1 H, it must be the single tertiary hydrogen at C2; the peak at d 5 1.90 ppm can then be assigned to the CH2 group at C6. This choice is also consistent with the general appearance of secondary hydrogen signals at higher field than those of tertiary hydrogens (Table 10-2). The remaining six hydrogens at d 5 1.42 ppm are not resolved, and all absorb at around the same place. 1
16.0
CH3
2.02 1.06
35.0
∑
H CH3 ∑/ Br 2 / H 3.37 3 1 4 1.42
5
6 1.90
33.1 30.2, 23.9, 23.6
A
A 1
H NMR assignments (ppm)
Br Š43.2
13
C assignments (ppm)
13
C NMR spectrum. Using the DEPT correlations in conjunction with the deshielding effect of bromine allows a ready assignment to the extent shown above.
b. To the student’s surprise, the optical activity of the sample decreased with time. Concurrently, the NMR spectrum of A diminished in intensity and new peaks appeared of an optically inactive isomer B. 1 H NMR: d 5 1.44 (m, 6 H), 1.86 (m, 4 H), 1.89 (s, 3 H) ppm; 13C NMR: d 5 20.8 (CH2), 26.7 (CH2), 28.5 (CH3), 37.6 (Cquat), 41.5 (CH2) ppm. Following the disappearance of A revealed a rate 5 k[A]. What is B, and how is it formed? SOLUTION Let us analyze the spectra of B, particularly in comparison to those of A. In the 1H NMR spectrum, we note that the number of signals has decreased from five to only three. Moreover, the low-field peak at d 5 3.37 ppm for the hydrogen next to bromine in A has disappeared and the methyl doublet is now a singlet, moved to lower field relative to its original position. In the 13C NMR spectrum, we
Important Concepts
recognize a similar simplification (hence symmetrization) from seven to five peaks. Moreover, the tertiary carbons (CH) have disappeared, there are only three types of CH2, and a quaternary carbon has appeared. The relatively deshielded CH3 carbon is also evident. Conclusions. The bromine atom must now be located at the same position as the CH3 substituent, as in the achiral 1-bromo-1-methylcyclohexane: 1.89
Br
28.5
CH3
Br
1.86
CH3 37.6
20.8, 26.7
1.44
B 1
41.5
B 13
H NMR assignments (ppm)
C NMR assignments (ppm)
How could that rearrangement have happened? Answer: We have a case of an SN1 reaction (Section 7-2) proceeding through a rearranging carbocation (Section 9-3): H CH3 ∑/ ŠBr
H CH3 ∑/ ⫹
CH3 ⫹
Br
CH3
ⴚ š ⫹ðBr ð
ⴚ š ⫺ðBr ð
A
B
The mechanism explains all of the observations: The loss of optical activity, the first-order disappearance of A, and the formation of B.
Important Concepts 1. NMR is the most important spectroscopic tool in the elucidation of the structures of organic molecules. 2. Spectroscopy is possible because molecules exist in various energetic forms, those at lower energy being convertible into states of higher energy by absorption of discrete quanta of electromagnetic radiation. 3. NMR is possible because certain nuclei, especially 1H and 13C, when exposed to a strong magnetic field, align with it (a) or against it (b). The a-to-b transition can be effected by radiofrequency radiation, leading to resonance and a spectrum with characteristic absorptions. The higher the external field strength, the higher the resonance frequency. For example, a magnetic field of 7.05 T causes hydrogen to absorb at 300 MHz, a magnetic field of 14.1 T causes it to absorb at 600 MHz. 4. High-resolution NMR allows for the differentiation of hydrogen and carbon nuclei in different chemical environments. Their characteristic positions in the spectrum are measured as the chemical shift, d, in ppm from an internal standard, tetramethylsilane. 5. The chemical shift is highly dependent on the presence (causing shielding) or absence (causing deshielding) of electron density. Shielding results in relatively high-field peaks [to the right, toward (CH3)4Si], deshielding in low-field ones. Therefore, electron-donor substituents shield, and electron-withdrawing components deshield. The protons on the heteroatoms of alcohols, thiols, and amines show variable chemical shifts and often appear as broad peaks because of hydrogen bonding and exchange. 6. Chemically equivalent hydrogens and carbons have the same chemical shift. Equivalence is best established by the application of symmetry operations, such as those using mirror planes and rotations. 7. The number of hydrogens giving rise to a peak is measured by integration. 8. The number of hydrogen neighbors of a nucleus is given by the spin – spin splitting pattern of its NMR resonance, following the N 1 1 rule. Equivalent hydrogens show no mutual spin – spin splitting.
Chapter 10
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436
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Using NMR Spectroscopy to Deduce Structure
9. When the chemical-shift difference between coupled hydrogens is comparable to their coupling constant, non-first-order spectra with complicated patterns are observed. 10. When the constants for coupling to nonequivalent types of neighboring hydrogens are different, the N 1 1 rule is applied sequentially. 11. Carbon NMR utilizes the low-abundance 13C isotope. Carbon – carbon coupling is not observed in ordinary 13C spectra. Carbon – hydrogen coupling can be removed by proton decoupling, thereby simplifying most 13C spectra to a collection of single peaks. 12. DEPT 13C NMR allows the assignment of absorptions to CH3, CH2, CH, and quaternary carbons, respectively.
Problems 25. Where on the chart presented in Figure 10-2 would the following be located: AM radio waves (n , 1 MHz 5 1000 kHz 5 106 Hz 5 106 s21, or cycles s21); FM broadcast frequencies (n , 100 MHz 5 108 s21)? 26. Convert each of the following quantities into the specified units. (a) 1050 cm21 into l, in mm; (b) 510 nm (green light) into n, in s21 (cycles s21, or hertz); (c) 6.15 mm into | n , in cm21; (d) 2250 cm21 into n, in s21 (Hz). 27. Convert each of the following quantities into energies, in kcal mol21. (a) A bond rotation of 750 wavenumbers (cm21); (b) a bond vibration of 2900 wavenumbers (cm21); (c) an electronic transition of 350 nm (ultraviolet light, capable of sunburn); (d) the broadcast frequency of the audio signal of TV channel 6 (87.25 MHz; before the advent of digital TV in 2009); (e) a “hard” X-ray with a 0.07-nm wavelength. 28. Calculate to three significant figures the amount of energy absorbed by a hydrogen when it undergoes an a-to-b spin flip in the field of (a) a 2.11-T magnet (n 5 90 MHz); (b) an 11.75-T magnet (n 5 500 MHz). 29. For each of the following changes, indicate whether it corresponds to moving to the right or to the left in an NMR spectrum. (a) Increasing radio frequency (at constant magnetic field strength); (b) increasing magnetic field strength (at constant radio frequency; moving “upfield”; Section 10-4); (c) increasing chemical shift; (d) increased shielding. 30. Sketch a hypothetical low-resolution NMR spectrum showing the positions of the resonance peaks for all magnetic nuclei for each of the following molecules. Assume an external magnetic field of 2.11 T. How would the spectra change if the magnetic field were 8.46 T? (a) CFCl3 (Freon 11)
(b) CH3CFCl2 (HCFC-141b)
Cl A (c) CF3 O C OH A Br
(Halothane)
31. If the NMR spectra of the molecules in Problem 30 were recorded by using high resolution for each nucleus, what differences would be observed? 32. The 1H NMR spectrum of CH3COCH2C(CH3)3, 4,4-dimethyl-2-pentanone, taken at 300 MHz, shows signals at the following positions: 307, 617, and 683 Hz downfield from tetramethylsilane. (a) What are the chemical shifts (d) of these signals? (b) What would their positions be in hertz, relative to tetramethylsilane, if the spectrum were recorded at 90 MHz? At 500 MHz? (c) Assign each signal to a set of hydrogens in the molecule. 33. Order the 1H NMR signals of the following compounds by chemical-shift position (lowest to highest). Which one is the most upfield? The most downfield? (a) H3C ¬ CH3 O B (d) H3C O C O CH3
(b) H2C “ CH2
(c) H3C ¬ O¬ CH3
(e)
CH3 A (f) H3C O N O CH3
Problems
34. Which hydrogens in the following molecules exhibit the more downfield signal relative to (CH3)4Si in the NMR experiment? Explain.
(a) (CH3)2O or (CH3)3N
O B (b) CH3COCH3
(c) CH3CH2CH2OH
or
(d) (CH3)2S or (CH3)2S“O
or
35. How many signals would be present in the 1H NMR spectrum of each of the cyclopropane derivatives shown below? Consider carefully the geometric environments around each hydrogen. Br Br
(a)
Br
(b)
Br ´
Br
(c)
Br
(d)
Br
Br
(e)
Br
Br
36. How many signals would be present in the 1H NMR spectrum of each of the following molecules? What would the approximate chemical shift be for each of these signals? Ignore spin – spin splitting.
(a) CH3CH2CH2CH3
CH3 A (c) HOCH2CCl A CH3
(b) CH3CHCH3 A Br
CH3 A (e) CH3CNH2 A CH3
(f ) CH3CH2CH(CH2CH3)2
O B (i) CH3CH2 O C A H
CH3 CH3O A A (j) CH3CH O C O CH3 A CH3
(g) CH3OCH2CH2CH3
CH3 A (d) CH3CHCH2CH3
H2C O CH2 A A (h) H2C O C M O
37. For each compound in each of the following groups of isomers, indicate the number of signals in the 1H NMR spectrum, the approximate chemical shift of each signal, and the integration ratios for the signals. Ignore spin – spin splitting. Indicate whether all the isomers in each group can be distinguished from one another by these three pieces of information alone. CH3 CH3 CH3 A A A (a) CH3CCH2CH3, BrCH2CHCH2CH3, CH3CHCH2CH2Br A Br CH2Cl CH3 A A (b) ClCH2CH2CH2CH2OH, CH3CHCH2OH, CH3CCH2OH A Cl CH3 CH3 CH3 A A A (c) ClCH2C CHCH3, ClCH2CH A Br
CH3 CH3 CH3 A A A CCH3, ClCH2C CHCH3, ClCH2CHCCH3 A A A A A CH3 Br Br CH3 Br
Chapter 10
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438
Using NMR Spectroscopy to Deduce Structure
Chapter 10
38. 1H NMR spectra for two haloalkanes are shown below. Propose structures for these compounds that are consistent with the spectra. (a) C5H11Cl, spectrum A; (b) C4H8Br2, spectrum B. 1H
1H
9H
NMR
NMR 6H
(CH3)4Si
2H
2H (CH3)4Si
4.0
3.5
3.0
300-MHz
2.5 1H
2.0
1.5
1.0
0.5
0.0
4.0
NMR spectrum ppm (δ) δ
3.5
3.0
300-MHz
2.5 1H
2.0
1.5
1.0
0.5
0.0
NMR spectrum ppm (δ) δ
B
A
39. The following 1H NMR signals are for three molecules with ether functional groups. All the signals are singlets (single, sharp peaks). Propose structures for these compounds. (a) C3H8O2, d 5 3.3 and 4.4 ppm (ratio 3 : 1); (b) C4H10O3, d 5 3.3 and 4.9 ppm (ratio 9 : 1); (c) C5H12O2, d 5 1.2 and 3.1 ppm (ratio 1: 1). Compare and contrast these spectra with that of 1,2-dimethoxyethane (Figure 10-15B). 40. (a) The 1H NMR spectrum of a ketone with the molecular formula C6H12O has d 5 1.2 and 2.1 ppm (ratio 3 : 1). Propose a structure for this molecule. (b) Each of two isomeric molecules related to the ketone in (a) has the molecular formula C6H12O2. Their 1H NMR spectra are described as follows: isomer 1, d 5 1.5 and 2.0 ppm (ratio 3 :1); isomer 2, d 5 1.2 and 3.6 ppm (ratio 3 : 1). All signals in these spectra are singlets. Propose structures for these compounds. To what compound class do they belong? 41. List the four important features of 1H NMR and the information you can derive from them. (Hint: See Sections 10-4 through 10-7.) 42. Describe in which ways the 1H NMR spectra of the compounds below would be similar and how they would differ. Address each of the four issues you listed in Problem 41. To which compound class does each of these compounds belong? O B CH3CH2 O C O O O CH3
O B CH3 O C O O O CH2CH3
O B CH3CH2 O C O CH3
O B CH3CH2CH2 O C O H
43. Below are shown three C4H8Cl2 isomers on the left and three sets of 1H NMR data that one would expect on application of the simple N 1 1 rule on the right. Match the structures to the proper spectral data. (Hint: You may find it helpful to sketch the spectra on a piece of scratch paper.) Cl Cl A A (a) CH3CH2CHCH2 Cl Cl A A (b) CH3CHCHCH3 Cl Cl A A (c) CH3CHCH2CH2
(i) d 5 1.5 (d, 6 H) and 4.1 (q, 2 H) ppm
(ii) d 5 1.6 (d, 3 H), 2.1 (q, 2 H), 3.6 (t, 2 H), and 4.2 (sex, 1 H) ppm (iii) d 5 1.0 (t, 3 H), 1.9 (quin, 2 H), 3.6 (d, 2 H), and 3.9 (quin, 1 H) ppm
Problems
Chapter 10
44. Predict the spin – spin splitting that you would expect to observe in the NMR spectra of each compound in Problem 36. (Reminder: Hydrogens attached to oxygen and nitrogen do not normally exhibit spin – spin splitting.) 45. Predict the spin – spin splitting that you would expect to observe in the NMR spectra of each compound in Problem 37. 46. The 1H NMR chemical shifts are given for each of the following compounds. As best you can, assign each signal to the proper group of hydrogens in the molecule and sketch a spectrum for each compound, incorporating spin – spin splitting whenever appropriate. (a) Cl2CHCH2Cl, d 4.0 and 5.8 ppm; (b) CH3CHBrCH2CH3, d 1.0, 1.7, 1.8, and 4.1 ppm; (c) CH3CH2CH2COOCH3, d 1.0, 1.7, 2.3, and 3.6 ppm; (d) ClCH2CHOHCH3, d 1.2, 3.0, 3.4, and 3.9 ppm. 47. 1H NMR spectra C through F (see below) correspond to four isomeric alcohols with the molecular formula C5H12O. Try to assign their structures.
1H
6H
NMR
1H
3H
NMR
6H
1H
2H 1.5
1.4
1.3
0.9
0.8
0.7 1.8 1.7 1.6 1.5 1.4
(CH3)4Si
3.0
2.5
2.0
1.5
1.0
0.5
2H
4.0
0.0
3.5
300-MHz 1H NMR spectrum ppm (δ) δ
(CH3)4Si
1H 1H
3.0
2.5
2.0
1.5
1.0
0.5
0.0
300-MHz 1H NMR spectrum ppm (δ) δ
D
C 1H
2H
1H
NMR
3H
NMR
3H
2H 4H
3H 4.0
2H 1H
3.8
3.6
4H
(CH3)4Si 1H
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.0
4.0
1H
3.5
3.0
2.5
2.0
1.5
1.0
300-MHz 1H NMR spectrum ppm (δ) δ
300-MHz 1H NMR spectrum ppm (δ) δ
E
(CH3)4Si
F
48. Sketch 1H NMR spectra for the following compounds. Estimate chemical shifts (see Section 10-4) and show the proper multiplets for peaks that exhibit spin – spin coupling. (a) CH3CH2OCH2Br; (b) CH3OCH2CH2Br; (c) CH3CH2CH2OCH2CH2CH3; (d) CH3CH(OCH3)2.
0.5
0.0
439
440
Chapter 10
1H
Using NMR Spectroscopy to Deduce Structure
49. A hydrocarbon with the formula C6H14 gives rise to 1H NMR spectrum G (margin). What is its structure? This molecule has a structural feature similar to that of another compound whose spectrum is illustrated in this chapter. What molecule is that? Explain the similarities and differences in the spectra of the two.
NMR 12 H
50. Treatment of the alcohol corresponding to NMR spectrum D in Problem 47 with hot concentrated HBr yields a substance with the formula C5H11Br. Its 1H NMR spectrum exhibits signals at d 5 1.0 (t, 3 H), 1.2 (s, 6 H), and 1.6 (q, 2 H) ppm. Explain. (Hint: See NMR spectrum C in Problem 42.) 51. The 1H NMR spectrum of 1-chloropentane is shown at 60 MHz (spectrum H) and 500 MHz (spectrum I). Explain the differences in appearance of the two spectra, and assign the signals to specific hydrogens in the molecule. 1.5
1.4
0.9
1H
0.8
3H
NMR 2H 6H
(CH3)4Si 2H (CH3)4Si 1.5
1.0
0.5
0.0
1H
300-MHz NMR spectrum ppm (δ) δ
4.0
G
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.0
60-MHz 1H NMR spectrum ppm (δ) δ H 1
H NMR
3H
2H
2H 2H2H
4.0
3.5
3.0
2.5
2.0
1.5
(CH3)4Si
1.0
0.5
0.0
500-MHz 1H NMR spectrum ppm (δ) δ
I 52. Describe the spin – spin splitting patterns that you expect for each signal in the 1H NMR spectra of the five bromocyclopropane derivatives illustrated in Problem 35. Note that in these compounds the geminal coupling constants (i.e., between nonequivalent hydrogens on the same carbon atom — Section 10-7) and trans vicinal coupling constants are smaller (ca. 5 Hz) than the cis vicinal coupling constants (ca. 8 Hz).
Problems
Chapter 10
53. Can the three isomeric pentanes be distinguished unambiguously from their broad-band proton-decoupled 13C NMR spectra alone? Can the five isomeric hexanes be distinguished in this way? 54. Predict the decoupling.
13
C NMR spectra of the compounds in Problem 36, with and without proton
55. Rework Problem 37 as it pertains to
13
C NMR spectroscopy.
56. How would the DEPT 13C spectra of the compounds discussed in Problems 35 and 37 differ in appearance from the ordinary 13C spectra? 57. From each group of three molecules, choose the one whose structure is most consistent with the proton-decoupled 13C NMR data. Explain your choices. (a) CH3(CH2)4CH3, (CH3)3CCH2CH3, (CH3)2CHCH(CH3)2; d 5 19.5 and 33.9 ppm. (b) 1-Chlorobutane, 1-chloropentane, 3-chloropentane; d 5 13.2, 20.0, 34.6, and 44.6 ppm. (c) Cyclopentanone, cycloheptanone, cyclononanone; d 5 24.0, 30.0, 43.5, and 214.9 ppm. (d) ClCH2CHClCH2Cl, CH3CCl2CH2Cl, CH2 5 CHCH2Cl; d 5 45.1, 118.3, and 133.8 ppm. (Hint: Consult Table 10-6.) 58. Propose a reasonable structure for each of the following molecules on the basis of the given molecular formula and of the 1H and proton-decoupled 13C NMR data. (a) C7H16O, spectra J and K (* 5 CH2 by DEPT); (b) C7H16O2, spectra L and M (the assignments in M were made by DEPT).
1
H NMR
8H
13
C NMR
3H
* * * * *
2H * (CH3)4 Si 2H 1H
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.0
60
40
300-MHz 1H NMR spectrum ppm (δ) δ
20
0
C NMR spectrum ppm (δ) δ
13
J
K
13
1
C NMR
H NMR 4H
CH2 CH2 6H
CH3
2H 4H Cquat
(CH3)4 Si
4
3.5
3
2.5
2
1.5
1
0.5
0
60
δ 300-MHz 1H NMR spectrum ppm (δ)
L
40
20
C NMR spectrum ppm (δ) δ
13
M
0
441
442
Using NMR Spectroscopy to Deduce Structure
Chapter 10
59.
1
The 1H NMR spectrum of cholesteryl benzoate (see Section 4-7) is shown as spectrum N. Although complex, it contains a number of distinguishing features. Analyze the absorptions marked by integrated values. The inset is an expansion of the signal at d 5 4.85 ppm and exhibits an approximately first-order splitting pattern. How would you describe this pattern? (Hint: The peak patterns at d 5 2.5, 4.85, and 5.4 ppm are simplified by the occurrence of chemical shift and/or coupling constant equivalencies.)
3H
H NMR
3H
6H
3H
2H
2H
5.0
4.9
4.8
4.7
2H
1H
(CH3)Si
1H 1H
8.0
7.5
7.0
6.5
6.0
5.5
5.0
4.5
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.0
300-MHz 1H NMR spectrum ppm (δ) δ
N
H CH3 /∑
CH3 %
CH3
CH3 H % ≥ % ≥ H H
O CO
CH3
)
Cholesteryl benzoate
60.
The terpene a-terpineol has the molecular formula C10H18O and is a constituent of pine oil. As the -ol ending in the name suggests, it is an alcohol. Use its 1H NMR spectrum (spectrum O, p. 443) to deduce as much as you can about the structure of a-terpineol. [Hints: (1) a-Terpineol has the same 1-methyl-4-(1-methylethyl)cyclohexane framework found in a number of other terpenes (e.g., carvone, Problem 43 of Chapter 5). (2) In your analysis of spectrum O, concentrate on the most obvious features (peaks at d 5 1.1, 1.6, and 5.4) and use chemical shifts, integrations, and the splitting of the d 5.4 signal (inset) to help you.]
Problems
1H
NMR
6H
3H
5.4
6.0
5.5
5.0
(CH3)4Si
2H
5.3
6H
4.5
4.0
3.5
300-MHz
1H
3.0
2.5
2.0
1.5
1.0
0.5
0.0
NMR spectrum ppm (δ) δ
O
61.
Study of the solvolysis of derivatives of menthol [5-methyl-2-(1-methylethyl)cyclohexanol] has greatly enhanced our understanding of these types of reactions. Heating the isomer of the 4-methylbenzenesulfonate ester shown below in 2,2,2-trifluoro-ethanol (a highly ionizing solvent of low nucleophilicity) leads to two products with the molecular formula C10H18. (a) The major product displays 10 different signals in its 13C NMR spectrum. Two of them are at relatively low field, about d 5 120 and 145 ppm, respectively. The 1H NMR spectrum exhibits a multiplet near d 5 5 ppm (1 H); all other signals are upfield of d 5 3 ppm. Identify this compound. (b) The minor product gives only seven 13C signals. Again, two are at low field (d , 125 and 140 ppm), but, in contrast with the 1H NMR data on the major isomer, there are no signals at lower field than d 5 3 ppm. Identify this product and explain its formation mechanistically. (c) When the solvolysis is carried out starting with the ester labeled with deuterium at C2, the 1 H spectrum of the resulting major product isomer in (a) reveals a significant reduction of the intensity of the signal at d 5 5 ppm, a result indicating the partial incorporation of deuterium at the position associated with this peak. How might this result be explained? [Hint: The answer lies in the mechanism of formation of the minor product in (b).]
`
CH3 CF3CH2OH, ⌬ 2
-
OSO2 @( (D)H CH(CH3)2
two C10H18 products
CH3
Team Problem 62. Your team is faced with a puzzle. Four isomeric compounds, A – D, with the molecular formula C4H9BrO react with KOH to produce E – G with the molecular formula C4H8O. Molecules A and B yield compounds E and F, respectively. The NMR spectra of compounds C and D are identical, and both furnish the same product, G. Although some of the starting materials are optically active, none of the products is. Moreover, each of E, F, and G displays only two 1H NMR signals of varying chemical shifts, none of them located between d 4.6 and 5.7 ppm. Both the respective resonances of E and G are complex, whereas F exhibits two singlets. Proton-decoupled 13C NMR spectra for E and G show only two peaks, whereas F exhibits three. Using this spectral information, work together to determine which isomers of C4H9BrO will yield the respective isomers of C4H8O. When you have matched reactant and product, divide the task of predicting the proton and carbon NMR spectra of E, F, and G among yourselves. Estimate the 1H and 13C chemical shifts for all, and predict the respective DEPT spectra.
Chapter 10
443
Using NMR Spectroscopy to Deduce Structure
Preprofessional Problems 63. The molecule (CH3)4Si (tetramethylsilane) is used as an internal standard in 1H NMR spectroscopy. One of the following properties makes it especially useful. Which one? (a) Highly paramagnetic (d) Highly nucleophilic
(b) Highly colored
(c) Highly volatile
64. One of the following compounds will show a doublet as part of its 1H NMR spectrum. Which one? (b) ClCH(CH3)2
(a) CH4
(c) CH3CH2CH3
G
Chapter 10
(d) H2C GCH2 ACA Br Br G
444
65. In the 1H NMR spectrum of 1-fluorobutane, the most deshielded hydrogens are those bound to (a) C4
(b) C3
(c) C2
(d) C1
66. One of the following compounds will have one peak in its 1H NMR spectrum and two peaks in its 13C NMR spectrum. Which one? (a)
(b)
(c) CH3 G CH3
Cl G (d) CH3CHCHCH3 G Cl
(e)
O F
O F
C H A P T E R
ELEVEN
Alkenes; Infrared Spectroscopy and Mass Spectrometry
O O
W
hat differentiates solid shortening from liquid cooking oil? Remarkably, the only significant structural difference is that the liquid contains carbon – carbon double bonds and the solid does not. Cooking oils are derivatives of alkenes, the simplest organic compounds containing multiple bonds. In this chapter and in Chapter 12, we shall investigate the properties, generation, and reactivity of alkenes. In the preceding several chapters, we learned that haloalkanes and alcohols, two major classes of compounds containing single-bonded functional groups, may undergo elimination under appropriate conditions to form alkenes. In this chapter we return to these processes and explore some additional features that affect their outcome. We shall then proceed in Chapter 12 to examine the reactions of alkenes, and we shall discover that they may be converted back into single-bonded substances by the process of addition. Thus, we shall see how alkenes can serve as intermediaries in many synthetic conversions. They are useful and economically valuable starting materials for plastics, synthetic fibers, construction materials, and many other industrially important substances. For example, addition reactions of many gaseous alkenes give oils as products, which is why this class of compounds used to be called “olefins” (from oleum facere, Latin, to make oil). Indeed, “margarine” is a shortened version of the original name,
i f CPC f i Alkene double bond
cis-9-Octadecenoic acid, also known as oleic acid, makes up more than 80% of natural olive oil extracted from the fruit of the European olive tree. It is acknowledged to be one of the most beneficial of all the food-derived fats and oils for human cardiovascular health. In contrast, the isomeric compound in which the double bond possesses trans instead of cis geometry has been found to have numerous adverse health effects.
446
Chapter 11
Alkenes; Infrared Spectroscopy and Mass Spectrometry
oleomargarine, for this product.* Because alkenes can undergo addition reactions, they are described as unsaturated compounds. In contrast, alkanes, which possess the maximum number of single bonds and thus are inert with respect to addition, are referred to as saturated. We begin with the names and physical properties of the alkenes and show how we evaluate the relative stability of their isomers. A review of elimination reactions allows us to further our discussion of alkene preparation. We also introduce two additional methods for determining molecular structure: a second type of spectroscopy — infrared (IR) spectroscopy — and a technique for determining the elemental composition of a molecule — mass spectrometry (MS). These methods complement NMR by ascertaining directly the presence or absence of functional groups and their characteristic bonds (O – H, CP C, etc.), as well as their arrangement in the overall structure.
11-1 Naming the Alkenes A carbon – carbon double bond is the characteristic functional group of the alkenes. Their general formula is CnH2n, the same as that for the cycloalkanes. Like other organic compounds, some alkenes are still known by common names, in which the -ane ending of the corresponding alkane is replaced by -ylene. Substituent names are added as prefixes. Common Names of Typical Alkenes CH D 3 CH2 P C G H
CH2 PCH2
Cl
Cl G D C PC D G Cl H Cl
Cl
Cl
Ethylene
Propylene
Trichloroethylene
(Fruit-ripening hormone in plants)
(Raw material for plastics)
(Common cleaning solvent)
In IUPAC nomenclature, the simpler ending -ene is used instead of -ylene, as in ethene and propene. More complicated systems require adaptations and extensions of the rules for naming alkanes (Section 2-5). Rule 1. To name the stem, find the longest chain that includes both carbons making up the double bond. The molecule may have longer carbon chains, but ignore them. CH3 A CH2 PCHCHCH2CH3 A methylpentene
CH2CH2CH3 A CH2 PCHCH(CH2)4CH3
H3C CH2CH3 A A CH3CH2CH2CH2C PCCH2CH2CH2CH3
A propyloctene
An ethylmethyldecene
(Not a hexene or a nonane derivative)
(Not a pentene or a heptene or an octene derivative)
Rule 2. Indicate the location of the double bond in the main chain by number, starting at the end closer to the double bond. (Cycloalkenes do not require the numerical prefix, but the carbons making up the double bond are assigned the numbers 1 and 2, unless another group takes precedence; see rule 6.) Alkenes that have the same molecular formula but differ in the *The name margarine itself originates indirectly from the Greek, margaron, pearl, and directly from margaric acid, the common name given to one of the constituent fatty acids of margarine, heptadecanoic acid, because of the shiny, “pearly” crystals it forms.
11-1 Naming the Alkenes
location of the double bond (such as 1-butene and 2-butene) are constitutional isomers that are also called double-bond isomers. A 1-alkene is also referred to as a terminal alkene; the others are called internal. Note that alkenes are easily depicted in line notation. 1
4
2 1
2
3
4
1
2
3
4
1
CH3CH PCHCH3
CH2 PCHCH2CH3
6
2
5
3
5
3
1-Butene
2-Butene
2-Pentene
(A terminal alkene; not 3-butene)
(An internal alkene and a double-bond isomer of 1-butene)
(Not 3-pentene)
4
Cyclohexene
Rule 3. Add substituents and their positions to the alkene name as prefixes. If the alkene stem is symmetric, begin from the end that gives the first substituent along the chain the lowest possible number. 2
H3C CH3
1
3A
2
1
3
4
5
CH3
1
CH2 PCHCHCH2CH3
2A
3
4
5
6
CH3CHCH PCHCH2CH3
3-Methyl-1-pentene
3-Methylcyclohexene
2-Methyl-3-hexene
(Not 6-methylcyclohexene)
(Not 5-methyl-3-hexene)
Exercise 11-1 Name the following two alkenes. CH3 (a)
CH3
(b) Br
CH3
Rule 4. Identify any stereoisomers. In a 1,2-disubstituted ethene, the two substituents may be on the same side of the molecule or on opposite sides. The first stereochemical arrangement is called cis and the second trans, in analogy to the cis-trans names of the disubstituted cycloalkanes (Section 4-1). Two alkenes of the same molecular formula differing only in their stereochemistry are called geometric or cis-trans isomers and are examples of diastereomers: stereoisomers that are not mirror images of each other. Same side of double bond
Opposite sides of double bond
cis
H3 C
CH G D 3 C PC D G H H cis-2-Butene
trans
H G D C PC D G H CH3
Cl A CH3CH CH3 G D C PC D G H H
trans-2-Butene
cis-4-Chloro-2-pentene
H 3C
Exercise 11-2 Name the following three alkenes. Cl (a)
Cl G D C PC D G H H
(b)
(c) Br
Chapter 11
447
Alkenes; Infrared Spectroscopy and Mass Spectrometry
In the smaller substituted cycloalkenes, the double bond can exist only in the cis configuration. The trans arrangement is prohibitively strained (as building a model reveals). However, in larger cycloalkenes, trans isomers are stable. CH3
CH3 CH2CH3 F
H3C trans-Cyclodecene
1-Ethyl-2,4-dimethylcyclohexene
3-Fluoro-1-methylcyclopentene
(In both cases, only the cis isomer is stable)
Rule 5. The labels cis and trans cannot be applied when there are three or four different substituents attached to the double-bond carbons. An alternative system for naming such alkenes has been adopted by IUPAC: the E, Z system. In this convention, the sequence rules devised for establishing priority in R,S names (Section 5-3) are applied separately to the two groups on each double-bonded carbon. When the two groups of higher priority are on opposite sides, the molecule is of the E configuration (E from entgegen, German, opposite). When the two substituents of higher priority appear on the same side, the molecule is a Z isomer (Z from zusammen, German, together). Z
Br
F G GC P C 2
G
1
CH2CH2CH3 G GC P C
CH3CH2
Higher priority on C2
ClCH2CH2
H
F
(Z)-1-Bromo-1,2-difluoroethene
G
Higher priority on C1
E
G
Chapter 11
G
448
CH3
(E)-1-Chloro-3-ethyl-4-methyl-3-heptene
Exercise 11-3 Name the following three alkenes. (a)
OCH3 D G C PC D G H3C CH2CH3
F
D D D G C PC D G H3C H
(b)
(c)
Cl
Rule 6. With the alkenes, we are introducing a new functional group after the alcohols. This rule addresses the problem that arises when we have both functions in a compound: Do we call it an alkene or an alcohol? The answer is that we give the hydroxy functional group precedence over the double bond. Therefore, we name alcohols containing double bonds alkenols, and the stem incorporating both functions is numbered to give the carbon bearing the OH group the lowest possible assignment. Note that the last e in alkene is dropped in the naming of alkenols.
CH2 PCHCH2OH
OH A2 CH3CH A3 Cl CHCH2CH3 G5 4D CPC D G H3C H
2-Propen-1-ol
(Z)-5-Chloro-3-ethyl-4-hexen-2-ol
3-Cyclohexen-1-ol
(Not 1-propen-3-ol)
(The two stereocenters are unspecified)
(Not 1-cyclohexen-4-ol)
OH
1
3
2
1
1 2
3 4
Exercise 11-4 Draw the structures of the following molecules: (a) trans-3-penten-1-ol; (b) 2-cyclohexen-1-ol.
11-2 Structure and Bonding in Ethene: The Pi Bond
Chapter 11
449
Rule 7. Substituents containing a double bond are named alkenyl; for example, ethenyl (common name, vinyl), 2-propenyl (allyl), and cis-1-propenyl. H CH2 PCH O
CH2 PCH O CH2 O
Ethenyl
2-Propenyl
(Vinyl)
(Allyl)
H3C
D G CPC D G
H
cis-1-Propenyl
As usual, the numbering of a substituent chain begins at the point of attachment to the basic stem. The example below constitutes an alkenol. However, we cannot incorporate both functional groups into the stem name. Therefore, the double bond is part of the substituent to the cyclooctanol. 1
3 2
5
4
3 2 1
¥ OH
trans-3-(4-Pentenyl)cyclooctanol
Exercise 11-5 (a) Draw the structure of trans-2-ethenylcyclopropanol. (b) Name the structure shown in the margin.
11-2 Structure and Bonding in Ethene: The Pi Bond The carbon – carbon double bond in alkenes has special electronic and structural features. This section reviews the hybridization of the carbon atoms in this functional group, the nature of its two bonds (s and p), and their relative strengths. We consider ethene, the simplest of the alkenes.
The double bond consists of sigma and pi components Ethene is planar, with two trigonal carbon atoms and bond angles close to 1208 (Figure 11-1). Therefore, both carbon atoms are best described as being sp2 hybridized (Section 1-8; Figure 1-21). Two sp2 hybrids on each carbon atom overlap with hydrogen 1s orbitals to form the four C – H s bonds. The remaining sp2 orbitals, one on each carbon, overlap with each other to form the carbon – carbon s bond. Each carbon also possesses a 2p orbital; these are aligned parallel to each other and are close enough to overlap, forming the P bond (Figure 11-2A). The electron density in a p bond is distributed over both carbons above and below the molecular plane, as indicated in Figure 11-2B. π bond
C
A
H C
H
120°
H
H
2p orbital
π
σ bond
H
H C
H
sp2 orbital
B
C
π
H
H 121.7⬚ H G D 116.6⬚ C P C D G H H
1.076 Å
1.330 Å
Figure 11-1 Molecular structure of ethene.
Figure 11-2 An orbital picture of the double bond in ethene. The s carbon – carbon bond is made by sp2 – sp2 overlap. The pair of p orbitals perpendicular to the ethene molecular plane overlap to form the additional p bond. For clarity, this overlap is indicated in (A) by the dashed green lines; the orbital lobes are shown artificially separated. Another way of presenting the p bond is depicted in (B), in which the “p-electron cloud” is above and below the molecular plane.
450
Chapter 11
Alkenes; Infrared Spectroscopy and Mass Spectrometry
The pi bond in ethene is relatively weak How much do the s and p bonds each contribute to the total double-bond strength? We know from Section 1-7 that bonds are made by overlap of orbitals and that their relative strengths depend on the effectiveness of this overlap. Therefore, we can expect overlap in a s bond to be considerably better than that in a p bond, because the sp2 orbitals lie along the internuclear axis (Figure 11-2). This situation is illustrated in energy-levelinteraction diagrams (Figures 11-3 and 11-4) analogous to those used to describe the bonding in the hydrogen molecule (Figures 1-11 and 1-12). Figure 11-5 summarizes our predictions of the relative energies of the molecular orbitals that make up the double bond in ethene.
Figure 11-3 Overlap between two sp2 hybrid orbitals (containing one electron each; shown in red) determines the relative strength of the s bond of ethene. In-phase interaction between regions of the wave function having the same sign reinforces bonding (compare inphase overlap of waves, Figure 1-4B) and creates a bonding molecular orbital. [Recall: These signs do not refer to charges; the 1 designations are chosen arbitrarily (see Figure 1-11).] Both electrons end up occupying this orbital and have a high probability of being located near the internuclear axis. The orbital stabilization energy, DEs, corresponds to the s-bond strength. The out-of-phase interaction, between regions of opposite sign (compare Figure 1-4C), results in an unfilled antibonding molecular orbital (designated s*) with a node.
Out-of-phase overlap Node H + H
C −
+ C
H − H
Negative overlap σ *
Antibonding molecular orbital of the σ bond: empty
H E
H
C
C
H
H ΔEσ
sp2
sp2 Bonding molecular orbital of the σ bond: filled
σ -Bond energy
Bonding electrons H
− C
+
+
H
C −
H H
Positive overlap σ
In-phase overlap
Thermal isomerization allows us to measure the strength of the pi bond How do these predictions of the p-bond strength compare with experimental values? We can measure the energy required to interconvert the cis form of a substituted alkene — say, 1,2-dideuterioethene — with its trans isomer. In this process, called thermal isomerization,
11-2 Structure and Bonding in Ethene: The Pi Bond
Chapter 11
Figure 11-4 Compare this picture of the formation of the p bond in ethene with Figure 11-3. Inphase interaction between two parallel p orbitals (containing one electron each; shown in blue) results in positive overlap and a filled bonding p orbital. The representation of this orbital indicates the probability of finding the electrons between the carbons above and below the molecular plane. Because p overlap is less effective than s, the stabilization energy, DEp, is smaller than DEs. The p bond is therefore weaker than the s bond. The out-of-phase interaction results in the antibonding molecular orbital p*.
Out-of-phase overlap Node + H H
− C
H
C
−
+
H
Negative overlap π *
Antibonding molecular orbital of the π bond: empty
H
E
C
C
H
H H
ΔEπσ
Bonding molecular orbital of the π bond: filled
p
π -Bond energy
451
p
Bonding electrons
H H
+
+
C
C
−
−
H H
* orbital
Positive overlap π
* orbital E
orbital
In-phase overlap
orbital
the two p orbitals making up the p bond are rotated 1808. At the midpoint of this rotation— 908 — the p (but not the s) bond has been broken (Figure 11-6). Thus, the activation energy for the reaction can be roughly equated with the p energy of the double bond. Thermal isomerization occurs, but only at high temperatures (.4008C). Its activation energy is 65 kcal mol21 (272 kJ mol21), a value we may assign to the strength of the p bond. Below 3008C, double bonds are configurationally stable; that is, cis stays cis and trans stays
Antibonding orbitals: *, * Bonding orbitals: ,
Figure 11-5 Energy ordering of the molecular orbitals making up the double bond. The four electrons occupy only bonding orbitals.
Broken πn bond
π bond
π bond
H H D
A
H C
C
H
H
Δ
C
D
D
B
‡ C
C
C
D H
D D C
Figure 11-6 Thermal isomerization of cis-dideuterioethene to the trans isomer requires breaking the p bond. The reaction proceeds from starting material (A) through rotation around the C – C bond until it reaches the point of highest energy, the transition state (B). At this stage, the two p orbitals used to construct the p bond are perpendicular to each other. Further rotation in the same direction results in a product in which the two deuterium atoms are trans (C).
Remember:
The symbol ‡ in Figure 11-6 denotes a transition state.
452
Alkenes; Infrared Spectroscopy and Mass Spectrometry
Chapter 11
111
65 π R
108 σ
C R′
H C CH3
100
Figure 11-7 Approximate bond strengths in an alkene (in kcal mol21). Note the relative weakness of the p bond.
trans. The strength of the double bond in ethene—the energy required for dissociation into two CH2 fragments—is 173 kcal mol21 (724 kJ mol21). Consequently the C–C s bond amounts to about 108 kcal mol21 (452 kJ mol21, Figure 11-7). Note that the other s bonds to the alkenyl carbon are also stronger than the analogous bonds in alkanes (Table 3-2). This effect is due largely to improved bonding overlap involving the relatively compact sp2 orbitals. As one consequence, the strongly bound alkenyl hydrogens are not abstracted in radical reactions. Instead, additions to the weaker p bond characterize the reactivity of alkenes (Chapter 12).
In Summary The characteristic hybridization scheme for the double bond of an alkene accounts for its physical and electronic features. Alkenes contain a planar double bond, incorporating trigonal carbon atoms. Their hybridization explains the strong s and weaker p bonds, stable cis and trans isomers, and the strength of the alkenyl-substituent bond. Alkenes are prone to undergo addition reactions, in which the weaker p bond but not the C – C s bond is broken.
11-3 Physical Properties of Alkenes
Table 11-1 Comparison of Melting Points of Alkenes and Alkanes Compound
Melting point (8C)
Butane trans-2-Butene cis-2-Butene
2138 2106 2139
Pentane trans-2-Pentene cis-2-Pentene
2130 2135 2180
Hexane trans-2-Hexene cis-2-Hexene trans-3-Hexene cis-3-Hexene
295 2133 2141 2115 2138
The carbon – carbon double bond alters many of the physical properties of alkenes relative to those of alkanes. Exceptions are boiling points, which are very similar, primarily a result of similar London forces (Figure 2-6C). Like their alkane counterparts, ethene, propene, and the butenes are gases at room temperature. Melting points, however, depend in part on the packing of molecules in the crystal lattice, a function of molecular shape. The double bond in cis-disubstituted alkenes imposes a U-shaped bend in the molecule that disrupts packing and reduces the melting point, usually below that of either the corresponding alkane or isomeric trans alkene (Table 11-1). A cis double bond is responsible for the sub-roomtemperature melting point of vegetable oil. Depending on their structure, alkenes may exhibit weak dipolar character. Why? Bonds between alkyl groups and an alkenyl carbon are polarized in the direction of the sp2-hybridized atom, because the degree of s character in an sp2 hybrid orbital is greater than that in an sp3. Electrons in orbitals with increased s character are held closer to the nucleus than those in orbitals containing more p character. This effect makes the sp2 carbon relatively electron withdrawing (although much less so than strongly electronegative atoms such as O and Cl) and creates a weak dipole along the substituent weak – alkenyl carbon bond. In cis-disubstituted alkenes, the two individual dipoles combine to give a net molecular dipole. These dipoles are opposed in trans-disubstituted alkenes and tend to cancel each other. The more polar cis-disubstituted alkenes often have slightly higher boiling points than their trans counterparts. The boiling-point difference is greater when the individual bond dipoles are larger, as in the case of the two 1,2-dichloroethene isomers, whose highly electronegative chlorine atoms also cause the direction of the bond dipoles to be reversed.
Polarization in Alkenes: Csp2 Pulls Electrons from Csp3 H3C Net dipole
CH D 3 G C PC D G H H b.p. 4ⴗC
H3C
H D G C PC D G H CH3 b.p. 1ⴗC
Cl No net dipole
Net dipole
Cl D G C PC D G H H b.p. 60ⴗC
Cl
H D G C PC D G Cl H
No net dipole
b.p. 48ⴗC
Another consequence of the electron-attracting character of the sp2 carbon is the increased acidity of the alkenyl hydrogen. Whereas ethane has an approximate pKa of 50,
11-4 Nuclear Magnetic Resonance of Alkenes
ethene is somewhat more acidic with a pKa of 44. Even so, ethene is a very weak acid compared with other compounds, such as the carboxylic acids or alcohols.
Acidity of the Ethenyl Hydrogen
CH2 P C
K 10⫺50
H
O
Relatively more acidic
O
CH3 O CH2 OH
H
K 10⫺44
š 2⫺ ⫹ CH3 O CH
H⫹
Ethyl anion
š ⫺ CH2 P CH
⫹
H⫹
Ethenyl (vinyl) anion
Exercise 11-6 Ethenyllithium (vinyllithium) is not generally prepared by direct deprotonation of ethene but rather from chloroethene (vinyl chloride) by metallation (Section 8-7). CH2 P CHCl
⫹
2 Li
(CH3CH2)2O
CH2 P CHLi 60%
⫹
LiCl
Upon treatment of ethenyllithium with acetone followed by aqueous work-up, a colorless liquid is obtained in 74% yield. Propose a structure.
In Summary The presence of the double bond does not greatly affect the boiling points of alkenes, compared with alkanes, but cis-disubstituted alkenes usually have lower melting points than their trans isomers, because the cis compounds pack less well in the solid state. Alkenyl hydrogens are more acidic than those in alkanes because of the electron-withdrawing character of the sp2-hybridized alkenyl carbon.
11-4 Nuclear Magnetic Resonance of Alkenes The double bond exerts characteristic effects on the 1H and 13C chemical shifts of alkenes (see Tables 10-2 and 10-6). Let’s see how to make use of this information in structural assignments.
The pi electrons exert a deshielding effect on alkenyl hydrogens Figure 11-8 shows the 1H NMR spectrum of trans-2,2,5,5-tetramethyl-3-hexene. Only two signals are observed, one for the 18 equivalent methyl hydrogens and one for the 2 alkenyl protons. The absorptions appear as singlets because the methyl hydrogens are too far away from the alkenyl hydrogens to produce detectable coupling. The low-field resonance of the alkenyl hydrogens (d 5 5.30 ppm) is typical of hydrogen atoms bound to alkenyl carbons. Terminal alkenyl hydrogens (RR9C P CH2) resonate at d 5 4.6 – 5.0 ppm, their internal counterparts (RCH P CHR9) at d 5 5.2 – 5.7 ppm. Why is deshielding so pronounced for alkenyl hydrogens? Although the electronwithdrawing character of the sp2-hybridized carbon is partly responsible, another phenomenon is more important: the movement of the electrons in the p bond. When subjected to an external magnetic field perpendicular to the double-bond axis, these electrons enter into a circular motion. This motion induces a local magnetic field that reinforces the external
Chapter 11
453
454
Chapter 11
Figure 11-8 300-MHz 1 H NMR spectrum of trans2,2,5,5-tetramethyl-3-hexene, illustrating the deshielding effect of the p bond in alkenes. It reveals two sharp singlets for two sets of hydrogens: the 18 methyl hydrogens at d 5 0.97 ppm and 2 highly deshielded alkenyl protons at d 5 5.30 ppm.
Alkenes; Infrared Spectroscopy and Mass Spectrometry
1H NMR
CH3 H3C
H
C C
CH3
C
18 H
H 3C
C
H CH3
H3C
(CH3)4Si Deshielded 2H
9
8
7
6
5
4
3
2
1
0
ppm (δ )
Figure 11-9 Movement of electrons in the p bond causes pronounced deshielding of alkenyl hydrogens. An external field, H0, induces a circular motion of the p electrons (shown in red) above and below the plane of a double bond. This motion in turn induces a local magnetic field (shown in green) that opposes H0 at the center of the double bond but reinforces it in the regions occupied by the alkenyl hydrogens.
Local magnetic field
Opposes H0 in this region of space
hlocal
H
H hlocal
C
C
Strengthens H0 in this region of space
H
H
-electron movement
hlocal
hlocal
External field, H0
Opposes H0 in this region of space
field at the edge of the double bond (Figure 11-9). As a consequence, the alkenyl hydrogens are strongly deshielded (Section 10-4). Exercise 11-7 The hydrogens on methyl groups attached to alkenyl carbons resonate at about d 5 1.6 ppm (see Table 10-2). Explain the deshielding of these hydrogens relative to hydrogens on methyl groups in alkanes. (Hint: Try to apply the principles in Figure 11-9.)
Cis coupling through a double bond is different from trans coupling When a double bond is not substituted symmetrically, the alkenyl hydrogens are nonequivalent, a situation leading to observable spin – spin coupling such as that shown in the spectra of cis- and trans-3-chloropropenoic acid (Figure 11-10). Note that the coupling constant
11-4 Nuclear Magnetic Resonance of Alkenes
1H NMR
1H
H
H C
10
11
1H1H
6.9 6.8 ppm
8
7
6
6.3 6.2 ppm
5
OH
O
J = 9 Hz
J = 9 Hz
9
C
Cl
ppm
Jcis is relatively small
C
(CH3)4Si
4
3
2
1
0
ppm (δ )
A
1H NMR
1H H
Cl 11
C
10 ppm
Jtrans is relatively large
C C
H
J = 14 Hz
J = 14 Hz
7.6 7.5 7.4 ppm
6.3 6.2 ppm
OH
O
(CH3)4Si
1H 1H 9
B
8
7
6
5
4
3
2
1
0
ppm (δ )
for the hydrogens situated cis (J 5 9 Hz) is different from that for the hydrogens arranged trans (J 5 14 Hz). Table 11-2 gives the magnitudes of the various possible couplings around a double bond. Although the range of Jcis overlaps that of Jtrans, within a set of isomers Jcis is always smaller than Jtrans. In this way cis and trans isomers can be readily distinguished. Coupling between hydrogens on adjacent carbon atoms, such as Jcis and Jtrans, is called vicinal. Coupling between nonequivalent hydrogens on the same carbon atom is referred to as geminal. In alkenes, geminal coupling is usually small (Table 11-2). Coupling to neighboring alkyl hydrogens (allylic, see Section 11-1) and across the double bond (1,4- or long-range) also is possible, sometimes giving rise to complicated spectral patterns. Thus,
Chapter 11
455
Figure 11-10 300-MHz 1 H NMR spectra of (A) cis-3chloropropenoic acid and (B) the corresponding trans isomer. The two alkenyl hydrogens are nonequivalent and coupled. The broad carboxylic acid proton ( – CO2H) resonates at d 5 10.80 ppm and is shown in the inset.
456
Chapter 11
Alkenes; Infrared Spectroscopy and Mass Spectrometry
Table 11-2
Coupling Constants Around a Double Bond J (Hz)
Type of coupling H
H
G D C PC D G
H
G D C PC D G
H
G D C PC D G
Name
Typical
Vicinal, cis
6 – 14
10
Vicinal, trans
11 – 18
16
Geminal
0–3
2
None
4 – 10
6
Allylic, (1,3)-cis or -trans
0.5 – 3.0
2
(1,4)- or long-range
0.0 – 1.6
1
H H
D GC G GH G C PC D G H H
Range
A A G C P C O CO H D A
H H A A O COC PC OCO A A A A
the simple rule devised for saturated systems, discounting coupling between hydrogens farther than two intervening atoms apart, does not hold for alkenes.
Further coupling leads to more complex spectra The spectra of 3,3-dimethyl-1-butene and 1-pentene illustrate the potential complexity of the coupling patterns. In both spectra, the alkenyl hydrogens appear as complex multiplets. In 3,3-dimethyl-1-butene (Figure 11-11A), Ha, located on the more highly substituted carbon atom, resonates at lower field (d 5 5.86 ppm) and in the form of a doublet of doublets with two relatively large coupling constants (trans Jab 5 18 Hz, cis Jac 5 10.5 Hz). Hydrogens Hb and Hc also absorb as a doublet of doublets each because of their respective coupling to Ha and their small mutual coupling (geminal Jbc 5 1.5 Hz). In the spectrum of 1-pentene (Figure 11-11B), additional coupling due to the attached alkyl group (see Table 11-2) creates a relatively complex pattern for the alkenyl hydrogens, although the two sets (terminal and internal) are clearly differentiated. In addition, the electron-withdrawing effect of the sp2 carbon and the movement of the p electrons (Figure 11-9) cause a slight deshielding of the directly attached (allylic) CH2 group. The magnitude of the coupling between these hydrogens and the neighboring alkenyl hydrogen is about the same (6–7 Hz) as the coupling with the two CH2 hydrogens on the other side. As a result, the multiplet for this allylic CH2 group appears as a quartet (with additional long-range couplings to the terminal alkenyl hydrogens), in accordance with the simple N 1 1 rule: N 5 (2 H from CH2) 1 D 1 H from PC ⫽ 3. GH
11-4 Nuclear Magnetic Resonance of Alkenes
1H NMR
CH3 C C
Ha Jab (trans) ⫽ 18 Hz
CH3 Hb Hc
Ha
Jab Jac
Jac (cis) ⫽ 10.5 Hz
9H
C
Hc
Jab
Jac
Jbc
Jbc
Jbc (geminal) ⫽ 1.5 Hz
(CH3)4Si 5.9 5.8 ppm
9
4.9 4.8 ppm
1 H 1 H1 H
8
7
6
5
4
3
2
1
0
ppm (δ )
A
1H NMR
CH2CH2CH3
H C
C
H
H
1.5 1.4 ppm 3H
Alkenyl hydrogens show non-first-order multiplets
1.0 0.9 ppm
5.0 4.9 ppm 5.9 5.8 ppm
(CH3)4Si
2.1 2.0 ppm 2H
2H
1H
9
B
8
7
6
5
4
457
Figure 11-11 300-MHz 1H NMR spectra of (A) 3,3-dimethyl-1-butene and (B) 1-pentene.
H3C Hb
Chapter 11
3
2
2H
1
0
ppm (δ )
Exercise 11-8
Working with the Concepts: Interpreting NMR Spectra of Alkenes Ethyl 2-butenoate (ethyl crotonate), CH3CH P CHCO2CH2CH3, in CCl4 has the following 1H NMR spectrum: d 5 6.95 (dq, J 5 16, 6.8 Hz, 1 H), 5.81 (dq, J 5 16, 1.7 Hz, 1 H), 4.13 (q, J 5 7 Hz, 2 H), 1.88 (dd, J 5 6.8, 1.7 Hz, 3 H), and 1.24 (t, J 5 7 Hz, 3 H) ppm; dd denotes a doublet of doublets, dq a doublet of quartets. Assign the various hydrogens and indicate whether the double bond is substituted cis or trans (consult Table 11-2). Strategy You are given both the compound’s structure and the NMR data. Therefore, this problem consists of deciphering the information. The 1H NMR spectrum shows five signals, one for each of the five
458
Chapter 11
Table 11-3 Comparison of 13C NMR Absorptions of Alkenes with the Corresponding Alkane Carbon Chemical Shifts (in ppm) H3C H3 C
G 122.8 D C PC D G
Alkenes; Infrared Spectroscopy and Mass Spectrometry
distinct hydrogen environments in the molecule. Each signal has an integration value telling us the number of hydrogens giving rise to that signal, and that is where we start. Then we analyze the location and splitting (if any) of each signal in turn. Solution • Beginning with the highest-field (smallest d ppm value) signal, at d 5 1.24 ppm, it integrates as 3 H; therefore, it must arise from one of the two CH3 groups in the molecule. Looking further, we see that it is split into a triplet, three lines. According to the N 1 1 rule for spin – spin splitting, this tells us that the CH3 group giving rise to this signal is adjacent to a carbon bearing two hydrogens, the CH2 group in the structure (2 neighboring hydrogens 1 1 5 3 lines). We can find the signal for this CH2 at d 5 4.13; as expected, it is a quartet (3 neighboring methyl hydrogens 1 1 5 4 lines). By a process of elimination, the CH3 group on the alkene carbon must be responsible for the signal at d 5 1.88. This signal is split as a doublet of doublets, indicative of splitting to two distinct single hydrogens: the two different alkene hydrogens. • We now turn to these final two hydrogens, which give signals at d 5 5.81 and 6.95 ppm, respectively. Both are doublets of quartets — in other words, eight-line patterns that appear as pairs of quartets. What does this imply? Each is evidently split into a quartet by the CH3 on the alkene carbon (3 1 1 5 4). Also, because the two alkene hydrogens are neighbors and in dissimilar environments, they split each other, causing the doublet splitting of each pattern (1 neighbor 1 1 5 2 lines). Which is which? The alkene hydrogen at d 5 6.95 ppm shows the larger quartet splitting, 6.8 Hz, suggesting that it is the one adjacent to the CH3 group. The other alkene hydrogen, at d 5 5.81 ppm, shows a much smaller quartet splitting of 1.7 Hz, consistent with its increased distance to the methyl. Notice that the order in which the information is presented allows us to assign each coupling constant to each splitting: given “dq, J 5 16, 6.8 Hz,” we infer that the first splitting (d) corresponds to the first J value (16 Hz), and the second (q) to the second J (6.8 Hz). • Finally, with J 5 16 Hz for the mutual splitting of the alkene hydrogens, we conclude from the data in Table 11-2 that they are trans to each other. In problems such as this, it is often instructive to reproduce all the information pictorially, as shown below. Notice that this representation is especially useful in revealing the mutual couplings by their identical J values, for example, 6.8 Hz for both the methyl hydrogens and the alkenyl hydrogen on the left. Identify the others on your own. ␦ 1.88 (dd, J 6.8, 1.7 Hz, 3 H) ␦ 6.95 (dq, J 16, 6.8 Hz, 1 H)
CH3
H ␦ 5.81 (dq, J 16, 1.7 Hz, 1 H) G D CP C ␦ 4.13 (q, J 7 Hz, 2 H) D G H CO2CH2CH3 ␦ 1.24 (t, J 7 Hz, 3 H)
H 3C
CH3
18.9
H G D 132.7 C PC D G H3C CH2CH3 H
123.7
12.3 20.5
14.0
Alkenes
H3 C H3C
G 34.0 D CHOCH D G
CH3
Try It Yourself O ‘ Ethenyl acetate, CH3 C OCH“CH2, displays the following 1H NMR data: d 5 7.23 (dd, J 5 14.4, 6.8 Hz, 1 H), 4.73 (dd, J 5 14.4, 1.6 Hz, 1 H), 4.52 (dd, J 5 6.8, 1.6 Hz, 1 H), 2.10 (s, 3 H). Interpret this spectrum.
CH3
19.2
Alkenyl carbons are deshielded in 13C NMR
22.2
CH3CH2CH2CH2CH3 13.5
Exercise 11-9
34.1
Alkanes
The carbon NMR absorptions of the alkenes also are highly revealing. Relative to alkanes, the corresponding alkenyl carbons (with similar substituents) absorb at about 100 ppm lower field (see Table 10-6). Two examples are shown in Table 11-3, in which the carbon chemical shifts of an alkene are compared with those of its saturated counterpart. Recall that, in broad-band decoupled 13C NMR spectroscopy, all magnetically unique carbons absorb as sharp single lines (Section 10-9). It is therefore very easy to determine the presence of sp2 carbons by this method.
11-5 Catalytic Hydrogenation: Stability of Double Bonds
Chapter 11
C H E MI C AL H I G H L I G H T 1 1 - 1 Prostaglandins
O
CO2H
^
@
HO
#
HO PGE1
O
CO2H
^
@
HO
#
HO PGE2
HO
Δ
HO
^
^
CO2H #
HO PGF2␣
O
CO2CH3
@
HO
HO CH3 ∑/
#
NMR analysis is widely used to determine the structures of complex molecules containing multiple functional groups. The first three compounds shown are members of the prostaglandin (PG) family of naturally occurring, biologically active substances. The 1H NMR spectra of these PGs reveal some aspects of their structures, but overall they are quite complicated, with many overlapping signals. In contrast, 13 C NMR permits rapid distinction between PG derivatives, merely by counting the peaks in three chemical-shift ranges. For example, PGE2 is readily distinguished by the presence of two signals near d 5 70 ppm for two hydroxy carbons, four alkene 13C resonances between d 5 125 and 140 ppm, and two carbonyls above d 5 170 ppm. Prostaglandins are extremely potent hormonelike substances with many biological functions, including muscle stimulation, inhibition of platelet aggregation, lowering of blood pressure, enhancement of inflammatory reactions, and induction of labor in childbirth. Indeed, the anti-inflammatory effects of aspirin (see Chemical Highlight 22-2) are due to its ability to suppress prostaglandin biosynthesis. An undesirable side effect of aspirin use is gastric ulceration, because some prostaglandins function to protect the stomach lining. The synthetic prostaglandin-like substance misoprostol exhibits a similar protective effect and is frequently administered together with aspirin or other anti-inflammatory agents to prevent ulcer formation.
Misoprostol
In Summary Hydrogen NMR is highly effective in establishing the presence of double bonds in organic molecules. Alkenyl hydrogens and carbons are strongly deshielded. The order of coupling is Jgem , Jcis , Jtrans. Coupling constants for allylic substituents display characteristic values as well. 13C NMR identifies alkenyl carbons by their unusually low-field chemical shifts, compared with those of alkane carbons.
11-5 Catalytic Hydrogenation of Alkenes: Relative Stability of Double Bonds When an alkene and hydrogen gas are mixed in the presence of catalysts such as palladium or platinum, two hydrogen atoms add to the double bond to give the saturated alkane (see Section 12-2). This reaction, which is called hydrogenation, is very exothermic. The heat released, called heat of hydrogenation, is typically about 230 kcal mol21 (2125 kJ mol21) per double bond. Hydrogenation of an Alkene D G C PC D G
ΔH° ⴚ30 kcal molⴚ1
HOH
Pd or Pt
A A OCOCO A A H H
459
460
Alkenes; Infrared Spectroscopy and Mass Spectrometry
Chapter 11
Heats of hydrogenation can be measured so accurately that they may be used to determine the relative energy contents, and therefore the thermodynamic stabilities, of alkenes. Let us see how this is done.
The heat of hydrogenation is a measure of stability
The fat molecules in butter and hard (stick) margarines are highly saturated, whereas those in vegetable oils have a high proportion of cis-alkene functions. Partial hydrogenation of these oils yields soft (tub) margarine.
In Section 3-10 we presented one method for determining relative stability: measuring heat of combustion. The less stable the molecule, the greater its energy content, and the more energy is released in this process. A very similar connection can be established using heats of hydrogenation. For example, what are the relative stabilities of the three isomers 1-butene, cis-2-butene, and trans-2-butene? Hydrogenation of each isomer leads to the same product: butane. If their respective energy contents are equal, their heats of hydrogenation should also be equal; however, as the reactions in Figure 11-12 illustrate, they are not. The most heat is evolved by hydrogenation of the terminal double bond, the next most exothermic reaction is that with cis-2-butene, and finally the trans isomer gives off the least heat. Therefore, the thermodynamic stability of the butenes increases in the order 1-butene , cis-2-butene , trans2-butene (Figure 11-12).
Highly substituted alkenes are most stable; trans isomers are more stable than cis isomers
CH3CH2CH
1-Butene
H2
cis-2-Butene
H2
trans-2-Butene
H2
Pt
Pt
Pt
Butane Butane
ΔH° 30.3 kcal mol1 (126.8 kJ mol1) ΔH° 28.6 kcal mol1 (119.7 kJ mol1) ΔH° 27.6 kcal mol1 (115.5 kJ mol1)
CH2
1-Butene
ΔE = 1.7 kcal mol−1
H3C
CH3 C
C
H
H3C
cis-2-Butene
+ H2
CH3
H
C
H
E
Butane
Increasing heat of hydrogenation
The results of the preceding hydrogenation reactions may be generalized: The relative stability of the alkenes increases with increasing substitution, and trans isomers are usually more stable than their cis counterparts. The first trend is due in part to hyperconjugation. Just as the stability of a radical increases with increasing alkyl substitution (Section 3-2),
C H
trans-2-Butene
+ H2 −30.3 kcal mol−1
ΔE = 1.0 kcal mol−1
+ H2 −28.6 kcal mol−1
−27.6 kcal mol−1
CH3CH2CH2CH3 Butane
Figure 11-12 The relative energy contents of the butene isomers, as measured by their heats of hydrogenation, tell us their relative stabilities. The diagram is not drawn to scale.
11-5 Catalytic Hydrogenation: Stability of Double Bonds
H C
C R
R
C
R
A
H
B
the p orbitals of a p bond can be stabilized by alkyl substituents. The second finding is easily understood by looking at molecular models. In cis-disubstituted alkenes, the substituent groups frequently crowd each other. Relative Stabilities of the Alkenes R CH2 P CH2
RCH P CH2
H
D G CPC D G (cis)
Least stable
R H
R D G CPC D G H R
H
R D G CPC D G H R
R
R D G CPC D G R R
R
(trans)
Increasing alkene stability
Most stable
Decreasing heat of hydrogenation
This steric interference is energetically unfavorable and absent in the corresponding trans isomers (Figure 11-13). Exercise 11-10
MODEL BUILDING
Rank the following alkenes in order of stability of the double bond to hydrogenation (order of DH 8 of hydrogenation): 2,3-dimethyl-2-butene, cis-3-hexene, trans-4-octene, and 1-hexene.
Cycloalkenes are exceptions to the generalization that trans alkenes are more stable than their cis isomers. In the medium-ring and smaller members of this class of compounds (Section 4-2), the trans isomers are much more strained (Section 11-1). The smallest isolated simple trans cycloalkene is trans-cyclooctene. It is 9.2 kcal mol21 (38.5 kJ mol21) less stable than the cis isomer and has a highly twisted structure. Exercise 11-11 Alkene A hydrogenates to compound B with an estimated release of 65 kcal mol21, more than double the values of the hydrogenations shown in Figure 11-12. Explain. H H H 65 kcal mol1
H 2, catalyst
A
461
Figure 11-13 (A) Steric congestion in cis-disubstituted alkenes and (B) its absence in trans alkenes explain the greater stability of the trans isomers.
R
H
H C
Chapter 11
B
In Summary The relative energies of isomeric alkenes can be estimated by measuring their heats of hydrogenation. The more energetic alkene has a higher DH8 of hydrogenation. Stability increases with increasing substitution because of hyperconjugation. Trans alkenes are more stable than their cis isomers because of steric hindrance. Exceptions are the smalland medium-ring cycloalkenes, in which cis substitution is more stable than trans substitution because of ring strain.
Molecular model of trans-Cyclooctene
462
Chapter 11
Alkenes; Infrared Spectroscopy and Mass Spectrometry
11-6 Preparation of Alkenes from Haloalkanes and Alkyl Sulfonates: Bimolecular Elimination Revisited With the physical aspects of alkene structure and stability as a background, let us now return to the various ways in which alkenes can be made. The most general approach is by elimination, in which two adjacent groups on a carbon framework are removed. The E2 reaction (Section 7-7) is the most common laboratory source of alkenes. Another method of alkene synthesis, the dehydration of alcohols, is described in Section 11-7.
General Elimination A A OCOCO A A A B
G G CPC D D
AB
Regioselectivity in E2 reactions depends on the base In Chapter 7 we discussed how haloalkanes (or alkyl sulfonates) in the presence of strong base can undergo elimination of the elements of HX with simultaneous formation of a carbon – carbon double bond. With many substrates, removal of a hydrogen can take place from more than one carbon atom in a molecule, giving rise to constitutional (double-bond) isomers. In such cases, can we control which hydrogen is removed — that is, the regioselectivity of the reaction (Section 9-9)? The answer is yes, to a limited extent. A simple example is the elimination of hydrogen bromide from 2-bromo-2-methylbutane. Reaction with sodium ethoxide in hot ethanol furnishes mainly 2-methyl-2-butene, but also some 2-methyl-1-butene. REACTION
E2 Reaction of 2-Bromo-2-methylbutane with Ethoxide Two types of hydrogens Primary Secondary
Reminder: “–HBr” under the
reaction arrow indicates the species removed from the starting material in the elimination reaction.
MECHANISM
CH3 A CH3CH2 O C O CH3 A ðBr ð 2-Bromo2-methylbutane
CH3CH2O Na , CH3CH2OH, 70°C HBr
H3C
CH G G 3 CP C D D CH3 H 70%
CH3CH2 G CP CH2 D H3C 30%
2-Methyl2-butene
2-Methyl1-butene
(More stable product)
(Less stable product)
In our example, the major product contains a trisubstituted double bond, so it is thermodynamically more stable than the minor product. Indeed, many eliminations are regioselective in this way, with the more stable product predominating. This result can be explained by analysis of the transition state of the reaction (Figure 11-14). Elimination of HBr proceeds through attack by the base on one of the neighboring hydrogens situated anti to the leaving group. In the transition state, there is partial C – H bond rupture, partial C – C double-bond formation, and partial cleavage at C – Br (compare Figure 7-8). The transition state leading to 2-methyl-2-butene is slightly more stabilized than that generating 2-methyl-1-butene (Figure 11-15A). The more stable product is formed faster because the structure of the transition state of the reaction resembles that of the products. Elimination reactions of this type that lead to the more highly substituted alkene are said to follow the Saytzev* rule: The
*Alexander M. Saytzev (also spelled Zaitsev or Saytzeff; 1841 – 1910), Russian chemist.
11-6 Preparation of Alkenes: Bimolecular Elimination
Baseδ − Baseδ −
H3C H
Baseδ − Baseδ −
H
H C
C
CH3 CH3
H3C
Br δ −
H3C
H
Newman projection
H
H
CH3
Br δ −
Partial double-bond character leading to trisubstituted double bond
Chapter 11
H H
C
C
H3C
CH3 CH2CH3 Br δ −
CH2CH3
H
H Br δ −
Partial double-bond character leading to terminal double bond
Newman projection
B
A
Figure 11-14 The two transition states leading to products in the dehydrobromination of 2-bromo-2-methylbutane. With unhindered base (CH3CH2O2Na1), transition state A is preferred over transition state B because there are more substituents around the partial double bond (Saytzev rule). With hindered base [(CH3)3CO2K1], transition state B is preferred over transition state A because there is less steric hindrance to abstraction of the primary hydrogens (Hofmann rule).
Use of a bulky base raises the energy of the transition state that would lead to the more stable, internal alkene product
E
+ Br NaOCH2CH3
With bulky base, lowerenergy transition state now Br + favors formation of less KOC(CH3)3 stable product
E
Lower-energy transition state favors formation of more stable product
Reaction coordinate
Reaction coordinate B
A
Figure 11-15 Potential energy diagrams for E2 reactions of 2-bromo-2-methylbutane with (A) sodium ethoxide (Saytzev rule) and (B) potassium tert-butoxide (Hofmann rule).
double bond forms preferentially between the carbon that contained the leaving group and the most highly substituted adjacent carbon that bears a hydrogen. A different product distribution is obtained when a more hindered base is used; more of the thermodynamically less favored terminal alkene is generated. E2 Reaction of 2-Bromo-2-methylbutane with tert-Butoxide, a Hindered Base More accessible primary hydrogens
(CH3)3COK, (CH3)3COH HBr
CH3 H3C G G GC P C G
CH3 A CH3CH2 O C O CH3 A ðBr ð
H
CH3
27%
CH3CH2 G GC P CH2 H3C 73%
REACTION
463
464
Chapter 11
Alkenes; Infrared Spectroscopy and Mass Spectrometry
To see why the terminal alkene is now favored, we again examine the transition state. Removal of a secondary hydrogen (from C3 in the starting bromide) is sterically more difficult than abstracting one of the more exposed methyl hydrogens. When the bulky base tert-butoxide is used, the energy of the transition state leading to the more stable product is increased by steric interference relative to that leading to the less substituted isomer; thus, the less substituted isomer becomes the major product (Figure 11-15B). An E2 reaction that generates the thermodynamically less favored isomer is said to follow the Hofmann rule,* named after the chemist who investigated a series of eliminations that proceeded with this particular mode of regioselectivity (Section 21-8).
Exercise 11-12
Working with the Concepts: Regioselectivity in Elimination When the following reaction is carried out with tert-butoxide in 2-methyl-2-propanol (tert-butyl alcohol), two products, A and B, are formed in the ratio 23 : 77. When ethoxide in ethanol is used, this ratio changes to 82 : 18. What are products A and B, and how do you explain the difference in their ratios in the two experiments? H G CCH3 G OH E B SB O O
G
H3C G CH3C G H
ECH3
Base, solvent
A
B
Strategy The difference between the two experiments is the use of a hindered base (tert-butoxide) in one and an unhindered base (ethoxide) in the other. Following the ideas presented in the text, we should expect A and B to be regioisomeric products of E2 reactions, arising from the removal of hydrogens from different carbon atoms adjacent to the one bearing the sulfonate leaving group. Solution • The bulky tert-butoxide is more likely to abstract a hydrogen atom from the less hindered adjacent (methyl) carbon atom, giving mostly B, the Hofmann-rule product: H i CH2 H≈ f )C (CH3)2HC i OO B SB O O
š (CH3)3CO ð
H G D C PC G D H (CH3)2HC H
CH3
B Major (Hofmann) product
• On the other hand, ethoxide preferentially removes a hydrogen from the tertiary carbon atom on the other side, because the result is the more stable trisubstituted alkene A, the Saytzev product:
CH3CH2š Oð
H H C[CH3 i }C O C i H3C & O3S CH3
H3C CH3
H D G C PC D G CH3 H 3C A
Major (Saytzev) product (More stable)
*Professor August Wilhelm von Hofmann (1818 – 1892), University of Berlin.
11-6 Preparation of Alkenes: Bimolecular Elimination
Chapter 11
Exercise 11-13
Try It Yourself (a) E2 reaction of 2-bromo-2,3-dimethylbutane, (CH3)2CBrCH(CH3)2, gives two products, A and B, in a 79 : 21 ratio using ethoxide in ethanol but a 27 : 73 ratio with tert-butoxide in 2-methyl-2propanol. What are A and B? (b) Use of (CH3CH2)3CO2 as the base gives an 8 : 92 ratio of A and B. Explain.
E2 reactions often favor trans over cis Depending on the structure of the alkyl substrate, the E2 reaction can lead to cis-trans alkene mixtures, in some cases with selectivity. For example, treatment of 2-bromopentane with sodium ethoxide furnishes 51% trans- and only 18% cis-2-pentene, the remainder of the product being the terminal regioisomer. The outcome of this and related reactions appears to be controlled again to some extent by the relative thermodynamic stabilities of the products, the more stable trans double bond being formed preferentially. Stereoselective Dehydrobromination of 2-Bromopentane
HBr
H CH3CH2 G G GC P C G
CH3CH2ONa, CH3CH2OH
H
CH3
51%
CH3 CH3CH2 G G GC P C G
CH G 3 CH3CH2CH2CBr ð G H
CH3CH2CH2CHPCH2
H
H 18%
31%
Unfortunately from a synthetic viewpoint, complete trans selectivity is rare in E2 reactions. Chapter 13 deals with alternative methods for the preparation of stereochemically pure cis and trans alkenes.
Some E2 processes are stereospecific Recall (Section 7-7) that the preferred transition state of elimination places the proton to be removed and the leaving group anti with respect to each other. Thus, before an E2 reaction takes place, bond rotation to an anti conformation occurs. This fact has additional consequences when reaction may lead to Z or E stereoisomers. For example, the E2 reaction of the two diastereomers of 2-bromo-3-methylpentane to give 3-methyl-2-pentene is stereospecific. Both the (R,R) and the (S,S) isomer yield exclusively the (E) isomer of the alkene. Conversely, the (R,S) and (S,R) diastereomers give only the (Z) alkene. (Build models!) As shown in the three-dimensional structures on the next page, anti elimination of HBr dictates the eventual configuration around the double bond. The reaction is stereospecific: One diastereomer (and its mirror image) produces only one stereoisomeric alkene, the other diastereomer furnishing the opposite configuration. Exercise 11-14 Which diastereomer of 2-bromo-3-deuteriobutane gives (E)-2-deuterio-2-butene, and which diastereomer gives the Z isomer?
In Summary Alkenes are most generally made by E2 reactions. Usually, the thermodynamically more stable internal alkenes are formed faster than the terminal isomers (Saytzev rule). Bulky bases may favor the formation of the products with thermodynamically less stable (e.g., terminal) double bonds (Hofmann rule). Elimination may be stereoselective, producing greater quantities of trans isomers than their cis counterparts from racemic starting materials. It also may be stereospecific, certain haloalkane diastereomers furnishing only one of the two possible stereoisomeric alkenes.
MODEL BUILDING
465
466
Chapter 11
Alkenes; Infrared Spectroscopy and Mass Spectrometry
Stereospecificity in the E2 Reaction of 2-Bromo-3-methylpentane Base:−
R R Br
−HBr
Br S
S
(E)-3-Methyl-2-pentene
Base:− Base:−
S R Br
−HBr
Br (Z)-3-Methyl-2-pentene
R S
Base:−
11-7 Preparation of Alkenes by Dehydration of Alcohols We have seen that treatment of alcohols with mineral acid at elevated temperatures results in alkene formation by loss of water, a process called dehydration, which proceeds by E1 or E2 pathways (Chapters 7 and 9). This section reviews this chemistry, now from the perspective of the alkene product. The usual way in which to dehydrate an alcohol is to heat it in the presence of sulfuric or phosphoric acid at relatively high temperatures (120 – 1708C). Acid-Mediated Dehydration of Alcohols A A OCOCO A A H ð OH
Acid,
G G CP C D D
HOH
The ease of elimination of water from alcohols increases with increasing substitution of the hydroxy-bearing carbon.
11-7 Preparation of Alkenes by Dehydration of Alcohols
Chapter 11
Relative Reactivity of Alcohols (ROH) in Dehydration Reactions R ⴝ primary secondary tertiary
Increasing ease of dehydration
CH3CH2 OH
Conc. H2SO4, 170C
CH2 P CH2
HOH
Primary alcohol
HOð H A A CH3C O CCH3 A A H H
50% H2SO4, 100C
CH3CH P CHCH3 80%
HOH
CH2 P CHCH2CH3 Trace
Secondary alcohol
(CH3)3COH
Dilute H2SO4, 50C HOH
Tertiary alcohol
CH3 E H2C P CE CH3 100%
Secondary and tertiary alcohols dehydrate by the unimolecular elimination pathway (E1), discussed in Sections 7-6 and 9-2. Protonation of the weakly basic hydroxy oxygen forms an alkyloxonium ion, now containing water as a good potential leaving group. Loss of H2O supplies the respective secondary or tertiary carbocations and deprotonation furnishes the alkene. The reaction is subject to all the side reactions of which carbocations are capable, particularly hydrogen and alkyl shifts (Section 9-3). Dehydration with Rearrangement H2SO4, H2O
H3C H E E C P CE E
CH3 ðOH A A CH3C O CH2 O CCH3 A A H H
H3C
CH2CH3 54%
CH3 A CH3CCHP CHCH3 A H 8%
Rearranged product
Exercise 11-15 Referring to Sections 7-6 and 9-3, write a mechanism for the preceding reaction. (Caution: As emphasized repeatedly, when writing mechanisms, use “arrow pushing” to depict electron flow; write out every step separately; formulate complete structures, including charges and relevant electron pairs; and draw explicit reaction arrows to connect starting materials or intermediates with their respective products. Don’t use shortcuts, and don’t be sloppy!)
Typically, the thermodynamically most stable alkene or alkene mixture results from unimolecular dehydration in the presence of acid. Thus, whenever possible, the most highly substituted system is generated; if there is a choice, trans-substituted alkenes predominate over the cis isomers. For example, acid-catalyzed dehydration of 2-butanol furnishes the equilibrium mixture of butenes, consisting of 74% trans-2-butene, 23% of the cis isomer, and only 3% 1-butene. Treatment of primary alcohols with mineral acids at elevated temperatures also leads to alkenes; for example, ethanol gives ethene and 1-propanol yields propene (Section 9-7). Conc. H2SO4,1808C
CH3CH2CH2OH uuuuuy CH3CH P CH2
other minor isomers
467
468
Alkenes; Infrared Spectroscopy and Mass Spectrometry
Chapter 11
The mechanism of this reaction begins with the initial protonation of oxygen. Then, attack by hydrogen sulfate ion or another alcohol molecule effects bimolecular elimination of a proton from one carbon atom and a water molecule from the other.
Exercise 11-16 (a) Propose a mechanism for the formation of propene from 1-propanol upon treatment with hot concentrated H2SO4. (b) Propene is also formed when propoxypropane (dipropyl ether) is subjected to the same conditions. Explain. Conc. H2SO4,1808C
CH3CH2CH2OCH2CH2CH3 uuuuuuy 2 CH3CH P CH2 1 H2O
In Summary Alkenes can be made by dehydration of alcohols. Secondary and tertiary systems proceed through carbocation intermediates, whereas primary alcohols can undergo E2 reactions from the intermediate alkyloxonium ions. All systems are subject to rearrangement and thus frequently give mixtures.
11-8 Infrared Spectroscopy The remaining sections in this chapter deal with two additional methods for the determination of the structures of organic compounds: infrared (IR) spectroscopy and mass spectrometry (MS). IR spectroscopy is a very useful tool because it is capable of detecting the characteristic bonds of many functional groups through their absorption of infrared light. IR spectroscopy measures the vibrational excitation of atoms around the bonds that connect them. The positions of the absorption lines associated with this excitation depend on the types of functional groups present, and the IR spectrum as a whole displays a pattern unique for each individual substance.
Absorption of infrared light causes molecular vibrations
B
A Frequency (v) Figure 11-16 Two unequal masses on an oscillating (“vibrating”) spring: a model for vibrational excitation of a bond.
At energies slightly lower than those of visible radiation, light causes vibrational excitation of the bonds in a molecule. This part of the electromagnetic spectrum is the infrared region (see Figure 10-2). The intermediate range, or middle infrared, is most useful to the organic chemist. IR absorption bands are described by either the wavelength, l, of the absorbed light in micrometers (1026 m; l < 2.5–16.7 mm; see Figure 10-2) or its reciprocal value, n (in units of cm21; , n 5 1yl). Thus, a typical infrared spectrum ranges called wavenumber, , , from n 5 600 to 4000 cm21, and the energy changes associated with absorption of this radiation range from 1 to 10 kcal mol21 (4 to 42 kJ mol21). Figure 10-3 described the general principles of a spectrometer, which apply also to an infrared instrument. Modern systems use sophisticated rapid-scan techniques and are linked with computers. This equipment allows for data storage, spectra manipulation, and computer library searches, so that unknown compounds can be matched with stored spectra. We can envision vibrational excitation simply by thinking of two atoms, A and B, linked by a flexible bond. Picture the atoms as two masses connected by a bond that stretches and compresses at a certain frequency, n, like a spring (Figure 11-16). In this picture, the frequency of the vibrations between two atoms depends both on the strength of the bond between them and on their atomic weights. In fact, it is governed by Hooke’s* law, just like the motion of a spring.
*Professor Robert Hooke (1635 – 1703), physicist at Gresham College, London.
11-8 Infrared Spectroscopy
Chapter 11
469
Figure 11-17 Various vibrational modes around tetrahedral carbon. The motions are labeled symmetric and asymmetric stretching or bending, scissoring, rocking, twisting, and wagging.
Symmetric stretching vibration (both outside atoms move away from or toward the center)
Symmetric bending vibration in a plane (scissoring)
Symmetric bending vibration out of a plane (twisting)
Asymmetric stretching vibration (as one atom moves toward the center, the other moves away)
Asymmetric bending vibration in a plane (rocking)
Asymmetric bending vibration out of a plane (wagging)
Hooke’s Law and Vibrational Excitation n| 5 k ,
n5 k5 f 5 m1, m2 5
(m1 1 m2 ) m1 m2 B f
vibrational frequency in wavenumbers (cm21) constant force constant, indicating the strength of the spring (bond) masses of attached atoms
This equation might lead us to expect every individual bond in a molecule to show one specific absorption band in the infrared spectrum. However, in practice, an interpretation of the entire infrared spectrum is considerably more complex and beyond the needs of the organic chemist. This is because molecules that absorb infrared light undergo not only stretching, but also various bending motions (Figure 11-17), as well as combinations of the two. The bending vibrations are mostly of weaker intensity, they overlap with other absorptions, and they may show complicated patterns. In addition, for a bond to absorb infrared light, its vibrational motion must cause a change in the molecular dipole. Therefore, vibrations of polar bonds give strong infrared absorption bands, whereas absorptions associated with nonpolar bonds may be weak or entirely absent. The practicing organic chemist can find good use for IR spectroscopy for two reasons: The vibrational bands of many functional groups appear at characteristic frequencies; furthermore, the entire infrared spectrum of any given compound is unique in its pattern and can be distinguished from that of any other substance.
Functional groups have typical infrared absorptions Table 11-4 lists the characteristic stretching wavenumber values for the bonds (shown in red) in some common structural units. Notice how most absorb in the region above 1500 cm21. We shall show the IR spectra typical of new functional groups when we introduce each of the corresponding compound classes in subsequent chapters. Figures 11-18 and 11-19 show the IR spectra of pentane and hexane. Above 1500 cm21, we see only the C – H stretching absorptions typical of alkanes, in the range from 2840 to 3000 cm21; no functional groups are present, and the two spectra are very similar in this
Table 11-4
Characteristic Infrared Stretching Wavenumber Ranges of Organic Molecules
Bond or Functional Group
n| (cm21)
RO O H (alcohols) O (carboxylic B RCO O H acids)
3200 – 3650
R2NO H (amines) RC qC O H (alkynes) G D (alkenes) C PC D G H A OC O H (alkanes) A
3250 – 3500
2500 – 3300 3260 – 3330 3050 – 3150
2840 – 3000
RC qCH (alkynes)
2100 – 2260
Bonds to Hydrogen
4000
Bond or Functional Group
n| (cm21)
RC q N (nitriles)
2220 – 2260
O O (aldehydes, B B RCH, RCR⬘ ketones) O B RCOR⬘ (esters) O (carboxylic B RCOH acids)
G D C PC (alkenes) D G (alcohols, A RC O OR⬘ ethers) A
COH OO H NO H
COC COO CON COX
Lighter atoms ⫽ higher frequency
Stronger bonds ⫽ higher frequency
Fingerprint region
3000
1735 – 1750
1710 – 1760 1620 – 1680 1000 – 1260
Single Bonds Triple Double Bonds Bonds CPC CPO CqC CqN CPN
3500
1690 – 1750
2500
2000 1500 Wavenumber
1000
600 cm⫺1
Increasing wavenumber (energy)
100
Transmittance (%)
Figure 11-18 IR spectrum of pentane. Note the format: Wavenumber is plotted (decreasing from left to right) against percentage of transmittance. A 100% transmittance means no absorption; therefore “peaks” in an IR spectrum point downward. The spectrum shows absorbances at n| C—H stretch 5 2960, 2930, and 2870 cm21; n| C—H bend 5 1460, 1380, and 730 cm21. The region between 600 and 1300 cm21 is also shown recorded at higher sensitivity (in red), revealing details of the pattern in the fingerprint region.
Csp3−H
H IR 0 4000 3500
3000
2500
CH2CHCHCHCH2
2000
Wavenumber 470
Recorded at higher sensitivity
H H H
H
1500
1000
600 cm−1
11-8 Infrared Spectroscopy
471
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Figure 11-19 IR spectrum of hexane. Comparison with that of pentane (Figure 11-18) shows that the location and appearance of the major bands are very similar, but the two fingerprint regions show significant differences at higher recorder sensitivity (in red).
100
Transmittance (%)
Csp3−H
Recorded at higher sensitivity H H H H CH2CHCHCHCHCH2
H
H
IR 0 4000
3500
3000
2500
2000
1500
1000
600 cm−1
Wavenumber
Figure 11-20 IR spectrum of 1-hexene: n| Csp 2H stretch 5 3080 cm21; n| C5C stretch 5 1640 cm21; n| Csp –H bend 5 995 and 915 cm21. 2
100
Transmittance (%)
2
Csp3−H
Csp2−H
C
C
3080 1640 H
H
H
H
H
C
C
C
C
H
H
H
H
995 C H IR 0 4000 3500
C
915
H 3000
2500
H
2000
1500
1000
600 cm−1
Wavenumber
region. However, below 1500 cm21, the spectra differ, as the scans at higher sensitivity show. This is the fingerprint region, in which absorptions due to C – C bond stretching and C – C and C – H bending motions overlap to give complicated patterns. Bands at approximately 1460, 1380, and 730 cm21 are common to all saturated hydrocarbons. Figure 11-20 shows the IR spectrum of 1-hexene. A characteristic feature of alkenes when compared with alkanes is the stronger Csp2 – H bond, which should therefore have a band at higher energy in the IR spectrum. Indeed, as Figure 11-20 shows, there is a sharp spike at 3080 cm21, due to this stretching mode, at a slightly higher wavenumber than the remainder of the C – H stretching absorptions. A useful rule of thumb is that Csp32H bonds give rise to peaks below 3000 cm21, whereas Csp22H bonds absorb above 3000 cm21. According to Table 11-4, the CPC stretching band should appear between about 1620 and 1680 cm21. Figure 11-20 shows a sharp band at 1640 cm21 assigned to this vibration. The other major peaks are the result of bending motions. For example, the two signals at 915 and 995 cm21 are typical of a terminal alkene.
472
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100
Transmittance (%)
Figure 11-21 IR spectrum of cyclohexanol: n| O2H stretch 5 3345 cm21; n| C2O 5 1070 cm21; Note the broad and very intense peak arising from the polar O – H bond.
Alkenes; Infrared Spectroscopy and Mass Spectrometry
O
H
H
O
C 3345
IR 0 4000
O
1070 3500
3000
2500
2000
1500
1000
600 cm−1
Wavenumber
Approximate Infrared Frequencies of Major Bending Modes for Alkenes H R G G GC P C G
H
H
915, 995 cmⴚ1
H R G G P C C G R H G
890 cmⴚ1
H R G G GC P C H R G
970 cmⴚ1
Two other strong bending modes may be used as a diagnostic tool for the substitution pattern in alkenes. One mode results in a single band at 890 cm21 and is characteristic of 1,1-dialkylethenes; the other gives a sharp band at 970 cm21 and is produced by the Csp2 – H bending mode of a trans double bond. The CPC stretching absorption of internal alkenes is usually less intense than that in terminal alkenes, because vibration of an internal CPC bond causes less change in the molecular dipole. In highly symmetric molecules, such as trans-3-hexene, the band for the CPC vibration may be too weak to be observed readily. However, the alkenyl C – H bend in trans-3-hexene still gives a very strong, sharp absorption at 970 cm21. In conjunction with NMR (Section 11-4), the presence or absence of such bands allows for fairly reliable structural assignments of specifically substituted double bonds. The O – H stretching absorption is the most characteristic band in the IR spectra of alcohols (Chapters 8 and 9), appearing as a readily recognizable intense, broad absorption over a wide range (3200 – 3650 cm21, Figure 11-21). The broadness of this band is due to hydrogen bonding to other alcohol molecules or to water. Dry alcohols in dilute solution exhibit sharp, narrow bands in the range from 3620 to 3650 cm21. In contrast, the C –X bonds of haloalkanes (Chapters 6 and 7) possess IR stretching frequencies at energies too low (,800 cm21) to be generally useful for characterization. Exercise 11-17 Three alkenes with the formula C4H8 exhibit the following IR absorptions: alkene A, 964 cm21; alkene B, 908 and 986 cm21; alkene C, 890 cm21. Assign a structure to each alkene.
In Summary The presence of specific functional groups can be ascertained by infrared spectroscopy. Infrared light causes the vibrational excitation of bonds in molecules. Strong bonds and light atoms vibrate at relatively high stretching frequencies measured in wavenumbers (reciprocal wavelengths). Conversely, weak bonds and heavy atoms absorb at lower wavenumbers, as would be expected from Hooke’s law. Highly polar bonds tend to exhibit more intense absorption bands. Because of the variety of stretching and bending modes, infrared spectra usually show complicated patterns. However, these patterns are diagnostic fingerprints for particular compounds. The presence of variously substituted alkenes may be detected by stretching signals at about 3080 (C – H) and 1640 (CPC) cm21 and by bending modes between 890 and 990 cm21. Alcohols show a characteristic band for the OH group in the range from 3200 to 3650 cm21. As a general rule, bands in the left half of the IR spectrum (above 1500 cm21) identify functional groups, whereas absorptions in the right half (below 1500 cm21) characterize specific compounds.
11-9 Mass Spectrometry
Chapter 11
11-9 Measuring the Molecular Mass of Organic Compounds: Mass Spectrometry In the examples and problems dealing with structure determinations of organic compounds, we have so far been given the molecular formula of the “unknown.” How is this information obtained? Elemental analysis (Section 1-9) gives us an empirical formula, which tells us the ratios of the different elements in a molecule. However, empirical and molecular formulas are not necessarily identical. For example, elemental analysis of cyclohexane merely reveals the presence of carbon and hydrogen atoms in a 1 : 2 ratio; it does not tell us that the molecule contains six carbons and twelve hydrogens. To determine molecular masses, the chemist turns to another important physical technique used to characterize organic molecules: mass spectrometry. This section begins with a description of the apparatus used and the physical principle on which it is based. Subsequently, we shall consider processes by which molecules fragment under the conditions required for molecular mass measurement, giving rise to characteristic recorded patterns called mass spectra. Mass spectra can help chemists distinguish between constitutional isomers and identify the presence of many functions such as hydroxy and alkenyl groups.
The mass spectrometer distinguishes ions by mass Mass spectrometry is not a form of spectroscopy in the conventional sense, because no radiation is absorbed (Section 10-2). A sample of an organic compound is introduced into an inlet chamber (Figure 11-22, upper left). It is vaporized, and a small quantity is allowed to leak into the source chamber of the spectrometer. Here the neutral molecules (M) pass through a beam of high-energy [usually 70 eV, or about 1600 kcal mol21 (6700 kJ mol21)] electrons. Upon electron impact, some of the molecules lose an electron to form the cor. responding radical cation, M1 , called the parent or molecular ion. Most organic molecules undergo only a single ionization.
Ionizing electrons 70-eV cathode (electron generator)
Leak into mass spectrometer
Volatilazation chamber (under vacuum)
– – –
M M M M M M M M M M M M
Sample M inlet
Source chamber
Magnet +• Accelerator plates (deflects M ) Strength H
M+• M+•
M+• M+•
M+•
M+•
M+• M+• M+• M+• M+• M+•
Anode (electron absorber)
M+• M+•
Relative abundance
100
MS
57
M+• M+•
50
Collector slit 29 41
Collector
M+•
Amplifier
0 0 20 40 60 80 Molecular weights of ions (as m/z)
Mass spectrum Recorder
Figure 11-22 Diagram of a mass spectrometer.
473
474
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Alkenes; Infrared Spectroscopy and Mass Spectrometry
Ionization of a Molecule on Electron Impact ⫹
M Neutral molecule
Molecular Masses of Organic Molecules CH4 m/z ⴝ 16 CH3OH m/z ⴝ 32
O B CH3COCH3 m/z ⴝ 74
e (70 eV)
M⫹j
Ionizing beam
Radical cation (Molecular ion)
⫹
2e
As charged particles, the molecular ions are next accelerated to high velocity by an electric field. (Molecules that are not ionized remain in the source chamber to be pumped . away.) The accelerated M1 ions are then subjected to a magnetic field, which deflects them from a linear to a circular path. The curvature of this path is a function of the strength of the magnetic field. Much as in an NMR spectrometer (Section 10-3), the strength of the magnetic field can be varied and therefore adjusted to give the exact degree of curvature to the path of the ions necessary to direct them through the collector slit to the collector, where they are detected and counted. Because lighter species are deflected more than heavier ones, the strength of the field necessary to direct the ions through the slit to the . collector is a function of the mass of M1 and therefore of the original molecule M. Thus, at a given magnetic field strength, only ions of a specific mass can pass through the collector slit. All others will collide with the internal walls of the instrument. Finally, the arrival of ions at the collector is translated electronically into a signal and recorded on a chart. The chart plots mass-to-charge ratio, myz (on the abscissa), versus peak height (on the ordinate); peak height measures the relative number of ions with a given myz ratio. Because only singly charged species are normally formed, z 5 1, and myz equals the mass of the ion being detected. Exercise 11-18 Three unknown compounds containing only C, H, and O give rise to the following molecular weights. Draw as many reasonable structures as you can. (a) myz 5 46; (b) myz 5 30; (c) myz 5 56.
High-resolution mass spectrometry reveals molecular formulas Consider substances with the following molecular formulas: C7H14, C6H10O, C5H6O2, and C5H10N2. All possess the same integral mass; that is, to the nearest integer, all four would be expected to exhibit a molecular ion at myz 5 98. However, the atomic weights of the elements are composites of the masses of their naturally occurring isotopes, which are not integers. Thus, if we use the atomic masses for the most abundant isotopes of C, H, O, and N (Table 11-5) to calculate the exact mass corresponding to each of the aforementioned molecular formulas, we see significant differences. Exact Masses of Four Compounds with m/z ⴝ 98
Table 11-5 Exact Masses of Several Common Isotopes Isotope 1
H C 14 N 16 O 32 S 35 Cl 37 Cl 79 Br 81 Br 12
Mass 1.00783 12.00000 14.0031 15.9949 31.9721 34.9689 36.9659 78.9183 80.9163
C7H14
C6H10O
C5H6O2
C5H10N2
98.1096
98.0732
98.0368
98.0845
Can we use mass spectrometry to differentiate these species? Yes. Modern high-resolution mass spectrometers are capable of distinguishing among ions that differ in mass by as little as a few thousandths of a mass unit. We can therefore measure the exact mass of any molecular ion. By comparing this experimentally determined value with that calculated for each species possessing the same integral mass, we can assign a molecular formula to the unknown ion. High-resolution mass spectrometry is the most widely used method for determining the molecular formulas of unknowns. Exercise 11-19 Choose the molecular formula that matches the exact mass. (a) myz 5 112.0888, C8H16, C7H12O, or C6H8O2; (b) myz 5 86.1096, C6H14, C4H6O2, or C4H10N2.
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C H E MI C AL H I G H L I G H T 1 1 - 2 Security in the 21st Century: Applications of IR and MS Spectroscopic methods are revolutionizing our ability to detect dangerous environmental substances in real time. Portable infrared imaging detectors (high-tech cameras) that can identify and pinpoint the location, extent, and movement of clouds of toxic gases are commercially available. These devices monitor the IR spectrum of the image of each pixel in the camera’s field of view. The detectors are programmed to alert a user to the presence of a variety of agents by matching the fingerprint regions of observed IR spectra with a customized database loaded into the device’s memory. Technological advances in the identification of chemical substances by their molecular masses have led to the development of the “puffer” machine that you may have seen at an airport security checkpoint. A passenger stands under the device’s archway while a puff of air is released, which passes his or her body and is wafted into a detector. The detector uses many of the basic features of the mass spectrometer: It ionizes molecules in the air stream and detects their masses. It differs in that masses are distinguished not by the amount their paths are curved by electric fields in a vacuum, but by how fast they drift through a series of charged rings at normal atmospheric pressure — thus the name ion-mobility spectrometry (IMS). Ion mobility, a function of the mass, shape, and size of a particle, allows
unambiguous identification of the original molecule by comparison with a standard database. In 10 s or less, these devices can detect both positive and negative ions deriving from minute amounts (less than 1029 g) of a variety of explosive substances, toxic industrial chemicals, illegal narcotics, and chemical warfare agents.
The Explosive Detection Trace Portal in operation at the San Francisco International Airport.
Molecular ions undergo fragmentation Mass spectrometry gives information not only about the molecular ion, but also about its component structural parts. Because the energy of the ionizing beam far exceeds that required to break typical organic bonds, some of the molecular ions break apart into virtually all possible combinations of neutral and ionized fragments. This fragmentation gives rise to a number of additional mass-spectral peaks, all of lower mass than the molecular ion from which they are derived. The spectrum that results is called the massspectral fragmentation pattern. The most intense peak in the spectrum is called the base peak. Its relative intensity is defined to be 100, and the intensities of all other peaks are described as a percentage of the intensity of the base peak. The base peak in a mass spectrum may be the molecular ion peak or it may be the peak of one of the fragment ions. For example, the mass spectrum of methane contains, in addition to the molecular ion . . peak, lines for CH31, CH21 , CH1, and C1 (Figure 11-23). These lines are formed by the processes shown in the margin. The relative abundances of these species, as indicated by the heights of the peaks, give a useful indication of the relative ease of their formation. It can be seen that the first C – H bond is cleaved readily, the myz 5 15 peak reaching 85% of the abundance of the molecular ion, which in this case is the base peak. The breaking of additional C – H bonds is increasingly difficult, and the corresponding ions have lower relative abundance. Section 11-10 considers the process of fragmentation in more detail and reveals how fragmentation patterns may be used as an aid to molecular structure determination.
Fragmentation of Methane in the Mass Spectrometer CH4⫹j ⫺H2
CH2⫹j ⫺H2
⫺Hj
CH3⫹ ⫺H2
C⫹j
CH⫹
Odd-electron (radical) cations
Even-electron cations
Chapter 11
Figure 11-23 Mass spectrum of methane. At the left is the spectrum actually recorded; at the right is the tabulated form, the largest peak (base peak) being defined as 100%. For methane, the base peak at myz 5 16 is due to the parent ion. Fragmentation gives rise to peaks of lower mass.
Alkenes; Infrared Spectroscopy and Mass Spectrometry
Fragment ions
MS 16 Base peak
100
Relative abundance
476
Tabulated Spectrum
m/z
Relative Abundance (%)
CH4
17 16
50
13CH 4
0 10
20
1.1 100.0 (base peak)
15
85.0
14
9.2
13
3.9
12
1.0
Molecular or Fragment ion
(M ⫹ 1)⫹1 M⫹ (parent ion)
1
(M ⫺ 1)⫹ (M ⫺ 2)⫹
1
(M ⫺ 3)⫹ (M ⫺ 4)⫹
1
m/z
Mass spectra reveal the presence of isotopes An unusual feature in the mass spectrum of methane is a small (1.1%) peak at myz 5 17; . it is designated (M 1 1)1 . How is it possible to have an ion present that has an extra mass unit? The answer lies in the fact that carbon is not isotopically pure. About 1.1% of natural carbon is the 13C isotope (see Table 10-1), giving rise to the additional peak. In the mass . spectrum of ethane, the height of the (M 1 1)1 peak, at myz 5 31, is about 2.2% that of the parent ion. The reason for this finding is statistical. The chance of finding a 13C atom
CHE M I C AL H I G H L I G H T 1 1 - 3 Detecting Performance-Enhancing Drugs Using Mass Spectrometry Recent and rather notorious cases of using illicit substances to enhance athletic performance have highlighted the technology that has made the detection of these substances possible (see Chapter 4 Opening). An instrument called a gas chromatograph (GC) separates a test sample into its individual components, each of which is analyzed by high-resolution mass spectrometry. The success of this method in detecting “cheating” rests in its exquisite sensitivity and extremely high quantitative precision. In cases of suspected administration of anabolic steroids such as testosterone (Section 4-7), two approaches are used. The first compares the ratio of testosterone (T) to its stereoisomer, epitestosterone (E; identical to T except that the hydroxy group on the five-membered ring is “down” instead of “up”). E and T occur naturally in humans in roughly equal amounts, but E, unlike T, does not enhance performance. Taking synthetic T alters the T : E ratio and is easily detected. Therefore, some athletes have taken synthetic T and E together, to maintain a T : E ratio within normal limits. Mass spectrometry identifies these situations, because of a quirk of biology: Synthetic steroids are made from plant-derived precursors, which have a slightly lower 13C content (relative to 12 C) than do steroids biosynthesized naturally in the human body. The differences are quite small (parts per thousand)
but readily detectable: After the steroids are separated from the samples by GC, they undergo combustion to CO2, and the mass spectrometer measures the ratio of 13CO2 to 12CO2. A 13C : 12C ratio that departs significantly from that found in normal human steroids and that approaches that found in plant-derived synthetics is considered to be strong evidence of “doping.”
American bicycle star Lance Armstrong leaving the doping test station during the 2001 Tour de France.
11-9 Mass Spectrometry
43
MS
100
base peak
Relative abundance
Two molecular ions
50
124 (C3H781Br) 122 (C3H779Br)
0 20
40
60
477
Figure 11-24 Mass spectrum of 1-bromopropane. Note the nearly equal heights of the peaks at myz 5 122 and 124, owing to the almost equal abundances of the two bromine isotopes.
(M⫺Br)⫹
CH3CH2CH2Br
0
Chapter 11
80
100
120
m/z
in a compound containing two carbons is double that expected of a one-carbon molecule. For a three-carbon moiety, it would be threefold, and so on. Other elements, too, have naturally occurring higher isotopes: Hydrogen (deuterium, 2H, about 0.015% abundance), nitrogen (0.366% 15N), and oxygen (0.038% 17O, 0.200% 18O) are examples. These isotopes also contribute to the intensity of peaks at masses higher than . M1 , but less so than 13C. Fluorine and iodine are isotopically pure. However, chlorine (75.53% 35Cl; 24.47% 37Cl) and bromine (50.54% 79Br; 49.46% 81Br) each exist as a mixture of two isotopes and give rise to readily identifiable isotopic patterns. For example, the mass spectrum of 1-bromopropane (Figure 11-24) shows two peaks of nearly equal intensity at myz 5 122 and 124. Why? The isotopic composition of the molecule is a nearly 1 : 1 mixture of CH3CH2CH279Br and CH3CH2CH281Br. Similarly, the spectra of monochloroalkanes exhibit ions two mass units apart in a 3 : 1 intensity ratio, because of the presence of about 75% R35Cl and 25% R37Cl. Peak patterns such as these are useful in revealing the presence of chlorine or bromine. Exercise 11-20 What peak pattern do you expect for the molecular ion of dibromomethane?
Exercise 11-21 Nonradical compounds containing C, H, and O have even molecular weights, those containing C, H, O, and an odd number of N atoms have odd molecular weights, but those with an even number of N atoms are even again. Explain.
In Summary Molecules can be ionized by an electron beam at 70 eV to give radical cations that are accelerated by an electric field and then separated by the different deflections that they undergo in a magnetic field. In a mass spectrometer, this effect is used to measure the molecular masses of molecules. At high resolution, the mass of the molecular ion permits the determination of the molecular formula. The molecular ion is usually accompanied by less massive fragments and isotopic “satellites,” owing to the presence of less abundant isotopes. In some cases, such as with Cl and Br, more than one isotope may be present in substantial quantities.
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11-10 Fragmentation Patterns of Organic Molecules Upon electron impact, molecules dissociate by breaking weaker bonds first and then stronger ones. Because the original molecular ion is positively charged, its dissociation typically gives one neutral fragment and one that is cationic. Normally the cationic fragment that forms bears its charge at the center that is most capable of stabilizing a positive charge. This section illustrates how the preference to break weaker bonds and to form more stable carbocationic fragments combines to make mass spectrometry a powerful tool for molecular structure determination.
Fragmentation is more likely at a highly substituted center The mass spectra of the isomeric hydrocarbons pentane, 2-methylbutane, and 2,2dimethylpropane (Figures 11-25, 11-26, and 11-27) reveal the relative ease of the several possible C – C bond-dissociation processes. In each case, the molecular ion produces a relatively weak peak; otherwise, the spectra of the three compounds are very different. Figure 11-25 Mass spectrum of pentane, revealing that all C – C bonds in the chain have been ruptured.
Relative abundance
43 (M − CH2CH3)+
MS
100
CH3CH2CH2CH2CH3
50
(CH2CH3)+ 27 29
(CH3)+
(M − CH3)+
15
57
M+• 72
0 20
0
40
60
80
100
120
m/z
Relative abundance
43 (M − CH2CH3)+
MS
100
57 (M − CH3)+
CH3 H3 C
50 29
C
CH2CH3
H
71 (M − H)+
Figure 11-26 Mass spectrum of 2-methylbutane. The peaks at myz 5 43 and 57 result from preferred fragmentation around C2 to give secondary carbocations.
72 M+•
15 0 0
20
40
60
80
m/z
100
120
11-10 Fragmentation Patterns of Organic Molecules
Relative abundance
479
Figure 11-27 Mass spectrum of 2,2-dimethylpropane. Only a very weak molecular ion peak is seen, because the fragmentation to give a tertiary cation is favored.
57 (M − CH3)+
MS
100
Chapter 11
CH3 H3C
50 29
41
C
CH3
CH3
15
72 M+•
0 0
20
40
60
80
100
120
m /z
Pentane fragments by C – C bond-breaking in four possible ways, each of which gives a carbocation and a radical. Only the positively charged cation is observed in the mass spectrum (see margin); the radical, being neutral, is “invisible.” For example, one process breaks the C1 – C2 bond to give a methyl cation and a butyl radical. The peak at myz 5 15 for CH31 in the mass spectrum (Figure 11-25) is very weak, consistent with the instability of this carbocation (Section 7-5). Similarly, the peak at myz 5 57 is weak, because it derives from fragmentation into a butyl cation and a methyl radical, and although the primary butyl cation is more stable than CH31, the methyl radical is a high-energy species. Thus this mode of fragmentation is not favored. The favored bond cleavages give peaks at myz 5 29 and 43, for the ethyl and propyl cations, respectively. Each of these fragmentations generates both a primary cation and a primary radical, and avoids formation of a methyl fragment. Each peak is surrounded by a cluster of smaller lines because of the presence of 13C, which results in a peak at one mass unit higher, and the loss of hydrogens, which produces peaks at masses one or more units lower. Note that loss of H? does not give a strong peak even if the carbocation left behind is stable: The hydrogen atom is a high-energy species (Section 3-1). The mass spectrum of 2-methylbutane (Figure 11-26) shows a pattern similar to that of pentane; however, the relative intensities of the various peaks differ. Thus, the peak at myz 5 71 (M 2 1)1 is larger, because loss of H? from C2 gives a tertiary cation. The signals at myz 5 43 and 57 are more intense, because both arise from loss of an alkyl radical from C2 to form a secondary carbocation.
Secondary carbocation
Preferred Fragmentation of 2-Methylbutane j CH3 H3C A D C O CH2CH3 H3C O C O CH2CH3 C H j CH j A G H H 3
m/z ⴝ 57
2
m/z ⴝ 72 Hj
CH3 Tertiary A carbocation C E H CH2CH3 H 3C m/z ⴝ 71
5
CH3 D H3C OC G H m/z ⴝ 43
Fragment Ions from Pentane CH3 m/z ⴝ 15
C3H7 m/z ⴝ 43
[CH3–CH2–CH2–CH2–CH3]j m/z ⴝ 72
C2H5
C4H9
m/z ⴝ 29
m/z ⴝ 57
Secondary carbocation
480
Chapter 11
j CH3 A H3C O C O CH3 A CH3 m/z ⴝ 72
Alkenes; Infrared Spectroscopy and Mass Spectrometry
The preference for fragmentation at a highly substituted center is even more pronounced in the mass spectrum of 2,2-dimethylpropane (Figure 11-27). Here, loss of a methyl radical from the molecular ion produces the 1,1-dimethylethyl (tert-butyl) cation as the base peak at myz 5 57. This fragmentation is so easy that the molecular ion is barely visible. The spectrum also reveals peaks at myz 5 41 and 29, the result of complex structural reorganizations, such as the carbocation rearrangements considered in Section 9-3.
Fragmentations also help to identify functional groups CH3 D H3C OC G CH3 m/z ⴝ 57
Particularly easy fragmentation of relatively weak bonds is also seen in the mass spectra of the haloalkanes. The fragment ion (M 2 X)1 is frequently the base peak in these spectra. A similar phenomenon is observed in the mass spectra of alcohols, which eliminate water . to give a large (M 2 H2O)1 peak 18 mass units below the parent ion (Figure 11-28). The bonds to the C – OH group also readily dissociate in a process called a cleavage, leading to resonance-stabilized hydroxycarbocations: A R OCOš OH A
70 eV Rj
G COš OH D
G CP OH D
The strong peak at myz 5 31 in the mass spectrum of 1-butanol is due to the hydroxymethyl cation, 1CH2OH, which arises from a cleavage. Alcohol Fragmentation by Dehydration and ␣ Cleavage ⴢ HO H A A R O C O CHR A H Mⴙⴢ Rj
H
HO A C E H
Relative abundance
R
(M ⴚ 18)ⴙ ⴢ
CH2R
R
HO A C E H
H
56 (M − H2O)+
MS
100
Figure 11-28 Mass spectrum of 1-butanol. The parent ion, at myz 5 74, gives rise to a small peak because of ready loss of water to give the ion at myz 5 56. Other fragment ions due to a cleavage are propyl (myz 5 43), 2-propenyl (allyl) (myz 5 41), and hydroxymethyl (myz 5 31).
CH2R
HO A C E H
H2O
RCH2j
Hj
[RCHP CHR]j
(CH2OH)+ 31
H 41 43
50
CH3CH2CHCH2OH
73 (M − H)+ 74 M+• 0 0
20
40
60
80
m/z
100
120
11-10 Fragmentation Patterns of Organic Molecules
Chapter 11
481
Exercise 11-22 Try to predict the appearance of the mass spectrum of 3-methyl-3-heptanol.
Alkenes fragment to give resonance-stabilized cations The fragmentation patterns of alkenes also reflect a tendency to break weaker bonds and to form more stable cationic species. The bonds one atom removed from the alkene function — the so-called allylic bonds—are relatively easily broken, because the result is a resonance-stabilized carbocation. For example, the mass spectra of terminal straight-chain alkenes such as 1-butene reveal formation of the 2-propenyl (allyl) cation, at myz 5 41, the base peak in the spectrum (Figure 11-29A). Allylic bond
CH2 P CH O CH2 O CH3
CH2 O CH O CH2
j
m/z ⴝ 56
CH3j
CH2 O CH O CH2 2-Propenyl (allyl) cation m/z ⴝ 41
Branched and internal alkenes fragment similarly at allylic bonds. Figure 11-29B shows the mass spectrum of 2-hexene, in which the base peak at myz 5 55 corresponds to formation of the resonance-stabilized 2-butenyl cation.
CH3 O CHP CH O CH2 O CH2 O CH3
CH3 O CH O CH O CH2
j
Figure 11-29 Mass spectra of (A) 1-butene, showing a peak at myz 5 41 from cleavage to give the resonance-stabilized 2-propenyl (allyl) cation; (B) 2-hexene, showing similar cleavage between C4 and C5 to give the 2-butenyl cation, with myz 5 55.
C2H5j
m/z ⴝ 84
CH3 O CH O CH O CH2 2-Butenyl cation m/z ⴝ 55
41 (M − CH3)+
MS
100
56 M+•
Relative abundance
Relative abundance
CH2 P CHCH2OCH3
50
CH3CH P CH CH2OCH2CH3
50 84 M+•
0
0 0
A
55 (M − CH2CH3)+
MS
100
20
40
m/z
0
60
B
20
40
60
m/z
80
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Exercise 11-23 The mass spectrum of 4-methyl-2-hexene shows peaks at myz 5 69, 83, and 98. Explain the origin of each.
In Summary Fragmentation patterns can be interpreted for structural elucidation. For example, the radical cations of alkanes cleave to form the most stable positively charged fragments, haloalkanes fragment by rupture of the carbon – halogen bond, alcohols readily dehydrate and undergo a cleavage, and alkenes break an allylic bond to form a resonancestabilized carbocation.
11-11 Degree of Unsaturation: Another Aid to Identifying Molecular Structure
Some C5H8 Hydrocarbons CH3 (Two bonds)
CHPCH2
CH3
(One bond ⴙ one ring)
(Two rings)
NMR and IR spectroscopy and mass spectrometry are important tools for determining the structure of an unknown. However, concealed in the molecular formula of every compound is an additional piece of information that can make the job easier. Consider a saturated, acyclic alkane: Its molecular formula is CnH2n12. In contrast, an alkene containing one double bond contains two hydrogens less: CnH2n; it is called unsaturated. Cycloalkanes also have the general formula CnH2n. You can see that hydrocarbons containing several double bonds and/or rings will deviate from the “saturated” formula CnH2n12 by the appropriate number of fewer hydrogens. The degree of unsaturation is defined as the sum of the numbers of rings and p bonds present in the molecule. Table 11-6 illustrates the relation between molecular formula, structure, and degree of unsaturation for several hydrocarbons. As Table 11-6 shows, each increase in the degree of unsaturation corresponds to a decrease of two hydrogens in the molecular formula. Therefore, starting with the general formula for acyclic alkanes (saturated; degree of unsaturation 5 0), CnH2n12 (Section 2-4), the degree of unsaturation may be determined for any hydrocarbon merely by comparing the actual number of hydrogens present with the number required for the molecule to be saturated, namely, 2n 1 2, where n 5 the number of carbon atoms present. For example, what is the degree of unsaturation in a hydrocarbon of the formula C5H8? A saturated compound with five carbons has the formula C5H12 (CnH2n12, with n 5 5). Because C5H8 is four hydrogens short of being saturated, the degree of unsaturation is 4y2 5 2. All molecules with this formula contain a combination of rings and p bonds adding up to two. Table 11-6 Degree of Unsaturation as a Key to Structure Formula
Representative Structures
C6H14
Degree of Unsaturation 0
;
C6H12
1 (one bond)
(one ring) ;
;
C6H10
2 (two bonds)
(one bond ⴙ one ring)
(two rings)
;
;
C6H8
3 (three bonds)
(two bonds ⴙ one ring)
(one bond ⴙ two rings)
11-11 Degree of Unsaturation
The presence of heteroatoms may affect the calculation. Let us compare the molecular formulas of several saturated compounds: ethane, C2H6, and ethanol, C2H6O, have the same number of hydrogen atoms; chloroethane, C2H5Cl, has one less; ethanamine, C2H7N, one more. The number of hydrogens required for saturation is reduced by the presence of halogen, increased when nitrogen is present, and unaffected by oxygen. We can generalize the procedure for determination of the degree of unsaturation from a molecular formula as follows. Step 1. Determine from the number of carbons (nC), halogens (nX), and nitrogens (nN) in the molecular formula the number of hydrogens required for the molecule to be saturated, Hsat. Hsat 5 2nC 1 2 2 nX 1 nN
(Oxygen and sulfur are disregarded.)
Step 2. Compare Hsat with the actual number of hydrogens in the molecular formula, Hactual, to determine the degree of unsaturation. Degree of unsaturation 5
Hsat 2 Hactual 2
Alternatively, these steps may be combined into one formula: Degree of unsaturation 5
2nC 1 2 1 nN 2 nH 2 nX 2
Exercise 11-24 Calculate the degree of unsaturation indicated by each of the following molecular formulas. (a) C5H10; (b) C9H12O; (c) C8H7ClO; (d) C8H15N; (e) C4H8Br2.
Exercise 11-25
Working with the Concepts: Degree of Unsaturation in Structure Determination Spectroscopic data for three compounds with the molecular formula C5H8 are given below; m denotes a complex multiplet. Assign a structure to each compound. (Hint: One is acyclic; the others each contain one ring.) (a) IR 910, 1000, 1650, 3100 cm21; 1H NMR d 5 2.79 (t, J 5 8 Hz), 4.8 – 6.2 (m) ppm, integrated intensity ratio of the signals 5 1 : 3. (b) IR 900, 995, 1650, 3050 cm21; 1 H NMR d 5 0.5–1.5 (m), 4.8–6.0 (m) ppm, integrated intensity ratio of the signals 5 5 : 3. (c) IR 1611, 3065 cm21; 1H NMR d 5 1.5–2.5 (m), 5.7 (m) ppm, integrated intensity ratio of the signals 5 3 : 1. Is there more than one possibility? Strategy Find the degree of unsaturation in order to set limits on the numbers of p bonds and rings. Use IR data to determine the presence or absence of p bonds. Then continue by using NMR data to define a reasonable structure possibility. • The molecular formula C5H8 corresponds to a degree of unsaturation 5 2. Therefore, all molecules must contain a combination of p bonds and rings adding up to two. Taking each in turn: (a) Four bands are listed in the IR spectrum; those at 1650 and 3100 cm21 correspond unambiguously to alkene C P C and alkenyl C – H stretching motions. The bands at 910 and 1000 cm21 are strongly suggestive of the presence of the terminal – CH P CH2 group (Section 11-8). The NMR spectrum displays two signals in an intensity ratio of 1 : 3. Because the molecule contains eight hydrogens total, this information implies two groups of hydrogens, one containing two and the other six. The two-hydrogen signal, showing a triplet at d 5 2.79, would appear to be due to a – CH2 – unit, coupled to two neighboring hydrogens on the surrounding carbon atoms. The signal for six hydrogens is in the range d 5 4.826.2, typical of alkenyl hydrogens and must therefore correspond to two – CH P CH2 groups. Combining these fragments, we obtain the structure CH2 P CH – CH2 – CH P CH2. (b) The IR data are essentially identical to those in (a), implying the presence of a – CH P CH2 unit. The NMR spectrum shows a five-hydrogen signal at high field and a three-hydrogen signal in the alkenyl region. The three-hydrogen signal is consistent with – CH P CH2, leaving us with a
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C3H5 fragment with NMR signals in the alkane region. Therefore, the molecule must contain a ring as the second degree of unsaturation, leaving
O CH CH2 as the only answer.
(c) The IR data are limited to C P C and alkenyl C – H bands, without additional information from absorptions below 1000 cm21. We again rely most heavily on the NMR spectrum, which again shows two signals, with the one at higher field three times as intense as the one at lower field. The molecule must contain six alkyl and two alkenyl hydrogens. There is no way to construct such a molecule with the correct formula and more than one double bond, so a ring must be present O CH2 CH3 as a second possibility. as well. The most straightforward option is , with The latter is less likely because its NMR would show a clear triplet for the – CH3 group, not a feature noted in the data.
Exercise 11-26
Try It Yourself Propose a structure for another compound with the molecular formula C5H8. Its IR spectrum, however, shows no absorption whatsoever in the region between 1600 and 2500 cm21.
In Summary The degree of unsaturation is equal to the sum of the numbers of rings and p bonds in a molecule. Calculation of this parameter makes solving structure problems from spectroscopic data easier.
THE BIG PICTURE In this chapter, we looked at alkenes, a compound class characterized by the carbon–carbon double bond. In Chapters 7 and 9, we learned that alkenes are prepared synthetically by elimination reactions of haloalkanes and alcohols. In this chapter, we examined these reactions in more depth. We saw that the structure of the base determines what products will form in E2 elimination from haloalkanes. Similarly, the structure of an alcohol undergoing acid-catalyzed dehydration determines what mechanism takes place and how easily it occurs. We also examined three additional methods that organic chemists use for determining molecular structure: calculation of the degree of unsaturation from a molecular formula, infrared (IR) spectroscopy, and mass spectrometry (MS). Mass spectrometry is the most important method for deriving the molecular formula; the degree of unsaturation tells us the total number of rings and p bonds in a molecule; IR spectra help identify the presence of the characteristic bonds of many functional groups. Combined with NMR spectra, this information is normally enough to determine a molecule’s structure. In later chapters, we shall cover ultraviolet (UV) spectroscopy, which provides information on multiple p bonds. In the next chapter, we turn to the reactions of alkenes, to see how their unsaturated structure defines their chemical behavior and the compounds into which they may be converted.
CHAPTER INTEGRATION PROBLEMS 11-27. The relative rates of E2 reaction of bromoethane, 2-bromopropane, and 2-bromo-2-methylpropane with sodium ethoxide in ethanol at 258C (at identical concentrations of substrate and base) are as follows: CH3CH2Br, 1; CH3CHBrCH3, 5; (CH3)3CBr, 40. Explain these data qualitatively. SOLUTION The E2 reaction is a one-step process. Therefore, its rate depends on the activation energy for this step, which in turn is related to the energy of the E2 transition state (Section 11-6). In addition, if some other reaction competes with the E2 process, the observed rate of elimination is diminished to the extent that some of the substrate molecules prefer to react via the alternative reaction pathway. For example, recall that bromoethane, a primary halide, reacts with ethoxide, a basic but nonbulky
Chapter Integration Problems
nucleophile, predominantly by the SN2 mechanism (Section 7-9). Comparison with 2-bromopropane, a secondary substrate, illustrates two points: (i) increased steric hindrance in 2-bromopropane interferes with substitution, reducing the competition from the SN2 mechanism; and (ii) the E2 transition state, which resembles ethene in the case of bromoethane, drops in energy in the case of 2-bromopropane, because it now resembles propene, a more stable alkene. Both arguments may be carried forward to the case of 2-bromo-2-methylpropane, a tertiary halide, because (i) the SN2 mechanism is sterically prohibited from taking place, and (ii) the E2 transition state now resembles 2-methylpropene, an even more stable alkene. Although unimolecular processes are now possible, at moderate to high concentrations of the strong base the tertiary substrate reacts preferentially via the E2 pathway. 11-28. Acid-catalyzed dehydration of 2-methyl-2-pentanol (dilute H2SO4, 508C) gives one major product and one minor product. Elemental analysis reveals that both contain only carbon and hydrogen in a 1 : 2 atom ratio, and high-resolution MS gives a mass of 84.0940 for the molecular ions of both compounds. Spectroscopic data for these substances are as follows: 1. Major product: IR 1660 and 3080 cm21; 1H NMR d 5 0.91 (t, J 5 7 Hz, 3 H), 1.60 (s, 3 H), 1.70 (s, 3 H), 1.98 (quin, J 5 7 Hz, 2 H), and 5.08 (t, J 5 7 Hz, 1 H) ppm. 2. Minor product: IR 1640 and 3090 cm21; 1H NMR d 5 0.92 (t, J 5 7 Hz, 3 H), 1.40 (sex, J 5 7 Hz, 2 H), 1.74 (s, 3 H), 2.02 (t, J 5 7 Hz, 2 H), and 4.78 (s, 2 H) ppm. Deduce the structures of the products, suggest mechanisms for their formation, and discuss why the major product forms in the greater amount. SOLUTION First, we write the structure of the starting material (alcohol nomenclature, Section 8-1) and what we know about the reaction: ðOH
Dilute H2SO4, 50°C
2-Methyl-2-pentanol
This is an acid-catalyzed reaction of a tertiary alcohol (Section 11-7). Even though we may know enough to be able to make a sensible prediction, let us proceed by interpretation of the spectra first, and see if the answer that we get is consistent with our expectations. Both compounds have the same molecular formula — they are isomers. The elemental analysis gives the empirical formula of CH2. The exact mass of CH2 is 12.000 1 2(1.00783) 5 14.01566. Thus, the MS data tell us that the molecular formula is 6(CH2) 5 C6H12, because 84.0940/14.01566 5 6. For the major product, peaks at 1660 and 3080 cm21 in the IR spectrum are in the ranges for the alkene CPC and C – H bond-stretching frequencies (1620 – 1680 and 3050 – 3150 cm21, Table 11-3). Armed with this information, we turn to the NMR spectrum and immediately look for signals in the region characteristic of alkene hydrogens, d 5 4.6 – 5.7 ppm (Table 10-2 and Section 11-4). Indeed, we find one at d 5 5.08 ppm. The information following the chemical shift position (t, J 5 7 Hz, 1 H) tells us that the signal is split into a triplet with a coupling constant of 7 Hz and has a relative integrated intensity corresponding to one hydrogen. The spectrum also shows three signals with integrations of 3 H each; it is usually reasonable to assume that simple signals with intensities of 3 between d 5 0 – 4 ppm indicate the presence of methyl groups, unless information to the contrary appears. Two of the methyl groups are singlets, and the other appears as a triplet. Finally, a signal integrating for 2 H at d 5 1.98 ppm is split into five lines. Assuming that a CH2 group gives rise to the latter signal, we have the following fragments to consider in piecing together a structure: CHPC, 3 CH3, and CH2. The sum of atoms in these fragments is C6H12, in agreement with the formula given for the product and giving us confidence that we are on the right track. There is only a limited number of ways in which to connect these pieces into a proper structure: CH3 A CH3 O CH P C O CH2 O CH3 (ignoring stereochemistry)
and
CH3 A CH3 O CH2 O CH P C O CH3
We can either use our chemical knowledge or, again, turn to the spectroscopy to determine which is correct. With the use of the splitting patterns in the NMR spectrum, a quick decision may be made. Using the N 1 1 rule (Section 10-7), we see that, under ideal conditions, an NMR signal will be split by N neighboring hydrogens into N 1 1 lines. In the first structure, the alkenyl hydrogen is neighbor to a CH3 group and should therefore appear with 3 1 1 5 4 lines, a quartet. The actual spectrum shows this signal as a triplet. Furthermore, this structure contains three methyl groups, but they have
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0, 1, and 2 neighboring hydrogens, respectively, and should appear as one singlet, one doublet, and one triplet — again, in disagreement with the actual spectrum. In contrast, the second structure fits: the alkenyl hydrogen has two neighbors, consistent with the observed triplet splitting, and two of the three CH3 groups are on an alkene carbon lacking a neighboring hydrogen and should be singlets. Using our chemical knowledge, we note that this correct structure has the same carbon connectivity as that of the starting material, whereas the incorrect one would have required a rearrangement. Turning to the minor product, we use the same logic: Again, the IR spectrum shows an alkene C P C stretch (at 1640 cm21) and an alkenyl C – H stretch at 3090 cm21. The NMR spectrum contains a singlet at d 5 4.78, integrating for 2 H: two alkenyl hydrogens. It also shows two methyl signals, as well as two others integrating 2 H for each. Therefore, we have 2 CH3, 2 CH2, and 2 alkenyl H, adding again (with the two alkenyl carbons) to C6H12. Even though many possible combinations can be devised at this point, we can make use of NMR splitting information to close in rapidly on the answer. One of the CH3 groups is a singlet, meaning that it must be attached to a carbon that lacks hydrogens. If we look at the preceding array of fragments, this latter carbon can only be an alkenyl carbon. Thus, we have CH3 – C P C, in which the bold carbon is not attached to hydrogen. Therefore, by process of elimination, both alkenyl hydrogens must be attached to the other alkenyl carbon: A CH3 – C P CH2. The remaining fragments can be attached to this piece in only one way, giving the final structure: CH3 OCH2 O CH2 A CH3 O C P CH2 Thus, we can complete the equation stated at the beginning of the problem as follows: ðOH
Dilute H2SO4, 50°C
Major product
2-Methyl-2-pentanol
Is this result in accord with our chemical expectations? Let us consider the mechanism (Section 11-7). Dehydration of a secondary or a tertiary alcohol under acidic conditions begins with protonation of the oxygen atom, giving a good potential leaving group (water). Departure of the leaving group results in a carbocation, and removal of a proton from a neighboring carbon atom (most likely by a second molecule of alcohol acting as a Lewis base) gives the alkene, overall an E1 process: > ROH p ðOH
H
)OH2
H
(or)
H2O
H > ROH p
The major product is the one with the more substituted double bond, the thermodynamically more stable alkene (Sections 11-5 and 11-7), as is typical in E1 dehydrations.
New Reactions 1. Hydrogenation of Alkenes (Section 11-5) G D C PC D G
H2
A A OC OC O A A H H
Pd or Pt
H 30 kcal mol1
Order of stability of the double bond R
G C P CH2 D R
R G D C PC D G H H
R
R
H G D C PC D G R H
more substituted alkene
Important Concepts
Preparation of Alkenes 2. From Haloalkanes, E2 with Unhindered Base (Section 11-6) CH G 3 G GC P C
CH3CH2ONa, CH3CH2OH
Saytzev rule
G
H A A O C O C O CH3 A A X
HX
More substituted (more stable) alkene
3. From Haloalkanes, E2 with Sterically Hindered Base (Section 11-6) H
HX
G
(CH3)3COK, (CH3)3COH
H
H A A O C O C O CH3 A A X
GC G GC P CH2
Hofmann rule
Less substituted (less stable) alkene
4. Stereochemistry of E2 Reaction (Section 11-6)
@
C G
CG C R ( R X
R R G G P C C G R R
Base
G
R R &
H
HX
G
Anti elimination
5. Dehydration of Alcohols (Section 11-7) A CG A OH
H2SO4, H2O
G G GC P C G
A G CG A H
Most stable alkene is major product Primary: E2 mechanism Secondary, tertiary: E1 mechanism Carbocations may rearrange
Order of reactivity: primary secondary tertiary
Important Concepts 1. Alkenes are unsaturated molecules. Their IUPAC names are derived from alkanes, the longest chain incorporating the double bond serving as the stem. Double-bond isomers include terminal, internal, cis, and trans arrangements. Tri- and tetrasubstituted alkenes are named according to the E,Z system, in which the R,S priority rules apply. 2. The double bond is composed of a s bond and a p bond. The s bond is obtained by overlap of the two sp2 hybrid lobes on carbon, the p bond by interaction of the two remaining p orbitals. The P bond is weaker (,65 kcal mol21) than its s counterpart (,108 kcal mol21) but strong enough to allow for the existence of stable cis and trans isomers. 3. The functional group in the alkenes is flat, sp2 hybridization being responsible for the possibility of creating dipoles and for the relatively high acidity of the alkenyl hydrogen. 4. Alkenyl hydrogens and carbons appear at low field in 1H NMR (d 5 4.6– 5.7 ppm) and 13C NMR (d 5 100– 140 ppm) experiments, respectively. Jtrans is larger than Jcis, Jgeminal is very small, and Jallylic is variable but small. 5. The relative stability of isomeric alkenes can be established by comparing heats of hydrogenation. It decreases with decreasing substitution; trans isomers are more stable than cis. 6. Elimination of haloalkanes (and other alkyl derivatives) may follow the Saytzev rule (nonbulky base, internal alkene formation) or the Hofmann rule (bulky base, terminal alkene formation). Trans alkenes as products predominate over cis alkenes. Elimination is stereospecific, as dictated by the anti transition state. 7. Dehydration of alcohols in the presence of strong acid usually leads to a mixture of products, with the most stable alkene being the major constituent.
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8. Infrared spectroscopy measures vibrational excitation. The energy of the incident radiation ranges from about 1 to 10 kcal mol21 (l < 2.5216.7 mm; n| < 600 – 4000 cm21). Characteristic peaks are observed for certain functional groups, a consequence of stretching, bending, and other modes of vibration, and their combination. Moreover, each molecule exhibits a characteristic infrared spectral pattern in the fingerprint region below 1500 cm21. 9. Alkanes show IR bands characteristic of C–H bonds in the range from 2840 to 3000 cm21. The CPC stretching absorption for alkenes is in the range from 1620 to 1680 cm21, that for the alkenyl C–H bond is about 3100 cm21. Bending modes sometimes give useful peaks below 1500 cm21. Alcohols are usually characterized by a broad peak for the O–H stretch between 3200 and 3650 cm21. 10. Mass spectrometry is a technique for ionizing molecules and separating the resulting ions magnetically by molecular mass. Because the ionizing beam has high energy, the ionized molecules also fragment into smaller particles, all of which are separated and recorded as the mass spectrum of a compound. High-resolution mass spectral data allow determination of molecular formulas from exact mass values. The presence of certain elements (such as Cl, Br) can be detected by their isotopic patterns. The presence of fragment-ion signals in mass spectra can be used to deduce the structure of a molecule. 11. Degree of unsaturation (number of rings 1 number of p bonds) is calculated from the molecular formula by using the equation Degree of unsaturation 5
Hsat 2 Hactual 2
where Hsat 5 2nC 1 2 2 nX 1 nN (disregard oxygen and sulfur).
Problems 29. Draw the structures of the molecules with the following names. (a) 4,4-Dichloro-trans-2-octene (c) 5-Methyl-cis-3-hexen-1-ol (e) (E)-3-Methoxy-2-methyl-2-buten-1-ol
(b) (Z)-4-bromo-2-iodo-2-pentene (d) (R)-1,3-Dichlorocycloheptene
30. Name each of the following molecules in accord with the IUPAC system of nomenclature. (a)
(c) Cl
(b)
OH Br
F (d)
(e)
HO
CF3
Cl
(f)
Cl
I
Cl CH3O (g)
OCH3
(h)
(i)
31. Name each of the compounds below. Use cis/trans and/or E/Z designations, if appropriate, to designate stereochemistry. Cl (a) H3C
G D C PC G D
H
H (b)
CH3
H3C
D G C PC G D
H3C
CH2Cl (c) H
H3C
D G C PC G D
H Cl
32. Of each pair of the following compounds, which one should have the higher dipole moment? The higher boiling point? (a) cis- or trans-1,2-Difluoroethene; (b) Z- or E-1,2-difluoropropene; (c) Z- or E-2,3-difluoro-2-butene 33. Draw the structures of each of the following compounds, rank them in order of acidity, and circle the most acidic hydrogen(s) in each: cyclopentane, cyclopentanol, cyclopentene, 3-cyclopenten-1-ol.
Preparation of Alkenes
G @ C O OH
G G CPC D D
O Cq C O
O Cq C O
O Cq C O
O Cq C O
O Cq CH
H2SO4,
1. X2 2. Base
H2, Lindlar catalyst
Na, liquid NH3
HX
X2
HBr, ROOR
Product: X G G CPC D D
Product: G G CPC D D
&
&
@
G
&
H G @ CO C
section number
X
Strong base (HO, RO, R2N)
7-7, 7-8 7-9, 11-6
7-6, 9-2, 9-3, 9-7, 11-7
13-4
H
H
H G G CPC D D
13-6
G G CP C D D
17-12
18-5, 18-6, 18-7
21-8
Product: O B CH G RCH2CHP C D R
OH,
489
G CPO D G C P PR3 D
O B 2 RCH2CH
Ag2O, H2O,
@
GNR3 GC O C C ( H
&
Product: G O CH P C D
Product:
13-6
H
Markovnikov addition
13-7
X
Product: X G G CPC D D
13-7
Product: O CHP CHBr X
anti-Markovnikov addition
13-8
490
Chapter 11
Alkenes; Infrared Spectroscopy and Mass Spectrometry
34. Assign structures to the following molecules on the basis of the indicated 1H NMR spectra A – E. Consider stereochemistry, where applicable. (a) C4H7Cl, NMR spectrum A; (b) C5H8O2, NMR spectrum B; (c) C4H8O, NMR spectrum C; (d) another C4H8O, NMR spectrum D (next page); (e) C3H4Cl2, NMR spectrum E (next page). 1
H NMR
3H
2H
1H
(CH3)4Si
1H
5.0
4.5
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.0
300-MHz 1H NMR spectrum ppm (δ ) A 1H
NMR 3H
6.0
5.5
5.0
4.5
(CH3)4Si
2H 1H 1H
1H
6.0
5.5
5.0
4.5
4.0
3.5
3.0
2.5
2.0
1.5
1.0
300-MHz 1H NMR spectrum ppm (δ ) B 1H
NMR 3H
1H
1H 4.3
1H
6.0
1H
5.5
5.0
4.5
1H
4.0
3.5
3.0
2.5
2.0
300-MHz 1H NMR spectrum ppm (δ ) C
1.5
(CH3)4Si
1.0
0.5
0.0
0.5
0.0
Problems
1H
Chapter 11
NMR 2H
5.9
5.8
5.7
2.4
2.3
2.2
2H
2H 1H
1H (CH3)4Si
6.0
5.5
5.0
4.5
4.0
3.5
300-MHz
1H
3.0
2.5
2.0
1.5
1.0
0.5
0.0
NMR spectrum ppm (δ )
D
1
H NMR 3H
1H
5.8
6.0
5.5
5.0
4.5
5.7
4.0
3.5
(CH3)4Si
3.0
2.5
2.0
1.5
1.0
0.5
0.0
300-MHz 1H NMR spectrum ppm (δ ) E
35. Explain the splitting patterns in 1H NMR spectrum D in detail. The insets are fivefold expansions. 36. For each of the pairs of alkenes below, indicate whether measurements of polarity alone would be sufficient to distinguish the compounds from one another. Where possible, predict which compound would be more polar.
CH2CH2CH3 H3C E E C P CE and E
H
H
H
H
CH2CH3 H E E C P CE E
E
(c)
E
H3C
H 3C CH2CH2CH3 CH2CH3 CH3CH2 E E E E C P CE C P CE (b) and
CH3CH2
H
H
E
(a)
H CH3 E E C P CE and CH3CH2CHPCH2
H
H
491
492
Chapter 11
Alkenes; Infrared Spectroscopy and Mass Spectrometry
37. Place the alkenes in each group in order of increasing stability of the double bond and increasing heat of hydrogenation. H3C
(a) CH2 PCH2
CH G D 3 C PC D G H3C CH3
H G D C PC D G H3C CH(CH3)2 H
(b)
H3C H3C
G C P CH2 D
CH(CH3)2 G D C PC D G H H3C H
H G D C PC D G (CH3)2CH CH(CH3)2 H
(c) CH3 (d) H3C
CH3
CH2 H 3C
H3C
(e)
38. Write the structures of as many simple alkenes as you can that, upon catalytic hydrogenation with H2 over Pt, will give as the product (a) 2-methylbutane; (b) 2,3-dimethylbutane; (c) 3,3dimethylpentane; (d) 1,1,4-trimethylcyclohexane. In each case in which you have identified more than one alkene as an answer, rank the alkenes in order of stability. 39. The reaction between 2-bromobutane and sodium ethoxide in ethanol gives rise to three E2 products. What are they? Predict their relative amounts. 40. What key structural feature distinguishes haloalkanes that give more than one stereoisomer on E2 elimination (e.g., 2-bromobutane, Problem 39) from those that give only a single isomer exclusively (e.g., 2-bromo-3-methylpentane, Section 11-6)? 41. Write the most likely major product(s) of each of the following haloalkanes with sodium ethoxide in ethanol or potassium tert-butoxide in 2-methyl-2-propanol (tert-butyl alcohol). (a) Chloromethane; (b) 1-bromopentane; (c) 2-bromopentane; (d) 1-chloro-1-methylcyclohexane; (e) (1-bromoethyl)-cyclopentane; (f) (2R,3R)-2-chloro-3-ethylhexane; (g) (2R,3S)-2-chloro-3ethylhexane; (h) (2S,3R)-2-chloro-3-ethylhexane. 42. Referring to the data in Chapter Integration Problem 11-27, predict how the rate of E2 reaction between 1-bromopropane and sodium ethoxide in ethanol would compare with those of the three substrates discussed in that problem under the same reaction conditions. 43. Draw Newman projections of the four stereoisomers of 2-bromo-3-methylpentane in the conformation required for E2 elimination. (See the structures labeled “Stereospecificity in the E2 Reaction of 2-Bromo-3-methylpentane” on p. 466.) Are the reactive conformations also the most stable conformations? Explain. 44. Referring to the answer to Problem 38 of Chapter 7, predict (qualitatively) the relative amounts of isomeric alkenes that are formed in the elimination reactions shown. 45. Referring to the answers to Problem 30 of Chapter 9, predict (qualitatively) the relative yields of all the alkenes formed in each reaction.
CH2
CH2
46. Compare and contrast the major products of dehydrohalogenation of 2-chloro-4-methylpentane with (a) sodium ethoxide in ethanol and (b) potassium tert-butoxide in 2-methyl-2-propanol (tert-butyl alcohol). Write the mechanism of each process. Next consider the reaction of 4-methyl-2-pentanol with concentrated H2SO4 at 1308C, and compare its product(s) and the mechanism of its (their) formation with those from the dehydrohalogenations in (a) and (b). (Hint: The dehydration gives as its major product a molecule that is not observed in the dehydrohalogenations.) 47. Referring to Problem 59 of Chapter 7, write the structure of the alkene that you would expect to be formed as the major product from E2 elimination of each of the chlorinated steroids shown.
A
B
48. 1-Methylcyclohexene is more stable than methylenecyclohexane (A, in the margin), but methylenecyclopropane (B) is more stable than 1-methyl-cyclopropene. Explain.
Problems
49. Give the products of bimolecular elimination from each of the following isomeric halogenated compounds. (b)
&
C6H5
C6H5
Br@ H &
Br@ H
(a)
C6H5
@& H CH3
C6H5 @& H3C H
One of these compounds undergoes elimination 50 times faster than the other. Which compound is it? Why? (Hint: See Problem 41.) 50. Explain in detail the differences between the mechanisms giving rise to the following two experimental results. CH3 ∑ Cl
CH3 ∑
%
Na OCH2CH3, CH3CH2OH
% CH(CH3)2
% CH(CH3)2 100%
CH3 ∑
CH3 ∑
CH3 ∑
Na OCH2CH3, CH3CH2OH
%
Cl % CH(CH3)2
% CH(CH3)2 25%
CH(CH3)2 75%
51. The molecular formulas and 13C NMR data (in ppm) for several compounds are given here. The type of carbon, as revealed from DEPT spectra, is specified in each case. Deduce a structure for each compound. (a) C4H6: 30.2 (CH2), 136.0 (CH); (b) C4H6O: 18.2 (CH3), 134.9 (CH), 153.7 (CH), 193.4 (CH); (c) C4H8: 13.6 (CH3), 25.8 (CH2), 112.1 (CH2), 139.0 (CH); (d) C5H10O: 17.6 (CH3), 25.4 (CH3), 58.8 (CH2), 125.7 (CH), 133.7 (Cquaternary); (e) C5H8: 15.8 (CH2), 31.1 (CH2), 103.9 (CH2), 149.2 (Cquaternary); (f ) C7H10: 25.2 (CH2), 41.9 (CH), 48.5 (CH2), 135.2 (CH). (Hint: This one is difficult. The molecule has one double bond. How many rings must it have?) 52. Data from both ordinary and DEPT 13C NMR spectra for several compounds with the formula C5H10 are given here. Deduce a structure for each compound. (a) 25.3 (CH2); (b) 13.3 (CH3), 17.1 (CH3), 25.5 (CH3), 118.7 (CH), 131.7 (Cquaternary); (c) 12.0 (CH3), 13.8 (CH3), 20.3 (CH2), 122.8 (CH), 132.4 (CH). 53. From the Hooke’s law equation, would you expect the C – X bonds of common haloalkanes (X 5 Cl, Br, I) to have IR bands at higher or lower wavenumbers than are typical for bonds between carbon and lighter elements (e.g., oxygen)? 54. Convert each of the following IR frequencies into micrometers. (a) 1720 cm21 (CPO) (d) 890 cm21 (alkene bend)
(b) 1650 cm21 (CPC) (e) 1100 cm21 (C – O)
(c) 3300 cm21 (O – H) (f) 2260 cm21 (C‚N)
55. Match each of the following structures with the IR data that correspond best. Abbreviations: w, weak; m, medium; s, strong; br, broad. (a) 905 (s), 995 (m), 1040 (m), 1640 (m), 2850 – 2980 (s), 3090 (m), 3400 (s, br) cm21; (b) 2840 (s), 2930 (s) cm21; (c) 1665 (m), 2890 – 2990 (s), 3030 (m) cm21; (d) 1040 (m), 2810 – 2930 (s), 3300 (s, br) cm21.
OH
OH A
B
C
D
56. You have just entered the chemistry stockroom to look for several isomeric bromopentanes. There are three bottles on the shelf marked C5H11Br, but their labels have fallen off. The NMR machine
Chapter 11
493
Alkenes; Infrared Spectroscopy and Mass Spectrometry
Chapter 11
is broken, so you devise the following experiment in an attempt to determine which isomer is in which bottle: You first treat a sample of the contents in each bottle with NaOH in aqueous ethanol, and then you determine the IR spectrum of each product or product mixture. Here are the results: NaOH
(i) C5H11Br isomer in bottle A uuy IR bands at 1660, 2850 ] 3020, and 3350 cm21 NaOH
(ii) C5H11Br isomer in bottle B uuy IR bands at 1670 and 2850 ] 3020 cm21 NaOH
(iii) C5H11Br isomer in bottle C uuy IR bands at 2850 2 2960 and 3350 cm21
(a) What do the data tell you about each product or product mixture? (b) Suggest possible structures for the contents of each bottle. 57. An organic compound exhibits IR spectrum F. From the group of structures below, choose the one that matches the spectrum best. OH
O
O OCH3
H
Transmittance (%)
100
IR 0 4000
3500
3000
2500
2000
1500
600 cm−1
1000
Wavenumber
F 58. The three compounds hexane, 2-methylpentane, and 3-methylpentane correspond to the three mass spectra shown below. Match each compound with the spectrum that best fits its structure on the basis of the fragmentation pattern. Relative abundance
494
100 50
100
MS
50
0 40
60
80
m /z
100
0 0 20
B
MS
50
0 0 20
A
100
MS
40
60 m /z
80
100
0 20 C
40
60
80
100
m /z
59. Assign as many peaks as you can in the mass spectrum of 1-bromopropane (Figure 11-24).
Problems
60. The following table lists selected mass-spectral data for three isomeric alcohols with the formula C5H12O. On the basis of the peak positions and intensities, suggest structures for each of the three isomers. A dash means that the peak is very weak or absent entirely. Relative Peak Intensities m/z 88 87 73 70 59 55 45 42
M1 (M 2 (M 2 (M 2 (M 2 (M 2 (M 2 (M 2
Isomer A
Isomer B
Isomer C
— 2 — 38 — 60 5 100
— 2 7 3 — 17 100 4
— — 55 3 100 33 10 6
1)1 15)1 18)1 29)1 15 2 18)1 43)1 18 2 28)1
61. Determine the molecular formulas corresponding to each of the following structures. For each structure, calculate the number of degrees of unsaturation from the molecular formula and evaluate whether your calculations agree with the structures shown.
(a)
Cl
(b)
(c)
(d) HO
N
(e)
(f )
O
(g)
(h)
62. Calculate the degree of unsaturation that corresponds to each of the following molecular formulas. (a) C7H12; (b) C8H7NO2; (c) C6Cl6; (d) C10H22O11; (e) C6H10S; (f) C18H28O2. 63. A hydrocarbon with an exact molecular mass of 96.0940 exhibits the following spectroscopic data: 1H NMR d 5 1.3 (m, 2 H), 1.7 (m, 4 H), 2.2 (m, 4 H), and 4.8 (quin, J 5 3 Hz, 2 H) ppm; 13 C NMR d 5 26.8, 28.7, 35.7, 106.9, and 149.7 ppm. The IR spectrum is shown below (spectrum G). Hydrogenation furnishes a product with an exact molecular mass of 98.1096. Suggest a structure for the compound consistent with these data.
Transmittance (%)
100
3072 1649
888 0 IR 4000
3500
3000
2500
2000
Wavenumber
G
1500
1000
600 cm−1
Chapter 11
495
496
Chapter 11
Alkenes; Infrared Spectroscopy and Mass Spectrometry
64. The isolation of a novel form of molecular carbon, C60, was reported in 1990. The substance has the shape of a soccer ball of carbon atoms and possesses the nickname “buckyball” (you don’t want to know the IUPAC name). Hydrogenation produces a hydrocarbon with the molecular formula C60H36. What is the degree of unsaturation in C60? In C60H36? Does the hydrogenation result place limits on the numbers of p bonds and rings in “buckyball”? (More on C60 is found in Chemical Highlight 15-1.) 65.
OH Geraniol
You have just been named president of the famous perfume company, Scents “R” Us. Searching for a hot new item to market, you run across a bottle labeled only C10H20O, which contains a liquid with a wonderfully sweet rose aroma. You want more, so you set out to elucidate its structure. Do so from the following data. (i) 1H NMR: clear signals at d 5 0.94 (d, J 5 7 Hz, 3 H), 1.63 (s, 3 H), 1.71 (s, 3 H), 3.68 (t, J 5 7 Hz, 2 H), 5.10 (t, J 5 6 Hz, 1 H) ppm; the other 8 H have overlapping absorptions in the range d 5 1.3 – 2.2 ppm. (ii) 13C NMR (1H decoupled): d 5 60.7, 125.0, 130.9 ppm; seven other signals are upfield of d 5 40 ppm. (iii) IR: | n 5 1640 and 3350 cm21. (iv) Oxidation with buffered PCC (Section 8-6) gives a compound with the molecular formula C10H18O. Its spectra show the following changes compared with the starting material: 1H NMR: signal at d 5 3.68 ppm is gone, but a new signal is seen at d 5 9.64 ppm; 13C NMR: signal at d 5 60.7 ppm is gone, replaced by one at d 5 202.1 ppm; IR: loss of signal at | n 5 3350 cm21; new peak at | n 5 1728 cm21. (v) Hydrogenation gives C10H22O, indentical with that formed on hydrogenation of the natural product geraniol (see margin).
66. Using the information in Table 11-4, match up each set of the following IR signals with one of these naturally occurring compounds: camphor; menthol; chrysanthemic ester; epiandrosterone. You can find the structures of the natural products in Section 4-7. (a) 3355 cm21; (b) 1630, 1725, 3030 cm21; (c) 1730, 3410 cm21; (d) 1738 cm21.
H ∑
OH ∑
67.
Identify compounds A, B, and C from the following information and explain the chemistry that is taking place. Reaction of the alcohol shown in the margin with 4methylbenzenesulfonyl chloride in pyridine produced A (C15H20O3S). Reaction of A with lithium diisopropylamide (LDA, Section 7-8) produces a single product, B (C8H12), which displays in its 1 H NMR a two-proton multiplet at about d 5 5.6 ppm. If, however, compound A is treated with NaI before the reaction with LDA, two products are formed: B and an isomer, C, whose NMR shows a multiplet at d 5 5.2 ppm that integrates as only one proton.
68.
The citric acid cycle is a series of biological reactions that plays a central role in cell metabolism. The cycle includes dehydration reactions of both malic and citric acids, yielding fumaric and aconitic acids, respectively (all common names). Both proceed strictly by enzymecatalyzed anti-elimination mechanisms.
{ H
CO2H H H
H* OH CO2H
Malic acid
OH A HO2CCHCHCH2CO2H A CO2H Isocitric acid
CO2H H Fumarase H2O
fumaric acid
HO
H* CH2CO2H
Aconitase H2O
aconitic acid
CO2H Citric acid
(a) In each dehydration, only the hydrogen identified by an asterisk is removed, together with the OH group on the carbon below. Write the structures for fumaric and aconitic acids as they are formed in these reactions. Make sure that the stereochemistry of each product is clearly indicated. (b) Specify the stereochemistry of each of these products, using either cis-trans or E,Z notation, as appropriate. (c) Isocitric acid (shown in the margin) also is dehydrated by aconitase. How many stereoisomers can exist for isocitric acid? Remembering that this reaction proceeds through anti elimination, write the structure of a stereoisomer of isocitric acid that will give on dehydration the same isomer of aconitic acid that is formed from citric acid. Label the chiral carbons in this isomer of isocitric acid, using R,S notation.
Problems
497
Chapter 11
Team Problem 69. The following data indicate that the dehydration of certain amino acid derivatives is stereospecific.
SO2Cl, pyridine
1. CH3
2
R@ GCO2CH2C6H5 HO & GC G C C ( H 1 R NHCO2CH2C6H5
R2 CO CH2C6H5 G D 2 C PC G D R1 NHCO2CH2C6H5
2. R3N base
1
a b c d
2
* /∑
N
NH H
O
H ∑
O
H ∑
O
!
OH
!
Divide the task of analyzing these data among yourselves to determine the nature of the stereocontrolled eliminations. Assign the absolute configuration (R,S) to compounds 1a – 1d and the E,Z configuration to compounds 2a – 2d. Draw a Newman projection of the active conformation of each starting compound (1a – 1d). As a team, apply your understanding of this information to determine the absolute configuration of the unassigned stereocenter (marked by an asterisk) in compound 3, which was dehydrated to afford compound 4, an intermediate in an approach to the synthesis of compound 5, an antitumor agent.
OCH3 O
SO2Cl, pyridine
1. CH3
N
NH
O
2. R3N base
H#
P1S
NH
H#
NH NHP2
2
NHP
O
/
O
/
H
H 3
4
!
(P1 and P2 are protecting groups)
H ∑
O
O
NH
NH
S
O O
#
H#
N
H
/
P 1S
H
S 5
OCH3 O
R1
R2
CH3 H CH(CH3)2 H
H CH3 H CH(CH3)2
498
Alkenes; Infrared Spectroscopy and Mass Spectrometry
Chapter 11
Preprofessional Problems
CH2
70. What is the empirical formula of compound A (see margin)? (a) C8H14; (b) C8H16; (c) C8H12; (d) C4H7 CH3
71. What is the degree of unsaturation in cyclobutane?
A
(a) Zero; (b) one; (c) two; (d) three 72. What is the IUPAC name for compound B (see margin)?
CH3CH2 H G D C PC G D H3C CH3
(a) (E )-2-Methyl-3-pentene; (b) (E)-3-methyl-2-pentene; (c) (Z)-2-methyl-3-pentene; (d) (Z )-3-methyl-2-pentene 73. Which of the following molecules would have the lowest heat of hydrogenation?
B
(a)
CH3 CH3
(b)
CH3
(c)
CH3
CH2
(d)
CH3
CH3
CH3
74. A certain hydrocarbon containing eight carbons was found to have two degrees of unsaturation but no absorption bands in the IR spectrum at 1640 cm21. The best structure for this compound is (a)
(b)
(c) CH2CH2CH P CH2
(d)
CH3
[
]
Interlude
Solving Problems in Organic Chemistry
With approximately one-third of the year’s course in organic chemistry behind you, let’s take stock of your ability to solve problems, and look to possible remedies for the difficulties you may have encountered. This interlude has the following structure: Understanding the Question Types of Problems in Organic Chemistry A General Approach to Problem Solving: The “WHIP” Strategy Solving Problems That Ask “What” Nomenclature Acidity Energy Stability Spectroscopy Solving Problems That Ask “How” and/or “Why” What is the product of a reaction? How does the product form? What reagent(s) do you need to convert one molecule into a specific other?
Understanding the question Perhaps the most common statement made by students coming to office hours is “I don’t understand!” When we ask them to elaborate, we frequently find that it is the question that they do not understand — never mind the solution. In fact, this situation is common in science and science-related classes. Exams not only test students’ knowledge; they also evaluate their ability to understand a written question and to rearrange it into a form that can be solved.
Types of problems in organic chemistry What kind of question is it? It is somewhat of an oversimplification, but there are essentially two categories of questions in organic chemistry. Categorizing the question is the first step in understanding it and defining what is being asked of you.
The “What?” question • • • • •
What is the IUPAC name? Which is the strongest acid? Calculate DH8 and determine which way the equilibrium lies. What is the most stable conformation? Interpret the following spectra of a molecule and suggest a possible structure for it. 499
500
Chapter 11
Solving Problems in Organic Chemistry
These are all “What?” questions. They ask you (1) to recall information and (2) to apply it to a relatively uncomplicated situation. As we shall soon see, approaches to solving “What” questions are usually straightforward (which does not necessarily mean that they are easy).
The “How” or “Why” question What about the second type of questions? These are the ones that focus on one or more chemical reactions from any of several perspectives: • What is the product? • What is the mechanism by which it forms? • What reagent(s) do you need to convert one molecule into a specific other? These may sound like “What” queries, but they really have a lot more to do with “How” and “Why” things happen in organic chemistry. A more sophisticated approach is usually needed, compared to the typical “What” enquiry. You can often recognize a “How” or a “Why” question by asking yourself: “Is something happening?” If the answer is yes, then the question is very likely asking you “How” or “Why.” Before we proceed to some examples, let us present a general approach to problem solving.
A general approach to problem solving: the “WHIP” strategy To succeed in organic chemistry requires time, commitment, and hard work. It also requires the development of organized problem-solving skills. To “WHIP” the challenge of solving a problem, proceed step by step: • What does the problem ask? Read the text describing the problem, and don’t start working on the answer until you understand everything it wants to know. Is it unclear? Read the question slowly and out loud to yourself. Look for key words that characterize the type of problem. Is it just looking for a piece of information (What?), or does it ask you to work out something more involved (How? Why?)? • How to start? What is your starting point? Where do you need to go? Ponder both of these questions to figure out what you need to accomplish. For example, to deduce the product of a reaction, you must identify both the relevant structural features of the starting material(s) (such as functional groups) and the chemical nature of the reagent that is acting on it (electrophile? nucleophile? oxidizing or reducing agent?). Evaluate in a general sense what might result. • Information needed? Have you forgotten something critical, such as the identity of a reagent? Or are you uncertain about a detail that might be important, such as primary vs secondary vs tertiary for a reacting carbon? Look it up — in the book, in your notes, or just ask — before you get too far into trying to work the problem. When you “solve” exercises incorrectly because of faulty information, it puts you in the position of having to “unlearn” a wrong approach in addition to still having to recognize the correct one. Under the stress of an exam situation, confusion becomes much more likely. • Proceed logically — no shortcuts! Do not skip steps in formulating a solution to a problem! Be pedantic in your accuracy: Write out each structure fully, if necessary in its Lewis form; complete your arrow pushing; add reaction arrows and reagents for each step; “doodle” on a separate piece of paper— organic chemistry is very visual and even a wrong structure, when drawn, may trigger your mind to lead you to the right one. Practice your problem-solving skills in front of others. Finally, trust the various processes that you have learned to lead you to the answers.
Solving Problems in Organic Chemistry
Solving problems that ask “What?” • Nomenclature: Nomenclature problems may ask you to give a name to a structure or to draw a structure that corresponds to a name. For example: Problem I-1. Provide the IUPAC name for the molecule given below.
A OH
A Br
Your What is straightforward: Name the molecule. How to start? Before naming a structure, you need to “read” it to determine connectivity and (if applicable) stereochemistry. Information needed? You’ll need to know the rules of choosing a parent chain (or ring), numbering it, and naming its substituents. Thus: A methyl substituent Another methyl substituent 1
2
3
A OH
An alcohol
4
5
A Br
6
“Read” the structure: 1. Six-carbon longest chain. 2. Four substituents (identified as shown). 3. Hydroxy function takes precedence over the others and defines numbering the stem.
A bromo substituent
Proceed, step by step. Apply the naming rules: (1) It is a “hexanol.” (2) Numbering begins at the end closer to the (highest-precedence) OH group. (3) Use alphabetical ordering to construct the name (“di-” is not counted — it is just a multiplier for “methyl”). The answer: 5-Bromo-2,2-dimethyl-3-hexanol • Acidity: Problem I-2. Which is the stronger acid, CH3NH2 or CH3OH? Again, a straightforward What. How to start? When you are asked to evaluate the strength of an acid, look at its conjugate base. The most stable conjugate base (on the basis of atom electronegativity as well as stabilizing inductive and resonance effects) corresponds to the strongest acid (Section 2-2). So, draw all the possible conjugate bases by removing a proton (H1) from every location in each molecule:
ðCH2NH2
CH3NHð
ðCH2OH
CH3Oð
Which is the most stable conjugate base? These four species have negative charges on three different atoms: C, N, and O. How do these atoms differ? Information needed! The three atoms differ in their respective electronegativity: O . N . C (Table 1-2). The more electronegative an atom is, the better it is at attracting electrons. The most electronegative atom therefore stabilizes best the extra electron pair and negative charge in the corresponding conjugate base. Finally, Proceed logically: (1) Oxygen is the most electronegative in the trio, so, with the extra electron pair and negative charge on O, CH3O :2 is the most stable of the four conjugate bases above. (2) Given that conclusion, CH3OH must be the stronger acid, and the hydrogen on the oxygen must be the most acidic of its four hydrogen atoms. • Energy: Energy change is one of the fundamental principles governing chemical reactions. Breaking bonds requires energy input; making bonds releases energy. DH8 5 (energy in) 2 (energy out). The equilibrium is favored in the direction of reaction for which DG8 is negative, which will often be determined by a negative DH8 (at least when DS8 is small).
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Problem I-3. Hydrogenation of alkenes typically displays DH8 5 230 kcal mol21. Which of the following statements are always true? (a) Mixing an alkene with H2 results in a fast reaction; (b) reaction of an alkene with H2 is exothermic; (c) all alkenes react equally favorably with H2. What? This is a three-part true – false question. You must think about each statement individually. (a) Start: Does energetic (thermodynamic) favorability necessarily determine how fast a reaction will go? Information: No. What does? Remember kinetics? Activation energies (Sections 3-3 and 11-5)? So, Proceed: False. (b) Start: Define exothermic. Information: Exothermic corresponds to a negative DH8 (Section 2-1). Proceed: This statement is true. (c) Start: Read carefully! The problem says equally favorably, which means that, if the statement is true, the hydrogenations of all alkenes will have exactly the same DH8. Information: Look at Figure 11-12. They don’t. Proceed: False! So the only true statement is (b). • Stability: Problem I-4. Rank the following four isomeric hydrocarbons in order of decreasing stability: (a) 1-cyclopropylpentane; (b) cis-1-ethyl-2-propylcyclopropane; (c) ethylcyclohexane; (d) cis-1,4-dimethylcyclohexane. What does the problem ask? To establish an approximate order of energy content for the four molecules and thus estimate relative stability. How to start? Careful structural drawings will help you recognize what features may stabilize or destabilize each hydrocarbon relative to the others. Notice that this problem includes a nomenclature component: Each name must be converted to a structure. Finally, to make the comparison fair, we need to draw the most stable possible conformation of each one. (a)
(b)
(c)
(d)
Information: The lower the energy content of a structure, the more stable it is. Steric crowding raises energy content and decreases stability (Section 2-8). So, a staggered conformation is more stable than an eclipsed one; anti is more stable than gauche; equatorial is more stable than axial (Section 4-4). In each case, the more stable option has less steric crowding. Note any other destabilizing effects (such as bond-angle strain). (a)
(b)
Serious bond-angle strain
(c)
(d) One axial substituent
Steric crowding (cis, eclipsed)
Proceed: As shown above, we identify the structural features that affect stability. How important is each one? Find the quantitative information: Bond-angle strain in cyclopropanes is the largest destabilizing factor (27.6 kcal mol21; Section 4-2). Next we add the enforced eclipsing in (b), which is at least as bad as the 4.9 kcal mol21 in the highest-energy conformation of butane (Figure 2-13). Cyclohexanes do not suffer from bond-angle strain, but (d) cannot avoid one axial substituent (1.7 kcal mol21, Table 4-3). With these data in hand, apply the logic: Higher energy content 5 less stable. The stability order is therefore (c) most stable, then (d), then (a), and finally (b) least stable. (If needed, you could even give approximate energy differences by adding the energy-content increments for each compound.)
Solving Problems in Organic Chemistry
• Spectroscopy: Spectroscopy is our most powerful tool for determining the structure of an organic compound. Solving spectroscopic problems is rather like identifying and combining the pieces of a puzzle to construct a final “picture.” As in solving any other sort of puzzle, there are efficiencies of strategy that you can learn to employ. Problem I-5. What is the structure of a molecule that displays the following spectroscopic data: 1 H NMR d 5 1.30 (broad s, 1 H), 3.89 (d, J 5 7.8 Hz, 2 H), 5.49 (t, J 5 7.8 Hz, 1 H) ppm; IR 1670 (weak), 3310 (strong, broad) cm21; MS m/z 5 88 (M1• ). A straightforward What. Start by recognizing what each type of spectrum reveals. Ideally, we would like to know: (a) Composition (molecular formula), either given or from mass spectrometry (b) Functional groups present, from an infrared spectrum (c) Number and location of hydrogens from the 1H NMR spectrum Information: (a) The composition of the molecule is not given. You do have mass data, a molecular weight of 88, and you could try to figure out by trial and error some combinations of common atoms that would add up to a molecular mass of 88, but that would be inefficient. Put this information aside for now and turn to the IR and 1H NMR spectra. (b) The IR spectrum is diagnostic of the OH functional group because of the strong, broad absorption at 3310 cm21; the weaker one at 1670 cm21 suggests a CPC double bond (Section 11-8). (c) Now for the 1H NMR spectrum: The broad singlet at d 5 1.30 ppm is almost certainly due to the OH hydrogen (Section 10-8). The signal at d 5 5.49 ppm gives away the presence of an alkenyl hydrogen. Finally, at d 5 3.89 ppm we have the signal for two hydrogens evidently deshielded by a neighboring OH group (Table 10-2). Proceed, logically, and step by step: We have some molecular fragments—our “puzzle pieces.” How do we put them together? The pieces are an OH—evidently attached to a CH2— and a double bond bearing a hydrogen: CPCH. On the basis of the integrations and spin–spin splittings of the 1H NMR signals (the alkenyl H is split into a triplet and the CH 2OH into a doublet, so they must be neighbors; Section 10-7), we must have something like the following substructure: CH2 OOH (No splitting from H on oxygen D because of fast exchange; Section 10-8) PC G Spin–spin splitting observed H This is obviously not a whole molecule. Furthermore, adding up the atomic masses gives us only 44. Atoms are clearly missing, but there are no additional signals in the 1H NMR spectrum. What to do? The molecular weight of this substructure gives us a strong hint: It is exactly half the value observed for the entire molecule. Thus, the molecule is symmetrical: It contains another set of hydrogen atoms (and carbons and oxygens) in identical environments as the ones already shown. This line of reasoning gets us to this structure:
HO O CH2 H
G D CPC D G
CH2 OOH
Three identical sets of hydrogens
H
Does this answer fit the data? The composition is C4H8O2, giving a mass of 48 1 8 1 32 5 88. The IR spectrum fits well. How about the 1H NMR spectrum? The integrations given were 1 H, 2 H, and 1 H, suggesting a molecule with only four hydrogens, but we saw above that this interpretation was a dead end. Recall, however, that NMR integrations give us only relative numbers of hydrogen atoms in each environment (Section 10-6). The values given above are one-half of the actual numbers of hydrogens responsible for each signal. It is not uncommon to be given a spectroscopic problem involving a simple organic molecule with such symmetry, so you should be aware of the possibility.
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At this stage, we have established that all the data fit our suggested structure. Are there any alternatives? The answer is yes, the trans isomer. However, careful scrutiny of Section 11-8 reveals that trans alkenes exhibit a sharp IR bending vibration at 970 cm21. Since such a band is not given, our answer stands as the best available. A final note: The problem would have been much easier if we had been given the molecular composition. Even an exact molecular mass from a high-resolution mass spectrum would have been helpful (Section 11-9): This molecule’s exact mass of 88.0524 tells us that its formula is C4H8O2 (by adding atomic masses from Table 11-5), as opposed to other alternatives with an approximate mass of 88, such as C4H12N2 (88.1002) or C5H12O (88.0889).
Solving problems that ask “How?” and/or “Why?” Let us turn now to the second type of question. These are the ones that address chemical reactions: the processes through which molecules change. • What is the product of a reaction? Even if you have memorized everything you can, you will still need to recognize a reaction type and apply the relevant mechanism to arrive at a product structure. This kind of question is rarely just “What?” For example, consider the following mock exam question and follow its analysis: Problem I-6. (5 points) Write the major product(s) of the following reaction. Br NaOH, CH3CH2OH
At first glance, this is a simple What-type problem. But is it? Let us see. How to start? Identify the starting organic molecule and the reagent by the compound categories to which they belong. See if that leads you to a solution. Hydroxide—good nucleophile—do you immediately think of nucleophilic substitution?
OH
Br NaOH, CH3CH2OH
Do you write
as your answer and quickly move on?
If so, you just lost most of those 5 points.
Haloalkane
Slow down! Take a second look: The substrate is tertiary, and 2OH is a strong base as well. Information: Table 7-4 tells you how a reaction between a tertiary substrate and a strong base proceeds: through elimination, by the E2 mechanism. But you are still not finished. More than one isomer can be formed. Section 11-6 covers the regiochemistry of elimination: Unhindered bases (such as hydroxide) give the more stable alkene (Saytzev rule). Now, finally, you can Proceed: Br NaOH, CH3CH2OH
The only full-credit answer
Hofmann (minor)
Saytzev (major)
• How does the product form? Why? Explain! These are the buzz words for “Write the mechanism.” If you have not addressed the reaction mechanism, you have not answered a “How” or a “Why” question, nor have you “explained” anything about a chemical reaction. Individual reaction steps are the words in the language of organic chemistry—the vocabulary. However, you need grammar in order to put the words together. The mechanisms constitute that grammar.
Solving Problems in Organic Chemistry
Chapter 11
Problem I-7. Which reaction below proceeds faster, (a) or (b)? What is the product? Explain. Cl
Cl
%
Excess NaOH, CH3CH2OH
(a)
´
(b)
Excess NaOH, CH3CH2OH
What does the question ask? “Explain!” There’s your cue: This is a mechanism problem. How to start? As in the preceding example, characterize the substrates: Here they are secondary chloroalkanes, with additional steric hindrance on the adjacent carbon. Again, the reagent is a strong nucleophile and base. Information: Displacement by SN2 is unlikely; as in Problem I-6, E2 should be favored (Table 7-4). What next? Proceed, logically: Use the question as a clue to how to proceed. Why should these two reactions proceed at different rates? How do the substrates differ? Answer: stereochemically. So, redraw both starting materials in the more instructive cyclohexane chair forms: H3C (a)
Cl
H3C Axial Cl is trans and anti to hydrogen atoms on the adjacent carbons
CH3
Cl
(b)
CH3
H H
Equatorial Cl is cis and gauche to hydrogen atoms on the adjacent carbons
H H
The E2 mechanism proceeds most readily from a geometry that aligns the leaving group and the hydrogen atom to be removed from an adjacent carbon atom in an anti manner (Section 11-6). Only the substrate in reaction (a) is capable of attaining this arrangement. Transformation (a) will therefore be the faster one, eliminating to give the cyclic alkene as the major product. Should you be asked specifically to show a mechanism for this transformation, base your answer on the Information relating to the E2 process presented in Section 7-7 and Figure 7-8. The carbon atom bearing the Cl is polarized d1. The reaction takes place in a single step, by simultaneous movement of three electron pairs. The first originates from hydroxide ion, the base, which attacks the hydrogen atom b to the leaving group. The second emanates from the breaking C2H bond and moves toward the d1 carbon atom to form the new p bond. The third leaves with the departing chloride ion, the leaving group. Using curved arrows to depict electron flow, we draw: H3 C
Cl δ+
CH3 H H
H3C q CH3
š HOð
• What reagent(s) do you need to effect the conversion of one molecule into a specific other? Synthesis requires you to have a command of your reaction vocabulary so that you know whether, and exactly how, one kind of compound can be transformed into another. The Reaction Road Maps preceding the end-of-chapter Problem sections in many of the chapters are designed to help you organize your studying for this purpose. Retrosynthetic analysis (Section 8-9) provides a framework for designing solutions for synthesis problems.
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O B
Problem I-8. Design a synthesis of (3-hexanone), beginning with any organic molecules containing no more than three carbons. What does the problem ask? Actually, several things. The product is a ketone, so you have to make that functional group. It has six carbons, and you are only allowed starting molecules with three, so you have to construct (at least) one carbon–carbon bond. You also have to decide what starting materials to use, and what reactions to do with them. That’s quite a lot! How to start (1): You need to make a ketone. Do you know any reactions that generate ketones? Information (1): So far, only one — oxidation of a secondary alcohol with a chromium(VI) reagent (Section 8-6). Proceed (1): Retrosynthetically, we say that the ketone derives from a secondary alcohol (below, left). An actual reaction that provides the ketone is shown below on the right. It must be the final step in your answer: Retrosynthetic Analysis O B
Final Synthesis Step (Forward Direction) OH A
OH A
O B
Na2Cr2O7, H2SO4
Make sure you include actual reagents above the reaction arrow
This arrow means “derives from”
What does the problem ask next? You are now faced with a new synthesis problem: making the starting material for this reaction (3-hexanol). It is a secondary alcohol. How to start (2): Ask yourself, do you know any ways to make secondary alcohols? Information (2): You know two ways, reduction of a ketone with a hydride reagent (Section 8-6) and addition of a Grignard reagent to an aldehyde (Section 8-8). Proceed (2): Check these two options for feasibility. Reduction of a ketone is not a useful method here, because the ketone we would need to reduce to make 3-hexanol is the one we are trying to make: 3-hexanone. We would just be going around in circles. Turning to the second route, consider combinations of Grignard reagents and aldehydes that will give us 3-hexanol. Thinking retrosynthetically, we find that there are two combinations, because we can make either the carbon – carbon bond to the left of the alcohol carbon (bond a), or the one to the right (bond b): Retrosynthetic Analysis of 3-Hexanol: Two Options
a
OH A
O B CH3CH2MgX HCCH2CH2CH3
a
Grignard reagent
b b
O B CH3CH2CH XMgCH2CH2CH3 Aldehyde
Aldehyde has four carbons
Aldehyde
Both are 3-carbon compounds
Grignard reagent
We can make bond a, but one of the starting materials is a four-carbon aldehyde. Because we are allowed to use only starting materials with three carbons or less, we would have to build that aldehyde from smaller molecules. On the other hand, we can construct bond b from two three-carbon components of the Grignard reaction. The required Grignard reagent is generated from a haloalkane, such as 1-bromopropane, and magnesium (Section 8-7). The final synthetic scheme can be presented now as follows:
CH3CH2CH2Br
1. Mg, (CH3CH2)2O 2. CH3CH2CHO 3. H, H2O
OH A
Na2Cr2O7, H2SO4
O B
C H A P T E R
[
Reactions of Alkenes
TWELVE
] F F F F F F F F
F F
F F
F F
F F
T
ake a look around your room. Can you imagine how different it would appear if every polymer-derived material (including everything made of plastic) were to be removed? Polymers have had an enormous effect indeed on modern society. The chemistry of alkenes underlies our ability to produce polymeric materials of diverse structure, strength, elasticity, and function. In the later sections of this chapter, we shall investigate the processes that give rise to such substances. They are, however, only a subset of the varied types of reactions that alkenes undergo. Additions constitute the largest group of alkene reactions and lead to saturated products. Through addition, we can take advantage of the fact that the alkene functional group bridges two carbons, and we can elaborate on the molecular structure at both carbons. Fortunately for us, most additions to the p bond are exothermic: They are almost certain to take place if a mechanistic pathway is available. Beyond our simple ability to add to the double bond, other features further enhance the usefulness and versatility of addition reactions. Many alkenes possess defined stereochemistry (E and Z) and, as we shall see in our discussions, many of their addition reactions proceed in a stereochemically defined manner. By combining these facts with the realization that additions to unsymmetric alkenes may also take place regioselectively, we shall find that we have the ability to exert a large measure of control over the course of these reactions and, consequently, the structures of the products that are formed. This control has been exquisitely refined in applications toward the synthesis of enantiomerically pure pharmaceuticals (see Chemical Highlights 5-4 and 12-2, and Section 12-2). We begin with a discussion of hydrogenation, focusing on the details of catalytic activation. Then we turn to the largest class of addition processes, those in which electrophiles such as protons, halogens, and metal ions are added to the alkene. Other additions that will contribute further to our synthetic repertoire include hydroboration, several oxidations (which can lead to complete rupture of the double bond if desired), and radical reactions. Each of these transformations takes us in a different direction; the Reaction Summary Road Map at the end of the chapter provides an overview of the interconversions leading to and from this versatile compound class.
12-1 Why Addition Reactions Proceed: Thermodynamic Feasibility The carbon – carbon p bond is relatively weak, and the chemistry of the alkenes is governed largely by its reactions. The most common transformation is addition of a reagent A – B to give a saturated compound. In the process, the A – B bond is broken, and A and B form
Modern electronic devices (for example, the iPhone shown in the photo) rely on batteries that can be recharged thousands of times. Polymerization of 1,1-difluoroethene gives a high-performance membrane (polyvinylidene fluoride) that allows charge to flow between the individual cells of lithiumion batteries but protects the battery from internal shortcircuits and catastrophic failure. Li-ion polymer batteries offer significant advantages in weight and energy capacity over earlier Li-ion and nickelbased designs. They are seeing increased use in consumer electronics such as cellular phones and laptop computers and are under development for applications in hybrid vehicles.
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single bonds to carbon. Thus, the thermodynamic feasibility of this process depends on the strength of the p bond, the dissociation energy DH A8 –B, and the strengths of the newly formed bonds of A and B. Addition to the Alkene Double Bond G D C PC D G
A
B
H ?
A
B
C
C
Recall that we can estimate the DH 8 of such reactions by subtracting the combined strength of the bonds made from that of the bonds broken (Section 3-4): ¢H° 5 (DH°p bond 1 DH°A2B ) 2 (DH°C2A 1 DH°C2B ) in which C stands for carbon. Table 12-1 gives the DH 8 values [obtained by using the data from Tables 3-1 and 3-4 and by equating the strength of the p bond to 65 kcal mol21 (272 kJ mol21)] and the
Table 12-1 Estimated DH 8 (all values in kcal mol21) for Additions to Ethenea CH2 P CH2
1
DH8P bond
AOB
uy
DH8A–B
A B A A HO C O C OH A A H H
DH8A–C
DH8B–C
,DH8
Hydrogenation
CH2 P CH2 65
1
HOH 104
uy
H H A A CH2 O CH2 101 101
233
Bromination
CH2 P CH2
1
65
.. .. : Br .. OBr .. :
uy
46
š ð ðBr šð ðBr A A H O C OO C O H A A H H 70 70
229
Hydrochlorination
CH2 P CH2
1
65
.. HOCl .. :
uy
103
šð H ðCl A A HO C O C OH A A H H 101 84
217
Hydration
CH2 P CH2 65 a
1
.. HOOH .. 119
uy
H ðš OH A A HO C O C OH A A H H 101 94
These values are only estimates: They do not take into account the changes in C – C and C – H s-bond strengths that accompany changes in hybridization.
211
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Chapter 12
509
estimated DH 8 values for various additions to ethene. In all the examples, the combined strength of the bonds formed exceeds, sometimes significantly, that of the bonds broken. Therefore, thermodynamically, additions to alkenes should proceed to products with release of energy. Exercise 12-1 Calculate the DH 8 for the addition of H2O2 to ethene to give 1,2-ethanediol (ethylene glycol) [DH 8HO–OH 5 49 kcal mol21 (205 kJ mol21)].
12-2 Catalytic Hydrogenation The simplest reaction of the double bond is its saturation with hydrogen. As discussed in Section 11-5, this reaction allows us to estimate the relative stability of substituted alkenes from their heats of hydrogenation. The process requires a catalyst, which may be either heterogeneous or homogeneous — that is, either insoluble or soluble in the reaction medium.
Hydrogenation takes place on the surface of a heterogeneous catalyst The hydrogenation of an alkene to an alkane, although exothermic, does not take place even at elevated temperatures. Ethene and hydrogen can be heated in the gas phase to 200 8C for prolonged periods without any measurable change. However, as soon as a catalyst is added, hydrogenation proceeds at a steady rate even at room temperature. The catalysts frequently are insoluble materials such as palladium (e.g., dispersed on carbon, Pd – C), platinum (Adams’s* catalyst, PtO2, which is converted into colloidal platinum metal in the presence of hydrogen), and nickel (finely dispersed, as in a preparation called Raney† nickel, Ra – Ni).
Scanning tunneling microscope (STM) image of a catalytic Pt surface. STM provides pictures at atomic resolution (the bar at the bottom right indicates 5 nm 5 50 Å). You can see the highly ordered patterns of parallel arrays of Pt atoms in brown. The yellow “crosslines” indicate steps on the surface, on which much of the catalytic activity takes place. (Courtesy Professor Gabor A. Somorjai and Dr. Feng Tao, University of California at Berkeley.)
*Professor Roger Adams (1889 – 1971), University of Illinois at Urbana – Champaign. † Dr. Murray Raney (1885 – 1966), Raney Catalyst Company, South Pittsburg, Tennessee.
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Reactions of Alkenes
First hydrogen transfers to carbon of surface-bound alkene Hydrogen atoms bound to metal atoms on catalyst surface
H
H
H
H
C
H HH
HH H
H C
CH2 CH2
H
CH2 CH2
Catalyst surface
Alkane product released from surface
H
Second hydrogen transfers, completing hydrogenation
H
H
H
H
H C
H
C
C
H
H
H
H
C
H
H O H has added syn
Figure 12-1 Mechanism of catalytic hydrogenation of ethene to produce ethane. The hydrogens bind to the catalyst surface and are delivered to the carbons of the surface-adsorbed alkene.
The major function of the catalyst is the activation of hydrogen to generate metal-bound hydrogen on the catalyst surface (Figure 12-1). Without the metal, thermal cleavage of the strong H – H bond is energetically prohibitive. Solvents commonly used in such hydrogenations include methanol, ethanol, acetic acid, and ethyl acetate.
Exercise 12-2
Working with the Concepts: Cis-trans Isomerization Under Alkene Hydrogenation Conditions During the catalytic hydrogenation of the cis double bond in oleic acid, a naturally occurring fatty acid, the formation of some of the trans isomer is observed. Explain. H
H H
H
R COOH
(CH2)7COOH R C8H17
Oleic acid
Strategy Examine the mechanism in Figure 12-1. Look for a pathway that could (1) allow rotation about the bond between the originally doublebonded carbons followed by (2) regeneration of a double bond with a trans configuration. Solution • The mechanism possesses two key features that are applicable to this problem. First, each of the first three steps is reversible. Second, the two hydrogen atoms add to the carbons of the original double bond one at a time. Let us see how we can make use of these characteristics to define an isomerization pathway.
12-2 Catalytic Hydrogenation
Chapter 12
511
• Write out the mechanism, beginning with the binding of the alkene to the catalyst surface. (Caution: Do not combine steps! Each step
in any mechanism should be written separately. Otherwise, you may miss a critical intermediate.) The mechanism, through the addition of the first hydrogen, goes as follows: /∑
H
R H /∑
H / (CH ) COOH 2 7
∑
Oleic acid
H H A A
H (CH2)7COOH
∑/
H R H H A A
H A
Single-hydrogen transfer intermediate
• In accordance with the general description in Figure 12-1, we know what happens next: The second hydrogen transfers over, and the hydrogenated product [stearic acid, CH3(CH2)16COOH] is released. But notice that this final step is irreversible. Once it occurs, there is no way back to any alkene, cis or trans. Therefore, the cis-trans isomerization pathway must involve only the species we have written in the partial mechanism above. • Look closely at the single-hydrogen transfer intermediate. It has a single bond between the two former alkene carbons that is conformationally flexible (Section 2-8). Carrying out a 1208 rotation and reversing the hydrogen transfer that led to this species gives us an alkene with a trans double bond: trans
/∑
R H H H A A
H / (CH ) COOH 2 7
H A
H (CH2)7COOH
∑/
Rotate 120
H
H R /∑
∑
H A
H / (CH ) COOH 2 7
∑
H
R H /∑
• Release of this alkene from the catalyst surface completes the isomerization. This transformation is exactly what occurs during the partial hydrogenation of vegetable oils, which is the commercial process for the formation of margarine and other partially saturated fats. The trans double bonds in these products — the so-called trans fatty acids — give rise to various adverse health effects that are described in more detail in Chemical Highlight 19-3.
Exercise 12-3
Try It Yourself During the course of the catalytic hydrogenation of 3-methyl-1-butene, some 2-methyl-2-butene is observed. Explain.
Hydrogenation is stereospecific An important feature of catalytic hydrogenation is stereospecificity. The two hydrogen atoms are added to the same face of the double bond (syn addition). For example, 1-ethyl-2-methylcyclohexene is hydrogenated over platinum to give specifically cis-1-ethyl2-methylcyclohexane. Addition of hydrogen can be from above or from below the molecular plane with equal probability. Therefore, each stereocenter is generated as both image and mirror image, and the product is racemic. H CH2CH3
H CH3
H CH3CH2
∑ %? /
CH3
H2, PtO2, CH3CH2OH, 25C
? /∑ š
CH2CH3
H H3C
82% 1-Ethyl-2-methylcyclohexene
cis-1-Ethyl-2-methylcyclohexane (Racemic)
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Reactions of Alkenes
Chiral catalysts permit enantioselective hydrogenation
P
P Rh O OCH3 BF4
Rh-(R,R)-DIPAMP BF4
HO CO2H NH2
%
∞
∞
H HO
%
COC f f
H
H
L-DOPA
(shown in the eclipsed conformation)
When steric hindrance inhibits hydrogenation on one face of a double bond, addition will take place exclusively to the less hindered face. This principle has been used to develop enantioselective or so-called asymmetric hydrogenation. The process employs homogeneous (soluble) catalysts, consisting of a metal, such as rhodium, and an enantiopure chiral phosphine ligand, which binds to the metal. A typical example is the Rh complex of the diphosphine (R,R)-DIPAMP (margin). After coordination of the alkene double bond and a molecule of H2 to rhodium, hydrogenation occurs via syn addition, just as in the case of insoluble metal catalysts. However, the asymmetrically arranged bulky groups in the chiral ligand prevent addition of hydrogen to one of the faces of the double bond, resulting in the formation of only one of the two possible enantiomers of the hydrogenation product (see also Chemical Highlight 5-4). This approach has proven to be a powerful method for synthesis of enantiomerically pure compounds of pharmaceutical significance. The industrial synthesis of l-DOPA (margin), the anti-Parkinson’s disease agent, employs in its key step the asymmetric hydrogenation of the alkene shown to give exclusively the necessary S stereoisomer of the reduction product.
/∑
H2, Rh-(R,R)-DIPAMP BF4 Enantiopure catalyst
/?
∑∑
O
#
P
H3CO Rh O H P
^
H!
OCCH3 B O
!
/∑
iP O
H3C
H3CO CH3CO B O
`
O HN Oli CH3
NH
H3CO H3CCO B O
CO2H O C N l COC H CH3 f (S) i H H H Δ
%
CP C
CPC
O
H3CO
CO2H
∑/
H
CO2H
∑/
H
%
∑/
/∑
H3COO
Single enantiomer (shown in the eclipsed conformation)
OCH3
Hydrogen is transferred from Rh to only one face of the double bond
Exercise 12-4 Catalytic hydrogenation of (S)-2,3-dimethyl-1-pentene gives only one optically active product. Show the product and explain the result. [Hint: Does addition of H2 either (1) create a new stereocenter or (2) affect any of the bonds around the stereocenter already present?]
In Summary Hydrogenation of the double bond in alkenes requires a catalyst. This transformation occurs stereospecifically by syn addition and, when there is a choice, from the least hindered side of the molecule. This principle underlies the development of enantioselective hydrogenation using chiral catalysts.
12-3 Nucleophilic Character of the Pi Bond: Electrophilic Addition of Hydrogen Halides As noted earlier, the p electrons of a double bond are not as strongly bound as those of a s bond. This p-electron cloud, located above and below the molecular plane of the alkene, is polarizable and capable of acting in a nucleophilic manner, just like the lone electron pairs of typical Lewis bases. The relatively high electron density of the double bond in 2,3-dimethylbutene is indicated (red) in its electrostatic potential map in the margin. In the
12-3 Electrophilic Addition of Hydrogen Halides
Chapter 12
513
upcoming sections, we shall discuss reactions between the nucleophilic alkene p bond and a variety of electrophiles. As in hydrogenation, the final outcome of such reactions is addition. However, there are several different mechanisms for these transformations, collectively called electrophilic additions, and they can be regioselective and stereospecific. We begin with the simplest electrophile, the proton.
Electrophilic attack by protons gives carbocations The proton of a strong acid may add to a double bond to yield a carbocation. This process is the reverse of the deprotonation step in the E1 reaction and has the same transition state (Figure 7-7). When a good nucleophile is present, particularly at low temperatures, the carbocation is intercepted to give the electrophilic addition product. For example, treatment of alkenes with hydrogen halides leads to the corresponding haloalkanes. The electrostatic potential map version of the general scheme shows the flow of electron density that occurs during this process. Mechanism of Electrophilic Addition of HX to Alkenes H G D CPC D G
Electrophilic attack
H A D OCOC G A
ðš Xð Nucleophilic trapping
H ðš Xð A A OCOCO A A
In the first step, an electron pair moves from the p bond (orange-red in the image at the left) toward the electrophilic (purple) proton to form a new s bond. The map of the carbocation product of this step (center) shows that electron deficiency is now centered on the cationic carbon. Next, addition of the negative (and, therefore, red) halide ion is shown. In the resulting haloalkane product at the right, the polarity of the new C – X bond is reflected by the orange-red color for the strongly d2 halogen atom, and the range of colors from greenish-blue to purple for the rest of the structure, indicating the distribution of d1 character among the remaining atoms. In a typical experiment, the gaseous hydrogen halide, which may be HCl, HBr, or HI, is bubbled through pure or dissolved alkene. Alternatively, HX can be added in a solvent, such as acetic acid. Aqueous work-up furnishes the haloalkane in high yield. š Ið
HI, 0°C
H 90% Cyclohexene
Iodocyclohexane
Exercise 12-5 Write out two step-by-step mechanisms for the addition of HI to cyclohexene shown above. In the first, use a free proton as the electrophile. In the second, use undissociated HI in the electrophilic addition step. Make sure to include all necessary curved arrows to depict electron-pair movement.
Although we write H1 as a reacting species in many mechanisms, we recognize that it is normally bound to a Lewis base in solution, such as X2 in this example.
514
Chapter 12
Reactions of Alkenes
The Markovnikov rule predicts regioselectivity in electrophilic additions Are additions of HX to unsymmetric alkenes regioselective? To answer this question, let us consider the reaction of propene with hydrogen chloride. Two products are possible: 2-chloropropane and 1-chloropropane. However, the only product observed is 2-chloropropane. Regioselective Electrophilic Addition to Propene
REACTION
CH3CH P CH2
CH3CHCH2 but no CH3CHCH2 A A A A H ðCl ð ðClð H 2-Chloropropane
Less substituted carbon: proton attaches here
1-Chloropropane
Similarly, reaction of 2-methylpropene with hydrogen bromide gives only 2-bromo2-methylpropane, and 1-methylcyclohexene combines with HI to furnish only 1-iodo-1methylcyclohexane. Two Further Examples of Regioselective Additions H3C
CH3
CH3 A CH3CCH2H A ðBrð
š HBrð
G C PCH2 D H3C
šð HI
CH3 Ið H
Less substituted
Less substituted
We can see from these examples that, if the carbon atoms participating in the double bond are not equally substituted, the proton from the hydrogen halide attaches itself to the less substituted carbon. As a consequence, the halogen ends up at the more substituted carbon. This phenomenon, referred to as the Markovnikov* rule, can be explained by what we know about the mechanism of electrophile additions of protons to alkenes. The key is the relative stability of the resulting carbocation intermediates. Consider the addition of HCl to propene. The regiochemistry of the reaction is determined in the first step, in which the proton attacks the p system to give an intermediate carbocation. Carbocation generation is rate determining; once it occurs, reaction with chloride proceeds quickly. Let us look at the crucial first step in more detail. The proton may attack either of the two carbon atoms of the double bond. Addition to the internal carbon leads to the primary propyl cation. Protonation of Propene at C2—More Substituted Carbon (Does Not Occur) CPC
H
‡
H
∑/
H3C
H HC≈ CO O 3 ZC Y H H
H
∑/
H
/∑
ANIMATED MECHANISM: Electrophilic addition of HCl to propene
'
MATION AN I
šð HCl
H
Transition state 1
CH3CH2CH2 Primary carbocation (Not observed)
Partial positive charge on primary carbon (unfavorable)
In contrast, protonation at the terminal carbon results in the formation of the secondary 1-methylethyl (isopropyl) cation.
*Professor Vladimir V. Markovnikov (1838 – 1904), University of Moscow, formulated his rule in 1869.
12-3 Electrophilic Addition of Hydrogen Halides
Chapter 12
515
Protonation of Propene at C1—Less Substituted Carbon CPC
H
H
H3C
H
Partial positive charge on secondary carbon (preferable)
/∑
/∑
H3C
H
∑/
H
H ≈' H CO ZC Y H
MECHANISM
‡
Transition state 2
CH3CHCH3 Secondary carbocation (Favored)
As we know, a primary carbocation is too unstable to be a reasonable reaction intermediate in solution. In contrast, secondary cations form relatively readily. In addition, notice that the transition states for the two possible modes of addition show positive charge building up on primary and secondary carbons, respectively. Thus, the energies, and stabilities, of the transition states will reflect the relative energies of the cations to which they lead. The energy of the transition state (and, therefore, the activation energy) leading to the secondary cation is lower, meaning that this cation forms much faster. Figure 12-2 is a potential-energy diagram showing the two competing pathways. We see that these are late transition states, whose energies closely resemble those of the product cations. High energy; resembles primary cation Primary cation TS-1
TS-2
Favored; resembles secondary cation
E
CH3CH
CH3CH2CH2 Cl− (Not observed)
Secondary cation CH3CHCH3 Cl−
CH2 CH3CH2CH2Cl
HCl
CH3CHClCH3 Reaction coordinate
On the basis of this analysis, we can rephrase the empirical Markovnikov rule: HX adds to unsymmetric alkenes in such a way that the initial protonation gives the more stable carbocation. For alkenes that are similarly substituted at both sp2 carbons, product mixtures are to be expected, because carbocations of comparable stability are formed. By analogy with other carbocation reactions (e.g., SN1, Section 7-3), when addition to an achiral alkene generates a chiral product, this product is obtained as a racemic mixture. Exercise 12-6 Predict the outcome of the addition of HBr to (a) 1-hexene; (b) trans-2-pentene; (c) 2-methyl2-butene; (d) 4-methylcyclohexene. How many isomers can be formed in each case?
Exercise 12-7 Draw a potential-energy diagram for the reaction in (c) of Exercise 12-6.
Figure 12-2 Potential-energy diagram for the two possible modes of HCl addition to propene. Transition state 1 (TS-1), which leads to the higher-energy primary propyl cation, is less favored than transition state 2 (TS-2), which gives the 1-methylethyl (isopropyl) cation.
Reactions of Alkenes
Carbocation rearrangements may follow electrophilic addition In the absence of a good nucleophile, carbocation rearrangements may occur following addition of an electrophile to the alkene double bond (Section 9-3). Rearrangements are favored in electrophilic additions of acids whose conjugate bases are poor nucleophiles. An example is trifluoroacetic acid, CF3CO2H. Its trifluoroacetate counterion is much less nucleophilic than are halide ions. Thus, addition of trifluoroacetic acid to 3-methyl-1-butene gives only about 43% of the normal product of Markovnikov addition. The major product results from a hydride shift that converts the initial secondary cation into a more stable tertiary cation before the trifluoroacetate can attach. Rearrangement Accompanying Addition of Trifluoroacetic Acid to 3-Methyl-1-butene
(CH3)2C O CH P CH2 A H Migrating hydride
C
CF3
P
A
A
A
A
ðOð P
O
A
H
ðOð H CF3 C Oð H CF3 C Oð A A A A (CH3)2C O CH O CH2 (CH3)2C O CH O CH2 A A H H 43% 57% P
ðOð
A
Chapter 12
A
516
3-Methyl-2-butyl trifluoroacetate
2-Methyl-2-butyl trifluoroacetate
Product of normal Markovnikov addition
Product resulting from carbocation rearrangement
Exercise 12-8 Write a detailed step-by-step mechanism for the reaction depicted above. Refer to Section 9-3 if necessary.
The extent of carbocation rearrangement is difficult to predict: It depends on the alkene structure, the solvent, the strength and concentration of the nucleophile, and the temperature. In general, rearrangements are favored under strongly acidic, nucleophile-deficient conditions.
In Summary Additions of hydrogen halides to alkenes are electrophilic reactions that begin with protonation of the double bond to give a carbocation. Trapping of the carbocation by halide ion gives the final product. The Markovnikov rule predicts the regioselectivity of hydrohalogenation to haloalkanes. As in any carbocation reaction, rearrangements may occur if good nucleophiles are absent.
12-4 Alcohol Synthesis by Electrophilic Hydration: Thermodynamic Control So far, we have seen attack on the double bond by a proton, followed by nucleophilic attachment of its counterion to the intermediate carbocation. Can other nucleophiles participate? Upon exposure of an alkene to an aqueous solution of sulfuric acid, which has a poorly nucleophilic counterion, water acts as the nucleophile to trap the carbocation formed by initial protonation. Overall, the elements of water add to the double bond, an electrophilic hydration. The addition follows the Markovnikov rule in that H1 adds to the less substituted carbon and the OH group ends up at the more substituted one. Because water is a poor nucleophile, carbocation rearrangements can intervene in the hydration process. This addition process is the reverse of the acid-induced elimination of water from alcohols (dehydration, Section 11-7). Its mechanism is the same in reverse, as illustrated in the hydration of 2-methylpropene, a reaction of industrial importance leading to 2-methyl2-propanol (tert-butyl alcohol).
12-4 Alcohol Synthesis by Electrophilic Hydration
Chapter 12
517
Electrophilic Hydration H3C H3 C
G C P CH2 D
š 50% HOH, H2SO4
REACTION
CH3 A H3C O C OCH2 OH A ðOH š 92%
2-Methylpropene
2-Methyl-2-propanol
Mechanism of the Hydration of 2-Methylpropene H3C H3C
G C P CH2 D
H
H
H3C
CH3 G D C A CH3
p HOH p p HOH p
CH3 A pD H CH3CO O A G H CH3
H H
CH3 A p CH3C O OH p A CH3
MECHANISM
Alkene hydration and alcohol dehydration are equilibrium processes In the mechanism of alkene hydration, all the steps are reversible. The proton acts only as a catalyst and is not consumed in the overall reaction. Indeed, without the acid, hydration would not occur; alkenes are stable in neutral water. The presence of acid, however, establishes an equilibrium between alkene and alcohol. This equilibrium can be driven toward the alcohol by using low reaction temperatures and a large excess of water. Conversely, we have seen (Section 11-7) that treating the alcohol with concentrated acid favors dehydration, especially at elevated temperatures. Alcohol
Conc. H2SO4, high temperature H2SO4, excess H2O, low temperature
alkene
H2O
Exercise 12-9 Treatment of 2-methylpropene with catalytic deuterated sulfuric acid (D2SO4) in D2O gives (CD3)3COD. Explain by a mechanism.
The reversibility of alkene protonation leads to alkene equilibration In Section 11-17 we explained that the acid-catalyzed dehydration of alcohols gives mixtures of alkenes in which the more stable isomers predominate. Equilibrating carbocation rearrangements, followed by E1, are partly responsible for these results. However, more important for the attainment of thermodynamic equilibration is that the E1 process is reversible: As we discussed above, alkenes can be protonated by acid to form carbocations. Let us illustrate this feature of reversible protonation by envisaging what can happen when 2-butanol is dehydrated under acidic conditions. To avoid complications arising from SN1, we will employ sulfuric acid bearing a nonnucleophilic counterion. The first thing to happen is protonation of the hydroxy group. Then, loss of water from the protonated alcohol gives the corresponding secondary carbocation. This species can undergo E1 in three different ways, to give the three observed products: 1-butene, cis-2-butene, and trans2-butene. The initial ratio of these isomers is controlled by the relative energies of the transition states leading to them: It is under kinetic control (Section 2-1). However, under the strongly acidic conditions, a proton can re-add to the double bond of any of the isomers. In the case of the two 2-butenes, this process engenders the corresponding secondary carbocations. In the case of 1-butene, and as noted earlier, regioselective Markovnikov
Hydration–Dehydration Equilibrium RCH P CH2 H2Oð Catalytic H
RCHCH3 A ð OH
518
Chapter 12
Reactions of Alkenes
addition leads to the same ion. Because this cation may again lose a proton to give any of the same three alkene isomers, the net effect is interconversion of the isomers to an equilibrium mixture, in which the thermodynamically most stable isomer is the major component. This system is therefore an example of a reaction that is under thermodynamic control (Section 2-1). Thermodynamic Control in the Acid-Catalyzed Dehydration of 2-Butanol ðOH A
H
H H E H O A
H2O
H
H
H 2O
H H H H H H
1-Butene
H H G D C
H
H
H H
cis-2-Butene trans-2-Butene
By this procedure, less stable alkenes may be catalytically converted into their more stable isomers (see below and margin). Acid-Catalyzed Equilibration of Alkenes (CH3)3C
C(CH3)3 G G CPC D D H H
Terminal Catalytic H
H A HO C OH
Catalytic H
Cis
(CH3)3C
H G G CPC D D H C(CH3)3 Trans
Exercise 12-10 Write a mechanism for this rearrangement. What is the driving force for the reaction? CH2CH3
Internal
H3 C
H 3C H
In Summary The carbocation formed by addition of a proton to an alkene may be trapped by water to give an alcohol, the reverse of alkene synthesis by alcohol dehydration. Reversible protonation equilibrates alkenes in the presence of acid, thereby forming a thermodynamically controlled mixture of isomers.
12-5 Electrophilic Addition of Halogens to Alkenes Halogens, although they do not appear to contain electrophilic atoms, add to the double bond of alkenes, giving vicinal dihalides. These compounds have uses as solvents for dry cleaning and degreasing and as antiknock additives for gasoline. Halogen addition works best for chlorine and bromine. The reaction of fluorine is too violent to be generally employed, and iodine addition is not normally favorable thermodynamically.
12-5 Electrophilic Addition of Halogens to Alkenes
Chapter 12
519
Halogenation of Alkenes ðš X Oš Xð
ðš Xð G
CO C
G
G D CPC D G
ðXð š Vicinal dihalide
X Cl, Br
Exercise 12-11 Calculate (as in Table 12-1) the DH 8 values for the addition of F2 and I2 to ethene. (For DH X8 2, see Section 3-5.)
Bromine addition is particularly easy to observe because bromine solutions immediately change from red to colorless when exposed to an alkene. This phenomenon is sometimes used as a color test for unsaturation. Halogenations are best carried out at or below room temperature in inert halogenated solvents such as tetrachloromethane (carbon tetrachloride).
Addition of bromine to an alkene results in almost immediate loss of the red-brown color of Br2 as the reaction takes place.
Electrophilic Halogen Addition of Br2 to 1-Hexene CH3(CH2)3CHP CH2
1-Hexene
Brð, CCl4 Br Oš ðš
CH3(CH2)3CHCH2Br ð A ðBrð 90% 1,2-Dibromohexane
Two Topologies of Alkene Addition Syn
Halogen additions to double bonds may seem to be similar to hydrogenations. However, their mechanism is quite different, as revealed by the stereochemistry of bromination; similar arguments hold for the other halogens.
or
Bromination takes place through anti addition What is the stereochemistry of bromination? Do the two bromine atoms add from the same side of the double bond (syn, as in catalytic hydrogenation) or from opposite sides (see margin)? Let us examine the bromination of cyclohexene. Addition on the same side should give cis-1,2-dibromocyclohexane; the alternative would result in trans-1,2dibromocyclohexane. The second pathway is borne out by experiment — only anti addition is observed. Because anti addition to the two reacting carbon atoms can take place with equal probability in two possible ways — in either case, from both above and below the p bond — the product is racemic. Anti Bromination of Cyclohexene Br ðBr ð ! ! Br , CCl ` ` Br ðBr ð 83% 2
4
Racemic trans-1,2-dibromocyclohexane
With acyclic alkenes, the reaction is also cleanly stereospecific. For example, cis-2-butene is brominated to furnish a racemic mixture of (2R,3R)- and (2S,3S)-2,3-dibromobutane; trans-2-butene results in the meso diastereomer.
Anti
or
REACTION
Reactions of Alkenes
2
G
CH3
4
cis-2-Butene
Racemic mixture of two enantiomers
MODEL BUILDING
(2R,3R)-2,3-Dibromobutane
C
trans-2-Butene
G
H
Br2, CCl4
G CH3
ðBr G C H3C ( S H
H CH3 C RG Br ð @
G GC P C
GH
&
H 3C
ð G Br CC S( H CH3
(2S,3S)-2,3-Dibromobutane
CH @ 3 G Br ð H& CC GC R S( H ðBr CH3 G
H 3C
GH
C
G GC P C
G
H
@
REACTION
Stereospecific 2-Butene Bromination H@ H ðBr G H3C & CH3 Br , CCl C C C G S H (R R G Br ðBr CH3 ð &
Chapter 12
G
520
Identical
meso-2,3-Dibromobutane
Cyclic bromonium ions explain the stereochemistry How does bromine attack the electron-rich double bond even though it does not appear to contain an electrophilic center? The answer lies in the polarizability of the Br – Br bond, which is prone to heterolytic cleavage upon reaction with a nucleophile. The p-electron cloud of the alkene is nucleophilic and attacks one end of the bromine molecule, with simultaneous displacement of the second bromine atom as bromide ion in an SN2-like process. What is the product of this process? We might expect a carbocation, in analogy with the proton additions discussed in Sections 12-3 and 12-4. However, if the first step of addition of bromine to cyclohexene were to give a carbocation, the bromide ion released in this process could attack the positively charged carbon atom from either the same or the opposite side of the ring, resulting in a mixture of cis- and trans-1,2-dibromocyclohexanes. However, as shown previously, only the trans product is obtained. How can we explain this result? The stereochemistry of bromination is explained if we propose that initial attack of bromine on the double bond gives a cyclic bromonium ion, in which the bromine bridges both carbon atoms of the original double bond to form a three-membered ring (Figure 12-3). The structure of this ion is rigid, and it may be attacked by bromide ion only on the side opposite the bridging bromine atom. The three-membered ring is thus opened stereospecifically (compare nucleophilic ring opening of oxacyclopropanes in Section 9-9). The leaving group is the bridging bromine. In symmetric bromonium ions, attack is equally probable at either carbon atom, thereby giving the racemic (or meso) products observed. δ−
δ−
Br
Br
−
δ+
Br
Br
+
C
A
C
C
δ+
C
C
Br
Br
Br
Br
C
C
−
+
C
B Figure 12-3 (A) Electron-pushing picture of cyclic bromonium ion formation. The alkene (red) acts as a nucleophile to displace bromide ion (green) from bromine. The molecular bromine behaves as if it were strongly polarized, one atom as a bromine cation, the other as an anion. (B) Orbital picture of bromonium ion formation.
12-6 The Generality of Electrophilic Addition
Chapter 12
521
Formation and Nucleophilic Opening of a Cyclic Bromonium Ion
MECHANISM
CPC
š Br f i COC or
SN2
š ðBr
COC
G
Leaving group
G
š ðBrð A ðBrð
š Brð
š GBrð O C C G
š ðBr
š ðBr ð Nucleophile
Exercise 12-12 Draw the intermediate in the bromination of cyclohexene, using the following conformational picture. Show why the product is racemic. What can you say about the initial conformation of the product? H
H H
H H
H H
H
H
H
H
H
H
H
H
H Conformational flip in cyclohexene
The halogenation of alkenes should not be confused with halogenation of alkanes (Sections 3-4 through 3-8). The addition to alkenes follows a mechanism in which a nucleophile (the alkene p bond) interacts with an electrophilic species (such as molecular Cl2 or Br2) via movement of electron pairs. Alkane halogenation is a radical process that requires an initiation step to generate halogen atoms. This step typically needs heat, light, or a radical initiator (such as a peroxide) and proceeds via a mechanism involving movement of single electrons.
In Summary Halogens add as electrophiles to alkenes, producing vicinal dihalides. The reaction begins with the formation of a bridged halonium ion. This intermediate is opened stereospecifically by the halide ion displaced in the initial step to give overall anti addition to the double bond. In subsequent sections, we shall see that other stereochemical outcomes are possible, depending on the electrophile.
12-6 The Generality of Electrophilic Addition Halogens are just one of many electrophile – nucleophile combinations that add to alkene double bonds. In this section we begin a survey of some of the most important of these processes, beginning with addition of halogens (the electrophile source) in the presence of water (the nucleophile). The products, 2-haloalcohols, commonly known as halohydrins, are widely used in a number of industrial and synthetic applications. In particular, they are important intermediates for the synthesis of oxacyclopropanes (epoxides, Section 9-9).
The bromonium ion can be trapped by other nucleophiles The creation of a bromonium ion in alkene brominations suggests that, in the presence of other nucleophiles, competition might be observed in the trapping of the intermediate. For example, bromination of cyclopentene in water as solvent gives the vicinal bromoalcohol (common name, bromohydrin). In this case, the bromonium ion is attacked by water, which is
MATION AN I
ANIMATED MECHANISM: Stereospecific bromination of 2-butenes
Reactions of Alkenes
Chapter 12
present in large excess. The net transformation is the anti addition of Br and OH to the double bond. The other product formed is HBr. The corresponding chloroalcohols (chlorohydrins) can be made from chlorine in water through a chloronium ion intermediate. Bromoalcohol (Bromohydrin) Synthesis
Br
H
OH2
H ^ ! > OH A H
H
Br
!
!
! !
^
`
`
`
Br2, H2O, 0C
> H )Br)
>) H )Br
> H )Br
H ! ^ )OH >
Br
Br O
O
O trans-2Bromocyclopentanol
Cyclopentene
Exercise 12-13 Write the expected product from the reaction of (a) trans-2-butene and (b) cis-2-pentene with aqueous chlorine. Show the stereochemistry clearly
Vicinal haloalcohols undergo intramolecular ring closure in the presence of base to give oxacyclopropanes (Section 9-9) and are therefore useful intermediates in organic synthesis.
HBr
Br2 HO OCH3
∑
H /∑ OCH3 ð Br H 76% trans-1-Bromo-2-methoxycyclohexane
Oxacyclopropane Formation from an Alkene Through the Haloalcohol ðClð ∑H Cl , H O, 0°C ! NaOH, H O, 25°C HCl ! OH ¨ H 70–73% 2
2
2
H ∑ ) [O ¨ H 70–73%
ð
Vicinal Haloether Synthesis
/
522
If alcohol is used as a solvent instead of water in these halogenations, the corresponding vicinal haloethers are produced, as shown in the margin.
Halonium ion opening can be regioselective In contrast with addition of two identical halogens, mixed additions to double bonds can pose regiochemical problems. Is the addition of Br and OH (or OR) to an unsymmetric double bond selective? The answer is yes. For example, 2-methylpropene is converted by aqueous bromine into only 1-bromo-2-methyl-2-propanol; none of the alternative regioisomer, 2-bromo-2-methyl-1-propanol, is formed.
12-6 The Generality of Electrophilic Addition
ðOH A CH3CCH2Br ð A CH3 82% yield
Br2, HO OH
H3C
G C PCH2 D H3C
HBr
ðBrð A OH CH3CCH2 A CH3
but no
1-Bromo-2-methyl2-propanol
2-Bromo-2-methyl1-propanol
The electrophilic halogen in the product always becomes linked to the less substituted carbon of the original double bond. The subsequently added nucleophile attaches to the more highly substituted center. How can this be explained? The situation is very similar to the acid-catalyzed nucleophilic ring opening of oxacyclopropanes (Section 9-9), in which the intermediate contains a protonated oxygen in the three-membered ring. In both reactions, the nucleophile attacks the more highly substituted carbon of the ring, because this carbon is more positively polarized than the other.
Greater ␦ here
š Br f i (CH3)2C O CH2
Regioselective Opening of the Bromonium Ion Formed from 2-Methylpropene
š HO OH
š (CH3)2C O CH2Br ð A OO O H H ð
Attack at more substituted carbon of bromonium ion
H
Exercise 12-14 What are the product(s) of the following reactions? Cl2, CH3OH
Br2, H2O
(b) H3C
Exercise 12-15
Working with the Concepts: Mechanism of Electrophilic Addition to Alkenes Write a mechanism for the reaction shown in Exercise 12-14(a). Strategy This problem is quite similar to the transformation shown on the preceding page. We have simply replaced bromine with chlorine. Solution • Initial attack of the alkene p bond on a molecule of Cl2 gives rise to a cyclic chloronium ion:
CH3
CHPCH2
š Clð š
š ðCl š
Cl
f i šð CH CH2 ðCl š • This species is not symmetric: In particular, of the two carbons bonded to the positive chlorine, the internal (secondary) carbon is more positively polarized. Attack of the nucleophilic solvent CH3
Recall:
Nucleophile — red Electrophile — blue Leaving group — green
MECHANISM CH3 A š CH3C O CH2Br ð A OH ð
A simple rule of thumb is that electrophilic additions of unsymmetric reagents of this type add in a Markovnikov-like fashion, with the electrophilic unit emerging at the less substituted carbon of the double bond. Mixtures are formed only when the two carbons are not sufficiently differentiated [see Exercise 12-14(b)].
(a) CH3CH P CH2
Chapter 12
523
Chapter 12
Reactions of Alkenes
methanol occurs preferentially at this center. Loss of a proton from the resulting oxonium ion completes the reaction:
šð ðCl f CH3 O CHO CH2 A O f CH3 f H
524
Cl
CH3
f i CH CH2
š CH3O OOH
H
šð ðCl f CH3 O CH O CH2 A Oð f CH3
Exercise 12-16
Try It Yourself Write a mechanism for the reaction shown in Exercise 12-14(b). [Caution: The presence of the methyl group on the cyclohexane ring has a significant stereochemical effect (compared with the other examples in the section)! Hint: The initial addition of halogen to the alkene p bond can occur either from the same face of the ring that contains the methyl group or from the opposite face. How many isomers do you expect to result from this addition?]
Exercise 12-17 What would be a good alkene precursor for a racemic mixture of (2R,3R)- and (2S,3S)-2-bromo3-methoxypentane? What other isomeric products might you expect to find from the reactions that you propose?
In general, alkenes can undergo stereo- and regiospecific addition reactions with reagents of the type A–B, in which the A–B bond is polarized such that A acts as the electrophile A1 and B as the nucleophile B2. Table 12-2 shows how such reagents add to 2-methylpropene. Table 12-2 Reagents A – B That Add to Alkenes by Electrophilic Attack H
CH3 G G CP C D D CH3 H
Name
1
d1
AOB
d2
Structure
Bromine chloride
.. .. : Br . . OCl .. :
Cyanogen bromide
.. : Br . . OCN:
Iodine chloride
. . .. :.I. OCl .. :
Sulfenyl chlorides
. . .. RS .. OCl .. :
Mercuric salts
.. XHgOXa, HO .. H
a
X here denotes acetate.
uy
H CH3 A A H O C O C O CH3 A A A B Addition product to 2-methylpropene
ðBrCH 2C(CH3)2 A ðClð š ðBrCH 2C(CH3)2 A CNð ICH2C(CH3)2 ð A ðClð š RSCH 2C(CH3)2 A ðClð š XHgCH2C(CH3)2 A ðOH š
12-7 Oxymercuration–Demercuration
Chapter 12
In Summary Halonium ions are subject to stereospecific and regioselective ring opening in a manner that is mechanistically very similar to the nucleophilic opening of protonated oxacyclopropanes. Halonium ions can be trapped by halide ions, water, or alcohols to give vicinal dihaloalkanes, haloalcohols, or haloethers, respectively. The principle of electrophilic additions can be applied to any reagent A – B containing a polarized or polarizable bond.
12-7 Oxymercuration – Demercuration: A Special Electrophilic Addition The last example in Table 12-2 is an electrophilic addition of a mercuric salt to an alkene. The reaction is called mercuration, and the resulting compound is an alkylmercury derivative, from which the mercury can be removed in a subsequent step. One particularly useful reaction sequence is oxymercuration – demercuration, in which mercuric acetate acts as the reagent. In the first step (oxymercuration), treatment of an alkene with this species in the presence of water leads to the corresponding addition product. REACTION
Oxymercuration
O O B B CH3COHgOCCH3
š HO OH
THF
CH3 š OH
/∑
CH3
/∑
3 H HgOCCH B O
Mercuric acetate
Alkylmercuric acetate
In the subsequent demercuration, the mercury-containing substituent is replaced by hydrogen through treatment with sodium borohydride in base. The net result is hydration of the double bond to give an alcohol. Demercuration NaBH4, NaOH, H2O
CH3 š OH
/∑
/∑
CH3 š OH HgOCCH 3 H B O
H
H
Hg
CH3CO B O
1-Methylcyclopentanol
/∑
Oxymercuration is anti stereospecific and regioselective. This outcome implies a mechanism similar to that for the electrophilic addition reactions discussed so far. The mercury reagent initially dissociates to give an acetate ion and a cationic mercury species. This mercury cation then attacks an alkene double bond, furnishing a mercurinium ion, with a structure similar to that of a cyclic bromonium ion. The water that is present attacks the more substituted carbon (Markovnikov rule regioselectivity) to give an alkylmercuric acetate intermediate. Replacement of mercury by hydrogen (demercuration) is achieved by sodium borohydride reduction through a complex and only incompletely understood mechanism. It is not stereospecific. The alcohol obtained after demercuration is the same as the product of Markovnikov hydration (Section 12-4) of the starting material. However, oxymercuration – demercuration is a valuable alternative to acid-catalyzed hydration, because no carbocation is involved; therefore oxymercuration – demercuration is not susceptible to the rearrangements that commonly occur under acidic conditions (Section 12-3). Its use is limited by the expense of the mercury reagent and its toxicity, which requires careful removal of mercury from the product and safe disposal.
O B CH3COH
525
526
Reactions of Alkenes
Chapter 12
CHE M I C AL H I G H L I G H T 1 2 - 1 Juvenile Hormone Analogs in the Battle against Insect-Borne Diseases Juvenile hormone (JH) is a substance that controls metamorphosis in insects. It is produced by the male wild silk moth, Hyalophora cecropia L., and its presence delays the maturation of insect larvae until the appropriate
developmental stage is reached. Exposure to JH causes insect metamorphosis to stall at the pupa stage: Mosquitoes exposed to JH fail to develop into biting, egg-laying adults. JH therefore has the potential for use in controlling mosquitoborne diseases such as malaria, yellow fever, and West Nile virus. This potential is limited by the instability of JH and the difficulty with which it can be isolated naturally or prepared synthetically. As a result, analogs have been sought that are more stable, suitably bioactive, and easy to prepare. The synthetic compound methoprene possesses all of these desirable features. Its synthesis utilizes several reactions we have studied, including oxymercuration – demercuration to give the tertiary methyl ether, ester hydrolysis (Section 8-5), and PCC oxidation of a primary alcohol to produce an aldehyde (Section 8-6). Whereas earlier attempts to prepare JH analogs gave compounds with very poor activity, methoprene is up to 1000 times more bioactive than JH toward a variety of pests. It is effective against fleas, mosquitoes, and fire ants and is now marketed under a number of trade names. Methoprene may be used indoors to eliminate flea infestations, greatly reducing the need for conventional insecticides. Although the product does not kill insects that have already reached adulthood, the eggs they lay after exposure do not develop into adults. Spreading methoprene as granules in areas where mosquitoes may breed prevents their survival past the pupal stage. Methoprene possesses relatively low toxicity toward vertebrates and, unlike chlorinated insecticides such as DDT (Chemical Highlight 3-2), it does not persist in the environment. Although it is stable enough to be effective for weeks to months after application, it is degraded over time by sunlight into innocuous smaller molecules. Methoprene and several other JH analogs have therefore become important new tools for pest management.
Juvenile hormone stops mosquito development at the pupa stage shown in the photo. Mosquito pupae live in water and breathe air from the surface through a pair of tubes on their back.
H3C
H3C
H3C {
O B COCH3
CH3
O % ≥ H
O
O O
Juvenile hormone
CH3
CH3
H2C
O B OCCH3
O B 1. Hg(OCCH3)2, CH3OH 2. NaBH4, KOH, H2O
H 3C
CH3
OCH3
H2C
OH
A
H PCC, CH2Cl2
H3C H3C
OCH3
CH3
O B CH H
Final steps
H 3C
OCH3
CH3
H3C Methoprene (JH analog)
CH3
O B COCH(CH3)2
12-7 Oxymercuration–Demercuration
527
Chapter 12
Mechanism of Oxymercuration – Demercuration
MECHANISM
Step 1. Dissociation O
O
O
O
CH3CO
CH3COHgOCCH3
HgOCCH3
Step 2. Electrophilic attack
O O
E
C
HgOCCH3
E
HgOCCH3
E
G
G
CP C G G
C
Mercurinium ion
Step 3. Nucleophilic opening (Markovnikov regioselectivity) O
O
HgOCCH3
CH3COHg
C
C
š OH
H
C
H
C OH ð
Alkylmercuric acetate
Step 4. Reduction
Ether Synthesis by Oxymercuration– Demercuration
O H
CH3COHg C
NaBH4, NaOH, H2O
C
C
C
CH3CO
Hg
OH ð
OH ð
1-Hexene
O
When the oxymercuration of an alkene is executed in an alcohol solvent, demercuration gives an ether, as shown in the margin.
O B 1. Hg(OCCH 3)2, CH3 OH 2. NaBH4, NaOH, H2O
Exercise 12-18 H
Working with the Concepts: Addressing a Difficult Mechanism Problem Explain the result shown below. O B 1. Hg(OCCH3)2 2. NaBH4, NaOH, H2O
OCH3 ð 65%
O
/∑
H2C
HOCH2 CH2OH
2-Methoxyhexane
CH2OH 42%
Strategy Start by addressing the apparent dissimilarity between the structure of the substrate and that of the product. You can clarify the pathway to a solution by numbering the carbon atoms in the substrate and then identifying the corresponding atoms in the product. Then you can begin the process of thinking mechanistically—following the steps of the process as we understand them to see where they lead. Solution • Begin with mercuration of the double bond:
O
5
/∑
HOCH2 CH2OH 8
7
/∑
4
7
Hg(O2CCH3)2
!
6
2 3
3
CH3CO2Hg!
1
HOCH2 CH2OH
H2C
2 4
1 5
CH2OH 8
6
Chapter 12
Reactions of Alkenes
• In all previous examples, the next step involved a molecule of solvent—either water or alcohol— which provided a nucleophilic oxygen atom to add to one carbon of the mercurinium ion and open the ring. In this case, however, we can see that an oxygen atom already present in the substrate (at C7) becomes attached to one of the original double-bond carbons (C2) in the product. This process is one of intramolecular bond formation. After removal of the mercury by NaBH4, we arrive at the final product:
!
CH3CO2Hg!
HgO2CCH3
NaBH4
O
O
/∑
š 2 CH2OH HOCH š
H2C
H2C
CH2OH
CH2OH
Exercise 12-19
Try It Yourself The reaction below proceeds to a cyclic product that is an isomer of the starting material. Suggest a structure for it. (Hint: Think mechanistically. Begin with the appropriate electrophilic attack. Then utilize a nucleophilic atom already present in the substrate in order to complete the addition process. Caution: There is a regioselectivity issue to address. Use the examples presented in the chapter section to guide you.)
EOH
1. Hg(O2CCH3)2 2. NaBH4, NaOH, H2O
In Summary Oxymercuration–demercuration is a synthetically useful method for converting alkenes regioselectively (following the Markovnikov rule) into alcohols or ethers. Carbocations are not involved; therefore, rearrangements do not occur.
12-8 Hydroboration – Oxidation: A Stereospecific Anti-Markovnikov Hydration So far in this chapter, we have seen two ways to add the elements of water to alkenes to give alcohols. This section presents a third method, which complements the other two synthetically by giving in a different regiochemical outcome. The process involves a reaction that lies mechanistically between hydrogenation and electrophilic addition: hydroboration of double bonds. The alkylboranes that result can be oxidized to alcohols.
The boron – hydrogen bond adds across double bonds Borane, BH3, adds to double bonds without catalytic activation, a reaction called hydroboration by its discoverer, H. C. Brown.* REACTION
Hydroboration of Alkenes G D
G CPC D
H G BOH D H
BH2 H G G C OC
Borane
An alkylborane
Repeat 2 G 2 CPC D
G D
528
H A A (O C O C)3B A A A trialkylborane
*Professor Herbert C. Brown (1912 – 2004), Purdue University, West Lafayette, Indiana, Nobel Prize 1979 (chemistry).
Borane (which by itself exists as a dimer, B2H6) is commercially available in ether and tetrahydrofuran (THF). In these solutions, borane exists as a Lewis acid-base complex with the ether oxygen (see Sections 2-2 and 9-5), an aggregate that allows the boron to have an electron octet (for the molecular-orbital picture of BH3, see Figure 1-17). How does the B – H unit add to the p bond? The p bond is electron rich and borane is electron poor. Therefore, it is reasonable to formulate an initial Lewis acid-base complex similar to that of a bromonium ion (Figure 12-3), requiring the participation of the empty p orbital on BH3. This shifts electron density from the alkene to boron. Subsequently, one of the hydrogens is transferred by means of a four-center transition state to one of the alkene carbons, while the boron shifts to the other. The stereochemistry of the addition is syn. All three B – H bonds are reactive in this way. The boron in the product alkylborane is again electron deficient. The electrostatic potential maps of the general scheme shown below (on a scale that maximizes the desired color changes) show how the boron in borane starts out as an electron-deficient species (blue), becomes more electron rich (red) in the complex, and then loses the electron density gained as it proceeds to product (blue).
529
Chapter 12
Lewis acid
BH3
š O Lewis base
H3B
12-8 Hydroboration–Oxidation
O
Borane–THF complex
Mechanism of Hydroboration C
C C
δ+
δ+
‡
C
C
MECHANISM
C C
H2B H B
H
H
H
B
δ
−
H H Borane–alkene complex
H
H2B
C H
Four-center transition state
MATION AN I
ANIMATED MECHANISM: Hydroboration – oxidation
Empty p orbital
‡
Regioselectivity of Hydroboration 3 RCH P CH2
BH3
Less hindered carbon
Hydroboration is not only stereospecific (syn addition), it is also regioselective. Unlike the electrophilic additions described previously, steric more than electronic factors primarily control the regioselectivity: The boron binds to the less hindered (less substituted) carbon. The reactions of the trialkylboranes resulting from these hydroborations are of special interest to us, as the next section will explain.
The oxidation of alkylboranes gives alcohols Trialkylboranes can be oxidized with basic aqueous hydrogen peroxide to furnish alcohols in which the hydroxy function has replaced the boron atom. The net result of the two-step sequence, hydroboration – oxidation, is the addition of the elements of water to a double bond. In contrast with the hydrations described in Sections 12-4 and 12-7, however, those using borane proceed with the opposite regioselectivity: In this sequence, the OH group ends up at the less substituted carbon, an example of anti-Markovnikov addition.
RCH2CH2
CH CH R H E 2 2 B A CH2CH2R
Chapter 12
Reactions of Alkenes
Hydroboration–Oxidation Sequence
REACTION 3 RCH P CHR
BH3, THF
(RCH2CHR)3B
(CH3)2CHCH2CH P CH2
R A š 3 RCH2CHOH š
H2O2, NaOH, H2O
1. BH3, THF 2. H2O2, NaOH, H2O
(CH3)2CHCH2CH2CH2š OH š 80%
4-Methyl-1-pentene
4-Methyl-1-pentanol
In the mechanism of alkylborane oxidation, the nucleophilic hydroperoxide ion attacks the electron-poor boron atom. The resulting species undergoes a rearrangement in which an alkyl group migrates with its electron pair — and with retention of configuration — to the neighboring oxygen atom, thus expelling a hydroxide ion in the process. Mechanism of Alkylborane Oxidation A OB Oš O Oš OH A R
ðš OOš OH
Hydroperoxide ion
R
š HO )
OB
OO R
O
O
OB
O
O
This process is repeated until all three alkyl groups have migrated to oxygen atoms, finally forming a triakyl borate (RO)3B. This inorganic ester is then hydrolyzed by base to the alcohol and sodium borate. (RO)3B
H2O
3 NaOH
Na3BO3
3 ROH
Because borane additions to double bonds and subsequent oxidation are so selective, this sequence allows the stereospecific and regioselective synthesis of alcohols from alkenes. The anti-Markovnikov regioselectivity of the hydroboration – oxidation sequence complements that of acid-catalyzed hydration and oxymercuration – demercuration. In addition, hydroboration, like oxymercuration, occurs without the participation of carbocations; therefore, rearrangements are not observed. A Stereospecific and Regioselective Alcohol Synthesis by Hydroboration–Oxidation H
1-Methylcyclopentene
D
H
CH3
š OH H š 86%
∑
∑
H
CH3 BD
H H2O2, NaOH, H2O
/∑
BH3, THF
/∑
CH3
/
MECHANISM
/
530
trans-2-Methylcyclopentanol
Exercise 12-20 Give the products of hydroboration – oxidation of (a) propene and (b) (E )-3-methyl-2-pentene. Show the stereochemistry clearly.
In Summary Hydroboration – oxidation constitutes another method for hydrating alkenes. The initial addition is syn and regioselective, the boron shifting to the less hindered carbon. Oxidation of alkyl boranes with basic hydrogen peroxide gives anti-Markovnikov alcohols with retention of configuration of the alkyl group.
12-9 Diazomethane, Carbenes, and Cyclopropane Synthesis
Chapter 12
12-9 Diazomethane, Carbenes, and Cyclopropane Synthesis Cyclopropanes make interesting synthetic targets. Their highly strained structures (Section 4-2) are fascinating to study, as are the functional contributions they make to a variety of naturally occurring biological compounds (Section 4-7). Cyclopropanes may be quite readily prepared by the addition of reactive species called carbenes to the double bond of alkenes. Carbenes have the general structure R2C : , in which the central carbon atom possesses an electron sextet. Although neutral, carbenes are electron deficient and act as electrophiles toward alkenes.
Diazomethane forms methylene, which converts alkenes into cyclopropanes The unusual substance diazomethane, CH2N2, is a yellow, highly toxic, and explosive gas. It decomposes on exposure to light, heat, or copper metal by loss of N2. The result is the highly reactive species methylene, H2C: , the simplest carbene.
C O N q Nð H2š
h or or Cu
Diazomethane
H2Cð ðNqNð Methylene
When methylene is generated in the presence of compounds containing double bonds, addition takes place to furnish cyclopropanes. The process is usually stereospecific, with retention of the original configuration of the double bond. Methylene Additions to Double Bonds
N2
`!H H C `! H H 40%
A A
CH2N2, h
CH3CH2
CH2CH3
Bicyclo[4.1.0]heptane
H
CH2N2, Cu, N2
H
} ~CH CH CH3CH2 2 3 50–70% cis-Diethylcyclopropane
Exercise 12-21
ð
Diazomethane is the simplest member of the class of compounds called diazoalkanes or diazo compounds, R2C P N2. When diazo compound A is irradiated in heptane solution at 278 8C, it gives a hydrocarbon, C4H6, exhibiting three signals in 1H NMR and two signals in 13C NMR spectroscopy, all in the aliphatic region. Suggest a structure for this molecule. CH2 P CHCH2CH P N P Nð A
Halogenated carbenes and carbenoids also give cyclopropanes Cyclopropanes may also be synthesized from halogenated carbenes, which are prepared from halomethanes. For example, treatment of trichloromethane (chloroform) with a strong base causes an unusual elimination reaction in which both the proton and the leaving group are removed from the same carbon. The product is dichlorocarbene, which gives cyclopropanes when generated in the presence of alkenes.
531
Reactions of Alkenes
Dichlorocarbene from Chloroform and Its Trapping by Cyclohexene CCl3
A
šð H (CH3)3CO š
šð ðCCl2 ðCl š Dichlorocarbene
ðCCl A 2
(CH3)3COH
ðClð š
CCl2 ! H
ðCCl2
H 59% In another route to cyclopropanes, diiodomethane is treated with zinc powder (usually activated with copper) to generate ICH2ZnI, called the Simmons-Smith* reagent. This species is an example of a carbenoid, or carbenelike substance, because, like carbenes, it also converts alkenes into cyclopropanes stereospecifically. Use of the Simmons-Smith reagent in cyclopropane synthesis avoids the hazards associated with diazomethane preparation. Simmons-Smith Reagent in Cyclopropane Synthesis
Zn–Cu, (CH3CH2)2O
CH2I2
–Metal iodide
H
C O
CH2
CH3
H CO C CH3 ( ( H H3C C
H 3C
G CPC D
O
H
An impressive example of the use of the Simmons-Smith reagent in the construction of natural products is the highly unusual, potent antifungal agent FR-900848, obtained in 1990 from a fermentation broth of Streptoverticillium fervens and first synthesized in 1996. Its most noteworthy feature is the fatty acid residue, which contains five cyclopropanes, four of which are contiguous and all of which were made by Simmons-Smith cyclopropanations. O HN O #
N
O
O
%
H N
%
%
%
%
%
%
%
%
%
%
Chapter 12
G D
532
` OH
≥ OH
FR-900848
In Summary Diazomethane is a useful synthetic intermediate as a methylene source for forming cyclopropanes from alkenes. Halogenated carbenes, which are formed by dehydrohalogenation of halomethanes, and the Simmons-Smith reagent, a carbenoid arising from the reaction of diiodomethane with zinc, also convert alkenes into cyclopropanes. Additions of carbenes to alkenes differ from other addition processes because a single carbon atom becomes bonded to both alkene carbons.
12-10 Oxacyclopropane (Epoxide) Synthesis: Epoxidation by Peroxycarboxylic Acids This section describes how an electrophilic oxidizing agent is capable of introducing a single oxygen atom to connect to both carbons of a double bond. This produces oxacyclopropanes, which may, in turn, be converted into vicinal anti diols. Sections 12-11 and 12-12 *Dr. Howard E. Simmons (1929 – 1997) and Dr. Ronald D. Smith (b. 1930), both with E. I. du Pont de Nemours and Company, Wilmington, Delaware.
12-10 Oxacyclopropane (Epoxide) Synthesis
will show methods for the attachment of oxygen atoms to each alkene carbon to give vicinal syn diols by partial double-bond cleavage or carbonyl compounds by complete double-bond cleavage.
Peroxycarboxylic acids deliver oxygen atoms to double bonds O B The OH group in peroxycarboxylic acids, R COOH, contains an electrophilic oxygen. These compounds react with alkenes by adding this oxygen to the double bond to form oxacyclopropanes. The other product of the reaction is a carboxylic acid. The transformation is of value because, as we know, oxacyclopropanes are versatile synthetic intermediates (Section 9-9). It proceeds at room temperature in an inert solvent, such as chloroform, dichloromethane, or benzene. This reaction is commonly referred to as epoxidation, a term derived from epoxide, one of the older common names for oxacyclopropanes. A popular peroxycarboxylic acid for use in the research laboratory is meta-chloroperoxybenzoic acid (MCPBA). For large-scale and industrial purposes, however, the somewhat shocksensitive (explosive) MCPBA has been replaced by magnesium monoperoxyphthalate (MMPP).
533
Chapter 12
Peroxycarboxylic Acids Electrophilic oxygen
ðOð B RO CO š OO š OH A peroxycarboxylic acid
ðOð B š CH3COOH š Peroxyethanoic (peracetic) acid
Cl
ðOð B COOH
meta-Chloroperoxybenzoic acid (MCPBA)
Oxacyclopropane Formation: Epoxidation of a Double Bond
ðOð
ý Ok
E
G
CP C G G
ðOð B OO H OO š RC O š
E
G
š RCOH
CO C
REACTION
ðOð B COOH
An oxacyclopropane
Electrophilic
The transfer of oxygen is stereospecifically syn, the stereochemistry of the starting alkene being retained in the product. For example, trans-2-butene gives trans-2,3dimethyloxacyclopropane; conversely, the cis-2-butene yields cis-2,3-dimethyloxacyclopropane.
O
O
H
CH2Cl2
H3C CO C H ( ( CH3 H 85% C
CH3
G D
Cl trans-2-Butene
meta-Chloroperoxybenzoic acid (MCPBA)
2
Magnesium monoperoxyphthalate (MMPP)
C
H
G CPC D
O
H 3C
O B COOH
CO ð B ðOð
Mg2
trans-2,3-Dimethyloxacyclopropane
What is the mechanism of this oxidation? It is related but not quite identical to electrophilic halogenation (Section 12-5). In epoxidation, we can write a cyclic transition state in which the electrophilic oxygen is added to the p bond at the same time as the peroxycarboxylic acid proton is transferred to its own carbonyl group, releasing a molecule of carboxylic acid, which is a good leaving group. The two new C – O bonds in the oxacyclopropane product are formally derived from the electron pairs of the alkene p bond and of the cleaved O – H linkage.
MECHANISM
Mechanism of Oxacyclopropane Formation
E
E
B E
R O ðš E E C CE E Ok O CE ð H
E
š OH ER t E C {O A B H {O t
E
E E C B CE
MATION AN I
ANIMATED MECHANISM: Oxacyclopropanation
E
Chapter 12
Reactions of Alkenes
Exercise 12-22 Outline a short synthesis of trans-2-methylcyclohexanol from cyclohexene. (Hint: Review the reactions of oxacyclopropanes in Section 9-9.)
In accord with the electrophilic mechanism, the reactivity of alkenes toward peroxycarboxylic acids increases with alkyl substitution, allowing for selective oxidations. For example, Epoxidation only at more electron-rich double bond O B CH3COOH (1 equivalent), CHCl3, 10°C
Disubstituted
E OH
Monosubstituted
86%
Hydrolysis of oxacyclopropanes furnishes the products of anti dihydroxylation of an alkene Treatment of oxacyclopropanes with water in the presence of catalytic acid or base leads to ring opening to the corresponding vicinal diols. These reactions follow the mechanisms described in Section 9-9: The nucleophile (water or hydroxide) attacks the side opposite the oxygen in the three-membered ring, so the net result of the oxidation – hydrolysis sequence constitutes an anti dihydroxylation of an alkene. In this way, trans-2-butene gives meso-2,3-butanediol, whereas cis-2-butene furnishes the racemic mixture of the 2R,3R and 2S,3S enantiomers. Vicinal Anti Dihydroxylation of Alkenes šð HO C OC
G
D
G C PC D
1. MCPBA, CH2Cl2 2. H, H2O
G
G
ð OH
Synthesis of Isomers of 2,3-Butanediol šð HO
H CH3 G C C H3C ( OH ð H
1. MCPBA, CH2Cl2 2. H, H2O
C G
G
šð HO
H CH3
@
C G
CG C H (R R OH CH3 ð G
D
G
cis-2-Butene
1. MCPBA, CH2Cl2 2. H, H2O
&
H G C PC D CH3 H3C H
meso-2,3-Butanediol
(2R,3R)-2,3-Butanediol
H@ H3 C & GC S
HO ð
G
D
trans-2-Butene
G
H G C PC D H CH3
H3C
@
MODEL BUILDING
&
534
ð OH Gš CC S( H CH3
(2S,3S)-2,3-Butanediol
(Racemic mixture)
Exercise 12-23 Give the products obtained by treating the following alkenes with MCPBA and then aqueous acid: (a) 1-hexene; (b) cyclohexene; (c) cis-2-pentene; (d) trans-2-pentene.
In Summary Peroxycarboxylic acids supply oxygen atoms to convert alkenes into oxacyclopropanes (epoxidation). Oxidation–hydrolysis reactions with peroxycarboxylic acids furnish vicinal diols in a stereospecifically anti manner.
12-11 Vicinal Syn Dihydroxylation with Osmium Tetroxide
Chapter 12
12-11 Vicinal Syn Dihydroxylation with Osmium Tetroxide Osmium tetroxide reacts with alkenes in a two-step process to give the corresponding vicinal diols in a stereospecifically syn manner. The process therefore complements the epoxidation – hydrolysis sequence described in the previous section, which proceeds with anti selectivity. REACTION Vicinal Syn Dihydroxylation with Osmium Tetroxide O O M J Os J M O O
G
D
G C PC D
OH ð Gš CO C
šð HO
1. OsO4, THF, 25°C 2. H2S
G
The process leads initially to an isolable cyclic ester, which is reductively hydrolyzed with H2S or bisulfite, NaHSO3. For example, ðOð ð O ð ðš B O š % Os O ðOð
J
H2S
OH ðOBððš % š OH
)
OsO4, THF, 25°C, 48 h
G
D
M
ðOð B
Osmium tetroxide
90% What is the mechanism of this transformation? The initial reaction of the p bond with osmium tetroxide constitutes a concerted (Section 6-4) addition in which three electron pairs move simultaneously to give a cyclic ester containing Os(VI). This process can be viewed as an electrophilic attack on the alkene: Two electrons flow from the alkene onto the metal, which is reduced [Os(VIII) y Os(VI)]. For steric reasons, the product can form only in a way that introduces the two oxygen atoms on the same face of the double bond — syn. This intermediate is usually not isolated but converted upon reductive work-up into the free diol.
VI O K Os N O O
) ) ) )
O
) )) )
O O MVIII J Os J M O O
) ) ) ) )
) ) ) )
) ) ) )
Mechanism of the Osmium Tetroxide Oxidation of Alkenes OH
H2S
OH
Because OsO4 is expensive and highly toxic, a commonly used modification calls for the use of only catalytic quantities of the osmium reagent and stoichiometric amounts of another oxidizing agent such as H2O2, which serves to reoxidize reduced osmium. An older reagent for vicinal syn dihydroxylation of alkenes is potassium permanganate, KMnO4. Although this reagent functions in a manner that is mechanistically similar to OsO4, it is less useful for synthesizing diols because of a tendency to give poorer yields owing to overoxidation. Potassium permanganate solutions, which are deep purple, are useful as a color test for alkenes, however: Upon reaction, the purple reagent is immediately converted into the brown precipitate of its reduction product, MnO2. Potassium Permanganate Test for Alkene Double Bonds ð OH Gš CO C
šð HO
KMnO4 Dark purple
G
D
G C PC D
MnO2 Brown precipitate
MECHANISM
535
G
CHE M I C AL H I G H L I G H T 1 2 - 2 Synthesis of Antitumor Drugs: Sharpless Enantioselective Oxacyclopropanation and Dihydroxylation In the decade of the 1990s, a significant shift occurred in the way in which new pharmaceuticals were synthesized. Before this time, most of the available methods for the preparation of chiral molecules in enantiomerically pure form were impractical on an industrial scale. Thus, typically, racemic mixtures were generated, although in many cases only one of the enantiomers in such mixtures possessed the desired activity (see Chemical Highlight 5-4). When it was medically and financially advantageous, such mixtures were resolved by classical methods (Section 5-8). However, fundamental conceptual advances in catalysis changed this situation. Some of the most useful examples are a series of highly enantioselective oxidation reactions of double bonds developed by K. B. Sharpless (see Chemical Highlight 5-4). The first such process is a variant of the oxacyclopropanation reaction discussed in Section 12-10, as applied specifically to 2-propenyl (allylic) alcohols. However, instead of
Sharpless Oxacyclopropanation Reagent
H 3C
a peroxycarboxylic acid, the reagent is tert-butyl hydroperoxide in the presence of titanium (IV) isopropoxide (“Sharpless epoxidation”), the function of the chiral auxiliary being assumed by tartaric acid diethyl ester (Chemical Highlight 5-3). The naturally occurring (1)-(2R,3R)-diethyl tartrate and its unnatural (2)-(2S,3S) mirror image are both commercial products. One delivers oxygen to one face of the double bond, the other to the opposite face, as shown below, giving either enantiomer of the oxacyclopropane product with high enantiomer excess (Section 5-2). The reaction works so well because the Ti tetraisopropoxide exchanges all four alkoxides for the four hydroxides derived from the other components of the mixture: two from the tartrate, one from the hydroperoxide, and one from the allylic alcohol. Once all assembled around the central metal, the transition state experiences maximally the stereocontrolling influence of the chiral auxiliary.
Reagent with R,R-tartrate
OH
95% ee
O H 3C H
S S
OH
OH
O tert-Butyl hydroperoxide
Ti[OCH(CH3)2]4
H 3C
Reagent with S,S-tartrate
OH
95% ee
O H 3C H
Titanium(IV) isopropoxide
R R
OH
Transition State for R,R-Tartrate
∑
H COOR ROOC / R R / H
∑
GCOOCH2CH3 R R} ( OH H3CH2COOC H or H GCOOCH2CH3 HO &GC S S} ( H H3CH2COOC OH
HO H &GC
O
O Ti
O
O O
Diethyl tartrate ligand
The role of the chiral coordinated ligand is to provide a pocket into which the substrate can enter in only one spatial orientation. In this respect, it bears the characteristics of many enzymes, biological catalysts that function in essentially the same way (see Chemical Highlight 5-5 and Chapter 26). In the absence of the chiral ligand, a racemic mixture forms. The Sharpless enantioselective oxacyclopropanation has been exploited in the synthesis of many chiral, enantiomerically pure building blocks for the construction of important drugs. An example is the key step of a route to the powerful antitumor agent aclacinomycin A (see also Chemical Highlight 7-1), depicted on top of the next page. 536
Sharpless applied the same principle of using a central metal that can hold a chiral directing group proximal to an alkene substrate in an enantioselective version of the OsO4-catalyzed alkene dihydroxylation (Section 12-11). Here, the essence of the chiral auxiliary is an amine derived from the family of natural alkaloids called the cinchona (Section 25-8). One of these amines is dihydroquinine, which is added in the linked dimeric form shown on p. 537. Instead of H2O2 as the stoichiometric oxidant (Section 12-11), Fe31 [as K3Fe(CN)6] is employed. This method has been applied by E. J. Corey (Section 8-9) to the enantioselective synthesis of ovalicin, a member of a class of fungal-derived natural products called antiangiogenesis agents: They inhibit the growth
CO2CH3 % ` OH
O
OH
A OH
OH
O
O
H3C
≈O ´
Reagent with S,S-tartrate
O
O
A OH
A OH
O
A OH
H3C
≥ O
O
N(CH3)2
O A OH
A OH
O
HO O
of new blood vessels and therefore have the capability to cut off the blood supply to solid tumors. The key to Corey’s synthetic route is the enantioselective syn dihydroxylation of the achiral cyclohexene derivative shown below.
Sharpless Dihydroxylation Reagent
H3C
O Aclacinomycin A
O
OH ´
ECH3
S
Reagent 97% ee
SECH3 ≥ OH
Catalytic OsO4
Transition State CH3
K3Fe(CN)6
B
O
B
N
OO C
H3CO O
OO N N
N
O
CH3CH2
B
O B O Os A O O L *
N
N
CH2CH3
O OCH3
Cinchona-based ligand
Key Step Toward Ovalicin
H3COH
H3CO E
O Dihydroquinine
O
/
O
99% ee
O Cinchona-based ligand
S
H3CO H3CO
OCH3
OCH3
of the molecule and, as of early 2009, it was hoped that this drug would eventually make it to market. O CH3 $ OH O % ´ ≥ H ~ OCH3 O
CH3 CH3
O CH3 $ OH O % ´ ≥ H ~ OCH3 H O N
CH3 CH3
~
A recently identified synthetic derivative of ovalicin, called TNP-470, possesses powerful antiangiogenesis activity, is chemically stable, nontoxic, noninflammatory, and potentially amenable to oral administration. Several preliminary studies revealed the effectiveness of TNP-470 against tumors in animals. By early 2004, TNP-470 had been entered in numerous human clinical trials to determine its applicability toward the treatment of cancers of the breast, brain, cervix, liver, and prostate, as well as AIDS-related Kaposi sarcoma, lymphoma, and leukemia. Unfortunately, as is often the case in drug testing, clinical trials were suspended, in this case because of neurological side effects. These problems were solved by attaching polymeric side chains at the chloride end
OH / OH S ∑ H
Reagent
()-Ovalicin
O
Cl
O TNP-470
537
538
Chapter 12
Reactions of Alkenes
Exercise 12-24 The stereochemical consequences of the vicinal syn dihydroxylation of alkenes are complementary to those of vicinal anti dihydroxylation. Show the products (indicate stereochemistry) of the vicinal syn dihydroxylation of cis- and trans-2-butene.
In Summary Osmium tetroxide, either stoichiometrically or catalytically together with a second oxidizing agent, converts alkenes into syn-1,2-diols. A similar reaction of purple potassium permanganate is accompanied by decolorization, a result that makes it a useful test for the presence of double bonds.
12-12 Oxidative Cleavage: Ozonolysis
REACTION
Ozonolysis Reaction of Alkenes
O3
ðš O š Oð G GC C D D ½O
Reduction [O]
G O CPš D
G
Ozonide
CH3 CH3CH2 G C PC D CH3 H
š O PC
G
D
G C PC D
D
Ozone is a blue gas that condenses to a dark blue, highly unstable liquid. Ozone is a powerful bacteriocide. As a result, ozonators are used for disinfecting water in pools and spas.
Although oxidation of alkenes with osmium tetroxide breaks only the p bond, other reagents may rupture the s bond as well. The most general and mildest method of oxidatively cleaving alkenes is through the reaction with ozone, ozonolysis. The products are carbonyl compounds. Ozone, O3, is produced in the laboratory in an instrument called an ozonator, in which an arc discharge generates 3–4% ozone in a dry oxygen stream. The gas mixture is passed through a solution of the alkene in methanol or dichloromethane. The first isolable intermediate is a species called an ozonide, which is reduced directly in a subsequent step by exposure to zinc in acetic acid or by reaction with dimethyl sulfide. The net result of the ozonolysis– reduction sequence is the cleavage of the molecule at the carbon–carbon double bond; oxygen becomes attached to each of the carbons that had originally been doubly bonded.
Carbonyl products
1. O3, CH2Cl2 2. Zn, CH3COOH
D
G
(Z)-3-Methyl-2-pentene
ðOð B CH3CH2CCH3 90%
ðOð B CH3CH
2-Butanone
Acetaldehyde
The mechanism of ozonolysis proceeds through initial electrophilic addition of ozone to the double bond, a transformation that yields the so-called molozonide. In this reaction, as in several others already presented, six electrons move in concerted fashion in a cyclic transition state. The molozonide is unstable and breaks apart into a carbonyl fragment and a carbonyl oxide fragment through another cyclic six-electron rearrangement. Recombination of the two fragments as shown yields the ozonide. Mechanism of Ozonolysis
MECHANISM
Step 1. Molozonide formation and cleavage C PC C
C MATION AN I
ANIMATED MECHANISM: Ozonolysis
kOk
> KO) k Hp O
O )p
> O p
O p)
A molozonide
C B )O)
C B O)
> )O p
A carbonyl oxide
12-12 Oxidative Cleavage: Ozonolysis
Chapter 12
Step 2. Ozonide formation and reduction
> O p
S ) 2> p (CH 3
O
O
B
D G > O) C B C O) D )> O p
D
G
CO OC O O k k k k
Zn,
2
G O C P> p D
(CH3)2> SP > O p
2
G O C P> p D
> ZnO p
O CH B C 3 O H
Ozonide
Exercise 12-25 An unknown hydrocarbon of the molecular formula C12H20 exhibited an 1H NMR spectrum with a complex multiplet of signals between 1.0 and 2.2 ppm. Ozonolysis of this compound gave two equivalents of cyclohexanone, whose structure is shown in the margin. What is the structure of the unknown?
Exercise 12-26 Give the products of the following reactions. O
(a)
H 3C [
1. O3, CH2Cl2 2. (CH3)2S
CH2
(b)
1. O3, CH2Cl2 O B 2. Zn, CH3COH
CH3
(c)
1. O3, CH2Cl2 2. (CH3)2S
{ CH2 P CH
Exercise 12-27
Working with the Concepts: Deducing the Structure of an Ozonolysis Substrate What is the structure of the following starting material? O C10H16
1. O3 2. (CH3)2S
O
Strategy Begin by counting atoms: How does the molecular formula of the product compare to that of the starting material? Then consider the reaction: What kind of reaction is it and what transformation does it achieve? Putting this information together should enable you to reconstruct the starting material. Solution • The molecular formula of the product is C10H16O2, which is identical to the starting material plus two oxygen atoms. This information simplifies the problem: How were these oxygens introduced? Can we imagine undoing that process in order to identify the original structure? • The reaction is ozonolysis, which achieves the overall transformation O
O
that is, the addition of two oxygen atoms to the original starting species, which is exactly the change observed in the problem.
O
539
540
Chapter 12
Reactions of Alkenes
• Reconstructing the starting molecule therefore requires nothing more than excising the two oxygen atoms and connecting the two carbonyl carbons by a double bond: Remove
O 9
10
8
10
1 2
9
3
8
1
2 3
Connect 7 6
Remove
5 4
4
6 7
5
O
On first sight, this seems easier said than done, because of the way in which we have written the dicarbonyl product. However, if we number the carbons as shown, and recall that carbon–carbon single bonds give rise to flexible molecules with multiple conformations, we find that connecting these two atoms is not so difficult.
Exercise 12-28
Try It Yourself Suggest a structure for a substance that, upon ozonolysis followed by treatment with (CH3)2S, gives as the sole product CH3COCH2CH2CH2CH2CHO. (Hint: Begin by writing out a bond-line formula of this product so that you can clearly see its structure and number its carbon atoms.)
In Summary Ozonolysis followed by reduction yields aldehydes and ketones. Mechanistically, the reactions presented in Sections 12-10 through 12-12 can be related in that initial attack by an electrophilic oxidizing agent on the p bond leads to its rupture. Unlike the reaction sequences studied in Sections 12-10 and 12-11, however, ozonolysis causes cleavage of both the p and the s bonds.
12-13 Radical Additions: Anti-Markovnikov Product Formation In this section, all radicals and single atoms are shown in green, as in Chapter 3.
REACTION
Markovnikov Addition of HBr CH3CH2CH P CH2 (Freshly distilled) HBr 24 h
šð ðBr A CH3CH2CHCH2H 90%
Radicals, lacking a closed outer shell of electrons, are capable of reacting with double bonds. However, a radical requires only one electron for bond formation, unlike the electrophiles presented in this chapter so far, which consume both electrons of the p bond upon addition. The product of radical addition to an alkene is an alkyl radical, and the final products exhibit anti-Markovnikov regiochemistry, similar to the products of hydroboration– oxidation (Section 12-8).
Hydrogen bromide can add to alkenes in anti-Markovnikov fashion: a change in mechanism When freshly distilled 1-butene is exposed to hydrogen bromide, clean Markovnikov addition to give 2-bromobutane is observed. This result is in accord with the ionic mechanism for electrophilic addition of HBr discussed in Section 12-3. Curiously, the same reaction, when carried out with a sample of 1-butene that has been exposed to air, proceeds much more quickly and gives an entirely different result. In this case, we isolate 1-bromobutane, formed by anti-Markovnikov addition. This change caused considerable confusion in the early days of alkene chemistry, because one researcher would obtain only one hydrobromination product, whereas another would obtain a different product or mixtures from a seemingly identical reaction. The mystery was solved by Kharasch* in the 1930s, when it was discovered that the culprits responsible for antiMarkovnikov additions were radicals formed from peroxides, ROOR, in alkene samples that had been stored in the presence of air. In practice, to effect anti-Markovnikov hydrobromination, radical initiators, such as peroxides, are added deliberately to the reaction mixture.
Markovnikov product (By ionic mechanism)
*Professor Morris S. Kharasch (1895 – 1957), University of Chicago.
12-13 Anti-Markovnikov Regiochemistry of Radical Additions
The mechanism of the addition reaction under these conditions is not an ionic sequence; rather, it is a much faster radical chain sequence. The reason is that the activation energies of the component steps of radical reactions are very small, as we observed earlier during the discussion of the radical halogenation of alkanes (Section 3-4). Consequently, in the presence of radicals, anti-Markovnikov hydrobromination simply outpaces the regular addition pathway. The initiation steps are
Chapter 12
Anti-Markovnikov Addition of HBr CH3CH2CH P CH2 (Exposed to oxygen) HBr 4h
1. the homolytic cleavage of the weak RO–OR bond [DH 8 < 39 kcal mol21 (163 kJ mol21)],
and 2. reaction of the resulting alkoxy radical with hydrogen bromide.
The driving force for the second (exothermic) step is the formation of the strong O–H bond. The bromine atom generated in this step initiates chain propagation by attacking the double bond. One of the p electrons combines with the unpaired electron on the bromine atom to form the carbon–bromine bond. The other p electron remains on carbon, giving rise to a radical. The halogen atom’s attack is regioselective, creating the relatively more stable secondary radical rather than the primary one. This result is reminiscent of the ionic additions of hydrogen bromide (Section 12-3), except that the roles of the proton and bromine are reversed. In the ionic mechanism, a proton attacks first to generate the more stable carbocation, which is then trapped by bromide ion. In the radical mechanism, a bromine atom is the attacking species, creating the more stable radical center. The alkyl radical subsequently reacts with HBr by abstracting a hydrogen and regenerating the chain-carrying bromine atom. Both propagation steps are exothermic, and the reaction proceeds rapidly. As usual, termination is by radical combination or by some other removal of the chain carriers (Section 3-4).
541
H A š CH3CH2CHCH2Br ð 65% Anti-Markovnikov product (By radical mechanism)
Mechanism of Radical Hydrobromination Initiation steps
MECHANISM Δ
š RO OR Oš š RO j
Δ
š HðBr ð
š 2 RO j š ROH
ΔH° 39 kcal mol1 (163 kJ mol1)
š ðBr j
ΔH° 17 kcal mol1 (71 kJ mol1)
MATION AN I
Propagation steps H G C P CH2 D CH3CH2 CH3CH2CHCH2Br
šð CH3CH2CH O CH2Br
š jBr ð
Secondary radical
ΔH° 5 kcal mol1 (21 kJ mol1)
H A š š ð ðBrj CH3CH2CHCH2Br
šð HðBr
ΔH° 11.5 kcal mol1 (48 kJ mol1)
Are radical additions general? Hydrogen chloride and hydrogen iodide do not give anti-Markovnikov addition products with alkenes; in both cases, one of the propagating steps is endothermic and consequently so slow that the chain reaction terminates. As a result, HBr is the only hydrogen halide that adds to an alkene under radical conditions to give anti-Markovnikov products. Additions of HCl and HI proceed only by ionic mechanisms to give normal Markovnikov products regardless of the presence or absence of radicals. Other reagents such as thiols, however, do undergo radical additions to alkenes. Radical Addition of a Thiol to an Alkene CH3CH P CH2
p CH3CH2p SH Ethanethiol
ROOR
p SCH2CH3 CH3CHCH2p A H Ethyl propyl sulfide
ANIMATED MECHANISM: Radical hydrobromination of 1-butene
542
Chapter 12
(CH3)3CH EOH O C(CH3)3 Bis(1,1-dimethylethyl) peroxide (Di-tert-butyl peroxide)
O H5C6
Reactions of Alkenes
In this example, the initiating alkoxy radical abstracts a hydrogen from sulfur to yield .. ., which then attacks the double bond. Bis(1,1-dimethylethyl) peroxide (di-tert-butyl CH3CH2 S .. peroxide) and dibenzoyl peroxide are commercially available initiators for such radical addition reactions. Exercise 12-29
EO O
C6H5 O
Dibenzoyl peroxide
Ultraviolet irradiation of a mixture of 1-octene and diphenylphosphine, (C6H5)2PH, furnishes 1-(diphenylphosphino)octane by radical addition. Write a plausible mechanism for this reaction. hv
(C6H5 ) 2PH 1 H2C“CH(CH2 ) 5CH3 vy (C6H5 ) 2POCH2OCH2 (CH2 ) 5CH3
As we saw in the case of hydroboration (Section 12-8), anti-Markovnikov additions are synthetically useful because their products complement those obtained from ionic additions. The ability to control regiochemistry is an important feature in the development of new synthetic methods.
In Summary Radical initiators alter the mechanism of the addition of HBr to alkenes from ionic to radical chain. The consequence of this change is anti-Markovnikov regioselectivity. Other species, most notably thiols, but not HCl or HI, are capable of undergoing similar reactions.
12-14 Dimerization, Oligomerization, and Polymerization of Alkenes Is it possible for alkenes to react with one another? Indeed it is, but only in the presence of an appropriate catalyst — for example, an acid, a radical, a base, or a transition metal. In this reaction the unsaturated centers of the alkene monomer (monos, Greek, single; meros, Greek, part) are linked to form dimers, trimers, oligomers (oligos, Greek, few, small), and ultimately polymers ( polymeres, Greek, of many parts), substances of great industrial importance. Polymerization
PC
D G D G D G CP C PC C PC G D G D G D
A A A A A A O C OC OC OC OC OC O A A A A A A
Monomers
Polymer
Carbocations attack pi bonds Treatment of 2-methylpropene with hot aqueous sulfuric acid gives two dimers: 2,4,4trimethyl-1-pentene and 2,4,4,-trimethyl-2-pentene. This transformation is possible because 2methylpropene can be protonated under the reaction conditions to furnish the 1,1-dimethylethyl (tert-butyl) cation. This species can attack the electron-rich double bond of 2-methylpropene with formation of a new carbon–carbon bond. Electrophilic addition proceeds according to the Markovnikov rule to generate the more stable carbocation. Subsequent deprotonation from either of two adjacent carbons furnishes a mixture of the two observed products. REACTION
Dimerization of 2-Methylpropene
CH2 P C(CH3)2
CH2 P C(CH3)2
H
CH3 CH3 A A CH3CCH2C P CH2 A CH3 2,4,4-Trimethyl1-pentene
CH3 A CH3CCH P C(CH3)2 A CH3 2,4,4-Trimethyl2-pentene
12-15 Synthesis of Polymers
Chapter 12
543
Mechanism of Dimerization of 2-Methylpropene CH3 D CH2 P C G CH3 CH3 A CH3C A CH3
H
CH3 D CH3C G CH3
H A b DCH3 CO C G A 2 H a CH A H
H
MECHANISM
CH3 D CH2 PC G CH3
CH3 CH3 A A CH3CCH2C P CH2 A CH3
From deprotonation a
CH3 A CH3CCHP C(CH3)2 A CH3 From deprotonation b
Repeated attack can lead to oligomerization and polymerization The two dimers of 2-methylpropene tend to react further with the starting alkene. For example, when 2-methylpropene is treated with strong acid under more stringent conditions, trimers, tetramers, pentamers, and so forth, are formed by repeated electrophilic attack of intermediate carbocations on the double bond. This process, which leads to alkane chains of intermediate length, is called oligomerization. Oligomerization of the 2-Methylpropene Dimers CH3 A CH3CCH P C(CH3)2 A CH3
CH3 CH2 A J CH3CCH2C G A CH3 CH3
CH3 CH3 CH3 A A D CH3CCH2CCH2C G A A CH3 CH3 CH3
H
CH3 CH3 A D CH3CCH2C G A CH3 CH3
CH3 D CH2 P C G CH3
CH2 PC
CH3 D G CH3
CH3 CH3 CH3 CH3 A A A D CH3CCH2CCH2CCH2C G A A A CH3 CH3 CH3 CH3
etc.
At higher temperatures, the oligomerization of alkenes continues to give polymers containing many subunits. Polymerization of 2-Methylpropene
n CH2 P C(CH3)2
H, 200°C
CH3 CH3 A A H O (CH2 O C)n1 CH2C P CH2 A CH3 Poly(2-methylpropene) (Polyisobutylene)
In Summary Catalytic acid causes alkene – alkene additions to occur, a process that forms dimers, trimers, oligomers containing several components, and finally polymers, which are composed of a great many alkene subunits.
12-15 Synthesis of Polymers Many alkenes are suitable monomers for polymerization. Polymerization is exceedingly important in the chemical industry, because many polymers have desirable properties, such as durability, inertness to many chemicals, elasticity, transparency, and electrical and thermal resistance. Although the production of polymers has contributed to pollution — many of them are not biodegradable — they have varied uses as synthetic fibers, films, pipes, coatings, and
This spectacular dress designed by Spanish born designer Paco Rabanne would not have been possible without synthetic polymers.
544
Table 12-3
Reactions of Alkenes
Chapter 12
Common Polymers and Their Monomers
Monomer
Structure
Polymer (common name)
Structure
Uses
Ethene
H2C P CH2
Polyethylene
O (CH2CH2)n O
Food storage bags, containers
Chloroethene (vinyl chloride)
H2C P CHCl
Poly(vinyl chloride) (PVC)
O (CH2CH)n O A Cl
Pipes, vinyl fabrics
Tetrafluoroethene
F2C P CF2
Teflon
O (CF2CF2)n O
Nonstick cookware
Polystyrene
O (CH2CH)n O
Foam packing material
Orlon
O (CH2CH)n O A CN
Clothing, synthetic fabrics
CHP CH2
Ethenylbenzene (styrene)
Propenenitrile (acrylonitrile)
Methyl 2-methylpropenoate (methyl methacrylate)
2-Methylpropene (isobutylene)
H2C P C
H G D CqN
CH G 3 H2C P C D COCH3 B O CH G 3 H2C P C D CH3
Plexiglas
CH3 A O (CH2C)n O A CO2CH3
Impact-resistant paneling
Elastol
CH3 A O (CH2C)n O A CH3
Oil-spill clean-up
molded articles. Polymers are also being used increasingly as coatings for medical implants. Names such as polyethylene, poly(vinyl chloride) (PVC), Teflon, polystyrene, Orlon, and Plexiglas (Table 12-3) have become household words. Acid-catalyzed polymerizations, such as that described for poly(2-methylpropene), are carried out with H2SO4, HF, and BF3 as the initiators. Because they proceed through carbocation intermediates, they are also called cationic polymerizations. Other mechanisms of polymerizations are radical, anionic, and metal catalyzed.
Radical polymerizations lead to commercially useful materials An example of radical polymerization is that of ethene in the presence of an organic peroxide at high pressures and temperatures. The reaction proceeds by a mechanism that, in its initial stages, resembles that of the radical addition to alkenes (Section 12-13). The peroxide initiators cleave into alkoxy radicals, which begin polymerization by addition to the double bond of ethene. The alkyl radical thus created attacks the double bond of another ethene molecule, furnishing another radical center, and so on. Termination of the polymerization can be by dimerization, disproportionation of the radical, or other radical-trapping reactions (Section 3-4). Mechanism of Radical Polymerization of Ethene Initiation steps š RO OR Oš š RO j
š RO j CH2 P CH2
j š ROCH 2 O CH2
12-15 Synthesis of Polymers
Chapter 12
545
C H E MI C AL H I G H L I G H T 1 2 - 3 Polymers in the Clean-Up of Oil Spills Poly-2-methylpropene (polyisobutylene) is the principal ingredient in a product called Elastol, which over two decades of use has been demonstrated to be an effective agent in the clean-up of oil spills. When Elastol is sprayed on an oil slick, its polymer chains, which normally wrap tightly around each other, unravel and mix with the oil. The oil is bound by the polymer into a viscous mat, which may be skimmed off the surface of a body of water. Elastol can absorb 100 times its volume of medium- to heavy-weight
petroleum oil. A valuable feature of this method is that the oil may be recovered from the polymer by processing through a special pump. Elastol is also effective for removing contaminants from groundwater, such as aromatic hydrocarbons and MTBE (methyl tert-butyl ether, which replaced lead as an octane enhancer for gasoline but whose seepage into groundwater supplies created environmental concerns). Elastol has been used to remediate petroleum spills in New Haven Harbor in Connecticut, on the Potomac and Neches (Texas) rivers, and around the oil pipelines in the Russian Arctic. Elastol has also proven to be useful in safely cleaning birds and other animals affected by oil spills. In particular, it selectively binds and removes the oil-based contaminants without affecting the natural skin oils of the animal.
This penguin is the victim of our oil-based economy: oil spill off the coast of South Africa.
n Poly(2-methylpropene) (Polyisobutylene)
Propagation steps š ROCH 2CH2j
CH2 P CH2
š ROCH 2CH2CH2CH2j
(n 1) CH2 P CH2
š ROCH 2CH2CH2CH2j š RO O (CH2CH2)n O CH2CH2j
Polyethene (polyethylene) produced in this way does not have the expected linear structure. Branching occurs by abstraction of a hydrogen along the growing chain by another radical center followed by chain growth originating from the new radical. The average molecular weight of polyethene is almost 1 million. Polychloroethene [poly(vinyl chloride), PVC] is made by similar radical polymerization. Interestingly, the reaction is regioselective. The peroxide initiator and the intermediate chain radicals add only to the unsubstituted end of the monomer, because the radical center formed next to chlorine is relatively stable. Thus, PVC has a very regular head-to-tail structure of molecular weight in excess of 1.5 million. Although PVC itself is fairly hard and brittle, it can be softened by addition of carboxylic acid esters (Section 20-4), called plasticizers (plastikos, Greek, to form). The resulting elastic material is used in “vinyl leather,” plastic covers, and garden hoses. CH2 P CHCl
ROOR
MECHANISM
O (CH2CH)nO A Cl Polychloroethene [Poly(vinyl chloride)]
Exposure to chloroethene (vinyl chloride) has been linked to the incidence of a rare form of liver cancer (angiocarcinoma). The Occupational Safety and Health Administration (OSHA) has set limits to human exposure of less than an average of 1 ppm per 8-h working day per worker.
Radical Polymerization of Ethene n CH2 P CH2 ROOR, 1000 atm, >100°C
O (CH2 O CH2)nO Polyethene (Polyethylene) (Plus branched isomers)
546
Chapter 12
Branching in Polyethene (Polyethylene) H A CH2CCH2CH2 A CH2 A CH2
Reactions of Alkenes
An iron compound, FeSO4, in the presence of hydrogen peroxide promotes the radical polymerization of propenenitrile (acrylonitrile). Polypropenenitrile (polyacrylonitrile), – (CH2CHCN)n – , also known as Orlon, is used to make fibers. Similar polymerizations of other monomers furnish Teflon and Plexiglas. Exercise 12-30 Prior to 2005, Saran Wrap was made by radical polymerization of 1,1-dichloroethene and chloroethene together. Propose a structure. Note: This is a “copolymerization,” in which the two monomers alternate in the final polymer.
Anionic polymerizations require initiation by bases Anionic polymerizations are initiated by strong bases such as alkyllithiums, amides, alkoxides, and hydroxide. For example, methyl 2-cyanopropenoate (methyl a-cyanoacrylate) polymerizes rapidly in the presence of even small traces of hydroxide. When spread between two surfaces, it forms a tough, solid film that cements the surfaces together. For this reason, commercial preparations of this monomer are marketed as Super Glue. What accounts for this ease of polymerization? When the base attacks the methylene group of a-cyanoacrylate, it generates a carbanion whose negative charge is located next to the nitrile and ester groups, both of which are strongly electron withdrawing. The anion is stabilized because the nitrogen and oxygen atoms polarize their multiple bonds in the sense d1C q Nd2 and d1C P Od2 and because the charge can be delocalized by resonance. Anionic Polymerization of Super Glue (Methyl -Cyanoacrylate)
š HO ð
ðOð B COCH3 D CH2 P C G C q Nð
Methyl 2-Cyanopropenoate
ðOð B COCH 3 D HOCH2 š C G C q Nð
ðOð B COCH3 D CH2 P C G C q Nð
ON EOCH3 ðOð ðš B C COCH 3 A D C HOCH2C CH 2š G A CNð CNð
etc.
(␣-Cyanoacrylate, Super Glue)
Metal-catalyzed polymerizations produce highly regular chains An important metal-catalyzed polymerization is that initiated by Ziegler-Natta* catalysts. They are typically made from titanium tetrachloride and a trialkylaluminum, such as triethylaluminum, Al(CH2CH3)3. The system polymerizes alkenes, particularly ethene, at relatively low pressures with remarkable ease and efficiency. Although we shall not consider the mechanism here, two features of Ziegler-Natta polymerization are the regularity with which substituted alkane chains are constructed from substituted alkenes, such as propene, and the high linearity of the chains. The polymers that result possess higher density and much greater strength than those obtained from radical polymerization. An example of this contrast is found in the properties of polyethene (polyethylene) prepared by the two methods. The chain branching that occurs during radical polymerization of ethene results in a flexible, transparent material (low-density polyethylene) used for food storage bags, whereas the Ziegler-Natta method produces a tough, chemically resistant plastic (high-density polyethylene) that may be molded into containers. *Professor Karl Ziegler (1898 – 1973), Max Planck Institute for Coal Research, Mülheim, Germany, Nobel Prize 1963 (chemistry); Professor Giulio Natta (1903 – 1979), Polytechnic Institute of Milan, Nobel Prize 1963 (chemistry).
12-16 Ethene: An Important Industrial Feedstock
In Summary Alkenes are subject to attack by carbocations, radicals, anions, and transition metals to give polymers. In principle, any alkene can function as a monomer. The intermediates are usually formed according to the rules that govern the stability of charges and radical centers.
12-16 Ethene: An Important Industrial Feedstock Ethene (ethylene) can serve as a case study for the significance of alkenes in industrial chemistry. This monomer is the basis for the production of polyethene (polyethylene), millions of tons of which are manufactured in the United States annually. The major source of ethene is the pyrolysis of petroleum, or hydrocarbons derived from natural gas, such as ethane, propane, other alkanes, and cycloalkanes (Section 3-3). Apart from its direct use as a monomer, ethene is the starting material for many other industrial chemicals. For example, acetaldehyde is obtained in the reaction of ethene with water in the presence of a palladium(II) catalyst, air, and CuCl2. The product formed initially, ethenol (vinyl alcohol), is unstable and spontaneously rearranges to the aldehyde (see Chapters 13 and 18). The catalytic conversion of ethene into acetaldehyde is also known as the Wacker* process. The Wacker Process CH2 P CH2
P
ðOð H2O, O2, catalytic PdCl2, CuCl2
š CH2 P CHOH
CH3CH
Ethenol (Vinyl alcohol)
Acetaldehyde
(Unstable)
Chloroethene (vinyl chloride) is made from ethene by a chlorination – dehydrochlorination sequence in which addition of Cl2 produces 1,2-dichloroethane. This compound is converted into the desired product by elimination of HCl. Chloroethene (Vinyl Chloride) Synthesis CH2 P CH2
Cl2
CH2 O CH2 A A Cl ðCl ð ð ð 1,2-Dichloroethane
HCl
šð CH2 P CHCl Chloroethene (Vinyl chloride)
Oxidation of ethene with oxygen in the presence of silver furnishes oxacyclopropane (ethylene oxide), the hydrolysis of which gives 1,2-ethanediol (ethylene glycol) (Section 9-11). Hydration of ethene gives ethanol (Section 9-11).
Oxacyclopropane (Ethylene oxide)
CH2
D
OH ðš ðš OH H, H2O
D
GýOk H2C CH2 G
CH2 P CH2
O2, catalytic Ag
CH2
1,2-Ethanediol (Ethylene glycol)
In Summary Ethene is a valuable source of various industrial raw materials, including ethanol, 1,2-ethanediol (ethylene glycol), and several important monomers for the polymer industry.
*Dr. Alexander Wacker (1846 – 1922), Wacker Chemical Company, Munich, Germany.
Chapter 12
547
548
Chapter 12
Reactions of Alkenes
12-17 Alkenes in Nature: Insect Pheromones Many natural products contain p bonds; several were mentioned in Sections 4-7 and 9-11. This section describes a specific group of naturally occurring alkenes, the insect pheromones (pherein, Greek, to bear; hormon, Greek, to stimulate).
Insect Pheromones O O
H
O B OCCH3
H
≥
H
H
H
H
H European vine moth
Japanese beetle
CHE M I C AL H I G H L I G H T 1 2 - 4 Metal-Catalyzed Alkene Metathesis for Constructing Medium and Large Rings One of the most startling examples of metal-catalyzed chemistry is alkene metathesis, a reaction in which two alkenes exchange their double-bonded carbon atoms, as shown in general form here. W2C P CX2
Catalyst
CW2
B
CY2
Y2C P CZ2
CX2
B
CZ2
The equilibrium in this reversible process may be driven by removal of one of the four components (Le Chatelier’s principle). This idea has been employed to form medium- and large-sized rings that are otherwise very difficult to construct, because of unfavorable strain and entropy factors. The example below illustrates ring closure of an acyclic starting compound with terminal double bonds. The products are a cyclic alkene and ethene, which, being a gas, evolves rapidly out of the reaction mixture, driving the equilibrium toward product.
i
Catalyst
i
i
C P CH2
C
i
i H H
i C B
H
i
C P CH2
Ciguatoxin is 100 times more poisonous than brevetoxin, the “red tide” toxin (Section 9-5), and is responsible for more human poisoning from seafood consumption than any other substance: More than 20,000 people become sick in this way each year, developing gastrointestinal, cardiovascular, and neurological abnormalities that can lead to paralysis, coma, and death. The tiny amounts of ciguatoxin in fish are far too small to have any effect on the flavor or odor of the food. Thus, a supply of this material is needed to develop sensitive detection methods. Ciguatoxin contains 13 ether rings and 30 stereocenters, making it a formidable synthetic target. Its synthesis was an amazing 12-year effort, culminating in the linkage of two polycyclic molecules containing five and seven rings, respectively, and closure of the final nine-membered “F” ring using alkene metathesis. The catalyst, developed by
iH
CH2
B
CH2
A recent, extraordinary application of methathesis is the synthesis of ciguatoxin by Hirama.* Ciguatoxin is produced by marine microorganisms associated with algae and is accumulated by some 400 species of warm-water reef fish. *Professor Masahiro Hirama (b. 1948), Tohoku University, Japan.
The coral grouper can turn lethal when ciguatoxin accumulates in its tissue.
12-17 Alkenes in Nature: Insect Pheromones
O $
O
OH H3 C
KO
/
B O
/
CH3
H
H3C CH2
B O
H
CH3
Male boll weevil
549
H
∑
CH3 H
CH3 A
Chapter 12
American cockroach
(Both enantiomers) Defense pheromones of larvae of chrysomelid beetle
This pheromone trap removes the males of the pea moth Cydia nigricana from circulation.
Pheromones are chemical substances used for communication within a living species. There are sex, trail, alarm, and defense pheromones, to mention a few. Many insect pheromones are simple alkenes; they are isolated by extraction of certain parts of the insect and separation of the resulting product mixture by chromatographic techniques. Often only minute quantities of the bioactive compound can be obtained, in which case the synthetic organic
BnO
A
≥
E
%
H O H
J
≥
H
H
I
H
≥ ≥ O H
≥ O ≥ H
G
CH3 H O H
%
H3C
)
L O
O M
(R cyclohexyl)
%
CH2
≥ ≥
(Bn “benzyl” protecting group,
%
H O H
PR3 Cl @ A Ru ) P CH Cl A PR3, CH2Cl2, 40C
OBn H
K
≥
OBn
% CH3 ´
≥ O ≥ O H ≥ H H
H
H
% O %
≥
H % O % % O % B C D H
H
H
´
% CH3 H % O %
0 H3C
HO
H % O %
H
H A
B
% O %
C
≥ O ≥ H
G
≥ ≥ O H H
F
H
I
%
H O H
J
H
≥
H O H
%
H
≥
0
O M
%
H3 C
H2CCH2
Grubbs,* contains a ruthenium atom linked by a double bond to carbon, an example of a so-called metal carbene complex. Metal carbenes were proposed (by Chauvin)† and confirmed as intermediates in alkene metathesis decades ago. Subsequently,
L O
≥
OH H
K
O H
%
Ciguatoxin
% CH3 ´
OH
E
≥
≥ O ≥ ≥ O H H ≥ H
D
% O %
≥
Na, NH3 (Removes Bn groups)
H
H
H
CH3
´
% CH3 H % O %
H3C
Grubbs and Schrock‡ prepared stable carbenes of Ru and Mo, respectively, that catalyze the process in a well-controlled manner. Alkene metathesis is now one of the most reliable and widely used methods for medium- and large-ring construction.
*Professor Robert H. Grubbs (b. 1942), California Institute of Technology, Nobel Prize 2005 (chemistry). † Professor Yves Chauvin (b. 1930), Institut Français du Pétrole, Rueil-Malmaison, France, Nobel Prize 2005 (chemistry). ‡ Professor Richard R. Schrock (b. 1945), Massachusetts Institute of Technology, Nobel Prize 2005 (chemistry).
550
Chapter 12
HO
H H
Reactions of Alkenes
chemist can play a very important role in the design and execution of total syntheses. Interestingly, the specific activity of a pheromone frequently depends on the configuration around the double bond (e.g., E or Z), as well as on the absolute configuration of any chiral centers present (R, S) and the composition of isomer mixtures. For example, the sex attractant for the male silkworm moth, 10-trans-12-cis-hexadecadien-1-ol (known as bombykol, margin), is 10 billion times more active in eliciting a response than the 10-cis-12-trans isomer, and 10 trillion times more active than the trans, trans compound. As is the case for the juvenile hormones (Chemical Highlight 12-1), pheromone research affords an important opportunity for achieving pest control. Minute quantities of sex pheromones can be used per acre of land to confuse male insects about the location of their female partners. These pheromones can thus serve as lures in traps to remove insects effectively without spraying crops with large amounts of other chemicals. It is clear that organic chemists in collaboration with insect biologists will make important contributions in this area in the years to come.
H
THE BIG PICTURE H Bombykol
We saw in Chapter 11 that in the double bonds of alkenes the p bond is weaker than the s bond. Thus, the general process of addition, in which p bonds are replaced by s bonds, is generally energetically favorable. We also found that although the CP C bond is nonpolar, it has a higher concentration of electrons than does a single bond. These two features are the basis of the reactivity of the alkene function. Thus, the alkene double bond acts as a nucleophile, and many of its reactions begin with attack by an electrophile on the p electrons. In this chapter, we found that alkene chemistry subdivides into several categories, including reactions with simple electrophiles, such as the proton, to give carbocations, followed by attachment of a nucleophile to give the final addition product. These processes are very much like the familiar acid-base reactivity we discussed in Chapter 2. Alkenes also undergo reactions that proceed through more complex pathways, often involving ring formation. Issues of regio- and stereochemistry apply, depending on the specific mechanism, but the underlying theme of nucleophile – electrophile interaction is almost always present. Many of the same ideas in alkene chemistry also apply to the carbon – carbon triple bond. In Chapter 13, we shall examine the behavior of alkynes and see that much of their chemistry is a direct extension of alkene reactivity to a system with two p bonds.
CHAPTER INTEGRATION PROBLEMS 12-31. Compare and contrast the addition reactions of each of the following reagents with (E )-3methyl-3-hexene: H2 (catalyzed by PtO2), HBr, dilute aqueous H2SO4, Br2 in CCl4, mercuric acetate in H2O, and B2H6 in THF. Consider regiochemistry and stereochemistry. Which of these reactions can be used to synthesize alcohols? In what respects do the resulting alcohols differ?
3
SOLUTION First, we need to identify the starting material’s structure. Recall (Section 11-1) that the designation “E” describes the stereoisomer in which the two highest-priority groups (according to Cahn-IngoldPrelog guidelines, Section 5-3) are on opposite sides of the double bond (i.e., trans to each other). In 3-methyl-3-hexene, the two ethyl groups are of highest priority. We therefore are starting with the compound shown in the margin. Now let us consider the behavior of this alkene toward our list of reagents. In each case, it may be necessary to choose between two different regiochemical and two different stereochemical modes of addition. We recognize this situation because the starting alkene is substituted differently at each carbon, a regiochemical characteristic, and it has a defined (E) stereochemistry. To answer such questions completely correctly, it is essential to consider the mechanism of each reaction — that is, to think mechanistically. Thus, the addition of H2 with PtO2 as a catalyst is an example of catalytic hydrogenation. Because the same kind of atom (hydrogen) adds to each of the alkene carbons, regiochemistry is not a consideration. Stereochemistry may be one, however. Catalytic hydrogenation is a syn addition, in which both hydrogen atoms attach to the same face of the alkene p bond (Section 12-2). If we view the
Chapter Integration Problems
Chapter 12
alkene in a plane perpendicular to the plane of the page (Figure 12-1), addition will be on the top face 50% of the time and on the bottom face 50% of the time:
H3C * CH3CH2
CH3CH2 H3C * C
H CH2CH3 H
H
H C CH2CH3
H3C ≈ ≈ CH2CH3 H CH3CH2 H OH
H
)
C
C
)
H
)
)
H OH H3C ≈ ≈ CH2CH3 H CH3CH2
Racemic mixture
The addition generates one stereocenter (marked in each of the two products in the center of the scheme by an asterisk); therefore each product molecule is chiral (Section 5-1). Because the products form in equal amounts, the consequence is a racemic mixture of (R)- and (S)-3-methylhexane. The next two reactions, with HBr and with aqueous H2SO4, begin with addition of the electrophile H1 (Sections 12-3 and 12-4). These processes generate carbocations and follow the regiochemical guideline known as Markovnikov’s rule: Addition is by attachment of H1 to the less substituted alkenyl carbon to give the more stable carbocation. The carbocation is trapped by any available nucleophile — Br2 in the case of HBr, and H2O in the case of aqueous H2SO4. Both steps proceed without stereoselectivity and, because the carbocation formed is already tertiary, rearrangement to a more stable carbocation is not possible. So we have the following result: H
)
≈ CH2CH3 H
H3C ≈ CH3CH2
C
)
)
H3C ≈ CH3CH2
Nucleophile (Br or H2O) adds to top or bottom
H
X A CH3CH2 O C O CH2CH2CH3 A CH3
CH2CH3 H
X Br (from HBr) or OH (from aqueous H2SO4)
Racemic mixture
The next two are examples of additions of electrophiles that form bridged cationic intermediates: a cyclic bromonium ion in the first case (Section 12-5) and a cyclic mercurinium ion in the second (Section 12-7). Additions therefore proceed stereospecifically anti, because the cyclic ion can be attacked only from the direction opposite the location of the bridging electrophile (Figure 12-3). With Br2, identical atoms add to both alkene carbons, so regiochemistry is not a consideration. In oxymercuration, the nucleophile is water, and it will add to the most substituted alkene carbon, because the latter is tertiary and possesses the greatest partial positive charge. Stereochemistry must be considered because addition of the electrophile occurs with equal probability from the top and from the bottom and stereocenters are created. So we have
)
H3C CH3CH2
C
Nucleophile (Br or H2O) adds only to bottom
CH2CH3 H
and only to C3 (tertiary position)
)Nu
E Br (from Br2) or Hg(O2CCH3) (from Hg[O2CCH3]2)
)
)
≈ CH2CH3 H
H C CH2CH3
CH3CH2 H3C C 3
E
Nu
E C ( CH2CH3 H
E Br (from Br2) or Hg(O2CCH3) Nu Br (from Br2) or OH (from H2O)
)Nu H3C ≈ CH3CH2
CH CH3CH2 & @ 3
Nucleophile (Br or H2O) adds only to top and only to C3 (tertiary position)
E
Nu
C
≈ CH2CH3 H
C
)
H3C ≈ CH3CH2
E 3
H3C ( CH3CH2
CH2CH3 H
&@
E
E
Racemic mixture
Equal amounts — a racemic mixture — of the two chiral, enantiomerically related products are formed. Finally, we come to hydroboration. Again, both regio- and stereoselectivity must be considered. As in hydrogenation, the stereochemistry is syn; unlike the preceding electrophilic additions, the regiochemistry is anti-Markovnikov: Boron attaches to the less substituted alkene carbon:
H3C CH3 CH2
C
BH2 CH2CH3 H
CH3 CH2 H3C C H
Racemic mixture
H C CH2CH3 BH2
H3C ≈ ≈ CH2CH3 H CH3CH2 HO BH2
)
C
)
H
)
)
HO BH2 H3C ≈ ≈ CH2CH3 H CH3CH2
551
Reactions of Alkenes
Chapter 12
For simplicity, addition of only one of the B – H bonds is shown. The picture strongly resembles that of hydrogenation, but, because of the unsymmetric nature of the reagent, both alkene carbons are now transformed into stereocenters. Three of the six reactions are well suited for the synthesis of alcohols: acid-catalyzed hydration, oxymercuration (after reduction of the C – Hg bond by NaBH4), and hydroboration (by oxidation of the C – B bond by H2O2). Compare the alcohol from hydration (shown earlier) with that from demercuration of the oxymercuration product: HO
CH2CH3 ? C H
C
H3C CH3 CH2
H3C CH3 CH2 ?Δ
C
HO
Hg(O2CCH3)
Hg(O2CCH3)
NaBH4, NaOH, H2O
CH2CH3 H
OH A CH3CH2 O C O CH2CH2CH3 A CH3 Racemic mixture
Racemic mixture
They are the same. Had a rearrangement taken place during hydration, this outcome might not have been the case (Section 9-3). Oxidation of the hydroboration product gives a different, regioisomeric alcohol having two stereocenters, also as a racemic mixture (the boron atoms are shown without the additional alkyl groups): B
CH2CH3 H
CH3CH2 H3C C H
H2O2, NaOH, H 2O
H C CH2CH3 iB D
H H3C CH3 CH2
Racemic mixture
C
OH CH2CH3 H
CH3 CH2 H3C C H
H C CH2CH3 OH
Racemic mixture
12-32. Thujene, which occurs in numerous plant oils, possesses the molecular formula C10H16 and is a monoterpene (Section 4-7). Several chemical and spectroscopic characteristics of thujene follow. What does each of these pieces of information tell you about the structure of thujene? (i) Thujene reacts instantly with one equivalent of KMnO4 in aqueous solution to discharge the purple color of permanganate and form a brown precipitate. Additional KMnO4 is not decolorized. (ii) Hydroboration–oxidation of thujene forms a compound C10H18O, called thujyl alcohol, whose 1H NMR spectrum shows a 1 H signal at d 5 3.40 ppm. (iii) Oxymercuration–demercuration of thujene forms a different alcohol C10H18O, whose 1 H NMR spectrum shows no signals downfield of d 5 3 ppm. (iv) Ozonolysis of thujene gives H O
?
CHO SOLUTION Beginning with the molecular formula, we may ascertain that thujene possesses three degrees of unsaturation (Section 11-11): [(2 3 10 1 2) 2 16]y2 5 6y2 5 3. (i) The fact that thujene reacts with only one equivalent of KMnO4 implies that only one of these unsaturations is a p bond (Section 12-11); the other two must be rings. (ii) The NMR spectrum of thujyl alcohol is consistent only H OH with a secondary alcohol: , because the signal in the region between d 5 3 and 4 integrates R R for only one hydrogen. (iii) In contrast, the oxymercuration product lacks an NMR signal in this region, meaning that this product cannot have a hydrogen on the HO-bearing carbon: In other words, it must be a tertiary alcohol. Taken together, these three pieces of information tell us that thujene has R1 R3 two rings and a trisubstituted double bond, . The result of ozonolysis completes the R2 H picture, because it gives a product with 10 carbons. Therefore, this product is simply thujene with its alkene function cleaved into two separate C P O units, one being an aldehyde and the other a ketone. Reversing this process, we have
H O
1. O3 2. Zn
?
C
H3C CH3 CH2
C
i
i
H
C
552
H
CHO Reconnect
Thujene
New Reactions
New Reactions 1. General Addition to Alkenes (Section 12-1) G D CPC D G
A B A A O CO CO A A
A OB
2. Hydrogenation (Section 12-2) H
GH COC
G
G D CPC D G
H2, catalyst
Syn addition Typical catalysts: PtO2, Pd–C, Ra–Ni
Electrophilic Additions 3. Hydrohalogenation (Section 12-3) R A HO C O CH3 A X
R
G C PCH2 D H
HX
Regiospecific (Markovnikov rule) Through more stable carbocation
4. Hydration (Section 12-4) D G CPC D G
H OH A A O CO CO A A
H, H2O
Through more stable carbocation
5. Halogenation (Section 12-5) X
G
CG C
X2, CCl4
G
G D CPC D G
X
Stereospecific (anti) X2 Cl2 or Br2, but not I2
6. Vicinal Haloalcohol Synthesis (Section 12-6) X X2, H2O
G
CG C
G
G D CPC D G
OH
OH attaches to more substituted carbon
7. Vicinal Haloether Synthesis (Section 12-6) X
G
X2, ROH
CG C
G
G D CPC D G
OR
OR attaches to more substituted carbon
8. General Electrophilic Additions (Section 12-6, Table 12-2)
E
A electropositive, B electronegative B attaches to more substituted carbon
CG C
G
COC
A B
G
AB
E
G D CPC D G
A
B
Chapter 12
553
Reactions of Alkenes
Chapter 12
9. Oxymercuration – Demercuration (Section 12-7)
G D CPC D G
O B 1. Hg(OCCH3)2, H2O 2. NaBH4, NaOH, H2O
H OH A A O CO CO A A
O B 1. Hg(OCCH3)2, ROH 2. NaBH4, NaOH, H2O
G D CPC D G
H OR A A O CO CO A A
OH or OR attaches to more substituted carbon
Initial addition is anti, through mercurinium ion
10. Hydroboration (Section 12-8) CH3
BH3
THF
(RCH2CH2)3B Regiospecific
THF
BH3
CH3 ∑ H H BO A
∑
H
G C PCH2 D
/
R
/
H
B attaches to less substituted carbon
Stereospecific (syn) and anti-Markovnikov
11. Hydroboration – Oxidation (Section 12-8) H OH A A O CO CO A A
1. BH3, THF 2. H2O2, HO
G D CPC D G
Stereospecific (syn) and anti-Markovnikov OH attaches to less substituted carbon
12. Carbene Addition for Cyclopropane Synthesis (Section 12-9) Using diazomethane: R
R
hv or or Cu
CH2N2
0 R
0 R
Stereospecific
Other sources of carbenes or carbenoids: CHCl3
Base
ðCCl2
CH2I2
Zn–Cu
ICH2ZnI
Oxidation 13. Oxacyclopropane Formation (Section 12-10) O B RCOOH, CH2Cl2
O B RCOH
E
O
E
G D CPC D G
COC
Stereospecific (syn)
14. Vicinal Anti Dihydroxylation (Section 12-10) HO
CG C
G
G D CPC D G
O B 1. RCOOH, CH2Cl2 2. H , H2O
G
O B RCOH
OH
15. Vicinal Syn Dihydroxylation (Section 12-11) G D CPC D G
HO 1. OsO4, 2. H2S; or catalytic OsO4, H2O2
Through cyclic intermediates
GOH COC
G
554
Important Concepts
16. Ozonolysis (Section 12-12) D G CPC D G
O 1. O3, CH3OH B 2. (CH3)2S; or Zn, CH3COH
G CPO D
OPC
D G
Through molozonide and ozonide intermediates
Radical Additions 17. Radical Hydrobromination (Section 12-13) G C P CH2 D
H Br A A O C O C OH A A H
HBr, ROOR
Anti-Markovnikov Does not occur with HCl or HI
18. Other Radical Additions (Section 12-13) D G CPC D G
RSH, ROOR
H SR A A O CO CO A A Anti-Markovnikov
Monomers and Polymers 19. Dimerization, Oligomerization, and Polymerization (Sections 12-14 and 12-15) n
G D CPC D G
H or ROjor B
A A O (C O C)n O A A
Important Concepts 1. The reactivity of the double bond manifests itself in exothermic addition reactions leading to saturated products. 2. The hydrogenation of alkenes is immeasurably slow unless a catalyst capable of splitting the strong H – H bond is used. Possible catalysts are palladium on carbon, platinum (as PtO2), and Raney nickel. Addition of hydrogen is subject to steric control, the least hindered face of the least substituted double bond frequently being attacked preferentially. 3. As a Lewis base, the p bond is subject to attack by acid and electrophiles, such as H1, X2, and Hg21. If the initial intermediate is a free carbocation, the more highly substituted carbocation is formed. Alternatively, a cyclic onium ion is generated subject to nucleophilic ring opening at the more substituted carbon. Carbocation formation leads to control of regiochemistry (Markovnikov rule); onium ion formation leads to control of both regio- and stereochemistry. 4. Mechanistically, hydroboration lies between hydrogenation and electrophilic addition. The first step is p complexation to the electron-deficient boron, whereas the second is a concerted transfer of the hydrogen to carbon. Hydroboration–oxidation results in the anti-Markovnikov hydration of alkenes. 5. Carbenes and carbenoids are useful for the synthesis of cyclopropanes from alkenes. 6. Peroxycarboxylic acids may be thought of as containing an electrophilic oxygen atom, transferable to alkenes to give oxacyclopropanes. The process is often called epoxidation. 7. Osmium tetroxide acts as an electrophilic oxidant of alkenes; in the course of the reaction, the oxidation state of the metal is reduced by two units. Addition takes place in a concerted syn manner through cyclic six-electron transition states to give vicinal diols. 8. Ozonolysis followed by reduction yields carbonyl compounds derived by cleavage of the double bond. 9. In radical chain additions to alkenes, the chain carrier adds to the p bond to create the more highly substituted radical. This method allows for the anti-Markovnikov hydrobromination of alkenes, as well as the addition of thiols and some halomethanes. 10. Alkenes react with themselves through initiation by charged species, radicals, or some transition metals to give polymers. The initial attack at the double bond yields a reactive intermediate that perpetuates carbon – carbon bond formation.
Chapter 12
555
Reactions of Alkenes
Reactions of Alkenes
&
@
H H A A A A G OCO CO OCO CO G CP C A A A A D D CH3 HO X H H
H
@
GC O C H H G
Chapter 12
&
556
12-3
11-5, 12-2, 18-9
HX
H2, catalyst
H
12-14
H or ROOR or base
12-15
AA H G A A O CH2CC P C D A O CO CO A A n dimer or oligomer
R H
G G CP C D D
H R
12-4
12-4
Substrate: G C P CH2 G D OC A H
Substrate: R G G CP C D D H H
H2SO4
H2SO4
12-4
H, H2O
@
GX & @ GX C C O O C C C C G G ( H ( H RO X
&
12-5
12-6
X2
X2, ROH
R
1. Br2 2. NaNH2, liquid NH3
NBS, h
Nu
RLi
Substrate: H G R CH P C D H(R)
Substrate: G G CP C D D CH2 O
Substrate: G G CP C D D CH2L
Substrate: G G CP C D D CH2 O
13-4
14-2
14-3
14-4
G R O C q C OH(R) G G G CP C CP C D D D D CH O CH2 Nu A Br
G G CP C D D
CH O A Li
Problems 33. With the help of the DH 8 values given in Tables 3-1 and 3-4, calculate the DH 8 values for addition of each of the following molecules to ethene, using 65 kcal mol21 for the carbon – carbon p bond strength. (a) (c) (e) (g)
Cl2 IBr (DH 8 5 43 kcal mol21) HI Br – CN (DH 8 5 83 kcal mol21; DH 8 for Csp3 – CN 5 124 kcal mol21)
(b) (d) (f ) (h)
IF (DH 8 5 67 kcal mol21) HF HO – Cl (DH 8 5 60 kcal mol21) CH3S – H (DH 8 5 88 kcal mol21; DH 8 for Csp3 – S 5 60 kcal mol21)
34. The bicyclic alkene car-3-ene, a constituent of turpentine, undergoes catalytic hydrogenation to give only one of the two possible stereoisomeric products. The product has the common name
Problems
Chapter 12
section number
G CPO D
O
&
@
G
2. NaBH4
G G CP C D D
O
32
)C
G D
R R
&
@
H
O
GC O C HO OH
12-10
12-11
C
C
H A A H OCO CO G A A OPC D H Br 12-12
12-13
H A A OCO CO A A H SR 12-13
O KMnO4, H2O or 1. O3 B HBr, ROOR RSH, ROOR RCOOH cat. OSO4, H2O2 2. Reduction
H
or h
, AlCl3
Substrate: or
14-8
H
G
1. BH3 2. H2O2, OH
O
O O
B 1. HgOCCH , ROH
@ G
12-9
&
@
12-8
C
&
12-7
H
@
O C C OH
@
G
R
GC O C H OH
&
&
R
H A A OCO CO A A OR H
14-9
or H
G G CP C D D
R2CuLi
H or
Substrate: O B C G G D CP C D D
Substrate: O B C G G D CP C D D
Substrate: O B C G G D CP C D D
18-8
18-9
18-10
18-11
OO O O A A O A A A A A J J J J B OC OC O C OC OC O C OC OC OC O C C C D D D D A A A A A A A G G D R H Nu H H A DC P C D OC A H
cis-carane, indicating that the methyl group and the cyclopropane ring are on the same face of the cyclohexane ring. Suggest an explanation for this stereochemical outcome.
H3C H C
(
H3C H C
100 atm H2, PtO2, CH3CH2OH, 25C
∞H
not H3C
H3C CH3
(
CH3
H3C
O
Substrate: G G O CP C B D D CHO C D A 15-11
A CCH2 O A
HNu, HO
H or OH
CH3
CH3
]H
557
558
Chapter 12
Reactions of Alkenes
35. Give the expected major product of catalytic hydrogenation of each of the following alkenes. Clearly show and explain the stereochemistry of the resulting molecules. CH3
H
(a)
š
CH3 % (b)
(c)
/∑ H2C
(CH3)2CH H
¨ H
36. Would you expect the catalytic hydrogenation of a small-ring cyclic alkene such as cyclobutene to be more or less exothermic than that of cyclohexene? (Hint: Which has more bond-angle strain, cyclobutene or cyclobutane?) 37. Give the expected major product from the reaction of each alkene with (i) peroxide-free HBr and (ii) HBr in the presence of peroxides. (a) 1-Hexene; (b) 2-methyl-1-pentene; (c) 2-methyl-2-pentene; (d) (Z)-3-hexene; (e) cyclohexene. 38. Give the product of addition to Br2 to each alkene in Problem 37. Pay attention to stereochemistry. 39. What alcohol would be obtained from treatment of each alkene in Problem 37 with aqueous sulfuric acid? Would any of these alkenes give a different product upon oxymercuration – demercuration? Upon hydroboration – oxidation? 40. Give the reagents and conditions necessary for each of the following transformations and comment on the thermodynamics of each (a) cyclohexanol y cyclohexene; (b) cyclohexene y cyclohexanol; (c) chlorocyclopentane y cyclopentene; (d) cyclopentene y chlorocyclopentane. (CH3)2CHCH P CHCO2CH3 4-Methyl-2-pentenoic acid methyl ester
41. Problem 51 of Chapter 6 presented a strategy for the synthesis of the amino acid (2S,3S)-3hydroxyleucine, requiring as the starting material a specific stereoisomer of 2-bromo-3-hydroxy-4methylpentanoic acid. Addition of bromine and water to the methyl ester of 4-methyl-2-pentenoic acid (margin) gives the corresponding ester of 2-bromo-3-hydroxy-4-methylpentanoic acid. (a) Which stereoisomer of the unsaturated ester, cis or trans, is needed for this addition to give the necessary stereoisomer of the product? (b) Can this strategy give the bromoalcohol as a single enantiomer or not? Explain, mechanistically. 42. Formulate the product(s) that you would expect from each of the following reactions. Show stereochemistry clearly. Cl
Br ,H O
2 (b) trans-3-Heptene uy
HCl
(a)
2 2 (c) 1-Ethylcyclohexene uuy
CH2CH3 CH3 NaOH, H O
2 (d) Product of (c) uuvuy
(g)
CH3 %
(e)
O B 1. Hg(OCCH3)2, CH3OH 2. NaBH4, CH3OH
Br , excess Na1N 2
2 3 (f) cis-2-Butene uuuuvuy
1. BH3, THF 2. H2O2, NaOH, H2O
43. Show how you would synthesize each of the following molecules from an alkene of appropriate structure (your choice). OH
O (b) Cl
(a)
(d)
H Br Δ$ @( H Br
(c)
Br H Δ$ @( Br H
Br H Δ$ @( H Br
(meso–4R,5S–isomer)
(Racemate of 4R,5R and 4S,5S isomers)
Problems
CH3 %
CH3 % (e)
Chapter 12
(f)
(More challenging. Hint: See Section 12-6.)
^
%
)O
~O
44. Propose efficient methods for accomplishing each of the following transformations. Most will require more than one step. Br (a)
HO H Δ$ Δ$
(b)
I
HO
(c)
Δ$ HO
H OH Δ$
HO H Δ$ Δ$
(meso–2R,3S–isomer)
H OH
(Racemate of 2R,3R and 2S,3S isomers)
HO H
H OH
H
O (d)
O
45. Reaction review. Without consulting the Reaction Road Map on pp. 556 – 557, suggest a reagent H D G to convert a general alkene, C P C , into each of the following types of compounds. D G
H OH A A (e) O C O COH A A
OH OH A A C O COH
(h)
H Br A A C O COH
(k)
Br H A A C O COH
(g)
CH3O H A A C O COH (j)
(m)
Cl H A A C O COH
(n)
( ) H A COC
E
(Markovnikov product)
(anti-Markovnikov product)
H (q)
C O COH
(f)
H H A A C O COH
CH2
(anti-Markovnikov product)
(Markovnikov product)
(polymer)
n
G D (p) C P O O P C D G
(c)
E
I H A A (d) O C O COH A A
O
E
(b)
OH H A A C O COH
E
(a)
Br Br A A C O COH
H SCH2CH3 A A C O COH
(anti-Markovnikov product)
(i)
C O COH
(l)
OH Br A A C O COH
CH3O Br A A (o) C O COH
559
560
Chapter 12
Reactions of Alkenes
46. Give the expected product of reaction of 2-methyl-1-pentene with each of the following reagents. (a) (d) (g) ( j)
H2, PtO2, CH3CH2OH HCl HI 1 peroxides ICl
(b) (e) (h) (k)
D2, Pd – C, CH3CH2OH HBr H2SO4 1 H2O Br2 1 CH3CH2OH
O B (p) Hg(OCCH3)2 H2 O, then NaBH 4
BH3, THF then NaOH 1 H2O2 HBr 1 peroxides Cl2 CH3SH 1 peroxides
O B (o) O 3, then Zn CH3 COH
(n) OsO4, then H2S
(m) MCPBA, CH2Cl2
(c) (f) (i) (l)
(q) Catalytic H2SO4 1 heat
47. What are the products of reaction of (E)-3-methyl-3-hexene with each of the reagents in Problem 46? 48. Write the expected products of reaction of 1-ethylcyclopentene with each of the reagents in Problem 46. 49. Write out detailed step-by-step mechanisms for the reactions in parts (c), (e), (f ), (h), (j), (k), (m), (n), (o), and (p) of Problem 46. 50. What alkene monomer gives the following polymer?
( ) CH3 H A A COC A A H H
n
51. Give the expected major product from reaction of 3-methyl-1-butene with each of the following reagents. Explain any differences in the products mechanistically. O B (a) 50% aqueous H2SO4; (b) Hg(OCCH3)2 in H2 O, followed by NaBH 4; (c) BH3 in THF, followed by NaOH and H2O2. 52. Answer the question posed in Problem 51 for cyclohexylethene. 53. Give the expected major product of reaction of magnesium monoperoxyphthalate (MMPP) with each alkene. In each case, also give the structure of the material formed upon hydrolysis in aqueous acid of the initial product. (a) 1-Hexene; (b) (Z)-3-ethyl-2-hexene; (c) (E)-3-ethyl-2-hexene; (d) (E)-3-hexene; (e) 1,2-dimethylcyclohexene. 54. Give the expected major product of reaction of OsO4, followed by H2S, with each alkene in Problem 53. 55. Give the expected major product of reaction of CH3SH in the presence of peroxides with each alkene in Problem 53. 56. Propose a mechanism for the peroxide-initiated reaction of CH3SH with 1-hexene. 57. Write the expected products of each of the following reactions. KOC(CH ) , (CH ) COH
3 3 3 3 (a) (E)-2-Pentene 1 CHCl3 uuuuuuuy
Zn-Cu, (CH CH ) O
3 2 2 (b) 1-Methylcyclohexene 1 CH2I2 uuuuuuy
Cu, D
(c) Propene 1 CH2N2 uuy KOC(CH ) , (CH ) COH
3 3 3 3 (d) (Z)-1,2-Diphenylethene 1 CHBr3 uuuuuuuy
Zn-Cu, (CH CH ) O
3 2 2 (e) (E)-1,3-Pentadiene 1 2 CH2I2 uuuuuuy
hv
(f) CH2 P CHCH2CH2CH2CHN2 uy
1
2H
H NMR
1H 1H 1H
(CH3)4Si 4.2
6.0
5.5
5.0
4.5
4.0
4.1
3.5
4.0
3.0
3.9
2.5
2.0
1.5
1.0
0.5
0.0
300-MHz 1H NMR spectrum ppm (δ ) A 58. 1H NMR spectrum A corresponds to a molecule with the formula C3H5Cl. The compound shows significant IR bands at 730 (see Problem 53 of Chapter 11), 930, 980, 1630, and 3090 cm21. (a) Deduce the structure of the molecule. (b) Assign each NMR signal to a hydrogen or group of hydrogens. (c) The “doublet” at d 5 4.05 ppm has J 5 6 Hz. Is this in accord with your assignment in (b)? (d) This “doublet,” upon fivefold expansion, becomes a doublet of triplets (inset, spectrum A), with J < 1 Hz for the triplet splittings. What is the origin of this triplet splitting? Is it reasonable in light of your assignment in (b)? 59. Reaction of C3H5Cl (Problem 58, spectrum A) with Cl2 in H2O gives rise to two products, both C3H6Cl2O, whose spectra are shown in B and C. Reaction of either of these products with KOH yields the same molecule C3H5ClO (spectrum D, next page). The insets show expansions of some of the multiplets. The IR spectrum reveals bands at 720 and 1260 cm21 and the absence of signals between 1600 and 1800 cm21 and between 3200 and 3700 cm21. (a) Deduce the structures of the compounds giving rise to spectra B, C, and D. (b) Why does reaction of the starting chloride compound with Cl2 in H2O give two isomeric products? (c) Write mechanisms for the formation of the product C3H5ClO from both isomers of C3H6Cl2O. 1
1
H NMR
H NMR 2H
2H
4H
4.1 4.1
4.0
3.7
4.0
3.8
3.7
3.6
1H
1H 1H
1H
4.0
3.5
3.0
(CH3)4Si
2.5
2.0
1.5
1.0
0.5
(CH3)4Si
0.0
4.0
300-MHz 1H NMR spectrum ppm (δ ) B
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.0
300-MHz 1H NMR spectrum ppm (δ ) C 561
562
Reactions of Alkenes
Chapter 12
1
H NMR
2H
1H
1H 3.3
1H
3.5
3.0
3.2
2.5
3.1
3.0
2.0
2.9
2.8
1.5
2.7
2.6
1.0
0.5
300-MHz 1H NMR spectrum ppm (δ ) D 1H
NMR 3H
6.00
5.95
5.90
5.85
5.80
1H1H 1H
6.0
1H
5.5
5.0
4.5
(CH3)4Si
1H
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.0
300-MHz 1H NMR spectrum ppm (δ ) E 60. 1H NMR spectrum E corresponds to a molecule with the formula C4H8O. Its IR spectrum has important bands at 945, 1015, 1665, 3095, and 3360 cm21. (a) Determine the structure of the unknown. (b) Assign each NMR and IR signal. (c) Explain the splitting patterns for the signals at d 5 1.3, 4.3, and 5.9 ppm (see inset for 10-fold expansion). 61. Reaction of the compound corresponding to spectrum E with SOCl2 produces a chloroalkane, C4H7Cl, whose NMR spectrum is almost identical with spectrum E, except that the broad signal at d 5 1.5 ppm is absent. Its IR spectrum shows bands at 700 (Problem 53, Chapter 11), 925, 985, 1640, and 3090 cm21. Treatment with H2 over PtO2 results in C4H9Cl (spectrum F, next page). Its IR spectrum reveals the absence of all the bands quoted for its precursor, except for the signal at 700 cm21. Identify these two molecules.
Problems
1H
NMR
Chapter 12
3H
3H
2H 1H
(CH3)4Si
4.0
3.5
3.0
2.5
1.5
2.0
300-MHz
1H
1.0
0.5
0.0
NMR spectrum ppm (δ )
F
62. The mass spectra of both of the compounds described in Problem 61 show two molecular ion peaks, two mass units apart, in an intensity ratio of about 3 : 1. Explain. 63. Give the structure of an alkene that will give the following carbonyl compounds upon ozonolysis followed by reduction with (CH3)2S. (a) CH3CHO only
64.
(c) (CH3)2C P O and H2C P O
(b) CH3CHO and CH3CH2CHO
O B (d) CH3CH2 CCH3 and CH3CHO
(e) Cyclopentanone and CH3CH2CHO
Plan syntheses of each of the following compounds, utilizing retrosynthetic-analysis techniques. Starting compounds are given in parentheses. However, other simple alkanes or alkenes also may be used, as long as you include at least one carbon – carbon bond-forming step in each synthesis.
O B (a) CH3CH2 CCHCH3 (propene) A CH3
(b) CH3CH2 CH2 CHCH2CH2CH3 (propene, again) A Cl
65. Show how you would convert cyclopentane into each of the following molecules. (a) cis-1,2-Dideuteriocyclopentane
(b) trans-1,2-Dideuteriocyclopentane O
(c)
! SCH2CH3
(d)
CH2
(e)
CH3
{ Cl
(f) 1,2-Dimethylcyclopentene
(g) trans-1,2-Dimethyl-1,2-cyclopentanediol
(c)
H3C≈
OH % (cyclohexene)
563
Chapter 12
Reactions of Alkenes
66. Give the expected major product(s) of each of the following reactions.
(a) CH3OCH2CH2CH P CH2
H H CH3CH2 D D (d) CP C G G H CH2CH2 (g) CH3CH P CH2
Catalytic HF
O B 1. Hg(OCCH3)2, CH3OH 2. NaBH4, CH3OH
CH3 D (b) H2CP C G CH2OH
1. Excess O3, CH2Cl2 2. (CH3)2S
O B 1. CH3COOH, CH2Cl2 2. H , H2O
H3C H D D (e) CP C G G CH3 CH3CH2
(h) CH2P CHNO2
CHPCH2 Conc. HI
(c)
Cl BrCN
(f)
1. OsO4, THF 2. NaHSO3
Catalytic KOH
(Hint: Draw Lewis structures for the NO2 group.)
67. (E)-5-Hepten-1-ol reacts with the following reagents to give products with the indicated formulas. Determine their structures and explain their formation by detailed mechanisms. (a) HCl, C7H14O (no Cl!); (b) Cl2, C7H13ClO (IR: 740 cm21; nothing between 1600 and 1800 cm21 and between 3200 and 3700 cm21). 68. When a cis alkene is mixed with a small amount of I2 in the presence of heat or light, it isomerizes to some trans alkene. Propose a detailed mechanism to account for this observation. 69. Treatment of a-terpineol (Chapter 10, Problem 60) with aqueous mercuric acetate followed by sodium borohydride reduction leads predominantly to an isomer of the starting compound (C10H18O) instead of a hydration product. This isomer is the chief component in oil of eucalyptus and, appropriately enough, is called eucalyptol. It is popularly used as a flavoring for otherwise foul-tasting medicines because of its pleasant spicy taste and aroma. Deduce a structure for eucalyptol on the basis of sensible mechanistic chemistry and the following proton-decoupled 13 C NMR data. (Hint: The IR spectrum shows nothing between 1600 and 1800 cm21 or between 3200 and 3700 cm21!) CH3
O B 1. Hg(OCCH3)2, H2O 2. NaBH4, H2O
eucalyptol, (C10H18O)
(CH3)2COH -Terpineol
CH3
H3C
CH2
13C
NMR: 22.8, 27.5, 28.8, 31.5, 32.9, 69.6, and 73.5 ppm
70. Both borane and MCPBA react highly selectively with molecules, such as limonene, that contain double bonds in very different environments. Predict the products of reaction of limonene with (a) one equivalent of BH3 in THF, followed by basic aqueous H2O2, and (b) one equivalent of MCPBA in CH2Cl2. Explain your answers. 71. Oil of marjoram contains a pleasant, lemon-scented substance, C10H16 (compound G). Upon ozonolysis, G forms two products. One of them, H, has the formula C8H14O2 and can be independently synthesized in the following way.
Limonene 1. Mg, (CH3CH2)2O O DD 2. H2C CHCH3
H3C
D
564
H3C
CH2Br
C8H16O I
PCC, CH2Cl2
C8H14O
1. BH3, THF 2. H2O2, NaOH 3. PCC, CH2Cl2 H
J
From this information, propose reasonable structures for compounds G through J. 72.
Humulene and a-caryophyllene alcohol are terpene constituents of carnation extracts. The former is converted into the latter by acid-catalyzed hydration in one step. Write a mechanism. (Hint: Follow the labeled carbon atoms retrosynthetically. The mechanism includes carbocation-induced cyclizations and hydrogen and alkyl-group migrations.)
Problems
*
* *
H
H, H2O
* *
* * *
Chapter 12
H *
OH
*
-Caryophyllene alcohol
Humulene
73. Predict the product(s) of ozonation of humulene (Problem 72), followed by reduction with zinc in acetic acid. If you had not known the structure of humulene ahead of time, would the identities of these ozonolysis products have enabled you to determine it unambiguously? 74.
Caryophyllene (C15H24) is an unusual sesquiterpene familiar to you as a major cause of the odor of cloves. Determine its structure from the following information. (Caution: The structure is totally different from that of a-caryophyllene alcohol in Problem 72.) Reaction 1
H , Pd–C
2 Caryophyllene uvvvy C15H28
Reaction 2 O l
H3C
Caryophyllene
1. O3, CH2Cl2 O B 2. Zn, CH3COH
CH3 CH2
O
/
H
μ
CH3 H H
l
O
l
O Reaction 3
H3C
Caryophyllene
C15H26O
1. O3, CH2Cl2 O B 2. Zn, CH3COH
CH3 H
μ
1. One equivalent of BH3, THF 2. H2O2, NaOH, H2O
O l CH3
/
H
/∑
H
H CH2OH
l
O
An isomer, isocaryophyllene, gives the same products as caryophyllene upon hydrogenation and ozonolysis. Hydroboration – oxidation of isocaryophyllene gives a C15H26O product isomeric to the one shown in reaction 3; however, ozonolysis converts this compound into the same final product shown. In what way do caryophyllene and its isomer differ? 75. Juvenile hormone (JH, Chemical Highlight 12-1) has been synthesized in several ways. Two intermediates are shown in (a) and (b). Propose completions for syntheses of JH that start with each of them. Your synthesis for (a) should be stereospecific. Also for (a), note that the double bond between C10 and C11 is the most reactive toward electrophilic reagents. H3C (a) H3C
H3C
H3C CH3
H3C
O COCH3
(b)
CH3
O COCH3
O Cl
Team Problem 76. The selectivity of hydroboration increases with increasing bulkiness of the borane reagent. (a) For example, 1-pentene is selectively hydroborated in the presence of cis- and trans-2-pentene when treated with bis(1,2-dimethylpropyl)borane (disiamylborane) or with 9-borabicyclo[3.3.1]nonane, 9-BBN. Divide the task of formulating the structure of the starting alkene used in preparing
565
Chapter 12
CH3 A (CH3)2CHCH BH 2
Bis(1,2-dimethylpropyl)borane (Disiamylborane)
both of these bulky borane reagents among yourselves. Make models to visualize the features of these reagents that direct the structural selectivity. (b) In an enantioselective approach to making secondary alcohols, two equivalents of one enantiomer of a-pinene are treated with BH3. The resulting borane reagent is treated with cis-2-butene followed by basic hydrogen peroxide to yield optically active 2-butanol. H3C
CH3
H 3C
CH3 H
BH3
2 CH3
2BH
1. 2. H2O2, OH
H CH3
-Pinene
9-BBN
/
EH B
Reactions of Alkenes
/∑
566
CH3 A HO OC O H A CH2CH3 Optically active
Share your model kits to make a model of a-pinene and the resulting borane reagent. Discuss what is directing the enantioselectivity of this hydroboration – oxidation reaction. What products besides 2-butanol result from the oxidation step?
Preprofessional Problems 77. A chiral compound, C5H8, upon simple catalytic hydrogenation, yields an achiral compound, C5H10. What is the best name for the former? (a) 1-Methylcyclobutene; (b) 3-methylcyclobutene; (c) 1,2-dimethylcyclopropene; (d) cyclopentene. 78. A chemist reacted 300 g of 1-butene with excess Br2 (in CCl4) at 25 8C. He isolated 418 g of 1,2-dibromobutane. What is the percent yield? (Atomic weights: C 5 12.0, H 5 1.00, Br 5 80.0.) (a) 26; (b) 36; (c) 46; (d) 56; (e) 66. 79. trans-3-Hexene and cis-3-hexene differ in one of the following ways. Which one? (a) Products of hydrogenation; (b) products of ozonolysis; (c) products of Br2 addition in CCl4; (d) products of hydroboration – oxidation; (e) products of combustion. 80. Which reaction intermediate is believed to be part of the following reaction? RCH P CH2
HBr, ROOR
RCH2CH2Br
(a) Radical; (b) carbocation; (c) oxacyclopropane; (d) bromonium ion. 81. When 1-pentene is treated with mercuric acetate, followed by sodium borohydride, which of the following compounds is the resulting product? (a) 1-Pentyne; (b) pentane; (c) 1-pentanol; (d) 2-pentanol.
THIRTEEN
C H A P T E R
Alkynes The Carbon – Carbon Triple Bond
S
N
O
O
A
lkynes are hydrocarbons that contain carbon – carbon triple bonds. It should not come as a surprise that their characteristics resemble the properties and behavior of alkenes, their double-bonded cousins. In this chapter we shall see that, like alkenes, alkynes find numerous uses in a variety of modern settings. For example, the polymer derived from the parent compound, ethyne (HC q CH), can be fashioned into electrically conductive sheets usable in lightweight, all-polymer batteries. Ethyne is also a substance with a relatively high energy content, a property that is exploited in oxyacetylene torches. A variety of alkynes, both naturally occurring and synthetic, have found use in medicine for their antibacterial, antiparasitic, and antifungal activities. OCq CO Alkyne triple bond
Because the O C q C O functional group contains two p linkages (which are mutually perpendicular; recall Figure 1-21), its reactivity is much like that of the double bond. For example, like alkenes, alkynes are electron rich and subject to attack by electrophiles. Many of the alkenes that serve as monomers for the production of polymeric fabrics, elastics, and plastics are prepared by electrophilic addition reactions to ethyne and other alkynes. Alkynes can be prepared by elimination reactions similar to those used to generate alkenes, and they are likewise most stable when the multiple bond is internal rather than terminal. A further, and useful, feature is that the alkynyl hydrogen is much more acidic than its alkenyl or alkyl counterpart, a property that permits easy deprotonation by strong bases. The resulting alkynyl anions are valuable nucleophilic reagents in synthesis. We begin with discussions of the naming, structural characteristics, and spectroscopy of the alkynes. Subsequent sections introduce methods for the synthesis of compounds in this class and the typical reactions they undergo. We end with an overview of the extensive industrial uses and physiological characteristics of alkynes.
Scanning tunneling microscopy (STM) is an indirect method for imaging individual atoms and molecules on a solid surface. The dialkyne whose structure is shown above glows brightly in the top STM image because it is in a conformation that is highly electrically conductive. In contrast, in the bottom image a change in molecular shape has “turned off” its conductivity, and it goes “dark” as a result. Such molecules are prototypes for “molecular switches,” which promise to revolutionize the fields of electronic components and computers in the 21st century.
Chapter 13
Common Names for Alkynes HC q CH
Alkynes
13-1 Naming the Alkynes A carbon–carbon triple bond is the functional group characteristic of the alkynes. The general formula for the alkynes is CnH2n22, the same as that for the cycloalkenes. The common names for many alkynes are still in use, including acetylene, the common name of the smallest alkyne, C2H2. Other alkynes are treated as its derivatives—for example, the alkylacetylenes. The IUPAC rules for naming alkenes (Section 11-1) also apply to alkynes, the ending -yne replacing -ene. A number indicates the position of the triple bond in the main chain.
HC q CH
CH3C q CCH3
CH3C q CCHCH2CH3
CH3 3A 2 1 CH3CC q CH
Ethyne
2-Butyne
4-Bromo-2-hexyne
3,3-Dimethyl-1-butyne
(An internal alkyne)
(A terminal alkyne)
Br
Acetylene
CH3C q CCH3
Dimethylacetylene
1
3 A4
2
5
6
4
A CH3
Alkynes having the general structure RC q CH are terminal, whereas those with the structure of RC q CR9 are internal. Substituents bearing a triple bond are alkynyl groups. Thus, the substituent – C q CH is named ethynyl; its homolog – CH2C q CH is 2-propynyl (propargyl). Like alkanes and alkenes, alkynes can be depicted in straight-line notation. [ ð
CH3CH2CH2C q CH
1
trans-1,2-Diethynylcyclohexane
Propylacetylene
2
3
CH2C q CH 2-Propynylcyclobutane (Propargylcyclobutane)
3
2 1
HCq CCH2OH 2-Propyn-1-ol (Propargyl alcohol)
In IUPAC nomenclature, a hydrocarbon containing both double and triple bonds is called an alkenyne. The chain is numbered starting from the end closest to either of the functional groups. When a double bond and a triple bond are at equidistant positions from either terminus, the double bond is given the lower number. Alkynes incorporating the hydroxy function are named alkynols. Note the omission of the final e of -ene in -enyne and of -yne in -ynol. The OH group takes precedence over both double and triple bonds in the numbering of a chain. OH 2
4 6
5
4
3
2
1
CH3CH2CH P CHC q CH
1
2
3
4
5
5
CH2 P CHCH2C q CH
3
1
6
3-Hexen-1-yne
1-Penten-4-yne
5-Hexyn-2-ol
(Not 3-hexen-5-yne)
(Not 4-penten-1-yne)
(Not 1-hexyn-5-ol)
Exercise 13-1 Give the IUPAC names for (a) all the alkynes of composition C6H10; H3C (b)
C G
568
CG C q CH H ( CH PCH2
(c) all butynols. Remember to include and designate stereoisomers.
13-2 Properties and Bonding in the Alkynes The nature of the triple bond helps explain the physical and chemical properties of the alkynes. In molecular-orbital terms, we shall see that the carbons are sp hybridized, and the four singly filled p orbitals form two perpendicular p bonds.
13-2 Properties and Bonding in the Alkynes
Chapter 13
569
Alkynes are relatively nonpolar Alkynes have boiling points very similar to those of the corresponding alkenes and alkanes. Ethyne is unusual in that it has no boiling point at atmospheric pressure; rather, it sublimes at 2848C. Propyne (b.p. 223.28C) and 1-butyne (b.p. 8.18C) are gases, whereas 2-butyne is barely a liquid (b.p. 278C) at room temperature. The medium-sized alkynes are distillable liquids. Care must be taken in the handling of alkynes: They polymerize very easily — frequently with violence. Ethyne explodes under pressure but can be shipped in pressurized gas cylinders that contain acetone and a porous filler such as pumice as stabilizers.
Dissociation Energies of C–C Bonds HC q CH
Ethyne is linear and has strong, short bonds
DH 229 kcal mol1 (958 kJ mol1)
In ethyne, the two carbons are sp hybridized (Figure 13-1A). One of the hybrid orbitals on each carbon overlaps with hydrogen, and a s bond between the two carbon atoms results from mutual overlap of the remaining sp hybrids. The two perpendicular p orbitals on each carbon contain one electron each. These two sets overlap to form two perpendicular p bonds (Figure 13-1B). Because p bonds are diffuse, the distribution of electrons in the triple bond resembles a cylindrical cloud (Figure 13-1C). As a consequence of hybridization and the two p interactions, the strength of the triple bond is about 229 kcal mol21, considerably stronger than either the carbon – carbon double or single bonds (margin). As with alkenes, however, the alkyne p bonds are much weaker than the s component of the triple bond, a feature that gives rise to much of its chemical reactivity. The C – H bond-dissociation energy of terminal alkynes is also substantial: 131 kcal mol21 (548 kJ mol21). π bond
p orbital
H2C P CH2
DH 173 kcal mol1 (724 kJ mol1) H3C O CH3
DH 90 kcal mol1 (377 kJ mol1)
π electron cloud
sp orbital
H
H
C
C
C
H
C
C
H
C
sp orbital π bond
p orbital A
σ bond
B
Figure 13-1 (A) Orbital picture of sp-hybridized carbon, showing the two perpendicular p orbitals. (B) The triple bond in ethyne: The orbitals of two sp-hybridized CH fragments overlap to create a s bond and two p bonds. (C) The two p bonds produce a cylindrical electron cloud around the molecular axis of ethyne. (D) The electrostatic potential map reveals the (red) belt of high electron density around the central part of the molecular axis.
Because both carbon atoms in ethyne are sp hybridized, its structure is linear (Figure 13-2). The carbon – carbon bond length is 1.20 Å, shorter than that of a double bond (1.33 Å, Figure 11-1). The carbon–hydrogen bond also is short, again because of the relatively large degree of s character in the sp hybrids used for bonding to hydrogen. The electrons in these orbitals (and in the bonds that they form by overlapping with other orbitals) reside relatively close to the nucleus and produce shorter (and stronger) bonds.
Alkynes are high-energy compounds The alkyne triple bond is characterized by a concentration of four p electrons in a relatively small volume of space. The resulting electron – electron repulsion contributes to the relative weakness of the two p bonds and to a very high energy content of the alkyne molecule itself. Because of this property, alkynes often react with the release of considerable amounts of energy. In addition to being prone to explosive decomposition, ethyne has a heat of combustion of 311 kcal mol21. As shown in the equation for ethyne combustion on the next page, this energy is distributed among only three product molecules, one of water and two
D
1.203 A
HO C q C O H 1.061 A
180
Linear ethyne
Figure 13-2 Molecular structure of ethyne.
570
Alkynes
Chapter 13
of CO2, causing each to be heated to extremely high temperatures (.25008C), sufficient for use in welding torches. Combustion of Ethyne HC q CH
2 CO2
2.5 O2
ΔH° 311 kcal mol1 (1301 kJ mol1)
H2O
As we found in our discussion of alkene stabilities (Section 11-5), heats of hydrogenation also provide convenient measures of the relative stabilities of alkyne isomers. In the presence of catalytic amounts of platinum or palladium on charcoal, the two isomers of butyne hydrogenate by addition of two molar equivalents of H2 to produce butane. Just as we discovered in the case of alkenes, hydrogenation of the internal alkyne isomer releases less energy, allowing us to conclude that 2-butyne is the more stable of the two. Hyperconjugation is the reason for the greater relative stability of internal compared with terminal alkynes.
The high temperatures required for welding are attained by combustion of ethyne (acetylene).
CH3CH2C q CH
2 H2
CH3C q CCH3
2 H2
Catalyst
Catalyst
CH3CH2CH2CH3
ΔH° 69.9 kcal mol1 (292.5 kJ mol1)
CH3CH2CH2CH3
ΔH° 65.1 kcal mol1 (272.4 kJ mol1)
Exercise 13-2 Are the heats of hydrogenation of the butynes consistent with the notion that alkynes are highenergy compounds? Explain. (Hint: Compare these values with the heats of hydrogenation of alkene double bonds.)
Terminal alkynes are remarkably acidic Relative Stabilities of the Alkynes
In Section 2-2 you learned that the strength of an acid, H – A, increases with increasing electronegativity, or electron-attracting capability, of atom A. Is the electronegativity of an atom the same in all structural environments? The answer is no: Electronegativity varies with hybridization. Electrons in s orbitals are more strongly attracted to an atomic nucleus than are electrons in p orbitals. As a consequence, an atom with hybrid orbitals high in s character (e.g., sp, with 50% s and 50% p character) will be slightly more electronegative than the same atom with hybrid orbitals with less s character (sp3, 25% s and 75% p character). This effect is indicated below in the electrostatic potential maps of ethane, ethene, and ethyne. The increasingly positive polarization of the hydrogen atoms is reflected in their increasingly blue shadings, whereas the carbon atoms become more electron rich (red) along the series. The relatively high s character in the carbon hybrid orbitals of terminal alkynes makes them more acidic than alkanes and alkenes. The pKa of ethyne, for example, is 25, remarkably low compared with that of ethene and ethane.
RC q CH , RC q CR9 More stable
Deprotonation of 1-Alkynes RC q C O H
ðB
RC q Cð
HB
Relative Acidities of Alkanes, Alkenes, and Alkynes H3C Hybridization: pKa:
CH3
H2C
CH2
sp3
sp2
50
44
Increasing acidity
HC
CH sp 25
13-3 Spectroscopy of the Alkynes
Chapter 13
This property is useful, because strong bases such as sodium amide in liquid ammonia, alkyllithiums, and Grignard reagents can deprotonate terminal alkynes to the corresponding alkynyl anions. These species react as bases and nucleophiles, much like other carbanions (Section 13-5). Deprotonation of a Terminal Alkyne pKa ⬇ 50
pKa ⬇ 25
CH3CH2C q CH (Stronger acid)
CH3CH2CH2CH2Li
(CH3CH2)2O
H A CH3CH2C q CLi CH3CH2CH2CH2
(Stronger base)
(Weaker base)
(Weaker acid)
Exercise 13-3
Working with the Concepts: Deprotonation of Alkynes What is the equilibrium constant, Keq, for the acid-base reaction shown above? Does its value explain why the reaction is written with only a forward arrow, suggesting that it is “irreversible”? Strategy Recall how pKa values relate to acid dissociation constants. Use this information to determine the value for Keq. Solution • The pKa is the negative logarithm of the acid dissociation constant. Dissociation of the alkyne therefore has a Ka < 10225, very unfavorable, at least in comparison with the more familiar acids. However, butyllithium is the conjugate base of butane, which has a Ka < 10250. As an acid, butane is 25 orders of magnitude weaker than is the terminal alkyne. Thus, butyllithium is that much stronger a base compared with the alkynyl anion. • The Keq for the reaction is found by dividing the Ka for the acid on the left by the Ka for the acid on the right: 10225y10250 5 1025. The reaction is very favorable in the forward direction, so much so that for all practical purposes it may be considered to be irreversible. (Caution: Use common sense to avoid major errors in solving acid-base problems, such as deciding that the equilibrium lies the wrong direction. Use this hint: The favored direction for an acid-base reaction converts the stronger acid/stronger base pair into the weaker acid/weaker base pair.)
Exercise 13-4
Try It Yourself Strong bases other than those mentioned here for the deprotonation of alkynes were introduced earlier. Two examples are potassium tert-butoxide and lithium diisopropylamide (LDA). Would either (or both) of these compounds be suitable for making ethynyl anion from ethyne? Explain, in terms of their pKa values.
In Summary The characteristic hybridization scheme for the triple bond of an alkyne controls its physical and electronic features. It is responsible for strong bonds, the linear structure, and the relatively acidic alkynyl hydrogen. In addition, alkynes are highly energetic compounds. Internal isomers are more stable than terminal ones, as shown by the relative heats of hydrogenation.
13-3 Spectroscopy of the Alkynes Alkenyl hydrogens (and carbons) are deshielded and give rise to relatively low-field NMR signals compared with those in saturated alkanes (Section 11-4). In contrast, alkynyl hydrogens have chemical shifts at relatively high field, much closer to those in alkanes. Similarly, the sp-hybridized carbons absorb in a range between that recorded for alkenes and alkanes. Alkynes, especially terminal ones, are also readily identified by IR spectroscopy. Finally, mass spectrometry can be a useful tool for identification and structure elucidation of alkynes.
571
572
Alkynes
Chapter 13
Figure 13-3 300-MHz 1H NMR spectrum of 3,3-dimethyl-1-butyne showing the high-field position (d 5 2.06 ppm) of the signal due to the alkynyl hydrogen.
1H NMR
9H CH3 CH3CC
CH
(CH3)4Si
CH3
1H
9
8
7
6
5
4
3
2
1
0
ppm (δ )
The NMR absorptions of alkyne hydrogens show a characteristic shielding Unlike alkenyl hydrogens, which are deshielded and give 1H NMR signals at d 5 4.6–5.7 ppm, protons bound to sp-hybridized carbon atoms are found at d 5 1.7 – 3.1 ppm (Table 10-2). For example, in the NMR spectrum of 3,3-dimethyl-1-butyne, the alkynyl hydrogen resonates at d 5 2.06 ppm (Figure 13-3). Why is the terminal alkyne hydrogen so shielded? Like the p electrons of an alkene, those in the triple bond enter into a circular motion when an alkyne is subjected to an external magnetic field (Figure 13-4). However, the cylindrical distribution of these electrons (Figure 13-1C) now allows the major direction of this motion to be perpendicular to
Local magnetic field
hlocal
Opposes H0 in this region of space
hlocal
Local magnetic field
R
H
H hlocal
C
hlocal
C
H
Strengthens H0 in this region of space
C
hlocal
hlocal C
H
H π electron movement
A
hlocal
External field, H0
Opposes H0 in this region of space
B
hlocal
π electron movement
External field, H0
Figure 13-4 Electron circulation in the presence of an external magnetic field generates local magnetic fields that cause the characteristic chemical shifts of alkenyl and alkynyl hydrogens. (A) Alkenyl hydrogens are located in a region of space where hlocal reinforces H0. Therefore, these protons are relatively deshielded. (B) Electron circulation in an alkyne generates a local field that opposes H0 in the vicinity of the alkynyl hydrogen, thus causing shielding.
13-3 Spectroscopy of the Alkynes
Chapter 13
573
that in alkenes and to generate a local magnetic field that opposes H0 in the vicinity of the alkyne hydrogen. The result is a strong shielding effect that cancels the deshielding tendency of the electron-withdrawing sp-hybridized carbon and gives rise to a relatively high-field chemical shift.
The triple bond transmits spin – spin coupling The alkyne functional group transmits coupling so well that the terminal hydrogen is split by the hydrogens across the triple bond, even though it is separated from them by three carbons. This result is an example of long-range coupling. The coupling constants are small and range from about 2 to 4 Hz. Figure 13-5 shows the NMR spectrum of 1-pentyne. The alkynyl hydrogen signal at d 5 1.94 ppm is a triplet (J 5 2.5 Hz) because of coupling to the two equivalent hydrogens at C3, which appear at d 5 2.16 ppm. The latter, in turn, give rise to a doublet of triplets, representing coupling to the two hydrogens at C4 (J 5 6 Hz) as well as that at C1 (J 5 2.5 Hz).
CH 1.1 1.0 0.9 ppm
1.6 1.5 ppm
3H 2.0 1.9 ppm
2H
2.2 2.1 ppm
9
8
7
6
J 2–4 Hz H A O C O C q COH A
Figure 13-5 300-MHz 1H NMR spectrum of 1-pentyne showing coupling between the alkynyl (green) and propargylic (blue) hydrogens.
1H NMR
CH3CH2CH2C
Long-Range Coupling in Alkynes
5
4
3
1H 2H
2
(CH3)4Si
1
0
ppm (δ )
Exercise 13-5
Working with the Concepts: Predicting an NMR Spectrum Predict the first-order splitting pattern in the 1H NMR spectrum of 3-methyl-1-butyne. Strategy First, write out the structure. Then identify groups of hydrogens within coupling distance of each other, both neighboring and long range. Finally, use information regarding approximate values of coupling constants (and the N 1 1 rule) to generate expected splitting patterns. Solution • The structure of the molecule is CH3 A CH3 O CHO C q CH 3
2
1
• The two methyl groups are equivalent and give one signal that is split into a doublet by the single hydrogen atom at C3 (N 1 1 5 2 lines). The coupling constant (J value) for this splitting is the typical 6 – 8 Hz found in saturated systems (Section 10-7).
574
Chapter 13
Alkynes
• The alkynyl O CqCH hydrogen at C1 experiences long-range coupling to the same H at C3, appearing also as a doublet, but J is smaller, about 3 Hz. • Finally, the signal for the hydrogen at C3 displays a more complex pattern. The 6 – 8-Hz splitting by the six hydrogens of the methyl groups gives a septet (N 1 1 5 7 lines). Each line of this septet is further split by the additional 3-Hz coupling to the alkynyl H. As the actual spectrum below shows, the outermost lines of this signal, a doublet of septets, are so small that they are barely visible (see Tables 10-4 and 10-5). (Caution: When interpreting 1H NMR spectra, be aware of the very low intensity of the outer lines in highly split signals. In fact, it is prudent to assume that such signals may consist of more lines than are readily visible.)
1H NMR
6H H C
H3C
C
CH 1.2 1.1
CH3
ppm
2.1 2.0 1.9
ppm
1H 2.6 2.5 ppm
9
8
7
6
(CH3)4Si
1H
5
4
3
2
1
0
ppm (δ )
Exercise 13-6
Try It Yourself Predict the first-order splitting pattern in the 1H NMR spectrum of 2-pentyne.
The 13C NMR chemical shifts of alkyne carbons are distinct from those of the alkanes and alkenes Carbon-13 NMR spectroscopy also is useful in deducing the structure of alkynes. For example, the triple-bonded carbons in alkyl-substituted alkynes resonate in the range of d 5 65–95 ppm, quite separate from the chemical shifts of analogous alkane (d 5 5 – 45 ppm) and alkene (d 5 100 – 150 ppm) carbon atoms (Table 10-6). Typical Alkyne 13C NMR Chemical Shifts HC q CH ␦ 71.9
HC q CCH2CH2CH2CH3 68.6
84.0
18.6
31.1
22.4
14.1
CH3CH2C q CCH2CH3 81.1 15.6
13.2 ppm
Terminal alkynes give rise to two characteristic infrared absorptions Infrared spectroscopy is helpful in identifying terminal alkynes. Characteristic stretching bands appear for the alkynyl hydrogen at 3260 – 3330 cm21 and for the CqC triple bond at 2100 – 2260 cm21. There is also a diagnostic n|Csp–H bending absorption at 640 cm21
13-3 Spectroscopy of the Alkynes
Chapter 13
575
Figure 13-6 IR spectrum of 1,7-octadiyne: n| Csp2H stretch 5 3300 cm21; n| CqC stretch 5 2120 cm21; n| Csp2H bend 5 640 cm21.
Transmittance (%)
100
2120 H H
C
C
H
C
C
C
H
C H
3300
C C
H H
IR 0 4000 3500
H
640
H H
3000
2500
2000
1500
1000
600 cm−1
Wavenumber
(Figure 13-6). Such data are especially useful when 1H NMR spectra are complex and difficult to interpret. However, the band for the CqC stretching vibration in internal alkynes is often weak, like that for internal alkenes (Section 11-8), thus reducing the value of IR spectroscopy for characterizing these systems.
Mass spectral fragmentation of alkynes gives resonance-stabilized cations The mass spectra of alkynes, like those of alkenes, frequently show prominent molecular ions. Thus high-resolution measurements can reveal the molecular formula and therefore the presence of two degrees of unsaturation derived from the presence of the triple bond. In addition, fragmentation at the carbon once removed from the triple bond is observed, giving resonance-stabilized cations. For example, the mass spectrum of 3-heptyne (Figure 13-7) shows an intense molecular ion peak at myz 5 96 and loss of both methyl
CH3CH2C
Relative abundance
67 (M − CH2CH3) 81 (M − CH3)
MS
100
CCH2CH2CH3
M • 96
50
0 0
20
40
60
m/z
80
100
Figure 13-7 Mass spectrum of . 3-heptyne, showing M1 at myz 5 96 and important fragments at myz 5 67 and 81 arising from cleavage of the C1 – C2 and C5 – C6 bonds.
576
Chapter 13
Alkynes
(cleavage a) and ethyl (cleavage b) fragments to give two different stabilized cations, with myz 5 81 and 67 (base peak), respectively: Fragmentation of an Alkyne in the Mass Spectrometer
CH2 O C O O C O CH2 O CH2CH3
a a
b
CH3 O CH2 O C q C O CH2 O CH2CH3
j
CH3j
m/z 96 C2H5j
CH2 O C O C O CH2 O CH2CH3 m/z 81
b
CH3 O CH2 O C O O C O CH2
CH3 O CH2 O C O C O CH2 m/z 67
Unfortunately, under the high energy conditions of the mass spectrometry experiment, migration of the triple bond can occur. Thus this fragmentation is not typically very useful for identifying the location of the triple bond in a longer-chain alkyne.
In Summary The cylindrical p cloud around the carbon – carbon triple bond induces local magnetic fields that lead to NMR chemical shifts for alkynyl hydrogens at higher fields than those of alkenyl protons. Long-range coupling is observed through the C q C linkage. Infrared spectroscopy provides a useful complement to NMR data, displaying characteristic bands for the C q C and qC – H bonds of terminal alkynes. In the mass spectrometer, alkynes fragment to give resonance-stabilized cations.
13-4 Preparation of Alkynes by Double Elimination The two basic methods used to prepare alkynes are double elimination from 1,2-dihaloalkanes and alkylation of alkynyl anions. This section deals with the first method, which provides a synthetic route to alkynes from alkenes; Section 13-5 addresses the second, which converts terminal alkynes into more complex, internal ones.
Alkynes are prepared from dihaloalkanes by elimination As discussed in Section 11-6, alkenes can be prepared by E2 reactions of haloalkanes. Application of this principle to alkyne synthesis suggests that treatment of vicinal dihaloalkanes with two equivalents of strong base should result in double elimination to furnish a triple bond. Double Elimination from Dihaloalkanes to Give Alkynes X
X
C
C
H
H
Base (2 equivalents) 2 HX
Cq C
Vicinal dihaloalkane
Indeed, addition of 1,2-dibromohexane (prepared by bromination of 1-hexene, Section 12-5) to sodium amide in liquid ammonia followed by evaporation of solvent and aqueous work-up gives 1-hexyne.
13-5 Preparation of Alkynes from Alkynyl Anions
577
Chapter 13
Example of Double Dehydrohalogenation to Give an Alkyne
A
CH3CH2CH2CH2CH A Br
CH2Br
1. 3 NaNH2, liquid NH3 2. H2O 2 HBr
CH3CH2CH2CH2C q CH
Three equivalents of NaNH2 are necessary in the preparation of a terminal alkyne because, as this alkyne forms, its acidic terminal hydrogen (Section 13-2) immediately protonates an equivalent amount of base. Eliminations in liquid ammonia are usually carried out at its boiling point, 2338C. Because vicinal dihaloalkanes are readily available from alkenes by halogenation, this sequence, called halogenation – double dehydrohalogenation, is a ready means of converting alkenes into the corresponding alkynes. A Halogenation–Double Dehydrohalogenation Used in Alkyne Synthesis 1. Br2, CCl4 2. NaNH2, liquid NH3 3. H2O
53% 1,5-Hexadiene
1,5-Hexadiyne
Exercise 13-7 Illustrate the use of halogenation – double dehydrohalogenation in the synthesis of the alkynes (a) 2-pentyne; (b) 1-octyne; (c) 2-methyl-3-hexyne.
Haloalkenes are intermediates in alkyne synthesis by elimination Dehydrohalogenation of dihaloalkanes proceeds through the intermediacy of haloalkenes, also called alkenyl halides. Although mixtures of E- and Z-haloalkenes are in principle possible, with diastereomerically pure vicinal dihaloalkanes only one product is formed because elimination proceeds stereospecifically anti (Section 11-6).
X
X
X
C
C
H
H
X
or H
H
Exercise 13-8 Give the structure of the bromoalkene intermediate in the bromination – dehydrobromination of cis-2-butene to 2-butyne. Do the same for the trans isomer. (Caution: There is stereochemistry involved in both steps. Hint: Refer to Section 12-5 for useful information, and use models.)
The stereochemistry of the intermediate haloalkene is of no consequence when the sequence is used for alkyne synthesis. Both E- and Z-haloalkenes eliminate with base to give the same alkyne.
In Summary Alkynes are made from vicinal dihaloalkanes by double elimination. Alkenyl halides are intermediates, being formed stereospecifically in the first elimination.
13-5 Preparation of Alkynes from Alkynyl Anions Alkynes can also be prepared from other alkynes. The reaction of terminal alkynyl anions with alkylating agents, such as primary haloalkanes, oxacyclopropanes, aldehydes, or ketones, results in carbon – carbon bond formation. As we know (Section 13-2), such anions are readily prepared from terminal alkynes by deprotonation with strong bases (mostly alkyllithium reagents, sodium amide in liquid ammonia, or Grignard reagents). Alkylation
Bð
Bð Newman projection Anti elimination
X G CP C
G H
An alkenyl halide
1 equivalent NaOCH3, CH3OH
BH X
578
Alkynes
Chapter 13
with methyl or primary haloalkanes is typically done in liquid ammonia or in ether solvents. The process is unusual, because ordinary alkyl organometallic compounds are unreactive in the presence of haloalkanes. Alkynyl anions are an exception, however. Alkylation of an Alkynyl Anion C q CH
C q CðLi
CH3CH2CH2CH2Li, THF
C q CCH2CH2CH3 CH3CH2CH2I, 65C LiI Alkylation by SN2 reaction
Deprotonation by strong base
85% 1-Pentynylcyclohexane
Attempted alkylation of alkynyl anions with secondary and tertiary halides leads to E2 products because of the strongly basic character of the nucleophile (recall Section 7-8). Ethyne itself may be alkylated in a series of steps through the selective formation of the monoanion to give mono- and dialkyl derivatives. Alkynyl anions react with other carbon electrophiles such as oxacyclopropanes and carbonyl compounds in the same manner as do other organometallic reagents (Sections 8-8 and 9-9). Reactions of Alkynyl Anions
LiNH2 (1 equivalent), liquid NH3
HC q CH
NH2H Deprotonation
HC q CLi
O f i 1. H2C O CH2 2. HOH LiOH Nucleophilic ring opening
OH A HC q CCH2CH2 92% 3-Butyn-1-ol
O
CH3C q CH
CH3CH2MgBr, (CH3CH2)2O, 20°C CH3CH2H Deprotonation
CH3C q CMgBr
1. 2. H2O Nucleophilic addition
OH Cq
CH
CH3
66% 1-(1-Propynyl)cyclopentanol
Exercise 13-9 Suggest efficient and short syntheses of these two compounds. (Hint: Review Section 8-9.) from
(a)
OH OH
(b)
from ethyne
Exercise 13-10 3-Butyn-2-ol is an important raw material in the pharmaceutical industry. It is the starting point for the synthesis of a variety of medicinally valuable alkaloids (Section 25-8), steroids (Section 4-7), and prostaglandins (Chemical Highlight 11-1 and Section 19-13), as well as vitamins E (Section 22-9) and K. Propose a short synthesis of 3-butyn-2-ol by using the techniques outlined in this section.
13-6 Reduction of Alkynes: Reactivity of Two Pi Bonds
Chapter 13
In Summary Alkynes can be prepared from other alkynes by alkylation with primary haloalkanes, oxacyclopropanes, or carbonyl compounds. Ethyne itself can be alkylated in a series of steps.
13-6 Reduction of Alkynes: The Relative Reactivity of the Two Pi Bonds Now we turn from the preparation of alkynes to the characteristic reactions of the triple bond. In many respects, alkynes are like alkenes, except for the availability of two p bonds. Thus, alkynes can undergo additions, such as hydrogenation and electrophilic attacks. Addition of Reagents A–B to Alkynes G D
R O C q C OR
B R R R G G or C PC C PC D D A B A R G D
A–B
A–B
R R R R A A A A A O C O C OB or A O C O C OB A A A A A B B A
In this section we introduce two new hydrogen addition reactions: step-by-step hydrogenation and dissolving-metal reduction by sodium to give cis and trans alkenes, respectively.
Cis alkenes can be synthesized by catalytic hydrogenation Alkynes can be hydrogenated under the same conditions used to hydrogenate alkenes. Typically, platinum or palladium on charcoal is suspended in a solution containing the alkyne and the mixture is exposed to a hydrogen atmosphere. Under these conditions, the triple bond is saturated completely. Complete Hydrogenation of Alkynes CH3CH2CH2C q CCH2CH3
H2, Pt
CH3CH2CH2CH2CH2CH2CH3 100% Heptane
3-Heptyne
Hydrogenation is a stepwise process that may be stopped at the intermediate alkene stage by the use of modified catalysts, such as the Lindlar* catalyst. This catalyst is palladium that has been precipitated on calcium carbonate and treated with lead acetate and quinoline. The surface of the metal rearranges to a less active configuration than that of palladium on carbon so that only the first p bond of the alkyne is hydrogenated. As with catalytic hydrogenation of alkenes (Section 12-2), the addition of H2 is a syn process. As a result, this method affords a stereoselective synthesis of cis alkenes from alkynes. Hydrogenation with Lindlar Catalyst H
H
Lindlar Catalyst 5% Pd–CaCO3, O B Pb(OCCH3)2,
H2, Lindlar catalyst, 25°C syn Addition of H2
100% 3-Heptyne
*Dr. Herbert H. M. Lindlar (b. 1909), Hoffman – La Roche Ltd., Basel.
cis-3-Heptene
N Quinoline
579
580
Chapter 13
Alkynes
Exercise 13-11 Write the structure of the product expected from the following reaction.
CH3 O O
H2, Lindlar catalyst, 25°C
O
Exercise 13-12
Some perfumes have star quality (from left to right): Jean Paul Gaultier MaDame Perfume, Paris Hilton Fairy Dust, Armani Prive Oranger Alhambra, and Jeanne Lanvin.
The perfume industry makes considerable use of naturally occurring substances such as those obtained from rose and jasmine extracts. In many cases, the quantities of fragrant oils available by natural product isolation are so small that it is necessary to synthesize them. Examples are the olfactory components of violets, which include trans-2-cis-6-nonadien-1-ol and the corresponding aldehyde. An intermediate in their large-scale synthesis is cis-3-hexen-1-ol, whose industrial preparation is described as “a closely guarded secret.” Using the methods in this and the preceding sections, propose a synthesis from 1-butyne.
With a method for the construction of cis alkenes at our disposal, we might ask: Can we modify the reduction of alkynes to give only trans alkenes? The answer is yes, with a different reducing agent and through a different mechanism.
Sequential one-electron reductions of alkynes produce trans alkenes When we use sodium metal dissolved in liquid ammonia (dissolving-metal reduction) as the reagent for the reduction of alkynes, we obtain trans alkenes as the products. For example, 3-heptyne is reduced to trans-3-heptene in this way. Unlike sodium amide in liquid ammonia, which functions as a strong base, elemental sodium in liquid ammonia acts as a powerful electron donor (i.e., a reducing agent). REACTION
Dissolving-Metal Reduction of an Alkyne 1. Na, liquid NH3 2. H2O
H
H 86% 3-Heptyne
trans-3-Heptene
In the first step of the mechanism of this reduction, the p framework of the triple bond accepts one electron to give a radical anion. This anion is protonated by the ammonia solvent (step 2) to give an alkenyl radical, which is further reduced (step 3) by accepting another electron to give an alkenyl anion. This species is again protonated (step 4) to give the product alkene, which is stable to further reduction. The trans stereochemistry of the final alkene is set in the first two steps of the mechanism, which give rise preferentially to the less sterically hindered trans alkenyl radical. Under the reaction conditions (liquid NH3, 2338C), the second one-electron transfer takes place faster than cis-trans equilibration of the radical. This type of reduction typically provides .98% stereochemically pure trans alkene.
13-6 Reduction of Alkynes: Reactivity of Two Pi Bonds
Chapter 13
581
Mechanism of the Reduction of Alkynes by Sodium in Liquid Ammonia
MECHANISM Step 1. One-electron transfer
R groups adopt trans-like geometry to minimize steric repulsion
− R RC
Na
CR
C
Na
C
R Alkyne radical anion
A
Step 2. First protonation − R C
R
H H−NH2
C
R
C
C −
R
NH2
Alkenyl radical
B
Step 3. Second one-electron transfer R
H C
R
H Na
C
Na
C
R
C
R
− Alkenyl anion
C
Step 4. Second protonation R
H C
C
R
R
H H−NH2
−
C
−
C
R
NH2
H Trans alkene
D
The equation below illustrates the application of dissolving-metal reduction in the synthesis of the sex pheromone of the spruce budworm, which is the most destructive pest to the spruce and fir forests of North America. The pheromone “lure” is employed at hundreds of sites in the United States and Canada as part of an integrated pest-management strategy (Section 12-17). The key reaction is reduction of 11-tetradecyn-1-ol to the corresponding trans alkenol. Subsequent oxidation to the aldehyde completes the synthesis. Primary alcohol
HO(CH2)10C q CCH2CH3 11-Tetradecyn-1-ol
Na, liquid NH3 Reduction
HO(CH2)10
H G G C PC D D CH2CH3 H
trans-11-Tetradecen-1-ol
The spruce budworm, a serious pest. Oxidized by PCC
PCC, CH2Cl2 Oxidation (Section 8-6)
to aldehyde O B H HC(CH2)9 G G C PC D D H CH2CH3
Sex pheromone of the spruce budworm
582
Chapter 13
Alkynes
Exercise 13-13
Working with the Concepts: Selectivity in Reduction When 1,7-undecadiyne (11 carbons) was treated with a mixture of sodium and sodium amide in liquid ammonia, only the internal bond was reduced to give trans-7-undecen-1-yne. Explain. (Hint: What reaction takes place between sodium amide and a terminal alkyne? Note that the pKa of NH3 is 35.) Strategy First, write out the equation for the reaction. Then consider the substrate’s functionality in the context of the reaction conditions. Solution • The equation is Na, NaNH2, NH3
• The conditions are strongly reducing (Na), but also strongly basic (NaNH2). We learned earlier in the chapter that the pKa of a terminal alkynyl hydrogen is about 25. Sodium amide, which is the conjugate base of the exceedingly weak acid ammonia, readily deprotonates the terminal alkyne, giving an alkynyl anion, RC q C:2. • The dissolving-metal reduction process requires electron transfer to the triple bond. However, the negative charge on the deprotonated terminal alkyne repels any attempt to introduce additional electrons, rendering that particular triple bond immune to reduction. Therefore, only the internal triple bond is reduced, producing a trans alkene.
Exercise 13-14
Try It Yourself What should be the result of the treatment of 2,7-undecadiyne with a mixture of excess sodium and sodium amide in liquid ammonia? Explain any differences between this outcome and that in Exercise 13-13.
In Summary Alkynes are very similar in reactivity to alkenes, except that they have two p bonds, both of which may be saturated by addition reactions. Hydrogenation of the first p bond, which gives cis alkenes, is best achieved by using the Lindlar catalyst. Alkynes are converted into trans alkenes by treatment with sodium in liquid ammonia, a process that includes two successive one-electron reductions.
13-7 Electrophilic Addition Reactions of Alkynes As a center of high electron density, the triple bond is readily attacked by electrophiles. This section describes the results of three such processes: addition of hydrogen halides, reaction with halogens, and hydration. The hydration is catalyzed by mercury(II) ions. As is the case in electrophilic additions to unsymmetrical alkenes (Section 12-3), the Markovnikov rule is followed in transformations of terminal alkynes: The electrophile adds to the terminal (less substituted) carbon atom.
Addition of hydrogen halides forms haloalkenes and geminal dihaloalkanes The addition of hydrogen bromide to 2-butyne yields (Z)-2-bromobutene. The mechanism is analogous to that of hydrogen halide addition to an alkene (Section 12-3).
13-7 Electrophilic Addition Reactions of Alkynes
Chapter 13
Addition of a Hydrogen Halide to an Internal Alkyne HBr, Br
CH3C q CCH3
CH D 3 G CP C D G Br H3 C 60% H
(Z)-2-Bromobutene
The stereochemistry of this type of addition is typically anti, particularly when excess halide ion is used. A second molecule of hydrogen bromide may also add, with regioselectivity that follows Markovnikov’s rule, giving the product with both bromine atoms bound to the same carbon, a geminal dihaloalkane. CH D 3 G CP C G D Br H3 C H
HBr
H Br A A CH3CHCCH3 A Br 90%
Both bromines add to the same carbon
2,2-Dibromobutane
The addition of hydrogen halides to terminal alkynes also proceeds in accord with the Markovnikov rule. Addition to a Terminal Alkyne
CH3C q CH
HI, 70C
H D G CP C D G H H3C 35% I
Both iodines add to the same carbon
I H A A CH3C O C O H A A I H 65%
Both hydrogens add to the same carbon
It is usually difficult to limit such reactions to addition of a single molecule of HX. Exercise 13-15 Write a step-by-step mechanism for the addition of HBr twice to 2-butyne to give 2,2-dibromobutane. Show clearly the structure of the intermediate formed in each step.
Halogenation also takes place once or twice Electrophilic addition of halogen to alkynes proceeds through the intermediacy of isolable vicinal dihaloalkenes, the products of a single anti addition. Reaction with additional halogen gives tetrahaloalkanes. For example, halogenation of 3-hexyne gives the expected (E)-dihaloalkene and the tetrahaloalkane. Double Halogenation of an Alkyne CH3CH2C q CCH2CH3
3-Hexyne
Br2, CH3COOH, LiBr
CH3CH2 Br D G CP C G D CH2CH3 Br 99% (E)-3,4-Dibromo-3-hexene
Exercise 13-16 Give the products of addition of one and two molecules of Cl2 to 1-butyne.
Br2, CCl4
Br Br A A CH3CH2C O CCH2CH3 A A Br Br 95% 3,3,4,4-Tetrabromohexane
583
584
Chapter 13
Alkynes
Mercuric ion – catalyzed hydration of alkynes furnishes ketones In a process analogous to the hydration of alkenes, water can be added to alkynes in a Markovnikov sense to give alcohols — in this case enols, in which the hydroxy group is attached to a double-bond carbon. As mentioned in Section 12-16, enols spontaneously rearrange to the isomeric carbonyl compounds. This process, called tautomerism, interconverts two isomers by simultaneous proton and double-bond shifts. The enol is said to tautomerize to the carbonyl compound, and the two species are called tautomers (tauto, Greek, the same; meros, Greek, part). We shall look at tautomerism more closely in Chapter 18 when we investigate the behavior of carbonyl compounds. Hydration followed by tautomerism converts alkynes into ketones. The reaction is catalyzed by Hg(II) ions. Hydration of Alkynes RC q CR
HOH, H, HgSO4
OH A RCH P CR
H O A B RC O CR A H
Tautomerism
Enol
Ketone
Hydration follows Markovnikov’s rule: Terminal alkynes give methyl ketones. Hydration of a Terminal Alkyne OH
OH O B
H2SO4, H2O, HgSO4
91% Exercise 13-17 Draw the structure of the enol intermediate in the reaction above.
Symmetric internal alkynes give a single carbonyl compound; unsymmetric systems lead to a mixture of ketones. Hydration of Internal Alkynes H2SO4, H2O, HgSO4
O 80% Only possible product
Example of Hydration of an Internal Alkyne That Gives a Mixture of Two Ketones CH3CH2CH2C q CCH3
H2SO4, H2O, HgSO4
O B CH3CH2CH2CCH2CH3 50%
O B CH3CH2CH2CH2CCH3 50%
Exercise 13-18 Give the products of mercuric ion – catalyzed hydration of (a) ethyne; (b) propyne; (c) 1-butyne; (d) 2-butyne; (e) 2-methyl-3-hexyne.
13-8 Anti-Markovnikov Additions to Triple Bonds
585
Chapter 13
Exercise 13-19
Working with the Concepts: Using Alkynes in Synthesis Propose a synthetic scheme that will convert compound A into B (see margin). [Hint: Consider OH A a route that proceeds through the alkynyl alcohol (CH3)2CC q CH.] Strategy The hint reveals to us a possible retrosynthetic analysis of the problem: HO
O
HO
O
Let us consider what we have learned so far in this chapter that can be helpful. This section has shown how alkynes can be converted to ketones by mercury ion-catalyzed hydration. Section 13-5 introduced a new strategy for the formation of carbon – carbon bonds through the use of alkynyl anions. Beginning with the three-carbon ketone A (acetone), our first task is to add a two-carbon alkynyl unit. Referring to Section 13-5, we can use any of several methods to convert ethyne into the corresponding anion. Solution • Adding the anion to acetone gives the necessary intermediate alcohol: O
HO
HC q CH
LiNH2 (1 equivalent), liquid NH3
HC q CLi
1. 2. H2O
Cq
CH
• Finally, hydration of the terminal alkyne function, as illustrated for the cyclohexyl derivative shown on the previous page, completes the synthesis: HO
HO H2SO4, H2O, HgSO4
O
Exercise 13-20
Try It Yourself Propose a synthesis of trans-3-hexene starting with 1-butyne.
In Summary Alkynes can react with electrophiles such as hydrogen halides and halogens either once or twice. Terminal alkynes transform in accord with the Markovnikov rule. Mercuric ion – catalyzed hydration furnishes enols, which convert into ketones by a process called tautomerism.
13-8 Anti-Markovnikov Additions to Triple Bonds Just as methods exist to permit anti-Markovnikov additions to double bonds (Sections 12-8 and 12-13), similar techniques allow additions to terminal alkynes to be carried out in an anti-Markovnikov manner.
Radical addition of HBr gives 1-bromoalkenes As with alkenes, hydrogen bromide can add to triple bonds by a radical mechanism in an anti-Markovnikov fashion if light or other radical initiators are present. Both syn and anti additions are observed.
HO O O A
B
586
Chapter 13
Alkynes
CH3(CH2)3C q CH
HBr, ROOR
CH3(CH2)3CH P CHBr 74% cis- and trans-1-Bromo-1-hexene
1-Hexyne
Aldehydes result from hydroboration – oxidation of terminal alkynes Terminal alkynes are hydroborated in a regioselective, anti-Markovnikov fashion, the boron attacking the less hindered carbon. However, with borane itself, this reaction leads ultimately to sequential hydroboration of both p bonds. To stop at the alkenylborane stage, bulky borane reagents, such as dicyclohexylborane, are used. Hydroboration of a Terminal Alkyne CH3(CH2)5C q CH
BH
THF
Anti-Markovnikov addition
H D G CP C G D B H
CH3(CH2)5
2
2
1-Octyne
94%
Dicyclohexylborane
Exercise 13-21 Dicyclohexylborane is made by a hydroboration reaction. What are the starting materials for its preparation?
Like alkylboranes (Section 12-8), alkenylboranes can be oxidized to the corresponding alcohols — in this case, to terminal enols that spontaneously rearrange to aldehydes. Hydroboration–Oxidation of a Terminal Alkyne CH3(CH2)5C q CH
1. Dicyclohexylborane 2. H2O2, HO Anti-Markovnikov hydroboration followed by oxidation
CH3(CH2)5
H D G CP C G D OH H Enol
Tautomerism
OH on less substituted carbon
H O A B CH3(CH2)5CO CH A H 70% Octanal
Exercise 13-22 Give the products of hydroboration – oxidation of (a) ethyne; (b) 1-propyne; (c) 1-butyne.
Exercise 13-23 Outline a synthesis of the following molecule from 3,3-dimethyl-1-butyne. O B (CH3)3CCH2CH
In Summary HBr in the presence of peroxides undergoes anti-Markovnikov addition to terminal alkynes to give 1-bromoalkenes. Hydroboration – oxidation with bulky boranes furnishes intermediate enols that tautomerize to the final product aldehydes.
13-9 Chemistry of Alkenyl Halides
13-9 Chemistry of Alkenyl Halides We have encountered haloalkenes — alkenyl halides — as intermediates in both the preparation of alkynes by dehydrohalogenation and also the addition to alkynes of hydrogen halides. Alkenyl halides have become increasingly important as synthetic intermediates in recent years as a result of developments in organometallic chemistry. These systems do not, however, follow the mechanisms familiar to us from our survey of the haloalkanes (Chapters 6 and 7). This section discusses their reactivity.
Alkenyl halides do not undergo SN2 or SN1 reactions Unlike haloalkanes, alkenyl halides are relatively unreactive toward nucleophiles. Although we have seen that, with strong bases, alkenyl halides undergo elimination reactions to give alkynes, they do not react with weak bases and relatively nonbasic nucleophiles, such as iodide. Similarly, SN1 reactions do not normally take place, because the intermediate alkenyl cations are species of high energy. H D CH2 PC G Br
I
H D CH2 PC Br G I
H D CH2 PC G Br
CH2 PCO H Br Ethenyl (vinyl) cation
Does not take place
Does not take place
Alkenyl halides, however, can react through the intermediate formation of alkenyl organometallics (see Exercise 11-6). These species allow access to a variety of specifically substituted alkenes. Alkenyl Organometallics in Synthesis Grignard addition to ketone gives tertiary alcohol
Br D CH2 PC G H
Mg
THF
1-Bromoethene (Vinyl bromide)
O B 1. CH3CCH3 2. H , H2O
MgBr D CH2 PC G H 90%
OH A C(CH3)2 D CH2 PC New G C–C H bond 65% 2-Methyl-3-buten-2-ol
Ethenylmagnesium bromide (A vinyl Grignard reagent)
Metal catalysts couple alkenyl halides to alkenes in the Heck reaction In the presence of soluble complexes of metals such as Ni and Pd, alkenyl halides undergo carbon – carbon bond formation with alkenes to produce dienes. In this process, called the Heck* reaction, a molecule of hydrogen halide is liberated. The Heck Reaction HCl is lost
Cl D PC H2C G H
H
D C P CH2 G H
Ni or Pd catalyst HCl
*Professor Richard F. Heck (b. 1931), University of Delaware.
A new C–C bond forms H A KCH KCH2 H2C C A H
Chapter 13
587
588
Alkynes
Chapter 13
CHE M I C AL H I G H L I G H T 1 3 - 1 Metal-Catalyzed Stille, Suzuki, and Sonogashira Coupling Reactions Three additional processes, the Stille, Suzuki, and Sonogashira* reactions, further broaden the scope of transition metal – catalyzed bond-forming processes. All utilize catalytic palladium or nickel; the differences lie
in the nature and functionality of the substrates commonly employed. In the Stille coupling, Pd catalyzes the direct linkage between alkenyl halides and alkenyltin compounds:
Stille Coupling Reaction O
O I
O
Pd catalyst, CuI, R3As
(CH3)3Sn
O O
O
93%
Copper(I) iodide and an arsenic-derived ligand, R3As, facilitate this very efficient process. The product shown was converted into a close relative of a microbially derived natural product that inhibits a factor associated with immune and
inflammation responses. This factor also affects HIV activation and cell-death processes that are disrupted in cancer. The Suzuki reaction replaces tin with boron and provides a different spectrum of utility. In particular,
Suzuki Coupling Reaction I
Ni catalyst, base
(HO)2B 63%
*Professor John K. Stille (1930–1990), Colorado State University; Professor Akira Suzuki (b. 1930), Kurashiki University, Japan; Professor Kenkichi Sonogashira (b. 1931), Osaka City University, Japan.
In common with other transition metal – catalyzed cross-couplings (see Chemical Highlight 8-3), assembly of the fragments around the catalyst precedes carbon – carbon bond formation. A simplified mechanism for the Heck reaction begins with reaction between the metal and the alkenyl halide to give an alkenylmetal halide (1). The alkene then complexes with the metal (2), and inserts itself into the carbon – metal bond, forming the new carbon – carbon linkage (3). Finally, elimination of HX in an E2-like manner gives the diene product and frees the metal catalyst (4). Mechanism of the Heck Reaction
(3) Alkene insertion
H2C
KCHH
H2C PCH O Pd O Cl
ECH2H ECl Pd CH A H
(4) Elimination
(2) H2CPCH2
H2C PCH O Pd O Cl ł H2C PCH2 —
H2C PCH O Cl
(1) Pd
H A KC H KCH2 C H2C A H
HCl
Pd
The growing popularity of the Heck reaction arises from both its versatility and its efficiency. In particular, it requires only a very small amount of catalyst compared with the quantity of the substrates; typically, 1% of palladium acetate in the presence of a phosphine ligand (R3P) is sufficient.
13-9 Chemistry of Alkenyl Halides
Chapter 13
589
The boron-containing substrate (a boronic acid) is efficiently prepared by hydroboration of a terminal alkyne with a special reagent, catechol borane:
Suzuki coupling succeeds with primary and secondary haloalkanes, which are poor Stille substrates. In the example below, Ni gives better results than does Pd.
Preparation of an Alkenylboronic Acid O
R
B OH
O
H, H2O
B O
O
(HO)2B
R
R H
Catechol borane
Alkenyl boronic acid
Boronic acids are prepared commercially in very large quantities, and the Suzuki coupling has become a major industrial process. Boronic acids are stable and easier to handle than organotin compounds, which are toxic and must be handled with great care. Finally, the Sonogashira reaction has a niche of its own as a preferred method for linking alkenyl and alkynyl
moieties. As in the Stille process, Pd, CuI, and ligands derived from nitrogen-group elements are employed. However, there is no need for tin; terminal alkynes react directly. The added base removes the HI by-product.
Sonogashira Coupling Reaction O
O I
H
Pd catalyst, CuI, R3P, base HI
89%
Examples of Heck Reactions Br
O B COCH3
New C–C bond
O B 1% Pd(OCCH3)2, R3P, 100C
72% ON EOH C
O B COCH3
O B 1% Pd(OCCH3)2, R3P, 100C
ON EOH C New C–C bond
Br 67%
COCH3 B O
Exercise 13-24 Write out a detailed step-by-step mechanism for the first of the two examples of Heck reactions above.
O B COCH3
590
Chapter 13
Alkynes
In Summary Alkenyl halides are unreactive in nucleophilic substitutions. However, they can participate in carbon – carbon bond-forming reactions after conversion to alkenyllithium or alkenyl Grignard reagents, or in the presence of transition-metal catalysts such as Ni and Pd.
13-10 Ethyne as an Industrial Starting Material Ethyne was once one of the four or five major starting materials in the chemical industry for two reasons: Addition reactions to one of the p bonds produce useful alkene monomers (Section 12-15), and it has a high heat content. Its industrial use has declined because of the availability of cheap ethene, propene, butadiene, and other hydrocarbons through oilbased technology. However, in the 21st century, oil reserves are expected to dwindle to the point that other sources of energy will have to be developed. One such source is coal. There are currently no known processes for converting coal directly into the aforementioned alkenes; ethyne, however, can be produced from coal and hydrogen or from coke (a coal residue obtained after removal of volatiles) and limestone through the formation of calcium carbide. Consequently, it may once again become an important industrial raw material.
Production of ethyne from coal requires high temperatures The high energy content of ethyne requires the use of production methods that are costly in energy. One process for making ethyne from coal uses hydrogen in an arc reactor at temperatures as high as several thousand degrees Celsius. Coal
H2
Δ
HC q CH
nonvolatile salts
33% conversion
The oldest large-scale preparation of ethyne proceeds through calcium carbide. Limestone (calcium oxide) and coke are heated to about 20008C, which results in the desired product and carbon monoxide. 3C
Coke
2000°C
CaO Lime
CaC2
CO
Calcium carbide
The calcium carbide is then treated with water at ambient temperatures to give ethyne and calcium hydroxide. Vivid demonstration of the combustion of ethyne, generated by the addition of water to calcium carbide.
CaC2
2 H2O
HC q CH
Ca(OH)2
Ethyne is a source of valuable monomers for industry Ethyne chemistry underwent important commercial development in the 1930s and 1940s in the laboratories of Badische Anilin and Sodafabriken (BASF) in Ludwigshafen, Germany. Ethyne under pressure was brought into reaction with carbon monoxide, carbonyl compounds, alcohols, and acids in the presence of catalysts to give a multitude of valuable raw materials to be used in further transformations. For example, nickel carbonyl catalyzes the addition of carbon monoxide and water to ethyne to give propenoic (acrylic) acid. Similar exposure to alcohols or amines instead of water results in the corresponding acid derivatives. All of these products are valuable monomers (see Section 12-15). Industrial Chemistry of Ethyne H HC q CH
CO
H2O
Ni(CO)4, 100 atm, >250°C
G C PCHCOOH D H Propenoic acid (Acrylic acid)
13-10 Ethyne as an Industrial Starting Material
Chapter 13
591
Polymerization of propenoic (acrylic) acid and its derivatives produces materials of considerable utility. The polymeric esters (polyacrylates) are tough, resilient, and flexible polymers that have replaced natural rubber (see Section 14-10) in many applications. Poly(ethyl acrylate) is used for O-rings, valve seals, and related purposes in automobiles. Other polyacrylates are found in biomedical and dental appliances, such as dentures. The addition of formaldehyde to ethyne is achieved with high efficiency by using copper acetylide as a catalyst. HC q CH
CH2 P O
Cu2C2–SiO2, 125°C, 5 atm
HC q CCH2OH
or
HOCH2C q CCH2OH
2-Propyn-1-ol (Propargyl alcohol)
2-Butyne-1,4-diol
The resulting alcohols are useful synthetic intermediates. For example, 2-butyne-1,4-diol is a precursor for the production of oxacyclopentane (tetrahydrofuran, one of the solvents most frequently employed for Grignard and organolithium reagents) by hydrogenation, followed by acid-catalyzed dehydration. Oxacyclopentane (Tetrahydrofuran) Synthesis
HOCH2C q CCH2OH
Catalyst, H2
HO(CH2)4OH
H3PO4, pH 2, 260–280°C, 90–100 atm
O 99%
H2O
Oxacyclopentane (Tetrahydrofuran, THF)
Several technical processes have been developed in which reagents d1A – Bd2 in the presence of a catalyst add to the triple bond. For example, the catalyzed addition of hydrogen chloride gives chloroethene (vinyl chloride), and addition of hydrogen cyanide produces propenenitrile (acrylonitrile). Addition Reactions of Ethyne H HC q CH
HCl
Hg2, 100–200°C
G C PCHCl D H Chloroethene (Vinyl chloride)
HC q CH
HCN
Cu, NH4Cl, 70–90°C, 1.3 atm
H
G C PCHCN D H 80–90% Propenenitrile (Acrylonitrile)
In 2007, the world produced 2.5 million tons of acrylic fibers, polymers containing at least 85% propenenitrile (acrylonitrile). Their applications include clothing (Orlon), carpets, and insulation. Copolymers of acrylonitrile and 10 – 15% vinyl chloride have fire-retardant properties and are used in children’s sleepwear.
In Summary Ethyne was once, and may again be in the future, a valuable industrial feedstock because of its ability to react with a large number of substrates to yield useful monomers and other compounds having functional groups. It can be made from coal and H2 at high temperatures, or it can be prepared from calcium carbide by hydrolysis. Some of the industrial reactions that it undergoes are carbonylation, addition of formaldehyde, and addition reactions with HX.
Poly(vinyl chloride) is extensively used in the construction industry for water and sewer pipes.
592
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Alkynes
13-11 Naturally Occurring and Physiologically Active Alkynes Although alkynes are not very abundant in nature, they do exist in some plants and other organisms. The first such substance to be isolated, in 1826, was dehydromatricaria ester, from the chamomile flower. More than a thousand such compounds are now known, and some of them are physiologically active. For example, some naturally occurring ethynylketones, such as capillin, an oil found in the chrysanthemum, have fungicidal activity. CH3C q C O C q CO C q C
H G G C PC D D H COCH3 B O
Dehydromatricaria ester
Capillin (Active against skin fungi)
The alkyne ichthyothereol is the active ingredient of a poisonous substance used by the Indians of the Lower Amazon River Basin in arrowheads. It causes convulsions in mammals. Two enyne functional groups are incorporated in the compound histrionicotoxin. It is one of the substances isolated from the skin of “poison arrow frog,” a highly colorful species of the genus Dendrobates. The frog secretes this compound and similar ones as defensive venoms and mucosal-tissue irritants against both mammals and reptiles. How the alkyne units are constructed biosynthetically is not clear.
O
O
HO HH G C J CH3C q C O C q CO C q C O C H
#
“Poison arrow” frog.
O B CH3C q C O C q C O C
H
N H
H
OH
O
O
N O
Ichthyothereol (A convulsant)
Histrionicotoxin
Many drugs have been modified by synthesis to contain alkyne substituents, because such compounds are frequently more readily absorbed by the body, less toxic, and more active than the corresponding alkenes or alkanes. For example, 3-methyl-1-pentyn-3-ol is available as a nonprescription hypnotic, and several other alkynols are similarly effective. Highly reactive enediyne ( – CqC – CH“CH – CqC – ) and trisulfide (RSSSR) functional groups characterize a class of naturally occurring antibiotic – antitumor agents discovered in the late 1980s, such as calicheamicin and esperamicin.
The Big Picture
H S
S
S
CH3
HO X≈ š O
K
H3COCNH B O
#
CH3 A HC q CCCH2CH3 A OH
OR
Enediyne group
3-Methyl-1-pentyn-3-ol
Calicheamicin (X = H) Esperamicin (X = OR)
(Hypnotic)
R and R = sugars (Chapter 24)
Ethynyl estrogens, such as 17-ethynylestradiol, are considerably more potent birth-control agents than are the naturally occurring hormones (see Section 4-7). The diaminoalkyne tremorine induces symptoms characteristic of Parkinson’s disease: spasms of uncontrolled movement. Interestingly, a simple cyclic homolog of tremorine acts as a muscle relaxant and counteracts the effect of tremorine. Compounds that cancel the physiological effects of other compounds are called antagonists (antagonizesthai, Greek, to struggle against). Finally, ethynyl analogs of amphetamine have been prepared in a search for alternative, more active, more specific, and less addictive central nervous system stimulants.
H
H %
}
}
CH3%OHC q CH % $ NCH2C q CCH2N
H
HO 17-Ethynylestradiol
Tremorine
NCH2C q CCH2N
NH2 A CH2CHC q CH
Tremorine antagonist
An amphetamine analog (Active in the central nervous system)
In Summary The alkyne unit is present in a number of physiologically active natural and synthetic compounds.
THE BIG PICTURE As we said at the end of the previous chapter, much of what we have encountered in our examination of the alkynes represents an extension of what we learned regarding alkenes. Addition reactions take place under very similar conditions, obeying the same rules of regio- and stereochemistry. Reagents such as the hydrogen halides and the halogens may add once or twice. Addition of the elements of water to one of the p bonds takes us in a new direction, however: The resulting alkenol (or enol, for short) undergoes rearrangement (tautomerism) to an aldehyde or a ketone. Finally, terminal alkynes display a form of reactivity that does not normally appear in the alkenes (or alkanes, for that matter): The – CqC – H hydrogen is unusually acidic. Its deprotonation gives rise to nucleophilic anions capable of forming new carbon – carbon bonds by reaction with a variety of functional groups possessing electrophilic carbon atoms.
Chapter 13
593
594
Chapter 13
Alkynes
In the next chapter we shall examine compounds containing multiple double bonds, including some made by the Heck process, a new organometallic reaction that we have just encountered. The same principles that have appeared repeatedly in Chapters 11 – 13 will continue to underlie the behavior of the systems that we cover next.
CHAPTER INTEGRATION PROBLEMS 13-25. Propose an efficient synthesis of 2,7-dimethyl-4-octanone, using organic building blocks containing no more than four carbon atoms. O
2,7-Dimethyl-4-octanone
SOLUTION We begin by analyzing the problem retrosynthetically (Section 8-9). What methods do we know for the preparation of ketones? We can oxidize alcohols (Section 8-6). Is this a productive line of analysis? If we look at the corresponding alcohol precursor, we can imagine synthesizing it by addition of an appropriate organometallic reagent to an aldehyde to form either bond a or bond b (Section 8-9):
OH
O
a
Target molecule
b
Alcohol precursor
Let us count carbon atoms in the fragments necessary for each of these synthetic pathways. To make bond a, we need to add a four-carbon organometallic to a six-carbon aldehyde. The bond b alternative would employ 2 five-carbon building blocks. Remember the restriction that only fourcarbon starting materials are allowed. From this point of view, neither of the preceding options is overly attractive. We shall shortly look again at route a but not route b. Do you see why? The latter would require initial construction of 2 five-carbon pieces, whereas the former needs formation of only 1 six-carbon unit from fragments containing four carbons or fewer. Now let us consider a second, fundamentally different ketone synthesis — hydration of an alkyne (Section 13-7). Either of two precursors, 2,7-dimethyl-3-octyne and 2,7-dimethyl-4-octyne, will lead to the target molecule. As shown here, however, only the latter, symmetric alkyne undergoes hydration to give just one ketone, regardless of the initial direction of addition. O
H2SO4, H2O, HgSO4
2,7-Dimethyl-3-octyne
O O H2SO4, H2O, HgSO4
2,7-Dimethyl-4-octyne
With 2,7-dimethyl-4-octyne appearing to be the precursor of choice, we continue by investigating its synthesis from building blocks of four carbons or fewer. The alkylation of terminal alkynes
New Reactions
Chapter 13
(Section 13-5) affords us a method of bond formation that divides the molecule into three suitable fragments, shown in the following analysis: Li
Li
Br
Br
LiNH2
The synthesis follows directly:
Br
LiC q CH, DMSO
LiNH2, liquid NH3
Br
Li
Although this three-step synthesis is the most efficient answer, a related approach derives from our earlier consideration of ketone synthesis with the use of an alcohol. It, too, proceeds through an alkyne. Construction of bond a of the target molecule, shown earlier, requires addition of an organometallic reagent to a six-carbon aldehyde, which, in turn, may be produced through hydroboration – oxidation (Section 13-8) of the terminal alkyne shown in the preceding scheme. O 1. Dicyclohexylborane 2. H2O2, NaOH, H2O
1. Li
, THF
OH
2. H, H2O
H
Oxidation of this alcohol by using a Cr(VI) reagent (Section 8-6) completes a synthesis that is just slightly longer than the optimal one described first. 13-26. Predict the product you would expect from the treatment of a terminal alkyne with bromine in water solvent, i.e., CH3CH2C q CH
Br2, H2O
SOLUTION Consider the problem mechanistically. Bromine adds to p bonds to form a cyclic bromonium ion, which is subject to ring opening by any available nucleophile. In the similar reaction of alkenes (Sections 12-5 and 12-6), nucleophilic attack is directed to the more substituted alkene carbon, namely, the carbon atom bearing the larger partial positive charge. Following that analogy in the case at hand and using water as the nucleophile, we may postulate the following as reasonable mechanistic steps:
O O
CH3CH2
O
CPC
H
CH3CH2 Br G G C PC D D H O Oð H
H
O
H2š O
O
CH3CH2C q CH
ðBrð
š ðš BrOBr ð
CH3CH2 Br G G C PC D D ð HO H
H
The product of the sequence is an enol, which, as we have seen (Section 13-8), is unstable and rapidly ðOð B tautomerizes into a carbonyl compound. In this case, the ultimate product is CH3CH2OC OCH2Br, a bromoketone.
New Reactions 1. Acidity of 1-Alkynes (Section 13-2) RC q CH
ðB
RC q Cð
BH
pKa ~ ~ 25 Base (B): NaNH2–liquid NH3; RLi–(CH3CH2)2O; RMgX–THF
, DMSO
2,7-dimethyl4-octyne
595
596
Chapter 13
Alkynes
Preparation of Alkynes 2. Double Elimination from Dihaloalkanes (Section 13-4) X X A A RC O CR A A H H
NaNH2, liquid NH3 2 HX
RC q CR
Vicinal dihaloalkane
3. From Alkenes by Halogenation – Dehydrohalogenation (Section 13-4)
RCH P CHR
1. X2, CCl4 2. NaNH2, liquid NH3
G RCH P C D
R NaNH2, liquid NH3
RC q CR
X
Alkenyl halide intermediate
Conversion of Alkynes into Other Alkynes 4. Alkylation of Alkynyl Anions (Section 13-5) 1. NaNH2, liquid NH3 2. RX
RC q CH
RC q CR
SN2 reaction: R must be primary
5. Alkylation with Oxacyclopropane (Section 13-5)
RC q CH
1. CH3CH2CH2CH2Li, THF O f f 2. H2C CH2 3. H, H2O
RC q CCH2CH2OH
Attack takes place at less substituted carbon in unsymmetric oxacyclopropanes
6. Alkylation with Carbonyl Compounds (Section 13-5)
RC q CH
1. CH3CH2CH2CH2Li, THF O B 2. RCR 3. H, H2O
OH A RC q CCR A R
Reactions of Alkynes 7. Hydrogenation (Section 13-6) RC q CR
Catalyst, H2
ΔH° ~ ~ 70 kcal mol1
RCH2CH2R
Catalysts: Pt, Pd–C
H RC q CR
H2, Lindlar catalyst
H G G C PC D D R R
ΔH° ~ ~ 40 kcal mol1
Cis alkene
8. Reduction with Sodium in Liquid Ammonia (Section 13-6)
RC q CR
1. Na, liquid NH3 2. H, H2O
H
R G G C PC D D R H Trans alkene
New Reactions
9. Electrophilic (and Markovnikov) Additions: Hydrohalogenation, Halogenation, and Hydration (Section 13-7) HX
RC q CR
HX
RCH P CXR
RCH2CX2R Geminal dihaloalkane
2 HX
RC q CH R
Br G G C PC D D Br R
Br2, Br
RC q CR
RCX2CH3 Br2
RCBr2CBr2R
Mainly trans
O B RCH2CR
Hg2, H2O
RC q CR
10. Radical Addition of Hydrogen Bromide (Section 13-8) HBr, ROOR
RC q CH
RCH PCHBr Anti-Markovnikov
Br attaches to less substituted carbon
11. Hydroboration (Section 13-8) R
RC q CH
H G G C PC D D BR2 H
R2BH, THF
Anti-Markovnikov and stereospecific (syn) addition B attaches to less substituted carbon )
Dicyclohexylborane (R
12. Oxidation of Alkenylboranes (Section 13-8)
H
B G G C PC D D R H
H2O2, HO
OH G G C PC D D R H
H
Tautomerism
O B RCH2CH
Enol
Organometallic Reagents 13. Alkenyl Organometallics (Section 13-9) R R
G G C PC D D
R
X Mg, THF
R
R
G G C PC D D
MgX R
14. Heck Reaction (Section 13-9)
R R
G G CP C D D
H
Cl
R
R
G G CP C D D 1
R3 R2
Ni or Pd catalyst HCl
R3 R A A RH KCH KC H C C R2 A A R R1
Chapter 13
597
598
Reactions of Alkynes
RCqC:M
13-2
OH A OH RCqCCR A A RCqCR RCqCCH2CHR R RCH2CH3
13-5
13-5
13-5
13-6
R H
G G CPC D D
H H
13-6
R H
G G CPC D D
section number
R
H H
13-6
RCX2CH3
RCX2CHX2
O B RCCH3
13-7
13-7
13-7, 17-4
H RCHP CHBr
13-8
G G CPC D D
H BR2
13-8
O B RCH2CH
13-8, 17-4
RCHP CH A RCH P CH
13-9
Via R Via RCXPCH2 RLi or RMgX or NaNH2
1. Base 2. RX
1. Base 2. O R
1. Base 2. RCR B O
H2, Pt or Pd–C
H2, Lindlar catalyst (syn-addition)
Na, liquid NH3 (anti-addition)
HX
RC q CH
X
G G CPC D D
X2
X Via RCHPCHBr
H
HgSO4, H2O, H2SO4
HBr, ROOR R2BH (R cyclohexyl)
1. HBr, ROOR 1. R2BH 2. H2O2, OH 2. RCHPCH2, Ni or Pd catalyst
Problems
Important Concepts 1. The rules for naming alkynes are essentially the same as those formulated for alkenes. Molecules with both double and triple bonds are called alkenynes, the double bond receiving the lower number if both are at equivalent positions. Hydroxy groups are given precedence in numbering alkynyl alcohols (alkynols). 2. The electronic structure of the triple bond reveals two p bonds, perpendicular to each other, and a s bond, formed by two overlapping sp hybrid orbitals. The strength of the triple bond is about 229 kcal mol21; that of the alkynyl C – H bond is 131 kcal mol21. Triple bonds form linear structures with respect to other attached atoms, with short C–C (1.20 Å) and C–H (1.06 Å) bonds. 3. The high s character at C1 of a terminal alkyne makes the bound hydrogen relatively acidic (pKa < 25). 4. The chemical shift of the alkynyl hydrogen is low (d 5 1.7 – 3.1 ppm) compared with that of alkenyl hydrogens because of the shielding effect of an induced electron current around the molecular axis caused by the external magnetic field. The triple bond allows for long-range coupling. IR spectroscopy indicates the presence of the CqC and qC – H bonds in terminal alkynes through bands at 2100 – 2260 cm21 and 3260 – 3330 cm21, respectively. 5. The elimination reaction with vicinal dihaloalkanes proceeds regioselectively and stereospecifically to give alkenyl halides. 6. Selective syn dihydrogenation of alkynes is possible with Lindlar catalyst, the surface of which is less active than palladium on carbon and therefore not capable of hydrogenating alkenes. Selective anti hydrogenation is possible with sodium metal dissolved in liquid ammonia because simple alkenes cannot be reduced by one-electron transfer. The stereochemistry is set by the greater stability of a trans disubstituted alkenyl radical intermediate. 7. Alkynes generally undergo the same addition reactions as alkenes; these reactions may take place twice in succession. Hydration of alkynes is unusual. It requires an Hg(II) catalyst, and the initial product, an enol, rearranges to a ketone by tautomerism. 8. To stop the hydroboration of terminal alkynes at the alkenylboron intermediate stage, modified dialkylboranes — particularly dicyclohexylborane — are used. Oxidation of the resulting alkenylboranes produces enols that tautomerize to aldehydes. 9. The Heck reaction links alkenes to alkenyl halides in a metal-catalyzed process.
Problems 27. Draw the structures of the molecules with the following names. (a) 1-Chloro-1-butyne
(b) (Z )-4-Bromo-3-methyl-3-penten-1-yne
(c) 4-Hexyn-1-ol
28. Name each of the compounds below, using the IUPAC system of nomenclature. Cl
OH
OH (b)
(d)
(e)
(c)
(f)
% %
(a)
29. Compare C – H bond strengths in ethane, ethene, and ethyne. Reconcile these data with hybridization, bond polarity, and acidity of the hydrogen. 30. Compare the C2 – C3 bonds in propane, propene, and propyne. Should they be any different with respect to either bond length or bond strength? If so, how should they vary? 31. Predict the order of acid strengths in the following series of cationic species: CH3CH2NH31, CH3CH“NH21, CH3CqNH1. [Hint: Look for an analogy among hydrocarbons (Section 13-2).]
Chapter 13
599
600
Chapter 13
Alkynes
32. The heats of combustion for three compounds with the molecular formula C5H8 are as follows: cyclopentene, DHcomb 5 21027 kcal mol21; 1,4-pentadiene, DHcomb 5 21042 kcal mol21; and 1-pentyne, DHcomb 5 21052 kcal mol21. Explain in terms of relative stability and bond strengths. 33. Rank in order of decreasing stability. (a) 1-Heptyne and 3-heptyne C q CCH3,
(b)
CH2Cq CH, and
(Hint: Make a model of the third structure. Is there anything unusual about its triple bond?) 34. Deduce structures for each of the following. (a) Molecular formula C6H10; NMR spectrum A; no strong IR bands between 2100 and 2300 or 3250 and 3350 cm21. (b) Molecular formula C7H12; NMR spectrum B; IR bands at about 2120 and 3330 cm21. (c) The percentage composition is 71.41% carbon and 9.59% hydrogen (the remainder is O), and the exact molecular mass is 84.0584; NMR and IR spectra C (next page). The inset in NMR spectrum C provides better resolution of the signals between 1.6 and 2.4 ppm. 1
6H
H NMR
4H
(CH3)4Si
2.0
1.5
1.0
0.5
0.0
300-MHz 1H NMR spectrum ppm (δ) δ
A 1
3H
H NMR
2H
1H
4H 2H (CH3)4Si
2.0
1.5
1.0
300-MHz H NMR spectrum ppm (δ) δ 1
B
0.5
0.0
Problems
Chapter 13
1
H NMR 2H
1H 2H
2.3
2H
2.0
1.7
1H
(CH3)4Si
4.0
3.5
3.0
2.0
2.5
1.5
1.0
0.5
0.0
δ 300-MHz 1H NMR spectrum ppm (δ)
Transmittance (%)
100
IR
0 4000 3500
3000
2500
2000
1500
1000
600 cm−1
Wavenumber
C 35. The IR spectrum of 1,8-nonadiyne displays a strong, sharp band at 3300 cm21. What is the origin of this absorption? Treatment of 1,8-nonadiyne with NaNH2, then with D2O, leads to the incorporation of two deuterium atoms, leaving the molecule unchanged otherwise. The IR spectrum reveals that the peak at 3300 cm21 has disappeared, but a new one is present at 2580 cm21. (a) What is the product of this reaction? (b) What new bond is responsible for the IR absorption at 2580 cm21? (c) Using Hooke’s law, calculate the approximate expected position of this new band from the structure of the original molecule and its IR spectrum. Assume that k and f have not changed. 36. Write the expected product(s) of each of the following reactions. CH3 A (a) CH3CH2CHCHCH2Cl A Cl
3 NaNH2, liquid NH3
CH3 Cl Cl CH3 A A A A (c) meso-CH3CHCH2CHCHCH2CHCH3
(b) CH3OCH2CH2CH2CHCHCH3 A A Br Br NaOCH3 (1 equivalent), CH3OH
2 NaNH2, liquid NH3
CH3 CH3 Cl Cl A A A A (d) (4R,5R)-CH3CHCH2CHCHCH2CHCH3
NaOCH3 (1 equivalent), CH3OH
601
602
Chapter 13
Alkynes
37. (a) Write the expected product of the reaction of 3-octyne with Na in liquid NH3. (b) When the same reaction is carried out with cyclooctyne (Problem 33b), the product is cis-cyclooctene, not trans-cyclooctene. Explain, mechanistically. 38. Write the expected major product of reaction of 1-propynyllithium, CH3C q C2Li1, with each of the following molecules in THF. Cl (b)
(a) CH3CH2Br
O B CH
(d)
(c) Cyclohexanone
CH3 %
O f i (e) CH3CH CH2
(f)
K K O
≥ H
39. Write the mechanism and final product for the reaction of 1-propynyllithium with trans-2,3dimethyloxacyclopropane. 40. Which of the following methods is best suited as a high-yield synthesis of 2-methyl3-hexyne,
?
Cl
H2, Lindlar's catalyst
Cl
NaNH2, liquid NH3
(b)
(a) 1. Cl2, CCl4 2. NaNH2, liquid NH3
(c)
Br
Br
Li
(e)
Li
(d)
41. Propose reasonable syntheses of each of the following alkynes, using the principles of retrosynthetic analysis. Each alkyne functional group in your synthetic target should come from a separate molecule, which may be any two-carbon compound (e.g., ethyne, ethene, ethanal). OH (b)
(a)
(c) OH
(d) (CH3)3CC‚CH [Be careful! What is wrong with (CH3)3CCl 1 2: C‚CH?]
42. Draw the structure of (R)-4-deuterio-2-hexyne. Propose a suitable retro-SN2 precursor of this compound. 43. Reaction review. Without consulting the Reaction Road Map on p. 598, suggest reagents to convert a general alkyne RCqCH into each of the following types of compounds.
R
(a) Br
G G CP C D D
Br
H
Br Br A A (b) R O COC O H A A Br Br
O H B A (c) R O COC O H A H
(Markovnikov product)
Problems
I H A A (d) R O COC O H A A I H
(g) R O CqCO R
R
(j) H
G G CP C D D
H
H
OH A (f) R O CqC O C OR A R
(e) R O CqCð M
(h) R O CqCO CH2 O CH2OH
H O A B (k) R O COC O H A H
H H A A (i) R O COC O H A A H H
(anti-Markovnikov product)
44. Give the expected major product of the reaction of propyne with each of the following reagents. (a) (b) (f) (i)
D2, Pd – CaCO3, Pb(O2CCH3)2, quinoline; Na, ND3; (c) 1 equivalent HI; (d) 2 equivalents HI; (e) 1 equivalent Br2; 1 equivalent ICl; (g) 2 equivalents ICl; (h) H2O, HgSO4, H2SO4; dicyclohexylborane, then NaOH, H2O2.
45. What are the products of the reactions of dicyclohexylethyne with the reagents in Problem 44? 46. Write the structures of the initially formed enol tautomers in the reactions of propyne and dicyclohexylethyne with dicyclohexylborane followed by NaOH and H2O2 (Problems 44, part i, and 45, part i). 47. Give the products of the reactions of your first two answers to Problem 45 with each of the following reagents. (a) H2, Pd – C, CH3CH2OH; (b) Br2, CCl4; (c) BH3, THF, then NaOH, H2O2; (d) MCPBA, CH2Cl2; (e) OsO4, then H2S. 48. Propose several syntheses of cis-3-heptene, beginning with each of the following molecules. Note in each case whether your proposed route gives the desired compound as a major or minor final product. (a) 3-Chloroheptane; (b) 4-chloroheptane; (c) 3,4-dichloroheptane; (d) 3-heptanol; (e) 4-heptanol; (f) trans-3-heptene; (g) 3-heptyne. 49. Propose reasonable syntheses of each of the following molecules, using an alkyne at least once in each synthesis. Br
I
Cl
(a)
I
(b)
(c) meso-2,3-Dibromobutane
CH3 (d) Racemic mixture of (2R,3R)and (2S,3S)-2,3-dibromobutane
(e)
O
Br
Cl
H
Cl
(f)
CH3
OH (g)
HO
H (h)
(i) O
Chapter 13
603
604
Chapter 13
Alkynes
50. Show how the Heck reaction might be employed to synthesize each of the following molecules.
O B COCH3
(a)
(b)
51. Propose a reasonable structure for calcium carbide, CaC2, on the basis of its chemical reactivity (Section 13-10). What might be a more systematic name for it? 52. Propose two different syntheses of linalool, a terpene found in cinnamon, sassafras, and orange flower oils. Start with the eight-carbon ketone shown here and use ethyne as your source of the necessary additional two carbons in both syntheses.
O
OH ?
Linalool
53. The synthesis of chamaecynone, the essential oil of the Benihi tree, requires the conversion of a chloroalcohol into an alkynyl ketone. Propose a synthetic strategy to accomplish this task.
?
Eventually
K K O
HO
Cq
Cl
K K O CH
Cq
CH
Chamaecynone
54.
Synthesis of the sesquiterpene bergamotene proceeds from the alcohol shown here. Suggest a sequence to complete the synthesis.
?
HOH2C ≈ 0 CH3
0 CH3
Bergamotene
55.
An unknown molecule displays 1H NMR and IR spectra D (next page). Reaction with H2 in the presence of the Lindlar catalyst gives a compound that, after ozonolysis and OO O BB B treatment with Zn in aqueous acid, gives rise to one equivalent of CH3CCH and two of HCH. What was the structure of the original molecule?
56.
Formulate a plausible mechanism for the hydration of ethyne in the presence of mercuric chloride. (Hint: Review the hydration of alkenes catalyzed by mercuric ion, Section 12-7.)
Problems
Chapter 13
1
H NMR 3H 1H
1H 1H
(CH3)4Si
6.0
5.5
5.0
4.5
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.0
1
300-MHz H NMR spectrum ppm (δ )
Transmittance (%)
100
IR
0 4000
3500
3000
2500
2000
1500
1000
600 cm−1
Wavenumber
D
57. A synthesis of the sesquiterpene farnesol requires the conversion of a dichloro compound into an alkynol, as shown below. Suggest a way of achieving this transformation. (Hint: Devise a conversion of the starting compound into a terminal alkyne.)
Cl
Cl ?
C q C O CH2OH
OH
Eventually
Farnesol
605
Alkynes
Chapter 13
Team Problem 58. Your team is studying the problem of an intramolecular ring closure of enediyne systems important in the total synthesis of dynemicin A, which exhibits potent antitumor activity.
%
H ≈ OH AA
≈ O≈
HN AA
O BB
COOH
%
OCH3
~
CH3 %
H BB O
AA OH
AA OH Dynemicin A
One research group tried the following approaches to effect this process. Unfortunately, all were unsuccessful. Divide the schemes among yourselves and assign structures to compounds A through D. (Note: R9 and R0 are protecting groups.)
H
%
LiNR2
A
CH3O
OCH3 OCH3
? ~
OH AA OCH3
š
Mild base
OCH3
Br
%
~
≥ OH
D
CH3 H ≈ % ≈ RN AA O ≈ H AA OCH3
´
%
3. RN O} A ≈
LiNR2
OH
OCH3 OCH3
?
AA OCH3
%
CH3 %
%
H
AA OCH3
H
%
C
CH3 H ≈ % ≈ RN AA O ≈ ~
~
~
OH
AA OCH3
OCH3 OCH3
LiNR2
%
RN A A
?
?
OCH3 OCH3
≈ O≈
1. (CH3CH2CH2CH2)4NF, THF (removes R) 2. CH3SO2Cl, (CH3CH2)3N
CH3 %
š
% B
OH
OCH3 OCH3
R
CH3 %
RN 2. AA
N AA
AA OCH3
R
%
CH3 %
O B
?
N A A
CH3SO2Cl, (CH3CH2)3N
š
1. CH3O
CH3 %
š
O B
H
š
606
Problems
A successful model study (shown here) provided an alternative strategy toward the completion of the total synthesis.
O
O
O
O
LDA
OH CHO
Discuss the advantages of this approach and apply it to the appropriate compound in approaches 1 through 3.
Preprofessional Problems 59. The compound whose structure is H – C q C(CH2)3Cl is best named (IUPAC) (a) 4-chloro-1-pentyne; (b) 5-chloropent-1-yne; (c) 4-pentyne-1-chloroyne; (d) 1-chloropent-4-yne. 60. A nucleophile made by deprotonation of propyne is (a) 2:CH2CH3; (b) 2:HC “ CH2; (c) 2:C q CH; (d) 2:C q CCH3; (e) 2:HC “ CHCH3. 61. When cyclooctyne is treated with dilute, aqueous sulfuric acid and HgSO4, a new compound results. It is best represented as
(a)
(b)
(c) OH
O
(d)
HO
OH
(e)
62. From the choices shown below, pick the one that best describes the structure of compound A. CH3 A
H3PO4, pH 2, 270°C, 100 atm
O (a) HOCH2CH(CH2)2OH A CH3
(b) HOCH2CHCH2OH A CH3
(c) HC q CCHCH2OH A CH3
(d) HCq CCH2CHCH2OH A CH3
Chapter 13
607
608
Chapter 13
Alkynes
63. From the choices shown below, pick the one that best describes the structure of compound A. Br Br A A CH3C O CCH2OH A A Br Br
Br2 (2 equivalents)
A (A primary alcohol; %C 68.6, %H 8.6, and %O 22.9)
(a) CH2 P CHCH2CH2OH
(b)
(d) CH3CH P CH O CH P CHOH
(e)
H2 (2 equivalents), Raney Ni
CH2OH
OH
1-butanol
(c) CH3Cq CCH2OH
C H A P T E R
FOURTEEN
Delocalized Pi Systems Investigation by Ultraviolet and Visible Spectroscopy
W
e live in a universe of color. Our ability to perceive and distinguish thousands of hues and shades of color is tied to the ability of molecules to absorb different frequencies of visible light. In turn, this molecular property is frequently a consequence of the presence of multiple p bonds. In the preceding three chapters we introduced the topic of compounds containing carbon–carbon p bonds, the products of overlap between two adjacent parallel p orbitals. We found that addition reactions to these chemically versatile systems provided entries both to relatively simple products, of use in synthesis, and to more complex products, including polymers—substances that have affected modern society enormously. In this chapter we expand further on all these themes by studying compounds in which three or more parallel p orbitals participate in p-type overlap. The electrons in such orbitals are therefore shared by three or more atomic centers and are said to be delocalized. A H C K CH C ÝE A A
A A H C K CH C K CH A A
Allylic radical
Conjugated diene
Our discussion begins with the 2-propenyl system — also called allyl — containing three interacting p orbitals. We then proceed to compounds that contain several double bonds: dienes and higher analogs. These compounds give rise to some of the most widely used polymers in the modern world, found in everything from automobile tires to the plastic cases around the computers used to prepare the manuscript for this textbook. The special situation of alternating double and single bonds gives rise to conjugated dienes, trienes, and so forth, possessing more extended delocalization of their p electrons. These substances illustrate new modes of reactivity, including thermal and photochemical cycloadditions and ring closures, which are among the most powerful methods for the synthesis of cyclic compounds such as steroidal pharmaceuticals. These processes exemplify a fundamentally new class of mechanisms: pericyclic reactions, which is the last type of mechanism we shall consider
b-Carotene is a pigment that is important for photosynthesis. It is part of the general family of carotenoids, produced in nature to the tune of 100 million tons per year and responsible for the orange color of carrots and many other fruits and vegetables. Its color is due to the presence of 11 contiguous double bonds, which cause the absorption of visible light.
610
Delocalized Pi Systems
Chapter 14
in this book. Following this chapter, we take the opportunity in an Interlude to review all the major types of organic processes covered in this course. We conclude the chapter itself with a discussion of light absorption by molecules with delocalized p systems. This process forms the basis of ultraviolet and visible spectroscopy.
14-1 Overlap of Three Adjacent p Orbitals: Electron Delocalization in the 2-Propenyl (Allyl) System What is the effect of a neighboring double bond on the reactivity of a carbon center? Three key observations answer this question. Observation 1. The primary carbon – hydrogen bond in propene is relatively weak, only 87 kcal mol21.
(CH3)2CHO H DH 98.5 kcal mol1 (412 kJ mol1) CH3CH2 O H DH 101 kcal mol1 (423 kJ mol1)
H G H2 C P C D CH2j
Propene
Hj
DH° 87 kcal mol1
2-Propenyl radical
A comparison with the values found for other hydrocarbons (see margin) shows that it is even weaker than a tertiary C – H bond. Evidently, the 2-propenyl radical enjoys some type of special stability. Observation 2. In contrast with saturated primary haloalkanes, 3-chloropropene dissociates relatively fast under SN1 (solvolysis) conditions and undergoes rapid unimolecular substitution through a carbocation intermediate. H G H 2C P C Q D CH2 O Cl ð QS
CH3OH, Δ Q Cl ð QS
H G H2C P C D CH2
Q CH3Q OH H
2-Propenyl cation
3-Chloropropene
H G H2C P C Q D OCH3 CH2Q ð ð
(CH3)3C O H DH 96.5 kcal mol1 (404 kJ mol1)
H G H2C P C D CH2 O H
ð ð
1
ð ð
DH° 87 kcal mol (364 kJ mol ) 1
ð ð
CH2 P CHCH2 O H
Decreasing bond strength
Dissociation Energies of Various C–H Bonds
3-Methoxypropene (SN1 product)
This finding clearly contradicts our expectations (recall Section 7-5). It appears that the cation derived from 3-chloropropene is somehow more stable than other primary carbocations. By how much? The ease of formation of the 2-propenyl cation in solvolysis reactions has been found to be roughly equal to that of a secondary carbocation. Observation 3. The pKa of propene is about 40. H2C P C
H G D CH2 O H
K ≈ 1040
H2C P C
H G H D CH2ð
2-Propenyl anion
Thus, propene is considerably more acidic than propane (pKa < 50), and the formation of the propenyl anion by deprotonation appears to be unusually favored. How can we explain these three observations?
Delocalization stabilizes 2-propenyl (allyl) intermediates Each of the preceding three processes generates a reactive carbon center — a radical, a carbocation, or a carbanion, respectively — that is adjacent to the p framework of a double bond. This arrangement seems to impart special stability. Why? The reason is electron delocalization: Each species may be described by a pair of equivalent contributing resonance forms (Section 1-5). These three-carbon intermediates have been given the name allyl (followed by the appropriate term: radical, cation, or anion). The activated carbon is called allylic.
14-1 Overlap of Three Adjacent p Orbitals
611
Chapter 14
Resonance Representation of Delocalization in the 2-Propenyl (Allyl) System
j CH2 O CH P CH2]
j [CH2 P CH O CH2
or
H A C E H b e H2C j CH2
or
H A C E H b e H2C CH2
or
H A C E H b e H2C CH2
Radical
[CH2 P CH O CH2
CH2 O CH P CH2] Cation
ð
Q CH2 O CH P CH2]
ð
Q [CH2 P CH O CH2
Anion
The 2-propenyl (allyl) pi system is represented by three molecular orbitals
Remember that resonance forms are not isomers but partial molecular representations. The true structure (the resonance hybrid) is derived by their superposition, better represented by the dotted-line drawings at the right of the classical picture.
MODEL BUILDING
The stabilization of the 2-propenyl (allyl) system by resonance can also be described in terms of molecular orbitals. Each of the three carbons is sp2 hybridized and bears a p orbital perpendicular to the molecular plane (Figure 14-1). Make a model: The structure is symmetric, with equal C – C bond lengths. π bond
Two nodes
π bond p orbital
C C
C
H σ bond σ bond
Figure 14-1 The three p orbitals in the 2-propenyl (allyl) group overlap, giving a symmetric structure with delocalized electrons. The s framework is shown as black lines.
Ignoring the s framework, we can combine the three p orbitals mathematically to give three p molecular orbitals. This process is analogous to mixing two atomic orbitals to give two molecular orbitals describing a p bond (Figures 11-2 and 11-4), except that there is now a third atomic orbital. Of the three resulting molecular orbitals, shown in Figure 14-2, one (p1) is bonding and has no nodes, one (p2) is nonbonding (in other words, it has the same energy as a noninteracting p orbital) and has one node, and one (p3) is antibonding, with two nodes. We can use the Aufbau principle to fill up the p molecular orbitals with the appropriate number of electrons for the 2-propenyl cation, radical, and anion (Figure 14-3).
π2 E
π1
+
•
−
+
−
+
−
Node
H H
π3
−
π3, antibonding
H H
+
Figure 14-3 The Aufbau principle is used to fill up the p molecular orbitals of 2-propenyl (allyl) cation, radical, and anion. In each case, the total energy of the p electrons is lower than that of three noninteracting p orbitals. Partial cation, radical, or anion character is present at the end carbons in these systems, a result of the location of the lobes in the p2 molecular orbital.
+
−
−
+
E
π2, nonbonding No node
+
+
+
−
−
−
π1, bonding Figure 14-2 The three p molecular orbitals of 2-propenyl (allyl), obtained by combining the three adjacent atomic p orbitals.
612
Chapter 14
Delocalized Pi Systems
The cation, with a total of two electrons, contains only one filled orbital, p1. For the radical and the anion, we place one or two additional electrons, respectively, into the nonbonding orbital, p2. The total p-electron energy of each system is lower (more favorable) than that expected from three noninteracting p orbitals — because p1 is greatly stabilized and filled with two electrons in all cases, whereas the antibonding level, p3, stays empty throughout. The resonance formulations for the three 2-propenyl species indicate that it is mainly the two terminal carbons that accommodate the charges in the ions or the odd electron in the radical. The molecular-orbital picture is consistent with this view: The three structures differ only in the number of electrons present in molecular orbital p2, which possesses a node passing through the central carbon; therefore, very little of the electron excess or deficiency will show up at this position. The electrostatic potential maps of the three 2-propenyl systems show their delocalized nature. (The cation and anion have been rendered at an attenuated scale to tone down the otherwise overwhelming intensity of color). To some extent, especially in the cation and anion, you can also discern the relatively greater charge density at the termini. Remember that these maps take into account all electrons in all orbitals, s and p. Partial Electron Density Distribution in the 2-Propenyl (Allyl) System H A C ee H2C e e CH2 −3
−3
Cation
H A C ee H2C e e CH2 −3j
−3j
Radical
H A C ee H2C e e CH2 −3
−3
Anion
In Summary Allylic radicals, cations, and anions are unusually stable. In Lewis terms, this stabilization is readily explained by electron delocalization. In a molecular-orbital description, the three interacting p orbitals form three new molecular orbitals: One is considerably lower in energy than the p level, another one stays the same, and a third is higher in energy. Because only the first two are populated with electrons, the total p energy of the system is lowered.
14-2 Radical Allylic Halogenation A consequence of delocalization is that resonance-stabilized allylic intermediates can readily participate in reactions of unsaturated molecules. For example, although halogens can add to alkenes to give the corresponding vicinal dihalides (Section 12-5) by an ionic mechanism, the course of this reaction is changed with added radical initiators (or on irradiation) and with the halogen present only in low concentrations. These conditions slow the ionic addition pathway sufficiently to allow a faster radical chain mechanism to take over, leading to radical allylic substitution.* *A complete explanation for this change requires a detailed kinetic analysis that is beyond the scope of this book. Suffice it to say that at low bromine concentrations the competing addition processes are reversible, and allylic substitution wins out.
14-2 Radical Allylic Halogenation
613
Chapter 14
Radical Allylic Halogenation X2 (low conc.), ROOR or hv
CH2 P CHCH3
CH2 P CHCH2X
REACTION HX
A reagent that is frequently used in allylic brominations in the laboratory is N-bromobutanimide (N-bromosuccinimide, NBS) suspended in tetrachloromethane. This species is nearly insoluble in CCl4 and is a steady source of very small amounts of bromine formed by reaction with trace impurities of HBr (see margin).
Mechanism of Br2 Generation from NBS @Ot ðN O š Brð Hðš Brð
NBS as a Source of Bromine O H2C H2C
{Ok
O
C N O Br
H2C
HBr
H2C
C
Protonation of oxygen (see also Exercise 2–11)
C NH
H ðO O
Br2
C
O
NBr
O
ðš Brð Br O š
{Ok Proton shift (tautomerism; Section 13-7)
NH
85%
O
{Ok
Nð
Br hv, CCl4
ðš Brð
H OO ðš
For example, NBS converts cyclohexene to 3-bromocyclohexene.
NOš Brð
{Ok
Butanimide
O
ðN O š Brð
O
N-Bromobutanimide (N-Bromosuccinimide, NBS)
š OH ðO
@Ot
O
ðN O H
3-Bromocyclohexene
{Ok
The bromine reacts with the alkene by a radical chain mechanism (Section 3-4). The process is initiated by light or by traces of radical initiators that cause dissociation of Br2 into bromine atoms. Propagation of the chain involves abstraction of a weakly bound allylic hydrogen by Br .. Mechanism of Allylic Bromination
MECHANISM
Initiation step ðš Br O š Brð š š
hv
H Abstraction
R
2 ðš Brj š
MATION AN I
Propagation steps R
R H
ðš Brj š
R
R
j
R j
HOš Brð š DH° 87 kcal mol1
DH° 87 kcal mol1
Allylic Radical R
R j
ANIMATED MECHANISM: Radical allylic halogenation
R
R j
Brð Br O š ðš š š
R
R Br
Brj ðš š
The resonance-stabilized radical may then react with Br2 at either end of the allylic system to furnish an allylic bromide and regenerate Br ., which continues the chain. Thus, alkenes
614
Chapter 14
Delocalized Pi Systems
that form unsymmetric allylic radicals can give mixtures of products upon treatment with NBS. For example, NBS, ROOR
CH2 P CH(CH2)5CH3
HBr
1-Octene
j CH2 P CHCH(CH2)4CH3
jCH2CHP CH(CH2)4CH3
Br2 Br . (from NBS)
Br2 Br . (from NBS)
Br A CH2 P CHCH(CH2)4CH3 28%
BrCH2CHP CH(CH2)4CH3 72%
3-Bromo-1-octene
1-Bromo-2-octene (Mixture of cis and trans)
Exercise 14-1 Ignoring stereochemistry, give all the isomeric monobromoheptenes that are possible from NBS treatment of trans-2-heptene.
Allylic chlorinations are important in industry because chlorine is relatively cheap. For example, 3-chloropropene (allyl chloride) is made commercially by the gas-phase chlorination of propene at 4008C. It is a building block for the synthesis of epoxy resin and many other useful substances. CH3CH P CH2
Cl2
400°C
ClCH2CH P CH2
HCl
3-Chloropropene (Allyl chloride)
Exercise 14-2 Predict the outcome of the allylic monochlorination of the following substrates. (a) Cyclohexene
(b)
(c) 1-Methylcyclohexene
The biochemical degradation of unsaturated molecules frequently involves radical abstraction of allylic hydrogens by oxygen-containing species. Such processes will be discussed in Section 22-9.
In Summary Under radical conditions, alkenes containing allylic hydrogens enter into allylic halogenation. A particularly good reagent for allylic bromination is N-bromobutanimide (N-bromosuccinimide, NBS).
14-3 Nucleophilic Substitution of Allylic Halides: SN1 and SN2 As our example of 3-chloropropene in Section 14-1 shows, allylic halides dissociate readily to produce allylic cations. These can be trapped at either end by nucleophiles in SN1 reactions. Allylic halides also readily undergo SN2 transformations.
Allylic halides undergo SN1 reactions The ready dissociation of allylic halides has important chemical consequences. Different allylic halides may give identical products upon solvolysis if they dissociate to the same
14-3 Nucleophilic Substitution of Allylic Halides
Chapter 14
allylic cation. For example, the hydrolysis of either 1-chloro-2-butene or 3-chloro-1butene results in the same alcohol mixture. The reason is that the same allylic cation is the intermediate. REACTION
Hydrolysis of Isomeric Allylic Chlorides 4
3
2
1
CH3CH P CHCH2 CH3CH P CHCH2Cl
Dissociation
Dissociation
Cl
Cl
Cl A CH3CHCH P CH2
MECHANISM
CH3CHCH P CH2 4
1-Chloro-2-butene
3
2
1
Allylic cation
3-Chloro-1-butene
Nucleophilic HOH trapping
CH3CH P CHCH2OH 2-Buten-1-ol
OH A CH3CHCH P CH2
H
3-Buten-2-ol
Exercise 14-3 Hydrolysis of (R)-3-chloro-1-butene gives racemic 3-buten-2-ol in addition to 2-buten-1-ol. Explain. (Hint: Review Section 7-3.)
Exercise 14-4
Working with the Concepts: Allylic Alcohols and Acid Treatment of 3-buten-2-ol with cold hydrogen bromide gives 1-bromo-2-butene and 3-bromo-1-butene. Explain by a mechanism. Strategy First we convert the text problem into a balanced equation. Then we examine the structure of 3-buten-2-ol, identifying its functional groups. Subsequently we consider the reaction conditions and decide how they will affect these functional groups, either individually or together. Solution Br OH A A HBr CH3CHCH P CH2 CH3CH P CHCH2Br H2O • CH3CHCH P CH2 • 3-Buten-2-ol is a secondary and allylic alcohol. • Recall (Section 9-2) that alcohols are protonated in the presence of strong acids and that, depending on structure, the resulting oxonium ions can react by SN2 or SN1 pathways. In the present case, clearly an SN1 reaction is indicated. • The resonance-stabilized allylic cation is trapped by bromide at either of the two allylic termini, producing the observed products.
OH A CH3CHCH P CH2
CH3CHCH P CH2 HBr
H2O Br CH3CH P CHCH2 Br A CH3CHCH P CH2 CH3CH P CHCH2Br H2O
615
616
Chapter 14
Delocalized Pi Systems
Exercise 14-5
Try It Yourself Write a mechanism for the following transformation: O
H, CH3OH
OCH3 HO
HO
OCH3
Allylic halides can also undergo SN2 reactions SN2 reactions of allylic halides with good nucleophiles (Section 6-8) are faster than those of the corresponding saturated haloalkanes. Two factors contribute to this acceleration. One is that the allylic carbon is attached to a relatively electron-withdrawing sp2 hybridized carbon (as opposed to sp3; Section 13-2), making it more electrophilic. The second is that overlap between the double bond and the p orbital in the transition state of the SN2 displacement (see Figure 6-4) is stabilizing, resulting in a relatively low activation barrier. The SN2 Reactions of 3-Chloro-1-propene and 1-Chloropropane Relative rate Acetone, 50°C
Acetone, 50°C
CH2 “CHCH2Cl 1 I2 -------------¡ CH2 “CHCH2I 1 Cl2 73 CH3CH2CH2Cl 1 I2 -------------¡
CH3CH2CH2I 1 Cl2 1
Exercise 14-6 The solvolysis of 3-chloro-3-methyl-1-butene in acetic acid at 258C gives initially a mixture containing mostly the structurally isomeric chloride with some of the acetate. After a longer period of time, no allylic chloride remains and the acetate is the only product present. Explain this result.
Cl
O O B B CH3COH, CH3COK
Cl
O O
3-Chloro-3-methyl1-butene
1-Chloro-3-methyl2-butene
3-Methyl-2-butenyl acetate
In Summary Allylic halides undergo both SN1 and SN2 reactions. The allylic cation intermediate of the SN1 reaction may be trapped by nucleophiles at either terminus, leading to product mixtures in the case of unsymmetrical systems. With good nucleophiles, allylic halides undergo SN2 reaction more rapidly than the corresponding saturated substrates.
14-4 Allylic Organometallic Reagents: Useful Three-Carbon Nucleophiles Propene is appreciably more acidic than propane because of the relative stability of the conjugated carbanion that results from deprotonation (Section 14-1). Therefore, allylic lithium reagents can be made from propene derivatives by proton abstraction by an alkyllithium. The process is facilitated by N,N,N9,N9-tetramethylethane-1,2-diamine (tetramethylethylenediamine, TMEDA), a good solvating agent.
14-5 Two Neighboring Double Bonds: Conjugated Dienes
617
Chapter 14
Allylic Deprotonation CH D 3 CH3CH2CH2CH2Li H2C P C G CH3
(CH3)2NCH2CH2N(CH3)2 (TMEDA)
CH D 3 H 2C P C CH3CH2CH2CH2 O H G CH2Li
Another way of producing an allylic organometallic is Grignard formation. For example, CH2 P CHCH2Br
Mg, THF, 0°C
3-Bromo-1-propene
CH2 P CHCH2MgBr 2-Propenylmagnesium bromide
The Simplest Conjugated and Nonconjugated Dienes CH2
CH
CH
CH2
Like their alkyl counterparts (Section 8-8), allylic lithium and Grignard reagents can react as nucleophiles. This is useful, because the embedded double bond allows for further functional-group transformations (Chapter 12). Exercise 14-7 Show how to accomplish the following conversion in as few steps as possible. (Hint: Allylic organometallic compounds react with ketones in the same manner as ordinary organometallic reagents.) O
HO
CH2CHO
1,3-Butadiene (Conjugated)
CH2
CHCH2
CH
In Summary Alkenes tend to be deprotonated at the allylic position, a reaction resulting in the corresponding delocalized anions. Allylic lithium or Grignard reagents can be made from the corresponding halides. Like their alkyl analogs, allyl organometallics function as nucleophiles.
14-5 Two Neighboring Double Bonds: Conjugated Dienes Now that we have caught a glimpse of the consequences of delocalization over three atoms, it should be interesting to learn what happens if we go one step further. Let us consider the addition of a fourth p orbital, which results in two double bonds separated by one single bond: a conjugated diene (conjugatio, Latin, union). In these compounds, delocalization again results in stabilization, as measured by heats of hydrogenation. Extended p overlap is also revealed in their molecular and electronic structures and in their chemistry.
Hydrocarbons with two double bonds are named dienes Conjugated dienes have to be contrasted with their nonconjugated isomers, in which the two double bonds are separated by saturated carbons, and the allenes (or cumulated dienes), in which the p bonds share a single sp hybridized carbon and are perpendicular to each other (Figure 14-4). You can see the contrasting p-electron distribution in conjugated and nonconjugated dienes in the electrostatic potential maps in the margin. In 1,3-butadiene, the regions of p-electron density (red) overlap, while in 1,4-pentadiene they are separated by a methylene group devoid of p electrons. In 1,2-propadiene (allene), these electrons are close, but are located in orthogonal (perpendicular) regions of space. The names of conjugated and nonconjugated dienes are derived from those of the alkenes in a straightforward manner. The longest chain incorporating both double bonds is
1,4-Pentadiene (Nonconjugated)
CH2
C
CH2
1,2-Propadiene (Allene, nonconjugated)
CH2
618
Chapter 14
Delocalized Pi Systems
π
Figure 14-4 The two p bonds of an allene share a single carbon and are perpendicular to each other.
π
H
H C H
C
σ
sp2
C
σ
H
sp2
π
π sp
CH2
C
CH2
found and then numbered to indicate the positions of the functional and substituent groups. If necessary, cis-trans or E,Z prefixes indicate the geometry around the double bonds. Cyclic dienes are named accordingly.
6
5
CH3
3G
G4
D H
7
5
CH2CH3 H 1 G4 5G P C C CH3 D G2 3G H CP C D D H H
H
CP C
D2 1 C P CH2 D H
trans-1,3-Pentadiene
H G4 3G CP C D D2 1 Br C P CH2 D H
CH3
cis-2-trans-4-Heptadiene
(Z)-4-Bromo-1,3-pentadiene
2
1 6 7
3 5
4
1
6
2
1
3
7
2 5 4
cis-1,4-Heptadiene
3 4 5 6
1,3-Cyclohexadiene
(A nonconjugated diene)
1,4-Cycloheptadiene (A nonconjugated cyclic diene)
Exercise 14-8 Suggest names or draw structures, as appropriate, for the following compounds. H3 C
(a)
CH3
(b) Br
(c) cis-3,6-Dimethyl-1,4-cyclohexadiene
CH3
(d) cis,cis-1,4-Dibromo-1,3-butadiene
Conjugated dienes are more stable than nonconjugated dienes The preceding sections noted that delocalization of electrons makes the allylic system especially stable. Does a conjugated diene have the same property? If so, that stability should be manifest in its heat of hydrogenation. We know that the heat of hydrogenation of a terminal alkene is about 230 kcal mol21 (see Section 11-5). A compound containing two noninteracting (i.e., separated by one or more saturated carbon atoms) terminal double bonds should exhibit a heat of hydrogenation roughly twice this value, about 260 kcal mol21. Indeed, catalytic hydrogenation of either 1,5-hexadiene or 1,4-pentadiene releases just about that amount of energy.
14-5 Two Neighboring Double Bonds: Conjugated Dienes
Chapter 14
619
Heat of Hydrogenation of Nonconjugated Alkenes Pt
CH3CH2CH “CH2 1 H2 ¡ CH3CH2CH2CH3 Pt
¢H° 5 230.3 kcal mol21 (2127 kJ mol21 )
CH2 “CHCH2CH2CH “CH2 1 2 H2 ¡ CH3 (CH2 )4CH3 ¢H° 5 260.5 kcal mol21 (2253 kJ mol21 )
Pt
CH2 “CHCH2CH “CH2 1 2 H2 ¡ CH3 (CH2 )3CH3
¢H° 5 260.8 kcal mol21 (2254 kJ mol21 )
When the same experiment is carried out with the conjugated diene, 1,3-butadiene, less energy is produced. Heat of Hydrogenation of 1,3-Butadiene Pt
CH2 “CHOCH “CH2 1 2 H2 ¡ CH3CH2CH2CH3 ¢H° 5 257.1 kcal mol21 (2239 kJ mol21 ) The difference, about 3.5 kcal mol21 (15 kJ mol21), is due to a stabilizing interaction between the two double bonds, as illustrated in Figure 14-5. Isolated double bonds 2 CH3CH2CH P CH2
2 H2
E
E 60.6 kcal mol1
Conjugated double bonds CH2P CHO CHP CH2
E 3.5 kcal mol1
Resonance stabilization
2 H2
Figure 14-5 The difference between the heats of hydrogenation of two molecules of 1-butene (a terminal monoalkene) and one molecule of 1,3-butadiene (a doubly terminal conjugated diene) reveals their relative stabilities. The value, about 3.5 kcal mol21, is a measure of the stabilization of 1,3-butadiene due to conjugation.
E 57.1 kcal mol1
2 CH3CH2CH2CH3
CH3CH2CH2CH3
Exercise 14-9 The heat of hydrogenation of trans-1,3-pentadiene is 254.2 kcal mol21, 6.6 kcal mol21 less than that of 1,4-pentadiene, even lower than that expected from the stabilizing energy of 1,3-butadiene. Explain. (Hint: See Section 11-5.)
Conjugation in 1,3-butadiene results from overlap of the pi bonds How do the two double bonds in 1,3-butadiene interact? The answer lies in the alignment of their p systems, an arrangement that permits the p orbitals on C2 and C3 to overlap (Figure 14-6A). The p interaction that results is weak but nevertheless amounts to a few kilocalories per mole because the p electrons are delocalized over the system of four p orbitals. Besides adding stability to the diene, this p interaction also raises the barrier to rotation about the single bond to more than 6 kcal mol21 (25 kJ mol21). An inspection of models shows that the molecule can adopt two possible extreme coplanar conformations. In one, designated s-cis, the two p bonds lie on the same side of the C2 – C3 axis; in the other, called s-trans, the p bonds are on opposite sides (Figure 14-6B). The prefix s refers to the fact that the bridge between C2 and C3 constitutes a single bond. The s-cis form is almost
MODEL BUILDING
620
Delocalized Pi Systems
Chapter 14
π bond
Weak π overlap
H
H
π bond
H
H H
1.34
1.47 Å 1.09 Å
122.9°
H
H H
H
Å
H s-cis Conformation: less stable
119.5°
H
H
H H
H H
H
ΔH° = −2.8 kcal mol−1 (−12 kJ mol−1)
H s-trans Conformation: more stable A Figure 14-6 (A) The structure of 1,3-butadiene. The central bond is shorter than that in an alkane (1.54 Å for the central C–C bond in butane). The p orbitals aligned perpendicularly to the molecular plane form a contiguous interacting array. (B) 1,3-Butadiene can exist in two planar conformations. The s-cis form shows steric hindrance because of the proximity of the two “inside” hydrogens highlighted in red.
B
3 kcal mol21 (12.5 kJ mol21) less stable than the s-trans conformation because of the steric interference between the two hydrogens on the inside of the diene unit.* Exercise 14-10 The dissociation energy of the central C – H bond in 1,4-pentadiene is only 77 kcal mol21. Explain. (Hint: See Sections 14-1 and 14-2 and draw the product of H atom abstraction.)
The p-electronic structure of 1,3-butadiene may be described by constructing four molecular orbitals out of the four p atomic orbitals (Figure 14-7). +
−
+
−
π4
Three nodes −
+
−
+
+
−
−
+
π3 −
+
+
−
+
+
−
−
π4 Antibonding molecular orbitals
Two nodes
π3
One node
π2
E
π2
Figure 14-7 A p-molecularorbital description of 1,3-butadiene. Its four electrons are placed into the two lowest p (bonding) orbitals, p1 and p2.
−
−
+
+
+
+
+
+
π1
Bonding molecular orbitals No nodes
−
−
−
π1
−
*The s-cis conformation is very close in energy to a nonplanar conformation in which the two double bonds are gauche (Section 2-8). Whether the s-cis or the gauche conformation is more stable remains a subject of controversy.
14-6 Electrophilic Attack on Conjugated Dienes
Chapter 14
621
In Summary Dienes are named according to the rules formulated for ordinary alkenes. Conjugated dienes are more stable than dienes containing two isolated double bonds, as measured by their heats of hydrogenation. Conjugation is manifested in the molecular structure of 1,3-butadiene, which has a relatively short central carbon – carbon bond with a small barrier to rotation of more than 6 kcal mol21 (25 kJ mol21). The two conformers, s-trans and s-cis, differ in energy by about 3 kcal mol21 (12.5 kJ mol21). The molecular-orbital picture of the p system in 1,3-butadiene shows two bonding and two antibonding orbitals. The four electrons are placed in the first two bonding levels.
14-6 Electrophilic Attack on Conjugated Dienes: Kinetic and Thermodynamic Control Does the structure of conjugated dienes affect their reactivity? Although more stable thermodynamically than dienes with isolated double bonds, conjugated dienes are actually more reactive kinetically in the presence of electrophiles and other reagents. 1,3-Butadiene, for example, readily adds 1 mol of cold hydrogen bromide. Two isomeric addition products are formed: 3-bromo-1-butene and 1-bromo-2-butene.
HBr
3-Bromo-1-butene
O
CH2 P CH O CH P CH2
Br
Br A HCH2 O CH O CH P CH2 70%
0°C
REACTION
HCH2 O CH P CH O CH2 30% 1-Bromo-2-butene
The generation of the first product is readily understood in terms of ordinary alkene chemistry. It is the result of a Markovnikov addition to one of the double bonds. But what about the second product? The presence of 1-bromo-2-butene is explained by the reaction mechanism. Initial protonation at C1 gives the thermodynamically most favored allylic cation.
MECHANISM
Protonation of 1,3-Butadiene H A CH2 O CH O CH P CH2
H
Primary nondelocalized cation not formed
Attack at C2
1
2
3
H A CH2 O CH O CHP CH2
H
4
CH2 P CH O CHP CH2
Attack at C1
H A CH2 O CHP CH O CH2
H3C is the same as
Delocalized allylic cation formed exclusively
This cation can be trapped by bromide in two possible ways to form the two observed products: At the terminal carbon, it yields 1-bromo-2-butene; and, at the internal carbon, it furnishes 3-bromo-1-butene. The 1-bromo-2-butene is said to result from 1,4-addition of hydrogen bromide to butadiene, because reaction has taken place at C1 and C4 of the original diene. The other product arises by the normal 1,2-addition. Nucleophilic Trapping of the Allylic Cation Formed on Protonation of 1,3-Butadiene CH3CHCH P CH2 A Br 1,2-Addition
Br Attack at internal carbon
H CH E be CH2 CH3CH
Br
CH3CHP CHCH2Br
Attack at terminal carbon
1,4-Addition
Many electrophilic additions to dienes give rise to product mixtures by both modes of addition, because of the intermediacy of allylic cations. For example, the bromination
H CH H E b e CH CH2
MATION AN I
ANIMATED MECHANISM: Addition of HBr to 1,3-butadiene
622
Delocalized Pi Systems
Chapter 14
CHE M I C AL H I G H L I G H T 1 4 - 1 “Face-to-Face” Interaction of Two Double Bonds connecting single bond. After this effect had been established, an interesting question was posed by organic chemists: What sort of interaction can we expect if we force two double bonds to come face to face to each other?
We have seen that the p interaction in conjugated dienes results in a small stabilization by delocalization of electrons (Section 14-5). In these systems, the two double bonds are neighbors, and orbital interaction occurs across the
H
/∑
E
Normal conjugated diene
H
H H
/∑
H
E
EE
H H H H
∑/
H
H H
HH
∑/
HH
Face-to-face “diene” Molecular model of D
The molecule designed to probe this question was D, synthesized from the highly strained A as shown. Thus, A is so reactive that it undergoes dimerization at 558C, first to the
diradical B, which then ring-closes to C. This species is not isolable, but spontaneously cleaves the newly formed fourmembered ring, now in the opposite direction, to give D.
55C
A
B
C
Diene D is reasonably stable at ambient temperature, but it decomposes eventually. Nevertheless, an X-ray structural analysis (Section 5-3) gave details of its carbon scaffold, which, as shown in the molecular model, is quite distorted. For starters, the substituent carbons to the double bonds are bent away from the p bond by 278. Moreover, the double bond is longer (1.354 Å) than normal (1.330 Å; Figure 11-1), and so are the bridging Csp3 – Csp3 linkages, 1.596 Å (normal value is 1.54 Å; Figure 2-4). The distance between the two double bonds is 2.395 Å. Two ethenes attracted by London forces (Figure 2-6C) would be expected to get as close as
D
3.4 Å (as in graphite, see Chemical Highlight 15-1), hence the carbon frame in D does enforce a face-to-face interaction, as designed. Is the p–p interaction in D stabilizing? The answer is complicated, because of the molecular distortions caused by the unusual carbon connectivity and because there is some (weak) hyperconjugation (Section 3-2) of the p orbitals with the adjacent s bonds. Nevertheless, calculations and spectroscopic measurements indicate that while there is some delocalization, electron repulsion seems to be dominant. Two double bonds appear to dislike staring into each others’ (p) faces.
of 1,3-butadiene proceeds through the (bromomethyl) allylic cation, instead of the cyclic bromonium ion encountered for normal alkenes (Section 12-5; see also Exercise 14-13). CCl4, 20C
H2 H EC H EC H C CH2 Br H
CH2 P CH O CH P CH2
Br OBr
CH2 O CH O CH P CH2 A A Br Br 54%
CH2 O CH P CHO CH2 A A Br Br 46%
3,4-Dibromo-1-butene
1,4-Dibromo-2-butene
Br
14-6 Electrophilic Attack on Conjugated Dienes
Chapter 14
Conjugated dienes also function as monomers in polymerizations induced by electrophiles, radicals, and other initiators (see Sections 12-14 and 12-15) and, as such, will be discussed in Section 14-10.
Exercise 14-11 Conjugated dienes can be made by the methods applied to the preparation of ordinary alkenes. Propose syntheses of (a) 2,3-dimethylbutadiene from 2,3-dimethyl-1,4-butanediol; (b) 1,3-cyclohexadiene from cyclohexane.
Exercise 14-12 Write the products of 1,2-addition and 1,4-addition of (a) HBr and (b) DBr to 1,3-cyclohexadiene. (Caution: What is unusual about the products of 1,2- and 1,4-addition of HX to unsubstituted cyclic 1,3-dienes?)
Changing product ratios: kinetic and thermodynamic control When the hydrobromination of 1,3-butadiene is carried out at 408C rather than 08C, a curious result is obtained: Instead of the original 70 : 30 mixture of 1,2- and 1,4-adducts, we now observe a 15 : 85 ratio of the corresponding bromobutenes: Hydrobromination of 1,3-Butadiene at 0C: Kinetic Control
HBr
Br A HCH2 O CH O CH P CH2 70%
Br
O
CH2 P CH O CH P CH2
0°C
HCH2 O CH P CH O CH2 30%
3-Bromo-1-butene
1-Bromo-2-butene
More
Less
Hydrobromination of 1,3-Butadiene at 40C: Thermodynamic Control
HBr
Br A HCH2 O CH O CH P CH2 15%
Br
O
CH2 P CH O CH P CH2
40°C
HCH2 O CH P CH O CH2 85%
Less
This ratio of bromides is obtained also by simply heating the initial 70 : 30 mixture or, more strikingly, either one of the pure isomers. How can this be explained? To understand these results, we have to go back to what we discussed in Section 2-1 regarding the kinetics and thermodynamics, or the rates and equilibria, that govern the outcome of reactions. From our observations, it is clear that at the higher temperature the two isomers are in equilibrium, their ratio reflecting their relative thermodynamic stability: 1-bromo-2-butene (with its internal double bond; Section 11-5) is a bit more stable than 3-bromo-1-butene. In general, a reaction whose ratio of products reflects their thermodynamic stability is said to be under thermodynamic control. Such is the case in the hydrobromination of 1,3-butadiene at 408C. What is happening at 08C? At that temperature, the two isomers are not interconverting and, hence, thermodynamic control is not attained. What, then, is the origin of the nonthermodynamic ratio? The answer lies in the relative rates at which the two products are generated from the intermediate allylic cation (drawn as the central starting point of the
More
623
624
Chapter 14
Delocalized Pi Systems
scheme below): 3-Bromo-1-butene, although thermodynamically less stable, is formed faster than 1-bromo-2-butene. In general, a reaction whose ratio of products reflects their relative rates of formation (hence, the relative heights of the respective activation barriers) is said to be under kinetic control. Such is the case in the hydrobromination of 1,3-butadiene at 08C. Kinetic Compared with Thermodynamic Control
Less stable product, predominates when reaction time is short or at low temperatures
H3C
H C ee CH e e CH2
1-Methyl-2-propenyl cation
CH3CH P CHCH2
Kinetic control (fast)
Br
Reverse reaction is slow at 0C, but fast at 40C
Thermodynamic control (slow)
Br A
Br A CH3CHCH P CH2
CH3CH P CHCH2 More stable product, predominates when reaction time is long or at higher temperatures
CH3CHCH P CH2
A potential energy diagram of our reaction (Figure 14-8) illustrates the lower activation barrier (and larger rate k1) associated with the formation of the less stable product and the higher barrier (and lower rate k2) leading to its more stable counterpart. The crucial feature is the reversibility (rate k21) of the kinetic step of trapping the intermediate cation. At 08C, the less stable, kinetic 3-bromo-1-butene product predominates, because reversal of its formation is relatively slow. At 408C, this product enters into rapid equilibrium with its cationic precursor and, eventually, with the thermodynamically more stable 1-bromo-2-butene. Why does the less stable product form with a lower activation barrier? Attack of the nucleophile (bromide, in this case) is faster at C3, the more substituted carbon, for a combination of reasons. When HBr protonates the terminal carbon atom of the diene, the freed bromide ion is initially relatively near to the adjacent carbon (C3) of the newly formed allylic cation. In addition, this cation is asymmetric; its positive charge is unequally distributed between carbons 1 and 3. More positive charge resides at the secondary carbon C3, as is reflected in the electrostatic-potential map shown in the margin. This position has less electron density (more blue) than the unsubstituted terminus. Thus, attack at C3 is favored kinetically by its greater partial positive charge and by its proximity to the newly released nucleophile. Kinetic control
Thermodynamic control Slow
Fast
k2 k1
k −2
k −1 Start here
H
H3C C
E
H
C +
CH2
+ Br−
CH3CHCH
CH2
ΔG°
Br CH3CH Figure 14-8 Kinetic control (to the left) as compared to thermodynamic control (to the right) in the reaction of 1-methyl-2-propenyl cation with bromide ion (middle).
(Less stable isomer)
CHCH2Br
(More stable isomer)
Reaction coordinate
Exercise 14-13
Working with the Concepts: Kinetic Versus Thermodynamic Control Bromination of 1,3-butadiene (see p. 622) at 608C gives 3,4-dibromo-1-butene and 1,4-dibromo-2-butene in a 10 : 90 ratio, whereas at 2158C this ratio is 60 : 40. Explain by a mechanism. Strategy To tackle problems of this type, you first need to formulate a mechanism that connects starting materials with products. Since we are dealing here with bromination of a double bond, refer back to Sections 12-5 and 12-6 for descriptions of the mechanism. Solution • Section 12-5 described how bromine attacks a double bond, namely, by forming a bromonium ion A initially, which is then subject to nucleophilic attack by bromide, leading to anti addition. • Section 12-6 revealed that when the double bond is substituted by alkyl groups, the bromonium ion is distorted, as in B, to reflect carbocation-like character on the more substituted carbon. • In the case of butadiene, the substituent is ethenyl, and the bromonium ion exists in its open form as a delocalized allylic cation C.
ðBrð ␦
R
šð ðBr
šð ðBr
ðBrð
šð ðBr
R
A, R H: symmetrical
B, R alkyl: unsymmetrical
C, R ethenyl: allylic cation
• Cation C can now be trapped by bromide ion at either its terminal or internal carbon. For reasons explained above, attack by bromide on the internal carbon appears slightly faster (kinetic control), giving the initial low temperature ratio of 60 : 40 of 3,4-dibromo-1-butene and trans-1,4-dibromo-2-butene. At 2158C, these two compounds are stable and do not undergo dissociation (the reverse of their formation). Kinetic control: šð ðBr
šð ðBr
š ð ðBr
ðBr ð
ðBr ð 60
šð ðBr
:
40
• Warming to 608C provides enough energy for the allylic bromide to function as a leaving group (by an SN1 reaction; Section 14-3) and thus to regenerate the allylic cation. At this temperature, the products and the cation are in rapid equilibrium, although the concentration of the higher-energy cation is very small. While the relative rates of bromide attack on the cation are still the same, this factor becomes irrelevant, because the product distribution is under thermodynamic control, reflecting relative stability: The 1,4-dibromo isomer is more stable than the 3,4-dibromo isomer, and the observed product ratio changes to 10 : 90. You can calculate by how much the two isomers differ in energy by applying the equation from Section 2-1. Thermodynamic control: šð ðBr
šð ðBr
šð ðBr
ðBr ð 60
ðBr ð
:
šð ðBr
šð ðBr
ðBr ð
ðBr ð 40
10
:
90
Exercise 14-14
Try It Yourself In the reaction shown below, which product is the result of 1,2- and 1,4-addition, respectively, and which is the kinetic and thermodynamic product, respectively? Cl
H3C HCl
H3C
H3C
Cl A
B
625
626
Chapter 14
Delocalized Pi Systems
In Summary Conjugated dienes are electron rich and are attacked by electrophiles to give intermediate allylic cations on the way to 1,2- and 1,4-addition products. These reactions may be subject to kinetic control at relatively low temperatures. At relatively higher temperatures, the kinetic product ratios may change to thermodynamic product ratios, when such product formation is reversible.
14-7 Delocalization Among More than Two Pi Bonds: Extended Conjugation and Benzene What happens when more than two conjugated double bonds are present in a molecule? Does its reactivity increase? And what happens if the molecule is cyclic? Does it react in the same way as its linear analog? In this section we begin to answer these questions.
Extended pi systems are thermodynamically stable but kinetically reactive When more than two double bonds are in conjugation, the molecule is called an extended P system. An example is 1,3,5-hexatriene, the next higher double-bond homolog of 1,3-butadiene. This compound is quite reactive and polymerizes readily, particularly in the presence of electrophiles. Despite its reactivity as a delocalized p system, it is also relatively stable thermodynamically. The increased reactivity of this extended p system is due to the low activation barriers for electrophilic additions, which proceed through highly delocalized carbocations. For example, the bromination of 1,3,5-hexatriene produces a substituted pentadienyl cation intermediate that can be described by three resonance structures. Bromination of 1,3,5-Hexatriene
BrCH2 O CH O CH
CH2 P CH O CH P CH O CHP CH2
CH O CH P CH2
Br2
BrCH2 O CH P CH O CH O CH
CH2
Br
1,3,5-Hexatriene
BrCH2 O CH P CH O CH P CHO CH2
Br A BrCH2CHCH P CHCH P CH2
Br A BrCH2CH P CHCHCHP CH2
BrCH2CH P CHCH P CHCH2Br
5,6-Dibromo-1,3-hexadiene
3,6-Dibromo-1,4-hexadiene
1,6-Dibromo-2,4-hexadiene
(A 1,2-addition product)
(A 1,4-addition product)
(A 1,6-addition product)
The final mixture is the result of 1,2-, 1,4-, and 1,6-additions, the last product being the most favored thermodynamically, because it retains an internal conjugated diene system. Exercise 14-15 Upon treatment with two equivalents of bromine, 1,3,5-hexatriene has been reported to give moderate amounts of 1,2,5,6-tetrabromo-3-hexene. Write a mechanism for the formation of this product.
14-7 Extended Conjugation and Benzene
627
Chapter 14
Some highly extended p systems are found in nature. An example is b-carotene, the orange coloring agent in carrots (see Chapter Opening), and its biological degradation product, vitamin A (retinol, below; see Chemical Highlight 18-3). Compounds of this type can be very reactive because there are many potential sites for attack by reagents that add to double bonds. In contrast, some cyclic conjugated systems may be considerably less reactive, depending on the number of p electrons (see Chapter 15). The most striking example of this effect is benzene, the cyclic analog of 1,3,5-hexatriene. H3 C CH3
CH3
CH3
H3C
CH3
CH3
CH3
CH3
CH3
-Carotene
CH3
CH3
CH3
CH2OH CH3 CH3 Vitamin A (Retinol)
Benzene, a conjugated cyclic triene, is unusually stable Cyclic conjugated systems are special cases. The most common examples are the cyclic triene C6H6, better known as benzene, and its derivatives (Chapters 15, 16, and 22). In contrast to hexatriene, benzene is unusually stable, both thermodynamically and kinetically, because of its special electronic makeup (see Chapter 15). That benzene is unusual can be seen by drawing its resonance forms: There are two equally contributing Lewis structures. Benzene does not readily undergo addition reactions typical of unsaturated systems, such as catalytic hydrogenation, hydration, halogenation, and oxidation. In fact, because of its low reactivity, benzene can be used as a solvent in organic reactions.
Benzene and Its Resonance Structures H C
H C C H
H C C
C H Benzene
Benzene Is Unusually Unreactive Exceedingly slow reaction
H2, Pd
OsO4, 25°C
No reaction
MODEL BUILDING H, H2O, 25°C
No reaction
Br2, 25°C
No reaction
In the chapters that follow, we shall see that the unusual lack of reactivity of benzene is related to the number of p electrons present in its cyclically conjugated array — namely, six. The next section introduces a reaction that is made possible only because its transition state benefits from six-electron cyclic overlap.
In Summary Acyclic extended conjugated systems show increasing thermodynamic stability but also kinetic reactivity because of the many sites open to attack by reagents and the ease of formation of delocalized intermediates. In contrast, the cyclohexatriene benzene is unusually stable and unreactive.
H
628
Chapter 14
Delocalized Pi Systems
14-8 A Special Transformation of Conjugated Dienes: Diels-Alder Cycloaddition Conjugated double bonds participate in more than just the reactions typical of the alkenes, such as electrophilic addition. This section describes a process in which conjugated dienes and alkenes combine to give substituted cyclohexenes. In this transformation, known as Diels-Alder cycloaddition, the atoms at the ends of the diene add to the alkene double bond, thereby closing a ring. The new bonds form simultaneously and stereospecifically.
The cycloaddition of dienes to alkenes gives cyclohexenes When a mixture of 1,3-butadiene and ethene is heated in the gas phase, a remarkable reaction takes place in which cyclohexene is formed by the simultaneous generation of two new carbon – carbon bonds. This is the simplest example of the Diels-Alder* reaction, in which a conjugated diene adds to an alkene to yield cyclohexene derivatives. The Diels-Alder reaction is in turn a special case of the more general class of cycloaddition reactions between p systems, the products of which are called cycloadducts. In the Diels-Alder reaction, an assembly of four conjugated atoms containing four p electrons reacts with a double bond containing two p electrons. For that reason, the reaction is also referred to as a [4 1 2]cycloaddition. The four-carbon component is simply called diene, the alkene is labeled dienophile “diene loving.” REACTION
Diels-Alder Cycloaddition of Ethene and 1,3-Butadiene
HC A HC
H2 C
CH J 2 M CH2
CH2 B CH2
200°C [4 2] Cycloaddition
CH2
HC HC C H2 20%
CH2
1,3-Butadiene
Ethene
Cyclohexene
(Four electrons)
(Two electrons)
(A cycloadduct)
What makes a good Diels-Alder reaction? Reactivity of the diene and dienophile
F3C
G C P CH2 D H
3,3,3-Trifluoro-1-propene (An electron-poor alkene)
The prototype reaction of butadiene and ethene actually does not work very well and gives only low yields of cyclohexene. It is much better to use an electron-poor alkene with an electron-rich diene. Substitution of the alkene with electron-attracting groups and of the diene with electron-donating groups therefore creates excellent reaction partners (see margin). The trifluoromethyl group, for example, is inductively (Section 8-3) electron attracting owing to its highly electronegative fluorine atoms. The presence of such a substituent enhances the Diels-Alder reactivity of an alkene. Conversely, alkyl groups are electron donating because of hyperconjugation (Sections 7-5, 11-3, and 11-5); their presence increases electron density and is beneficial to dienes in the Diels-Alder reaction. The electrostatic potential maps in the margin illustrate these effects. The double bond with the trifluoromethyl substituent has less electron density (yellow) than the double bonds bearing methyl groups (red). Other alkenes have substituents that interact with double bonds by resonance. For example, carbonyl-containing groups and nitriles are good electron acceptors by this effect. Double bonds bearing such substituents are electron poor because of the contribution of resonance forms that place a positive charge on an alkene carbon atom.
*Professor Otto P. H. Diels (1876–1954), University of Kiel, Germany, Nobel Prize 1950 (chemistry); Professor Kurt Alder (1902–1958), University of Köln, Germany, Nobel Prize 1950 (chemistry).
14-8 Diels-Alder Cycloaddition
Chapter 14
629
Groups That Are Electron Withdrawing by Resonance ðš Oð A C H H R C G H
ðOð C H H R H2 C P C G H
H2C
ðš Oð A CH N R H2C O C G H
š Nð
Nð
J
C H H2C P C G H
C H C G H
H2C
C N
š Nð J
H2C OC G H
Some examples of the trend in reactivity of dienophiles and dienes are H3 C
F 3C
NC
NC
NC Dienophile
H3C
H3C
H3CO
H3C Diene
Increasing reactivity
Exercise 14-16 Classify each of the following alkenes as electron poor or electron rich, relative to ethene. Explain your assignments. F O (a) H2C P CHCH2CH3
F
(b)
(c)
O
(d)
F F
O
Exercise 14-17 The double bond in nitroethene, H2C P CHNO2, is electron poor, and that in methoxyethene, H2C P CHOCH3, is electron rich. Explain, using resonance structures.
Examples of reaction partners that undergo efficient Diels-Alder cycloaddition are 2,3-dimethyl-1,3-butadiene and propenal (acrolein). O B B
H3C
H3C H
H
100°C, 3 h
H3C
H3C 2,3-Dimethyl1,3-butadiene
O B B
Propenal (Acrolein)
90% Diels-Alder cycloadduct
H3C CH3 G G C OC M J CH2 H2C 2,3-Dimethyl-1,3-butadiene (An electron-rich diene)
630
Delocalized Pi Systems
Chapter 14
CHE M I C AL H I G H L I G H T 1 4 - 2 Conducting Polyenes: Novel Functional Materials Can you imagine replacing all the copper wire in our electrical power lines and appliances with an organic polymer? A giant step toward achieving this goal was made in the late 1970s by Heeger, MacDiarmid, and Shirakawa,* for which they received the Nobel Prize in 2000. They synthesized a polymeric form of ethyne (acetylene) that conducts electricity as metals do. This discovery caused a fundamental change in how organic polymers (“plastics”) were viewed. Indeed, normal plastics are used to insulate and protect us from electrical currents. What is so special about polyethyne (polyacetylene)? For a material to be conductive, it has to have electrons that are free to move and sustain a current, instead of being localized, as in most organic compounds. In this chapter, we have seen how such delocalization is attained by linking sp2 hybridized carbon atoms in a growing chain: conjugated polyenes. We
have also learned how a positive charge, a single electron, or a negative charge can “spread out” along the p network, not unlike a molecular wire. Polyacetylene has such a polymeric structure, but the electrons are still too rigid to move with the facility required for conductivity. To achieve this goal, the electronic frame is “activated” by either removing electrons (oxidation) or adding them (reduction), a transformation called doping. The electron hole (positive charge) or electron pair (negative charge) delocalize over the polyenic structure in much the same way as that shown for extended allylic chains in Section 14-6. In the original breakthrough experiment, polyacetylene, made from acetylene by transition metal – catalyzed polymerization (see Section 12-15), was doped with iodine, resulting in a spectacular 10-million-fold increase in conductivity. Later refinement improved this figure to 1011, essentially organic copper!
H
H
H
H 1e
H
H
H
H
Oxidation (doping)
trans-Polyacetylene
Conducting polyacetylene
The black, shiny, flexible foil of polyacetylene (polyethyne) made by polymerization of gaseous ethyne on the walls of the reaction vessel shown.
*Professor Alan J. Heeger (b. 1936), University of California at Santa Barbara, California; Professor Alan G. MacDiarmid (1927–2007), University of Pennsylvania, Philadelphia, Pennsylvania; Professor Hideki Shirakawa (b. 1936), University of Tsukuba, Japan.
14-8 Diels-Alder Cycloaddition
Because of its sensitivity to air and moisture, polyacetylene is difficult to use in practical applications. However, the idea of using extended p systems to impart organic conductivity can be exploited with a range of
Chapter 14
631
materials, all of which have proven utility. Many of these contain especially stabilized cyclic 6-p-electron units, such as benzene (Section 15-2), pyrrole, and thiophene (Section 25-4).
Organic Conductors and Applications
H N
H N N H
Poly( p-phenylene vinylene)
Polyaniline
(Electroluminescent displays, such as in mobile phones)
(Conductor; electromagnetic shielding of electronic circuits; antistatic, such as in carpets)
H N N H
H N N H
S
S S
S
Polypyrrole
Polythiophene
(Electrolyte; screen coating; sensing devices)
(Field-effect transistors, such as in supermarket checkouts; antistatic, such as in photographic film)
Apart from these applications in electronics, conducting polymers can be made to “light up” when excited by an electric field, a phenomenon called electroluminescence that has gained enormous utility in the form of organic light-emitting diodes (OLEDs). Simply put, such organic materials can be viewed as organic light “bulbs.” They
S
are relatively light and flexible and encompass a broad spectrum of colors. Because organic polymers are, in principle, readily processed into any shape or form, they provide novel, flexible displays, such as books, luminous cloth, and wall decorations. The future, indeed, looks bright!
These bright colors are thanks to OLEDs.
632
Delocalized Pi Systems
Chapter 14
The carbon – carbon double bond in the cycloadduct is electron rich and sterically hindered. Thus, it does not react further with additional diene. The parent 1,3-butadiene, without additional substituents, is electron rich enough to undergo cycloadditions with electron-poor alkenes.
O B COCH2CH3
160C, 15 h
O B COCH2CH3
94%
Ethyl propenoate (Ethyl acrylate)
Many typical dienes and dienophiles have common names, owing to their widespread use in synthesis (Table 14-1).
Exercise 14-18 Formulate the products of [4 1 2]cycloaddition of tetracyanoethene with (a) 1,3-butadiene; (b) cyclopentadiene; (c) 1,2-dimethylenecyclohexane (see margin).
1,2-Dimethylenecyclohexane
Table 14-1
Typical Dienes and Dienophiles in the Diels-Alder Reaction Dienes
CH3 H 3C H3C 1,3-Butadiene
CH3 1,3-Cyclopentadiene
trans, trans2,4-Hexadiene
2,3-Dimethyl1,3-butadiene
1,3-Cyclohexadiene
Dienophiles
NC
CN
NC
CN
Tetracyanoethene
H
CN
H
CN
cis-1,2-Dicyanoethene
O O O 2-Butenedioic anhydride (Maleic anhydride)
CO2CH3 A C c C A CO2CH3 Dimethyl butynedioate (Dimethyl acetylenedicarboxylate)
H
CO2CH3
H
CO2CH3
Dimethyl cis-2-butenedioate (Dimethyl maleate)
H
CH3O2C
CO2CH3 H
Dimethyl trans-2-butenedioate (Dimethyl fumarate)
O B H2C P CHCH
O B H2C P CHCOCH3
Propenal (Acrolein)
Methyl propenoate (Methyl acrylate)
14-8 Diels-Alder Cycloaddition
Chapter 14
The Diels-Alder reaction is concerted The Diels-Alder reaction takes place in one step. Both new carbon – carbon single bonds and the new p bond form simultaneously, just as the three p bonds in the starting materials break. As mentioned earlier (Section 6-4), one-step reactions, in which bond breaking happens at the same time as bond making, are concerted. The concerted nature of this transformation can be depicted in either of two ways: by a dotted circle, representing the six delocalized p electrons, or by electron-pushing arrows. Just as six-electron cyclic overlap stabilizes benzene (Section 14-7), the Diels-Alder process benefits from the presence of such an array in its transition state. Two Pictures of the Transition State of the Diels-Alder Reaction New C–C bond
.. . ..
. .. ..
.. . ... .... . .. .
MECHANISM
‡ or
Dotted-line picture
New C–C bond
Electron-pushing picture
Six-membered ring
An orbital representation (Figure 14-9) clearly shows bond formation by overlap of the p orbitals of the dienophile with the terminal p orbitals of the diene. While these four carbons rehybridize to sp3, the remaining two internal diene p orbitals give rise to the new p bond. The mechanism of the Diels-Alder reaction requires that both ends of the diene point in the same direction to be able to reach the dienophile carbons simultaneously. This means that the diene has to adopt the energetically slightly less favorable s-cis conformation, relative to the more stable s-trans form (Figure 14-6).
Will be a full π bond
Diene (1,3-Butadiene)
Full π bond sp3
HC HC
CH2
HC HC
CH2
CH2 CH2
Full σ bond
Will be a σ bond
sp3
sp3
CH C 2
Will be a bond
CH2 H2C
Dienophile (Ethene)
Full bond
H2C
sp3
Cycloadduct (Cyclohexene)
Figure 14-9 Orbital representation of the Diels-Alder reaction between 1,3-butadiene and ethene. The two p orbitals at C1 and C4 of 1,3-butadiene and the two p orbitals of ethene interact, as the reacting carbons rehybridize to sp3 to maximize overlap in the two resulting new single bonds. At the same time, p overlap between the two p orbitals on C2 and C3 of the diene increases to create a full double bond.
633
Delocalized Pi Systems
Chapter 14
CH3
CH3 Slightly unfavorable
More unfavorable
H3C
CH3 Particularly pronounced steric hindrance
H s-trans
s-cis (Reacting conformation)
s-trans
s-cis (Reacting conformation)
This necessity affects the rates of the cycloaddition: When the s-cis form is particularly hindered or impossible, the reaction slows down or does not occur. Conversely, when the diene is constrained to s-cis, the transformation is accelerated. Particularly Reactive Dienes
Unreactive Dienes
The Diels-Alder reaction is stereospecific As a consequence of the concerted mechanism, the Diels-Alder reaction is stereospecific. For example, reaction of 1,3-butadiene with dimethyl cis-2-butenedioate (dimethyl maleate, a cis alkene) gives dimethyl cis-4-cyclohexene-1,2-dicarboxylate. The stereochemistry at the original double bond of the dienophile is retained in the product. In the complementary reaction, dimethyl trans-2-butenedioate (dimethyl fumarate, a trans alkene) gives the trans adduct. In the Diels-Alder Reaction, the Stereochemistry of the Dienophile Is Retained
H
150–160°C, 20 h
CO2CH3 H
/∑
∑
H
COCH3 B O
CO2CH3
H 68%
Dimethyl cis-2-butenedioate (Dimethyl maleate)
Dimethyl cis-4-cyclohexene-1,2-dicarboxylate (Cis product)
(Cis starting material)
200–205°C, 3.5 h
CO2CH3 H
/∑
H
O B COCH3
/
MODEL BUILDING
O B COCH3
/
634
H
Dimethyl trans-2-butenedioate (Dimethyl fumarate)
H CO2CH3
∑
H3COC B O
95% Dimethyl trans-4-cyclohexene-1,2-dicarboxylate (Trans product)
(Trans starting material)
Similarly, the stereochemistry of the diene also is retained. Note that the cycloadducts depicted here contain stereocenters and may be either meso or chiral. However, since we begin with achiral starting materials, the ensuing products, when chiral, are formed as racemates via two equal-energy transition states (see, for example, Sections 5-7 and 12-5).
14-8 Diels-Alder Cycloaddition
Chapter 14
In other words, the stereospecificity exhibited by the Diels-Alder process refers to relative, not absolute, stereochemistry. As always, we are depicting only one enantiomer of a chiral (but racemic) product, as in the case of the cycloadditions of dimethyl trans-2-butenedioate (above) and cis,trans-2,4-hexadiene (below). In the Diels-Alder Reaction, the Stereochemistry of the Diene Is Retained CH3 NC H H
CN CN
CH3
/∑ CN H CH3
CN
NC trans,trans-2,4-Hexadiene
Tetracyanoethene (Methyls end up cis)
(Both methyls “outside”)
H NC CH3 H
MODEL BUILDING
H CH3 /∑ CN
CN
H3C H CN /∑
CN
CN CN
NC
/∑ CN H CH3
CN
CH3 cis, trans-2,4-Hexadiene
(Methyls end up trans)
(One methyl “inside”; one methyl “outside”)
Exercise 14-19
Working with the Concepts: Diels-Alder Reactions Draw the product of the following Diels-Alder cycloaddition: H3C
CN
? NC
H3C
Strategy In dealing with problems that feature Diels-Alder reactions, it is useful to remind yourself of the spatial approach of diene and dienophile, ideally using molecular models. Once you feel comfortable with visualizing how the reactants align in the transition state, you can practice drawing it. Solution • Using Figure 14-9 as a guide, draw, in perspective, the two components of the reaction before cycloaddition. • Complete your drawing by moving the six participating electrons appropriately to generate the two new s bonds and the new p bond. • If your drawing looks distorted, straighten it out to a regular cyclohexene ring, adding the substituents, including their relative stereochemistry. H3 C
H3C
H3C
H3C
H3C CN NC
CN NC
H3C
] CN ~ CN
635
Delocalized Pi Systems
Exercise 14-20
Try It Yourself Add structures of the missing products or starting materials to the following Diels-Alder reaction schemes. H3C
] CN
?
(a)
- CN
H3C
H3C
H3C
(b) ?
CH3 % F F F
? ^
H3C
CH3
F
Exercise 14-21 cis,trans-2,4-Hexadiene reacts very sluggishly in [4 1 2]cycloadditions; the trans,trans isomer does so much more rapidly. Explain. [Hint: The Diels-Alder reaction requires the s-cis arrangement of the diene (Figure 14-9, and see Figure 14-6).]
Diels-Alder cycloadditions follow the endo rule The Diels-Alder reaction is highly stereocontrolled, not only with respect to the substitution pattern of the original double bonds, but also with respect to the orientation of the starting materials relative to each other. Consider the reaction of 1,3-cyclopentadiene with dimethyl cis-2-butenedioate. Two products are conceivable, one in which the two ester substituents on the bicyclic frame are on the same side (cis) as the methylene bridge, the other in which they are on the side opposite (trans) from the bridge. The first is called the exo adduct, the second the endo adduct (exo, Greek, outside; endo, Greek, within). The terms refer to the position of groups in bridged systems. Exo substituents are placed cis with respect to the shorter bridge; endo substituents are positioned trans to this bridge. In general, in an exo addition, the substituents on the dienophile point away from the diene. Conversely, in an endo addition they point toward the diene. Exo and Endo Cycloadditions to Cyclopentadiene
Hydrogens point toward the diene
H
Y Y Y Y Y Y YH Y Y
H
H
Shorter bridge (methylene)
H
Exo addition
Y Y Y Y Y Y H3CO2C Y Y Y H3CO2C H
CO2CH3
¥H !CO2CH3 ¥ H
CO2CH3
CO2CH3
Substituents point toward the diene
H
!
Chapter 14
Exo groups
Substituents point away from the diene
H
H H
H
Endo addition
H !
636
H Hydrogens point away from the diene
¥ CO2CH3 ! H ¥ CO2CH3
Endo groups
14-8 Diels-Alder Cycloaddition
Chapter 14
637
C H E MI C AL H I G H L I G H T 1 4 - 3 The Diels-Alder Reaction is “Green” In the Diels-Alder reaction, both starting materials are consumed to give a new product without generating any additional materials. Such transformations are called “atom economical,” because all of the atoms of starting materials are incorporated in the product. Atom-economical reactions are a key ingredient of green chemistry (see Chemical Highlight 3-1), and the Diels-Alder reaction adheres to several of its principles. Thus, it generates no (or little) waste, incorporates all starting materials in the product(s), and retains functionality. It normally does not require protecting groups; it can be run with neat reagents, thus avoiding solvents; and it is often efficient enough to allow crystallization or distillation of pure product, thus avoiding column chromatography. Many Diels-Alder reactions need some heating to proceed at reasonable rates, but this problem can be overcome by the application of catalysts that allow room-temperature transformations. For example, Lewis acids (Section 2-2) accelerate cycloadditions greatly. A case in which this effect has been quantified is shown below. Such catalysis also affects exo/endo ratios and can result in enantioselectivity when optically active catalysts are employed (see, for example, Chemical Highlights 5-4 and 9-3, and Section 12-2). The effect of the Lewis acid is to activate the dienophile by complexation to the carbonyl oxygen lone pairs, rendering the carbonyl group even more electron withdrawing.
O
Water, the greenest solvent
Sometimes solvents are unavoidable, as in intramolecular reactions (Exercise 14-24), when high dilution is important (Section 9-6). The green solvent of choice is, naturally, water. In fact, as shown in the example, water is not only feasible as a solvent, it can speed up Diels-Alder reactions on its own, in addition to improving stereoselectivity, especially when combined with Lewis acid catalysis. This effect of water has been ascribed to hydrogen bonding and hydrophobic effects (Section 8-2) on the transition state.
Solvent, catalyst
H
N
H O
N
H H
O N
Endo krel krel krel krel
(in (in (in (in
CH3CN, uncatalyzed) CH3CN, catalyzed by Cu21) H2O, uncatalyzed) H2O, catalyzed by Cu21)
1 158,000 287 232,000
67 94 84 93
Exo : : : :
The Diels-Alder reaction usually proceeds with endo selectivity, that is, the product in which the activating electron-withdrawing group of the dienophile is located in the endo position is formed faster than the alternative exo isomer. This occurs even though the exo product is often more stable than its endo counterpart. This observation is referred to as the endo rule. The preference for endo cycloaddition has its origin in a variety of steric and electronic influences on the transition state of the reaction. Although
33 6 16 7
MODEL BUILDING
638
Chapter 14
Delocalized Pi Systems
the endo transition state is only slightly lower in energy, this is sufficient to control the outcome of most Diels-Alder reactions we shall encounter. Mixtures may ensue in the case of highly substituted systems or when several different activating substituents are present.
The Endo Rule
Y Y H Y Y
80C
COCH3 B O
CH3O
Y Y Y Y Y
H H
Endo addition
¥! H COCH3 B O 91%
H
H
O
O
O O
O O
Methyl propenoate
The relative stereochemistry of the product of a general Diels-Alder reaction that follows the endo rule is illustrated below. As an aid to keeping track of the substituents and where they are going, we can use the general labels “o” (for outside, as in outside the halfcircle marked by the diene carbon sequence) and “i” (for inside) for the two possible stereochemical orientations of groups attached to the end of the diene. We then label the substituents on the dienophile with respect to their orientation in the transition state of the reaction as either endo or exo. The structure of the expected product with all substituents in place is shown on the right of the equation. You can see that “o” is always cis to “endo.” This scheme allows you to get the structure of your product quickly, without the need for stereodrawings. However, it does not replace a thorough understanding of the principles that led to it!
o i i o o “outside” i “inside”
endo
exo
i o * C exo ~endo
endo
exo
~ exo & ¥ endo i o
)
ANIMATED MECHANISM: Diels-Alder cycloaddition (endo rule)
Endo product
)
MATION AN I
O
14-8 Diels-Alder Cycloaddition
Exercise 14-22
Working with the Concepts: The Endo Rule Draw the product of the reaction of trans,trans-2,4-hexadiene with methyl propenoate (show the stereochemistry clearly). Strategy First, we draw the structures of the two reagents, trans,trans-2,4-hexadiene and methyl propenoate. Next, to get the stereochemistry of the product right, we need to line up the reagents as in the transition state picture in Figure 14-9, one above the other. The ester function of the dienophile is, according to the endo rule, expected to be positioned endo. Solution • Following the prescribed steps gives us CH3
H3C ∑/ H
CH3
`
H
O
H
H3C
`
^
H3C
%
H
H3CO2C
H
CH3
H
H3CO2C
CO2CH3
• We can check the result, using our general schematic formalism above. For this purpose, we label all substituents in the reagents: The two methyl groups in the diene are located “outside” and therefore are designated “o.” The ester function in the dienophile is labeled endo. Plugging in these labeled groups into our generalized product structure confirms our solution. CH3
o
`
o
`
o
^
CH3O2C
endo
o
CH3 endo
Exercise 14-23
Try It Yourself Predict the products of the following reactions (show the stereochemistry clearly): (a) trans-1,3pentadiene with 2-butenedioic anhydride (maleic anhydride); (b) 1,3-cyclopentadiene with dimethyl trans-2-butenedioate (dimethyl fumarate).
Exercise 14-24 The Diels-Alder reaction can also occur in an intramolecular fashion. Draw the two transition states leading to products in the following reaction. O B COCH3 H H 65
^
75%
%
^
180°C, 5 h
O B COCH3 % H
`
`
O B COCH3
:
H 35
Chapter 14
639
640
Delocalized Pi Systems
Chapter 14
In Summary The Diels-Alder reaction is a concerted cycloaddition that proceeds best between electron-rich 1,3-dienes and electron-poor dienophiles to furnish cyclohexenes. It is stereospecific with respect to the stereochemistry of the double bonds and with respect to the arrangements of the substituents on diene and dienophile: It follows the endo rule.
14-9 Electrocyclic Reactions The Diels-Alder reaction couples the ends of two separate p systems. Can rings be formed by the linkage of the termini of a single conjugated di-, tri-, or polyene? Yes, and this section will describe the conditions under which such ring closures (and their reverse), called electrocyclic reactions, take place. Cycloadditions and electrocyclic reactions belong to a class of transformations called pericyclic (peri, Greek, around), because they exhibit transition states with a cyclic array of nuclei and electrons.
Electrocyclic transformations are driven by heat or light
Photochemical reactions are used increasingly in “green” technology. This reactor on the roof of Complutense University of Madrid is used for water disinfection. A polymer-supported dye absorbs sunlight to convert oxygen to a more reactive state (“singlet oxygen”), which destroys harmful bacteria in water (picture courtesy of Professor Guillermo Orellana, UCM).
Let us consider first the conversion of 1,3-butadiene into cyclobutene. This process is endothermic because of ring strain. Indeed, the reverse reaction, ring opening of cyclobutene, occurs readily upon heating. However, ring closure of cis-1,3,5-hexatriene to 1,3-cyclohexadiene is exothermic and takes place thermally. Is it possible to drive these transformations in the thermally disfavored directions? We know that in a thermal reaction this is a difficult task, because equilibrium is governed by thermodynamics (Section 2-1). However, the problem can be surmounted in some cases by using light, so-called photochemical reactions. In these, absorption of a photon by the starting material excites the molecule into a higher energy state. We have seen how such absorptions form the basis for spectroscopy (Section 10-2; see also Section 14-11). Molecules can relax from such excited states to furnish thermodynamically less stable products than starting material(s). We shall not deal with the details of photochemistry in this text, but we do note that it allows electrocyclic reaction equilibria to be driven in the energetically unfavorable direction. Therefore, irradiation of 1,3-cyclohexadiene with light of appropriate frequency will cause its conversion to its triene isomer. Similarly, irradiation of 1,3-butadiene effects ring closure to cyclobutene. Electrocyclic Reactions CH2
Δ
CH2
CH2
hv
CH2
cis-1,3,5-Hexatriene
ΔH° 14.5 kcal mol1 (60.7 kJ mol1) Ring closure to six-membered ring is exothermic
1,3-Cyclohexadiene
CH2
Δ
CH2
hv
Cyclobutene
CH2 ΔH° 9.7 kcal mol1 (40.6 kJ mol1) Ring opening of four-membered ring is exothermic CH2 1,3-Butadiene
Exercise 14-25 Give the products obtained on heating the following compounds.
(a)
(b)
14-9 Electrocyclic Reactions
Chapter 14
Electrocyclic reactions are concerted and stereospecific Like the Diels-Alder cycloaddition, electrocyclic reactions are concerted and stereospecific. Thus, the thermal isomerization of cis-3,4-dimethylcyclobutene gives only cis,trans-2,4hexadiene. H
CH3 [ ~H [ CH ~
CH3 H
3
H
CH3
cis-3,4-Dimethylcyclobutene
cis, trans-2,4-Hexadiene
Heating its isomer, trans-3,4-dimethylcyclobutene, provides only trans,trans-2,4-hexadiene. CH3
CH3 [ ~H [H ~ CH3
Δ
H H CH3
trans-3,4-Dimethylcyclobutene
trans, trans-2,4-Hexadiene
Figure 14-10 takes a closer look at these processes. As the bond between carbons C3 and C4 in the cyclobutene is broken, these carbon atoms must rehybridize from sp3 to sp2 and rotate to permit overlap between the emerging p orbitals and those originally present. In such thermal cyclobutene ring openings, the carbon atoms are found to rotate in the same direction, either both clockwise or both counterclockwise. This mode of reaction is called a conrotatory process. In the case of cis-3,4-dimethylcyclobutene, both the clockwise and counterclockwise courses result in the same product, cis,trans-2,4-hexadiene. However, for trans-3,4-dimethylcyclobutene, two products are possible. The counterclockwise mode leads
MODEL BUILDING
641
642
Delocalized Pi Systems
Chapter 14
CH3 moves inward
Clockwise
‡
CH3
H3C H
H 3C
Δ
H
CH3
H
H Clockwise
A
cis-3,4-Dimethylcyclobutene
CH3
H H3 C
H
CH3 moves outward
Conrotatory
cis,trans-2,4-Hexadiene
Counterclockwise
H3C
H CH3
H Counterclockwise
B
Δ
H3C
Conrotatory
trans-3,4-Dimethylcyclobutene
‡ H
H
Δ
CH3 H
H
CH3
H3 C
trans,trans-2,4-Hexadiene
Conrotatory
to the observed trans,trans-2,4-hexadiene. Rotation in the opposite direction would form the corresponding cis,cis isomer but is sterically encumbered and not observed. Fascinatingly, the photochemical closure (photocyclization) of butadiene to cyclobutene proceeds with stereochemistry exactly opposite that observed in the thermal opening. In this case, the products arise by rotation of the two reacting carbons in opposite directions. In other words, if one rotates clockwise, the other does so counterclockwise. This mode of movement is called disrotatory (Figure 14-11). Can these observations be generalized? Let us look at the stereochemistry of the cis1,3,5-hexatriene – cyclohexadiene interconversion. Surprisingly, the six-membered ring is formed thermally by the disrotatory mode, as can be shown by using derivatives. For example, heated trans,cis,trans-2,4,6-octatriene gives cis-5,6-dimethyl-1,3-cyclohexadiene, and cis,cis,trans-2,4,6-octatriene converts into trans-5,6-dimethyl-1,3-cyclohexadiene, both disrotatory closures. Counterclockwise
Clockwise
hv
H
H3 C Counterclockwise
CH3
H3C
H
H
H
Clockwise
hv
H3 C
CH3 H
H
Disrotatory
H H3C Figure 14-11 Disrotatory photochemical ring closure of cis,transand trans,trans-2,4-hexadiene. In the disrotatory mode, one carbon rotates clockwise, the other counterclockwise.
H CH3
Both CH3 groups move inward
trans-3,4-Dimethylcyclobutene
Figure 14-10 (A) Conrotatory ring opening of cis-3,4-dimethylcyclobutene. Both reacting carbons rotate clockwise. The sp3 hybrid lobes in the ring change into p orbitals, the carbons becoming sp2 hybridized. Overlap of these p orbitals with those already present in the cyclobutene starting material creates the two double bonds of the cis,trans diene. (B) Similar conrotatory opening of trans-3,4-dimethylcyclobutene in a counterclockwise fashion proceeds to the trans,trans diene. (C) The alternative clockwise conrotatory opening of trans-3,4-dimethylcyclobutene does not occur because of steric encumbrance in the transition state.
H
H3 C
Clockwise
C
H
trans,trans-2,4-Hexadiene
Clockwise
H3C
CH3 H
H
Disrotatory
H3C
CH3
14-9 Electrocyclic Reactions
Stereochemistry of the Thermal 1,3,5-Hexatriene Ring Closure
MODEL BUILDING
CH3
%
CH3
Chapter 14
Δ
%
Disrotatory
CH3
CH3
Disrotatory
trans, cis, trans-2,4,6-Octatriene
cis-5,6-Dimethyl-1,3-cyclohexadiene
%
CH3
Δ
CH3
~CH
Disrotatory
CH3
3
Disrotatory
cis, cis, trans-2,4,6-Octatriene
trans-5,6-Dimethyl-1,3-cyclohexadiene
In contrast, the corresponding photochemical reactions occur in conrotatory fashion. Stereochemistry of the Photochemical 1,3,5-Hexatriene Ring Closure CH3
%
CH3
hv Conrotatory
CH3
~CH
3
This stereocontrol is observed in many other electrocyclic transformations and is governed by the symmetry properties of the relevant p molecular orbitals. The WoodwardHoffmann* rules describe these interactions and predict the stereochemical outcome of all electrocyclic reactions as a function of the number of electrons taking part in the process and whether the reaction is carried out photochemically or thermally. A complete treatment *Professor Robert B. Woodward (1917 – 1979), Harvard University, Cambridge, Massachusetts, Nobel Prize 1965 (chemistry); Professor Roald Hoffmann (b. 1937), Cornell University, Ithaca, New York, Nobel Prize 1981 (chemistry).
MODEL BUILDING
643
644
Delocalized Pi Systems
Chapter 14
CHE M I C AL H I G H L I G H T 1 4 - 4 An Electrocyclization Cascade in Nature: Immunosuppressants from Streptomyces Cultures Streptomyces is a group of bacteria found predominantly in soil and in decaying vegetation. You may have noticed them in action by the “earthy” odor of compost heaps. These bacteria are prolific in producing medicinally useful, biologically active compounds and have therefore been investigated thoroughly by organic chemists. Among the components of some Streptomyces cultures are polyenes, including the conjugated tetraene spectinabilin. In 2001, the promising potent immunosuppressants called SNF 4435 C and D (an acronym referring to the biological screening method) were discovered in the same cultures. Chemists noticed that these compounds were isomers of spectinabilin and speculated that they were formed by a cascade of two electrocyclizations.
O H3C CH3 O CH3
O2N
O
CH3
CH3
Surgeons prepare to insert a new heart into a transplant patient. Immunosuppressants will be essential to avoid rejection by the immune system of the recipient.
OCH3
Spectinabilin
H3C
O2N
O
H3C
O2N
O
O
O CH3
CH3
O CH3
O OCH3
CH3 SNF 4435 C
Spectinabilin is an all-trans tetraene and therefore incapable of undergoing ring closure. It was found, however, that sunlight induced cis-trans isomerization of the two internal double bonds furnishing the cis,cis isomer that is perfectly set up to enter into an 8p conrotatory cyclization. Because
H3C
[
≈
H3C
CH3
OCH3
CH3 SNF 4435 D
there is a stereocenter (*), there are two modes of conrotatory movement (clockwise and counterclockwise), giving two isomeric cyclooctatrienes. These compounds then undergo subsequent 6p disrotatory closure to the two SNF isomers.
14-9 Electrocyclic Reactions
645
Chapter 14
Formation of SNF 4435 Isomers from Spectinabilin O OCH3
O 2N O
O Spectinabilin
Sunlight
O2N O
O
q
*
O OCH3
8 Conrotatory ring closure
O
NO2 O
O
NO2
O
O
OCH3
O
OCH3
6 Disrotatory ring closure
H3C
O 2N
O
H3C
O2N
O
O
O CH3
≈
CH3
OCH3
CH3 SNF 4435 C
The proposed pathway was confirmed in 2004 and 2005 by so-called biomimetic syntheses that were inspired by the biogenetic proposal. These approaches centered on the direct construction of the cis,cis isomer of spectinabilin using a
O H 3C
[
O H3C
CH3
CH3
OCH3
CH3 SNF 4435 D
Stille coupling (Chemical Highlight 13-1) of the two halves of the molecule. In fact, the resulting tetraene converted to the two SNF isomers spontaneously under the conditions of the reaction.
Delocalized Pi Systems
Table 14-2
Stereochemical Course of Electrocyclic Reactions (Woodward-Hoffmann Rules)
Number of pairs of participating electrons Even Odd
Thermal process
Photochemical process
Conrotatory Disrotatory
Disrotatory Conrotatory
of this subject is best left to a more advanced course in organic chemistry. However, the predicted stereochemical course of electrocyclic reactions can be summarized in the simple manner shown in Table 14-2. Exercise 14-26 The cyclic polyene A (an “annulene”; see Section 15-6) can be converted to either B or C by a sequence of electrocyclic ring closures, depending on whether light or heat are used. Identify the conditions necessary to effect either transformation and identify each step as either con- or disrotatory. H
H
~
%
H H
?
H
?
H H
B
%
~
~
%
H
H
%
~
H
H
H
A
C
Exercise 14-27
Working with the Concepts: An Electrocyclic Reaction with a Twist
/∑
Heating cis-3,4-dimethylcyclobutene, A, in the presence of dienophile B gave exclusively the diastereomer C. Explain by a mechanism. H CH3 /∑ H CH3 CN ? CN H ' /∑ CN CH3 CN H 3C H
∑
A
B
C
Strategy This reaction looks like a cycloaddition. We can confirm this idea by checking the atom stoichiometry: C6H10(A) 1 C4H2N2(B) 5 C10H12N2(C), so the reaction is atom economical. What kind of cycloaddition? To determine this, we need to do some retrosynthetic analysis on C. Solution • Working backward, we see that cyclohexene C resembles a Diels-Alder addition product of B to a 2,4-hexadiene isomer. Since the two methyl groups in C are trans to each other, the diene cannot be symmetrical: the only choice is cis,trans-2,4-hexadiene D: CH3 CN
/∑
H CH3 /∑ H CN
CH3
H CN
∑
/∑
/
Chapter 14
/
646
CN
H3C H C
D
B
Y
CH3
‡
CH3
Y NC Y Y Y Y NC Y E
• D must be derived from isomer A by thermal, conrotatory electrocyclic ring opening. • Is the stereochemistry of the cycloaddition of B to D exo or endo? Inspection of the relative positioning of the substituents at the four contiguous stereocenters shows it to be endo, as in E.
14-10 Polymerization of Conjugated Dienes: Rubber
Chapter 14
647
Exercise 14-28
Try It Yourself Irradiation of ergosterol gives provitamin D2, a precursor of vitamin D2 (a deficiency of which causes softening of the bones, especially in children). Is the ring opening conrotatory or disrotatory? (Caution: The product is written in a more stable conformation than that obtained upon ring opening.) CH3 %R %
CH3 % ≥ H
CH3 hv
CH3
OH ´
≥ H
) HO
CH3 %R % ≥ H
CH3 Provitamin D2
Ergosterol
≥ H
´
CH3 ´
H H3C≈% CH3% %
HO
Vitamin D2
In Summary Conjugated dienes and hexatrienes are capable of (reversible) electrocyclic ring closures to cyclobutenes and 1,3-cyclohexadienes, respectively. The diene – cyclobutene system prefers thermal conrotatory and photochemical disrotatory modes. The triene – cyclohexadiene system reacts in the opposite way, proceeding through thermal disrotatory and photochemical conrotatory rearrangements. The stereochemistry of such electrocyclic reactions is governed by the Woodward-Hoffmann rules.
14-10 Polymerization of Conjugated Dienes: Rubber Like simple alkenes (Section 12-15), conjugated dienes can be polymerized. The elasticity of the resulting materials has led to their use as synthetic rubbers. The biochemical pathway to natural rubber features an activated form of the five-carbon unit 2-methyl-1,3-butadiene (isoprene, see Section 4-7), which is an important building block in nature.
1,3-Butadiene can form cross-linked polymers When 1,3-butadiene is polymerized at C1 and C2, it yields a polyethenylethene (polyvinylethylene). 1,2-Polymerization of 1,3-Butadiene Unit of polymerization
2n CH2 P CH O CHP CH2
Initiator
CH2 CH2 B B CH CH A A O (CH O CH2 O CH O CH2)n O
Cross-linking
Alternatively, polymerization at C1 and C4 gives either trans-polybutadiene, cis-polybutadiene, or a mixed polymer. 1,4-Polymerization of 1,3-Butadiene n CH2 P CH O CH P CH2
Initiator
O (CH2 O CH P CHO CH2)n O cis- or trans-Polybutadiene
Butadiene polymerization is unique in that the product itself may be unsaturated. The double bonds in this initial polymer may be linked by further treatment with added chemicals, such as radical initiators, or by radiation. In this way, cross-linked polymers arise (Figure 14-12), in
Figure 14-12 Cross-linking underlies the elasticity of polybutadiene chains in rubber.
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which individual chains have been connected into a more rigid framework. Cross-linking generally increases the density and hardness of such materials. It also greatly affects a property characteristic of butadiene polymers: elasticity. The individual chains in most polymers can be moved past each other, a process allowing for their molding and shaping. In crosslinked systems, however, such deformations are rapidly reversible: The chain snaps back to its original shape (more or less). Such elasticity is characteristic of rubbers.
Synthetic rubbers are derived from poly-1,3-dienes Elasticity in action: This bouncing ball was photographed using colored strobe lights, flashing 50 times per second, giving about 75 separate images. The sequence shows that the ball’s trajectory is a parabola. The ball moves fastest near the ground and slowest at the top of an arc. The horizontal speed is approximately constant, and only the vertical speed is changing (due to gravity). The height of successive bounces is less as the ball loses energy by hitting the ground.
Polymerization of 2-methyl-1,3-butadiene (isoprene, Section 4-7) by a Ziegler-Natta catalyst (Section 12-15) results in a synthetic rubber (polyisoprene) of almost 100% Z configuration. Similarly, 2-chloro-1,3-butadiene furnishes an elastic, heat- and oxygen-resistant polymer called neoprene, with trans chain double bonds. Several million tons of synthetic rubber are produced in the United States every year. CH3 A n H2C P C O CH P CH2
TiCl4, AlR3
H3C
H G G C PC D D CH2)n O O (H2C
2-Methyl-1,3-butadiene
Cl A n H2C P C O CH P CH2
(Z)-Polyisoprene
CH2)n O G G C PC D D H O (H2C Cl
TiCl4, AlR3
2-Chloro-1,3-butadiene
Latex, the precursor of natural rubber, is drained from the bark of Hevea brasiliensis.
Neoprene
Natural Hevea rubber is a 1,4-polymerized (Z)-poly(2-methyl-1,3-butadiene), similar in structure to polyisoprene. To increase its elasticity, it is treated with hot elemental sulfur in a process called vulcanization (Vulcanus, Latin, the Roman god of fire), which creates sulfur cross-links. This reaction was discovered by Goodyear* in 1839. One of the earliest and most successful uses of the product, Vulcanite, was the manufacture of dentures that could be molded for good fit. Before the 1860s, false teeth were embedded in animal bone, ivory, or metal. The swollen appearance of George Washington’s lips (visible on U.S. currency) is attributed to his ill-fitting ivory dentures. Nowadays, dentures are made from acrylics (Section 13-10). Rubber remains an indispensable component of many commercial products, including tires (the major use), shoes, rainwear, and other clothing incorporating elastic fibers. Copolymers in which the double bonds of 1,3-butadiene undergo polymerization with the double bonds of other alkenes have assumed increasing importance in recent years. By varying the proportions of the different monomers in the polymerization mix, the properties of the final product may be “tuned” over a considerable range. One such substance is a three-component copolymer of propenenitrile, 1,3-butadiene, and ethenylbenzene, known as ABS (for acrylonitrile/butadiene/styrene copolymer). The diene imparts the rubberlike property of flexibility, whereas the nitrile hardens the polymer. The result is a highly versatile material that may be fashioned into sheets or molded into virtually any shape. Its strength and ability to tolerate deformation and stress have allowed its utilization in everything from clock mechanisms to camera and computer cases to automobile bodies and bumpers.
Polyisoprene is the basis of natural rubber How is rubber made in nature? Plants construct the polyisoprene framework of natural rubber by using as a building block 3-methyl-3-butenyl pyrophosphate (isopentenyl pyrophosphate). This molecule is an ester of pyrophosphoric acid and 3-methyl-3-buten-1-ol. An enzyme equilibrates a small amount of this material with the 2-butenyl isomer, an allylic pyrophosphate. The radiator guard and the bumper corners of this massive truck are made of ABS copolymer.
*Charles Goodyear (1800 – 1860), American inventor, Washington, D.C.
14-10 Polymerization of Conjugated Dienes: Rubber
Chapter 14
Biosynthesis of the Two Isomers of 3-Methylbutenyl Pyrophosphate CH3 A H2C P CCH2CH2OH
O B HO O P O OH O A OH
3-Methyl-3-buten-1-ol
O B HO O PO OH A OH
REACTION
Phosphoric acid HOH
O O B B HO O P O O O P O OH A A OH OH HOH
Pyrophosphoric acid
Enzyme
O O B B CPCHCH2O O P O O O P O OH A A H3C OH OH H3C
O
O
CH3 O O A B B H2C P CCH2CH2O O P O O O P O OH A A OH OH 3-Methyl-3-butenyl pyrophosphate
3-Methyl-2-butenyl pyrophosphate
Although the subsequent processes are enzymatically controlled, they can be formulated simply in terms of familiar mechanisms (OPP 5 pyrophosphate). Mechanism of Natural Rubber Synthesis Step 1. Ionization to stabilized (allylic) cation
OPP
OPP Step 2. Electrophilic attack
MECHANISM
OPP
OPP H
Step 3. Proton loss
OPP
H Step 4. Second oligomerization
OPP
H
Geranyl pyrophosphate
OPP
OPP
OPP
OPP
H
OPP Farnesyl pyrophosphate
In the first step, ionization of the allylic pyrophosphate gives an allylic cation. Attack by a molecule of 3-methyl-3-butenyl pyrophosphate, followed by proton loss, yields a dimer called geranyl pyrophosphate. Repetition of this process leads to natural rubber.
Many natural products are composed of 2-methyl-1,3-butadiene (isoprene) units Many natural products are derived from 3-methyl-3-butenyl pyrophosphate, including the terpenes first discussed in Section 4-7. Indeed, the structures of terpenes can be dissected into
649
650
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five-carbon units connected as in 2-methyl-1,3-butadiene. Their structural diversity can be attributed to the multiple ways in which 3-methyl-3-butenyl pyrophosphate can couple. The monoterpene geraniol and the sesquiterpene farnesol, two of the most widely distributed substances in the plant kingdom, form by hydrolysis of their corresponding pyrophosphates.
OH
OH
Geraniol
Farnesol
Coupling of two molecules of farnesyl pyrophosphate leads to squalene, a biosynthetic precursor of the steroid nucleus (Section 4-7).
steroids Squalene
Bicyclic substances, such as camphor, a chemical used in mothballs, nasal sprays, and muscle rubs, are built up from geranyl pyrophosphate by enzymatically controlled electrophilic carbon – carbon bond formations. Camphor Biosynthesis from Geranyl Pyrophosphate
OPP
Cis-trans isomerization OPP
OPP
Geranyl pyrophosphate
is the same as
O Camphor
Other higher terpenes are constructed by similar cyclization reactions.
In Summary 1,3-Butadiene polymerizes in a 1,2 or 1,4 manner to give polybutadienes with various amounts of cross-linking and therefore variable elasticity. Synthetic rubber can be made from 2-methyl-1,3-butadiene and contains varied numbers of E and Z double bonds. Natural rubber is constructed by isomerization of 3-methyl-3-butenyl pyrophosphate to the 2-butenyl system, ionization, and electrophilic (step-by-step) polymerization. Similar mechanisms account for the incorporation of 2-methyl-1,3-butadiene (isoprene) units into the polycyclic structure of terpenes.
14-11 Electronic Spectra: Ultraviolet and Visible Spectroscopy In Section 10-2, we explained that organic molecules may absorb radiation at various wavelengths. Spectroscopy is possible because absorption is restricted to quanta of defined energies, hv, to effect specific excitations with energy change DE. ¢E 5 hv 5
hc (c 5 velocity of light) l
14-11 Electronic Spectra: UV and Visible Spectroscopy
Figure 10-2 divided the range of electromagnetic radiation into various subsections, from high-energy X-rays to low-energy radiowaves. Among these subsections, one stood out by its representation in colors, the visible spectrum. Indeed, this is the only region of the electromagnetic spectrum that the human body, with our eyes as the “spectrometer,” is able to resolve. The effect of other forms of radiation on us is less well defined: X-rays and ultraviolet light (sunburn!) are destructive, infrared radiation is perceived as heat, and micro- and radio waves are undetectable. To understand how colors come about, we have to go back to an experiment by Isaac Newton, who showed that white light, when passed though a prism, can be dispersed into the full color spectrum, much like a rainbow, for which water droplets fulfil the function of the prism. Thus, what we perceive as white light is actually the effect of “broad-band irradiation” (to borrow a term from NMR spectroscopy; Section 10-9) on the light receptors in our retina (see Chemical Highlight 18-3). Consequently, we see an object (or compound) as colored when it absorbs light from part of the visible spectrum, reflecting the remainder. For example, when an object absorbs blue light, we see it as orange; when it absorbs green light, we see it as purple. Orange absorption turns the object blue and purple absorption turns it green. In organic chemistry, such colored compounds are most frequently those containing an array of conjugated double bonds. The excitation of electrons in these bonds happens to match the energy of visible light (and, as it turns out, also much of the ultraviolet spectrum), giving them their color. Recall the orange color of b-carotene (Section 14-7) in the Chapter Opening, and you are certainly familiar with the effects of indigo (margin) on blue jeans. In this section, we shall quantify the color of organic compounds by spectroscopy in the wavelength range from 400 to 800 nm, called visible spectroscopy (see Figure 10-2 and margin below). We also examine the range from 200 to 400 nm, called ultraviolet spectroscopy. Because these two wavelength ranges are so close to each other, they are typically measured simultaneously with the same spectrometer. Both techniques are particularly useful for investigating the electronic structures of unsaturated molecules and for measuring the extent of their conjugation. A UV-visible spectrometer is constructed according to the general scheme in Figure 10-3. As in NMR, samples are usually dissolved in solvents that do not absorb in the spectral region under scrutiny. Examples are ethanol, methanol, and cyclohexane, none of which has absorption bands above 200 nm. The events triggered by electromagnetic radiation at UV and visible wavelengths lead to the excitation of electrons from filled bonding (and sometimes nonbonding) molecular orbitals to unfilled antibonding molecular orbitals. These changes in electron energies are recorded as electronic spectra. As in NMR, FT instruments have greatly improved sensitivity and ease of spectral acquisition.
Chapter 14
651
Sunlight is split into its component colors (the visible spectrum) by raindrops to give a rainbow.
O
N H
H N
O Indigo
Ultraviolet and visible light give rise to electronic excitations Consider the bonds in an average molecule: We can safely assume that, except for lone pairs, all electrons occupy bonding molecular orbitals. The compound is said to be in its ground electronic state. Electronic spectroscopy is possible because ultraviolet radiation and visible light have sufficient energy to transfer many such electrons to antibonding orbitals, thereby creating an excited electronic state (Figure 14-13). Dissipation of the absorbed energy may occur in the form of a chemical reaction (cf. Section 14-9), as emission of light (fluorescence, phosphorescence), or simply as emission of heat.
The blue in blue jeans is due to the presence of indigo.
The Visible Spectrum Antibonding MO E
E hv
E
Electron movement
Bonding MO Ground state
Excited state
Figure 14-13 Electronic excitation by transfer of an electron from a bonding to an antibonding orbital converts a molecule from its ground electronic state into an excited state.
Color
Wavelength
violet blue green yellow orange red
380–450 nm 450–495 nm 495–570 nm 570–590 nm 590–620 nm 620–750 nm
652
Chapter 14
Delocalized Pi Systems
Figure 14-14 Electronic transitions in a simple p system. The wavelength of radiation required to cause them is revealed as a peak in the ultraviolet or visible spectrum.
* Antibonding E
n
The orange-red b-carotene and deep blue azulene differ in their p-electronic structure.
*
*
n
Nonbonding [as found in lone electron pairs or the propenyl (allyl) radical]
Bonding
Organic s bonds have a large gap between bonding and antibonding orbitals. To excite electrons in such bonds requires wavelengths much below the practical range (,200 nm). As a result, the technique has found its major uses in the study of p systems, in which filled and unfilled orbitals are much closer in energy. Excitation of such electrons gives rise to PnP* transitions. Nonbonding (n) electrons are even more readily promoted through nnP* transitions (Figure 14-14). Because the number of p molecular orbitals is equal to that of the component p orbitals, the simple picture presented in Figure 14-14 is rapidly complicated by extending conjugation: The number of possible transitions skyrockets, and with it the complexity of the spectra. A typical UV spectrum is that of 2-methyl-1,3-butadiene (isoprene), shown in Figure 14-15. The position of a peak is defined by the wavelength at its maximum, a lmax value (in nanometers). Its intensity is reflected in the molar extinction coefficient or molar absorptivity, e, that is characteristic of the molecule. The value of e is calculated by dividing the measured peak height (absorbance, A) by the molar concentration, C, of the sample (assuming a standard cell length of 1 cm). A e5 C The value of e can range from less than a hundred to several hundred thousand. It provides a good estimate of the efficiency of light absorption. Electronic spectral absorption bands are frequently broad, as in Figure 14-15, not the sharp lines typical of many NMR spectra.
Figure 14-15 Ultraviolet spectrum of 2-methyl-1,3-butadiene in methanol, lmax 5 222.5 nm (e 5 10,800). The indentations at the sides of the main peak are called shoulders.
10 9
Shoulders
8
l max = 222.5 mm CH3
Shoulder
Absorbance A
7 CH2
6
C
CH
CH2
5 4 3 2 1
0.0
UV 200
220
240
260
280
300
320
340
Wavelength (nm)
Electronic spectra tell us the extent of delocalization Electronic spectra often indicate the size and degree of delocalization in an extended p system. The more double bonds there are in conjugation, the longer is the wavelength for the lowest-energy excitation (and the more peaks will appear in the spectrum). For example, ethene absorbs at lmax 5 171 nm, and an unconjugated diene, such as 1,4-pentadiene, absorbs at lmax 5 178 nm. A conjugated diene, such as 1,3-butadiene, absorbs at much lower energy (l max 5 217 nm). Further extension of the conjugated system leads to corresponding incremental increases in the lmax values, as shown in Table 14-3. The hyperconjugation of alkyl groups and the improved p overlap in rigid, planar cyclic systems
14-11 Electronic Spectra: UV and Visible Spectroscopy
Table 14-3
653
Chapter 14
lmax Values for the Lowest Energy Transitions in Ethene and Conjugated Pi Systems
Alkene structure
Name
e
lmax (nm)
Ethene
171
15,500
1,4-Pentadiene
178
Not measured
1,3-Butadiene
217
21,000
2-Methyl-1,3-butadiene
222.5
10,800
trans-1,3,5-Hexatriene
268
36,300
trans,trans-1,3,5,7-Octatetraene
330
Not measured
2,5-Dimethyl-2,4-hexadiene
241.5
13,100
1,3-Cyclopentadiene
239
4,200
1,3-Cyclohexadiene
259
10,000
A steroid diene
282
Not measured
A steroid triene
324
Not measured
A steroid tetraene
355
Not measured
b-Carotene (vitamin A precursor)
497 (orange appearance)
133,000
Azulene, a cyclic conjugated hydrocarbon
696 (blue-violet appearance)
150
CH3 % CH3 %
≥ H
CH3 % CH3 %
≥ H
CH3 % CH3 %
≥ H
(For structure, see Section 14-7.)
654
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Delocalized Pi Systems
Figure 14-16 The energy gap between the highest occupied and lowest unoccupied molecular orbitals (abbreviated HOMO and LUMO, respectively) decreases along the series ethene, the 2-propenyl (allyl) radical, and butadiene. Excitation therefore requires less energy and is observed at longer wavelengths.
* *
LUMO
hv
n
hv
4*
LUMO
3* HOMO
hv
2 HOMO
E
LUMO
HOMO
1 Decreasing HOMO–LUMO gap
H2C P CH2 Ethene
H C ee H2C e e CH2
H H COC M J CH2 H2C
2-Propenyl radical
1,3-Butadiene
both seem to contribute. Beyond 400 nm (in the visible range), molecules become colored: first yellow, then orange, red, violet, and finally blue-green. For example, the contiguous array of 11 double bonds in b-carotene (Section 14-7) is responsible for its characteristic intense orange appearance (lmax 5 480 nm). Why should larger conjugated p systems have more readily accessible and lower-energy excited states? The answer is pictured in Figure 14-16. As the overlapping p-orbital array gets longer, the energy gap between filled and unfilled orbitals gets smaller, and more bonding and antibonding orbitals are available to give rise to additional electronic excitations. Finally, conjugation in cyclopolyenes is governed by a separate set of rules, to be introduced in the next two chapters. Just compare the electronic spectra of benzene, which is colorless (Figure 15-6), with that of azulene, which is deep blue (Figure 14-17), and then compare both with the data in Table 14-3. Exercise 14-29 Arrange the following compounds in the order of their increasing lmax values. (a) 1,3,5-Cycloheptatriene
(b) 1,5-Hexadiene
(c) 1,3-Cyclohexadiene
(e) Polyacetylene
(d)
CH3 %
(f)
Absorbance (log ε)
5
Figure 14-17 UV – visible spectrum of azulene in cyclohexane. The absorbance is plotted as log e to compress the scale. The horizontal axis, representing wavelength, also is nonlinear.
4 3 2 1 UV-VIS 0 220
250
300
Wavelength (nm)
350 400
500 600 700
The Big Picture
Chapter 14
655
C H E MI C AL H I G H L I G H T 1 4 - 5 The Contributions of IR, MS, and UV to the Characterization of Viniferone 1
H and 13C NMR data enabled the elucidation of much of the structure of the grape-seed-derived antioxidant viniferone (Chemical Highlight 10-5), shown again below. In particular, the presence of two carbonyl carbons and a total of eight alkene or benzene carbons was revealed. Also, a sixmembered ether ring was strongly indicated. Sorting out the rest of the structure was greatly aided by complementary information derived from mass, IR, and UV measurements. Although 13C NMR gave the number of carbon atoms, high-resolution mass spectrometry revealed the entire
molecular formula as C15H14O8 and thus the presence of nine degrees of unsaturation (15 carbons 3 2 5 30; 30 1 2 5 32; 32 2 14 hydrogens 5 18; 18y2 5 9). IR spectroscopy is very characteristic for five-membered rings such as the one in viniferone: Bands at 1700, 1760, and 1790 cm21 for various individual and combined vibrational modes of the CPO groups, and a very broad, strong absorption between 2500 and 3300 cm21 for the carboxylic acid OOH also appear in the IR spectra of the simpler model compound shown below.
O B
E
/ H H H O
∑/
H
O
O
/∑ A OH OH H H Viniferone
The site of the linkage between this ring and the rest of the structure of viniferone was apparent from the 1H NMR spectrum. In the simpler model, the alkene hydrogens appear at d 5 6.12 (H2) and 7.55 (H3) ppm; viniferone shows only one signal at d 5 6.19. The absence of any signal in the vicinity of d 5 7.5 – 7.6 ppm implies that the point of connection is C3.
O B
H
%? %
HO H
H A
HH
O
2
H 3
OH O OH Model compound
Finally, UV spectroscopy helps to identify the oxygensubstitued benzene ring by the presence of absorptions in the vicinity of 275 nm for the cyclic conjugated moiety. This assignment also completes the identification of the nine degrees of unsaturation: the two CPO groups, the CPC double bond in the five-membered ring, the three double bonds in the benzene ring, and the three rings themselves.
In Summary UV and visible spectroscopy can be used to detect electronic excitations in conjugated molecules. With an increasing number of molecular orbitals, there is an increasing variety of possible transitions and hence number of absorption bands. The band of longest wavelength is typically associated with the movement of an electron from the highest occupied to the lowest unoccupied molecular orbital. Its energy decreases with increasing conjugation.
THE BIG PICTURE We have progressed from a single electron in a single p orbital, as in radicals (Sections 3-1 and 3-2), to a pair of electrons in neighboring p orbitals, as in double bonds (Chapter 12), to two pairs of electrons in neighboring p orbitals, as in triple bonds (Chapter 13), to the present chapter, which extends this theme. We now know that an infinite number of p orbitals can line up in carbon sequences that exhibit the phenomenon of conjugation. Conjugation delocalizes electrons and therefore charges, as in a molecular wire. It can also be viewed as a way by which one end of a molecule may “communicate” with the other chemically. For example, protonation of a conjugated polyene has a profound effect on the molecule along the entire chain of sp2-hybridized carbons. Even more impressive are the pericyclic reactions, following the last new kind of mechanism that we shall encounter in this book. For example, in the Diels-Alder cycloaddition, the respective termini of the diene and the dienophile add to each other in concert, bond formation and bond breaking occurring at the same time. Similar communication is evident in spectral data, particularly in UV–visible spectroscopy. Where does this lead us? Returning to the analogy of a conjugated polyene as a molecular wire, we can ask a simple question: What happens if we touch the ends of that wire to complete a closed circuit? A lot, as the next two chapters will reveal.
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Delocalized Pi Systems
CHAPTER INTEGRATION PROBLEMS 14-30. a. Propose a reasonable mechanism for the transformation of trans,trans-2,4-hexadien-1-ol (sorbyl alcohol) into trans-5-ethoxy-1,3-hexadiene. H, CH3CH2OH
OH
OCH2CH3 Sorbyl alcohol
trans-5-Ethoxy-1,3-hexadiene
SOLUTION This process takes place in acidic ethanol solution (otherwise known as wine). We begin by inspecting the structures of the starting and final molecules to find that (1) an alcohol has become an ether and (2) the double bonds have moved. Let us consider relevant information that we have seen concerning these functional groups. The conversion of two alcohols into an ether in the presence of acid was introduced in Section 9-7. Protonation of one alcohol molecule gives an alkyloxonium ion, whose leaving group (water) may be replaced in either an SN2 or an SN1 fashion by a molecule of a second alcohol:
CH3CH2OH2 CH3 A CH3COH2 A CH3
SN2
CH3CH2OH
SN1
CH3CH2OH
CH3CH2OCH2CH3 CH3 A CH3COCH2CH3 A CH3
Can either of these processes be adapted to the present question? Sorbyl alcohol is an allylic alcohol, and we have just seen (Section 14-3) that allylic halides readily undergo both SN2 and SN1 displacement reactions. A good deal of Chapter 9 dealt with comparing and contrasting the behavior of haloalkanes with that of alcohols: Protonation of the OH group of an alcohol opens the way to both substitution and elimination chemistry. We should therefore expect that allylic alcohols will be similarly reactive. We need next to consider whether the SN2 or the SN1 mechanism is more appropriate. The fact that the double bonds have moved in the course of the reaction is a helpful piece of information. Referring again to Section 14-3 and also 14-6, we note that the allylic cation that results from dissociation of a leaving group in the first step of an SN1 reaction is delocalized, and nucleophiles can therefore attach at more than one site. Let us explore this line of reasoning by examining the carbocation derived from protonation and loss of water from sorbyl alcohol: OH
H
H2O
OH2
Sorbyl alcohol
Like an allylic cation, this carbocation is delocalized. However, the initial system is a more extended conjugated one, and the resulting cation is a hybrid of three, rather than two, contributing resonance forms. As in electrophilic addition to a conjugated triene (Section 14-7), an incoming nucleophile has the option of attaching to any of three positions. In this particular case, attachment of ethanol to a secondary carbon, where a relatively high fraction of the positive charge resides (Section 14-6), is the predominant (kinetic) result. H OO
CH3CH2š OH
O BB OCH2CH3 Ethyl trans, trans-2,4-hexadienoate (Ethyl sorbate)
O
b. Esters of sorbic acid undergo Diels-Alder reactions. Predict the major product to arise from the heating of ethyl sorbate with 2-butenedioic anhydride (maleic anhydride, see structure, Table 14-1). Consider carefully all stereochemical aspects of the process.
Chapter Integration Problems
657
Chapter 14
SOLUTION The Diels-Alder reaction is a cycloaddition between a diene, in this case the ethyl sorbate, and a dienophile, usually an electron-poor alkene (Section 14-8). We should first have a look at the two reaction partners to visualize the new bonding connections that this process will effect. To do so, we need to rotate about the single bond between carbon 3 and carbon 4 of ethyl sorbate to place its double bonds in the necessary conformation. We do this carefully, so as not to mistakenly change the stereochemistry of the double bonds: They both start out trans and must be that way after we have finished. Next, we connect the ends of the diene system to the carbon atoms of the dienophile’s alkene function (dotted lines below), giving us the connectivity of the product, but without stereochemistry specified: CH3CH2O2C bb
O B OCH2CH3
bb
CH3CH2O2C
O
O
O O
CH3
O
CH3
O
To complete the problem, we must finally take into account two details of the Diels-Alder reaction: (1) Stereochemical relations in the components are retained in the course of the reaction; and (2) unsaturated substituents on the double bond of the dienophile prefer to locate themselves underneath the diene system (in endo positions) in the course of the cycloaddition. Making use of illustrations similar to those in the text, we can visualize the process as follows: CO2CH2CH3 H H
H3C
O H
O O
CH3CH2O2C H O Δ$ H B $ O $ B $ @ H O HC H 3
H
It takes some careful examination to see how the spatial arrangement of the groups in the sketch of CH CH O C 3 2 2 the transition state translates into the final positions in the product. As you work on this aspect of the problem, look in particular at each carbon taking part in the formation of the two new single bonds [dotted lines in transition state (left side of reaction)]. Examine the positions of the substituents on these carbons relative to the two new bonds. A view that may be useful for this purpose is one in which the whole picture is rotated by 908 (see margin). It is reasonably clear from this perspective that the four hydrogen atoms will end up all cis to each other in the newly formed cyclohexene ring. Try Problems 56 and 68 for more practice with mechanisms. 14-31. Because of its superb stereoselectivity, the Diels-Alder reaction is often used as a key step in the construction of acyclic building blocks containing several defined stereocenters. The strategy is to retrosynthetically convert the target into a cyclohexane derivative that is accessible by [4 1 2]cycloaddition. Suggest a synthesis of (racemic) compound A from starting materials containing four carbons or less.
H
CN ≥
H
#
O
CN
O
A
SOLUTION On first sight, this problem seems impossible. The trick is not to focus on the immediate solution, but on the process of retrosynthetic analysis. Thus, the first task is to find a precursor to A that is an appropriately substituted cyclohexene. We can worry about how to make it (by a Diels-Alder reaction) later.
H
H H
H
O O
CH3
O
658
Chapter 14
Delocalized Pi Systems
We note that A is a substituted hexane containing aldehyde functions at the two termini. Is there a retrosynthetic step that will link these two functional carbons to make a six-membered ring? The answer is in Section 12-12: (reverse) ozonolysis, a process that will connect the two carbonyl carbons through a C – C double bond. To draw the resulting cyclohexene B with the correct stereochemistry, it helps to first redraw A in an angular fashion, by rotation around the indicated bonds, before removing the two oxygens and closing the ring. O O O
H
CN H
H H
H
NC CN
CN
O
CN CN
CN
O
CN
O
B
H
This, like any, cyclohexene can be dismantled by a retro-endo Diels-Alder process (Section 14-8), providing the two components C and D of the forward reaction in the required stereochemical form. Dienophile D is readily available (Table 14-1). cis-1,3-Hexadiene, C, will have to be made, and there are many ways by which this task can be accomplished. The crucial feature of CN
B
CN C
D
its construction is the stereochemistry of the double bond. How do we synthesize cis alkenes? The answer is in Section 13-6: the Lindlar-catalyzed hydrogenation of alkynes. Hence, a good precursor of C is enyne E, which can be derived from F and iodoethane (Section 13-5). CH3CH2I C
E
F
But-1-en-3-yne (“vinylacetylene”), F, is made industrially by the CuCl-catalyzed dimerization of ethyne. How would you make it? We learned that alkenes can be converted to alkynes by halogenationdouble dehydrohalogenation (Section 13-4). This approach suggests 1,3-butadiene as a starting material. We have seen, however, that halogenation of 1,3-butadiene, such as bromination, is not as straightforward as that of a monoalkene: 1,2- and 1,4-addition occurs (Section 14-6). Is that a problem? Answer: No. The 1,2-dihalobutene will eliminate normally, first to 2-bromobutadiene (via deprotonation at the more acidic allylic position) and then to F: Br
Br Br
base
base
HBr
HBr
F
The 1,4-dihalobutene will also eliminate, but in a manner that uses the conjugative ability of the double bond, producing 1-bromo-1,3-butadiene as an intermediate, which then goes on to product. ðB H base
Br Br
HBr
Br
HBr
F
On the basis of this analysis, write a forward scheme for the synthesis of A, starting with 1,3butadiene, bromoethane, and D as the sources of carbon.
New Reactions
New Reactions 1. Radical Allylic Halogenations (Section 14-2) Br A RCHCH P CH2
NBS, CCl4, hv
RCH2CH P CH2
RCH P CHCH2Br
DH of allylic C–H bond ~ 87 kcal mol1
2. SN2 Reactivity of Allylic Halides (Section 14-3) CH2 P CHCH2X
Acetone
Nuð
CH2 P CHCH2Nu
X
Faster than ordinary primary halides
3. Allylic Grignard Reagents (Section 14-4) Mg, (CH CH ) O
3 2 2 CH2 P CHCH2Br uvvuuuy CH2 P CHCH2MgBr
Can be used in additions to carbonyl compounds
4. Allyllithium Reagents (Section 14-4) 2 .. CH3CH2CH2CH2Li, TMEDA RCH2CH P CH2 uuuuuuuuy RCHCH P CH2 Li1
pKa of allylic C – H bond < 40
5. Hydrogenation of Conjugated Dienes (Section 14-5) H , Pd–C, CH CH OH
DH° 5 257.1 kcal mol21
2 3 2 CH2 P CH OCH P CH2 uuuuuuuy CH3CH2CH2CH3
but compare H , Pd–C, CH CH OH
DH° 5 260.8 kcal mol21
2 3 2 CH3(CH2)3CH3 CH2 P CH OCH2 OCH P CH2 uuuuuuuy
6. Electrophilic Reactions of 1,3-Dienes: 1,2- and 1,4-Addition (Section 14-6) X A CH2 P CHCHCH3
XCH2CH P CHCH3
X A CH2 P CHCHCH2X
XCH2CH P CHCH2X
HX
CH2 P CH O CH P CH2
X2
CH2 P CH O CH P CH2
7. Thermodynamic Compared with Kinetic Control in SN1 Reactions of Allylic Derivatives (Section 14-6) X A Fast Slow CH3CH P CHCH2 X CH3CHCH P CH2 CH3CH P CHCH2X More stable product
Less stable product
Stability of primary allylic cation ~ ordinary secondary cation
(Formation reversible at higher temperature)
8. Diels-Alder Reaction (Concerted and Stereospecific, Endo Rule) (Section 14-8) R %
R A Δ
A R
0 R
[A )
A
A electron acceptor Requires s-cis diene; better with electron-poor dienophile
Chapter 14
659
660
Chapter 14
Delocalized Pi Systems
9. Electrocyclic Reactions (Section 14-9) Δ
Δ
Conrotatory
Disrotatory
hv
hv
Disrotatory
Conrotatory
10. Polymerization of 1,3-Dienes (Section 14-10) 1,2-Polymerization
Initiator
2n CH2 P CHO CHP CH2
CH2 CH2 B B CH CH A A O (CHO CH2 O CHO CH2)n O
1,4-Polymerization Initiator
n CH2 P CHO CHP CH2
O (CH2 O CH P CH O CH2)n O Cis or trans
11. 3-Methyl-3-butenyl Pyrophosphate as a Biochemical Building Block (Section 14-10) CH3 A CH2 P C O CH2CH2OPP
Enzyme
(CH3)2C PCHCH2OPP
(CH3)2C P CHCH2 Allylic cation
3-Methyl-3-butenyl pyrophosphate
OPP
Pyrophosphate ion
C – C bond formation
OPP
OPP
H
OPP
(OPP)
etc.
Important Concepts 1. The 2-propenyl (allyl) system is stabilized by resonance. Its molecular-orbital description shows the presence of three p molecular levels: one bonding, one nonbonding, and one antibonding. Its structure is symmetric, any charges or odd electrons being equally distributed between the two end carbons. 2. The chemistry of the 2-propenyl (allyl) cation is subject to both thermodynamic and kinetic control. Nucleophilic trapping may occur more rapidly at an internal carbon that bears relatively more positive charge, giving the thermodynamically less stable product. The kinetic product may rearrange to its thermodynamic isomer by dissociation followed by eventual thermodynamic trapping. 3. The stability of allylic radicals allows radical halogenations of alkenes at the allylic position. 4. The SN2 reaction of allylic halides is accelerated by orbital overlap in the transition state. 5. The special stability of allylic anions allows allylic deprotonation by a strong base, such as butyllithium – TMEDA.
Problems
Chapter 14
6. 1,3-Dienes reveal the effects of conjugation by their relative stability (compared to nonconjugated systems) and a relatively short internal bond (1.47 Å). 7. Electrophilic attack on 1,3-dienes leads to the preferential formation of allylic cations. 8. Extended conjugated systems are reactive because they have many sites for attack and the resulting intermediates are stabilized by resonance. 9. Benzene has special stability because of cyclic delocalization. 10. The Diels-Alder reaction is a concerted stereospecific cycloaddition reaction of an s-cis diene to a dienophile; it leads to cyclohexene derivatives. It follows the endo rule. 11. Conjugated dienes and trienes equilibrate with their respective cyclic isomers by concerted and stereospecific electrocyclic reactions. 12. Polymerization of 1,3-dienes results in 1,2- or 1,4-additions to give polymers that are capable of further cross-linking. Synthetic rubbers can be synthesized in this way. Natural rubber is made by electrophilic carbon – carbon bond formation involving biosynthetic five-carbon cations derived from 3-methyl-3-butenyl pyrophosphate. 13. Ultraviolet and visible spectroscopy gives a way of estimating the extent of conjugation in a molecule. Peaks in electronic spectra are usually broad and are reported as Lmax (nm). Their relative intensities are given by the molar absorptivity (extinction coefficient) e.
Problems 32. Draw all resonance forms and a representation of the appropriate resonance hybrid for each of the following species. CH3
%
(a)
(b)
j
(c)
š
š
(d)
(e)
33. For each species in Problem 32, indicate the resonance form that is the major contributor to the resonance hybrid. Explain your choices. 34. Illustrate by means of appropriate structures (including all relevant resonance forms) the initial species formed by (a) breaking the weakest C – H bond in 1-butene; (b) treating 4-methylcyclohexene with a powerful base (e.g., butyllithium – TMEDA); (c) heating a solution of 3-chloro-1-methylcyclopentene in aqueous ethanol. 35. Rank primary, secondary, tertiary, and allylic radicals in order of decreasing stability. Do the same for the corresponding carbocations. Do the results indicate something about the relative ability of hyperconjugation and resonance to stabilize radical and cationic centers? 36. Give the major product(s) of each of the following reactions. OH Conc. HBr
Br
H2O
(a)
(b)
CH3
(c)
Cl CH3 (d) CH2I
O B CH3COH
CH3 (e)
CH3CH2OH
CHP CH2 OH
KSCH3, DMSO
(f) Cl
CH2I
37. Formulate detailed mechanisms for the reactions in Problem 36 (a, c, e, f ). 38. Rank primary, secondary, tertiary, and (primary) allylic chlorides in approximate order of (a) decreasing SN1 reactivity; (b) decreasing SN2 reactivity.
CH3NO2, Δ
661
Delocalized Pi Systems
Chapter 14
39. Rank the following six molecules in approximate order of decreasing SN1 reactivity and decreasing SN2 reactivity. Cl Cl
(b)
(a)
Cl
(c)
Cl (d)
(e)
(f)
Cl
Cl
40. How would you expect the SN2 reactivities of simple saturated primary, secondary, and tertiary chloroalkanes to compare with the SN2 reactivities of the compounds in Problem 39? Make the same comparison for SN1 reactivities. 41. Give the major product(s) of each of the following reactions. H3C I /∑
CH3 ∑ H2O
(a)
NBS, CCl4, ROOR
(b)
∑
∑
H
H
CH3 A (c) (S)-CH3CH2CHCH P CH2 O B 1. CH3CH, THF 2. H , H2O
(d)
(CH3)2C P CH (f) H3C
i f
H
∑/
(e) Product of (d)
GCH2CH3 CP C G G H H
CH3CH2 NBS, CCl4, ROOR
G
662
C
CH3CH2CH2CH2Li, TMEDA
KSCH3, DMSO
Br
42. Write a detailed step-by-step mechanism to show how each of the products arises from the reaction in Problem 41, part (a). 43. The following reaction sequence gives rise to two isomeric products. What are they? Explain the mechanism of their formation. Cl 1. Mg 2. D2O
OH CH3
44. Starting with cyclohexene, propose a reasonable synthesis of the cyclohexene derivative shown in the margin. 45. Give a systematic name to each of the following molecules.
(a)
(b)
OH
Br [ (c)
Br }
(d)
46. Compare the allylic bromination reactions of 1,3-pentadiene and 1,4-pentadiene. Which should be faster? Which is more energetically favorable? How do the product mixtures compare? NBS, ROOR, CCl
4 CH2 P CH O CH P CH O CH3 uuuuuvy
NBS, ROOR, CCl
4 CH2 P CH O CH2 P CH O CH2 uuuuuvy
Problems
663
Chapter 14
47. We learned in Section 14-6 that electrophilic additions to conjugated dienes at low reaction temperatures give kinetic product ratios. Furthermore, these kinetic mixtures may change to mixtures with thermodynamic product ratios when the temperature is raised. Do you think that cooling the thermodynamic product mixture back to the original low reaction temperature will change it back to the original kinetic ratio? Why or why not? 48. Compare the addition of H1 to 1,3-pentadiene and 1,4-pentadiene (see Problem 46). Draw the structures of the products. Draw a qualitative reaction profile showing both dienes and both proton addition products on the same graph. Which diene adds the proton faster? Which one gives the more stable product? 49. What products would you expect from the electrophilic addition of each of the following reagents to 1,3-cycloheptadiene? (a) HI; (b) Br2 in H2O; (c) IN3; (d) H2SO4 in CH3CH2OH. (Hint: For b and c, see Exercise 14-13, specifically drawings A – C). 50. Give the products of the reaction of trans-1,3-pentadiene with each of the reagents in Problem 49. 51. What are the products of reaction of 2-methyl-1,3-pentadiene with each of the reagents in Problem 49? 52. Write a detailed step-by-step mechanism to show how each of the products arises from the reactions in Problem 51, above. 53. Give the products expected from reaction of deuterium iodide (DI) with (a) 1,3-cycloheptadiene; (b) trans-1,3-pentadiene; (c) 2-methyl-1,3-pentadiene. In what way does the observable result of reaction of DI differ from that of reaction of HI with these same substrates [compare with Problems 49(a), 50(a), and 51(a)]? 54. Arrange the following carbocations in order of decreasing stability. Draw all possible resonance forms for each of them. 1
(a) CH2 “ CHO CH2
1
1
(b) CH2 “ C H
1
(d) CH3 OCH “ CHO CHO CH3
(c) CH3 C H2
1
(e) CH2 “ CHO CH “ CHO CH2
55. Sketch the molecular orbitals for the pentadienyl system in order of ascending energy (see Figures 14-2 and 14-7). Indicate how many electrons are present, and in which orbitals, for (a) the radical; (b) the cation; (c) the anion (see Figures 14-3 and 14-7). Draw all reasonable resonance forms for any one of these three species. 56. Dienes may be prepared by elimination reactions of substituted allylic compounds. For example, CH3 A H3C O C P CH O CH2OH
Catalytic H2SO4, Δ
CH3 A H2C P C O CH P CH2
LDA, THF
CH3 A H3C O C P CH O CH2Cl
OH OH
Propose detailed mechanisms for each of these 2-methyl-1,3-butadiene (isoprene) syntheses. 57. Give the structures of all possible products of the acid-catalyzed dehydration of vitamin A (Section 14-7). 58. Propose a synthesis of each of the following molecules by Diels-Alder reactions. O B (a)
[CN ~ CN
H3C (b)
CH3 %
H (c)
≥ CH3
O B [COCH3 )
OH Br Bromoconduritol
O O O
COCH3 B O
59. The haloconduritols are members of a class of compounds called glycosidase inhibitors. These substances possess an array of intriguing biological activity, from antidiabetic and antifungal to activity against HIV virus and cancer metastasis. Stereoisomeric mixtures of bromoconduritol (margin) are commonly used in studies of these properties. One recent synthesis of these compounds proceeds via the bicyclic ethers A and B. (a) Identify the starting materials for a one-step preparation of both of these ethers via a Diels-Alder reaction. (b) Which of the starting molecules in your answer is the diene, and which is the dienophile? (c) This Diels-Alder reaction gives an 80 : 20 mixture of B and A. Explain.
A
O
O O B
O
O
Chapter 14
Delocalized Pi Systems
60. Give the product(s) of each of the following reactions. (a) (b) (c) (d) (e) (f ) (g)
OH
61.
N Pš N G
COCH3 B O
Dimethyl azodicarboxylate
Propose an efficient synthesis of the cyclohexenol in the margin, beginning exclusively with acyclic starting materials and employing sound retrosynthetic analysis strategy. [Hint: A Diels-Alder reaction may be useful, but take note of structural features in dienes and dienophiles that permit Diels-Alder reactions to work well (Section 14-8).]
62. Dimethyl azodicarboxylate (see margin) takes part in the Diels-Alder reaction as a dienophile. Write the structure of the product of cycloaddition of this molecule with each of the following dienes. (a) 1,3-Butadiene; (b) trans,trans-2,4-hexadiene; (c) 5,5-dimethoxycyclopentadiene; (d) 1,2-dimethylenecyclohexane. Ignore the stereochemistry at nitrogen in the products (amines undergo rapid inversion, as we shall see in Section 21-2). 63. Bicyclic diene A reacts readily with appropriate alkenes by the Diels-Alder reaction, whereas diene B is totally unreactive. Explain.
A
B
64. Formulate the expected product of each of the following reactions.
H3CO
/
(b)
(a)
H3C hv
(c)
∑
∑
D
H3C
H
CH3O
CH3 ∑ H
/
D ∑ H
hv
/
O B CH3OC G
3-Chloro-1-propene (allyl chloride) 1 NaOCH3 cis-2-Butene 1 NBS, peroxide (ROOR) 3-Bromo-1-cyclopentene 1 LDA trans,trans-2,4-Hexadiene 1 HCl trans,trans-2,4-Hexadiene 1 Br2, H2O 1,3-Cyclohexadiene 1 methyl propenoate (methyl acrylate) 1,2-Dimethylenecyclohexane 1 methyl propenoate (methyl acrylate)
/
664
Δ
Δ
(d)
CH3
H
65. Which of the reactions shown below will occur under the influence of heat? Light?
(a)
H %
(b)
H ´
(c)
0 H
0 H
66. Explain the following reaction sequence. O B 1. Pd(OCCH3)2, R3P, K2CO3
Br
O B 2. H2C P CH O COCH2CH3
O
B
COCH2CH3
67. Give abbreviated structures of each of the following compounds: (a) (E)-1,4-poly-2-methyl-1,3butadiene [(E)-1,4-polyisoprene]; (b) 1,2-poly-2-methyl-1,3-butadiene (1,2-polyisoprene); (c) 3,4poly-2-methyl-1,3-butadiene (3,4-polyisoprene); (d) copolymer of 1,3-butadiene and ethenylbenzene
Problems
Chapter 14
(styrene, C6H5CH P CH2, SBR, used in automobile tires); (e) copolymer of 1,3-butadiene and propenenitrile (acrylonitrile, CH2 P CHCN, latex); (f ) copolymer of 2-methyl-1,3-butadiene (isoprene) and 2-methylpropene (butyl rubber, for inner tubes).
CH3
68. The structure of the terpene limonene is shown in the margin (see also Exercise 5-29). Identify the two 2-methyl-1,3-butadiene (isoprene) units in limonene. (a) Treatment of isoprene with catalytic amounts of acid leads to a variety of oligomeric products, one of which is limonene. Devise a detailed mechanism for the acid-catalyzed conversion of two molecules of isoprene into limonene. Take care to use sensible intermediates in each step. (b) Two molecules of isoprene may also be converted into limonene by a completely different mechanism, which takes place in the strict absence of catalysts of any kind. Describe this mechanism. What is the name of the reaction? 69.
The carbocation derived from geranyl pyrophosphate (Section 14-10) is the biosynthetic precursor of not only camphor, but also limonene (Problem 68) and a-pinene (Chapter 4, Problem 46). Formulate mechanisms for the formation of the latter two compounds.
70. What is the longest-wavelength electronic transition in each of the following species? Use molecular-orbital designations such as nyp*, p 1yp 2, in your answer. (Hint: Prepare a molecular-orbital energy diagram, like that in Figure 14-16, for each.) (a) 2-Propenyl (allyl) cation; (b) 2-propenyl (allyl) radical; (c) formaldehyde, H2C P O; (d) N2; (e) pentadienyl anion (Problem 55); (f) 1,3,5-hexatriene. 71. Ethanol, methanol, and cyclohexane are commonly used solvents for UV spectroscopy because they do not absorb radiation of wavelength longer than 200 nm. Why not? 72. The ultraviolet spectrum of a 2 3 1024 M solution of 3-penten-2-one exhibits a pyp* absorption at 224 nm with A 5 1.95 and an nyp* band at 314 nm with A 5 0.008. Calculate the molar absorptivities (extinction coefficients) for these bands. 73. In a published synthetic procedure, acetone is treated with ethenyl (vinyl) magnesium bromide, and the reaction mixture is then neutralized with strong aqueous acid. The product exhibits the 1 H NMR spectrum shown below. What is its structure? When the reaction mixture is (improperly) allowed to remain in contact with aqueous acid for too long, an additional new compound is observed. Its 1H NMR spectrum has peaks at d 5 1.70 (s, 3 H), 1.79 (s, 3 H), 2.25 (broad s, 1 H), 4.10 (d, J 5 8 Hz, 2 H), and 5.45 (t, J 5 8 Hz, 1 H) ppm. What is the structure of the second product, and how did it get there?
1H
6H NMR
5.9
5.8
5.0
4.9
4.8
(CH3)4Si
1H
1H
1H
6.0
5.1
1H
5.5
5.0
4.5
4.0
3.5
3.0
2.5
2.0
1.5
300-MHz 1H NMR spectrum ppm (δ )
1.0
0.5
0.0
H 3C
C E N
CH2
Limonene
665
Delocalized Pi Systems
Chapter 14
74.
CH3
Farnesol is a molecule that makes flowers smell good (lilacs, for instance). Treatment with hot concentrated H2SO4 converts farnesol first into bisabolene and finally into cadinene, a compound of the essential oils of junipers and cedars. Propose detailed mechanisms for these conversions. CH3
CH3
H 3C
CH3
H3C
H %
OH Farnesol
^
CH3
^
666
(CH3)2CH H3C
H
CH3
CH3 Cadinene
Bisabolene
75. The ratio of 1,2- to 1,4-addition of Br2 to 1,3-butadiene (Section 14-6) is temperature dependent. Identify the kinetic and thermodynamic products, and explain your choices. O CH3 CH3O O
76. Diels-Alder cycloaddition of 1,3-butadiene with the cyclic dienophile shown in the margin takes place at only one of the two carbon – carbon double bonds in the latter to give a single product. Give its structure and explain your answer. Watch stereochemistry. This transformation was the initial step in the total synthesis of cholesterol (Section 4-7), completed by R. B. Woodward (see Section 14-9) in 1951. This achievement, monumental for its time, revolutionized synthetic organic chemistry.
Team Problem 77. As a team, consider the following historic preparation of a tris(1,1-dimethylethyl) derivative of Dewar benzene, B, by the photochemical isomerisation of 1,2,4-tris(1,1-dimethylethyl)benzene by van Tamelen and Pappas (1962). B does not revert to A via either a thermal or a photochemical electrocyclic mechanism. Formulate a mechanism for the conversion of A to B and explain the kinetic robustness of B with respect to the regeneration of A.
Dewar benzene
B
A
Preprofessional Problems 78. How many nodes are present in the LUMO (lowest unoccupied molecular orbital) of 1,3butadiene? (a) Zero; (b) one; (c) two; (d) three; (e) four 79. Arrange the following three chlorides in decreasing order of SN1 reactivity. CH3CH2CH2Cl
H2C P CHCHCH3 A Cl
CH3CH2CHCH3 A Cl
A
B
C
(a) A . B . C; (b) B . C . A; (c) B . A . C; (d) C . B . A.
Problems
80. When cyclopentadiene is treated with tetracyanoethene, a new product results. Its most likely structure is
CN CN A A (a) HOC O C A A CN CN
CN A H2C O C O CN A A (c) H2C O C O CN A CN
CN (b)
CN CN CN
CN (d)
CN NC CN
81. Which common analytical method will most clearly and rapidly distinguish A from B? CH3
CH3 A
B
(a) IR spectroscopy; (b) UV spectroscopy; (c) combustion analysis; (d) visible spectroscopy.
Chapter 14
667
[
]
Interlude
A Summary of Organic Reaction Mechanisms Although we are only just past the halfway point in our survey of organic chemistry, with the completion of Chapter 14 we have in fact now seen examples of each of the three major classes of organic transformations: radical, polar, and pericyclic processes. This section summarizes all of the individual mechanism types that we have so far encountered in each of these reaction classes.
Radical reactions follow chain mechanisms Radical reactions begin with the generation of a reactive odd-electron intermediate by means of an initiation step and convert starting materials into products through a chain of propagation steps. We have seen both radical substitutions (Chapter 3) and radical additions (Chapter 12). Substitution is capable of introducing a functional group into a previously unfunctionalized molecule; radical addition is an example of functional-group interconversion. These individual subcategories are summarized in Table 1.
Polar reactions constitute the largest class of organic transformations Interactions of polarized, or charged, species lead to the greatest variety in organic chemistry and to the largest number of reaction mechanism types: the typical chemistry of the organic functional groups. Two mechanisms each for substitution and elimination were first presented in Chapters 6 and 7. Both unimolecular and bimolecular pathways were found to be possible for each mechanism, depending on the structure of the substrate and, in some cases, the reaction conditions. With the introduction of functional groups containing p bonds, we encountered polar addition reactions in two distinct forms: nucleophilic in Chapter 8 and electrophilic in Chapter 12. These processes are summarized in Table 2.
Pericyclic reactions lack intermediates The final class of reactions is those characterized by cyclic transition states in which there is continuous overlap of a cyclic array of orbitals. These processes take place in a single step, without the intervention of any intermediate species. They may combine multiple components to introduce new rings, as in the Diels-Alder and other cycloaddition reactions, or they may take the form of ring-opening and ring-closing processes—the electrocyclic reactions. Examples are presented in Table 3. 668
A Summary of Organic Reaction Mechanisms
Table 1 1.
Chapter 14
669
Types of Radical Reactions
Radical Substitution
Mechanism: Radical chain
(Section 3-4)
initiation X OX
Δ or hv
2 Xj
propagation CO H ⫹ Xj
HX ⫹
Cj ⫹ X O X
Cj
CO X ⫹ Xj
termination Examples:
Cj ⫹ Xj
CO X hv
(Alkanes) RH 1 X2 -----¡ RX 1 HX
(Sections 3-4 through 3-8) hv
(Allylic systems) CH2 “CHCH3 1 X2 -----¡ CH2 “CHCH2X 1 HX 2.
(Sections 14-2 and 22-9)
Radical Addition
Mechanism: Radical chain
(Section 12-13)
Examples: Peroxides
(Alkenes) RCH“CH2 1 HBr --------¡ RCH2CH2Br
(Sections 12-13 and 12-15)
Anti-Markovnikov product Peroxides
(Alkynes) RC‚CH 1 HBr --------¡ RCH“CHBr
(Section 13-8)
670
Types of Polar Reactions
Table 2 1.
Delocalized Pi Systems
Chapter 14
Bimolecular Nucleophilic Substitution
Mechanism: Concerted backside displacement (SN2) Nuð⫺
Nu O C
CO X
(Sections 6-2, 6-4, and 6-5)
⫹ X⫺
Example: HO2 1 CH3Cl ¡ CH3OH 1 Cl 2
(Sections 6-2 through 6-9)
100% Inversion at a stereocenter
2.
Unimolecular Nucleophilic Substitution
Mechanism: Carbocation formation–nucleophilic attack (SN1; usually accompanied by E1)
C⫹ ⫹ X⫺
CO X
C⫹
Nuð⫺
(Section 7-2)
Nu O C
Example: H2O 1 (CH3 ) 3CCl ¡ (CH3 ) 3COH 1 HCl
(Sections 7-2 through 7-5)
Racemization at a stereocenter
3.
Bimolecular Elimination
Mechanism: Concerted deprotonation–p-bond formation–expulsion of leaving group (E2) Bð⫺
&¨ C
H C &¨
C PC
X
(Section 7-7)
⫹ HB ⫹ X⫺
Example: CH3CH2O 2 1 CH3CHClCH3 ¡ CH3CH2OH 1 CH3CH“CH2 1 X 2
(Sections 7-7 and 11-6)
Anti transition state preferred
4.
Unimolecular Elimination
Mechanism: Carbocation formation–deprotonation and p-bond formation (E1) accompanies SN1 HO C O C OX
Bð⫺
H O C O C⫹
(Section 7-6)
H O C O C⫹ ⫹ X⫺
C PC
⫹ HB
Example: H2O
(CH3 ) 3CCl ¡ (CH3 ) 2C“CH2 1 HCl
(Sections 7-6 and 11-7)
A Summary of Organic Reaction Mechanisms
671
Types of Polar Reactions [continued]
Table 2 5.
Chapter 14
Nucleophilic Addition
Mechanism: Nucleophilic addition–protonation C PO
Nu)⫺
NuO C O O⫺
(Sections 8-6 and 8-8)
NuO C O O⫺
H⫹
NuO C O OH
Examples: (Hydride reagents) NaBH4 1 (CH3 ) 2C“O ¡ (CH3 ) 2CHOH CH3 A R O C O OH A CH3
(Organometallic reagents) RMgX ⫹ (CH3)2C P O
6.
(Section 8-6)
(Sections 8-8 and 14-4)
Electrophilic Addition
Mechanism: Electrophilic addition–nucleophilic attack
E O C O C⫹
E O C O C⫹
⫺
)Nu
or
ð
C PC
E⫹ D G C C ( (
ð
E⫹
(Sections 12-3 and 12-5)
E O C O CO Nu
Example: Br ƒ (Alkenes) RCH“CH2 1 HBr ¡ RCHCH3 Markovnikov product
(Sections 12-3 through 12-7)
672
Table 3 1.
Delocalized Pi Systems
Chapter 14
Types of Pericyclic Reactions
Cycloaddition
Mechanism: Concerted, through cyclic array of electrons
(Section 14-8)
Example:
CN (Diels-Alder reaction)
⌬
⫹
¥ CN
CN
¥ CN
Stereospecific, endo product preferred
2.
Electrocyclic Reactions
Mechanism: Concerted, through cyclic array of electrons
(Section 14-9)
Examples:
(Cyclobutene
butadiene)
CH ? 3 CH ? 3
CH3
⌬
CH3 Conrotatory thermally CH3 (Hexatriene
cyclohexadiene) CH3
⌬
CH ? 3 CH ? 3 Disrotatory thermally
(Section 14-9)
C H A P T E R
FIFTEEN
Benzene and Aromaticity Electrophilic Aromatic Substitution O O
uring the early nineteenth century, the oil used to illuminate the streets of London and other cities came from rendered whale fat (called “blubber”). Eager to determine its components, in 1825 the English scientist Michael Faraday* heated whale oil to obtain a colorless liquid (b.p. 80.18C, m.p. 5.58C) that had the empirical formula CH. This compound posed a problem for the theory that carbon had to have four valences to other atoms, and it was of particular interest because of its unusual stability and chemical inertness. The molecule was named benzene and was eventually shown to have the molecular formula C6H6. Annual production in the United States alone is 8 million liters per year.
D
Benzene
Benzene has four degrees of unsaturation (see Table 11-6), satisfied by a 1,3,5cyclohexatriene structure (one ring, three double bonds) first proposed by Kekulé† and Loschmidt,‡ yet it does not exhibit the kind of reactivity expected for a conjugated triene. In other words, its structure does not seem to correlate with its observed function. However, benzene is not completely unreactive. For example, it reacts with bromine, albeit only in the *Professor Michael Faraday (1791 – 1867), Royal Institute of Chemistry, London. † Professor F. August Kekulé; see Section 1-4. ‡ Professor Josef Loschmidt (1821 – 1895), University of Vienna, Austria.
O
The structure of vanillin is that of a substituted benzene ring, highlighted in red here and in benzene on the left. Vanillin is the essential component of oil of vanilla, extracted from the fermented bean pods of the vanilla orchid (Vanilla fragrans). It is the world’s second most expensive spice (after saffron), and global demand is estimated at 2000 tons per year.
674
Chapter 15
Benzene and Aromaticity
presence of catalytic amounts of a Lewis acid, such as iron tribromide, FeBr3 (Section 15-9). Surprisingly, the result is not addition but substitution to furnish bromobenzene. H
H
H
H
H
H
H
HBr H
H
H
H
Benzene
Bromobenzene
∑
H
Br OBr
Br ?H ? Br H
H
Br
FeBr3
∑
H
H
(Substitution product)
Addition product not formed
The formation of only one monobromination product is nicely consistent with the sixfold symmetry of the structure of benzene. Further reaction with bromine introduces a second halogen atom to give the three isomeric 1,2-, 1,3-, and 1,4-dibromobenzenes. Br Br
Br
Br
Br2, FeBr3
Br
HBr
Br Br
Br Br 1,2-Dibromobenzene (the same as 1,6-dibromobenzene)
Kekulé’s original model of benzene.
1,3-Dibromobenzene
1,4-Dibromobenzene
Historically, the observation of only one 1,2-dibromobenzene product posed another puzzle. If the molecule has a cyclohexatriene topology with alternating single and double bonds, two isomers should have been formed: 1,2- and 1,6-dibromobenzene, in which the two substituents are connected by double or single carbon – carbon bonds, respectively. Kekulé solved this puzzle brilliantly by boldly proposing that benzene should be viewed as a set of two rapidly equilibrating (he used the word “oscillating”) cyclohexatriene isomers, which would render 1,2- and 1,6-dibromobenzene indistinguishable. We now know that this notion was not quite right. In terms of modern electronic theory, benzene is a single compound, best described by two equivalent cyclohexatrienic resonance forms (Section 14-7). Why does the cyclic array of p electrons in benzene impart unusual stability? How can we quantify this observation? Considering the diagnostic power of NMR (Section 11-4) and electronic spectroscopy (Section 14-11) in establishing the presence of delocalized p systems, can we expect characteristic spectral signals for benzene? In this chapter we shall answer these questions. We look first at the system of naming substituted benzenes, second at the electronic and molecular structure of the parent molecule, and third at the evidence for its unusual stabilizing energy, the aromaticity of benzene. This aromaticity and the special structure of benzene affect its spectral properties and reactivity. We shall see what happens when two or more benzene rings are fused to give more extended p systems. Similarly, we shall compare the properties of benzene with those of other cyclic conjugated polyenes. Finally, we shall study the special mechanism by which substituents are introduced into the benzene ring: electrophilic aromatic substitution.
15-1 Naming the Benzenes is the same as
Benzene and its derivatives were originally called aromatic compounds because many of them have a strong aroma. Benzene, even though its odor is not particularly pleasant, is viewed as the “parent” aromatic molecule. Whenever the symbol for the benzene ring with its three double bonds is written, it should be understood to represent only one of a pair of contributing resonance forms. Alternatively, the ring is sometimes drawn as a regular hexagon with an inscribed circle.
15-1 Naming the Benzenes
Many monosubstituted benzenes are named by adding a substituent prefix to the word “benzene.” F
NO2
Fluorobenzene
H3C
Nitrobenzene
CH3 AA CO CH3
(1,1-Dimethylethyl)benzene (tert-Butylbenzene)
There are three possible arrangements of disubstituted benzenes. These arrangements are designated by the prefixes 1,2- (ortho, or o-) for adjacent substituents, 1,3 (meta-, or m-) for 1,3-disubstitution, and 1,4 (para-, or p-) for 1,4-disubstitution. The substituents are listed in alphabetical order.
Br
CH2CH3
Cl NO2
Cl
CH3CH A CH3 1,2-Dichlorobenzene (o-Dichlorobenzene)
1-Bromo-3-nitrobenzene (m-Bromonitrobenzene)
1-Ethyl-4-(1-methylethyl)benzene ( p-Ethylisopropylbenzene)
To name tri- and more highly substituted derivatives, the six carbons of the ring are numbered so as to give the substituents the lowest set of numbers, as in cyclohexane nomenclature.
CH3 3
CH3
4 1
5
C
4
HC
NO2 1-Bromo-2,3-dimethylbenzene
2
4
3
5
Br
NO2
2
c
6
CH2CH3 3
6
2
5
NO2 1
1,2,4-Trinitrobenzene
1 6
CHP CH2
1-Ethenyl-3-ethyl-5-ethynylbenzene
The following benzene derivatives will be encountered in this book.
CH3
CH3
CHP CH2
CH3
OCH3
CH3 H3C Methylbenzene (Toluene)
1,2-Dimethylbenzene (o-Xylene)
CH3
1,3,5-Trimethylbenzene (Mesitylene)
(Common industrial and laboratory solvents)
Ethenylbenzene (Styrene)
Methoxybenzene (Anisole)
(Used in polymer manufacture)
(Used in perfume)
Chapter 15
675
676
Benzene and Aromaticity
Chapter 15
OH
NH2
O B CCH3
CHO
COOH
Benzenol (Phenol)
Benzenamine (Aniline)
(An antiseptic and anesthetic used in sore throat sprays)
Benzenecarbaldehyde (Benzaldehyde)
1-Phenylethanone (Acetophenone)
Benzenecarboxylic acid (Benzoic acid)
(A hypnotic drug)
(A food preservative)
(Used in dye manufacture) (Artificial almond flavoring)
We shall employ IUPAC nomenclature for all but three systems. In accord with the indexing preferences of Chemical Abstracts, the common names phenol, benzaldehyde, and benzoic acid will be used in place of their systematic counterparts. Ring-substituted derivatives of such compounds are named by numbering the ring positions or by using the prefixes o-, m-, and p-. The substituent that gives the compound its base name is placed at carbon 1.
CH3
CHP CH2
OH Br
I
Br Br Br
1-Iodo-2-methylbenzene (o-Iodotoluene)
2,4,6-Tribromophenol
1-Bromo-3-ethenylbenzene (m-Bromostyrene)
A number of the common names for aromatic compounds refer to their fragrance and natural sources. Several of them have been accepted by IUPAC. As before, a consistent logical naming of these compounds will be adhered to as much as possible, with common names mentioned in parentheses.
Aromatic Flavoring Agents OH
CH2OH
OH
CH3 OCH3
COOCH3
OH CHO
Phenylmethanol (Benzyl alcohol)
Br
∑
CH3
trans-1-(4-Bromophenyl)2-methylcyclohexane
Methyl 2-hydroxybenzoate (Methyl salicylate, oil of wintergreen flavor)
4-Hydroxy-3-methoxybenzaldehyde (Vanillin, vanilla flavor)
CH(CH3)2 5-Methyl-2-(1-methylethyl)phenol (Thymol, thyme flavor)
The generic term for substituted benzenes is arene. An arene as a substituent is referred to as an aryl group, abbreviated Ar. The parent aryl substituent is phenyl, C6H5. The C6H5CH2 – group, which is related to the 2-propenyl (allyl) substituent (Sections 14-1 and 22-1), is called phenylmethyl (benzyl) (see examples in margin).
15-2 Structure and Resonance Energy of Benzene
Chapter 15
677
Exercise 15-1 Write systematic and common names of the following substituted benzenes. OH
CH3
Cl
NO2
D
(b)
(a) NO2
(c) NO2
Exercise 15-2 Draw the structures of (a) (1-methylbutyl)benzene; (b) 1-ethenyl-4-nitrobenzene (p-nitrostyrene); (c) 2-methyl-1,3,5-trinitrobenzene (2,4,6-trinitrotoluene — the explosive TNT; see also Chemical Highlight 16-1).
Exercise 15-3 The following names are incorrect. Write the correct form. (a) 3,5-Dichlorobenzene; (b) o-aminophenyl fluoride; (c) p-fluorobromobenzene.
In Summary Simple monosubstituted benzenes are named by placing the substituent name before the word “benzene.” For more highly substituted systems, 1,2-, 1,3-, and 1,4- (or ortho-, meta-, and para-) prefixes indicate the positions of disubstitution. Alternatively, the ring is numbered, and substituents labeled with these numbers are named in alphabetical order. Many simple substituted benzenes have common names.
15-2 Structure and Resonance Energy of Benzene: A First Look at Aromaticity Benzene is unusually unreactive: At room temperature, benzene is inert to acids, H2, Br2, and KMnO4 — reagents that readily add to conjugated alkenes (Section 14-6). The reason for this poor reactivity is that the cyclic six-electron arrangement imparts a special stability in the form of a large resonance energy (Section 14-7). We shall first review the evidence for the structure of benzene and then estimate its resonance energy by comparing its heat of hydrogenation with those of model systems that lack cyclic conjugation, such as 1,3-cyclohexadiene.
The benzene ring contains six equally overlapping p orbitals If benzene were a conjugated triene, a “cyclohexatriene,” we should expect the C–C bond lengths to alternate between single and double bonds. In fact, experimentally, the benzene molecule is a completely symmetrical hexagon (Figure 15-1), with equal C–C bond lengths of 1.39 Å, between the values found for the single bond (1.47 Å) and the double bond (1.34 Å) in 1,3-butadiene (Figure 14-6). Figure 15-2 shows the electronic structure of the benzene ring. All carbons are sp2 hybridized, and each p orbital overlaps to an equal extent with its two neighbors. The resulting delocalized electrons form circular p clouds above and below the ring. The
1.09 Å
1.39 Å 120°
H
H 120°
120°
Figure 15-1 The molecular structure of benzene. All six C – C bonds are equal; all bond angles are 1208.
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Benzene and Aromaticity
Chapter 15
Benzene π bond
H1s−Csp2
H
π Cloud
H H H
H
H
C H
Csp2−Csp2
H
C
C
C
C
H C H
H
H B
A
π Cloud
C
Figure 15-2 Orbital picture of the bonding in benzene. (A) The s framework is depicted as straight lines except for the bonding to one carbon, in which the p orbital and the sp2 hybrids are shown explicitly. (B) The six overlapping p orbitals in benzene form a p-electron cloud located above and below the molecular plane. (C) The electrostatic potential map of benzene shows the relative electron richness of the ring and the even distribution of electron density over the six carbon atoms.
symmetrical structure of benzene is a consequence of the interplay between the s and p electrons in the molecule. The symmetrical s frame acts in conjunction with the delocalized p frame to enforce the regular hexagon.
Benzene is especially stable: heats of hydrogenation A way to establish the relative stability of a series of alkenes is to measure their heats of hydrogenation (Sections 11-5 and 14-5). We may carry out a similar experiment with benzene, relating its heat of hydrogenation to those of 1,3-cyclohexadiene and cyclohexene. These molecules are conveniently compared because hydrogenation changes all three into cyclohexane. The hydrogenation of cyclohexene is exothermic by 228.6 kcal mol21, a value expected for the hydrogenation of a cis double bond (Section 11-5). The heat of hydrogenation of 1,3-cyclohexadiene (DH 8 5 254.9 kcal mol21) is slightly less than double that of cyclohexene because of the resonance stabilization in a conjugated diene (Section 14-5); the energy of that stabilization is (2 3 28.6) 2 54.9 5 2.3 kcal mol21 (9.6 kJ mol21).
H2
2 H2
Catalytic Pt
H 28.6 kcal mol1 (120 kJ mol1)
Catalytic Pt
H 54.9 kcal mol1 (230 kJ mol1)
Armed with these numbers, we can calculate the expected heat of hydrogenation of benzene, as though it were simply composed of three double bonds like that of cyclohexene, but with the extra resonance stabilization of conjugation as in 1,3-cyclohexadiene.
3 H2
Catalyst
H 3 (H of hydrogenation of
H ? ) 3 (resonance correction in
(3 28.6) (3 2.3) kcal mol 85.8 6.9 kcal mol1 78.9 kcal mol1 (330 kJ mol1)
1
)
15-3 Pi Molecular Orbitals of Benzene
Chapter 15
+ 3 H2 “1,3,5-Cyclohexatriene” + 2 H2
3 H2 +
1,3-Cyclohexadiene
Benzene
E + H2 Cyclohexene Measured heat of hydrogenation 28.6 kcal mol–1 (120 kJ mol–1)
Measured heat of hydrogenation 54.9 kcal mol–1 (230 kJ mol–1)
Resonance energy = 29.6 kcal mol–1 (124 kJ mol–1)
Estimated heat of hydrogenation 78.9 kcal mol–1 (330 kJ mol–1)
Measured heat of hydrogenation 49.3 kcal mol–1 (206 kJ mol–1)
Cyclohexane Figure 15-3 Heats of hydrogenation provide a measure of benzene’s unusual stability. Experimental values for cyclohexene and 1,3-cyclohexadiene allow us to estimate the heat of hydrogenation for the hypothetical “1,3,5-cyclohexatriene” (see Chemical Highlight 15-2). Comparison with the experimental DH8 for benzene gives a value of approximately 29.6 kcal mol21 for the aromatic resonance energy.
Now let us look at the experimental data. Although benzene is hydrogenated only with difficulty (Section 14-7), special catalysts carry out this reaction, so the heat of hydrogenation of benzene can be measured: DH 8 5 249.3 kcal mol21, much less than the 278.9 kcal mol21 predicted. Figure 15-3 summarizes these results. It is immediately apparent that benzene is much more stable than a cyclic triene containing alternating single and double bonds. The difference is the resonance energy of benzene, about 30 kcal mol21 (126 kJ mol21). Other terms used to describe this quantity are delocalization energy, aromatic stabilization, or simply the aromaticity of benzene. The original meaning of the word aromatic has changed with time, now referring to a thermodynamic property rather than to odor.
In Summary The structure of benzene is a regular hexagon made up of six sp2-hybridized carbons. The C – C bond length is between those of a single and a double bond. The electrons occupying the p orbitals form a p cloud above and below the plane of the ring. The structure of benzene can be represented by two equally contributing cyclohexatriene resonance forms. Hydrogenation to cyclohexane releases about 30 kcal mol21 less energy than is expected on the basis of nonaromatic models. This difference is the resonance energy of benzene.
15-3 Pi Molecular Orbitals of Benzene We have just examined the atomic orbital picture of benzene. Now let us look at the molecular orbital picture, comparing the six p molecular orbitals of benzene with those of 1,3,5-hexatriene, the open-chain analog. Both sets are the result of the contiguous overlap of six p orbitals, yet the cyclic system differs considerably from the acyclic one. A comparison of the energies of the bonding orbitals in these two compounds shows that cyclic conjugation of three double bonds is better than acyclic conjugation.
679
680
Benzene and Aromaticity
Chapter 15
+ −
−
+
− +
+ −
+
+ − − +
−
− +
+ +
− −
−
+
+
+
−
+
ψ6
−
−
−
+ −
−
− +
− +
+
+
+ −
+
+
−
+ −
ψ5
ψ4
−
−
−
+
+
+
E Node
Node
Node
−
−
+ +
+ +
+ −
−
−
+
−
+ −
− +
−
+
− − +
−
ψ3
+ −
+ + − −
+ + − −
ψ1 Benzene
−
−
+
+
Node
ψ2
+
+
+
−
−
+ + −
−
+
+
−
−
−
+
+
−
− π6
+ +
−
−
+
+ π5
− +
+
−
−
− π4
+ −
+
+
−
+ π3
− −
−
+
+
− π2
+ +
+
−
−
+ π1
− 1,3,5-Hexatriene
Figure 15-4 Pi molecular orbitals of benzene compared with those of 1,3,5-hexatriene. The orbitals are shown at equal size for simplicity. Favorable overlap (bonding) takes place between orbital lobes of equal sign. A sign change is indicated by a nodal plane (dashed line). As the number of such planes increases, so does the energy of the orbitals. Note that benzene has two sets of degenerate (equal energy) orbitals, the lower energy set occupied (c2, c3), the other not (c4, c5), as shown in Figure 15-5.
Cyclic overlap modifies the energy of benzene’s molecular orbitals Figure 15-4 compares the p molecular orbitals of benzene with those of 1,3,5-hexatriene. The acyclic triene follows a pattern similar to that of 1,3-butadiene (Figure 14-7), but with two more molecular orbitals: The orbitals all have different energies, the number of nodes increasing in the procession from p1 to p6. The picture for benzene is different in all respects: different orbital energies, two sets of degenerate (equal energy) orbitals, and completely different nodal patterns. Is the cyclic p system more stable than the acyclic one? To answer this question, we have to compare the combined energies of the three filled bonding orbitals in both. Figure 15-5 shows the answer: The cyclic p system is stabilized relative to the acyclic one. On going from 1,3,5-hexatriene to benzene, two of the bonding orbitals (p1 and p3) are lowered in energy and one is raised; the effect of the one raised in energy is more than offset by the two that are lowered.
15-3 Pi Molecular Orbitals of Benzene
Cumulative lowering of orbital energies
Antibonding molecular orbitals
E
Bonding molecular orbitals
Inspection of the signs of the wave functions at the terminal carbons of 1,3,5-hexatriene (Figure 15-4) explains why the orbital energies change in this way: Linkage of C1 and C6 causes p orbital overlap that is in phase for p1 and p3 but out of phase for p2.
Some reactions have aromatic transition states The relative stability of six cyclically overlapping orbitals accounts in a simple way for several reactions that proceed readily by what has seemed to be a complicated, concerted movement of three electron pairs: the Diels-Alder reaction (Section 14-8), osmium tetroxide addition to alkenes (Section 12-11), and the first step of ozonolysis (Section 12-12). In all three processes, a transition state occurs with cyclic overlap of six electrons in p orbitals (or orbitals with p character). This electronic arrangement is similar to that in benzene and is energetically more favored than the alternative, sequential bond breaking and bond making. Such transition states are called aromatic.
Aromatic Transition States C O VIII O M J Os J M O O
O
C
GJ O
O
Diels-Alder reaction
Osmium tetroxide addition
Ozonolysis
Exercise 15-4 If benzene were a cyclohexatriene, 1,2-dichloro- and 1,2,4-trichlorobenzene should each exist as two isomers. Draw them.
Chapter 15
681
Figure 15-5 Energy levels of the p molecular orbitals in benzene and 1,3,5-hexatriene. In both, the six p electrons fill the three bonding molecular orbitals. In benzene, two of them are lower in energy and one is higher than the corresponding orbitals in 1,3,5-hexatriene. Overall, energy is reduced and stability increased in going from the acyclic to the cyclic system.
682
Chapter 15
Benzene and Aromaticity
Exercise 15-5 A
B
The thermal ring opening of cyclobutene to 1,3-butadiene is exothermic by about 10 kcal mol21 (Section 14-9). Conversely, the same reaction for benzocyclobutene, A, to compound B (shown in the margin) is endothermic by the same amount. Explain.
In Summary Two of the filled p molecular orbitals of benzene are lower in energy than are those in 1,3,5-hexatriene. Benzene is therefore stabilized by considerably more resonance energy than is its acyclic analog. A similar orbital structure also stabilizes aromatic transition states.
H ~ 10 kcal mol1
15-4 Spectral Characteristics of the Benzene Ring The electronic arrangement in benzene and its derivatives gives rise to characteristic ultraviolet spectra. The hexagonal structure is also manifest in specific infrared bands. What is most striking, cyclic delocalization causes induced ring currents in NMR spectroscopy, resulting in unusual deshielding of hydrogens attached to the aromatic ring. Moreover, the different coupling constants of 1,2- (ortho), 1,3- (meta), and 1,4- (para) hydrogens in substituted benzenes indicate the substitution pattern.
The UV – visible spectrum of benzene reveals its electronic structure
Suntan lotions are applied to protect the skin from potential cancercausing high-energy UV radiation. They contain substances, such as PABA, that absorb light in the damaging region of the electromagnetic spectrum.
COOH
NH2
Cyclic delocalization in benzene gives rise to a characteristic arrangement of energy levels for its molecular orbitals (Figure 15-4). In particular, the energetic gap between bonding and antibonding orbitals is relatively large (Figure 15-5). Is this manifested in its electronic spectrum? As shown in Section 14-11, the answer is yes: Compared with the spectra of the acyclic trienes, smaller lmax values are expected for benzene and its derivatives. You can verify this effect in Figure 15-6: The highest-wavelength absorption for benzene occurs at 261 nm, closer to that of 1,3-cyclohexadiene (259 nm, Table 14-3) than to that of 1,3,5hexatriene (268 nm). The ultraviolet and visible spectra of aromatic compounds vary with the introduction of substituents; this phenomenon has been exploited in the tailored synthesis of dyes (Section 22-11). Simple substituted benzenes absorb between 250 and 290 nm. For example, the water-soluble 4-aminobenzoic acid ( p-aminobenzoic acid, or PABA, see margin), 4-H2N – C6H4 – CO2H, has a lmax at 289 nm, with a rather high extinction coefficient of 18,600. Because of this property, it is used in suntan lotions, in which it filters out the dangerous ultraviolet radiation emanating from the sun in this wavelength region. Some individuals cannot tolerate PABA because of an allergic skin reaction, and many recent sun creams use other compounds for protection from the sun (“PABA-free”).
4-Aminobenzoic acid ( p-Aminobenzoic acid, PABA)
Benzene
Absorbance (A)
Figure 15-6 The distinctive ultraviolet spectra of benzene: lmax(e ) 5 234(30), 238(50), 243(100), 249(190), 255(220), 261(150) nm; and 1,3,5-hexatriene: lmax(e) 5 247(33,900), 258(43,700), 268(36,300) nm. The extinction coefficients e of the absorptions of 1,3,5-hexatriene are very much larger than those of benzene; therefore the spectrum at the right was taken at lower concentration.
Absorbance (A)
1,3,5-Hexatriene
UV
UV
230 240 250 260 270 280 Wavelength (nm)
230 240 250 260 270 280 Wavelength (nm)
15-4 Spectral Characteristics of the Benzene Ring
Chapter 15
683
The infrared spectrum reveals substitution patterns in benzene derivatives The infrared spectra of benzene and its derivatives show characteristic bands in three regions. The first is at 3030 cm21 for the phenyl – hydrogen stretching mode. The second ranges from 1500 to 2000 cm21 and includes aromatic ring C – C stretching vibrations. Finally, a useful set of bands due to C – H out-of-plane bending motions is found between 650 and 1000 cm21. Typical Infrared C–H Out-of-Plane Bending Vibrations for Substituted Benzenes (cm1) R
R
R
R
R R R 690–710 730–770
735–770
690–710 750–810
790–840
Their precise location indicates the specific substitution pattern. For example, 1,2dimethylbenzene (o-xylene) has this band at 738 cm21, the 1,4 isomer at 793 cm21; the 1,3 isomer (Figure 15-7) shows two absorptions in this range, at 690 and 765 cm21. Figure 15-7 The infrared spectrum of 1,3-dimethylbenzene (m-xylene). There are two C – H stretching absorptions, one due to the aromatic bonds (3030 cm21), the other to saturated C – H bonds (2920 cm21). The two bands at 690 and 765 cm21 are typical of 1,3-disubstituted benzenes.
Transmittance (% )
100
H
H H C
H
H H
H H
IR 0 4000 3500
C
H
H 3000
2500 2000 1500 Wavenumber
1000
600 cm −1
The mass spectrum of benzene indicates stability Figure 15-8 depicts the mass spectrum of benzene. Noticeable is the lack of any significant fragmentation, attesting to the unusual stability of the cyclic six-electron framework (Section 15-2). The (M 1 1)1• peak has a relative height of 6.8%, as expected for the relative abundance of 13C in a molecule made up of six carbon atoms.
The NMR spectra of benzene derivatives show the effects of an electronic ring current 1
H NMR is a powerful spectroscopic technique for the identification of benzene and its derivatives. The cyclic delocalization of the aromatic ring gives rise to unusual deshielding, which causes the ring hydrogens to resonate at very low field (d < 6.5 – 8.5 ppm), even lower than the already rather deshielded alkenyl hydrogens (d < 4.6 – 5.7 ppm, see Section 11-4).
Chemical Shifts of Allylic and Benzylic Hydrogens CH2 P CHO CH3 Allylic: 1.68 ppm
CH3 Benzylic: 2.35 ppm
684
Chapter 15
Benzene and Aromaticity
Figure 15-8 The mass spectrum of benzene reveals very little fragmentation.
M⫹• 78
MS
Relative abundance
100
Molecular ion
50
0 0
20
40
60
80
100
120
m/z
The 1H NMR spectrum of benzene, for example, exhibits a sharp singlet for the six equivalent hydrogens at d 5 7.27 ppm. How can this strong deshielding be explained? In a simplified picture, the cyclic p system with its delocalized electrons may be compared to a loop of conducting metal. When such a loop is exposed to a perpendicular magnetic field (H0), the electrons circulate (called a ring current), generating a new local magnetic field (hlocal). This induced field opposes H0 on the inside of the loop (Figure 15-9), but it reinforces H0 on the outside — just where the hydrogens are located. Such reinforcement results in deshielding. The effect is strongest close to the ring and diminishes rapidly with increasing distance from it. Thus, benzylic nuclei are deshielded only about 0.4 to 0.8 ppm more than their allylic counterparts, and hydrogens farther away from the p system have chemical shifts that do not differ much from each other and are similar to those in the alkanes. Opposes H0 in this region of space hlocal
Ring current of circulating electrons
Induced local magnetic field, hlocal
hlocal
H
H
Strengthens H0 in this region of space
Figure 15-9 The p electrons of benzene may be compared to those in a loop of conducting metal. Exposure of this loop of electrons to an external magnetic field H0 causes them to circulate. This “ring current” generates a local field, reinforcing H0 on the outside of the ring. Thus, the hydrogens resonate at lower field.
hlocal
Strengthens H0 in this region of space
hlocal Opposes H0 in this region of space
External field, H0
15-4 Spectral Characteristics of the Benzene Ring
Whereas benzene exhibits a sharp singlet in its NMR spectrum, substituted derivatives may have more complicated patterns. For example, introduction of one substituent renders the hydrogens positioned ortho, meta, and para nonequivalent and subject to mutual coupling. An example is the NMR spectrum of bromobenzene, in which the signal for the ortho hydrogens is shifted slightly downfield relative to that of benzene. Moreover, all the protons are coupled to one another, thus giving rise to a complicated spectral pattern (Figure 15-10). Figure 15-11 shows the NMR spectrum of 4-(N,N-dimethylamino)benzaldehyde. The large chemical shift difference between the two sets of hydrogens on the ring results in a near-first-order pattern of two doublets. The observed coupling constant is 9 Hz, a typical splitting between ortho protons.
1H NMR
Br H
H
H
H H
7.3 7.2 ppm
7.6 7.5 7.4 ppm
2H
1H NMR
685
Chapter 15
3H
CHO H
H 9
8
7
6
ppm (δ ) H
H
6H
N(CH3)2
7.8 7.7 ppm
9.8 9.7 ppm
(CH3)4Si 2H
1H
9
Figure 15-10 A part of the 300-MHz 1H NMR spectrum of bromobenzene, a non-first-order spectrum.
8
6.8 6.7 6.6 ppm
2H
7
6
5
4
3
2
1
0
ppm (δ )
Figure 15-11 300-MHz 1H NMR spectrum of 4-(N,N-dimethylamino) benzaldehyde ( p-dimethylaminobenzaldehyde). In addition to the two aromatic doublets (J 5 9 Hz), there are singlets for methyl (d 5 3.09 ppm) and the aldehydic hydrogens (d 5 9.75 ppm).
A first-order spectrum revealing all three types of coupling is shown in Figure 15-12. 1-Methoxy-2,4-dinitrobenzene (2,4-dinitroanisole) bears three ring hydrogens with different chemical shifts and distinct splittings. The hydrogen ortho to the methoxy group appears at
1H NMR
OCH3
8.5 8.4 ppm 8.8 8.7 ppm
7.3 7.2 ppm
1H 1H 9
8
H
NO2
H
H NO2
3H
(CH3)4Si
1H
7
6
5
4
ppm (δ )
3
2
1
0
Figure 15-12 300-MHz 1H NMR spectrum of 1-methoxy-2,4dinitrobenzene (2,4-dinitroanisole).
686
Chapter 15
Aromatic Coupling Constants Jortho 5 6 – 9.5 Hz Jmeta 5 1.2 – 3.1 Hz Jpara 5 0.2 – 1.5 Hz 13C
NMR Data of Two Substituted Benzenes (ppm) H3 C
21.3
137.8
129.3
Benzene and Aromaticity
d 5 7.23 ppm as a doublet with a 9-Hz ortho coupling. The hydrogen flanked by the two nitro groups (at d 5 8.76 ppm) also appears as a doublet, with a small (3-Hz) meta coupling. Finally, we find the remaining ring hydrogen absorbing at d 5 8.45 ppm as a doublet of doublets because of simultaneous coupling to the two other ring protons. Para coupling between the hydrogens at C3 and C6 is too small (,1 Hz) to be resolved; it is evident as a slight broadening of the resonances of these protons. In contrast with 1H NMR, 13C NMR chemical shifts in benzene derivatives are dominated by hybridization and substituent effects. Because the induced ring current flows directly above and below the aromatic carbons (Figure 15-9), they are less affected by it. Moreover, the relatively large 13C chemical shift range, about 200 ppm, makes ring current contributions (only a few ppm) less noticeable. Therefore, benzene carbons exhibit chemical shifts similar to those in alkenes, between 120 and 135 ppm when unsubstituted (see margin). Benzene itself exhibits a single line at d 5 128.7 ppm. Exercise 15-6
128.5
Can the three isomeric trimethylbenzenes be distinguished solely on the basis of the number of peaks in their proton-decoupled 13C NMR spectra? Explain.
125.6
NO2 148.3
Exercise 15-7
Working with the Concepts: Using Spectral Data to Assign the Structure of a Substituted Benzene
123.4 129.5 134.7
A hydrocarbon shows a molecular-ion peak in the mass spectrum at myz 134. The associated [M 1 1] ion at myz 135 has about 10% of the intensity of the molecular ion. The largest peak in the spectrum is a fragment ion at myz 119. The other spectral data for this compound are as follows: 1H NMR (90 MHz) d 5 7.02 (broad s, 4 H), 2.82 (septet, J 5 7.0 Hz, 1 H), 2.28 (s, 3 H), and 1.22 ppm (d, J 5 7.0 Hz, 6 H); 13C NMR d 5 21.3, 24.2, 38.9, 126.6, 128.6, 134.8, and 145.7 ppm; IR n˜ 5 3030, 2970, 2880, 1515, 1465, and 813 cm–1; UV lmax(e) 5 265(450) nm. What is its structure? Strategy The spectra supply a lot of information, so your first problem is Where do I start? Typically, researchers look at the most important data first — the mass and the 1H NMR spectrum — and then use the other spectra as corroborating evidence. The following solution employs only one of several strategies that you could apply. It is often not necessary to dwell in such depth on the information embedded in the individual spectra, and you may find it more useful first to identify the most important pieces, such as the molecular ion in the mass spectrum, the spectral regions and/or splitting patterns in the 1H NMR spectrum, the number of 13C signals, and a characteristic IR frequency or UV absorption. Whenever you are ready to formulate a (sub)structure, you should do so, and keep validating (or discarding) it as you analyze additional data. Trial and error is the key! Solution • The mass spectrum is already quite informative. There are not many hydrocarbons with a molecular ion of myz 134 for which a sensible molecular formula can be derived. For example, the maximum number of carbons has to be less than C12 (mass 144, too large). If it were C11, the molecular formula would have to be C11H2, ruled out by a quick look at the 1H NMR spectrum, which shows more than two hydrogens in the molecule. (Besides, how many structures can you formulate with the composition C11H2?) The next choice, C10H14, looks good, especially since going down once again in the number of carbons to C9 would require the presence of 26 hydrogens, an impossible proposition, because the maximum number of hydrogens is dictated by the generic formula CnH2n12 (Section 2-4). • C10H14 means four degrees of unsaturation (Section 11-11), suggesting the presence of a combination of rings, double bonds, and triple bonds. The last option is less likely because of the absence of a band at ,2200 cm–1 in the IR spectrum (Section 13-3). This leaves us with double bonds and/or rings. • A check of the 1H NMR spectrum does not show signals for normal double bonds, but instead an aromatic peak. A benzene ring has the postulated four degrees of unsaturation. Hence we are dealing with an aromatic compound.
15-5 Polycyclic Aromatic Hydrocarbons
• Looking at the fragmentation pattern in the mass spectrum, the mass ion at m/z 119 points to the loss of a methyl group, which must therefore be present in our molecule. Turning to the 1H NMR spectrum in more detail and remembering that we have 14 hydrogens in our structure, we can now probe how these hydrogens are distributed. The 1H NMR spectrum reveals 4 H in the aromatic region, all very close in chemical shift, giving rise to a broad singlet. The remainder of the hydrogen signals appear to be due to alkyl groups: One of them is clearly a methyl substituent at d 5 2.28 ppm; the others are part of a more complex array, containing two sets of mutually coupled nuclei. Closer inspection of this array shows it to be a septet (indicating six equivalent neighbors) and a doublet (indicating a single hydrogen as a neighbor), pointing to the presence of a 1-methylethyl substituent (Table 10-5). The relatively large d values for the single methyl and the tertiary hydrogen of the 1-methylethyl groups suggest that they are close to the aromatic ring (Table 10-2). • Moving on to the 13C NMR spectrum, the preceding analysis requires the presence of (at least) three sp3-hybridized carbon peaks (Table 10-6). Indeed, there are three signals at d 5 21.3, 24.2, and 38.9 ppm, and no additional peaks in this region. Instead, we see four resonances in the aromatic range. Since a benzene ring has six carbons, there must be some symmetry to the molecule. • The IR spectrum reveals the presence of Caromatic2H units (n˜ 5 3030 cm21), and the signal at n˜ 5 813 cm21 indicates a para disubstituted benzene. • Finally, the electronic spectrum shows the presence of a conjugated system, obviously a benzene ring. • Putting all of this information together suggests a benzene ring with two substituents, a methyl and a 1-methylethyl group. The 13C NMR spectrum rules out ortho or meta disubstitution, leaving only 1-methyl-4-(1-methylethyl)benzene as the solution (see margin).
Exercise 15-8
Try It Yourself Deduce the structure of a compound C16H16Br2 that exhibits the following spectral data: UV lmax (log e) 5 226sh(4.33), 235sh(4.55), 248(4.78), 261(4.43), 268(4.44), 277(4.36) nm (sh 5 shoulder); IR n˜ 5 3030, 2964, 2933, 2233, 1456, 1362, 1060, 892, 614 cm21; 1H NMR (300 MHz) d 5 7.75 (s, 1 H), 7.45 (s, 1 H), 2.41 (t, J 5 7.0 Hz, 4 H), 1.64 (sextet, J 5 7.0, 4 H), 1.06 (t, J 5 7.4 Hz, 6 H); 13C NMR d 5 13.51, 21.54, 21.89, 78.26, 96.68, 124.4, 125.2, 135.2, 136.8 ppm. (Caution: Look closely at the NMR spectra with reference to the molecular formula. Moreover, the IR spectrum contains important information. Hint: There is symmetry in the molecule.)
In Summary Benzene and its derivatives can be recognized and structurally characterized by their spectral data. Electronic absorptions take place between 250 and 290 nm. The infrared vibrational bands are found at 3030 cm21 (Caromatic – H), from 1500 to 2000 cm21 (C – C), and from 650 to 1000 cm21 (C – H out-of-plane bending). Most informative is NMR, with low-field resonances for the aromatic hydrogens and carbons. Coupling is largest between the ortho hydrogens, smaller in their meta and para counterparts.
15-5 Polycyclic Aromatic Hydrocarbons When several benzene rings are fused together to give more extended p systems, the molecules are called polycyclic benzenoid or polycyclic aromatic hydrocarbons (PAHs). In these structures, two or more benzene rings share two or more carbon atoms. Do these compounds also enjoy the special stability of benzene? The next two sections will show that they largely do. There is no simple system for naming these structures, so we shall use their common names. The fusion of one benzene ring to another results in a compound called naphthalene. Further fusion can occur in a linear manner to give anthracene, tetracene, pentacene, and so on,
Chapter 15
(CH3)2CH A
A CH3 1-Methyl-4(1-methylethyl)benzene
687
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Chapter 15
Benzene and Aromaticity
CHE M I C AL H I G H L I G H T 1 5 - 1 The Allotropes of Carbon: Graphite, Diamond, and Fullerenes Elements can exist in several forms, called allotropes, depending on conditions and modes of synthesis. Thus, elemental carbon can arrange in more than 40 configurations, most of them amorphous (i.e., noncrystalline), such as coke (Sections 3-3 and 13-10), soot, carbon black (as used in printing ink), and activated carbon (as used in air and water filters). You probably know best two crystalline modifications of carbon: graphite and diamond. Graphite, the most stable carbon allotrope, is a completely fused polycyclic benzenoid p system, consisting of layers arranged in an open honeycomb pattern and 3.35 Å apart. The fully delocalized nature of these sheets (all carbons are sp2 hybridized) gives rise to their black color and conductive capability. Graphite’s lubricating property is the result of the ready mutual sliding of its component planes. The “lead” of pencils is graphitic carbon, and the
⬚ 3.35〈
⬚ 3.35〈
Graphite
Diamond
black pencil marks left on a sheet of paper consist of rubbed-off layers of the element. In the colorless diamond, the carbon atoms (all sp3 hybridized) form an insulating network of cross-linked cyclohexane chair conformers. Diamond is the densest and hardest (least deformable) material known. It is also less stable than graphite, by 0.45 kcal/g C atom, and transforms into graphite at high temperatures or when subjected to high-energy radiation, a little-appreciated fact in the jewelry business. A spectacular discovery was made in 1985 by Curl, Kroto, and Smalley* (for which they received the Nobel Prize in 1996): buckminsterfullerene, C60, a new, spherical allotrope of carbon in the shape of a soccer ball. They found that laser evaporation of graphite generated a variety of carbon clusters in the gas phase, the most abundant of which contained 60 carbon atoms. The best way of assembling such a cluster while satisfying the tetravalency of carbon is to formally “roll up” 20 fused benzene rings and to connect the dangling valencies in such a way as to generate 12 pentagons: a so-called truncated icosahedron with 60 equivalent vertices — the shape of a soccer ball. The molecule was named after Buckminster Fuller† because its shape is reminiscent of the “geodesic domes” designed by him. It is soluble in organic solvents, greatly aiding in the proof of its structure and the exploration of its chemistry. For example, the 13 C NMR spectrum shows a single line at d 5 142.7 ppm, in the expected range (Sections 15-4 and 15-6). Because of its curvature, the constituent benzene rings in C60 are strained and the energy content relative to graphite is 10.16 kcal/g C atom. This strain is manifested in a rich chemistry, including electrophilic, nucleophilic, radical, and concerted addition reactions (Chapter 14). The enormous interest spurred by the discovery of C60 rapidly led to a number of exciting developments, such as the design of multigram synthetic methods (commercial material sells for as low as $1 per gram); the isolation of many other larger carbon clusters, dubbed “fullerenes,”
*Professor Robert F. Curl (b. 1933), Rice University, Houston, Texas; Professor Harold W. Kroto (b. 1939), University of Sussex, England; Professor Richard E. Smalley (1943–2005), Rice University, Houston, Texas. † Richard Buckminster Fuller (1895 – 1983), American architect, inventor, and philosopher.
Buckminsterfullerene C60
15-5 Polycyclic Aromatic Hydrocarbons
Chapter 15
C 70
Chiral C84
such as the rugby-ball – shaped C70; chiral systems (e.g., as in C84); isomeric forms; fullerenes encaging host atoms, such as He and metal nuclei (“endohedral fullerenes”); and the synthesis of conducting salts (e.g., Cs3C60, which becomes superconducting at 40 K). Moreover, reexamination of the older literature and newer studies have revealed that C60 and other fullerenes are produced simply upon incomplete combustion of organic matter under certain conditions or by varied heat treatments of soot and therefore have probably been “natural products” on our planet since early in its formation. From a materials point of view, perhaps most useful has been the synthesis of graphitic tubules, so-called nanotubes, based on the fullerene motif. Nanotubes are even harder than diamond, yet elastic, and show unusual magnetic and electrical (metallic) properties. There is the real prospect that nanotubes may replace the computer chip as we currently know it in the manufacture of a new generation of faster and smaller computers (see also Chemical Highlight 14-2). Nanotubes also function as a molecular “packaging material” for other structures, such as metal catalysts and even biomolecules. Thus, carbon in the fullerene modification has taken center stage in the new field of nanotechnology, aimed at the construction of devices at the molecular level.
An example of a geodesic dome (of the type whose design was pioneered by Buckminster Fuller) forms part of the entrance to EPCOT Center, Disney World, Florida. It is 180 feet high with a diameter of 165 feet.
Carbon nanotube
689
690
Benzene and Aromaticity
Chapter 15
a series called the acenes. Angular fusion (“annulation”) results in phenanthrene, which can be further annulated to a variety of other benzenoid polycycles. 6 7
5 1
8 8a
2 3
4a 4
5
1 7
2
6
3
8 8a
4a 4
Naphthalene
MODEL BUILDING
9 9a
10a 10
1 7
2
6
3
5
12
4a 4
Anthracene
12a
11a
11
10a
5a 5
10
6a 6
7
Tetracene (Naphthacene)
4 9
3
8
2
4a
4b
8 8a 9
10a 1
10
Phenanthrene
Each structure has its own numbering system around the periphery. A quaternary carbon is given the number of the preceding carbon in the sequence followed by the letters a, b, and so on, depending on how close it is to that carbon.
Exercise 15-9 Name the following compounds or draw their structures. (a) 2,6-Dimethylnaphthalene (b) 1-Bromo-6-nitrophenanthrene (c) 9,10-Diphenylanthracene Br
(d)
NO2
(e) HO3S
Naphthalene is aromatic: a look at spectra In contrast with benzene, which is a liquid, naphthalene is a colorless crystalline material with a melting point of 80 8C. It is probably best known as a moth repellent and insecticide, although in these capacities it has been partly replaced by chlorinated compounds such as 1,4-dichlorobenzene (p-dichlorobenzene). The spectral properties of naphthalene strongly suggest that it shares benzene’s delocalized electronic structure and thermodynamic stability. The ultraviolet and NMR spectra are particularly revealing. The ultraviolet spectrum of naphthalene (Figure 15-13) shows a pattern typical of an extended conjugated system, with peaks at wavelengths as long as 320 nm. On the basis of this observation, we conclude that the electrons are delocalized more extensively than in benzene (Section 15-2 and Figure 15-6). Thus, the added four p electrons enter into efficient overlap with those of the attached benzene ring. In fact, it is possible to draw several resonance forms. Resonance Forms of Naphthalene
Alternatively, the continuous overlap of the 10 p orbitals and the fairly even distribution of electron density can be shown as in Figure 15-14. According to these representations, the structure of naphthalene should be symmetric, with planar and almost hexagonal benzene rings and two perpendicular mirror planes bisecting the
15-5 Polycyclic Aromatic Hydrocarbons
691
Chapter 15
Figure 15-13 Extended p conjugation in naphthalene is manifest in its UV spectrum (measured in 95% ethanol). The complexity and location of the absorptions are typical of extended p systems.
5
Absorbance (log ε )
4
3
2
1
UV 200
220
240
260
280
300
320
340
Wavelength (nm)
Figure 15-14 (A) Orbital picture of naphthalene, showing its extended overlap of p orbitals. (B) Electrostatic potential map, revealing its electron density distribution over the 10 carbon atoms. 1.42 Å 1.37 Å
A
B 1.39 Å
molecule. X-ray crystallographic measurements confirm this prediction (Figure 15-15). The C–C bonds deviate only slightly in length from those in benzene (1.39 Å), and they are clearly different from pure single (1.54 Å) and double bonds (1.33 Å). Further evidence of aromaticity is found in the 1H NMR spectrum of naphthalene (Figure 15-16). Two symmetric multiplets can be observed at d 5 7.49 and 7.86 ppm, characteristic of aromatic hydrogens deshielded by the ring-current effect of the p-electron loop (see Section 15-4, Figure 15-9). Coupling in the naphthalene nucleus is very similar to that in substituted benzenes: Jortho 5 7.5 Hz, Jmeta 5 1.4 Hz, and Jpara 5 0.7 Hz. The 13C NMR spectrum shows three lines with chemical shifts that are in the range of other benzene derivatives (see margin). Thus, on the basis of structural and spectral criteria, naphthalene is aromatic.
121°
Figure 15-15 The molecular structure of naphthalene. The bond angles within the rings are 1208.
13C NMR Data of Naphthalene (ppm) 134.4
Exercise 15-10 A substituted naphthalene, C10H8O2, gave the following spectral data: 1H NMR d 5 5.20 (broad s, 2 H), 6.92 (dd, J 5 7.5 Hz and 1.4 Hz, 2 H), 7.00 (d, J 5 1.4 Hz, 2 H), and 7.60 (d, J 5 7.5 Hz, 2 H) ppm; 13C NMR d 5 107.5, 115.3, 123.0, 129.3, 136.8, and 155.8 ppm; IR n˜ 5 3300 cm21 (broad). What is its structure? [Hints: Review the values for Jortho, meta, para in benzene (Section 15-4). The number of NMR peaks is less than maximum.]
1.40 Å
128.5
126.5
692
Chapter 15
Figure 15-16 The 300-MHz 1 H NMR spectrum of naphthalene reveals the characteristic deshielding due to a p-electronic ring current.
Benzene and Aromaticity
1H NMR
H H
H
H
H H
7.9 7.8 ppm
H
7.6 7.5 7.4 ppm
4H
9
H
8
(CH3)4Si
4H
7
6
5
4
3
2
1
0
ppm (δ )
Most fused benzenoid hydrocarbons are aromatic These aromatic properties of naphthalene hold for most of the other polycyclic benzenoid hydrocarbons. It appears that the cyclic delocalization in the individual benzene rings is not significantly perturbed by the fact that they have to share at least one p bond. Linear and angular fusion of a third benzene ring onto naphthalene result in the systems anthracene and phenanthrene. Although isomeric and seemingly very similar, they have different thermodynamic stabilities: Anthracene is about 6 kcal mol21 (25 kJ mol21) less stable than phenanthrene, even though both are aromatic. Enumeration of the various resonance forms of the molecules explains why. Anthracene has only four, and only two contain two fully aromatic benzene rings (red in the structures shown here). Phenanthrene has five, three of which incorporate two aromatic benzenes, one even three benzenes. Resonance in Anthracene
Resonance in Phenanthrene
Exercise 15-11 Draw all the possible resonance forms of tetracene (naphthacene, Section 15-5). What is the maximum number of completely aromatic benzene rings in these structures?
15-6 Other Cyclic Polyenes: Hückel’s Rule
Chapter 15
In Summary The physical properties of naphthalene are typical of an aromatic system. Its UV spectrum reveals extensive delocalization of all p electrons, its molecular structure shows bond lengths and bond angles very similar to those in benzene, and its 1 H NMR spectrum reveals deshielded ring hydrogens indicative of an aromatic ring current. Other polycyclic benzenoid hydrocarbons have similar properties and are considered aromatic.
15-6 Other Cyclic Polyenes: Hückel’s Rule The special stability and reactivity associated with cyclic delocalization is not unique to benzene and polycyclic benzenoids. Thus, we shall see that other cyclic conjugated polyenes can be aromatic, but only if they contain (4n 1 2) p electrons (n 5 0, 1, 2, 3, . . .). In contrast, 4n p circuits may be destabilized by conjugation, or are antiaromatic. This pattern is known as Hückel’s* rule. Nonplanar systems in which cyclic overlap is disrupted sufficiently to impart alkenelike properties are classified as nonaromatic. Let us look at some members of this series, starting with 1,3-cyclobutadiene.
1,3-Cyclobutadiene, the smallest cyclic polyene, is antiaromatic 1,3-Cyclobutadiene, a 4n p system (n 5 1), is an air-sensitive and extremely reactive molecule in comparison to its analogs 1,3-butadiene and cyclobutene. Not only does the molecule have none of the attributes of an aromatic molecule like benzene, it is actually destabilized through p overlap by more than 35 kcal mol21 (146 kJ mol21) and therefore is antiaromatic. As a consequence, its structure is rectangular, and the two diene forms represent isomers, equilibrating through a symmetrical transition state, rather than resonance forms.
Cyclobutadiene: antiaromatic
1,3-Cyclobutadiene Is Unsymmetrical ‡
1 Ea ~ ~ 3–6 kcal mol (13–26 kJ mol1)
Transition state
Free 1,3-cyclobutadiene can be prepared and observed only at very low temperatures. The reactivity of cyclobutadiene can be seen in its rapid Diels-Alder reactions, in which it can act as both diene (shown in red) and dienophile (blue). H
O
MODEL BUILDING
H CH3O CH3O CH3O CH3O
O
O
O
*Professor Erich Hückel (1896 – 1984), University of Marburg, Germany.
693
694
Chapter 15
Benzene and Aromaticity
CHE M I C AL H I G H L I G H T 1 5 - 2 Juxtaposing Aromatic and Antiaromatic Rings in Fused Hydrocarbons What happens when we fuse aromatic to antiaromatic rings? The general answer is that the resulting system tries to distort its antiaromatic substructures to minimize the destabilizing contributions of cyclic 4n p circuits.
A Biphenylene
A case in point is the juxtaposition of benzene with cyclobutadiene, as in the series of molecules shown below, which was studied by one of your authors (P. V.).
B Angular [3]phenylene
In these hydrocarbons, the contribution of antiaromaticity due to the cyclobutadiene rings is reduced by enforcing bond alternation in the neighboring benzene rings, thus
C C3-symmetric [4]phenylene
minimizing the relative importance of resonance forms in which a double bond is “inside” the four-membered ring. This effect is illustrated by the specific resonance forms of
Resonance Forms of Biphenylene
Intermediate
Intermediate
Minor
Exercise 15-12 1,3-Cyclobutadiene dimerizes at temperatures as low as 22008C to give the two products shown. Explain mechanistically.
2
0 0 H H
H H %
`
200°C
H H % %
0 H H
^
Major
Substituted cyclobutadienes are less reactive because of steric protection, particularly if the substituents are bulky; they have been used to probe the spectroscopic features of the cyclic system of four p electrons. For example, in the 1H NMR spectrum of 1,2,3-tris(1,1dimethylethyl)cyclobutadiene (1,2,3-tri-tert-butylcyclobutadiene), the ring hydrogen resonates at d 5 5.38 ppm, at much higher field than expected for an aromatic system. This and other properties of cyclobutadiene show that it is quite unlike benzene.
15-6 Other Cyclic Polyenes: Hückel’s Rule
molecules A – C, which contribute strongly to the respective overall structures. As shown for biphenylene A below, other resonance contributors are less (or un-) important. Here, because of symmetry, the extent of bond alternation is equal on both sides of the cyclobutadiene core. However,
Chapter 15
695
in phenylenes B and C, the outside rings, in order to preserve their own aromaticity, “gang up” to maximize bond alternation in the center benzene, rendering it increasingly cyclohexatriene-like (see Section 15-2). An abbreviated, but quite descriptive, resonance picture is shown below.
Illustrative Resonance Forms of C
Major
These conclusions are based on the observed spectral, X-ray structural, and chemical properties of A – C. For example, the d value for the center “aromatic” hydrogens in B is 6.13 ppm, reflecting the presence of a diminished benzenoid ring current (Section 15-4). The extent of bond alternation between single (1.50 Å) and double (1.34 Å) bonds is maximal in the center cyclohexatriene of C, and both B and C undergo addition reactions typical of alkenes (Chapter 12). Most convincingly, the DH8hydrogenation of C [283.0 to 284.2 kcal mol21 (2347 to 2352 kJ mol21)]
Minor
is that expected for three minimally interacting double bonds, as in the “hypothetical” cyclohexatriene discussed in Section 15-2! Hydrocarbons like the phenylenes are not only of fundamental interest in efforts to understand aromaticity, but also constitute potential building blocks for a new generation of organic electronic materials (see also Chemical Highlight 14-2). They are flat, like graphite (Chemical Highlight 15-1), but they have activated electronic structures, a prerequisite for electron mobility.
1,3,5,7-Cyclooctatetraene is nonplanar and nonaromatic Let us now examine the properties of the next higher cyclic polyene analog of benzene, 1,3,5,7-cyclooctatetraene, another 4n p cycle (n 5 2). Is it antiaromatic, like 1,3cyclobutadiene? First prepared in 1911 by Willstätter,* this substance is now readily available from a special reaction, the nickel-catalyzed cyclotetramerization of ethyne. It is a yellow liquid (b.p. 152 8C) that is stable if kept cold but that polymerizes when heated. It is oxidized by air, catalytically hydrogenated to cyclooctane, and subject to electrophilic additions and to cycloaddition reactions. This chemical reactivity is diagnostic of a normal polyene (Section 14-7). Spectral and structural data confirm the ordinary alkene nature of cyclooctatetraene. Thus, the 1H NMR spectrum shows a sharp singlet at d 5 5.68 ppm, typical of an alkene. The molecular-structure determination reveals that cyclooctatetraene is actually nonplanar
*Professor Richard Willstätter (1872 – 1942), Technical University, Munich, Germany, Nobel Prize 1915 (chemistry).
4 HC q CH Ni(CN)2, 70°C, 15–25 atm
70% 1,3,5,7-Cyclooctatetraene: nonaromatic
696
Chapter 15
Benzene and Aromaticity
Figure 15-17 The molecular structure of 1,3,5,7cyclooctatetraene. Note the alternating single and double bonds of this nonplanar, nonaromatic molecule.
H
126.1°
117.6° 1.48 Å 1.34 Å
and tub-shaped (Figure 15-17). The double bonds are nearly orthogonal (perpendicular) and not conjugated. Conclusion: The molecule is nonaromatic.
Exercise 15-13 On the basis of the molecular structure of 1,3,5,7-cyclooctatetraene, would you describe its double bonds as conjugated (i.e., does it exhibit extended p overlap)? Would it be correct to draw two resonance forms for this molecule, as we do for benzene? (Hint: Build molecular models of the two forms of 1,2-dimethylcyclooctatetraene.)
Exercise 15-14 Cyclooctatetraene A exists in equilibrium with less than 0.05% of a bicyclic isomer B, which is trapped by Diels-Alder cycloaddition to 2-butenedioic anhydride (maleic anhydride, Table 14-1) to give compound C. O
H
O
B
O
[
[H
O
O O A
C
What is isomer B? Show a mechanism for the A y B y C interconversion. (Hint: Work backward from C to B and review Section 14-9.)
Only cyclic conjugated polyenes containing (4n 1 2) pi electrons are aromatic Unlike cyclobutadiene and cyclooctatetraene, certain higher cyclic conjugated polyenes are aromatic. All of them have two properties in common: They contain (4n 1 2) p electrons, and they are sufficiently planar to allow for delocalization. The first such system was prepared in 1956 by Sondheimer*; it was 1,3,5,7,9,11,13,15,17cyclooctadecanonaene, containing 18 p electrons (4n 1 2, n 5 4). To avoid the use of such cumbersome names, Sondheimer introduced a simpler system for naming cyclic conjugated
*Professor Franz Sondheimer (1926 – 1981), University College, London.
15-6 Other Cyclic Polyenes: Hückel’s Rule
Chapter 15
polyenes. He named completely conjugated monocyclic hydrocarbons (CH)N as [N]annulenes, in which N denotes the ring size. Thus, cyclobutadiene would be called [4]annulene; benzene, [6]annulene; cyclooctatetraene, [8]annulene. The first almost unstrained aromatic system in the series after benzene is [18]annulene. H
H
H H
H
H H
H H
H
H
H
H
H
H H
H
H
H H
H
H
H H H
H H
H
H
H
H
H
H H
H
H
[18]Annulene (1,3,5,7,9,11,13,15,17-Cyclooctadecanonaene)
Exercise 15-15 The three isomers of [10]annulene shown here have been prepared and exhibit nonaromatic behavior. Why? (Hint: Build models!)
cis,cis,cis,cis,cis[10]Annulene
trans,cis,cis,cis,cis[10]Annulene
trans,cis,trans,cis,cis[10]Annulene
[18]Annulene is fairly planar and shows little alternation of the single and double bonds. The extent of the delocalization of its p electrons is pictured in the electrostatic potential map in the margin. Like benzene, it can be described by a set of two equivalent resonance forms. In accord with its aromatic character, the molecule is relatively stable and undergoes electrophilic aromatic substitution. It also exhibits a benzenelike ring-current effect in its 1 H NMR spectrum (see Problems 64 and 65). Since the preparation of [18]annulene, many other annulenes have been made. As long as they are planar and delocalized, those with (4n 1 2) p electrons, such as benzene and [18]annulene, are aromatic, whereas those with 4n p electrons, such as cyclobutadiene and [16]annulene, are antiaromatic. When cyclic delocalization is prohibited by angle or steric strain, such as in cyclooctatetraene or [10]annulene (Exercise 15-15), the systems are nonaromatic. Of course, cyclic polyenes in which there is no contiguous array of p orbitals are not annulenes and therefore also nonaromatic.
[18]Annulene: aromatic
The Annulenes and Other Cyclic Polyenes
Cyclobutadiene: planar, antiaromatic
Benzene: planar, aromatic
Cyclooctatetraene: nonplanar nonaromatic
[10]Annulene: nonplanar nonaromatic
[12]Annulene: planar antiaromatic
697
698
Chapter 15
[14]Annulene: planar aromatic
Benzene and Aromaticity
[16]Annulene: planar antiaromatic
[18]Annulene: planar aromatic
1,3-Cyclopentadiene: noncyclically delocalized nonaromatic
1,4-Cyclohexadiene: nondelocalized nonaromatic
The alternating behavior of the annulenes between aromatic and antiaromatic had been predicted earlier by the theoretical chemist Hückel, who formulated this (4n 1 2) rule in 1931. Hückel’s rule expresses the regular molecular-orbital patterns calculated for planar, cyclic conjugated polyenes. The p orbitals mix to give an equal number of p molecular orbitals, as shown in Figure 15-18. For example, the four p orbitals of cyclobutadiene result in four molecular orbitals, the six p orbitals of benzene in six molecular orbitals, and so on. These orbitals occur in symmetrically disposed degenerate pairs, except for the lowest bonding and highest antibonding ones, which are unique. Thus, cyclobutadiene has one such degenerate pair of molecular orbitals, benzene has two, and so on. A closed-shell system is possible only if all bonding molecular orbitals are occupied (see Section 1-7), that is, only if there are (4n 1 2)p electrons. On the other hand, 4n p cycles always contain a pair of singly occupied orbitals, an unfavorable electronic arrangement. Figure 15-18 (A) Hückel’s 4n 1 2 rule is based on the regular pattern of the p molecular orbitals in cyclic conjugated polyenes. They are equally spaced and degenerate, except for the highest and lowest ones, which are unique. (B) Molecular-orbital levels in 1,3cyclobutadiene. Four p electrons are not enough to result in a closed shell (in other words, doubly filled molecular orbitals), so the molecule is not aromatic. (C) The six p electrons in benzene produce a closed-shell configuration, so benzene is aromatic.
Highest antibonding molecular orbital
E
)
)
Pairs of degenerate molecular orbitals
x
Lowest bonding molecular orbital Cyclopolyene with x 1 conjugated double bonds
Cyclobutadiene (x 1)
Benzene (x 2)
A
B
C
Exercise 15-16 On the basis of Hückel’s rule, label the following molecules as aromatic or antiaromatic. (a) [30]Annulene; (b) [20]annulene; (c) trans-15,16-dihydropyrene; (d) the deep blue (see Figure 14-17) azulene; (e) S-indacene. 2 1
3
H ´
10
4
15 16
0 H
9
8
1
5
8
7
2
6
6 7
trans-15,16-Dihydropyrene
3
4
Azulene
5
S-indacene
15-7 Hückel’s Rule and Charged Molecules
Chapter 15
In Summary Cyclic conjugated polyenes are aromatic if their p electron count is 4n 1 2. This number corresponds to a completely filled set of bonding molecular orbitals. Conversely, 4n p systems have open-shell, antiaromatic structures that are unstable, are reactive, and lack aromatic ring-current effects in 1H NMR. Finally, when steric constraints impose nonplanarity, cyclic polyenes behave as nonaromatic alkenes.
15-7 Hückel’s Rule and Charged Molecules Hückel’s rule also applies to charged molecules, as long as cyclic delocalization can occur. Their aromaticity is reflected in relative thermodynamic and kinetic stability, the observation of ring currents in the NMR experiment, and the absence of bond alternation in crystal structures. This section shows how charged aromatic systems can be prepared.
The cyclopentadienyl anion and the cycloheptatrienyl cation are aromatic 1,3-Cyclopentadiene is unusually acidic [pKa < 16; comparable to alcohols (Section 8-3)] because the cyclopentadienyl anion resulting from deprotonation contains a delocalized, aromatic system of six p electrons. The negative charge is equally distributed over all five carbon atoms. For comparison, the pKa of propene is 40. An electrostatic potential map of the molecule is shown in the margin, on a scale that attenuates the otherwise overwhelming effect of the negative charge.
Cyclopentadienyl anion: aromatic
Aromatic Cyclopentadienyl Anion H
H
š
š
H
etc.
pKa 16
or
In contrast, the cyclopentadienyl cation, a system of four p electrons, can be produced only at low temperature and is extremely reactive. When 1,3,5-cycloheptatriene is treated with bromine, a stable salt is formed, cycloheptatrienyl bromide. In this molecule, the organic cation contains six delocalized p electrons, and the positive charge is equally distributed over seven carbons (as shown in the electrostatic potential map in the margin). Even though it is a carbocation, the system is remarkably unreactive, as is expected for an aromatic system. In contrast, the cycloheptatrienyl anion is antiaromatic, as indicated by the much lower acidity of cycloheptatriene (pKa 5 39) compared with that of cyclopentadiene.
Cycloheptatrienyl cation: aromatic
Aromatic Cycloheptatrienyl Cation H
H
Br2, HBr
Br
Exercise 15-17 Draw an orbital picture of (a) the cyclopentadienyl anion and (b) the cycloheptatrienyl cation (consult Figure 15-2).
etc.
or
699
700
Benzene and Aromaticity
Chapter 15
Exercise 15-18 The rate of solvolysis of compound A in 2,2,2-trifluoroethanol at 258C exceeds that of compound B by a factor of 1014. Explain. CH3
C(CH3)3 (CH3)3C
OCCF3 B O
H 3C
H 3C
A
OCCF3 B O
B
Exercise 15-19 On the basis of Hückel’s rule, label the following molecules aromatic or antiaromatic. (a) Cyclopropenyl cation; (b) cyclononatetraenyl anion; (c) cycloundecapentaenyl anion.
Nonaromatic cyclic polyenes can form aromatic dianions and dications Cyclic systems of 4n p electrons can be converted into their aromatic counterparts by two-electron oxidations and reductions. For example, cyclooctatetraene is reduced by alkali metals to the corresponding aromatic dianion. This species is planar, contains fully delocalized electrons, and is relatively stable. It also exhibits an aromatic ring current in 1H NMR. Nonaromatic Cyclooctatetraene Forms an Aromatic Dianion K, THF
Eight electrons, nonaromatic
=
Reduction by two electrons
2 K
Ten electrons, aromatic
Similarly, [16]annulene can be either reduced to its dianion or oxidized to its dication, both products being aromatic. On formation of the dication, the configuration of the molecule changes. Aromatic [16]Annulene Dication and Dianion from Antiaromatic [16]Annulene
+ +
CF3SO3H, SO2, CH2Cl2, –80°C
K, THF
Oxidation by two electrons
Reduction by two electrons
=
[16]Annulene Fourteen π electrons, aromatic
Sixteen electrons, antiaromatic
Eighteen electrons, aromatic
15-8 Electrophilic Aromatic Substitution
Chapter 15
701
Exercise 15-20
Working with the Concepts: Recognizing Aromaticity in Charged Molecules Azulene (see Exercise 15-16d) is readily attacked by electrophiles at C1, by nucleophiles at C4. Explain. Strategy We need to formulate first the various resonance forms of the species resulting from these two modes of attack. Inspection of these structures might then provide the answer. Solution • Attack by electrophiles at C1: E
H
E
E
H
H etc.
A cycloheptatrienyl cation: aromatic
Attack by E1 at C1 generates a fused, aromatic cycloheptatrienyl cation framework. • Attack by nucleophiles at C4: Nuð H
Nu
ð
H etc.
A cyclopentadienyl anion: aromatic
Attack by Nu2 at C4 generates a fused, aromatic cyclopentadienyl anion framework.
Exercise 15-21
Try It Yourself The triene A can be readily deprotonated twice to give the stable dianion B. However, the neutral analog of B, the tetraene C (pentalene), is extremely unstable. Explain. H
Base
ð ð
H H A
H B
C
In Summary Charged species may be aromatic, provided they exhibit cyclic delocalization and obey the 4n 1 2 rule.
15-8 Synthesis of Benzene Derivatives: Electrophilic Aromatic Substitution In this section we begin to explore the reactivity of benzene, the prototypical aromatic compound. The aromatic stability of benzene makes it relatively unreactive, despite the presence of three formal double bonds. As a result, its chemical transformations require forcing
E
Nu
702
Benzene and Aromaticity
Chapter 15
conditions and proceed through new pathways. Not surprisingly, however, most of the chemistry of benzene features attack by electrophiles. We shall see in Section 22-4 that attack by nucleophiles is rare but possible, provided that a suitable leaving group is present.
Benzene undergoes substitution reactions with electrophiles Benzene is attacked by electrophiles, but, in contrast to the corresponding reactions of alkenes, this reaction results in substitution of hydrogens — electrophilic aromatic substitution — not addition to the ring. REACTION
Electrophilic Aromatic Substitution H
E
H
H
H
H
H
EX
Electrophile
H
H
HX
H
H
H
Under the conditions employed for these processes, nonaromatic conjugated polyenes would rapidly polymerize. However, the stability of the benzene ring allows it to survive. Let us begin with the general mechanism of electrophilic aromatic substitution.
Electrophilic aromatic substitution in benzene proceeds by addition of the electrophile followed by proton loss The mechanism of electrophilic aromatic substitution has two steps. First, the electrophile E1 attacks the benzene nucleus, much as it would attack an ordinary double bond. The resonance stabilized cationic intermediate thus formed then loses a proton to regenerate the aromatic ring. Note two important points in the formulation of this general mechanism. First, always show the hydrogen at the site of the initial electrophilic attack. Second, the positive charge in the resulting cation is indicated by three resonance forms and is located ortho and para to the carbon that has been attacked, a result of the rules for drawing resonance forms (Sections 1-5 and 14-1).
MECHANISM
Mechanism of Electrophilic Aromatic Substitution Step 1. Electrophilic attack
E
E
H
E
H
H
E
or
H E
Step 2. Proton loss
E H
H
E
E
H
H
E
The first step in this mechanism is not favored thermodynamically. Although charge is delocalized in the cationic intermediate, the formation of the C – E bond generates an sp3-hybridized carbon in the ring, which interrupts cyclic conjugation: The intermediate is
15-8 Electrophilic Aromatic Substitution
E E +
H
A
703
Figure 15-19 (A) Orbital picture of the cationic intermediate resulting from attack by an electrophile on the benzene ring. Aromaticity is lost because cyclic conjugation is interrupted by the sp3-hybridized carbon. The four electrons in the p system are not shown. (B) Dotted-line notation to indicate delocalized nature of the charge in the cation.
+
sp3
Chapter 15
H
B
not aromatic (Figure 15-19). However, the next step, loss of the proton at the sp3-hybridized carbon, regenerates the aromatic ring. This process is more favored than nucleophilic trapping by the anion that accompanies E1. Such trapping would give a nonaromatic addition product. The overall substitution is exothermic, because the bonds formed are stronger than the bonds broken. Figure 15-20 depicts a potential energy diagram in which the first step is rate determining, a kinetic finding that applies to most electrophiles that we shall encounter. The subsequent loss of proton is much faster than initial electrophilic attack because it leads to the aromatic product in an exothermic step, which furnishes the driving force for the overall sequence. The following sections look more closely at the most common reagents employed in this transformation and the details of the mechanism. +
E␦
‡
␦+ H Rate-determining transition state
+
E H
Slow
Ea
Intermediate cation Not aromatic, endothermic
E
+ E+
‡ ␦+
E +
H␦
Fast
ΔH˚
Aromatic
E + H+ Aromatic, exothermic
Reaction coordinate
Exercise 15-22 We learned in Section 11-5 that the DH8 of the hydrogenation of cis-2-butene is 228.6 kcal mol–1. Taking this system as a model for a double bond in benzene, estimate the corresponding DH8 of the hydrogenation of benzene to 1,3-cyclohexadiene. What is the difference between these two double bonds? (Hint: Consult Figure 15-3.)
In Summary The general mechanism of electrophilic aromatic substitution begins with electrophilic attack by E1 to give an intermediate, charge-delocalized but nonaromatic cation in a rate-determining step. Subsequent fast proton loss regenerates the (now substituted) aromatic ring.
Figure 15-20 Potential-energy diagram describing the course of the reaction of benzene with an electrophile. The first transition state is rate determining. Proton loss is relatively fast. The overall rate of the reaction is controlled by Ea; the amount of exothermic energy released is given by DH8.
704
Chapter 15
Bromination of Benzene
Benzene and Aromaticity
15-9 Halogenation of Benzene: The Need for a Catalyst REACTION
Br2, FeBr3
Br
An example of electrophilic aromatic substitution is halogenation. Benzene is normally unreactive in the presence of halogens, because halogens are not electrophilic enough to disrupt its aromaticity. However, the halogen may be activated by Lewis acidic catalysts, such as ferric halides (FeX3) or aluminum halides (AlX3), to become a much more powerful electrophile. How does this activation work? Lewis acids have the ability to accept electron pairs. When a halogen such as bromine is exposed to FeBr3, the two molecules combine in a Lewis acid-base reaction.
MECHANISM
Activation of Bromine by the Lewis Acid FeBr3
Br
š š ðBr OBr OFeBr3
š OBr š ðBr ð FeBr3
Bromobenzene
š Br
š Br OFeBr3
In this complex, the Br – Br bond is polarized, thereby imparting electrophilic character to the bromine atoms. Electrophilic attack on benzene is at the terminal bromine, allowing the other bromine atom to depart with the good leaving group FeBr42. In terms of electron flow, you can also view this process as a nucleophilic substitution of [Br2FeBr3] by the benzene double bond, not unlike an SN2 reaction (see also Figure 15-21). Electrophilic Attack on Benzene by Activated Bromine
Figure 15-21 X-ray structure of the carbocation arising from attack of bromine on hexamethylbenzene. This cation can be isolated because it is unusually stabilized by the methyl substituents and because it lacks a proton on the sp3-hybridized carbon, thus disabling bromoarene formation.
H
F Cl Br I F Cl Br I C6H5 F Cl Br I
F Cl Br I C6H5 C6H5 C6H5 C6H5 H H H H H
FeBr4
H
The FeBr42 formed in this step now functions as a base, abstracting a proton from the cyclohexadienyl cation intermediate. This transformation not only furnishes the two products of the reaction, bromobenzene and hydrogen bromide, but also regenerates the FeBr3 catalyst. Bromobenzene Formation
Dissociation Energies [DH 8] of Bonds A – B [kcal mol21 (kJ mol21)] B
Br
Table 15-1
A
š š ðBr OBr OFeBr3
š O FeBr3 Br ðBr
H
HBr
FeBr3
Br
DH8 38 58 46 36 126 96 81 65 112 135.8 103.2 87.5 71.3
(159) (243) (192) (151) (527) (402) (339) (272) (469) (568) (432) (366) (298)
A quick calculation confirms that electrophilic bromination of benzene is exothermic. A phenyl – hydrogen bond (approximately 112 kcal mol21, Table 15-1) and a bromine molecule (46 kcal mol21) are lost in the process. Counterbalancing this loss is the formation of a phenyl – bromine bond (DH 8 5 81 kcal mol21) and an H – Br bond (DH 8 5 87.5 kcal mol21). Thus, the overall reaction is exothermic by 158 2 168.5 5 210.5 kcal mol21 (43.9 kJ mol21). As in the radical halogenation of alkanes (Section 3-7), the exothermic nature of aromatic halogenation decreases down the periodic table. Fluorination is so exothermic that direct reaction of fluorine with benzene is explosive. Chlorination, on the other hand, is controllable but requires the presence of an activating catalyst, such as aluminum chloride or ferric chloride. The mechanism of this reaction is identical with that of bromination. Finally, electrophilic iodination with iodine is endothermic and thus not normally possible. Much like the radical halogenation of alkanes, electrophilic chlorination and bromination of benzene (and substituted benzenes, Chapter 16) introduces functionality that can be
15-10 Nitration and Sulfonation of Benzene
Chapter 15
utilized in further reactions, in particular C – C bond formations through organometallic reagents (see Problem 54, Section 13-9, and Chemical Highlight 13-1). Exercise 15-23 When benzene is dissolved in D2SO4, its 1H NMR absorption at d 5 7.27 ppm disappears and a new compound is formed having a molecular weight of 84. What is it? Propose a mechanism for its formation. (Caution: In all mechanisms of electrophilic aromatic substitution, always draw the H atom at the site of electrophilic attack.)
Exercise 15-24 Professor G. Olah* and his colleagues exposed benzene to the especially strong acid system HF – SbF5 in an NMR tube and observed a new 1H NMR spectrum with absorptions at d 5 5.69 (2 H), 8.22 (2 H), 9.42 (1 H), and 9.58 (2 H) ppm. Propose a structure for this species.
In Summary The halogenation of benzene becomes more exothermic as we proceed from I2 (endothermic) to F2 (exothermic and explosive). Chlorinations and brominations are achieved with the help of Lewis acid catalysts that polarize the X – X bond and activate the halogen by increasing its electrophilic power.
15-10 Nitration and Sulfonation of Benzene In two other typical electrophilic substitutions of benzene, the electrophiles are the nitronium ion (NO21), leading to nitrobenzene, and sulfur trioxide (SO3), giving benzenesulfonic acid.
Benzene is subject to electrophilic attack by the nitronium ion To bring about nitration of the benzene ring at moderate temperatures, it is not sufficient just to treat benzene with concentrated nitric acid. Because the nitrogen in HNO3 has little electrophilic power, it must somehow be activated. Addition of concentrated sulfuric acid serves this purpose by protonating the nitric acid. Loss of water then yields the nitronium ion, NO21, a strong electrophile, with much of its positive charge residing on nitrogen, as shown in the electrostatic potential map.
REACTION
Nitration of Benzene
HNO3, H2SO4
NO2 Activation of Nitric Acid by Sulfuric Acid
> N HOO p
> O) l i
> O) l
> ON HO A i > O) p H
HO OSO3H
> O) p
HSO4
Nitric acid
> O) l
> N HOO A i > O) p H
H2> O)
> > O PNP p O p Nitronium ion
*Professor George A. Olah (b. 1927), University of Southern California, Los Angeles, Nobel Prize 1994 (chemistry).
Nitrobenzene
705
706
Chapter 15
Benzene and Aromaticity
The nitronium ion, with its positively charged nitrogen, then attacks benzene.
MECHANISM Mechanism of Aromatic Nitration
OPNP O
OSO3H
NO2
H
HOSO3H
NO2
H
Nitrobenzene
Aromatic nitration is the best way to introduce nitrogen-containing substituents into the benzene ring. The nitro group functions as a directing group in further substitutions (Chapter 16) and as a masked amino function (Section 16-5), as unraveled in benzenamines (anilines; Section 22-10).
Sulfonation is reversible Concentrated sulfuric acid does not sulfonate benzene at room temperature. However, a more reactive form, called fuming sulfuric acid, permits electrophilic attack by SO3. Commercial fuming sulfuric acid is made by adding about 8% of sulfur trioxide, SO3, to the concentrated acid. Because of the strong electron-withdrawing effect of the three oxygens, the sulfur in SO3 is electrophilic enough to attack benzene directly. Subsequent proton transfer results in the sulfonated product, benzenesulfonic acid. Mechanism of Aromatic Sulfonation
MECHANISM
H
H
š Oð
S
B
š O
Oð B SO š O ð B HO ð
ðO
B
H
š O
ð
ðOð B S K N
š OH
Benzenesulfonic acid
Aromatic sulfonation is readily reversible. The reaction of sulfur trioxide with water to give sulfuric acid is so exothermic that heating benzenesulfonic acid in dilute aqueous acid completely reverses sulfonation. REACTION
Sulfonation of Benzene
Hydration of SO3 ðOð B S K N ð O Oð
HO H
ðOð O PS OH
heat
ð OH
Reverse Sulfonation: Hydrolysis SO3, H2SO4
SO3H
H2O, catalytic H 2 SO 4, 100C
H
HOSO3H
SO3H
Benzenesulfonic acid
The reversibility of sulfonation may be used to control further aromatic substitution processes. The ring carbon containing the substituent is blocked from attack, and electrophiles are directed to other positions. Thus, the sulfonic acid group can be introduced to serve as a directing blocking group and then removed by reverse sulfonation. Synthetic applications of this strategy will be discussed in Section 16-5.
15-10 Nitration and Sulfonation of Benzene
Chapter 15
707
Benzenesulfonic acids have important uses Sulfonation of substituted benzenes is used in the synthesis of detergents. Thus, long-chain branched alkylbenzenes are sulfonated to the corresponding sulfonic acids, then converted into their sodium salts. Because such detergents are not readily biodegradable, they have been replaced by more environmentally acceptable alternatives. We shall examine this class of compounds in Chapter 19. Aromatic Detergent Synthesis SO3, H2SO4
R
SO3H
R
NaOH H2O
SO3Na
R
R branched alkyl group
Another application of sulfonation is to the manufacture of dyes, because the sulfonic acid group imparts water solubility (Chapter 22). Sulfonyl chlorides, the acid chlorides of sulfonic acids (see Section 9-4), are usually prepared by reaction of the sodium salt of the acid with PCl5 or SOCl2. Preparation of Benzenesulfonyl Chloride O S O ONa
SO2Cl POCl3 NaCl
PCl5
O
Sulfonyl chlorides are frequently employed in synthesis. For example, recall that the hydroxy group of an alcohol may be turned into a good leaving group by conversion of the alcohol into the 4-methylbenzenesulfonate (p-toluenesulfonate, tosylate; Sections 6-7 and 9-4). Sulfonyl chlorides are important precursors of sulfonamides, many of which are chemotherapeutic agents, such as the sulfa drugs discovered in 1932 (Section 9-11). Sulfonamides are derived from the reaction of a sulfonyl chloride with an amine. Sulfa drugs specifically contain the 4-aminobenzenesulfonamide (sulfanilamide) function. Their mode of action is to interfere with the bacterial enzymes that help to synthesize folic acid (Chemical Highlight 25-3), thereby depriving them of an essential nutrient and thus causing bacterial cell death. Sulfa Drugs N RHN
H2N
SO2NHR′
SO2NH
O CH3
General structure
Sulfamethoxazole (Gantanol) (Antibacterial, used to treat urinary infections)
CH3O N
N H2N
SO2NH
H2N
SO2NH
N Sulfalene (Kelfizina) (Antiparasitic)
N Sulfadiazine (Antimalarial)
MATION AN I
ANIMATED MECHANISM: Electrophilic aromatic sulfonation of benzene
708
Chapter 15
Benzene and Aromaticity
About 15,000 sulfa derivatives have been synthesized and screened for antibacterial activity; some have become new drugs. With the advent of newer generations of antibiotics, the medicinal use of sulfa drugs has greatly diminished, but their discovery was a milestone in the systematic development of medicinal chemistry.
Exercise 15-25 Formulate mechanisms for (a) the reverse of sulfonation; (b) the hydration of SO3.
In Summary Nitration of benzene requires the generation of the nitronium ion, NO21, which functions as the active electrophile. The nitronium ion is formed by the loss of water from protonated nitric acid. Sulfonation is achieved with fuming sulfuric acid, in which sulfur trioxide, SO3, is the electrophile. Sulfonation is reversed by hot aqueous acid. Benzenesulfonic acids are used in the preparation of detergents, dyes, compounds containing leaving groups, and sulfa drugs.
15-11 Friedel-Crafts Alkylation None of the electrophilic substitutions mentioned so far has led to carbon – carbon bond formation, one of the primary challenges in organic chemistry. In principle, such reactions could be carried out with benzene in the presence of a sufficiently electrophilic carbon-based electrophile. This section introduces the first of two such transformations, the Friedel-Crafts* reactions. The secret to the success of both processes is the use of a Lewis acid, usually aluminum chloride. In the presence of this reagent, haloalkanes attack benzene to form alkylbenzenes. In 1877, Friedel and Crafts discovered that a haloalkane reacts with benzene in the presence of an aluminum halide. The resulting products are the alkylbenzene and hydrogen halide. This reaction, which can be carried out in the presence of other Lewis acid catalysts, is called the Friedel-Crafts alkylation of benzene. REACTION
Friedel-Crafts Alkylation H
R
RX
AlX3
HX
The reactivity of the haloalkane increases with the polarity of the C – X bond in the order RI , RBr , RCl , RF. Typical Lewis acids are BF3, SbCl5, FeCl3, AlCl3, and AlBr3. Friedel-Crafts Alkylation of Benzene with Chloroethane CH2CH3
H CH3CH2Cl
AlCl3, 25°C
HCl
28% Ethylbenzene
*Professor Charles Friedel (1832 – 1899), Sorbonne, Paris; Professor James M. Crafts (1839 – 1917), Massachusetts Institute of Technology, Cambridge, MA.
15-11 Friedel-Crafts Alkylation
Chapter 15
With primary halides, the reaction begins with coordination of the Lewis acid to the halogen of the haloalkane, much as in the activation of halogens in electrophilic halogenation. This coordination places a partial positive charge on the carbon bearing the halogen, rendering it more electrophilic. Attack on the benzene ring is followed by proton loss in the usual manner, giving the observed product. Mechanism of Friedel-Crafts Alkylation with Primary Haloalkanes Step 1. Haloalkane activation
MECHANISM
p X) p
RCH2
+ _ p X)AlX3 RCH2)p
AlX3
Step 2. Electrophilic attack
f
H2C
R
CH2R
i+ > X OAlX3 p
+
AlX4
H
Step 3. Proton loss
CH2R
ðX p O AlX3
HX
AlX3
CH2R
H
With secondary and tertiary halides, free carbocations are usually formed as intermediates; these species attack the benzene ring in the same way as the cation NO21.
Exercise 15-26 Write a mechanism for the formation of (1,1-dimethylethyl)benzene (tert-butylbenzene) from 2-chloro-2-methylpropane (tert-butyl chloride), benzene, and catalytic AlCl3.
Intramolecular Friedel-Crafts alkylations can be used to fuse a new ring onto the benzene nucleus. An Intramolecular Friedel-Crafts Alkylation Cl AlCl3, CS2 and CH3NO2 (solvents), 25C, 72 h HCl
H 31% Tetralin (Common name)
Friedel-Crafts alkylations can be carried out with any starting material that functions as a precursor to a carbocation, such as an alcohol or alkene (Sections 9-2 and 12-3).
709
710
Chapter 15
Benzene and Aromaticity
Friedel-Crafts Alkylations Using Other Carbocation Precursors H
CH3CH2CHCH3 OH
CH3CH2CHCH3
BF 3 , 60C, 9 h HOH
36%
Forms CH3CH2CHCH3
(1-Methylpropyl)benzene
H
H HF, 0C
62% Cyclohexylbenzene
Forms
H Unreactive Halides in the FriedelCrafts Alkylation Br
Friedel-Crafts vinylations or arylations employing haloalkenes or haloarenes do not work (margin). The reason is that the corresponding cations are energetically inaccessible (Sections 13-9 and 22-10). Exercise 15-27
Cl
In 2007, more than 3.7 million tons of (1-methylethyl)benzene (isopropylbenzene or cumene), an important industrial intermediate in the manufacture of phenol (Section 22-4), was synthesized in the United States from propene and benzene in the presence of phosphoric acid. Write a mechanism for its formation in this reaction.
Exercise 15-28
Working with the Concepts: Reversible Friedel-Crafts Alkylations Heating any of the three isomeric dimethylbenzenes with HF in the presence of BF3 to 80oC leads to the equilibrium mixture shown below. Formulate a mechanism for these isomerizations, starting with 1,2-dimethylbenzene and simply using H1 to represent the acid. Why is the equilibrium concentration of the 1,2-isomer the lowest? CH3
CH3
CH3
CH3 CH3 18%
58%
CH3 24%
1,2-Dimethylbenzene (o-Xylene)
1,3-Dimethylbenzene (m-Xylene)
1,4-Dimethylbenzene (p-Xylene)
Strategy Topologically, these isomerizations are reminiscent of alkyl shifts, which we studied earlier in connection with carbocation rearrangements (Section 9-3). To generate a carbocation from a benzene derivative, we protonate with acid (Section 15-8, Exercises 15-23 and 15-24). Protonation can occur anywhere on the ring and is reversible.
15-11 Friedel-Crafts Alkylation
Solution • There are three possible distinct delocalized carbocations (write them), but the one of interest results from protonation at the methyl-bearing carbon, shown below as its resonance forms. CH3
CH3
/∑
H CH3
/∑
H
CH3 H CH3
/∑
CH3
CH3 H CH3
A
• Inspection of the resonance contributors reveals that one of them, form A, looks exactly like the type of carbocation drawn in Section 9-3 for H and alkyl shifts. • Execution of a methyl shift from A gives a new carbocation, which can deprotonate to give 1,3-dimethylbenzene. CH3
CH3 H CH3
CH3
CH3
/∑
H
H
H
CH3 H
H
H
H
∑/
∑/
CH3 H
/∑ H H CH3
H
A
B
C
H
H
1,3-Dimethylbenzene
1,4-Dimethylbenzene
Alternatively, another methyl shift (best visualized from resonance form B) gives C, which then loses a proton to give the 1,4-isomer. Therefore, the mechanism is indeed a sequence of simple alkyl shifts occurring in protonated benzene. • Why is 1,2-dimethylbenzene the minor component of the mixture? You have probably guessed the answer: steric hindrance, like that seen in cis alkenes (Figure 11-13).
Exercise 15-29
Try It Yourself Intramolecular Friedel-Crafts alkylation of A provided the expected B, but also 10% of C, suggesting the operation of a competing pathway to the normal mechanism. Propose such a pathway. D D D
D D A
OH
BF3
D B
C
In Summary The Friedel-Crafts alkylation produces carbocations (or their equivalents) capable of electrophilic aromatic substitution by formation of aryl–carbon bonds. Haloalkanes, alkenes, and alcohols can be used to achieve aromatic alkylation in the presence of a Lewis or mineral acid.
Chapter 15
711
712
Benzene and Aromaticity
Chapter 15
15-12 Limitations of Friedel-Crafts Alkylations The alkylation of benzenes under Friedel-Crafts conditions is accompanied by two important side reactions: One is polyalkylation; the other, carbocation rearrangement. Both cause the yield of the desired products to diminish and lead to mixtures that may be difficult to separate. Consider first polyalkylation. Benzene reacts with 2-bromopropane in the presence of FeBr3 as a catalyst to give products of both single and double substitution. The yields are low because of the formation of many by-products. CH(CH3)2
H
(CH3)2CHBr
CH(CH3)2
FeBr3
HBr Overalkylation
H
H 25%
CH(CH3)2 15%
(1-Methylethyl)benzene (Isopropylbenzene)
1,4-Bis(1-methylethyl)benzene ( p-Diisopropylbenzene)
The electrophilic aromatic substitutions that we studied in Sections 15-9 and 15-10 can be stopped at the monosubstitution stage. Why do Friedel-Crafts alkylations have the problem of multiple electrophilic substitution? It is because the substituents differ in electronic structure (a subject discussed in more detail in Chapter 16). Bromination, nitration, and sulfonation introduce an electron-withdrawing group into the benzene ring, which renders the product less susceptible than the starting material to electrophilic attack. In contrast, an alkylated benzene is more electron rich than unsubstituted benzene and thus more susceptible to electrophilic attack. Exercise 15-30 Treatment of benzene with chloromethane in the presence of aluminum chloride results in a complex mixture of tri-, tetra-, and pentamethylbenzenes. One of the components in this mixture crystallizes out selectively: m.p. 5 808C; molecular formula 5 C10H14; 1H NMR d 5 2.27 (s, 12 H) and 7.15 (s, 2 H) ppm; 13C NMR d 5 19.2, 131.2, and 133.8 ppm. Draw a structure for this product.
The second side reaction in aromatic alkylation is skeletal rearrangement (Section 9-3). For example, the attempted propylation of benzene with 1-bromopropane and AlCl3 produces (1-methylethyl)benzene. CH(CH3)2
H REACTION
CH3CH2CH2Br
AlCl3 HBr Rearranged alkyl group
The starting haloalkane rearranges by a hydride shift to the thermodynamically favored 1-methylethyl (isopropyl) cation in the presence of the Lewis acid. Rearrangement of 1-Bromopropane to 1-Methylethyl (Isopropyl) Cation
MECHANISM H
CH3CHO CH2 OBr
AlCl3
CH3CHCH3 1-Methylethyl (isopropyl) cation
BrAlCl3
15-12 Limitations of Friedel-Crafts Alkylations
Exercise 15-31
Working with the Concepts: Rearrangements in Friedel-Crafts Alkylations Attempted alkylation of benzene with 1-chlorobutane in the presence of AlCl3 gave not only the expected butylbenzene, but also, as a major product, (1-methylpropyl)benzene. Write a mechanism for this reaction. Strategy First write an equation for the described transformation:
AlCl3
Cl
HCl
Butylbenzene
(1-Methylpropyl)benzene
Consider the mechanism for each product separately. Solution • The first product is derived readily by a normal Friedel-Crafts alkylation:
H
š ðCl O AlCl3
š Cl O AlCl3
HCl
AlCl3
• The second product contains a rearranged butyl group, the result of electrophilic aromatic substitution by a secondary butyl cation. The required rearrangement is brought about by a Lewis acid – catalyzed hydride shift: H š CH3CH2CHO CH2 OClð
CH3CH2CH O CH3
AlCl3
AlCl
4
H
CH3CH2CHCH3
Exercise 15-32
Try It Yourself Write a mechanism for the following reaction:
Cl
AlCl3
H
Chapter 15
713
714
Chapter 15
Benzene and Aromaticity
Because of these limitations, Friedel-Crafts alkylations are rarely used in synthetic chemistry. Can we improve this process? A more useful reaction would require an electrophilic carbon species that could not rearrange and that would, moreover, deactivate the ring to prevent further substitution. There is such a species — an acylium cation — and it is used in the second Friedel-Crafts reaction, the topic of the next section.
In Summary Friedel-Crafts alkylation suffers from overalkylation and skeletal rearrangements by both hydrogen and alkyl shifts.
15-13 Friedel-Crafts Acylation (Alkanoylation) The second electrophilic aromatic substitution that forms carbon–carbon bonds is FriedelCrafts alkanoylation (as in butanoylation, pentanoylation, and so on). A more popular alternative systematic name for this process is acylation, which is the term we shall use. IUPAC O O B B retains the common naming of formyl for HC– and acetyl for CH3C– (Section 17-1), hence the corresponding acylations are called formylation and acetylation. Acylation proceeds through the intermediacy of acylium cations, with the general structure RCqO :1. This section describes how these ions readily attack benzene to form ketones. Friedel-Crafts Acylation O B C
O B RCX, AlX3
R
HX
Friedel-Crafts acylation employs acyl chlorides Benzene reacts with acyl halides in the presence of an aluminum halide to give 1-phenylalkanones (phenyl ketones). An example is the preparation of 1-phenylethanone (acetophenone) from benzene and acetyl chloride, by using aluminum chloride as the Lewis acid. Friedel-Crafts Acylation of Benzene with Acetyl Chloride
REACTION
H
O B CH3CCl
O B C
1. AlCl3 2. H2O, H
CH3
HCl
61% 1-Phenylethanone (Acetophenone)
Acyl chlorides are reactive derivatives of carboxylic acids. They are readily formed from carboxylic acids by reaction with thionyl chloride, SOCl2. (We shall explore this process in detail in Chapter 19.) Preparation of an Acyl Chloride Even though it is unstable and reactive, the formyl cation, H –C q O1, is a fundamental small organic molecule that is (relatively) abundant in such diverse environments as hot flames and cold interstellar space. It has been detected in the gas surrounding the comet Hale-Bopp, a spectacular visitor to Earth’s skies in 1997.
O B RCOH
SOCl2
O B RCCl
SO2
HCl
Acyl halides react with Lewis acids to produce acylium ions The key reactive intermediates in Friedel-Crafts acylations are acylium cations. These species can be formed by the reaction of acyl halides with aluminum chloride. The Lewis acid initially
15-13 Friedel-Crafts Acylation (Alkanoylation)
715
Chapter 15
coordinates to the carbonyl oxygen because of resonance (see Exercise 2-11). This complex is in equilibrium with an isomer in which the aluminum chloride is bound to the halogen. Dissociation then produces the acylium ion, which is stabilized by resonance and, unlike alkyl cations, is not prone to rearrangements. As shown in the electrostatic potential map of the acetyl cation in the margin, most of the positive charge (blue) resides on the carbonyl carbon. Acylium Ions from Acyl Halides
AlCl3
ðOð B Xð RC O š
AlCl3
D ðO B RC O š Xð
D ðOð A RC P š X
AlCl3
ðOð B š RC O X OAlCl3
Acetyl cation
[RC q Oð Acylium ion
RC P š O] ðš XAlCl3
Sometimes carboxylic anhydrides are used in acylation in place of the halides. These molecules react with Lewis acids in a similar way. Acylium Ions from Carboxylic Anhydrides
ðOð ðOð B B RC O š O O CR
AlCl3 D ðOð ðOð A B RC P O O CR
AlCl3 D ðO ðOð B B RC O š O O CR
AlCl3
ðOð ðOð B B RC O š O O CR A AlCl3
RC P š O]
[RC q Oð
Acylium ions undergo electrophilic aromatic substitution The acylium ion is sufficiently electrophilic to attack benzene by the usual aromatic substitution mechanism. Electrophilic Acylation
MECHANISM )O)
H
H R
C
Ok
R
C R
C
k
O)
Because the newly introduced acyl substituent is electron withdrawing (see Sections 14-8 and 16-1), it deactivates the ring and protects it from further substitution. Therefore polyacylation does not occur, whereas polyalkylation does (Section 15-12). The effect is accentuated by the formation of a strong complex between the aluminum chloride catalyst and the carbonyl function of the product ketone. Lewis Acid Complexation with 1-Phenylalkanones
D ðO B C
)O) R
D
D
C
AlCl3
AlCl3
R
H
ðOð B š Cl3AlOCR
716
Chapter 15
Benzene and Aromaticity
This complexation removes the AlCl3 from the reaction mixture and necessitates the use of at least one full equivalent of the Lewis acid to allow the reaction to go to completion. Aqueous work-up is necessary to liberate the ketone from its aluminum chloride complex, as illustrated by the following examples.
H
O B CH3CH2CCl
1. AlCl3 (1.7 equivalents) 2. H2O, H
O B CCH2CH3
HCl
84% 1-Phenyl-1-acetone (Propiophenone)
H
O O B B CH3COCCH3
1. AlCl3 (2.4 equivalents) 2. H2O, H
O B CCH3
CH3COOH
85% 1-Phenylethanone (Acetophenone)
The selectivity of the Friedel-Crafts acylation for single substitution allows for the selective introduction of carbon chains into the benzene nucleus, a task that proved difficult to accomplish by Friedel-Crafts alkylation (Section 15-12). Since we know how to convert the carbonyl function into an alcohol by hydride reduction (Section 8-6) and the hydroxy substituent into a leaving group that can be further reduced by hydride (Section 8-7), we can synthesize the corresponding hydrocarbon. This sequence of acylation – reduction constitutes a roundabout alkylation protocol that occurs selectively. We shall encounter more direct “deoxygenations” of carbonyl groups later (Sections 16-5 and 17-10). Preparation of Hexylbenzene by Hexanoylation–Reduction of Benzene O O
1. NaBH4 2. HBr
AlCl3
Cl
HCl
Br LiAlH4
Hexylbenzene
Exercise 15-33 O B The simplest acyl chloride, formyl chloride, H – C – Cl, is unstable, decomposing to HCl and CO upon attempted preparation. Therefore, direct Friedel-Crafts formylation of benzene is impossible. An alternative process, the Gattermann-Koch reaction, enables the introduction of the formyl group, – CHO, into the benzene ring by treatment with CO under pressure, in the presence of HCl and Lewis acid catalysts. For example, methylbenzene (toluene) can be formylated at the para position in this way in 51% yield. The electrophile in this process was observed directly for the first time in 1997 by treating CO with HF – SbF5 under high pressure: 13C NMR: d 139.5 ppm; IR: ˜n 5 2110 cm21. What is the structure of this species and the mechanism of its reaction with
Chapter Integration Problems
methylbenzene? Explain the spectral data. (Hints: Draw the Lewis structure of CO and proceed by considering the species that may arise in the presence of acid. The comparative spectral data for free CO are 13C NMR: d 5 181.3 ppm; IR: n˜ 5 2143 cm21.) CH3
CH3
CO
HCl
H
AlCl3, CuCl
CHO
In Summary The problems of Friedel-Crafts alkylation (multiple substitution and carbocation rearrangements) are avoided in Friedel-Crafts acylations, in which an acyl halide or carboxylic acid anhydride is the reaction partner, in the presence of a Lewis acid. The intermediate acylium cations undergo electrophilic aromatic substitution to yield the corresponding aromatic ketones.
THE BIG PICTURE The concept of aromaticity (and antiaromaticity) may seem strange and new to you. However, it is really just an extension of other electronic effects we have encountered earlier, starting with Coulomb’s law (Section 1-2), the octet rule (Section 1-3), and the Aufbau principle (Section 1-6). Other examples of orbital overlap and electron delocalization have been shown to be either stabilizing or destabilizing, including the stability ordering of radicals (Section 3-2) and carbocations (Section 7-5) due to hyperconjugation, and the general phenomenon of (de)stabilization by p-electron delocalization (Chapter 14). In this context, aromaticity is simply another important kind of electronic effect in organic chemistry. We shall explore additional implications of aromaticity in Chapter 16, showing how single substituents on the benzene ring affect further substitutions. As a unit, the benzene ring occurs in organic molecules ranging from polystyrene to aspirin; learning how to incorporate benzene rings into other molecules and changing the substituents on a benzene ring are important aspects of many branches of organic synthesis.
CHAPTER INTEGRATION PROBLEMS 15-34. Compound A, C8H10, was treated with Br2 in the presence of FeBr3 to give product B, C8H9Br. The spectral data for A and B are given below. Assign structures to A and B. A: 1H NMR (CDCl3): d 5 2.28 (s, 6 H), 6.95 (m, 3 H), 7.11 (td, J 5 7.8, 0.4 Hz, 1 H) ppm. 13 C NMR (CDCl3): d 5 21.3, 126.1, 128.2, 130.0, 137.7 ppm. IR (neat) selected values: n˜ 5 3016, 2946, 2921, 769, 691 cm21. UV (CH3OH): lmax 5 261 nm. B: 1H NMR (CDCl3): d 5 2.25 (s, 3 H), 2.34 (s, 3 H), 6.83 (dd, J 5 7.9, 2.0 Hz, 1 H), 7.02 (dd, J 5 2.0, 0.3 Hz, 1 H), 7.36 (dd, J 5 7.9, 0.3 Hz, 1 H) ppm. 13 C NMR (CDCl3): d 5 20.7, 22.7, 121.6, 128.1, 131.6, 132.1, 136.9, 137.4 ppm. IR (neat) selected values: n˜ 5 3012, 2961, 2923 cm21. UV (CH3OH): lmax 5 265 nm. SOLUTION A cursory glance at the molecular formulas and the spectral data confirms that bromination of a substituted benzene is taking place: One hydrogen in A (C8H10) is replaced by a bromine atom (C8H9Br), and both compounds show aromatic peaks in both the 1H and 13C NMR spectra. The IR spectra confirm this assignment by the presence of a Caromatic – H stretching signal for A and B, and the UV spectra are consistent with a phenyl chromophore. A closer look reveals that both compounds contain two methyl groups attached to the aromatic system at d < 2.3 ppm (Table 10-2). Subtracting 2 3 CH3 from C8H10 (A) leaves C6H4, a phenyl fragment: Compound A must be a dimethylbenzene isomer. Consequently, B must be a bromo(dimethyl) benzene isomer. The question for both is: Which isomer? To find the answers, it is useful to write down all the possible options. For A, there are three isomers: ortho-, meta-, and para-dimethylbenzene (xylene; Section 15-1).
Chapter 15
717
718
Chapter 15
Benzene and Aromaticity
The Three Possible Structures for A CH3
CH3
CH3 CH3
CH3 CH3
Ortho
Para
Meta
A
Is it possible to distinguish between them on the basis of NMR spectroscopy? The answer is yes, because of the varying degrees of symmetry inherent to each ring. Thus, the ortho isomer should exhibit only two types of aromatic hydrogens (two each) and three phenyl carbon signals in the respective NMR spectra. The para isomer is even more symmetrical, containing only one type of benzene hydrogen and two types of phenyl carbons. Both of these predictions are incompatible with the observed data for A. Thus, while some of the aromatic hydrogens are unresolved and appear as a multiplet (3 H), the presence of a single unique proton at d 5 7.11 ppm is consistent only with meta-dimethylbenzene. (Which one of the two possible unique hydrogens gives rise to this signal? Hint: Look at the coupling pattern of the 1H NMR signal.) Similarly, this isomer contains four distinct ring carbons, as seen in the 13C NMR spectrum. Armed with the knowledge of the structure of A, we can now formulate the possible products of its electrophilic aromatic bromination: The Three Possible Structures for B CH3
CH3
CH3
Br CH3
CH3
Br
CH3
Br B
Which one is B? Symmetry (or lack thereof ) provides the answer. The 1H NMR spectrum of B reveals two distinct methyl groups and three separate ring hydrogen resonances, whereas the 13C NMR spectrum exhibits two methyl carbon peaks, in addition to six aromatic carbon absorptions. The combined data are only compatible with the center structure above, 1-bromo-2,4-dimethylbenzene. The reaction of A to provide B is therefore: CH3
CH3 Br2, FeBr3 HBr
CH3
CH3 Br A
B
Why does single bromination of A give only isomer B? Read on to Chapter 16! But first try Problem 43 for more practice with spectroscopy. 15-35. The insecticide DDT (see Chemical Highlight 3-2) has been made in ton quantities by treatment of chlorobenzene with 2,2,2-trichloroacetaldehyde in the (required) presence of concentrated H2SO4. Formulate a mechanism for this reaction. Cl ðOð 2
Cl3CCH
CCl3 99% H2SO4, 15°C, 5 h
Cl
C H 98%
Chlorobenzene
2,2,2-Trichloroacetaldehyde
DDT
Cl
Chapter Integration Problems
Chapter 15
SOLUTION Let us first take an inventory of what is given: 1. The product is composed of two subunits derived from chlorobenzene and one subunit derived from the aldehyde. 2. In the process, the starting materials — two chlorobenzenes and one trichloroacetaldehyde, which amount to a combined atomic total of C14H11Cl5O — turn into DDT, with the molecular formula C14H9Cl5. Conclusion: The elements of H2O are extruded. 3. Topologically, the transformation constitutes a replacement of an aromatic hydrogen by an alkyl carbon substituent, strongly implicating the occurrence of a Friedel-Crafts alkylation (Section 15-12). We can now address the details of a possible mechanism. Friedel-Crafts alkylations require positively polarized or cationic carbon electrophiles. In our case, inspection of the product suggests that the carbonyl carbon in the aldehyde is the electrophilic aggressor. This carbon is positively polarized to begin with, because of the presence of the electron-withdrawing oxygen and chlorine substituents. A nonoctet dipolar resonance form (Section 1-5) illustrates this point. Activation of 2,2,2-Trichloroacetaldehyde as an Electrophile
Cl3C
ðOð
Oð ðš
ECH
ECH
H
Cl3C
H
H
Cl 3C
ECH
ðOH H
Cl 3C
Hydroxycarbocation
Dipolar nonoctet
Octet
OH ðš
EC H
H
Protonated aldehyde
In the presence of strong acid, protonation of the negatively polarized oxygen of the neutral species generates a positively charged intermediate in which the electrophilic character of the carbonyl carbon is further accentuated in the resonance form depicting a hydroxy carbocation. The stage is now set for the first of two electrophilic aromatic substitution steps (Section 15-11). First Electrophilic Aromatic Substitution Step
Cl
H
Cl3C
ðš OH A ECH
Cl H
HO H A A Z Z C yy G y CCl3 H ZZ
OH A C H A CCl3
Cl
H
The product of this step is an alcohol, which can be readily converted into the corresponding carbocation by acid (Section 9-2), partly because the ensuing charge is resonance stabilized by the adjacent benzene ring (benzylic resonance, Section 22-1, related to allylic resonance, Sections 14-1 and 14-3). The cation subsequently enables the second electrophilic aromatic substitution step to provide DDT. Another more advanced mechanism is the subject of Problem 66. Alcohol Activation and Second Electrophilic Aromatic Substitution Step H H2O
Cl
f
C
i
H
H
Cl
C
etc. CCl3
CCl3 Cl
Cl3C Cl
C A H
H
ZZ y ZZ y
yy
Cl
OH A C H A CCl3
H
Cl
DDT
719
720
Chapter 15
Benzene and Aromaticity
New Reactions 1. Hydrogenation of Benzene (Section 15-2) H2, catalyst
H° 49.3 kcal mol1 Resonance energy: ~30 kcal mol1
Electrophilic Aromatic Substitution 2. Chlorination, Bromination, Nitration, and Sulfonation (Sections 15-9 and 15-10) X2, FeX3
C6H6
HNO3, H2SO4
C6H6
SO3, H2SO4
C6H6
C6H5X
HX
X Cl, Br
C6H5NO2
H2O
C6H5SO3H
H2SO4, H2O, Δ
Reversible
3. Benzenesulfonyl Chlorides (Section 15-10) C6H5SO3Na
PCl5
C6H5SO2Cl
POCl3
NaCl
4. Friedel-Crafts Alkylation (Section 15-11) C 6H 6
RX
AlCl3
C6H5R
HX
overalkylated product
R is subject to carbocation rearrangements
Intramolecular Cl AlCl3
HCl
Alcohols and alkenes as substrates
C6H6
C6H6
OH A RCHR
BF3, 60°C
RCHP CH2
H2O
HF, 0°C
R A C6H5CHR R A C6H5CHCH3
5. Friedel-Crafts Acylation (Section 15-13) Acyl halides
C6H6
O B RCCl
1. AlCl3 2. H2O
O B C6H5CR
HCl
Requires at least one full equivalent of Lewis acid
Anhydrides
C6H6
O O B B CH3COCCH3
1. AlCl3 2. H2O
O B C6H5CCH3
CH3COOH
R
R
OR
22-1
HE
HE
X
15-11
15-11
Nu
22-1
)CH2
22-1
COOH
22-2
ON R
22-2
CH3
R
15-11
(RL) )Nu
Nu
22-4, 26-5
Especially with e-poor arenes
R OH
22-2
(RR CHL) (RCH2L) (RCH3) (RR CH2) (RR CHOH) (RCH2OR ) R OH, MnO2 R Li KMnO4, OH )Nu, SN2 H2, Pd–C solvolysis
H
22-1
15-10
H3C ON R
15-13
ON R
15-13
OH
22-4
(RN2) H2O,
R
R
E (Rdonor or halogen)
16-2, 16-3
E
E
R
H E
16-5
NH2
16-5
NO2
E HCl, Zn(Hg) CF3CO3H (Racceptor) (RNO2) (RNH2)
16-2, 16-3
R
HO COOH
22-6
HO
22-7
X(CN)
22-10
F
22-10
R
16-5
22-10
N
CH2Cl
R
OH
NN
26-7
(Ralkyl) ClCH2OCH2CH3, SnCl4
22-11
(RN2) OH
HCl, Zn(Hg) H2O, O B (RSO3H) (RR C)
16-5
(RN2) (RN2BF4) (RN2) (ROCH2CHPCH2) (ROH) H3PO2 CuX or CuCN, NaOH, CO2
O O O Sec. or tert. B B X2, FeX3 HNO3, H2SO4 SO3, H2SO4 R X, AlX3 R CHPCH2, HF B ROH, H or BF3 R CCl, AlCl3 R COCR , AlCl3 (RH) (RH) (RH) (RH) (RH) (RH) (RH) (RH)
15-10
H
(RR CH2) X2, h
H2, Pt (RH)
15-9
R
H
14-7, 15-2
R
HE
SO3H
H
NO2
H
X
section number
H
Reactions of Benzene and Substituted Benzenes H
721
722
Chapter 15
Benzene and Aromaticity
Important Concepts 1. Substituted benzenes are named by adding prefixes or suffixes to the word benzene. Disubstituted systems are labeled as 1,2-, 1,3-, and 1,4- or ortho, meta, and para, depending on the location of the substituents. Many benzene derivatives have common names, sometimes used as bases for naming their substituted analogs. As a substituent, an aromatic system is called aryl; the parent aryl substituent, C6H5, is called phenyl; its homolog C6H5CH2 is named phenylmethyl (benzyl). 2. Benzene is not a cyclohexatriene but a delocalized cyclic system of six p electrons. It is a regular hexagon of six sp2-hybridized carbons. All six p orbitals overlap equally with their neighbors. Its unusually low heat of hydrogenation indicates a resonance energy or aromaticity of about 30 kcal mol21 (126 kJ mol21). The stability imparted by aromatic delocalization is also evident in the transition state of some reactions, such as the Diels-Alder cycloaddition and ozonolysis. 3. The special structure of benzene gives rise to unusual UV, IR, and NMR spectral data. 1H NMR spectroscopy is particularly diagnostic because of the unusual deshielding of aromatic hydrogens by an induced ring current. Moreover, the substitution pattern is revealed by examination of the o, m, and p coupling constants. 4. The polycyclic benzenoid hydrocarbons are composed of linearly or angularly fused benzene rings. The simplest members of this class of compounds are naphthalene, anthracene, and phenanthrene. 5. In these molecules, benzene rings share two (or more) carbon atoms, whose p electrons are delocalized over the entire ring system. Thus, naphthalene shows some of the properties characteristic of the aromatic ring in benzene: The electronic spectra reveal extended conjugation, 1 H NMR exhibits deshielding ring-current effects, and there is little bond alternation. 6. Benzene is the smallest member of the class of aromatic cyclic polyenes following Hückel’s (4n 1 2) rule. Most of the 4n p systems are relatively reactive anti- or nonaromatic species. Hückel’s rule also extends to aromatic charged systems, such as the cyclopentadienyl anion, cycloheptatrienyl cation, and cyclooctatetraene dianion. 7. The most important reaction of benzene is electrophilic aromatic substitution. The ratedetermining step is addition by the electrophile to give a delocalized hexadienyl cation in which the aromatic character of the original benzene ring has been lost. Fast deprotonation restores the aromaticity of the (now substituted) benzene ring. Exothermic substitution is preferred over endothermic addition. The reaction can lead to halo- and nitrobenzenes, benzenesulfonic acids, and alkylated and acylated derivatives. When necessary, Lewis acid (chlorination, bromination, Friedel-Crafts reaction) or mineral acid (nitration, sulfonation) catalysts are applied. These enhance the electrophilic power of the reagents or generate strong, positively charged electrophiles. 8. Sulfonation of benzene is a reversible process. The sulfonic acid group is removed by heating with dilute aqueous acid. 9. Benzenesulfonic acids are precursors of benzenesulfonyl chlorides. The chlorides react with alcohols to form sulfonic esters containing useful leaving groups and with amines to give sulfonamides, some of which are medicinally important. 10. In contrast with other electrophilic substitutions, including Friedel-Crafts acylations, FriedelCrafts alkylations activate the aromatic ring to further electrophilic substitution, leading to product mixtures.
Problems 36. Name each of the following compounds by using the IUPAC system and, if possible, a reasonable common alternative. (Hint: The order of functional group precedence is – COOH . 2CHO . 2 OH . 2 NH2.) OCH3
COOH
OH CHO
(a)
(b)
(c)
Cl NO2
Problems
NH2
NH2
Chapter 15
CH3 CH3
(d)
(e)
(f) CH3
COOH CH2CH3
Br
OH (g)
CH2CH2OH
(h) CH3O
(i)
OCH3
CH3
Br
O
37. Give a proper IUPAC name for each of the following commonly named substances.
CH3
OH
OH CH3
OCH3 (b)
(a)
(c)
H3C
OH CH3
CH2CHP CH2
CH2(CH2)4CH3 Hexylresorcinol
Durene
Eugenol
38. Draw the structure of each of the following compounds. If the name itself is incorrect, give a correct systematic alternative. (a) o-Chlorobenzaldehyde; (b) 2,4,6-trihydroxybenzene; (c) 4-nitro-o-xylene; (d) m-isopropylbenzoic acid; (e) 4,5-dibromoaniline; (f) p-methoxy-mnitroacetophenone. 39. The complete combustion of benzene is exothermic by approximately 2789 kcal mol21. What would this value be if benzene lacked aromatic stabilization? 40. The 1H NMR spectrum of naphthalene shows two multiplets (Figure 15-16). The upfield absorption (d 5 7.49 ppm) is due to the hydrogens at C2, C3, C6, and C7, and the downfield multiplet (d 5 7.86 ppm) is due to the hydrogens at C1, C4, C5, and C8. Explain why one set of hydrogens is deshielded more than the other. 41. Complete hydrogenation of 1,3,5,7-cyclooctatetraene is exothermic by 2101 kcal mol21. Hydrogenation of cyclooctene proceeds with DH 8 5 223 kcal mol21. Are these data consistent with the description of cyclooctatetraene presented in the chapter? 42. Which of the following structures qualify as being aromatic, according to Hückel’s rule?
(a)
j
CHP CH2 (b)
(c)
(d)
2
(e)
(f )
2 K
(g)
723
724
Benzene and Aromaticity
Chapter 15
43. Following are spectroscopic and other data for several compounds. Propose a structure for each of them. (a) Molecular formula 5 C6H4Br2. 1H NMR spectrum A. 13C NMR: 3 peaks. IR: n˜ 5 745 (s, broad) cm21. UV: lmax(e) 5 263(150), 270(250), and 278(180) nm. (b) Molecular formula 5 C7H7BrO. 1H NMR spectrum B. 13C NMR: 7 peaks. IR: n˜ 5 765(s) and 680(s) cm21. (c) Molecular formula 5 C9H11Br. 1H NMR spectrum C. 13C NMR: d 5 20.6 (CH3), 23.6 (CH3), 124.2 (Cquaternary), 129.0 (CH), 136.0 (Cquaternary), and 137.7 (Cquaternary) ppm.
1H
NMR 2H
2H
7.6
7.2
7.1
(CH3)4Si 1H
3H
NMR
8.0 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 300-MHz 1H NMR spectrum ppm (δ ) A
7.2
7.1
7.0
6.9
6.8
(CH3)4Si
3H 1H
10.0 9.5 9.0 8.5 8.0 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 300-MHz 1H NMR spectrum ppm (δ ) B 1H
6H
NMR
3H 2H (CH3)4Si
7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 300-MHz 1H NMR spectrum ppm (δ ) C
44. Both methylbenzene (toluene) and 1,6-heptadiyne have molecular formulas of C7H8 and molecular masses of 92. Which of the two mass spectra shown below corresponds to which compound? Explain your reasoning.
Problems
100
MS Relative abundance
Relative abundance
100
50
0
Chapter 15
MS
50
0 20
40
60 m/z
80
100
0
20
40
60
80
100
m/z
45. (a) Is it possible to distinguish the three isomers of dimethoxybenzene solely on the basis of the number of peaks in their proton-decoupled 13C NMR spectra? Explain. (b) How many different isomers of dimethoxynaphthalene exist? How many peaks should each one exhibit in the protondecoupled 13C NMR spectrum? 46. The species resulting from the addition of benzene to HF – SbF5 (Exercise 15-24) shows the following 13C NMR absorptions: d 5 52.2(CH2), 136.9(CH), 178.1(CH), and 186.6(CH) ppm. The signals at d 5 136.9 and d 5 186.6 are twice the intensity of the other peaks. Assign this spectrum. 47. Reaction review. Without consulting the Reaction Road Map on p. 721, suggest a reagent to convert benzene into each of the following compounds.
C(CH3)3
Cl
(a)
(b)
ON ECH3 C
A
(c)
CH2CH3
A
(e)
NO2
A
A
SO3H
A
(f)
(d)
Br
A
A
(g)
(h)
48. Give the expected major product of addition of each of the following reagent mixtures to benzene. (Hint: Look for analogies to the reactions presented in this chapter.) (a) (b) (c) (d) (e)
Cl2 1 AlCl3 T2O 1 T2SO4 (T 5 tritium, 3H) (CH3)3COH 1 H3PO4 N2O5 (which tends to dissociate into NO21 and NO32) (CH3)2C P CH2 1 H3PO4 (f) (CH3)3CCH2CH2Cl 1 AlCl3
Br Br A A (g) (CH3)2CCH2CH2C(CH3)2 AlBr3
(h) H3C
COCl AlCl3
49. Write mechanisms for reactions (c) and (f) in Problem 48.
O B Cl O S O OH B O
SO3H
50. Hexadeuteriobenzene, C6D6, is a very useful solvent for 1H NMR spectroscopy because it dissolves a wide variety of organic compounds and, being aromatic, is very stable. Suggest a method for the preparation of C6D6. 51. Propose a mechanism for the sulfonation of benzene using chlorosulfuric acid, ClSO3H (in the margin).
HCl
725
726
Chapter 15
Benzene and Aromaticity
52. Benzene reacts with sulfur dichloride, SCl2, in the presence of AlCl3 to give diphenyl sulfide, C6H5 – S – C6H5. Propose a mechanism for this process. 53. (a) 3-Phenylpropanoyl chloride, C6H5CH2CH2COCl, reacts with AlCl3 to give a single product with the formula C9H8O and an 1H NMR spectrum with signals at d 5 2.53 (t, J 5 8 Hz, 2 H), 3.02 (t, J 5 8 Hz, 2 H), and 7.2 – 7.7 (m, 4 H) ppm. Propose a structure and a mechanism for the formation of this product. (b) The product of the process described in (a) is subjected to the following reaction sequence: (1) NaBH4, CH3CH2OH; (2) Conc. H2SO4, 100 8C; (3) H2, Pd – C, CH3CH2OH. The final product exhibits five resonance lines in its 13C NMR spectrum. What is the structure of the substance formed after each of the steps in this sequence? 54. The text states that alkylated benzenes are more susceptible to electrophilic attack than is benzene itself. Draw a graph like that in Figure 15-20 to show how the energy profile of electrophilic substitution of methylbenzene (toluene) would differ quantitatively from that of benzene. 55. Like haloalkanes, haloarenes are readily converted into organometallic reagents, which are sources of nucleophilic carbon.
Br
Mg, (CH3CH2)2O, 25°C
MgBr
Phenylmagnesium bromide Grignard reagents
Cl
Mg, THF, 50°C
MgCl
Phenylmagnesium chloride
The chemical behavior of these reagents is very similar to that of their alkyl counterparts. Write the main product of each of the following sequences.
(a) C6H5Br
1. Li, (CH3CH2)2O 2. CH3CHO 3. H, H2O
(b) C6H5Cl
1. Mg, THF O D G 2. H2C O CH2 3. H, H2O
56. Give efficient syntheses of the following compounds, beginning with benzene. (a) 1-Phenyl-1heptanol; (b) 2-phenyl-2-butanol; (c) 1-phenyloctane. (Hint: Use a method from Section 15-14. Why will Friedel-Crafts alkylation not work?) H
O
57. Vanillin, whose structure is shown in the margin and is the subject of the Chapter Opening, is a benzene derivative with several functional groups, each one of which displays its characteristic reactivity. What would you expect to be the products of reaction of vanillin with each of the following reagents? (a) NaBH4, CH3CH2OH (b) NaOH, then CH3I
CH3O OH Vanillin
Vanillin is obtained by extraction from the seed pods of plants in the Vanilla genus in a process that dates back at least 500 years: The Mexican Aztecs used it to flavor xocoatl, a chocolate drink. Cortez discovered it in the court of Montezuma and was responsible for introducing it to Europe. Increasing demand for vanillin necessitated development of synthetic procedures, which involve extraction of related compounds from other plant-derived sources. One of the most important of these sources is the wood-derived waste from the manufacture of paper. The conversion of eugenol (an extract from cloves) into vanillin is very similar chemically. First, treatment
Problems
727
Chapter 15
of eugenol with KOH at 150 8C in a high-boiling solvent causes positional isomerization of its side-chain double bond: CH2CHPCH2
CHPCHCH3 KOH, 150°C, 1.5 h
CH3O
CH3O OH
OH Eugenol
Subsequently, oxidative cleavage, (see Section 12-12) completes the synthesis of vanillin. (c) Propose a mechanism for the isomerization depicted in the scheme. 58. Because of cyclic delocalization, structures A and B shown here for o-dimethylbenzene (o-xylene) are simply two resonance forms of the same molecule. Can the same be said for the two dimethylcyclooctatetraene structures C and D? Explain. CH3
CH3
CH3 CH3
CH3
A
B
CH3 CH3
C
CH3
D
59. The energy levels of the 2-propenyl (allyl) and cyclopropenyl p systems (see margin) are compared qualitatively in the diagram below. (a) Draw the three molecular orbitals of each system, using plus and minus signs and dotted lines to indicate bonding overlap and nodes, as in Figure 15-4. Does either of these systems possess degenerate molecular orbitals? (b) How many p electrons would give rise to the maximum stabilization of the cyclopropenyl system, relative to 2-propenyl (allyl)? (Compare Figure 15-5, for benzene.) Draw Lewis structures for both systems with this number of p electrons and any appropriate atomic charges. (c) Could the cyclopropenyl system drawn in (b) qualify as being “aromatic”? Explain.
2-Propenyl (Allyl)
Antibonding molecular orbital
Nonbonding molecular orbital E Cyclopropenyl
Bonding molecular orbital
Cyclopropenyl
60.
61.
O
2-Propenyl (Allyl)
2,3-Diphenylcyclopropenone (see structure in the margin) forms an addition product with HBr that exhibits the properties of an ionic salt. Suggest a structure for this product and a reason for its existence as a stable entity. Is cyclobutadiene dication (C4H421) aromatic according to Hückel’s rule? Sketch its p molecular orbital diagram to illustrate.
62. All the molecules shown below are examples of “fulvenes,” or methylenecyclopentadienes. N(CH3)2
5-Methylene1,3-cyclopentadiene “Fulvene”
6-Dimethylaminofulvene
H3C
CH3
6,6-Dimethylfulvene
C6H5
C6H5
6,6-Diphenylfulvene
C6H5
C6H5
2,3-Diphenylcyclopropenone
Chapter 15
Benzene and Aromaticity
(a) One of these structures is considerably more acidic than the others, with a pKa around 20. Identify it and its most acidic hydrogen(s), and explain why it is an unusually strong acid for a molecule with only carbon-hydrogen bonds. (b) None of the 7-methylene-1,3,5-cycloheptatrienes that correspond to the fulvene structures above show any unusual acidity. Explain.
H
H
63. A characteristic reaction of fulvenes is nucleophilic addition. To which carbon in a fulvene would you expect nucleophiles to add, and why? 64.
1,6-Methano[10]annulene
(a) The 1H NMR spectrum of [18]annulene shows two signals, at d 5 9.28 (12 H) and 22.99 (6 H) ppm. The negative chemical shift value refers to a resonance upfield (to the right) of (CH3)4Si. Explain this spectrum. (Hint: Consult Figure 15-9.) (b) The unusual molecule 1,6-methano[10]annulene (shown in the margin) exhibits two sets of signals in the 1H NMR spectrum at d 5 7.10 (8 H) and 20.50 (2 H) ppm. Is this result a sign of aromatic character?
65. The 1H NMR spectrum of the most stable isomer of [14]annulene shows two signals, at d 5 20.61 (4 H) and 7.88 (10 H) ppm. Two possible structures for [14]annulene are shown here. How do they differ? Which one corresponds to the NMR spectrum described?
B
A
66. Explain the following reaction and the indicated stereochemical result mechanistically. (H HOCH2 H ∞CH3 C
O DG H2C C C ( CH3 H O B HgOCCH3
Phenylmercury acetate
1. AlCl3 2. H, H2O
67. Metal-substituted benzenes have a long history of use in medicine. Before antibiotics were discovered, phenylarsenic derivatives were the only treatment for a number of diseases. Phenylmercury compounds continue to be used as fungicides and antimicrobial agents to the present day. On the basis of the general principles explained in this chapter and your knowledge of the characteristics of compounds of Hg21 (see Section 12-7), propose a sensible synthesis of phenylmercury acetate (shown in the margin).
Team Problem 68. As a team, discuss the following complementary experimental results as they pertain to the mechanism of electrophilic aromatic substitution. (a) A solution of HCl and benzene is colorless and does not conduct electricity, whereas a solution of HCl and AlCl3 and benzene is colored and does conduct electricity. (b) The following are 13C NMR chemical shifts for the species below (see also Exercise 15-24):
1
yZZy y y
728
2
H 6
y y
3
4
5
H
Cl and C5: 186.6 ppm C3: 178.1 ppm C2 and C4: 136.9 ppm C6: 52.2 ppm
Problems
Chapter 15
(c) The relative rates of chlorination of the following compounds are as shown. Compound
Relative Rate
Benzene Methylbenzene 1,4-Dimethylbenzene 1,2-Dimethylbenzene 1,2,4-Trimethylbenzene 1,2,3-Trimethylbenzene 1,2,3,4-Tetramethylbenzene 1,2,3,5-Tetramethylbenzene Pentamethylbenzene
0.0005 0.157 1.00 2.1 200 340 2000 240,000 360,000
(d) When 1,3,5-trimethylbenzene is treated with fluoroethane and one equivalent of BF3 at 280 8C, an isolable solid salt with a melting point of 215 8C is produced. Heating the salt results in 1-ethyl-2,4,6-trimethylbenzene.
Preprofessional Problems 69. o-Iodoaniline is the common name of which of the following compounds?
CH3
NH2
NH2
NH2
I
I (b)
(a)
(c)
(d)
CH3
I I
70. The species that is not aromatic according to Hückel’s rule is
(a)
(b)
(d)
(c)
71. When compound A (shown in the margin) is treated with dilute mineral acid, an isomerization takes place. Which of the following compounds is the new isomer formed?
CH3
CH2CH3
CH2CH2CH3
CH2CH3
CH3 CH2CH3 (c)
(d) CH3 CH2CH3
CHCH3 A
(b)
(a)
CHCH3
CH2CH3
729
730
Chapter 15
Benzene and Aromaticity
72. Which set of reagents will best carry out the conversion shown?
Br (a) HBr, peroxides; (b) Br2, FeBr3; (c) Br2 in CCl4; (d) KBr. 73. One of the compounds shown here contains carbon – carbon bonds that are 1.39Å long. Which one? CH3 (a)
(b)
(c)
CH3
(e) H3CC q CCH3
(d) CH3
SIXTEEN
C H A P T E R
[
]
Electrophilic Attack on Derivatives of Benzene
O O
Substituents Control Regioselectivity
O O
t some time in your life you have probably ingested at least one of the painkillers aspirin, acetaminophen, naproxen, or ibuprofen, perhaps better known under one of their respective brand names, aspirin, Tylenol, Naprosyn, and Advil. Aspirin, acetaminophen, and ibuprofen are ortho- or paradisubstituted benzenes; naproxen is a disubstituted naphthalene. How are such compounds synthesized? The answer is by electrophilic aromatic substitution.
A
OH
COOH O B OCCH3
OH
HNCCH3 B O 2-Acetyloxybenzoic acid (Aspirin)
H 3C
O
N-(4-Hydroxyphenyl)acetamide (Acetaminophen)
CH3 OH CH3O
O 2-[2-(6-Methoxynaphthyl)]propanoic acid (Naproxen)
2-[4-(2-Methylpropyl)phenyl]propanoic acid (Ibuprofen)
Aspirin, prepared industrially by selective electrophilic aromatic substitution of phenol, is arguably the blockbuster drug of all times. Its active metabolite, 2-hydroxybenzoic acid (salicylic acid), obtained from the bark of the white willow tree, has been used for four millennia for the treatment of inflammation and to relieve pain or discomfort caused by arthritis, soft-tissue injuries, and fever. Aspirin was discovered by the German company Bayer in the late 19th century and ironically marketed together with another drug, heroin, whose addictive side effects were not recognized then.
732
Chapter 16
Electrophilic Attack on Derivatives of Benzene
Chapter 15 described the use of this transformation in the preparation of monosubstituted benzenes. In this chapter we analyze the effect of such a first substituent on the reactivity and regioselectivity (orientation) of a subsequent electrophilic substitution reaction. Specifically, we shall see that substituents on benzene can be grouped into (1) activators (electron donors), which generally direct a second electrophilic attack to the ortho and para positions, and (2) deactivators (electron acceptors), which generally direct electrophiles to the meta positions. We will then devise strategies toward the synthesis of polysubstituted arenes, such as the analgesics depicted on the previous page. Acceptor
Donor ortho
ortho
meta
meta
para Activated ring
Deactivated ring
16-1 Activation or Deactivation by Substituents on a Benzene Ring In Section 14-8 we discussed the effect that substituents have on the efficiency of the DielsAlder reaction: Electron donors on the diene and acceptors on the dienophile are beneficial to the outcome of the cycloaddition. Chapter 15 revealed another manifestation of these effects: Introduction of electron-withdrawing substituents into the benzene ring (e.g., as in nitration) caused further electrophilic aromatic substitution (EAS) to slow down, whereas the incorporation of donors, as in the Friedel-Crafts alkylation, caused substitution to accelerate. What are the factors that contribute to the activating or deactivating nature of substituents in these processes? How do they make a monosubstituted benzene more or less susceptible to further electrophilic attack? NO2
CH3
Increasing rate of EAS
The electronic influence of any substituent is determined by an interplay of two effects that, depending on the structure of the substituent, may operate simultaneously: induction and resonance. Induction occurs through the s framework, tapers off rapidly with distance, and is mostly governed by the relative electronegativity of atoms and the resulting polarization of bonds (Tables 1-2 and 8-2). Resonance takes place through p bonds, is therefore longer range, and is particularly strong in charged systems (Section 1-5, Chapter 14). Let us look at both of these effects of typical groups introduced by electrophilic aromatic substitution, starting with inductive donors and acceptors. Thus, simple alkyl groups, such as methyl, are donating by virtue of their hyperconjugating s frame, a phenomenon that we encountered earlier (Sections 7-5 and 11-5). On the other hand, trifluoromethyl (by virtue of its electronegative fluorines) is electron withdrawing. Similarly, directly bound heteroatoms, such as N, O, and the halogens (by virtue of their relative electronegativity), as well as positively polarized atoms, such as those in carbonyl, cyano, nitro, and sulfonyl functions, are inductively electron withdrawing.
16-1 Activation or Deactivation by Substituents
Chapter 16
Inductive Effects of Some Substituents on the Benzene Ring Donors D
Acceptors A
D
A
D –CH3, other alkyl groups
A –CF3, –NR2, –OR, –X (–F, –Cl, –Br, –I),
ðOð O š O B B K O CR, O C q Nð, ONH , O S O OH B šð O ðOð
Now we turn to substituents that resonate with the aromatic p system. Resonance donors bear at least one electron pair capable of delocalization into the benzene ring. Therefore, such groups as –NR2, –OR, and the halogens belong in that category. You will note that, inductively, these groups are electron withdrawing; in other words, here the two phenomena, induction and resonance, are opposing each other. Which one wins out? The answer depends on the relative electronegativity of the heteroatoms (Table 1-2) and on the ability of their respective p orbitals to overlap with the aromatic p system. For amino and alkoxy groups, resonance overrides induction. For the halogens, induction outcompetes resonance, making them weak electron acceptors. Resonance Donation to Benzene ðD
D
D
D
ð , –Cl , –Br ð D –NR2, –OR, –F ð, –I ð
Finally, groups bearing a polarized double or triple bond, whose positive (d1) end is attached to the benzene nucleus, such as carbonyl, cyano, nitro, and sulfonyl, are electron withdrawing through resonance. Resonance Acceptance from Benzene K B
A
A)
E
A) E B
B
A) E B
A) E B
)O) )O) > O B B A Kp > B O CR,O C q N), ON O OH p H , OS B B > O p) )O)
Note that, here, resonance reinforces induction. Electrostatic potential maps indicate the presence of electron-donating substituents by depicting the benzene ring as shaded relatively red; electron-withdrawing groups make the benzene ring appear shaded relatively blue (green).
733
734
Chapter 16
Electrophilic Attack on Derivatives of Benzene
Benzene
Methylbenzene (Toluene)
Benzenamine (Aniline)
Nitrobenzene
Exercise 16-1 Explain the 1H NMR spectral assignments in Figures 15-11 and 15-12. (Hint: Draw resonance structures involving the substituents on the benzene ring.)
Exercise 16-2 The 13C NMR spectrum of phenol, C6H5OH, shows four lines at d 5 116.1 (C2), 120.8 (C4), 130.5 (C3), and 155.6 (C1) ppm. Explain these assignments. (Hint: The 13C chemical shift for benzene is d 5 128.7 ppm.)
How do we know whether a substituent functions as a donor or acceptor? In electrophilic aromatic substitution, the answer is simple. Because the attacking species is an electrophile, the more electron rich the arene, the faster the reaction. Conversely, the more electron poor the arene, the slower the reaction. Hence, electron donors activate the ring, whereas electron acceptors deactivate. Relative Rates of Nitration of C6H5X CF3 . NO2 X 5 NH(C6H5) . OH . CH3 . H . Cl . CO2CH2CH3 . 106 1000 25 1 0.033 0.0037 2.6 3 1025 6 3 1028 Increasing rate of nitration
Exercise 16-3 Specify whether the benzene rings in the compounds below are activated or deactivated. CO2H
CH2CH3
OCH3
NO2
(a)
(b)
(c)
(d)
CH3 CH2CH3
CF3
N(CH3)2
In Summary When considering the effect of substituents on the reactivity of the benzene nucleus, we have to analyze the contributions that occur by induction and resonance. We can group these substituents into two classes: (1) electron donors, which accelerate electrophilic aromatic substitutions relative to benzene, and (2) electron acceptors, which retard them.
16-2 Directing Inductive Effects of Alkyl Groups We are now ready to tackle the question of the regioselectivity (orientation) of electrophilic aromatic substitution of substituted benzenes. What controls the position along the benzene ring at which an electrophile will attack? We begin with the electrophilic substitution reactions of alkyl-substituted benzenes, such as methylbenzene (toluene), in which the methyl group is electron donating by induction.
16-2 Directing Inductive Effects of Alkyl Groups
Chapter 16
Groups that donate electrons by induction are activating and direct ortho and para Electrophilic bromination of methylbenzene (toluene) is considerably faster than the bromination of benzene itself. The reaction is also regioselective: It results mainly in para (60%) and ortho (40%) substitutions, with virtually no meta product. Electrophilic Bromination of Methylbenzene (Toluene) Gives Ortho and Para Substitution CH3
CH3
CH3
CH3
REACTION
Br BrO Br, FeBr3, CCl4
HBr
Br
39%
Br 60%
HOH p
p
Nu
H2> O p
> C OOH p D Nu
MECHANISM
p HO p
p
p
Alkoxide ion
Note that the new Nu – C bond is made up entirely of the electron pair of the nucleophile. The entire transformation is reminiscent of an SN2 reaction. In that process, a leaving group is displaced. Here, an electron pair is displaced from a shared position between carbon and oxygen to a position solely on the oxygen atom. Additions of strongly basic nucleophiles to carbonyl groups typically follow the nucleophilic addition – protonation pathway. The second mechanism predominates under acidic conditions and begins with electrophilic attack. In the first step, protonation of the carbonyl oxygen occurs, facilitated by the polarization of the CPO bond and the presence of lone electron pairs on the oxygen atom. The oxygen is only weakly basic, as shown by the pKa of the conjugate acid, which ranges from about 27 to 28. Thus, in a dilute acidic medium, in which many carbonyl addition reactions are carried out, most of the carbonyl compound will stay unprotonated. However, the small amount of protonated material behaves like a very reactive carbon electrophile. Nucleophilic attack by the nucleophile completes the addition process and shifts the first, unfavorable equilibrium. Electrophilic Protonation–Addition (Acidic Conditions)
O C P> p
H
C POH p
)NuH )NuH
CO> OH p D HNu
H H
MECHANISM CO> OH p D Nu
Protonated carbonyl group pKa
8
The electrophilic protonation–addition mechanism is best suited for reactions of relatively weakly basic nucleophiles. The acidic conditions are incompatible with strongly basic nucleophiles, because protonation of the nucleophile occurs.
In Summary There are three regions of reactivity in aldehydes and ketones. The first two are the two atoms of the carbonyl group and are the subject of the remainder of this chapter. The third is the adjacent carbon. The reactivity of the carbonyl group is governed by addition processes. Nucleophilic organometallic and hydride reagents result in irreversible alcohol formation (after protonation). Ionic additions of NuH (Nu 5 OH, OR, SR, NR2) are reversible and may begin with nucleophilic attack at the carbonyl carbon, followed by electrophilic trapping of the alkoxide anion so generated. Alternatively, in acidic media, protonation precedes the addition of the nucleophile.
17-6 Addition of Water to Form Hydrates This section and Sections 17-7 and 17-8 introduce the reactions of aldehydes and ketones with water and alcohols. These compounds attack the carbonyl group through the mechanisms just outlined, utilizing either acid or base catalysis.
789
790
Aldehydes and Ketones
Chapter 17
Water hydrates the carbonyl group Water is capable of attacking the carbonyl function of aldehydes and ketones. The transformation can be catalyzed by either acid or base and leads to equilibration with the corresponding geminal diols, RC(OH)2R9, also called carbonyl hydrates. REACTION
Hydration of the Carbonyl Group C PO
HOH
H or HO
C O OH D HO
K
Geminal diol
In the base-catalyzed mechanism, the hydroxide functions as the nucleophile. Water then protonates the intermediate adduct, a hydroxy alkoxide, to give the product diol and to regenerate the catalyst. Mechanism of Base-Catalyzed Hydration
MECHANISM C
> O p
)> OH
p
O) GC G > p > HO p
> HOH p
Hydroxy alkoxide
OH GC G > p > HO p
> HO p)
Geminal diol
In the acid-catalyzed mechanism, the sequence of events is reversed. Here, initial protonation facilitates nucleophilic attack by the weak nucleophile water. Subsequently, the catalytic proton is lost to reenter the catalytic cycle.
MECHANISM Mechanism of Acid-Catalyzed Hydration O C P> p
H
OH C P O >
H2> O p
Protonated carbonyl
OH GC O > p HO) A H
OH GC O > p > HO p
H
Geminal diol
Hydration is reversible
š CP O
š CO Oð Carbocationlike
As these equations indicate, hydrations of aldehydes and ketones are reversible. The equilibrium lies to the left for ketones and to the right for formaldehyde and aldehydes bearing inductively electron-withdrawing substituents. For ordinary aldehydes, the equilibrium constant approaches unity. How can these trends be explained? Look again at the resonance structures of the carbonyl group, as described in Section 17-2 and repeated here in the margin. In the dipolar resonance form, the carbon possesses carbocation character. Therefore, alkyl groups, which stabilize carbocations (Section 7-5), stabilize carbonyl compounds as well. Conversely, inductively electron-withdrawing substituents, such as CCl3 and CF3, increase positive charge at the carbonyl carbon and destabilize carbonyl compounds. The stabilities of the product diols are affected to a lesser extent by substituents. As a net result, therefore, relative to hydration of formaldehyde, the hydration reactions of aldehydes and ketones are progressively more endothermic, whereas hydrations of carbonyl compounds containing electronwithdrawing groups are more exothermic. These thermodynamic effects are paralleled by differences in kinetic reactivity: Carbonyl compounds containing electron-withdrawing groups are the most electrophilic and the most reactive, followed in turn by formaldehyde, other aldehydes, and finally ketones. We shall see that the same trends also govern other addition reactions.
17-7 Addition of Alcohols to Form Hemiacetals and Acetals
Cl3C
H
H
O B C E H
H3 C
H
O B C E H
H3C
H
O B C E H
CH3
O B Cl3CCH
K 104
O B HCH
K 103
O B CH3CH Electron-withdrawing CCl3 group generates additional positive charge at carbonyl carbon
Electron-donating CH3 groups reduce positive charge at carbonyl carbon
Increasing reactivity
O B CH3CCH3
K
1
K 102
Increasing favorability of addition
Equilibrium Constants K for the Hydration of Typical Carbonyl Compounds
Relative Reactivities of Carbonyl Groups: Aldehydes Are More Reactive than Ketones O B C E H
791
Chapter 17
Exercise 17-7 O O O B B B (a) Rank in order of increasing favorability of hydration: Cl3CCH, Cl3CCCH3, Cl3CCCCl3. 18 O B 18 (b) Treatment of acetone with H2 O results in the formation of labeled acetone, CH3CCH3. Explain.
Despite the fact that hydration is energetically favorable in some cases, it is usually not possible to isolate carbonyl hydrates as pure substances: They lose water too easily, regenerating the original carbonyl compound. However, they may participate as intermediates in subsequent chemistry, such as the oxidation of aldehydes to carboxylic acids under aqueous conditions (Sections 8-6 and 17-4).
In Summary The carbonyl group of aldehydes and ketones is hydrated by water. Aldehydes are more reactive than ketones. Electron-withdrawing substituents render the carbonyl group more electrophilic. Hydration is an equilibrium process that may be catalyzed by acids or bases.
17-7 Addition of Alcohols to Form Hemiacetals and Acetals In this section we shall see that alcohols add to carbonyl groups in much the same manner as water does. Both acids and bases catalyze the process. In addition, acids catalyze the further transformation of the initial addition product to furnish acetals by replacement of the hydroxy group with an alkoxy substituent.
Aldehydes and ketones form hemiacetals reversibly Not surprisingly, alcohols also undergo addition to aldehydes and ketones, by a mechanism virtually identical to that outlined for water. The adducts formed are called hemiacetals (hemi, Greek, half) because they are intermediates on the way to acetals.
The English artist Damien Hirst became famous for a series of works in which dead animals are preserved in aqueous formaldehyde, such as “The Physical Impossibility of Death in the Mind of Someone Living,” a 14-foot (4.3 m) tiger shark immersed in a glass tank.
792
Aldehydes and Ketones
Chapter 17
Hemiacetal Formation
R
O B DC
DH
H O A RO C OOR A H
ROH
R
O B DC
DR
H O A RO C OOR A R
ROH
A hemiacetal
A hemiacetal
Like hydration, these addition reactions are governed by equilibria that usually favor the starting carbonyl compound. Hemiacetals, like hydrates, are therefore usually not isolable. Exceptions are those formed from reactive carbonyl compounds such as formaldehyde or 2,2,2trichloroacetaldehyde. Hemiacetals are also isolable from hydroxy aldehydes and ketones when cyclization leads to the formation of relatively strain-free five- and six-membered rings. Intramolecular Hemiacetal Formation: Cyclic Hemiacetals Are More Stable than Acyclic Hemiacetals
)O) B C H
H
> O)
H
H 2C H2C
Glucose CH2OH
H H C PO
H C
H OH
H
C
C
HO
C H2
O
C
)> O
C H2C
H
H
H 2C
CH2
C H2
> O) CH2
O
O
O
OH H 0.003% Aldehyde form H or HO
CH2OH O
H
C
C
H OH
H
C
C
H
OH 99%
HO
H C OH New stereocenter
Cyclic hemiacetal (Two stereoisomers)
5-Hydroxypentanal
A cyclic hemiacetal, stable
Intramolecular hemiacetal formation is common in sugar chemistry (Chapter 24). For example, glucose, the most common simple sugar in nature, exists as an equilibrium mixture of an acyclic pentahydroxyaldehyde and two stereoisomeric cyclic hemiacetals. The cyclic forms constitute more than 99% of the mixture in aqueous solution.
Acids catalyze acetal formation In the presence of excess alcohol, the acid-catalyzed reaction of aldehydes and ketones proceeds beyond the hemiacetal stage. Under these conditions, the hydroxy function of the initial adduct is replaced by another alkoxy unit derived from the alcohol. The resulting compounds are called acetals. (Ketal is an older term for an acetal derived from a ketone.) Acetal Synthesis
REACTION
O B RCR
H
2 ROH
OR A RO COOR A R An acetal
H2O
17-8 Acetals as Protecting Groups
Chapter 17
793
The net change is the replacement of the carbonyl oxygen by two alkoxy groups and the formation of one equivalent of water. Let us examine the mechanism of this transformation for an aldehyde. The initial reaction is the ordinary acid-catalyzed addition of the first molecule of alcohol. The resulting hemiacetal can be protonated at the hydroxy group, changing this substituent into water, a good leaving group. Upon loss of water, the resulting carbocation is stabilized by resonance with a lone electron pair on oxygen. A second molecule of alcohol now adds to the electrophilic carbon, initially giving the protonated acetal, which is then deprotonated to the final product. Mechanism of Acetal Formation
MECHANISM
Step 1. Hemiacetal generation: acid-catalyzed addition of first molecule of alcohol
ðOð RCH
ðš OH
ðOH H H
C
C H
R
R
š ROH
H
ðš OH
ðš OH
š OH R
R
C
H
H
R
H
H
ð OR
OR H
C
Hemiacetal
Step 2. Acetal generation: acid-catalyzed SN1 displacement of water by second molecule of alcohol H
ðš OH R
C OR ð
ðOH
H
H H
R
C OR ð
š O H2
H
R
š H2 O
H
C ðOR
R
š OH R
H C
š ROH
OR
H R
C OR ð
R
Oð H
H H
H
R
OR
C OR ð
Acetal
Each step is reversible; the entire sequence, starting from the carbonyl compound and ending with the acetal, is an equilibrium process. In the presence of the acid catalyst, the equilibrium may be shifted in either direction: toward acetal by using excess alcohol or removing water; toward starting aldehyde or ketone by adding excess water, a process called acetal hydrolysis. However, in contrast with hydrates and hemiacetals, acetals may be isolated as pure substances by neutralizing the acid catalyst used in their formation. Reversal of acetal formation cannot occur in the absence of acid. Therefore, acetals may be prepared and put to use synthetically, as described in the next section.
In Summary Alcohols react with aldehydes and ketones to form hemiacetals. This process, like hydration, is reversible and is catalyzed by both acids and bases. Hemiacetals are converted by acid and excess alcohol into acetals. Acetals are stable under neutral or basic conditions but are hydrolyzed by aqueous acid.
17-8 Acetals as Protecting Groups In the conversion of an aldehyde or ketone to an acetal, the reactive carbonyl function is transformed into a relatively unreactive ether-like moiety. Because acetal formation is reversible, this process amounts to one of masking, or protecting, the carbonyl group. Such protection is necessary when selective reactions (for example, with nucleophiles) are required at one functional site of a molecule and the presence of an unprotected carbonyl group
MATION AN I
ANIMATED MECHANISM: Acetal formation
794
Aldehydes and Ketones
Chapter 17
elsewhere in the same molecule might interfere. This section describes the execution of such protecting strategies.
Cyclic acetal formation protects carbonyl groups from attack by nucleophiles Diols, such as 1,2-ethanediol, are particularly effective reagents compared to normal alcohols for forming acetals. Use of diols converts aldehydes and ketones into cyclic acetals, which are generally more stable than their acyclic counterparts. The reason lies in part in their relatively more favorable (or, better, less unfavorable) entropy of formation (Section 2-1). Thus, as shown below, two molecules of starting material (carbonyl compound and diol) are converted into two molecules of product (acetal and water) in this process. In contrast, acetal formation with a normal alcohol (Section 17-7) uses up three molecules (carbonyl compound and two equivalents of alcohol), resulting in the same number of product molecules. Acetaldehyde undergoes acidcatalyzed cyclotrimerization to the acetal commonly known as paraldehyde, a hypnotic sedative that is also used to prevent the growth of mold on leather, 3 CH3CHO
Cyclic Acetalization O B CH3CH
HOCH2CH2OH
H
O
H3C[ O
H H2O
H2C
CH2
O
O
ACA
H3C H 73%
]CH3
O 0 CH3
O
Paraldehyde
O O
O
O
A cyclic acetal
Cyclic acetals are readily hydrolyzed in the presence of aqueous acid, but are not attacked by many basic, organometallic, and hydride reagents. These properties make them most useful as protecting groups for the carbonyl function in aldehydes and ketones. An example is the alkylation of an alkynyl anion with 3-iodopropanal 1,2ethanediol acetal.
ICH2CH2 O CHO 3-Iodopropanal
HOCH2CH2OH, H Protection
Use of a Protected Aldehyde in Synthesis CH3(CH2)3C q CLi HO O 1-Hexynyllithium C SN2 reaction ICH2CH2O O LiI 3-Iodopropanal 1,2-ethanediol acetal
HO O C CH3(CH2)3C q C O CH2CH2O O 70% 4-Nonynal 1,2-ethanediol acetal
H, H2O HOCH2CH2OH
CH3(CH2)3C q C O CH2CH2 O CHO
Deprotection
90% 4-Nonynal
When the same alkylation is attempted with unprotected 3-iodopropanal, the alkynyl anion attacks the carbonyl group.
17-8 Acetals as Protecting Groups
Chapter 17
795
Exercise 17-8
Working with the Concepts: Using an Acetal Protecting Group in Synthesis Suggest a convenient way of converting compound A into compound B. (Caution: How many carbon atoms are present in A and B, respectively?) O O OH
Br A
B
Strategy Begin by identifying the structural change that the problem asks you to achieve. Notice that the target molecule B contains one more carbon atom in the chain than does the starting compound A. Therefore, the required transformation is RBr y RCH2OH. We can prepare primary alcohols by addition of Grignard reagents to formaldehyde (Section 8-8): 1. H2CPO, ether 2. H, H2O
RMgBr
RCH2OH
The problem with substrate A is that it already contains a carbonyl group. This group is likely to be incompatible with the preparation and use of a Grignard function at the opposite end of the molecule. How can we prevent interference by this carbonyl group? Solution • Before doing anything else, protect the carbonyl group as an acetal. Being similar in function to ethers, acetals are inert to Grignard reagents. • We follow acetal formation with the reactions needed to complete our synthesis: Convert the bromoalkane functional group to the Grignard reagent, react with formaldehyde to add the necessary carbon atom, and protonate the resulting alkoxide with aqueous acid. • The final step, treatment with aqueous acid, conveniently also hydrolyzes the acetal group and restores the original ketone function. This is the deprotection step referred to in the example in the text above. O HOCH2CH2OH, H
O
O
Mg, ether
Br
H2C P O
O
O
O
Br
O
MgBr O
O MgBr
H, H2O
OH
Exercise 17-9
Try It Yourself Propose a synthetic scheme to convert bromobenzene into HOCH2 O (Hint: See Sections 15-13 and 16-3.)
OC
O l i
.
CH3
Thiols react with the carbonyl group to form thioacetals Thiols, the sulfur analogs of alcohols (see Section 9-10), react with aldehydes and ketones by a mechanism identical with the one described for alcohols. Instead of a proton catalyst, a Lewis acid, such as BF3 or ZnCl2, is often used in ether solvent. The reaction produces the sulfur analogs of acetals, thioacetals, again, especially well for cyclic systems. Cyclic Thioacetal Formation from a Ketone HSCH2CH2SH, ZnCl2, (CH3CH2)2O, 25°C H2O
O
S
S
95% A cyclic thioacetal
RS
SR
Thioacetal
796
Chapter 17
Aldehydes and Ketones
These sulfur derivatives are stable in aqueous acid, a medium that hydrolyzes ordinary acetals. The difference in reactivity may be useful in synthesis when it is necessary to differentiate two different carbonyl groups in the same molecule. Hydrolysis of thioacetals is carried out using mercuric chloride in aqueous acetonitrile. The driving force is the formation of insoluble mercuric sulfides. Thioacetal Hydrolysis S
H2O, HgCl2, CaCO3, CH3CN
S
O
Exercise 17-10
Working with the Concepts: Differentiating Carbonyl Groups How would you accomplish the following transformation? (Hint: See Section 8-8.)
S
S
S
S
O O
OH
Strategy Formation of a secondary alcohol may be achieved by addition of a Grignard reagent to an aldehyde. Choose a deprotection method that will release the aldehydic carbonyl function but will leave the thioacetal protected cyclic ketone part untouched. Solution • Hydrolyze the acetal with aqueous acid. • React the resulting aldehyde with the appropriate Grignard reagent to give the desired alcohol.
S
S
S O
S
H, H2O
CHO
O
1. CH3CH2CH2MgCl 2. H, H2O
S
S
OH
Neither of these steps affects the thioacetal moiety.
Exercise 17-11 OH
Try It Yourself O O
Show how the starting material for Exercise 17-10 may be converted into the alcohol in the margin.
17-9 Nucleophilic Addition of Ammonia and Its Derivatives
797
Chapter 17
Thioacetals are desulfurized to the corresponding hydrocarbon by treatment with Raney nickel (Section 12-2). Thioacetal generation followed by desulfurization is used to convert a carbonyl into a methylene group under neutral conditions.
S
Raney Ni, H2
S
H
H
65%
Exercise 17-12 Suggest a possible synthesis of cyclodecane from
.
(Caution: Simple hydrogenation will not do. Hint: See Section 12-12.)
In Summary Acetals and thioacetals are useful protecting groups for aldehydes and ketones. Acetals are formed under acidic conditions and are stable toward bases and nucleophiles. They are hydrolyzed by aqueous acid. Thioacetals are usually prepared using Lewis acid catalysts and are stable to both aqueous base and acid. Mercuric salts are required for thioacetal hydrolysis. Thioacetals may also be desulfurized to hydrocarbons with Raney nickel.
17-9 Nucleophilic Addition of Ammonia and Its Derivatives H
Ammonia and the amines may be regarded as nitrogen analogs of water and alcohols. Do they add to aldehydes and ketones? In fact, they do, giving products corresponding to those just studied. We shall see one important difference, however. The products of addition of amines and their derivatives lose water, furnishing either of two new derivatives of the original carbonyl compounds: imines and enamines.
GšD N A H
H
Ammonia
R
GšD N A H
H
Primary amine
Ammonia and primary amines form imines Upon exposure to an amine, aldehydes and ketones initially form hemiaminals, the nitrogen analogs of hemiacetals. Hemiaminals of primary amines readily lose water to form a carbon– nitrogen double bond. This function is called an imine (an older name is Schiff* base) and is the nitrogen analog of a carbonyl group.
REACTION
> H2 RO N
C
> O p
Addition
H
H A A > N O) E H Op R C
Deprotonation at N, reprotonation at O
H A > N OH p E H Op C R A hemiaminal
The mechanism of the elimination of water from the hemiaminal is the same as that for the decomposition of a hemiacetal to the carbonyl compound and alcohol. It begins with protonation of the hydroxy group. (Protonation of the more basic nitrogen just leads back to the original carbonyl compound.) Dehydration follows, and then deprotonation of the intermediate iminium ion. *Professor Hugo Schiff (1834 – 1915), University of Florence, Italy.
Elimination of water H2O H2O
C
> N
)
Imine Formation from Amines and Aldehydes or Ketones
An imine
R
798
Aldehydes and Ketones
Chapter 17
Mechanism of Hemiaminal Dehydration
MECHANISM R
H GšD N A CO š OH
R
H
H
H GšD N A CO š OH H
R
š HOH š HOH
GšD N A C
R
H
Hemiaminal
G D N B C
R
H H H
Iminium ion
G k N B C
Imine
Processes such as imine formation from a primary amine and an aldehyde or ketone, in which two molecules are joined with the elimination of water (or other small molecules such as alcohols), are called condensations. Condensation of a Ketone with a Primary Amine R R f f RNH2 O P C H2O RNP C i i R R Imines are useful compounds: For example, they are commonly employed in the synthesis of more complex amines (Section 21-6). However, the condensation reaction above is reversible, making imines very susceptible to hydrolysis and often difficult to isolate as pure substances. As a result, when a synthetic sequence requires an imine as an intermediate, the appropriate amine and carbonyl compound are usually mixed in the presence of the reagent that is needed to convert with the imine further. Thus, the imine reacts as soon as it is formed. In this way, the condensation equilibrium is constantly shifted to produce more imine, continuing the reaction until it finally goes to completion. If care is taken to remove the water that is formed in the condensation process (for example, by continuous distillation from the reaction mixture), imines may be isolated, often in high yield. For example, NH2 O B CH3CCH3
H
H3C H 3C
i C PN f
H2O
95% Exercise 17-13 Reagent A has been used with aldehydes to prepare crystalline imidazolidine derivatives, such as B, for the purpose of their isolation and structural identification. Write a mechanism for the formation of B. [Hint: Notice that the product is a nitrogen analog of an acetal. Develop a mechanism analogous to that for acetal formation (Section 17-7), using the two amine groups in the starting diamine in place of two alcohol molecules.] H NH N H
C6H5 C E 6H5
O B
H3C
H
CH3OH, H
H2O
CH A6 5 N CH3 H N A C6H5
A
B
N,Nⴕ-Diphenyl-1,2-ethanediamine
2-Methyl-1,3-diphenyl-1,3-diazacyclopentane (2-Methyl-1,3-diphenylimidazolidine) (m.p. 102ⴗC)
17-9 Nucleophilic Addition of Ammonia and Its Derivatives
Chapter 17
799
C H E MI C AL H I G H L I G H T 1 7 - 1 Imines in Biology interconversion of amino acids and 2-oxocarboxylic acids (ketoacids). Imines are the key intermediates in this transformation, by which amino acids are broken down biologically (a process called catabolism).
Imines play pivotal roles in the biological transformations of 2-aminocarboxylic acids (amino acids), the building blocks of proteins (Chapter 26). Pyridoxal and pyridoxamine, two derivatives of vitamin B6 (pyridoxine), mediate the
CH2OH HOCH2
OH N
CH3
Pyridoxine (Vitamin B6)
H CHO NH2 A RCH2CHCO2H
HOCH2
OH
HOCH2
CO H D 2 NO CH G J G C CH2R OH
N
Amino acid
N
CH3
Pyridoxal
H 2C
D
Imine tautomerization
CH3
Imine 1
CO H D 2 G CH2R OH
NP C
HOCH2
CH2NH2 HOCH2
OH
N
CH3
Imine 2
The sequence begins with formation of an imine from the amino acid and the oxidized form of the vitamin, pyridoxal. This imine converts to an isomer via tautomerism: simultaneous proton and double-bond shift (See Sections 13-7 and 13-8). Hydrolysis of the new imine gives pyridoxamine and the ketoacid.
N
O B RCH2CCO2H
CH3
Pyridoxamine
Keto acid
Depending on the body’s metabolic needs, the pyridoxamine thus formed may proceed to react with other ketoacids to produce required amino acids (the scheme shown above run in reverse), or it may serve to facilitate the ultimate disposal of the nitrogen by excretion.
Special imines aid in the identification of aldehydes and ketones Several amine derivatives condense with aldehydes and ketones to form imine products that are highly crystalline and have sharp melting points. Among these are hydroxylamine, derivatives of hydrazine, H2NNH2, such as 2,4-dinitrophenylhydrazine, and semicarbazide. These imine products, called oximes, hydrazones, and semicarbazones, respectively, are more stable than the simple imines described above. As shown in the margin for oximes, each of these derivatives is stabilized by resonance delocalization of a lone pair on the oxygen or nitrogen that is attached to the imine nitrogen atom. Table 17-5 illustrates the structures of these substances. Melting points have been tabulated for imine derivatives of thousands of aldehydes and ketones. Before spectroscopic methods became routinely available, chemists used comparisons
ðš OH f C P N
š OH
l
ðC ON
Resonance stabilization in oximes
800
Chapter 17
Aldehydes and Ketones
Table 17-5 Imine Derivatives of Aldehydes and Ketones i
Product of reaction with CPO
Reagent
f
i C f
H2NOH Hydroxylamine
NOH
Oxime
O2N
O 2N H H 2N
NH
NO2
i C f
2,4-Dinitrophenylhydrazine
N
NO2
N
2,4-Dinitrophenylhydrazone
O
O H 2N
NHCNH2
NHCNH2
i C f
Semicarbazide
N Semicarbazone
of these melting points to determine whether an aldehyde or ketone of unknown structure was identical with a previously characterized compound. For example, the distillation of castor oil yields a compound with the formula C7H14O. Its boiling point of about 1508C is similar to those recorded for heptanal, 2-heptanone, and 4-heptanone. The unknown is readily identified as heptanal by preparing and determining the melting points of its semicarbazone (1098C) and 2,4-dinitrophenylhydrazone (1088C). These values match those tabulated for the derivatives of the aldehyde and are quite different from those recorded for the two ketones: The respective derivatives of 2-heptanone exhibit melting points of 1278 and 898C, and those of 4-heptanone, 1338 and 758C.
Condensations with secondary amines give enamines R H EH š N A R Secondary amine
The condensations of amines described so far are possible only for primary derivatives, because the nitrogen of the amine has to supply both of the hydrogens necessary to form water. Reaction with a secondary amine, such as azacyclopentane (pyrrolidine), therefore takes a different course. After the initial addition, water is eliminated by deprotonation at carbon to produce an enamine. This functional group incorporates both the ene function of an alkene and the amino group of an amine. Enamine Formation
O B CH3CH2CCH2CH3
H A N
Addition
OH A CH3CH2 O C ON A CH2CH3
HOH H2O
CH3CH P C
i
N
f CH2CH3
90% Azacyclopentane (Pyrrolidine)
Hemiaminal
An enamine
Enamine formation is reversible and hydrolysis occurs readily in the presence of aqueous acid. Enamines are useful substrates in alkylations (Section 18-4).
17-9 Nucleophilic Addition of Ammonia and Its Derivatives
Exercise 17-14
Working with the Concepts: Enamine Formation Formulate a detailed mechanism for the acid-catalyzed enamine formation shown above. Strategy Begin in a manner identical to the mechanisms of addition of ammonia and primary amines to give imines. As you proceed step by step, look for the point at which the structure of the secondary amine forces the mechanism to diverge from imine formation. Solution • Add the nucleophilic amine nitrogen as described earlier in this section. The negatively charged oxygen picks up a proton and the positively charged nitrogen loses one to give the usual hemiaminal intermediate. H O
O
N
H
N H H HO
N
H
HO
N
• Using the catalytic acid, protonate the oxygen, converting it into a better leaving group (water). The water departs, leaving an iminium ion, just as in the case of imine formation. • At this point, however, the nitrogen lacks another hydrogen to lose: Imine formation is impossible. Thus the system must take an alternative path. • Examine the structure of the actual enamine product (above): One hydrogen is missing from the carbon adjacent to the one bearing the nitrogen (the a-carbon). This hydrogen can be lost as a proton from the iminium intermediate, giving the final product.
HO
N
H
H2O
N
N
H2O
H
N
H
Exercise 17-15
Try It Yourself Write the product, in addition to the mechanism of its formation, for the following acid-catalyzed reaction. O
O H
H2O
N H
Chapter 17
801
Chapter 17
Aldehydes and Ketones
Exercise 17-16 Write the products of the following reactions, under acid-catalyzed conditions. O
(a)
O
(c)
OH
NH2 A (b) CH3CHCHCH3 A NH2
H N
O O B B CH3CH2C O CCH3
C6H5NHNH2
(Hint: See Section 17-7.)
In Summary Primary amines attack aldehydes and ketones to form imines by condensation. Hydroxylamine gives oximes, hydrazines lead to hydrazones, and semicarbazide results in semicarbazones. Secondary amines react with aldehydes and ketones to give enamines.
17-10 Deoxygenation of the Carbonyl Group In Section 17-5 we reviewed methods by which carbonyl compounds can be reduced to alcohols. Reduction of the CPO group to CH2 (deoxygenation) also is possible. Two ways in which this may be achieved are Clemmensen reduction (Section 16-5) and thioacetal formation followed by desulfurization (Section 17-8). This section presents a third method for deoxygenation—the Wolff-Kishner reduction.
Strong base converts simple hydrazones into hydrocarbons Condensation of hydrazine itself with aldehydes and ketones produces simple hydrazones: Synthesis of a Hydrazone O B CH3CCH3
H2N ONH2 Hydrazine
D N B C
CH3CH2OH H2O
H3C
NH2
G D
802
CH3
Acetone hydrazone
Hydrazones undergo decomposition with evolution of nitrogen when treated with base at elevated temperatures. The product of this reaction, called the Wolff-Kishner* reduction, is the corresponding hydrocarbon. REACTION
Wolff-Kishner Reduction NH2 D N B RCR
NaOH
(HOCH2CH2)2O, 180–200°C
RCH2R
N2
The mechanism of nitrogen elimination includes a sequence of base-mediated hydrogen shifts. The base first removes a proton from the hydrazone to give the corresponding delocalized anion. Reprotonation may occur on nitrogen, regenerating starting material, or on carbon, thereby furnishing an intermediate azo compound (Section 22-11) leading on to the product. The base removes another proton from the azo nitrogen to generate a new anion, *Professor Ludwig Wolff (1857 – 1919), University of Jena, Germany; Professor N. M. Kishner (1867 – 1935), University of Moscow.
17-10 Deoxygenation of the Carbonyl Group
Chapter 17
which rapidly decomposes irreversibly, with the extrusion of nitrogen gas. The result is a strongly basic alkyl anion (carbanion), which is immediately protonated by the aqueous medium to give the final hydrocarbon product. Mechanism of Nitrogen Elimination in the Wolff-Kishner Reduction
G D
R
šð HO
H2O Deprotonation
G D
š N OH
R
H
R
OH
Reprotonation
š Nð š N A RCR A H
B
B
š N A RCR A H
š N A C
G D
HOH
š NH
š NH D š N B C R R
B
H
š N B C
R
H
š N
MECHANISM
šð HO
HOH H2O Deprotonation
RCR A H
Irreversible loss of N2
H
OH
Reprotonation
Azo intermediate
In practice, the Wolff-Kishner reduction is carried out without isolating the intermediate hydrazone. An 85% aqueous solution of hydrazine (hydrazine hydrate) is added to the carbonyl compound in a high-boiling alcohol solvent such as diethylene glycol (HOCH2CH2OCH2CH2OH, b.p. 2458C) or triethylene glycol (HOCH2CH2OCH2CH2OCH2CH2OH, b.p. 2858C) containing NaOH or KOH, and the mixture is heated. Aqueous work-up yields the pure hydrocarbon. O CH3
1. H2NNH2, H2O, diethylene glycol, NaOH, 2. H2O
CH3
O 69% Wolff-Kishner reduction complements the Clemmensen and thioacetal desulfurization methods of deoxygenating aldehydes and ketones. Thus, the Clemmensen reduction is unsuitable for compounds containing acid-sensitive groups, and hydrogenation of multiple bonds can accompany desulfurization with hydrogen and Raney nickel. Such functional groups are generally not affected by Wolff-Kishner conditions.
Wolff-Kishner reduction aids in alkylbenzene synthesis We have already seen that the products of Friedel-Crafts acylations may be converted into alkyl benzenes by using Clemmensen reduction. Wolff-Kishner deoxygenation also is frequently employed for this purpose and is particularly useful for acid-sensitive, base-stable substrates. Wolff-Kishner Reduction of a Friedel-Crafts Acylation Product O B CCH3
NH2NH2, H2O, KOH, triethylene glycol,
CH2CH3
95% Exercise 17-17 Propose a synthesis of hexylbenzene from hexanoic acid.
RCH2R
HO
803
804
Chapter 17
Aldehydes and Ketones
In Summary The Wolff-Kishner reduction is the decomposition of a hydrazone by base, the second part of a method of deoxygenating aldehydes and ketones. It complements Clemmensen and thioacetal desulfurization procedures.
17-11 Addition of Hydrogen Cyanide to Give Cyanohydrins Besides alcohols and amines, several other nucleophilic reagents may attack the carbonyl group. Particularly important are carbon nucleophiles, because new carbon–carbon bonds can be made in this way. Section 8-8 explained that organometallic compounds, such as Grignard and alkyllithium reagents, add to aldehydes and ketones to produce alcohols. This section and Section 17-12 deal with the behavior of carbon nucleophiles that are not organometallic reagents—the additions of cyanide ion and of a new class of compounds called ylides. Hydrogen cyanide adds reversibly to the carbonyl group to form hydroxyalkanenitrile adducts commonly called cyanohydrins. The equilibrium may be shifted toward the adduct by the use of liquid HCN as solvent. However, it is dangerous to use such large amounts of HCN, which is volatile and highly toxic. Therefore, slow addition of HCl to an excess of the strongly basic NaCN is typically employed to generate HCN in a moderately alkaline mixture. Cyanohydrin Formation REACTION
O
HO
NaCN
CN
Conc. HCl NaCl
60% 1-Hydroxycyclohexanecarbonitrile (Cyclohexanone cyanohydrin)
MECHANISM
Cyanohydrin formation requires the presence of both free cyanide ion and undissociated HCN, a condition satisfied by maintaining a moderately basic pH. Cyanide adds to the carbonyl carbon in the slow, reversible first step of the mechanism. The HCN then protonates the negatively charged oxygen in the fast second step. This protonation is necessary to shift the overall equilibrium toward the product. Mechanism of Cyanohydrin Formation
ðNq Cð
C
Oð NC
Oð COš D
HO CN NC
COš OH D
ðCN
Exercise 17-18 Rank the following carbonyl compounds in order of thermodynamic favorability of HCN addition: acetone, formaldehyde, 3,3-dimethyl-2-butanone, acetaldehyde. (Hint: See Section 17-6.)
We shall see in subsequent chapters (Sections 19-6 and 26-2) that cyanohydrins are useful intermediates because the nitrile group can be modified by further reaction.
In Summary The carbonyl group in aldehydes and ketones can be attacked by carbon-based nucleophiles. Organometallic reagents give alcohols and cyanide gives cyanohydrins.
17-12 Addition of Phosphorus Ylides: The Wittig Reaction
R
G š C OP(C6H5)3 D H Ylide
Another useful reagent in nucleophilic additions contains a carbanion that is stabilized by an adjacent, positively charged phosphorus group. Such a species is called a phosphorus ylide, and its attack on aldehydes and ketones is called the Wittig* reaction. The Wittig reaction is a powerful method for the selective synthesis of alkenes from aldehydes and ketones. *Professor Georg Wittig (1897 – 1987), University of Heidelberg, Germany, Nobel Prize 1979 (chemistry).
17-12 Addition of Phosphorus Ylides: The Wittig Reaction
Chapter 17
Deprotonation of phosphonium salts gives phosphorus ylides Phosphorus ylides are most conveniently prepared from haloalkanes by a two-step sequence; the first step is the nucleophilic displacement of halide by triphenylphosphine to furnish an alkyltriphenylphosphonium salt. Phosphonium Salt Synthesis R (C6H5)3Pð Triphenylphosphine
G š CH2 O Xð
C6H6
š RCH2P(C6H5)3 ð Xð
An alkyltriphenylphosphonium halide
The positively charged phosphorus atom renders any neighboring proton acidic. In the second step, deprotonation by bases, such as alkoxides, sodium hydride, or butyllithium, gives the ylide. Ylides can be isolated, although they are usually only generated in solutions that are subsequently treated with other reagents. Ylide Formation
> OP(C6H5)3 RCH
H A RC O P(C6H5)3 X– A H
CH3CH2CH2CH2 O Li
THF
CH3CH2CH2CH2H
RCHP P(C6H5)3 Ylide
Notice that we may formulate a second resonance structure for the ylide by delocalizing the negative charge onto the phosphorus. In this form, the valence shell on phosphorus has been expanded and a carbon – phosphorus double bond is present.
The Wittig reaction forms carbon – carbon double bonds When an ylide is exposed to an aldehyde or ketone, their reaction ultimately furnishes an alkene by coupling the ylide carbon with that of the carbonyl. The other product of this transformation is triphenylphosphine oxide. REACTION
The Wittig Reaction CPO
(C6H5)3 P P C
Aldehyde or ketone
O B CH3CH2CH2CH
Ylide
CH3 A CH3CH2C P P(C6H5)3
CPC
Alkene
(CH3CH2)2O, 10°C (C6H5)3PO
(C6H5)3P P O Triphenylphosphine oxide
CH3 A CH3CH2CH2CH P CCH2CH3 66% 3-Methyl-3-heptene
The Wittig reaction is a valuable addition to our synthetic arsenal because it forms carbon – carbon double bonds. In contrast with eliminations (Sections 11-6 and 11-7), it gives rise to alkenes in which the position of the newly formed double bond is unambiguous. Compare, for example, two syntheses of 2-ethyl-1-butene, one by the Wittig reaction, the other by elimination.
LiX
805
806
Chapter 17
Aldehydes and Ketones
Comparison of Two Syntheses of 2-Ethyl-1-butene By Wittig reaction O B CH3CH2CCH2CH3
CH2 B CH3CH2CCH2CH3
CH2 P P(C6H5)3
(C6H5)3P P O
(Only one isomer)
By elimination CH3 A CH3CH2CCH2CH3 A Br
Base
CH2 B CH3CH2CCH2CH3
CH3 A CH3CH2C P CHCH3
(Mixture of isomers)
What is the mechanism of the Wittig reaction? The negatively polarized carbon in the ylide is nucleophilic and can attack the carbonyl group. The result is a phosphorus betaine,* a dipolar species of the kind called a zwitterion (Zwitter, German, hybrid). The betaine is short lived and rapidly forms a neutral oxaphosphacyclobutane (oxaphosphetane), characterized by a four-membered ring containing phosphorus and oxygen. This substance then decomposes to the product alkene and triphenylphosphine oxide. The driving force for the last step is the formation of the very strong phosphorus – oxygen double bond.
Mechanism of the Wittig Reaction
MECHANISM p RCH OP(C6H5)3
R A RCH OC O R A A (C6H5)3P )O p)
R p G O GC P p R
MATION AN I
ANIMATED MECHANISM: The Wittig reaction
A phosphorus betaine
R C A R O p)
R A O
A O
RCH A (C6H5)3P
RCH P C
(C6H5)3P PO
R
An oxaphosphacyclobutane (Oxaphosphetane)
Wittig reaction can be carried out in the presence of ether, ester, halogen, alkene, and alkyne functions. Many display useful stereoselectivity. For example, reactions between nonconjugated ylides and aldehydes typically result in cis (or Z) alkenes with good selectivity. A Stereoselective Wittig Reaction Giving Mostly cis Product CH3CH2CH2 CH3CH2CH2CH P P(C6H5)3
CH3(CH2)4CHO, THF
i
(C6H5)3PO
f
H
CPC 70%
f i
(CH2)4CH3 H
(cis : trans ratio = 6)1) 1
*Betaine is the name of an amino acid, (CH3)3N CH2COO2, which is found in beet sugar (beta, Latin, beet) and exists as a zwitterion.
17-12 Addition of Phosphorus Ylides: The Wittig Reaction
Chapter 17
In contrast, the presence of conjugation in the ylide frequently results in trans products, as shown in the example below, taken from the commercial synthesis of vitamin A1 (Section 14-7) by the German chemical company BASF (Badische Anilin and Soda Fabriken). A Highly trans-Selective Wittig Reaction: BASF Vitamin A1 Synthesis O
P(C6H5)3X
CH3O , CH3OH
P(C6H5)3
O B OCCH3
CH3 A
A H
HO, H2O
OH
Ester hydrolysis Vitamin A1
Exercise 17-19 Propose syntheses of 3-methylenecyclohexene from (a) 2-cyclohexenone and (b) 3-bromocyclohexene, using Wittig reactions.
Exercise 17-20 Propose a synthesis of the following dienone from the indicated starting materials. [Hint: Make use of a protecting group (Section 17-8).] O B CH3CCH2CH2CHP CHCHP CH2
from
O B CH3CCH2CH2CH2Br
and
O B HCCHP CH2
Exercise 17-21 Develop concise synthetic routes from starting material to product. You may use any material in addition to the given compound (more than one step will be required). (a)
(b)
CH2PCH(CH2)4CH PCH2
H
O B OCCH3
(Hint: See Section 12-12.)
(Hint: See Section 14-8.)
In Summary Phosphorus ylides add to aldehydes and ketones to give betaines that decompose by forming carbon – carbon double bonds. The Wittig reaction affords a means of synthesizing alkenes from carbonyl compounds and haloalkanes by way of the corresponding phosphonium salts.
807
Aldehydes and Ketones
Chapter 17
17-13 Oxidation by Peroxycarboxylic Acids: The Baeyer-Villiger Oxidation When ketones are treated with peroxycarboxylic acids (Section 12-10), the result is an oxidation of the carbonyl function to an ester, a transformation called Baeyer-Villiger* oxidation. The mechanism of this reaction starts by nucleophilic addition of the hydroperoxy end of the peracid to the carbonyl group to generate a reactive peroxide analog of a hemiacetal. This unstable adduct decomposes through a cyclic transition state in which an alkyl group shifts from the original carbonyl carbon to oxygen to give an ester. REACTION
Baeyer-Villiger Oxidation O B CH3CCH2CH3
O B CF3COOH, CH2Cl2
2-Butanone
A
R
R
Peroxycarboxylic acid
)O) B CO O
>) O A R
H
A
O B RCOOH
A
Ketone
H > )O )> O B RO A C C ) OOO R )p p A
MECHANISM
O B RCR
O B CH3CO CH2CH3 72% Ethyl acetate
O A
808
> O) i C OR l O
Ester
Cyclic ketones are converted into cyclic esters. Attack is at the carbonyl rather than at the carbon – carbon double bond (Section 12-10). Unsymmetric ketones, such as that in the following reaction, can in principle lead to two different esters. Why is only one observed? The answer is that some substituents migrate more easily than others. Experiments have established their relative ease of migration, or migratory aptitude.
O B CH3COOH, CHCl3
O O 56%
O
The ordering suggests that the migrating carbon possesses carbocationic character in the transition state for rearrangement. Migratory Aptitudes in the Baeyer-Villiger Reaction Methyl , primary , phenyl , secondary , tertiary Exercise 17-22 Predict the outcome of the following oxidations with a peroxycarboxylic acid. CH3 (a)
(b) O
O
(c) O
*Professor Johann Friedrich Wilhelm Adolf von Baeyer (1835 – 1917), University of Munich, Nobel Prize 1905 (chemistry); Victor Villiger (1868 – 1934), BASF, Ludwigshafen, Germany.
The Big Picture
Chapter 17
809
In Summary Ketones can be oxidized with peroxycarboxylic acids to give esters; with unsymmetric ketones, the esters can be formed selectively by migration of only one of the substituents.
17-14 Oxidative Chemical Tests for Aldehydes Although the advent of NMR and other spectroscopy has made chemical tests for functional groups a rarity, they are still used in special cases in which other analytical tests may fail. Two characteristic simple tests for aldehydes will again turn up in the discussion of sugar chemistry in Chapter 24; they make use of the ready oxidation of aldehydes to carboxylic acids. The first is Fehling’s* test, in which copper(II) ion (such as in CuSO4) is the oxidant. In a basic medium, the precipitation of red copper(I) oxide indicates the presence of an aldehyde function. Fehling’s Test O B RCH
Cu2
NaOH, tartrate (see Chemical Highlight 5-3), H2O
Cu2O
O B RCOH
Brick-red
The second is Tollens’s† test, in which a solution of silver ion (such as in AgNO3) precipitates a silver mirror on exposure to an aldehyde.
Addition of a molecule containing an aldehyde group to a solution of blue copper(II) sulfate (right) causes the formation of a brickred precipitate of copper(I) oxide in the Fehling test.
Tollens’s Test O B RCH
Ag
NH3, H2O
Ag
O B RCOH
Mirror
The Fehling and Tollens tests are not commonly used in large-scale syntheses. However, the Tollens reaction is employed industrially to produce shiny silver mirrors on glass surfaces, such as the insides of Thermos bottles.
THE BIG PICTURE We have begun an extended, systematic study of compounds containing the carbonyl group. Nine chapters remain in this book; carbonyl chemistry will be a significant presence in every one of them and will be a main focus of six of the nine. Why do these compounds play such a prominent role in organic chemistry? Recall that in Chapter 2 we introduced the properties of electrophilicity and nucleophilicity and discussed how they underlie almost all of the chemical reactivity of organic compounds. In the chapter just completed, we explored the carbonyl group as an electrophilic function, illustrating a number of processes that involve the addition of nucleophiles to the carbonyl carbon. In the next chapter, we shall see the other side of the picture of carbonyl chemistry: The carbon atom adjacent to the carbonyl group—the a-carbon—is a potential site of nucleophilic reactivity. It is this capacity for both electrophilic and nucleophilic behavior at adjacent sites in the same molecule that sets carbonyl compounds apart from almost any other class of organic substances. It is no wonder that the formation of carbon – carbon bonds in nature is in large measure based on carbonyl chemistry.
*Professor Hermann C. von Fehling (1812 – 1885), Polytechnic School of Stuttgart, Germany. † Professor Bernhard C. G. Tollens (1841 – 1918), University of Göttingen, Germany.
The Tollens test detects the presence of readily oxidized functional groups in organic molecules, such as aldehydes. Aldehyde addition to a solution of silver(I) in aqueous ammonia (right) causes the rapid deposition of a silver mirror on the glass walls of the tube (left).
810
Aldehydes and Ketones
Chapter 17
In the next chapter, we shall find out how to generate nucleophilic reactivity at the a carbons of aldehydes and ketones and how to use that reactivity in the formation of bonds with a variety of electrophiles, including the carbon atoms of other molecules.
CHAPTER INTEGRATION PROBLEMS 17-23. Oxidation of 4-hydroxybutanal with PCC (pyH1 CrO3Cl2, Section 8-6) does not produce the dialdehyde that one might expect. Instead a cyclic compound called a lactone is formed: O
O
PCC, CH2Cl2
HO
H
O
O ( No
H
H) O
Explain this result. SOLUTION Let us first examine the type of compound with which we are dealing. It contains both alcohol and aldehyde functions. Section 17-7 described the reversible addition of alcohols to aldehydes and ketones to give hemiacetals; furthermore, it showed that the equilibrium of the process lies strongly on the product side when it leads to a five- or six-membered ring. For example:
H O HO
H
O H
H
HO
O
H
O OH
H
H OH
If we compare this hemiacetal with the product of the actual oxidation reaction shown earlier, we see that it is the hemiacetal whose oxidation produces the observed product. Notice that, in general, aldehyde hemiacetals possess the same structural feature that is present in ordinary alcohols that can H A be oxidized to carbonyl compounds: OCOOOH. A 17-24. a. Propose a synthesis of norethynodrel, the major component in one common oral contraceptive preparation (Enovid, compare Chemical Highlight 4-3). Use as your starting compound the following nortestosterone derivative (left-hand structure). OH CH3% % `C q CH
CH3 %OH %
17
H ≥ % ≥ H H O
H ≥ % ≥ H H O
Starting compound
Norethynodrel
SOLUTION Analyze the problem retrosynthetically (Section 8-9). The target molecule is identical with the starting compound in all but one respect: the additional substituent on the cyclopentane ring (at C17). We know that we cannot add groups directly to secondary alcohols to make tertiary alcohols. However, addition reactions of organometallic reagents to ketones afford tertiary alcohols (Section 8-8). The necessary organometallic reagent for our purpose is Li12CqCH, an alkynyl anion (Section 13-5). Let us see if we can devise a plan based on this chemistry. The final step in the synthesis would appear to be addition of LiCqCH to a C17 ketone, but, as the following reaction shows, there is a major problem: The precursor molecule contains a second carbonyl group in the six-membered ring at the lower left (at C3). In such a process, it would be impossible to prevent the alkynyl anion from adding indiscriminately to C3 as well as C17.
Chapter Integration Problems
CH3 %
3
O
Chapter 17
CH3 %OH % ` Cq CH
17 1. LiC q CH 2. H , H2O
H ≥ % ≥ H H
H ≥ % ≥ H H
HCq C
O
HO
What to do? A similar problem arose in the synthesis of testosterone described in Chemical Highlight 9-2. It was solved by the use of protecting groups. We can do something similar here. The underlying conceptual problem is that we need to avoid having both C3 and C17 exist as carbonyls at the same time in any molecule in our synthesis. Failure to do so will doom a synthetic plan. With this concept in mind, we can imagine a modified ending to our synthesis, in which the carbon–oxygen double bond at C3 is protected in a form unreactive toward organometallic reagents during the previously shown step. After the addition reaction has been completed, the protecting group may be removed. An acetal derived from 1,2-ethanediol should do nicely (Section 17-8): CH3 %
O
3
H ≥ % ≥ H H
O
CH3 %OLi % ` Cq CH
17
H ≥ % ≥ H H
LiC q CH
O
H+, H2O
Norethynodrel
O
O
In this scheme, the aqueous acid does double duty: It protonates the alkoxide that results from alkynyllithium addition, and it hydrolyzes the acetal to a ketone. How does the synthesis begin? The C17 hydroxy group in the original starting material must be oxidized to a carbonyl. However, we cannot perform that oxidation in the presence of the carbonyl function at C3, because we would then violate the aforedescribed principle: We would create a molecule with both carbonyls present at the same time and the need to later react one and not the other with a reagent that cannot discriminate between them. With these considerations in mind, it becomes clear that protection of C3 must come first, before oxidation at C17: CH3 %OH % Starting compound
H ≥ % ≥ H H
H+, HOCH2CH2OH
O O
CH3 % PCC, CH2Cl2
H ≥ % ≥ H H
O O
b. Why is pyridinium chlorochromate (PCC) in CH2Cl2 a better choice than K2Cr2O7 for an oxidant in the second step? SOLUTION What could go wrong with K2Cr2O7? Consider the conditions under which it is used: typically in aqueous sulfuric acid (Section 8-6). Therefore, this reagent carries the risk of hydrolysis of the acetal group under the acidic conditions. PCC is a neutral, nonaqueous reagent, ideal for substrates containing acid-sensitive functions. c. Addition of alkynyllithium to the carbonyl at C17 is stereoselective to give the tertiary alcohol shown. Why? Would addition to a carbonyl at C3 also be stereoselective? SOLUTION What is the usual origin of stereoselectivity in reactions such as this? Steric hindrance in the immediate vicinity of the reacting center is the most common cause. The methyl substituent at the position adjacent to C17 is axial with respect to the six-membered ring and sterically hinders approach of the alkynyl reagent to the top face of the C17 carbonyl carbon (see Section 4-7). Addition from below is therefore strongly favored. There is no comparable steric hindrance in the vicinity of C3 to give rise to any preference for addition from either side.
O
811
Aldehydes and Ketones
New Reactions Synthesis of Aldehydes and Ketones 1. Oxidation of Alcohols (Section 17-4) R A RCHOH
O B RCR
CrO3, H2SO4
O B RCH
PCC, CH2Cl2
RCH2OH
Stable to oxidizing agent
Allylic oxidation OH H
MnO2, CHCl3
C
C
B
C
B
C
O B C
C
2. Ozonolysis of Alkenes (Section 17-4)
C
O
O
C
B
1. O3, CH2Cl2 2. Zn, CH3CO2H
B
C
B
C
3. Hydration of Alkynes (Section 17-4) O H2O, Hg2+, H2SO4
RC q CH
RCCH3
4. Friedel-Crafts Acylation (Section 17-4) O
B
C6H6
1. AlCl3 2. H+, H2O
RCCl
O B
C6H5CR
HCl
Reactions of Aldehydes and Ketones 5. Reduction by Hydrides (Section 17-5) O B RCH
NaBH4, CH3CH2OH
O B RCR
RCH2OH
1. LiAlH4, (CH3CH2)2O 2. H+, H2O
Selectivity O B CHCR
1. LiAlH4, (CH3CH2)2O 2. H+, H2O
RCH
B
RCH
B
OH A CHCR A H
6. Addition of Organometallic Compounds (Section 17-5) RLi or RMgX
CH2
THF
O
B
Chapter 17
B
812
Formaldehyde
RLi or RMgX
O B RCH
THF
Aldehyde
RLi or RMgX
O B RCR Ketone
RCH2OH Primary alcohol
OH A RCR A H Secondary alcohol
THF
OH A RCR A R Tertiary alcohol
OH A RCR A H
New Reactions
Chapter 17
7. Addition of Water and Alcohols—Hemiacetals (Sections 17-6 and 17-7) O B RCR
OH A RCR A OH
H2O, H+ or HO–
OH A RCR A OR
ROH, H or HO
Carbonyl hydrate
O O B B Keq: ROCOR < ROCOH < H2C
B
O B RCR
Hemiacetal
(A geminal diol)
Intramolecular addition HO
R
O O
H or HO
R
HO
Cyclic hemiacetal
8. Acid-Catalyzed Addition of Alcohols—Acetals (Sections 17-7 and 17-8) O B RCR 2 ROH
H
OR A RCR A OR
H2O
Acetal
Cyclic acetals O B RCR
O
H
HOCH2CH2OH
O
C
D G
R
R
Ketone, protected as cyclic acetal (Stable to base, LiAlH4, RMgX)
Ketone
9. Thioacetals (Section 17-8) Formation RS
A
O B RCR 2 RSH
BF3 or ZnCl2, (CH3CH2)2O
A SR
CO RO R (Stable to aqueous acid, base, LiAlH4, RMgX)
Hydrolysis RS
A
R
ASR
C O O
H2O, HgCl2, CaCO3, CH3CN
R
10. Raney Nickel Desulfurization (Section 17-8)
S
S C
D G
R
R
Raney Ni, H2
RCH2R
O B RCR
H2O
O
813
Aldehydes and Ketones
11. Addition of Amine Derivatives (Section 17-9)
O B RCR
R D N B C D G R R
RNH2, H
H2O
Imine
12. Enamines (Section 17-9)
O B RCH2CR
R R
G NH D
RCHP C
Secondary amine
D G
R A N OR
H2O
R
Enamine
13. Wolff-Kishner Reduction (Section 17-10) O B RCR
H2NNH2, H2O, HO,
RCH2R
14. Cyanohydrins (Section 17-11) O B RCR
HO
ACN
CO RO R
HCN
Cyanohydrin
15. Wittig Reaction (Section 17-12) RCH2X
C6H6
P(C6H5)3
RCH2P(C6H5)3 X Phosphonium halide
Triphenylphosphine Works with primary or secondary haloalkanes
Base
RCH2P(C6H5)3 X
RCH P P(C6H5)3 Ylide
R C
CHR
(C6H5)3P
A
R (Not always stereoselective)
16. Baeyer-Villiger Oxidation (Section 17-13) O B RCR
O B RCOOH
CH2Cl2
Ketone
O B RCOR
O B RCOH
Ester
Migratory aptitudes in Baeyer-Villiger oxidation Methyl , primary , phenyl , secondary , cyclohexyl , tertiary
B
THF
B
O B RCR RCH P P(C6H5)3
A
Chapter 17
A
814
O
Problems
Important Concepts 1. The carbonyl group is the functional group of the aldehydes (alkanals) and ketones (alkanones). It has precedence over the hydroxy, alkenyl, and alkynyl groups in the naming of molecules. 2. The carbon – oxygen double bond and its two attached nuclei in aldehydes and ketones form a plane. The CPO unit is polarized, with a partial negative charge on oxygen and a partial positive charge on carbon. 3. The 1H NMR spectra of aldehydes exhibit a peak at d ¯ 9.8 ppm. In 13C NMR, the carbonyl carbon absorbs at ,200 ppm. Aldehydes and ketones have strong infrared bands in the region 1690 – 1750 cm21; this absorption is due to the stretching of the CPO bond. Because of the availability of low-energy n → p* transitions, the electronic spectra of aldehydes and ketones have relatively long wavelength bands. This class of compounds displays characteristic mass spectral fragmentations around the carbonyl function. 4. The carbon – oxygen double bond undergoes ionic additions. The catalysts for these processes are acids and bases. 5. The reactivity of the carbonyl group increases with increasing electrophilic character of the carbonyl carbon. Therefore, aldehydes are more reactive than ketones. 6. Primary amines undergo condensation reactions with aldehydes and ketones to imines; secondary amines condense to enamines. 7. The combination of Friedel-Crafts acylation and Wolff-Kishner or Clemmensen reduction allows synthesis of alkylbenzenes free of the limitations of Friedel-Crafts alkylation. 8. The Wittig reaction is an important carbon – carbon bond-forming reaction that produces alkenes directly from aldehydes and ketones. 9. The reaction of peroxycarboxylic acids with the carbonyl group of ketones produces esters.
Problems 25. Draw structures and provide IUPAC names for each of the following compounds. (a) Methyl ethyl ketone; (b) ethyl isobutyl ketone; (c) methyl tert-butyl ketone; (d) diisopropyl ketone; (e) acetophenone; (f ) m-nitroacetophenone. 26. Name or draw the structure of each of the following compounds.
O B (a) (CH3)2CHCCH(CH3)2
O
O (b)
H
(c)
O O Cl (d) H
G G CPC D D
H (e) CH2CHO
(g) (Z )-2-Acetyl-2-butenal
(f) Br
(h) trans-3-Chlorocyclobutanecarbaldehyde
O
Chapter 17
815
816
Chapter 17
Aldehydes and Ketones
Preparation of Aldehydes and Ketones RCH2OH, RCHR A OH CrVI
R
G G CP C D D 1. O3 2. (CH3)2S or Zn, CH3COOH
OC q CO
OC q CO
Hg2, H2O
1. R2 BH 2. H2O2, HO
O B RCX, AlX3
CrVI
Product: O
Product: O
Products: G G CPO O P C D D
8-6
R
12-12
13-7
13-8
16-5, 22-2
CHO A OH
MnO2
R
15-13
G G CP C D D
Product: G G CP C D D G C B O
17-4, 22-2
O O O B B HC H E ECE R R H RH
18-4
20-2
20-2
20-4
20-6
20-8
22-8
23-1
Product:
Product: O B EC H R R
Product: O B EC H R H
Products: O B EC H R H
Products: O B EC H R H
Products: O B EC H R H
Product: O
Products: O O B B RCH2CCHCOR A R
O A J OCOC G A R
ROH
RNH2
NH3
O
ROH
1. R 2. HH2O
G G CP C D D
A NO
1. R2CuLi 2. H, H2O
O B RCX
CH3 CH3 CH3 A A A 1. LiAl OC(CH3)3 3H CHCH 1. CH AlH AlH CHCH CHCH 1. CH 1. CH 3 2 2AlH 3 2 2 3 2 2 2. H , H2O 2. H, H2O 2. H, H2O 2. H, H2O O B RCX
O B RCOR
O B RCNHR
RC q N
CrVI
OH
OH
1. RO, ROH 2. H, H2O
O B RCH2COR
1. RC 2. H
Problems
nes O
817
Chapter 17
section number &@ GC O OH(R)
(R)HO H or OH, (H2O)
&@ GC OSR
G RO N P C D
RS
R R(H)
G G CP C D D
H or OH, H2O
HgCl2, CaCO3, CH3CN, H2O
&@ GCOOH
NR2
H or OH, H2O
NC
O G G CP C D D
O G G CP C D D
H
R
OH, H2O
Product: O A l OCOC A i R
G
17-6, 17-7, 17-8
17-8
Other product: R2
Other product: R2
17-9
17-9
17-11
18-1
18-4
O B C E H R R
23-1
23-2
23-4
23-4
Products: O O B B RCCHCCH2R A R
Products: O B EC H R CH2R
Product: OH O A B RO C O CH A R
Product: OH O A l RO C O C A i R H
ROH
24-5
24-9
24-9
Product: CHO A H OO A OH R
Product: O R
25-4
H
Product: O R
R O
CO2
ROH O B 1. RCH2CCH2R, RO, ROH 2. H, H2O O B RCOR
1. OH, H2O 2. H, H2O
O O B B RCCHCOR A R
1. CH3CH2CH2CH2Li O B 2. RCR 3. HgCl2, HgO, CH3OH
S
S
R G N X HIO4 S catalyst O B RCH
A O C O OH A O C O OH A
H2, Pd-BaSO4, Fe3, H2O2, H2O H, H2O
CqN A H OO A OH R
COOH A H OO A OH R
H, H2O
RH
O
H R
818
Reactions of Aldehydes and Ketones OH OH A A RO C O H(R) RO C O H(R) A A H R
8-6, 17-5
8-8, 17-5
CH2H(R)
section number
OH OH OR SR H R A A A A N NR2 G B RO C O H(R) RO C O H(R) RO C O H(R) RO C O H(R) RCH P C A A A A C D E H R OH OR OR SR R H(R)
16-5
17-6
17-7
17-7, 17-8
17-8
17-9
Substrate: RC6H5
NaBH4 or LiAlH4
RLi or RMgX
17-9
17-10
17-11
17-12
H2NNH2, H2O HO,
NaCN, H
RCHP(C6H5)3
HO, H2O,
H2, PdOC
H or
Substrate: O B C H E G CP C D D
Substrate: O B RCH2CR
Zn(Hg), HCl,
H2O, H or HO
ROH, H or HO
ROH, H
R
O B C E H
RSH, H
RNH2, H or HO
R2 NH, H or HO
O B C E H R HR
Oxidant (i.e., CrO3, KMnO4, Ag, Cu2, H2O2, HNO3, O2-Rh3, Br2)
CH3 A (CH3CH)2NLi (LDA)
H or HO, H2O
X2, H
X2, HO
1. NaH or LDA 2. RX
Substrate: O B C E H R H
Substrate: O A l OCO C A i H
Substrate: O A l OCO C A i H
Substrate: O A l OCO C A i H
Substrate: H O A l OCO C A i H
Substrate: O A l OCO C A i H
Substrate: O B RCH2CR
17-13
17-14, 19-5, 19-6, 24-4, 24-9
18-1
18-2
18-3
18-3
18-4
18-5
O B EC H O R O H(R)
O B RCOH
O G G CP C D D
O A J OCOC G A X
X O A J OCOC G A X
O A J OCOC G A R
RCOOH B O
RCH2H(R)
OH (R)H H HR(H) A C B RO C O H(R) A EC H CN R H(R)
G G CP C D D
OH
18-8
NuH, HO
1. R2 CuLi 2. EX (EH, R)
Substrate: O B C H E G CP C D D
Substrate: O B C H E G CP C D D
18-9
18-10
O R R O A A A A O J J A A A A J OC OCOC OC OCOC RCH2CPCCR O C O C O C G G A A A A G A A B R E Nu H O H H (EH, R)
Reactions of Aldehydes and Ketones (continued)
R
HO,
H2O
R2NH, NaBH3CN, H2O or CH3OH CH2O, R2NH or CH3CH2OH
Substrates: O A J O CO C D A H O B C H E G CP C D D
NuH,
H
O B EC H
O B 1. RCOCH2CH3, NaOCH2CH3, CH3CH2OH 2. H, H2O
section number
O B EC H HR R D
R
N
X S catalyst
(NH4)2CO3 or P2O5 or P2S5
C6H5NHNH2, CH3CH2OH,
Substrate: O B Substrate: O B RCCH2R
B O
Substrate: O B RCH2CR
22-8
23-1
18-11
21-6
21-9
O O A A A l l CO C O C O C O C A A A i i H
NR2 A RO C O H(R) A H
O B RCCHR A CH2NR2
O O B B RCCHCR A R
OH
Nu
819
OH
Substrate: O B EC H R H
Substrate: CHO A H OO A OH R
23-4
24-7
OH O HHC K N O NHC6H5 A J A RO C O C G A HC K R N O NHC6H5 H R
Substrate: O R
R O
1. NH3, CH2O 2. HNO3
1. NH3, HCN 2. H, H2O,
Substrate: O O B B RCCH2COCH2CH3
Substrate: O B EC H R H
25-5
26-2
25-3
CO2CH2CH3
CH3CH2O2C E H R R X XNH, O, S
R
N
R
NH 3
A RO CO CO A B H O
820
Aldehydes and Ketones
Chapter 17
27. The following spectroscopic data are for two carbonyl compounds with the formula C8H12O. Suggest a structure for each compound. The letter “m” stands for the appearance of this particular part of the spectrum as an uninterpretable multiplet. (a) 1H NMR: d 5 1.60 (m, 4 H), 2.15 (s, 3 H), 2.19 (m, 4 H), and 6.78 (t, 1 H) ppm. 13C NMR: d 5 21.8, 22.2, 23.2, 25.0, 26.2, 139.8, 140.7, and 198.6 ppm. (b) 1H NMR: d 5 0.94 (t, 3 H), 1.48 (sex, 2 H), 2.21 (q, 2 H), 5.8 – 7.1 (m, 4 H), and 9.56 (d, 1 H) ppm. 13C NMR: d 5 13.6, 21.9, 35.2, 129.0, 135.2, 146.7, 152.5, and 193.2 ppm. 28. The compounds described in Problem 27 have very different ultraviolet spectra. One has lmax(e) 5 232(13,000) and 308(1450) nm, whereas the other has lmax(e) 5 272(35,000) nm and a weaker absorption near 320 nm (this value is hard to determine accurately because of the intensity of the stronger absorption). Match the structures that you determined in Problem 27 to these UV spectral data. Explain the spectra in terms of the structures. 29. Following are spectroscopic and analytical characteristics for an unknown compound. Propose a structure. Empirical formula: C8H16O. 1H NMR: d 5 0.90 (t, 3 H), 1.0–1.6 (m, 8 H), 2.05 (s, 3 H), and 2.25 (t, 2 H) ppm. IR: 1715 cm21. UV: lmax(e) 5 280(15) nm. MS: myz 5 128 (M1• ); intensity of (M 1 1)1 peak is 9% of M1• peak; important fragments are at myz 5 113 (M 2 15)1 myz 5 85 (M 2 43)1, myz 5 71 (M 2 57)1, myz 5 58 (M 2 70)1 (the second largest peak), and myz 5 43 (M 2 85)1 (the base peak). 30. Reaction review. Without consulting the Reaction Road Map on pp. 816 –817, suggest reagents to convert each of the starting materials below into 3-hexanone. OH (a)
(b)
O
O
(c)
S
N
S
N
(d)
(e)
(g)
(h)
(f) HO
CN
31. Indicate which reagent or combination of reagents is best suited for each of the following reactions. OH
O
O (b)
(a)
(c)
O B CH3CCH2CH2CH2CHO
H
OH
O (d)
O (e)
(f) O
32. Write the expected products of ozonolysis (followed by mild reduction—e.g., by Zn) of each of the following molecules. (a) CH3CH2CH2CHPCH2
(b)
(c)
(d)
Problems
Chapter 17
33. For each of the following groups, rank the molecules in decreasing order of reactivity toward addition of a nucleophile to the most electrophilic sp2-hybridized carbon. OOO O OO B BB BBB (a) (CH3)2CPO, (CH3)2CPNH, (CH3)2CPO H (b) CH3CCH3, CH3CCCH3, CH3CCCCH3 (c) BrCH2COCH3, CH3COCH3, CH3CHO, BrCH2CHO 1
34. Give the expected products of reaction of butanal with each of the following reagents. (a) LiAlH4, (CH3CH2)2O, then H1, H2O (b) CH3CH2MgBr, (CH3CH2)2O, then H1, H2O (c) HOCH2CH2OH, H1 35. Give the expected products of reaction of 2-pentanone with each of the reagents in Problem 34. 36. Give the expected products of reaction of 4-acetylcyclohexene with each of the reagents in Problem 34. 37. Give the expected product(s) of each of the following reactions. O
O excess CH3OH
(a)
OH
excess CH3OH
(b)
H
O O CH3
(c)
H 3C
S
OH
O NHNH2
H
(d) CH3CCH3 HOCH2CHCH2CH2CH3
O O
CH3 % 2 CH3CH2SH
(e)
BF3, (CH3CH2)2O
(f)
(CH3CH2)2NH
O 38. Formulate detailed mechanisms for (a) the formation of the hemiacetal of acetaldehyde and methanol under both acid- and base-catalyzed conditions and (b) the formation of the intramolecular hemiacetal of 5-hydroxypentanal (Section 17-7), again under both acid- and base-catalyzed conditions. 39. Formulate the mechanism of the BF3-catalyzed reaction of CH3SH with butanal (Section 17-8). 40.
Overoxidation of primary alcohols to carboxylic acids is caused by the water present in the usual aqueous acidic Cr(VI) reagents. The water adds to the initial aldehyde product to form a hydrate, which is further oxidized (Section 17-6). In view of these facts, explain the following two observations. (a) Water adds to ketones to form hydrates, but no overoxidation follows the conversion of a secondary alcohol into a ketone. (b) Successful oxidation of primary alcohols to aldehydes by the water-free PCC reagent requires that the alcohol be added slowly to the Cr(VI) reagent. If, instead, the PCC is added to the alcohol, a new side reaction forms an ester. This is illustrated for 1-butanol. O B CH3CH2CH2CH2OH uuuvvy CH3CH2CH2COCH2CH2CH2CH3 PPC, CH2Cl2
(c) Give the products expected from reaction of 3-phenyl-1-propanol and water-free CrO3 (1) when the alcohol is added to the oxidizing agent and (2) when the oxidant is added to the alcohol.
H
821
822
Chapter 17
Aldehydes and Ketones
41. Explain the results of the following reactions by means of mechanisms. ðOð
š Oð š ðOH
H
O
š (a) HO
O
O
H
O
OH
(b) HO (c) Explain why hemiacetal formation may be catalyzed by either acid or base, but acetal formation is catalyzed only by acid, not by base. 42. The two isomeric compounds below are naturally occurring insect pheromones. The isomer on the left attracts the male olive fruit fly; the one on the right, the female. (a) What kind of isomeric relationship exists between these two structures? (b) What functional group is contained in these molecules? (c) Both of these compounds are hydrolyzed under aqueous acidic conditions. Draw the products. Are the product molecules from the two starting isomers the same or different?
O
O
O
O A N
CH2OH
N
CH3
A O
O
43. Formulate a plausible mechanism for the following reaction. The product is a precursor of mediquox (shown in the margin), an agent used to treat respiratory infections in chickens (no, we are not making this up). NH2
Mediquox
H3C
NH2
O B C G
Benzene-1,2-diamine
GC G B O
CH3
N
CH3
N
CH3
CH3CH2OH
2,3-Dimethylquinoxaline
44. The formation of imines, oximes, hydrazones, and related derivatives from carbonyl compounds is reversible. Write a detailed mechanism for the acid-catalyzed hydrolysis of cyclohexanone semicarbazone to cyclohexanone and semicarbazide.
N
O B NHCNH2 D
O
H2O
H
O B NH2NHCNH2
45. Reaction review II. Without consulting the Reaction Road Map on p. 818, suggest reagents to convert cyclohexanone into each of the compounds below. OH (a)
S (b)
O
HO
S
CH2CH3
HO
(c)
(d)
(g)
(h)
N(CH2CH3)2
O (e)
(f)
CN
Problems
CH3O
OCH3
N
HO
(i)
(j)
Chapter 17
(k)
(l)
46. Propose reasonable syntheses of each of the following molecules, beginning with the indicated starting material. HO (a)
OH
O B B
from
OH
(b) C6H5NPC(CH2CH3)2 from 3-pentanol
OH (c)
from 1,5-pentanediol HO
O
(d)
from O
OH
47. The UV absorptions and colors of 2,4-dinitrophenylhydrazone derivatives of aldehydes and ketones depend sensitively on the structure of the carbonyl compound. Suppose that you are asked to identify the contents of three bottles whose labels have fallen off. The labels indicate that one bottle contained butanal, one contained trans-2-butenal, and one contained trans-3-phenyl-2propenal. The 2,4-dinitrophenylhydrazones prepared from the contents of the bottles have the following characteristics. Bottle 1: m.p. 187 – 1888C; lmax 5 377 nm; orange color Bottle 2: m.p. 121 – 1228C; lmax 5 358 nm; yellow color Bottle 3: m.p. 252 – 2538C, lmax 5 394 nm; red color Match up the hydrazones with the aldehydes (without first looking up the melting points of these derivatives), and explain your choices. (Hint: See Section 14-11.) 48. Indicate the reagent(s) best suited to effect these transformations. O (a) O O B (b) CH3CH P CHCH2CH2CH
O B CH3CH2CH2CH2CH2CH
O B (c) CH3CH P CHCH2CH2CH
CH3CH P CHCH2CH2CH2OH
]OH
%
-OH
%
(d)
O O
49. The molecule bombykol, whose structure is shown below, is a powerful insect pheromone, the sex attractant of the female silk moth (see Section 12-17). It was initially isolated in the amount of 12 mg from extraction of 2 tons of moth pupae. Propose a synthesis of bombykol from two simpler molecules based on the Wittig reaction. CH3CH2CH2CPCCHPCH(CH2)8CO2CH2CH3 Bombykol
O
823
824
Chapter 17
Aldehydes and Ketones
50. For each of the following molecules, propose two methods of synthesis from the different precursor molecules indicated. (a) CH3CHPCHCH2CH(CH3)2 from (1) an aldehyde and (2) a different aldehyde CH3 (b)
from (1) a dialdehyde and (2) a diketone CH3
51. Three isomeric ketones with the molecular formula C7H14O are converted into heptane by Clemmensen reduction. Compound A gives a single product upon Baeyer-Villiger oxidation; compound B gives two different products in very different yields; compound C gives two different products in virtually a 1 : 1 ratio. Identify A, B, and C. 52. Give the product(s) of reaction of hexanal with each of the following reagents. (a) HOCH2CH2OH, H1
(b) LiAlH4, then H1, H2O
(c) NH2OH, H1
(d) NH2NH2, KOH, heat
(e) (CH3)2CHCH2CHPP(C6H5)3
(f)
(g) Ag1, NH3, H2O
(h) CrO3, H2SO4, H2O
(i) HCN
N H
, H
53. Give the product(s) of reaction of cycloheptanone with each of the reagents in Problem 52. 54. Formulate in full detail the mechanism for the Wolff-Kishner reduction of 1-phenylethanone (acetophenone) to ethylbenzene (see p. 803). 55. The general equation for the Baeyer-Villiger oxidation (see p. 808) begins with a reaction between a ketone and a peroxycarboxylic acid to form a peroxy analog of a hemiacetal. Formulate a detailed mechanism for this process.
O
56. Formulate a detailed mechanism for the Baeyer-Villiger oxidation of the ketone shown in the margin (refer to Exercise 17-22). 57. Give the two theoretically possible Baeyer-Villiger products from each of the following compounds. Indicate which one is formed preferentially.
(a)
O r CO
O CH3
O CH3
(b)
O
(c)
(d)
O B (e) C6H5CCH3
H
58. Propose efficient syntheses of each of the following molecules, beginning with the indicated starting materials.
from
(a) H3C
O (c)
OH
O
CHO
CHO
(b) H
OH
OH CH3
from ClCH2CH2CH2OH
rC f
CH3
CHO from H
OH
Problems
59.
Explain the fact that, although hemiacetal formation between methanol and cyclohexanone is thermodynamically disfavored, addition of methanol to cyclopropanone goes essentially to completion: O
HO
OCH3
CH3OH 60. The rate of the reaction of NH2OH with aldehydes and ketones is very sensitive to pH. It is very low in solutions more acidic than pH 2 or more basic than pH 7. It is highest in moderately acidic solution (pH , 4). Suggest explanations for these observations. 61. Compound D, formula C8H14O, is converted by CH2PP(C6H5)3 into compound E, C9H16. Treatment of compound D with LiAlH4 yields two isomeric products F and G, both C8H16O, in unequal yield. Heating either F or G with concentrated H2SO4 produces H, with the formula C8H14. Ozonolysis of H produces a ketoaldehyde after Zn–H1, H2O treatment. Oxidation of this ketoaldehyde with aqueous Cr(VI) produces O CO2H Identify compounds D through H. Pay particular attention to the stereochemistry of D. 62. In 1862, it was discovered that cholesterol (for structure, see Section 4-7) is converted into a new substance named coprostanol by the action of bacteria in the human digestive tract. Make use of the following information to deduce the structure of coprostanol. Identify the structures of unknowns J through M as well. (i) Coprostanol, upon treatment with Cr(VI) reagents, gives compound J, UV lmax(e) 5 281(22) nm and IR 5 1710 cm21. (ii) Exposure of cholesterol to H2 over Pt results in compound K, a stereoisomer of coprostanol. Treatment of K with the Cr(VI) reagent furnishes compound L, which has a UV peak very similar to that of compound J, lmax(e) 5 258(23) nm, and turns out to be a stereoisomer of J. (iii) Careful addition of Cr(VI) reagent to cholesterol produces M: UV lmax(e) 5 286(109) nm. Catalytic hydrogenation of M over Pt also gives L. 63.
Three reactions that include compound M (see Problem 62) are described here. Answer the questions that follow. (a) Treatment of M with catalytic amounts of acid in ethanol solvent causes isomerization to compound N: UV lmax(e) 5 241(17,500) and 310(72) nm. Propose a structure for N. (b) Hydrogenation of compound N (H2–Pd, ether solvent) produces compound J (Problem 62). Is this the result that you would have predicted, or is there something unusual about it? (c) Wolff-Kishner reduction of compound N (H2NNH2, H2O, HO2, D) leads to 3-cholestene. Propose a mechanism for this transformation. H / CH3 CH3 ∑ % $ CH3 %
CH3 CH3
H %
≥ H 3-Cholestene
Team Problem 64. In acidic methanol, 3-oxobutanal is transformed into a compound with the molecular formula C6H12O3. O
O CH3OH, H
H
C6H12O3
3-Oxobutanal
As a group, analyze the following 1H NMR and IR spectral data: 1H NMR (CCl4): d 5 2.19 (s, 3 H), 2.75 (d, 2 H), 3.38 (s, 6 H), 4.89 (t, 1 H) ppm; IR: 1715 cm21.
Chapter 17
825
826
Chapter 17
Aldehydes and Ketones
Consider the chemical shifts, the splitting patterns, and the integrations of the signals in the NMR spectrum and discuss possible fragments that could give rise to the observed multiplicities. Use the IR information to assign the functional group that exists in the new molecule. Present an explanation for your structural determination, including reference to the spectral data, and suggest a detailed mechanism for the formation of the new compound.
Preprofessional Problems 65. In the transformation shown here, which of the following compounds is most likely to be compound A (use IUPAC name)? (a) 5-Octyn-7-one; (b) 5-octyn-2-one; (c) 3-octyn-2-one; (d) 2-octyn-3-one. CrO , H SO , acetone
3 2 4 3-Octyn-2-ol uuuuvvu uvy A
O B 66. The reaction CO O H3C H
CH2 P C
f
H
i
demonstrates (a) resonance; OH
(b) tautomerism; (c) conjugation; (d) deshielding. 67. Which of the following reagents converts benzenecarbaldehyde (benzaldehyde) into an oxime? (a) H2NNHC6H5; (b) H2NNH2; (c) O3; (d) H2NOH; (e) CH3CH(OH)2. 68. Which of the following statements is correct? In the IR spectrum of 3-methyl-2-butanone, the most intense absorption is at (a) 3400 cm21, owing to an OH stretching mode; (b) 1700 cm21, owing to a CPO stretching mode; (c) 2000 cm21, owing to a CH stretching mode; (d) 1500 cm21, owing to the rocking of an isopropyl group.
C H A P T E R
EIGHTEEN
Enols, Enolates, and the Aldol Condensation a,b-Unsaturated Aldehydes and Ketones
O
H
ave another look at the Chapter Opening above. The very fact that you can see it is made possible by the chemistry alluded to in the caption: Photons impinge on the p system (Section 14-11) of cis-retinal bound to a protein to cause cis-trans isomerization. The associated conformational changes result, within picoseconds, in a nerve impulse that is translated by your brain into “vision” (Chemical Highlight 18-3). The crucial chemical feature of retinal that makes this process possible is the (in this case extended) conjugative communication between the carbonyl group and the adjacent p system. This chapter will show you that the carbonyl group (much like an ordinary carbon – carbon double bond; Section 14-1) activates adjacent C–H and CPC bonds even in much simpler systems, because of resonance. After you have absorbed the material that follows, you will be able to “look” at this page with quite a different perspective! In the last chapter, we saw how the structure of the carbonyl group — a multiple bond that is also highly polar — gives rise to a characteristic combination of functional behaviors: addition reactions mediated by electrophilic attack (usually by protons) on the Lewis basic oxygen and attack by nucleophiles on the carbon. We turn now to a third site of reactivity in aldehydes and ketones, the carbon next to the carbonyl group, known as the A-carbon. The carbonyl group induces enhanced acidity of hydrogens on the a-carbon. Moving these A-hydrogens may lead to either of two electron-rich species: unsaturated alcohols called enols or their corresponding conjugate bases, known as enolate ions. Both enols and enolate ions are important nucleophiles, capable of attacking electrophiles such as protons, halogens, haloalkanes, and even other carbonyl compounds.
␣-Hydrogen
C A H
ðOð B C E ␣-Carbon
D C PC Enol
OH
Photomicrograph of rod and cone cells in the retina. All known image-resolving eyes and, indeed, all known visual systems in nature use a single molecule, cis-retinal, for light detection. Absorption of a photon isomerizes the cis double bond to trans. The accompanying massive change in overall molecular geometry is responsible for triggering the nerve impulse that is perceived as vision.
Enols, Enolates, and the Aldol Condensation
Chapter 18
We begin by introducing the chemistry of enolates and enols. Especially important is a reaction between enolate ions and carbonyl compounds, called the aldol condensation. This process is widely used to form carbon – carbon bonds both in the laboratory and in nature. Among the possible products of aldol condensation are a,b-unsaturated aldehydes and ketones, which contain conjugated carbon – carbon and carbon – oxygen p bonds. As expected, electrophilic additions may take place at either p bond. However, more significantly, a,b-unsaturated carbonyl compounds are also subject to nucleophilic attack, a reaction that may involve the entire conjugated system.
18-1 Acidity of Aldehydes and Ketones: Enolate Ions The pKa values of aldehyde and ketone a-hydrogens range from 16 to 21, much lower than the pKa values of ethene (44) or ethyne (25), but comparable with those of alcohols (15 – 18). Strong bases can therefore remove an a-hydrogen. The anions that result are known as enolate ions or simply enolates. Deprotonation of a Carbonyl Compound at the ␣-Carbon ␣-Carbon
ð
C OC
B
šð O J
š
C OC
š
šð O š
šð O
CPC š
BH
Enolate ion
pKa 16–18 (aldehydes) 19–21 (ketones)
Acetone enolate
O
C OC A H
šð O J
O
828
Why are aldehydes and ketones relatively acidic? We know that acid strength is enhanced by stabilization of the conjugate base (Section 2-2). In the enolate ion, the inductive effect of the positively polarized carbonyl carbon strongly stabilizes the negative charge at the a-position. Aldehydes are stronger acids than ketones because their carbonyl carbon bears a larger partial positive charge (Section 17-6). Further strong stabilization is provided by delocalization of charge onto the electronegative oxygen, as described by the resonance forms just pictured. The effect of delocalization is also reflected in the electrostatic potential map of the acetone enolate shown in the margin (on an attenuated scale), which exhibits negative charge (red) on the a-carbon as well as on the oxygen. An example of enolate formation is the deprotonation of cyclohexanone by lithium diisopropylamide (LDA; Section 7-8). Enolate Preparation O
H H
␣
H ␣
H
CH3 A (CH3CH)2NLi LDA
H THF
O Li H
H
CH3 A (CH3CH)2NH
Cyclohexanone enolate ion
Exercise 18-1 Identify the most acidic hydrogens in each of the following molecules. Give the structure of the enolate ion arising from deprotonation. (a) Acetaldehyde; (b) propanal; (c) acetone; (d) 4-heptanone; (e) cyclopentanone.
O aD OC CZ
Resonance hybrid
Each resonance form contributes to the characteristics of the enolate ion and thus to the chemistry of carbonyl compounds. The resonance hybrid possesses partial negative charges on both carbon and oxygen; as a result, it is nucleophilic and may attack electrophiles at
1 8 - 2 K e t o –E n o l E q u i l i b r i a
829
Chapter 18
either position. A species that can react at two different sites to give two different products is called ambident (“two fanged”: from ambi, Latin, both; dens, Latin, tooth). The enolate ion is thus an ambident anion. Its carbon atom is normally the site of reaction, undergoing nucleophilic substitution with SN2 substrates such as suitable haloalkanes. Because this reaction attaches an alkyl group to the reactive carbon, it is called alkylation (more specifically, C-alkylation). As we shall see in Section 18-4, alkylation is a powerful method for carbon – carbon bond formation of ketones. For example, alkylation of cyclohexanone enolate with 3-chloropropene takes place at carbon. Alkylation at oxygen (O-alkylation) is uncommon, although oxygen is typically the site of protonation. The product of protonation is an unsaturated alcohol, called an alkenol (or enol for short). Enols are unstable and rapidly isomerize back to the original ketones (recall Section 13-7). Ambident Behavior of Cyclohexanone Enolate Ion
ðOð
ðOð CH2CHP CH2
CH2 P CHCH2Cl, THF
k
␣-Carbon: usual site of alkylation by SN2 reaction
ðš OH
ðš Oð
ðOð H
H, H2O
C-alkylation, Cl
O-protonation
62% Cyclohexanone enolate ion
2-(2-Propenyl)cyclohexanone
Cyclohexanone enol
Cyclohexanone
Exercise 18-2 Give the products of reaction of cyclohexanone enolate with (a) iodoethane (reacts by C-alkylation) and (b) chlorotrimethylsilane, (CH3)3Si – Cl (reacts by O-silylation).
In Summary The hydrogens on the carbon next to the carbonyl group in aldehydes and ketones are acidic, with pKa values ranging from 16 to 21. Deprotonation leads to the corresponding enolate ions, which may attack electrophilic reagents at either oxygen or carbon. Protonation at oxygen gives enols.
18-2 Keto – Enol Equilibria We have seen that protonation of an enolate at oxygen leads to an enol. The enol, an unstable isomer of an aldehyde or ketone, rapidly converts into the carbonyl system: It tautomerizes (Section 13-7). These isomers are called enol and keto tautomers. We begin by discussing factors affecting their equilibria, in which the keto form usually predominates. We then describe the mechanism of tautomerism and its chemical consequences.
An enol equilibrates with its keto form in acidic or basic solution Enol – keto tautomerism proceeds by either acid or base catalysis. Base simply removes the proton from the enol oxygen, reversing the initial protonation. Subsequent (and slower) C-protonation furnishes the thermodynamically more stable keto form.
MECHANISM
Base-Catalyzed Enol–Keto Equilibration
CP C
š O D
Enol form
H ðB
Deprotonation
CP C
š Oð D Enolate ion
š CO C
š Oð J
BH
Reprotonation
H š Oð A J ðB COC Keto form
830
Chapter 18
Enols, Enolates, and the Aldol Condensation
In the acid-catalyzed process, the enol form is protonated at the double-bonded carbon away from the hydroxy-bearing neighbor. The electrostatic potential map of ethenol (margin) shows that this carbon bears more negative charge (red). Moreover, the resulting cation is resonance-stabilized by the attached hydroxy group, and inspection of the corresponding resonance form reveals it to be simply the protonated carbonyl compound. Deprotonation then gives the product. Ethenol
Acid-Catalyzed Enol–Keto Equilibration š OH Dš C PC
H š A DOH C OC š
H
H šð O A J C OC
š H OH A J C OC
Protonated carbonyl system
Enol form
H
Keto form
Both the acid- and base-catalyzed enol – keto interconversions occur rapidly in solution whenever there are traces of the required catalysts. Remember that although the keto form (usually) predominates, the enol-to-keto conversion is reversible and the mechanisms by which the keto form equilibrates with its enol counterpart are the exact reverse of the above two schemes.
Substituents can shift the keto – enol equilibrium The equilibrium constants for the conversion of the keto into the enol forms are very small for ordinary aldehydes and ketones, only traces of enol (which is less stable by ca. 8 – 12 kcal mol21) being present. However, relative to its keto form, the enol of acetaldehyde is about a hundred times more stable than the enol of acetone, because the less substituted aldehyde carbonyl is less stable than the more substituted ketone carbonyl. REACTION
Keto – Enol Equilibria O B HO CH2CH
K 6 107
D H2C P C G
OH H O
O
Acetaldehyde
O B HO CH2CCH3 Acetone
G 8.5 kcal mol1 (35.5 kJ mol1)
Ethenol (Vinyl alcohol)
K 5 109
D H2C P C G
OH CH3
G 11.3 kcal mol1 (47.3 kJ mol1)
2-Propenol
Enol formation leads to deuterium exchange and stereoisomerization What are some consequences of enol formation by tautomerism? One is that treatment of a ketone with traces of acid or base in D2O solvent leads to the complete exchange of all the a-hydrogens.
1 8 - 2 K e t o –E n o l E q u i l i b r i a
Chapter 18
Hydrogen–Deuterium Exchange of Enolizable Hydrogens ␣-Hydrogens (exchange in D2O)
Not ␣-hydrogens (do not exchange)
O B CH3CCH2CH3 2-Butanone
O B CD3CCD2CH3
D2O, DO
1,1,1,3,3-Pentadeuterio-2-butanone
This reaction can be conveniently followed by 1H NMR, because the signal for these hydrogens slowly disappears as each one is sequentially replaced by deuterium. In this way, the number of a-hydrogens present in a molecule can be readily determined.
Exercise 18-3 Formulate mechanisms for the base- and acid-catalyzed replacement of a single a-hydrogen in acetone by deuterium from D2O.
Exercise 18-4 Write the products (if any) of deuterium incorporation by the treatment of the following compounds with D2O – NaOD. (a) Cycloheptanone (c) 3,3-Dimethyl-2-butanone
(b) 2,2-Dimethylpropanal (d) O CHO
Exercise 18-5
Working with the Concepts: Assigning NMR Signals of a Cyclic Ketone The 1H NMR spectrum of cyclobutanone consists of a quintet at d 5 2.00 ppm and a triplet at d 5 3.13 ppm. Assign the signals in this spectrum to the appropriate hydrogens in the molecule. Strategy Using the information in Section 17-3 regarding chemical shifts of hydrogens in ketones, apply the tools of NMR analysis that you acquired in Chapter 10. Solution • The structure of cyclobutanone implies an 1H NMR spectrum displaying two signals: One for the four a-hydrogens (on C2 and C4) and another for the two b-hydrogens (on C3). • We learned in Section 17-3 that a-hydrogens in carbonyl compounds are more deshielded than b-hydrogens. Furthermore, in accordance with the (N 1 1) rule for spin – spin splitting (Section 10-7), the signal for the a-hydrogens should appear as a triplet as a result of splitting by the two b-hydrogen neighbors; thus we may assign the triplet signal at d 5 3.13 ppm to the four a-hydrogens. • Conversely, splitting by four neighboring a-hydrogens should cause the signal for the b-hydrogens to appear as a quintet, consistent with the signal at d 5 2.00 ppm.
Exercise 18-6
Try It Yourself What would you expect to observe to change in the 1H NMR spectrum of cyclobutanone upon treatment with D2O – NaOD? (Hint: Deuterium atoms do not display signals in an 1H NMR spectrum.)
O B C D G CH2 H2C G D
CH2
831
832
Chapter 18
Enols, Enolates, and the Aldol Condensation
Another consequence of enol formation, or enolization, is the ease with which stereoisomers at a-carbons interconvert. For example, treatment of cis-2,3-disubstituted cyclopentanones with mild base furnishes the corresponding trans isomers. The trans isomers are more stable for steric reasons. Base-Catalyzed Isomerism of an ␣-Substituted Ketone
MODEL BUILDING
OK CH2CHP CH2
/
CH3
H
Reprotonation
H CH2CHP CH2
∑
∑
CH3
H
CH3 95%
The reaction proceeds through the enolate ion, in which the a-carbon is planar and, hence, no longer a stereocenter. Reprotonation from the side cis to the 3-methyl group results in the trans diastereomer (Sections 4-1 and 5-5). Another consequence of enolization is the difficulty of maintaining optical activity in a compound whose only stereocenter is an a-carbon. Why? As the (achiral) enol converts back into the keto form, a racemic mixture of R and S enantiomers is produced. For example, at room temperature, optically active 3-phenyl-2-butanone racemizes with a half-life of minutes in basic ethanol. Racemization of Optically Active 3-Phenyl-2-butanone O B i CO CCH3 H ( H3C
H5C6
O G D C PC D G H3C CH3
CH3CH2O, CH3CH2OH
H5C6
C
(S)-3-Phenyl2-butanone
CH3CH2O, CH3CH2OH
Enolate ion
O B i CO CCH3 H 3C ( H H5C6
C
/
∑ H
Deprotonation 10% KOH, CH3CH2OH
/∑
/∑
CH2CHP CH2 H
O
/
O
(R)-3-Phenyl2-butanone
(Achiral)
Exercise 18-7 Bicyclic ketone A rapidly equilibrates with a stereoisomer upon treatment with base, but ketone B does not. Explain. O
O H %
CH3 %
0 H
0 H
A
B
In Summary Aldehydes and ketones are in equilibrium with their enol forms, which are roughly 10 kcal mol21 less stable. Keto – enol equilibration is catalyzed by acid or base. Enolization allows for easy H – D exchange in D2O and causes isomerization at stereocenters next to the carbonyl group.
18-3 Halogenation of Aldehydes and Ketones This section examines a reaction of the carbonyl group that can proceed through the intermediacy of either enols or enolate ions — halogenation. Aldehydes and ketones react with halogens at the a-carbon. In contrast with deuteration, which proceeds to completion with either acid or base, the extent of halogenation depends on whether acid or base catalysis has been used.
18-3 H a l o g e n a t i o n o f A l d e h y d e s a n d K e t o n e s
Chapter 18
In the presence of acid, halogenation usually stops after the first halogen has been introduced, as shown in the following example. Acid-Catalyzed -Halogenation of Ketones O B HO CH2CCH3
O B BrCH2CCH3 44%
Br–Br, CH3CO2H, H2O, 70C
REACTION HBr
Bromoacetone
The rate of the acid-catalyzed halogenation is independent of the halogen concentration, an observation suggesting a rate-determining first step involving the carbonyl substrate. This step is enolization. The halogen then rapidly attacks the double bond to give an intermediate oxygen-stabilized halocarbocation. Subsequent deprotonation of this species furnishes the product. Mechanism of the Acid-Catalyzed Bromination of Acetone
MECHANISM
Step 1. Enolization (rate determining) ðOð B CH3CCH3 H
H š OH A HOCOC G A CH3 H P
š OH D H2C P C H G CH3
Step 2. Halogen attack š OH D H2C P C G CH3
š OH J H2C O C G A CH3 Br
š OH D H2C O C G A CH3 Br
Br
Br O Br Step 3. Deprotonation H D O B BrCH2CCH3
O B BrCH2CCH3
H
Why is further halogenation retarded? The answer lies in the requirement for enolization. To repeat halogenation, the halo carbonyl compound must enolize again by the usual acid-catalyzed mechanism. However, the electron-withdrawing power of the halogen makes protonation, the initial step in enolization, more difficult than in the original carbonyl compound. Halogenation Slows Down Enolization Because Protonation Is Disfavored Less basic than unsubstituted ketone
O B BrCH2CCH3 Electron withdrawing
H D O B BrCH2CCH3
H
833
834
Chapter 18
Mechanism of Halogenation of an Enolate Ion
O G C P CH2 Br OBr D R
O B RCCH2Br Br
Enols, Enolates, and the Aldol Condensation
Therefore, the singly halogenated product is not attacked by additional halogen until the starting aldehyde or ketone has been used up. Base-mediated halogenation is entirely different. It proceeds instead by the formation of an enolate ion, which then attacks the halogen. Here the reaction continues until it completely halogenates the same a-carbon, leaving unreacted starting material (when insufficient halogen is employed). Why is base-catalyzed halogenation so difficult to stop at the stage of monohalogenation? The electron-withdrawing power of the halogen increases the acidity of the remaining a-hydrogens, accelerating further enolate formation and hence further halogenation.
Exercise 18-8 Write the products of the acid- and base-catalyzed bromination of cyclohexanone.
More acidic than unsubstituted ketone
In Summary Halogenation of aldehydes and ketones in acid can proceed selectively to the monohalocarbonyl compounds. In base, all a-hydrogens are replaced before another molecule of starting material is attacked.
18-4 Alkylation of Aldehydes and Ketones We have seen that enolates may be generated from ketones and aldehydes by treatment with bases, such as LDA (Sections 7-8 and 18-1). The nucleophilic a-carbon of the enolate may participate in SN2 alkylation reactions with suitable haloalkanes, forming a new carbon – carbon bond in the process. This section presents some features of enolate alkylation and compares it with a related process, alkylation of a species called an enamine.
Alkylation of enolates can be difficult to control In principle, the alkylation of an aldehyde or a ketone enolate is nothing more than a nucleophilic substitution. Alkylation of an Aldehyde or Ketone ␣-Carbon
AC H
O B C E
␣-Hydrogen
Base (e.g., LDA)
O B C š CE
R–X SN2
AC R
O B C E
Enolate ion New carbon–carbon bond
In practice, however, the reaction may be complicated by several factors. The enolate ion is a fairly strong base. Therefore, alkylation is normally feasible using only halomethanes or primary haloalkanes; otherwise, E2 elimination converting the haloalkane into an alkene (Section 7-8) becomes a significant process. Other side reactions can present additional difficulties. Alkylation of aldehydes usually fails because their enolates undergo a highly favorable condensation reaction, which is described in the next section. Even alkylation of ketones may be problematic, because the product of monoalkylation may lose another a-hydrogen under the reaction conditions and become alkylated a second time. Furthermore, if the starting ketone is unsymmetrical, alkylation may occur at either of the two a-carbons, leading to two regioisomeric products. The reaction of 2-methylcyclohexanone with iodomethane illustrates both of these problems.
18-4 A l k y l a t i o n o f A l d e h y d e s a n d K e t o n e s
Chapter 18
Products of Alkylation of 2-Methylcyclohexanone O H H
O H
CH3 H
LDA, CH3I, THF
O CH3
H
CH3
H3C
CH3
H
H
Regioisomeric monoalkylation products
O H3C
H
O CH3 CH3
H3C
H3C
CH3 CH3
Polyalkylation products
Under some circumstances, ketones may be alkylated successfully. In the virtually ideal example below, the ketone possesses only one a-hydrogen and the primary allylic halide is an excellent SN2 substrate (Section 14-3). The base in this example is sodium hydride, NaH (Section 8-6). Successful Alkylation of a Ketone O B C6H5CCH(CH3)2 Only ␣-hydrogen in the molecule
2-Methyl-1-phenyl1-propanone
1. NaH, THF 2. (CH3)2C P CHCH2Br H–H, NaBr
O B C6H5CC(CH3)2 A CH2CHP C(CH3)2 88% 2,2,5-Trimethyl-1-phenyl4-hexen-1-one
Exercise 18-9 O The reaction with base of the compound shown in the margin gives three isomeric products C8H12O. What are they? (Hint: Try intramolecular alkylations. Caution: As stated in Section 18-1, enolate alkylation “normally” proceeds only at carbon. Is the present case “normal”?)
(CH2)3Br
Exercise 18-10 C-alkylation of cyclohexanone enolate with 3-chloropropene (Section 18-1) is much faster than the corresponding reaction with 1-chloropropane. Explain. (Hint: See Section 14-3.) What product(s) would you expect from reaction of cyclohexanone enolate with (a) 2-bromopropane and (b) 2-bromo-2-methylpropane? (Hint: See Chapter 7.)
Enamines afford an alternative route for the alkylation of aldehydes and ketones Section 17-9 showed that the reaction of secondary amines such as azacyclopentane (pyrrolidine) with aldehydes or ketones produces enamines. As the following resonance forms indicate, the nitrogen substituent renders the enamine carbon – carbon double bond electron rich. Furthermore, the dipolar resonance contribution gives rise to significant nucleophilicity at the remote carbon, even though the enamine is neutral. As a result, electrophiles may attack at this position. Let us see how this attack may be used to synthesize alkylated aldehydes and ketones.
š N H Azacyclopentane (Pyrrolidine)
835
836
Chapter 18
Enols, Enolates, and the Aldol Condensation
Resonance in Enamines
C PC
D
N
CO C
N J
Carbon is nucleophilic
Ethenamine
Exposure of enamines to haloalkanes results in alkylation at the nucleophilic carbon to produce an iminium salt. Upon aqueous work-up, iminium salts hydrolyze by a mechanism that is the reverse of the one formulated for imine formation in Section 17-9. The results are a new alkylated aldehyde or ketone and the original secondary amine. Alkylation of an Enamine Alkylated at -carbon
O B CH3CH2CCH2CH3
N H, H H2O
N G GC P CHCH3 CH3CH2
CH3–I I
Enamine
3-Pentanone
CH3 N A M COCCH3 D A CH3CH2 H
HOH H,
O B CH3CH2CCH(CH3)2
– N H
An iminium salt
2-Methyl-3-pentanone
Exercise 18-11 Formulate the mechanism for the final step of the sequence just shown: hydrolysis of the iminium salt.
How does the alkylation of an enamine compare with the alkylation of an enolate? Enamine alkylation is far superior, because it minimizes double or multiple alkylation: Under the conditions of the process, the iminium salt formed after the first alkylation is relatively stable and unable to react with additional haloalkane. Enamines can also be used to prepare alkylated aldehydes, as shown here. (We shall see in the next section that aldehyde enolates undergo a new reaction called the aldol condensation and, therefore, cannot be alkylated readily.)
O B (CH3)2CHCH
N 1. H 2. CH3CH2Br 3. H, H2O
O B (CH3)2CCH A CH3CH2 67% 2,2-Dimethylbutanal
2-Methylpropanal
Exercise 18-12 Alkylations of the enolate of ketone A are very difficult to stop before dialkylation occurs, as illustrated here. Show how you would use an enamine to prepare monoalkylated ketone B. O
A
1. LDA, THF 2. BrCH2CO2C2H5
CH2CO2C2H5
C2H5O2CCH2 CH2CO2C2H5 O
94%
O
B
18-5 The Aldol Condensation
Chapter 18
In Summary Enolates give rise to alkylated derivatives upon exposure to haloalkanes. In these reactions, control of the extent and the position of alkylation, when there is a choice, may be a problem. Enamines derived from aldehydes and ketones undergo alkylation to the corresponding iminium salts, which can hydrolyze to the alkylated carbonyl compounds.
18-5 Attack by Enolates on the Carbonyl Function: Aldol Condensation We have seen the dual functional capability of carbonyl compounds: electrophilic at the carbonyl carbon, potentially nucleophilic at the adjacent a-carbon. In this section we introduce one of the most frequently employed carbon – carbon bond-forming strategies: attack of an enolate ion at a carbonyl carbon. The product of this process is a b-hydroxy carbonyl compound. Subsequent elimination of water may occur, leading to a,b-unsaturated aldehydes and ketones. The next three sections describe these reactions in detail, and Chemical Highlights 18-1 and 18-2 illustrate them in a biological context.
Aldehydes undergo base-catalyzed condensations Addition of a small amount of dilute aqueous sodium hydroxide to acetaldehyde at low temperature initiates the conversion of the aldehyde into a dimer, 3-hydroxybutanal, with the common name aldol (from aldehyde alcohol). Upon heating, this hydroxyaldehyde dehydrates to give the final product, the a,b-unsaturated aldehyde trans-2-butenal. This reaction is an example of the aldol condensation. The aldol condensation is general for aldehydes and, as we shall see, sometimes succeeds with ketones as well. We first describe its mechanism before turning to its uses in synthesis. REACTION
Aldol Condensation Between Two Molecules of Acetaldehyde New bond
H H3C
G C PO D
O B H2CCH A H
NaOH, H2O, 5C Aldol addition
OH O B A CH3C O CHCH A A H H
Dehydration
3-Hydroxybutanal
O B H CH G G CPC D D H3C H
H2O
trans-2-Butenal (An , -unsaturated aldehyde)
The aldol condensation highlights the two most important facets of carbonyl group reactivity: enolate formation and attack of nucleophiles at a carbonyl carbon. The base (hydroxide) is not strong enough to convert all of the starting aldehyde into the corresponding enolate ion, but it does bring about an equilibrium between the aldehyde and a small amount of enolate. Because the ion is formed in the presence of a large excess of aldehyde, its nucleophilic a-carbon can attack the carbonyl group of another acetaldehyde molecule. Protonation of the resulting alkoxide furnishes 3-hydroxybutanal. Mechanism of Aldol Formation
MECHANISM
Step 1. Enolate generation
ðš OH
A
Oð ðš H2C P C H A
ðOð B HC O CH2 O H
Small equilibrium concentration of enolate
š HOH
MATION AN I
ANIMATED MECHANISM: Aldol condensation – dehydration
837
Chapter 18
Enols, Enolates, and the Aldol Condensation
Step 2. Nucleophilic attack Electrophilic carbonyl carbon
š Oð
A
ðOð B CH3CH
CH2 P C
A
838
H
ðš Oð
ðOð B A CH3C O CH2CH A H
New carbon–carbon bond
Nucleophilic ␣-carbon
Step 3. Protonation šð ðOð ðO A B š OH CH3C O CH2CH H O A H
ðš OH ðOð B A CH3C O CH2CH A H 50–60%
šð HO
3-Hydroxybutanal (Aldol)
Note that hydroxide ion functions as a catalyst in this reaction. The last two steps of the sequence drive the initially unfavorable equilibrium toward product, but the overall reaction is not very exothermic. The aldol is formed in 50 – 60% yield and does not react further when its preparation is carried out at low temperature (58C). Exercise 18-13 Give the structure of the hydroxyaldehyde product of aldol condensation at 58C of each of the following aldehydes: (a) propanal; (b) butanal; (c) 2-phenylacetaldehyde; (d) 3-phenylpropanal.
Exercise 18-14 Can benzaldehyde undergo aldol condensation? Why or why not?
At elevated temperature, the aldol is converted into its enolate ion. Elimination of hydroxide ion, normally a poor leaving group, is driven thermodynamically by the formation of the conjugated, relatively stable final product. The net result of this transformation is a hydroxide-catalyzed dehydration of the aldol. We describe the overall process of aldol reaction followed by dehydration as aldol condensation. Recall (Section 17-9) that a condensation is a reaction that combines two molecules into one with the elimination of (typically) a molecule of water. MATION AN I
ANIMATED MECHANISM: Aldol condensation – dehydration
Mechanism of Aldol Dehydration ðš OH H ðOð A A B CH3C O C O CH A A H H
ðš OH
OH ðš Oð ðš A D CH3C OCHP C G A H H
š HOH
Enolate š HO ð as a leaving group is unusual
OH ðš š Oð A D CH3C OCHP C G A H H
ðOð B CH3CHP CHCH
š HO ð
What makes the aldol condensation synthetically useful? It couples two carbonyl compounds, with carbon – carbon bond formation, to furnish a product with a carbonyl group
18-5 T h e A l d o l C o n d e n s a t i o n
839
Chapter 18
and either an alcohol or an alkene function. Thus, at low temperatures, its outcome is a b-hydroxycarbonyl compound: New bond
O
O O B 2 CH3CHCH A CH3
NaOH, H2O, 5°C Aldol addition
CH3 OH CH3 O A A A B CH3CH O C O C O C OH A A H CH3 85%
2-Methylpropanal
O
3-Hydroxy-2,2,4-trimethylpentanal
At higher temperatures, however, it furnishes an a,b-unsaturated carbonyl compound, which has considerable synthetic utility (Sections 18-9 through 18-11): H
NaOH, H2O,
2
Aldol addition and dehydration
O
H
H
New double bond
O 80% Heptanal
(Z)-2-Pentyl-2-nonenal
Exercise 18-15 Give the structure of the a,b-unsaturated aldehyde product of aldol condensation of each of the aldehydes in Exercise 18-13.
Ketones can undergo aldol condensation So far, we have discussed only aldehydes as substrates in the aldol condensation. What about ketones? Treatment of acetone with base does indeed lead to some 4-hydroxy-4methyl-2-pentanone, but the conversion is poor because of an unfavorable equilibrium with starting material. Aldol Formation from Acetone New bond
O B CH3CCH3 94%
HO
OH O A B CH3C O CH2CCH3 A CH3 6% 4-Hydroxy-4-methyl-2-pentanone
The lesser driving force of the aldol reaction of ketones is due to the greater stability of a ketone than an aldehyde [about 3 kcal mol21 (12.5 kJ mol21)]. As a result, the aldol addition of ketones is endothermic. To drive the reaction forward, we can extract the product alcohol continuously from the reaction mixture as it is formed. Alternatively, under more vigorous conditions, dehydration and removal of water move the equilibrium toward the a,b-unsaturated ketone (see margin).
OH O A B CH3C O CH2CCH3 A CH3 NaOH, H2O,
H3C H3C
O B G C P CHCCH3 D 80%
4-Methyl-3-penten-2-one
H2O (removed)
840
Enols, Enolates, and the Aldol Condensation
Chapter 18
Exercise 18-16
Working with the Concepts: Practicing Mechanisms of Aldol Reactions Formulate the mechanism for the aldol reaction of acetone. Strategy As we have recommended previously as a strategy for learning to solve problems, look for a pattern to follow elsewhere in the text or in your notes — in this case, the mechanism for the aldol reaction of acetaldehyde given earlier in this section. Replace the earlier substrate with the new one and follow the identical sequence of steps, making sure that all your bond-forming and bond-cleavage processes occur at the correct locations with respect to the functional groups in each reacting molecule. Solution • In order, (1) use the base to deprotonate an a-carbon of one acetone molecule to form an enolate ion. • (2) Add the a-carbon of the negatively charged enolate to the carbonyl carbon of a second molecule of the ketone. • (3) Protonate the resulting alkoxide oxygen to give the final hydroxyketone product:
ðOð B CH3C O CH2 O H
ðOð B CH3CCH3
š Oð Dš CH2 P C G CH3
ðš OH
Step 1
ðš Oð D H2C P C G CH3
Step 2
Oð ðš ðOð A B CH3C O CH2CCH3 A CH3
š H O OH
Step 3
š HOH
ðš OH ðOð A B CH3C O CH2CCH3 A CH3
š HO ð
Exercise 18-17
Try It Yourself The aldol reaction above is reversible. Propose a mechanism for the conversion of its product, 4-hydroxy-4-methyl-2-pentanone, back into two molecules of acetone in the presence of 2OH.
In Summary Treatment of enolizable aldehydes with catalytic base leads to b-hydroxy aldehydes at low temperature and to a,b-unsaturated aldehydes upon heating. The reaction proceeds by enolate attack on the carbonyl function. Aldol addition to a ketone carbonyl group is energetically unfavorable. To drive the aldol condensation of ketones to product, special conditions have to be used, such as removal of the water or the aldol formed in the reaction.
18-6 Crossed Aldol Condensation What happens if we try to carry out an aldol condensation between the enolate of one aldehyde and the carbonyl carbon of another? In such a situation, called crossed aldol condensation, mixtures ensue, because enolates of both aldehydes are present and may react with the carbonyl groups of either starting compound. For example, a 1 : 1 mixture of acetaldehyde and propanal gives the four possible aldol addition products in comparable amounts.
18-6 Crossed Aldol Condensation
841
Chapter 18
C H E MI C AL H I G H L I G H T 1 8 - 1 Enzyme-Catalyzed Stereoselective Aldol Condensations in Nature Glucose (C6H12O6, Section 24-1), the most common sugar in nature, is absolutely essential to all life on Earth. In many species it is the sole energy source for the major organs, including the entire central nervous system. All species synthesize glucose from smaller molecules. The process is called gluconeogenesis, and it employs a crossed aldol condensation (catalyzed by enzymes appropriately named aldolases) to construct a carbon – carbon bond between two three-carbon precursors. A primary amine in the enzyme (specifically, a substituent in the amino acid lysine, Section 26-1) condenses with the carbonyl group of the monophosphate ester of 1,3dihydroxyacetone to form an iminium salt. Enzyme-catalyzed
Enzyme ONH2
CH2OPO32 f Enzyme ON P C Base H i H C H f HO
H2O
∑/
∑/
CH2OPO32 f OP C H i C H f HO
deprotonation gives an enamine (Section 17-9), whose nucleophilic carbon (red) attacks the aldehyde carbonyl carbon (blue) of 2,3-dihydroxypropanal (glyceraldehyde) 3-phosphate. This nitrogen analog of a crossed aldol condensation proceeds with 100% stereoselectivity: The enzyme causes the two substrate molecules to approach one another in a single, highly ordered geometry. Bond formation occurs exclusively to link the bottom faces of both the (red) enamine carbon and the carbonyl carbon (blue) to each other, resulting in the illustrated stereochemistry in the product. Hydrolysis of the resulting iminium salt gives the phosphate ester of the six-carbon sugar fructose, which is converted to glucose in a subsequent step.
1,3-Dihydroxyacetone monophosphate
CH2OPO32 f Enzyme ON O C ii H C OH f H HO H EO C Enamine
CH2OPO32 A Enzyme ON P C H HO H
∑/ HO
H CH2OPO32
OH
H
OH
H
OH
CH2OPO32
CH3CH
CH3C H
O
O
CHCH CH3
3-Hydroxy-2-methylbutanal
2. Acetaldehyde enolate adds to propanal. O O ðOð OH A CH3CH2CH CH2 P CH CH3CH2C CH2CH H 3-Hydroxypentanal
CH2OPO32 Fructose1,6-diphosphate
1. Propanal enolate adds to acetaldehyde. O
H
H
(All four reactions occur simultaneously)
OH
HO
OH
Result of Nonselective Crossed Aldol Reaction Between Acetaldehyde and Propanal
ðOð A CH3CHP CH
CH2OPO32 A OP C
H
(R)-Glyceraldehyde3-phosphate
O
H 2O
O
O
842
Enols, Enolates, and the Aldol Condensation
Chapter 18
CHE M I C AL H I G H L I G H T 1 8 - 2 Enzymes in Synthesis: Stereoselective Crossed Aldol Condensations The same enzymes that catalyze aldol condensations in nature may also be employed in the synthetic laboratory. In nature, the enzyme 2-deoxyribose-5-phosphate aldolase (DERA) catalyzes the condensation of acetaldehyde with 2O
3PO
OH %
O
CH3CHO
glyceraldehyde 3-phosphate to produce the phosphate ester of 2-deoxyribose, the sugar component of the deoxyribonucleic acid (DNA) present in all living things (Chapter 26). 2O
DERA
OH %
3PO
H
OH
(R)-Glyceraldehyde 3-phosphate
OH %
O
H
2-Deoxyribose 5-phosphate
DERA is employed to carry out laboratory syntheses of crossed aldol products in which both 100% stereoselectivity and 100% selectivity in the roles of the components are observed. An example is its catalysis of the condensation of propanal with (2R)-3-azido-2-hydroxypropanal. The condensation proceeds in only one of the four possible ways, via reaction of the a-carbon of propanal with the carbonyl carbon of the substituted aldehyde. In addition, the two new
N3
O
≥
CH3CH2CHO
stereocenters are formed in single configurations: R at the new C2, and S at C3. This particular application is significant, because the azido group is readily reduced to an amino (NH2) group, and the resulting compound is a member of a class of highly biologically active compounds called amino sugars. This class includes both antibiotics and anticancer agents (Section 24-12), as well as drugs with antihypertensive activity. DERA
OH %
N3
CH3 %
≥
H
OH
(2R)-3-Azido2-hydroxypropanal
O
H
(2R,3S,4R)-5-Azido3,4-dihydroxypentanal
3. Acetaldehyde enolate adds to acetaldehyde. ðOð OH O O O A A B B B CH3C O CH2CH CH3CH CH2 P CH CH3CH2CH A Not involved H
O
O
3-Hydroxybutanal
O CH3CH2CH
4. Propanal enolate adds to propanal. O O ðOð OH A CH3CHP CH CH3CH CH3CH2C CHCH Not involved
H
O
O
CH3
3-Hydroxy-2-methylpentanal
Can we ever efficiently synthesize a single aldol product from the reaction of two different aldehydes? We can, when one of the aldehydes has no enolizable hydrogens, because two of the four possible condensation products cannot form. We add the enolizable aldehyde slowly to an excess of the nonenolizable reactant in the presence of base. As soon as the enolate of the addend is generated, it is trapped by the other aldehyde.
18-7 Intramolecular Aldol Condensation
843
Chapter 18
A Successful Crossed Aldol Condensation CH3
CH3
CH3CCHO
CH3CH2CHO
NaOH, H2O,
CH3CCH
Added slowly
CH3 2,2-Dimethylpropanal
CH3 CCHO
H2O
CH3
Propanal
2,4,4-Trimethyl-2-pentenal
(No -hydrogens)
Exercise 18-18 Show the likely products of the following aldol condensations. CHO
CHO
(a)
CH3CHO
(b) 2
(reacts with itself)
(c) CH2PCHCHO 1 CH3CH2CHO
In Summary Crossed aldol condensations furnish product mixtures unless one of the reaction partners cannot enolize.
18-7 Intramolecular Aldol Condensation
The product of reaction (a) of Exercise 18-18 derives its common name—cinnamaldehyde—from the flavor that it imparts to these yummy items.
It is possible to carry out aldol condensations between two carbonyl groups located in the same molecule. Such a reaction is called an intramolecular aldol condensation and is important in the synthesis of cyclic compounds, especially five- and six-membered rings. For example, heating a dilute solution of hexanedial with aqueous base results in the formation of the cyclic product 1-cyclopentenecarbaldehyde. Here, one end of the molecule functions as the nucleophilic enolate component (after deprotonation), the other as the electrophilic carbonyl partner. After the initial aldol addition, subsequent dehydration furnishes the product.
Unfavorable Strained Ring Formation from 2,5-Hexanedione O
New bond
O O B B HCCH2CH2CH2CH2CH Site of enolate formation
Hexanedial
KOH, H2O, H2O Aldol addition
O B CH
C H C OH H
O B CH
CH3
H3C O
H2O Dehydration
H 62% OH
1-Cyclopentenecarbaldehyde
H3C Why does the intramolecular aldol condensation take place in preference to the intermolecular alternative? For one, attack at the carbonyl carbon of a different molecule is minimized by high dilution of the dialdehyde in the reaction mixture. More important, however, intermolecular reactions must overcome the loss of entropy arising from the conversion of two molecules into one as the initial hydroxyaldehyde product forms. In contrast, their intramolecular variant merely converts the acyclic substrate into a cyclic one, a process that costs much less in entropic terms and therefore is more favorable both kinetically (Section 9-6) and thermodynamically (Section 17-8). The entropic advantages of intramolecularity are particularly noticeable in intramolecular ketone condensations, transformations that provide a ready source of cyclic and bicyclic a,b-unsaturated ketones. Since ketones can have two enolizable a-carbons, in principle,
H CH3 O
H3 C
CH3 O
844
Enols, Enolates, and the Aldol Condensation
Chapter 18
CHE M I C AL H I G H L I G H T 1 8 - 3 Reactions of Unsaturated Aldehydes in Nature: The Chemistry of Vision H3C
CH3
CH3
H
CH3
O
H3C H
CH3
H
CH3
CH3
H H
Retinal isomerase
H
H
CH3
H3C H
trans-Retinal
O
cis-Retinal
Vitamin A (retinol, see Section 14-7 and the Chapter Opening) is an important nutritional factor in vision. Enzymecatalyzed oxidation converts it into trans-retinal. Trans-retinal is present in the light-receptor cells of the human eye, but before it can fulfill its biological function, it has to be isomerized by an enzyme, retinal isomerase, to give cis-retinal. cis-Retinal behaves like a “loaded spring” by virtue of the steric hindrance associated with the cis double bond (Section 11-5). The molecule fits well into the active site of a protein called opsin (approximate molecular weight 38,000). As shown on the facing page, cis-retinal reacts with one of the amine substituents of opsin to form the imine rhodopsin, the light-sensitive chemical unit in the eye. The electronic spectrum of rhodopsin, with a lmax at 506 nm (e 5 40,000), indicates the presence of a protonated imine group. When a photon strikes rhodopsin, the cis-retinal part isomerizes extremely rapidly, in only picoseconds (10212 s), to the trans isomer. This isomerization induces a geometric change, which appears to severely disrupt the snug fit of the original molecule in the protein cavity. Within nanoseconds (1029 s), a series of new intermediates form from this photoproduct, accompanied by conformational changes in the protein structure, followed by eventual hydrolysis of the illfitting retinal unit. This sequence initiates a nerve impulse
perceived by us as light. The trans-retinal is then reisomerized to the cis form by retinal isomerase and re-forms rhodopsin, ready for another photon. What is extraordinary about this mechanism is its sensitivity, which allows the eye to register as little as one photon impinging on the retina. Studies using nerve cells have shown that the geometric change brought about by isomerization of rhodopsin-bound retinal physically opens a “pore” in a special kind of protein called an ion channel, which is located in the cell membrane. Positively charged ions flow through the open pore into the cell, constituting an electrical current. The difference between this process and that of vision is that the geometric change in rhodopsin does not directly open the ion-channel pore in visual system cells. Rather, the dissociation of trans-retinal from rhodopsin causes the activation of a substance called G protein, discovered by Gilman* and Rodbell.† G protein is bound to the inner surface of the cell membrane and, upon activation, opens the ion channel to initiate the electrical signal of a nerve impulse. *Professor Alfred G. Gilman (b. 1941), University of Texas Southwestern Medical School, Nobel Prize 1994 (physiology or medicine). † Dr. Martin Rodbell (1925–1998), National Institutes of Health, Nobel Prize 1994 (physiology or medicine).
several aldol products may be possible. However, the reversibility of the reaction ensures that usually the least strained ring is generated, typically one that is five or six membered. Thus, reaction of 2,5-hexanedione results in condensation between an enolate at C1 and the C5 carbonyl group. An alternative pathway, attack of a C3 enolate on C5 to give a strained three-membered ring, is not observed.
MODEL BUILDING
Intramolecular Aldol Condensation of a Dione O O O B1 5B 3 CH3CCH2CH2CCH3
NaOH, H2O, New bond
CH3 42% 2,5-Hexanedione
3-Methyl-2-cyclopentenone
H2O
18-7 Intramolecular Aldol Condensation
H3C
H3C
CH3
845
Chapter 18
H H
H3C
CH3
H
H3C
CH3
H3C
H
H + H2N
CH3
Protein
H3C H
+
N
H
H+, H2O − H2O
O Protein
cis-Retinal
Opsin
Rhodopsin
hυ
Protein
trans-Retinal
cis-Retinal
Protein Rhodopsin after photon absorption
Rhodopsin before photon absorption
Exercise 18-19 Predict the outcome of intramolecular aldol condensations of the following compounds. O O B B (a) Cyclodecane-1,5-dione (b) C6H5C(CH2)2CCH3 (c) O
CH2C(CH2)3CH3 B O
(d) 2,7-Octanedione
Exercise 18-20 The intramolecular aldol condensation of 2-(3-oxobutyl)cyclohexanone (margin) can, in principle, lead to four different compounds (ignoring stereochemistry). Draw them and suggest which one would be the most likely to form. (Hint: Build models!)
O
O CH3
2-(3-Oxobutyl)cyclohexanone
846
Chapter 18
O
Enols, Enolates, and the Aldol Condensation
Exercise 18-21 Prepare the following compounds from any starting material, using aldol reactions in the crucial step. (Hint: The second preparation requires a double aldol addition.) OH Rotundone
(a)
O B CHP CHCCH3
(b)
O
OH
In Summary Intramolecular aldol condensation succeeds with both aldehydes and ketones. It can be highly regioselective and gives the least strained cycloalkenones.
18-8 Properties of A,B-Unsaturated Aldehydes and Ketones We have seen that the products of aldol condensation are a,b-unsaturated aldehydes and ketones. How do the properties of a,b-unsaturated aldehydes and ketones compare with the individual characteristics of their two types of double bonds? We shall find that their chemistry in some situations is a simple composite of the two and under other circumstances involves the a,b-unsaturated carbonyl, or enone, functional group as a whole. As later chapters will indicate, this complex reactivity is quite typical of molecules with two functional groups, or difunctional compounds.
Conjugated unsaturated aldehydes and ketones are more stable than their unconjugated isomers Like conjugated dienes (Section 14-5), a,b-unsaturated aldehydes and ketones are stabilized by resonance, resulting in the delocalization of the partial positive charge on the carbonyl carbon. As a consequence, and as shown in the electrostatic potential map of 2-butenal in the margin, the double bond is relatively electron poor (green, not red; compare Section 12-3), and the b-carbon is electrophilic (blue). Resonance Forms of 2-Butenal O
2-Butenal
REACTION
O
O
O B CH3CHP CHO CH
CH3CHP CHO CH
O
Black pepper (shown in the photo as the fruit before drying) owes its characteristic aroma to the presence of the sesquiterpene (Section 4-7) rotundone, a bicyclic a,b-unsaturated ketone. Rotundone is also present in basil, oregano, and thyme, and imparts a peppery aroma to some red wines. A possible synthesis of rotundone making use of the intramolecular aldol condensation is the subject of Problem 46.
CH3CHO CHP CH
Thus, unconjugated b,g-unsaturated carbonyl compounds rearrange readily to their conjugated isomers. The carbon – carbon double bond is said to “move into conjugation” with the carbonyl group, as the following example shows. Isomerization of a ,␥-Unsaturated Carbonyl Compound to a Conjugated System O B CH2 P CHCH2CH
3-Butenal
H or HO, H2O
O B CH3CH P CHCH
2-Butenal: more stable because of conjugation
The isomerization can be acid or base catalyzed. In the base-catalyzed reaction, the intermediate is the conjugated dienolate ion, which is reprotonated at the carbon terminus.
1 8 - 8 P r o p e r t i e s o f a,b -U n s a t u r a t e d A l d e h y d e s a n d K e t o n e s
Mechanism of Base-Mediated Isomerization of , -Unsaturated Carbonyl Compounds
MECHANISM
H
O O
O B CH2 P CHC O CH
847
Chapter 18
HO
Deprotonation
Protonation here gives conjugated aldehyde
H O
O B CH2 O CHP CHO CH
O
O B CH2 P CHO CHO CH
CH2 P CHO CH P CH
Reprotonation
HO OH
Dienolate ion
O B CH3CH P CHCH
Exercise 18-22 Propose a mechanism for the acid-catalyzed isomerization of 3-butenal to 2-butenal. (Hint: An intermediate is 1,3-butadien-1-ol.)
a,b-Unsaturated aldehydes and ketones undergo the reactions typical of their component functional groups a,b-Unsaturated aldehydes and ketones undergo many reactions that are perfectly predictable from the known chemistry of the carbon – carbon and carbon – oxygen double bonds. For example, hydrogenation by palladium on carbon gives the saturated carbonyl compound. H2, Pd–C, CH3CO2CH2CH3 (ethyl acetate solvent)
O
O 95%
The carbon – carbon p system also undergoes electrophilic additions. For example, bromination furnishes a dibromocarbonyl compound (compare Section 12-5). O B CH3CH P CHCCH3
Br–Br, CCl4
3-Penten-2-one
Br O A B CH3CHCHCCH3 A Br 60% 3,4-Dibromo-2-pentanone
The carbonyl function undergoes the usual addition reactions (Section 17-5). Thus, nucleophilic addition of amines (Section 17-9) results in the expected condensation products (see, however, the next section). OH D N B CH P CHCCH3
O B CH P CHCCH3 NH2OH, H H2O
4-Phenyl-3-buten-2-one
Oxime (m.p. 115C)
Exercise 18-23 Propose a synthesis of 3-phenyl-2-methyl-1-propanol starting from propanal.
HO
Enols, Enolates, and the Aldol Condensation
In Summary a,b-Unsaturated aldehydes and ketones are more stable than their nonconjugated counterparts. Either base or acid catalyzes interconversion of the isomeric systems. Reactions typical of alkenes and carbonyl compounds are also characteristic of a,bunsaturated aldehydes and ketones.
18-9 Conjugate Additions to A,B-Unsaturated Aldehydes and Ketones We now show how the conjugated carbonyl group of a,b-unsaturated aldehydes and ketones can enter into reactions that involve the entire functional system. These reactions are 1,4-additions of the type encountered with conjugated dienes, such as 1,3-butadiene (Section 14-6). The reactions proceed by acid-catalyzed, radical, or nucleophilic addition mechanisms, depending on the reagents.
The entire conjugated system takes part in 1,4-additions Addition reactions in which only one of the p bonds of a conjugated system takes part are classified as 1,2-additions (compare Section 14-6). Examples are the additions of Br2 to the carbon – carbon double bond and NH2OH to the carbon – oxygen double bond of a,bunsaturated aldehydes and ketones (Section 18-8).
1,2-Addition of a Polar Reagent A–B to a Conjugated Enone O
ON C
C A–B
CP C
A D B OH A C
COC O A A B A
or
CP C
However, several reagents add to the conjugated p system in a 1,4-manner, a result also called conjugate addition. In these transformations, the nucleophilic part of a reagent attaches itself to the b-carbon, and the electrophilic part (most commonly, a proton) binds to the carbonyl oxygen. As the resonance forms and the resonance hybrid below illustrate, the b-carbon in an a,b-unsaturated carbonyl compound possesses a partial positive charge and is therefore a second electrophilic site in these molecules, in addition to the carbonyl carbon itself (recall Section 17-2). Addition of nucleophiles to the b-carbon is therefore not a surprising process. Resonance Forms of ␣,-Unsaturated Carbonyl Compounds
B
C
CP C
O␦
ðš Oð
ðOð
A
Chapter 18
B
848
CO C
C
␦
C
C C
␦
Resonance hybrid: Both the carbonyl and -carbons are electrophilic.
The initial product of conjugate addition to an a,b-unsaturated carbonyl compound is an enol, which subsequently rapidly tautomerizes to its keto form. Thus the end result appears to be that of 1,2-addition of Nu – H to the carbon – carbon double bond.
1 8 - 9 C o n j u g a t e A d d i t i o n s t o a,b-U n s a t u r a t e d C o m p o u n d s
Chapter 18
1,4-Addition of a Polar Reagent A–B to a Conjugated Enone A OO O A C O C
l A–B AH CO C CP C C OC OC A A A B B H B
N
New bond to carbon
Oxygen and nitrogen nucleophiles undergo conjugate additions Water, alcohols, amines, and similar nucleophiles undergo 1,4-additions, as the following examples (hydration and amination) show. Although these reactions can be catalyzed by acid or base, the products are usually formed faster and in higher yields with base. These processes are readily reversed at elevated temperatures. O B CH2 P CHCCH3
HOH, Ca(OH)2, 5°C
REACTION
O B HOCH2 O CHCCH3 A H
-Carbon: site of nucleophilic attack
3-Buten-2-one
O H3 C B G C P CHCCH3 D H3C
4-Hydroxy-2-butanone H D CH3N , H2O G H
O CH3NH A B (CH3)2C O CHCCH3 A H 75%
Site of nucleophilic attack
4-Methyl-3-penten-2-one
4-Methyl-4-(methylamino)-2-pentanone
What determines whether 1,2- or 1,4-additions will take place? With the nucleophiles shown, both processes are reversible. Usually the 1,4-products form because they are carbonyl compounds and are more stable than the species arising from 1,2-addition (hydrates, hemiacetals, and hemiaminals, Sections 17-6, 17-7, and 17-9). Exceptions include amine derivatives such as hydroxylamine, semicarbazide, or the hydrazines, for which 1,2-addition eventually leads to an imine product whose precipitation out of solution drives the equilibrium. The mechanism of the base-catalyzed addition to conjugated aldehydes and ketones is direct nucleophilic attack at the b-carbon to give the enolate ion, which is subsequently protonated. Mechanism of Base-Catalyzed Hydration of , -Unsaturated Aldehydes and Ketones
š HO ð
ðO ð B C CP C
š HOC O CP C
ðš Oð D
H Oš OH
ðOð B š HOC O CHC
Exercise 18-24 Treatment of 3-chloro-2-cyclohexenone with sodium methoxide in methanol gives 3-methoxy-2cyclohexenone. Write the mechanism of this reaction. (Hint: Start with a conjugate addition.)
The ability to form new bonds with nucleophiles at the b carbon makes a,b-unsaturated aldehydes and ketones extremely useful in synthesis. This ability extends to carbon nucleophiles, permitting the construction of new carbon–carbon bonds. The remainder of the chapter illustrates processes of this type.
MECHANISM
ðš OH
849
Chapter 18
O GG C6H5CCH
GG
850
Enols, Enolates, and the Aldol Condensation
Hydrogen cyanide also undergoes conjugate addition CH2
1-Phenylpropenone KCN, H
O GG C6H5CCHCH2CN G H
Treatment of a conjugated aldehyde or ketone with cyanide in the presence of acid may result in attack by cyanide at the b-carbon, in contrast with cyanohydrin formation (Section 17-11). This transformation proceeds through a 1,4-addition pathway. The reaction includes protonation of the oxygen, then nucleophilic b-attack, and finally enol–keto tautomerization. Exercise 18-25 Formulate the mechanism of the acid-catalyzed 1,4-addition of cyanide to 1-phenylpropenone (see margin).
67% 4-Oxo-4-phenylbutanenitrile
In Summary a,b-Unsaturated aldehydes and ketones are synthetically useful building blocks in organic synthesis because of their ability to undergo 1,4-additions. Hydrogen cyanide addition leads to b-cyano carbonyl compounds; oxygen and nitrogen nucleophiles also can add to the b-carbon.
18-10 1,2- and 1,4-Additions of Organometallic Reagents Organometallic reagents may add to the a,b-unsaturated carbonyl function in either 1,2- or 1,4-fashion. Organolithium reagents, for example, react almost exclusively by direct nucleophilic attack at the carbonyl carbon. Exclusive 1,2-Addition of an Organolithium H3C H3C
O B G C P CHCCH3 D
4-Methyl-3-penten-2-one
OH A G C P CHCCH3 D A H3C CH3 81% H3C
1. CH3Li, (CH3CH2)2O 2. H, H2O
2,4-Dimethyl-3-penten-2-ol
Reactions of Grignard reagents with a,b-unsaturated aldehydes and ketones are not generally very useful, because Grignards may give 1,2-addition, 1,4-addition, or both, depending on the structures of the reacting species and conditions. Fortunately, another type of organometallic reagent, an organocuprate, is very effective in giving rise only to products of conjugate addition. Organocuprates have the empirical formula R2CuLi, and they may be prepared by adding two equivalents of an organolithium reagent to one of copper(I) iodide, CuI. Example of Preparation of an Organocuprate 2 CH3Li 1 CuI y (CH3)2CuLi 1 LiI Lithium dimethylcuprate (An organocuprate reagent)
Organocuprates are highly selective in undergoing 1,4-addition reactions: Exclusive 1,4-Addition of a Lithium Organocuprate -Carbon: site of nucleophilic attack
O B CH3(CH2)5CHP CCH A CH3 2-Methyl-2-nonenal
1. (CH3)2CuLi, THF, 78°C, 4 h 2. H, H2O
CH3 O A B CH3(CH2)5CHCHCH A CH3 40% 2,3-Dimethylnonanal
The copper-mediated 1,4-addition reactions are thought to proceed through complex electron-transfer mechanisms. The first isolable intermediate is an enolate ion, which can be
1 8 - 1 0 1, 2 - a n d 1 , 4 - A d d i t i o n s o f O r g a n o m e t a l l i c R e a g e n t s
trapped by alkylating species, as shown in Section 18-4. Conjugate addition followed by alkylation constitutes a useful sequence for a,b-dialkylation of unsaturated aldehydes and ketones.
, -Dialkylation of Unsaturated Carbonyl Compounds O B C
O J COC OC A A R R
1. R2CuLi 2. R X
CP C
The following example illustrates this reaction. O
O
O ŠCH3
CH3
)
1. (CH3CH2CH2CH2)2CuLi, THF 2. CH3I
) CH2CH2CH2CH3 84%, 4:1
) CH2CH2CH2CH3
trans- and cis-3-Butyl-2-methylcyclohexanone
Exercise 18-26
Working with the Concepts: Conjugate Addition in Synthesis O
Show how you might synthesize
from 3-methyl-2-cyclohexenone.
CH3 Strategy The fact that your starting material is an a,b-unsaturated ketone and your product has an additional bond to its b-carbon makes conjugate addition a reasonable place to start. However, for this problem, retrosynthetic analysis (working backward) is especially useful. The presence of an additional ring should suggest an intramolecular aldol condensation as a synthetic possibility. Analyze retrosynthetically first; then proceed in the forward direction. Solution • In the target molecule (below, left), consider the CPC double bond of the a,b-unsaturated ketone (arrow) to be the result of an intramolecular aldol condensation between the a-carbon of a cyclohexanone and a side-chain aldehyde function (center structure). • Consider the bond connecting this side chain to the b-carbon of the cyclohexanone (center, arrow) as deriving from a conjugate addition to 3-methyl-2-cyclohexenone, the specified starting material: O
O
O
O
H
CH3
CH3
CH3 Target molecule
Ketoaldehyde
3-Methyl-2-cyclohexenone
• Before translating this retrosynthetic analysis into a synthesis in the forward direction, notice that the side chain, which we intend to use as a nucleophile for 1,4-addition, will have to be introduced as an organocuprate reagent, which in turn has to come from a halide via an organolithium species. However, it also contains a carbonyl function, which will need to be protected, so that the organolithium reagent can be prepared successfully (recall Section 17-8): O HOCH2CH2OH, H
Br
H
O Br
O H
1. Li 2. CuI
O LiCu
O H
2
Chapter 18
851
852
Enols, Enolates, and the Aldol Condensation
Chapter 18
• The synthesis is completed by addition of the cuprate to the cyclohexenone, hydrolysis of the acetal to free the aldehyde group, and intramolecular aldol condensation: O
O
O
H
O O
CH3
O
LiCu
H
2
CH3 O
O
H
O
H, H2O
OH,
H2O,
CH3
CH3
Exercise 18-27
Try It Yourself O CH3
Design a synthesis of
CH3
from 3-methyl-2-cyclohexenone.
CH3
In Summary Organolithium reagents add 1,2 and organocuprate reagents add 1,4 to a,bunsaturated carbonyl systems. The 1,4-addition process initially gives rise to a b-substituted enolate, which upon exposure to haloalkanes furnishes a,b-dialkylated aldehydes and ketones.
18-11 Conjugate Additions of Enolate Ions: Michael Addition and Robinson Annulation Like other nucleophiles, enolate ions undergo conjugate additions to a,b-unsaturated aldehydes and ketones, in a reaction known as the Michael* addition.
REACTION
Michael Addition O
O CH3
O B CH2 P CHCC6H5
CH3
O B CH2CH2CC6H5
CH3CH2O K , CH3CH2OH, (CH3CH2)2O
64% The mechanism of the Michael addition includes nucleophilic attack by the enolate ion on the b-carbon of the unsaturated carbonyl compound (the Michael “acceptor”), followed by protonation of the resulting enolate.
MECHANISM
O G CPC
Mechanism of the Michael Addition: 1,4-Addition of an Enolate
CPC
O B C E
O O M D C OC O CO C PC
H O OH HO
O O M J C OC O C O C O C A H
*Professor Arthur Michael (1853 – 1942), Harvard University, Cambridge, MA.
18-11 Michael Addition and Robinson Annulation
Chapter 18
853
C H E MI C AL H I G H L I G H T 1 8 - 4 Alexander Borodin: Composer, Chemist, and Pioneering Teacher Alexander Borodin (1833 – 1887) was one of the most prominent Russian composers of the 19th century. He is known primarily for several orchestral works and an opera, Prince Igor, from which a number of enduringly popular excerpts emerged, including the Polovetsian Dances and the music used in the 1950s Broadway musical Kismet (in 1954 Borodin was actually awarded a posthumous Tony Award for the music in this show). However, Borodin was doing chemistry experiments before he was composing music, generating foul odors in the family apartment as a young teen. He attended school in St. Petersburg and graduated with a medical degree in 1856. His career in chemistry began with a postdoctoral appointment in Heidelberg, Germany, in the laboratory of Professor Emil Erlenmeyer — a name that should ring a bell with almost all of you because of the flask named after him. During the ensuing years Borodin made some significant contributions to organic chemistry. He demonstrated the first synthesis of an organofluorine compound, contributed to the development of carboxylic acid chemistry, and invented an analytical method for measuring nitrogen concentration that allowed determination of the amount of urea excreted in urine. He is perhaps most noted for his work on aldehydes: The historical record is a bit muddled, but Borodin may have been the discoverer of the aldol reaction. The German August Kekulé (of benzene fame) and the Frenchman Adolph Wurtz were also working on reactions of aldehydes under basic conditions around the same time, but Borodin almost certainly started earlier and carried his studies further. He described aldol reactions and condensations of a number of straight-chain aldehydes in the 1860s; his principal failing was a certain degree of laziness in properly writing up his work for publication, thus leading to still-debated issues of priority in discovery. Borodin’s scientific legacy took a step further in the 1870s, when he turned more of his attention to teaching.
He introduced the first advanced medical course for women and championed the cause of educating women to pursue careers in science and medicine. He also capitalized on his musical fame to organize charity concerts to raise funds to help poor students. Although his musical accomplishments fall short of those of his contemporaries Tchaikovsky and Rimsky-Korsakov, Borodin was and continues to be held in the highest esteem for a lifetime of achievement in both art and science.
Tomb of the composer and chemist Alexander Borodin in St. Petersburg, Russia.
As the mechanism indicates, the reaction works because of the nucleophilic potential of the a-carbon of an enolate and the electrophilic reactivity of the b-carbon of an a,b-unsaturated carbonyl compound. With some Michael acceptors, such as 3-buten-2-one, the products of the initial addition are capable of a subsequent intramolecular aldol condensation, which creates a new ring. Michael Addition Followed by Intramolecular Condensation H3 C
O B CH2 PCHCCH3
CH3CH2ONa, CH3CH2OH, (CH3CH2)2O, 10C Michael addition
3-Buten-2-one
O CH3
CH3
CH3 , HO
O
O CH3
Intramolecular aldol condensation
HOH
O
OH 54%
O 86%
854
Chapter 18
Enols, Enolates, and the Aldol Condensation
The synthetic sequence of a Michael addition followed by an intramolecular aldol condensation is also called a Robinson* annulation. MATION AN I
ANIMATED MECHANISM: Robinson annulation
Robinson Annulation Michael addition
CH O
C C H2 O
O
H2O
Aldol condensation
The Robinson annulation has found extensive use in the synthesis of polycyclic ring systems, including steroids and other natural products containing six-membered rings. Exercise 18-28
Working with the Concepts: Using Michael and Robinson Reactions in Synthesis O
Propose a synthesis of
using Michael or Robinson reactions. C6H5
Strategy Use retrosynthetic analysis to identify reasonable starting materials. (Hint: Always make your first “disconnection” at the CPC double bond in the a,b-unsaturated ketone. Then proceed in a forward direction, following the steps presented in this section.) Solution • The CPC double bond in an a,b-unsaturated ketone may be formed from an aldol condensation. Therefore, we begin with its “disconnection” to identify a compound that can be converted to the desired final product. Break the bond, and place a carbonyl group (“PO”) at the former b-carbon. Why? In the forward direction, aldol condensation converts the carbonyl group in the starting compound into the b-carbon of the product. The result of this disconnection is the ketoaldehyde below: O O Disconnect CH3 O C6H5
C6H5
H
In the retrosynthetic analysis, the original -carbon becomes a carbonyl group
• This compound is an example of a “1,5-dicarbonyl” compound: If you begin counting at either one of the two carbonyl carbons, the other one is the fifth atom away. This type of structure may be prepared by Michael addition of an enolate to an a,b-unsaturated compound. We identify possible starting materials retrosynthetically, by “disconnecting” at either a – b bond between the carbonyl carbons. There are two possible disconnections:
*Sir Robert Robinson (1886 – 1975), Oxford University, England, Nobel Prize 1947 (chemistry).
The Big Picture
Second Possible Disconnection
First Possible Disconnection O
O 
␣
H3 C
O
O
␣
H
␣
H3C

␣
O
O

CH3
H3C
H
C 6H 5
C6H5
O
␣
H 2C
H
H3C
O

CH2
␣
H
C6H5
C 6 H5
• We have two options from which to choose: Michael addition of the enolate of acetone to 2phenylpropenal (above, left), and Michael addition of the enolate of 2-phenylacetaldehyde to 3buten-2-one (above, right). Which is better? Michael additions are normally carried out under basic conditions, and we know (Section 18-5) that aldehydes undergo aldol condensation with base much more readily than do simple ketones. Thus, the second option above risks interference from the aldol reaction between two molecules of 2-phenylacetaldehyde. The first option is better because the enolate of acetone will undergo Michael addition to the aldehyde much more easily than aldol condensation with itself. Thus, we have all the pieces we need to put the synthesis together in the forward direction — it is a Robinson annulation:
H3C
O
O
O CH3
H2C
H C6H5
Intramolecular aldol condensation
CH3CH2ONa, CH3CH2OH Michael addition
O
O
H3C
H C6H5 C6H5
Exercise 18-29
Try It Yourself Propose syntheses of the following compounds by Michael or Robinson reactions.
(a)
O
O
O
(b)
In Summary The Michael addition results in the conjugate addition of an enolate ion to give dicarbonyl compounds. The Robinson annulation reaction combines a Michael addition with a subsequent intramolecular aldol condensation to produce new cyclic enones.
THE BIG PICTURE The aldol condensation and the synthetic sequences that can be built around it provide us with powerful methods for synthesizing complex organic molecules. These reactions provide ways to form carbon–carbon bonds at three sites: the a-, the b-, and the carbonyl carbons. We can construct elaborate acyclic compounds using these processes, and we can also close rings. In the problems that follow, take note of the remarkable variety of compounds that can be formed just using the chemical properties of the carbonyl group and its adjacent atoms. Continuing our survey of carbonyl chemistry, we turn next to the carboxylic acids, to see how these more highly oxidized systems compare. The principle of enhanced acidity for hydrogens attached to atoms adjacent to a carbonyl carbon is illustrated by the acidity
Chapter 18
855
856
Chapter 18
Enols, Enolates, and the Aldol Condensation
of the O – H hydrogen in these compounds. In addition, the OH portion of the carboxylic acid function is a potential leaving group. The next chapter presents a new method by which substitution reactions can occur, one that is central to the chemistry of all compounds in the carboxylic acid family: the addition – elimination pathway.
CHAPTER INTEGRATION PROBLEMS
O H %
18-30. Propose as many syntheses as you can of the bicyclic ketone in the margin, using at least one aldol condensation in each. You may use any starting materials you like.
% H
SOLUTION To approach this problem, it is useful to evaluate the information we have been given. The target compound is a saturated ketone, but aldol condensations give rise to a,b-unsaturated carbonyl compounds. Therefore, we can infer that any synthesis we develop must finish with hydrogenation of a CPC double bond in conjugation with the carbonyl group. Based on the target structure, three a,b-unsaturated ketones can be identified as possible precursors. O
O H %
␣ 
␣


␣
␣

% H
O H %

% H
% H
O ␣ 
Having identified these three structures, we can proceed to formulate the compounds that, upon undergoing intramolecular aldol condensation, will produce each one. Retrosynthetically, disconnect each CPC double bond; each disconnected b-carbon derives from a carbonyl group. That gives the following structures, shown in the same left-to-right order as their corresponding aldol enone products above. ␣
CH3 O H

O H %
O
O
O
% H
␣
␣
H
O 

All that remains is to complete the syntheses in the forward direction, including necessary reagents. We do so below for the first of these three starting compounds; the other two can be completed in exactly the same way. ␣
CH3 O H
OCH3
O H %
O ␣ OH,
% H
H2O,

O H %
H2, Pd–C
% H
H % % H
18-31. The Robinson annulation sequence is a powerful method for the construction of six-membered rings. It is therefore no surprise that this sequence has been widely applied in steroid synthesis. Beginning with the bicyclic ketone shown in the margin, propose a synthesis that makes use of one or more Robinson annulations for the steroid shown here.
O
OCH3 CH3
O
Chapter Integration Problems
Chapter 18
SOLUTION The Robinson annulation combines Michael addition to an a,b-unsaturated ketone with intramolecular aldol condensation (Section 18-11) to afford a cyclohexenone. Retrosynthetic analysis (Section 8-9) of the target molecule leads to the disconnection of two bonds in ring A: the carbon – carbon double bond, by a retro-aldol condensation, and a single bond by a retro-Michael addition. The Robinson annulation for the construction of ring A is closely related to the example in Section 18-11, which condenses 2-methylcyclohexanone with 3-buten-2-one. OCH3
OCH3 CH3 “ “ A B
H 3C B
O
O
O
CH3
At this stage, we have simplified the target from a tetracyclic to a tricyclic molecule. However, the latter is a b,g-unsaturated ketone, not an obvious product of Robinson annulation for the formation of ring B. Postponing consideration of this problem for the moment, can we formulate a Robinson annulation that transforms the initial ketone into a tricyclic product with the same molecular skeleton as that of the product of the last reaction? Yes we can: In fact, we can construct a doublebond isomer: OCH3
OCH3 O B CH3CH2CCH P CH2
CH3ONa, CH3OH
Michael addition
NaO
O
OCH3
OCH3
H2O Aldol condensation
O
H3C O
H3C
O
We are quite close to a complete solution. All that remains is to connect the tricyclic ketone that we have just prepared with the necessary second Robinson sequence. To do so, we think mechanistically, asking ourselves, “What is the structure of the enolate needed to initiate the annulation?” It is shown at the right in the following reaction, and we note that it is an allylic enolate ion (Section 14-4), for which we can write a second resonance contributor. It therefore may be derived by deprotonation of the tricyclic ketone in the preceding reaction at the allylic (g) position. OCH3
H3C O
H
OCH3
OCH3
Base
H3C O
H3C
O Resonance-stabilized allylic enolate anion
What does this realization mean in practical terms? We can carry out the needed second Robinson annulation starting directly from the preceding a,b-unsaturated ketone; it is not necessary to first prepare the b,g-unsaturated ketone revealed in our initial retrosynthetic disconnection, because they give the same enolate upon deprotonation. The complete synthesis is shown on the following page.
857
Chapter 18
Enols, Enolates, and the Aldol Condensation
OCH3
OCH3
OCH3
O B CH3CH2CCH P CH2 CH3ONa, CH3OH
H2O
O
H3C
O
O H3C
O OCH3
O B CH3CCH P CH2 CH3ONa, CH3OH
OCH3
CH3
O
CH3
H2O
O CH3
O
A final note: You may have recognized that the allylic anion discussed above is also benzylic and hence enjoys additional resonance stabilization by conjugation with the benzene ring. Benzylic resonance will be discussed in detail in Section 22-1. For more practice with Robinson annulations, go to Problems 59, 63, and 64.
New Reactions Synthesis and Reactions of Enolates and Enols 1. Enolate Ions (Section 18-1) O B RCH2CR
O D RCH P C G R
LDA or KH or (CH3)3COK or other strong base, 78°C
Enolate ion
2. Keto – Enol Equilibria (Section 18-2) O B RCH2CR
Catalytic H or HO Tautomerism
OH D RCH P C G R
3. Hydrogen – Deuterium Exchange (Section 18-2) O B RCH2CR
D2O, DO or D
O B RCD2CR
4. Stereoisomerization (Section 18-2) O B C G CR H }( R R
H or HO
O B C G CR } R ( H R
G
G
858
5. Halogenation (Section 18-3) O B RCH2CR
X2, H HX
O B RCHCR A
New Reactions
6. Enolate Alkylation (Section 18-4)
D RCH P C G
O R
O B RCHCR A R
RX X
SN2 reaction: R0X must be methyl or primary halide
7. Enamine Alkylation (Section 18-4) R A N O R J X CO C A R
R X
N
D CP C
R A N OR
H, H2O
CO C A R
R2NH
SN2 reaction: RX must be methyl, primary, or secondary halide
8. Aldol Condensations (Sections 18-5 through 18-7) O B 2 RCH2CH
OH O A B RCH2C O CHCH A A H R
HO
CHO D RCH2CH P C G R
HO,
Aldol adduct
Condensation product
Crossed aldol condensation (one aldehyde not enolizable) O B RCH
O B R CH2CH
HO, H2O
D RCH P C G
R CHO
Ketones O B RCCH2R
O OH R B A A RC O CH O CR A CH2R
HO
R O A B RC P CO CR A CH2R
, H2O Drive equilibrium
Intramolecular aldol condensation O
(CH2)n
O CH2R O
R HO
(CH2)n
H2O
R
R
Unstrained rings preferred
Reactions of A,B-Unsaturated Aldehydes and Ketones 9. Hydrogenation (Section 18-8)
C PC
H2, Pd, CH3CH2OH
N
O B C
CO CO C A A H H
H2O
Chapter 18
859
Enols, Enolates, and the Aldol Condensation
10. Addition of Halogen (Section 18-8)
O B C C PC
N
X2, CCl4
CO CO C A A X X
11. Condensations with Amine Derivatives (Section 18-8)
N B C
O B C ZNH2
C PC
D
Z
C PC Z = OH, RNH, etc.
Conjugate Additions to A,B-Unsaturated Aldehydes and Ketones 12. Hydrogen Cyanide Addition (Section 18-9)
O B C C PC
KCN, H
CO CO C A A CN H
13. Water, Alcohols, Amines (Section 18-9) N
ROH
CO CO C A A OR H
RNH2
N
C PC
CO CO C A A OH H
N
O B C
H2O
CO CO C A A RNH H
14. Organometallic Reagents (Section 18-10)
1. RLi, (CH3CH2)2O 2. H, H2O
O B C C PC
HO R D D C D C PC 1,2-Addition
1. R2CuLi, THF 2. H, H2O
N
Chapter 18
N
860
CO CO C A A R H 1,4-Addition
Additions of RMgX may be 1,2 or 1,4, depending on reagent and substrate structure.
Problems
Cuprate additions followed by enolate alkylations
O B C
C PC
N
1. R2CuLi, THF 2. R X
CO CO C A A R R
15. Michael Addition (Section 18-11)
O B C
O
D C PC D R
R
C PC
O H O A B B RC O C O C O C O CR
H
16. Robinson Annulation (Section 18-11)
O
CH2
CH A C K O
HO
O
Important Concepts 1. Hydrogens next to the carbonyl group (A-hydrogens) are acidic because of the electron-withdrawing nature of the functional group and because the resulting enolate ion is resonance stabilized. 2. Electrophilic attack on enolates can occur at both the a-carbon and the oxygen. Haloalkanes usually prefer the a-carbon. Protonation of the oxygen leads to enols. 3. Enamines are neutral analogs of enolates. Resonance donation of the nitrogen lone pair imparts nucleophilic character on the remote double bond carbon, which can be alkylated to give iminium cations that hydrolyze to aldehydes and ketones on aqueous work-up. 4. Aldehydes and ketones are in equilibrium with their tautomeric enol forms; the enol – keto conversion is catalyzed by acid or base. This equilibrium allows for facile a-deuteration and stereochemical equilibration. 5. A-Halogenation of carbonyl compounds may be acid or base catalyzed. With acid, the enol is halogenated by attack at the double bond; subsequent renewed enolization is slowed down by the halogen substituent. With base, the enolate is attacked at carbon, and subsequent enolate formation is accelerated by the halogens introduced. 6. Enolates are nucleophilic and reversibly attack the carbonyl carbon of an aldehyde or a ketone in the aldol condensation. They also attack the b-carbon of an a,b-unsaturated carbonyl compound in the Michael addition. 7. a,b-Unsaturated aldehydes and ketones show the normal chemistry of each individual double bond, but the conjugated system may react as a whole, as revealed by the ability of these compounds to undergo acid- and base-mediated 1,4-additions. Cuprates add in 1,4-manner, whereas alkyllithiums normally attack the carbonyl function.
Problems 32. Underline the a-carbons and circle the a-hydrogens in each of the following structures. O O B (a) CH3CH2CCH2CH3
O B (b) CH3CCH(CH3)2
O H3C ≈ (d)
O [CH3 (e)
H3C[ (c)
CH3 CH3
[CH3
CHO (f )
Chapter 18
861
862
Chapter 18
Enols, Enolates, and the Aldol Condensation
O B (g) (CH3)3CCH
O B (h) (CH3)3CCH2CH
33. Write the structures of every enol and enolate ion that can arise from each of the carbonyl compounds illustrated in Problem 32. 34. What product(s) would form if each carbonyl compound in Problem 32 were treated with (a) alkaline D2O; (b) 1 equivalent of Br2 in acetic acid; (c) excess Cl2 in aqueous base? 35. Describe the experimental conditions that would be best suited for the efficient synthesis of each of the following compounds from the corresponding nonhalogenated ketone. Br O A B (a) C6H5CHCCH3
O
Cl Cl Cl
Cl
Cl
(c)
(b) O
36. Propose a mechanism for the following reaction. (Hint: Take note of all of the products that are formed and base your answer on the mechanism for acid-catalyzed bromination of acetone shown in Section 18-3.) O
O Cl SO2Cl2
Catalytic HCl, CCl4
SO2 HCl
37. Give the product(s) that would be expected on reaction of 3-pentanone with 1 equivalent of LDA, followed by addition of 1 equivalent of
(a) CH3CH2Br
(b) (CH3)2CHCl
O B (c) (CH3)2CHCH2OS B O
CH3
(d) (CH3)3CCl
38. Give the product(s) of the following reaction sequences.
N 1. H, H
(a) CH3CHO
N 1. H, H 2. (CH3)2C P CHCH2Cl 3. H, H2O
2.
(b)
CH2CHO
CH2Br
3. H, H2O
39. The problem of double compared with single alkylation of ketones by iodomethane and base is mentioned in Section 18-4. Write a detailed mechanism showing how some double alkylation occurs even when only one equivalent each of the iodide and base is used. Suggest a reason why the enamine alkylation procedure solves this problem.
N H 2O
40. Would the use of an enamine instead of an enolate improve the likelihood of successful alkylation of a ketone by a secondary haloalkane? 41. Formulate a mechanism for the acid-catalyzed hydrolysis of the pyrrolidine enamine of cyclohexanone (shown in the margin).
H
42. Write the structures of the aldol condensation products of (a) pentanal; (b) 3-methylbutanal; (c) cyclopentanone.
O N H
43. Write the structures of the expected major products of crossed aldol condensation at elevated temperature between excess benzaldehyde and (a) 1-phenylethanone (acetophenone—see Section 17-1 for structure); (b) acetone; (c) 2,2-dimethylcyclopentanone. 44. Formulate a detailed mechanism for the reaction that you wrote in (c) of Problem 43.
Problems
Chapter 18
45. Give the likely products for each of the following aldol addition reactions. (a) 2
CH2CHO
NaOH, H2O
NaOH, H2O
CHO (CH3)2CHCHO
(b)
CHO
O CH3
(c) H H3C
CH3
NaOH, H2O NaOH, H2O
(d)
O
O
CHO
46. Rotundone (margin) is the natural product responsible for the peppery aroma in peppers, many herbs, and red wines (p. 846). What cyclic diketone will give rotundone upon intramolecular aldol condensation? 47. Write all possible products of the base-catalyzed crossed aldol reactions between each pair of reaction partners given below. (Hint: Multiple products are possible in every case; be sure to include thermodynamically unfavorable as well as favorable ones.)
Rotundone
(a) Butanal and acetaldehyde (b) 2,2-Dimethylpropanal and acetophenone (c) Benzaldehyde and 2-butanone 48. For each of the three crossed aldol reactions described in Problem 47, indicate which, if any, of the multiple possible products should predominate in the reaction mixture and explain why. 49. Aldol condensations may be catalyzed by acids. Suggest a role for H1 in the acid-catalyzed version. (Hint: Consider what kind of nucleophile might exist in acidic solution, where enolate ions are unlikely to be present.) 50. Reaction review. Without consulting the Reaction Road Maps on pp. 818 – 819, suggest reagents to convert butanal into each of the following compounds. O (a)
O
O (b)
H Cl
H Br
(c)
H
Br
O H (d)
O
O (e)
H
(f)
O (g)
H OH
51. Reaction review II. Without consulting the Reaction Road Maps on pp. 818 – 819, suggest reagents to convert acetophenone into each of the following compounds. O
O (a) C6H5
(b) C6H5 O
(d) C6H5
O CCl3
(c) C6H5
O (e) C6H5
OH
O (f) C6H5
CH2Br
52. Reaction review III. Without consulting the Reaction Road Maps on pp. 818 – 819, suggest reagents to convert 3-buten-2-one into each of the following compounds. O O O O (a) (b) (c) (d) C6H5
H
863
864
Chapter 18
Enols, Enolates, and the Aldol Condensation
O O
(e)
OH
O
(f)
OH
(g)
O
(h)
OH
53. A number of highly conjugated organic compounds have found use as sunscreens. One of the more widely used is 4-methylbenzylidene camphor (4-MBC), whose structure is shown in the margin. This compound is effective in absorbing so-called UV-B radiation (with wavelengths between 280 and 320 nm, responsible for most sunburns). Suggest a simple synthesis of this compound using a crossed aldol condensation. 4-MBC
54.
The distillate from sandalwood is one of the oldest and most highly valued fragrances in perfumery. The natural oil is in short supply and, until recently, synthetic substitutes have been difficult to prepare. Polysantol® (margin) is the most successful of these substitutes. Its synthesis in 1984 involved the aldol condensation shown below.
OH O
O
Polysantol®
O KOH, H2O, CH3OH
H
(a) This synthetic step, although usable, had a significant drawback that was responsible for its modest yield (60%). Identify this problem in detail. (b) A solution that avoids conventional aldol condensation was published in 2004. A bromoketone is prepared and, upon reaction with Mg metal, gives a magnesium enolate. This enolate then reacts with the aldehyde selectively to give a hydroxyketone that may be dehydrated to the desired product. Discuss how this approach solves the problem outlined in part (a). O
O MgBr
O Br2, CH3CO2H, CCl4
Mg, ether
Br O
O
H
HO 55. Write the expected major product of reaction of each of the carbonyl compounds (i) – (iii) with each of the reagents (a) – (h). O (i)
(a) (c) (e) (g)
O
CH2CH2CH3 D (ii) CH3CHP C G CH B O H2, Pd, CH3CH2OH Cl2, CCl4 CH3Li, (CH3CH2)2O NH2NHCNH2, CH3CH2OH B O
(b) (d) (f) (h)
(iii)
LiAlH4, (CH3CH2)2O KCN, H1, H2O (CH3CH2CH2CH2)2CuLi, THF (CH3CH2CH2CH2)2CuLi, followed by treatment with CH2PCHCH2Cl in THF
56. Give the expected product(s) of each of the following reactions. O B (a) C6H5CCH2CH2CH3
1. LDA, THF 2. CH3CH2Br, HMPA
H (b)
O H
1. LDA, THF 2. BrCH2COCH3 B O
Problems
H3C
Chapter 18
O
CH3 1. (CH3)2CuLi, THF 2. C6H5CH2Cl
CH3
LDA, THF
(d)
(c) O
(CH2)4Br
57. Write the products of each of the following reactions after aqueous work-up. O O O B B (a) C6H5CCH3 CH2 P CHCC6H5
LDA, THF
(b)
1. (CH2 P CH)2CuLi, THF O B 2. CH2 P CHCCH3
O (c)
O B (CH3)2C P CHCH
NaOH, H2O
1. (CH3)2CuLi, THF O B 2. (CH3)2C P CHCCH3
CH3 (d) O
(e) Write the results that you expect from base treatment of the products of reactions (c) and (d). 58. Write the final products of the following reaction sequences.
O B O CH2 P CHCCH3
(a)
NaOCH3, CH3OH,
O
O O CH3 B CH2 P CHCCH3
H3C (b)
KOH, CH3OH,
(c)
1. LDA, THF 2. HC q CCCH3 B O
(d) Write a detailed mechanism for reaction sequence (c). (Hint: Treat the 3-butyn-2-one reagent as a Michael acceptor in the first step.) 59. Propose syntheses of the following compounds by using Michael additions followed by aldol condensations (i.e., Robinson annulation). Each of the compounds shown has been instrumental in one or more total syntheses of steroidal hormones.
O
H3C (a)
OCH3
(b) CO2CH2CH3
O CH3
(c) CH3O
O B EC
O CHO O
CH3
(d) O
60. Would you expect addition of HCl to the double bond of 3-buten-2-one (shown in the margin) to follow Markovnikov’s rule? Explain your answer by a mechanistic argument.
O B CH3CCHP CH2 3-Buten-2-one
865
866
Enols, Enolates, and the Aldol Condensation
Chapter 18
61. Using the following information, propose structures for each of these compounds. (a) C5H10O, NMR spectrum A, UV lmax(e) 5 280(18) nm; (b) C5H8O, NMR spectrum B, UV lmax(e) 5 220(13,200), 310(40) nm; (c) C6H12, NMR spectrum C, UV lmax(e) 5 189(8,000) nm; (d) C6H12O, NMR spectrum D, UV lmax(e) 5 282(25) nm. (See the next page for spectrum D.)
3H 3H
3H
2.4
1.6
0.9
3H
2H 2H 6.9 6.8 6.7 6.6 6.5 6.4 6.3 6.2 6.1
(CH3)4Si
3.0
A
2.5
2.0
1.5
1.0
0.5
0.0
7.0
300-MHz 1H NMR spectrum ppm (δ )
(CH3)4Si
1H
1H
6.5
6.0
5.5
5.0
4.5
300-MHz
B
4.0 1H
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.0
NMR spectrum ppm (δ )
Next, for each of the following reactions, name an appropriate reagent for the indicated interconversion. (The letters refer to the compounds giving rise to NMR spectra A through D.) (e) A y C; (f) B y D; (g) B y A.
3H 3H
1H
5.0
1H
4.5
2H
4.0
3.5
3.0
2.5
2.0
2H
1.5
300-MHz 1H NMR spectrum ppm (δ ) C
1.0
0.5
Problems
6H
4H
2H
2.5
3.0
(CH3)4Si
2.0
1.5
1.0
0.5
0.0
300-MHz 1H NMR spectrum ppm (δ ) D
62.
Treatment of cyclopentane-1,3-dione with iodomethane in the presence of base leads mainly to a mixture of three products.
O
O
O CH3
NaOH, CH3I, CH3CH2OH
O
O CH3
H
H
CH3
O
O
A
OCH3
B
C
(a) Give a mechanistic description of how these three products are formed. (b) Reaction of product C with a cuprate reagent results in loss of the methoxy group. For example,
O
O 1. (CH3CH2CH2CH2)2CuLi, THF 2. H, H2O
OCH3 C
CH2CH2CH2CH3 D
Suggest a mechanism for this reaction, which is another synthetic route to enones substituted at the b-carbon. (Hint: See Exercise 18-24.) 63.
A somewhat unusual synthesis of cortisone-related steroids includes the following two reactions.
Chapter 18
867
Enols, Enolates, and the Aldol Condensation
Chapter 18
CH3 OC(CH3)3 A OC(CH3)3 CH3CHCH2CH2Br
(CH3)3COK, benzene
O CH3 OC(CH3)3 CH3I, (CH3)3COK, benzene
OC(CH3)3 O
CH3 H3C OC(CH3)3 (CH3)3CO O
(a) Propose mechanisms for these two transformations. Be careful in choosing the initial site of deprotonation in the starting enone. The alkenyl hydrogen, in particular, is not acidic enough to be the one initially removed by base in this reaction. (b) Propose a sequence of reactions that will connect the carbons marked by arrows in the third structure shown to form another six-membered ring. 64. The following steroid synthesis contains modified versions of two key types of reactions presented in this chapter. Identify these reaction types and give detailed mechanisms for each of the transformations shown. H3C %
O KOC(CH3)3, THF
O
O
^
868
CO2CH3 H3C % O
O
H3C %
K2CO3, CH3OH
) CO2CH3
O
) CO2CH3
O
O
65. Devise reasonable plans for carrying out the following syntheses. Ignore stereochemistry in your strategies. CH3 (a)
, starting from cyclohexanone O
H 3C O (b)
CH3
CN CH3 O
O
, starting from 2-cyclohexenone
(Hint: Prepare
CH3 CH2
in your first step.)
Problems
Chapter 18
66. Write reagents (a, b, c, d, e) where they have been omitted from the following synthetic sequence. Each letter may correspond to one or more reaction steps. This sequence is the beginning of a synthesis of germanicol, a naturally occurring triterpene. The diol used in the step between (a) and (b) provides selective protection of the more reactive carbonyl group. [Hint: See Problem 63 when formulating (b).]
(a)
O
O
H3C
CH3 O
CH3
H3C
H , HOCH2CH2OH
O (b)
O
O
O
CH3
H3C %
1. Li, liquid NH3* 2. (c)
O
H CH3
H3C
H3C %
^
CH3
CH3 %
CH3CO B O H3C
O
H3C %
(e)
%
H
^
%
H
CH3CO B O H3C
H CH3 H3C
CH3
CH3
^
^
CH3 % CH3 H
^
%
HO H3C
H CH3
%
H % H3C %
H CH3 Germanicol
Team Problem 67. When 2-methylcyclopentanone is treated with the bulky base triphenylmethyllithium under the two sets of conditions shown, the two possible enolates are generated in differing ratios. Why is this so? OLi
O CH3
(C6H5)3CLi
Conditions A: Ketone added to excess base 72% Conditions B: Excess ketone added to base 6%
OLi CH3
(d)
CH3
^
O
CH3 O % O
^
CH3 O % O
^
H3C %
28% 94%
To tackle this problem, you have to invoke the principles of kinetic versus thermodynamic control (review Sections 11-6, 14-6, and 18-2); that is, which enolate is formed faster and which one is more stable? Divide your team so that one group considers conditions A and the other conditions B. Use curved arrows to show the flow of electrons leading to each enolate. Then assess whether your set of conditions is subject to enolate equilibration (thermodynamic control) or not (kinetic control). Reconvene to discuss these issues and draw a qualitative potential-energy diagram depicting the progress of deprotonation at the two a sites. *An electron-transfer process (compare alkyne reduction, Section 13-6) that is the equivalent of adding hydride, H:2, to the b-carbon. The product is the enolate of the saturated ketone.
CH3 %
O
869
870
Chapter 18
Enols, Enolates, and the Aldol Condensation
Preprofessional Problems 68. When 3-methyl-1,3-diphenyl-2-butanone is treated with excess D2O in the presence of catalytic acid, some of its hydrogens are replaced by deuterium. How many? (a) One; (b) two; (c) three; (d) six; (e) eight. 69. How would you best classify the following reaction? (a) Wittig reaction; (b) cyanohydrin formation; (c) conjugate addition; (d) aldol addition. H E H2C P CHCHPNH H O B H3CC
CH3
NC
NCCH2CHP CHNH2
70. The aqueous hydroxide-promoted reaction of the compound shown in the margin with (CH3)3CCHO yields exclusively one compound. Which one is it? O B (a) H3CC
CHP C
(c) (CH3)3CCHP CH
f
OCH3 A (b) H3C O C A OCH3
C(CH3)3
i H
CH3
O B (d) (CH3)3CCHP CHO C
CH3
CH3
71. The 1H NMR spectrum of 2,4-pentanedione indicates the presence of an enol tautomer of the dione. What is its most likely structure? OH (a)
OH
O (b)
OH
O (c)
OH
EOH C P C C P E H (d) HO CH3 H3C H
C H A P T E R
NINETEEN
Carboxylic Acids O N O O
R, S O
W
hat types of compounds are the end products of biological metabolism? We inhale oxygen and exhale carbon dioxide, one product of the oxidation of organic compounds (Chapter 3). Carboxylic acids also are major products of biochemistry occurring under oxidizing conditions and are thus widespread throughout living systems on Earth. You are probably familiar with the bottle of wine that has turned to vinegar, the result of enzymatic oxidation of ethanol to acetic acid. Carboxylic acids are characterized by the presence of the carboxy group, a functional group containing a hydroxy unit attached to a carbonyl carbon. This substituent is written COOH or CO2H; both representations are used in this chapter. Carboxylic acids not only are widely distributed in nature, they are also important industrial chemicals. For example, besides being the most important building block in the assembly of complex biological molecules, acetic acid is an industrial commodity that is produced in very large quantities. Much of the functional behavior of carboxylic acids can be anticipated if we view them structurally as hydroxy carbonyl derivatives. Thus, the hydroxy hydrogen is acidic, the oxygens are basic and nucleophilic, and the carbonyl carbon is subject to nucleophilic attack. O B CH
)O) Basic Electrophilic B OH DC Acidic OOH R G> Carboxy group > In this chapter, we first introduce the system of naming carboxylic acids and then list some of their physical and spectroscopic characteristics. We then examine their acidity and basicity, two properties that are strongly influenced by the interaction between the electron-withdrawing carbonyl group and the hydroxy function. Methods for the preparation of the carboxy group are considered next, followed by a survey of its reactivity. Reactions of carboxylic acids will feature a new two-step substitution pathway, addition – elimination, for the replacement of the hydroxy group by other nucleophiles, such as halide, alkoxide, and amide. The chemistry of the carboxylic acid derivatives, which result from these transformations, is the subject of Chapter 20.
The carboxylic acid fexofenadine (Allegra) is a top-selling nonsedating antihistamine drug for the treatment of allergy symptoms. It contains a stereocenter, but both enantiomers are equally effective, hence it is sold as a racemate.
872
Carboxylic Acids
Chapter 19
19-1 Naming the Carboxylic Acids Like other organic compounds, many carboxylic acids have common names that are used frequently in the literature. These names often indicate the natural sources from which the acids were first isolated (Table 19-1). Chemical Abstracts retains the common names for the two simplest, formic and acetic acids. The carboxy function and the functional groups of its derivatives take precedence in naming over any other group discussed so far. The overall “order of precedence” of the most common functional groups is given below. In any molecule containing more than one functional group, the group highest in precedence is entered as a suffix in the name, and all other substituents as prefixes. For example, consider HOCH2CH2COCH2CHO, a molecule containing alcohol, ketone, and aldehyde functions. In its name, the first two groups appear as prefixes (hydroxy- and oxo-, respectively), while the aldehyde group, being of highest precedence, is signified by the suffix -al: This compound is 5-hydroxy-3-oxopentanal. Alkenes and alkynes are exceptions. They rank below the amines in order of precedence, but when a carbon – carbon double or triple bond is part of the parent chain or ring of a molecule, we insert -en(e)- or -yn(e)- just before the suffix for the highest-ranking functional group. The following examples of names for carboxylic acids illustrate these principles. Order of Precedence of Functional Groups O B RCOH Carboxylic acid
O O O B B B RCOCR RCOR Anhydride
Ester
O B RCX Acyl halide
O B RCNR2 RCN Amide
Nitrile
O B RCH
O B RCR ROH RSH RNH2
Aldehyde
Ketone
Alcohol
Thiol
Amine
Increasing precedence in naming
Table 19-1 Names and Natural Sources of Carboxylic Acids Structure
IUPAC name
Common name
Natural source
HCOOH
Methanoic acid
Formic acida
From the “destructive distillation” of ants ( formica, Latin, ant)
CH3COOH
Ethanoic acid
Acetic acida
Vinegar (acetum, Latin, vinegar)
CH3CH2COOH
Propanoic acid
Propionic acid
CH3CH2CH2COOH
Butanoic acid
Butyric acid
CH3(CH2)3COOH CH3(CH2)4COOH
Pentanoic acid Hexanoic acid
Valeric acid Caproic acid
Dairy products ( pion, Greek, fat) Butter (particularly if rancid) (butyrum, Latin, butter) Valerian root Odor of goats (caper, Latin, goat)
O O O O
O O
a
Used by Chemical Abstracts.
19-1 Naming the Carboxylic Acids
Chapter 19
873
The IUPAC system derives the names of the carboxylic acids by replacing the ending -e in the name of the alkane by -oic acid. The alkanoic acid stem is numbered by assigning the number 1 to the carboxy carbon and labeling any substituents along the longest chain incorporating the CO2H group accordingly. CH 0 3
Br H /∑ C E H H3 C COOH
OH
0
CH2P CHCOOH
CH3
O
Br O O
O
O
O
O
Propenoic acid (Acrylic acid)
(R)-2-Bromopropanoic acid ( -Bromopropionic acid)
(2R,3S)-Dimethylpentanoic acid ( R, S-Dimethylvaleric acid)
In multiply functionalized carboxylic acids, the main chain is chosen to include other functional groups as much as possible. Saturated cyclic acids are named as cycloalkanecarboxylic acids. Their aromatic counterparts are the benzoic acids. In these compounds, the carbon attached to the carboxy functional group is Cl.
OH O 5-Butyl-6-heptenoic acid (Better than 5-ethenylnonanoic acid)
O B CH3CCHCH2CH2COOH A CH2CH2CH3 5-Oxo-4-propylhexanoic acid
COOH
Cl Br
1
HO 2
2 1
COOH
1-Bromo-2-chlorocyclopentanecarboxylic acid
2-Hydroxybenzoic acid (o-Hydroxybenzoic acid, salicylic acid)
2-Hydroxybenzoic (salicylic) acid has been known for over 2000 years as an analgesic (pain-killing) folk remedy present in willow bark (salix, Latin, willow). The acetic acid ester of salicylic acid is better known as aspirin (Chemical Highlight 22-2). Dicarboxylic acids are referred to as dioic acids. OO BB HOCCOH
O O B B HOCCH2COH
HOOCCH2CH2COOH
Ethanedioic acid (Oxalic acid)
Propanedioic acid (Malonic acid)
Butanedioic acid (Succinic acid)
HOOC(CH2)3COOH
HOOC(CH2)4COOH
HO2CCH P CHCO2H
Pentanedioic acid (Glutaric acid)
Hexanedioic acid (Adipic acid)
cis-2-Butenedioic acid (Maleic acid) or trans-2-Butenedioic acid (Fumaric acid)
Their common names again derive from their natural origins. For example, butanedioic (succinic) acid was discovered in the distillate from amber (succinum, Latin, amber), and trans-2-butenedioic (fumaric) acid is found in the plant Fumaria, which was burned in ancient times to ward off evil spirits ( fumus, Latin, smoke).
Human beings have known since prehistoric times that chewing on the bark of the white willow relieves pain.
874
Chapter 19
Carboxylic Acids
Exercise 19-1 Give systematic names or write the structure, as appropriate, of the following compounds. COOH OH
(a) Br
Cl
(b)
COOH (c)
O
CH3O O
(d) 2,2-Dibromohexanedioic acid (f) 4-(1,1-Dimethylethyl)benzoic acid
NO2
(e) 4-Hydroxypentanoic acid
In Summary The systematic naming of the carboxylic acids is based on the alkanoic acid stem. Cyclic derivatives are called cycloalkanecarboxylic acids, their aromatic counterparts are referred to as benzoic acids, and dicarboxylic systems are labeled alkanedioic acids.
19-2 Structural and Physical Properties of Carboxylic Acids What is the structure of a typical carboxylic acid? What are the characteristic physical properties of carboxylic acids? This section answers these questions, beginning with the structure of formic acid. We shall see that carboxylic acids exist mainly as hydrogen-bonded dimers.
Formic acid is planar
1.097 Å
H 1.343 Å
111.0
O 0.972 Å
124.1
1.202 Å
Ok
C
124.9
The molecular structure of formic acid is shown in Figure 19-1. It is roughly planar, as expected for a hydroxy-substituted carbonyl compound, with an approximately trigonal carbonyl carbon. (Compare the structure of methanol, Figure 8-1, with that of acetaldehyde, Figure 17-2.) These structural characteristics are found in carboxylic acids in general.
106.3
H
The carboxy group is polar and forms hydrogen-bonded dimers
Figure 19-1 The molecular structure of formic acid is planar, with a roughly equilateral trigonal arrangement about the carbonyl carbon.
The carboxy function derives strong polarity from the carbonyl double bond and the hydroxy group, which forms hydrogen bonds to other polarized molecules, such as water, alcohols, and other carboxylic acids. Not surprisingly, therefore, the lower carboxylic acids (up to butanoic acid) are completely soluble in water. As neat liquids and even in fairly dilute solutions (in aprotic solvents), carboxylic acids exist to a large extent as hydrogen-bonded dimers, each O–H∙ ∙ ∙O interaction ranging in strength from about 6 to 8 kcal mol21 (25 to 34 kJ mol21). Carboxylic Acids Form Dimers Readily
2 RCOOH
OZHOO G J ROC COR G J O O HZ O Two hydrogen bonds
O OH (E )-3-Methyl-2-hexenoic acid
Table 19-2 shows that carboxylic acids have relatively high melting and boiling points, because they form hydrogen bonds in the solid as well as the liquid state. Carboxylic acids, especially those possessing relatively low molecular weights and correspondingly high volatility, exhibit characteristically strong odors. For example, the presence of butanoic acid contributes to the characteristic strong flavor of many cheeses, and (E )-3-methyl-2-hexenoic acid is one of the principal compounds responsible for the smell of human sweat.
19-3 Spectroscopy and Mass Spectrometry of Carboxylic Acids
Table 19-2
Melting and Boiling Points of Functional Alkane Derivatives with Various Chain Lengths
Derivative
Melting point (8C)
Boiling point (8C)
2182.5 297.7 297.8 292.0 8.4 2183.3 2136.4 2114.7 2121.0 16.7 2187.7 2122.8 2126.5 295.0 281.0 220.8
2161.7 224.2 65.0 221.0 100.6 288.6 12.3 78.5 20.8 118.2 242.1 46.6 97.4 56.5 48.8 141.8
CH4 CH3Cl CH3OH HCHO HCOOH CH3CH3 CH3CH2Cl CH3CH2OH CH3CHO CH3COOH CH3CH2CH3 CH3CH2CH2Cl CH3CH2CH2OH CH3COCH3 CH3CH2CHO CH3CH2COOH
In Summary The carboxy function is planar and contains a polarizable carbonyl group. Carboxylic acids exist as hydrogen-bonded dimers and exhibit unusually high melting and boiling points.
19-3 Spectroscopy and Mass Spectrometry of Carboxylic Acids The polarizable double bond and the hydroxy group also shape the spectra of carboxylic acids. In this section, we show how both NMR and IR methods are used to characterize the carboxy group. We also illustrate how carboxylic acids fragment in the mass spectrometer.
The carboxy hydrogen and carbon are deshielded As in aldehydes and ketones, the hydrogens positioned on the carbon next to the carbonyl group are slightly deshielded in 1H NMR spectra. The effect diminishes rapidly with increasing distance from the functional group. The hydroxy proton resonates at very low field (d 5 10 – 13 ppm). As in the NMR spectra of alcohols, its chemical shift varies strongly with concentration, solvent, and temperature, because of the strong ability of the OH group to enter into hydrogen bonding. The 1H NMR spectrum of pentanoic acid is shown in Figure 19-2. 1
H NMR Chemical Shifts of Alkanoic Acids
CH3COOH
2.08 ppm
CH3CH2COOH
1.16 2.36 ppm
(CH3)2CHCOOH
1.21
2.56 ppm
HCOOH
8.08 ppm
The 13C NMR chemical shifts of carboxylic acids also are similar to those of the aldehydes and ketones, with moderately deshielded carbons next to the carbonyl group and the typically low-field carbonyl absorptions. However, the amount of deshielding is smaller because the positive polarization of the carboxy carbon is somewhat attenuated by the presence of the extra OH group.
Chapter 19
875
876
Chapter 19
Carboxylic Acids
1H NMR
O
3H
CH3CH2CH2CH2COH
1.4 1.3 ppm
1.7 1.6 ppm
1.0 0.9 ppm
2H (CH3)4Si 2.4 2.3 ppm
12.0 11.5 ppm
2H each
1H 20 19 18 17 16 15 14 13 12 11 10 9
8
7
6
5
4
3
2
1
0
ppm (δ ) Figure 19-2 300-MHz 1H NMR spectrum of pentanoic acid. The scale has been expanded to 20 ppm to allow the signal for the acid proton at d 5 11.75 ppm to be shown. The methylene hydrogens at C2 absorb at the next lowest field as a triplet (d 5 2.35 ppm, J 5 7 Hz), followed by a quintet and a sextet, respectively, for the next two sets of methylenes. The methyl group appears as a distorted triplet at highest field (d 5 0.92 ppm, J 5 6 Hz).
Typical 13C NMR Chemical Shifts of Alkanoic Acids CH3COOH
CH3CH2COOH
9.0 27.8
21.1 177.2 ppm
CH3CH2CHO
compare
5.2 36.7
180.4 ppm
201.8 ppm
This attenuation is best understood by writing resonance forms. Recall from Section 17-2 that aldehydes and ketones are described by two such representations. The dipolar Lewis structure indicates the polarization of the CPO bond. Even though it is a minor contributor (because of the lack of an octet on carbon), it explains the deshielding of the carbonyl carbon. In carboxylic acids, however, the corresponding dipolar form contributes less to the resonance hybrid: The hydroxy oxygen can donate an electron pair to give a third arrangement in which the carbon and both oxygen atoms have octets. The degree of positive charge on the carbonyl carbon and therefore its deshielding are greatly reduced. The changes in electron density distribution around the carbonyl function when going from acetone to acetic acid are shown in their respective electrostatic potential maps. The strongly positive polarization (blue) of the carbonyl carbon in the ketone is diminished in the acid (green). Resonance in Aldehydes and Ketones O) )>
)O)
Acetone
H3C
DCG
H3C
R
DC G
R
The contribution of the second resonance form, though minor, explains the strong deshielding of the carbonyl and adjacent carbons
Resonance in Carboxylic Acids )O)
Acetic acid
DCG > R OH >
O) )>
DC R G> OH >
O) )>
DC > R OH
The third resonance form explains the attenuated deshielding of the carbonyl carbon relative to that of aldehydes and ketones
19-3 Spectroscopy and Mass Spectrometry of Carboxylic Acids
Exercise 19-2
Working with the Concepts: Deducing the Structure of Carboxylic Acids from NMR Data A foul-smelling carboxylic acid with b.p. 1648C gave the following NMR data: 1H NMR: d 5 1.00 (t, J 5 7.4 Hz, 3 H), 1.65 (sextet, J 5 7.5 Hz, 2 H), 2.31 (t, J 5 7.4 Hz, 2 H), and 11.68 (s, 1 H) ppm: 13C NMR: d 5 13.4, 18.5, 36.3, and 179.6 ppm. Assign a structure to it. Strategy A good place to start is the 13C NMR spectrum, because it gives you, in addition to confirming the presence of a carboxy carbon, the number of additional unique carbon atoms in the molecule. The 1H NMR spectrum will expand on that information through the number of signals (hence the number of different hydrogens), their chemical shift (hence possibly the presence of other functional groups), integration (hence the relative abundance of each type of hydrogen), and spin–spin splitting (hence the number of equivalent neighbors of the hydrogen giving rise to a particular signal). Solution • The 13C NMR spectrum shows a total of four signals, one of which, at d 5 179.6 ppm, is due to the carboxy carbon. This leaves us with three additional peaks, all in the saturated carbon region (Section 10-9). One of them (d 5 36.3 ppm) is slightly deshielded relative to the other two, likely a carbon atom next to the carboxy group. • Turning to the 1H NMR spectrum, we can identify the carboxy hydrogen at d 5 11.68 ppm, in addition to three additional signals. Again, all of the latter appear in the saturated region (Table 10-2), one of them (d 5 2.31 ppm) slightly deshielded relative to the other two, likely attached to the carbon next to the carboxy group. • Focusing next on the integrated values for the hydrogen signals, we note that their ratios relative to the single carboxy proton are 3 : 2 : 2. The highest-field absorption, at d 5 1.00 ppm, integrates for 3 H and is likely due to the presence of a methyl group. The signal at d 5 2.31 ppm, suspected to be relatively deshielded by the attached carboxy function, integrates for 2 H, pointing to the presence of a –CH2COOH substructure. • An analysis of the splitting patterns suggests that the methyl group, a triplet, has two equivalent hydrogen neighbors, indicating the presence of CH3 – CH2 – . Moreover, the signal at d 5 2.31 ppm is also a triplet, extending the suggested substructure above to – CH2CH2COOH. • Putting all of this information together leads to the solution: CH3CH2CH2COOH (butanoic or “butyric” acid, with the smell of rancid butter). We can confirm this assignment by looking at the multiplicity of the 1H NMR signal for the CH2 fragment flanked by CH3 on one, and CH2 on the other side, CH3CH2CH2COOH. Assuming that the coupling constants in both directions will be very similar, the (N 1 1) rule predicts a sextet, as is indeed observed.
Exercise 19-3
Try It Yourself Assign the structure of a carboxylic acid on the basis of the following NMR data: 1H NMR: d 5 1.20 (d, J 5 7.5 Hz, 6 H), 2.58 (septet, J 5 7.5 Hz, 1 H), and 12.17 (s, 1 H) ppm; 13C NMR: d 5 18.7, 33.8, and 184.0 ppm. (Caution: The 13C NMR spectrum reveals only the number of unique carbon atoms in the molecule. Hint: The presence of a signal integrating for 6 H in the 1 H NMR spectrum is significant. One carbon cannot be attached to six hydrogens.)
The carboxy group shows two important IR bands The carboxy group consists of a carbonyl group and an attached hydroxy substituent. Consequently, both characteristic stretching frequencies are seen in the infrared spectrum. The O – H bond gives rise to a very broad band at a lower wavenumber (2500 – 3300 cm21) than is observed for alcohols, largely because of strong hydrogen bonding. The IR spectrum of propanoic acid is shown in Figure 19-3.
Chapter 19
877
100
O
3100 (broad) IR 0 4000 3500
CH2CHCO
H
H 3000
2500
H
1715
2000 1500 Wavenumber
1000
600 cm−1
Mass spectra of carboxylic acids show three modes of fragmentation The mass spectra of carboxylic acids typically show fairly weak molecular ions, because fragmentation occurs readily in several ways. In common with aldehydes and ketones, both a cleavage and McLafferty rearrangement are observed. In addition, cleavage at the C3 –C4 bond occurs together with migration of an a-hydrogen to oxygen, because a resonancestabilized carbocation results:
>
> >
O c C A O A H
> >
Cleavage RCH2CH2CH2j
>
j O B > RCH2CH2CH2 O C O OH >
O B C B O A H
> >
> >
Mass Spectral Fragmentation of Carboxylic Acids O B C A O A H
m/z 45 Protonated CO2
RH H CH O H 2C
CH OH CH2
j
McLafferty rearrangement RCHPCH2
j O H A B RCH2 O CH2 O CH O C O OH
> >
Figure 19-3 IR spectrum of propanoic acid: n˜O–H stretch 5 3000 cm21; n˜CPO stretch 5 1715 cm21. The peaks associated with these stretching vibrations are broad because of hydrogen bonding.
Carboxylic Acids
> >
Chapter 19
Transmittance (%)
878
j OH A KC H H2C OH m/z 60 Acetic acid enol
C3–C4 Cleavage and -H shift RCH2j
OH A CH2 P CHO C OOH m/z 73 Protonated propenoic acid
Figure 19-4 shows the mass spectrum of butanoic acid. Exercise 19-4 Figure 19-5 shows the mass spectrum of pentanoic acid. Identify the peaks resulting from the three cleavage modes illustrated on the following page.
19-4 Acidic and Basic Character of Carboxylic Acids
+•
OH H2C
C
879
Figure 19-4 Mass spectrum of butanoic acid. The molecular ion and peaks arising from the cleavage modes described in the text are identified.
60
MS
100
Chapter 19
CH3CH2CH2COOH
Relative abundance
OH
50
CH2 (HOC
+
CH
OH
C OH
O )+
73
45 88 M+• 0 0
20
40
60
80
100
m/z
MS
100
Relative abundance
Figure 19-5 Mass spectrum of pentanoic acid. (See Exercise 19-4.) Note the relatively low intensity of the molecular ion peak.
CH3CH2CH2CH2COOH
50
102 M+• 0 0
20
40
60
80
100
120
m/z
In Summary The NMR signals for carboxylic acids reveal highly deshielded acid protons and carbonyl carbons and moderately deshielded nuclei next to the functional group. The infrared spectrum shows characteristic bands at about 2500 –3300 (O –H) and 1710 (CPO) cm21. The mass spectra of carboxylic acids reflect three characteristic modes of fragmentation.
19-4 Acidic and Basic Character of Carboxylic Acids Like alcohols (Section 8-3), carboxylic acids exhibit both acidic and basic character: Deprotonation to carboxylate ions is relatively easy, protonation more difficult.
Carboxylic acids are relatively strong acids Carboxylic acids have much lower pKa values than alcohols do, even though the relevant hydrogen in each case is that of a hydroxy group.
Chapter 19
Carboxylic Acids
Carboxylic Acids Dissociate Readily O B > RCOH p
O B > RCO p)
> H2O p
104 –105 4–5
Ka pKa
> 2 HOH
Carboxylate ion
Why do carboxylic acids dissociate to a greater extent than do alcohols? The difference is that the hydroxy substituent of a carboxylic acid is attached to a carbonyl group, whose positively polarized and sp2-hybridized carbonyl carbon exerts a powerful electron-withdrawing inductive effect. In addition, the carboxylate ion is significantly stabilized by resonance, much as is the enolate ion formed by deprotonation of the a-carbon in aldehydes and ketones (Section 18-1). Resonance in Carboxylate and Enolate Ions (B base)
BH
)> O)
)O) B > O) RC O p
O
B)
)O) B > RCOH p
> O) RC O p
)> O) O
Carboxylate ion
> O RC P p
pKa
4–5
Enolate ion
BH
)> O)
)O) B CHR RC O >
CHR RC O >
)> O) O
)O) B B) RCCH2R
O
880
RC P CHR
pKa
19–21
In contrast with enolates, two of the three resonance forms in carboxylate ions are equivalent (Section 1-5). As a result, carboxylates are symmetric, with equal carbon – oxygen bond lengths (1.26 Å), in between the lengths typical of the carbon– oxygen double (1.20 Å) and single (1.34 Å) bonds in the corresponding acids (see Figure 19-1). The electrostatic potential map of acetate ion in the margin shows the equal distribution of negative charge (red) over the two oxygens.
Electron-withdrawing substituents increase the acidity of carboxylic acids Acetate ion
We saw previously (Section 8-3) that the inductive effect of electron-withdrawing groups close to the hydroxy function increases the acidity of alcohols. A similar phenomenon is observed in carboxylic acids. Table 19-3 shows the pKa values of selected acids. Note that two or three electron-withdrawing groups on the a-carbon can result in carboxylic acids whose strength approaches that of some inorganic (mineral) acids. Exercise 19-5 In the following sets of acids, rank the compounds in order of decreasing acidity.
(a) CH3CH2COOH F A (b) CH3CHCH2COOH COOH
F
Br A CH3CHCOOH
CH3CBr2COOH
Br A CH3CHCH2COOH COOH
COOH
(c) F
19-4 Acidic and Basic Character of Carboxylic Acids
Table 19-3 pKa Values of Various Carboxylic and Other Acids Compound
pKa
Alkanoic acids HCOOH CH3COOH ClCH2COOH Cl2CHCOOH Cl3CCOOH F3CCOOH CH3CH2CH2COOH CH3CH2CHClCOOH CH3CHClCH2COOH ClCH2CH2CH2COOH
pKa
Dioic acids 3.55 4.76 2.82 1.26 0.63 0.23 4.82 2.84 4.06 4.52
Benzoic acids 4-CH3C6H4COOH C6H5COOH 4-ClC6H4COOH
Compound
HOOCCOOH HOOCCH2COOH HOOCCH2CH2COOH HOOC(CH2)4COOH
1.27, 2.83, 4.20, 4.35,
4.19 5.69 5.61 5.41
Other acids H3PO4 HNO3 H2SO4 HCl H2O CH3OH
2.15 (first pKa) 21.4 23.0 (first pKa) 28.0 15.7 15.5
4.36 4.20 3.98
The dioic acids have two pKa values, one for each CO2H group. In ethanedioic (oxalic) and propanedioic (malonic) acids, the first pKa is lowered by the electron-withdrawing inductive effect of the second carboxy group on the first. In the higher dioic acids, the two pKa values are close to those of monocarboxylic acids. The relatively strong acidity of carboxylic acids means that their corresponding carboxylate salts are readily made by treatment of the acid with base, such as NaOH, Na2CO3, or NaHCO3. These salts are named by specifying the metal and replacing the ending -ic acid in the acid name by -ate. Thus, HCOO2Na1 is called sodium formate; CH3COO2Li1 is named lithium acetate; and so forth. Carboxylate salts are much more water soluble than the corresponding acids because the polar anionic group is readily solvated.
Carboxylate Salt Formation CH3 A CH3CCH2CH2COOH A CH3
NaOH, H2O
CH3 A CH3CCH2CH2COONa A CH3
4,4-Dimethylpentanoic acid
Sodium 4,4-dimethylpentanoate
(Slightly water soluble)
(Highly water-soluble salt)
HOH
Carboxylic acids may be protonated on the carbonyl oxygen The lone electron pairs on both of the oxygen atoms in the carboxy group can, in principle, be protonated, just as alcohols are protonated by strong acids to give alkyloxonium ions (Sections 8-3 and 9-2). It has been found that it is the carbonyl oxygen that is more basic. Why? Protonation at the carbonyl oxygen gives a species whose positive charge is delocalized by resonance. The alternative possibility, resulting from protonation at the hydroxy group, does not benefit from such stabilization (see also Exercise 2-9).
Chapter 19
881
Chapter 19
Carboxylic Acids
Protonation of a Carboxylic Acid Positive charge is stabilized by resonance
)O) B D C G> D H R O A H
Not observed
)O) B D C G> R OH >
D
H
D O )>
> O
H
K
106
B C D G> R OH >
H
D
882
pKa
6
D C M R OH >
Resonance stabilization is not possible
Note that protonation is nevertheless very difficult, as is shown by the very high acidity (pKa < 26) of the conjugate acid. We shall see, however, that such protonations are important in many reactions of the carboxylic acids and their derivatives.
Exercise 19-6 The pKa of protonated acetone is 27.2 and that of protonated acetic acid is 26.1. Explain.
In Summary Carboxylic acids are acidic because the polarized carbonyl carbon is strongly electron withdrawing and deprotonation gives resonance-stabilized anions. Electronwithdrawing groups increase acidity, although this effect wears off rapidly with increasing distance from the carboxy group. Protonation is difficult, but possible, and occurs on the carbonyl oxygen to give a resonance-stabilized cation.
19-5 Carboxylic Acid Synthesis in Industry Carboxylic acids are useful reagents and synthetic precursors. The two simplest ones are manufactured on a large scale industrially. Formic acid is employed in the tanning process in the manufacture of leather and in the preparation of latex rubber. It is synthesized efficiently by the reaction of powdered sodium hydroxide with carbon monoxide under pressure. This transformation proceeds by nucleophilic addition followed by protonation. Formic Acid Synthesis NaOH
CO
150°C, 100 psi
H, H2O
HCOONa
HCOOH
There are three important industrial preparations of acetic acid: ethene oxidation through acetaldehyde (Section 12-16); air oxidation of butane; and carbonylation of methanol. The mechanisms of these reactions are complex. Acetic Acid by Oxidation of Ethene CH2 P CH2
O2, H2O, catalytic PdCl2 and CuCl2 Wacker process
CH3CHO
O2, catalytic Co3
CH3COOH
Acetic Acid by Oxidation of Butane CH3CH2CH2CH3
O2, catalytic Co3, 15–20 atm, 180°C
CH3COOH
Acetic Acid by Carbonylation of Methanol CH3OH
CO, catalytic Rh3, I2, 30–40 atm, 180°C Monsanto process
CH3COOH
19-6 Methods for Introducing the Carboxy Functional Group
Global demand of acetic acid exceeds 6.5 million tons (6.5 3 109 kg) annually. It is used to manufacture monomers for polymerization, such as methyl 2-methylpropenoate (methyl methacrylate; Table 12-3), as well as pharmaceuticals, dyes, and pesticides. Two dicarboxylic acids in large-scale chemical production are hexanedioic (adipic) acid, used in the manufacture of nylon (see Chemical Highlight 21-4), and 1,4-benzenedicarboxylic (terephthalic) acid, whose polymeric esters with diols are fashioned into plastic sheets, films, and bottles for soft drinks.
Chapter 19
883
COOH
COOH 1,4-Benzenedicarboxylic acid
19-6 Methods for Introducing the Carboxy Functional Group
(Terephthalic acid)
The oxidation of primary alcohols and aldehydes to carboxylic acids by aqueous Cr(VI) was described in Sections 8-6 and 17-4. This section presents two additional reagents suitable for this purpose. It is also possible to introduce the carboxy function by adding a carbon atom to a haloalkane. This transformation can be achieved in either of two ways: the carbonation of organometallic reagents or the preparation and hydrolysis of nitriles.
Oxidation of primary alcohols and of aldehydes furnishes carboxylic acids Primary alcohols oxidize to aldehydes, which in turn may readily oxidize further to the corresponding carboxylic acids. Polyethylene terephthalate is Dacron, used in the Jarvik artificial heart. As a thin film, it is Mylar.
Carboxylic Acids by Oxidation
RCH2OH
Oxidation
O B RCH
O B RCOH
Oxidation
In addition to aqueous CrO3, KMnO4 and nitric acid (HNO3) are frequently used in this process. Because it is one of the cheapest strong oxidants, nitric acid is often chosen for large-scale and industrial applications.
2 HNO3
O B ClCH2CH2CH
25°C
3-Chloropropanal
O B ClCH2CH2COH 79%
2 NO2
H2 O
3-Chloropropanoic acid
Exercise 19-7 Give the products of nitric acid oxidation of (a) pentanal; (b) 1,6-hexanediol; (c) 4-(hydroxymethyl)cyclohexanecarbaldehyde.
Organometallic reagents react with carbon dioxide to give carboxylic acids Organometallic reagents attack carbon dioxide (usually in the solid form known as “dry ice”) much as they would attack aldehydes or ketones. The product of this carbonation process is a carboxylate salt, which upon protonation by aqueous acid yields the carboxylic acid. Carbonation of Organometallics
R O MgX
ðOð B C B ðOð 2
THF
RLi
CO2
THF
š OðMgX D ROC M Oð RCOOLi
H, HOH XMgOH
H, HOH LiOH
RCOOH
RCOOH
884
Chapter 19
Carboxylic Acids
Recall that an organometallic reagent is usually made from the corresponding haloalkane: RX 1 Mg y RMgX. Hence, carbonation of the organometallic reagent allows the two-step transformation of RX into RCOOH, the carboxylic acid with one more carbon. For example, Cl A CH3CH2CHCH3
THF
Mg
MgCl A CH3CH2CHCH3
COOH A CH3CH2CHCH3 86%
1. CO2 2. H, H2O
2-Chlorobutane
2-Methylbutanoic acid
Nitriles hydrolyze to carboxylic acids Another method for converting a haloalkane into a carboxylic acid with an additional carbon atom is through the preparation and hydrolysis of a nitrile, RC q N. Recall (Section 6-2) that cyanide ion, 2: C q N :, is a good nucleophile and may be used to synthesize nitriles through SN2 reactions. Hydrolysis of the nitrile in hot acid or base furnishes the corresponding carboxylic acid (and ammonia or ammonium ion). Carboxylic Acids from Haloalkanes Through Nitriles
CN
RX
X
RC q N
1. HO 2. H, H2O NH3 or NH4
RCOOH
The mechanism of this reaction will be described in detail in Section 20-8. Carboxylic acid synthesis with the use of nitrile hydrolysis is preferable to Grignard carbonation when the substrate contains other groups that react with organometallic reagents, such as the hydroxy, carbonyl, and nitro functionalities. CH2CN
CH2COOH H2SO4, H2O, 15 min,
NO2
NO2 95%
(4-Nitrophenyl)ethanenitrile
(4-Nitrophenyl)acetic acid
Exercise 19-8
Working with the Concepts: Carboxylic Acids from Haloalkanes Suggest a method for accomplishing the conversion shown below. Several steps may be necessary. O
O ?
Br H HOOC H Strategy Before you start, inspect both structures carefully. Notice that the target molecule has an additional carbon atom that must be introduced. The last two methods in the section above provide possible options, but neither one can be applied directly to the starting material. The carbonyl function will interfere with any attempt to proceed using organometallic reagents and will also react with cyanide ion. Let us look at each of these options. We know from Section 17-11 that cyanide addition is reversible and can be avoided under basic conditions. Can we therefore displace bromide by cyanide and follow with hydrolysis under basic conditions? Look more closely at the substrate—Br is attached to a tertiary carbon. The necessary SN2 displacement will not occur; only elimination will take place (Section 7-8). Fortunately, organometallic reagents can be prepared from almost any type of halogenated compound, including tertiary haloalkanes, provided that interference from other functional groups can be avoided. This realization gives us the framework for a solution.
19-6 Methods for Introducing the Carboxy Functional Group
Chapter 19
885
Solution • Protect the aldehyde group in a way that will allow formation and use of a Grignard reagent: Form the cyclic acetal using 1,2-ethanediol and acid catalysis. • Following the sequence in the section above, prepare the Grignard reagent and add it to CO2. • Protonate the resulting carboxylate under acidic, aqueous conditions. The acetal hydrolyzes as well, deprotecting the aldehyde function: O Br
O
HOCH2CH2OH, H
Br
H
1. Mg, THF 2. CO2 3. H, H2O
O
O HOOC
H
Exercise 19-9
Try It Yourself Using chemical equations, show how you would convert each of the following halogenated compounds into a carboxylic acid with one additional carbon atom. If more than one method can be used successfully, show all. If not, give the reason behind your choice of approach. (a) 1-Chloropentane; (b) iodocyclopentane; (c) 4-bromobutanoic acid (Hint: See Sections 8-7 and 8-9); (d) chloroethene (Hint: See Section 13-9); (e) bromocyclopropane (Hint: See Problem 60 of Chapter 6).
Hydrolysis of the nitrile group in a cyanohydrin, prepared by addition of HCN to an aldehyde or a ketone (Section 17-11), is a general route to 2-hydroxycarboxylic acids, which possess antiseptic properties. OH OH A A CHO HO C O CN HO C O COOH NaCN, NaHSO3, H2O
HCl, H2O, 12 h
46% Benzaldehyde
2-Hydroxy-2-phenylethanenitrile (Mandelonitrile)
2-Hydroxy-2-phenylacetic acid (Mandelic acid)
Exercise 19-10 Suggest ways to effect the following conversions (more than one step will be required). CHO
HOCHCOOH
H3 C
COOH
(b)
(a) Br %
COOH ´
(c)
(Caution: Pay attention to the stereochemistry!) 0 OCH3
0 OCH3
In Summary Several reagents oxidize primary alcohols and aldehydes to carboxylic acids. A haloalkane is transformed into the carboxylic acid containing one additional carbon atom either by conversion into an organometallic reagent and carbonation or by displacement of halide by cyanide followed by nitrile hydrolysis.
H
886
Chapter 19
Carboxylic Acids
19-7 Substitution at the Carboxy Carbon: The Addition – Elimination Mechanism The carbonyl group in carboxylic acids shows reactivity similar to that in aldehydes and ketones: It is subject to attack by nucleophiles at carbon and electrophiles at oxygen. However, the presence of the carboxy OH group in the structure adds another dimension to the chemical function of carboxylic acids: Just as in alcohols, this OH may be converted into a leaving group (Section 9-2). As a result, after nucleophilic addition to the carbonyl carbon takes place, the leaving group may depart, resulting in a net substitution process and a new carbonyl compound. This section introduces this process and the general mechanisms by which it takes place. )O) B C D G R L Carboxylic acid derivative
The carbonyl carbon is attacked by nucleophiles As we have learned from our study of aldehydes and ketones, carbonyl carbons are electrophilic and can be attacked by nucleophiles. This type of reactivity is observed in the carboxylic acids and the carboxylic acid derivatives, substances with the general formula RCOL (L stands for leaving group). Carboxylic Acid Derivatives O O
O
O
O
RCX
RCOCR
RCOR
RCNR2
Acyl halide
Anhydride
Ester
Amide
In contrast to the addition products of aldehydes and ketones (Sections 17-5 through 17-7), the intermediate formed upon attack of a nucleophile on a carboxy carbon can decompose by eliminating a leaving group. The result is substitution of the nucleophile for the leaving group via a process called addition – elimination.
Substitution at the carboxy carbon occurs by addition – elimination The carbonyl carbon is trigonal and sp2-hybridized. As we have seen in the reactions of alkenes and benzenes (Chapters 12 and 15), the substitution mechanisms associated with saturated, tetrahedral, sp3-hybridized carbon atoms do not typically occur in unsaturated, trigonal planar systems. Their planarity renders backside, SN2-type displacement geometrically difficult, and sp2-hybridized carbon atoms form poor carbocations, disfavoring SN1. Substitution by the addition – elimination sequence combines two simple mechanistic processes. Addition of a nucleophile to the carbon of a carbonyl group is a reaction we know well. Elimination is just the mechanistic reverse of the addition process. Mechanism type 4(a) from Section 2-2 may be used to illustrate both, as follows. Forward direction — addition:
Xð
A P B
Reverse direction — elimination: X O A O Bð
X O A O Bð Xð
A PB
When the site of attack (atom A) already contains a potential leaving group (Y), then the two steps above can occur one after the other to give net substitution of X for Y: General Mechanistic Pattern for an Addition–Elimination Sequence
Xð
Y i
A PB
Addition of X
Elimination of Y
X i
A PB
Yð
i
i
Y A X O A O Bð A
As this general scheme shows, both steps in an addition – elimination sequence may be reversible. The scheme below shows addition – elimination as applied to a carboxylic acid derivative. The intermediate species formed after addition but before elimination (in contrast
19-7 Substitution at the Carboxy Carbon
Chapter 19
to both starting material and product) contains a tetrahedral carbon center. It is therefore called a tetrahedral intermediate. REACTION
Nucleophilic Substitution by Addition–Elimination )O) B C D G R L
Trigonal carbon
Nucleophile
O OH A RO C OL A Nu
HONu
Leaving group
O R
Tetrahedral carbon
O B C D G Nu R
Nu
R
L
Nu
R
Nu
Carboxylic acid derivative
L OH
O
O L
Trigonal carbon
Tetrahedral intermediate
Substitution product
Substitution by addition – elimination is the most important pathway for the formation of O O B B carboxylic acid derivatives as well as for their interconversion — that is, RCL y RCL9. The remainder of this section and those that follow will describe how such derivatives are prepared from carboxylic acids, and Chapter 20 will explore their properties and chemistry.
L
Color code reminder: Nucleophile — red Electrophile — blue Leaving group — green
Addition – elimination is catalyzed by both base and acid Addition – elimination reactions may be carried out under either basic or acidic conditions. We have seen how additions of nucleophiles to aldehydes and ketones (Sections 17-5 through 17-9; Table 17-4) may be catalyzed by either bases or acids. The same is true for additions of nucleophiles to carboxylic acid derivatives. Eliminations from the tetrahedral intermediate are similarly catalyzed: Recall that this process is mechanistically just the reverse of addition; therefore, the same catalytic effects are observed. Let us examine the roles of both base and acid in detail. All nucleophiles are Lewis bases. However, a nucleophile with a removable proton (Nu–H) may be deprotonated by a strong base (denoted as :B2) to give the negatively charged and thus more strongly nucleophilic : Nu2, which is the attacking species (Step 1). Addition – elimination occurs (Step 2), and the base may be regenerated (Step 3), making it a catalyst for the overall process. Typical nucleophiles in sequences such as this include water and alcohols, which upon deprotonation give hydroxide and alkoxide ions, respectively. Mechanism of Base-Catalyzed Addition – Elimination
MECHANISM
Step 1. Deprotonation of NuH Nu OH
ðB
ðNu
BH
Step 2. Addition – elimination )O) B DCG R L
O) )> A R O C OL A Nu
)Nu
)O) B DCG Nu R
Step 3. Regeneration of catalyst ðL
H OB
LH
ðB
(Alternatively, ðL may act as a base in step 1)
)L
887
888
Chapter 19
Carboxylic Acids
In cases where the nucleophile is already strong, base catalysis is unnecessary, and the entire mechanism may be described by Step 2, above. Addition – elimination reactions may also benefit from acid catalysis. The acid functions in two ways: First, it protonates the carbonyl oxygen (Step 1), activating the carbonyl group toward nucleophilic attack (Section 17-5). Second, protonation of L (Step 2) makes it a better leaving group (recall Section 6-7 and 9-2). Mechanism of Acid-Catalyzed Addition – Elimination Step 1. Protonation )O) B C D G R L
D
H
)O
B C D G R L
H
Step 2. Addition – elimination D
H
)O
B C D G R L
)NuH
H D O )> A RO C OLH A Nu
H D O )> A RO C OL A NuOH
D
H
)O
B C )LH D G Nu R
Step 3. Deprotonation; regeneration of catalytic proton D
H
)O
B C D G Nu R
)O) B C D G R Nu
H
Substitution in carboxylic acids is inhibited by a poor leaving group and the acidic proton We can apply the general addition–elimination process to the conversion of carboxylic acids into their derivatives. However, we must first overcome two problems. First, we know that the hydroxide ion is a poor leaving group (Section 6-7). Second, the carboxy proton is acidic, and most nucleophiles are bases (Section 6-8). Therefore, the desired nucleophilic attack (path a in the following equation) can encounter interference from an acid-base reaction (path b). Indeed, if a nucleophile is very basic, such as alkoxide, formation of the carboxylate ion will be essentially irreversible, and nucleophilic addition to the carbonyl carbon becomes very difficult. Competing Reactions of a Carboxylic Acid with a Nucleophile )>) O A > R O CO OH p A Nu Tetrahedral intermediate
Esterification RCOOH
ROH
H
RCOOR
H 2O
a
Nu)
a
Reversible nucleophilic addition (path a)
)O) b B b C D G > R O p OH
)Nu Reversible acid-base reaction (path b)
)O) B DC G > R O) p
NuH
Carboxylate ion
On the other hand, with less basic nucleophiles, especially under acidic conditions, the ready reversibility of carboxylate formation may permit nucleophilic addition to compete and ultimately lead to substitution through the addition – elimination mechanism. A typical example is the esterification of a carboxylic acid (Section 9-4), in which an alcohol and a carboxylic acid react to yield an ester and water. The nucleophile, an alcohol, is a weak base, and acid is present to protonate both the carbonyl oxygen, activating it toward nucleophilic addition, and the carboxy OH, converting it into a better leaving group, water.
19-8 Preparation of Acyl Halides and Anhydrides
Chapter 19
Acid Catalysis in Esterification Step 1. Protonation of the carboxy group activates it toward addition of an alcohol (R9OH).
H )O) B DC G > R OH p
)OH B DC G > R OH p
Starting carboxylic acid
Protonated carboxylic acid
Step 2. Tetrahedral intermediate forms by addition of R9OH and loss of H1. Step 3. Protonation of an OH oxygen makes it a better leaving group, allowing its departure. )> OH A > R O CO OH p A )OR > Tetrahedral
H
)> OH A R O CO OH p 2 A )OR > Protonated OH:
intermediate
H2O
better leaving group
EH )O B C D G > R OR p
)O) B C D G > R OR p
Protonated ester
Final ester product
The sections that follow will examine this and other carboxy substitutions in detail. Exercise 19-11 Write out Step 2 of the above sequence in full.
In Summary Nucleophilic attack on the carbonyl group of carboxylic acid derivatives is a key step in substitution by addition – elimination. Either acid or base catalysis may be observed. For carboxylic acids, the process is complicated by the poor leaving group (hydroxide) and competitive deprotonation of the acid by the nucleophile, acting as a base. With less basic nucleophiles, addition can occur.
19-8 Carboxylic Acid Derivatives: Acyl Halides and Anhydrides With this section, we begin a survey of the preparation of carboxylic acid derivatives. Replacement of the hydroxy group in RCOOH by halide gives rise to acyl halides; substitution by alkanoate (RCOO2) furnishes carboxylic anhydrides. Both processes first require transformation of the hydroxy functionality into a better leaving group.
Acyl halides are formed by using inorganic derivatives of carboxylic acids O B RCX
The conversion of carboxylic acids into acyl halides employs the same reagents used in the synthesis of haloalkanes from alcohols (Section 9-4): SOCl2 and PBr3. The problem to be solved in both cases is identical — changing a poor leaving group (OH) into a good one.
Acyl halide
Acyl Halide Synthesis O B CH3CH2CH2COH
O B ClSCl, reflux O P S P O HCl
O B CH3CH2CH2CCl 85%
Butanoic acid
REACTION O Cl
Butanoyl chloride
O B CBr
O B COH 3
PBr3 H3PO3
3 90%
889
Carboxylic Acids
(These reactions fail with formic acid, HCOOH, because formyl chloride, HCOCl, and formyl bromide, HCOBr, are unstable. See Exercise 15-33.) These transformations begin with the conversion of the carboxylic acid into an inorganic derivative in which the substituent on the carbonyl carbon is a good leaving group: O B C E H EH R O
O B S E H Cl Cl
ðOð is a good B S leaving group šE H š ðO š Clð š š
OH is a poor ðš š leaving group
O B C E H EH R O
O O B B C S E H E H R O Cl
HCl
Br
Br A P E H
Br
HBr
O Br B A C P E H E H R O Br
Brð ðš is a good A P leaving group šE H š ðO š Brð š š
OH is a poor ðš š leaving group
Exercise 19-12
Working with the Concepts: Acyl Chlorosulfite Formation Propose a mechanism for the reaction of SOCl2 with a carboxylic acid to give the inorganic derivative (an acyl chlorosulfite) shown above. Strategy It is tempting to employ the same steps for this problem as we used in the conversion of an alcohol into the corresponding alkyl chlorosulfite, ROSOCl, upon reaction with SOCl2 (Section 9-4), namely, displacement of chloride from sulfur by the hydroxy group of the alcohol:
> >
RCH2OH
O B ClSCl
O B RCH2OSCl
H
Cl
However, direct application of this mechanism to a carboxylic acid is not likely to be correct. We saw in Section 19-5 that protonation of a carboxylic acid occurs at the carbonyl oxygen atom, not on the oxygen of the hydroxy group. The reason is straightforward: Protonation of the carbonyl oxygen gives a resonance-stabilized intermediate, whereas reaction at OH does not. The same can be said of reaction of a carboxylic acid with any electrophile, such as the sulfur in SOCl2 or the phosphorus in PBr3. Solution • It is more likely that the transformation in question proceeds as follows:
>
Cl
O B ES H > O Cl B EC H> O OH R
>
O B ES H > Cl O A EC N O R
H
>
O B C E H> OH R
Cl i S PO f Cl
>
> >
Chapter 19
> >
890
>
• This reaction pathway converts the carbonyl oxygen, rather than the hydroxy, into the leaving group. • To our knowledge, this specific mechanism has not been tested experimentally. We present it as an example of a mechanistic hypothesis that is proposed for a set of reacting species, based on experimental observations of other similar species.
19-8 Preparation of Acyl Halides and Anhydrides
Chapter 19
Exercise 19-13
Try It Yourself Propose a mechanism for the reaction of PBr3 with a carboxylic acid to give the inorganic derivative (an acyl dibromophosphite) shown above Exercise 19-12.
Acid-catalyzed addition – elimination follows. Mechanism of Acyl Chloride Formation with Thionyl Chloride
MECHANISM
Step 1. Addition )O) B )O) C D G B > >) OSCl R > >
D
> H )Cl) >
H D )> O )O) B A >) RO CO> OSCl A > > )Cl) > Tetrahedral
H
)O B )O) >) )Cl C D G B > >) R OSCl > > >
intermediate
Step 2. Elimination H D > )O >) O B A >OS ROCOO A > G >) Cl )Cl) > >
)O) B C D G >) R Cl >
>PSP O > O > >
> ) H )Cl >
The mechanism of acyl bromide formation using phosphorus tribromide (PBr3) is similar (see Section 9-4).
Acids combine with acyl halides to produce anhydrides The electronegative power of the halogens in acyl halides activates the carbonyl function to attack by other, even weak, nucleophiles (Chapter 20). For example, treatment of acyl halides with carboxylic acids results in carboxylic anhydrides. O B CH3CH2CH2COH Butanoic acid
O B ClCCH2CH2CH3 Butanoyl chloride
O O B B CH3CH2CH2COCCH2CH2CH3 85%
, 8 h HCl
O
O
O
Butanoic anhydride
As the name indicates, carboxylic anhydrides are formally derived from the corresponding acids by loss of water. Although carboxylic acid dehydration is not a general method for anhydride synthesis, cyclic examples may be prepared by heating dicarboxylic acids. A condition for the success of this transformation is that the ring closure lead to a five- or six-membered ring product.
O O B B RCOCR Carboxylic anhydride
891
892
Chapter 19
Carboxylic Acids
Cyclic Anhydride Formation
H2C A H2C
D
COOH 300°
G
H2O
H2C
O
O B C O
O
H2C
CB O 95%
COOH
O
Butanedioic anhydride (Succinic anhydride)
Butanedioic acid (Succinic acid)
Because the halogen in the acyl halide and the RCO2 substituent in the anhydride are good leaving groups and because they activate the adjacent carbonyl function, these carboxylic acid derivatives are useful synthetic intermediates for the preparation of other compounds, a topic to be discussed in Sections 20-2 and 20-3. Exercise 19-14 Suggest two preparations, starting from carboxylic acids or their derivatives, for each of the following compounds. O O B B (a) CH3COCCH2CH3
H3C O A B (b) CH3CHCCl
Exercise 19-15 Propose a mechanism for the thermal formation of butanedioic anhydride from the dioic acid.
In Summary The hydroxy group in COOH can be replaced by halogen by using the same reagents used to convert alcohols into haloalkanes—SOCl2 and PBr3. The resulting acyl halides are sufficiently reactive to be attacked by carboxylic acids to generate carboxylic anhydrides. Cyclic anhydrides may be made from dicarboxylic acids by thermal dehydration.
19-9 Carboxylic Acid Derivatives: Esters
O B Esters have the general formula RCOR9. Their widespread occurrence in nature and many practical uses make them perhaps the most important of the carboxylic acid derivatives. This section describes the mineral acid–catalyzed reaction of carboxylic acids with alcohols that forms esters.
Carboxylic acids react with alcohols to form esters When a carboxylic acid and an alcohol are mixed together, no reaction takes place. However, upon addition of catalytic amounts of a mineral acid, such as sulfuric acid or HCl, the two components combine in an equilibrium process to give an ester and water (Section 9-4). This method of ester formation was first described by the legendary German chemist Emil Fischer and his coworker Arthur Speier in 1895. It is therefore called Fischer-Speier or (more commonly) Fischer esterification. REACTION
Acid-Catalyzed (Fischer) Esterification O B RCOH Carboxylic acid
ROH
O B RCOR
Alcohol
Ester
H
H2O
19-9 Preparation of Esters
Chapter 19
Esterification is not very exothermic: DH8 is usually close to 0. The associated entropy change is also small; therefore DG8 < 0 and K < 1. How can the equilibrium be shifted toward the ester product? One way is to use an excess of either of the two starting materials; another is to remove the ester or the water product from the reaction mixture. In practice, esterifications are most often achieved by using the alcohol as a solvent. O B CH3COH
Acetic acid
H2SO4,
CH3OH
H2O
O B CH3COCH3 85%
O
O
Methyl acetate
Solvent
Exercise 19-16 Show the products of the acid-catalyzed reaction of each of the following pairs of compounds. (a) Methanol 1 pentanoic acid; (b) formic acid 1 1-pentanol; (c) cyclohexanol 1 benzoic acid; (d) 2-bromoacetic acid 1 3-methyl-2-butanol.
The opposite of esterification is ester hydrolysis. This reaction is carried out under the same conditions as esterification, but, to shift the equilibrium, an excess of water is used in a water-miscible solvent. CH3 A CH3CH2CH2CH2CCOOCH2CH3 A CH3
CH3 A CH3CH2CH2CH2CCOOH A CH3 85%
H2SO4, HOH, acetone,
CH3CH2OH
2,2-Dimethylhexanoic acid
Ethyl 2,2-dimethylhexanoate
Exercise 19-17 Give the products of acid-catalyzed hydrolysis of each of the following esters. O
(a) CH3(CH2)3C q CCH2CH2COOCH(CH3)2
O
(b)
O
(c) CH3CH2CHCH2OOC
A
CH3
Esterification proceeds through acid-catalyzed addition – elimination The presence of the acid catalyst features prominently in the mechanism of ester formation: It causes the carbonyl function to undergo nucleophilic attack by the alcohol (methanol in this example, step 2) and the hydroxy group to leave as water (step 3). We previewed this mechanism earlier (Section 19-7). Here we present it with full details. Mechanism of Acid-Catalyzed Esterification
MECHANISM
Step 1. Protonation of the carboxy group )O) B C D G > O OH R >
D
> O
H H
H
B C D G > O OH R >
H D )> O A C D G > OOH R >
H D > )O A C D M > O OH R
893
894
Carboxylic Acids
Chapter 19
MATION AN I
ANIMATED MECHANISM: Esterification
Step 2. Attack by methanol D
H D > )O A O R O C O> A >G H O) D G H3C H
H
> O
B DCG > O OH R >
OH CH3> >
H D > )O A R O C O> O A >G H O) D> H3C
H H
Tetrahedral intermediate Relay point: Can go back to starting material Can go forward to product
Step 3. Elimination of water H D O )> A > ROCOO A >G H O) D> H3C D
H
> O
R
B C D
GO > O CH3 >
H H
H D )> O A C D > O CH3 R GO >
H D > )O A DH ROCOO A >G H O) D> H3C
H2O H2O
H D > )O A C D M > R O O CH3
H H
ðOð B C D G > O CH3 R O >
Initially, protonation of the carbonyl oxygen gives a delocalized carbocation (step 1). Now the carbonyl carbon is susceptible to nucleophilic attack by methanol. Proton loss from the initial adduct furnishes the tetrahedral intermediate (step 2). This species is a crucial relay point, because it can react in either of two ways in the presence of the mineral acid catalyst. First, it can lose methanol by the reverse of steps 1 and 2. This process, beginning with protonation of the methoxy oxygen, leads back to the carboxylic acid. The second possibility, however, is protonation at either hydroxy oxygen, leading to elimination of water and to the ester (step 3). All the steps are reversible; therefore, either addition of excess alcohol or removal of water favors esterification by shifting the equilibria in steps 2 and 3, respectively. Ester hydrolysis proceeds by the reverse of the sequence and is favored by aqueous conditions. Exercise 19-18 Give the mechanism of esterification of a general carboxylic acid RCOOH with methanol in which the alcohol oxygen is labeled with the 18O isotope (CH318OH). Does the labeled oxygen appear in the ester or in the water product?
Hydroxy acids may undergo intramolecular esterification to lactones When hydroxy carboxylic acids are treated with catalytic amounts of mineral acid, cyclic esters — or lactones — may form. This process is called intramolecular esterification and is favorable for formation of five- and six-membered rings. Unlike the intermolecular esterification, the equilibrium is now favored by entropy and therefore particularly facile: Water is expelled from the starting material.
19-9 Preparation of Esters
Formation of a Lactone
Chapter 19
895
O
O HOCH2CH2CH2CH2COOH
O
O
H2SO4, H2O H2O
90%
10%
A lactone
Exercise 19-19
Working with the Concepts: A Multistep Mechanism Explain the following result mechanistically. H
HO
O H, H2O
O
O
H
H2O
O
CH2COOH Strategy The starting compound possesses a five-membered ring hemiacetal group, which, as we saw in Section 17-7, can be a reasonably stable structure. Hemiacetals equilibrate with their open-chain hydroxycarbonyl isomers in the presence of aqueous acid or base: šð HO
šð HO
H
H
šð O CH2COOH
ðOð
H ðO OH
H
Hš OH CH2COOH
CH2COOH
OH
OH
H
H
OH
OH2
O
H2O
H
OH
Tetrahedral intermediate
Exercise 19-20
Try It Yourself If the starting compound in Exercise 19-19 were labeled with the atom, where would the 18O label end up in the product?
18
O isotope at its ring oxygen
In Summary Carboxylic acids react with alcohols to form esters, as long as a mineral acid catalyst is present. This reaction is only slightly exothermic, and its equilibrium may be shifted in either direction by the choice of reaction conditions. The reverse of ester formation is ester hydrolysis. The mechanism of esterification is acid-catalyzed addition of alcohol to the carbonyl group followed by acid-catalyzed dehydration. Intramolecular ester formation results in lactones, favored only when five- or six-membered rings are produced.
O
>
OH
> >
>
> >
O
H D O
>
> >
> >
> >
> > >
> >
>
H
O
H
O
OH
H
> > >
> > > >
O
H
>
> >
O
OH O
>
H
OH
H
O
> >
O
> > > > >
> >
OH
> >
O
E
> >
The structure that results contains three functional groups: an aldehyde carbonyl, an alcohol hydroxy, and a carboxylic acid carboxy. Solution • Intramolecular reaction of the hydroxy group with the carboxylic acid function gives the cyclic ester (lactone) product shown above. • We complete the solution to the exercise by writing out the mechanism, step by step.
OH
H
product
896
Chapter 19
Carboxylic Acids
19-10 Carboxylic Acid Derivatives: Amides O RCNR2 Carboxylic amide
As we saw earlier (Section 17-9), amines also are capable of attacking the carbonyl function. When the carbonyl is that of a carboxylic acid, the product is a carboxylic amide,* the last major class of carboxylic acid derivatives. The mechanism is again addition – elimination but is complicated by acid-base chemistry.
Amines react with carboxylic acids as bases and as nucleophiles Nitrogen lies to the left of oxygen in the periodic table. Therefore, amines (Chapter 21) are better bases as well as better nucleophiles than alcohols (Section 6-8). To synthesize carboxylic amides, therefore, we must address the problem discussed in Section 19-7: interference from competing acid-base reaction. Indeed, exposure of a carboxylic acid to an amine initially forms an ammonium carboxylate salt, in which the negatively charged carboxylate is very resistant to nucleophilic attack. Ammonium Salts from Carboxylic Acids ðOð B š RCO OH
ðNH3
ðOð B š RCO ð H NH3
Ammonia
An ammonium carboxylate salt
Notice that salt formation, though very favorable, is nonetheless reversible. Upon heating, a slower but thermodynamically favored reaction between the acid and the amine can take place. The acid and the amine are removed from the equilibrium, and eventually salt formation is completely reversed. In this second mode of reaction, the nitrogen acts as a nucleophile and attacks the carbonyl carbon. Completion of an addition – elimination sequence leads to the amide. Although it is convenient, this method suffers from the high temperatures required to reverse ammonium carboxylate formation. Therefore, better procedures rely on the use of activated carboxylic acid derivatives, such as acyl chlorides (Chapter 20). REACTION
O B CH3CH2CH2COH
Formation of an Amide from an Amine and a Carboxylic Acid
(CH3)2NH
155C H2O
O B CH3CH2CH2CN(CH3)2 84%
O N
N, N-Dimethylbutanamide
MECHANISM
R
)O) B DCG
> OH >
)NH3
Mechanism of Amide Formation O) )> A OH RO C O > A > NH 3
Stepwise proton transfer: deprotonation at N, reprotonation at O
Tetrahedral intermediates
O) )> A RO C O > OH2 A NH2 >
H2> O)
)O) B DCG NH2 R > An amide
In the mechanism shown above, the elimination step is shown to proceed from a zwitterionic tetrahedral intermediate. However, the prevalence of such a species is likely to be highly pH dependent. Thus, an alternative possibility is to eliminate water from a neutral species, *Do not confuse the names of carboxylic amides with those of the alkali salts of amines, also called amides (e.g., lithium amide, LiNH2).
19-11 Reduction of Carboxylic Acids by LiAlH4
Chapter 19
as shown in the margin. Amide formation is reversible. Thus, treatment of amides with hot aqueous acid or base regenerates the component carboxylic acids and amines.
Alternative Mechanism of Amide Formation OH )> A OH RO C O > A > NH2 >
Dicarboxylic acids react with amines to give imides Dicarboxylic acids may react twice with the amine nitrogen of ammonia or of a primary amine. This sequence gives rise to imides, the nitrogen analogs of cyclic anhydrides (cf. p. 892). O B O C H2C CH2COOH CH2COO NH4 NH 290C )NH O, C H CH2COOH CH2COONH4 2H 2 NH CB N Butanedioic acid O 83% O Butanimide
H
OH )> A OH2 RO C O > A NH2 >
3
2
3
H2O
(Succinimide)
EH )O B C D G NH2 R >
Recall the use of N-halobutanimides in halogenations (Section 14-2).
Amino acids cyclize to lactams In analogy to hydroxycarboxylic acids, some amino acids undergo cyclization to the corresponding cyclic amides, called lactams (Section 20-6). O O O O B B )NH šCH2 CH2 CH2 COH HN H2NCH2 CH2 CH2 CO H2O
H
H
86% A lactam
The penicillin class of antibiotics derives its biological activity from the presence of a lactam function (see Chemical Highlight 20-2). Lactams are the nitrogen analogs of lactones (Section 19-9). Exercise 19-21 Formulate a detailed mechanism for the formation of butanimide from butanedioic acid and ammonia.
In Summary Amines react with carboxylic acids to form amides by an addition–elimination process that begins with nucleophilic attack by the amine on the carboxy carbon. Amide formation is complicated by reversible deprotonation of the carboxylic acid by the basic amine to give an ammonium salt.
19-11 Reduction of Carboxylic Acids by Lithium Aluminum Hydride Lithium aluminum hydride is capable of reducing carboxylic acids all the way to the corresponding primary alcohols, which are obtained upon aqueous acidic work-up. Reduction of a Carboxylic Acid RCOOH
1. LiAlH4, THF 2. H, H2O
RCH2OH
897
H
N
)O) B DCG NH2 R >
898
Chapter 19
Carboxylic Acids
Example: 1. LiAlH4, THF 2. H, H2O
COOH Double Hydride Additions to Carboxylate Ion )O) B C D G
> O) Li > H3Al O H
R
CH2OH 65%
Although the exact mechanism of this transformation is not completely understood, it is clear that the hydride reagent first acts as a base, forming the lithium salt of the acid and hydrogen gas. Carboxylate salts are generally resistant to attack by nucleophiles. Yet, despite the negative charge, lithium aluminum hydride is so reactive that it is capable of donating two hydrides to the carbonyl function of the carboxylate, possibly assisted by aluminum (margin; see also Section 8-6). The product of this sequence, a simple alkoxide, gives the alcohol after protonation. Exercise 19-22
H A Al E H
O H )> A O) Li RO C O > A > H
Propose synthetic schemes that produce compound B from compound A. (a) CH3CH2CH2CN
CH3CH2CH2CH2OH
A
B
(b)
CH2COOH
CH2CD2OH
A
B
In Summary The nucleophilic reactivity of lithium aluminum hydride is sufficiently great to effect the reduction of carboxylates to primary alcohols.
HO AlH3 OAlH2
> O) Li GC D >
/∑
R
H H
19-12 Bromination Next to the Carboxy Group: The Hell-VolhardZelinsky Reaction Like aldehydes and ketones, alkanoic acids can be monobrominated at the a-carbon by exposure to Br2. The addition of a trace amount of PBr3 is necessary to get the reaction started. Because the highly corrosive PBr3 is difficult to handle, it is often generated in the reaction flask (in situ). This is achieved by the addition of a little elemental (red) phosphorus to the mixture of starting materials; it is converted into PBr3 instantaneously by the bromine present.
REACTION
A Hell-Volhard-Zelinsky* Reaction
CH3CH2CH2CH2COOH
Br–Br, trace P
Br A CH3CH2CH2CHCOOH 80%
HBr
2-Bromopentanoic acid
The acyl bromide, formed by the reaction of PBr3 with the carboxylic acid (Section 19-8), is subject to rapid acid-catalyzed enolization. The enol is brominated to give the 2-bromoacyl bromide. This derivative then undergoes an exchange reaction with unreacted acid to furnish the product bromoacid and another molecule of acyl bromide, which reenters the reaction cycle.
MECHANISM
Mechanism of the Hell-Volhard-Zelinsky Reaction Step 1. Acyl bromide formation (see Exercise 19-12 and Section 19-8) O B 3 RCH2COH
PBr3
O B 3 RCH2CBr
H3PO3
Acyl bromide
*Professor Carl M. Hell (1849–1926), University of Stuttgart, Germany; Professor Jacob Volhard (1834 –1910), University of Halle, Germany; Professor Nicolai D. Zelinsky (1861–1953), University of Moscow.
19-13 Biological Activity of Carboxylic Acids
O B RCH2CBr
H
B A O
Step 2. Enolization
RCH
OH
C
Br
Enol
B A O
Step 3. Bromination
RCH
OH
C
Br–Br
Br
O B RCHCBr A Br
HBr
Step 4. Exchange O B RCHCBr A Br
O B RCH2COH
O B RCHCOH A Br
O B RCH2CBr Reenters step 2
Exercise 19-23 Formulate detailed mechanisms for steps 2 and 3 of the Hell-Volhard-Zelinsky reaction. (Hints: Review Sections 18-2 for step 2 and 18-3 for step 3.)
The bromocarboxylic acids formed in the Hell-Volhard-Zelinsky reaction can be converted into other 2-substituted derivatives. For example, treatment with aqueous base gives 2-hydroxy acids, whereas amines yield a-amino acids (Chapter 26). An example of amino acid synthesis is the preparation of racemic 2-aminohexanoic acid (norleucine, a naturally occurring but rare amino acid) shown here.
CH3(CH2)4COOH Hexanoic acid
Br2, trace PBr3, 70°–100°C, 4 h
Br A CH3(CH2)3CHCOOH 86%
NH3, H2O, 50°C, 30 h
2-Bromohexanoic acid
NH2 A CH3(CH2)3CHCOOH 64% 2-Aminohexanoic acid (Norleucine)
In Summary With trace amounts of phosphorus (or phosphorus tribromide), carboxylic acids are brominated at C2 (the Hell-Volhard-Zelinsky reaction). The transformation proceeds through 2-bromoacyl bromide intermediates.
19-13 Biological Activity of Carboxylic Acids Considering the variety of reactions that carboxylic acids can undergo, it is no wonder that they are very important, not only as synthetic intermediates in the laboratory, but also in biological systems. This section will provide a glimpse of the enormous structural and functional diversity of natural carboxylic acids. A discussion of amino acids will be deferred to Chapter 26. As Table 19-1 indicates, even the simplest carboxylic acids are abundant in nature. Formic acid is present not only in ants, where it functions as an alarm pheromone, but also in plants. For example, one reason why human skin hurts after it touches the stinging nettle is that formic acid is deposited in the wounds. Acetic acid is formed through the enzymatic oxidation of ethanol produced by fermentation. Vinegar is the term given to the dilute (ca. 4 – 12%) aqueous solution thus generated in ciders, wines, and malt extracts. Louis Pasteur in 1864 established the involvement of bacteria in the oxidation stage of this ancient process.
Chapter 19
899
900
Carboxylic Acids
Chapter 19
CHE M I C AL H I G H L I G H T 1 9 - 1 Soaps from Long-Chain Carboxylates The sodium and potassium salts of long-chain carboxylic acids have the property of aggregating as spherical clusters called micelles in aqueous solution. In such aggregates, the hydrophobic alkyl chains (Section 8-2) of the acids come together in space because of their attraction to
one another by London forces (Section 2-6) and their tendency to avoid exposure to polar water. As shown above, the polar, water-solvated carboxylate “head groups” form a spherical shell around the hydrocarbonlike center. Polar “head”
Hydrophobic “tail”
O C
R
Water
Na⫹
O
O⫺ Na⫹
O⫺
Na⫹
Na⫹
O C
C
O O⫺
C
O⫺
O C
O⫺
Na⫹
Na⫹
O C
O⫺
O C
Na⫹
O⫺ O C
Na⫹
O⫺ O C
O⫺ O C
Na⫹
O⫺
O C
Na⫹
O⫺
O C
O⫺
Na⫹
O C
O⫺
O C
Na⫹
O C
O
Na⫹
O
O
C
C
O⫺
C O⫺
Na⫹
O⫺
Na⫹
O⫺
O⫺
Na⫹
Na⫹
Fatty acids are derived from coupling of acetic acid Acetic acid exhibits diverse biological activities, ranging from a defense pheromone in some ants and scorpions to the primary building block for the biosynthesis of more naturally occurring organic compounds than any other single precursor substance. For example, 3-methyl-3-butenyl pyrophosphate, the crucial precursor in the buildup of the terpenes (Section 14-10), is made by the enzymatic conversion of three molecules of CH3COOH into an intermediate called mevalonic acid. Further reactions degrade the system to the five-carbon (isoprene) unit of the product.
3 CH3 O COOH
Enzymes
CH2 O COOH A CH3O COH A CH2O CH2OH Mevalonic acid
Enzymes
CH3O C
CH2 J i
O O B B CH2O CH2OO P OO O P OOH A A OH OH
3-Methyl-3-butenyl pyrophosphate
19-13 Biological Activity of Carboxylic Acids
H HO ^
H % ^
H
A conceptually more straightforward mode of multiple coupling is found in the biosynthesis of fatty acids. This class of compounds derives its name from its source, the natural fats, which are esters of long-chain carboxylic acids (see Section 20-4). Hydrolysis or saponification (so called because the corresponding salts form soaps; sapo, Latin, soap—see Chemical Highlight 19-1) yields the corresponding fatty acids. The most important of them are between 12 and 22 carbons long and may contain cis carbon – carbon double bonds. Fatty Acids (CH2)7COOH D G CP C G D H H
CH3(CH2)7
Hexadecanoic acid (Palmitic acid)
cis-9-Octadecenoic acid (Oleic acid)
In accord with their biosynthetic origin, fatty acids consist mostly of even-numbered carbon chains. A very elegant experiment demonstrated that linear coupling occurred in a highly
^
COOH CH3
^
H3C %
^
`
H HO H3C %
False-color scanning electron micrographs of the collar of a pure cotton shirt before washing (left) and after washing (right). Before washing, the threads and subthreads are covered with grime and skin flakes. (Magnified 70 times.)
CH3(CH2)14COOH
901
to the need for detergents of very high solubility and cleansing power but reduced foaming (sudsing). Novel sulfonates containing ether linkages have been developed to meet this demand. Certain steroid bile acids such as cholic acid (Section 4-7) also have surfactant or detergentlike properties and are found in the bile duct. These substances are released into the upper intestinal tract, where they emulsify water-insoluble fats through the formation of micelles. Hydrolytic enzymes can then digest the dispersed fat molecules.
%
Because these carboxylate salts also create films on aqueous surfaces, they act as soaps. The polar groups stick into the water while the alkyl chains assemble into a hydrophobic layer. This construction reduces the surface tension of the water, permitting it to permeate cloth and other fabrics and giving rise to the foaming typical of soaps. Cleansing is accomplished by dissolving ordinarily water-insoluble materials (oils, fats) in the hydrocarbon interior of the micelles. One problem with carboxylate soaps is that they form curdlike precipitates with the ions present in hard water (e.g., Mg21 and Ca21). Detergents based on alkanesulfonates, RSO32 Na1, and alkyl sulfates, ROSO32 Na1, avoid this drawback but have caused pollution problems in lakes and streams because branches in their alkyl chains made early versions nonbiodegradable. The microorganisms associated with normal sewage treatment procedures are capable of breaking down only straight-chain systems. The recent widespread introduction of energy- and water-saving front-loading washing machines has given rise
Chapter 19
H OH
Cholic acid
902
Carboxylic Acids
Chapter 19
NH2
Mercapto group O
O HS
CH2
CH2
H N
C
CH2
CH2
H N
H
C
C
O−
CH3 R
C
HO
CH2
O
P
O− O
P
O
CH3
N CH2
O
N
O
H
O
N
H
H
−O
N
H O
OH
P
O−
O
2-Aminoethanethiol part
Pantothenic acid part
Adenosine diphosphate (ADP) part
Figure 19-6 Structure of coenzyme A. For this discussion, the important part is the mercapto function. For convenience, the molecule may be abbreviated HSCoA.
REACTION
CH314COOH
regular fashion. In it, singly labeled radioactive (14C) acetic acid was fed to several organisms. The resulting fatty acids were labeled only at every other carbon atom. Organism
CH314CH2CH214CH2CH214CH2CH214CH2CH214CH2CH214CH2CH214CH2CH214COOH Labeled hexadecanoic (palmitic) acid
The mechanism of chain formation is very complex, but the following scheme provides a general idea of the process. A key player is the mercapto group of an important biological relay compound called coenzyme A (abbreviated HSCoA; Figure 19-6). This function binds acetic acid in the form of a thiol ester called acetyl CoA. Carboxylation transforms some of this thiol ester into malonyl CoA. The two acyl groups are then transferred to two molecules of acyl carrier protein. Coupling occurs with loss of CO2 to furnish a 3-oxobutanoic thiol ester. Coupling of Acetic Acid Units
MECHANISM Step 1. Thiol ester formation
CH3COOH Acetic acid
HSCoA Coenzyme A
O B CH3CSCoA
HOH
Acetyl coenzyme A
Step 2. Carboxylation O B CH3CSCoA
CO2
Acetyl CoA carboxylase
O O B B HOCCH2CSCoA Malonyl CoA
Step 3. Acetyl and malonyl group transfers O B CH3CSCoA
HS O protein
O B CH3CS O protein
HSCoA
Acyl carrier protein
O O B B HOCCH2CSCoA
HS O protein Acyl carrier protein
O O B B HOCCH2CSO protein
HSCoA
19-13 Biological Activity of Carboxylic Acids
Chapter 19
903
C H E MI C AL H I G H L I G H T 1 9 - 2 Trans Fatty Acids and Your Health More than 90% of the double bonds in naturally occurring unsaturated fatty acids possess the cis configuration, contributing to the reduced melting temperatures of vegetable oils compared with saturated fats (Section 11-3). Exposure of vegetable oil to catalytic hydrogenation conditions produces solid margarine. However, this process does not hydrogenate all of the double bonds: As we saw in Exercise 12-2, a significant fraction of cis double bonds are merely isomerized to trans by the catalyst and remain in the final solid product. For example, the fat in one U.S. brand of synthetic hard (stick) margarine consists of about 18% saturated fatty acids (SFAs) and 23% trans fatty acids (TFAs). The fat in soft (tub) margarines, whose exposure to catalytic hydrogenation conditions is less, has about the same level of SFAs but far less (5 – 10%) TFAs. For comparison, natural butter is 50 – 60% SFAs but only 3 – 5% TFAs. What, if any, are the health consequences of TFAs in the human diet? It has long been suspected that TFAs are not metabolized by the body in the same way as their cis counterparts; in the 1960s and 1970s, this suspicion was confirmed by studies that indicated that TFAs in foods greatly affect lipid metabolism. Perhaps the most alarming finding was that TFAs accumulate in cell membranes and increase the levels of low-density lipoproteins (LDLs, popularly if imprecisely known as “bad cholesterol”) in the bloodstream while reducing high-density lipoprotein levels (HDLs, the so-called good cholesterol, see Chemical Highlight 4-2). Studies beginning in the 1990s implicated dietary TFAs in increased risk of breast cancer and heart disease. It is now generally believed that the health effects of TFAs are far more adverse than even those of SFAs. While TFAs in general make up a minor component of a healthy diet, the use of partially hydrogenated (and, therefore, TFA-rich) oils in commercial baked goods, and of shortenings (18% TFA) for frying in restaurants, renders the TFA content of cakes, cookies, and pastries, as well as French fries and other fried
fast foods, alarmingly high. In 2005 the American Heart Association (AHA) recommended that fat consumption make up no more than 30% of caloric intake, and that the total of trans and saturated fats be limited to less than 10%. In 2006 the AHA amended that recommendation to limit TFAs to no more than 1% of total calories. A typical “fast food” meal could exceed this amount by many multiples. As a result, the U.S. Food and Drug Administration required all food products to list TFA content beginning in 2006. Since that time, numerous communities across the United States, including major cites such as New York and Philadelphia, and the entire state of California, have enacted bans on the use of partially hydrogenated, high-TFA oils for restaurant cooking. Public pressure became so intense that by the end of 2008 most major fast-food chains had switched to low- or no-TFA oils for all purposes.
The new French fries: now without trans fatty acids.
Step 4. Coupling O O B B HOCCH2CSO protein
O B CH3CSO protein CO2
O O B B CH3CCH2CS O protein A 3-oxobutanoic thiol ester
Reduction of the ketone function to a methylene group follows. The resulting butanoic thiol ester repeatedly undergoes a similar sequence of reactions elongating the chain, always by two carbons. The eventual product is a long-chain acyl group, which is removed from the protein by hydrolysis. O B CH3CH2 O(CH2CH2)O n CH2C OSO protein
H2O
O B CH3CH2 O(CH2CH2)O n CH2C OOH Fatty acid
HSO protein
904
Chapter 19
Carboxylic Acids
Arachidonic acid is a biologically important unsaturated fatty acid Naturally occurring unsaturated fatty acids are capable of undergoing further transformations leading to a variety of unusual structures. An example is arachidonic acid, which is the biological precursor to a multitude of important chemicals in the human body, such as prostaglandins (Chemical Highlight 11-1), thromboxanes, prostacyclins, and leukotrienes.
OH %
OH %
COOH
Arachidonic acid
COOH
Leukotriene B4 (Potent chemotactic factor; e.g., causes cell migrations)
O HO O
COOH
;
O
HO
OH
O
O
Prostaglandin F2 (Induces labor, abortion, menstruation)
COONa
O COOH
O O Human red blood cells, erythrocytes, and activated platelets, thrombocytes (blue), trapped in a fibrin (yellow) blood clot.
O B CH3CO
COOH
Aspirin
OH
HO
OH
Thromboxane A2
Prostacyclin I2, sodium salt
(Contracts smooth muscles; aggregates blood platelets)
(The most potent natural inhibitor of platelet aggregation; vasodilator, used in heart-bypass operations and in kidney transplant patients)
Each of these compounds possesses powerful biological properties. For example, some of the prostaglandins are responsible for the tissue inflammation associated with rheumatoid arthritis. The salicylate derivative aspirin (Chemical Highlight 22-2) is capable of combating such symptoms by inhibiting the conversion of arachidonic acid into prostaglandins. In the early 1990s, asthma, a condition that afflicts some 15 million Americans, was recognized to be a chronic inflammatory disease. Recent research has shown that the leukotrienes mediate airway inflammation, and the thromboxanes contribute to
19-13 Biological Activity of Carboxylic Acids
905
Chapter 19
C H E MI C AL H I G H L I G H T 1 9 - 3 Plastics, Fibers, and Energy from Biomass-Derived Hydroxyesters five-carbon acid, stiffer with more four-carbon acid. PHBV is stable at temperatures up to 1408C, but it is fully degraded to innocuous small organic molecules in 6 months when exposed to microorganisms in soil, compost, or water. One application of PHBV is in controlled drug release: The polymer encapsulates a pharmaceutical, which is released only after the coating has degraded sufficiently by ester hydrolysis to form the two original hydroxyacids, which are natural products of human metabolism and quite harmless. Of dramatic longer-term interest is the 2008 announcement of the development of a genetically engineered switchgrass that produces a significant amount of polyhydroxyalkanoic acid – derived polyesters (PHAs) in its cell walls. The PHA polymer can be used directly in the manufacture of a biodegradable plastic (Mirel), and the plant residues that remain are a potential source of biomass energy.
UN 19-20 t/k
UN 19-21 t/k
Ingeo garments made from corn sugar fibers hit the world of fashion.
Biodegradable plastic bottles would hopefully transform this heap into compost.
n
冣 冢
冣
B OCH(CH2)xC A R n
A
B OCHCH2C A R n
A
A
B OCH2C
A
冢 冣 冢
A
The persistent problem of conventional plastic waste disposal grows ever more severe as landfills approach capacity with these highly degradation-resistant substances (Section 12-15). Biodegradable plastics provide an option for nonreusable items such as plastic bags, wraps, and bottles. A recently developed and commercialized biobased and biodegradable plastic is poly(b-hydroxybutyrate-co-b-hydroxyvalerate) (PHBV), a copolymer of 3-hydroxybutanoic acid and 3-hydroxypentanoic acid. PHBV is a polyester that is produced by bacterial fermentation of mixtures of acetic and propanoic acids. The ratio of the two hydroxy acids controls the properties of the plastic—it is more flexible with more
A
The first decade of the 21st century has seen dramatic developments in the use of naturally occurring but non – petrochemical-derived raw materials for numerous purposes. Synthetic fibers for fabric manufacture are almost entirely petroleum-derived. Alternative polyesters such as poly(lactic acid) (Ingeo) can be made from vegetable matter — corn initially, but eventually from the starch that may be extracted from otherwise unusable plant residues (stalks, straws, etc.). The preparation of such polyesters is estimated to use two-thirds less fossil fuel than conventional fiber manufacture, and the process emits 80 – 90% less greenhouse gas.
PLA
PHBV
PHA
(Ingeo)
(R = – CH3, – CH2CH3)
(Mirel)
bronchoconstriction. These findings have led to the use of anti-inflammatory agents such as the corticosteroids (Section 4-7) as antiasthmatics, because they are capable of blocking the biosynthesis of arachidonic acid itself.
Exercise 19-24 Identify the arachidonic acid backbone in each of its four derivatives just shown.
906
Chapter 19
Carboxylic Acids
Nature also produces complex polycyclic carboxylic acids Many biologically active natural products that have carboxy groups as substituents of complex polycyclic frames have physiological potential that derives from other sites in the molecule. In these compounds the function of the carboxy group may be to impart water solubility, to allow for salt formation or ion transport, and to enable micellar-type aggregation. Two examples are gibberellic acid, one of a group of plant growth-promoting substances manufactured by fermentation, and lysergic acid, a major product of hydrolyzed extracts of ergot, a fungal parasite that lives on grasses, including rye. Many lysergic acid derivatives possess powerful psychotomimetic activity. In the Middle Ages, thousands who ate rye bread contaminated by ergot experienced the poisonous effects characteristic of these compounds: hallucinations, convulsions, delirium, epilepsy, and death. The synthetic lysergic acid diethylamide (LSD) is one of the most powerful hallucinogens known. The effective oral dose for humans is only about 0.05 mg.
O
O O ) HO
HOOC [
H %
C
0 0 H 0 H3C HO2C
ECH3 N [H
O N
OH
N H
N
Gibberellic acid
Lysergic acid
In Summary The numerous naturally occurring carboxylic acids are structurally and functionally diverse. The condensation of multiple acetyl units gives rise to the straight-chain fatty acids. The fatty acids in turn are converted into a wide array of substances exhibiting varied biological and, often, useful medicinal properties.
THE BIG PICTURE The carboxylic acid functional group combines the carbonyl group of the aldehydes and ketones with the hydroxy group of the alcohols. The carbonyl group enhances the acidity of the hydroxy hydrogen, and the hydroxy group gives rise to the addition – elimination pathway for substitution about the carbonyl carbon. Such substitution reactions result in the formation of the carboxylic acid derivatives known as halides, anhydrides, esters, and amides. These derivatives are important starting points for the synthesis of a variety of other classes of compounds. The next chapter will reveal the remarkable chemical versatility that has made these derivatives mainstays of laboratory chemistry as well as countless naturally occurring biological processes. After we have completed our presentation of carbonyl chemistry, the only simple compound class left to be discussed will be the amines (Chapter 21). The final five chapters of the text will deal with compounds combining multiple functional groups, with an emphasis on those of synthetic and biochemical importance.
Chapter Integration Problems
Chapter 19
CHAPTER INTEGRATION PROBLEMS 19-25. Propose a mechanism for the exchange between an acyl bromide and a carboxylic acid, as it occurs in step 4 of the Hell-Volhard-Zelinsky reaction. O B RCHCBr A Br
O B RCH2COH
O B RCHCOH A Br
O B RCH2CBr
SOLUTION First we must consider exactly what changes have occurred and then by which pathways these changes could have taken place. The two starting compounds have exchanged Br and OH substituents on their carbonyl carbons. Given the main chapter theme, addition – elimination pathways for carbonyl substitution reactions, let us look for a logical way to enable that process. The a-bromoacyl bromide has a strongly electrophilic (d1) carbonyl carbon, because of the electron-withdrawing effects of the two bromines. This acyl bromide is therefore a good candidate to be attacked by the most nucleophilic atom in the two compounds, the carbonyl oxygen of the carboxylic acid: Recall that the carbonyl oxygen is the most favorable site for attachment of electrophiles, because the species that results is resonance stabilized:
O B C E H OH RCH2
Br f RCH O A A EC H O Br A + EC M O OH RCH2
Br f RCH O A A C OE HBr B C E H O OH RCH2
Br f RCH i CPO f Br
Elimination of bromide ion from this tetrahedral intermediate gives a protonated anhydride. Readdition of bromide to the activated carbonyl group in the anhydride gives a new tetrahedral intermediate. At this stage the exchange sequence can be completed by elimination of the carboxylic acid. This process may be viewed as involving the shift of a proton from one oxygen to another using a six-membered-ring transition state, similar to those we have seen on several previous occasions. This step results directly in the formation of the exchanged product molecules, an a-bromocarboxylic acid and an acyl bromide unsubstituted at the a-carbon: E Br RCH A KC H O OH >
>
>
> >
> >
A
> >
>
>
>
>
O > B C E H Br RCH2
> >
>
>
> >
> >
>
>
E Br RCH A C E N O O A RCH2 C H A HOE Br
> >
> >
> > Br
19-26. Lactones with rings of more than six members may be synthesized if ring strain and transannular interactions are minimized (see Sections 4-2 through 4-5 and 9-6). One of the most important commercial musk fragrances is a dilactone, ethylene brassylate. Its synthesis, illustrated here, begins with acid-catalyzed reaction of tridecanedioic acid with 1,2-ethanediol, which yields a substance A with the molecular formula C15H28O5. Under the conditions in which it forms, substance A converts into a polymer (lower left structure). Strong heating reverses polymer formation, regenerating substance A. In a slower process, A transforms into the final macrocyclic product
> >
>
> >
E Br RCH A C OE N O B EC H O OH RCH2
>
Br f RCH O A A C OE HBr B EC H O OH RCH2
907
908
Chapter 19
Carboxylic Acids
(lower right structure), which is distilled from the reaction mixture, thereby shifting the equilibria in its favor.
Commercial Synthesis of a Macrocyclic Musk O O B B C O (CH2)11O C G D HO OH
HOCH2CH2OH
Tridecanedioic acid (Brassylic acid)
1,2-Ethanediol (Ethylene glycol)
H
O
O O
O O B B O C(CH2)11COCH2CH2O n O
H
C15H28O5
Polymer
O
H,
Ethylene brassylate
A
a. Suggest a structure for compound A. SOLUTION What information do we possess regarding this compound? We have its molecular formula and the structures both of its immediate precursors and of two of its products. It is possible to arrive at the solution by using any of these clues. For example, the polymeric product is constructed of repeating monomer units, each of which contains an ethylene glycol ester of brassylic acid, just missing the OH from the carboxy group at the left and the H from the hydroxy group at the right. Indeed, the formula of this unit is C15H26O4, or structure A less one molecule of water. If we simply write its structure and then add the missing elements of a molecule of water, we get A, the monoester shown.
O
O OH O
HO Ethylene glycol monoester of brassylic acid (A)
To use another approach, the molecular formulas of the two starting materials add up to C15H30O6, or A 1 H2O. Therefore, A is the product of a reaction that combines brassylic acid and ethylene glycol, and releases a molecule of water in the process, exactly what is observed in esterification. b. Propose mechanisms for the reactions that interconvert structure A with both the polymer and the macrocycle. SOLUTION Both reactions are examples of esterifications. Polymer formation begins with ester formation between the hydroxy group of one molecule of structure A and the carboxy group of another. We follow the pattern of addition – elimination (Section 19-9), making use of acid catalysis as specified in the original equation. Remember: The answers to mechanism problems should make use of only those chemical species actually specified in the reaction. The mechanistic sequence of steps is adapted directly from the text examples: (1) protonation of the carbonyl carbon to be attacked, (2) nucleophilic addition of a hydroxy oxygen to form the tetrahedral intermediate, (3) protonation of one hydroxy group of the intermediate to form a good leaving group, (4) departure of water, and (5) proton loss from oxygen to yield the final product.
Chapter Integration Problems
H
>
> > O
O O
HO
HO
OH
O
H OH
> HO
>
HO
O
O
O OH
O
O
O
OH
O
O
O
O
>
HO
OH
OH
Chapter 19
Tetrahedral intermediate
O HO
HO
O
O
OH2
O
O
OH
O
H2O
H
G
O
O
O HO
O
O
O OH
O
H
O HO
O
O
O O
O
OH
O
The product of this process is a dimer; repetition of such ester formation many times at both ends of this molecule eventually gives rise to the polyester observed in the reaction. The same steps take place in reverse to convert the polymer back into monoester A upon heating. What about the mechanism for macrocyclic lactone formation? We realize that the process must be intramolecular to form a ring (compare Sections 9-6 and 17-7). We can write exactly the same sequence of steps as we did for esterification, but we use the free hydroxy group at one end of molecule A to attack the carboxy carbon at the other end.
O
O
H
OH O
O OH
HO O
O
O
)p OH OH
H
O
> O pH
> O p
O
HO
O
OH p
O OH
O
OH2 p
O H2O
O
O
O O
H
O
909
Carboxylic Acids
Macrocyclic lactone synthesis is a topic of considerable interest to the pharmaceutical industry because this function constitutes the basic framework of many medicinally valuable compounds. Examples include erythromycin A, a widely used macrolide antibiotic (Chemical Highlight 20-2), and tacrolimus (FK-506), a powerful immunosuppressant that shows great promise in controlling the rejection of transplanted organs in human patients. O
%
`
O `
O
H3C
O
O
OH O
N
CH3
O
O
OCH3 O %
O
%
O
O
OH % O
OCH3
%
OH
^
`
HO^
OH
O
`
%
H3CO
N(CH3)2 %
OH
%
HO
CH3
H3CO `
Chapter 19
^
910
CH3 Erythromycin A
`
CH3
Tacrolimus (FK-506)
New Reactions 1. Acidity of Carboxylic Acids (Section 19-4) ðOð B C E Hš R OH
ðš Oð A C E Nš O R
ðOð B C E Hš Oð R
H2š O
Resonance-stabilized carboxylate ion Ka 104–105, pKa
4–5
Salt formation RCOOH
RCOONa
NaOH
H 2O
Also with Na2CO3, NaHCO3
2. Basicity of Carboxylic Acids (Section 19-4) š Oð l ROC i š OH
š O OH
H
l ROC i š O OH
ROC
š O OH f r š
O O H
Resonance-stabilized protonated carboxylic acid
Preparation of Carboxylic Acids 3. Oxidation of Primary Alcohols and Aldehydes (Section 19-6) RCH2OH
Oxidizing agent
RCOOH
Oxidizing agents: aqueous CrO3, KMnO4, HNO3
RCHO
Oxidizing agent
RCOOH
Oxidizing agents: aqueous CrO3, KMnO4, Ag, H2O2, HNO3
H3š O
OH
New Reactions
Chapter 19
4. Carbonation of Organometallic Reagents (Section 19-6) RMgX
CO2
RLi
CO2
THF
RCOOMgX
THF
RCOOLi
H, H2O
H, H2O
RCOOH RCOOH
5. Hydrolysis of Nitriles (Section 19-6) H2O, , H or HO
RC q N
RCOOH
NH4
or
NH3
Reactions of Carboxylic Acids 6. Nucleophilic Attack at the Carbonyl Group (Section 19-7) Base-catalyzed addition – elimination O B RCL
ðNu
O A R O C OL A Nu
Addition
L leaving group
O B RCNu
Elimination
L
Tetrahedral intermediate
Acid-catalyzed addition – elimination
O B RCL
D
H
O
H
B RCL
OH A R O C OL A NuH
ðNuH Addition
OH A R O C OL A Nu
H
H
D
HL Elimination
Tetrahedral intermediate
Derivatives of Carboxylic Acids 7. Acyl Halides (Section 19-8)
RCOOH
O B RCCl
SOCl2
SO2
H3PO3
HCl
Acyl chloride
3 RCOOH
O B 3 RCBr
PBr3
Acyl bromide
8. Carboxylic Anhydrides (Section 19-8)
RCOOH
O B RCCl
O O B B RCOCR Anhydride
H
O
OH A R O C OL G A H Nu
HCl
B RCNu
H
O B RCNu
911
912
Chapter 19
Carboxylic Acids
Cyclic anhydrides
COOH
C
(CH2)n
(CH2)n
H2O
COOH
C
O l i O f r O
Best for five- or six-membered rings
9. Esters (Section 19-9) Acid-catalyzed esterification
RCO2H
O B RCOR
H
ROH
K
1
H2O
Cyclic esters (lactones) CH2OH
CH2 i (CH2)n O f C r O
H
(CH2)n
K
COOH
H2O
Lactone K 1 for five- and six-membered rings
10. Carboxylic Amides (Section 19-10)
RCOOH
RCOO
RNH2
RNH3
O B RCNHR
Imides
C
COOH
(CH2)n
RNH2
(CH2)n C
COOH
O l i NR f r O
Cyclic amides (lactams) CH2NH2 (CH2)n COOH
CH2 i (CH2)n NH f C r O
H2O
Lactam
11. Reduction with Lithium Aluminum Hydride (Section 19-11)
RCOOH
1. LiAlH4, (CH3CH2)2O 2. H, H2O
RCH2OH
12. Bromination: Hell-Volhard-Zelinsky Reaction (Section 19-12)
RCH2COOH
Br2, trace P
Br A RCHCOOH
2 H2O
H2O
Problems
Chapter 19
Important Concepts 1. Carboxylic acids are named as alkanoic acids. The carbonyl carbon is numbered 1 in the longest chain incorporating the carboxy group. Dicarboxylic acids are called alkanedioic acids. Cyclic and aromatic systems are called cycloalkanecarboxylic and benzoic acids, respectively. In these systems the ring carbon bearing the carboxy group is assigned the number 1. 2. The carboxy group is approximately trigonal planar. Except in very dilute solution, carboxylic acids form dimers by hydrogen bonding. 3. The carboxylic acid proton chemical shift is variable but relatively high (d 5 10 – 13), because of hydrogen bonding. The carbonyl carbon is also relatively deshielded but not as much as in aldehydes and ketones, because of the resonance contribution of the hydroxy group. The carboxy function shows two important infrared bands, one at about 1710 cm21 for the C P O bond and a very broad band between 2500 and 3300 cm21 for the O–H group. The mass spectrum of carboxylic acids shows facile fragmentation in three ways. 4. The carbonyl group in carboxylic acids undergoes nucleophilic displacement via the addition – elimination pathway. Addition of a nucleophile gives an unstable tetrahedral intermediate that decomposes by elimination of the hydroxy group to give a carboxylic acid derivative. 5. Lithium aluminum hydride is a strong enough nucleophile to add to the carbonyl group of carboxylate ions. This process allows the reduction of carboxylic acids to primary alcohols.
Problems 27. Name (IUPAC or Chemical Abstracts system) or draw the structure of each of the following compounds. O Br H3C Cl D G P C C (a) (d) (b) (c) OH OH D G CO2H (CH3)2CH O COOH $OH
H G (f) CPC D D CO2H HO2C
0
(e) CO2H
CH3
HO
Cl D
CH2CO2H
COOH (h)
(g)
COOH Phthalic acid
OH o-Orsellinic acid
(i) 4-Aminobutanoic acid (also known as GABA, a critical participant in brain biochemistry); ( j) meso-2,3-dimethylbutanedioic acid; (k) 2-oxopropanoic acid (pyruvic acid); (l) trans-2-formylcyclohexanecarboxylic acid; (m) (Z)-3-phenyl-2-butenoic acid; (n) 1,8-naphthalenedicarboxylic acid. 28. Name each of the following compounds according to IUPAC or Chemical Abstracts. Pay attention to the order of precedence of the functional groups. COOH
O
O (a) HO
(b) Cl
OH H
(c)
COOH
(d) NO2
COOH
O
29. Rank the group of molecules shown here in decreasing order of boiling point and solubility in water. Explain your answers. COOH
CH3
CHO
CH2OH
30. Rank each of the following groups of organic compounds in order of decreasing acidity. O B (a) CH3CH2CO2H, CH3CCH2OH, CH3CH2CH2OH
(b) BrCH2CO2H, ClCH2CO2H, FCH2CO2H
913
914
Chapter 19
Carboxylic Acids
Cl Cl A A (c) CH3CHCH2CO2H, ClCH2CH2CH2CO2H, CH3CH2CHCO2H
(d) CF3CO2H, CBr3CO2H, (CH3)3CCO2H NO2
(e)
COOH, O2N
COOH, O2N
COOH, CH3O
COOH
31. Propose a structure for a compound that displays the spectroscopic data that follow. The molecular ion in the mass spectrum appears at m/z 5 116. IR n˜ 5 1710 (s) and 3000 (s, broad) cm21. 1H NMR: d 5 0.94 (t, J 5 7.0 Hz, 6 H), 1.59 (m, 4 H), 2.36 (quin, J 5 7.0 Hz, 1 H), and 12.04 (broad s, 1 H) ppm; 13C NMR: 11.7, 24.7, 48.7, and 183.0 ppm.
Preparation of Carboxylic Acids O B RCOR H or OH, H2O
section number
O B RCH
RCH2OH
Oxidant (i.e., CrO3, KMnO4, Ag, Cu2, H2O2, HNO3, O2–Rh3, Br2)
K2Cr2O7, H, H2O
RMgBr
RC q N
CO2
H or OH, H2O
Other product: ROH 8-5, 19-9, 20-1, 20-4
17-14, 19-5, 19-6, 24-4, 24-9
8-6, 19-6
19-6
19-6, 20-8, 26-2
O B C E H R OH
20-1, 20-2
20-1, 20-3
20-1, 20-6, 26-5
22-2
24-5
26-6
Product: COOH
Products: O O B B HCOH RCH
Other product: CH3
H or OH, H2O
H or OH, H2O
H or OH, H2O,
KMnO4, OH,
HIO4, H2O
H2, Pd–C
O B RCCl
O O B B RCOCR
O B G R RCN D R
R
O OH B A HC O C O R A H
O B RCOCH2
Reactions of Carboxylic Acids O B RCOR
O B RCO
OH A C G D OH R
O B RCX (X P Cl, Br)
O O B B RCOCR
9-4, 19-9
19-4
19-4
19-8
19-8
ROH, H
OH
H
SOCl2 or PBr3
O B RCCl
section number
O B G R RCN D R
19-10, 26-4, 26-6, 26-7
HN
G D
RCH2OH
O B RCHCOH A Br
O B RCCH2R
RCH2COOH
O B RCH
19-11
19-12
23-2
23-2
24-9
Substrate: O B RCH2COH
Substrate: O B RCCHCOOH A R
Substrate: HOOCCHCOOH A R
Substrate: RCHCOOH A OH
Br2, P
Fe3, H2O2
R
R , or DCC
O B EC H R OH
LiAlH4
915
916
Chapter 19
Carboxylic Acids
32. (a) An unknown compound A has the formula C7H12O2 and infrared spectrum A (p. 917). To which class does this compound belong? (b) Use the other spectra (NMR-B, p. 917, and F, p. 919; IR-D, E, and F, pp. 918 – 919) and spectroscopic and chemical information in the reaction sequence to determine the structures of compound A and the other unknown substances B through F. References are made to relevant sections of earlier chapters, but do not look them up before you have tried to solve the problem without the extra help. (c) Another unknown compound, G, has the formula C8H14O4 and the NMR and IR spectra labeled G (p. 920). Propose a structure for this molecule. (d) Compound G may be readily synthesized from B. Propose a sequence that accomplishes this efficiently. (e) Propose a completely different sequence from that shown in (b) for the conversion of C into A. (f) Finally, construct a synthetic scheme that is the reverse of that shown in (b); namely, the conversion of A into B. O B 1. Hg(OCCH3)2, H2O 2. NaBH4
C6H10
CrO3, H2SO4 , acetone, 0 C
C6H12O
Section 12-7
B
Section 8-6
C
C NMR: 22.1 24.5 126.2 ppm 1 H NMR-B
C NMR: 24.4 25.9 35.5 69.5 ppm
13
13
CH2PP(C6H5)3
C6H10O
C7H12
Section 17-12
D
E
C NMR: 23.8 26.5 40.4 208.5 ppm IR-D
13
1. BH3, THF 2. HO, H2O2 Section 12-8
IR-E
C7H14O F
Na2Cr2O7, H2O, H2SO4
A
Section 8-6
IR-F H NMR-F
1
33. Give the products of each of the following reactions. (a) (CH3)2CHCH2CO2H 1 SOCl2 uy COOH
(c)
CH3CH2OH
(e) Product of (d), heated strongly
H
(b) (CH3)2CHCH2CO2H 1 CH3COBr uy COOH NH3
(d) CH3O
(f) Phthalic acid [Problem 27(h)], heated strongly
34. When 1,4- and 1,5-dicarboxylic acids, such as butanedioic (succinic) acid (Section 19-8), are treated with SOCl2 or PBr3 in attempted preparations of the diacyl halides, the corresponding cyclic anhydrides are obtained. Explain mechanistically. 35. Reaction review. Without consulting the Reaction Road Map on p. 914, suggest reagents to convert each of the following starting materials into hexanoic acid: (a) hexanal; (b) methyl hexanoate; (c) 1-bromopentane; (d) 1-hexanol: (e) hexanenitrile. 36. Reaction review II. Without consulting the Reaction Road Map on p. 915, suggest reagents to convert hexanoic acid into each of the following compounds: (a) 1-hexanol; (b) hexanoic anhydride; (c) hexanoyl chloride; (d) 2-bromohexanoic acid; (e) ethyl hexanoate; (f) hexanamide. 37. Fill in suitable reagents to carry out the following transformations. (a) (CH3)2CHCH2CHO uy (CH3)2CHCH2CO2H
Br
(b)
(c)
CHO OH
OH
CO2H (d)
OH A CHCO2H
OH
Cl O
O (e)
O
O OH
O
(f) (CH3)3CCO2H uy (CH3)3CCO2CH(CH3)2
Problems
Chapter 19
Transmittance (%)
100
3040
1704
IR
0 4000
3500
3000
2500
2000
1500
600 cm−1
1000
Wavenumber
IR-A
1
H NMR
4H 2H
4H
5.68
5.5
5.0
5.66
5.64
2.0
4.0
4.5
3.5
1.8
3.0
δ 300-MHz H NMR spectrum ppm (δ) 1
B
NHCH3 (g)
CH3 A NH ECH3 C
B
O
1.6
2.5
2.0
1.5
917
Carboxylic Acids
100
Transmittance (%)
Chapter 19
1715 IR 0 4000
3500
3000
2500
2000
1500
1000
600 cm−1
Wavenumber IR-D
100
Transmittance (%)
918
3072
1649
IR 0 4000
888 3500
3000
2500
2000
1500
1000
600 cm−1
Wavenumber
IR-E
38. Propose syntheses of each of the following carboxylic acids that employ at least one reaction that forms a carbon – carbon bond.
(a) CH3CH2CH2CH2CH2CH2CO2H
OH A (b) CH3CHCH2CO2H
CH3 A (c) H3C O C O CO2H A CH3
Problems
1
H NMR
2H
5H
4H 2H 1H (CH3)4Si
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.0
300-MHz 1H NMR spectrum ppm (δ) δ
F
Transmittance (%)
100
3328 IR 0 4000
3500
1035
3000
2500
2000
1500
1000
600 cm−1
Wavenumber IR-F
39. (a) Write a mechanism for the esterification of propanoic acid with 18O-labeled ethanol. Show clearly the fate of the 18O label. (b) Acid-catalyzed hydrolysis of an unlabeled ester with 18 O-labeled water (H218O) leads to incorporation of some 18O into both oxygens of the carboxylic acid product. Explain by a mechanism. (Hint: You must use the fact that all steps in the mechanism are reversible.)
Chapter 19
919
Carboxylic Acids
Chapter 19
1
H NMR 6H
2.4
2.2
2.0
1.8
1.6
4H 4H
4.0
3.5
3.0
(CH3)4Si
2.5
2.0
1.5
1.0
0.5
0.0
300-MHz 1H NMR spectrum ppm (δ) δ
G 100
Transmittance (%)
920
1742
IR
0 4000
3500
3000
2500
2000
1500
1000
600 cm−1
Wavenumber
IR-G
40. Give the products of reaction of propanoic acid with each of the following reagents. (a) SOCl2 (e)
CH2NH2
(b) PBr3
(c) CH3CH2COBr 1 pyridine
(d) (CH3)2CHOH 1 HCl
(f) Product of (e), heated strongly
(g) LiAlH4, then H1, H2O
(h) Br2, P
41. Give the product of reaction of cyclopentanecarboxylic acid with each of the reagents in Problem 40.
Problems
42.
921
Chapter 19
When methyl ketones are treated with a halogen in the presence of base, the three hydrogen atoms on the methyl carbon are replaced to give a CX3-substituted ketone. This product is not stable under the basic conditions and proceeds to react with hydroxide, ultimately furnishing the carboxylic acid (as its conjugate base) and a molecule of HCX3, which has the common name haloform (i.e., chloroform, bromoform, and iodoform, for X 5 Cl, Br, and I, respectively). For example: ðOð B RCCBr3
ðOð B š RCO ð
š ð OH
HCBr3
Propose a series of mechanistic steps to convert the tribromoketone into the carboxylate. What is the leaving group? Why do you think this species is capable of acting as a leaving group in this process? 43. Interpret the labeled peaks in the mass spectrum of 2-methylhexanoic acid (MS-H).
CH3CH2CH2CH2CHCOOH O
Relative abundance
74
MS
100
50
CH3
87
45 130 0 0
20
40
60
80 m/z
100
120
140
160
MS-H
44. Suggest a preparation of hexanoic acid from pentanoic acid. 45. Give reagents and reaction conditions that would allow efficient conversion of 2-methylbutanoic acid into (a) the corresponding acyl chloride; (b) the corresponding methyl ester; (c) the corresponding ester with 2-butanol; (d) the anhydride; (e) the N-methylamide; CH3 A (f) CH3CH2CHCH2OH
Br A (g) CH3CH2CCO2H A CH3
46. Treatment of 4-pentenoic acid (margin) with Br2 in the presence of dilute aqueous base yields a nonacidic compound with the formula C5H7BrO2. (a) Suggest a structure for this compound and a mechanism for its formation. (b) Can you find a second, isomeric product whose formation is also mechanistically reasonable? (c) Discuss the issues that contribute to determining which of the two is the major product in this reaction. (Hint: Review Section 12-6.) 47. Show how the Hell-Volhard-Zelinsky reaction might be used in the synthesis of each of the following compounds, beginning in each case with a simple monocarboxylic acid. Write detailed mechanisms for all the reactions in one of your syntheses.
O OH 4-Pentenoic acid
922
Carboxylic Acids
Chapter 19
O (a) CH3CH2CHCO2H A NH2
(b)
(d) HO2CCH2SSCH2CO2H
(e) (CH3CH2)2NCH2CO2H
(c)
CHCO2H A CO2H
OH OH
(f) (C6H5)3PCHCO2H Br A CH3
48. The Hell-Volhard-Zelinsky reaction produces only bromocarboxylic acids. However, modified procedures have been developed to convert acyl chlorides into a-chloro- and a-bromoacyl chlorides by reaction with N-chloro- and N-bromobutanimide (N-chloro- and N-bromosuccinimide, NCS and NBS: Section 14-2), respectively. Reaction with I2 gives a-iodoacyl chlorides. (SOCl2 is used as the solvent to maintain the acyl chloride functional group.) Suggest a mechanism for any one of these processes. NCS, HCl, SOCl2, 70°C
C6H5CH2CH2COCl
C6H5CH2CHClCOCl 84%
NBS, HBr, SOCl2, 70°C
C6H5CH2CHBrCOCl 71%
I2, HI, SOCl2, 85°C
O B CH3CNH2
49. How would you expect the acidity of acetamide to compare with that of acetic acid? With that of acetone? Which protons in acetamide are the most acidic? Where would you expect acetamide to be protonated by very strong acid?
Acetamide
O
C6H5CH2CHICOCl 75%
O
50. Attempted CrO3 oxidation of 1,4-butanediol to butanedioic acid results in significant yields of “g-butyrolactone.” Explain mechanistically.
␥-Butyrolactone
51. Following the general mechanistic scheme given in Section 19-7, write detailed mechanisms for each of the following substitution reactions. (Note: These transformations are part of Chapter 20, but try to solve the problem without looking ahead.)
(a)
B CCl CH3CH2OH
B (b) CH3CNH2 H2O
H
B COCH2CH3
HCl
B CH3COH NH4
52. Suggest structures for the products of each reaction in the following synthetic sequence.
H ´ O ≥ H IR: 1745 cm1
N 1. H, H 2. (CH3)2C P CHCH2Br 3. H, H2O
1. O3, CH2Cl2 2. Zn, CH3COOH 3. KMnO4, NaHCO3
HOCH2CH2OH, H
C14H22O
C16H26O2
H
I
IR: 1675 and 1745 cm1
IR: 1670 cm1
C13H20O4
1. H, H2O 2. NaBH4
C11H18O3
Catalytic H, Δ
C11H16O2
J
K
L
IR: 1715 and 3000 (broad) cm1
IR: 1715, 3000 (broad), and 3350 cm1
IR: 1770 cm1
Problems
Chapter 19
53. SN2 reactions of simple carboxylate ions with haloalkanes (Section 8-5) in aqueous solution generally do not give good yields of esters. (a) Explain why this is so. (b) Reaction of 1-iodobutane with sodium acetate gives an excellent yield of ester if carried out in acetic acid (as shown here). Why is acetic acid a better solvent for this process than water? B CH3CH2CH2CH2I CH3CONa 1-Iodobutane
CH3CO2H, 100°C
Sodium acetate
B CH3CH2CH2CH2OCCH3 NaI 95% Butyl acetate
(c) The reaction of 1-iodobutane with sodium dodecanoate proceeds surprisingly well in aqueous solution, much better than the reaction with sodium acetate (see the following equation). Explain this observation. (Hint: Sodium dodecanoate is a soap and forms micelles in water. See Chemical Highlight 19-1.)
CH3CH2CH2CH2I CH3(CH2)10CO2 Na 54.
H 2O
B CH3(CH2)10COCH2CH2CH2CH3
The iridoids are a class of monoterpenes with powerful and varied biological activities. They include insecticides, agents of defense against predatory insects, and animal attractants. The following reaction sequence is a synthesis of neonepetalactone, one of the nepetalactones, which are primary constituents of catnip. Use the information given to deduce the structures that have been left out, including that of neonepetalactone itself.
C10H16O2
CHO
Base
CrO3, H2SO4, 0 C
M
N IR: 890, 1630, 1640, 1720, and 3000 (very broad) cm1
IR: 890, 1645, 1725 (very strong), and 1705 cm1
1. Dicyclohexylborane, THF 2. HO , H2O2
C11H16O2
55.
CH3OH, H
C10H14O2
C11H18O3
H , H2O,
O
P
Neonepetalactone
IR: 890, 1630, 1640 and 1720 cm1
IR: 1630, 1720, and 3335 cm1
IR: 1645 and 1710 cm1 UV max 241 nm
Propose two possible mechanisms for the following reaction. (Hint: Consider the possible sites of protonation in the molecule and the mechanistic consequences of each.) Devise an isotope-labeling experiment that might distinguish your two mechanisms. HCl,
CH A OH COOH
CH A OO
C B O
56. Propose a short synthesis of 2-butynoic acid, CH3CqCCO2H, starting from propyne. (Hint: Review Sections 13-2 and 13-5.) 57. The benzene rings of many compounds in nature are prepared by a biosynthetic pathway similar to that operating in fatty acid synthesis. Acetyl units are coupled, but the ketone functions are not reduced. The result is a polyketide thiolester, which forms rings by intramolecular aldol condensation.
923
924
Chapter 19
Carboxylic Acids
O O O O B B B B CH3CCH2CCH2CCH2C O S O protein Polyketide thiolester
o-Orsellinic acid [for structure, see Problem 27(g)] is a derivative of salicylic acid and is prepared biosynthetically from the polyketide thiolester shown. Explain how this transformation might take place. Hydrolysis of the thiolester to give the free carboxylic acid is the last step.
Team Problem 58. Section 19-9 showed you that 4- and 5-hydroxy acids can undergo acid-catalyzed intramolecular esterification to produce the corresponding lactone in good yield. Consider the following two examples of lactonization. Divide the analysis of the reaction sequences among your group. Propose reasonable mechanisms to explain the respective product formation. O ONa
H3C^
Br2, CH2Cl2
Br ?
O O (Note stereochemistry!)
O
O 2
OH
H , H2O
O OH
OH
H, H2O
OH O
(Two diastereomers)
H3C
O
O
O
O
CH3
(Cis and trans isomers)
Reconvene to present your mechanisms to each other.
Preprofessional Problems 59. What is the IUPAC name of the compound shown? (a) (b) (c) (d)
(E )-3-Methyl-2-hexenoic acid (Z )-3-Methyl-2-hexenoic acid (E )-3-Methyl-3-hexenoic acid (Z )-3-Methyl-3-hexenoic acid
CO2H i f CP C f i H CH3CH2CH2 H 3C
60. Select the acid with the highest Ka (i.e., lowest pKa). CO2H (a) H3CCO2H
(b)
CO2H (c)
I HO2C
(d) Cl2CHCO2H Cl
61. The acid whose structure is shown in the margin can best be prepared via one of these sequences. Which one? C(CH3)3
(a) H3CBr Br3CCO2H (b) (CH3)3CI
Mg, ether
Benzene
K
H, H2O (work-up)
CO2
CH3 KMnO4
(c)
CH3 (d) H3C
C(CH3)3
C(CH3)3
KMnO4
C H A P T E R
TWENTY
Carboxylic Acid Derivatives
O
O
N
N
N
O
O
W
hen complex organic systems have evolved in nature, they are frequently made by the combination of carboxy carbonyl groups and the heteroatoms of alcohols, amines, or thiols. Why? The addition– elimination processes introduced in Chapter 19 provide mechanistic pathways of relatively low activation energy for the interconversion of variously substituted carboxylic acid derivatives, many of which play central roles in biology (Chapter 26). Chemists find these compounds similarly useful, as we shall see in this chapter, which deals with the chemistry of four major carboxylic acid derivatives: halides, anhydrides, esters, and amides. Each has a substituent, L, that can function as a leaving group in substitution reactions. We already know, for instance, that displaceO B ment of the halide in RCX by a carboxy group leads to anhydrides. ðOð B C E H R L O B L = X, OCR, OR, NR2
Carboxylic acid derivatives
We begin with a comparison of the structures, properties, and relative reactivities of carboxylic acid derivatives. We then explore the chemistry of each type of compound. Halides and anhydrides are valuable reagents in the synthesis of other carbonyl compounds. Esters and amides are enormously important in nature; for example, the esters include common flavoring agents, waxes, fats, and oils; among the amides we find urea and penicillin. The alkanenitriles, RC q N, are also treated here because they have similar reactivity.
The sturdiness of the amide linkage is amply demonstrated by the properties of poly( pphenylene terephthalamide), which is used in bulletproof vests and body armor under the commercial name of Kevlar. Developed by DuPont chemist Stephanie Kwolek, Kevlar is 16 times as stiff as nylon, the polymer formerly used in such applications. Kevlar’s properties derive from the planarity of its benzene rings combined with restricted rotation about the amide linkages (Section 20-6). Thousands of law enforcement officers owe their lives to this remarkable material, which can stop the bullet from a 9-mm handgun traveling at 1200 feet per second.
N
Chapter 20
Carboxylic Acid Derivatives
20-1 Relative Reactivities, Structures, and Spectra of Carboxylic Acid Derivatives Carboxylic acid derivatives undergo substitution reactions with nucleophiles, such as water, organometallic compounds, and hydride reducing agents. These transformations proceed through the familiar (often acid- or base-catalyzed) addition–elimination sequence (Section 19-7). Addition–Elimination in Carboxylic Acid Derivatives ðOð B EC H R L
šð ðO A RO C O L A NuH
ðNuH
ðOð B EC H R Nu
H
ðL
Tetrahedral intermediate
The relative reactivities of the substrates follow a consistent order: Acyl halides are most reactive, followed by anhydrides, then esters, and finally the amides, which are least reactive. Relative Reactivities of Carboxylic Acid Derivatives in Nucleophilic Addition – Elimination with Water O B RCX
HOH
Acyl halide
Increasing reactivity
926
O O B B RCOCR
HOH
Anhydride
O B RCOR
HOH
Ester
O B RCNR2
HOH
Amide
20C Fast
20C Slow
RCOOH
RCOOH
Very slow: Needs catalyst and heat
RCOOH
HX
RCOOH
RCOOH
Exceedingly slow: Needs catalyst and prolonged heating
ROH
HNR2
The order of reactivity depends directly on the structure of L: how good a leaving group it is and what effect it has on the adjacent carbonyl function. This effect can be understood by looking at the degree to which the various resonance forms A – C contribute to the structure of the carboxylic acid derivative. In this picture, B is a very minor contributor, because it is an electron sextet structure. We have used such forms only to emphasize the polarity of the carbonyl function. This leaves us with A and the dipolar alternative C. A will always be major, because it has no charge separation (Section 1-5). The question is to what extent C helps to stabilize the functional group by delocalization of the lone pairs on the substituent L onto the carbonyl oxygen. Resonance in Carboxylic Acid Derivatives ðOð B GCG š R L
ðš Oð A GCG š R L
ðš Oð A G CP R L
A
B Nonoctet, minor
C
20-1 Relative Reactivities of Carboxylic Acid Derivatives
Chapter 20
The answer to this question lies in the electronegativity of L: Decreasing electronegativity goes with increasing contribution of C. As we saw in Section 6-7, decreasing electronegativity also goes hand in hand with decreasing leaving-group ability, and in turn with decreasing acidity of the conjugate acid. L
X
O B OCR
OR
NR2
Decreasing electronegativity Decreasing leaving-group ability Decreasing acidity of conjugate acid HL Increasing contribution of resonance form C
Thus, dipolar resonance form C is most important when L is NR2, because nitrogen is the least electronegative atom in the series. Consequently, amides are the least reactive carboxylic acid derivatives. In esters, the contribution of C is somewhat diminished, because oxygen is more electronegative than nitrogen. Nonetheless, resonance remains strong (see also Section 16-1) and, while esters are more reactive than amides, they are still rather resistant to nucleophilic attack. On the other hand, in anhydrides, the lone pairs on the central oxygen are shared by two carbonyl groups, making them less available for resonance as in C. For that reason, they are more reactive than esters. Finally, acyl halides are the most reactive for two reasons: relative electronegativity (F, Cl; Table 1-2) and poor overlap of the large halogen p orbitals (Cl, Br, I; Section 1-6) with the relatively small 2p lobes of the carbonyl carbon. The electrostatic potential maps in the margin depict examples of the two extremes: acetyl chloride and acetamide. In the chloride, the carbonyl carbon is much more positively charged (blue) than in the amide (green). Simultaneously, the extent of electron donation by the L group (Cl versus NH2) is greater in the amide, as indicated by the greater negative charge on the carbonyl oxygen (red).
Acetyl chloride
Acetamide
Exercise 20-1
Working with the Concepts: Relative Reactivity of Carboxylic Acid Derivatives Phosgene, phenylmethyl chloroformate, bis(1,1-dimethylethyl) dicarbonate, dimethyl carbonate, and urea are derivatives of carbonic acid H2CO3 (made by dissolving CO2 in water). Rank them in order of decreasing reactivity in nucleophilic addition – eliminations. O B ClCCl
O B C6H5CH2OCCl
O O B B (CH3)3COCOCOC(CH3)3
O B CH3OCOCH3
O B H2NCNH2
Phosgene
Phenylmethyl chloroformate
Bis(1,1-dimethylethyl) dicarbonate
Dimethyl carbonate
Urea
Strategy All of these compounds possess heteroatom (O, N, or halogen)-containing substituents on both
O B
sides of the carbonyl carbon, as does carbonic acid, HOCOH. The arguments about relative reactivity of carboxylic acid derivatives made in this section apply here in the same way. Solution • To a first approximation, the effects of the substituents on either side of the carbonyl carbon may be considered to be additive. The only difference is that the effect of the second substituent on the carbonyl is somewhat reduced, because the two groups have to compete with each other for resonance with the C P O p bond. • Thus phosgene, with halogens on both sides, may be considered as a “double acyl halide” and is the most reactive. • Reactivity decreases in the order in which the compounds are illustrated: The chloroformate is “half halide/half ester”; both carbonyls in the dicarbonate are “half anhydride/half ester”; the carbonate is a “double ester”; and urea is a “double amide” and the least reactive.
927
928
Chapter 20
Carboxylic Acid Derivatives
Exercise 20-2
Try It Yourself Where would you place the compounds below in the reactivity sequence of Exercise 20-1? (Hint: Consider steric effects.) O B (CH3)3COCOC(CH3)3
O B (CH3)3COCCl
O O B B CH3OCOCOCH3
Bis(1,1-dimethylethyl) carbonate
1,1-Dimethylethyl chloroformate
Dimethyl dicarbonate
The greater the resonance, the shorter the C – L bond The extent of resonance can be observed directly in the structures of carboxylic acid derivatives. In the progression from acyl halides to esters and amides, the C – L bond becomes progressively shorter, owing to increased double-bond character (Table 20-1). The NMR spectra of amides reveal that rotation about this bond has become restricted. For example, N,N-dimethylformamide at room temperature exhibits two singlets for the two methyl groups, because rotation about the C – N bond is very slow on the NMR time scale. The evidence points to considerable p overlap between the lone pair on nitrogen and the carbonyl carbon, as a result of the increased importance of the dipolar resonance form in amides. The measured barrier to this rotation is about 21 kcal mol21 (88 kJ mol21). Slow Rotation in N,N-Dimethylformamide H3C i f H3C
šO C N
ýO l i
H
ýOk H 3C f i NP C f i H H3C sp2 hybridized
Slow rotation
Ea = 21 kcal mol1
H 3C i f H3C
šO C N
ýO l i
H
ýOk H3C f i NP C f i H H3C
O N
Note that these observations are not consistent with the structure for ammonia shown in Section 1-8, Figure 1-20. Ammonia (and simple amines, as we shall see in Section 21-2) adopt a geometry based on sp3 hybridization, in order to minimize repulsion between electron pairs. In contrast, the lone pair on nitrogen in an amide resides in a p orbital, to permit p overlap with the p orbital on the carbonyl carbon atom. Consequently, the amide nitrogen atom possesses trigonal planar geometry and sp2 hybridization: The stabilization due to p overlap in the planar form overrides the normal preference for a tetrahedral arrangement of electron pairs. We shall see in Section 26-4 that the planarity of the nitrogen in amides is the single most important determinator of structure, and therefore function, in peptides and proteins. Exercise 20-3 O B CH3CNHNHC6H5 1-Acetyl 2-phenylhydrazide
The methyl group in the 1H NMR spectrum of 1-acetyl 2-phenylhydrazide, shown in the margin, exhibits two singlets at d 5 2.02 and 2.10 ppm at room temperature. Upon heating to 1008C in the NMR probe, the same compound gives rise to only one signal in that region. Explain.
Table 20-1
L Cl OCH3 NH2
O B C – L Bond Lengths in RC – L Compared with R – L Single Bond Distances
Bond length (Å) in R – L
O B Bond length (Å) in RC – L
1.78 1.43 1.47
1.79 (not shorter) 1.36 (shorter by 0.07 Å) 1.36 (shorter by 0.11 Å)
20-1 Relative Reactivities of Carboxylic Acid Derivatives
Table 20-2
O B Carbonyl Stretching Frequencies of Carboxylic Acid Derivatives RC – L n˜C“O (cm21)
Cl
1790 – 1815
O B OCR OR NR92
Increasing n˜C“O
L
1740 – 1790 1800 – 1850
Two bands are observed, corresponding to asymmetric and symmetric stretching motions
1735 – 1750 1650 – 1690
Infrared spectroscopy can also be used to probe resonance in carboxylic acid derivatives. The dipolar resonance structure weakens the C P O bond and causes a corresponding decrease in the carbonyl stretching frequency (Table 20-2). The IR data for carboxylic acids reported in Section 19-3 refer to the common dimeric form, in which hydrogen bonding reduces the stretching frequencies of both the O – H and C P O bonds to about 3000 and 1700 cm21, respectively. A special technique — vapor deposition at very low temperature — allows the IR spectra of carboxylic acid monomers to be measured, for direct comparison with the spectra of carboxylic acid derivatives. Monomeric acetic acid displays n˜C P O at 1780 cm21, similar to the value for carboxylic anhydrides, higher than that for esters, and lower than that of halides, consistent with the degree of resonance delocalization in carboxylic acids. In addition to infrared bands due to their carbonyl groups, amides display bands resulting from the stretching of their N – H bonds. In the region of 3100 to 3400 cm21, amides with two NH bonds show two bands, and those with one NH bond display one (see Section 20-3). The 13C NMR signals of the carbonyl carbons in carboxylic acid derivatives are less sensitive to polarity differences and fall in a narrow range near 170 ppm. 13C
NMR Chemical Shifts of the Carbonyl Carbon in Carboxylic Acid Derivatives
O B CH3CCl
O O B B CH3COCCH3
O B CH3COH
O B CH3COCH3
O B CH3CNH2
␦ = 170.3
166.9
177.2
170.7
172.6 ppm
In common with other carbonyl compounds, the mass spectra of carboxylic acid derivatives typically show peaks resulting from both a-cleavage and McLafferty rearrangement. Exercise 20-4 The mass spectrum of methyl pentanoate (molecular weight 5 116) shows fragmentation peaks at m/z 5 85, 74, and 57. Assign these peaks.
Carboxylic acid derivatives are basic and acidic The extent of resonance in carboxylic acid derivatives is also seen in their basicity (protonation at the carbonyl oxygen) and acidity (enolate formation). In all cases, protonation requires strong acid, but it gets easier as the electron-donating ability of the L group increases. Protonation is important in acid-catalyzed nucleophilic addition – elimination reactions.
Chapter 20
929
930
Carboxylic Acid Derivatives
Chapter 20
Protonation of a Carboxylic Acid Derivative ðOð B GCG š R L
H O B GCG š R L
EH ðš O A GCG š R L
šE
H
EH O ðš A G CP R L A relatively strong contribution of this resonance form stabilizes the protonated species
Exercise 20-5 The oxygen in acetyl chloride is much less basic than the one in acetamide. Explain, using resonance structures.
For related reasons, the acidity of the hydrogens next to the carbonyl group increases along the series. The acidity of a ketone lies between those of an acyl chloride and an ester. Acidities of ␣-Hydrogens in Carboxylic Acid Derivatives in Comparison with Acetone O B CH3CCl
O O O O B B B B CH3CN(CH3)2 CH3COCH3 CH3CCH3 CH3CCl pKa
30
25
20
16
O Exercise 20-6 Cl
Give an example of a reaction from Chapter 19 that takes advantage of the relatively high acidity of the a-hydrogens in acyl halides.
Acetyl chloride
O B In Summary The relative electronegativity and the size of L in RCL controls the extent of resonance of the lone electron pair(s) and the relative reactivity of a carboxylic acid derivative in nucleophilic addition – elimination reactions. This effect manifests itself structurally and spectroscopically, as well as in the relative acidity and basicity of the a-hydrogen and the carbonyl oxygen, respectively.
CH3 O A B CH3CHCH2CBr 3-Methylbutanoyl bromide
O F
20-2 Chemistry of Acyl Halides
Pentanoyl fluoride
O B CH
O B The acyl halides, RCX, are named after the alkanoic acid from which they are derived. The halides of cycloalkanecarboxylic acids are called cycloalkane carbonyl halides. Acyl halides undergo addition – elimination reactions in which nucleophiles displace the halide leaving group. These compounds are so reactive that catalysts are usually not necessary for their conversion.
F
O F
Cyclohexanecarbonyl fluoride
Addition–Elimination Reactions of Acyl Halides ðOð B šð ðNu RCX š
Oð ðš A RO CO š Xð A š Nu Tetrahedral intermediate
ðOð B RCNu
Xð ðš š
20-2 Chemistry of Acyl Halides
Chapter 20
O B RCX O B RCOH
HOH
O O B B RCOCR HX
O B RCOH HX
H A RNH
ROH
O B RCOR HX
R2CuLi
LiAl(OR)3H
O B RCR LiX CuR
O B RCNHR HX
O B RCH LiX Al(OR)3
Figure 20-1 Nucleophilic addition – elimination reactions of acyl halides.
Figure 20-1 shows a variety of nucleophilic reagents and the corresponding products. It is because of this wide range of reactivity that acyl halides are useful synthetic intermediates. Let us consider these transformations one by one (except for anhydride formation, which was covered in Section 19-8). Examples will be restricted to acyl chlorides, which are the most readily accessible, but their transformations can be generalized to a considerable extent to the other acyl halides.
Water hydrolyzes acyl chlorides to carboxylic acids Acyl chlorides react with water, often violently, to give the corresponding carboxylic acids and hydrogen chloride. This transformation is a simple example of addition – elimination and is representative of most of the reactions of this class of compounds. REACTION
Acyl Chloride Hydrolysis O B CH3CH2CCl
O B CH3CH2COH 100%
HOH
Propanoyl chloride
HCl
Propanoic acid
MECHANISM Mechanism of Acyl Chloride Hydrolysis ðOð B GCG
CH3CH2
šð Cl š
O H2š š
Oð ðš A šð CH3CH2OCO Cl A š ðOH2
Chloride: good leaving group
šð ðCl
O
O
ðOð B GCG G H š O CH3CH2 A H
H
O
O
Cl
ðOð B š CH3CH2COH š
O Cl
O
− Cl
O
Tetrahedral intermediate
Alcohols convert acyl chlorides into esters The analogous reaction of acyl chlorides with alcohols is a highly effective way of producing esters. A base such as an alkali metal hydroxide, pyridine, or a tertiary amine is usually added to neutralize the HCl by-product. Because acyl chlorides are readily made from the corresponding carboxylic acids (Section 19-8), the sequence RCOOH y RCOCl y RCOOR9 is a good method for esterification. By maintaining neutral or basic conditions, this preparation avoids the equilibrium problem of acid-catalyzed ester formation (Fischer esterification, Section 19-9).
O
−
931
932
Carboxylic Acid Derivatives
Chapter 20
Ester Synthesis from Carboxylic Acids Through Acyl Chlorides O B RCOH O B CH3CCl
1,1-Dimethylethyl acetate (tert-Butyl acetate)
HCl
O B RCOR
ROH, base HCl
O B CH3COCH2CH2CH3 75%
N(CH2CH3)3 (See Problem 35)
HN(CH2CH3)3 Cl
Propyl acetate
1-Propanol
Acetyl chloride
O B C E H OC(CH3)3 H3C
HOCH2CH2CH3
O B RCCl
SOCl2
Triethylammonium chloride
Exercise 20-7 You have learned that 2-methyl-2-propanol (tert-butyl alcohol) dehydrates in the presence of acid (Section 9-2). Suggest a synthesis of 1,1-dimethylethyl acetate (tert-butyl acetate, shown in the margin) from acetic acid. Avoid conditions that might dehydrate the alcohol.
Amines convert acyl chlorides into amides Secondary and primary amines, as well as ammonia, convert acyl chlorides into amides. As shown in the first example below, aqueous ammonia works quite well for the synthesis of simple amides: NH3, being a much stronger nucleophile than water, reacts preferentially with the carbonyl derivative. As in ester formation, the HCl formed is neutralized by added base (which can be excess amine). Amides from Acyl Halides
REACTION
O B CNH2
O B CCl
NH3
H2O
NH4 Cl
Excess
86% Benzoyl chloride
O B CH2 P CHCCl
Benzamide
2 CH3NH2
O B CH2 P CHCNHCH3 68%
CH3NH3 Cl
N-Methylpropenamide
Propenoyl chloride
MECHANISM
Benzene, 5°C
The mechanism of this transformation is, again, addition – elimination, beginning with attack of the nucleophilic amine nitrogen at the carbonyl carbon. Mechanism of Amide Formation from Acyl Chlorides
NR H2š
ðš Oð A š RO C O Cl ð A R N A HH H A
ðOð B C E Hš R Cl ð
š ð ðCl
ðOð B C E H EH R N A A R H
ðOð B C E H š ER R N A H
H
Tetrahedral intermediate
Note that in the last step, a proton must be lost from nitrogen to give the amide. Consequently, tertiary amines (which have no hydrogens on nitrogen) cannot form amides. Instead, they
20-2 Chemistry of Acyl Halides
Chapter 20
convert acyl halides to acyl ammonium salts. In these species, the nitrogen has no lone electron pair to stabilize the carbonyl function by resonance. On the contrary, the positive charge activates the carbonyl carbon to nucleophilic attack. Consequently, acyl ammonium salts show reactivity that is similar to those of acyl halides. A useful consequence of this behavior is that acylations of other functions can be carried out in the presence of tertiary amino substituents. Acetylation of a tert-Amino Alcohol O OH H3C
N(CH3)2
O B CH3CCl
OH H 3C
O
Cl CH3 N CH3 H3C
H3C
CH3 N(CH3)2
O Reactive intermediate acyl ammonium salt
Exercise 20-8 Some amide preparations from acyl halides require a primary or secondary amine that is too expensive to use also as the base to neutralize the hydrogen halide. Suggest a solution to this problem.
Organometallic reagents convert acyl chlorides into ketones Organometallic reagents (RLi and RMgX) attack the carbonyl group of acyl chlorides to give the corresponding ketones. However, these ketone products are themselves prone to further attack by the relatively unselective organolithium (RLi) and Grignard (RMgX) reagents to give alcohols (see Section 8-8). Ketone formation is best achieved by using diorganocuprates (Section 18-10), which are more selective than RLi or RMgX and do not add to the product ketone. Formation of a Ketone from an Acyl Halide O M
O M
CCl
CCHP C(CH3)2
[(CH3)2C P CH]2CuLi
LiCl
(CH3)2C P CHCu
70%
Reduction of acyl chlorides results in aldehydes We can convert an acyl chloride into an aldehyde by hydride reduction. In this transformation, we again face a selectivity problem: Sodium borohydride and lithium aluminum hydride convert aldehydes into alcohols. To prevent such overreduction, we must modify LiAlH4 by letting it react first with three molecules of 2-methyl-2-propanol (tert-butyl alcohol; see Section 8-6). This treatment neutralizes three of the reactive hydride atoms, leaving one behind that is nucleophilic enough to attack an acyl chloride but not the resulting aldehyde. Reductions by Modified Lithium Aluminum Hydride Preparation of reagent LiAlH4
3 (CH3)3COH
LiAl[OC(CH3)3]3H Lithium tri(tert-butoxy)aluminum hydride
3 HOH
933
934
Chapter 20
Carboxylic Acid Derivatives
Reduction O B RCCl
LiAl[OC(CH3)3]3H
1. Ether solvent 2. H, H2O
O B RCH
LiCl
Al[OC(CH3)3]3
Reduction of an Acyl Halide to an Aldehyde O B CH3CHP CHCCl
1. LiAl[OC(CH3)3]3H, (CH3OCH2CH2)2O, 78°C 2. H, H2O
O B CH3CHP CHCH 48%
2-Butenoyl chloride
2-Butenal
Exercise 20-9 Prepare the following compounds from butanoyl chloride. O
O OH
(a) O
O
O N(CH3)2
(c)
O
(b)
(d)
(e)
H
In Summary Acyl chlorides are attacked by a variety of nucleophiles, the reactions leading to new carboxylic acid derivatives, ketones, and aldehydes by addition – elimination mechanisms. The reactivity of acyl halides makes them useful synthetic relay points on the way to other carbonyl derivatives.
20-3 Chemistry of Carboxylic Anhydrides
O O B B Carboxylic anhydrides, RCOCR, are named by simply adding the term anhydride to the acid name (or names, with regard to mixed anhydrides). This method also applies to cyclic derivatives. O O O
O
O O 1,2-Benzenedicarboxylic anhydride (Phthalic anhydride)
O O B B CH3COCCH3
O O O B B CH3COCCH2CH3 Acetic propanoic anhydride (A mixed anhydride)
O O
O
O O
O
O O
O
O O
Acetic anhydride
2-Butenedioic anhydride (Maleic anhydride)
Pentanedioic anhydride (Glutaric anhydride)
The reactions of carboxylic anhydrides with nucleophiles, although less vigorous, are completely analogous to those of the acyl halides. The leaving group is a carboxylate instead of a halide ion.
20-3 Chemistry of Carboxylic Anhydrides
Chapter 20
935
Nucleophilic Addition–Elimination of Anhydrides ðOð ðOð B B š RCO O O CR ðNuH
Carboxylate ion: good leaving group
ðš Oð ðOð B A RC O š O OCR A NuH
ðOð B RC O Nu O H
ðOð B ðOCR š
Tetrahedral intermediate
Typical Reactions of Anhydrides O O B B CH3COCCH3
HOH Hydrolysis
O O B B CH3COH HOCCH3 100%
Acetic anhydride
O
Acetic acid
O
O CH3OH
O
OCH3 83%
Alcoholysis (ester formation)
Propanoic anhydride
O HO
Methyl propanoate
Propanoic acid
In every addition – elimination reaction except hydrolysis, the carboxylic acid side product is usually undesired and is removed by work-up with aqueous base. Cyclic anhydrides undergo similar nucleophilic addition – elimination reactions that lead to ring opening. Nucleophilic Ring Opening of Cyclic Anhydrides O O
CH3OH, 100C (See Problem 38)
O O B B HOCCH2CH2COCH3 96%
O Butanedioic (succinic) anhydride
Exercise 20-10 Treatment of butanedioic (succinic) anhydride with ammonia at elevated temperatures leads to a compound C4H5NO2. What is its structure? (Hint: Consult Section 19-10.)
Exercise 20-11 Formulate the mechanism for the reaction of acetic anhydride with methanol in the presence of sulfuric acid.
Although the chemistry of carboxylic anhydrides is very similar to that of acyl halides, anhydrides have some practical advantages. Acyl halides are so reactive that they are difficult to store for extended periods without some hydrolysis occurring due to exposure to atmospheric moisture. As a result, chemists usually prepare acyl halides immediately before they are to be used. Anhydrides, being slightly less reactive toward nucleophiles, are more stable, and several (including all the examples illustrated in this section) are commercially available. Consequently, carboxylic anhydrides are often the preferred reagents for the preparation of many carboxylic acid derivatives.
ðOð B RCNu
ðOð B š HOCR
936
Carboxylic Acid Derivatives
Chapter 20
In Summary Anhydrides react with nucleophiles in the same way as acyl halides do, except that the leaving group is a carboxylate ion. Cyclic anhydrides furnish dicarboxylic acid derivatives.
20-4 Chemistry of Esters
O B As mentioned in Section 19-9, esters, RCOR9, constitute perhaps the most important class of carboxylic derivatives. They are particularly widespread in nature, contributing especially to the pleasant flavors and aromas of many flowers and fruits. Esters undergo typical carbonyl chemistry but with reduced reactivity relative to that of either acyl halides or carboxylic anhydrides. This section begins with a discussion of ester nomenclature. Descriptions follow of the transformations that esters undergo with a variety of nucleophilic reagents.
Esters are alkyl alkanoates O B Esters are named as alkyl alkanoates. The ester grouping, –COR, as a substituent is called alkoxycarbonyl. A cyclic ester is called a lactone (the common name, Section 19-9); the systematic name is oxa-2-cycloalkanone (Section 25-1). Its common name is preceded by a-, b-, g-, d-, and so forth, depending on ring size. O B CH3COCH3
O
O B CH3CH2COCH2CH3
O CH3 A B CH3COCH2CH2CHCH3
Ethyl propanoate (Ethyl propionate)
3-Methylbutyl acetate (Isopentyl acetate)
O Methyl acetate
(A component of banana flavor)
O
O B COCH3
O
O
O
NH2
2-Methylpropyl propanoate (Isobutyl propionate)
3-Methylbutyl pentanoate (Isopentyl valerate)
Methyl 2-aminobenzoate (Methyl anthranilate)
(A component of rum flavor)
(A component of apple flavor)
(A component of grape flavor)
O
O
O
O
H3C
O
O
g-Valerolactone (Systematic: 5-methyloxa-2-cyclopentanone)
O O -Propiolactone (This compound is a carcinogen and is systematically called oxa-2-cyclobutanone; see Section 25-1)
Lactones: Cyclic esters
O O
(Z)-7-Dodecenyl acetate
CH3
O
O
␥ -Butyrolactone (Systematic: oxa-2-cyclopentanone)
In addition to their prevalence in plants, esters play biological roles in the animal kingdom. Section 12-17 included several examples of esters that function as insect pheromones. Perhaps the most bizarre of these esters is (Z )-7-dodecenyl acetate, a component of the pheromone mixture in several species of moths. This same compound was recently found to also be the mating pheromone of the elephant (who said nature has no sense of humor?). Section 20-5 will describe a number of more conventional biological functions of esters.
20-4 Chemistry of Esters
Chapter 20
937
Lactones (cyclic esters) are also widely distributed in nature. The ready availability of g-valerolactone from renewable plant-derived sources has led to its serious consideration as a potential “green” biofuel. Problem 46(d) describes this development in more detail. Exercise 20-12 Name the following esters. O
(a)
O O B B (b) CH3OCCH2CH2COCH3
O
This couple is blissfully ignorant of the fact that they are attracted to each other by the same ester pheromone as moths.
(c) CH2 P CHCO2CH3
In industry, lower esters, such as ethyl acetate (b.p. 778C) and butyl acetate (b.p. 1278C), are used as solvents. For example, butyl butanoate has replaced the ozone-depleting trichloroethane as a cleaning solvent in the manufacture of electronic components such as computer chips. Higher nonvolatile esters are used as softeners (called plasticizers; see Section 12-15) for brittle polymers — in flexible tubing (e.g., Tygon tubing), rubber pipes, and upholstery.
Esters hydrolyze to carboxylic acids; the leaving-group problem Esters undergo nucleophilic substitution reactions by means of addition–elimination pathways, albeit with reduced reactivity relative to halides and anhydrides. Thus, catalysis by acid or base becomes a frequent necessity. For example, esters are cleaved to carboxylic acids and alcohols in the presence of excess water and strong acid, and the reaction requires heating to proceed at a reasonable rate. The mechanism of this transformation is the reverse of acid-catalyzed esterification (Section 19-9). As in esterification, the acid serves two purposes: It protonates the carbonyl oxygen to make the ester more reactive toward nucleophilic attack, and it protonates the alkoxy oxygen in the tetrahedral intermediate to make it a better leaving group. Exercise 20-13
Working with the Concepts: Ester Hydrolysis in Acid Formulate a mechanism for the acid-catalyzed hydrolysis of g-butyrolactone. Strategy Find the structure of g-butyrolactone (on the previous page). It is a cyclic ester. Find the general mechanism for addition – elimination under acidic conditions (Section 19-7). Substitute g-butyrolactone for the general starting material in that mechanism, and proceed. Solution • Step by step: Begin with protonation of the carbonyl oxygen, H ðOð ðO ðOð
H
ðOð
• Continue with addition – elimination. Water is the nucleophile. Use protonation to turn the ring oxygen into a good leaving group. In the tetrahedral intermediate, cleave the bond between the (former) carbonyl carbon and the ring oxygen.
ðO H2š O
H
H H
ðOð
O k
H š O ðOð
H ýO k ýOk H H ðO
H H
š OH š OH
• Deprotonate the carbonyl oxygen. The mechanism is completed. You have your product. H
ðOð
Oð š HO
š OH
š HO
š OH
Oð
H
938
Chapter 20
Carboxylic Acid Derivatives
Strong bases also promote ester hydrolysis through an addition – elimination mechanism (Sections 19-7 and 20-1). The base (B) converts the poor nucleophile water into the negatively charged and more highly nucleophilic hydroxide ion. Bð
ðOH
HO OH
B OH
Ester hydrolysis is frequently achieved by using hydroxide itself, in at least stoichiometric amounts, as the base. REACTION
Example of Ester Hydrolysis Using Aqueous Base O
O 1. KOH, H2O, CH3OH, 2. H, H2O
OCH3
OH
CH3OH
100% Methyl 3-methylbutanoate
3-Methylbutanoic acid
Mechanism of Base-Mediated Ester Hydrolysis
MECHANISM
Step 1. Addition – elimination ðOð B š RCOCH 3
Oð ðš A RO COš OCH3 A ð OH
ðš OH
ðOð B š RCOH
Methoxide: poor leaving group
ðš OCH3
Tetrahedral intermediate
Step 2. Deprotonation ðOð B š RCO OH Carboxylic acid
ðš OCH3
ðOð B š RCO ð
Methoxide (Strong base)
Carboxylate ion
š HOCH 3
Favorable acid-base reaction of step 2 drives equilibrium of step 1
Ester hydrolysis in base (saponification, see Section 19-13) differs from the acid-catalyzed version in several ways. Consider the addition – elimination (step 1) first: The tetrahedral intermediate can revert to starting material by release of a hydroxide ion, or it can go forward toward product through expulsion of methoxide. Either way, the process requires loss of a poor leaving group. Both hydroxide and methoxide are strong bases, and strong bases are poor leaving groups (Section 6-7). How can these transformations take place at all? The answer lies in part in the strength of the C P O bond and the associated stability of carbonyl compounds. The tetrahedral intermediate is so much higher in energy that its conversion to a carbonyl compound is typically exothermic even when a poor leaving group is expelled. In addition, the base-mediated process benefits from a very favorable acid-base reaction in step 2: The strong base methoxide, having been released at the addition – elimination stage, rapidly deprotonates the acid to give carboxylate ion. This step is very favorable and drives the entire hydrolysis sequence essentially to completion. Subsequent acidic aqueous work-up gives the carboxylic acid product.
Exercise 20-14
Try It Yourself Formulate a mechanism for the base-mediated hydrolysis of g-butyrolactone.
20-4 Chemistry of Esters
Transesterification takes place with alcohols
Transesterification
Esters react with alcohols in an acid- or base-catalyzed transformation called transesterification. This allows for the direct conversion of one ester into another without proceeding through the free acid. Like esterification, transesterification is a reversible reaction. To shift the equilibrium, a large excess of the alcohol is usually employed, sometimes in the form of solvent.
O B RCOR
Ethyl octadecanoate
H or OCH3
CH3OH Methanol (Solvent)
O B C17H35COCH3 90%
O B RCOR
CH3CH2OH
Methyl octadecanoate
Alkoxy groups trade places
Lactones are opened to hydroxy esters by transesterification. Conversion of a Lactone into an Open-Chain Ester O O
Br
H
OH
O
HO
Br
O 80% ␥-Butyrolactone
3-Bromopropanol
3-Bromopropyl 4-hydroxybutanoate
The mechanisms of transesterification by acid and base are analogous to the mechanisms of the corresponding hydrolyses to the carboxylic acids. Thus, acid-catalyzed transesterification begins with protonation of the carbonyl oxygen, followed by nucleophilic attack of the alcohol on the carbonyl carbon. In contrast, under basic conditions the alcohol is first deprotonated, and the resulting alkoxide ion then adds to the ester carbonyl group. Exercise 20-15 Formulate mechanisms for the acid- and base-catalyzed transesterifications of g-butyrolactone by 3-bromo-1-propanol.
Amines convert esters into amides Amines, which are more nucleophilic than alcohols, readily transform esters into amides. No catalyst is needed, but heating is required. Amide Formation from Methyl Esters O B RCOCH3
H RNH
O B RCNHR
CH3OH
Uncatalyzed Conversion of an Ester into an Amide O B CH3(CH2)7CHP CH(CH2)7COCH3 Methyl 9-octadecenoate
H CH3(CH2)11NH 1-Dodecanamine
R OH
H or OR
Conversion of an Ethyl Ester into a Methyl Ester O B C17H35COCH2CH3
939
Chapter 20
230°C (See Problem 43)
O B CH3(CH2)7CHP CH(CH2)7CNH(CH2)11CH3 69% N-Dodecyl-9-octadecenamide
CH3OH
ROH
940
Chapter 20
Carboxylic Acid Derivatives
Grignard reagents transform esters into alcohols Esters can be converted into alcohols by using two equivalents of a Grignard reagent. In this way, ordinary esters are transformed into tertiary alcohols, whereas formate esters furnish secondary alcohols. REACTION Alcohols from Esters and Grignard Reagents O B CH3CH2COCH2CH3
2 CH3CH2CH2MgBr
Ethyl propanoate
O B HCOCH3
Methyl formate
1. (CH3CH2)2O 2. H, H2O CH3CH2OH
Propylmagnesium bromide
2 CH3CH2CH2CH2MgBr
OH A CH3CH2CCH2CH2CH3 A CH2CH2CH3 69%
Two identical alkyl groups, derived from Grignard reagent
4-Ethyl-4-heptanol
1. (CH3CH2)2O 2. H, H2O CH3OH
Butylmagnesium bromide
OH A HCCH2CH2CH2CH3 A CH2CH2CH2CH3 85% 5-Nonanol
The reaction begins with addition of the organometallic to the carbonyl function in the usual manner to give the magnesium salt of a hemiacetal (Section 17-7). At room temperature, rapid elimination results in the formation of an intermediate ketone (or aldehyde, from formates). The resulting carbonyl group then immediately adds a second equivalent of Grignard reagent (recall Section 8-8). Subsequent acidic aqueous work-up leads to the observed alcohol.
MECHANISM
Mechanism of the Alcohol Synthesis from Esters and Grignard Reagents ðOð B C E H R OCH3
O O MgBr A RO C O OCH3 A R
ROMgBr
Ester
Tetrahedral intermediate
ðOð B EC H R R Ketone (Reaction cannot be stopped at this stage)
O B C6H5COCH3 Methyl benzoate
Fast CH3OMgBr
ROMgBr Fast
OMgBr A RO C O R A R Alkoxide
H , H2 O
OH A RO C O R A R Alcohol (Final product)
Exercise 20-16 Propose a synthesis of triphenylmethanol, (C6H5)3COH, beginning with methyl benzoate (shown in the margin) and bromobenzene.
20-4 Chemistry of Esters
Chapter 20
Esters are reduced by hydride reagents to give alcohols or aldehydes The reduction of esters to alcohols by LiAlH4 requires 0.5 equivalent of the hydride, because only two hydrogens are needed per ester function. Reduction of an Ester to an Alcohol O B NCHCOCH2CH3 A CH3
1. LiAlH4 (0.5 equivalent), (CH3CH2)2O 2. H , H2O
NCHCH2OH A CH3 90%
CH3CH2OH
A milder reducing agent allows the reaction to be stopped at the aldehyde oxidation stage. Such a reagent is bis(2-methylpropyl)aluminum hydride (diisobutylaluminum hydride) when used at low temperatures in toluene. Reduction of an Ester to an Aldehyde H3C O A B CH3CHCOCH2CH3 Ethyl 2-methylpropanoate
CH3 A (CH3CHCH2)2AlH
1. Toluene, 60C 2. H , H2O
Bis(2-methylpropyl)aluminum hydride (Diisobutylaluminum hydride, DIBAL)
CH3CH2OH
CH3 A CH3CHCHO 2-Methylpropanal
Esters form enolates that can be alkylated The acidity of the a-hydrogens in esters is sufficiently high that ester enolates are formed by treatment of esters with strong base at low temperatures. Ester enolates react like ketone enolates, undergoing alkylations. Alkylation of an Ester Enolate O B CH3COCH2CH3 pKa
LDA, THF, 78C
25 Nucleophilic carbon
OðLi ðš f CH2 P C i OCH2CH3 Ethyl acetate enolate ion
CH2 P CHCH2 OBr, HMPA LiBr
O B CH2 P CHCH2CH2COCH2CH3 97% Ethyl 4-pentenoate
The pKa of esters is about 25. Consequently, ester enolates exhibit the typical side reactions of strong bases: E2 processes (especially with secondary, tertiary, and b-branched halides) and deprotonations. The most characteristic reaction of ester enolates is the Claisen condensation, in which the enolate attacks the carbonyl carbon of another ester. This process will be considered in Chapter 23. Exercise 20-17 Give the products of the reaction of ethyl cyclohexanecarboxylate with the following compounds or under the following conditions (and followed by acidic aqueous work-up, if necessary). (a) H1, H2O; (b) HO2, H2O; (c) CH3O2, CH3OH; (d) NH3, D; (e) 2 CH3MgBr; (f) LiAlH4; (g) 1. LDA, 2. CH3I.
In Summary Esters are named as alkyl alkanoates. Many of them have pleasant odors and are present in nature. They are less reactive than acyl halides or carboxylic anhydrides and therefore often require the presence of acid or base to transform. With water, esters hydrolyze to the corresponding carboxylic acids or carboxylates; with alcohols, they undergo transesterification and, with amines at elevated temperatures, furnish amides.
941
942
Chapter 20
Carboxylic Acid Derivatives
Grignard reagents add twice to give tertiary alcohols (or secondary alcohols, from formates). Lithium aluminum hydride reduces esters all the way to alcohols, whereas bis(2methylpropyl)aluminum (diisobutylaluminum) hydride allows the process to be stopped at the aldehyde stage. With LDA, it is possible to form ester enolates, which can be alkylated by electrophiles.
20-5 Esters in Nature: Waxes, Fats, Oils, and Lipids Esters are essential components in the cells of all living organisms. This section introduces several of the most common types of natural esters with brief descriptions of their biological functions.
Waxes are simple esters, whereas fats and oils are more complex
CH2OH A CHOH A CH2OH 1,2,3-Propanetriol (Glycerol)
O B CH2OCR O B HCOCR O B CH2OCR 1,2,3-Propanetriol triester (Triglyceride)
Esters made up of long-chain carboxylic acids and long-chain alcohols constitute waxes, which form hydrophobic (Section 8-2) and insulating coatings on the skin and fur of animals, the feathers of birds, and the fruits and leaves of many plants. Spermaceti and beeswax are liquids or very soft solids at room temperature and are used as lubricants. Sheep’s wool provides wool wax, which when purified yields lanolin, a widely used cosmetic base. The leaves of a Brazilian palm are the source of carnauba wax, a tough and water-resistant mixture of several solid esters. Carnauba wax is highly valued for its ability to take and maintain a high gloss and is used as a floor and automobile wax. O B CH3(CH2)14CO(CH2)15CH3
O B CH3(CH2)nCO(CH2)mCH3 n ⴝ 24, 26; m ⴝ 29, 31
Hexadecyl hexadecanoate (Cetyl palmitate)
Beeswax
(Wax from the sperm whale)
Triesters of 1,2,3-propanetriol (glycerol) with long-chain carboxylic acids constitute fats and oils (Sections 11-3 and 19-13). They are also called triglycerides. The acids in fats and oils (fatty acids) are typically unbranched and contain an even number of carbon atoms; if the fats are unsaturated, the double bonds are usually cis. Fats are biological energy reserves, which are stored in the body’s tissues until their metabolism leads ultimately to CO2 and water. Oils have a similar function in plant seeds. As food components, fats and oils serve as solvents for food flavors and colors and contribute to a sense of “fullness” after eating, because they leave the stomach relatively slowly. Saturated fats containing hexadecanoic (palmitic), tetradecanoic (myristic), and dodecanoic (lauric) acids have been implicated as dietary factors in atherosclerosis (hardening of the arteries; Chemical Highlight 4-2). Fortunately for lovers of chocolate, cocoa butter is mainly a low-melting triglyceride that contains two molecules of octadecanoic (stearic) acid — which, despite being saturated, does not contribute to elevated low-density lipoprotein (LDL) levels or atherosclerosis — and one molecule of (Z)-9-octadecenoic (oleic) acid. The latter is also the chief fatty acid component of olive oil, the major source of dietary fat among the Greeks, who have very low rates of heart disease.
Lipids are biomolecules soluble in nonpolar solvents A “chocoholic’s” dream.
Extraction of biological material with nonpolar solvents gives a wide assortment of compounds that includes terpenes and steroids (Section 4-7), fats and oils, and a variety of other low-polarity substances collectively called lipids (lipos, Greek, fat). Lipid fractions include phospholipids, important components of cell membranes, which are derived from carboxylic acids and phosphoric acid. In the phosphoglycerides, glycerol is esterified with two adjacent fatty acids and a phosphate unit that bears another substituent derived from a
20-5 Esters in Nature: Waxes, Fats, Oils, and Lipids
943
Chapter 20
low-molecular-weight alcohol such as choline, [HOCH2CH2N(CH3)3]1 2OH. The substance shown here is an example of a lecithin, a lipid found in the brain and nervous system. Hexadecanoic (palmitic) acid unit
O B H2 C O O O C O R O A B RO C O O O C OH O A B H2C O O O P O OR A O
O B O CH2OC(CH2)14CH3 cis B A CH3(CH2)7CHP CH(CH2)7COCH O A B cis-9-Octadecenoic O CH (CH2)2N(CH3)3 PO 2 (oleic) acid unit A O Choline unit
A phosphoglyceride
Palmitoyloleoylphosphatidylcholine (A lecithin)
Compounds such as these play roles in the transmission of nerve impulses and exhibit other biological effects. For example, a lipid containing the amide analog of oleic acid [(Z)-9octadecenamide] was recently identified as an essential sleep-inducing agent in the brain; indeed, persons with lipid-restricted diets have difficulty experiencing deep sleep. Because these molecules carry two long hydrophobic fatty acid chains and a polar head group (the phosphate and choline substituent), they are capable of forming micelles in aqueous solution (see Chemical Highlight 19-1). In the micelles, the phosphate unit is solubilized by water, and the ester chains are clustered inside the hydrophobic micellar sphere (Figure 20-2A). Phosphoglycerides can also aggregate in a different way: They may form sheets called a lipid bilayer (Figure 20-2B). This capability is significant because, whereas micelles are usually limited in size (,200 Å in diameter), bilayers may be as much as 1 mm (107 Å) in length. This property makes them ideal constituents of cell membranes, which act as permeability barriers regulating molecular transport into and out of the cell. Lipid bilayers are relatively stable molecular assemblies. The forces that drive their formation are similar to those at work in micelles: London interactions (Section 2-6) between the hydrophobic alkane chains, and coulombic and solvation forces between the polar head groups and between these polar groups and water.
In Summary Waxes, fats, and oils are naturally occurring, biologically active esters. Lipids are a broad class of biological molecules soluble in nonpolar solvents. They include glycerolderived triesters, which are components of cell membranes. Polar head O−
Inner aqueous compartment
+
O H2C
O Nonpolar tails
CH
O
O
C
C
R
R'
P
OCH2CH2N(CH3)3
CH2 O
O
Bilayer membrane
O A C E H R L
E )> O A C E N R L
B
C
A
L 5 leaving group Basicity increases with increasing contribution of resonance structure C.
3. Enolate Formation (Section 20-1) ðOð B EC H
RCH2
ðš Oð
Base
RCH
Lð
B š HC RCH
C
š Oð
Lð
š HC RCH B
Lð
A
B
C
Acidity of the neutral derivative generally increases with decreasing contribution of resonance structure C in the anion.
Reactions of Acyl Halides 4. Water (Section 20-2) O B RCX
O B RCOH
H2O
HX
Carboxylic acid
5. Carboxylic Acids (Sections 19-8 and 20-2) O B RCX
O O B B RCOCR
RCO2H
HX
Carboxylic anhydride
6. Alcohols (Section 20-2) O B RCX
ROH
O B RCOR Ester
ðš Oð
E
O
E
%
%
O
g. The ring carbonyl group has been deprotected, and the carboxy group transformed into a ketone. For the first time in the synthesis, a new carbon – carbon bond has been introduced, between the carboxy carbon and an ethyl group. This process will most likely require the use of an organometallic reagent, from which the ring carbonyl has been protected in sequence (f). From which carboxylic acid derivatives can ketones be prepared? Acyl halides (Section 20-2) and nitriles (Section 20-8). Since you don’t know how to convert a carboxylic acid into its nitrile, we choose the following sequence for (g): 1. SOCl2 (gives halide), 2. (CH3CH2)2CuLi, ether (converts halo- into ethylcarbonyl), and 3. H1, H2O (deprotects ring carbonyl, Section 17-8). h. Another carbon–carbon bond is made, forming a ring. We trace the pattern of carbon atoms before and after this step to identify which ones were connected (see margin). They are the ring carbonyl carbon and the CH2 of the ethyl group, a to the side-chain carbonyl. We have here an intramolecular aldol condensation (Section 18-7), readily achieved by aqueous base. So (h), the final step, is 2OH, H2O.
B
Analysis of Step (h) Aldol Condensation
Carboxylic Acid Derivatives
E H
Chapter 20
%
960
HX (Removed with pyridine, triethylamine, or other base)
L
New Reactions
7. Amines (Section 20-2) O B RCX
O B RCNHR
RNH2
Amide
HX (Removed with pyridine, triethylamine, excess RⴕNH2, or other base)
8. Cuprate Reagents (Section 20-2) O B RCX
O B RCR
1. R2CuLi, THF 2. H, H2O
RCu
LiX
Ketone
9. Hydrides (Section 20-2) O B RCX
O B RCH
1. LiAl[OC(CH3)3]3H, (CH3CH2)2O 2. H, H2O
LiX
Al[OC(CH3)3]3
Aldehyde
Reactions of Carboxylic Acid Anhydrides 10. Water (Section 20-3) O O B B RCOCR
O B 2 RCOH
H2O
Carboxylic acid
11. Alcohols (Section 20-3) O O B B RCOCR
O B RCOR
ROH
O B RCOH
Ester
12. Amines (Section 20-3) O O B B RCOCR
O B RCOH
O B RCNHR
RNH2
Amide
Reactions of Esters 13. Water (Ester Hydrolysis) (Sections 19-9 and 20-4) Acid catalysis O B RCOR
O B RCOH
Catalytic H
H2O
ROH
ROH
Carboxylic acid
Base catalysis O B RCOR
O B RCO
H2O
OH
Carboxylate ion
1 equivalent
14. Alcohols (Transesterification) and Amines (Section 20-4) O B RCOR
ROH
H or OR
O B RCOR
ROH
Ester
O B RCOR
RNH2
O B RCNHR Amide
ROH
Chapter 20
961
Chapter 20
Carboxylic Acid Derivatives
15. Organometallic Reagents (Section 20-4) O B RCOR
1. 2 RMgX, (CH3CH2)2O 2. H, H2O
OH A R O C O R A R
ROH
Tertiary alcohol
Methyl formate O B HCOCH3
1. 2 RMgX, (CH3CH2)2O 2. H, H2O
OH A HO C O R A R
CH3OH
Secondary alcohol
16. Hydrides (Section 20-4) O B RCOR
1. LiAlH4, (CH3CH2)2O 2. H, H2O
RCH2OH Alcohol
O B RCOR
CH3 A 1. (CH3CHCH2)2AlH, toluene, 60 C 2. H, H2O
O B RCH Aldehyde
17. Enolates (Section 20-4) O B RCH2COR
šð ðO A RCH COR
ðOð B š H COR RCH
LDA, THF
RX
B
962
Ester enolate ion
Reactions of Amides 18. Water (Section 20-6) O B RCNHR
H2O
H,
O B RCOH
RNH3
Carboxylic acid
O B RCNHR
H 2O
O B RCO
HO,
RNH2
19. Hydrides (Section 20-6) O B RCNHR
1. LiAlH4, (CH3CH2)2O 2. H, H2O
RCH2NHR Amine
O B RCNHR
CH3 A 1. (CH3CHCH2)2AlH, (CH3CH2)2O 2. H , H2O
O B RCH Aldehyde
20. Enolates and Amidates (Section 20-7) ðOð B RCH2CNR2 pKa
Base
30
Amide enolate ion
ðOð B š RCH2CNHR pKa
š Oð D RCHP C G NR2
22
Base
ðš Oð A NR RCH2C Pš Amidate ion
R O A B RCHCOR
Important Concepts
21. Hofmann Rearrangement (Section 20-7) O B RCNH2
Br2, NaOH, H2O, 75°C
RNH2
CO2
Amine
Reactions of Nitriles 22. Water (Section 20-8)
RC q N
H 2O
O B RCNH2
H or HO,
H or HO,
Amide
O B RCOH Carboxylic acid
23. Organometallic Reagents (Section 20-8) 1. RMgX or RLi 2. H, H2O
RC q N
O B RCR Ketone
24. Hydrides (Section 20-8) RC q N
1. LiAlH4 2. H, H2O
RCH2NH2 Amine
RC q N
CH3 A 1. (CH3CHCH2)2AlH 2. H, H2O
O B RCH Aldehyde
25. Catalytic Hydrogenation (Section 20-8) RC q N
H2, PtO2, CH3CH2OH
RCH2NH2 Amine
Important Concepts 1. The electrophilic reactivity of the carbonyl carbon in carboxylic acid derivatives is weakened by good electron-donating substituents. This effect, measurable by IR spectroscopy, is responsible not only for the decrease in the reactivity with nucleophiles and acid, but also for the increased basicity along the series: acyl halides – anhydrides – esters – amides. Electron donation by resonance from the nitrogen in amides is so pronounced that there is hindered rotation about the amide bond on the NMR time scale. 2. Carboxylic acid derivatives are named as acyl halides, carboxylic anhydrides, alkyl alkanoates, alkanamides, and alkanenitriles, depending on the functional group. 3. Carbonyl stretching frequencies in the IR spectra are diagnostic of the carboxylic acid derivatives: Acyl chlorides absorb at 1790 – 1815 cm21, anhydrides at 1740 – 1790 and 1800 – 1850 cm21, esters at 1735 – 1750 cm21, and amides at 1650 – 1690 cm21. 4. Carboxylic acid derivatives generally react with water (under acid or base catalysis) to hydrolyze to the corresponding carboxylic acid; they combine with alcohols to give esters and with amines to furnish amides. With Grignard and other organometallic reagents, they form ketones; esters may react further to form the corresponding alcohols. Reduction by hydrides gives products in various oxidation states: aldehydes, alcohols, or amines. 5. Long-chain esters are the constituents of animal and plant waxes. Triesters of glycerol are contained in natural oils and fats. Their hydrolysis gives soaps. Triglycerides containing phosphoric acid ester subunits belong to the class of phospholipids. Because they carry a highly polar head group and hydrophobic tails, phospholipids form micelles and lipid bilayers. 6. Transesterification can be used to convert one ester into another.
Chapter 20
963
964
Carboxylic Acid Derivatives
Chapter 20
7. The functional group of nitriles is somewhat similar to that of the alkynes. The two component atoms are sp hybridized. The IR stretching vibration appears at about 2250 cm21. The hydrogens next to the cyano group are deshielded in 1H NMR. The 13C NMR absorptions for nitrile carbons are at relatively low field (d , 112–126 ppm), a consequence of the electronegativity of nitrogen.
Problems 30. Name (IUPAC system) or draw the structure of each of the following compounds. COCl
O (a)
(b)
I O O B B COCCH2CH3
(d)
CH3
(g) Propyl butanoate ( j) N, N-Dimethylbenzamide
O O B B (c) CF3COCCF3
O B (e) (CH3)3CCOCH2CH3
O B (f) CH3CNH
(h) Butyl propanoate (k) 2-Methylhexanenitrile
(i) 2-Chloroethyl benzoate (l) Cyclopentanecarbonitrile
31. Name each of the following compounds according to IUPAC or Chemical Abstracts rules. Pay attention to the order of precedence of the functional groups. O (a)
OCH3
(b)
CONH2
O
OH CO2CH2CH3
(c)
Cl
NHCH3
(d) H
O
CONH2
O 32. (a) Use resonance forms to explain in detail the relative order of acidity of carboxylic acid derivatives, as presented in Section 20-1. (b) Do the same, but use an argument based on inductive effects. O B CH3CN(CH3)2 i
O O B B CH3CNCCH3 A CH3 ii
33. In each of the following pairs of compounds, decide which possesses the indicated property to the greater degree. (a) Length of C – X bond: acetyl fluoride or acetyl chloride. (b) Acidity of the boldface H: CH2(COCH3)2 or CH2(COOCH3)2. (c) Reactivity toward addition of a nucleophile: (i) an amide or (ii) an imide (as shown in the margin). (d) High-energy infrared carbonyl stretching frequency: ethyl acetate or ethenyl acetate. 34. Give the product(s) of each of the following reactions. O B (a) CH3CCl 2
(b)
冢
冢
NH2
O CuLi
THF
Cl
2
H3C O A B (c) H3C O C O CCl A H 3C
LiAl[OC(CH3)3]3H, THF, 78°C
OCH3 H3C
(e) H3C
LiAl[OC(CH3)3]3H, THF, 78°C
CCl B O
O B CH2 O CCl (d) A CH2 O CCl B O
2
CH2OH,
N
Problems
Chapter 20
35. Formulate a mechanism for the reaction of acetyl chloride with 1-propanol shown on p. 932. 36. Give the product(s) of the reactions of acetic anhydride with each of the following reagents. Assume in all cases that the reagent is present in excess. (b) NH3
(a) (CH3)2CHOH (c)
MgBr, THF; then H, H2O
(d) LiAlH4, (CH3CH2)2O; then H1, H2O
37. Give the product(s) of the reaction of butanedioic (succinic) anhydride with each of the reagents in Problem 36. 38. Formulate a mechanism for the reaction of butanedioic (succinic) anhydride with methanol shown on p. 935. 39. Give the products of reaction of methyl pentanoate with each of the following reagents under the conditions shown. (a) NaOH, H2O, heat; then H1, H2O (c) (CH3CH2)2NH, heat (e) LiAlH4, (CH3CH2)2O; then H1, H2O
(b) (CH3)2CHCH2CH2OH (excess), H1 (d) CH3MgI (excess), (CH3CH2)2O; then H1, H2O (f) [(CH3)2CHCH2]2AlH, toluene, low temperature; then H1, H2O
40. Give the products of reaction of g-valerolactone (5-methyloxa-2-cyclopentanone, Section 20-4) with each of the reagents in Problem 39. 41. Draw the structure of each of the following compounds: (a) b-butyrolactone; (b) b-valerolactone; (c) d-valerolactone; (d) b-propiolactam; (e) a-methyl-d-valerolactam; (f) N-methyl-g-butyrolactam. 42. Formulate a mechanism for the acid-catalyzed transesterification of ethyl 2-methylpropanoate (ethyl isobutyrate) into the corresponding methyl ester. Your mechanism should clearly illustrate the catalytic role of the proton. 43. Formulate a mechanism for the reaction of methyl 9-octadecenoate with 1-dodecanamine shown on p. 939. 44. Reaction review. Suggest reagents to convert each of the following starting materials into the indicated product: (a) hexanoyl chloride into acetic hexanoic anhydride; (b) methyl hexanoate into N-methylhexanamide; (c) hexanoyl chloride into hexanal; (d) hexanenitrile into hexanoic acid; (e) hexanamide into hexanamine; (f) hexanamide into pentanamine; (g) ethyl hexanoate into 3-ethyl-3-octanol; (h) hexanenitrile into 1-phenyl-1-hexanone [C6H5CO(CH2)4CH3]. 45. Give the product of each of the following reactions. H3C
0 COOCH3
O
(CH3)2CHNH2, CH3OH,
CH3CH2
O B (c) CH3COCH3 excess
(d)
O
(b) ´
(a)
1. KOH, H2O 2. H, H2O
%
COOCH3 %
MgBr
CH D 3 CH G COOCH2CH3
1. (CH3CH2)2O, 20°C 2. H, H2O
1. LDA, THF, 78°C 2. CH3I, HMPA 3. H, H2O
COOCH3 (e)
CH3 A 1. (CH3CHCH2)2AlH, toluene, 60°C 2. H , H2O
46. For each of the naturally occurring lactones below, draw the structure of the compound that would result from hydrolysis using aqueous base.
Sedanenolide, major contributor to the flavor of celery (a)
O O
965
966
Chapter 20
Carboxylic Acid Derivatives
H ´ (b)
O
Nepetalactone, major active component in catnip (Caution: Do you notice anything unusual about one of the functional groups after hydrolysis?
≥ H O O
(c)
CH3 O (CH3CH2)2N
CO2CH3
´
N H
0
DEET
g-Valerolactone, impotant in the fragrance industry and a potential biofuel
O
(d) A novel “green” process converts levulinic acid (4-oxopentanoic acid, CH3COCH2CH2COOH), which is readily obtained from waste biomass, into g-valerolactone by treatment with gaseous H2 in the presence of a special hydrogenation catalyst. Speculate on how this reaction may take place. 47. N,N-Diethyl-3-methylbenzamide (N,N-diethyl-m-toluamide), marketed as DEET, is perhaps the most widely used insect repellent in the world and is especially effective in interrupting the transmission of diseases by mosquitoes and ticks. Propose one or more preparations of DEET (see structure, margin) from the corresponding carboxylic acid and any other appropriate reagents. 48. Like many chiral pharmaceuticals, Ritalin, widely used to treat attention deficit disorder, is synthesized and marketed as a racemic mixture. In 2003, a new and potentially practical synthetic route to the active enantiomer (see structure, margin) was described. The first step is shown below, the reaction of dimethyl sulfate with a six-membered ring lactam. Propose a mechanism for this reaction.
C6H5
O B CH3O O S O OCH3 B O
Ritalin (Active stereoisomer)
N H
O
N
OCH3
O B 49. Formulate a mechanism for the formation of acetamide, CH3CNH2, from methyl acetate and ammonia. 50. Give the products of the reactions of pentanamide with the reagents given in Problem 39(a, e, f). Repeat for N,N-dimethylpentanamide. 51. Formulate a mechanism for the acid-catalyzed hydrolysis of 3-methylpentanamide shown on p. 948. (Hint: Use as a model the mechanism for general acid-catalyzed addition – elimination presented in Section 19-7.) 52. What reagents would be necessary to carry out the following transformations? (a) Cyclohexanecarbonyl chloride y pentanoylcyclohexane; (b) 2-butenedioic (maleic) anhydride y (Z)-2-butene1,4-diol; (c) 3-methylbutanoyl bromide y 3-methylbutanal; (d) benzamide y 1-phenylmethanamine; (e) propanenitrile y 3-hexanone; (f) methyl propanoate y 4-ethyl-4-heptanol. 53. Nylon-6,6 was the first completely synthetic fiber, initially prepared in the mid-1930s at DuPont. It is a copolymer whose subunits are alternating molecules of 1,6-hexanediamine and hexanedioic (adipic) acid, linked by amide functions. (a) Write a structural representation of Nylon-6,6. (b) Hexanedinitrile (adiponitrile) may be used as a common precursor to both 1,6-hexanediamine and hexanedioic acid. Show the reactions necessary to convert the dinitrile into both the diamine and the diacid. (c) Formulate a mechanism for the conversion into the diacid under the reaction conditions that you chose in (b). 54. Upon treatment with strong base followed by protonation, compounds A and B undergo cis-trans isomerization, but compound C does not. Explain. [ COOCH3
[ CON(CH3)2
[ CONH2
)
)
)
COOCH3
A
CON(CH3)2 B
C
CONH2
Problems
967
Chapter 20
55. 2-Aminobenzoic (anthranilic) acid is prepared from 1,2-benzenedicarboxylic anhydride (phthalic anhydride) by using the two reactions shown here. Explain these processes mechanistically.
O
O B COH
O NH3, 300°C
O
1. NaOH, Br2, 80°C 2. H, H2O
NH
NH2 O
O
1,2-Benzenedicarboxylic anhydride (Phthalic anhydride)
1,2-Benzenedicarboximide (Phthalimide)
2-Aminobenzoic acid (Anthranilic acid)
56. On the basis of the reactions presented in this chapter, write reaction summary charts for esters and amides similar to the chart for acyl halides (Figure 20-1). Compare the number of reactions for each of the compound classes. Is this information consistent with your understanding of the relative reactivity of each of the functional groups? 57. Show how you might synthesize chlorpheniramine, a powerful antihistamine used in several decongestants, from each of carboxylic acids A and B. Use a different carboxylic amide in each synthesis.
N
N
N
CHCH2COOH
Cl
CHCH2CH2COOH
Cl
CHCH2CH2N(CH3)2
Cl B
A
Chlorpheniramine
58. Although esters typically have carbonyl stretching frequencies at about 1740 cm21 in the infrared spectrum, the corresponding band for lactones can vary greatly with ring size. Three examples are shown in the margin. Propose an explanation for the IR bands of these smaller ring lactones. 59. Upon completing a synthetic procedure, every chemist is faced with the job of cleaning glassware. Because the compounds present may be dangerous in some way or have unpleasant properties, a little serious chemical thinking is often beneficial before “doing the dishes.” Suppose that you have just completed a synthesis of hexanoyl chloride, perhaps to carry out the reaction in Problem 34(b); first, however, you must clean the glassware contaminated with this acyl halide. Both hexanoyl chloride and hexanoic acid have terrible odors. (a) Would cleansing the glassware with soap and water be a good idea? Explain. (b) Suggest a more pleasant alternative, based on the chemistry of acyl halides and the physical properties (particularly the odors) of the various carboxylic acid derivatives. 60.
Show how you would carry out the following transformation in which the ester function at the lower left of the molecule is converted into a hydroxy group but that at the upper right is preserved. (Hint: Do not try ester hydrolysis. Look carefully at how the ester groups are linked to the steroid and consider an approach based on transesterification.) CH3 CH3 %
O B C E H } H3C O
OCH3 D C M O
CH3 ?
CH3 %
CH3 %
CH3 % } HO
OCH3 D C M O
O
O
O
O
1735 cmⴚ1
1770 cmⴚ1
O O 1840 cmⴚ1
968
Chapter 20
Carboxylic Acid Derivatives
61. The removal of the C17 side chain of certain steroids is a critical element in the synthesis of a number of hormones, such as testosterone, from steroids in the relatively readily available pregnane family. CH3 H3C %
O
CH3 %OH 17 %
17
H3C H % ≥ % ≥ H H
H 3C H % ≥ % ≥ H H
Several steps
} HO
H3C
O
O Pregnan-3␣ -ol-20-one
OH
Testosterone
How would you carry out the comparable transformation, shown in the margin, of acetylcyclopentane into cyclopentanol? (Note: In this and subsequent synthetic problems, you may need to use reactions from several areas of carbonyl chemistry discussed in Chapters 17 – 20.)
?
62. Propose a synthetic sequence to convert carboxylic acid A into the naturally occurring sesquiterpene a-curcumene. CH3
CH3 ?
CO2H
CH3
CH3 H 3C
CH3 ␣ -Curcumene
A
63. Propose a synthetic scheme for the conversion of lactone A into amine B, a precursor to the naturally occurring monoterpene C. CH3
CH3 CH3 H3C
CH3
?
CH3
H3C
H 3C
CH3
HO
O
CH3
HO
O
N(CH3)2 A
64.
CH3
CH3
B
C
Propose a synthesis of b-selinene, a member of a very common family of sesquiterpenes, beginning with the alcohol shown here. Use a nitrile in your synthesis. Inspection of a model may help you choose a way to obtain the desired stereochemistry. (Is the 1-methylethenyl group axial or equatorial?) CH3 %
CH3 %
?
'OH
≥ H H2C
CH2
≥ H H2C
CH3  -Selinene
65. Give the structure of the product of the first of the following reactions, and then propose a scheme that will ultimately convert it into the methyl-substituted ketone at the end of the scheme. This example illustrates a common method for the introduction of methyl groups into synthetically prepared steroids. (Hint: It will be necessary to protect the carbonyl function.) O HCN
≥ H
C11H15NO IR: 1715, 2250 cmⴚ1
?
CH3 % ≥ H
O
Problems
66.
1H
Spectroscopic data for two carboxylic acid derivatives are given in NMR-A and NMR-B. Identify these compounds, which may contain C, H, O, N, Cl, and Br but no other elements. (a) 1H NMR: spectrum A (one signal has been amplified to reveal all peaks in the multiplet). IR: 1728 cm21. High-resolution mass spectrum: myz for the molecular ion is 116.0837. See table for important MS fragmentation peaks. (b) 1H NMR: spectrum B. IR: 1739 cm21. Highresolution mass spectrum: The intact molecule gives two peaks with almost equal intensity: myz 5 179.9786 and 181.9766. See table for important MS fragmentation peaks. 1H
NMR
NMR 3H
6H
3H
969
Chapter 20
Mass Spectrum of Unknown A m/z
Intensity relative to base peak (%)
116 101 75 57 43 29
0.5 12 26 100 66 34
3H 4.4 4.3 4.2
2H (CH3)4Si
2H
5.1 5.0 4.9
1H
1H
(CH3)4Si
Mass Spectrum of Unknown B 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0
5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0
300-MHz 1H NMR spectrum ppm (δ )
300-MHz 1H NMR spectrum ppm (δ ) B
A
Team Problem 67. Friedel-Crafts acylations are best carried out with acyl halides, but other carboxylic acid derivatives undergo this process, too, such as carboxylic anhydrides or esters. These reagents may have some drawbacks, however, the subject of this problem. Before you start, discuss as a group the mechanisms for forming acylium ions from acyl halides and carboxylic anhydrides in Section 15-13. Then divide your group in two and analyze the outcome of the following two reactions. Use the NMR spectral data given to confirm your product assignments. (Hint: D is formed via C.) O
O AlCl3
O
A (C8H8O) B (C10H12O)
O
AlCl3
O
A (C8H8O) C (C8H10) D (C10H12O)
Compound A: 1H NMR: d 2.60 (s, 3H), 7.40–7.50 (m, 2H), 7.50–7.60 (m, 1H), 7.90–8.00 (m, 2H). Compound B: 1H NMR: d 2.22 (d, 6H), 3.55 (sep, 1H), 7.40 – 7.50 (m, 2H), 7.50 – 7.60 (m, 1H), 7.90 – 8.00 (m, 2H). Compound C: 1H NMR: d 1.20 (t, 3H), 2.64 (q, 2H), 7.10 – 7.30 (m, 5H). Compound D: 1H NMR: d 1.25 (t, 3H), 2.57 (s, 3H), 2.70 (q, 2H), 7.20 (d, 2H), 7.70 (d, 2H).
m/z
Intensity relative to base peak (%)
182 180 109 107 101 29
13 13 78 77 3 100
970
Chapter 20
Carboxylic Acid Derivatives
Reconvene to share your solutions. Then specifically address the nature of the complications ensuing when using the reagents shown. Finally, all together, consider the following reaction sequence. Again, use a mechanistic approach to arrive at the structures of the products. O
O
AlCl3
C10H10O3
Zn(Hg), HCl,
C10H12O2
SOCl2
C10H11ClO
AlCl3
C10H10O
O
Preprofessional Problems F A (CH3)2CCO2CH(CH3)2
68. What is the IUPAC name of the compound shown in the margin? (a) Isopropyl 2-fluoro3-methylbutanoate; (b) 2-fluoroisobutanoyl 2-propanoate; (c) 1-methylethyl 2-fluorobutyrate; (d) 2-fluoroisopropyl isopropanoate; (e) 1-methylethyl 2-fluoro-2-methylpropanoate. O B 69. Saponification of (CH3)2CHC18OCH2CH2CH3 with aqueous NaOH will give (a) (CH3)2CHCO22Na1 1 CH3CH2CH218OH; O B (b) (CH3)2CHC18O2Na1 1 CH3CH2CH2OH; (c) (CH3)2CHOCH2CH2CH3 1 C q 18O; (d) (CH3)2CHCHO 1 CH3CH2CH218OH.
O J H 2C O C A A H2C O O A
70. The best description for compound A (see margin) is (a) an amide; (b) a lactam; (c) an ether; (d) a lactone. 71. Which of the three compounds below would be the most reactive toward hydrolysis with aqueous base? O B COCH3 (a)
O B CCl (b)
O B CNH2 (c)
C H A P T E R
[
TWENTY ONE
]
Amines and Their Derivatives
N
Functional Groups Containing Nitrogen
O O
O
ur atmosphere consists of about one-fifth oxygen, O2, and nearly four-fifths nitrogen, N2. We are fully aware of the importance of the oxygen part: We need it to breathe and nature uses it abundantly in water, alcohols, ethers, and many other organic and inorganic molecules. What about the nitrogen component? Unlike O2, which is ultimately the reactive ingredient in biological oxidations, N2 itself is relatively inert. However, in its reduced form of ammonia, NH3, and its organic derivatives, the amines, it plays as active a role in nature as oxygen. Thus, amines and other nitrogen-bearing compounds are among the most abundant organic molecules. As components of the amino acids, peptides, proteins, and alkaloids, they are essential to biochemistry. Many amines, such as the neurotransmitters, possess powerful physiological activity; related substances have found medicinal uses as decongestants, anesthetics, sedatives, and stimulants (Chemical Highlight 21-1). Similar activity is found in cyclic amines in which nitrogen is part of a ring, the nitrogen heterocycles (Chapter 25). In many respects, the chemistry of the amines is analogous to that of the alcohols and ethers (Chapters 8 and 9). For example, all amines are basic (although primary and secondary amines can also behave as acids), they form hydrogen bonds, and they act as nucleophiles in substitution reactions. However, there are some differences in reactivity, because nitrogen is less electronegative than oxygen. Thus, primary and secondary amines are less acidic and form weaker hydrogen bonds than alcohols, and they are more basic and more nucleophilic. This chapter will show that these properties underlie their physical and chemical characteristics and give us a variety of ways to synthesize amines.
Anxiety, by Edvard Munch (1863 – 1944). Effexor (venlafaxine; 1-[2(dimethylamino)-1-(4-methoxyphenyl)ethyl]cyclohexanol) was the top-selling drug in 2007 for the treatment of depression and anxiety. As shown in the molecular model, the OH proton forms a hydrogen bond to the lone pair of the amine nitrogen.
972
Chapter 21
Amines and Their Derivatives
21-1 Naming the Amines Amines are derivatives of ammonia, in which one, two, or three of the hydrogens have been replaced by alkyl or aryl groups. Therefore, amines are related to ammonia in the same sense as ethers and alcohols are related to water. Note, however, that the designations primary, secondary, and tertiary (see below) are used in a different way. In alcohols, ROH, the nature of the R group defines this designation; in amines, the number of R substituents on nitrogen determines the amine classification. H
D Dš N A H
H
Ammonia
Although higher organisms cannot activate nitrogen by reduction to ammonia, some microorganisms do. Thus, the nodules in the root system of the soybean are the sites of nitrogen reduction by Rhizobium bacteria.
R
D Dš N A H
R
H
Primary amine
D Dš N A R⬘
R
H
Secondary amine
D Dš N A R⬘
R⬙
Tertiary amine
The system for naming amines is confused by the variety of common names in the literature. Probably the best way to name aliphatic amines is that used by Chemical Abstracts — that is, as alkanamines, in which the name of the alkane stem is modified by replacing the ending -e by -amine. The position of the functional group is indicated by a prefix designating the carbon atom to which it is attached, as in the alcohols (Section 8-1).
CH3NH2 Methanamine
CH3 A CH3CHCH2NH2 2-Methyl-1-propanamine
H2N H *C (R)-trans-3-Penten-2-amine
Substances with two amine functions are diamines, two examples of which are 1,4butanediamine and 1,5-pentanediamine. Their contribution to the smell of dead fish and rotting flesh leads to their descriptive common names, putrescine and cadaverine, respectively. NH2
H2N
H2N
NH2 1,5-Pentanediamine (Cadaverine)
1,4-Butanediamine (Putrescine)
The aromatic amines, or anilines, are called benzenamines (Section 15-1). For secondary and tertiary amines, the largest alkyl substituent on nitrogen is chosen as the alkanamine stem, and the other groups are named by using the letter N-, followed by the name of the additional substituent(s). NH2 H A CH3NCH2CH3 The smell of rotting fish is due largely to putrescine.
CH3 A CH3NCH2CH2CH3
N
N
Benzenamine (Aniline) (A primary amine)
N-Methylethanamine
N,N-Dimethyl-1-propanamine
(A secondary amine)
(A tertiary amine)
21-2 Structural and Physical Properties of Amines
An alternative way to name amines treats the functional group, called amino-, as a substituent of the alkane stem. This procedure is analogous to naming alcohols as hydroxyalkanes. It is also used when other functional groups are present in the molecule, because the amine function has the lowest order of precedence of all functional groups discussed in this text (Section 19-1).
CH3CH2NH2
(CH3)2NCH2CH2CH3
Aminoethane
N,N-Dimethylaminopropane
973
Chapter 21
CH3NH2 Methylamine (CH3)3N Trimethylamine
N H
HO
3-(N-Ethylamino)-1-butanol
Many common names are based on the term alkylamine (see margin), as in the naming of alkyl alcohols.
N H3C
CH2
Benzylcyclohexylmethylamine
Exercise 21-1 Name each of the following molecules twice, first as an alkanamine, then as an alkyl amine. N(CH3)2 NH2 A (a) CH3CHCH2CH3 (b)
H3C H *C
(c) Br
NH2
Exercise 21-2 Draw structures for the following compounds (common name in parentheses): (a) 2-propynamine (propargylamine); (b) (N-2-propenyl)phenylmethanamine (N-allylbenzylamine); (c) N,2-dimethyl2-propanamine (tert-butylmethylamine).
In Summary There are several systems for naming amines. Chemical Abstracts uses names of the type alkanamine and benzenamine. Alternatives for alkanamine are the terms aminoalkane and alkylamine. The common name for benzenamine is aniline.
21-2 Structural and Physical Properties of Amines Nitrogen often adopts a tetrahedral geometry, similar to carbon, but with one lone electron pair instead of a fourth bond. Thus, amines generally have tetrahedral shapes around the heteroatom. However, this arrangement is not rigid because of a rapid isomerization process called inversion. Nitrogen is not as electronegative as oxygen, so while amines form hydrogen bonds, they are weaker than the hydrogen bonds formed by alcohols. As a result, amines have relatively low boiling points compared to alcohols.
Lone electron pair
1.01 Å
1.47 Å
The alkanamine nitrogen is tetrahedral In contrast to amides (Section 20-1), the nitrogen orbitals in amines are very nearly sp3 hybridized (see Section 1-8), forming an approximately tetrahedral arrangement. Three vertices of the tetrahedron are occupied by the three substituents, the fourth by the lone electron pair. As we shall see, it is this electron pair that is the source of the basic and nucleophilic properties of the amines. The term pyramidal is often used to describe the geometry adopted by the nitrogen and its three substituents. Figure 21-1 depicts the structure of methanamine (methylamine).
N
H 105.9⬚
H
CH3 112.9⬚
Figure 21-1 Nearly tetrahedral structure of methanamine (methylamine).
974
Chapter 21
Amines and Their Derivatives
CHE M I C AL H I G H L I G H T 2 1 - 1 Physiologically Active Amines and Weight Control A large number of physiologically active compounds owe their activity to the presence of amino groups. Several simple examples are well-known prescription or illegal drugs, such as adrenaline, Benzedrex, Urotropine, amphetamine, mescaline, and Effexor (see Chapter Opening). A recurring pattern in many (but not all) of these compounds is the 2-phenylethanamine (b-phenethylamine) unit — that is, a nitrogen connected by a two-carbon chain to a benzene
HNCH3
∑
H
nucleus (highlighted below in green where applicable). This structural feature appears to be crucial for the binding to brain receptor sites responsible for neurotransmitter action at certain nerve terminals. Such amines can affect a wide variety of behavior, from controlling appetite and muscular activity to creating potentially addictive euphoric stimulation.
HO /
N NHCH3 N
HO CH3
OH
N N
Epinephrine (Adrenaline)
Propylhexedrine (Benzedrex)
Hexamethylenetetramine (Urotropine)
(Adrenergic stimulant)
(Nasal decongestant)
(Antibacterial agent)
NH2 A CH2 O C O CH3 A H
CH2CH2NH2
CH3O
CH2CH2NH2
CH3O OCH3 Amphetamine
Mescaline
2-Phenylethanamine
(Antidepressant, central nervous system stimulant. The N-methyl derivative, methamphetamine, is a dangerous, addictive drug known as speed, crank, crystal, or ice.)
(Hallucinogen)
(-Phenethylamine)
The problem of selective targeting of these sites by molecular design has played a major role in the development of drugs that control weight and combat obesity. This effort has taken several directions. Thus, whereas many of the “diet pills” rely on their anorectic (appetite-suppressing) effect, newer products, such as sibutramine (approved by the FDA in 1997), which has no stimulating side effects, act by increasing satiety. In other words, you feel hungry, but you are more quickly satisfied after you start eating, or feel full longer. A recent compound, rimonabant, works on a different principle, and its discovery illustrates one of the ways chosen by researchers to find new drugs. It is known that smoking cannabis (marijuana; Sections 9-11 and 20-6) can induce hunger pangs (“the munchies”).
Controlling weight by keeping fit: aerobics class.
Therefore, it was hypothesized that blocking the corresponding cannabinoid receptors in the brain would eliminate such craving. Such an effect then could be useful in weight control. Indeed, extensive screening showed that rimonabant is very active in this respect, and the drug entered the European market in 2006. However, the U.S. FDA refused approval in 2007, because of concerns over potential depressive side effects. In 2008 the drug was pulled off the European market and several companies discontinued developing related compounds, putting a stop to this promising approach, at least for now. Another approach to controlling weight is to increase metabolic rate (and hence body temperature), a goal most simply and commonly attained by exercising. An arguably
21-2 Structural and Physical Properties of Amines
less taxing alternative is a thermogenic drug. Caffeine is such a compound, but its enhancing effect on the metabolic
Chapter 21
975
rate is short-lived and is followed by a period of actually depressing it.
H % % C N ECH3
A
Cl
CH3
O
CH3
E
H3C
E
O
N
N
NO
Cl Cl
N ON H
N
A
CH3
Cl Rimonabant
Sibutramine (Meridia)
N
N
O
Whereas thermogenic drugs help control body weight by “burning off fat,” the exact opposite — namely, inhibition of calorific intake by preventing metabolism — is being attempted with yet another class of drugs. An example is orlistat (approved by the FDA in 1999), a molecule that retards the enzymatic breakdown of fat in the gut, thus allowing food to be excreted undigested. Initially sold only by prescription, the FDA approved an over-the-counter formulation in 2007 (marketed under the name all¯l ), a first for a weight-loss drug. Amines are not the only agents that are being investigated for their potential in weight control. For example, the completely nondigestible fat olestra contains no nitrogens, and its development is based on the finding that molecules with more than five ester functions (recall that “normal” fats are triglycerides; Section 20-5) are not
Caffeine
metabolized by the body’s enzymes. Although the structure of olestra looks forbidding, it is actually a simple octaester of common table sugar, sucrose (shown in green; Chapter 24). Olestra is currently used in some foods, especially savory snacks such as potato chips, cheese puffs, and crackers. You may question the validity of this type of drug research, because it appears to cater to the glutton or the “fashionable” in the search of the perfect body. Not so: Do not forget that obesity is a major health hazard, causing chronic diseases, such as cardiovascular and respiratory problems, hypertension, diabetes, and certain cancers, and that it shortens the life span of those afflicted. It appears to be mostly the result of a metabolic (perhaps in some cases genetic) disposition that cannot otherwise be controlled.
O O
O CH3(CH2)n
(CH2)8CH3
O
O O
O
O
O
% @ CH3(CH2)5
O
%
CH3(CH2)n
H N
H
(CH2)nCH3
O
O
O O
O
CH3(CH2)n O
O
O
H3C
(CH2)nCH3 O O
O
(CH2)nCH3
O
CH3 Orlistat (Xenical)
CH3(CH2)n
O
Olestra
(CH2)nCH3
O
976
Chapter 21
Amines and Their Derivatives
Exercise 21-3 The bonds to nitrogen in methanamine (methylamine; see Figure 21-1) are slightly longer than the bonds to oxygen in the structure of methanol (Figure 8-1). Explain. (Hint: See Table 1-2.)
MODEL BUILDING
The tetrahedral geometry around an amine nitrogen suggests that it should be chiral if it bears three different substituents, the lone electron pair serving as the fourth. The image and mirror image of such a compound are not superimposable, by analogy with carbonbased stereocenters (Section 5-1). This fact is illustrated with the simple chiral alkanamine N-methylethanamine (ethylmethylamine).
Image and Mirror Image of N-Methylethanamine (Ethylmethylamine) CH2CH3
CH3CH2 N
N
H H3 C
H CH3 Mirror plane
However, samples of the amine prove not to be optically active. Why? Amines are not configurationally stable at nitrogen, because of rapid isomerization by a process called inversion. The molecule passes through a transition state incorporating an sp2-hybridized nitrogen atom, as illustrated in Figure 21-2. This transition state is similar to that depicted in Figure 6-4 for the inversion observed in SN2 reactions (Section 6-5). The barrier to this motion in ordinary small amines has been measured spectroscopically and found to be between 5 and 7 kcal mol21 (ca. 20 – 30 kJ mol21). It is therefore impossible to keep an enantiomerically pure, simple di- or trialkylamine from racemizing at room temperature when the nitrogen atom is the only stereocenter.
Transition State of Inversion CH3
sp 3
CH3CH2 Figure 21-2 Inversion at nitrogen rapidly interconverts the two enantiomers of N-methylethanamine (ethylmethylamine). Thus, the compound exhibits no optical activity.
CH2
N H H3 C Chiral
sp 2 sp 3
‡
N
CH2CH3 N H
H
CH3
Achiral
CH3 Chiral
Amines form weaker hydrogen bonds than alcohols do Because alcohols readily form hydrogen bonds (Section 8-2), they have unusually high boiling points. In principle, so should amines, and indeed Table 21-1 bears out this
21-3 Spectroscopy of the Amine Group
Table 21-1
977
Chapter 21
Physical Properties of Amines, Alcohols, and Alkanes Melting point (8C)
Boiling point (8C)
CH4 CH3NH2 CH3OH
2182.5 293.5 297.5
2161.7 26.3 65.0
CH3CH3 CH3CH2NH2 CH3CH2OH
2183.3 281 2114.1
288.6 16.6 78.5
CH3CH2CH3 CH3CH2CH2NH2 CH3CH2CH2OH
2187.7 283 2126.2
242.1 47.8 97.4
Compound
Compound
Melting point (8C)
(CH3)2NH (CH3)3N
293 2117.2
7.4 2.9
(CH3CH2)2NH (CH3CH2)3N
248 2114.7
56.3 89.3
(CH3CH2CH2)2NH (CH3CH2CH2)3N
240 294
NH3 H2O
277.7 0
Boiling point (8C)
110 155 233.4 100
expectation. However, because amines form weaker hydrogen bonds* than alcohols do, their boiling points are lower and their solubility in water less. In general, the boiling points of the amines lie between those of the corresponding alkanes and alcohols. The smaller amines are soluble in water and in alcohols because they can form hydrogen bonds to the solvent. If the hydrophobic part of an amine exceeds six carbons, the solubility in water decreases rapidly; the larger amines are essentially insoluble in water.
In Summary Amines adopt an approximately tetrahedral structure in which the lone electron pair occupies one vertex of the tetrahedron. They can, in principle, be chiral at nitrogen but are difficult to maintain in enantiomerically pure form because of fast inversion at the nitrogen. Amines have boiling points higher than those of alkanes of similar size. Their boiling points are lower than those of the analogous alcohols because of weaker hydrogen bonding, and their water solubility is between that of comparable alkanes and alcohols.
21-3 Spectroscopy of the Amine Group Primary and secondary amines can be recognized by infrared spectroscopy because they exhibit a characteristic broad N – H stretching absorption in the range between 3250 and 3500 cm21. Primary amines show two strong peaks in this range, whereas secondary amines give rise to only a single line (see also IR spectra of amides, Section 20-1). Primary amines also show a band near 1600 cm21 that is due to a scissoring motion of the NH2 group (Section 11-8, Figure 11-17, and margin). Tertiary amines do not give rise to such signals, because they do not have a hydrogen that is bound to nitrogen. Figure 21-3 shows the infrared spectrum of cyclohexanamine. Nuclear magnetic resonance spectroscopy also may be useful for detecting the presence of amino groups. Amine hydrogens often resonate to give broadened peaks, like the OH signal in the NMR spectra of alcohols. Their chemical shift depends mainly on the rate of exchange of protons with water in the solvent and the degree of hydrogen bonding.
*Whereas all amines can act as proton acceptors in hydrogen bonding, only primary and secondary amines can function as proton donors, because tertiary amines lack such protons.
Scissoring Motion of the –NH2 Hydrogens
H
H N
Chapter 21
Amines and Their Derivatives
Figure 21-3 (A) Infrared spectrum of cyclohexanamine. The molecule exhibits two strong peaks between 3250 and 3500 cm21, characteristic of the N–H stretching absorptions of the primary amine functional group. The broad band near 1600 cm21 results from scissoring motions of the N–H bonds. (B) N-Methylcyclohexanamine shows only one N–H peak at 3300 cm21. (C) N,N-Dimethylcyclohexanamine has no peaks between 3250 and 3500 cm21.
Transmittance (%)
100
H
N
H NH2 scissoring
3358
3283
2 N–H peaks IR 0 4000
3500
3000
2500
A
2000 1500 Wavenumber
600 cm−1
1000
100
Transmittance (%)
100
H
N
Transmittance (%)
978
CH3
No N–H peak
H3C
N
CH3
1 N–H peak
B
IR 0 4000 3500 3000 2500 Wavenumber
C
IR 0 4000 3500 3000 2500 Wavenumber
Figure 21-4 shows the 1H NMR spectrum of azacyclohexane (piperidine), a cyclic secondary amine. The amine hydrogen peak appears at d 5 1.22 ppm, and there are two other sets of signals, multiplets at d 5 1.34 and 2.61 ppm. The absorption at lowest field can be assigned to the hydrogens neighboring the nitrogen, which are deshielded by the effect of its electronegativity. Exercise 21-4 Would you expect the hydrogens next to the heteroatom in an amine, RCH2NH2, to be more or less deshielded than those in an alcohol, RCH2OH? Explain. (Hint: See Exercise 21-3.)
The 13C NMR spectra of amines show a similar trend: Carbons bound directly to nitrogen resonate at considerably lower field than do the carbon atoms in alkanes. However, as in 1H NMR (Exercise 21-4), nitrogen is less deshielding than oxygen.
21-3 Spectroscopy of the Amine Group
6H
ppm
H
H
H
4H
H
H H
H H N
H
1H
H
H
(CH3)4Si
2.7 2.6 2.5 ppm
9
8
7
6
5
4
3
2
1
0
ppm (δ )
13
C Chemical Shifts in Various Amines (ppm)
25.5
23.6 27.2
25.1 26.6
25.7
68.0
36.7
compared with 47.5
N H
O
50.4
(CH3CH2)2NH 15.4 44.2
NH2
H A CH3NCH(CH3)2 33.6
50.9 22.6
Mass spectrometry readily establishes the presence of nitrogen in an organic compound. Unlike carbon, which is tetravalent, nitrogen is trivalent. Because of these valence requirements and because nitrogen has an even atomic weight (14), molecules incorporating one nitrogen (or any odd number of nitrogens) have an odd molecular weight (review Exercise 11-21). You can readily verify this count by considering that the molecular formula for an amine containing a nitrogen is CnH2n13N. The mass spectral fragmentation patterns of amines also aid in structural assignments. For example, the mass spectrum of N,N-diethylethanamine (triethylamine) shows the peak for the molecular ion at myz 5 101 (Figure 21-5). More prominent, however, the base peak is at myz 5 86 and is caused by the loss of a methyl group by a-cleavage (Section 11-10). Such fragmentation is favored because it results in a resonance-stabilized iminium ion. Mass-Spectral Fragmentation of N,N-Diethylethanamine +j (CH3CH2)2NCH2O CH3 m/z ⴝ 101 N,N-Diethylethanamine (Triethylamine)
CH3j
979
Figure 21-4 300-MHz 1H NMR spectrum of azacyclohexane (piperidine). Like the OH hydrogen signal in alcohols, the amine NH peak may appear almost anywhere in the normal hydrogen chemicalshift range. Here the NH absorption occurs at d 5 1.22 ppm, and the signal is fairly sharp because of the use of dry solvent (CDCl3).
1H NMR 1.5 1.4 1.3 1.2
Chapter 21
(CH3CH2)2N P CH2
š CH2 (CH3CH2)2NO
m/z ⴝ 86 Iminium ion
The rupture of the C–C bond next to nitrogen is frequently so easy that the molecular ion cannot be observed. For example, in the mass spectrum of 1-hexanamine, the molecular ion (myz 5 101) is barely visible; the dominating peak corresponds to the methyleneiminium fragment [CH2PNH2]1 (myz 5 30).
Chapter 21
Figure 21-5 Mass spectrum of N,N-diethylethanamine (triethylamine), showing a molecular ion peak at myz 5 101. In general, molecules incorporating one nitrogen atom have an odd molecular weight. The base peak is due to loss of a methyl group, resulting in an iminium ion with myz 5 86.
Amines and Their Derivatives
86 (M − CH3)+
MS
100
(CH3CH2)3N Relative abundance
980
50
.
M+ 101
0 0
20
40
60
80
100
120
m/z
Exercise 21-5
Working with the Concepts: Assigning the Structure of an Amine A compound with the molecular formula C8H11N exhibited the following spectral data: Mass spectrum: myz (relative intensity) 5 121 (M1, 6), 91 (15), 30 (100). 1H NMR: d 5 1.16 (s, 2 H), 2.73 (t, J 5 7 Hz, 2 H), 2.93 (t, J 5 7 Hz, 2 H), 7.20 (m, 5 H) ppm. 13C NMR: d 40.2 (CH2), 43.6 (CH2), 126.1 (CH), 128.4 (CH), 128.8 (CH), 139.9 (Cquaternary) ppm. IR (neat): selected | n 5 3366, 3284, 3062, 3026, 2932, 2850, 1603, 746, 700 cm21. What is it? Strategy The combined spectral data give us more than enough information to elucidate the structure of this unknown. The following analyzes them in order of appearance in the question, starting with the mass and 1H NMR spectra to reach a tentative assignment. The remaining data are then used to support it. However, in problems like this, you can start with whatever spectral information you deem most relevant, jump from one type of spectroscopy to another, and piece together the solution that way. Solution • The mass spectrum shows a molecular ion (myz 5 121), which corresponds to the molecular formula (as it should), C8H11N. • More instructive are the fragment peaks: the first, at myz 5 91, signifies the loss of a piece with mass 5 30, exactly the value of the second fragment ion. In other words, the molecule breaks up very efficiently into two segments of myz 5 91 and 30. How do we dissect C8H11N to provide such fragments? It should not take you long to realize that myz 5 30 must be due to [CH2 PNH2]1, as there is no other sensible alternative. (The only other molecule with this mass that can be written is C2H6, ethane.) Therefore, myz 5 91 has the composition C7H7. • Turning to the 1H NMR spectrum, we note that the hydrogens are divided into four groups, in the ratio 2 : 2 : 2 : 5. The first group gives rise to a singlet and is therefore due to a pair of hydrogens without immediate hydrogen neighbors or attached to N, as in NH2 (decoupled by rapid proton exchange). The next two sets of hydrogens appear as triplets and thus must be due to mutually coupled (hence neighboring) CH2 groups, suggesting the presence of substructure A – CH2CH2 – B. • Subtracting the unit C2H4 from C8H11N leaves C6H7N for A 1 B. • Finally, we observe an aromatic multiplet of 5 H at d 5 7.20 ppm, strongly indicative of a phenyl group, C6H5. If this group were A, then B would have to be NH2, and a possible structure would be 2-phenylethanamine, C6H5CH2CH2NH2. Such a solution would certainly be in accord with the
21-4 Acidity and Basicity of Amines
mass spectrum, the dominant fragments being methyleneiminium and benzyl cation [C6H5CH2]1. [Parenthetically, benzyl cation rapidly rearranges under mass spectral conditions to its more stable, aromatic annulene isomer, cycloheptatrienyl cation (Section 15-7)]. • Let us see whether the remainder of the spectral data supports our tentative assignment. The 13 C NMR spectrum, indeed, exhibits the expected six lines with the appropriate chemical shifts (Table 10-6, Section 21-3). • The IR spectrum contains bands for the NH2 group, | n 5 3366, 3284, and 1603, very similar to those shown in Figure 21-3(A), in addition to the aromatic and saturated C – H absorptions (Figure 15-7), all fully consistent with the proposed structure, shown in the margin.
Chapter 21
NH2
2-Phenylethanamine
Exercise 21-6 H A N
Try It Yourself What approximate spectral data (IR, NMR, myz) would you expect for N-ethyl-2,2dimethylpropanamine, shown in the margin?
In Summary The IR stretching absorption of the N – H bond ranges between 3250 and 3500 cm21; the corresponding 1H NMR peak is often broad and can be found at variable d. The electron-withdrawing nitrogen deshields neighboring carbons and hydrogens, although to a lesser extent than the oxygen in alcohols and ethers. The mass spectra of simple alkanamines that contain only one nitrogen atom have odd-numbered molecular ion peaks, because of the trivalent character of nitrogen. Fragmentation occurs in such a way as to produce resonance-stabilized iminium ions.
21-4 Acidity and Basicity of Amines Like the alcohols (Section 8-3), amines are both basic and acidic. Because nitrogen is less electronegative than oxygen, the acidity of amines is about 20 orders of magnitude less than that of comparable alcohols. Conversely, the lone pair is much more available for protonation, thereby causing amines to be better bases.
Acidity and Basicity of Amines š Amine acting as an acid: RNH A H
ðB
Ka
Acid
š 2 Amine acting as a base: RNH Base
RNH
HB
Conjugate base
Kb
HA
H A RNH2
ðA
Conjugate acid
Amines are very weak acids We have seen evidence that amines are much less acidic than alcohols: Amide ions, R2N2, are used to deprotonate alcohols (Section 9-1). The equilibrium of this proton transfer lies strongly to the side of the alkoxide ion. The high value of the equilibrium constant, about 1020, is due to the strong basicity of amide ions, which is consistent with the low acidity of amines. The pKa of ammonia and alkanamines is of the order of 35.
N-Ethyl-2,2-dimethylpropanamine
981
982
Chapter 21
Amines and Their Derivatives
Acidity of Amines H2 O
š RNH A H
Ka
RNH
Amine
Amide ion
(Weak acid)
(Strong base)
H2OH
[RNH][H 2OH]
Ka
š [RNH] A H
pKa
10
35
35
The deprotonation of amines requires extremely strong bases, such as alkyllithium reagents. For example, lithium diisopropylamide, the sterically hindered base used in some bimolecular elimination reactions (Section 7-8), is made in the laboratory by treatment of N-(1-methylethyl)-2-propanamine (diisopropylamine) with butyllithium. Preparation of LDA H3C CH3 A A CH3CHNCHCH3 H
CH3CH2CH2CH2Li CH3CH2CH2CH2H
N-(1-Methylethyl)2-propanamine (Diisopropylamine)
Preparation of Sodium Amide 2 Na
2 NH3 Catalytic Fe3
2 NaNH2
H2
Li H3C CH3 A A CH3CHNCHCH 3 Lithium diisopropylamide, LDA
An alternative synthesis of amide ions is the treatment of amines with alkali metals. Alkali metals dissolve in amines (albeit relatively slowly) with the evolution of hydrogen and the formation of amine salts (much as they dissolve in water and alcohol, furnishing H2 and metal hydroxides or alkoxides, Section 9-1). For example, sodium amide can be made in liquid ammonia from sodium metal in the presence of catalytic amounts of Fe31, which facilitates electron transfer to the amine. In the absence of such a catalyst, sodium simply dissolves in ammonia (labeled “Na, liquid NH3”) to form a strongly reducing solution (Section 13-6).
Amines are moderately basic, ammonium ions are weakly acidic Amines deprotonate water to a small extent to form ammonium and hydroxide ions. Thus, amines are more strongly basic than alcohols but not nearly as basic as alkoxides. Protonation occurs at the site of the free electron pair as pictured in the electrostatic potential map of N,N-dimethylmethanamine (trimethylamine) in the margin. Basicity of Amines š 2 HO š RNH OH N,N-Dimethylmethanamine (Trimethylamine)
Kb
Amine
H A RNH2
HO ð
Ammonium ion
The resulting ammonium salts can be primary, secondary, or tertiary, depending on the number of substituents on nitrogen.
RNH3 Cl
R2NH2 Br
R3NH I
Primary ammonium chloride
Secondary ammonium bromide
Tertiary ammonium iodide
21-4 Acidity and Basicity of Amines
Ammonium salts are named by attaching the substituent names to the ending -ammonium followed by the name of the anion. It is useful to view the basicity of the amines as a measure of the acidity of their conjugate acids (Section 2-2), the ammonium ions. These species are stronger acids than water (pKa 5 15.7; Table 2-2) or alcohols but much weaker than carboxylic acids (pKa 5 4 – 5; Section 19-4).
H2 O
Conjugate acid
Ka
RNH2
H2OH
Ka
[RNH2][H2OH]
Base
Reminder: The weaker the conjugate acid, the higher its pKa and the stronger the corresponding base
pKa
[RNH2] A H
1010
10
Any factor, such as substituents or hybridization, that increases the electron density of the amine nitrogen increases the basicity of the amine and therefore the pKa of the corresponding ammonium salt. Conversely, decreasing the electron density of the amine nitrogen decreases its basicity and the pKa of the corresponding ammonium salt. For example, alkyl1 ammonium salts are slightly less acidic than ammonium ion, NH4, and therefore the corresponding amines are more basic. The reason is the electron-donating character of alkyl groups (Sections 7-5 and 16-1). However, as shown below, the pKa values do not increase in a regular way with increasing alkyl substitution. In fact, tertiary amines are typically less basic than secondary systems. Solvation is responsible for this observation. Thus, increasing the number of alkyl groups on the amine nitrogen increases unfavorable steric disruption of the solvent shell (Section 8-3). At the same time, it decreases the number of hydrogens attached to the nitrogen capable of entering into favorable hydrogen bonds (Section 21-2). Both phenomena counteract the inductive donor properties of the alkyl groups in solution. Indeed, in the gas phase, in which there is no solvent, the trend is as expected: 1 1 1 1 (pKa ) NH4 , CH3NH3 , (CH3 )2NH2 V (CH3 )3N H.
pKa Values of a Series of Simple Ammonium Ions*
NH4
CH3NH3
(CH3)2NH2
(CH3)3NH
N N
N
Increasing steric hindrance pKa ⴝ 9.24
10.62
10.73
1
CH3NH3 Cl2
Methylammonium chloride
CH3 f NE H I i CH2CH3
Cyclopentylethylmethylammonium iodide
Acidity of Ammonium Ions H A RNH2
Chapter 21
9.79
*A confusing practice in the literature is to refer to the pKa value of an ammonium ion as being that of the neutral amine. In the statement “the pKa of methanamine is 10.62,” what is meant is the pKa of the methylammonium ion. The pKa of methanamine is actually 35.
983
984
Chapter 21
Amines and Their Derivatives
CHE M I C AL H I G H L I G H T 2 1 - 2 Separation of Amines from Other Organic Compounds by Aqueous Extraction Techniques The isolation of natural products or the steps of organic synthesis frequently involve the problem of separating basic, neutral, and acidic components of a mixture. We can solve this problem by exploiting the basicity of amines and, more generally, the acid-base properties of a number of functional groups in their purification by aqueous extraction of the corresponding salts. These procedures make use of the fact that salts are, in general, water soluble, whereas their neutral conjugate acids or bases are not. Thus, for example, the basic amines can be extracted from organic solvents into an acidic aqueous phase in the form of their ammonium salts. Conversely, acidic compounds, such as carboxylic acids, are removed from an organic medium by extraction into a basic aqueous phase, in which they are soluble as carboxylates. These processes can then be reversed by respective basification or acidification of the aqueous layers, ultimately allowing the separation of basic, neutral, and acidic components of a mixture. A flow chart that describes such a separation procedure is shown on page 985. First, a solution of the mixture in an organic solvent is treated with mild aqueous base, such as NaHCO3, thereby converting any carboxylic acid that may be present into its sodium salt. This salt, being water soluble, moves from the organic into the aqueous layer. After the layers are separated, the carboxylic acid may be brought out of the water
Extraction of solutes from an organic into an aqueous layer (and the reverse) is accomplished by shaking the mixture in a separating funnel and then letting the two phases separate, as shown. Physical separation of the layers occurs by careful draining through the stopcock opening at the bottom.
The nitrogen lone pairs in arenamines, carboxamides, and imines are less available for protonation In contrast to the effect of alkyl groups, the basicity of amines is decreased by electron-withdrawing substituents on the nitrogen. For example, the pKa of 2-[bis(2hydroxyethyl)amino]ethanol is only 7.75, a consequence of the presence of three inductively HO
š N
OH
ðNH2
ðNH2 ðOð
2-[Bis(2-hydroxyethyl)amino]ethanol
H N
HO
š NH2
CH3
OH
OH
Benzenamine (Aniline) Resonance diminishes the availability of the N lone pair
NH3
Cyclohexanamine
Acetamide Resonance diminishes the availability of the N lone pair
NH3
E H
ðO CH3
OH pKa ⴝ 7.75
4.63
10.66
š NH2
0.63 Protonation occurs on oxygen
21-4 Acidity and Basicity of Amines
solution by acidification with strong mineral acid, such as HCl. Next, the organic solution that remains is treated with acid, thus converting any amine present into the corresponding water-soluble ammonium salt, which moves into the
Chapter 21
water phase. Separation of the layers gives an organic solution, which contains the neutral substances. The amine is removed from water solution by treatment with strong base, such as NaOH.
Separation of Acidic (Carboxylic Acid), Basic (Amine), and Neutral (Haloalkane) Organic Compounds by Aqueous Extraction B RCOH RNH2 RX Extract with aqueous NaHCO3 Aqueous layer
Organic layer
B RCO Na
RNH2 RX Extract with aq. HCl
Acidify with HCl
Aqueous layer
B RCOH
RNH3 Cl Basify with NaOH
RNH2
electron-withdrawing oxygens. Similarly, benzenamine (aniline) is considerably less basic (pKa 5 4.63) than its saturated analog cyclohexanamine (pKa 5 10.66) and other primary amines. There are two reasons for this effect. One is the sp2 hybridization of the aromatic carbon attached to the nitrogen in benzenamine, which renders that carbon relatively electron withdrawing (Sections 11-3 and 13-2), in this way making the nitrogen lone pair less available for protonation. The second reason is the resonance stabilization of the system by delocalization of the electron pair into the aromatic p system (Section 16-1). This resonance is lost upon protonation. A similar example is acetamide, in which the appended acetyl group ties up the nitrogen lone pair by induction (the carbonyl carbon is positively polarized) and resonance (Section 20-1). As might be expected, the. .hybridization at nitrogen itself also drastically affects basicity, in the order : NH3 . R2C“N R¿ . RC‚N: , a phenomenon that we already encountered in the discussion of the relative acidity of alkanes, alkenes, and alkynes (Section 13-2). Thus, iminium ions (Section 17-9) have pKa values estimated to be of the order of 7 to 9; N-protonated nitriles (Section 20-8) are even more acidic (pKa , 25). Table 21-2 summarizes the pKa values of the conjugate acids of some representative amines.
Exercise 21-7 Explain the decreasing pKa values of the following (protonated) amines: NH2 pKa ⴝ
10.7
NH2 9.5
NH2 8.2
985
Organic layer
RX
986
Chapter 21
Amines and Their Derivatives
Table 21-2
pKa Values of Various Amines
Compound
pKa of conjugate acid
:NH3 .. CH3 N H2 .. (CH3)2 N H (CH3)3N:
9.24 10.62 10.73 9.79 7.75
(HOCH2CH2)3N: ðNH2
Compound
pKa of conjugate acid
ðNH2 4.63
ðOð 0.63 CH3
10.66
NH2
R A EC N E R R N
7–9 , 25
R O C q Nð
In Summary Amines are poor acids and require alkyllithium reagents or alkali metal treatment to form amide ions. In contrast, they are moderately good bases, leading to the relatively weakly acidic ammonium salts.
21-5 Synthesis of Amines by Alkylation Amines can be synthesized by alkylating nitrogen atoms. Several such procedures take advantage of an important property of the nitrogen in many compounds: It is nucleophilic.
Amines can be derived from other amines As nucleophiles, amines react with haloalkanes to give ammonium salts (Section 6-2). Unfortunately, this reaction is not clean, because the resulting amine product usually undergoes further alkylation. How does this complication arise? Consider the alkylation of ammonia with bromomethane. When this transformation is carried out with equimolar quantities of starting materials, the weakly acidic product (methylammonium bromide), as soon as it is formed, (reversibly) donates a proton to the starting, weakly basic ammonia. The small quantities of methanamine generated in this way then compete effectively with the ammonia for the alkylating agent, and this further methylation generates a dimethylammonium salt. The process does not stop there, either. This salt can donate a proton to either of the other two nitrogen bases present, furnishing N-methylmethanamine (dimethylamine). This compound constitutes yet another nucleophile competing for bromomethane; its further reaction leads to N,N-dimethylmethanamine (trimethylamine) and, eventually, to tetramethylammonium bromide, a quaternary ammonium salt. The final outcome is a mixture of alkylammonium salts and alkanamines. Methylation of Ammonia First Alkylation. Two steps give primary amine Step 1. Nucleophilic Substitution:
H3Nð
CH3 O Br
CH3NH3 Br Methylammonium bromide
Step 2. Deprotonation:
CH3NH2 Br A H
ðNH3
š 2 CH3NH Methanamine (Methylamine)
HNH3 Br
21-5 Synthesis of Amines by Alkylation
Chapter 21
Subsequent Alkylation. Gives secondary, tertiary, and quaternary amines or ammonium salts, respectively NH2 CH3š
CH3Br
CH3Br
(CH3)2š NH
Amine hydrobromide
CH3Br
(CH3)3Nð
Amine hydrobromide
N-Methylmethanamine (Dimethylamine)
N,N-Dimethylmethanamine (Trimethylamine)
(CH3)4N Br Tetramethylammonium bromide
The mixture of products obtained upon treatment of haloalkanes with ammonia or amines is a serious drawback that limits the usefulness of direct alkylation in synthesis. As a result, indirect alkylation methods are frequently applied, particularly in the preparation of primary amines. Exercise 21-8 Like other amines, benzenamine (aniline) can be benzylated with chloromethylbenzene (benzyl chloride), C6H5CH2Cl. In contrast with the reaction with alkanamines, which proceeds at room temperature, this transformation requires heating to between 90 and 958C. Explain. (Hint: See Section 21-4.)
Indirect alkylation leads to primary amines The synthesis of primary amines requires a nitrogen-containing nucleophile that will undergo reaction only once and that can be converted subsequently into the amino group. For example, cyanide ion, 2CN, turns primary and secondary haloalkanes into nitriles, which are then reduced to the corresponding amines (Section 20-8). This sequence allows the conversion RX n RCH2NH2. Note, however, that this method introduces an additional carbon into the haloalkane framework, because cyanide is alkylated at carbon and not nitrogen. Conversion of a Haloalkane into the Homologous Amine by Cyanide Displacement–Reduction
RC qN CH3(CH2)8Br
NaCN
RC qN
CN
RX
LiAlH4 or H2, metal catalyst
DMSO
X RCH2NH2
CH3(CH2)8CN 93%
NaBr
1-Bromononane
H2, Raney Ni, 100 atm
Decanenitrile
CH3(CH2)8CH2NH2 92% 1-Decanamine
To transform a haloalkane selectively into the corresponding amine without additional carbons requires a modified nitrogen nucleophile that is unreactive after the first alkylation. Such a nucleophile is the azide ion, N32, which reacts with haloalkanes to furnish alkyl azides. These azides in turn are reduced by catalytic hydrogenation (Pd–C) or by lithium aluminum hydride to the primary amines.
R — N3 Alkyl azide
Azide Displacement–Reduction
Br
Naš NPNPš N
CH3CH2OH NaBr SN2 reaction
NPNPN
1. LiAlH4, (CH3CH2)2O 2. H, H2O
NH2
91%
89%
3-Cyclopentylpropyl azide
3-Cyclopentylpropanamine
987
988
Amines and Their Derivatives
Chapter 21
A nonreductive approach to synthesizing primary amines uses the (commercially available) anion of 1,2-benzenedicarboximide (phthalimide), the imide of 1,2-benzenedicarboxylic (phthalic) acid. This process is also known as the Gabriel* synthesis. Because the nitrogen in the imide is adjacent to two carbonyl functions, the acidity of the NH group (pKa 5 8.3) is much greater than that of an ordinary amide (pKa 5 22, Section 20-7). Deprotonation can therefore be achieved with as mild a base as carbonate ion, the resulting anion being monoalkylated in good yield. The amine subsequently can be liberated by acidic hydrolysis, initially as the ammonium salt. Base treatment of the salt then produces the free amine. Gabriel Synthesis of a Primary Amine O
O OH
ðNH3, 300° C
OH
H2O
O K2CO3, H2O
ðNH
ðNðK
O
O
KBr
O Relatively acidic
97% 1,2-Benzenedicarboxylic acid (Phthalic acid)
HC q CCH2OBr, DMF, 100°C
1,2-Benzenedicarboximide (Phthalimide)
NH2 HC q CCH2š 73% O ðNCH2C q CH
H2SO4, H2O, 120°C
HC q CCH2NH3 HSO4
NaOH, H 2O
2-Propynamine (Propargylamine)
O
O
O 93%
OH
ONa
N-2-Propynyl1,2-benzenedicarboximide (N-Propargylphthalimide)
OH
ONa
O
O Removed in aqueous work-up
Exercise 21-9 The cleavage of an N-alkyl-1,2-benzenedicarboximide (N-alkyl phthalimide) is frequently carried out with base or with hydrazine, H2NNH2. The respective products of these two treatments are the 1,2-benzenedicarboxylate A or the hydrazide B. Write mechanisms for these two transformations. (Hint: Review Section 20-6.) O
O COO
COO
OH
RNH2
NR
NH A NH
H2NNH2 RNH2
O A
*Professor Siegmund Gabriel (1851 – 1924), University of Berlin.
O B
21-6 Synthesis of Amines by Reductive Amination
Chapter 21
989
Exercise 21-10 Show how you would apply the Gabriel method to the synthesis of each of the following amines: (a) 1-hexanamine; (b) 3-methylpentanamine; (c) cyclohexanamine; (d) H2NCH2CO2H, the amino acid glycine. (Caution: In the synthesis of glycine, you need to watch out for the carboxylic acid function. Can you see why?) For each of these four syntheses, would the azide displacement – reduction method be equally good, better, or worse?
In Summary Amines can be made from ammonia or other amines by simple alkylation, but this method gives mixtures and poor yields. Primary amines are better made in stepwise fashion, employing nitrile and azide groups, or protected systems, such as 1,2-benzenedicarboxylic imide (phthalimide) in the Gabriel synthesis.
21-6 Synthesis of Amines by Reductive Amination A more general method of amine synthesis, called reductive amination of aldehydes and ketones, allows the construction of primary, secondary, and tertiary amines. In this process, the carbonyl compound is exposed to an amine containing at least one N – H bond (NH3, primary, secondary amines) and a reducing agent to furnish a new alkylated amine directly (a primary, secondary, or tertiary amine, respectively). The new C – N bond is formed to the carbonyl carbon of the aldehyde or ketone.
General Reductive Amination ␦ f
i
R
i ðN O H f
Reduction
H i f C O Nð R~ i R ;
O PC
R
The sequence begins by the initial condensation of the amine with the carbonyl component (Section 17-9) to produce the corresponding imine (for NH3 and primary amines) or iminium ion (secondary amines). Similar to the carbon – oxygen double bond in aldehydes and ketones, the carbon – nitrogen double bond in these intermediates is then reduced by simultaneous catalytic hydrogenation or by added special hydride reagents. Reductive Amination of a Ketone with a Primary Amine R G CPO D R
H2NR
R R G D C PN D R
Condensation
Reduction
R R G D C PN D R
H2O
R A RO C ONHR A H
This reaction succeeds because of the selectivity of the reducing agents: either hydrogen gas in the presence of a catalyst (Section 12-2) or sodium cyanoborohydride, Na12BH3CN. Both react faster with the imine double bond than with the carbonyl group under the conditions employed. With Na12BH3CN, the conditions are relatively acidic (pH 5 2 – 3), activating the imine double bond by protonation on nitrogen and thus facilitating hydride attack at carbon. As shown in the margin, the relative stability of the modified borohydride reagent
MATION AN I
ANIMATED MECHANISM: Reductive amination
N c ␦ C Electron withdrawing A E B~ H & H H Diminished ability of H to leave as :H, hence reagent is less sensitive to H
Amines and Their Derivatives
Chapter 21
at such low pH (at which NaBH4 hydrolyzes; Section 8-6) is due to the presence of the electron-withdrawing cyano group, which renders the hydrogens less basic (hydridic). In a typical procedure, the carbonyl component and the amine are allowed to equilibrate with the imine and water in the presence of the reducing agent. In this way, ammonia furnishes primary amines, and primary amines result in secondary amines. Amine Synthesis by Reductive Amination O K C
NH3
NOH K C
H2O
H H A f C ON A i H H 89%
H2, Ni, CH3CH2OH, 70°C, 90 atm
G
G
990
H2O
H
H
Benzaldehyde
Phenylmethanamine (Benzylamine)
Not isolated
O
NCH3
H NCH3
NaBH3CN, CH3CH2OH, pH 2–3
H2O
CH3NH2
H
H2O
78% Cyclohexanone
N-Methylcyclohexanamine
Not isolated
Similarly, reductive aminations with secondary amines give tertiary amines.
H3C
H
O
H3C
N(CH3)2
(CH3)2NH, NaBH3CN, CH3OH
H3 C
CH3
H3C
CH3 89%
Secondary amines cannot form imines with aldehydes and ketones, but are in equilibrium with the corresponding N,N-dialkyliminium ions (Section 17-9), which are reduced by addition of H2 from cyanoborohydride.
CH2
O
NCH2C6H5
NCH2C6H5
H
CH2
N-(Phenylmethyl)cyclopentanamine (Benzylcyclopentylamine)
Iminium ion
NaBH3CN, CH3OH
NCH2C6H5 HCH2 100% N-Methyl-N-(phenylmethyl)cyclopentanamine (Benzylcyclopentylmethylamine)
Exercise 21-11 Formulate a mechanism for the reductive amination with the secondary amine shown in the preceding example.
21-6 Synthesis of Amines by Reductive Amination
Chapter 21
Exercise 21-12
Working with the Concepts: Applying Reductive Amination Explain the following transformation by a mechanism. O
O NaBH3CN, CH3OH
H
H
N
NH2
35%
Strategy As usual, you need to take an inventory of the functionalities present in the starting material, the reagent, and the product (an amine). The result is that we are dealing with an intramolecular variant of reductive animation. This realization should be the starting point for retrosynthetic analysis. Solution • First, write down the generic process retrosynthetically: CPO
CO N A H
HON
• Thus, the retrosynthetic disconnection from the product is O N
H
N O H
H
O H NH2
• The mechanism of the forward reaction follows accordingly (not all intermediates are shown): C H
KO
C H
H
NH2
C i H
CH B O
ðNH
NaBH3CN
NOH
H2O
CH B O
KO
H
N
H2O
NaBH3CN
N
Exercise 21-13
Try It Yourself Treatment of aldehyde A with the modified sodium cyanoborohydride reagent shown gave the natural product buflavine, a bioactive constituent of the bulbs of the Boophane flava plant from southern Africa. Its mass spectral molecular ion occurs at myz 5 283. Suggest a structure. H3CO H
H3CO
O
O B NaBH(OCCH3)3
H N
Buflavine
35%
CH3 A
The Boophane plant.
991
992
Chapter 21
Amines and Their Derivatives
In Summary Reductive amination furnishes alkanamines by reductive condensation of amines with aldehydes and ketones.
21-7 Synthesis of Amines from Carboxylic Amides Carboxylic amides can be versatile precursors of amines by reduction of the carbonyl unit (Section 20-6). Since amides are in turn readily available by reaction of acyl halides with amines (Section 20-2), the sequence acylation – reduction constitutes a controlled monoalkylation of amines. Utility of Amides in Amine Synthesis O B RCCl
H2NR
Base HCl
O B RCNHR
RCH2NHR
LiAlH4, (CH3CH2)2O
Primary amides can also be turned into amines by oxidation with bromine or chlorine in the presence of sodium hydroxide — in other words, by the Hofmann rearrangement (Section 20-7). Recall that in this transformation the carbonyl group is extruded as carbon dioxide, so the resulting amine bears one carbon less than the starting material. Amines by Hofmann Rearrangement RNH2
O
C
O
B
Br2, NaOH, H2O
B
O B RCNH2
Exercise 21-14 Suggest synthetic methods for the preparation of N-methylhexanamine from hexanamine (two syntheses) and from N-hexylmethanamide (N-hexylformamide).
In Summary Amides can be reduced to amines by treatment with lithium aluminum hydride. The Hofmann rearrangement converts amides into amines with loss of the carbonyl group.
21-8 Reactions of Quaternary Ammonium Salts: Hofmann Elimination Much like the protonation of alcohols, which turns the – OH into the better leaving group –1OH2 (Section 9-2), protonation of amines might render the resulting ammonium salts subject to nucleophilic attack. In practice, however, amines are not sufficiently good leaving groups (they are more basic than water) to partake in substitution reactions. Nevertheless, they can function as such in the Hofmann* elimination, in which a tetraalkylammonium salt is converted to an alkene by a strong base. Section 21-5 described how complete alkylation of amines leads to the corresponding quaternary alkylammonium salts. These species are unstable in the presence of strong base, because of a bimolecular elimination reaction that furnishes alkenes (Section 7-7). The base attacks the hydrogen in the b-position with respect to the nitrogen, and a trialkylamine departs as a neutral leaving group.
*This is the Hofmann of the Hofmann rule for E2 reactions (Section 11-6) and the Hofmann rearrangement (Section 20-7).
21-8 Hofmann Elimination
Chapter 21
Bimolecular Elimination of Quaternary Ammonium Ions
GNR3 GCO C H
C
C
ðNR3
HOH
Alkene
š ðOH
In the procedure of Hofmann elimination, the amine is first completely methylated with excess iodomethane (exhaustive methylation) and then treated with wet silver oxide (a source of HO2) to produce the ammonium hydroxide. Heating degrades this salt to the alkene. When more than one regioisomer is possible, Hofmann elimination, in contrast to most E2 processes, tends to give less substituted alkenes as the major products. Recall that this result adheres to Hofmann’s rule (Section 11-6) and appears to be caused by the bulk of the ammonium group, which directs the base to the less hindered protons in the molecule. Hofmann Elimination of 1-Butanamine CH3CH2CH2CH2NH2
Excess CH3I, K2CO3, H2O
1-Butyltrimethylammonium iodide
1-Butanamine
š HOð H
CH3CH2CH2CH2N(CH3)3 I
H H G C C H ( N(CH3)3 CH3CH2
CH3CH2CH P CH2
&
@
G
C G
1-Butyltrimethylammonium hydroxide
HOH
Ag2O, H2O AgI
N(CH3)3
1-Butene
Exercise 21-15 Give the structures of the possible alkene products of the Hofmann elimination of (a) N-ethylpropanamine (ethylpropylamine) and (b) 2-butanamine. (Caution: In the Hofmann elimination of 2-butanamine, you need to consider several products. Hint: Review Section 11-6).
The Hofmann elimination of amines has been used to elucidate the structure of nitrogencontaining natural products, such as alkaloids (Section 25-8). Each sequence of exhaustive methylation and Hofmann elimination cleaves one C – N bond. Repeated cycles allow the heteroatom to be precisely located, particularly if it is part of a ring. In this case, the first carbon – nitrogen bond cleavage opens the ring. H
H N A CH3
N-Methylazacycloheptane
1. CH3I 2. Ag2O, H2O 3.
1. CH3I 2. Ag2O, H2O 3.
H
N(CH3)2 N,N-Dimethyl-5-hexenamine
1,5-Hexadiene
Exercise 21-16 Why is exhaustive methylation and not, say, ethylation used in Hofmann eliminations for structure elucidation? (Hint: Look for other possible elimination pathways.)
N(CH3)3
993
994
Chapter 21
Amines and Their Derivatives
Exercise 21-17 3-Ethenyl-1,4-pentadiene
An unknown amine of the molecular formula C7H13N has a 13C NMR spectrum containing only three lines of d 5 21.0, 26.8, and 47.8 ppm. Three cycles of Hofmann elimination are required to form 3-ethenyl-1,4-pentadiene (trivinylmethane; margin) and its double-bond isomers (as side products arising from base-catalyzed isomerization). Propose a structure for the unknown.
In Summary Quaternary methyl ammonium salts, synthesized by amine methylation, undergo bimolecular elimination in the presence of base to give alkenes.
21-9 Mannich Reaction: Alkylation of Enols by Iminium Ions In the aldol reaction, an enolate ion attacks the carbonyl group of an aldehyde or ketone (Section 18-5) to furnish a b-hydroxycarbonyl product. A process that is quite analogous is the Mannich* reaction. Here, however, it is an enol that functions as the nucleophile and an iminium ion, derived by condensation of a second carbonyl component with an amine, as the substrate. The outcome is a b-aminocarbonyl product. To differentiate the reactivity of the three components of the Mannich reaction, it is usually carried out with (1) a ketone or aldehyde; (2) a relatively more reactive aldehyde (often formaldehyde, CH2 PO, see Section 17-6); and (3) the amine, all in an alcohol solvent containing HCl. These conditions give the hydrochloride salt of the product. The free amine, called a Mannich base, can be obtained upon treatment with base. REACTION
Mannich Reaction O
O
CH2 P O
(CH3)2NH
HCl, CH3CH2OH,
CH2N(CH3)2 Cl A H 85%
Ketone
More reactive aldehyde
CH3 A CH3CHCHP O
CH2 P O
Amine
Salt of Mannich base
CH3NH2
1. HCl, CH3CH2OH, 2. HO, H2O
2-Methylpropanal
Aldehyde
MATION AN I
ANIMATED MECHANISM: The Mannich reaction
More reactive aldehyde
Amine
CH3 A CH3C O CHP O A CH2NHCH3 70% 2-Methyl-2-(N-methylaminomethyl)propanal Mannich base
The mechanism of this process starts with iminium ion formation between the aldehyde (e.g., formaldehyde) and the amine, on the one hand, and enolization of the ketone, on the other. The electrostatic potential maps in the margin (p. 995) illustrate the electron deficiency of the iminium ion (blue) and the contrasting relative electron abundance (red and yellow) of the enol. One can also recognize that the electron density is greater at the a-carbon (yellow) next to the hydroxy-bearing carbon (green). As soon as the enol is formed, it undergoes nucleophilic attack on the electrophilic iminium carbon, and the resulting species converts to the Mannich salt by proton transfer from the carbonyl oxygen to the amino group. *Professor Carl U. F. Mannich (1877 – 1947), University of Berlin.
21-9 Mannich Reaction
Chapter 21
995
Mechanism of the Mannich Reaction
MECHANISM
Step 1. Iminium ion formation
CH2 P O
(CH3)2NH2 Cl
CH2 PN(CH3)2 Cl
H2O
Step 2. Enolization šð HO
ðOð H
Step 3. Carbon–carbon bond formation Iminium ion
HH
šð HO
š O
Cl
Cl
N(CH3)2 CH2š
CH2 PN(CH3)2 Step 4. Proton transfer H H Cl š O
ðOð CH2š N(CH3)2
Enol
CH2N(CH3)2 Cl A H Salt of Mannich base
The following example shows the Mannich reaction in natural product synthesis. In this instance, one ring is formed by condensation of the amino group with one of the two carbonyl functions. Mannich reaction of the resulting iminium salt with the enol form of the other carbonyl function follows. The product has the framework of retronecine (see margin), an alkaloid that is present in many shrubs and is hepatotoxic (causes liver damage) to grazing livestock.
CHO N A H
CH2OH
H ´ N
O
Mannich Reaction in Synthesis CHO
HO *
O
CHO
CHO
H H2O
N
H
N 52%
N Retronecine
Exercise 21-18
Working with the Concepts: Practicing the Mannich Reaction Write the product of the Mannich reaction of ammonia, formaldehyde, and cyclopentanone. Strategy The Mannich reaction starts by condensation of the amine with the more reactive carbonyl component. We have to identify this constituent and convert it into the iminium species. We then couple the iminium intermediate with the enol form of the less reactive carbonyl component.
996
Chapter 21
Amines and Their Derivatives
Solution • The more reactive carbonyl species is formaldehyde. H
H i f • Formaldehyde and ammonia condense to CPN . f i H H ðš OH • The enol tautomer of cyclopentanone is
.
• Reaction of the two species occurs by attack of the electron-rich enol on the positively polarized iminium ion, and deprotonation on basic work-up finishes the job. ðš OH
H H
š O
H
i f CPN f i H H
ðOð š N
H
H
š N
H
H
Exercise 21-19
Try It Yourself Write the products of each of the following Mannich reactions: (a) 1-hexanamine 1 formaldehyde 1 2-methylpropanal; (b) N-methylmethanamine 1 formaldehyde 1 acetone; (c) cyclohexanamine 1 formaldehyde 1 cyclohexanone.
Exercise 21-20 b-Dialkylamino alcohols and their esters are useful local anesthetics. Suggest a synthesis of the anesthetic Tutocaine hydrochloride, beginning with 2-butanone. O O NH2
(H3C)2N H
Cl
Tutocaine hydrochloride
In Summary Condensation of aldehydes (e.g., formaldehyde) with amines furnishes iminium ions, which are electrophilic and may be attacked by the enols of ketones (or other aldehydes) in the Mannich reaction. The products are b-aminocarbonyl compounds.
21-10 Nitrosation of Amines Amines react with nitrous acid, through nucleophilic attack on the nitrosyl cation, NO1. The product depends very much on whether the reactant is an alkanamine or a benzenamine (aniline) and on whether it is primary, secondary, or tertiary. This section deals with alkanamines; aromatic amines will be considered in the next chapter. To generate NO1, we must first prepare the unstable nitrous acid by the treatment of sodium nitrite with aqueous HCl. In such an acid solution, an equilibrium is established with the nitrosyl cation. (Compare this sequence with the preparation of the nitronium cation from nitric acid; Section 15-10.)
21-10 Nitrosation of Amines
997
Chapter 21
Nitrosyl Cation from Nitrous Acid O Oš Naðš NP š O
HCl
O Oš Hš NP š O
NaCl
H
H
H
Sodium nitrite
G š NP š O O Oš D
ðNP š O
Nitrous acid
ðNq Oð
H2š O
Nitrosyl cation
The nitrosyl cation is electrophilic and is attacked by amines to form an N-nitrosammonium salt. Nð
š NO NP š O
š NP š O
N-Nitrosammonium salt
The course of the reaction now depends on whether the amine nitrogen bears zero, one, or two hydrogens. Tertiary N-nitrosammonium salts are stable only at low temperatures and decompose upon heating to give a mixture of compounds. Secondary N-nitrosammonium salts are simply deprotonated to furnish the relatively stable N-nitrosamines as the major products.
(CH3)2 NH
H A NPO Cl (CH3)2 NO
NaNO2, HCl, H2O, 0 C
HCl
R2N O NPO N-Nitrosamine
H3C i NONPO f H3C 88–90% N-Nitrosodimethylamine
Similar treatment of primary amines initially gives the analogous monoalkyl-N-nitrosamines. However, these products are unstable because of the remaining proton on the nitrogen. By a series of hydrogen shifts, they first rearrange to the corresponding diazohydroxides. Then protonation, followed by loss of water, gives highly reactive diazonium ions, R–N21. When R is a secondary or a tertiary alkyl group, these ions lose molecular nitrogen, N2, and form the corresponding carbocations, which may rearrange, deprotonate, or undergo nucleophilic trapping (Section 9-3) to yield the observed mixtures of compounds. Mechanism of Decomposition of Primary N-Nitrosamines Step 1. Rearrangement to a diazohydroxide R G š NOš NP š O D H
R G š NOš NP š OH D H
H H
R G OH NO š NPš D H
H H
ROš NPš NO š OH Diazohydroxide
Step 2. Loss of water to give a diazonium ion
H
H2O H2O
RONq Nð Diazonium cation
Step 3. Nitrogen loss to give a carbocation
RONq Nð
N2
R
CH3CH2CH2 O C
H D
G
ROš NPš NO š OH2
@
H
&
ROš NPš NO š OH
NH2
Pure R enantiomer NaNO2, HCl, H2O
product mixtures
Exercise 21-21 The mechanism just shown pertains to the decomposition of diazonium ions in which the alkyl groups R are secondary or tertiary. The result depicted in the margin was reported in 1991 and addresses the pathway chosen in the case of R 5 primary alkyl. Is the mechanism the same?
H@ D& GC OCH2 CH2CH3 HO 100% Pure S enantiomer
998
Chapter 21
Amines and Their Derivatives
CHE M I C AL H I G H L I G H T 2 1 - 3 N-Nitrosodialkanamines and Cancer N-Nitrosodialkanamines are notoriously potent carcinogens in a variety of animals. Although there is no direct evidence, they are suspected of causing cancer in humans as well. Most nitrosamines appear to cause liver cancer, but some of them are very organ specific in their carcinogenic potential
(bladder, lungs, esophagus, nasal cavity, etc.). Their mode of carcinogenic action appears to begin with enzymatic oxidation of one of the a-carbons to a hemiaminal (Section 17-9), which allows eventual formation of an unstable monoalkylN-nitrosamine, as shown here for N-nitrosodiethylamine:
Enzymatic oxidation
NO f N
HO
NO f N
Monoalkyl-N-nitrosamines are sources of carbocations, as described in this section. Carbocations are powerful electrophiles that are thought to attack one of the bases in DNA to inflict the kind of genetic damage that seems to lead to cancerous cell behavior. In the early 1970s, farm animals in Norway that had been fed herring meal preserved using relatively large doses of sodium nitrite began to show a high incidence of liver disorders, including cancer. Fish meal contains amines (for example, trimethylamine is primarily responsible for its odor) and it did not take long to realize that these amines reacted with the added nitrite to produce nitrosamines. This finding was disturbing because amines also occur in meats, and sodium nitrite is often added during the meat-curing process. Sodium nitrite inhibits the growth of bacteria that cause botulism, retards the development of rancidity and off-odors during storage, and preserves the flavor of added spices and smoke (if the meat is smoked). In addition, it gives meat an appetizing pink coloration and special flavor. The color has its origin in nitrite-derived nitric oxide, NO, which forms a red complex with the iron in myoglobin. (The oxygen complex of myoglobin gives blood its characteristic color; see Section 26-8). Does the practice of curing meat lead to nitrosamine contamination? The answer is yes: Nitrosamines have been detected in a variety of cured meats, such as smoked fish, frankfurters (N-nitrosodimethylamine), and fried bacon [N-nitrosoazacyclopentane (N-nitrosopyrrolidine); see Exercise 25-7]. This finding poses an environmental dilemma: Removing sodium nitrite as a food additive would prevent it from becoming a source of nitrosamines, but it would also cause a significant rise in botulism poisoning. Moreover, human exposure to nitrosamines occurs not only through meat; nitrosamines are also present in beer, nonfat dry milk, tobacco products, rubber, some cosmetics, and the gastric juices of the stomach. Their occurrence in the stomach can be called “natural,” as it is traced to bacteria in the mouth that reduce nitrate (which is, in turn, prevalent in vegetables such as spinach, beets, radishes, celery, and cabbages) to
O
H
NO f HN
nitrite, which then reacts with amines present in other foods we ingest. Because of the potential carcinogenicity of nitrosamines in humans, the level of nitrite in foods is closely regulated (,200 ppm). During the past 20 years, a number of measures, such as the use of additives that inhibit nitrosamine formation in curing (for example, vitamin C; Section 22-9) or changes in production processes (for example, the making of beer), have helped to significantly reduce human exposure to nitrosamines, estimated to be about 0.1 mg per day. It can be argued that we have evolved to tolerate such small quantities, but definitive data are still lacking (see also Chemical Highlight 25-4, on natural pesticides).
N A NO
N N O
N-Nitrosoazacyclopentane (N-Nitrosopyrrolidine)
Nitrosamines are found in fried bacon.
21-10 Nitrosation of Amines
Chapter 21
999
The nitrosyl cation also attacks the nitrogen of N-methylamides. The products, Nmethyl-N-nitrosamides, are precursors to useful synthetic intermediates. Nitrosation of an N-Methylamide O B RCNHCH3
O B RCNCH3 A NO
NO
H
N-Methylamide
N-Methyl-N-nitrosamide
Exercise 21-22
Working with the Concepts: Applying Basic Principles: A New Reaction of NO1 The ketone shown below undergoes nitrosation to the corresponding oxime (for oximes, see Section 17-9). Explain by a mechanism. O
O NaNO2, HCl
NH
OH
Strategy This is a new reaction, not discussed in the text. To try to understand it, you need to identify the changes that have occurred in going from starting material to product and to recall the function of the reagent. Solution • Topologically, we have attached the NO unit to the a-position of the starting ketone. This becomes clearer if we consider the tautomer of the product: O
N H N O
• Turning to the reagent, this section has shown that sodium nitrite in the presence of acid is a source of the electrophilic nitrosyl cation. In our example, it appears that attack has occurred a to the carbonyl function. Do we know anything about how electrophiles might attack ketones in this manner? Or, to put it differently, how do we make the a-carbonyl carbon nucleophilic? • Recall from Chapter 18 that ketones are in equilibrium with their enol tautomers, an equilibrium that is established rapidly in the presence of catalytic acid. These enols feature electron-rich double bonds that are ready targets for electrophiles, such as iminium ions (Mannich reaction, Section 21-9) or halogens (Section 18-3). Let us formulate a similar attack by NO1.
EH ðO
š ðOH
ðOð
ðNqO ð
H
NN
O ð
• We have now established the connectivity of the final product. To form it, all we need is to deprotonate the carbonyl function and tautomerize the nitrosyl part to the oxime. ðOð
ðOð
ðOð H
H NN
Oð
For the follow-up Exercise 21-23, see p. 1002.
NN
Oð
NN
H
N O
H
i
EH ðO
ðOH
1000
Amines and Their Derivatives
Chapter 21
CHE M I C AL H I G H L I G H T 2 1 - 4 Amines in Industry: Nylon In addition to their significance in medicine (Chemical Highlight 21-1), the amines have numerous industrial applications. An example is 1,6-hexanediamine (hexamethylenediamine, HMDA), needed in the
manufacture of nylon. The amine is polycondensed with hexanedioic (adipic) acid to produce Nylon 6,6, out of which hosiery, gears, and millions of tons of textile fiber are made.
Polycondensation of Adipic Acid with HMDA O OH
HO
O
O
Hexanedioic acid (Adipic acid)
O O O
H 2N
NH2 H
Double salt
1,6-Hexanediamine (Hexamethylenediamine)
O
H2O Polymerization
H H2N
NH2
270 C, 250 psi
O
O
O
Nylon 6,6
The high demand for nylon has stimulated the development of several ingenious cheap syntheses of the monomeric precursors. Thus, hexanedioic (adipic) acid is currently produced from benzene by three different multistep routes, all culminating in the last step in the salt-catalyzed oxidation of cyclohexanol (or cyclohexanol/cyclohexanone mixtures) with nitric acid. A “green” approach was disclosed in 2006 by chemists at Rhodia Chimie, France, in which air is used as the oxidizing agent instead of the toxic and corrosive (and more expensive) HNO3.
Adipic Acid Production Current industrial process
A “green” synthesis
OH HNO3, 60 C, V, Cu salts
COOH COOH Hexanedioic acid (Adipic acid)
Air, 120 C Mn, Co salts
The left leg of movie star Marie Wilson (who can be seen dangling from a crane) advertising Nylon stockings (Hollywood, 1949).
21-10 Nitrosation of Amines
The hexanediamine component of nylon was originally made by Wallace Carothers* at E. I. du Pont de Nemours from hexanedioic acid, which was turned into
Chapter 21
1001
hexanedinitrile (adiponitrile) by treatment with ammonia. Finally, catalytic hydrogenation furnished the diamine.
Hexanediamine from Adipic Acid O O B B HOC(CH2)4COH
NH3, 4 H2O
Hexanedioic acid (Adipic acid)
H2, Ni, 130 C, 2000 psi
NqC(CH2)4C q N Hexanedinitrile (Adiponitrile)
Later, a still shorter hexanedinitrile synthesis was discovered that used 1,3-butadiene as a starting material. Chlorination of butadiene furnished a mixture of 1,2-and 1,4-dichlorobutene (Section 14-6). This mixture can be
H2N(CH2)6NH2 1,6-Hexanediamine (Hexamethylenediamine)
converted directly into only the required 3-hexenedinitrile with sodium cyanide in the presence of cuprous cyanide. Selective hydrogenation of the alkenyl double bond then furnishes the desired product.
Hexanedinitrile (Adiponitrile) from 1,3-Butadiene
CH2 P CHO CH P CH2
Cl O Cl
ClCH2CH P CHCH2Cl
Cl A ClCH2CHCH P CH2
NCCH2CH P CHCH2CN
CuCN, NaCN NaCl H2, catalyst
3-Hexenedinitrile
In the mid-1960s, Monsanto developed a process that uses a more expensive starting material but takes just one
NC(CH2)4CN Hexanedinitrile
step: the electrolytic hydrodimerization of propenenitrile (acrylonitrile).
Electrolytic Hydrodimerization of Propenenitrile (Acrylonitrile) 2 CH2P CHC qN
2e
2 H
N q CCHCH2 O CH2CHC q N A A H H
Propenenitrile (Acrylonitrile)
To counter Monsanto’s challenge, du Pont devised yet another synthesis, again starting with 1,3-butadiene, but now eliminating the consumption of chlorine, removing the toxicwaste problems of the disposal of copper salts, and using cheaper hydrogen cyanide rather than sodium cyanide. The
synthesis is based on the conceptually simplest approach: direct regioselective anti-Markovnikov addition of two molecules of hydrogen cyanide to butadiene. A transition-metal catalyst, such as iron, cobalt, or nickel, is needed to effect this regiocontrol.
Hydrogen Cyanide Addition to 1,3-Butadiene CH2PCHCH P CH2 2 HCN
*Dr. Wallace H. Carothers (1896 – 1937), E. I. du Pont de Nemours and Company, Wilmington, Delaware.
Catalyst
NCCH2CHCHCH2CN A A H H
Amines and Their Derivatives
Chapter 21
Exercise 21-23
Try It Yourself The following reaction has been explored as a possible first step in the preparation of acetaminophen (Chapter 16 Opening). What is its mechanism and what is the mechanistic connection to the process described in Exercise 21-22? OH
OH
OH NO
NaNO2, HCl
NO 90%
10%
Base treatment of N-methyl-N-nitrosamides gives diazomethane N-Methyl-N-nitrosamides are converted into diazomethane, CH2N2, upon treatment with aqueous base. O C
A
N A
OP
Making Diazomethane
B
A
H3C
A
1002
NH2
KOH, H2O, (CH3CH2)2O, 0°C
N
N-Methyl-N-nitrosourea
Nð CH2 PNPš
NH3
K2CO3
H2O
Diazomethane
Diazomethane is used in the synthesis of methyl esters from carboxylic acids. However, it is exceedingly toxic and highly explosive in the gaseous state (b.p. 2248C) and in concentrated solutions. It is therefore usually generated in dilute ether solution and immediately allowed to react with the acid. This method is very mild and permits esterification of molecules possessing acid- and base-sensitive functional groups, as shown in the following example. H N
O
H N
O
OH
OCH3
CH2N2, (CH3CH2)2O, CH3OH
O O
O NH2
O
O
O NH2
75% When it is irradiated or exposed to catalytic amounts of copper, diazomethane evolves nitrogen to generate the reactive carbene methylene, H2C: Methylene reacts with alkenes by addition to form cyclopropanes stereospecifically (Section 12-9). Exercise 21-24 Formulate a mechanism for the methyl esterification by diazomethane. (Hint: Review the resonance structures for CH2N2 in Section 1-5.)
In Summary Nitrous acid attacks amines, thereby causing N-nitrosation. Secondary amines give N-nitrosamines, which are notorious for their carcinogenicity. N-Nitrosamines derived
Chapter Integration Problems
from primary amines decompose through SN1 or SN2 processes to a variety of products. N-Nitrosation of N-methylamides results in the corresponding N-nitrosamides, which liberate diazomethane upon treatment with hydroxide. Diazomethane is a reactive substance used in the methylesterification of carboxylic acids and as a source of methylene for the cyclopropanation of alkenes.
THE BIG PICTURE We have seen the role of the amine function as a base or nucleophile at various points of the text. As early as in Sections 1-3 and 1-8, we learned about its valence electron count and hybridization, using NH3 as an example. In Section 2-2, we saw its first use as a base or nucleophile, and a resulting practical application in the resolution of ammonium tartrates was featured in Section 5-8. Amine nucleophilicity was an essential part of Chapters 6 and 7, elaborated in its synthetic aspects in Sections 17-9, 18-4, 18-9, 19-10, 20-6, and 20-7. The present chapter summarizes, reinforces, and extends this material. In particular, it points out the similarities and differences between the amines and the alcohols. Thus, nitrogen is located immediately to the left of oxygen in the periodic table. It therefore has one valence electron less and, correspondingly, lesser positive nuclear charge. Hence, it is trivalent, and not divalent, and less electronegative. However, chemically, it shows many similarities, and the amine – imine functionalities strongly resemble their oxygen counterparts — namely, the alcohol – carbonyl functionalities. In the laboratory, as well as in nature, the two systems complement each other, as described in several Chemical Highlights and problems. Once again, the similarities and differences in how these systems function can be traced back to the similarities and differences in their electronic structures. We have now completed our survey of the basic organic chemistry compound classes, as defined by simple functional groups. In the remainder of the book, we shall examine compounds with multiple functional groups, whether they be the same or different. We shall see the amino group and amine derivatives again in these chapters, particularly when studying heterocycles, proteins, and nucleic acids. We shall learn that the combination of the amino and carbonyl groups underlies the molecular structure of life itself.
CHAPTER INTEGRATION PROBLEMS 21-25. On the basis of the synthetic methods for amines provided in this chapter, carry out retrosynthetic analyses of the antidepressant Prozac, with 4-trifluoromethylphenol and benzene as your starting materials. OH O
H NCH3
, F3C CF3 Prozac [R,S-N-Methyl-3-(4-trifluorophenoxy)-3-phenyl-1-propanamine]
SOLUTION As for any synthetic problem, you can envisage many possible solutions. However, the constraints of given starting materials, convergence, and practicality rapidly narrow the number of available options. Thus, it is clear that the 4-trifluoromethylphenoxy group is best introduced by Williamson ether synthesis (Section 9-6) with an appropriate benzylic halide, compound A (Section 22-1), with the use of our first starting compound, 4-trifluoromethylphenol.
Chapter 21
1003
1004
Chapter 21
Amines and Their Derivatives
Retrosynthetic Step to A
O
H NCH3
H NCH3
X O
F3C
F3C A
The task therefore reduces to devising routes to A, utilizing the second starting material, benzene, and suitable building blocks. We know that the best way to introduce an alkyl chain into benzene is by Friedel-Crafts acylation (Section 15-13). This reaction would also supply a carbonyl function, as in compound B, which could be readily transformed [by reduction to the alcohol (Section 8-6) and conversion of the hydroxy into a good leaving group, X (Chapter 9)] into compound A (or something O B similar). On paper, an attractive acylating reagent would be ClCCH2CH2NHCH3. However, this reagent carries two functional groups that would react with each other either intermolecularly to produce a polyamide (see Chemical Highlight 21-4) or intramolecularly to produce a b-lactam (Section 20-6). This problem could be circumvented by using a leaving group in place of the amino function, the latter to be introduced by one of the methods in Section 21-5. Retrosynthesis of A O O X
A
Cl
X
B
Consideration of cyanide as a building block (Section 21-5) opens up another avenue of retrosynthetic disconnection of compound B (and therefore compound A) through compound C. Retrosynthesis of B O
O CN
X
B C
CN
D
The required compound D could be envisaged to arise from acetylbenzene (made by Friedel-Crafts acetylation of benzene), followed by acid-catalyzed halogenation of the ketone (Section 18-3). Reduction of the nitrile group in compound C could be carried out with concomitant carbonyl conversion to give a primary amine version of compound A, namely, E. N-Methylation might be most conveniently accomplished by reductive amination (although, as we shall see in Chapter 22, the benzylic position may be sensitive to the reductive conditions employed). Retrosynthesis of E OH
O CN
C
O NH2
E
Phenyloxacyclopropane
CN
Chapter Integration Problems
Intermediate E can be approached by using nucleophilic cyanide in a different manner — namely, by attack at the less hindered position of phenyloxacyclopropane. The latter would arise from phenylethene (styrene) by oxacyclopropanation (Section 12-10), and phenylethene could in turn be readily made from acetylbenzene by reduction – dehydration. Inspection of the synthetic methods described in Sections 21-5 through 21-8 may give you further ideas about how to tackle the synthesis of Prozac, but they are all variations on the schemes formulated so far. However, consideration of the Mannich reaction (Section 21-9) provides a more fundamental alternative, attractive because of its more highly convergent nature to a derivative of compound B with the intact amino function in place. Indeed, this is the commercial route used by Eli Lilly to the final drug. Retro-Mannich Synthesis of B (XPCH3NH) O
O NCH3 H
CH2 P O
CH3NH2
B
21-26. Formulate a mechanism for the following reaction.
H2O (trace),
N
N A
B
SOLUTION As always, let us first analyze what is given, before delving into a possible solution. We note that C8H13N on the left turns into the same on the right: We are dealing with an isomerization. There are no reagents, only a catalytic amount of water (a proton source). The topology of a seven-membered ring changes to that of a five-membered one, and the functionality of a (di)enamine (Section 17-9) is transformed into that of an a,b-unsaturated imine (Section 18-8). Our task is to find a way by which water will open the seven-membered ring and eventually close some acyclic intermediate to the fivemembered ring product. What does an enamine do in the presence of water? Answer: it hydrolyzes (Section 17-9). Mechanism of Hydrolysis HOš OH
š N
OH ðš
H H
š N
N
A
š ðOH
H
C
ðOH
H
H2š O
N
OH ðš
OH ð D
N Hš OH E
H
N Hš ½Ok F
Let us formulate the mechanism of this step as applied to our starting material A. Crucial in dealing with enamines is the increased basicity (and nucleophilicity) of the b-carbon, which leads to its preferred protonation (Section 17-9) [and alkylation (Section 18-4)]. Thus, protonation of A gives B,
Chapter 21
1005
1006
Chapter 21
Amines and Their Derivatives
which is attacked by hydroxide to furnish C, thus completing the initial hydration sequence in enamine hydrolysis (Section 17-9). The next step, breaking the C – N bond, has a “quirk” in our system: The nitrogen, usually activated to leave by direct protonation, is part of an enamine unit, which is (again) protonated at carbon to generate D containing an iminium leaving group. Ring opening therefore proceeds through E to F. Now that we have accomplished ring opening, what is next? Simply looking at the topology of F and the desired B, we can see that we need to form a (double) bond between the b-carbon of the imine and the carbonyl carbon. If the imine were a ketone, we would accomplish this by an intramolecular aldol condensation (Section 18-7). Here, we utilize the imine as a masked ketone, via its enamine form G as the nucleophile. The individual steps are completely analogous.
An “Imine Version” of the Intramolecular Aldol Condensation H
H
O
HOš OH
N š
š ðOH
N f H
ðš OH
H
O
š N f H
H2š O
ð O
H
F
H G
H2š O
H
š N f H
ðš OH
ð OH
HO
N f H
H
H2š O
š N f H
ð OH
N B
Note that the rapid imine F to enamine G conversion obviates an alternative pathway of an intramolecular Mannich reaction (Section 21-9). Formulate this alternative!
New Reactions 1. Acidity of Amines and Amide Formation (Section 21-4) RNH2 R2NH
K
H2O
RNH
H3O
R2NLi
CH3CH2CH2CH2Li
1035
Ka
CH3CH2CH2CH3
Lithium dialkylamide
2 NH3
2 Na
Catalytic Fe3
2 NaNH2
H2
Ka
1010
2. Basicity of Amines (Section 21-4)
RNH2
RNH3
H2O
H2O
RNH3
HO
RNH2
H3O
K
New Reactions
Salt formation
RNH2
RNH3 Cl
HCl
Alkylammonium chloride General for primary, secondary, and tertiary amines
Preparation of Amines 3. Amines by Alkylation (Section 21-5)
š 2 RNH
R A RNH2 X
RX
General for primary, secondary, and tertiary amines
Drawback: Multiple alkylation R A RNH2 X
RNR3 X
RX
4. Primary Amines from Nitriles (Section 21-5)
RX
DMSO SN2
CN
X
RCN
LiAlH4 or H2, catalyst
RCH2NH2
R limited to methyl, primary, and secondary alkyl groups
5. Primary Amines from Azides (Section 21-5)
RX
CH3CH2OH SN2
N3
X
RN3
1. LiAlH4, (CH3CH2)2O 2. H2O
RNH2
R limited to methyl, primary, and secondary alkyl groups
6. Primary Amines by Gabriel Synthesis (Section 21-5) O
O
NH
1. K2CO3, H2O 2. RX, DMF
1. H2SO4, H2O, 120°C 2. NaOH, H2O
NOR
O
O
R limited to methyl, primary, and secondary alkyl groups
7. Amines by Reductive Amination (Section 21-6) O B RCR
NH3, NaBH3CN, H2O or alcohol
NH2 A R O C OR A H
Reductive methylation with formaldehyde R2NH
CH2 P O
NaBH3CN, CH3OH
R2NCH3
RNH2
Chapter 21
1007
1008
Chapter 21
Amines and Their Derivatives
8. Amines from Carboxylic Amides (Section 21-7) O R B D RCN G R
1. LiAlH4, (CH3CH2)2O 2. H2O
D RCH2N G
R R
9. Hofmann Rearrangement (Section 21-7) O B RCNH2
Br2, NaOH, H2O
RNH2
CO2
Reactions of Amines 10. Hofmann Elimination (Section 21-8) RCH2CH2NH2
Excess CH3I, K2CO3
Ag2O, H2O
RCH2CH2N(CH3)3 I
RCH2CH2N(CH3)3
OH
AgI
RCHP CH2
N(CH3)3
H2O
11. Mannich Reaction (Section 21-9) O B RCCH2R
CH2 P O
O B RCCHR A CH2N(CH3)2
1. HCl 2. HO
(CH3)2NH
12. Nitrosation of Amines (Section 21-10) Tertiary amines
NaNO2, HX
R3N
R3NO NO X Tertiary N-nitrosammonium salt
Secondary amines R
G NH D R
R
G NO NP O D R
NaNO2, H
N-Nitrosamine
Primary amines RNH2
NaNO2, H
RNPNOH
H
RN2
H2O
N2
R
mixture of products
13. Diazomethane (Section 21-10) O B CH3NCR A NP O
KOH
Nð CH2 PNPš
Esterification with diazomethane O B RCOH
CH2N2
O B RCOCH3
N2
Preparation of Amines
RNO2 Zn(Hg), HCl or H2, Ni or Fe, HCl
section number
O B EC H R R
R R G G CPN D R
O B C G G D CP C D D
O B C E H EH(R) R N A H(R)
RNH2
H, H2O
RNH2
H orOH, H2O,
Substrate: Raryl
Product: OH A RO C ONHR A R
Other product: O B EC H R R
Product: O A A J OCO CO C D A A RHN H
Other product: RCOOH
16-5
17-9
17-9
18-9
20-6
O B C E H EH(R) R N A H(R)
LiAlH4
O B RCNH2
RC q N
X2, NaOH
LiAlH4
RBr
HO N
G D
H(R) H(R)
Product: RNH2 20-6
20-7
20-8
21-5
&
H(R) G G @ CON D H(R)
21-5
21-6
21-9
Product: R2 H(R) A G R1O C O N D A H(R) H
Product: O B R1CCHR2 R A G CH2N D H(R)
22-4, 26-5
22-4
Product: RNH
Product: NH2
25-2
&
@
Product: GNHR
GC G C C ( Nu
NO2
25-6
26-2
Product:
Product: NH2 A RO C O CN A H
N
NH2
NO2
LiAlH4
RN3
HO N
G D
H(R)
, H(R) NaBH3CN O B EC H 2 1 R R
CH2 PO, H, R G HN D H(R) O B
RNH2
X NO2
R1CCH2R2
1009
NO2
NH 2
ð Nu
X
R A N
NaNH2, liquid NH3
NH3, HCN
N
O B C E H R H
21-9
O B R1CCHR2 A G CH2N D
R2 A G R1 O C O N D A H
O B 1 R CCH2R2, CH2 P O, H
R1 G C P O, H D 2 R
17-9
21-7
R2, R1 NaBH3CN
O B EC H
O B CF3COOH
A R
Substrate: NH2
16-5
A R
/∑
R1 C G G G CPN D R2
R3NO NO or R2NO NO
21-10
NaNO2, H
NO2
E E N
A R
N2
22-10
A R
Substrate: NH2
NaNO2, H
(
O B RCX
O B RCOH,
ON A
C GC
25-2
R A N
NHR
20-2
R
25-2
O
R
O B RCOR
Other product: ROH
20-4
O B CE E N A
G N(CH2)3OH D
O B CE E N A
19-10
R
O B CE E N A
section number
H(R) G G @ CON D H(R)
O B C G G D CP C D D
18-9
O A A J OCO CO C A A N H D G
NO2
NO2
22-4, 26-5
NO2
X
1. R X 2. H, H2O
Substrate: R2N A CN C
Substrate: R2NH
O B C E EH C
18-4
C
O B C E E R C
17-9
R2N A CN
@&
G
∑/
NO2
Reactions of Amines
C G
∑/
1010
&
R
R O
N A
25-3
RLi
21-4
O
G Nð D
R
R
G & @ H
21-4
H A N
RBr
21-5
or
O B G RCN D
26-6
O B RCOH, N P CPN
R A N D G
R A N
G & @
Problems
Important Concepts 1. Amines can be viewed as derivatives of ammonia, just as ethers and alcohols can be regarded as derivatives of water. 2. Chemical Abstracts names amines as alkanamines (and benzenamines), alkyl substituents on the nitrogen being designated as N-alkyl. Another system is based on the label aminoalkane. Common names are based on the label alkylamine. 3. The nitrogen in amines is sp3 hybridized, the nonbonding electron pair functioning as the equivalent of a substituent. This tetrahedral arrangement inverts rapidly through a planar transition state. 4. The lone electron pair in amines is less tightly held than in alcohols and ethers, because nitrogen is less electronegative than oxygen. The consequences are a diminished capability for hydrogen bonding, higher basicity and nucleophilicity, and lower acidity. 5. Infrared spectroscopy helps to differentiate between primary and secondary amines. Nuclear magnetic resonance spectroscopy indicates the presence of nitrogen-bound hydrogens; both hydrogen and carbon atoms are deshielded in the vicinity of the nitrogen. Mass spectra are characterized by iminium ion fragments. 6. Indirect methods, such as displacements with azide or cyanide, or reductive amination, are superior to direct alkylation of ammonia for the synthesis of amines. 1
7. The NR3 group in a quaternary amine, R¿2NR3, is a good leaving group in E2 reactions; this enables the Hofmann elimination to take place. 8. The nucleophilic reactivity of amines manifests itself in reactions with electrophilic carbon, as in haloalkanes, aldehydes, ketones, and carboxylic acids and their derivatives.
Problems 27. Give at least two names for each of the following amines. NH2 NH2 (b)
(a)
CH3 A NCH2CH2CH3
(d)
(g)
N
N H
(e) (CH3)3N
(c) Cl O B (f) CH3CCH2CH2N(CH3)2
Cl (h) (CH3CH2)2NCH2CHPCH2
28. Give structures that correspond to each of the following names. (a) N,N-Dimethyl-3-cyclohexenamine; (b) N-ethyl-2-phenylethylamine; (c) 2-aminoethanol; (d) m-chloroaniline. 29. Name each of the following compounds pictured in Chemical Highlight 21-1 according to IUPAC or Chemical Abstracts. Pay attention to the order of precedence of the functional groups. (a) Propylhexedrine; (b) amphetamine; (c) mescaline; (d) epinephrine. 30. As mentioned in Section 21-2, the inversion of nitrogen requires a change of hybridization. (a) What is the approximate energy difference between pyramidal nitrogen (sp3 hybridized) and trigonal planar nitrogen (sp2 hybridized) in ammonia and simple amines? (Hint: Refer to the Ea of inversion.) (b) Compare the nitrogen atom in ammonia with the carbon atom in each of the following species: methyl cation, methyl radical, and methyl anion. Compare the most stable geometries and the hybridizations of each of these species. Using fundamental notions of orbital energies and bond strengths, explain the similarities and differences among them.
Chapter 21
1011
1012
Chapter 21
Amines and Their Derivatives
31. Use the following NMR- and mass-spectral data to identify the structures of two unknown compounds, A and B. A: 1H NMR: d 5 0.92 (t, J 5 6 Hz, 3 H), 1.32 (broad s, 12 H), 2.28 (broad s, 2 H), and 2.69 (t, J 5 7 Hz, 2 H) ppm. Mass spectrum myz (relative intensity) 5 129(0.6) and 30(100). B: 1H NMR: d 5 1.00 (s, 9 H), 1.17 (s, 6 H), 1.28 (s, 2 H), and 1.42 (s, 2 H) ppm. Mass spectrum myz (relative intensity) 5 129(0.05), 114(3), 72(4), and 58(100). 32. The following spectroscopic data (13C NMR and IR) are for several isomeric amines of the formula C6H15N. Propose a structure for each compound. (a) 13C NMR: d 5 23.7 (CH3) and 45.3 (CH) ppm. IR: 3300 cm21. (b) 13C NMR: d 5 12.6 (CH3) and 46.9 (CH2) ppm. IR: no bands in 3250 – 3500 cm21 range. (c) 13C NMR: d 5 12.0 (CH3), 23.9 (CH2), and 52.3 (CH2) ppm. IR: 3280 cm21. (d) 13C NMR: d 5 14.2 (CH3), 23.2 (CH2), 27.1 (CH2), 32.3 (CH2), 34.6 (CH2), and 42.7 (CH2) ppm. IR: 1600 (broad), 3280, and 3365 cm21. (e) 13C NMR: d 5 25.6 (CH3), 38.7 (CH3), and 53.2 (Cquaternary) ppm. IR: no bands in 3250 – 3500 cm21 region. 33. The following mass-spectral data are for two of the compounds in Problem 32. Match each mass spectrum with one of them. (a) myz (relative intensity) 5 101(8), 86(11), 72(79), 58(10), 44(40), and 30(100). (b) myz (relative intensity) 5 101(3), 86(30), 58(14), and 44(100). 34. Is a molecule whose conjugate acid has a high pKa value a stronger or a weaker base than a molecule whose conjugate acid has a low pKa value? Explain using a general equilibrium equation. 35. In which direction would you expect each of the following equilibria to lie? (a) NH3 OH
NH2 H2O CH3NH3 OH
(b) CH3NH2 H2O (c) CH3NH2 (CH3)3NH
CH3NH3 (CH3)3N
36. How would you expect the following classes of compounds to compare with simple primary amines as bases and acids? (a) Carboxylic amides; for example, CH3CONH2
(b) Imides; for example, CH3CONHCOCH3
(c) Enamines; for example, CH2PCHN(CH3)2
(d) Benzenamines; for example,
NH2
37. Several functional groups containing nitrogen are considerably stronger bases than are ordinary amines. One is the amidine group found in DBN and DBU, both of which are widely used as bases in a variety of organic reactions. N
N
N
C N Amidine group
N
N 1,5-Diazabicyclo[4.3.0]non-5-ene (DBN)
1,8-Diazabicyclo[5.4.0]undec-7-ene (DBU)
NH
B
Another unusually strong organic base is guanidine, H2NCNH2. Indicate which nitrogen in each of these bases is the one most likely to be protonated and explain the enhanced strength of these bases relative to simple amines. 38. Reaction review: Without consulting the Reaction Road Map on page 1009, suggest reagents that convert each of the following starting materials into the indicated product: (a) (chloromethyl)benzene (benzyl chloride) into phenylmethanamine (benzylamine); (b) benzaldehyde into phenylmethanamine (benzylamine); (c) benzaldehyde into N-ethylphenylmethanamine (benzylethylamine); (d) (chloromethyl)benzene (benzyl chloride) into 2-phenylethanamine (phenethylamine); (e) benzaldehyde into N,N-dimethylphenylmethanamine (benzyldimethylamine); (f) 1-phenylethanone (acetophenone) into 3-amino-1-phenyl-1-propanone; (g) benzonitrile into phenylmethanamine (benzylamine); (h) 2-phenylacetamide into phenylmethanamine (benzylamine).
Problems
39. The following syntheses are proposed for amines. In each case, indicate whether the synthesis will work well, poorly, or not at all. If a synthesis will not work well, explain why.
(a) CH3CH2CH2CH2Cl
(b) (CH3)3CCl
1. KCN, CH3CH2OH 2. LiAlH4, (CH3CH2)2O
1. NaN3, DMSO 2. LiAlH4, (CH3CH2)2O
CONH2
CH3CH2CH2CH2NH2
(CH3)3CNH2 NH2
Br2, NaOH, H2O
(c)
(d)
CH3NH2
Br
NHCH3 O NK, DMF
1.
H ´
(e)
0 H
2. H, H2O,
[ CH Br 2 ≥ CH3
2. LiAlH4, (CH3CH2)2O
(f)
(g)
CHO
冢
冣
CH2 O NH 2
1. (CH3)3CNH2 2. NaBH3CN, CH3CH2OH
(h) H2NCH2CH2CHO
CH2NHC(CH3)3 NH
NaBH3CN, CH3CH2OH
Br
[ CH NH 2 2 ≥ CH3
0 H CH2NH2, HO
1.
COCl
(i)
H ´
O
Br 1. HNO3, H2SO4 2. Fe, H
( j)
1. NaH, THF 2. CH3I 3. LiAlH4, (CH3CH2)2O
NH
CH D 3 N
O NH2 40. For each synthesis in Problem 39 that does not work well, propose an alternative synthesis of the final amine, starting either with the same material or with a material of similar structure and functionality. 41. Give the structures of all possible nitrogen-containing organic products that might be expected to form on reaction of chloroethane with ammonia. (Hint: Consider multiple alkylations.) 42. Phenylpropanolamine (PPA) has long been an ingredient in many cold remedies and appetite suppressants. In late 2000, the FDA asked manufacturers to remove products containing this compound from the market because of evidence of an increased risk of hemorrhagic stroke. This action has caused a switch to the safer pseudoephedrine as the active component of such medications. OH NH2 A A CH O CH O CH3 Phenylpropanolamine
OH NHCH3 A A CH O CH O CH3 Pseudoephedrine
Chapter 21
1013
Chapter 21
HN
Amines and Their Derivatives
Suppose that you are the director of a major pharmaceutical laboratory with a huge stock of phenylpropanolamine on hand and the president of the company issues the order, “Pseudoephedrine from now on!” Analyze all your options, and propose the best solution that you can find for the problem. 43. Apetinil, an appetite suppressant (i.e., diet pill; see Chemical Highlight 21-1), has the structure shown in the margin. Is it a primary, a secondary, or a tertiary amine? Propose an efficient synthesis of Apetinil from each of the following starting materials. Try to use a variety of methods. (a) C6H5CH2COCH3
CH3 A (c) C6H5CH2CHCOOH
44. Suggest the best syntheses that you can for the following amines, beginning each with any organic compounds that do not contain nitrogen. (a) Butanamine; (b) N-methylbutanamine; (c) N,Ndimethylbutanamine. 45. Give the structures of the possible alkene products of Hofmann elimination of each of the following amines. If a compound can be cycled through multiple eliminations, give the products of each cycle.
(a)
H 3C
NH2 A CHCH2CH3
NH2
(b)
(d)
(e) N H
(c) H3C
N A CH3
N
46. What primary amine(s) would give each of the following alkenes or alkene mixtures upon Hofmann elimination? (a) 3-Heptene; (b) mixture of 2- and 3-heptene; (c) 1-heptene; (d) mixture of 1- and 2-heptene. 47. Formulate a detailed mechanism for the Mannich reaction between 2-methylpropanal, formaldehyde, and methanamine shown on page 994. 48. Reaction of the tertiary amine tropinone with (bromomethyl)benzene (benzyl bromide) gives not one but two quaternary ammonium salts, A and B. H3 C
G
N
CH2Br
A B
[C15H20NO]ⴙ Brⴚ
O Tropinone (C8H13NO)
Compounds A and B are stereoisomers that are interconverted by base; that is, base treatment of either pure isomer leads to an equilibrium mixture of the two. (a) Propose structures for A and B. (b) What kind of stereoisomers are A and B? (c) Suggest a mechanism for the equilibration of A and B by base. (Hint: Think “reversible Hofmann elimination.”) 49. Attempted Hofmann elimination of an amine containing a hydroxy group on the b-carbon gives an oxacyclopropane product instead of an alkene.
O
NH2
H2C O CH2
1. Excess CH3I 2. Ag2O, H2O 3.
O
O
HO
O
Apetinil
Br A (b) C6H5CH2CHCH3
O
1014
H2C O CH2 (CH3)3N
Problems
Chapter 21
1015
(a) Propose a sensible mechanism for this transformation. (b) Pseudoephedrine (see Problem 42) and ephedrine are closely related, naturally occurring compounds, as the similar names imply. In fact, they are stereoisomers. From the results of the following reactions, deduce the precise stereochemistries of ephedrine and pseudoephedrine.
Ephedrine
1. CH3I 2. Ag2O, H2O 3.
} & H
1. CH3I 2. Ag2O, H2O 3.
O (}H CH3
Pseudoephedrine
H}&
O (}H CH3
50. Show how each of the following molecules might be synthesized by Mannich or Mannich-like reactions. (Hint: Work backward, identifying the bond made in the Mannich reaction.) N(CH3)2
O (a)
(b)
N
(c) H3CCHCN A NH2
O
O
O (d)
O N
(e)
N
51. Tropinone (Problem 48) was first synthesized by Sir Robert Robinson (famous for the Robinson annulation reaction; Section 18-11), in 1917, by the following reaction. Show a mechanism for this transformation. H3C H
G
N
f
H3C f CP i H2C O CH3NH2 iC PO A H2C f P O H3C C f H
O
52. Illustrate a method for achieving the transformation shown in the margin, using combinations of reactions presented in Sections 21-8 and 21-9. 53. Give the expected product(s) of each of the following reactions. H3C (a)
HE CH
NH2
NaNO2, HCl, H2O
NaNO2, HCl, 0 C
(b) N H
54. Tertiary amines add readily as nucleophiles to carbonyl compounds, but, lacking a hydrogen on nitrogen, they cannot deprotonate to give a stable product. Instead, their addition gives an intermediate that is highly reactive toward other nucleophiles. Thus, tertiary amines are occasionally used as catalysts for the addition of weak nucleophiles to carboxylic acid derivatives. (a) Fill in the missing structures in the scheme below. (CH3 ) 3N 1 CH3COCl Δ
2Cl2
Intermediate A C5H9ClNO
Δ
Intermediate B (C5H9NO)1
O
O CH2
1016
Amines and Their Derivatives
Chapter 21
(b) Intermediate B readily enters into reactions with weak nucleophiles such as phenol (margin). Formulate the mechanism for this process and give the structure of the product.
OH
55. Phenol
Tertiary amines undergo reversible conjugate addition to a,b-unsaturated ketones (see Chapter 18). This process is the basis for the Baylis-Hillman reaction, which is catalyzed by tertiary amines, that resembles a crossed aldol reaction. An example is shown below. O
OH
O H
O
(CH3)3N
(a) Formulate a mechanism for this process. Begin with conjugate addition of the amine to the enone. Give the products of each of the following Baylis-Hillman reactions: O
(b) CH3CHO
(CH3)3N
O
(c)
56.
O H
(CH3)3N
Reductive amination of excess formaldehyde with a primary amine leads to the formation of a dimethylated tertiary amine as the product (see the following example). Propose an explanation. 2 CH2 P O
(CH3)3CCH2NH2
NaBH3CN, CH3OH
2,2-Dimethylpropanamine
(CH3)3CCH2N(CH3)2 84% N,N,2,2-Tetramethylpropanamine
57. Several of the natural amino acids are synthesized from 2-oxocarboxylic acids by an enzymecatalyzed reaction with a special coenzyme called pyridoxamine. Use electron-pushing arrows to describe each step in the following synthesis of phenylalanine from phenylpyruvic acid.
HOCH2
OH N
CO H D 2 G OH CH2
CH2N P C
CH2NH2 O B CH2CCO2H
HOCH2
CH3
Pyridoxamine
N
CH3
Phenylpyruvic acid
CO H D 2 H H E NP C G C CH2 HOCH2 OH
CHO
CO H D 2
HOCH2
HH N O CH K G C CH2 HOCH2 OH
H2O
N A H
CH3
N
CH3
OH N
CH3
Pyridoxal
NH2 A CH2CHCO2H Phenylalanine
Problems
58.
Chapter 21
1017
From the following information, deduce the structure of coniine, a toxic amine found in poison hemlock, which, deservedly, has a very bad reputation. IR: 3330 cm21. 1H NMR: d 5 0.91 (t, J 5 7 Hz, 3 H), 1.33 (s, 1 H), 1.52 (m, 10 H), 2.70 (t, J 5 6 Hz, 2 H), and 3.0 (m, 1 H) ppm. Mass spectrum: myz (relative intensity) 5 127 (M1, 43), 84(100), and 56(20).
Coniine
1. CH3I 2. Ag2O, H2O 3.
1. CH3I 2. Ag2O, H2O 3.
mixture of three compounds
(CH3)3N
mixture of 1,4-octadiene and 1,5-octadiene
59. Pethidine, the active ingredient in the narcotic analgesic Demerol, was subjected to two successive exhaustive methylations with Hofmann eliminations, and then ozonolysis, with the following results:
C15H21NO2
1. CH3I 2. Ag2O, H2O 3.
C16H23NO2
1. CH3I 2. Ag2O, H2O 3.
Pethidine
(CH3)3N
1. O3, CH2Cl2 2. Zn, H2O
C14H16O2
2 CH2O OHC } C O CO2CH2CH3 ( CHO
(a) Propose a structure for pethidine based on this information. (b) Propose a synthesis of pethidine that begins with ethyl phenylacetate and cis-1,4-dibromo-2butene. (Hint: First prepare the dialdehyde ester shown in the margin, and then convert it into pethidine.)
C O CO2CH2CH3 OHCCH2}( CH2CHO
60. Skytanthine is a monoterpene alkaloid with the following properties. Analysis: C11H21N. 1H NMR: two CH3 doublets (J 5 7 Hz) at d 5 1.20 and 1.33 ppm; one CH3 singlet at d 5 2.32 ppm; other hydrogens give rise to broad signals at d 5 1.3 – 2.7 ppm. IR: no bands $ 3100 cm21. Deduce the structures of skytanthine and degradation products A, B, and C from this information. Cl 1.
Skytanthine
1. CH3I 2. Ag2O, H2O 3.
C12H23N
1. O3, CH2Cl2 2. Zn, H2O
CH2 P O
2. KOH, H2O
C11H21NO
A
B
IR: v˜ ⴝ 1646 cmⴚ1
IR: v˜ ⴝ 1715 cmⴚ1
CH3COOH
C9H19NO
CO3H, CH2Cl2
Careful oxidation
O
C IR: v˜ ⴝ 3620 cmⴚ1
CH3 CH2N(CH3)2
IR: v˜ ⴝ 1745 cmⴚ1
61. Many alkaloids are synthesized in nature from a precursor molecule called norlaudanosoline, which in turn appears to be derived from the condensation of amine A with aldehyde B. Formulate a mechanism for this transformation. Note that a carbon – carbon bond is formed in the process. Name a reaction presented in this chapter that is closely related to this carbon – carbon bond formation.
CH2CH2NH2
HO
O B CH2CH
HO
HO
HO
HO
HO
alkaloids
NH OH
OH A
B
Norlaudanosoline
Chapter 21
Amines and Their Derivatives
Team Problem 62. Quaternary ammonium salts catalyze reactions between species dissolved in two immiscible phases, a phenomenon called phase-transfer catalysis. For example, heating a mixture of 1-chlorooctane dissolved in decane with aqueous sodium cyanide shows no sign of the SN2 product, nonanenitrile. On the other hand, addition of a small amount of (phenylmethyl)triethylammonium chloride results in a rapid, quantitative reaction.
N
Cl
CH3(CH2)7Cl NaCN 1-Chlorooctane
CH3(CH2)7CN NaCl 100% (Phenylmethyl)triethylammonium chloride
Nonanenitrile
As a team, discuss possible answers to the following questions: (a) What is the solubility of the catalyst in the two solvents? (b) Why is the SN2 reaction so slow without catalyst? (c) How does the ammonium salt facilitate the reaction?
Preprofessional Problems 63. One of the following four amines is tertiary. Which one? (a) Propanamine; (b) N-methylethanamine; (c) N,N-dimethylmethanamine; (d) N-methylpropanamine. 64. Identify the best conditions for the following transformation: O B CH3CH2CNH2
CH3CH2NH2 CO2
(a) H2, metal catalyst; (b) excess CH3I, K2CO3; (c) Br2, NaOH, H2O; (d) LiAlH4, ether; (e) CH2N2, ether. 65. Rank the basicities of the following three nitrogen-containing compounds (most basic first): NH3
CH3NH2
(CH3)4N1 NO32
A
B
C
(a) A . B . C; (b) B . C . A; (c) C . A . B; (d) C . B . A; (e) B . A . C. 66. Which of the following formulas best represents diazomethane?
š (a) CH2 P NPN
1018
(d) ðCH2 O Nq Nð
šOH N PCPN (b) HOš
(c)
EH š N P CPN H H
(e) CH2 O Nq Nð
67. Use the following partial IR- and mass-spectral data to identify one of the structures among the selection given. IR spectrum: 3300 and 1690 cm21; mass spectrum: myz 5 73 (parent ion). O O B (a) HCN(CH3)2 O B (d) CH3CH2CNH2
NHCH3 (b)
(c) H2NCH2Cq CCH2NH2
H O
(e)
NH2 H
C H A P T E R
TWENTY TWO
Chemistry of Benzene Substituents Alkylbenzenes, Phenols, and Benzenamines
N N
N
O F O enzene (Chapter 15) used to be a common O laboratory solvent until OSHA (U.S. Occupational Safety and Health Administration) placed it on its list of carcinogens. Chemists now use methylbenzene (toluene) instead, which has very similar solvating power but is not carcinogenic. Why not? The reason is the relatively high reactivity of the benzylic hydrogens that renders methylbenzene subject to fast metabolic degradation and extrusion from the body, unlike benzene, which can survive many days embedded in fatty and other tissues. Thus, the benzene ring, though itself quite unreactive because of its aromaticity, appears to activate neighboring bonds or, more generally, affects the chemistry of its substituents. You should not be too surprised by this finding, because it is complementary to the conclusions of Chapter 16. There we saw that substituents affect the behavior of benzene. Here we shall see the reverse. How does the benzene ring modify the behavior of neighboring reactive centers? This chapter takes a closer look at the effects exerted by the ring on the reactivity of alkyl substituents, as well as of attached hydroxy and amino functions. We shall see that the behavior of these groups (introduced in Chapters 3, 8, and 21) is altered by the occurrence of resonance. After considering the special reactivity of aryl-substituted (benzylic) carbon atoms, we turn our attention to the preparation and reactions of phenols and benzenamines (anilines). These compounds are found widely in nature and are used in synthetic procedures as precursors to substances such as aspirin, dyes, and vitamins.
B
Ciprofloxacin (“Cipro”) is a synthetic tetrasubstituted benzene derivative used widely to treat bacterial infections, especially of the urinary tract. The photo shows a laboratory worker producing tablets of the drug.
1020
Chapter 22
Sole site of reactivity
Phenylmethyl (benzyl) system Sites of reactivity
2-Propenyl (allyl) system
Chemistry of Benzene Substituents
22-1 Reactivity at the Phenylmethyl (Benzyl) Carbon: Benzylic Resonance Stabilization Methylbenzene is readily metabolized because its methyl C – H bonds are relatively weak with respect to homolytic and heterolytic cleavage. When one of these methyl hydrogens has been removed, the resulting phenylmethyl (benzyl) group, C6H5CH2 (see margin), may be viewed as a benzene ring whose p system overlaps with an extra p orbital located on an attached alkyl carbon. This interaction, generally called benzylic resonance, stabilizes adjacent radical, cationic, and anionic centers in much the same way that overlap of a p bond and a third p orbital stabilizes 2-propenyl (allyl) intermediates (Section 14-1). However, unlike allylic systems, which may undergo transformations at either terminus and give product mixtures (in the case of unsymmetrical substrates), benzylic reactivity is regioselective and occurs only at the benzylic carbon. The reason for this selectivity lies in the disruption of aromaticity that goes with attack on the benzene ring.
Benzylic radicals are reactive intermediates in the halogenation of alkylbenzenes We have seen that benzene will not react with chlorine or bromine unless a Lewis acid is added. The acid catalyzes halogenation of the ring (Section 15-9). Br Br–Br
Br–Br, FeBr3
No reaction
–HBr
In contrast, heat or light allows attack by chlorine or bromine on methylbenzene (toluene) even in the absence of a catalyst. Analysis of the products shows that reaction takes place at the methyl group, not at the aromatic ring, and that excess halogen leads to multiple substitution. REACTION
CH2H
CH2Br
Br2
HBr
(Bromomethyl)benzene
CH3
CH2Cl
CHCl2
CCl3
Cl2, hv
Cl2, hv
Cl2, hv
HCl
HCl
HCl
(Chloromethyl)benzene
(Dichloromethyl)benzene
(Trichloromethyl)benzene
Each substitution yields one molecule of hydrogen halide as a by-product. As in the halogenation of alkanes (Sections 3-4 through 3-6) and the allylic halogenation of alkenes (Section 14-2), the mechanism of benzylic halogenation proceeds through radical intermediates. Heat or light induces dissociation of the halogen molecule into atoms. One of them abstracts a benzylic hydrogen, a reaction giving HX and a phenylmethyl (benzyl) radical. This intermediate reacts with another molecule of halogen to give the product, a (halomethyl)benzene, and another halogen atom, which propagates the chain process.
22-1 Benzylic Resonance Stabilization
Chapter 22
1021
Mechanism of Benzylic Halogenation
MECHANISM ðš XOš Xð
Initiation: CH2 O H
or h
2ðš Xj
jCH2
CH2X
Xj
Propagation:
XOX
HX
Xj
Phenylmethyl (benzyl) radical
Remember: Single-headed (“fishhook”) arrows denote the movement of single electrons.
What explains the ease of benzylic halogenation? The answer lies in the stabilization of the phenylmethyl (benzyl) radical by the phenomenon called benzylic resonance (Figure 22-1). As a consequence, the benzylic C – H bond is relatively weak (DH8 5 87 kcal mol21, 364 kJ mol21); its cleavage is relatively favorable and proceeds with a low activation energy. Inspection of the resonance structures in Figure 22-1 reveals why the halogen attacks only the benzylic position and not an aromatic carbon: Reaction at any but the benzylic carbon would destroy the aromatic character of the benzene ring.
CH2
CH2
CH2
δ CH2
CH2 or
δ
B
A
δ
δ
H C H C
Exercise 22-1 For each of the following compounds, draw the structure and indicate where radical halogenation is most likely to occur upon heating in the presence of Br2. Then rank the compounds in approximate descending order of reactivity under bromination conditions. (a) Ethylbenzene; (b) 1,2-diphenylethane; (c) 1,3-diphenylpropane; (d) diphenylmethane; (e) (1-methylethyl)benzene.
Benzylic cations delocalize the positive charge Reminiscent of the effects encountered in the corresponding allylic systems (Section 14-3), benzylic resonance can affect strongly the reactivity of benzylic halides and sulfonates in nucleophilic displacements. For example, the 4-methylbenzenesulfonate (tosylate) of 4-methoxyphenylmethanol (4-methoxybenzyl alcohol) reacts with solvent ethanol rapidly via an SN1 mechanism. This reaction is an example of solvolysis, specifically ethanolysis, which we described in Chapter 7.
Figure 22-1 The benzene p system of the phenylmethyl (benzyl) radical enters into resonance with the adjacent radical center. The extent of delocalization may be depicted by (A) resonance structures, (B) dotted lines, or (C) orbitals.
1022
Chemistry of Benzene Substituents
Chapter 22
O B CH2OS B O
CH3O
CH3
SN1
CH3CH2OH
Ethanolysis
(4-Methoxyphenyl)methyl 4-methylbenzenesulfonate
REACTION
(A primary benzylic tosylate)
CH3O
CH2OCH2CH3
HO3S
CH3
1-(Ethoxymethyl)-4-methoxybenzene
The reason is the delocalization of the positive charge of the benzylic cation through the benzene ring, allowing for relatively facile dissociation of the starting sulfonate.
MECHANISM
Mechanism of Benzylic Unimolecular Nucleophilic Substitution CH3š O
CH2 OL
CH2
L
CH2
CH2
CH2
CH2
CH3CH2OH
product
Oð CH3 MATION AN I
CH3 Oð
CH3 Oð
CH3 O
CH3 Oð
Octet form
ANIMATED MECHANISM: Benzylic nucleophilic substitution
Benzylic cation
Several benzylic cations are stable enough to be isolable. For example, the X-ray structure of the 2-phenyl-2-propyl cation (as its SbF62 salt) was obtained in 1997 and shows the phenyl – C bond (1.41 Å) to be intermediate in length between those of pure single (1.54 Å) and double bonds (1.33 Å), in addition to the expected planar framework and trigonal arrangement of all sp2 carbons (Figure 22-2), as expected for a delocalized benzylic system.
Exercise 22-2 Which one of the two chlorides will solvolyze more rapidly: (1-chloroethyl)benzene, C6H5CHCl, or chloro(diphenyl)methane, (C6H5)2CHCl? Explain your answer. A CH3
H3C 114°
+ 123° 122° 1.41 Å
116°
H3C
1.40 Å
1.48 Å 1.43 Å 1.37 Å
Figure 22-2 Structure of the 2-phenyl-2-propyl cation.
The above SN1 reaction is facilitated by the presence of the para methoxy substituent, which allows for extra stabilization of the positive charge. In the absence of this substituent, SN2 processes may dominate. Thus, the parent phenylmethyl (benzyl) halides and sulfonates undergo preferential and unusually rapid SN2 displacements, even under solvolytic conditions, and particularly in the presence of good nucleophiles. As in allylic SN2 reactions (Section 14-3), two factors contribute to this acceleration. One is that the benzylic carbon is made relatively more electrophilic by the neighboring sp2-hybridized phenyl carbon
22-1 Benzylic Resonance Stabilization
X
−
Chapter 22
Figure 22-3 The benzene p system overlaps with the orbitals of the SN2 transition state at a benzylic center. As a result, the transition state is stabilized, thereby lowering the activation barrier toward SN2 reactions of (halomethyl)benzenes.
‡
H C H Nu
1023
−
(as opposed to sp3-hybridized ones; Section 13-2). The second is stabilization of the SN2 transition state by overlap with the benzene p system (Figure 22-3). MATION AN I
CH2Br CH2CN
SN2
CN
Br
ANIMATED MECHANISM: Benzylic nucleophilic substitution
81% (Bromomethyl)benzene (Benzyl bromide) (
Phenylethanenitrile (Phenylacetonitrile)
100 times faster than SN2 reactions of primary bromoalkanes)
Exercise 22-3 Phenylmethanol (benzyl alcohol) is converted into (chloromethyl)benzene in the presence of hydrogen chloride much more rapidly than ethanol is converted into chloroethane. Explain. Phenylmethyl (benzyl) cation
Resonance in benzylic anions makes benzylic hydrogens relatively acidic A negative charge adjacent to a benzene ring, as in the phenylmethyl (benzyl) anion, is stabilized by conjugation in much the same way that the corresponding radical and cation are stabilized. The electrostatic potential maps of the three species (rendered in the margin at an attenuated scale for optimum contrast) show the delocalized positive (blue) and negative (red) charges in the cation and anion, respectively, in addition to the delocalized electron (yellow) in the neutral radical. Resonance in Benzylic Anions CH3
ðCH 2
CH2 k
CH2
Phenylmethyl (benzyl) radical
CH2 k
H
pKa
41
The acidity of methylbenzene (toluene; pKa < 41) is therefore considerably greater than that of ethane (pKa < 50) and comparable to that of propene (pKa < 40), which is deprotonated to produce the resonance-stabilized 2-propenyl (allyl) anion (Section 14-4). Consequently, methylbenzene (toluene) can be deprotonated by butyllithium to generate phenylmethyllithium.
Phenylmethyl (benzyl) anion
1024
Chemistry of Benzene Substituents
Chapter 22
Deprotonation of Methylbenzene CH3
CH2Li
(CH3)2NCH2CH2N(CH3)2, THF,
CH3CH2CH2CH2Li
Methylbenzene (Toluene)
CH3CH2CH2CH2H
Phenylmethyllithium (Benzyllithium)
Exercise 22-4 Which molecule in each of the following pairs is more reactive with the indicated reagents, and why? (a) (C6H5)2CH2 or C6H5CH3, with CH3CH2CH2CH2Li CH2Br
(b)
CH2Cl or
OCH3
OCH3
CH3CHOH
(c)
, with NaOCH3 in CH3OH
CH3CHOH or
, with HCl NO2
In Summary Benzylic radicals, cations, and anions are stabilized by resonance with the benzene ring. This effect allows for relatively easy radical halogenations, SN1 and SN2 reactions, and benzylic anion formation.
22-2 Benzylic Oxidations and Reductions Because it is aromatic, the benzene ring is quite unreactive. While it does undergo electrophilic aromatic substitutions (Chapters 15 and 16), reactions that dismantle the aromatic six-electron circuit, such as oxidations and reductions, are much more difficult to achieve. In contrast, such transformations occur with comparative ease when taking place at benzylic positions. This section describes how certain reagents oxidize and reduce alkyl substituents on the benzene ring.
Oxidation of alkyl-substituted benzenes leads to aromatic ketones and acids Reagents such as hot KMnO4 and Na2Cr2O7 may oxidize alkylbenzenes all the way to benzoic acids. Benzylic carbon – carbon bonds are cleaved in this process, which usually requires at least one benzylic C – H bond to be present in the starting material (i.e., tertiary alkylbenzenes are inert). Complete Benzylic Oxidations of Alkyl Chains CH2CH2CH2CH3
1. KMnO4, HO, 2. H, H2O
COOH
3 CO2
HOOC
H3C
80% 1-Butyl-4-methylbenzene
1,4-Benzenedicarboxylic acid (Terephthalic acid)
22-2 Benzylic Oxidations and Reductions
The reaction proceeds through first the benzylic alcohol and then the ketone, at which stage it can be stopped under milder conditions (see margin and Section 16-5). The special reactivity of the benzylic position is also seen in the mild conditions required for the oxidation of benzylic alcohols to the corresponding carbonyl compounds. For example, manganese dioxide, MnO2, performs this oxidation selectively in the presence of other (nonbenzylic) hydroxy groups. (Recall that MnO2 was used in the conversion of allylic alcohols into a,b-unsaturated aldehydes and ketones; see Section 17-4.)
Chapter 22
1,2,3,4-Tetrahydronaphthalene (Tetralin) CrO3, CH3COOH , H2O, 21°C
Selective Oxidation of a Benzylic Alcohol with Manganese Dioxide H CH3O
OH
O
OH CH3O
OH
OH
MnO2, acetone, 25°C, 5 h
CH3O
Not isolated
CH3O 94%
O
Benzylic ethers are cleaved by hydrogenolysis Exposure of benzylic alcohols or ethers to hydrogen in the presence of metal catalysts results in rupture of the reactive benzylic carbon – oxygen bond. This transformation is an example of hydrogenolysis, cleavage of a s bond by catalytically activated hydrogen.
CH2 O OR CH2H H2, Pd–C, 25°C
HOR
Exercise 22-5 Write synthetic schemes that would connect the following starting materials with their products. CH3
CH2CH3
(a)
O H 3C
O H 3C
O
CH3
(b)
O
CH3 O
71% 1-Oxo-1,2,3,4-tetrahydronaphthalene (1-Tetralone)
Cleavage of Benzylic Ethers by Hydrogenolysis
O
Hydrogenolysis is not possible for ordinary alcohols and ethers. Therefore, the phenylmethyl (benzyl) substituent is a valuable protecting group for hydroxy functions. The following scheme shows its use in part of a synthesis of a compound in the eudesmane class of essential oils, which includes substances of importance in both medicine and perfumery.
1025
1026
Chemistry of Benzene Substituents
Chapter 22
Phenylmethyl Protection in a Complex Synthesis HO % CH3 %
1. NaH, THF 2. C6H5CH2Br
O
O
≥ H
O O
RO % CH3 %
Protection of OH (Section 9-6)
O
CH3COOH, H2O
O
≥ H
Deprotection of carbonyl group (Section 17-8)
O 80%
(R C6H5CH2)
RO % CH3 %
Wittig reaction (Section 17-12)
O
RO % CH3 % ≥ H
1. BH3, THF 2. Oxidation (to alcohol) 3. Oxidation (to ketone)
CHCH3
RO % CH3 %
Hydroboration–oxidation* (Section 12-8) Oxidation (Section 8-6)
≥ H
94%
%
≥ H 93%
CH3CHP P(C6H5)3, DMSO
COCH3 99%
(Mixture of E and Z isomers)
1. CH3Li, (CH3CH2)2O 2. H, H2O
98%
H2, Pd–C, CH3CH2OH Deprotection of OR
HO % CH3 % CH3 A CCH3 A OH
≥ H
O
%
CH3 A CCH3 A OH
≥ H
%
(Section 8-8)
RO % CH3 %
O
98%
*While the literature reports the use of special oxidizing agents, in principle 2. H2O2, 2OH and 3. CrO3 would have been satisfactory.
Because the hydrogenolysis of the phenylmethyl (benzyl) ether in the final step occurs under neutral conditions, the tertiary alcohol function survives untouched. A tertiary butyl ether would have been a worse choice as a protecting group, because cleavage of its carbon – oxygen bond would have required acid (Section 9-8), which may cause dehydration (Section 9-2).
In Summary Benzylic oxidations of alkyl groups take place in the presence of permanganate or chromate; benzylic alcohols are converted into the corresponding ketones by manganese dioxide. The benzylic ether function can be cleaved by hydrogenolysis in a transformation that allows the phenylmethyl (benzyl) substituent to be used as a protecting group for the hydroxy function in alcohols.
22-3 Names and Properties of Phenols Arenes substituted by hydroxy groups are called phenols (Section 15-1). The p system of the benzene ring overlaps with an occupied p orbital on the oxygen atom, a situation resulting in delocalization similar to that found in benzylic anions (Section 22-1). As one result of this extended conjugation, phenols possess an unusual, enolic structure. Recall that enols are usually unstable: They tautomerize easily to the corresponding ketones because of the relatively strong carbonyl bond (Section 18-2). Phenols, however, prefer the enol to the keto form because the aromatic character of the benzene ring is preserved.
22-3 Names and Properties of Phenols
Chapter 22
Keto and Enol Forms of Acetone and Phenol ðš OH
ðOð H H3C
K ~ 109
C H2 Acetone
ðOð
š ðOH
H
CH2
H3C
2-Propenol
šH ðO
š
š
OH
OH
ð
H K ~ 10
š
OH
ð
13
2,4-Cyclohexadienone
š
Phenol
Phenols and their ethers are ubiquitous in nature; some derivatives have medicinal and herbicidal applications, whereas others are important industrial materials. This section first explains the names of these compounds. It then describes an important difference between phenols and alkanols — phenols are stronger acids because of the neighboring aromatic ring.
Phenols are hydroxyarenes Phenol itself was formerly known as carbolic acid. It forms colorless needles (m.p. 418C), has a characteristic odor, and is somewhat soluble in water. Aqueous solutions of it (or its methyl-substituted derivatives) are applied as disinfectants, but its main use is for the preparation of polymers (phenolic resins; Section 22-6). Pure phenol causes severe skin burns and is toxic; deaths have been reported from the ingestion of as little as 1 g. Fatal poisoning may also result from absorption through the skin. Substituted phenols are named as phenols, benzenediols, or benzenetriols, according to the system described in Section 15-1, although some common names are accepted by IUPAC (Section 22-8). These substances find uses in the photography, dyeing, and tanning industries. The compound bisphenol A (shown in the margin; see also Chemical Highlight 22-1) is an important monomer in the synthesis of epoxyresins and polycarbonates, materials that are widely employed in the manufacture of durable plastic materials, food packaging, dental sealants, and coatings inside beverage cans (Chemical Highlight 22-1). Phenols containing the higher-ranking carboxylic acid functionality are called hydroxybenzoic acids. Many have common names. Phenyl ethers are named as alkoxybenzenes. As a substituent, C6H5O is called phenoxy.
OH
COOH
OH
OH
OH
H3C O C O CH3
OH Bisphenol A
OH OH
NO2 CH3 4-Methylphenol (p-Cresol)
Cl 4-Chloro3-nitrophenol
OH
OH OH 3-Hydroxybenzoic acid (m-Hydroxybenzoic acid)
1,4-Benzenediol (Hydroquinone)
1,2,3-Benzenetriol (Pyrogallol)
Many examples of phenol derivatives, particularly those exhibiting physiological activity, are depicted in this book (e.g., see Chemical Highlights 5-4, 9-1, 21-1, 22-1, and 22-2, and
1027
1028
Chemistry of Benzene Substituents
Chapter 22
Sections 4-7, 9-11, 15-1, 22-9, 24-12, 25-8, and 26-1). You are very likely to have ingested without knowing it the four phenol derivatives shown here.
OH
O H3C
EO
N A H
HO
Capsaicin (Active ingredient in hot pepper, as in jalapeño or cayenne pepper)
Chili peppers.
HO
OH
OH
Resveratrol
OH
(Cancer chemopreventive from grapes; see also Chemical Highlight 22-1)
HO
O
OH
~O HO
OH
O
OH OH O
HO
4-(4-Hydroxyphenyl)-2-butanone
Epigallocatechin-3 gallate (Cancer chemopreventitive from green tea)
(Flavor of raspberries)
Green tea.
Raspberries.
Phenols are unusually acidic Phenols have pKa values that range from 8 to 10. Even though they are less acidic than carboxylic acids (pKa 5 3 – 5), they are stronger than alkanols (pKa 5 16 – 18). The reason is resonance: The negative charge in the conjugate base, called the phenoxide ion, is stabilized by delocalization into the ring. Acidity of Phenol OH ðš
Oð ðš H
ðOð
ðOð
k
pKa
10
Phenoxide ion
ðOð k
22-3 Names and Properties of Phenols
1029
Chapter 22
The acidity of phenols is greatly affected by substituents that are capable of resonance. 4-Nitrophenol (p-nitrophenol), for example, has a pKa of 7.15.
OH ðš
Oð ðš
ðOð
H
ðOð
k
ðOð etc.
N
N
K H k ð O ½O
K H k ð O ½O
N
N H k ð O ½O
K H k ð O ½O
pKa 7.15
N H k ½ ½O
ýO E
Negative charge delocalized into the nitro group
The 2-isomer has similar acidity (pKa 5 7.22), whereas nitrosubstitution at C3 results in a pKa of 8.39. Multiple nitration increases the acidity to that of carboxylic or even mineral acids. Electron-donating substituents have the opposite effect, raising the pKa.
OH
OH NO2
NO2
O 2N
OH NO2
NO2
CH3
2,4-Dinitrophenol
2,4,6-Trinitrophenol (Picric acid)
4-Methylphenol ( p-Cresol)
pKa 4.09
pKa 0.25
pKa 10.26
As Section 22-5 will show, the oxygen in phenol and its ethers is also weakly basic, in the case of ethers giving rise to acid-catalyzed cleavage.
Exercise 22-6 Why is 3-nitrophenol (m-nitrophenol) less acidic than its 2- and 4-isomers but more acidic than phenol itself?
Exercise 22-7 Rank in order of increasing acidity: phenol, A; 3,4-dimethylphenol, B; 3-hydroxybenzoic (m-hydroxybenzoic) acid, C; 4-(fluoromethyl)phenol [p-(fluoromethyl)phenol], D.
In Summary Phenols exist in the enol form because of aromatic stabilization. They are named according to the rules for naming aromatic compounds explained in Section 15-1. Those derivatives bearing carboxy groups on the ring are called hydroxybenzoic acids. Phenols are acidic because the corresponding anions are resonance stabilized.
1030
Chemistry of Benzene Substituents
Chapter 22
CHE M I C AL H I G H L I G H T 2 2 - 1 Two Phenols in the News: Bisphenol A and Resveratrol This section mentions two phenols as examples of chemicals to which you have potentially frequent exposure: bisphenol A (p. 1027) and resveratrol (p. 1028). Bisphenol A is the essential ingredient of the familiar polycarbonate plastics used in clear baby bottles, instant-formula cans, dental sealants, the linings of food and beverage containers, and reusable plastic bottles. More than 2 billion pounds are made each year in the United States alone, and three times as much are made worldwide. Yet there is continuing controversy surrounding this monomer, first made in 1891— a controversy that illustrates the difficulty of interpreting scientific data in terms of human risk assessment. Carbonate unit
O
CH3 A C A CH3
O B O O C OO n
Bisphenol A unit Polycarbonate plastics
The problem is that bisphenol A behaves as an estrogen mimic in animals. The monomer has been found to leach from the plastic, at increasing rates when heated, such as in a microwave oven. For example, a study carried out in 2003 showed that even very low levels of bisphenol A (about 20 ppb) caused chromosome aberrations in developing mouse eggs. These amounts are in the same range as are encountered in human blood and urine. Because the processes through which human and mouse eggs are prepared for fertilization are very similar, the study is a cause for concern, although it does not prove that humans are at risk.
A polycarbonate plastic bottle in action.
Moreover, in other studies carried out with adult rats, bisphenol A appeared to have no adverse effects on reproduction and development. A critique of these findings, dated 2008, pointed out that fetuses and newborns lack the liver enzyme needed to detoxify the chemical, again raising the specter of adverse effects on children. As a report by the National Institute of Environmental Health Sciences stresses, the discrepancies in studies of this type may be due to the use of different strains of animals, varying degrees of exposure and differing background levels of estrogenic pollutants, dosing regimens, and housing of the animals
22-4 Preparation of Phenols: Nucleophilic Aromatic Substitution Phenols are synthesized quite differently from the way in which ordinary substituted benzenes are made. Direct electrophilic addition of OH to arenes is difficult because of the scarcity of reagents that generate an electrophilic hydroxy group, such as HO1. Instead, phenols are prepared by formal nucleophilic displacement of a leaving group from the arene ring by hydroxide, HO2, reminiscent of, but mechanistically quite different from, the synthesis of alkanols from haloalkanes. This section considers the ways in which this transformation may be achieved.
Nucleophilic aromatic substitution may follow an addition – elimination pathway Treatment of 1-chloro-2,4-dinitrobenzene with hydroxide replaces the halogen with the nucleophile, furnishing the corresponding substituted phenol. Other nucleophiles, such
22-4 Nucleophilic Aromatic Substitution
(singly versus group). Remember that we are talking about extremely small concentrations of a biologically active compound (parts per billion!) that affect only a certain percentage of animals and to differing degrees. And then there are the big questions of the extent to which animal studies are relevant to humans and whether there is a threshold level of exposure that humans can tolerate because of the presence of evolutionary natural detoxification mechanisms. In 2007 and 2008, two government-convened groups reiterated that human exposure is at a level that causes adverse effects in humans and that there is concern about prenatal and earlychildhood exposure. As a result, both Canada and the U.S. Congress began moving toward a ban of bisphenol A in children’s products, and some retailers, such as Wal-Mart and Toys“R”Us, are phasing out their sale. Concurrently, industrial producers are replacing polycarbonate plastics with other polymers. The issues surrounding the potentially harmful effects of chemicals in the environment are equally relevant with respect to their potentially beneficial effects. An example is resveratrol. This compound has been used in traditional medicine to treat conditions of the heart and liver, and recently scientists took an interest in its physiological properties. It is present in various plants and foods, such as eucalyptus, lily, mulberries, peanuts, and, most prominently, in the skin of white and, especially, red grapes, where it is found in concentrations of 50–100 mg/g. The compound is a chemical weapon against invading organisms, such as fungi. Grapes are used for wine making, so resveratrol occurs in red wine, at levels of up to 160 mg per ounce. Studies have suggested that the regular consumption of red wine reduces the incidence of coronary heart disease, a finding described as the “French paradox,” namely, the low incidence of heart problems in France despite the relatively high-fat diet.
Chapter 22
Resveratrol may be the active species, as recent research has indicated its potentially beneficial cardiovascular effects, including its action as an antioxidant, which inhibits lipid peroxidation (Section 22-9), and as an antiplatelet agent (Chemical Highlight 22-2), which prevents atherosclerosis. Other investigations have shown that the molecule is also an active antitumor agent, involved in retarding the initiation, promotion, and progression of certain cancers, with seemingly minimal toxicity. Perhaps most interesting, it was discovered that resveratrol significantly extended the lifespan of certain species of yeast, worm, the fruit fly, and fish. Along similar lines, it canceled the life-shortening effects of a high-fat diet in mice. As a result of these promising discoveries, resveratrol has been hailed by manufacturers as the “French paradox in a bottle” and pushed for wide-scale consumption. However, experts advise caution. For example, little is known about its metabolism and how it affects the liver, the above results are derived in large part from in vitro experiments, and, like bisphenol A, it has estrogen-like physiological effects, enhancing the growth of breast cancer cells. For the time being, an occasional glass of red wine may be the best course of action, if any!
Resveratrol protects grapes from fungus, such as that shown here.
as alkoxides or ammonia, may be similarly employed, forming alkoxyarenes and arenamines, respectively. Processes such as these, in which a group other than hydrogen is displaced from an aromatic ring, are called ipso substitutions (ipso, Latin, on itself). The products of these reactions are intermediates in the manufacture of useful dyes. REACTION
Nucleophilic Aromatic Ipso Substitution Ipso position
Cl
Nu NO2
NO2 ðNu
NO2
Electron withdrawing
NO2
1031
Cl
1032
Chemistry of Benzene Substituents
Chapter 22
Cl
OH NO2
NO2 Na2CO3, HOH, 100°C
NO2
NaCl
NO2 90%
1-Chloro-2,4dinitrobenzene
2,4-Dinitrophenol
Cl
NH2 NO2
NO2
ðNH3,
NH4Cl
NO2 85%
NO2
2,4-Dinitrobenzenamine (2,4-Dinitroaniline)
The transformation is called nucleophilic aromatic substitution. The key to its success is the presence of one or more strongly electron-withdrawing groups on the benzene ring located ortho or para to the leaving group. Such substituents stabilize an intermediate anion by resonance. In contrast with the SN2 reaction of haloalkanes, substitution in these reactions takes place by a two-step mechanism, an addition – elimination sequence similar to the mechanism of substitution of carboxylic acid derivatives (Sections 19-7 and 20-2). Mechanism of Nucleophilic Aromatic Substitution Step 1. Addition (facilitated by resonance stabilization)
MECHANISM
Nu
Cl
Nu
Cl
NO2
Nu
Cl
NO2
k
k
ðOð N
Eš Oð
Nuð
E ðš O
Cl
N N
ðOð B N Eš Oð
NO2
N Hš Oð
ð O
Nu Oð
NO2
Cl NO2
Nu
Cl
Oð ðš N
Eš Oð
MATION AN I
ANIMATED MECHANISM Nucleophilic aromatic substitution
N Hš Oð
E ðš O
NO2
The negative charge is strongly stabilized by resonance involving the ortho- and para-NO2 groups.
22-4 Nucleophilic Aromatic Substitution
Chapter 22
1033
Step 2. Elimination (only one resonance structure is shown) ðOð š Clð B NE š Oð k
Nu
ðOð B NE š Oð
Nu
NO2
š ðCl ð
NO2
In the first and rate-determining step, ipso attack by the nucleophile produces an anion with a highly delocalized charge, for which several resonance structures may be written, as shown. Note the ability of the negative charge to be delocalized into the electron-withdrawing groups. In contrast, such delocalization is not possible in 1-chloro-3,5-dinitrobenzene, in which these groups are located meta; so this compound does not undergo ipso substitution under the conditions employed. Nuð
Cl
Nu
O2N
NO2
Cl
Nu
k
O2N
NO2
Cl
Cl k
O2N
Nu
NO2
O2N
NO2
Meta-NO2 groups do not provide resonance stabilization of the negative charge.
In the second step, the leaving group is expelled to regenerate the aromatic ring. The reactivity of haloarenes in nucleophilic substitutions increases with the nucleophilicity of the reagent and the number of electron-withdrawing groups on the ring, particularly if they are in the ortho and para positions. Exercise 22-8 Write the expected product of reaction of 1-chloro-2,4-dinitrobenzene with NaOCH3 in boiling CH3OH.
Exercise 22-9
Working with the Concepts: Using Nucleophilic Aromatic Substitution in Synthesis Ofloxacin, an antibiotic of the quinolone class (Section 25-7), is used in the treatment of infections of the respiratory and urinary tract, the eyes, the ears, and skin tissue. The quinolone antibiotics are yet another alternative in the ongoing battle against penicillin (and other drug)-resistant bacteria. The final three steps in the synthesis of ofloxacin are shown below, the last of which is a simple ester hydrolysis (Section 20-4). Formulate mechanisms for the other two transformations from A to B and then to C. O
F
O
O
F
O
CH3 A N , toluene, 100C N H
O
O
NaH, 70–90°C
F F
N
F
F HN
OH
O
CH3
H3C A
B
O
O
F
O
F
O
O B COH
NaOH, H2O
N
N EN
O
CH3
CH3
C
N
N EN
O
CH3
Ofloxacin
CH3
Strategy Taking an inventory of the topological changes and of the nature of the bonds formed and bonds broken, we note that B and C are the result of two successive intra- and one intermolecular nucleophilic aromatic substitutions, respectively. The first steps are ring closures and therefore especially favored for entropic and enthalpic reasons. (Formation of six-membered rings is a favorable reaction, see Sections 9-6 and 17-7.) The aromatic ring in A should be highly activated with respect to nucleophilic attack, because it bears four inductively electron-withdrawing fluorine substituents, in addition to the resonance-based electron-withdrawing carbonyl group. To propose a mechanism, let’s look at the individual steps. Solution • In the conversion of A to B, the strong base Na1H2 is employed. Where are the acidic sites in A? Most obvious is the hydroxy function (pKa , 15–16, Section 8-3, Table 8-2), which will certainly be deprotonated. • What about the amino group? Closer inspection of structure A shows that the nitrogen is connected via a double bond to two carbonyl moieties. Thus, the function is resonance stabilized in much the same way as an ordinary amide (but more so): Resonance in A ðOð
F
ðOð
ðš Oð
F
ðOð
O
O
F
O
F š F HN
F
ðOð ðš Oð
F F
OH
F HN
F
H3C
OH
F
H3C
F HN
OH
H3C
Consequently, it should be relatively acidic, even more acidic than an amide (pKa < 22, Section 20-7). It is therefore likely that A is doubly deprotonated before attacking the benzene nucleus. Nucleophilic aromatic substitution is then readily formulated to assemble the first new ring. In the scheme below, only one—the dominating—resonance form of the anion resulting from nucleophilic attack on the arene is shown, that in which the negative charge is delocalized onto the carbonyl oxygen. First Ring Closure ðOð
F
O
ðš Oð
O F
F ðš N
F
ðOð
O
F
š Oð š
F
O
F
ðFð F š HC
H3C
š N
ðš Fð š
O š N
F
š Oð š
3
O
š Oð š
F H3C
The second nucleophilic substitution by alkoxide is unusual, inasmuch as it has to occur at a position meta to the carbonyl function. Therefore, the intermediate anion is not stabilized by resonance, but solely by inductive effects. The regioselectivity is controlled by strain: The alkoxide cannot “reach” the more favorable carbon para to the carbonyl. Second Ring Closure to B O
O
F
F
O
F
F
F
ðFð š ðO š
ðš Fð š
N
O
O N
F O
CH3 B ðOð š The conversion of B to C features an intermolecular substitution with an amine nucleophile. Of the two carbons bearing the potential fluoride leaving groups, the one para to the carbonyl is picked, because of the resonance stabilization of the resulting intermediate anion. The initial product is formed as the ammonium salt, from which the free base C is liberated by basic work-up (Chemical Highlight 21-2).
H N š
ðOð
ðš Oð
O
F
O
N
F O
B 1034
CH3
CH3
Nð
A
š N A CH3
CH3
O
O ð
N
F
O
O
CH3
F H A N ðFð š O
O
O HF (on basic work-up)
N CH3
C
22-4 Nucleophilic Aromatic Substitution
Chapter 22
Exercise 22-10
Try It Yourself Propose a mechanism for the following conversion. Considering that the first step is rate determining, draw a potential-energy diagram depicting the progress of the reaction. (Hint: This is a nucleophilic aromatic substitution.) NO2 O H3C
O SO2
H3C
SO2 NO2
Haloarenes undergo substitution through benzyne intermediates Haloarenes devoid of electron-withdrawing substituents do not undergo simple ipso substitution. Nevertheless, when haloarenes are treated with nucleophiles that are also strong bases, if necessary at highly elevated temperatures, they convert to products in which the halide has been replaced by the nucleophile. For example, if exposed to hot sodium hydroxide followed by neutralizing work-up, chlorobenzene furnishes phenol. Cl
REACTION
OH 1. NaOH, H2O, 340°C, 150 atm 2. H, H2O
Chlorobenzene
NaCl
Phenol
Treatment with potassium amide results in benzenamine (aniline). Cl
NH2 1. KNH2, liquid NH3 2. H, H2O
KCl
Benzenamine (Aniline)
It is tempting to assume that these substitutions follow a mechanism similar to that formulated for nucleophilic aromatic ipso substitution earlier in this section. However, when the last reaction is performed with radioactively labeled chlorobenzene (14C at C1), a very curious result is obtained: Only half of the product is substituted at the labeled carbon; in the other half, the nitrogen is at the neighboring position. 14
C label
Cl
NH2 NH2 KNH2, liquid NH3
KCl
Chlorobenzene-1-14C
50%
50%
Benzenamine-1-14C
Benzenamine-2-14C
1035
1036
Chapter 22
Chemistry of Benzene Substituents
Direct substitution does not seem to be the mechanism of these reactions. What, then, is the answer to this puzzle? A clue is the attachment of the incoming nucleophile only at the ipso or at the ortho position relative to the leaving group. This observation can be accounted for by an initial base-induced elimination of HX from the benzene ring, a process reminiscent of the dehydrohalogenation of haloalkenes to give alkynes (Section 13-4). In the present case, elimination is not a concerted process, but rather takes place in a sequential manner, deprotonation preceding the departure of the leaving group (step 1 of the mechanism shown). Both stages in step 1 are difficult, with the second being worse than the first. Why is that? With respect to the initial anion formation, recall (Section 11-3) that the acidity of Csp2 – H is very low (pKa < 44), and the same is true in general for phenyl hydrogens. The presence of the adjacent p system of benzene does not help, because the negative charge in the phenyl anion resides in an sp2 orbital that is perpendicular to the p frame and is therefore incapable of resonance with the double bonds in the six-membered ring. Thus, deprotonation of the haloarene requires a strong base. It takes place ortho to the halogen, because the halogen’s inductive electron-withdrawing effect acidifies this position relative to the others. Although deprotonation is not easy, the second stage of step 1, subsequent elimination of X2, is even more difficult because of the highly strained structure of the resulting reactive species, called 1,2-dehydrobenzene or benzyne. Mechanism of Nucleophilic Substitution of Simple Haloarenes
MECHANISM
Step 1. Elimination occurs stepwise X
−
X NH2
H pKa ~ 44
MATION AN I
ANIMATED MECHANISM: Nucleophilic aromatic substitution via benzynes
+ X
−
− HNH2
−
Highly strained triple bond
sp 2 - Hybrid orbital is perpendicular to aromatic π system
Phenyl anion
Benzyne
(Intermediate)
(Reactive intermediate; not isolated)
Step 2. Addition occurs to both strained carbons −
NH2
NH2
− − NH2
−
−
NH2
H — NH2
−
H — NH2
NH2
− − NH2
NH2 H H NH2
Why is benzyne so strained? Recall that alkynes normally adopt a linear structure, a consequence of the sp hybridization of the carbons making up the triple bond (Section 13-2). Because of benzyne’s cyclic structure, its triple bond is forced to be bent, rendering it unusually reactive. Thus, benzyne exists only as a reactive intermediate under these conditions, being rapidly attacked by any nucleophile present. For example (step 2), amide ion or even ammonia solvent can add to furnish the product benzenamine (aniline). Because the two ends of the triple bond are equivalent, addition can take place at either carbon, explaining the label distribution in the benzenamine obtained from 14C-labeled chlorobenzene. Benzyne is too reactive to be isolated and stored in a bottle, but it can be observed spectroscopically under special conditions. Irradiation of benzocyclobutenedione at 77 K
22-4 Nucleophilic Aromatic Substitution
Chapter 22
(21968C) in frozen argon (m.p. 5 21898C) produces a species whose IR and UV spectra are assignable to benzyne, formed by loss of two molecules of CO. Generation of Benzyne, a Reactive Intermediate O hv, 77 K
2 CO
O Benzocyclobutene-1,2-dione
Although benzyne is usually represented as a cycloalkyne (Figure 22-4A), its triple bond exhibits an IR stretching frequency of 1846 cm21, intermediate between the values for normal double (cyclohexene, 1652 cm21) and triple (3-hexyne, 2207 cm21) bonds. The 13C NMR values for these carbons (d 5 182.7 ppm) are also atypical of pure triple bonds (Section 13-3), indicating a considerable contribution of the cumulated triene (Section 14-5) resonance form (Figure 22-4B). The bond is weakened substantially by poor p orbital overlap in the plane of the ring.
Poor overlap 1.42 Å 1.24 Å 1.43 Å
1.40 Å
A
B
C
Figure 22-4 (A) The orbital picture of benzyne reveals that the six aromatic p electrons are located in orbitals that are perpendicular to the two additional hybrid orbitals making up the distorted triple bond. These hybrid orbitals overlap only poorly; therefore, benzyne is highly reactive. (B) Resonance in benzyne. (C) The electrostatic potential map of benzyne shows electron density (red) in the plane of the six-membered ring at the position of the distorted sp-hybridized carbons.
Exercise 22-11 1-Chloro-4-methylbenzene (p-chlorotoluene) is not a good starting material for the preparation of 4-methylphenol (p-cresol) by direct reaction with hot NaOH, because it forms a mixture of two products. Why does it do so, and what are the two products?
Exercise 22-12 Explain the regioselectivity observed in the following reaction. (Hint: Consider the effect of the methoxy group on the selectivity of attack by amide ion on the intermediate benzyne.) Br
NH2 H2N
OCH3
OCH3
KNH2, liquid NH3
OCH3
KBr Major
Minor
1037
1038
Chapter 22
Chemistry of Benzene Substituents
Phenols are produced from arenediazonium salts The most general traditional laboratory procedure for making phenols is from arenamines through their arenediazonium salts, ArN21X2. Recall that primary alkanamines can be N-nitrosated but that the resulting species rearrange to diazonium ions, which are unstable — they lose nitrogen to give carbocations (Section 21-10). In contrast, primary benzenamines (anilines) are attacked by cold nitrous acid, in a reaction called diazotization, to give relatively stable, isolable, although still reactive arenediazonium salts. Compared to their alkanediazonium counterparts, these species enjoy resonance stabilization and are prevented from undergoing immediate N2 loss by the high energy of the resulting aryl cations (to be discussed in more detail in Section 22-10).
Diazotization š N c N
NH2 NaNO2, H, H2O, 0°C
R
R Arenediazonium ion
When arenediazonium ions are gently heated in water, nitrogen is evolved and the resulting aryl cations are trapped extremely rapidly by the solvent to give phenols. Decomposition of Arenediazonium Salts in Water to Give Phenols N N+
+
OH
Δ −N2
R
HOH −H +
R
R
Aryl cation
In these reactions, the “super” leaving group N2 accomplishes what halides are only able to do when attached to a highly electron-deficient benzene nucleus (nucleophilic aromatic substitution) or under extreme conditions (through benzyne intermediates), namely, replacement by hydroxide. The three mechanisms are completely different. In nucleophilic aromatic substitution, the nucleophile attacks prior to departure of the leaving group. In the benzyne mechanism, the nucleophile acts initially as a base, followed by extrusion of the leaving group and subsequent nucleophilic attack on the strained triple bond. In the arenediazonium-ion decomposition, the leaving group exits first, followed by trapping by water. The utility of this phenol synthesis is apparent when you recall that arenamines are derived from nitroarenes by reduction, and nitroarenes are made from other arenes by electrophilic aromatic substitution (Chapters 15 and 16). Therefore, retrosynthetically (Section 8-9), we can picture the hydroxy group in any position of a benzene ring that is subject to electrophilic nitration.
22-4 Nucleophilic Aromatic Substitution
Chapter 22
Retrosynthetic Connection of Phenols to Arenes D(X)
D(X) OH
D(X) NH2
NO2
OH
NH2
NO2
A
A
A
OH
D(X)
Donor-activated or halogen-bearing arene
A
NH2
NO2
Acceptor-deactivated arene
Two examples are depicted below. Br
Br
Br
HNO3, H2SO4, 20C
Br NaNO2, HCl, H2O, 0C
H2, Ni
NO2
Br 80C
N2Cl 80%
NH2 98%
OH 87%
70% (separated from the ortho isomer by crystallization)
O
CH3
O
O
CH3
HNO3, H2SO4
4-Bromophenol ( p-Bromophenol)
O
CH3 NaNO2, H2SO4, H2O, 0C
CH3COOH, Fe
NO2
CH3
80C
N2 HSO4
NH2
85%
70%
O
CH3
OH
82% 1-(3-Hydroxyphenyl)ethanone (m-Hydoxyacetophenone)
Exercise 22-13 ortho-Benzenediazoniumcarboxylate A (made by diazotization of 2-aminobenzoic acid, Problem 20-59) is explosive. When warmed in solution with trans,trans-2,4-hexadiene, it forms compound B. Explain by a mechanism. (Hint: Two other products are formed, both of which are gases.) Nð Nq
CH3 %
H3C
CH3
š Oð E C B ðOð A
0 CH3 B
Exercise 22-14 Propose a synthesis of 4-(phenylmethyl)phenol ( p-benzylphenol) from benzene. (Caution: Remember that Friedel-Crafts reactions do not work with deactivated arenes.)
1039
1040
Chemistry of Benzene Substituents
Chapter 22
Phenols can be made from haloarenes by Pd catalysis While we have seen that ordinary halobenzenes are resilient to reaction with hydroxide, they undergo such nucleophilic displacements in the presence of Pd salts and added phosphine ligands PR3. Pd-Catalyzed Phenol Synthesis from Haloarenes X
OH KOH, Pd catalyst, PR3, 100C
The reaction is general for substituted benzenes, providing a complement to the diazonium method described above. OCH3
O
OCH3
O
CH3
KOH, Pd catalyst, PR3, 100C
CH3
KOH, Pd catalyst, PR3, 100C
Br Cl
OH
OH 90%
98%
4-Methoxyphenol ( p-Methoxyphenol)
1-(3-Hydroxyphenyl)ethanone (m-Hydoxyacetophenone)
The mechanism is related to that of the Heck and other Pd-catalyzed reaction (Section 13-9; Chemical Highlight 13-1). As shown in a simplified manner below, it begins by insertion of the metal into the aryl halide bond, exchange of the halide for hydroxide, and extrusion of the final product with regeneration of the catalyst. Mechanism of the Pd-Catalyzed Phenol Synthesis from Haloarenes X X
OH
Pd
Pd
OH
HO
Pd
X
Pd
Similar substitutions can be carried out with alkoxides to give phenol ethers, and with amines, including ammonia, to furnish benzenamines. OCH3
OCH3 Pd catalyst, PR3, 100C
N H 98%
NH2
Br
3-Methoxy-N-(2-methylpropyl)benzenamine 3-Methoxy-N-(2-methylpropyl)aniline NH3 (14 atm), Pd catalyst, PR3, 90C
Br
NH2
89% 2-(1-Methylethyl)benzenamine (o-Isopropylaniline)
22-5 Alcohol Chemistry of Phenols
Although the preceding methods are valuable in the preparation of specifically substituted phenols, the parent compound is made industrially by the air oxidation of (1-methylethyl)benzene (isopropylbenzene or cumene; see also Exercise 15-27) to the benzylic hydroperoxide and its subsequent decomposition with acid (margin). The “by-product” acetone is valuable in its own right and makes this process highly cost effective, quite apart from the environmentally benign use of air as an oxidant.
1041
Chapter 22
A “Green” Industrial Phenol Synthesis Cheap starting materials
Friedel-Crafts H PO 3 4 alkylation
Exercise 22-15 How would you make the following phenols from the given starting materials? (Hint: Consult Chapters 15 and 16.) CF3
CF3
OCH3 HO
(a)
NO2
(b)
from
OCH3 O2 (air), ROOR
from OOH
OH NO2 OH
(c)
H2SO4
from
OH
In Summary When a benzene ring bears enough strongly electron-withdrawing substituents, nucleophilic addition to give an intermediate anion with delocalized charge becomes feasible, followed by elimination of the leaving group (nucleophilic aromatic ipso substitution). Phenols result when the nucleophile is hydroxide ion, arenamines (anilines) when it is ammonia, and alkoxyarenes when alkoxides are employed. Very strong bases are capable of eliminating HX from haloarenes to form the reactive intermediate benzynes, which are subject to nucleophilic attack to give substitution products. Phenols may also be prepared by decomposition of arenediazonium salts in water and by Pd-catalyzed hydroxylations of haloarenes.
22-5 Alcohol Chemistry of Phenols The phenol hydroxy group undergoes several of the reactions of alcohols (Chapter 9), such as protonation, Williamson ether synthesis, and esterification.
The oxygen in phenols is only weakly basic Phenols are not only acidic but also weakly basic. They (and their ethers) can be protonated by strong acids to give the corresponding phenyloxonium ions. Thus, as with the alkanols, the hydroxy group imparts amphoteric character (Section 8-3). However, the basicity of phenol is even less than that of the alkanols, because the lone electron pairs on the oxygen are delocalized into the benzene ring (Sections 16-1 and 16-3). The pKa values for phenyloxonium ions are, therefore, lower than those of alkyloxonium ions. pKa Values of Methyl- and Phenyloxonium Ion H
f O CH3š i
H E š O
H
OH ðš
H OH CH3š
H
H
pKa 2.2
pKa 6.7
H
O
Valueadded products
1042
Chapter 22
Chemistry of Benzene Substituents
Unlike secondary and tertiary alkyloxonium ions derived from alcohols, phenyloxonium derivatives do not dissociate to form phenyl cations, because such ions have too high an energy content (see Section 22-10). The phenyl – oxygen bond in phenols is very difficult to break. However, after protonation of alkoxybenzenes, the bond between the alkyl group and oxygen is readily cleaved in the presence of nucleophiles such as Br2 or I2 (e.g., from HBr or HI) to give phenol and the corresponding haloalkane. COOH
COOH HBr,
OCH3
CH3Br
OH 90%
3-Methoxybenzoic acid (m-Methoxybenzoic acid)
3-Hydroxybenzoic acid (m-Hydroxybenzoic acid)
Exercise 22-16 Why does cleavage of an alkoxybenzene by acid not produce a halobenzene and the alkanol?
Alkoxybenzenes are prepared by Williamson ether synthesis The Williamson ether synthesis (Section 9-6) permits easy preparation of many alkoxybenzenes. The phenoxide ions obtained by deprotonation of phenols (Section 22-3) are good nucleophiles. They can displace the leaving groups from haloalkanes and alkyl sulfonates. OH
OCH2CH2CH3
CH3CH2CH2Br
NaOH, H2O NaBr, HOH
Cl
Cl 63%
3-Chlorophenol (m-Chlorophenol)
1-Chloro-3-propoxybenzene (m-Chlorophenyl propyl ether)
Esterification leads to phenyl alkanoates The reaction of a carboxylic acid with a phenol (Section 19-9) to form a phenyl ester is endothermic. Therefore, esterification requires an activated carboxylic acid derivative, such as an acyl halide or a carboxylic anhydride. O B OCCH2CH3
OH
O B CH3CH2CCl
CH3
CH3 4-Methylphenol ( p-Cresol)
NaOH, H2O NaCl, HOH
Propanoyl chloride
4-Methylphenyl propanoate ( p-Methylphenyl propanoate)
Exercise 22-17 Explain why, in the preparation of acetaminophen (Chemical Highlight 22-2), the amide is formed rather than the ester. (Hint: Review Section 6-8.)
22-5 Alcohol Chemistry of Phenols
Chapter 22
1043
C H E MI C AL H I G H L I G H T 2 2 - 2 Aspirin: A Phenyl Alkanoate Drug COOH OH
O O B B CH3COCCH3, H,
COOH O B OCCH3
CH3COOH
2-Acetyloxybenzoic acid (o-Acetoxybenzoic acid, acetylsalicylic acid, aspirin)
2-Hydroxybenzoic acid (o-Hydroxybenzoic acid, salicylic acid)
Aspirin prevents heart attacks by minimizing the formation of blood clots in the coronary arteries. The photograph shows such a blood clot (orange) in the left main pulmonary artery of a patient (see also p. 904).
The year 1997 marked the 100th birthday of the synthesis of the acetic ester of 2-hydroxybenzoic acid (salicylic acid); namely, 2-acetyloxybenzoic acid (acetylsalicylic acid), better known as aspirin (see Section 19-13 and Chapter 16 Opening). Aspirin is the first drug that was clinically tested before it was marketed, in 1899. More than 100 billion tablets of aspirin per year are taken by people throughout the world to relieve headaches and rheumatoid and other pain, to control fever, and to treat gout and arthritis. Production capacity in the United States alone is 10,000 tons per year. Salicylic acid (also called spiric acid, hence the name aspirin [“a” for acetyl]) in extracts from the bark of the willow tree or from the meadowsweet plant had been used since ancient times (see Chapter 16 Opening) to treat pain, fever, and swelling. This acid was first isolated in pure form in 1829, then synthesized in the laboratory, and finally produced on a large scale in the nineteenth century and prescribed as an analgesic, antipyretic, and anti-inflammatory drug. Its bitter taste and side effects, such as mouth irritation and gastric bleeding, prompted the search for better derivatives that resulted in the discovery of aspirin. In the body, aspirin functions as a precursor of salicylic acid, which irreversibly inhibits the enzyme cyclooxygenase. OH
This enzyme causes the production of prostaglandins (see Chemical Highlight 11-1 and Section 19-13), molecules that in turn are inflammatory and pain producing. In addition, one of them, thromboxane A2, aggregates blood platelets, necessary for the clotting of blood when injury occurs. This same process is, however, undesirable inside arteries, causing heart attacks or brain strokes, depending on the location of the clot. Indeed, a large study conducted in the 1980s showed that aspirin lowered the risk of heart attacks in men by almost 50% and reduced the mortality rate during an actual attack by 23%. Many other potential applications of aspirin are under investigation, such as in the treatment of pregnancy-related complications, viral inflammation in AIDS patients, dementia, Alzheimer’s disease, and cancer. Despite its popularity, aspirin can have some serious side effects: It is toxic to the liver, prolongs bleeding, and causes gastric irritation. It is suspected as the cause of Reye’s syndrome, a condition that leads to usually fatal brain damage. Because of some of these drawbacks, many other drugs compete with aspirin, particularly in the analgesics market, such as naproxen, ibuprofen, and acetaminophen (see the beginning of Chapter 16). Acetaminophen, better known as Tylenol, is prepared from 4-aminophenol by acetylation. OH
O O B B CH3COCCH3, CH3COOH CH3COOH
NH2 4-Aminophenol ( p-Aminophenol)
HNCCH3 B O N-(4-Hydroxyphenyl)acetamide [N-( p-Hydroxyphenyl)acetamide, acetaminophen, Tylenol]
In Summary The oxygen in phenols and alkoxybenzenes can be protonated even though it is less basic than the oxygen in the alkanols and alkoxyalkanes. Protonated phenols and their derivatives do not ionize to phenyl cations, but the ethers can be cleaved to phenols and haloalkanes by HX. Alkoxybenzenes are made by Williamson ether synthesis, aryl alkanoates by acylation.
1044
Chemistry of Benzene Substituents
Chapter 22
OCH3 O B CH3CCl
22-6 Electrophilic Substitution of Phenols The aromatic ring in phenols is also a center of reactivity. The interaction between the OH group and the ring strongly activates the ortho and para positions toward electrophilic substitution (Sections 16-1 and 16-3). For example, even dilute nitric acid causes nitration. OH
OH
H Methoxybenzene (Anisole)
OH NO2
HNO3, CHCl3, 15C
HCl AlCl3, CS2
OCH3
NO2 61%
26% 2-Nitrophenol (o-Nitrophenol)
4-Nitrophenol ( p-Nitrophenol)
Friedel-Crafts acylation of phenols is complicated by ester formation and is better carried out on ether derivatives of phenol (Section 16-5), as shown in the margin. Phenols are halogenated so readily that a catalyst is not required, and multiple halogenations are frequently observed (Section 16-3). As shown in the following reactions, tribromination occurs in water at 208C, but the reaction can be controlled to produce the monohalogenation product through the use of a lower temperature and a less polar solvent.
KC H O CH3 70% 1-(4-Methoxyphenyl)ethanone (p-Methoxyacetophenone)
Halogenation of Phenols OH
OH Br
Br
3 Br–Br, H2O, 20C
HBr
CH3
Br 100% 2,4,6-Tribromophenol
Br
Br–Br, CHCl3, 0C
but
3 HBr
Phenol
OH
OH
CH3 80% 2-Bromo-4-methylphenol
4-Methylphenol ( p-Cresol)
Electrophilic attack at the para position is frequently dominant because of steric effects. However, it is normal to obtain mixtures resulting from both ortho and para substitutions, and their compositions are highly dependent on reagents and reaction conditions. Exercise 22-18 Friedel-Crafts methylation of methoxybenzene (anisole) with chloromethane in the presence of AlCl3 gives a 2;1 ratio of ortho; para products. Treatment of methoxybenzene with 2-chloro2-methylpropane (tert-butyl chloride) under the same conditions furnishes only 1-methoxy-4(1,1-dimethylethyl) benzene (p-tert-butylanisole). Explain. (Hint: Review Section 16-5.)
Exercise 22-19
Working with the Concepts: Devising Synthetic Strategies Starting with Substituted Phenols Proparacaine is a local anesthetic that is used primarily to numb the eye before minor surgical procedures, such as the removal of foreign objects or stitches. Show how you would make it from 4-hydroxybenzenecarboxylic acid. O
O OH HO 4-Hydroxybenzenecarboxylic acid
H2 N
O
O Proparacaine
N
22-6 Electrophilic Substitution of Phenols
Chapter 22
1045
Strategy Inspection of the structures of both the starting material and the product reveals three structural modifications: (1) An amino group is introduced into the benzene core, suggesting a nitration-reduction sequence (Section 16-5) that relies on the ortho-directing effect of the hydroxy substituent: (2) the phenol function is etherified, best done by Williamson synthesis (Section 22-5): (3) the carboxy function is esterified with the appropriate amino alcohol (Section 20-2). What is the best sequence in which to execute these manipulations? To answer this question, consider the possible interference of the various functions with the suggested reaction steps. Solution • Modification 1 could be carried out with the starting material, resulting in the corresponding amino acid. However, “unmasking” the amino group (from its precursor nitro) early bodes trouble with steps 2 and 3, because amines are better nucleophiles than alcohols. Therefore, attempted etherification of the aminated phenol will lead to amine alkylation (Section 21-5). Similarly, attempts to effect ester formation in the presence of an amine function will give rise to amide generation (Section 19-10). However, early introduction of the nitro group and maintaining it as such until the other functions are protected looks good. • For modification 2, Williamson ether synthesis in the presence of a carboxy group should be fine, because the carboxylate ion is a poorer nucleophile than phenoxide ion (Section 6-8). • The preceding protection of the phenolic function is important to the success of modification 3, ester formation with the amino alcohol, especially if we want to activate the carboxy group as an acyl chloride. • Putting it all together, a possible (and, indeed, the literature) synthesis of proparacaine is as follows:
O
O OH
HO
HNO3
O2N
O OH
O2N
NaOH, CH3CH2CH2Cl
HO
OH
O O
O2N
1. SOCl2 2. HOCH2CH2N(CH2CH3)2
O
N H2, Pd-C
proparacaine
O
• Why does the amino alcohol not react with the acyl halide at nitrogen in the last step? After all, amines are more nucleophilic than alcohols. The answer is: It does, but because the amine function is tertiary, it can only form an acyl ammonium salt. This function has reactivity similar to that of an acyl chloride. Thus, attack by the hydroxy group on the acyl ammonium carbonyl carbon wins out thermodynamically to give the ester in the end (Section 20-2).
R
O B C E H
š Clð š
ðNR3
O B C E H š R NR3 ðClð š
š ROH
ðNR3
R
Exercise 22-20
Try It Yourself Device an alternative route to proparacaine from 4-hydroxybenzenecarboxylic acid, using Pd catalysis.
Under basic conditions, phenols can undergo electrophilic substitution, even with very mild electrophiles, through intermediate phenoxide ions. An industrially important application is the reaction with formaldehyde, which leads to o- and p-hydroxymethylation. Mechanistically, these processes may be considered enolate condensations, much like the aldol reaction (Section 18-5).
O B C E H
OR
1046
Chemistry of Benzene Substituents
Chapter 22
Hydroxymethylation of Phenol Oð ðš
š ðOH
ðOð
HO
ðOð
k
š O CH2 P š
š š HOH
ðOð
OH ðš
ðOð
H š šð CH2O
š š HOH š š HO
š
ðš š OCH2
OH ðš
š OH CH2š
š CH2š OH
H
The initial aldol products are unstable: They dehydrate on heating, giving reactive intermediates called quinomethanes. OH
O CH2OH
O
OH CH2
HO,
HO, HOH
HOH
CH2
CH2OH o-Quinomethane
p-Quinomethane
Because quinomethanes are a,b-unsaturated carbonyl compounds, they may undergo Michael additions (Section 18-11) with excess phenoxide ion. The resulting phenols can be hydroxymethylated again and the entire process repeated. Eventually, a complex phenol– formaldehyde copolymer, also called a phenolic resin (e.g., Bakelite), is formed. Their major uses are in plywood (45%), insulation (14%), molding compounds (9%), fibrous and granulated wood (9%), and laminates (8%).
Phenolic Resin Synthesis O
O CH2
O CH2
H
O
O
OH CH2
CH2 P O
k
H OH
OH
OH
CH2
OH CH2 polymer
CH2OH
H2 C OH
n H2O
22-6 Electrophilic Substitution of Phenols
1047
Chapter 22
In the Kolbe*-Schmitt† reaction, phenoxide attacks carbon dioxide to furnish the salt of 2-hydroxybenzoic acid (o-hydroxybenzoic acid, salicylic acid, precursor to aspirin; see Chemical Highlight 22-2). OH ðOð B C B ðOð
NaOH, H2O, pressure
OH ðOð B C Hš šðNa O
Exercise 22-21
Working with the Concepts: Recognizing Phenol as an Enol Formulate a mechanism for the Kolbe-Schmitt reaction. Strategy As always, we take an inventory of the components of the reaction: starting materials, other reagents and reaction conditions, and products. Phenol is an electron-rich arene (see this and Section 16-3) and also acidic. Carbon dioxide has an electrophilic carbon that is attacked by nucleophilic carbon atoms, such as in Grignard reagents (Section 19-6). The reaction conditions are strongly basic. Finally, the product looks like that of an electrophilic ortho substitution. Solution • Under basic conditions, phenol will exist as phenoxide, which, like an enolate ion (Section 18-1), can be described by two resonance forms (see margin). • Enolate alkylations occur at carbon (Section 18-10). In analogy, you can formulate a phenoxide attack on the electrophilic carbon of CO2. Alternatively, you can think of this reaction as an electrophilic attack of CO2 on a highly activated benzene ring. • Finally, deprotonation occurs to regenerate the aromatic arene. p p )O) )OH )O) O p B )O) H O) B CHp Ep OH C C p B B )O) )O) As you will have noticed, the selectivity for ortho attack by CO2 in this process is exceptional. Although not completely understood, it may involve direction of the electrophile by the Na1 ion in the vicinity of the phenoxide negative charge.
Exercise 22-22
Try It Yourself Phentolamine (as a water-soluble methanesulfonic acid salt) is an antihypertensive that has recently been introduced into dentistry: It cuts in half the time taken to recover from the numbing effect of local anesthetics. The key step in its preparation dates from 1886, the reaction shown below. What is its mechanism? (Caution: This is not a nucleophilic aromatic substitution. Hint: Think keto – enol tautomerism). CH3
OH
OH
OH CH3
CH3
160C
OH
NH2
N H
N N
70% HN
Phentolamine
*Professor Adolph Wilhelm Hermann Kolbe (1818–1884), University of Leipzig, Germany. † Professor Rudolf Schmitt (1830–1898), University of Dresden, Germany.
Various wood products incorporating Bakelite are used in the construction of houses.
p )O)
)O) ½
Phenoxide ion
1048
Chemistry of Benzene Substituents
Chapter 22
OH
OH
Exercise 22-23 Cl
Cl
Hexachlorophene (margin) is a skin germicide formerly used in soaps. It is prepared in one step from 2,4,5-trichlorophenol and formaldehyde in the presence of sulfuric acid. How does this reaction proceed? (Hint: Formulate an acid-catalyzed hydroxymethylation for the first step.)
Cl Cl Cl
Cl Hexachlorophene
In Summary The benzene ring in phenols is subject to electrophilic aromatic substitution, particularly under basic conditions. Phenoxide ions can be hydroxymethylated and carbonated.
22-7 An Electrocyclic Reaction of the Benzene Ring: The Claisen Rearrangement At 2008C, 2-propenyloxybenzene (allyl phenyl ether) undergoes an unusual reaction that leads to the rupture of the allylic ether bond: The starting material rearranges to 2-(2-propenyl)phenol (o-allylphenol). REACTION
O
E CH2CHP CH2
OH CH2CHP CH2
H
75% 2-Propenyloxybenzene (Allyl phenyl ether)
2-(2-Propenyl)phenol (o-Allylphenol)
This transformation, called the Claisen* rearrangement, is another concerted reaction with a transition state that accommodates the movement of six electrons (Sections 14-8 and 15-3). The initial intermediate is a high-energy isomer, 6-(2-propenyl)-2,4-cyclohexadienone, which enolizes to the final product (Sections 18-2 and 22-3). Mechanism of the Claisen Rearrangement
MECHANISM O
O
O
H
OH
O O
6-(2-Propenyl)2,4-cyclohexadienone
The Claisen rearrangement is general for other systems. With the nonaromatic 1-ethenyloxy2-propene (allyl vinyl ether), it stops at the carbonyl stage because there is no driving force for enolization. This is called the aliphatic Claisen rearrangement. *Professor Ludwig Claisen (1851–1930), University of Berlin, Germany.
22-7 Claisen Rearrangement
Chapter 22
1049
Aliphatic Claisen Rearrangement H EC N
H 2C CH2 A OH KCH2 C H
H2C
H KC H
CH2 A ON ECH2 C H 50%
255°C
1-Ethenyloxy-2-propene (Allyl vinyl ether)
4-Pentenal
The carbon analog of the Claisen rearrangement is called the Cope* rearrangement; it takes place in compounds containing 1,5-diene units. Cope Rearrangement
178°C
72% 3-Phenyl-1,5-hexadiene
trans-1-Phenyl-1,5-hexadiene
Note that all of these rearrangements are related to the electrocyclic reactions that interconvert cis-1,3,5-hexatriene with 1,3-cyclohexadiene (see margin and Section 14-9). The only difference is the absence of a double bond connecting the terminal p bonds.
Electrocyclic Reaction of cis-1,3,5-Hexatriene
Exercise 22-24 Explain the following transformation by a mechanism. (Hint: The Cope rearrangement can be accelerated greatly if it leads to charge delocalization.) HO NaOH, H2O
OHC
Exercise 22-25
Working with the Concepts: Applying Claisen and Cope Rearrangements Citral, B, is a component of lemon grass and as such used in perfumery (lemon and verbena scents). It is also an important intermediate in the BASF synthesis of vitamin A (Section 14-7, Chemical Highlight 18-3). The last step in the synthesis of citral requires simply heating the enol ether A. How do you get from A to B? O
O
A
H
B Citral
*Professor Arthur C. Cope (1909–1966), Massachusetts Institute of Technology, Cambridge.
Strategy As always, we take an inventory of the components of the reaction: starting materials, other reagents and reaction conditions, and products. Here, this is straightforward: There are no reagents, we simply apply heat, and it appears that the reaction is an isomerization. You need to confirm this suspicion, by determining the molecular formulas of A and B. Indeed, it is C10H16O for both. What thermal reactions could you envisage for A? Solution • You note that A contains a diene unit connected to an isolated double bond. Hence, in principle, an intramolecular Diels-Alder reaction to C might be feasible (Section 14-8; Exercise 14-24): Potential Diels-Alder Reaction of A
O
O
A
C
• It is apparent that this pathway is unfavorable for two reasons: (1) Both the diene and the dienophile are electron rich and therefore not good partners for cycloaddition (Section 14-8), and, even more obvious, (2) a strained ring is formed in C. • An alternative is based on the recognition of a 1,5-hexadiene unit, the prerequisite for a Cope rearrangement. In A, this diene unit contains an oxygen, hence we can write a Claisen rearrangement to see where it leads us. Claisen Rearrangement of A H
O
O
A
D
• The product D contains a 1,5-diene substructure, capable of a Cope rearrangement, leading to citral B. Cope Rearrangement of D O
H O
H
D
B
Exercise 22-26
Try It Yourself The ether A provides B upon heating to 2008C. Formulate a mechanism. (Caution: The terminal alkenyl carbon cannot reach the para position of the benzene ring. Hint: Start with the first step of a Claisen rearrangement.)
O CH3
H 3C
A
1050
OH
CH3
H3C
B
22-8 Oxidation of Phenols: Benzoquinones
In Summary 2-Propenyloxybenzene rearranges to 2-(2-propenyl)phenol (o-allylphenol) by an electrocyclic mechanism that moves six electrons (Claisen rearrangement). Similar concerted reactions are undergone by aliphatic unsaturated ethers (aliphatic Claisen rearrangement) and by hydrocarbons containing 1,5-diene units (Cope rearrangement).
22-8 Oxidation of Phenols: Benzoquinones Phenols can be oxidized to carbonyl derivatives by one-electron transfer mechanisms, resulting in a new class of cyclic diketones, called benzoquinones.
Benzoquinones and benzenediols are redox couples The phenols 1,2- and 1,4-benzenediol (for which the respective common names catechol and hydroquinone are retained by IUPAC) are oxidized to the corresponding diketones, ortho- and para-benzoquinone, by a variety of oxidizing agents, such as sodium dichromate or silver oxide. Yields can be variable when the resulting diones are reactive, as in the case of o-benzoquinone, which partly decomposes under the conditions of its formation. Benzoquinones from Oxidation of Benzenediols OH
O Ag2O, (CH3CH2)2O
O Low yield
OH
O
O O
O Catechol
o-Benzoquinone
OH
O Na2Cr2O7, H2SO4
O 92%
OH
O
O
O
O
Hydroquinone
p-Benzoquinone
The redox process that interconverts hydroquinone and p-benzoquinone can be visualized as a sequence of proton and electron transfers. Initial deprotonation gives a phenoxide ion, which is transformed into a phenoxy radical by one-electron oxidation. Proton
Chapter 22
1051
1052
Chemistry of Benzene Substituents
Chapter 22
dissociation from the remaining OH group furnishes a semiquinone radical anion, and a second one-electron oxidation step leads to the benzoquinone. All of the intermediate species in this sequence benefit from considerable resonance stabilization (two forms are shown for the semiquinone). We shall see in Section 22-9 that redox processes similar to those shown here occur widely in nature. Redox Relation Between p-Benzoquinone and Hydroquinone š ðOH
Oð ðš
šj ðO
H
ð OH
ðš Oj
e
Oð ðš
ðOð
e
H
ð OH
ð OH
Phenoxide ion
Phenoxy radical
ð Oð
ðOð
ðO j Semiquinone radical anion
Exercise 22-27 Give a minimum of two additional resonance forms each for the phenoxide ion, phenoxy radical, and semiquinone radical anion shown in the preceding scheme.
The enone units in p-benzoquinones undergo conjugate and Diels-Alder additions p-Benzoquinones function as reactive a,b-unsaturated ketones in conjugate additions (see Section 18-9). For example, hydrogen chloride adds to give an intermediate hydroxy dienone that enolizes to the aromatic 2-chloro-1,4-benzenediol. O
OH
OH
HCl
H Cl
Cl O
O
p-Benzoquinone
OH
6-Chloro-4-hydroxy2,4-cyclohexadienone
2-Chloro-1,4-benzenediol
The double bonds also undergo cycloadditions to dienes (Section 14-8). The initial cycloadduct to 1,3-butadiene tautomerizes with acid to the aromatic system. Diels–Alder Reactions of p-Benzoquinone O C6H6, 20C, 48 h
OH
O H %
HCl,
% H O
O
OH 88% overall
22-9 Oxidation-Reduction Processes in Nature
Chapter 22
1053
C H E MI C AL H I G H L I G H T 2 2 - 3 Chemical Warfare in Nature: The Bombardier Beetle The oxidizing power of p-benzoquinones is used by some arthropods, such as millipedes, beetles, and termites, as chemical defense agents. Most remarkable among these species is the bombardier beetle. Its name is descriptive of its defense mechanism against predators, usually ants, which involves firing hot corrosive chemicals from glands in its posterior, with amazing accuracy. At the time of an attack (simulated in the laboratory by pinching the beetle with finetipped forceps, see photo), two glands located near the end of the beetle’s abdomen secrete mainly hydroquinone and hydrogen peroxide, respectively, into a reaction chamber. This chamber contains enzymes that trigger the explosive oxidation of the diol to the quinone and simultaneous decomposition of hydrogen peroxide to oxygen gas and water. This cocktail is audibly expelled at temperatures up to 1008C in the direction of the enemy from the end of the beetle’s abdomen, aided for aim by a 2708 rotational
capability. In some species, firing occurs in pulses of about 500 per second, like a machine gun.
The bombardier beetle in action.
Exercise 22-28 Explain the following result by a mechanism. (Hint: Review Section 18-9.) O H3 C
O CH3
CH3O
OCH3 O
CH3CH2ONa, CH3CH2OH
H3C
CH3
CH3CH2O
OCH2CH3 O
In Summary Phenols are oxidized to the corresponding benzoquinones. The diones enter into reversible redox reactions that yield the corresponding diols. They also undergo conjugate additions and Diels-Alder additions to the double bonds.
22-9 Oxidation-Reduction Processes in Nature This section describes some chemical processes involving hydroquinones and p-benzoquinones that occur in nature. We begin with an introduction to the biochemical reduction of O2. Oxygen can engage in reactions that cause damage to biomolecules. Naturally occurring antioxidants inhibit these transformations, as do several synthetic preservatives.
Ubiquinones mediate the biological reduction of oxygen to water Nature makes use of the benzoquinone – hydroquinone redox couple in reversible oxidation reactions. These processes are part of the complicated cascade by which oxygen is used in biochemical degradations. An important series of compounds used for this purpose are the ubiquinones (a name coined to indicate their ubiquitous presence in nature), also collectively called coenzyme Q (CoQ, or simply Q). The ubiquinones are substituted p-benzoquinone derivatives bearing a side chain made up of 2-methylbutadiene units (isoprene; Sections 4-7 and 14-10). An enzyme system that utilizes NADH (Chemical Highlights 8-1 and 25-2) converts CoQ into its reduced form (QH2).
1054
Chemistry of Benzene Substituents
Chapter 22
CH3O
CH3 A (CH2CHP CCH2)nH
CH3O
CH3
O
CH3O
CH3 A (CH2CHP CCH2)nH
CH3O
CH3
OH Enzyme, reducing agent
O
OH
Ubiquinones (n 6, 8, 10) (Coenzyme Q)
Reduced form of coenzyme Q (Reduced Q, or QH2)
QH2 participates in a chain of redox reactions with electron-transporting iron-containing proteins called cytochromes (Chemical Highlight 8-1). The reduction of Fe31 to Fe21 in cytochrome b by QH2 begins a sequence of electron transfers involving six different proteins. The chain ends with reduction of O2 to water by addition of four electrons and four protons. O2 1 4 H1 1 4 e2 uy 2 H2O
Phenol derivatives protect cell membranes from oxidative damage The biochemical conversion of oxygen into water includes several intermediates, including superoxide, O22. , the product of one-electron reduction, and hydroxy radical, . OH, which arises from cleavage of H2O2. Both are highly reactive species capable of initiating reactions that damage organic molecules of biological importance. An example is the phosphoglyceride shown here, a cell-membrane component derived from the unsaturated fatty acid cis,cis-octadeca-9,12-dienoic acid (linoleic acid). Initiation step
H CH3(CH2)4 13
H 11
C
O B CH2O O C(CH2)14CH3 O A B CH2(CH2)6CO O CH O A A 9 CH2O OPO(CH2)2N(CH3)3 B O
jOH
R
R j
R
R j
HOH
R R j
R
Pentadienyl radical
The doubly allylic hydrogens at C11 are readily abstracted by radicals such as . OH (Section 14-2). The resonance-stabilized pentadienyl radical combines rapidly with O2 in the first of two propagation steps. Reaction occurs at either C9 or C13 (shown here), giving either of two peroxy radicals containing conjugated diene units. Propagation step 1 R j
R
R
O2
ðš O A ðO j
Peroxy radical
R
22-9 Oxidation-Reduction Processes in Nature
1055
Chapter 22
In the second propagation step this species removes a hydrogen atom from C11 of another molecule of phosphoglyceride, or more generally, lipid (Section 20-5), thereby forming a new dienyl radical and a molecule of lipid hydroperoxide. The dienyl radical may then reenter propagation step 1. In this way, a large number of lipid molecules may be oxidized following just a single initiation event. Propagation step 2 R
R
ðš O A ðO j
R H
H R
R
R
11
C
š ðO A ðOH
R
R j
Lipid hydroperoxide
Numerous studies have confirmed that lipid hydroperoxides are toxic, their products of decomposition even more so. For example, loss of . OH by cleavage of the relatively weak O – O bond gives rise to an alkoxy radical, which may decompose by breaking a neighboring C – C bond (b-scission), forming an unsaturated aldehyde.
-Scission of a Lipid Alkoxy Radical R
R O
šð O
H
R Oj
jOH
ðOH
R
H ð
H
O
jR
ð
R
Alkoxy radical
Through related but more complex mechanisms, certain lipid hydroperoxides decompose to give unsaturated hydroxyaldehydes, such as trans-4-hydroxy-2-nonenal, as well as the dialdehyde propanedial (malondialdehyde). Molecules of these general types are partly responsible for the smell of rancid fats. Both propanedial and the a,b-unsaturated aldehydes are extremely toxic, because they are highly reactive toward the proteins that are present in close proximity to the lipids in cell membranes. For example, both dials and enals are capable of reacting with nucleophilic amino and mercapto groups from two different parts of one protein or from two different protein molecules, and these reactions produce cross-linking (Section 14-10). Cross-linking severely inhibits protein molecules from carrying out their biological functions (Chapter 26).
HO H CH3(CH2)4 O trans-4-Hydroxy-2-nonenal
O H
protein1O S
protein1O S
O
R
H
NO protein2
Protein2ONH2
Protein1OSH
R
H
R
Processes such as these are thought by many to contribute to the development of emphysema, atherosclerosis (the underlying cause of several forms of heart disease and stroke), certain chronic inflammatory and autoimmune diseases, cancer, and, possibly, the process of aging itself. Does nature provide the means for biological systems to protect themselves from such damage? A variety of naturally occurring antioxidant systems defend lipid molecules inside cell membranes from oxidative destruction. The most important is vitamin E, a collection
H
Propanedial (Malondialdehyde)
Cross-linking of Proteins by Reaction with Unsaturated Aldehydes O
O
H
1056
Chemistry of Benzene Substituents
Chapter 22
of eight compounds with very similar structures, commonly represented by one of them, a-tocopherol (margin). They all possess a long hydrocarbon chain (see Problem 46 of Chapter 2), a feature that makes them lipid soluble. Their reducing qualities stem from the presence of the hydroquinone-like aromatic ring (Section 22-8). The corresponding phenoxide ion is an excellent electron donor. The protective qualities of vitamin E rest on its ability to break the propagation chain of lipid oxidation by the reduction of radical species.
CH3 HO CH3 O
H3C
R
CH3 Vitamin E (-Tocopherol)
Reactions of Vitamin E with Lipid Hydroperoxy and Alkoxy Radicals
R branched C16H33 chain
CH3
e-Transfer
CH3
š HO
ðO š
H
CH3 H3C
O CH3
CH3 H3C
R
O CH3
Vitamin E
R
š lipid O O j or š š lipid O O j O O
Vitamin E phenoxide
Lipid radicals
CH3 š jO CH3 H3C
O CH3
R
š lipid O O ð or š š lipid O O ð O O
H
š lipid O OH or š š lipid O O O OH
-Tocopheroxy radical
Lipid peroxidation has been implicated in retinal diseases, and clinicians prescribe antioxidants to help in the treatment of diabetic retinopathy. This condition is characterized by a compromised blood supply that causes the development of fatty exudates and fibrous tissue on the retina (yellow areas in photo of the eye).
In this process, lipid radicals are reduced and protonated. Vitamin E is oxidized to an a-tocopheroxy radical, which is relatively unreactive because of extensive delocalization and the steric hindrance of the methyl substituents. Vitamin E is regenerated at the membrane surface by reaction with water-soluble reducing agents such as vitamin C.
Regeneration of Vitamin E by Vitamin C CH3
OH ðO š
O
H# HO
O H
CH2OH
CH3
š jO
OH
ðO š
CH3
CH3 O
H3C CH3
Vitamin C
R
H3C
O CH3
R
š jO
O
H# HO
O H
CH2OH Semidehydroascorbic acid
The product of vitamin C oxidation eventually decomposes to lower-molecular-weight watersoluble compounds, which are excreted by the body. Exercise 22-29 Vitamin C is an effective antioxidant because its oxidation product semidehydroascorbic acid is stabilized by resonance. Give other resonance forms for this species.
22-9 Oxidation-Reduction Processes in Nature
1057
Chapter 22
Benzoquinones consume glutathione, an intracellular reducing agent Virtually all living cells contain the substance glutathione, a peptide that incorporates a mercapto functional group (see Sections 9-10 and 26-4). The mercapto function serves to reduce disulfide linkages in proteins to SH groups and to maintain the iron in hemoglobin in the 21 oxidation state (Section 26-8). Glutathione also participates in the reduction of oxidants, such as hydrogen peroxide, H2O2, that may be present in the interior of the cell. COO O HSCH2 O A B B A H3NCHCH2CH2 O C ONHCHO C ONHCH2COOH Glutathione
The molecule is converted into a disulfide (Section 9-10) in this process but is regenerated by an enzyme-mediated reduction. 2 Glutathione O SH
H2O2
Glutathione peroxidase
2 H2O
glutathione O SO SO glutathione
Glutathione reductases
Benzoquinones and related compounds react irreversibly with glutathione in the liver by conjugate addition. Cell death results if glutathione depletion is extensive. Acetaminophen is an example of a substance that exhibits such liver toxicity at very high doses. Cytochrome P-450, a redox enzyme system in the liver, oxidizes acetaminophen to an imine derivative of a benzoquinone, which in turn consumes glutathione. Vitamin C is capable of reversing the oxidation. OH
OH
O
SO glutathione
Cytochrome P-450
Glutathione–SH
Vitamin C
HNCCH3 B O
NCCH3 B O
HNCCH3 B O
OH C(CH3)3
Acetaminophen
Synthetic analogs of vitamin E are preservatives Synthetic phenol derivatives are widely used as antioxidants and preservatives in the food industry. Perhaps two of the most familiar are 2-(1,1-dimethylethyl)-4-methoxyphenol (butylated hydroxyanisole, or BHA) and 2,6-bis(1,1-dimethylethyl)-4-methylphenol (butylated hydroxytoluene, or BHT; see Exercise 16-14). For example, addition of BHA to butter increases its storage life from months to years. Both BHA and BHT function like vitamin E, reducing oxygen radicals and interrupting the propagation of oxidation processes.
In Summary Oxygen-derived radicals are capable of initiating radical chain reactions in lipids, thereby leading to toxic decomposition products. Vitamin E is a naturally occurring phenol derivative that functions as an antioxidant to inhibit these processes within membrane lipids. Vitamin C and glutathione are biological reducing agents located in the intraand extracellular aqueous environments. High concentrations of benzoquinones can bring
OCH3 2-(1,1-Dimethylethyl)4-methoxyphenol (BHA)
OH (CH3)3C
C(CH3)3
CH3 2,6-Bis(1,1-dimethylethyl)4-methylphenol (BHT)
1058
Chapter 22
Chemistry of Benzene Substituents
about cell death by consumption of glutathione; vitamin C can protect the cell by reduction of the benzoquinone. Synthetic food preservatives are structurally designed to mimic the antioxidant behavior of vitamin E.
22-10 Arenediazonium Salts As mentioned in Section 22-4, N-nitrosation of primary benzenamines (anilines) furnishes arenediazonium salts, which can be used in the synthesis of phenols. Arenediazonium salts are stabilized by resonance of the p electrons in the diazo function with those of the aromatic ring. They are converted into haloarenes, arenecarbonitriles, and other aromatic derivatives through replacement of nitrogen by the appropriate nucleophile.
Arenediazonium salts are stabilized by resonance The reason for the stability of arenediazonium salts, relative to their alkane counterparts, is resonance and the high energy of the aryl cations formed by loss of nitrogen. One of the electron pairs making up the aromatic p system can be delocalized into the functional group, which results in charge-separated resonance structures containing a double bond between the benzene ring and the attached nitrogen. Resonance in the Benzenediazonium Cation š N
ð Nð B N
N
N N
ð Nð B N
ð Nð B N
At elevated temperatures (.508C), nitrogen extrusion does take place, however, to form the very reactive phenyl cation. When this is done in aqueous solution, phenols are produced (Section 22-4). Why is the phenyl cation so reactive? After all, it is a carbocation that is part of a benzene ring. Should it not be resonance stabilized, like the phenylmethyl (benzyl) cation? The answer is no, as may be seen in the molecular-orbital picture of the phenyl cation (Figure 22-5). The empty orbital associated with the positive charge is one of the sp2 hybrids aligned in perpendicular fashion to the p framework that normally produces aromatic resonance stabilization. Hence, this orbital cannot overlap with the p bonds, and the positive p orbitals
+
+
Phenyl cation
A
Empty sp2 orbital
B
C
Figure 22-5 (A) Line structure of the phenyl cation. (B) Orbital picture of the phenyl cation. The alignment of its empty sp2 orbital is perpendicular to the six-p-electron framework of the aromatic ring. As a result, the positive charge is not stabilized by resonance. (C) The electrostatic potential map of the phenyl cation, shown at an attenuated scale for better contrast, pinpoints much of the positive charge (blue edge on the right) as located in the plane of the six-membered ring.
22-10 Arenediazonium Salts
charge cannot be delocalized. Moreover, the cationic carbon would prefer sp hybridization, an arrangement precluded by the rigid frame of the benzene ring. We used a similar argument to explain the difficulty in deprotonating benzene to the corresponding phenyl anion (Section 22-4).
Arenediazonium salts can be converted into other substituted benzenes When arenediazonium salts are decomposed in the presence of nucleophiles other than water, the corresponding substituted benzenes are formed. For example, diazotization of arenamines (anilines) in the presence of hydrogen iodide results in the corresponding iodoarenes. CHO
O O
CHO
O
CH3COOH, HI, NaNO2
O
NH2
N2
I 53%
Attempts to obtain other haloarenes in this way are frequently complicated by side reactions. One solution to this problem is the Sandmeyer* reaction, which makes use of the fact that the exchange of the nitrogen substituent for halogen is considerably facilitated by the presence of cuprous [Cu(I)] salts. The detailed mechanism of this process is complex, and radicals are participants. Addition of cuprous cyanide, CuCN, to the diazonium salt in the presence of excess potassium cyanide gives aromatic nitriles. Sandmeyer Reactions CH3
CH3
CH3 Cl
NH2
N2 HCl, NaNO2, 0°C
Cl CuCl, 60°C N2
79% overall 2-Methylbenzenamine (o-Methylaniline)
1-Chloro-2-methylbenzene (o-Chlorotoluene)
Cl
Cl
1. HBr, NaNO2, 0°C 2. CuBr, 100°C
NH2
Br 73%
2-Chlorobenzenamine (o-Chloroaniline)
CH3
1-Bromo-2-chlorobenzene (o-Bromochlorobenzene)
CH3 1. HCl, NaNO2, 0°C 2. CuCN, KCN, 50°C
NH2
CN 70% 4-Methylbenzonitrile ( p-Tolunitrile)
*Dr. Traugott Sandmeyer (1854 – 1922), Geigy Company, Basel, Switzerland.
Chapter 22
1059
1060
Chapter 22
Chemistry of Benzene Substituents
Exercise 22-30 Propose syntheses of the following compounds, starting from benzene.
OH
CN
(b)
(a) I
(c) SO3H
CN
The diazonium group can be removed by reducing agents. The sequence diazotization – reduction is a way to replace the amino group in arenamines (anilines) with hydrogen. The reducing agent employed is aqueous hypophosphorous acid, H3PO2. This method is especially useful in syntheses in which an amino group is used as a removable directing substituent in electrophilic aromatic substitution (Section 16-5). Reductive Removal of a Diazonium Group CH3
CH3
CH3 NaNO2, H, H2O
H3PO2, H2O, 25C
Br
Br
Br
NH2
N2
H 85% 1-Bromo-3-methylbenzene (m-Bromotoluene)
Another application of diazotization in synthetic strategy is illustrated in the synthesis of 1,3-dibromobenzene (m-dibromobenzene). Direct electrophilic bromination of benzene is not feasible for this purpose; after the first bromine has been introduced, the second will attack ortho or para. What is required is a meta-directing substituent that can be transformed eventually into bromine. The nitro group is such a substituent. Double nitration of benzene furnishes 1,3-dinitrobenzene (m-dinitrobenzene). Reduction (Section 16-5) leads to the benzenediamine, which is then converted into the dihalo derivative. Synthesis of 1,3-Dibromobenzene by Using a Diazotization Strategy NO2
Br
NH2
HNO3, H2SO4,
1. NaNO2, H, H2O 2. CuBr, 100C
H2, Pd
NO2
NH2
Br
Exercise 22-31 Propose a synthesis of 1,3,5-tribromobenzene from benzene.
In Summary Arenediazonium salts, which are more stable than alkanediazonium salts because of resonance, are starting materials not only for phenols, but also for haloarenes,
22-11 Electrophilic Substitution with Arenediazonium Salts
Chapter 22
1061
arenecarbonitriles, and reduced aromatics by displacement of nitrogen gas. The intermediates in some of these reactions may be aryl cations, highly reactive because of the absence of any electronically stabilizing features, but other, more complicated mechanisms may be followed. The ability to transform arenediazonium salts in this way gives considerable scope to the regioselective construction of substituted benzenes.
22-11 Electrophilic Substitution with Arenediazonium Salts: Diazo Coupling Being positively charged, arenediazonium ions are electrophilic. Although they are not very reactive in this capacity, they can accomplish electrophilic aromatic substitution when the substrate is an activated arene, such as phenol or benzenamine (aniline). This reaction, called diazo coupling, leads to highly colored compounds called azo dyes. For example, reaction of N,N-dimethylbenzenamine (N,N-dimethylaniline) with benzenediazonium chloride gives the brilliant orange dye Butter Yellow. This compound was once used as a food coloring agent but has been declared a suspect carcinogen by the Food and Drug Administration. Diazo Coupling
Dyes are important additives in the textile industry. Azo dyes, while still in widespread use, are becoming less attractive for this purpose because some have been found to degrade to carcinogenic benzenamines.
Cl
N
J
H, H2O pH 3–5
HCl
N(CH3)2 Cl
)N(CH3)2
Azo function
J
N(CH3)2
4-Dimethylaminoazobenzene ( p-Dimethylaminoazobenzene, Butter Yellow)
Dyes used in the clothing industry usually contain sulfonic acid groups that impart water solubility and allow the dye molecule to attach itself ionically to charged sites on the polymer framework of the textile. Industrial Dyes
(CH3)2N
O
J
N
SO3Na
N
Methyl Orange pH 3.1, red pH 4.4, yellow
SO3Na
NaO3S N P
P
NH2
N N
N Congo Red pH 3.0, blue-violet pH 5.0, red
H2N
O S
O
Na
1062
Chemistry of Benzene Substituents
Chapter 22
CHE M I C AL H I G H L I G H T 2 2 - 4 William Perkin and the Origins of Industrial and Medicinal Chemistry In 1851, William Perkin* was a 13-year-old pupil at City of London School. His teacher, Thomas Hall, recognized Perkin’s interest and aptitude in science but also realized his own limitations. Hall therefore encouraged Perkin to attend lectures at the Royal Institution, including a series presented by the legendary scientist Michael Faraday.† A few years earlier, Faraday and others had sponsored the founding of the Royal College of Chemistry in response to Justus Liebig’s‡ scathing criticism of the sorry state of chemical research in England. Liebig’s brilliant young student, August von Hofmann,§ was hired as its first director. London in the 1850s was perhaps the filthiest, most polluted city in the world. It was lit by gas lamps, with the gas extracted from coal. The residue of the extraction was massive quantities of coal tar, most of which was simply dumped into London’s streams and rivers. However, coal tar was rich in aromatic amines, and Hofmann, who found these compounds fascinating, could obtain coal tar in London by the barrelful, just by asking. By 1856, Perkin had begun studying in Hofmann’s laboratory. Malaria was rampant in England (indeed, in most of the world) at the time, and the only remedy was quinine, the single source of which was the bark of the South American cinchona tree. Perkin knew that the molecular formula of quinine was C20H24N2O2 and that Hofmann had isolated several 10-carbon amines, including naphthalenamine,
C10H9N. Lacking any knowledge of molecular structure, Perkin decided to try to synthesize quinine by oxidative dimerization of such coal tar – derived 10-carbon aromatic amines. What could be more useful than to convert a major environmental pollutant into a desperately needed medicine? Perkin’s experiments gave black crude products, out of which could be extracted brightly colored powders. Pursuing these experiments with more amines and, eventually, amine mixtures, Perkin stumbled onto a procedure that gave a brilliant purple substance to which was given the name “mauve.” Within a short time, dyes based on Perkin’s mauve had transformed the world of fashion — not exactly what he had in mind originally, but the discovery did make him very rich. Previously, the only source of purple dye (so-called Royal Purple, because only royalty could afford it) was from the shell lining of a species of mollusk found in only two locations on the Mediterranean coast. Thus was born the synthetic dye industry, the first of the large-scale industries based on chemical science. Remarkably, the correct basic chemical structure of Perkin’s mauve was not determined and published until 1994, by Professor Otto Meth-Cohn and undergraduate Mandy Smith, at the University of Sunderland, England. The reaction that forms the major constituent of the dye is shown below.
N
H 3C NH2
NH2
NH2 CH3
2
K2Cr2O7, H2SO4, H2O
N H
N
CH3 NH2
CH3
*Sir William Henry Perkin (1838 – 1907), Royal College of Chemistry, London, England. † This the same Faraday as the discoverer of benzene (Chapter 15 Opening).
‡
Baron Justus von Liebig (1803 – 1873), University of Munich, Germany. § This is the same Hofmann as that of the Hofmann rule (Section 11-6) and the Hofmann rearrangement (Section 20-7).
Exercise 22-32 Write the products of diazo coupling of benzenediazonium chloride with each of the following molecules. (a) Methoxybenzene; (b) 1-chloro-3-methoxybenzene; (c) 1-(dimethylamino)-4(1,1-dimethylethyl)benzene. (Hint: Diazo couplings are quite sensitive to steric effects.)
In Summary Arenediazonium cations attack activated benzene rings by diazo coupling, a process that furnishes azobenzenes, which are often highly colored.
The Big Picture
Chapter 22
1063
The process involves oxidation of some of the amine starting material to nitrosobenzenes, such as C6H5NO, followed by acid-catalyzed electrophilic aromatic substitution reactions between the nitroso compounds and other molecules of amine.
Scanning electron micrograph of Mycobacterium smegmatis found in soil. The species is used as a model for studies of tuberculosis caused by Mycobacterium tuberculosis. Image width: 8.38 3 1026 m.
Linda Evangelista models a lacy mauve dress from Christian Dior’s autumn collection: l max 5 550 nm.
Perkin’s discovery did not help cure malaria, but it nevertheless had an enormous impact on medical science. In the 1860s, Robert Koch¶, one of the founders of the field ¶
Professor H. H. Robert Koch (1843 – 1910), University of Göttingen, Germany, Nobel Prize 1905 (physiology or medicine).
of bacteriology, began to follow up on reports that anthrax was caused by microscopic rod-shaped bodies found in the blood of infected sheep. He developed practical techniques for culturing and growing colonies of bacteria, a concept introduced by Pasteur a generation earlier but never demonstrated. Koch went on to discover the microorganisms that caused tuberculosis, conjunctivitis, amebic dysentery, and cholera, publishing superb micrographs to support his findings. By the last decades of the nineteenth century, Paul Ehrlich** became aware of reports that Perkin’s dyes could be used to stain cells, and he applied these ideas to help better visualize Koch’s tuberculosis bacilli. He realized that the staining phenomena were in fact chemical reactions and that the dyes stained different microbiological entities differently. Ehrlich found that some dyes could be used as therapeutic agents to combat the effects of bacterial toxins, thus pioneering the field of chemotherapy and making great strides in the development of immunology. In 1910, Ehrlich developed the first effective drug to combat syphilis, a disease that had become a worldwide scourge. His work helped give birth to the fields of biochemistry, cell biology, and medicinal chemistry. Thus, Perkin’s discoveries of a halfcentury earlier had changed more than the world of fashion: They changed the world.
**Professor Paul Ehrlich, (1854 – 1915), Royal Institute for Experimental Therapy, Frankfurt, Germany, Nobel Prize 1908 (physiology or medicine).
THE BIG PICTURE This chapter completes our understanding of the interplay between the benzene ring and its alkyl, hydroxy, and amino or modified amino substituents. Just as much as the electronic character of such substituents may activate or deactivate the ring with respect to substitution (Chapter 16; Sections 22-4 and 22-6) and the way in which substitutents are positioned controls the location of attack by electrophiles and nucleophiles, the ring imparts special reactivity on attached nuclei because of its resonating capacity. In other words, and to repeat the underlying theme of the text, the structure of the substituted aromatic compound
1064
Chapter 22
Chemistry of Benzene Substituents
determines its function. In many respects, this behavior is a simple extension of the chemistry of delocalized systems (Chapter 14) with the added feature of aromaticity (Chapter 15). In the next chapter, we shall continue to examine the effect that two functional groups in the same molecule have on each other, now turning to carbonyl groups. In Chapter 24, we shall study molecules containing the carbonyl and several hydroxy groups and their biological relevance. Then, in Chapters 25 and 26, we conclude with biologically important compounds containing an array of functionality.
CHAPTER INTEGRATION PROBLEMS 22-33. 5-Amino-2,4-dihydroxybenzoic acid A is a potential intermediate in the preparation of natural products of medicinal value (Section 22-3). Propose syntheses, starting from methylbenzene (toluene). COOH
CH3 HO
NH2 OH A
SOLUTION This problem builds on the expertise that you gained (Chapter 16) in controlling the substitution patterns of target benzenes, but now with a greatly expanded range of reactions. The key is, again, recognition of the directing power of substituents — ortho, para or meta (Section 16-2) — and their interconversion (Section 16-5). Retrosynthetic analysis of compound A reveals one carbon-based substituent, the carboxy group, which can be readily envisaged to be derivable from the methyl group in the starting material (by oxidation, Section 22-2). In the starting material, the carbon-based substituent is ortho, para directing, suggesting its use (retrosynthesis 1) in the introduction of the two hydroxy functions (as in compound B, through nitration – reduction – diazotization – hydrolysis; Sections 22-4 and 22-10). In compound A, it is meta directing and potentially utilizable (retrosynthesis 2) for the amination at C3 (as in compound C, through nitration – reduction). Retrosynthesis 1 COOH
COOH A
E
HO
O2N
OH
CH3
CH3
O2N
NO2
NO2
B
Retrosynthesis 2 COOH A
COOH
COOH
CH3
E
NH2
NO2
C
The question is, Are compounds B and C effective precursors of compound A? The answer is yes. Nitration of compound B should take place at the desired position (C3 in the product), ortho and para, respectively, to the two hydroxy substituents, thus placing the nitrogen at its position in compound A. Electrophilic attack between the OH groups would be expected to be sterically hindered (Section 16-5).
Chapter Integration Problems
Conversely, the amino group in compound C, especially when protected as an amide, should direct electrophilic substitution to the less hindered ortho carbon and the para carbon, again yielding the desired pattern. The actual proposed synthetic schemes would be then as follows. Synthesis 1 CH3
CH3
COOH
O 2N
O2N
Na2Cr2O7, H, H2O
HNO3, H2SO4
H2, Ni
NO2
NO2 COOH
COOH H2N
1. NaNO2, HCl 2. H2O,
HO
1. HNO3, H2SO4 2. H2, Ni
A
OH
NH2
Synthesis 2 CH3
COOH
COOH Na2Cr2O7, H, H2O
HNO3, H2SO4
H2, Ni
NO2 COOH
COOH
O B 1. CH3CCl 2. HNO3, H2SO4
1. H2, Ni 2. NaNO2, HCl 3. H2O, 4. H, H2O
O2N
NH2 NO2
O B NCCH3 H
A
22-34. A key reaction in the early development of the “pill” (Section 4-7; Chemical Highlight 4-3) was the “dienone-phenol” rearrangement shown below. Write a mechanism. CH3 OH % C
H
CH3
H 3C H % ≥ % ≥ H H O
CH3 OH % C H ≥ % ≥ H H
HO
SOLUTION This is a mechanistic problem involving an acid-catalyzed rearrangement featuring a migrating alkyl (methyl) group. Sound familiar? Review Section 9-3 on carbocation rearrangements! How do we obtain a suitable carbocation from our starting material? Answer: Protonation of the carbonyl group, which furnishes a resonance-stabilized hydroxypentadienyl cation (Section 14-6, 14-7, and Exercise 18-22). Protonation of Dienone Function CH3 % O ð
CH3 %
H
HO
CH3 % etc.
š HO A
Chapter 22
1065
1066
Chapter 22
Chemistry of Benzene Substituents
Two of the resonance structures (write all of them) place a positive charge next to the methyl group. Only form A is “productive,” inasmuch as it leads to product by aromatization, which is the major driving force for the whole process. Methyl Shift and Phenol Formation H3 C H %C
CH3 %
š HO
CH3 OH % C
CH3
H ≥ % ≥ H H
H
š HO
HO
A
New Reactions Benzylic Resonance 1. Radical Halogenation (Section 22-1) RCH2
RCHX
RCHj
X2
HX
RCH
through
etc. j Benzylic radical
Requires heat, light, or a radical initiator
2. Solvolysis (Section 22-1) RCHOSO2R
RCHOR
ROH
RCH
SN 1
RSO3H
RCH
through
etc.
Benzylic cation
3. SN2 Reactions of (Halomethyl)benzenes (Section 22-1) CH2X
CH2Nu )Nu
X
Through delocalized transition state
4. Benzylic Deprotonation (Section 22-1) CH3
CH2Li
pKa
RLi
RH
Phenylmethyllithium (Benzyllithium)
41
Oxidation and Reduction Reactions on Aromatic Side Chains 5. Oxidation (Section 22-2) R
O
RCH2 CrO3
COOH 1. KMnO4, HO, 2. H, H2O
New Reactions
Benzylic alcohols R
RCHOH
O
MnO2, acetone
6. Reduction by Hydrogenolysis (Section 22-2) CH2OR
CH3 H2, Pd–C, ethanol
ROH
C6H5CH2 is a protecting group for ROH.
Phenols and Ipso Substitution 7. Acidity (Section 22-3) p )O)
OH
O
H
etc.
p
pKa
Phenoxide ion
10
Much stronger acid than simple alkanols
8. Nucleophilic Aromatic Substitution (Section 22-4) Cl
Nu
Nu
Cl NO2
NO2 Nu)
NO2
NO2
NO2
Cl
NO2
Nucleophile attacks at ipso position.
9. Aromatic Substitution Through Benzyne Intermediates (Section 22-4) Cl
NH2 NaNH2, liquid NH3
NH3
NaCl Nucleophile attacks at both ipso and ortho positions.
X
OH 1. NaOH, 2. H , H2O
10. Arenediazonium Salt Hydrolysis (Section 22-4) N2
NH2 NaNO2, H, 0C
OH H2O,
Benzenediazonium cation
N2
Chapter 22
1067
1068
Chapter 22
Chemistry of Benzene Substituents
11. Substitution with Pd Catalysis X
OH KOH, Pd catalyst, PR3, 100C
Reactions of Phenols and Alkoxybenzenes 12. Ether Cleavage (Section 22-5) OR
OH HBr,
RBr
Aryl C–O bond is not cleaved.
13. Ether Formation (Section 22-5) OH
OR RX
NaOH, H2O
Alkoxybenzene Williamson method (Section 9-6)
14. Esterification (Section 22-5) O B OCR
OH
O B RCCl
Base
Phenyl alkanoate
15. Electrophilic Aromatic Substitution (Section 22-6) OCH3
OCH3
OCH3 E
E
H
E
16. Phenolic Resins (Section 22-6) OH
OH CH2OH
CH2 P O
NaOH
HO H 2O
OH
O
OH
OH
CH2 , NaOH
polymer
New Reactions
17. Kolbe Reaction (Section 22-6) OH
OH
CO2
COOH
1. NaOH, pressure 2. H , H2O
18. Claisen Rearrangement (Section 22-7) Aromatic Claisen rearrangement
O
O
OH
H
Aliphatic Claisen rearrangement
O
O
H
19. Cope Rearrangement (Section 22-7)
R
R
20. Oxidation (Section 22-8)
OH
O Na2Cr2O7, H
OH
O 2,5-Cyclohexadiene-1,4-dione (p-Benzoquinone)
21. Conjugate Additions to 2,5-Cyclohexadiene-1,4-diones (p-Benzoquinones) (Section 22-8) OH
O
HCl Cl
O
OH
Chapter 22
1069
1070
Chemistry of Benzene Substituents
Chapter 22
22. Diels-Alder Cycloadditions to 2,5-Cyclohexadiene-1,4-diones (p-Benzoquinones) (Section 22-8) O
O H %
0 H
O
O
23. Lipid Peroxidation (Section 22-9) H
H C R
OOH
H
O2, radical chain reaction
C R
R
Fragmentation to alkoxy radicals, followed by -scission
toxic substances such as 4-hydroxy-2-alkenals
R
24. Inhibition by Antioxidants (Section 22-9) š Oj
H R
C R
R
š OOj
H
R
CH3
C
HO
R or
OH
H
CH3
C
O
OOH
H
R
CH3
R or
CH3 H3C
š jO CH3 H3C
C R
CH3
R
Vitamin E (or BHA or BHT)
25. Vitamin C as an Antioxidant (Section 22-9) OH
OH
ðš O
O
H#
e
H
HO
jš O
O
H#
O
O H
HO
CH2OH
CH2OH Semidehydroascorbic acid
Arenediazonium Salts 26. Sandmeyer Reactions (Section 22-10) N2X
X CuX,
X Cl, Br, I, CN
27. Reduction (Section 22-10) N2
H H3PO2
O
N2
N2
R
Problems
28. Diazo Coupling (Section 22-11) N c N J
OH
H
OH Azo compound Occurs only with strongly activated rings
Important Concepts 1. Phenylmethyl and other benzylic radicals, cations, and anions are reactive intermediates stabilized by resonance of the resulting centers with a benzene p system. 2. Nucleophilic aromatic ipso substitution accelerates with the nucleophilicity of the attacking species and with the number of electron-withdrawing groups on the ring, particularly if they are located ortho or para to the point of attack. 3. Benzyne is destabilized by the strain of the two distorted carbons forming the triple bond. 4. Phenols are aromatic enols, undergoing reactions typical of the hydroxy group and the aromatic ring. 5. Benzoquinones and benzenediols function as redox couples in the laboratory and in nature. 6. Vitamin E and the highly substituted phenol derivatives BHA and BHT function as inhibitors of the radical-chain oxidation of lipids. Vitamin C also is an antioxidant, capable of regenerating vitamin E at the surface of cell membranes. 7. Arenediazonium ions furnish reactive aryl cations whose positive charge cannot be delocalized into the aromatic ring. 8. The amino group can be used to direct electrophilic aromatic substitution, after which it is replaceable by diazotization and substitution, including reduction.
Problems 35. Give the expected major product(s) of each of the following reactions. CH2CH3
(a)
Cl2(1 equivalent), hv
(b)
NBS (1 equivalent), hv
36. Formulate a mechanism for the reaction described in Problem 35(b). 37. Propose syntheses of each of the following compounds, beginning in each case with ethylbenzene. (a) (1-Chloroethyl)benzene; (b) 2-phenylpropanoic acid; (c) 2-phenylethanol; (d) 2-phenyloxacyclopropane. 38. Predict the order of relative stability of the three benzylic cations derived from chloromethylbenzene (benzyl chloride), 1-(chloromethyl)-4-methoxybenzene (4-methoxybenzyl chloride), and 1-(chloromethyl)-4-nitrobenzene (4-nitrobenzyl chloride). Rationalize your answer with the help of resonance structures. 39. By drawing appropriate resonance structures, illustrate why halogen atom attachment at the para position of phenylmethyl (benzyl) radical is unfavored compared with attachment at the benzylic position. 40. Triphenylmethyl radical, (C6H5)3C?, is stable at room temperature in dilute solution in an inert solvent, and salts of triphenylmethyl cation, (C6H5)3C1, can be isolated as stable crystalline solids. Propose explanations for the unusual stabilities of these species.
Chapter 22
1071
1072
Chapter 22
Chemistry of Benzene Substituents
41. Give the expected products of the following reactions or reaction sequences. CH2Cl
(a) BrCH2CH2CH2
H2O,
1. KCN, DMSO 2. H, H2O,
(b)
1. CH3CH2CH2CH2Li, (CH3)2NCH2CH2N(CH3)2, THF 2. C6H5CHO 3. H, H2O,
(c)
Fluorene
CH2Br
C16H14
42. The hydrocarbon with the common name fluorene is acidic enough (pKa < 23) to be a useful indicator in deprotonation reactions of compounds of greater acidity. Indicate the most acidic hydrogen(s) in fluorene. Draw resonance structures to explain the relative stability of its conjugate base. 43. Outline a straightforward, practical, and efficient synthesis of each of the following compounds. Start with benzene or methylbenzene. Assume that the para isomer (but not the ortho isomer) may be separated efficiently from any mixtures of ortho and para substitution products. CH2CH2Br
CONH2
COOH
O Br
(a)
(b)
(c)
Br
(d) COOCH3
Cl
44. Rank the following compounds in descending order of reactivity toward hydroxide ion. Br
Br
Br
Br
Br
NO2
NO2 NO2
NO2 NO2
NO2
NO2
45. Predict the main product(s) of the following reactions. In each case, describe the mechanism(s) in operation. Cl
Cl
Cl Cl
NO2 H2NNH2
(a)
NaOCH3, CH3OH
(b) O2N
LiN(CH2CH3)2, (CH3CH2)2NH
(c)
NO2 CH3
NO2
46. Starting with benzenamine, propose a synthesis of aklomide, an agent used to treat certain exotic fungal and protozoal infections in veterinary medicine. Several intermediates are shown to give you the general route. Fill in the blanks that remain; each requires as many as three sequential reactions. (Hint: Review the oxidation of amino- to nitroarenes in Section 16-5.) NH2
NH2 (a)
NO2
NO2
(b)
(c)
(d)
Cl Br
NO2
Br
Cl CN
Cl CONH2 Aklomide
Problems
47. Explain the mechanism of the following synthetic transformation. (Hint: Two equivalents of butyllithium are used.) OCH3
1. CH3CH2CH2CH2Li 2. H2C P O 3. H, H2O
F
OCH3 CH2OH CH2CH2CH2CH3
48. In nucleophilic aromatic substitution reactions that proceed by the addition–elimination mechanism, fluorine is the most easily replaced halogen in spite of the fact that F2 is by far the worst leaving group among halide ions. For example, 1-fluoro-2,4-dinitrobenzene reacts much more rapidly with amines than does the corresponding chloro compound. Suggest an explanation. (Hint: Consider the effect of the identity of the halogen on the rate-determining step.) 49. Based on the mechanism presented for the Pd-catalyzed reaction of a halobenzene with hydroxide ion, write out a reasonable mechanism for the Pd-catalyzed reaction of 1-bromo-3-methoxybenzene with 2-methyl-1-propanamine shown in Section 22-4. 50. Give the likely products of each of the following reactions. Each one is carried out in the presence of a Pd catalyst, a phosphine, and heat.
(a) F3C
Cl
(b)
Br
CH3CH2CH2SH
N H O
(c) CH3O
I
Na CN
(d)
Br
š
51. A very efficient short synthesis of resveratrol (Chemical Highlight 22-1) was reported in 2006. Fill in reasonable reagents for steps (a) through (d). Refer to the referenced text sections as needed. O B CH3CO
HO
HO CHO
(a)
CH P CH2
Section 17-12
(b)
HO
HO
O B CH3CO (c)
CH P CH
Section 13-9
CH P CH2
Sections 19-9, 20-3
CH3CO B O
O B OCCH3
(d) Section 20-4
resveratrol
CH3CO B O
52. The reaction sequence shown below illustrates the synthesis of 2,4,5-trichlorophenoxyacetic acid (2,4,5-T), a powerful herbicide. A 1 : 1 mixture of the butyl esters of 2,4,5-T and its dichlorinated analog 2,4-D was used between 1965 and 1970 as a defoliant during the Vietnam War
Chapter 22
1073
1074
Chapter 22
Chemistry of Benzene Substituents
under the code name Agent Orange. Propose mechanisms for the reactions in the synthesis of this substance, whose effects on the health of those exposed to it remain topics of considerable controversy.
Cl
Cl
Cl
Cl
1. NaOH, 150°C 2. H, H2O
Cl
OH
Cl
Cl
ClCH2COOH, NaOH, H2O,
NaCl
Cl
OCH2COOH
NaCl
Cl
2,4,5-Trichlorophenol (2,4,5-TCP)
Cl 85%
2,4,5-Trichlorophenoxyacetic acid (2,4,5-T)
53. Give the expected major product(s) of each of the following reactions and reaction sequences. 1. KMnO4, OH, 2. H, H2O
(a)
1. (CH3)2CHCl, AlCl3 2. HNO3, H2SO4 3. KMnO4, NaOH, 4. H, H2O
CH3 CH2OH
(b)
O B CH2CCH3
1. MnO2, acetone 2. KOH, H2O,
(c)
54. Rank the following compounds in order of descending acidity. OH
(a) CH3OH
(b) CH3COOH
OH
(c)
(d)
SO3H
OH
OH
(e)
(f)
OCH3
CF3
55. Design a synthesis of each of the following phenols, starting with either benzene or any monosubstituted benzene derivative. OH
OH Br
OH Br
Cl
(b)
(a)
(c) The three benzenediols
NO2
(d) NO2
CH3
56. Starting with benzene, propose syntheses of each of the following phenol derivatives.
OCH2COOH
O B OCCH3
NHCOCH3
Cl
COOH
(a)
(b)
(c) Br
Cl The herbicide 2,4-D
OCH2CH3
Br
Phenacetin
Dibromoaspirin
(The active ingredient in Midol)
(An experimental drug for the treatment of sickle-cell anemia)
Problems
Chapter 22
57. Name each of the following compounds.
OH
OH
OH
Cl
HO
(a)
(b)
O
OH
(c)
O
(d)
Br
(e)
SO3H
SCH3 O
CH2OH
58. Give the expected product(s) of each of the following reaction sequences.
O
OH 1. 2 CH2PCHCH2Br, NaOH 2.
(a)
CH3 1. 2. O3, then Zn, H 3. NaOH, H2O,
(b)
Cl Cl
OH Ag2O
(c) OH
Cl Cl
OH
OH H3C
O
O CH3 Ag2O
(d)
CH3CH2SH
(e)
OH
O
(two possibilities)
(f) O
59. As a children’s medicine, acetaminophen (Tylenol) has a major marketing advantage over aspirin: Liquid Tylenol preparations (essentially, acetaminophen dissolved in flavored water) are stable, whereas comparable aspirin solutions are not. Explain. 60. Chili peppers contain significant quantities of vitamins A, C, and E, as well as folic acid and potassium. They also contain small quantities of capsaicin, the fiery essence responsible for the “hot” in hot peppers. The pure substance is in fact quite dangerous: Chemists working with capsaicin must work in special filtered-air enclosures and wear full body protection. One milligram placed on the skin will cause a severe burn. Although capsaicin has no odor or flavor of its own, its pungency — in the form of detectable stimulation of the nerves in the mucous membranes of the mouth — can be detected even when the substance is diluted to one part in 17 million of water. The hottest peppers exhibit about 1y20th of this level of pungency. The structure of capsaicin is shown on p. 1028. Some of the data that were used in its elucidation are presented below. Interpret as much of this information as you can. MS: mYz 5 122, 137 (base peak), 195 (tricky!), 305. IR: n~ 5 972, 1660, 3016, 3445, 3541 cm21. 1 H NMR: d 5 0.93 (6H, d, J 5 8 Hz), 1.35 (2H, quin, J 5 7 Hz), 1.62 (2H, quin, J 5 7 Hz), 1.97 (2H, q, J 5 7 Hz), 2.18 (2H, t, J 5 7 Hz), 2.20 (1H, m), 3.85 (3H, s), 4.33 (2H, d, J 5 6 Hz), 5.33 (2H, m), 5.82 (2H, broad s), 6.73 (1H, d, J 5 9 Hz), 6.78 (1H, s), 6.85 (1H, d, J 5 9 Hz) ppm.
61.
Biochemical oxidation of aromatic rings is catalyzed by a group of liver enzymes called aryl hydroxylases. Part of this chemical process is the conversion of toxic aromatic
1075
1076
Chapter 22
Chemistry of Benzene Substituents
hydrocarbons such as benzene into water-soluble phenols, which can be easily excreted. However, the primary purpose of the enzyme is to enable the synthesis of biologically useful compounds, such as the amino acid tyrosine.
NH2 A CH2CHCOOH
NH2 A CH2CHCOOH O2, hydroxylase
OH Phenylalanine
Tyrosine
(a) Extrapolating from your knowledge of benzene chemistry, which of the following three possibilities seems most reasonable: The oxygen is introduced by electrophilic attack on the ring; the oxygen is introduced by free-radical attack on the ring; or the oxygen is introduced by nucleophilic attack on the ring? (b) It is widely suspected that oxacyclopropanes play a role in arene hydroxylation. Part of the evidence is the following observation: When the site to be hydroxylated is initially labeled with deuterium, a substantial proportion of the product still contains deuterium atoms, which have apparently migrated to the position ortho to the site of hydroxylation.
R
R
R
O2, hydroxylase
as an intermediate
through H
D D
D
OH
O
Suggest a plausible mechanism for the formation of the oxacyclopropane intermediate and its conversion into the observed product. (Hint: Hydroxylase converts O2 into hydrogen peroxide, HO – OH.) Assume the availability of catalytic amounts of acids and bases, as necessary. Note: In victims of the genetically transmitted disorder called phenylketonuria (PKU), the hydroxylase enzyme system described here does not function properly. Instead, phenylalanine in the brain is converted into 2-phenyl-2-oxopropanoic (phenylpyruvic) acid, the reverse of the process shown in Problem 57 of Chapter 21. The buildup of this compound in the brain can lead to severe retardation; thus people with PKU (which can be diagnosed at birth) must be restricted to diets low in phenylalanine. 62. A common application of the Cope rearrangement is in ring-enlargement sequences. Fill in the reagents and products missing from the following scheme, which illustrates the construction of a 10-membered ring.
O (a)
O O
(d)
(e)
(f)
200C
O (b)
O O
1. LiAlH4, (CH3CH2)2O 2. H, H2O
(c)
Excess PCC, CH2Cl2
Problems
Chapter 22
63. As mentioned in Section 22-10, the Sandmeyer reactions, in which copper(I) ion catalyzes substitution of the nitrogen in arenediazonium salts by Cl, Br, or CN, take place by complex mechanisms involving radicals. Explain why these substitutions do not follow either the SN1 or SN2 pathway. 64. Formulate a detailed mechanism for the diazotization of benzenamine (aniline) in the presence of HCl and NaNO2. Then suggest a plausible mechanism for its subsequent conversion into iodobenzene by treatment with aqueous iodide ion (e.g., from K1I2). Bear in mind your answer to Problem 63. 65. Show how you would convert 3-methylbenzenamine into each of the following compounds: (a) methylbenzene; (b) 1-bromo-3-methylbenzene; (c) 3-methylphenol; (d) 3-methylbenzonitrile; (e) N-ethyl-3-methylbenzenamine. 66. Devise a synthesis of each of the following substituted benzene derivatives, starting from benzene.
CN
Cl
OH
OH O2N
(a)
(b)
(c)
(d)
COOH
Br
Cl CN
NO2
COOH
Cl
COOH
Br
(e)
Br
(f)
Br
(g) Cl Br
I
Br
67. Write the most reasonable structure of the product of each of the following reaction sequences.
1. NaNO2, HCl, 5C 1. NaNO2, HCl, 5C
NH2
OH
NH2 2. HO
(a)
Golden Yellow
(b)
NH 2.
SO3H
SO3H
For the following reaction, assume that electrophilic substitution occurs preferentially on the most activated ring.
1. NaNO2, HCl, 5°C OH
NH2 2.
(c) SO3H
Orange I
Metanil Yellow
1077
1078
Chapter 22
Chemistry of Benzene Substituents
68. Show the reagents that would be necessary for the synthesis by diazo coupling of each of the following three compounds. For structures, see Section 22-11. (a) Methyl Orange
(b) Congo Red
J N
(c) Prontosil, H2N
SO2NH2
N
, which is
NH2
converted microbially into sulfanilamide, H2N
SO2NH2
(The accidental discovery of the antibacterial properties of prontosil in the 1930s led indirectly to the development of sulfa drugs as antibiotics in the 1940s.) 69.
(a) Give the key reaction that illustrates the inhibition of fat oxidation by the preservative BHT. (b) The extent to which fat is oxidized in the body can be determined by measuring the amount of pentane exhaled in the breath. Increasing the amount of vitamin E in the diet decreases the amount of pentane exhaled. Examine the processes described in Section 22-9 and identify one that could produce pentane. You will have to do some extrapolating from the specific reactions shown in the section.
70.
The urushiols are the irritants in poison ivy and poison oak that give you rashes and make you itch upon exposure. Use the following information to determine the structures of urushiols I (C21H36O2) and II (C21H34O2), the two major members of this family of unpleasant compounds. H2, Pd–C, CH3CH2OH
Urushiol II
urushiol I
Excess CH3I, NaOH
Urushiol II
1. O3, CH2Cl2 2. Zn, H2O
C23H38O2 Dimethylurushiol II
CH3CH2CH2CH2CH2CH2CHO
C16H24O3 Aldehyde A
Synthesis of Aldehyde A OCH3 1. SO3, H2SO4 2. HNO3, H2SO4
C7H7NSO6
H , H2O,
B
C7H8O2
C7H7NO3
1. H2, Pd, CH3CH2OH 2. NaNO2, H,H2O 3. H2O,
C
1. CO2, pressure, KHCO3, H2O 2. NaOH, CH3I 3. H, H2O
D
C9H10O4
1. LiAlH4, (CH3CH2)2O 2. H, H2O 3. MnO2, acetone
E
C9H10O3
1. C6H5CH2O(CH2)6CH P P(C6H5)3 2. Excess H2, Pd–C, CH3CH2OH 3. PCC, CH2Cl2
aldehyde A
F
71. Is the site of reaction in the biosynthesis of norepinephrine from dopamine (see Chapter 5, Problem 65) consistent with the principles outlined in this chapter? Would it be easier or more difficult to duplicate this transformation nonenzymatically? Explain.
Problems
Chapter 22
Team Problem 72. As a team, consider the following schemes that outline steps toward the total synthesis of taxodone D, a potential anticancer agent. For the first scheme, divide your team into two groups, one to discuss the best option for A to effect the initial reduction step, the second to assign a structure to B, using the partial spectral data provided. O B OCCH3
O B CH3CO
O B CH3CO
H3C %
H3C %
A
≥ H
O B OCCH3
H, toluene,
≥ H
O
B
NMR: 5.99 (dd, 1 H), 6.50 (d, 1 H) ppm. IR: ˜ 1720 cm1. MS: m/z 384 (M). 1H
OH
Reconvene to discuss both parts of the first scheme. Then, as a group, analyze the remainder of the synthesis shown in the second scheme below. Use the spectroscopic data to help you determine the structures of C and taxodone, D. O B COOH
B
Cl
OH, H2O
C NMR: 3.51 (dd, 1 H), 3.85 (d, 1 H) ppm. MS: m/z 400 (M).
1H
D
Taxodone NMR: 6.55 (d, 1 H), 6.81 (s, 1 H) ppm, no other alkenyl or aromatic signals. IR: ˜ 3610, 3500, 1628 cm1. UV-Vis: max ( ) 316 (20,000) nm. MS: m/z 316 (M). 1H
Propose a mechanism for the formation of D from C. (Hint: After ester hydrolysis, one of the phenoxide oxygens can donate its electron pair through the benzene ring to effect a reaction at the para position. The product contains a carbonyl group in its enol form. For the interpretation of some of the spectral data, see Section 17-3.)
Preprofessional Problems 73. After chlorobenzene has been boiled in water for 2 h, which of the following organic compounds will be present in greatest concentration? O OH
(a) C6H5OH
(b)
OH
(c) Cl
(d) C6H5Cl
(e)
Cl Cl
H
HI, D
74. What are the products of the following reaction? C6H5OCH3 uy? OCH3
(a) C6H5I 1 CH3OH
(b) C6H5OH 1 CH3I
(c) C6H5I 1 CH3I
(d) I
H2
1079
1080
Chemistry of Benzene Substituents
Chapter 22
75. The transformation of 4-methylbenzenediazonium bromide to toluene is best carried out by using: (a) H1, H2O
(c) H2O, 2OH
(b) H3PO2, H2O
(d) Zn, NaOH
76. What is the principal product after the slurry obtained on treating benzenamine (aniline) with potassium nitrite and HCl at 08C has been added to 4-ethylphenol? OH NPNC6H5
NHC6H5
(b)
(a)
OH
OH
OH
(c)
(d) NPNC6H5
CH2CH3
CH2CH3
CH2CH3
CH2CH3
C6H5
77. Examination of the 1H NMR spectra of the following three isomeric nitrophenols reveals that one of them displays a hydroxy (phenolic) proton at substantially lower field than do the other two. Which one? NO2
NO2
NO2 OH
(a)
(b)
(c) OH OH
C H A P T E R
Ester Enolates and the Claisen Condensation
TWENTY THREE
O
Synthesis of b-Dicarbonyl Compounds; Acyl Anion Equivalents
O
I
n Chapter 18, during our survey of carbonyl chemistry, we learned that many techniques developed by the synthetic organic chemist are in fact based on natural processes for the construction of carbon – carbon bonds in biological systems. The aldol condensation (Sections 18-5 through 18-7) is one such process, a powerful method for converting aldehydes and ketones into b-hydroxycarbonyl compounds. In this chapter we shall examine first the related Claisen condensation, in which attack of an ester enolate on a carbonyl group generates a new carbon – carbon bond. We have already seen its use in the biosynthesis of long-chain carboxylic acids (Section 19-13). The products of Claisen condensations are 1,3-dicarbonyl compounds, more commonly known as b-dicarbonyl compounds, which are important for their versatility in synthesis. Examples of -Dicarbonyl Compounds O O B B CH3CCH2CCH3 2,4-Pentanedione (Acetylacetone) (A -diketone)
O O B B CH3CCH2COCH3
O
O O
Methyl 3-oxobutanoate (Methyl acetoacetate) (A -ketoester)
O O B B HOCCH2COH Propanedioic acid (Malonic acid) (A -dicarboxylic acid)
Another natural carbon – carbon bond-forming process that has been adapted for use by synthetic chemists involves thiamine, or vitamin B1, shown on page 1100. Thiamine plays an essential role in several biochemical processes, including the biosynthesis of sugars, as we shall see in Chapter 24. Chemical Highlight 23-2 describes how thiamine also mediates sugar metabolism by converting pyruvic acid, a product of sugar metabolism, into acetyl CoA (Section 19-13). The relevant carbon – carbon bond-forming process makes use of a new kind of nucleophile derived from aldehydes and ketones, an acyl anion equivalent.
The diketone 1,2-indanedione was developed by Professor Madeleine Joullié at the University of Pennsylvania for the detection of latent fingerprints on porous surfaces such as paper. The reactive carbonyl groups undergo condensation with traces of amino acids deposited by human touch, and the adducts exhibit fluorescence on exposure to light (as shown in the photo on the left, courtesy of Professor Joullié). This advance in forensic science was spectacularly successful in pinpointing the perpetrator in the assassination of the Israeli Minister of Tourism in 2001, who was shot to death in the Jerusalem Hyatt Hotel. Initial investigations led to a room, in which a newspaper produced clear fingermarks (see photo on the right, courtesy of the division of identification and forensic science of the Israel Police, with the help of Professor Joseph Almog, The Hebrew University of Jerusalem), which were successfully compared by the AFIS (automated fingerprint identification system) to point out the assassin.
1082
Chapter 23
Ester Enolates and the Claisen Condensation
These species illustrate how a normally electrophilic carbonyl carbon atom can be temporarily made nucleophilic and used to form a new carbon – carbon bond with the carbonyl carbon of another aldehyde or ketone molecule, giving a-hydroxyketones as the ultimate products.
23-1 B-Dicarbonyl Compounds: Claisen Condensations Ester enolates undergo addition – elimination reactions with ester functions, furnishing b-ketoesters. These transformations, known as Claisen* condensations, are the ester analogs of the aldol reaction (Section 18-5).
Claisen condensations form B-dicarbonyl compounds Ethyl acetate reacts with a stoichiometric amount of sodium ethoxide to give ethyl 3-oxobutanoate (ethyl acetoacetate). REACTION
Claisen Condensation of Ethyl Acetate O B CH3COCH2CH3
O B CH3COCH2CH3
NaOCH2CH3, CH3CH2OH CH3CH2OH
O O B B CH3CCH2COCH2CH3 75% Ethyl 3-oxobutanoate (Ethyl acetoacetate)
Ethyl acetate
Unlike imine formation (Section 17-9) or the aldol condensation (Section 18-5), in which two molecules are joined with elimination of water, the Claisen condensation proceeds by elimination of a molecule of alcohol. The Claisen condensation begins with formation of the ester enolate ion (step 1). Addition – elimination of this species to the carbonyl group of another ester molecule furnishes a 3-ketoester (steps 2 and 3). These steps are reversible; the overall process is endothermic through this stage. Indeed, the initial deprotonation of the ester (pKa < 25) by ethoxide is particularly unfavorable (pKa of ethanol 5 15.9). Mechanism of the Claisen Condensation
MECHANISM
Step 1. Ester enolate formation O B CH3COCH2CH3
MATION AN I
ANIMATED MECHANISM: The Claisen condensation
Na ðš OCH2CH3 š
š Oð Na ðCH2 O C i
CH2 P C š OCH2CH3
š Oð f i
š OCH2CH3
š OH CH3CH2
Step 2. Nucleophilic addition ðOð B CH3COCH2CH3
O B ðCH COCH CH 2 2 3
Oð ðš A CH3 O C O OCH2CH3 A CH2COCH2CH3 New B carbon–carbon bond O
*This is the Claisen of the Claisen rearrangement (Section 22-7).
23-1 b-Dicarbonyl Compounds: Claisen Condensations
Chapter 23
Step 3. Elimination Oð ðš A CH3CO š OCH2CH3 A š CH2COCH2CH3 B O
O O B B CH3CCH2COCH2CH3
ðš OCH2CH3 š
3-Ketoester
The product of step 3, a 3-ketoester (or b-ketoester), possesses hydrogens on its central carbon atom that are unusually acidic (pKa < 11) due to the presence of carbonyl groups on both sides. As a result, this ketoester immediately loses a proton to ethoxide, converting the base into ethanol (step 4). This is the reason that a stoichiometric rather than a catalytic amount of ethoxide is necessary for a Claisen condensation to succeed: Each molecule of 3-ketoester formed protonates a molecule of ethoxide, thereby removing the base needed to catalyze steps 1 to 3. The use of a stoichiometric amount of ethoxide assures that the base will not be completely depleted until all of the starting ester has been converted to product. Furthermore, the highly favorable deprotonation in step 4 also serves to drive the unfavorable equilibria of steps 1 to 3 in the forward direction. Work-up with aqueous acid (step 5) restores the proton to the central carbon atom of the ketoester, furnishing the final product and completing the process. Step 4. Deprotonation of ketoester drives equilibrium O O B B CH3CCH2COCH2CH3 Acidic, pKa
11
O O B B š CH3COCHOCOCH 2CH3
ðš OCH2CH3 š
O Oð ðš B A CH3CPCHOCOCH2CH3
O Oð ðš B A CH3COCHPCOCH2CH3
Step 5. Protonation upon aqueous work-up O O B B š CH3CCHCOCH 2CH3
H, H2O
O O B B CH3CCH2COCH2CH3
To prevent transesterification, both the alkoxide and the ester are usually derived from the same alcohol. Exercise 23-1 Give the products of Claisen condensation of (a) ethyl propanoate; (b) ethyl 3-methylbutanoate; (c) ethyl pentanoate. For each, the base is sodium ethoxide, the solvent ethanol.
Protons flanked by two carbonyl groups are acidic How does the structure of the 3-ketoester make its deprotonation so favorable? The acidity of the hydrogens flanked by the two carbonyl groups is much enhanced by resonance stabilization of the corresponding anion. Table 23-1 lists the pKa values of several b-dicarbonyl compounds and related systems, such as methyl cyanoacetate and propanedinitrile (malonodinitrile). The importance of the final deprotonation step in the Claisen condensation is made clear when condensation is attempted with an ester bearing only one a-hydrogen. The product of a
OH CH3CH2š š
1083
Ester Enolates and the Claisen Condensation
Chapter 23
Table 23-1
pKa Values for B-Dicarbonyl and Related Compounds
Name
Structure
pKa
2,4-Pentanedione (Acetylacetone)
O O B B CH3CCH2CCH3
9
Methyl 2-cyanoacetate
O B NCCH2COCH3
9
Ethyl 3-oxobutanoate (Ethyl acetoacetate) Propanedinitrile (Malonodinitrile)
More acidic
1084
Diethyl propanedioate (Diethyl malonate)
O O B B CH3CCH2COCH2CH3
11
NCCH2CN
13
O O B B CH3CH2OCCH2COCH2CH3
13
reaction with such an ester would be a 2,2-disubstituted 3-ketoester lacking the acidic protons necessary to drive the equilibrium. Hence, no Claisen condensation product is observed. REACTION
Failure of a Claisen Condensation
O B 2 (CH3)2CHCOCH2CH3
Na OCH2CH3, CH3CH2OH Equilibrium lies to the left
Ethyl 2-methylpropanoate
O CH3 O B A B O CH3CH2OH (CH3)2CHCOCOOCOCH 2CH3 A CH3 Note lack of acidic hydrogens Ethyl 2,2,4-trimethyl-3-oxopentanoate (Not observed)
That this result is due to an unfavorable equilibrium can be demonstrated by treating a 2,2disubstituted 3-ketoester with base: Complete reversal of the Claisen condensation process (called retro-Claisen condensation) ensues, giving two molecules of simple ester through a mechanism that is the exact reverse of the forward reaction.
MECHANISM Reversal of a Claisen Condensation (Retro-Claisen Condensation) š CH3CH2O ð
ðOð CH3 ðOð B A O š (CH3)2CHCOCOOCOCH 2CH3 A CH3
ðOð B š (CH3)2CHCOCH 2CH3
Oð ðš A š (CH3)2CPCOCH 2CH3
OH CH3CH2š
OH CH3CH2š
Oð CH3 ðOð ðš A A O O (CH3)2CHCOOCOOCOCH 2CH3 A A CH3CH2O ð CH3 ðOð B š (CH3)2CHCOCH 2CH3
š CH3CH2O ð
Exercise 23-2 Explain the following observation. O CH3 B A 2 CH3COCOCOOCH3 A CH3
1. CH3ONa, CH3OH 2. H, H2O
O B CH3CCH2COOCH3
2 (CH3)2CHCOOCH3
Claisen condensations can have two different esters as reactants Mixed Claisen condensations start with two different esters. Like crossed aldol condensations (Section 18-6), they are typically unselective and furnish product mixtures. However,
23-1 b-Dicarbonyl Compounds: Claisen Condensations
Chapter 23
1085
a selective mixed condensation is possible when one of the reacting partners has no ahydrogens, as in ethyl benzoate. A Selective Mixed Claisen Condensation ON EOCH2CH3 C
O B CH3CH2COCH2CH3
O O B B COCHCOCH2CH3 A CH3
1. CH3CH2ONa, CH3CH2OH 2. H, H2O
71%
Ethyl benzoate
Ethyl 2-methyl-3-oxo3-phenylpropanoate
Exercise 23-3 Give all the Claisen condensation products that would result from treatment of a mixture of ethyl acetate and ethyl propanoate with sodium ethoxide in ethanol.
Exercise 23-4 Is the mixed Claisen condensation between ethyl formate and ethyl acetate likely to afford one major product? Explain and give the structure(s) of the product(s) you expect.
Intramolecular Claisen condensations result in cyclic compounds The intramolecular version of the Claisen reaction, called the Dieckmann* condensation, produces cyclic 3-ketoesters. As expected (Section 9-6), it works best for the formation of five- and six-membered rings. O O O CO2CH2CH3 1. CH CH O Na , CH CH OH B B 2. H , H O CH3CH2OCCH2CH2CH2CH2CH2COCH2CH3 3
2
3
MODEL BUILDING
2
2
60% O O
O
O O
O
O Diethyl heptanedioate
Ethyl 2-oxocyclohexanecarboxylate
Exercise 23-5
Working with the Concepts: Predicting a Successful Claisen Condensation Two cyclic products are possible from the Dieckmann (intramolecular Claisen) condensation shown below, but only one of them actually forms. Describe and explain briefly the outcome of this reaction. O O B B CH3CH2OCCH2CH2CH2CH2CHCOCH2CH3 A CH3
*Professor Walter Dieckmann (1869 – 1925), University of Munich, Germany.
1. CH3CH2ONa, CH3CH2OH 2. H, H2O
1086
Ester Enolates and the Claisen Condensation
Chapter 23
Strategy To draw the possible products, use the mechanistic pattern described throughout this section. Solution • Form a new carbon– carbon bond between the carbon atom a to one of the ester carbonyl groups and the carbonyl carbon atom of the other. The products of the two possible options are shown below:
O O O B ␣ B CH3CH2OCCH2CH2CH2CH2CHCOCH2CH3 A CH3
O O B␣ B CH3CH2OCCH2CH2CH2CH2CHCOCH2CH3 A CH3
1. CH3CH2ONa, CH3CH2OH 2. H, H2O
CH3 ␣
CO2CH2CH3
H
O
1. CH3CH2O Na , CH3CH2OH 2. H, H2O
CH3CH2O2C
CH3
␣
• Notice that the product of the first condensation lacks an acidic a-hydrogen between its two carbonyl groups. This molecule is not isolated under Claisen condensation conditions, because the equilibrium associated with its formation cannot be driven to completion by deprotonation (step 4 of the mechanism presented earlier in this section; see also “Failure of a Claisen Condensation”). Only the product of the second condensation process is obtained.
CHE M I C AL H I G H L I G H T 2 3 - 1 Claisen Condensations in Biochemistry O B CH3CSCoA
CO2
Acetyl CoA carboxylase
O O B B HOCCH2CSCoA Malonyl CoA
Acetyl coenzyme A
The methylene group in the carboxylated species is much more reactive than the methyl group in acetyl thioesters and participates in a wide variety of Claisenlike condensations. Although these processes require enzyme catalysis, they may be written in simplified form, shown below.
The coupling processes that build fatty acid chains from thioesters of coenzyme A (Section 19-13) are forms of Claisen condensations. The carboxylation of acetyl CoA into malonyl CoA (shown above) is a variant in which the carbon of CO2 rather than that of an ester carbonyl group is the site of nucleophilic attack.
O B ECH H3C SR H A RSE HCH2E HO C C B B O
CO2
ðš Oð R A f H3C O C O Sð A i RSE HCH2 H C B
RSH
RSE HCH2E HCH3 C C B B O
O
O
O (RSH acyl carrier protein; see Section 19-13.)
The product of the reaction above, the acetoacetyl derivative of the acyl carrier protein, is the starting point for the biosynthesis of other compounds besides fatty acids. Steroids derive from a sequence of enzyme-catalyzed Claisen-like
condensations that produce the branched five-carbon skeleton of 2-methyl-1,3-butadiene (isoprene), the building block of the terpenes (Section 4-7). In this variation, the enolate of acetyl CoA adds to the ketocarbonyl of acetoacetyl CoA,
23-1 b-Dicarbonyl Compounds: Claisen Condensations
Chapter 23
Exercise 23-6
Try it Yourself Formulate a mechanism for the following reaction.
O
1. CH3CH2ONa, CH3CH2OH 2. H, H2O
CO2CH2CH3 CH3CO2CH2CH3
CO2CH2CH3
CO2CH2CH3 O 60–80%
Diethyl 1,2-benzenedicarboxylate (Diethyl phthalate)
Ketones undergo mixed Claisen reactions Ketones can participate in the Claisen condensation. Because they are more acidic than esters, they are deprotonated before the ester has a chance to undergo self-condensation. The products (after acidic work-up) may be b-diketones, b-ketoaldehydes, or other b-dicarbonyl compounds. The reaction can be carried out with a variety of ketones and esters both interand intramolecularly. O B CH3COCH2CH3
O B CH3CCH3
1. NaH, (CH3CH2)2O 2. H, H2O
dehydration, and phosphorylation finally gives 3-methyl3-butenyl (isoprenyl) pyrophosphate.
giving b-hydroxy-b-methylglutaryl CoA (HMG CoA). A sequence of enzyme-catalyzed reduction, decarboxylation, O O B B C C E H E H H3C CH2 SCoA
O O B B CH3CCH2CCH3 85%
OH O B A A HO2C H EC H EC H CH2 SCoA CH2 H3C
HMG synthase, thiolase
CoASH ECH2H C H B O OH A A HO2C H EC H ECH2OH CH2 CH2 H 3C
HMG CoA reductase
-Hydroxy--methylglutaryl CoA (HMG CoA)
Adenosine triphosphate (ATP, Chemical Highlight 23-2) CO2, H2O
Mevalonic acid
CH3 O O B B A O O O P O O KC H ECH2 O O O P A A CH2 CH2 O O Isoprenyl pyrophosphate
In animals, six molecules of isoprenyl pyrophosphate are used to construct the hydrocarbon squalene, which contains a 24-carbon chain and six methyl substituents. Squalene, in
turn, undergoes a complex sequence of cyclization, rearrangement, and bond cleavage to give steroids, such as cholesterol (Sections 4 – 7 and 14 – 10).
Squalene
1087
1088
Ester Enolates and the Claisen Condensation
Chapter 23
O
Exercise 23-7 1,3-Cyclohexanedione (margin) can be prepared by an intramolecular mixed Claisen condensation between the ketone carbonyl and ester functions of a single molecule. What is the structure of this substrate molecule?
O 1,3-Cyclohexanedione
Retrosynthetic analysis clarifies the synthetic utility of the Claisen condensation
O
Having seen a variety of types of Claisen condensations, we can now ask how this process may be logically analyzed for synthetic use. Three facts are available to help us: (1) Claisen condensations always form 1,3-dicarbonyl compounds; (2) one of the reaction partners in a Claisen condensation must be an ester, whose alkoxide group is lost in the course of the condensation; and (3) the other reaction partner (the source of the nucleophilic enolate) must contain at least two acidic hydrogens on an a-carbon. In addition, if a mixed condensation is being considered, one reaction partner should be incapable of self-condensation (e.g., it should lack a-hydrogens). If we are given the structure of a target molecule and wish to determine whether (and, if so, how) it can be made by a Claisen condensation, we must analyze it retrosynthetically with the preceding points in mind. For example, let us consider whether 2-benzoylcyclohexanone can be made by a Claisen condensation. It is a 1,3-dicarbonyl compound, meeting the first requirement. What bond forms in a Claisen condensation? By examining all the examples in this section, we find that the new bond in the product always connects one of the carbonyl groups of the 1,3-dicarbonyl moiety to the carbon atom between them. Our target molecule contains two such bonds, which we label a and b. As we continue our analysis by disconnecting each of these strategic bonds in turn, we must employ the second point: The carbonyl group at which the new carbon – carbon bond forms starts out as part of an ester function. Thus, working backward, we must imagine reattaching an alkoxy group to this carbonyl carbon. Thus:
O a
b
2-Benzoylcyclohexanone
Disconnect here Reattach O OR here
O
O
O
O
Disconnect here Reattach O OR here
O
O
OR a
a
b
b
⫹
RO
Disconnection of bond a reveals a ketoester, which undergoes intramolecular Claisen condensation in the forward direction, whereas disconnection of bond b gives cyclohexanone and a benzoic ester. Both condensations are quite feasible; however, the second is preferable because it constructs the target from two smaller pieces: O
O
CH3CH2O
1. CH3CH2ONa, CH3CH2OH 2. H , H2O
O
O
Exercise 23-8
Working with the Concepts: Claisen Retrosynthetic Analysis O
Show a synthesis of
O B CCO2CH2CH3
using a Claisen or a Dieckmann condensation.
23-1 b-Dicarbonyl Compounds: Claisen Condensations
Strategy Begin by identifying the strategic bonds — the ones that may be made in a Claisen-type condensation. Once you have found them (there will be two), disconnect each one in turn and attach an alkoxy group to the separated carbonyl carbon. This procedure will give you the two possible starting points for the Claisen-based synthesis. Solution • The strategic bonds are those connecting the two carbonyl groups to the shared atom between them (arrows). Label them a and b. • (1) Break each of these bonds, and (2) attach an ethoxy group to the carbonyl fragment (use an ethoxy group because the target molecule is an ethoxy ester). You should come up with something like this, which constitutes your retrosynthetic analysis: a O
O B CCO2CH2CH3
b Disconnect at a
O
Disconnect at b
O B CCO2CH2CH3
O B CCO2CH2CH3
O
Attach O OCH2CH3 to separated carbonyl carbons (arrows)
O B CH3CH2OC
O B CCO2CH2CH3
O O B CH3CH2OCCO2CH2CH3
• At this stage you have identified the two possible starting points. To complete the exercise, write the condensation reactions in the forward direction: Synthesis a: O B CH3CH2OC
␣
O B CCO2CH2CH3
O
O B CCO2CH2CH3
O
O B CCO2CH2CH3
1. CH3CH2ONa, CH3CH2OH 2. H , H2O
Synthesis b: O
O B CH3CH2OCCO2CH2CH3
1. CH3CH2ONa, CH3CH2OH 2. H , H2O
Synthesis a is a Dieckmann condensation featuring attack of the a-carbon of the ketone enolate on the ester carbonyl at the other end of the molecule. Synthesis b is a mixed Claisen condensation in which cyclohexanone enolate attacks one of the two ester groups of diethyl ethanedioate (diethyl oxalate).
Chapter 23
1089
1090
Chapter 23
Ester Enolates and the Claisen Condensation
Exercise 23-9
Try It Yourself Suggest syntheses of the following molecules by Claisen or Dieckmann condensations. O O O B B (a) CH3CCH2CH
O
CO2CH2CH3
(b)
O
(c) H3C
In Summary Claisen condensations are endothermic and therefore would not take place without a stoichiometric amount of base strong enough to deprotonate the resulting 3ketoester. Mixed Claisen condensations between two esters are nonselective, unless they are intramolecular (Dieckmann condensation) or one of the components is devoid of a-hydrogens. Ketones also participate in selective mixed Claisen reactions because they are more acidic than esters.
23-2 B-Dicarbonyl Compounds as Synthetic Intermediates Having seen how to prepare b-dicarbonyl compounds, let us explore their synthetic utility. This section will show that the corresponding anions are readily alkylated and that 3-ketoesters are hydrolyzed to the corresponding acids, which can be decarboxylated to give ketones or new carboxylic acids. These transformations open up versatile synthetic routes to other functionalized molecules.
B-Dicarbonyl anions are nucleophilic The unusually high acidity of the methylene hydrogens of b-ketocarbonyl compounds can be used to synthetic advantage. Their unusually low pKa values (ca. 9 – 13, see Table 23-1) allow alkoxide bases to remove a proton from this methylene group essentially quantitatively, giving an enolate ion that may be alkylated to produce substituted derivatives. For example, treatment of ethyl 3-oxobutanoate (ethyl acetoacetate) with NaOCH2CH3 effects complete deprotonation to the enolate, which reacts via the SN2 mechanism with iodomethane to afford the methylated derivative shown. The remaining acidic hydrogen at this position can be abstracted with the somewhat stronger base KOC(CH3)3. The resulting anion may undergo another SN2 alkylation, shown here with benzyl bromide, to yield the doubly substituted product.
-Ketoester Alkylations
O O B B CH3CCHCOCH2CH3 A H
1. NaOCH2CH3, CH3CH2OH 2. CH3 CH3CH2OH NaI
1. KOC(CH3)3, (CH3)3COH CH2Br
O O B B CH3CCHCOCH2CH3 A CH3 New bond
Ethyl 3-oxobutanoate (Ethyl acetoacetate)
2. (CH3)3COH KBr
O CH3 O B A B CH3C O C OO COCH2CH3 A CH2 New bond
65%
77%
Ethyl 2-methyl3-oxobutanoate
Ethyl 2-methyl2-(phenylmethyl)3-oxobutanoate
23-2 b-Dicarbonyl Compounds as Synthetic Intermediates
Chapter 23
Other b-dicarbonyl compounds undergo similar reactions.
O O B B CH3CH2OCCHCOCH2CH3 A H
O O B B CH3CH2OCCHCOCH2CH3 A CH3CH2CH A CH3 84%
1. NaOCH2CH3, CH3CH2OH CH3 A 2. CH3CH2CHBr CH3CH2OH NaBr
Diethyl propanedioate (Diethyl malonate)
Diethyl 2-(1-methylpropyl)propanedioate
Exercise 23-10 Give a synthesis of 2,2-dimethyl-1,3-cyclohexanedione from methyl 5-oxohexanoate.
3-Ketoacids readily undergo decarboxylation Hydrolysis of 3-ketoesters furnishes 3-ketoacids, which in turn readily undergo decarboxylation under mild conditions. The products — ketones and carboxylic acids — contain the alkyl groups introduced in prior alkylation steps.
REACTION
Formation and Decarboxylation of 3-Ketoacids O O B B RCCHCOCH2CH3 A R
Hydrolysis
O O B B RCCHCOH A R
O B RCCHR A H
3-Ketoacid
O O B B CH3CH2OCCHCOCH2CH3 A R
O O B B HOCCHCOH A R
Hydrolysis
CO2
Ketone
O B RCHCOH A H
Carboxylic acid
O O B B CH3CCHCOCH2CH3 A (CH2)3CH3
1. NaOH, H2O 2. H2SO4, H2O, 100C CH3CH2OH CO2
2-Heptanone
Ethyl 2-butyl-3-oxobutanoate
O O B B CH3CH2OCCHCOCH2CH3 A CH3CH2CH A CH3 Diethyl 2-(1-methylpropyl)propanedioate
O B CH3CCH(CH2)3CH3 A H 60%
H2SO4, H2O, CH3CH2OH CO2
CH3 A CH3CH2CHCHCOOH A H 65% 3-Methylpentanoic acid
CO2
1091
Ester Enolates and the Claisen Condensation
Chapter 23
Decarboxylation, or loss of CO2, is not a typical reaction of carboxylic acids under ordinary conditions. However, b-ketoacids are unusually prone to decarboxylation for two reasons. First, the Lewis basic oxygen of the 3-keto function is ideally positioned to bond with the carboxy hydrogen by means of a cyclic six-atom transition state. Second, this transition state has aromatic character (Section 15-3), because three electron pairs shift around the cyclic six-atom array. The species formed in decarboxylation are CO2 and an enol, which tautomerizes rapidly to the final ketone product.
Mechanism of Decarboxylation of 3-Ketoacids
MECHANISM ðOð B C
H
C C
H 3C H
ðš OH
ðOð Oð
H
C H3C
CH2
ðOð B C B ðOð
O B C CH2
H3C
CO2
H
Note that only carboxylic acids with a second carbonyl group in the 3- or b-position are structurally capable of reacting in this way. Carboxylic acids lacking a b-carbonyl function do not decarboxylate, regardless of the presence of C P O groups elsewhere in the molecule. Loss of CO2 occurs readily only from the neutral carboxylic acid. If the ester is hydrolyzed with base, acid must be added to protonate the resulting carboxylate salt in order for decarboxylation to occur. Decarboxylation of substituted propanedioic (malonic) acids follows the same mechanism.
Exercise 23-11 Formulate a detailed mechanism for the decarboxylation of CH3CH(COOH)2 (methylmalonic acid).
The acetoacetic ester synthesis leads to methyl ketones The combination of alkylation followed by ester hydrolysis and finally decarboxylation allows ethyl 3-oxobutanoate (ethyl acetoacetate) to be converted ultimately into 3substituted or 3,3-disubstituted methyl ketones. This strategy is called the acetoacetic ester synthesis.
Acetoacetic Ester Synthesis O R O B A B CH3C O CO COCH2CH3 A R
O R B CH3CCH R A
O O B B CH3CCH2COCH2CH3
A
1092
3,3-Disubstituted methyl ketone
Methyl ketones with either one or two substituent groups on C3 can be synthesized by using the acetoacetic sequence.
23-2 b-Dicarbonyl Compounds as Synthetic Intermediates
Chapter 23
Syntheses of Substituted Methyl Ketones O O B B CH3CCH2COCH2CH3
1. NaOCH2CH3, CH3CH2OH 2. CH3CH2CH2CH2Br
O O B B CH3CCHCOCH2CH3 A CH2CH2CH2CH3 72%
1. NaOH, H2O 2. H2SO4, H2O, 100C
O B CH3CCH2 O CH2CH2CH2CH3 60% 2-Heptanone
1. KOC(CH3)3, (CH3)3COH 2. CH3CH2CH2CH2I
1. KOH, H2O, 100C 2. HCl, H2O, 100C
O CH2CH2CH2CH3 B A CH3C O CHOCH2CH2CH2CH3
A
A
O O B B CH3CCCOCH2CH3
CH3CH2CH2CH2 CH2CH2CH2CH3 80%
64% 3-Butyl-2-heptanone
Exercise 23-12 Propose syntheses of the following ketones, beginning with ethyl 3-oxobutanoate (ethyl acetoacetate). (a) 2-Hexanone; (b) 2-octanone; (c) 3-ethyl-2-pentanone; (d) 4-phenyl-2-butanone.
The malonic ester synthesis furnishes carboxylic acids Diethyl propanedioate (malonic ester) is the starting material for preparing 2-alkylated and 2,2-dialkylated acetic acids, a method called the malonic ester synthesis.
Malonic Ester Synthesis O O B B CH3CH2OCCH2COCH2CH3
O R O B A B CH3CH2OC O CO COCH2CH3 A R
R A H O CO COOH A R 2,2-Dialkylated acetic acid
Like the acetoacetic ester route to ketones, the malonic ester synthesis can lead to carboxylic acids with either one or two substituents at C2. Synthesis of a 2,2-Dialkylated Acetic Acid O O B B CH3CH2OCCHCOCH2CH3 A CH3
1. NaOCH2CH3, CH3CH2OH 2. CH3(CH2)9Br, 80°C 3. KOH, H2O, CH3CH2OH, 80°C 4. H2SO4, H2O, 180°C
CH3 A CH3(CH2)9CHCOOH 74%
Diethyl 2-methylpropanedioate (Diethyl methylmalonate)
2-Methyldodecanoic acid
1093
1094
Chapter 23
Ester Enolates and the Claisen Condensation
Exercise 23-13 (a) Give the structure of the product formed after each of the first three steps in the preceding synthesis of 2-methyldodecanoic acid. (b) How would you make the starting material, diethyl 2-methylpropanedioate?
The rules and limitations governing SN2 reactions apply to the alkylation steps. Thus, tertiary haloalkanes exposed to b-dicarbonyl anions give mainly elimination products. However, the anions can successfully attack acyl halides, a-bromoesters, a-bromoketones, and oxacyclopropanes. Exercise 23-14 The first-mentioned compound in each of the following parts is treated with the subsequent series of reagents. Give the final products. (Note: The choice of the conditions for the last step(s), either direct acid-catalyzed hydrolysis – decarboxylation or stoichiometric base hydrolysis – acidification – decarboxylation, are arbitrary. The conditions given below are those that gave the best yields experimentally.) (a) CH3CH2O2C(CH2)5CO2CH2CH3: NaOCH2CH3; CH3(CH2)3I; NaOH; and H1, H2O, D (b) CH3CH2O2CCH2CO2CH2CH3: NaOCH2CH3; CH3I; KOH; and H1, H2O, D O O B B (c) CH3CCHCO2CH3: NaH, C6H6; C6H5CCl; and H, H2O, A CH3 O B (d) CH3CCH2CO2CH2CH3: NaOCH2CH3; BrCH2CO2CH2CH3; NaOH; and H, H2O,
(e) CH3CH2CH(CO2CH2CH3)2: NaOCH2CH3; BrCH2CO2CH2CH3; and H1, H2O, D O O B B (f) CH3CCH2CO2CH2CH3: NaOCH2CH3; BrCH2CCH3; and H, H2O,
Exercise 23-15
Working with the Concepts: Syntheses Using B-Dicarbonyl Compounds Propose a synthesis of cyclohexanecarboxylic acid from diethyl propanedioate (malonate), CH2(CO2CH2CH3)2, and 1-bromo-5-chloropentane, Br(CH2)5Cl. Strategy Begin by comparing the structures of the two starting materials with that of the target molecule. Identify the strategic bonds in the product — the ones that may be made in a b-dicarbonyl-based carboxylic acid synthesis. Once you have done so, find the carbon atoms in the starting materials that give rise to these bonds. Solution • The strategic bonds are those at the a-carbon of the target carboxylic acid (arrows). Line up the product structure with those of the starting molecules in a way that clarifies how they are to be linked. You should come up with something like this, which constitutes your retrosynthetic analysis: Br ␣
␣
H2C
CO2H
Cyclohexanecarboxylic acid
CO2CH2CH3 CO2CH2CH3
Cl 1-Bromo-5chloropentane
Diethyl propanedioate (Diethyl malonate)
23-3 b-Dicarbonyl Anion Chemistry: Michael Additions
Chapter 23
• What makes this particular synthesis challenging to analyze? To construct a cyclic product, we must connect the malonate a-carbon to two functionalized atoms of a single substrate molecule, in this case the 1-bromo-5-chloropentane. The five carbon atoms of this compound combine with the a-carbon of the malonate to furnish the desired six-membered ring. • The actual synthetic sequence can be depicted as follows. Deprotonation of the malonate a-carbon followed by SN2 displacement at the C – Br bond (better leaving group; recall Section 6-7) of the substrate gives a 5-chloropentyl malonate derivative. Subsequent treatment with another equivalent of base removes the remaining a-hydrogen, and the resulting enolate ion displaces chloride intramolecularly to form the ring. Hydrolysis and decarboxylation complete the synthesis:
H2C(CO2CH2CH3)2
1. NaOCH2CH3, CH2CH3OH 2. Br(CH2)5Cl
NaOCH2CH3, CH3CH2OH
Cl(CH2)5 O CH(CO2CH2CH3)2
CO2CH2CH3
1. NaOH, H2O 2. H2SO4, H2O,
CO2H
CO2CH2CH3
Exercise 23-16
Try It Yourself How would you modify the synthesis in Exercise 23-15 in order to use it to prepare heptanedioic acid, HOOC(CH2)5COOH? (Hint: The same starting materials may be used.)
In Summary b-Dicarbonyl compounds such as ethyl 3-oxobutanoate (acetoacetate) and diethyl propanedioate (malonate) are versatile synthetic building blocks for elaborating more complex molecules. Their unusual acidity makes it easy to form the corresponding anions, which can be used in nucleophilic displacement reactions with a wide variety of substrates. Their hydrolysis produces 3-ketoacids that are unstable and undergo decarboxylation on heating.
23-3 B-Dicarbonyl Anion Chemistry: Michael Additions Reaction of the stabilized anions derived from b-dicarbonyl compounds and related analogs (Table 23-1) with a,b-unsaturated carbonyl compounds leads to 1,4-additions. This transformation, an example of Michael addition (Section 18-11), is base-catalyzed and works with a,b-unsaturated ketones, aldehydes, nitriles, and carboxylic acid derivatives, all of which are termed Michael acceptors. Michael Addition
(CH3CH2O2C)2CH2
O B CH2 P CHCCH3
O O O
O B (CH3CH2O2C)2CHOCH2CH2CCH3 71% O O
O
+
O
O O Diethyl propanedioate (Diethyl malonate)
Catalytic CH3CH2ONa, CH3CH2OH, 10 to 25°C
O 3-Buten-2-one (Methyl vinyl ketone) (Michael acceptor)
Diethyl 2-(3-oxobutyl)propanedioate
1095
1096
Ester Enolates and the Claisen Condensation
Chapter 23
Why do stabilized anions undergo 1,4- rather than 1,2-addition to Michael acceptors? 1,2-Addition occurs, but is reversible with relatively stable anionic nucleophiles, because it leads to a relatively high-energy alkoxide. Conjugate addition is favored thermodynamically because it produces a resonance-stabilized enolate ion.
Exercise 23-17 Formulate a detailed mechanism for the Michael addition process just depicted. Why is the base required in only catalytic amounts?
Exercise 23-18 Give the products of the following Michael additions [base in square brackets].
(a) CH3CH2CH(CO2CH2CH3)2 O
O B CH2PCHCH
[NaOCH2CH3]
O CH2PCHCq N
(b)
[NaOCH3]
O
(c)
CO2CH2CH3
H3C
CH3CH P CHCO2CH2CH3
[KOCH2CH3]
Exercise 23-19 Explain the following observation. (Hint: Consider proton transfer in the first Michael adduct.) NC O
O
2 CH2 P CHC q N
NaOCH3, CH3OH
CN O
O
81%
␣
The following reaction is a useful application of Michael addition of anions of b-ketoesters to a,b-unsaturated ketones. The addition gives initially the diketone shown in the margin. Deprotonation of the a-methyl group in the side chain gives an enolate that is perfectly positioned to react with the carbonyl carbon of the cyclohexanone ring. This intramolecular aldol condensation forms a second six-membered ring. Recall (Section 18-11) that the synthesis of six-membered rings by Michael addition followed by aldol condensation is called Robinson annulation.
O
H3C O CO2CH2CH3
O O CO2CH2CH3
O B CH2 P CHCCH3
Na OCH2CH3, CH3CH2OH
CO2CH2CH3
23-3 b-Dicarbonyl Anion Chemistry: Michael Additions
Chapter 23
1097
Exercise 23-20
Working with the Concepts: Mechanism of Robinson Annulation Formulate a detailed mechanism for the preceding transformation. Strategy You need to show every step of the process in sequence, starting with the Michael addition and following with the intramolecular aldol condensation. Review the details of these reactions in Section 18-11. Solution • Following the cues in the text, we begin with a step-by-step description of the Michael addition: deprotonation of the a-carbon between the ketone and ester carbonyl groups in the cyclic b-ketoester, addition to the b-carbon of the a,b-unsaturated ketone, and protonation of the resulting enolate by a molecule of alcohol. CH3 H3C O
O
H
CO2CH2CH3
O
O
O
ðš OCH2CH3
H 3C O
O
HOOCH2CH3
CO2CH2CH3 CO2CH2CH3
CO2CH2CH3
• The intramolecular aldol reaction follows: Ethoxide deprotonates the a-methyl group, giving the corresponding enolate, which adds to the ring carbonyl carbon. The resulting alkoxide is protonated by alcohol. Ethoxide then initiates loss of a molecule of water by removal of an a-hydrogen and expulsion of hydroxide from the b-carbon, completing the aldol condensation and giving the final product. ðš OCH2CH3
O
O
HOOCH2CH3
CH2 O
H i H2C ðOð
CO2CH2CH3
O
ðš O
CO2CH2CH3
ðš OCH2CH3 O
CO2CH2CH3
O HO
O
H HO
CO2CH2CH3
CO2CH2CH3
CO2CH2CH3
Exercise 23-21
Try It Yourself O
Illustrate a Robinson annulation that ensues upon treatment of a mixture of
O
and
3-buten-2-one with sodium ethoxide in ethanol. Draw the intermediate of the initial Michael addition.
Chapter 23
Ester Enolates and the Claisen Condensation
In Summary b-Dicarbonyl anions, like ordinary enolate anions, undergo Michael additions to a,b-unsaturated carbonyl compounds. Addition of a b-ketoester to an enone gives a diketone, which can generate six-membered rings by intramolecular aldol condensation (Robinson annulation).
23-4 Acyl Anion Equivalents: Preparation of A-Hydroxyketones
O B C E R Acyl anion
Throughout our study of carbonyl compounds, we have been used to thinking of the carbonyl carbon atom as being electrophilic and its a-carbon, in the form of an enol or an enolate, as being nucleophilic. These tendencies govern the rich chemistry of the functional groups built around the C P O unit. However, although the array of transformations available to carbonyl compounds is large, it is not without limits. For example, we have no way of making a direct connection between two carbonyl carbon atoms, because both are electrophilic: Neither can serve as a nucleophilic electron source to attack the other. One may imagine a hypothetical carbonyl-derived nucleophilic species, such as an acyl anion (margin), which might add to an aldehyde or ketone to give an a-hydroxyketone, as follows. A Plausible (but Unfeasible) Synthesis of -Hydroxyketones šð ðOððO B A RC O CHR
ðOð B HCR
ðOð B RCð
š ðOððOH B A RC O CHR
HOH HO
Acyl anion (cannot be made)
The ability to generate such an anion would expand the versatility of carbonyl chemistry immensely, making a wide variety of 1,2-difunctionalized systems available, in analogy to the 1,3-difunctional products that we can prepare by using aldol and Claisen condensations. Unfortunately, acyl anions are high-energy species and cannot be generated readily for synthetic applications. Consequently, chemists have explored the construction of other chemical species containing negatively charged carbon atoms that can undergo addition reactions and later be transformed into carbonyl groups. These special nucleophiles, called masked acyl anions or acyl anion equivalents, are the subject of this section. Exercise 23-22 Why are acyl anions not formed by reaction of a base with an aldehyde? (Hint: See Sections 17-5 and 18-1.)
Cyclic thioacetals are masked acyl anion precursors Cyclic thioacetals are formed by the reaction of dithiols with aldehydes and ketones (Section 17-8). The hydrogens on the carbon positioned between the two sulfur atoms in thioacetals are acidic enough (pKa < 31) to be removed by suitably strong bases, such as alkyllithiums. The negative charge in the conjugate base is stabilized inductively by the highly polarizable sulfur atoms. Deprotonation of 1,3-Dithiacyclohexane, a Cyclic Thioacetal, by Butylithium
CH3CH2CH2CH2Li, THF
S
S
S
S pKa 31.1
H
1,3-Dithiacyclohexane
H
CH3CH2CH2CH2H
S
)
1098
H
S Li
S
S
Li
23-4 Acyl Anion Equivalents
1099
Chapter 23
The anion of 1,3-dithiacyclohexane, as well as the anions of its substituted derivatives, are nucleophilic and add to aldehydes and ketones, furnishing alcohols with an adjacent thioacetal group. The example below shows a sequence that begins with the formation of a substituted 1,3-dithiacyclohexane (also known as a 1,3-dithiane) from an aldehyde. Deprotonation gives the masked acyl anion, which adds to the carbonyl group of 2-cyclohexenone to give an alcohol. Finally, hydrolysis of the thioacetal function in this substance returns the original carbonyl group, now as part of an a-hydroxyketone product. O 1.
CH3CHO
SH, HS CHCl3, HCl H2O (Section 17-8)
Electrophilic
S
S
CH3CH2CH2CH2Li, THF
H 3C H 91% 2-Methyl1,3-dithiacyclohexane
S
š
2. H, H2O
S
Li CH3
Nucleophilic
Masked acyl anion
S
S
OH
H3C
H2O, HgCl2, CaCO3, CH3CN
O OH
H3 C
New bond
70%
93% 1-Acetyl2-cyclohexen-1-ol
In this synthesis, the electrophilic carbonyl carbon of the starting aldehyde is transformed into a nucleophilic atom, the negatively charged C2 of a 1,3-dithiacyclohexane anion. After this anion is added to the ketone, hydrolysis of the thioacetal function regenerates the original electrophilic carbonyl group. The sequence therefore employs the reversal of the polarization of this carbon atom to form the carbon – carbon bond. Reagents exhibiting reverse polarization greatly increase the strategies available to chemists in planning syntheses. We have in fact seen this strategy before: Conversion of a haloalkane into an organometallic (e.g., Grignard) reagent (Section 8-7) reverses the polarity of the functionalized carbon from electrophilic (d1C – Xd2) to nucleophilic (d2C – Md1). While this section describes the application of dithiacyclohexane anions as masked acyl anions to the preparation of a-hydroxyketones only, it is readily apparent that their alkylation with other electrophilic reagents allows for a general ketone synthesis (see Chapter Integration Problem 23-26). Exercise 23-23 N
Formulate a synthesis of 2-hydroxy-2,4-dimethyl-3-pentanone, beginning with simple aldehydes and ketones and using a 1,3-dithiacyclohexane anion.
š S
Masked acyl anions can be generated catalytically in the reaction of aldehydes with thiazolium salts. These salts are derived from thiazoles by alkylation at nitrogen. Thiazole (margin) is a heteroaromatic compound (Section 25-4) containing sulfur and nitrogen. Thiazolium salts have an unusual feature — a relatively acidic proton located between the two heteroatoms (at C2).
Thiazole
R
O
Thiazolium salts catalyze aldehyde coupling
N
X
š S Thiazolium salt
1100
Ester Enolates and the Claisen Condensation
Chapter 23
CHE M I C AL H I G H L I G H T 2 3 - 2 Thiamine: A Natural, Metabolically Active Thiazolium Salt NH2 CH2O
H 3C
Thiamine (A H)
N
CH2CH2OA S
N
adenosine triphosphate (ATP), the main energy source in cells. (For the structure of adenosine, see Section 26-9.) Acetyl CoA is the only molecule capable of entering the TCA cycle; therefore, for any food-derived compound to serve as an energy source, it must first be converted into acetyl CoA. The metabolism of glucose leads to pyruvic (2oxopropanoic) acid. The role of thiamine is to catalyze the conversion of pyruvic acid into acetyl CoA by first facilitating loss of a molecule of CO2 and then activating the remaining acetyl fragment as an acyl anion equivalent. The first transformation begins with addition of the thiamine conjugate base to the a-keto carbon of pyruvate. The adduct thus formed decarboxylates readily, generating CO2 and a resonance-stabilized zwitterion. This product is an acetyl anion equivalent, which participates in nucleophilic attack on a sulfur atom of a substance called lipoamide. The result is a tetrahedral intermediate from which the thiamine catalyst, having done its job, is eliminated. Finally, in a transesterificationlike process (Section 20-4), the thiol group of coenzyme A replaces the dihydrolipoamide (DHLPA) function at the acetyl carbonyl carbon, furnishing acetyl CoA. Under nonoxidative (anaerobic, or “oxygen debt”) conditions, such as those found in muscle tissue undergoing extreme exertion, an alternative process drives the formation of energy-rich ATP: Pyruvic acid is reduced enzymatically
The catalytic activity of thiazolium salts in aldehyde dimerization has an analogy in nature: the action of thiamine, or vitamin B1. Thiamine, in the form of its pyrophosphate, is a coenzyme* for several biochemical transformations, including the breakdown of the sugar glucose (Chapter 24). In living systems, the tricarboxylic acid (TCA) cycle, also known as the citric acid or Krebs† cycle, generates two-thirds of the energy derived from food in higher organisms. The TCA cycle combines acetyl CoA
OPCOCO2H A H2COCO2H
H2COCO2H A HOOCOCO2H A H2COCO2H
CH3COSCoA, Citric acid synthetase
Citric acid
2-Oxobutanedioic (Oxaloacetic) acid
O O B B O O O O P POOH Thiamine pyrophosphate (TPP) A A A OH OH
(Section 19-13) with 2-oxobutanedioic (oxaloacetic) acid to form citric acid. In each turn of the cycle, two carbons of citric acid are oxidized to CO2, regenerating oxaloacetic acid and giving rise in a coupled process to a molecule of
*A molecule required by an enzyme for its biological function. † Sir Hans Adolf Krebs, 1900 – 1981, Oxford University, Nobel Prize 1953 (physiology or medicine.)
Pyruvate Adduct Formation R H3C E E C C i i R ONP ES C B
B
R H3 C E E C C i i R ONP ES C
ðO š O C O C O CH3
B O
B O
H
ðO š O C O C O CH
Deprotonated thiamine pyrophosphate (TPP)
Pyruvate
O O
N
CH3
B O
OH
Addition compound
3
23-4 Acyl Anion Equivalents
Chapter 23
Decarboxylation R H3C E E C C i i R ONP ES C
B
C EE CH3 HO
OH
O
O O
B
B
ðO š O C O C O CH3
B O
R H3C E E C C i i R ONE ES C B C EE CH3 HO
R H3C E E C C i i R ONP ES C
CO2
Acetyl anion equivalent
Acetyl Transfer and Formation of Acetyl CoA R H3C E E C C i i R ON ES C
R H3C E E C C i i R ONP ES C
S
S
B
C EE CH3 HO
O O
O
H
B
B
R
SO C O CH3
HS
R
Lipoamide
TPP
OH Tetrahedral intermediate
(R amide linkage to enzyme)
SO C O CH3 P
HS R
DHLPA
P
O
ACoSH
CH3 O C O SCoA
O
S-Acetyl dihydrolipoamide
to (S)-(1)-2-hydroxypropanoic (lactic) acid. Excessive buildup of lactic acid in muscle tissue causes fatigue and cramps. Removal of the lactic acid from the muscle occurs both by slow diffusion into the bloodstream and by enzymecatalyzed conversion back into pyruvic acid, once the condition of oxygen debt has been relieved. This need to alleviate oxygen debt is the reason why you breathe hard during and after physical exercise.
O B CH3CCOOH
Lactic acid dehydrogenase
HO H [Δ COCOOH D H 3C (S)-()-2-Hydroxypropanoic (lactic) acid
The final event in the women’s heptathlon at the 2004 Olympic games in Athens, the 800-m run, took place in 408C (1008F) heat. Here, the gold medalist (left) and another competitor show the combined effects of oxygen debt and lactic acid buildup in their muscles.
1101
Ester Enolates and the Claisen Condensation
Chapter 23
Thiazolium Salts Are Acidic
N
š ð
2
H
1S
N k ð š S
ð
š S
R O
O
O
N3
5
R
R
O
R
4
N k S
ð
HOH
17–19
pKa
Thiazolium salt
Resonance-stabilized thiazolium salt conjugate base
In the presence of thiazolium salts, aldehydes undergo conversion into a-hydroxyketones. An example of this process is the conversion of two molecules of butanal into 5-hydroxy-4-octanone. The catalyst is N-dodecylthiazolium bromide, which contains a longchain alkyl substituent to improve its solubility in organic solvents. REACTION
Aldehyde Coupling (CH2)11CH3
New bond
O
N
O B 2 CH3CH2CH2CH
Br S N-Dodecylthiazolium bromide, NaOH, H2O
O OH B A CH3CH2CH2COCHCH2CH2CH3 76% 5-Hydroxy-4-octanone
Butanal
The mechanism of this reaction begins with reversible addition of C2 in the deprotonated thiazolium salt to the carbonyl function of an aldehyde. Mechanism of Thiazolium Salt Catalysis in Aldehyde Coupling
MECHANISM
Step 1. Deprotonation of thiazolium salt Step 2. Nucleophilic attack by catalyst
š S
ð
H
R
N ðš Oð A O R C š S A H
R
R
O
N
ðOð B CH E
O
R
O
N ðš OH A O R C š S A H
HOH
Step 3. Masked acyl anion formation
š S pKa
R
O
N
R
OH A C O R A H
N
HO
š S
O
R
O
N
f OH C i R
f
OH
C i R
š S
Acyl anion equivalent
17–18
Step 4. Nucleophilic attack on second aldehyde
N
C i R
š S
OH f
H
ðOð B CH E
R
R
N HO ðš Oð A A C O C O R š S A A R H
R
O
R
O
O
1102
HOH
N HO OH A A C O C O R š S A A R H
The Big Picture
Chapter 23
Step 5. Liberation of a-hydroxyketone
O
N HO OH A A C O C O R š S A A R H
R
HO
N ðš Oð OH A A O C O R C š S A A R H
O OH B A RCOCR A H
HOH
The product alcohol of step 2 contains a thiazolium unit as a substituent. This group is electron withdrawing and increases the acidity of the adjacent proton. Deprotonation leads to an unusually stable masked acyl anion. Nucleophilic attack by this anion on another molecule of aldehyde, followed by loss of the thiazolium substituent, liberates the a-hydroxyketone. Comparison of the thiazolium method for synthesis of a-hydroxyketones with the use of dithiacyclohexane anions is instructive. Thiazolium salts have the advantage in that they are needed in only catalytic amounts. However, their use is limited to the synthesis of O OH B A molecules R–C–CH–R in which the two R groups are identical. The dithiacyclohexane method is more versatile and can be used to prepare a much wider variety of substituted a-hydroxyketones.
Exercise 23-24 Which of the following compounds can be prepared by using thiazolium ion catalysts, and which are accessible only from 1,3-dithiacyclohexane anions? Formulate syntheses of at least two of these substances, one by each route. OH
(a)
OH
(b) O
O O
(c)
(d) HO
O OH B A COCH
R
O
OH
(e) O
In Summary a-Hydroxyketones are available from addition of masked acyl anions to aldehydes and ketones. The conversion of aldehydes into the anions of the corresponding 1,3-dithiacyclohexanes (1,3-dithianes) illustrates the method of reverse polarization. The electrophilic carbon changes into a nucleophilic center, thereby allowing addition to an aldehyde or ketone carbonyl group. Thiazolium ions catalyze the dimerization of aldehydes, again through the transformation of the carbonyl carbon into a nucleophilic atom.
THE BIG PICTURE Returning to the chemistry of the C P O functional group, we have seen further evidence of the central role that carbonyl compounds play in synthesis, both in the laboratory and in nature. While there are a variety of reactions introduced in this chapter that appear to be new, they have close analogs in processes we have seen before. The most prominent of these is the Claisen condensation, which is the ester counterpart to the aldol condensation of aldehydes and ketones and similarly gives a product with multiple functional groups. Notice how this topic again brings us back to the concepts of acid-base chemistry, especially with regard to the deprotonation of a-carbons whose hydrogens are rendered acidic by either
N š S
R
O
H2O
1103
1104
Chapter 23
Ester Enolates and the Claisen Condensation
one or (in the case of b-dicarbonyl compounds) two neighboring C P O functions. Decarboxylation is an elimination reminiscent of other reactions with aromatic transition states, including several described in Chapter 15, as well as the rearrangements just discussed in Section 22-7. Finally, the two types of acyl anion equivalents illustrate very different ways to modify a carbonyl carbon such that it can support a negative charge and thus behave as a nucleophile. This is a more sophisticated form of functional group modification, but conceptually it relies on nothing more complicated than an application of the principles discussed at the beginning of Chapter 2: To make a hydrogen acidic, stabilize the negative charge associated with its conjugate base. The innovation in this chemistry is the use of reversible modification of the C P O group to accomplish this task, such that the carbonyl function can be restored after its modified, nucleophilic “alter ego” has done its job. Chapter 24 begins the final phase of our introductory course in organic chemistry, wherein we have a closer look at some of the organic compounds of nature. Fittingly, we start with the carbohydrates, compounds combining the familiar features of alcohols and carbonyl compounds.
CHAPTER INTEGRATION PROBLEMS 23-25. a. Consider the process shown below. Although it is a mixed Claisen condensation, it forms a single product with the molecular formula C11H14O6 in 80% yield. Propose a structure for the product and rationalize the pathway by which it forms. The first component, diethyl ethanedioate (diethyl oxalate), is used in excess. OO O O BB B B CH3CH2OCCOCH2CH3 CH3CH2OCCH2CH2CH2COCH2CH3 Diethyl ethanedioate (Diethyl oxalate)
1. CH3CH2ONa, CH3CH2OH 2. H, H2O
Diethyl pentanedioate
SOLUTION We begin by noticing that diethyl oxalate lacks a-hydrogens and therefore cannot undergo Claisen condensation on its own. Next we see that the formula of the product equals the sum of the formulas of the two starting materials (C15H26O8) minus two molecules of ethanol (C4H12O2) — a change that would be expected for the product of a process involving two Claisen condensations. Because the oxalate is present in excess, it is logical to assume that the first reaction is a Claisen condensation between one of its ester carbonyl groups and an enolate ion derived from the pentanedioate ester: O
O
CH3CH2O CH3CH2O
O OCH2CH3
O
CH3CH2O
OCH2CH3
OCH2CH3 OCH2CH3
O O
O
O
The formula of this intermediate product is C13H20O7 — one molecule of ethanol has been eliminated so far. A second condensation must still take place, but it cannot involve addition to another molecule, because the product we are aiming for does not contain enough atoms in its formula. The only option is a second intramolecular Claisen condensation, between the remaining ester group of the oxalate-derived portion of this intermediate and the other a-carbon of the pentanedioatederived moiety: O
O
O ␣
CH3CH2O
OCH2CH3
O
CH3CH2O
OCH2CH3
OCH2CH3
O
O O
O
Chapter Integration Problems
b. Propose a synthesis of the following compound:
O
Chapter 23
CO2CH2CH3
H 3C H 3C CO2CH2CH3
O
SOLUTION Using the approach illustrated in (a), we can analyze this target retrosynthetically in the context of ring-formation by double Claisen condensation: O
CO2CH2CH3
CO2CH2CH3
H3C H3C O
H3C
CO2CH2CH3
H3C
CO2CH2CH3
CO2CH2CH3
CO2CH2CH3
What makes this mixed Claisen condensation feasible? Although both diesters possess a-hydrogens and can therefore give rise to enolate ions upon treatment with base, the first (dimethyl substituted) ester will not undergo successful condensation with a second identical molecule, because the product will lack the necessary additional a-hydrogen that needs to be removed to drive the equilibrium forward. Thus, this diester may be used in excess and will participate only as the carbonyl partner in reaction with enolate ions derived from the pentanedioate. Reaction of the ester mixture with excess ethoxide in ethanol, followed by treatment with aqueous acid, gives rise to the desired product. 23-26. Propose syntheses of the two ketones in (a) and (b), using the methodology introduced in this chapter. Ground rules for these problems: You may use any organic compounds provided that they contain no more than six carbons each. Any inorganic reagents are allowable. O
a.
(an intermediate in the synthesis of the widely used perfume essence linalool — see Problem 52 of Chapter 13)
SOLUTION Where do we start? We note that the target molecule is a 2-alkanone — a methyl ketone. As a result, we may be able to utilize the acetoacetic ester chemistry introduced in Section 23-2, which is specific in the preparation of methyl ketones. Recall that the acetoacetic ester synthesis affords ketones of the general formulas RCH2COCH3 and RR9CHCOCH3, depending on whether we alkylate the starting ester once or twice. The desired product fits the first of these general formulas, so let us examine a retrosynthetic analysis along these lines. The strategic bond (Section 8-9) is the one between the R group and the C3 a-carbon of the ketone: O
O
Cl
CO2CH2CH3
Strategic bond
With a general strategy in hand, let us look at its details: A b-ketoester alkylation is necessary and we know (Section 18-4) that alkylations of enolates in general follow the SN2 mechanism. Is the proposed substrate suitable for this process? We note that the leaving group, Cl, is located on a carbon atom that is both primary and allylic; hence this substance should be an excellent SN2 substrate (Sections 6-9, 7-9, and 14-3). The final synthesis can therefore be laid out in a manner virtually identical with that of 2-heptanone on page 1091, replacing 1-bromobutane with the allylic chloride: O
CO2CH2CH3
1. NaOH, H2O 2. H2SO4, H2O,
Cl
CO2CH2CH3 O
b.
O
O NaOCH2CH3, CH3CH2OH
(an intermediate in the synthesis of the sex attractant of the bark beetle Ips confusus)
1105
1106
Ester Enolates and the Claisen Condensation
Chapter 23
SOLUTION We note immediately that this target molecule is not a methyl ketone; thus the acetoacetic ester synthesis is not conveniently applicable. We have made simple ketones before by adding organometallic reagents to aldehydes to give secondary alcohols, followed by oxidation (Section 8-9). By this route, the strategic bonds are those to the corresponding alcohol carbon, giving us two potential retrosynthetic disconnections, as shown here. OH
O MgCl
a
a
H
b
b
O H
BrMg
While the preceding provides perfectly good solutions, the problem requires us to employ methodology from the current chapter. Having ruled out acetoacetic ester chemistry, we have an alternative in the use of acyl anion equivalents, for which we saw examples applied to the synthesis of a-hydroxyketones (Section 23-4). Can simple aldehydes and ketones be prepared with these equivalents as well? The answer is yes, by simple alkylation with RX, followed by hydrolysis. Thus,
“ S
š
S is the synthetic equivalent of
O ”
H
š
H “ and S
š
S is the synthetic equivalent of
O ”
R
š
R
Turning to the problem at hand, we analyze retrosynthetically, identifying the two possible strategic bonds in the disubstituted 1,3-dithiacyclohexane corresponding to our target. The resulting monosubstituted dithianes are readily derived from the corresponding aldehydes by thioacetalization (Section 17-8): O
Br
S
a
S
O
S
S
a
b
b
S
SH
Br
O
HS
S
H
H
HS
SH
We have now to decide which strategic bond construction, a or b, is better before designing our final answer. Two considerations are important in this decision: (1) ease of bond formation and (2) size and structural and functional complexity of starting materials. Bond a forms in an SN2 displacement of a branched, primary haloalkane — not the best situation, especially with strongly basic nucleophiles (Section 7-9), and dithiacyclohexanes have pKa values in the 30s. Bond b employs a primary, allylic substrate—a far better alternative. Approach b is also preferable on the basis of our second consideration:
New Reactions
two even-sized (five-carbon-unit) pieces with separated functional groups. Our synthetic scheme for the solution is therefore as shown. 1. CH3CH2CH2CH2Li HS SH, CHCl3, HCl
O
S
2. Br
S H
H
S
H2O, HgCl2, CaCO3, CH3CN
S
O
New Reactions Synthesis of B-Dicarbonyl Compounds 1. Claisen Condensation (Section 23-1) O B 2 CH3COR
1. NaOR, ROH 2. H, H2O
O O B B CH3CCH2COR ROH
Endothermic reaction; equilibrium driven to anion of product by excess base.
Acidic
2. Dieckmann Condensation (Section 23-1) CO2R (CH2)n
1. NaOR, ROH 2. H, H2O
O l C i CHCO2R
ROH
O O B B RCCH2CCH3
ROH
(CH2)n
CH2CO2R
3. B-Diketone Synthesis (Section 23-1) O B RCOR
O B CH3CCH3
1. NaOR, ROH 2. H, H2O
Intramolecular O B CCH3 (CH2)n
C
1. Na OR, ROH 2. H, H2O
(CH2)n
CO2R
O J
i CH2 f C M O
ROH
3-Ketoesters as Synthetic Building Blocks 4. Enolate Alkylation (Section 23-2) O B RCCH2CO2R
1. NaOR, ROH 2. RX
O B RCCHCO2R A R
5. 3-Ketoacid Decarboxylation (Section 23-2) O O B B RCCH2COR
HO RO
O O B B RCCH2CO
H
O O B B RCCH2COH
CO2
O B RCCH3
Chapter 23
1107
1108
Chapter 23
Ester Enolates and the Claisen Condensation
6. Acetoacetic Ester Synthesis of Methyl Ketones (Section 23-2) 1. NaOR, ROH 2. RX 3. HO 4. H,
O O B B CH3CCH2COR O
O B CH3CCH2R
O
R alkyl, acyl, CH2COR, CH2CR RX oxacyclopropane
7. Malonic Ester Synthesis of Carboxylic Acids (Section 23-2) 1. NaOR, ROH 2. RX 3. HO 4. H,
O O B B ROCCH2COR
O
O B RCH2COH
O
R alkyl, acyl, CH2COR, CH2CR RX oxacyclopropane
8. Michael Addition (Section 23-3) O B CH2 P CHCR
H2C
CO2CH3 f
NaOCH3, CH3OH
i CO2CH3
O CO2CH3 B f RCCH2CH2CH i CO2CH3
Michael acceptor
Acyl Anion Equivalents 9. 1,3-Dithiacyclohexane (1,3-Dithiane) Anions as Acyl Anion Equivalents (Section 23-4) A A R O Cð A A
O B Cð RO
synthetic equivalent of
A electron-withdrawing, conjugating, or polarizable group
1. CH3CH2CH2CH2Li, THF O B 2. RCR
S
S
S
HgCl2, H2O
HO H C H f i R R
S
S
S
1. CH3CH2CH2CH2Li, THF 2. RX 3. HgCl2, H2O
A R
O B C f i R R
10. Thiazolium Salts in Aldehyde Coupling (Section 23-4)
N
O B 2 RCH
(CH2)11CH3 D
Br S N-Dodecylthiazolium bromide
HO O A B RC O CR A H
HO O A B RC O CH A R
Problems
Chapter 23
Important Concepts 1. The Claisen condensation is driven by the stoichiometric generation of a stable B-dicarbonyl anion in the presence of excess base. 2. B-Dicarbonyl compounds contain acidic hydrogens at the carbon between the two carbonyl groups because of the inductive electron-withdrawing effect of the two neighboring carbonyl functions and because the anions resulting from deprotonation are resonance stabilized. 3. Although mixed Claisen condensations between esters are usually not selective, they can be so with certain substrates (nonenolizable esters, intramolecular versions, ketones). 4. 3-Ketoacids are unstable; they decarboxylate in a concerted process through an aromatic transition state. This property, in conjunction with the nucleophilic reactivity of 3-ketoester anions, allows the synthesis of substituted ketones and acids. 5. Because acyl anions are not directly available by deprotonation of aldehydes, they have to be made as masked reactive intermediates or stoichiometric reagents by transformations of functional groups.
Problems 27. Arrange the following compounds in order of increasing acidity. Estimate pKa values for each. O O
O
(b) CH3CO2H
(a)
O
(e) CH3CHO
(c) CH3OH
O B COCH3
(d)
O
(f)
(g) CH3O2CCH2CO2CH3
(h) CH3O2CCO2CH3
28. Give the expected results of the reaction of each of the following molecules (or combinations of molecules) with excess NaOCH2CH3 in CH3CH2OH, followed by aqueous acidic work-up.
(a) CH3CH2CH2COOCH2CH3
CH3 A (b) C6H5CHCH2COOCH2CH3
CH3 A (c) C6H5CH2CHCOOCH2CH3
O O B B (d) CH3CH2OC(CH2)4COCH2CH3
O O B B (e) CH3CH2OCCH(CH2)4COCH2CH3 A CH3
(f) C6H5CH2CO2CH2CH3 1 HCO2CH2CH3
[
CO2CH2CH3
(g) C6H5CO2CH2CH3 1 CH3CH2CH2CO2CH2CH3
(h)
[
CO2CH2CH3
CH2CO2CH2CH3
(i) CH2CO2CH2CH3
O B CH3CH2OC
O B COCH2CH3
O O B B CH3CH2OCCH2CH2COCH2CH3
1109
1110
Ester Enolates and the Claisen Condensation
Chapter 23
29. The following mixed Claisen condensation works best when one of the starting materials is present in large excess. Which of the two starting materials should be present in excess? Why? What side reaction will compete if the reagents are present in comparable amounts?
O O B B CH3CH2COCH3 (CH3)2CHCOCH3
NaOCH3, CH3OH
O O B B (CH3)2CHCCHCOCH3 A CH3
30. Suggest a synthesis of each of the following b-dicarbonyl compounds by Claisen or Dieckmann condensations.
(a)
(c) H3C
O O B B CH2CCHCOCH2CH3
O O B B (b) C6H5CCHCOCH2CH3 A C6H5
O
OO O BB B (d) HC CCH2COCH2CH3
CO2CH2CH3
O O B B (f) CH3CH2OCCH2COCH2CH3
O O B B (e) C6H5CCH2CC6H5
O O B B CCH2CCH3
(g)
31. Do you think that propanedial, shown in the margin, can be easily prepared by a simple Claisen condensation? Why or why not?
O O B B HCCH2CH Propanedial
32. Devise a preparation of each of the following ketones by using the acetoacetic ester synthesis.
O B
O O
(b)
(a)
(c)
(d) B O
OCH2CH3 O
33. Devise a synthesis for each of the following four compounds by using the malonic ester synthesis.
(a)
OH
(b) O
OH
O
(c)
H2CCOOH A H2CCOOH
(d)
COOH
34. Use the methods described in Section 23-3, with other reactions if necessary, to synthesize each of the following compounds. In each case, your starting materials should include one aldehyde or ketone and one b-dicarbonyl compound.
Problems
O
O
(a)
O
(b) O
O
(c)
CH(CO2CH2CH3)2
(Hint: A decarboxylation is necessary.)
O
Carbonic acid, H2CO3,
35.
Chapter 23
冢
冣
O B C
, is commonly assumed to be an unstable
HO OH compound that easily decomposes into a molecule of water and a molecule of carbon dioxide: H2CO3 y H2O 1 CO2x. Indeed, the evidence of our own experience in opening a container of any carbonated beverage supports this apparently obvious notion. However, in 2000 it was discovered that this assumption is not quite correct: Carbonic acid is actually a perfectly stable, isolatable compound in the complete absence of water. Its decomposition, which is a decarboxylation reaction, is strongly catalyzed by water. It is exceedingly difficult to completely exclude water without the use of special techniques, which explains why carbonic acid has been such an elusive species to obtain in pure form. Based on the discussion of the mechanism of decarboxylations of 3-ketocarboxylic acids in Section 23-2, suggest a role for a molecule of water to play in catalyzing the decarboxylation of carbonic acid. (Hint: Try to arrange one water and one carbonic acid molecule to give a sixmembered ring stabilized by hydrogen bonds, and then see if a cyclic aromatic transition state for decarboxylation exists.)
36. Based on your answer to Problem 35, predict whether or not water should catalyze the decarboxylation of each of the following compounds. For each case in which your answer is yes, draw the transition state and the final products. O B C
(a) RO
(c) H2N
(b)
(a monoester of carbonic acid) OH
O B C
O B C RO
(carbamic acid)
O B C
(d) H2N
OH
(a diester of carbonic acid) OR
(a carbamate ester) OR
37. Write out, in full detail, the mechanism of the Michael addition of malonic ester to 3-buten-2-one in the presence of ethoxide ion. Be sure to indicate all steps that are reversible. Does the overall reaction appear to be exo- or endothermic? Explain why only a catalytic amount of base is necessary. 38. Give the likely products of each of the following reactions. All are carried out in the presence of a Pd catalyst, a ligand for the metal such as a phosphine, and heat. (Hint: Refer to Section 22-4 for related Pd-catalyzed reactions of halobenzenes with a nucleophile.)
O
(a)
Br
ðCH(CO CH CH ) 2 2 3 2
(b) O2N
Cl
39. Based on the mechanism presented for the Pd-catalyzed reaction of a halobenzene with hydroxide ion (Section 22-4), write out a reasonable mechanism for the Pd-catalyzed reaction of Problem 38(a).
O
1111
1112
Chapter 23
Ester Enolates and the Claisen Condensation
40. Using the methods described in this chapter, design a multistep synthesis of each of the following molecules, making use of the indicated building blocks as the sources of all the carbon atoms in your final product.
O O B , from CH3CO2CH2CH3 and CH3CCHPCH2
(a) CH3
O O
O B , from CH3I, CH2(CO2CH2CH3)2 and CH3CCHPCH2HH(Hint: First make
(b)
.) O
O O
(c)
O B O , from CH3I, CH2(CO2CH2CH3)2 and BrCH2CCH3HH(Hint: First make
O .) O
41. Give the products of reaction of the following aldehydes with catalytic N-dodecylthiazolium bromide. (a) (CH3)2CHCHO; (b) C6H5CHO; (c) cyclohexanecarbaldehyde; (d) C6H5CH2CHO. 42. Give the products of the following reactions. BF3
(a) C6H5CHO 1 HS(CH2)3SH uy
THF
(b) Product of (a) 1 CH3CH2CH2CH2Li uy
What are the results of reaction of the substance formed in (b) with each aldehyde in Problem 41, followed by hydrolysis in the presence of HgCl2? 43. (a) On the basis of the following data, identify unknowns A, found in fresh cream prior to churning, and B, possessor of the characteristic yellow color and buttery odor of butter. A: MS: myz (relative abundance) 5 88(M1, weak), 45(100), and 43(80). 1 H NMR: d 5 1.36 (d, J 5 7 Hz, 3 H), 2.18 (s, 3 H), 3.73 (broad s, 1 H), 4.22 (q, J 5 7 Hz, 1 H) ppm. IR: n˜ 5 1718 and 3430 cm21. B: MS: myz (relative abundance) 5 86(17) and 43(100). 1 H NMR: d 5 2.29 (s) ppm. IR: n˜ 5 1708 cm21. (b) What kind of reaction is the conversion of compound A into compound B? Does it make sense that this should take place in the churning of cream to make butter? Explain. (c) Outline laboratory syntheses of compounds A and B, starting only with compounds containing two carbons. (d) The UV spectrum of compound A has a lmax at 271 nm, whereas that of compound B has a lmax at 290 nm. (Extension of the latter absorption into the violet region of the visible spectrum is responsible for the yellow color of compound B.) Explain the difference in lmax. 44. Write chemical equations to illustrate all primary reaction steps that can occur between a base such as ethoxide ion and a carbonyl compound such as acetaldehyde. Explain why the carbonyl carbon is not deprotonated to any appreciable extent in this system. 45. The nootkatones are bicyclic ketones responsible for the flavor and aroma of grapefruits. Nootkatones also repel termites and are used commercially to protect wood structures from termite infestation. Fill in the missing reagents in the partial synthetic sequence for the preparation of isonootkatone shown on the next page. More than one synthetic step may be needed for each transformation.
Problems
Chapter 23
O Br
%
CO2CH3
CO2H
HO2C
Br
CO2CH3 %
O
% %
% %
O
O Isonootkatone
Nootkatone
46. b-Dicarbonyl compounds condense with aldehydes and ketones that do not undergo self-aldol reaction. The products are a,b-unsaturated dicarbonyl compounds, and the process goes by the colorful name of Knoevenagel condensation. (a) An example of a Knoevenagel condensation is given below. Propose a mechanism.
O
O B CH3CCH2CO2CH2CH3
NaOCH2CH3, CH3CH2OH
O B CCH3 CO2CH2CH3
(b) Give the product of the Knoevanagel condensation shown below.
CHO
CH2(CO2CH2CH3)2
NaOCH2CH3, CH3CH2OH
(c) The diester shown in the margin is the starting material for the dibromide used in the synthesis of isonootkatone (Problem 45). Suggest a preparation of this diester using the Knoevenagel condensation. Propose a sequence to convert it into the dibromide of Problem 45. 47. The following ketones cannot be synthesized by the acetoacetic ester method (why?), but they can be prepared by a modified version of it. The modification includes the preparation (by Claisen O O B B condensation) and use of an appropriate 3-ketoester, RCCH2COCH2CH3, containing an R group that appears in the final product. Synthesize each of the following ketones. For each, show the structure and synthesis of the necessary 3-ketoester as well. O
(a)
(b) O C6H5CH2
O
O
(c)
(d) (Hint: Use a Dieckmann condensation.)
O C6H5CH2 (Hint: Use a double Claisen condensation.)
CO2CH2CH3 CO2CH2CH3
1113
1114
Chapter 23
Ester Enolates and the Claisen Condensation
48. Some of the most important building blocks for synthesis are very simple molecules. Although cyclopentanone and cyclohexanone are readily available commercially, an understanding of how they can be made from simpler molecules is instructive. The following are possible retrosynthetic analyses (Section 8-9) for both of these ketones. Using them as a guide, write out a synthesis of each ketone from the indicated starting materials.
O
O B BrCH2CCH3
O O O B B HCCH2CH2CCH3
O B CH3CCH3
O O B B HCCH2COCH2CH3
Cyclopentanone
O
O B CH3COCH2CH3
O B CH2PCHCCH3
O O O B B HCCH2CH2CH2CCH3
O B CH3CCH3
O O B B HCCH2COCH2CH3
Cyclohexanone
O B CH3COCH2CH3
49. A short construction of the steroid skeleton (part of a total synthesis of the hormone estrone) is shown here. Formulate mechanisms for each of the steps. (Hint: A process similar to that taking place in the second step is presented in Problem 48 of Chapter 18.)
CH3 OK
O
O
KOH, CH3OH,
CH3O CH3 O OK
H 3C C6H6,
OK
CH3O
50.
CH3 O
SO3H,
CH3O
Using methods described in Section 23-4 (i.e., reverse polarization), propose a simple synthesis of each of the following molecules. HO O A B (a) CH2PCHCHCCH2C6H5
HO CH3 A A (c) CH3CHCHCHO
(b) O
51.
Propose a synthesis of ketone C, which was central in attempts to synthesize several antitumor agents. Start with aldehyde A, lactone B, and anything else you need.
CHO
AO
H 2C
O
O
AO
O
A O
H2C
O
A O
O A
B
C
Problems
Team Problem 52. Split your team in two, each group to analyze one of the following reaction sequences by a mechanism (13C 5 carbon-13 isotope).
O O B B (a) 13CH3CCH2COCH2CH3
O O B B (b) CH3CCH2COCH2CH3
1. NaH 2. CH3CH2CH2CH2Li 3. CH3I 4. OH, H2O 5. H , H2O,
1. CH3CH2O, CH3CH2OH 2. (CH3)3CCl 3. OH, H2O 4. H , H2O,
O B CH313CH2CCH3
O B CH3CCH3
CH3 f H2CPC i CH3
Reconvene and discuss your results. Specifically, address the position of the 13C label in the product of (a) and the failure to obtain alkylation in (b). As a complete team, also discuss the mechanism of the following transformation. (Hint: For the first step, a minimum of three equivalents of KNH2 is required.)
O Cl
N A H
O
1. K NH2, liquid NH3 2. H, H2O (work-up)
H O N A H 78%
O
Preprofessional Problems 53. Two of the following four compounds are more acidic than CH3OH (i.e., two of these have Ka greater than methanol). Which ones?
CH3CH2OCH2CH3
O
O B CH3CCH2CHO
A
B
C
CF3CH2OH D
(a) A and B; (b) B and C; (c) C and D; (d) D and A; (e) D and B.
54. The reaction of ethyl butanoate with sodium ethoxide in CH3CH2OH gives
OH A (a) CH3CH2CH2CHCHCO2CH2CH3 A CH2CH3
O B (b) CH3CH2CH2CCHCO2CH2CH3 A CH2CH3
O B (c) CH3CH2CCH2CH2CO2CH2CH3
OH A (d) CH3CH2CH2CHCH2CO2CH2CH3
Chapter 23
1115
Chapter 23
HO2C(CH2)2CH A
CO H D 2 G CO2H
Ester Enolates and the Claisen Condensation
55. When acid A (margin) is heated to 2308C, CO2 and H2O are evolved and a new compound is formed. Which one? CO H D 2 (a) HO2CCH2CHPC G CO2H
(b) HO2CCH2CH2CH2CH3
O
O
O
(c)
(d) CH3CH2CH(CO2H)2
O
(e) O
O
56. A compound with m.p. 5 2228C has a parent peak in its mass spectrum at myz 5 113. The 1H NMR spectrum shows absorptions at d 5 1.2 (t, 3 H), 3.5 (s, 2 H), and 4.2 (q, 2 H). The IR spectrum exhibits significant bands at n˜ 5 3000, 2250, and 1750 cm21. What is its structure? O O O
(a)
P
1116
CH
(b)
(c)
C
O
H
NH2
OCH2CH2CH3 CN
H N
(e) NCCH2CO2CH2CH3
(d) OCH(CH3)2
H N
CH2
C H A P T E R
TWENTY FOUR
Carbohydrates Polyfunctional Compounds in Nature O
O
O O O
T
ake a piece of bread and place it in your mouth. After a few minutes it will begin to taste distinctly sweet, as if you had added sugar to it. Indeed, in a way, this is what happened. The acid and enzymes in your saliva have cleaved the starch in the bread into its component units: glucose molecules. You all know glucose as dextrose or grape sugar. The polymer, starch, and its monomer, glucose, are two examples of carbohydrates. Carbohydrates are most familiar to us as major contributors to our daily diets, in the form of sugars, fibers, and starches, such as bread, rice, and potatoes. In this capacity, they function as chemical energy-storage systems, being metabolized to water, carbon dioxide, and heat or other energy. Members of this class of compounds give structure to plants, flowers, vegetables, and trees. They also serve as building units of fats (Sections 19-13 and 20-5) and nucleic acids (Section 26-9). Carbohydrates are considered to be polyfunctional, because they possess multiple functional groups. Glucose, C6(H2O)6, and many related simple members of this compound class form the building blocks of the complex carbohydrates and have the empirical formulas Cn(H2O)n, essentially hydrated carbon. We shall first consider the structure and naming of the simplest carbohydrates — the sugars. We shall then turn our attention to their chemistry, which is governed by the presence of carbonyl and hydroxy functions along carbon chains of various lengths. Sections 24-1 through 24-3 will address aspects of their properties and chemical behavior. Beginning with Section 24-4, we shall discuss reactions of carbohydrates that are useful preparatively, both for elucidating their structure as well as converting them into other substances. We have already seen an example of the biosynthesis of carbohydrates (Chemical Highlight 18-1). Finally, we shall describe a sampling of more complex types of carbohydrates found in nature.
24-1 Names and Structures of Carbohydrates The simplest carbohydrates are the sugars, or saccharides. As chain length increases, the increasing number of stereocenters gives rise to a multitude of diastereomers. Fortunately for chemists, nature deals mainly with only one of the possible series of enantiomers. Sugars are polyhydroxycarbonyl compounds and many form stable cyclic hemiacetals, which affords additional structural and chemical variety.
O
O O
O
O
O
In everyday life, when we say the word “sugar,” we usually refer to sucrose, the most widely occurring disaccharide in nature. Obtained from sugar beet and sugar cane (illustrated here), sucrose is prepared commercially in pure form in greater quantities than any other chemical substance. In addition, the fibrous “waste” material of the sugar cane plant, composed of more complex carbohydrates, is now widely exploited as a “green” biofuel.
1118
Carbohydrates
Chapter 24
CHO A HO C O OH A CH2OH
Sugars are classified as aldoses and ketoses Carbohydrate is the general name for the monomeric (monosaccharides), dimeric (disaccharides), trimeric (trisaccharides), oligomeric (oligosaccharides), and polymeric (polysaccharides) forms of sugar (saccharum, Latin, sugar). A monosaccharide, or simple sugar, is an aldehyde or ketone containing at least two additional hydroxy groups. Thus, the two simplest members of this class of compounds are 2,3-dihydroxypropanal (glyceraldehyde) and 1,3-dihydroxyacetone (margin). Complex sugars (Section 24-11) are those formed by the linkage of simple sugars through ether bridges. Aldehydic sugars are classified as aldoses; those with a ketone function are called ketoses. On the basis of their chain length, we call sugars trioses (three carbons), tetroses (four carbons), pentoses (five carbons), hexoses (six carbons), and so on. Therefore, 2,3-dihydroxypropanal (glyceraldehyde) is an aldotriose, whereas 1,3-dihydroxyacetone is a ketotriose. Glucose, also known as dextrose, blood sugar, or grape sugar (glykys, Greek, sweet), is a pentahydroxyhexanal and hence belongs to the class of aldohexoses. It occurs naturally in many fruits and plants and in concentrations ranging from 0.08 to 0.1% in human blood. A corresponding isomeric ketohexose is fructose, the sweetest natural sugar (some synthetic sugars are sweeter), which also is present in many fruits ( fructus, Latin, fruit) and in honey. Another important natural sugar is the aldopentose ribose, a building block of the ribonucleic acids (Section 26-9).
O
O
O 2,3-Dihydroxypropanal (Glyceraldehyde) (An aldotriose)
CH2OH A CPO A CH2OH
O
Constitutional isomers
CHO A HO C O OH A HO O C OH A HO C O OH A HO C O OH A CH2OH
O
O 1,3-Dihydroxyacetone
O
(A ketotriose)
CH2OH A C PO A HO O C OH A HO C O OH A HO C O OH A CH2OH
O
O
O
CHO A HO C O OH A HO C O OH A HO C O OH A CH2OH
O
O O
O
O
O
O
O
O O O
O
O
Glucose
Fructose
Ribose
(An aldohexose)
(A ketohexose)
(An aldopentose)
Exercise 24-1 To which classes of sugars do the following monosaccharides belong?
CHO A (a) HCOH A HCOH A CH2OH Erythrose
Diabetics need to carefully monitor the glucose levels in their blood, as with the simple device shown.
CHO A HOCH A (b) HOCH A HCOH A CH2OH Lyxose
CH2OH A C PO A (c) HOCH A HCOH A CH2OH Xylulose
A disaccharide is derived from two monosaccharides by the formation of an ether (usually, acetal) bridge (Sections 17-7 and 24-11). Hydrolysis regenerates the monosaccharides. Ether formation between a mono- and a disaccharide results in a trisaccharide, and repetition of this process eventually produces a natural polymer (polysaccharide). Polysaccharides constitute the framework of cellulose and starch (Section 24-12).
24-1 Names and Structures of Carbohydrates
Chapter 24
Most sugars are chiral and optically active With the exception of 1,3-dihydroxyacetone, all the sugars mentioned so far contain at least one stereocenter. The simplest chiral sugar is 2,3-dihydroxypropanal (glyceraldehyde), with one asymmetric carbon. Its dextrorotatory form is found to be R and the levorotatory enantiomer S, as shown in the Fischer projections of the molecule. Recall that, by convention, the horizontal lines in Fischer projections represent bonds to atoms above the plane of the page (Section 5-4). MODEL BUILDING
Fischer Projections of the Two Enantiomers of 2,3-Dihydroxypropanal (Glyceraldehyde) CHO ´ H OH is the same as H# C ! OH ≥ CH2OH CH2OH
CHO ´ H is the same as HO # C !H HO ≥ CH2OH CH2OH
(R)-()-2,3-Dihydroxypropanal [D-()-Glyceraldehyde]
(S)-()-2,3-Dihydroxypropanal [L-()-Glyceraldehyde]
CHO
CHO
([]D25C 8.7)
25C
([]D
8.7)
Even though R and S nomenclature is perfectly satisfactory for naming sugars, an older system is still in general use. It was developed before the absolute configuration of sugars was established, and it relates all sugars to 2,3-dihydroxypropanal (glyceraldehyde). Instead of R and S, it uses the prefixes d for the (1) enantiomer of glyceraldehyde and l for the (2) enantiomer (Chemical Highlight 5-2). Those monosaccharides whose highest-numbered stereocenter (i.e., the one farthest from the aldehyde or keto group) has the same absolute configuration as that of d-(1)-2,3-dihydroxypropanal [d-(1)-glyceraldehyde] are then labeled d; those with the opposite configuration at that stereocenter are named l. Two diastereomers that differ only at one stereocenter are also called epimers. Designation of a D and an L Sugar CHO H
OH
O
H
H
OH
H
OH
H
OH
H
OH
HO
HO
Highest-numbered stereocenter
CH2OH
CH2OH D-Aldose
H CH2OH
Highest-numbered stereocenter
L-Ketose
The d,l nomenclature divides the sugars into two groups. As the number of stereocenters increases, so does the number of stereoisomers. For example, the aldotetrose 2,3,4-trihydroxybutanal has two stereocenters and hence may exist as four stereoisomers: two diastereomers, each as a pair of enantiomers. The next higher homolog, 2,3,4,5tetrahydroxypentanal, has three stereocenters, and therefore eight stereoisomers are possible: four diastereomeric pairs of enantiomers. Similarly, 16 stereoisomers (as eight enantiomeric pairs) may be formulated for the corresponding pentahydroxyhexanal. Like many natural products, these diastereomers have common names that are often used, mainly because the complexity of these molecules leads to long systematic names. This chapter will therefore deviate from our usual procedure of labeling molecules systematically. The isomer of 2,3,4-trihydroxybutanal with 2R,3R configuration is called erythrose; its diastereomer, threose. Note that each of these isomers has two enantiomers, one belonging to the family of the d sugars, its mirror image to the l sugars. The sign of the optical rotation is not correlated with the d and l label (just as in the R,S notation: (2) Does not necessarily correspond to S, and (1) does not necessarily correspond to R; see Section 5-3). For example, d-glyceraldehyde is dextrorotatory, but d-erythrose is levorotatory.
1119
1120
Carbohydrates
Chapter 24
MODEL BUILDING
Diastereomeric 2,3,4-Trihydroxybutanals: Erythrose (2 Enantiomers) and Threose (2 Enantiomers) Diastereomers Enantiomers
Enantiomers
CHO H
R
H
R
OH OH
CHO
CHO
CHO
HO
S
HO
S
H
R
OH
HO
S
H
R
HO
S
H
CH2OH
O
O
O
D-()-Erythrose
O
O R
S
S
R
O
L-()-Erythrose
Mirror plane
CH2OH 2R,3S
O
O
O S
OH
2S,3R
O
R
H
CH2OH
2S,3S
O R
H
CH2OH
2R,3R
O
H
O
O
O
D-()-Threose
S
O
L-()-Threose
Mirror plane
As mentioned previously, an aldopentose has three stereocenters and hence 23 5 8 stereoisomers. There are 24 5 16 such isomers in the group of aldohexoses. Why then use the d,l nomenclature even though it designates the absolute configuration of only one stereocenter? Probably because almost all naturally occurring sugars have the D configuration. Evidently, somewhere in the structural evolution of the sugar molecules, nature “chose” only one configuration for one end of the chain. The amino acids are another example of such selectivity (Chapter 26). Figure 24-1 shows Fischer projections of the series of d-aldoses up to the aldohexoses. To prevent confusion, chemists have adopted a standard way to draw these projections: The carbon chain extends vertically and the aldehyde terminus is placed at the top. In this convention, the hydroxy group at the highest-numbered stereocenter (at the bottom) points to the right in all d sugars. Figure 24-2 shows the analogous series of ketoses.
Exercise 24-2 Give a systematic name for (a) d-(2)-ribose and (b) d-(1)-glucose. Remember to assign the R and S configuration at each stereocenter.
HO H H OH *C *C *C HOH2C CHO H OH A
Exercise 24-3 Redraw the hashed-wedged line structure of sugar A (shown in the margin) as a Fischer projection and find its common name in Figure 24-1.
In Summary The simplest carbohydrates are sugars, which are polyhydroxy aldehydes (aldoses) and ketones (ketoses). They are classified as d when the highest-numbered stereocenter is R, l when it is S. Sugars related to each other by inversion at one stereocenter are called epimers. Most of the naturally occurring sugars belong to the d family.
24-1 Names and Structures of Carbohydrates
CHO H
OH CH2OH
CHO H
OH
H
OH
CHO
D-()-Glyceraldehyde
HO H
CH2OH
CHO
OH
HO
H
OH
H
OH
HO
H
OH
H
OH
H
H
CH2OH
H
CH2OH
D-()-Ribose
CHO H
CHO
OH
HO
H
H
HO
H
OH
CHO
H
H
OH
H
OH
HO
H
OH
H
OH
H
OH
H
OH
HO
H
OH
H
OH
H
OH
H
OH
H
CH2OH
CHO
HO
H
H
OH
HO
H
HO
H
H
OH
H
CH2OH
CH2OH
H
HO
OH
H
CH2OH
D-()-Mannose
CH2OH
H
CHO H
OH
HO
H
HO
OH CH2OH
D-()-Gulose
OH
D-()-Lyxose
CHO
OH
D-()-Glucose
H
D-()-Xylose
OH HO
D-()-Altrose
OH
CH2OH
D-()-Arabinose
H
CH2OH
H
CHO
H
D-()-Allose
Figure 24-1 D-Aldoses (up to the aldohexoses), their signs of rotation, and their common names.
D-()-Threose
CHO
CHO
1121
CH2OH
D-()-Erythrose
CHO
Chapter 24
D-()-Idose
H
CHO
OH
HO
H
H
HO
H
H
HO
H
OH
H
OH
CH2OH
CH2OH
D-()-Galactose
D-()-Talose
CH2OH O CH2OH 1,3-Dihydroxyacetone
CH2OH O H
OH CH2OH
D-()-Erythrulose
CH2OH
CH2OH O
O
H
OH
HO
H
OH
H
CH2OH
D-()-Xylulose
CH2OH
O
OH CH2OH
D-()-Ribulose
CH2OH
H
CH2OH
O H
CH2OH
O
H
OH
HO
H
H
OH
H
OH
HO
H
OH
H
OH
H
O
OH
HO
H
H
HO
H
OH
H
OH
CH2OH
CH2OH
CH2OH
CH2OH
D-()-Psicose
D-()-Fructose
D-()-Sorbose
D-()-Tagatose
Figure 24-2 D-Ketoses (up to the ketohexoses), their signs of rotation, and their common names.
1122
Chapter 24
Carbohydrates
24-2 Conformations and Cyclic Forms of Sugars Sugars are molecules with multiple functional groups and multiple stereocenters. This structural complexity gives rise to a variety of chemical properties. To enable chemists to focus on the portion of a sugar molecule involved in any given chemical process, several ways of depicting sugars have been developed. We have seen the Fischer representation in Section 24-1; this section shows how to interconvert Fischer and hashedwedged representations. In addition, it introduces the cyclic isomers that exist in solutions of simple sugars.
Fischer projections depict all-eclipsed conformations Recall (Section 5-4) that the Fischer projection represents the molecule in an all-eclipsed arrangement. It can be translated into an all-eclipsed hashed-wedged line picture. MODEL BUILDING
Fischer Projection and Hashed-Wedged Line Structures for D-(ⴙ)-Glucose
6
CH2OH
4
6
H #O 5 CH2OH H H OH
All-eclipsed hashed-wedged line structure
A
A
/ A
∑
∑/ A
5
H OH H OH ∑/ ∑/ HOH2C 5 C 3 C 1 6 C 4 C 2 CHO
A
H#! OH
180⬚ rotation of C3 and C5
A
OH
1
H OH ∑/ C 1 HO C 3 2 CHO A 6 4 HO C 5 CH2OH C H /∑ HO H H
A
H
Fischer projection
!H
#
H#! OH
OH CHO
#
OH CH2OH
HO#3
4
H
2
!
3
HO #! H
H
A
H
H#!OH
A
HO
OH
CHO
2
/∑
H
1
/∑
CHO
HO H H OH
All-staggered hashed-wedged line structure
A molecular model can help you see that the all-eclipsed form actually possesses a roughly circular shape, as illustrated by the “scrolled” rendition above (center picture). In the subsequent all-eclipsed hashed-wedged line structure, notice that the groups on the right of the carbon chain in the original Fischer projection now project upward (wedged bonds). From this conformer, we can reach the all-staggered form by 1808 rotations of C3 and C5.
Sugars form intramolecular hemiacetals Sugars are hydroxycarbonyl compounds that should be capable of intramolecular hemiacetal formation (see Section 17-7). Indeed, glucose and the other hexoses, as well as the pentoses, exist as an equilibrium mixture with their cyclic hemiacetal isomers, in which the hemiacetals strongly predominate. In principle, any one of the five hydroxy groups could add to the carbonyl group of the aldehyde. However, three- and fourmembered rings are too strained, and five- and six-membered rings are the products, the latter usually dominating. The six-membered ring structure of a monosaccharide is called a pyranose, a name derived from pyran, a six-membered cyclic ether (see Sections 9-6 and 25-1). Sugars in the five-membered ring form are called furanoses, from furan (Section 25-3).
24-2 Conformations and Cyclic Forms of Sugars
Chapter 24
1123
Five- and Six-Membered Cyclic Hemiacetal Formation New stereocenter
A
B
A
H2C 4 A H2C
O
2
3
A
A
3
A1 CH2
A 1C
H O
A
C
A
H2C 4 A H2C
O
H⫹
A
OH HA
OH
CH2 2
Furanose ring
Furan
New stereocenter
O
A
3
1
O
O
CH2
A
O
H2C
O
B
C
H⫹
A
O
H2C 4
H2C5 OA H O 1C H2C 4 A A H2C CH2 OH A
O OH H H2C 5
2
3
2
Pyranose ring
Pyran
To depict a d-series sugar correctly in its cyclic form, draw the hashed-wedged line representation of the all-eclipsed structure on the previous page and flip it upside down, as shown below. Rotation of C5 places its hydroxy group in position to form a six-membered cyclic hemiacetal by addition to the C1 aldehyde carbon. Similarly, a five-membered ring can be made by rotation of C4 to place its OH group in position to bond to C1. The resulting drawings depict the ring in a somewhat artificial way: It is shown as if it were planar and from a perspective in which the bottom-most ring bonds (between C2 and C3 in the structures below) are interpreted as being in front of the plane of the page. As we shall see shortly, a slightly modified version of this type of drawing is quite commonly used to depict the cyclic forms of sugars.
MODEL BUILDING
MATION AN I
Cyclic Hemiacetal Formation by Glucose
C
OH HO
/
/
H
C
H
Leads to pyranose
/∑
/∑
/ A
∑
∑/ A
1 CHO 3 2
∑
A
H
Rotate C5 by 120⬚ around C4–C5 bond
5 6
C4
/∑
HO
H HOCH2 ∑ / 6 C OH 5 HO 1 CHO C4 H 3 2 C C H H OH HO
∑
Flip upside down
Leads to furanose OH H ∑ /C CH OH 2
/∑
A
H OH H ∑/ C 1 HO C 3 2 CHO A 6 4 HO C 5 CH2OH C H /∑ HO H
ANIMATED MECHANISM: Cyclic hemiacetal formation by glucose
D-Glucose
6
6
CH2OH A HO O C O H O A C 4 OH A A3 H C A H 5
H
H A 2 C A OH
1C
D-Glucofuranose
(Less stable)
D G OH
New stereocenter
H A C4 A HO
CH2OH A O C5 H A D H 1C OH H G A3 A OH 2 C C A A New H OH stereocenter D-Glucopyranose
(More stable) (Groups on the right in the original Fischer projection [circled] point downward in the cyclic hemiacetal except at C5, which has been rotated)
Chapter 24
Carbohydrates
Exercise 24-4 Draw the Fischer projection of l-(2)-glucose and illustrate its transformation into the corresponding six-membered cyclic hemiacetal.
In contrast with glucose, which exists primarily as the pyranose, fructose forms both fructopyranose and fructofuranose in a rapidly equilibrating 68 : 32 mixture. Cyclic Hemiacetal Formation by Fructose
MODEL BUILDING
/∑
O
2 C O CH2OH 1
4
/
C5
O
∑
H
O
HO
OH H2C6O O
/∑
1124
H
H
HO
3
CO C
OH
D-Fructose
6
O HOCH2 A C5 H HO A A4 A 3 H C C A A OH H 32%
CH2OH D1 2C G OH New stereocenter (anomeric carbon)
H A C5 A HO
CH2 6
O
H HO A4 A 3 C C A A OH H
2C
CH2OH D1 G OH
New stereocenter (anomeric carbon)
68%
D-Fructofuranose
D-Fructopyranose
(Furanose five-membered cyclic sugar)
(Pyranose six-membered cyclic sugar)
Note that, upon cyclization, the carbonyl carbon turns into a new stereocenter. As a consequence, hemiacetal formation leads to two new compounds, two diastereomers differing in the configuration of the hemiacetal group. If that configuration is S in a d-series sugar, that diastereomer is labeled a; when it is R in a d sugar, the isomer is called b. Hence, for example, d-glucose may form a- or b-d-glucopyranose or -furanose. Because this type of diastereomer formation is unique to sugars, such isomers have been given a separate name: anomers. The new stereocenter is called the anomeric carbon.
Exercise 24-5
Working with the Concepts: Relationships Between Sugar Stereoisomers The anomers a- and b-glucopyranose should form in equal amounts because they are enantiomers. True or false? Explain your answer. Strategy Be sure that you understand the meanings of enantiomer and diastereomer, and how physical properties of molecules relate to those stereoisomeric relationships, before you answer this question about anomers. Solution False! Anomers differ in configuration at only the anomeric carbon (C1). The configurations at the remaining stereocenters (C2 through C5) are the same in a-glucopyranose as they are b-glucopyranose. Anomers are therefore diastereomers—not enantiomers—and should not form in equal amounts. Enantiomers differ from each other in configuration at every stereocenter.
24-2 Conformations and Cyclic Forms of Sugars
Chapter 24
1125
Exercise 24-6
Try It Yourself How is a-l-glucopyranose related to a-d-glucopyranose? What is the configuration of the anomeric carbon in a-l-glucopyranose, R or S? Answer the same question for b-l-glucopyranose and b-dglucopyranose.
Fischer, Haworth, and chair cyclohexane projections help depict cyclic sugars How can we represent the stereochemistry of the cyclic forms of sugars? One approach uses Fischer projections. We simply draw elongated lines to indicate the bonds formed upon cyclization, preserving the basic “grid” of the original formula. Adapted Fischer Projections of Glucopyranoses S
O M COH H HO
R
H O COOH
OH
H Cyclization
H
OH
HO
H
OH
H
H
OH
H
CH2OH
HO O COH
H
H O
OH
OH
HO
O
H OH
H H
CH2OH
In the Fischer projection of the a form of a d sugar, the anomeric OH points toward the right. In the Fischer projection of the b form, the anomeric OH is on the left.
CH2OH
-D-()-Glucopyranose
-D-()-Glucopyranose
(m.p. 146C)
(m.p. 150C)
Haworth* projections more accurately represent the real three-dimensional structure of the sugar molecule. The cyclic ether is written in line notation as a pentagon or a hexagon, the anomeric carbon (in a d sugar) placed on the right, and the ether oxygen put on top. The substituents located above or below the ring are attached to vertical lines. In relating the Haworth projection to a three-dimensional structure, we interpret the ring bond at the bottom (between C2 and C3) to be in front of the plane of the paper, and the ring bonds containing the oxygen are understood to be in back. Haworth Projections
H A4 A H H A3 A OH
O
H A 1 H A OH 2A A OH
-D-()-Erythrofuranose
HOCH2 6A H A5 A4 H A OH HO A3 A H
O
H A 1 H A OH 2A A OH
-D-()-Glucopyranose
HOCH2 6A H A5 A4 H A OH HO A3 A H
O 1
H
2A
A OH
OH A A H
-D-()-Glucopyranose
In a Haworth projection, the a anomer has the OH group at the anomeric carbon pointing down, whereas the b anomer has it pointing up.
*Sir W. Norman Haworth (1883–1950), University of Birmingham, England, Nobel Prize 1937 (chemistry).
Groups on the right in the Fischer projection point downward in the Haworth formula.
1126
Carbohydrates
Chapter 24
Exercise 24-7 Draw Haworth projections of (a) a-d-fructofuranose; (b) b-d-glucofuranose; and (c) b-darabinopyranose.
Haworth projections are used extensively in the literature, but here, to make use of our knowledge of conformation (Sections 4-3 and 4-4), the cyclic forms of sugars will be presented mostly as envelope (for furanoses) or chair (for pyranoses) conformations (Figures 4-4 and 4-6). As in Haworth notation, the ether oxygen will be placed usually top right and the anomeric carbon at the right vertex of the envelope or chair. Conformational Pictures of Glucofuranose and -pyranose CH2OH A HOCH
H O H A
H HO
H CH2OH
HO OH
A H
HO
CH2OH
O H A
OH
HO H
A H
HO
O H A
OH H
H
Envelope conformation
OH OH
H
H
H
Chair conformation
Chair conformation
O
O
O
OH
O O
O O
O
O
O O
O
O O
O
O
O
O -D-Glucofuranose
-D-Glucopyranose (One axial –OH group)
-D-Glucopyranose (All substituents equatorial)
Although there are exceptions, most aldohexoses adopt the chair conformation that places the bulky hydroxymethyl group at the C5 terminus in the equatorial position. For glucose, this preference means that, in the a form, four of the five substituents can be equatorial, and one is forced to lie axial; in the b form, all substituents can be equatorial. This situation is unique for glucose; the other seven d aldohexoses (see Figure 24-1) contain one or more axial substituents.
Exercise 24-8 Using the values in Table 4-3, estimate the difference in free energy between the all-equatorial conformer of b-d-glucopyranose and that obtained by ring flip (assume that DG8CH2OH 5 DG8CH3 5 1.7 kcal mol21 and that the ring oxygen mimics a CH2 group).
In Summary Hexoses and pentoses can take the form of five- or six-membered cyclic hemiacetals. These structures rapidly interconvert through the open-chain polyhydroxyaldehyde or ketone, with the equilibrium usually favoring the six-membered (pyranose) ring.
24-3 Anomers of Simple Sugars: Mutarotation of Glucose
Chapter 24
1127
24-3 Anomers of Simple Sugars: Mutarotation of Glucose Glucose precipitates from concentrated solutions at room temperature to give crystals that melt at 1468C. Structural analysis by X-ray diffraction reveals that these crystals contain only the a-d-(1)-glucopyranose anomer (Figure 24-3). When crystalline a-d-(1)-glucopyranose 8C 5 1112 is dissolved in water and its optical rotation measured immediately, a value [a]25 d is obtained. Curiously, this value decreases with time until it reaches a constant 152.7. The process that gives rise to this effect is the interconversion of the a and b anomers. Figure 24-3 Structure of a-D-(1)-glucopyranose, with selected bond lengths and angles.
H 1.102 Å
O H
0.971 Å
1.427 Å
CH2
1.425 Å
O H
110.6
113.8
O H
A HO
A
H
1.525 Å
H
1.426 Å 1.095 Å
109.8
H OH
1.389 Å
O
H
0.969 Å
In solution, the a-pyranose rapidly establishes an equilibrium (in a reaction that is catalyzed by acid and base; see Section 17-7) with a small amount of the open-chain aldehyde, which in turn undergoes reversible ring closure to the b anomer. Interconversion of Open-Chain and Pyranose Forms of D-Glucose CHO
CH2OH HO
O
H or HO
H
HO OH OH 36.4% -D-()-Glucopyranose ([ ]D25C 112)
Axial
H HO
OH H
H
OH
H
OH
CH2OH 0.003% Aldehyde form
H or HO
CH2OH HO
O OH
HO HO
H Equatorial
63.6% -D-()-Glucopyranose ([ ]D
25C
18.7)
The b form has a considerably lower specific rotation (118.7) than its anomer; therefore, the observed a value in solution decreases. Similarly, a solution of the pure b anomer (m.p. 1508C, obtainable by crystallizing glucose from acetic acid) gradually increases its specific rotation from 118.7 to 152.7. At this point, a final equilibrium has been reached, with 36.4% of the a anomer and 63.6% of the b anomer. The change in optical rotation observed when a sugar equilibrates with its anomer is called mutarotation (mutare, Latin, to change). Interconversion of a and b anomers is a general property of sugars. This includes all monosaccharides capable of existing as cyclic hemiacetals. Exercise 24-9 An alternative mechanism for mutarotation bypasses the aldehyde intermediate and proceeds through oxonium ions. Formulate it.
1128
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Carbohydrates
Exercise 24-10
Working with the Concepts: Calculating Specific Rotations of Diastereomeric Sugar Mixtures Calculate the equilibrium ratio of a- and b-glucopyranose (which has been given in the text) from the specific rotations of the pure anomers and the observed specific rotation at mutarotational equilibrium. Strategy The specific rotation of an equilibrium mixture is the average of the rotations of the contributing isomers — however, weighted by the respective mole fractions in the mixture. All the raw data you need are given in the problem statement. Solution • Let’s designate the mole fraction of the a form as xa and the mole fraction of its b isomer as xb. Their respective specific rotations are given as 11128 (a) and 118.78 (b), whereas the equilibrium mixture exhibits a value of 152.78. Thus, we have 152.7 5 (1112)(xa) 1 (118.7)(xb) • By the definition of mole fraction, xa 1 xb 5 1, so we can substitute one for the other in the above equation. Solving gives xa 5 0.364 and xb 5 0.636. Therefore the equilibrium ratio is xa y xb 5 (0.636) y (0.364) 5 1.75.
Exercise 24-11
Try It Yourself The pure a and b forms of d-galactose exhibit [a]d values of 1150.7 and 152.8, respectively. The equilibrium mixture after mutarotation in water shows a specific rotation of 180.2. Calculate the composition of the equilibrium mixture.
Exercise 24-12 By using Table 4-3, estimate the difference in energy between a- and b-glucopyranose at room temperature (258C). Then calculate it by using the equilibrium percentage.
In Summary The hemiacetal carbon (anomeric carbon) can have two configurations: a or b. In solution, the a and b forms of the sugars are in equilibrium with each other. The equilibration can be followed by starting with a pure anomer and observing the changes in specific rotation, a phenomenon also called mutarotation.
24-4 Polyfunctional Chemistry of Sugars: Oxidation to Carboxylic Acids Simple sugars exist as isomers: the open-chain structure and the a and b anomers of various cyclic forms. Because all of these isomers equilibrate rapidly, the relative rates of their individual reactions with various reagents determine the product distribution of a particular transformation. We can therefore divide the reactions of sugars into two groups, those of the linear form and those of the cyclic forms, because the two structures contain different functional groups. Although the two forms may sometimes react competitively, we see in this section that reactions of aldoses with oxidizing agents take place at the aldehyde moiety of the open-chain form, not the hemiacetal function of the cyclic isomers.
Fehling’s and Tollens’s tests detect reducing sugars Because they are polyfunctional compounds, the open-chain monosaccharides undergo the reactions typical of each of their functional groups. For example, aldoses contain the oxidizable
24-4 Polyfunctional Chemistry of Sugars
Chapter 24
formyl group and therefore respond to the standard oxidation tests such as exposure to Fehling’s or Tollens’s solutions (Section 17-14). The a-hydroxy substituent in ketoses is similarly oxidized. Results of Fehling’s and Tollens’s Tests on Aldoses and Ketoses COOH
CHO H
OH H
HO H
OH
H
OH
H Blue Cu2 complex, HO, H2O (Fehling’s solution)
Cu2O
OH
HO
Brick-red precipitate
H
H
OH
H
OH
CH2OH
CH2OH
D-Glucose
D-Gluconic
acid
(An aldonic acid)
OOH BA RC CHR
Ag, NH4OH, H2O (Tollens’s solution)
Ketose
OO BB RC CR
Ag
-Dicarbonyl compound
Silver mirror
In these reactions, the aldoses are transformed into aldonic acids, ketoses into a-dicarbonyl compounds. Sugars that respond positively to these tests are called reducing sugars. All ordinary monosaccharides are reducing sugars.
Oxidation of aldoses can give mono- or dicarboxylic acids Aldonic acids are made on a preparative scale by oxidation of aldoses with bromine in buffered aqueous solution (pH 5 526). For example, d-mannose yields d-mannonic acid in this way. Upon subsequent evaporation of solvent from the aqueous solution of the aldonic acid, the g-lactone (Section 20-4) forms spontaneously. Aldonic Acid Preparation and Subsequent Dehydration to Give an Aldonolactone
HO
H
HO
H
H
OH
H
OH CH2OH
O
M COH
M C
HO
H
HO
H
Br2, H2O
HO
H
HO
H
2 HBr
H
OH
H
OH CH2OH
H2O
H H
OH
HO
O or
CH2OH A C H A
O
CHO
A
Br2 oxidizes only here
O OH A
H HO
H
CH2OH
H 83%
D-Mannose
D-Mannonic
acid
D-Mannono-␥ -lactone
Exercise 24-13 What would the product be if the oxidation of d-mannose took place at the hemiacetal hydroxy group of the cyclic pyranose form, rather than at the carbonyl group of the open-chain isomer, as shown above?
More vigorous oxidation of an aldose leads to attack at the primary hydroxy function as well as at the formyl group. The resulting dicarboxylic acid is called an aldaric, or
M O
1129
1130
Chapter 24
Carbohydrates
saccharic, acid. This oxidation can be achieved with warm dilute aqueous nitric acid (see Section 19-6). For example, d-mannose is converted into d-mannaric acid under these conditions. Preparation of an Aldaric Acid HNO3 oxidizes only here
CHO HO
H
HO
H
COOH
HNO3, H2O, 60C
HO
H
HO
H
H
OH
H
OH
H
OH
H
OH
CH2OH
COOH 44%
D-Mannose
D-Mannaric
acid
Exercise 24-14 The two sugars d-allose and d-glucose (Figure 24-1) differ in configuration only at C3. If you did not know which was which and you had samples of both, a polarimeter, and nitric acid at your disposal, how could you distinguish the two? (Hint: Write the products of oxidation.)
In Summary The chemistry of the sugars is largely that expected for carbonyl compounds containing several hydroxy substituents. Oxidation (by Br2) of the formyl group of aldoses gives aldonic acids; more vigorous oxidation (by HNO3) converts sugars into aldaric acids.
24-5 Oxidative Cleavage of Sugars The methods for oxidation of sugars discussed so far leave the basic skeleton intact. A reagent that leads to C – C bond rupture is periodic acid, HIO4. This compound oxidatively degrades vicinal diols to give carbonyl compounds. REACTION
Oxidative Cleavage of Vicinal Diols with Periodic Acid
/
OH ∑ H
HIO4, H2O
/∑
OH
O B CH CH B O 77%
H cis-1,2-Cyclohexanediol
Hexanedial
The mechanism of this transformation proceeds through a cyclic periodate ester, which decomposes to give two carbonyl groups. Mechanism of Periodic Acid Cleavage of Vicinal Diols
MECHANISM C O OH A C O OH A
HIO4
H2O
OH C O OH D A EI P O C O O JO Cyclic periodate ester
2 CPO
HIO3
24-6 Reduction of Monosaccharides to Alditols
Chapter 24
Because most sugars contain several pairs of vicinal diols, oxidation with HIO4 can give complex mixtures. Sufficient oxidizing agent causes complete degradation of the chain to one-carbon compounds, a technique that has been applied in the structural elucidation of sugars. For example, treatment of glucose with five equivalents of HIO4 results in the formation of five equivalents of formic acid and one of formaldehyde. Similar degradation of the isomeric fructose consumes an equal amount of oxidizing agent, but the products are three equivalents of the acid, two of the aldehyde, and one of carbon dioxide. Periodic Acid Degradation of Sugars 1 CHO
H HO H H
2 3 4 5
1 CH2OH
OH H OH OH
6
CH2OH D-Glucose
5 HIO4
O B 5 HCOH From C1–C5
O B HCH From C6
2
HO H H
3 4 5
O H OH OH
5 HIO4
O B 3 HCOH
From C3–C5
6
CH2OH D-Fructose
It is found that (1) the breaking of each C – C bond in the sugar consumes one molecule of HIO4, (2) each aldehyde and secondary alcohol unit furnishes an equivalent of formic acid, and (3) the primary hydroxy function gives formaldehyde. The carbonyl group in ketoses gives CO2. The number of equivalents of HIO4 consumed reveals the size of the sugar molecule, and the ratios of products are important clues to the number and arrangement of hydroxy and carbonyl functions. Notice in particular that, after degradation, each carbon fragment retains the same number of attached hydrogen atoms as were present in the original sugar. Exercise 24-15 Write the expected products (and their ratios), if any, of the treatment of the following compounds with HIO4. (a) 1,2-Ethanediol (ethylene glycol); (b) 1,2-propanediol; (c) 1,2,3-propanetriol; (d) 1,3-propanediol; (e) 2,4-dihydroxy-3,3-dimethylcyclobutanone; (f) d-threose.
Exercise 24-16 Would degradation with HIO4 permit the following sugars to be distinguished? Explain. (For structures, see Figures 24-1 and 24-2.) (a) d-Arabinose and d-glucose; (b) d-erythrose and d-erythrulose; (c) d-glucose and d-mannose.
In Summary Oxidative cleavage with periodic acid degrades the sugar backbone to formic acid, formaldehyde, and CO2. The ratio of these products depends on the structure of the sugar.
24-6 Reduction of Monosaccharides to Alditols Aldoses and ketoses are reduced by the same types of reducing agents that convert aldehydes and ketones into alcohols. The resulting polyhydroxy compounds are called alditols. For example, d-glucose gives d-glucitol (older name, d-sorbitol) when treated with sodium borohydride. The hydride reducing agent traps the small amount of the open-chain form of the sugar, in this way shifting the equilibrium from the unreactive cyclic hemiacetal to the product.
O B 2 HCH From C1 and C6
CO2 From C2
1131
1132
Carbohydrates
Chapter 24
Preparation of an Alditol CHO CH2OH HO
H O OH H
Red seaweed contains relatively large amounts of D-glucitol.
OH
HO OH
HO
HCHOH
H
H
OH
HO
NaBH4, CH3OH
H
H
OH
H
OH
H
OH
H
OH
CH2OH
CH2OH
D-Glucose
D-Glucitol (D-Sorbitol)
Many alditols are found in nature. d-Glucitol is present in red seaweed in concentrations as high as 14%, as well as in many berries (but not in grapes), in cherries, in plums, in pears, and in apples. It is prepared commercially from d-glucose by high-pressure hydrogenation or by electrochemical reduction. Glucitol is widely used as a sweetener in products such as mints, cough drops, mouthwashes, and chewing gum, where it is often identified by its alternative name, sorbitol. Glucitol is similar in caloric content to glucose. However, the types of bacteria present in the mouth that cause dental caries are less capable of metabolizing glucitol than glucose. Exercise 24-17 (a) Reduction of d-ribose with NaBH4 gives a product without optical activity. Explain. (b) Similar reduction of d-fructose gives two optically active products. Explain.
In Summary Reduction of the carbonyl function in aldoses and ketoses (by NaBH4) furnishes alditols.
24-7 Carbonyl Condensations with Amine Derivatives As might be expected, the carbonyl function in aldoses and ketoses undergoes condensation reactions with amine derivatives (Section 17-9). For example, treatment of d-mannose with phenylhydrazine gives the corresponding hydrazone, d-mannose phenylhydrazone. Surprisingly, the reaction does not stop at this stage but can be induced to continue with additional phenylhydrazine (two extra equivalents). The final product is a double phenylhydrazone, also called an osazone (here, phenylosazone). In addition, one equivalent each of benzenamine (aniline), ammonia, and water is generated. Phenylhydrazone and Phenylosazone Formation H
HO
H H
H
OH
H
OH CH2OH
D-Mannose
C6H5NHNH2 CH3CH2OH, , 30 min H2O
HO HO
H
NHC6H5
H H
H
OH
H
OH CH2OH 75% D-Mannose phenylhydrazone
N
NHC6H5
CPN
NHC6H5
C
A
HO
N C
A
O C
A
H
2 C6H5NHNH2, CH3CH2OH, C6H5NH2, NH3, H2O
HO
H
H
OH
H
OH CH2OH 95% A phenylosazone
The mechanism of osazone synthesis involves tautomerism to a 2-ketoamine, followed by a complex sequence of eliminations and condensations. Once formed, the osazones do
24-8 Ester and Ether Formation: Glycosides
not continue to react with excess phenylhydrazine but are stable under the conditions of the reaction. Historically, the discovery of osazone formation marked a significant advance in the practical aspects of sugar chemistry. Sugars are well known for their reluctance to crystallize from syrups. Their osazones, however, readily form yellow crystals with sharp melting points, thus simplifying the isolation and characterization of many sugars, particularly if they have been formed as mixtures or are impure. Exercise 24-18 Compare the structures of the phenylosazones of d-glucose, d-mannose, and d-fructose. In what way are they related?
In Summary One equivalent of phenylhydrazine converts a sugar into the corresponding phenylhydrazone. Additional hydrazine reagent causes oxidation of the center adjacent to the hydrazone function to furnish the osazone.
24-8 Ester and Ether Formation: Glycosides Because of their multiple hydroxy groups, sugars can be converted into alcohol derivatives. This section explores the formation of simple esters and ethers of monosaccharides and also addresses selective reactions at the anomeric hydroxy group in their cyclic isomers.
Sugars can be esterified and methylated Esters can be prepared from monosaccharides by standard techniques (Sections 19-9, 20-2, and 20-3). Excess reagent will completely convert all hydroxy groups, including the hemiacetal function. For example, acetic anhydride transforms b-d-glucopyranose into the pentaacetate. Complete Esterification of Glucose O CH2OH HO
O OH
HO HO
O O B B 5 CH3COCCH3, pyridine, 0C, 24 h 5 CH3COOH
B
O
CH2OCCH3 O
B
CH3CO
B
OCCH3
CH3CO B
O
H
O
CH3CO B
O
H
91% -D-Glucopyranose
-D-Glucopyranose pentaacetate
Williamson ether synthesis (Section 9-6) allows complete methylation. Complete Methylation of a Pyranose O
HO
OH HO OH
H
-D-Ribopyranose
O B 4 CH3OSOCH3, NaOH B O 4 CH3OSO3Na 4 HOH
CH3O
O OCH3 CH3O H OCH3 70%
-D-Ribopyranose tetramethyl ether
Notice that the hemiacetal function at C1 is converted into an acetal group. The acetal function can be selectively hydrolyzed back to the hemiacetal (see Section 17-7).
Chapter 24
1133
1134
Carbohydrates
Chapter 24
Selective Hydrolysis of a Sugar Acetal O
CH3O
OCH3
O
CH3O
8% HCl, HOH,
CH3O H OCH3
CH3O OCH3 67% D-Ribopyranose
CH3OH
OH
trimethyl ether
(Mixture of and forms)
MATION AN I
ANIMATED MECHANISM: Methyl glycoside formation
It is also possible to convert the hemiacetal unit of a sugar selectively into the acetal. For example, treatment of d-glucose with acidic methanol leads to the formation of the two methyl acetals. Sugar acetals are called glycosides. Thus, glucose forms glucosides. Selective Preparation of a Glycoside (Sugar Acetal)
CH2OH HO
CH3OH, 0.25% HCl, H2O
O
HO HO
OH
HOH
- or -D-Glucopyranose
CH2OH HO
CH2OH
O H
HO HO
HO
O OCH3
HO HO
OCH3
H
Methyl -D-glucopyranoside
Methyl -D-glucopyranoside
(m.p. 166C, [␣]25C 158) D
25C (m.p. 105C, [␣]D 33)
Because glycosides contain a blocked anomeric carbon atom, they do not show mutarotation in the absence of acid, they test negatively to Fehling’s and Tollens’s reagents (they are nonreducing sugars), and they are unreactive toward reagents that attack carbonyl groups. Such protection can be useful in synthesis and in structural analysis (see Exercise 24-19). Exercise 24-19 The same mixture of glucosides is formed in the methylation of d-glucose with acidic methanol, regardless of whether you start with the a or b form. Why?
Exercise 24-20 Draw the structure of methyl a-d-arabinofuranoside.
Exercise 24-21 ∑
∑
H HOCH2 DOG / OHC / C C CHO A A OCH3 H A
Methyl a-d-glucopyranoside consumes two equivalents of HIO4 to give one equivalent each of formic acid and dialdehyde A (shown in the margin). An unknown aldopentose methyl furanose reacted with one equivalent of HIO4 to give dialdehyde A, but no formic acid. Suggest a structure for the unknown. Is there more than one solution to this problem?
Neighboring hydroxy groups in sugars can be linked as cyclic ethers The presence of neighboring pairs of hydroxy groups in the sugars allows for the formation of cyclic ether derivatives. For example, it is possible to synthesize five- or six-membered cyclic sugar acetals from the vicinal (and also from some b-diol) units by treating them with carbonyl compounds (Section 17-8).
24-8 Ester and Ether Formation: Glycosides
1135
Chapter 24
C H EMIC A L H IG H LIG H T 24 - 1 Protecting Groups in Vitamin C Synthesis H3C O
H
O B H⫹, 2 CH3CCH3
HO H OH H 1 ∑/ ∑/ 6 2 CH2OH 5 /4∑ 3 HOCH2 H OH O
5
CH3 C
H
⫺2 H2O
CH2OH 4
1
3
O O C H 3C CH3
OH
O
O
O
O
O O
2
O
H2C 6
Sorbose
O
O
O O
O
O
Virtually all vitamin C sold is synthetic material. At an early stage of its synthesis from the sugar sorbose, the primary hydroxy group at C1 of sorbose must be oxidized selectively to the carboxy function, without affecting the OH groups at carbons 3 through 6. Sorbose exists mostly as the cyclic hemiacetal derived by addition of the hydroxy group at C5 to the C2 carbonyl. Upon acid-catalyzed reaction with excess acetone, the hydroxy groups at C2 and C3 are blocked as a five-membered cyclic acetal and those at C4 and C6 are protected as a six-membered acetal ring. Potassium permanganate then converts the
unprotected CH2OH (C1) into CO2H. Deprotection follows (see Problem 63 for more details.) OH A CO
K
B
O HO H C C A A HO COO A HO O C OH A CH2OH Vitamin C
Cyclic Acetal Formation from Vicinal Diols
OH OH C
C
⫹
O B CH3CCH3
H3C A CH3 A C O O H⫹
C
C
⫹
H2O
Acetone acetal
Such processes work best when the two OH groups are positioned cis to allow a relatively unstrained five- or six-membered ring to form. Cyclic acetal and ether formation is often employed to protect selected alcohol functions. The remaining hydroxy moieties can then be converted into leaving groups, transformed by elimination, or oxidized to carbonyl compounds, as is done in the commercial synthesis of Vitamin C (Chemical Highlight 24-1). HO
Exercise 24-22 Suggest a synthesis of the compound shown in the margin from d-galactose. (Caution: You will need protecting groups. Hint: Consider a strategy that uses cyclic acetals for protection.)
CH2Br
O
HO HO
OH
1136
Chapter 24
Carbohydrates
In Summary The various hydroxy groups of sugars can be esterified or converted into ethers. The hemiacetal unit can be selectively protected as the acetal, also called a glycoside. Finally, the various diol units in the sugar backbone can be linked as cyclic acetals, depending on steric requirements.
24-9 Step-by-Step Buildup and Degradation of Sugars Larger sugars can be made from smaller ones and vice versa, by chain lengthening and chain shortening. These transformations can also be used to structurally correlate various sugars, a procedure applied by Fischer to prove the relative configuration of all the stereocenters in the aldoses shown in Figure 24-1.
Cyanohydrin formation and reduction lengthens the chain To lengthen the chain of a sugar, an aldose is first treated with HCN to give the corresponding cyanohydrin (Section 17-11). Because this transformation forms a new stereocenter, two diastereomers appear. Separation of the diastereomers and partial reduction of the nitrile group by catalytic hydrogenation in aqueous acid then gives the aldehyde groups of the chain-extended sugars. Sugar Chain Extension Through Cyanohydrins Step 1. Cyanohydrin formation New stereocenter
CN HC P O H HO
OH
HCN
H
CN
H
OH
H
OH
HO
H
HO
H OH
H HO
H
Two new diastereomeric nitriles
Step 2. Reduction and hydrolysis (only one diastereomer is shown) C qN H H HO
OH OH H
HC PNH H2, Pd–BaSO4, H, H2O, 4 atm
H
HC P O
OH OH
H HO
H Imine
H2O NH2H
H
OH
H
OH
HO
H
Extended sugar
In this hydrogenation, a modified palladium catalyst (similar to the Lindlar catalyst, Section 13-6) allows selective reduction of the nitrile to the imine, which hydrolyzes under the reaction conditions. The special catalyst is necessary to prevent the hydrogenation from proceeding all the way to the amine (Section 20-8). The preceding chain-lengthening sequence is an improved and shortened version of what is known as the Kiliani-Fischer* synthesis of chain-extended sugars. In the late 1800s, Kiliani demonstrated that addition of cyanide to an aldose gives the cyanohydrin, which, upon hydrolysis (Sections 19-6 and 20-8), is converted into the chain-extended aldonic acid. Fischer subsequently succeeded in converting this aldonic acid into an aldose by lactone formation (Sections 19-9 and 24-4) followed by reduction (Section 20-4; for Fischer’s reduction method, see Problem 62 at the end of this chapter). *Professor Heinrich Kiliani (1855 – 1945), University of Freiburg, Germany; Professor Emil Fischer (see Section 5-4).
24-9 Step-by-Step Buildup and Degradation of Sugars
Chapter 24
Exercise 24-23
Working with the Concepts: Sugar Synthesis What are the products of chain extension of d-erythrose? Strategy As the text above indicates, Kiliani-Fischer chain extension adds a new formyl group to C1 of an aldose, converting the original formyl function into that of a secondary alcohol. Find d-erythrose in Figure 24-2 and identify the sugar(s) that should result from operation of this process. Solution • d-Erythrose is a four-carbon aldose. Addition of a formyl substituent to C1 leaves C2, C3, and C4 unchanged, while changing the original carbonyl group to that of a secondary alcohol. The emerging new stereocenter may form with either the R or S configuration. • Figure 24-2 shows that both d-ribose and d-arabinose have five carbons, and in both of them the “bottom” three carbons (C3, C4, and C5) are identical in structure and stereochemistry to the corresponding three carbon atoms (C2, C3, and C4) in d-erythrose. d-Ribose and d-arabinose differ only at C2, the new stereocenter. They are the products of d-erythrose chain extension.
Exercise 24-24
Try It Yourself What are the chain-extension products of d-arabinose?
Ruff degradation shortens the chain Whereas the preceding approach synthesizes higher sugars, complementary strategies degrade higher sugars to lower sugars one carbon at a time. One of these strategies is the Ruff* degradation. This procedure removes the carbonyl group of an aldose and converts the neighboring carbon into the aldehyde function of the new sugar. The Ruff degradation is an oxidative decarboxylation. The sugar is first oxidized to the aldonic acid by aqueous bromine. Exposure to hydrogen peroxide in the presence of ferric ion then leads to the loss of the carboxy group and oxidation of the new terminus to the aldehyde function of the lower aldose. REACTION
Ruff Degradation of Sugars HC P O H
OH
COOH Br2, H2O
OH
H
Fe3, H2O2, H2O
HC P O
CO2
The oxidative decarboxylation takes place by means of two one-electron oxidations. The first gives a carboxy radical, which is unstable and rapidly loses CO2. The resulting hydroxysubstituted radical is oxidized again to give the aldehyde. Mechanism of Oxidative Decarboxylation
e, H Fe3
Fe2
š
š
ON EOj C A HO C O OH A HO C O OH
š
CO2H A HO C O OH A HO C O OH
MECHANISM
CO2
H Ej EOH C A HO C O OH
*Professor Otto Ruff (1871 – 1939), University of Danzig, Germany.
e, H Fe3
Fe2
CHO A HO C O OH
1137
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Chapter 24
Carbohydrates
C H E M I CA L H IG H LIG H T 24 -2 Sugar Biochemistry In nature, carbohydrates are produced primarily by a reaction sequence called photosynthesis (see also Section 9-11). In this process, sunlight impinging on the chlorophyll of green plants is absorbed. The photochemical energy thus obtained is used to convert carbon dioxide and water into oxygen and the polyfunctional structure of carbohydrates.
The detailed mechanism of this transformation is complicated and takes many steps, the first of which is the absorption of one quantum of light by the extended p system (Section 14-11) of chlorophyll (chloros, Greek, green, and phillon, Greek, leaf).
Photosynthesis of Glucose in Green Plants Sunlight, chlorophyll
6 CO2 1 6 H2O uuuy C6(H2O)6 1 6 O2
Mg N ½N
Glucose
/
Released metabolic energy
Nk
N ý
H
∑
H
/ H
O
O
O CO2CH3
Chlorophyll a
Chemical Highlight 18-1 described an aldol-based biosynthesis of carbohydrates, and Chemical Highlight 23-2 introduced thiamine pyrophosphate (TPP) in its role in the metabolism of glucose to produce biochemical energy. Here we illustrate another biochemical sequence that uses thiamine catalysis to interconvert ketoses. In ketose interconversion, the enzyme transketolase employs a molecule of thiamine to catalyze the transfer of a (hydroxymethyl)carbonyl unit from xylulose to erythrose, producing fructose in the process. Mechanistically, the deprotonated thiazolium ion first attacks the carbonyl group of the donor sugar (xylulose) to form an addition product, in a way completely analogous to addition to aldehydes (Section 23-4). Because the donor
sugar contains a hydroxy group next to the reaction site, this initial product can decompose by the reverse of the addition process to an aldehyde (glyceraldehyde) and a new thiamine intermediate. This new intermediate attacks another aldehyde (erythrose) to produce a new addition product. The catalyst then dissociates as TPP, releasing the new sugar molecule (fructose). The cycle of photosynthesis, carbohydrate interconversion, and carbohydrate metabolism is a beautiful example of how nature reuses its resources. First, CO2 and H2O are consumed to convert solar energy into chemical energy and O2. When an organism needs some of the stored energy, it is generated by conversion of carbohydrate into CO2 and H2O, using up roughly the same amount of O2 originally liberated.
Ruff degradation gives low yields because of the sensitivity of the products to the reaction conditions. Nevertheless, the procedure is useful in structural elucidations (Exercise 24-25). Fischer originally carried out such studies to establish the relative configurations of the monosaccharides (the Fischer proof). The next section will describe some of the logic behind his approach to the problem. Exercise 24-25 Ruff degradation of two d pentoses, A and B, gave two new sugars, C and D. Oxidation of C with HNO3 gave meso-2,3-dihydroxybutanedioic (tartaric) acid, that of D resulted in an optically active acid. Oxidation of either A or B with HNO3 furnished an optically active aldaric acid. What are compounds A, B, C, and D?
24-10 Relative Configurations of the Aldoses
Chapter 24
Transketolase-Catalyzed Biosynthesis of Fructose from Erythrose Using Thiamine Pyrophosphate (Hydroxymethyl)carbonyl group R H3C D D (Thiazolium ion CP C of TPP), S D NP R C H Transketolase
Erythrose
Glyceraldehyde
Sugar Activation
H3C R E E C C i i RO NP ES HO C
B
B
H3C R E E C C i i R ONP ES C
RO C O C O CH2OH H
Deprotonated thiamine pyrophosphate
H
O O
RO C O C O CH2OH B H O
H
O O
O O
Fructose (New sugar)
Removal of Old Aldehyde
HO
B
H3C R E E C C i i S R ONP E C
Xylulose (Donor sugar)
CHO A H O COOH A CH2OH
O B RCH
O
CHO A H O COOH A H O CO OH A CH2OH
C EE CH2OH HO
CH2OH A P CPO A HO O COH A H O COOH A CH2OH
CH2OH A CP O A HO O COH A H O COOH A H O COOH A CH2OH
OH
Addition compound
Donor sugar
Introduction of New Aldehyde
H
P
RO C O C O CH2OH
HO
O O
CH2OH
O O
C EH
HO
H
O O
B
H3C R E E C C i i R O NP ES HO C B
B
H3C R E E C C i i R ON ES C O
O B RCH
H
O
RO C O C O CH2OH
OH
New sugar
In Summary Sugars can be made from other sugars by step-by-step one-carbon chain lengthening (cyanohydrin formation and reduction) or shortening (Ruff degradation).
24-10 Relative Configurations of the Aldoses: An Exercise in Structure Determination Imagine that we have been presented with 14 jars, each filled with one of the aldoses in Figure 24-1, each labeled with a name, but no structural formula. How would we establish the structure of each substance without using spectroscopy? This was the challenge faced by Fischer over a century ago. In an extraordinary display of scientific logic, Fischer solved this problem by interpreting the results of a combination of carefully thought out synthetic manipulations, designed to convert the aldoses he had
TPP
1139
1140
H
Chapter 24
Carbohydrates
CHO
available to him into one another and into related substances. Fischer made one assumption: That the dextrorotatory enantiomer of 2,3-dihydroxypropanal (glyceraldehyde) had the d (and not the l) configuration. This assumption was proved correct only after a special kind of X-ray analysis was developed in 1950, long after Fischer’s time. Fischer’s was a lucky guess; otherwise, all the structures of the d sugars in Figure 24-1 would have had to be changed into their mirror images. Fischer’s goal was to establish the relative configurations of all the stereoisomers — that is, to associate each unique sugar with a unique relative sequence of stereocenters.
R
OH OH
H
CH2OH Structure 1 (D-Erythrose) HNO3
The structures of the aldotetroses and aldopentoses can be determined from the optical activity of the corresponding aldaric acids
COOH H
OH
H
OH
Beginning with the structure of d-glyceraldehyde, we shall now set out to prove unambiguously the structures of the higher d aldoses. We shall use Fischer’s logic, although not his procedures, because most of these aldoses were unavailable in his time. (Consult Figure 24-1 as required while following these arguments.) We begin with chain extension of d-glyceraldehyde, obtaining a mixture of d-erythrose and d-threose. Oxidation of d-erythrose with nitric acid gives optically inactive meso-tartaric acid, which tells us that d-erythrose must have structure 1, with both – OH groups on the same side in the Fischer representation. In contrast, oxidation of d-threose forms an optically active acid. Thus, d-threose must have the opposite stereochemistry at C2 (structure 2). Both must have the same (R) configuration at C3, the carbon derived from the C2 stereocenter of our starting material, d-glyceraldehyde. The difference is at C2: d-Erythrose is 2R; d-threose, 2S. Now that we know the structures of d-erythrose and d-threose, we continue as follows. Chain extension of d-erythrose (Exercise 24-23) gives a mixture of two pentoses: d-ribose and d-arabinose. Their configurations at C3 and C4 must be the same as those of C2 and C3 in d-erythrose; they differ at C2, the new stereocenter created in the lengthening procedure. Nitric acid oxidation of d-ribose gives an optically inactive (meso) acid; thus d-ribose has structure 3. d-Arabinose oxidizes to an optically active acid; it must have structure 4.
COOH meso-Tartaric acid (Not optically active)
CHO HO H
S
H OH
CH2OH Structure 2 (D-Threose) HNO3
COOH HO H
H OH COOH
COOH
()-Tartaric acid (Optically active)
CHO
OH
H
OH
H
H HNO3
OH
H
H H
COOH
OH
HO
OH OH
CHO
D-Erythrose
Chain extension
OH
H
OH
HNO3
H
H
OH
H
OH COOH
CH2OH
Structure 3 (D-Ribose)
(Not optically active)
HO
H
H
CH2OH
Meso
COOH
Structure 4 (D-Arabinose)
Optically active
Proceeding in the same way from d-threose, chain extension gives a mixture of d-xylose and d-lyxose. The oxidation product of d-xylose is meso; therefore d-xylose has structure 5. The optically active oxidation product of d-lyxose confirms its structure as 6. COOH H HO H
CHO
OH H OH COOH Meso
(Not optically active)
H HNO3
HO H
OH
HO HO
H OH CH2OH
Structure 5 (D-Xylose)
CHO
D-Threose
Chain extension
H
COOH
H H OH CH2OH
Structure 6 (D-Lyxose)
HNO3
HO
H
HO
H
H
OH COOH
Optically active
24-10 Relative Configurations of the Aldoses
Chapter 24
Symmetry properties also define the structures of the aldohexoses We now know the structures of the four aldopentoses and can extend the chain of each of them. This process gives us four pairs of aldohexoses, each pair distinguished from the others by the unique sequence of stereocenters at C3, C4, and C5. The members of each pair differ only in their configuration at C2. The structural assignment for the four sugars obtained from d-ribose and d-lyxose, respectively, is again accomplished by oxidation to the corresponding aldaric acids. Both d-allose and d-galactose give optically inactive oxidation products, in contrast with d-altrose and d-talose, which give optically active dicarboxylic acids. CHO
CHO
H
OH
H
H
OH
HO
H
H
OH
HO
H
OH
H
CH2OH
CHO
OH
HO
CHO
H
HO
H
H
OH
HO
H
H
H
OH
HO
H
OH
H
OH
H
CH2OH
D-Allose
D-Galactose
(From D-ribose)
(From D-lyxose)
(Both give meso dicarboxylic acids)
CH2OH
OH CH2OH
D-Altrose
D-Talose
(From D-ribose)
(From D-lyxose)
(Both give optically active dicarboxylic acids)
The structural assignments of the four remaining hexoses (with structures 7 – 10) cannot be based on the approach taken so far, because all give optically active diacids upon oxidation. We join the actual strategy devised by Fischer to solve a problem that had become substantially more complicated. He found that chain extension of d-arabinose gave a mixture of d-glucose and d-mannose, one of which must have structure 7 and the other structure 8. Displaying synthetic creativity decades ahead of his time, he developed a procedure to exchange the functionalities at C1 and C6 of a hexose — converting C1 to a primary alcohol and C6 to an aldehyde. Carrying out this procedure starting with both d-glucose and d-mannose, he found that the result of C1/C6 exchange in glucose was a new sugar, which did not exist in nature and was given the name gulose. On the other hand, exchange of C1 and C6 in mannose returned mannose, unchanged. Examination of structures 7 and 8 reveals why this is so and gives us a glimpse of Fischer’s genius. If we perform a C1/C6 exchange in 8 and then rotate the resulting Fischer projection by 1808, we realize that the initial and final structures are identical: The exchange procedure gives the original sugar back. Thus 8 is identified as the structure of d-mannose, and 7, in which C1/C6 exchange gives a new structure, must correspond to d-glucose. CHO H HO
CH2OH
OH H
H C1/C6 exchange
HO
OH H
CHO HO HO
CH2OH HO
H H
C1/C6 exchange
HO
CHO
H H
Rotate by 180
HO
H
HO
H
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
CH2OH 7 (D-Glucose)
CHO New sugar (Gulose)
CH2OH
CHO
8 (D-Mannose)
Exercise 24-26 Was the gulose that Fischer synthesized from d-glucose a d-sugar or an l-sugar? (Caution: In a very uncharacteristic oversight, Fischer himself got the wrong answer at first, and that confused everybody for years.)
CH2OH Same as 8
1141
1142
Chapter 24
CHO H
OH
H
OH
HO
Carbohydrates
Exercise 24-27 Fischer proved that his structural reasoning was correct by comparing the nitric acid oxidation of d-glucose to that of his synthetic gulose. What do these two reactions produce?
H OH
H
CH2OH 9
CHO HO
H OH
H HO
H OH
H
CH2OH 10
With the synthesis of gulose, one of the two remaining hexoses (the products of chain lengthening of d-xylose) was identified as structure 9. By process of elimination, the other, d-idose, could be assigned structure 10, thus completing the exercise. Exercise 24-28 In the preceding discussion, we assigned the structures of d-ribose and d-arabinose by virtue of the fact that, upon oxidation, the first gives a meso diacid, the second an optically active diastereomer. Could you arrive at the same result by 13C NMR spectroscopy?
In Summary Step-by-step one-carbon chain lengthening or shortening, in conjunction with the symmetry properties of the various aldaric acids, allows the stereochemical assignments of the aldoses.
24-11 Complex Sugars in Nature: Disaccharides A substantial fraction of the natural sugars occur in dimeric, trimeric, higher oligomeric (between 2 and 10 sugar units), and polymeric forms. The sugar most familiar to us, so-called table sugar, is a dimer.
Sucrose is a disaccharide derived from glucose and fructose
Sucrose is a pure chemical compound and forms large crystals from saturated aqueous solutions.
Sucrose, ordinary table sugar (see Chapter Opening), is one of the few natural chemicals consumed in unmodified form (water and NaCl are examples of others). Its average yearly consumption in the United States is about 175 pounds per person. Sugar is isolated from sugar cane and sugar beets, in which it is particularly abundant (about 14 – 20% by weight), although it is present in many plants in smaller concentrations. World production is about 150 million tons a year. Sucrose has not been considered in this chapter so far, because it is not a simple monosaccharide; rather, it is a disaccharide composed of two units, glucose and fructose. The structure of sucrose can be deduced from its chemical behavior. Acidic hydrolysis splits it into glucose and fructose. It is a nonreducing sugar. It does not form an osazone. It does not undergo mutarotation. These findings suggest that the component monosaccharide units are linked by an acetal bridge (Section 17-7) connecting the two anomeric carbons; in this way, the two cyclic hemiacetal functions block each other. X-ray structural analysis confirms this hypothesis: Sucrose is a disaccharide in which the a-d-glucopyranose form of glucose is attached to b-d-fructofuranose in this way. CH2OH HO
Both anomeric carbons in sucrose are parts of acetal functional groups.
HO HOCH2 HO
CH2OH
O H HO
H HO
O CH2OH
Acetal
HO or
O
OH O
OH
O
HO
O
Acetal
CH2OH
OH CH2OH H
Sucrose, a -D-fructofuranosyl-␣-D-glucopyranoside
Sucrose has a specific rotation of 166.5. Treatment with aqueous acid decreases the rotation until it reaches a value of 220. The same effect is observed with the enzyme invertase. The phenomenon, known as the inversion of sucrose, is related to mutarotation
24-11 Complex Sugars in Nature: Disaccharides
Chapter 24
of monosaccharides. It includes three separate reactions: hydrolysis of the disaccharide to the component monosaccharides a-d-glucopyranose and b-d-fructofuranose; mutarotation of a-d-glucopyranose to the equilibrium mixture with the b form; and mutarotation of b-d-fructofuranose to the slightly more stable b-d-fructopyranose. Because the value for the specific rotation of fructose (292) is more negative than the value for glucose (152.7) is positive, the resulting mixture, sometimes called invert sugar, has a net negative rotation, inverted from that of the original sucrose solution. Inversion of Sucrose CH2OH
O
HO
H
HO HO
OH O
O CH2OH
OH CH2OH
CH2OH
H, H2O or invertase
HO
O H
HO HO
H
OH
18%
Sucrose
␣-D-Glucopyranose
CH2OH
HOCH2
O
HO
OH
HO HO
HO
H
O HO
HO
OH
OH OH
OH
CH2OH 16%
32% -D-Glucopyranose
O
CH2OH
34%
-D-Fructofuranose
-D-Fructopyranose
The noncaloric, edible fat substitute olestra (Chemical Highlight 21-1) consists of a mixture of sucroses esterified with seven or eight fatty acids obtained from vegetable oils, such as hexadecanoic (palmitic) acid. The fatty acids shield the sucrose core of olestra so effectively that it is completely immune to attack by digestive enzymes and passes through the digestive tract unchanged. Exercise 24-29
Working with the Concepts: Practicing Sugar Chemistry Write the products (if any) of the reaction of sucrose with excess (CH3)2SO4, NaOH. Strategy The chemistry of disaccharides and other more complex carbohydrates is largely the same as that of monosaccharides. Identify the functional groups present and consider processes comparable to those described earlier for simple sugars. Solution • Dimethyl sulfate is a reagent that is commonly used to convert alcohols into methyl ethers via the Williamson ether synthesis (Sections 9-6 and 22-8). (Iodomethane would be equally effective but is more costly.) • As illustrated by the example in Section 24-8, dimethyl sulfate methylates all free – OH groups in a molecule, converting them into – OCH3 groups by SN2 processes. Carrying out this reaction with sucrose gives as the product CH3O CH3O
CH2OCH3 O H OCH
3
CH3O
O
CH3OCH2 For the follow-up Exercise 24-30, see next page.
O
OCH3 CH2OCH3
1143
1144
Carbohydrates
Chapter 24
C H E M I CA L H IG H LIG H T 24 -3 Carbohydrate-Derived Sugar Substitutes
CH2OH OH
H HO
HO
H OH
H
CH2OH Xylitol
H
CH2OH O H OH
H
H
OH
H
OH H
O H
OH H
H
H OH CH2OH
Lactitol
In spite of the commercial success of the synthetic artificial sweeteners saccharin and aspartame (Section 26-4), these substances cannot completely replace sugar in products such as chocolate, because their physical properties are very different from those of sucrose. As a result, confections containing these sweeteners must be supplemented with other materials to give them perceived characteristics acceptable to the consumer. The formulation of low-calorie chocolate is an unusual challenge, because much of the caloric content and appealing “melt-in-your-mouth” character of chocolate comes from its fat content. Replacement of the fat in chocolate with lowcalorie sweeteners has been attempted. Various mixtures of alditols such as xylitol and lactitol, together with low-calorie synthetic glucopyranose polymers, have been used in the past with some success. More recently, maltitol has been employed with superior results. This disaccharide alditol is almost as sweet as sucrose but possesses only two-thirds of its caloric content. In addition, maltitol gives chocolate a rich creamy texture, mimicking the effect of fat. One of the most hotly debated issues associated with the search for low- and noncaloric sweeteners is the case
CH2OH
HO
CH2OH O H OH
H
H O H
H
OH
H
OH
OH
H
H
CH2OH
OH CH2OH
Maltitol
of stevioside, found in the leaves of the Paraguayan herb Stevia rebaudiana. Stevioside is 300 times as sweet as sucrose, and its hybrid terpene – carbohydrate structure renders it relatively resistant to the digestive process and therefore virtually noncaloric. Nonetheless, although stevioside is widely used in China, Japan, and Brazil, it is banned as a food additive in the United States, Canada, and the European Union because of concerns over certain metabolic degradation products that may lead to cancer or sterility. One of the newest low-calorie sweetener candidates is the naturally occurring ketose d-tagatose (Figure 24-2). This substance is virtually indistinguishable in taste from sucrose — note the similarity in structure between tagatose and fructose — but it has far fewer calories, a minimal effect on blood-sugar levels, and does not support the bacteria that cause tooth decay. Moreover, its preparation is relatively “green,” involving simple base treatment of the normally discarded lactose-rich waste from milk production (whey). Marketing of tagatose as a sweetener in jams and chocolates began in Europe in late 2007.
Exercise 24-30
Try It Yourself Write the products (if any) of the reaction of sucrose with (a) 1. H1, H2O, 2. NaBH4; and (b) NH2OH.
Acetals link the components of complex sugars Sucrose contains an acetal linkage between the anomeric carbons of the component sugars. One could imagine other acetal linkages with other hydroxy groups. Indeed, maltose (malt sugar), which is obtained in 80% yield by enzymatic (amylase) degradation of starch (to be discussed in the next section), is a dimer of glucose in which the hemiacetal oxygen of one glucose molecule (in the a anomeric form) is bound to C4 of the second.
24-11 Complex Sugars in Nature: Disaccharides
Chapter 24
OH
HO
HO
OH
O
HO
O
O
O
HO
OH
OH
O
HO
O O
HO OH
Stevioside
Leaves of the Paraguayan herb Stevia rebaudiana.
The most recent successful entrant into the artificial sweetener marketplace is sucralose (marketed in the United States as Splenda®). A trichlorinated derivative of sucrose, sucralose is 600 times as sweet as sucrose. It has the benefit of sufficient heat stability to be usable in cooking and baking, and it is available in a variety of forms appropriate to its diverse potential uses. Sucralose is absorbed to a limited extent by the digestive tract, but there is no evidence of any ill effects in humans. Sucralose has been approved for safe consumption in dozens of countries, including the United States and Canada.
CH2OH HO HO
1145
Cl
CH2OH
O H
HO HO
OH
O
O H H CH OH 2 OH O O 4 OH HO 2 3 OH 1 H
-Maltose, an ␣-D-glucopyranosyl--D-glucopyranose
In this arrangement, one glucose retains its unprotected hemiacetal unit with its distinctive chemistry. For example, maltose is a reducing sugar; it forms osazones, and it undergoes mutarotation. Maltose is hydrolyzed to two molecules of glucose by aqueous acid or by the enzyme maltase. It is about one-third as sweet as sucrose. Exercise 24-31 Draw the structure of the initial product of b-maltose when it is subjected to (a) Br2 oxidation; (b) phenylhydrazine (3 equivalents); (c) conditions that effect mutarotation.
O CH2Cl
Sucralose
OH CH2Cl
H
1146
Chapter 24
Carbohydrates
Another common disaccharide is cellobiose, obtained by the hydrolysis of cellulose (to be discussed in the next section). Its chemical properties are almost identical with those of maltose, and so is its structure, which differs only in the stereochemistry at the acetal linkage — b instead of a. CH2OH HO
H
O
CH2OH
O
HO OH H
O OH
HO OH H
-Cellobiose, a -D-glucopyranosyl--D-glucopyranose
Aqueous acid cleaves cellobiose into two glucose molecules just as efficiently as it hydrolyzes maltose. However, enzymatic hydrolysis requires a different enzyme, b-glucosidase, which specifically attacks only the b-acetal bridge. In contrast, maltase is specific for a-acetal units of the type found in maltose. After sucrose, the most abundant natural disaccharide is lactose (milk sugar). It is found in human and most animal milk (about 5% solution), constituting more than one-third of the solid residue remaining on evaporation of all volatiles. Its structure is made up of galactose and glucose units, connected in the form of a b-d-galactopyranosyl-a-d-glucopyranose. Crystallization from water furnishes only the a anomer. Lactose contains an unprotected hemiacetal unit and is therefore a reducing sugar that undergoes mutarotation. The first step of its degradation in the body is hydrolysis of the glycosidic bond by the enzyme lactase to give galactose and glucose. Lack of this enzyme causes lactose intolerance, a condition that is associated with abdominal cramps and diarrhea after the consumption of milk and other dairy products. OH CH2OH
H
O O
HO OH H
CH2OH
O H
HO OH OH
Crystalline ␣-lactose, a -D-galactopyranosyl-␣-D-glucopyranose
Got lactose?
In Summary Sucrose is a dimer derived from linking a-d-glucopyranose with b-dfructofuranose at the anomeric centers. It shows inversion of its optical rotation on hydrolysis to its mutarotating component sugars. The disaccharide maltose is a glucose dimer in which the components are linked by a carbon – oxygen bond between an a anomeric carbon of a glucose molecule and C4 of the second. Cellobiose is almost identical with maltose but has a b configuration at the acetal carbon. Lactose has a b-d-galactose linked to glucose in the same manner as in cellobiose.
24-12 Polysaccharides and Other Sugars in Nature Polysaccharides are the polymers of monosaccharides. Their possible structural diversity exceeds that of alkene polymers (Sections 12-14 and 12-15), particularly in variations of chain length and branching. Nature, however, has been remarkably conservative in its construction of such macromolecules. The three most abundant natural polysaccharides — cellulose, starch, and glycogen — are derived from the same monomer: glucose.
Cellulose and starch are unbranched polymers Cellulose is a poly-b-glucopyranoside linked at C4, containing about 3000 monomeric units and having a molecular weight of about 500,000. It is largely linear.
24-12 Polysaccharides and Other Sugars in Nature
Chapter 24
1147
C4
H
CH2OH
O
H
CH2OH
O
H O
O HO
HO OH H
CH2OH
O
O HO OH H
C1
O OH H
Cellulose
Individual strands of cellulose tend to align with one another and are connected by multiple hydrogen bonds. The development of so many hydrogen bonds is responsible for the highly rigid structure of cellulose and its effective use as the cell-wall material in organisms. Thus, cellulose is abundant in trees and other plants. Cotton fiber is almost pure cellulose, as is filter paper. Wood and straw contain about 50% of the polysaccharide. Like cellulose, starch is a polyglucose, but its subunits are connected by a-acetal linkages. It functions as a food reserve in plants and (like cellulose) is readily cleaved by aqueous acid into glucose. Major sources of starch are corn, potatoes, wheat, and rice. Hot water swells granular starch and allows the separation of the two major components: amylose (,20%) and amylopectin (,80%). Both are soluble in hot water, but amylose is less soluble in cold water. Amylose contains a few hundred glucose units per molecule (molecular weight, 150,000–600,000). Its structure is different from that of cellulose, even though both polymers are unbranched. The difference in the stereochemistry at the anomeric carbons leads to the strong tendency of amylose to form a helical polymer arrangement (not the straight chain shown in the formula). Note that the disaccharide units in amylose are the same as those in maltose. H
CH2OH
O
O
C4
H H
HO HO
CH2OH
O
O H H
HO
Cl
HO Maltose unit
CH2OH
O
O H
HO HO
Cell walls (like those depicted in a leaf of moss) rely on cellulose for rigidity.
O
Amylose
In contrast with amylose, amylopectin is branched, mainly at C6, about once every 20 to 25 glucose units. Its molecular weight runs into the millions. H
CH2OH
O
O H
HO HO H O
CH2OH
Although we normally associate cellulose with plants, this sea squirt produces cellulose for its outer wall, one of very few animals to do so.
O
O H H
HO HO
O
CH2
Branching point
O H H
HO HO
O
CH2OH
O H
HO HO Amylopectin
O
1148
Chapter 24
Carbohydrates
Glycogen is a source of energy
Michael Phelps degrading glycogen.
Another polysaccharide similar to amylopectin but with greater branching (1 per 10 glucose units) and of much larger size (as much as 100 million molecular weight) is glycogen. This compound is of considerable biological importance because it is one of the major energystorage polysaccharides in humans and animals and because it provides an immediate source of glucose between meals and during (strenuous) physical activity. It is accumulated in the liver and in rested skeletal muscle in relatively large amounts. The manner in which cells make use of this energy storage is a fascinating story in biochemistry. A special enzyme, glycogen phosphorylase, first breaks glycogen down to give a derivative of glucose, a-d-glucopyranosyl 1-phosphate. This transformation takes place at one of the nonreducing terminal sugar groups of the glycogen molecule and proceeds step by step — one glucose molecule at a time. Because glycogen is so highly branched, there are many such end groups at which the enzyme can “nibble” away, making sure that, at a time of high-energy requirements, a sufficient amount of glucose becomes available quickly.
Nonreducing terminus
CH2OH HO
O H
HO CH2OH HO
HO O H
HO HO Nonreducing terminus
O
(
CH2OH
O
O H
HO CH2OH
HO O
O H
HO HO
O
CH2
)n
O H
HO HO
CH2OH
O
O H
HO HO
OO
Glycogen
CH2OH HO
H3PO4, glycogen phosphorylase
O H
HO HO
CH2OH
O
CH2OH
O H
HO HO
(
CH2OH
HO
HO HO
HO O
O H
HO HO
O
)n
O
-D-Glucopyranosyl 1-phosphate
CH2
O B OPOH B O
O H
HO HO
O
CH2OH
O H
HO HO
OO
24-12 Polysaccharides and Other Sugars in Nature
O 1
O
O
O 4
3
2
O 5
O 6
O
i
8
7
9 17
O
16
O
15
O
14
O
13
O
12
10 11
i i O
O
O
O
O
Phosphorylase (Eight -D-glucopyranosyl 1-phosphates released)
O
O
O
6
7
8
14
13
12
O
O
9 10 11
i i O
O
O
O
O
O
O
O
O
O
Transferase
9
6
O
7
8
O
O
14
O
13
O
12
10 11
i i O
O
-1,6-Glucosidase (One glucose released)
10 6
O
7
8
O
O
14
O
13
O
12
11
i O
O
O
Phosphorylase cannot break a-1,6-glycosidic bonds. As soon as it gets close to such a branching point (in fact, as soon as it reaches a terminal residue four units away from that point), it stops (Figure 24-4). At this stage, a different enzyme comes into play, transferase, which can shift blocks of three terminal glucosyl residues from one branch to another. One glucose substituent remains at the branching point. Now a third enzyme is required to remove this last obstacle to obtaining a new straight chain. This enzyme is specific for the kind of bond at which cleavage is needed; it is a-1,6-glucosidase, also known as the debranching enzyme. When this enzyme has completed its task, phosphorylase can continue degrading the glucose chain until it reaches another branch, and so forth.
Cell-surface carbohydrates mediate cell-recognition processes For multicellular organisms to function, each of the great variety of cells must be able to recognize and interact specifically with other cells and with various chemical species. These specific interactions are described as cell-recognition processes. Cell recognition usually involves noncovalent binding, commonly by means of hydrogen bonds, with molecules on the exterior surface of the cell. After binding takes place, movement of extracellular material into the interior of the cell may occur. Cell recognition is central to virtually every cell function. Examples of recognition in cell – cell interactions include the response of the immune system to invasion by “foreign” cells, infection of cells by bacteria or viruses, and fertilization of ova by sperm. Certain bacteria, such as cholera and pertussis (“whooping cough”), do not invade cells directly, but the toxins they produce bind to cell surfaces and are subsequently “escorted” inside. These recognition processes are based on carbohydrates that are present on cell surfaces. These carbohydrates are linked either to lipids (glycolipids; Section 20-5) or to proteins (glycoproteins; Chapter 26) that are embedded in the cell wall (Figure 24-5). Glucosyl cerebroside is the simplest of the glycolipids and illustrates key features of this class of
Chapter 24
1149
Figure 24-4 Steps in the degradation of a glycogen side chain. Initially, phosphorylase removes glucose units 1 through 5 and 15 through 17 step by step. The enzyme is now four sugar units away from a branching point (10). Transferase moves units 6 through 8 in one block and attaches them to unit 14. A third enzyme, a-1,6-glucosidase, debranches the system at glucose unit 10, by removing glucose 9. A straight chain has been formed and phosphorylase can continue its degradation job.
Chapter 24
Carbohydrates
Figure 24-5 Schematic drawing illustrating carbohydrate recognition at the cell surface, enabling cell – cell binding, in addition to adhesion of other species, such as microbes (bacteria, viruses, and bacterial toxins). The sugar chains can be linked to proteins (red ribbons).
Virus Cell adhesion Recognition site
Oligosaccharide chain Docking cell
Recognition site Bacterial toxin
Oligosaccharide chains Oligosaccharide chains
Bacteria
Cell
Recognition site
compounds: a polar carbohydrate “head” that resides on the cell surface, attached to two virtually nonpolar “tails” that embed in the cell membrane. Differences in carbohydrate composition at the “head” give rise to the different functions that these species display. For example, cholera toxin binds specifically to a glycolipid called GM1 pentasaccharide. This substance is found naturally at the surface of the cells of the intestine that are responsible for water absorption. The result is disruption of this function and severe diarrhea. O
H HO HO H
H
H NH /∑
H OH
H O CH2OH
O
CH2
OH
/∑
1150
H OH
Glucosyl cerebroside
The human blood groups—O, A, B, and AB—are each characterized by different oligosaccharides on the red blood cell surface. When an incompatible blood type is introduced by transfusion, the body’s immune system produces antibodies that bind to the oligosaccharides on the surfaces of the foreign red cells, causing them to clump together. The result is widespread clogging of blood vessels and death. Type O cells are “coated” with a hexasaccharide whose structure is common to all four blood groups; therefore, type O cells are not recognized as “foreign” by any recipient, regardless of blood type. In type A and type B blood cells, two different additional saccharides are linked to the type O hexasaccharide. It is this “seventh” saccharide that triggers the immune response. Therefore, persons of types A and B cannot donate to each other, and neither can donate to someone of type O. By contrast, type AB cells
24-12 Polysaccharides and Other Sugars in Nature
Chapter 24
contain both the type A and type B saccharides, and therefore individuals with type AB blood can receive transfusions from any donor without risk of immune system reaction. Oddly enough, the biological purpose (if any) of the blood cell surface oligosaccharides is a mystery. Some scientists speculate that the presence of different blood groups (and the associated antibodies) may play a role in inhibiting viral infection: Viruses transmitted between persons with different blood groups might have a greater likelihood of being recognized as “foreign” and destroyed by the recipient’s immune system. Nevertheless, people with a rare genetic abnormality that prevents their blood cells from forming any surface oligosaccharides are still capable of living normal, healthy lives.
Modified sugars may contain nitrogen Many of the naturally occurring sugars have a modified structure or are attached to some other organic molecule. In one large class of sugars, at least one of the hydroxy groups has been replaced by an amine function. They are called glycosylamines when the nitrogen is attached to the anomeric carbon and amino deoxy sugars when it is located elsewhere. CH2OH HO
CH2OH
O
HO NH2
HO
O OH
HO
OH
NH2
-D-Glucopyranosylamine
2-Amino-2-deoxy-D-glucopyranose
(A glycosylamine)
(An amino deoxy sugar)
When a sugar is attached by its anomeric carbon to the hydroxy group of another complex residue, it is called a glycosyl group. The remainder of the molecule (or the product after removal of the sugar by hydrolysis) is called the aglycon. An example is adriamycin, a member of the anthracycline family of antibiotics. Adriamycin and its deoxy analog daunomycin have been remarkably effective in the treatment of a wide variety of human cancers. They now constitute a cornerstone of combination cancer chemotherapy. The aglycon part of these systems is a linear tetracyclic framework incorporating an anthraquinone moiety (derived from anthracene; see Section 15-5). The amino sugar is called daunosamine. Substituted anthraquinone
#´
OH
O
O
OH
H3C
HO
OH
≥ O
H3CN H
Sugar daunosamine
Adriamycin (R OH) Daunomycin (R H)
O
H HN
OH HN A C
NH B NHCNH2
OH NH2
CH2OH OH
O H2N
O O
Aglycon
CH3O
OH
H3C OHC
A
OH
B
O
Streptose unit
R
O
HO 2-Deoxy-2-methylaminoL-glucose unit
Streptomycin
An unusual group of antibiotics, the aminoglycoside antibiotics, is based on oligosaccharide structures almost exclusively. Of particular therapeutic importance is streptomycin (an antituberculosis agent), isolated in 1944 from cultures of the mold Streptomyces griseus. The molecule consists of three subunits: the furanose streptose, the glucose derivative 2-deoxy-2-methylamino-l-glucose (an example of the rare l form), and streptidine, which is actually a hexasubstituted cyclohexane.
In Summary The polysaccharides cellulose, starch, and glycogen are polyglucosides. Cellulose consists of repeating dimeric cellobiose units. Starch may be regarded as a polymaltose derivative. Its occasional branching poses a challenge to enzymatic degradation, as does
Streptidine unit
1151
1152
Chapter 24
Carbohydrates
C H E M I CA L H IG H LIG H T 24 -4 Sialic Acid, “Bird Flu,” and Rational Drug Design In 1918 an influenza strain of unprecedented virulence jumped from its normal avian hosts to humans. By the time the outbreak subsided a year later, this “flu” killed as many as 100 million people — the numbers were so staggering that only estimates are possible — nearly 5% of the world’s population. The flu virus is a simple entity, its biochemistry governed by a half-dozen or so segments of RNA (Chapter 26) that control the synthesis of a handful of proteins. Two proteins in particular, hemagglutinin (HA) and neuraminidase (NA), determine the virus’s ability to bind to and infect cells and to reproduce by release of new virus particles from infected cells, respectively. Influenza viruses mutate rapidly, giving rise to strains that generate modified versions of HA and NA. The flu responsible for the 1918–1919 pandemic is labeled H1N1. In all, 16 forms of HA and 9 of NA have been identified. The H5N1 strain is of most current concern: It infects vast populations of birds worldwide, especially in Southeast Asia. At present, H5N1 rarely jumps to humans, but, like H1N1, it is highly lethal when it does. The fear that a mutant variety of H5N1 will gain the ability of human-to-human transmission has sparked considerable research into vaccines to prevent infection, as well as antiviral drugs that inhibit the function of viral proteins such as NA once infection has occurred. The vaccine route is made more difficult by the tendency of the virus to mutate in almost every reproductive cycle. The antiviral strategy relies on the known biochemistry of a carbohydrate that occurs naturally in the blood and tissues of all vertebrate animals: sialic acid. Sialic acid describes not one but a group of compounds, N- or O-substituted derivatives of neuraminic acid, a monosaccharide with a nine-carbon backbone. These compounds have essential biological functions in the blood, in the brain (especially for learning and memory), and elsewhere. The term “sialic acid” is often used to refer to the most common member of the family, N-acetylneuraminic acid (Neu5Ac). Sialic acid also protects organisms from invading pathogens;
Emergency hospital ward set up during the 1918 Spanish flu pandemic at the US Army’s Camp Funston in Kansas, USA.
paradoxically, however, its presence on cell surfaces facilitates the binding of HA of the influenza virus, which leads to infection. In the 1970s, studies identified a sialic acid derivative, DANA, that bound to NA and inhibited viral reproduction. An X-ray crystal structure study of the DANA – NA complex showed that the hydroxy group at C4 of the bound derivative was located close to a negatively charged carboxylate ion that was essential for NA activity. Carboxylic acids have a pKa of about 4 (Section 19-4); therefore, at the nearly neutral pH of the body’s fluids, an acid will be in the form of its conjugate base. In an effort to make this binding stronger, a group of Australian chemists converted the C4 hydroxy into an amino group, which would be mostly protonated at pH ¯ 7 (Section 21-4). The amine analog was found to bind to and inhibit NA 100 times more strongly. In a final modification, this substituent was converted to a guanidino function,
glycogen. Metabolism of these polymers first gives monomeric glucose, which is then further degraded. Finally, many sugars exist in nature in modified form or as simple appendages to other structures. Examples include cell-surface carbohydrates and amino sugars. The aminoglycoside antibiotics consist entirely of saccharide molecules, modified and unmodified.
THE BIG PICTURE The carbohydrates are a broad and diverse class of compounds, responsible for countless functions in biological chemistry. Nonetheless, their chemical behavior and their biological activity arise directly from their structures as polyfunctional molecules with well-defined shapes, which in turn result from their stereochemical characteristics. Their study provides an excellent opportunity to review the properties of the carbonyl and hydroxy functional groups, and to explore how such simple structural components can give rise to a rich array of molecular properties. We continue in Chapter 25 with a look at heterocyclic compounds, also powerful contributors to biological function, but with structures built around rings. These systems contain oxygen, sulfur, or nitrogen atoms in addition to carbon. In Chapter 26, we will return to systems in which carbohydrates play an important role when we discuss the nucleic acids.
Chapter Integration Problem
NH B O NH O C O NH2, which is a far stronger base than a simple amine, because the positive charge of its conjugate acid is highly delocalized by resonance (Section 21-6). This compound became the drug zanamivir, now marketed as Relenza by the GlaxoSmithKline pharmaceutical company. Relenza is quite likely to be the front-line antiviral compound in the worldwide effort to prepare for a bird flu outbreak. Relenza is the result of a process now commonly followed by pharmaceutical companies called rational drug design. Sophisticated software packages model both the three-dimensional shape and the surface charge distribution
Chapter 24
of target molecules, such as NA. These models are in turn employed to “design” drug “candidates” that have complementary shapes and charge distributions, so that they will “dock” with the target in a way that inhibits its function. This strategy enables chemists to identify optimal structural features of a potential drug candidate before actually going into the lab to synthesize the molecule. The graphic below displays schematically the process that led from sialic acid to the drug Relenza, based on the experimental and computational evaluations of the attraction of each drug-candidate structure to the carboxylate ion and other groups in the critical region of the neuraminidase (NA) target molecule.
Rational Design of the Neuraminidase (NA) Inhibitor Zanamivir (Relenza) Sialic acid (Neu5Ac)
HO
OH $ H $ O
@ HO
OH CO⫺ 2
RNH #
/∑
RNH # @ HO
4
HO
OH $ H $ O
@ HO
@ HO
(
O R ⴝ CH3C
4-AminoDANA
DANA
HO
1153
RNH #
CO⫺ 2 4
Zanamivir (Relenza)
HO
OH $ H $ O
@ HO
) OH $ H $ O
@ HO CO⫺ 2
RNH #
4
@ H3N⫹
@ HN
CO⫺ 2 4
NH2 H2N⫹ ON EO⫺ C
ON EO⫺ C
ON EO⫺ C
ON EO⫺ C
NA
NA
NA
NA
Weak inhibitor (hydrogen bonding)
Slightly better inhibitor (hydrogen bonding)
100ⴛ better inhibitor (ionic attraction)
Final pharmaceutical (strong ionic attraction)
CHAPTER INTEGRATION PROBLEM 24-32. Rutinose is a sugar that is part of several bioflavonoids, compounds found in many plants that have significant therapeutic value in maintaining cardiovascular health in general and the strength of the walls of blood vessels in particular. Rutin is one rutinose-containing bioflavonoid found in buckwheat and eucalyptus. Hesperidin is another, derived from the peels of lemons and oranges. Each contains rutinose bound to a tricyclic aglycon (Section 24-12). OH
OH OH
HO
rutinose
O O OH
O Rutin
OCH3 O
O
rutinose OH
O
Hesperidin
Use the following information to deduce a structure for the sugar rutinose.
The peels of these citrus fruits are rich in bioflavonoids.
1154
Carbohydrates
Chapter 24
a. Rutinose is a reducing sugar that, upon acid hydrolysis, gives one equivalent each of d-glucose and a sugar A with the formula C6H12O5. Sugar A reacts with four equivalents of HIO4 to give four equivalents of formic acid and one equivalent of acetaldehyde. What can we conclude regarding sugar A at this stage? SOLUTION What does the result of HIO4 degradation tell us? Each equivalent of HIO4 cleaves a bond between two oxygen-bearing carbon atoms. As the examples in Section 24-5 illustrate, formic acid can arise from either a terminal formyl group or an internal secondary hydroxy group. Acetaldehyde is an unusual degradation product. Its formation implies a terminal methyl substituent, attached to a secondary hydroxy carbon. A logical reconstruction of sugar A from this information may look like this: HCOOH
CHO A CHOH A CHOH A CHOH A CHOH A CH3
HCOOH HCOOH HCOOH
L-()-Mannose
CHO A CH3
1. HSCH2CH2SH, ZnCl2 2. Raney Ni (Section 17-8) 3. O2, Pt* 4. (H2O) (Section 24-4) 5. [(CH3)2CHCH2]2AlH (Section 20-4)
As a check, we note that our structure does indeed have the molecular formula C6H12O5. b. Sugar A can be synthesized from l-(2)-mannose as shown in the margin. Step 3 (noted by an asterisk) is a special reaction that converts the terminal primary alcohol into a carboxylic acid group. What does this result reveal about the stereocenters in sugar A? SOLUTION We follow the reaction sequence step by step to obtain the full open-chain structure of sugar A:
Sugar A
S H
OH
H
OH
HO HO
OS C OH
O
CHO HSCH2CH2SH, ZnCl2
CH3
H
OH
H
OH
H
HO
H
HO
CH2OH
Raney Ni
H
OH
H
OH
H
HO
H
HO
CH2OH
L-Mannose
H
Na–Hg
H H
HO
H
OH
H
OH
H2O
H
HO
H H CO2H
Sugar A Aldonic acid
CH3
OH
HO
H
HO
Sugar A Alditol
CH3
O2, Pt
H CH2OH
Thioacetal
CH3
O
C K O Sugar A Aldonolactone
H
OH
H
OH
HO
H
HO
H CHO
Sugar A
Sugar A has the name 6-deoxy-l-mannose. c. Complete methylation of rutinose with the use of excess dimethyl sulfate (Section 24-8) gives a heptamethylated derivative. On subsequent mild acid hydrolysis, one equivalent of 2,3,4-tri-O-methyld-glucose and one equivalent of the 2,3,4-tri-O-methyl derivative of sugar A are obtained. What possible structure(s) of rutinose is consistent with these data? SOLUTION Dimethyl sulfate treatment converts all free – OH groups into – OCH3 moieties (Section 24-8). Therefore, we may conclude that rutinose possesses seven hydroxy substituents, and (at least) one must be
Chapter Integration Problem
Chapter 24
part of a cyclic hemiacetal function. Recall that rutinose is a reducing sugar (Section 24-4). We may also conclude that both monosaccharide components of rutinose are in cyclic forms. Why? The openchain forms of glucose and sugar A contain a total of nine – OH groups. For rutinose to possess only seven, two of the original nine must be built into a cyclic glycosidic acetal function linking the sugars, similar to the linkages in the disaccharides maltose, cellobiose, and lactose (Section 24-11). Acid hydrolysis of the heptamethylated rutinose gives two trimethylated monosaccharides. Acid cleaves the glycoside linkage between the two sugars (Section 24-11), but where did the seventh methyl group go? It must have been attached to a hemiacetal oxygen, in the form of a methyl glycoside — an acetal — which we know is cleaved by acid much more easily than an ordinary methyl ether (Section 24-8). The six methyl groups (at carbons 2, 3, and 4 in the two resulting monosaccharides) must be bound to oxygen atoms that are not part of either the glycosidic linkage between the sugars or their rings. Only the oxygen atoms left unmethylated by dimethyl sulfate treatment can constitute these parts of rutinose. Thus, we may conclude that the oxygens at C5 in both methylated monosaccharide products are contained in pyranose (six-membered) rings, because five-membered rings would have contained the oxygens at C4 instead. We are almost finished. There remain three hydroxy groups that are candidates for the linkage between the sugars: the hemiacetal – OH at C1 of sugar A and either the C1 hemiacetal hydroxy or the C6 primary – OH of the glucose. To summarize: 1
1
HOCH
CHOH
H H
O
HO
2 3 4 5
OH
H Candidates for glycosidic linkage
OH
HO H
H H
H
6
2 3 4
OH H
O
OH
5 6
CH2OH
CH3 6-Deoxy-L-mannopyranose Sugar A
D-Glucopyranose
In fact, we already have the answer: Rutinose is a reducing sugar and must therefore possess at least one hemiacetal function. The only option therefore is to assign the latter to the glucose moiety and the glycoside bond to C6 – OH and C1 of sugar A. We conclude by redrawing the structures in more descriptive Haworth projections and chair conformations (the stereochemistry at C1 of sugar A is not defined) by following the procedures in Section 24-2. In the following three drawings, the glucopyranose moieties are below the sugar A plane. Because A is an l sugar, care is necessary to preserve the proper absolute stereochemistry. One method is to follow the same procedure as for d sugars: Place the ether linkage in back and rotate the Fischer projection 908 clockwise, such that groups originally on the right in the Fischer projection now point downward and those on the left point upward. Indicators d and l refer to the configuration at C5. Whereas the substituent (C6) at this position points upward in a d sugar, with C5 inverted, this group must now point downward. The two structures I and II below result. An equally correct alternative is to rotate the Fischer projection of an l sugar the opposite way — counterclockwise. In this manner structure III at the right is obtained. Notice that the C6 methyl group now points upward, the anomeric carbon (C1) is to the left, and the structure of sugar A is presented in a view rotated by 1808 about an axis perpendicular to the page, relative to the drawings at the right. Build models! 6
O
Rutinose
H 5
HO 4
H
O O
6CH3
H 3
OH
1
1
H 2
OH
6
H 3C
O O CH2 5
H 4
HO
H OH 3
H I
H
HO
HO 1
OH
O
O
HO
O
HO
CH2
OH
4
HO
2
OH
2
5
H O
H OH
H
H OH
HO
3
H II
H
H OH HO
6
CH2
O
CH3
5
3
2
III
OH
H 1
OH
4
OH
1155
1156
Chapter 24
Carbohydrates
New Reactions 1. Cyclic Hemiacetal Formation in Sugars (Section 24-2) CHO H
OH
HO
CH2OH
H
H
OH
H
OH
HO
H2O
O
HO OH
OH
␣-and -Glucopyranoses
CH2OH
2. Mutarotation (Section 24-3) CHO CH2OH HO
H O
HO
H2O
HO
H OH OH
H
OH
CH2OH
H
HO
H2O
OH
HO
OH
OH
OH CH2OH
anomer
anomer
(Equilibrium [] D25C 52.7)
([] D25C 112)
O
([] D25C 18.7)
3. Oxidation (Section 24-4, most nonessential H and OH substituents have been omitted) Tests for reducing sugars CHO
COOH
2
Cu , OH , H2O (Fehling’s solution) or Ag, NH4OH, H2O (Tollens’s solution)
H
OH
H
CH2OH
Cu2O or Red
OH CH2OH
Aldonic acid synthesis
CHO
O B C
COOH
O
Br2, H2O H2O
H
OH
H
CH2OH
H
OH
CH2OH
CH2OH
␥-Lactone
Aldonic acid
Aldaric acid synthesis CHO HO H
H OH CH2OH
COOH HNO3, H2O
HO H
H OH COOH
Aldaric acid
Ag Silver mirror
New Reactions
4. Sugar Degradation (Section 24-5) R A H OC O OH A H O C OOH A R R A C PO A CH2OH
HIO4, H2O
R A C PO A OH
HIO4, H2O
R A H OC O OH A C PO A H
R A 2 CPO A H
CH2O
R A C PO A H
HIO4, H2O
R A H OC O OH A COOH
HIO4, H2O
HCOOH
R A C PO A H
CO2
5. Reduction (Section 24-6) CH2OH
CHO NaBH4, CH3OH
OH
H
H
OH CH2OH
CH2OH
Alditol
6. Hydrazones and Osazones (Section 24-7) CHO H
OH
H
OH
C6H5NHNH2 (1 equivalent), CH3CH2OH
HCPNONHC6H5 A CPNONHC6H5
CHPNONHC6H5 H
OH
H
OH
C6H5NHNH2, CH3CH2OH,
H2O
C6H5NH2, NH3, H2O
H
CH2OH
CH2OH
CH2OH
OH
Phenylhydrazone
Osazone
7. Esters (Section 24-8)
CH2OH HO
O O B B 5 RCOCR, pyridine
O
HO OH
OH
O B O CH2OCR B O RCO RCO B RCO O B O
␣ and  anomers
O B OCR
5 RCOOH
H2O
␣ and  anomers
8. Glycosides (Section 24-8) CH2OH HO
O
CH3OH, H
HO
H2O, H
OH
CH2OH HO
O
HO OH
OH
␣ and  anomers
OCH3
␣ and  anomers
9. Ethers (Section 24-8) CH2OH HO
O
5 (CH3)2SO4, Na Na2SO4
HO OH ␣ and  anomers
OH
CH3O CH3O
CH2OCH3 O CH3O
␣ and  anomers
OCH3
Chapter 24
1157
1158
Chapter 24
Carbohydrates
10. Cyclic Acetals (Section 24-8) R⬘
O B CH3CCH3, H⫹
O
R
OH
O
⫺H2O
OH HO
R⬘ R
OH O
H
O H3C
H
CH3
11. Chain Extension Through Cyanohydrins (Section 24-9) CN A HOCOOH
CHO
H2, Pd–BaSO4, H⫹, H2O
HCN
H
OH
H
CHO A HOCOOH
OH
Sugar
OH
H
CH2OH
CH2OH
CH2OH
Cyanohydrin
Extended sugar
12. Ruff Degradation (Section 24-9) CHO H
COOH
OH
H
OH
H
OH
Br2, H2O
CHO Fe3⫹, H2O2 ⫺CO2
H
OH CH2OH
CH2OH
H
OH CH2OH
Important Concepts 1. Carbohydrates are naturally occurring polyhydroxycarbonyl compounds that can exist as monomers, dimers, oligomers, and polymers. 2. Monosaccharides are called aldoses if they are aldehydes and ketoses if they are ketones. The chain length is indicated by the prefix tri-, tetr-, pent-, hex-, and so forth. 3. Most natural carbohydrates belong to the d family; that is, the stereocenter farthest from the carbonyl group has the same configuration as that in (R)-(1)-2,3-dihydroxypropanal [d-(1)glyceraldehyde]. 4. The keto forms of carbohydrates exist in equilibrium with the corresponding five-membered (furanoses) or six-membered (pyranoses) cyclic hemiacetals. The new stereocenter formed by cyclization is called the anomeric carbon, and the two anomers are designated A and B. 5. Haworth projections of d sugars depict the cyclic ether in line notation as a pentagon or a hexagon, the anomeric carbon placed on the right and the ether oxygen at the top. The substituents located above or below the ring are attached to vertical lines. The ring bond at the bottom (between C2 and C3) is understood to be in front of the plane of the paper, and the ring bonds containing the oxygen are understood to be in back. The a anomer has the OH group at the anomeric carbon pointing downward, whereas the b anomer has it pointing upward. 6. Equilibration between anomers in solution gives rise to changes in the measured optical rotation called mutarotation. 7. The reactions of the saccharides are characteristic of carbonyl, alcohol, and hemiacetal groups. They include oxidation of the aldehyde to the carboxy function of aldonic acids, double oxidation to aldaric acids, oxidative cleavage of vicinal diol units, reduction to alditols, condensations, esterifications, and acetal formations.
Problems
8. Sugars containing hemiacetal functions are called reducing sugars, because they readily reduce Tollens’s and Fehling’s solutions. Sugars in which the anomeric carbon is acetalized are nonreducing. 9. The synthesis of higher sugars is based on chain lengthening, the new carbon being introduced by cyanide ion. The synthesis of lower sugars relies on Ruff chain shortening, a terminal carbon being expelled as CO2. 10. The Fischer proof uses the techniques of chain lengthening and shortening together with the symmetry properties of aldaric acids to determine the structures of the alsoses. 11. Di- and higher saccharides are formed by the combination of hemiacetal and alcohol hydroxy groups of two sugars to give an acetal linkage. 12. The change in optical rotation observed in acidic aqueous solutions of sucrose, called the inversion of sucrose, is due to the equilibration of the starting sugar with the various cyclic and anomeric forms of its component monomers. 13. Many sugars contain modified backbones. Amino groups may have replaced hydroxy groups, there may be substituents of various complexity (aglycons), the backbone carbon atoms of a sugar may lack oxygens, and (rarely) the sugar may adopt the l configuration.
Problems 33. The designations d and l as applied to sugars refer to the configuration of the highest-numbered stereocenter. If the configuration of the highest numbered stereocenter of d-ribose (Figure 24-1) is switched from d to l, is the product l-ribose? If not, what is the product? How is it related to d-ribose (i.e., what kind of isomers are they)? 34. To which classes of sugars do the following monosaccharides belong? Which are d and which are l? CH2OH O
CHO
CHO
H
OH
HO
H
H
OH
HO
H
H
OH
HO
H
(a) HOCH2
OH
(b) HO
H
CH2OH
(c)
CH3
OH
H
OH CH2OH
()-Rhamnose
()-Apiose
H
()-Mannoheptulose
35. Draw open-chain (Fischer-projection) structures for l-(1)-ribose and l-(2)-glucose (see Exercise 24-2). What are their systematic names? 36. Identify the following sugars, which are represented by unconventionally drawn Fischer projections. (Hint: It will be necessary to convert these projections into more conventional representations without inverting any of the stereocenters.) CH2OH CH2OH H
(a) HOCH2
CHO
(b) HOCH2
OH
OH
H
OH
(e) HOCH2
(d) HOCH2 H
H
OH H CHO
CHO H OH
H
(c) HO
HO
H
HO
H
H OH
H
O OH
OHC
HO
Chapter 24
1159
1160
Chapter 24
Carbohydrates
37. Redraw each of the following sugars in open-chain form as a Fischer projection, and find its common name. OH H OH * C * C C C (a) OHC G G CH2OH C C &¥ & ¥ H OH HO H H
HO
(c)
H
CH2OH O H H OH
HO
HOCH2 (b) HH HO
H
O
OH OH CH2OH
HO
H
O OH
(d) HO
OH
CH2OH OH
H
38. For each of the following sugars, draw all reasonable cyclic structures, using either Haworth or conformational formulas; indicate which structures are pyranoses and which are furanoses; and label a and b anomers. (a) (2)-Threose; (b) (2)-allose; (c) (2)-ribulose; (d) (1)-sorbose; (e) (1)-mannoheptulose (Problem 34). 39. Are any of the sugars in Problem 38 incapable of mutarotation? Explain your answer. 40. Draw the most stable pyranose conformation of each of the following sugars. (a) a-d-Arabinose; (b) b-d-galactose; (c) b-d-mannose; (d) a-d-idose. 41. Ketoses show positive Fehling’s and Tollens’s tests not only by oxidation to a-dicarbonyl compounds, but through a second process: Ketoses isomerize to aldoses in the presence of base. The aldose then undergoes oxidation by the Fehling’s or Tollens’s solution. Using any ketose in Figure 24-2, propose a base-catalyzed mechanistic pathway to the corresponding aldose. [Hint: Review Section 18-2.] 42. What are the products (and their ratios) of periodic acid cleavage of each of the following substances: (a) 1,3-dihydroxyacetone; (b) rhamnose (Problem 34); (c) glucitol. 43. Write the expected products of the reaction of each of the following sugars with (i) Br2, H2O; (ii) HNO3, H2O, 608C; (iii) NaBH4, CH3OH; and (iv) excess C6H5NHNH2, CH3CH2OH, D. Find the common names of all the products. (a) d-(2)-Threose; (b) d-(1)-xylose; (c) d-(1)galactose. 44. Draw the Fischer projection of an aldohexose that will give the same osazone as (a) d-(2)-idose and (b) l-(2)-altrose. 45. (a) Which of the aldopentoses (Figure 24-1) would give optically active alditols upon reduction with NaBH4? (b) Using d-fructose, illustrate the results of NaBH4 reduction of a ketose. Is the situation more complicated than reduction of an aldose? Explain. 46. Which of the following glucoses and glucose derivatives are capable of undergoing mutarotation? (a) a-d-Glucopyranose; (b) methyl a-d-glucopyranoside; (c) methyl a-2,3,4,6-tetra-O-methyl-dglucopyranoside (i.e., the tetramethyl ether at carbon 2, 3, 4, and 6); (d) a-2,3,4,6-tetra-O-methyld-glucopyranose; (e) a-d-glucopyranose 1,2-acetone acetal. 47. (a) Explain why the oxygen at C1 of an aldopyranose can be methylated so much more easily than the other oxygens in the molecule. (b) Explain why the methyl ether unit at C1 of a fully methylated aldopyranose can be hydrolyzed so much more easily than the other methyl ether functions in the molecule. (c) Write the expected product(s) of the following reaction. D-Fructose
CH3OH, 0.25% HCl, H2O
-----------------------------¡
48. Of the four aldopentoses, two form diacetals readily when treated with excess acidic acetone, but the other two form only monoacetals. Explain.
Problems
1161
Chapter 24
49. d-Sedoheptulose is a sugar intermediate in a metabolic cycle (the pentose oxidation cycle) that ultimately converts glucose into 2,3-dihydroxypropanal (glyceraldehyde) plus three equivalents of CO2. Determine the structure of d-sedoheptulose from the following information. B B D-Sedoheptulose -----¡ 4 HCOH 2 HCH CO2 6 HIO4
C H5NHNH2 an osazone identical with that formed d-Sedoheptulose ----6--------¡ by another sugar, aldoheptose A Ruff degradation
Aldoheptose A -----------------¡ aldohexose B HNO3, H2O, ¢
Aldohexose B -------------¡ an optically active product Ruff degradation
Aldohexose B -----------------¡ D-ribose 50. Is it possible for two different diastereomeric aldoses to give the same product upon KilianiFischer chain elongation? Why or why not? 51. In each of the following groups of three d aldoses, identify the two that give the same product upon Ruff degradation. (a) Galactose, gulose, talose; (b) glucose, gulose, idose; (c) allose, altrose, mannose. 52. Illustrate the results of chain elongation of d-talose through a cyanohydrin. How many products are formed? Draw them. After treatment with warm HNO3, does the product(s) give optically active or inactive dicarboxylic acids? 53.
(a) Write a detailed mechanism for the isomerization of b-d-fructofuranose from the hydrolysis of sucrose into an equilibrium mixture of the b-pyranose and b-furanose forms. (b) Although fructose usually appears as a furanose when it is part of a polysaccharide, in the pure crystalline form, fructose adopts a b-pyranose structure. Draw b-d-fructopyranose in its most stable conformation. In water at 208C, the equilibrium mixture contains about 68% b-dpyranose and 32% b-d-furanose. (c) What is the free-energy difference between the pyranose and furanose forms at this temperature? (d) Pure b-d-fructopyranose has [a]d208C 5 2132. The equilibrium pyranose – furanose mixture has [a]d208C 5 292. Calculate [a]d208C for pure b-dfructofuranose.
54. Classify each of the following sugars and sugar derivatives as either reducing or nonreducing. (a) d-Glyceraldehyde; (b) d-arabinose; (c) b-d-arabinopyranose 3,4-acetone acetal; (d) b-darabinopyranose acetone diacetal; (e) d-ribulose; (f) d-galactose; (g) methyl b-d-galactopyranoside; (h) b-d-galacturonic acid (as shown in the margin); (i) b-cellobiose; ( j) a-lactose. 55. Is a-lactose capable of mutarotation? Write an equation to illustrate your answer. 56. Trehalose, sophorose, and turanose are disaccharides. Trehalose is found in the cocoons of some insects, sophorose turns up in a few bean varieties, and turanose is an ingredient in low-grade honey made by bees with indigestion from a diet of pine tree sap. Identify among the following structures those that correspond to trehalose, sophorose, and turanose on the basis of the following information: (i) Turanose and sophorose are reducing sugars. Trehalose is nonreducing. (ii) Upon hydrolysis, sophorose and trehalose give two molecules each of aldoses. Turanose gives one molecule of an aldose and one molecule of a ketose. (iii) The two aldoses that constitute sophorose are anomers of each other.
(a)
HOCH2 HO
(b) HO
O
HO HOCH2 HO
HO
O
OH
HOCH2
O O
HO
O OH
CH2OH
OH
O OH
HO OH
HO H
COOH O H OH
H
H
OH
OH H
-D-Galacturonic acid
1162
Carbohydrates
Chapter 24
(c)
HOCH2
(d)
O
HO
HOCH2
O
HO
HO
HO HO
HO
HO
O O
O CH2OH
O CH2OH
OH
HO OH
OH
OH
57. Identify and, using the information in Sections 24-1 and 24-11, provide names for the carbohydrates that are contained in the structure of stevioside (Chemical Highlight 24-3). 58. The disaccharide at the terminus of the oligosaccharide attached to the surface of type B red blood cells is an a-d-galactopyranosyl-b-d-galactopyranose. The linkage is an acetal from C1 of the first galactose to the C3 hydroxy group of the second. In other words, the full name is 3-(a-dgalactopyranos-l-yl)-b-d-galactopyranose. Draw the structure of this disaccharide, using conformationally accurate chairs for the six-membered rings. 59. Glucose reacts with ammonia in the presence of a trace of acid to give predominantly b-dglucopyranosylamine (Section 24-12). Propose a reasonable mechanism for this transformation. Why is only the hydroxy group at C1 replaced?
H
OH
HOCH2
OH
60. (a) A mixture of (R)-2,3-dihydroxypropanal (d-glyceraldehyde) and 1,3-dihydroxyacetone that is treated with aqueous NaOH rapidly yields a mixture of three sugars: d-fructose, d-sorbose, and racemic dendroketose (only one enantiomer is shown here). Explain this result by means of a detailed mechanism. (b) The same product mixture is also obtained if either the aldehyde or the ketone alone is treated with base. Explain. [Hint: Closely examine the intermediates in your answer to (a).]
CH2OH
61. Write or draw the missing reagents and structures (a) through (g). What is the common name of (g)?
CH2OH O
Dendroketose D-()-Xylose
(a)
(c)
(b)
NH3,
(d)
D-Xylonic
Methyl
acid
D-xylonate
CO2
Br2, NaOH
C5H11NO5 (e)
C4H11NO4
NH3
(f )
C4H8O4 (g)
The preceding sequence (called the Weerman degradation) achieves the same end as what procedure described in this chapter? 62.
Fischer’s solution to the problem of sugar structures was actually much more difficult to achieve experimentally than Section 24-10 implies. For one thing, the only sugars that he could readily obtain from natural sources were glucose, mannose, and arabinose. (Erythrose and threose were, in fact, not then available at all, either naturally or synthetically.) His ingenious solution required a way to exchange the functionalities at C1 and C6 of glucose and mannose in order to make the critical distinction described at the end of the section. (Of course, had gulose existed in nature, all this effort would have been unnecessary, but Fischer wasn’t so lucky.) Fischer’s plan led to unexpected difficulties, because at a key stage he got a troublesome mixture of products. Nowadays we solve the problem in the manner shown below. Fill in the missing reagents and structures (a) through (g). Use Fischer projections for all structures. Follow the instructions and hints in parentheses.
(a)
H, H2O
O2, Pt
D-()-Glucose
(b) Methyl D-glucoside
(Both isomers; write only one)
(d) D-Glucuronic
acid (Write the open-chain form only)
(c)
(A special reaction that oxidizes only the primary hydroxy at C6 into a carboxylic group)
NaBH4
(e) Gulonic acid
Methyl D-glucuronoside
H2O
Na–Hg
(f) Gulonolactone
(Reduces lactones to aldehydes)
(g) Gulose (Write the open-chain form only)
Problems
63.
Chapter 24
Vitamin C (ascorbic acid, Section 22-9) is present almost universally in the plant and animal kingdoms. (According to Linus Pauling, mountain goats biosynthesize from 12 to 14 g of it per day.) Animals produce it from d-glucose in the liver by the four-step sequence d-glucose n d-glucuronic acid (see Problem 62) n d-glucuronic acid g-lactone n l-gulonic acid g-lactone n vitamin C. CH2OH A H OC O OH A COH A C O OH B O C O OH A CPO
HO is the same as
OH A C C
C
O
H OC O O A HO OC O H A CH2OH
Vitamin C
The enzyme that catalyzes the last reaction, l-gulonolactone oxidase, is absent from humans, some monkeys, guinea pigs, and birds, presumably because of a defective gene resulting from a mutation that may have occurred some 60 million years ago. As a result, we have to get our vitamin C from food or make it in the laboratory. In fact, the ascorbic acid in almost all vitamin supplements is synthetic. An outline of one of the major commercial syntheses follows. Draw the missing reagents and products (a) through (f ). (Hint: See Chemical Highlight 24-1.)
D-Glucose
(a)
D-glucitol
Microbial oxidation at C5 (by Gluconobacter oxydans)
(b)
(c)
L-Sorbose
L-Sorbofuranose
(d) (Two steps)
(Open chain)
CH3 O O
CH3 O (e)
O H3C
O
CH3
COOH
CH2OH A H OC O OH A COH HO OC A H OC O OH A CPO A COOH 2-Keto-L-gulonic acid
(f)
O
CH2OH A H OC O OH A COH A H OC O OH A CPO A CPO
Keto form of vitamin C
Team Problem 64. This problem is meant to encourage you to think as a team about how you might establish the structure of a simple disaccharide, with some additional information at your disposal. Consider d-lactose (Section 24-11) and assume you do not know its structure. You are given the knowledge that it is a disaccharide, linked in a b manner to the anomeric carbon of only one of the sugars, and you are given the structures of all of the aldohexoses (Table 24-1), as well as all of their possible methyl ethers. Deal with the following questions as a team or by dividing the work, before joint discussion, as appropriate. (a) Mild acid hydrolyzes your “unknown” to d-galactose and d-glucose. How much information can you derive from that result? (b) Propose an experiment that tells you that the two sugars are not connected through their respective anomeric centers. (c) Propose an experiment that tells you which one of the two sugars contains an acetal group used to bind the other. (Hint: The functional group chemistry of the monosaccharides described in the chapter can be applied to higher sugars as well. Specifically, consider Section 24-4 here.) (d) Making use of the knowledge of the structure of all possible methyl ethers of the component monosaccharides, design experiments that will tell you which (nonanomeric) hydroxy group is responsible for the disaccharide linkage. (e) Similarly, can you use this approach to distinguish between a furanose and pyranose structure for the component of the disaccharide that can mutarotate?
vitamin C
1163
1164
Chapter 24
CHO H
OH CH2OH
OH O A B HOCH2CHCHCH2CH A OH
Carbohydrates
Preprofessional Problems 65. Most natural sugars have a stereocenter that is identical to that in (R)-2,3-dihydroxypropanal, shown as a Fischer projection in the margin. What is the (very popular) common name for this compound? (a) d-(1)-Glyceraldehyde; (b) d-(2)-glyceraldehyde; (c) l-(1)-glyceraldehyde; (d) l-(2)-glyceraldehyde 66. What kind of sugar is the compound shown in the margin? (a) An aldopentose; (b) a ketopentose; (c) an aldohexose; (d) a ketohexose 67. Which one of the following statements is true about the oxacyclohexane conformer of the sugar b-d-(1)-glucopyranose? (a) One OH group is axial, but all remaining substituents are equatorial. (b) The CH2OH group is axial, but all remaining groups are equatorial. (c) All groups are axial. (d) All groups are equatorial. 68. The methyl glycoside of mannose is made by treating the sugar with (a) AlBr3, CH3Br; (b) dilute aqueous CH3OH; (c) CH3OCH3 and LiAlH4; (d) CH3OH, HCl; (e) oxacyclopropane, AlCl3 69. One of the statements below is correct about the sugar shown. Which one? H CH2OH
H
O
HOCH2 O
HO
O
H
H
OH CH2OH
OH H
(a) It is a nonreducing sugar. (c) It exists in two anomeric forms.
H
OH
H
(b) It forms an osazone. (d) It undergoes mutarotation.
TWENTY FIVE
C H A P T E R
Heterocycles O
Heteroatoms in Cyclic Organic Compounds
N
N
O
N
N
L
ook at the list of the world’s top ten prescription drugs (Table 25-1). What do their structures have in common? Apart from the presence of heteroatoms, such as oxygen, nitrogen, and sulfur, they all contain at least two rings. Moreover, in all but two cases, these rings are made up not solely of carbon, called carbocycles (Chapter 4), but also of heteroatoms and are thus called heterocycles. This chapter describes the naming, syntheses, and reactions of some saturated and aromatic heterocyclic compounds in order of increasing ring size, starting with the heterocyclopropanes. Some of this chemistry is a simple extension of transformations presented earlier for carbocycles. However, the heteroatom often causes heterocyclic compounds to exhibit special chemical behavior. Most physiologically active compounds owe their biological properties to the presence of heteroatoms, mainly in the form of heterocycles. A majority of the known natural products are heterocyclic. It is therefore not surprising that more than half the published chemical literature deals with such compounds — their synthesis, isolation, and interconversions. Indeed, we have already encountered many examples — the cyclic ethers (Section 9-6), acetals (Sections 17-8, 23-4, and 24-8), carboxylic acid derivatives (Chapters 19 and 20), and amines (Chapter 21). The bases in DNA, whose sequence stores hereditary information, are heterocycles (Section 26-9); so are many vitamins, such as B1 (thiamine, Chemical Highlight 23-2), B2 (riboflavin, Chemical Highlight 25-3), B6 (pyridoxine), the spectacularly complex B12, and vitamins C and E (Section 22-9). The structures of vitamins B6 and B12, as well as additional examples of heterocyclic systems and their varied uses, are depicted here.
How many of you can do without that morning cup of coffee? The stimulating ingredient in coffee is the heterocycle caffeine, 135 mg of it in a normal 8-oz serving. Of course, there is also Roman espresso, which has 100 mg of caffeine in less than an ounce of liquid.
1166
Heterocycles
Chapter 25
Table 25-1
Top Ten U.S. Prescription Drugs (Ranked by 2007 Sales) by Generic and, in Parentheses, Selected Brand Namesa
1. Atorvastatin (Lipitor)
6. Venlafaxine (Effexor) CH3
CH3
N H
OH %
O
H3C
OH %
COOH
N
H3C
H3CE
(Racemate) Antidepressant
Cholesterol reducer
7. Clopidogrel (Plavix) O l šS /
CH3
O
N
N HN
CH3
N
Stomach acid reducer
S
3. Lansoprazole (Prevacid)
Antiplatelet agent
O l S
N
8. Simvastatin (Sivastin, Zocor) N
O CF3
O
(Racemate) Antiulcerative and stomach acid reducer
Antihypercholesterolemic
≥
OH
≥
HO HO
H N O
9. Amlodipine (Norvasc) H 3C
H N
O
CH3O2C
Cl Salmeterol (Racemate)
Fluticasone
Antihypertensive
Antiasthmatic
10. Escitalopram (Lexapro) NC
5. Montelucast (Singulair)
O
S ´
N
CO2H H3C H3C
H3C
ENE CH3
F HO
Antiasthmatic a
NH2
CO2CH2CH3
≥ F
Cl
/ CH3
≥
O
H /
H3C
CH3
CH3
H3C
4. Fluticasone and salmeterol mixture (Advair Diskus) F S O
O
≥
H3C
O
HO /
N H
HO / H H3C / ≥ / ≥ F H
Cl
O
CH3
O H3 C /
O
O
E
O A CH3
OH
O
F
2. Esomeprazole (Nexium) H3C
EN
Total U.S. sales of pharmaceuticals reached $295 billion in 2007, with lipitor leading at $6.2 billion.
Antidepressant
25-1 Naming the Heterocycles
CONH2
´
/
H
OH
CONH2
H3C
HOCH2
CH3 / H
´
´
E
H R
´
!
´
! CH3 CH3
H H
H O
H 2C
H
N
N N
N
E
š N OH
HOCH2
CH3
N H
H O
CH3 H Vitamin B12 (Cobalamin)
(Catalyzes biological rearrangements and methylations)
N H3C
HN
N
HOOC
HN
N N
O
COO COOH Viagra (Sildenafil citrate)
N
Cl
O HO
O
N
N O
HO
CH3 A N
CH3
∑
H
O
CH3
0
O l O O l l S
N
(Enzyme cofactor vitamin with multiple functions)
OH
HO H
NH2
EPE O l O O H
CH3
Pyridoxine, vitamin B6
H2NOC
NH O H3C / H
HOCH2
CONH2
´
H2NOC / ýN R ½N H3C E H3 C Co H / Nk ýN H2NOC H 0 CH3 H3C O
1167
Chapter 25
N3 Zidovudine (AZT)
Diazepam (Valium)
(Antiviral AIDS drug, see Chemical Highlight 26-3)
(Tranquilizer)
(Treats erectile dysfunction; see also Chapter Integration Problem 25-29)
25-1 Naming the Heterocycles Like all other classes of compounds we have encountered, the heterocycles contain many members bearing common names. Moreover, there are several competing systems for naming heterocycles, which is sometimes confusing. We shall adhere to the simplest system. We regard saturated heterocycles as derivatives of the related carbocycles and use a prefix to denote the presence and identity of the heteroatom: aza- for nitrogen, oxa- for oxygen, thia- for sulfur, phospha- for phosphorus, and so forth. Other widely used names will be given in parentheses. The location of substituents is indicated by numbering the ring atoms, starting with the heteroatom. In the following examples, space-filling renditions of the most representative ring sizes, from three to six membered, give you a perspective of their actual shape.
Heterocycles
O
F
N A CH3
O
Oxacyclopropane (Oxirane, ethylene oxide)
2 1S
N-Methylazacyclopropane (N-Methylaziridine)
Br&
A
4
2
1
H3C & CH3
NH
3-Ethylazacyclobutane (3-Ethylazetidine)
S
N H
5 1O
trans-3,4-Dibromooxacyclopentane (trans-3,4-Dibromotetrahydrofuran)
Azacyclopentane (Pyrrolidine)
3
6
N H
2 1O
S
3-Methyloxacyclohexane (3-Methyltetrahydropyran)
S
Br
4∑
CH3
5
S
3
2
2,2-Dimethylthiacyclobutane (2,2-Dimethylthietane) 4
O
O
Oxacyclobutane (Oxetane)
2-Fluorothiacyclopropane (2-Fluorothiirane)
CH3CH2 3
3
A
Chapter 25
∑
1168
3-Cyclopropylthiacyclohexane (3-Cyclopropyltetrahydrothiopyran)
Thiacyclopentane (Tetrahydrothiophene)
N Azacyclohexane (Piperidine)
The common names of unsaturated heterocycles are so firmly entrenched in the literature that we shall use them here. NH2 5
4 3
4
3
4
3
4
3
5
4
4
6
3
3
5 2
2
5
2
5
N1 H
O
Pyrrole
Furan
2
5
S
1
2
6
1
2
7
N
N
8
1
1
6
N1 H
7
Quinoline
N
N N
N H
Adenine (See Section 26-9)
S
N
Thiophene
Pyridine
N Indole
Exercise 25-1 Name or draw the following compounds. (a) trans-2,4-Dimethyloxacyclopentane (trans-2,4dimethyltetrahydrofuran); (b) N-ethylazacyclopropane; Br
(c)
(d) O 2N
N
NO2
N H
25-2 Nonaromatic Heterocycles As illustrated by the chemistry of the oxacyclopropanes (Section 9-9), ring strain allows the three- and four-membered heterocycles to undergo nucleophilic ring opening readily. In contrast, the larger, unstrained systems are relatively inert to attack.
25-2 Nonaromatic Heterocycles
Chapter 25
1169
Ring strain makes heterocyclopropanes and heterocyclobutanes reactive Heterocyclopropanes are relatively reactive because ring strain is released by nucleophilic ring opening. Under basic conditions, this process gives rise to inversion at the less substituted center (Section 9-9). MODEL BUILDING Strained ring
š O
C6H5
Attack at less substituted carbon
2-Phenyloxacyclopropane
70% CH3CH2š NH2, H2O, 120C, 16 days
2-Methoxy-1-phenylethanol
CH3CH2š NH ≥H ! CH3 G CO C G H3C NHCH2CH3 H 55% ! ≥
CH2CH3 A N `H H3C ^ & ( CH3 H
Strained ring
CH3OH
Oð CH3š
OH ðš A C6H5CHCH2š OCH3 85%
Attack with inversion
N-Ethyl-(2S,3S)-trans-2,3dimethylazacyclopropane
meso-N, Nⴕ-Diethyl-2,3-butanediamine
Exercise 25-2 Explain the following result by a mechanism. (Caution: This is not an oxidation. Hints: Calculate the molecular formulas of starting material and product. Try a ring opening catalyzed by the Lewis acid and consider the options available to the resulting intermediate. See Section 9-3.)
^
H ? O ?
MgBr2, (CH3CH2)2O
O
H
^
100%
Exercise 25-3 2-(Chloromethyl)oxacyclopropane reacts with hydrogen sulfide ion (HS2) to give thiacyclobutan3-ol. Explain by a mechanism.
Exercise 25-4
Working with the Concepts: Practicing Mechanisms That Involve Heterocycles Isomeric cylindricines A and B, isolated in 1993, are the two main alkaloids (Section 25-8) present in extracts from the Australian marine plant Clavelina cylindrica. The two compounds equilibrate to a 3 : 2 mixture. Formulate a mechanism for this process. (Hint: Check Exercise 9-25.)
O N
O Cl
)
N
Cl Cylindricine A
Cylindricine B
1170
Heterocycles
Chapter 25
CHE M I C AL H I G H L I G H T 2 5 - 1 Smoking, Nicotine, Cancer, and Medicinal Chemistry Nicotine is an azacyclopentane derivative that is responsible for the addictive character of cigarette smoking. Like other drugs of abuse, its presence activates the brain’s pleasure centers by stimulating the release of the neurotransmitter dopamine, while its absence activates the body’s stress systems (“withdrawal”). The causal link between smoking and cancer is well established, and there are over 40 known carcinogens among the thousands of compounds contained in cigarette smoke, among them benz[a]pyrene (Section 16-7).
N
NO
(Section 21-10)
N
ANA H3C NO
(Chemical Highlight 21-3)
Nicotine
NO A š N
N
Upon protonation of the oxygen in the nitroso group, these substances become reactive alkylating agents, capable of transferring methyl groups to nucleophilic sites in biological molecules such as DNA, as shown below. The diazohydroxide that remains decomposes through a diazonium ion to a carbocation, which may inflict additional molecular damage (Section 21-10).
š KOð
ðN A ðNH E CH3 R
Ring opening and oxidation
š KOH
H
ðN A ðNH E CH3 R
ðš OH E ðN B H EN CH3 R
Nu)(in DNA)
The ability of nicotine to act as a tumor promoter has surfaced in several studies. For example, certain human lung cancer cells show 50% acceleration in their rate of division if nicotine is added to the culture medium. Nicotine also inhibits the “suicide” of malignant cells, one of the main repair and clean-up mechanisms of the body in case of cell damage. Thus, nicotine actually helps cells with genetic damage to
A
H3C
A
š N A H3C
H /
4-(N-Methyl-N-nitrosamino)1-(3-pyridyl)-1-butanone
CH3
NO
A
H /
Nicotine appears to play a dual contributory role, because its metabolites are outright carcinogens and because the parent system itself, while not causing cancer, is a tumor promoter. The metabolic pathway has as the initial step the N-nitrosation of the azacyclopentane (pyrrolidine) nitrogen. Oxidation and ring opening (compare Chemical Highlight 21-3) then take place, giving a mixture of two N-nitrosodialkanamines (N-nitrosamines), each of which is a known powerful carcinogen. š N
H
O N 4-(N-Methyl-N-nitrosamino)4-(3-pyridyl)butanal
ðš OH E ðN B Nu O CH3 E Nð R
Nicotine patches are used to wean individuals from their addiction to cigarette smoking.
Strategy Following the hint, Exercise 9-25 points out that a 2-chloroethylsulfide can undergo intramolecular SN2 reaction to an intermediate sulfonium ion, which is subject to opening by external nucleophilic attack. Recognizing that the “action” part of cylindricine A is a 2-chloroethylamine substructure, we can apply the same principle here. Solution • Intramolecular attack by the amino nitrogen on the chlorine-bearing carbon produces reactive intermediate ammonium ion A (next page). • The activated azacyclopropane ring may be attacked at either of its two carbons by the chloride ion liberated in the first step. • Ring opening via the primary carbon regenerates starting material (a). • However, attack on the neighboring position (b) unravels A to cylindricine B. • Build a model!
25-2 Nonaromatic Heterocycles
survive and, presumably, proliferate. This finding has caused concern, because nicotine is the crucial ingredient in a number of smoking-cessation therapies that feature gums, patches, lozenges, and inhalators. There are about 1.3 billion smokers worldwide, a staggering number, to which the United States contributes 60 million. Among them, the majority wants to and has tried to quit. Because tobacco-related illnesses are the second leading cause of death (after heart disease), there is a tremendous need for new therapies. A new concept in this vein is the
Chapter 25
new drug varenicline (Chantix), which came on the market in 2006. It is a nicotine receptor “partial agonist,” that is, this molecule targets the same receptors site of the central nervous system as nicotine, but with only partial efficacy. As such, it both reduces the craving for “a smoke” and decreases the pleasure associated with it, but not completely, thus minimizing the chances of a relapse. The synthesis of varenicline is a showcase of reactions that you have encountered in this book, as practiced by medicinal chemists.
Synthesis of Varenicline Br
Br
1. O3 2. H2, Pt
Li LiBr
Section 22-4
Br
Section 14-8
Section 12-12
Li
CHO C
C6H5CH2NH2, H, H2, Pt
H2, Pd
NCH2C6H5
Section 21-6
NH
Section 22-2
O O B B CF3COCCF3
Section 20-3
¥ CHO O B NCCF3
HNO3, H
O B NCCF3
O2N
Section 16-4
O B NCCF3
H2N H2, Pd Section 16-5
O2N
OO BB HCCH
H2N
N
N Na2CO3, H2O
O B NCCF3
Section 17-9
N
Section 20-6
NH
N Varenicline
O
O
O N R H
N
a
R Cl
b
Cl
N H
R
H Cl Cylindricine A
A
1171
Cylindricine B
1172
Chapter 25
Heterocycles
Exercise 25-5
Try It Yourself Treatment of thiacyclobutane with chlorine in CHCl3 at 2708C gives ClCH2CH2CH2SCl in 30% yield. Suggest a mechanism for this transformation. [Hint: The sulfur in sulfides is nucleophilic (Section 9-10).]
The reactivity of the four-membered heterocycloalkanes bears out expectations based on ring strain: They undergo ring opening, as do their three-membered cyclic counterparts, but more stringent reaction conditions are usually required. The reaction of oxacyclobutane with CH3NH2 is typical.
O ð
š 2 CH3NH
150°C
š š CH3NH(CH OH 2)3 45% N-Methyl-3-amino-1-propanol
The b-lactam antibiotic penicillin functions through related ring-opening processes (see Chemical Highlight 20-2). Exercise 25-6 2-Methyloxacyclobutane reacts with HCl to give two products. Write their structures.
Heterocyclopentanes and heterocyclohexanes are relatively unreactive The unstrained heterocyclopentanes and heterocyclohexanes are relatively inert. Recall that oxacyclopentane (tetrahydrofuran, THF) is used as a solvent. However, the heteroatoms in aza- and thiacycloalkanes allow for characteristic transformations (see Sections 9-10, 17-8, 18-4, and Chapter 21). In general, ring opening occurs by conversion of the heteroatom into a good leaving group. Exercise 25-7 Treatment of azacyclopentane (pyrrolidine) with sodium nitrite in acetic acid gives a liquid, b.p. 99 – 1008C (15 mm Hg), that has the composition C4H8N2O. Propose a structure for this compound. (Hint: Review Section 21-10.)
Aromatic Heterocyclopentadienes
š N A H
š O
Pyrrole
Furan
In Summary The reactivity of heterocyclopropanes and heterocyclobutanes results from the release of strain by ring opening. The five- and six-membered heterocycloalkanes are less reactive than their smaller-ring counterparts.
25-3 Structure and Properties of Aromatic Heterocyclopentadienes Pyrrole, furan, and thiophene are 1-hetero-2,4-cyclopentadienes. Each contains a butadiene unit bridged by an sp2-hybridized heteroatom bearing lone electron pairs. These systems contain delocalized p electrons in an aromatic six-electron framework. This section considers the structures and methods of preparation of these compounds.
Pyrrole, furan, and thiophene contain delocalized lone electron pairs š S Thiophene
The electronic structure of the three heterocycles pyrrole, furan, and thiophene is similar to that of the aromatic cyclopentadienyl anion (Section 15-7). The cyclopentadienyl anion may
25-3 Aromatic Heterocyclopentadienes
H
H C
sp 2 Hybridized
H C
H
H
C
C N H
X
C
C H
H p Orbital
p Orbital
Pyrrole
Furan (X = O) Thiophene (X = S)
A
B
be viewed as a butadiene bridged by a negatively charged carbon whose electron pair is delocalized over the other four carbons. The heterocyclic analogs contain a neutral atom in that place, again bearing lone electron pairs. One of these pairs is similarly delocalized, furnishing the two electrons needed to satisfy the 4n 1 2 rule (Section 15-7). To maximize overlap, the heteroatoms are hybridized sp2 (Figure 25-1), the delocalized electron pair being assigned to the remaining p orbital. In pyrrole, the sp2-hybridized nitrogen bears a hydrogen substituent in the plane of the molecule. For furan and thiophene, the second lone electron pair is placed into one of the sp2 hybrid orbitals, again in the plane and therefore with no opportunity to achieve overlap. This arrangement is much like that in the phenyl anion (Section 22-4). As a consequence of delocalization, pyrrole, furan, and thiophene exhibit properties that are typical of aromatic compounds, such as unusual stability, deshielded protons in the 1H NMR spectra due to the presence of ring currents, and the ability to undergo electrophilic aromatic substitution (Section 25-4). The delocalization of the lone pair in the 1-hetero-2,4-cyclopentadienes can be described by charge-separated resonance forms, as shown for pyrrole.
k
N H
k
k
N H
N H
N H
k
N H
Notice that there are four dipolar forms in which a positive charge is placed on the heteroatom and a negative charge successively on each of the carbons. This picture suggests that the heteroatom should be relatively electron poor and the carbons relatively electron rich. The electrostatic potential maps below confirm this expectation. Thus (on the same scale), the nitrogen in pyrrole is less electron rich (orange) than that in its saturated counterpart azacyclopentane (red), whereas the diene portion in pyrrole is more electron rich (red) than that in 1,3-cyclopentadiene (yellow).
Pyrrole
Azacyclopentane
š C A H Cyclopentadienyl anion
MODEL BUILDING
Resonance Forms of Pyrrole k
1173
Figure 25-1 Orbital pictures of (A) pyrrole and (B) furan (X 5 O), and thiophene (X 5 S). The heteroatom in each is sp2 hybridized and bears one delocalized lone electron pair.
sp 2 Hybrid orbital: perpendicular to aromatic π frame
C
H C
Chapter 25
1,3-Cyclopentadiene
Heterocycles
As this description suggests, the extent of aromaticity in these systems depends on the relative ability of the heteroatom to donate its lone electron pair, in turn described by its respective electronegativity (Table 1-2). Consequently, the aromatic character increases along the series furan , pyrrole , thiophene, a trend that is reflected in relative reactivity and stability. Aromaticity in Heterocyclopentadienes
N H
S
k k
O
k
Chapter 25
k k
1174
Increasing aromaticity
Exercise 25-8 Azacyclopentane and pyrrole are both polar molecules. However, the dipole vectors in the two molecules point in opposite directions. What is the sense of direction of this vector in each structure? Explain your answer.
Pyrroles, furans, and thiophenes are prepared from G-dicarbonyl compounds Syntheses of the heterocyclopentadienes use a variety of cyclization strategies. A general approach is the Paal-Knorr* synthesis (for pyrroles) and its variations (for the other heterocycles). The target molecule is made from an enolizable g-dicarbonyl compound that is treated with an amine derivative (for pyrroles) or P2O5 (for furans) or P2S5 (for thiophenes). Cyclization of a G-Dicarbonyl Compound to a 1-Hetero-2,4-Cyclopentadiene
R
R
RNH2, or P2O5, or P2S5
O O
H2O
R
R
X X ⴝ NRⴕ, O, S
ðOð ðOð B B CH3CCH2CH2CCH3
š 2 (CH3)2CHNH
CH3COOH, , 17 h
š N A (CH3)2CH 70%
H3 C
CH3
N-(1-Methylethyl)2,5-dimethylpyrrole
ðOð
C6H5 C6H5
P2O5, 150C
ðOð
C6H5
C6H5
ðOð
62% *Professor Karl Paal (1860 – 1935), University of Erlangen, Germany; Professor Ludwig Knorr (1859 – 1921), University of Jena, Germany.
25-4 Reactions of the Aromatic Heterocyclopentadienes
Chapter 25
O O
ðOð ðOð B B CH3CCH2CH2CCH3
P2S5, 140–150C
S
H3C
CH3 S
60% 2,5-Dimethylthiophene
Exercise 25-9 Formulate a possible mechanism for the acid-catalyzed dehydration of 2,5-hexanedione to 2,5dimethylfuran. (Hint: The crucial ring closure is accomplished by the oxygen of one carbonyl group attacking the carbon of the second. See also Section 18-2.)
Exercise 25-10 4-Methylpyrrole-2-carboxylic acid (compound B) is the trail pheromone of the ant species Atta texana. A third of a milligram has been estimated to be sufficient to mark a path around Earth, and each ant carries only 3.3 ng (1029 g). Propose a synthesis starting from 3-methylcyclobutene1-carboxylic acid (compound A). (Hint: What dione is the retrosynthetic precursor to compound B, and how can you make it from compound A?) H3C
H3C
COOH
N H
COOH
B
A
Exercise 25-11 The following equation is an example of another synthesis of pyrroles. Write a mechanism for this transformation. (Hint: Refer to Section 17-9.) O B CH3CCHCO2CH2CH3 A NH2 Ethyl 2-amino3-oxobutanoate
O B CH3CCH2CO2CH2CH3 Ethyl 3-oxobutanoate
H3C
CO2CH2CH3
CH3CH2O2C
N H
CH3
Diethyl 3,5-dimethylpyrrole2,4-dicarboxylate
In Summary Pyrrole, furan, and thiophene contain delocalized aromatic p systems analogous to that of the cyclopentadienyl anion. A general method for the preparation of 1-hetero2,4-cyclopentadienes is based on the cyclization of enolizable g-dicarbonyl compounds.
25-4 Reactions of the Aromatic Heterocyclopentadienes The reactivity of pyrrole, furan, and thiophene and their derivatives is governed largely by their aromaticity and is based on the chemistry of benzene. This section describes some of their reactions, particularly electrophilic aromatic substitution, and introduces indole, a benzofused analog of pyrrole.
1175
1176
Heterocycles
Chapter 25
Pyrroles, furans, and thiophenes undergo electrophilic aromatic substitution As expected for aromatic systems, the 1-hetero-2,4-cyclopentadienes undergo electrophilic substitution. There are two sites of possible attack, at C2 and at C3. Which one should be more reactive? An answer can be found by the same procedure used to predict the regioselectivity of electrophilic aromatic substitution of substituted benzenes (Chapter 16): enumeration of all the possible resonance forms for the two modes of reaction. Consequences of Electrophilic Attack at C2 and C3 in the Aromatic Heterocyclopentadienes Attack at C2
MECHANISM
E
X
H
X
E
H
X
H X
E
E
Strongly contributing all-octet form Three resonance forms
Attack at C3 H
H E
X
E
X
E
X
Strongly contributing all-octet form Two resonance forms
Both modes benefit from the presence of the resonance-contributing heteroatom, but attack at C2 leads to an intermediate with an additional resonance form, thus indicating this position to be the preferred site of substitution. Indeed, such selectivity is generally observed. However, because C3 also is activated to electrophilic attack, mixtures of products can form, depending on conditions, substrates, and electrophiles. REACTION
Electrophilic Aromatic Substitution of Pyrrole, Furan, and Thiophene NO2
O B CH3CONO2, –10C
N H
CH3COOH
NO2
N H 50%
N H 13%
2-Nitropyrrole
3-Nitropyrrole O B CH3CCl, SnCl4
Cl – Cl, CH2Cl2
O
HCl
Cl O 64% 2-Chlorofuran
H3C
S
HCl
H3C
S 64%
CCH3 B O
2-Acetyl5-methylthiophene
25-4 Reactions of the Aromatic Heterocyclopentadienes
Chapter 25
1177
The relative reactivity of benzene and the three heterocycles in electrophilic substitutions is the result of contributions from the aromaticity of the respective rings and the stabilization of the intermediate cations. It increases in the order benzene ! thiophene , furan , pyrrole. Exercise 25-12
Working with the Concepts: Predicting the Position of Electrophilic Aromatic Substitution of Substituted Thiophene The monobromination of thiophene-3-carboxylic acid gives only one product. What is its structure, and why is it the only product formed? Strategy The two preferred sites of reaction should be at the carbon atoms next to sulfur, in our unsymmetrical thiophene C2 and C5. To see which one of these is more active, we need to compare the respective resonance forms of the intermediate cation resulting from electrophilic attack at either position. Solution • Attack at C2 COOH
COOH Br2
H
Br
š S
COOH
COOH
H
H
š S
Br
š S
Br
S
Br
A
• Attack at C5 COOH
COOH
COOH
COOH
Br2 Br
š S
Br
Br H
š S
H
š S
Br H
S
• Result: Attack on C5 avoids placing the positive charge on C3, as in A, bearing the electron-withdrawing carboxy function. Therefore, the only product is 5-bromo-3-thiophenecarboxylic acid.
Exercise 25-13
Try It Yourself The monobromination of 3-methylfuran gives only one product. What is its structure, and why is it the only product formed?
Pyrrole is extremely nonbasic compared to ordinary amines (Section 21-4), because the lone electron pair on nitrogen is tied up by conjugation. Very strong acid is required to effect protonation, and it takes place on C2, not on the nitrogen. The Protonation of Pyrrole Occurs on Carbon N H
H
N H
H
pKa 4.4
H
Pyrroles are not only very nonbasic, they are, in fact, relatively acidic. Thus, while azacyclopentane has a pKa 5 35 (normal for an amine, Section 21-4), the corresponding value for pyrrole is 16.5! The reasons for this increase in acidity are the change in hybridization from sp3 to sp2 (see Section 11-3) and the delocalization of the negative charge (as in cyclopentadienyl anion, Section 15-7).
Chapter 25
Heterocycles
Pyrrole Is Relatively Acidic
sp3
H pKa 35 Azacyclopentane
š N
š N
sp2
N
N
H
š N
Base
N
N
Five resonance forms
pKa 16.5
Exercise 25-14 Explain why pyrrole is protonated at C2 rather than on the nitrogen.
1-Hetero-2,4-cyclopentadienes can undergo ring opening and cycloaddition reactions Furans can be hydrolyzed under mild conditions to g-dicarbonyl compounds. The reaction may be viewed as the reverse of the Paal-Knorr-type synthesis of furans. Pyrrole polymerizes under these reaction conditions, whereas thiophene is stable. Hydrolysis of a Furan to a G-Dicarbonyl Compound
H3 C
O
CH3
CH3COOH, H2SO4, H2O,
O O B B CH3CCH2CH2CCH3 90% 2,5-Hexanedione
Raney nickel desulfurization (Section 17-8) of thiophene derivatives results in sulfurfree acyclic saturated compounds.
S
CH(OCH2CH3)2
Raney Ni, (CH3CH2)2O,
CH3(CH2)3CH(OCH2CH3)2 50%
NiS
Being least aromatic (Section 25-3), the p system of furan (but not of pyrrole or thiophene) possesses sufficient diene character to undergo Diels-Alder cycloadditions (Section 14-8). O O O
NH
(CH3CH2)2O, 25C
[H O [H NH
O
H2N H ? H2C OC i CO2H
O
O O
N
O 95%
Indole is a benzopyrrole
∑
1178
N H Tryptophan
Indole is the most important benzannulated (fused-ring) derivative of the 1-hetero-2,4cyclopentadienes. It forms part of many natural products, including the amino acid tryptophan (Section 26-1). Indole is related to pyrrole in the same way that naphthalene is related to benzene. Its electronic makeup is indicated by the various possible resonance forms that can be formulated for the molecule. Although those resonance forms that disturb the cyclic six-p-electron
25-5 Structure and Preparation of Pyridine: An Azabenzene
Chapter 25
system of the fused benzene ring are less important, they indicate the electron-donating effect of the heteroatom. Resonance in Indole 4
3
6 7
5
2 N1 i H
N i H
N i H
etc.
N i H
Exercise 25-15 Predict the preferred site of electrophilic aromatic substitution in indole. Explain your choice.
Exercise 25-16 Irradiation of compound A in ethoxyethane (diethyl ether) at 21008C generates the enol form, B, of acetylbenzene and a new compound, C, which isomerizes to indole upon warming to room temperature.
N
O
A
h , 100C
OH f H2CP C
C
B
The 1H NMR spectrum of compound C shows signals at d 5 3.79 (d, 2 H) and 8.40 (t, 1 H) ppm in addition to four aromatic absorptions. Indole has peaks at d 5 6.34 (d, 1 H), 6.54 (broad d, 1 H), and 7.00 (broad s, 1 H) ppm. What is compound C? (Hint: This reaction proceeds by a mechanism similar to the mass spectral McLafferty rearrangement, Section 17-3.)
In Summary The donation of the lone electron pair on the heteroatom to the diene unit in pyrrole, furan, and thiophene makes the carbon atoms in these systems electron rich and therefore more susceptible to electrophilic aromatic substitution than the carbons in benzene. Electrophilic attack is frequently favored at C2, but substitution at C3 is also observed, depending on conditions, substrates, and electrophiles. Some rings can be opened by hydrolysis or by desulfurization (for thiophenes). The diene unit in furan is reactive enough to undergo DielsAlder cycloadditions. Indole is a benzopyrrole containing a delocalized p system.
25-5 Structure and Preparation of Pyridine: An Azabenzene Pyridine can be regarded as a benzene derivative — an azabenzene — in which an sp2hybridized nitrogen atom replaces a CH unit. The pyridine ring is therefore aromatic, but its electronic structure is perturbed by the presence of the electronegative nitrogen atom. This section describes the structure, spectroscopy, and preparation of this simple azabenzene.
Pyridine is aromatic Pyridine contains an sp2-hybridized nitrogen atom like that in an imine (Section 17-9). In contrast to pyrrole, only one electron in the p orbital completes the aromatic p-electron arrangement of the aromatic ring; as in the phenyl anion, the lone electron pair is located
sp2
N Pyridine
1179
1180
Chapter 25
Heterocycles
Figure 25-2 (A) Orbital picture of pyridine. The lone electron pair on nitrogen is in an sp2-hybridized orbital and is not part of the aromatic p system. (B) The electrostatic potential map of pyridine reveals the location of the lone electron pair on nitrogen (red) in the molecular plane and the electron-withdrawing effect of the electronegative nitrogen on the aromatic p system (green; compare to the electrostatic potential map of pyrrole in Section 25-3).
p Orbital
H
H
Perpendicular to aromatic π frame
N
H H
H
sp 2 Orbital
A
B
in one of the sp2 hybrid atomic orbitals in the molecular plane (Figure 25-2). Therefore, in pyridine, the nitrogen does not donate excess electron density to the remainder of the molecule. Quite the contrary: Because nitrogen is more electronegative than carbon (Table 1-2), it withdraws electron density from the ring, both inductively and by resonance. Resonance in Pyridine
N
N
N
N
N
Exercise 25-17 Azacyclohexane (piperidine) is polarized because of the presence of the relatively electronegative nitrogen. The polarization in pyridine is twice as large. Explain.
Aromatic delocalization in pyridine is evident in the 1H NMR spectrum, which reveals the presence of a ring current. The electron-withdrawing capability of the nitrogen is manifest in larger chemical shifts (more deshielding) at C2 and C4, as expected from the resonance picture. 1
H NMR Chemical Shifts (ppm) in Pyridine and Benzene 7.46
H H
7.06
H
8.50
Pyridine Is a Weak Base N
H
H 7.27
N
N A H Pyridinium ion pKa 5.29
Because the lone pair on nitrogen is not tied up by conjugation (as it is in pyrrole, Exercise 25-14), pyridine is a weak base. (It is used as such in numerous organic transformations.) Compared with alkanamines (pKa of ammonium salts < 10; Section 21-4), the pyridinium ion has a low pKa, because the nitrogen is sp2 and not sp3 hybridized (see Section 11-3 for the effect of hybridization on acidity). Pyridine is the simplest azabenzene. Some of its higher aza analogs are shown here. They behave like pyridine but show the increasing effect of aza substitution — in particular, increasing electron deficiency. Minute quantities of several 1,4-diazabenzene (pyrazine) derivatives are responsible for the characteristic odors of many vegetables. One drop of 2-methoxy-3-(1-methylethyl)-1,4-diazabenzene (2-isopropyl-3-methoxypyrazine) in a large
25-5 Structure and Preparation of Pyridine: An Azabenzene
Chapter 25
swimming pool would be more than adequate to give the entire pool the odor of raw potatoes.
1,3-Diazabenzene (Pyrimidine)
1,4-Diazabenzene (Pyrazine)
ð
š N
N
N
1,3,5-Triazabenzene (1,3,5-Triazine)
N
N
ð
ð
N
N
ð
ð
N
ð
N
ð
1,2-Diazabenzene (Pyridazine)
N
1,2,4,5-Tetraazabenzene (1,2,4,5-Tetrazine)
N
N
1,2,3-Triazabenzene (1,2,3-Triazine)
š N
CH(CH3)2
N
OCH3
N
ð
ð
N
N
š N
N
ð
ð
š N
N
1,2,4-Triazabenzene (1,2,4-Triazine)
š N
CH2CH(CH3)2
N
OCH3
2-Methoxy-3-(1-methylethyl)1,4-diazabenzene
2-Methoxy-3-(2-methylpropyl)1,4-diazabenzene
(Potatoes)
(Green peppers)
Pyridines are made by condensation reactions Pyridine and simple alkylpyridines are obtained from coal tar. Many of the more highly substituted pyridines are in turn made by both electrophilic and nucleophilic substitution of the simpler derivatives. Pyridines can be made by condensation reactions of acyclic starting materials such as carbonyl compounds with ammonia. The most general of these methods is the Hantzsch* pyridine synthesis. In this reaction, two molecules of a b-dicarbonyl compound, an aldehyde, and ammonia combine in several steps (Chapter Integration Problem 25-28) to give a substituted dihydropyridine, which is readily oxidized by nitric acid to the aromatic system. When the b-dicarbonyl compound is a 3-ketoester, the resulting product is a 3,5pyridinedicarboxylic ester. Hydrolysis followed by pyrolysis of the calcium salt of the acid causes decarboxylation. Hantzsch Synthesis of 2,6-Dimethylpyridine CH2 B ðOð
H CO2CH2CH3
A
A
CH3CH2O2C
ð
CH3CH2O2C
3 H2O
H3C
ð
ð
A
A
ð
CH2 H2 C A A C C M MO CH3 H 3C O ðNH3
H CO2CH2CH3
š N H 89%
CH3
Diethyl 1,4-dihydro-2,6-dimethyl3,5-pyridinedicarboxylate
H CH3CH2O2C
H CO2CH2CH3
1. KOH, H2O 2. CaO, 2 CH3CH2OH, 2 CaCO3
H3C
N š 65%
CH3
Diethyl 2,6-dimethyl-3,5pyridinedicarboxylate
*Professor Arthur R. Hantzsch (1857 – 1935), University of Leipzig, Germany.
H3C
N š 65%
CH3
2,6-Dimethylpyridine
HNO3, H2SO4
1181
1182
Heterocycles
Chapter 25
The first step of the Hantzsch pyridine synthesis is an example of a four-component reaction: Four molecules combine in a specific fashion to form a single product. This is remarkable considering the number of possible condensation pathways that are available to the starting materials. The reason for its success is the reversibility of channels that do not lead to the observed dihydropyridine and the careful tuning of reactivity of the participating molecules: ammonia (or ammonium acetate); a relatively reactive aldehyde; and a b-dicarbonyl constituent that is used twice. (For a stepwise tour through the mechanism, see Chapter Integration Problem 25-28). Multicomponent reactions of this type, in which water is the only by-product, are by their very nature atom economical and hence “green” (Chemical Highlight 3-1). Even more environmentally friendly is the use of water as a solvent and, in the aromatization step to the substituted pyridine, simply oxygen in the presence of activated carbon (a form of highly porous carbon derived from charcoal). A “Super Green” Hantzsch Pyridine Synthesis CH3CHO O O B B 2 CH3CCH2COCH2CH3 O B NH OCCH 4 3
O H2O
CH3
O
O
O O2, C
O
80%
N H
O
O
93%
H3C
CH3
CH3
O
H3C
N
CH3
Exercise 25-18 What starting materials would you use in the Hantzsch synthesis of the following pyridines? CH3
C(CH3)3 NO2
(a) CH3CH2O2C
CN
NC (b)
(c) H 3C
H3C
N
CN N
CH3
CH3CH2
N
CH2CH3
CH3
Exercise 25-19
Working with the Concepts: Practicing Mechanisms of Pyridine Syntheses The Hantzsch synthesis of pyridines features 1,4-dihydropyridines in the first step. A variant of the method uses hydroxylamine (Table 17-5), which can be regarded as an oxidized version of ammonia. With this reagent, pyridines are formed directly from 1,5-dicarbonyl compounds, in turn readily made by Michael additions of enolates to a,b-unsaturated aldehydes and ketones (Section 18-11). Formulate the mechanisms of the two steps in the pyridine synthesis shown. NaOCH2CH3, CH3CH2OH
O
O
O O
HONH2, HCl, CH3CH2OH
N
25-5 Structure and Preparation of Pyridine: An Azabenzene
Chapter 25
1183
Strategy Looking at the first step of the reaction, you should recognize a Michael addition of cyclohexanone to an unsaturated ketone. (You can review this reaction in Section 18-11). The transformation in the second step looks similar to the Paal-Knorr synthesis of pyrroles (Exercise 25-9), except that the result is the construction of a six-membered (not a five-membered) ring. The mechanism you are looking for should be based on the mechanisms for these two reactions. Solution • In the first step, the product is formed by C–C bond construction between the carbon next to the carbonyl in cyclohexanone and the terminal position of the enone; in other words, cyclohexanone is alkylated at the a-position. Remember that ketones are alkylated through their enolates (Section 18-4) and, in this case, alkylation occurs by attack of cyclohexanone enolate on the positively polarized enone b-carbon, as sketched below. ␦
H
␦
ðB BH
Oð
, bh
O ðB
š Oð
O
O
)
• For the second step, recall that primary amines react with ketones reversibly to furnish imines by loss of water (Section 17-9) and that, in fact, hydroxylamine makes oximes (Table 17-5). These condensations proceed through the intermediacy of hemiaminals by nucleophilic attack of the amine nitrogen on the carbonyl carbon.
H2NOH
)
NH 2
C6H5
)O
)
))
O) )O
)
O)
C6H5
G HO
C6H5
H2O
N A HO
)
)
)O
NH
G HO Hemiaminal
)O
C6H5
)
OH
)
))
H
Oxime
• Imines (like carbonyl compounds that form enols) are in equilibrium with their tautomeric forms, enamines (Section 17-9). If we write this form for our oxime, we arrive at an aminocarbonyl compound that is poised to undergo fast intramolecular formation of another hemiaminal. Dehydration produces a new cyclic enamine, which, upon inspection, is nothing but a hydrated pyridine. Finally, the driving force of aromaticity facilitates the rapid loss of water and generation of the pyridine product.
Tautomerism
HO
)
Oxime
Intramolecular hemiaminal
Enamine
H
H
)
H2O
C6H5
N
)
N A OH
C6H5 N A )OH OH
)
HO
H2O
C6H5
)
)
NH )O
)
)O
)
N
C6H5
C6H5
1184
Heterocycles
Chapter 25
Exercise 25-20
Try It Yourself Sketch a plausible mechanism by which the following condensation reaction occurs. [Hints: Start with an aldol condensation of the aldehyde with the enamine. (Review enamine alkylation, Section 18-4). Then use the resulting product as a Michael acceptor of the second enamine.] C(CH3)3 H3C
CN
N H
O
N H
H3C
OH
O
CH3
O
N
HN N H
CN
1,3-Diazabenzene-2,4,6-triol prefers the triketo tautomeric form (Section 22-3) by about 29 kcal mol21. It is commonly known as barbituric acid (pKa 5 7.4) and constitutes the basic frame of a group of sedatives and sleep inducers called barbiturates, of which veronal and phenobarbital are two examples (margin).
Veronal
O
NC H
Exercise 25-21
HN O
H O
H
H2N
O
(H3C)3C
2
HO
O
Tautomerism
N
OH
HN O
N H
O
Barbituric acid
Phenobarbital
Propose a synthesis of veronal from diethyl propanedioate (malonic ester; Section 23-2) and urea (Section 20-6). (Hint: In the presence of a base, such as an alkoxide, amides are in equilibrium with the corresponding amidates; Section 20-7.)
In Summary Pyridines are aromatic but electron poor. The lone pair on nitrogen makes the heterocycle weakly basic. Pyridines are prepared by condensation of a b-dicarbonyl compound with ammonia and an aldehyde.
25-6 Reactions of Pyridine The reactivity of pyridine derives from its dual nature as both an aromatic molecule and a cyclic imine. Both electrophilic and nucleophilic substitution processes may occur, leading to a variety of substituted derivatives.
Pyridine undergoes electrophilic aromatic substitution only under extreme conditions Because the pyridine ring is electron poor, the system undergoes electrophilic aromatic substitution only with great difficulty, several orders of magnitude more slowly than benzene, and at C3 (see Section 15-8). Electrophilic Aromatic Substitution of Pyridine NO2
NaNO3, fuming H2SO4, 300C
Br Br–Br, H2SO4, SO3
H2O
N
HBr
N
4.5%
3-Nitropyridine
N
N
86%
3-Bromopyridine
25-6 Reactions of Pyridine
1185
Chapter 25
Exercise 25-22 Explain why electrophilic aromatic substitution of pyridine occurs at C3.
Activating substituents allow for milder conditions or improved yields. NO2
KNO3, fuming H2SO4, 100C H2O
N
H3C
CH3
N 81%
H3C
2,6-Dimethylpyridine
CH3
2,6-Dimethyl-3-nitropyridine
Br Br–Br, CH3COOH, 20C HBr
N
N 90%
NH2
2-Aminopyridine
NH2
2-Amino-5-bromopyridine
Pyridine undergoes relatively easy nucleophilic substitution Because the pyridine ring is relatively electron deficient, it undergoes nucleophilic substitution much more readily than does benzene (Section 22-4). Attack at C2 and C4 is preferred because it leads to intermediates in which the negative charge is on the nitrogen. An example of nucleophilic substitution of pyridine is the Chichibabin* reaction, in which the heterocycle is converted into 2-aminopyridine by treatment with sodium amide in liquid ammonia. Chichibabin Reaction
ðNH 2
1. NaNH2, liquid NH3 2. H, H2O
N
N
H
Mechanism of the Chichibabin Reaction
H OH
N
H
NH2 Addition
70% 2-Aminopyridine
This reaction proceeds by the addition – elimination mechanism. The first step is attack by 2: NH2 at C2, a process that resembles 1,2-addition to an imine function. Expulsion of a hydride ion, H :2, from C2 is followed by deprotonation of the amine nitrogen to give H2 and a resonance-stabilized 2-pyridineamide ion. Protonation by aqueous work-up furnishes the final product. Note the contrast with electrophilic substitutions, which include proton loss, not expulsion of hydride as a leaving group. Transformations related to the Chichibabin reaction take place when pyridines are treated with Grignard or organolithium reagents.
N
H
LiH
Li
N 49% 2-Phenylpyridine
*Professor Alexei E. Chichibabin (1871 – 1945), University of Moscow, Russia.
NH2 H
ðH Elimination
N
Methylbenzene (toluene), 110C, 8 h
N
NH2
1186
Heterocycles
Chapter 25
CHE M I C AL H I G H L I G H T 2 5 - 2 Pyridinium Salts in Nature: Nicotinamide Adenine Dinucleotide, Dihydropyridines, and Organocatalysis NH2 O B CNH2
Reactive site
C
Nicotinamide
N
C
HC
C
H
H
H Pyrophosphate
Ribose
H
Ribose
H
H HO
OH
N
O
H
H HO
O PP A O
OO CH2
N
O
Adenine
CH N
OO CH2
N
OH
P PO A O
O Nicotinamide adenine dinucleotide
A complex pyridinium derivative, nicotinamide adenine dinucleotide (NAD1) is an important biological oxidizing agent. The structure consists of a pyridine ring [derived from 3-pyridinecarboxylic (nicotinic) acid], two ribose molecules (Section 24-1) linked by a pyrophosphate bridge, and the heterocycle adenine (Section 26-9). Most organisms derive their energy from the oxidation (removal of electrons) of fuel molecules, such as glucose or fatty acids; the ultimate oxidant (electron acceptor) is oxygen, which gives water. Such biological oxidations proceed through a cascade of electron-transfer reactions requiring the intermediacy of special redox reagents. NAD1 is one such molecule. In the oxidation of a substrate, the pyridinium ring in NAD1 undergoes a two-electron reduction with simultaneous protonation. NAD1 is the electron acceptor in many enzymatic oxidations of alcohols to aldehydes (including the conversion of vitamin A into retinal, Chemical Highlight 18-3; see also Chemical Highlight 8-1). This reaction can be seen as a transfer of hydride from C1 of the alcohol to C4 of the pyridinium nucleus with simultaneous deprotonation to give
Reduction of NADⴙ H
H
H CONH2
CONH2
H
2
e š N A R
N A R NADⴙ
NADH
the aldehyde and the dihydropyridine, NADH. With other enzymes, the reverse is achieved, namely, the reduction of aldehydes and ketones to alcohols with NADH (see Chapter 8, Problems 58 and 59, and Section 22-9). The “action” part of NADH constitutes a simple dihydropyridine of the type readily accessible in the first step of the Hantzsch pyridine synthesis (Section 25-5). Therefore chemists have probed whether such compounds can be used as metal-free alternatives to hydride reagents, such as
In most nucleophilic substitutions of pyridines, halides are leaving groups, the 2- and 4-halopyridines being particularly reactive. OCH3
Cl NaOCH3, CH3OH NaCl
N
N 75% 4-Methoxypyridine
25-6 Reactions of Pyridine
LiAlH4, in carbonyl reductions, or as surrogates for catalytically activated (e.g., Pd or Pt) hydrogen in hydrogenations. Indeed, such so-called biomimetic reactions (because they mimic the principles applied by nature) have been achieved
Chapter 25
1187
with “Hantzsch esters” to reduce 2-oxoesters to the corresponding alcohols. Similarly, conjugate hydride addition to a,b-unsaturated aldehydes, followed by protonation provides the saturated aldehydes.
Hantzsch Esters in Reductions
O
H OCH3
O
OH
H
H3CH2CO2C
CO2CH2CH3
O
90%
N H
H3C
H
CH3
H
H3CH2CO2C
CO2CH2CH3
cat. R2NH2
77%
H3C
O
H3C
OCH3
Cu2
H
N H
CH3
H
The first process is not quite metal free, as it requires catalytic amounts of Lewis acidic Cu21 to activate the carbonyl function. This has the advantage that in the presence of chiral ligands (see Section 12-2 and Chemical Highlights 5-4 and 12-2) to the Cu21, the hydride adds enantioselectively from only one side of the molecule to generate only one enantiomer of the alcohol, much as nature does it. In the second process, similar activation takes place using catalytic amounts of an ammonium salt, which converts intermittently the aldehyde function to the corresponding
iminium ion, in which the positive charge serves to render the b-carbon a better hydride acceptor. Traditionally, such activations are performed with H1 or Lewis acids, and hydrogenation of alkenes normally uses heterogeneous metal catalysts (Section 12-2). To contrast this and related transformations from their metal-catalyzed variants, they have been termed “organocatalytic.” Because they provide great selectively and avoid potentially toxic metal ingredients, they also fall into the category of “green” chemistry.
The NADⴙ–NADH Redox Couple H A RC O O OH A H
H 4
N A R NADⴙ
CONH2
O
H3C
Enzyme
O B RCH
H 4
CONH2 H
š N A R NADH
Exercise 25-23 Propose a mechanism for the reaction of 4-chloropyridine with methoxide. [Hint: Think of the pyridine ring as containing an a,b-unsaturated imine function (see Sections 17-9 and 18-9).]
Exercise 25-24 The relative rates of the reactions of 2-, 3-, and 4-chloropyridine with sodium methoxide in methanol are 3000 : 1 : 81,000. Explain.
1188
Heterocycles
Chapter 25
In Summary Pyridine undergoes slow electrophilic aromatic substitution preferentially at C3. Nucleophilic substitution reactions occur more readily to expel hydride or another leaving group from either C2 or C4.
25-7 Quinoline and Isoquinoline: The Benzopyridines We can imagine the fusion of a benzene ring to pyridine in either of two ways, giving us quinoline and isoquinoline (1- and 2-azanaphthalene, according to our systematic nomenclature). Both are liquids with high boiling points. Many of their derivatives are found in nature or have been synthesized in the search for physiological activity. Like pyridine, quinoline and isoquinoline are readily available from coal tar. As might be expected, because pyridine is electron poor compared with benzene, electrophilic substitutions on quinoline and isoquinoline take place at the benzene ring. As with naphthalene, substitution at the carbons next to the ring fusion predominates. NO2
N Quinoline
H2SO4, SO3, fuming HNO3, 15–20C, 5 h
N
H2O
N
N
Isoquinoline
NO2 43%
35% Quinoline
N
5-Nitroquinoline
8-Nitroquinoline
CHE M I C AL H I G H L I G H T 2 5 - 3 Folic Acid, Vitamin D, Cholesterol, and the Color of Your Skin Folic acid is essential to human life. Its structure incorporates a 1,3,5,8-tetraazanaphthalene (pteridine) ring system, together with 4-aminobenzoic acid (Section 15-4) and (S)aminopentanedioic (glutamic) acid (Section 26-1). Folic acid is necessary for the proper development of the nervous system in the very early stages of pregnancy. One of its functions is the transfer of one-carbon fragments between
H2N
N
O
B
N O
H
A
N
N
B
CH2 ON
H
A
CH2CH2COH
A
≥
C ON O CHCOH
B
OH 1,3,5,8-Tetraazanaphthalene part
biomolecules (Chapter 9, Problems 69 and 70). A deficiency of this substance, which must be obtained from the diet, is associated with crippling and often fatal birth defects such as spina bifida (“open spine”) and anencephaly (a failure of the brain to develop normally). The U.S. Public Health Service recommends that all women of child-bearing age take 400 mg (0.4 mg) of folic acid daily.
H
O 4-Aminobenzoic acid part
(S)-2-Aminopentanedioic (glutamic) acid part
Folic acid
∑
HO
Vitamin D2
Vitamin D is also an essential nutrient for good health. It supports the growth of healthy bones in children, and its lack leads to a deforming condition known as rickets. Throughout our lives, vitamin D plays a crucial role in maintaining proper levels of calcium and phosphorus in the body. Unlike folic acid, your body can manufacture
vitamin D. The starting material is cholesterol (Chemical Highlight 4-2), and the other indispensable ingredient is sunlight — specifically, the ultraviolet B radiation (the range from 280 to 315 nm) that falls on the Earth’s surface when the sun is high in the sky. From these facts derives a truly remarkable story.
25-7 Quinoline and Isoquinoline: The Benzopyridines
Chapter 25
1189
NO2 H2SO4, HNO3, 0C, 30 min H2O
N
N
ð
ð
ð
N
NO2 8%
72% Isoquinoline
5-Nitroisoquinoline
8-Nitroisoquinoline
In contrast with electrophiles, nucleophiles prefer reaction at the electron-poor pyridine nucleus. These reactions are quite analogous to those with pyridine.
The Peruvian fire stick secretes quinoline from glands behind the head as a chemical defense against predators, such as frogs, cockroaches, spiders, and ants.
Chichibabin Reaction of Quinoline and Isoquinoline 1. NaNH2, liquid NH3, 20C, 20 days 2. H, H2O
N
N 80%
NH2
2-Aminoquinoline
The shades of human skin range from nearly jet black in some tropical populations to the palest of skin tones in redheads from northern Europe (the United Kingdom in particular). How and why is this the case? The extended aromatic p system in the tetraazanaphthalene ring of folic acid absorbs ultraviolet light strongly (Section 15-5) and undergoes subsequent structural changes; in short, sunlight, which is necessary for the production of vitamin D, destroys the body’s stores of folic acid. The variations in human skin color around Earth represent the response of human evolution to the varying amounts of UV light to which humans are exposed. At the evolutionary dawn of humanity, as the amount of protective hair covering the body diminished, individuals who possessed greater amounts of skin-darkening pigments had a higher likelihood of giving birth to healthy babies. Their slightly darker skin provided some protection for the folic acid in their bodies, while still allowing the synthesis of adequate amounts of vitamin D. What about lighter-skinned populations? Assuming the accuracy of the “out of Africa” hypothesis, humans moving northward to latitudes with less direct sunlight would have faced vitamin D deficiencies as a function of the darkness of their skin. In this situation, individuals with slightly lighter skin would have had the advantage, their increased vitamin D production increasing the odds of reaching reproductive age. Over the generations of northward human migration, the lightening of the skin would have reached the present balance that allows both the preservation of necessary
amounts of folic acid and the synthesis of adequate quantities of vitamin D. How do the darkest-skinned individuals synthesize enough vitamin D, when their skin permits only very little UV light to penetrate? An adaptation developed in this population, which favored individuals who produced higher levels of cholesterol, the precursor of vitamin D, in their blood. We see the consequences of this modification in the higher incidence of cholesterol-related heart disease and stroke in this group.
The diversity of the interplay between folic acid and vitamin D production: Thuram (top), Henry, Lama, and Petit of the French national soccer team on the way to winning the World Cup in 1998.
1190
Heterocycles
Chapter 25
COOH
COOH 1. KNH2, liquid NH3 2. CH3COOH
N
ð
ð
N
ðNH2 71% 1-Aminoisoquinoline4-carboxylic acid
Exercise 25-25 Quinoline and isoquinoline react with organometallic reagents exactly as pyridine does (Section 25-6). Give the products of their reaction with 2-propenylmagnesium bromide (allylmagnesium bromide).
ð
ð
The following structures are representative of higher aza analogs of naphthalene. N
N
N
ð
N
ð
N
1,2-Diazanaphthalene (Cinnoline)
2,3-Diazanaphthalene (Phthalazine)
š N
1,3-Diazanaphthalene (Quinazoline)
š N
N
ð
N
N 1,4-Diazanaphthalene (Quinoxaline)
š N
ð
š N
N
1,3,8-Triazanaphthalene (Pyrido[2,3-d]pyrimidine)
N
N
1,3,5,8-Tetraazanaphthalene (Pteridine)
Exercise 25-26
Working with the Concepts: Recognizing Retrosynthetic Disconnections in the Azanaphthalenes As described in Section 25-5, some pyridine syntheses are based on a variety of condensation reactions, typically aldol (Section 18-5) and imine condensations (Section 17-9). Bearing this in mind, how would you dissect the pyridine ring of quinoline A retrosynthetically to two appropriate starting materials for its construction? CH3 CH3 N
CH3
A
Strategy The most obvious disconnection is the imine linkage, which could arise from the corresponding amine and carbonyl component. The latter reveals an a,b-unsaturated ketone that might be assembled by an aldol condensation. Solution • First, we disconnect the imine double bond (step 1). This produces a benzenamine on one side and a carbonyl group on the other. • Second, we apply a retroaldol step to the remaining double bond (step 2) to generate 2-acetylbenzeneamine and 2,4-pentanedione. • Are these two materials suitable starting materials for the synthesis of A? The answer is yes: This reaction is an example of the so-called Friedländer synthesis of quinolines, and it proceeds with either acid or base catalysis. While one can envisage several competing condensations of the starting materials, they are reversible and there is considerable driving force toward the aromatic cyclization product. Moreover, the aminoketone component contains the two functions in conjugation, thus preventing its self-condensation (write the dipolar ammonium enolate resonance form). CH3 CH3 O CH3 O
ð
CH3 N
CH3
CH3
Step 1
H2N O
CH3
Step 2
O NH2
CH3
+ O
CH3
25-8 Alkaloids: Nitrogen Heterocycles in Nature
Exercise 25-27
Try It Yourself Sketch plausible retrosynthetic disconnections for the following molecules. [Hints: For (b), see Section 17-10; for (c), see Problem 17-43.] CH3
C6H5 N
H3C
(a)
N
(b)
(c)
N
H3C
N
N
C6H5
In Summary The azanaphthalenes quinoline and isoquinoline may be regarded as benzopyridines. Electrophiles attack the benzene ring, nucleophiles attack the pyridine ring.
25-8 Alkaloids: Physiologically Potent Nitrogen Heterocycles in Nature The alkaloids are bitter-tasting, natural nitrogen-containing compounds found particularly in plants. The name is derived from their characteristic basic properties (alkali-like), which are induced by the lone electron pair of nitrogen. As with acyclic amines (Chapter 21), the (Lewis) basic nature of the alkaloids, in conjunction with their particular three-dimensional architecture, gives rise to often potent physiological activity. We have already noted some examples of this behavior in the narcotics morphine and heroin (Section 9-11), the psychoactive lysergic acid and LSD (Section 19-13), and the antibiotic penicillins (Chemical Highlight 20-2).
X
O B C E
/
RO
N
ECH3
/
H
/ /
∑
/
N
N H
Morphine (R ⴝ H) O B Heroin (R ⴝ CH3C)
H ´/ S
O
Lysergic acid (X ⴝ OH) Lysergic acid diethylamide, LSD [X ⴝ (CH3CH2)2N]
∑
RO H
H
NCH3
!
O B H RCNH /´
O
CH ∑ ! 3 CH3
H CO2H
Penicillin
Nicotine (see also Chemical Highlight 25-1), present in dried tobacco leaves in 2 – 8% concentration, is the stimulating ingredient in cigarettes and other tobacco products. O
H /
N
N A H3C
Nicotine
O
N N A CH3
N
Caffeine (R ⴝ CH3) Theobromine (R ⴝ H)
H3CH
N
CO CH $ H2 3
∑
RH
CH3 i N
O B ∑ ! OC H Cocaine
Chapter 25
1191
CHE M I C AL H I G H L I G H T 2 5 - 4 Nature Is Not Always Green: Natural Pesticides green potatoes (exposed to sunlight, which increases their toxin level), the drinking of herbal teas, the consumption of “poison” mushrooms, and so forth. Abraham Lincoln’s mother died from drinking milk from a cow that had grazed on the toxic snakeroot plant. The issue strikes at the heart of the argument that “organic” food, especially fruits and vegetables, is more healthy than conventional produce. Proponents advocate that organic products are better because they are grown without added synthetic pesticides, and critics claim that this very fact makes them prone to be contaminated with higher levels of bacteria and natural toxins. H3C H H3C
H ´
!
!
effective concentration of a drug to be greatly increased, reaching dangerous levels, or to be decreased, making the treatment less effective. Another example is the herbal antidepressant hypericum (St. John’s wort), which can cause abortions and also interferes with the action of the birth control pill (Chemical Highlight 4-3).
O
O
OH
HO
CH3
HO
CH3
CH3 CH3
Bergamottin
What is the purpose of these compounds in plant life? Plants cannot run away from predators and invading organisms, such as fungi, insects, animals, and humans, and they have no organs with which to defend themselves. Instead, they have developed an array of chemical weapons, “natural pesticides,” with which to mount an effective defense strategy. Tens of thousands of these chemicals are now known. They are either already present in the existing plant or are generated in a primitive “immune response” to external damage, such as by caterpillars or herbivorous insects. For example, in the tomato plant, a small polypeptide (Section 26-4) containing 18 amino acids, systemin, is the chemical alarm *Professor Bruce N. Ames (b. 1928), University of California at Berkeley.
1192
CH3
Solanine
OH
O
N
(R ⴝ sugar)
In addition to the problem of immediate potential toxicity of plant chemicals, there is a growing body of evidence that points to adverse interactions between foods and pharmaceuticals. For example, bergamottin in grapefruit juice is responsible for the “grapefruit juice effect” on the bioavailability of several prescription drugs. This can cause the
O
≥
!
≥
[ RO
Green potatoes are toxic because of the presence of the alkaloid solanine.
≥ H
CH3 H ≥ ≥ H H
≥
Many people believe that everything synthetic is somehow suspect and “bad,” and that all of nature’s chemicals are benign. As pointed out by Ames* and others, this is a misconception. While we have seen that, indeed, many manufactured chemicals have problems with toxicity and adverse effects on the environment, nature’s chemicals are not any different from synthetic ones. Nature has its own highly productive laboratory, which puts out compounds by the millions, many of which are highly toxic, such as quite a few of the alkaloids found in plants. Consequently, there are numerous (sometimes lethal) cases of poisoning (especially of children) due to the accidental ingestion of plant material, the eating of
CH3
OH
O
OH
Hypericin (Active component of hypericum)
signal for external attack. The molecule travels rapidly through the plant, initiating a cascade of reactions that produce chemical poisons. The effect is either to fend off attackers completely or to slow them down sufficiently so that other predators will consume them. Interestingly, one of these compounds is salicylic acid, the core of aspirin (Chemical Highlight 22-2), which prevents the point of damage (much like a wound) from being infected. Plants in distress have evolved to use chemicals as alarm pheromones (Section 12-17), activating the chemical weapons complex of (as yet) undamaged neighbors by air- or waterborne molecular signals. They may also develop resistance (immunity) by chemical pathways. Americans consume about 1.5 g of natural pesticides per person per day, in the form of vegetables, fruit, tea, coffee, and so forth — 10,000 times more than their intake of
Natural Plant Pesticides Compound
Plant food (concentration in ppm) Apple, carrot, celery, grapes, lettuce, potato (50 – 200); basil, dill, sage, thyme, and other herbs (.1000); coffee (roasted beans, 1800)
HO COOH HO Caffeic acid (Carcinogen)
N
C
Cabbage (35 – 590); cauliflower (12 – 66), Brussels sprouts (110 – 1560); brown mustard (16,000 – 72,000); horseradish (4500)
S
Allyl isothiocyanate (Carcinogen)
Orange juice (31); black pepper (8000)
CH3
Š
CH3
H
(R)-Limonene (Carcinogen)
Carrot (10 – 20)
OH
(CH2)5CH3 Carotatoxin (Neurotoxin)
O
O
O
Parsley (11,000 – 112,000); celery (1,300 – 46,000)
Psoralen (Carcinogen)
NH2
Banana (15,000)
HO N H Serotonin (Neurotransmitter, vasoconstrictor)
synthetic pesticide residues. The concentration of these natural compounds ranges in the parts per million (ppm), orders of magnitude above the levels at which water pollutants (e.g., chlorinated hydrocarbons) and other synthetic pollutants (e.g., detergents, Chemical Highlight 19-1) are usually measured (parts per billion, or ppb). Few of these plant toxins have been tested for carcinogenicity but, of those tested (in rodents), roughly half are carcinogenic, the same proportion as that of synthetic chemicals. Many have proven toxicity. The table gives examples of some (potentially) toxic pesticides in common foods. Why, then, have we all not been exterminated by these poisons? One reason is that the level of our exposure to any one of these natural pesticides is very small. More important,
we, like plants, have evolved to defend ourselves against this barrage of chemical projectiles. Thus, for starters, our first line of defense, the surface layers of the mouth, esophagus, stomach, intestine, skin, and lungs, is discarded once every few days as “cannon fodder.” In addition, we have multiple detoxifying mechanisms, rendering ingested poisons nontoxic; we excrete a lot of material before it does any harm; our DNA has many ways of repairing damage; and finally, our ability to smell and taste “repugnant” substances (such as the “bitter” alkaloids, rotten food, milk that is “off,” eggs that smell of “sulfur”) serves as an advance warning signal. In the final analysis, we each must judge what we put into our bodies, but the age-old wisdoms still hold: Avoid anything in excess, and maintain variety in your diet.
1193
Chapter 25
An Anopheles stephensi mosquito lunching on a human host through its pointed feeding tube. Engorgement has proceeded to the point that a droplet of blood emanates from the abdomen.
Heterocycles
Even more stimulating than nicotine are caffeine and theobromine, present in coffee and tea or cocoa (chocolate), respectively. Perhaps the most dangerous stimulant is cocaine, extracted from the leaves of the coca shrub, which is cultivated mainly in South America for the purpose of illegal drug trafficking. Cocaine is shipped and sold in the form of the water-soluble hydrochloride salt (“street cocaine”), which may be ingested through the nasal passages by “snorting” or orally and intravenously. The actual alkaloid is known as “freebase” or “crack” and is inhaled by smoking. There are severe physical and psychological side effects of the drug, such as brain seizures, respiratory collapse, heart attack, paranoia, and depression. The compound has some good uses, nevertheless. For example, it functions as a very effective topical anesthetic in eye operations. Quinine, isolated from cinchona bark (as much as 8% concentration), is the oldest known effective antimalarial agent. A malaria attack consists of a chill accompanied or followed by a fever, which terminates in a sweating stage. Such attacks may recur regularly. The name “malaria” is derived from the Italian malo, bad, and aria, air, referring to the old belief that the disease was caused by noxious effluent gases from marshland. The actual culprit is a protozoan parasite (Plasmodium species) transmitted by the bite of an infected female mosquito of the genus Anopheles (see Chemical Highlight 3-2). It is estimated that from 300 to 500 million people are affected by this disease, which kills over 2 million each year, more than half of them children. Strychnine is a powerful poison (the lethal dose in animals is about 5 – 8 mg kg21), the lethal ingredient of many a detective novel. H
N
/∑
H HO
H R
H
/
R
N
≥
CH3O
N
0
H
≥
1194
H
H ≥ H
H H
NH
/
≥ HO
H
N Quinine
O
Strychnine (RⴝH) Brucine (RⴝCH3O)
H
H
1,2,3,4-Tetrahydroisoquinoline
The isoquinoline and 1,2,3,4-tetrahydroisoquinoline nuclei are abundant among the alkaloids, and their derivatives are physiologically active, for example, as hallucinogens, central nervous system agents (depressants and stimulants), and hypotensives. Note that the pharmacophoric 2-phenylethanamine unit (see Chemical Highlight 21-1) is part of these nuclei and is also present in most of the other alkaloids considered in this section. (Find it in morphine, lysergic acid, quinine — there is a quirk here — and strychnine.)
In Summary The alkaloids are natural nitrogen-containing compounds, many of which are physiologically active.
THE BIG PICTURE The discussion of saturated and aromatic heterocycles pulls together several concepts and applications in this book — in particular, the ring opening of strained heterocycles; the extension of the principles of aromaticity to heteroaromatic systems, including electrophilic and nucleophilic aromatic substitution; the use of condensation methodology in the construction of heteroaromatic systems; and a glimpse at their structural diversity in nature and as synthetic drugs. The field of heterocyclic chemistry is vast, and this chapter has only highlighted a few of its aspects. We shall see in the next and final chapter that heterocycles are integral parts of the nucleic acids DNA and RNA.
Chapter Integration Problems
Chapter 25
CHAPTER INTEGRATION PROBLEMS 25-28. As we have seen for the heterocyclopentadienes (Section 25-3) and pyridine (Section 25-5), heteroaromatic compounds can almost invariably be made by condensation reactions of carbonyl substrates with appropriate heterofunctions. a. Write a plausible mechanism for the first step of the Hantzsch synthesis of 2,6-dimethylpyridine (Section 25-5), shown again here. CH2 B O
H CO2CH2CH3
A
A
CH3CH2O2C
CH2 H2C A A C C M MO D CH H3C O 3 NH3
H
CH3CH2O2C
CO2CH2CH3
A
3 H2O
N H
H3C
CH3
SOLUTION By following the fate of the four components in the starting mixture, you can see that the ammonia has reacted at the two keto carbonyl carbons, presumably by imine and then enamine formation (Section 17-9), whereas the formaldehyde component has made bonds to the acidic methylene of the 3-oxobutanoates (Section 23-2), presumably initially by an aldol-like condensation process (Section 18-5). Let us formulate these steps one at a time. Step 1. Aldol-like condensation of formaldehyde with ethyl 3-oxobutanoate
H 2C P O
O B CH3CCH2CO2CH2CH3
CO CH CH G 2 2 3 H2CP C D CCH3 B O
H2O
Michael acceptor
Step 2. Enamine formation of ammonia with ethyl 3-oxobutanoate O B CH3CCH2CO2CH2CH3
NH3
NH2 A CH3CP CHCO2CH2CH3
H2O
Enamine
We note that step 1 furnishes a Michael acceptor, whereas step 2 results in an enamine. The latter can react with the former, in analogy to the reaction of enolates, by Michael addition, in this case to a neutral oxoenamine. Step 3. Michael addition of the enamine CH3CH2O2C H3C
H NH2
H2C O ð
CO2CH2CH3 CH3 H
CO2CH2CH3
CH3CH2O2C H3C
H
CH3CH2O2C
CO2CH2CH3
H shift O N ð H2
CH3
H3C
N O H2 Oxoenamine
This species is now perfectly set up to undergo an intramolecular imine condensation of the 3,4dihydropyridine, which tautomerizes (Sections 13-7 and 18-2) to the more stable 1,4-dihydro product.
CH3
1195
1196
Heterocycles
Chapter 25
Step 4. Intramolecular imine formation and tautomerization CH3CH2O2C H3C
CO2CH2CH3 N O H2
H
CH3CH2O2C
CO2CH2CH3
H2O
CH3
H 3C
Oxoenamine
N
CH3
Diethyl 3,4-dihydro-2,6-dimethyl3,5-pyridinedicarboxylate
CO2CH2CH3
CH3CH2O2C H3C
N H
CH3
Diethyl 1,4-dihydro-2,6dimethyl-3,5-pyridinedicarboxylate
b. On the basis of the discussion of (a) and Exercises 25-11 and 25-14, suggest some simple retrosynthetic (retrocondensation) reactions to indole, quinoline, and 1,4-diazanaphthalene (quinoxaline) from ortho-disubstituted benzenes. SOLUTION You can view indole as a benzofused enamine. The enamine part is retrosynthetically connected to the corresponding enolizable carbonyl and amino function (Section 17-9). H N H
N H2
O
Indole
Quinoline can be viewed as a benzofused a,b-unsaturated imine. The imine nucleus is retrosynthetically opened to the corresponding amine and a,b-unsaturated carbonyl fragments (Section 17-9), which can be constructed by an aldol condensation (Section 18-5) employing acetaldehyde. O O B CH3CH
H
NH2
N Quinoline
1,4-Diazanaphthalene can be dissected by retrosynthetic hydrolysis of the two imine functions (Section 17-9) to 1,2-benzenediamine and ethanedione (glyoxal). N
NH2
OO BB HCCH
NH2
N 1,4-Diazanaphthalene (Quinoxaline)
25-29. Viagra (sildenafil citrate) was introduced in 1998 as an effective way to treat male erectile dysfunction (MED). It was discovered accidentally during clinical trials of the compound as a treatment for coronary heart disease and works by enhancing the production of nitric oxide (NO) in erectile tissue, ultimately leading to vasodilation (see Chemical Highlight 26-1). The route by which Viagra was made originally in the research laboratory is shown below.
O
O
O O
1. H2NNH2, H2O 62%
O
O
H N
O B 2. CH3OSOCH3 B O 3. NaOH, H2O
N
71%
Chapter Integration Problems
O
4. HNO3, H2SO4 5. SOCl2 6. NH3, H2O 7. H2, Pd–C
CH3 A N N
O HO
H2N
42%
O Cl
CH3 A N N
O
Chapter 25
8.
, pyridine
88%
H2N
CH3 A N N
O H2N O
O
9. NaOH, H2O, CH3CH2OH
CH3 A N N
HN
81%
10. ClSO3H (excess) 11. NCH3 HN
, CH3CH2OH 88%
N NH O O O O
O
HN
N
O N
12. Citric acid
N
O Sildenafil
N
A
H3C
100%
CH3N
HN
O BB S
A
N
A
BB S
O
CH3 A N N
A H
O COO
HO
HOOC
COOH Viagra (Sildenafil citrate)
With your current knowledge of synthetic chemistry, you should be able to understand each of the 12 steps in this synthesis, even though some of the functional units are new to you. Identify the essential features of each step and rationalize its outcome. SOLUTION This problem is typical of what the practitioner of organic synthesis might encounter while reading the literature. He or she may not be versed in the particular class of substances described, or the specific reagents employed, but will nevertheless be able to follow the narrative by extrapolating from basic principles. Step 1. We can see topologically that the N2 unit in the reagent hydrazine, H2NNH2, has been added across the b-dicarbonyl function (Chapter 23) in some way. Comparison of the respective molecular formulas reveals that two molecules of water are lost on going from starting materials to product, suggestive of a double condensation. Substituted hydrazines undergo just that with ketones (Section 17-10; see also Exercise 25-27b), normally only once; here we do it twice. Application to a 1,3-dione produces a diazacyclopentadiene, which aromatizes by deprotonation (to an aromatic anion, see Section 15-7), followed by reprotonation. The resulting 2-azapyrrole (Section 25-4) is called pyrazole, and this is one way of making it. General Pyrazole Synthesis
R
O
1. H 2. H
N N R H2NNH2
2 H2O
R
R H
H
HN N
O
R
R Pyrazole derivative
Step 2. Dimethylsulfate, (CH3O)2SO2, is a methylating agent (Section 6-7) and serves to alkylate the pyrazole nitrogen. Step 3. This step is a simple base-mediated ester hydrolysis (Section 20-4).
CH3 A N N
1197
1198
Chapter 25
Heterocycles
Step 4. The product of the sequence 4 – 7 contains an additional amino substituent on the pyrazole ring. The requisite nitrogen is introduced by way of an electrophilic nitration (Sections 15-10 and 25-4). Steps 5 and 6. This sequence converts the carboxylic acid to the carboxylic amide via the acid chloride (Sections 19-8 and 20-2). Step 7. We now convert the nitro substituent introduced in step 4 into the amino function by catalytic hydrogenation (Section 16-5). Step 8. Next, we introduce the 2-ethoxybenzoyl fragment by amide formation (Section 20-2). Step 9. This condensation is unusual, as it involves deactivated amino and carbonyl groups. It is facilitated by its intramolecular nature, much like that of cyclic imide formation (Section 19-10), and the dipolar (Section 20-1) aromatic resonance stabilization of the product, shown in a general way below.
šð ðO
ðOð N N
R
A
H
A
H
N
R
N
Step 10. This step is a variant of aromatic sulfonation (Section 15-10) and simultaneous sulfonyl chloride generation, using chlorosulfonic acid, ClSO3H, the acid chloride of sulfuric acid. Step 11. We arrive at sildenafil, the active ingredient of Viagra, by sulfonamide production (Section 15-10). Step 12. To allow for water solubility, sildenafil is administered in the protonated form of the ammonium salt of citric acid, the drug Viagra itself.
New Reactions 1. Reactions of Heterocyclopropanes (Section 25-2) X
1. ðNu or HNu 2. H2O
HX
Nu
2. Ring Opening of Heterocyclobutanes (Section 25-2) 1. ðNu or HNu 2. H2O
X
HX
Nu
Less reactive than heterocyclopropanes
3. Paal-Knorr Synthesis of 1-Hetero-2,4-cyclopentadienes (Section 25-3) P2S5,
R
S
R
R
R
P2O5,
R
O O
R
O
(NH4)2CO3,
R
N H
R
4. Reactions of 1-Hetero-2,4-cyclopentadienes (Section 25-4) Electrophilic substitution E
X
X
E
Main product
E
through X
H
S
O
Relative reactivity
N H
New Reactions
Ring opening O CH3
H , H2 O
H3C
H3C
CH3
O
O
Cycloaddition O O
O
NH
[H O [H
(CH3CH2)2O, 25C
NH O
O
5. Hantzsch Synthesis of Pyridines (Section 25-5) RCHO
O B COR
R
A
H 2C A C OP
HNO3
A
CH2 A CP R O
CO2R
RO2C
A
A
O B ROC
R
R
R
N H
NH3 R
R CO2R
RO2C R
1. KOH 2. CaO,
R
N
R
R
N
6. Reactions of Pyridine (Sections 25-5 and 25-6) Protonation (Section 25-5) H
N H
N
Pyridinium ion (pKa ⴝ 5.29)
Electrophilic substitution (Section 25-6)
E
H E
E H
N
N
N
Ring is deactivated relative to benzene.
Nucleophilic substitution (Section 25-6) H, H2O
NaNH2, liquid NH3
Chichibabin reaction
N
N
NH2
H
N
H X
Nu ðNu
RLi
Organometallic reagent
N
N
R
N Halopyridine (X ⴝ Br, Cl)
NH2
N
X
Chapter 25
1199
Chapter 25
Heterocycles
7. Reactions of Quinoline and Isoquinoline (Section 25-7) Electrophilic substitution
Nucleophilic substitution
N
N
N
N
Important Concepts 1. The heterocycloalkanes can be named using cycloalkane nomenclature. The prefix aza- for nitrogen, oxa- for oxygen, thia- for sulfur, and so forth, indicates the heteroatom. Other systematic and common names abound in the literature, particularly for the aromatic heterocycles. 2. The strained three- and four-membered heterocycloalkanes undergo ring opening with nucleophiles easily. 3. The 1-hetero-2,4-cyclopentadienes are aromatic and have an arrangement of six p electrons, similar to that in the cyclopentadienyl anion. The heteroatom is sp2 hybridized, the p orbital contributing two electrons to the p system. As a consequence, the diene unit is electron rich and reactive in electrophilic aromatic substitutions. 4. Replacement of one (or more) of the CH units in benzene by an sp2-hybridized nitrogen gives rise to pyridine (and other azabenzenes). The p orbital on the heteroatom contributes one electron to the p system; the lone electron pair is located in an sp2 hybrid atomic orbital in the molecular plane. Azabenzenes are electron poor, because the electronegative nitrogen withdraws electron density from the ring by induction and by resonance. Electrophilic aromatic substitution of azabenzenes is sluggish. Conversely, nucleophilic aromatic substitution occurs readily; this is shown by the Chichibabin reaction, substitutions by organometallic reagents next to the nitrogen, and the displacement of halide ion from halopyridines by nucleophiles. 5. The azanaphthalenes (benzopyridines) quinoline and isoquinoline contain an electron-poor pyridine ring, susceptible to nucleophilic attack, and an electron-rich benzene ring that enters into electrophilic aromatic substitution reactions, usually at the positions closest to the heterocyclic unit.
Problems 30. Name or draw the following compounds. (a) cis-2,3-Diphenyloxacyclopropane; (b) 3-azacyclobutanone; (c) 1,3-oxathiacyclopentane; (d) 2-butanoyl-1,3-dithiacyclohexane; COOH (e)
O
(f)
CHO
N A CH3
(g)
CH3 (h) S
N
CH3
31. Identify by name (either IUPAC or common) as many of the heterocyclic rings contained in the structures shown in Table 25-1 as you can. 32. Give the expected product of each of the following reaction sequences. H ´
) [
1200
(a)
≥ H
O
1. LiAlH4, (CH3CH2)2O 2. H, H2O
OCH2CH3 H3C (c)
O
Dilute HCl, H2O
CH3 CH3
NaOCH2CH3, CH3CH2OH,
(b)
N G CH3
Problems
1201
Chapter 25
33. The penicillins are a class of antibiotics containing two heterocyclic rings that interfere with the construction of cell walls by bacteria (Chemical Highlight 20-2). The interference results from reaction of the penicillin with an amino group of a protein that closes gaps that develop during construction of the cell wall. The insides of the cell leak out, and the organism dies. (a) Suggest a reasonable product for the reactions of penicillin G with the amino group of a protein (protein – NH2). (Hint: First identify the most reactive electrophilic site in penicillin.) O B H C6H5CH2CN & C H
H C S
&
&
O
Protein–NH2
a “penicilloyl” protein derivative
& H CO2H
C
N
C CH3 CH3
Penicillin G
(b) Penicillin-resistant bacteria secrete an enzyme (penicillinase) that catalyzes hydrolysis of the antibiotic faster than the antibiotic can attack the cell-wall proteins. Propose a structure for the product of this hydrolysis and suggest a reason why hydrolysis destroys the antibiotic properties of penicillin. H2O penicillinase
Penicillin G uuuuuuy
penicilloic acid (Hydrolysis product; no antibiotic activity)
34. Propose reasonable mechanisms for the following transformations. 1. SnCl4 (a Lewis acid), CH2Cl2 2. H, H2O
(a) O (b)
CH2OH 1. CH3CH2CH2CH2Li, BF3–O(CH2CH3)2, THF 2. H, H2O
O
C6H5
C6H5
OH
(Hint: See Section 15-13.)
CH3CO B O
Br (
}
(
}
(
(c) H3C } }H CH3 O H
H (CH3
O O B B MgBr2, CH3COCCH3
H3C H
35. Rank the following compounds in increasing order of basicity: water, hydroxide, pyridine, pyrrole, ammonia. 36. Each of the heterocyclopentadienes in the margin contains more than one heteroatom. For each one, identify the orbitals occupied by all lone electron pairs on the heteroatoms and determine whether the molecule qualifies as aromatic. Are any of these heterocycles a stronger base than pyrrole? 37. Give the product of each of the following reactions.
N H
N H
Pyrazole
Imidazole
O
O
N CH3NH2
(a)
N N
(b)
O
N
P2O5,
O 38. 1-Hetero-2,4-cyclopentadienes can be prepared by condensation of an a-dicarbonyl compound and certain heteroatom-containing diesters. Propose a mechanism for the following pyrrole synthesis. H5C6
A
OO O O BB B B C6H5CCC6H5 CH3OCCH2NCH2COCH3 A CP O H3C
C6H5
NaOCH3, CH3OH,
CH3OOC
N A H
COOCH3
How would you use a similar approach to synthesize 2,5-thiophenedicarboxylic acid?
S
O
Thiazole
Isoxazole
1202
Heterocycles
Chapter 25
39. Give the expected major product(s) of each of the following reactions. Explain how you chose the position of substitution in each case.
Cl2
(a)
HNO3, H2SO4
(b)
COOCH3
O
S
(c)
Cl A CH3CHCH3, AlCl3
CH3
CH3
H C B O
S
Br (d)
N2Cl, NaOH, H2O
N
Br2
(e) N H
S
(Hint: basic conditions. First deprotonate nitrogen.)
40. Give the products expected of each of the following reactions. H3C
G D N
CH3 O Fuming H2SO4, 270°C
(a)
(b)
O
S
O
, pressure
KSH, CH3OH,
(c)
N
Br
N O 1. C6H5COCl, SnCl4 2. Raney Ni,
(d)
(CH3)3CLi, THF,
(e)
S
N 41. Propose a synthesis of each of the following substituted heterocycles, using synthetic sequences presented in this chapter. H3C (a)
H3C
CH3
O
CH3
H3C (b)
CH3
CH3
(c) H5C6
N
C6H5
(d) N H
H3C
S
42. The structures of caffeine, the principal stimulant in coffee, and theobromine, its close relative in chocolate, are given in Section 25-8. Propose an efficient synthetic method to convert theobromine into caffeine. Cl N Cl
N Cl
N
Cyanuric chloride NH3
NH2 N H2N
44. Chelidonic acid, a 4-oxacyclohexanone (common name, g-pyrone), is found in a number of plants and is synthesized from acetone and diethyl ethanedioate. Formulate a mechanism for this transformation.
N N
Melamine
43. Melamine, or 2,4,6-triaminotriazine, is a toxic heterocyclic compound that has been implicated in the illnesses and deaths of both house pets and humans who ingested melamine-contaminated food. As its formula shows, melamine has a very high nitrogen content. Proteins (Chapter 26) are the main natural source of nitrogen in foods, and nitrogen analysis is commonly used to determine protein content in foods. The illegal addition of melamine to packaged foods increases their nitrogen content and makes them appear richer in protein; in reality, they are deadly. (a) A typical protein in food contains about 15% nitrogen. What is the % N in melamine? Does this result explain the motivation behind the melamine “doping” of packaged foods? (b) Melamine is synthesized by addition of ammonia to cyanuric chloride (2,4,6-trichlorotriazine, see equation in the margin). What kind of reaction is this? Formulate a mechanism and explain why cyanuric chloride displays this type of chemistry.
NH2
O OO O BB B CH3CCH3 2 CH3CH2OCCOCH2CH3
1. NaOCH2CH3, CH3CH2OH 2. HCl,
HOOC
O
COOH
Chelidonic acid
Problems
45. Porphyrins are polyheterocyclic constituents of hemoglobin and myoglobin, the molecules that transport molecular oxygen in living systems (Section 26-8), of the cytochromes, which also play central roles in biological redox processes (Section 22-9), and of chlorophyll (Chemical Highlight 24-2), which mediates photosynthesis in all green plants. Porphyrins are the products of a remarkable reaction between pyrrole and an aldehyde in the presence of acid:
P
RD
C
GH
N H
N
H
R
R N
N
ð
N H
H
ð
ð
ðOð
ð
ð
R
R A porphyrin
2
ð
N H
ð
Oð GC P
ð
H
ð
This reaction is complicated and has many steps. The simpler condensation of one molecule of benzaldehyde and two of pyrrole to give the product shown below, a dipyrrylmethane, is illustrative of the first stage in porphyrin formation. Propose a step-by-step mechanism for this process.
H
N H
N H
A dipyrrylmethane
46. Isoxazoles (Problem 36) have taken on increased significance as synthetic targets, because they are found in the structures of some recently discovered naturally occurring molecules that show promise as antibiotics (see Chemical Highlight 20-2). Isoxazoles may be prepared by reaction of alkynes with compounds containing the unusual nitrile oxide functional group:
RC q CR RC q N O š Oð
š N
R
šð O
Nitrile oxide
R
R
Isoxazole
Suggest a mechanism for this process. 47. Reserpine is a naturally occurring indole alkaloid with powerful tranquilizing and antihypertensive activity. Many such compounds possess a characteristic structural feature: one nitrogen atom at a ring fusion separated by two carbons from another nitrogen atom.
N ) H H
N H
} H ) CH3OOC
}
CH3O
) OOC ≥ OCH3
OCH3 OCH3 OCH3
Reserpine
A series of compounds with modified versions of this structural feature have been synthesized and also shown to have antihypertensive activity, as well as antifibrillatory properties. One such synthesis is shown here. Name or draw the missing reagents and products (a) through (c).
Chapter 25
1203
1204
Chapter 25
Heterocycles
H N f i H2C O CH2,
(a)
CH3CH2OH
N
CO2CH2CH3
C8H14N2O (b)
CO2CH2CH3
N H
LiAlH4
C8H16N2 (c)
48. Starting with benzenamine (aniline) and pyridine, propose a synthesis for the antimicrobial sulfa drug sulfapyridine. N
N H Indole
Sulfapyridine
N H
49. Derivatives of benzimidazole possess biological activity somewhat like that of indoles and purines (of which adenine, Section 25-1, is an example). Benzimidazoles are commonly prepared from benzene-1,2-diamine. Devise a short synthesis of 2-methylbenzimidazole from benzene-1,2-diamine. NH2
N N H
N
NH2
N
Benzimidazole
N
O B NHS B O
N
?
CH3 N H
NH2 Benzene-1,2-diamine
Purine
2-Methylbenzimidazole
50. The Darzens condensation is one of the older methods (1904) for the synthesis of three-membered heterocycles. It is most commonly the reaction of a 2-halo ester with a carbonyl derivative in the presence of base. The following examples of the Darzens condensation show how it is applied to the synthesis of oxacyclopropane and azacyclopropane rings. Suggest a reasonable mechanism for each of these reactions. Cl A (a) C6H5CHO C6H5CHCOOCH2CH3
KOC(CH3)3, (CH3)3COH
(b) C6H5CHPNC6H5 ClCH2COOCH2CH3 H3C H 3C
Br A N N
O
O i
Br
O f i H5C6 O C O C O C6H5 D G H COOCH2CH3
KOC(CH3)3, CH3OCH2CH2OCH3
C6H5 A N f i C6H5CHO CHCOOCH2CH3
51. (a) The compound shown in the margin, with the common name 1,3-dibromo-5,5-dimethylhydantoin, is useful as a source of electrophilic bromine (Br1) for addition reactions. Give a more systematic name for this heterocyclic compound. (b) An even more remarkable heterocyclic compound (ii) is prepared by the following reaction sequence. Using the given information, deduce structures for compounds i and ii, and name the latter. H3C
CH3
D G C PC G D CH3 H3C
1,3-Dibromo-5,5-dimethylhydantoin, 98% H2O2
O B Ag OCCH3
C6H13BrO2 i
AgBr, CH3COOH
C6H12O2 ii
Heterocycle ii is a yellow, crystalline, sweet-smelling compound that decomposes upon gentle heating into two molecules of acetone, one of which is formed directly in its n y p* excited state (Sections 14-11 and 17-3). This electronically excited product is chemiluminescent.
ii
O O B B CH3CCH3 CH3CCH3
n
*
O B h 2 CH3CCH3
Heterocycles similar to compound ii are responsible for the chemiluminescence produced by a number of species [e.g., fireflies (see Chemical Highlight 9-1) and several deep-sea fish]; they also serve as the energy sources in commercial chemiluminescent products.
Problems
1205
Chapter 25
52. Azacyclohexanes (piperidines) can be synthesized by reaction of ammonia with cross-conjugated dienones: ketones conjugated on both sides with double bonds. Propose a mechanism for the following synthesis of 2,2,6,6-tetramethylaza-4-cyclohexanone. O B
O B (CH3)2C P CHCCH P C(CH3)2
NH3
H3C H3C
CH3 N H
CH3
53. Quinolines (Section 25-7) are heterocycles that are widely used in medicinal chemistry because of the diversity of biological activity — including anticancer utility — that their derivatives display. A short synthesis of 3-acyldihydroquinolines (which may be converted into 3-acylquinolines by mild oxidation) is shown below. Propose a mechanism for this process. (Hint: Review Section 18-10.) O
O H
CH3
CH3
Base catalyst
O
NH2
N H
O CH3
54.
At which position(s) do you expect 3-acetylquinoline (margin) to undergo nitration in the presence of a mixture of sulfuric and fuming nitric acids? Will this reaction be faster or slower than nitration of quinoline itself?
N 3-Acetylquinoline
55. Compound A, C8H8O, exhibits 1H NMR spectrum A. Upon treatment with concentrated aqueous HCl, it is converted almost instantaneously into a compound that exhibits spectrum B (p. 1206). What is compound A, and what is the product of its treatment with aqueous acid?
1
H NMR 5H
1H 7.4
7.3
3.9
1H
3.8
3.2
3.1
2.8
1H
(CH3)4Si
7.5
7.0
6.5
6.0
5.5
5.0
4.5
4.0
3.5
3.0
2.5
δ 300-MHz H NMR spectrum ppm (δ) 1
A
2.0
1.5
1.0
0.5
0.0
1206
Heterocycles
Chapter 25
1
H NMR
5H
7.4
7.3
4.9
4.8
3.8
3.7
3.6
(CH3)4Si 1H
7.5
7.0
6.5
6.0
5.5
5.0
1H
4.5
4.0
1H 2H
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.0
300-MHz 1H NMR spectrum ppm (δ) δ B
56. Heterocycle C, C5H6O, exhibits 1H NMR spectrum C and is converted by H2 and Raney nickel into compound D, C5H10O, with spectrum D (p. 1207). Identify compounds C and D. (Note: The coupling constants of the compounds in this problem and the next one are rather small; they are therefore not nearly as useful in structure elucidation as those around a benzene ring.)
1H
NMR 3H
7.4
7.3
6.3
6.2
6.1
2.4
6.0
2.3
1H 1H 1H
(CH3)4Si
7.5
7.0
6.5
6.0
5.5
5.0
4.5
4.0
3.5
3.0
2.5
2.0
300-MHz 1H NMR spectrum ppm (δ ) C
1.5
1.0
0.5
0.0
Problems
1
Chapter 25
H NMR 3H
3.9 3.8 3.7 3.6 3.5
1.3
1.9
1.8
1.2
1.1
1.7
2H 1H
3H 1H
4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 300-MHz 1H NMR spectrum ppm (δ) δ
D
57.
1
(CH3)4Si
The commercial synthesis of a useful heterocyclic derivative requires treatment of a mixture of aldopentoses (derived from corncobs, straw, etc.) with hot acid under dehydrating conditions. The product, E, has 1H NMR spectrum E, shows a strong IR band at 1670 cm21, and is formed in nearly quantitative yield. Identify compound E and formulate a mechanism for its formation.
H NMR
1H
1H 1H 9.6
1H
9.5
7.6
7.2
7.1
6.5
(CH3)4Si
9.5
9.0
8.5
8.0
7.5
7.0
6.5
6.0
5.5
5.0
4.5
4.0
3.5
δ 300-MHz 1H NMR spectrum ppm (δ)
E
3.0
2.5
2.0
1.5
1.0
0.5
0.0
1207
1208
Heterocycles
Chapter 25
H1 , D
Aldopentoses uuy C5H4O2 E
Compound E is a valuable synthetic starting material. The following sequence converts it into furethonium, which is useful in the treatment of glaucoma. What is the structure of furethonium?
1. NH3, NaBH3CN 2. Excess CH3I, (CH3CH2)2O
E uuuuuuuuuy furethonium
58.
Treatment of a 3-acylindole with LiAlH4 in (CH3CH2)2O reduces the carbonyl all the way to a CH2 group. Explain by a plausible mechanism. (Hint: Direct SN2 displacement of alkoxide by hydride is not plausible.)
O B CR
CH2R Excess LiAlH4, (CH3CH2)2O,
O B CR
(Compare
OH A C O R) A H
LiAlH4
N H
N H
59. The sequence in the margin is a rapid synthesis of one of the heterocycles in this chapter. Draw the structure of the product, which has 1H NMR spectrum F.
1. O3, CH2Cl2 2. (CH3)2S 3. NH3
1
H NMR
1H 1H 1H
3H
9.2
8.5
8.4
7.9
7.8
7.7
7.6
7.5
(CH3)4Si
1H
9.5
9.0
8.5
8.0
7.5
7.0
6.5
6.0
5.5
5.0
4.5
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.0
300-MHz H NMR spectrum ppm (δ) δ 1
F
Team Problem 60. This problem introduces two literature syntheses of indole derivatives, and you are asked to come up with plausible mechanisms for them. Divide your team in two, each group concentrating on one of the methods.
Problems
1209
Chapter 25
Fischer Indole Synthesis of 2-Phenylindole
H3C
H
N A H
NH2 H3C
H2O
N A H
O
C
H
N
NH4
N i H 2-Phenylindole
In this procedure, a hydrazone of an enolizable aldehyde or ketone is heated in strong acid, causing ring closure with simultaneous expulsion of ammonia to furnish the indole nucleus. [Hints: The mechanism of the reaction proceeds in three stages: (1) an imine – enamine tautomerization (recall Section 17-9); (2) an electrocyclic reaction (a “diaza-Cope” rearrangement; recall Section 22-7); (3) another imine – enamine (in this case, benzenamine) tautomerization; (4) ring closure to the heterocycle; and (5) elimination of NH3.]
Reissert Indole Synthesis of Ethyl Indole-2-carboxylate O
O
CH3
O
O
O
KOCH2CH3
O
NO2
H2, Pt
O
NO2
2-Methylnitrobenzene (o-Nitrotoluene)
A 2-oxopropanoate ester (A pyruvate ester)
CO2CH2CH3 N i H Ethyl indole-2-carboxylate
In this sequence, a 2-methylnitrobenzene (o-nitrotoluene) is first converted into an ethyl 2-oxopropanoate (pyruvate, Chemical Highlight 23-2) ester, which, on reduction, is transformed into the target indole. [Hints: (1) The nitro group is essential for the success of the first step. Why? Does this step remind you of another reaction? Which one? (2) Which functional group is the target of the reduction step (recall Section 16-5)? (3) The ring closure to the heterocycle requires a condensation reaction.]
Preprofessional Problems 61. The proton decoupled 13C NMR spectrum of pyridine will display how many peaks? (a) One; (b) two; (c) three; (d) four; (e) five. 62. Pyrrole is a much weaker base than azacyclopentane (pyrrolidine) for which of the following reasons? (a) The nitrogen in pyrrole is more electropositive than that in pyrrolidine; (b) pyrrole is a Lewis acid; (c) pyrrole has four electrons; (d) pyrrolidine can give up the proton on the nitrogen atom more readily than can pyrrole; (e) pyrrole is aromatic. 63. Which of the following compounds would you expect to be the major organic product of the two-step sequence shown here? O CH3 A (CH3CH)2NLi O
THF
CH3CH2I
š N A H
š N A H
Pyrrole
Azacyclopentane (Pyrrolidine)
1210
Heterocycles
Chapter 25
NH
NH ECH2CH3
(a)
O
O OCH2CH3
(b)
O
O
H CH2CH3
(c)
CH2CH3 (d) O
64. This reaction yields one main organic product. Which of the following compounds is it?
2-Phenylthiophene C6H5 (a)
C6H5 (b)
S
CCH3 B O
O B SnCl4, CH3CCl
(c) H3CC B O
S
O
(d) H 3C B O
S
C6H5 S
C H A P T E R
TWENTY SIX
Amino Acids, Peptides, Proteins, and Nucleic Acids N
Nitrogen-Containing Polymers in Nature
O
O
O
n page 1 of this text we defined organic chemistry as the chemistry of carbon-containing compounds. We then went on to point out that organic molecules constitute the chemical bricks of life. Indeed, a historical definition of organic chemistry restricts it to living organisms. What is life and how do we, as organic chemists, approach its study? A functional definition of life refers to it as a condition of matter manifested by growth, metabolism, reproduction, and evolution. The underlying basic processes are chemical, and researchers hope to decipher their complexity by investigating specific reactions or reaction sequences. The “whole,” however, is much more complex than these individual pieces, because they interact by multiple feedback loops in a manner that makes it impossible to show that each effect has a simple cause. This final chapter will provide a glimpse of that complexity by taking you from amino acids to their polymers, the polypeptides — in particular, the large natural polypeptides called proteins — and to their biological origin, DNA. Proteins have an astounding diversity of functions in living systems. As enzymes, they catalyze transformations ranging in complexity from the simple hydration of carbon dioxide to the replication of entire chromosomes — great coiled strands of DNA, the genetic material in living cells. Enzymes can accelerate certain reactions many millionfold. We have already encountered the protein rhodopsin, the photoreceptor that generates and transmits nerve impulses in retinal cells (Chemical Highlight 18-3). Other proteins serve for transport and storage. Thus, hemoglobin carries oxygen; iron is transported in the blood by transferrin and stored in the liver by ferritin. Proteins play a crucial role in coordinated motion, such as muscle contraction. They give mechanical support to skin and bone; they are the antibodies responsible for our immune protection; and they control growth and differentiation — that is, which part of the information stored in DNA is to be used at any given time. We begin with the structure and preparation of the 20 most common amino acids, the building blocks of proteins. We then show how amino acids are linked by peptide bonds in the three-dimensional structure of hemoglobin and other polypeptides. Some proteins contain thousands of amino acids, but we shall see how to determine the sequence of amino acids in many polypeptides and synthesize these molecules in the laboratory. Finally, we consider how other polymers, the nucleic acids DNA and RNA, direct the synthesis of proteins in nature.
Pregabalin [Lyrica; (S)-3(aminomethyl)-5-methylhexanoic acid] is an unnatural amino acid that is effective in the treatment of chronic pain. The molecule is shown in its zwitterionic ammonium carboxylate form.
1212
Amino Acids, Peptides, Proteins, and Nucleic Acids
Chapter 26
26-1 Structure and Properties of Amino Acids
␣-Carbon
COOH H2N
H R
Basic
Amino acids are carboxylic acids that bear an amine group. The most common of these in nature are the 2-amino acids, or A-amino acids, which have the general formula RCH(NH2)COOH; that is, the amino function is located at C2, the a-carbon. The R group can be alkyl or aryl, and it can contain hydroxy, amino, mercapto, sulfide, carboxy, guanidino, or imidazolyl groups. Because of the presence of both amino and carboxy functions, amino acids are both acidic and basic.
Acidic S
␣-Amino acid
The stereocenter of common 2-amino acids has the S configuration More than 500 amino acids exist in nature, but the proteins in all species, from bacteria to humans, consist mainly of only 20. Adult humans can synthesize all but eight, and two only in insufficient quantities. This group is often called the essential amino acids because they must be included in our diet. Although amino acids can be named in a systematic manner, they rarely are; so we shall use their common names. Table 26-1 lists the 20 most common amino acids, along with their structures, their pKa values, and the three- and (the newer) one-letter codes that abbreviate their names. We shall see later how to use these codes to describe peptides conveniently. COOH H
H 2N
Table 26-1 Natural (2S)-Amino Acids
R R
Three-letter code
One-letter code
pKa of A-COOH
pKa of A-1NH3
pKa of acidic function in R
Isoelectric point, pI
Glycine
Gly
G
2.3
9.6
—
6.0
Alanine Valinea Leucinea Isoleucinea
Ala Val Leu Ile
A V L I
2.3 2.3 2.4 2.4
9.7 9.6 9.6 9.6
— — — —
6.0 6.0 6.0 6.0
Phenylalaninea
Phe
F
1.8
9.1
—
5.5
Proline
Pro
P
2.0
10.6
—
6.3
Serine Threoninea
Ser Thr
S T
2.2 2.1
9.2 9.1
— —
5.7 5.6
Tyrosine
Tyr
Y
2.2
9.1
10.1
5.7
Asparagine
Asn
N
2.0
8.8
—
5.4 Continued
Name
H Alkyl group CH3 CH(CH3)2 CH2CH(CH3)2
CHCH2CH3 (S) A CH3 H2C COOHb H
HN
CH2 Hydroxy containing CH2OH
CHOH (R) A CH3 H2C
OH
Amino containing
O B CH2CNH2
26-1 Structure and Properties of Amino Acids
1213
Chapter 26
Table 26-1 Natural (2S)-Amino Acids (continued) Three-letter code
One-letter code
pKa of A-COOH
pKa of A-1NH3
pKa of acidic function in R
Isoelectric point, pI
Glutamine Lysinea
Gln Lys
Q K
2.2 2.2
9.1 9.0
— 10.5c
5.7 9.7
Argininea
Arg
R
2.2
9.0
12.5c
10.8
Tryptophana
Trp
W
2.8
9.4
—
5.9
Histidinea
His
H
1.8
9.2
6.1c
7.6
Cysteined Methioninea
Cys Met
C M
2.0 2.3
10.3 9.2
8.2 —
5.1 5.7
Aspartic acid Glutamic acid
Asp Glu
D E
1.9 2.2
9.6 9.7
3.7 4.3
2.8 3.2
R
Name
Amino containing (continued)
O B CH2CH2CNH2 (CH2)4NH2
NH B (CH2)3NHCNH2 H2C
N H H2C NH N Mercapto or sulfide containing CH2SH CH2CH2SCH3 Carboxy containing CH2COOH CH2CH2COOH a
Essential amino acids. bEntire structure. cpKa of conjugate acid. dThe stereocenter is R because the CH2SH substituent has higher priority than the COOH group.
MODEL BUILDING
Amino acids may be depicted by either hashed-wedged line structures or by Fischer projections. How to Draw l-Amino Acids and Their Relation to the l-Sugars C2, or ␣-carbon
≥
R
O
H2N A C O COOH R H #
N
O
COOH ´ H2N # C ! H ≥ R
S (L)
O
H2N
N
CHO
COOH
O
H R
R
Hashed-wedged line structures
In all but glycine, the simplest of the amino acids, C2 is a stereocenter and usually adopts the S configuration. Other stereocenters located in the substituent R may have either R (as in threonine) or S configuration (as in isoleucine). As in the names of the sugars (Section 24-1), an older amino acid nomenclature uses the prefixes d and l, which relate all the l-amino acids to (S)-2,3-dihydroxypropanal (lglyceraldehyde). As emphasized in the discussion of the natural d sugars, a molecule belonging to the l family is not necessarily levorotatory. For example, both valine ([a]258C 5 113.9) d and isoleucine ([a]d258C 5 111.9) are dextrorotatory.
HO
H CH2OH
(S)-2,3-Dihydroxypropanal (L-Glyceraldehyde) Fischer projections
1214
Chapter 26
Amino Acids, Peptides, Proteins, and Nucleic Acids
Exercise 26-1 Give the systematic names of alanine, valine, leucine, isoleucine, phenylalanine, serine, tyrosine, lysine, cysteine, methionine, aspartic acid, and glutamic acid.
Exercise 26-2 Draw hashed-wedged line structures for (S )-alanine, (S )-phenylalanine, (R)-phenylalanine, and (S)-proline.
Exercise 26-3 COOH H2N
H
Among the amino acids in which R 5 alkyl (Table 26-1), proline can be readily distinguished from the others by IR spectroscopy. How? (Hint: Review Section 21-3.)
R Proton transfer
Amino acids are acidic and basic: zwitterions COO
H2N H A H R Zwitterion
Glycine zwitterion
Because of their two functional groups, the amino acids are both acidic and basic; that is, they are amphoteric (Section 8-3). The carboxylic acid group protonates the amine function, thus forming a zwitterion. This ammonium carboxylate form is favored because an ammonium ion is much less acidic (pKa < 10 – 11) than a carboxylic acid (pKa < 2 – 5). The highly polar zwitterionic structure allows amino acids to form particularly strong crystal lattices. Most of them therefore are fairly insoluble in organic solvents, and they decompose rather than melt when heated. The electrostatic potential map of glycine in the margin depicts its highly dipolar nature, originating from the juxtaposition of the electron-rich (red) carboxylate function with the electron-poor (blue) ammonium group. The structure of an amino acid in aqueous solution depends on the pH. Consider, for example, the simplest member of the series, glycine. The major form in neutral solution is the zwitterion. However, in strong acid (pH , 1), glycine exists predominantly as the cationic ammonium carboxylic acid, whereas strongly basic solutions (pH . 13) contain mainly the deprotonated 2-aminocarboxylate ion. These forms interconvert by acid-base equilibria (Section 2-2). HO
H2NCH2COOH A H
H2NCH2COO A H
H
An ammonium ion predominates at pH < 1
A neutral zwitterion predominates at pH ? 6
HO H
H2NCH2COO A carboxylate ion predominates at pH > 13
Increasing pH
Table 26-1 records pKa values for each functional group of the amino acids. For glycine, the first value (2.3) refers to the equilibrium
H3NCH2COOH
H2O
H3NCH2COO
H2OH
pKa ⴝ 2.3
K1
[H3NCH2COO][H2OH]
102.3
[H3NCH2COOH] Note that this pKa is more than two units less than that of an ordinary carboxylic acid (pKa CH3COOH 5 4.74), an observation that is true for all the other a-aminocarboxy groups in
26-1 Structure and Properties of Amino Acids
Table 26-1. This difference is a consequence of the electron-withdrawing effect of the protonated amino group. The second pKa value (9.6) describes the second deprotonation step:
H3NCH2COO
H2NCH2COO
H2O
H2OH
pKa ⴝ 9.6
K2
[H2NCH2COO][H2OH]
109.6
[H3NCH2COO ]
At the isoelectric point, the net charge is zero The pH at which the extent of protonation equals that of deprotonation is called the isoelectric pH or the isoelectric point (pI; Table 26-1). At this pH, the amount of positive charge balances that of negative charge and the concentration of the charge-neutralized zwitterionic form is at its greatest. For an amino acid without any additional acidic or basic groups, such as glycine, the value of its pI is the average of its two pKa values. pI 5
pKa-cooh 1 pKa-1 n h2h 2
5 (for glycine)
pH at which concentration of zwitterion is at its maximum
2.3 1 9.6 5 6.0 2
When the side chain of the acid bears an additional acidic or basic function, the pI is either decreased or increased, respectively, as one would expect. Table 26-1 shows seven amino acids in which this is the case. Specifically, for the four amino acids with an acidic side chain, the pI is the average of its two lowest pKa values. Conversely, for the three amino acids with a basic side chain, the pI is the average of its two highest pKa values. Assignment of pKa Values in Selected Amino Acids 1.9
H3N
2.2
COOH
COOH
H H2N
H
H
H H2N
H
COOH
H H2N
CH2 9.0
9.1
2.2
COOH
CH2
CH2 A 9.6 C OK HOH
CH2
9.0
CH2 CH2
CH2 OH
10.1
Tyrosine
NH
2H
10.5
Lysine
H CH2
CH2 3.7
Aspartic acid
2.2
HN H
N H
C
NH2 12.5
Arginine
Why is this so? Picture these amino acids in their fully protonated form and then raise the pH (in other words, add base) to observe what happens to their net charge. For example, the amino dicarboxylic acid, aspartic acid, will be positively charged at low pH due to the presence of the ammonium substituent. To reach a charge-neutral form, on average one of the carboxy functions has to be deprotonated. This will happen at a pH midpoint between the two respective pKa values (1.9 and 3.7), at pI 5 2.8. The similar glutamic acid has the corresponding value at 3.2. At physiological pH, both of the carboxy functions are deprotonated, and the molecules exist as the zwitterionic anions aspartate and glutamate. (Monosodium glutamate, MSG, is used as a flavor enhancer in various foods.) Tyrosine, which bears the relatively nonacidic neutral phenol substituent (at low pH), has a pI 5 5.7, which is midway between the pKa’s of the other two more acidic groups.
Chapter 26
1215
1216
Amino Acids, Peptides, Proteins, and Nucleic Acids
Chapter 26
CH EMICAL H IG H L I G H T 2 6 - 1 Arginine and Nitric Oxide in Biochemistry and Medicine
NH2 B (CH2)3NHO C ONH2
OOC
O2, nitric oxide synthase
NH3 OOC
H
H
NG-Oxo-L-arginine
>
>
In the late 1980s and early 1990s, scientists, among them the 1998 Nobelists (medicine) Furchgott, Ignarro, and Murad,* made a series of startling discoveries. The simple but highly reactive and exceedingly toxic molecule nitric oxide, j O , is synthesized in a wide variety of cells in NP> mammals, including humans. In the body, nitric oxide performs several critical biological functions as diverse as control of blood pressure, inhibition of platelet aggregation, cell differentiation, neurotransmission, and penile erection. It also plays a major role in the activity of the immune system. For example, macrophages (cells associated with the body’s immune system) destroy bacteria and tumor cells by exposing them to nitric oxide. Nitric oxide is synthesized by the enzyme-catalyzed oxidation of arginine, as shown above. The enzyme involved, nitric oxide synthase, may well be essential for cell survival. Thus, a recent study shows that fruit flies, Drosophila, genetically modified to be incapable of making this enzyme, die as embryos. *Professor Robert F. Furchgott (b. 1916), State University of New York, Brooklyn; Professor Louis J. Ignarro (b, 1941), University of California at Los Angeles; Professor Ferid Murad (b. 1936), University of Texas, Dallas.
š NH2
P
H
O
>
O2, H
H
Guanidine
O N NH3 A OOC (CH2)3NHO C ONH2 A H O H j
A
Oj N NH3 B OOC (CH2)3NHO C ONH2
N H2š
O2 H 2O
NG-Hydroxy-L-arginine
L -Arginine
š NH B C E H
OH N B (CH2)3NHOCONH2 A
NH3
O B NHO C ONH2 jNP O (CH2)3>
NH3 OOC
H L-Citrulline
Nitric oxide
Nitric oxide is released by cells on the inner walls of blood vessels and causes adjacent muscle fibers to relax. This 1987 discovery explains the effectiveness of nitroglycerin and other organic nitrates as treatments for angina and heart attacks, a nearly century-old mystery: These substances are metabolically converted into NO, which dilates the blood vessels (see also Chapter Integration Problem 25-29). Paradoxically, considering the essential benefits of NO, its overproduction leads to septic shock. Uncontrolled release may also be the cause of brain damage after a stroke and disorders such as Alzheimer’s and Huntington’s diseases. The occurrence of high levels of nitrite ion, NO22 (the product of NO oxidation), in the joints of people with rheumatoid arthritis indicates overproduction of NO as a response to inflammation. Similar associations have been established for schizophrenia, urinary disorders, and multiple sclerosis. The rapidly evolving story of NO is an example of how little we still know about the functions of the body, but also how rapidly a discovery leads to the evolution of an entire field.
In contrast to the above examples, lysine bears an additional basic amino group that is protonated in a strongly acidic medium to furnish a dication. When the pH of the solution is raised, deprotonation of the carboxy group occurs first, followed by respective proton loss from the nitrogen at C2 and the remote ammonium function. The isoelectric point is located halfway between the last two pKa values, at pI 5 9.7. Arginine carries a substituent NH B new to us: the relatively basic guanidino group, –NHCNH2, derived from the molecule guanidine (margin). The pKa of its conjugate acid is 12.5, almost three units greater than that of the ammonium ion (Section 21-4). The pI of arginine lies midway between that of the guanidinium and ammonium groups, at 10.8.
26-2 Synthesis of Amino Acids
1217
Chapter 26
Exercise 26-4 Guanidine is found in turnip juice, mushrooms, corn germ, rice hulls, mussels, and earthworms. Its basicity is due to the formation of a highly resonance-stabilized conjugate acid. Draw its resonance forms. (Hint: Review Section 20-1.)
Histidine (margin) contains another new subsituent, the basic imidazole ring (see Problem 6 of Chapter 25). In this aromatic heterocycle, one of the nitrogen atoms is hybridized as in pyridine and the other is hybridized as in pyrrole.
COOH
HH2N
1.8
H CH2
9.2
Exercise 26-5
HNð
Draw an orbital picture of imidazole. (Hint: Use Figure 25-1 as a model.) 6.1
N G H Histidine
The imidazole ring is relatively basic because the protonated species is stabilized by resonance.
Nk
Resonance in Protonated Imidazole D
Nk š N H
H ½N
N
H
š N A H
D
š N H
H
Imidazole
N A H pKa ⴝ 7.0
This resonance stabilization is related to that in amides (Sections 20-1 and 26-4). Imidazole is significantly protonated at physiological pH (pI 5 7.6). It can therefore function as a proton acceptor and donor at the active site of a variety of enzymes (see, e.g., chymotrypsin, Section 26-4). The amino acid cysteine bears a relatively acidic mercapto substituent (pKa 5 8.2, pI 5 5.1). Recall that, apart from their acidic character, thiols can be oxidized to disulfides under mild conditions (Section 9-10). In nature, various enzymes are capable of oxidatively coupling and reductively decoupling the mercapto groups in the cysteines of proteins and peptides, thereby reversibly linking peptide strands (Section 9-10).
In Summary There are 20 elementary l-amino acids, all of which have common names. Unless there are additional acid-base functions in the side chain, their acid-base behavior is governed by two pKa values, the lower one describing the deprotonation of the carboxy group. At the isoelectric point, the number of amino acid molecules with net zero charge is maximized. Some amino acids contain additional acidic or basic functions, such as hydroxy, amino, guanidino, imidazolyl, mercapto, and carboxy.
26-2 Synthesis of Amino Acids: A Combination of Amine and Carboxylic Acid Chemistry Chapter 21 treated the chemistry of amines, Chapters 19 and 20 that of carboxylic acids and their derivatives. We use both in preparing 2-amino acids.
Hell-Volhard-Zelinsky bromination followed by amination converts carboxylic acids to 2-amino acids What would be the quickest way of introducing a 2-amino substituent into a carboxylic acid? Section 19-12 pointed out that simple 2-functionalization of an acid is possible by
COOH
HH2N
2.0
H CH2SH
10.3
8.2
Cysteine
1218
Amino Acids, Peptides, Proteins, and Nucleic Acids
Chapter 26
the Hell-Volhard-Zelinsky bromination. Furthermore, the bromine in the product can be displaced by nucleophiles, such as ammonia. In these two steps, propanoic acid can be converted into racemic alanine.
CH3CH2COOH
Br A CH3CHCOOH
Br–Br, trace PBr3 HBr
Propanoic acid
ðNH3, H2O, 25C, 4 days HBr
NH
3
A CH3CHCOO
80%
56%
2-Bromopropanoic acid
Alanine
Unfortunately, this approach suffers frequently from relatively low yields. A better synthesis utilizes Gabriel’s procedure for the preparation of primary amines (Section 21-5).
The Gabriel synthesis can be adapted to produce amino acids Recall that N-alkylation of 1,2-benzenedicarboxylic imide (phthalimide) anion followed by acid hydrolysis furnishes amines (Section 21-5). To prepare an amino acid instead, we can use diethyl 2-bromopropanedioate (diethyl 2-bromomalonate) in the first step of the reaction sequence. This alkylating agent is readily available from the bromination of diethyl propanedioate (malonate). Now the alkylation product can be hydrolyzed and decarboxylated (Section 23-2). Hydrolysis of the imide group then furnishes an amino acid. Gabriel Synthesis of Glycine O
O
ðNð K O Potassium 1,2-benzenedicarboxylic imide (Potassium phthalimide)
COOCH2CH3 A HC O Br A COOCH2CH3
COOCH2CH3 D N O CH G COOCH2CH3 O 85%
KBr
Diethyl 2-bromopropanedioate (Diethyl 2-bromomalonate)
H, H2O, 2 CH3CH2OH
O COOH D NO CH G COOH O
H, H2O CO2
ECOOH H
H3NCH2COO
COOH
85% Glycine
One of the advantages of this approach is the versatility of the initially formed 2-substituted propanedioate. This product can itself be alkylated, thus allowing for the preparation of a variety of substituted amino acids. O NO CH(CO2CH2CH3)2 O
1. CH3CH2ONa, CH3CH2OH 2. RX 3. H, H2O,
NH
3
A RCHCOO
26-2 Synthesis of Amino Acids
Exercise 26-6 Propose Gabriel syntheses of methionine, aspartic acid, and glutamic acid.
Exercise 26-7
Working with the Concepts: Practicing Variants of the Gabriel Amino Acid Synthesis A variation of the Gabriel synthesis uses diethyl N-acetyl-2-aminopropanedioate (acetamidomalonic ester), A. Propose a synthesis of racemic serine from this reagent. O
H
NH
CO2CH2CH3
N H
3
A HOCH2CHCOO
CO2CH2CH3 A
erine
Strategy We are starting with a substituted malonic ester. You should be asking yourself the following questions: What synthesis should I be looking at? What substituent needs to be introduced? How do I achieve this substitution? Solution • We apply the principles of the malonic ester synthesis of substituted carboxylic acids (Section 23-2). • The substituent to be introduced is hydroxymethyl. • Hydroxymethylation is achieved by crossed aldol addition reaction (Section 18-6) of the malonic ester enolate with formaldehyde. • Decarboxylation (Section 23-2) and amide hydrolysis (Section 20-6) is achieved with aqueous acid. HO
O A
CH2
HO CO2CH2CH3
O, NaOH
N H
H
H , H2O,
CO2CH2CH3
H3N
COO
Exercise 26-8
Try It Yourself In the search for lead compounds for the treatment of certain neurodegenerative disorders, medicinal chemists used diethyl N-acetyl-2-aminopropanedioate (acetamidomalonic ester, reagent A in Exercise 26-7) for the synthesis of the heterocyclic amino acid B from starting compound A. Suggest how they might have done so. [Caution: You need to introduce a leaving group into A. Hint: The heterocycle is aromatic (Section 25-3) and therefore attached carbons are reactive (Section 22-1).] H3CO
OCH3
š O
HO
N
š O H2N
A
OH
COOH B
N
Chapter 26
1219
1220
Chapter 26
Amino Acids, Peptides, Proteins, and Nucleic Acids
Amino acids are prepared from aldehydes by the Strecker synthesis The crucial step in the Strecker* synthesis is a variation of the cyanohydrin formation from aldehydes and hydrogen cyanide (Section 17-11). O B RCH
HCN
OH A R O C O CN A H Cyanohydrin
When the same reaction is carried out in the presence of ammonia or ammonium cyanide, H4N1CN2, it is the intermediate imine that undergoes addition of hydrogen cyanide, to furnish the corresponding 2-amino nitriles. Subsequent acidic or basic hydrolysis (Section 20-8) results in the desired amino acids. MATION AN I
ANIMATED MECHANISM: The Strecker synthesis
Strecker Synthesis of Alanine O B CH3CH
NH3 H2O
Acetaldehyde
NH B CH3CH Imine intermediate
HCN
NH2 A H3C O C O CN A H
NH
H, H2O,
2-Aminopropanenitrile
3
A CH3CHCOO 55% Alanine
Exercise 26-9 Propose Strecker syntheses of glycine (from formaldehyde) and methionine (from 2-propenal). (Hint: Review Section 18-9.)
In Summary Racemic amino acids are made by the amination of 2-bromocarboxylic acids, applications of the Gabriel synthesis of amines, and the Strecker synthesis. The last method proceeds through an imine variation of the preparation of cyanohydrins, followed by hydrolysis.
26-3 Synthesis of Enantiomerically Pure Amino Acids
H
≥
CH3O
N
]H
≥
N
H
O Brucine
≥ H
≥ HO
&
CH3O
All the methods of the preceding section produce amino acids in racemic form. However, we noted that most of the amino acids in natural polypeptides have the S configuration. Thus, many synthetic procedures — in particular, peptide and protein syntheses — require enantiomerically pure compounds. To meet this requirement, either the racemic amino acids must be resolved (Section 5-8) or a single enantiomer must be prepared by enantioselective reactions. A conceptually straightforward approach to the preparation of pure enantiomers of amino acids would be resolution of their diastereomeric salts. Typically, the amine group is first protected as an amide and the resulting product is then treated with an optically active amine, such as the inexpensive alkaloid brucine (Section 25-8 and margin). The two diastereomers formed can be separated by fractional crystallization. Unfortunately, in practice, this method can be tedious and can suffer from poor yields.
*Professor Adolf Strecker (1822 – 1871), University of Würzburg, Germany.
26-3 Synthesis of Enantiomerically Pure Amino Acids
Chapter 26
Resolution of Racemic Valine NH
O B HNCH A (CH3)2CHCHCOOH
3
A (CH3)2CHCHCOO
HCOOH
Protection
Racemate (R,S)-Valine
HOH
Racemate
80% (R,S)-N-Formylvaline Brucine (abbreviated B*), CH3OH, 0C
O COOB*H B S HCN H H CH(CH3)2
HB*OOC R
H
(CH3)2HC Diastereomers Separate by fractional crystallization NaOH, H2O, 0C
O B NCH H
NaOH, H2O, 0C
Remove brucine and hydrolyze amide
COO S
OOC R
H
H
CH(CH3)2 70%
(CH3)2HC 70%
H3N
Enantiomer (S)-Valine
NH3
Enantiomer (R)-Valine
In an alternative approach, the stereocenter at C2 is formed enantioselectively, such as in enantioselective hydrogenations of a,b-unsaturated amino acids (Section 12-2). Nature makes use of this strategy in the biosynthesis of amino acids. Thus, the enzyme glutamate dehydrogenase converts the carbonyl group in 2-oxopentanedioic acid into the amine substituent in (S)-glutamic acid by a biological reductive amination (for chemical reductive aminations, see Section 21-6). The reducing agent is NADH (Chemical Highlight 25-2). O B HOOCCH2CH2CCOOH
NH
NH3
H
NADH, glutamate dehydrogenase
2-Oxopentanedioic acid
NAD
(S)-Glutamic acid
(S)-Glutamic acid is the biosynthetic precursor of glutamine, proline, and arginine. Moreover, it functions to aminate other 2-oxo acids, with the help of another type of enzyme, a transaminase, making additional amino acids available. NH
3
A H O C O COO A R
O B RCCOO
Transaminase
O B RCCOO
NH
3
A H O C O COO A R
In Summary Optically pure amino acids can be obtained by resolution of a racemic mixture or by enantioselective formation of the C2 stereocenter.
3
A HOOCCH2CH2CHCOO
H2O
1221
1222
Chapter 26
Amino Acids, Peptides, Proteins, and Nucleic Acids
CH EMICAL H IG H L I G H T 2 6 - 2 Enantioselective Synthesis of Optically Pure Amino Acids: Phase-Transfer Catalysis A better alternative to the preparation of pure enantiomers by resolution of a-amino acid racemates is their direct generation from achiral precursors. For this purpose, we can use enantiopure chiral catalysts in the synthetic step that engenders chirality at C2. An example is the catalytic asymmetric (“enantioselective”) hydrogenation of a-enamino acids (Chemical Highlight 5-4 and Section 12-2). An alternative is the alkylation of suitable derivatives of glycine in the presence of optically active ammonium salts. For example, the imine-ester derivative of glycine shown on the facing page was benzylated to give 66% enantiomeric excess (Section 5-2) of the R product, using a biphasic aqueous CH2Cl2 system and 0.1 equivalent of an added ammonium salt of cinchonine. Cinchonine is an alkaloid obtained cheaply from the South American cinchona tree. Crystallization and hydrolysis of the product’s imine and ester functions gave optically pure phenylalanine. The first step of this process is carried out in a rapidly stirred mixture of dichloromethane, which contains the organic starting materials, and basic water. The ammonium salt, composed of hydrophobic substituents around the positively charged nitrogen, is soluble in both phases and shuttles back and forth between them, carrying with it either chloride or hydroxide as counterion. Hydroxide in the organic phase is responsible for a-deprotonation of the
A cinchona tree in the Amazon rainforest.
protected glycine. The enolate so formed is not free, but is ion paired with the chiral ammonium ion. As a result, alkylation by SN2 occurs preferentially from one side (by diastereomeric transition states; Section 5-7) to generate an excess of one enantiomer of protected alanine. In other words, the handedness of the optically pure catalyst ensures preferential formation of one enantiomer (here R), purified subsequently by crystallization. Quaternary ammonium salts, such as tetrabutylammonium, have been used more generally in such biphasic
26-4 Peptides and Proteins: Amino Acid Oligomers and Polymers Amino acids are very versatile biologically because they can be polymerized. In this section, we describe the structure and properties of such polypeptide chains. Long polypeptide chains are called proteins (somewhat arbitrarily defined as .50 amino acids) and are one of the major constituents of biological structures. Proteins serve an enormous variety of biological functions; these functions are often facilitated by the twists and folds of the component chains.
Amino acids form peptide bonds 2-Amino acids are the monomer units in polypeptides. The polymer forms by repeated reaction of the carboxylic acid function of one amino acid with the amine group in another to make a chain of amides (Section 20-6). The amide linkage joining amino acids is also called a peptide bond. Peptide bond
R O A B 2n HN O C O COH A A H H 2-Amino acid (␣-Amino acid)
R O R O A B A B O (NH O C O C O NH O C O C)n A A H H
2n H2O
Polyamino acid (Polypeptide)
The oligomers formed by linking amino acids in this way are called peptides. For example, two amino acids give rise to a dipeptide, three to a tripeptide, and so forth. The individual amino acid units forming the peptide are referred to as residues. In some proteins, two or more polypeptide chains are linked by disulfide bridges (Sections 9-10 and 26-1).
26-4 Peptides and Proteins
Chapter 26
1223
O
(C6H5)2C PNCH2COOC(CH3)3
CH2Br
N
Cinchonine salt (0.1 equiv.), NaOH, H2O, CH2Cl2
} H
75%
H O
1. Crystallize 2. H, H2O,
H3N } H
OH ´
O N
O
83% R, 17% S
Cl
N
H Cl
(R)-Phenylalanine N-4-Chlorobenzylcinchonium chloride (Cinchonine marked red)
systems to enable normally insoluble counterions to enter the organic phase and react with organic substrates. The observed reactivity is called phase-transfer catalysis, of which the above is an example. Other applications are the SN2 reaction (Team Problem 62 in Chapter 21), MnO42 oxidations (Section 12-11), dichlorocarbene additions
(Section 12-9), and hydride reductions (Section 8-6). The employment of minimal quantities of organic solvents in conjunction with water, the simplicity of the reaction procedure, the catalytic nature of the process, and its selectivity makes phase-transfer catalysis a “greener” alternative to many homogeneous transformations.
Of great importance for the structure of polypeptides is the fact that the peptide bond is fairly rigid at room temperature and planar, a result of the conjugation of the amide nitrogen lone electron pair with the carbonyl group (Section 20-1). The N–H hydrogen is almost always located trans to the carbonyl oxygen, and rotation about the C–N bond is slow because the C–N bond has partial double-bond character. The result is a relatively short bond (1.32 Å), between the length of a pure C–N single bond (1.47 Å, Figure 21-1) and that of a C–N double bond (1.27 Å). On the other hand, the bonds adjacent to the amide function enjoy free rotation. Thus, polypeptides are relatively rigid but nevertheless sufficiently mobile to adopt a variety of conformations. Hence, they may fold in many different ways. Most biological activity is due to such folded arrangements; straight chains are usually inactive. Planarity Induced by Resonance in the Peptide Bond Rotation fast
ðOð H R H ≥$ B A N H EC H EC H š C N C ≥( B A R H H O
ðš H Oð H R ≥$ A A N H EC N EC H N C C ≥( B A R H H O
O N
1.51 Å
R
R
1.24 Å
1.32 Å
N
1.46 Å
Rotation slow
Polypeptides are characterized by their sequence of amino acid residues In drawing a polypeptide chain, the amino end, or N-terminal amino acid, is placed at the left. The carboxy end, or C-terminal amino acid, appears at the right. The configuration at the C2 stereocenters is usually presumed to be S.
O
1224
Amino Acids, Peptides, Proteins, and Nucleic Acids
Chapter 26
How to Draw the Structure of a Tripeptide R O R O R A B A B A H3N O C O C O N O C O C O N O C O COO A A A A A H H H H H C-terminus
N-terminus
Amino acid 1
Amino acid 2
Amino acid 3
The chain incorporating the amide (peptide) bonds is called the main chain, the substituents R, R9, and so forth, are the side chains. The naming of peptides is straightforward. Starting from the amino end, the names of the individual residues are simply connected in sequence, each regarded as a substituent to the next amino acid, ending with the C-terminal residue. Because this procedure rapidly becomes cumbersome, the three-letter abbreviations listed in Table 26-1 are used for larger peptides.
O CH3 A B H3NCH2 O C O NHCHCOO
CH3 O B H3NCH O C O NHCH2COO A
CH3 (CH3)2CH A A HCOH CH2 O C6H5CH2 O A A B B A H3NCH O C O NHCH O C O NHCHCOO O O
O
N
O
O N
N
O O
O
Glycylalanine Gly-Ala
N
N
O N
Alanylglycine Ala-Gly
O
N
Phenylalanylleucylthreonine Phe-Leu-Thr
Let us look at some examples of peptides and their structural variety. A dipeptide ester, aspartame, is a low-calorie artificial sweetener (NutraSweet; see Chapter Integration Problem 26-28). In the three-letter notation, the ester end is denoted by OCH3. Glutathione, a tripeptide, is found in all living cells, and in particularly high concentrations in the lens of the eye. It is unusual in that its glutamic acid residue is linked at the g-carboxy group (denoted g-Glu) to the rest of the peptide. It functions as a biological reducing agent by being readily oxidized enzymatically at the cysteine mercapto unit to the disulfide-bridged dimer. MODEL BUILDING
COO A CH2 O C6H5CH2 O A B B A H3NCH O C O NH O CH O COCH3
COO O HSCH2 O A B B A H3NCHCH2CH2 O C O NHCHO C O NHCH2COOH
Aspartylphenylalanine methyl ester Asp-Phe-OCH3 (Aspartame)
␥-Glutamylcysteinylglycine ␥-Glu-Cys-Gly (Glutathione)
26-4 Peptides and Proteins
1225
Chapter 26
Orn
Gramicidin S is a cyclic peptide antibiotic constructed out of two identical pentapeptides that have been joined head to tail. It contains phenylalanine in the R configuration and a rare amino acid, ornithine [Orn, a lower homolog (one less CH2 group) of lysine]. In the short notation in which gramicidin S is shown in the margin, the sense in which the amino acids are linked (amino-to-carboxy direction) is indicated by arrows.
Leu
Val
(R)-Phe
Pro
Pro
(R)-Phe
Val
Leu
Val Orn
Gramicidin S
Orn
O
Pro
Leu
N O
(R)-Phe
N N N O
O N O N
N
N
N
O
N
N
O N
O
Gramicidin S. (Computer-generated model, courtesy Professor Evan R. Williams and Dr. Richard L. Wong, University of California at Berkeley. The red dashed lines represent hydrogen bonds.)
Insulin illustrates the three-dimensional structure adopted by a complex sequence of amino acids (Figure 26-1). This protein hormone is an important drug in the treatment of diabetes because of its ability to regulate glucose metabolism. Insulin contains 51 amino acid residues incorporated into two chains, denoted A and B. The chains are connected by two disulfide bridges, and there is an additional disulfide linkage connecting the cysteine residues at positions 6 and 11 of the A chain, causing it to loop. Both chains fold up in a way that minimizes steric interference and maximizes electrostatic, London, and hydrogenbonding attractions. These forces give rise to a fairly condensed three-dimensional structure (Figure 26-2). Because synthetic methods gave only low yields, insulin used to be isolated from the pancreas of slaughtered cows and pigs, purified, and sold as such. In the 1980s, the development of genetic engineering methods (Section 26-11) allowed the cloning of the human gene that codes for insulin. The gene was placed into bacteria modified to produce the drug continuously. In this way, enough material is generated to treat millions of diabetics throughout the world. Figure 26-1 Bovine (cattle) S S A A insulin is made up of two amino Gly-Ile-Val-Glu-Gln-Cys-Cys-Ala-Ser-Val-Cys-Ser-Leu-Tyr-Gln-Leu-Glu-Asn-Tyr-Cys-Asn acid chains, linked by disulfide A A 21 5 10 15 bridges. The amino (N-terminal) S S end is at the left in both chains. A D S S B chain A A Phe-Val-Asn-Gln-His-Leu-Cys-Gly-Ser-His-Leu-Val-Glu-Ala-Leu-Tyr-Leu-Val-Cys-Gly-Glu-Arg-Gly-Phe-Phe-Tyr-Thr-Pro-Lys-Ala A chain
5
10
15
20
25
30
1226
Chapter 26
Amino Acids, Peptides, Proteins, and Nucleic Acids
Figure 26-2 Three-dimensional structure of insulin. Residues in chain A are blue, those in B green. The disulfide bridges are indicated in red. (After Biochemistry, 6th ed., by Jeremy M. Berg, John L. Tymoczko, and Lubert Stryer. W. H. Freeman and Company. Copyright © 1975, 1981, 1988, 1995, 2002, 2007.)
Exercise 26-10 Vasopressin, also known as antidiuretic hormone, controls the excretion of water from the body. Draw its complete structure. (Caution: There is a disulfide bridge between the two cysteine residues.) S
A
S
A
Cys-Tyr-Phe-Gln-Asn-Cys-Pro-Arg-Gly-NH2
Proteins fold into pleated sheets and helices: secondary and tertiary structure
ðš O O
š N
ðš O
Hydrogen bond
š N
O
H
H
Insulin and other polypeptide chains adopt well-defined three-dimensional structures. Whereas the sequence of amino acids in the chain defines the primary structure, the folding pattern of the chain induced by the spatial arrangement of close-lying amino acid residues gives rise to the secondary structure of the polypeptide. The secondary structure results mainly from the rigidity of the amide bond and from hydrogen (and other noncovalent) bonding (margin) along the chain(s). Two important arrangements are the pleated sheet, or b configuration, and the a helix. In the pleated sheet (also called b sheet; Figure 26-3), two chains line up with the amino groups of one peptide opposite the carbonyl groups of a second, thereby allowing hydrogen bonds to form. Such bonds can also develop within a single chain if it loops back on itself. Multiple hydrogen bonding of this type can impart considerable rigidity to a system. The planes of adjacent amide linkages form a specific angle, a geometry that produces the observed pleated-sheet structure, in which the R groups protrude above and below at each kink. The A helix, as shown in Figure 26-4, allows for intramolecular hydrogen bonding between nearby amino acids in the chain: The carbonyl oxygen of each amino acid is interacting with the amide hydrogen four residues ahead. There are 3.6 amino acids per turn of the helix, two equivalent points in neighboring turns being about 5.4 Å apart. The CPO and NOH bonds point in opposite directions and are roughly aligned with the helical axis. On the other hand, the (hydrophobic) R groups point away from the helix. Not all polypeptides adopt idealized structures such as these. If too much charge of the same kind builds up along the chain, charge repulsion will enforce a more random orientation. In addition, the rigid proline, because its amino nitrogen is part of the substituent ring and has no N2H available for hydrogen bonding, can cause a kink or bend in an a helix. Further folding, coiling, and aggregation of polypeptides is induced by distant residues in the chain end and give rise to their tertiary structure. A variety of forces, all arising from the R group, come into play to stabilize such molecules, including disulfide bridges, hydrogen
26-4 Peptides and Proteins
H
O C
R C
C
H
H
C
C
H
H
O
H
C
R
C H
C
R
C
C
R
H
C
H
C
R
H H
H
C
R
N C
N
H
N
C
O
O
H
O
R
N C
R
N C
H
O
C N
H
H
H
H
H
N
C
C
O
O
R
N C
H
O
H
C
R
H
O
R
N C
H
H
N
C
C
H
H
C N
O
O
R
N C
H
C
R
O
R
N C
H
H
N
C
C
H
H
N
O
O
R C
C
R
O
R
N C
N H
O
H
H
Chapter 26
R
C
N
C
H
O
~6.7 Å
N C
H
O
R
∑
N
H C
R
H C
C
N C
N
R
∑
∑
R
C
H C
R
H C
O
H
O
H
O
C
H C
N H
O
H
R
H C
Figure 26-3 (A) The pleated sheet, or b configuration, which is held in place by hydrogen bonds (dotted lines) between two polypeptide strands. (After “Proteins,” by Paul Doty, Scientific American, September 1957. Copyright © 1957, Scientific American, Inc.) (B) The peptide bonds define the individual pleats (shaded in yellow); the positions of the side chains, R, are alternately above and below the planes of the sheets. The dotted lines indicate hydrogen bonds to a neighboring chain or to water.
H
R H
C O
C N O
R N C
O
H
N
H
C
N
C C
C N CH R R
O
O
O C
R
O
H
N H
N H
H
O O
O
H
R H
N C H
R N C
C
C
C C
C
H
N C
H C
N
H
C H
C H
H
O
C
R
H
R
H
H
C
H R C H
N
N O
H O
C
C
R
C R
~5.4Å Figure 26-4 The a helix, in which the polymer chain is arranged as a right-handed spiral held rigidly in shape by intramolecular hydrogen bonds (red dotted lines). (After “Proteins,” by Paul Doty, Scientific American, September 1957. Copyright © 1957, Scientific American, Inc.)
bonds, London forces, and electrostatic attraction and repulsion. There are also micellar effects (Chemical Highlight 19-1 and Section 20-5): The polymer adopts a structure that maximizes exposure of polar groups to the aqueous environment while minimizing exposure of hydrophobic groups (e.g., alkyl and phenyl), the “hydrophobic effect” (Section 8-2, Chemical Highlight 19-1). Pronounced folding is observed in the globular proteins, many of which perform chemical transport and catalysis (e.g., myoglobin and hemoglobin, Section 26-8, Figure 26-8). In the
H
1227
1228
Amino Acids, Peptides, Proteins, and Nucleic Acids
Chapter 26
fibrous proteins, such as myosin (in muscle), fibrin (in blood clots), and a-keratin (in hair, nails, and wool), several a helices are coiled to produce a superhelix (Figure 26-5). The tertiary structures of enzymes and transport proteins (proteins that carry molecules from place to place) usually give rise to three-dimensional pockets, called active sites or binding sites. The size and shape of the active site provide a highly specific “fit” for the substrate or ligand, the molecule on which the protein carries out its intended function. The inner surface of the pocket typically contains a specific arrangement of the side chains of polar amino acids that attracts functional groups in the substrate by hydrogen bonding or ionic interactions. In enzymes, the active site aligns functional groups and additional molecules in a way that promotes their reactions with the substrate. An example is the active site of chymotrypsin, a mammalian digestive enzyme responsible for the degradation of proteins in food. Chymotrypsin accomplishes the hydrolysis of peptide bonds at body temperature and at physiological pH. Recall that ordinary amide hydrolysis requires much more drastic conditions (Section 20-6). Moreover, the enzyme also recognizes specific peptide linkages that are targeted for selective cleavage, such as the carboxy end of phenylalanine residues (see Section 26-5, Table 26-2). How does it do that? A simplified picture of the active part of this large molecule (of dimensions 51 3 40 3 40 Å) in this process is shown in the following scheme. Peptide Hydrolysis in the Active Site of Chymotrypsin
Aspartic acid
Hydrophobic pocket
O
Substrate peptide cleavage
O −
Aspartic acid
O O − H
H Histidine
Histidine
N
H
NH
O
O
N
H NH
O
C
H NH
C
H
C
C
O
N
N
O
H Serine
Substrate peptide
Chymotrypsin chain
Serine
NH2
Aspartic acid
O O −
Bound ester hydrolysis
Note: In this scheme, the red arrows depict only the respective addition steps of the two addition–elimination sequences.
H Histidine
N
H O
C
C
O
N
OH
Serine
H NH
26-4 Peptides and Proteins
Chapter 26
1229
The enzyme has four important, close-lying parts, all of which work together to facilitate the hydrolysis reaction: a hydrophobic pocket, and the residues of aspartic acid, histidine, and serine. The hydrophobic pocket (see Section 8-2) helps to bind the polypeptide to be “digested” by attraction of the hydrophobic phenyl substituent of one of its component phenylalanine residues. With the phenyl group held in this pocket, the three amino acid residues cooperate in a proton-transfer relay sequence to effect nucleophilic addition–elimination (Sections 19-7 and 20-6) of the serine hydroxy group to the carbonyl function of Phe, releasing the amine part of the cleaved polypeptide. The remainder of the substrate is held by an ester linkage to the enzyme, positioned to undergo ester hydrolysis (Section 20-4) by a water molecule. This reaction is aided by a tandem proton-transfer sequence similar to that used for hydrolyzing the peptide bond. With the link to the enzyme broken, the carboxy segment of the original substrate is now free to leave the intact active site of chymotrypsin, making room for another polypeptide. Exercise 26-11
20 Å
The scheme just shown omits the respective elimination steps of the two nucleophilic addition– elimination reactions. Show how the enzyme aids in accelerating them as well. [Hint: Draw the result of the “electron pushing” depicted in the first (or second) picture of the scheme and think about how a reversed electron and proton flow might help.]
Denaturation, or breakdown of the tertiary structure of a protein, usually causes precipitation of the protein and destroys its catalytic activity. Denaturation is caused by exposure to excessive heat or extreme pH values. Think, for example, of what happens to clear egg white when it is poured into a hot frying pan or to milk when it is added to lemon tea. Some molecules, such as hemoglobin (Section 26-8), also adopt a quaternary structure (see Figure 26-9), in which two or more polypeptide chains, each with its own tertiary structure, combine to form a larger assembly. A simplified picture of the progression from primary to quaternary structures is given below. The Progression from Primary to Secondary, Tertiary, and Quaternary Polypeptide Structures Amino acids
Pleated sheet
Alpha helix
Figure 26-5 Idealized picture of a superhelix, a coiled coil.
Pleated sheet
α-helix Primary
Secondary
Tertiary
The three-dimensional constellation of polypeptides is a direct consequence of primary structure: In other words, the amino acid sequence specifies in which way the chain will coil, aggregate, and otherwise interact with internal and external molecular units. Therefore, knowledge of this sequence is of paramount importance in the understanding of protein structure and function. How to obtain this knowledge is the subject of the next section.
In Summary Polypeptides are polymers of amino acids linked by amide bonds. Their amino acid sequences can be described in a shorthand notation using the three- or one-letter abbreviations compiled in Table 26-1. The amino end group is placed at the left, the carboxy end at the right. Polypeptides can be cyclic and can also be linked by disulfide and hydrogen bonds. The sequence of amino acids is the primary structure of a polypeptide, folding gives rise to its secondary structure, further folding and coiling produce its tertiary structure, and aggregation of several polypeptides results in the quaternary structure.
Quaternary
1230
Chapter 26
Amino Acids, Peptides, Proteins, and Nucleic Acids
26-5 Determination of Primary Structure: Amino Acid Sequencing Biological function in polypeptides and proteins requires a specific three-dimensional shape and arrangement of functional groups, which, in turn, necessitate a definite amino acid sequence. One “monkey wrench” residue in an otherwise normal protein can completely alter its behavior. For example, sickle-cell anemia, a potentially lethal condition, is the result of changing a single amino acid in hemoglobin (Section 26-8). The determination of the primary structure of a protein, called amino acid or polypeptide sequencing, can help us to understand the protein’s mechanism of action. In the late 1950s and early 1960s, amino acid sequences were discovered to be predetermined by DNA, the molecule containing our hereditary information (Section 26-9). Thus, through a knowledge of protein primary structure, we can learn how genetic material expresses itself. Functionally similar proteins in related species should, and do, have similar primary structures. The closer their sequences of amino acids, the more closely the species are related. Polypeptide sequencing therefore strikes at the heart of the question of the evolution of life itself. This section shows how chemical means, together with analytical techniques, allow us to obtain this information.
First, purify the polypeptide
Electrophoresis of proteins on polyacrylamide. Each column of blue bands illustrates the separation of a mixture of proteins into their individual components.
The problem of polypeptide purification is an enormous one, and attempts at its solution consume many days in the laboratory. Several techniques can separate polypeptides on the basis of size, solubility in a particular solvent, charge, or ability to bind to a support. Although detailed discussions are beyond the scope of this book, we shall briefly describe some of the more widely used methods. In dialysis, the polypeptide is separated from smaller fragments by filtration through a semipermeable membrane. A second method, gel-filtration chromatography, uses a carbohydrate polymer in the form of a column of beads as a support. Smaller molecules diffuse more easily into the beads, spending a longer time on the column than large ones do; thus, they emerge from the column later than the large molecules. In ion-exchange chromatography, a charged support separates molecules according to the amount of charge that they carry. Another method based on electric charge is electrophoresis. A spot of the mixture to be separated is placed on a plate covered with a thin layer of chromatographic material (such as polyacrylamide) that is attached to two electrodes. When the voltage is turned on, positively charged species (e.g., polypeptides rich in protonated amine groups) migrate toward the cathode, negatively charged species (carboxy-rich peptides) toward the anode. The separating power of this technique is extraordinary. More than a thousand different proteins from one species of bacterium have been resolved in a single experiment. Finally, affinity chromatography exploits the tendency of polypeptides to bind very specifically to certain supports by hydrogen bonds and other attractive forces. Peptides of differing sizes and shapes have differing retention times in a column containing such a support.
Second, determine which amino acids are present When the polypeptide strand has been purified, the next step in structural analysis is to establish its composition. To determine which amino acids and how much of each is present in the polypeptide, the entire chain is degraded by amide hydrolysis (6 N HCl, 1108C, 24 h) to give a mixture of the free amino acids. The mixture is then separated and its composition recorded by an automated amino acid analyzer. This instrument consists of a column bearing a negatively charged support, usually containing carboxylate or sulfonate ions. The amino acids pass through the column in slightly acidic solution. They are protonated to a greater or lesser degree, depending on their structure, and therefore are more or less retained on the column. This differential retention separates the amino acids, and they come off the column in a specific order, beginning with the most acidic and ending with the most basic. At the end of the column is a reservoir containing a special indicator. Each amino acid produces a violet color (see Chapter Integration Problem 26-29) whose intensity is proportional to the amount of that
26-5 Amino Acid Sequencing
Methionine
Absorbance
Threonine
Histidine Valine
Aspartic acid
Chapter 26
Serine Glutamic acid
Glycine
Alanine
Isoleucine Leucine
Phenylalanine Lysine NH3
Tyrosine Cysteine
Proline
Volume of solvent eluted pH
3.25
4.25
Figure 26-6 A chromatogram showing the presence of various amino acids separated by an amino acid analyzer using a polysulfonated ion-exchange resin. The more acidic products (e.g., aspartic acid) are generally eluted first. Ammonia is included for comparison.
acid present and is recorded in a chromatogram (Figure 26-6). The area under each peak is a measure of the relative amount of a specific amino acid in the mixture. The amino acid analyzer can readily establish the composition of a polypeptide. For example, the chromatogram of hydrolyzed glutathione (Section 26-4) gives three equal-sized peaks, corresponding to Glu, Gly, and Cys. Exercise 26-12 Hydrolysis of the A chain in insulin (Figure 26-1) gives a mixture of amino acids. What are they and what is their relative abundance in the mixture?
Sequence the peptide from the amino (N-terminal) end Once we know the gross makeup of a polypeptide, we can determine the order in which the individual amino acids are bound to one another — the amino acid sequence. Several different methods can reveal the identity of the residue at the amino end. Most exploit the uniqueness of the free amino substituent, which may enter into specific chemical reactions that serve to “tag” the N-terminal amino acid. One such procedure is the Edman* degradation, and the reagent used is phenyl isothiocyanate, C6H5NPCPS (a sulfur analog of an isocyanate; Section 20-7). Recall (Section 20-7) that isocyanates are very reactive with respect to nucleophilic attack, and the same is true of their sulfur analogs. In the Edman degradation, the terminal amino group adds to the isothiocyanate reagent to give a thiourea derivative (refer to Section 20-6 for the urea function). Mild acid causes extrusion of the tagged amino acid as a phenylthiohydantoin, leaving the remainder of the polypeptide unchanged (see also Exercise 26–13). The phenylthiohydantoins of all amino acids are well known, so the N-terminal end of the original polypeptide can be readily identified. The new chain, carrying a new terminal amino acid, is now ready for another Edman degradation to tag the next residue, and so forth. The entire procedure has been automated to allow the routine identification of polypeptides containing 50 or more amino acids. Beyond that number, the buildup of impurities becomes a serious impediment. The reason for this drawback is that each degradation round, though proceeding in high yield, is not completely quantitative, thus leaving small quantities of incompletely reacted peptide admixed with the new one. You can readily envisage that, with each step, this problem increases until the mixtures become intractable. *Professor Pehr V. Edman (1916 – 1977), Max Planck Institute for Biochemistry, Martinsried, Germany.
5.28
Arginine
1231
1232
Chapter 26
Amino Acids, Peptides, Proteins, and Nucleic Acids
Edman Degradation of the A Chain of Insulin
š N PCPS
CH2CH3 CH3 A A O CH3CH O CH3CH O A A B B B H2š NCH2C O N O CH O C O N O CH O C O N H H H
the A chain of insulin
H2O
CH3 CH2CH3 A A S O CH3CH O CH3CH O A A B B B B NHCNHCH2C O N O CH O C O N O CH O C O N H H H š Thiourea
the A chain of insulin
H, H2O
S N
NH CH2
CH3 CH2CH3 A A CH3CH O CH3CH O A A B B H2NCH O C O N O CH O C O N H H
O
the A chain of insulin
1. C6H5NPCPS 2. H, H2O
Phenylthiohydantoin derived from glycine
S N O
NH CHCHCH3 A CH2CH3
CH3 A CH3CH O B A H2NCH O C O N H
the A chain of insulin
Phenylthiohydantoin derived from isoleucine
etc. Exercise 26-13 H C6H5
N
N
C
S
O
R H
The pathway of phenylthiohydantoin formation is not quite as straightforward as indicated by the red arrow in step 2 of the preceding scheme, but rather proceeds through the intermediacy of an isomeric thiazolinone (margin), which then rearranges under acidic conditions to the more stable phenylthiohydantoin. Formulate a mechanism for the formation of this compound (R 5 H) from the acid treatment of the phenylthiourea of glycine amide, C6H5NHC(5S)NHCH2C(5O)NH2. [Hint: Sulfur is more nucleophilic than nitrogen (Section 6-8).]
Exercise 26-14 Polypeptides can be cleaved into their component amino acid fragments by treatment with dry hydrazine. This method reveals the identity of the carboxy end. Explain.
The chopping up of longer chains is achieved with enzymes The Edman procedure allows for the ready sequencing of only relatively short polypeptides. For longer ones (e.g., those with more than 50 residues), it is necessary to cleave the larger chains into shorter fragments in a selective and predictable manner. These cleavage methods
26-5 Amino Acid Sequencing
Table 26-2
Chapter 26
Specificity of Hydrolytic Enzymes in Polypeptide Cleavage
Enzyme
Site of cleavage
Trypsin Clostripain Chymotrypsin Pepsin Thermolysin
Lys, Arg, carboxy end Arg, carboxy end Phe, Trp, Tyr, carboxy end Asp, Glu, Leu, Phe, Trp, Tyr, carboxy end Leu, Ile, Val, amino end
rely mostly on hydrolytic enzymes. For example, trypsin, a digestive enzyme of intestinal liquids, cleaves polypeptides only at the carboxy end of arginine and lysine. Selective Hydrolysis of the B Chain of Insulin by Trypsin Phe-Val-Asn-Gln-His-Leu-Cys-Gly-Ser-His-Leu-Val-Glu-Ala-Leu-Tyr-Leu-Val-Cys-Gly-Glu-Arg-Gly-Phe-Phe-Tyr-Thr-Pro-Lys-Ala 5
10
15
20
25
30
Trypsin, H2O
Phe-Val-Asn-Gln-His-Leu-Cys-Gly-Ser-His-Leu-Val-Glu-Ala-Leu-Tyr-Leu-Val-Cys-Gly-Glu-Arg 1 Gly-Phe-Phe-Tyr-Thr-Pro-Lys 1 Ala
A more selective enzyme is clostripain, which cleaves only at the carboxy end of arginine. In contrast, chymotrypsin, which, like trypsin, is found in mammalian intestines, is less selective and cleaves at the carboxy end of phenylalanine (see Section 26-4), tryptophan, and tyrosine. Other enzymes have similar selectivity (Table 26-2). In this way, a longer polypeptide is broken down into several shorter ones, which may then be sequenced by the Edman procedure. After a first enzymatic cleavage, sequences of segments of the polypeptide under investigation are determined, but the order in which they are linked is not. For this purpose, selective hydrolysis is carried out a second time, by using a different enzyme that provides pieces in which the connectivities broken under the first conditions are left intact, so-called overlap peptides. The solution is then found by literally “piecing” together the available information like a puzzle. Exercise 26-15
Working with the Concepts: Determining Amino Acid Sequences A polypeptide containing 21 amino acids was hydrolyzed by thermolysin. The products of this treatment were Gly, Ile, Val-Cys-Ser, Leu-Tyr-Gln. Val-Glu-Gln-Cys-Cys-Ala-Ser, and Leu-GluAsn-Tyr-Cys-Asn. When the same polypeptide was hydrolyzed by chymotrypsin, the products were Cys-Asn, Gln-Leu-Glu-Asn-Tyr, and Gly-Ile-Val-Glu-Gln-Cys-Cys-Ala-Ser-Val-Cys-SerLeu-Tyr. Give the amino acid sequence of this molecule. Strategy Piecing together puzzles of this type is best done visually. First lay out all the individual products, best in order of decreasing length. Starting with the largest oligopeptide resulting from the hydrolysis of the starting material by one enzyme, look for matching partial sequences of pieces obtained in the hydrolysis by the second enzyme. One of the remaining oligopeptides should provide an overlap sequence with a second by lining it up against one of the termini of the starting large fragment, thus allowing the gradual lengthening of it, until all degradation products have been reassembled to the original polypeptide. Solution • Product tabulation: Thermolysin Hydrolysis
Chymotrypsin Hydrolysis
Val-Glu-Gln-Cys-Cys-Ala-Ser (a) Leu-Glu-Asn-Tyr-Cys-Asn (b) Leu-Tyr-Gln (c) Val-Cys-Ser (d) Gly (e) Ile (f)
Gly-Ile-Val-Glu-Gln-Cys-Cys-Ala-SerVal-Cys-Ser-Leu-Tyr (g) Gln-Leu-Glu-Asn-Tyr (h) Cys-Asn (i)
1233
Amino Acids, Peptides, Proteins, and Nucleic Acids
• The largest fragment is (g), which arises from the chymotrypsin hydrolysis, and we can readily line up several of the pieces of the thermolysin treatment against it: Gly-Ile-Val-Glu-Gln-Cys-Cys-Ala-Ser-Val-Cys-Ser-Leu-Tyr-
⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎨ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ⎜ ⎨ ⎜ ⎝ ⎛ ⎜ ⎨ ⎜ ⎝
(e) (f)
(a)
(d)
(g)
Leu-Tyr-Gln (c)
• This process establishes overlap peptide (c) as the link between (g) and (h), which furnishes the larger substructure (g) 1 (h): Gly-Ile-Val-Glu-Gln-Cys-Cys-Ala-Ser-Val-Cys-Ser-Leu-Tyr-Gln-Leu-Glu-Asn-Tyr(g)
(h)
⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎨ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝
Leu-Glu-Asn-Tyr-Cys-Asn (b) ⎛ ⎜ ⎨ ⎜ ⎝
Chapter 26
⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎨ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ⎜ ⎜ ⎜ ⎜ ⎨ ⎜ ⎜ ⎜ ⎜ ⎝
1234
(i) • Inspection of the right terminus of (g) 1 (h) points to (b) as the overlap peptide to the remainder of the chain and (i). The polypeptide sequence is therefore Gly-Ile-Val-Glu-Gln-Cys-Cys-Ala-Ser-Val-Cys-Ser-Leu-Tyr-Gln-Leu-Glu-Asn-Tyr-Cys-Asn • If you check Figure 26-1, you will recognize the A chain of insulin.
Exercise 26-16
Try It Yourself What are the polypeptide products obtained in the hydrolysis of the A chain of insulin by pepsin?
Protein sequencing is made possible by recombinant DNA technology Despite the success of the techniques for polypeptide sequencing described so far, allowing for the structure elucidation of hundreds of proteins, their application to large systems (i.e., those containing more than 1000 residues) is an expensive, laborious, and time-consuming business. Progress in this field was significantly impeded until the advent of recombinant DNA technology (Section 26-11). As we shall see (Section 26-10, Table 26-3), the sequences of the four bases in DNA — adenine, thymine, guanine, and cytosine — are directly correlated to the amino acid sequences of the proteins encoded by genes or the corresponding messenger RNA. Modern developments have led to the rapid automated analysis of DNA, and the knowledge gained can be immediately translated into primary protein structure. In this way, tens of thousands of proteins have been sequenced in the past few years.
In Summary The structure of polypeptides is established by various degradation schemes. First, the polymer is purified; then the kind and relative abundance of the component amino acids are determined by complete hydrolysis and amino acid analysis. The N-terminal residues can be identified by Edman degradation. Repeated Edman degradation gives the sequence of shorter polypeptides, which are made from longer polypeptides by specific enzymatic hydrolysis. Finally, recombinant DNA technology has made primary structure analysis of larger proteins relatively easy.
26-6 Synthesis of Polypeptides: A Challenge in the Application of Protecting Groups In a sense, the topic of peptide synthesis is a trivial one: Only one type of bond, the amide linkage, has to be made. The formation of this linkage was described in Section 19-10. Why discuss it further? This section shows that, in fact, achieving selectivity poses great problems, for which specific solutions have to be found. Consider even as simple a target as the dipeptide glycylalanine. Just heating glycine and alanine to make the peptide bond by dehydration would result in a complex mixture of di-, tri-, and higher peptides with random sequences. Because the two starting materials can form bonds either to their own kind or to each other, there is no way to prevent random oligomerization.
26-6 Synthesis of Polypeptides
Chapter 26
1235
An Attempt at the Synthesis of Glycylalanine by Thermal Dehydration
Gly Ala
H2O
Gly-Gly Ala-Gly Gly-Ala Ala-Ala Gly-Gly-Ala Ala-Gly-Ala etc. Desired product
Selective peptide synthesis requires protecting groups To form peptide bonds selectively, the functional groups of the amino acids have to be protected. There are several amino- and carboxy-protecting strategies available, two of which are described on the following page (see also Problem 26-55). The amino end is frequently blocked by a phenylmethoxycarbonyl group (abbreviated carbobenzoxy or Cbz), introduced by reaction of an amino acid with phenylmethyl chloroformate (benzyl chloroformate). Protection of the Amino Group in Glycine
H3NCH2COO
O B CH2OCCl
Glycine
O B CH2OCNHCH2COOH 80%
NaOH NaCl HOH
Phenylmethyl chloroformate (Benzyl chloroformate)
Phenylmethoxycarbonylglycine (Carbobenzoxyglycine, Cbz-Gly)
The amino function is deprotected by hydrogenolysis (Section 22-2), which initially furnishes the carbamic acid as a reactive intermediate (Section 20-7). Decarboxylation occurs instantly to restore the free amine. Deprotection of the Amino Group in Glycine CH2H O B CH2OCNHCH2COOH
H2, Pd–C
O B HOCNHCH2COOH
CO2
Carbamic acid function
Another amino-protecting group is 1,1-dimethylethoxycarbonyl (tert-butoxycarbonyl, Boc), introduced by reaction with bis(1,1-dimethylethyl) dicarbonate (di-tert-butyl dicarbonate). Protection of the Amino Group in Amino Acids as the Boc Derivative R
A
H3NCHCOO
O O B B (CH3)3COCOCOC(CH3)3
(CH3CH2)3N CO2, (CH3)3COH
O R B A (CH3)3COCNHCHCOOH 70–100%
Bis(1,1-dimethylethyl) dicarbonate (Di-tert-butyl dicarbonate)
1,1-Dimethylethoxycarbonylamino acid (tert-Butoxycarbonylamino acid, Boc-amino acid)
Deprotection in this case is achieved by treatment with acid under conditions mild enough to leave peptide bonds untouched. Deprotection of Boc-Amino Acids O R A B (CH3)3COCNHCHCOOH
R
HCl or CF3COOH, 25°C
A
H3NCHCOO
CO2
CH2 P C(CH3)2
H3NCH2COO 95%
1236
Chapter 26
Amino Acids, Peptides, Proteins, and Nucleic Acids
Exercise 26-17 The mechanism of the deprotection of Boc-amino acids is different from that of the normal ester hydrolysis (Section 20-4): It proceeds through the intermediate 1,1-dimethylethyl (tert-butyl) cation. Formulate this mechanism.
The carboxy terminus of an amino acid is protected by the formation of a simple ester, such as methyl or ethyl. Deprotection results from treatment with base. Phenylmethyl (benzyl) esters can be cleaved by hydrogenolysis under neutral conditions (Section 22-2).
Peptide bonds are formed by using carboxy activation With the ability to protect either end of the amino acid, we can synthesize peptides selectively by coupling an amino-protected unit with a carboxy-protected one. Because the protecting groups are sensitive to acid and base, the peptide bond must be formed under the mildest possible conditions. Special carboxy-activating reagents are used. Perhaps the most general of these reagents is dicyclohexylcarbodiimide (DCC). The electrophilic reactivity of this molecule is similar to that of an isocyanate (Section 20-7); it is ultimately hydrated to N,N9-dicyclohexylurea. Peptide Bond Formation with Dicyclohexylcarbodiimide (DCC)
š ðOð N B B C C E HšE HšEH R O N
An O-acyl isourea
D RN G
H
O B RCNHR
H N
OPC
O
N B C B N
O
RCOOH
H
N H
Dicyclohexylcarbodiimide (DCC)
N,Nⴕ-Dicyclohexylurea
The role of DCC is to activate the carbonyl group of the acid to nucleophilic attack by the amine. This activation arises from the formation of an O-acyl isourea, in which the carbonyl group possesses reactivity similar to that in an anhydride (Section 20-3). Armed with this knowledge, let us solve the problem of the synthesis of glycylalanine. We add amino-protected glycine to an alanyl ester in the presence of DCC. The resulting product is then deprotected to give the desired dipeptide. Preparation of Gly-Ala
O B (CH3)3COCNHCH2COOH Boc-Gly
H 3C O A B H2NCHCOCH2C6H5
DCC
O O H3C O B B A B (CH3)3COCNHCH2CNHCHCOCH2C6H5
Ala-OCH2C6H5
1. H, H2O 2. H2, Pd–C
Boc-Gly-Ala-OCH2C6H5
O CH3 B A H3NCH2CNHCHCOO
C6H5CH3
CO2
CH2 P C(CH3)2
Gly-Ala
For the preparation of a higher peptide, deprotection of only one end is required, followed by renewed coupling, and so forth.
26-6 Synthesis of Polypeptides
Chapter 26
1237
Exercise 26-18
Working with the Concepts: Writing Mechanisms of DCC Coupling Reactions The mechanism of peptide coupling with DCC proceeds through the O-acyl isourea shown in this section. What happens next depends on reaction conditions: In some cases amide formation occurs by direct attack of amine; in other cases a carboxylic anhydride is an intermediate, which is then attacked by the amine to generate the amide product. Formulate the two competing pathways. Strategy Since the O-acyl isourea is an analog of an anhydride, you should review the mechanisms of nucleophilic attack on this functional group (Section 20-3). We can then extrapolate to the O-acyl isourea, using either the amine or the carboxylic acid as the respective nucleophiles. Solution • The O-acyl isourea is activated by protonation at the imine nitrogen. This site is unusually basic because of resonance stabilization in the protonated species, not unlike the situation encountered in the case of the guanidino group (Section 26-1). The resulting protonated urea substituent is an excellent leaving group. We can then write an addition–elimination sequence for the attack of the amine on the carbonyl carbon to eventually furnish the amide.
ð
C6H11E HE
N
ðOð H
š š N O A C6H11
R
C6H11E EH š N Oð HE š š N O R A C6H11
RNH2
ðOð N, Nⴕ-dicyclohexylurea
R
C6H11E EH N ðš Oð NR H2 HE š š N O R A C6H11
ðOð
H
NR H2
R
š NHR
• The same protonated O-acyl isourea as above can also be attacked by carboxylate ion. Addition–elimination results in the carboxylic anhydride. ðOð ðš O
C6H11E EH O Oð š N ðš O HE R R š š O N A C6H11
ðOð ðOð
ð
ð
C6H11E EH š N Oð HE š š N O R A C6H11
R
N, Nⴕ-dicyclohexylurea
R
š O
R
• The carboxylic anhydride is attacked by the amine through the usual addition–elimination mechanism to produce the amide (Section 20-3).
Exercise 26-19
Try It Yourself DCC is also used for the synthesis of esters from carboxylic acids and alcohols. Write a mechanism.
Exercise 26-20 Propose a synthesis of Leu-Ala-Val from the component amino acids.
In Summary Polypeptides are made by coupling an amino-protected amino acid with another in which the carboxy end is protected. Typical protecting groups are readily cleaved esters and related functions. Coupling proceeds under mild conditions with dicyclohexylcarbodiimide as a dehydrating agent.
1238
Chapter 26
Amino Acids, Peptides, Proteins, and Nucleic Acids
26-7 Merrifield Solid-Phase Peptide Synthesis Polypeptide synthesis has been automated. This ingenious method, known as the Merrifield* solid-phase peptide synthesis, uses a solid support of polystyrene to anchor a peptide chain. Polystyrene is a polymer (Section 12-15) whose subunits are derived from ethenylbenzene (styrene). Although beads of polystyrene are insoluble and rigid when dry, they swell considerably in certain organic solvents, such as dichloromethane. The swollen material allows reagents to move in and out of the polymer matrix easily. Thus, its phenyl groups may be functionalized by electrophilic aromatic substitution. For peptide synthesis, a form of Friedel-Crafts alkylation is used to chloromethylate a few percent of the phenyl rings in the polymer. Electrophilic Chloromethylation of Polystyrene CH2 O CHO CH2 O CH
CH2 O CHO CH2 O CH ClCH2OCH2CH3, SnCl4 CH3CH2OH
CH2Cl Polystyrene
Functionalized polystyrene
Exercise 26-21 Formulate a plausible mechanism for the chloromethylation of the benzene rings in polystyrene. (Hint: Review Section 15-11.)
A dipeptide synthesis on chloromethylated polystyrene proceeds as follows. MATION AN I
ANIMATED MECHANISM: Merrified synthesis of peptides
Solid-Phase Synthesis of a Dipeptide O R B A (CH3)3COCNHCHCOO 1. Attachment of protected amino acid
ClCH2 Cl
O R O B A B (CH3)3COCNHCHCOCH2 2. Deprotection of amino terminus
polystyrene chain
CF3CO2H, CH2Cl2
R O A B H2NCHCOCH2 3. Coupling to protected second amino acid
polystyrene chain
polystyrene chain
O R A B (CH3)3COCNHCHCOOH, DCC
O R O R O B A B A B (CH3)3COCNHCHCNHCHCOCH2
polystyrene chain
*Professor Robert B. Merrifield (1921–2006), Rockefeller University, New York, Nobel Prize 1984 (chemistry).
26-8 Polypeptides in Nature
4. Deprotection of amino terminus
CF3CO2H, CH2Cl2
R O R O A B A B H2NCHCNHCHCOCH2 5. Disconnection of dipeptide from polymer R O R B A H3NCHCNHCHCOO A
polystyrene chain
HF
FCH2
polystyrene chain
Dipeptide
First an amino-protected amino acid is anchored on the polystyrene by nucleophilic substitution of the benzylic chloride by carboxylate. Deprotection is then followed by coupling with a second amino-protected amino acid. Renewed deprotection and final removal of the dipeptide by treatment with hydrogen fluoride complete the sequence. The great advantage of solid-phase synthesis is the ease with which products can be isolated. Because all the intermediates are immobilized on the polymer, the products can be purified by simple filtration and washing. Obviously, it is not necessary to stop at the dipeptide stage. Repetition of the deprotection–coupling sequence leads to larger and larger peptides. Merrifield designed a machine that would carry out the required series of manipulations automatically, each cycle requiring only a few hours. In this way, the first total synthesis of the protein insulin was accomplished. More than 5000 separate operations were required to assemble the 51 amino acids in the two separate chains; thanks to the automated procedure, this took only several days. Automated protein synthesis has opened up exciting possibilities. First, it is used to confirm the structure of polypeptides that have been analyzed by chain degradation and sequencing. Second, it can be used to construct synthetic proteins that might be more active and more specific than natural ones. Such proteins could be invaluable in the treatment of disease or in understanding biological function and activity.
In Summary Solid-phase synthesis is an automated procedure in which a carboxyanchored peptide chain is built up from amino-protected monomers by cycles of coupling and deprotection.
26-8 Polypeptides in Nature: Oxygen Transport by the Proteins Myoglobin and Hemoglobin Two natural polypeptides function as oxygen carriers in vertebrates: the proteins myoglobin and hemoglobin. Myoglobin is active in the muscle, where it stores oxygen and releases it when needed. Hemoglobin is contained in red blood cells and facilitates oxygen transport. Without its presence, blood would be able to absorb only a fraction (about 2%) of the oxygen needed by the body. How is the oxygen bound in these proteins? The secret of the oxygen-carrying ability of myoglobin and hemoglobin is a special nonpolypeptide unit, called a heme group, attached to the protein. Heme is a cyclic organic molecule (called a porphyrin) made of four linked, substituted pyrrole units surrounding an iron atom (Figure 26-7). The complex is red, giving blood its characteristic color. The iron in the heme is attached to four nitrogens but can accommodate two additional groups above and below the plane of the porphyrin ring. In myoglobin, one of these groups is the imidazole ring of a histidine unit attached to one of the a-helical segments of the protein (Figure 26-8A). The other is most important for the protein’s function—bound oxygen. Close to the oxygen-binding site is a second imidazole of a histidine unit, which protects this side of the heme by steric hindrance. For example, carbon monoxide, which also binds to the
Chapter 26
1239
1240
Amino Acids, Peptides, Proteins, and Nucleic Acids
Chapter 26
O CH3 O
NOH
O
N
N HO N
N
N
N
N O
Fe
N H3C
N
CH3 N
N
CH3 eme
Porphine
Figure 26-7 Porphine is the simplest porphyrin. Note that the system forms an aromatic ring of 18 delocalized p electrons (indicated in red). A biologically important porphyrin is the heme group, responsible for binding oxygen. Two of the bonds to iron are dative (coordinate covalent), indicated by arrows.
Fresh meat rapidly assumes a dark, sometimes grayish appearance. However, when it is treated with carbon monoxide, it retains its bright red color, a feature that appeals to consumers. The reason is the formation of a stable Fe–CO complex of the heme in hemoglobin.
iron in the heme group, and thus blocks oxygen transport, is prevented from binding as strongly as it normally would because of the presence of the second imidazole group. Consequently, CO poisoning can be reversed by administering oxygen to a person who has been exposed to the gas. The two imidazole substituents in the neighborhood of the iron atom in the heme group are brought into close proximity by the unique folding pattern of the protein. The rest of the polypeptide chain serves as a mantle, shielding and protecting the active site from unwanted intruders and controlling the kinetics of its action (Figure 26-8B and C). Myoglobin and hemoglobin offer excellent examples of the four structural levels in proteins. The primary structure of myoglobin consists of 153 amino acid residues of known sequence. Myoglobin has eight a-helical segments that constitute its secondary structure, the longest having 23 residues. The tertiary structure has the bends that give myoglobin its three-dimensional shape. Hemoglobin contains four protein chains: two a chains of 141 residues each, and two b chains of 146 residues each. Each chain has its own heme group and a tertiary structure similar to that of myoglobin. There are many contacts between the chains; in particular, a1 is closely attached to b1, as is a2 to b2. These interactions give hemoglobin its quaternary structure (Figure 26-9).
Histidine residue HC Histidine residue
N
HN N
C CH2
C H
N
50
C
H C Fe
O O
60
20
CH2
30
C H
40
70
Heme
Imidazole 110
10
Heme plane 90
130 80 1
140
153
150
C Figure 26-8 (A) Schematic representation of the active site in myoglobin, showing the iron atom in the heme plane bound to a molecule of oxygen and to the imidazole nitrogen atom of one histidine residue. (B) Schematic representation of the tertiary structures of myoglobin and its heme. (C) Secondary and tertiary structure of myoglobin. (After “The Hemoglobin Molecule,” by M. F. Perutz, Scientific American, November 1964. Copyright © 1964, Scientific American, Inc.)
26-9 Biosynthesis of Proteins: Nucleic Acids
β
1241
Chapter 26
Figure 26-9 The quaternary structure of hemoglobin. Each a and b chain has its own heme group. (After R. E. Dickerson and
β
I. Geis, 1969, The Structure and Action of Proteins, Benjamin-Cummings, p. 56. Copyright © 1969 by Irving Geis.)
α
Heme group
α
The folding of the hemoglobin and myoglobin of several living species is strikingly similar even though the amino acid sequences differ. This finding implies that this particular tertiary structure is an optimal configuration around the heme group. The folding allows the heme to absorb oxygen as it is introduced through the lung, hang on to it as long as necessary for transport, and release it when required.
A
O O P P O
5' end
A
O
A
H2C H
26-9 Biosynthesis of Proteins: Nucleic Acids
Base
O
H
H
How does nature assemble proteins? The answer to this question is based on one of the most exciting discoveries in science, the nature and workings of the genetic code. All hereditary information is embedded in the deoxyribonucleic acids (DNA). The expression of this information in the synthesis of all proteins, including the many enzymes necessary for cell function, is carried out by the ribonucleic acids (RNA). After the carbohydrates and polypeptides, the nucleic acids are the third major type of biological polymer. This section describes their structure and function.
H H
O
A
O O P P O
A
O
A
H2C
Base
O H
H
H
Four heterocycles define the structure of nucleic acids
H H
O
A
Considering the structural diversity of natural products, the structures of DNA and RNA are simple. All their components, called nucleotides, are polyfunctional, and it is one of the wonders of nature that evolution has eliminated all but a few specific combinations. Nucleic acids are polymers in which phosphate units link sugars, which bear various heterocyclic nitrogen bases (Figure 26-10). In DNA, the sugar units are 2-deoxyriboses, and only four bases are present: cytosine (C), thymine (T), adenine (A), and guanine (G). The sugar characteristic of RNA is ribose, and again there are four bases, but the nucleic acid incorporates uracil (U) instead of thymine.
O O P P O
A
O
A
H2C
Base
O H
H
H 3' end
H O
A
H
Nucleic Acid Sugars and Bases HOCH2
OH
O
H
HOCH2
H
H H
H HO
H
HO
2-Deoxyribose
NH2 H3C
N H
O
Cytosine (C)
OH
Ribose
NH2
O
N
H
H
H
NH N H
Thymine (T)
O
Figure 26-10 Part of a DNA chain. The base is a nitrogen heterocycle. The sugar is 2-deoxyribose.
OH
O
N N H
N N
Adenine (A)
O N N H
O NH
N Guanine (G)
NH NH2
N H
O
Uracil (U)
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Chapter 26
Exercise 26-22 Even though the preceding structures do not (except for adenine) indicate so, cytosine, thymine, guanine, and uracil are aromatic, albeit somewhat less so than the corresponding azapyridines. Explain. (Hint: Recall the discussions about amide resonance in Sections 20-1 and 26-4 and Chapter Integration Problem 25-9.)
We construct a nucleotide from three components. First, we replace the hydroxy group at C1 in the sugar with one of the base nitrogens. This combination is called a nucleoside. Second, a phosphate substituent is introduced at C5. In this way, we obtain the four nucleotides of both DNA and RNA. The positions on the sugars in nucleosides and nucleotides are designated 19, 29, and so forth, to distinguish them from the carbon atoms in the nitrogen heterocycles.
Human male chromosomes, the structures in the cell that carry the genetic information encoded in their component DNA.
NH2 N HOCH2
N
O
N N
OH
OH
Adenosine (A nucleoside)
Nucleotides of DNA O
NH2 O O
N
O O
N
N
O
O B 5 HOPOCH2 A OH 4
N
N N
O
N
N
O B HOPOCH2 A OH
N
1
O
N
N N
N 2
3
O
OH
OH 29-Deoxyadenylic acid
2ⴕ-Deoxyguanylic acid
NH2
O H3C
N H2O3POCH2
O
N
OH 2ⴕ-Deoxycytidylic acid
O
H2O3POCH2
O
OH Thymidylic acid
N N
EH O
EH NH2
26-9 Biosynthesis of Proteins: Nucleic Acids
Chapter 26
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CH EMICAL H I G H L I G H T 2 6 - 3 Synthetic Nucleic Acid Bases and Nucleosides in Medicine O
O
H N
O NH
F
H2N HO
O 5-Fluorouracil (Fluracil)
HO
N
HN
O O
N N
HO
N
A
O
H2N
CH3
HN
N
O
N
N
N3
O
9-[(2-Hydroxyethoxy)methyl]guanine (Acyclovir)
OH OH
3ⴕ-Azido-3ⴕ-deoxythymidine (Zidovudine, or AZT)
The central role played by nucleic acid replication in biology has been exploited in medicine. Many hundreds of synthetically modified bases and nucleosides have been prepared and their effects on nucleic acid synthesis investigated. Some of them in clinical use include 5-fluorouracil (fluracil), an anticancer agent, 9-[(2-hydroxyethoxy)methyl]guanine (acyclovir), which is active against two strains of herpes simplex virus, 39-azido-39-deoxythymidine (zidovudine, or AZT),
1--D-Ribofuranosyl-1,2,4,-triazole-3-carboxamide (Ribavirin)
the first approved drug (in 1987) that combats the AIDS virus, and 1-b-d-ribofuranosyl-1,2,4-triazole-3-carboxamide (ribavirin), used to treat hepatitis C and respiratory viral infections. Substances such as these may interfere with nucleic acid replication by masquerading as legitimate nucleic acid building blocks. The enzymes associated with this process are fooled into incorporating the drug molecule, and synthesis of the biological polymer cannot continue.
Nucleotides of RNA O
NH2 O O
N
O O
N
O
N
N
H2O3POCH2
N
N
N
O
N H2O3POCH2
N
N
O
N O
O
OH
Adenylic acid
OH
OH
NH2
O
N H2O3POCH2
OH
N
O
OH
Cytidylic acid
OH
Guanylic acid
N O
H2O3POCH2
OH
N
O
EH O
OH
Uridylic acid
The polymeric chain shown in Figure 26-10 is then readily derived by repeatedly forming a phosphate ester bridge from C59 (called the 59 end) of the sugar unit of one nucleotide to C39 (the 39 end) of another (Figure 26-10). In this polymer, the bases adopt the same role as that of the 2-substituent in the amino acids of a polypeptide: Their sequence varies from one nucleic acid to another and determines the fundamental biological properties of the system. Viewed from the perspective of storing
N N
EH NH2
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Amino Acids, Peptides, Proteins, and Nucleic Acids
information, polypeptides do so by using an amide polymer backbone along which differing side chains are attached, like the letters in a word. In the case of polypeptides there are 20 such letters, for the 20 natural amino acids (Table 26-1). The same goal is achieved in nucleic acids through a sugar–phosphate polymeric array that bears a sequence of amine bases, except that now we have only four letters: C, T, A, G for DNA and C, U, A, G for RNA. Information Storage in Polypeptides and Nucleic Acids A polypeptide R1
R3
R2
R4
NHO C O C A B H O A nucleic acid Base 2 Base 1
H
H H(DNA) OH(RNA) Base 3 Base 4 O O B H O O OP O O A H O
Nucleic acids form a double helix Nucleic acids, especially DNA, can form extraordinarily long chains (as long as several centimeters) with molecular weights of as much as 150 billion. Like proteins, they adopt secondary and tertiary structures. In 1953, Watson and Crick* made their ingenious proposal that DNA is a double helix composed of two strands with complementary base sequences. A crucial piece of information was that, in the DNA of various species, the ratio of adenine to thymine, like that of guanine to cytosine, is always one to one. Watson and Crick concluded that two chains are held together by hydrogen bonding in such a way that adenine and guanine, respectively, in one chain always face thymine and cytosine in the other, and vice versa (Figure 26-11). Thus, if a piece of DNA in one strand has the sequence H N O HZ O
N
Sugar
D
N
N Z HON N O
Figure 26-11 Hydrogen bonding between the base pairs adenine – thymine and guanine – cytosine. The components of each pair are always present in equal amounts. The electrostatic potential maps of the respective pairs adenine – thymine and guanine – cytosine show how regions of opposite charges (red and blue) come to lie face to face.
CH3
Adenine–thymine
N G Sugar
H O Z HO N
N
Sugar
D
N
N O HZN N N O HZ O H
N G Sugar
Guanine–cytosine
*Professor James D. Watson (b. 1928), Harvard University, Cambridge, Massachusetts, Nobel Prize 1962 (medicine); Professor Sir Francis H. C. Crick (1916–2004), Cambridge University, England, Nobel Prize 1962 (medicine).
26-9 Biosynthesis of Proteins: Nucleic Acids
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G
C A
T
Sugar
C
G
Chapter 26
Phosphorus atom
34 Å T
A A
T C
A
G
B
Figure 26-12 (A) The two nucleic acid strands of a DNA double helix are held together by hydrogen bonding between the complementary sets of bases. Note: The two chains run in opposite directions and all the bases are on the inside of the double helix. The diameter of the helix is 20 Å; base–base separation across the strands is ,3.4 Å; the helical turn repeats every 10 residues, or 34 Å. (B) Space-filling model of DNA double helix (green and red strands). In this picture, the bases are shown in lighter colors than the sugar–phosphate backbone. (C) The DNA double helix, in a view down the axis of the molecule. The color scheme in (A) and (C) matches that in Figure 26-10. (After Biochemistry, 6th ed., by Jeremy M. Berg, John L. Tymoczko, and Lubert Stryer, W. H. Freeman and Company. Copyright © 1975, 1981, 1988, 1995, 2002, 2007.)
-A-G-C-T-A-C-G-A-T-C-, this entire segment is hydrogen bonded to a complementary strand running in the opposite direction, -T-C-G-A-T-G-C-T-A-G-, as shown. O A O G O C O T O A O C O G O A O T O CO O T O C O G O A O T O G O C O T O A O GO Because of other structural constraints, the arrangement that maximizes hydrogen bonding and minimizes steric repulsion is the double helix (Figure 26-12). The cumulative hydrogenbonding energies involved are substantial considering that just single base pairing, as in Figure 26-11, is favorable by 5.8 kcal mol21 (24 kJ mol21) for G–C and 4.3 kcal mol21 (18 kJ mol21) for A–T.
DNA replicates by unwinding and assembling new complementary strands There is no restriction on the variety of sequences of the bases in the nucleic acids. Watson and Crick proposed that the specific base sequence of a particular DNA contained all genetic information necessary for the duplication of a cell and, indeed, the growth and development of the organism as a whole. Moreover, the double-helical structure suggested a way in which DNA might replicate — make exact copies of itself — and so pass on the genetic code. In this mechanism, each of the two strands of DNA functions as a template. The double helix partly unwinds, and enzymes called DNA polymerases then begin to assemble the new DNA by coupling nucleotides to one another in a sequence complementary to that in the template, always juxtaposing C to G and A to T (Figure 26-13). Eventually, two complete double helices are produced from the original. This process is at work throughout the entire
Transmission electron microscope picture of replicating DNA. Unwinding of the two complementary strands generates a “bubble” that enlarges to form a Y-shaped molecule termed a replication fork.
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Figure 26-13 Simplified model of DNA replication. The double helix initially unwinds to two single strands, each of which is used as a template for reconstruction of the complementary nucleic acid sequence.
New nucleic acid strand 5'
Unwound double helix C A
T
5'
C G
A C G
A
A T
T
T
G T C G
T
G
A
C T G
A T
G C
A
A
C
A
G A C
G C
C
T G C
T
3'
G
5' G C
A C T
5' A 3'
3'
G
T
T
G
T
Old DNA
A
G C
C
C T G
A T
A
A
A
G A C
G C
C
T G C
T G
3'
New nucleic acid strand
human genetic material, or genome — some 2.9 billion base pairs — with an error frequency of less than 1 in 10 billion base pairs.
In Summary The nucleic acids DNA and RNA are polymers containing monomeric units called nucleotides. There are four nucleotides for each, varying only in the structure of the base: cytosine (C), thymine (T), adenine (A), and guanine (G) for DNA; cytosine, uracil (U), adenine, and guanine for RNA. The two nucleic acids differ also in the identity of the sugar unit: deoxyribose for DNA, ribose for RNA. DNA replication and RNA synthesis from DNA is facilitated by the complementary character of the base pairs A – T, G – C, and A – U. The double helix partly unwinds and functions as a template for replication.
26-10 Protein Synthesis Through RNA The mechanism of duplicating the entire nucleotide sequence in DNA replication is used by nature and by chemists to obtain partial copies of the genetic code for various purposes. In nature, the most important application is the assembly of RNA, called transcription, which transcribes the parts of the DNA that contain the information (the genes) necessary to synthesize proteins in the cell. The process by which this transcribed information is decoded and used to construct proteins is called translation. The three key players in protein synthesis are the “DNA transcript” messenger RNA (mRNA), the “delivery unit” for the specific amino acids to be connected by peptide bonds, transfer RNA (tRNA), and the catalyst that enables amide bond formation — the ribosome. Protein synthesis starts with mRNA, a transcript of a piece of a single strand of partly unwound DNA (Figure 26-14). Its chain is much shorter than that of DNA, and it does not stay bound to the DNA but breaks away as its synthesis is finished. The mRNA is the template responsible for the correct sequencing of the amino acid units in proteins. How does mRNA do that? Each sequence of three bases, called a codon, specifies a particular amino acid (Table 26-3). Simple permutation of this three-base code with a total of four
DNA
Figure 26-14 Simplified picture of messenger RNA synthesis from a single strand of (partly unwound) DNA.
OAO GO CO T O AO C O
Y
Y
Y
Y
Y
Y
U O C O G OA O U O G O mRNA
OAO CO T O
Y
Y
Y
O U O G OA
26-10 Protein Synthesis Through RNA
Table 26-3
Three-Base Code for the Common Amino Acids Used in Protein Synthesis
Amino acid
Base sequence
Amino acid
Base sequence
Amino acid
Base sequence
Ala (A)
GCA GCC GCG GCU
His (H)
CAC CAU
Ser (S)
Ile (I)
AUA AUC AUU
AGC AGU UCA UCG UCC UCU
Leu (L)
CUA CUC CUG CUU UUA UUG
Thr (T)
ACA ACC ACG ACU
Trp (W)
UGG
Lys (K)
AAA AAG
Tyr (Y)
UAC UAU
Val (V)
GUA GUG GUC GUU
Chain initiation
AUG
Chain termination
UGA UAA UAG
Arg (R)
AGA AGG CGA CGC CGG CGU
Asn (N)
AAC AAU
Asp (D)
GAC GAU Met (M)
AUG
Cys (C)
UGC UGU
Phe (F)
UUU UUC
CAA CAG
Pro (P)
CCA CCC CCG CCU
Gln (Q)
Glu (E)
GAA GAG
Gly (G)
GGA GGC GGG GGU
bases gives 43 5 64 possible distinct sequences. That number is more than enough, because only 20 different amino acids are needed for protein synthesis. This might seem like overkill, but consider that the next lower alternative — namely, a two-base code — would give only 42 5 16 combinations, too few for the number of different amino acids found in natural proteins. Codons do not overlap; in other words, the three bases specifying one amino acid are not part of another preceding or succeeding codon. Moreover, the “reading” of the base sequence is consecutive; each codon immediately follows the next, uninterrupted by genetic “commas” or “hyphens.” Nature also makes full use of all 64 codons by allowing for several of them to describe the same amino acid (Table 26-3). Only tryptophan and methionine are characterized by single three-base codes. Some codons act as signals to initiate or terminate production of a polypeptide chain. Note that the initiator codon (AUG) is also the codon for methionine. Thus, if the codon AUG appears after a chain has been initiated, methionine will be produced. The complete base sequence of the DNA in a cell defines its genetic code. Mutations in the base sequence of DNA can be caused by physical (radiation) or chemical (carcinogens; see, e.g., Section 16-7) interference. Mutations can either replace one base with another or can add or delete one base or more. Here is some of the potential value of redundant codons. If, for example, the RNA sequence CCG (proline) were changed as a result of a DNA mutation to the RNA sequence CCC, proline would still be correctly synthesized.
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Exercise 26-23 What is the sequence of an mRNA molecule produced from a DNA template strand with the following composition? 59-ATTGCTCAGCTA-39
Exercise 26-24 (a) What amino acid sequence is encoded in the following mRNA base sequence (starting at the left end)? A-A-G-U-A-U-G-C-A-U-C-A-U-G-C-U-U-A-A-G-C (b) Identify the mutation that would have to occur for Trp to be present in the resulting peptide.
Figure 26-15 Representation of the biosynthesis of the tripeptide Gly-Ala-Asn. The tRNAs carrying their specific amino acids line up their anticodons along the codons of mRNA, before ribosomal enzymes accomplish amide linkage.
O G O G O A O G O C O A OAOAO C O
Y
Y
Y
Y
Y
Y
Y
Y
Y
CO COU
C O G OU
U O UO G
O
O
O
OPC
OPC
OPC
H2NCH2
H2š NCH
H2š NCH
A A
A A A
CH3
mRNA chain (with codons) tRNAs (with anticodons)
A A A
CH2
Amino acids delivered by tRNAs
A
OPC
A
NH2 Gly
Ala
Asn
With a copy of the requisite codons in hand, proteins are then synthesized along the mRNA template with the help of a set of other important nucleic acids, the tRNAs. These molecules are relatively small, containing from 70 to about 90 nucleotides. Each tRNA is specifically designed to carry one of the 20 amino acids for delivery to the mRNA in the course of protein buildup. The amino acid sequence encoded in mRNA is read codon by codon by a complementary three-base sequence on tRNA, called an anticodon. In other words, the individual tRNAs, each carrying its specific amino acid, line up along the mRNA strand in the correct order. At this stage, catalytic ribosomes (very large enzymes) containing their own RNA facilitate peptide-bond formation (Figure 26-15). As the polypeptide chain grows longer, it begins to develop its characteristic secondary and tertiary structure (a helix, pleated sheets, etc.), helped by enzymes that form the necessary disulfide bridges. All of this happens with remarkable speed. It is estimated that a protein made up of approximately 150 amino acid residues can be biosynthesized in less than 1 minute, and that without protecting groups (Section 26-6)! Clearly, nature still has the edge over the synthetic organic chemist, at least in this domain.
In Summary RNA is responsible for protein biosynthesis; each three-base sequence, or codon, specifies a particular amino acid. Codons do not overlap, and more than one codon can specify the same amino acid.
26-11 DNA Sequencing and Synthesis: Cornerstones of Gene Technology The sheep Dolly, the world’s first clone of an adult mammal cell, died in 2003 at the age of 7.
Molecular biology is undergoing a revolution because of our capability to decipher, reproduce, and alter the genetic code of organisms. Individual genes or other DNA sequences from a genome can be copied (cloned), often on a large scale. Genes of higher organisms
26-11 DNA Sequencing and Synthesis
can be expressed (i.e., they can be made to start protein synthesis) in lower organisms, or they can be modified to produce “unnatural” proteins. Modified genes have been reintroduced successfully into the organism of their origin, causing changes in the physiology and biochemistry of the “starting” host. Much of this development is due to biochemical advances such as the discovery of enzymes that can selectively cut, join, or replicate DNA and RNA. For example, restriction enzymes cut long molecules into small defined fragments that can be joined by DNA ligases to other DNAs (recombinant DNA technology). Polymerases catalyze DNA replication, and selected DNA sequences are made on a large scale by the polymerase chain reaction (described later in this section). The list goes on and on, and its contents are beyond the scope of this text. The essential foundation for these discoveries has been the knowledge of the primary structure of nucleic acids and the development of methods for their synthesis.
Rapid DNA sequencing has deciphered the human genome The sequencing of DNA can be accomplished by using both chemical (introduced by Gilbert* with his student Maxam) and enzymatic methods, especially the Sanger† dideoxynucleotide protocol. In both approaches, and in analogy to protein analysis (Section 26-5), the unwieldy DNA chain is first cleaved at specific points by enzymes called restriction endonucleases. There are more than 200 such enzymes, providing access to a multitude of overlapping sequences. In the Maxam–Gilbert procedure, described here in a simplified manner, the pieces of DNA obtained are then tagged at their 59-phosphate ends (the starting end) with a radioactive phosphorus (32P) label for analytical detection in the next step (oligonucleotides are colorless and cannot be identified visually). This step consists of subjecting multiple copies of the pure, labeled fragments in four separate experiments to chemicals that cleave only next to specific bases (i.e., only at A, G, T, and C, respectively). The conditions of cleavage are controlled so that cleavage occurs essentially only once on each copy, but, overall, at all possible places at which the particular base occurs. The products of the individual experiments are mixtures of oligonucleotides whose ends are known: Half of them contain the radioactive label at the beginning end, while the other half do not have any label and are ignored. The position of this end nucleotide in the original DNA is deduced from the length of the radioactive fragment, a measurement accomplished by electrophoresis (Section 26-5), which separates mixtures of oligonucleotides by size. The patterns of products that are made and identified in this way reveal the base sequence. For a simple analogy, picture a measuring tape that has been marked randomly by spots of four different colorless labels. You can make multiple copies of this tape, and you have four special automatic scissors that cut the tape at only one specific label. You divide your collection of tapes into four batches and let each type of scissors do its work. The result from each experiment will be a collection of tapes of different lengths indicating where on the original a specific label occurs. In the Sanger method, the piece of DNA to be sequenced, the template strand, is subjected to an enzyme, DNA polymerase, that replicates it many times (as the complementary strand, Section 26-9), starting from one end (the 39 end). Replication requires addition of the four ingredient nucleotides (abbreviated as N below) A, G, T, and C [in the form of their reactive triphosphates (TP)], respectively, for chain buildup. The trick is to add very small amounts of a modified nucleotide to the mixture, which causes termination of chain growth at the very spot at which the normal nucleotide would have been incorporated. The modification is simple: Instead of the usual 29-deoxyribose (d), we use 29,39-dideoxyribose (dd). Moreover, to visualize the end result, the base attached to the dideoxyribose is labeled with a fluorescent dye molecule that can be detected by exposure to light of the appropriate wavelength. *Professor Walter Gilbert (b. 1932), Harvard University, Nobel Prize 1980 (chemistry). † Professor Frederick Sanger (b. 1918), Cambridge University, England, Nobel Prizes 1958 (chemistry; for the structure of insulin) and 1980 (chemistry; for nucleotide sequencing).
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The Nucleotide Ingredients in the Sanger DNA Sequencing Method O O O B B B O O P O O O P O O O P O O A A A O O O
Base
O H H
H 2
3
OH
O O O B B B O O P O O O P O O O P O O A A A O O O
H H
H
3
H
H
“Normal” 2ⴕ-deoxyribonucleoside triphosphate (dNTP)
Base
O
Dye
H 2
H
H
Fluorescent-dye-labeled 2ⴕ,3ⴕ-dideoxyribonucleoside triphosphate (ddNTP)
The stepwise procedure is as follows (Figure 26-16). In the first step, multiple copies of the template strand are made by a procedure that also adds a short known nucleotide sequence to its 39-end. This auxiliary is necessary to bind to a complementary section of the template, making a short double strand that serves as a primer for DNA polymerase; in other words, it tells the enzyme where to start replicating (namely, where the known double strand ends and the unknown single strand starts). We then divide our supply of template strands to be sequenced into four portions and carry out polymerase reactions on each of them separately. In the first, we use dATP, dTTP, dGTP, and dCTP, in addition to 1% of ddATP, and let the reaction run its course. The polymerase will begin to synthesize a complementary strand to the template; that is, when it encounters a T, it will add an A; when it encounters a C, it will add a G, and vice versa. The key is that 1% of the time, when it finds a T, it will add Fragments ending with ddA
Template strand 5' end ATGACTAGGCCTGC
3' end
Primer
Step 1
DNA polymerase dATP, dTTP, dGTP, dCTP, 1% ddATP
ddACG ddATCCGGACG ddACTGATCCGGACG
Electrophoresis Step 2
5' A T G A C T A G G C C T G C
+
3' T A C T G A T C C G G A C G
Electrical poles
3' 5' Template Replicate strand strand Repeat with ddT
_
Electrophoresis plate
Repeat with ddG Repeat with ddC Figure 26-16 Sanger dideoxynucleotide sequencing of a sample oligonucleotide, called the template strand. The attached known primer sequence tells the enzyme DNA polymerase where to start replication. Replication stops when a 29,39-dideoxyribonucleotide (in the experiment shown, it is ddA) is added to the growing chain. Electrophoresis of the resulting mixture reveals the length of the fragments with the corresponding end nucleotide and, therefore, the position of this nucleotide in the replicate strand and its complement in the template strand.
26-11 DNA Sequencing and Synthesis
a defunct ddA, lacking the OH group necessary for further chain lengthening. Replication will stop and the incomplete strand will fall off the template. Note that, since 99% of the time T will be matched by dA and therefore replication will continue uninterrupted, all Ts along the sequence have essentially the same chance of being “hit” by ddA. In this way, all Ts are tagged and a collection of oligonucleotides of different lengths is assembled, all of which end with ddA. In step 2, this mixture is then separated by electrophoresis and the position of the bands visualized by fluorescent light emission of the dye that is part of ddA. The position of the bands gives the length of the fragment strand and therefore the positions of the nucleotide A in the replicate strand and T in the template strand. Having established the location of all the Ts in the template strand, the same experiment is carried out three additional times, with ddT, ddG, and ddC, respectively, as the chainstopping additives, thus giving away the position of the other three nucleotides in the sequence. The Sanger method has been automated such that oligonucleotides with 3 million bases can be reliably sequenced in a day or less. The keys to automation have been the attachment of dyes of differing color to the ddNTP building blocks, for example, red for T, green for A, yellow for G, and blue for C; the use of laser detection; parallel analysis; and the everincreasing power of computers. The sequencing machine produces chromatographic printouts providing the sequence under investigation on simple sheets of paper (Figure 26-17). The technique has become so convenient that DNA sequencing, rather than peptide sequencing, has become the method of choice for structural elucidation of the encoded proteins (see Sections 26-5 and 26-10). Moreover, it has allowed the determination of the sequence of entire genomes with billions of bases, including our own. The first genome of a live organism to be sequenced (in 1995) was that of the bacterium Haemophilus influenzae Rd. (which causes ear infections, such as meningitis), with 1.8 million base pairs (Mb). After this milestone, larger genomes were tackled, including those of yeast (12 Mb; 1996), the worm Caenorhabditis elegans (97 mB; 1998), and the fruit fly (160 Mb; 1999), as a preamble for the “real thing,” our own blueprint of hereditary information. In 1990, the Human Genome Project was begun — an international effort aimed at establishing the sequence of the 2.9 billion bases in human DNA and hence identifying the associated genes. The project was originally expected to take about 15 years, but thanks to the above advances in nucleotide sequencing, it was finished in April 2003. Only about 1% of the sequence codes for proteins (Section 26-10), which are responsible for what we are and how we function. Most of our genome (about 95%) consists of long chains of oftenrepeating segments that do not code for amino acids and that seem to be active in regulating gene expression, affecting cellular machinery in the reading of nearby genes and protein synthesis, and functioning as the “glue” of chromosomal structure. A big surprise is that we have only about 27,000 genes, many fewer than the expected 100,000, and a rather sobering number, considering that it is in the ballpark of that of the worm Caenorhabditis elegans (19,500 genes) or of the first two other mammalian genomes to be unraveled, those of the mouse (2002) and rat (2004), both with about 29,000 genes. It now appears that the
Chapter 26
The ordinary soil-dwelling worm Caenorhabditis elegans has about as many genes as we do. Surprised?
The genetic code of the soybean was determined in 2008. It has about 66,000 genes, more than twice that of the human genome. Surprised again?
GTGTGAAAT TGT TAT CGC TGT T T CC TCATGGT CATAT TGGCGTAAT 10 20 30 40
Figure 26-17 The color of the peaks in a printout from an automated DNA sequencer tells us the sequence of an oligonucleotide.
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quality of genes is as important as their quantity: Human complexity may arise from the fact that our genes can produce not only more, but also more sophisticated proteins, capable of fulfilling more than one function. Moreover, complexity is also dependent on the interplay of proteins, evidently more varied in humans. The development of automated DNA sequencing in conjunction with several ingenious time-saving strategies, including sophisticated computer analyses, has dramatically shortened the time required to decipher genomes. In 2008, the sequencing of the genes of the first (two) females was reported, accomplished in only a few months in each case. One of the donors was afflicted with cancer, and the study allowed the detection of all the mutations that might have been responsible for the onset of the disease. Such information might lead to new therapies and help doctors in their choices among existing treatments. The deciphering of the human genome has sparked the development of several new research areas attempting to grapple with the enormous quantity of data becoming available on a daily basis. For example, functional genomics aims to find out what genes do, especially as it pertains to human health. Comparative genomics is establishing similarities in the genomes of a variety of species, from pigs to Japanese puffer fish. Equally or arguably even more important is proteomics, the identification and functional study of the abundant new proteins that are being discovered as a consequence of the knowledge of the encoding gene. Compared to the genome, the corresponding proteome may be much more complex because proteins exist in a multitude of substitutionally “decorated” forms, such as phosphorylated, glycosylated, sulfated, and other variants. Some estimates place their number at 1 million. The Protein Structure Initiative is a U.S. national program devoted to determining the three-dimensional shapes of these molecules. Started in the year 2000 and greatly aided by robotic technology, it had determined over 3200 structures of proteins derived from plants, mice, yeast, and bacteria by mid-2008. Finally, bioinformatics is the application of computer technology to the discovery of more deep-seated information in biological data. To use language as an analogy: The genome has provided words; now is the time to decipher sentences and meanings. Among the tasks that will occupy the attention of scientists for decades are the determination of the number of genes, their exact locations, and their regulation and function. Unraveling these aspects of genetics may help us understand an organism’s susceptibility to disease; the details of gene expression and protein synthesis; the cooperation between proteins to elicit complex function; the evolutionary relationship among the species on our planet; the relationship of single point mutations to disease; and the details of multiple gene expression for the development of complex traits.
Like sequencing, DNA synthesis is automated You can order your custom-made oligonucleotide today and receive it tomorrow. The reason for this efficiency is the use of automated nucleotide coupling by so-called DNA synthesizers that operate on the same principle as that of the Merrifield polypeptide synthesis (Section 26-7): solid-phase attachment of the growing chain and the employment of protected nucleotide building blocks. Protection of the bases cytosine, adenine, and guanine is required at their amino functions (absent in thymine, which needs no protection), in the form of their amides. Protected (Except for Thymine) DNA Bases C6H5 A C E N O HN
N H
O H3C
N O
Cytosine (C)
C6H5 A C E N O HN NH
N H Thymine (T)
O
N N H
N N Adenine (A)
O N N H
O
NH N
N H
Guanine (G)
26-11 DNA Sequencing and Synthesis
Chapter 26
The sugar moiety is blocked at C59 as a dimethoxytrityl [di(4-methoxyphenyl)phenylmethyl, DMT] ether, readily cleaved by mild acid through an SN1 mechanism (Section 22-1 and Problem 40 in Chapter 22), much like 1,1-dimethylethyl (tert-butyl) ethers (Section 9-8, Chemical Highlight 9-2). To anchor the first protected nucleoside on the solid support, the C39-OH is attached to an activated linker, a diester. Unlike the Merrifield medium of polystyrene, the solid used in oligonucleotide synthesis is surface-functionalized silica (SiO2) bearing an amino substituent as a “hook.” Coupling to the anchor nucleoside is by amide formation. Anchoring the Protected Nucleoside on SiO2 Protected base 1
5
DMTO OO CH2
O
H H
3
DMTO OO CH2
H
H2š NCH2CH2CH2 O SiO2
Protected base 1
O
H
H H
H
H
H O A KC H O CH2CH2CO B O
H O A KC H O CH2CH2CNH(CH2)3 O SiO2 B O
NO2
Silica-anchored first nucleoside
DMT CH3O
C
OCH3
RO
With the first nucleoside in place, we are ready to attach to it the second. For this purpose, the point of attachment, the 59-OH, is deprotected with acid. Subsequent addition of a 39-OH activated nucleoside effects coupling. The activating group is an unusual phosphoramidite [containing P(III)], which, as we shall see shortly, also serves as a masked phosphate [P(V)] for the final dinucleotide and is subject to nucleophilic substitution, not unlike PBr3 (recall Sections 9-4 and 19-8). The displacement reaction is catalyzed and furnishes a phosphite derivative; the catalyst is the, again unusual, aromatic heterocycle tetrazole, a tetrazacyclopentadiene related to pyrrole (Section 25-3) and imidazole (Section 26-1). Finally, the phosphorus is oxidized with iodine to the phosphate oxidation state.
RO
DMTOCH2
š N ðN
Cl2CHCO2H, CH2Cl2
H
H
O H A KC H O CH2CH2CNH(CH2)3 O SiO2 B O
H
H
O A EPH
H
H
Nq CCH2CH2O
N H ,N
N[CH(CH3)2]2
OR
š N š N H Tetrazole
Protected base 2
O
OR A P E H
A phosphite
Dinucleotide Synthesis: Deprotection, Coupling, and Oxidation
Protected SiO2-anchored nucleoside
NR2
A phosphoramidite
Dimethoxytrityl [Di(4-methoxyphenyl)phenylmethyl]
Protected base 1 š HOCH 2 O H H
OR A P E H
N N H
1253
1254
Amino Acids, Peptides, Proteins, and Nucleic Acids
Chapter 26
DMTOCH2 H
Protected base 2
O
DMTOCH2
H
Protected base 2
O
H
H
I2, H2O, THF, 2,6-dimethylpyridine
H
H O A P Nq CCH2CH2OE HOCH2
H
H
H
H
H O A Protected O PP H base 1 f OCH2 O NCCH2CH2O H H
Protected base 1
O
H
H
H
H
H
H O A KC H O CH2CH2CNH(CH2)3 O SiO2 B O
H O A KC H O CH2CH2CNH(CH2)3 O SiO2 B O
The sequence (1) DMT hydrolysis, (2) coupling, and (3) oxidation can be repeated multiple times in the synthesizer machine until the desired oligonucleotide is assembled in its protected and immobilized form. The final task is to remove the product from the silica and deprotect the DMT-bearing terminal sugar, all the bases, and the phosphate group, without cleaving any of the other bonds. Remarkably, this task can be done in just two steps, first treatment with acid and then with aqueous ammonia, as shown here for the dinucleotide made in the preceding scheme. Liberation of the Solid-Phase-Supported and Protected Dinucleotide
DMTOCH2 H H
Protected base 2
O
HOCH2 H
H H
H
H O A Protected O PP H base 1 f OCH2 O NCCH2CH2O H H H H
Cl2CHCO2H, CH2Cl2
ðNH3, H2O
DMTOH
H2CPCHCN
H O A KC H O CH2CH2CNH(CH2)3 O SiO2 B O
Base 2
O H
H
O H A O PP H Base 1 f OCH2 O HO H H H H HO
H
Exercise 26-25
Working with the Concepts: Exploring Mechanistic Features of Solid-Phase Dinucleotide Synthesis As shown in this section, oligonucleotide synthesis features as the key step a tetrazole-catalyzed coupling of an alcohol (the first nucleotide) with a phosphoramidite (the second nucleotide). On first sight (focusing on all the nitrogen lone pairs), the function of tetrazole appears to be that of a base (to deprotonate the alcohol), but this heterocycle is actually quite acidic (pKa 5 4.8), about as much as acetic acid (Section 19-4). Moreover, kinetic experiments indicate a ratedetermining step involving both the tetrazole and phosphoramidite. Explain these observations and suggest a plausible mechanism for the coupling process. (Hint: Review pyrrole in Sections 25-3 and 25-4.)
26-11 DNA Sequencing and Synthesis
N
A
A
ROH
OR' A P E H NC(CH2)2O N[CH(CH3)2]2
N B N
N H
HN[CH(CH3)2]2
OR' A EPH NC(CH2)2O OR
Strategy We follow the hint and try to extrapolate the chemistry of pyrrole to that of tetrazole, in particular its acidity and, considering the kinetic evidence that points to attack of phosphoramidite on the heterocycle, its most nucleophilic site. Solution • Why is tetrazole so acidic? This molecule is obviously no ordinary amine, which would exhibit a pKa of about 35 (Section 21-4). A clue is provided by the observation of similar relatively high acidity in pyrrole, with a pKa of 16.5 (Section 25-4). The explanation was delocalization of the charge in the anion and the sp2 hybridization of the substituent carbons. The same reasoning holds here, accentuated by the presence of the additional three electron-withdrawing nitrogen atoms on the ring. • Turning to the mechanism of the coupling reaction and considering the acidity of tetrazole, it is reasonable to assume that the most basic amine nitrogen on the phosphoramidite is protonated rapidly (reversibly), changing the substituent into an excellent leaving group (the first step in the scheme below). • Formulation of this protonated species suggests the nature of the rate-determining second step: nucleophilic displacement of the protonated amine from the phosphorus. • Which nitrogen of tetrazole will attack? Recalling the electrophilic aromatic substitution chemistry of pyrrole (Section 25-4), we note that optimal resonance stabilization of the resulting cation is attained by involving the position at the terminus of the diene portion, in the present case N2. • Rapid proton loss (step 3) then furnishes a new intermediate, in which the amino substituent on the original phosphorus reagent has been replaced by tetrazolyl. 3
šð R'O A PH E š N[CH(CH3)2]2 O NC(CH2)2š
šð R'O A PH š N E š š NC(CH2)2 O NO B Nð A HN
H H
H H
šð R'O A PH E N[CH(CH3)2]2 NC(CH2)2š O D H
š N
š N
4
2
ðN
š H
1N
HN[CH(CH3)2]2
šð R'O A PH š N E š š O NC(CH2)2 NO B Nð A ðN
• Tetrazolyl anion is a much better leaving group than the original amide, because it is a much weaker base. Thus, the actual coupling with the alcohol takes place through this intermediate by nucleophilic displacement, as shown next.
š ROH
šð R'O A PH š N E š NC(CH2)2š O NO B Nð A ðN
N
N
N
šð R'O A PH E š O NC(CH2)2š OR
N H
Exercise 26-26
Try It Yourself Formulate mechanisms for all the hydrolysis reactions that effect deprotection of the solid-phase supported dinucleotide in the scheme on page 1254. [Caution: The cyanoethyl protecting group is not removed by nucleophilic hydrolysis of the phosphate ester, because such a process would cleave all P – O bonds indiscriminately. Can you imagine another way? Hint: Like other carboxylic acid derivatives (Section 20-1), the a hydrogen in nitriles is acidic.]
Chapter 26
1255
1256
Chapter 26
Amino Acids, Peptides, Proteins, and Nucleic Acids
CH EMICAL H IG H L I G H T 2 6 - 4 DNA Fingerprinting The human genome is unique and unalterable for each individual and is the same for every cell in his or her body. Therefore, in principle, it should be usable to establish identity unambiguously, much like conventional fingerprinting. However, sequencing 3 billion base pairs per person is still too time consuming and costly to be practicable. Instead, DNA fingerprinting applies abbreviated and much faster techniques. An older but instructive approach, called Southern blotting (after its inventor, E. M. Southern), is probably best known from its application in crime laboratories and in paternity cases in the late 1980s and early 1990s. It is based on the finding that those pieces of DNA that seemingly have no coding function contain repeating sequences (from 20 to 100 base pairs) called variable-number tandem repeats or VNTR. The key is that the number of repeats, from 1 to 30 in a row, at certain places (also called loci) differs among individuals. The size of the corresponding DNA section will differ accordingly. The DNA fingerprint then is the visualization of the number of VNTR at a number of loci. In practice, the sample of DNA (for example, from the blood at the scene of a crime or from the suspected father of a child) was cut into smaller pieces by restriction enzymes and the smaller fragments sorted by size using electrophoresis through a slab of agarose gel. To save the electrophoresis pattern on a more stable surface, it was literally “blotted” onto a nylon membrane (hence the “Southern blot”). To locate a specific VNTR sequence on one of the DNA fragments, a DNA sequence complementary to that of a VNTR locus was generated, bearing a radioactive or chemiluminescent label. This probe was then allowed to bind to complementary DNA sequences on the membrane and the pattern visualized on a photographic film (see photo). Standard forensic analysis used between 3 and 5 loci, and analysis of the combined data placed the odds of finding two individuals with the same observed pattern at 1 to about 10–100 million, unless they were biologically related. A similar procedure can be followed to establish genealogy, even over several generations. Thus, your VNTR could be derived from the DNA of either or both of your parents, but you can’t have any VNTR that are not present in your mother or father. A hypothetical case is shown in the simplified Southern blot patterns on the facing page. You can see that the parents’ sequences (and no others) show up in part in their joint biological offspring (red and blue). However, the son from another marriage exhibits none of the current father’s traits (no blue), but those of his biological father (green). Even more strikingly, the adopted daughter has a completely incongruent pattern (orange). In paternity
A bloodstain on a murder victim is suspected to belong to the perpetrator. Which one of the suspects is it? (Image credited to Cellmark Diagnostics, Abingdon, England.)
disputes, the identity of the father can be ascertained with a probability of up to 100,000 to 1. Currently, forensic DNA analysis uses the polymerase chain reaction to amplify 13 specific DNA loci. The VNTR at these loci are only four base pairs, called short tandem repeats (STR). The amplification primers are chemically labeled with differently colored fluorescent dyes so that the PCR products can be subjected to electrophoresis into the path of a camera, through a capillary or a thin slab of polyacrylamide gel, and detected by their fluorescence. Elapsed electrophoresis time to detection enables measurement of the molecular weight of the PCR products, from which the number of STR at any of the loci can be determined. The odds of obtaining a “false positive” with this method can be as low as 1 in 100 billion. A case of an application of DNA fingerprinting that had historical impact is the identification of the remains of Tsar Nicholas II of Russia and some of his family members. It had been believed that the Tsar’s entire family, along with the Royal Physician and three servants, had been executed by the Bolshevik revolutionaries during the night of July 16, 1918, and their remains hastily buried in a makeshift grave. This story could not be confirmed until
26-11 DNA Sequencing and Synthesis
Chapter 26
VNTR Patterns of a Family
Mother
Father
Son
the grave was found in 1991 and nine bodies exhumed. Mitochondrial DNA (mtDNA) sequencing using bone samples enabled identification of the Tsar, his wife (the Tsarina), their three daughters, and the other four skeletons belonging to the non-family members. Each of the three purported daughters revealed an mtDNA sequence consistent with the purported parents, confirming their family relationship. Moreover, the sequences relating the three daughters to the Tsarina were also found in Prince Philip, the Duke of Edinburgh, who is a distant relative of the late Tsarina through a maternal line. The Tsar’s remains were similarly analyzed by comparison to the DNA fingerprints of two living relatives of the Tsar. The fate of the youngest, fourth daughter and the hemophiliac son (and Crown Prince) of the Tsar remained unknown until 2008, when their burned and shattered skeletons were found in a forest near the grave of the other members of the family. Despite the poor condition of the bone fragments recovered, sufficient amounts of DNA could be extracted to confirm their identity.
Son from mother’s previous marriage
Daughter
Adopted daughter
Tsar Nicholas II of Russia and his family were executed in 1918 during the Russian revolution, as was verified in 1991 and 2008 by DNA fingerprinting.
1257
1258
Chapter 26
Amino Acids, Peptides, Proteins, and Nucleic Acids
Exercise 26-27 In the dinucleotide synthesis on pages 1253 and 1254, the attachment of the first nucleoside to silica employs a 4-nitrophenyl ester as a leaving group. Why would the use of this substituent be advantageous?
The polymerase chain reaction (PCR) makes multiple copies of DNA
Is this Queen Hatshepsut?
DNA cloning—the preparation of large numbers of identical DNA molecules for the study of their sequence, expression, and regulation—revolutionized molecular biology in the mid-1970s. Cloning needs living cells to amplify inserted DNA. Therefore, the discovery of a procedure by Mullis* in 1984 that reproduces DNA segments in vitro by the millions, without the necessity of living cells, marked a stunning advance: the polymerase chain reaction (PCR). The key to this reaction is the ability of some DNA-copying enzymes to remain stable at temperatures as high as 958C. The original enzyme used in PCR is Taq DNA polymerase, found in a bacterium, Thermus aquaticus, in a hot spring in Yellowstone National Park (now other improved enzymes have become available). As described earlier in the Sanger dideoxyribose method of DNA sequencing, polymerases need a supply of the four nucleotides and a short primer to start their job. In nature, a primase enzyme takes care of the priming and the entire DNA sequence behind the primer is reproduced. In PCR, the primer consists of a short (20 bases) oligonucleotide strand that is complementary to a short piece of the DNA to be copied. In practice, the PCR is carried out as shown schematically in Figure 26-18. The reaction flask is charged with the (double-stranded) DNA to be copied, the four nucleotides, the primer, and Taq polymerase. In the first step, the mixture is heated to 90–958C to induce the DNA to separate into two strands. Cooling to 548C allows the primers to attach themselves to the individual DNA molecules. Raising the temperature to 728C provides optimum (in fact, “native”) conditions for Taq polymerase to add nucleotides to the primer along the attached DNA strand until the end, resulting in a complementary copy of the template. All of this takes only a few minutes and can be repeated in automated temperature-regulated vessels multiple times. Because the products of each cycle are again separated into their component strands to function as additional templates in the subsequent cycle, the amount of DNA produced with time increases exponentially: After n cycles, the quantity of DNA is 2n21. For example, after 20 cycles (less than 3 h), it can be about 1 million; after 32 cycles, 1 billion. There are many practical applications of this technique. In medical diagnostics, bacteria and viruses (including HIV) are readily detected through the use of specific primers. Early cancer detection is made possible by identifying mutations in growth-control genes. Inherited disorders can be pinpointed, starting from unborn babies to dead bodies. For example, blood stains on the clothing of Abraham Lincoln have been analyzed to show that he had a genetic disorder called Marfan’s syndrome, which affects the structure of the connective tissue. In forensics and legal medicine, PCR has been used to identify the origin of blood, saliva, and other biological clues left by the perpetrator of a crime. Paternity and genealogy can be proved fairly conclusively (Chemical Highlight 26-4). In molecular evolution, ancient DNA can be multiplied or even reconstructed by amplification of isolated fragments. In this way, the agent of the bubonic plague was successfully identified to be the bacterium Yersinia pestis, isolated from the teeth of victims buried in French mass graves dating from 1590 to 1722. Similarly, some of the genes of a 2400-year-old Egyptian mummy have been deciphered, leading to a broad effort to use this technology to help identify the genealogy of other mummies found in ancient Egyptian graves. For example, the purported identity of a mummy excavated from a tomb found in Egypt’s Valley of the Kings as that of the female ruler Hatshepsut in 2007 is being confirmed by DNA analysis. Similarly, the possibility that two embalmed fetuses found in King Tutankhamun’s tomb are his daughters is being established. What will be next? *Dr. Kary B. Mullis (b. 1945), La Jolla, California, Nobel Prize 1993 (chemistry).
The Big Picture
DNA target sequence a
b
c
d
e
5'
3'
3'
5' a'
b'
c'
d'
e'
Separate strands by heating to 95°C 5'
3'
3'
5' Attach primers by cooling to 54°C d
5'
3' 5'
3'
Primer
d'
b 5'
Primer
3'
3'
5' b'
Synthesize DNA by extending the primers at 72°C
5'
3'
3'
5' a'
b' b 5'
3'
c' c
d' d
e 3' 5'
THE BIG PICTURE This concluding chapter drives home the message we have expressed in many places: Life is based on organic molecules. From a chemical standpoint, life at the level of amino acids and nucleic acids is astonishingly simple. It is the structures and properties of the polymeric arrays that add the more complex dimensions and functions typical of living organisms. To use an analogy, think of a modern desktop computer. At the level of individual chips, the interactions are well defined and understood, and can be analyzed in terms of on and off pulses of current. However, when we add some controlling software, guiding the microcircuits to act in unison, we achieve the complex problem-solving ability we desire from a computer. Similarly, at the molecular level of biology, all molecules, no matter how complicated, react according to the same chemical principles we have studied in this book. However, when the molecules and cells act in unison, the unpredictable behavior of life occurs. For those of you who will go on to specialize in the life sciences, as biochemists, molecular cell biologists, doctors, and so on, this chapter will have whetted your appetite for more and will also have laid a molecular foundation for your future studies. For others, it will have provided a glimpse of the molecular basis for all life-related biochemical processes, including protein synthesis, function, and biological origin in the nucleic acids. In today’s society, as we grapple with issues involving genetic engineering, development of drugs to fight disease, environmental dangers, energy resources — an entire range of issues involving science and technology — it is useful to understand the chemical principles that underlie these developments and make them possible for good and for ill.
Chapter 26
1259
Figure 26-18 Polymerase chain reaction (PCR). A cycle consists of three steps: strand separation, attachment of primers, and extension of primers by DNA synthesis. The reactions are carried out in a closed vessel. The cycle is driven by changes in temperature. Sequences on one strand of the original DNA are denoted by abcde and those on the complementary strand by a9b9c9d9e9. The primers are shown in blue and the new DNA extending from the primers in the respective complementary color to the old strands (red or green). (After Biochemistry, 6th ed., by Jeremy M. Berg, John L. Tymoczko, Lubert Stryer, W. H. Freeman and Company. Copyright © 1975, 1981, 1988, 1995, 2002, 2007.)
Amino Acids, Peptides, Proteins, and Nucleic Acids
CHAPTER INTEGRATION PROBLEMS 26-28. Aspartame (NutraSweet), Asp-Phe-OCH3, appears to be a simple target of synthesis. That this is not so becomes apparent when you try to devise routes to it, starting simply from aspartic acid (Asp) and phenylalanine (Phe). Analyze the problem and formulate various approaches to the synthesis of this compound. O H2N H * }
H N *
HO
O
OCH3 H CH2C6H5 }
O
Aspartame
SOLUTION Simple retrosynthetic analysis dissects the molecule into its two amino acid components, Asp and Phe-OCH3 (made by methyl esterification of Phe), which might be envisaged to be coupled with DCC (Section 26-6). There is a problem, however: Asp has an additional b-carboxy group, bound to interfere. We are therefore facing the task of preparing a selectively carboxy-protected Asp. We can think about accomplishing it by two strategies: (A) the direct protection of the Asp nucleus or, if strategy A is not successful, (B) a total synthesis of an appropriately derivatized Asp from simpler starting materials that avoids Asp altogether. The following strategies focus on practical solutions to our problem found in the literature, although you may come up with alternatives that might be even better. A. Selective Protection of the Asp Nucleus Are the two carboxy groups chemically sufficiently different to warrant the investigation of selective monoprotection of one of the carboxy groups in Asp? The answer is yes, perhaps. Thus, we have learned that an a-amino group causes an increase in the acidity of a carboxylic acid by about 2 pK units (Section 26-1). For Asp, pKa (a-COOH) 5 1.9, whereas pKa (b-COOH) 5 3.7 (Table 26-1). One might therefore be tempted to try a selective ammonium salt formation with the a-carboxylate function at carefully controlled pH, followed by thermolysis (Section 19-10). The problem is that, under the thermal conditions of the relatively slow amide formation, fast proton exchange takes place. It is better to hope for differing reactivity of the two carbonyl groups in addition – elimination reactions of an appropriate derivative. The electron-withdrawing effect of the a-amino group should manifest itself here in increased electrophilicity of the adjacent carbonyl carbon (Sections 17-6, 19-4, and 20-1), rendering the corresponding ester function more susceptible to basic hydrolysis. Indeed, this approach has been realized successfully with the N-Cbz-protected (phenylmethyl) diester of Asp (prepared by standard procedures, see Section 26-6), as shown below.
O
H NHCbz * O
H5C6
O
O C6H5
H2O, acetone, LiOH, 0°C
H NCbz * OH }
Chapter 26
}
1260
H5C6
O
O
O 67%
Cbz-Asp-(OCH2C6H5)2
The product was coupled with Phe-OCH3 and DCC, and then deprotected by catalytic hydrogenolysis (Section 26-6) to give aspartame. A simplified synthesis relies on the potential to protect difunctional compounds as cyclic derivatives. For example, 1,2-diols are “masked” as cyclic acetals (Section 24-8), hydroxy acids as lactones (Section 19-9), amino acids as lactams (Section 19-10), and dicarboxylic acids as anhydrides (Section 19-8). The last two possibilities merit consideration as applied to Asp. However, direct lactam formation can be quickly ruled out because of the complications of ring strain (although b-lactams have been used in the preparation of aspartame). This problem is absent with respect to dehydration to the five-membered ring anhydride. Because anhydrides are activated carboxylic acid derivatives (Section 20-3), the Asp anhydride can be coupled directly with Phe-OCH3 without the help of added DCC. Nucleophilic attack of the amino end of Phe-OCH3 occurs preferentially at the desired position, albeit not completely so; 19% of the product derives from peptide-bond formation at the b-carboxy group of Asp.
Chapter Integration Problems
O
OH
CH3COOH
H H3N
O Phe–OCH3
*
*
OH
O O B B H , CH3COCCH3
}
}
H H 2N
Chapter 26
O
aspartame
O
O 93%
61%
Asp anhydride
B. Total Syntheses of Carboxy-Differentiated Asp Derivatives The alternative to approach A is to start from scratch and build up the Asp framework by using the methods of Section 26-2 in such a way as to provide as a final product a selectively monoprotected carboxy derivative. There are many possible strategies for the solution of this problem, but, as you contemplate them, you will discover that finding an appropriate masked b-carboxy group that is not unmasked during the manipulations of the a-amino acid part is not easy. In the following Gabriel synthesis, recourse was made to the relatively innocuous 2-propenyl substituent, which is eventually elaborated by oxidative cleavage (Sections 12-12 and 24-5) to the free b-carboxy moiety. O NO CH(CO2CH2CH3)2
1. CH3CH2O, CH3CH2OH Br 2.
O
O NO C(CO2CH2CH3)2 A CH2CHP CH2 O
H2N H ∑*
H, H2O,
OH
1. Cbz protection 2. Phe–OCH3, DCC 3. Oxidative cleavage
O 26-29. The amino acid analyzer (Section 26-5) “detects” the elution of any amino acid by the presence of an indicator that turns a deep violet color. This indicator is ninhydrin A, which reacts with amino acids to the purple compound B (“Ruhemann’s purple”), generating as by-products an aldehyde and CO2. O
O 2 OH
P
O
NaOH
H2NCHCOOH
RCH CO2
N
H2O
O
OH
NaO
R O
O A Ninhydrin
O
B Ruhemann’s purple
Amino acid
Write a reasonable mechanism for this process. SOLUTION Let’s take an inventory: (1) Ninhydrin is a carbonyl hydrate (Section 17-6) of the corresponding trione, whose central carbonyl group is activated by its two neighbors. (2) Two molecules of the indicator disassemble one molecule of the amino acid into B (containing the amino nitrogen), an aldehyde (accounting for the RCH part), and CO2 (derived from the carboxy function). (3) Decarboxylations of carboxylic acids occur readily in the presence of 3-oxo (or similar) functions (Section 23-2). (4) The highly delocalized (hence colored; Section 14-11) B is the enolate (Section 18-1) of an imine (Section 17-9). The recognition of the last point allows us to simplify the problem somewhat, as B must be derived by a condensation of the dehydrated A with the corresponding amine C. Simplifying B O
O O H2N
B O
O C
aspartame
1261
Amino Acids, Peptides, Proteins, and Nucleic Acids
Coupling this realization with point 1 suggests that the first step of the mechanism is a base-catalyzed dehydration (Section 17-6) to the trione form of ninhydrin, which condenses with the amino acid to the corresponding imine D. Steps 1 and 2 O
A
O
R
OH
H2NCHCOOH
O
H2O
N
H2O
COOH
O
O
R D
Point 3 encourages us to look for a 3-oxocarboxylic acid-type intermediate that would spontaneously lose CO2 via an aromatic transition state (Section 23-2). Indeed, D fits the bill: It is an imine derivative of such a species, electronically quite equivalent, and smoothly rendering E. Step 3 O
O R N
N
R
H
CO2
H
O O
O O
E
Now we can see the step that provides a connection to C and also the aldehyde product: simple hydrolysis of E. Step 4 O
P
2O, HO
E
C
RCH
As already discussed, C then condenses with ninhydrin under the basic conditions to generate the salt B.
New Reactions 1. Acidity of Amino Acids (Section 26-1) R
R
A
A
H3NCHCOO
H3NCHCOOH pKa
2–3
pKa
Isoelectric point pI
9–10
pKCOOH pKNH
3
2
2. Strongly Basic Guanidino Group in Arginine (Section 26-1)
H
š NH2 A C Dš NH2 RHN
š NH2 A C D š RHN NH2
NH2 B C D D š š RHN NH2
B
š NH B C D D š š RHN NH2
pKa
B
Chapter 26
O
1262
13
3. Basicity of Imidazole in Histidine (Section 26-1)
R
H N N
H
R
H N
R
H N NH
NH
pKa 7.0
New Reactions
Chapter 26
Preparation of Amino Acids 4. Hell-Volhard-Zelinsky Bromination Followed by Amination (Section 26-2)
1. Br2, trace PBr3 2. NH3, H2O
RCH2COOH
NH3 A RCHCOO
5. Gabriel Synthesis (Section 26-2) O
O NK BrCH(CO2CH3)2
NCH(CO2CH3)2
KBr
1. RONa 2. RX 3. H, H2O,
NH3 A RCHCOO
O
O
6. Strecker Synthesis (Section 26-2) O B RCH
NH2 A RCHCN
HCN, NH3
H , H2O,
NH3 A RCHCOO
Polypeptide Sequencing 7. Hydrolysis (Section 26-5) 6 N HCl, 1108C, 24 h
Peptide uuuuuuuy amino acids 8. Edman Degradation (Section 26-5) S NP C P S
R O A B H2NCHCO N O H
H, H2O
H2O
N
NH
H2N O
R
O Lower polypeptide
Phenylthiohydantoin
Preparation of Polypeptides 9. Protecting Groups (Section 26-6)
NH3 A RCHCOO
O B C6H5CH2OCCl Phenylmethyl chloroformate (Benzyl chloroformate)
NH3 A RCHCOO
NaOH NaCl
O O B B (CH3)3COCOCOC(CH3)3
O R B A C6H5CH2OCNHCHCOOH Cbz-protected amino acid
(CH3CH2)3N
H2, Pd–C Deprotection C6H5CH3 CO2
O R B A (CH3)3COCNHCHCOOH
Bis(1,1-Dimethylethyl) dicarbonate (Di-tert-butyl dicarbonate)
Boc-protected amino acid
10. Peptide-Bond Formation with Dicyclohexylcarbodiimide (DCC; Section 26-6) Cbz-Gly
Ala-OCH2C6H5
C6H11NPCPNC6H11 DCC
Cbz-Gly-Ala-OCH2C6H5
O B C6H11NHCNHC6H11
NH3 A RCHCOO
H , H2O Deprotection CO2 CH2PC(CH3)2
NH3 A RCHCOO
1263
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Amino Acids, Peptides, Proteins, and Nucleic Acids
Chapter 26
11. Merrifield Solid-Phase Synthesis (Section 26-7)
P
ClCH2OCH2CH3, SnCl4 CH3CH2OH
P O CH2Cl
O R A B 1. (CH3)3COCNHCHCOO 2. H , H2O
O R B A P O CH2OC O CHNH2
O R A B 1. (CH3)3COCNHCHCOOH, DCC 2. H, H2O
P = polystyrene
O R O R B A B A P O CH2OC O CHNHC OCHNH2
HF
P O CH2F
R O R A B H3NCHCNHCHCOO A
Important Concepts 1. Polypeptides are poly(amino acids) linked by amide bonds. Most natural polypeptides are made from only 19 different l-amino acids and glycine, all of which have common names and threeand one-letter abbreviations. 2. Amino acids are amphoteric; they can be protonated and deprotonated. 3. Enantiomerically pure amino acids can be made by classical fractional crystallization of diastereomeric derivatives or by enantioselective reactions of appropriate achiral precursors. 4. The structures of polypeptides are varied; they can be linear, cyclic, disulfide bridged, pleated sheet, a-helical or superhelical, or disordered, depending on size, composition, hydrogen bonding, and electrostatic and London forces. 5. Amino and nucleic acids are separated mainly by virtue of their size- or charge-based differences in ability to bind to solid supports. 6. Polypeptide sequencing entails a combination of selective chain cleavage and amino acid analysis of the resulting shorter polypeptide fragments. 7. Polypeptide synthesis requires end-protected amino acids that are coupled by dicyclohexylcarbodiimide. The product can be selectively deprotected at either end to allow for further extension of the chain. The use of solid supports, as in the Merrifield synthesis, can be automated. 8. The proteins myoglobin and hemoglobin are polypeptides in which the amino acid chain envelops the active site, heme. The heme contains an iron atom that reversibly binds oxygen, allowing for oxygen uptake, transport, and delivery. 9. The nucleic acids are biological polymers made of phosphate-linked base-bearing sugars. Only four different bases and one sugar are used for DNA and RNA, respectively. Because the base pairs adenine–thymine (uracil for RNA) and guanine–cytosine are held by particularly favorable hydrogen bonding, a nucleic acid can adopt a dimeric helical structure containing complementary base sequences. In DNA, this arrangement unwinds and functions as a template during DNA replication and RNA synthesis. In protein synthesis, each amino acid is specified by a set of three consecutive RNA bases, called a codon. Thus, the base sequence (genetic code) in a strand of RNA translates into a specific amino acid sequence in a protein. 10. DNA sequencing relies on restriction enzymes, radioactive labeling, and specific chemical cleavage reactions to small fragments, analyzed by electrophoresis. 11. DNA synthesis employs silica as a support on which the growing oligonucleotide sequence is built up with the help of base, alcohol, and phosphite – phosphate protecting groups. 12. The polymerase chain reaction (PCR) makes multiple copies of DNA.
Problems 30. Draw stereochemically correct structural formulas for isoleucine and threonine (Table 26-1). What is a systematic name for threonine? 31. The abbreviation allo means diastereomer in amino acid terms. Draw allo-l-isoleucine, and give it a systematic name. 32. Draw the structure that each of the following amino acids would have in aqueous solution at the indicated pH values. (a) Alanine at pH 5 1, 7, and 12; (b) serine at pH 5 1, 7, and 12; (c) lysine
Problems
Chapter 26
at pH 5 1, 7, 9.5, and 12; (d) histidine at pH 5 1, 5, 7, and 12; (e) cysteine at pH 5 1, 7, 9, and 12; (f) aspartic acid at pH 5 1, 3, 7, and 12; (g) arginine at pH 5 1, 7, 12, and 14; (h) tyrosine at pH 5 1, 7, 9.5, and 12. 33. Group the amino acids in Problem 32 according to whether they are (a) positively charged, (b) neutral, or (c) negatively charged at pH 5 7. 34. Show how the pI values for the amino acids in Problem 32 (see Table 26-1) are derived. For each amino acid that possesses more than two pKa values, give the reason for your choice when calculating pI. 35. Show how Hell-Volhard-Zelinsky bromination followed by amination can be used to synthesize each of the following amino acids in racemic form: (a) Gly; (b) Phe; (c) Ala. 36. Show how the use of a Strecker synthesis can give each of the following amino acids in racemic form: (a) Gly; (b) Ile; (c) Ala. 37. What amino acid would result from carrying out the following synthetic sequence on
NOCH(CO2CH2CH3)2?
1. CH3CH2O2Na1, CH3CH2OH; 2. Br(CH2)4NHCOCH3; 3. H1, H2O, D. 38. Using either one of the methods in Section 26-2 or a route of your own devising, propose a reasonable synthesis of each of the following amino acids in racemic form. (a) Val; (b) Leu; (c) Pro; (d) Thr; (e) Lys. 39. (a) Illustrate the Strecker synthesis of phenylalanine. Is the product chiral? Does it exhibit optical activity? (b) It has been found that replacement of NH3 by an optically active amine in the Strecker synthesis of phenylalanine leads to an excess of one enantiomer of the product. Assign the R or S configuration to each stereocenter in the following structures, and explain why the use of a chiral amine causes preferential formation of one stereoisomer of the final product.
}
}
H2O
1. H, H2O 2. H2, Pd–C
O
Mainly
40. The antibacterial agent in garlic, allicin (Chemical Highlight 9-4, Problem 73 in Chapter 9), is synthesized from the unusual amino acid alliin by the action of the enzyme allinase. Because allinase is an extracellular enzyme, this process takes place only when garlic cells are crushed. Propose a reasonable synthesis for the amino acid alliin. (Hint: Begin by designing a synthesis of an amino acid from Table 26-1 that is structurally related to alliin.)
O NH3 B A H2C P CHCH2SCH2CHCOO Alliin
41. Devise a procedure for separating a mixture of the four stereoisomers of isoleucine into its four components: (1)-isoleucine, (2)-isoleucine, (1)-alloisoleucine, and (2)-alloisoleucine (Problem 31). (Note: Alloisoleucine is much more soluble in 80% ethanol at all temperatures than is isoleucine.)
CH2 H [ C ONH 3 }
H3C H [ C CH2 H [ C O ND H NC
O
NH2 E C , HCN ) H H 3C
}
O B CH2CH
COO
Mainly
1265
1266
Amino Acids, Peptides, Proteins, and Nucleic Acids
Chapter 26
42. Identify each of the following structures as a dipeptide, tripeptide, and so forth, and point out all the peptide bonds. (CH3)2CH O HSCH2 CH3 O A A A B B (a) H3N O CH O C O NH O CH O C O NH O CH O COO
HOCH2 O CH2COO A B A (b) H3N O CH O C O NH O CH O COO
NH
OH A O CH2 O CH2(CH2)3NH3 CH3CH O A B B A B A H3N O CH O C O NH O CH O C O N O CH O C O NH O CH O COO N
(c)
OH
CH2 O O O CH2 O CH2CH(CH3)2 B B B A A B A (d) H3N O CH O C O NH O CH2 O C O NH O CH2 O C O NH O CH O C O NH O CH O COO
43. Using the three-letter abbreviations for amino acids, write the peptide structures in Problem 42 in short notation. 44. Indicate which of the amino acids in Problem 32 and the peptides in Problem 42 would migrate in an electrophoresis apparatus at pH 5 7 (a) toward the anode or (b) toward the cathode. 45. Silk is composed of b sheets whose polypeptide chains consist of the repeating sequence GlySer-Gly-Ala-Gly-Ala. What characteristics of amino acid side chains appear to favor the b-sheet configuration? Do the illustrations of b-sheet structures (Figure 26-3) suggest an explanation for this preference? 46. Identify as many stretches of a helix as you can in the structure of myoglobin (Figure 26-8C). Prolines are located in myoglobin at positions 37, 88, 100, and 120. How does each of these prolines affect the tertiary structure of the molecule? 47. Of the 153 amino acids in myoglobin, 78 contain polar side chains (i.e., Arg, Asn, Asp, Gln, Glu, His, Lys, Ser, Thr, Trp, and Tyr). When myoglobin adopts its natural folded conformation, 76 of these 78 polar side chains (all but those of two histidines) project outward from its surface. Meanwhile, in addition to the two histidines, the interior of myoglobin contains only Gly, Val, Leu, Ala, Ile, Phe, Pro, and Met. Explain. 48. Explain the following three observations. (a) Silk, like most polypeptides with sheet structures, is water insoluble. (b) Globular proteins such as myoglobin generally dissolve readily in water. (c) Disruption of the tertiary structure of a globular protein (denaturation) leads to precipitation from aqueous solution. 49. In your own words, outline the procedure that might have been followed by the researchers who determined which amino acids were present in vasopressin (Exercise 26-10). 50. Write the products of a single Edman degradation of the peptides in Problem 42. 51. What would be the outcome of reaction of gramicidin S with phenyl isothiocyanate (Edman degradation)? (Hint: With which functional group does this substance react?) 52. The polypeptide bradykinin is a tissue hormone that can function as a potent pain-producing agent. By means of a single treatment with the Edman reagent, the N-terminal amino acid in bradykinin is identified as Arg. Incomplete acid hydrolysis of the intact polypeptide causes random cleavage of many bradykinin molecules into an assortment of peptide fragments that includes Arg-Pro-ProGly, Phe-Arg, Ser-Pro-Phe, and Gly-Phe-Ser. Complete hydrolysis followed by amino acid analysis indicates a ratio of 3 Pro, 2 Phe, 2 Arg, and one each of Gly and Ser. Deduce the amino acid sequence of bradykinin. 53. The amino acid sequence of met-enkephalin, a brain peptide with powerful opiatelike biological activity, is Tyr-Gly-Gly-Phe-Met. What would be the products of step-by-step Edman degradation of met-enkephalin?
Problems
The peptide shown in Problem 42(d) is leu-enkephalin, a relative of met-enkephalin with similar properties. How would the results of Edman degradation of leu-enkephalin differ from those of met-enkephalin? 54. Secreted by the pituitary gland, corticotropin is a hormone that stimulates the adrenal cortex. Determine its primary structure from the following information. (i) Hydrolysis by chymotrypsin produces six peptides: Arg-Trp, Ser-Tyr, Pro-Leu-Glu-Phe, Ser-Met-Glu-His-Phe, Pro-Asp-AlaGly-Glu-Asp-Gln-Ser-Ala-Glu-Ala-Phe, and Gly-Lys-Pro-Val-Gly-Lys-Lys-Arg-Arg-Pro-ValLys-Val-Tyr. (ii) Hydrolysis by trypsin produces free lysine, free arginine, and the following five peptides: Trp-Gly-Lys, Pro-Val-Lys, Pro-Val-Gly-Lys, Ser-Tyr-Ser-Met-Glu-His-Phe-Arg, and Val-Tyr-Pro-Asp-Ala-Gly-Glu-Asp-Gln-Ser-Ala-Glu-Ala-Phe-Pro-Leu-Glu-Phe. 55. Propose a synthesis of leu-enkephalin [see Problem 42(d)] from the component amino acids. 56. The following molecule is thyrotropin-releasing hormone (TRH). It is secreted by the hypothalamus, causing the release of thyrotropin from the pituitary gland, which, in turn, stimulates the thyroid gland. The thyroid produces hormones, such as thyroxine, that control metabolism in general. NH O
N CH2 O O O A B B B HNO CHO C ONHO CHO C ONO CHO C ONH2 Pyroglutamic acid
Histidine
Prolinamide
The initial isolation of TRH required the processing of 4 tons of hypothalamic tissue, from which 1 mg of the hormone was obtained. Needless to say, it is a bit more convenient to synthesize TRH in the laboratory than to extract it from natural sources. Devise a synthesis of TRH from Glu, His, and Pro. Note that pyroglutamic acid is the lactam of Glu and may be readily obtained by heating Glu to 1408C. 57. (a) The structures illustrated for the four DNA bases (Section 26-9) represent only the most stable tautomers. Draw one or more alternative tautomers for each of these heterocycles (review tautomerism, Sections 13-7 and 18-2). (b) In certain cases, the presence of a small amount of one of these less stable tautomers can lead to an error in DNA replication or mRNA synthesis due to faulty base pairing. One example is the imine tautomer of adenine, which pairs with cytosine instead of thymine. Draw a possible structure for this hydrogen-bonded base pair (see Figure 26-11). (c) Using Table 26-3, derive a possible nucleic acid sequence for an mRNA that would code for the five amino acids in met-enkephalin (see Problem 53). If the mispairing described in (b) were at the first possible position in the synthesis of this mRNA sequence, what would be the consequence in the amino acid sequence of the peptide? (Ignore the initiation codon.) 58. Factor VIII is one of the proteins participating in the formation of blood clots. A defect in the gene whose DNA sequence codes for Factor VIII is responsible for classic hemophilia. Factor VIII contains 2332 amino acids. How many nucleotides are needed to code for its synthesis? 59. In addition to the conventional 20 amino acids in Tables 26-1 and 26-3, two others, selenocysteine (Sec) and pyrrolysine (Pyl), are incorporated into proteins using the nucleic acid–based cell machinery described in Section 26-10. The three-base codes for Sec and Pyl are UGA and UAG, respectively. These codes normally serve to terminate protein synthesis. However, if they are preceded by certain specific base sequences, they instead cause incorporation of these unusual amino acids and continued growth of the peptide chain. Pyl is rare, occurring only in some ancient bacteria. Sec, containing the side chain CH2SeH, is widespread; in fact, it occurs in at least two dozen human proteins that rely on the reactivity of the essential trace element selenium for their function. The pKa for the SeH group in Sec is 5.2. (For comparison, the pKa for the SH in Cys is 8.2.) (a) Draw the structure of Sec at pH 7; compare it with that of Cys at the same pH. (b) Determine pI for Sec. (c) Given the relationship between sulfur and selenium in the periodic table, and the pKa differences between SH and SeH, how do you expect the chemical reactivities of Sec and Cys to compare? 60.
Hdroxyproline (Hyp), like many other amino acids that are not “officially” classified as essential, is nonetheless a very necessary biological substance. It constitutes about 14% of the amino acid content of the protein collagen. Collagen is the main constituent of skin and connective tissue. It is also present, together with inorganic substances, in nails, bones, and teeth. (a) The systematic name for hydroxyproline is (2S,4R)-4-hydroxyazacyclopentane-2-carboxylic
Chapter 26
1267
Amino Acids, Peptides, Proteins, and Nucleic Acids
Chapter 26
acid. Draw a stereochemically correct structural formula for this amino acid. (b) Hyp is synthesized in the body in peptide-bound form from peptide-bound proline and O2, in an enzymecatalyzed process that requires vitamin C. In the absence of vitamin C, only a defective, Hyp-deficient collagen can be produced. Vitamin C deficiency causes scurvy, a condition characterized by bleeding of the skin and swollen, bleeding gums. In the following reaction sequence, an efficient laboratory synthesis of hydroxyproline, fill in the necessary reagents (i) and (ii), and formulate detailed mechanisms for the steps marked with an asterisk. O O
ClCH2
i
O
NO CH2
NaCH(COOCH2CH3)2*
O O
O
Cl
COOCH2CH3 SO2Cl2
NCH2 O
COOCH2CH3
NCH2 O
O
ii*
hydroxyproline
O
O
O
(c) Gelatin, which is partly hydrolyzed collagen, is rich in hydroxyproline and, as a result, is often touted as a remedy for split or brittle nails. Like most proteins, however, gelatin is almost completely broken down into individual amino acids in the stomach and small intestine before absorption. Is the free hydroxyproline thus introduced into the bloodstream of any use to the body in the synthesis of collagen? (Hint: Does Table 26-3 list a three-base code for hydroxyproline?) The biosynthesis of oligosaccharides (Chapter 24) employs the chemical constituents of proteins and nucleic acids as well as carbohydrates. In the example shown, a disaccharide linkage is created between a molecule of galactose and a molecule of N-acetylgalactosamine. The galactose (the “donor” sugar) is carried into the process as a uridine diphospate ester, and the “acceptor” galactosamine is held in place by a glycoside linkage to the hydroxyl group of a serine residue in a protein. The enzyme galactosyl transferase specifically forms a disaccharide linkage between C1 of the donor and C3 of the acceptor:
6
CH2OH 5
O
OH 4
H OH
H 3
H
O O B B H OOP OOOP OO N O A A 2 H H O O OH H H 1
2OH
5
H NO O
O
OH 4
H
H OH 3
H
H
A C O A H Serine O O CH2 O CH A 2 NH NH A A C O A CH3 1
B
H
6 CH
O
B
61.
OH
OH Uridine diphosphate–galactose
N-Acetylgalactosamine–protein
CH2OH H
CH2OH
H H
O
OH H OH
O H
H
H
H H
OH
H
A C O A O O CH2 O CH A NH A
O
B
O
OH
NH A C O A CH3 B
1268
Galactose 1 → 3 N-acetylgalactosamine–protein
O O B B OO P O O O P O O N O A A H H O O H H OH
OH
Uridine diphosphate (UDP)
H NO O
Problems
What type of basic mechanistic process does this reaction resemble? Discuss the roles played by the various participants of the reaction. 62.
Sickle-cell anemia is an often fatal genetic condition caused by a single error in the DNA gene that codes for the b chain of hemoglobin. The correct nucleic acid sequence (read from the mRNA template) begins with AUGGUGCACCUGACUCCUGAGGAGAAG . . . , and so forth. (a) Translate this into the corresponding amino acid sequence of the protein. (b) The mutation that gives rise to the sickle-cell condition is replacement of the boldface A in the preceding sequence by U. What is the consequence of this error in the corresponding amino acid sequence? (c) This amino acid substitution alters the properties of the hemoglobin molecule — in particular, its polarity and its shape. Suggest reasons for both these effects. (Refer to Table 26-1 for amino acid structures and to Figure 26-8C for the structure of myoglobin, which is similar to that of hemoglobin. Note the location of the amino acid substitution in the tertiary structure of the protein.)
Team Problem 63. Amino acids can be used as enantiomerically pure starting materials in organic synthesis. Scheme I depicts the first steps in the synthesis of a reagent employed in the preparation of enantiomerically pure b-amino acids, such as that occurring in the side chain of taxol (Section 4-7). Scheme II features an ester of the same amino acid for the preparation of an unusual heterocyclic dipeptide used in the study of polypeptide conformations. Scheme I: Synthesis of an enantiomerically pure reagent O
O
NH2 H NH 2 *∑ O CO2K
1. NaHCO3, Cl 2. H, H2O
H
A C9H15N2O3ⴚ Kⴙ Six-membered nitrogen heterocyle
Potassium salt of asparagine
OCH3
B C11H18N2O5
Consider the following questions: 1. There are two diastereomers of A formed in a 90 : 10 ratio. The major isomer is the one that can attain the most stable chair conformation. Are the two substituents on the ring cis or trans to each other? Label their positions as equatorial or axial. 2. Which nitrogen is nucleophilic and yields the carbamic ester (Section 20-6) B? Scheme II: Synthesis of an unusual heterocyclic dipeptide
O
O
NH2 H NH 2 *∑ O
H
C
Fmoc–amino acid chloride
D
O 1,1-Dimethylethyl (tert-butyl) ester of asparagine
O O
O
∑
H A N
C15H20N2O3 (Acyclic compound)
& R H
Fluorenylmethyloxycarbonyl(Fmoc)– amino acid chloride
Cl
Heterocyclic dipeptide
The Fmoc protecting group (highlighted in the box) is used instead of Cbz or Boc, with which you are familiar, because the amino acid chloride is necessary to make the new amide bond. Neither the Cbz nor the Boc group is stable under these conditions.
Consider the following questions: 1. What functional group is generated in C? 2. Where is the peptide bond in D? Circle it. Reconvene to discuss the answers to the questions posed in each scheme and the structures you proposed for A – D.
Chapter 26
1269
1270
Chapter 26
Amino Acids, Peptides, Proteins, and Nucleic Acids
Preprofessional Problems COOH H
H2N
CH3 A
64. Structure A (shown in the margin) is that of a naturally occurring a-amino acid. Select its name from the following list. (a) Glycine; (b) alanine; (c) tyrosine; (d) cysteine. 65. The primary structure of a protein refers to: (a) cross-links with disulfide bonds; (b) presence of an a helix; (c) the a-amino acid sequence in the polypeptide chain; (d) the orientation of the side chains in three-dimensional space. 66. Which one of the following five structures is a zwitterion? O B (a) O2CCH2CNH2
(b)
2
1
(d) CH3(CH2)16CO2 K
O2CCH2CH2CO22
2
O l (e) HO C i š Oð
1
(c) H3NCH2CO22 Oð iš HO Cl O
67. When an a-amino acid is dissolved in water and the pH of the solution adjusted to 12, which of the following species is predominant? O B (a) RCHCOH A NH2
O B (b) RCHCOH A NH 3
O B (c) RCHCO A NH 3
O B (d) RCHCO A NH2
68. How many stereocenters are present in the small, naturally occurring protein glycylalanylalanine? (a) Zero; (b) one; (c) two; (d) three.
MC AT ® Q uestions The section that follows includes material from previously administered MCAT® items and is reprinted with permission of the Association of American Medical Colleges (AAMC).
Passage Number I The addition of hydrogen halides to double or triple carbon-carbon bonds can result in more than 1 product, as illustrated in Reactions 1 and 2. HX
C5H11
⫹ C5H11
C5H11
X
X Product A
Product B
Reaction 1
C6H5O C q C O CH3
HX
H H X X H H ⫹ CPC CPC E E C6H5 C6H5 CH3 CH3 Product C
Product D
Reaction 2
A chemist postulated that the formation of Products A and C involves a carbocation intermediate, whereas Products B and D result from a competing radical-chain addition. It was discovered that the results of these additions were strongly affected by the reaction conditions. For example, when Reactions 1 and 2 were carried out in the presence of silica gel (SiO2) or alumina (Al2O3) surfaces, the yields and product distributions were dramatically altered. Tables 1 and 2 summarize the effects of the reaction conditions on the results of Reactions 1 and 2. 1. From Tables 1 and 2 it can be concluded that the reaction of HCl with 1-octene or 1-phenylpropyne in the absence of silica or alumina: (a) results in a mixture of products. (b) results in a 100% yield of the products. (c) requires 3 hr of reaction time. (d) does not occur within 1 hr. 2. Which of the following is most likely the predominant competing reaction in Reaction 2? (a) Rearrangement of carbocationic intermediates (b) Migration of the double bond in the product alkenes
TABLE 1 Unsaturated compound 1-Octene 1-Octene 1-Octene 1-Octene
Experimental Results of Reaction 1 under Various Conditions Reaction conditions
Product A (%)
Product B (%)
Starting material left over (%)
HCl, 1 hr HCl, Al2O3, 1 hr HBr, 0.3 hr HBr, SiO2, 0.7 hr
0 60 11 96
0 0 83 0
100 11 6 trace MCAT-1
MCAT® Questions
TABLE 2
Experimental Results of Reaction 2 under Various Conditions
Unsaturated compound
Reaction conditions
Product C (%)
C6H5CqCCH3 C6H5CqCCH3
HCl, 1 hr HCl, SiO2, 1 hr HBr, 3 hr
0 E: 39 Z: 11 E: 29 Z: 8 E: 14 Z: 79
C6H5CqCCH3 C6H5CqCCH3
HBr, Al2O3, 0.3 hr
Starting material left over (%)
Product D (%) 0
100
0 E: 18 Z: 29
44
0
0
0
Table adapted from P. J. Kropp, K. A. Daus, S. D. Crawford, M. W. Tubergen, K. D. Kepler, S. L. Craig, and V. P. Wilson, “Surface-Mediated Reactions. 1. Hydrohalogenation of Alkenes and Alkynes.” ©1990 by the American Chemical Society.
(c) Reduction of the product alkenes with H2 (d) Addition of HX to the product alkenes 3. Which of the following is the structure of the carbocation intermediate that will result in the formation of Product C? 1 q COCH3 (a) C6H5OC 1 (b) C6H5OCP CHOCH 3 1 (c) C6H5OC q COCH 3 1 (d) C6H5OCH P COCH3 4. The transformation of 1-octene to Product A involves the conversion of: (a) sp2 hybrid orbitals to sp hybrid orbitals. (b) sp2 hybrid orbitals to sp3 hybrid orbitals. (c) sp hybrid orbitals to sp2 hybrid orbitals. (d) sp hybrid orbitals to sp3 hybrid orbitals. 5. According to the results of Table 1, which of the following should be the major product of the reaction of 1-butene with HBrySiO2 after 0.7 hr? (a) 1-Bromo-1-butene (b) 2-Bromo-2-butene (c) 2-Bromobutane (d) 1-Bromobutane For further information, see Chapters 12 and 13.
Passage Number II A variety of highly functionalized compounds (Compounds 1–4) were recently isolated from marine sources in British Columbia. O
O
H O
OH
O Compound 1
O
O
O
O
O
OH
O
MCAT-2
O Compound 2
Compound 3
Compound 4
MCAT® Questions
The marine material was extracted with methanol, and the methanol was removed from the resulting organic material by evaporation. The residue was subsequently extracted from a saturated NaCl solution with ethyl acetate. The ethyl acetate fractions were combined, and the ethyl acetate was removed by evaporation. The remaining material contained Compounds 1–4, which were separated and purified by chromatography. The structures of the four compounds were determined by NMR and IR spectroscopy. Furthermore, the structure of Compound 3 was confirmed by treatment with acetic anhydride to give a previously known ester, Compound 5 (Reaction 1).
Compound 3
O B (CH3C)2O
Compound 5
Acetic anhydride
Reaction 1
1. Compounds 1 and 3 can be distinguished from each other by their IR spectra because the spectrum of: (a) Compound 1 will exhibit a COO stretch, and that of Compound 3 will not. (b) Compound 1 will exhibit an OOO stretch, and that of Compound 3 will not. (c) Compound 3 will exhibit a COH stretch, and that of Compound 1 will not. (d) Compound 3 will exhibit an OOH stretch, and that of Compound 1 will not. 2. Conjugated double bonds can be found in which of the compounds described in the passage? (a) Compounds 1 and 2 only (b) Compound 4 only (c) Compounds 1, 2, and 3 only (d) Compounds 1, 2, 3, and 4 3. Compound 6 is shown below. What is the stereochemical relationship between Compound 6 and Compound 3?
O
#
OH
O
H
(a) (b) (c) (d)
They They They They
are are are are
O identical. structural isomers. nonsuperimposable and are mirror images of each other. nonsuperimposable and are not mirror images of each other.
4. Which of the following reaction conditions can be used to convert Compound 2 into Compound 4? (a) H2SO4 yH2O (b) NaOH yCH3CH2OH (c) H2 yPdyC (d) Cr2O722yH1 For further information, see Chapter 11.
Passage Number III A forensic chemist was given a liquid sample and was asked to identify its components. A small portion of the liquid was subjected to gas chromatographic (GC) analysis, and it was observed that the sample contained two volatile compounds, one of which was diethylamine.
MCAT-3
MCAT-4
MCAT® Questions
The sample was dissolved in methylene chloride (CH2Cl2), and the diethylamine was removed by extraction. The methylene chloride was subsequently removed by evaporation, leaving the other volatile component, Compound A. Compound A was neutral to litmus but dissolved in refluxing sulfuric acid. Workup of the sulfuric acid solution yielded a solid, Compound B (melting point [MP] 5 133°C), and a liquid, Compound C. Possible structures for Compound B are listed in Table 1. The identity of Compound B was then determined from its neutralization equivalent, which is equal to the molecular weight of the acid divided by the number of carboxy groups in the molecule. The IR spectrum of Compound C indicated that it was an alcohol, and the proton NMR data allowed the chemist to assign its structure. Ultimately, the structure of Compound A was deduced from the identities of Compounds B and C. The structure of Compound A was confirmed by comparison with an authentic reference sample. TABLE 1
Possible Structures of Compound B
Compound
Structure
MP
Molecular weight (g/mole)
C6H5OCHPCHCOOH
133°C
148
133°C
112
133°C 133°C 133°C
104 152 202
Cinnamic acid
O Furoic acid Malonic acid Mandelic acid Sebacic acid
COOH
HOOCOCH2OCOOH C6H5OCH(OH)COOH HOOC(CH2)8OCOOH
1. Which of the following solutions would be suitable for the extraction of diethylamine from the methylene chloride solution? (a) Benzene (b) Ethyl alcohol (c) 10% aqueous HCl (d) 10% aqueous NaHCO3 2. Based on the observation that Compound A was determined to be a neutral substance, which of the following types of compounds should have been considered as possible structures? (a) RC(O)H and RC(O)OH (b) RC(O)OR and ROOOR (c) RONH2 and RC(O)OR (d) RC(O)OH and ROOOR 3. Which absorption band from the IR spectrum of Compound C is most characteristic of an alcohol? (a) A broad absorption band at approximately 3,300 cm21 (b) A broad absorption band at approximately 2,000 cm21 (c) A strong, sharp absorption band at approximately 1,600 cm21 (d) A strong, sharp absorption band at approximately 800 cm21 4. Under the proper conditions, any of the compounds in Table 1 could react with diethylamine to form the corresponding: (a) nitrile (b) amine (c) imine (d) amide For further information, see Chapter 19.
MCAT® Questions
Passage Number IV Two chemists proposed methods to synthesize esters. CHEMIST A Chemist A suggested that the best method for preparing an ester involves the reaction of a carboxylic acid with an alcohol, as shown in Reaction A. O B CH3COH ⫹ HOCH2CH3
O B CH3COCH2CH3 ⫹ H2O
H2SO4
Reaction A
Chemist A proposed the following mechanism for Reaction A. O B CH3COH
⫹
H⫹
OH B CH3COH
CH3CH2OH
OH A CH3COH A HOCH2CH3 ⫹
⫹
OH B H2O ⫹ CH3COCH2CH3
OH A⫹ CH3COH2 A OCH2CH3
OH A CH3COH ⫹ H⫹ A OCH2CH3
O B CH3COCH2CH3 ⫹ H⫹ CHEMIST B Chemist B pointed out that Reaction A often gives low yields because the equilibrium constant is relatively low. As an alternative, Chemist B suggested that a higher yielding method to synthesize esters involves the reaction of an acid chloride (such as acetyl chloride) with an alcohol, as shown in Reaction B. O B CH3COH
SOCl2
O B CH3CCl Et3N HOCH2CH3
O B CH3COCH2CH3 ⫹ HCl Reaction B
1. According to Chemist A, the role of the sulfuric acid in Reaction A is to: (a) provide energy for the nucleophilic attack of the alcohol. (b) act as a solvating medium. (c) make the hydroxy group of the alcohol more nucleophilic. (d) make the carbonyl group of the carboxylic acid more electrophilic. 2. Chemist A separated the three organic compounds in Reaction A on a thin-layer chromatography plate that separated the compounds according to polarity. If the least polar compound traveled the farthest up the plate, what was the order of the distances traveled by the three compounds? (a) Carboxylic acid , alcohol , ester (b) Ester , alcohol , carboxylic acid (c) Alcohol , carboxylic acid , ester (d) Ester , carboxylic acid , alcohol
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3. Which of the following is most likely a role of the weak base Et3N in Reaction B? (a) It neutralizes the acetic acid. (b) It neutralizes the HCl by-product. (c) It protonates the acid chloride. (d) It protonates the ethanol. 4. If (R)-2-phenylpropanoic acid is substituted for acetic acid in Reaction A, what will Chemist A predict about the stereochemistry of the product ester? (a) The ester will have an R configuration. (b) The ester will have an S configuration. (c) The product will be a mixture of the R and S esters. (d) The product will not contain a chiral center. For further information, see Chapter 20.
Passage Number V The preparation, isolation, and characterization of highly unstable compounds remain a challenge for synthetic organic chemists. To circumvent these problems, chemists have developed the method of matrix isolation, which typically involves trapping an unstable compound by freezing it in a solid matrix. Two high-energy compounds that have been successfully isolated by this technique are benzyne and a propellane.
Benzyne Benzyne is a highly unstable intermediate that is thought to be formed from the base-promoted elimination of a halide and a hydrogen from adjacent carbon atoms in an aryl halide. It has been proposed that the structure of benzyne is highly strained and contains a very reactive triple bond. Although benzyne cannot be routinely isolated, evidence for the existence of this intermediate has been provided by the isolation of its decomposition by-product, triphenylene, which is presumably formed from a reaction between three molecules of benzyne. In addition, benzyne can now be trapped (and detected spectroscopically) by the photochemical decomposition of a high-energy reactant in a frozen argon matrix (see Reaction 1). O O
h
O
Ar, 8 K
Benzyne
O Reaction 1
Propellanes Propellanes are extremely reactive compounds in which two carbons that are directly connected are also connected by three other bridges. The propellane shown in Reaction 2 readily reacts with itself to form the corresponding dimer.
2 Propellane
Reaction 2
MCAT® Questions
Under controlled reaction conditions, however, this propellane can be isolated. As shown in Reaction 3, when the diiodo compound is treated with potassium at relatively high temperatures, the resulting propellane may be trapped by rapidly freezing the vapor into a nitrogen matrix.
I K
N2
395 K
29 K
I Reaction 3
1. The high reactivity of the propellane described in the passage is most likely a result of: (a) its bent shape. (b) forced planarity. (c) ring strain. (d) steric repulsion between substituents. 2. Which of the following represents a resonance structure of benzyne? (a)
(b)
(c)
(d)
3. On the basis of the information in the passage, what is the temperature at which the product of Reaction 3 is frozen in a matrix? (a) 2273°C (b) 2244°C (c) 120°C (d) 151°C 4. According to the passage, the solid matrix in which an unstable compound is to be trapped should: (a) be composed of a substance with a high freezing point. (b) be composed of a substance with a low freezing point. (c) react with the starting material to give the unstable product. (d) react with the unstable product to give a new, stable product. 5. In addition to benzyne, which of the following compounds will most likely be formed in Reaction 1? (a) 4 O2 (b) 2 CO (c) 2 CO2 (d) H2CO3 For further information, see Chapter 22.
Passage Number VI Recently, it has been reported that the mushroom tyrosinase enzyme catalyzes an oxidative coupling reaction of 2,6-disubstituted phenols (Compounds 1a–f ) to the corresponding diphenoquinones (Compounds 2a–f ) and bisphenols (Compounds 3a–f ), as shown below. It was discovered that the outcome of the reaction was affected by the presence of acetonitrile, as well as the substituents on the phenol starting materials (see Table 1). It was also observed that the reaction did not proceed under an argon atmosphere.
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TABLE 1
Results of Oxidative Coupling Reactions Product yield (%)
Compound
Structure OH
1a
1b
OH
OH
1c
OH
1d
OH
1e
1f
OCH3
H3CO
OH
Cl
2a–f (diphenoquinones)
3a–f (bisphenols)
9
96 (70*)
0 (20*)
9
98 (72*)
0 (20*)
55
0 (50*)
0 (24*)
58
0 (46*)
0 (24*)
60
0 (40*)
0 (20*)
72
0
0
Rxn. time (hr)
Cl
*Reaction carried out in acetonitrile solvent.
1. The reaction of Compound 1f is most likely unsuccessful because the chloride substituent: (a) is too sterically bulky. (b) contains lone electron pairs. (c) is electron donating. (d) is electron withdrawing. 2. From the data in Table 1, what can be determined about the effect of reaction time on the product distribution? (a) Longer reaction times result in lower yields. (b) Longer reaction times result in higher yields. (c) Longer reaction times result in increased bisphenol formation. (d) Nothing can be determined without further data. 3. Which of the following compounds from the passage can be most obviously distinguished from each other by their IR spectra? (a) Compounds 1a and 1d (b) Compounds 2d and 2e (c) Compounds 3b and 3e (d) Compounds 3d and 3e 4. Which of the phenols in Table 1 has the most acidic hydrogen? (a) Compound 1a (b) Compound 1b (c) Compound 1e (d) Compound 1f For further information, see Chapter 22.
MCAT® Questions
Passage Number VII Six unlabeled vials were discovered in a laboratory and were each thought to contain one of the following sweeteners.
H H H SO3H N
O
⫹
H3N ⫺
O2C
Compound A
O NH SO2 Compound C
H
O
Compound B
⫹
H3N ⫺
O2C
O
CH3 N A H
CHO H
OH
OH
H
OH
CH2OH
S
O
H
Compound E
H A NH CH
Compound D
CHO HO
OH CH3
N A H
CH2OH Compound F
A chemist performed a variety of experiments in order to determine the identity of the unknown compounds. 1. HNO3 oxidizes primary alcohols and aldehydes to the corresponding carboxylic acid. One of the sweeteners was found to be optically active; however, its optical activity vanished upon treatment with HNO3. Which of the following compounds could it be? (a) Compound B (b) Compound C (c) Compound E (d) Compound F 2. Phenylketonuria (PKU) is a genetic disease, the victims of which are unable to metabolize phenylalanine [C6H5CH2CH(CO22)(NH31)]. Which of the sweeteners in the passage should someone with PKU avoid ingesting? (a) Compound A (b) Compound B (c) Compound C (d) Compound D 3. The trade name of one stereoisomer of Compound B is Aspartame. How many total stereoisomers of Compound B exist? (a) 2 (b) 4 (c) 6 (d) 8
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4. Which of the following observations best supports the argument that the physiological sensation of taste involves a receptor site that is chiral? (a) Compound A is 2 times as sweet as Compound B. (b) Compound A is 1y10 as sweet as Compound C. (c) Two stereoisomers of Compound E are sweet. (d) One stereoisomer of Compound C is sweet and another is tasteless. For further information, see Chapter 24.
Passage Number VIII A chemist tested the proteins in a snake venom for hemorrhagenic activity (the ability to cause bleeding). To separate the proteins, the chemist used two methods. The first method was gel-filtration chromatography, which separates proteins according to size. In this technique, a column is filled with porous beads, and a mixture of proteins is allowed to pass through the beads. Proteins that are too large to enter the beads pass through the column at a faster rate than do smaller proteins; therefore, the larger proteins emerge from the column earlier. In the second method, ion-exchange chromatography, proteins are separated on the basis of their net charge. If a column is loaded with positively charged beads, negatively charged proteins bind to the beads, and positively charged proteins pass through the column. The negatively charged proteins are subsequently released from the beads after the addition of a sodium chloride solution. EXPERIMENT 1 The snake venom was subjected to gel-filtration chromatography, and Fractions 1, 2, 3, and 4 were collected in order as they passed out of the column (each fraction contains either a single protein or a group of proteins). The chemist discovered that only the material in the last fraction (Fraction 4) showed hemorrhagenic activity. EXPERIMENT 2 When the material in Fraction 4 was subjected to ion-exchange chromatography, Fraction A passed out of the column. Addition of a sodium chloride solution resulted in the elution of Fractions B and C. None of the three materials exhibited hemorrhagenic activity; however, activity was restored when Fractions A, B, and C were recombined. EXPERIMENT 3 The chemist subsequently determined that Fractions B and C each contained a single, pure protein (Proteins B and C, respectively), whereas Fraction A contained two different proteins (Proteins A1 and A2). 1. Which of the following modifications to Experiment 1 might result in a better separation of only the materials found in Fraction 4? (a) Using beads with smaller pores (b) Using beads with larger pores (c) Using a shorter column (d) Filling the column with fewer beads 2. In Experiment 2, which of the following is the most reasonable explanation for the effect of the sodium chloride solution? (a) Sodium and chloride ions increase the charge interactions between the proteins and the charged beads. (b) Sodium and chloride ions clean out the column by passing through the column immediately. (c) Sodium and chloride ions bind to the charged proteins, and the resulting complex is retained in the column. (d) Sodium and chloride ions competitively bind to the beads, and the charged proteins are released from the column.
MCAT® Questions
3. The beads in a gel-filtration chromatography column should have which of the following characteristics? (a) Neutral and inert (b) Neutral and reactive (c) Charged and inert (d) Charged and reactive 4. Which of the following statements accurately describes the relative size and charge of Proteins A1 and A2? (a) They are approximately the same size and have the same charge. (b) They are approximately the same size and have opposite charges. (c) They are very different in size and have the same charges. (d) They are very different in size and have opposite charges. For further information, see Chapter 26.
Passage Number IX Cyclophosphamide (Compound 1) is an anticancer drug that is effective against a broad range of human tumors. The metabolic pathway of cyclophosphamide has been the subject of much research. The metabolism as it was understood in 1990 is shown in Scheme A. Scheme A Inactive metabolite Step 2
H
HO
N
O Cl
P N
O
H N
[O]
O Cl
P
Step 1
O
Cl
N Cl
Cyclophosphamide
Compound 2
Compound 1
Step 3 Inactive metabolite
Step 4
O H
Base Step 5
O
NH2 O P N
Cl Cl
Compound 3
O H
⫹
NH2 A HO O P P O A N
Cl Cl
Phosphoramide mustard
Compound 4
Cyclophosphamide (Compound 1) is enzymatically oxidized in the liver to Compound 2 (Step 1). Oxidation of Compound 2 to an inactive carbonyl-containing metabolite can occur. This is also an enzyme-catalyzed reaction, but the enzyme is different from the enzyme causing the oxidation in Step 1. If this second oxidation does not occur, an equilibrium between Compound 2 and Compound 3 is established, favoring Compound 3.
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At this point, the pathway splits into two irreversible reactions. In Step 4, oxidation by a third enzyme forms another inactive metabolite. Alternatively, a base can remove an a–proton from Compound 3. Loss of the leaving group followed by neutralization of all charges produces propenal or acrolein and phosphoramide mustard (Compound 4), the anticancer active metabolite. 1. Which of the following functional groups best describes the portion of cyclophosphamide shown below? (R represents an alkyl group.) RO
(a) (b) (c) (d)
P
O
Carboxylic acid derivative Phosphate ester Phosphate ether Alkyl phosphorylase
2. The aldehyde, (E)-2-butenal, is an analog of propenal, the by-product produced in Step 5. O H
(E)-2-Butenal
If (E)-2-butenal were treated with bromine in carbon tetrachloride, the dibromide product would be the result of: (a) anti addition. (b) cis addition. (c) syn addition. (d) cycloaddition. 3. Enzymes are naturally occurring proteins which catalyze biological reactions. Because they are catalysts, enzymes: (a) change the energy of the reactants. (b) change the energy of the products. (c) change the energy difference (DH) between products and reactants. (d) change the reaction pathway. 4. What is the notation for the reaction mechanism for Step 5? (a) E1 (b) E2 (c) SN1 (d) SN2 5. Step 1 produces only one of the two possible isomeric secondary alcohols next to the N atom. The enzymatic reaction which produces the single isomer is called: (a) regioselective. (b) regiospecific. (c) stereoselective. (d) stereospecific. For further information, see Chapters 8, 12, and 20.
Passage Number X Many organic compounds contain two functional groups, both of which may react with certain reagents. When attempting to react only one of the groups, it is often necessary to protect the other group by temporarily changing it into a functional group that is inert to the reagent to be used. This temporary group is known as a protecting group and it is removed at the end of the reaction.
MCAT® Questions
ALCOHOLS Alcohols can be protected by treatment with acetic anhydride, an enol ether, or an alkyl halide (see Reactions 1–3). Reaction 1 O
O CH3
CH2COCH3
HO
O O
CH3
Pyridine
Compound I
O
O
CH3CO
CH2COCH3 Compound II
Reaction 2
OH Br
⫹
Br
Compound III
O
O
O Compound IV
Compound V
Reaction 3 OH
⫹
CH3l
O
Base
CH3
AMINES As with alcohols, amines can be protected with acetic anhydride (see Reaction 4). Reaction 4 O H3C
O NH2
⫹
O
O
H N
Pyridine
H3C
H3C
CH3
O
CH3 O
ACIDS Carboxylic acids can be protected by treatment with thionyl chloride and potassium t-butoxide, as illustrated in Reaction 5. Reaction 5 O BrCH2CH2COH Compound VI
O 1) SOCl2 2) K⫹⫺OC(CH3)3
BrCH2CH2COC(CH3)3
⫹
KCl
Compound VII
1. Which of the following functional groups are present in the product shown for Reaction 4? (a) Two ketones and an amide (b) A ketone and an amide (c) A ketone and an ester (d) An ester and an amide
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MCAT® Questions
2. What is the product of the first step of Reaction 5? O (a)
ClOSCH2CH2COH O
(b)
ClCH2CH2COH O
(c)
BrCH2CH2CCl O
(d)
BrCH2CH2COSOCl 3. If an ester, an ether, and an acid are contained in the same molecule, in what order will they react with aqueous base? (a) The ether, the acid, then the ester (b) The acid, the ester, then the ether (c) The acid, the ether, then the ester (d) The ester, the acid, then the ether 4. When Compound VII is subjected to acid hydrolysis, Compound VI is regenerated and 2-methylpropene is formed. Which of the following transformations accounts for the formation of 2-methylpropene? (a) Dehydration of an alcohol (b) Nucleophilic displacement of a halide (c) Cleavage of an ether (d) Oxidation of an acid 5. An (a) (b) (c) (d)
important characteristic of a protecting group is that it must be: small, to prevent steric hindrance of other reactions. very reactive toward most reagents. very inert to all reagents. easy to remove at the end of a reaction.
For further information, see Chapters 9 and 20.
Passage Number XI A chemist conducted a series of experiments in which styrene and para-substituted styrenes were allowed to react with anhydrous nitric acid in dichloromethane. Figure 1 shows the structures of some of the reactants the chemist used and the products obtained in the study.
ONO2 CH P CH2 R
CH R
Compound 1
Figure 1 Reactants and products.
ONO2 CH3
CH R
Compound 2
Compound 3
CH2NO2
MCAT® Questions
Table 1 shows the results of five experiments in the study.
TABLE 1
Experimental Results
Experiment 1 2 3 4 5
Organic substrate (Compound 1)
Major product
Minor product(s)
Styrene (R 5 H) 4-Methylstyrene (R 5 2CH3) 4-Nitrostyrene (R 5 2NO2) 4-Trifluoromethylstyrene (R 5 2CF3) 4-Chlorostyrene (R 5 Cl)
Compound 2 (R 5 H) Compound 2 (R 5 2CH3) Compound 3 (R 5 2NO2) Compound 3 (R 5 2CF3) Compound 2 and Compound 3 (R 5 Cl)
Yes No No No Yes
Yes in the last column of Table 1 means that minor products do form during the experiment, and no means that minor products do not form. 1. In Experiment 1, the major product results from what type of reaction? (a) An addition reaction (b) A substitution reaction (c) An elimination reaction (d) A cleavage reaction 2. The minor products of Experiment 1 result from mononitration of the aromatic ring. These by-products are related structurally in that they are (a) diastereomers. (b) resonance forms. (c) constitutional isomers. (d) enantiomers. 3. Is the hypothetical explanation that Compound 3 (R 5 2NO2) arises from an SN2 reaction between Compound 2 and HNO3 a plausible explanation? (a) Yes, because the nitro group bonds to a primary carbon atom, which favors an SN2 reaction (b) Yes, because the 2ONO2 group activates the site of substitution (c) No, because SN2 reactions do not occur on aromatic substrates (d) No, because hydride ion cannot serve as a leaving group 4. Which of the following structures shows the major product of a reaction between 4-ethylstyrene and anhydrous nitric acid in dichloromethane? (a)
NO2 CH
CH3
CH3CH2 (b)
ONO2 CH CH3CH2
CH3
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MCAT® Questions
ONO2
(c)
CH
CH2NO2
CH3CH2 (d)
CH2
CH2NO2
CH3CH2 5. A reaction between Compound 1, where R is an acetyl group (2COCH3), and nitric acid would most closely mimic which experiment shown in Table 1? (a) Experiment 1 (b) Experiment 2 (c) Experiment 3 (d) Experiment 5 For further information, see Chapters 5, 12, and 22.
Passage Number XII Two chlorinated compounds, Compound 1 and Compound 2, have been isolated from a culture medium of the fungal strain Bjerkandera fumosa. The structures of these compounds (Figure 1) were proven by synthesis. The synthesis of Compound 1 is shown in Figure 2. O
OH %
OH OH MeO
MeO Cl
Cl
Compound 1
Compound 2
Figure 1 Structures of Compounds 1 and 2.
OH
O
CHO a
b
MeO
MeO
MeO
Cl
Cl
Compound 3
Compound 4
Cl Compound 5
O
c, d
OH
MeO
e
Compound 1
Cl Compound 6
Figure 2 Synthesis of Compound 1; key: (a) standard procedure; (b) PDC, CH2Cl2, rt.; (c) TMSCl, Et3N, rt.; (d) cat. OsO4, NMMO, acetone, 08C; (e) Zn(BH4)2, ether, 08C.
MCAT® Questions
A common procedure converted Compound 3 into Compound 4, which was allowed to react with PDC to produce Compound 5. The latter compound was converted into its trimethylsilyl enol ether and treated with OsO4 in the presence of N-methylmorpholine N-oxide (NMMO) to produce Compound 6, an a-ketol. When treated with NaBH4, Compound 6 gave a mixture of erythro and threo diols but yielded the desired erythro product exclusively on treatment with Zn(BH4)2. The synthetic Compound 1 was shown to be identical to the natural compound by gas chromatography (GC) and mass spectroscopy (MS). Compound 2 was synthesized by a known procedure, starting with Compound 7. The structure of the synthetic Compound 2 was also shown to be identical to the natural compound by GC/MS. COOCH3 MeO Cl Compound 7
1. If the chemists wished to cleave the methyl-oxygen bond of the methoxy ether in Compound 2, which of the following reagents would accomplish the task? (a) HI (b) I2/NaOH (c) NaI (d) CH3I 2. Step a of Figure 2 was accomplished by reacting Compound 3 with: (a) ethyl iodide in dry ether. (b) ethylmagnesium iodide in dry ether. (c) ethanol and a catalytic amount of HCl. (d) sodium borohydride in ethanol. 3. PDC functioned as what type of reagent in the synthetic scheme of Figure 2? (a) A reducing agent, because a carbonyl group was converted into a secondary alcohol (b) A reducing agent, because a secondary alcohol was converted into a ketone (c) An oxidizing agent, because a carbonyl group was converted into a secondary alcohol (d) An oxidizing agent, because a secondary alcohol was converted into a ketone 4. Compound 5 can be named as an aryl-substituted: (a) propan-1-one. (b) propan-1-o1. (c) propan-2-one. (d) propan-3-one. 5. What difference did the researchers use to distinguish the erythro and threo isomers by gas chromatography? (a) Retention times (b) Peak heights (c) Peak areas (d) Injection times For further information, see Chapters 8 and 22.
Passage Number XIII A research group conducted two experiments, summarized below, in which a substrate underwent elimination by an E2 mechanism. The factors that influence the orientation of the carbon-carbon double bond in the product were evaluated. The orientation was either Hofmann or Zaitsev, depending on the location of the double bond.
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EXPERIMENT 1 Several 2-halohexanes were subjected to elimination. Either methoxide or tert-butoxide was the base and the corresponding alcohol was the solvent. The general reaction is shown in Equation 1, and the results are recorded in Table 1. Equation 1 Base
CH3CH2CH2CH2CHCH3 A X
CH3CH2CH2CH2CHPCH2
Solvent
Hofmann
⫹ (E)- and (Z)-CH3CH2CH2CH P CHCH3 Zaitsev
TABLE 1
Results of Experiment 1 2-Hexenes
Halogen X
Base/solvent
1-Hexene (%)
E (%)
Z (%)
I CI F I Cl F
CH3O2/CH3OH CH3O2/CH3OH CH3O2/CH3OH tert-BuO2/tert-BuOH tert-BuO2/tert-BuOH tert-BuO2/tert-BuOH
19 33 70 78 91 97
63 50 21 15 5 2
18 17 9 7 4 1
EXPERIMENT 2 A single substrate, 2-iodobutane, was subjected to elimination by a variety of bases in the solvent dimethyl sulfoxide (DMSO). The general reaction is shown in Equation 2, and the results are recorded in Table 2. Equation 2 C
CH3CH2CHCH3 A I
F
K Base DMSO
CH3CH2CH PCH2
⫹
(E)- and (Z)-CH3CH PCHCH3
Hofmann
TABLE 2
Zaitsev
Results of Experiment 2 Base
pKa (Base-H)
1-Butene (%)
p-Nitrobenzoate Benzoate Phenoxide Methoxide tert-Butoxide
8.9 11.0 16.4 29.0 32.2
5.8 7.2 11.4 17.0 20.7
1. How does the nature of the leaving group correlate with the orientation of the product produced in Experiment 1? (a) The poorer the leaving group, the higher the proportion of Hofmann orientation. (b) The poorer the leaving group, the higher the proportion of Zaitsev orientation. (c) The nature of the leaving group has little effect on orientation. (d) The poorer the leaving group, the higher the proportion of (E)-2-hexene formed.
MCAT® Questions
2. A by-product was observed in the reaction shown by Equation 1 when the base/solvent pair was CH3O2/CH3OH. What is the most likely byproduct? (a) CH3CH2CH2CH2CH2CH2OCH3 (b) CH3CH2CH2CH2CH(OCH3)CH3 (c) CH3CH2CH2CH2CH(OH)CH3 (d) CH3CH2CH2CH2CH2CH2OH 3. If the leaving group in Experiment 1 were Br2, approximately what percentage of 1-hexene would be expected from the reaction when the base/solvent pair is CH3O2/CH3OH? (a) 15% (b) 25% (c) 50% (d) 70% 4. A student wished to prepare Compound 2.
CH2
Compound 2
Which of the following alkyl halides and base/solvent pairs would produce the highest yield of the desired product? (a) 2-Fluoro-1-methylcyclohexane and CH3O2/CH3OH (b) 2-Fluoro-1-methylcyclohexane and tert-BuO2/tert-BuOH (c) 1-Chloro-1-methylcyclohexane and CH3O2/CH3OH (d) 1-Fluoro-1-methylcyclohexane and tert-BuO2/tert-BuOH For further information, see Chapters 6 and 7.
Passage Number XIV The three-step reaction sequence shown in Figure 1 is a proven method for the synthesis of Compound 3, a very useful reagent for the asymmetric synthesis of medicinally important organic compounds. Step 1 Mg
PhBr
Solvent
PhMgBr
Step 2
H Ph
H Ph b) H3O⫹
HO
OH
#
CO2CH3
#
# HO
a) PhMgBr
Ph Ph Compound 1
Compound 2
Step 3
H Ph Pyridine CH2Cl2
O H Ph CH3CO
OH
#
# Ph Ph
Compound 2
CH3COCl
#
# HO
OH
Ph Ph Compound 3
Figure 1 Synthesis of Compound 3 (Ph 5 phenyl, C6H52).
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MCAT® Questions
The following facts relate to the first two steps shown in Figure 1. Fact 1: The reaction shown in Step 1 does not occur if methanol is the solvent. Fact 2: The transformation shown in Step 2 requires a minimum of three moles of phenylmagnesium bromide per mole of Compound 1 to effectively produce Compound 2. Fact 3: If the reaction shown in Step 2a is conducted with one equivalent of PhMgBr and followed by a workup (Step 2b) in aqueous acid, there is no net consumption of Compound 1.
1. If only one equivalent of PhMgBr is added in Step 2a, the PhMgBr is converted into what organic product? (a) Phenol (b) Benzene (c) Bromobenzene (d) Benzoic acid 2. What is the function of pyridine in Step 3? (a) To serve as the principal solvent (b) To capture HCl as the hydrochloride (c) To remove a proton from CH3COCl (d) To neutralize acetic acid as it forms 3. Which of the following compounds can replace PhMgBr in Step 2a? (a) PhOMgBr (b) PhI (d) PhLi (c) Ph2Cu 4. What functional groups are present in Compound 3? (a) Ether and ester (b) Alcohol and ether (c) Alcohol and ester (d) Alcohol, ether, and ketone 5. What product is obtained from a reaction between methyl benzoate and an excess of PhMgBr? (a) Triphenylmethanol (b) Diphenylmethanol (c) Diphenyl ketone (d) Phenyl benzoate For further information, see Chapters 8 and 20.
Passage Number XV Scientists have proposed three different mechanisms for the Diels-Alder reaction between a diene and a dienophile. Mechanism 1 involves a cyclic 6-centered transition state in which no intermediate is formed. Mechanism 2 results in a dipolar intermediate from the stepwise attack of a pair of electrons from the diene on the dienophile. Mechanism 3 is a stepwise process in which the intermediate is a diradical. Mechanism 1
⫹
⫺ ⫹ Mechanism 2
Mechanism 3
Observations 1–6 have been compiled from experimental data.
MCAT® Questions
OBSERVATIONS 1. The reaction is facilitated by electron-withdrawing groups on the dienophile. 2. If more than one product is possible, the substituents in the favored products are 1,2 or 1,4 to each other, rather than 1,3. 3. If two stereoisomers are possible, the addition is predominantly endo; that is, the larger substituents of the dienophile are under the ring. 4. Only the s-cis conformation of the diene undergoes Diels-Alder reactions. 5. Both the diene and the dienophile retain their stereochemistry in the product. 6. The rates of reaction depend very little on the nature of the solvent. B
B
A
A
s-trans
s-cis
C
C D
D
1. Observation 6 supports Mechanism 1 over Mechanism 2 because if a charged intermediate were formed: (a) nonpolar solvents would increase the rate of reaction. (b) nonpolar solvents would not affect the rate of reaction. (c) polar solvents would increase the rate of reaction. (d) polar solvents would not affect the rate of reaction. 2. Which of the following dienophiles would best facilitate a Diels-Alder reaction? CH3 (a) H
H
H
(b) H
CH3
H
CH3
(c) H
H
H
COOCH3
(d) H
COOCH3
H
COOCH3
3. The progress of the following Diels-Alder reaction was monitored by IR spectroscopy. O
O
⫹
H
CH3
CH3
Which of the following spectroscopic features would be the most useful in determining when the reaction is complete? (a) The appearance of the product C2C stretch (b) The appearance of the product alkene C2H stretch
MCAT-21
MCAT® Questions
(c) The disappearance of the dienophile CPO stretch (d) The disappearance of the diene CPC stretch 4. What is the name of the Diels-Alder product of the reaction shown below? H3C ⫹ H3C
(a) (b) (c) (d)
Cl
H
4-Chloro-1,2-dimethylcyclohexene 1-Chloro-3,4-dimethylcyclohexene 3-Chloro-1,6-dimethylcyclohexene 5-Chloro-1,2-dimethylcyclohexene
5. According to Observations 3 and 5, what should be the major product of the following reaction? Cl
H
H
COOCH3
⫹
H Cl
O
#
(b)
Cl H H
#
(a)
#
#
COCH3
COCH3 H
O O
(d)
COCH3
H Cl
#
Cl H
#
(c)
#
#
MCAT-22
H
H
COCH3 O
For further information, see Chapter 14.
Passage Number XVI A research study of a soft coral of the genus Lobophytum collected in the Philippines revealed a new diterpene Compound 3 and two related but previously known compounds, Compounds 1 and 2. The compounds (Figure 1) were isolated from a freeze-dried aqueous extract of the invertebrate by successive fractionations. Fractions that displayed HIV-inhibitory activity were purified by HPLC by elutions with increasing concentrations of acetonitrile in water to yield Compounds 1–3. Compounds 1 and 2 were identified as the known cembranolides, lobohedleolide and its isomer (7E)-lobohedleolide, respectively. High-resolution FABMS of Compound 3 gave a (MH)1 at m/z 376.2496 (C22H33NO4). The IR spectrum of Compound 3 displayed significant bands at 3400–3100, 1760, and 1674 cm21. The 1H and 13C NMR spectra of Compound 3 were similar to those of Compound 1 except that the resonances due to the exo-methylene group at C-17 in Compound 1 were replaced by dimethylaminomethyl (2CH2NMe2) resonances. The methylene protons of the dimethylaminomethyl group gave 1H NMR signals at 3.36 and 3.45 ppm. These signals were coupled to each other and to that of the proton at C-15. Compound 3 was isolated as a white gum with a specific rotation of 113.18 when measured in chloroform at a concentration of 0.25 g/100 mL. The UV spectrum of Compound 3 was taken in ethanol and showed absorptions at lmax 222 (log e 5 3.56) and 218 (log e 5 3.60) nm. All three compounds exhibited moderate HIV-inhibitory activity in a standard anti-HIV assay.
MCAT® Questions
20 11
13
H
19
COOH
17
1 15 16
3
O
H
O
18
Compound 1
H HOOC O
H
O
Compound 2
H NMe2
COOH H
O
O
Compound 3
Figure 1 Naturally occurring diterpenes.
1. What kind of isomers are Compounds 1 and 2? (a) Constitutional (b) Enantiomers (c) Diastereomers (d) Anomers 2. Excluding carbon-carbon double bonds, what functional groups are present in Compound 2? (a) Acid and lactone (b) Acid, ketone, and ether (c) Peroxide and lactone (d) Acid and hemiacetal 3. If Compound 1 were to react with dimethylamine to form Compound 3, the reaction would be described as what type of addition? (a) 1,2 (b) 1,4 (c) 2,3 (d) 3,4 4. Which carbon atoms in Compound 3 are stereogenic? (The numbering system used for carbon atoms in Compound 3 is analogous to that used in Compound 1.) (a) 1, 2, and 15 (b) 1, 2, and 17 (c) 3, 7, and 11 (d) 1, 2, 3, 7, 11, and 15
MCAT-23
MCAT-24
MCAT® Questions
5. How many degrees of unsaturation are present in Compound 3? (a) 2 (b) 3 (c) 6 (d) 7 For further information, see Chapters 5, 11, and 20.
Passage Number XVII Albuterol (a9-[tert-butylamino)methyl]-4-hydroxy-m-xylene-a,a9-diol), a drug used for the treatment of asthma, is synthesized according to the following pathway. O B CH3COCH2
O B CCH2N
CH2 C4H9
O B CH3CO Step 1 H⫹, H2O
HOCH2
O B CCH2 O N
CH2 C4H9
HO NaBH4, Step 2 CH CH OH 3 2
HOCH2
OH A CHCH2N
CH2 C4H9
HO Step 3 H2, Pd/C
HOCH2
OH A CHCH2NHC4H9
HO Albuterol
Sodium borohydride (NaBH4) is utilized in Step 2 because it is much less reactive than LiAlH4 and is consequently more selective. Typically, only aldehydes and ketones react with NaBH4. Albuterol is water-soluble and is a long-acting bronchodilator. 1. Which of the following shows the correct structure of the C4H9 group that is bonded to the nitrogen of albuterol (a9-[tert-butylamino)methyl]-4-hydroxy-m-xylene-a,a9-diol)? (a) CH3 H CH E CH3 (b) CH3CH2CH A CH3
MCAT® Questions
(c) CH3CHCH2 A CH3 CH3 A (d) CH3 O C A CH3 2. When albuterol is dissolved in water, which of the following hydrogen-bonded structures does NOT contribute to its water solubility? (a) HOCH2 HO O O H CHCH2N O C4H9 A A OH H
HO
H A OOH
(b) HOCH2
H A CHCH2N O C4H9 A OH
HO
(c)
H A OOH
HOCH2
H A C O CH2NHC4H9 A OH
HO
(d)
HOCH2 H OO A H
CHCH2NHC4H9 A OH
HO
3. In the synthesis of albuterol, Step 2 is which of the following types of reactions? (a) An oxidation reaction (b) A reduction reaction (c) A hydrolysis reaction (d) A substitution reaction 4. Which of the following structures is formed first when albuterol is dissolved in a dilute aqueous solution of NaOH? (a) HOCH2 HO
(b)
HOCH2 HO
F
B
CHCH2NH( O C4H8)Na A OH
NaB F
CHCH O NHC4H9 A OH
MCAT-25
MCAT-26
MCAT® Questions
(c)
HOCH2
NaB F
HO
CHCH2NC4H9 A OH HOCH2
(d) B
F NaO
CHCH2NHC4H9 A OH
5. What type of reaction is illustrated by Step 3? (a) Heterogeneous catalysis (b) Homogeneous catalysis (c) Halophilic substitution (d) Bimolecular elimination 6. What is the product of the following reaction?
COOCH3 NaBH4 CH3OH
O Br
H
(a) COOH
O Br
H
(b) CH2OH
O Br
H
(c) CH2OH
{
HO
Br
H
MCAT® Questions
(d) COOCH3
{
HO
Br
H
7. Step 3 represents the conversion of a tertiary amine to: (a) a primary amine. (b) a secondary amine. (c) an amide. (d) an imine. 8. How many chiral carbons exist in synthetic albuterol? (a) 0 (b) 1 (c) 2 (d) 3 9. Which of the following by-products is formed in Step 1? (a)
O B CH3C O CH3
(b) CH3OH O B (c) CH3C O OH (d) CH3OCH3 For further information, see Chapters 8, 21, and 22.
Passage Number XVIII An organic chemistry class was assigned to prepare a variety of alkyl halides from the corresponding alcohols. The students carried out several experiments, 3 of which are summarized below. EXPERIMENT 1 Sodium bromide was refluxed with ethanol for 8 hours. No reaction occurred. C2H5OH 1 NaBr uvy no reaction EXPERIMENT 2 Hydrogen bromide was refluxed with ethanol for 8 hours to produce ethyl bromide and water. C2H5OH 1 HBr uvy C2H5Br 1 H2O EXPERIMENT 3 3-Methyl-2-butanol was treated with hydrogen bromide to produce only 2-bromo-2-methylbutane. OH A (CH3)2CHCHCH3 ⫹ HBr
Br A (CH3)2CCH2CH3
MCAT-27
MCAT-28
MCAT® Questions
CONCLUSIONS The students ultimately deduced that Experiment 2 proceeded via protonation of the hydroxy group followed by displacement of water by bromide. In contrast, the class concluded that Experiment 3 involved the rearrangement of an intermediate carbocation followed by reaction with bromide. 1. What type of reaction mechanism is exemplified by Experiment 2? (a) E1 (b) E2 (c) SN1 (d) SN2 2. Why is Experiment 2 much more efficient than Experiment 1? (a) OH2 is a better leaving group than water is. (b) Water is a better leaving group than OH2 is. (c) NaBr is more soluble in ethanol than HBr is. (d) NaBr is more volatile than HBr is. 3. According to the results in the passage, which of the following reactions should produce the highest yield of CH3CH2Br? (a) CH3CH2OH 1 Br2 (b) CH3CH2OH 1 NaBr 1 Br2 (c) CH3CH2OH 1 NaBr 1 H2SO4 (d) CH3CH2OH 1 KBr 4. Which of the following carbocations is the most stable? 1 (a) CH3CH2CH2C H2 1
(b) CH3CH2CH2C HCH3 1
(c) (CH3)3C 1
(d) C H3 5. If the proposed mechanism for the reaction in Experiment 3 is correct, which of the following assumptions can be made? (a) The rate of rearrangement is less than the rate of reaction of the carbocation with bromide. (b) The rate of rearrangement is greater than the rate of reaction of the carbocation with bromide. (c) The rate of rearrangement is equal to the rate of reaction of the carbocation with bromide. (d) The rate of rearrangement is directly proportional to the rate of reaction of the carbocation with bromide. 6. To synthesize a pure secondary alkyl halide, which of the following alcohols should the students treat with HBr? (a) CH3CH2OH (b) (CH3)2CHCH(OH)CH(CH3)2 (c) (CH3)3COH (d) (CH3)2CHOH 7. In a further experiment, some students heated 2-methyl-2-butanol with concentrated sulfuric acid instead of hydrogen bromide. Which of the following products was most likely formed in the greatest amount? (a) CH3 O C P CH O CH3 A CH3 (b) CH2 P C O CH2 O CH3 A CH3
MCAT® Questions
(c)
Br A CH3 O C O CH2 O CH3 A CH3
(d)
Br A CH3 O CH O CH O CH3 A CH3
8. During the course of the experiments, the students were surprised to discover that 2-methyl-2-propanol had a significantly lower boiling point than 1-butanol. What is the reason for this difference? (a) 2-Methyl-2-propanol has a lower molecular weight than 1-butanol. (b) 2-Methyl-2-propanol has a higher density than 1-butanol. (c) The branched carbon chain in 2-methyl-2-propanol interferes with hydrogen bonding. (d) The branched carbon chain in 2-methyl-2-propanol destabilizes the carbocation intermediate. 9. The students tried to synthesize an alkyl chloride by substituting HCl for HBr in the procedure described in Experiment 2, but the reaction was sluggish. What is the reason for this decreased reactivity? (a) Chloride is a weaker nucleophile than bromide is. (b) Chloride is a weaker electrophile than bromide is. (c) Chloride stabilizes the intermediate carbocation that is produced. (d) Chloride destabilizes the intermediate carbocation that is produced. For further information, see Chapters 9 and 11.
Passage Number XIX A new carbon-carbon bond can be created by alkylating a carbanion with an alkyl bromide (RBr). Typically the carbanion is formed by the abstraction of an acidic proton by butyllithium (BuLi). One of the groups that has been utilized to activate the proton toward abstraction is the sulfone substituent that is illustrated in methyl phenyl sulfone, Compound I, in the following reaction. O B C6H5SCH3 B O Compound I
O BF B C6H5SCH2Li B O
BuLi, THF 0⬚C
RBr
C6H5SO2CH2R ⫹ C6H5SO2CHR2 ⫹ C6H5SO2CR3 Product A
Product B
Product C
Monoalkylations, dialkylations, and trialkylations are possible, as indicated by Products A–C. The products were distinguished by gas chromatography because the retention times increased in the following order: Compound I , Product A , Product B , Product C. Table 1 shows the distribution of Products A, B, and C and Compound I obtained when methyl phenyl sulfone was alkylated with a variety of alkyl halides.
MCAT-29
MCAT® Questions
TABLE 1 Products (%) Trial
A
B
C
Compound I (%)
71 66 40 36
19 19 25 43
0 0 2.5 0
11 16 33 21
RBr
1 2 3 4
CH3CH2CH2Br CH3CH2CH2CH2Br CH2PCHCH2Br C6H5CH2Br
In another experiment, the reactions were repeated using potassium hydride (KH) instead of butyllithium to determine the effect on monoalkylation versus dialkylation. The results of the comparison between BuLi and KH are compiled in Table 2. TABLE 2 Ratio of monoalkylation to dialkylation RBr
BuLi
KH
CH3CH2CH2Br CH3CH2CH2CH2Br CH2PCHCH2Br C6H5CH2Br
3.7 :1 3.5 :1 1.6 :1 0.8 :1
29 :1 21 :1 2.5 :1 1.3 :1
1. The sulfone group activates the hydrogen toward abstraction by BuLi or KH because the resulting: (a) carbocation is stabilized by resonance. (b) carbocation is destabilized by resonance. (c) carbanion is stabilized by resonance. (d) carbanion is destabilized by resonance.
Intensity
Intensity
2. Which of the following gas chromatograms most accurately represents the mixture of products and Compound I obtained from Trial 4 in Table 1? (b) (a)
0
0
Time
Time
(d) Intensity
(c) Intensity
MCAT-30
0
Time
0
Time
3. According to Table 1, which of the following compounds represents the monoalkylated Product A that was formed with the LEAST amount of contamination by the dialkylated Product B? (a) C6H5SO2CH2C6H4Br (b) C6H5SO2CH2CH2C6H5 (c) C6H5SO2CH2CH2CH2CH2Br (d) C6H5SO2CH2CH2CH2CH3
MCAT® Questions
4. Product B is most likely formed by which of the following reaction mechanisms? F
B
(a) C6H5SO2CH2Li
BuLi
⫺2
C6H5SO2CHLi2 2 RBr
C6H5SO2CHR2 F
B
(b) C6H5SO2CH2Li
RBr
C6H5SO2CH2R F
B
C6H5SO2CH2Li
C6H5SO2CHR2 F
B
(c) C6H5SO2CH2Li
RBr
RBr
F
C6H5SO2CHR C6H5SO2CH2R BrF
C6H5SO2CHR2 F
B
(d) C6H5SO2CH2Li
RF
BrF
C6H5SO2CHRBr C6H5SO2CHBrLi 2 RF
C6H5SO2CHR2 5. According to Table 2, if chemists want to obtain the highest ratio of a dialkylated product, they should use: (a) an unsaturated alkyl halide and BuLi. (b) an unsaturated alkyl halide and KH. (c) a saturated alkyl halide and BuLi. (d) a saturated alkyl halide and KH. 6. Which of the following structures corresponds to the major product of Trial 4 in Table 1? (a)
O B C6H5S O(CH2C6H5)2 B O
(b)
O H B A C6H5CH2S O CCH2C6H5 B A O CH2C6H5
(c)
O H B A C6H5S O CC6H5 B A O CH2C6H5
(d)
O H B A C6H5S O CCH2C6H5 B A O CH2C6H5
7. If methyl ethyl sulfone were utilized in the alkylation reaction instead of Compound I, which of the following problems would occur with the alkylation reaction? (a) A greater number of products could be formed. (b) The methyl hydrogen would no longer be acidic.
MCAT-31
MCAT-32
MCAT® Questions
(c) The products would not be volatile enough to analyze by gas chromatography. (d) The carbanion would be too sterically hindered to undergo alkylation. For further information, see Chapter 18.
Passage Number XX The most common glyceryl trialkanoates (commonly known as fats or glycerides) are triesters of glycerol and long-chain carboxylic acids. Compound 1 and Compound 2 are two naturally occurring glyceryl trialkanoates. (Note: In Compound 1, R 5 oleic acid; in Compound 2, R 5 stearic acid.) CH2OH A CHOH A CH2OH
CH2OCOR A CHOCOR A CH2OCOR
Glycerol
Glyceryl trialkanoate
CH3(CH2)7 (CH2)7COOH H D CPC G E H H
CH3(CH2)16COOH
Oleic acid
Stearic acid
The long-chain carboxylic acids, called fatty acids, may be saturated or unsaturated. Oils are fats that are made up of predominantly cis unsaturated fatty acids; they are liquid at room temperature. Fats with little unsaturation are solid at room temperature. Fats undergo the typical reactions of esters. An important commercial reaction of fats is alkaline hydrolysis, or saponification. The products of saponification are glycerol and the salts of the corresponding fatty acids, which are often used as soaps. Phosphoglycerides are fats in which glycerol is esterified with 2 different fatty acids and phosphoric acid. These monophosphate esters are called phosphatidic acids. Naturally occurring free phosphatidic acids are rare. Usually the phosphoric acid substituent is esterified to a second alcohol such as choline, as exemplified by phosphatidylcholine. O B CH2OC(CH2)14CH3
O 1 B CH3(CH2)14CO # C ! H 2
O B ⫹ CH2OPOCH2CH2N(CH3)3 3 B O⫺ Phosphatidylcholine
1. The reduction of Compound 1 to Compound 2 represents the conversion of: (a) a solid to an oil. (b) an oil to a solid. (c) a fatty acid to a triglyceride. (d) a triglyceride to a fatty acid. 2. What is the product when Compound 2 undergoes saponification with NaOH? (a) CH3(CH2)16CH2OH (b) CH3(CH2)16CH2O2Na1 (c) CH3(CH2)16COOH (d) CH3(CH2)16COO2Na1
MCAT® Questions
3. Based on the structure of phosphatidylcholine, which of the following structures accurately represents choline? (a) (CH3)3N1CH2CH2OH (b) (CH3)3N1CH2CH2OPO32 (c) (CH3)3N1CH2CH2COO2 (d) (CH3)3N1CH2CH2OPO4CH2OH 4. The salts of fatty acids are used as soaps because the salts: (a) have a polar region and a nonpolar region and are thus insoluble in water. (b) have a polar region and a nonpolar region and thus help organic materials become water-soluble. (c) are exclusively polar and thus dissolve in aqueous solutions. (d) are exclusively nonpolar and thus dissolve organic materials. 5. Which of the following properties is characteristic of the phosphorus-containing substituent on Carbon 3 of phosphatidylcholine? (a) It is a primary amine. (b) It is hydrophobic. (c) It is polar. (d) It is nonpolar. 6. What is the most likely explanation for the observation that unsaturated glyceryl alkanoates are liquids? (a) The cis double bonds do not allow the carbon chain to fit into an orderly crystal structure. (b) The unsaturated glyceryl alkanoates are more susceptible to the effects of hydrogen bonding than the corresponding saturated compounds are. (c) The molecular weights of unsaturated glyceryl alkanoates are much higher than those of the corresponding saturated compounds. (d) Unsaturated glyceryl alkanoates have higher densities than the corresponding saturated compounds. 7. Which of the following methods can be used to increase the melting point of a natural fat? (a) Heating the fat (b) Cooling the fat (c) Esterifying the fat (d) Hydrogenating the fat 8. How many chiral carbons does phosphatidylcholine have? (a) 0 (b) 1 (c) 2 (d) 3 9. A chemist wants to monitor the conversion of glycerol to a glyceryl trialkanoate by infrared spectroscopy. Which one of the following glycerol stretches will NOT be present in the infrared spectrum of a glyceryl trialkanoate? (a) The C—C stretch (b) The C—H stretch (c) The O—H stretch (d) The C—O stretch 10. What functional group is formed when a saponification product (fatty acid salt) is treated with HCl? (a) An alcohol group (b) An aldehyde group (c) An ester group (d) A carboxylic acid group For further information, see Chapters 19 and 20.
MCAT-33
MCAT-34
MCAT® Questions
A nswer Key to MCAT
®
Questions
Complete solutions available online at www.whfreeman.com/vollhardtschore6E
Passage Number I
Passage Number IX
Passage Number XVI
1. 2. 3. 4. 5.
1. 2. 3. 4. 5.
1. 2. 3. 4. 5.
d d b b c
Passage Number II 1. 2. 3. 4.
d b d a
Passage Number III 1. 2. 3. 4.
c b a d
Passage Number IV 1. 2. 3. 4.
d a b a
Passage Number V 1. 2. 3. 4. 5.
c b b b c
Passage Number VI 1. 2. 3. 4.
d d c d
Passage Number VII 1. 2. 3. 4.
d b b d
Passage Number VIII 1. 2. 3. 4.
a d a a
b a d b d
Passage Number X 1. 2. 3. 4. 5.
b c b a d
Passage Number XI 1. 2. 3. 4. 5.
a c d b c
Passage Number XII 1. 2. 3. 4. 5.
a b d a a
Passage Number XIII 1. 2. 3. 4.
c b b d
Passage Number XIV 1. 2. 3. 4. 5.
b b d c a
Passage Number XV 1. 2. 3. 4. 5.
c d c a a
c a b a d
Passage Number XVII 1. 2. 3. 4. 5. 6. 7. 8. 9.
d c b d a d b b c
Passage Number XVIII 1. 2. 3. 4. 5. 6. 7. 8. 9.
d b c c b d a c a
Passage Number XIX 1. 2. 3. 4. 5. 6. 7.
c b d b a d a
Passage Number XX 1. b 2. d 3. a 4. b 5. c 6. a 7. d 8. b 9. c 10. d
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