Organic Chemistry, Sixth Edition

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Library of Congress Cataloging-in-Publication Data Wade, L. G. Organic chemistry / L. G. Wade, Jr. p. cm. Includes bibliographical references and index. ISBN 0-13-147871-0 I. Chemistry, Organic-Textbooks. l. Title. QD251.3.w33 2006 547-dc22

2004060047

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Brief Contents Preface xxiii

1

Introduction and Review

2

Structure and Properties of Organic Molecules 39

3

Structure and Stereochemistry of Alkanes 81

4

The Study of Chemical Reactions 125

5

Stereochemistry 167

6

Alkyl Halides: Nucleophilic Substitution and Elimination 212

7

Structure and Synthesis of Alkenes 279

8

Reactions of Alkenes 321

9

Alkynes 382

10

Structure and Synthesis of Alcohols 417

11

Reactions of Alcohols 460

12

Infrared Spectroscopy and Mass Spectrometry 508

13

Nuclear Magnetic Resonance Spectroscopy 559

14

Ethers, Epoxides, and Sulfides 623

15

Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy 663

16

Aromatic Compounds 705

17

Reactions of Aromatic Compounds 749

18

Ketones and Aldehydes 805

19

Amines 836

20

Carboxylic Acids 870

21

Carboxylic Acid Derivatives 978

22

Condensations and Alpha Substitutions of Carbonyl Compounds 1041

23

Carbohydrates and Nucleic Acids 1097

24

Amino Acids, Peptides, and Proteins 1153

25

Lipids 1200

26

Synthetic Polymers 1222 Appendices 1243 Answers to Selected Problems A1 Photo Credits PC1 Index 11

vi

Contents Preface xxiii About the Author xli

1

Introduction and Review 1 1-1

The Origins of Organic Chemistry

1-2

Principles of Atomic Structure

1-3

Bond Formation: The Octet Rule

1-4

Lewis Structures

1-5

Multiple Bonding

3

7 8

1

6

Summary: Common Bonding Patterns (Uncharged) 1-6

Electronegativity and Bond Polarity

1-7

11 Ionic Structures 12

1-8

9

9

Formal Charges

Summary: Common Bonding Patterns in Organic Compounds and Ions 1-9

Resonance

13

17

1-10

Structural Formulas

1-11

Molecular Formulas and Empirical Formulas

21

1-12

Arrhenius Acids and Bases

1-13

Br¢nsted-Lowry Acids and Bases

1-14

Lewis Acids and Bases

1 Glossary 32 Study Problems 34

13

20

22

29

Chapter

2

Structure and Properties of Organic Molecules 39 2-1

39

Wave Properties of Electrons in Orbitals

41

2-2

Molecular Orbitals

2-3

Pi Bonding

2-4

Hybridization and Molecular Shapes

2-5

Drawing Three-Dimensional Molecules

2-6

General Rules of Hybridization and Geometry

2-7

Bond Rotation

44

56

45

49

54

2-8

Isomerism

2-9

Polarity of Bonds and Molecules

61

58

2-10

Intermolecular Forces

2-11

Polarity Effects on Solubilities

2-12

Hydrocarbons

2-13

Organic Compounds Containing Oxygen

2-14

Organic Compounds Containing Nitrogen Chapter

68

2 Glossary 75

Study Problems

77

50

65 71

73

vii

viii

Contents

3

Structure and Stereochemistry of Alkanes 81 3-1

Classification of Hydrocarbons (Review)

3-2

Molecular Formulas of Alkanes

3-3

Nomenclature of Alkanes

83

Summary: Rules for Naming Alkanes

89 3-5 Uses and Sources of Alkanes 91 3-6 Reactions of Alkanes 93 3-4

81

82 88

Physical Properties of Alkanes

3-7

Structure and Conformations of Alkanes

3-8

Conformations of Butane

3-9

Conformations of Higher Alkanes

98

94

100

100

3-10

Cycloalkanes

3-11

cis-trans Isomerism in Cycloalkanes

3-12

Stabilities of Cycloalkanes; Ring Strain

3-13

Cyclohexane Conformations

107

103 103

Problem-Solving Strategy: Drawing Chair Conformations 110

111 3-15 Conformations of Disubstituted Cyclohexanes 114 3-14

Conformations of Monosubstituted Cyclohexanes Problem-Solving Strategy: Recognizing cis and trans Isomers 116

3-16

117 Chapter 3 Glossary 119

Bicyclic Molecules

Study Problems 122

4

The Study of Chemical Reactions 1 25 4-1

Introduction

125

125

4-2

Chlorination of Methane

4-3

The Free-Radical Chain Reaction

126

Key Mechanism: Free-Radical Halogenation 128

130

4-4

Equilibrium Constants and Free Energy

4-5

Enthalpy and Entropy

4-6

Bond-Dissociation Enthalpies

4-7

Enthalpy Changes in Chlorination

4-8

Kinetics and the Rate Equation

4-9

Activation Energy and the Temperature Dependence of Rates

139

133

140

134

135 137

4-10

Transition States

4-11

Rates of Multistep Reactions

4-12

Temperature Dependence of Halogenation

142

144 149

4-13

Selectivity in Halogenation

4-14

The Hammond Postulate

143

Problem-Solving Strategy: Proposing Reaction Mechanisms 151

153

4-15

Radical Inhibitors

4-16

Reactive Intermediates

155

Summary: Reactive Intermediates Chapter 4 Glossary Study Problems

160 163

160

Contents

5

Stereochemistry 167 5-1

Introduction 167

5-2

Chirality 168

5-3

(R) and (S) Nomenclature of Asymmetric Carbon Atoms 174

5-4

Optical Activity 179

5-5

Biological Discrimination of Enantiomers 184

5-6

Racemic Mixtures 185

5-7

Enantiomeric Excess and Optical Purity 186

5-8

Chirality of Conformationally Mobile Systems 187

5-9

Chiral Compounds without Asymmetric Atoms 189

5-10

Fischer Projections 191

5-11

Diastereomers 196

Summary : Fischer Projections and Their Use 196 Summary: Types of Isomers 197 5-12

Stereochemistry of Molecules with Two or More Asymmetric Carbons 198

5-13

Meso Compounds 199

5-14

Absolute and Relative Configuration 201

5-15

Physical Properties of Diastereomers 203

5-16

Resolution of Enantiomers 204 Chapter 5 Glossary 207 Study Problems 209

6

Alkyl Halides: Nucleophilic Substitution and Elimination 212 6-1

Introduction 212

6-2

Nomenclature of Alkyl Halides 213

6-3

Common Uses of Alkyl Halides 215

6-4

Structure of Alkyl Halides 217

6-5

Physical Properties of Alkyl Halides 218

6-6

Preparation of Alkyl Halides 220 Summary: Methods for Preparing Alkyl Halides 223

6-7

Reactions of Alkyl Halides: Substitution and Elimination 225

6-8

Second-Order Nucleophilic Substitution: The SN2 Reaction 226

6-9

Generality of the SN2 Reaction 228

Key Mechanism: The SN2 Reaction 227

Summary: SN2 Reactions of Alkyl Halides 228 6-10

Factors Affecting SN2 Reactions: Strength of the Nucleophile 230 Summary: Trends in Nucleophilicity 231

6-11

Reactivity of the Substrate in SN2 Reactions 234

6-12

Stereochemistry of the SN2 Reaction 238

6-13

First-Order Nucleophilic Substitution: The SN1 Reaction 240 Key Mechanism: The SN1 Reaction 241

6-14

Stereochemistry of the SNI Reaction 244

6-15

Reanangements in SN1 Reactions 246

6-16

Comparison of SN1 and SN2 Reactions 249 Summary: Nucleophilic Substitutions 251

6-17

First-Order Elimination : The E l Reaction 252 Key Mechanism: The E1 Reaction 252

Summary: Carbocation Reactions 256 6-18

Positional Orientation of Elimination: Zaitsev's Rule 257

ix

X

Contents 6-19

Second-Order Elimination: The E2 Reaction 258

6-20

Stereochemistry of the E2 Reaction 261

6-21

Comparison of EI and E2 Elimination Mechanisms 262

Key Mechanism: The E2 Reaction 259

Summary: Elimination Reactions 264 Problem-Solving Strategy: Predicting Substitutions and Eliminations 264

Summary: Reactions of Alkyl Halides 267 Chapter 6 Glossary 270 Study Problems 273

7

Structure and Synthesis of Alkenes 279 7-1

Introduction 279

7-2

The Orbital Description of the Alkene Double Bond 280

7-3

Elements of Unsaturation 281

7-4

Nomenclature of Alkenes 283

7-5

Nomenclature of Cis-Trans Isomers 285

7-6

Commercial Importance of Alkenes 288

Summary: Rules for Naming Alkenes 287 7-7

Stability of Alkenes 290

7-8

Physical Properties of Alkenes 296

7-9

Alkene Synthesis by Elimination of Alkyl Halides 298

7-10

Alkene Synthesis by Dehydration of Alcohols 306 Key Mechanism: Acid-Catalyzed Dehydration of an Alcohol 307

7-11

Alkene Synthesis by High-Temperature Industrial Methods 309 Problem-Solving Strategy: Proposing Reaction Mechanisms 310

Summary: Methods for Synthesis of Alkenes 314 Chapter 7 Glossary 316 Study Problems 318

8

Reactions of Alkenes 321 8-1

Reactivity of the Carbon-Carbon Double Bond 321

8-2

Electrophilic Addition to Alkenes 322 Key Mechanism: Electrophilic Addition to Alkenes 322

8-3

Addition of Hydrogen Halides to Alkenes 324

8-4

Addition of Water: Hydration of Alkenes 330

8-5

Hydration by Oxymercuration-Demercuration 333

8-6

Alkoxymercuration-Demercuration 335

8-7

Hydroboration of Alkenes 336

8-8

Addition of Halogens to Alkenes 342

8-9

Formation of Halohydrins 345

8-10

Catalytic Hydrogenation of Alkenes 348

8-11

Addition of Carbenes to Alkenes 350

8-12

Epoxidation of Alkenes 353

8-13

Acid-Catalyzed Opening of Epoxides 355

8-14

Syn Hydroxylation of Alkenes 358

8-15

Oxidative Cleavage of Alkenes 360

8-16

Polymerization of Alkenes 363 Problem-Solving Strategy: Organic Synthesis 367

Contents Summary: Reactions of Alkenes 370 Chapter 8 Glossary 374 Study Problems 376

9

Alkynes 382 9-1

Introduction 382

9-2

Nomenclature of Alkynes 383

9-3

Physical Properties of Alkynes 384

9-4

Commercial Importance of Alkynes 384

9-5

Electronic Structure of Alkynes 386

9-6

Acidity of Alkynes; Formation of Acetylide Ions 387

9-7

Synthesis of Alkynes from Acetylides 389

9-8

Synthesis of Alkynes by Elimination Reactions 393 Summary: Syntheses of Alkynes 396

9-9 9-10

Addition Reactions of Alkynes 396 Oxidation of Alkynes 406 Problem-Solving Strategy: Multistep Synthesis 408

Summary: Reactions of Alkynes 409 Chapter 9 Glossary 412 Study Problems 413

10

Structure and Synthesis of Alcohols 41 7 10-1

Introduction 417

10-2

Structure and Classification of Alcohols 417

10-3

Nomenclature of Alcohols and Phenols 419

10-4

Physical Properties of Alcohols 423

10-5

Commercially Important Alcohols 425

10-6

Acidity of Alcohols and Phenols 427

10-7

Synthesis of Alcohols: Introduction and Review 430 Summary: Previous Alcohol Syntheses 430

10-8

Organometallic Reagents for Alcohol Synthesis 432

10-9

Addition of Organometallic Reagents to Carbonyl Compounds 435 Key Mechanism: Gri g n ard Reactions 435

Summary: Grignard Reactions 442 10-10

Side Reactions of Organometallic Reagents: Reduction of Alkyl Halides 443

10-11

Reduction of the Carbonyl Group: Synthesis of 1° and 2° Alcohols 445 Summary: Reactions of LiAlH4 and NaBH4 448 Summary: Alcohol Syntheses 449

10-12

Thiols (Mercaptans) 451 Chapter

10 Glossary

454

Study Problems 455

11

Reactions of Alcohols 460 11-1

Oxidation States of Alcohols and Related Functional Groups 460

11-2

Oxidation of Alcohols 462

11-3

Additional Methods for Oxidizing Alcohols 465

11-4

Biological Oxidation of Alcohols 467

xi

xii

Contents 11-5

Alcohols as NucJeophiles and Electrophiles; Formation of Tosylates 469 Summary: SN2 Reactions of Tosylate Esters 47l

11-6

Reduction of Alcohols 472

11-7

Reactions of Alcohols with Hydrohalic Acids 472

11-8

Reactions of Alcohols with Phosphorus Halides 477

11-9

Reactions of Alcohols with Thionyl Chloride 478

11-10

Dehydration Reactions of Alcohols 480

11-11

Unique Reactions of Diols 488

11-12

Esterification of Alcohols 490

11-13

Esters of Inorganic Acids 491

11-14

Reactions of Alkoxides 494

Problem-Solving Strategy: Proposing Reaction Mechanisms 484

Key Mechanism: The Williamson Ether Synthesis 494 Problem-Solving Strategy: Multistep Synthesis 496

Summary: Reactions of Alcohols 499 Chapter 11 Glossary 502 Study Problems 503

12

Infrared Spectroscopy and Mass Spectrometry 508 12-1

Introduction 508

12-2

The Electromagnetic Spectrum 509

12-3

The Infrared Region 510

12-4

Molecular Vibrations 511

12-5

IR-Active and IR-Inactive Vibrations 513

12-6

Measurement of the IR Spectrum 514

12-7

Infrared Spectroscopy of Hydrocarbons 517

12-8

Characteristic Absorptions of Alcohols and Amines 522

12-9

Characteristic Absorptions of Carbonyl Compounds 523

12-10

Characteristic Absorptions of C- N Bonds 529

12-11

Simplified Summary of IR Stretching Frequencies 530

12-12

Reading and Interpreting IR Spectra (Solved Problems) 532

12-13

Introduction to Mass Spectrometry 537

12-14

Determination of the Molecular Formula by Mass Spectrometry 541

12-15

Fragmentation Patterns in Mass Spectrometry 544 Summary: Common Fragmentation Patterns 549 Chapter 12 Glossary 551 Study Problems 552

13

Nuclear Magnetic Resonance Spectroscopy 559 13-1

Introduction 559

13-2

Theory of Nuclear Magnetic Resonance 559

13-3

Magnetic Shielding by Electrons 562

13-4

The NMR Spectrometer 564

13-5

The Chemical Shift 565

13-6

The Number of Signals 572

13-7

Areas of the Peaks 573

13-8

Spin-Spin Splitting 576 Problem-Solving Strategy: Drawin g an NMR Spectrum 581

Contents 13-9

Complex Splitting 585

13-10

Stereochemical Nonequivalence of Protons 588

13-11

Time Dependence of NMR Spectroscopy 591 Problem-Solving Strategy: Interpreting Proton NMR Spectra 594

13-12

Carbon-13 NMR Spectroscopy 599

13-13

Interpreting Carbon NMR Spectra 607

13-14

Nuclear Magnetic Resonance Imaging 609 Problem-Solving Strategy: Spectroscopy Problems 610

Chapter 13 Glossary 614 Study Problems 616

14

Ethers, Epoxides, and Sulfides 623 14-1

Introduction 623

14-2

Physical Properties of Ethers 623

14-3

Nomenclature of Ethers 628

14-4

Spectroscopy of Ethers 631

14-5

The Williamson Ether Synthesis 633

14-6

Synthesis of Ethers by Alkoxymercuration-Demercuration 634

14-7

Industrial Synthesis: Bimolecular Dehydration of Alcohols 635

14-8

Cleavage of Ethers by HBr and HI 636

Summary: Syntheses of Ethers 636 14-9

Autoxidation of Ethers 639 Summary: Reactions of Ethers 639

14-10

Sulfides (Thioethers) 640

14-11

Synthesis of Epoxides 642

14-12

Acid-Catalyzed Ring Opening of Epoxides 645

14-13

Base-Catalyzed Ring Opening of Epoxides 649

Summary: Epoxide Syntheses 645

14-14

Orientation of Epoxide Ring Opening 650

14-15

Reactions of Epoxides with Grignard and Organolithium Reagents 652

14-16

Epoxy Resins: The Advent of Modern Glues 653 Summary: Reactions of Epoxides 655 Chapter 14 Glossary 656 Study Problems 658

15

Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy 663 15-1

Introduction 663

15-2

Stabilities of Dienes 663

15-3

Molecular Orbital Picture of a Conjugated System

15-4

Allylic Cations 669

15-5

1,2- and lA-Addition to Conjugated Dienes 670

15-6

665

Kinetic versus Thermodynamic Control in the Addition of HBr to 1,3-Butadiene 672

15-7

Allylic Radicals 674

15-8

Molecular_ Orbitals of the Allylic System 676

15-9

Electronic Configurations of the Allyl Radical, Cation, and Anion 678

15-10

SN2 Displacement Reactions of Allylic Halides and Tosylates 679

xii i

xiv

Contents 15-11

The Diels-Alder Reaction 680

15-12

The Diels-Alder as an Example of a Pericyclic Reaction 689

Key Mechanism: The Diels-Alder Reaction 680

15-13

Ultraviolet Absorption Spectroscopy 692 Chapter 15 Glossary 699 Study Problems 701

16

Aromatic Compounds 705 16-1

Introduction: The Discovery of Benzene 705

16-2

The Structure and Properties of Benzene 705

16-3

The Molecular Orbitals of Benzene 709

16-4

The Molecular Orbital Picture of Cyclobutadiene 712

16-5

Aromatic, Antiaromatic, and Nonaromatic Compounds 714

16-6

Hiickel's Rule 714

16-7

Molecular Orbital Derivation of Htickel's Rule 716

16-8

Aromatic Ions 717

16-9

Heterocyclic Aromatic Compounds 723

16-10

Polynuclear Aromatic Hydrocarbons 727

16-11

Aromatic Allotropes of Carbon 729

16-12

Fused Heterocyclic Compounds 731

16-13

Nomenclature of Benzene Derivatives 732

16-14

Physical Properties of Benzene and Its Derivatives 734

16-15

Spectroscopy of Aromatic Compounds 735 Chapter 16 Glossary 738 Study Problems 740

17

Reactions of Aromatic Compounds 749 17-1

Electrophilic Aromatic Substitution 749

17-2

Halogenation of Benzene 751

17-3

Nitration of Benzene 753

Key Mechanism: Electrophilic Aromatic Substitution 750

17-4

Sulfonation of Benzene 755

17-5

Nitration of Toluene: The Effect of Alkyl Substitution 757

17-6

Activating, Ortho, Para-Directing Substituents 759

17-7

Deactivating, Meta-Directing Substituents 763

Summary: Activating, Ortho, Para-Directors 762 Summary: Deactivating, Meta-Directors 766 17-8

Halogen Substituents: Deactivating, but Ortho, Para-Directing 766

17 -9

Effects of Multiple Substituents on Electrophilic

Summary: Directing Effects of Substituents 768 Aromatic Substitution 768 17-10 17-11

The Friedel-Crafts Alkylation 771 The Friedel-Crafts Acylation 775 Summary: Comparison of Friedel-Crafts Alkylation and Acylation 778

17-12

Nucleophilic Aromatic Substitution 780

17-13

Addition Reactions of Benzene Derivatives 785

17-14

Side-Chain Reactions of Benzene Derivatives 787

17 -15

Reactions of Phenols 791

Contents Summary: Reactions of Aromatic Compounds 794 Chapter 17 Glossary 797 Study Problems 800

18

Ketones and Aldehydes 805 18-1

Carbonyl Compounds 805

18-2

Structure of the Carbonyl Group 806

18-3

Nomenclature of Ketones and Aldehydes 806

18-4

Physical Properties of Ketones and Aldehydes 809

18-5

Spectroscopy of Ketones and Aldehydes 811

18-6

Industrial Importance of Ketones and Aldehydes 818

18-7

Review of Syntheses of Ketones and Aldehydes 818

18-8

Synthesis of Ketones and Aldehydes Using 1,3-Dithianes 822

18-9

Synthesis of Ketones from Carboxylic Acids 823

18-10 18-11

Synthesis of Ketones from Nitriles 824 Synthesis of Aldehydes and Ketones from Acid Chlorides 825 Summary: Syntheses of Ketones and Aldehydes 826

18-12

Reactions of Ketones and Aldehydes: Nucleophilic Addition 829

18-13

The Wittig Reaction 832

Key Mechanism: Nucleophilic Additions to Carbonyl Groups 831

18-14

Hydration of Ketones and Aldehydes 836

18-15

Formation of Cyanohydrins 838

18-16

Formation of Imines 840 Key Mechanism: Formation of Imines 840

18-17

Condensations with Hydroxylamine and Hydrazines 843

18-18

Formation of Acetals 845

Summary: Condensations of Amines with Ketones and Aldehydes 844 Key Mechanism: For m ation of Acetals 846 Problem-Solving Strategy: Proposing Reaction Mechanisms 848

18-19

Use of Acetals as Protecting Groups 850

18-20

Oxidation of Aldehydes 852

18-21

Reductions of Ketones and Aldehydes 853 Summary: Reactions of Ketones and Aldehydes 855 Glossary 858 Study Problems 861

19

Amines 870 19-1

Introduction 870

19-2

Nomenclature of Amines 871

19-3

Structure of Amines 873

19-4

Physical Properties of Amines 875

19-5

Basicity of Amines 877

19-6

Effects on Amine Basicity 878

19-7

Salts of Amines 880

19-8

Amine Salts as Phase-Transfer Catalysts 882

19-9

Spectroscopy of Amines 884

19-10

Reactions of Amines with Ketones and Aldehydes (Review) 888

19-11

Aromatic Substitution of Arylamines and Pyridine (Review) 888

19-12

Alkylation of Amines by Alkyl Halides 892

XV

xvi

Contents 19-13

Acylation of Amines by Acid Chlorides 893

19-14

Formation of Sulfonamides 895

19-15

Amines as Leaving Groups: The Hofmann Elimination 897

19-16

Oxidation of Amines; The Cope Elimination 900

19-17

Reactions of Amines with Nitrous Acid 902

19-18

Reactions of Arenediazonium Salts 904

19-19

Synthesis of Amines by Reductive Amination 911

19-20

Synthesis of Amines by Acylation-Reduction 913

19-21

Syntheses Limited to Primary Amines 915

Summary: Reactions of Amines 908

Summary: Synthesis of Amines 921 Chapter 19 Glossary 923 Study Problems 926

20

Carboxylic Acids 935 20-1

Introduction 935

20-2

Nomenclature of Carboxylic Acids 935

20-3

Structure and Physical Properties of Carboxylic Acids 939

20-4

Acidity of Carboxylic Acids 940

20-5

Salts of Carboxylic Acids 944

20-6

Commercial Sources of Carboxylic Acids 947

20-7

Spectroscopy of Carboxylic Acids 948

20-8

Synthesis of Carboxylic Acids 952

20-9

Reactions of Carboxylic Acids and Derivatives;

Summary: Syntheses of Carboxylic Acids 955 Nucleophilic Acyl Substitution 957 20-10

Condensation of Acids with Alcohols: The Fischer Esterification 958 Key Mechanism: Fischer Esterification 959

20-11

Esterification Using Diazomethane 962

20-12

Condensation of Acids with Amines: Direct Synthesis of Amides 963

20-13

Reduction of Carboxylic Acids 963

20-14

Alkylation of Carboxylic Acids to Form Ketones 965

20-15

Synthesis and Use of Acid Chlorides 966 Summary: Reactions of Carboxylic Acids 968 Chapter 20 Glossary 970 Study Problems 971

21

Carboxylic Acid Derivatives 978 21-1

Introduction 978

21-2

Structure and Nomenclature of Acid Derivatives 979

21-3

Physical Properties of Carboxylic Acid Derivatives 985

21-4

Spectroscopy of Carboxylic Acid Derivatives 988

21-5

Interconversion of Acid Derivatives by Nucleophilic Acyl Substitution 994 Key Mechanism: Addition-El i m i n ation Mechan ism of Nucleophilic Acyl Substitution 995

21-6

Transesterification 1003

21-7

Hydrolysis of Carboxylic Acid Derivatives 1006

21-8

Reduction of Acid Derivatives 1011

Problem-Solving Strategy: Propos i n g Reaction Mechan isms 1004

Contents 21-9

Reactions of Acid Derivatives with Organometallic Reagents 1013

21-10

Summary of the Chemistry of Acid Chlorides 1014

21-11

Summary of the Chemistry of Anhydrides 1016

21-12

Summary of the Chemistry of Esters 1019

21-13

Summary of the Chemistry of Amides 1022

21-14

Summary of the Chemistry of Nitriles 1025

21-15

Thioesters 1026

21-16

Esters and Amides of Carbonic Acid 1028 Chapter 21 Glossary 1030 Study Problems 1032

22

Condensations and Alpha Substitutions of Carbonyl Compounds 1041 22-1

Introduction 1041

22-2

Enols and Enolate Ions 1042

22-3

Alpha Halogenation of Ketones 1046

22-4

a

22-5

Alkylation of Enolate Ions 1052

22-6

Formation and Alkylation of Enamines 1053

22-7

The Aldol Condensation of Ketones and Aldehydes 1056

22-8

Dehydration of Aldol Products 1060

Bromination of Acids: The HVZ Reaction 1051

Key Mechanism: Base-Catalyzed Aldol Condensation 1056 Key Mechanism: Base-Catalyzed Dehydration of an Aldol 1060

22-9

Crossed Aldol Condensations 1061 Problem-Solving Strategy: Proposing Reaction Mechanisms 1062

22-10

Aldol Cyclizations 1064

22-11

Planning Syntheses Using Aldol Condensations 1065

22-12

The Claisen Ester Condensation 1067

22-13

The Dieckmann Condensation: A Claisen Cyclization 1070

22-14

Crossed Claisen Condensations 1071

Key Mechanism: The Claisen Ester Condensation 1067

22-15

Syntheses Using ,B-Dicarbonyl Compounds 1074

22-16

The Malonic Ester Synthesis 1076

22-17

The Acetoacetic Ester Synthesis 1079

22-18

Conjugate Additions: The Michael Reaction 1081

22-19

The Robinson Annulation 1085 Problem-Solving Strategy: Proposing Reaction Mechanisms 1086

Summary: Enolate Additions and Condensations 1088 Chapter 22 Glossary 1090 Study Problems 1092

23

Carbohydrates and Nucleic Acids 1097 23-1

Introduction 1097

23-2

Classification of Carbohydrates 1098

23-3

Monosaccharides 1099

23-4

Erythro and Threo Diastereomers 1102

23-5

Epimers 1103

23-6

Cyclic Structures of Monosaccharides 1104

xvii

xviii

Contents 23-7

Anomers of Monosaccharides; Mutarotation 1108

23-8

Reactions of Monosaccharides: Side Reactions in Base 1110

23-9

Reduction of Monosaccharides 1112

23-10

Oxidation of Monosaccharides; Reducing Sugars 1113

23-11

Nonreducing Sugars: Formation of Glycosides 1115

23-12

Ether and Ester Formation 1117

23-13

Reactions with Phenylhydrazine: Osazone Formation 1119

23-14

Chain Shortening: The Ruff Degradation 1120

23-15

Chain Lengthening: The Kiliani-Fischer Synthesis 1121

23-16

Fischer's Proof of the Configuration of Glucose 1124

23-17

Determination of Ring Size; Periodic Acid Cleavage of Sugars 1 127

23-18

Disaccharides 1129

23-19

Polysaccharides 1134

23-20

Nucleic Acids: Introduction 1137

Summary: Reactions of Sugars 1122

23-21

Ribonucleosides and Ribonucleotides 1139

23-22

The Structure of Ribonucleic Acid 1141

23-23

Deoxyribose and the Structure of Deoxyribonucleic Acid 1141

23-24

Additional Functions of Nucleotides 1145 Chapter 23 Glossary 1147 Study Problems 1149

24

Amino Acids, Peptides, and Proteins 1 1 53 24-1

Introduction 1153

24-2

Structure and Stereochemistry of the a-Amino Acids 1154

24-3

Acid-Base Properties of Amino Acids 1158

24-4

Isoelectric Points and Electrophoresis 1160

24-5

Synthesis of Amino Acids 1161

24-6

Resolution of Amino Acids 1167

24-7

Reactions of Amino Acids 1167

Summary: Syntheses of Amino Acids 1166

Summary: Reactions of Amino Acids 1170 24-8

Structure and Nomenclature of Peptides and Proteins 1170

24-9

Peptide Structure Determination 1174

24-10

Solution-Phase Peptide Synthesis 1180

24-11

Solid-Phase Peptide Synthesis 1183

24-12

Classification of Proteins 1188

24-13

Levels of Protein Structure 1189

24-14

Protein Denaturation 1191 Chapter 24 Glossary 1193 Study Problems 1196

25

lipids 1 200 25-1

Introduction 1200

25-2

VVaxes 1200

25-3

Triglycerides 1201

25-4

Saponification of Fats and Oils; Soaps and Detergents 1204

Contents 25-5

Phospholipids 1208

25-6

Steroids 1209

25-7

Prostaglandins 1212 Terpenes 1213

25-8

Chapter 25 Glossary 1217 Study Problems 1218

26

Synthetic Polymers 1 222 26-1

Introduction 1222

26-2

Addition Polymers 1223

26-3

Stereochemistry of Polymers 1229

26-4

Stereochemical Control of Polymerization; Ziegler-Natta Catalysts 1230

26-5

Natural and Synthetic Rubbers 1230

26-6

Copolymers of Two or More Monomers 1232

26-7

Condensation Polymers 1232

26-8

Polymer Structure and Properties 1236 Chapter 26 Glossary 1238 Study Problems 1240

Appendices 1 243 lA

NMR Absorption Positions of Protons in Various Structural Environments 1244

lC

Spin-Spin Coupling Constants 1246 I3 C Chemical Shifts in Organic Compounds 1247

2A

Characteristic Infrared Group Frequencies 1248

IB

2B

IR: Characteristic Infrared Absorptions of Functional Groups 1251

3

UV: The Woodward-Fieser Rules for Predicting UV-Visible

4A

Methods and Suggestions for Proposing Mechanisms 1257

4B

Suggestions for Developing Multistep Syntheses 1260

Spectra 1253

5

pKa Values for Representative Compounds 1261

Mechanism Boxes CHAPTER 6

Allylic Bromination 222 Inversion of Configuration in the SN2 Reaction 238 Racemization in the SN1 Reaction 245 Hydride Shift in an SN1 Reaction 247 Methyl Shift in an SN1 Reaction 248 Rearrangement in an E1 Reaction 255

CHAPTER 7

Dehydrohalogenation by the E2 Mechanism 298 Stereochemistry of the E2 Reaction 300 E2 Debromination of a Vicinal Dibromide 304

CHAPTER 8

Ionic Addition of HX to an Alkene 325 Free-Radical Addition of HBr to Alkenes 327

xix

XX

Contents

Acid-Catalyzed Hydration of an Alkene 331 Oxymercuration of an Alkene 333 Hydroboration of an Alkene 338 Addition of Halogens to Alkenes 343 Formation of Halohydrins 345 Epoxidation of Alkenes 354 Acid-Catalyzed Opening of Epoxides 355 CHAPTER 9

Metal-Ammonia Reduction of an Alkyne 399 Acid-Catalyzed Keto-Enol Tautomerism 403 Base-Catalyzed Keto-Tautomerism 405

CHAPTER 10

Hydride Reduction of a Carbonyl Group 446

CHAPTER 11

Reaction of a Tertiary Alcohol with HBr (SN1) 473 Reaction of a Primary Alcohol with HBr (SN2) 473 Reaction of Alcohols with PBr3 477 (Review): Acid-Catalyzed Dehydration of an Alcohol 480 The Pinacol Rearrangement 488

CHAPTER 14

Cleavage of an Ether by HBr or H I 637 Acid-Catalyzed Opening of an Epoxide in Water 646 Acid-Catalyzed Opening of an Epoxide in an Alcohol Solution 647 Base-Catalyzed Opening of Epoxides 649

CHAPTER 15

1,2- and 1,4-Addition to a Conjugated Diene 671 Free-Radical Allylic Bromination 674

CHAPTER 17

Bromination of Benzene 751 Nitration of Benzene 754 Sulfonation of Benzene 755 Friedel-Crafts Alkylation 772 Friedel-Crafts Acylation 776 Nucleophilic Aromatic Substitution (Addition-Elimination) 781 Nucleophilic Aromatic Substitution (Benzyne Mechanism) 784 The Birch Reduction 786

CHAPT ER 18

Nucleophilic Additions to Carbonyl Groups 831 The Wittig Reaction 834 Hydration of Ketones and Aldehydes 837 Formation of Cyanohydrins 838 Formation of Imines 840 Formation of Acetals 846 Wolff-Kishner Reduction 855

CHAPTER 19

Electrophilic Aromatic Substitution of Pyridine 890 Nucleophilic Aromatic Substitution of Pyridine 891 Acylation of an Amine by an Acid Chloride 893 Hofmann Elimination 897 The Cope Elimination of an Amine Oxide 901 Diazotization of an Amine 902 The Hofmann Rearrangement of Amides 919

Contents

CHAPTER 20

Nucleophilic Acyl Substitution in the Basic Hydrolysis of an Ester 957

CHAPTER 21

Addition-Elimination Mechanism of Nucleophilic Acyl Substitution 995

Esterification Using Diazomethane 962

Conversion of an Acid Chloride to an Anhydride 998 Conversion of an Acid Chloride to an Ester 998 Conversion of an Acid Chloride to an Amide 999 Conversion of an Acid Anhydride to an Ester 999 Conversion of an Acid Anhydride to an Amide 1000 Conversion of an Ester to an Amide (Ammonolysis of an Ester) 1000 Transesterification 1005 Saponification of an Ester 1007 Basic Hydrolysis of an Amide 1009 Acidic Hydrolysis of an Amide 1009 Base-Catalyzed Hydrolysis of a Nitrile 1010 Hydride Reduction of an Ester 1011 Reaction of an Ester with 2 Moles of a Grignard Reagent 1014 CHAPTER 22

Alpha Substitution 1041 Addition of an Enolate to Ketones and Aldehydes (a Condensation) 1042 Substitution of an Enolate on an Ester (a Condensation) 1042 Base-Catalyzed Keto-Enol Tautomerism 1042 Acid-Catalyzed Keto-Enol Tautomerism 1043 Base-Promoted Halogenation 1046 Fina I Steps of the Haloform Reaction 1048 Acid-Catalyzed Alpha Halogenation 1050 Acid-Catalyzed Aldol Condensation 1059 1,2-Addition and 1,4-Addition (Conjugate Addition) 1082

CHAPTER 23

Formation of a Cyclic Hemiacetal 1104 Base-Catalyzed Epimerization of Glucose 1111 Base-Catalyzed Enediol Rearrangement 1111

CHAPTER 26

Free-Radical Polymerization 1225 Cationic Polymerization 1226 Anionic Polymerization 1228

na�

Key Mechanism Boxes

CHAPTER 4

Free-Radical Halogenation 128

CHAPTER 6

The SN2 Reaction 227 The SN1 Reaction 241 The E1 Reaction 252 The E2 Reaction 259

CHAPTER 7

Acid-Catalyzed Dehydration of an Alcohol 307

CHAPTER 8

Electrophilic Addition to Alkenes 322

xxi

xxii

Contents

CHAPTER 10

Grignard Reactions 435

CHAPTER 11

The Williamson Ether Synthesis 494

CHAPTER 15

The Diels-Alder Reaction 680

CHAPTER 17

Electrophilic Aromatic Substitution 750

CHAPTER 20

Fischer Esterification 959

CHAPTER 22

Base-Catalyzed Aldol Condensation 1056 Base-Catalyzed Dehydration of an Aldol 1060 The Ciaisen Ester Condensation 1067

Answers to Selected Problems A1 Photo Credits PC1 Index 11

Preface As you begin your study of organic chemistry, you might feel overwhelmed by the number of compounds, names, reactions, and mechanisms that confront you. You might even wonder whether you can learn all this material in a single year. The most important function of a textbook is to organize the material to show that most of organic chemistry consists of a few basic principles and many extensions and applications of these principles. Relatively little memorization is required if you grasp the major concepts and develop flexibility in applying those concepts. Frankly, I have a poor memory, and I hate memorizing lists of information. I don't remember the specifics of most of the reactions and mechanisms in this book, but I can work them out by remembering a few basic principles, such as "alcohol dehydrations usually go by E 1 mechanisms." Still, you ' ll have to learn some facts and fundamental principles to serve as the work­ ing "vocabulary" of each chapter. As a student, I learned this the hard way when I made a D on my second organic chemistry exam. I thought organic would be like general chemistry, where I could memorize a couple of equations and fake my way through the exams. For example, in the ideal gas chapter, I would memorize PV nRT, and I was good to go. When I tried the same approach in organic, I got a D. We learn by making mistakes, and I learned a lot in organic chemistry. In writing this book, I've tried to point out a small number of important facts and princi­ ples that should be learned to prepare for solving problems. For example, of the hundreds of reaction mechanisms shown in this book, about 20 are the fundamental mechanistic steps that combine into the longer, more complicated mechanisms. I've highlighted these fundamental mechanisms in Key Mechanism boxes to alert you to their importance. Spectroscopy is anoth­ er area where a student might feel pressured to memorize hundreds of facts, such as NMR chemical shifts and infrared vibration frequencies. I couldn't do that, so I've always gotten by with knowing about a dozen NMR chemical shifts and about a dozen IR vibration frequencies, and knowing how they are affected by other influences. I've listed those important infrared frequencies in Table 12-2 and the important NMR chemical shifts in Table 13-3. Don't try to memorize your way through this course. It doesn't work; you have to know what's going on so you can apply the material. Also, don't think (like I did) that you can get by without memorizing anything. Read the chapter, listen carefully to the lectures, and work the problems. The problems will tell you whether or not you know the material. If you can do the problems, you should do well on the exams. If you can't do the problems, you prob­ ably won ' t be able to do the exams, either. If you keep having to look up an item to do the problems, that item is a good one to learn. Here are some hints I give my students at the beginning of the course:

To the Student

=

1.

2. 3.

4.

Read the material in the book before the lecture (expect 13-15 pages per lecture). Know­ ing what to expect and what is in the book, you can take fewer notes and spend more time listening and understanding the lecture. After the lecture, review your notes and the book, and do the in-chapter problems. Also, read the material for the next lecture. If you are confused about something, visit your instructor during office hours immediate­ ly, before you fall behind. Bring your attempted solutions to problems with you to show the instructor where you are having trouble. To study for an exam, begin by reviewing each chapter and your notes, then concentrate on the end-of-chapter problems. Also use old exams for practice, if available. Remember the two "golden rules" of organic chemistry.

1. Don't Get Behind! The course moves too fast, and it's hard to catch up. 2. Work Lots of Problems.

need more work.

Everyone needs the practice, and the problems show w here you xxiii

A Student/s Guide to Using This Text SEVERAL KINDS OF STUDY AIDS are provided to emphasize and review

the most important points. The text uses a color scheme that should help you to identify each study aid and its purpose.

Getting Oriented As you read through a chapter, the features in B LUE should help to organize the material. Some of these features are First Exposure Icons, Rules, Summary Tables, Reaction Summaries, and Glossaries. First-Exposure Icons

Hundreds of reactions appear in the text, and many types of reactions appear several times. When you study these reactions, it will help to recognize when you are seeing a new reaction for the first time. A blue pointing hand indicates the introduction of an important general type of reaction. In most cases, specific examples and variations of that type of reaction will follow the general reaction. Rules

MARKOVNIKOV' S RULE: The addition of a proton acid to the double bond of an alkene results in a product with the acid proton bonded to the carbon atom that already holds the greater number of hydrogen atoms.

I

SUMMARY

Nucleophilic Substitutions Promoting factors

strong nueleophile needed

substrate (RX)

3° > 2°

CH3X

solvent

good ionizing solvent needed

wide variety of solvents

leaving group

good one required

good one required

other

AgN03 forces ionization

Characteristics

xxiv

Summary Tables

weak nucleophiles are OK

nueleophile

/

Well-known rules, imp�rtant insights, and key definitions are often set apart and printed in/blue type. These rules, insights, and definitions are central to understanding the material in their respective chapters.

> 1° > 2°

kinetics

iirst order, k,[RX]

stereochemistry

mixture of inversion and retention

complete inversion

rearrangements

common

impossible

second order, k,[RX][Nuc:-]

Whenever a large amount of material lends itself to a concise summary, a summary table is provided to compare and contrast this material. For example, this summary table compares the factors affecting SN1 and SN2 reactions.

Reaction Summaries

At the conclusion of each section on syntheses or reactions of a functional group ("Reactions of Alkenes," for example), a summary table is provided for efficient review. Each summary, highlighted by a blue background, includes cross-references to reactions that are discussed elsewhere.

I,

SUMMARY

Reactions of Alkenes

1. Electrophilic Additions a.

" /

Addition of hydrogen halides (Section 8-3) C=C

/ "

+

I I

H

(HX

=

Hel, HEr, or HI)

I

- C-C -

H-X

I

X

Markovnikov orientation (anti·Markovnikov with HEr and peroxides)

Example CH3

I

CH3-C-CH 3 no p roxldes

�

CH3

I

CH3 - C=CH2 2·methylpropene

+

HBr

I

EI

t·b utyl bromide (Mar kovnlkov onentation)

P�

TH3

CH3- CH- CH2Br isobutyl bromide

(anti-Markovnikovorientation)

Glossaries

Each chapter ends with a glossary that defines and explains technical terms introduced in that chapter. New terms defined in the glossary are printed in boldface the first time they appear in the chapter. Don't think of the glossaries as simply dictionaries for looking up words. The Index works better for that. The real purpose of the glossaries is to act as a study aid for reviewing the material. Read through them after acid derivatives Compounds that are related to carboxylic acids but have other electron· Chapter 1 0 you read each chapter, and withdrawing groups in place of the -OH group of the acid. Three examples are acid chlo· rides, esters, and amides. (p. 439) Glossary they will help to jog your memory as you go over the o o o o definitions and make sure II II II II R-C-O-R' R-C-NHZ R-C-OH R-C-Cl you understand and can amide carboxylic acid acid chloride ester use all the new terms. alcohol A compound in which a hydrogen atol11 of a hydrocarbon has been replaced by a hydroxyl group, -OH. (p. 417) Alcohols are classitied as primary, secondary, or tertiary depending on whether the hydroxyl group is bonded to a primary, secondary, or tertiary carbon atom. (p. 4 1 7) OH

I

R-C-H

I

OH

I

R-C-R

I

OH

I

R-C-R

I

H

H

R

primary alcohol

secondary alcohol

tertiary alcohol

xxv

Understanding H ow Reactions Occu r M echanism Boxes and Key M echanism Boxes

These boxes are provided to help you find the important mechanisms easily when you review a chapter. The mechanism boxes (about 1 00 total) have large blue headings that make them easy to see as you thumb through the chapter. The format of the Mechanism Boxes should help you focus on the individual steps of each reaction and how those steps Formation of Halohydrins contribute to the overall reaction.

MECHANISM 8-8

Step 1: Electrophilic attack forms a halonium ion.

. .. C=C + :X. , X: � / .. � " CI. I) ,, (X

=

' x '" ! \

-C - C /

:X:-

\

halonium ion

Br, or

Step 2: Water opens the hal onium ion; deprotonation gives the balohydrin. :x:

I

I

I

I+

:x:

back-side aUac

t�==

I

I

I

I

+

-C-C -

-C-C-

H p+

x-

:0

KEY M ECHANISM 8- 1

Electrophilk Acldition to Alkenes

A wide variety of eleetrophilic additions involve similar mechanisms. First, a strong electrophiJe attracts the loosely held electrons from the pi bond of an alkene. The elec­ trophile forms a sigma bond to one of the carbons of the (former) double bond, while the other carbon becomes a carbocation. The carbocation (a strong electrophile) reacts with a nucleophile (often a weak nucleophile) to form another sigma bond. Step 1: Attac k of th e

pi bond on t he electrophile for ms

----'>

a carbocation.

I

/

I

"

-C-C+ E

Step 2: Attack by a nucleophile gives the addition product.

I

/� _ + Nue :

-C-C + I

E

"

----'>

I

I

-C-C I

E

I

Nue

EXAMPLE: Ionic addition of HBr to 2-butene

This example shows what happens when gaseous HBr adds to 2-butene. The proton in HBr i s e lectrophilic; it reacts with the alkene to form a carbocation. Bromide ion reacts rapidly with the carbocation to give a stable product in which the elements of HBr have added to the ends of the double bond.

Step 1: Protonation of the double bond forms a carbocation.

H H I I CH3 -C=C-CH \H . . 3 v�.r :

2° »

1°

Rearrangements are common.

Essential Problem-Solving Skills

This list is provided at the end of each chapter to remind you of the kinds of skills needed to solve typical problems associated with the material in that chapter. When you finish a chapter, this list can point out concepts you might need to review, or it might suggest types of problems and solutions you have not considered. Reviewing the problem-solving skills is often a good prelude to doing the end-of-chapter problems.

I

Essenti a l Problem-Solving Skills i n Chapter 8

1. Predict the products of additions, oxidations, reductions, and cleavages of alkenes, including

(a) (b)

orientation of reaction (regiochemistry), stereochemistry.

2. Propose logical mechanisms to explain the observed products of alkene reactions, including regiochemistry and stereochemistry.

3. Use retrosynthetic analysis to solve multistep synthesis problems with alkenes as reagents, intermediates, or products.

4. When more than one method i s usable for a chemical transformation, choose the better method and explain its advantages .

5. Use clues provided b y products of reactions such a s ozonolysis t o determine the struc­ ture of an unknown alkene. In studying these reaction-intensive chapters, students ask whether they should "memorize" all the reactions . Doing organic chemistry is Eke speaking a foreign language, and the reactions are our vocabulary. Without knowing the words, how can you construct sentences? Making flash cards often helps. In organic chemistry, the mechanisms, regiochemistry, and stereochemistry are our

grammar. You must develop facility with the reactions, as you develop facility with the

words and grammar you use in speaking. Problems and multistep syntheses are the sen­ tences of organic chemistry. You must practice combining all aspects of your vocabulary in solving these problems.

Students who fail exams often do so because they have memorized the vocabulary, but they have not practiced doing problems. Others fail because they think they can do problems, but they lack the vocabulary. If you understand the reactions and can do the end-of-chapter problems without looking back, you should do well on your exams.

xxix

Lea rning to Use Organic Chemistry by Working Problems Reading through a chapter can be deceptively easy. Everything might make sense, yet you may not be ready to apply the principles to new and different cases. Solving problems gives you an opportunity to practice using the material, which is how you really learn it anyway. Problems also tell you whether or not you actually understand what you are doing. In most cases, if you can do well on the problems, you should do well on the exams. Problems

The in-chapter problems appear right after the relevant sections of the text. These problems provide immediate review and reinforcement of the material as you learn it, helping to make sure you understand each section well enough before moving on to the next. Later, end-of-chapter problems promote additional review and practice. Your instructor may choose to Chromic acid oxidation of an alcohol (Section 1 1 -2A) occurs i n two steps: formation of the chromate ester, followed by assign specific problems that an elimination of H + and chromium. Which step do you expect to be rate-limiting? Careful kinetic studies have shown reflect the emphasis of the that Compound A undergoes chromic acid oxidation over 10 times as fast as Compound B. Explain this large di fference lectures. Problems with red in rates. stars (*) are more difficult problems that require extra ,H2Cr04 H2S04 , thought and perhaps some (slower) "O (faster) extension of the material H o H OH H Compound 8 Compound A presented in the chapter.

* 1 1 -61

�

�

�

S O LV E D P R O B L E M 7 - 2 Which o f the following alkenes are stable?

(a)

8

9

,

'0

5

,

cis

, ,,,,, , \. .,- { � Lrans

(b)

� '

(d)

6

,

5

Solved Problems

,

£ � . 6

lrans

jJ.� ,

� 8 �'

tra n s

SO LUTI O N Compound (a) is stable. Although the double bond is at a bridgehead, it is not a bridged bicyclic system. The trans double bond is in a ten-membered ring. Compound (b) is a Bredt's rule violation and is not stable. The largest ring contains six carbon atoms, and the trans double bond cannot be stable in this bridgehead position. Compound (c) (norbomene) is stable. The (cis) double bond is not at a bridgehead carbon. Compound (d) is stable. Although the double bond is at the bridgehead of a bridged bicyclic system, there is an eight-membered ring to accommodate the trans double bond.

PROBLEM 7- 1 1 Explain why each of the following alkenes is stable or unstable. (a) 1 ,2-dimethylcyclobutene (b) tralls- I ,2-dimethylcyciobutene (c) tralls-3,4-dimethylcyclobutene (el) tralls- I ,2-dimethylcyclodecene

xxx

Where appropriate, solved problems (highlighted by a beige background) are provided to show how you might approach a particular type of problem and what kind of answer is expected. For example, a solved problem might work through a mechanism to show how it is broken down into individual steps and how red curved arrows show movement of electrons.

Follow Up Problems

Solved problems are often followed by another problem to give students an immediate opportunity to practice the principles covered in the solved problems.

Preface I am always interested to hear from students using this book. If you have any suggestions about how the book might be made better, or if you've found an error, please let me know. (L. G. Wade, Whitman College, Walla Walla, WA, 99362: E-mail [email protected] ). I take students' suggestions seriously, and hundreds of them now appear in this book. For example, Whitman student Brian Lian suggested Figure 2 1 -9, and University of Minnesota student (and race-car driver) Jim Coleman gave me the facts on the use of methanol at Indianapolis. Good luck with your study of organic chemistry. I ' m certain you will enjoy this course, especially if you let yourself relax and develop an interest in how organ­ ic compounds influence our lives. My goal in writing this book has been to make the process a little easier: to build the concepts logically on top of each other, so they flow naturally from one to the next. The hints and suggestions for problem-solving have helped my students in the past, and I hope some of them will help you to learn and use the material. Even if your memory is worse than mine (highly unlikely), you should be able to do well in organic chemistry. I hope this will be a good learning experience for all of us.

Student Resou rces Solutions M a n u a l (0- 1 3- 1 47882-6) The Solutions Manual, prepared by Jan W. Simek of California Polytechnic S tate University, contains complete solutions to all the problems. The Solutions Manual also gives helpful hints on how to approach each kind of problem. This supplement is a useful aid for any student, and it i s particularly valuable for students w h o feel they understand the material but need more help with problem solving. Appendix 1 of the Solutions Manual summarizes the IUPAC system of nomenclature. Appendix 2 reviews and demonstrates how acidity varies with structure in organic molecules, and how one can predict the direction of an acid-base equilibrium. Brief answers to many of the in-chapter prob­ lems are given at the back of this book. These answers are sufficient for a student on the right track, but they are of limited use to one who is having difficulty working the problems.

Molecular Model Kits

Every organic chemistry student needs a set of molecular models. These models are used to demonstrate a multitude of principles, including stereochemistry, ring strain, conformations of cyclic and acyclic systems, isomerism, and many others. Model kits allow students to construct their own molecules and see the three­ dimensional aspects of organic chemistry that can only be imagined on a two­ dimensional drawing. Prentice Hall Molecular Model Kit (ISBN: 0-205-08136-3) This model kit allows students to build space-filling and ball-and-stick models of organic molecules. This durable kit is often recommended for students who are likely to use the models for a full-year course, and especially for those who intend to go on in chemistry or biochemistry. Brumlik Framework Molecular Model Kit (ISBN: 0-13-330076-5) Models constructed with this kit allow students to see the relationship between atoms in organ­ ic molecules, including precise interatomic distances and bond angles. The flexible bonds can form strained systems, with the amount of bend in the bonds giving a qual­ itative idea of the amount of strain. Brumlik Universal Molecular Model Kit (ISBN: 0-13-931700-7) This scien­ tifically accurate molecular model set demonstrates the framework of a molecule, the space-filling capacity of a molecule, and molecular orbitals. This kit features color­ coded atomic valence spheres and connectors. Its parts are fully interchangeable with the Brumlik Framework Molecular Model Kit.

xxxi

xxxii

Preface Com puter Software ChemOffice Software (ISBN: 0-1 3-032956-8) Issued on CD, this software package includes student versions of Cambridge Software's popular molecular modeling software, ChemDraw and Chem3D. The package is functionally similar to ChemOffice Software. A workbook featuring chapter-specific exercises written especially for Organic Chemistry 6/e is available for download on the Companion Web site. Molecu lar Modeling Workbook plus CD Featuring SpartanView and Spartan­ Build software, this workbook includes a software tutorial and numerous challenging exercises students can tackle to solve problems involving structure building and analy­ sis, using the tools included in the two pieces of Spartan software. Available free when packaged with the text; please ask your Prentice Hall representative for details, or send e-mail to [email protected].

Internet Resou rces Wade Companion Web site http://chem.prenhall .com/wade

For students who want additional practice, the Companion Web site provides a useful resource. Students can test their knowledge of the chapter with a Self Quiz in multiple-choice format. Hints for each question are provided, and feedback is immediately available when students submit the quiz for online grading. The Molecule Gallery features hundreds of 3D molecular models that the student can explore by rotating, changing the orientation and color of, and magnifying the image. Other features of the Web site include Current Topics modules that highlight recent research and developments in topics related to each chapter, and for students who are using ChemOffice Ltd., the Companion Web site also features ChemOffice Activities modules, where the student can practice structure drawing.

To the I n structor

In writing the first edition of this text, my goal was to produce a modern, readable text that uses the most effective techniques of presentation and review. Subsequent editions extended and refined that goal, with substantial rewriting and reorganization and with the addition of several new features. This sixth edition incorporates even more refine­ ments than the fifth, with revisions in the organization, writing, and graphics. Some of the modifications made in the most recent editions are: About 1 00 of the most important mechanisms have been organized into mechanism boxes, with large blue headings for easy review. In this sixth edition, these boxes have been refined to make the individual steps clearer to students. I've tried to choose most of the standard mechanisms that nearly everyone teaches; yet, in some cases, it seems that other mechanisms would be good candidates. If there are additional mechanisms that should be boxed, or some that should not be boxed, please let me know what you think. In choosing the Key Mechanisms, I've used two major criteria. If the mecha­ nism is one of the fundamental mechanisms that make up the longer, more complex mechanisms, then it must be a Key Mechanism. Examples are S N I , SN2, E l , E2, nucleophilic acyl substitution, electrophilic aromatic substitution, and so forth. The other criterion is more subjective. If the mechanism is one of the ones I routinely ex­ pect students to do on exams, then it is a Key Mechanism. Examples are formation of imines and acetals, aldol and Claisen condensations, and so on. If you feel I have left one out or included one that should not be a Key Mechanism, please let me know. Updated Coverage. In this sixth edition, I've updated several terms to those that have gradually received acceptance among organic chemists. Examples are bond­ dissociation enthalpy to replace the more ambiguous bond-dissociation energy and the newer transliteration Zaitsev to replace the older SaytzeJf. I've continued

1. Mechanism Boxes.

2.

Preface the gradual transition to the newer IUPAC names with the revised locations of numbers, such as in hexa- l ,3-diene instead of 1 ,3-hexadiene. I've also begun a transition from kcal to kJ as the primary energy units, since kJ units are used in all general chemistry texts at this time. I've added several sections to the fifth and sixth editions to cover new material or material of current interest. Chapter 4: A section on free-radical inhibitors was added to show students how some of the common inhibitors break the free-radical chain reaction, and their importance in chemistry and biochemistry. Chapter 5: Using the Mislow and Siegel definition (1. Am. Chem. Soc. 1984, 106, 3 3 1 9), I introduce the popular (but often incorrectly defined) term stereocenter and explain the differences between this term and the IUPAC telms chirality center and asymmetric carbon atom (or chiral carbon atom). The term stereocenter is much broader than the more precise term asymmetric carbon atom, and it assumes that one already knows the stereochemical properties of the molecule (to know which bonds will give rise to stereoisomers upon their interchange). Therefore, I have continued to encourage students to identify the (immediately apparent) asymmetric carbon atoms to use as tools in examining a molecule to determine its stereochemistry. Chapter 8: The Nobel Prize-winning asymmetric reduction work by NOyOIi and Knowles is discussed, together with its implications for enantioselective drug synthesis. Chapter 14: The Nobel Prize-winning Sharpless asymmetric epoxidation is discussed, together with the factors that selectively enhance the formation of one enantiomer of the product. Chapter 12: Fourier-transform IR spectroscopy is discussed, together with the reasons why this technique gives improved sensitivity and resolution over the dispersive method. Chapter 13: The NMR spectra have been convelted to high-field (300 MHz) spectra from the excellent Aldrich collection. The expansion boxes have been re­ fined and clarified to make sure that the individual splittings are visible. Chapter 16: A section has been added that discusses the aromaticity of Fullerenes and their relationship to other allotropes of carbon. Chapter 24: A section has been added that discusses prions: proteins which are thought to be infectious because of misfolding, resulting in clumping and formation of plaques. This topic relates the topic at hand (protein conformations) directly to the ongoing concern about mad cow disease. 3. Electrostatic Potential Maps. Electrostatic potential maps are used in cases where they might help students to visualize the charge distribution of a species in a way that helps to explain the electrophilic or nucleophilic nature of a compound. In introducing EPMs, I've emphasized their qualitative nature without stressing their mathematical derivation. As a result, I've explained and used EPMs much like they are introduced in the general chemistry textbooks. Several new EPMs have been added in the sixth edition. The entire book has been edited, with many sections reorganized and rewritten to enhance clarity. As in the first edition, each new topic is introduced carefulJy and explained thoroughly. Many introductory sections have been rewritten to update them and make them more approachable for students. Whenever possible, illustrations have been added or mod­ ified to help students visualize the physical concepts. The emphasis continues to be on chemical reactivity. Chemical reactions are introduced as soon as possible, and each functional group is considered in view of its reactivity toward electrophiles, nucleophiles, oxidants, reductants, and other reagents. "Electron-pushing" mechanisms are stressed throughout as a means of explaining and

xxxi ii

xxxiv

Preface predicting this reactivity. Structural concepts such as stereochemistry and spectroscopy are thoroughly treated as useful techniques that enhance the fundamental study of chemical reactivity.

O rg a n izati on

This book maintains the traditional organization that concentrates on one functional group at a time while comparing and contrasting the reactivity of different functional groups. Reactions are emphasized, beginning with Lewis acid-base reactions in Chapter 1 , continuing with thermodynamics and kinetics in Chapter 4, and covering most of the important substitution, addition, and elimination reactions in the three chapters following stereochemistry. Spectroscopic techniques (IR, MS, and NMR) are covered in Chapters 1 2 and 13, so that they can be included in the first semester. This early coverage is needed to allow effective use of spectroscopy in the laboratory. Still, a large amount of organic chemistry has been covered before this digression into structure determination. The principles of spectroscopy are practiced and reinforced in later chapters, where the characteristic spectral features of each functional group are summarized and reinforced by practice problems.

Key Features

F lexi b i l ity of Coverage

No two instructors teach organic chemistry exactly the same way. This book covers all the fundamental topics in detail, building each new concept on those that come before. Many topics may be given more or less emphasis at the discretion of the instructor. Examples of these topics are 1 3 C NMR spectroscopy, ultraviolet spectroscopy, conser­ vation of orbital symmetry, amino acids and proteins, nucleic acids, and the special top­ ics chapters, lipids and synthetic polymers. Another area of flexibility is in the problems. The wide-ranging problem sets review the material from several viewpoints, and more study problems are provided than most students are able to complete. This large variety allows the instructor to select the most appropriate problems for the individual course. U p-to-Date Treatment

In addition to the classical reactions, this book covers many techniques and reactions that have more recently gained wide use among practicing chemists. Molecular-orbital theory is introduced early and used to explain electronic effects in conjugated and aro­ matic systems, pericyclic reactions, and ultraviolet spectroscopy. Carbon- 1 3 NMR spec­ troscopy is treated as the routine tool it has become in most research laboratories, and the DEPT technique is introduced in this edition. Many of the newer synthetic tech­ niques are also included, such as asymmetric hydrogenation and epoxidation, use of sodium triacetoxyborohydride, Birch reduction, Swern oxidations, alkylation of 1 ,3-dithianes, and oxidations using pyridinium chlorochromate. Reactio n Mechanisms

Reaction mechan isms are important in all areas of organic chemistry, but they are dif­ ficult for many students. Students fall into the trap of memorizing a mechanism while not understanding why it proceeds as it does. This book stresses the principles used to predict mechanisms. Problem-solving sections develop basic techniques for approach­ ing mechanism problems, and they work to minimize rote memorization. These tech­ niques emphasize deciding whether the reaction is acidic, basic, or free radical in nature, then breaking it down into Lewis acid-base interactions and using "electron pushing arrows" to illustrate these individual steps. Important mechanisms are highlighted by placing them in the Mechanism and Key Mechanism boxes.

Preface Introduction to M echanisms U sing Free-Radical H alogenation

The advantages and disadvantages of using free-radical halogenation to introduce reaction mechanisms have been debated for many years. The principal objection to free-radical halogenation is that it is not a useful synthetic reaction. But useful reactions such as nucleophilic substitution and additions to alkenes are complicat­ ed by participation of the solvent and other effects . Gas-phase free-radical halogenation allows a clearer treatment of kinetics and thermodynamics, as long as its disadvantages as a synthetic reaction are discussed and students are aware of the limitations.

Organic Synthesis

Organic synthesis is stressed throughout this book, with progressive discussions of the process involved in developing a synthesis. Retrosynthetic analysis is emphasized, and the student learns to work backward from the target compound and forward from the starting materials to find a common intermediate. Typical yields have been provided for many synthetic reactions, although I hope students will not misuse these numbers. Too often students consider the yield of a reac­ tion to be a fixed characteristic just as the melting point of a compound is fixed. In prac­ tice, many factors affect product yields, and literature values for apparently similar reactions often differ by a factor of 2 or more. The yields given in this book are typical yields that a good student with excellent technique might obtain.

Spectroscopy

Spectroscopy is one of the most important tools of the organic chemist. This book devel­ ops the theory for each type of spectroscopy and then discusses the characteristic spec­ tral features. The most useful and dependable characteristics are summarized into a small number of rules of thumb that allow the student to interpret most spectra without looking up or memorizing large tables of data. For reference use, extensive tables of NMR and IR data and a more complete version of the Woodward-Fieser rules for UV are provided as appendices. This approach is particularly effective with IR and NMR spectroscopy, and with mass spectrometry. Practical rules are given to help students see what information is available in the spectrum and what spectral characteristics usually correspond to what structural features. Sample problems show how the informa­ tion from various spectra is combined to propose a structure. The emphasis is on helping students develop an intuitive feel for using spectroscopy to solve structural problems.

Nomenclatu re

IUPAC nomenclature is stressed throughout the book, but common nomenclature is also discussed and used to develop students' familiarity. Teaching only the IUPAC nomen­ clature might be justifiable in theory, but such an approach would handicap students in their further study and use of the literature. Much of the literature of chemistry, biology, and medicine uses common names such as methyl ethyl ketone, isovaleric acid, methyl t-butyl ether, y-aminobutyric acid, and s-caprolactam. This book emphasizes why sys­ tematic nomenclature is often preferred, yet it encourages familiarity with common names as well. I've enjoyed working on this new edition, and I hope that it is an improved fine-tuning of the fifth edition. I've tried to make this book as error-free as possible, but I ' m sure some errors have slipped by. If you find errors, or have suggestions

xxxv

xxxvi

Preface about how the book might be made better, please let me know (L. G. Wade, Whitman College, Walla Walla, WA, 99362; E-mail wadelg@ whitman.edu). Errors can be fixed quickly in the next printing. I ' ve already started a file of possible changes and improvements for the seventh edition, and I hope many of the current users will contribute suggestions to this file. I hope this book makes your job easier and helps more of your students to succeed. That's the most important reason why I wrote i t. Instructor Resou rces I n stru ctor's Resou rce on CD/ DVD (0- 1 3-1 47875-3) Prepared by Rizalia Klausmeyer of Baylor University and Jo B lackburn of Richland College. This lec­ ture resource provides a fully searchable and integrated collection of resources to help you make efficient and effective use of your lecture preparation time, as well as to enhance your classroom presentations and assessment efforts. This resource features almost all the art from the text, including tables; two pre-built Power­ Point™ presentations ; PDF files of the art, animations, movies, molecules, and the Instructor's Resource Manual Word™ files. This CD also features a search­ engine tool that enables you to find relevant resources via a number of different parameters, such as key terms, learning objectives, figure numbers, and resource type (e.g., Media Activities) . This CD/DVD set also contains the TestGen, a computer­ ized version of the Test Item File that enables professors to create and tailor exams to their needs. One Key Cou rse Management OneKey offers the best teaching and learning resources all in one place. OneKey for Organic Chemistry, Sixth Edition is all your stu­ dents need for anytime anywhere access to your course materials. OneKey is all you need to plan and administer your course. Conveniently organized by textbook chapter, these compiled resources help you save time and help your students reinforce and apply what they have learned in class. Available resources include Concept Review with Key Terms, Research Navigator, and Self-Quiz. Assessment content includes Quiz, Master Quiz, MCAT Study Guide, and the Test Item File. Resources from the Instructor's Resource Center on CDIDVD are also included. All the content in OneKey is also available in WebCT and Blackboard. Prentice Hall offers content cartridges for these text-specific Course Management Systems. Visit www.prenhall.comldemo for details. These courses also offer just the Test Item File separately. Organic Chemistry On-Line Homework System Prentice Hall, in conjunction with Robert B. Grossman and Raphael A. Finkel and a team of programmers at the Uni­ versity of Kentucky, has developed a homework system for organic chemistry that can finally support the types of problems assigned in organic chemistry. The homework system contains hundreds of organic chemistry structure-drawing problems, allows stu­ dents to draw structures, recognizes correct and incorrect answers, and provides struc­ ture-specific feedback. This drawing activity with immediate feedback results in a better learning experience for students. The homework system also grades student responses automatically and stores their activity in a gradebook, providing you a better way to track student learning while saving you time. Please contact your local Prentice Hall repre­ sentative for more information. Transparency Set (0- 1 3-1 47878-8)

This package comprises 275 four-color acetates of the most useful images, computer art, and line drawings from the text. The Trans­ parency Pack is available at no charge to adopters of Organic Chemistry, Sixth Edition.

Test Item File (0- 1 3-1 47876-1 ) by Gary Hollis, Roanoke College. This is a print­ ed version of all questions in the TestGen software found on the IReD. The new Sixth Edition contains over 2 1 00 questions.

Preface I am pleased to thank the many talented people who helped with this revision. More than anyone else, Jan Simek, author of the Solutions Manual, has consistently provided me with excellent advice and sound j udgment through several editions of this book. In this edition, Jan co-authored the section on DEPT spectroscopy, most of the new problems, and all of the Answers to Selected Problems. Particular thanks are also due to Carol Pritchard-Martinez and Ray Mullaney, who both made thousands of useful suggestions throughout the writing and revision process, and who helped to shape this new edition. I would like to thank the reviewers for their valuable insight and commentary. Although I did not adopt all their suggestions, most of them were helpful and con­ tributed to the quality of the final product. Sixth Edition Prescriptive Reviewers

Bill Baker Barry Coddens Barbara Colonna Chris Gorman Geneive Henry William Jenks Przemyslaw Maslak Rabi Musah Allan Pinhas Suzanne Ruder Maria de Graca Vicente

University of South Florida Northwestern University University of Miami North Carolina State University Susquehanna University Iowa State University Pennsylvania State University University at Albany University of Cincinnati Virginia Commonwealth University Louisiana State University

Sixth Edition Manuscript Reviewers

David Alonso Dan Becker John Berger Bob Bly Mary Boyd Hindy Bronstein Philip Brown Christine Brzezowski Patrick Buick David Cantillo Dee Ann Casteel Amber Charlebois Cai Chengzhi Jamie Lee Cohen Richard Conley Robert Crow William Donaldson Aouad Emmanuel Malcolm Forbes Anne Gaquere Rainer Glaser Steven Graham Fathi Halaweish Julius Harp Christine Hermann Kathy Hess Steve Holmgren Angela King Vera Kolb Paul Kropp Scott Lewis

Andrews University Loyola University Montclair State University University of South Carolina Loyola University, Chicago Fordham College at Lincoln Center North Carolina State University University of Alberta Florida Atlantic University Hillsborough Community College B ucknell University William Paterson University University of Houston Pace University Middlesex County College St. Louis College of Pharmacy Marquette University Rutgers University, Newark Campus University of North Carolina, Chapel Hill State University of West Georgia University of Missouri, Columbia St. John's University, Jamaica South Dakota State University North Carolina A&T University Radford University Cypress College Montana State University Wake Forest University University of Wisconsin, Parkside University of North Carolina, Chapel Hill James Madison University

xxxvii

Acknowledg ments

xxxv i i i

Preface Guigen Li Helena Malinakova Mark Mascal John Masnovi John McBride Martin McClinton James R. McKee Gary Miracle Gholam Mirafzal Tom Mitzel David Modarelli Andrew Morehead Richard Morrison Thomas N alIi Michael Nee Mark Niemczyk Glenn Nomura Patrick O'Connor Cyril Parkanyi Anthony Pearson John Penn James Poole Owen Priest John Rainier Kirk Schanze David Shultz Joseph Sloop Luise Strange John Struss Joseph Tufariello Kent Voelkner Dan Von Riesen Sheild Wallace Lisa Whalen

Texas Tech University University of Kansas University of California, Davis Cleveland State University Northwest Vista College Brevard Community College University of the Sciences in Philadelphia Texas Tech University Drake University Trinity College University of Akron East Carolina University University of Georgia Winona State University University of California, Berkeley Wheaton College Georgia Perimeter College Rutgers University Florida Atlantic University Case Western Reserve University West Virginia University Ball State University Northwestern University University of Utah University of Florida North Carolina State University United States Military Academy Georgia Perimeter College University of Tampa University at Buffalo Lake Superior State College Roger Williams University Albuquerque Technical Vocational Institute University of New Mexico

Sixth Edition Accuracy Reviewers

Thomas Nalli Susan ScheIble

Winona State University University of Colorado at Denver

Fifth Edition Prescriptive Reviewers

Simon Bott Weldon Burnham Robert S. Coleman Sergio 1. Cortes Steven Todd Deal Gary Earl K. R. Fountain Christopher M. Hadad Christopher Allen Hansen Eric Miller Nicholas R. Natale Tony E. Nicolas John G. O'Brien Michael W. Pelter William R. Pool Erik E. Simanek Donald G. Slavin

University of Houston Richland College Ohio State University University of Texas at Dallas Georgia Southern University Augustana College Truman State University Ohio State University Washington State University San Juan College University of Idaho New York City Technical College Truman State University Purdue University Calumet Westminster College Texas A&M University Community College of Philadelphia

Preface Greg Slough Kenneth W. Stagliano Maria Vogt

Kalamazoo College Illinois Institute of Technology Bloomfield College

Fifth Edition Manuscript Reviewers

Neil T. Allison Jeff Charonnat William Dailey John DiCesare John Farrar John F. Helling Thomas G. Jackson Madeline Jouille Stephen Kass Eugene A. Kline William Mancini Jerrold Meinwold Nicholas R. Natale Snorri T. Sigurdsson Maria Vogt Stephen A. Woski Catherine Woytowicz

University of Arkansas California State University, Northridge University of Pennsylvania University of Tulsa University of St. Francis University of Florida University of Alabama University of Pennsylvania University of Minnesota Tennessee Technological University Paradise Valley Community College Cornell University University of Idaho University of Washington B loomfield College University of Alabama George Washington University

Fifth Edition Accuracy Reviewers

Donnie Byers Jeff Charonnat Sergio J. Cortes Giovanna Ghirlanda Christopher M. Hadad Catherine Woytowicz

Johnson County Community College California State University, Northridge University of Texas at Dallas University of Pennsylvania Ohio State University George Washington University

Fifth Edition Application Notes Contributor

Christian P. Whitman

The University of Texas at Austin

Fourth Edition Reviewers

Mohammed Ali Paul T. Buonora Jeff Charonnat Arnold Craig Rhoda E. R. Craig William Dailey S. Todd Deal Roger D. Frampton Catherine Franklin Philip Hampton Catherine Hagen Howard Norman R. Hunter T. G. Jackson Francis M. Klein Eugene A. Kline N. Dale Ledford Clifford C. Leznoff R. Daniel Libby James W. Long James G. Macmillan Donald S. Matteson Robert McClelland James C. McKenna

Southeast Missouri State University University of Scranton California State University, Northridge Montana State University Kalamazoo College University of Pennsylvania Georgia Southern University Tidewater Community College SUNY, University at Albany University of New Mexico Texarkana College University of Manitoba University of South Alabama Creighton University Tennessee Technological University University of South Alabama York University Moravian College University of Oregon University of Northern Iowa Washington State University University of Toronto Oklahoma City Community College

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Preface

Gary W. Morrow Richard Narske Nicholas R. Natale Bjorn Olesen Michael Pena Bryan W. Roberts Joseph M. Ross Melvin L. Rueppel William N. Setzer Warren V. Sherman Tami Spector Kenneth W. Stagliano David H. Thompson Maria Vogt

University of Dayton Augustana College University of Idaho Southeast Missouri State University Arizona State University University of Pennsylvania Rueppel Consulting University of Alabama in Huntsville Chicago State University University of San Francisco Illinois Institute of Technology Purdue University Bloomfield College

Finally, I want to thank the people at Prentice Hall, whose dedication and flexibility contributed to the completion of this project. As Executive Editor, Nicole Folchetti kept the project moving, ensured the needed resources were available, and made many useful comments and suggestions. Production editor Donna King kept the production process organized, on track, and on schedule. It has been a pleasure working with aU these thoroughly professional and competent people. L. G. Wade, Jr. Walla Walla, Washington

About the Author L. G. "Skip" Wade decided to become a chemistry major during his sophomore year at Rice University, while taking organic chemistry from Professor Ronald M. Magid. After receiving his B.A. from Rice in 1 969, Wade went on to Harvard University, where he did research with Professor James D. White. While at Harvard, he served as the Head Teaching Fellow for the organic laboratories and was strongly influenced by the teach­ ing methods of two master educators, Professors Leonard K. Nash and Frank H. Westheimer. After completing his Ph.D. at Harvard in 1 974, Dr. Wade joined the chemistry faculty at Colorado State University. Over the course of fifteen years at Colorado State, Dr. Wade taught organic chemistry to thousands of students working toward careers in all areas of biology, chemistry, human medicine, veterinary medicine, and environ­ mental studies. He also authored research papers in organic synthesis and in chemical education, as well as eleven books reviewing current research in organic synthesis. Since 1 989, Dr. Wade has been a chemistry professor at Whitman College, where he teaches organic chemistry and pursues research interests in organic synthesis and foren­ sic chemistry. Dr. Wade received the A. E. Lange Award for Distinguished Science Teaching at Whitman in 1 993. Dr. Wade's interest in forensic science has led him to testify as an expert witness in court cases involving drugs and firearms, and he has worked as a police firearms instructor, drug consultant, and boating safety officer. He also enjoys repairing and restoring old violins and bows, which he has done professionally for many years.

xli

1 Introd uction and Revievv

The modern definition of organic chemistry is the chemistry of carbon compounds. What is so special about carbon that a whole branch of chemistry is devoted to its compounds? Unlike most other elements, carbon forms strong bonds to other car­ bon atoms and to a wide variety of other elements. Chains and rings of carbon atoms can be built up to form an endless variety of molecules. It is this diversity of carbon compounds that provides the basis for life on Earth. Living creatures are composed largely of complex organic compounds that serve structural, chemical, or genetic functions. The term organic literally means "derived from living organisms." Originally, the science of organic chemistry was the study of compounds extracted from living organisms and their natural products. Compounds such as sugar, urea, starch, waxes, and plant oils were considered "organic," and people accepted Vitalism, the belief that natural products needed a "vital force" to create them. Organic chemistry, then, was the study of compounds having the vital force. Inorganic chemistry was the study of gases, rocks, and minerals, and the compounds that could be made from them. In the nineteenth century, experiments showed that organic compounds could be synthesized from inorganic compounds. In 1 828, the German chemist Friedrich Wohler converted ammonium cyanate, made from ammonia and cyanic acid, to urea simply by heating it in the absence of oxygen.

1 -1

The Origins o f O rg a n ic C h e m i stry

o

NHt -OCN

heat

-----7

II

H2N -C - NH2

ammonium cyanate

urea

(inorganic)

(organic)

Urea had always come from living organisms and was presumed to contain the vital force, yet ammonium cyanate is inorganic and thus lacks the vital force. Some chemists claimed that a trace of vital force from Wohler's hands must have contami­ nated the reaction, but most recognized the possibility of synthesizing organic com­ pounds from inorganics. Many other syntheses were carried out, and the vital force theory was eventually discarded. Since Vitalism was disproved in the early nineteenth century, you'd think it would be extinct by now. And you'd be wrong! Vitalism lives on today in the minds of those who believe that "natural" (plant-derived) vitamins, flavor compounds, etc. are somehow different and more healthful than the identical "artificial" (synthesized) compounds.

The

Jarvik 7 artificial heart, composed

largely of synthetic organic materials.

1

2

Chapter 1:

Introduction and Review As chemists, we know that plant-derived compounds and the synthesized com­

founds are identical. Assuming they are pure, the only way to tell them apart is through

One

of

nicotine's

i ncrease

the

effects

is

concentration

to of

dopamine, a chemical i n the brain's reward

system.

Release

of

this

chemical makes smokers feel good and reinforces the need to smoke.

4C dating: Compounds synthesized from petrochemicals have a lower content of radioactive 1 4C and appear old because their 1 4C has decayed over time. Plant-derived compounds are recently synthesized from CO2 in the air. They have a higher content of radioactive 1 4c. Some large chemical suppliers provide isotope ratio analyses to show that their "naturals" have high 14C content and are plant-derived. Such a sophisticated analysis lends a high-tech flavor to this twenty-first-century form of Vitalism. Even though organic compounds do not need a vital force, they are still distin­ guished from inorganic compounds. The distinctive feature of organic compounds is that they all contain one or more carbon atoms. Still, not all carbon compounds are organic; substances such as diamond, graphite, carbon dioxide, ammonium cyanate, and sodium carbonate are derived from minerals and have typical inorganic properties. Most of the millions of carbon compounds are classified as organic, however. We ourselves are composed largely of organic molecules, and we are nourished by the organic compounds in our food. The proteins in our skin, the lipids in our cell mem­ branes, the glycogen in our livers, and the DNA in the nuclei of our cells are all organic compounds. Our bodies are also regulated and defended by complex organic compounds.

nicotine

nt

OH

CH20H

I

Q

HC

H3

N

carmine

vitamin C

H

0 COOH

O

-

-

HO

morphine

OH

OH

HO OH

0

H

Four examples of organic compounds in living organisms. Tobacco contains nicotine, an addictive alkaloid. Rose hips contain vitamin C, essential for preventing scurvy. The red dye carmine comes from cochineal insects, shown on prickly pear cactus. Opium poppies contain morphine, a pain-relieving, addictive alkaloid.

One

of

the

reasons

chemists

synthesize derivatives of complex organic compounds l i ke morp h i n e is t o discover new d r u g s that reta i n t h e good properties (potent pain­ relieving) but not the bad proper­ ties (highly addictive).

Chemists have learned to synthesize or simulate many of these complex mole­ cules. The synthetic products serve as drugs, medicines, plastics, pesticides, paints, and fibers. Many of the most important advances in medicine are actually advances in organic chemistry. New synthetic drugs are developed to combat disease, and new polymers are molded to replace failing organs. Organic chemistry has gone full circle. It began as the study of compounds derived from "organs," and now it gives us the drugs and materials we need to save or replace those organs.

1-2

Before we begin our study of organic chemistry, we must review some basic prin­ ciples. Many of these concepts of atomic and molecular structure are crucial to your understanding of the structure and bonding of organic compounds. 1 -2A

Electronic Structu re of the Atom

element's chemical properties are determined by the number of protons in the nucleus and the corresponding number of electrons around the nucleus. The electrons form bonds and determine the structure of the resulting molecules. Because they are small and light, electrons show properties of both particles and waves; in many ways, the electrons in atoms and molecules behave more like waves than like particles. Electrons that are bound to nuclei are found in orbitals. The Heisenberg uncer­ tainty principle states that we can never determine exactly where the electron is; nevertheless, we can determine the electron density, the probability of finding the electron in a particular part of the orbital. An orbital, then, is an allowed energy state for an electron, with an associated probability function that defines the distribution of electron density in space. Atomic orbitals are grouped into different "shells" at different distances from the nucleus. Each shell is identified by a principal quantum number n, with n I for the lowest-energy shell closest to the nucleus. As n increases, the shells are farther from the nucleus, higher in energy, and can hold more electrons. Most of the common elements in organic compounds are found in the first two rows of the periodic table, indicating that their electrons are found in the first two electron shells. The first shell (n 1 ) can hold two electrons, and the second shell ( n 2) can hold eight. The first electron shell contains just the Is orbital. All s orbitals are spherically symmetrical, meaning that they are nondirectional. The electron density is only a func­ tion of the distance from the nucleus. The electron density of the Is orbital is graphed in Figure 1 -2. Notice how the electron density is highest at the nucleus and falls off exponentially with increasing distance from the nucleus. The I s orbital might be imag­ ined as a cotton boll, with the cottonseed at the middle representing the nucleus. The density of the cotton is highest nearest the seed, and it becomes less dense at greater distances from this "nucleus." The second electron shell consists of the 2s and 2p orbitals. The 2s orbital is spher­ ically symmetrical like the I s orbital, but its electron density is not a simple exponential function. The 2s orbital has a smaller amount of electron density close to the nucleus. An

=

=

1 -2

Pri n c i p l es of Ato m i c Struct u re

Structure of the Atom

Atoms are made up of protons, neutrons, and electrons. Protons are positively charged and are found together with (uncharged) neutrons in the nucleus. Electrons, which have a negative charge that is equal in magnitude to the positive charge on the proton, occu­ py the space surrounding the nucleus (Figure 1-1). Protons and neutrons have similar masses, about 1 800 times the mass of an electron. Almost all the atom's mass is in the nucleus, but it is the electrons that take part in chemical bonding and reactions. Each element is distinguished by the number of protons in the nucleus (the atom­ ic number). The number of neutrons is usually similar to the number of protons, although the number of neutrons may vary. Atoms with the same number of protons but different numbers of neutrons are called isotopes. For example, the most common kind of carbon atom has six protons and six neutrons in its nucleus. Its mass number (the sum of the protons and neutrons) is 1 2, and we write its symbol as 1 2c. About 1 % of carbon atoms have seven neutrons; the mass number is 1 3 , written 1 3c. A very small fraction of carbon atoms have eight neutrons and a mass number of 14. The 1 4C iso­ tope is radioactive, with a half-life (the time it takes for half of the nuclei to decay) of 5730 years. The predictable decay of 1 4C is used to determine the age of organic materials up to about 50,000 years old. 1 -2B

Principles of Atomic Structure

=

cloud of electrons

\

nucleus (protons and neutrons)

... Figure 1 -1

Basic atomic structure. An atom has a dense, positively charged nucleus surrounded by a cloud of electrons.

3

4

Chapter 1: Introduction and Review

Is

t

electron density

-- distance from the

distance --

nucleus

� Fig u re

1 -2

nu cl eu s

Graph and diagram of the Is atomic orbital. The electron density is highest at the nucleus and drops off exponentially with increasing distance from the nucleus in any direction.

Most of the electron density is farther away, beyond a region of zero electron density called a node. Because most of the 2s electron density is farther from the nucleus than that of the Is, the 2s orbital is higher in energy. Figure 1 -3 shows a graph of the 2s orbital. In addition to the 2s orbital, the second shell also contains three 2p atomic orbitals, one oriented in each of the three spatial directions. These orbitals are called the 2px, the 2py, and the 2pz, according to their direction along the x, y, or z axis. The 2p orbitals are slightly higher in energy than the 2s, because the average location of the electron in a 2p orbital is farther from the nucleus. Each p orbital consists of two lobes, one on either side of the nucleus, with a nodal plane at the nucleus. The nodal plane is a flat (planar) region of space, including the nucleus, with zero electron density. The three 2p orbitals differ only in their spatial orientation, so they have identical energies. Orbitals with identical energies are called degenerate orbitals. Figure 1 -4 shows the shapes of the three degenerate 2p atomic orbitals.

2s

distance __

from the nucleus

� Fig u re

1 -3

Graph and diagram of the 2s atomic orbi tal. The 2s orbital has a small region of high electron density close to the nucleus, but most of the electron density is farther from the nucleus, beyond a node, or region of zero electron density.

t

electron density

-- distance from the nucleus

5

1-2 Principles of Atomic Structure -

-- ------

1 /£ , , distance from : the nucleus , , , , , ,

electron density

,

directions of axes

\

nucleus

(z comes out toward us)

nodal plane

.... F i g u re

1 -4

The 2p orbitals. There are three 2p the

2px orbital

the

2px, 2py, and 2pz

orbitals superimposed at 90° angles

orbitals, oriented at right angles to each other. Each is labeled according

to its orientation along the x, y, or z axis.

The Pauli exclusion principle tells us that each orbital can hold a maximum of 2 electrons, provided that their spins are paired. The first shell (one I s orbital) can accommodate 2 electrons. The second shell (one 2s orbital and three 2p orbitals) can accommodate 8 electrons, and the third shell (one 3s orbital, three 3p orbitals, and five 3d orbitals) can accommodate 1 8 electrons. 1 -2C

E lectronic Config u rations of Atoms

Aujbau means "building up" in German, and the aujbau principle tells us how to build up the electronic configuration of an atom's ground (most stable) state. Starting with the lowest-energy orbital, we fill the orbitals in order until we have added the proper number of electrons. Table 1 - 1 shows the ground-state electronic configurations of the elements in the first two rows of the periodic table. TABLE 1 -1

E l ectro n i c Configurations of the Elements of the Fi rst and Second Rows Element H

He Li Be B C

N

0 F

Ne

Configuration

Valence Electrons

Is' ls2 I s22s ' ls22s2 I s22s22p.� Is22s22p1-2p� Is22s22p.�2p�2p� Is22s22p;2p�2pi I s22s22p;2p;2Pi I s22s22p;2p;2p�

I 2

Relative orbital energies

2

- 2 p,

3

4

5

6 7 8

energy

-2s - Is

- 2 p ,.

- 2 P:

6

C hap te r ] : Introduction and Review

� Fig u re

1 -5

Partial periodic table

noble

First three rows of the periodic table.

gases

The organization of the periodic table

(VIII)

IA

results from the fi l l i ng of atomic

-

-

orbitals i n order of increasing energy.

H

IIA

IlIA

IVA

VA

VIA

VIlA

He

For these representative elements, the

Li

Be

B

C

N

0

F

Ne

Na

Mg

Al

Si

P

S

Cl

Ar

number of the column corresponds to the number of valence electrons.

--

Lith i u m carbonate, a salt of lithium, is a mood-stabilizing agent used to treat the psychiatric disorder known as mania. Mania is characterized by behaviors such as elated mood, feel­ ings of greatness, racing thoughts, and an inabil ity to sleep. We don't know how lith ium carbonate helps to stabilize these patients' moods.

----

Two additional concepts are illustrated in Table 1 - 1 . The valence electrons are those electrons that are in the outermost shell. Carbon has four valence elec­ trons, nitrogen has five, and oxygen has six. Helium has two valence electrons, and neon has eight, corresponding to a fil led first shell and second shell, respectively. In general (for the representative elements), the column or group number of the peri­ odic table corresponds to the number of valence electrons (Figure 1 -5). Hydrogen and lithium have one valence electron, and they are both in the first column (group IA) of the periodic table. Carbon has four valence electrons, a n d it is in group IVA of the periodic table. Notice in Table I - I that carbon's third and fourth valence electrons are not paired; they occupy separate orbitals . Although the Pauli exclusion principle says that two electrons can occupy the same orbital, the electrons repel each other, and pairing requires additional energy. Hund's rule states that when there are two or more orbitals of the same energy, electrons will go into different orbitals rather than pair up i n the same orbital. The first 2p electron (boron) goes into one 2p orbital, the second (carbon) goes into a different orbital, and the third (nitrogen) occupies the l ast 2p orbital . The fourth, fifth, and sixth 2p electrons must pair up with the first three electrons. PROBLEM

1-1

Write the electronic configurations of the third-row elements shown in the partial periodic table in Figure 1 - 5 .

1 -3

Bond Formati o n : T h e Octet R u l e

In 1 9 1 5 , G. N. Lewis proposed several new theories describing how atoms bond together to form molecules. One of these theories states that a filled shell of electrons is especially stable, and atoms transfer or share electrons in such a way as to attain a filled shell of electrons. A filled shell of electrons is simply the electron configuration of a noble gas, such as He, Ne, or Ar. This principle has come to be called the octet rule because a filled shell implies eight valence electrons for the elements in the second row of the periodic table. 1 -3A

Ion i c Bond ing

There are two ways that atoms can interact to attain noble-gas configurations. Some­ times atoms attain noble-gas configurations by transferring electrons from one atom to another. For example, lithium has one electron more than the helium configuration, and fluorine has one electron less than the neon configuration. Lithium easily loses its valence electron, and fluorine easily gains one:

1 -4 Lewis Structures ..

Li ·..---·--... F :

:F:

+

electron transfer

He configuration

Ne configuration

..

Li+ : F : ionic bond

A transfer of one electron gives each of these two elements a noble-gas configu­ ration. The resulting ions have opposite charges, and they attract each other to form an ionic bond. Ionic bonding usually results in the formation of a large crystal lattice rather than individual molecules. Ionic bonding is common in inorganic compounds but relatively uncommon in organic compounds. 1 -3 B

Covalent Bonding

Covalent bonding, in which electrons are shared rather than transferred, is the most common type of bonding in organic compounds. Hydrogen, for example, needs a sec­ ond electron to achieve the noble-gas configuration of helium. If two hydrogen atoms come together and form a bond, they "share" their two electrons, and each atom has two electrons in its valence shell.

H·

+

H :H

H'

each H shares two electrons (He configuration)

We will study covalent bonding in more detail in Chapter 2.

One way to symbolize the bonding in a covalent molecule is to use Lewis structures. In a Lewis structure, each valence electron is symbolized by a dot. A bonding pair of electrons is symbolized by a pair of dots or by a dash ( - ) . We try to arrange all the atoms so they have their appropriate noble-gas configurations: two electrons for hydrogen, and octets for the second-row elements. Consider the Lewis structure of methane ( CH4 ) ' H H :C:H H

H or

I I

H-C-H H

methane

Carbon contributes four valence electrons, and each hydrogen contributes one, to give a total of eight electrons. All eight electrons surround carbon to give it an octet, and each hydrogen atom shares two of the electrons. The Lewis structure for ethane ( C2H6 ) is more complex. H H H :C:C:H H H

or

H

H

H

H

I I H-C-C-H I I

ethane

Once again, we have computed the total number of valence electrons ( 1 4) and distrib­ uted them so that each carbon atom is surrounded by 8 and each hydrogen by 2. The only possible structure for ethane is the one shown, with the two carbon atoms sharing a pair of electrons and each hydrogen atom sharing a pair with one of the carbons. The ethane structure shows the most important characteristic of carbon-its ability to form strong carbon-carbon bonds.

1 -4

Lewi s Struct u res

7

Ch apter 1: Introd uction and Review

8

Valence-shell electrons that are not shared between two atoms are called A pair of nonbonding electrons is often called a lone pair. Oxygen atoms, nitrogen atoms, and the halogens (F, Cl, Br, I) usually have non bonding electrons in their stable compounds. These lone pairs of nonbonding electrons help to determine the reactivity of their parent compounds. The following Lewis structures show one lone pair of electrons on the nitrogen atom of methyl­ amine and two lone pairs on the oxygen atom of ethanol . Halogen atoms usually have three lone pairs, as shown in the structure of chloromethane. nonbonding electrons.

H

I

./

H

H

H-C-N-H

I

H

lone pair

I

H

lone pairs

I

.// :

H

H

H

I .. H-C -Cl.:.::::

_

H-C-C-O

I

I

H

methylamine

I

I

I

H

ethanol

..

lone pairs

chloromethane

A correct Lewis structure should show any lone pairs. Organic chemists often draw structures that omit most or all of the lone pairs. These are not true Lewis struc­ tures because you must imagine the correct number of nonbonding electrons.

PROBLEM-SOLVING

H?ftl;

Lewis structures are the way we write orga n i c chemistry. Learn ing now to draw them q u ickly and correctly will help you throughout this course.

PRO BLEM

1 -2 Y,

Draw Lewis structures for the following compounds.

(a) (c) (e) (g) (i)

ammonia,

NH3

H30+ ethylamine, CH3CH2NH2 fluoroethane, CH3CH2F borane, BH3

hydronium ion,

H20 C3Hg

(b) (d)

propane,

(h)

2-propanol,

water,

CH30CH3 CH3CH(OH)CH3 (j) boron trifluoride, BF3 (1)

dimethyl ether,

Explain what i s unusual about the bonding in compounds in parts (i) and 0).

1 -5

M u lti ple B on d i ng

In drawing Lewis structures in Section 1 -4, we placed just one pair of electrons between any two atoms. The sharing of one pair between two atoms is called a single bond. Many molecules have adjacent atoms sharing two or even three electron pairs. The sharing of two pairs is called a double bond, and the sharing of three pairs is called a triple bond. Ethylene ( C2H 4 ) is an organic compound with a double bond. When we draw a Lewis structure for ethylene, the only way to show both carbon atoms with octets is to draw them sharing two pairs of electrons. The following examples show organic compounds with double bonds. In each case, four electrons (two pairs) are shared between two atoms to give them octets. A double dash ( ) symbolizes a double bond. =

H. .H :C : : C: H' 'H

H. : C : : o": H'

H. : C : : N: H' 'H

or

or

or

Acetylene, in combination with oxy­

H ,"",,/H / C = C ,"",,H H

gen, burns with an intense flame

ethylene

that has diverse applications. It can be

used for welding

parts of a

bridge underwater and for repairing an oil pipeline in Siberia.

H H

� C = O::

formaldehyde

H '"""/ C = N''"""H H formaldimine

Acetylene (C2H2) has a triple bond. Its Lewis structure shows three pairs of electrons between the carbon atoms to give them octets. The following examples show organic compounds with triple bonds. A triple dash ( - ) symbolizes a triple bond.

1-6

H : C ::: C : H

H H H : C : C ::: C : C : H H H

H H : C : C ::: N: H

or

or

or

H H I I H - C - C== C - C - H I I H H dimethylacetylene

H I H - C - C== N: I H acetonitrile

..

H - C== C - H acetylene

..

Electronegativity and Bond Polarity

9

All these Lewis structures show that carbon normally forms four bonds in neutral or­ ganic compounds. Nitrogen generally forms three bonds, and oxygen usually forms two. Hy­ drogen and the halogens usually form only one bond. The number of bonds an atom usually forms is called its valence. Carbon is tetravalent, nitrogen is trivalent, oxygen is divalent, and hydrogen and the halogens are monovalent. By remembering the usual number of bonds for these cornmon elements, we can write organic structures more easily. If we draw a structure with each atom having its usual number of bonds, a correct Lewis structure usually results.

�

SUMMARY

Common Bonding Patterns (Uncharged)

valence: lone pairs:

I - CI

-NI

-0-

-H

- Cl:

carbon

nitrogen

oxygen

hydrogen

halogens

2

a

3

4 a

..

..

2

3

1

P R O B L E M 1-3

Write Lewis structures for the following molecular formulas. 9» HCN KG.) HONO yt) N2 ¥ H2CNH .Jta HC02H � CO2 �) HNNH ji), C3H6 00 C2H3Cl C3H4 (two double bonds) � C3H4 (one triple bond)

�

PROBLEM 1 -4

Circle any lone pairs (pairs of nonbonding electrons) in the structures you drew for Problem 1-3.

A bond with the electrons shared equally between the two atoms is called a nonpolar covalent bond. The bond in H2 and the C - C bond in ethane are nonpolar covalent bonds. In most bonds between two different elements, the bonding electrons are at­ tracted more strongly to one of the two nuclei. An unequally shared pair of bonding electrons is called a polar covalent bond. : CI: .. ICD··

-C I nonpolar covalent bond

polar covalent bond

Na+

:cC

ionic bond

When carbon is bonded to chlorine, for example, the bonding electrons are attracted more strongly to the chlorine atom. The carbon atom bears a small partial positive charge, and the chlorine atom bears an equal amount of negative charge. Figure 1-6 shows the polar carbon-chlorine bond in chloromethane. We symbolize the bond polarity by an

..

1

PROBLEM-SOLVING

HiltZ;

These "usual numbers of bonds" might be single bonds, or they m ight be combined into double and triple bonds. For example, three bonds to nitrogen m i ght be three single bonds, one sing le bond and one double bond, or one triple bond (:N=N:). In working problems, consider all possibil ities.

1 -6 E l ectroneg ativity and Bond Pola rity

10

Chapter 1: Introduction and Review

chloromethane

chloromethane ..... Figure 1 -6

Bond polarity. Chloromethane contains a polar carbon--chlorine bond with a partial negative charge on chlorine and a partial positive charge on carbon. The electrostatic potential map shows a red region (electron-rich) around the partial negative charge and a blue region (electron-poor) around the paJtial positive charge. Other colors show intermediate values of electrostatic potential.

arrow with its head at the negative end of the polar bond and a plus sign at the positive end. The bond polarity is measured by its dipole moment (fL), defined to be the amount of charge separation (8+ and 8-) multiplied by the bond length. The symbol 8+ means "a small amount of positive charge"; 8- means "a small amount of negative charge." Figure 1 -6 also shows an electrostatic potential ml:tp (EPM) for chloromethane, using color to represent the calculated charge distribution in a molecule. Red shows electron-rich regions, and blue shows electron-poor regions. Orange, yellow, and green show intermediate levels of electrostatic potential. In chloromethane, the red region shows the partial negative charge on chlorine, and the blue region shows the partial pos­ itive charges on carbon and the hydrogen atoms. We often use electronegativities as a guide in predicting whether a given bond will be polar and the direction of its dipole moment. The Pauling electronegativity scale, most com­ monly Llsed by organic chemists, is based on bonding properties, and it is useful for predict­ ing the polarity of covalent bonds. Elements with higher electTonegativities generally have more attraction for the boncling electrons. Therefore, in a bond between two cli fferent atoms, the atom with the higher electronegativity is the negative end of the dipole. Figure 1 -7 shows Pauling electronegativities for some of the important elements in organic compounds. Notice that the electronegativities increase from left to right across the periodic table. Nitrogen, oxygen, and the halogens are all more electronegative than carbon; sodium, lithium, and magnesium are less electronegative. Hydrogen's electronegativi­ ty is si milar to that of carbon, so we usually consider C - H bonds to be nonpolar. We will consider the polarity of bonds and molecules in more detail in Section 2-9. P R O B L E M 1 -5

Use electronegativities to predict the direction of the dipole moments of the following bonds. (a) C-Cl C-N �) c-s (e) C - B o(r c-o f'd.l (;f) N - C I N a (g) QJ{) N -s (:it B - Cl N B til v p

'""fI 2.2 Li

1.0

Na

0.9

� Figure 1 -7

The Pauling electronegativities of some of the elements found in organic compounds.

K

�

Be

B

0

F

1.6

1.8

2.5

3.0

3.4

4.0

1.3

l .6

1.9

2.2

2.6

3.2

Mg

AI

C

Si

N P

S

CI

Br

3.0

�

2.7 -'---

1-7 Formal Charges

In polar bonds, the partial charges (8+ and 8-) on the bonded atoms are real. Formal charges provide a method for keeping track of electrons, but they may or may not correspond to real charges. In most cases, if the Lewis structure shows that an atom has a formal charge, it actually bears at least part of that charge. The concept of formal charge helps us determine which atoms bear most of the charge in a charged molecule, and it also helps us to see charged atoms in molecules that are neutral overall . To calculate formal charges, count how many electrons contribute to the charge of each atom and compare that number with the number of valence electrons in the free, neutral atom (given by the group number in the periodic table). The electrons that contribute to an atom's charge are all its unshared (nonbonding) electrons; plus 2. half the (bonding) electrons it shares with other atoms, or one electron of each bonding pair. 1.

The formal charge of a g iven atom can be calculated by the formula formal charge ( FC )

=

[group number] - [nonbonding electrons] - 1 [sharecl electrons]

S O LV E D P R O B L E M 1 - 1

Compute the formal charge (FC) on each atom in the following structures. (a) Methane ( CH4 ) H H :C:H H S O LU T I O N

Each of the hydrogen atoms in methane has one bonding pair of electrons (two shared electrons). Half of two shared electrons is one electron, and one valence electron is what hydrogen needs to be neutral. Hydrogen atoms with one bond are formally neutral: FC 1 - 0 - 1 O. The carbon atom has four bonding pairs of electrons (eight electrons). Half of eight shared electrons is four electrons, and four electrons are what carbon (groupIVA) needs to be neutral. Carbon is formally neutral whenever it has four bonds: FC 4 - 0 - �(8) O. (b) The hydronium ion, H30+ =

=

=

=

e�two nonbonding electrons)

H

MH

�------------------,

H

three bonds, six bonding electrons

SOLUTION

In drawing the Lewis structure for this ion, we use eight electrons: six from oxygen plus three from the hydrogens, minus one because the ion has a positive charge. Each hydro­ gen has one bond and is formally neutral. Oxygen is surrounded by an octet, with six bonding electrons and two nonbonding electrons. Half the bonding electrons plus all the nonbonding electrons contribute to its charge: � + 2 5; but oxygen (group VIA) needs six valence electrons to be neutral. Consequently, the oxygen atom has a formal charge + l. of + 1: FC 6 - 2 - 1( 6) =

=

=

(c) H 3 N - BH3 Boron

has fOLlr bonds, eight bonding electrons

H H + .. .. H: N: B : H H H Nitrogen has four bonds. eight bonding electrons

1 -7 Formal Cha rges

11

12

Chapter

1:

Introduction and Review

SOLUTION This is a neutral compound where the individual atoms are formally charged. The Lewis struc­ ture shows that both nitro en and boron have four shared bonding pairs of electrons. Both boron and nitrogen have "2 = 4 electrons contributing to their charges. Nitrogen (group V) needs five valence electrons to be neutral, so it bears a formal charge of +1 . Boron (group ill) needs only three valence electrons to be neutral, so it bears a formal charge of 1

/

-

Nitrogen:

FC

=

Boron:

FC

=

5 3

-

-

-1(8) !(8) 0

0

-

=

.

+1

=-1

S O LUTI O N In this structure, both carbon and nitrogen have four shared pairs of bonding electrons. With four bonds, carbon is formally neutral; however, nitrogen is in group V, and it bears a formal positive charge: FC 0 4 +1 . This compound might also be drawn with the following Lewis structure: =

5

-

-

=

H

,,+ / C-N / "

H

..

H

H

In this structure, the carbon atom has three bonds with six bonding electrons. We calculate that 3 electrons, so carbon is one short of the four needed to be formally neutral:

g

=

!(6)

FC = 4 0 = +l . Nitrogen has six bonding electrons and two nonbonding electrons. We calculate that + 2 so the nitrogen is uncharged in this second structure: -

g

=

-

5,

FC

=

5

-

2

-1(6)

=0

The significance of these two Lewis structures is discussed in Section 1-9.

Most organic compounds contain only a few common elements, usually with complete octets of electrons. The summary table on the facing page shows the most commonly occurring bonding structures, using dashes to represent bonding pairs of electrons . Use the rules for calculating formal charges to verify the charges shown on these structures. A good understanding of the structures shown here will help you to draw organic compounds and their ions quickly and correctly.

1- 8 Ionic Structu res

Some organic compounds contain ionic bonds. For example, the structure of methylam­ monium chloride ( CH3NH3Cl) cannot be drawn using just covalent bonds. That would require nitrogen to have five bonds, implying ten electrons in its valence shell. The cor­ rect structure shows the chloride ion ionically bonded to the rest of the structure. H H I I H - C - N LH I I H H methylammonium chloride

H H H I I/ H-C-N I I "" H H 9:

too many el ectrons around nitrogen

cannot be drawn covalently

1-9

I

SUMMARY

Resonance

Common Bonding Patterns in Organic Compounds and Ions

Atom

Val ence Electrons

B

3

C

4

-C -

N

5

-N-

0

6

hal ogen

7

Pos itively C harged

Neutral

Negativel y C harged

-B-

- B=--

-C -

-C =-

(no octet) /'

+

1

1 1

1

1 .. -N1 ..

1 � (no octet) 1+

1

-OL 1

1

-N=- 0:

- 0-

..

- el L

-Cl:

-

:Cl :-

PROBLEM·SOLVING

••

les s common

more common

P R O B L E M 1 ·6 Draw Lewis structures for the following compounds and ions, showing appropriate formal charges.

(a) [CH30H2]+ (d) NaOCH3 (g) NaBH4

(c) ( CH3 hNH2Ci

(b) NH4Cl

(e) +CH3 (h) NaBH3CN (k) KOC( CH3 h

(j) [HONH3]+

(0 TH3

(i) ( CH3hO- BF3 (I) [H2C= OH]+

Reso n a nce Hybrids

Some compounds ' structures are not adequately represented by a single Lewis structure. When two or more valence-bond structures are possible, differing only in the placement of electrons, the molecule will usually show characteristics of both structures. The different structures are called resonance structures or resonance forms because they are not different compounds, just different ways of drawing the same compound. The actual molecule is said to be a resonance hybrid of its resonance forms. In Solved Problem 1 - I (d) we saw that the ion [H2CNH2t might be represented by either of the following resonance forms:

f

H H

�+ t? / C-N /

H

�

H

H H

� +/ C=N /

res onance forms of a resonance hybrid

H

�

H

J

H H

�15+

/

15'/

H

C=-=-=N

�

H

combined representation

HritZ;

This is a very i mportant table. Work enough problems to become fami liar with these bonding patterns so you can recognize other patterns as being either u n usual or wrong.

Some molecules can be drawn either covalently or ionically. For example, sodium acetate (NaOCOCH3) may be drawn with either a covalent bond or an ionic bond be­ tween sodium and oxygen. Because sodium generally forms ionic bonds with oxygen (as in NaOH), the ionic ally bonded structure is usually preferred. In general, bonds be­ tween atoms with very large electronegativity differences (about 2 or more) are usually drawn as ionic. ·0· H tj" H II I II I .. .. Na-O-C-C-H Na+ -:O C CH .. I I H H

1 -9A

13

1 -9 Reso nance

14

Chapter 1 : Introduction and Review The actual structme of this ion is a resonance hybrid of the two structures. In the actual molecule, the positive charge is delocalized (spread out) over both the carbon atom and the nitrogen atom. In the left resonance form, the positive charge is on carbon, but carbon does not have an octet. Nitrogen's nonbonding electrons can move into the bond (as indicated by the red arrow) to give the second structme with a double bond a positive charge on nitrogen and an octet on carbon. The combined representation attempts to combine the two reso­ nance forms into a single pictme with the charge shared by carbon and nitrogen. Spreading the positive charge over two atoms makes the ion more stable than it would be if the entire charge were localized only on the carbon or only on the nitrogen. We call this a resonance-stabilized cation. Resonance is most important when it allows a charge to be delocalized over two or more atoms, as in this example. Resonance stabilization plays a crucial role in organic chemistry, especially in the chemistry of compounds having double bonds. We will use the concept of reso­ nance frequently throughout this course. For example, the acidity of acetic acid (following) is enhanced by resonance effects. When acetic acid loses a proton, the resulting acetate ion has a negative charge delocalized over both of the oxygen atoms. Each oxygen atom bears half of the negative charge, and this delocalization stabilizes the ion. Each of the carbon-oxygen bonds is halfway between a single bond and a double bond, and they are said to have a bond order of 1 �. .. H 0: I / H-C-C "' . I ..O�H H

'

"

+

acetic acid

PROBLEM-SOLVING

HiltZ;

Second-row elements (8, C, N, 0, F) can not have more than eight electrons in their valence shel ls. The following is NOT a valid Lewis structure:

II

H

.. /0:

�. ··'

H-C-N H

J

H O I / ., H - C- C '\. I 0: .. H acetate ion

+

We use a single double-headed arrow between resonance forms (and often en­ close them in brackets) to indicate that the actual structure is a hybrid of the Lewis structures we have drawn. By contrast, an equilibrium is represented by two arrows in opposite directions. Some uncharged molecules actually have resonance-stabilized structures with equal positive and negative formal charges. We can draw two Lewis structures for ni­ tromethane (CH3N02) , but both of them have a formal positive charge on nitrogen and a negative charge on one of the oxygens. Thus, nitromethane has a positive charge on the nitrogen atom and a negative charge spread equally over the two oxygen atoms. The N - ° bonds are midway between single and double bonds, as indicated in the combined representation:

l

C"

H I +?/,O: H-C-N "'� I ..0..H

H 0- + I +/ H - C-N ' �' _.L I 0 ' H resonance forms

combined representation

Remember that individual resonance forms do not exist. The molecule does not "resonate" between these structures. It is a hybrid with some characteristics of both. An analogy is a mule, which is a hybrid of a horse and a donkey. The mule does not "resonate" between looking like a horse and looking like a donkey; it looks like a mule all the time, with the broad back of the horse and the long ears of the donkey. 1 -9 8

M ajor and M i nor Resonance Contri butors

Two or more correct Lewis structures for the same compound may or may not repre­ sent electron distributions of equal energy. Although separate resonance forms do not

1 -9 Resonance

exist, we can estimate their relative energies as if they did exist. More stable resonance forms are closer representations of the real molecule than less stable ones. The two resonance forms shown earlier for the acetate ion have similar bonding, and they are of identical energy. The same is true for the two resonance forms of nitromethane. The following resonance forms are bonded differently, however. H

/ ,,+ C-N / "

H

..

H major contributor

H

l

minor contributor

These structures are not equal in estimated energy. The first structure has the positive charge on nitrogen. The second has the positive charge on carbon, and the car­ bon atom does not have an octet. The first structure is more stable because it has an additional bond and all the atoms have octets. Many stable ions have a positive charge on a nitrogen atom with four bonds (see Summary Table, page 13). We call the more stable resonance form the major contributor, and the less stable form is the minor contributor. The structure of the actual compound resembles the major contributor more than it does the minor contributor. Many organic molecules have major and minor resonance contributors. Formaldehyde (H2C=0 ) can be written with a negative charge on oxygen, balanced by a positive charge on carbon. This polar resonance form is higher in estimated ener­ gy than the double-bonded structure because it has charge separation, fewer bonds, and a positively charged carbon atom without an octet. The charge-separated structure is only a minor contributor, but it helps to explain why the formaldehyde C = 0 bond is very polar, with a partial positive charge on carbon and a partial negative charge on oxygen. The electrostatic potential map (EPM) also shows an electron-rich region (red) around oxygen and an electron-poor region (blue) around carbon in formaldehyde. O IJ II C IJ+ / " H H all octets no charge separation (major contributor)

no octet on C charge separation (minor contributor)

IfL

dipole moment

In drawing resonance forms, we try to draw structures that are as low in energy as possible. The best candidates are those that have the maximum number of octets and the maximum number of bonds. Also, we look for structures with the minimum amount of charge separation. Only electrons can be delocalized. Unlike electrons, nuclei cannot be delocalized. They must remain in the same places, with the same bond distances and angles, in all the resonance contributors. The following general rules will help us to draw realistic resonance structures: 1.

All the resonance structures must be valid Lewis structures for the compound. 2. Only the placement of the electrons may be shifted from one structure to another. (Electrons in double bonds and lone pairs are the ones that are most commonly shifted.) Nuclei cannot be moved, and the bond angl es must remain the same.

EPM of formaldehyde

15

Chapter 1: Introduction and Review

16

H

H

H

I H '-... �C'-...C/ C :-:::H / H H

I H '-...+/C� / C C / "H H

I H +/ '-... �C'-...C C / "H H

H

H

H

HiltZ;

H H CI / �C/ .::C H / "H H

NOT resonance

resonance forms

PROBLEM-SOLVING

H

3.

The number of unpaired electrons (if any) must remain the same. Most stable compounds have no unpaired electrons, and all the electrons must remain paired in all the resonance structures. 4. The major resonance contributor is the one with the lowest energy. Good con ­ tributors generally have all octets satisfied, as many bonds as possible, and as lit­ tle charge separation as possible. Negative charges are more stable on more electronegative atoms, such as 0, N, and S. 5. Resonance stabilization is most i mportant when it serves to delocalize a charge over two or more atoms.

Resonance forms can be compared using the fol lowing criteria, beginning with the most i mportant: 1. As many octets as possible 2. As many bonds as possible 3. Any negative charges on electronegative atoms 4. As little charge separation as possible

S O LV E D P R O B L E M 1 - 2

For each of the following compounds, draw the important resonance forms. Indicate which structures are major and minor contributors or whether they would have the same energy. (a) [CH30CH2]+ SOLUTION

rI

I . "J

H .. / H-C-O-C+ .. " H HI

H H 1 + / H-C-O=C H H

minor contributor

major contributor

H

The first (minor) structure has a carbon atom with only six electrons around it. The second (major) structure has octets on all atoms and an additional bond. 0 (b) H "-" ,f" C-C "/ H H SOLUTION

f

H H

"-":-1

/

O:/ . "C=C / " H H

0:

H

/J

C-C

"-

H

minor contributor

.

.

l

major contributor

Both of these structures have octets on oxygen and both carbon atoms, and they have the same number of bonds. The first structure has the negative charge on carbon; the second has it on oxygen. Oxygen is the more electronegative element, so the second structure is the major contributor. (c) H2S04

r

SOLUTION

·· ·· 0 .. II .. H-O-S -O-H .. II ..

..0..

I

.. :0: .. 1 + .. H-O-S-O-H ..0..

0 .. t ·· H-O-S-O.. .. H I :0: .

�

.

-

�

J

: 0 :.. 1+2 H-O-S-O-H .. .. I :0:••

The first structure, with more bonds and less charge separation, is possible because sulfur is a third-row element with accessible d orbitals, giving it an expandable valence. For example, SF6 is a stable compound with 12 electrons around sulfur. Theoretical calculations suggest that the last structure, with octets on all atoms, may be the major resonance contributor, however. We cannot always predict the major con­ tributor of a resonance hybrid.

1 - 1 0 Structural Formulas P R O B L E M 1 -7

Draw the important resonance forms for the following molecules and ions. (a) CO�(b) NO)" (c) N02" (d) H2C = CH -cHi (e) H2C = CH - CH2" (I) SO�- (g) [CH3C(OCH3h ]+ P R O B L E M 1 -8

For each of the following compounds, draw the i mportant resonance forms. Indicate which structures are major and minor contributors or whether they have the same energy. (a) [H2CN02r (b) H2C = CH- N02 (c) [H2COH]+ + (d) H2CNN (e) [H2CCN](I) H2N - CH-CH=CH - NH2 0 o 0

I I

I

(g) H-C - CH-C - H

Condensed Structura l Formu las

Condensed structural formulas (Table 1 -2) are written without showing all the indi­ v idual bonds. In a condensed structure, each central atom is shown together w ith the atoms that are bonded to it. The atoms bonded to a central atom are often l isted after the central atom (as in CH3CH3 rather than H3C-CH3) even if that is not their actu­ al bonding order. In many cases, if there are two or more identical groups, parentheses and a subscript may be used to represent all the identical groups. Nonbonding elec­ trons are rarely shown in condensed structural formulas. TABLE 1-2 Compound

ethane

isobutane

Examples of Condensed Structural Formulas lewis Structure

H H I I H---C ---C - H I I H H H H H I I I H--c-C-C-H

h

I

h

Condensed Structural Formula

CH3CH3

(CH3)3CH

H-C-H I H

n-hexane

HinZ;

In drawing resonance forms for ions, see how you can delocalize the charge over several atoms. Try to spread a negative charge over electronegative elements like oxygen and nitrogen. Try to spread a positive charge over as many carbons as possible, but especial ly over any atoms that can bear the positive charge and still have an octet; for example, oxygen (with three bonds) or nitrogen (with four bonds).

(h) H-C-NH2

Several kinds of formulas are used by organic chemists to represent organic com­ pounds. Some of these formulas involve a shorthand notation that requires some explanation. Structural formulas actually show which atoms are bonded to which. There are two types of structural formulas, complete Lewis structures and condensed structural formulas. In addition, there are several ways of drawing condensed structur­ al formulas. As we have seen, a Lewis structure symbolizes a bonding pair of electrons as a pair of dots or as a dash ( - ) . Lone pairs of electrons are shown as pairs of dots. 1-10A

PROBLEM-SOLVING

17

H H H H H H I I I I I I H-C-C-C-C-C-C-H I I I I I I H H H H H H

CH/CH2)4CH3 ( Continued)

1 - 10 Structural Fo rmu las

18

Chapter 1: Introduction and Review TABLE 1-2

Continued Lewis Structure

Compound

H

I

H

I

H

H

H

H

H

H

.. I I H-C-C-O-C-C-H I I .. I I

diethyl ether

H

I

H

H

H

I

Condensed Structural Formula

CH3CHPCH2CH3 or CH3CH2-O-CH2CH3 or (CH3CH2)P

.. .

H-C-C-O-H .

ethanol

I

I

I

.. I

H

H

H :O-H H H-C-C

isopropyl alcohol

I

I

I

H

I

..

H

H

H

H

H

I

C-H

-

I

H-C-N-C-H

dimethylamine

I

I

I

When a condensed structural formula is written for a compound containing double or triple bonds, the multiple bonds are often drawn as they would be in a Lewis structure. Table 1-3 shows examples of condensed structural formulas containing multiple bonds. Notice that the - CHO group of an aldehyde and the - COOH group of a carboxylic acid are actually bonded differently from what the condensed notation suggests. TABLE 1-3

Condensed Structural Formulas for Double and Triple Bonds Lewis Structure

Compound

H 2-butene

I

H

Condensed Structural Formula

H

I

I

H-C-C=C-C-H

I

I

H

H

I

H

H acetonitrile

I

H-C-C=N:

I

H H acetaldehyde

I

0 II

"

'

H-C-C-H

I

H H acetone

I

II

H

I

H-C-C-C-H

I

H

I

H

"0' II .. H-C-C-O-H .. I H

acetic acid

" 0"

I

H

o II

CH3COOH or CH3C-OH orCH3C 02H

1-10 Structural Fo rmulas

As you can see from Tables 1-2 and 1 -3, the distinction between a complete Lewis structu ral formula and a condensed structural formula can be blurry. Chemists often draw formulas with some parts condensed and other parts completely draw n out. You should work with these different types of formulas so that you understand what all of them mean. P R O B L E M 1 -9 Draw complete Lewis structures for the foll owing condensed structural formulas. (a) CH3(CH2hCH(CH3h (b) (CH3 hCHCH2Ci (c) CH3CH 2COCHCH2 (d) CH3CH 2 CHO (e) CH3COCN (0 (CH3hCCOOH (g) (CH3CH2hCO

1-108

Line-Ang l e Form u las

Another kind of shorthand used for organic structures is the line-angle formula, sometimes called a skeletal structure or a stick figure. Line-angle formulas are often used for cyclic compounds and occasionally for noncyclic ones. In a stick figure, bonds are represented by lines, and carbon atoms are assumed to be present wherever two lines meet or a line begins or ends. Nitrogen, oxygen, and halogen atoms are shown, but hydrogen atoms are not usually drawn unless they are bonded to an atom that is drawn. Each carbon atom is assumed to have enough hydrogen atoms to give it a total of four bonds. Nonbonding elec trons are rarely shown. Table 1 -4 shows some examples of line-angle drawings.

TABLE 1-4

E xamples of Line-Angle Drawings

Compound

hexane 2-hexene 3-hexanol

2-cyclohexenone

2-methylcyclohexanol

nicotinic acid (a vitamin, also called niacin)

Condensed Structure

Line-Angle Formula

CH3(CHJ4CH3 CH3CH=CHCH2CH2CH3

� �

� OH

CH3CH2CH(OH)CH2CH2CH3 0 /CH2,,- � "",, CH2 C I I CH2 CH '- � CH 4

/CH2,,CHOH CH? I I C�2 /CHCH3 CH2 H I H,,--- /C� /COOH C C I II C C / -:9' "- N "-H H

a

a I

h

N

C( C(

H

OH

CH3

or

(i' 0

COOH or

N

oH

19

20

Chapter

1:

Introduction and Review PROBLEM 1-10

Give Lewis structures corresponding to the following line-angIe structures.

(.l

(e)

V H

(bl

rY CHO

cJ

(c)

o

0

( l g

PROBLEM 1-1 1

O� N

H

(d)

o

if

/\ �OH

o (h) �

H � 9 H OH �

Draw condensed structural formulas corresponding to the following line-angle structures. (a)

1- 1 1 Molecu l a r Fo rmulas a nd Empiri ca l Fo rm u l as

�

�

(b)

�

(c)

I

(d)

Before we can write possible structural formulas for a compound, we need to know its molecular formula. The molecular formula simply gives the number of atoms of each element in one molecule of the compound. For example, the molecular formula for I-butanol is C4H 100. CH3CH 2CH 2CH 20H I-butanol, molecular formula C4H 10 0

Calculation of the Empirical Formula

Molecular formulas can be determined by a two-step process. The first step is the determination of an empirical formula, simply the relative ratios of the elements present. Suppose, for example, that an unknown compound was found by quantitative elemental analysis to contain 40.0% carbon and 6.67% hydrogen. The remainder of the weight (53.3%) is assumed to be o xygen. To convert these numbers to an empirical formula, we can follow a simple procedure. 1.

Assume the sample contains 1 00 g, so the percent value gives the number of grams of each element. Divide the number of grams of each element by the atomic weight to get the number of moles of that atom in the 1 00-g sample. 2. Divide each of these numbers of moles by the smallest one. This step should give recognizable ratios. For the unknown compound, we do the following computations: 40.0 g C 1 2.0 g/ mol

=

6.67 g H 1.o 1 g/ mol

=

53.3 g O

1 6 . 0 g/mol

=

3.33 mol C;

3.33 mol 3.33 mol

6.60 mol H;

6.60 mol 3.33 mol

3.33 mol 0;

3.33 mol 3.33 mol

1 .98

=

2

1

The first computation divides the number of grams of carbon by 12, the number of grams of hydrogen by 1 , and the number of grams of o xygen by 16. We

1-12

Arrhenius Acids and Bases

21

compare these numbers by dividing them by the smallest number, 3.33. The final result is a ratio of one carbon to two hydrogens to one oxygen. This result gives the empirical formula C,H20, or CH20, which simply shows the ratios of the ele­ ments. The molecular formula can be any multiple of this empirical formula, because any multiple also has the same ratio of elements. Possible molecular for­ mulas are CH20, C2H402, C3H603, C4H s04, etc. Calculation of the Molecular Formula

How do we know the correct molecular formula? We can choose the right multiple of the empirical formula if we know the molecular weight. Molecular weights can be determined by methods that relate the freezing-point depression or boiling-point elevation of a solvent to the molal concen­ tration of the unknown. If the compound is volatile, we can convert it to a gas and use its volume to determine the number of moles according to the gas law. Newer methods include mass spectrometry, which we will cover in Chapter Ii. For our example (empirical formula CH20), let's assume that the molecular weight is determined to be about 60. The weight of one CH20 unit is 30, so our un­ known compound must contain twice this many atoms. The molecular formula must be C2H402. The compound might be acetic acid. In Chapters 1 2, 13, and 15 we will use spectroscopic techniques to determine the complete structure for a compound once we know its molecular formula.

°

II CH3-C-OH acetic acid, C2H402

P R O B LEM 1-1 2

Compute the empirical and molecular formulas for each of the following elemental analyses. In each case, propose at least one structure that fits the molecular fonnula. C H N CI MW (a) 40.0% (b) 32.0% (c) 37.2% (d) 38.4%

6.67% 6.67% 7.75% 4.80%

0 18.7% 0 0

0 0 55.0% 56.8%

90 75 64 125

PROBLEM-SOLVING

The properties and reactions of acids and bases are central to our study of organic chemistry. We need to consider exactly what is meant by the terms acid and base. Most people would agree that H2S04 is an acid and NaOH is a base. Is BF3 an acid or a base? Is ethylene ( H2C = CH2) an acid or a base? To answer these questions, we need to understand the three di fferent definitions of acids and bases: the Arrhenius def­ inition, the Brj2jnsted-Lowry definition, and the Lewis definition. Acidic compounds were first classified on the basis of their sour taste. The Latin terms acidus (sour) and acetum (vinegar) gave rise to our modem terms acid and acetic acid. Alkaline compounds (bases) were substances that neutralize acids, such as limestone and plant ashes (al kalai in Arabic). The Arrhenius theory, developed at the end of the nineteenth century, defined acids as substances that dissociate in water to give H30 + ions. The stronger acids, such as sulfuric acid (H2S04) , were assumed to dissociate to a greater degree than weaker acids, such as acetic acid ( CH3COOH) . H2SO4 su I furic

+

H zO



1- 14 Lewis Acids and B ases Resona nce Sta b i l ization The negative charge of a conjugate base may be delocalized over two or more atoms by resonance. Depending on how electronega­ tive those atoms are, and how many share the charge, resonance delocalization is often the dominant effect helping to stabilize an anion. Consider the following conjugate bases. Conjugate Base

�

Acid

CH 3CH2- Q:­

CH 3CH 2-OH

15.9

ethoxide ion

ethanol

(weak acid)

·0· · II .. CH 3-C-O:-

4.74

••

l

: 0:-

�

I

o

CH 3- S =0:.

..0 0..

(moderate acid)

acetic acid

acetate ion

·· · II .. CH 3- S-O:II ..

II

..0..

.

II

CH 3- S-OH

�

- 1.2

II

(strong acid)

o methanesulfonic acid

methanesulfonate ion

Ethoxide ion is the strongest of these three bases. Ethoxide has a negative charge localized on one oxygen atom; acetate ion has the negative charge shared by two oxygen atoms; and the methanesulfonate ion has the negative charge spread over three oxygen atoms. The pKa values of the conjugate acids of these anions show that acids are stronger if they deprotonate to give resonance-stabilized conjugate b as e s . PROBLEM 1 - 1 8

Write equations for the following acid-base reactions. Label the conjugate acids and bases, and show any resonance stabilization. Predict whether the equibbrium favors the reactants or products. (a) (c) (e) (g)

CH3CH20H + CH3NH­ CH30H + H 2S04 CH3NH! + CH30CH3S03 + CH3COOH

(b) CH3CH2COOH + CH3NHCH3 (d) NaOH + H2S (0 CH30- + CH3COOH

The Brl1lnsted-Lowry definition of acids and bases depends on the transfer of a proton from the acid to the base . The base uses a pair of nonbonding electrons to form a bond to the proton. G. N. Lewi s reasoned that this kind of reaction does not need a proton. Instead, a base could use its lone pair of electrons to bond to some other electron-deficient atom. In effect, we can look at an acid-base reaction from the viewpoint of the bonds that are formed and broken rather than a proton that is transferred. The following reaction shows the proton transfer, with emphasis on the bonds being broken and formed. Organic chemists routinely use curved arrows to show the movement of the participating electrons. B :� H : A V

B :H

+

:A

Lewis bases are species with non bonding electrons that can be donated to form new bonds. Lewis acids are species that can accept these electron pairs to form new bonds. Since a Lewis acid accepts a pair of electrons, it is called an

1 - 14 Lewis Aci d s a n d Bases

29

30

Chapter 1: Introduction and Review

PROBLEM-SOLVING

Htnp

A nucleoph i l e donates e lectrons An electroph ile accepts electrons. Acidic protons may serve as electron acceptors.

electrophile, from the Greek words meaning "lover of electrons." A Lewis base is called a nucleophile, or "lover of nuclei," because it donates electrons to a nucleus with an empty (or easily vacated) orbital. In this book, we sometimes use colored type for emphasis: blue for nucleophiles, green for electrophiles, and occasionally red for acidic protons. The Lewis acid-base definitions include reactions having nothing to do with protons. Following are some examples of Lewis acid-base reactions. Notice that the common Br0nsted-Lowry acids and bases also fall under the Lewis definition, with a proton serving as the electrophile. Curved arrows (red) are used to show the movement of electrons, generally from the nucleophile to the electrophile.

B :-==---- H+

nucleophile

electrophile

(Lewis base)

(Lewis acid)

H-

H 1

F 1

� : �� - F H

nucleophi Ie

F

B-H bond formed

H F 1 1_ H - N-B -F I 1 H F +

electrophile

H I .. .. H Cl : CH3- 0 :� _C " � I 1\""- " H

bond formed

H I .. CH3 -O- C - H I H ••

+

nucleophi Ie

electrophile

nucleophile base

electrophile

bond formed

acid

(conjugated acid)

'0' II � �.. H -.l.... q - C- CH3 -----7 H3N : +

.. : CI :

bond formed

(conjugated base)

Some of the terms associated with acids and bases have evolved specific mean­ ings in organic chemistry. When organic chemists use the term base, they usually mean a proton acceptor (a Brpnsted-Lowry base). Similarly, the term acid usually means a proton donor (a Br0nsted-Lowry acid). When the acid-base reaction involves forma­ tion of a bond to some other element (especially carbon), organic chemists refer to the electron donor as a nucleophile (Lewis base) and the electron acceptor as an electrophile (Lewis acid). The following illustration shows electrostatic potential maps for the preceding second example, the reaction of NH3 with BF3 . The electron-rich (red) region of NH3 attacks the electron-poor (blue) region of B F3. The product shows high electron densi­ ty on the boron atom and its three fluorine atoms and low electron density on nitrogen and its three hydrogen atoms.

1 - 14 Lewis Acids and B ases

31

The curved-arrow formalism is used to show the flow of an electron pair from the electron donor to the electron acceptor. The movement of each pair of electrons in­ volved in making or breaking bonds is indicated by its own separate an·ow, as shown in the preceding set of reactions. In this book, these curved arrows are always printed in red. In the preceding reaction of CH30- with CH3Cl, one curved arrow shows the lone pair on oxygen forming a bond to carbon. Another curved arrow shows that the C - CI bonding pair detaches from carbon and becomes a lone pair on the Cl- product.

I

.

H

CH3 -(Xi� H _CpCl: . . -----'--"r I H nucleophile electrophile • •

.I . H

• •

�

CH3-OE>C- H .. I H

+

�Cl: ..

The curved-arrow formalism is universally used for keeping track of the flow of electrons in reactions. We have also used this device (in Section 1-9, for example) to keep track of electrons in resonance structures as we imagined their "flow" in going from one resonance structure to another. Remember that electrons do not "flow" in resonance structures: They are simply delocalized. Still, the curved-arrow formalism helps our minds flow from one resonance structure to another. We will find ourselves constantly using these (red) curved arrows to keep track of electrons, both as reactants change to products and as we imagine additional resonance structures of a hybrid.

PROBLEM-SOLVING

Hi-nv

Use one curved arrow for each pair of electrons participating in the reaction.

P R O B L E M 1 - 1 9 ( PA RT I A L LY S O LV E D )

In the following acid-base reactions, 1. determine which species are acting as electrophiles (acids) and which are acting as nu­ cleophiles (bases). 2. use the curved-arrow formalism to show the movement of electron pairs in these reac­ tions, as well as the imaginary movement in the resonance hybrids of the products. 3. indicate which reactions are best termed Br0nsted-Lowry acid-base reactions. (a)

I

o

CH3-C-H

+

acetaldehyde

HCl

This reaction is a proton transfer from HCl to the C = O group of acetaldehyde. Therefore, it is a Br0nsted-Lowry acid-base reaction, with HCl acting as the acid (pro­ ton donor) and acetaldehyde acting as the base (proton acceptor). Before drawing any curved arrows, remember that arrows must show the movement of electrons from the electron-pair donor (the base) to the electron-pair acceptor (the acid). An arrow must go from the electrons on acetaldehyde that form the bond to the hydrogen atom, and the bond to chlorine must break, with the chloride ion taking these electrons. Drawing these arrows is easier once we draw good Lewis structures for all the reactants and products.

�

:O: II CH 3 -C-H

.. H -Cl l.- . . :

base

acid

[

+/ 0 :

H

PROBLEM-SOLVING

�II CH3 -C-H minor

maior

The resonance forms of the product show that a pair of electrons can be moved be­ tween the oxygen atom and the C = O pi bond. The positive charge is delocalized over the carbon and oxygen atoms, with most of the positive charge on oxygen because all octets are satisfied in that resonance structure. 0o I (b) CH3-C-H + CH3- OCH -C-H 3 acetaldehyde O-CH3

I

I

Hi-nv

T h e curved arrows we use in mechanisms show the flow of electrons and not the movement of atoms. We wil l use these curved arrows constantly throughout this course.

32

Chap ter 1 : Introduction and Review

In this case, no proton has been transferred, so this is not a Br¢nsted-Lowry acid-base reaction. Instead, a bond has formed between the C = O carbon atom and the oxygen of the CH3 - 0- group. Drawing the Lewis structures helps to show that the CH3 - 0- group (the nucleophile in this reaction) donates the electrons to form the new bond to acetaldehyde (the electrophile). This result agrees with our intuition that a negative­ ly charged ion is likely to be electron-rich and therefore an electron donor. ..

:0: I CH3 - C- H I : O-CH3 ••

:O II J CH3 -C -H

.. - : O-CH3 � ••

electroph i le

nucleoph ile

Notice that acetaldehyde acts as the nucleophile (Lewis base) in part (a) and as the electrophile (Lewis acid) in part (b). Like most organic compounds, acetaldehyde is both acidic and basic. It acts as a base if we add a strong enough acid to make it donate electrons or accept a proton. It acts as an acid if the base we add is strong enough to donate an elec­ tron pair or abstract a proton. B H3 I CH3 - O - CH3 -

+

o II (d) CH3-C - H

+

-OH

0I CH3 - C - H I OH

�

I H 0 LH - � - � -H

o

Chapte r 1 G l ossa ry

II (e) CH3-C-H

+

(f) CH3-NH2

CH3-Cl

+

-OH

+

CH3-NH2-CH3

�

+

Cl-

Each chapter ends with a glossary that summarizes the most important new terms in the chapter. These glossaries are more than just a dictionary to look up unfamiliar terms as you encounter them (the index serves that purpose). The glossary is one of the tools for reviewing the chapter. You can read carefully through the glossary to see if you understand and remember all the terms and associated chemistry mentioned there. Anything that seems unfamiliar should be reviewed by turning to the page number given in the glossary listing. acid-dissociation constant (Ka) The equilibrium constant for the reaction of the acid with

water to generate H30+. (p. 23) HA acid

t+ or

\- distance

wave function (lji) with positive sign

node

�

... Figure 2-3

First harmonic of a guitar string. The two halves of the string are separated by a node, a point with zero displacement. The two halves vibrate out of phase with each other.

nucleus

/- distance represented by nucleus

probability function 1112 (electron density)

wave function (lji) with negative sign

at any point is given by 0/2, the square of the wave function at that point. Notice that the plus sign and the minus sign of these wave functions are not charges. The plus or minus sign is the instantaneous sign of the constantly changing wave func­ tion. The Is orbital is spherically symmetrical, and it is often represented by a cir­ cle (representing a sphere) with a nucleus in the center and with a plus or minus sign to indicate the instantaneous sign of the wave function (Figure 2-2). If you gently place a finger at the center of a guitar string while plucking the string, your finger keeps the midpoint of the string from moving. The displacement ( movement + or ) at the midpoint is always zero; this point is a node. The string now vibrates in two parts, with the two halves vibrating in opposite directions. We say that the two halves of the string are out of phase: When one is displaced upward, the other is displaced downward. Figure 2-3 shows this first harmonic of the guitar string. The first harmonic of the guitar string resembles the 2p orbital (Figure 2-4). We have drawn the 2p orbital as two "lobes," separated by a node (a nodal plane). The two lobes of the p orbital are out of phase with each other. Whenever the wave function has a plus sign in one lobe, it has a minus sign in the other lobe. -

nodal plane

�� �'" nucleus

represented by

wave function (instantaneous picture)

� Figure 2-4

The 2p orbital. The 2p orbital has two lobes, separated by a nodal plane. The two lobes are out of phase with each other. When one has a plus sign, the other has a minus si gn .

nodal plane

/ nodal plane

�/ nucleus � equivalent wave function (instantaneous picture)

represented by nodal plane

2-2

2-1 A

Molecular Orbitals

41

Linear Com b i nation of Ato m ic O rbita l s

Atomic orbitals can combine and overlap to give more complex standing waves. We can add and subtract their wave functions to give the wave functions of new orbitals. This process is called the linear combination of atomic orbitals (LCAO). The num­ ber of new orbitals generated always equals the number of orbitals we started with. When orbitals on different atoms interact, they produce molecular orbitals (MOs) that lead to bonding (or antibonding). 2. When orbitals on the same atom interact, they give hybrid atomic orbitals that define the geometry of the bonds. 1.

We begin by looking at how atomic orbitals on different atoms interact to give molecular orbitals. Then we consider how atomic orbitals on the same atom can inter­ act to give hybrid atomic orbitals. The stability of a covalent bond results from a large amount of electron density in the bonding region, the space between the two nuclei (Figure 2-5). In the bonding region, the electrons are close to both nuclei, resulting in a lowering of the overall energy. The bonding electrons also mask the positive charges of the nuclei, so the nuclei do not repel each other as much as they would otherwise. There is always an optimum distance for the two bonded nuclei. If they are too far apart, their attraction for the bonding electrons is diminished. If they are too close together, their electrostatic repulsion pushes them apart. The internuclear distance where attraction and repUlsion are balanced, which also gives the minimum energy (the strongest bond), is called the bond length.

2-2 M o lecu l a r O rbita ls

bonding region • +

nucleus 1

2-2A

/

•

+

electrons in this region "'nucleus 2 attract both nuclei and mask the positive charges from repelling each other

The Hyd rogen Molecule; S i g m a Bon d i n g

The hydrogen molecule is the simplest example of covalent bonding. As two hydrogen atoms approach each other, their Is wave functions can add constructively so that they reinforce each other, or destructively so that they cancel out where they overlap. Figure 2-6 shows how the wave functions interact constructively when they are in phase and have the same sign in the region between the nuclei. The wave functions reinforce each other and increase the electron density in this bonding region. The result is a bonding molecular orbital (bonding MO). The bonding MO depicted in Figure 2-6 has most of its electron density centered along the line connecting the nuclei. This type of bond is called a cylindrically sym­ metrical bond or a sigma bond (0" bond). Sigma bonds are the most common bonds in organic compounds. All single bonds in organic compounds are sigma bonds, and every double or triple bond contains one sigma bond. The electrostatic potential map of H 2 shows its cylindrically symmetrical sigma bond, with the highest electron density (red) in the bonding region between the two protons.

� Figure 2-5

Electron density in a molecular orbital. A bonding molecular orbital places a large amount of electron density in the bonding region, the space between the two nuclei.

42

Chapter 2 : Structure and Properties of Organic Molecules Constructive Interaction: The two Is orbitals are in phase and have the same sign. add

�

Is

Is

Is

Is

�

� Figure 2-6

Formation of a cr-bonding MO. When the Is orbitals of two hydrogen atoms overlap in phase, they interact constructively to form a bonding MO. The electron density in the bonding region (between the nuclei) is increased. The result is a cylindrically symmetrical bond, or sigma bond, designated cr.

bonding molecular orbital represented by:

cr-bonding MO

When two hydrogen I s orbitals overlap out of phase with each other, an antibonding molecular orbital results (Figure 2-7). The two I s wave functions have opposite signs, so they tend to cancel out where they overlap. The result is a node (actually a nodal plane) separating the two atoms. The presence of a node separating the two nuclei usually indicates that the orbital is antibonding. Destructive interaction: The two I s orbitals are out of phase.

�--�

add

�

node

�

� Figure 2-7

Formation of a cr* antibonding MO. When two I s orbitals overlap out of phase, they interact destructively to form an antibonding MO. The positive and negative values of the wave functions tend to cancel out in the region between the nuclei, and a node separates the nuclei. We use an asterisk ( * ) to designate anti bonding orbitals such as this sigma anti bonding orbital, cr*.

antibonding m�lecular orbital I

represented by:

node cr* anti bonding MO

2-2 Molecular Orbitals ------- -----

43

-

node antibonding cr*

energy

.... Fig ure 2-8

I

atomic orbital

Relative energies of atomic and molecular orbitals. When the two hydrogen Is orbitals overlap, a sigma bonding MO and a sigma anti bonding MO res u lt. The bonding MO is lower in energy than the atomic Is orbital, and the antibonding orbital is higher in energy. Two electrons (represented by arrows) go into the bonding MO with opposite spins, forming a stable H2 molecule. The antibonding orbital is vacant.

atomic orbital

bonding molecular orbital

Figure 2-8 shows the relative energies of the atomic orbitals and the molecu­ lar orbitals of the H2 system. When the I s orbitals are in phase, the resulting molecular orbital is a sigma bonding MO, with lower energy than that of a I s atomic orbital. When two I s orbitals overlap out of phase, they form an antibond­ ing (0"*) orbital with higher energy than that of a Is atomic orbital. The two elec­ trons in the H2 system are found with paired spins in the sigma bonding MO, giving a stable H2 molecule. In stable molecules, the anti bonding orbitals (such as cr* ) are usually vacant. 2-28

S i g m a Overlap I nvolv i n g p O rbitals

When two p orbitals overlap along the line between the nuclei, a bonding orbital and an antibonding orbital result. Once again, most of the electron density is centered along the line between the nuclei. This linear overlap is another type of sigma bonding MO. The constructive overlap of two p orbitals along the line joining the nuclei forms a cr bond represented as follows: •

+ r__'", .

-

•

(J

•

bonding MO

S O LV E D P R O B L E M 2 - 1

Draw the cr':' antibonding orbital that results from the destructive overlap of the two Px orbitals just shown. S O LU TI O N

This orbital results from the destructive overlap of lobes of the two p orbitals with opposite phases. If the signs are reversed on one of the orbitals, adding the two orbitals gives an anti bonding orbital with a node separating the two nuclei: node +

Px

-

Px

•

•

cr* anti bonding MO

44

Chapter 2:

S tructure

and Properties of Org an ic Molecules Overlap of an s orbital with a p orbital also gives a bonding MO and an anti­ bonding MO, as shown in the following illustration. Constructive overlap of the s orbital with the Px orbital gives a sigma bonding MO with its electron density centered along the line between the nuclei. Destructive overlap gives a (]"* antibonding orbital with a node separating the nuclei.

s

Px

Ci

bonding MO node

+

Px

2-3 Pi Bo n d i ng

-

(-) s

=CHCH3

HiltZ;

S i m i l a r groups on the sa me side of the double bond: cis. Similar g roups o n opposite sides of the double bond: trans.

I -butene

Cis and trans isomers are only one type of stereoisomerism. The study of the structure and chemistry of stereoisomers is called stereochemistry. We will encounter stereochemistry throughout our study of organic chemistry, and Chapter 5 is devoted entirely to this field. Cis-trans isomers are also called geometric isomers because they differ in the geometry of the groups on a double bond. The cis isomer is always the one with simi­ lar groups on the same side of the double bond, and the trans isomer has similar groups on opposite sides of the double bond. To have cis-trans isomerism, there must be two d(fferent groups on each end of the double bond. For example, I -butene has two identical hydrogens on one end of the double bond. Reversing their positions does not give a different compound. Similarly, 2-methyl-2-butene has two identical methyl groups on one end of the double bond. Reversing the methyl groups does not give a different compound.

(a)

57

F2C=CH2 C>=CHCHCH3

(0

crCHCH;

PRO B LE M 2 - 1 1

Give the relationship between the following pairs of structures. The possible relation­ ships are same compound

constitutional isomers (structural isomers)

cis-trans isomers

not isomers (different molecularformula)

PROBLEM-SOLVING

HiltZ;

Identical groups on one of the double-bonded carbons implies no cis-trans isomerism.

58

Chapter 2 : Structure and Properties of Organic Molecules CH3CH,CHCH,CH,CH3 I CH3 -

-

Br (b)

"-

/

C=C

H

/ "-

Br (d)

(I)

(g) (h) (i) (j)

H

/ "C=C / "-

-

Br

H and

H

Br

/ "C=C / "-

-

Br

Br (c)

Br

H

"/

C=C

/ "-

H H

Cl Cl H Cl 1 1 1 1 H-C-C-H (e) H-C-C-H and 1 1 1 1 H H Cl H

H and Br

and

-

H CH3 CH3 H 1 1 1 1 H-C C - H and H - C C - CH3 1 1 1 1 CH3 H H H CH3- CH2 - CH2 - CH3 and CH3 - CH = CH - CH3 CH2 =CH - CH2CH2CH3 and CH3 - CH = CH - CH2CH3 CH2 = CHCH2CH2CH3 and CH3CH2CH2CH =CH2 CH 3 CH 3 and (k) CH 3

(X

y

2- 9 Pola rity of Bonds and Molecu les

(y CH 3

0

In Section 1 -6, we reviewed the concept of polar covalent bonds between atoms with different electronegativities. Now we are ready to combine this concept with molecu­ lar geometry to study the polarity of entire molecules. 2-9A

Bon d D i pole Moments

Bond polarities can range from nonpolar covalent, through polar covalent, to totally ionic. In the following examples, ethane has a nonpolar covalent C-C bond. Methyl­ amine, methanol, and chloromethane have increasingly polar (C-N, C-O, and C-Cl) covalent bonds. Methylammonium chloride (CH3NHI Cn has an ionic bond between the methylammonium ion and the chloride ion. �

�

�

H3C - CH3

H3 C -NH2

H3C-OH

H3 C -Cl

ethane

methylamine

methanol

chloromethane

nonpolar

+

H3 CNHPmethylammonium ch loride

I on ic

increasing polarity

The polarity of an individual bond is measured as its bond dipole moment, jL, defined as jL = 15 x d where 8 is the amount of charge at either end of the dipole and d is the distance between the charges. Dipole moments are expressed in units of the debye (D), where 1 debye 3.34 X 1 0-30 coulomb meters. If a proton and an electron (charge l .60 X 10- 19 coulomb) were 1 A apart (distance 1 0- 1 0 meter), the dipole moment would be jL ( l.60 X 1 0- 19 coulomb) X ( 10- 1 0 meter) 1 .60 X 10-29 coulomb meter =

=

=

2-9

Expressed in debyes,

fL

=

1.60 X 10-29 e· m 3.34 X 10-30 e mi D

=

Polarity of Bonds and Molecules

4 .8 0

.

A simple rule of thumb, using common units, is that fL

( in debyes )

=

4.8 X 8 ( electron charge ) X d ( in angstroms )

Dipole moments are measured experimentally, and they can be used to calculate other information such as bond lengths and charge separations. Bond dipole moments in organic compounds range from zero in symmetrical bonds to about 3.6 D for the strongly polar e-N: triple bond. Table 2-1 shows typi­ cal dipole moments for some of the bonds common in organic molecules. Recall that the positive end of the crossed arrow corresponds to the less electronegative (partial positive charge) end of the dipole. TABLE 2-1

Bond D ipole M oments (D ebye) for Some Common Covalent Bonds

Bond

Bond

Dipole Moment, I.t

+-
!l = 1.9 D

+-> !l = 1.0 D

chloromethane

chloroform

H

� Figure 2-21 Molecular dipole moments. A molecular dipole moment is the vector sum of the individual bond dipole moments.

� +-+

�

NF3 J.L = 0.2 D

=0

CHpH o

OH (j) o

hydrocortisone

vitamin E

acid chloride An acid derivative with a chlorine atom in place of the hydroxyl group. (p. 73) o

II

R-C-Cl alcohol A compound that contains a hydroxyl group; R -OH. (p. 71) aldehyde A carbonyl group with one alkyl group and one hydrogen; (p. 72) o

II

R-C-H. alkanes Hydrocarbons containing only single bonds. (p. 68) alkenes Hydrocarbons containing C=C double bonds. (p. 69) alkynes Hydrocarbons containing C - C triple bonds. (p. 70) alkyl group A hydrocarbon group with only single bonds; an alkane with one hydrogen removed, to allow bonding to another group; symbolized by R. (p. 69) amide An acid derivative that contains an amine instead of the hydroxyl group of the acid. (p. 74) o

II

R-C-NH2

o

o

II

R-C-NHR'

II

R-C-NR;

amine An alkylated analogue of ammonia; R -NH2, R2NH, or R3N. (p. 73) aromatic hydrocarbons (arenes) Hydrocarbons containing a benzene ring, a six-membered ring with three double bonds. (p. 70) bond dipole moment A measure of the polruity of an individual bond in a molecule, defined as 10 f.L = (4.8 X d X 8), where f.L is the dipole moment in debyes (10- esu A), d is the bond

-

length in angstroms, and 8 is the effective amount of charge separated, in units of the electronic charge. (p. 58) carbonyl group The functional group, as in a ketone or aldehyde. (p. 72) " C=O ./ carboxyl group The -COOH functional group, as in a carboxylic acid. (p. 72) carboxylic acid A compound that contains the carboxyl group; (p. 72) o

II

R-C-OH

Chapter 2 Glossary

75

76

Chapter 2 : S tructure and Properties of Organic Molecules cis-trans isomers (geometric isomers) Stereoisomers that differ in their cis-trans arrangement

on a ring or a double bond. The cis isomer has similar groups on the same side, and the trans isomer has similar groups on opposite sides. (p. 57) constitutional isomers (structural isomers) Isomers whose atoms are connected differently; they differ in their bonding sequence. (p. 56) cyano group The -C = N functional group, as in a nitrile. (p. 74) dipole-dipole forces Attractive intermolecular forces resulting from the attraction of the positive and negative ends of the molecular dipole moments of polar molecules. (p. 62) dipole moment See bond dipole moment and molecular dipole moment. (p. 58) double bond A bond containing four electrons between two nuclei. One pair of electrons forms a sigma bond, and the other pair forms a pi bond. (p. 44) ester An acid derivative with an alkyl group replacing the acid proton. (p. 73) o

II

R-C-OR' ether A compound with an oxygen bonded between two alkyl groups; R -0-R'. (p. 7 1 ) functional group The reactive, nonalkane part of an organic molecule. (p. 69) geometric isomers See cis-trans isomers. (p. 57) hybrid atomic orbital A directional orbital formed from a combination of s and p orbitals on 2

the same atom. Two orbitals are formed by sp hybridization, three orbitals by sp hybridization, 3 and four orbitals by sp hybridization. (p. 45) sp hybrid orbitals give a bond angle of 1 800 (linear geometry). 2 sp hybrid orbitals give bond angles of 1 200 (trigonal geometry). 3 sp hybrid orbitals give bond angles of 1 09.50 (tetrahedral geometry). hydrocarbons Compounds composed exclusively of carbon and hydrogen. alkanes: Hydrocarbons containing only single bonds. (p. 68) alkenes: Hydrocarbons containing C=C double bonds. (p. 69) alkynes: Hydrocarbons containing C- C triple bonds. (p. 70) cycloalkanes, cycloalkenes, cycloalkynes: Alkanes, alkenes, and alkynes in the form of a ring. (p. 68) aromatic hydrocarbons: Hydrocarbons containing a benzene ring, a six-membered ring with three double bonds. (p. 70)

o

benzene hydrogen bond A particularly strong attraction between a nonbonding pair of electrons and an electrophilic 0-H or N - H hydrogen. Hydrogen bonds have bond energies of about 5 kcal/mol (2 1 kJ/mol), compared with about 1 00 kcallmol (about 400 kJ/mol) for typical C-H bonds. (p. 63) hydroxyl group The -OH functional group, as in an alcohol. (p. 71) isomers Different compounds with the same molecular formula. (p. 56) constitutional isomers (structural isomers) are connected differently; they differ in their bonding sequence. stereoisomers differ only in how their atoms are oriented in space. cis-trans isomers (geometric isomers) are stereoisomers that differ in their cis-trans arrangement on a ring or a double bond. stereochemistry is the study of the structure and chemistry of stereoisomers. ketone A carbonyl group with two alkyl groups attached. (p. 72) o

II

R-C-R' linear combination of atomic orbitals (LeAO) Wave functions can add to each other to pro­ duce the wave functions of new orbitals. The number of new orbitals generated equals the orig­ inal number of orbitals. (p. 4 1 ) London dispersion forces Intermolecular forces resulting from the attraction of correlated temporary dipole moments induced in adjacent molecules. (p. 62) molecular dipole moment The vector sum of the bond dipole moments (and any nonbonding pairs of electrons) in a molecule; a measure of the polarity of a molecule. (p. 60)

Study Problems molecular orbital (MO) An orbital formed by the overlap of atomic orbitals on different atoms. MOs can be either bonding or anti bonding, but only the bonding MOs are filled in most stable molecules. (p. 4 1 ) A bonding molecular orbital places a large amount of electron density in the bonding re­ gion between the nuclei. The energy of an electron in a bonding MO is lower than it is in an atomic orbital . An anti bonding molecular orbital places most of the electron density outside the bond­ ing region. The energy of an electron in an antibonding MO is higher than it is in an atomic orbital. nitrile A compound containing a cyano group, - C= N. (p. 74) node In an orbital, a region of space with zero electron density. (p. 40) pi bond (7T bond) A bond fonned by sideways overlap of two p orbitals. A pi bond has its elec­ tron density in two lobes, one above and one below the line joining the nuclei. (p. 44) sigma bond (0- bond) A bond with most of its electron density centered along the line joining the nuclei; a cylindrically symmetrical bond. Single bonds are normally sigma bonds. (p. 4 1 ) stereochemistry The study of the structure and chemistry of stereoisomers. (p. 57) stereoisomers Isomers that differ only in how their atoms are oriented in space. (p. 57) structural isomers (IUPAC term: constitutional isomers) Isomers whose atoms are connect­ ed differently; they differ in their bonding sequence. (p. 56) triple bond A bond containing six electrons between two nuclei. One pair of electrons forms a sigma bond and the other two pairs form two pi bonds at right angles to each other. (p. 52) valence-shell electron-pair repulsion theory (VSEPR theory) Bonds and lone pairs around a central atom tend to be separated by the largest possible angles: about 1 800 for two, 1 200 for three, and 109.50 for four. (p. 45) van der Waals forces The attractive forces between neutral molecules, including dipole-di­ pole forces and London forces. (p. 62) dipole-dipole forces: The forces between polar molecules resulting from attraction of their permanent dipole moments. London forces: Intermolecular forces resulting from the attraction of correlated temporary dipole moments induced in adjacent molecules. wave function (.p) The mathematical description of an orbital. The square of the wave function (1//) is proportional to the electron density. (p. 39)

C

Essential Probl e m-Solvi ng Ski l ls in Chapter 2

1.

Draw the structure of a single bond, a double bond, and a triple bond.

2. Predict the hybridization and geometry of the atoms in a molecule. 3. Draw a three-dimensional representation of a given molecule. 4.

Identify constitutional isomers and stereoisomers.

S.

Identify polar and nonpolar molecules, and predict which ones can engage in hydro­ gen bonding.

6.

Predict general trends in the boiling points and solubilities of compounds, based on their size, polarity, and hydrogen-bonding ability.

7.

Identify the general classes of hydrocarbons, and draw structural fonnulas for examples.

8.

Identify the classes of compounds containing oxygen or nitrogen, and draw structural formulas for examples.

Study Problems 2-23

Define and give examples of the fol lowing tenns: (b) antibonding MO (e) pi bond (h) constitutional isomers (k) bond dipole moment (j) stereoisomers

(a) bonding M O (d) sigma bond (g) triple bond

(c) hybrid atomic orbital

(f)

(i)

(I)

double bond cis-trans isomers molecular dipole moment

77

78

Chapter 2: Structure and Properties of Organic Molecules (m) dipole-dipole forces (p) miscible liquids (s) functional group 2-24

2-25

H

H

H - C - CI

CI -C-CI

CI

H

I

I

2-27

2-28

I

Cyclopropane ( C3 H6 , a three-membered ring) is more reactive than most other cycloalkanes. (a) Draw a Lewis structure for cyclopropane. (b) Compare the bond angles of the carbon atoms in cyclopropane with those in an acyclic ( noncyclic) alkane. (c) Suggest why cyclopropane is so reactive. For each of the following compounds, 1. Give the hybridization and approximate bond angles around each atom except hydrogen . 2 . Draw a three-dimensional diagram, including any lone pairs of electrons. (c) NH2NH2 (b) -OH (a) H3 0+ (d) ( CH3hN (f) CH3COOH (e) (CH3)4 N+ (i) CH20 (h) CH3 0H (g) CH3CH=NH For each of the following compounds, 1. Draw a Lewis structure. 2. Show the kinds of orbitals that overlap to form each bond. 3. Give approximate bond angles around each atom except hydrogen. (a) ( CH3)zNH (b) CH3 CH20H (e) CH3 - C= C - CHO (d) CH3 - CH=C( CH3h (g)

2-29

(r) alkyl group

Give a defi nition and an example for each class of organic compounds. (c) alkyne (b) alkene (a) alkane (d) aIcohol (e) ether (f) ketone (i) carboxylic acid (h) aromatic hydrocarbon (g) aldehyde (k) amine (I) amide (j) ester (m) nitrile If the carbon atom in CH2CI2 were flat, there would be two stereoisomers. The carbon atom in CH2CI2 is actually tetra­ hedral. Make a model of this compound, and determine whether there are any stereoisomers of CH2CI2.

I

2-26

(0) hydrogen bonding

(n) London forces (q) hydrocarbons

o

II

CH3 - C - CH3

(h)

(Oy O U

3 In most amines, the nitrogen atom is sp hybridized, with a pyramidal structure and bond angles close to 1 09°. In formamide, the nitrogen atom is found to be planar, with bond angles close to 120°. Explain this surprising finding. (Hint: Consider resonance forms and the overlap needed in them.) o

II

..

H-C-NH2 formamide 2-30

Predict the hybridization and geometry of the carbon and nitrogen atoms in the following ions. (Hint: Resonance.) o

2-31

2-32

+ II ;: (a) CH3 - C-CH2 (b) H2N -CH=CH- CH2 Draw orbital pictures of the pi bonding in the following compounds: (a) CH3COCH3 (b) HCN (c) CH2=CH - CH=CH - CN (d) CH3 - C= C - CHO (a) Draw the structure of cis- CH3 - CH= CH - CH2CH3 showing the pi bond with its proper geometry. (b) Circle the six coplanar atoms in this compound. (c) Draw the trans isomer, and circle the coplanar atoms. Are there still six? (d) Circle the coplanar atoms in the following structure.

Study Proble ms 2-33 2-34

In 2-pen tyne (CH3 CCCH2CH3 ) there are four atoms in a straight line . Use dashed lines an d wedges to draw a three­ dimen sional re pre sen tation of this mole cule, an d circle the four atoms that are in a straight line . Which of the fol lowin g compoun ds show cis-tran s i somerism? Draw the cis an d tran s i somers of the ones that do. (c) CH2=C(CH3 h (b) CH3 - C= C - CH3 (a) CH3CH= CHCH3 (d) cyclopen tene,

2-35

0

(e) CH3 -CH= C - CH -CH 2

I

II

CH2CH2CH3

3

Give the re lation ships between the following pairs of struct ures. The possible relation ships are : same compoun d, cis-trans i somers, con sti tution al (structural) isomers, n ot isomers (di fferen t molecular formula). (b) CH2= CH - CH2Cl an d CHCl=CH- CH3 (a) CH3 CH2CH2CH3 an d ( CH3hCH CH3 (c)

CH3

\:=!

�

CH3

CH3 and

(d)

CH3

CH3

f

CH3

\:=!

an d

= CH2

CH3

(f)

an d

(h) 2-36 2-37

(e)

NC NC

" /

C= C

/

"

CN (f)

CN

()

0

II

(d) CH3 - C - CH3

(c) CBr4

(b) CH3 -CN

0

(g)

N

Cl

I

H Diethyl ethe r and I -butan ol are isomers, and they have simi lar solubilitie s in water. Their boiling points are very different, however. Explain why these two compoun ds have simi lar solubil ity propertie s but dramatically differen t boiling poin ts. CH3 CH2 - O - CH2CH3 diethyl ether, bp 35°C 8 .4 mL dissolves in 100 mL H20

2-39

� � O

Sulfur dioxide has a dipole moment of 1.60 D. Carbon dioxide has a dipole momen t of zero, even though C - O bonds are more polar than S - 0 bonds. Explain this apparent contradicti on . For e ach of the following compoun ds, 1 . Draw the Lewis structure . 2. Show how the bond dipole momen ts (an d those of an y nonbon din g pai rs of e lectron s) con tribute to the mole cular dipole moment. 3 . Pre dict whether the compound will have a large ( > I D ) , small, or zero dipole moment. (a) CH3 - CH= N - CH3

2-38

79

CH3CH2CH2CH 2 - OH I-butanol, bp 118°C 9.1 mL dissolves in 100 mL H 20

N-methylpyrrolidine has a boi lin g point of 8 1 °C, an d piperidine has a boilin g poin t of 106°C. (a) Explain this large differen ce (25°C) in boilin g point for the se two i somers. (b) Te trahydropyran has a boiling poin t of 88°C, an d cyclopentan ol has a boiling point of 141°C. These two i somers have a boiling poin t diffe ren ce of 53°C. Explain why the two oxygen -contain in g i somers have a much l arge r boilin g poin t difference than the two amine isomers.

C

N-CH3

N-methylpyrrolidine , bp 81 °C

Co

tetrahydrop yran, bp 88 °C

C

N-H

piperidine , bp 106°C

o-

OH

cyclopentanol, bp 141 °C

Chapter 2: Structure and Properties of Organic Molecules

80 2-40

Which of the following pure compounds can form hydrogen bonds? Which can form hydrogen bonds with water? (3) (CH3 CH2hNH (b) (CH3CH2hN (c) CH3 CH2CH20H (e) CH3 (CH2hCH3 (f) CH2=CH - CH2CH3 (d) ( CH3 CH2CH2hO (i) CH3CH2CHO (h) CH3CH2COOH (g) CH3COCH3 O 0 � yO (j)

2-41

2-42

0

(k)

(I)

U

�

CH3 - C-NH2

Predict which compound in each pair has the higher boiling point. Explain your prediction. (3) CH3 CH20CH3 or CH3 CH ( OH) CH3 (b) CH3 CH2CH2CH3 or CH3 CH2CH2CH2CH3 (c) CH3 CH2CH2CH2CH3 or (CH3hCHCH2CH3 (d) CH3 CH2CH2CH2CH3 or CH3 CH2CH2CH2CH2Cl Circle the functional groups in the following structures. State to which class (or classes) of compounds the structure belongs. (3)

(d)

()

(b)

0

a

COOH

I

(c)

0

0

~ �

(e)

CH'OCH;

6

CHO

0

(f)

NH2

I

(h) CH3 - CH - COOCH3

(g)

�

Cr

-H

CN 2-43

Dimethyl sulfoxide (DMSO) has been used as an anti-inflammatory rub for racehorses. DMSO and acetone seem to have similar structures, but the C= O carbon atom in acetone is planar, while the S=O sulfur at om in DMSO is pyramidal. Draw correct Lewis structures for DMSO and acetone, predict the hybridiz ations, and expl ain these observations. o

o

CH3- S - CH3

CH3- C - CH3

DMSO

acetone

II

2-44

II

Many naturally occurring compounds contain more than one functional group. Identify the functional groups in the fol­ lowing compounds: (3) Penicillin G is a naturally occurring antibiotic. (b) Dopamine is the neurotransmitter that is deficient in Parkinson's disease. (c) Thyroxine is the principal thyroid hormone. (d) Testosterone is a male sex hormone. o

II

CH -C -NH

6 J=r?a O

OH

penicillin G

HO

hh y Y '

0

_

I

_

�

I

thyroxine

dopamine OH NH2

t

CH2 - H - COOH o

testosterone

3

Structure and Stereochem istry of Alkanes

W

henever possible, we will study organic chemistry using families of compounds to organize the material. The properties and reactions of the compounds in a family are similar, just as their structures are similar. B y considering how the structural features o f a class of compounds determine their prop­ erties, we can predict the properties and reactions of similar new compounds. This organization elevates organic chemistry from a catalog of many individual com­ pounds to a systematic study of a few types of compounds. Families of organic molecules are classified according to their reactive parts, called functional groups . . We considered some of the common functional groups in Sections 2- 1 2 through 2- 1 4. An alkane is a hydrocarbon that contains only single bonds. The alkanes are the sim­ plest and least reactive class of organic compounds because they contain only hydrogen 3 and sp hybridized carbon, and they have no reactive functional groups. Although alkanes undergo reactions such as cracking and combustion at high temperatures, they are much less reactive than other classes of compounds having functional groups.

We classify hydrocarbons according to their bonding (Section 2- 1 2), as shown in Table 3- 1 . Alkanes have only single bonds. A hydrocarbon with a carbon-carbon dou­ ble bond (such as ethylene) is an alkene. If a hydrocarbon has a carbon-carbon triple bond (like acetylene), it is an alkyne. Hydrocarbons with aromatic rings (resembling benzene) are called aromatic hydrocarbons. TABLE 3-1

Classification of H yd roca rbo ns (Revi ew)

Summary of Hydrocarbon Classification Functional Group

Example

alkanes

none (no double or triple bonds)

CH3-CH2-CH3, propane

alkenes

""'C =C / ./ '-

CH2=CH-CH3, propene

al\cynes

-C=C - triple bond

H-C=C-CH3, propyne

Compound Type

aromatics

3-1

I

benzene ring

C ""'C � ""'C /

I

II

/C � /C "'" C

I

81

82

Chapter 3: Structure and Stereochemistry of Alkanes

A hydrocarbon with no double or triple bonds is said to be saturated because it has the maximum number of bonded hydrogens. Another way to describe alkanes, then, is as the class of saturated hydrocarbons.

3-2 Molecu l a r Fo rmu las of A l ka n es

Table 3-2 shows the structures and formulas of the first 20 unbranched alkanes. Any isomers of these compounds have the same molecular formulas even though their structures are different. Notice how the molecular formulas increase by two hydrogen atoms each time a carbon atom is added. The structures of the alkanes in Table 3-2 are purposely written as chains of -CH2 - groups (methylene groups), terminated at each end by a hydrogen atom. This is the general formula for the unbranched (straight-chain) alkanes. These alkanes differ only by the number of methylene groups in the chain. If the molecule contains n carbon atoms, it must contain (2n + 2) hydrogen atoms. Figure 3-1 shows how this pattern appears in structures and how it leads to formu­ las of the form CnH2n+2 . A series of compounds, like the n-alkanes, that differ only by the number of -CH2 - groups, is called a homologous series, and the individual members of the series are called homologs. For example, butane is a homolog of propane, and both of these are homologs of hexane and decane. Although we have derived the CnH2n+2 formula using the unbranched n-alkanes, it applies to branched alkanes as well. Any isomer of one of these n-alkanes has the same molecular formula. Just as butane and pentane follow the CnH2n+2 rule, their branched isomers isobutane, isopentane, and neopentane also follow the rule. PROBLEM 3- 1 Using the general molecular formula for alkanes, (a) Predict the molecular formula of the C2S straight-chain alkane. (b) Predict the molecular formula of 4,6-diethyl- 1 2-(3,5-dimethyloctyl)triacontane, an alkane containing 44 carbon atoms.

TABLE 3-2 Alkane methane ethane propane butane pentane hexane heptane octane nonane decane undecane dodecane tridecane tetradecane pentadecane hexadecane heptadecane octadecane nonadecane eicosane triacontane a

Formu las and Physical Properties of the Unbranched A l kanes, Called the n-Al kanes Number of Carbons

Structure

Formula

Boiling Point (0C)

Melting Point (OC)

Densitya

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 30

H-CH2-H H-(CH2h-H H-(CH2h-H H-(CH2)4 -H H-(CH2)s-H H-(CH2)6-H H-(CH2h-H H-(CH2)g-H H-(CH2)9-H H-(CH2)lo-H H-(CH2)II-H H-(CH2)12-H H-(CH2)I3-H H-(CH2)14-H H-(CH2)ls-H H-(CH2)16-H H-(CH2)I7-H H-(CH2)ls-H H-(CH2)19-H H-(CH2ho-H H-(CH2)30-H

CH4 C2H6 C3 Hs C4HIO CSHI2 C6HI4 C7HI6 CsHls C9H20 C1o H22 C11 H24 CI2H26 C13 H2S CI4 H30 C1sH32 CI6H34 C17H36 C1sH3S CI9H40 C2oH42 C30H62

- 1 64 -89 -42 0 36 69 98 1 26 151 174 1 96 216 235 254 271 287 303 317 330 343 >450

- 1 83 - 1 83 - 1 89 -138 - 1 30 -95 -9 1 -57 -5 1 -30 -26 -10 -5 6 10 18 23 28 32 37 66

0.55 0.51 0.50 0.58 0.63 0.66 0.68 0.70 0.72 0.73 0.74 0.75 0.76 0.76 0.77 0.77 0.76 0.76 0.78 0.79 0.81

Densities are given i n g/mL at 20°C, except for methane and ethane, whose densities are given at their boiling points.

3-3 Nomenclature of Alkanes H

H

I

H-C-H

I

H

I

H

I

H-C-C-H

I

H

I

H

I

H

H

I

H

I

I

H-C-C-C-H

H

I

H

I

I

H

I

H

I

H-C-C-C-C -H

I

H

H

I

H

H

I

H

I

H

I

H

methane, CH4

CH3

CH3

I

I

CH3 -CH2-CH2-CH2-CH3

CH3 -CH-CH2-CH3

CH3 -C -CH3

I

or H ---( CH2Js H

CH3 isopentane, CSHI2

neopentane, CSHI2

... Figure 3-1 Examples of the general alkane molecular formula, CnH2n+2.

The names methane, ethane, propane, and butane have histori cal roots. From pentane on, alkanes are named using the Greek word for the number of carbon atoms, plus the suffix -ane to identify the molecule as an alkane. Table 3-2 gives the names and physical propertie s of the n-alkanes up to 20 carbon atoms.

3-3A

Com mon Na mes

If all alkane s had unbran ched ( straight-chain) structure s, their nomen clature would be simple. Most alkanes have structural i somers, however, and we need a way of naming all the different i somers. For example, there are two i somers of formula C4H 10. The unbranched isomer is simply called butane (or n-butane, meaning "normal" butane), and the bran ched i somer i s called isobutane, meaning an "i somer of butane." CH3

I

CH3-CH2-CH2-CH3

CH3-CH-CH3

butane (n-butane)

isobutane

The three isomers of CsH 12 are called pentane (or n-pentane), isopentane, and

neopentane.

CH3

I

CH3-CH2-CH2-CH2-CH3

CH3- CH-CH2-CH3

pentane ( n-pentane)

isopentane

CH3

I

CH3-C-CH3

I

CH3 neopentane

Isobutane, isopentane, and neopentane are common names or trivial names, meaning hi storical names arising from common u sage. Common names cannot easily describe the larger, more compli cated molecules having many i somers, however. The number of i somers for any molecular formula grows rapidly as the number of carbon atoms increases. For example, there are 5 stru ctural i somers of hexane, 1 8 i somers of octane, and 75 i somers of decane! We need a sy stem of nomenclature that enables u s to name compli cated molecules without having to memorize hundreds of these hi stor­ i cal common names.

3-3 No menclature of Alka n es

83

84

Chapter 3: Structure and Stereochemi stry of Alkanes 3-3 8

IUPAC or Systematic Na mes

A group of chemists representing the countries of the world met in 1892 to devise a sy stem for naming compounds that would be simple to use, require a minimum of memorization, and yet be flexible enough to name even the most compli cated organi c compounds. This was the first meeting of the group that came to be known as the International Union of Pure and Applied Chemistry, abbreviated IUPAC. This interna­ tional group has developed a detailed system of nomenclature that we call the IUPAC rules. The IUPAC rules are accepted throughout the world as the standard method for naming organic compounds. The names that are generated using this sy stem are called IUPAC names or systematic names. The IUPAC sy stem works con si stently to name many di fferent families of com­ pounds. We will consider the nanling of alkanes in detail, and later extend these rules to other kinds of compounds as we encounter them. The IUPAC sy stem u ses the longest chain of carbon atoms as the main chain, whi ch i s numbered to give the loca­ tions of side chains. Four rules govern this process. Rule 1: The M ain Chain the compound. I

Find the longest continuous chain of carbon atoms, and use the name of this chain as the base name of the compound.

CH2CH3 CH3

I

CH-CH2 -CH2 -CH3 3

4

The first rule of nomen clature gives the base name of

5

6

3-methylhexane

For example, the longest chain of carbon atoms in the compound at left contain s six carbons, so the compound is named as a hexane derivative. The longest chain i s rarely drawn in a straight line; look carefully to find it. The groups attached to the main chain are called substituents because they are sub stituted (in place of a hydrogen atom) on the main chain. When there are two

longest chains of equal length, use the chain with the greater number of substituents as the main chain. The following compound contain s two different seven- carbon chains and i s named as a heptane. We choose the chain on the right as the main chain becau se it has more sub stituents (in red) attached to the chain. CH3

I

CH-CH1-

I

CH - CH-CH2CH3 CH3

CH-CH3

I

CH3 wrong seven-carbon chain, but only three substituents

PROBLEM-SOLVING

HinZ;-

When looking for the lon gest

continuous chain (to give the base name), look to find a l l the different chains of that length. Often, the

correct seven-carbon chain, four substituents

Rule 2: Numbering the Main Chain To give the location s of the sub stituents, assign a number to each carbon atom on the main chain.

Number the longest chai n , begi nning with the end o f the chain neare st a sub sti tuent.

longest chain with the most su bstituents is not obvious.

We start the n umbering from the end nearest a branch so the numbers of the sub­ stituted carbons will be as low as possible. In the preceding heptane stru cture on the right, numbering from top to bottom gives the first bran ch at C3 ( carbon atom 3), but numbering from bottom to top gives the first bran ch at C2. Numbering from bottom to top i s correct. (If each end had a substituent the same di stance in, we would start at the end nearer the second bran ch point.)

iCH CH3

I

�

(meth

3

'CH ----1 C H 2

I

�CH ---.J C H CH3

6

CH3 CH2CH3

I

CH

7CH3

I

-

5CH ----D CH?

I

�CH -----.JC H

CH3

I

3-3 No menc l at ure of A lkanes

CH3

yl)

(meth

7CH 3

incorrect

I

2CH i

d

�

CH2CH3 CH3

(;

)

ethYI

H3

correct 3-ethyl-2,4,S-trimethylheptane

Rule 3: Naming Alkyl Groups

Next, name the substituent groups.

Name the substituent groups attached to the longest chain as alkyl groups. Give the location of each alkyl group by the number of the main-chain carbon atom to which it is attached.

Alkyl groups are named by replacing the -ane suffix of the alkane name with -yl. Methane becomes methyl; ethane becomes ethyl. CH4

methane

CH3-

methyl group

CH3-CH3

ethane

CH3-CH2-

ethyl group

CH3-CH2-CH3

propane

CH3-CH2-CH2-

propyl group

The following alkanes show the use of alkyl group nomenclature. 7 CH? --.JlCH?- --.2CH 3

CH3-CHz i

I

-

CH3 ---1 CH2 ----.J C H ----:! C H2 ---2 C H2 ----D C H

3-medlylhexane

CH3

3-ethy l-6-medlylnonane

Figure 3-2 gives the names of the most common alkyl groups, those having up to four carbon atoms . The propyl and butyl groups are s imply unbranched three­ and four-carbon alkyl groups. These groups are often named as "n-propyl " and "n-butyl " groups, to distinguish them from other kinds of (branched) propyl and butyl groups.

One carbon

CH3-

methyl group

Three carbons

Two carbons

CH3

I

CH3-CH2-

CH3-CH2-CH2-

CH3-CH-

ethyl group

propyl group (or "n-propyl group")

isopropyl group

Four carbons

CH3

I

CH3

I

CH3-CH2-CH2-CH2-

CH3-CH-CH2-

CH3-CH2-CH-

butyl group (or "n-butyl group")

isobutyl group

sec-butyl group

CH3

I

CH 3 -C-

I

CH3

tert-butyl group (or "I-butyl group") � Figure 3-2 Some common alkyl groups.

85

86

Ch apter 3: Structure and Stereochemistry of Alkan es The simple branched alkyl groups are usually known by common names. The i sopropyl and i sobutyl groups have a characteristic "iso" ( CH3 h CH grouping, just as in i sobutane.

�

�

�

C 3

C 3

C 3

CH­ / CH3

CH-CH2 / CH3

CH-CH3 / CH3

isopropyl group

isobutyl group

isobutane

The names of the secondary-butyl (sec-butyl) and tertiary-butyl ( tert-butyl or t-butyl ) groups are based on the degree of alkyl substitution of the carbon atom attached to the main chain. In the sec-butyl group, the carbon atom bonded to the main chain i s secondary (2°), or bonded to two other carbon atoms. In the t-butyl group, it is tertiary (30), or bonded to three other carbon atoms. In both the n-butyl group and the i sobutyl group, the carbon atoms bonded to the main chain are primary (10), bonded to only one other carbon atom. H

R

R

I

I

I R - CI

R -C -

I

H

R -C-

I

H

a primary ( 1 0) carbon

R

a tertiary (3°) carbon

a secondary (2°) carbon

H

I

CHJ. CH CH?-C2 I

CH3

CH3

CH CH -C-

C H3- C -

3

H

2

I

I I

1

CH 3

H

sec- b utyl group (2°)

n -butyI group ( 1 0)

t-butyl group (3°)

SOLVED PROBLEM 3-1

Give the structures of 4-isopropyloctane and 5-t-butyldecane. SOLUTION

4-Isopropyloctane has a chain of eight carbon s, with an isopropyl group on the fourth carbon . 5-t-Butyldecane has a chain of ten carbons, with a t -butyl group on the fifth. CH3 - CH - CH3

I

CH3 - CH2 -CH2 -CH -CH2 - CH2 - CH2 -CH3 4-isopropyloctane

CH3

I

CH3 - C - CH3

I

CH3 -CH2 - CH2 -CH2 - CH- CH2 - CH2 - CH2 -CH2 - CH3 5-I-butyldecane

Haloalkanes can be named just like alkanes, with the halogen atom treated as a sub stituent. Halogen sub stituents are named.fluoro-, chloro-, bromo-, and iodo-.

Br

I

CH3-CH -CH2CH3 2-bromobutane

CH3 CI

I

I

CH3-CH -CH - CH2CH3 3-chloro-2-methylpentane

F

I

CH3-CH-CH2F 1 ,2-difluoropropane

3-3 Nomenclature of Alkanes PROBLEM 3-2

PROBLEM-SOLVING

Name the following alkanes and haloalkanes. CHo- CH3

87

Hinp

Please remember the a l kyl g roups in Figure 3-2. You will encounter them

I -

many times throughout this course.

(a) CH3- CH-CH2- CH3

CH3- CHo- CH2 CH(CH3 )o-

I

I

(c) CH3 - CH2-CH- CH- CH2- CH2-CH3

CH3 CH-CH3

I

CHo- CH3 I I (d) CH3 -CH2- CH2 - CH2 - CH2 -CH2 - CH -CH - CH3 Rule 4: Organizing M ultiple Groups pounds with more than one sub stituent.

The final rule deal s with naming com­

When two or more sub stituents are pre sent, l i st them in alphabetical order. When two or more of the same alkyl sub stituent are present, use the prefixes di-, tri-, tetra-, etc. to avoid having to name the alkyl group twi ce. di- mean s 2

penta- means 5 hexa- means 6

tri- mean s 3

tetra- means 4

Using this rule, we can con struct names for some compli cated structures. Let's finish naming the heptane on p. 84. Thi s compound (at right) has an ethyl group on C3 and three methyl groups on C2, C4, and CS . The ethyl group is listed alphabetically before the methyl groups.

3-ethyl-2,4,5-trimethylheptane

SOLVED PROBLEM 3-2

Give a systematic (IUPAC) name for the following compound. CH3

I

CH- CH3

I

CH2CH3

I

CH3 -CH- CH-CH2 -CH -CH3

I

CH3 - C - CH3

I

CH3 SOLUTION

The longest carbon chain contains eight carbon atoms, so this compound is named as an octane. Numbering from left to right gives the first branch on C2; numbering from right to left gives the first branch on C3, so we number from left to right.

PROBLEM-SOLVING

Hinp

When su bstituents are al phabetized,

CH3 There are four methyl groups: two on C2, one on C3, and one on C6. These four groups will be listed as "2,2,3,6-tetramethyl . . . ." There is an isopropyl group on C4. Listing the isopropyl group and the methyl groups alphabetically, we have 4-isopropyl-2,2,3,6-tetramethyloctane

iso- is used as part of the a l kyl group name, but the hyphenated prefixes are not. Thus isobutyl is a l phabetized with i, but n-butyl, t-butyl, and

sec- butyl are al phabetized with b.

The number prefixes di-, tri-, tetra-, etc. are ig nored in al phabetizing.

88

I,

Chapter 3 : Structure and Stereochemistry of Alkanes SUMMARY

Rules for Naming Alkanes

To name an alkane, we follow four rules: 1. Find the longest continuous chain of carbon atoms, and use this chain as the base name. 2. Number the longest chain, beginning with the end nearest a branch. 3. Name the substituents on the longest chain (as alkyl groups). Give the location of each substituent by the number of the main­ chain carbon atom to which it is attached. 4. When two or more substituents are present, list them in alphabetical order. When two or more of the same alkyl substituent are present, use the prefixes di-, trio, tetra-, and so on (ignored in alphabetizing) to avoid hav ing to name the alkyl group twice.

PROBLEM-SOLVING

HiltP

PROBLEM 3-3

Always com pare the total n u m be r of

Write structures for the following compounds. (b) 3-methyl-S-propylnonane (a) 3-ethyl-3-methylpentane (d) 5-isopropyl-3,3,4-trimethyloctane (c) 4-t-butyl-2-methylheptane

carbon atoms in the name with the n u m ber in the structure to make sure they match. For example, an

isopropyldimethyloctane should have

PROBLEM 3-4

3 + 2 + 8 carbon atoms.

Provide IUPAC names for the following compounds. (b) CH3 - C(CH3 h - CH3

(a) (CH3 hCHCH2CH3

CH2CH3

I

(c) CH3CH2CH2CH-CH(CH3)2

CCCH3)3

I

(e) CH3CH2CHCHCH3

I

CH(CH3)2 PROBLEM 3-5

Give structures and names for (a) the five isomers of C6H 14

(b) the nine isomers of C7 H 16

Complex Substituents Complex alkyl groups are named by a systematic method using the longest alkyl chain as the base alkyl group. The base alkyl group is numbered beginning with the carbon atom (the "head carbon") bonded to the main chain. The substituents on the base alkyl group are listed with appropriate numbers, and parentheses are used to set off the name of the complex alkyl group. The following examples i l lustrate the systematic method for naming complex alkyl groups .

I

CH2CH3 2

1

3

-CH-CH-CH3

1

CH3

a ( l -ethyl-2-methylpropyl) group

I

CH3

1

2

CH3

3

1

4

-C-CH2-CH-CH3

1

CH3

a (l, 1 ,3-trimethylbutyl) group

3 CH3

1

2CH-CH3 CH3CH,

1

-

I

I

CH-CH?CH3 -

1

8 2 2 ICH32CH2 --.JCH---.:t CH2 --.?CH-2CH7CH 2CH9CH 3

3-ethyl-5-( I-ethy1-2-methylpropyl)nonane

1 , I -dimethyl-3-(I, 1 ,3-trimethylbutyl)cyclooctane

3-4 Physical Propert ies of Alkanes

89

PROBLEM 3 - 6

Draw the structures of the following groups, and give their more common names. (b) the (2-methylpropyl) group (d) the ( I , I-dimethylethyl) group

(a) the ( l -methylethyl) group (c) the ( l -methylpropyl) group

PROBLEM 3 - 7

Draw the structures of the following compounds. (a) 4-( 1-methylethyl)heptane (b) 5-(1 ,2,2-trimethylpropyl)nonane PROBLEM 3-8

Without looking at the structures, give a molecular formula for each compound in Problem 3-7. Use the names of the groups to determine the number of carbon atoms, then use the ( 2n + 2 ) rule.

Alkanes are used primarily as fuels, solvents, and lubricants. Natural gas, gasoline, kerosene, heating oil, lubricating oil, and paraffin "wax" are all composed primarily of alkanes, with different physical properties resulting from different ranges of molecular weights.

3-4A

3-4 Physica l Properties of A l ka n es

Sol ubilities and Densities of Al kanes

Alkanes are nonpolar, so they dissolve in nonpolar or weakly polar organic solvents. Alkanes are said to be hydrophobic ("water hating") because they do not dissolve in water. Alkanes are good lubricants and preservatives for metals because they keep water from reaching the metal surface and causing corrosion. Densities of the n-alkanes are listed in Table 3-2 (p. 82). Alkanes have densities around O.7g/mL, compared with a density of l .Og/mL for water. Because alkanes are less dense than water and insoluble in water, a mixture of an alkane (such as gasoline or oil) and water quickly separates into two phases, with the alkane on top.

3-4B

Boiling Points of A l kanes

Table 3-2 also gives the boiling points and melting points of the unbranched alkanes. The boiling points increase smoothly with increasing numbers of carbon atoms and increasing molecular weights. Larger molecules have larger surface areas, resulting in increased intermolecular van der Waals attractions. These increased attractions must be overcome for vaporization and boiling to occur. Thus, a larger molecule, with greater surface area and greater van der Waals attractions, boils at a higher temperature. A graph of n-alkane boiling points versus the number of carbon atoms (the blue line in Figure 3-3) shows the increase in boiling points with increasing molecular weight. Each additional CHz group increases the boiling point by about 30°C up to about ten carbons, and by about 20°C in higher alkanes. The green line in Figure 3-3 represents the boiling points of some branched alkanes. In general, a branched alkane boils at a lower temperature than the n-alkane with the same number of carbon atoms. This difference in boiling points arises because branched al­ kanes are more compact, with less surface area for London force interactions.

3-4C

Melting Points of A l kanes

The blue line in Figure 3-4 is a graph of the melting points of the n-alkanes. Like their boiling points, the melting points increase with increasing molecular weight. The melting point graph is not smooth, however. Alkanes with even numbers of carbon atoms pack better into a solid structure, so that higher temperatures are needed to melt them. Alkanes

Oil floats on water. Note how the oil slick (from the leaking Exxon Valdez) spreads across the top of the water. Oil recovery booms, containing nonpolar fibers, are used to soak up and contain the spilled oil. Note how most of the oil slick ends at the oil recovery booms.

90

Ch apter

3: Struct ure and Stereochemistry of Alkanes

400

G � Figure

0 C '0

200

0..

3-3

on

Alkane boiling points. The boiling points of the unbranched alkanes (blue) are compared with those of some branched alkanes (green). Because of their smaller surface areas, branched alkanes have lower boiling points than unbranched alkanes.

100

;§

.D

'0

CH3-(CH2)n-CH3

300

a

-100 -200

isoal ka nes L a

CH3 '" CH-(CH2)nCH3 / CH3

l5

10

5

20

number of carbon atoms

with odd numbers of carbon atoms do not pack as well, and they melt at lower tempera­ tures. The sawtooth-shaped graph of melting points is smoothed by drawing separate lines (green and red) for the alkanes with even and odd numbers of carbon atoms. Branching of the chain also affects an alkane ' s melting point. A branched alkane generally melts at a higher temperature than the n-alkane with the same number of carbon atoms. Branching of an alkane gives it a more compact three-dimensional structure, which packs more easily into a solid structure and increases the melting point. The boiling points and melting points of three isomers of formula C6H 14 show that the boiling points decrease and the melting points increase as the shape of the molecule becomes more highly branched and compact.

�3

CH3

3 CH3 / �CH-CH

C

CH-CH2-CH 2 -CH 3

/ CH3

I

3 CH-C-CH 3 2-CH

'" CH3

/ CH3

I

CH3

bp 500e mp -98°e

bp 58°e mp -135°e

bp 6Qoe mp -154°e PROBLEM 3-9

List each set of compounds in order of increasing boiling point. (a) hexane, octane, and decane (b) octane, (CH 3hC-C (CH3h and CH 3CH 2C(CH 3 hCH2CH2 CH 3 PROBLEM 3-10

Repeat Problem 3-9, listing the compounds in order of increasing melting point.

50 u o

."§o

0..

� Figure 3-4

Alkane melting points. The melting point curve for n-alkanes with even numbers of carbon atoms is slightly higher than the curve for alkanes with odd numbers of carbons.

�n

'.:j

a

even numbers

-50

� �

-lOa

Q) a -150

odd numbers

-200 a

5

10

15

number of carbon atoms

20

3-5 Uses and Sources of Alkanes Distillation separates alkanes into fractions with similar boiling points. These fractions are suited for different uses based on their physical properties, such as volatility and viscosity. 3-SA

Major Uses of A l kanes

91

3-5 Uses a n d Sources of Al ka nes

(1-(2 The first four alkanes (methane, ethane, propane, and butane) are gases at room temperature and atmospheric pressure. Methane and ethane are difficult to li que­ fy, so they are usually handled as compressed gases. Upon cooling to cryogenic (very low) temperatures, however, methane and ethane become liquids. Liquefied natural gas, mostly methane, can be transported in special refrigerated tankers more easily than it can be transported as a compressed gas.

(r(4 Propane and butane are easily liquefied at room temperature under modest pressure. These gases, often obtained along with l i quid petroleum, are stored in low­ pressure cylinders of liquefied petroleum gas (LPG). Propane and butane are good fuels, both for heating and for internal combustion engines. They burn cleanly, and pollution-control equipment is rarely necessary. In many agricultural areas, propane and butane are more cost-effective tractor fuels than gasoline and diesel fuel. Propane ® and butane have l argely replaced Freons (see Section 6-3D) as propellants in aerosol cans. Unlike alkanes, the chlorofluorocarbon Freon propellants are suspected of dam­ aging the earth ' s protective ozone l ayer.

(5-(8 The next four alkanes are free-flowing, volatile liquids. Isomers of pentane, hexane, heptane, and octane are the primary constituents of gasoline. Their volatility is crucial for this use because the injection system simply s quirts a stream of gasoline into the intake air as it rushes through. If gasoline did not evaporate easily, it would reach the cylinder in the form of droplets. Droplets cannot burn as efficiently as a vapor, so the engine would smoke and give low mileage. In addition to being volatile, gasoline must resist the potentially damaging explo­ sive combustion known as knocking. The antiknock properties of gasoline are rated by an octane number that is assigned by comparing the gasoline to a mixture of n -heptane (which knocks badly) and isooctane (2,2,4-trimethylpentane, which is not prone to knocking). The gasoline being tested is used in a test engine with variable compression ratio. Higher compression ratios induce knocking, so the compression ratio is increased until knocking begins. Tables are available that show the percentage of isooctane in an isooctaneiheptane blend that begins to knock at any given compression ratio. The octane number assigned to the gasoline is simply the percentage of isooctane in an isooctaneiheptane mixture that begins to knock at that same compression ratio. CH3

I

Clean-burning propane-powered vehicles help to reduce air pollution in urban areas.

CH3

I

CH3-C-CH2-CH-CH3

I

CH3 n-heptane (0 octane) prone to knocking

2,2,4-trimethylpentane CLOO octane) "isooctane," resists knocking

(9-(16 The nonanes ( C9 ) through about the hexadecanes ( C 16) are higher-boiling liq­ uids that are somewhat viscous. These alkanes are used in kerosene, jet fuel, and diesel fuel. Kerosene, the lowest-boiling of these fuels, was once widely available but is now harder to find. It is less volatile than gasoline and less prone to forming explosive m ixtures. Kerosene was used in kerosene lamps and heaters, which use wicks to allow this heavier fuel to bum. Jet fuel is similar to kerosene, but more highly refined and less odorous. Diesel fuel is not very volati le, so it does not evaporate in the intake air. In a diesel engine, the fuel is sprayed directly into the cylinder right at the top of the compression stroke. The hot, highly compressed air in the cylinder causes the fuel to burn quickly, swirling and vaporizing as it burns. Some of the alkanes in diesel fuel have fairly high freezing points, and they may solidify in cold weather. This partial

Incomplete combustion of gasoline and

other

motor

fuels

releases

significant quantities of volatile or­ ganic compounds (VOCs) into the atmosphere. VOCs are composed of short-chained alkanes, alkenes, aromatic compounds, and a variety of other hydrocarbons. VOCs are components of air pollution and contribute to cardiac and respira­ tory diseases.

92

Chapter 3: Structure and Stereochemistry of Alkanes solidifi cation causes the diesel fuel to turn into a waxy, semisolid mass. Owners of diesel engines in cold climates often mix a small amount of gasoline with their diesel fuel in the winter. The added gasoline dissolves the frozen alkanes , diluting the slush and allowing it to be pumped to the cylinders. C16 and Up A lkanes with more than 1 6 carbon atoms are most often used as lubri­ cating and heating oils. These are sometimes called "mineral" oils because they come from petroleum, which was once considered a mineral. Paraffin "wax" is not a true wax, but a purified mixture of high-molecular­ weight alkanes with melting points well above room temperature. The true waxes are long-chain esters, discussed in Chapter 25. 3-58

The large distillation tower at left is used to separate petroleum into fractions based on their boiling points. The "cat cracker" at right uses catalysts and high temperatures to crack large molecules into smaller ones.

Al kane Sources; Petroleum Refining

Alkanes are derived mostly from petroleum and petroleum by-products. Petroleum, often called crude oil, is pumped from well s that reach into pockets containing the remains of prehistori c plants. The principal constituents of crude oil are alkanes, some aromati cs, and some undesirable compounds containing sulfur and nitrogen. The com­ position of petroleum and the amounts of contaminants vary from one source to another, and a refinery must be carefully adjusted to process a parti cul ar type of crude oil. Because of their different qualities , different prices are paid for light Arabian crude, West Texas crude, and other classes of crude petroleum. The first step in refining petroleum is a careful fractional distillation. The prod­ u cts of that distillation are not pure alkanes but mixtures of alkanes with useful ranges of boiling points. Table 3-3 shows the major fractions obtained from distillation of crude petroleum. After distillation, catalytic cracking converts some of the less valuable fractions to more valuable products. Catalyti c cracking involves heating alkanes in the presen ce of materials that catalyze the cleavage of l arge molecules into smaller ones. Cracking is often used to convert higher-boiling fractions into mixtures that can be blended with gasoline. When cracking is done in the presence of hydrogen (hydrocracking), the produ ct i s a mixture of alkanes free of sulfur and nitrogen impurities. The following reaction shows the catalytic hydrocracking of a molecule of tetradecane into two molecules of heptane. CH3 - ( CH2 ) 1 2 -CH3

3-5C

+

H2

heat

S'O I 2 or AI 2O3 cataI yst )

2 CH3 - ( CH2 ) s-CH3

Natural Gas; Methane

Natural gas was once treated as a waste product of petroleum produ ction and

destroyed by flaring it off. Now natural gas is an equally valuable natural resource, pumped and stored throughout the world. Natural gas is about 70% methane, 10% ethane, and 1 5 % propane, depending on the source of the gas. Small amounts of other hydro carbons and contaminants are also present. Natural gas is often found above pockets of petroleum or coal, although it is also found in places where there is little TABLE 3-3

Major Fractions Obtained from Distillation of Crude Petroleum

Boiling Range (DC)

Methane hydrate, containi ng methane molecules surrounded by water molecules, is formed under high pressure on the cold seafloor. When brought to the surface, it quickly melts and releases the methane.

Number of Carbons

Fraction

Use

2-4

petroleum gas gasoline kerosene diesel heavy oil petroleum "jelly," paraffin "wax" asphalt

LP gas for heating

under 30° 30°- 1 80° 1 60°-230° 200°-3200 300°-450° > 300° (vacuum)

8-1 6 10- 1 8 1 6-30 >25

residue

>35

4-9

motor fuel heat, jet fuel motor fuel heating, lubrication

3-6 Reactions of Alkanes

93

or no recoverable petroleum or coal. Natural gas is used primarily as a fuel to heat buildings and to generate electricity. It is also important as a staIting material for the production of fertilizers. Although the methane we burn as natural gas is millions of years old, another 300 million tons per year (estimated) of new methane is synthesized by microbes in diverse places such as the stomachs of plant-eating animals and the mud under the seafloor. Most of the undersea methane is eaten by other microbes, but some escapes at methane seeps. Under the sea, cold, high-pressure conditions may allow formation of methane hydrate, with individual methane molecules trapped inside cages of water molecules. When methane hydrate is brought to the surface, it quickly melts and the methane escapes. We currently have no practical methods for capturing and using microbial methane or methane hydrate. Much of this methane escapes to the atmos­ phere, where it acts as a greenhouse gas and contributes to global warming.

Alkanes are the least reactive class of organic compounds. Their low reactivity is reflected in another term for alkanes: paraffins. The name paraffin comes from two Latin terms, parum, meaning "too little," and affinis, meaning "affinity." Chemists found that alkanes do not react with strong acids or bases or with most other reagents. They attributed this low reactivity to a lack of affinity for other reagents, so they coined the name "paraffins." Most useful reactions of alkanes take place under energetic or high-temperature conditions. These conditions are inconvenient in a laboratory because they require speciali zed equipment, and the rate of the reaction is difficult to control. Alkane reac­ tions often form mixtures of products that are difficult to separate. These mixtures may be of commercial importance for an industry, however, where the products may be separated and sold separately. Newer methods of selective functionalization may even­ tually change this picture. For now, however, the following alkane reactions are rarely seen in laboratory applications, but they are widely used in the chemical industry and even in your home and car. 3-6A

3-6 Reactions of Al kanes

Combustion

Combustion is a rapid oxidation that takes place at high temperatures, converting alkanes to carbon dioxide and water. Little control over the reaction is possible, except for moderating the temperature and controlling the fuel/air ratio to achieve efficient burning.

CIlH(21l + 2)

+

excess 0 2

Example

�

heat

n C O2

+

(n

+

1)

Hp

�

heat

Unfortunately, the burning of gasoline and fuel oil pollutes the air and depletes the petroleum resources needed for lubricants and chemical feedstocks. Solar and nuclear heat sources cause less pollution, and they do not deplete these important natural resources. Facilities that use these more environment-friendly heat sources are, how­ ever, more expensive than those that rely on the combustion of alkanes. 3-68

Cracking and Hydrocracking

As discussed in Section 3 - S B , catalytic cracking of large hydrocarbons at high temperatures produces smaller hydrocarbons. The cracking process usually oper­ ates under conditions that give the maximum yields of gasoline. In hydrocracking, hydrogen is added to give saturated hydrocarbons; cracking without hydrogen gives mixtures of alkanes and alkenes.

Combustion is the most common reaction of alkanes. Lightning initiated this fire in a tank containing 3 million gallons of gasoline at the Shell Oil storage facility in Woodbridge, NJ (June l l , 1 996).

94

Chapter 3: Structure and Stereochemistry of Alkanes Catalytic hydrocracking

H" -

C I 2H26

heat

catalyst

long-chain alkane

)

C7 H I 6

shorter-chain alkanes Catalytic cracking

heal

C I 2H26

long-chain alkane

catalyst

)

C 7H I 6

shorter-chain alkanes and alkenes

3-6C

Halogenation

Alkanes can react with halogens ( F2 C1 2 , Brb 12 ) to form alkyl halides. For example, ' methane reacts with chlorine ( CI 2 ) to form chloromethane (methyl chloride), dichloromethane (methylene chloride), trichloromethane (chloroform), and tetra­ chloromethane (carbon tetrachloride).

Heat or light is usually needed to initiate this halogenation. Reactions of alkanes with chlorine and bromine proceed at moderate rates and are easily controlled. Reactions with fluorine are often too fast to control, however. Iodine reacts very slowly or not at all. We will discuss the halogenation of alkanes in Chapter 4.

3-7 Structu re a n d Conformations of Alka n es

Although alkanes are not as reactive as other classes of organic compounds, they have many of the same structural characteristics. We will use simple alkanes as examples to study some of the properties of organic compounds, including the structure of sp3 hybridized carbon atoms and properties of C-C and C- H single bonds.

3-7A

Structure of Methane

The simplest alkane is methane, CH4. Methane is perfectly tetrahedral, with the 1 09.5° bond angles predicted for an sp3 hybrid carbon. Four hydrogen atoms are covalently bonded to the central carbon atom, with bond lengths of 1 .09 A.

3-7 Structure and Conformations of Alkanes 3-7 B

Conformations of Ethane

Ethane, the two-carbon alkane, is composed of two methyl groups with overlapping sp3 hybrid orbitals forming a sigma bond between them.

ethane

ethane

ethane

The two methyl groups are not fixed in a single position but are relatively free to rotate about the sigma bond connecting the two carbon atoms. The bond maintains its linear bonding overlap as the carbon atoms turn. The different arrangements formed by rotations about a single bond are called conformations, and a specific conforma­ tion is called a conformer ( "conformational isomer"). Pure conformers cannot be isolated in most cases, because the molecules are constantly rotating through all the possible conformations. rotate

overlap maintained

linear overlap of sigma bond

rotate

In drawing conformations, we often use Newman projections, a way of drawing a molecule looking straight down the bond connecting two carbon atoms (Fig. 3-5). The front carbon atom is represented by three lines (three bonds) coming together in a

H

-----

H

viewed from the end

perspective drawing

Newman projection

.... Figure 3-5

The Newman projection looks straight down the carbon-carbon bond.

95

96

Chapter 3 : Structure and Stereochemistry of Alkanes Newman projections:

(}

(\

=

0°

(}

,:\

H�H �-4HH H H H HH H H H H HH H H H H H H

Sawhorse structures: � Fig u re 3-6

Ethane conformations. The eclipsed conformation has a dihedral angle e 0°, and the staggered conformation has e 60°. Any other conformation is called a skew conformation. =

=

eclipsed, (} 0°

staggered, (} 60°

=

=

skew, e

=

anything else

Y shape. The back carbon is represented by a circle with three bonds pointing out from it. Until you become familiar with Newman projections, you should make models and compare your models with the drawings. An infinite number of conformations are possible for ethane, because the angle between the hydrogen atoms on the front and back carbons can take on an infinite number of values. Figure 3-6 uses Newman projections and sawhorse structures to illustrate some of these ethane conformations. Sawhorse structures picture the mole­ cule looking down at an angle toward the carbon-carbon bond. Sawhorse structures can be misleading, depending on how the eye sees them. We will generally use per­ spective or Newman projections to draw molecular conformations. Any conformation can be specified by its dihedral angle (e), the angle between the C bonds on the front carbon atom and the C bonds on the back carbon in the Newman projection. Two of the conformations have special names. The conforma­ tion with e 0° is called the eclipsed conformation because the Newman projection shows the hydrogen atoms on the back carbon to be hidden (eclipsed) by those on the front carbon. The staggered conformation, with e 60°, has the hydrogen atoms on the back carbon staggered halfway between the hydrogens on the front carbon. Any other intermediate conformation is called a skew conformation. In a sample of ethane gas at room temperature, the ethane molecules are rotating and their conformations are constantly changing. These conformations are not all equally favored, however. The lowest-energy conformation is the staggered conforma­ tion, with the electron clouds in the C bonds separated as much as possible. The eclipsed conformation places the C - electron clouds closer together; it is about 1 2.6 kJ/mol (3.0 kcal/mol) higher in energy than the staggered conformation. Three kilocalories is not a large amount of energy, and at room temperature, most molecules have enough kinetic energy to overcome this small rotational barrier. Figure 3-7 shows how the potential energy of ethane changes as the carbon-carbon bond rotates. The y axis shows the potential energy relative to the most stable (staggered) conformation. The x axis shows the dihedral angle as it increases from 0° (eclipsed) through 60° (staggered) and on through additional eclipsed and staggered conformations as e continues to increase. As ethane rotates toward an eclipsed conformation, its poten­ tial energy increases, and there is resistance to the rotation. This resistance to twisting (torsion) is called torsional strain, and the 1 2.6 kJ/mol (3.0 kcal/mol) of energy required is called torsional energy. -

H

-

=

=

HH

-

H

3-7 Structure and Conformations of Alkanes

97

L-----�----���--L---� e

$H�OH H H

dihedral angle .... Figure 3-7

The torsional energy of ethane is lowest in the staggered conformation. The eclipsed conformation is about 1 2.6 kllmol (3.0 kcaJ/mol) higher in energy. At room temperature, this barrier is easily overcome and the molecules rotate constantly.

H

H

Conformational analysis is the study of the energetics of different confor­ mations. Many reactions depend on a molecule's ability to twist into a particular conformation; conformational analysis can help to predict which conformations are favored and which reactions are more likely to take place. We will apply conformational analysis to propane and butane first, and later to some interesting cycloalkanes .

3-7C

Conformations of Propa ne

Propane is the three-carbon alkane, with formula C3 HS. Figure 3-8 shows a three­

dimensional representation of propane and a Newman projection looking down one of the carbon-carbon bonds. Figure 3-9 shows a graph of the torsional energy of propane as one of the car­ bon-carbon bonds rotates. The torsional energy of the eclipsed conformation is about 1 3 . 8 kJ/mol (3.3 kcal/mol), only 1 .2 kJ (0.3 kcal) more than that requ ired for ethane. Apparently, the torsional strain resulting from eclipsing a carbon-hydrogen bond with a carbon-methyl bond is only 1 .2 kJ (0.3 kcaJ) more than the strain of eclipsing two carbon-hydrogen bonds.

PROBLEM-SOLVING

Htni::

A C - H bond eclipsed with another C - H bond contributes 4.2 kJlmol

PROBLEM 3- 1 1

(1 .0 kcal/mol) torsional energy

Draw a graph, similar to Figure 3-9, of the torsional strain of 2-methylpropane as it rotates about the bond between C l and C2. Show the dihedral angle and draw a Newman projection for each staggered and eclipsed conformation.

(one-third of eclipsed ethane). A C - H bond ecli psed with a C - CH3 bond contributes 5.4 kJ/mol ( 1 .3 kcal/mol).

H

-

:$: CH3

viewed from the end

---- -------

perspective drawing ---

-

Newman projection

.... Figure 3-8

Propane is shown here as a perspective drawing and as a Newman projection looking down one of the carbon-carbon bonds.

98

Chapter 3 : Structure an d S tere oc he mis try of Alkanes

�----�----��- e

0°

60°

1 20°

dihedral angle

� Fig u re 3-9

Torsional energy of propane. When a C - C bond of propane rotates, the torsional energy varies much like it does in ethane, but with 13.8 kllmol (3.3 kcal/mol) torsional energy in the eclipsed conformation.

3- 8

Butane is the four-carbon alkane, with molecular formula C4H 10. We refer to n-butane

as a straight-chain alkane, but the chain of carbon atoms is not really straight. The angles between the carbon atoms are close to the tetrahedral angle, about 1 09.5°. Rotations about any of the carbon-carbon bonds are possible; Figure 3 -1 0 shows Newman projections, looking along the central C2 -C3 bond, for four conformations of butane. Construct butane with your molecular models, and sight down the C2 -C3 bond. Notice that we have defined the dihedral angle e as the angle between the two end methyl groups. Three of the conformations shown in Figure 3 -1 0 are given special names. When the methyl groups are pointed in the same direction (e 0° ) , they eclipse each other. This conformation is called totally eclipsed, to distinguish it from the

Conformati o n s o f B uta ne

=

.-

_

. =:::::-=---=====--=========-===-==-===============

H H

�b: y

H

3

H

H

CH3

H

totally eclipsed (0°) .& Figure

gauche (60°)

$ CH3

eclipsed ( 1 20°)

anti ( 1 80°)

3-10

Butane conformations. Rotations about the center bond in butane give different molecular shapes. Three of these conformations have specific names. ======

H 1 800

H

3-8 Conformations of Butane other eclipsed conformations like the one at e 1 20°. At e 60°, the butane mole­ cule is staggered and the methyl groups are toward the left and right of each other. This 60° conformation is called gauche (pronounced gosh), a French word meaning "left" or "awkward." Another staggered conformation occurs at e 1 80°, with the methyl groups pointing in opposite directions. This conformation is called anti because the methyl groups are "opposed." =

=

=

3-8A

Torsional Energy of B uta ne

A graph of the relative torsional energies of the butane conformations is shown in Figure 3 - 1 1 . All the staggered conformations (anti and gauche) are lower in energy than any of the eclipsed conformations. The anti conformation is lowest in energy because it places the bulky methyl groups as far apart as possible. The gauche confor­ mations, with the methyl groups separated by just 60°, are 3 . 8 kJ (0.9 kcal) higher in energy than the anti conformation because the methyl groups are close enough that their electron clouds begin to repel each other. Use your molecular models to compare the crowding of the methyl groups in these conformations .

3-8B

Steric Strai n

The totally eclipsed conformation is about 6 kJ 0 .4 kcal) higher in energy than the other eclipsed conformations because it forces the two end methyl groups so close together that their electron clouds experience a strong repulsion. This kind of interfer­ ence between two bulky groups i s called steric strain or steric hindrance. The fol­ lowing structure shows the interference between the methyl groups in the totally eclipsed conformation.

I S kJ

-2 1 kJ

(3.6 kcal ) (5 kcal)

1 20°

1 80°

I S kJ

lowest

eclipsed

anti

(5 kcal)

60° 3.8 kJ (0.9 kcal)

(3.6 kcal)

totally eclipsed

gauche

0° - 2 1 kJ

0.00

H��:3 "Q):,� 1-1 I-I I-I Y H H

240° 15 kJ (3.6 kcal)

300° 3.8 kJ (0.9 kcal)

(5 kcal)

eclipsed

gauche

totally eclipsed

360° -21 kJ

1-1'3C �1-I H YI H -l

.. Figure 3-1 1

Torsional energy of butane. The anti conformation is lowest in energy, and the totally eclipsed conformation is highest in energy.

()

99

1 00

Ch ap te r 3: Structu re and Ste reochemist ry of Alkanes

Totally eclipsed conformation of butane

PROBLEM-SOLVING

HiltZ;-

A C - CH3 bond ecli psed with another C - CH3 bond contributes a bout 1 2-1 3 kJ/mol (3 kcal/mol) torsional

Rotating the totally eclipsed conformation 60° to a gauche conformation releas­ es most, but not all, of this steric strain. The gauche conformation is still 3. 8 kJ (0.9 kcal) h igher in energy than the most stable anti conformation. What we have learned about the conformations of butane can be appl ied to other alkanes. We can predict that carbon-carbon s ingle bonds w il l assume staggered con­ formations whenever possible to avoid ecl ipsing of the g roups attached to them. Among the staggered conformations, the anti conformation is p referred because it has the lowest tors ional energy. We must remember, however, that there is enough thermal energy p resent at room temperature for the molecules to rotate rapidly among all the d ifferent conformations. The relative stabil ities are important because more molecules w il l be found in the more stable conformations than in the less stable ones.

energy. (Tot a l l y ecl i psed butane is

PROBLEM 3-1 2

about 21 kJ/mol with 4.2 kJ/mol for each of the two interactions between C - H bonds, leaving 1 2.6 kJ/mol for the methyl-methyl ecl i psing.)

The h igher alkanes resemble butane in the ir p reference for anti and gauche conforma­ tions about the carbon-carbon bonds. The lowest-energy conformation for any straight-chain alkane is the one w ith all the internal carbon-carbon bonds in its anti conformations. These anti conformations g ive the chain a z igzag shape. At room tem­ perature, the internal carbon-carbon bonds undergo rotation, and many molecules contain gauche conformations. Gauche conformations make kinks in the z igzag struc­ ture. Nevertheless, we frequently d raw alkane chains in a z igzag structure to rep resent the most stable arrangement.

3-9 Conformatio n s o f H i g h e r A l ka nes

H H

� � R

H

H

H

H

R

anti conformation

Draw a graph, similar to Figure 3 - 1 1 , of the torsional energy of 2-methylbutane as it rotates about the C2 - C3 bond.

�

R

� H

H

gauche conformation

octane, all anti conformation

PRO BLEM 3- 1 3 Draw a perspective representation of the most stable conformation of 3-methylhexane .

. 3-1 0 Cycl oa l ka n es

Many organic compounds are cyclic: They contain rings of atoms. The carbohydrates we eat are cyclic, the nucleotides that make up our DNA and RNA are cycl ic, and the antibiotics we use to treat diseases are cyc l ic. In this chapter, we use the cycloalkanes to illustrate the properties and stabil ity of cyc l ic compounds.

1 01

3 - 1 0 Cycloalkanes H H \ / /H H C \ H-C / "-- C -H \ ../"H H,,-- / C C H../" \ / "-- H c-c H ../" I I "-- H H H

H-C-C-H

H H \ / H /C" /H \ ' C C / "-- H H/ \ H-C- C-H

or

or

or

or

or

cyclopropane

D

cyclobutane

o

cyclopentane

cyclohexane

cycloheptane

C3H6

C4Hg

Cs H , o

o C6H '2

o C7H ' 4

H

H

I

I

H

H

I

I

H-C-C-H

I

I

I

H

I

H

... Figure 3- 1 2 Structures of some cycloalkanes.

Cycloalkanes are alkanes that contain rings of carbon atoms. Simple cycloalkanes are named like acyclic (noncyclic) alkanes, with the prefix cyclo- indicating the presence of a ring. For example, the cycloalkane with four carbon atoms in a ring is called cyclobutane. The cycloalkane with seven carbon atoms in a ring is cycloheptane. Line-angle formulas are often used for drawing the rings of cycloalkanes (Figure 3 - 1 2).

3-1 0A

General Molecular Form ulas of Cycloa l kanes

Simple cycloalkanes are rings of CH2 groups (methylene groups). Each one has exactly twice as many hydrogen atoms as carbon atoms, giving the general molecular formula CnH21l" This general formula has two fewer hydrogen atoms than the ( 2n + 2) formula for an acyclic alkane because a ring has no ends, and no hydrogens are needed to cap off the ends of the chain.

Cyclopropane was once used as general

anesthetic

because

its

vapors, like those of other simple alkanes

and

cycloalkanes,

cause

sleepiness and a loss of conscious­ ness.

After

inhalation

into

the

lungs, cyclopropane goes into the blood. Due to its nonpolar nature,

3-1 0 8

it rapidly leaves the blood

Physical Properties of Cycloa l kanes

Most cycloalkanes resemble the acyclic (noncyclic), open-chain alkanes in their phys­ ical properties and in their chemistry. They are nonpolar, relatively inert compounds with boiling points and melting points that depend on their molecular weights. The cycloalkanes are held in a more compact cyclic shape, so their physical properties are similar to those of the compact, branched alkanes. The physical properties of some common cycloalkanes are listed in Table 3-4. TABLE 3-4

Physica l Properties o f Some Simple Cycloalkanes

Cycloalkane

Formula

Boiling Point (0C)

Melting Point (OC)

Density

cyclopropane cyclobutane cyclopentane cyclohexane cycloheptane cyclooctane

C3 H6 C4Hg C s H IO C6 H ' 2 C7 H '4 CgH16

-33 -12 49 81 1 18 1 48

- 1 28 -50 -94 7 - 12 14

0.72 0.75 0.75 0.78 0.8 1 0.83

and

passes through the nonpolar mem­ branes

surrounding

the

central

nervous system, where it produces anesthesia. longer

Cyclopropane

used

as

an

is

no

anesthetic

because it is highly flammable (like ether) and

can cause explosions

when mixed with air.

102

Chapter 3 : Structure and Stereochemistry of Alkanes 3- 10C

Nomen clature of Cycloalkanes

Cycloalkanes are named much l ike acycl ic alkanes. Substituted cycloalkanes use the cycloalkane for the base name, w ith the alkyl groups named as substituents. If there is j ust one substituent, no numbering is needed.

a

CH3 CH3

I

C-C-CH3

I

H

I

H

( 1 ,2-dimethylpropyl)cyclohexane

t-butylcycloheptane

methylcyclopentane

\

If there are two or more s ubstituents on the ring, the r ing carbons are num­ bered to g ive the lowest poss ible numbers for the substituted carbons. The number­ ing beg ins w ith one of the substituted r ing carbons and continues in the d irection that g ives the lowest pos s ible numbers to the other substituents. In the name, the subst ituents are l isted in alphabetical order. When the numbering could begin w ith e ither of two alkyl groups (as in a d isubst ituted cycloalkane), begin w ith the one that is alphabetically first.

q

H

H'

H

CH3

l -ethyl-2-methylcyclobutane

1 , 1 ,3-trimethylcyclopentane

1 , I -dieth y 1-4-isopropy lcyclohexane

When the acyclic portion of the molecule contains more carbon atoms than the cyc l ic portion (or when it contains an important functional group), the cycl ic portion is sometimes named as a cycloalkyl substituent.

I

H-C

AC CH' CH,IUP4�;' CH3

4-cycl opropy 1- 3-meth y loctane

3

4

C-CH2-CH2-CH2 5

cyclopentylcyclohexane

5-cyclobutyl- l -pentyne

PROBLEM 3-14 Give

(al

names for the fol lowing compounds.

(c)

PRO B L E M 3- 1 5

Draw the structure and give the molecular formula for each of the fol lowing compounds. (b) propylcyclohexane (a) cyclododecane (d) 3-ethyl - l , l -dimethylcyclohexane (c) cyclopropylcyclopentane

3- 1 2 Stabilities of CycIoalkanes; Ring Strain Open-chain alkanes undergo rotations about their carbon-carbon single bonds, so they are free to assume any of an infinite number of conformations. Alkenes have rigid double bonds that prevent rotation, giving rise to cis and trans isomers with dif­ ferent orientations of the groups on the double bond (Section 2-8). Cycloalkanes are similar to alkenes in this respect. A cycloalkane has two distinct faces. If two sub­ stituents point toward the same face, they are cis. If they point toward opposite faces, they are trans. These geometric isomers cannot interconvert without breaking and re-forming bonds. Figure 3- 1 3 compares the cis-trans isomers of 2-butene with those of 1 ,2-dimethylcyclopentane. Make models of these compounds to convince yourself that cis- and trans- l ,2-dimethylcyclopentane cannot interconvert by simple rotations about the bonds.

1 03

3-1 1 cis-tra n s Isomerism i n Cycloa l ka n es

PROBLEM 3-1 6 Which of the following cycloalkanes are capable of geometric (cis-trans) isomerism? Draw the cis and trans isomers. (b) l ,4-dimethylcyclohexane (a) 3-ethyl- l , l -dimethylcyclohexane (c) l -ethyl-3-methylcyclopentane (d) l -cyclopropyl-2-methylcyclohexane

PROBLEM 3- 1 7 Give

(a)

H

� � 'II

names for the following cycJoalkanes.

IUPAC

�

. ."P-

C(CH3 ) 3

CH3

(b)

CH2CH2CH3

CH2CH3

H

H '" / C=C / '" CH3 H3 C

H3 C H ", / C=C '" / CH3 H

cis-2-butene

trans-2-butene

... Figure 3-1 3

() - � � 'H'

H H3 C

�H CH3

CH3

CH3

cis- l ,2-di methy IcycIopentane

H3 C

() = � ++ R H

H CH3

H

CH3

trans- l ,2-dimethylcyclopentane

Although all the simple cycloalkanes (up to about C20) have been synthesized, the most common rings contain five or six carbon atoms. We will study the stabilities and conformations of these rings in detail because they help to determine the properties of many important organic compounds. Why are five-membered and six-membered rings more common than the other sizes? Adolf von B aeyer first attempted to explain the relative stabilities of cyclic molecules in the late nineteenth century, and he was awarded a Nobel Prize for this work in 1 905. Baeyer reasoned that the carbon atoms in acyclic alkanes have bond an­ 3 gles of 1 09.5°. (We now explain this bond angle by the tetrahedral geometry of the sp hybridized carbon atoms.)

Cis-trans isomerism in cycJoalkanes. Like alkenes, cycloalkane rings are restricted from free rotation. Two substituents on a cycloalkane must be either on the same side (cis) or on opposite sides (trans) of the ring.

3-1 2 Sta b i l ities of Cycloa l kan es; R i n g Stra i n

104

Ch apter 3: S tru ctu re and S tereoche mi stry of Alk anes

1 09.5°

� Figure 3-14

tetrahedral angle

The ring strain of a planar cyclobutane results from two factors: angle strain from the compressing of the bond angles to 90° rather than the tetrahedral angle of 1 09.5°, and torsional strain from eclipsing of the bonds.

____

!.. _ _ _ _ _

I I

/

I

�/

---...

1 9.5°

n� angle I On �/ compress

Newman projection of planar cyclobutane

If a cycloalkane requires bond angles other than 1 09.5°, the orbitals of its carbon-carbon bonds cannot achieve optimum overlap, and the cycloalkane must have some angle strain (sometimes called Baeyer strain) associated with it. Figure 3-14 shows that a planar cyclobutane, with 90° bond angles, is expected t o have significant angle strain. In addition to this angle strain, the Newman projection in Figure 3 - 1 4 shows that the bonds are eclipsed, resembling the totally eclipsed conformation of butane (Section 3-7). This eclipsing of bonds gives rise to torsional strain. Together, the angle strain and the torsional strain add to give what we call the ring strain of the cyclic compound. The amount of ring strain depends primarily on the size of the ring. Before we discuss the ring strain of different cycloalkanes, we need to consider how ring strain is measured. In theory, we should measure the total amount of energy in the cyclic compound and subtract the amount of energy in a similar, strain-free ref­ erence compound. The difference should be the amount of extra energy due to ring strain in the cyclic compound. These measurements are commonly made using heats

of combustion.

3-12A

Heats of Combustion

The heat of combustion is the amount of heat released when a compound is burned with an excess of oxygen in a sealed container called a bomb calorimeter. If the com­ pound has extra energy as a result of ring strain, that extra energy is released in the combustion. The heat of combustion is usually measured by the temperature rise in the water bath surrounding the "bomb." A cycloalkane can be represented by the molecular formula (CH 2) m so the general reaction in the bomb calorimeter is:

n C02

+

n H20

+

n(energy per CH2) heat of combustion

The molar heat of combustion of cyclohexane is nearly twice that of cyclo­ propane, simply because cyclohexane contains twice as many methylene (CH2) groups per mole. To compare the relative stabilities of cycloalkanes, we divide the heat of combustion by the number of methylene (CH2) groups . The result is the energy per CH2 g roup. These normalized energies allow us to compare the relative amounts of ring strain (per methylene g roup) in the cycloalkanes.

3- 1 2 Stabi l i ti es of Cycloalkanes; Ring Strain TABLE 3-5

1 05

Heats of Combustion (per Mole) for Some Simple Cycloalkanes

Ring Size

Cycloalkane

3 cyclopropane 4 cyclobutane 5 cyclopentane 6 cyclohexane 7 cycloheptane 8 cyclooctane reference: long-chain alkane

Molar Heat of Combustion

Heat of Combustion per CH2 Group

Ring Strain per CH2 Group

Total Ring Strain

209 1 kJ 2744 kJ 3320 kJ 395 1 kJ 4637 kJ 5309 kJ

697 . 1 kJ 686 . 1 kJ 664.0 kJ 658.6 kJ 662.4 kJ 663.6 kJ 658.6 kJ

38.5 kJ 27.5 kJ 5.4 kJ 0.0 kJ 3.8 kJ 5 . 1 kJ 0.0 kJ

1 1 5 kJ (27.6 kcal) 1 10 kJ (26.3 kcal) 27 kJ (6.5 kcal) 0.0 kJ (0.0 kcal) 27 kJ (6.4 kcal) 4 1 kJ (9.7 kcal) 0.0 kJ (0.0 kcal)

Table 3-5 shows the heats of combustion for some simple cycloalkanes. The refer­ ence value of 658.6 kJ ( 1 57.4 kcal) per mole of CH2 groups comes from an unstrained long-chain alkane. The values show large amounts of ring strain in cyclopropane and cyclobutane. Cyclopentane, cycloheptane, and cyclooctane have much smaller amounts of ring strain, and cyclohexane has no ring strain at all. We will discuss several of these rings in detail to explain this pattern of ring strain. 3-1 2 8

Cyclopropane

Table 3-5 shows that cyclopropane bears more ring strain per methylene group than any other cycloalkane. Two factors contribute to this l arge ring strain. First is the angle strain required to compress the bond angles from the tetrahedral angle of 1 09.5° to the 3 60° angles of cyclopropane. The bonding overlap of the carbon--carbon sp orbitals i s 3 weakened when the bond angles differ s o much from the tetrahedral angle. The sp orbitals cannot point directly toward each other, and they overlap at an angle to form weaker "bent bonds" (Fig. 3 - 1 5). Torsional strain is the second factor i n cyclopropane ' s large ring strain. The three-membered ring i s planar, and all the bonds are eclipsed. A Newman projection of one of the carbon--carbon bonds (Fig. 3- 1 6) shows that the conformation resembles the totally eclipsed conformation of butane. The torsional strain in cyclopropane is not as great as its angle strain, but it helps to account for the large total ring strain. Cyclopropane i s generally more reactive than other alkanes. Reactions that open the cyclopropane ring release 1 1 5 kJ (27.6 kcal) per mole of ring strain, which pro­ vides an additional driving force for these reactions. P R O B L E M 3-1 8

The heat of combustion of cis- l ,2-dimethy Icyclopropane is larger than that of the trans isomer. Which isomer is more stable? Use drawings to explain this difference in stability.

�--

1 09.5° tetrahedral

,

�---. ,'

,

,

,

angle

,

..... Figure 3-1 5

,

&' ,

- - - - _ - - - _ _ _ _

600

" 49.5° angle . , compressIOn

"bent bonds" nonlinear overlap

Angle strain i n cyclopropane. The bond angles are compressed to 60° from the usual 1 09.5° bond angle of 3 sp hybridized carbon atoms. This severe angle strain leads to nonlinear 3 overlap of the sp orbitals and "bent bonds."

106

Chapter 3: Structure and Stereochemistry of Alkanes

Newman projection of cyclopropane

A Figure 3- 1 6

Torsional strain i n cyclopropane. All the carbon-carbon bonds are eclipsed, generating torsional strain that contributes to the total ring strain. 3-12C

Cyclobutane

The total ring strain in cyclobutane is almost as great as that in cyclopropane, but is distributed over four carbon atoms. If cyclobutane were perfectly planar and square, it would have 90° bond angles. A planar geometry requires eclipsing of all the bonds, however, as in cyclopropane. To reduce this torsional strain, cyclobutane actually assumes a slightly folded form, with bond angles of 88°. These smaller bond angles require slightly more angle strain than 90° angles, but the relief of some of the tor­ sional strain appears to compensate for a small increase in angle strain (Figure 3- 17). PROBLEM 3-1 9

trans- l ,2-Dimethylcyclobutane is more stable than cis- l ,2-dimethylcyclobutane, but cis­ I ,3-dimethylcyclobutane is more stable than trans- l ,3-dimethylcyclobutane. Use drawings to explain these observations. 3-1 2 0 The conformation of cyclopentane

If cyclopentane had the shape of a planar, regular pentagon, its bond angles would be 1 08°, close to the tetrahedral angle of 1 09.5°. A planar structure would require all the bonds to be eclipsed, however. Cyclopentane actually assumes a slightly puckered "envelope" conformation that reduces the eclipsing and lowers the torsional strain (Figure 3-1 8). This puckered shape is not fixed, but undulates by the thermal up -and­ down motion of the five methylene groups. The "flap" of the envelope seems to move around the ring as the molecule undulates.

is important because ribose and deoxyribose,

the

sugar

compo­

nents of RNA and DNA, respective­ ly, assume cyclopentane-like ring conformations.

These

Cyclopentane

conforma­

tions are crucial to the properties and reactions of R N A and DNA.

not quite eclipsed

H

� ,

H slightly folded conformation A Figure 3-1 7

H

U

t quite eclipsed

Newman projection of one bond

The conformation of cyclobutane is slightly folded. Folding gives partial relief from the eclipsing of bonds, as shown in the Newman projection. Compare this actual structure with the hypothetical planar structure in Figure 3- 1 4.

3- 1 3 Cyc l oh exane Conformations H

H

H

H

... F i g u re 3 - 1 8

H

"flap" folded upward

The conformation o f cyclopentane i s slightly folded, like the shape o f an envelope. This puckered conformation reduces the eclipsing of adjacent CH2 groups.

Newman projection showing relief of eclipsing of bonds

viewed from

We will cover the conformations of cyclohexane in more detail than other cycloalkanes because cyclohexane ring systems are particularly common. Carbohy­ drates, steroids, plant products, pesticides, and many other important compounds contain cyclohexane-like rings whose conformations and stereochemistry are criti­ cally important to their reactivity. The abundance of cyclohexane rings in nature i s probably due t o both their stability and the selectivity offered b y their predictable conformations. The combustion data (Table 3-5) show that cyclohexane has no ring strain. Cy­ clohexane must have bond angles that are near the tetrahedral angle (no angle strain) and also have no eclipsing of bonds (no torsional strain). A planar, regular hexagon would have bond angles of 1 20° rather than 1 09.5°, implying some angle strain. A pla­ nar ring would also have torsional strain because the bonds on adjacent CH2 groups would be eclipsed. Therefore, the cyclohexane ring cannot be planar. 3- 1 3A

1 07

3-1 3 Cyclohexa ne Confo rmations

Chair and Boat Confor m ations

Cyclohexane achieves tetrahedral bond angles and staggered conformations by assum­ ing a puckered conformation. The most stable conformation is the chair conforma­ tion shown in Figure 3- 1 9. Build a molecular model of cyclohexane, and compare its shape with the drawings in Figure 3- 1 9. In the chair conformation, the angles between

H

chair conformation

� H

H

H H

H

---

I H

H

H

�

"k 'ir _ l:f -

- - - - - . - -

H

- - -

H

_ _

viewed along the "seat" bonds

H H

H

H

yfy yfy � � H

CH2 CH2

H

Newman projection

... Figure 3- 1 9

H H

The chair conformation of cyclohexane has one methylene group puckered upward and another puckered downward. Viewed fro m the Newman projection, the chair has no eclipsing of the carbon-carbon bonds. The bond angles are 1 09.5°.

1 08

Chapter 3 : Structure and Stereochemistry of Alkanes

"flagpole" hydrogens

H H

�H

� �ii : � : = H

H

symmetrical boat

boat conformation

� Figure 3-20

eclipsed

In the symmetrical boat conformation of cyclohexane, eclipsing of bonds results in torsional strain. In the actual molecule, the boat is skewed to give the twist boat, a conformation with less eclipsing of bonds and less interference between the two flagpole hydrogens.

The

conformations

of

biological

molecules are critical for their ac­ tivities. For example, steroids will fit into their receptors in only one conformation. The correct fit acti­ vates the receptor, reSUlting in a biological activity.

� 'Jr

H

t

H H

CH' C H,

HH

1Yf 'j(

H H

HH

Newman projection

"twist" boat

the carbon-carbon bonds are all 1 09.5°. The Newman p rojection looking down the "seat" bonds shows all the bonds in staggered conformations. The boat conformation of cyclohexane (Figure 3-20) also has bond angles of 1 09.5° and avoids angle strain. The boat conformation resembles the chair conformation except that the "footrest" methylene group is folded upward. The boat conformation suffers from torsional strain, however, because there is eclipsing of bonds. This eclipsing forces two of the hydrogens on the ends of the "boat" to interfere with each other. These hydrogens are called flagpole hydrogens because they point upward from the ends of the boat like two flagpoles. The Newman p rojection in Figure 3-20 shows this eclipsing of the carbon-carbon bonds along the sides of the boat. A cyclohexane molecule in the boat conformation actually exists as a slightly skewed twist boat conformation, also shown in Figure 3-20. If you assemble your molecular model in the boat conformation and twist it slightly, the flagpole hyd rogens move away from each other and the eclipsing of the bonds is reduced. Even though the twist boat is lower in energy than the symmetrical boat, it is still about 23 kJ/mol (5.5 kcal/mol) higher in energy than the chair conformation. When someone refers to the "boat conformation," the twist boat (or simply twist) conformation is often intended. At any instant, most of the molecules in a cyclohexane sample are in chair con­ formations. The energy barrier between the boat and chai r is sufficiently low, howev­ e r, that the conformations interconvert many times each second. The interconversion from the chai r to the boat takes place by the footrest of the chair flipping upward and forming the boat. The highest-energy point in this process is the conformation where the footrest is planar with the sides of the molecule. This unstable arrangement is called the half-chair conformation. Figure 3-2 1 shows how the energy of cyclohex­ ane varies as it interconverts between the boat and chai r forms.

3-1 3 8

Axial and Eq uatorial Positions

If we could freeze cyclohexane in the chair conformation, we would see that there are two different kinds of carbon-hydrogen bonds. Six of the bonds (one on each carbon atom) are directed up and down, parallel to the axis of the ring. These are called axial bonds. The other six bonds point out from the ring, along the "equator" of the ring.

3- 1 3 Cyc10hexane Conformations

4 L==?6 I 3 half-chair 2

�

-r

23 k] twist (5 .5 kcal)

6 �l chair

6

I

2

\- - - - - - /

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

4

half-chair

4� 3 S

S

1 09

S

j

42 k] ( 1 0 kcal) 29 k] (6.9 kcal )

L______________

S 6 I 4� chair -

.... Figure 3-21

Conformational energy of cyclohexane. The chair conformation is most stable, followed by the twist boat. To convert between these two conformations, the molecule must pass through the unstable half-chair conformation.

I These are called equatorial bonds. The axial bonds and hydrogens are shown in red in Figure 3 -22, and the equatorial bonds and hydrogens are shown in green. Each carbon atom in cyclohexane is bonded to two hydrogen atoms, one directed upward and one downward. As the carbon atoms are numbered in Figure 3 -22, C l has an axial bond upward and an equatorial bond downward. C2 has an equatorial bond upward and an axial bond downward. The pattern alternates. The odd-numbered carbon atoms have axial bonds up and equatorial bonds down, like Cl . The even-numbered carbons have equatorial bonds up and axial bonds down, like C2. Thi s pattern of alternating axial and equatorial bonds is helpful for predicting the conformations of substituted cyc1ohexanes, as we see in Sections 3 - 1 3 and 3 - 1 4.

axis

H

axial

H

.... Fig u re 3-22

seen from the side

seen from above

Axial bonds are directed vertically, parallel to the axis of the ring. Equatorial bonds are directed outward, toward the equator of the ring. As they are numbered here, the odd-numbered carbons have their upward bonds axial and their downward bonds equatorial. The even-numbered carbons have their downward bonds axial and their upward bonds equatorial.

1 10

Chapter 3 : Structure and Stereochemistry of Alkanes

P R O B L E M - S O LV I N G S T R AT E G Y D rawi n g C h a i r Co nfo r m a t i o n s

Drawing realistic pictures of cyclohexane conformations is not difficult, but certain rules should be fol lowed to show the actual positions and angles of the substituents on the ring. Make a cyc10hexane ring with your models, put it in a chair conformation, and use it to follow along with this discussion. When you hold your model at the angle that corresponds to a draw­ ing, the angles of the bonds in the model should correspond to the angles in the drawing. To draw the carbon-carbon bond framework, first draw two parallel lines, slightly slanted and slightly offset. The atoms at the ends of these bonds lie in a plane, and they define what will be the "armrests" of our chair. ..------

------

Draw the headrest and footrest carbons, and draw the lines connecting them to the armrests. The two lines connecting the headrest carbon should be parallel to the two lines connecting the footrest. headrest

in back

�t::=\ = � /"

i

in front

footrest

Notice that the carbon-carbon bond framework uses lines with only three different slopes, labeled a, b, and c. Compare this drawing with your model, and notice the pairs of carbon-carbon bonds with three distinct slopes. slop�

slope b

\

slope c

------

We can draw the chair with the headrest to the left and the footrest to the right, or vice versa. Practice drawing it both ways.

Now fill in the axial and equatorial bonds. The axial bonds are drawn vertically, either up or down. When a vertex of the chair points upward, its axial bond also points upward. If the vertex points downward, its axial bond points downward. C l is a downward-pointing vertex, and its axial bond also points downward. C2 points upward, and its axial bond points upward. axis

4�1 I

3

6

II

2

I

The equatorial bonds take more thought. Each carbon atom is represented by a vertex formed by two lines (bonds), having two of the possible slopes a, b, and c. Each equatorial bond should have the third slope: the slope that is not represented by the two lines forming the vertex.

Look at your model as you add the equatorial bonds. The vertex C l is formed by lines of slopes b and c, so its equatorial bond should have slope a. The equatorial bond at C2 should have slope b, and so on. Notice the W- and M-shaped patterns that result when these bonds are drawn correctly.

� �

3 - 1 4 Conformations of Monosubstituted Cyc10hexanes

111

PRO B L E M 3-20

The cyclohexane chair just drawn has the headrest to the left and the footrest to the right. Draw a cyclohexane chair with its axial and equatorial bonds, having the headrest to the right and the footrest to the left. PRO B L E M 3-2 1

Draw 1 ,2,3 ,4,5,6-hexamethyIcycIohexane with all the methyl groups (b) in equatorial positions. If your cycIohexane rings look awkward or slanted when using the analytical approach just shown, then try the artistic approach:* Draw a wide M, and draw a wide W below it, dis­ placed about half a bond length to one side or the other. Connect the second atoms and the fourth atoms to give the cyclohexane ring with four equatorial bonds. (a) in axial positions.

� \

W

\

W

displaced to the right

displaced to the left

The other two equatorial bonds are drawn parallel to the ring connections. The axial bonds are then drawn vertically.

': � --ZJ=tt �

e

I

e

e

:

a

*See

,

�

a

, ,

"

a

V. Dragojlovic, J. Chem. Educ.

I

:

,

'e

a,

e

:

e

e

I

,

a

a

"

e

,

"

a,

"e

: I

a

I

,

e

a

2 001 , 78, 923 .

A substituent on a cyclohexane ring (in the chair confonnation) can occupy either an axial or an equatorial position. In many cases, the reactivity of the substituent depends on whether its position is axial or equatorial. The two possible chair conformations for methylcyclohexane are shown in Figure 3-23. These confonnations are in equilibrium because they interconvert at room temperature. The twist boat serves as an intennediate in this chair-chair interconversion, sometimes called a "ring-flip." Place different-colored

axial

�� �

4

eqUatorial

2

--

4 d� 4 K - 6 --

3

boat

2

3

2

1

H

3-1 4 Co nformati ons of Mo nosu bstituted Cyclohexanes

CH 3

equatorial

axial

-ring-flip

CH3 axial

� Figure 3-23

CH3 equatorial

Chair--chair interconversion of methyIcyclohexane. The methyl group is axial in one conformation, and equatorial in the other.

1 12

Chapter 3 : Structure and Stereochemistry of Alkanes

*�::$ '

CH

H H

1

4

3 6

H

5

CH2

H H

H

H

Newman projection (a)

$$ H

H

� Figure 3-24

(a) When the methyl substituent is i n an axial position on Cl , it is gauche to C3. (b) The axial methyl group on C1 is also gauche to C5 of the ring.

H

3

/--

ga

tH2 2

6

CH 3

I

CH2

H H

H

H

Newman projection ( b)

atoms in the axial and equatorial positions of your cyclohexane model, and notice that the chair-chair interconversion changes axial to equatorial and equatorial to axial. The two chair conformations of methylcyclohexane interconvert at room tem­ perature, so the one that is lower in energy predominates. Careful measurements have shown that the chair with the methyl group in an equatorial position is the most stable conformation. It is about 7.6 kJ/mol ( 1 .8 kcaVmol) lower in energy than the conforma­ tion with the methyl group in an axial position. Both of these chair conformations are lower in energy than any boat conformation. We can show how the 7.6 kJ energy dif­ ference between the axial and equatorial positions arises by examining molecular models and Newman projections of the two conformations. First, make a model of methylcyclohexane and use it to follow this discussion. Consider a Newman projection looking along the armrest bonds of the conforma­ tion with the methyl group axial (Figure 3 -24a): The methyl group is on C l , and we are looking from C l toward C2. There is a 60° angle between the bond to the methyl group and the bond from C2 to C3, and the methyl substituent and C3 are in a gauche relation­ ship. In our analysis of torsional strain in butane, we saw that a gauche interaction raises the energy of a conformation by 3.8 kJ/mol (0.9 kcal/mol) relative to the anti conforma­ tion. This axial methyl group is also gauche to C5, as you will see if you look along the CI -C6 bond in your model. Figure 3 -24b shows this second gauche relationship. The Newman projection for the conformation with the methyl group equatorial shows that the methyl group has an anti relationship to both C3 and C5 . Figure 3 -25

� Figure 3-25

-

Looking down the C 1 C2 bond of the equatori al conformation. Notice that the methyl group is anti to C3.

f;t� �

H3C

H2

H

CH2

H

y Y H

H

Newman projection

3 - 1 4 Conformations of Monosubstituted Cyclohexanes H HL

�

ax ial

\-t--\ � H

H

___

H equatorial

.... Figure 3-26

C " ' ''H "H l al

equatorial

I ,3-diaxial interactions

more stable by 1 .7 kcallmol (7. 1 kllmol)

shows the Newman projection along the C I - C2 bond, with the anti relationship of the methyl group to C3 . PRO B L E M 3-22

Draw a Newman projection, similar to Figure 3-2S, down the C I -C6 bond i n the equatori­ al conformation of methylcyclohexane. Show that the equatorial methyl group is also anti to CS. (Using your models will help.) The axial methylcyclohexane conformation has two gauche interactions, each representing about 3 . 8 kJ (0.9 kcal) of additional energy. The equatorial methyl group has no gauche interactions. Therefore, we predict that the axial conformation is higher in energy by 7 . 6 kJ ( 1 . 8 kcal) per mole, in good agreement with the experimental value. Figure 3 -26 shows that the gauche relationship of the axial methyl group with C3 and C5 places the methyl hydrogens close to the axial hydrogens on these carbons, and their electron clouds begin to interfere. This form of steric hindrance is called a 1,3-diaxial interaction because it involves sub­ stituents on the carbon atom that would be numbered C3 if the carbon bearing the methyl group was numbered C 1 . These 1 ,3 -diaxial interactions are not present in the equatorial conformation. A larger group usually has a larger energy difference between the axial and equa­ torial positions, because the 1 ,3 -diaxial interaction shown in Figure 3 -26 is stronger for larger groups. Table 3-6 shows the energy differences between the axial and equa­ torial positions for several alkyl groups and functional groups. The axial position is higher in energy in each case.

TABLE 3-6

Energy Differences Between the Axial and Equatorial Conformations of Monosubstituted Cyclohexanes !::J. G (axial-equatorial) X

q X

axial

H

H

113

�

dx equatorial

-F

-CN -Cl -Br

-OH -COOH -CH 3 -CH 2CH 3 -CH (CH 3h -C(CH 3 h

(kJ/mol)

(kcal/mol)

0.8 0.8 2. 1 2.5

0.2 0.2 O.S 0.6 1 .0

4. 1

5.9 7.6 7. 9 8.8 23

1 .4

1 .8 1 .9 2. 1 5.4

The axial substituent interferes with the axial hydrogens on C3 and CS. This interference is called a 1 ,3-diaxial interaction.

1 14

Chapter 3: Structure and Stereochemistry of Alkanes PRO B L E M 3-23

Table 3-6 shows that the axial-equatorial energy difference for methyl, ethyl, and isopropyl groups increases gradually: 7.6, 7.9, and 8.8 kJ/mol ( l .8, l .9, and 2. 1 kcaUmol). The t-butyl group jumps to an energy difference of 23 kJ/mol (5.4 kcal/mol), over twice the value for the isopropyl group. Draw pictures of the axial conformations of isopropylcyclohexane and t-butylcyclohexane, and explain why the t-butyl substituent experiences such a large increase in axial energy over the isopropyl group. PROBLEM 3-24

Draw the most stable conformation of (a) ethylcyclohexane (b) isopropylcyclohexane

3-1 5 Conformati ons of D i s u bstituted Cycl o hexa nes

(c) t-butylcyclohexane

The steric interference between substituents in axial positlOns is particularly severe when there are l arge groups on two carbon atoms that bear a 1 ,3-diaxial relationship (cis on C l and C3 , or C l and C5), as in the two chair conformations of cis- l , 3-dimethylcyclohexane shown here. The less stable conformation has both methyl groups in axial position s. The more stable conformation has both methyl groups in equatorial positions. Note the strongly unfavorable 1 ,3 -diaxial i nteraction between the two methyl groups in the diaxial conformation. The mole­ cule can relieve this 1 ,3-diaxial interference by flipping to the diequatorial confor­ mation. Use your models to compare the diaxial and di equatorial forms of cis- l ,3 -dimethy lcyclohexane.

�

1 ,3 - dlaxial

H

,H \\" C . --H

"-...

'-.....

mteraction

.H \\\' C --H

� feqUatorial ).T

: � ial

L------I

aX

H

H . I equatorial H H.::."C

C

H

H

diaxial-very unfavorable

H

�

'H H

diequatorial-much more stable

Either of the chair conformations of trans- l ,3-dimethylcyclohexane has one methyl group in an axial position and one in an equatorial position. These conforma­ tions have equal energies, and they are present in equal amounts. Chair conformations of trans- i, 3 -dimethylcyclohexane a xi al

CH3

H

equatorial

H

-L lL--:::.J equatorial CH3

same

energy

n

H

3

axial

Now we can compare the relative stabilities of the cis and trans i somers of 1 ,3-dimethylcyclohexane. The most stable conformation of the cis isomer has both methyl groups in equatorial positions. Either conformation of the trans isomer places one methyl group in an axial position. The trans isomer is therefore higher in energy than the cis isomer by about 7.6 kJ/mol (1 .8 kcal/mol), the energy difference between axial and equatorial methyl groups. Remember that the cis and trans isomers cannot interconvert, and there is no equilibrium between these isomers.

3 - 1 5 Conformations of Disubstituted Cyclohexanes S O LV E D P R O B L E M 3 - 3

(a) Draw both chair conformations of cis- I ,2-dimethylcyclohexane, and determine which conformer is more stable. (b) Repeat for the trans isomer. (c) Predict which isomer (cis or trans) is more stable. SOLUTION

PROBLEM-SOLVING

Htltp

R i ng-flips change the axial or

equatori a l positi o n i n g of groups, but they cannot change their cis-trans relationships. Converting cis into trans would req u i re brea king and re-for m i n g bonds.

(a) There are two possible chair conformations for the cis isomer, and these two conforma­ tions interconvert at room temperature. Each of these conformations places one methyl group axial and one equatorial, giving them the same energy.

i

equato r a l

CH3 �I -\ �CH3 H

aXIal

() same energy

PROBLEM-SOLVING

�H axial CH3

H H3C H CH,

3t-

equatorial

higher energy (diaxial)

equatonal

conformer is more stable. (e) Predict which isomer (cis or trans) is more stable. PROBLEM 3-26

Use your results from Problem 3-25 to complete the following table. Each entry shows the positions of two groups arranged as shown. For example, two groups that are trans on adja­ cent carbons (trans- I ,2) must be both equatorial (e,e) or both axial (a,a).

(e,e) or (a,a)

bond a x i a l and their up bond equatoria l . For example, cis-1 , 3 (both

odd, one even) w i l l be one axial, one

equatorial

(b) Repeat for the trans isomer.

(e,a) or (a,e)

equatorial, the even-n u m bered carbons wi l l a l l have their down

up, both odd) w i l l be both axial or

(a) Draw both chair conformations of cis- l ,4-dimethylcyclohexane, and determine which

1 ,2 1 ,3 1 ,4

up bond axial and their down bond

both equatorial; cis- 1 , 2 (both up, one

PROBLEM 3-25

trans

carbons are s i m i l a r, as are the odd-num bered carbons a l l have their

(c) The trans isomer is more stable. The most stable conformation of the trans isomer is diequatorial and therefore about 7.6 kJ/mol ( 1 .8 kcal/mol) lower in energy than either conformation of the cis isomer, each having one methyl axial and one equatorial. Remember that cis and trans are distinct isomers and cannot interconvert.

cis

cyclohexane, the odd-n u m bered even-numbered carbons. If the

lower energy (diequalorial )

Positions

Htltp

If you n u m be r the carbons in a

(b) There are two chair conformations of the trans isomer that interconvert at room temper­ ature. Both methyl groups are axial in one, and both are equatorial in the other. The diequatorial conformation is more stable because neither methyl group occupies the more hindered axial position.

3-1 SA

11 S

Substit uents of Different Sizes

In many substituted cycIohexanes, the substituents are different sizes. As shown in Table 3-6 (p. 1 1 3), the energy difference between the axial and equatorial positions for a larger group is greater than that for a smalIer group. In general, if both groups cannot be equatorial , the most stable conformation has the larger group equatorial and the smaller group axial.

equatorial. This tip a l lows you to pred ict the a nswers before you draw them.

116

Chapter 3 : Structure and Stereochemistry of Alkanes S O LV E D P R O B L E M 3 - 4

Draw the most stable conformation of trans- l -ethyl-3-methylcyclohexane. S O LUTION

4=l:

First, we draw the two conformations. axial

H, H

CH3

n H

CH3 axial

equatorial

CH2CH3 equatorial

more stable

less stable

Both of these conformations require one group to be axial while the other is equatorial. The ethyl group is bulkier than the methyl group, so the conformation with the ethyl group equato­ rial is more stable. These chair conformations are in equilibrium at room temperature, and the one with the equatorial ethyl group predominates. P RO B L E M 3-27

Draw the two chair conformations of each of the following substituted cyclohexanes. In each case, label the more stable conformation. (b) trans- l -ethyl-2-methylcyclohexane (a) cis- I -ethyl-2-methylcyclohexane (c) cis- I -ethyl-4-isopropylcyclohexane (d) trans- l -ethyl-4-methylcyclohexane

P R O B L E M - S O LV I N G S T R AT E G Y Recog n i z i n g cis a n d tra n s I s o m e rs

Some students find it difficult to look at a chair conformation and tell whether a disubstitut­ ed cyclohexane is the cis isomer or the trans isomer. In the following drawing, the two methyl groups appear to be oriented in similar directions. They are actually trans but are often mistaken for cis.

C �!�: � H down

up

CX· CH3 CH3 1

\\\'

2

trans- l ,2-dimethylcyclohexane

This ambiguity is resolved by recognizing that each of the ring carbons has two avail­ able bonds, one upward and one downward. In this drawing, the methyl group on C l is on the downward bond, and the methyl on C2 is on the upward bond. Because one is down and one is up, their relationship is trans. A cis relationship would require both groups to be upward or both to be downward. PROBLEM 3-28

Name the following compounds. Remember that two cis; one up bond and one down bond are trans.

up

bonds are cis; two down bonds are

3- 1 6 B icyclic Molecules

3-15 B

Extremely B u l ky Groups

Some groups, such as tertiary-butyl groups, are so bulky that they are extremely strained in axial positions. Regardless of the other groups present, cyclohexanes with t-butyl substituents are most stable when the t-butyl group is in an equatorial position. The following figure shows the severe steric interactions in a chair conformation with a t-butyl group axial.

H H-tF=\H C, , " H C H 'CH �H-C H 33

extremely crowded

�

/

!�£,' i 'l)

_

/�" '.

strongly preferred conformation

If two t-butyl groups are attached to the ring, both of them are much less strained i n equatorial position s . When neither chair conformation allows both bulky groups to be equatorial, they may force the ri ng into a twist boat conforma­ tion. For example, either chair conformation of cis- l ,4-di-t-butylcyclohexane re­ quires one of the bulky t-butyl groups to occupy an axial position. This compound is more stable in a twist boat conformation that allows both bulky groups to avoid axial positions .

I-butyl group moves out o f the axial position

twist boat

PRO B L E M 3-29 Draw the most stable conformation of (a) cis- l -t-butyl-3-ethylcycIohexane (b) trans- l -t-butyl-2-methylcycIohexane (c) trans- l -t-butyl-3-( l , I -dimethyJpropyl)cycIohexane

Two or more rings can be joined into bicyclic or polycyclic systems. There are three ways that two rings may be joined. Fused rings are most common, sharing two adj a­ cent carbon atoms and the bond between them. Bridged rings are also common, shar­ ing two nonadjacent carbon atoms (the bridgehead carbons) and one or more carbon atoms (the bridge) between them. Spirocyclic compounds, in which the two rings share only one carbon atom, are relatively rare.

3-1 6 Bicycl i c Mo lecu les

117

118

Chapter 3 : Structure and Stereochemistry of Alkanes

_IW a:;

bridgehead "' bO",

co

bicyclo[2.2. 1 ] heptane (norbornane)

bicyclo[4.4.0]decane (decal in)

3- 16A

spirocyclic

bridged bicyclic

fused bicyclic

spiro[4.4]nonane

Nomen clature of 8icyclic Alkanes

The name of a bicyclic compound is based on the name of the alkane having the same number of carbons as there are in the ring system. This name fol lows the prefix bicyclo and a set of brackets enclosing three numbers. The following examples contain eight car­ bon atoms and are named bicyclo[4.2.0]octane and bicyclo[3 .2. 1 ]octane, respectively.

J;

zero-carbon bridge Bicyclic

molecules

are

found

in

many natural product structures. A derivative of bicyclo[3.2.1 joctane,

four-carbon bndge

in which nitrogen replaces the car­

that produces euphoria.

two-carbon bndge

two-carbon bndge

three-carbon bridge bicyclo[3.2. l loctane

bicyclo[4.2.0]octane

bon at the one-carbon bridge, is found in cocaine, an addictive drug

� �

ne-carbon bJ idge

All fused and bridged bicyclic systems have three bridges connecting the two bridge­ head atoms (red circles) where the rings connect. The numbers in the brackets give the number of carbon atoms in each of the three bridges connecting the bridgehead carbons, in order of decreasing size. PROBLEM 3-30

Name the following compounds. (')

(b) �

0>

3-16 8

(0)

CO

(d)

0

cis- and trans-Decalin

Decalin (bicyclo[4.4.0]decane) is the most common example of a fused-ring system. Two geometric isomers of decalin exist, as shown in Figure 3-27. In one isomer the

H CO H c

H

CO H

is-decali n

trans-decal in bridgehead carbons

bridgehead carbons

� Figure 3-27

cis-Decalin has a ring fusion where the second ring is attached by two cis bonds. trans-Decal in is fused using two trans bonds. (The other hydrogens are omitted for cl arity.)

cis-decal in

trans- decali n

Chapter 3 Glossary rings are fused using two cis bonds, while the other is fused using two trans bonds. You should make a model of decalin to follow this discussion. If we consider the left ring in the drawing of cis-decalin, the bonds to the right ring are both directed downward (and the attached hydrogens are directed upward). These bonds are therefore cis, and this is a cis ring fusion. In trans-decalin, one of the bonds to the right ring is directed upward and the other downward. These bonds are trans, and this is a trans ring fusion. The six-membered rings in both isomers assume chair conformations, as shown in Figure 3-27. The conformation of cis-decalin is somewhat flexible, but the trans isomer is quite rigid. If one of the rings in the tran s isomer did a chair-chair interconversion, the bonds to the second ring would both be­ come axial and would be directed 1 800 apart. This is an impossible conformation, and it prevents any chair-chair interconversion in trans-decalin. PRO B L E M 3 - 3 1

Use your models to do a chair-chair interconversion on each ring of the conformation of cis­ decalin shown in Figure 3-27. Draw the conformation that results.

C2H2n+2.

acyclic Not cyclic. (p. 1 0 1 ) alkane A hydrocarbon having only single bonds; a saturated hydrocarbon; general formula:

(p. 8 1 )

alkyl group The group o f atoms remaining after a hydrogen atom i s removed from an alkane;

Cha pter 3 G lossary

an aIkanelike substituent. Symbolized by R. (p. 85)

angle strain or Baeyer strain The strain associated with compressing bond angles to smaller

(or larger) angles. (p. 1 04) anti conformation A conformation with a 1 800 dihedral angle between the largest groups. Usually the lowest-energy conformation. (p. 99) aromatic hydrocarbon A hydrocarbon having a benzene-like aromatic ring. (p. 70) axial bond One of six bonds (three up and three down) on the chair conformation of the cyclo­ hexane ring that are parallel to the "axis" of the ring. (p. 1 08) bridged bicyclic compound A compound containing two rings joined at nonadjacent carbon atoms. (p. 1 1 7)

bridged bicyclic systems (bridgeheads circled)

bridgehead carbons The carbon atoms shared by two or more rings. Three chains of carbon atoms (bridges) connect the bridgeheads. (p. 1 1 7 ) chair-chair interconversion (ring-flip) The process o f one chair conformation of a cyclo­ hexane flipping into another one, with all the axial and equatorial positions reversed. The boat conformation is an intermediate for the chair-chair interconversion. (p. I l l )

H

H)()C�/ H

� I

chair (methyls axial)

boat

chair (methyls equatorial)

or trans arrangement on a ring or double bond. (p. 1 03) cis: Having two similar groups directed toward the same face of a ring or double bond. trans: Having two similar groups directed toward opposite faces of a ring or double bond. cis-trans isomers (geometric isomers) Stereoisomers that differ only with respect to their cis

119

120

Chapter 3 : Structure and Stereochemistry of Alkanes

H)C

"

/ C=C " /

CH)

H)C

"

/ C=C / "

H CH) trans-2-butene

H H cis-2-butene

~

H H)C

CH)

H

CH)

trans- l ,2-dimethylcyclopentane

cis- l ,2-dimethylcyclopentane

combustion A rapid oxidation at high temperatures in the presence of air or oxygen. (p. 93) common names The names that have developed historically, generally with a specific name for each compound; also called trivial names. (p. 83) conformational analysis The study of the energetics of different conformations. (p. 97) conformations or conformers Structures that are related by rotations about single bonds. In

most cases, conformations interconvert at room temperature, and they are not true isomers. (p. 95) e = 0° A

H H

�

e

=

� � �

60°

CH

H H H

H

CH)

H

H

H

totally eclipsed conformation

gauche conformation

conformations of cyclohexanes (p. 1 07)

H

� t chair

H

>

y boat

half-chair

�

H

�

e

=

1 80°

H CH) anti co nfo rmati on

H

L �;1 twi st boat

chair conformation: The most stable conformation of cydohexane, with one part puckered

upward and another part puckered downward. boat conformation: The less stable puckered conformation of cyclohexane, with both parts puckered upward. The most stable boat is actually the twist boat (or simply twist) confor­ mation. Twisting minimizes torsional strain and steric strain. flagpole hydrogens: Two hydrogens (blue) in the boat conformation point upward like flagpoles. The twist boat reduces the steric repulsion of the flagpole hydrogens. half-chair conformation: The unstable conformation halfway between the chair conforma­ tion and the boat conformation. Part of the ring is flat in the half-chair conformation. constitutional isomers (structural isomers) Isomers whose atoms are connected differently; they differ in their bonding sequence. (p. 55) cracking Heating large alkanes to cleave them into smaller molecules. (p. 92) catalytic cracking: Cracking in the presence of a catalyst. hydrocracking: Catalytic cracking in the presence of hydrogen to give mixtures of alkanes. cyclic Containing a ring of atoms. (p. 1 00) cycloalkane An alkane containing a ring of carbon atoms; general formula: C"H21l' (p. 1 0 1 ) degree of alkyl substitution The number of alkyl groups bonded to a carbon atom i n a com­ pound or in an alkyl group. (p. 86) H

I R -C - H I H

primary

(1°)

carbon atom

H

I R-C -H I R

secondary

(2°)

carbon atom

R

I R -C - H I R

tertiary

(3°)

carbon atom

R

I I

R-C -R R quaternary

(4°)

carbon atom

1,3-diaxial interaction The strong steric strain between two axial groups on cydohexane car­ bons with one carbon between them. (p. 1 1 3 ) dihedral angle (8) (see also conformations) The angle between two specified groups in a Newman projection. (p. 96) .

Chapter 3 Glossary eclipsed conformation Any conformation with bonds directly lined up with each other, one behind the other, in the Newman projection. The conformation with e 0° is an eclipsed con­ formation. See also staggered conformation. (p. 96) equatorial bond One of the six bonds (three down and three up) on the cyclohexane ring that are directed out toward the "equator" of the ring. The equatorial positions are shown in green in the following drawing. (p. 1 09) =

axis

e

e

axial bonds in red; equatorial bonds in green fused-ring system A molecule in which two or more rings share two adjacent carbon atoms.

(p. 1 1 7)

fused-ring systems gauche conformation A conformation with a 60° dihedral angle between the l argest groups.

(p. 99) geometric isomers See cis-trans isomers, the IUPAC term. (p. 1 03) halogenation The reaction of alkanes with halogens, in the presence of heat or light, to give

products with halogen atoms substituted for hydrogen atoms. (p. 94) hea l or ligh

ll x F, Cl, Br R - X + XH R - H + X2 heat of combustion The heat given off when a mole of a compound is burned with excess oxygen to give CO2 and H20 in a bomb calorimeter. A measure of the energy content of a molecule. (p. 1 04) homologs Two compounds that differ only by one or more -CH2- groups. (p. 82) hydrophilic Attracted to water, soluble in water. hydrophobic Repelled by water; insoluble in water. (p. 89) IUPAC names The systematic names that follow the rules adopted by the International Union of Pure and Applied Chemistry. (p. 84) kerosene A thin, volatile oil distilled from petroleum, with a boiling range higher than that of gasoline and lower than that of diesel fuel. Kerosene was once used in lanterns and heaters, but now most of this petroleum fraction is further refined for use as jet fuel. (p. 9 1 ) methane hydrate An icelike substance consisting of individual methane molecules trapped inside cages of water molecules. (p. 93) =

I

methine group The - CH - group. methylene group The - CH 2- groups. (p. 82) methyl group The - CH3 group. (p. 85) Il-alkane, normal alkane, or straight-chain alkane An alkane with all its carbon atoms in a

single chain, with no branching or alkyl substituents. (p. 83) Newman projections A way of drawing the conformations of a molecule by looking straight

�

down the bond connecting two carbon atoms. (p. 95) H

�

1 800 dihedral angle

H CH3 a Newman projection of butane in the anti conformation H

octane number A rating of the antiknock properties of a gasoline blend. Its octane number is the percentage of isooctane (2,2,4-trimethylpentane) in an isooctane/heptane blend that begins to knock at the same compression ratio as the gasoline being tested. (p. 9 1 )

121

122

Chapter 3 : Structure and Stereochemistry of Alkanes paraffins Another term for alkanes. (p. 93) ring strain The extra strain associated with the cyclic structure of a compound, as compared

with a similar acyclic compound; composed of angle strain and torsional strain. (p. 1 04) angle strain or Baeyer strain: The strain associated with compressing bond angles to smaller (or larger) angles. torsional strain: The strain associated with eclipsing of bonds in the ring. saturated Having no double or triple bonds. (p. 8 1 ) sawhorse structures A way of picturing conformations by looking down at an angle toward the carbon-carbon bond. (p. 96) skew conformation Any conformation that is not precisely staggered or eclipsed. (p. 96) spirocyclic compounds Bicyclic compounds in which the two rings share only one carbon atom. (p. 1 1 7) staggered conformation Any conformation with the bonds equally spaced in the Newman projection. The conformation with e 60° is a staggered conformation. (p. 96) =

nli

HH

H H

=

0°

�

H H H

eclipsed conformation of ethane

H

�

yty � H

H

Ii = 600

H

H

staggered conformation of ethane

steric strain The interference between two bulky groups that are so close together that their electron clouds experience a repUlsion. (p. 99) substituent A side chain or appendage on the main chain. (p. 84) systematic names Same as IUPAC names, the names that follow the rules adopted by the International Union of Pure and Applied Chemistry. (p. 84) torsional energy or conformational energy The energy required to twist a bond into a specific conformation. (p. 96) torsional strain The resistance to twisting about a bond. (p. 96) totally eclipsed conformation A conformation with a 0° dihedral angle between the largest groups. Usually the highest-energy conformation. (p. 99)

I

Essential Prob lem-Solving Ski l ls in Chapter 3

1. Explain and predict trends in physical properties of alkanes.

2. Correctly name alkanes, cycloalkanes, and bicyclic alkanes. 3. Given the name of an alkane, draw the structure and give the molecular formula. 4.

Compare the energies of alkane conformations and predict the most stable confOlmation.

5.

Compare the energies of cycloalkanes, and explain ring strain.

6.

Identify and draw cis and trans stereoisomers of cycloalkanes.

7.

Draw accurate cyclohexane conformations, and predict the most stable conformations of substituted cyclohexanes.

Study Problems 3-32

Define and give an example for each term. (a) alkane (d) saturated (g) hydrophilic

U) methyl group

(b) (e) (h) (k)

alkene hydrophobic n-alkane common name

(c) alkyne

(f)

aromatic methylene group (I) systematic name (i)

Study Problems eclipsed (0) Newman projection gauche (r) anti conformation (8) an acyclic alkane cis-trans isomers on a ring (u) chair conformation (v) boat conformation twist boat (x) half-chair conformation (z) equatorial position (A) catalytic cracking (y) axial position (D) bridged bicyclic compound (B) chair--chair interconversion (C) fused ring system (E) bridgehead carbon atoms (F) combustion Which of the following Lewis structures represent the same compound? Which ones represent different compounds? CH3 CH3 H CH3 H CH. H H H H H H H

(m) conformers (p) staggered

3-33

123

I

(n) (q) (t) (w)

I

I

(a) H - C - C - C - H

I

H H

"

/ CH3 (b) CH3 " H

/

I

I

H

C=C

C=C

I I

H

H

"

� CH2,f CH3 "

CH3 CH3

/ "

H

H

CH3

I

I

I

H

I I

CH3 -"

/ CH2 - C ,\-

CH2

I

H

H

CH3 H

I J

H-C-C-H

C - CH?

"

/ H H " H

/

CH3

C=C

C=C

/ "

/

I

I

H-C-C-H

I

I

H

H '

'" .r

H

I

I

I

CH3 H

H

"

CH3 CH3 CH3

H

U V � LttI ~ "' H

I

H-C-C-H

H

CH3

CH3

(e)

I

I

H

CH3 / "

I

H-C-C-C-C-H

H ° CH3

�

CH3

H

CH3

H

CH3

CH3

3-34

CH3 H

Draw the structure that corresponds with each name. (b) 4-isopropyldecane (e) 2,2,4,4-tetramethylhexane (h) isobutylcyclopentane (j) pentylcyclohexane (k) cyclobutylcyclohexane Each of the following descriptions applies to more than one alkane. In each case, match the description. (a) a methyl heptane (b) a diethyldecane (d) a trans-dimethylcyclopentane (e) a (2,3-dimethylpentyl)cycloalkane

(a) 3-ethyloctane (d) 2,3-dimethyl-4-propylnonane (g) cis- l -ethyl-4-methylcyclohexane

3-35

CH3 H

(c) sec-butylcycloheptane

(f)

trans- l ,3-diethylcyclopentane t-butylcyclohexane cis- l -bromo-3-chlorocyclohexane draw and name two structures that (i) (I)

(c) a cis-diethylcycloheptane

(f)

a bicyclodecane

1 24 3-36 3-37

Chapter 3 : Structure and Stereochemistry of Alkanes Write structures for a homologous series of alcohols (R -OH) having from one to six carbons. Give the IUPAC names of the following alkanes. (a) CH3C(CH3 hCH(CH2CH3)CH2CH2CH(CH3h (b) CH3CH2 - H- CH2CH2 - H-CH3

T

CH3CHCH3

� (,) (:r>

T

CH3CHCH3

(')

3-38 3-39

3-40 3-41

3-42 3-43

3-44

3-45 3-46

Draw and name eight isomers of molecular formula CSH 1S. The following names are all incorrect or incomplete, but they represent real structures. Draw each structure and name it correctly. (b) 3-isopropylhexane (a) 2-ethylpentane (c) 5-chloro-4-methylhexane (0 2,3-diethylcyclopentane (e) 2-cyclohexylbutane (d) 2-dimethylbutane In each pair of compounds, which compound has the higher boiling point? Explain your reasoning. (a) octane or 2,2,3-trimethylpentane (b) heptane or 2-methylnonane (c) 2,2,5-trimethylhexane or nonane There are eight different five-carbon alkyl groups. (a) Draw them. (b) Give them systematic names. (c) In each case, label the degree of substitution (primary, secondary, or tertiary) of the head carbon atom, bonded to the main chain. Use a Newman projection, about the indicated bond, to draw the most stable conformer for each compound. (a) 3-methylpentane about the C2 - C3 bond (b) 3,3-dimethylhexane about the C3 - C4 bond (a) Draw the two chair conformations of cis- l ,3-dimethylcyclohexane, and label all the positions as axial or equatorial. (b) Label the higher-energy conformation and the lower-energy conformation. (c) The energy difference in these two conformations has been measured to be about 23 kJ (5.4 kcal) per mole. How much of this energy difference is due to the torsional energy of gauche relationships? (d) How much energy is due to the additional steric strain of the 1 ,3-diaxial interaction? Draw the two chair conformations of each compound, and label the substituents as axial and equatorial. In each case, determine which conformation is more stable. (a) cis- l -ethyl-2-isopropylcyclohexane (b) trans-l -ethyl-2-isopropylcyclohexane (d) trans- l -ethyl-3-methylcyclohexane (c) cis- l -ethyl-3-methylcyclohexane (e) cis- l -ethyl-4-methylcyclohexane Using what you know about the conformational energetics of substituted cyclohexanes, predict which of the two decalin isomers is more stable. Explain your reasoning. The most stable form of the common sugar glucose contains a six-membered ring in the chair conformation with all the substituents equatorial. Draw this most stable conformation of glucose.

y Oi CHPH HO � OH

HO

OH glucose

4

vacant

p orbital

\

The Study of Chem ical Reactions

carbocation

The most interesting and useful aspect of organic chemistry i s the study of reactions. We cannot remember thousands of specific organic reactions, but we can organize the reactions into logical groups based on how the reactions take place and what interme­ diates are involved. We begin our study by considering the halogenation of alkanes, a relatively simple reaction that takes place in the gas phase, without a solvent to com­ plicate the reaction. In practice, alkanes are so unreactive that they are rarely used as starting materials for most organic syntheses. We start with them because we have already studied their structure and properties, and their reactions are relatively uncom­ plicated. Once we have used alkanes to introduce the tools for studying reactions, we will apply those tools to a variety of more useful reactions. The overall reaction, with the reactants on the left and the products on the right, is only the first step in our study of a reaction. If we truly want to understand a reaction, we must also know the mechanism, the step-by-step pathway from reactants to products. To know how well the reaction goes to products, we study its thermo­ dynamics, the energetics of the reaction at equilibrium. The amounts of reactants and products present at equilibrium depend on their relative stabilities. Even though the equilibrium may favor the formation of a product, the reaction may not take place at a useful rate. To use a reaction in a realistic time period (and to keep the reaction from becoming violent), we study its kinetics, the variation of reac­ tion rates with different conditions and concentrations of reagents. Understanding the reaction ' s kinetics helps us to propose reaction mechanisms that are consistent with the properties we observe.

4-1

The reaction of methane with chlorine produces a mixture of chlorinated products, whose composition depends on the amount of chlorine added and also on the reaction conditions. Either light or heat is needed for the reaction to take place at a useful rate. When chlorine is added to methane, the first reaction is

4-2

H

I

H-C-H

I

I ntroduction

Ch lori nation of M eth ane

H +

C] - Cl

heat or light

)

I

H - C - Cl

I

+

H - Cl

H

H methane

alleyl group

chlorine

chloromethane (methyl chloride)

hydrogen chloride

125

1 26

Chapter 4: The Study of Chemical Reactions This reaction may continue; heat or light is needed for each step: H

I H-C-Cl I H

Cl

I H-C-Cl I

CI

I

CI

CI-C-Cl

C]-C-Cl

H

CI

I

H

+

HCl

+

HC]

I I

+

HCl

This sequence raises several questions about the chlorination of methane. Why is heat or light needed for the reaction to go? Why do we get a mixture of products? Is there any way to modify the reaction to get just one pure product? Are the observed products formed because they are the most stable products possible? Or are they fa­ vored because they are formed faster than any other products? The answers to these questions involve three aspects of the reaction: the mecha­ nism, the thermodynamics, and the kinetics. 1 . The mechanism is the complete, step-by-step description of exactly which bonds break and which bonds form in what order to give the observed products.

2. Thermodynamics is the study of the energy changes that accompany chemical and physical transformations. It allows us to compare the stability of reactants and products and predict which compounds are favored by the equilibrium. 3. Kinetics is the study of reaction rates, determining which products are formed fastest. Kinetics also helps to predict how the rate will change if we change the reaction conditions. We will use the chlorination of methane to show how we study a reaction. Before we can propose a detailed mechanism for the chlorination, we must learn everything we can about how the reaction works and what factors affect the reaction rate and the product distribution. A careful study of the chlorination of methane has established three important characteristics: 1. The chlorination does not occur at room temperature in the absence of light. The reaction begins when light falls on the mixture or when it is heated. Thus, we know this reaction requires some form of energy to initiate it.

2. The most effective wavelength of light is a blue color that is strongly absorbed by chlorine gas. This finding implies that light is absorbed by the chlorine mole­ cule, activating chlorine so that it initiates the reaction with methane. 3. The light-initiated reaction has a high quantum yield. This means that many molecules of the product are formed for every photon of light absorbed. Our mechanism must explain how hundreds of individual reactions of methane with chlorine result from the absorption of a single photon by a single mole­ cule of chlorine.

4-3 The Free-Ra d i ca l Cha i n Reaction

A chain reaction mechanism has been proposed to explain the chlorination of methane. A chain reaction consists of three kinds of steps: 1. The initiation step, which generates a reactive intermediate.

2. Propagation steps, in which the reactive intermediate reacts with a stable mole­ cule to form another reactive intermediate, allowing the chain to continue until the supply of reactants is exhausted or the reactive intermediate is destroyed. 3. Termination steps, side reactions that destroy reactive intermediates and tend to slow or stop the reaction.

4-3 The Free-Radical Chain Reaction

1 27

In studying the chlorination of methane, we will consider just the first reaction to form chloromethane (common name methyl chloride). This reaction is a substitution: Chlorine does not add to methane, but a chlorine atom substitutes for one of the hydrogen atoms, which becomes part of the HCl by-product. H I H-C -@ I H

+

heat or light (hv)

@- Cl

H I H-C I H

)

+

@- Cl

chloromethane

chlorine

methane

--@

(methyl chloride)

4-3A

The I n itiation Step: Generation of Radicals

Blue light, absorbed by chlorine but not by methane, promotes this reaction. There­ fore, initiation probably results from the absorption of light by a molecule of chlorine. Blue light has about the right energy to split a chlorine molecule ( CI2) into two chlorine atoms (242 kj/mol or 58 kcal/mol).* The splitting of a chlorine molecule by absorption of a photon is shown as follows: .. r--

: CI �: CI .. :

+

photon (hv)

.

..

.

:CI· + · CI :

Notice the fishhook-shaped half-arrows used to show the movement of single unpaired electrons. Just as we use curved arrows to represent the movement of electron pairs, we use these curved half-arrows to represent the movement of single electrons. These half-arrows show that the two electrons in the CI - Cl bond separate, and one leaves with each chlorine atom. The splitting of a Cl2 molecule is an initiation step that produces two highly reactive chlorine atoms. A chlorine atom is an example of a reactive intermediate, a short-lived species that is never present in high concentration because it reacts as quickly as it is formed. Each CI' atom has an odd number of valence electrons (seven), one of which is unpaired. The unpaired electron is called the odd electron or the radical electron. Species with unpaired electrons are called radicals or free radicals. Radicals are electron-deficient because they lack an octet. The odd elec­ tron readily combines with an electron in another atom to complete an octet and form a bond. Figure 4-1 shows the Lewis structures of some free radicals. Radicals are often represented by a structure with a single dot representing the unpaired odd electron.

--- ---

Lewis structures

:C I ·

:B r·

H :O ·

Wrillen

------

H H :C · H

course of everyday life, reactive oxygen species are encountered in the environment and produced in the body. These compounds break down

into

short-lived

hydroxyl

radicals, which can react with the body's

proteins

and

DNA.

The

reSUlting damage accumulates and may result in heart disease, cancer, and premature aging.

-----

HH H :C :C · HH

CI ·

Br·

HO ·

CH3·

CH3 CH2·

chlorine atom

bromine atom

hydroxyl radical

methyl radical

ethyl radical

*The energy of a photon of light is related to its frequency v by the relationship E = hv, where h is Planck's constant. Blue light has an energy of about 50 kJ (60 kcal) per einstein (an einstein is a mole of photons).

2

Free radicals may play a role in dis­ eases and accelerate aging. In the

... Figure 4- 1

Free radicals. Free radicals are reactive species with odd numbers of electrons. The unpaired electron is rep resented by a dot in the formula.

128

Chapter 4: The Study of Chemical Reactions P RO B L E M 4 - 1

Draw Lewis structures for the following free radicals. (a) The n-propyl radical, CH3-CH2-CH2 (c) The isopropyl radical 4-3 8

(b) The t-butyl radical, (CH3hC' (d) The iodine atom

Propagation Steps

When a chlorine radical collides with a methane molecule, it abstracts (removes) a hydro­ gen atom from methane. One of the electrons in the C -H bond remains on carbon while the other combines with the odd electron on the chlorine atom to form the H - CI bond. First propagation step

y�o� + Cl ' H-C-H

H I H-C ' I H

I H

methane

chlorine atom

+

methyl radical

H-Cl hydrogen chloride

This step forms only one of the final products: the molecule of HCl. A later step must form chloromethane. Notice that the first propagation step begins with one free radical (the chlorine atom) and produces another free radical (the methyl radical). The regen­ eration of a free radical is characteristic of a propagation step of a chain reaction. The reaction can continue because another reactive intermediate is produced. In the second propagation step, the methyl radical reacts with a molecule of chlorine to form chloromethane. The odd electron of the methyl radical combines with one of the two electrons in the CI - CI bond to give the CI- CH 3 bond, and the chlo­ rine atom is left with the odd electron. Secon.d propagation step

Y/'1�� + Cl-Cl H-C

H I H-C -Cl I H

I H

methyl radical

+

chloromethane

chlorine molecule

Cl ' chlorine atom

In addition to forming chloromethane, the second propagation step produces anoth­ er chlorine atom. The chlorine atom can react with another molecule of methane, giving HCI and a methyl radical, which reacts with Cl2 to give chloromethane and regenerate yet another chlorine atom. In this way, the chain reaction continues until the supply of the re­ actants is exhausted or some other reaction consumes the radical intermediates. The chain reaction explains why many molecules of methyl chloride and HCI are formed by each photon of light that is absorbed. We can summarize the reaction mechanism as follows. •

� KEY MECHANISM 4-1

Free-Radical Halogenation

Like many other radical reactions, free-radical halogenation is a chain reaction. Chain reactions usually require one or more initiation steps, to form radicals, followed by propagation steps that produce products and regenerate radicals. Initiation: Radicals are formed. Light supplies the energy to split a chlorine molecule. CI-Cl

+

hv (light)

2 Cl ·

4-3 The Free-Radical Chain Reaction Propagation: A radical reacts to generate another radical. Step 1:

A chlorine radical abstracts a hydrogen to generate an alkyl radical.

H I

H-C-H

I H

+

H I H -C' I H

Cl'

+

H-Cl

methyl radical Step 2:

continues the chain

The alkyl radical reacts with Cl2 to generate the product and a chlorine radical.

H I H-C ' I H

+

-----7

Cl - Cl

H I H - C -CI I H

+

CI '

chloromethane

The chlorine radical generated in Step 2 goes on to react in Step I, continuing the chain . The overall reaction is simply the s um of the propagation steps:

H I H-C-H I H Question:

+

C]- C!

H I H - C -Cl I H

+

H - CI

What factor s characterize the propagation steps of a chain reaction?

P RO B L E M 4 - 2

(a) Write the propagation steps leading to the formation of dichloromethane ( CH2Cl2 ) from chloromethane. (b) Explain why free-radical halogenation usually gives mixtures of products. (c) How could an industrial plant control the proportions of methane and chlorine to favor production of CCI4? To favor CH3Cl?

4-3C

Term i nation Reactions

If anything happens to consume some of the free-radical intermediates without gener­ ating new ones, the chain reaction will slow or stop. Such a side reaction is called a termination reaction: a step that produces fewer reactive intermediates (free radi­ cals) than it consumes. The following are some of the possible termination reactions in the chlorination of methane: H 1r1� H-C- + Cl ' I H Cl r1

0·

H I H - C -Cl I H Cl-Cl

129

130

Chapter 4: The Study of Chemical Reactions

H H 1r1� I H-C + ·C-H I I H H

H H I I H-C-C-H I I H H H

H

I

H-C· collides with wall

I H

Cl · collides with wall

Some anti-cancer agents act by generating highly reactive hydrox­ yl

radicals,

which

damage

and

degrade the DNA of the rapidly dividing tumor cells. As a result, the cells die and the tumor shrinks. One example of a radical generator is bleomycin, which is used for the treatment of testicular cancer.

PROBLEM-SOLVING

H?ltp

In a free-radical chain reaction,

initiation steps generally create new free radicals. Propagation steps usually combine a free radical and a reactant to give a product and another free radical. Termination steps generally decrease the number of free radicals.

H-

? -B H

CI

-1M�al l J

The combination of any two free radicals is a termination step because it decreases the number of free radicals. Other termination steps involve reactions of free radicals with the walls of the vessel or other contaminants. Although the first of these termination steps gives chloromethane, one of the products, it consumes the free radi­ cals that are necessary for the reaction to continue, thus breaking the chain. Its contri­ bution to the amount of product obtained from the reaction is small compared with the contribution of the propagation steps. While a chain reaction is in progress, the concentration of radicals is very low. The probability that two radicals will combine in a termination step is lower than the probability that each will encounter a molecule of reactant and give a propagation step. The termination steps become important toward the end of the reaction, when there are relatively few molecules of reactants available. At this point, the free radicals are less likely to encounter a molecule of reactant than they are to encounter each other (or the wall of the container). The chain reaction quickly stops. P RO B L E M 4 - 3

Each o f the following proposed mechanisms for the free-radical chlorination o f methane is wrong. Explain how the experimental evidence disproves each mechanism. (a) Cl2 + hv � Cl ; (an "activated" form of C12) Cl ; + CH4 � HCI + CH3CI (b) CH4 + hv � ·CH3 + H· · CH3 + Cl2 � CH3Cl + CI· CI · + H- � HCI P R OB L E M 4 - 4

Free-radical chlorination of hexane gives very poor yields of l -chlorohexane, while cyclo­ hexane can be converted to chlorocyclohexane in good yield. (a) How do you account for this difference? (b) What ratio of reactants (cyclohexane and chlorine) would you use for the synthesis of chlorocyclohexane?

4-4 Equilibri u m Constants and Free Energy

N ow that we have determined a mechanism for the chlorination of methane, we can consider the energetics of the individual steps. Let's begin by reviewing some of the principles needed for this discussion. Thermodynamics is the branch of chemistry that deals with the energy changes accompanying chemical and physical transformations. These energy changes are most u seful for describing the properties of systems at equilibrium. Let's review how ener­ gy and entropy variables describe an equilibrium.

4-4 Equilibrium Constants and Free Energy

13 1

The equilibrium concentrations of reactants and products are governed by the equilibrium constant of the reaction. For example, if A and B react to give C and D, then the equilibrium constant Keg i s defined by the following equation: ) C + D [products] [C][D] Ke g [reactants] [A][B] The value of Keg tells us the position of the equilibrium: whether the products or the reactants are more stable, and therefore energetically favored. If Keg is larger than 1, the reaction is favored as written from left to right. If Keg is less than 1, the reverse re­ action is favored (from right to left as written). The chlorination of methane has a large equilibrium constant of about 1.1 X 1 0 1 9 . A

+

B

(

= ----

CH4 Keq

+

CI 2

(

)

CH3C1

[CH3Cl][ HCI]

=

[CH4] [CI2 ]

=

+

HCI

1 . 1 X 1 019

The equilibrium constant for chlorination is so large that the remaining amounts of the reactants are close to zero at equilibrium. Such a reaction is said to go to completion, and the value of Keg is a measure of the reaction's tendency to go to completion. From the value of Keq we can calculate the change in free energy (sometimes called Gibbs free energy) that accompanies the reaction. Free energy is represented by G, and the change (Ll) in free energy associated with a reaction is represented by LlG, the difference between the free energy of the products and the free energy of the reactants. LlG is a measure of the amount of energy available to do work.

LlG

=

(free energy of products ) - (free energy of reactants)

If the energy levels of the products are lower than the energy levels of the reactants (a "downhill" reaction), then the reaction i s energetically favored; and this equation gives a negative value of LlG, corresponding to a decrease in the energy of the system. The standard Gibbs free energy change, LlGo, is most commonly used. The symbol ° designates a reaction involving reactants and products in their standard states (pure substances in their most stable states at 25°C and 1 atm pressure). The relation­ ship between LlGo and Keg is given by the expression K eq

=

e -t:J.GOjRT

or, conversely, by

where

R = 8 . 3 1 4 Jlkelvin-mol (l.987 cal/kelvin-mol), the gas constant T absolute temperature, in kelvins* e 2.7 1 8, the base of natural logarithms The value of RT at 25°C is about 2.48 kJ/mol (0.592 kcal/mol). =

=

The formula shows that a reaction is favored ( Ke q > 1) if it has a negative value of LlGo (energy is released). A reaction that has a positive value of LlGo (energy must be added) is unfavorable. These predictions agree with our intuition that reac­ tions should go from higher-energy states to lower-energy states, with a net decrease in free energy. *Absolute temperatures (in kelvins) are correctly given without a degree sign, as in the equation 25°C

298 K. We

will often include the degree sign, however, to distinguish absolute temperatures (K) from equilibrium constants (K) as in 25°C

=

298°K.

=

PROBLEM-SOLVING

H?nl/

A reaction with a negative t:J.G is favored.

A reaction with a positive unfavorable.

t:J.G is

132

Chapter 4: The Study of Chemical Reactions S O LV E D P RO B L E M 4- 1

Calculate the value of !:!.Go for the chlorination of methane. S O L UTION

-2.303RT(log Keq) Keq for the chlorination is l . 1 X 1 019, and log Keq !:!'Go

=

=

1 9 .04

At 25°C (about 298° Kelvin), the value of RT is RT

=

( 8 . 3 1 4 llkelvin-mol) ( 298 kelvins)

=

2478 llmol, or 2.48 kJ/mol

Substituting, we have !:!'Go

=

(-2.303)(2.478 kllmol) ( 1 9.04)

=

- 108.7 kJ/mo\ ( -25 .9 kcaIjmol)

This is a large negative value for !:!.Go, showing that this chlorination has a large driving force that pushes it toward completion.

In general, a reaction goes nearly to completion (>99%) for values of !1Go that are more negative than about 1 2 kJ /mol or 3 kcalj mol. Table 4- 1 shows what per­ centages of the starting materials are converted to products at equilibrium for reactions with various values of !1Go. -

-

P RO B L E M 4 - 5

The following reaction has a value of !:!'Go

=

-

2 . 1 kJ/mol (-0.5 kcaIjmol ) .

(a) Calculate Keq a t room temperature (25° C ) for this reaction a s written. (b) Starting with aiM solution of CH3Br and H 2 S, calculate the final concentrations of all four species at equilibrium. P RO B L E M 4-6

At room temperature (25° C), the reaction of two molecules of acetone to form diacetone alco­ hol proceeds to an extent of about 5%. Determine the value of !:!'Go for this reaction.

OH II I CH3-C-CH2-C(CH3)2 o

o

II 2CH3-C-CH3 acetone

TABLE 4-1

diacetone alcohol

Product Composition as a Function of !:!'GO at 25°C

100% !:!'GO kJ/mol

+4.0 +2.0 0.0 -2.0 -4.0 -8.0 -12.0 -16.0 -20.0

kcal/mol

K

Conversion to Products

( + 1 .0) ( +0.5) (0.0) ( -0.5) ( - 1 .0) ( -1.9) ( -2.9) ( -3.8) ( -4.8)

0.20 0.45 1 .0 2.2 5.0 25 1 27 638 3200

17% 31% 50% 69% 83% 96% 99.2% 99.8% 99.96%

t)

90%

c...

80%

.u; ....

70%

u

60%

'"

::l '0 0 ....

B

a 0

(\) >a 0

50%

0.0

-4.0

-8.0 -12.0 !:!'Go, kllmol

-1 6.0

-20.0

4-5 Enthalpy and Entropy Two factors contribute to the change in free energy: the change in enthalpy and the change in entropy multiplied by the temperature. tlGo tlGO tlHo tlso

=

=

=

=

tlHo - Ttlso

( free energy of products) - ( free energy of reactants) ( enthalpy of products) - (enthalpy of reactants ) ( entropy of products ) - (entropy of reactants)

At low temperatures, the enthalpy term (tlHO) is usually much larger than the entropy term (-TtlSO), and the entropy term is sometimes ignored. 4-5A

E nthalpy

The change in enthalpy (tlHO) is the heat of reaction-the amount of heat evolved or consumed in the course of a reaction, usually given in kilojoules (or kilocalories) per mole. The enthalpy change is a measure of the relative strength of bonding in the prod­ ucts and reactants. Reactions tend to favor products with the lowest enthalpy (those with the strongest bonds). If weaker bonds are broken and stronger bonds are formed, heat is evolved and the reaction is exothermic (negative value of tlHO). In an exothermic reaction, the enthalpy term makes a favorable negative contribution to tlGo. If stronger bonds are broken and weaker bonds are formed, then energy is consumed in the reaction, and the reaction is endothermic (positive value of tlHO). In an endothermic reaction, the enthalpy term makes an unfavorable positive contribution to tlGo. The value of tlHo for the chlorination of methane is about - 105. 1 kJ/mol ( - 25.0 kcaIjmol) . This is a highly exothermic reaction, with the decrease in enthalpy serving as the primary driving force. 4-5 8

E ntropy

Entropy is often described as randomness, or freedom of motion. Reactions tend to favor products with the greatest entropy. Notice the negative sign in the entropy term (-TtlSO) of the free-energy expression. A positive value of the entropy change (tlSO), indicating that the products have more freedom of motion than the reactants, makes a favorable (negative) contribution to tlGo. In many cases, the enthalpy change (tlHO) is much larger than the entropy change (tlSO), and the enthalpy term dominates the equation for tlGO. Thus, a nega­ tive value of tlso does not necessarily mean that the reaction has an unfavorable value of tlGo. The formation of strong bonds (the change in enthalpy) is usually the most important component in the driving force for a reaction. In the chlorination of methane, the value of tlso is + 1 2. 1 J/kelvin-mole. The -Ttlso term in the free energy is -TtlSo

=

=

The value of tlGo tlGo

=

=

=

- ( 298°K) ( 1 2. 1 J/Kelvin-mol)

=

-3610 J/mol

-3. 6 1 kJ/mol ( -0.86 kcaIjmol )

- 1 08.7 kJ/mol i s divided into enthalpy and entropy terms: tlHo - TtlSo

=

- 105. 1 kJ/mol - 3.61 kJ/mol

- 108.7 kJ/mol ( -25.9 kcaIjmol)

The enthalpy change is the largest factor in the driving force for chlorination. This is the case in most organic reactions: The entropy term is often small in relation to the enthalpy term. When we discuss chemical reactions involving the breaking and forming of bonds, we can often use the values of the enthalpy changes (tlHO), under the assumption that tlGo == tlHo. We must be cautious in making this approximation, however, because some reactions have relatively small changes in enthalpy and larger changes in entropy.

4-5 Entha l py and Entropy

133

134

Chapter 4: The Study of Chemical Reactions S O LV E D P RO B L E M 4 - 2

Predict whether the value o f I::!.so for the dissociation o f Cl2 i s positive (favorable) o r nega­ tive (unfavorable). What effect does the entropy term have on the sign of the value of I::!.Go for this reaction? SOL UTION

Two isolated chlorine atoms have much more freedom of motion than a single chlorine molecule. Therefore, the change in entropy is positive, and the entropy term (-TI::!.S O) is negative. This negative (favorable) value of (-TI::!. SO ) is small, however, compared with the much larger, positive (unfavorable) value of I::!.Ho. The chlorine molecule is much more sta­ ble than two chlorine atoms, showing that the positive enthalpy term predominates. PROBLEM-SOLVING

Hi-ltv

In general, two smaller molecules (or

fragments, such as radicals) have more freedom of motion (greater entropy) than one larger molecule.

P RO B L E M 4 - 7

When ethene i s mixed with hydrogen i n the presence of a platinum catalyst, hydrogen adds across the double bond to form ethane. At room temperature, the reaction goes to completion. Predict the signs of I::!.Ho and I::!.So for this reaction. Explain these signs in terms of bonding and freedom of motion. Pt catal yst �

)

H H I I H-C-C-H I I H H

ethene

ethane

P RO B L E M 4 - 8

For each reaction, estimate whether I::!. So for the reaction is positive, negative, or impossible to predict.

CloH22

heat

C3H6 + C7HI6 heptane propene n-decane (b) The formation of diacetone alcohol: 0 0

(a)

catalyst

II 2 CH3-C-CH3

-OH

0

II (c) CH3-C-OH + CHpH

4-6 Bond-Dissociation Enthal pies

OH II I CH3-C-C�-C(CH3)2 H+

0

II CH3-C-OCH3

+

HzO

We can put known amounts of methane and chlorine into a bomb calorimeter and use a hot wire to initiate the reaction. The temperature rise in the calorimeter is used to calculate the precise value of the heat of reaction, fl.Ho. This measurement shows that 105 kJ (25 kcal) of heat is evolved (exothermic) for each mole of methane converted to chloromethane. Thus, fl.Ho for the reaction is negative, and the heat of reaction is given as fl.Ho - 1 05 kli mol ( -25 kcaljmol ) In many cases, we want to predict whether a particular reaction will be endother­ mic or exothermic, without actually measuring the heat of reaction. We can calculate an approximate heat of reaction by adding and subtracting the energies involved in the breaking and forming of bonds. To do this calculation, we need to know the energies of the affected bonds. The bond-dissociation enthalpy (BDE, also called bond-dissociation energy) is the amount of enthalpy required to break a particular bond homolytically, that is, in such a way that each bonded atom retains one of the bond's two electrons. In contrast, when a bond is broken heterolytically, one of the atoms retains both electrons. =

4-7 Enthalpy Changes in Chlorination Homolytic cleavage (free radicals result) r

A' +

A:B �

.. f..

..

: CI : CI: U '

'B

2 : Cl '

"

6.HO

=

6.HO

=

bond-dissociation enthalpy 242 kJ/mol (58 kcal/mol)

Heterolytic cleavage (ions result) r-

A:B

-----;.

A+

+

:B

n· (CH3)3C - �1 :

-----;.

(CH3hC+

+

: Cl :-

(6.H o varies with solvent)

Homolytic cleavage (radical cleavage) forms free radicals, while heterolytic cleavage (ionic cleavage) forms ions. Enthalpies for heterolytic (ionic) cleavage de­ pend strongly on the solvent's ability to solvate the ions that result. Homolytic cleav­ age is used to define bond-dissociation enthalpies because the values do not vary so much with different solvents or with no solvent. Note that a curved arrow is used to show the movement of the electron pair in an ionic cleavage, and curved half-arrows are used to show the separation of individual electrons in a homolytic cleavage. Energy is released when bonds are formed, and energy is consumed to break bonds. Therefore, bond-dissociation enthalpies are always positive (endothermic). The overall enthalpy change for a reaction is the sum of the dissociation enthalpies of the bonds broken minus the sum of the dissociation en thaI pies of the bonds formed. 6.Ho 2:(BDE of bonds broken ) - 2: ( BDE of bonds formed) By studying the heats of reaction for many different reactions, chemists have developed reliable tables of bond-dissociation enthalpies. Table 4-2 gives the bond-dissociation enthalpies for the homolysis of bonds in a variety of molecules. =

We can use values from Table 4-2 to predict the heat of reaction for the chlorination of methane. This reaction involves the breaking (positive values) of a CH3 - H bond and a CI - CI bond, and the formation (negative values) of a CH3 -CI bond and a H - CI bond. Overall reaction

Bonds broken CI- Cl CH3 - H Total 6.Ho

6.Ho (per mole) +242 kJ (+58 kcal ) +435 kJ (+ 104 kcal ) +677 kJ (+162 kcal ) =

Bonds formed 6.Ho (per mole) H - Cl -431 kJ (-103 kcal ) CH3 - CI -351 kJ (-84 kcal) Total -782 kJ (-1 87 kcal )

+677 kJ + (-782) kJ

=

- 1 05 kJ/mol (-25 kcal/mol )

The bond-dissociation enthalpies also provide the heat of reaction for each indi­ vidual step: First propagation step

CI · + CH4 Breaking a CH3 - H bond Forming an H - CI bond Step total

�

' CH3 + HCI +435 kJ/mol (+ 1 04 kcaljmol ) -43 1 kJ/mol (-103 kcaljmol ) +4 kJ/mol (+ 1 kcal/mol )

4-7 Enthal py Changes in Chlorination

135

136

Chapter 4: The Study of Chemical Reactions

TABLE 4-2

Bond-Dissociation Enthalpies for Homolytic Cleavages A:B � A· + ·B Bond-Dissociation Enthalpy

Bond-Dissociation Enthalpy kJ/mol

kcal/mol

F-F CI-CI Br-Br I-I H-F H-CI H-Br H-I HO-H HO-OH

435 444 159 242 192 151 569 431 368 297 498 213

104 106 38 58 46 36 136 103 88 71 119 51

CH3-H CH3-F CH3-Cl CH3-Br CH3-1 CH3-OH

435 456 351 293 234 381

104 109 84 70 56 91

CH3CH2-H CH3CH2-F CH3CH2-Cl CH3CH2-Br CH3CH2-1 CH3CH2-OH CH3CH2CH2-H CH3CH2CH2-F CH3CH2CH2-CI CH3CH2CH2-Br CH3CH2CH2-I CH3CH2CH2-OH

410 448 339 285 222 381 410 448 339 285 222 381

98 107 81 68 53 91 98 107 81 68 53 91

Bond

D-D

kJ/mol

kcal/mol

397 444 335 285 222 381

95 106 80 68 53 91

381 444 331 272 209 381

91 106 79 65 50 91

356 364 464 473

85 87 111 113

368 356 343 351 339

88 85 82 84 81

Bonds to secondary carbons

H - X bonds and X - X bonds

H-H

Bond

Methyl bonds

Bonds to primary carbons

(CH3hCH-H (CH3hCH-F (CH3hCH-CI (CH3hCH-Br (CH3hCH-I (CH3hCH-OH

Bonds to tertiary carbons

(CH3hC-H (CH3hC-F (CH3hC-CI (CH3hC-Br (CH3hC-I (CH3)3C-OH

Other C- H bonds

PhCH2-H (benzylic)

CH2=CHCH2-H (allylic) CH2=CH-H (vinyl) Ph-H (aromatic) C-C bonds

CH3-CH3 CH3CH2-CH3 CH3CH2-CH2CH3 (CH3hCH-CH3 (CH3)3C-CH3

Second propagation step

PROBLEM-SOLVING

H ?nv

Bond-dissociation enthalpies are for

breaking bonds, which costs energy. In

calculating values of t'lHo, use positive BDE values for bonds that are broken

and negative values for bonds that are formed.

'CH3 + Cl2 � CH3Cl + CI ' +243 klimol (+58 kcal/ mol ) Breaking a Cl - Cl bond -352 klimol ( - 84 kcal/ mol ) Forming a CH3 - Cl bond - 1 09 klimol ( -26 kcal/ mol ) Step total Grand total

=

+4 kli mol + ( - 1 09 kli mol )

=

- 105 klimol ( -25 kcalimol )

The sum of the values of I1Ho for the individual propagation steps gives the overall enthalpy change for the reaction. The initiation step, Cl2 - 2 CI ' , is not added to give the overall enthalpy change because it is not necessary for each molecule of prod­ uct formed. The first splitting of a chlorine molecule simply begins the chain reaction, which generates hundreds or thousands of molecules of chloromethane. P R OB L E M 4 - 9

(a) Propose a mechanism for the free-radical chlorination of ethane,

CH3 - CH3 + Cl2

hv

�

CH3 - CH2Cl + HCI

4-8 Kinetics and the Rate Equation

137

(b) Calculate 6Ho for each step in this reaction.

(c) Calculate the overall value of 6Ho for this reaction. Alter native M echanis m

The mechanism we have used is not the only one that might be proposed to explain the reaction of methane with chlorine. We know that the initiating step must be the splitting of a molecule of C12, but there are other propaga­ tion steps that would form the correct products: (a) CI' + CH3 - H � CH3 - Cl + H- 6.Ho +435 kJ - 351 kJ =

(b) H'

+ Cl - CI � H - Cl + CI'

6.Ho

=

+242 kJ - 431 kJ

=

+84 kJ ( +20 kcal)

-189 kJ ( -45 kcal)

Total -105 kJ ( -25 kcal) This alternative mechanism seems plausible, but Step (a) is endothermic by 84 kJ/mol (20 kcal/mol). The previous mechanism provides a lower-energy alternative. When a chlorine atom collides with a methane molecule, it will not react to give methyl chlOIide and a hydrogen atom (6.Ho +84 kJ +20 kcal ) ; it will react to give HCI and a methyl radical (6.Ho +4 kJ +1 kcal) , the first propagation step of the correct mechanism. =

=

=

=

P R OB L E M 4 - 1 0 (a) Using bond-dissociation enthalpies from Table 4-2 (page 1 36), calculate the heat of reaction

for each of the steps in the free-radical bromination of methane.

B r2 + CH 4

heat o r light

) CH 3 B r + HB r

(b) Calculate the overall heat of reaction.

Kinetics is the study of reaction rates. How fast a reaction goes is just as important as the position of its equilibrium. Just because thermodynamics favors a reaction (nega­ tive 6.GO) does not necessarily mean the reaction will actually occur. For example, a mixture of gasoline and oxygen does not react without a spark or a catalyst. Similarly, a mixture of methane and chlorine does not react if it is kept cold and dark. The rate of a reaction is a measure of how fast the products appear and the reac­ tants disappear. We can determine the rate by measuring the increase in the concentrations of the products with time, or the decrease in the concentrations of the reactants with time. Reaction rates depend on the concentrations of the reactants. The greater the con­ centrations, the more often the reactants collide and the greater the chance of reaction. A rate equation (sometimes called a rate law) is the relationship between the concen­ trations of the reactants and the observed reaction rate. Each reaction has its own rate equation, determined experimentally by changing the concentrations of the reactants and measuring the change in the rate. For example, consider the general reaction A + B � C + D The reaction rate is usually proportional to the concentrations of the reactants ([A] and [B]) raised to some powers, a and b. We can use a general rate expression to represent this relationship as rate kr[AY'[B]b where kr is the rate constant, and the values of the powers (a and b) must be deter­ mined experimentally. We cannot guess or calculate the rate equation from just the stoichiometry of the reaction. The rate equation depends on the mechanism of the reaction and on the rates of the individual steps in the mechanism. In the general rate equation, the power a is called the order of the reaction with respect to reactant A, and b is the order of the reaction with respect to B. The sum of these powers, (a+b), is called the overall order of the reaction. The following reaction has a simple rate equation: =

CH3 - Br + -OH

H,O/acetone

-

) CH3 - OH + Br-

4-8 Kinetics and the Rate Equation

138

Chapter 4: The Study of Chemical Reactions

Experiments show that doubling the concentration of methyl bromide, [ CH3Br], dou­ bles the rate of reaction. Doubling the concentration of hydroxide ion, [-OH], also doubles the rate. Thus, the rate is proportional to both [ CH3Br] and [-OH], so the rate equation has the following form: rate kr[CH3Br][-OH] This rate equation is first order in each of the two reagents because it is proportional to the first power of their concentrations. The rate equation is second order overall be­ cause the sum of the powers of the concentrations in the rate equation is 2; that is, ( first order ) + ( first order ) = second order overall. Reactions of the same overall type do not necessarily have the same form of rate equation. For example, the following similar reaction has a different kinetic order: =

( CH3 h C - Br + -OH

H20/acetone

'>

( CH3 h C - OH + Br-

Doubling the concentration of t buty l bromide [ ( CH3 h C - Br] causes the rate to double, but doubling the concentration of hydroxide ion [-OH] has no effect on the rate of this particular reaction. The rate equation is rate kr[ ( CH3 ) 3C -Br] This reaction is first order in t buty l bromide, and zeroth order in hydroxide ion (pro­ portional to [-OH] to the zeroth power). It is first order overall. The most important fact to remember is that the rate equation must be deter­ mined experimentally. We cannot predict the form of the rate equation from the stoi­ chiometry of the reaction. We determine the rate equation experimentally, then use that information to propose consistent mechanisms. -

=

-

S O LV ED P R OB L E M 4 - 3

Chloromethane reacts with dilute sodium cyanide ( Na+ -C=N) according to the follow­ ing equation: CH3-CI + T=N � CH3 - C - N + Clchloride acetonitrile cyanide chloromethane When the concentration of chloromethane is doubled, the rate is observed to double. When the concentration of cyanide ion is tripled, the rate is observed to triple. (a) What is the kinetic order with respect to chloromethane? (b) What is the kinetic order with respect to cyanide ion? (c) What is the kinetic order overall? (d) Write the rate equation for this reaction. SOLUTIO N

(a) When [CH3CI] is doubled, the rate doubles, which is 2 to the first power. The reaction is first order with respect to chloromethane. (b) When [-CN] is tripled, the reaction rate triples, which is 3 to the first power. The reaction is first order with respect to cyanide ion. (c) First order plus first order equals second order overall. (d) rate kr[CH3CI][TN] =

P R OB L E M 4 - 1 1

The reaction of t-butyl chlOlide with methanol ( CH3hC - Cl t-butyl chloride

+

CH3 - OH methanol

is found to follow the rate equation rate

=

�

( CH3hC - OCH3 methyl t-butyl ether

kr[ ( CH3hC - CI]

(a) What is the kinetic order with respect to t-butyl chloride? (b) What is the kinetic order with respect to methanol? (c) What is the kinetic order overall?

+

HCI

4-9 Activation Energy and the Temperature Dependence of Rates P R OB L E M 4 - 1 2

Under certain conditions, the bromination of cyclohexene follows an interesting rate law: H Br

G

Br H

(a) What is the kinetic order with respect to cyclohexene? (b) What is the kinetic order with respect to bromine? (c) What is the overall kinetic order? P R OB L E M 4 - 1 3

When a small piece of platinum is added to a mixture of ethene and hydrogen, the following reaction occurs: H Pt catalyst

ethene

hydrogen

)

I

H

I

H-C-C-H

I

H

I

H

eth ane

Doubling the concentration of hydrogen has no effect on the reaction rate. Doubling the con­ centration of ethene also has no effect. (a) What is the kinetic order of this reaction with respect to ethene? With respect to hydro­ gen? What is the overall order? (b) Write the unusual rate equation for this reaction. (c) Explain this strange rate equation, and suggest what one might do to accelerate the reaction.

Each reaction has its own characteristic rate constant, kr. Its value depends on the conditions of the reaction, especially the temperature. This temperature dependence is expressed by the Arrhenius equation, kr Ae -E,,/RT =

where

A

=

Ea

R T

a constant (the "frequency factor") activation energy the gas constant, 8 .3 1 4 llkelvin-mole ( 1 .987 callkelvin-mole) the absolute temperature

=

=

=

The activation energy, E a, is the minimum kinetic energy the molecules must have to overcome the repulsions between their electron clouds when they collide. The exponential term e -Ea/RT corresponds to the fraction of collisions in which the particles have the minimum energy Ea needed to react. We can calculate Ea for a reaction by measuring how kr varies with temperature, and substituting into the Arrhenius equation. The frequency factor A accounts for the frequency of collisions and the fraction of collisions with the proper orientation for the reaction to occur. In most cases, only a small fraction of collisions occur between molecules with enough speed and with just the right orientation for reaction to occur. Far more collisions occur without enough kinetic energy or without the proper orientation, and the molecules simply bounce off each other. The Arrhenius equation implies that the rate of a reaction depends on the fraction of molecules with kinetic energy of at least E a. Figure 4-2 shows how the distribution of kinetic energies in a sample of a gas depends on the temperature. The black curved line shows the molecular energy distribution at room temperature, and the dashed lines show the energy needed to overcome barriers of 4 kllmol ( 1 kcal/mol), 40 kllmol ( 1 0 kcal/mol), and 80 kl ( 1 9 kcal/mol). The area under the curve to the right of each barri­ er corresponds to the number of molecules with enough energy to overcome that

4-9 Activation Energy and the Tem perature Dependence of Rates

1 39

140

Chapter 4: The Study of Chemical Reactions

i

\

room temperature (3000K)

� Figure 4-2

The dependence of kinetic energies on temperature. This graph shows how the number of molecules with a given activation energy decreases as the activation energy increases. At a higher temperature (red curve), more collisions have the needed energy.

40 kJ/mol 80 kllmal

�

energy (E)

barrier. The red curve shows how the energy distribution is shifted at 1 00°C. At 100°C, many more molecules have the energy needed to overcome the energy barriers, espe­ cially the 80 kJ/mol barrier. For smaller temperature changes, chemists often use an approximation: For reactions with typical activation energies of about 40 to 60 kJ/mol ( 1 0 to 15 kcal/mol), the reaction rate approximately doubles when the temperature is raised by lo°C, as from 27°C (near room temperature) to 37°C (body temperature). Because the relative rate constant, kreb increases quickly when the temperature is raised, it might seem that raising the temperature would always be a good way to save time by making reactions go faster. The problem with raising the temperature is that all reactions are accelerated, including all the unwanted side reactions. We try to find a temperature that allows the desired reaction to go at a reasonable rate without producing unacceptable rates of side reactions.

4-10 Transition States

The activation energy Ea represents the energy difference between the reactants and the transition state, the highest-energy state in a molecular collision that leads to reaction. In effect, the activation energy is the barrier that must be overcome for the reaction to take place. The value of Ea is always positive, and its magnitude depends on the relative energy of the transition state. The term transition state implies that this configuration is the transition between the reactants and products, and the molecules can either go on to products or return to reactants. Unlike the reactants or products, a transition state is unstable and cannot be iso­ lated. It is not an intermediate, because an intermediate is a species that exists for some finite length of time, even if it is very short. An intermediate has at least some stability, but the transition state is a transient on the path from one intermediate to another. The transition state is often symbolized by a superscript double dagger ( :j: ) , and the changes in variables such as free energy, enthalpy, and entroPl involved in achieving the transition state are symbolized IlG:j:, IlH:j:, and IlS:j: . IlG is similar to Ea, and the symbolllG:j: is often used in speaking of the activation energy. Transition states have high energies because bonds must begin to break before other bonds can form. The following equation shows the reaction of a chlorine radical with methane. The transition state shows the C - H bond partially broken and the H - Cl bond partially formed. Transition states are often enclosed by brackets to emphasize their transient nature.

H I H-C-H I H

. /H H - C ...... H

+ 'Cl transition state

+

H-Cl

4-10 Transition States

t

transition state �

.... Figure 4-3

:j:

(��'�t0�:� �['f{O

Reaction-energy diagram for a one-step exothermic reaction. The reactants are toward the left, and the products are toward the right. The vertical axis represents the potential energy. The transition state is the highest point on the graph, and the activation energy is the energy difference between the reactants and the transition state.

_________

C+D (products) reaction coordinate __

Reactio n-Energy D i agrams The concepts of transition state and activation ener­ gy are easier to understand graphically. Figure 4-3 shows a reaction-energy diagram for a one-step exothennic reaction. The vertical axis of the energy diagram represents the total potential energy of all the species involved in the reaction. The horizontal axis is called the reaction coordinate. The reaction coordinate symbolizes the progress of the reaction, going from the reactants on the left to the products on the right. The tran­ sition state is the highest point on the graph, and the activation energy is the energy difference between the reactants and the transition state. The heat of reaction (!::..HO) is the difference in energy between the reactants and the products. If a catalyst were added to the reaction in Figure 4-3, it would create a transition state of lower energy, thereby lowering the activation energy. Addition of a catalyst would not change the energies of the reactants and products, however, so the heat of reaction and the equilibrium constant would be unaffected. S O LV E D P R OB L E M 4 - 4

Consider the following reaction:

This reaction has an activation energy (Ea) of + 1 7 kJImol ( +4 kcal/mol ) and a !:!.Ho of +4 kJ/ mol (+ 1 kcal/mol ) . Draw a reaction-energy diagram for this reaction. SOLUTION

We draw a diagram that shows the products to be 4 kJ higher i n energy than the reactants. The barrier is made to be 17 kJ higher in energy than the reactants.

+"

- --

i

E,

r

[H3C ---H- -- C1)"

--- ---- ---- --. kl _ _ _

_

CH4 + CI·

141

transition state

��:k: ���

reaction coordinate __ P R OB L E M 4 - 1 4 (a) Draw the reaction-energy diagram for the reverse reaction:

·CR3 + RCI � CR4 + CI·

(b) What is the activation energy for this reverse reaction? (c) What is the heat of reaction (!:!.HO) for this reverse reaction?

kJ

Enzymes serve as biological cata­ lysts.

They

speed

up

reactions

without changing the energies of the

reactants

(called

substrates)

and products. Without enzymes, most of the reactions in our cells would not go fast enough to keep us alive.

142

Chapter 4: The Study of Chemical R e actions PROBLEM 4-1 5 (a) Draw a reaction-energy diagram for the following reaction:

The activation energy is 4 kllmol (1 kcal/mol), and the overall !J.Ho for the reaction is 1 09 kllmol ( -26 kcal/ mol ) . (b) Give the equation for the reverse reaction. (c) What is the activation energy for the reverse reaction? -

4-11 Rates of M u ltiste p Reactions

Many reactions proceed by mechanisms involving several steps and several intermedi­ ates. As we saw in Section 4.7, for example, the reaction of methane with chlorine goes through two propagation steps. The propagation steps are shown here, along with their heats of reaction and their activation energies. Just the propagation steps are shown because the rate of the initiation step is controlled by the amount of light or heat available to split chlorine molecules. Step

CH4 + CI ' � 'CH3 + HCI 'CH3 + Cl2 � CH3C1 + CI ·

-

!::. HO (per mole) +4 kJ (+ 1 kcal ) 1 09 kJ ( -26 kcal )

Ea (per mole) l 7 kJ (4 kcal ) 4 kJ ( 1 kcaJ )

In this reaction, CI· and CH3· are reactive intermediates. Unlike transition states, reactive intermediates are stable as long as they do not collide with other atoms or molecules. As free radicals, however, CI · and CH3· are quite reactive toward other molecules. Figure 4-4 shows a single reaction-energy profile that includes both propa­ gation steps of the chlorination. The energy maxima (high points) are the unstable transition states, and the energy minima (low points) are the intermediates. This com­ plete energy profile provides most of the important information about the energetics of the reaction. The Rate- lim iting Step

In a multistep reaction, each step has its own characteris­ tic rate. There can be only one overall reaction rate, however, and it is controlled by the rate-limiting step (also called the rate-determining step). In general, the highest­ energy step of a multistep reaction is the "bottleneck," and it determines the overall rate. How can we tell which step is rate limiting? If we have the reaction-energy dia­ gram, it is simple: The highest point in the energy diagram is the transition state with the highest energy-the transition state for the rate-limiting step. The highest point in the energy diagram of the chlorination of methane (Figure 4-4) is the transition state for the reaction of methane with a chlorine radical.

rate-limiting transition state

� Figure 4-4

Combined reaction-energy diagram for the chlorination of methane. The energy maxima are transition states, and the energy minima are intermediates. (Units are kllmo\.)

I

cD

l�-� �;-�w =:a �

. . ���;;:�51� . . � : . .. C

:

H3

�

_ _ _

C

..

I

.

..

reaction coordinate

.

-----+-

C

H3

W"

"

Cl + Clo

- I ® kl/m o

4- 1 2 Temperature Dependence of Halogenation This step must be rate limiting. If we calculate a rate for this slow step, it will be the rate for the overall reaction. The second, faster step will consume the products of the slow step as fast as they are formed.

We now apply what we know about rates to the reaction of methane with halogens. The rate-limiting step for chlorination is the endothermic reaction of the chlorine atom with methane to form a methyl radical and a molecule of HCI. Rate- limiting step

CH4 + Cl'

�

'CH3 + HCl

The activation energy for this step is 17 kJ/mol (4 kcal/mol). At room temperature, the value of e -EJRT is 1 300 X 1 0-6 . This value represents a rate that is fast but controllable. In a free-radical chain reaction, every propagation step must occur quickly, or the free radicals will undergo unproductive collisions and become involved in termination steps. We can predict how quickly the various halogen atoms react with methane given relative rates based on the measured activation energies of the slowest steps: Relative Rate (e-Ea/RT x 1 0 6) Reaction

p. Cl' Br' I'

+ + + +

CH4 CH4 CH4 CH4

--

--

---

HF + 'CH3 HCl + 'CH3 HBr + 'CH3 HI + 'CH3

fa (per mole)

27°C(3000K)

5 kJ ( 1 .2 kcal) (4 kcal) 75 kJ ( 1 8 kcal) 1 40 kJ (34 kcal)

1 40,000 1 300 9 X 1 0-8 2 X 1 0- 19

17 kJ

227°C(5000K)

300,000 18,000 0.0 1 5 2 X 1 0-9

These relative rates suggest how easily and quickly methane reacts with the differ­ ent halogen radicals. The reaction with fluorine should be difficult to control because its rate is very high. Chlorine should react moderately at room temperature, but it may be­ come difficult to control if the temperature lises much (the rate at SOooK is rather high). The reaction with bromine is very slow, but heating might give an observable rate. Iodi­ nation is probably out of the question because its rate is exceedingly slow, even at SOO°K. Laboratory halogenations show that our predictions are right. In fact, fluorine reacts explosively with methane, and chlorine reacts at a moderate rate. A mixture of bromine and methane must be heated to react, and iodine does not react at all. PRO B L E M 4- 1 6

The bromination of methane proceeds through the following steps:

(a)

(b) (c) (d)

t:.Ho (Ee r mole1 hv Br2 � 2 Br' + 1 92 kJ (46 kcal ) CH4 + Br' --3> 'CH3 + HBr 67 kJ ( 16 kcal ) --3> ' CH3 + Br2 CH3Br + Br' - 1 0 1 kJ ( -24 kcal ) Draw a complete reaction-energy diagram for this reaction. Label the rate-limiting step. Draw the structure of each transition state. Compute the overall value of t:.Ho for the bromination.

P R O B L E M 4- 1 7

Ea (Eer mole1 (46 kcal ) 75 kJ ( 1 8 kcal ) 4 kJ ( 1 kcal )

192 kJ

(a) Using the B DEs in Table 4-2 (p. 1 36), compute the value of t:.Ho for each step in the

iodination of methane. (b) Compute the overall value of t:.Ho for iodination. (c) Suggest two reasons why iodine does not react well with methane.

4-1 2 Tem peratu re Dependence of Halogenation

143

144

Chapter 4: The Study of Chemical Reactions

4-1 3

Up to now, we have limited our discussions to the halogenation of methane. Beginning our study with such a simple compound allowed us to concentrate on the thermody­ namics and kinetics of the reaction. Now we consider halogenation of the "higher" alkanes, meaning those of higher molecular weight.

Selectivity in Halogenation

4-13A

Chlori nation of Propa ne: Prod uct Ratios

Halogenation is a substitution, where a halogen atom replaces a hydrogen. R - H + X2

�

R-X + H-X

In methane, all four hydrogen atoms are identical, and it does not matter which hydro­ gen is replaced. In the higher alkanes, replacement of different hydrogen atoms leads to different products. In the chlorination of propane, for example, two monochlorinated (just one chlorine atom) products are possible. One has the chlorine atom on a primary carbon atom, and the other has the chlorine atom on the secondary carbon atom. I°

carbon 2° carbon \ / CH3- CH2-CH3

hv, 2 5°C

+

propane

)

Cl I CH2-CH2-CH3

+

l -chloropropane, 40% (n-propyl chloride)

Cl I CH3-CH - CH3 2-chloropropane, 60% (isopropyl chloride)

The product ratio shows that replacement of hydrogen atoms by chlorine is not random. Propane has six primary hydrogens (hydrogens bonded to primary carbons) and only two secondary hydrogens (bonded to the secondary carbon), yet the major product results from substitution of a secondary hydrogen. We can calculate how reac­ tive each kind of hydrogen is by dividing the amount of product observed by the num­ ber of hydrogens that can be replaced to give that product. Figure 4-5 shows the definition of primary, secondary, and tertiary hydrogens and the calculation of their relative reactivity. Replacing either of the two secondary hydro­ gens accounts for 60% of the product, and replacing any of the six primary hydrogens accounts for 40% of the product. We calculate that each secondary hydrogen is 4.5 times

H I R- C -H I H

R I R- C -H I H

R I R- C -H I R

primary ( 1 0) hydrogens

secondary (2°) hydrogens

tertiary (3°) hydrogen

Six primary ( J O) hydrogens

) replacement

� Figure 4-5

Definitions of primary, secondary, and tertiary hydrogens. There are six primary hydrogens in propane and only two secondary hydrogens, yet the major product results from replacement of a secondary hydrogen.

relative reactivily

CH3 -CH2-CH2- Cl primary chloride

40%

6 hydrogens

=

6.67% per H

Two secondary (2°) hydrogens

C12, ilv

replacemen�

The 2° hydrogens are

��6�

=

Cl I CH3 -CH- CH 3 secondary chloride

60%

2 hydrogens

4.5 times as reactive as the 1 ° hydrogens.

=

30.0% per H

4-13 Selectivity in Halogenation Initiation: Splitting of the chlorine molecule +

Cl 2

hv

------c>

2 CI·

First propagation step: Abstraction (removal) of a primary or secondary hydrogen

pri mary radical

secondary radical

Second propagation step: Reaction with chlorine to form the alkyl chloride

' CH2- CH2- CH3

+

Cl2

------c>

or

CH3- CH-CH3 secondary radical

Cl-CH2- CH2-CH3

+

CI·

primary chloride ( l -chloropropane)

primary radical

+

CI2

------c>

CI I CH3-CH- CH3

+

Cl·

secondary chloride (2-chl oroprop ane)

� Figure 4-6

The mechanism for free-radical chlorination of propane. The rust propagation step forms either a primary radical or a secondary radical. This radical determines whether the final product will be the primary chloride or the secondary chloride.

as reactive as each primary hydrogen. To explain this preference for reaction at the sec­ ondary position, we must look carefully at the reaction mechanism (Figure 4-6). When a chlorine atom reacts with propane, abstraction of a hydrogen atom can give either a primary radical or a secondary radical. The structure of the radical formed in this step determines the structure of the observed product, either l -chloropropane or 2-chloropropane. The product ratio shows that the secondary radical is formed prefer­ entially. This preference for reaction at the secondary position results from the greater stability of the secondary free radical and the transition state leading to it. P R O B L E M 4- 1 8

What would be the product ratio in the chlorination of propane if all the hydrogens were abstracted at equal rates? P R O B L E M 4- 1 9

Classify each hydrogen atom in the following compounds as primary ( 1 0), secondary (2°), or tertiary (30). (a) butane (c) 2-methylbutane (b) isobutane (d) cyclohexane (e) norbornane (bicyclo[2.2. 1 Jheptane) 4- 1 3 8

Free-Rad ical Sta b i lities

Figure 4-7 shows the energy required (the bond-dissociation enthalpy ) to form a free radical by breaking a bond between a hydrogen atom and a carbon atom. This energy is greatest for a methyI carbon, and it decreases for a primary carbon, a secondary carbon, and a tertiary carbon. The more highly substituted the carbon atom, the less energy is required to form the free radical. From the information in Figure 4-7, we conclude that free radicals are more sta­ ble if they are more highly substituted. The following free radicals are listed in de­ creasing order of stability.

1 45

1 46

Chapter 4: The Study of Chemical Reactions Formation of a methyl radical

Bond-dissociation enthalpy

/ili0 Formation of a primary

(JO) radical �

CH3 - CH2- CH3

H'

=

435 kJ ( 1 04 kcal)

+

CH3 - CH2- CH2

t"W = 4 1 0 kJ (98 kcal)

+

CH3 - CH-CH3

t"W = 397 kJ (95 kcal)

Formation of a secondary (2 °) radical �

CH3 - CH2 - CH3

� Figure 4-7

Formation of a tertiary

H·

(30) radical

CH3

Enthalpy required to form a free radical. Bond-dissociation enthalpies show that more highly substituted free radicals are more stable than less highly substituted ones.

I

�

CH 3 - C -H

I

H-

t"W = 38 1 kJ (9 1 kcal)

CH3

R

I

R-C '

I

R

I

>

R-C'

I

R

>

H

I

R-C

I

H >

te rti ary

H

s ec ondary

>

I

H-C'

I

H

H >

>

>

>

primary

>

Me ' methyl

In the chlorination of propane, the secondary hydrogen atom is abstracted more often because the secondary radical and the transition state leading to it are lower in energy than the primary radical and its transition state. Using the bond-dissociation enthalpies in Table 4-2 (page 136), we can calculate D.Ho for each of the possible reaction steps. Abstraction of the secondary hydrogen is 3 kcallmol ( 1 3 kllmol) more exothermic than abstraction of the primary hydrogen. 1°

H:

CH3- CH2-CH3

+

Cl ·

- -;.

CH3- CH2- CH2 '

+

H -Cl

Energy required to break the CH3CH2CH2+ H bond

+ 4 1 0 kllmol ( +98 kcal/mol)

Energy released in forming the H + Cl bond

-43 1 kllmol ( - 1 03 kcal/mol)

Total energy for reaction at the primary position:

- 2 1 kllmol (

- 5 kcal/mol)

CH3

I Energy required to break the CH3 - CH + H bond

+ 397 kllmol ( + 95 kcal/mol)

Energy released in forming the H + Cl bond

-43 1 kllmol ( - 1 03 kcal/mol)

Total energy for reaction at the secondary position:

- 34 kllmol (

- 8 kcal/mol)

A reaction-energy diagram for this rate-limiting first propagation step appears in Figure 4-8. The activation energy to form the secondary radical is slightly lower, so the secondary radical is formed faster than the primary radical.

4- 1 3 Selectivity in Halogenation

-

difference in activation energies (about 4 kJ)

1-----------

I

1 ° Ea 2 ° E-al.- -- >-> � CHFH2CH3 + CI ·

-

I ° radical � Figure 4-8

�

1 3 kJ difference

CH3

R I R-C' I H

>

2°

>

H I R-C ' I H

>

>

1°

>

Vitamin B12 is an essential dietary factor and a deficiency results in anemia and neurological damage. The vitamin assists two different enzymes in the production and the stabilization These

of

methyl

methyl

radicals.

radicals are

then

used for the synthesis of important cellular components.

H I H-C' I H least stable

Stability of radicals 3°

methyl

Like carbocations, radicals can be stabilized by resonance. Overlap with the p orbitals of a 1T bond allows the odd electron to be delocalized over two carbon atoms. Resonance delocalization is particularly effective in stabilizing a radical. H I H CH3 C " � '-.... / C C· I I H H

�

H

H I C

,, / � /

'C I H

C I H

CH3

H

p 1IIi-+--

top view

side view

CH3

/ � / ' """'' fj .

fj.

or

H I C

C I H

C I H

orbital

odd electron ..... Figure 4- 1 5

Orbital diagram of the methyl radical. The structure of the methyl radical is like that of the methyl cation (Figure 4- 1 3), except there is an additional electron. The odd electron is in the p orbital perpendicular to the plane of the three C-H bonds.

1 58

Chapter 4: The Study of Chemical Reactions P R O B L E M 4-30

Rank the following radicals i n decreasing order of stability. Classify each as primary, second­ ary, or tertiary. (a) The isopentyl radical, (CH3hCHCH2 - CH2 (b) The 3 -methyl-2-butyl radical, CH3 - CH - CH(CH3h (c) The 2-methyl-2-butyl radical, CH3 - C(CH3)CH 2 CH3

4-16C

Ca rbanions

A carbanion has a trivalent carbon atom that bears a negative charge. There are eight electrons around the carbon atom (three bonds and one lone pair), so it is not electron deficient; rather, it is electron rich and a strong nucleophile (Lewis base). A carbanion has the same electronic structure as an amine. Compare the structures of a methyl carbanion and ammonia: H I H-C:I H

methyl anion

H I

H-N:

methyl anion

I H ammonia

The hybridization and bond angles of a simple carbanion also resemble those of 3 an amjne. The carbon atom is sp hybridized and tetrahedral. One of the tetrahedral positions is occupied by an unshared lone pair of electrons. Figure 4- 1 6 compares the orbital structures and geometry of ammonia and the methyl anion. Like amines, carbanions are nucleophilic and basic. A carbanion has a negative charge on its carbon atom, however, making it a more powerful base and a stronger nucleophile than an amine. For example, a carbanion is sufficiently basic to remove a proton from ammonia.

ammonia .. Figure 4- 1 6

Comparison of orbital structures of methyl anion and ammonia. Both the methyl anion and ammonia have an 3 sp hybridized central atom, with a nonbonding pair of electrons occupying one of the tetrahedral positions.

Carbanions that occur as intermediates in organic reactions are almost always bonded to stabilizing groups. They can be stabilized either by inductive effects or by resonance. For example, halogen atoms are electron withdrawing, so they stabilize carbanions through the inductive withdrawal of electron density. Resonance also plays an important role in stabilizing carbanions. A carbonyl group (C 0) stabilizes an adj acent carbanion by overlap of its 1T bond with the nonbonding electrons of the carb­ anion. The negative charge is delocalized onto the electronegative oxygen atom of the carbonyl group. =

.' 0 �

H

I

C - C -H / IV H H

�. + - : OH

l·o� H

/

H

�'"-

C-

. - �P2

:0:

'" /

C=C

H H resonance-stabilized carbanion

/ '"

H H

1

+

Hp

2 This resonance-stabilized carbanion must be sp hybridized and planar for effective delocalization of the negative charge onto oxygen (Section 2-6). Resonance­ stabilized carbanions are the most common type of carbanions we will encounter in organic reactions.

4- 1 6 Reactive Intermediates PRO B L E M 4-3 1

Acetylacetone (2,4-pentanedione) reacts with sodium hydroxide to give water and the sodium salt of a carbanion. Write a complete structural formula for the carbanion, and use resonance forms to show the stabilization of the carbanion. 0

o

II

II

H3C - C -CH2-C - CH3 acety lacetone (2,4-pentanedione) PRO B L E M 4-32

Acetonitrile (CH3 C = N ) is deprotonated by very strong bases. Write resonance forms to show the stabilization of the carbanion that results.

4-1 6 D

Carbenes

Carbenes are uncharged reactive intermediates containing a divalent carbon atom. The simplest carbene has the formula : CH2 and is called methylene, just as a - CH 2 group i n a molecule i s called a methylene group. One way of generating carbenes i s to form a carbanion that can expel a halide ion. For example, a strong base can abstract a proton from tribromomethane ( CHBr3 ) to give an inductively stabilized carbanjon. This carbanion expels bromide ion to give dibromocarbene. Br 1 "1 Y\ Br- C - H -OH I Br

�

H2O

+

tribromomethane

��

Br "

Br-C : I Br

Br /

C:

dibromocarbene

a carbanion

The electronic structure of dibromocarbene is shown next. The carbon atom has only six electrons in its valence shell. It is sp 2 hybridized, with trigonal geometry. An unshared pair of electrons occupies one of the sp 2 hybrid orbitals, and there is an empty p orbital extending above and below the plane of the atoms. A carbene has both a lone pair of electrons and an empty p orbital, so it can react as a nucleophile or as an electrophile. empty p orbital

-:\) / G u

nonbonding e l ectrons pai,;d in this sp- orbItal

Br " """C Bf

,, ;

hybr'd'"d

Methylene itself is formed when diazomethane (CH2N2) is heated or irradiated with light. The diazomethane molecule splits to form a stable nitrogen molecule and the very reactive carbene. H,' '- + C� N- N : l,(

H/

diazomethane

heat or light

)

methylene

nitrogen

The most common synthetic reaction of carbenes is their addition to double bonds to form cyclopropane rings. For example, dibromocarbene adds to cyclohexene to give an interesting bicyclic compound.

+

Be

1 59

1 60

Chapter 4: The Study of Chemical Reactions

:C

/

Br

-------c'>

'"

Br dibromocarbene

cyclohexene

O

C

/

"'

Br

Br

No simple carbenes have never been purified or even made in a high concentra­ tion, because when two carbenes c ol l i de, th ey immediately dimerize (two of them bond together) to give an alkene. R ".. C : / R

/

+ : C "..

R

very fast

)

R

Carbenes and carbenoids (carbene-like reagents) are useful both for the synthe­ sis of other compounds and for the investigation of reaction mechanisms. The carbene intermediate is generated in the presence of its target compound, so that it can react immediately, and the concentration of the carbene is always low. Reactions using car­ benes are discussed in Chapter 8. PRO B L E M 4-33

When i t i s strongly heated, ethyl diazoacetate decomposes to give nitrogen gas and a carbene. Draw a Lewis structure of the carbene. + : N=N

o -

CH

II

-

C

-

O

-

C � CH 3

ethyl diazoacetate

(I

S U M MARY

Reactive Intermediates

Structure

-C+

radicals

-C,

carbanions carbenes

Chapte r 4 G l ossa ry

I

carbocations

I

I I

-

I -

C .·

I

Stability

Properties

3° > 2° > 1 ° > + CH3

electrophilic strong acids

3° > 2° >

electron deficient

-: CH3

>

1° >

' CH3

nucleophilic strong bases

1 ° > 2° > 3°

both nucleophilic and electrophilic

""

C: ../

I

I

activation energy (Ea) The energy difference between the reactants and the transition state; the minimum energy the reactants must have for the reaction to occur. (p. 1 39) bond-dissociation enthalpy (BDE) The amount of enthalpy required to break a particular bond homolytically, to give radicals. (p. 1 34)

A:B

---->

A'

+

B'

carbanion A strongly nucleophilic species with a negatively charged carbon atom having only three bonds. The carbon atom has a nonbonding pair of electrons. (p. 1 58) carbene A highly reactive species with only two bonds to an u ncharged carbon atom and with a nonbonding pair of electrons. The simplest carbene is methylene, : CH 2 . (p. 1 59)

Chapter 4 Glossary carbocation (carbonium ion, carbenium ion) A strongly electrophilic species with a posi­ tively charged carbon atom having only three bonds. (p. 1 55) catalyst A substance that increases the rate of a reaction (by lowering Ea) without being con­ sumed in the reaction. (p. 1 4 1 ) chain reaction A multistep reaction where a reactive intermediate formed in one step brings about a second step that generates the intermediate needed for the following step. (p. 1 26) initiation step: The preliminary step in a chain reaction, where the reactive intermediate is first formed. propagation steps: The steps in a chain reaction that are repeated over and over to form the product. The sum of the propagation steps should give the net reaction. termination steps: Any steps where a reactive intermediate is consumed without another one being generated. enthalpy (heat content; H) A measure of the heat energy in a system. In a reaction, the heat absorbed or evolved is called the heat of reaction, t:.Ho. A decrease in enthalpy (negative t:.HO) is favorable for a reaction. (p. 1 33) endothermic: Consuming heat (having a positive t:.HO). exothermic: Giving off heat (having a negative t:.HO). entropy (S) A measure of disorder or freedom of motion. An increase in entropy (positive t:.SO) is favorable for a reaction. (p. 1 33) equilibrium A state of a system such that no more change is taking place; the rate of the for­ ward reaction equals the rate of the reverse reaction. (p. 1 3 1 ) equilibrium constant A quantity calculated from the relative amounts of the products and reactants present at equilibrium. (p. 1 3 1 ) For the reaction aA

+

bB

�

cC

+

dD

the equilibrium constant is

free energy (Gibbs free energy; G) A measure of a reaction's tendency to go in the direction written. A decrease in free energy (negative t:. G) is favorable for a reaction. (p. 1 3 1 ) t:. G

Free-energy change i s defined:

=

t:. H - Tt:.S

standard Gibbs free energy change: ( t:. GO ) The free-energy change corresponding to reac­ tants and products in their standard states (pure substances in their most stable states) at 25°C and 1 atm pressure. t:. Go is related to Keq by

Keq

= e -!:J.Go/RT

(p. 1 3 1 )

Hammond postulate Related species (on a reaction-energy diagram) that are closer in energy are also closer in structure. In an exothermic reaction, the transition state is closer to the reac­ tants in energy and in structure. In an endothermic reaction, the transition state is closer to the products in energy and in structure. (p. 1 49) heterolytic cleavage (ionic cleavage) The breaking of a bond in such a way that one of the atoms retains both of the bond's electrons. A heterolytic cleavage forms two ions. (p. 1 34)

homolytic cleavage (radical cleavage) The breaking of a bond in such a way that each atom retains one of the bond's two electrons. A homolytic cleavage produces two radicals. (p. 1 34)

A·

+

·B

inductive effect A donation (or withdrawal) of electron density through sigma bonds. (p. 1 56) intermediate A molecule or a fragment of a molecule that is formed in a reaction and exists for a finite length of time before it reacts in the next step. An intermediate corresponds to a rel­ ative minimum (a low point) in the reaction-energy diagram. (p. 140) reactive intermediate: A short-lived species that is never present in high concentration because it reacts as quickly as it is formed. (p. 1 55)

1 61

1 62

Chapter 4: The Study of Chemical Reactions kinetics The study of reaction rates. (p. 1 37) mechanism The step-by-step pathway from reactants to products showing which bonds break and which bonds form in what order. The mechanism should include the structures of all inter­ mediates and arrows to show the movement of electrons. (p. 1 26) potential-energy diagram See reaction-energy diagram. (p. 1 4 1 ) radical (free radical) A highly reactive species i n which one o f the atoms has an odd number of electrons. Most commonly, a radical contains a carbon atom with three bonds and an "odd" (unpaired) electron. (p. 1 57) radical inhibitor A compound added to prevent the propagation of free-radical chain reactions. In most cases, the inhibitor reacts to form a radical that is too stable to propagate the chain. (p. 1 53) rate equation (rate law) The relationship between the concentrations of the reagents and the observed reaction rate. (p. 1 37) A general rate law for the reaction A + B � C + D is

rate

= kr[A]O[B]

b

kinetic order: The power of a concentration term in the rate equation. The preceding rate equation is ath order in [A], bth order in [B], and a+bth order overall . rate constant: The proportionality constant k r i n the rate equation. rate-limiting step (rate-determining step) The slowest step in a multistep sequence of reactions. In general, the rate-limiting step is the step with the highest-energy transition state. (p. 1 42) rate of a reaction The amount of product formed or reactant consumed per unit of time. (p. 1 37) reaction-energy diagram (potential-energy diagram) A plot of potential-energy changes as the reactants are converted to products. The vertical axis is potential energy (usually free ener­ gy, but occasionally enthalpy). The horizontal axis is the reaction coordinate, a measure of the progress of the reaction. (p. 1 4 1 )

:t:/ transition state

products reaction coordinate

�

resonance stabilization Stabilization that takes place by delocalization of electrons in a pi­ bonded system. Cations, radicals, and anions are often stabilized by resonance delocaliza­ tion. (p. 1 56)

substitution A reaction in which one atom replaces another, usually as a substituent on a car­ bon atom. (p. 1 27) thermodynamics The study of the energy changes accompanying chemical transformations. Thermodynamics is generally concerned with systems at equilibrium. (p. 1 30) transition state (activated complex) The state of highest energy between reactants and prod­ ucts. A relative maximum (high point) on the reaction-energy diagram. (p. 1 40)

Study Problems

I 1.

2.

3.

4.

5. 6. 7. 8.

1 63

Essential Problem-Solvi ng Skills i n Chapter 4

Explain the mechanism and energetics of the free-radjcal halogenation of alkanes. Based on the selectivity of halogenation, predict the products of halogenation of an alkane. Calculate free-energy changes from equilibrium constants. Calculate enthalpy changes from bond-dissociation enthalpies. Determine the order of a reaction, and suggest a possible mechanism based on its rate equation. Use energy diagrams to discuss transition states, activation energies, intermediates, and the rate-determining step of a multistep reaction. Use the Hammond postulate to predict whether a transition state will be reactant-like or product-like. Describe the structures of carbocations, carbanions, free radicals, and cat·benes and the structural features that stabilize them. Explain which are electrophilic and which are nucleophilic.

Study Problems 4-34

4-35

Define and give an example for each term. (a) homolytic cleavage (b) heterolytic cleavage (d) carbocation (e) carbanion ( g) carbonium ion (h) intermediate (k) rate equation (j) transition state (m) rate constant (0) reaction mechanism (p) substitution reaction (q) activation energy (s) rate-limiting step (t) Hammond postulate Consider the following reaction-energy diagram.

(c) free radical carbene (i) catalyst (I) equilibrium constant (0) chain reaction (r) bond-dissociation enthalpy (u) resonance stabilization (f)

(a) (b) (c) (d)

4-36 4-37

4-38

Label the reactants and the products. Label the activation energy for the first step and the second step. Is the overall reaction endothermic or exothermic? What is the sign of !:J.HO? Which points in the curve correspond to intermediates? Which correspond to transition states? Label the transition state of the rate-limiting step. Does its structure resemble the reactants, the products, or an intermediate? Draw a reaction-diagram profile for a one-step exothermic reaction. Label the parts that represent the reactants, products, transition state, activation energy, and heat of reaction. Draw a reaction-energy diagram for a two-step endothermic reaction with a rate-limiting second step. Treatment of t-butyl alcohol with concentrated HCI gives t-butyl chloride. CH3

CH3

I

CH - C - OH 3

I

CH3

I-butyl alcohol

+

W +

Cl-

I

CH3 - C - Cl

I

+

H20

CH3

I-butyl chloride

When the concentration of H + is doubled, the reaction rate doubles. When the concentration of t-butyl alcohol is tripled, the reaction rate triples. When the chloride ion concentration is quadrupled, however, the reaction rate is unchanged. Write the rate equation for this reaction.

1 64 4-39

Chapter 4: The Study of Chemical Reactions Label each hydrogen atom in the following compounds as primary ( 1 0), secondary (2°), or tertiary (3°). (b) (CH3 hCCH2C(CH3h

(a) CH3CH2CH(CH3h

(c) 4-40

4-41

4-42

4-43

O

CH3 d)

(

cb

Use bond-dissociation enthalpies (Table 4-2, p. 1 36) to calculate values of !1Ho for the following reactions. (a) CH3 - CH 3 + 12 -- CH3CH2I + HI (b) CH3CH2CI + HI -- CH3CH2I + HCl (c) ( CH3 h C - OH + HCl -- (CH 3 h C - CI + H20 (d) CH3CH2CH3 + H2 -- CH3CH3 + CH4 (e) CH3CH20H + HBr -- CH3CH2-Br + H20 Use the information in Table 4-2 (p. l 36) to rank the following radicals in decreasing order of stability.

For each of the following alkanes, 1 . Draw all the possible monochlorinated derivatives. 2. Determine whether free-radical chlorination would be a good way to make any of these monochlorinated derivatives. 3. Which monobrominated derivatives could you form in good yield by free-radical bromination? (a) cyclopentane (b) methylcyclopentane d) 2,2,3,3-tetramethylbutane (c) 2,3-dimethylbutane ( Write a mechanism for the light-initiated reaction of cyclohexane with chlorine to give chlorocyclohexane. Label the initi­ ation and propagation steps. hv -----7

cyclohex ane

4-44

+

HCl

chlorocyclohexane

Draw the important resonance structures of the following free radicals.

(a) CH2 = CH - CH 2

(d) * 4-45

O

CI

0

(b)

(e)

< /-

0 CH2

(c)

II

CH3-C-O ·

0

In the presence of a small amount of bromine, the following light-promoted reaction has been observed.

H 3C

-U

C H3

+

� V hv

H 3C

C H3 Br

+

(Hint: Notice which H atom has been lost in both products.) (b) Explain why only this hydrogen atom has been replaced, in preference to any of the other hydrogen atoms in the starting material. (a) Write a mechanism for this reaction. Your mechanism should explain how both products are formed.

Study Problems 4-46

4-47

4-48

4-49

4-50

4-51 " '4-52

For each compound, predict the major product of free-radical bromination. Remember that bromination is highly selective, and only the most stable radical will be formed. (a) cyclohexane (b) methylcyclopentane (c) 2,2,3-trimethylbutane (0 hexane (e) 3-methyloctane (d) decalin

ethylbenzene

When exactly I mole of methane is mixed with exactly 1 mole of chlorine and light is shone on the mixture, a chlorina­ tion reaction occurs. The products are found to contain substantial amounts of di-, tri-, and tetrachloromethane. (a) Explain how a mixture is formed from this stoichiometric mixture of reactants, and propose mechanisms for the formation of these compounds from chloromethane. (b) How would you run this reaction to get a good conversion of methane to CH3Cl? Methane to CCI4? The chlorination of pentane gives a mixture of three monochlorinated products. (a) Draw their structures. (b) Predict the ratios in which these monochlorination products will be formed, remembering that a chlorine atom abstracts a secondary hydrogen about 4.5 times as fast as it abstracts a primary hydrogen. (a) Draw the structure of the transition state for the second propagation step in the chlorination of methane.

'CH3 + Cl 2 � CH3Cl + Cl' Show whether the transition state is product-like or reactant-like, and which of the two partial bonds is stronger. (b) Repeat for the second propagation step in the bromination of methane. Peroxides are often added to free-radical reactions as initiators because the oxygen-oxygen bond is homolytically cleaved rather easily. For example, the bond-dissociation enthalpy of the 0 - 0 bond in hydrogen peroxide ( H - 0 - 0 - H) is only 2 1 2 kllmol (5 1 kcal/mol). Give a mechanism for the hydrogen peroxide-initiated reaction of cyclopentane with chlorine. When dichloromethane is treated with strong NaOH, an intermediate is generated that reacts like a carbene. Draw the structure of this reactive intermediate, and propose a mechanism for its formation. When ethene is treated in a calorimeter with H2 and a Pt catalyst, the heat of reaction is found to be - 1 37 kllmol ( - 32.7 kcal/mol ) , and the reaction goes to completion. When the reaction takes place at 1400° K, the equilibrium is found to be evenly balanced, with Keq = l . Compute the value of D.. S for this reaction. CH 2 - CH 2 + H 2

" '4-53

*4-54

1 65

Pl catalyst +---

D.. H = - 1 37 kllmol

l

When a small amount of iodine is added to a mixture of chlorine and methane, it prevents chlorination from occurring. Therefore, iodine is afree-radical inhibitor for this reaction. Calculate D.. Ho values for the possible reactions of iodine with species present in the chlorination of methane, and use these values to explain why iodine inhibits the reaction. (The I - Cl bond-dissociation enthalpy is 2 1 1 kllmol or 50 kcal/mol.) Tributyltin hydride (Bu3SnH ) is used synthetically to reduce alkyl halides, replacing a halogen atom with hydrogen. Free­ radical initiators promote this reaction, and free-radical inhibitors are known to slow it or stop it. Your job is to develop a mechanism, using the following reaction as the example.

+

trace Br2,

h.v

+

The following bond-dissociation enthalpies may be helpful:

o

H 397 kJ/moi

o

Br 285 kJ/moi

Br -Br

192 kJ/mol

H -B r

368 kJ/mo!

B U3Sn -H

3 10 kJ/mo!

BU3Sn - Br

552 kJ/mo!

(a) Propose initiation and propagation steps to account for this reaction. (b) Calculate values of D.. H for your proposed steps to show that they are energetically feasible. (Hint: A trace of Br2 and light suggests it's there only as an initiator, to create Br' radicals. Then decide which atom can be abstracted most favorably from the starting materials by the Br' radical. That should complete the initiation. Now decide what energetically favored propagation steps will accomplish the reaction.)

1 66 * 4-55

Chapter 4: The Study of Chemical Reactions When healthy, Earth's stratosphere contains a low concentration of ozone ( 03 ) that absorbs poten­ tially harmful ultraviolet (UV) radiation by the cycle shown at right. Chlorofluorocarbon refrigerants, such as Freon 1 2 ( CF2 CI 2 ) , are stable in the lower atmos­ phere, but in the stratosphere, they absorb high-energy UV radiation to generate chlorine radicals.

hv

�o �+

The presence of a small number of chlorine radicals appears to lower ozone concentrations dramati- heat cally. The following reactions are all known to be exothermic (except the one requiring light) and to have high rate constants. Propose two mechanisms to explain how a small number of chlorine radicals can destroy large numbers of ozone molecules. Which of the two mechanisms is more likely when the concentration of chlorine atoms is very smaIl? CI - O - O - CI * 4-56

hv -----,>

CI - O ' 2 CI - 0'

+ 0 �

�

O 2 + CI'

CI - O - O - Cl

Deuterium (D) is the hydrogen isotope of mass number 2, with a proton and a neutron in its nucleus. The chemistry of deuterium is nearly identical to the chemistry of hydrogen, except that the C - D bond is slightly stronger than the C - H bond by 5 . 0 kllmol ( 1 2 kcal/mol). Reaction rates tend to be slower if a C - D bond (as opposed to a C - H bond) is broken in a rate- limiting step. This effect, called a kinetic isotope effect, is clearly seen in the chlorination of methane. Methane undergoes free­ radical chlorination 1 2 times as fast as tetradeuteriomethane ( CD4 ) . Faster:

CH4 + CI '

Slower:

CD4 + CI'

�

�

CH3Cl + HCI

relative rate

CH3CI + DCI

relative rate

= =

12

(a) Draw the transition state for the rate-limiting step of each of these reactions, showing how a bond to hydrogen or deuterium is being broken in this step. (b) Monochlorination of deuterioethane ( C2HsD ) leads to a mixture containing 93% C2H4DCl and 7% C 2 HsCI. Calculate the relative rates of abstraction per hydrogen and deuterium in the chlorination of ethane. (c) Consider the thermodynamics of the chlorination of methane and the chlorination of ethane, and use the Hammond postulate to explain why one of these reactions has a much larger isotope effect than the other.

5 Ste reochem i stry

Stereochemistry is the study of the three-dimensional structure of molecules. No one can understand organic chemistry, biochemistry, or biology without using stereochem­ istry. Biological systems are exquisitely selective, and they often discriminate between molecules with subtle stereochemical differences. We have seen (Section 2-8) that iso­ mers are grouped into two broad classes: constitutional isomers and stereoisomers. Constitutional isomers (structural isomers) differ in their bonding sequence; their atoms are connected differently. Stereoisomers have the same bonding sequence, but they differ in the orientation of their atoms in space. Differences in spatial orientation might seem unimportant, but stereoisomers often have remarkably different physical, chemical, and biological properties. For example, the cis and trans isomers of butenedioic acid are a special type of stereoiso­ mer called cis-trans isomers (or geometric isomers). Both compounds have the formu­ la HOOC - CH CH - COOH, but they differ in how these atoms are arranged in space. The cis isomer i s called maleic acid, and the trans i somer is caUedfumaric acid. Fumaric acid is an essential metabolic intermediate in both plants and animals, but maleic acid is toxic and irritating to tissues. =

o H HO - C

II

II

" /

o

C=C

/ "

C - OH H

fumaric acid, mp 287°C essential metabolite

A double bond in rhodopsin, a visu­ al pigment found in your eyes that enables you to see at night, is con­ verted from the cis isomer to the trans isomer when light strikes the eye. As a result, a nerve impulse travels to the brain and you see the source of the light.

N-opsin

/;

II

II

H

Introd u ction

0

o HO -C

5-1

" /

C=C

/

C - OH

"

H

maleic acid, mp 1 38°C toxic irritant

The discovery of stereochemistry was one of the most important break­ throughs in the structural theory of organic chemistry. Stereochemistry explained why several types of i somers exist, and it forced scientists to propose the tetrahedral carbon atom. In this chapter, we study the three-dimensional structures of molecules to understand their stereochemical relationships. We compare the various types of stereoisomers and study ways to differentiate among stereoisomers. In future chap­ ters, we will see how stereochemistry plays a major role in the properties and reac­ tions of organic compounds.

Ij /; /;

rhodopsin

1 67

1 68

Chapter 5: Stereochemistry

5-2 Chirality

What is the difference between your left hand and your right hand? They look simi­ lar, yet a left-handed glove does not fit the right hand. The same principle applies to your feet. They look almost identical, yet the left shoe fits painfully on the right foot. The relationship between your two hands or your two feet is that they are nonsuper­ imposable (nonidentical) mirror images of each other. Objects that have left-handed and right-handed forms are called chiral (ki'rel, rhymes with "spiral"), the Greek word for "handed." We can tell whether an object is chiral by looking at its mirror image (Figure 5 - 1 ). Every physical object (with the possible exception of a vampire) has a mirror image, but a chiral object has a mirror image that is different from the original object. For example, a chair and a spoon and a glass of water all look the same in a mirror. Such objects are called achiral, meaning "not chiral." A hand looks different in the mirror. If the original hand were the right hand, it would look like a left hand in the mirror.

� Figure 5-1

Use of a mirror to test for chirality. An object is chiral if its mirror image is different from the original object.

right hand

Besides shoes and gloves, we encounter many other chiral objects every day (Figure 5-2). What is the difference between an English car and an American car? The English car has the steering wheel on the right-hand side, while the American car has it on the left. To a first approximation, the English and American cars are nonsuperimpos­ able mirror images. Most screws have right-hand threads and are turned clockwise to tighten. The mirror image of a right-handed screw is a left-handed screw, turned coun­ terclockwise to tighten. Those of us who are left-handed realize that scissors are chiral. Most scissors are right-handed. If you use them in your left hand, they cut poorly, if at all. A left-handed person must go to a well-stocked store to find a pair of left-handed scissors, the mirror image of the "standard" right-handed scissors.

mirror

� Figure 5-2

Common chiral objects. Many objects come in "left-handed" and "right-handed" versions.

mirror

mirror

5-2 Chirality

1 69

PRO B L E M 5 - 1

Determine whether the following objects are chiral or achiral.

•

5-2A

C h i ral ity and Ena ntiomerism i n O rganic Molecu les

Like other objects, molecules are either chiral or achiral. For example, consider the two geometric isomers of 1 ,2-dichlorocyclopentane (Figure 5-3). The cis isomer is achiral because its mirror image is superimposable on the original molecule. Two molecules are said to be superimposable if they can be placed on top of each other and the three-dimensional position of each atom of one molecule coincides with the equivalent atom of the other molecule. To draw the mirror image of a molecule, sim­ ply draw the same structure with left and right reversed. The up-and-down and front­ and-back directions are unchanged. These two mirror-image structures are identical (superimposable), and cis- l ,2-dichlorocyclopentane is achiral. The mirror image of trans- l ,2-dichlorocyclopentane is different from (nonsuper­ imposable with) the original molecule. These are two different compounds, and we should expect to discover two mirror-image isomers of trans- l ,2-dichlorocyclopen­ tane. Make models of these isomers to convince yourself that they are different no matter how you twist and tum them. Nonsuperimposable mirror-image molecules are called enantiomers. A chiral compound always has an enantiomer (a nonsuperimpos­ able mirror image). An achiral compound always has a mirror image that is the same as the original molecule. Let's review the definitions of these words.

chiral: achiral:

same compound

di fferent compounds

/ 1: ""

/ 1: H ""

cis- I ,

CI

Htn!/

mirror image the same or different? Different: The object is chiral. Same: The object is achiral.

mirror-image isomers; pairs of compounds that are nonsuperimposable minor images ("handed") different from its mirror image; having an enantiomer ("not handed") identical with its mirror image; not chiral

enantiomers:

CI

PROBLEM-SOLVING

Every object has a mi rror image. Is its

'

CI

CI

2-dichlorocyclopentane (achiral)

H

lrans- l ,

CI

'

Cl

H

2-dichlorocyclopentane (chiral)

� Figure 5-3

Stereoisomers of 1 ,2-dichlorocyclopen tane. The cis isomer has no enantiomers; it is achiral. The trans isomer is chiral; it can exist in either of two nonsuperimposable enantiomeric forms.

1 70

Chapter 5: Stereochemistry

Any compound that is chiral must have an enantiomer. Any compound that is achiral cannot have an enantiomer. PROBLEM-SOLVING

HtnZ;

Stereochemistry is a difficult topic for many students. Use your models to help you see the rel ationships between structures. Once you have experience working with these three-dimensional rel ationships, you

P R O B L E M 5-2

Make a model and draw a three-dimensional structure for each compound. Then draw the mirror image of your original structure and determine whether the mirror image is the same compound. Label each structure as being chiral or achiral, and label pairs of enantiomers. (b) trans- l ,2-dimethylcyclobutane (a) cis- l ,2-dimethylcyclobutane (d) 2-bromobutane (c) cis- and trans- ! , 3 -dimethylcyclobutane

may (or may not) be able to visualize them without constructing models.

(0)

5-2B

dJo

(f)

jOyO/ LJ"-'

Asymmetric Carbon Atoms, C h i ral ity Centers, and Stereocenters

The three-dimensional drawing of 2-bromobutane in Figure 5-4 shows that 2bromobutane cannot be superimposed on its mirror image. Thi s simple molecule is chiral, with two distinct enantiomers. What is i t about a molecule that makes it chi­ ra!? The most common feature (but not the only one) that lends chirality is a carbon atom that is bonded to four different groups. Such a carbon atom is called an asymmetric carbon atom or a chiral carbon atom, and is often designated by an asterisk ( * ) . Carbon atom 2 of 2-bromobutane is bonded to a hydrogen atom, a bromine atom, a methyl group, and an ethyl group. It is an asymmetric carbon atom, and it is responsible for the chirality of 2-bromobutane. An asymmetric carbon atom is the most common example of a chirality center, the IUPAC term for any atom holding a set of ligands in a spatial arrangement that is not superimposable on its mirror image. Chirality centers belong to an even broader group called stereocenters. A stereocenter (or stereogenic atom) is any atom at which the interchange of two groups gives a stereoisomer. 1 Asymmetric carbons and the double-bonded carbon atoms in cis-trans isomers are the most common types of stereocenters. Figure 5-5 compares these successively broader definitions.

mirror

� Figure 5-4

2-Bromobutane is chiral by virtue of an asymmetric carbon atom (chiral carbon atom), marked by an *.

--:.---._ ----.. _._.- ---- .=-=== ==='- . --:... - -l The term slereocenler (slereogenic atom) is not consistently defined. The original (Mislow) definition is given here. Some sources simply detine it as a synonym for an asymmetric carbon (chiral carbon) or for a chirality cente!:

5-2

asym metric carbon

Chirality

171

� Figure 5-5 Asymmetric carbon atoms are examples of chirality centers, which are examples of stereocenters.

chiral ity centers ( * ) stereocenters (circled)

Make a model of an asymmetric carbon atom, bonded to four different-colored atoms. Also make its mirror image, and try to superimpose the two (Figure 5-6). No matter how you twist and turn the models, they never look exactly the same. If two of the four groups on a carbon atom are the same, however, the arrange­ ment usually is not crural. Figure 5-7 shows the mirror image of a tetrahedral structure with only three different groups ; two of the four groups are the same. If the structure on the right is rotated 1 800, it can be superimposed on the left structure. We can generalize at this point, but keep in mind that the ultimate test for chirality is always whether the molecule's mirror image is the same or different. 1. If a compound has no asymmetric carbon atom, it is usually achiral. 2. If a compound has just one asymmetric carbon atom, it is chiral. 3. If a compound has more than one asymmetric carbon, it may or may not be

chiral. (We will see examples in Section 5- 1 2.)

mirror

� Figure 5-6 Enantiomers of an asymmetric carbon atom. These two mirror images are nonsuperimposable.

m irror

rotate

1 800

same as the fust structure --- _ ._---

� Figure 5-7 A carbon atom bonded to just three different types of groups is not chiral.

1 72

Chapter 5 : Stereochemistry S O LV E D P R O B L E M 5 - 1

Star ( * ) each asymmetric carbon atom i n the following structure:

0 5 �

H OH

4

H

SOLUTION

6

j

1 ",

:;:

2

H -CH3

CI

There are three asymmetric carbons, starred i n red. 1 . The (CHOH) carbon of the side chain is asymmetric. Its four substituents are the ring, a hydrogen atom, a hydroxyl group, and a methyl group. 2. Carbon atom C ' of the ring is asymmetric. Its four substituents are the side chain, a hydrogen atom, the part of the ring closer to the chlorine atom ( - CH 2 - CHCI- ) , and the part of the ring farther from the chlorine atom ( -CH 2 - CH 2 CH 2 - CHCI - ) . 3. The ring carbon bearing the chlorine atom i s asymmetric. Its four substituents are the chlorine atom, a hydrogen atom, the part of the ring closer to the side chain, and the part of the ring farther from the side chain. Notice that different groups might be different in any manner. For example, the ring carbon bearing the chlorine atom is asymmetric even though two of its ring substituents initially appear to be - CH 2 - groups. These two parts of the ring are different because one is closer to the side chain and one is farther away. The entire structure of the group must be considered. PROBLEM-SOLVING

Htltv

To draw the mirror image of a

structure, keep up-and-down and front-and-back aspects as they are in the original structure, but reverse left

PROB L E M 5-3

Draw a three-dimensional structure for each compound, and star all asymmetric carbon atoms. Draw the mirror image for each structure, and state whether you have drawn a pair of enantiomers or just the same molecule twice. Build molecular models of any of these exam­ ples that seem difficult to you.

and right.

OH

� Br (a) l-bromobutane

� OH

a

(b) l -pentanol

OH

�

(d) 3-pentanol

(c) 2-pentanol

CI

(e) chlorocyc1ohexane

(g) cis- I ,2-dichlorocyclobutane PROBLEM-SOLVING

�

(h) trans- I ,2-dichlorocyclobutane

Htltv

To determine whether a ring carbon is asymmetric, see if there is a difference in the path around the ring in each

(i) trans- I ,3-dichlorocyclobutane

direction. If there is, then the two ring bonds are "different groups. "

PRO B L E M 5-4

For each of the stereocenters (circled) i n Figure 5-5, (a) draw the compound with two of the groups on the stereocenter interchanged. (b) give the relationship of the new compound to the original compound.

5-2 Chirality

r:f

of symmetry (a)

internal mirror plane

5-2C

M i rror Planes of Sym metry

In Figure 5-3 we saw that cis-l ,2-dichlorocyclopentane is achiral. Its mirror image was found to be identical with the original molecule. Figure 5-8 shows a shortcut that often shows whether a molecule is chiral. If we draw a line down the middle of cis1 ,2-dichlorocyclopentane, bisecting a carbon atom and two hydrogen atoms, the part of the molecule that appears to the right of the line is the mirror image of the part on the left. This kind of symmetry is called an internal mirror plane, sometimes sym­ bolized by the Greek lowercase letter sigma ((]'). Since the right-hand side of the molecule is the reflection of the left-hand side, the molecule's mirror image is the same as the original molecule. Notice in the following figure that the chiral trans isomer of 1 ,2-dichlorocy­ clopentane does not have a mirror plane of symmetry. The chlorine atoms do not reflect into each other across our hypothetical mirror plane. One of them is directed up, the other down. not a

plane

of symmetry

H

el1antiol11ers

/\

Cl

Cl

�

does not correspond

H

different from the structure at left

We can generalize from these and other examples to state the following principle: even though it may contain asymmetric carbon atoms. The converse is not true, however. When we cannot find a mirror plane of symmetry, that does not necessarily mean that the molecule must be chiral. The following exam­ ple has no internal mirror plane of symmetry, yet the mirror image is superimposable on the original molecule. You may need to make models to show that these mirror im­ ages are just two drawings of the same compound. ",·CI Br CI "" Br """Br CI Br""" CI Any molecule that has an internal mirror plane of symmetl)) cannot be chiral,

CC

Internal mirror plane. cis1 ,2-Dichlorocyclopentane has an internal mirror plane of symmetry. Any compound with an internal mirror plane of symmetry cannot be chiral. .... Figure 5-8

�CI tl

t

D

(hydrogens are omitted for clarity)

173

174

r hm

Chapter 5: S te eoc e is try

� Figure 5-9

A

carbon atom with two identical substituents (only three different substituents) usually has an internal mirror plane of symmetry. The structure is not chiral.

viewed from

this angle

Using what we know about mirror planes of symmetry, we can see why a chiral (asymmetric) carbon atom is special. Figure 5-4 showed that an asymmetric carbon has a mirror image that is nonsuperimposable on the original structure. If a carbon atom has only three different kinds of substituents, however, it has an internal mirror plane of symmetry (Figure 5-9). Therefore, it cannot contribute to chirality in a molecule. PRO BLEM 5-5

For each compound, determine whether the molecule has an internal milTor plane of symmetry. If it does, draw the rnin-or plane on a three-dimensional drawing of the molecule. If the mole­ cule does not have an internal milTor plane, determine whether or not the structure is chiral. (b) cis- l ,2-dibromocyclobutane (a) methane t cop,", u,, m

::: :�::�:�::�:: ::::� ;:

:: ;;s

�CHO

OH

NH,

I -

(h) CH3-CH-COOH alanine

5-3 (R) a n d (5) Nomenclature of Asymmetric Ca rbon Atoms

Alanine, from Problem 5-5(h), is one of the amino acids found in common proteins. Alanine has an asymmetric carbon atom, and it exists in two enantiomeric forms. O'\- / OH HO",f'° C C 1*

/ ,

C'I"""H

CH3

NH2

natural alanine

J

/ "

' H""" C H2N

CH3

unnatural alanine

These mirror images are different, and this difference is reflected in their biochemistry. Only the enantiomer on the left can be metabolized by the usual enzyme; the one on the right is not recognized as a useful amino acid. Both are named alanine, however, or 2-aminopropanoic acid in the IUPAC system. We need a simple way to distinguish between enantiomers and to give each of them a unique name.

and (S) Nomenclature of Asymmetlic Carbon Atoms The difference between the two enantiomers of alanine lies in the three­ dimensional arrangement of the four groups around the asymmetric carbon atom. Any asymmetric carbon has two possible (mirror-image) spatial arrangements, which we call configurations. The alanine enantiomers represent the two possible arrangements of its four groups around the asymmetric carbon atom. If we can name the two configurations of any asymmetric carbon atom, then we have a way of specifying and naming the enantiomers of alanine or any other chiral compound. The Cahn-Ingold-Prelog convention is the most widely accepted system for naming the configurations of chirality centers. Each asymmetric carbon atom is assigned a letter (R) or (S) based on its three-dimensional configuration. To deter­ mine the name, we follow a two-step procedure that assigns "priorities" to the four substituents and then assigns the name based on the positions of these substituents. Here is the procedure: 1. Assign a "priority" to each g roup bonded to the asymmetric carbon. We speak of group 1 as having the highest priority, group 2 second, group 3 third, and group 4 as having the lowest priority. (a) Look at the first atom of the group-the atom bonded to the asymmetric carbon. Atoms with higher atomic numbe rs receive higher priorities. For CD CH example, if the four groups bonded to an asymmetric carbon atom were H, I .,l. 3 CH3, NH2, and F, the fluorine atom (atomic number 9) would have the C"';"' H @ highest priority, followed by the nitrogen atom of the NH2 group (atomic / CD F NH number 7), then by the carbon atom of the methyl group (atomic number (j) 6). Note that we look only at the atomic number of the atom directly attached to the asymmetric carbon, not the entire group. Hydrogen would have the lowest priority. With different isotopes of the same element, the heavier isotopes have higher priorities. For example, tritium eH) receives a higher priority than deuterium eH), followed by hydrogen eH). 5-3 (R)

2

Examples of priority for atoms bonded to a n asymmetric carbon:

I> Br> CI > S > F> 0> N> l3C> 1 2C> Li> 3H> 2H> IH (b) In case of ties, use the next atoms along the chain of each group as tiebreak­ ers. For example, we assign a higher priority to isopropyl -CH(CH3h than to ethyl -CH2CH3. The first carbon atom in the ethyl group is bond­ ed to two hydrogens and one carbon, while the first carbon in the isopropyl group is bonded to two carbons and one hydrogen. An ethyl group and a -CH2CH2Br have identical first atoms and second atoms, but the bromine atom in the third position gives -CH2CH2Br a higher priority than -CH2CH3. One high-priority atom takes priority over any number of lower-priority atoms. Examples

H H CH3 CH2Br 1 1 1 1 -C-Br > -C-Cl > -C-CH3 > -C-CH 3 > -CH2CH2CH2CH3 1 1 1 1 H CI CH3 H For this method, imagine that each pi bond is broken and the atoms at both

(c) Treat double a nd triple bonds as if each were a bond to a separate atom.

175

176

Chapter 5: Stereochemistry

ends duplicated. Note that when you break a bond, you always add two imaginary atoms. H H H H / I I I R-C=C"- becomes R-C-C-H H ©©

break and duplicate

H H H I /H I / R-C=N becomes R-C-N

break and duplicate

R-C -C-H becomes break and duplicate

OH I R-C=O break and duplicate

®© �� R-C-C-H ©©

OH I becomes R-C-O

�©

2. Using a three-dimensional drawing or a model, put the fourth-priority group in back and view the molecule along the bond from the asymmetric carbon to the fourth-priority group. Draw an arrow from the first-priority group, through the second, to the third. (R) rectus, sinister,

If the arrow points clockwise, the asymmetric carbon atom is called (Latin, "upright"). If the arrow points counterclockwise, the chiral carbon atom is called (S) (Latin, "left").

rotate

rotate

(R) enantiorner

(S) en antiorner

and (S) Nomenclature of Asymmetric Carbon Atoms 177 Alternatively, you can draw the arrow and imagine turning a car's steering wheel in that direction. If the car would go to the left, the asymmetric carbon atom is designated (S). If the car would go to the right, the asymmetric carbon atom is designated (R). Let's use the enantiomers of alanine as an example. The naturally occur­ ring enantiomer is the one on the left, determined to have the (S) configuration. Of the four atoms attached to the asymmetric carbon in alanine, nitrogen has the largest atomic number, giving it the highest priority. Next is the -COOH car­ bon atom, since it is bonded to oxygen atoms. Third i s the methyl group, fol­ lowed by the hydrogen atom. When we position the natural enantiomer with its HtnZ; hydrogen atom pointing away from us, the arrow from -NH2 to -COOH to Until you become comfortable with drawings, use models to -CH3 points counterclockwise. Thus, the naturally occurring enantiomer of working you assign (R) and (5) alanine has the (S) configuration. Make models of these enantiomers to illustrate help configurations. how they are named (R) and (S). 5-3 (R)

PROBLEM-SOLVING

G)

G)

COOH clockwise I H"�",IIC "(i) H2N CH3 (j)

COOH I 0 CII ' ' 'H / , (j) H3C NH2 (i)

counterclockwise

@

natural (S)-alanine

unnatural (R )-alanine

S O LV E D P R O B L E M 5 - 2

Draw the enantiomers of 1,3-dibromobutane and label them as (R) and (S). (Making a model is particularly helpful for this type of problem.) *

CH -CH -CH-CH

I

Br

2

2

I

Br

3

S O L U TIO N

The third carbon atom in 1 ,3-dibromobutane is asymmetric. The bromine atom receives fust priority, the (-CH2CH2Br) group second priority, the methyl group third, and the hydro­ gen fourth. The following mirror images are drawn with the hydrogen atom back, ready to assign (R) or (S) as shown.

S O LV E D P R O B L E M 5 - 3

The structure of one of the enantiomers of carvone i s shown here. Find the asymmetric carbon atom, and determine whether it has the (R) or the (S) configuration. o

(Continued)

PROBLEM-SOLVING

HtnZ;

In assigning priorities for a ring carbon, go around the ring in each direction until you find a point of difference; then use the difference to determine which ring carbon has higher priority than the other.

178

Chapter 5: Stereochemistry

Scientists frequently use the iso­ topes of hydrogen to assign the configuration of the products of biological reactions. Ethanol, made

chiral by the presence of a deuteri­ um at (2, is one of the early examples.

eH)

S O L U TI O N

The asymmetric carbon atom is one of the ring carbons, as indicated by the asterisk in the following structure. Although there are two - CH 2 - groups bonded to the carbon, they are different -CH2 - groups. One is a -CH2-CO - group, and the other is a - CH 2 - CH=C group. The groups are assigned priorities, and this is found to be the (S) enantiomer. Group CD:

��

C* - C-CH2

I

CH3 Group (1):

��

C* - CH -C-O

I

2

Group G):

C-

��

C* - CH -C-C-C=O 2

PROBLEM-SOLVING

HiltZ;-

If the lowest-priority atom (usually H) is oriented toward you, you don't need to turn the structure around. You can leave it as it is with the H toward you and apply the RI5 rule backward.

II I

H

I

CH3

I

PRO B L E M 5-6

Star ( * ) each asymmetric carbon atom in the following examples, and determine whether it has the (R) or (S) configuration. CH3

I

(a) HO :;C "

H

(b)

CH2CH3

(d)

(,)

Q Cl

Q H

CI

Cl

(h)

PROBLEM-SOLVING

HiltZ;-

Interchanging any two substituents on an asymmetric carbon atom inverts its (R) or (5) configuration. If there is only one asymmetric carbon in a molecule, inverti ng its configuration gives the enantiomer.

PROBLEM 5-7

In Problem 5-3, you drew the enantiomers for a number of chiral compounds. Now go back and designate each asymmetric carbon atom as either (R) or (S) .

5-4 Optical Activity

Mirror-image molecules have nearly identical physical properties. Compare the following properties of (R)-2-bromobutane and (S)-2-bromobutane. {R}-2-Bromobutane

{S}-2-Bromobutane

9 1 .2 - 1 12 1 .436 1 .253

9 1 .2 - 1 12 1 .436 1.253

boiling point (0 C) melting point (0 C) refractive i ndex density

179

5-4 Optica l Activity

Differences in enantiomers become apparent in their interactions with other chiral molecules, such as enzymes. Still, we need a simple method to distinguish between enantiomers and measure their purity in the laboratory. Polarimetry is a common method used to distinguish between enantiomers, based on their ability to rotate the plane of polarized light in opposite directions. For example, the two enantiomers of thyroid hormone are shown below. The (S) enantiomer has a powerful effect on the metabolic rate of all the cells in the body. The (R) enantiomer is useless. In the labora­ tory, we distinguish between the enantiomers by observing that the active one rotates the plane of polarized light to the left. HO

� }- � }-C:;c{ O

I

COOH

I

ri ri OR � O� _ _ "

�

I

I

wrong enantiomer (R) rotates polarized light to the right

thyroid hormone (5) rotates polarized Light to the left

5-4A

H,N, l : HOOC /) "' CH'

Plane- Po l a rized Light

Most of what we see is unpolarized light, vibrating randomly in all directions. Plane­ is composed of waves that vibrate in only one plane. Although there are other types of "polarized light," the term usually refers to plane-polarized light. When unpolarized light passes through a polarizing filter, the randomly vibrating light waves are filtered so that most of the light passing through is vibrating in one direction (Figure 5-10). The direction of vibration is called the axis of the filter. Polariz­ ing filters may be made from carefully cut calcite crystals or from specially treated plas­ tic sheets. Plastic polarizing filters are often used as lenses in sunglasses, because the axis of the filters can be positioned to filter out reflected glare. When light passes first through one polarizing filter and then through another, the amount of light emerging depends on the relationship between the axes of the two

polarized light

waves vibrating in a single plane

waves vibrating in all directions

-·m light source

polarizing filter

plane-polarized light

"'ill Figure 5-1 0

Function of a polarizing filter. The waves of plane-polarized light vibrate primarily in a single plane.

180

Chapter 5: Stereochemistry second polarizing filter, axis parallel (0 the first

first polarizing filter

-(D- t> /

observer sees

maximum light plane-polarized

first polarizing filter � Figure 5-1 1

second polarizing filter, axis perpendicular to the first

-1'\j)\

Crossed poles. When the axis of a second polarizing filter is parallel to the first, a maximum amount of light passes through. When the axes of the filters are perpendicular (crossed poles), no light passes through.

light

plane-polarized

� �

-O-k � L [ [

Y

J -,

ob""

,

no light

light

filters (Figure 5-1 1). If the axes of the two filters are lined up (parallel), then nearly all the light that passes through the first filter also passes through the second. If the axes of the two filters are perpendicular (crossed poles), however, all the polarized light that emerges from the first filter is stopped by the second. At intermediate angles of rota­ tion, intermediate amounts of light pass through. You can demonstrate this effect for yourself by wearing a pair of polarized sun­ glasses while looking at a light source through another pair (Figure 5-12). The second pair seems to be transparent, as long as its axis is lined up with the pair you are wear­ ing. When the second pair is rotated to 90°, however, the lenses become opaque, as if they were covered with black ink. 5-4B

Rotation of Plane-Polarized Light

When polarized light passes through a solution containing a chiral compound, the chi­ ral compound causes the plane of vibration to rotate. Rotation of the plane of polarized

� Figure 5- 1 2

Using sunglasses to demonstrate parallel axes of polarization and crossed poles. When the pairs of sunglasses are parallel, a maximum amount of light passes through. When they are perpendicular, very little light passes through.

light is called optical activity, and substances that rotate the plane of polarized light are said to be optically active. Before the relationship between chirality and optical activity was known, enan­ tiomers were called optical isomers because they seemed identical except for their oppo­ site optical activity. The term was loosely applied to more than one type of isomerism among optically active compounds, however, and this ambiguous term has been replaced by the well-defined term enantiomers. Two enantiomers have identical physical properties, except for the direction they rotate the plane of polarized light. Enantiomeric compounds rotate the plane of polarized light by exactly the same amount but in opposite directions. If the (R) isomer rotates the plane 30° clockwise, the (S) isomer will rotate it 30° coun­ terclockwise. If the (R) enantiomer rotates the plane 5° counterclockwise, the (S) enantiomer will rotate it 5° clockwise. We cannot predict which direction a particular enantiomer [either (R) or (S)] will rotate the plane of polarized light. (R) and (S) are simply names, but the direction and magnitude of rotation are physical properties that must be measured. 5-4C

Polari metry

A polarimeter measures the rotation of polarized light. It has a tubular cell filled with a solution of the optically active material and a system for passing polarized light through the solution and measuring the rotation as the light emerges (Figure 5-13). The light from a sodium lamp is filtered so that it consists of just one wavelength (one color), because most compounds rotate different wave­ lengths of light by different amounts. The wavelength of light most commonly used for polarimetry is a yellow emission line in the spectrum of sodium, called the sodium D line. Monochromatic (one-color) light from the source passes through a polarizing filter, then through the sample cell containing a solution of the optically active com­ pound. On leaving the sample cell, the polarized light encounters another polarizing filter. This filter is movable, with a scale allowing the operator to read the angle between the axis of the second (analyzing) filter and the axis of the first (polarizing) filter. The operator rotates the analyzing filter until the maximum amount of light is

sodium monochromator filter lamp

polarizing filter

sample cell

analyzing filter

detector

... Figure 5-1 3

Schematic diagram of a polarimeter. The light originates at a source (usually a sodium lamp) and passes through a polarizing filter and the sample cell. An optically active solution rotates the plane of polarized light. The analyzing filter is another polarizing filter equipped with a protractor. It is turned until a maximum amount of light is observed, and the rotation is read from the protractor.

5-4

Optical Activity

PROBLEM-SOLVING

181

Hi-Iti:;-

Don't confuse the process for naming a structure (R) or (5) with the process for measuring an optical rotation. Just because we use the terms clockwise and counterclockwise in naming (R) and (5) does not mean that light follows our naming rules.

182

Chapter 5: Stereochemistry �

ana l yzing filter

chiral solution

transmitted, then the observed rotation is read from the protractor. The observed rota­ tion is symbolized by a, the Greek letter alpha. Compounds that rotate the plane of polarized light toward the right (clock­ wise) are called dextrorotatory, from the Greek word dexios, meaning "toward the right." Compounds that rotate the plane toward the left (counterclockwise) are called levorotatory, from the Latin word laevus, meaning "toward the left." These terms are sometimes abbreviated by a lowercase d or I. Using IUPAC notation, the direc­ tion of rotation is specified by the ( +) or ( ) sign of the rotation: Dextrorotatory (clockwise) rotations are ( + ). Levorotatory (counterclockwise) rotations are ( ). For example, the isomer of 2-butanol that rotates the plane of polarized light clockwise is named ( + )-2-butanol or d-2-butanol. Its enantiomer, ( )-2-butanol or 1-2-butanol, rotates the plane counterclockwise by exactly the same amount. You can see the principle of polarimetry by using two pairs of polarized sunglasses, a beaker, and some corn syrup or sugar solution. Wear one pair of sun­ glasses, look down at a light, and hold another pair of sunglasses above the light. Notice that the most light is transmitted through the two pairs of sunglasses when their axes are parallel. Very little light is transmitted when their axes are perpendicular. Put syrup into the beaker, and hold the beaker above the bottom pair of sunglass­ es so the light passes through one pair of sunglasses (the polarizing filter), then the beaker (the optically active sample), and then the other pair of sunglasses (the analyz­ ing filter); see Figure 5-14. Again, check the angles giving maximum and minimum light transmission. Is the syrup solution dextrorotatory or levorotatory? Did you notice the color variation as you rotated the filter? You can see why just one color of light should be used for accurate work. -

� Figure 5- 1 4

Apparatus for using a l ightbulb and two pairs of polarized sunglasses as a simple polarimeter.

-

-

5-4D

Specific Rotation

The angular rotation of polarized light by a chiral compound is a characteristic phys­ ical property of that compound, just like the boiling point or the density. The rota­ tion (a) observed in a polarimeter depends on the concentration of the sample solution and the length of the cell, as well as the optical activity of the compound. For example, twice as concentrated a solution would give twice the original rotation. Similarly, a 20-cm cell gives twice the rotation observed using a similar concentra­ tion in a lO-cm cell. To use the rotation of polarized light as a characteristic property of a com­ pound, we must standardize the conditions for measurement. We define a com­ pound's specific rotation [a] as the rotation found using a lO-cm ( I-dm) sample cell and a concentration of 1 g/mL. Other cell lengths and concentrations may be used, as long as the observed rotation is divided by the path length of the cell (l) and the concentration (c). [a] a(observed) =

where

--'­ --'---

c·l

a( observed) rotation observed in the polarimeter c concentration in grams per mL I length of sample cell (path length) in decimeters (dm) =

=

=

5-4 Optical Activity S O LV E D P R O B L E M 5 - 4

When one of the enantiomers of 2-butanol i s placed i n a polarimeter, the observed rotation is 4.05° counterclockwise. The solution was made by diluting 6 g of 2-butanol to a total of 40 mL, and the solution was placed into a 200-mm polarimeter tube for the measurement. Determine the specific rotation for this enantiomer of 2-butanol. S O L U TIO N

Since it is levorotatory, this must be (- )-2-butanol. The concentration is 6 g per 40 mL = 0. 1 5 g/mL, and the path length is 200 mm = 2 dm. The specific rotation is ?5 [ ] (l'D

=

-4.05° = -13 .50 (0.15 ) (2 )

A rotation depends on the wavelength of light used and also on the temperature, so these data are given together with the rotation. In Solved Problem 5-4, the "25" means that the measurement was made at 25° C, and the "D" means that the light used was the D line of the sodium spectrum. Without even measuring it, we can predict that the specific rotation of the other enantiomer of 2-butanol will be [alE? = + 13.5° where the ( ) refers to the clockwise direction of the rotation. This enantiomer would be called ( + ) -2-butanol. We could refer to this pair of enantiomers as ( + ) -2-butanol and ( - ) -2-butanol or as (R)-2-butanol and (S)-2-butanol. Does this mean that (R)-2-butanol is the dextrorotatory isomer because it is named (R), and (S)-2-butanol is levorotatory because it is named (S)? Not at all! The rotation of a compound, ( + ) or ( - ) , is something that we measure in the polarimeter, depending on how the molecule interacts with light. The (R) and (S) nomenclature is our own artificial way of describing how the atoms are arranged in space. In the laboratory, we can measure a rotation and see whether a particular substance is ( +) or ( - ) . On paper, we can determine whether a particular drawing is named (R) or (S). But it is difficult to predict whether a structure we call (R) will rotate polarized light clockwise or counterclockwise. Similarly, it is difficult to predict whether a dextrorotatory substance in a flask has the (R) or (S) configuration. +

PRO B L E M 5-8

A solution o f 2.0 g of ( + )-glyceraldehyde, HOCH2 - CHOH - CHO, i n 1 0 mL of water was placed in a 1 00-mm cell. Using the sodium D line, a rotation of + l .74° was found at 25° C. Determine the specific rotation of (+ ) -glyceraldehyde. PRO B L E M 5-9

A solution of 0.5 g of ( -)-epinephrine (see Fig. 5- 1 5) dissolved in 1 0 mL of dilute HCI was placed in a 20-cm polarimeter tube. Using the sodium D line, the rotation was found to be -5.0° at 25° C. Determine the specific rotation of epinephrine. PROB L E M 5-1 0

A chiral sample gives a rotation that is close to 1 80°. How can one tell whether this rota­ tion is + 1 80° or - 1 800?

183

184

h er 5: Stereochemistry

C apt

5-5 Biologica l Discrimi nation of Ena nti ome rs

If the direction of rotation of polarized light were the only difference between enan­ tiomers, one might ask whether the difference was important. Biological systems com­ monly distinguish between enantiomers, and two enantiomers may have totally different biological properties. In fact, any chiral probe can distinguish between enantiomers, and a polarimeter is only one example of a chiral probe. Another example is your hand. If you needed to sort a box of gloves into right-handed gloves and left-handed gloves, you could distinguish between them by checking to see which ones fit your right hand. Enzymes in living systems are chiral, and they are capable of distinguishing be­ tween enantiomers. Usually, only one enantiomer of a pair fits properly into the chiral active site of an enzyme. For example, the levorotatory form of epinephrine is one of the principal hormones secreted by the adrenal medulla. When synthetic epinephrine is given to a patient, the ( ) form has the same stimulating effect as the natural hor­ mone. The ( + ) form lacks this effect and is mildly toxic. Figure 5-15 shows a simpli­ fied picture of how only the ( ) enantiomer fits into the enzyme's active site. Biological systems are capable of distinguishing between the enantiomers of many different chiral compounds. In general, just one of the enantiomers produces the characteristic effect; the other either produces no effect or has a different effect. Even your nose is capable of distinguishing between some enantiomers. For example, ( - ) -carvone is the fragrance associated with spearmint oil; ( + ) -carvone has the tangy odor of caraway seed. The receptor sites for the sense of smell must be chiral, therefore, -

-

o (R)-( )-epinephrine

enzyme active site

-

enzyme-substrate complex

natural epinephrine

OH (y0H

Y

*

� Figure 5 - 1 5

Chiral recognition of epinephrine by an enzyme. Only the levorotatory enantiomer fits into the active site of n ym

the e z e

.

(S)-(+)-epinephrine unnatural epinephrine

does not fit the enzyme's active site

just as the active sites in most enzymes are chiral. In general, enantiomers do not inter­ act identically with other chiral molecules, whether or not they are of biological origin.

O�

CH3

Q

O

/�CH2 H3C ( + )-carvone (caraway

H2C

Mixtures

185

Enzymes can exist as two enan­ tiomers, although only one enan­ tiomer is found in nature. In 1992,

tiomers of an enzyme that cuts pep­ tide substrates, and showed for the first time that each enzyme acts only on the corresponding enantiomeric peptide substrate.

/" C

seed)

Racemic

Stephen Kent and co-workers reported the synthesis of both enan­

�

C

5-6

CH3

(- )-carvone (spearmint)

PRO B L E M 5 - 1 1

If you had the two enantiomers of carvone in unmarked bottles, could you use just your nose and a polarimeter to determine (a) whether it is the ( + ) or ( ) enantiomer that smells l ike spearmint? (b) whether it is the (R) or (S) enantiomer that smells like spearmint? (c) With the information given in the drawings of carvone above, what can you add to your answers to (a) or (b)? -

Suppose we had a mixture of equal amounts of ( + ) -2-butanol and ( - ) -2-butanol. The ( +) isomer would rotate polarized light clockwise with a specific rotation of + 13.5°, and the ( ) isomer would rotate the polarized light counterclockwise by exactly the same amount. We would observe a rotation of zero, just as though 2-butanol were achiral. A solution of equal amounts of two enantiomers, so that the mixture is opti­ cally inactive, is called a racemic mixture. Sometimes a racemic mixture is called a racemate, a (±) pair, or a (d,f) pair. A racemic mixture is symbolized by placing (±) or (d,l) in front of the name of the compound. For example, racemic 2-butanol would be symbolized by (± ) -2-butanol" or "(d,l)-2-butanol." -

5-6 Racemi c Mixtures

"

CH3

,

H"'

HO

C

CH3

and

I

\ CH2CH3

(S )-( + )-2-butanol + 1 3.5° rotation

I

/ CH3CH2

A racemic mixture contains eq u al amounts of the two enantiomers.

\'''H

C

OH

(R )-( - )-2-butanol - 1 3 .5° rotation

You might think that a racemic mixture would be unusual, since it requires exactly equal amounts of the two enantiomers. This is not the case, however. Many reactions lead to racemic products, especially when an achiral molecule is converted to a chiral molecule. A reaction that uses optically inactive reactants and catalysts cannot produce a product that is optically active. Any chiral product must be formed as a racemic mixture. For example, hydrogen adds across the C 0 double bond of a ketone to produce an alcohol.

Many drugs currently on the mar­ ket are racemic mixtures. Keta­ mine, for example, is a potent anesthetic agent, but its use is lim­ ited because it is hallucinogenic

responsible

for

the

" C=O

/ R'

H,. Ni

o

R

I

R'-C-O

I

H

I

H

isomer is anesthetic

effects, and the (R) isomer causes the hallucinogenic effects.

=

R

(5)

(making it a drug of abuse widely known as "K"). The

CI

ketamine

186

Chapter 5: Stereochemistry -

---_._--- .-..

--.---

acid H2 from top

� Figure 5- 1 6

Hydrogenation of 2-butanone fOnTIS racemic 2-butanol. Hydrogen adds to either face of the double bond. Addition of H2 to one face gives the (R) product, while addition to the other face gives the (S) product.

add H 2 from bottom

)

CH3CH2·" CH3 ..../ 4.8 x 0 x d H .· /"' ' 8C+ X8where 0 is the charge and d is the bond length. /L

/L =

--

H

The eiectronegativities of the halogens increase in the order CI I F Br electronegalivily:




1°

>

:j:

+

!\ .. ---

transition state

substrate

The o rder of reactivity for substrates is CH3X

I

�

---

p roduct

..

:X:

leaving g roup

2°. (3° alkyl h alide s cannot react by this mechanism.)

( Continued)

228

Chapter 6:

Alkyl Halides: Nucleophilic Substitution and Elimination

EXAMPLE : Reaction of 1 -bromobutane with so dium methoxi de gives 1 -methoxybutane . +

CH3CH2CH2CH2Br

-------7)

NaBr

I -methoxybutane

I -bromobutane

sodium methoxide

+

CH3CH2CH2CH20CH3

:Br:

nucleophile

product

transition state

electrophile (substrate)

leaving group

PR OBLEM 6-1 3: Under certain conditions, the reaction of 0.5 M I-bromobutane with 1 .0 M sodium methoxide forms I -methoxybutane at a rate of 0.05 mollL per second. What would be the rate if 0.1 M I -bromobutane and 2.0 M NaOCH3 were used?

6-9

Many useful reactions take place by the SN2 mechanism. The reaction of an alkyl halide, such as methyl iodide, with hydroxide ion gives an alcohol. Other nucleophiles convert alkyl halides to a wide variety of functional groups. The following table summarizes some of the types of compounds that can be formed by nucleophilic displacement of alkyl halides.

General ity of the SN2 Reaction

I

S U M M ARY

SN 2 Reactions of A lky l Ha li des Nuc:-

+

Nucleophile R-X

+

-:1:

R-X

+

-:O H

R-X

+

-:O R' ..

R-X

+

R-X

+

-: SR'

R-X

+

R- X

+

: N H3 .. .. + -: N= N= N:-

R-X

+

-: SH

-: C- C - R'

R-X

+

R-X

+

R' - COO:-

R-X

+

:P(Ph)3

-: C - N:

R-X -----3> -----3> -----3> -----3> -----3> -----3> -----3> -----3> -----3> -----3> -----3>

-----3>

Nuc- R

+

Product

..

X-

Class of Product

R-I:

alkyl halide

R-OH

alcohol

R-O R'

ether

R-SH

thiol (mercaptan)

R- SR'

thioether (s ulfide)

R-NH; X.. .. + R- N= N= N:-

amine azide

R-C- C - R'

alkyne

R- C - N:

nitrile

R'- COO -R

ester

[R-PPh3 ]+ -X

phosphonium s alt

..

..

6-9 Generality of the SN2 Reaction

Examples

U

+

U u

- : OH

CH 2 C1

CH2OH

hydroxide

benzyl chloride

benzyl alcohol

0'

U··

+

CH3I iodomethane

methoxybenzene

phenoxide

(methyl iodide)

CH3CH2CH2CH2CH2Br

+

CH3CH2CH2CH2CI

+

I-bromopentane

..

: SH

(methyl phenyl ether) -----?

CH3CH2CHzCH2CHzSH I-pentanethiol

: NH3

l-chlorobutane

ammonia

(n-butyl chloride)

(excess)

+

CH3CH2B r

Na+ -: C==C-H

bromoethane

-----?

CH3CH2CH2CH2NH2 I-butanamine

(n-butylamine) -----?

CH3CH2 - C==C -H l-butyne

sodium acetylide

(ethyl bromide)

+

CH3CH2CH}

- : C==N

1-iodopropane

(ethylacetylene) -----?

CH3CH2CH2- C==N butaneni trile

cyanide

(n-propyl iodide)

(butyronitrile)

Ha lo gen Ex chan ge Rea ctions The SN2 reaction provides a useful method for synthesizing alkyl iodides and fluorides, which are more difficult to make than alkyl chlorides and bromides. Halides can be converted to other halides by halogen exchange reactions, in which one halide displaces another. Iodide is a good nucleophile, and many alkyl chlorides react with sodium iodide to give alkyl iodides. Alkyl fluorides are difficult to synthesize directly, and they are often made by treating alkyl chlorides or bromides with KF under conditions that use a crown ether and an aprotic solvent to enhance the normally weak nucleophilicity of the fluoride ion (see Section 6- 1 0).

R - X + 1R - X + KF

R - I + x-

-----?

IS-crown-6

CH3CN

)

R -F + KX

Examples

H2C=CH - CH2Cl

+

NaI

CH3CH2Cl

+

KF

allyl chloride

ethyl chloride

OCH1

+

H2C=CH - CH2I allyl iodide

l8-crown-6

CH3CN

CH3CH2F ethyl fluoride

+

NaCl

KCl

229

230

Chapter 6: Alkyl Halides: Nucleophilic Substitution and Elimination

P R O B L E M 6- 1 4 Predict the major products of the following substitutions.

(a)

CH3CH2Br

(b)

HC=e:-+Na

(CH 3 hCO- +K-

+

ethyl bromide

potassium l-butoxide

CH3CH2CH2CH2Cl-

+

I -chlorobutal1e

sodium acetylide

(c) (CH 3 hCHCH2Br + excess NH3 (d) CH3CH21 + NaCN(e) I-chloropentane + NaI-

(f) l-chloropentane

+

CH3CN

18-crowl1-6

KF

P R O B L E M 6- 1 5 Show how you would convert l-chlorobutane into the following compounds. I-butanol (b) I-fluorobutane

(a) (c) (e) (g)

6-10

We will use the SN2 reaction as an example o f how w e study the properties o f the species that participate in the reaction. Both the nucleophile and the substrate (the alkyl halide) are important, as well as the type of solvent used. We begin by consider­ ing what makes a good nucleophile. The nature of the nucleophile strongly affects the rate of the SN2 reaction. A strong nucleophile is much more effective than a weak one in attacking an electrophilic carbon atom. For example, both methanol (CH 30H) and methoxide ion (CH 30-) have easily shared pairs of nonbonding electrons, but methoxide ion reacts with electrophiles in the SN2 reaction about 1 million times faster than methanol. It is generally true that a species with a negative charge is a stronger nucleophile than a similar, neutral species. Methoxide ion has nonbonding electrons that are readily available for bonding. In the transition state, the negative charge is shared by the oxygen of methoxide ion and by the halide leaving group. Methanol, however, has no negative charge; the transition state has a partial negative charge on the halide but a partial positive charge on the methanol oxygen atom. We can generalize the case of methanol and the methoxide ion to say that

Facto rs A ffecti ng SN2 Reactions: Stren gth of the Nucleophi le

A

H

�C -". H ''' J. ' H

C H -" g.. 3

base is always a stronger nucleophile than its conjugate acid.

-

conjugate base (stronger nucleophile)

C H 3 -O· I H "

.

H

�\

conjugate acid (weaker nucleophile)

(d) CH3 - (CH2h- CN (f) CH3CH2- O - (CHzh- CH3

l-iodobutane

CH3 - (CH2h-C=CH CH3 - (CH2h- NH2

",c

H7 H

l,"}"

.

-

l �

,:�

H

1

J 'J

��

C H 3-O---C ---I: " /\" H H lower E"

1 CH -O---C --I: !

3

1 /\" H H H

higher E"

=1=

-

H . / + : I: CH 3-O-C �.,. HH ..

.

••

-

..+

H

/

+ :1:CH 3 -O-C � .. . I H HH

6-10 TABLE 6-3

Some Common Nucleophiles. listed in Decreasing Order of Nucleophilicity in Hydroxylic Solvents Such as Water and Alcohols

( CH3 CH2)3P:

strong nucleophiles

moderate nucleophiles

The "nitrogen mustard" anticancer drugs are believed to alkylate DNA using two SN2 reactions. First, the

: B r:-

nitrogen

: NH3

-:S - H :1:-

..

ride portion to generate a reactive

o

gen atom of DNA. The process is re­

..

II

peated, linking the two strands of the double-helix DNA, and thereby

CH3C -O:-

preventing replication of the DNA.

CH2CH2Cl

••

:F:-

weak nucleophiles

CH3 - 0 :-

I

H3C - N:

H-O - H

••

displaces

intermediate that alkylates a nitro­

: Cl:-

(CH3 CH2 )2NH -: C - N (CH3 CH2 )3N:

nucleophile

chloride on the primary alkyl chlo­

CH3-�- CH3

H- O:-

23 1

Facto rs Affecting SN2 Reacti o n s: S trength of the Nucleophile

CH3 -Q- H

� CH 2

CH 2 0Cl

nitrogen mustard

We might be tempted to say that methoxide is a much better nucleophile because it is much more basic. This would be a mistake because basicity and nucleophilicity are different properties. Basicity is defined by the eq uilibrium constant for abstracting a proton. Nucleophilicity is defined by the rate of attack on an electrophilic carbon atom. In both cases, the nucleophile (or base) forms a new bond. If the new bond is to a proton, it has reacted as a base; if the new bond is to carbon, it has reacted as a nuc­ leophile. Predicting which way a species will react may be difficult; most (but not all) good nucleophiles are also strong bases, and vice versa.

-

CH2CH2Cl

J

H 3C - N-CH2 \ / CH2

intermediate

CI-

Basicity B-H

+

A:-

Nucleophilicity

I

�I

B:-

+

-C-X

B -C -

1(;

I

+

X :-

Table 6-3 lists some common nucleophiles in decreasing order of their nucle­ ophilicity in common solvents such as water and alcohols. The strength of nucle­ ophiles shows three major trends:

C

S U M MARY

Tren ds i n Nucleoph ilicity

1. A species with a negative c harge i s a stronger nucleophile than a similar neu tral species. In particular, a base i s a stronger nucle­ ophile than its conjugate acid.

- : SH

>

..

>

- : NH2

2. Nucleophil icity decreases from left to right in the periodic table, following the increase in elec tronegativity from left to right. The more electronegative elements have more tightly held non bonding electrons that are less reactive toward forming new bonds. ..

> : OH ..

..

..

:F:..

(CH3CH2) 3P : > (CH3 CH2) 2S :

3. Nucleophilicity increases down the periodic table, following t he increase i n size and polari zability. ..

..

:1:- > :Br:- > : CI : - > : F : -

..

-: SeH

..

> - : SH > - : OH

232

Chapter 6:

Alkyl Halides: Nucleophilic Substitution and Elimination Sp3 orbital

\

back lobe

H

H "

I

C----- X

.� - X H"'J

F-

F\

H H

H

"hard," small valence shell

� FIGURE 6-5

transition state

more bonding

Comparison of fluoride ion and iodide ion as nucleophiles in the SN2 reaction. Fluoride has tightly bound electrons that cannot begin to form a C F bond until the atoms are close together. Iodide has more loo sel y bound outer electrons that begin bonding earlier in the reaction.

\

-

"soft," large valence shell

8-

:j:

H

I

8-

:j:

C----- X

F\

H H

transition state

The third trend (size and polarizability) reflects an atom's ability to engage in partial bonding as it begins to attack an electrophilic carbon atom. As we go down a column in the periodic table, the atoms become larger, with more electrons at a greater distance from the nucleus. The electrons are more loosely held, and the atom is more polarizable: Its electrons can move more freely toward a positive charge, resulting in stronger bonding in the transition state. The increased mobility of its electrons enhances the atom's ability to begin to form a bond at a relatively long distance. Figure 6-5 illustrates this polarizability effect by comparing the attack of iodide ion and fluoride ion on a methyl halide. The outer shell of the fluoride ion is the second shell. These electrons are tightly held, close to the nucleus. Fluoride is a "hard" (low­ polarizability) nucleophile, and its nucleus must approach the carbon nucleus quite closely before the electrons can begin to overlap and form a bond. In the transition state, there is little bonding between fluorine and carbon. In contrast, the outer shell of the iodide ion is the fifth shell. These electrons are loosely held, making the iodide ion a "soft" (high-polarizability) nucleophile. The outer electrons begin to shift and overlap with the carbon atom from farther away. There is a great deal of bonding between iodine and carbon in the transition state, which lowers the energy of the transition state. 6 - 1 0A

Steric Effects on Nucleophi l i city

To serve as a nucleophile, an ion or molecule must get in close to a carbon atom to attack it. Bulky groups on the nucleophile hinder this close approach, and they slow the reaction rate. For example, the t-butoxide ion is a stronger base (for abstracting protons) than ethoxide ion, but t-butoxide ion has three methyl groups that hinder any close approach to a carbon atom. Therefore, ethoxide ion is a stronger nucleophile than t-butoxide ion. When bulky groups interfere with a reaction by virtue of their size, we call the effect steric hindrance. three methyl groups hinder attack at a carbon atom

CH3

I

..

CH3 -C -O: .. -

I

CH3

t-butoxide (hindered) stronger base, yet weaker nucleophile

.

CH3 -CH2 -0: .

••

ethoxide (unhindered) weaker base, yet stronger nucleophile

6-10 Factors Affecting SN2 Reactions: Strength of the NucIeophile

233

Steric hindrance has little effect on bas icity because basicity involves attack on an unhindered proton. When a nucleophilic attack at a carbon atom is involved, however, a bulky base cannot approach the carbon atom so easily. Most bases are also nucleophiles, capable of attacking either a proton or an electrophilic carbon atom. If we want a species to act as a base, we use a bulky reagent like t­ butoxide ion. If we want it to react as a nucleophile, we use a less hindered reagent, like ethoxide. PRO B L E M 6-1 6 F or each pair, predict the stronger nucIeophile in the solvent). Explain your prediction.

(a) (c) (e) (g)

(CH3CH2hN or (CH3CH2hNH NH3 or PH3 (CH3hN or (CH3hO (CH3hCHO- or CH3CH2CH20-

6- 1 08

(b) (d) (0 (h)

SN2

reaction (using an alcohol as the

(CH3hO or (CH3hS CH3S- or H2S CH3S- or CH30H

1 - or Cl-

Another factor in the nucleophilicity of these ions is their solvation, particularly i n protic solvents. A protic solvent is one that has acidic protons, usually in the form of 0- H or N -H groups. These groups form hydrogen bonds to negatively charged nucleophiles. Protic solvents, especially alcohols, are convenient solvents for nucleophilic substitutions because the reagents (alkyl halides, nucleophiles, etc.) tend to be quite soluble. Small anions are solvated more strongly than large anions in a protic solvent because the solvent approaches a small anion more closely and forms stronger hydrogen bonds. When an anion reacts as a nucleophile, energy is required to "strip off' some of the solvent molecules, breaking some of the hydrogen bonds that sta­ bilized the solvated anion. More energy is required to strip off solvent from a small, strongly solvated ion such as fluoride than from a large, diffuse, less strongly solvated ion like iodide.

/

Hooo : F : �.

H-O

'"

H 0

R

'" / o

R

R

C -X

H --; \..,; H

@� o '"

0 0

Hiltv

Steric hindrance (bulki ness) h i nders n uc!eophil icity (SN2) more than it does

Solvent Effects o n Nucleophilicity

�

PROBLEM-SOLVING

solvent partially stripped off in the transition state

The enhanced solvation of smaller anions in protic solvents, requiring more energy to strip off their solvent molecules, reduces their nucleophilicity. This trend reinforces the trend in polarizability: The polarizability increases with increasing atomic number, and the solvation energy (in protic solvents) decreases with increas­ ing atomic number. Therefore, nucleophilicity (in protic solvents) generally increas­ es down a column in the periodic table, as long as we compare similar species with similar charges.

basicity.

234

Chapter 6: Alkyl Halides: Nucleophilic Substitution and Elimination In contrast with protic solvents, aprotic solvents (solvents without 0- H or N - H groups) enhance the nucleophilicity of anions. An anion is more reactive in an aprotic solvent because it is not so strongly solvated. There are no hydrogen bonds to be broken when solvent must make way for the nucleophile to approach an elec­ trophilic carbon atom. The relatively weak solvating ability of aprotic solvents is also a disadvantage: Most polar, ionic reagents are not soluble in simple aprotic solvents such as alkanes. Polar aprotic solvents have strong dipole moments to enhance solubility, yet they have no 0 - H or N H groups to form hydrogen bonds with anions. Examples of useful polar aprotic solvents are acetonitrile, dimethylformamide, and acetone. We can add specific solvating reagents to enhance solubility without affecting the reactivity of the nucleophile. For example, the "crown ether" 1 8-crown-6 solvates potassium ions. Using the potassium salt of a nucleophile and solvating the potassium ions causes the nucleophilic anion to be dragged along into solution. -

acetonitrile

d imethylformamide

acetone

(DMF)

18-crown-6 solvates

K+ ions

The following example shows how fluoride ion, normally a poor nucleophile in hydroxylic (protic) solvents, can be a good nucleophile i n an aprotic solvent. Although KF is not very soluble in acetonitrile, 18-crown-6 solvates the potassium ions, and the poorly solvated (and therefore nucleophilic) fluoride ion follows. KF, 18-crown-6 CH3CN

6-11 Reactivity of the Substrate i n SN2 Reactions

)

Just as the nucleophile is important in the SN2 reaction, the structure of the alkyl halide is equally important. We will often refer to the alkyl halide as the substrate: literally, the compound that is being attacked by the reagent. Besides alkyl halides, a variety of other types of compounds serve as substrates in SN2 reactions. To be a good substrate for SN2 attack by a nucleophile, a molecule must have an electrophilic car­ bon atom with a good leaving group, and that carbon atom must not be too sterically hindered for a nucleophile to attack. 6-11A

Leavi ng-Group Effects on the Substrate

A leaving group serves two purposes in the SN2 reaction: It polarizes the C- X bond-making the carbon atom electrophilic-and it leaves with the pair of electrons that once bonded it to the electrophilic carbon atom. To fill these roles, a good leaving group should be 1. electron withdrawing, to polarize the carbon atom,

stable (not a strong base) once it has left, 3. polarizable, to stabilize the transition state. 2.

Reactivity of the Substrate in SN2 Reactions

6- 1 1

235

The leaving group must be elect ron withd rawing to create a partial positive charge on the carbon atom, making the carbon electrophilic. An electron-withdrawing leaving group also stabilizes the negatively charged transition state. Halogen atoms are strong­ ly electronegative, so alkyl halides are common substrates for SN2 reactions. Oxygen, nitrogen, and sulfur also form strongly polarized bonds with carbon; given the right substituents, they can form the basis for excellent leaving groups. 1.

Strongly polarized

C +-,> X ( X

2.

halogen)

=

The leaving group must be stable once it has left with the pair of electrons that bonded it to carbon. A stable leaving group is needed for favorable energetics. The leaving group is leaving in the transition state; a reactive leaving group would raise the energy of the transition state, slowing the reaction. Also, the energy of the leaving group is reflected in the energy of the products. A reactive leaving group would raise the energy of the products, driving the equilibrium toward the reactants. bond forming

Nuc :

bond breaking

r:�A��

�\ + C -X /j�

/ Nuc -C �

+

transition state

Good leaving groups should be weak bases; therefore, they are the conjugate bases of strong acids. The hydrohalic acids HCl, HBr, and HI are strong, and their con­ jugates (Cl-, Br-, and 1-) are all weak bases. Other weak bases, such as sulfate ions, sulfonate ions, and phosphate ions, can also serve as good leaving groups. Table 6-4 lists examples of good leaving groups. Hydroxide ion, alkoxide ions, and other strong bases are poor leaving groups for SN2 reactions. For example, the - OH group of an alcohol is a poor leaving group because it would have to leave as hydroxide ion.

*

+

Br-CH3

- : OH

PROBLEM-SOLVING

hydroxide ions, al koxide ions, or other strong bases servi ng as leaving groups.

(strong base)

Ions that are strong bases and poor leaving groups:

-: OH

hydroxide

-: OR

-: NH2

alko·�ide

amide

Table 6-4 also lists some neutral molecules that can be good leaving groups. A neutral molecule often serves as the leaving group from a posit ively charged species. TABLE 6-4

Weak Bases That Are Common Leaving Groups

Ions:

Neutral molecules:

..0".

..

..

:CI:

H

I

:Br:

halides

: I:

..

H

I

-

II

:O-S-R

..

II -.0..

sulfonate

R

I

-

..0".

.

II

:O-S-OR

..

..

II

I

-

sulfate

phosphate

R

I

:N-R

:P-R

water

alcohols

amines

phosphines

I

R

.

.

:O-P-OR

II ..0..

:O-R

I

.

:0:

..0..

:O-H

R

-

..

Hiltv

Do not write SN2 reactions that show

236

Chapter 6: Alkyl Halides: Nucleophilic Substitution and Elimination For example, if an alcohol is placed in an acidic solution, the hydroxyl group is pro­ tonated. Water then serves as the leaving group. Note that the need to protonate the alcohol (requiring acid) limits the choice of nucleophiles to those few that are weak bases, such as bromide and iodide. A strongly basic nucleophile would become pro­ tonated in acid.

..

1+

: Br:;:;'--- CH3 -O-H '••

H

H •.

protonated alcohol

+

I

: O -H water

3.

Finally, a good leaving group should be polarizable, to maintain partial bonding with the carbon atom in the transition state. This bonding helps stabilize the transition state and reduce the activation energy. The departure of a leaving group is much like the attack of a nucleophile, except that the bond is breaking rather than forming. Polarizable nucleophiles and polarizable leaving groups both stabilize the transition state by engaging in more bonding at a longer distance. Iodide ion, one of the most polarizable ions, is both a good nucleophile and a good leaving group. In contrast, flu­ oride ion is a small, "hard" ion. Fluoride is both a poor nucleophile (in protic solvents) and a poor leaving group in SN2 reactions. PROBLEM 6-1 7

When dimethyl ether (CH3 - O - C H 3 ) is treated with concentrated HBr, the initial prod­ ucts are CH3Br and CH30H. Propose a mechanism to account for this reaction.

6-11 B

Steric Effects on the Substrate

Different alkyl halides undergo SN2 reactions at vastly different rates. The structure of the substrate is the most important factor in its reactivity toward SN2 displacement. The reaction goes rapidly with methyl halides and with most primary substrates. It is more sluggish with secondary halides. Tertiary halides fai l to react at all by the SN2 mecha­ nism. Table 6-5 shows the effect of alkyl substitution on the rate of SN2 displacements. For simple alkyl halides, the relative rates for SN2 displacement are Relative rates for SN2: The physical explanation for this order of reactivity is suggested by the information in Table 6-5. All the slow-reacting compounds have one property in common: The back side of the electrophilic carbon atom is crowded by the presence of bulky groups. Tertiary halides are more hindered than secondary halides, which are more hindered than primary halides. Even a bulky primary halide (like neopentyl bromide) undergoes SN2 reaction at a rate similar to that of a tertiary halide. The relative rates show that TABLE 6-5

Effect of Substituents on the Rates of SN2 Reactions

Class of Halide methyl primary (10) secondary (2°) tertiary (3°) n-butyl (1°) isobutyl (10) neopentyl ( l0)

Example

Relative Rate

CH3 -Br CH3CH2 -Br (CH3hCH-Br (CH3hC -Br CH3CH2CH2CH2 -Br (CH3hCHCH2 -Br (CH3hCCH2 -Br

>1000 50 1

PROBLEM-SOLVING

(R)

and

(5)

HtnJ/

are j ust names. Don't

rely o n n a m es to dete r m i n e the

CH3CHFOCH3 (S)

+

NaBr

(a) W h y i s bromide rather than fluoride replaced? (b) Draw perspective structures (as shown on the previous page for 2-bromobutane) for the starting material , the transition state, and the product.

(c) Does the product show retention or inversion of configuration? (d) Is thi s result consistent with reaction by the SN2 mechanism?

stereoche m i stry of a react i o n .

240

Chapter 6: Alkyl Halides: Nucleophilic Substitution and Elimination

6-1 3 Fi rst-O rder Nucleoph i l ic Substituti o n : The SN 1 Reaction

When t-butyl bromide is placed in boiling methanol, methyl t-butyl ether can be isolat­ ed from the reaction mixture. Because this reaction takes place with the solvent acting as the nucleophile, it is called a solvolysis (solvo for "solvent," plus lysis, meaning "cleavage"). (CH3 ) 3 C - Br I-butyl bromide

+

CH3 - OH

�

boil

+

( CH3)3C -O - CH3

methanol

methyl I-butyl ether

HBr

This solvolysis is a substitution because methoxide has replaced bromide on the t-butyl group. It does not go through the SN2 mechanism, however. The SN2 requires a strong nucleophile and a substrate that is not too hindered. Methanol is a weak nucleophile, and t-butyl bromide i s a hindered tertiary halide-a poor SN2 substrate. If this substitution cannot go by the SN2 mechanism, what kind of mechanism might be involved? An important clue is kinetic: Its rate does not depend on the con­ centration of methanol, the nucleophile. The rate depends only on the concentration of the substrate, t-butyl bromide.

This rate equation is first order overall : first order in the concentration of the alkyl halide and zeroth order in the concentration of the nucleophile. Because the rate does not depend on the concentration of the nucleophile, we infer that the nucleophile is not present in the transition state of the rate-limiting step. The nucleophile must react after the slow step. This type of substitution is called the SNI reaction, for Substit ut ion, N ucle­ ophilic, unimolecular. The term unimolecular means there is only one molecule in­ volved in the transition state of the rate-limiting step. The mechanism of the SNI reaction of t-butyl bromide with methanol is shown here. Ionization of the alkyl halide (fIrst step) is the rate-limiting step. Step 1: Formation of carbo cation

(rate limiting) (CH3hC +

.. : Br :

+

Step 2: Nucleophilic attack on the carbocation

(CH3)3C +

� ..

?

: -CH3

(slow)

?

.. +

(CH3)3C- -CH3

H

(fast)

H

Final Step: Loss of proton to solvent

..

(CH3)3C - Q -CH3

+

.. +

CH3 - O - H

I

(fast)

H The SN1 mechanism is a multistep process. The first step is a slow ionization to form a carbocation. The second step is a fast attack on the carbocation by a nucle­ ophile. The carbocation is a strong electrophile; it reacts very fast with nucleophiles, including weak nucleophiles. The nucleophile is usually weak, because a strong nucleophile would be more likely to attack the substrate and force some kind of second-order reaction. If the nucleophile is an uncharged molecule like water or an

et..> ,.,....

6- 1 3

KEY M ECHAN ISM 6-4

The

S

N1

First-Order Nucleophilic Substitution: The SN I Reaction

241

Reaction

The SN1 reaction involves a two-step mechanism. A slow ionization gives a carbocation that reacts quickly with a (usually weak) nucleophile. Reactivity: 3° > 2° > 1 °. Step 1.

Step 2.

Formation of the carbocation (rate-limiting). R- X : U· Nucleophilic attack on the carbocation (fast). R

+�

R - Nuc

If the nucleophile is water or an alcohol, a third step is needed to deprotonate the product. EXAM PLE: Solvo ly sis of 1-io do-1-methylcyclohexane in methanol.

Step 1:

Formation of a carbocation (rate-limiting).

Step 2:

Nucleophilic attack by the solvent (methanol).

Step 3:

Deprotonation to form the product.

(j-

CH3 QCHl

product

+

+/ H CH3 -Q ", H

(protonated methanol)

PROBLEM 6-22: Propose an SN 1 mechanism for the solvolysis of 3-bromo2,3-dimethylpentane in ethanol.

alcohol, the positively charged product must lose a proton to give the final uncharged product. The general mechanism for the SN1 reaction is summarized in the Key Mechanism box. The reaction-energy diagram of the SN1 reaction (Figure 6-8) shows why the rate does not depend on the strength or concentration of the nucleophile. The ioniza­ tion (first step) is highly endothermic, and its large activation energy determines the overall reaction rate. The nucleophilic attack (second step) is strongly exothermic, with a lower-energy transition state. In effect, a nucleophile reacts with the carboca­ tion almost as soon as it forms. The reaction-energy diagrams of the SN1 mechanism and the SN2 mechanism are compared in Figure 6-8. The SN 1 has a true intermediate, the carbocation. The

P ROBLEM-SOLVING

HtltP

Never show a proton f a l l i n g off into thin a i r. Show a poss i b l e base (often the solvent) a bstracting the proton.

242

Chapter 6: Alkyl Halides: Nucleophilic Substitution an d Elimin ati on

rate l i miting transition state

� Figure 6-8

Reaction-energy di agrams of the SN 1 and SN2 reactions . The SN 1 is a two­ step mechanism with two transition states ( :j: 1 and :j: 2) and a carbocation intermediate. The SN2 has only one transition state and no intermediate.

t

�tl

/ intermediate �

single transition state

R+ + X -

+ Nuc: -

R - X + Nuc : -

R-Nuc + X -

R- Nuc + X -

-=- - -.= - ===-=========== --

intermediate appears as a relative minimum (a low point) in the reaction-energy dia­ gram. Reagents and conditions that favor formation of the carbocation (the slow step) accelerate the SN 1 reaction; reagents and conditions that hinder its formation retard the reaction. 6- 1 3A

Su bstituent Effects

The rate-limiting step of the SN 1 reaction is ionization to form a carbocation, a strongly endothermic process. The transition state resembles the carbocation (Hammond postulate, Section 4-14); consequently, rates of SN 1 reactions depend strongly on carbo­ cation stability. In Section 4- 1 6A, we saw that alkyl groups stabilize carbocations by donating electrons through sigma bonds (the inductive effect) and through overlap of filled orbitals with the empty p orbital of the carbocation (hyperconjugation). Highly substituted carbocations are therefore more stable. p orbital

vacant

\

weak

8-t

carbocation stability: 3° > 2° > 1 ° > +CH3

8+ CH3 0- + ,/;CH3 � .C 0 +1 8+ CH3 inductive effect carbocation

alkyl group

hyperconjugation

Reactivity toward SN 1 substitution mechanisms follows the stability of carbocations: SN 1 reactivity: This order is oppos ite that of the SN2 reaction. Alkyl groups hinder the SN2 by block­ ing attack of the strong nucleophile, but alkyl groups enhance the SN 1 by stabilizing the carbocation intermediate. Resonance stabilization of the carbocation can also promote the SN 1 reaction. For example, allyl bromide is a primary halide, but it undergoes the SN I reaction

243

6- 1 3 First-Order N u cleoph i l i c Substitution: The S N 1 Reaction

about as fast as a secondary halide. The carbocation formed by ionization is reso­ nance stabilized, with the positive charge spread equally over two carbon atom s .

Nuc :

Be resonance-stabilized carbocation

Vinyl and aryl halides generally do not undergo SN I or SN2 reactions. An SN I reaction would require ionization to form a vinyl or aryl cation, either of which is less stable than most alkyl carbocations. An SN2 reaction would require back-side attack by the nucleophile, which is made impossible by the repulsion of the electrons in the double bond or aromatic ring.

a vinyl halide

6-1 3 8

an aryl halide

Leavi ng-Gro u p Effects

The leaving group is breaking its bond to carbon in the rate-limiting ionization step of the SN1 mechanism. A highly polarizable leaving group helps stabilize the rate-limiting transition state through partial bonding as it leaves. The leaving group should be a weak base, very stable after it leaves with the pair of electrons that bonded it to carbon. Figure 6-9 shows the transition state of the ionization step of the SN1 reaction. Notice how the leaving group is taking on a negative charge while it stabilizes the new carbocation through partial bonding. The leaving group should be stable as it takes on this negative charge, and it should be polarizable to engage in effective partial bonding as it leaves. A good leaving group is just as necessary in the SN 1 reaction as it is in the SN2, and similar leaving groups are effective for either reaction. Table 6-4 (page 235) lists some common leaving groups.

Hi-ltv

PRO B L E M 6-23

PROBLEM-SOLVING

Choose the member of each pair that w i l l react faster by the S N 1 mechan ism.

Pri m a ry cations a re rarely formed in

(a) (b)

s o l ution u n less they are resonance­

I -bromopropane or 2-bromopropane

sta b i l i zed.

2-bromo-2-methylbutane or 2-bromo-3-methylbutane

(c) n-propyl bromide or allyl bromide

(d) (e) (f)

I -bromo-2,2-dimethylpropane or 2-bromopropane 2-iodo-2-methylbutane or t-butyl chloride 2-bromo-2-methylbutane or ethyl iodide

X is taking on a partial negative charge

: x-

.... Figure 6-9 In the transition state of the SN 1 ionization, the leaving group is taking on a negative charge. The C

-

X

bond is breaking, and a polarizable leaving group can still maintain

partial bonding in the transition state

substantial overlap.

244

Chapter 6: Alkyl Halides: Nucleophilic Substitution

and Eli mi natio n

P R O B L E M 6-24 3-Bromocyclohexene i s a secondary halide, and benzyl bromide i s a primary halide. Both halides undergo SN 1 substitution about as fast as most tertiary halides. Use resonance struc­ tures to explain this enhanced reactivity.

Q-

Br

3-bromocyclohexene

6- 1 3C

2° good ionizing solvent needed

kinetics

first order, k,.[RX]

stereochemistry rearrangements

mixture of inversion and retention common

good one required AgN03 forces ionization

Characteristics

strong nucleophile needed CH3X > 1 ° > 2° wide variety of solvents good one required

second order, kr[RX][Nuc:-] complete inversion impossible

PROBLEM-SOLVING

P RO B L E M 6-27 For each reaction, give the expected substitution product, and predict whether the mechanism will be predominantly first order or second order. (a) 2-chloro-2-methylbutane + C H 3COOH (b) isobutyl bromide + sodium methoxide (c) l -iodo- I -methylcyclohexane + ethanol (d) cyclohexyl bromide + methanol (e) cyclohexyl bromide + sodium ethoxide

PRO B L E M 6-28 When (R)-2-bromobutane i s heated with water, the S N 1 substitution proceeds twice as fast as the SN2. Calculate the e.e. and the specific rotation expected for the product. The specific rotation of (R)-2 butanol is - 1 3.5°.

PRO BLEM 6-29 A reluctant first-order substrate can be forced to ionize by adding some silver nitrate (one of the few soluble silver salts) to the reaction. Silver ion reacts with the halogen to form a silver halide (a highly exothermic reaction), generating the cation of the alkyl group. R-X

+

Ag +

�

R+

+

AgX

!

Give mechanisms for the following silver-promoted rearrangements. CH3

CH3

I

I

(a) C H3 -C -CH 2 - I

C H3 -C -CH2 -CH3

I

I

C H3

OH

AgN03, HPICH3CHPH

)

Hi-ltv

The strength of the nucleophile (or base) usua l ly determ ines the order of the reaction. Strong nucleophiles encourage second-order reactions, a n d weak nucleophi les more commonly react by fi rst-order

on 30 h a l i des, a nd SN 1 i s u n l i kely reactions. Also, SN2 is u n l ikely

on 1 0 h a l ides.

252

Chapter 6 : Alkyl Halides: Nucleophilic Substitution and Elimination A n elimination involves the loss of two atoms or groups from the substrate, usually with formation of a pi bond. Elimination reactions frequently accompany and compete with substitutions. By varying the reagents and conditions, we can often modify a reaction to favor substitution or to favor elimination. First we will discuss eliminations by them­ selves. Then we consider substitutions and eliminations together, trying to predict what products and what mechanisms are likely with a given set of reactants and conditions. Depending on the reagents and conditions involved, an elimination might be a first-order (E l ) or second-order (E2) process. The following examples illustrate the types of eliminations we cover in this chapter.

6-1 7 Fi rst-Order E l i m i n atio n : The E 1 Reaction

E1:

(

r--- H CH?CH3 1 ---, 1 H-C-C-CH CH3 1 �. CH3 1?

CH3 OH .. )

+

:

r:

-

.. +

CH3 -O-H 1 H

E2:

..

CHP-H H",

CH2CH3 / C=C '" / CH2CH3 H3C .. : Br :

6-1 7A

Mechanism and Kinetics of the E 1 Reaction

The abbreviation El stands for Elimination, unimolecular. The mechanism is called unimolecular because the rate-limiting transition state involves a single molecule rather than a collision between two molecules. The slow step of an E l reaction is the same as in the SN 1 reaction: unimolecular ionization to form a carbocation. In a fast second step, a base abstracts a proton from the carbon atom adj acent to the C+. The electrons that once formed the carbon-hydrogen bond now form a pi bond between two carbon atoms. The general mechanism for the E l reaction is shown in the following Key Mechanism box .

v-� KEY M ECHAN ISM 6-8

•.

The E1 Reaction

The E l reaction requires ionization to a carbocation intermediate like the SN 1 , so it fol­ lows the same order of reactivity: 3° > 2° > > 1 ° A base (usually weak) deprotonates the carbocation to give an alkene. Step 1: U nimolecular ionization to give a carbocation (rate-li miting).

1

I

-C-C-

I +/ -C-C

(�

� ('t

� '"

..

:x:

+

Step 2: Deprotonation by a weak base (often the solvent) gives the alkene (fast).

B :-

+/ 1 -C-C

��J

'"

+-)

B-H

+

'" /

C=C

/ '"

6- 1 7 First-Order Elimination: The E l Reaction

EXAM PLE: E1 e l i m i n ation of bromocyclohexa ne in metha n o l .

Step 1: Ionization gives a carbocation and bromide ion.

O=B.· . H

r·

I '-.-

(

CH3 0H, heat )

Step 2: Methanol abstracts a proton to give cyclohexene.

Gt

(X

H H CH30H H�·

H

+

H

PROBLEM: Show what happens in step 2 of the Example, (El elimination of bromo­ cyclohexane in methanol) if the solvent acts as a nucleophile rather than as a base.

6-1 7 B

Com petition with the S N 1 Reaction

The E l reaction almost always occurs together with the S N I . Whenever a carbocation is formed, it can undergo either substitution or elimination, and mixtures of products often result. The following reaction shows the formation of both elimination and sub­ stitution products in the reaction of t-butyl bromide with boiling ethanol . CH3

I

CH3 - C - Br

I

CH3

�

+

heat

+

(-butyl bromide

ethanol

2-methylpropene (E I product)

CH3

I

CH 3 - C - O -CH2 - CH3

I

CH3

ethyl (-butyl ether (SN I product)

The 2-methylpropene product results from dehydrohalogenation, an elimina­ tion of hydrogen and a halogen atom. Under these first-order conditions (the absence of a strong base), dehydrohalogenation takes place by the El mechanism: Ionization of the alkyl halide gives a carbocation intermediate, which loses a proton to give the alkene. Substitution results from nucleophilic attack on the carbocation. Ethanol serves as a base in the elimination and as a nucleophile in the substitution. Step 1: ionization to form a carbocation.

.

: Br :

.

: Br :

CH 3 - C - CH 3 +

IJ

CH3 - C - CH 3

I

CH3

I

CH3

Step 2: (by the Ei mechanism): Basic attack by the solvent abstracts a proton to give an alkene.

CH3CH� - 0 : ""----"'" H

.

-

I

H

It

+

I

I

H - C - C - CH 3 H

CH3

+

253

2 54

Chapter 6:

Alkyl H al ide s :

Nucleophilic S ubstitution an d E li min ati o n or,

+

(by the SNi mechanism):

�. CHFH2-Q-H

CH3 -C-CH3 I CH3

Nucleophilic attack by the solvent on the carbocation .

?

..

: CH2CH3 )

CH3 -C-CH 3 + I CH3 CH3CHPH2

Under ideal conditions, one of these first-order reactions provides a good yield of one product or the other. Often, however, carbocation intermediates react in two or more ways to give mixtures o f products. For this reason, SN 1 and E l reactions of alkyl halides are not often used for organic synthesis. They have been studied in great detail to learn about the properties of carbocations, however.

PROBLEM 6-30

Some S N 1 substitution probably accompanies the E l elimination of bromocyclohexane shown in Key Mechanism Box

(a) (b)

6-8.

Show the mechanism and product of the corresponding S N1 reaction. C ompare the function of the solvent (methanol) in the E l and SN1 reactions.

6-1 7 C

O rbitals and Energetics

In the second step of the E l mechanism, the carbon atom next to the C + must rehy­ 2 bridize to sp as the base attacks the proton and electrons flow into the new pi bond.

B- H

11:

bond

The potential-energy diagram for the E 1 reaction (Figure 6- 1 2) is similar to that for the SN 1 reaction. The ionization step is strongly endothermic, with a rate-limiting transition state. The second step is a fast exothermic deprotonation by a base. The base tJ

/transition state rate-limiting t2

1

EI rate

1

=

kr [R - X l

-C-C­

I:'; +

H B : -J"

1

� Figure 6-1 2

1

-C-C-

Reaction-energy diagram of the E l reaction. The first step is a rate-limiting ionization. C ompare this energy profile with that of the SN 1 reaction, Figure

: X :-

6-8.

1

H

1

:X:

reaction coordinate

�

'\ / C=C / '\ : X :B -H

6- 1 7 First-Order Elimination: The E 1

Reaction

255

is not involved in the reaction until after the rate-limiting step, so the rate depends only on the concentration of the alkyl halide. Weak bases are common in E 1 reactions. Like other carbocation reactions, the E1 may be accompanied by rearrangement. Compare the following E1 reaction (with rearrangement) with the SN I reaction of the same substrate, shown in Mechanism 6-6. Note that the solvent acts as a base in the E 1 reaction and a nucleophile in the SN 1 reaction. M E C H A N I S M 6-9

Rea rra ngement in an E 1 Reaction

Like other reactions involving carbocations, the E1 may be accompanied by rearrangement.

Step 1:

Ionization to form a carbocation.

: Br : H ..

CI

1

CH3 - C - C - CH3

1

Step 2: A

1

CH3

1

H

2-bromo-3-methylbutane

CH3

2° carbocation

hydride shift forms a more stable carbocation. H

... 2° > 1 °.

PROBLEM 6-33 Give the substitution and elimination products you would expect from the following reactions. (a) 3-bromo-3-ethylpentane heated in methanol (b) l-iodo- I -methylcyclopentane heated in ethanol (c) 3-bromo-2,2-dimethylbutane heated in ethanol (d) iodocyclohexane + silver nitrate in water (see Problem 6-29)

6- 1 8 Positional Orientation of Elimination: Zaitsev's

Many compounds can eliminate in more than one way, to give mixtures of products. In many cases, we can predict which elimination product will predominate. jn the exam­ ple shown in Mechanism Box 6-9, the carbocation can lose a proton on eitht;Y' of t;vo adjacent carbon atoms.

CH3CHi�H / � H H l :t

rl

+

CH -C-C-C-H 3 I I I H CH3 H

Positional O rientation o f E l i m i n ati o n : Zaitsev's R u le

2-methyl - l -butene disubstituted, minor (10%)

The first product has a trisubstituted double bond, with three substituents (circled) on the doubly bonded carbons. It has the general formula R2C=CHR. The second prod­ uct has a disubstituted double bond, with general formula R2C=CH2 (or R -CH=CH - R ) . In most E l and E2 eliminations where there are two or more possible elimination products, the product with the most substituted double bond will predominate. This general principle is called Zaitsev's rule,* and reactions that give the most substituted alkene are said to follow Zaitsev orientation. ZAITSEV'S RULE :

In elimination reactions, the most substituted alkene

usually predominates.

R2C=CR2 tetrasubstituted

>

R2C=CHR trisubstituted

>

RHC=CHR

disubstituted

and

R2C=CH2

>

This order of preference is the same as the order of stability of alkenes. We consider the stability of alkenes in more detail in Section 7-7, but for now, it is enough j ust to know that more substituted alkenes are more stable. In Chapter 7, we will study some unusual reactions where Zaitsev's rule does not apply. PROBLE M 6-34 When I -bromo- l -methylcyciohexane i s heated i n ethanol for an extended period of time, three products result: one ether and two alkenes. Predict the products of this reaction, and propose a mechanism for their formation. Predict which of the two alkenes is the major elim­ ination product.

S O LV E D P R O B L E M 6 - 2 When 3-iodo-2,2-dimethylbutane is treated with silver nitrate i n ethanol, three elimina­ tion products are formed. Give their structures, and predict which ones are formed in larger amounts.

SO LUTI O N Silver nitrate reacts with the alkyl iodide to give silver iodide and a cation.

CH3

I

CH3 - C - CHI -CH3

I

CH3

+ Ag+

CH3 ----?

I

CH3 - C -CH-CH3 + AgI

I

CH3

+

I

{-

( Continued ) *Zaitsev is transliterated from the Russian, and may also be spelled Saytzeff.

257

6-1 8

+ 2-methyl-2-butene trisubstituted, maior (90%)

Rule

RHC=CH2 monosubstituted

/

258

Chapter 6 : Alkyl Halides: Nucleophilic S ubstitution a n d Elimination

PROBLEM-SOLVING

Htnl/

Whenever a carbocation is formed next to a more h i g h ly s u bstituted carbon, consider whether a

This secondary carbocation can lose a proton t o give an unrearranged alkene (A), o r i t can rearrange to a more stable tertiary cation.

Loss of a proton CH3

rearrangement m i g ht occur.

I I

3

/

H

�

CH - C - C CH3

C-H

I

H

+

Product (A) + CH3CHPH2

Rearrangement CH3

1

CH3

H

I

1

CH3- C - C -CH3

1--------'" +

H

I

CH3 - C -C -CH3 +

(methyl shift)

1

CH3

CH}

3° carbocation

2° carbocation

The tertiary cation can lose a proton in either of two positions. One of the products (B) is a tetrasubstituted alkene, and the other (C) is disubstituted.

Formation of a tetrasubstituted alkene CH3

1

H � ..

HOCH,CH3

y-I

..

CH3 - C -- C -CH3

I

+

-

)

H}C H}C

CH3

,,

/

/

CH3

C=C

"

+

+

CH3CHPH2

CH}

(D)

(tetrasubstituted)

Formation of a disubstituted alkene

�

C 3

.. CH3 CH?OH _ ..

GH

H

1

r' C - C - CH3 ( j' + 1

C / " H H

CH3

H

1

�

C 3

C - C - CH3

H - C.I"

1

H

1

C H3 (e)

(disubstituted) Product B predominates over product C because the double bond in B is more substitued. Whether product A is a major product will depend on the specific reaction conditions and whether proton loss or rearrangement occurs faster.

PROBLEM 6-35 Each of the two carbocations in Solved Problem 6-2 can also react with ethanol to give a substi­ tution product. Give the structures of the two substitution products formed in this reaction.

6- 1 9 Seco nd-Ord e r E l i m i n ati o n : T h e E2 Reaction

Eliminations can also take place under second-order conditions with a strong base present. As an example, consider the reaction of t-butyl bromide with methoxide ion in methanol. This is a second-order reaction because methoxide ion is a strong base as well as a strong nucleophile. It attacks the alkyl halide faster than the halide can ionize to give a first-order reaction. No substitution product (methyl t-butyl ether) is observed, how­ ever. The SN2 mechanism is blocked because the tertiary alkyl halide is too hindered.

6- 1 9

Second-Order Elimination: The E2 Reaction

259

The observed product is 2-methylpropene, resulting from elimination of HBr and formation of a double bond.

..

CH3- O- H .•

E2

-----7

: Br : The rate of this elimination is proportional to the concentrations of both the alkyl halide and the base, giving a second-order rate equation. This is a bimolecular process, with both the base and the alkyl halide participating in the transition state, so this mechanism is abbreviated E2 for Elimination, bimolecular. In the E2 reaction just shown, methoxide reacts as a base rather than as a nucleophile. Most strong nucIeophiles are also strong bases, and elimination commonly results when a strong base/nucIeophile is used with a poor SN2 substrate such as a 30 or hindered 20 alkyl halide. Instead of attacking the back side of the hindered electrophilic carbon, methoxide abstracts a proton from one of the methyl groups. This reaction takes place in one step, with bromide leaving as the base is abstracting a proton. In the general mechanism of the E2 reaction, a strong base abstracts a proton on a carbon atom adjacent to the one with the leaving group As the base abstracts a proton, a double bond forms and the leaving group leaves. Like the SN2 reaction, the E2 is a concerted reaction in which bonds break and new bonds form at the same time, in a single step . •":':,�

KEY M ECHAN ISM 6- 1 0

The E2 Reaction

The concerted E2 reaction takes place in a single step. A strong base abstracts a proton on a carbon next to the leaving group, and the leaving group leaves. The product is an alkene.

+

B -H

transition state

EXAM PLE: E2 elimination of 3-bromopentane with sodium ethoxide.

The order of reactivity for alkyl halides in E2 reactions is 30 > 20 > 1 0. PROBLEM 6-36: Under second-order conditions (strong base/nucIeophile), SN2 and E2 reactions may occur simultaneously and

compete with each other. Show what products might be expected from the hindered 2° alkyl halide) with sodium ethoxide.

reaction

of 2-bromo-3-methylbutane (a moderately

260

Chapter 6: Alkyl Halides: Nucleophilic Substitution and Elimination

Reactivity of the Substrate in the E2 The order of reactivity of alkyl halides to­ ward E2 dehydrohalogenation is found to be

This reactivity order reflects the greater stability of highly substituted double bonds. Elimination of a tertiary halide gives a more substituted alkene than elimination of a secondary halide, which gives a more substituted alkene than a primary halide. The stabilities of the alkene products are reflected in the transition states, giving lower activation energies and higher rates for elimination of alkyl halides that lead to highly substituted alkenes. M ixtu res of Products in the E2 The E2 reaction requires abstraction of a proton on a carbon atom next to the carbon bearing the halogen. If there are two or more possibilities, mixtures of products may result. In most cases, Zaitsev's rule predicts which of the possible products will be the major product: the most substituted alkene. For example, the E2 reaction of 2-bromobutane with potassium hydroxide gives a mixture of two products, I -butene (a monosubstituted alkene) and 2-butene (a disub­ stituted alkene). As predicted by Zaitsev's rule, the disubstituted isomer 2-butene is the major product.

..

HO :

�

H

H

(

..

: OH

H

I � 1 (, I H-C-C-C - CH3 1 1 1 H (Br H

KOH

+ 8 1 % 2-butene

1 9% I -butene

2-bromobutane

monosubstituted

disubstituted

Similarly, the reaction of I -bromo- I -methylcyclohexane with sodium ethoxide gives a mixture of a disubstituted alkene and a trisubstituted alkene. The trisubstituted alkene is the major product.

NaBr NaOCH2CH3 CH3CHPH

)

+ methy lenecyc10hexane

I -methylcyclohexene

disubstituted, minor

trisubstituted, major

PROBLEM 6-37 1. Predict the elimination products of the following reactions. When two alkenes are possi­ ble, predict which one will be the major product. Explain your answers, showing the degree of substitution of each double bond in the products. 2. Which of these reactions are likely to produce both elimination and substitution products? (a) 2-bromopentane + NaOCH3

(b) 3-bromo-3-methylpentane

(c) 2-bromo-3-ethylpentane

+ +

NaOMe

(Me

=

methyl, CH3 )

NaOH

(d) cis- I -bromo-2-methykycIohexane

+

NaOEt

(Et

=

ethyl, CH2CH 3 )

6-20

Stereochemistry of the

Like the SN2 reaction, the E2 follows a concerted mechanism: Bond breaking and bond formation take place at the same time, and the partial formation of new bonds lowers the energy of the transition state. Concerted mechanisms require specific geo­ metric arrangements so that the orbitals of the bonds being broken can overlap with those being formed and the electrons can flow smoothly from one bond to another. The geometric arrangement required by the SN2 reaction is a back-side attack; with the E2 reaction, a coplanar arrangement of orbitals is needed. E2 elimination requires partial formation of a new pi bond, with its parallel p orbitals, in the transition state. The electrons that once formed a C - H bond must begin to overlap with the orbital that the leaving group is vacating. Formation of this new pi bond implies that these two sp 3 orbitals must be parallel so that pi overlap is possible as the hy­ drogen and halogen leave and the orbitals rehybridize to the p orbitals of the new pi bond. Figure 6-13 shows two conformations that provide the necessary coplanar align­ ment of the leaving group, the departing hydrogen, and the two carbon atoms. When the hydrogen and the halogen are anti to each other (e 180° ), their orbitals are aligned. This is called the anti-coplanar conformation. When the hydrogen and the halogen eclipse each other (e 0 ° ) , their orbitals are once again aligned. This is called the syn-coplanar conformation. Make a model corresponding to Figure 6-13, and use it to follow along with this discussion. Of these possible conformations, the anti-coplanar arrangement is most com­ monly seen in E2 reactions. The transition state for the anti-coplanar arrangement is a staggered conformation, with the base far away from the leaving group. In most cases, this transition state is lower in energy than that for the syn-coplanar elimination. The transition state for syn-coplanar elimination is an eclipsed conformation. In addition to the higher energy resulting from eclipsing interactions, the transition state suffers from interference between the attacking base and the leaving group. To abstract the proton, the base must approach quite close to the leaving group. In most cases, the leaving group is bulky and negatively charged, and the repulsion between the base and the leaving group raises the energy of the syn-coplanar transition state.

E2 Reaction

261

6-20 Ste reoch e m i stry of the E2 Reaction

=

=

H

R R

$: X

R

R

)

��

Base

··

R'

IC

/ \)

R

Enzyme-catalyzed eliminations gen­ erally proceed by E2 mechanisms and produce only one stereoisomer. Two catalytic groups are involved: One abstracts the hydrogen, and the other assists in the departure of the leaving group. The groups are positioned appropriately to allow an anti-coplanar elimination.

Base- H

�

C1C:::'/

�R

�

•

•

8

------7

R " II R""'-

G X

anti-coplanar transition state

(staggered conformation- l ower energy)

.... Figure 6-1 3 Concerted transition states of the E2 reaction. The orbitals of the hydrogen atom and the halide must be aligned syn-coplanar transition state (eclipsed conformati on -hi gher energy)

so they can begin to form a pi bond in the transition state.

262

Chapter 6: Alkyl Halides: Nucleophihc Substitution and Elimination Some molecules are rigidly held in eclipsed (or nearly eclipsed) conformations, with a hydrogen atom and a leaving group in a syn-coplanar arrangement. Such com­ pounds are likely to undergo E2 elimination by a concerted syn-coplanar mechanism. Deuterium labeling (using D, the hydrogen isotope with mass number 2) is used in the following reaction to show which atom is abstracted by the base. Only the hydrogen atom is abstracted, because it is held in a syn-coplanar position with the bromine atom. Remember that syn-coplanar eliminations are unusual, however; anti�coplanar elimi­ nations are more common. H

D

D

PROBLEM-SOLVING

Htnv

Models a re h e l pful whenever complex stereochemistry is involved.

+

NaBr

PROBLEM 6-38 When the first compound shown here i s treated with sodium methoxide, the only elimjnation product is the trans isomer. The second diastereomer (blue) gives only the cis product. Use your models and careful drawings of the transition states to explain these results.

H

(

H

Ph Ph

=

/

" CH3

�)

Ph

phenyl group,

H " H -.,." c -C/ '"'''' CH 3 / " Br "

Ph

Co m pa r i son of E 1 a n d E 2 E l i m i n ation Mech a n isms

HOCH3

The E2 is a stereospecific reaction, because different stereoisomers of the start­ ing material react to give different stereoisomers of the product. This stereospecificity results from the anti-coplanar transition state that is usually involved in the E2. We consider more of the implications of the anti-coplanar transition state in Chapter 7. For now, Problem 6-38 will give you an opportunity to build models and see how the stereochemistry of an E2 elimination converts different stereoisomers into different stereoisomers of the product.

/ " "" B r -.,.C -C II"".

6-2 1

+

Ph

NaOCH) CH)OH

NaOCH, CH,oH

)

)

Ph

H

Ph

H

/ C=C / "

"

trans

"

/

C=C cis

/

CH3 Ph

"

Ph

CH3

Let's summarize the major points to remember about the E l and E2 reactions, fo­ cusing on the factors that help us predict which of these mechanisms will operate under a given set of experimental conditions. Then we will organize these factors into a short table. Effect of the Base The nature of the base is the single most important factor in determining whether an elimination will go by the E l or E2 mechanism. If a strong base is present, the rate of the bimolecular reaction will be greater than the rate of ion­ ization, and the E2 reaction will predominate (perhaps accompanied by the SN2). If no strong base is present, with a good solvent a unimolecular ionization is likely, followed by loss of a proton to a weak base such as the solvent. Under these conditions, the E l reaction usually predominates (always accompanied by the SN 1 ).

El : E2:

Base strength is unimportant (usually weak). Strong bases are required.

6-21 Comparison of El and E2 Elimination Mechanisms Effect of the Solvent The slow step of the E l reaction is the formation of two ions. Like the SNl, the E l reaction critically depends on polar ionizing solvents such as water and the alcohols. In the E 2 reaction, the transition state spreads out the negative charge of the base over the entire molecule. There is no more need for solvation in the E 2 transition state than in the reactants. The E 2 is therefore less sensitive to the solvent; in fact, some reagents are stronger bases in less polar solvents. E l: Good ionizing solvent required. E 2 : Solvent polarity i s not so important. Effect of the S ubstrate

For both the E l and the E2 reactions, the order of reactivity is ( 10 usually

will not go E I )

In the E l reaction, the rate-limiting step i s formation of a carbocation, and the reactivity order reflects the stability of carbocations. In the E2 reaction, the more sub­ stituted halides generally form more substituted, more stable alkenes. Ki netics The rate of the E l reaction is proportional to the concentration of the alkyl halide [RX] but not to the concentration of the base. It follows a first-order rate equation. The rate of the E 2 reaction is proportional to the concentrations of both the alkyl halide [RX] and the base [B:-]. It follows a second-order rate equation. E 1 rate kr[RX] E 2 rate kr[RX][B:-] =

=

In most E l and E 2 eliminations with two or more possible products, the product with the most substituted double bond (the most stable product) predominates. This principle is called Zaitsev's rule, and the most highly substituted product is called the Zaitsev product. O rientation of E l imi nation

� Br

EI or E2

)

E l , E 2: Usually Zaitsev orientation.

major (cis + trans)

minor

The E l reaction begins with an ionization to give a flat carboca­ tion. No particular geometry is required. The E 2 reaction takes place through a concerted mechanism that requires a coplanar arrangement of the bonds to the atoms being eliminated. The transition state is usually anti-coplanar, although it may be syn-coplanar in rigid systems. Stereochemistry

E l : No particular geometry required for the slow step. E 2 : Coplanar arrangement (usually anti) required for the transition state. Rea rra n g ements The El reaction involves a carbocation intermediate. This inter­ mediate can rearrange, usually by the shift of a hydride or an alkyl group, to give a more stable carbocation. The E2 reaction takes place in one step with no intermediates. No rearrangement is possible in the E 2 reaction. E l : Rearrangements are common. E 2 : No rearrangements.

263

t

264

Chapter

6:

SUM MARY

Alkyl Halides: Nucleophilic Substitution and Elimination

E l i mination Reactions E1

Promoting factors

E2

base

weak bases work

solvent

good ionizing solvent

wide variety of solvents

substrate

3° > 2°

3° > 2° > 1°

leaving group Characteristics

strong base required

good one required

good one required

orientation

most substituted alkene

first order, kr[RX]

second order, kr[RX][B:-]

stereochemistry

no special geometry

coplanar transition state required

rearrangements

common

impossible

kinetics

most substituted alkene

P R O B L E M - S O LV I N G S T R AT E G Y

Predicting Substitutions and Eliminations

.

. R I...-� :

SNI

R SN2

+� �

+

Nuc:-

/

� C� : I

EI

I

-C-C-

��3

B:-

I

+/

- C-C

l et

�

I� I

-C-C-

R+

--7

R - Nuc

..

C-X: \ 1...-

+

-

+

�

I

+

I

(slow) (fast)

Nuc - C -

+/ -C-C �

(�

:x:

/

--7

•.

.

:X: .

-:X:

(slow)

H

B -H

�)

E2(H

B:-

�

--7

'-.... /' = C C

/'

'-....

+

+

/ � C=C / �

B -H

(fast)

+

:X:-

Given a set of reagents and solvents, how can we predict what products will result and which mechanisms will be involved? Should you memorize all this theory about substitutions and eliminations? Students sometimes feel overwhelmed at this point. Memorizing is not the best way to approach this material because it is too much and there are too many factors. Besides, the real world with its real reagents and solvents is not as clean as our equations on paper. Most nucleophiles are also basic, and most bases are also nucleophilic. Most solvents can solvate ions or react as nucleophiles, or both. We will review the most important factors that determine the reaction pathway, organized in a sequence that allows you to predict as much as can be predicted. The first principle you must understand is that we cannot always predict one unique product or one unique mechanism. Often, the best we can do is to eliminate some of the possibilities and make some good predictions. Remembering this limitation, here are some general guidelines:

6-2 1 Comparison of E l and E2 Elimination Mechanisms 1.

The strength of the base or nucleophile determines the order of the reaction. If a strong nucleophile (or base) is present, it will force second-order kinetics, either SN2 or E2. A strong nucleophile attacks the electrophilic carbon atom or abstracts a proton faster than the molecule can ionize for first-order reactions. If no strong base or nucleophile is present, the fastest reaction will probably be a first-order reaction, either SNI or E l . Addition of silver salts to the reaction can force some kind of ionization. This is the most important of these guidelines. Consider the following examples:

�

�

Br

Br

E2

(no SN2 on 3° carbon)

2. Primary halides usually undergo the SN2 reaction, occasionally the E2 reaction. Primary halides rarely undergo first-order reactions, since primary carbocations are rarely formed. With good nucleophiles, SN2 substitution is usually observed. With a strong base, E2 elimination may also be observed. Sometimes silver salts or high temperatures are used to force a primary halide to ionize, usually with rearrangement to give a more stable carbocation. In such a case, the rearranged SNI and E l products may be observed.

� � 3.

SN2

Br Br

AgN03, heat) CH30H

(and possibly E2)

SN 1 and E 1

(both with rearrangement)

Tertiary halides usually undergo the E2 reaction (strong base) or a mixture of SNI and El (weak base). Tertiary halides cannot undergo the SN2 reaction. A strong base forces second-order kinetics, resulting in elimination by the E2 mechanism. In the absence of a strong base, tertiary halides react by first-order processes, usually a mixture of SN I and E l . The spe­ cific reaction conditions determine the ratio of substitution to elimination.

E2

(no SN2 on 3° carbon)

CHpH) heat 4.

The reactions of secondary halides are the most difficult to predict. With a strong base, either the SN2 or the E2 reaction is possible. With a weak base and a good ionizing solvent, either the SN1 or the E l reaction is possible. Mixtures of products are common. Figure 6- 14 shows these possibilities with a secondary halide under second­ order and first-order conditions.

� Br

�

NaOCH3) CH30H

SN2 and E2

CH3OH) heat

SNI and El

Br

5.

Some nucleophiles and bases favor substitution or elimination. To promote elimination, the base should readily abstract a proton but not readily attack a carbon atom. A bulky strong base, such as t-butoxide [-OC ( CH 3 h), enhances elirnination. ( Continued)

265

266

Chapter 6: Alkyl Halides: Nucleophilic Substitution and Elimination

second-order conditions (strong base/nucleophile)

first-order conditions (weak base/nucleophile)

.. Figu re 6-1 4 Under second-order conditions (strong base/nucleophile), a secondary alkyl halide might undergo either substitution ( SN2) or elimination (E2). Under first-order conditions (weak base/nucleophile), SN 1 and E l are possible. To promote substitution, we need a good nucleophile with limited basicity: a highly polar­ izable species that is the conjugate base of a strong acid. Bromide ( Br -) and iodide (1-) are examples of good nucleophiles that are weak bases and favor substitution.

� Br

NaI

Br

�

NaOCCCH3)3 ) (CH3hCOH

mostly E2

P RO B LE M 6-39: Give the structures of the products expected from the indicated mech­ anisms in the preceding examples. S O LV E D P R O B L E M 6 - 3 Predict the mechanisms and products o f the following reactions.

U Br

(a)

CH,

heat

l-bromo-l -methylcyclohexane Br

I

(b) CH3-CH-CH2CH2CH2CH3 2-bromohexane

NaOCH3 )

SO LUT I O N (a) There is no strong base or nucleophile present, so this reaction must be first order, with an ionization of the alkyl halide as the slow step. Deprotonation of the carbocation gives either of two elimination products, and nucleophilic attack gives a substitution product.

carbocation

m

aj or

minor

E I elimination products

substitution product (SN1)

6-21 Comparison of E l and E2 Elimination Mechanisms (b) This reaction takes place with a strong base, so it is second order. This secondary halide can undergo both SN2 substitution and E2 elimination . Both products wil l be formed, with the relative proportions of substitution and elimination depending on the reaction conditions. major

PROBLEM-SOLVING

267

HinZ;

The strength of the base/n ucleophile usually determines the order of the reaction.

minor

�------�v�--� E2 products

OCH3

I

CH3 -CH -CH2CH2CH2CH3 SN2 product P RO B L E M 6 -40 Predict the products and mechanisms o f the following reactions. When more than one product or mechanism is possible, explain which are most likely. (a) l -bromohexane sodium ethoxide in ethanol (b) 2-chlorohexane + NaOCH3 in methanol (c) t-butyl bromide NaOCH2CH3 in ethanol (d) t-butyl bromide heated in ethanol (e) isobutyl iodide KOH in ethanol/water (0 isobutyl chloride + AgN03 in ethanol/water (g) neopentyl bromide + AgN03 in ethanol/water (h) I-bromo- l -methylcyclopentane heated in methanol (i) I-bromo- l -methylcyclopentane + NaOEt in ethanol

+

+

+

C

SUMMARY

PROBLEM-SOLVING

HinZ;

Don't try to memorize your way t h ro u g h this cha pter. Try to

u n derstand what h a p pens i n the different reactions. Some memorizing i s necessary, but s i m p l y memorizing everyth ing won't a l low you to predict new reactions.

Reactions of A l kyl Halides

Some of these reactions have not yet been covered, but they are included here for completeness and for later reference. Notice the section numbers, indicating where each reaction is covered.

1. Nucleophilic substitutions ( Secti o n 6-9) a. Alcohol formation R-X

+

�

- : OH

+

R - OH

:X -

Example

CH3CH2- OH

CH3CH2 -Br + NaOH

b. Halide exchange R-X + R -Cl

+

NaBr

ethyl alcohol

ethyl bromide

+

R-I

: 1:lS-crown-6

KF

)

+

R-F

: X-

+

KCI

Example

H2C=CH - CH2Cl

+

Nal

H2C=CH -CH2I

+

NaCl

allyl iodide

allyl chloride c. Williamson ether synthesis

R -X

+

R'O:-

R -0-R'

+

: X-

ether synthesis

R-X

+

R'S:-

R - S-R'

+

: X-

thioether synthesis

268

Chapter 6: Alkyl Halides: Nucleophilic Substitution and Elimination

Example

+

CH3-1

-----7

CH3CH2-O- Na+

methyl iodide

sodium ethoxide

CH3-O-CH2CH3

Na+ 1-

+

methyl ethyl ether

d. Amine synthesis

R-X

+

excess

-----7

: NH3

R-NH2

R - NHt X-

+

NHt : X­

amine

Example

n-propyl bromide

n-propylamine

e. Nitrile synthesis

+

R-X

- : C==N :

R -C==N :

cyanide

nitrile

+

:X­

Example

l-chloro-3-methylbutane f.

4-methylpentanenitrile

Alleyne synthesis

R - C==C :-

R'-X

+

R - C==C - R'

acetylide ion

+

: X­

alkyne

Example

CH3-C==C - H propyne

CH3-C==C:- Na + propynide ion 2.

+

-----7

NaNH2

sodium amide +

CH3-C==C :- Na+

+

NH3

sodium propynide

CH3CH2- 1 -----7

CH3-C==C - CH2CH3

ethyl iodide

+

NaI

2-pentyne

Eliminations a. Dehydrohalogenation (Sections 6-18 and 7-9A)

I

H

I

-C-C-

I

Example

I

---7

KOH

� /

/ �

+

: X-

X

� Br

�+�

2-bromohexane

b.

C=C

2-hexene

}-hexene

Dehalogenation (Section 7-9D)

Br I

I

�

I

I

/

- C-C -

Br

C=C

/ �

+

I-Br

+

KBr

6-21 Comparison of El and E2 Elimination Mechanisms

ex:

Example

trans-l ,2-dibromocyclohexane

3.

cyclohexene

Formation of organometallic reagents (Section 10-8) a. Grignard reagents

R-X ex

=

+ Cl, Br, or I)

C�CH2-O-CH2CH3

Mg

)

R-Mg -X organomagnesium halide eGrignard reagent)

Example

0:'

+

O:

ether -----7

Mg

bromocyclohexane

b.

cyclohexylmagnesium bromide

Organolithium reagents

R -X ex

=

+

gB,

2 Li

Cl, Br, or I)

------7

+

R -Li

Li+ X­

organolithium

Example

CH3CHzCH2CH2 -Br

+

2 Li

....:.=�) .:..:

hexane

CH3CHzCH2CHz -Li n-butyllithium

n-butyl bromide

4. Coupling of organocopper reagents (Section 10-9) +

2 R -Li

R2 CuLi

+

CuI

R'-X

------7

------7

� CuLi

+

LiI

R - R'

+

R-Cu

+

LiX

Example CuI �

4Li �

n-CgH 17 /'

5.

H

", /

(CH3)2 CuLi

C=C

/ '"

Reduction (Section 10-10)

R-X

(1) Mg orLi

)

R-H

Example

C9H'9-CH2-B r n-decyl bromide

(1) Mg, ether (2)HP

)

C9H '9- CH3 n-decane

H CH3

+

LiBr

269

270

Chapter 6: Alkyl Halides: Nucleophilic Substitution and Elimination

Chapter 6 Glossary

acid A species that can donate a proton. acidity (acid strength): The thermodynamic reactivity of an acid, expressed quantitatively by the acid-dissociation constant Ka. A derivative of an alkane in which one (or more) of the hydrogen atoms has been replaced by a halogen. (p. 212) allylic The saturated position adjacent to a carbon-carbon double bond. (p. 221) allylic halogenation Substitution of a halogen for a hydrogen at the allylic position. (p. 221)

alkyl halide (haloalkane)

Br

I

H2C=CH-CH-CH3 a\lylic bromide

+

+ Br

I

H2C-CH=CH-CH3

N-bromosuccinimide

rearranged allylic bromide

+

�

N-H

o

succinimide

(product of an allylic shift)

allylic shift A rearrangement that results from reaction at either end of a resonance-stabilized ally[jc intermediate. (p. 222) anti Adding to (or eliminating from) opposite faces of a molecule. (p. 261) anti-coplanar: Having a dihedral angIe of 180°. syn-coplanar: Having a dihedral angle of 0°. H

HX

*� x

anti-coplanar

syn-coplanar

aprotic solvent A solvent that has no acidic protons; a solvent with no 0-H or N-H 234) aryl halide An aromatic compound (benzene derivative) in which a halogen is bonded to one of the carbon atoms of the aromatic ring. (p. 212) base An electron-rich species that can abstract a proton. (p. 231) basicity (base strength): The thermodynamic reactivity of a base, expressed quantitatively

groups. (p.

by the base-dissociation constant Kb. A reaction in which the breaking of bonds and the formation of new bonds occur at the same time (in one step). (pp. 227, 259) dehydrohalogenation An elimination in which the two atoms lost are a hydrogen atom and a halogen atom. (pp. 225,253) electrophile (Lewis acid) A species that can accept an electron pair from a nucleophile, form­ ing a bond. (p. 226) electrophilicity (electrophile strength) The kinetic reactivity of an electrophile. elimination A reaction that involves the loss of two atoms or groups from the substrate, usually resulting in the formation of a pi bond. (p. 225, 252) El reaction (elimination, unimolecular) A multistep elimination where the leaving group is lost in a slow ionization step, then a proton is lost in a second step. Zaitsev orientation is generally preferred. (p. 252)

concerted reaction

B-H ..

..

:CI:

:Cl:

E2 reaction (elimination, bimolecular) A concerted elimination involving a transition state where the base is abstracting a proton at the same time that the leaving group is leaving. The anti-coplanar transition state is generally preferred. (p. 258)

Chapter 6 Glossary

B-H

------i>

E2

H H

"/

C=C

/

"-

CH,

0

CH3

.. :CI:

A generic name for a group of chlorofluorocarbons used as refrigerants, propellants, and solvents. Freon-12® is C F2CI2, and Freon-22® is CHCIF2. (p. 216) geminal dihalide A dihalide with both halogens on the same carbon atom. (p. 214)

freons

CH3-CH2-CBr2-CH3 a geminal dibromide

A substitution where one halogen atom replaces another; com­ monly used to form fluorides and iodides. (p. 229) hydride shift (symbolized � H) Movement of a hydrogen atom with a pair of electrons from one atom (usually carbon) to another. Hydride shifts are examples of rearrangements that con­ vert carbocations into more stable carbocations. (p. 246) hydroxylic solvent A solvent containing OH groups (the most common type of protic sol­ vents). (p. 234) inversion of configuration (see also Walden inversion) A process in which the groups around a chit'al carbon atom are changed to the opposite spatial configuration, usually as a result of back-side attack. (p. 238, 244)

halogen exchange reaction

+

RI "--"""C-Br: .. H / R2

HO-C"'''.. H

(S)

(R)

..

/

RI

, R2

+

:Br:­

The SN2 reaction goes with inversion of configuration. leaving group The atom or group of atoms that departs during a substitution or elimination. The leaving group can be charged or uncharged, but it leaves with the pair of electrons that orig­ inally bonded the group to the remainder of the molecule. (p. 225) methyl shift (symbolized �CH3) Rearrangement of a methyl group with a pair of electrons from one atom (usually carbon) to another. A methyl shift (or any alkyl shift) in a carbocation generally results in a more stable carbocation. (p. 248) nucleophile (Lewis base) An electron-rich species that can donate a pair of electrons to form a bone!. (p. 226, 231) nucleophilicity (nucleophile strength) The kinetic reactivity of a nucleophile; a measure of the rate of substitution in a reaction with a standard substrate. nucleophilic substitution A reaction where a nucleophile replaces another group or atom (the leaving group) in a molecule. (p. 225) organic synthesis The preparation of desired organic compounds from readily available start­ ing materials. polarizable Having electrons that are easily displaced toward a positive charge. Polarizable atoms can begin to form a bond at a relatively long distance. (p_ 232) primary halide, secondary halide, tertiary halide These terms specify the substitution of the halogen-bearing carbon atom (sometimes called the head carbon). If the head carbon is bonded to one other carbon, it is primary; if it is bonded to two carbons, it is secondary; and if bonded to three carbons, it is tertiary. (p. 214) CH3 CH3 -C-Br

CH3 a secondary halide (2°)

CH3 a tertiary halide (3°)

I

a primary halide (1°)

I

CH3-CH-Br

I

protic solvent A solvent containing acidic protons, usually 0 -H or N -H groups. (p. 233) racemization The loss of optical activity that occurs when a reaction shows neither clean retention of configuration nor clean inversion of configuration. (p_ 244) reagent The compound that serves as the attacking species in a reaction.

27 1

272

Chapter 6: Alkyl Halides: Nucleophilic Substitution and Elimination

rearrangement A reaction involving a change in the bonding sequence within a molecule. Rearrangements are common in reactions such as the SN 1 and El involving carbocation inter­ mediates. (p. 246) retention of configuration Formation of a product with the same configuration as the reactant. In a nucleophilic substitution, retention of configuration occurs when the nucleophile assumes the same stereochemical position in the product as the leaving group occupied in the reactant. (p. 244) solvolysis A nucleophilic substitution or elimination where the solvent serves as the attacking reagent. Solvolysis literally means "cleavage by the solvent." (p. 240)

An atom that gives rise to stereoisomers when its groups are interchanged. Asym­ metric carbon atoms and double-bonded carbons in cis-trans alkenes are the most common stereocenters. stereospecific reaction A reaction in which different stereoisomers react to give different stereoisomers of the product. (pp. 239, 262) steric hindrance (steric strain) Interference by bulky groups when they approach a position where their electron clouds begin to repel each other. (p. 237) substitution (displacement) A reaction in which an attacking species (nucleophile, elec­ trophile, or free radical) replaces another group. (p. 225) SN2 reaction (substitution, nucleophilic, bimolecular) The concerted displacement of one 3 nucleophile by another on an sp hybrid carbon atom. (p. 227) SNI reaction (substitution, nucleophilic, unimolecular) A two-step interchange of nucle­ ophiles, with bond breaking preceding bond formation. The first step is ionization to form a carbocation. The second step is the reaction of the carbocation with a nucleophile. (p. 240) substrate The compound that is attacked by the reagent. (p. 226) syn Adding to (or eliminating from) the same face of a molecule. (p. 261) syn-coplanar Having a dihedral angle of 0°. See anti-coplanar for a diagram. transition state In each individual step of a reaction, the state of highest energy between reac­ tants and products. The transition state is a relative maximum (high point) on the reaction-energy diagram. (p. 227) vicinal dihalide A dihalide with the halogens on adjacent carbon atoms. (p. 214)

stereocenter

Br CH3 -CHBr-CHBr-CH) a vicinal dibromide

I

CH)-CH=C -CH) a vinyl bromide

vinyl halide A derivative of an alkene in which one (or more) of the hydrogen atoms on the double-bonded carbon atoms has been replaced by a halogen. (p. 212) Walden inversion (see also inversion of configuration) A step in a reaction sequence in which an asymmetric carbon atom undergoes inversion of configuration. (p. 238) Zaitsev's rule (Saytzeff rule) An elimination usually gives the most substituted alkene prod­ uct. Zaitsev's rule does not always apply, but when it does, the reaction is said to give Zaitsev orientation. (p. 257)

I

1.

2.

Essential Problem-Solving Skills in Chapter 6 Correctly name alkyl halides, and identify them as

1°,2°,or 3°.

Predict the products of SN1,SN2,E l , and E2 reactions, including stereochemistry.

3.

Draw the mechanisms and energy profiles of

SN1, SN2, El, and E2 reactions.

S.

Predict which substitutions or eliminations will be faster, based on differences in substrate, base/nucleophile, leaving group, or solvent.

6.

Predict whether a reaction will be first-order or second-order.

S.

Use Zaitsev's rule to predict major and minor elimination products.

4. Predict and explain the rearrangement of cations in first-order reactions.

7. When possible, predict predominance of substitution or elimination.

Study Problems

273

Study Problems 6-41

Define and give an example for each term. nucleophile (b) electrophile (d) substitution (e) SN2 reaction (g) solvolysis (h) elimination (j) El reaction (k) rearrangement (m) steric hindrance (n) alkyl halide (p) vinyl halide (q) allybc halide (s) secondary halide (t) tertiary halide (v) syn elimination (w) stereospecific reaction

6-42

Draw the structures of the following compounds. sec-butyl chloride (b) isobutyl bromide (e) trans-l,4-diiodocyclohexane (d) 2,2,2-trichloroethanol (g) chloroform (h) l -chloro- l-isopropylcyclopentane Give systematic (IUPAC) names for the following compounds.

6-43

(a)

(c) leaving group (0 SN 1 reaction (i) E2 reaction (I) base (0) aryl halide (r) primary habde (u) anti elimination

(c)

(a)

(a)

�

r

(b)

(0

(i)

a

(c)

7

6-44

�

(a)�CI

or

Cl

(b)

(0

�

Cl

�

or

Br

�

Br

or

(0

CH"

a

Cl ""CH3 \. CI H

�I

� Br

�

or

Br

�

Predict the compound in each pair that will undergo solvolysis (in aqueous ethanol) more rapidly.

(a) (CH3CH2hCH-CI

or

(CH3hC-Cl

(b)

� Cl

or

Cl

X

r

or

Br 6-46

1

F

Predict the compound in each pair that will undergo the SN2 reaction faster.

(d)

6-45

q� C

CI CH,

Cl Cl

(O)

1,2-dibromo-3-methylpentane methylene chloride t-amyl iodide

Br

Show how each compound might be synthesized by the SN2 displacement of an alkyl halide.

(c)

0 o

274

6-47

6-48

Chapter 6: Alkyl Halides: N ucleophilic Substitution and Elimination

(g) (a) (b)

H-C - C-CH2CH2CH3 Give two syntheses for (CH3hCH-O-CH2CH3, and explain which synthesis is better. A student wanted to synthesize methyl t-butyl ether, CH3-O-C(CH3h He attempted the synthesis by adding sodium methoxide (CH30Na) to t-butyl chloride, but he obtained none of the desired product. Show what product is formed in this reaction, and give a better synthesis for methyl t-butyl ether.

When ethyl bromide is added to potassium t-butoxide, the product is ethyl t-butyl ether. CH3CH2-Br

(CH3hC-O-+K

+

potassium I-butoxide

ethyl bromide

�

(CH3hC-O-CH2CH3

ethyl I-butyl either

What happens to the reaction rate if the concentration of ethyl bromide is doubled? What happens to the rate if the concentration of potassium t-butoxide is tripled and the concentration of ethyl bromide is doubled? (c) What happens to the rate if the temperature is raised? When t-butyl bromide is heated with ethanol, one of the products is ethyl t-butyl ether. (a) What happens to the reaction rate if the concentration of ethanol is doubled? (b) What happens to the rate if the concentration of t-butyl bromide is tripled and the concentration of ethanol is doubled? (c) What happens to the rate if the temperature is raised?

(a) (b)

6-49

6-50

6-51

Chlorocyclohexane reacts with sodium cyanide (NaCN) in ethanol to give cyanocyclohexane. The rate of formation of cyanocyclohexane increases when a small amount of sodium iodide is added to the solution. Explain this acceleration in the rate. Give the solvolysis products expected when each compound is heated in ethanol.

(a) 6-52

�

Br

(b)

a

CI CH3

(')

Allylic halides have the structure

" /

(a) (b) (c)

� Br

I

(d) �

/\

Br

I

C=C-C-x

I

Show how the first-order ionization of an allylic halide leads to a resonance-stabilized cation. Draw the resonance structures of the allylic cations formed by ionization of the following halides. Show the products expected from SN 1 solvolysis of these halides in ethanol.

(i)

(l Br A/

(ii)

(y CH2Br

0

(iii)

� Br

(iv)

�

Br

6-53

List the following carbocations in decreasing order of their stability.

6-54

Two of the carbocations in Problem 6-53 are prone to rearrangement. Show how they might rearrange to more stable carbocations.

6-55

Draw perspective structures or Fischer projections for the substitution products of the following reactions.

(b) -------'>

acetone

f

CH3

Br H

H CH3

CH2CH3

+ NaOH

water/acetone )

EtOH, heat)

275

Study Problems 6-56

Predict the products of the following SN2 reactions.

(a) CH3CH2ONa

(c)


2° » 1 °

Rearrangements are common.

SO LUTI O N The first step is protonation of the hydroxyl group, which converts it to a good leaving group.

CH3 � \ 1 .( CH3 -C-O-H H SO I .. CH3 +

2

4

CH3 H 1 / CH3 -C-O+ 1 "H CH 3 .•

+

The second step is ionization of the protonated alcohol to give a carbocation.

CH3 H 1 (" / CH3 -C-O+ 1 .. "H CH 3

Abstraction of a proton completes the mechanism.

HS

04-

Alkene Synthesis by High-Temperature Industrial Methods

7-1 1

309

PROBLEM 7-25

Propose mechanisms for the following reactions.

o

OH � (XoH CH3

cyclopentanol

(b)

cyclopentene H2S04

heat

2-pentanol

(c)

�

H2S04�

�

+

I -pentene

2-pentene

UCH3 QCH3 etCH, +

heat

2-methylcyclohexanol

7-1 1 A

�

I -methylcyclohexene

+

methylenecyclohexane

3-methylcyclohexene

7-1 1

Catalytic Cracking of A l k a nes

The least expensive way to make alkenes on a large scale is by the catalytic cracking of petroleum: heating a mixture of alkanes in the presence of a catalyst (usually aluminosilicates). Alkenes are formed by bond cleavage to give an alkene and a short­ ened alkane.

I I I I I I I I H-C-C - C - C - C - C - C - C - H I I I I I I I I H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

heat � catalyst

I I I I I H-C-C - C - C - C - H I I I I I H

H

H

long-chain alkane

H

H

H

H

H

H

H

H

shorter alkane

Cracking is used primarily to make small alkenes, up to about six carbon atoms. Its value depends on having a market for all the different alkenes and alkanes produced. The average molecular weight and the relative amounts of alkanes and alkenes can be controlled by varying the temperature, catalyst, and concentration of hydrogen in the cracking process. A careful distillation on a huge column separates the mixture into its pure components, ready to be packaged and sold. Because the products are always mixtures, catalytic cracking is not suitable for laboratory synthesis of alkenes. B etter methods are available for synthesizing relatively pure alkenes from a variety of other functional groups. Several of these methods are discussed i n Sections 7-9, 7- 10, and later sections listed in the sum­ mary on page 3 1 4. 7-1 1 8

Dehyd rog enation o f Alkanes

Dehydrogenation is the removal of Hz from a molecule, just the reverse of hydro­

genation. Dehydrogenation of an alkane gives an alkene. This reaction has an unfavor­ able enthalpy change but a favorable entropy change.

I I -C-CI I H

H

heat ) catalyst

'" /

C=C

+ 80 to + 1 20 kl ( + 20 to +30 kcal)

/ '"

+

/1S0

A l ke n e Synth esis by H i g h-Te m pe ratu re I n d u stri a l M ethods

H2 =

+ 1 25 l/kelvin-mol

H + H

",

/

I

H C=C-C-H

I

H alkene

I

H

310

Chapter 7 : Structure and Synthesis o f Alkenes

PI, 500°C )

The hydrogenation of alkenes (Section 7-7) is exothermic, with values of b.. H o around - 80 to - 1 20 kJ/mol ( - 20 to -30 kcal/mol). Therefore, dehydrogenation is endother­ mic and has an unfavorable (positive) value of b.. Ho. The entropy change for dehydro­ genation is strongly favorable ( L� S o = + 1 20 J/kelvin-mol ) , however, because one alkane molecule i s converted into two molecules (the alkene and hydrogen), and two molecules are more disordered than one. The equilibrium constant for the hydrogenation-dehydrogenation equilibrium depends on the change in free energy, b.. G b.. H - Tb.. S. At room temperature, the enthalpy term predominates and hydrogenation is favored. When the temperature is raised, however, the ( - Tb.. S ) entropy term becomes larger and eventually dominates the expression. At a sufficiently high temperature, dehydrogenation is favored. =

PROB L E M 7-26

The dehydrogenation of butane to trans-2-butene has I1HO = + 1 1 6 kl/mol ( + 27.6 kcaIjmol ) and I1So = + 1 1 7 J/kelvin-mol ( + 28.0 caIjkelvin-mol ) . (a) Compute the value o f I1Go for dehydrogenation at room temperature (25°C or 298°K). Is dehydrogenation favored or disfavored? (b) Compute the value of I1G for dehydrogenation at 1 000°C, assuming I1S and I1H are constant. Is dehydrogenation favored or disfavored?

In many ways, dehydrogenation is similar to catalytic cracking. In both cases, a catalyst is used to lower the activation energy, and both reactions use high temperatures to increase a favorable entropy term ( - Tb.. S ) and overcome an unfavorable enthalpy term ( b.. H ) . Unfortunately, dehydrogenation and catalytic cracking also share a tenden­ cy to produce mixtures of products, and neither reaction is well suited for laboratory synthesis of alkenes.

P R O B L E M - S O LV I N G S T R AT E G Y

Proposing Reaction M echanisms At this point, we have seen examples of three major classes of reaction mechanisms: Those involving strong bases and strong nucleophiles Those involving strong acids and strong electrophiles Those involving free radicals Many students have difficulty proposing mechanisms. We can use some general principles to approach this process, however, by breaking it down into a series of logical steps. Using a sys­ tematic approach, we can usually come up with a mechanism that is at least possible and that explains the products, without requiring any unusual steps. Appendix 4 contains more com­ plete methods for approaching mechanism problems. First, Classify the Reaction

Before you begin to propose a mechanism, you must determine what kind of reaction you are dealing with. Examine what you know about the reactants and the reaction conditions: A free-radical initiator such as chlorine, bromine, or a peroxide (with heat or light) sug­ gests that a free-radical chain reaction is most likely. Free-radical reactions were discussed in detail in Chapter 4.

7- 1 1 A l kene S ynthesis by H i gh-Temperature Industrial Methods Strong acids or strong electrophiles (or a reactant that can dissociate to give a strong electrophile) suggest mechanisms such as the SN 1 , E l , alcohol dehydration, etc. that involve carbocations and other strongly acidic intermediates. Strong bases or strong nucleophiles suggest mechanisms such as the SN2 or E2, involv­ ing attack by the strong base or nucleophile on a substrate.

General Princi ples for Drawi ng Mechanisms Once you have decided which type of mechanism i s most likely (acidic, basic, or free-radical), some general principles can guide you in proposing the mechanism. Some principles for free­ radical reactions were discussed in Chapter

4.

Now we consider reactions that involve either

strong nucleophiles or strong electrophiles as intermediates. In later chapters, we will apply these principles to more complex mechanisms. Whenever you start to work out a mechanism,

draw all the bonds and all the sub­

stituents of each carbon atom affected throughout the mechanism. Three-bonded carbon atoms are likely to be the reactive intermediates. If you attempt to draw condensed formulas or line-angle formulas, you will likely misplace a hydrogen atom and show the wrong carbon atom as a radical, cation, or anion.

Show only one step at a time; never combine steps, unless two or more bonds really do change position in one step (as in the E2 reaction, for example) . Protonation of an alco­ hol and loss of water to give a carbocation, for example, must be shown as two steps. You must not simply circle the hydroxyl and the proton to show water fal ling off.

Use curved arrows to show the movement of electrons in each step of the reaction. This movement is alway s /rom the nucleophile (electron donor) to the electrophile (electron acceptor). For example, protonation of an alcohol must show the arrow going from the elec­ trons of the hydroxyl oxygen to the proton-never from the proton to the hydroxyl group. Don 't use curved arrows to try to "point out" where the proton (or other reagent) goes.

Reactions I nvolving Strong N ucleophiles When a strong base or nucleophile is present, we expect to see intermediates that are also strong bases and strong nucleophiles; anionic intermediates are common. Acids and elec­ trophiles in such a reaction are generally weak. Avoid drawing carbocations, H3 0+, and other strong acids. They are unlikely to coexist with strong bases and strong nucleophi les. Functional groups are often converted to alkoxides, carbanions, or other strong nucleophiles by deprotonation or reaction with a strong nucleophile. Then the carbanion or other strong nucleophile reacts with a weak electrophile such as a carbonyl group or an alkyl halide. Consider, for example, the mechanism for the dehydrohalogenation of 3-bromopentane.

CH 3 - CH,_ - C H - CH 2 - CH 3

I

Br Someone who has not read Chapter

6 or these guidelines for classifying mechanisms might

propose an ionization, followed by loss of a proton:

y y)

In.correct mechan.ism

H

I

H

I

CH - C - C - C H - CH 3

I

H

h

Br

2

3

Bc

unlikely)

3

CH - C - C -CH, - C H 3

I

H

+

( unlikelyr- H +

This mechanism would violate several general principles of proposing mechanisms. First, in the presence of ethoxide ion ( a strong base), both the carbocation and the H+ ion are unlike­ ly. Second, the mechanism fails to explain why the strong base is required; the rate of ion­ ization would be unaffected by the presence of ethoxide ion. Also, H+ doesn ' t just fall off (even in an acidic reaction); it must be removed by a base. ( Continued)

Bc

311

312

Chapter 7 : Structure and Synthesis o f Alkenes The presence of ethoxide ion (a strong base and a strong nucleophile) in the reaction sug­ gests that the mechanism involves only strong bases and nucleophiles and not any strongly acidic intermediates. As shown in Section 7-9A, the reaction occurs by the E2 mechanism, an example of a reaction involving a strong nucleophile. In this concerted reaction, ethoxide ion removes a proton as the electron pair left behind forms a pi bond and expels bromide ion. Correct mechanism

H I CH3 -C=C-CH -CH3 I H :Br: 2

..

Reactions I nvolving Strong Electrophi les

When a strong acid or electrophile is present, expect to see intermediates that are also strong acids and strong electrophiles. Cationic intermediates are common, but avoid draw­ ing any species with more than one + charge. Bases and nucleophiles in such a reaction are generally weak. Avoid drawing carbanions, alkoxide ions, and other strong bases. They are unlikely to coexist with strong acids and strong electrophiles. Functional groups are often converted to carbocations or other strong electrophiles by protonation or by reaction with a strong electrophile; then the carbocation or other strong electrophile reacts with a weak nucleophile such as an alkene or the solvent. For example, consider the dehydration of 2,2-dimethyl-l -propanol:

CH 3 I

CH 3-C-CH 2 -OH I

CH 3 The presence of sulfuric acid indicates that the reaction is acidic and should involve strong electrophiles. The carbon skeleton of the product is different from the reactant. Under these acidic conditions, formation and rearrangement of a carbocation would be likely. The hydroxyl group is a poor leaving group; it certainly cannot ionize to give a carbocation and -OH (and we do not expect to see a strong base like -OH in this acidic reaction). The hydroxyl group is weakly basic, however, and it can become protonated in the presence of a strong acid.

r

Step 1:

Protonation of the hydroxyl group

CH3 I .� CH3- -CH2-9-H + H2S04 CH3 .

starting alcohol

------7

+

CH 3 . H 1 I + CH 3-C-CH?-O-H I CH 3 .

.

+

HSO-4

protonated alcohol

The protonated hydroxyl group -OH2 is a good leaving group. A simple ionization to a carbocation would form a primary carbocation. Primary carbocations, however, are very unstable. Thus, a methyl shift occurs as water leaves, so a primary carbocation is never formed. A tertiary carbocation results. (You can visualize this as two steps if you prefer.) Step 2:

Ionization with rearrangement

CH 3 I

CH 3-C-CH 2 I�

CH 3

CH 3

H

1+ -O-H

U·

protonated alcohol

I

with CH3 shift (- CH3 )

CH 3-C-CH +

I

2

CH 3

tertiary carbocation

+

H20 :

7 -11 Alkene Synthesis by High-Temperature Industrial Methods

The final step is loss of a proton to a weak base, such as HSO; or H20 (but not -OH, which is incompatible with the acidic solution). Either of two types of protons, labeled I and 2 in the following figure, could be lost to give alkenes. Loss of proton 2 gives the required product. Step 3:

Abstraction of a proton to form the required product -----7

H

CH3 / C=C / " H CH2-CH3 "

abstract proton

1

"

CH3

or

/

H

C=C / " CH3 CH3

abstract proton 2 observed product

Because abstraction of proton 2 gives the more highly substituted (therefore more sta­ ble) product, Zaitsev's rule predicts it will be the major product. Note that in other problems, however, you may be asked to propose mechanisms to explain unusual compounds that are only minor products.

P ROBLEM 7 - 2 7

For practice i n recognizing mechanisms, classify each reaction according to the type of mechanism you expect: 1. Free-radical chain reaction 2. Reaction involving strong bases and strong nucleophiles 3. Reaction involving strong acids and strong electrophiles

Ba(OH)2

�)C;:

o

0-11 f ' _

(c) styrene

11-0

o

0

c-o-o-c

heat

(d) ethylene

f ' _

)

polystyrene

polyethylene

P ROBLEM 7 - 2 8

Propose mechanisms for the following reactions: (a) CH3 - CH2-CH2-CH2-OH

(Hint: Hydride shift)

(b)

U

Br

CH3-CH=CH -CH3

31 3

314

Chapter

7:

PROBLEM-SOLVING

Structure and Synthesis of Alkenes

Htnl/

Alcohol dehydrations usually go

Propose mechanisms for the following reactions.

through EI elimination of the protonated alcohol, with a carbocation intermediate.

0 0H oH eS

P ROBL EM 7 - 2 9

(a)

Rearrangements are common.

(b)

heat

c5 0+0 + + � � II

H2SO4

+

heat

OH

~

(c)

0

H 3P04

H2SO4

heat

OH

(d)

I

Summ ary

00

H:��'.

+

00 CO OJ +

Methods for Synthesis of Alkenes

1. Dehydrohalogenation of alkyl halides (Section 7-9)

H

I

I

-C-C-

I

X

I

base, heat (loss of HX)

"

/

"

/

C=C

Example

d� H

cyclooctene

chlorocyclooctane 2. Dehalogenation of vicinal dibromides (Section 7-9D)

Br

I

I

-C-C-

I

NaI or

I

Br

Example

J-I " .;,;Ph ""C-C " Ph"'/' Br Br

H

NaI acetone

"

/

/

C=C

"

7 - 1 1 Alkene Synthesis by High-Temperature Industrial Methods

315

3. Dehydration of alcohols (Section 7- 1 0)

I I

I I

conc. H2S04 or H 3P04

- C- C-

Examp"cj: H

heat

)

/ '" C=C / '"

+

Hp

OH

ex:

H cyclohexanol

+

H,O

cyclohexene

4. Dehydrogenation of alkanes (Section 7-11B)

(Useful only for small alkenes; com monly gives mixtures.)

I I

I I

heat, catalyst

-C -C H

)

H

(Useful o nly for small al ke nes; commo nly gives mixtures.)

{

Example Pt,500°C

5. Hofmann and Cope eliminations (Sections 19-1 5 and

H

I I

I-bute ne + cis- a nd trans-2-bute ne I,3-butadie ne + H 2

+

1 9-16)

I I

-C-C+N(CH3h

( Usually gives the least highly substituted al ke ne.)

Example

CH3-CH -CH-CH3 2

I

+N(CH3)3

6. Reduction of alkynes (Section 9-9)

R- C=C-R'

R-C=C-R'

R

H2, Pd/BaS04 quinoline

H

'"

/

R'

C=C / '"

R H ", / C=C H

/

'"

cis-alkene

H

trans-alkene

R'

Examples H2, PdIBaS04 quinoline

(Continued)

316

Chapter 7 : Structure and Synthesis of Alkenes

7. Wittig reaction

(Section 18-13) "-

R' "C=CHR" / R

R' R

C=O /

C

+

Ph3P=CHR"

+

Ph3P=O

Example C=O

+

Ph3P=CHCH3

cyc!opentanone

Chapter 7 Glossary

alkene (olefin) A hydrocarbon with one or more carbon-carbon double bonds. (p. 279) diene: A compound with two carbon-carbon double bonds. (p. 284) triene: A compound with three carbon-carbon double bonds. (284) tetraene: A compound with four carbon-carbon double bonds. (284) allyl group A vinyl group plus a methylene group: CH2=CH-CH2- (p. 285) Bredt's rule A stable bridged bicyclic compound cannot have a double bond at a bridgehead

position unless one of the rings contains at least eight carbon atoms. (p. 295) bicycIic: Containing two rings. bridged bicycIic: Having at least one carbon atom in each of the three links connecting the bridgehead carbons.

a bridged bicyC!ic compound

a Bredt's rule violation

bridgehead carbons: Those carbon atoms that are part of both rings, with three bridges of bonds connecting them. catalytic cracking The heating of petroleum products in the presence of a catalyst (usually an aluminosilicate mineral), causing bond cleavage to form alkenes and alkanes of lower molecu­ lar weight. (p. 309) cis-trans isomers (geometric isomers) Isomers that differ in their cis-trans arrangement on a ring or double bond. Cis-trans isomers are a subclass of diastereomers. (p. 285) cis: Having similar groups on the same side of a double bond or a ring. trans: Having similar groups on opposite sides of a double bond or a ring. Z: Having the higher-priority groups on the same side of a double bond. E: Having the higher-priority groups on opposite sides of a double bond. dehalogenation The eJjmination of a halogen (X2) from a compound. Dehalogenation is for­ mally a reduction. (p. 303)

X

X

I

I

I

/ + C=C " / "

-C-C-

I

(or NaI in acetone)

dehydration The elimination of water from a compound; usually acid-catalyzed. (p. 306)

H

I

OH

I

-C-C-

I

I

/ C=C + " / "

Hp

dehydrogenation The elimination of hydrogen (H2) from a compound; usually done in the

presence of a catalyst. (p. 309) H

I

H

I

-C-C-

I

I

E

Pt, high temperature

)

" /

C=C

/

"

+

Chapter 7 Glossary dehydrohalogenation The elimination of a hydrogen halide (HX) from a compound; usually base-promoted. (p. 298)

H

I

X

I

- C - C-

I

I

�

KO H

double-bond isomers Constitutional isomers that differ only in the position of a double bond. Double-bond isomers hydrogenate to give the same alkane. (p. 291) element of unsaturation A structural feature that results in two fewer hydrogen atoms in the

molecular formula. A double bond or a ring is one element of unsaturation; a triple bond is two elements of unsaturation. (p. 281) geminal dihalide A compound with two halogen atoms on the same carbon atom. geometric isomers See cis-trans isomers. (p. 285) heteroatom Any atom other than carbon or hydrogen. (p. 282) Hofmann product The least highly substituted alkene product. (p. 299) hydrogenation Addition of hydrogen to a molecule. The most common hydrogenation is the addition of H2 across a double bond in the presence of a catalyst (catalytic hydrogenation). The value of ( -/:::'.H0) for this reaction is called the heat of hydrogenation. (p. 290) '" /

C=C

/

'"

+

H H2

�

Pt

I

H

I

-C-C-

I

I

heat of hydrogenation

olefin An alkene (p. 279) polymer A substance of high molecular weight made by linking many small molecules, called monomers. (p. 289) addition polymer: A polymer formed by simple addition of monomer units. polyolefin: A type of addition polymer with an olefin serving as the monomer. saturated Having only single bonds; incapable of undergoing addition reactions. (p. 281) Saytzeff Alternate spelling of Zaitsev. stereospecific reaction A reaction in which different stereoisomers react to give different stereoisomers of the product. (p. 300). trans-diaxial An anti and coplanar arrangement allowing E2 elimination of two adjacent sub­

stituents on a cyclohexane ring. The substituents must be trans to each other, and both must be in axial positions on the ring. (p. 301) unsaturated Having multiple bonds that can undergo addition reactions. (p. 281) vicinal dihalide A compound with two halogens on adjacent carbon atoms. (p. 303) vinyl group An ethenyl group, CH2=CH - . (p. 285) Zaitsev's rule (Saytzeff's rule): An elimination usually gives the most stable alkene product, commonly the most substituted alkene. Zaitsev's rule does not always apply, especially with a bulky base or a bulky leaving group. (p. 291) Zaitsev elimination: An elimination that gives the Zaitsev product. Zaitsev product: The most substituted alkene product.

I

E ssential Problem-Solving Skills in Chapter 7

1. Draw and name all alkenes with a given molecular formula.

2.

Use the E-Z and cis-trans systems to name geometric isomers.

3.

Use heats of hydrogenation to compare stabilities of alkenes.

4.

Predict relative stabilities of alkenes and cycloalkenes, based on structure and stereo­ chemistry.

5.

Predict the products of dehydrohalogenation of alkyl halides, dehalogenation of dibro­ mides, and dehydration of alcohols, including major and minor products.

6.

Propose logical mechanisms for dehydrohalogenation, dehalogenation, and dehydra­ tion reactions.

7.

Predict and explain the stereochemistry of E2 eliminations to form alkenes. Predict E2 reactions on cyclohexane systems.

8.

Propose effective single-step and multistep syntheses of alkenes.

317

31 8

Chapter

7:

Structure and Synthesis of Alkenes

Study Problems 7-30

Define each term and give an example.

(b) Zaitsev elimination

(g)

dehydrogenation

(h) dehydrohalogenation

(j)

dehalogenation

(k)

7-32

(f)

(c) element of unsaturation

Bredt's rule violation geminal dihalide

(I) (i)

hydrogenation dehydration vicinal dihalide

(n) polymer

(m) heteroatom 7-31

(e)

(a) double-bond isomers (d) Hofmann product

Draw a structure for each compound.

(e)

(a) 3-methyl-l-pentene (d) 1,3-cyclohexadiene

(b) cis-3-methyl-3-hexene

(g)

(h) (Z)-2-bromo-2-pentene

vinylcyclopropane

cycloocta- l,4-diene

(f)

(Z)-3-methyl-2-octene

(i)

(3Z,6E)-1,3,6-octatriene

(c)

3,4-dibromobut- l -ene

Give a correct name for each compound.

(a

)

CH3-CH2 -C-CH?-CH?- -CH3

II

(c)

(b) ( CH3CH2hC= CHCH 3

-

�

CH2

(d)

7-33

(e)

'1

�

Label each structure as Z, E, or neither.

(c)

(b)

7-34

(a)

00

Draw and name all five isomers of formula C3HSF.

(b) Draw all 12 acyclic (no rings) isomers of formula C4H7Br. Include stereoisomers. (c) Cholesterol, C27H460, has only one pi bond. With no additional information, what else can you say about its structure?

7-35

Draw and name all stereo isomers of 3-methylhexa-2,4-diene

(a )

Using the cis-trans nomenclature.

(b) Using the E-Z nomenclature. 7-36

Determine which compounds show cis-trans isomerism. Draw and label the isomers, using both the cis-trans and E-Z nomenclatures where applicable.

( a)

(e)

I-pentene

7-37

(f)

(b) 2-pentene

(d) l , l-dibromopropene

(c) 3-hexene

1,2-dibromopropene

hexa-2,4-diene

For each alkene, indicate the direction of the dipole moment. For each pair, determine which compound has the larger dipole moment.

( a)

cis-l ,2-difluoroethene or trans- l ,2-difluoroethene

(b) cis-l ,2-dibromoethene or trans-2,3-dibromo-2-butene (c) cis- l,2-dibromo- l,2-dichloroethene or cis- l ,2-dichloroethene

7-38

c/

Predict the products of the following reactions. When more than one product is expected, predict which will be the major product.

(a) (c 7-39

)

H

H2SO4

(b)

heat

H, Br

cf

NaOCH3

)

(d)

Write a balanced equation for each reaction.

(a

)

CH3-CH2 -CH-CH3

1

OH

H2S0 4, heat )

�

H 3P04 heat

OH H; Br

cf

NaOC(CH3J3

�)CC H

Br

)

Study Problems

(c)

7-40

Br

Br

CH3-CH-CH-CH3 I

I

Zn, CH3COOH

-----7

(c)

(d)

(b) cyclopentanol

cyclopentyl bromide

cyclopentane (not by dehydrogenation)

Predict the products formed by sodium hydroxide-promoted dehydrohalogenation of the following compounds. In each case, predict which will be the major product.

(f)

7-43

I

)

Show how you would prepare cyclopentene from each compound.

(e)

(a) l -bromobutane (d) I-bromo- l -methylcyclohexane 7-42

NaOH, heat

Br

(a) trans-I, 2-dibromocyclopentane 7-41

(d)

CH3 CH3 I I CH3 -CH-C-CH 3

319

(c)

(b) 2-chlorobutane

3-bromopentane

cis- l -bromo-2-methylcyclohexane

trans- l -bromo-2-methylcyclohexane

(a) I-butene (d) methylenecyclohexane

What halides would undergo dehydrohalogenation to give the following pure alkenes?

(e)

(b) isobutylene

(c)

2-pentene

4-methylcyclohexene

In the dehydrohalogenation of alkyl halides, a strong base such as t-butoxide usually gives the best results via the E2

(a )

mechanism. Explain why a strong base such as t-butoxide cannot dehydrate an alcohol through the E2 mechanism.

(b) Explain why strong acid, used in the dehydration of an alcohol, is not effective in the dehydrohalogenation of an 7-44

(a ) (c)

7-45

alkyl halide.

Predict the major products of acid-catalyzed dehydration of the following alcohols.

(b) l -methylcyclopentanol (d) 2,2-dimethyl-l -propanol

2-pentanol 2-methylcyclohexanol

Dehydration of 2-methylcyclopentanol gives a mixture of three aLkenes. Propose mechanisms to account for these three products.

C(�H

H2SO4

heat

0+ d major

2-methylcyclopentanol 7-46

CH,

CH3

H'

minor

minor

Predict the dehydrohalogenation product(s) that result when the following alkyl halides are heated in alcoholic When more than one product is formed, predict the major and minor products.

(a)

(CH3),-CH-C(CH3),-

(b)

I

Br

(d)

7-47

cf

+

(c)

(CH3),C-CH,-CH 3 I

Br

Br

(XCH3 C

(CH3)2CH-TH-CH3

KOH.

l

EI eliminations of alkyl haLides are rarely useful for synthetic purposes because they give mixtures of substitution and elimina­ tion products. Explain why the sulfuric acid-catalyzed dehydration of cyclohexanol gives a good yield of cyclohexene even though the reaction goes by an E l mechanism. (Hint: What are the nucleophiles in the reaction mixture? What products are formed if these nucleophiles attack the carbocation? What further reactions can these substitution products undergo?)

*7-48

The following reaction is called the pinacol rearrangement. The reaction begins with an acid-promoted ionization to give a carbocation. This carbocation undergoes a methyl shift to give a more stable, resonance-stabilized cation. Loss of a pro­ ton gives the observed product. Propose a mechanism for the pinacol rearrangement.

H3C CH3 I I CH3-C-C-CH 3 I I HO OH pinacol

CH3 I CH3-C-C-CH 3 I CH3 o

II

pi nacolone

320 7-49

Chapter 7 : Structure and Synthesis of Alkenes

Propose a mechanism to explain the formation of two products in the following reaction. NBS, hv

7-50

7-51 7-52

A chemist allows some pure (2S,3R)-3-bromo-2,3-diphenylpentane to react with a solution of sodium ethoxide (NaOCH2CH3) in ethanol. The products are two alkenes: A (cis-trans mixture) and B, a single pure isomer. Under the same conditions, the reaction of (2S,3S)-3-bromo-2,3-diphenylpentane gives two alkenes, A (cis-trans mixture) and C. Upon catalytic hydrogenation, all three of these alkenes (A, B, and C) give 2,3-diphenylpentane. Determine the structures of A, B, and C, give equations for their formation, and explain the stereospecificity of these reactions.

The energy difference between cis- and trans-2-butene is about 4 kllmol; however, the trans isomer of 4,4-dimethyl2-pentene is nearly 16 kllmol more stable than the cis isomer. Explain this large difference.

A double bond in a six-membered ring is usually more stable in an endocyclic position than in an exocyclic position. Hydrogenation data on two pairs of compounds follow. One pair suggests that the energy difference between endocyclic and exocyclic double bonds is about 9 kllmol. The other pair suggests an energy difference of about 5 kllmo\. Which number do you trust as being more representative of the actual energy difference? Explain your answer.

Ou

endocyclic 7-53

exocyclic

107

(c)

(XBr

"'"

(d)

'Br

(XBr

Br

� H

,

H One of the following dichloronorbornanes undergoes elimination much faster than the other. Determine which one reacts faster, and explain the large difference in rates. (CH 3) 3CO-K (CH 3)3COH

cis *7-55

Cl

trans

+

d::r

CI

H

CI

A graduate student wanted to make methylenecyclobutane, and he tried the following reaction. Propose structures for the other products, and give mechanisms to account for their formation.

()=-

*7-56

110

Predict the products of the following eliminations of vicinal dibromides with potassium iodide. Remember to consider the geometric constraints of the E2 reaction.

(b)

*7-54

116 105 heats of hydrogenation (kJ/mol)

JS

+

methylenecyc10butane (minor)

other product

Give a mechanism to explain the formation of the following product. In your mechanism, explain the cause of the rearrangement, and explain the failure to form the Zaitsev product. OH

8

Reactions of Alkenes

A

ll al ke nes have a commo n feature : a carbo n-carbo n double bo nd. The reac ­ tio ns of al ke nes arise from the reactivity of the carbon-carbo n double bo nd. O nce agai n, the co ncept of the fu nctio nal group helps to organize a nd sim­ plify the study of chemical reactio ns. By studyi ng the characteristic reactio ns of the double bo nd, we can predict the reactio ns of al ke nes we have never see n before.

Because si ngle bo nds (sigma bo nds) are more stable than pi bonds, the most co mmon re­ actio ns of double bo nds transform t he pi bo nd i nto a sigma bo nd. For example, catalytic hydrogenation converts the C C pi bo nd and the H - H sigma bo nd i nto two C - H sig ma bo nds (Sectio n 7-7). The reaction is exothermic (i�.H0 about -80 to -120 k limol or about -20 to -30 kcal jmol), showi ng that the product is more stable than the reactants. =

=

'"

C=C

/

/

'"

+

H -H

catalyst

)

I I -C-CI I H H

+

8-1 Reactivity of the Ca rbon-Ca rbon Double Bond

energy

Hydroge natio n of an al ke ne is a n example of an addition, one of the three major reactio n types we have studied: addition, elimi nation, and substitutio n. I n an additio n, two molecules combi ne to form o ne product molecule. Whe n an al ke ne u ndergoes addition, two groups add to the carbo n atoms of the double bo nd and the carbo ns become saturated. I n ma ny ways, additio n is the reverse of elimination, i n which o ne molecule splits i nto two fragme nt molecules. I n a substitution, one fragme nt replaces another fragme nt i n a molecule. Addition

'"

/

C=C

/ '"

+

Elimination

Substitution

I I -C-CI I X

-

C -X

I

I

�

y

+

�

X-Y

y-

'"

/

I

X

I

C=C

�

I

-C-C-

/

+

'"

-C I

I

-

Y

I

X-Y

y

+

X321

322

Chapter 8: Reactio ns of Alkenes Addition is the most commo n reactio n of alke nes, a nd i n this chapter we co nsid­ er additio ns to alke nes i n detail. A wide variety of fu nctio nal groups can be formed by addi ng suitable reage nts to the double bo nds of alke nes.

8-2 El ectrop h i l i c Add ition to Al kenes

... Figure 8-1

The elec trons i n the pi bo nd are s pread farther from the carbon nucle i and are more loosel y hel d than the si gma electro ns.

In pri nciple, many differe nt reage nts could add to a double bo nd to form more stable products ; that is , the reactio ns are e nergetically favorable. Not all of these reactio ns have co nve nie nt rates, howe ver. For example, the reactio n of ethyle ne with hydroge n (to gi ve ethane) is stro ngly exothermic, but the rate is very slow. A mixture of ethyle ne a nd hydroge n ca n remai n for years without appreciable reactio n. Addi ng a catalyst such as p lati num, palladium, or nickel allows the reactio n to take place at a rapid rate. Some reage nts react with carbo n-carbo n double bo nds without the aid of a cat ­ alyst. To u nderstand what types of reage nts react with double bo nds, co nsider the structure of the pi bo nd . Although the electro ns i n the sigma bo nd framework are tightly held, the pi bo nd is delocalized abo ve a nd below the sigma bo nd (Figure 8-1 ). The pi-bo ndi ng electro ns are spread farther from the carbon nuclei, a nd they are more loosely held. A stro ng electrophile has a n affinity for these loosely held electro ns; it can pull them away to form a new bo nd (Figure 8-2), leavi ng o ne of the carbon atoms with o nly three bo nds a nd a positive charge: a carbocatio n. I n effect, the double bo nd has reacted as a nucleophile, do nati ng a pair of electro ns to the electrophile. Most additio n reactio ns i nvolve a seco nd step i n which a nucleophile attacks the carbocatio n (as i n the seco nd step of the S N 1 reactio n), formi ng a stable addition prod­ uct. I n the product, both the electrophile a nd the nucleophile are bo nded to the carbo n atoms that were co nnected by the double bo nd. This reactio n is outli ned i n Key Mech­ a nism Box 8-1 , ide ntifyi ng the electrophile as E+ a nd the nucleophile as Nuc: -. This type of reactio n requires a stro ng electrophile to attract the electro ns of the pi bo nd a nd ge nerate a carbocatio n i n the rate-limiti ng step . Most al ke ne reactio ns fall i nto this large class of electrophilic additions to alke nes.

� Figure 8-2

The pi bo nd as a nucleophile. A strong electro phile attracts the electrons out of the pi bo nd to form a ne w si gma bo nd, ge nerati ng a carbocatio n. The (re d) c urve d arro w sho ws the mo vement of e lectro ns, fro m the e lectro n-rich pi bo nd to the elec tron- poor elec tro ph ile.

Q o

.

....... C+-- C "1/,,,.

'1/"'1/

� '" empt y p orbital ."'t-:"--KEY MECHANISM 8-1 " A wide variety of electrophilic additio ns i nvol ve similar mechanisms. First, a stro ng electrophile attracts the loosely hel d electro ns from the pi bo nd of an alke ne. The e1ec­ trophile forms a sigma bo nd to o ne of the carbo ns of the (former) double bo nd, while the other carbo n becomes a carbocation. The carbocation (a stro ng electro phile) reacts with a nucleo phile (o ften a weak nucleophile) to form another sigma bo nd.

Step 1:

Attack of the pi bond on the electrophile for ms a car bocation. �

+

I

/

I

"-

- C - C+

E on the more substituted carbon

8-2 Electrophilic Addition to Alkenes Step 2: Attack by a nuc1eophile gives the addition product.

I

+

/�_

-C-C+ '" I

Nuc:

I I -C-CI I

-----0>

E

E

Nuc

EXAMPLE: Ionic addition of HBr to 2-butene This example shows what happens when gaseous HEr adds to 2-butene. The proton in HEr is electrophilic; it reacts with the alkene to form a carbocation. Bromide ion reacts rapidly with the carbocation to give a stable product in which the elements of HBr have added to the ends of the double bond.

Step 1: Protonation of the double bond forms a carbocation.

H H I I CH -C=C-CH 3 \H 3 u�!:

�

•.

Step 2: Bromide ion attacks the carbocation.

H H I I .. CH -C-C-CH + :Br: .. 3 I + 3 H�

H H I I CH -C-C-CH 3 3 I + H

+ :Br:

H H I I CH -C-C-CH 3 I I 3 H :Br:

�)

PROBLEM: Explain why the + charge of the carbocation always appears at the carbon of the (former) double bond that has NOT bonded to the electrophile. We will consider several types of additions to alkenes, using a wide variety of reagents: water, borane, hydrogen, carbenes, halogens, oxidizing agents, and even other alkenes. Most, but not all, of these will be electrophilic additions. Table

TABLE 8-1

"

/

Type of Addition ) [Elements Added] a

Types of Additions to Alkenes /

c=c "

H

x

I

halogenation

-C-C-

rH 20]

hydrogenation

[H2], a reduction

I

I

H

H

I I

)

[X2], an oxidation

halohydrin formation

I

[HOX], an oxidation

OH OH hydroxylation

oxidative cleavage

)

[02], an oxidation

I

I

HX addition

I

I

[HX]

-C-C-

[HOOHJ, an oxidation

/

"

" C=O

/ O=C

/\

)

[0], an oxidation

-C-C-

I

[CH2]

I

"These are not the reagents used but simply the groups that appear in the product.

X

I

-C-C-

I

I

I

OH

I

) -C-C-

I

I

H

X

I

I

-C-C-

I

H cyclopropanation )

°

epoxidation

I

X

I

-C-C-

)

summarizes

Product

OH

I

hydration )

8 -1

I

H

/\

" / C

-C-C-

I

I

323

324

Chapter 8: Reactio ns of Alkenes the classes of additions we will cover. Note that the table shows what ele me nts have added across the double bond i n the fi nal product, but it says nothi ng about reage nts or mechanis ms. As we study these reactio ns, you should note the regiochemistry of each reactio n, also called the orientation of addition, meaning which part of the reage nt adds to which e nd of the double bo nd. Also note the stereochemistry if the reaction is stereospecific.

8-3

Orientation of Addition: Markovnikov's Rule

S-3A

Add ition of Hyd rogen H a l i des to Al kenes

The si mple mechanism show n for additio n of HBr to 2-butene applies to a large nu mber of electrophilic additio ns. We can use this mechanism to predict the outcome of some fairly complicated reactio ns. For example, the addition of RBr to 2- methyl-2-butene could lead to either of two products , yet only one is observed. CH3

CH3

I

H - Br

+

CH3 - C -CH-CH3

I

I

I

or CH3 - C - CH -CH3

I

I

H

Br H observed

Br

not observed

The first step is proto natio n of the double bo nd. If the proton adds to the seco ndary carbo n, the product will be diffe re nt from the o ne formed if the proto n adds to the tertiary carbo n.

add H+ to secondary carbon

)

CH3

I

CH3 - C - CH -CH3

+

I

H

Be

tertiary carbocation

CH3

I

CH3 - C

)CH- CH3

add H+ to tertiary carbon

)

CH3

I

CH3 - C - CH -CH3

I

H - Br

H

'-..:;.

PROBLEM-SOLVING

Hi-ltv

Stability of carbocations:

3° > 2° > 1 ° > +CH3

+

Br -

secondary carbocation

When the proto n adds to the seco ndary carbo n, a tertiary carbocatio n results. Whe n the proto n adds to the tertiary carbo n atom , a secondary carbocatio n results. The tertiary carbocatio n is more stable (see Sectio n 4- 1 6A) , so the first reactio n is favored. The seco nd half of the mechanism produces the fi nal product of the additio n of HBr to 2- methyl-2-butene. CH3

CH3

I

CH3 - C - CH - CH3 +

Br :

J

I

H

I I

CH3 - C -CH - CH3

I

Br H

Note that proto nation of o ne carbo n ato m of a double bo nd gives a carbocatio n o n the carbo n atom that was not proto nated. Therefore , the proto n adds to the e nd of the dou­ ble bond that is less substituted to give the more substituted carbocation (the more sta ­ ble carbocatio n).

8-3 Addition of Hydrogen Halides to Alkenes

MECHANISM 8-2 Step 1:

Ionic Addition of HX to an Alkene

Protonatio n of the pi bond forms a c arbo catio n. H

� .. '"C=C + H X: '"

/ Step 2:

'" I C- C / I

\ �

+

:X:-

+ on the more substituted carbon Attack by the h alide io n gives the additio n product .

I '" -C:X:�C .. / I . .

H

+

:X: H

I

I

I

I

-C-C-

EXAMPLE:

The ionic additio n o f HBr to prope ne shows proto nation of the less substituted c ar­ bon to give the more substituted c arboc atio n. Reaction with bromide ion completes the additio n. CH3 H �

I

I

I

I

H-C-C-H :1;I.r= H product

*-

CH3

I /H H-C-C I +"H H

Positive charge on less substituted carbon. Less stable; 1101 formed.

There are m any ex amples o f reactio ns where the proto n adds to the less substituted c arbo n atom o f the double bo nd i n order to produce the more substituted c arbocatio n. The additio n of HBr ( and other hydroge n h alides) is s aid to be regioselective bec ause i n e ach c ase, o ne of the two possible orie nt atio ns of additio n results prefere nti ally over the other. A Russi an chemist, Vl adimir M arkov nikov, first showed the orie nt atio n o f additio n o f HBr to alke nes i n 186 9. M arkov nikov stated:

M arkovnikov's Rule

MARKO VN IKO V'S RULE: The additio n of a proto n acid to the double bo nd o f an alke ne re sults i n a product with the acid proto n bo nded to the c arbo n atom th at alre ady holds the gre ater number o f hydroge n atoms. This is the origi nal stateme nt o f Markovnikov's rule. Re actions th at fo llow this ru le are s aid to follow Markovnikov orientation and give the Markovnikov product. We are o fte n i nterested i n addi ng electrophiles other th an proto n acids to the double bonds o f alke nes. M arkov nikov ' s rule c an be exte nded to i nclude a wide variety o f other additio ns, b ased o n the additio n o f the electrophile i n such a w ay as to produce the most stable c arboc atio n. MARKO VNIKO V' S RULE (exte nded) : I n an electrophilic additio n to an alke ne , the electrophile add s i n such a w ay as to ge nerate the most st able i ntermedi ate.

325

326

Chapter

8:

Reactions of Alkenes

a�H 3 �: QCH3 .. : r: \ � H I �r: -------7 � '-...A ' H H H .

� Figure 8-3

-------7

aC�� . •

H H

product

An electrophile adds to the less

Positive charge

substituted end of the double bond to

notformed.

on less substituted carbon.

give the more substituted (and

Less stable;

therefore more stable) carbocation.

Figure 8-3 shows how HBr adds to I-methylcyclohexene to give the product with an additional hydrogen bonded to the carbon that already had the most bonds to hydrogen (one) in the alkene. Note that this orientation results from addition of the proton in the way that generates the more stable carbocation. Like HBr, both HCI and HI add to the double bonds of alkenes, and they also fol­ low Markovnikov's rule; for example, CH3

I

CH -C-CH-CH CH3 3

I

I

Cl H

2

H H'-..., / C ;'H

HI

0

CQ

HCl

H

PROBLEM 8-1 Predict the major products of the following reactions.

(c)

(a) CH3-CH=CH2

+

I-methylcyclohexene

(b) 2-methylpropene + HCl (d) 4-methylcyclohexene + HBr

HBr

+

HI

PROBLEM 8 - 2 When 1,3-butadiene reacts with 1 mol of HBr, both 3-bromo-l -butene and J-bromo-2-butene are formed. Propose a mechanism to account for this mixture of products.

8-3B

Free-Radical Addition of HBr: Anti-Markovnikov Addition

In 1 933, M. S. Kharasch and F. W. Mayo showed that anti-Markovnikov products result from addition of HBr (but not HCl or HI) in the presence of peroxides. Perox­ ides give rise to free radicals that act as catalysts to accelerate the addition, causing it to occur by a different mechanism. The oxygen-oxygen bond in peroxides is rather weak. It can break to give two radicals.

' :' .. - R -O R-O �. ..

-------7

heat

..

R- O '

..

+ ' O-R

+ 150 kJ (+ 36

kcal)

8-3

Addition of Hydrogen Halides to Alkenes

Alkoxy radicals (R - 0·) catalyze the anti-Markovnikov addition of HBr. The mechanism of this free-radical chain reaction is shown next. MECHANISM 8-3

Free-Radical Addition of HBr to Alkenes

Initiation: Radicals are formed.

R-O-O-R � R-O· + ·O-R R-O· + H-Br � R-O-H + Br·

Propagation: A radical reacts to generate another radical.

Step 1:

A bromine radical adds to the double bond to generate an alkyl radical on the more substituted carbon atom. / '" C=C '" /

Step 2:

Br·

+

----7

The alkyl radical abstracts a hydrogen atom from HEr to generate the product and a bromine radical. Br H Br I I I / -C-C- + Br· -C-C· + H-Br '"

I

I

I

The bromine radical generated in Step 2 goes on to react in Step 1, continuing the chain. EXAMPLE: Free-radical addition of HBr to propene.

Initiation: Radicals are formed.

R -O- O-R � R-O· + ·O-R R-O· + H-Br � R-O-H + Br·

Propagation: A radical reacts to generate another radical.

Step 1:

A bromine radical adds to the double bond to generate an alkyl radical on the secondary carbon atom. H

Br I

"'.

C-C-H / I H 3C H

. on the 2° carbon

Step 2:

The alkyl radical abstracts a hydrogen atom from HBr to generate the product and a bromine radical. H

"'.

Br I

C-C - H / I H 3C H

+

H-Br

H I

Br I

H-C-C-H

I

I

CH3 H

The bromine radical generated in Step 2 goes on to react in Step

+

Br·

1, continuing the chain.

327

328

Chapter 8: Reactions of Alkenes

Let's consider the individual steps. In the initiation step, free radicals generated from the peroxide react with HBr to form bromine radicals.

G· .. + H-Br: R-ol .. � ..

jjHO

.. .. · R-O-H + :Br

-63

=

kJ (-15 kcal)

'" an octet of electrons in its valence shell, making it electron­ The bromine radical lacks � deficient and electrophilic. It adds to a double bond, forming a neW free radical with � the odd electron on a carbon atom. �'"C=C/

I I

. . :Br· + ..

jjHD

/

-C-C · :Br:

'"

=

-12 kJ ( -3

kcal)

This free radical reacts with an HBr molecule to form a C-H bond and generate another bromine radical. .. I I -C-C- + :Br·

I

I

jjHO

=

-25

Br H

kJ (-6 kcal)

The regenerated bromine radical reacts with another molecule of the alkene, continuing the chain reaction. Note that each propagation step starts with one free radical and ends with another free radical. The number of free radicals is constant, until free radicals come together and terminate the chain reaction.

Radical Addition of HBr to Unsymmetrical Alkenes Now we must explain the anti-Markovnikov orientation found in the products of the peroxide-catalyzed reaction. When the alkene is unsymmetrical, adding the bromine radical to the second­ ary end of the double bond forms a tertiary radical.

CH3

I

CH3 -C-CH-CH3 •

I

Br

tertiary radical (more stable)

CH3

I

but not

CH3 -C-CH-CH3 '

I

Br

secondary radical (less stable)

As we saw in the protonation of an alkene, the electrophile (in this case, Br ) adds to the less substituted end of the double bond, and the unpaired electron appears on the more substituted carbon to give the more stable free radical. This intermediate reacts with HBr to give the anti-Markovnikov product, in which H has added to the more substituted end of the double bond: the end that started with fewer hydrogens. CH3 I CH3-C-CH-CH3 + H-Br .

PROBLEM-SOLVING Stability of radicals:

30 > 20 > 10 >

Hi-ltv . CH3

I

Br

CH3

I I

CH3-C-CH-CH3 + Br ·

I

H Br

anti-Markovnikov product

Note that both mechanisms for the addition of HBr to an alkene (with and with­ out peroxides) follow our extended statement of Markovnikov's rule: In both cases, the electrophile adds to the less substituted end of the double bond to give the more stable carbocation or free radical. In the ionic reaction, the electrophile is H+. In the peroxide-catalyzed free-radical reaction, Br· is the electrophile. Many students wonder why the reaction with Markovnikov orientation does not take place in the presence of peroxides, together with the free-radical chain reaction. It actually does take place, but the peroxide-catalyzed reaction is faster. If just a tiny

8-3 Addition of Hydrogen Halides to Alkenes

329

bit of peroxide is present, a mixture of Markovnikov and anti-Markovnikov products results. If an appreciable amount of peroxide is present, the radical chain reaction is so much faster than the uncatalyzed ionic reaction that only the anti-Markovnikov product is observed. The reversal of orientation in the presence of peroxides is called the peroxide effect. It occurs only with the addition of HBr to alkenes. The reaction of an alkyl rad­ ical with HCl is strongly endothermic, so the free-radical chain reaction is not effec­ tive for the addition of HC\.

I / C] - C - C ' � I

I

I

C] - C - C - H I I

H - Cl

+

+

t1HO

CI'

=

+42 kJ ( + 10 kcal)

S imilarly, the reaction of an iodine atom with an alkene is strongly endothermic, and the free-radical addition of HI is not observed. Only HBr has just the right reactiv­ ity for each step of the free-radical chain reaction to take place. I·

+

� /

C=C

I

/ �

/

t1HO

I - C - C' � I

+ 54kJ ( +13kcal)

=

PROBLEM-SOLVING

PROBLEM 8-3

Predict the major products of the following reactions, and propose mechanisms to support your predictions.

o (a) 2-methylpropene

0

I

+ HBr + CH3CH2 - O - O-CH2CH3

(

(c) I-phenylpropene + HBr + di-t-butyl peroxide

PhenYl

=

Ph

=

0-)

SOLVED PROBLEM 8-1

Show how you would accomplish the following synthetic conversions. (a) Convert I-methylcyclohexene to I-bromo- l -methylcyclohexane. SOLUTION

This synthesis requires the addition of HBr to an alkene with Markovnikov orientation. Ionic addition of HBr gives the correct product.

+ I-methylcyclohexene

HBr I-bromo-I-methylcyclohexane

(b) Convert I-methylcyclohexanol to I-bromo-2-methylcyclohexane. SOLUTION

This synthesis requires the conversion of an alcohol to an alkyl bromide with the bromine atom at the neighboring carbon atom. This is the anti-Markovnikov product, which could be formed by the radical-catalyzed addition of HBr to I-methylcyclohexene.

+ I-methylcyclohexene

HBr

R-O-O-R ) heat

structures, including all bonds and charges, when writing a mechanism or determining the course of a reaction.

I

+ HBr + CH3 - C -O-O-C - CH3

(b) I-methylcyclopentene

Hi-ltv

Remember to write out complete

(X

CH3 Br

l-bromo-2-methylcyclohexane ( Continued )

330

Chapter

8:

Reactions of Alkenes

l-Methylcyclohexene is easily synthesized by the dehydration of I-methylcyclohexa­ nol. The most substituted alkene is the desired product.

+ l -methylcyc1ohexanol

l-methylcyc1ohexene

The two-step synthesis is summarized as follows:

HEr ROaR

I-methylcyclohexanol

)

(XCH3

Br l-bromo-2-methylcyclohexane

I-methy lcyc10hexene

PROBLEM 8-4

(a) I-butene � I-bromobutane (b) I-butene � 2-bromobutane (c) 2-l11ethylcyclohexanol � I-bromo-l-methylcyclohexane Show how you would accomplish the following synthetic conversions.

(d)

8-4 Add ition of Water: Hyd rati on of A l ke n es

2-l11ethyl-2-butanol

�

2-broI110-3-l11ethylbutane

An alkene may react with water in the presence of a strongly acidic catalyst to form an alcohol. Formally, this reaction is a hydration (the addition of water), with a hydrogen atom adding to one carbon and a hydroxyl group adding to the other. Hydration of an alkene is the reverse of the dehydration of alcohols we studied in Section 7- 10. Hydration of an alkene

H I -

c

I

alken e

OH I -

c

-

I

alcohol

(Markovnikov orientation) Dehydration of an alcohol

H OH I I - C - CI I alcohol

alkene

For dehydrating alcohols, a concentrated dehydrating acid (such as H2S04 or H3P04) is used to drive the equilibrium to favor the alkene. Hydration of an alkene, on the other hand, is accomplished by adding excess water to drive the equilibrium toward the alcohol. 8-4A

Mechanism of Hydration

The principle of microscopic reversibility states that a forward reaction and a reverse reac­ tion taking place under the same conditions (as in an equi]jbrium) must follow the same re­ action pathway in microscopic detail. The hydration and dehydration reactions are the two complementary reactions in an equi]jblium; therefore, they must follow the same reaction pathway. It makes sense that the lowest-energy transition states and intermediates for the reverse reaction are the same as those for the forward reaction, except in reverse order. According to the principle of microscopic reversibility, we can write the hydra­ tion mechanism by reversing the order of the steps of the dehydration (Section 7-10).

8-4 Addition of Water: Hydration of Alkenes

Protonation of the double bond forms a carbocation. Nucleophilic attack by water, fol­ lowed by loss of a proton, gives the alcohol. MECHANISM 8-4 Step 1:

Protonation of the double bond forms a carbocation.

�

/

Step 2:

Acid-Catalyzed Hydration of an Alkene

Cb LH

"c-L I



I

..

R-B -O :

I

R borate ester

+

:OH

8-7 Hydroboration of Alkenes Hydrolysis of the borate ester

..

: O-R

I

I C ..

O-R

I

..

I

?-�-vT R :OH R

: O-B - O :

I

R : OH R

O-R

O-R

..

O-B

I

I

R :OH

:0:

I

R

(The other two OR groups hydrolyze similarly)

I

O- B

I

R

I

O-H + R- OH + -OH

Hydroboration of alkenes is another example of a stereospecific reaction, in which different stereoisomers of the starting compound react to give different stereoisomers of the product. Problem 8- 14considers the different products formed by the hydroboration-oxidation of two acyclic diastereomers. SOLVED PROBLE M 8-4

A norbornene molecule labeled with deuterium is subjected to hydroboration-oxidation. Give the structures of the intermediates and products.

J):

"'

(

B H;

THF

D

elida (inside) face deuterium- labeled norbornene

alkylborane

alcohol (racemic mixture)

SOLUT ION

The syn addition of BH3 across the double bond of norbornene takes place mostly from the more accessible outside (exo) face of the double bond. Oxidation gives a product with both the hydrogen atom and the hydroxyl group in exo positions. (The less accessible inner face of the double bond is called the endo face.) PROBLE M 8-1 2

In the hydroboration of I-methylcyclopentene shown in Solved Problem 8-3, the reagents are achi.ral, and the products are chiral. The product is a racemic mixture of trans-2-methylcy­ clopentanol, but only one enantiomer is shown. Show how the other enantiomer is formed. PROBLE M 8-1 3

Predict the major products of the following reactions. Include stereochemistry where applicable.

(a) I -methylcycloheptene + BH3 . THF, then H202> OH(b) trans-4,4-dimethyl-2-pentene + BH3 ' THF, then H202, OH-

(c)

�/, �

H

+ BH3 ' THF, then HP2' OH-

CH3

PROBLE M 8-1 4

(a) When (Z)-3-methyl-3-hexene undergoes hydroboration-oxidation, two isomeric prod­ ucts are formed. Give their structures, and label each asymmetric carbon atom as (R) or

(S). What is the relationship between these isomers?

(b) Repeat part (a) for (E)-3-methyl-3-hexene. What is the relationship between the products

formed from (Z)-3-methyl-3-hexene and those formed from (E)-3-methyl-3-hexene?

341

342

Chapter 8: Reactions of Alkenes PROBLEM 8- 1 5

Show how you would accomplish the following transformations. (.

l

(b

l

eo eo

CO

�

OH

CQ

OH (c) I -methylcycloheptanol --7 2-methylcycloheptanol PROBLEM 8- 1 6

When HBr adds across the double bond of 1 ,2-dimethylcyclopentene, the product is a mix­ ture of the cis and trans isomers. Show why this addition is not stereospecific.

8-8

Halogens add to alkenes to form vicinal dihalides.

Add ition of H a logens to Al kenes

I I

I I

X

"-....

/'

C=C '-... /'

+

X2

-C-C-

--7

X

usually anti addition

S-SA

Mechanism of Halogen Addition

A halogen molecule ( Br2, C12, or 12) is electrophilic; a nucleophile can react with a halogen, displacing a halide ion: Nuc : -

� +

: Br-Br: .. � ..

..

..

Nuc- Br:

..

: Br : -

+

In this example, the nucleophile attacks the electrophilic nucleus of one bromine atom, and the other bromine serves as the leaving group, departing as bromide ion. Many re­ actions fit this general pattern; for example: � . + : Br - Br: + : Br : HO Br: HO : .. .. \:;: ..

..

..

..

.

..

: Cl :-

+

"-.... � ..

C=C "-.... /'

+

..

: BrBr: .. \:;:

: Br·· /+ \ - C- C I I

+

..

: Br : -

bromoniulll ion

In the last reaction, the pi electrons of an alkene attack the bromine molecule, expelling bromide ion. A bromonium ion results, containing a three-membered ring with a positive charge on the bromine atom. This bromonium ion is similar in structure to the mercurinium ion discussed in Section 8-5. Similar reactions with other halogens form other halonium ions. The structures of a chloronium ion, a bromonium ion, and an iodonium ion are shown next.

8-8 Addition of Halogens to Alkenes Examples

'" Cl " /+ \ - C -C -

'" Br" /+ \ - C -C -

." ( 1+ \ - C- C -

chloronium ion

bromonium ion

iodonium ion

I

I

I

I

I

343

I

Unlike a nonnal carbocation, all the atoms in a halonium ion have filled octets. The three-membered ring has considerable ring strain, however, which, combined with a positive charge on an electronegative halogen atom, makes the halonium ion strongly electrophilic. Attack by a nucleophile, such as a halide ion, opens the halonj­ um ion to give a stable product.

MECHANISM 8-7 Step 1:

Addition of Halogens to Alkenes

Electrophilic attack forms a halonium ion. '" x " /+\ -C-C-

,, C =� .. .. C + : X-X :

/'

"

I

.. \l.:

+

I

: X:-

hal onium ion

The halide ion opens the halonium ion.

Step 2:

:X:

..

I I

I I

-C -C -

X-

Step 1:

:X:

attacks from the back side

EXAMPLE: Addition of Br2 to propene.

Electrophilic attack forms a bromonium ion.

f:)l\ .. ..

.

H ", :Br- Br: C=C . . �. "" / H3 C H

I

H H3C

propene

+

When a solution of bromine (red-brown) is added to cyclohexene, the bromine color quickly disappears because bromine adds across the double bond. When bromine is added to cyclohexane (at right), the color persists.

' Br' I \ C -C .

\H H

bromonium ion

Bromide ion opens the bromonium ion

Step 2:

H

Br

�C - C/ / \H H Br

H3C

1 ,2-dibromopropane

Chlorine and bromine commonly add to alkenes by the halonium ion mechanism. Iodination is used less frequently because diiodide products decompose easily. Any solvents used must be inert to the halogens; methylene chloride CH2CI2 , chlorofonn CHCI3 , and carbon tetrachloride CCI4 are the most frequent choices.

(

)

(

)

(

)

344

Chapter 8: Reactions of Alkenes 8-88

Stereochemistry of Halogen Addition

The addition of bromine to cyclopentene is a stereospecific anti addition.

i? �H

H

d�

Br

but not

H

cyc10pentene

Br

trans- l ,2-dibromocyc1opentane (92%)

cis- I ,2-dibromocyclopentane (not formed)

Anti stereochemistry results from the bromonium ion mechanism. When a nucleophile attacks a halonium ion, it must do so from the back side, in a manner similar to the S N 2 displacement. This back-side attack assures anti stereochemistry of addition.

H Br

enantiomer

+

H

trans

Halogen addition is another example of a stereospecific reaction, in which different stereoisomers of the starting material give different stereoisomers of the prod­ uct. Figure 8-4 shows additional examples of the anti addition of halogens to alkenes. The addition of bromine has been used as a simple chemical test for the pres­ ence of olefinic double bonds. A solution of bromine in carbon tetrachloride is a clear, deep red color. When this red solution is added to an alkene, the red bromine color disappears (we say it is "decolorized"), and the solution becomes clear and

(JC

H H

cyc10hexene

H

+ Cl 2

H

� CI CI + CI Cl -f------J � � H

Br H3 C H / ,%0 C-C ' �H / CH3 Br (+ enantiomer)

cis-2-butene � Figure 8-4

Ex amples of the anti addition of halogens to alkenes. The stereospecific anti addition gives predictable stereoisomers of the products.

H

racemic trans- l ,2-dichlorocyc1ohexane

CH3 H Br

+ Br + H CH3

+

CH3 Br H

+ H + Br

(±)-2,3-dibromobutane

= :f:: CH3

trans-2-butene

CH3

meso-2,3-dibromobutane

8-9 Formation of Halohydrins

345

colorless. (Although there are other functional groups that decolorize bromine, few do it as quickly as alkenes.) PROBLEM 8-1 7

Give mechanisms to account for the stereochemistry of the products observed from the addi­ tion of bromine to cis- and trans-2-butene (Figure 8-4). Why are two products formed from the cis isomer but only one from the trans? (Making models will be helpful.) PROBLEM 8-1 8

Propose mechanisms and predict the major products of the following reactions. Include stere­ ochemistry where appropriate. (a) cycloheptene + Br2 in CH2Cl2 ( E)-3-decene + Br2 in CCI4 (b) (0)

CD

+

CI, ;n CHCl,

(d)

---0---


H

BH2

OH

anti-Markovnikov orientation (syn stereochemistry) Example

(1) B H3

. THF

(2) H2 02, OH

f. Polymerization

" /

C= C

(Section 8-16) '\..

I / R- C - C+ "I

/ "

/

C=C'\.. /

----'»

I I I / R -C-C -C-C+ I

I

(also radical and anionic polymerization)

I

"

---;.

pol y mer

Example

CH3 -CH= CH2

n

BF3

)

l�-�l lk t J H3

propylene

polypropylene

(Section 8- 10)

2. Reduction: Catalytic Hydrogenation

" /

C=C

/ "

n

+ H2

Pt, Pd, or Ni

I I

I I

- C -C-

)

H

H

(syn addition)

(Section 8-11)

3. Addition ofCarbenes: Cyclopropanation

" /

C=C

I

\ /

/

/ \

C

"-

y

(X,Y

o

=

X

H, Cl, Br, I, or -COOEt)

Example

cyclohexene

I

-C-C-

+

CHBr,

�B r �Br

(Continued)

372

Chapter 8: Reactions of Alkenes

4. Oxidative Additions

a.

Addition of halogens (Section 8-8) x

I

"-

/ C=C "/

I

- C-C-

I

I X (anti addition)

Example

�

ilr,

"

Br trans-l,2-dibromocyclohexane

cyclohexene

NBS provides a trace of Br2 that (with light as initiator) allows radical substitution to proceed faster than the ionic addition. (Section 6-6B)

NBS, hv ) (trace Br2) cyclohexene b.

3-bromocyclohexene (Section 8-9)

Halohydrin formation

CH3 ""'OH

+ anti addition (Markovnikov orientation) c.

Epoxidation

"-

/

C=C

/

"-

Q

"\ Br if

(Section 8-12)

+

0

R- C-O-O-H

�

II

0 O-0

Example

"

"-

/

Anti hydroxylation

/

"-

C=C

C-OOH

II

Cl

cyclohexene

d.

-

+

0

I

\/

+

syn addition

peroxyacid

alkene

I

-C - C -

�

0

R -C-O-H

II

00 �O>-r +

OH

epoxycyclohexane Cl (cyclohexene oxide)

(Section 8- 1 3)

R-C-O-O-H II o

I

I

-C-C\ / o

OH I I -C-CI I OH

8-16 Polymerization of Alkenes Example

II

o

o

H-C-OOH, H30+ )

Q

em

trans-cyclohexane-l,2-diol

cycIohexene

e.

Ji / OH '\ H

Syn hydroxylation

"-C=C / / "-

(Section 8-14) +

+

I

-OH,Hp

I

-C-CI I OH OH (syn addition)

Example

fI "' OH '\ OH

Q

o

II

cycIohexene

cis-cycIohexane- l,2-diol

5. Oxidative Cleavage ofAlkenes

a. R R

(Section 8-15)

Ozonolysis

" /

C=C

R'

/ "-

+

03

+

-4

H ozonide

I

CH3

I

CH 3 - C = C - CH 3 2-methyl-2-butene

acetaldehyde

acetone

b. Potassium permanganate

R

"-C=C / R' ,,/ R H

+

KMn04

warm �

R R

"-

/ C=O

+

/ R' O=C"­ OH

ketones and acids (aldehydes are oxidized) Example

H CH3 I I CH3-C=C-CH3 + KMn04 2-methyl-2-butene

"

ketones and aldehydes

Example

H

O=C

/

warm � acetic acid

acetone

R'

H

373

374

Chapter

8: Reactions of Alkenes

Chapte r 8 G l ossa ry

addition A reaction involving an increase in the number of groups attached to the alkene and a decrease in the number of elements of unsaturation. (p. 321) anti addition: An addition in which two groups add to opposite faces of the double bond (as in addition of Br2). (p. 344) electrophilic addition: An addition in which the electrophile (electron-pair acceptor) bonds to one of the double-bonded carbons first, followed by the nucleophile. (p. 322) syn addition: An addition in which two groups add to the same face of the double bond (as in osmium tetroxide hydroxylation). (p. 348) addition polymer (chain-growth polymer) A polymer that results from rapid addition of one molecule at a time to a growing polymer chain, usually with a reactive intermediate (cation, radical, or anion) at the growing end of the chain. (p. 363) alkoxy group (alkoxyl group) ( - 0 - R) An alkyl group bonded through an oxygen atom, as in an ether. alkoxymercuration The addition of mercuric acetate to an alkene in an alcohol solution, forming an aLkoxymercurial intermediate. Demercuration gives an ether. (p. 335)

"

/

C=C

/

"

+

R -OH

Hg(OAc)2

R-O

R- O

I I -C -C I I

I

I

-C - C-

I

HgOAc

I

H

alpha elimination (a elimination) The elimination of two atoms or groups from the same car­ bon atom. Alpha eliminations are frequently used to form carbenes. (p. 352)

CHBr3 + KOH

----'>

:CBr2 + H20 + KEr

anionic polymerization The process of forming an addition polymer by chain-growth poly­ merization involving an anion at the end of the growing chain. (p. 366) asymmetric induction (enantioselective synthesis) The formation of an optically active product from an optically inactive starting material. Such a process requires the use of an opti­ cally active reagent or catalyst. (p. 349) beta elimination (f3 elimination) The elimination of two atoms or groups from adjacent carbon atoms. This is the most common type of elimination. (p. 352)

H

I

Br

I

- C-C -

I

I

+

"

/

KOH

C=C

/ "

+

HzO

+

KEf

carbene A reactive intermediate with a neutral carbon atom having only two bonds and two non bonding electrons. Methylene ( :CH2 ) is the simplest carbene. (p. 350) cationic polymerization The process of forming an addition polymer by chain-growth poly­ merization involving a cation at the end of the growing chain. (p. 363) chain-growth polymer See addition polymer. (p. 363) demercuration The removal of a mercury species from a molecule. Demercuration of the products of oxymercuration and alkoxymercuration is usually accomplished using sodium borohydride. (p. 333) epoxide (oxirane) A three-membered cyclic ether. (p. 353) epoxidation: Formation of an epoxide, usually from an alkene. A peroxyacid is generally used for alkene epoxidations. free-radical polymerization The process of forming an addition polymer by chain-growth polymerization involving a free radical at the end of the growing chain. (p. 365) glycol A 1,2-diol. (p. 355) halogenation The addition of a halogen (X2) to a molecule, or the free-radical substitution of an X for an H (p. 344) halohydrin A beta-haloalcohol, with a halogen and a hydroxyl group on adjacent carbon atoms. (p. 345)

" /

C=C

/ "

I I

I I

- C -C CI

OH

a chlorohydrin

+

HCI

Chapter

halonium ion A reactive, cationic intermediate with a three-membered ring containing a halo­ gen atom; usually, a chloronium ion, a bromonium ion, or an iodonium ion. (p. 342) heterogeneous catalysis Use of a catalyst that is in a separate phase from the reactants. For example, a platinum hydrogenation catalyst is a solid, a separate phase from the liquid alkene. (p. 348) homogeneous catalysis Use of a catalyst that is in the same phase as the reactants. For exam­ ple, the acid catalyst in hydration is in the liquid phase with the alkene. (p. 348) hydration The addition of water to a molecule. Hydration of an alkene forms an alcohol. (p. 330) H

OH

I I -C -C I I hydroboration The addition of borane (BH3 ) or one of its derivatives (BH3 ' THF, for exam­ ple) to a molecule. (p. 336) hydrogenation The addition of hydrogen to a molecule. The most common hydrogenation is the addition of H2 across a double bond in the presence of a catalyst (catalytic hydrogenation or catalytic reduction). (p. 348) hydroxylation The addition of two hydroxyl groups, one at each carbon of the double bond; formally, an oxidation. (p. 358) HO

OH

I I - C-C I I Markovnikov's rule (original statement) When a proton acid adds to the double bond of an alkene, the proton bonds to the carbon atom that already has more hydrogen atoms. (extended statement) In an electrophilic addition to an alkene, the electrophile adds in such a way as to generate the most stable intermediate. (p. 324)

+

HCI

c(:'

H Markovnikov product

Markovnikov orientation: An orientation of addition that obeys the original statement of Markovnikov's rule; one that gives the Markovnikov product. (p. 325) anti-Markovnikov orientation: An orientation of addition that is the opposite of that pre­ dicted by the original statement of Markovnikov's rule; one that gives the anti-Markovnikov product. (p. 326) monomer One of the small molecules that bond together to form a polymer. (p. 363) organic synthesis The preparation of desired organic compounds from readily available materials. (p. 367) oxidative cleavage The cleavage of a carbon-carbon bond through oxidation. Carbon-carbon double bonds are commonly cleaved by ozonolysis/reduction or by warm, concentrated per­ manganate. (p. 360) oxymercuration The addition of aqueous mercuric acetate to an alkene. (p. 333) ""

/

/

""

C=C

HO +

Hg(OAc)2

I I

I I

-C -C -

+

HOAc

HgOAc

ozonolysis The use of ozone, usually followed by reduction, to cleave a double bond. (p. 366) peroxide effect The reversal of orientation of HBr addition to aLkenes in the presence of per­ oxides. A free-radical mechanism is responsible for the peroxide effect. (p. 329) peroxyacid (peracid) A carboxylic acid with an extra oxygen atom and a peroxy ( - 0 - 0 - ) linkage. The general formula is RC03H. (p. 353) polymer A high-molecular-weight compound composed of many molecules of a smaller, simpler compound called the monomer. (p. 363) polymerization: The reaction of monomer molecules to form a polymer.

8 Glossary

375

376

Chapter

8: Reactions of Alkenes regioselective reaction A reaction in which one direction of bond making or bond breaking occurs preferentially over all other directions. For example, the addition of HCl is regioselec­ tive, predicted by Markovnikov's rule. Hydroboration-oxidation is regioselective because it consistently gives anti-Markovnikov orientation. (p. 325) retrosynthetic analysis A method of working backward to solve multistep synthetic prob­ lems. (p. 367) Simmons-Smith reaction A cyclopropanation of an alkene using the carbenoid reagent generated from diiodomethane and the zinc-copper couple. (p. 35 1 )

o

CH212, Zn(Cu))

(J>

Simmons-Smith reaction

stereospecific reaction A reaction that converts different stereoisomers of the starting material into different stereoisomers of the product. (p.

I

341 )

Essentia l Problem-Solving Skil l s i n Chapter 8

1. Predict the products of additions, oxidations, reductions, and cleavages of alkenes, including (a) orientation of reaction (regiochemistry), (b) stereochemistry. 2. Propose logical mechanisms to explain the observed products of alkene reactions, including regiochemistry and stereochemistry. 3. Use retrosynthetic analysis to solve multistep synthesis problems with alkenes as reagents, intermediates, or products. 4. When more than one method is usable for a chemical transformation, choose the better method and explain its advantages. 5. Use clues provided by products of reactions such as ozonolysis to determine the struc­ ture of an unknown alkene. In studying these reaction-intensive chapters, students ask whether they should "memorize" all the reactions. Doing organic chemistry is like speaking a foreign language, and the reactions are our vocabulary. Without knowing the words, how can you construct sentences? Making flash cards often helps. In organic chemistry, the mechanisms, regiochemistry, and stereochemistry are our grammar. You must develop facility with the reactions, as you develop facility with the words and grammar you use in speaking. Problems and multistep syntheses are the sen­ tences of organic chemistry. You must practice combining all aspects of your vocabulary in solving these problems. Students who fail exams often do so because they have memorized the vocabulary, but they have not practiced doing problems. Others fail because they think they can do problems, but they lack the vocabulary. If you understand the reactions and can do the end-of-chapter problems without looking back, you should do well on your exams.

Study Problems 8-46

Define each term, and give an example. (a) dimerization (d) stereospecific addition (g) Markovnikov addition (j) hydrogenation (m) heterogeneous catalysis (p) hydroxylation (s) hydroboration (v) oxymercuration-demercuration (y) alkoxymercuration-demercuration

(b) polymerization (e) syn addition (h) anti-Markovnikov addition (k) hydration (n) halogenation (q) epoxidation (t) alpha elimination (w) carbene addition (z) monomer

(c) electrophilic addition (f) anti addition (i) peroxide effect (I) (0) (r) (u)

homogeneous catalysis halohydrin oxidative cleavage beta elimination (x) cationic polymerization (aa) addition polymer

377

Study Problems 8-47

Predict the major products of the fol lowing reactions, and give the structures of any intermediates. Include stereochem­ istry where appropriate.

Y'"

(oj

�

�

(d)

( I ) 03

(2)

m

(OJ

8-48

J

'

J

(h)

;.

U (:( eo

CH3C03H

GJ

(

(CH3)2S

PhC03H

(g) �

(

(b)

H +, HP

(

)

H2

(

----?

Pt

kJ

nJ

( I ) Hg(OAc)2' Hp

(2) NaBH4

0--

Br2 CCI4

V

0--

(c)

------0>

(2) H202, -OH

HP2

KMoO,, -OH

,

H', H,O

eo

(cold, dil)

( I ) 03

(2) (CH3)2S

( IJ

Cl2

------0>

H20

Propose mechanisms consistent with the following reactions.

(a) Y'" (c)

�

(e)

�

HBr ROOR

Br

~

"

(b)

Br

HBr

~ �

HCl CH30H

)

� OCH)

+

� CI

Br

(f)

(g)

Br2

�

2

H H

LiCI in CH30H

0

�c=c� H

'\..

(d)

+

------;>

�

OCH3

Q I �

H+

�

+

H CH -c-c( C HI HI 3

�

H2S04 Hp

() I �

CHEr3 NaOH

JyCI Jy +

Br

-o, -

"

)

�

()1 �

)

HCI ) ROOR

KMn04' -OH

05°4 -----?

(P J

( I ) BH3 . THF

(O V (;l U U

HBr

R�

U eo

0--

B' Br

"

378 8-49

Chapter 8: Reactions of Alkenes Show how you would synthesize each compound using methylenecyclohexane as your starting material .

(a) (d)

(g) 8-50

(fo U et

H

()B' � (c) V (fo � � Ck�'

methylenecyclohexane

(b)

O

(,)

(h)

OH

CH;

OH

Cl

OH

(n

0)

Limonene is one of the compounds that give lemons their tangy odor. Show the structures of the products expected when limonene reacts with an excess of each of these reagents.

limonene

8-51

(a) borane in tetrahydrofuran, followed by basic hydrogen peroxide (b) m-chloroperoxybenzoic acid (c) ozone, then dimethyl sulfide (d) a mixture of osmic acid and hydrogen peroxide (e) hot, concentrated potassium permanganate (1) peroxyacetic acid in water hydrogen and a platinum catalyst (g) (h) hydrogen bromide gas (i) hydrogen bromide gas in a solution containing dimethyl peroxide (j) bromine water (k) chlorine gas (I) mercuric acetate in methanol, followed by sodium borohydride (m) methylene iodide pretreated with the zinc-copper couple The structures of three monomers are shown. In each case, show the structure of the polymer that would result from polymerization of the monomer. Vinyl chloride is polymerized to "vinyl" plastics and PVC pipe. Tetrafluoroethylene polymerizes to Teflon ®, used as non-stick coatings and PTFE valves and gaskets. Acrylonitrile is polymerized to Orlon®, used in sweaters and carpets.

F" F

/F

/ C = C"

F

tetraftuoroethylene

*8-52

8-53 8-54

acrylonitrile

When styrene (vinylbenzene) is commercially polymerized, about 1-3% of 1 ,4-divinylbenzene is often added to the styrene. The incorporation of some divinylbenzene gives a polymer with more strength and better resistance to organ­ ic solvents. Explain how a very small amount of divinylbenzene has a marked effect on the properties of the polymer. The cationic polymerization of isobutylene (2-methylpropene) is shown in Section 8- 1 6A. Isobutylene is often polymer­ ized under free-radical conditions. Propose a mechanism for the free-radical polymerization of isobutylene. Poly(ethyl acrylate) has the formula

Give the structure of the ethyl acrylate monomer.

Study Problems 8-55

8-56

8-57

8-58

Draw the structures of the following compounds, and determine which member of each pair is more reactive toward the addition of HBr. (a) propene or 2-methylpropene (b) cyclohexene or I -methylcyclohexene (e) I -butene or 1 ,3-butadiene Cyclohexene is dissolved in a solution of l ithium chloride in chloroform. To this solution is added one equivalent of bromine. The material isolated from this reaction contains primarily a mixture of trans- l ,2-dibromocyclohexane and trans- l -bromo-2-chlorocyclohexane. Propose a mechanism to show how these compounds are formed. Draw a reaction-energy diagram for the propagation steps of the free-radical addition of HBr to isobutylene. Draw curves representing the reactions leading to both the Markovnikov and the anti-Markovnikov products. Compare the values of fl eD and Ea for the rate-limiting steps, and explain why only one of these products is observed. Give the products expected when the following compounds are ozonized and reduced.

ca)

8-59

ef

(b)

�

Cd)

(d)

a a

CH3 OH

O CQ

OH

(b)

OH H C1

�..

(e)

(e)

c::( CO Br

OH

OH

(f)

OCH3

OH

8-61

JS)

Show how you would make the following compounds from a suitable cyclic alkene. (a)

8-60

379

Unknown X, CSH9Br, does not react with bromine or with dilute KMn04. Upon treatment with potassium t-butoxide, X gives only one product, Y, CSH8. Unlike X, Y decolorizes bromine and changes KMn04 from purple to brown. Catalytic hydrogenation of Y gives methylcyclobutane. Ozonolysis-reduction of Y gives dialdehyde Z, CSH802. Propose consistent structures for X, Y, and Z. Is there any aspect of the structure of X that is still unknown? One of the constituents of turpentine is a-pinene, formula CJOH 1 6. The fol lowing scheme (called a "road map") gives some reactions of a-pinene. Determine the structure of a-pinene and of the reaction products A through E. E

I

A

C JOH 1S02 H 3 O'

D

C 1oH l 6O

1

C JOH l6Brz

PhC0 3 H

Be, CC14

j -flr a-pinene C 1 oH l6

( I ) 03 (2) (CH3)2S

Br2 Hp

1

B

C J OH 1 7OBr H,SO, heat

c

C IOH l SBr

CH3

8-62

The sex attractant of the housefly has the formula C23H46. When treated with warm potassium permanganate, this pheromone gives two products: CH3 ( CH2 ) 1 2COOH and CH3 (CH2hCOOH. Suggest a structure for this sex attractant. Explain which PaIt of the structure is uncertain.

380 8-63

Chapter 8: Reactions of Alkenes In contact with a platinum catalyst, an unknown alkene reacts with 3 equivalents of hydrogen gas to give l -isopropyl4-methylcyclohexane. When the unknown alkene is ozonized and reduced, the products are the following: 0

o

0

I

II

o

0

II

I

H - C - CH2 -C-C-CH3

H-C-H

0

I

II

CH3- C - CH2- C - H

Deduce the structure of the unknown alkene. *8-64

Propose a mechanism for the following reaction.

o 8-65

The two butenedioic acids are called fumaric acid (trans) and maleic acid (cis). 2,3-Dihydroxybutanedioic acid is called tartaric acid. H HOOC

/ " C=C / "

COOH H

HOOC H

/ " C=C " /

COOH

HOOC-CH-CH-COOH

I

OH

H

maleic acid

fumaric acid

I

OH

tartaric acid

Show how you would convert (a) fumaric acid to ( ± ) -tartaric acid. (b) fumaric acid to meso-tartaric acid. (c) maleic acid to ( ± ) -tartaric acid. (d) maleic acid to meso-tartaric acid. 8-66

The compound BD3 is a deuterated form of borane. Predict the product formed when l -methylcyc1ohexene reacts with BD3 . THF, followed by basic hydrogen peroxide.

8-67

A routine addition of HBr across the double bond of a vinylcyclopentane gave an unexpected rearranged product. Propose a mechanism for the formation of this product, and explain why the rearrangement occurs.

HBr

�

8-68

An unknown compound decolorizes bromine in carbon tetrachloride, and it undergoes catalytic reduction to give decalin. When treated with warm, concentrated potassium permanganate, this compound gives cis-cyclohexane- I ,2-dicarboxylic acid and oxalic acid. Propose a structure for the unknown compound.

co (X - -,-

� � 1J (1¥�

unknown compound

/J o"

J,

Co /JCd.)

decal in

II

o

COOH +

COOH cis cycl ohexane l 2 d icarboxyl ic acid

II

0

HO - C - C - OH oxalic acid

(�

further oxidation

)

Study Problems *8-69

381

Many enZymes catalyze reactions that are similar to reactions we might use for organic synthesis. Enzymes tend to be stereospecific in their reactions, and asymmetric induction is common. The following reaction, part of the tricarboxylic acid cycle of cell respiration, resembles a reaction we might use in the laboratory ; however, the enzyme-catalyzed reac­ tion gives only the (S) enantiomer of the product, malic acid. H

" / C

HO COOH H� / C

COOH

II

I

fumarase

C

CH2COOH (S)-malic acid

/ " HOOC H fumaric acid

COOD product in D20

(a) What type of reaction does fumarase catalyze? (b) Is fumaric acid crural? Is malic acid crural? In the enzyme-catalyzed reaction, is the product (malic acid) optically active?

"'8-70

(c) If we could run the preceding reaction in the laboratory using sulfuric acid as the catalyst, would the product (malic acid) be optically active? (d) Do you expect the fumarase enzyme to be a chiral molecule? (e) When the enzyme-catalyzed reaction takes place in D20, the only product is the stereoisomer just pictured. No enan­ tiomer or diastereomer of this compound is formed. Is the enzyme-catalyzed reaction a syn or anti addition? (0 Assume we found conditions to convert fumaric acid to deuterated malic acid using hydroboration with BD3 . THF, followed by oxidation with D202 and NaOD. Use Fischer projections to show the stereoisomer(s) of deuterated malic acid you would expect to be formed. (a) The following cyclization has been observed in the oxymercuration-demercuration of this unsaturated alcohol. Propose a mechanism for this reaction.

( I) Hg(OAc)2

(2) NaBH4

(b) Predict the product of formula C7HI3BrO from the reaction of this same unsaturated alcohol with bromine. Propose *8-71

8-72

a mechanism to support your prediction. An inexperienced graduate student treated 5-decene with borane in THF, placed the flask in a refrigerator, and left for a party. When he returned from the party, he discovered that the refrigerator was broken, and it had gotten quite warm inside. Although all the THF had evaporated from the flask, he treated the residue with basic hydrogen peroxide. To his surprise, he recovered a fair yield of I -decanol. Use a mechanism to show how this reaction might have occurred. (Hint: The addition of BH3 is reversible.) We have seen many examples where halogens add to alkenes with anti stereochemistry via the halonium ion mechanism. However, when l -phenylcyclohexene reacts with chlorine in carbon tetrachloride, a mixture of the cis and trans isomers of the product is recovered. Propose a mechanism, and explain this lack of stereospecificity.

+

I -phenylcycJohexene

transand cis1 ,2-dichloro- l -phenylcycJohexane

9

H

Al kynes

9- 1

Introduction

H

are hydrocarbons that contain carbon-carbon triple bonds. Alkynes are also called acetylenes because they are derivatives of acetylene, the simplest alkyne.

Alkynes

CH3CH 2 - C - C - H

CH 3 - C = C - CH3

acetylene

ethyl acetylene

dimethy lacetylene

ethyne

I -b utyne

H-C=C-H

2-butyne

The chemistry of the carbon-carbon triple bond is similar to that of the double bond. In this chapter, we see that alkynes undergo most of the same reactions as alkenes, especially the additions and the oxidations. We also consider reactions that are specific to alkynes: some that depend on the unique characteristics of the C - C triple bond, and others that depend on the unusual acidity of the acetylenic C - H bond. A triple bond gives an alkyne four fewer hydrogens than the corresponding alkane. Its molecular formula is like that of a molecule with two double bonds: C n H 2n 2 . There­ fore, the triple bond contributes two elements of unsaturation (eu) (Section 7-3). H '-..... /H H-C-C-H '-..... / H H

H '-..... /H /C = C '-..... H H

ethane, C 2H6 o e u, C Il H21l + 2

ethene, C 2H4 l eu, C Il H2 1l

H - C == C - H ethyne, C2 H2

2 e u, C n H21l

_

2

Alkynes are not as common in nature as alkenes, but some plants do use alkynes to protect themselves against disease or predators. Cicutoxin is a toxic compound found in water hemlock, and capillin protects a plant against fungal diseases. The alkyne func­ tional group is not common in drugs, but parsalmide is used as an analgesic, and ethynyl estradiol (a synthetic female honnone) is a common ingredient in birth control pills. Dynemicin A is an antibacterial compound that is being tested as an antitumor agent. HOCH2CH2CH2- C == C - C == C - CH = CH - CH=CH-CH=CH - CHCH2CH2CH3 cicu toxin

I

o

11 CH3-C ==C-C == C - C capillin

382

-0 � j

OH

9-2 Nomenclature of Alkynes

I

o COH

HO

OH

parsalmide

o

ethynyl estradiol

OH

dynernicin A

PROBLEM 9 - 1 Draw structural formulas of at least two alkynes of each molecular formula.

(a)

(c)

(b) C S H l 2

C6H IO

C7H I O

IUPAC

I U PAC Names The nomenclature for alkynes is similar to that for alkenes. We find the longest continuous chain of carbon atoms that includes the triple bond and change the -ane ending of the parent alkane to -yne. The chain is numbered from the end closest to the triple bond, and the position of the Iliple bond is designated by its lower­ numbered carbon atom. Substituents are given numbers to indicate their locations.

9-2

Nomenclature of Alkynes

CH3

IUPAC name:

H - C-C - H

C H3 - C - C - H

ethyne (acetylene)

propyne

I

Br I

CH3 - C - C - CH3

CH3- C H - C - C - C H2 - C H - CH3

2-butyne

6-bromo-2-111ethy1-3-heptyne

but-2-yne

6-bromo-2-methylhept-3-yne

When additional functional groups are present, the suffixes are combined to pro­ duce the compound names of the alkenynes (a double bond and a triple bond), alkynols (a triple bond and an alcohol), and so on. The new system (placing the number right before the group) helps to clarify these names. The rules give alkenes and alcohols higher priority than alkynes, so the numbering begins at the end closer to these higher-priority groups.

IUPAC IUPAC

H2C=C-C==C-CH3 I CH3

CH3-CH-C==C-H CH3-C-C - CH-CH2CH3 I I OH OCH3

IUPAC name:

2-methyl- l -penten-3-yne

3-butyn-2-o1

4-methoxy-2-hexyne

new IUPAC name:

2-methylpent- l -en-3-yne

but-3-yn-2-o1

4-methoxyhex-2-yne

The common names of alkynes describe them as derivatives of acetylene. Most alkynes can be named as a molecule of acetylene with one or two alkyl substituents. This nomenclature is like the common nomenclature for ethers, where we name the two alkyl groups bonded to oxygen. Common Names

H-C=C-H acetylene

CH3-C=C-H methylacetylene

(CH3hCH -C - C-CH(CH3h diisopropylacetylene

R-C=C-H an alkylacetylene

R-C-C-R' a dialkylacetylene

Ph-C=C-H

CH3-C C-CH2CH3

Ph -C - C - Ph

H-C-C-CH20H

phenyl acetylene

diphenylacetylene

elh y I melhy lacety lene

hydroxymethylacetylene ( propargyl alcohol )

383

384

Chapter 9:

Alkynes

Many of an alkyne's chemical properties depend on whether there is an acetylenic hydrogen ( H - C = C ) , that is, whether the triple bond comes at the end of a carbon chain. Such an alkyne is called a terminal alkyne or a terminal acetylene. If the triple bond is located somewhere other than the end of the carbon chain, the alkyne is called an internal alkyne or an internal acetylene.

[Ej---- C =C- CH2CH3

I -butyne, a terminal alkyne

(no acetylenic hydrogen)

CH3 -C=C-CH3 2-butyne, an internal alkyne

PROBLEM 9-2

For each molecular formula, draw all the isomeric alkynes, and give their IUPAC names. Cir­ cle the acetylenic hydrogen of each terminal alkyne. (a) C4H6 (two isomers) (b) CSH8 (three isomers)

9-3

Physical Properties of Alkynes

The physical properties of alkynes (Table 9-1 ) are similar to those of alkanes and alkenes of similar molecular weights. Alkynes are relatively nonpolar and nearly insoluble in water. They are quite soluble in most organic solvents, including acetone, ether, methyl­ ene chloride, chloroform, and alcohols. Many alkynes have characteristic, mildly offen­ sive odors. Acetylene, propyne, and the butynes are gases at room temperature, just like the corresponding alkanes and alkenes. In fact, the boiling points of alkynes are nearly the same as those of alkanes and alkenes with similar carbon skeletons. TABLE 9-1

Physical Properties of Selected Alkynes

Name

Structure

ethyne (acetylene) propyne 1 -butyne 2-butyne 1 -pentyne 2-pentyne 3-methyl- 1 -butyne I -hexyne 2-hexyne 3-hexyne 3,3-dimethyl-l-butyne I -heptyne l -octyne 1 -nonyne 1 -decyne

9-4

Commercial I mportance of Alkynes

9-4A

mp (0C)

bp (0C)

Density (g /cm 3)

-82 - 10 1 - 1 26 -32 -90 - 101

- 84 - 23 8 27 40 55 28 71 84 82 38 1 00 1 25 151 1 74

0.62 0.67 0.67 0.69 0.70 0.7 1 0.67 0.72 0.73 0.73 0.67 0.73 0.75 0.76 0.77

H - C-C - H H -C = C - CH3 H - C - C - CH2CH3 CH3 - C =C - CH3 H - C - C - CH2CH2CH3 CH3 - C = C - CH2CH3 CH3 - CH(CH3 ) - C - C - H H - C =C - (CH2h - CH3 CH3 - C - C -CH2CH2CH3 CH3CH2-C - C-CH2CH3 (CH3hC-C - C - H H - C == C - (CH2)4CH3 H -C = C - (CH2)sCH3 H - C = C - (CH2)6CH3 H - C = C - (CH2hCH3

- 1 32 -90 - 10 1 -81 -81 -79 -50 -36

Uses of Acetylene and Methylacetylene

Acetylene is by far the most important commercial alkyne. Acetylene is an important industrial feedstock, but its largest use is as the fuel for the oxyacetylene welding torch. Acetylene is a colorless, foul-smelling gas that burns in air with a yellow, sooty flame. When the flame is supplied with pure oxygen, however, the color turns to light blue, and flame temperature increases dramatically. A comparison of the heat of

9-4 Commercial I mport ance of Alkynes

385

combustion for acetylene with those of ethene and ethane shows why this gas makes an excellent fuel for a high-temperature flame. !::.. HO

- 1 561 kJ divided by 5 moles of products H2C = CH2

3 O2 � 2 CO2 +2 H20 !::.. Ho - 1 4 1 0 kJ divided by 4 moles of products

+

HC = CH +

� O2 � 2 CO2 + I H20 !::.. Ho - 1 326 kJ divided by 3 moles of products

=

=

=

=

- 1 56 1 kJ ( -373 kcal) - 3 1 2 kJ/mol of products ( -75 kcal/mol)

- 1 4 1 0 kJ ( - 337 kcal ) -352 kJ/mol of products ( - 84 kcal/mol )

=

=

- 1 326 kJ ( - 3 1 7 kcal ) -442 kJ/mol of products ( - 1 06 kcal/mol )

If we were simply heating a house by burning one of these fuels, we might choose ethane as our fuel because it produces the most heat per mole of gas consumed. In the welding torch, we want the highest possible temperature of the gaseous prod­ ucts. The heat of reaction must raise the temperature of the products to the flame tem­ perature. Roughly speaking, the increase in temperature of the products is proportional to the heat given off per mole of products formed. This rise in temperature is largest with acetylene, which gives off the most heat per mole of products. The oxyacetylene flame reaches temperatures as high as 2800°C. When acetylene was first used for welding, it was considered a dangerous, explosive gas. Acetylene is thermodynamically unstable. When the compressed gas is subjected to thermal or mechanical shock, it decomposes to its elements, releas­ ing 234 kJ (56 kcal) of energy per mole. This initial decomposition often splits the container, allowing the products (hydrogen and finely divided carbon) to burn in the air. H-C=C-H

�

2C

+

H2

!::.. HO !::.. HO

=

=

-234 kJ/mol ( -56 kcal/mol ) - 1 090 kJ/mol 2 ( -26 1 kcal/mol )

Acetylene is safely stored and handled in cylinders that are filled with crushed firebrick wet with acetone. Acetylene dissolves freely in acetone, and the dissolved gas is not so prone to decomposition. Firebrick helps to control the decomposition by minimizing the free volume of the cylinder, cooling and controlling any decomposi­ tion before it gets out of control. Methylacetylene also is used in welding torches. Methylacetylene does not decompose as easily as acetylene, and it burns better in air (as opposed to pure oxygen). Methylacetylene is well suited for household soldering and brazing that requires higher temperatures than propane torches can reach. The industrial synthe­ sis of methylacetylene gives a mixture with its isomer, propadiene (allene). This mixture is sold commercially under the name MAPP® gas (MethylAcetylene­ ProPadiene).

9-8

CH3 - C == C - H

H2C = C = CH2

methylacetylene

propadiene (allene)

M a n ufacture of Acetylene

Acetylene, one of the cheapest organic chemicals, is made from coal or from natural gas. The synthesis from coal involves heating lime and coke (roasted coal) in an

An oxygen-acetylene flame is hot enough to melt steel for welding. A cutting torch uses an extra jet of oxygen to burn away hot steel.

386

Chapter 9: Alkynes

electric furnace to produce calcium carbide. Addition of water to calcium carbide pro­ duces acetylene and hydrated lime. 3 C + CaO l i me coke

electric furnace, 25000e

CaC2

CO

+

calcium carbide

H - C = C - H + Ca ( OH h hydrated lime

acelylene

This second reaction once served as a light source in coal mines until battery­ powered lights became available. A miner's lamp works by allowing water to drip slowly onto some calcium carbide. Acetylene is generated, feeding a small flame where the gas burns in air with a yellow flickering light. Unfortunately, this flame ignites the methane gas commonly found in coal seams, causing explosions. Battery-powered miner's lamps provide better light and reduce the danger of methane explosions. The synthesis of acetylene from natural gas is a simple process. Natural gas con­ sists mostly of methane, which forms acetylene when it is heated for a very short peri­ od of time. 2 CH4

l .soooe 0.01 sec

)

H-C-C-H

+

3 H2

Although this reaction is endothermic, there are twice as many moles of products as reactants. The increase in the number of moles results in an increase in entropy, and the ( - Tt:.S) term in the free energy ( t:.G t:.H - Tt:.S) predominates at this high temperature. =

PROBLEM 9 -3

Carbide lamp on a miner's cap, used around 1 920.

9 -5

Electronic Structure of Alkynes

What reaction would acetylene likely undergo if it were kept at 1 500°C for too long?

In Section 2-4, we studied the electronic structure of a triple bond. Let's review this structure, using acetylene as the example. The Lewis structure of acetylene shows three pairs of electrons in the region between the carbon nuclei: H:C:: :C:H Each carbon atom is bonded to two other atoms, and there are no nonbonding valence electrons. Each carbon atom needs two hybrid orbitals to form the sigma bond frame­ work. Hybridization of the s orbital with one p orbital gives two hybrid orbitals, directed 1 80° apart, for each carbon atom. Overlap of these sp hybrid orbitals with each other and with the hydrogen s orbitals gives the sigma bond framework. Experi­ mental results have conftrmed this linear ( 1 80°) structure. H

---

C

-----

C

---

H

Two pi bonds result from overlap of the two remaining unhybridized p orbitals on each carbon atom. These orbitals overlap at right angles to each other, forming one pi bond with electron density above and below the C - C sigma bond, and the other with electron density in front and in back of the sigma bond. The shape of these pi bonds is such that they blend to form a cylinder of electron density encircling the sigma bond between the two carbon atoms.

9-6

H

Acidity of Alkynes; Formation of Acetylide Ions

387

H overlap of p orbitals

cylinder of electron density

The carbon-carbon bond length in acetylene is 1.20 A, and each carbon-­ hydrogen bond is 1.06 A. Both bonds are shorter than the corresponding bonds in ethane and in ethene. 54

H. r/l .H A H� f_ /C-C"". H \ \� H 1 .09

A

ethane

r l .33 A

H '-.. f _ / H /.C=C H � '-... H 1 .08

I.20 A

C H-C=C-H

(

A

ethene

,MA

ethyne

The triple bond is relatively short because of the attractive overlap of three bonding pairs of electrons and the high s character of the sp hybrid orbitals. The sp hybrid orbitals are about one-half s character (as opposed to one-third s character of sp 2 hybrids and one-fourth of sp 3 hybrids), using more of the closer, tightly held s orbitals. The sp hybrid orbitals also account for the slightly shorter C-H bonds in acetylene compared with ethylene. Terminal alkynes are much more acidic than other hydrocarbons. Removal of an 9-6 acetylenic proton forms an acetylide ion, which plays a central role in a1kyne chemistry. The acidity of an acetylenic hydrogen stems from the nature of the sp Acidity of Alkynes; hybrid =C-H bond. Table 9-2 shows how the acidity of a C-H bond Formation of varies with its hybridization, increasing with the increasing s character of the Acetylide Ions TABLE 9-2 Compound

Conjugate Base

H H I I H-C-C- H I I H H /H H '-. . /C=C '-... H H :NH3

H H I I H -C-C<JI I H H H '-... C=C 0/ '-... H H : NH;

H -C=:C- H R-OH

H -C=:C<JR-O:-

Hybridization

s

Character

p Ka

sp3

25%

50

sp2

33%

44 35

(ammonia)

sp (alcohols)

weakest acid

50%

25 16-18

stronger acid

388

Chapter 9: Alkynes

orbitals : sp 3 < sp 2 < sp. (Remember that a smaller value of pKa corresponds to a stronger acid.) The acetylenic proton is about 10 19 times as acidic as a vinyl proton. Abstraction of an acetylenic proton gives a carbanion that has the lone pair of electrons in the sp hybrid orbital. Electrons in this orbital are close to the nucleus, and there i s less charge separation than in carbanions with the lone pair in sp 2 or sp 3 hybrid orbitals. Ammonia and alcohols are included for comparison; note that acetylene can be deprotonated by the amide CNH2) ion, but not by an alkoxide ion COR ) . Very strong bases (such as sodium amide) deprotonate terminal acetylenes to form carbanions called acetylide ions (or alkynide ions). Hydroxide ion and alkoxide ions are not strong enough bases to deprotonate alkynes. Internal alkynes do not have acetylenic protons, so they do not react.

CH3CH2- C=C- H

O

CH3CH2 -C - C : - +Na

---7

NaNH2

+

O

---7

NH3

C=C ,

+Na

+

NH3

sodium cyclohexylacetylide

sodium amide

cyclohexylacetylene

+

sodium butynide

sodium amide

I -butyne, a termi nal alkyne

C=C-H

NaNH2

+

NaNHo

CH3-C=C-CH3

-)

no reaction

(no acetylenic proton) 2-butyne, an internal alkyne

Sodium amide ( Na+ - : N H 2 ) is frequently used as the base in forming acetylide salts. The amide ion CNH 2 ) is the conjugate base of ammonia, a compound that is itself 3 35). One a base. Ammonia is also a very weak acid, however, with Ka 10- 5 ( pKa of its hydrogens can be reduced by sodium metal to give the sodium salt of the amide ion, a very strong conjugate base. =

=

H

I

H - N-H

+

Na

ammonia

R - C - C- H

+

Na+ - : NH 2

Fe

3

H + catalyst

)

I

Na+ -: N - H

+

-t H2 i

sodium amide ("sodamide")

R - C == C : - +Na

+

: NH3

a sodium acetylide

Acetylide ions are strong micleophiles. In fact, one of the best methods for syn­ thesizing substituted alkynes is a nucleophilic attack by an acetylide ion on an unhin­ dered alkyl halide. We consider this displacement reaction in detail in Section 9-7 A.

P R O B L E M 9-4

The boiling points of I -hexene (64°C) and I -hexyne (7 1 °C) are sufficiently close that it is difficult to achieve a clean separation by distillation. Show how you might use the acidity of I -hexyne to remove the last trace of it from a sample of I -hexene.

9-7 Synthesis of Alkynes from Acetylides PROBLEM 9-5

Predict the products of the following acid-base reactions, or indicate if no significant reaction would take place. (b) H - C = C - H + CH3Li

(a) H - C = C - H + NaNH2 (c) H - C = C - H + NaOCH3 (e) H - C = C:- +Na + CH30H

(d) H - C = C - H + NaOH

(g) H - C = C:- +Na + H2C = CH2

(h) H 2 C = CH2 + NaNH2

(0 H - C = e:- �a + H20

(i) CH30H + NaNH2

Two different approaches are commonly used for the synthesis of alkynes. In the first, an appropriate electrophile undergoes nucleophilic attack by an acetylide ion. The electrophile may be an unhindered primary alkyl halide (undergoes SN2), or it may be a carbonyl compound (undergoes addition to give an alcohol). Either reaction joins two fragments and gives a product with a lengthened carbon skeleton. This approach is used in many laboratory syntheses of alkynes. The second approach forms the triple bond by a double dehydrohalogenation of a dihalide. This reaction does not enlarge the carbon skeleton. Isomerization of the triple bond may occur (see Section 9-8), so dehydrohalogenation is useful only when the desired product has the triple bond in a thermodynamically favored position.

9-7A

9-7

Synthesis of Alkynes from Acetylides

A l kylatio n of Acetyl ide Ions

An acetylide ion is a strong base and a powerful nucleophile. It can displace a halide ion from a suitable substrate, giving a substituted acetylene. �

- :R- C=C

LA

+ R '- X (R '- X must be

R- C = C-R'

+

X-

a primary alkyl halide)

If this SN2 reaction is to produce a good yield, the alkyl halide must be an excellent SN2 substrate: It must be primary, with no bulky substituents or branches close to the reaction center. In the following examples, acetylide ions displace primary halides to form elongated alkynes. H- C sodium acetyl ide

O

l -bromobutane

C - CH2CH2C H2CH3 I -hexyne (buty lacety lene) (75%)

C=C-H

ethynylcyclohexane (eyclohexylacetylene)

(I)

NaNH2

(2) ethyl bromide

l -cyclohexyl-l -butyne (ethyJcyclohexy Jaeety lene) (70%)

+

N aBr

389

390

Ch apter 9: Alkynes

If the back-side approach is hindered, the acetylide ion may abstract a proton, giving elimination by the E2 mechanism. Br

CH3CH2- C ==C : butynide ion

PROBLEM-SOLVING

+

I

E2

H3C -CH - CH3

�

butyne

isopropyl bromide

Htnp

Alkylation of acetyl ide ions is an excel l ent way of length ening a carbon cha in. The triple bond can l ater be reduced (to an a l kane or an a l k ene) if needed.

propene

S O LV E D P R O B L E M 9 - 1

Show how to synthesize 3-decyne from acetylene and any necessary alkyl halides. SOLUTI O N

Another name for 3-decyne is ethyl n-hexylacetylene. It can be made by adding an ethyl group and a hexyl group to acetylene. This can be done in either order; we begin by adding the hexyl group. H - C =C - H acetylene CH3 (CH2)s - C - C - H l - octyne

( I ) NaNH2

( J )NaNH2 ) (2)CH3CH2Br

CH3 (CH2)s - C = C - H l - octyne CH3 (CH2)s - C = C - CH2CH3 3- decyne

P R O B L E M 9-6

Show the reagents and intermediates involved in the other order of synthesis of 3-decyne, by adding the ethyl group first and the hexyl group last. PRO BLEM 9-7

Show how you might synthesize the following compounds, using acetylene and any suitable alkyl halides as your starting materials. If the compound given cannot be synthesized by this method, explain why. (b) 2-hexyne (a) I -hexyne (c) 3-hexyne (d) 4-methyl-2-hexyne (e) 5-methyl-2-hexyne (f) cyclodecyne

9-7B

Addition of Acetylide Ions to Carbonyl G roups

Like other carbanions, acetylide ions are strong nucleophiles and strong bases. In ad­ dition to displacing halide ions in SN2 reactions, they can add to carbonyl ( C = O ) groups. Figure 9- 1 shows the structure of the carbonyl group. Because oxygen i s more electronegative than carbon, the C = 0 double bond is polarized. The oxygen atom has a partial negative charge balanced by an equal amount of positive charge on the carbon atom.

� F i g u re 9-1

The C = O double bond of a carbonyl group resembles the C = C double bond of an alkene; however, the carbonyl double bond is strongly polarized. The oxygen atom bears a partial negative charge, and the carbon atom bears a partial positive charge.

R .. , 8+ 8" 'IIC = O: "'" +------>' R � = 2.4 D

9-7 Synthesis of Alkynes from Acetylides

The positively charged carbon i s electrophilic; attack by a nucIeophile places a negative charge on the electronegative oxygen atom.

I I

.. : Nuc -C - O

I

alkoxide

I I

l) '

"

.. ..

carbonyl group is used in the syn· to cause

..

Nuc - C-OH

The addition of an acetyl ide ion to a thesis of ethchlorvynol, a drug used sleep.

drowsiness and induce

Ethchlorvynol

is

relatively

nonpolar, enhancing its distribution

The product of this nucleophilic attack is an alkoxide ion, a strong base. (An alkoxide ion is the conjugate base of an alcohol, a weak acid.) Addition of water or a dilute acid protonates the alkoxide to give the alcohol.

I .. "" .. Nuc -C - O : - H - O - H

391

+

into the fatty tissue of the central nervous system .

-: O .. H

An acetylide ion can serve as the nucIeophile in this addition to a carbonyl group. The acetylide ion adds to the carbonyl group to form an alkoxide ion. Addition of dilute acid (in a separate step) protonates the alkoxide to give the alcohol. R

�

R'-C == C: -

+

� .,

R

acetyli de

R

" 'C = O'

I I

R'- C == C -C-O: .. -

----7

..

R

aldehyde or ketone

R

I .. R '-C == C - C - O: I ..

R

H 30+

I I

R'-C==C- C - OH

�

R

R an acetylenic alcohol

An acetylide adds to formaldehyde ( H2 C = O ) to give (after the protonation step) a primary alcohol with one more carbon atom than there was in the acetyl ide. H R - C == C: ,

+

H

'"

H

I I

.. R'-C == C - C-O:

C = O. .' �

'

H

� formaldehyde

..

H

I . R'-C - C - C - O. :I ..

H

I I

R'-C==C- C- OH

H

Example

H

a 1 0 alcohol

( I ) NaNH2

CH3- C =C - H propyne

(2) H ?- C = O

PROBLEM-SOLVING

)

CH3-C - C -CH2-OH 2-butyn- l -ol ( 1 °)

An acetylide adds to an aldehyde to give, after protonation, a secondary alcohol. The two groups of the secondary alcohol are the acetylide and the alkyl group that was bonded to the carbonyl group of the aldehyde.

HtltP

N umbers ( 1 ), (2), (3), etc. are used to show a sequence of separate reactions over a single arrow. If the num bers were omitted, it would incorrect ly imply mixing all these reagents together, rather than adding them in separate steps.

392

Chapter 9: Alkynes

R '- C == C :

-

R I .. R'-C==C-C - O :-

.. " C = O. . \.....; H R

+

..

I

H

an aldehyde

�

R

I .. R'-C==C-C - O: I ..

R'-C

R I C- C-OH

I

H

H

a 2° alcohol

Example

CH3 I CH3- CH - C = C - H

Ph

CH3

I

( I ) NaNH? (2) PhCHO

I

CH3-CH -C=C- CH - OH

(3) H 30+

3-methyl- l -butyne

4-methyl- l -phenyl-2-pentyn- l -ol (2°)

A ketone has two alkyl groups bonded to its carbonyl carbon atom. Addition of an acetylide, followed by protonation, gives a tertiary alcohol. The three alkyl groups bond­ ed to the carbinol carbon atom (the carbon bearing the - OH group) are the acetylide and the two alkyl groups originally bonded to the carbonyl group in the ketone.

R '- C == C:

-

"

R I .. R '- C==C- C - O : -

R

+

.. C = O.. \.....; R

----7

..

I

R

a ketone

�

R

I

.. R'-C==C- C - O: I ..

R H 3 0+

I

)

R '- C == C - C - OH

I

R

R a 3° alcohol

Example

0

OH

( I ) Na+ - : C=C - H

(2) HP+

cyclohexanone

)

C=C-H

l -ethynylcyclohexanol (3°)

S O LV E D P R O B L E M 9 - 2

Show how you would synthesize the following compound, beginning with acetylene and any necessary additional reagents.

9-8 Synthesis o f A l kynes by Elimination Reactions

393

S O LUT I O N

We need to add two groups to acetylene: an ethyl group and a six-carbon aldehyde (to form the secondary alcohol). If we formed the alcohol group first, the weakly acidic - OH group would interfere with the alkylation by the ethyl group. Therefore, we should add the less reactive ethyl group first, and add the alcohol group later in the synthesis.

The ethyl group is not acidic, and it does not i nterfere with the addition of the second group:

NaNH,

-)

P RO B L E M 9-8

Show how you would synthesize each compound, beginning with acetylene and any neces­ sary additional reagents. (a) 2-propyn- l -ol (propargyl alcohol) (b) 2-phenyl-3-butyn-2-ol

OH

I

CH3-C-C=C-H I Ph

PROBLEM-SOLVING

(d) 3-methyl-4-hexyn-3-01

(c) 2-heptyn-4-ol

OH

I

CH3CH2 -C-C=C-CH3 I CH3

In some cases, we can generate a carbon-carbon triple bond by eliminating two mole­ cules of HX from a dihalide. Dehydrohalogenation of a geminal or vicinal dihalide gives a vinyl halide. Under strongly basic conditions, a second dehydrohalogenation may occur to form an alkyne.

H H I I R-C-C-R'

I

X

I

base - HX

(fast)

X

a geminal dihalide

base - HX

(fast)

H ", / R' /C=C ", R X vinyl halide

9-8

Synthesis of Alkynes by Elimination Reactions

base - HX

(slow)

)

R - C = C -R' alkyne

vinyl halide

a vicinal dihalide

H X I I R - C -C-R' I I H X

H ", / R' /C = C ", R X

base - HX

(slow)

)

Htlti/

If a synthesis req u i res both a l kylation of a n acetylide a nd addition to a carbonyl, add the less reactive group fi rst: a l kylate, then add to a carbonyl. In genera l, you should add reactive functio n a l groups late in a synthesis.

R - C = C - R' alkyne

394

Chapter 9: Alkynes Con d itions for E l i m ination We have already seen (Section 7-9A) many examples of dehydrohalogenation of alkyl halides. The second step is new, however, because it involves dehydrohalogenation of a vinyl halide to give an alkyne. This second dehydro­ halogenation occurs only under extremely basic conditions-for example, fused (molten) KOH or alcoholic KOH in a sealed tube, usually heated to temperatures close to 200°C. Sodium amide is also used for the double dehydrohalogenation. Since the amide ion C: N H2) is a much stronger base than hydroxide, the amide reaction takes place at a lower temperature. The following reactions are carefully chosen to form products that do not rearrange (see below).

Br

I

Br

I

CH3 -CH 2 - CH -CH - CH3 2,3-dibrol1lopentane

CH3 -CH2 - CH2 -CH2 -CHCl2 l , l -dichloropentane

KOH (fused) ) 200°C

( I ) NaNH2 , I SO° C (2) Hp

CH3- CH2-C

C-CH3

2-pentyne (4S%)

CH3 -CH2 -CH2 - C == C - H l -pentyne (SS%)

Unfortunately, the double dehydrohalogenation is limited by the severe conditions. Any functional groups that are sensitive to strong bases cannot survive; also, the alkyne products may rearrange under these extremely basic conditions. Figure 9-2 shows how the loss of protons at one carbon atom and their replacement elsewhere leads to isomerization of the triple bond. This ability to isomerize implies that all the possible triple-bond isomers will equilibrate, and the most stable iso­ mer will predominate. The most stable alkyne isomer is usually the internal alkyne or a mixture of internal alkynes. Any of several isomers of dibromopentane give mostly 2-pentyne on dehydrohalo­ genation with fused KOH at 200°C. In each case, the alkyne initially formed rearranges to the most stable isomer, 2-pentyne.

Base-Catalyzed Rearrangements

�f� C-R B:-

R-C - C

+

I

H resonance-stabi lized carbanlon

an acetylene

R-C

Yf�B : -

C-C-R

I

+

H an isomerized acetylene � Fig u re 9-2

resonance-stabilized carbanion

Under extremely basic conditions, an acetylenic triple bond can migrate along the carbon chain by repeated deprotonation and reprotonation.

3

9-8 Synthesis of A l kynes by Elimination Reaction s

r

CH -CH -CH - CH 2CH3 I I Br Br

Br- H - CH2CH2CH2CH3

2,3-dibromopentane

I , I -dibromopentane

Br

Br

I CH3- C - CH,CH,CH3 - I

rH2- rH - CH2CH2CH3

2,2-dibromopentane

1 ,2-dibromopentane

Br

395

Br

KOH, 200°C

CH3-C

C -CH2CH3

2-pentyne

Br

PROB L E M 9-9

Give a mechanism to show how l , l -dibromopentane reacts with fused KOH at 200°C to give 2-pentyne.

P RO B L E M 9 - 1 0

'

( 1 7 kllmol) more stable than their corresponding tenninal alkynes. Calculate the ratio of Using heats of hydrogenation, we can show that most internal alkynes are about 4 kcallmol

terminal alkyne to internal alkyne present at equilibrium at 200°C.

Isomerization also results when sodium amide is used as the base in the double dehydrohalogenation. All possible triple-bond isomers are formed, but sodium amide is such a strong base that it deprotonates the terminal acetylene, removing it from the equilibrium. The acetylide ion becomes the favored product. When water is added to quench the reaction, the acetylide ion is protonated to give the terminal alkyne . R - C == C - H

..

+

R-C

R - C == C : ­

: NH2

CH3(CH2 )4- C - C - CH3 2-octyne

( I ) NaNH 2 , I SO°C (2) H 20

C-H

major product

acetylide ion (major component)

one component of the mixture

CHiCH2 )4-CH2- C == C - H l -octyne (80%)

P RO B LE M 9 - 1 1

(a) Propose a mechanism to show how 2-pentyne reacts with sodium amide to give I -pentyne. (b) Explain how this reaction converts a more stable isomer (2-pentyne) to a less stable isomer ( l -pentyne). (c) How would you accomplish the opposite reaction, converting l -pentyne into 2-pentyne? PROBLEM 9- 1 2

Show which of the fol lowing compounds could be synthesized in good yield by a double dehydrohalogenation from a dihalide. In each case, 1. Show which base you would use (KOH or NaNH2)' 2. Show how your starting material might be synthesized from an alkene. (a) 2-butyne (b) l -octyne (d) cyclodecyne (c) 2-octyne

PROBLEM-SOLVING R"

\1 )

C =C

/

H

I

KOH, 200°C

most stable internal alkyne

/X

H?nP"

"

R'

NaNH2, I S00C (2) HP

terminal alkyne

396

Chapter 9 : Alkynes

I

SUMMARY

Syntheses of Alkynes

1. Alkylation of acetylide ions (Section 9-7 A) R - C = C - R' + R - C = C:- + R' - X (R' - X must be an unhindered primary halide or tosylate)

X-

Example

2-hexyne

I-bromopropane

sodium propynide

2. Additions to carbonyl groups (Section 9-7B)

+

R' R'

" /

R'

R'

I I R'

I .. R-C==C-C-O:I ..

C= d.:

R-C=C-C-OH

R'

o

Example

I

+ CH3CH2-C-H

H-C==C-CH-CH2CH3 I -pentyn-

propanal

sodium acetylide

3. Double dehydrohalogenation ofalkyl dihalides (Section 9-8) X

X

I I R- C - C - R ' I I H

H

or

X

I I R-C-C-R' I I H

H

3-01

OH

I

fused KOH

)

R - C == C - R'

X

(KOH forms internal alkynes; NaNH2 forms terminal alkynes.) Examples

CH3CH2-CHz-CCl2- CH3

KOH, 200°C (fused)

) ----'....:. --' ..�

2,2-dichloropentane

CH3CH2 - CH2 - CCl2 - CH3 2,2-dichloropentane

9-9

Addition Reactions of Alkynes

NaNH2 , 1 50°C ----�)

CH3CH2 - C = C - CH3 2-pentyne CH3CHz- CH2- C = C - H I -pentyne

We have already discussed some of the most important reactions of alkynes. The nucleophilic attack of acetylide ions on electrophiles, for example, is one of the best methods for making more complicated alkynes (Section 9-7) . Now we consider reac­ tions that involve transformations of the carbon-carbon triple bond itself. Many of the reactions of alkynes are similar to the corresponding reactions of alkenes because both involve pi bonds between two carbon atoms. Like the pi bond of an alkene, the pi bonds of an alkyne are electron-rich, and they readily undergo addi­ tion reactions. Table 9-3 shows how the energy differences between the kinds of carbon-carbon bonds can be used to estimate how much energy it takes to break a par­ ticular bond. The bond energy of the alkyne triple bond is only about 226 kJ (54 kcal) more than the bond energy of an alkene double bond. This is the energy needed to break one of the pi bonds of an alkyne. Reagents add across the triple bonds of alkynes just as they add across the dou­ ble bonds of alkenes. In effect, this reaction converts a pi bond and a sigma bond into

9-9

TABLE 9-3

Approximate Bond Energies of Carbon-Carbon Bonds

Bond

Total Energy

Class of Bond

Approximate Energy

c-c c=c c=c

347 kJ (83 kcal) 6 1 1 kJ ( 146 kcal) 837 kJ (200 kcal)

alkane sigma bond alkene pi bond second alkyne pi bond

347 kJ (83 kcal) 264 kJ (63 kcal) 226 kJ (54 kcal)

Addition Reactions of Alkynes

two sigma bonds. S ince sigma bonds are generally stronger than pi bonds, the reaction is usually exothermic. Alkynes have two pi bonds, so up to two molecules can add across the triple bond, depending on the reagents and the conditions.

�

R - C=C - R'

+

t � bon

�

A

A-B

R

"

/

)'B

C=C

"'-

A-B

A

B

A

B

I I R - C - C -R' I I

)

R'

We must consider the possibility of a double addition whenever a reagent adds across the triple bond of an alkyne. Some conditions may allow the reaction to stop after a single addition, while other conditions give double addition. 9-9A

Catalytic H yd rogen ation to A l ka nes

In the presence of a suitable catalyst, hydrogen adds to an alkyne, reducing it to an alkane. For example, when either of the butyne isomers reacts with hydrogen and a platinum catalyst, the product is n-butane. Platinum, palladium, and nickel catalysts are commonly used in this reduction.

R - C = C - R'

Pt, Pd,

+ 2 H2

or

Ni

)

H

H

H

H

I I R - C - C - R' I I

Examples

H - C = C - CH2CH3

+

2 H2

Pt

�

I -butyne

CH3 - C - C - CH3 2-butyne

H - CH2 - CH2 - CH2CH3 butane ( 1 00%)

+

2 H2

Pt

�

CH3 - CH2 - CH2 - CH3 butane ( 1 00%)

Catalytic hydrogenation takes place in two stages, with an alkene intermediate. With efficient catalysts such as platinum, palladium, or nickel, it is usually impossible to stop the reaction at the alkene stage.

R - C=C-R'

H

H

H

H

I I R - C - C - R' I I

397

398

Chapter 9 : Alkynes

R- C== C - R

,

� F i g u re 9-3

,

H H - - -'± '

Catalytic hydrogenation of alkynes using the L i nd lar cataly s t.

Pd (partially poisoned)

9-9 8

Cata lytic Hydrogenation to cis Al kenes

Hydrogenation of an alkyne can be stopped at the alkene stage by using a "poisoned" (partially deactivated) catalyst made by treating a good catalyst with a compound that makes the catalyst less effective. Lindlar's catalyst is a poisoned palladium catalyst, composed of powdered barium sulfate coated with palladium, poisoned with quinoline. Nickel boride ( Ni2B ) is a newer alternative to Lindlar's catalyst that is more easily made and often gives better yields. R R-C=C-R' alkyne

H

'" /

C=C

/

'"

R' H

cis alkene

quinoline (Lindlar' s catalyst)

The catalytic hydrogenation of alkynes is sirnilar to the hydrogenation of alkenes, and both proceed with syn stereochemistry. In catalytic hydrogenation, the face of a pi bond contacts the solid catalyst, and the catalyst weakens the pi bond, allowing two hydrogen atoms to add (Figure 9-3). This simultaneous (or nearly simultaneous) addition of two hydrogen atoms on the same face of the alkyne ensures syn stereochemistry. In an internal alkyne, syn addition gives a cis product. For example, when 2-hexyne is hydrogenated using the Lindlar catalyst, the product is cis-2-hexene.

Lincllar' s catal y st .

)

cis-2-hexene

2-hexyne Lindlar's catalyst .

l -hexyne

9-9C

)

H'-. / (CH2)3CH3 , /C = C '-., H H l -hexene

M etal-Ammonia Red ucti on to trans Al kenes

To form a trans alkene, two hydrogens must be added to the alkyne with anti stereo­ chemistry. Sodium metal in liquid ammonia reduces alkynes with anti stereochem­ istry, so this reduction is used to convert alkynes to trans alkenes. R-C=C-R' alkyne

+

NalNH3 trans alkene

9-9 Addition Reactions of Alkynes Example

H3C ",--/H /C = C ",--H (CH2)4CH3 2-octyne

trans-2-octene (80%)

Ammonia (bp -33°C) is a gas at room temperature, but it is kept liquid by using dry ice to cool the reaction vessel. A s sodium dissolves in liquid ammonia, it gives up electrons, which produce a deep blue color. It is these solvated electrons that actually reduce the alkyne. NH3

+

Na

NH3 · e-(deep blue solution)

�

+

Na+

solvated electron

The metal-ammonia reduction proceeds by addition of an electron to the alkyne to form a radical anion, followed by protonation to give a neutral radical. Protons are provided by the ammonia solvent or by an alcohol added as a cosolvent. Addition of another electron, followed by another proton, gives the product. MECHANIS M 9-1

Metal-Am monia Red uction of an Al kyne

This mechanism involves addition of an electron, followed by a proton, then addition of a second electron, followed by a second proton.

Step

1:

An electron adds to the alkyne, forming a radical anion.

�

R - C"-" := C - R'

e

R ",---

�

alkyne

0

C=C "'---

-0

R'

radical anion

Step 2: The radical anion is protonated to give a radical.

R ",--0 C=C

/H H-N :

-O-----Y v "H Step

3:

+

vinyl radical

An electron adds to the radical, forming an anion. R" 0\e /C = C " R' H

�

vinyl anion

Step

4:

Protonation of the anion gives an alkene. R" G� /H H- N " : /C = C " H R' H

v

+

�

trans alkene

399

400

Chapter 9: Alkynes The anti stereochemistry of the sodium-ammonia reduction appears to result from the greater stability of the vinyl radical in the trans configuration, where the alkyl groups are farther apart. An electron is added to the trans radical to give a trans vinyl anion, which is quickly protonated to the trans alkene. PROBLEM 9- 1 3

Show how you would convert (a) 3-octyne to cis-3-octene (c) cis-cyc1odecene to trans-cyc1odecene

9-9 0

(b) 2-pentyne to trans-2-pentene (d) trans-2-pentene to cis-2-pentene

Addition of Ha logens

Bromine and chlorine add to alkynes just as they add to alkenes. If 1 mole of halogen adds to an alkyne, the product is a dihaloalkene. The stereochemistry of addition may be either syn or anti, and the products are often mixtures of cis and trans isomers.

R-C=C-R'

+

(X2

=

� X2 Cl2 or Br2)

+

Example

If 2 moles of halogen add to an alkyne, a tetrahalide results. Sometimes it is dif­ ficult to keep the reaction from proceeding all the way to the tetrahalide even when we want it to stop at the dihalide. x R-C=C-R'

+ (X2

=

2 X2 Cl2 or Br2)

I

X

X

X

I

R-C-C-R'

I

I

Example

CH (CH ) 3

2 3

CI

Cl

Cl

Cl

I I -C-C-H I I

( 1 00%) PROBLEM 9- 1 4

In the addition of j ust 1 mole of bromine to I -hexyne, should the I -hexyne be added to a bromine solution or should the bromine be added to the I -hexyne? Explain your answer.

9-9 E

Addition of Hydrogen H a l i des

Hydrogen halides add across the triple bond of an alkyne in much the same way they add across the alkene double bond. The initial product is a vinyl halide. When a hydrogen halide adds to a terminal alkyne, the product has the orientation predicted by Markovnikov's rule. .A second molecule of HX can add, usually with the same orientation as the first. / .

/

/

9-9 Addition Reactions of Alkynes

[]}-

R- C = C- H + X (HX HCI, HBr, or HI) =

[EJ- x

)

1 �

R-C-C-H I X L!:!J

ill

For example, the reaction of I -pentyne with HBr gives the Markovnikov product. In an internal alkyne such as 2-pentyne, however, the acetylenic carbon atoms are equally substituted, and a mixture of products results.

+

H - C = C -CH2CH2CH3

HBr

H '-..... / CH2CH2CH3 /C = C '-..... Br H

-----7

2-bromo- l -pentene (Markovnikov product)

I -pentyne

CH3-C ==C-CH2CH3

+

HBr

-----7

Br H I I CH3-C=C- CH2CH3

+

H Br I I CH3 - C = C - CH2CH3 3-bromo-2-pentene (E and Z isomers)

2-bromo-2-pentene (E and Z isomers)

2-pentyne

The mechanism is similar to the mechanism of hydrogen halide addition to alkenes. The vinyl cation formed in the first step is more stable with the positive charge on the more highly substituted carbon atom. Attack by halide ion completes the reaction. +

R-C=C

/H

+

.. :X:

�.

vinyl cation

Markovnikov orientation

When 2 moles of a hydrogen halide add to an alkyne, the second mole usually adds with the same orientation as the first. This consistent orientation leads to a gemi­ nal dihalide. For example, a double Markovnikov addition of HBr to I -pentyne gives 2,2-dibromopentane.

H - C = C - CH2CH2CH3

H Br

HBr

------7

------7

I -pentyne

H Br I I H - C -C - CH2CH2 CH 3 I I H Br

2-bromo- l -pentene

PROBLEM 9 -1 5

Propose a mechanism for the entire reaction of I -pentyne with 2 moles of RBr. Show why Markovnikov's rule should be observed in both the first and second additions of HBr.

2,2-dibromopentane

40 1

402

Chapter 9: Alkynes PRO B L E M 9-1 6

The reaction of 2-octyne with 2 equivalents of HCl gives a mixture of two products. (a) Give the structures of the two products. (b) Show why the second equivalent of HCl adds with the same orientation as the first in each case.

In Section 8-3B, we saw the effect of peroxides on the addition of HBr to alkenes. Peroxides catalyze a free-radical chain reaction that adds HBr across the dou­ ble bond of an alkene in the anti-Markovnikov sense. A similar reaction occurs with alkynes, with HBr adding with anti-Markovnikov orientation.

H-C

C-CH2CH2CH3 l-pentyne

+

[!!}- B r

ROaR

)

I -bromo- l -pentene (mixture of E and Z isomers)

PRO BLEM 9-1 7

Propose a mechanism for the reaction of I -pentyne with RBr in the presence of peroxides. Show why anti-Markovnikov orientation results. P R O B L E M 9- 1 8

Show how l -hexyne might be converted to (b) I -bromo- l -hexene (a) 1 ,2-dichloro- l -hexene (c) 2-bromo- l -hexene (d) 1 , l ,2,2-tetrabromohexane (e) 2-bromohexane (f) 2,2-dibromohexane

9-9F

Hyd ration of Alkynes to Ketones and Aldehydes

Mercuric lon-Catalyzed Hydration Alkynes undergo acid-catalyzed addition of water across the triple bond in the presence of mercuric ion as a catalyst. A mixture of mercuric sulfate in aqueous sulfuric acid is commonly used as the reagent. The hydration of alkynes is similar to the hydration of alkenes, and it also goes with Markovnikov ori­ entation. The products are not the alcohols we might expect, however.

R-C=C-H

+

H R" I � C-C-H O

Hp

alkyne a vinyl alcohol (enol)

�

ketone

Electrophilic addition of mercuric ion gives a vinyl cation, which reacts with water and loses a proton to give an organomercurial alcohol.

�Hg2+

R-C - C - H alkyne

�

+ �6:

R-C = C "

Hg+

vinyl cation

�

/ /

organomercurial alcohol

9-9 Addition Reactions of Alkynes

403

Under the acidic reaction conditions, mercury is replaced by hydrogen to give a vinyl alcohol, called an enol.

"- I� /n H+

HO:

/ R

C=C

"-

Hg +

(

)

H

HO:

/ R

H HO: / "c=c + Hg2+ / "R H

1

C - C-H

"-+

organomercurial alcohol

\..:1

Hg+

vinyl alcohol (enol)

resonance-stabilized intermediate

Enols tend to be unstable and isomerize to the ketone form. As shown next, this isomerizat ion involves the shift of a proton and a double bond. The (boxed) hydroxyl proton is lost, and a proton is regained at the methyl position, while the pi bond shifts from the C C position to the C 0 position. This type of rapid equilibrium is called a tautomerism . The one shown is the keto-enol tautomerism, which is covered i n more detail in Chapter 2 2 . The keto form usually predominates. =

=

A number of biological reactions involve the formation of an enol. Researchers are focusing on ways to use these reactions for thera­ peutic purposes. Several investiga­ tors have synthesized stable enols by

placing

bulky

substituents

around the double bond.

enol

keto

keto-enol tautomerism

In acidic solut ion, the keto-enol tautomerism takes place by addition of a proton to the adjacent carbon atom, followed by loss of the hydroxyl proton from oxygen.

Acid-Catalyzed Keto-Enol Tautomerism

MECHANISM 9-2

Under acidic conditions, the proton first adds at its new position on the adjacent carbon atom, and then is removed from its old position in the hydroxyl group. Step 1: Addition of a proton at the methylene group. H

� �

R ",

I

.,::9 C-c

L.!!J O.. +

I

H

�

---f!!J

resonance-stabilized intermediate

enol form

Step 2: Loss of the hydroxyl proton.

�

[illH

R

::::

0: ..

H

I

c -c

+

I

H

---f!!J

�

� � . I 1 �I �: R � ", )c -c ---f!!J

[ill0 H

0+ ..

resonance-stabilized intermediate

I

PROBLEM-SOLVING

H?ItP

To move a proton (as in a

�)

tautomerism) u nder acidic conditions,

H

try ad ding a proton in the new

keto form

position, then removing it from the old position.

404

Chapter 9: Alkynes

For example, the mercuric-catalyzed hydration of I-butyne gives I -buten-2-01 as an intermediate. In the acidic solution, the intermediate quickly equilibrates to its more stable keto tautomer, 2-butanone.

I -buten-2-o1

I-butyne

PROBLEM 9-19

When 2-pentyne reacts with mercuric sulfate in dilute sulfuric acid, the product is a mixture of two ketones. Give the structures of these products, and use mechanisms to show how they are formed. Hy drobor ation -Oxidation In Section 8-7 we saw that hydroboration-oxidation adds water across the double bonds of alkenes with anti-Markovnikov orientation. A similar reaction takes place with alkynes, except that a hindered dialkylborane must be used to prevent addition of two molecules of borane across the triple bond. Di(sec­ ondary isoamyl)borane, called "disiamylborane," adds to the triple bond only once to give a vinylborane. (Amyl is an older common name for penty! .) In a terminal alkyne, the boron atom bonds to the terminal carbon atom.

Sia

R-C-C-H terminal alkyne

H3C ",

/

CH-CH-

I

H3C

disiamylborane a vinylborane

CH 3 "sec-isoamyl" or "siamyl"

Oxidation of the vinylborane (using basic hydrogen peroxide) gives a vinyl alco­ hol (enol), resulting from anti-Markovnikov addition of water across the triple bond. This enol quickly tautomerizes to its more stable carbonyl (keto) form. In the case of a terminal alkyne, the keto product is an aldehyde. This sequence is an excellent method for converting terminal alkynes to aldehydes.

R

I

H

H-C-C

,.L, l!!J

vinylborane

unstable enol form

/

�0

aldehyde

Under basic conditions, the keto-enol tautomerism operates by a different mech­ anism than it does in acid. In base, the proton is first removed from its old position in the OR group, and then replaced on carbon. In acid, the proton was first added on car­ bon, and then removed from the hydroxyl group.

9-9 Addition Reactions of Alkynes

405

Base-Catalyzed Keto-Enol Tautomerism

MECHANISM 9-3

Under basic conditions, the proton is first removed from its old pos ition i n the enol, and then replaced in its new position on the adj acent carbon atom of the ketone or aldehyde. Step 1: Loss of the hydroxyl proton.

stabilized "enolate" ion

enol form

Step 2: Reprotonation on the adjacent carbon atom.

�� ] HeJ , H

/

� :0:

R

I

/ H-C-C

LkJ

stabilized "enol ate" ion

PROBLEM-SOLVING

H

�O:

+ -OH

position, then adding it to the new position.

Hydroboration of I -hexyne, for example, gives the vinylborane with boron on the less highly substituted carbon. Oxidation of this i ntermediate gives an enol that quickly tautomerizes to hexanal .

CH3(CH2)3 - C = C - H

+

Sia2BH

H

I-hexyne

/

C=C

/

'"

H B Sia2

a vinylborane

vinylborane

enol

enol

tautomerism) u nder basic conditions, try removing the proton from its old

keto form

CH3(CH2)3 '\

Ht/tp

To move a proton (as in a

hexanal

(65%)

PROBLEM 9-20

The hydroboration-oxidation of internal alkynes produces ketones. (a) When hydroboration-oxidation is applied to 2-butyne, a single pure product is obtained. Determine the structure of this product, and show the intermediates in its formation. (b) When hydroboration-oxidation is applied to 2-pentyne, two products are obtained. Show why a mixture of products should be expected with any unsymmetrical internal alkyne.

406

Chapter 9: Alkynes PROBLEM 9-21

For each compound, give the product(s) expected from (1) HgS04/H2S04-catalyzed hydra­ tion and (2) hydroboration-oxidation. (b) 2-hexyne (a) I-hexyne (d) cyclodecyne (c) 3-hexyne PROBLEM 9-22

Disiamylborane adds only once to alkynes by virtue of its two bulky secondary isoamyl groups. Disiamylborane is prepared by the reaction of BH3'THF with an alkene. (a) Draw the structural formulas of the reagents and the products in the preparation of disiamylborane. (b) Explain why the reaction in part (a) goes only as far as the dialkylborane. Why is Sia3B not formed?

9-10 Oxidation of Alkynes

9-1 0A

Permanganate Oxidations

Under mild conditions, potassium permanganate oxidizes alkenes to glycols, com­ pounds with two - OH groups on adj acent carbon atoms (Section 8- 1 4B). Recall that this oxidation involves adding a hydroxyl group to each end of the double bond (hydroxylation). A similar reaction occurs with alkynes. If an alkyne is treated with aqueous potassium permanganate under nearly neutral conditions, an a-diketone results. This is conceptually the same as hydroxylating each of the two pi bonds of the alkyne, then losing two molecules of water to give the diketone. o

0

II

II

R -C -C - R'

R - C==C - R '

diketone For example, when 2-pentyne is treated with a cold, dilute solution of neutral perman­ ganate, the product is 2,3-pentanedione. o

II

0

II

CH3- C - C - CH2CH3

2,3-pentanedione

2-pentyne

(90%)

Terminal alkynes probably give a keto-aldehyde at first, but the aldehyde quickly oxi­ dizes to an acid under these conditions. o

R-C

C-H

II

0

II

R - C - C - OH

KMn04 ) H20, neutral

keto-acid

keto-aldehyde

terminal alkyne

If the reaction mixture becomes warm or too basic, the diketone undergoes ox­ idative cleavage. The products are the salts of carboxylic acids, which can be convert­ ed to the free acids by adding dilute a ci d. 0

R-C - C - R '

KMn04, KOH H20, heat

)

0

I

II

R - C - O- + - O - C - R ' Hel

-----'.>

HzO

0

II

R-C - OH

0 +

II

HO-C-R'

9-10 Oxidation of Alkynes

For example, warm, basic permanganate cleaves the triple bond of 2-pentyne to give acetate and propionate ions. Acidification reprotonates these anions to acetic acid and propionic acid. o

CH3-C==C - CH2 CH3

KMn04' KOH HP, heat

2-pentyne

o

I

CH3- C -O-

+

I

-O - C - CH2CH3

acetate

propionate

o

o

II

CH3- C - OH

+

"

HO - C - CH2CH3 propionic acid

acetic acid

Terminal alkynes are cleaved similarly to give a carboxylic acid and CO2. o

( I ) KMn04' KOH, H20 (2) H+

l-hexyne

9-10B

I

CH3(CH2 )3 - C - OH

+

pentanoic acid

Ozonolysis

Ozonolysis of an alleyne, followed by hydrolysis, cleaves the triple bond and gives two carboxylic acids. Either permanganate cleavage or ozonolysis can be used to deter­ mine the position of the triple bond in an unknown alkyne (see Problem 9 -24). (I) 0) ) H 20

R - C - C - R'

( 2)

R - COO H

+

R' - COO H

Examples

CH3 - C=C - CH 2CH3

+

2-pentyne

propionic acid

acetic acid

(I) 0)

(2) H20

I-hexyne

)

pentanoic acid

formic acid

PROBLEM 9-23

Predict the product(s) you would expect from treatment of each compound with (1) dilute, neutral KMn04 and (2) warm, basic KMn04, then dilute acid. (a) I-hexyne (b) 2-hexyne (c) 3-hexyne (d) 2-methyl-3-hexyne (e) cyclodecyne PROBLEM 9-24

Oxidative cleavages can help to determine the positions of the triple bonds in alkynes. (a) An unknown alleyne undergoes oxidative cleavage to give adipic acid and two equivalents of acetic acid. Propose a structure for the alleyne. (I) 0) ) H20

unknown alkyne ( 2)

HOOC - ( CH2 ) 4 - COOH adipic acid

+

2 CH3COOH

(b) An unknown alkyne undergoes oxidative cleavage to give the following triacid plus one equivalent of propanoic acid. Propose a structure for the alkyne.

unknown alkyne a triacid

propionic acid

CO2 t

407

408

Chapter 9: Alkynes P R OB L E M-SOLVIN G ST R AT E GY

Multistep Synthesis

Multistep synthesis problems are useful for exercising your knowledge of organic reac­ tions, and in Chapter 8 we illustrated a systematic approach to synthesis. Now we apply this approach to a fairly difficult problem emphasizing alkyne chemistry. The compound to be synthesized is cis-2-methylhex-4-en-3-01. (The "3-01" means there is an alcohol -OH group on C3.) H "

H / C=C " / H3C CH-CH-CH3

I

I

OH

CH3

cis-2-methylhex-4-en-3-o1

The starting materials are acetylene and compounds containing no more than four carbon atoms. In this problem, it is necessary to consider not only how to assemble the car­ bon skeleton and how to introduce the functional groups, but also when it is best to put in the functional groups. We begin with an examination of the target compound, and then we examine possible intermediates and synthetic routes. 1. Review the functional groups and carbon skeleton of the target compound.

The target compound contains seven carbon atoms and two functional groups: a cis car­ bon-carbon double bond and an alcohol. The best method for generating a cis double bond is the catalytic hydrogenation of a triple bond (Section 9-9B). OH

I

H3C-C=C-CH-CH(CH3)2

Lindlar's catalyst

Using this hydrogenation as the final step simplifies the problem to a synthesis of this acetyleruc alcohol. We know how to form carbon-carbon bonds next to triple bonds, and we have seen the formation of acetylenic alcohols (Section 9-7B).

2. Review the functional groups and carbon skeletons of the starting materials, and see how their skeletons might fit together in the target compound.

Acetylene is listed as one of the starting materials, and we have good methods (Section 9-7) for making carbon-carbon bonds next to triple bonds, by using acetylide ions as nucJe­ ophiles. We can break the target structure into three pieces, each containing no more than four carbon atoms. OH

I

CH3

I

-C=C-

-CH-CH-CH3

acetylene

4 carbons (functionalized)

3. Compare methods for assembling the carbon skeleton of the target compound. Which ones provide a key intermediate with the correct carbon skeleton and func­ tional groups correctly positioned for conversion to the functionality in the target molecule?

Acetylenic alcohols result when acetylides add to ketones and aldehydes (Section 9-7B). Reaction of the acetylide ion with 2-methylpropanal gives one of the groups needed on the triple bond. o

H-C=C-

+

II

H-C-CH(CH3)2 2-methylpropanal

H30+ ----7

--7

9-1 0 Oxidation of Alkynes

409

A methyl group is needed on the other end of the double bond of the target compound. Methylation requires formation of an acetylide, however (Section 9-7 A): CH3I

+

-:C=C-R

�

H3C-C=C-R +

1-

Since the hydroxyl group in the acetylenic alcohol is much more acidic than the acetylenic proton, any attempt to form the acetylide would fail.

This problem can be overcome by adding the methyl group first and then the alcohol portion. In general, we try to add less reactive groups earlier in a synthesis, and more reactive groups later. In this case, we make the alcohol group after adding the alkyl group because the alkyl group is less likely to be affected by subsequent reactions. H-C=C-H

(I) NaN� (2) CH31

o

H3C-C=e:-

+

NaNH2

) H3C-C=C-H

H3C-C=e:- Na+ OH

II

I

H30+

H-C-CH(CH3h � -----7 H3C-C=C-CH-CH(CH3)2

4. Working backward through as many steps as necessary, compare methods for syn­ thesizing the reactants needed for assembly of the key intermediate with the correct carbon skeleton and functionality.

These compounds are all allowed as starting materials. Later, when we have covered more synthetic reactions, we will encounter problems that require us to evaluate how to make the compounds needed to assemble the key intermediates.

5. Summarize the complete synthesis in the forward direction, including all steps and all reagents, and check it for errors and omissions.

Thjs final step is left to you as an exercise. Try to do it without looking at this solution, reviewing each thought process as you summarize the synthesis. Now practice using a systematic approach with the syntheses in Problem 9-25. PROBLEM 9-25: Develop syntheses for the following compounds, using acetylene and compounds containing no more than four carbon atoms as your organic starting materials. (a) 3-methylnon-4-yn-3-ol ("3-01" means there is an OH group on C3.) (b) cis- l -ethyl-2-methylcyclopropane

(c) CH3CH2

.A H

H

I

SUMMARY

CH2CH2CH3

Reactions of Alkynes

I. ACETYLIDE CHEMISTRY 1. Formation of acetylide anions (alkynides) (Section 9-6)

R-C=C-H R-C=C-H R-C=C-H

+

+

NaNH2 R'Li

+ R'MgX

� � �

R-C=e:- �a + NH3 R-C-CLi + R'H R-C-CMgX + R'-H ( Continued)

41 0

Chapter 9: Alkynes Example

2.

sodium propyni de (propynyl sodi um)

sodium amide

propyne

Alkylation of acetylide ions (Section 9-7A) R-C=e:-

+

R'-X

-----+

R - C = C -R'

(R'-X must be an unhindered pri mary halide or tosylate.)

Example +

CH3CH2-C=e:- �a sodium butynide 3.

CH3CH2CH2-Br l-bromopropane

-----+

CH3CH2-C=C-CH2CH2CH3 3-heptyne

Reactions with carbonyl groups (Section 9-7B)

R-C=C:

+

R'

R' " · C=0..· / R'

R'

I

.. R-C=C-C-O:.. I

I

R-C=C-C-OH

I

R'

Example

R'

°

II

CH3-C=C:-

( 1 ) CH3CH2-C-CH3

Na+

(2) Hp

sodium propynide

OH

I

CH3-C=C-C-CH2CH3

)

I

CH3

3-methyl-4-hexyn-3-ol

I I. ADDITIONS TO THE TRIPLE BOND 1. Reduction to alkanes (Section 9-9A) H

+

R-C=C-R'

Pt, Pd, or Ni

2 H2

I

H

I

R-C-C-R'

)

I

H

I

H

Example +

CH3CH2-C=C-CH2-OH

2 H2

Pt

�

CH3CH2-CH2-CH2-CH2-0H

I-pentanol

2-pentyn - l -ol

2.

Reduction to alkenes (Sections 9-9B and 9-9C) R-C=C-R'

+

H2

PdlBaS04, quinoline

R )

R' " / C=C / " H H

cis

R-C=C-R'

Na,NH3

)

trans

Examples

�

CH2CH3 / C=C " / H H

CH3C 2 CH3CH2 -C=C-CH2CH3

3-hexyne

quinoline

cis-3-hexene

9-10 Oxidation of Alkynes Na, NH3

CH3CH2 -C=C-CH2CH3

)

3-hexyne

411

CH3C 2 H � / C=C "'/ H CH2CH3 trans-3-hexene

3.

Addition of halogens (X2

=

Cl2, Br2) (Section 9-9D)

x

R-CX=CX-R'

R-C-C-R'

X

I

I

R-C-C-R'

I

I

X Example

4.

X

Br

CH3C=CCH2CH3

CH3CBr=CBrCH2CH3

2-pentyne

cis- and trans2,3-dibromo-2-pentene

Addition of hydrogen halides (where HX

I

CH3-C-C-CH CH3 I I 2 Br Br 2,2,3,3-tetrabromopentane

HCl, HBr, or HI) (Section 9-9E)

=

H- X

R-C=C-R'

Br

I

R-CH=CX-R'

H- X

H

X

I

I

R-C-C-R'

)

I

(Markovnikov orientation)

I

H

X

Example

CI CH3CH2 -C=C- H

HCI

HCI

�

�

I

CH3CH2-C-CH3 I

CI

I-butyne 2-chloro-l-butene 5.

2,2-dichlorobutane

Addition of water (Section 9-9F) a. Catalyzed by HgS041H2S04

H

I

R-C-C- H

II

I

H

° (Markovnikov orientation)

vinyl alcohol (unstable)

ketone (stable)

Example °

CH3-C=C-H

+

II

CH3-C-CH3

H20

propyne

2-propanone (acetone)

b. Hydroboration-oxidation

H R-C-C- H

(I) Sia2 BH . THF

I

R-C-C- H

,

I

H (anti-Markovnikov orientation)

CH3-C-C-H propyne

aldehyde (stable)

vinyl alcohol (unstable)

Example

o

(I) Si�BH . THF (2) HP2 ' NaOH

I

0

,

II

CH3-CH2-C-H propanal

( Continued)

41 2

Chapter 9: Alkynes

I I I. OXIDATION OF ALKYNES (SECTION 9-10) 1.

Oxidation to a-diketones (Section 9-lOA)

° R-C=C- R'

Example

°

II

II

H2 0,

R-C-C- R'

n u ra

e t l

° CH3-C=C-CH2CH3

2-pentyne 2.

0

II

Hz0'

II

CH3-C-C-CH2CH3

neu tral

pentane-2,3-dione

Oxidative cleavage (Section 9-lOB)

R-C=C-R'

(1) KMn04' -OH (2) H+

o

°

II

II

+

R-C-OH

)

HQ-C-R'

Examples

(1) KMn04' NaOH

(1) KMn04' NaOH

Chapter 9 Glossary

)

)

acetylene The simplest alkyne, H -C=C-H. Also used as a synonym for alkyne, a gener­ ic term for a compound containing a C=C triple bond. (p. 382) acetylide ion (alkynide ion) The anionic salt of a terminal alkyne. Metal acetylides are

organometallic compounds with a metal atom in place of the weakly acidic acetylenic hydrogen of a terminal alkyne. (p. 387) R - C=C-H

+

N a+ -:NH2

�

R - C=CJ- +N a a sodium acetylide

+

:NH3

alkoxide ion R - 0-, the conjugate base of an alcohol. (p. 387) R-O:.. alkoxide

+

H20

� �

R-O-H .. alcohol

+

-OH

alkyne Any compound containing a carbon-carbon triple bond. (pp. 382, 384) A terminal alkyne has a triple bond at the end of a chain, with an acetylenic hydrogen. An internal alkyne has the triple bond somewhere other than at the end of the chain.

acetylenic hydrogen

[E}-- C=C-CH2CH3 I-butyne,

a

terminal alkyne

(no acelylenic hydrogen) CH3-C=C-CH3

2-butyne, an internal alkyne

amyl An older common name for pentyl. (p. 404) enol An alcohol with the hydroxyl group bonded to a carbon atom of a carbon-carbon double

bond. Most enols are unstable, spontaneously isomerizing to their carbonyl tautomers, called the keto form of the compound. See tautomers. (p. 403) Lindlar's catalyst A heterogeneous catalyst for the hydrogenation of alkynes to cis alkenes. In its most common form, it consists of a thin coating of palladium on barium sulfate, with quino­ line added to decrease the catalytic activity. (p. 398) s character The fraction of a hybrid orbital that corresponds to an s orbital; about one-half for 2 3 sp hybrids, one-third for sp hybrids, and one-fourth for sp hybrids. (p. 387)

Study Problems

413

siamyl group A contraction for secondary isoamyl, abbreviated "Sia." This is the l ,2-dimethyl­ propyl group. Disiamylborane is used for hydroboration of terminal alkynes because this bulky borane adds only once to the triple bond. (p. 404) H3 C Sia

"" CH-CH-

/

H3C

�

I

CH 3

" sec - i soamyl " or "siamyl"

alkyne

disiamylborane

R'

H

H

BSi�

/ "" C=C / ""

a vinylborane

tautomers Isomers that can quickly interconvert by the movement of a proton (and a dou­ ble bond) from one site to another. An equilibrium between tautomers is called a tautomerism. (p.403)

enol form

keto form

The keto-enol tautomerism is the equilibrium between these two tautomers. vinyl cation A cation with a positive charge on one of the carbon atoms of a C=C double bond. The cationic carbon atom is usually sp hybridized. Vinyl cations are often generated by the addition of an electrophile to a carbon-carbon triple bond. (p. 401) R R - C-C-R'

pJ

""

+

C=C-R'

/ \

E

Sp2

\

sp

a vinyl cation

I

Essential Problem-Solving Skills in Chapter 9

1. Name alkynes, and draw the structures from their names. 2. Explain why alkynes are more acidic than alkanes and alkenes. Show how to generate

nucleophilic acetylide ions and heavy-metal acetylides. 3. Propose effective single-step and multistep syntheses of alkynes. 4. Predict the products of additions, oxidations, reductions, and cleavages of alkynes, includ­

ing orientation of reaction (regiochemistry) and stereochemistry. 5. Use alkynes as starting materials and intermediates in one-step and multistep syntheses. 6. Show how the reduction of an alkyne leads to an alkene or alkene derivative with the

desired stereochemistry.

Study Problems 9-26

Briefly define each term, and give an example. (a) alkyne (b) acetylide ion (e) Lindlar's catalyst (d) tautomerism

(g) vinyl cation (i) hydration of an alkyne 9-27

(c) enol

( 0 disiamylborane

(h) oxidative cleavage of an alkyne (j)

hydroboration of an alkyne

Write structural formulas for the following compounds. (a) 3-nonyne (b) methyl-n-pentylacetylene (e) S-methyl-3-octyne (d) cyclohexylacetylene

(c) ethynylbenzene

(0

trans-3,S-dibromocyclodecyne

(g) 3-octyn-2-ol

(i)

l ,4-heptadiyne

(j)

vinylacetylene

(h) cis-6-ethyl-2-octen-4-yne (k) (S)-3-methyl- l -penten-4-yne

414

C hapter 9: Alkynes

9-28

Give common names for the following compounds. (a) CH3-C=C-CH2CH3 (b) Ph-C=C-H (c) 3-methyl-4-octyne (d) (CH3hC-C=C-CH(CH3)CH2CH3

9-29

Give IUPAC names for the following compounds. Ph

I

(a) CH3-C=C-CH-CH3

(d)

9-30

(a) Draw and name the seven alkynes of formula C6H 10. (b) Which compounds in part (a) will react with sodium amide? Show the products that result.

9-31

A marginal note on page 391 states, "The addition of an acetylide ion to a carbonyl group is used in the synthesis of ethchlorvynol, a drug used to cause drowsiness and induce sleep." Show how you would accomplish this synthesis from acetylene and a carbonyl compound.

OH

I

CH3CH2-C -C-CH

I

C

H

/�

CHCI

ethchlorvynol 9-32

Muscalure, the sex attractant of the common housefly, is cis-9-tricosene. Most syntheses of alkenes give the more stable trans isomer as the major product. Devise a synthesis of muscalure from acetylene and other compounds of your choice. Your synthesis must give specifical ly the cis isomer of muscalure.

CH3(CH2)7

'"

H

C=C

/

/

"

(CH?)I?CH3 - H

cis-9-tricosene, "muscalure" 9-33

9-34

Predict the products of reaction of I-pentyne with the following reagents. (a) I equivalent of HCI (b) 2 equivalents of HCl (e) 1 equivalent of Br2 (d) H2, Pd/ BaS04, quinoline (h) warm, concd. KMn04, NaOH (g) cold, dilute KMn04 (j) NaNH2 (k) H2S04/HgS04, H20

(c) excess H2, Ni (f) 2 equivalents of Br2 (i) Na, liquid ammonia (I) Sia2BH, then H 2 02 , -OH

Show how you would accomplish the following synthetic transformations. Show all intermediates. (a) 2,2-dibromobutane � I-butyne (b) 2,2-dibromobutane � 2-butyne (c) I-butyne � 3-octyne (d) trans-2-hexene � 2-hexyne (e) cis-2-hexene � l-hexyne (f) cyclodecyne � cis-cyclodecene (g) cyclodecyne � trans-cyclodecene (h) l -hexyne � 2-hexanone, CH3COCH2CH2CH2CH3 (i) l-hexyne � hexanal, CH3(CH2)4CHO (j) trans-2-hexene � cis-2-hexene

Study Problems 9-35

415

Potassiu m hydroxide is mixed with I ,2-dibromo-4,4-dimethylpentane, and the mixture is heated at 200°C in a sealed tube for 1 hour. The product mixture is distilled, and the distil late is collected over a boil i ng range of 7S-SSoC to give a mixture (A) of two compounds. This mixture is treated with sodium amide, followed by acetone, then by dilute acid, to give another product mixture (B). Mixture B is distilled, with one component (C) collected between SO-S4°C and another component (D) collected between 1 40-1S0°C under a good vacuum. Further analysis shows that product D con­ tains a hydroxyl group. Identify the compounds in m ixtures A and B, as well as pure compounds C and D. 1 ,2-dibromohexane

(I) KOH, 200°C

Mixture A

(2) distilled

( 1 ) NaNH2 o

II

(2) CH3-C-CH3

f �

(3) H30+

pure C, b.p. SO-S4°

9-36

d i s til led

Mixture B

ure D, b.p. 140- 1 S00/vacuul1l

Predict the products formed when CH3CH2-C=C:- Na+ reacts with the following compounds. (a) ethyl bromide (b) t-butyl bromide (c) formaldehyde (d) cyclohexanone (e) CH3CH2CH2CHO (f) cyclohexanol (g) 2-butanone, CH3CH2COCH3

9-37

Show how you would synthesize the following compounds, starting with acetylene and any compounds containing no more than four carbon atoms. (a) I-hexyne (b) 2-hexyne (c) cis-2-hexene (d) trans-2-hexene (e) hexane (f) 2,2-dibromohexane (g) pentanal, CH3CH2CH2CH2CHO (h) 2-pentanone, CH3-CO-CH2CH2CH3 (i) (± )-3, 4-dibromohexane U) meso-2,3-butanediol

9-38

When treated with hydrogen and a platinum catalyst, an unknown compound (X) absorbs 5 equivalents of hydrogen to give n-butylcyclohexane. Treatment of X with an excess of ozone, followed by dimethyl sulfide and water, gives the fol­ lowing products: o

II

0

II

o

0

II

H-C-CH2-CH2-C-C-H

II

o

0

II

H-C-C-H

II

o

0

II

H-C-C-OH

II

H-C-OH

Propose a structure for the unknown compound (X). Is there any uncertainty in your structure? 9-39

When compound (Z) is treated with ozone, followed by dimethyl sulfide and washing with water, the products are formic acid, 3-oxobutanic acid, and hexanal. o

o

(Z)

II

H-C-OH formic acid

+

II

o

0

II

CH3-C-CH2-C-OH 3-oxobutanoic acid

+

II

CH3(CH2)4-C-H hexan al

Propose a structure for compound (Z). What uncertainty is there in the structure you have proposed?

41 6 9-40

Chapter 9: Alkynes Show how you would synthesize the following compounds from acetylene and any other needed reagents: (a) 6-phenylhex-l -en-4-yne (b) cis-l-phenyl-2-pentene (c) trans- l-phenyl-2-pentene

f: '

Cd)

H H

H

C"d i" '"'''''om,,)

OH

CH2CH3 *9-41

The following functional-group interchange is a useful synthesis of aldehydes. o

I

R-C- C-H

R-CH2-C-H

terminal alkyne

aldehyde

(a) What reagents were used in this chapter for this transformation? Give an example to i llustrate this method. (b) This functional-group interchange can also be accomplished using the following sequence.

R-C=C-H

NaOCH2CH3 CH3CHPH

Propose mechanisms for these steps.

(c) Explain why a nucleophilic reagent such as ethoxide adds to an alkyne more easily than it adds to an alkene. *9 - 42

Using any necessary inorganic reagents, show how you would convert acetylene and isobutyl bromide to (a ) meso-2,7-dimethyl-4,5-octanediol, (CH3hCHCH2CH( OH)CH(OH)CH2CH(CH3 h (b) (± )-2, 7-dimethyl-4,5-octanediol

*9-43

Show how you would convert the following starting materials into the target compound. You may use any additional reagents you need.

other reagents as needed

________7�________7

VX� 0

- C -CH2-

0' _

10

Structure and Synthesis of Alcohols

1 0-1 Introduction

Alcohols are organic compounds containing hydroxyl ( - OH ) groups. They are some of the most common and useful compounds in nature, in industry, and around the house. The word alcohol is one of the oldest chemical terms, derived from the early Arabic al-kuhl. Originally it meant "the powder," and later "the essence." Ethyl alcohol, distilled from wine, was considered to be "the essence" of wine. Ethyl alcohol (grain alcohol) is found in alcoholic beverages, cosmetics, and drug preparations. Methyl alcohol (wood alcohol) is used as a fuel and solvent. Isopropyl alcohol (rubbing alco­ hol) is used as a skin cleanser for injections and minor cuts.

OH

I

CH3-CH2 - OH

CH3- OH

CH3- CH - CH3

ethyl alcohol ethanol

methyl alcohol methanol

isopropyl alcohol 2-propanol

Alcohols are synthesized by a wide variety of methods, and the hydroxyl group may be converted to most other functional groups. For these reasons, alcohols are ver­ satile synthetic i ntermediates. In this chapter, we discuss the physical properties of alcohols and summarize the methods used to synthesize them. In Chapter 1 1 ( Reac­ tions of Alcohols), we continue our study of the central role that alcohols play in organic chemistry as reagents, solvents, and synthetic intermediates.

10-2 Structure and Classification of Alcohols

The structure of an alcohol resembles the structure of water, with an alkyl group replacing one of the hydrogen atoms of water. Figure 1 0- 1 compares the structures of water and methanol. Both have sp3 -hybridized oxygen atoms, but the C - 0 - H bond angle in methanol ( 1 08.9°) is considerably l arger than the H - O - H bond angle in water (104.5°) because the methyl group is much larger than a hydrogen atom. The bulky methyl group counteracts the bond angle compression caused by oxy­ gen's nonbonding pairs of electrons. The O - H bond lengths are about the same in water and methanol (0.96 A), but the C - 0 bond is considerably longer ( 1 .4 A), reflecting the l arger covalent radius of carbon compared to hydrogen. One way of organizing the alcohol family is to classify each alcohol according to the type of carbinol carbon atom: the one bonded to the - OH group. If this carbon atom is primary (bonded to one other carbon atom), the compound is a primary alcohol. A secondary alcohol has the - OH group attached to a secondary carbon atom, and a 417

418

Chapter 1 0: Structure and Synthesis of Alcohols

1 .4 A

0.96 A

H

� Figure 10-1

��

H

1 04.50 H

y�

p\

� 1 08.90

H H

water

water

Comparison of the structures of water and methyl alcohol.

"

0.96 A

H methyl alcohol

methyl alcohol

tertiary alcohol has it bonded to a tertiary carbon. When we studied alkyl halides (Chapter 6), we saw that primary, secondary, and tertiary halides react differently. The same is true for alcohols. We need to Jearn how these classes of alcohols are similar and under what conditions they react differently. Figure 10-2 shows examples of primary, secondary, and tertiary alcohols. Compounds with a hydroxyl group bonded directly to an aromatic (benzene) ring are called phenols. Phenols have many properties similar to those of alcohols, while other properties derive from their aromatic character. In this chapter, we consid­ er the properties of phenols that are simjJar to those of alcohols and note some of the differences. In Chapter 16, we consider the aromatic n ature of phenols and the reac­ tions that result from that aromaticity. Type

Examples

Structure

CH3

H Primary alcohol

I

R - C - OH

I

CH3 HCH2 - 0H

ethanol

2-methyl-l-propanol

H

CH3

I

CH - OH

R' Secondary alcohol

I

R - C - OH

I

H

I

CH2

I

CH3 2-butanol

R' Tertiary alcohol

�

CH3CH2 - OH

(jH

R - C - OH

I

o-

I

I

I

CH3

OH

6

phenol

cholesterol

Ph - C - OH Ph

2-methyl-2-propanol

Phenols

HO Ph

CH 3 - C - OH

OH

benzyl alcohol

cyclohexanol

I

R"

CH2 - OH

OH

CH3

I

o-

triphenylmethanol

I-methylcyclopentanol

OH

&

CH3

3-methylphenol

HO

f) I�

0H

hydroquinone

... Figure 10-2

Classification of alcohols. Alcohols are classified according to the type of carbon atom (primary, secondary, or tertiary) bonded to the hydroxyl group. Phenols have a hydroxyl group bonded to a carbon atom in a benzene ring.

10-3 Nomenclat ure of Alcohols and Phenols

10-3A

IUPAC Names ("Alkanol" Names)

The IUPAC system provides unique names for alcohols, based on rules that are similar to those for other classes of compounds. In general, the name carries the -ol suffix, together with a number to give the location of the hydroxyl group. The formal rules are summarized in the fol lowing three steps: 1. Name the longest carbon chain that contains the carbon atom bearing the - OR group. Drop the final -e from t h e alkane name and add the suffix -ol to

give the root name.

2. Number the longest carbon chain starting at the end nearest the hydroxyl group, and use the appropriate number to indicate the position of the - OR group. (The hydroxyl group takes precedence over double and triple bonds.)

3. Name all the substituents and give their numbers, as you would for an alkane or an alkene.

In the following example, the longest carbon chain has four carbons, so the root name is butanol. The - OR group is on the second carbon atom, so this is a 2-butanol . The complete IUPAC name is I-bromo-3 , 3-dimethyl-2-butanol. The new IUPAC positioning of numbers would place the 2 next to the group it locates (-01), giving the name I -bromo-3,3-dimethylbutan-2-01. CH OH 3

I

I

4CH _3C -2CH -ICH2 -Br 3

I

CH3

Cyclic alcohols are named using the prefix cyclo-; the hydroxyl group is assumed to be on C l .

Cts 4

IUPAC name: new IUPAC name:

6

3

H

i�

2

�

Br

OH H

trans-2-bromocyclohexanol trans-2 -bromocyclohexan-I-ol

I-ethylcyclopropanol

I -eth ylcyclopropan- l -ol

SOLVED PROBLEM 10-1

Give the systematic (IUPAC) name for the following alcohol.

CH?I

I

-

CH3-CH2-CH

-

CH?-OH

1-

CH-CH-CH3

I

CH3 SOLUTION The longest chain contains six carbon atoms, but it does not contain the carbon bonded to the hydroxyl group. The longest chain containing the carbon bonded to the -OH group is the one outlined by the green box, containing five carbon atoms. This chain is numbered from right to left in order to give the hydroxyl-bearing carbon atom the lowest possible number.

CH,I iCH?

r-:------:----::+-- - ----'

I -

5CH3-4CH2- 3CH-2CH

The correct name for this compound is 3-(iodomethyl)-2-isopropylpentan- l -ol.

1 0-3 Nomenclature of Alcohols and Phenols

419

420

Chapter

10:

Structure and Synthesis of Alcohols In naming alcohols containing double and triple bonds, use the -al suffix after the alkene or alkyne name. The alcohol functional group takes precedence over double and triple bonds, so the chain is numbered in order to give the lowest possible number to the carbon atom bonded to the hydroxyl group. The position of the - OH group is given by putting its number before the -al suffix. Numbers for the multiple bonds were once given early in the name, but the 1 997 revision of the IUPAC rules puts them next to the -en or -yn suffix they describe. B oth the new and old placements of the numbers are shown in the following figure.

I

HO IUPAC name:

3

H3

5

�

� -4

p

F


eLher . LI ------'>

1 0-9 Addition of Organometallic Reagents to Carbonyl Compounds Because they resemble carbanions, Grignard and organolithium reagents are strong nucleophiles and strong bases. Their most useful nucleophilic reactions are additions to carbonyl ( C = O ) groups, much like we saw with acetylide ions (Section 9-7 B ) . The carbonyl group is polarized, with a partial positive charge on carbon and a partial negative charge on oxygen. The positively charged carbon is electrophilic; attack by a nucleophile places a negative charge on the electronegative oxygen atom.

I .. I

R- C - O : -

1 0-9 Addition of Org anometall i c Reag ents to Carbonyl Compounds

The product of this nucleophilic attack is an alkoxide ion, a strong base. Addition of water or a dilute acid protonates the alkoxide to give the alcohol.

I ..

R- C - OH

I

+

.. : OH

Either a Grignard reagent or an organolithium reagent can serve as the nucle­ ophile in this addition to a carbonyl group. The following discussions refer to Grignard reagents, but they also apply to organolithium reagents. The Grignard reagent adds to the carbonyl group to form an alkoxide ion. Addition of dilute acid (in a separate step) protonates the alkoxide to give the alcohol. We are i nterested primarily in the reactions of Grignard reagents with ketones and aldehydes. Ketones are compounds with two alkyl groups bonded to a carbonyl group. Aldehydes have one alkyl group and one hydrogen atom bonded to the car­ bonyl group. Formaldehyde has two hydrogen atoms bonded to the carbonyl group. R R

'" /

H

'"

c=o

R

/

C=O

an aldehyde

a ketone

H H

'" /

C=O

formaldehyde

The electrostatic potential map (EPM) of formaldehyde shows the polarization of the carbonyl group, with an electron-rich (red) region around oxygen and an electron-poor (blue) region near carbon.

..;�t�'-:� KEY MECHANISM

1 0- 1

G rignard Reactions

Grignard and organolithium reagents provide some of the best methods for assembling a carbon skeleton. These strong nucleophiles add to ketones and aldehydes to give alkoxide ions, which are protonated to give alcohols. Formation of the Grignard reagent: Magnesium reacts with an alkyl halide in an

ether solution .

R'-X

+

Mg

cthe7

R ' - MgX

Reaction }: The Glignard reagent attacks a carbonyl compound to form an alkoxide s alt.

R'

-------7 ether

� I ..

R '- - C - O: -

I ..

R'

magnesium alkoxjde salt

(Continued)

EPM of formaldehyde

435

436

Chapter 1 0: Structure and Synthesis of Alcohols Reaction 2: After the f1rst reaction is complete, water or dilute acid is added to proto­

nate the alkoxide and give the alcohol . R'

R

� I .�H rO - H � I ... . I I .. R '- - C - O : -

..

+MgX

R '- - C -O - H

)

XMgOH

+

R

R'

alcohol

magnesium alkoxide salt

EXA M PLE: Addition of phenylmagnesium bromide to acetone.

Formation of the Grignard reagent: Magnesium reacts with bromobenzene in an

ether solution to give phenylmagnesium bromide. +

Mg

----i>

ether

� o-

MgBr

phenylmagnesium bromide

Reaction 1: The Grignard reagent attacks a carbonyl compound to form an

alkoxide salt.

o� -�-:o:CH3

-

I

CH3

..

magnesium alkoxide salt

Reaction 2: After the first reaction is complete, water or dilute acid is added to proto­

nate the alkoxide and give the alcohol.

o� - '\

CH3

�

I / C - O: I "

CH3

magnesium alkoxide salt

o��

CH3

r-

�H�O. . -H + MgBr

)

-

I

- OH

+

BrMgOH

CH3

2-phenyl-2-propanol

Q U E STIO N : What would be the result if water were accidentally added in the first reaction with the Grignard reagent and the carbonyl compound?

1 0-9A

Addition to Formaldehyde: Formation of Primary Alcohols

Addition of a Grignard reagent to formaldehyde, fol lowed by protonation, gives a primary alcohol with one more carbon atom than i n the Grignard reagent.

(jfr

§

H

MgX

Grignard reagent

----i>

ether

§ II

C - O - +MgX H

formaldehyde

primary alcohol

1 0-9 Addition of Organometallic Reagents to Carbonyl Compounds

437

For example, H H

butylmagne�ium bromide

",

/

C=O

H

( I ) ether solvent ) (2) Hp+

CH3CH2CH2CH2 -

.

(b)

1 0-98

�OH

(c)

CH?OH or -

PROBLEM-SOLVING Note the use of

a r row.

Grignard reagents add to aldehydes to give, after protonation, secondary alcohols.

MgX

+

R'

R' " H

G- II

C - O - +MgX

C=O /

H

aldehyde

Grignard reagent

R'

R'

G- I

I

H 0+ �

C - O - + MgX

G- II

C - OH H

H

secondary alcohol

The two alkyl groups of the secondary alcohol are the alkyl group from the Grig­ nard reagent and the alkyl group that was bonded to the carbonyl group of the aldehyde.

+

H3C H

"

/

CH3 C=O

�

ether

I

3

acetaldehyde

CH3

I

"

CH3 -CH - C - O - +Mo-Br 2

I

H

"

CH - CH -C-O- +Mo-Br 2

I

H

H 0+ �

CH3

I

CH CH? - C-OH 3

_

I

H

2 bu tanol -

(85%)

*

HiltZ; to show

separate reactions with one reaction

Addition to Aldehydes: Formation of Secondary Alcohols

G-

- OH

I -pentanol (92%)

Show how you would synthesize the following alcohols b y adding an appropriate Grignard reagent to formalde h yde

(a) V

T

H

formaldehyde

PRO B L E M 10-13

rY CH20H

I

438

Chapter 10 : Structure and Synthesis of Alcohols

PROBLEM-SOLVING

H?nJ/

A secondary a l cohol has two groups

o n the carbinol carbon atom. Consider

P R O B L E M 10-14

Show two ways you could synthesize each of the following alcohols by adding a n appropri­ ate Grignard reagent to an aldehyde.

two poss i b l e reactions, with either

OH

group added as the G r i g n a rd reagent.

(a)

(b)

� '-H}O+

448

I

Chapter 1 0: Structure and Synthesis of Alcohols S U M MARY

Reacti ons of LiAIH4 a nd N a B H4

NaBH4

LiAlH4

I

R-CH2-OH

I R-C-R'

I R-CH-R'

""C=C/' /' ""

no reaction

no reaction

I

no reaction

R-CH2-OH

I

no reaction

R-CH2-OH

0

aldehyde

R-C-H 0

ketone alkene

R-CH2-OH

I

OH

OH R-CH-R'

0

R-C-O-

acid anion

anion in base

0

ester

R-C-OR'

Note: The products shown are the final products, after hydrolysis of the alkoxide.

PROBLEM-SOLVING

Htni::-

When m a k i n g a pri mary or secondary alcohol, you can consider a d d i n g a n a l kyl group last (as a Grignard reagent) or a d d i n g a hydrogen last (by reduci ng a ketone or aldehyde).

P RO B L E M 10-24

Predict the products you would expect from the reaction of NaBH4 with the following compounds. (c) Ph - COOH

(d)

I C

(yO

(e)

V

H

/,

I C

o

0

°

yy y

"-

(f)

OCH3

°

°

P RO B L E M 10-25

D

I

C 'H

Repeat Problem 1 0-24 using LiAlH4 (followed by hydrolysis) as the reagent. P RO B L E M 10-26

Show how you would synthesize the following alcohols by reducing appropriate car­ bonyl compounds. O O (a) I -heptanol

(b) 2-heptanol

(c) 2-methyl-3-hexanol

(d)

[y l � OH

10-11 C

Cata lytic Hydrogenation of Ketones and Aldehydes

Reducing a ketone or an aldehyde to an alcohol involves adding two hydrogen atoms across the C = O bond. This addition can be accomplished by catalytic hydrogenation, commonly using Raney nickel as the catalyst. OR o

I

-C-

+

R2

Raney N i

)

I - CH-

10-11 Reduction of the Carbonyl Group: Synthesis of 1 ° and 2° Alcohols

449

Raney nickel is a finely divided hydrogen-bearing form of nickel made by treating a ruckel-aluminum alloy with a strong sodium hydroxide solution. The aluminum in the alloy reacts to form hydrogen, leaving behind a finely divided ruckel powder saturated with hydrogen. Raney nickel is an effective catalyst for the hydrogenation of ketones and alde­ hydes to alcohols. Carbon-carbon double bonds are also reduced under these conditions, however, so any alkene double bonds in the starting material will also be reduced. In most cases, sodium borohydride is more convenient for reducing simple ketones and aldehydes.

0 / H2C=C H - CH 2 - C - C"'CH3

I

r �::: I

2.2-d;m"hY l

CH 3

CH3

Raney Ni )

+

CH3 -CH2 - CH2 -

H

rI

- CH 20H

CH3

2,2-dimethyl - l -pentanol (94%)

pe"

CH1 ,

(for comparison)

I .

)

H 2 C=C H - CH 2 - C - CH2 OH

I

CH3 2,2-dimethylpent-4-en- l -ol

I

S U M MARY

Alcoh ol Synt h eses

I. FROM ALKENES 1. Hydration (S ecti on s 8-4 through

8-7)

a. Acid catalyzed: forms Markovnikov alcohols; b. Oxymercuration-demercuration: forms Markovnikov alcohols

{ c.

CH3

I

H3C-C-CH3

( l ) Hg(OAclz, H20 ( 2) NaBH4

I

OH

Hydroboration-oxidation: forms anti-Markovnikov alcohols (1) BH3· THF (2)

H202, NaOH

CH3

I

HO -CH2-C-CH3

)

I

H

2. Hydroxylation: forms vicinal diols (glycols) (Sections 8- 1 3 and 8- 1 4) a. Syn hydroxylation, using KMn041NaOH or using OsO/H202

o

H

or

/l- OH � OH

KMnOiNaOH

.

'

cycIopentene b. Anti hydroxylation, using peracids

o

cycIopentene

ff

cis-cycIopentane-l ,2-diol

B

/+ OH �H cm

(+ enantiomer)

trans-cycIopentane- l,2-diol ( Continued)

Chapter 10: Structure and Synthesis of Alcohols

450

II. FROM ALKYL HALIDES: NUCLEOPHILIC SUBSTITUTION

1. Second-order substitution: primary (and some secondary) halides

(SECTIONS 6-9 AND 6-13)

KOH Hp

�

2. First-order substitution: tertiary (and some secondary) halides

I CH - C - CH CH 3

3

I

acetone/water beat

3

Cl I-butyl chloride

)

3 r CH - C -CH H

3

I

3

OH -butyl alcohol

isobutylene

t

(Section 10-9)

III. FROM CARBONYL COMPOUNDS: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 1. Addition of a Grignard or organolithium reagent o

I -C-

I -CI R

O- +MgX

+

------7

ether

R -MgX

a. Addition to formaldehyde gives a primmy alcohol ethyl magnesium bromide

+

H 0+

�

I

OH -C-

I

R

(1) ether solvent (2) Hp+

I-propanol

b. Addition to an aldehyde gives a secondary alcohol

c.

< !-

MgBr

(1) ether solvent (2) H30+

+

phenyl magnesium bromide

o-�

acetaldehyde

(1) ether ) (2) Hp+

cyclohexanone d. Addition to an acid halide or an ester gives a tertimy alcohol

OCH,CH, OR

l -ethylcyclohexanol

o

I

o

I

CH3-C -OCH3 methyl acetate

0) 2

d -OH I CH3

I -phenylethanol

Addition to a ketone gives a tertiary alcohol

CH3-C - C l acety1 chloride or

H

Q-MgBr

cyc10hexylmagnesium bromide ) ( 2) Hp+

1,1-dicyclohexylethanol

10-12 Thiols ( Mercaptans)

451

e. Addition to ethylene oxide gives a primary alcohol (with two carbon atoms added)

2-cyclohexylethanol

cyclohexylmagnesium bromide

2. Reduction of carbonyl compounds

(Section 1 0- 1 1 ) a . Catalytic hydrogenation of aldehydes and ketones o

OH

II

+

-C-

Raney Ni

H2

)

I

-CH-

This method is usually not as selective or as effective as the use of hydride reagents. b. Use of hydride reagents (1) Reduction of an aldehyde gives a primary alcohol o

< )- �

-H

benzyl alcohol

benzaldehyde

(2) Reduction of a ketone gives a secondary alcohol

cyclohexanol

cyclohexanone (3) Reduction of an acid or ester gives a primary alcohol o

II

CH3-(CH2)8-C-OH decanoic acid o

II

(I) LiAlH4

)

CH3-(CH2)8-CH2-0H I-decanol

CH3-(CH2)8-C-OCH3 methyl decanoate

IV. SYNTHESIS OF PHENOLS (CHAPTER 17)

Thiols are sulfur analogues of alcohols, with an -SH group in place of the alcohol - OH group. Oxygen and sulfur are in the same column of the periodic table (group VIA), with oxygen in the second row and sulfur in the third. IUPAC names for thiols are derived from the alkane names, using the suffix -thiol. Thiols are also caJled mercaptans ("captures mercury") because they form stable heavy-metal derivatives. Common names are formed like those of alcohols, using the name of the alkyl group with the word mercaptan . The - SH group itself is caJled a mercapto group. IUPAC name: common name:

methanethiol methyl mercaptan

l-butanethiol II-butyl mercaptan

2-butene-l-thiol

10-12 Th iols (Mercaptans)

2-mercaptoethanol

452

C hap te r 10: Structure and Synthesis of Alcohols Thiols' ability to complex heavy metals has proved useful for making antidotes to heavy-metal poisoning. For example, in World War II the Allies were concerned that the Germans would use lewisite, a volatile arsenic compound, as a chemical warfare agent. Thiols complex strongly with arsenic, and British scientists developed dimercaprol (2,3-dimercapto- l -propanol) as an effective antidote. The Allies came to refer to this compound as "British anti-lewisite" (BAL), a name that is still used. Dimercaprol is useful against a variety of heavy metals, including arsenic, mercury, and gold.

CI H

'" /

C=C

/

'"

CH2 -CH-CH2

H

I

SH As Cl2

Garlic has served throughout history as a remedy for numerous diseases. Movies have often depicted its power to repel werewolves and vampires. The characteristic garlic odor derives from the many sulfur compounds it contains. One of the major constituents is allicin. a compound with antibacterial properties.

The odor of thiols is their strongest characteristic. Skunk scent is composed mainly of 3-methyl- l -butanethiol and 2-butene- l -thiol, with small amounts of other thiols. Ethanethiol is added to natural gas (odorless methane) to give it the characteris­ tic "gassy" odor for detecting leaks. Although oxygen is more electronegative than sulfur, thiols are more acidic than alcohols. Their enhanced acidity results from two effects: First, S - H bonds are gen­ erall y weaker than 0 - H bonds, making S - H bonds easier to break. Second, the thiolate ion (R - S-) has its negative charge on sulfur, which allows the charge to be delocalized over a larger region than the negative charge of an alkoxide ion, borne on a smaller oxygen atom. Thiolate ions are easily formed by treating the thiol with aque­ ous sodium hydroxide.

CH3 - CH2 -SH

+

-OH

�

(2) H30+

( 1)

C 6H 1 zO A

H2SO4 heat

C 6H IO B

)

,

t K Mn04 I warm, coned.

H2S04 t heat I

F

P) Mg, ether (2)

(yO

(3) Hp+ * 10-50

10-51

C5H9Br

0

HBr

H

I

R- C - R'

I

I

[0]

I

I

R- C - R'

I

H

----i>

R"

alkane no bonds to 0

II

R-C-H

+

[0]

----i>

o

II

R - C - OH

H20

aldehyde two bonds to 0

carboxylic acid three bonds to 0

o

�

secondary alcohol one bond to 0

[0]

o

OH

----i>

alkane no bonds to 0

R - C - R'

I

----i>

primary alcohol one bond to 0

H

H

R-C- H

[ 0]

H

alkane no bonds to 0 H

I

R-

+

�

- R'

H2 0

(no further oxidation)

ketone two bonds to 0 .... Figure 11-1 Oxidation states of alcohols. An alcohol is more oxidized than an alkane, yet less oxidized than carbonyl compounds such as ketones, aldehydes, and acids. Oxidation of a primary alcohol leads to an aldehyde, and further oxidation leads to an acid. Secondary alcohols are oxidized to ketones. Tertiary alcohols cannot be oxidized without breaking carbon-carbon bonds.

OH

I

R - C - R'

I

461

(usually no further oxidation)

R"

tertiary alcohol one bond to 0

We can tell that an oxidation or a reduction of an alcohol has taken place by counting the number of C - 0 bonds to the carbon atom. For example, in a primary alcohol, the carbinol ( C - OH ) carbon atom has one bond to oxygen; in an aldehyde, the carbonyl carbon has two (more oxidized); and in an acid, it has three. Oxidation of an alcohol usu­ ally converts C - H bonds to C - 0 bonds. If we convert an alcohol to an alkane, the carbinol carbon loses its bond to oxygen and gains another bond to hydrogen. Figure 1 1- 1 compares the oxidation states of primary, secondary, and tertiary alcohols with those obtained by oxidation or reduction. The symbol [0] indicates an unspecified oxidizing agent. Notice that oxidation of a primary or secondary alcohol forms a carbonyl ( C = O ) group by the removal of two hydrogen atoms: one from the carbinol carbon and one from the hydroxyl group. A tertiary alcohol cannot easily oxidize because there is no hydrogen atom available on the carbinol carbon.

P R O B L E M 1 1- 1

Classify each reaction as an oxidation, a reduction, or neither.

o

o

II

H-C-H

---7

II

HO-C-OH

462

(c)

Chapter H3C CH3 C H3 CH3-HOC-COH UC"-H

1 1 : Reactions of Alcohols

I

I I

I

H+

--?

II

°

(e)

0 (i) 0 0

(g)

(k)

------7

HEr

Os04

------?

H 2 02

a:Br

rY0H

V

(h)

0H (XOH BH2 ()

�

O + H20

OH (X",.'OH

BR, · THF

11-2 Oxidation of Alcohols

Primary and secondary alcohols are easily oxidized b y a variety of reagents, including chromium reagents, permanganate, nitric acid, and even household bleach (NaOCl, sodium hypochlorite). The choice of reagent depends on the amount and value of the alcohol. We use cheap oxidants for large-scale oxidations of simple, inexpensive alco­ hols. We use the most effective and selective reagents, regardless of cost, for delicate and valuable alcohols. In this chapter, we study only the oxidants that have the widest range of uses and the best selectivity. An understanding of the most common oxidants can later be extended to include additional reagents.

1 1-2A

Oxidation of Secondary Alcohols

Secondary alcohols are easily oxidized to give excellent yields of ketones. The chromic acid reagent is often best for laboratory oxidations of secondary alcohols. OH

I

R - CH- R '

Na2Cr20/ H2 S04

)

o

I

R-C-R'

secondary alcohol

ketone

Example

cyclohexanol

cyclohexanone

(90%)

The chromic acid reagent is prepared by dissolving sodium dichromate, ( Na2Cr207 ) in a mixture of sulfuric acid and water. The active species in the mixture is probably chromic acid, H2Cr04, or the acid chromate ion, HCrO;. Adding chromium trioxide ( Cr03 ) to dilute sulfuric acid achieves the same result.

+

N�Crz07

+

HzO

o

2 H2S04

2 HO - Cr - OH o

I

o

+

+

I

sodium dichromate

Cr03

1 1 -2 Oxidation of Alcohols

H+

o

II

I II

+

- O - Cr - OH o

acid chromate ion

chromic acid

The mechanism of chromic acid oxidation probably involves the formation of a chromate ester. Elimination of the chromate ester gives the ketone. In the elimination, the carbinol carbon retains its oxygen atom but loses its hydrogen and gains the second bond to oxygen. Formation of the chromate ester

R'

I

R-C-O-H

I

+

o

R - C - O - Cr - OH

o

II

I

I

+

0

H

chromate ester

chromic acid

alcohol

I

I

H - O - Cr - OH

H

0

R'

I

Elimination of the chromate ester and oxidation of the carbinol carbon

R'

"0-

I

..

1--'

" V II�

R'

II

�

R - C - O - Cr - OH

" .

HzO

J

I

.

R - C = O.:

+

"0-

..0..

.

� '. /Cr- OH

- ..0., '

Cr (IV)

Cr (VI)

The chromium(IV) species formed reacts further to give the stable reduced form, chromium(III). Both sodium dichromate and chromic acid are orange, while chromic 3 ion (Cr + ) is a deep blue. One can follow the progress of a chromic acid oxidation by observing the color change from orange through various shades of green to a greenish blue. In fact, the color change observed with chromic acid can be used as a test for the presence of an oxidizable alcohol.

1 1 -2 8

Oxidation of Pri mary Alcohols

Oxidation of a primary alcohol initially forms an aldehyde. Unlike a ketone, however, an aldehyde is easily oxidized further to give a carboxylic acid. OH

I

R - CH - H primary alcohol

[0]

-7

o

II

R - C-H aldehyde

[ 0]

-7

2 HSO:;-

o

HO - Cr - OH

chromium trioxide

+

chromic acid (H2Cr04)

I

H zO

2 Na+

o

I

R - C - OH carboxylic acid

Obtaining the aldehyde is often difficult, since most oxidizing agents strong enough to oxidize primary alcohols also oxidize aldehydes. Chromic acid generally oxidizes a primary alcohol all the way to the carboxylic acid.

463

C ha

464

pte

2° > 1 °. As in other carbocation reactions, realTangements are common. With primary alcohols, reanangement and isomerization of the products are so common that acid-catalyzed dehydration is rarely a good method for converting them to alkenes. The following mechanism shows how I -butanol undergoes dehydration with reanangement to give a mixture of I -butene and 2-butene. The more highly sub­ stituted product, 2-butene, is the major product, in accordance with the Zaitsev rule (Section 6- 1 9).

HHHH H-C-C-C-C-H H-O: H+ H H \.... H

HI )=+J IHl H H H-C-C-C-C-H H-O·+ H H H H

ionization of the protonated alcohol, with rearrangement I

I

I

I

I

I

I

I

I

( I.

I

I

I

I

I

H20 leaves H : - migrates

I

)

H H � + H-C-C-C-C-H HHHH I

I

I

I

I

I

secondary carbocation

>, 01)

� i3

I

I

-C-C-

I

H

I

OH H2S04

I

I

+

-C-C -

I

H

I

OH2 .... Figure 11-2

HS04-

reaction coordinate --------------- --_ . - ---

Reaction-energy diagram for dehydration of an alcohol.

481

482

Chapter 1 1 : Reactions of Alcohols Loss of either proton to give two products

H H H + I I I H-C - C - C - C - H I Hb

� chJ�

�

� B :-�

secondary carbocation

H H H H H H I I I I I I H-C-C=C-C-H + H-C=C-C - C - H I I I H Hb H

[ loss of H: [

2-butene (major, 70%) a disu bsti tuted alkene

� chJ�

loss of H: I-butene (minor, 30%) a monosubstituted alkene

Let's review the utility of dehydration and give guidelines for predicting the products :

1. Dehydration usually goes by the E l mechanism. Rearrangements may occur to form more stable carbocations.

2. Dehydration works best with tertiary alcohols and almost as well with secondary alcohols. Rearrangements and poor yields are common with primary alcohols.

3. (Zaitsev 's rule) If two or more alkenes might be formed by deprotonation of the carbocation, the most highly substituted alkene usually predominates.

Solved Problem 1 1 -3 shows how these rules are used to predict the products of dehydrations. The carbocations are drawn to show how rearrangements occur and how more than one product may result. PROBLEM-SOLVING

HinZ;

Most alcohol dehydrations go by E I mechanisms i nvolving protonation of the OH group, fo l l owed by loss of water.

S O LV E D P R O B L E M 1 1- 3

Predict the products of sulfuric acid-catalyzed dehydration of the following alcohols. (b) neopentyl alcohol (a) I -methyIcyclohexanol SOLUTION

(a) I -Methylcyclohexanol reacts to form a tertiary carbocation. A proton may be abstract­ ed from any one of three carbon atoms. The two secondary atoms are equivalent, and abstraction of a proton from one of these carbons leads to the trisubstituted double bond of the major product. Abstraction of a methyl proton leads to the disubstituted double bond of the minor product. H

(XC",,-1 / +

H

l -methylcyclohexanol

H

cation

protonated

loss of Ha major product (trisubstituted)

H

H

loss of Hb minor product (disubstituted)

(b) Neopentyl alcohol cannot simply ionize to form a primary cation. Rearrangement occurs as the leaving group leaves, giving a tertiary carbocation. Loss of a proton from the adjacent secondary carbon gives the trisubstituted double bond of the major

1 1 - 1 0 Dehydration Reactions of Alcohols

483

product. Loss of a proton from the methyl group gives the disubstituted double bond of the minor product.

3

CH3

I

. :- w

CH -C-CH2OH

I

•.

CH3

�

CH3

(" +

I�

+

CH 3 -C-CH,-OH, - .. -

I

CH3

CH3

neopenlyl alcohol

ionization with

(2,2-dimcthyl- l -propanol )

rearrangement

3°

+

loss of

I

CH3 -C -CH2

H

I

CH3 cation

CH,CH3

""/ /"" C =C

H;

H

Hb

CH3

loss of

minor product

major product

(disubstitllted)

(trisubstituted)

PROBLEM-SOLVING

P R O B L E M 1 1-22

Predict the products of the sulfuric acid-catalyzed dehydration of the following alcohols. When more than one product is expected, label the major and minor products. (a) 2-methyl-2-butanol (b) I -pentanol (c) 2-pentanol (d) l -isopropylcyclohexanol (e) 2-methylcyclohexanol P R O B L E M 1 1-23

Some alcohols undergo rearrangement or other unwanted side reactions when they dehydrate in acid. Alcohols may be dehydrated under mildly basic conditions lIsing phosphorus oxy­ chloride ( POCI3 ) in pyridine. The alcohol reacts with phosphorus oxychloride much like it reacts with tosyl chloride (Section 1 1 -5), displacing a chloride ion from phosphorus to give an alkyl dichlorophosphate ester. The dichlorophosphate group is an outstanding leaving group. Pyridine reacts as a base with the dichlorophosphate ester to give an E2 elimination. Propose a mechanism for the dehydration of cyclohexanol by POCI3 in pyridine. a

II

CI

/P"

I

CI

CI

phosphorus oxychloride

1 1 -10B

o N

pyridine

Bimolecu l a r Dehydration to Form Ethers (Ind ustrial)

In some cases, a protonated primary alcohol may be attacked by another molecule of the alcohol and undergo an SN2 displacement. The net reaction is a bimolecular dehy­ dration to form an ether. For example, the attack by ethanol on a protonated molecule of ethanol gives diethyl ether. CH3

/ CH3CH?� 0 - C, .

nucleophilic

electrophilic

'

+

.· 0!·

. - �k� protonated ether

H

"'H water

Bimolecular dehydration can be used to synthesize symmetrical dialkyl ethers from simple, unhindered primary alcohols. This method is used for the industrial synthesis of

HtftJ/

Draw the carbocation, look for poss i b l e rearra ngements, then consider all the ways that the orig inal carbocation a n d any rearranged carbocation m i ght lose protons to g ive a l kenes. Za itsev's rule usually pred i cts the major product.

diethyl ether

484

Chapter 1 1 : Reactions of Alcohols diethyl ether ( CH3CH2 - O - CH2CH 3 ) and dimethyl ether (CH3 - O - CH 3 ) . Under the acidic dehydration conditions, two reactions compete: Elimination (to give an alkene) competes with substitution (to give an ether). Substitution to give the ethel; a bimolecular dehydration H2S04, 140°C

)

CH3CH2 - O - CH2CH3

diethyl ether

ethanol

+

H20

Elimination to give the alkene, a unimolecular dehydration

CH2 = CH2 +

H20

ethylene P R O B L E M 1 1-24

Contrast the mechanisms of the two preceding dehydrations of ethanol. How can we control these two competing dehydrations? The ether synthesis (substitution) shows two molecules of alcohol giving two product molecules: one of diethyl ether and one of water. The elimination shows one molecule of alcohol giv­ ing two molecules: one of ethylene and one of water. The elimination results in an increase in the number of molecules and therefore an increase in the randomness (entropy) of the system. The elimination has a more positive change in entropy ( ilS) than the substitution, and the -TilS term in the Gibbs free energy becomes more favorable for the elimination as the temperature increases. Substitution (to give the ether) is favored around 1 40aC and below, and elimination is favored around I 80aC and above. Diethyl ether is produced industrially by heating ethanol with an acidic catalyst at around 1 40aC. P R O B L E M 1 1-25

Explain why the acid-catalyzed dehydration is a poor method for the synthesis of an unsym­ metrical ether such as ethyl methyl ether, CH3CH2 - O - CH3. P R O B L E M 1 1 -26

Propose a detailed mechanism for the following reaction. H2S04, heat

)

Propos i n g Reaction Mech a n isms

P R O B L E M - S O LV I N G S T R AT E G Y

In view of the large number of reactions we've covered, proposing mechanisms for reactions you have never seen may seem nearly impossible. As you gain experience in working mech­ anism problems, you will start to see similarities to known reactions. Let's consider how an organic chemist systematically approaches a mechanism problem. (A more complete ver­ sion of this method appears in Appendix 4.) Although this stepwise approach cannot solve all mechanism problems, it should provide a starting point to begin building your experience and confidence. Determ i ning the Type of Mechanism

First, determine what kinds of conditions and catalysts are involved. In general, reactions may be classified as involving (a) strong electrophiles (including acid-catalyzed reactions),

1 1 - 1 0 Dehydration Reactions of Alcohols (b) strong nucleophiles (including base-catalyzed reactions), or (c) free radicals. These three types of mechanisms are quite distinct, and you should first try to determine which type is involved. (a) In the presence of a strong acid or a reactant that can dissociate to give a strong elec­

trophile, the mechanism probably involves strong electrophiles as intermediates. Acid­ catalyzed reactions and reactions involving carbocations (such as the SNl , the E l , and most alcohol dehydrations) fall in this category. (b) In the presence of a strong base or a strong nucleophile, the mechanism probably involves

strong nucleophiles as intermediates. Base-catalyzed reactions and those depending on base strength (such as the SN2 and the E2) generally fall in this category. (c) Free-radical reactions usually require a free-radical initiator such as chlorine, bromine,

NBS, or a peroxide. In most free-radical reactions, there is no need for a strong acid or base.

Once you have determined which type of mechanism you will write, there are gener­ al methods for approaching the problem. At this point, we consider mostly the electrophilic reactions covered in recent chapters. Suggestions for drawing the mechanisms of reactions involving strong nucleophiles and free-radical reactions are collected in Appendix 4. Reactions I nvolving Strong E lectroph iles

When a strong acid or electrophile is present, expect to see intermediates that are strong acids and strong electrophiles; cationic intermediates are common. Bases and nucle­ ophiles in such a reaction are generally weak, however. Avoid drawing carbanions, alkox­ ide ions, and other strong bases. They are unlikely to coexist with strong acids and strong electrophiles. Functional groups are often converted to carbocations or other strong electrophiles by protonation or reaction with a strong electrophile; then the carbocation or other strong elec­ trophile reacts with a weak nucleophile such as an alkene or the solvent. 1. Consider the carbon skeletons of the reactants and products, and decide which car­ bon atoms in the products are most likely derived from which carbon atoms in the reactants. 2. Consider whether any of the reactants is a strong enough electrophile to react with­ out being activated. If not, consider how one of the reactants might be converted to a strong electrophile by protonation of a Lewis basic site (or complexation with a Lewis acid).

Protonation of an alcohol, for example, converts it to a strong electrophile, which can undergo attack or lose water to give a carbocation, an even stronger electrophile. Protona­ tion of an alkene converts it to a carbocation.

3. Consider how a nucleophilic site on another reactant (or, in a cyclization in another part of the same molecule) can attack the strong electrophile to form a bond needed in the product. Draw the product of this bond formation.

If the intermediate is a carbocation, consider whether it is likely to rearrange to form a bond in the product. If there isn't any possible nucleophilic attack that leads in the direction of the product, consider other ways of converting one of the reactants to a strong electrophile. 4. Consider how the product of a nucleophilic attack might be converted to the final product (if it has the right carbon skeleton) or reactivated to form another bond needed in the product.

To move a proton from one atom to another (as in an isomerization), try adding a proton to the new position, then removing it from the old position. 5. Draw out all steps of the mechanism using curved arrows to show the movement of electrons.

Be careful to show only one step at a time. Com mon M istakes to Avoid in Draw i n g Mechan isms

1.

Do not use condensed or line-angle formulas for reaction sites. Draw all the bonds and all the substituents of each carbon atom affected throughout the mechanism. In ( Continued)

485

486

Chapter 1 1 : Reactions of Alcohols reactions involving strong electrophiles and acidic conditions, three-bonded carbon atoms are likely to be carbocations. If you draw condensed formulas or line-angle for­ mulas, you will likely misplace a hydrogen atom and show a reactive species on the wrong carbon. 2.

Do not show more than one step occurring at once. Do not show two or three bonds chang­ ing position in one step unless the changes really are concerted (take place simultaneous­ ly). For example, protonation of an alcohol and loss of water to give a carbocation are two steps. You must not show the hydroxyl group "jumping" off the alcohol to join up with an anxiously waiting proton.

3.

Remember that curved arrows show movement of electrons, always from the nucleophile (electron donor) to the electrophile (electron acceptor). For example, protonation of a double bond must show the arrow going from the electrons of the double bond to the pro­ ton- never from the proton to the double bond. Resist the urge to use an arrow to "point out" where the proton (or other reagent) goes.

SAMPLE PROBLEM

To illustrate the stepwise method for reactions involving strong electrophiles, we will devel­ op a mechanism to account for the following cyclization:

OR The cyclized product is a minor product in this reaction. Note that a mechanism problem is different from a synthesis problem: In a mechanism problem, we are limited to the reagents given and are asked to explain how these reactants form these products under the conditions shown. Also, a mechanism problem may deal with how an unusual or unexpected minor product is formed. In the presence of sulfuric acid, this is clearly an acid-catalyzed mechanism. We ex­ pect strong electrophiles, cationic intermediates (possibly carbocations), and strong acids. Carbanions, alkoxide ions, and other strong bases and strong nucleophiles are unlikely. 1. Consider the carbon skeletons of the reactants and products, and decide which carbon atoms in the products are most likely derived from which carbon atoms ' in the reactants.

Drawing the statting material and the product with all the substituents of the affected carbon atoms, we see the major changes shown here. A vinyl hydrogen must be lost, a C - C bond must be formed, a methyl group must move over one carbon atom, and the hydroxyl group must be lost.

=

2. Consider whether any of the reactants is a strong enough electrophile to react with­ out being activated. If not, consider how one of the reactants might be converted to a strong electrophile by protonation of a Lewis basic site (or complexation with a Lewis acid).

The starting material is not a strong electrophile, so it must be activated. Sulfuric acid could generate a strong electrophile either by protonating the double bond or by proto­ nating the hydroxyl group. Protonating the double bond would form the tertiary carboca­ tion, activating the wrong end of the double bond. Also, there is no good nucleophilic site

1 1 - 1 0 Dehydration Reactions of Alcohols on the side chain to attack this carbocation to form the correct ring. Protonating the dou­ ble bond is a dead end.

does not lead toward product

The other basic site is the hydroxyl group. An alcohol can protonate on the hydroxyl group and lose water to form a carbocation.

3. Consider how a nucleophilic site on another reactant (or, in a cyclization, in anoth­ er part of the same molecule) can attack the strong electrophile to form a bond needed in the product. Draw the product of this bond forrnation.

The carbocation can be attacked by the electrons in the double bond to form a ring; but the positive charge is on the wrong carbon atom to give a six-membered ring. A favorable rearrangement of the secondary carbocation to a tertiary one shifts the positive charge to the correct carbon atorn and accomplishes the methyl shift we identified in Step 1 . Attack by the (weakly) nucleophilic electrons in the double bond gives the correct six-membered ring.

4. Consider how the product of nucleophilic attack might be converted to the final product (if it has the right carbon skeleton) or reactivated to forrn another bond needed in the product. Loss of a proton (to HS04 or H20, but not to -OH, which is not compatible!) gives the

observed product.

5.

Draw out all steps of the mechanism using curved arrows to show the movement of electrons.

Combining the equations written immediately above gives the complete mechanism for this reaction. The following problems require proposing mechanisms for reactions involving strong electrophiles. Work each one by completing the five steps just described. ( Continued)

487

488

Ch a

t

11:

p er Reactions of Alcohols Propose�O a mechaniH sm for each re c io 0 P R O B L E M 1 1-27

a t n.

(a)

(b)

(c)

1 1 -1 1 U n iq u e Reactions of Diols

H2S04, heat

00CH3 U

)

W

Hp

I

�

ero CHpH o Q--CH, II

+

+

1 1-1 1 A

+

(f

The Pinacol Rea rrangement

Using our knowledge of alcohol reactions, we can explain results that seem strange at first glance. The following dehydration is an example of the pinacol rearrangement:

H3C

CH3

HO

OH

1 1 H3 C - C - C - CH3 1 1 pinacol (2,3-dimethyl-2,3-butanediol)

�

CH3

1

H3C - - - CH3 + H20 o

T

CH3

pinacolone (3,3-dimethyl-2-butanone)

The pinacol rearrangement is formally a dehydration. The reaction is acid­ catalyzed, and the first step is protonation of one of the hydroxyl oxygens. Loss of water gives a tertiary carbocation, as expected for any tertiary alcohol. Mi gration of a methyl group places the positive charge on the carbon atom bearing the second - OH group, where oxygen' s nonbonding electrons help to stabilize the charge through resonance. This extra stability i s the driving force for the rearrangement. Deprotonation of the resonance-stabilized cation gives the product, pinacolone.

MECHANISM 1 1-5 Step 1:

The PililaGol Rearrangement

Protonation of a hydroxyl group.

Step 2: Loss of water gives a carbocation.

H3C

CH3

1 1 H3 C - C - C - CH3

+

H+

�/

HQI : : QI TT

H3C CH3 I I H3C-C-C-CH3

1

I)

HO : +OH 2 .. ••

H 3C CH3 1 / H3C -C-C+ "1 CH3 HO :

+

H20

-

[

Step 3: Methyl migration fOnTIS a resonance-stabilized carbocation.

H3

C

(methyl migration)

)

l

1 1- 1 1 Unique Reactions of Diols

�

CH 3

CH 3

I H 3C-C-C-CH 3 I I

489

H 3C-C- -CH 3 II I H-O+ CH3

+

H-Q: CH 3

••

resonance-stabilized carbocation

Step 4: Deprotonation gives the product.

I

H 3C-C-C-CH 3

)0:I

H

CH 3

CH 3

CH 3 +

I

..

CH 3

I H 3C-C-C-CH 3 I I



""""' C = O ,../

+

Hinz;

By analogy with the pinacol rearrangement, watch for carbocation rearrangements that place the + charge on a carbinol carbon atom.

"'/ O=C.... ...

ketones and aldehydes

490

C h apter 1 1 : Reactions of Alcohols

1 ,2-Diols (glycols), such as those formed b y hydroxylation o f alkenes, are cleaved by periodic aci d ( HI04 ) . The products are the same ketones and aldehydes that would be formed by ozonolysis-reduction of the alkene. Hydroxylation followed by periodic acid cleavage serves as a useful alternative to ozonolysis, and the periodate cleavage by i tself is useful for determining the structures of sugars (Chapter 23). Periodic acid cleavage of a glycol probably involves a cyclic periodate interme­ diate like that shown here.

� FOO =
H (d) �� H H ___

To an organic chemist, the term ester normally means an ester of a carboxylic acid, unless some other kind of ester is specified. Replacing the -OH group of a car­ boxylic acid with the - OR group of an alcohol gives a carboxylic ester. The follow­ ing reaction, called the Fischer esterification, shows the relationship between the alcohol and the acid on the left and the ester and water on the right.

Este r ificat i o n o f Alco h o l s

R -O

-EJ

alcohol

I H -O+� - R' o

+

acid

o

II

R -O-C-R'

+

ester

I H-O-H I

For example, if we mix isopropyl alcohol with acetic acid and add a drop of sul­ furic acid as a catalyst, the following equilibrium results.

i -EJ

CH3

H- - o CH3

isopropyl al cohol

acetic acid

�H3 � H - C -O - C -CH I CH3 isopropyl acetate

3

+

H-O-H water

1 1 - 1 3 Esters of Inorganic Acids

Because the Fischer esterification is an equilibrium (often with an unfavorable equilibri­ um constant), clever techniques are often required to achieve good yields of esters. For example, we can use a large excess of the alcohol or the acid. Adding a dehydrating agent removes water (one of the products), dJiving the reaction to the right. There is a more powerful way to form an ester, however, without having to deal with an unfavor­ able equilibrium. An alcohol reacts with an acid chloride in an exothermic reaction to give an ester.

R-O

-EJ

§� - R'

�

o

+

alcohol

o

R-O- -R'

acid chloride

+

ester

491

The alcohol groups of unpleasant­ tasting drugs are often converted to esters in order to mask the taste. In most cases, the ester has a less unpleasant taste than the free alcohol.

�

The mechanisms of these reactions that form acid derivatives are covered with similar mechanisms in Chapter 21. PROBLEM 1 1 - 3 1

II

Show the alcohol and the acid chloride you would use to make the following esters. 0 o

II

(b) CHiCH2)3-0- C - CH2CH3 II-butyl propionate

(a) CH3CH2CH2C -OCH2CH2CH, n-propyl butyrate

(c) H3C

--
-O-!--< > o

o (d)

O- - CH(CH3)2

cyclopropyl benzoate

p-tolyl isobutyrate

11-13

In addition to forming esters with carboxylic acids, alcohols form inorganic esters with inorganic acids such as nitric acid, sulfuric acid, and phosphoric acid. In each type of ester, the alkoxy (-OR) group of the alcohol replaces a hydroxyl group of the acid, with loss of water. We have already studied tosylate esters, composed of para-toluenesulfonic acid and alcohols (but made using tosyl chloride, Section 11-5). Tosylate esters are analogous to sulfate esters (Section 1 1 - 1 3A), which are composed of su lfuric acid and alcohols.

R-O

-EJ

+

-o

�L II 0

--CH3

_

0 para-tollleneslllfonic acid (TsOH)

alcohol

(

�II

-o-

Este rs of I n orga n ic Acids

0 R-O-

)

CH3

I H20 I

+

0 para-tollleneslllfonate ester (ROTs) _

Made using tosyl chloride

R -O

-EJ

alcohol

+

@3-- �II

-o-

0

_

�I

-o-

0 CH3

0 para-toillenesllifonyl chloride (TsCI)

pyridine

�

R-O-

CH3

_

0 tosylate ester (ROTs)

+

I H CI I

492

Chap ter 11: Reactions of Alcohols 11-13A

Su lfate Esters

A sulfate ester is like a sulfonate ester, except there is no alkyl group directly bonded to the sulfur atom. In an alkyl sulfate ester, alkoxy groups are bonded to sulfur through oxygen atoms. Using methanol as the alcohol,

I CH3 -O-S-OH I

sulfuric acid

methyl sulfate

o

o

The body converts the hydroxyl groups of some drugs to their sul­ fate derivatives in order to produce water-soluble compounds that are readily excreted. The reaction is not as common as it might be because of limited availability of inorganic sulfate in the body.

o

o

o

� II � S-OH II

+

II II

1 H20 1

CH3 -O- S-O-CH3 o

+

Hp

dimethyl sulfate

Sulfate ions are excellent leaving groups. Like sulfonate esters, sulfate esters are good electrophiles. Nucleophiles react with sulfate esters to give alkylated products. For example, the reaction of dimethyl sulfate with ammonia gives a sulfate salt of methylamine, CH3NHt CH30S03". ·

0

·

·

H"" � I CH3 -O-S-O-CH3 H-N: H/ 10 · I ..0.. ..

ammonia

dimethyl sulfate

H

"0

.

.

1+ 1

:O-S-O-CH3

methylammonium ion

methylsulfate ion

.. ..

H-N-CH3 H

I II

..0. .

PROB LEM 1 1 - 3 2

Use resonance structures to show that the negative charge in the methylsulfate anion is shared equally by three oxygen atoms.

11-13 B

Nitrate Esters

Nitrate esters

are formed from alcohols and nitric acid. R-O- Nt"

°

""0-

alcohol

nitric acid

+

alkyl nitrate ester

I H- O - H I ·

.

The best known nitrate ester is "nitroglycerine," whose systematic name is glyceryl trinitrate. Glyceryl trinitrate results from the reaction of glycerol ( 1 ,2,3-propanetriol) with three molecules of nitric acid.

iI H-0-N02

CH -O-NO2 Z

CHz- 0- N02 Illustration of Alfred Nobel operating the apparatus he used to make nitroglycerine. The temperature must be monitored and controlled carefully during this process; therefore, the operator's stool has only one leg to ensure that he stays awake.

glycerol (glycerine)

nitric acid

+

1

3 H20

1

glyceryl trinitrate (nitroglycerine)

Nitroglycerine was first made in 1847 and was found to be a much more pow­ erful explosive than black powder, which is a physical mixture of potassium nitrate, sulfur, and charcoal. In black powder, potassium n itrate i s the oxidizer, and sulfur and charcoal provide the fuel to be oxidized. The rate of a black powder explosion is

1 1 - 1 3 Esters of I norganic Acids limited by how fast oxygen from the grains of heated potassium nitrate can diffuse to the grains of sulfur and charcoal. A black powder explosion does its work by the rapid increase in pressure resulting from the reaction. The explosion must be con­ fined, as in a cannon or a firecracker, to be effective. In nitroglycerine, the nitro groups are the oxidizer and the CH and CH 2 groups are the fuel to be oxidized. This intimate association of fuel and oxidizer allows the explosion to proceed at a much faster rate, forming a shock wave that propagates through the explosive and initiates the reaction. The explosive shock wave can shatter rock or other substances without the need for confinement. Because of its unprecedent­ ed explosive power, nitroglycerine was called a high explosive. Many other high explo­ sives have been developed, including picric acid, TNT (trinitrotoluene), PETN (pentaerythritol tetranitrate), and RDX (research department explosive). Nitroglycerine and PETN are nitrate esters.

¢

493

Nitroglycerin i s used t o relieve angina, a condition where the heart is not receiving enough oxygen. Angina is characterized by severe pain in the chest, often triggered by stress or exercise. Workers with angina at the Nobel dynamite plant discovered this effect. They noticed that their symptoms improved when they went to work. In the mitochondria, nitroglycerin is me­ tabolized to NO (nitric oxide), which regulates and controls many meta­ bolic processes.

OH

02N

N02

N02

RDX

PETN

picric acid

Pure nitroglycerine is hazardous to make, use, and transport. Alfred Nobel's family were experts at making and using nitroglycerine, yet his brother and several workers were killed by an explosion. In 1 866, Nobel found that nitroglycerine soaks into diatomaceous earth to give a pasty mixture that can be molded into sticks that do not detonate so easily. He called the sticks dynamite and founded the firm Dynamit Nobel, which is still one of the world's leading ammunition and explosives manufac­ turers. The Nobel prizes are funded from an endowment that originated with Nobel's profits from the dynamite business.

1 1 -1 3C

Pho s p h ate Ester s

Alkyl phosphates are composed of 1 mole of phosphoric acid combined with moles of an alcohol. For example, methanol forms three phosphate esters. o

� II � P - OH I

I

I

1, 2, or 3

"

o

CH 3-O -P - OH

+

OH

phosphoric acid

"

HzO

I

OH

mono methyl phosphate

"

o

I

o

CH} - O-P - OH

CH3 - O -P-O-CH3

+

+

HzO

O - CH}

dimethyl phosphate

HzO

I

O - CH3

trimethyl phosphate

Phosphate esters play a central role in biochemistry. Figure 1 1 -3 shows how phosphate ester linkages compose the backbone of the nucleic acids RNA (ribonucleic acid) and DNA (deoxyribonucleic acid). These nucleic acids, which carry the genetic informa­ tion in the cell, are discussed in Chapter 23.

By controlling the formation of phosphate esters on key proteins, the body is able to regulate many cellular processes. Any disruption of these processes can result in numerous health problems, includ­ ing cancer, diabetes, and obesity.

494

Chapter 11: Reactions of Alcohols

� Figure 11-3

P hosphate ester groups bond the individual nucleotides together in DNA. The "base" on each of the nucleotides corresponds to one of the four heteroxyclic bases of DNA (see Section 23-20).

11-14

Reactions of Alkoxides

In Section 1O-6B, we learned to remove the hydroxyl proton from an alcohol by reduc­ tion with an "active" metal such as sodium or potassium. This reaction generates a sodium or potassium salt of an alkoxide ion and hydrogen gas.

.. ..

R-O-H

+

Na

R-O-H

+

K

The reactivity of alcohols toward sodium and potassium decreases in the order: methyl > 1 ° > 2° > 3°. Sodium reacts quickly with primary alcohols and some secondary alcohols. Potassium is more reactive than sodium and is commonly used with tertiary alcohols and some secondary alcohols. Some alcohols react sluggishly with both sodium and potassium. In these cases, a useful alternative is sodium hydride, usually in tetrahydrofuran solution. Sodium hydride reacts quickly to form the alkoxide, even with difficult compounds .

..

R-O-H

alcohol

Sodium metal reacts vigorously with simple primary alcohols such as ethanol.

+

NaH

sodium hydride

THF

hydrogen

sodium alkoxide

The alkoxide ion is a strong nucleophile as well as a powerful base. Unlike the alcohol itself, the alkoxide ion reacts with primary alkyl halides and tosylates to form ethers. This general reaction, called the Williamson ether synthesis, is an SN2 dis­ placement. The alkyl halide (or tosylate) must be primary so that a back-side attack is not hindered. When the alkyl halide is not primary, elimination usually results.

"N�-::"''''''' KEY MECHANISM 11-6

This is the most important method for making ethers. Step

1:

�

Form the alkoxide of the alcohol having the more hindered group .

R-O-H

+ Na (or NaH or K)

R - O: - Na+

alkoxide ion

+

+H2

!

1 1 - 14 Reactions of Alkoxides Step 2: The alkoxide displaces the leaving group of a good SN2 substrate. .. .. (' R - Q -CH2 -R' R -O:,-:Na R'--:CH -X ____ ___ 2 ••

alkoxide ion

primary halide or tosylate

+

495

NaX

ether

EXAMPLE: Synthesis of cyclopentyl ethyl ether

Step

1:

Form the alkoxide of the alcohol with the more h i n dered gro up .

Step

2:

The alkoxide displaces the leaving group of a good SN2 substrate.

� O -CH2 -CH3 ------? �

+

Na+ Br-

PROBLEM: Why is the cyclohexyl group chosen for the alkoxide and the ethyl group chosen for the halide? Why not use cyclohexyl bromide and sodium ethoxide to make cyclopentyl ethyl ether?

In using the Williamson ether synthesis, one must remember that the alkyl halide (or tosylate) must be a good SN2 substrate, usually primary. In proposing a Williamson synthesis, we usually choose the less hindered alkyl group to be the halide (or tosylate) and the more hindered group to be the a l ko xi de i on .

PROBLEM 1 1 - 3 3

A good Williamson synthesis of ethyl methyl ether would be sodium ethoxide

melhyl iodide

ethyl melhyl ether

What is wrong with the following proposed synthesis of ethyl methyl ether? First, ethanol i s treated with acid to protonate the hydroxyl group (making it a good leaving group), and then methoxide is added to displace water.

(incorrect synthesis of ethyl methyl ether) PROBLEM 1 1 - 3 4

(a)

Show how ethanol and cyclohexanol may b e used to synthesize cyclohexyl ethyl ether (tosylation followed by the Williamson ether synthesis). (b) Why can't we synthesize this product simply by mixing the two alcohols, adding some sulfuric acid, and heating? PROBLEM 1 1 - 3 5

A student wanted to make (R)-2-ethoxybutane, using the Williamson ether synthesis. He remembered that the Williamson synthesis involves an SN2 displacement, which takes place with inversion of configuration. He ordered a bottle of (S)-2-butanol for his chiral starting material. He also remembered that the SN2 goes best on primary halides and tosylates, so he

PROBLEM-SOLVING

Hi-ItZ;

In using the Williamson ether synthesis to make R- 0 -R', choose the less hindered al kyl group to serve as the alkyl halide (R' -X), because it will make a better SN2 substrate. Choose the more hindered alkyl group to form the alkoxide (R-O-), because it is less sensitive to steric hindrance in the reaction. R-O-

+

R'-X

---'>

R-O-R'

496

Chapter 1 1 : Reactions of Alcohols made ethyl tosylate and sodium (S)-2-butoxide. After warming these reagents together, he obtained an excellent yield of 2-ethoxybutane. (a) What enantiomer of 2-ethoxybutane did he obtain? Explain how this enantiomer results from the SN2 reaction of ethyl tosylate with sodium (S)-2-butoxide. (b) What would have been the best synthesis of (R)-2-ethoxybutane? (c) How can this student convert the rest of his bottle of (S)-2-butanol to (R)-2-ethoxybutane? P R O B L E M 11-36

The anions of phenols (phenoxide ions) may be used in the Williamson ether synthesis, espe­ cially with very reactive alkylating reagents such as dimethyl sulfate. Using phenol, dimethyl sulfate, and other necessary reagents, show how you would synthesize methyl phenyl ether.

PRO B LEM-SOLVING STRATEGY

Multistep Synthesis

Chemists use organic syntheses both to make larger amounts of useful natural compounds and to invent totally new compounds in search of improved properties and biological effects. Synthesis also serves as one of the best methods for developing a firm command of organic chemistry. Planning a practical multistep synthesis requires a working knowledge of the applications and the limitations of a variety of organic reactions. We will often use synthesis problems for reviewing and reinforcing the reactions we have covered. We use a systematic approach to solving multistep synthesis problems, working back­ ward, in the "retrosynthetic" direction. We begin by studying the target molecule and con­ sidering what final reactions might be used to create it from simpler intermediate compounds. Most syntheses require comparison of two or more pathways and the interme­ diates involved. Eventually, this retrosynthetic analysis should lead back to starting materi­ als that are readily available or meet the requirements defined in the problem. We can now extend our systematic analysis to problems involving alcohols and the Grignard reaction. As examples, we consider the syntheses of an acyclic diol and a disubsti­ tuted cyclohexane, concentrating on the crucial steps that assemble the carbon skeletons and generate the final functional groups. SAMPLE PROBLEM

Our first problem is to synthesize 3-ethyl-2,3-pentanediol from compounds containing no more than three carbon atoms.

I

I -

CH,CH3

I

CH3 -CH-C-CH 2 -CH OH

OH

3

3-ethyl-2,3-pentanediol

1. Review the functional groups and carbon skeleton of the target compound.

The compound is a vicinal diol (glycol) containing seven carbon atoms. Glycols are com­ monly made by hydroxylation of alkenes, and this glycol would be made by hydroxyla­ tion of 3-ethyl-2-pentene, which effectively becomes the target compound. CH2-CH3 I CH3- CH=C-CH2 -CH3 3-ethyl-2-pentene

KMn04 cold, dilute (or other methods)

CH2-CH3 I CH3-CH-C-CH2-CH3 I I OH OH 3 -ethyl-2,3-pentanediol

2. Review the functional groups and carbon skeletons of the starting materials (if spec­

ified), and see how their skeletons might fit together into the target compound.

The limitation is that the starting materials must contain no more than three carbon atoms. To form a 7-carbon product requires at least three fragments, probably a 3-carbon fragment and two 2-carbon fragments. A functional group that can be converted to an

1 1 - 1 4 Reactions of Alkoxides alkene will be needed on either C2 or C3 of the chain, since 3-ethyl-2-pentene has a dou­ ble bond between C2 and C3.

I -

CH?-CH3

CH3-CH=C-CH2-CH3

3. Compare methods for assembling the carbon skeleton of the target compound. Which

ones provide a key intermediate with the correct carbon skeleton and functional groups correctly positioned for conversion to the functionality in the target molecule? At this point, the Grignard reaction is our most powerful method for assembling a carbon skeleton, and Grignards can be used to make primary, secondary, and tertiary alcohols (Section 1 0-9). The secondary alcohoI3-ethyl-2-pentanol has its functional group on C2, and the tertiary alcohol 3-ethyl-3-pentanol has it on C3. Either of these alcohols can be synthesized by an appropriate Grignard reaction, but 3-ethyl-2-pentanol may dehydrate to give a mixture of products. Because of its symmetry, 3-ethyl-3-pentanol dehydrates to give only the desired alkene, 3-ethyl-2-pentene. It also dehydrates more easily because it is a tertiary alcohol.

!

I

CH2-CH3

CH3-CH-CH-CH2-CH3

I

OH

3-ethyl-2-pentanol

CH2-CH3

2:

CH3-CH= -CH2-CH3

(major) 3-ethyl-2-pentene

PrefelTed synthesis:

I

CH2-CH3

CH3-CH2- -CH2-CH3

r

+

CH3=CH-C-CH2-CH3

I

CH2-CH3

CH3-CH2=C-CH2-CH3

H2S04

OH

(only product) 3-ethy 1-2-pentene

3-ethyl-3-pentanol 4. Working backward through as many steps as necessary, compare methods for syn­

thesizing the reactants needed for assembly of the key intermediate. (This process may require writing several possible reaction sequences and evaluating them, keep­ ing in mind the specified starting materials.) The key intermediate, 3-ethyl-3-pentanol, is simply methanol substituted by three ethyl groups. The last step in its synthesis must add an ethyl group. Addition of ethyl magne­ sium bromide to 3-pentanone gives 3-ethyl-3-pentanol.

II

CH3-CH2-C-CH2 -CH3

( 1 ) CH3CH2-MgBr (2) H30+

)

I

CH2-CH3

I

CH3-CH2-C-CH2-CH3 OH

o 3-pentanone

3-ethyl-3-pentanol

The synthesis of 3-pentanone from a three-carbon fragment and a two-carbon fragment requires several steps (see Problem 11-37). Perhaps there is a better alternative, consider­ ing that the key intermediate has three ethyl groups on a carbinol carbon atom. Two sim­ ilar alkyl groups can be added in one Grignard reaction with an acid chloride or an ester (Section 1 O-9D). Addition of 2 moles of ethyl magnesium bromide to a three-carbon acid chloride gives 3-ethyl-3-pentanol.

II

o

CH3-CH2-C-CI

propionyl chloride

(1) 2 CH3CH2-MgBr (2) H30+

I

CH2-CH3

T

I

CH2-CH3

CH3-CH2- -CH2-CH3 OH

3-ethyl-3-pentanoI ( Continued)

(minor) 3-ethyl-l-pentene

497

498

Chapter 1 1 : Reactions of Alcohols 5. Summarize the complete synthesis in the forward direction, including all steps and

all reagents, and check it for errors and omissions.

CH?-CH3 -

I

-C-CH2-CH3 CH3 -CT-f "2

I

OH

propionyl chloride

3-ethyl-3-pentanol KMn04 (cold, dilute)

CH?-CH3 -

)

I

CH3-CH -C-CH2 -CH3

I

I

OH

OH

3-ethyl - 2,3-pentanediol

3-ethyl - 2- pentene PROBLEM 1 1 - 3 7

To practice working through the early parts of a multistep synthesis, devise syntheses of (a) 3-ethyl-2-pentanol from compounds containing no more than three carbon atoms. (b) 3-pentanone from alcohols containing no more than three carbon atoms. SAMPLE PROBLEM

As another example of the systematic approach to multistep synthesis, let's consider the synthesis of l-bromo-2-methylcyclohexane from cyclohexanol. H

G

OH

�

Ct� H

H3

Br

1. Review the functional groups and carbon skeleton of the target compound.

The skeleton has seven carbon atoms: a cyclohexyl ring with a methyl group. It is an alkyl bromide, with the bromine atom on a ring carbon one atom removed from the methyl group.

2. Review the functional groups and carbon skeletons of the starting materials (if specified), and see how their skeletons might fit together into the target compound.

The stru.ting compound has only six carbon atoms; clearly, the methyl group must be added, presumably at the functional group. There are no restrictions on the methylating reagent, but it must provide a product with a functional group that can be converted to an adjacent halide. 3. Compare methods for assembling the carbon skeleton of the target compound to determine which methods provide a key intermediate with the correct carbon skele­ ton and functional groups at the correct positions for being converted to the func­ tionality in the target molecule.

Once again, the best choice is a Grignard reaction, but there are two possible reactions that give the methylcyclohexane skeleton. A cycJohexyl Grignard reagent can add to formaldehyde, or a methyl Grignard reagent can add to cyclohexanone. (There ru.-e other possibilities, but none that are more direct.)

6� CH3MgBr

H + H

+

'" /

C=O

(10

( 1 ) ether

(2) H3O+

CH?OH

)

Ct

H

cf

to '" "m'tiom""'"

alcohol C

(1) et her ) (2) Hp+

OH

�

to be function ali zed

alcohol D

Neither product has its alcohol functional group on the carbon atom that is functionalized in the target compound. Alcohol C needs its functional group moved two carbon atoms,

1 1 -14 Reactions of Alkoxides

499

but alcohol D needs it moved only one carbon atom. Converting alcohol D to an alkene functionalizes the correct carbon atom. Anti-Markovnikov addition of HBr converts the alkene to an alkyl halide with the bromine atom on the correct carbon atom. ROOR

HEr

rf

CH3

�Br H

)

alcohol D

target compound

4. Working backward through as many steps as necessary, compare methods for syn­ thesizing the reactants needed for assembly of the key intermediate.

(t

All that remains is to make cyc1ohexanone by oxidation of cyciohexanol. OH H

5. Summarize the complete synthesis in the forward direction, including all steps and all reagents, and check it for errors and omissions.

ROOR

HEr

)

ri-

VBr

CH3

H

Problem 11-38 provides practice intermediates.

111

multistep syntheses and using alcohols as

PROBLEM 11-38:

Develop syntheses for the following compounds. As starting materials, you may use cyc1opentanol, alcohols containing no more than four carbon atoms, and any common reagents and solvents. (a) trans-cyciopentane-l,2-diol (b) l -chloro-l-ethylcyciopentane OCH2CH3 (d) (e)

(,) � i:�3

C 1.

SUMMARY

o

=="\1

CH'CH3

Reactions of Alcohols

Oxidation-reduction reactions a. Oxidation of secondary alcohols to ketones (Section 11-2A) OH

I

R-CH-R'

Example

I

o

NazCrz07, H2S04

)

OH CH3 -CH -CH2CH3 2-butanol

II R - C - R' o

Na2Cr207, H2S04

)

II

CH3-C -CH2CH3 2-butanone

( Continued)

500 b.

Chapter 11: Reactions of Alcohols

Oxidation of primary alcohols to carboxylic acids (Section 11-2B) Na2Cr207, H2S04

R-CH2-OH

o

II

l

R-C-OH o

II

Example CH3(CH2)4-CH2-0H

CH3(CH2)4-C-OH

I-hexanol

hexanoic acid

c. Oxidation of primary alcohols to aldehydes (Section 11-2B)

II R-C-H o

PCC

R-CH2-OH

�

Example I-hexanol

hexanal

d. Reduction of alcohols to alkanes (Section 1 1 -6)

R-OH

(1) TsCl / pyridine ) R-H (2) LiAIH4

Example (1)

TsCI/pyridine )

(2) LiAl H 4

2.

cycIohexanol

cycIohexane

Cleavage of the alcohol hydroxyl group -C+O-H Conversion of alcohols to alkyl halides (Sections 11-7 through 11-9)

a.

2ipyridine

HCl or SOCI

R-Cl -' R-OH ----------=------>l HBror PBr3

R-OH -----'---l--> R-Br R-OH

Examples

HI or

(CH3hC-OH t-butyl alcohol

(CH3hCH - CH2OH isobutyl alcohol

CH3(CH2)4-CH20H I-hexanol b.

piI 2

-'- -->l R - I --------=-HCl

�

PBr3

piI2

(CH3hC-CI t-butyl chloride

)

(CH3hCH-CH2Br isobutyl bromide

)

CH3(CH2)4 -CH21 l-iodohexane

Dehydration of alcohols to form alkenes (Section I I - I DA) H OH H2 S04 or H3P04 I I ) -C-C-

I

I

Example

cycIohexanol

" /

C=C

/ "

+

o

cycIohexene

HzO

11-14 Reactions of Alkoxides

c. Industrial dehydration of alcohols to form ethers (Section 11-lOB) R-O-R +

+

�

2 R-OH

__

Example

CH3CH 2-O-CH 2CH 3 + H 20 diethyl ether

2 CH3CH20H ethanol

3.

H 20

I Cleavage of the hydroxyl proton -C -O+ H I a. Tosylation (Section 11-5)

' CH3 R-OH CI-�-oo

II

+

alcohol

_

o

(CH 3hCH -OH isopropyl alcohol

TsCl / pyridine

R-OH

II

(CH 3hCH -OTs isopropyl tosylate

o

II R'-C-Cl (acyl chloride)

R-O-C-R' HCI o

)

II

+

ester

II

('I

a �

0

Br

11-43

d� d

Chapter 1 1 : Reactions of Alcohols

I

o-

H

(d)

H

11-47

�

-----'>

�

�

0

�

Predict the major products of dehydration catalyzed by sulfuric acid. (a) I-hexanol (b) 2-hexanol (d) I -methylcyclopentanol (e) cyclopentylmethanol

I

OH CHCH,CH,

(c) 3-pentanol

� OH

0

II

° +

CH3 -C -OH

Show how you would make the methanesulfonate ester of cyclohexanol, beginning with cyclohexanol and an appropriate acid chloride. o

II ('f- O-�S-CH 3 V H

Show how you would convert (S)-2-hexanol to (b) ( R)-2-bromohexane (a) (S)-2-chlorohexane

(

c) (R)-2-hexanol

When l -cyclohexylethanol is treated with concentrated aqueous HBr, the major product is I -bromo-l-ethylcyclohexane.

~

HBr Hp

------?

(a) Give a mechanism for this reaction. (b) How would you convert l -cyclohexylethanol to ( l-bromoethyl)cyclohexane in good yield?

11-48

�

(f) 2-methylcyclopentanol

Predict the esterification products of the following acid/alcohol pairs.

cyclohexyl methanesulfonate: 11-46

CH,B'.

CH20H

CH,CH,CH,

(b)

11-45

�

d

if

Br

?

if

Show how you would make each compound, beginning with an alcohol of your choice.

(Y 0CH3

HO

(a)

(

c

)

V

°

II

(f)

� C -OH LJ

Study Problems

505

o (i)

(g)

H H ,r\;

�;�" OTs I

CH3 11-49 11-50

11-51

Predict the major products (including stereochemistry) when cis-3-methylcyclohexanol reacts with the following reagents, (c) Lucas reagent (b) SOCl2 (a) PBr3 (e) TsClIpyridine, then NaBr (d) concentrated HBr Show how you would use simple chemical tests to distinguish between the following pairs of compounds, In each case, describe what you would do and what you would observe, (a) I-butanol and 2-butanol (b) 2-butanol and 2-methyl-2-butanol (c) cyclohexanol and cyclohexene (d) cyclohexanol and cyclohexanone (e) cyclohexanone and l-methylcyclohexanol H",

(a) 11-52

0-

H

/H

/C=C'"

(b)

�

C

B

)

II o

}----.

E

3,4-dimethyl-3-hexanol

Give the structures of the intermediates and products V through Z, Mg, ether , cyclopelltanol

11-55

II

(c) -O -S-CH3

�

Mg, ether -""--------,-;. D (Grignard reagent)

N�Crp7' H2S04

11-54

I

o

Compound A is an optically active alcohoL Treatment with chromjc acid converts A into a ketone, B. In a separate reac­ tion, A is treated with PBr3, converting A into compound C. Compound C is purified, and then it is allowed to react with magnesium in ether to give a Grignard reagent, D, Compound B is added to the resulting solution of the Grignard reagent After hydrolysis of the initial product (E), this solution is found to contain 3,4-dimethyl-3-hexanoL Propose structures for compounds A, B, C, D, and K A

11-53

0� c6

Write the important resonance structures of the following anions,

x

y

Z

V

Under acid catalysis, tetrahydrofurfuryl alcohol reacts to give surprisingly good yields of dihydropyran, Propose a mecha­ nism to explain this useful synthesis,

o o

tetrahydrofurfuryl alcohol

dihydropyran

Propose mechanisms for the following reactions, In most cases, more products are formed than are shown here, You only need to explain the formation of the products shown, however. CI

HCI, ZnClz

) O (a rrunor product)

506

Chapter 11: Reactions of Alcohols OH

«) 11-56

en

H���',

OH

CO CO +

OH

11-58

u:

Show how you would synthesize the following compounds. As starting materials, you may use any alcohols containing four or fewer carbon atoms, cyclohexanol, and any necessary solvents and inorganic reagents.

(')

11-57

+

�

U

OH

Show how you would synthesize the following compound. As starting materials, you may use any alcohols containing five or fewer carbon atoms and any necessary solvents and inorganic reagents.

The following pseudo-syntheses (guaranteed not to work) exemplify a common conceptual error. heat

�r-'H?S04 ) OH

11-59

11-60

(a) What is the conceptual error implicit in these syntheses? (b) Propose syntheses that are more likely to succeed.

Two unknowns, X and Y, both having the molecular formula C4HsO, give the following results with four chemical tests. Propose structures for X and Y consistent with this information. Compound X Compound Y

Bromine

Na Metal

Chromic Acid

Lucas Reagent

decolorizes no reaction

bubbles no reaction

orange to green no reaction

no reaction no reaction

The Williamson ether synthesis involves the displacement of an alkyl halide or tosylate by an alkoxide ion. Would the synthesis shown be possible by making a tosylate and displacing it? If so, show the sequence of reactions. If not, explain why not and show an alternative synthesis that would be more likely to work.

� �

OH

make the tosylate and displace?

� �

OCH3

)

*11-61

Study Problems

Chromic acid oxidation of an alcohol (Section 1 1 -2A) occurs in two steps: formation of the chromate ester, followed by an elimination of H+ and chromium. Which step do you expect to be rate-limiting? Careful kinetic studies have shown that Compound A undergoes chromic acid oxidation over I a times as fast as Compound B. Explain this large difference in rates.

~

OH

*11-62

*11-63

507

(H2Cr04

(slower)

H Compound A

HO

� H Compound B

Many alcohols undergo dehydration at aoc when treated with phosphorus oxychloride ( POCI 3 ) in the basic solvent pyri­ dine. (Phosphorus oxychloride is the acid chloride of phosphoric acid, with chlorine atoms in place of the hydroxyl groups of phosphoric acid.) ( Propose a mechanism for the dehydration of cyclopentanol using POCl3 and pyridine. The first half of the mecha­ a) nism, formation of a dichlorophosphate ester, is similar to the first half of the mechanism of reaction of an alcohol with thionyl chloride. Like a tosylate, the dichlorophosphate group is a good leaving group. The second half of the mechanism might be either first order or second order; draw both alternatives for now. (b) When trans-2-methylcyclopentanol undergoes dehydration using POCl3 in pyridine, the major product is 3-methyl­ cyclopentene, and not the Zaitsev product. What is the stereochemistry of the dehydration? What does this stereo­ chemistry imply about the correct mechanism in part (a)? Explain your reasoning. Alcohols combine with ketones and aldehydes to form interesting derivatives, which we will discuss in Chapter 1 8. The following reactions show the hydrolysis of two such derivatives. Propose mechanisms for these reactions.

a)

(

(b)

C) CXJ

OCH,

:,�

,

HO

f)

+

H

CHpH

12

fixed mirror

I nfra red S pectrosco py a n d M a ss S pectrometry

laser calibration beam sample

� 12-1

I n trod ucti o n

detector

One of the most important tasks of organic chemistry is the determination of organic structures. When an interesting compound is isolated from a natural source, its struc­ ture must be completely determined before a synthesis is begun. Whenever we run a reaction, we must determine whether the product has the desired structure. The struc­ ture of an unwanted product must be known so the reaction conditions can be altered to favor the desired product. In many cases, a compound can be identified by chemical means. We find the molecular formula by analyzing the elemental composition and determining the molecular weight. If the compound has been characterized before, we can compare its physical properties (melting point, boiling point, etc.) with the published values. Chemical tests can suggest the functional groups and narrow the range of possible structures before the physical properties are used to make an identification. These procedures are not sufficient, however, for complex compounds that have never been synthesized and characterized. They are also impractical with compounds that are difficult to obtain, because a relatively large sample is required to complete the elemental analysis and all the functional group tests. We need analytical techniques that work with tiny samples and that do not destroy the sample. Spectroscopic techniques often meet these requirements. Absorption spec­ troscopy is the measurement of the amount of light absorbed by a compound as a func­ tion of the wavelength of light. In general, a sample is irradiated by a light source, and the amount of light transmitted at various wavelengths is measured by a detector and plotted on a graph. Unlike chemical tests, most spectroscopic techniques are nondes­ tructive; that is, the sample is not destroyed. Many different kinds of spectra can be measured with little or no loss of sample. In this book, we cover four spectroscopic or related techniques that serve as poweIful tools for structure determination in organic chemistry: Infrared (IR) spectroscopy, covered in this chapter, observes the vibrations of bonds and provides evidence of the functional groups present. Mass spectrometry (MS), also covered in this chapter, bombards molecules with electrons and breaks them apart. Analysis of the masses of the fragments gives the molecular weight, possibly the molecular formula, and clues to the structure and functional groups. Less than a milligram of sample is destroyed in this analysis.

508

Nuclear magnetic resonance (NMR) spectroscopy, covered in Chapter 13, observes the chemical environments of the hydrogen atoms (or the carbon atoms) and provides evidence for the structure of the alkyl groups and clues to the functional groups.

12-2 The Electromagnetic Spectrum

509

Ultraviolet (UV) spectroscopy, covered in Chapter 1 5 , observes electronic tran­ sitions and provides information on the electronic bonding in the sample. These spectroscopic techniques are complementary, and they are most powerful when used together. In many cases, an unknown compound cannot be completely iden­ tified from one spectrum without additional information, yet the structure can be deter­ mined with confidence using two or more different types of spectra. In Chapter 1 3, we consider how clues from different types of spectroscopy are combined to provide a reliable structure.

Visible light, infrared light, ultraviolet light, microwaves, and radio waves are exam­ ples of electromagnetic radiation. They all travel at the speed of light, about 3 X 1 0 10 cm/ second but they differ in frequency and wavelength. The frequency of a wave is the number of complete wave cycles that pass a fixed point in a second. Frequency, represented by the Greek letter v (nu), is usually given in hertz (Hz), mean­ ing cycles per second. The wavelength, represented by the Greek letter A (lambda), is the distance between any two peaks (or any two troughs) of the wave.

wavelength 1-+----- 1, ---��I

I�---- A ----I-� The wavelength and frequency, which are inversely proportional, are related by the equation VA = e

or

A

e = v

speed of light (3 X 1 0 1 0 cm/ sec ) v = frequency in hertz A = wavelength in centimeters Electromagnetic waves travel as photons, which are massless packets of energy. The energy of a photon is proportional to its frequency and inversely proportional to its wavelength. A photon of frequency v (or wavelength A) has an energy given by where

e =

E

he = hv = -A

where h is Planck's constant, 6.62 X 1 0- 7 kJ ' sec or 1 .5 8 X 1 0- 7 kcal . sec. Under certain conditions, a molecule struck by a photon may absorb the photon'S energy. In this case, the molecule's energy is increased by an amount equal to the photon'S energy, hv. For this reason, we often represent the irradiation of a reaction mixture by the symbol hv. The electromagnetic spectrum is the range of all possible frequencies, from zero to infinity. In practice, the spectrum ranges from the very low radio frequencies used to communicate with submarines to the very high frequencies of gamma rays. Figure 1 2- 1 shows the wavelength and energy relationships of the various parts of the electromagnetic spectrum. The electromagnetic spectrum is continuous, and the exact positions of the dividing lines between the different regions are somewhat arbitrary. Toward the top of the spectrum in Figure 1 2- 1 are the higher frequencies, shorter wavelengths, and

3

3

12-2

The Electromagnetic S pectrum

510

Chapter

1 2:

Infrared Spectroscopy a n d Mass Spectrometry

higher frequency

Wavelength (A)

Region

C lll

gamma rays

shorter wavelength

10- 9 10- 7

>< 0 �

!@ 1J.l � Figure 1 2-1

longer wavelength

107 105

vacuum UV

103

near UV

10- 5

i nfrared

10- 3

Molecular effects ionization electroni c transitions

visible

10- 4

(IR)

10

radio

10- 2 10- 4 10- 6

microwave

102 104

lower frequency

kllmol

X rays

10- 1

The electromagnetic spectrum.

Energy

molecular vibrations rotational Ill otion nuclear spin transitions

----- ------

higher energies. Toward the bottom are the lower frequencies, longer wavelengths, and lower energies. X rays (very high energy) are so energetic that they excite electrons past all the energy levels, causing ionization. Energies in the ultraviolet-visible range excite electrons to higher energy levels within molecules. Infrared energies excite molecular vibrations, and microwave energies excite rota­ tions. Radio-wave frequencies (very low energy) excite the nuclear spin transitions observed in NMR spectroscopy.

12-3

The I n fra red Reg i o n

The infrared (from the Latin, infra, meaning "below" red) region of the spectrum cor­ responds to frequencies from just below the visible frequencies to just above the high­ est microwave and radar frequencies: wavelengths of about 8 X 1 0-5 cm to 1 X 1 0-2 cm. Common infrared spectrometers operate in the middle of this region, at wavelengths between 2.5 X 1 0-4 cm and 25 X 1 0-4 cm, corresponding to energies of about 4.6 to 46 kllmol 0 . 1 to 11 kcal/mol). Infrared photons do not have enough energy to cause electronic transitions, but they can cause groups of atoms to vibrate with respect to the bonds that connect them. Like electronic transitions, these vibra­ tional transitions correspond to distinct energies, and molecules absorb infrared radia­ tion only at certain wavelengths and frequencies. The position of an infrared band is specified by its wavelength ( A ) , measured in microns (/Lm ) . A micron (or micrometer) corresponds to one millionth ( 1 0-6) of a meter, or 1 0-4 cm. A more common unit, however, is the wavenumber (v), which cOlTesponds to the number of cycles (wavelengths) of the wave in a centimeter. The wavenumber is the reciprocal of the wavelength (in centimeters). Since 1 cm = 1 0,000 /Lm, the wavenum­ ber can be calculated by dividing 1 0,000 by the wavelength in microns. The units of the wavenumber are cm- 1 (reciprocal centimeters). v (cm- I )

1 O,000 /Lm/cm A (/Lm)

1 A (cm)

= ---

or

A (/Lm) -

1 O,000 /Lm/cm ---'--v (cm- I )

-

-

For example, an absorption at a wavelength of 4 /Lm corresponds to a wavenumber of 2500 cm- 1 • v =

1 O,000 /Lm/cm 4 /Lm

=

2500 cm- I

or

A

=

1 0,000 /Lm/ cm 2500 cm- I

=

4r 'I.m

1 2-4 Molecular Vibrations

51 1

Wavenumbers ( in cm- I ) have become the most common method for specifying absorptions, and we will use wavenumbers throughout this book. The wavenumber is proportional to the frequency ( v ) of the wave, so it is also proportional to the energy of a photon of this frequency (E = h v ) . Some references still use microns, however, so you should know how to convert these units.

IR

PROB L E M 1 2 -1

Complete the following conversion table.

v(cm- I ) A (/-Lin )

1 700

4000 2.50

3.33

3.03

4.55

1 640

1 600

400 25.0

Before discussing characteristic infrared absorptions, it's helpful to understand some theory about the vibrational energies of molecules. The following drawing shows how a covalent bond between two atoms acts like a spring. If the bond is stretched, a restor­ ing force pulls the two atoms together toward their equilibrium bond length. If the bond is compressed, the restoring force pushes the two atoms apart. If the bond is stretched or compressed and then released, the atoms vibrate. --

spring force

spring force

stretched

com pressed

�

�

1 2-4

M o lecu l a r V i b rations

-+-

eq u i l ib ri u Jll bond length

The frequency of the stretching vibration depends on the masses of the atoms and the stiffness of the bond. Heavier atoms vibrate more slowly than lighter ones; for exam­ ple, the charactel;stic frequency of a C -D bond is lower than that of a C - H bond. In a group of bonds with similar bond energies, the frequency decreases with increasing atomic weight.

Stronger bonds are generally stiffer, requiring more force to stretch or compress them. Thus, stronger bonds usually vibrate faster than weaker bonds (assum.ing the atoms have similar masses). For example, 0 -H bonds are stronger than C -H bonds, and 0 -H bonds vibrate at higher frequencies. Triple bonds are stronger than double bonds, so triple bonds vibrate at higher frequencies than double bonds. Simi­ larly, double bonds vibrate at higher frequencies than single bonds. In a group of bonds having atoms of similar masses, the frequency increases with bond eneJgy. Table 1 2- 1 lists some common types of bonds, together with their stretching frequencies to show how frequency varies with the masses of the atoms and the strength of the bonds. An infrared spectrum is a graph of the energy absorbed by a molecule as a func­ tion of the frequency or wavelength of light. The IR spectrum of methanol is shown in Figure 1 2-2. In the infrared region, absorptions generally result from exciting the vibra­ tional modes of the bonds in the molecule. Even with simple compounds, infrared spec­ tra contain many different absorptions, not just one absorption for each bond. The methanol spectrum (Figure 1 2-2) is a good example. We can see the broad 0 - H stretch around 3300 cm- I , the C - H stretch just below 3000 cm- I , and the C - 0 stretch just above 1 000 cm- I . We also see absorptions resulting from bending vibra­ tions, including scissoring and twisting vibrations. In a bending vibration, the bond lengths stay constant, but the bond angles vibrate about their equilibrium values.

PROBLEM-SOLVING

H?np

Use spectrum (singular) and spectra (pl ural) correctly: "This spectruJll is . "These spectra are . . . . "

512

Chapter 1 2: Infrared Spectroscopy

and Mass Spectrometry

TABLE 1 2-1

Bond Stretching Frequencies.

In a group of bonds with similar bond energies, the frequency decreases with increasing atomic

weight. In a group of bonds between similar atoms, the frequency The bond energies and frequencies l isted here are approximate. Bond

C-H C-D C-C

C-C C=C C-C

j

increases with bond energy.

Stretching Frequency (cm- 1 )

Bond Energy [kJ (kcal)] 420 ( l aO ) <note these are trs>

Frequency decreases with increasing atomic mass

heavier atoms

j j j

3000

420 ( 1 00)

2 1 00

350 (83)

1 200

Frequency increases with bond energy

j j

350 (83) 6 1 1 ( 1 46) 840 (200)

C-N C=N C=N

8 9 1 (2 1 3)

C-O c=o

745 ( 1 78)

305 (73) 6 1 5 ( 1 47)

stronger bond

1 200 1 660 2200 1 200 1 650 2200

1

360 (86)

v decreases

v increases

1

1 1 00 1 700

Consider the fundamental vibrational modes of a water molecule in the following diagram. The two 0 - H bonds can stretch in phase with each other (symmetric stretch­ ing), or they can stretch out of phase (antisymmetric stretching). The H - 0 - H bond angle can also change in a bending vibration, making a scissoring motion. H

H

H

H

,/

symmetric stretching

anti symmetric stretching

bending (scissoring)

A nonlinear molecule with n atoms generally has 3n - 6 fundamental vibrational modes. Water (3 atoms) has 3 ( 3 ) - 6 = 3 fundamental modes, as shown in the preced2.5 3 100 11 illll! " il ll 1IIIlIilI I I JI I

""'

I I 1n111IT iti'\i;'l I' , ' ' ' :lI tfi ' l ",,ii+ii+ii+ ,**H+++\�,**H+I80

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1

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3500

I

4.5

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wavelength (p,m)

III

5.5

5

7

6

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8

9

10

11

12

13

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1 4 15 1 6

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I II I

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l

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3000

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i ll;

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���� ='�'S

· �o �, . .�,� .� . � l I �����_� i� I �llll�l� I� l i� I �llllllll���lll� I ! !� l �� il� 2500 2000 1 800 1600 1 400 1 200 1000 800

wavenumber (em-I)

..

���

600

... Figure 1 2-2 The infrared spectrum of methanol shows 0 - H, C - H, and C - 0 stretching absorptions, together with absorptions from several bending modes.

12-5

IR-Active and IR-Inactive Vibrations

513

ing figure. Methanol has 3 ( 6 ) - 6 1 2 fundamental modes, and ethanol has 3 ( 9 ) - 6 = 2 1 fundamental modes. We also observe combinations and multiples (over­ tones) of these simple fundamental vibrational modes. As you can see, the number of absorptions in an infrared spectrum can be quite large, even for simple molecules. It is highly unlikely that the IR spectra of two different compounds (except enan­ tiomers) will show the same frequencies for all their various complex vibrations. For this reason, the infrared spectrum provides a "fingerprint" of a molecule. In fact, the region of the IR spectrum containing most of these complex vibrations (600 to 1 400 cm -I ) is commonly called the fingerprint region of the spectrum. The simple stretching vibrations in the 1 600 to 3500 cm- I region are the most characteristic and predictable; our study of infrared spectroscopy will concentrate on them. Although our introductory study of IR spectra will largely ignore bending vibra­ tions, you should remember that these absorptions generally appear in the 600 to 1 400 cm- I region of the spectrum. Experienced spectroscopists can tell a great deal about the structure of a molecule from the various kinds of bending vibrations known as "wagging," "scissoring," "rocking," and "twisting" that appear in the finger­ print region (see Figure 1 2-2). The reference table of IR frequencies (Appendix 2) lists both stretching and bending characteristic frequencies. =

Not all molecular vibrations absorb infrared radiation. To understand which ones do and which do not, we need to consider how an electromagnetic field interacts with a molecular bond. The key to this interaction lies with the polarity of the bond, measured as its dipole moment. A bond with a dipole moment can be visualized as a positive charge and a nega­ tive charge separated by a spring. If this bond is placed in an electric field (Figure 1 2-3), it is either stretched or compressed, depending on the direction of the field. Qpe of the components of an electromagnetic wave is a rapidly reversing electric field ( E ) . This field alternately stretches and compresses a polar bond, as shown in Figure 1 2-3. When the electric field is in the same direction as the dipole moment, the bond is compressed and its dipole moment decreases. When the field is opposite the dipole moment, the bond stretches and its dipole moment increases. If this alternate stretching and compressing of the bond occurs at the frequency of the molecule's nat­ ural rate of vibration, energy may be absorbed. Vibrations of bonds with dipole moments generally result in IR absorptions and are said to be IR active. If a bond is symmetrical and has zero dipole moment, the electric field does not interact with the bond. For example, the triple bond of acetylene ( H - C - C - H ) has zero dipole moment, and the dipole moment remains zero i f the bond i s stretched or compressed. B ecause the vibration produces no change in the dipole moment, there is no absorption of energy. This vibration is said to be IR inactive, and it produces no +---->

®------0

j !I E

eIectnc . field

+

represented as

- t

I R-Active a nd I R- I na ctive Vi b rations

� t

force on +

t in direction of field

molecule compressed dipole moment decreased force on opposite direction of field

12- 5

electric field

force 011 + i ll directioll of field

molecule stretched dipole moment increased

+

force 011 opposite direction of field

« Figure 1 2-3 Effect of an electric field o n a polar bond. A bond with a dipole moment (as in HF, for example) is either stretched or compressed by an electric field, depending on the direction of the field. Notice that the force on the positive charge is in the direction of the electric field (E) , and the force on the negative charge is in the opposite direction.

514

Chapter 1 2 :

Infrared Spectroscopy and Mass Spectrometry absorption in the IR spectrum. The key to an IR-active vibration is that the

vibration

must change the dipole moment of the molecule.

In general, if a bond has a dipole moment, its stretching frequency causes an absorption in the IR spectrum. If a bond is symmetrically substituted and has zero dipole moment, its stretching vibration is weak or absent in the spectrum. Bonds with zero dipole moments sometimes produce absorptions (usually weak) because molecu­ lar collisions, rotations, and vibrations make them unsymmetrical part of the time. PROBLEM 1 2-2

Which of the bonds shown in red are expected to have IR-active stretching frequencies?

H -C - C -H

H - C =C - H

H - C= C - CH3

H3C - C - C -CH3

I

H H C-C-H 3 I H H3C H

12-6

Measu rement of the I R S pectru m

C =C

/

/ "

CH3 H

Infrared spectra can be measured using liquid, solid, or gaseous samples that are placed in the beam of infrared light. A drop of a l iquid can be placed as a thin film between two salt plates made of NaCI or KEr, which are transparent to infrared light at most important frequencies. A solid can be ground with KEr and pressed into a disk that is placed in the light beam. Alternatively, a solid sample can be ground into a pasty mull with paraffin oil. As with a liquid, the mull is placed between two salt plates. Solids can also be dissolved in common solvents such as CH2CI2, CCI4, or CS2 that do not have absorptions in the areas of interest. Gases are placed in a longer cell with polished salt windows. These gas cells often contain mirrors that reflect the beam through the cell several times for stronger absorption. An infrared spectrometer measures the frequencies of infrared light absorbed by a compound. In a simple infrared spectrometer (Figure 1 2-4), two beams of light are used. The sample beam passes through the sample cell, while the reference beam passes through a reference cell that contains only the solvent. A rotating mirror alternately allows light from each of the two beams to enter the monochromator.

motor

� Figure 1 2-4

Block diagram of a dispersive infrared spectrometer. The sample beam passes through the sample cell while the reference beam passes through a reference cell that contains only the solvent. A rotating mirror alternately allows light from each of the two beams to enter the monochromator where they are compared. The chart recorder graphs the difference in light transmittance from the two beams.

"

rotating segmented mirror detector

�

glowing wIre source

VV � �.n� t .

� v

chart recorder

light transmitted

1 2-6

Measurement of the IR Spectrum

51 5

The monochromator uses prisms or diffraction gratings to allow only one fre­ quency of light to enter the detector at a time. It scans the range of infrared frequencies as a pen moves along the conesponding frequencies on the x axis of the chart paper. Higher frequencies (shorter wavelengths) appear toward the left of the chart paper. The detector signal is proportional to the difference in the intensity of light in the sam­ ple and reference beams, with the reference beam compensating for any absorption by air or by the solvent. The detector signal controls movement of the pen along the y axis, with 100% transmittance (no absorption) at the top of the paper, and 0% trans­ mittance (absorption of all the light) at the bottom. The spectrometer shown in Figure 1 2-4 is called a dispersive instrument because it disperses light into all the different frequencies and measures them individually. Dis­ persive instruments require expensive prisms and diffraction gratings, and they must be manually aligned and calibrated on a regular basis. Since only one frequency is observed at a time, dispersive instruments require strong IR sources, and they require 2 to 10 minutes to scan through a complete spectrum. Dispersive infrared spectrometers are being replaced by Fourier transform infrared (FT-IR) spectrometers. A Fourier transform infrared spectrometer (FT-IR) uses an interferometer, like that shown in Figure 1 2-5, to measure an IR spectrum. The infrared light goes from the glowing source to a beamsplitter, usually made of polished KBr, placed at a 45° angle. Part of the beam passes through the beam splitter, and part is reflected at a right angle. The reflected beam strikes a stationary mirror, while the transmitted beam strikes a mirror that moves at a constant speed. The beams return from the mirrors to recombine at the beamsplitter. The beam from the moving mirTOr has traveled a different distance than the beam from the fixed mirror, and the two beams combine to create an interfer­ ence pattem called an interferogram. This interferogram, which simultaneously con­ tains all frequencies, passes through the sample compartment to reach the detector. The interferogram shown in the upper half of Figure 1 2-6 contains all the infor­ mation contained in the spectrum shown in the lower half. The intelferogram is said to be in the time domain, cOlTesponding to the energy seen by the detector as the mirror moves through the signal. A standard computer algorithm called a Fourier transform converts the time domain to the frequency domain spectrum that allows us to see the strength of absorption as a function of the frequency (or wavelength). Figure 1 2-6 shows both the interferogram and the IR spectrum of n-octane.

fixed mirror

IR source IR source

moving mirror

laser calibration beam sample

i::J

detector

.... Figure 1 2-5 Block diagram of an interferometer in an Ff-IR spectrometer. The light beams reflected from the fixed and moving mirrors are combined to fom1 an interferogram, which passes through the sample to enter the detector.

516

Chapter 1 2 : Infrared Spectroscopy

and Mass Spectrometry

+4 -0

on

+2

-

2

-

;>

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.... 8 u 8

0

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-

-

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ii ! i !.i I! 'I sc issor ++++++ 1 +++ 1 -+++1 -+++-1+1+1+11+ : 1 1-++-+ ,+++-+--iHI +++++-+-+-H-+--l I

CH3 1 +-+:+1+-+--1 rock -+++ 1 11-+ KlH+flllt+H ll HIHlH HH++++HHHHHHiH++H-III-H 1 - 1H++ 1-1I11I+ H�I+� 11 rock :

I

,, !

II I

3500 3000

CH2

++- I+I+I+I+I+I+Hf+Hf+HH-H-H-H-H�,I CH3(CH2)6CH3 t-t-t-+-++-t-t--H--+-l H -+++--+++--+I+I+ +HHt+HHHHt+HHf+++

II

1 600 1400 1 200 1000

, I i II

wavenumber (cm - 1 )

1

III

I11I

17.-oc tane

800

fingerprint region

600 1

... F i g u re 1 2-6 (a) Interferogram generated by n-octane. (b) Infrared spectrum of n-octane. Notice that the frequencies shown in a routine IR spectrum range from about 600 em- 1 to about 4000 em- 1 .

----- - -----.-===================================================

--

._-_._--- ----

The FT- IR spectrometer has several major advantages over the dispersive instrument: Its sensitivity is better because it measures all frequencies simultaneously rather than scanning through the individual frequencies. Less energy is needed from the source, and less time (typically 1 to 2 seconds) is needed for a scan. Several scans can be completed in a few seconds and averaged to improve the signal. Resolution and accuracy are also improved because a laser beam is used alongside the IR beam to con­ trol the speed of the moving mirror and to time the collection of data points. The laser beam is a precise frequency reference that keeps the spectrometer accurately calibrated. In the infrared spectrum of n-octane [Figure 1 2-6(b)] there are fom major absorption bands. The broad band between 2800 and 3000 cm- I results from C - H stretching vibrations, and the band at 1 467 cm- I results from a scissoring vibration of the CH2 groups. The absorptions at 1 378 and 722 cm- I result from the bending vibrations (rock­ ing) of CH 3 and CH2 groups, respectively. Since most organic compounds contain at least some saturated C - H bonds and some CH 2 and CH 3 groups, all these bands are com­ mon. In fact, without an authentic spectrum for comparison, we could not look at this spec­ trum and conclude that the compound is octane. We could be fairly certain that it is an alkane, however, because we see no absorption bands corresponding to functional groups. Another characteristic in the octane spectrum is the absence of any identifiable C - C stretching absorptions. (Table 1 2- 1 shows that C - C stretching absorptions occur around 1 200 cm- 1 .) Although there are seven C - C bonds in octane, their di­ pole moments are small, and their absorptions are weak and indistinguishable. This re­ sult is common for alkanes with no functional groups to polarize the C - C bonds.

12-7 Infrared Spectroscopy of Hydrocarbons Hydrocarbons contain only carbon--carbon bonds and carbon-hydrogen bonds. An infrared spectrum does not provide enough i nformation to identify a structure conclu­ sively (unless an authentic spectrum is available to compare "fingerprints"), but the absorptions of the carbon--carbon and carbon-hydrogen bonds can indicate the pres­ ence of double and triple bonds.

1 2-7A

Carbon-Carbon Bond Stretch i n g

Stronger bonds generally absorb a t higher frequencies because of their greater stiff­ ness. Carbon-carbon single bonds absorb around 1 200 cm- I , C = C double bonds absorb around 1 660 cm- 1 , and C - C triple bonds absorb around 2200 cm-1 . Carbon-carbon bond stretching frequencies

1 200 cm- I 1 660 cm- I < 2200 cm- I

C-C C=C C=C

As discussed for the octane spectrum, C - C single bond absorptions (and most other absorptions in the fingerprint region) are not very reliable. We use the fingerprint region primarily to confirm the identity of an unknown compound by comparison with an authentic spectrum. The absorptions of C = C double bonds, however, are useful for structure determi­ nation. Most unsymmetrically substituted double bonds produce observable stretching absorptions in the region of 1 600 to 1 680 cm- 1 . The specific frequency of the double­ bond stretching vibration depends on whether there is another double bond nearby. When two double bonds are one bond apart (as in 1 ,3-cyclohexadiene on the following figure) they are said to be conjugated. As we will see in Chapter 1 5 , conjugated double bonds are slightly more stable than isolated double bonds because there is a small amount of pi bonding between them. This overlap between the pi bonds leaves a little less electron den­ sity in the double bonds themselves; as a result, they are a little less stiff and vibrate a lit­ tle more slowly than an isolated double bond. Isolated double bonds absorb around 1 640 to 1 680 cm- I , and conjugated double bonds absorb around 1 620 to 1 640 cm- 1 .

o

O

------- 'ome p; o,,,',p

-------

1 645 c m- I cyclohexene (isolated)

less pi overlap than an isolated double bond

1 620 cm- I 1 ,3-cyc1ohexadiene (conjugated)

The effect of conjugation is even more pronounced in aromatic compounds, which have three conjugated double bonds in a six-membered ring. Aromatic C = C bonds are more like 1 � bonds than true double bonds, and their reduced pi bonding results in less stiff bonds with lower stretching frequencies, around 1 600 cm- I .

Characteristic C

[0

=

OJ

� �

'

�

1 600 em- I

bond order = I t

C stretching jr-equencies

isolated C = C conjugated C = C aromatic C = C

1 640- 1 680 cm- I 1 620- 1 640 cm- I approx. 1 600 cm- 1

12-7

I nfra red Spectroscopy of Hyd roca rbons

517

518

Chapter 1 2: Infrared Spectroscopy and Mass Spectrometry Carbon-carbon triple bonds in alkynes are stronger and stiffer than carbon-carbon single or double bonds, and they absorb infrared light at higher frequencies. Most alkyne C - C triple bonds have stretching frequencies between 2 1 00 and 2200 cm- L . Terminal alkynes usually give sharp C - C stretching signals of moderate intensity. The C - C stretching absorption of an internal alkyne may be weak or absent, however, due to the symmetry of the disubstituted triple bond with a very small or zero dipole moment.

�

R-C

C-H

terminal alkyne

R-C= C-R' internal alkyne

12-7B

C - C stretch observed around 2 10 0 to 2200 cm -

C = C stretch may be weak or absent

Carbon-Hydrogen Bond Stretc h i ng

Alkanes, alkenes, and alkynes also have characteristic C - H stretching frequencies. Car­ bon-hydrogen bonds involving sp 3 hybrid carbon atoms generally absorb at frequencies 2 1 just below (to the right of) 3000 cm- ; those involving sp hybrid carbons absorb just t above (to the left of) 3000 cm- . We explain this difference by the amount of s character in the carbon orbital used to form the bond. The s orbital is closer to the nucleus than the p orbitals, and stronger, stiffer bonds result from orbitals with more s character. 2 3 An sp orbital is one-fourth s character, and an sp orbital is one-third s charac­ 2 ter. We expect the bond using the sp orbital to be slightly stronger, with a higher vi­ bration frequency. The C- H bond of a terminal alkyne is formed using an sp-hybrid orbital, with about one-half s character. This bond is stiffer than a C - H bond using 2 an sp 3 or sp hybrid carbon, and it absorbs at a higher frequency: about 3300 cm- I . C - H bond stretching frequencies: sp > sp

I

I

I

I

-C-C-H "/

C=C

-C

12-7C

PROBLEM-SOLVING

Htni::

The unsaturated C H stretch, to the left of 3000 cm -1 , should alert you to look for a weak ( ( stretch. =

-

=

/ "-

2

> sp

3

Sp3 hybridized, one-fourth s character

2800-3000 cm- I

Sp2 hybridized, one-third s character

3000 3 100 cm- I -

sp hybridized, one-half s character

3300 cm- I

H

C-H

I nterpretation of the IR Spectra of Hyd rocarbons

Figure 1 2-7 compares the IR spectra of n-hexane, I -hexene, and cis-2-octene. The hexane spectrum is similar to that of n-octane (Figure 1 2-6). The C - H stretching fre­ quencies form a band between 2800 and 3000 cm -1, and the bands in the fingerprint region are due to the bending vibrations discussed for Figure 1 2-6. This spectrum sim­ ply indicates the absence of any IR-active functional groups. The spectrum of I -hexene shows additional absorptions characteristic of a dou­ ble bond. The C - H stretch at 3080 cm - I corresponds to the alkene = C - H bonds 2 1 involving sp hybrid carbons. The absorption at 1 642 cm- results from stretching of the C C double bond. (The small peak at 1 820 cm -I is likely an overtone at double the frequency of the intense peak at 9 10 cm- I . ) The spectrum o f cis-2-octene (Figure 1 2-7c) resembles the spectrum o f I -hex­ ene, except that the C = C stretching absorption at 1 660 cm- I is very weak in cis2-octene because the disubstituted double bond has a very small dipole moment. =

1 2-7 Infrared Spectroscopy of Hydrocarbons

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H

'"

/

/

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H

C =C

H3C

(c )

(CH2)4CH3

cis-2-octene

1600 1 400 1 200 1 000

wavenumber (em- I )

600 L I

16

I

600

I

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1 2 13 14 15 1 6

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1 800 1 600 1400 1 200 1000

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stretch

6

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2500

i

5.5

, ,

2500

3.5

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wavelength (J.Lm)

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3023 = C-H iii II st retch i i i i I ! i i !i I I! I

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, I!

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,

2000 1 800 1 600 1400 1 200 1 000

2500

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�

I f lil

II , I I

I

i

"

'"

·

II , ! i

CH3(CH2hCH

I

III

ill' III111

lit! I ii

i ll !

ill

I! Ii Iii I II :qI ;d! C -H slTet�h ! I i ,

.1.

'I

'I

.I

i! I '.

II! III "

,

.Ii

,

5;5

I

4

1::11

3080 = C -H stretch

,

.� .� 20 1 1 11

3000

1

.� ,I

40 f

. i I,

II

60 ; ! I ·

I II , �.

3500

,

I

I

' 1 I i, I I jI i Ii I i

!

wavelength (J.Lm)

;5

II

i

I

I,

800

... Figure 12-7

Comparison of the IR spectra of (a) n-hexane, (b) I-hexene, and (c) cis-2-octene. The most characteristic absorptions in the I-hexene spectrum are the C = C stretch at 1 642 cm- I and the unsaturated = C - H stretch at 3080 em- I . The nearly symmetrically substituted double bond in cis-2-octene gives a weak C = C absorption at 1660 cm - I . The unsaturated = C - H stretch at 3023 cm - I is still apparent, however.

600

51 9

520

Ch apter 1 2: Infrared Spectroscopy and Mass Spectrometry Even without the weak C C stretching absorption, the unsaturated C-H stretching absorption j ust above 3000 em- I suggests the presence of an alkene double bond. Figure 1 2-8 compares the IR spectra of l -octyne and 4-octyne. In addition to the alkane absorptions, the l -octyne spectrum shows sharp peaks at 33 1 3 and 2 1 1 9 em- I . I The absorption at 33 1 3 em- results from stretching of the stiff - C - H bond formed by the sp hybrid alkyne carbon. The 2 11 9 em- I absorption results from stretching of the C = C triple bond. The spectrum of 4-octyne is not very helpful. Since there is no acetylenic hydro­ gen, there is no = C - H stretching absorption around 3300 cm- I . There is no visible C - C stretching absorption around 2 1 00 to 2200 cm- 1 either, because the disubstituted triple bond has a very small dipole moment. This spectrum fails to alert us to the pres­ ence of a triple bond. =

2.5 1 00

3

3.5

4

4.5

=

wavelength (j.Lm)

5 I

I:

5.5 1

II

6

7

II

8

10

II

12

80 60

40

stretch

20 Itf, " , ,++l+tttttH+ tH+ tttH+1tt 4000

o

3500

2.5 1 00

3000

3

2500

3.5

4

80

40

II

4.5

wavenumber (cm - I )

2000

II I

5

1 800

1 600

1 400

wavelength (j.Lm)

II

5.5 I II

6

1 200

1 000

800

7

II

600

12

II I

II

Ht-H+H+IHtI+H+Ht+tt+H1+H+Ht- HUt+I-H-I+t-+t 1

II

20

• Figure 1 2-8

II

3

=

(a) l -octyne

+t ++tH+tH+tHCf'thH-H -+ -t-f-"'t-t+H-+-H-t-H-t+++-l I s t retch visible !+++t+H-t+H-++t+++t+++-H1l1H

60

0 llillJ.ll.lJ.JJ.!.Ll.l!.ill.ll.L 4000 3500

mlmlmlmlmmtutlmlmJmJm:j±{;H=: - ::������1tUtl+� C C ( CH2 )s CH + I ++ +-HI-HH

ffHIIH-H+tl-H-I+ I+t+Ht+tH-H-H-I+H-ti+H H+t.Ht�1+t-1-t1+t+H- 1

3000

2500

2000

I.

1 800

1 600

wavenumber (cm - I )

1 400

C H 3(CH 2h - C = C - (CH 2hC H 3

1 200

(b)

1 000

4-octyne

800

600

Comparison of the IR spectra of l -octyne and 4-octyne (a) The IR spectrum of l -octyne shows characteristic absorptions at 33 1 3 em - I (alkynyl = C - H stretch) and at 2 1 1 9 cm- I (C = C stretch). (b) We cannot tel1 that 4-octyne is an alkyne from its IR spectrum because it displays neither of the characteristic absorptions seen in (a). There is no alkynyl = C - H bond, and its symmetrically substituted triple bond has too smal1 a dipole moment to produce the C = C stretching absorption seen in the spectrum of l-octyne.

521

1 2-7 Infrared Spectroscopy of Hydrocarbons P R O B L E M 1 2 -3

For each hydrocarbon spectrum, determine whether the compound is an alkane, an alkene, an alleyne, or an aromatic hydrocarbon. More than one unsaturated group may be present. 5 .,

wavelength (J.Lm)

I,

I ii;

I! I , II Ii Ii

I

II "

1I i '

i

!, !. I ,

, il Ii i

I ii , !! :i j Ii Ii I I I Ii

Ii

'i '

!i l il

;;

L.

11

1 600 1 800 2000 wavenumber (em - I )

1

3 " ' :" " 1 "

i

25 1 00

:1

II

'1I IIil l.1 I

2.5 1 00 1'

!

III'

1tttfttttl ft llllTi tf lTI1 tl t+t1t+t1ffi1 +ttH+rt HHtt+I-f+tHttHti o,"++I+H-fHJJ±l"'ll+t l t

� tHttHHtHtHl *ii: !

40 ' : :� .�

00: IE

2

4000

ttlttHHtHltl i'ttl1 II iill .ill

I : : I!I "

III ' .111.1.

ii Hq iii! i Ii!! iii!

3500

1 400

1 200

1 000

I I

I

600

800

IIIHit tHlltH:>nf H m "'iT ' IFiT '

1

I

! "

I,

I

! i

.1

il

::11 1

2500

I I

I

I

I I

ii ,

I III·

II

I

IIi ! II i

I II !

II I I i

j

I ; I II

II, II ii ' '1

I

o

I'

: 1 I

'j 1

'i l ! j j Ii

I,

HI

3000

, ,

.., 1 1 11

iill ,. , 111111 1I 1HHttH IttHHt.tt! l+rtflH

nn I ill! i I I

III "

I

: : 1" ' I

wavelength (J.Lm) 55

45 ;

!illl!! '111 ' 1 lilt" 1111 !! I II . ' II i I I I I )i i ' . 11 1 .11 Iii 1 ' 11 1 1

i.

I

I

I.

II I

I i I

3.5

35

(c ) • !. . � tttt� tt 'i; 6 0 • � HttHt++t+tt+t t I

80

I

I.

,

II

Ii i Ii 1' 1 '

I I I

I. '

II

II .

I

,

I

I iii:

Ii

Ii i 'l

14 15 16

12

II

Ii I

i1

11

:,

I !

II

,

I

II

I

I

II i i i , I i:

II I II

II I

11

12 ,

II

14

1,5 1,6

{

I' I

13 ,

,

II

I I

II ' II

!il . Ii

I 800 1 600 2000 wavenumber (em - I )

i ll i .1 I i I I II

I I I!

1 400

1 200

1 000

800

600

522

Chapter 1 2 : Infrared Spectroscopy and Mass Spectrometry

12-8

The 0 - H bonds of alcohols and the N - H bonds of amines are strong and stiff. The vibration frequencies of 0 - H and N - H bonds therefore occur at higher frequen­ cies than those of most C - H bonds (except for alkynyl = C - H bonds).

Cha racteristic Absorptions of Alcohols a n d Am i n es

H

H

R"

R -N - H

R -N -R'

R-N-R'

secondary

tertiary

I

R-O-H

pri m ary

alcohol

I

I

�------�vr--� a

0 - H and N - H stretching frequencies

m i n es

3 300 c m- I , broad 3000 cm - I , broad

alcohol O - H acid O - H

3 3 00 cm- I , broad with spikes

amine N - H

Alcohol 0 - H bonds absorb over a wide range of frequencies, centered around I 3 300 cm- . Alcohol molecules are involved in hydrogen bonding, with different mol­ ecules having different instantaneous arrangements. The 0 - H stretching frequencies reflect this diversity of hydrogen-bonding arrangements, resulting in very broad absorptions. Notice the broad 0 - H absorption centered around 3 300 cm -[ in the infrared spectrum of I -butanol (Figure 1 2-9). Figure 1 2-9 also shows a strong C - 0 stretching absorption centered near I 1 050 cm- . Compounds with C - O bonds (alcohols and ethers, for example) gener­ ally show strong absorptions in the range of 1 000 to 1 200 cm - I ; however, there are other functional groups that also absorb in this region. Therefore, a strong peak between 1 000 and 1 200 cm- I does not necessarily imply a c - o bond, but the absence of an absorption in this region suggests the absence of a C - 0 bond. For simple ethers, this unreliable C - O absorption is usually the only clue that the com­ pound might be an ether. Amine N - H bonds also have stretching frequencies in the 3 300 cm- I region, or even slightly higher. Like alcohols, amines participate in hydrogen bonding that can

2 .5 1 00 frn-rn"",,,. ",

, 80

1lW. I!I!! ff+l4"+H+H+t+i-HB

IIi

I

0 4000

II

i' l

3500

� Figure 1 2-9

5

5.5

II

1 ' : ::,', ",

ill

7 TTlTT l

, Ir7

1

10

11

1 2 1 3 1 4 15 1 6

· 1 -'-T"'T'rr...,.-;--r-,-,-r"rrl...,.-;--T-i-T-l r +-;:TT:TTII.,mTIT1 'rrrTTTT,

! �H+H-l-i-i+I+itt +!H .J-+'. ii !i !-!iH ' + 1-H-l-i-HI-H1I 1-l-i-H l iH-H-1 'i I+ NHHtH--+-H+-+v++I+!+-1=-f"-H-++++-H- 1 l ii-H I -Hi1-l-i-i-lff i l' !lH i' I i f-!" i! ' i; i j 1+1H i i · -r-�-1\ -H-1+hdH --fI-I-I-H-\fI+-H-t-� H+ H- I+ l l + ! j -l Hf H + ++l l-mH ' H I I I H + ! + W l 1i -! -H I -I+H 1 /F'�rl; -i\;i-4 , i ii I ,It , ! i ii i i i! i

i

+f+++j-f+l-!,A'H++ ' ' l tw + iiil i i ' ItHHiHHj,1IlHt i Hi' 1 butanol it l l l 'i ttHt ++.!+++ 4ff+HJ 1 1 +++-i .!: dil I I l '! CH 3 (CH?)3 0 H

I!'.' :, ','" " 1,,1 II I

,

I �:::mmit, ;:tl i t

. T

'

! ;!IT :",""",+:"'i:rm-mr:rT:1fT iTrr,.,. I wavelength (/-Lm)

4.5

l , tn mTTT mTTT

'Ht H!tl-l+H H, t,Ipttib

60 • � * I, I *i!UHiI* i1I tw . I M ::il: 'ii:. iii! 40 I IIH i6��11

20: E I

4

I

'' �: +i+++� ++I

:�

: II"

3.5

-

:II

:1: :. ' H;H-!: ; ;f+

I!

.

I

,

I l' I

.1 i i i ! ! !I l

1+1H-i+H-i+H-i+1H-i+H-i+i+H-'l+HHHffi1'iffIi1-H-Hi I -Hf-H+-t--+-H1 ++++-H+H-+++-I I : : I: ; i���:�h -I--l---+-!-+-++++-H i!

I

:

ill

I

I

: !

, i

i" I

:

; 11

(\

°

1 H-H-H- HH+I HHI+H+ I-H-H-H-H , ,+ I-H-H-H+H-I+1;+11iH+h saturated II+I+ Yl+ �-H-HH-++-H-- H+J-t- H--H-+ 1 1 H-!+H-I+ffil---js t r e tc h �C HHIi-H-l i+HH-H' H I ·l+H -H-H + Hi -H-H -H-H -I-H-1 H+ + I+ fiI --i--i-i--I--+++-+-I--l-t--J-t- +-++-+- H 1 1 111 III " I i l l l l l ll l

:,'

,,

I I "

: 1

''''II '' II -l-i-!-H+!-H-lI+'HI-\-Ii+HH-HH+H+-I-H-ft l-!-�-f�-· I-�-�-I--I·-t--r+1-1-+-I'-H-1 �Hii+l H-H+!-H-HII+I+!-l-i-HfH-HI I .u1..LL!. HJHJJ.j2+� �H= I -'---'--'--'l00ll l LL I LLww OO . Ot6f lillll.l l'll.2l.LL 0ll 1l.60 .! ll..l0ll..Ll..lL ll..l �000��Ht+i�'J .LLLO'-':-0!..!.LLl.LL OCLO--':�--'-'---L-l----'---' -'-J :80:-:0L 600--'-:-: 40:'-'::0.!..!.!..l..L.12 - --'--'---'I8 wavenumber (em - I )

-

The IR spectrum of I -butanol shows a broad, intense 0 H stretching absorption centered around 3 300 em- I . The broad shape is due to the diverse nature of the hydrogen-bonding interactions of alcohol molecules.

1 2-9 Characteristic Absorptions of Carbonyl Compounds

wavelength (fLm)

1111

1.1 II , ,

I, .

3500

3000

2500

2000

II

1 800

II

10

11

12

13

14

15 16 ,

,

i

I II '

1.1

.

iii , II

II I I ' I

I I

Ii

!

,

Ii I, i

I ! ,

9

I

! ;

.

8

dl 1 600

wavenumber (em - I )

1 400

II I' II II

I, 1 200

1 000

... Figure 1 2-10 The IR spectrum of dipropylamine shows a broad N - H stretching absorption centered around in this broad absorption.

800

3300 em - I .

Notice the spike

broaden the N - H absorptions. With amines, however, the absorption is somewhat weaker, and there may be one or more sharp spikes superimposed on the broad N - H stretching absorption: often one N - H spike for a secondary amine ( R 2 NH) and two N - H spikes for a primary amine ( RNH 2 ) . These sharp spikes, combined with the presence of nitrogen in the molecular formula, help to distinguish amines from alco­ hols. Tertiary amines ( R 3 N ) have no N - H bonds, and they do not give rise to N - H stretching absorptions in the IR spectrum. Figure 1 2- 1 0 shows the spectrum of dipropylamine, a secondary amine.

Because it has a large dipole moment, the C = O double bond produces intense in­ frared stretching absorptions. Carbonyl groups absorb at frequencies around 1 700 cm- I , but the exact frequency depends on the specific functional group and the rest of the molecule. For these reasons, infrared spectroscopy is often the best method for detecting and identifying the type of carbonyl group in an unknown compound. To simplify our discussion of carbonyl absorptions, we first consider a "normal" stretch­ ing frequency for simple ketones, aldehydes, and carboxylic acids, then we examine the types of carbonyl groups that deviate from this frequency.

12-9A

12-9

Cha racte ristic Absorptions of Carbonyl Compo u n ds

Si m p l e Ketones, Alde hydes, a n d Acids

The C = O stretching vibrations of simple ketones, aldehydes, and carboxylic acids occur at frequencies around 1 7 1 0 cm I These frequencies are higher than those for C C double bonds because the C = 0 double bond is stronger and stiffer. -

.

=

o

11 /

1 7 1 0 em - 1

R-C-R' ketone

o

11 /

1 7 1 0 cm - 1

R-C-H al de hyd

� 2700, 2800 cm- 1

�/

1 7 1 0 cm- 1

R-C-O-H acid

\broad, 2500 - 3500 cm - 1

600

523

524

Chapter 1 2: Infrared Spectroscopy and Mass Spectrometry

Ii

5 100�: 1 1 1 111 80

1�

i Of_

i T'

20

o

I�

T

II! ,I

I

" 'I

saturated

��r

I I

h

,r I

lil,1

I

III

3500

,'I ii i

II'

II

3

, I 1II1I

, :,

::

'il '1, 111l 1 ,

' I

' q,, '

I

i � i II saturated 60 ! � ii i ' C - H ����tchlll! '

jM

I� :

j �� " 1 20 11�

1

0) 4000

I

I

Ii I i

'iii!! Ili ';II ' '

l li , l

, ,

I

,

i ': 1 ;' " �iii:! ii , , iii

:1 , II

I"

3500

i!!i

iii

i i i Hi!" 1 :! ill ! i i i

i i i i 1 if ! t

T

i'

I I, I '

i

.I

I II

" I

, I, I

I

" ·1 ++-f H·lljiI·ft.1H++-l-H+I 1'1'H,I H, +H++t+I+HIIIH C=O II + 1 H+ Hr-+-H- ++-H-HH-+-++++t-HH stretch III II I II 0 I i

i

Ii' II "

4

II I

,

I

III

'!i

11

(a) 2-heptanone

! I I

2000 1800 1600 1400 1200 1000 wavenumber (cm - I )

2500

I I I ',

I

'I! I !i

' I !,II iii, I l � f � �AA+t�/ +ttt/·hrtHI1++l*H·H+I +I+Iill ttttHm im,!+!Htttl+�'IT IT , ltr */-�i·+t�/-r+-+rr+I+t-r/-rt-+r / I l ! I

3000

3,5

I i

I'

l +'f!+t 1+t+H+I++1 H+H! lt HliH ftttttttH tH ttttH+tHt+t+

! !! I

II �I, I

I I I

i�

I'

" !l

I

'I

iii I

::; ::;,

II ,

'I

I Ii

I,

,

' I

25 100 . ��,:I;;

40

' ++H++H++H+:i+i+�i ::::: ;1

I

il � - H C 1 Htltt , , !I ' mtH-1t t ltt iI 1 +t++ 1 !+!+ IH!H'H+I :11 II ' il i: '11 111 I

I �

4000

80

,

ll

!� I I i I

i 1" " ' ,! I I I , iii I i Iii I : , II! i i i

:n iii

I'

60 I � I 40

II

III I I 1111 ''1

wavelength (jl.m) 5 5.5 6 7 8 9 10 II 1 2 1,3 4 15 16 I ; i H+f-H+f-'H+f-IttrlJl+H'N-++-tft-Hr-+-+-+++1-ITt-t+-1r-+-+-+\i ' f.h +tfI!+i+i+i1,!4-H-j'i>kif.h

4.5

35

4.5

II! II

wavelength (jl.m ) 5 5:5 III , ;1

7

I ll !

8

9

10

Illi

800

11

600

1 2 1 3 1 4 1 5 16

i, , I iii! / f , ,i · ' / " , II I , ,I O i ·/+I+I· H\H+/·H+H� / +l+I H /H L \ t;H-1 H1 1 + t+HtFW /+ HI' tI ' I+\�-/+-H-+++t-H' I'-r+-++++t-h stretch +/ H+/+I+I 'I+i+li · I,ll , , I I ' I II Iii i I I [II , 'Ii il I i ll I l ! II ' a l l tt ' -r-i-n , ti ' -Hii+t++liilii ' t+t+Ht' Imti H! +t++++ttt-tHTt+t+tt m+l+H+tttiitiTH-mm!--H+t II , II I

1 1 1I111 '

I" I i / , ' l l· \ I

� � Utttt��� lltt!ttl1ttt�tttt��tt.u�����t�t·ttt·tttttt�t1-tt+�;+� i ' " ' I / iI �· ·Il '

II I '

' I

' I I' ' i Ii i ' I CH}CH2CH2- C - H I I II I H+ I+lI'HI -!' ' li::�H'J2729 �'LL, Uq!.i1 HH II +l I I I' H-f+ I+t+t:!' H'H -H++ C = 2. l : , 'H+t+tI·H+I+I. ' I : 0 I , I I I I l + H+ " -ItI I+t-H· I+H+I fH+i+HHII !! I i ! t++I H++HII+ I ·t+t+I ' H!+I+H!+I+mm l l: 1 I I+-i l t'+t'I r++ I_ L -H +r-+L -I J ' H t iJ + . + 1-+ , IH IT ' +� IT I IT + ' , T I H-H +m I I I I , " 1 1:

3000

:I ;

ili 'H 2500

I

it ' I� : lmlm ttl:w:U=r=r�(�b�) b�lI�ty�ra�l�d�eh�Y�d� e r4=t:1 U U ':i : JllQ 2000 1 800 [600 1400 1200 1000 800 600 wavenumber (cm - I ) stretch

... Figure 12-11 Infrared spectra of (a) 2-heptanone and (b) butyraldehyde. Both the ketone and the aldehyde show intense carbonyl 1 absorptions near 1 7 1 0 cm - I . In the aldehyde spectrum, there are two peaks (2720 and 2820 cm- ) characteristic of the aldehyde C - H stretch.

PROBLEM-SOLVING

H?np

Real spectra are rarely perfect, Samples often contain traces of water, g iving weak absorptions in the O-H region. Many compounds oxidize in air. For example, alcohols often g ive weak ( = 0 absorptions from oxidized impurities.

In addition to the strong C = O stretching absorption, an aldehyde shows a char­ acteristic set of two low-frequency C - H stretching frequencies around 2700 and 2800 cm- I . Neither a ketone nor an acid produces these absorptions. Figure 1 2- 1 1 compares the IR spectra of a ketone and an aldehyde. Notice the characteristic carbonyl stretching absorptions in both spectra, as well as the aldehyde C - H absorptions at 2720 and 2820 cm -1 in the butyraldehyde spectrum. A carboxylic acid produces a characteristic broad 0 - H absorption in addi­ tion to the intense carbonyl stretching absorption (Figure 1 2- 1 2) . B ecause of the unusually strong hydrogen bonding in carboxylic acids, the broad 0 - H stretch­ ing frequency is shifted to about 3000 cm- I , centered on top of the usual C - H absorption. This broad 0 - H absorption gives a characteristic overinflated shape to the peaks in the C- H stretching region. Participation of the aci d carbonyl group in hydrogen bonding frequently results in broadening of the strong carbonyl absorption as well .

1 2-9 Characteristic Absorptions of Carbonyl Compounds

106 ,II: III I I

80

!I,II I 1111 11

I '"

1 . 11 II II

I I I" h III I I

'! !�

I

60 II ;�

20

I ,

II

M

40

I, :�

I

I

stretch

I 'I

I

II

1 111

I

III I

Iii

I III

1 1. 1

,,! , id

I

I ' :II! II i Ii I " I"

I

I IHiil1

2500

I !i !

i

I! ,

I

Ii J!

I'

I

i '

i I'

I!

II

ii

stretch

C=O

III

I

7 , I ! l II ,I II ; "Ij

6

I II );

I

II

I

I I I1111 I 1 1I

1 1 , 11 I

I

I IIII

II,

II I1 I !I

IllUUUlljl

I

,I

I

5.5

I II II

I ,! 1'1 I llli!!1

I Hi!

3000

wavelength (/Lm)

4.5

I

, �ret��

I :

I

3500

4

I i i II ll i i i i I I I III' I i ' 1111 111 I l i ! 1 " l j• 'I� , !1 "I l' . i. I t l ' , " I If 1 I I; � I I I O - H I I,,'I ! I

I

IIII IIII I I ',I

Ii

0 4000

I,ll!! 1111 !III 3.5

3

Ii :1 II

'II

I

"i i

I

I

:

"" I

!

II

!

I I!

14 15 1 6

il!

I I II 1III � I II I

I I I

I i II

1 400

1 800

1,3

I

I

Ii!!

i

I '

, ':!lllllir 1 600 wavenumber (cm - I )

2000

i!

ill iii

" I,

I

12

11

I; i'

Ii

I

iii

I

I

9 10 I I ii!i II I

II

I! I

II

8,

525

,I

1 200

hexanoic acid

CH3(CH2)4COOH

II

I IIII 1 000

I

600

800

t

O· · -

\

R- C

H

\

O --....

... Figure 12-12 Infrared spectrum of hexanoic acid. Carboxylic acids show a broad 0 - H absorption from about 2500 to 3500 em - I . This broad absorption gives the entire C -H stretching region a broad appearance, punctuated by sharper C -H stretching absorptions.

\

--.... O

;f-

C-R

H

-..O

\

1 7 1 0 cm - I

broad, about 3000

cm- I

SOLVED PROBLEM 1 2 - 1

Determine the functional group(s) in the compound whose IR spectrum appears here.

wavelength (/Lm) ill

, I II

', III I I

iii III ! � ! II

!

II

I \I :i i

,I , 'Ii t

o

4000

: 11 .1

Ii! 'I

I ,

III I

i lt

3500

3000

2500

I I

I I II

ii !' i

!

i

I

I

Iii

,I

I

I I

ii

I Ii I

I

: II ! II Ii

",

II!

II , ! I i !!

!

1 600 wavenumber (cm - I )

2000

IJ!iI I

I iii

,I ,I

I I

I Ii I I

II i ll l liii I 1 11 , I! I iii

II

1 800

1 400

1 200

:I

1 000

I

800

600

SOL U T I O N First, look at the spectrum and see what peaks (outside the fingerprint region) don't look like alkane peaks: a weak peak around 3400 em - I , a strong peak about 1 720 em - I , and an unusual C -H stretching region. The C -H region has two additional peaks around 2720 and 2820 em - I . The strong peak at 1 725 em- I must be C = 0, and the peaks at 2720 and 2820 c m- I suggest I an aldehyde. The weak peak around 3400 em - might be mistaken for an alcohol 0 - H . From experience, we know alcohols give much stronger 0 - H absorptions. This small peak might be from an impurity of water or from a small amount of the hydrate of the aldehyde (see Chapter 1 8). Many IR spectra show small, unexplained absorptions in the 0- H region.

a

526

Chapter 12: Infrared Spectroscopy and Mass Spectrometry

PRO B l E M 1 2 - 4

Spectra are given for three compounds, Each compound has one or more of the following functional groups: alcohol, amine, ketone, aldehyde, and carboxylic acid, Determine the functional group(s) in each compound,

'

I

wavelength (porn)

" Iillilll lUI I ll+�lllll Jilll llllllllll�III,1 I_'" 1,1I 1I1 J -l-U IIi1_ I iII I tHLLIUII 11I 1 1 I I I I I I I I 'I 1-L � iii: I "mtl:�*m:m: Tl [ , "1[11 'llI '1 �lll_'i l�LU I lm"tl I 11,1 � ml I I 1111' ,'1111111111111 I rlvl I 1111 II l Imit I, I illriII I II : 4-� I I -l±# :LH � I (a) H- � H HI Milll +, 1: l fltfiiiiii �-+±tf ' t+tt\ +-fH-ttttt# 1 1 1 1 I�ttn I + I -ltl, t lWn; ll lil l ,i lJifltfiliH $" ffi : illlL �i I llJ�HI! IH'I I,I IIi"IHI HI III HIl ijlHI mH H I lml- 1+itl+- +I--r -HH- H+ , 1./ H 1+1--hl+t� --�1+ +H-h++-h--+Ht illl! J

1,5 3 I IIUUI!I IIII 1111111..ill! J III iii"

" I

I

II

3,5

4

,

4,5

5

'

'

5,5

I

I

JlI I AN

lill

9

8

7

,

1m ,' jlJl J I1 , I IillIIIII fTT, ,I ItMI I I tt cE illl l I\Wj, 1l �� , I'IIfTilII 1m-,I,iil + I' I! ' 1WI!lII 'll I,� JII..ljj1JJllj .!IIJ - , 'IU l i n

I

6

10

11

h'J

,

I

12

13

14 1 5 1 6 I

,

"

I I 1\1.1 II ILlIII I II I

'.I ll ,II

_1+H-I --t-++-t-H'H -l,/+I-t-H-+I -, H,II- H---+,-I+++-I-+-I-+-I-J+--+I++

' _'LU , I I I II I 1' II, IT I r1111 I , III ll I +++'-l III , iU11!1JILLL UL+ Iln I ,ilL! 1111 tH-'r II TIll I mr1111 rrr -IT i l IlnllTIT1111 11 I lTITllllTIT I lilllllll I !!II !m Iii, Mt I nITH I I I 1II I , HflliHI �H H' ' / H1' HlHifh11HH1111 ' 11 1.1 1 1 HIHHIII" � H-+HI+HH+ t� +H I 1-+1-1+ tiff -H-It LfI- I -H tm+-+ --+1-++ - -t++�HI lL,T I nll

1 1 ,1 1 1 I

,

4000

TfTTn II I l' i !, I, I II!l1 1 ,n, 1:111111111 I ' 11111111

[m'

II

'

3500

, 1 , 1 II I I1I1 mr

!

,

I

I,

3000

2000

1 800

u I TITII ' li iiilil l'llUJtl Ii�W I 3

II

3,5

1 600

-

1 400

1 200

wavelength (porn)

4.5 5 5,5 " ,*:i" I I I!:!!JJJJIlIIIiI I�IIH�illl II II , ': I I mllttlt,IlI' ITt 'Nil lid ," , nT11_11 II IT:I I!, .' ' II !. ," "" I , I IJUIJIl/ lllWl1ill lll' 1$. ' JJJJ IIi I, ,llil liLi Ili! ' , ililll -I, , II flIT 80 , IllrnT �i " . I

2,5 1 00 [,::

"

wavenumber (em - ] )

2500

I , II ! '

4

I II

1 000

!

800

ilL

o

M

T

4000

l i li

(b)

I

1 I

,

I

, iii

!

"

, !i!, , ;! I

"

Ii

'Ii

3500

,. i,'

3000

'

I

2000

]J..l.PIl

I

I

wavenumber (em - I )

2500

4

I

Ii

1 800

1 600

1 400

1 200

wavelength (porn) 5.5

4.5 ' I i

'II

nn

11

L OOO

,0

'1 '1

II

m

111 1 1

II

'I

600

I I I �_ I I I I--l- II II�_

Jt 1 1;TTmr rrr rrn�:m ll: , If lir WI l'tfHHi ' hm Wj ,�, , , tt�-Htt I�II fit Htt $ � I I Hrmt-j '-1 tiltH i i' 60 - � � ill ' I ' Jjl �y;i!llhfHI �H WM �- � jJnIJ.lIl J' f Il II,w IlITIl liIf";HNli. Hl,T+',IIH'Utj' ' 1j�Ul:1!fii IIlJlUlrl- tlilitl '-llil 1l1! 1lrTTTllTl ' lII _ -JilI IJtH:, ITI 1 IT, IT:! I Tn m Trrr II 11 1111-I-mUtlt ill,l i ii, m !!' 'It" [Ilj"iJPM,I�1,IlI I ii' III I !Ii L IlUIJ1H4LL11!N I III ml- II'�llUJ�-l- I v ,11mlTIT , I I �rn. �I l.!-!-!-Im -W- TLU iWU-W-!- lijF 40 me q jj * I IW InL! III Jll ! I U I,' i I '' II I n II! I I iHi ' 1mTITlII rn ,n! I !n l iT � �, m ' llll \ ilt JHlt/�Hi II'HI1 1 '1U\" -mtmt l_4iit H_rtt ,Hrtt :mi'iit lHiit Hl!l+ffili+t-M-i tt_\IHtiI_tffiHtffi�Hm-i+rtt +m - iitIi +iit1iitffiffi :II�-t-t+ttt111UI 20 IIlllillJ I111 1 II W[I, ' Iij iii! UtlllJUJJU' � fw.tlilllJJJJlWJUUillJ illI I i i I fIT1Ifm IIfIT ImTfll lm n,nlillll� 1 1 111111I11!! ' TflTITITlnTrl " 1111 I lTI III i lll ll! I II Im-' lilia " 1111111 llItH JIL ill Uti l.wlilE1 Jilil JUJ lll IUm TtU"rf 11thH',jlIJtlllljilllllllflll i litII ,Illlllmt lifillillUl!,tllnill ll 11- ill, IIImrmr JlILt'l 1IJil JjJJ1lTfTtiTil III1 l J11T T

I

I

7 8 9 10 I ! LLLLJlllLlliIlJ!-jJiIII I � " mrmrmrlll 'M IJill _D I 1 -11111 , :I

6

I I !

I

11 ,

12

I

13

I 'I

14 1 5 1 6

I I800 I 12 ,

1,3

I

600

14 1 5 1 6

J-,

+- +H--+- I-++---h- I +-I+ I -i-+f.-I IHHI++I+f1++j-H1tHtttfl+HtH1fHHH-HIrl+1H+Httf+l H+i-l-I+l-/- -+ I l ', I I I I i i I I I I I I I I I i i , I il llIlJJIITIrrffifITiTITWfITIITlllillllil!iI, \W rUIJJIllI I:m+ H,iliLU liliJJIIUJllI _ I JI -:4 ' , . . jii

m

i l ll l lllll l l l l l l l l l l! ,

"

1I111111 , 1 1I11! I IIIIIII 11 ill

wavenumber (em - I )

2000

1 800

1 600

I

.

.

'

,

. , , '

"

,

, ,

"

,I

1 400

1 200

1 000

800

600

1 2-9 Characteristic Absorptions of Carbonyl Compounds 1 2-98

527

Resona n ce lowe r i n g of Carbonyl Frequencies

In Section 1 2-7A we saw that conjugation of a C=C double bond lowers its stretch­ ing frequency. This is also true of conj ugated carbonyl groups, as shown next. Delo­ calization of the pi electrons reduces the electron density of the carbonyl double bond, weakening it and lowering the stretching frequency from about 1 7 1 0 cm- 1 to about 1 685 em- I for conj ugated ketones, aldehydes, and acids.

0

' 685 ,m

2-cyclohexenone

O 1 690 cm - l �

I

C H/ � 2-butenal

benzoic acid

The C=C absorption of a conj ugated carbonyl compound may not be ap­ parent in the IR spectrum because it is so much weaker than the C=O absorption. The presence of the C=C double bond is often inferred from its effect on the C = 0 frequency and the presence of unsaturated =C- H absorptions above 3000 em- I . The carbonyl groups of amides absorb at particularly low IR frequencies: about 1 640 to 1 680 em- I (Figure 1 2- 1 3 ) . The dipolar resonance structure (shown next) places part of the pi bond between carbon and nitrogen, leaving less than a full C = 0 double bond.

The IR absorption frequency of the amide N - H group is sensitive to the strength of hydrogen bonding. Therefore, IR spectroscopy provides structural information about pep­ tide and protein conformations, which are stabilized by the hydro­ gen bonding of amide groups.

��

about 1 640 c m -

CH3CH2CH2-C- NH2

The very low frequency of the amide carbonyl might be mistaken for an alkene C=C stretch. For example, consider the spectra of butyramide (C= O I about 1 640 em- ) and I -methylcyclopentene (C=C at 1 65 8 em- I ) in Figure 1 2- 1 3 . Three striking differences are evident in these spectra: ( 1 ) The amide carbonyl absorption is much stronger than the absorption of the alkene double bond; (2) there are prominent N - H stretching absorptions in the amide spectrum ; I and (3) there is a n unsaturated C- H stretching (j ust t o the left o f 3000 em- ) in the alkene spectrum. These examples show that we can distinguish between C=O and C=C absorptions, even when they appear in the same region of the spectrum . Like primary amines, most rimary ami des show two spikes i n the N - H stretching region (about 3 300 cm- R), as in the butyral11ide spectrum (Figure 1 2- 13). Secondary amides (like secondary amines) generally show one N - H spike.

o

II

R -C-NH2 primary amide

R - NH2 primary amine

o

I

R-C-NH- R secondary amide

R - NH- R ' secondary amine

528

Chapter 1 2: I nfrare d Spectroscopy and Mass Spectrometry

0

1

6 11;, 80 I

60

40

I

'�

: 1:

.. HlJ.LIHU' L IH1H II I I I

. i] : I I ,. II Ii

I

!M

I

I' ll I II , 11� l[l l l l ii 1 IIIili l l l l l l I

ii " I

.� , � :�

1

35 11 : 1 .

I

lill

1 4

1 ;'!�'11 I

I

55 6 ':: , . 1,:

wavelength (j.Lm)

·�::1 1

, "

Ii! I i i

' 11 1 1

o

"'�

1

"'" I

2.5 100 80

'

I'

T

II

1 :: ,,,

' "

1 1+'

I I '

'

� �

3

I l � II i!; .i !R ,II ,i i '

,!,il Ii ! , '

Ii

60 !I �S II ".

1 M i ,I'

I 1II 'I I. II, ! I i i

III1

, ,II

I

o

I

4000

l ltl:

, ;,,1

I

I '�

Ii ! I ' d

I

3500

I

..

3000

5

5.5

I II II I

iill l , \1 Ii Ii

! II I

I

!I

6

I

C

I

I

\I

I I

II

1Llllll . I iliilill

.

I

I

I

I

1

-H-+I-:I++-I+:+1_1

6 II· II

I

1400 1200 1000 8

wavelength (j.Lm)

!I'!

2500

ilil

12, 13 14 15 16

I

1 ill I ! 1 1I I L.L II I LL I I� ' · -:'-':Ll" !1 l lw lI l w JJ. 1I-,::,;1 1.u ill..L i II� I !!I,::-:l,:! L.LL.L�:-!--.L..L.L..L-,--I-,---,--,::: �ill.LIL.L .L!:,; J

, III

I I I l iliilil

I

stretch

4.5

II ! Ii I 1. 1 I I I. I ! , (b)

IIi'

"

wavenumber (cm- I )

In !;

I

I

"

I ,,

III lii.i !

!Ii

1111, . , 40 1 i I tl!1 1.1 . i: �� Ii '!!H : II 20 I ' alkene = C - H I I

I II

4

3.5

Ii '

III1 I i i I I 1.1 I

II I

I

11

10

9

'11+ lj��II+I-! \++'f:II+tH-I+t'fW'I+H : H!+,1+t-H'+IHI- H+f H-i-l+H--H-f-H:IIII 1 '1 I I I -I

CH

1111 1 1 ',

III I i i!

1 1

' I I Ii

!I I , 'I I I 1 ++-f+ti + H lfI+trn l +tHIH-++t+t-+-+ + fffi 1 +-+-nrl, I'H'+++++, i ttH I +-+-+-r+-++'I-+-+ I I t++-'l++++-+tHH+l' I H-+H-H i I I I II ,I ' I I I II I I I I , II I

: i ': : ::' HM+lI+lHII.H:' ,:'! " c ��: ,L : � Iffi11 : 1 1 ITIlTII 1 I ' � ' · I i'"'; ' ' ' l..!J.l.!� I 3500 3000 2500 2000�Htl..!J.l.!I�.L!1800' 1600 4000

20

8

; 1 1 17

I I _ H-H-H +-H-I+ H+++I+i-H+H n lu+t-I� . . cNH--+-1 -+-r- I--+T hr��'-Y�-rt-r r-r+�1 ' +1·H++H+H·I+++H -I+H-H� I i ++-+I-+-HyI'-++!' -+-1--+-+++ +++-H+ +++++++ H-H'+I 'H ' 1+1ffif-H--+-ft +-+-+-r---H I I '++++ I I H-H I ++++ I H-++ I +t+H-H I I +++++

I , " 'l 'II

I

5

II III

7

I ,! "" i

1658

II!, i

I , ! ,

"

I il i Ii I

I

I ii ,i

,I i " Iii I' II

:1

,

!

, I I ,I, I !i i!

I '

i

I

!

II iii

C stretch

Ili

I, !i!

II I!'

, II I ill ,

I"

800

11

10

9

'I I I ," i , .1 !Ii

,I

600

12 1 3 1 4 15 16

,

i I

!I J I Ii

, I, I

I

,

!

III I

2000 1 800 1 600 1400 1200 1000 II

i I

I

I

wavenumber (cm - I )

800

600

i

.. Figure 12-13 Characteristic IR spectra of amides. The carbonyl group of butyramide (a)

and the C = C double bond of I -methylcyclopentene (b) absorb in the same region, but three clues distinguish the alkene from the amide: (1) The C = 0 absorption is much stronger than the C = C; (2) there are N H absorptions (near 3300 em- I ) in the amide; and (3) there is an unsaturated = C - H absorption in the alkene.

-

12-9C

IR spectroscopy can also be used to monitor the progress of biological reactions. For example, the hydrol­ ysis of complex lipids (esters of glycerol) causes a characteristic de­ crease in intensity of the ester car­ bonyl absorption at 1 73 5 cm- " with a corresponding appearance of a carboxylic acid absorption near 1 7 1 0 cm- ' .

Carbonyl Absorptions Above 1710 cm-

1

Some carbonyl groups absorb at frequencies higher than 1 7 1 0 cm- I . For example, sim­ ple carboxylic esters absorb around 1 735 cm - I . These higher-frequency absorptions are also seen in strained cyclic ketones (in a five-membered ring or smaller). In a small ring, the angle strain on the carbonyl group forces more electron density into the C = O double bond, resulting i n a stronger, stiffer bond.

o

I

/

about

1 735 cm - I

R-C-O-R' a carboxylic ester

)

o�1 738 c m-I

II

)

CH3(CH2)6C - OCH2CH3 ethyl octanoate

o

II

)1785

C / " CH2 CH?

\

/

CH2

-

cyclobutanone

cm- I

)

1 2- 1 0

Characteristic Absorptions of C -N Bonds

Infrared absorptions of carbon-nitrogen bonds are similar to those of carbon-carbon bonds, except that carbon-nitrogen bonds are more polar and give stronger absorp­ tions. Carbon-nitrogen single bonds absorb around 1 200 cm- I , in a region close to many C - C and C - O absorptions. Therefore, the C - N single bond stretch is rarely useful for structure determination. Carbon-nitrogen double bonds absorb in the same region as C = C double bonds, around 1 660 cm - I ; however, the C N bond gives rise to stronger absorptions because of its greater dipole moment. The C = N stretch often resembles a carbonyl absorption in intensity. The most readily recognized carbon-nitrogen bond is the triple bond of a nitrile (Figure 1 2- 1 4). The stretching frequency of the nitrile C = N bond is close to that of an acetylenic C = C triple bond, about 2200 cm- \ however, nitriles generally absorb I I above 2200 cm - (2200 to 2300 cm - I ), while alkynes absorb below 2200 cm - . Also, nitrile triple bonds are more polar than C C triple bonds, so nitriles usually produce stronger absorptions than alkynes.

12-10

C h a racteristic Absorptions of C - N Bonds

=

}

C - N bond stretching frequencies

for comparison :

" , ll l i I, I111 1I l i l l

100..5

40 20 o

;

: II I � , II1 Ii .11.1 .M I I

. .

: r "i 'l ,�

I

I, I" i I', 'III

1 200 c m- I

C=N

1 660 cm-

I

C=N

> 2200 cm- I

C=C

< 2200 cm- I

usually stmng

(usually moderate or weak)

ill II! " III! iili l l " I I "II I I II I II i I I " I ii , I IllliliiNl 11, !1i liii 'I I II Ii ! ; I I , " I i 1!II�iil!1i I iii! 1 I 11,! ,r! i' ,11I1 f.! iII II II II I !i 'ii' I " I '" ' 1i l' II I I I , j 1 1 n!! . . wm mmmmmmLm !;+H-i'++Hm-t-'lt +t I IItHittt--r-t-r-t--r--t-t- -r-t-r--H -t ---t-t-t--r-t--H 11 '111Ofititit++++++++C++++;-;-!N 1 l1 1t 11 '1 H-H-ittttitHttil\-t+++++t I ,' l l i l lin,l I I, ,"

" ii ii' I ' i l l l iil

I'

4;5

5 3;!!"HI

I'! 1111 ' , i ! 'I" I

,

80 �I ' I'I I 60

C-N

I

I I ' 1i

litl1,ftlmHrriiltlffi'm l l li" Ti:lIl,iR- r

1I11I

I

'Iii

wavelength (JLm)

';5

stretch

1,0

1,1

1,2 1,3 14 1 5 1 6

i, II I I lil ,! ++ II I HH+Hltt l t ltH- I+t+1 +H-i1 -1 11-i-illi H+t-t-t-+' i-t--H+t-++++++r-1 2249 -Ht+ +++

: I ! I!' !I' II I tttmJUJUJ.JJrlltml�!m !m!�!!�.tI;-lt-lt-ltrltrltrlt!It!ttrlttrttrttrltltUIttiUirUrUrUJWmrUrUrU���-;:wdd±�bl�'t�yr�O�n�it�ri �l e='='tU+t+�rn HtfttIHHtHHtltt tHHtIHH ffi' ltftt ,ltftt IHtt, it,!t, t;H+:

I i

I

i i

'ii

, I ! :I I'I

IIIL , , I I I

4000

h!

i Iiii"

I I .

i

'

i

!

t+, I-i-i-i-i +t+I+l+lt �t-:, I-I+t-t' Iffitmtt l! ff I it lilil' ttttttttttt·1+t-t-1 C H C H CH C - N 3 l l �����mml m q W I 'I � ' � � �� � � � ' I tt tt tt tt tt tt tt tt i i tt fF 9 � I-tl--t+--1+-1+-t +-t +t -t-tl _ 2 ""FF2'FF"Ff.+-t i � ltttt�ttltttttttttt�:tr-lTnFf"'Ff I ! J

,

j

• Figure 1 2- 1 4

Nitrile triple bond stretching absorptions are at slightly higher frequencies (and usually more intense) than those of alkyne triple bonds. Compare this spectrum of butyronitrile with that of l -octyne in Figure 1 2-8.

PROBLEM 1 2 - 5

The infrared spectra for three compounds are provided. Each compound has one or more of the following functional groups: conjugated ketone, ester, amide, nitrile, and alkyne. Deter­ mine the functional group(s) in each compound. ( Continued)

529

530

1 2:

Chapter

2 .5 1 00

mr 1111

80 1111

11111,1 ,

3

:1 1

20

II

I

:

I

I

Ii II I

I

I

I I

"

I I

I

3000

3500

II

I

I I

. 1 11

I

II

I

20

"

I

I

E,

o I 4000

" , "(

.

I

':1 4000

12-11

! i i

I

Ii

i. I ! ! III III ! i

,I

'I

I

, ,,

I

i

I

,

I

II

II 1 600 wavenumber (em I )

2000

1 800

,

"

I

I i i

:

1 1 1 1 1 11 1 1 1 1 i l ; II I 1 685

II! ii ,,

I

II

I

,

I

I

I I

I

12

I

I !

I I

11,1 Iii

Iii 1 200

ill 1000

I

I

i

I

:

I

I

I

I

: il I'

:

I

I

I

I

I

II I. i ll

14 15 16 , I I

13

I

I

I ,

I I

i i' I , ! I , I

1400

II

I

II II I I I

I

I

I

I

I

I

!

i I

,

!

800

I

I

I I

600

11 ,

1,2

13 ,

l4 1 5 1 6 I

I I

', ,I

. I

, " I 'I

I,,'

�' , '11 1. 1 ,n I

:!, I' ,II ,

III , I I I I Ii

3000

I II 1 I 1 11 , -tl ii:-t, iTI 1 738 mtm'TItIttl

II "

2500

'

II" I ! I i ' II

"

'Ii :

II

I

3500

I I I

I

I

I

I

II

'I

I I

'II II

I

3000

1111

1111 11

I

II

S i m p l ified S u m m a ry of I R Stretch i ng Freq uencies

' I:

"

iii

"

I 2500

III

'I I II.

,

Ii

I

,l lli

II I

' I

I '

1 600

5:5

I, ; j 11 ,I' iii

,I I ,

r

III

II I

II

, I 1 1'1

I, I

Ii , ! ,'II

I

I I

, ,iI I I

I" I i

,I

1 800

II I

I

1400

wavelength (JLm)

4;5 III I I I II I

II III,

I II

2000

I! , ! iii i ! II

wavenumber (em - I )

3

111I l ' iI I I I I ii i ! II, : I I Ii

1,1

10

II Ii;

i l l ' I i

Ii I

I

I ' Il ;
i '

I ii

3500

i

J i.i I I 40 ' i

I

I

,

! II,

5;5

",!II I' Hj ii' Iliil l l , Ii ' l

II I i , I "iii '

!; 60 I I �

I

I

II I

wavelength (JLm)

II ! I1 1 II

I , I

1I �

I

ti: 1l

10

1111

I

'

5 1 00.?:"1I1 1 i 1 I 80 ' I

!

I

I

ll i' �

A

,

, II

.

!T

i

I

I

'1 1'1'iH'II lllIiI I II I1I1 , , 80 I % (b)I I II ,': I ,� i, "Wlf+lf+l -H+ 60 iI, ��N m 11 11,' M I ':

'i �

,I,

1

-

1 00

40

I,

I

I

Ii I I

2500

55 I

I

I I !I I ,l l ! I

II II

I I I II I I

I

wavelength (JLm)

45 : I I�I I I

. 1 . IUWill illilLiI ,I I I ,

"

I

II

o 4000

Ii

!

"

'"

'

40

3;5 l iill III

I ! I (a)I I , i: I j; M� II 1.I 1 , I W, 1± I ' I i j! I ], �E I

60

Infrared Spectroscopy and Mass Spectrometry

!

I II

Ii.

Iii

I,

2000

I

I. I

oom 1 600

wavenumber (em - I )

I

II

1 200

1 000 9

8

1:1

I

I Ii i

I i" 1 800

II

II

I

I

I

12

13

1,4 1 5 1 6

'11 1 I

I

, II ,

I'I

1 400

11

600

I

I

"

10

800

:1

1 200

I

, II , II

, 1000

800

600

It may seem there are too many numbers to memorize in infrared spectroscopy. There are hundreds of characteristic absorptions for different kinds of compounds, and a detailed reference table of characteristic frequencies is given in Appendix 2. Please glance at Appendix 2, and note that Appendix 2A is organized visually, while Appendix 2B is or­ ganized by functional groups. For everyday use, we can get by with only a few stretching

12-11 Simplified Summary oflR Stretching Frequencies

I

25 1 00
35

3

80

.�t:

,

3500

2500

3000

55

I

2000

I

65

6

C=C c-lo CIN

T

I

20

o 4000

7

wavelength (11m)

1 800

I

II

1 600

8

9

10

II

cTc c-o C-LN

I

I

1400

I

14

16

f

r

I

12

531

fi n erpri t regi n

1 200

1 000

800

600

wavenumber (em-I)

TABLE 12-2

Summary of lR Stretching Frequencies

Frequency (cm-1) 3300

3000

Functional Group

Comments

alcohol amine, amide alkyne

O-H N-H = C -H

alkane

-C-H

just below 3000 Clll I

=C

....--- H

alkene

just above 3000 CIll I

I I

2200

alkyne nitrile

1710

carbonyl

"

O-H

acid

- C =C- C =N "C=O ....---

(very strong)

1660

PROBLEM-SOLVING

always broad may be broad, sharp, or broad with spikes always sharp, usually strong

very broad just below 2200 cm-I just above 2200 cm-I ketones, aldehydes, acids esters higher, about 1735 Clll-I conjugation lowers frequency amides lower, about 1650 CIll-1

alkene

"C=C ""--....--"

conjugation lowers frequency aromatic C=C about 1600 cm-I

imine

"C = N""---

stronger than C=C

amide

" C=O

....---

....---

stronger than C=C (see above)

Ethers, esters, and alcohols also show C - O stretching between 1000 and 1200 cm-I.

frequencies, shown in Table 12-2. In using this table, remember that the numbers are approximate and they do not give ranges to cover all the unusual cases. Also, remember how frequencies change as a result of conjugation, ring strain, and other factors. Strengths a nd Lim itations of I nf ra red Spectroscopy The most useful aspect of infrared spectroscopy is its ability to identify functional groups, but IR does not provide much information about the carbon skeleton or the alkyl groups in the com­ pound. These aspects of the structure are more easily determined by NMR, as we will see in Chapter 13. Even an expert spectroscopist can rarely determine a structure based only on the IR spectrum.

Htnp

Ta b l e 12-2 provides the numbers but not the understa n d i n g and practice needed to work most IR problems.

Learn to use the material in this table, then practice doing problems until you feel confident.

532

Chapter

12:

Infrared Spectroscopy and Mass Spectrometry

Ambiguities often arise in the interpretation of IR spectra. For example, a strong absorption at 1680 em-I might arise from an amide, an isolated double bond, a conjugated ketone, a conjugated aldehyde, or a conjugated carboxylic acid. Famil­ iarity with other regions of the spectrum usually enables us to determine which of these functional groups is present. In some cases, we cannot be entirely certain of the functional group without additional information, usually provided by other types of spectroscopy. Infrared spectroscopy can provide conclusive proof that two compounds are either the same or different. The peaks in the fingerprint region depend on complex vibrations involving the entire molecule, and it is impossible for any two compounds (except enantiomers) to have precisely the same infrared spectrum. 1. It indicates the functional groups in the compound.

To summarize, an infrared spectrum is valuable in three ways:

2. It shows the absence of other functional groups that would give strong absorp­

tions if they were present.

3. It can confirm the identity of a compound by comparison with a known sample.

You have an unknown with an absorption at 1680 em- I ; it might be an amide, an isolated double bond, a conjugated ketone, a conjugated aldehyde, or a conjugated carboxylic acid. Describe what spectral characteristics you would look for to help you determine which of these possible functional groups might be causing the 1680 peak.

SOLVED PROB LE M 12-2

SO L U TION Amide: (1680 peak is strong) Look for N

3300 em- I .

-

H absorptions (with spikes) around

Isolated double bond: (1680 peak is weak or moderate) Look for = C - H absorp­ I

tions just above 3000 em- . Conjugated ketone: (1680 peak is strong) There must be a double bond nearby, conjugated with the C = O, to lower the C=O frequency toI 1680 em-I . Look for the C = C of the nearby double bond (moderate, 1620 to 1640 em- ) and its = C - H above 3000 em- I . Conjugated aldehyde: (1680 peak is strong) Look for the aldehyde C - H stretch I about 2700 and 2800 emand C - H of the nearby double I . Also look for the C = C bond (1620 to 1640 em- and just above 3000 em- I ). Conjugated carboxylic acid: (1680 peak is strong) Look for the characteristic I acid 0- H stretch centered on top of the C - H stretch around 3000 cm- . Also look for the C = C and = C - H of the nearby double bond (1620 to 1640 em- I and just above 3000 cm-I).

12-12

Reading and I nte rpret i n g I R S pectra (Solved Problems)

Many students are unsure how much information they should be able to obtain from an infrared spectrum. In Chapter 13 , we will use IR together with NMR and other infor­ mation to determine the entire structure. For the present, concentrate on getting as much information as you can from the IR spectrum by itself. Several solved spectra are included in this section to show what information can be inferred. An experienced spectroscopist could obtain more information from these spectra, but we will concen­ trate on the major, most reliable, features. Study this section by looking at each spectrum and writing down the i mportant frequencies and your proposed functional groups. Then look at the solution and

12-12

R eading and I nter pr eting IR S pec rt a (So lved Pro blems) co mpar e it with y our solutio n. Th e ac tual str uc tur es of th esec ompounds ar e sh own at th e end of th is sec tion. Th ey ar e not given with th e so lutions bec ause you cannot determine these structures using only the infrared spectra, so a c omplete str uc tur e is not a par t of ar ealistic solution. mm�ft1� ' ' ft+IIl+!It!1"4'mu+mrnftm!!m1 mrrr1l)1,!l1] 1m 'illl Ht ,. IttiH: I : : H t HttiHt I 80 � ri IIlilll , ; t+H+Htm , lffi,ritJiH,,

2.5 I 00

•

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wavelength (/-Lm) 5 5.5 6

7

8

.

Ht

9

11

10

12

:,�flrIIIHlm! H#i�11 II lli lmmHlV mH+H -H+H·Ht*H±llii+H-ill*JjJ+lli�-llillili}:lWkl-1 iI �11 ffIf�tfIlllfml!Ht�tHfjHftHltl ::#itMil IlffIf-H#�#�H1 +H � ::: . u HI!;� ,HilIYrNll4�ittill +R-H+ftt-H# iH11 -*1+Itii l�f Mmfiilimu -Ht U 111�II +ttt ,fjlKU"lllU ! . t . 11 1 11 _

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4000

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2500

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1 000

,

800

10 ! I I !

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4000

1 1 1

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r--r---t-t- -j-i-.-r-

�uu

600

5 54

Chapter 1 2: Infrared Spectroscopy and Mass Spectrometry

1 00

I-ll

-

80: . �

60 40 20

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0 4000

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(c )

'1 11111

IIIHII

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,

f l "il

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, Ii i

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12-17

12-18

12-19

12-20

2500

14

1 ,5 1 6

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1 400

1 200

7

8

1 000

9

800

10

11

12

13

600

14

15 16

, i

I 1. 1

II '

3000

13

'" i l l

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3500

12

, II , 'I II

! ';

,

II i I I

wavelength (p,m) 5

I

I '

II

1 800 1 600 2000 wavenumber (em- I )

2500

11 ,

Iit II III

,

I i1 i'

II

II

!

, '1 1 ' 1 1 ; L il li

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3000

3500

I I' II

i ll ' !i I ii ,

1m

II

i l.

11 1. "

II

wavelength (p,m) 55 ,

45 ,

II

, ,' I I II

,� '�

. �, 'i .�

3

I i III 1 600 1 800 2000 I wavenumber (em- ) I I

III ! !!!

'Iii i '

1 400

1 200

1 000

800

600

Predict the masses and the structures of the most abundant fragments observed in the mass spectra of the following compounds. (c) 4-methyl-2-pentanol ) (b) 3-methyl-2-hexene (a) 2-methylpentane (e) cyclohexyl isopropyl ether [cyclohexyl - O - CH(CH3 zJ (d) 2-methyl- I -phenylpropane Give logical fragmentation reactions to account for the following ions observed in these mass spectra. (b) methylcyclohexane: 98, 83 (a) n-octane: 1 14, 85, 7 1 , 57 )(c) 2-methyl-2-pentene: 84, 69 (d) 1 -pentanol : 70, 55, 4 1 , 3 1 (e) N-ethylaniline (PhNHCH2CH3 : 1 2 1 , 1 06, 77 A common lab experiment is the dehydration of cyclohexanol to cyclohexene. (a) Explain how you could tell from the IR spectrum whether your product was pure cyclohexene, pure cyclohexanol, or a mixture of cyclohexene and cyclohexanol. Give approximate frequencies for distinctive peaks. (b) Explain why mass spectrometry might not be a good way to distinguish cyclohexene from cyclohexanol. (A true story.) While organizing the undergraduate stockroom, a new chemistry professor found a half-gallon jug contain­ ing a cloudy liquid (bp 1 00- 1 05°C), marked only "STUDENT PREP." She ran a quick mass spectrum, which is printed below. As soon as she saw the spectrum (without even checking the actual mass numbers), she said, "I know what it is." 1 00

80

g 60 0)

oj -0 t:

ii 40 oj

20 o

-----

5r ,

20

I

t I

I I

1----1----

10

I

30

40

50

Jrtlll 60

70

80

mlz

1 011

I ! I

1 36

I

93

.1.1.

90

1 00

1 10

1 20

1 30

1 40

1 50

1 60

* 12-21

12-23

(a) What compound is the "student prep"? Any uncertainty in the structure?

(b) Suggest structures for the fragments at 1 36, 1 07, and 93. Why is the base peak (at mlz 57) so strong? A C - D (carbon-deuterium) bond is electronically much like a C - H bond, and it has a similar stiffness, measured by the spring constant, k. The deuterium atom has twice the mass of a hydrogen atom, however. (a) The infrared absorption frequency is proportional to Vk/m. Use this relationship to calculate the IR absorption fre­ quency of a typical C - D bond. ) (b) A chemist dissolves a sample in deuterochloroform (CDCI3 , then decides to take the IR spectrum and simply evap­ orates most of the CDCI3. What functional group will appear to be present in this IR spectrum as a result of the CDCI3 impurity? The mass spectrum of n-octane shows a prominent molecular ion peak (mlz 1 1 4). There is also a large peak at mlz 57, but it is not the base peak. The mass spectrum of 3,4-dimethylhexane shows a smaller molecular ion, and the peak at mass 57 is the base peak. Explain these trends in abundance of the molecular ions and the ions at mass 57, and predict the intensi­ ties of the peaks at masses 57 and 1 1 4 in the spectrum of 2,2,3,3-tetramethylbutane. An unknown, foul-smelling hydrocarbon gives the mass spectrum and infrared spectrum shown. (a) Use the mass spectrum to propose a molecular formula. How many elements of unsaturation are there? (b) Use the IR spectrum to determine the functional group(s), if any. (c) Propose one or more structures for this compound. What parts of the structure are uncertain? If you knew that hydro­ genation of the compound gives n-octane, would the structure still be uncertain? (d) Propose structures for the major fragments at 39, 67, 8 1 , and 95 in the mass spectrum. Explain why the base peak is so strong. 1 .00 80

1 1 / r 'I

/

KOH heat

•

)-C=C-
i 1 0

A signal around 83 to 04 suggests protons on a carbon bearing an electronegative element such as oxygen or a halogen. Protons that are more distant from the electronegative atom will be less strongly deshielded.

I

- O - C -H

I

I

I

Cl- C -H

Br-C -H

I

I

I

I - C -H

I

around 83-84 for hydrogens on carbons bearing oxygen or halogen

4.

Signals around 87 to 88 suggest the presence of an aromatic ring. If some of the aromatic absorptions are farther downfield than 87.2, an electron-withdrawing substituent may be attached.

�Y x

around

87 - 88

H

13- 1 1

Time Dependence of NMR Spectroscopy

595

5. Signals around 85 to 86 suggest vinyl protons. Splitting constants can differentiate cis and trans isomers. H

'" / / '" C=C C=C / / '" '" H H H around 85- 86; J 10 H z around 85 - 86; J 1 5 Hz 6. Learn to recognize ethyl groups and isopropyl groups (and structures that resemble these groups) by their characteristic splitting patterns. =

=

isopropyl group

ethyl group 7.

Signals around 82. 1 to 82.5 may suggest protons adjacent to a carbonyl group or next to an aromatic ring. A singlet at 82. 1 often results from a methyl group bonded to a carbonyl group. o

II I - C - C -B I

around 82. 1 - 82.5 8.

!( r

CB3

singlet, 82.3

singlet, 82.1

Signals in the range 89 to 8 1 0 suggest an aldehyde. o

II

C / "aldehyde,

9.

H 89 -810

A sharp singlet around 82.5 suggests a terminal alkyne. - C=C- H around 82.5

These hints are neither exact nor complete. They are simple methods for making edu­ cated guesses about the major features of a compound from its NMR spectrum. The hints can be used to draw partial structures to examine all the possible ways they might be com­ bined to give a molecule that corresponds with the spectrum. Figure 13-38 gives a graphic presentation of some of the most common chemical shifts. A more complete table of chem­ ical shifts appears in Appendix l . -COOH

8 1 I -8 1 2

�

Q(

II

o -C-H

10

9

X

'-

/'

I

C=C

H

I 8

I

X-C-H

I 7

/'

'-

I

I

6

5 8 (ppm)

x =0, hal

I

4

II

o

I I

-c-c

I D

I

3

I

2

o

.... Figure 13-38

Common chemical shifts in the I H NMR spectrum. ( Continued)

596

Chapter 1 3 : Nuclear Magnetic Resonance Spectroscopy

!" !

I

10

! j' i

"

"

i: :

9

8

7

5

6

3

4

2

o

i5 (ppm) .... Figure 13-39

Proton NMR spectrum for a compound of formula C4Hg02'

SAMPLE PROBLEM

Consider how you might approach the NMR spectrum shown in Figure 1 3-39, The molecu­ lar formula is known to be C4Hg02, implying one element of unsaturation (the saturated formula would be C4H 10 02)' Three types of protons appear in this spectrum. The signals at 04.1 and 8 1 .3 resemble an ethyl group-confirmed by the 2:3 ratio of the integrals. partial structure: -CH2-CH3 The ethyl group is probably bonded to an electronegative element, since its methylene (-CH2- ) protons absorb close to 04. The molecular formula contains oxygen, so an ethoxy group is suggested. partial structure: - 0 - CH2 -CH3 The singlet at 82. 1 ( area 3) might be a methyl group bonded to a carbonyl group. A carbonyl group would also account for the element of unsaturation. =

°

II

partial structure: -C-CH3 We have accounted for all eight hydrogen atoms in the spectrum. Putting together all the clues, we arrive at a proposed structure. °

II

CH�-CH�-O -C-CH� !ethyl acetate

At this point, the structure should be rechecked to make sure it is consistent with the molec­ ular formula, the proton ratios given by the integrals, the chemical shifts of the signals, and the spin-spin splitting. In ethyl acetate, the Ha protons give a triplet (split by the adjacent CH2 group, J 7 Hz) of area 3 at 81.3 ; the Hb protons give a quartet (split by the adjacent CH3 group, J 7 Hz) of area 2 at 04. 1 ; and the He protons give a singlet of area 3 at 82. 1 . =

=

PROBLEM

13-22

Draw the expected NMR spectrum of methyl propionate, and point out how it differs from the spectrum of ethyl acetate. °

II

CH3-O-C-CH2-CH3 m ethyl propionate

597

1 3- 1 1 Time Dependence of NMR Spectroscopy SOLVED PROBLEM 13-4

Propose a structure for the compound of molecular formula C4 H 1 00 whose proton NMR spectrum follows. OHz

50Hz

I""I

",ii,' :'

I! i! ,ll; i I

: 1

;:, ; "

,I

111111111111 1.08 0.98

i'.I.II'1 10

i·i, :'1' I: i i "I , , , " " , , 1 , , ' 9

8

7

4

5

6

"

'"

'

3

2

1

1

'

0

8 (ppm)

SOLUTION

The molecular formula C4 H 1 00 indicates no elements of unsaturation. Four types of hydrogens appear in this spectrum, in the ratio 2: 1 : 1:6. The singlet (one proton) at 82.4 might be a hydroxyl group, and the signal (two protons) at 83.4 corresponds to protons on a carbon atom bonded to oxygen. The 83.4 signal is a doublet, implying that the adjacent carbon atom bears one hydrogen. partial slruCllIre:

H

I

H-O-C H,-C-

-

I

(Since we cannot be certain that the 82.4 absorption is actually a hydroxyl group, we rnight consider shaking the sample with D 20. If the 2.4 ppm absorption represents a hydroxyl group, it will shrink or vanish after shaking with D 20.) The signals at 8 1 .8 and 80.9 resemble the pattern for an isopropyl group. The integral ratio of 1 :6 supports this assumption. Since the methine (-C H-) proton of the isopropyl group absorbs at a fairly high field, the isopropyl group must be bonded to a carbon

I

atom rather than an oxygen.

partial structure:

Our two partial structures add to a total of six carbon atoms (compared with the four in the molecular formula) because two of the carbon atoms appear in both partial structures. Drawing the composite of the partial structures, we have isobutyl alcohol: /

3

CWI

HC H"- O- C Hb2 -C"

C H� This structure must be rechecked to make sure that it has the con'ect molecular formula and that it accounts for all the structural evidence provided by the spectrum (Problem 1 3-23). PROBLEM 1 3-23

Give the spectral assignments for the protons in isobutyl alcohol (Solved Problem 13.4). For example, Ha is a singlet, area

=

1,

at 82.4

Chapter 13: Nuclear M agnetic Resonance Spectroscopy

598

PROBLEM 13- 2 4

Five proton NMR spectra are given here, together with molecular formulas. I n each case, propose a structure that is consistent with the spectrum.

!Offseti

om,:

i �! I

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f! f! I.:

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7

8

9

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, : : • I :: : , .•

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1 3-12 Cat'bon-1 3 NMR Spectroscopy

!

;

T

, ,

;"' . j

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to

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'

i 8

'I'

I , l' fT! i I;

-

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:

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.

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,

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4

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2

1

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o (ppm)

10

9

8

7

6

5

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o (ppm)

Where does a carbonyl group absorb in the NMR? Where does an internal alkyne absorb? In the proton NMR, both of these groups are invisible, Sometimes we can infer their presence: If the carbonyl group has a proton attached (an aldehyde proton), the peak between 89 and 10 alerts us to its presence, If the adjacent carbon atom has hydrogens, their signals between 82. 1 and 2.5 are suggestive, but we still can't see the carbonyl group. An internal alkyne is even more diffic ult, because there are no distinc­ tive absorptions in the proton NMR and usually none in the IR either. The development of Fourier transform NMR spectroscopy made carbon NMR (!3C NMR or CMR) possible, and high-field superconducting spectrometers allowed it to become nearly as convenient as proton NMR eH NMR). Carbon NMR deter­ mines the magnetic environments of the carbon atoms themselves. Carbonyl carbon atoms, alkyne carbon atoms, and aromatic carbon atoms all have chat'acteristic chemi­ cal shifts in the I3C NMR spectrum.

ii i f b 13-12A Sens t v ty o Car on NMR

Cat'bon NMR took longer than proton NMR to become a routine technique because carbon NMR signals are much weaker than froton signals. About 99% of the carbon atoms in a natural sample are the isotope l-c. This isotope has an even number of

1 3- 1 2

Carbon-13 NMR Spectroscopy

600

Chapter 13: Nuclear M agnetic Resonance Spectroscopy

protons and an even number of neutrons, so it has no magnetic spin and cannot give rise to NMR signals. The less abundant isotope 1 3C has an odd number of neutrons, giving it a magnetic spin of �, just like a proton. Because only 1 % of the carbon atoms in a sample are the magnetic 1 3C isotope, the sensitivity of I3C NMR is decreased by a factor of 100. In addition, the gyromagnetic ratio of l 3C is only one-fourth that of the proton, so the I3C resonance frequency (at a given magnetic field) is only one-fourth of that for I H NMR . The smaller gyromagnetic ratio leads to a further decrease in sensitivity. Because I 3C NMR is less sensitive than lH NMR, special techniques are needed to obtain a spectrum. If we simply operate the spectrometer in a normal (called continuous wave or CW) manner, the desired signals are very weak and become lost in the noise. When many spectra are averaged, however, the random noise tends to cancel while the desired signals are reinforced. If several spectra are taken and stored in a computer, they can be averaged and the accumulated spectrum plotted by the computer. Since the 1 3C NMR technique is much less sensitive than the 1 H NMR technique, hundreds of spectra are commonly averaged to produce a usable result. Several minutes are required to scan each CW spectrum, and this averaging procedure is long and tedious. Fortunately, there is a better way.

i f 13-1 2B Four er Trans orm NMR Spectroscopy

When magnetic nuclei are placed in a magnetic field and irradiated with a pulse of radio frequency close to their resonant frequency, the nuclei absorb some of the energy and precess like little tops at their resonant frequencies (Figure 1 3-40). This precession of many nuclei at slightly different frequencies produces a com­ plex signal that decays as the nuclei lose the energy they gained from the pulse. This signal is called a free induction decay (or transient) and it contains all the information needed to calculate a spectrum . The free induction decay (FID) can be recorded by a radio receiver and a computer in 1 to 2 seconds, and many FIDs can be averaged in a few minutes. A computer converts the averaged transients into a spectrum. A Fourier transform is the mathematical technique used to compute the spec­ trum from the free induction decay, and this technique of using pulses and collect­ ing transients is called Fourier transform spectroscopy. A Fourier transform spectrometer is usually more expensive than a continuous wave spectrometer, since it must have fairly sophisticated electronics capable of generating precise pulses and accurately receiving the complicated transients. A good I3C NMR instrument usually has the capability to do I H NMR spectra as well. When used with proton spectroscopy, the Fourier transform technique produces good spectra with very small amounts (less than a milligram) of sample.

v

V V --­ /,

C)

FT

0

---

sp ectrum

o

pulse

.... Figure 13-40

nuclei precess

free induction decay ( FID)

Fourier transform NMR spectroscopy. The FT NMR spectrometer delivers a radio-frequency pulse close to the resonance frequency of the nuclei. Each nucleus precesses at its own resonance frequency, generating a free induction decay (FID). Many of these transient Fills are accumulated and averaged in a short period of time. A computer does a Fourier transform (FT) on the averaged FID, producing the spectrum printed on the recorder.

1 3- 1 2 C arbon- 1 3 NMR Spectroscopy

h i l hif b 13-12C Car on C em ca S ts

601

Figure 1 3-4 1 gives typical ranges of chemical shifts for carbon atoms in organic mol­ ecules. A more detailed table of carbon chemical shifts is provided as Appendix 1 C. Carbon chemical shifts are usually about 1 5 to 20 times larger than comparable proton chemical shifts, which makes sense because the carbon atom is one atom closer to a shielding or deshielding group than its attached hydrogen. For example, an aldehyde proton absorbs around 89 .4 in the I H NMR spectrum, and the carbonyl carbon atom absorbs around 1 80 ppm downfield from TMS in the l 3C spectrum. Figure 1 3-42 com­ pares the proton and carbon spectra of a complex aldehyde to show this relationship between proton and carbon chemical shifts.

I I I

�

C=o

aromatic C

�

"---_ --

I -B I / � / / -{-o- / � C - Cl, C - Br

C=C

50

1 00

1 50

200

o TMS

ppm down field from

.... Figure 13-41

Table of approximate chemical shift values for 1 3C NMR. Most of these values for a carbon atom are about 1 5 to 20 times the chemical shift of a proton if it were bonded to the carbon atom.

TMS

The proton (lower) and carbon (upper) spectra in Figure 1 3-42 are calibrated so the full width of the proton spectrum is 1 0 ppm, while the width of the 13C spectrum is 200 ppm (20 times as large). Notice how the corresponding peaks in the two spectra almost line up vertically. This proportionality between l3C NMR and I H NMR chemical shifts is an approximation that allows us to make a first estimate of a carbon atom's chemical shift. For example, the peak for the aldehyde proton is at 89.5 in the proton spectrum, so we expect the peak for the aldehyde carbon to appear at a chemical shift between 1 5 and 20 times as large (between 8 1 44 and 8 1 92) in the carbon spectrum. The actual position is at 8 1 80.

(

,

I3C NMR

I

����

1 60

1 80

200

al deh

- .�

1 20

1 40

1 00

80

I

' H NMR

T i

!"

II ,

l'

,

.

7

8

9

1 3C NMR

s p ectru m ,

' -, , ' ,

1 10

o

20

40

60

r:�=�?l j�

arbon

17

I: ' :

-

5

6

4

:

i

:

3

2

"I o

o (ppm)

... Figure 13-42

I H NMR (lower) and 1 3C NMR (upper) spectra of a heterocyclic aldehyde. Notice the correlation between the chemical shifts in the two spectra. The proton spectrum has a sweep width of 1 0 ppm, and the carbon spectrum has a width of 200 ppm. :-::===

-

-=.--=-

-

---

.

--- --- - --

602

Chapter 1 3: Nuclear Magnetic Resonance S p ectro scopy

NMR spectroscopy is an important method for determ i n i ng the three­ dimensional structures of proteins i n solution. This technique com­ bines specialized experiments that use both

13e and 'H NMR spec­

troscopy.

Notice also the triplet at 877 in the l 3C NMR spectrum in Figure 1 3-42. This is the carbon signal for deuterated chloroform ( CDCI 3 ) , split into three equal-sized peaks by coupling with the deuterium atom. Chloroform-d ( CDCI3 ) is a common sol­ vent for I3C NMR because the spectrometer can "lock" onto the signal from deuteri­ um at a different frequency from carbon. The CDCl3 solvent signal is a common feature of carbon NMR spectra. Because chemical shift effects are larger in l 3C NMR, an electron-withdrawing group has a substantial effect on the chemical shift of a carbon atom beta (one carbon removed) to the group. For example, Figure 1 3-43 shows the ' H NMR and l 3C NMR spectra of 1 ,2,2-trichloropropane. The methyl ( CH3 ) carbon absorbs at 33 ppm down­ field from TMS because the two chlorine atoms on the adjacent - CCl2- carbon have a substantial effect on the methyl carbon. The chemical shift of this methyl carbon is about 1 5 times that of its attached protons ( 82. 1 ) , in accordance with our prediction. Similarly, the chemical shift of the - CH 2 C l carbon (56 ppm) is about 1 5 times that of its protons ( 84.0 ) . Although the CCl2 carbon has no protons, the proton in a - CHCI2 group generally absorbs around 85 .8. The carbon absorption at 87 ppm is about 1 5 times this proton shift.

iff h i b d 1 3-120 Important D erences Between Proton an Car on Tec n ques

M ost of the characteristics of I3C NMR spectroscopy are similar to those of the ' H NMR technique. There are some important differences, however. Operating Frequency The gyromagnetic ratio for l3C is about one-fourth that of the proton, so the resonance frequency is also about one-fourth. A spectrometer with a 70,459-gauss magnet needs a 300-MHz transmitter for protons and a 75.6-MHz trans­ mitter for l3c . A spectrometer with a 1 4,092-gauss magnet needs a 60-MHz transmit­ 3 ter for k protons and a 1 5 . 1 -M Hz transmitter for l c. Pea Areas The areas of l3C NMR peaks are not necessarily proportional to the

number of carbons giving rise to the peaks. Carbon atoms with two or three protons attached usually give the strongest absorptions, and carbons with no protons tend to give weak absorptions. Newer spectrometers have an integrating mode that uses gated ---- ----

�

---.

--

200

1 80

1 60

1 40

1 20

1 00

80

60

40

20

I

'

I

I

I

I

I

I

I

I

I

11

\

-�-

--

I

I

I

L.

Itt,

-

-

o

TMS

/

r--

I

Cl

T tiS

�

Figure 13-43

Proton and J 3 C NMR spectra of 1 ,2,2-trichloropropane.

1 0.0

9.0

8.0

7.0

1

,I

3.0

2.0

" I ,I 1 .0

o

13- 1 2

Carbon- I 3 NMR Spectroscopy

decoupling to equalize the absorptions of different carbon atoms. This mode makes peak integrals nearly proportional to the relative numbers of carbon atoms.

603

i li i 13-12E Sp n-Spin Sp tt ng

I3C NMR splitting patterns are quite different from those observed in 1 H NMR. Only 1 % of the carbon atoms in the I 3C NMR sample are magnetic, so there is only a small probability that an observed I3C nucleus is adjacent to another I 3C nucleus. Therefore, carbon-carbon splitting can be ignored. Carbon-hydrogen coupling is common, however. Most carbon atoms are bonded directly to hydrogen atoms or are sufficiently close to hydrogen atoms for carbon-hydrogen spin-spin coupling to be observed. Extensive carbon-hydrogen coupling produces splitting patterns that can be compli­ cated and difficult to interpret. l Proton Spin Oecoup ing To simplify l 3C NMR spectra, they are commonly recorded using proton spin decoupling, where the protons are continuously irradiat­ ed with a broadband ("noise") proton transmitter. As a result, all the protons are con­ tinuously in resonance, and they rapidly flip their spins. The carbon nuclei see an average of the possible combinations of proton spin states. Each carbon signal appears as a single, unsplit peak because any carbon-hydrogen splitting has been eliminated. The spectra in Figures 1 3-42 and 1 3-43 were generated in this manner.

PROBLEM

13-25

Draw the expected broadband-decoupled I 3 C NM R spectra o f the following compounds. Use Figure 13-4 1 to estimate the chemical shifts. (a)

O

(b)

o

(d)

0 H '" H

/

o

/

C=C'"

II

C-H H

ff l O -Resonance Decoup ing

Proton spin decoupling produces spectra that are very simple, but some valuable information is lost in the process. Off-resonance decoupling simplifies the spectrum but allows some of the splitting information to be retained (Figure 1 3-44). With off-resonance decoupling, the 13C nuclei are split only

.... Figure 13-44

200

1 80

1 60

140

1 20

1 00 8 (ppm)

80

60

40

Off-resonance-decoupled 13C NMR spectrum of 1 ,2,2-trichloropropane. The CCl2 group appears as a singlet, the CH2CI group as a triplet, and the CH 3 group as a quartet. Compare this spectrum with Figure 13-43.

604

Chapter 1 3: Nuclear Magnetic Resonance Spectroscopy

J 3C NMR

TMS

2-butanone

TMS

240

220

200

1 80

1 60

1 40

1 20

1 00

8 (ppm)

.... F i g u re 1 3-45

Off-resonance-decoupled (upper) and broadband-decoupled (lower)

3 1 C

NMR spectra of 2-butanone.

by the protons directly bonded to them. The N + 1 rule applies, so a carbon atom with one proton (a methine) appears as a doublet, a carbon with two attached protons (a methylene) gives a triplet, and a methyl carbon is split into a quartet. Off­ resonance-decoupled spectra are easily recognized by the appearance of TMS as a quartet at 0 ppm, split by the three protons of each methyl group. The best procedure for obtaining a l3C NMR spectrum is to run the spectrum twice: The singlets in the broadband-decoupled spectrum indicate the number of non­ equivalent types of carbon atoms and their chemical shifts. The multiplicities of the signals in the off-resonance-decoupled spectrum indicate the number of hydrogen atoms bonded to each carbon atom. 1 3C spectra are often given with two traces, one broadband decoupled and the other off-resonance decoupled. If j ust one trace is given, it is usually broadband decoupled. Figure 1 3-45 shows both spectra for 2-butanone.

PROBLEM

13-2 6

PROBLEM

13-2 7

(a) Show which carbon atoms correspond with which peaks in the

3 1 C

NMR spectrum of 2-butanone (Figure 1 3-45). (b) Draw the proton NMR spectrum you would expect for 2-butanone. How well do the proton chemical shifts predict the carbon chemical shifts, using the " 1 5 to 20 times as large" rule of thumb?

Repeat Problem 13-25, sketching the off-resonance--decoupJed

1 3C

spectra of the compounds.

1 3-12F DEPT 1 3e NMR

DEPT (Distortionless Enhanced Polarization Transfer) is a more recent technique

that provides the same information as off-resonance decoupling. DEPT is easier to run on modern, computer-controlled Fourier-transform spectrometers. DEPT gives better sensitivity, and it avoids overlapping multiplets because all the peaks remain decou­ pled singlets.

1 3- 1 2 Carbon- 1 3 NMR Spectroscopy

Each l 3 C nucleus is magnetically coupled to the protons bonded to it. Under the right circumstances, this magnetic coupling allows the transfer of polarization from the protons to the carbon nucleus. The number of protons bonded to the l 3 C nucleus deter­ mines how this polarization transfer occurs. A DEPT experiment usually includes three spectral scans: 1. The nonnal decoupled scan, in which each type of 1 3 C nucleus appears as a singlet. The DEPT-90 scan, in which only the CH (methine) carbons bonded to exactly one proton appear. 3. The DEPT- 1 35 scan, in which the CH 3 (methyl) groups and CH (methine) groups appear normally, and the CH2 groups give negative peaks. Carbons that are bonded to no protons do not appear. 2.

As shown graphically in Table 1 3-4, this information allows us to distinguish among carbons bonded to 0, 1 , 2, or 3 hydrogen atoms: Carbons with no H's appear only in the nOlmal spectrum, but not in either DEPT spectrum. Methine carbons (CH) give normal positive peaks in all three spectra. Methylene ( CH2 ) carbons give normal peaks in the nonnal spectrum, no peaks in the DEPT-90 spectrum, and negative peaks in the DEPT- 1 35 spectrum. Methy I ( CH 3 ) carbons give normal peaks in the normal spectrum, no peaks in the DEPT-90 spectrum, and normal peaks in the DEPT- 1 35 spectrum. Figure 1 3-46 shows the normal decoupled l 3 C NMR spectrum of but-3-en-2-one (a), plus the DEPT-90 spectrum (b), and the DEPT- 1 35 spectrum (c). Note that the car­ bonyl carbon (Cb, no protons) appears only in the regular spectrum. Cc, with 1 proton, appears nonnally in all the spectra. Cd, with two protons, appears as a negative peak in the DEPT- 1 35 spectrum. Ca, the methyl carbon with three protons, vanishes in the DEPT-90 spectrum but appears as a nonnal peak in the DEPT- 1 35 spectrum. •

•

•

•

TABLE 1 3-4

SUMMARY

S u m mary of DEPT Spectra

DEPT S pectra

quaternary

methine

Type oJ J3e

Protons

-C-

I

O

I

1

I

-C-H

I

methylene

- CH 2

I

2

methyl

- CH3

3

Normal J3e NMR

DEPT-90

DEPT-135

� �~� ~ T � ~

605

Chapter 1 3: Nuclear Magnetic Resonance Spectroscopy

606

I3

C NMR

'

,

a H d .:;:::-C "b /CH 3 H 2C C C

,

I ($J �

II

,

,

,

I

($J CH 3

CI-I

S (a)

(c)

,

CH2

0

(b)

I

L

'----

DEPT-90

DEPT- 1 35

200

I

I

I

I

1 80

1 60

1 40

1 20

I

I

I

I

I

I

1 00

80

60

40

20

o

.... F i g u re 13-46

1 3 C NMR spectrum and DEPT spectra of but-3-en-2-one.

PROBLEM 13-2 8

The standard 1 3 C NMR spectrum of phenyl propanoate is shown here. Predict the appearance of the DEPT-90 and DEPT- 1 35 spectra.

.

' !

.

: .

i : ,

200

80

60

1 40

1 20

00

80

60

40

20

.�

0

1 3- 1 3 Interpreting Carbon NMR Spectra

Interpreting 13C N M R spectra uses the same principles as interpreting I H N M R spec­ tra. In fact, carbon spectra are often easier to interpret. The I 3C N M R spectrum pro­ vides the following information: 1. The number of different signals implies how many different types of carbons are

607

1 3- 1 3

Interpreting Carbon NMR Spectra

present. The chemical shifts of those signals suggest what types of functional groups con­ tain those carbon atoms. 3. The peak areas (in the integrating mode) imply how many carbons of each type are present. 4. The splitting of signals in the off-resonance-decoupled spectrum or the DEPT-90 and DEPT- 1 35 spectra indicate how many protons are bonded to each carbon atom. 2.

For example, in the I 3C N M R spectrum of o-valerolactone (Figure 1 3-47), the CH2 groups in the upper (off-resonance-decoupled) spectrum are split into triplets, but they appear as singlets in the lower (broadband-decoupled) spectrum. Let's consider how we might solve this structure, given only the 13C N M R spec­ trum and the molecular formula. As we have seen in Figures 1 3-4 1 and 1 3-42, the sig­ nal at 173 ppm is appropriate for a carbonyl carbon. The off-resonance-decoupled spectrum shows a singlet at 1 73 ppm, implying that no hydrogens are bonded to the carbonyl carbon. The chemical shift of the next absorption is about 70 ppm. This is about 20 times the chemical shift of a proton on a carbon bonded to an electronegative element. The molecular formula implies that the electronegative element must be oxygen. Since the signal at 70 ppm is a triplet in the off-resonance-decoupled spectrum, this carbon must be a methylene ( - CH 2 - ) group. o

I I I3 C N M R

I

I

I

I

I

I

I

- CH2- O -

partial structures:

I

I

I

I

I

I

c= o

/ \

1 73

ppm

� 83 - 04

H

I (I

- c- o 70 ppm

� 82. 1 H

-C-

I

�

0 I II 30 ppm -C -C� I I

I

I

I

I

C SH S0 2

1

aO

lui

C S H S0 2

.1111

aO

I I I I I I I I I I 200 1 90 1 80 1 70 1 60 1 5 0 1 40 1 30 1 20 1 1 0 1 00

8

I 90

(ppm)

I 80

I 70

I 60

I 50

I 40

I 30

I 20

I 10

l

I 0

.... F i g u re 13-47

Off-resonance-decoupled and broadband-decoupled spectra of o-valerolactone, molecular formula CS H S 02·

608

Chapter 1 3 : Nuc l ear Magnetic Resonance Spectroscopy

The signal at 30 ppm corresponds to a carbon atom bonded to a carbonyl group. Remember that a proton on a carbon adjacent to a carbonyl group absorbs around 2 . 1 ppm, and we expect the carbon to have a chemical shift about 1 5 to 20 times as large. This carbon atom is a methylene group, as shown by the triplet in the off-resonance-decoupled spectrum. o

partial structures:

II -CH2 -C -

- CH2 - O -

The two signals at 1 9 and 22 ppm are from carbon atoms that are not directly bonded to any deshielding group, although the carbon at 22 ppm is probably closer to one of the oxygen atoms. These are also triplets in the off-resonance-decoupled spec­ trum and they correspond to methylene groups. We can propose:

19

3� / 1 73

""'C(O

22 --"

70 )

0

I partial structures: -C - CH2 - CH 2 - CH2 - 0 I

o

II - CH2 - C-

The molecular formula C5H802 implies the presence oftwo elements of unsatura­ tion. The carbonyl ( C 0 ) group accounts for one, but there are no more carbonyl groups and no double-bonded alkene carbon atoms. The other element of unsaturation must be a ring. Combining the partial structures into a ring gives the complete structure. In the following problems, only the broadband-decoupled spectra are provided. In cases where off-resonance-decoupled spectra are available, the off-resonance mul­ tiplicity of each peak is indicated: ( s ) singlet, ( d ) doublet, ( t ) triplet, and (q) quartet. =

=

=

=

=

PROB LEM

13-29

A bottle of allyl bromide was found to contain a large amount of an impurity. A careful dis­ tillation separated the i mpurity, which has the molecular formula C 3 H60. The following 13 C NMR spectrum of the impurity was obtained: I 1 3C N M R

I

I

I

I

I

I

I

I

I

(t)

C3 H 6 0

( t) (d)

200

I 1 80

I 1 60

I 1 40

I 1 20

I 1 00

1 80

1 60

1 40

,I 20

1 1 o

8 (ppm)

(a) Propose a structure for this impurity. 13 (b) Assign the peaks in the C NMR spectrum to the carbon atoms in the structure. (c) Suggest how this impurity arose in the allyl bromide sample.

PROBLEM 13-30

An inexperienced graduate student was making some 4-hydroxybutanoic acid. He obtained an excellent yield of a different compound, whose 1 3C NMR spectrum is shown here.

1 3- 1 4 Nuclear Magnetic Resonance I maging I

1 3C NMR

I

I

I

I

I

C4H602

I

I

(t)

I

609

I

(t)

(t)

(s)

200

I

1 80

I

1 60

j

I

1 40

I

1 20

I

(pp m)

1 00

[)

I

80

,I

60

I 20

,I

40

I o

(a) Propose a structure for this product. (b) Assign the peaks in the l3C NMR spectrum to the carbon atoms in the structure. PRO BLEM 13- 3 1

A laboratory student was converting cyclohexanol to cyclohexyl bromide by using one equiv­ alent of sodium bromide in a large excess of concentrated sulfuric acid. The major product she recovered was not cyclohexyl bromide, but a compound of formula C6H 0 that gave the 1 following 1 3 C NMR spectrum : I

I

I

I

I

I

I

I

I

I

1 3C NMR

(t)

C 6H I O

(t)

(d)

200

I

1 80

,I

1 60

I 1 40

I

1 20

1 I 1 00

[)

(ppm)

I 80

,I 60

I

40

I 20

I

o

(a) Propose a structure for this product. (b) Assign the peaks in the 1 3 C NMR spectrum to the carbon atoms in the structure.

(c) Suggest modifications in the reaction to obtain a better yield of cyclohexyl bromide.

When chemists use NMR spectroscopy, they take great pains to get the most uniform magnetic field possible (often homogeneous to within one part per billion) . They place small tubes of homogeneous solutions in the magnetic field and spin the tubes to aver­ age out any remaining variations in the magnetic field. Their goal is to have the sam­ ple behave as if it were all at a single point in the m ag netic field, with every molecule subjected to exactly the same external magnetic field. Nuclear magnetic resonance imaging uses the same physical effect, but its goals are almost the opposite of chemical NMR. In NMR imaging, a heterogeneous sample (commonly a living human body) is placed in the magnetic field of a large-bore superconducting magnet. The magnetic field is purposely nonuniform, with a gradient that allows just the protons in one plane of the sample to be in resonance at any one time. By using a combination of field gradients and sophisticated Fourier transform

1 3- 1 4

Nuclear Magnetic Resonance Imaging

610

Chapter 1 3: N uc1ear Magnetic Resonance Spectroscopy

� Figure 13·48

(a) MRl scan of a human brain showing a metastatic tumor in one hemisphere. (b) MRI image of the pelvic region showing severe damage in an arthritic hip.

The 31p signa l of phosphates in cells and tissues can be followed by

31p NMR

spectroscopy.

This

technique has been used to study the effects of exercise and oxygen starvat ion on the meta bolism of phosphate esters such as ATP.

techniques, the instrument can look selectively at one point within the sample, or a line within the sample, or a plane within the sample. The computer generates an image of a two-dimensional slice through the sample. A succession of slices can be accumulated in the computer to give a three-dimensional plot of the proton resonances within the bulk of the sample. Medical NMR imaging is commonly called magnetic resonance imaging (MRI) to avoid the common fear of the word nuclear and the misconception that "nuclear" means "radioactive." There is nothing radioactive about an NMR spectrometer. In fact, MRI is the least invasive, least hazardous method available for imaging the interior of the body. The only common side effect is claustrophobia from being confined within the ring of the wide-bore magnet. The MRI image can easily distinguish watery tissues, fatty tissues, bone, air spaces, blood, etc. by their differences in composition and movement. By using proton relaxation times, the technique becomes even more useful. In a strong magnetic field, slightly more proton spins are aligned with the field (the lower-energy state) than against it. A radio-frequency pulse of j ust the right duration inverts some spins, in­ creasing the number of spins oriented against the magnetic field. The spins gradually relax to their normal state over a period of a few seconds. By following the free-in­ duction decay, the spectrometer measures how quickly spin relaxation occurs in each pixel of the sample. Differing relaxation times are coded by color or intensity in the image, giving valuable information about the tissues involved. For example, cancerous tissues tend to have longer relaxation times than the corresponding normal tissues, so tumors are readily apparent in the NMR image. Figure 13-48 shows two actual MRI images: The first image is a slice through a patient's head showing a brain tumor. The second image is a slice through another patient's pelvic region showing an arthritic hip.

P R O B L E M - S O LV I N G STRAT E G Y

Spectroscopy Pro b l em s

We have now learned to use both IR and NMR spectroscopy, as well as mass spectrometry, to determine the structures of unknown organic compounds. These techniques usually pro­ vide a unique structure with little chance of error. A major part of successful spectral inter­ pretation is using an effective strategy rather than simply looking at the spectra, hoping that something obvious will jump out. A systematic approach should take into account the strengths and weaknesses of each technique. The following table summarizes the informa. tion provided by each spectroscopic technique.

1 3 - 1 4 Nuclear Magnetic Resonance Imaging Summary of the Information Provided by Each Type of Spectroscopy

molecular weight molecular formula heteroatoms functional groups alkyl substituents

�

MS

.j (HRMS) J

S S

IR

NMR

H

S H

�

.j

NOles: ,I, usually provides this information. H, usually provides helpful i nformation. S, sometimes provides helpful information.

We can summarize how you might go about identifying an unknown compound, but the actual process depends on what you already know about the chemistry of the compound and what you learn from each spectrum. Always go through the process with scratch paper and a pencil so you can keep track of mass numbers, formulas, possible functional groups, and carbon skeletons. Look for a molecular ion, and determine a tentative molecular weight. Remember that some compounds (alcohols, for example) may fail to give a visible mo­ lecular ion. If the molecular weight is odd, consider a nitrogen atom. If an HRMS is available, compare the "exact" mass with the tables to find a molecular formula with a mass close to the experimental value. Look for anything unusual or characteristic about the mass spectrum: Does the M + 2 peak of the parent ion look larger than the M + 1 peak? It might contain S, Cl, or Br. Is there a large gap and a peak at 1 27 characteristic of iodine? Although you might look at the MS fragmentation patterns to help determine the structure, this is more time-consuming than going on to other spectra. You can verify the fragmentation patterns more easily once you have a proposed structure. 2. Infrared spectrum. Look for 0- H, N - H, or = C - H peaks in the 3300 cm- I region. Are there saturated C - H peaks to the right of 3000 cm- I ? Unsaturated = C - H peaks to the left of 3000 cm - I ? Also look for C = C or C = N stretch around 2200 cm- I , and for C = O, C = C, or C = N stretch between 1 600 and 1 800 cm- I . The 1. Mass spectrum.

exact position of the peak, plus other characteristics (intensity, broadening), should help to determine the functional groups. For example, a broad 0 -H band centered over the C - H stretch at 3000 em -] might imply a carboxylic acid group, - COOH. The combination of IR and an odd molecular ion in the mass spectrum should con­ fil"m amines, amides, and nitriles. A strong alcohol - OH absorption in the IR might suggest that the apparent molecular ion in the mass spectrum could be low by 1 8 units from loss of water.

First look for strongly deshielded protons, such as carboxylic acids ( 8 1 0 to 8 12), aldehydes (89 to 8 1 0), and aromatic protons (87 to 88). Moderately deshielded peaks might be vinyl protons (85 to 86) or protons on a carbon bonded to an electronegative atom such as oxygen or halogen (83 to 84). A peak around 82. 1 to 82.5 might be an acetylenic proton or a proton on a carbon next to a carbonyl group, a benzene ring, or a vinyl group. These possibilities should be checked to see which are consistent with the IR spec­ trum. Finally, the spin-spin splitting patterns should be analyzed to suggest the structures of the alkyl groups present.

3. Nuclear magnetic resonance spectrum.

Once you have considered all the spectra, there should be one or two tentative struc­ tures. Each structure should be checked to see whether it accounts for the major characteris­ tics of all the spectra. •

•

Are the molecular weight and formula of the tentative structure consistent with the appearance (or the absence) of the molecular ion in the mass spectrum? Are there peaks in the mass spectrum corresponding to the expected fragmentation products? Does the tentative structure explain each of the characteristic stretching frequencies in the infrared spectrum? Does it account for any shifting of frequencies from their usual positions? ( Continued)

611

Chapter 1 3: Nuclear Magnetic Resonance Spectroscopy

612

•

Does the tentative structure account for each proton (or carbon) in the NMR spec­ trum? Does it also account for the observed chemical shifts and spin-spin splitting patterns?

If the tentative structure successfully accounts for all these features of the spectra, you can be confident that it is correct.

PROBLEM ( PART I ALLY SOLVED ) 13-32

Sets of spectra are given for two compounds. For each set, ( 1 ) Look at each spectrum individually, and list the structural characteristics you can determine from that specu·um. (2) Look at the set of spectra as a group, and propose a tentative structure. (3) Verify that your proposed structure accounts for the major features of each spectrum. The solution for compound 1 is given after the problem, but go as far as you can before looking at the solution. 1 00

I COl�lpound I

80

g

I

-

II)

60

OJ "0 c:

.E OJ

I

Ib6

I

1 2 M+

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40

--

!

20 -

o 10

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20

,

30

1

,

40

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50

,

60

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70

,

,

80

90

,

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,

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120

,

1 30

,

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,

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,

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f �

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III 11 1 111 1 1 ,

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3500

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I ii !i

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1 800

1 600

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wavenumber (em - I )

II

l [

7

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1000

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2500

I

II I , ,

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9

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I

,[ 10

I

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I

lill i , II I

,I

wavelength (/Lm)

III lliI1 IIII

4

II

,

o 4000

,

3 .5

3

2 .5 1 00

5

a (ppm)

4

3

2

o

1 3- 1 4 Nuclear Magnetic Resonance Imaging

80

g '"

-

60

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40

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20

II-+-

I

1 00

7

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I

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70

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80

90

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1 10

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--- I ?- l�-

1 00

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1 30

1 40

1 50

1 60

wavelength (J.Lm) 3

2.5 I 0 0 imrrmm";r ,,ITT , r ,,;n,

80

60

40

20

�

I1 11

3.5 " I

'. II . Ii!

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1 80

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1 60

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III

i ii! i i l . ,',,/

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, 1N , , � , I,

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55 ,

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I ,I!! • I, I I I i, , Compollnd 2 IH+H+tIH " , :- I+H+ H1 ' + 1 H+ttt H lili i ll H ! 1IHH-I + I tHI-+ + +H++!-H-f-f -I-+-H --H+H--H-t-I-f--Hl -- IH ' I , 1 , "I 'II I I , ! ', , I-H+H I-I+- 1+tt+Hhll+tt+ +tt+,, ttl-H ,,,I,' H,-ti ,, I+H- H-H+H-l-i ;ti+ +i ' ++I++H-H+IH F +I +'' IIIH-f-H+ II'H+I-H+++++ + ++-I-I+-' I-++-+-+I-+++ +Ht-\-! ,ll! iii! 1 1 '1 ji i ! 1 l i l l , "I ,I I ,, : I, I , II" I , " I I ! I ! Ii I' I 1I 11 ,II 1 ,1 I ;ii i I I I ! ! Iii! iil II

II II

2500

1 40

1 800

2000

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1 600

1 20

1 00

1 200

1 400

wavenu m ber ( cm - I )

40

60

80

CDCI3

800

1 000

solvent ') "

I""I

I H NMR !

OHz

50Hz

600

o

20

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Compollnd 2 j

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[) (ppm)

SOLUTION TO COM POU N D I

The MS shows an odd molecular weight at 1 2 1 and a large even-numbered fragment at 1 06. These features may indicate the presence of a nitrogen atom. Infrared spectrum: The IR shows a sharp peak around 3400 em-I , possibly the N - H of an arnine or the = C - H of a terminal alkyne. Because the MS suggests a nitrogen atom, and there is no other evidence for an alkyne (no C - C stretch around 2200 em- I ), the 3400 cm-I absorption is probably an N - H bond. The unsaturated C -H absorptions above 3000 cm -I, combined with an aromatic C = C stretch around 1 600 em- I , indicate an aromatic ring_

Mass spectrum :

=

(Continued)

Chapter 13: Nuclear Magnetic Resonance Spectroscopy

614

NMR spectrum: The NMR

shows complex splitting in the aromatic region, probably from a benzene ring: The total integral of 5 suggests the ring is monosubstituted. Part of the aromatic absorption is shifted upfield of 87.2, suggesting that the substituent on the benzene ring is a pi electron-donating group Like an amine or an ether. An ethyl group (total ar·ea 5) is seen at 8 1 .2 and 83 . 1 , appropriate for protons on a carbon atom bonded to nitrogen. A broad singlet of area I appears at 83.5, probably resulting from the N - H seen in the IR spectrum. Combining this information, we propose a nitrogen atom bonded to a hydrogen atom, a benzene ring, and an ethyl group. The total mo­ lecular weight for this structure would be 1 2 1 , in agreement with the molecular ion in the mass spectrum.

Proposed structure for compound J

H H-C

",

/ C=C " / '

H .

� C-C "-/'

H

/

,

.

2

C-N-CH -CH H

I

H

3

The proposed structure shows an ar·omatic ring with 5 protons, which explains the aromatic signals in the NMR and the C = C at 1 600 cm- I and the C - H above 3000 cm- I in the IR. The aromatic ring is bonded to an electron-donating - NHR group, which explains the odd molecular weight, the N - H absorption in the IR, and the aromatic signals shifted above 87.2 in the NMR. The ethyl group bonded to nitrogen explains the ethyl signals in the NMR, deshielded to 83 . 1 by the nitrogen atom. The base peak in the MS (M 15 1 06) is explained by the loss of a methyl group to give a resonance-stabilized cation: =

-

=

�

.. / Ph - N - C +

I

H

H

"

�

H

/ + Ph - N = C

I

H

'"

J

H H

+

· CH3

loss of

15

IIIlz 1 06

Cha pter 1 3 G lossa ry

accidentally equivalent nuclei Nuclei that are nonequivalent by NMR yet absorb at nearly the same chemical shift and ar·e not resolved. Nuclei that absorb at the same chemical shift can­ not split each other, whether they are chemically equivalent or accidentally equivalent. (p. 573) chemically equivalent atoms Atoms that cannot be distinguished chemically. The replace­ ment test for chemically equivalent atoms gives identical compounds. (p. 572) chemical shift The difference (in ppm) between the resonance frequency of the proton (or car­ bon nucleus) being observed and that of tetramethylsilane (TMS). Chemical shifts are usually given on the 8 (delta) scale, in parts per million downfield from TMS. (p. 565) complex splitting Splitting by two or more different kinds of protons with different coupling constants. (p. 585) coupling constant (J) The distance (in hertz) between two adjacent peaks of a multiplet. (p. 582) 13 DEPT (Distortionless Enhanced Polarization Transfer) A method of running several C spectra with different pulse sequences so that the carbon atoms appear differently depending on whether they are bonded to 0, 1 , 2, or 3 protons. (p. 604) deshielded Bonded to a group that withdraws part of the electron density from around the nu­ cleus. The absorptions of deshielded nuclei are moved downfield, resulting in larger chemical shifts. (p. 563) diastereotopic atoms Nuclei that occupy diastereomeric positions. The replacement test for diastereotopic atoms gives diastereomers. Diastereotopic nuclei can be distinguished by NMR, and they can split each other unless they are accidentally equivalent. (p. 589) downfield At a lower value of the applied magnetic field, toward the left (higher values of 8) on the NMR spectrum. The more deshielded a nucleus is, the farther downfield it absorbs. (p. 564) Fourier transform spectroscopy Spectroscopy that involves collecting transients (containing all the different resonance frequencies) and converting the averaged transients into a spectrum using the mathematical Fourier transform. (p. 600) transient (free induction decay, or FID): The signal that results when many nuclei are irradiated by a pulse of energy and precess at their resonance frequencies. gyromagnetic ratio (y) A measure of the magnetic properties of a nucleus. The resonance frequency (v) is given by the equation v = yBeff/27r, where Beff is the effective magnetic field at the nucleus. The gyromagnetic ratio of a proton is 26,753 sec- I gauss- I . The gyromagnetic ratio of a 13C nucleus is 6728 sec- I gauss - I . (p. 56 1 )

Chapter 1 3 Glossary induced magnetic field The magnetic field set up by the motion of electrons in a molecule (or in a wire) in response to the application of an external magnetic field. (p. 562) integration The measurement of the area under a peak, proportional to the number of protons giving rise to that peak. (p. 574) magnetically coupled Nuclei that are close enough that their magnetic fields influence each other, resulting in spin-spin splitting. (p. 576) magnetic moment The magnitude of a nuclear magnetic field, characterized by the gyromag­ netic ratio 1'. (p. 559) magnetic resonance imaging (MRI) The medical term for NMR imaging, avoiding the word nuclear. Use of field gradients in a large-bore magnet to scan two-dimensional slices of a patient's body. (p. 609) multiplet: A group of peaks resulting from the spin-spin splitting of the signal from a single type of nucleus. A doublet has two peaks, a triplet has three peaks, a quartet has four peaks, etc. (p. 578) N + 1 rule: A signal that is being split by N equivalent protons is split into a multiplet with N+ I individual peaks. (p. 578) nuclear magnetic resonance spectroscopy (NMR) A form of spectroscopy that measures the absorption of radio-frequency energy by nuclei in a magnetic field. The energy absorbed caus­ es nuclear spin transitions. (p. 559) carbon magnetic resonance ( 1 3 C NMR, CMR): NMR of the 1 3 C isotope of carbon. (p. 599) proton magnetic resonance ( I H NMR, PMR): NMR of protons. (p. 559) off-resonance decoupling A technique used with I 3C NMR in which only the protons directly bonded to a carbon atom cause spin-spin splitting. (p. 603) relaxation time A measure of how slowly the nuclear spins return to their normal state after an RF pulse near their resonance frequency. Alternatively, the evening after a chemistry exam. (p. 6 1 0) shielded Surrounded by electrons whose induced magnetic field opposes the externally applied magnetic field. The effective magnetic field at the shielded nucleus is less than the applied magnetic field. (p. 562) spin decoupling Elimination of spin-spin splitting by constantly irradiating one type of nuclei at its resonance frequency. (p. 603) spin-spin splitting (magnetic coupling) The interaction of the magnetic fields of two or more nuclei, usually through the bonds connecting them. Spin-spin splitting converts a single signal to a multiplet, a set of smaller peaks. (p. 576) TMS Tetramethylsilane, an NMR standard whose absorption is defined as 00.00 (p. 565) upfield At a higher value of the applied magnetic field, toward the right (lower values of 0) on the NMR spectrum. The more shielded a nucleus is, the farther upfield it absorbs. (p. 564)

I

Essential Problem..:Solvi ng Ski l ls i n Chapter 1 3

1.

Given a structure, determine which protons are equivalent and which are nonequivalent; predict the number of signals and their approximate chemical shifts.

2.

Given the chemical shifts of absorptions, suggest likely types of protons.

3.

Use the integral trace to determine the relative numbers of different types of protons.

4.

Predict which protons in a structure will be magnetically coupled, and predict the num­ ber of peaks and approximate coupling constants of their multiplets.

5.

Use proton spin-spin splitting patterns to determine the structure of alkyl and other groups.

6.

Draw the general features of the NMR spectrum of a given compound.

Predict the approximate chemical shifts of carbon atoms in a given compound. Given the chemical shifts of 1 3C absorptions, suggest likely types of carbons. I3 8. Use the off-resonance decoupled or DEPT C spectrum to determine the number of hydrogens bonded to a given carbon.

7.

9.

Combine the chemical shifts, integrals, and spin-spin splitting patterns in the NMR spectrum with information from infrared and mass spectra to determine the structures of organic compounds.

615

Chapter 1 3 : Nuclear M agnetic Resonance Spectroscopy

61 6

Study Problems

An unknown compound has the molecular formula C9H J J Br. Its proton NMR spectrum shows the following absorptions: singlet, 07. 1 , integral 4.4 cm singlet, 02.3, integral 1 3.0 cm singlet, 02.2, integral 6.7 cm Propose a structure for this compound. Predict the multiplicity (the number of peaks as a result of splitting) for each shaded proton in the following compounds.

13-33

13-34

(b) CH3 - C H-O H

(a) CH3 - CH 2 - CCl2 - CH 3

* H

(d) H

(c) CH3 - C H-C H 3

I

I

CH3

H

CH'

Q-

(e)

CH3

CH3

0

H H Predict the approximate chemical shifts of the protons in the following compounds. (a) benzene (b) cyclohexane (c) CH 3 -O - CH2 -CH2 -CH2Cl

1 3-35

0

II

(d) CH 3 - CH 2 - C = C - H

< )-- �

(f)

(e) CH3-CH2-C-CH3

( CH3 ) 2CH - 0 -CH2 -CH2 - OH

0

(g)

0

-H H

O
LI��iI'-" -

'L 1

11111111 1 .68

1

1111 1111 11 3.68

6

50Hz

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I

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3.58

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5

i5 (ppm )

4

3

2

o

Study Problems * 13-45

When 2-chloro-2-methylbutane is treated with a variety of strong bases, the products always seem to contain two isomers (A and B) of formula Cs H 1 0. When sodium hydroxide is used as the base, isomer A predominates. When potassium t-butoxide is used as the base, isomer B predominates. The ' H and I 3 C NMR spectra of A and B are given below. (a) Determine the structures of isomers A and B . (b) Explain why A is the major product when using sodium hydroxide as the base and why B is the major product when using potassium t-butoxide as the base. 200

1 80

140

1 60

1 20

1 00

' 3C NMR

� 11 80

C DC I 3

: :

60

OHz

OHz

50Hz

� � '" �

i somer A

.�

T

�

�

IIt

I I leT

10

9

8

7

6

200

1 80

1 60

140

1 20

ri-

I

-ll�l

:

v:

i

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IIIIItII

5.28

5.18

' H NMR

50Hz

jl

;1

20

,

I

I

i,

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�

=t=; -7-

1

1

'--

1

-

1

11 1111 1111 11

! : ! !

1 .78

1.58

1 .68

4

3

2

1 00

80

60

40

I

(

T

i

;

0

20

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C DC I3 .

,

,

isomer B

j

I

I

I

,

:

,

10

13-47

1 :

,

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9

8

7

6

5 o (ppm)

:

:

4

0

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,

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,-

H-

i

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f I T

o (ppm)

0

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5

' 3C NMR

,

40

IIIIII

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' H NMR

13-46

619

3

l

lJ -f , 2

:

:l

o

(A true story.) A major university was designated as a national nuclear magnetic resonance center by the National Science Foundation. Several large superconducting instruments were being installed when a government safety inspec­ tor appeared and demanded to know what provisions were being made to handle the nuclear waste produced by these instruments. Assume you are the manager of the NMR center, and offer an explanation that could be understood by a nonscientist. A compound was isolated as a minor constituent in an extract from garden cress. Its spectra are shown here. ( 1 ) Look at each spectrum individually, and list the structural characteristics you can determine from that spectrum. (2) Look at the set of spectra as a group, and propose a tentative structure. (3) Verify that your proposed structure accounts for the major features of each spectrum.

Chapter 13: Nuclear Magnetic Resonance Spectroscopy

620

1 00 80

g 60 oj -OCH3

CXOH

(0

"'"

OCH3

f li h 14-3C Nomenclature o Cyc c Et ers

Cyclic ethers are our first examples of heterocyclic compounds, containing a ring in which a ring atom is an element other than carbon. This atom, called the heteroatom, is numbered 1 in numbering the ring atoms. Heterocyclic ethers are an especially important and useful class of ethers.

id ( i ) Epox es Ox ranes

We have already encountered some of the chemistry of epox­ ides (Section 8- 1 2). Epoxides are three-membered cyclic ethers, usually fonned by peroxyacid oxidation of the corresponding aikenes. The common name of an epoxide is formed by adding "oxide" to the name of the alkene that is oxidized. The following reactions show the synthesis and common names of two simple epoxides. o

H2C =CH2 ethylene

+

II Ph - C -OOH

o

II Ph -C- OH

�

peroxybenzoic acid

ex:

ethylene oxide

benzoic acid

Ethylene oxide has been used as a fumigant for foods, textiles, and soil, and for sterilizing biomedical instruments.

p eroxybenzoic aCid )

probably due to its ability to alky­

H cyclohexene oxide

H

O H

4

3

�\

CH3

trans- l ,2-epoxy-4-methylcyclohexane

H

/ 2 0�3

1)

CH3

H

\:?H - CH -CH3 6

2

OCH3 cis-2,3-epoxy-4-methoxyhexane

Another systematic method names epoxides as derivatives of the parent compound, ethylene oxide, using "oxirane" as the systematic name for ethylene oxide. In this sys­ tem, the ring atoms of a heterocyclic compound are numbered starting with the het­ eroatom and going in the direction to give the lowest substituent numbers. 1

oxirane

diffuses

late critical cel lular enzymes.

One systematic method for naming epoxides is to name the rest of the molecule and use the term "epoxy" as a substituent, giving the numbers of the two carbon atoms bonded to the epoxide oxygen.

5

readily

ing them. Its antibacterial effect is

cyclohexene

'� '�· H

It

through materials without damag­

H /O � CH3 3( H CH3 0)2

trans-2-methoxy-3-methyloxirane

630

Chapter 1 4: Ethers, Epoxides, and Sulfides The least common cyclic ethers are the four-membered oxetanes. Because these four-membered rings are strained, they are more reactive than larger cyclic ethers and open-chain ethers. They are not as reactive as the highly strained oxi­ ranes (epoxides), however.

Oxetanes

CH3

f+1 �

2

CH2 CH3

CH3 H

2-ethyl-3,3-dimethyloxetane

oxetane

Furans (Oxolanes) The five-membered cyclic ethers are commonly named after an aromatic member of this group, furan. We consider the aromaticity of furan and other heterocycles in Chapter 16. The systematic term oxolane is also used for a five­ membered ring containing an oxygen atom.

4 HH )( X

H H H 3-methoxyfuran

furan

0_

H

tetrahydrofuran (THF) (oxolane)

The saturated five-membered cyclic ether resembles furan but has four additional hy­ drogen atoms. Therefore, it is called tetrahydrofuran (THF). One of the most polar ethers, tetrahydrofuran is an excellent nonhydroxylic organic solvent for polar reagents. Grignard reactions sometimes succeed in THF even when they fail in diethyl ether.

Pyrans (Oxanes) The six-membered cyclic ethers are commonly named as deriva­ tives of pyran, an unsaturated ether. The saturated compound has four more hydrogen atoms, so it is called tetrahydropyran (THP). The systematic term oxane is also used for a six-membered ring containing an oxygen atom. H H

i) H

H

I

I

0

H H

H H

U

H

H

H H

H

pyran

4-methylpyran

0

H H

tetrahydropyran (THP) (oxane)

Dioxanes Heterocyclic ethers with two oxygen atoms in a six-membered ring are called dioxanes. The most common form of dioxane is the one with the two oxygen atoms in a l ,4-relationship. l ,4-Dioxane is miscible with water, and it is widely used as a polar solvent for organic reactions.

:

()o :

1 ,4-dioxane

6

I O ? '

l� : O A CH3 3

4-methyl- l ,3-dioxane

ryo� �O �

dibenzo- l ,4-dioxane (dioxin)

Dioxin is a common name for dibenzodioxin, 1 ,4-dioxane fused with two ben­ zene rings. The name "dioxin" is often used incorrectly in the news media for 2,3,7,8tetrachlorodibenzodioxin (TCDD), a toxic contaminant in the synthesis of the herbicide called 2,4,5-T or Agent Orange. Surprisingly, TCDD has been in the environment for many millions of years because it is also formed in forest fires. Most dioxins are toxic

14-4 Spectroscopy of Ethers

and carcinogenic (cause cancer) because they associate with DNA and cause a mis­ reading of the genetic code.

Cl Cl

u 6

51 4

0",

J

�?

3

°

Cl

II CH2-C-OH

Cl

Cl

2,4,5-trichlorophenoxyacetic acid (2,4,5-T or Agent Orange)

yY yY 0

CI

�O� CI

2,3,7,8-tetrachlorodibenzodioxin (TCDD, incolTectly "dioxin")

PROBLEM 14-5 l A-Dioxane is made commercially by the acid-catalyzed dehydration of an alcohol. Show what alcohol will dehydrate to give lA-dioxane. (a) (b) Propose a mechanism for this reaction.

PROBLEM 14-6 Name the following heterocyclic ethers.

(a)

O

(bl

o

(l o

CI

Br

( l e

6 o

ocHCH 2 3

Infrared Spectroscopy of Ethers Infrared spectra do not show obvious or reli­ able absorptions for ethers. Most ethers give a moderate to strong C-O stretch around 1000 to 1200 cm-I (in the fi ngerprint region), but many compounds other than ethers give similar absorptions. Nevertheless, the IR spectrum can be useful because it shows the absence of carbonyl (C=O) groups and hydroxyl (0-H) groups. If the molecular formula contains an oxygen atom, the lack of carbonyl or hydroxyl absorp­ tions in the I R suggests an ether.

14-4

Spectroscopy of Ethers

Mass Spectrometry of Ethers

The most common fragmentation of ethers is cleavage next to one of the carbon atoms bonded to oxygen. Because this carbon is alpha to the oxygen atom, this fragmentation is called a cleavage. The resulting oxonium ion (oxygen with three bonds and a positive charge) is resonance-stabilized by the nonbonding electrons on oxygen. ll'

Cleavage

[R

fCH2-0-R,]t

R·

+

not observed oxonium ion

Another common cleavage is the loss of either of the two alkyl groups to give another oxonium ion or an alkyl cation. Loss of an alkyl group

+

·R'

not observed oxonium ion

631

632

Chapter 14: Ethers, Epoxides, and Sulfides

or

[R-CH2-0fR't

R-CH2-O'

+R'

+

alkyl cation

not observed

The mass spectrum of diethyl ether appears in Figure 14-5. The four most abun­ dant ions correspond to the molecular ion, loss of an ethyl group, a cleavage, and loss of an ethylene molecule combined with a cleavage. All these modes of cleavage form resonance-stabilized oxonium ions. PROBLEM 14-7 Propose a fragmentation to account for each numbered peak in the mass spectrum of n-butyl isopropyl ether. -.---,.--.---.--�--�--,- --,---,---,--,---,---,--� 100 ,-�� 43

II

::1,---+-+--+-!

' I I � i 2: 1---1--+---IIIt\I---+ II' IHI I---iIj

57

1-

40

73

'-r--

10

I

20

30

40

50

60

101

I--+--I---+

70

80

90

I

I

I

I

l�o�11 I

I!�:

100

110

I

120

130

140

150

160

mlz

100 80 Q) u 0: OJ '" 0: ::> .D '"

60 40

L

31

, , --+-, ---,

-_._.

201---l---

20

30

40

50

60

70

80

90

100

110

120

130

140

150

m/z Loss of an ethyl group

+

H - O=CH-CH3

+

111/z45 IX

Cleavage

+

CH3- CH2-O=CH2 +

� Figure 14-5 The mass spectrum of diethyl ether shows major peaks for the molecular ion, loss of an ethyl group, a cleavage, and (l' cleavage combined with loss of a molecule of ethylene.

m/z 59 (l'

·CH2CH3 loss of 29

Cleavage combined with loss of an ethylene molecule +

CH3-CH2-O=CH2 /I1/z 59

----.>

+

H-O=CH2 III/Z 31

+

CH2=CH2 loss of 28

·CH3 loss or 15

160

14-5 The Williamson Ether Synthesis In the l3C NMR spectrum, a carbon atom bond­ ed to oxygen generally absorbs between 865 and 890. Protons on carbon atoms bonded to oxygen usually absorb at chemical shifts between 83.5 and 84 in the I H NMR spectrum. Both alcohols and ethers have resonances in this range. See, for example, the NMR spectra of methyl t-butyl ether (page 574) and ethanol (page 592). If a compound containing C, H, and 0 has resonances in the correct range, and if there is no 0 H stretch or C 0 stretch in the IR spectrum, an ether is the most likely functional group. NMR Spectroscopy of Ethers

-

I �13C -C-OI

865-890

H�'H 83.5 - 84

=

We have already seen most of the common methods for synthesizing ethers. We re­ view them at this time, looking more closely at the mechanisms to see which methods are most suitable for preparing various kinds of ethers. The Williamson ether synthesis (Section 11-14) is the most reliable and versatile ether synthesis. This method involves the SN2 attack of an alkoxide ion on an unhindered primary alkyl halide or tosylate. Secondary alkyl halides and tosylates are occasionally used in the Williamson synthesis, but elimination competes, and the yields are often poor.

R -O-R'

The alkoxide is commonly made by adding Na, 11-14).

K,

+

..

:X:

or NaH to the alcohol (Section

Examples (I) Na

ethoxycyc!ohexane (92%)

cyclohexanol

(I) NaH

(2)

CH3-1

)

2-methoxy-3,3-dimethylpentane (90%)

3,3-dimethyl-2-pentanol

SOLVED PROBLEM 14-1 (a) Why is the following reaction a poor method for the synthesis of t-butyl propyl ether? (b) What would be the major product from this reaction? (c) Propose a better synthesis of t-butyl propyl ether.

H3

T CH3-C-Br I

CH3 sodium propoxide

H

I-butyl bromide

does not give

X )

CH3

I

CH3-C-O-CH?-CH-?CH3

I

CH3 I-butyl propyl ether

14-5

The Williamson Ether Synthesis

633

634

Chapter 14: Ethers, Epoxides, and Sulfides

SOLUTION (a) The desired SN2 reaction cannot occur on the tertiary alkyl halide. (b) The alkoxide ion is a strong base as well as a nuc1eophile, and elimination prevails.

�H

.... . �

CH3CH?CH?-O· - -

Na

hl

+

CH3

E2

--

H-C-C-CH3

I

H

sodium propoxide

�I r

isobutylene

I-butyl bromide

(c) A better synthesis would use the less hindered alkyl group as the SN2 substrate and the alkoxide of the more hindered alkyl group.

r

HJ

�7 +N a CH CH?-C-H ) - I �r

CH -c-o3

I

CH)

+

SN2 --

CH)

I

CHJ I-butyl propyl ether

l-bromopropane

sodium I-butoxide

I

CH)-C-O -CH?-CH?- CH)

PROBLEM 14-8 Propose a Williamson synthesis of 3 -butoxy- l , l -dimethylcyclohexane from 3 , 3-dimethyl­ cyc1ohexanol and butanol.

Synthesis of Phenyl Ethers A phenol (aromatic alcohol) can be used as the alkoxide fragment (but not the halide fragment) for the Williamson ether synthesi s . Phenols are more acidic than aliphatic alcohols (Section 1 0-6), and sodium hydroxide is sufficiently basic to form the phenoxide ion. As with other alkoxides, the elec­ trophile should have an unhindered primary alkyl group and a good leaving group.

(I) NaOH

2-butoxynitrobenzene

2-nitrophenol

(80%) PROBLEM-SOLVING

HiltZ;

To convert two alcohols to an ether,

convert the more hindered alcohol to its alkoxide. Convert the less hindered alcohol to its tosylate (or an alkyl halide). Make sure the tosylate (or halide) is a good SN2 substrate.

14-6

Synthesis of Ethers by Alkoxymercuration­ Demercuration

PROBLEM 14-9 Show how you would use the Williamson ether synthesis to prepare the following ethers. You may use any alcohols or phenols as your organic starting materials. (a) cyclohexyl propyl ether (b) isopropyl methyl ether (c) I-methoxyA-nitrobenzene (d) ethyl n-propyl ether (two ways)

(e) benzyl t-butyl ether ( benzyl

=

Ph - CH 2 - )

The alkoxymercuration-demercuration process adds a molecule of an alcohol across the double bond of an alkene (Section 8-6). The product is an ether, as shown here. Hg(OAc)2 ROH

)

I

-

C

-

I

I

C

-

I

AcOHg :q-R mercurial ether

I

-

I

C

C

H

OR

-

I

-

I

14-7 Industrial Synthesis: Bimolecular Dehydration of Alcohols

635

Example

(I) Hg(OAc)2' CH 30H ) (2) NaBH4

CH3( CH)3-CH-CH3

I

OCH3 2-methoxyhexane, 80% (Markovnikov product)

l-hexene

PROBLEM-SOLVING

PROBLEM 14- 10 Show how the following ethers rrught be synthesized using ( l ) alkoxymercuration-demercuration and (2) the Williamson synthesis. (When one of these methods cannot be used for the given ether, point out why it will not work.) (a) 2-methoxybutane (b) ethyl cyclohexyl ether (d) l-methyl-I-methoxycyclopentane (e) 2-methyl- l -methoxycyclopentane (f) t-butyl phenyl ether (e) I-methyl- l -isopropoxycyclopentane

group of the alcohol to the more substituted carbon atom of the C=C double bond.

The least expensive method for synthesizing simple symmetrical ethers is the acid-cat­ alyzed bimolecular dehydration, discussed in Section I I- lO B. Unimolecular dehydra­ tion (to give an alkene) competes with bimolecular dehydration. To form an ether, the alcohol must have an unhindered primary alkyl group, and the temperature must be kept low. If the alcohol is hindered or the temperature is too high, the delicate balance between substitution and elimination shifts in favor of elimination, and very little ether is formed. Bimolecular dehydration is used in industry to make symmetrical ethers from primary alcohols. Because the dehydration is so limited in its scope, it finds little use in the laboratory synthesis of ethers. Bimolecular dehydration

R-O-R

+

H20

2 CH30 H

CH3-O-CH3

+

H20

methyl alcohol

dimethyl ether (100%)

2 R-OH Examples

2 CH3CH20H

CH3CH2-O-CH2CH3

ethyl alcohol

diethyl ether (88%)

+

2 CH3CH2CHPH

CH3CH2CH2-0 -CH2CH2CH3

n-propyl alcohol

n-propyl ether

H20

+

Hp

(75%)

CH3-CH-CH3

H2C=CH-CH3

OH

unimolecular dehydration (no ether is formed)

I

isopropyl alcohol

+

Hp

If the conditions are carefully controlled, bimolecular dehydration is a cheap synthesis of diethyl ether. In fact, this is the industrial method used to produce millions of gal­ lons of diethyl ether each year.

HiltZ;

Alkoxymercuration adds the -OR

14-7

Industrial Synthesis: Bimolecular Dehydration of Alcohols

636

Chapter 14: Ethers, Epoxides, and Sulfides

PROBLEM-SOLVING

H?nP"

PROBLEM 14- 1 1 Explain why bimolecular dehydration is a poor method for making unsymmetrical ethers such as ethyl methyl ether.

Bimolecular dehydration of alcohols is generally a poor synthetic method.

PROBLEM 14- 12 Propose a mechanism for the acid-catalyzed dehydration of n-propyl alcohol to n-propyl ether, as shown above. When the temperature is allowed to rise too high, propene is formed. Propose a mechanism for the formation of propene, and explain why it is favored at higher temperatures. PROBLEM 14- 13 Which of the following ethers can be formed in good yield by dehydrating the correspon­ ding alcohols? For those that cannot be formed by dehydration, suggest an alternative method that will work. (a) dibutyl ether (b) ethyl n-propyl ether (e) di- sec-butyl ether

I.

SUMMARY

Syntheses of Ethers

1. The Williamson ether synthesis (Sections 11-14 and 14-5)

+

R-O: x

=

�

R'-X

+

R-O-R'

X-

R' must be primary

CI, Br, I, OTs, etc.

2. Addition of an alcohol across a double bond: alkoxymercuration-demercuration (Sections 8-6 and 14-6)

"

C=C

/

/

"

+

Hg(OAc)2

R-OH

1 1

1

1 1

-C-C-

1

-C-C-

)

1

AcOHg

H

OR

1

OR

Markovnikov orientation 3. Bimolecular dehydration of alcohols: industrial synthesis (Section l 1- lOB and 14-7)

2 R-OH

(

H+

)

R-O-R +

H20

R must be primary

14-8

Unlike alcohols, ethers are not commonly used as synthetic intermediates because they do not undergo many reactions. This unreactivity makes ethers attractive as sol­ vents. Even so, ethers do undergo a limited number of characteristic reactions. Ethers are cleaved by heating with HBr or H I to give alkyl bromides or alkyl iodides.

Cleavage of Ethers by HBr and HI

R-O-R'

excess

(X

=

HX

Br or I)

R-X

+

R'-X

Ethers are unreactive toward most bases, but they can react under acidic conditions. A pro­ tonated ether can undergo substitution or elimination with the expulsion of an alcohol. Ethers react with concentrated HBr and H I because these reagents are sufficiently acidic to protonate the ether, while bromide and iodide are good nUcleophiles for the substitution.

r:jj='

X..�

R-O-R' + H + - X ether

(X

=

Br or I)

( )

1

H

� R-O1+ v ..

-

H R'

protonated ether

�

X-R + :O-R' alkyl halide

alcohol

HX

----7

X-R +

X-R'

14-8

Cleavage of Ethers by HEr and HI

In effect, this reaction converts a dialkyl ether into two alkyl halides. The conditions are very strong, however, and the molecule must not contain any acid-sensitive func­ tional groups. Iodide and bromide ions are good nucleophiles but weak bases, so they are more likely to substitute by the SN2 mechanism than to eliminate by the E2 mechanism. Mechanism 1 4- 1 shows how bromide ion cleaves the protonated ether by displacing an alcohol. In most cases, the alcohol reacts further with HBr to give the alkyl bromide (see Section 1 1 -7). The reaction of cyclohexyl ethyl ether with HBr is an example of this displacement. The cyclohexanol produced by the cleavage reacts further with BBr, and the final products are ethyl bromide and bromocyclohexane.

MECHANISM 14-1

Cleavage of an Ether by HBr or HI

------,

Ethers are cleaved by a nucleophilic substitution of Br- or 1- on the protonated ether. Step

1:

Protonation of the ether to form a good leaving group.

R

R'

R

.. " .� H � 1?( 0: /. 0

R'

,,+ /

H

O.

/.

Step 2: SN2 cleavage of the protonated ether.

(R H .. C+O./ /. R'

:Br: .. +

-----7

./ :0 ." R' H

:�!-R

+

Step 3: Conversion of the alcohol fragment to the alkyl halide.

(Does not occur with phenols.)

R'-O-H �

R'-Br

+

H20

This conversion can occur by either of the two mechanisms shown in Section 11-7, depending on the structure of the alcohol and the reaction conditions. The protonated alcohol undergoes either SNI or SN2 substitu­ tion by bromide ion. EXAMPLE: Cleavage of cyclopentyl ethyl ether by HBr.

Step

1:

Proto nation of the ether to form a good leaving group,

U

.� .. HcJr= R-CH2-CH3

cyclopentyl ethyl ether

Step 2: Cleavage of the protonated ether

+ (Continued)

637

638

Chapter 14: Ethers,

Epoxides, and Sulfides Step 3: Conversion of the alcohol fragment to the alkyl halide.

First, the alcohol is protonated to form a good leaving group.

The protonated ether undergoes SNI or SN2 substitution by bromide ion. H 20 :

H I. rt-H

G

O �: H

: r :-

H

�

a

HBr

»

Hel

PROBLEM 14- 14 Propose a mechanism for the following reaction.

Q

excess HBr

)

Br

tetrahydrofuran

�Br

1,4-dibromobutane

Phenyl Ethers Phenyl ethers (one of the groups bonded to oxygen is a benzene ring) react with RBr or HI to give alkyl halides and phenols. Phenols do not react further to give halides because the sp2-hybridized carbon atom of the phenol cannot undergo the SN2 (or SN 1 ) reaction needed for conversion to the halide.

C) ,,:::::

ethyl phenyl ether

PROBLEM-SOLVING

HiltZ:-

HBr and HI convert both alkyl groups

(but not aromatic groups) of an ether

protonated ether

I�

phenol (no further reaction)

ethyl bromide

PROBLEM 14- 15 Predict the products of the following reactions. An excess of acid is available in each case. + HBr (b) tetrahydropyran + HI

(a) ethoxycyclohexane

to alkyl halides; phenols are unreactive, however.

..

'OH

'

(e) anisole (methoxybenzene) + HBr

r

(e) Ph -O-CH2CH2 - H -CH2-O-CH2CH) CH)

(d) +

COl �

HBr

0

+

HI

14-9 Aut oxi dation of Ethers PROBLEM 14-16 Boron tribrornide (BBr3) cleaves ethers to give alh)'l halides and alcohols. R-O-R' R-O-BBr2

+

+

BBr3

3 H20

�

�

R-O-BBr2

ROH

+

+

R'Br

B(OHh

+

2 HBr

The reaction is thought to involve attack by a bromide ion on the Lewis acid-base adduct of the ether with BBr3 (a strong Lewis acid). Propose a mechanism for the reaction of butyl methyl ether with BBr3 to give (after hydrolysis) butanol and bromomethane.

When ethers are stored in the presence of atmospheric oxygen, they slowly oxidize to produce hydroperoxides and dialkyl peroxides, both of which are explosive. Such a spontaneous oxidation by atmospheric oxygen is called an autoxidation. excess

R-O-CH 2 R'

OOH

02

I

(slow)

ether

+

R-O-CH-R'

14-9

Autoxidation of Ethers

R-O-O-CH 2-R'

hydroperoxide

dialkyl peroxide

Example

OOH

H3 C

excess 02

"

(weeks or months)

/

I

CH-O-C-CH o + 0

I

H 3C

CH 3

diisopropy\ ether

hydroperoxide

diisopropy\ peroxide

Organic chemists often buy large containers of ethers and use small quantities over several months. Once a container has been opened, it contains atmospheric oxygen, and the autoxidation process begins. After several months, a large amount of peroxide may be pres­ ent. Distillation or evaporation concentrates the peroxides, and an explosion may occur. Such an explosion may be avoided by taking a few simple precautions. Ethers should be bought in small quantities, kept in tightly sealed containers, and used promptly. Any procedure requiring evaporation or distillation should use only perox­ ide-free ether. Any ether that might be contaminated with peroxides should be discard­ ed or treated to destroy the peroxides.

t

SUMMARY

Reactions of Ethers

1. Cleavage by HBr and HI (Section 14-8)

excess

R-O-R'

(X

=

excess

Ar-O-R Ar

=

(X

=

HX

Br, I)

R-X

HX

Br,

+

Ar-OH

I)

R'-X +

R-X

aromatic ring

Example

excess

CH3CH2-O-CH3

HI ethyl iodide

ethyl methyl ether

methyl iodide

2. Autoxidation (Section 14-9)

OOH R-O-CH2-R' ether

excess O2

---->J

(slow)

I

R-O-CH-R' hydroperoxide

+

R-O-O-CH2-R' dialkyJ peroxide

639

640

Chapter

14:

Ethers, Epoxides, and Sulfides

14-10

Sulfides are also called thioethers because they are the sulfur analogues of ethers.

Like thiols, sulfides have strong characteristic odors: The odor of dimethyl sulfide is

Sulfides (Thioethers) reminiscent of oysters that have been kept in the refrigerator for too long. Sulfides are named like ethers, with "sulfide" replacing "ether" in the common names. In the IUPAC (alkoxy alkane) names, "alkylthio" replaces "alkoxy."

Q'

CH3-S-CH3 dimethyl sulfide

�

SCH3

methyl phenyl sulfide

2CH3

4-ethylthio-2-methyl-2-pentene

Sulfides are easily synthesized by the Williamson ether synthesis, using a thiolate ion as the nucleophile. CH3CH2-Sethanethiolate

� + CH3CH2CH2 � l-bromopropane

+

CH3CH2CH2-S-CH2CH3 ethyl propyl sulfide

Thiols are more acidic than water. Therefore, thiolate ions are easily generated by treating thiols with aqueous sodium hydroxide. + CH 3CH2-SH + Na -OH CH3CH2-S- �a + H20 �) pKa 10.5 sodium ethanethiolate pKa 15.7 Because sulfur is larger and more polarizable than oxygen, thiolate ions are even better nucleophjles than alkoxide ions. Thjolates are such effective nucleophiles that secondary alkyl halides often react to give good yields of SN2 products. =

=

Br

H

\!'

/ "-

CH3

H SCH3 \ --' C / "-

CH3

CH2CH3

(R)-2-bromobutane

CH2CH3

(S )-2-(methylthio )butane

PROBLEM 14-17 Show how you would synthesize butyl isopropyl sulfide using I-butanol, 2-propanol, and any solvents and reagents you need.

Sulfides are much more reactive than ethers. In a sulfide, sulfur's valence is not necessarily filled: Sulfur can form additional bonds with other atoms. Sulfur forms particularly strong bonds with oxygen, and sulfides are easily oxidized to sulfoxides and sulfones. Sulfoxides and sulfones are drawn using either hypervalent double­ bonded structures or formally charged single-bonded structures as shown here.

R-S-R' sulfide

sulfoxide

'0' II R-S-R' sulfoxide

[

'0II R-S- R' � .Q.

�

I :9:

_

sulfone

]

1 R-S-R' :cF 2+

14-10 Sulfides ( Thioethers) The hydrogen peroxide/acetic acid combination is a good oxidant for sulfides. One equivalent of peroxide gives the sulfoxide, and a second equivalent further oxidizes the sulfoxide to the sulfone. This reagent combination probably reacts via the peroxy­ acid, which is formed in equilibrium with hydrogen peroxide. o

o

II

CH3 -C-OH

+

II

H-O-O-H

+

CH3 -C-O-O-H

H-O-H

peroxyacid

acid

Because they are easily oxidized, sulfides are often used as mild reducing agents. For example, we have used dimethyl sulfide to reduce the potentially explosive ozonides that result from ozonolysis of alkenes (Section 8-1 5).

(X

CH3

CH3

��

°3 ------0>

H

CH3 -S-CH3 dimethyl sulfide

H

ozonide

,

Q

CH3 +

0 II

CH3 -S -CH3 dimethyl sulfoxide

H

Sulfur compounds are more nucleophilic than the corresponding oxygen com­ pounds, because sulfur is larger and more polarizable and its electrons are less tightly held in orbitals that are farther from the nucleus. Although ethers are weak nucle­ ophiles, sulfides are relatively strong nucleophiles. Sulfides attack unhindered alkyl halides to give sulfonium salts.

.�

R-S-R ..

R'-X �

alkyl halide

sulfide

Example

CH3-� .-CH3

X-

+

CH31

dimethyl sulfide

�

sulfonium salt

1+

CH3 1-

CH3-�:-CH3

trimethylsulfonium iodide

Sulfonium salts are good alkylating agents because the leaving group is an uncharged sulfide. Sulfur's polarizability enhances partial bonding in the transition state, lower­ ing its energy.

+

Xnucleophile

R- S-R

+

X-

sulfide

sulfonium salt

Example

1pyridine

trimethylsulfonium iodide

N-methylpyridinium iodide

dimethyl sulfide

641

642

Chapter 14: Ethers, Epoxides, and Sulfides

Sulfonium salts are common alkylating agents in biological systems. For example, ATP activation of methionine forms the sulfonium salt S-adenosylmethionine (SAM), a biological methylating agent. +

N H3

CH3 o

II

0

0

II

(

II

- O-P -O-P -O-P -O-CH2

I

OH

I

I

�

I

I

:S -CH CH -CH-COO-

�

..

2 me hionine

N H2

(=¢;

OH

OH

H

H OH

OH

OH

OH

S-adenosylmethionine (SAM)

adenosine triphosphate (ATP)

SAM converts norepinephrine to epinephrine (adrenaline) in the adrenal glands. OH

HO

-bf , -

I

OH

+

'"

H

adenosine

SAM

the nitrogen mustards. which are less reactive alkylating agents that are used as anticancer drugs. Nitro­ gen mustards alkylate DNA. which prevents its reproduction and ulti­ mately kills the cells.

/

CH 3 -N: '"

CH2CH2Cl CH2CH2Cl

HO

h CHCH2-�±-CH3 I ---«J- OH I H epinephrine

norepinephrine

The sulfur mustards gave rise to

OH

N H3

C-/ ��� CH2 CH2-2: H-COO,H

CHCH2 -N

( CH3

PROBLEM 14-18 Mustard gas, Cl-CH2CH2 -S-CH2CH2-C1, was used as a poisonous chemical agent in World War I. Mustard gas is much more toxic than a typical primary alkyl chloride. Its tox­ icity stems from its ability to alkylate amino groups on important metabolic enzymes, render­ ing the enzymes inactive. (a) Propose a mechanism to explain why mustard gas is an exceptionally potent alkylating agent. (b) Bleach (sodium hypochlorite, NaOCI, a strong oxidizing agent) neutralizes and inacti­ vates mustard gas. Bleach is also effective on organic stains because it oxidizes colored compounds to colorless compounds. Propose products that might be formed by the reac­ tion of mustard gas with bleach.

nitrogen mustard

14-11

Synthesis of Epoxides

Epoxides are easily made from alkenes, and (unlike other ethers) they undergo a vari­ ety of useful synthetic reactions. For these reasons, epoxides are valuable synthetic intermediates. Here we review the epoxidation techniques already covered (Section 8-12) and consider in more detail the useful syntheses and reactions of epoxides. 1 4-1 1 A

Peroxyacid Epoxidation

Peroxyacids (sometimes called peracids) are used to convert alkenes to epoxides. If

the reaction takes place in aqueous acid, the epoxide opens to a glycol. Therefore, to make an epoxide, we use a weakly acidic peroxyacid that is soluble in aprotic solvents such as CH2CI 2. Because of its desirable solubility properties, meta-chloroperoxyben­ zoic acid (MCPBA) is often used for these epoxidations.

""

/

/

""

c=c

°I R-C-O-O-H

+

alkene

°

1 4-11

/\ -C-CI

peroxyacid

I

Synthesis of Epoxides

° I R -C-O-H

+

epoxide

acid

Example

Cl

ex:

MCPBA

< }- L

MCPBA

)

cyclohexene

epoxycyclohexane

O-O-H

meta-chloroperoxybenzoic acid

(100%)

The epoxidation takes place in a one-step, concerted reaction that maintains the stereochemistry of any substituents on the double bond. R ° / C .... ../ / ° \ .0I ).. 8 1 /c

"C/

-----7

H--'

"

alkene

O� / C I /0 H

,,/ C" 1/ 0 /c \

peroxyacid

epoxide

R

acid

The peroxyacid epoxidation is quite general, with electron-rich double bonds reacting fastest. The following reactions are difficult transformations made possible by this se­ lective, stereospecific epoxidation procedure. The second example uses magnesium monoperoxyphthalate (MMPP), a relatively stable water-soluble peroxyacid often used in large-scale epoxidations.

(X

MCPBA (I equiv)

H

""

/

Lk.

0

)

CH3

CH3

1 ,2-dimethyl-l ,4-cyclohexadiene

Ph

�CH3

CH,

/

CH3

C=C

""

cis-4,S-epoxy-4,S-dimethylcyclohexene I

»� �2(

Ph , /0,, : CH3

MMPP

N02

H

N02

(E)-2-methy 1-2-nitro- 3-phenyIox i rane

(E)-2-nitro-I-phenylpropene

2

1 4-1 1 B

Base-Promoted Cyclization of Halohydrins

A second synthesis of epoxides and other cyclic ethers involves a variation of the Williamson ether synthesis. If an alkoxide ion and a halogen atom are located in the same molecule, the alkoxide may displace a halide ion and form a ring. Treatment of a halohydrin with a base leads to an epoxide through this intemal SN2 attack. .. :O� H

II -C-CIX I (X

=

�-:O-H ..

Ct, Br, I)

+

I /?: -C-CI c*

-----7

X-

:0: /\ -C-CI

I

643

644

Chapter 14: Ethers, Epoxides, and Sulfides

Halohydrins are easily generated by treating alkenes with aqueous solutions of halogens. Bromine water and chlorine water add across double bonds with Markovnikov orientation (Section 8-11). The following reaction shows the reaction of cyclopentene with chlorine water to give the chlorohydrin. Treatment of the chlorohy­ drin with aqueous sodium hydroxide gives the epoxide. Formation of the chlorohydrin

~ H

cyc\opentene

chlorine water

Cl

trans-chlorohydrin (mixture of enantiomers)

chloronium ion

Displacement of the chlorohydrin

-OH

(---7

H

CI

�C2 H

trans-chlorohydrin

alkoxide

--

W H

H Cl-

epoxide (50% overall)

This reaction can be used to synthesize cyclic ethers with larger rings. The difficulty lies in preventing the base (added to deprotonate the alcohol) from attacking and dis­ placing the halide. 2,6-Lutidine, a bulky base that cannot easily attack a carbon atom, can be used to deprotonate the hydroxyl group to give a five-membered cyclic ether. Five-, six-, and seven-membered (and occasionally four-membered) cyclic ethers are formed this way.

n

J:\:�

CH3

CI

O-.J H

••

chloro-alcohol

+

D

CH3

CH3

2,6-lutidine (2,6-dimethylpyridine)

alkoxide

Cl-

0

2-methyltetrahydrofuran (85%)

PROBLEM 14-19 Show how you would accomplish the following transformations ..Some of these examples require more than one step. (a) 2-methylpropene ---7 2,2-dimethyloxirane (b) I-phenylethanol ---7 2-phenyloxirane (c) 5-chloro- l -pentene ---7 tetrahydropyran (d) 5-chloro-l-pentene ---7 2-methyltetrahydrofuran (e) 2-chloro-I-hexanol ---7 1,2-epoxyhexane

PROBLEM 14-20 The 2001 Nobel Prize in chemistry was awarded to three organic chemists who have developed methods for catalytic asymmetric syntheses. An asymmetric (or enantioselec­ tive) synthesis is one that converts an achiral starting material into mostly one enantiomer of a chiral product. K. Barry Sharpless (The Scripps Research Institute) developed an asymmetric epoxidation of allylic alcohols that gives excellent chemical yields and greater than 90% enantiomeric excess. The Sharpless epoxidation uses tert butyl hydroperoxide, titanium (IV) isopropoxide, and a dialkyl tartrate ester as the reagents. The following epoxidation of geraniol is typical. -

14-12 Acid-Catalyzed Ring Opening of Epoxides

�

) OH

' .. 0 I � OH 80% yield, > 90% ee

geraniol

+::

H2CH3

H

Reagents:

HO

+

H

COOCH2CH3

titanium (IV) isopropoxide

tert-butyl hydroperoxide

diethyl L-tartrate

(a) Which of these reagents is most likely to be the actual oxidizing agent? That is, which reagent is reduced in the reaction? What is the likely function of the other reagents?

(b) When achiral reagents react to give a chiral product, that product is normally formed as a racemic mixture of enantiomers. How can the Sharpless epoxidation give just one nearly pure enantiomer of the product? (c) Draw the other enantiomer of the product. What reagents would you use if you wanted to epoxidize geraniol to give this other enantiomer?

I 1.

SUMMARY

Epoxide Syntheses

Peroxyacid epoxidation (Section 14-11A) o " /

2.

/

C=C

"

/\

R- C- OOH

+

o

o

II

- C-C-

I

I

II

R - C- OH

+

Base-promoted cyclization of halohydrins (Section 14-1lB)

I

X

I

- C-C-

I

OH

0, _

b as

I

- C- C -

\/

I

X Example

I

e �

o

=

CI, Br, I, OTs, etc.

0, �

H

CH-CH2 CI

I

OR

2-chloro-l -phenylethanol

NaOH,Hp

)

_

- CH2

\/ o

2-phenyloxirane

Epoxides are much more reactive than common dialkyl ethers because of the large strain energy (about 105 kllmol or 25 kcal/mol) associated with the three-membered ring. Unlike other ethers, epoxides react under both acidic and basic conditions. The products of acid-catalyzed opening depend primarily on the solvent used. In Water In Section 8-13 we saw that acid-catalyzed hydrolysis of epoxides gives glycols with anti stereochemistry. The mechanism of this hydrolysis involves protonation of oxygen (forming a good leaving group), followed by SN2 attack by

14-12

Acid-Catalyzed Ring Opening of Epoxides

645

646

Chapter

14:

Ethers, Epoxides, and Sulfides water. Anti stereochemistry results from the back-side attack of water on the protonated epoxide.

MECHANISM 14-2

Acid-Catalyzed Opening of Epoxides in Water

Epoxides open in acidic solutions to form glycols. Step

1:

Protonation of the epoxide to form a strong electrophile.

w- 0t '

0

H-O-H

(

'

"

)

W

H

H

H

H

1 ,2-epoxycyclopentane

Step 2: Water attacks and opens the ring.

� �H H

U~ Hi:):

Step 3: Deprotonation to give the diol.

H. )

�

�

HO'

0+ / -G"' H H

H

0+ / '" H H

H

H

H

+

OH

trans-cyclopentane-I,2-diol (mixture of enantiomers)

Direct anti hydroxylation of an alkene (without isolation of the epoxide interme­ diate) is possible by using an acidic aqueous solution of a peroxyacid. As soon as the epoxide is formed, it hydrolyzes to the glycol. Peroxyacetic acid (CH3C03H) and peroxyformic acid (HC03H) are often used for the anti hydroxylation of alkenes.

H3C" H

H /

,C=C / '"

CH3

trans-2-butene

o

II CH3-C-OOH )

HO

",

JI

..,CH 3

""C -c ." '" H3C� OH

meso-butane-2,3-diol

PROBLEM 14-21 Propose mechanisms for the epoxidation and ring-opening steps of the epoxidation and hydrol­ ysis of trans-2-butene shown above. Predict the product of the same reaction with cis-2-butene.

When the acid-catalyzed opening of an epoxide takes place with an alcohol as the solvent, a molecule of alcohol acts as the nucleophile. This reaction pro­ duces an alkoxy alcohol with anti stereochemistry. This is an excellent method for making compounds with ether and alcohol functional groups on adjacent carbon atoms. For example, the acid-catalyzed opening of 1,2-epoxycyclopentane in a methanol solution gives trans-2-methoxycyclopentanol. In Alcohols

14-12 Acid-Catalyzed Ring Opening of Epoxides

MECHANISM 14-3

Acid-Catalyzed Opening of an Epoxide in an Alcohol Solution

Epoxides open in acidic alcohol solutions to form 2-alkoxy alcohols.

Step 1: Protonation of the epoxide to form a strong electrophile.

��� � . � H

H

H

H

1,2-epoxycyclopentane

Step 2: The alcohol (solvent) attacks and opens the ring.

�, H

�

:O-H

/

CH3

Step 3: Deprotonation to give the product, a 2-alkoxy alcohol.

)O.:J H

H

CH3

CH3 iiH

/

� H

+

CH

,(lH,

' Q -CH,

trans-2-methoxycyclopentanol (82%) (mixture of enantiomers)

PROBLEM 14-22 Cellosolve® is the trade name for 2-ethoxyethanol, a common industrial solvent. This com­ pound is produced in chemical plants that use ethylene as their only organic feedstock. Show how you would accomplish this industrial process.

Using Hydrohalic Acids

When an epoxide reacts with a hydrohalic acid ( HC], HBr, or H I), a halide ion attacks the protonated epoxide. This reaction is analogous to the cleavage of ethers by HBr or HI. The halohydrin initially formed reacts further with HX to give a 1,2-dihalide. This is rarely a useful synthetic reaction, because the 1,2-dihalide can be made directly from the alkene by electrophilic addition of X2 .

/O\� -C-C- + H- X I

I

�

..

:OH

I

I

-C-C-

I

X ---c">

HX

I

I

X (several steps)

PROBLEM 14-23 When ethylene oxide is treated with anhydrous RBr gas, the major product is 1,2-dibro­ moethane. When ethylene oxide is treated with concentrated aqueous RBr, the major product is ethylene glycol. Use mechanisms to explain these results.

I

I

-C-C-

I

X

647

648

Chapter 14: Ethers, Epoxi des , and Sulfides

squalene epoxidase (enzyme)

)

4

cyclized intermediate

squalene-2,3-epoxide

squalene

HO HO cholesterol

lanosterol

... Figure 14-6 Role of squalene in the biosynthesis of steroids. The biosynthesis of steroids starts with epoxidation of squalene to squalene2,3-epo xide. The opening of this epoxide promotes cyclization of the carbon skeleton under the control of an enzyme. The cyclized intermediate is converted to lanosterol, then to other steroids.

Inhibitors of squalene epoxidase are used in antifungal drugs to treat ath­ lete's foot, jock itch, ring worm, and ™

nail infections. The drug Tinactin

(tolnaftate) inhibits squalene epoxi­ dase, which blocks the synthesis of the steroids the fungus needs to make its cell membrane. The defec­ tive cell membrane kills the fungus.

The Opening of Squalene-2,3-Epoxide

Steroids are tetracyclic compounds that serve a wide variety of biological functions, including hormones (sex hormones), emulsi­ fiers (bile acids), and membrane components (cholesterol). The biosynthesis of steroids is believed to involve an acid-catalyzed opening of squalene-2,3-epoxide (Figure 14-6). Squalene is a member of the class of natural products called terpenes (see Section 25-8). The enzyme squalene epoxidase oxidizes squalene to the epoxide, which opens and forms a carbocation that cyclizes under the control of another enzyme. The cyclized intermedi­ ate rearranges to lanosterol, which is converted to cholesterol and other steroids. Although cyclization of squalene-2,3-epoxide is controlled by an enzyme, its mechanism is similar to the acid-catalyzed opening of other epoxides. The epoxide oxygen becomes protonated and is attacked by a nucleophile. In this case, the nucle­ ophile is a pi bond. The initial result is a tertiary carbocation (Figure 14-7).

-----7

HO protonated squalene-2,3-epoxide

HO tertiary carbocation

tertiary carbocation

... Figure 14-7 Cyclization of squalene epoxide begins with the acid-catalyzed opening of the epoxide. Each additional cyclization step forms another carbocation.

. . .

14-13

Base-Catalyzed Ring Opening of Epoxides

649

This initial carbocation is attacked by another double bond, leading to the for­ mation of another ring and another tertiary carbocation. A repetition of this process leads to the cyclized intermediate shown in Figure 14-6. Note that this sequence of steps converts an achiral, acyclic starting material (squalene) into a compound with four rings and seven asymmetric carbon atoms. The enzyme-catalyzed sequence takes place with high yields and complete stereospecificity, providing a striking example of asymmetric induction in a biological system. PROBLEM 14 -24 Show the rest of the mechanism for formation of the cyclized intermediate in Figure 14-6.

Most ethers do not undergo nucleophilic substitutions or eliminations under basic con­ ditions, because an alkoxide ion is a poor leaving group. Epoxides have about 1 05 kllmol (25 kcal/mol) of ring strain that is released upon ring opening, however, and this strain is enough to compensate for the poor alkoxide leaving group. Figure 14-8 compares the energy profiles for nucleophilic attack on an ether and on an epoxide. The starting epoxide is about 1 05 kllmol (25 kcal/mol) higher in energy than the ether, and its displacement has a lower activation energy.

1 4- 1 3

Base-Cata Iyzed Ring Opening of Epoxides

lower

Ea

r

2?

1 05 kJ

;;., ring strain

t yO " -C-- C " / .. )

:0 : I I - C- CI I

HO : -

OH

..,c ..,

--------------

:§ �

�c.

HO . . : -'----'" R VO - R' reaction coordinate

-

The reaction of an epoxide with hydroxide ion leads to the same product as the acid-catalyzed opening of the epoxide: a 1 ,2-diol (glycol), with anti stereochemistry. In fact, either the acid-catalyzed or base-catalyzed reaction may be used to open an epox­ ide, but the acid-catalyzed reaction takes place under milder conditions. Unless there is an acid-sensitive functional group present, the acid-catalyzed hydrolysis is preferred.

MECHANISM 14-4

B ase-Catalyzed Openi ng of Epoxi des

Strong bases and nucleophiles do not attack and cleave most ethers. Epoxides are more reac­ tive, however, because opening the epoxide relieves the strain of the three-membered ring. Strong bases can attack and open epoxides, even though the leaving group is an alkoxide. Step

1: A

strong base attacks and opens the ring to an alkoxide.

H

(fa

(n)

(A true story.) An inexperienced graduate student moved into a laboratory and began work. He needed some diethyl ether for a reaction, so he opened an old, rusty I-gallon can marked "ethyl ether" and found there was half a gallon left. To puri­ fy the ether, the student set up a distillation apparatus, started a careful distillation, and went to the stockroom for the other reagents he needed. While he was at the stockroom, the student heard a muffled "boom." He quickly returned to his lab to find a worker from another laboratory putting out a fire. Most of the distillation apparatus was embedded in the ceiling. (a) Explain what probably happened. (b) Explain how this near-disaster might have been prevented. (a) Show how you would synthesize the pure (R) enantiomer of 2-butyl methyl sulfide starting with pure (R)-2-butanol and any reagents you need. (b) Show how you would synthesize the pure (S) enantiomer of the product. (a) Predict the values of mlz and the structures of the most abundant fragments you would observe in the mass spectrum of di-n-propyl ether. (b) Give logical fragmentations to account for the foUowing ions observed in the mass spectrum of 2-methoxypentane: 102, 87,7 1 , 5 9, 3 1 . The following reaction resembles the acid-catalyzed cyclization of squalene oxide. Propose a mechanism for this reaction.

Show how you would accomplish the following synthetic transformations in good yield. --i> I-phenyl-2-hexanol --i> I-methoxy-2-hexanol --i> 2-methoxy- l -hexanol Give the structures of intermediates A through H in the following synthesis of trans- l-cyclohexyl-2-methoxycyclohexane. (a) I-hexene (b) I-hexene (c) l -hexene

B

A 14-40

(i)

� Wv �

potassium t-butoxide + n-butyl bromide

659

o

(h) propylene oxide

(k)

14-34

Study Problems

coned. HBr heat

MCPBA

)

C (gas)

F

/

+

(I) E

D

Mg , ether

)

G

Na

---i>

H

(Another true story.) An organic lab student carried out the reaction of methylmagnesium iodide with acetone (CH3COCH3 ), followed by hydrolysis. During the distillation to isolate the product, she forgot to mark the vials she used

660

Chapter 14: Ethers, Epoxides, and S ulfides

to collect the fractions. She turned in a product of formula C4 H 100 that boiled at 35°C. The IR spectrum showed only a weak 0 - H stretch around 3300 cm-I, and the mass spectrum showed a base peak at mlz 59. The NMR spectrum showed a quartet (J 7 Hz) of area 2 at 83.5 and a triplet (J 7 Hz) of area 3 at 81.3 . Propose a structure for this product, explain how it corresponds to the observed spectra, and suggest how the student isolated this compound. Show how you would synthesize the following ethers in good yield from the indicated starting materials and any addi­ tional reagents needed. (a) cyclopentyl n-propyl ether from cyclopentanol and I-propanol (b) n-butyl phenyl ether from phenol and I -butanol (c) 2-methoxydecane from a decene (d) I -methoxydecane from a decene (e) l -ethoxy- l -methylcyclohexane from I-methylcyclohexene (f) trans-2,3-epoxyoctane from trans-2-octene There are two different ways of making 2-ethoxyoctane from 2-octanol using the Williamson ether synthesis. When pure ( - )-2-octanol of specific rotation - 8.24° is treated with sodium metal and then ethyl iodide, the product is 2-ethoxyoc­ tane with a specific rotation of - 1 5.6° When pure ( -)-2-octanol is treated with tosyl chloride and pyridine and then with sodium ethoxide, the product is also 2-ethoxyoctane. Predict the rotation of the 2-ethoxyoctane made using the tosy­ lationlsodium ethoxide procedure, and propose a detailed mechanism to support your prediction. Under base-catalyzed conditions, several molecules of propylene oxide can react to give short polymers. Propose a mech­ anism for the base-catalyzed formation of the following trimer. =

14-41

14-42

14-43

=

CH3

CH3

C�

HO -CH2 -CH -0- CH2 -CH -0-CH2 -CH - OH

I

I .

I .

Under the right conditions, the following acid-catalyzed double cyclization proceeds in remarkably good yields. Propose a mechanism. Does this reaction resemble a biological process you have seen?

14-44

OH

Propylene oxide is a chiral molecule. Hydrolysis of propylene oxide gives propylene glycol, another chiral molecule. Draw the enantiomers of propylene oxide. Propose a mechanism for the acid-catalyzed hydrolysis of pure (R)-propylene oxide. Propose a mechanism for the base-catalyzed hydrolysis of pure (R)-propylene oxide. Explain why the acid-catalyzed hydrolysis of optically active propylene oxide gives a product with a rotation opposite that of the product of the base-catalyzed hydrolysis. An acid-catalyzed reaction was carried out using methyl cellosolve (2-methoxyethanol) as the solvent. When the 2-methoxyethanol was redistilled, a higher-boiling fraction (bp 1 62°C) was also recovered. The mass spectrum of this fraction showed the molecular weight to be 1 34. The IR and NMR spectra are shown here. Determine the structure of this compound, and propose a mechanism for its formation.

*14-45

(a) (b) (c) (d)

*14-46

4 1 6 3 2 3 m 5 � 5 mm� m ' "m "II�m : 5 mmwrnUJm ' � � l i 1 0GII' I" ====�==� = ;m � � : � :U l rnuU[mnu m�� ' 1 ' 4 1l � ° t l �I� � l t �W I , I � � , �,5� 1 , " Ii ,; ! I I" I wavelength ([Lm)

80 60

40 20

I ;�

I

� .�

II

ill

I

' " " ;

II

:111

I,ll

li!1

A

III

5

IIIII

. 1" 1 1 111

E 1'litH l itHlt lttHl l, tlrHiIiHtl" +i Hl li i IH l

II 3ilm �rt+ffi3000H+,i+\,1fH++ 00 4000 o

,I

,Ii

II

,I,

2 00 5

1

III

! !1

Problem 14-46 !:,, '' !I' Iii'

I",

I i i!

;iii

II!

,,!

II

'

I

III I

'I'

I

,I

I',

i

I

II ,I

wavenumber (em-I)

II

I II

2000 1800 1600 1400 '

'Ii ,Iii

I II

j

I'

I,' II I

i'

1200

'I

11

i, II ii 11

000

800

600

Study Problems

200

18 0

140

160

1 20

i

Problem

:

,

10 14-47

:

iI :

100

I

i

+-T-+++4 + +-T-++-++ -+ 14-46 H--+++-++ � ',

:

,

; ,

,

i

:

i

�r

§ �

7

6

20

o

, I

I ;

3,68

:

3,58

) 5 (ppm)

,

11,

I

4

3

:

2

o

A compound of molecular formula CsH sO gives the IR and NMR spectra shown here. Propose a structure, and show how it is consistent with the observed absorptions.

4

4,5

,II'

5

5;5

6

7

wavelength (/Lm) I

I!

jH-H;iH' H-l-H+ +H;I'

Ii

i

,iii ! I i

!I I

i ! , iI I

1 60

I'II

120

140 I

II il ': III : i 1I 1;:·

'

ii I' !

I!

_,

i

'

i

,

I

I !

,

9

8

i

\

100

iii 0'H' i Z

I'�'

'OLI

�I

0 r Z

t I

i I �

IIlilI' i

7

6

5

5 (ppm)

II

I,

II I

, II

, , I

I II ,i,

1400 1200

1000

800

60

80

40

It

i

, ,

t-+++-t+--H i '1: � f-+-i-++--i--t"C

t\

�

i� :t "

600 20

01Hz �

'

'' ' ' , ' ' ' L 3 8, " 3.9, . u u

I

Ii

l! 1

wavenumber (em - I)

I : . I 'ir-',', I ' P oble m' _l,. 4-47 1+-t-+++-+-+-+-HIH+++-+-++-+1m , +-+-H-t-++i , ' "

'

I

III ,11 II, II

1j

2000 1800 1600

180

12 13 14 15 16

II

,!

I, if;

200

10

9

Iii"

di

il!!

8

I.

II j 1-1+1+11++, Ii I '

10

40

50Hz

I""I

3,78

8

9

OHz

60

I � ���I�������'"��

i

'I

80

661

,

,

510HZ

o

... . . . . .

if T

11 �I" I

�

WJI-----;�_

4

3

2

-+-

-:,' '-

' I""I" 'I""I'",I8""I' Lf' 3.lu, 3.0u, 2.9u, 2. u, 2.7u, ,- +, '-tl'i+.', l-+-+-+--+I---l7"F7H 3.2u

.' i

L-i-

--L �

,;

' , "";"

o

662 *14-48

Chapter ]4: Ethers, Epoxides, and S ulfides

A new graduate student was studying the insecticidal properties of a series of polycyclic epoxides. He epoxidized alkene

A using two different methods. First he used MCPBA, which gave an excellent yield of an epoxide that he labeled B. Then he treated alkene A with bromine water to form the bromohydrin, followed by 2,6-dimethylpyridine (see page 299)

to form an epoxide in fair yield. To his surprise, the second method produced an epoxide (C) with different physical and chemical properties from the first. In particular, C reacts with strong nucleophiles much faster than B. Propose structures for B and C, and propose mechanisms to show why different products are formed. Explain why C reacts so much faster with strong nucleophiles. MCPBA

alkene A

(2) 2,6-dimethylpyridine

epoxide B

epoxide C

15

Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy Double bonds can interact with each other if they are separated by just one single bond. Such interacting double bonds are said to be conjugated. Double bonds with two or more single bonds separating them have little interaction and are called isolated double bonds. For example, 1 , 3-pentadiene has conjugated double bonds, and 1 ,4-pentadiene has isolated double bonds.

1 5-1 Introduction

conjugated double bonds (more stable than isolated double bonds) --"""'c=c/' � /'

--"""'C=C/'

/

CH2

--.....

isolated double bonds

Because of the interaction between the double bonds, systems containing conju­ gated double bonds tend to be more stable than similar systems with isolated double bonds. In this chapter, we consider the unique properties of conjugated systems, the theoretical reasons for this extra stability, and some of the characteristic reactions of molecules containing conjugated double bonds. We also study ultraviolet spec­ troscopy, a tool for determining the structures of conj ugated systems.

In Chapter 7, we used heats of hydrogenation to compare the relative stabilities of alkenes. For example, the heats of hydrogenation of 1 -pentene and trans-2-pentene show that the disubstituted double bond in trans-2-pentene is 10 kJ/mol (2.5 kcal/mol) more stable than the monosubstituted double bond in 1 -pentene.

� J -pentene

trans-2-pentene

t).HO tlW

=

=

1 5-2 Stabilities of Dienes

-]26 kJ (- 30. 1 kcal)

- 1 1 6 kJ ( - 27.6 kcaJ) 663

664

Chapter 1 5 : Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy

When a molecule has two isolated double bonds, the heat of hydrogenation is close to the sum of the heats of hydrogenation for the individual double bonds. For example, the heat of hydrogenation of 1 ,4-pentadiene is -252 kJimol ( -60.2 kcal/mol), about twice that of I-pentene. /).HO

1 ,4-pentadiene

=

-252 kJ ( - 60.2 kca\)

For conjugated dienes, the heat of hydrogenation is less than the sum for the individual double bonds. For example, trans-l,3-pentadiene has a monosubstituted double bond like the one in I -pentene and a disubstituted double bond like the one in 2-pentene. The sum of the heats of hydrogenation of I -pentene and 2-pentene is -242 kJ (-57.7 kcal), but the heat of hydrogenation of trans- l ,3-pentadiene is -225 kJimol ( -53.7 kcal/mol), showing that the conjugated diene has about 1 7 kJ/mol (4.0 kcal/mol) extra stability.

trans- [,3-pentadiene

2 H2

�

Pt

�

+

( - 1 26 kJ)

+

I-pentene

�

�

2-pentene

?

�

predicted

- 242 kJ ( -57.7 kcal)

(- 1 1 6 kJ)

- 225 kJ ( -53.7 kcal) actual value 17 kJ (4.0 kcal) more stable by

What happens if two double bonds are even closer together than in the conju­ gated case? Successive double bonds with no intervening single bonds are called cumulated double bonds. Consider 1 ,2-pentadiene, which contains cumulated dou­ ble bonds. Such 1 ,2-diene systems are also called allenes, after the simplest member of the class, 1 ,2-propadiene or "allene," H2C=C=CH2. The heat of hydrogenation of 1 ,2-pentadiene is -292 kJimol (-69.8 kcal/mol).

H '-----

/H =C=C C / '----H CH 2CH 3

I ,2-pentadiene

(ethylallene)

pentane

/).HO

I-pentene + 2-pentene

/).HO

CH3CH2CH2CH2CH3

sum of

�

�

=

=

-292 kJ ( - 69.8 kcal)

-242 kJ (-57.7 kcal)

1 ,2-pentadiene is less stable by 50 kJ

( 1 2.1 kcal)

Because 1 ,2-pentadiene has a larger heat of hydrogenation than 1 ,4-pentadiene, we conclude that the cumulated double bonds of allenes are less stable than isolated dou­ ble bonds and much less stable than conjugated double bonds. Figure 1 5- 1 summarizes the relative stability of isolated, conjugated, and cumulated dienes and compares them with alkynes. Rank each group of compounds in order of increasing heat of hydrogenation. (a) 1 ,2-hexadiene; 1 ,3 ,5-hexatriene; 1 ,3-hexadiene; 1 ,4-hexadiene; 1 ,5-hexadiene; 2,4-hexadiene P R O BLEM 15-1

cumulated diene

=c�

1,2-pentadiene

1 5-3 Molecular Orbital Picture of a Conjugated System alkyne

665

terminal =-----/""-..

internal

I-pentyne

alkyne =

292 kJ

,-

2-pentyne

291 kJ 275

t

isolated diene

isolated diene

�

kJ

I ,4 -pentadiene

252 kJ

t

alkane (pentane or hexane)

�

lmns-I,4-hexadiene

=r t

conjugated diene

�

I t

Imns -I ,3-pentadiene

225 kJ

... Figure 15-1 Relative energies of conjugated, isolated, and cumulated dienes compared with alkynes, based on heats of hydrogenation (kllmol).

In a strongly acidic solution, 1,4-cyclohexadiene tautomerizes to 1,3-cyclohexadiene. Propose a mechanism for this rearrangement, and explain why it is energetically favorable. P R O BLEM 15-2

(Review) The central carbon atom of an allene is a member of two double bonds, and it has an interesting orbital arrangement that holds the two ends of the molecule at right angles to each other. (a) Draw an orbital diagram of allene, showing why the two ends are perpendicular. (b) Draw the two enantiomers of 1 ,3-dichloroallene. A model may be helpful. P R O BLEM 15-3

Figure 1 5- 1 shows that the compound with conjugated double bonds i s 1 7 kllmol (4.0 kcal/mo1) more stable than a similar compound with isolated double bonds. This 17 kllmol of extra stability in the conjugated molecule is called the resonance energy of the system. (Other terms favored by some chemists are con­ jugation energy, de localization energy, and stabilization energy.) We can best explain thi s extra stab i li ty of conjugated systems by examining their molecular orbitals. Let's begin with the molecular orbitals of the simplest conjugated diene, 1 ,3-butadiene.

1S-3A

Structure a n d Bon d i n g of 1,3-Butadiene

The heat of hydrogenation of 1 ,3-butadiene is about 17 kllmol (4.0 kcal/mol) less than twice that of I -butene, showing that 1 ,3-butadiene has a resonance energy of 1 7 kllmol.

15-3 Molecular Orbital Picture of a Conjugated System

666

C h apter 1 5 :C onj ugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy

� Figure 15-2 Structure of I ,3-butadiene in its most stable conformation. The 1 .48 A central carbon-carbon single bond is shorter than the 1.54 A bonds typical of alkanes because of its partial double-bond character.

small amount of overlap

H

H

1,3-butadiene

I:1HO

I-butene

I:1W

resonance energy of I ,3-butadiene

=

=

-237 kJ ( -56.6 kcal )

= -127 kJ (-30.3 kcal) X 2 = -254 kJ (-60.6 kcal)

254 kJ - 237 kJ

=

1 7 kJ (

4 .0 kcal )

Figure 1 5-2 shows the most stable conformation of I ,3-butadiene. Note that this con­ formation is planar, with the p orbitals on the two pi bonds aligned. The C2 - C3 bond in I ,3-butadiene ( 1 .48 A) is considerably shorter than a carbon-carbon single bond in an alkane ( 1 .54 A). This bond is shortened slightly by the increased s character of the sp2 hybrid orbitals, but the most important cause of this bond shortening is its pi bonding overlap and pmtial double-bond character. The planar confor­ mation, with the p orbitals of the two double bonds aligned, allows overlap between the pi bonds. In effect, the electrons in the double bonds are delocalized over the entire molecule, creating some pi overlap and pi bonding in the C2 - C3 bond. The length of this bond is intermediate between the normal length of a single bond and that of a double bond. Lewis structures are not adequate to represent delocalized molecules such as I ,3-buta­ diene. To represent the bonding in conjugated systems accurately, we must consider molec­ ular orbitals that represent the entire conjugated pi system, and not just one bond at a time. 15-38

Constructi ng the Molecular Orbitals of 1,3-8utadiene

All four carbon atoms of I ,3-butadiene are sp2 hybridized, and (in the planar confor­ mation) they all have overlapping p orbitals. Let's review how we constructed the pi molecular orbitals (MOs) of ethylene from the p atomic orbitals of the two carbon atoms (Figure 1 5-3). Each p orbital consists of two lobes, with opposite phases of the wave function in the two lobes. The plus and minus signs used in drawing these or­ bitals indicate the phase of the wave function, not electrical charges. To minimize confusion, we will color the lobes of the p orbitals to emphasize the phase difference. In the pi bonding molecular orbital of ethylene, the lobes that overlap in the bonding region between the nuclei are in phase; that is, they have the same sign (+ overlaps with + , and - overlaps with -). We call this reinforcement constructive overlap. Constructive overlap is an important feature of all bonding molecular orbitals. In the pi antibonding molecular orbital (marked by *), on the other hand, lobes of opposite phase (with opposite signs, + with -) overlap in the bonding region. This de­ structive overlap causes cancelling of the wave function in the bonding region. Midway between the nuclei, this anti bonding MO has a node: a region of zero electron density where the positive and negative phases exactly cancel. Electrons have lower energy in the bonding MO than in the original p orbitals, and higher energy in the antibonding MO. In the ground state of ethylene, two electrons are in the bonding MO, but the antibonding MO is vacant. Stable molecules tend to have filled bonding MOs and empty antibonding MOs.

node

destructi ve overlap energy

1 5-3 Molecular Orbital Picture of a Conj ugated System (antibonding) 11:*

=

energy of the isolated orbitals on C I and C2

p

(bonding) 11:

constructive overlap

667

=

Several important principles are illustrated in Figure 15-3. Constructive overlap results in a bonding interaction; destructive overlap results in an antibonding interac­ tion. Also, the number of pi molecular orbitals is always the same as the number of p orbitals used to form the Mas. These molecular orbitals have energies that are sym­ metrically distributed above and below the energy of the starting p orbitals. Half are bonding Mas, and half are antibonding MOs. Now we are ready to construct the molecular orbitals of 1 ,3-butadiene. The p orbitals on C l through C4 overlap, giving an extended system of four p orbitals that form four pi molecular orbitals. Two MOs are bonding, and two are antibonding. To represent the four p orbitals, we draw four p orbitals in a line. Although 1 ,3-butadiene is not linear, this sim­ ple straight-line representation makes it easier to draw and visualize the molecular orbitals.

.... Figure 15-3 The pi molecular orbitals of ethylene. The pi bonding orbital of ethylene is formed by constructive overlap of unhybridized p orbitals on the 2 sp -hyb rid carbon atoms. Destructive overlap of these two orbitals forms the antibonding pi orbital. Combination of two p orbitals must give exactly two molecular orbitals. PROBLEM-SOLVING

HiltZ;

Stable molecules tend to have filled bonding Mas and empty anti bonding Mas.

represented by

The lowest-energy molecular orbital always consists entirely of bonding interac­ tions. We indicate such an orbital by drawing all the positive phases of the p orbitals overlapping constructively on one face of the molecule, and the negative phases over­ lapping constructively on the other face. Figure 1 5-4 shows the lowest-energy MO for I,3-butadiene. This MO places electron density on all four p orbitals, with slightly more on C2 and C3. (In these figures, larger and smaller p orbitals are used to show which atoms bear more of the electron density in a particular MO.) This lowest-energy orbital is exceptionally stable for two reasons: There are three bonding interactions, and the electrons are delocalized over four nuclei. This orbital bonding bonding bonding

.... Figure 15-4 The 1Tj bonding MO of 1 ,3-butadiene. The lowest-energy orbital of l ,3-butadiene has bonding interactions between all adjacent carbon atoms. This orbital is labeled 1Tj because it is a pi bonding orbital and i t has the lowest energy.

Chapter 1 5 : Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy

668

� Figure 1 5 -5 The 7T2 bonding MO of 1,3-butadiene. The second MO of I,3-butadiene has one node in the center of the molecule. There are bonding interactions at the C I - C2 and C3 - C4 bonds, and there is a (weaker) antibonding interaction between C2 and C3. This 7T2 orbital is bonding, but is not as strongly bonding as 7T J .

bonding

anti bonding

bonding

helps to illustrate why the conjugated system is more stable than two isolated double bonds. It also shows some pi-bond character between C2 and C3, which lowers the en­ ergy of the planar conformation and helps to explain the short C2 - C3 bond length. As with ethylene, the second molecular orbital (7T2) of butadiene (Figure 1 5-5) has one vertical node in the center of the molecule. This MO represents the classic pic­ ture of a diene. There are bonding interactions at the C 1 - C2 and C3 - C4 bonds, and there is a (weaker) antibonding interaction between C2 and C3. The 7T2 orbital has two bonding interactions and one antibonding interaction, so we expect it to be a bonding orbital ( 2 bonding - 1 antibonding 1 bonding ) . It is not as strongly bonding nor as low in energy as the all-bonding 7TI orbital. Adding and subtracting bonding and antibonding interactions is not a reliable method for calculat­ ing energies of molecular orbitals, but it is useful for predicting whether a given orbital is bonding or anti bonding and for ranking orbitals in order of their energy. The third butadiene MO (7T;) has two nodes (Figure 1 5-6). There is a bonding interaction at the C2 - C3 bond, and there are two antibonding interactions, one be­ tween C l and C2 and the other between C3 and C4. This is an antibonding orbital ( * ) , and it is vacant in the ground state. The fourth, and last, molecular orbital (7T;) of 1 ,3-butadiene has three nodes and is totally antibonding (Figure 1 5-7). This MO has the highest energy and is unoccupied in the molecule's ground state. This highest-energy MO (7T;) is typical: For most systems, the highest-energy MO has antibonding interactions between all pairs of adjacent atoms. Butadiene has four pi electrons (two electrons in each of the two double bonds in the Lewis structure) to be placed in the four MOs just described. Each MO can accommodate two electrons, and the lowest-energy MOs are filled first. Therefore, the four pi electrons go into 7TI and 7T2. Figure 1 5-8 shows the electronic configuration of 1 ,3-butadiene. Both bonding MOs are filled, and both antibonding MOs are empty. Most stable molecules have this arrangement of filled bonding orbitals and vacant antibonding orbitals. Figure 1 5-8 also compares the relative energies of the ethylene MOs with the butadiene MOs to show that the conjugated butadiene system is slightly more stable than two ethylene double bonds. The partial double-bond character between C2 and C3 in 1 ,3-butadiene explains why the molecule is most stable in a planar conformation. There are actually two planar conformations that allow overlap between C2 and C3. These conformations arise by ro­ tation about the C2 - C3 bond, and they are considered single-bond analogues of trans and cis isomers about a double bond. Thus, they are named s-trans ("single" -trans) and if s-cis ("single" -cis) conformations. =

an ti bo ndi ng

anti bonding *

3

I I

node

I

node

.A. Figure 1 5-6 The 7T�' antibonding MO of 1 ,3-butadiene. The third MO of I ,3-butadiene has two nodes, giving two antibonding interactions and one bonding interaction. This is an anti bonding orbital, and it is vacant in the ground state.

The 7Tt anti bonding molecular orbital of I ,3-butadiene. The highest­ energy MO of I ,3-butadiene has three nodes and three anti bonding interactions. It is strongly antibonding, and it is vacant in the ground state. � Figure 1 5 -7

all anti bonding

butadiene

eth y lene

EJ EJ

re * 3

1 5 -4 Allylic Cations

EJ

IT*

energy of isolated p orbital

ami bodnd bo� �gi ng

- - - - - - - - - - - - - - - - - - - -

GB

- - - - - - - - -

¥.

�

------

H

H

.... Figure 15-8 The electronic configurations of ethylene and 1,3-butadiene. In both 1,3-butadiene and ethylene, the bonding MOs are filled and the anti bonding MOs are vacant. The average energy of the electrons is slightly lower in butadiene. This lower energy is the resonance stabilization of the conjugated diene.

GB

GB

re I

-

H

H

H (

H

H

s-trans

)

H

�

� rJ:I'

m i ld

i nterference

H

s-cis

The s-trans conformation is 1 2 kJ/mol (2.8 kcal/mol) more stable than the s-cis conformation, which shows interference between the two nearby hydrogen atoms. The rotational barrier for these conformers (rotation about the C2 - C3 bond) is only about 20 kllmol (about 5 kcal/mol) compared with about 250 kllmol (60 kcal/mol) for rotation of a double bond in an alkene. The s-cis and s-trans conformers of butadiene (and all the skew conformations in between) easily interconvert at room temperature. Conjugated compounds undergo a variety of reactions, many of which involve inter­ mediates that retain some of the resonance stabilization of the conjugated system. Common intermediates include allylic systems, particularly allylic c.ations and radi­ cals. Allylic cations and radicals are stabilized by delocalization. First, we consider some reactions involving allylic cations and radicals, then (Section 1 5-8) we derive the molecular orbital picture of their bonding. In Chapter 7, we saw that the - CH2 - CH = CH2 group is called the allyl group. Many common names use this terminology. /' ....... ' C = C ....... /' /

I T .......

allylic position

allyl bromide

669

all yl alcohol

15-4 Allylic Cations

allylbenzene

When allyl bromide is heated with a good ionizing solvent, it ionizes to the allyl cation, an allyl group with a positive charge. More-substituted analogues are called

670

Chapter 1 5 : Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy

allylic cations. All allylic cations are stabilized by resonance with the adjacent double

· : 13 r :

bond, which delocalizes the positive charge over two carbon atoms.

H2C = CH - CH2 £$:r : allyl bromide

[H2C = CH - CH2

� H

allyl cation

(�

H

H substituted allylic cations

H

�

W +

H

i -CH=CH2]

+

H H

H

Draw another resonance form for each of the substituted allylic cations shown in the preced­ ing figure, showing how the positive charge is shared by another carbon atom. In each case, state whether your second resonance form is a more important or less important resonance contributor than the first structure. (Which structure places the positive charge on the more­ substituted carbon atom?) P R O B L E M 1 5 -4

When 3-bromo- I-methylcyclohexene undergoes solvolysis i n hot ethanol, two products are formed. Propose a mechanism that accounts for both of these products. CH3 Br CH3 0CH2CH3 CH3 CH3CH)OH ) + OCH 2CH3 heat P R O BLE M 1 5 - 5

0

0

a

We can represent a delocalized ion such as the allyl cation either by resonance forms, as shown on the left in the following figure, or by a combined structure, as shown on the right. Although the combined structure is more concise, it is sometimes confusing because it attempts to convey all the information provided by two or more resonance forms.

[ �

�

H2C = C - CH2 1

2

3 +

resonance forms

� ]

+ H2C - C = CH2 1

2

3

or

�

I+ I+ H2C = C = CH2

combined representation I

3

Because of its resonance stabilization, the (primary) allyl cation is about as stable as a simple secondary carbocation, such as the isopropyl cation. Substituted allylic cations generally have at least one secondary carbon atom bearing part of the positive charge; they are about as stable as simple tertiary carbocations such as the t-butyl cation. H3C+ < 1 ° < 2°, allyl < 3°, substituted allylic

Stability of carbocations

�+

*+

H;C ==CH ==C H2 8+

8+

CH3 -CH == CH == CH2

1 5-5

+

is about as stable as CH3-CH-CH3 / CH3 is about as stable as CH3 - C "--+ CH3

Electrophilic additions to conjugated dienes usually involve allylic cations as intermedi­ ates. Unlike simple carbocations, an allylic cation can react with a nucleophile at either 1 ,2- and 1 ,4-Addition of its positive centers. Let's consider the addition of HBr to 1 ,3-butadiene, an elec­ to Conjugated Dienes trophilic addition that produces a mixture of two constitutional isomers. One product, 3-bromo- l -butene, results from Markovnikov addition across one of the double bonds. In the other product, I -bromo-2-butene, the double bond shifts to the C2 - C3 position.

1 5-5 1 ,2- and l ,4-Addition to Conjugated Dienes

3-bromo- 1 -butene l ,2-addition

I -bromo-2-butene l ,4-addition

The first product results from electrophilic addition of HBr across a double bond. This process is called a l,2-addition whether or not these two carbon atoms are numbered I and 2 in naming the compound. In the second product, the proton and bro­ mide ion add at the ends of the conjugated system to carbon atoms with a 1 ,4-relation­ ship. S uch an addition is called a l,4-addition whether or not these carbon atoms are numbered 1 and 4 in naming the compound.

'"

/

1

C=C-C =C

1

/

'"

21

1

- C-C-C=C

I I

A -B

1

A

1

B

/

'"

+

1 ,2-addition

l ,4-addition

The mechanism is similar to other electrophilic additions to alkenes. The proton is the electrophile, adding to the alkene to give the most stable carbocation. Protonation of 1 ,3butadiene gives an allylic cation, which is stabilized by resonance delocalization of the positive charge over two carbon atoms. Bromide can attack this resonance-stabilized in­ termediate at either of the two carbon atoms sharing the positive charge. Attack at the secondary carbon gives 1 ,2-addition; attack at the primary carbon gives l ,4-addition. 1 ,2- a nd 1 ,4-Ad d itien t o a C(£)I'lj ugated Diene

M E C H A N I S M 1 5- 1

Step 1: Protonation of one of the double bonds forms a resonance-stabilized allylic cation. Br---J- H

� H

/

C=C H

/

'"

/

H

C=C

/

H

'"

H

H

H / H H - C - C+ / '" 1 C=C H / '" H H

1

ally lie cation

Step 2: A nucleophile attacks at either electrophilic carbon atom. H

H3C - C + '"

1

H

/

C =C

/

'"

H

: BI : �

H

1

C=C

H3C -C-Br

1 ,4

H

/

'"

l ,2-addition

/

'"

H

H

The key to formation of these two products is the presence of a double bond in position to form a stabilized allylic cation. Molecules having such double bonds are likely to react via resonance-stabilized intermediates. Treatment of an alkyl halide with alcoholic AgN03 often promotes ionization. + Ag + R - CI � AgCl + R+ PRO B L E M 15-6

When 3-chloro- l -methylcyclopentene reacts with AgN03 in ethanol, two isomeric ethers are formed. Suggest structures, and propose a mechanism for their formation.

H

1

/

H

H - C -C H / ,\1 C - C+ H / '" H H

H

and

1 H3C - C,\H

/

H

1 1

C-C-Br H

1 ,4-addition

671

672

C hapter 1 5:C onjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy

Propose a mechanism for each reaction, showing explicitly how the observed mixtures of products are formed. (a) 3-methyl-2-buten- l -01 + HBr -7 I -bromo-3-methyl-2-butene + 3-bromo-3-methyl- l -butene (b) 2-methyl-3-buten-2-01 + HBr -7 I-bromo-3-methyl-2-butene + 3-bromo-3-methyl- l -butene (c) 1 ,3-butadiene + Br2 -7 3,4-dibromo- l -butene + 1 ,4-dibromo-2-butene (d) l -chloro-2-butene + AgN03 , H 20 -7 2-buten-I -ol + 3-buten-2-01 (e) 3-chloro- l -butene + AgN03 , H 2 0 -7 2-buten - l -ol + 3-buten-2-01 P R O B L E M 1 5 -7

1 5-6 Kinetic versus Therm odynamic Control in the Addition of H Br to 1 , 3-Butadiene

One of the interesting peculiarities of the reaction of 1 ,3-butadiene with HBr is the effect of temperature on the products. If the reagents are allowed to react briefly at - 80°C, the 1 ,2-addition product predominates. If this reaction mixture is later allowed to warm to 40° C, however, or if the original reaction is carried out at 40° C, the com­ position favors the l ,4-addition product. (80%)

H2C -CH -CH=C I-I2 I I H Br

( 1 ,2-product)

(20%)

H?C - - CH = CH- CH?I I Br H

( l ,4-product)

H2C -CH -CH=CH? I I H Br

( l ,2-product)

H2C - CH = CH- CH? I I Br H

( l ,4-product)

1400C

(85%)

This variation in product composition reminds us that the most stable product is not always the major product. Of the two products, we expect I -bromo-2-butene (the l ,4-product) to be more stable, since it has the more substituted double bond. This prediction is supported by the fact that this isomer predominates when the reaction mixture is warmed to 40°C and allowed to equilibrate. A reaction-energy diagram for the second step of this reaction (Figure 15-9) helps to show why one product is favored at low temperatures and another at higher tempera-

� Figure 15-9 Reaction-energy diagram for the second step of the addition of 1 ,3 butadiene to HBr. The allylic carbocation (center) can react at either of its electTophilic carbon atoms. The transition state (t) leading to I ,2-addition has a lower energy than that leading to the 1,4 product, so the 1 ,2-product is formed faster (kinetic product). The 1 ,2-product is not as stable as the 1 ,4-product, however. If equilibrium is reached, the 1 ,4-product predominates (thermodynamic product).

:1= ( l ,4)

:1= (1,2)

r1

CH3 -

--""'-- - -

"'H�.2

--------------

CH , - CH - CH = CH?

.

I

Br 1 ,2-product

�C

H2

intermediate +

(formed faster)

T

- - - - - - - - - - - ---"'�-

CH3 - CH = CH - CH 2B r

(more stable) 1 ,4-product

reaction coordinate

1 5-6 Kinetic versus Thermodynamic Control in the Addition of HBr to 1 ,3-Butadiene

tures. The allylic cation is in the center of the diagram; it can react toward the left to give the 1 ,2-product or toward the right to give the l A-product. The initial product depends on where bromide attacks the resonance-stabilized allylic cation. Bromide can attack at either of the two carbon atoms that share the positive charge. Attack at the sec­ ondary carbon gives 1 ,2-addition, and attack at the primary carbon gives 1 ,4-addition.

[

H +') H2C CH-CH=CH2

H3C-(C�-CH=CH2 +

: Br .. : -

I H3C-CH-CH = CH?

aLlack at secondary carbon

I Br

I ,2-addition product

+ H3C-CH=CH-C�2 ( Br .. :-

delocalized allylic cation

-

�

\ a ac \rimary carbon �

•

at

I Br

1 A-addition product

Thermodynamic Control at 40°C

At 40°C, a significant fraction of molecular colli­ sions have enough energy for reverse reactions to occur. Notice that the activation energy for the reverse of the 1 ,2-addition is less than that for the reverse of the 1 ,4-addition. Although the 1 ,2-product is still formed faster, it also reverts to the allylic cation faster than the l A-product does. At 40°C, an equilibrium is set up, and the relative energy of each species determines its concentration. The l A-product is the most stable species, and it predominates. Since thermodynamics determine the results, this situation is called thermodynamic control (or equilibrium control) of the reaction. The l A-product, favored under these conditions, is called the thermodynamic product. We will see many additional reactions whose products may be determined by kinetic control or by thermodynamic control, depending on the conditions. In general, reactions that do not reverse easily are kinetically controlled because an equilibrium is rarely established. In kinetically controlled reactions, the product with the lowest-energy transition state predominates. Reactions that are easily reversible are thermodynamical­ ly controlled unless something happens to prevent equilibrium from being attained. In thermodynamically controlled reactions, the lowest-energy product predominates. When Br2 is added to l ,3-butadiene at - 15°C, the product mixture contains 60% of product A and 40% of product B. When the same reaction takes place at 60°C, the product ratio is 1 0% A and 90% B. (a) Propose structures for products A and B . (Hint: In many cases, an allylic carbocation is more stable than a bromonium ion.) (b) Propose a mechanism to account for formation of both A and B. ( Continued)

1 5 -8

k

HoC-CH = CH -CH?

Kinetic Control at - 80°C The transition state for 1 ,2-addition has a lower energy than the transition state for l A-addition, giving the 1 ,2-addition a lower activation energy ( Ea ) . This is not surprising, because 1 ,2-addition results from bromide attack at the more substituted secondary carbon, which bears more of the positive charge because it is better stabilized than the primary carbon. Because the 1 ,2-addition has a lower activation ener­ gy than the l A-addition, the 1 ,2-addition takes place faster (at all temperatures). Attack by bromide on the allylic cation is a strongly exothermic process, so the reverse reaction has a large activation energy. At - 80°C, few collisions take place with this much energy, and the rate of the reverse reaction is practically zero. Under these conditions, the product that is formed faster predominates. Because the kinetics of the reaction determine the results, this situation is called kinetic control of the reaction. The 1 ,2-product, favored under these conditions, is called the kinetic product.

P R O BL E M

tt

]

673

Chapter

674

15:

Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy

(c) Show why A predominates at - 15°C, but B predominates at 60°C. (d) If you had a solution of pure A, and its temperature was raised to 60°C, what would you expect to happen? Propose a mechanism to support your prediction.

Like allylic cations, aUylic radi cal s are s t ab i li z ed b y resonance delocalizatio n . For example, the mechanism of free-radical bromination of cyclohexene is shown next. Substitution occurs entirely at the alJylic position, where abstraction of a hydrogen gives a resonance-stabilized allylic radical as the i ntermedi ate .

1 5-7 Allylic Radicals

Free-Rad ica l Al lylic B ro m i nation

M EC H A N I S M 1 5-2

Initiation:

Formation of radicals.

Propagation:

hlJ �

2 Br '

Each step consumes a radical and forms another radical leading to products.

First Propagation Step: The bromine radical abstracts an aUylic hydrogen to produce an allylic radical. H H

¢c: H

+

Br'

�

H

allyJic hydrogens

[ye: err:] �

Second Propagation Step: The aUylic radical i n turn reacts

+

RBr

an allylic radical

with a bromine molecule to form an allyl bromide and a new

bromine atom, which continues the chain.

[ye: err:] �

Regeneration of Br2: NBS

aUylic radical

+

Br2

Q(

�

H Br

allylic bromide

+

Br'

continues the chain

reacts with HBr to regenerate the molecule of bromjne used in the allylic bromination step.

HBr

+

�

N - Br

o

N-bromosllccinimide (NBS)

Stabil ity of Allylic Radicals

+

�

N-H

o

sllccinimide

Why is it that (in the first propagation step) a bromine radical abstracts only an allylic hydrogen atom, and not one from another secondary site? Abstraction of allylic hydrogens is preferred because the allylic free radical that results is resonance-stabilized. The bond-dissociation enthalpies required to generate several free radicals are compared next. Notice that the allyl

1 5-7 Allylic Radicals

radical (a primary free radical) is actually 1 3 kJ/mol (3 kcal/mol) more stable than the tertiary butyl radical. CH3CH2 - H ( CH3hCH - H (CH3 ) 3 C - H H2C = CH - CH2 - H

� � � �

CH3CH2• (CH3hCH· ( CH3 hC· H2C = CH - CH2•

+ + + +

H· H· H· H·

D.. H

D.. H

D.. H

D.. H

=

=

=

=

+4 1 0 k J + 393 k J +381 kJ +368 kJ

( +98 kcal) ( primary ) ( + 94 kcal ) ( secondary ) ( + 9 1 kcal) ( tertiary ) ( + 88 kcal ) ( allyl)

The allylic 2-cyclohexenyl radical has its unpaired electron delocalized over two secondary carbon atoms, so it is even more stable than the un substituted allyl radical. The second propagation step may occur at either of the radical carbons, but in this symmetrical case, either position gives 3-bromocyclohexene as the product. Less-symmetrical compounds often give mixtures of products resulting from an allylic shift: In the product, the double bond can appear at either of the positions it occupies in the resonance forms of the aUylic radical. An allylic shift in a radical reaction is similar to the l ,4-addition of an electrophilic reagent such as HBr to a diene (Section 1 5-5). The following propagation steps show how a mixture of products results from the free-radical allylic bromination of I -butene. +

CH3 - CH - CH = CH �

Br·

� �

�

[CH3 - CH-CH=CH2

�

]

CH3 - CH = CH - CH2 + HBr

resonance-stabilized allylic radical

i

i

CH3 -CH-CH =CH1- + CH3 - C H = CH -CH1 + B r · I I Br Br (mixture)

When methylenecyclohexane is treated with a low concentration of bromine under iuadiation by a sunlamp, two substitution products are formed.

P R O B L E M 15-9

+ methylenecyclohexane

two substitution products

+

HBr

(a) Propose structures for these two products. (b) Propose a mechanism to account for their formation.

Bromination Using NBS At higher concentrations, bromine adds across double bonds (via a bromonium ion) to give saturated dibromides (Section 8-8). In the allylic bromination just shown, bromine substitutes for a hydrogen atom. The key to getting substitution is to have a low concentration of bromine, together with light or free radi­ cals to initiate the reaction. Free radicals are highly reactive, and even a small concen­ tration of radicals can produce a fast chain reaction. S imply adding bromine might make the concentration too high, resulting i n ionic addition o f bromine across the double bond. A convenient bromine source for allylic bromination is N-bromosuccinimi de (NBS), a brominated deriv­ ative of succinimide. S uccinimide is a cyclic amide of the four-carbon diacid succinic acid.

675

676

Chapter 1 5 : Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy

� -H � N

N-B'

o

0

succinimide

succinic acid

N-bromosuccinimide (NBS)

NBS provides a fairly constant, low concentration of Br2 because it reacts with RBr liberated in the substitution, converting it back into Br2. This reaction also removes the RBr byproduct, preventing it from adding across the double bond by its own free-rad­ ical chain reaction. Step 1 :

Step 2:

Free-radical aUylic substitution (mechanism on p. 674)

R - R + Br2

hv

R - Br + RBr

NBS converts the RBr byproduct back i nto B r2

Q

�

0

c(-H 0

N-B,

+

RBr

-------7

0

+

Br2

0

NBS

succinimide

The NBS reaction is carried out in a clever way. The allylic compound is dis­ solved in carbon tetrachloride, and one equivalent of NBS is added. NBS is denser than CCl4 and not very soluble in it, so it sinks to the bottom of the CC14 solution. The reaction is initiated using a sunlamp for illumination or a radical initiator such as a per­ oxide. The NBS gradually appears to rise to the top of the CCl4 layer. It is actually converted to succinimide, which is less dense than CCI4. Once all the solid succin­ imide has risen to the top, the sunlamp is turned off, the solution is filtered to remove the succinimide, and the CCl4 is evaporated to recover the product. Devise a complete mechanism for the light-initiated reaction of I-hexene with NBS in carbon tetrachloride solution.

PRO BLEM 1 5- 1 0

Predict the product(s) of light-initiated reaction with NBS in CCl4 for the following start­ ing materials.

PRO BLEM 1 5- 1 1

(a)

1 5-8 Molecular Orbitals of the Allylic System

cyc!opentene

(b) trans-2-pentene

toluene

Let's take a closer look at the electronic structure of allylic systems, using the allyl radical as our example. One resonance form shows a pi bond between C2 and C3 with the radical electron on C l , and the other shows a pi bond between C l and C2 with the radical electron on C3. These two resonance forms imply that there is half a pi bond between C l and C2 and half a pi bond between C2 and C3, with the radical electron half on C l and half on C3.

1 5-8 M olecular Orbitals of the Al l yli c System

1t bonding

--------_._-_. -.-

677

.... Figure 1 5-1 0 Geometric structure of the allyl cation, allyl radical, and allyl anion.

H

12

/C� 3 /H 'C C

H'" I

H H I �C1 2� 3 H '" /

1 H

1 H

C 1

H

resonance forms

H I,

'" I -?," C � 3 /H o· C C o'

H

C' 1

f H

H

f H

combined representation

Remember that no resonance form has an independent existence: A compound has characteristics of all its resonance forms at the same time, but it does not "res­ onate" among them. The p orbitals of all three carbon atoms must be parallel to have pi bonding overlap simultaneously between C l and C2 and between C2 and C3. The geometric structure of the allyl system is shown in Figure 1 5- l O. The allyl cation, the allyl radical, and the allyl anion all have this same geometric structure, differing only in the number of pi electrons. Just as the four p orbitals of 1 ,3-butadiene overlap to form four molecular orbitals, the three atomic p orbitals of the allyl system overlap to form three molecular orbitals, shown in Figure 1 5- 1 1 . These three MOs share several important features with the MOs of the butadiene system. The first MO is entirely bonding, the second has one node, and the third has two nodes and (because it is the highest-energy MO) is entirely antibonding. As with butadiene, we expect that half of the MOs will be bonding, and half anti­ bonding; but with an odd number of MOs, they cannot be symmetrically divided. One anti bonding antibonding

antibonding 1t:; energy of isolated p orbital

2

nodes

nonbonding nonbonding

- nonbonding 1t 2

bonding 1t I

bonding

bonding

I node

0 nodes

PROBLEM-SOLVING

Htltp

In drawing pi MO's, several p orbitals

combine to give the same n umber of MOs: half bonding and half antibonding. If there are an odd number of MOs, the middle one is nonbonding. The lowest-energy MO has no nodes; each higher MO has one more node.

The highest-energy MO is entirely antibonding, with a node at each overlap. In a stable system, the bonding MOs are filled, and the antibonding MOs are empty.

electrons in the allyl radical

.... Figure 1 5-1 1 The three molecular orbitals of the allyl system. The lowest-energy MO ( 7T I ) has no nodes and is entirely bonding. The intermediate orbital ( 7T2 ) is nonbonding, having one symmetrical node that coincides with the center carbon atom. The highest-energy MO ( 7Tj') has two nodes and is entirely anti bonding. In the allyl radical, 7T I is filled. The unpaired electron is in 7T2, having its electron density entirely on C l and C3.

678

Chapter 1 5 : Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy

of the MOs must appear at the middle of the energy levels, neither bonding nor anti­ bonding: It is a nonbonding molecular orbital. Electrons in a nonbonding orbital have the same energy as in an isolated p orbital. The structure of the nonbonding orbital (7T2) may seem strange because there is zero electron density on the center p orbital (C2). This is the case because 7T2 must have one node, and the only symmetrical position for one node is in the center of the molecule, crossing C2. We can tell from its structure that 7T2 must be nonbonding, because C l and C3 both have zero overlap with C2. The total is zero bonding, or a non bonding orbital.

1 5-9 Electronic Configurations of the Allyl Radical, Cation, and Anion

The right-hand column of Figure 1 5 - 1 1 shows the electronic structure for the allyl rad­ ical, with three pi electrons in the lowest available molecular orbitals. Two electrons are in the all-bonding MO (7Tj ) , representing the pi bond shared between the C I - C2 bond and the C2 - C3 bond. The unpaired electron goes into 7T2 with zero electron density on the center carbon atom (C2). This MO representation agrees with the reso­ nance picture showing the radical electron shared equally by C l and C3, but not C2. Both the resonance and MO pictures successfully predict that the radical will react at either of the end carbon atoms, C l or C3. The electronic configuration of the ally I cation (Figure 1 5 - 1 2) differs from that of the allyl radical; i t lacks the unpaired electron in 7T2, which has half of its electron density on C l and half on C3. In effect, we have removed half an electron from each of C l and C3, while C2 remains unchanged. This MO picture is consistent with the resonance picture showing the positive charge shared by C l and C 3 . H I H H '" / C � / C +C I I H H

H I H' H " � C """ / C+ C I I H H

resonance forms

H I ,C ,

H" -I +" C .,Y 2 I H

�

/I

H

C -+ I 2 H

combined representation

Figure 1 5 - 1 2 also shows the electronic configuration of the allyl anion, which differs from the allyl radical in having an additional electron in 7T2, the nonbonding orbital

antibonding n

� Fig u re 1 5- 1 2 Comparison o f the electronic structure of the allyl radical with the allyl cation and the allyl anion. The allyl cation has no electron in 7T2, leavi n g half a positive charge on each of C l and C3. The allyl anion has two electrons in 7Tb giving half a negative charge to each of C 1 and C3.

energy of isolated p orbital

� K--+-T-+-�

allyl cation (2 n electrons)

3D 0 -

allyl radical allyl anion (3 n electrons) (4 n electrons)

n*

additional electron here - non bonding

bonding

n2

nl

1 5- 1 0 SN2 Displacement Reactions of Al ly lic Halides and Tosylates

679

with its electron density divided between Cl and C3. In agreement with the reso­ nance picture, this electron's negative charge is divided equally between C l and C3. H

H

I H H " """...... C � / -: C "c I I H

H

H

H

I C , /H " #' C:C� I I H

I H , / " 7, C � LC L C 2 I I 2

H

H

H

resonance forms

H

combined representation

The molecular orbital representation shows the allyl anion with a pair of electrons in 7T2, a nonbonding orbital. This picture i s also consistent with the resonance forms shown earlier, with a lone pair of nonbonding electrons evenly divided between C I and C3. PRO B L E M 1 5- 1 2 When I -bromo-2-butene is added to magnesium metal in dry ether, a Grignard reagent is formed. Addition of water to this Grignard reagent gives a mixture of I -butene and 2-butene (cis and trans). When the Grignard reagent is made using 3-bromo- l -butene, addition of water produces exactly the same mixture of products in the same ratios. Explain this curious result.

Allylic halides and tosylates show enhanced reactivity toward nucleophilic displace­ ment reactions by the SN2 mechanism. For example, allyl bromide reacts with nucle­ ophiles by the SN2 mechanism about 40 times faster than n-propyl bromide. Figure 15- 13 shows how this rate enhancement can be explained by allybc delo­ calization of electrons in the transition state. The transition state for the SN2 reaction looks like a trigonal carbon atom with a p orbital perpendicular to the three sub­ stituents. The electrons of the attacking nucleophile are forming a bond using one lobe of the p orbital while the leaving group's electrons are leaving from the other lobe.

1 5- 1 0 SN2 Displacement Reactions of A l l ylic H alides and Tosylates

S N 2 reaction on n-propyl bromide

transition state :j:

" C H

�

,,· H

Nuc

/

H

:Br: transition state

.... Fig u re 1 5-1 3 Allylic delocalization in the SN2 transition state. The transition state for the SN2 reaction of allyl bromidc with a nucIeophile is stabilized by conjugation of the double bond with the p orbital that is momentarily present on the reacting carbon atom. The resulting overlap lowers the energy of the transition state, increasing the reaction rate.

680

Chapter

15:

Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy When the substrate is allylic, the transition state receives resonance stabilization through conjugation with the p orbitals of the pi bond. This stabilization lowers the en­ ergy of the transition state, resulting in a lower activation energy and an enhanced rate. The enhanced reactivity of allylic halides and tosylates makes them particularly attractive as electrophiles for SN2 reactions. Allylic halides are so reactive that they couple with Grignard and organolithium reagents, a reaction that does not work well with unactivated halides.

H2C = CH - CH2Br + CH3 - ( CH2) 3 - Li allyl bromide

/1-

�

H2C = CH - CH2 - (CH2h - CH3 + LiBr l-heptene (85%)

butylithium

P R O B L E M 15 - 1 3 Show how you might synthesize the fol lowing compounds starting with alkyl or alkenyl halides containing four carbon atoms or fewer. (a) l -heptene (b) 5-methylhex-2-ene *(c) 5-decene

1 5-1 1

In 1 928, German chemists OUo Diels and Kurt Alder discovered that alkynes and alkenes with electron-withdrawing groups add to conjugated dienes to form six-membered rings. The Diels-Alder reaction has proven to be a useful synthetic tool, providing one of the best ways to make six-membered rings with diverse functionality and controlled stereo­ chemistry. Diels and Alder were awarded the Nobel Prize in 1 950 for their work. The Diels-Alder reaction is also called a [4 + 2] cycloaddition because a ring is formed by the interaction of four pi electrons in the diene with two pi electrons of the alkene or alkyne. Since the electron-poor alkene or alkyne is prone to react with a diene, it is called a dienophile ("lover of dienes"). In effect, the Diels-Alder reaction converts two pi bonds into two sigma bonds. We can symbolize the Diels-Alder reac­ tion by using three arrows to show the movement of three pairs of electrons. This elec­ tron movement is concerted, with three pairs of electrons moving simultaneously. The electron-withdrawing groups ( - W ) are usually carbonyl-containing (C = 0) groups o r cyano ( - C = N) groups. The Diels-Alder reaction is like a nucJeophile-electrophile reaction. The diene is electron-rich, and the dienophile is electron-poor. Simple dienes such as 1 ,3 -butadiene are sufficiently electron-rich to be effective dienes for the Diels-Alder reaction. The presence of electron-releasing groups such as alkyl groups or alkoxy ( - OR ) groups may further enhance the reactivity of the diene. Simple alkenes and alkynes such as ethene and ethyne are poor dienophiles, however. A good dienophile generally has one or more electron-withdrawing groups ( - W ) pulling electron density away from the pi bond. Dienophiles commonly have carbonyl-containing ( C = O ) groups or cyano ( - C - N) groups to enhance their Diels-Alder reactivity. Figure 1 5 - 1 4 shows some representative Diels-Alder reactions involving a variety of different dienes and dienophiles .

The Diels-Alder Reaction The precursor to vitamin D under­ goes a reverse cycloaddition reac­ tion (that is, the ring opens) in the skin

after exposure to the

sun's

ultraviolet rays. In cold climates, chil­ dren who have l ittle exposure to sunlight often cannot synthesize or consume enough vitamin D, and as a result they develop rickets. Fish liver oils are rich sources of vitamin D.

•

�

KEY M E C H AN I S M 1 5-3

The Diels-Alder Reaction

The Diels-Alder is a one-step, concerted mechanism. A diene reacts with an electron-poor alkene to give a new cyclohexene ri ng. H ",,- /W

6 �� �

Ll (heat)

)

C

H

diene electron-rich

/ ""-

H

dienophi le electron-poor

a cyclohexene ring

1 5 - 1 1 The Diels-Alder Reaction

681

A diene reacts with an electron-poor alkyne to give a cyclohexadiene. w

I

6 �ffi C:c-C H I

diene

EXAMPLES:

dienophile

a cyclohexa- l , 4-diene ring

0

(

0

� ex?

+

-----0>

COOCH3 C ( CCOOCH3 0

0

I

III

+

I

-----0>

COOCH3 Q COOCH3 I

I

H3C �C""'-C/ H H3C�b1H=� H3CX H/C'---H H3C H 0C O�C /OCH3 3 �C/ 'OCH C C C-OCHl 0 CC O� ""'-OCH3 0 O CH30J: ��

---

-

-

-

_.. --. -�-��-��--�-

Diels-Alder adduct

dienophile

cliene

+

-

N

+

II

I

III

-----7

I

+

-- ---_.

N

I

-----7

I

\

I

II

\

!

o

.

.... Figure 1 5-14 Examples o f the Diels-Alder reaction. Electron-releasing substituents activate the diene; electron-withdrawing substituents activate the dien oph i le .

682

Chapter 1 5 : Conjugated S ystems, Orbital Symmetry, and Ultraviolet Spectroscopy

PROBLEM-SOLVING

HinZ:;

A Diels-Alder product always contains

PROBLEM 1 5- 1 4

two ends of the diene form new bonds to the ends of the dienophile. The center (formerly single) bond of

(

the diene becomes a double bond. The dienophile's double bond

b)

becomes a single bond (or its triple

� �

CH3

bond becomes a double bond) .

(d )

PROBLEM-SOLVING

HinZ:;

To deconstruct a Diels-Alder product,

groups. (If a single bond, the

a ( )

dienoph ile had a double bond; if double, the dienophile had a triple

0

I

C

+

III

C

O

I

�C "

(f) OCH 3

0, �

C H3 O

+ O

;:: �

a

0

o

I

C -CH,

(ly" II

the two double bonds of the diene dienophile.

(e)

O � /OCH3 C

(b)

(d)

C - OCH3

� C - OCH3 II

o

CHP cap

o

bond.) Break the two bonds that join the diene and dienophile, and restore and the double (or triple) bond of the

I

I

C

III

C

I

�C "

+

(

OCH 3

CN CN

What dienes and dienophiles would react to give the following Diels-Alder products?

center of what was the diene. Directly usually with electron-withdrawing

+

O � /OCH3 C

PRO B L E M 1 5- 1 5

look for the double bond at the across the ring is the dienophile bond,

¢o o

Predict the products of the following proposed Diels-Alder reactions.

one more ring than the reactants. The

D

(e)

CHP

I

C - OCH,CH,

D2� I

CN

CN

(e)

(\::y CN

lV H

0

(n UW H

0

1S- 11A Stereochem ical Req u i re ments of the Diels-Alder Tra n sition State

The mechanism of the Diels-Alder reaction is a concerted cyclic movement of six electrons: four in the diene and two in the dienophile. For the three pairs of electrons to move simultaneously, the transition state must have a geometry that allows overlap of the two end p orbitals of the diene with those of the dienophile. Figure 1 5- 1 5 shows the required geometry of the transition state. The geometry of the DieJs-Alder transi­ tion state explains why some isomers react differently from others, and it enables us to predict the stereochemistry of the products. Three stereochemical features of the Diels-Alder reaction are controlled by the requirements of the transition state: s-cis Conformation of the Diene

The diene must be in the s-cis conformation to react. When the diene is in the s-trans conformation, the end p orbitals are too far apart to overlap with the p orbitals of the dienophile. The s-trans conformation usually has a lower energy than the s-cis, but thi s energy difference is not enough to prevent most dienes from undergoing Diels-Alder reactions. For example, the s-trans conformation of butadiene is only 9.6 kllmol (2.3 kcal/mol) lower in energy than the s-cis conformation.

1 5- 1 1 The Diels-Alder Reaction :j:

"�V Q QY{Q (\ 6� � ,

C-C

H/

UH

� c-

overlap begins as these orbitals come together

reactants

/H "- H

H \

H-C

- , .. \J /H H ---. C

683

H

/H""H C

C-C---/\ H H WI

/D

C� H w./

U

product blue diene green dienophile red new bonds =

=

transition state

=

... Figure 15-15 The geometry of the Diels-Alder transition state. The Diels-Alder reaction has a concerted mechanism, with all the bond making and bond breaking occurring in a single step. Three pairs of electrons move simultaneously, requiring a transition state with overlap between the end p orbitals of the diene and those of the dienophile.

S-CIS

s-trans J 2 kJ/mol more stable

Structural features that aid or hinder the diene in achieving the s-cis conforma­ tion affect its ability to participate in Diels-Alder reactions. Figure 1 5- 1 6 shows that dienes with functional groups that hinder the s-cis conformation react more slowly than butadiene. Dienes with functional groups that hinder the s-trans conformation react faster than butadiene. Because cyclopentadiene is fixed in the s-cis conformation, it is highly reactive in the Diels-Alder reaction, It is so reactive, in fact, that at room temperature, cyclopentadiene slowly reacts with itself to form dicyclopentadiene, Cyclopentadiene

Diels-Alder rate compared with that of i,3-butadiene - slower

(no Diels-Alder)

similar to butadiene

faster

----+

o

.... Figure 15-16 Dienes that easily adopt the s-cis conformation undergo the DieJs-Alder reaction more readily.

684

Chapter 1 5 : Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy

is regenerated by heating the dimer above 200°C. At this temperature, the Diels-Alder reaction reverses, and the more volatile cyclopentadiene monomer distills over into a cold flask. The monomer can be stored indefinitely at dry-ice temperatures. H

at>

--

--

H

syn Stereochemistry The Diels-Alder reaction is a syn addition with respect to both the diene and the dienophile. The dienophile adds to one face of the diene, and the diene adds to one face of the dienophile. As you can see from the transition state in Figure 1 5 - 1 5 , there is no opportunity for any of the substituents to change their stereochemical positions during the course of the reaction. Substituents that are on the same side of the diene or dienophile will be cis on the newly formed ring. The following examples show the results of this syn addition.

( (

0

+

�

II

-OCH, -----?

C-OCH3 II 0

( +

Q

"''' C-OCH3 H 0II

[jC

cis (meso)

cis

CH3 I ' H � /H C I CH3

H 0 II " ' C-OCH,

COOCH3 I C III C I COOCH3

-----?

1

H

""

1

COOCH COOCH

CH3

c is (meso)

The E ndo Rule

:

(+

f

0

II

CHP-C II

-----?

0

( + CH3 I 'H

�

C I H

/CH3

'\

COOCH,

H COOCH3

[jC

trans (racemic)

trans

COOCH} I C III C I COOCH3

a B

-OCH,

-----?

1 ,. 1

H 3C

COOCH COOCH

H

trans (racemic)

:

When the dienophile has a pi bond in its electron-withdrawing group (as in a carbonyl group or a cyano group), the p orbitals in that electron-withdrawing group approach one of the central carbon atoms (C2 or C3) of the diene. This proximity results in secondary overlap: an overlap of the p orbitals of the electron-withdrawing group with the p orbitals of C2 and C3 of the diene (Figure 1 5- 1 7). Secondary overlap helps to stabilize the transition state. The influence of secondary overlap was first observed in reactions using cyclopentadiene to form bicyclic ring systems. In the bicyclic product (called norbornene), the electron-withdrawing substituent occupies the stereochemical position closest to the central atoms of the diene. This position is called the endo position because the substituent seems to be i nside the pocket formed by the six-membered ring of norbornene. This stereochemical preference for the electron-withdrawing substituent to appear i n the endo position is called the endo rule.

1 5- 1 1

The D iels-Alder Reaction

transition state

... Figure 1 5- 1 7 In most Die1s-Alder reactions, there i s secondary overlap between the p orbitals of the electron-withdrawing group and one of the central carbon atoms of the diene. Secondary overlap stabilizes the transition state, and it favors products having the electron-withdrawing groups i n endo positions.

� �

0

+

cxo

H

" / C

II

H

/C" /H H C

------?

II

0

endo exo endo

�o

tit; o =c " endo H

0

0

stereochemical positions of norbornene

+

------?

en do

H C=O

.-1'C P"

0

I

/0

o

The endo rule is useful for predicting the products of many types of Diels-Alder reactions, regardless of whether they use cyclopentadiene to form norbornene sys­ tems. The following examples show the use of the endo rule with other types of Diels-Alder reactions.

c

+

H

"'C

/H

/C '" /H C H II

II

0

------?

d::t�

O = C '"endo H

but not

d=+c:o H

685

686

Chapter 15: Conjugated S ystems, Orbital Symmetry, and Ultraviolet Spectroscopy

�o o

+

S O LV E D P R O B L E M

o

1 5-1

�

o

endo

/'m",'"' "pl"'''g w',,, CH,

Use the endo rule to predict the product of the following cycloaddition.

�+l OCH3

C - CH,

II

0

SOLUTION Imagine this diene to be a substituted cyclopentadiene; the endo product will be formed.

�� '+l OCH3

an

C - CH,

II

0

i m gi e CH2 replacing I-I ' s

endo product

(A

In the imaginary reaction, we replaced the two inside hydrogens with the rest of the cyclopentadiene ring. Now we put them back and have the actual product.

CH3 0 H \

I

H

/ ···· C -CH3 CHP H II '"

o

endo product PROBLEM

1 5- 1 6

Predict the major product for each proposed Diels-Alder reaction. Include stereochemistry where appropriate.

H

H

1 5-1 1 B

C -N "- / C

II

/ "­

C

H

Diels-Alder Reactions Using U nsym metrical Reagents

Even when the diene and dienophile are both unsymmetrically substituted, the Diels-Alder reaction usually gives a single product (or a maj or product) rather than a

1 5 - 1 1 The Diels-Alder Reaction

random mixture. We can usually predict the major product by considering how the substituents polarize the diene and the dienophile in their charge-separated resonance forms. If we then arrange the reactants to connect the most negatively charged carbon in the (electron-rich) diene with the most positively charged carbon in the (electron­ poor) dienophile, we can usually predict the correct orientation. The following exam­ ples show that an electron-donating substituent (D) on the diene and an electron-withdrawing substituent (W) on the dienophile usually show either a 1 ,2- or I ,4-relationship in the product. Formation of 1,4-product

l

w

D

�

U

but not W

D

l ,4-product

("I

�W l ,3-product

Predicting this product

dienophile

diene

Formation of 1,2-product

�

+

0

l

W

�

�w D

1 ,2-prodllct

charge-separated resonance forms

l ,4-product

D

but not

Ow

1 , 3-prodllct

Predicting this product

diene

dienophile

charge-separated resonance forms

1 ,2-product

687

688 Stable

Chapter 1 5 : Conj ugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy compounds

that

m im ic

these charge-separated structures are used to generate protein cata­ lysts or catalytic antibodies to cat­ a l yze

the

Diels-Alder

reaction.

These catalysts can be designed to produce the disfavored isomer.

In most cases, we don't even need to draw the charge-separated resonance forms to determine which orientation of the reactants is preferred. We can predict the major products of unsymmetrical Diels-Alder reactions simply by remembering that the electron-donating groups of the diene and the electron-withdrawing groups of the dienophile usually bear either a 1 ,2-relationship or a l ,4-relationship in the products, but not a l ,3-relationship. S O LV E D P R O B L E M 1 5 - 2

OCH3 ) ri CN � +I

Predict the products of the following proposed D iels-Alder reactions.

(a)

�

CH� 3

(b)

S O LUTI O N (a) The methyl group i s weakly electron-donating to the diene, and the carbonyl group is electron-withdrawing from the dienophile. The two possible orientations place these groups in a l ,4-relationship or a 1 ,3-relationship. We select the l ,4-relationship for our predicted product. (Experimental results show a 70:30 preference for the l ,4-product. )

CI ,,-CH; o

D

CH3

1 ,4-relationship (major) (70% )

(-C=N)(-OCH3)

1 ,3-relationship (rillnor) (3 0%)

(b) The methoxyl group i s strongly electron-donating to the diene, and the cyano group is electron-withdrawing from the dienophile. Depending on the orientation of addition, the product has either a 1 ,2- or a 1 ,3-relationship of these two groups. We select the 1,2-relationship, and the endo rule predicts cis stereochem­ istry of the two substituents.

CH'

1 ,2-relationship (product)

�

H , VCN

1 ,3-relationship (not formed)

PROBLEM 1 5- 1 7 In Solved Problem 1 5-2, we simply predicted that the products would have a 1 ,2- or l ,4-rela­ tionship of the proper substituents. Draw the charge-separated resonance forms of the reactants to support these predictions.

PRO BLEM 1 5-18

COOCH3 CIII CI H

'O

Predict the products of the following Diels-Alder reactions.

(a)

CHP D

+

CHP '

(b)

�

+

Q o

1 5- 1 2 The Diels-Alder as an Example of a Pericyclic Reaction

CR3

"' (d)

) A

CR3

+

l

CN

To understand why the Diels-Alder reaction takes place, we must consider the molecu­ lar orbitals involved. The Diels-Alder is a cycloaddition: Two molecules combine in a one-step, concerted reaction to form a new ring. Cycloadditions such as the Diels-Alder are one class of pericyclic reactions, which involve the concerted forming and breaking of bonds within a closed ring of interacting orbitals. Figure IS- I S (page 683) shows the closed loop of interacting orbitals i n the Diels-Alder transition state. Each carbon atom of the new ring has one orbital involved in this closed loop. A concerted peri cyclic reaction has a single transition state, whose activation en­ ergy may be supplied by heat (thermal induction) or by ultraviolet light (photochemi­ cal induction). Some pericyclic reactions proceed only under thermal induction, and others proceed only under photochemical induction. Some pericyclic reactions take place under both thermal and photochemical conditions, but the two sets of conditions give different products. For many years, pericyclic reactions were poorly understood and unpredictable. Around 1 965, Robert B. Woodward and Roald Hoffmann developed a theory for predict­ ing the results of pericyclic reactions by considering the symmetry of the molecular orbitals of the reactants and products. Their theory, called conservation of orbital sym­ metry, says that the MOs of the reactants must flow smoothly into the MOs of the prod­ ucts without any drastic changes in symmetry. In that case, there will be bonding interactions to help stabilize the transition state. Without these bonding interactions in the transition state, the concerted cyclic reaction cannot occur. Conservation of symmetry has been used to develop "rules" to predict which pericyclic reactions are feasible and what products will result. These rules are often called the Woodward-Hoffmann rules. 1S-12A

689

Conservation of Orbital Symmetry i n the Diels-Alder Reaction

We will not develop all the Woodward-Hoffmann rules, but we will show how the molecular orbitals can indicate whether a cycloaddition will take place. The simple Diels-Alder reaction of butadiene with ethylene serves as our first example. The molecular orbitals of butadiene and ethylene are represented in Figure 1 5 - 1 8 . B uta­ diene, with four atomic p orbitals, has four molecular orbitals: two bonding MOs (filled) and two antibonding MOs (vacant). Ethylene, with two atomic p orbitals, has two MOs: a bonding MO (filled) and an antibonding MO (vacant). In the Diels-Alder reaction, the diene acts as the electron-rich nucleophile, and the dienophile acts as the electron-poor electrophile. If we imagine the diene con­ tributing a pair of electrons to the dienophile, the highest-energy electrons of the diene require the least activation energy for such a donation. The electrons in the highest-energy occupied orbital, called the Highest Occupied Molecular Orbital (HOMO), are the important ones because they are the most weakly held. The HOMO of butadiene is 1T 2 , and its symmetry determines the course of the reaction. The orbital in ethylene that receives these electrons is the lowest-energy orbital available, the Lowest Unoccupied Molecular Orbital (LUMO). In ethylene, the LUMO is the 1T * anti bonding orbital. If the electrons in the HOMO of butadiene can flow smoothly into the LUMO of ethylene, a concerted reaction can take place. Figure 1 5 - 1 9 shows that the HOMO of butadiene has the correct symmetry to overlap in phase with the LUMO of ethylene. Having the correct symmetry means the

15-12 The Diels-Alder as an Exam ple of a Pericyclic Reaction Pericyclic reactions are very useful for organic synthesis, but they are rarely seen i n nature. One excep­ tion

is

the

enzyme

chorismate

mutase. This enzyme plays a cen­ tral role in the formation of aro­ matic compounds in bacteria, fungi, and plants.

690

Chapter 1 5 : Conjugated S ystems, Orbital Symmetry, and Ultraviolet Spectroscopy

---�- -----

n*1 I LUMO

n�1 ILUMO

n21-H-1 HOMO

Molecular orbitals of butadiene and ethylene. � Figure 15-18

n I H I HOMO ethylene

butadiene

orbitals that form the new bonds can overlap constructively: plus with plus and minus with m i nus. These bonding interactions stabilize the transition state and promote the concerted reaction. This favorable result shows the reaction is symmetry-allowed. The Diels-Alder reaction is common, and this theory correctly predicts a favorable transition state. 1 5-1 2 B

The "Forbidden" [2

+

2] Cycloaddition

If a cycloaddition produces an overlap of positive-phase orbitals with negative­ phase orbitals (destructive overlap), antibonding interactions are generated. Anti­ bonding interactions raise the activation energy; thus the reaction is classified as symmetry-forbidden. The thermal [2 + 2] cycloaddition of two ethylenes to give cyclobutane is a symmetry-forbidden reaction. butadiene

HOMO

-���-�-- 1T* transition absorbs at a wavelength of 2 1 7 nm (540 kJ/mol) compared with 1 7 1 nm (686 kJ/mol) for ethylene. This longer wavelength (lower-energy) absorption results from a smaller energy difference between the HOMO and LUMO in butadiene than in ethylene.

hv

2 1 7 nm

[±]

------7

11 2

� Figure 1 5-24

EJ

IHjfOI

[±J

-

2 1 7 nm (540 kJ)

I ethylene I 11

ground state ,3-butadiene

I

excited state

absorption. The principal 7T � 1T* transition in 1 ,3,5-hexatriene occurs at 258 nm (452 kJ/mol, or 1 08 kcal/mol). We can summarize the effects of conjugation on the wavelength of UV absorp­ tion by stating a general rule: A compound that contains a longer chain of conjugated double bonds absorbs light at a longer wavelength. p-Carotene, which has 1 1 conju-

11 * 6

B

>-> bfj '-



ground excited state state 1 ,3,5-hexatriene

.--- --��.�--.�.-- , --�. ---- "-- ------

I

258 om (452 kJ)

-----

15-13

U l travio let Absorption Spectroscopy

gated double bonds in its pi system, absorbs at 454 nm, well into the visible region of the spectrum, corresponding to absorption of blue light. White light from which blue has been removed appears orange. f3-Carotene is the principal compound responsible for giving carrots their orange color. Carotene derivatives provide many of the colors we see in fruits, vegetables, and autumn leaves.

695

{3-carotene

Because they have no interaction with each other, isolated double bonds do not contribute to shifting the UV absorption to longer wavelengths. Both their reactions and their UV absorptions are like those of simple alkenes. For example, 1 ,4-penta­ diene absorbs at 1 78 nm, a value that is typical of simple alkenes rather than conjugat­ ed dienes.

1 5 - 1 3C

I -pentene, l 76 nm

isolated

1 ,4-pentadiene, 1 78 nm

conjugated

I

,3-pentadiene, 223 nm

Obta i n i n g an U ltraviolet Spectrum

To measure the ultraviolet (or UV-visible) spectrum of a compound, the sample is dis­ solved in a solvent (often ethanol) that does not absorb above 200 nm. The sample solution is placed in a quartz cell, and some of the solvent is placed in a reference cell. An ultraviolet spectrometer operates by comparing the amount of light transmitted through the sample (the sample beam) with the amount of light in the reference beam. The reference beam passes through the reference cell to compensate for any absorption of light by the cell and the solvent. The spectrometer (Figure 1 5 -26) has a source that emits all frequencies of UV light (above 200 nm). This light passes through a monochromator, which uses a dif­ fraction grating or a prism to spread the light into a spectrum and select one wave­ length. This single wavelength of light is split into two beams, with one beam passing through the sample cell and the other passing through the reference (solvent) cell. The detector continuously measures the intensity ratio of the reference beam (Ir) com­ pared with the sample beam (Is ) . As the spectrometer scans the wavelengths in the UV region, a printer draws a graph (called a spectrum) of the absorbance of the sample as a function of the wavelength. reference cell with solvent

source

�

\ �

monochromator

sample dissolved 1Il solvent

� �= /

--- - -- -

� - � �';�l �b�;� � = dete"",! I II

reference beam

I

printer plot of log (1/1) versus A

Carotene derivatives absorb different wavelengths of light, depending on the length of the conjugated system and the presence of other functional groups.

.... F i g u re 1 5-26 Diagram of an ultraviolet spectrometer. In the ultraviolet spectrometer, a monochromator selects one wavelength of light, which is split into two beams. One beam passes through the sample cell, while the other passes through the reference cell. The detector measures the ratio of the two beams, and the printer plots this ratio as a function of wavelength.

Chapter

696

15:

Conj ugated S ystems Orbital S ymmetry, and Ultraviolet Spectroscopy ,

The absorbance, A, of the sample at a particular wavelength is governed by Beer's law.

A = IOg where

(�)

= eel

c = sample concentration in moles per liter

I = path length of light through the cell in centimeters e = the molar absorptivity (or molar extinction coefficient) of the sample

The molar extinction coefficient

e,

associated with a wavelength of

maximum absorbance Amax, is par­ ticularly

useful

for

determining

drug concentrations. For example, the concentration of tetracycl ine is measured at 380 n m where the molar absorptivity va lue is 1 6,200.

Molar absorptivity (e) is a measure of how strongly the sample absorbs light at that wavelength. If the sample absorbs light at a particular wavelength, the sample beam (Is) is less intense than the reference beam (11')' and the ratio Ir/Is is greater than 1 . The ratio is equal to 1 when there is no absorption. The absorbance (the logarithm of the ratio) is therefore greater than zero when the sample absorbs, and is equal to zero when it does not. A UV spectrum is a plot of A, the absorbance of the sample, as a function of the wavelength. UV-visible spectra tend to show broad peaks and valleys. The spectral data that are most characteristic of a sample are as follows: 1. The wavelength(s) of maximum absorbance, called

2. The value of the molar absorptivity

A m ax

e at each maximum

Since UV-visible spectra are broad and lacking in detail, they are rarely printed as actual spectra. The spectral information is given as a list of the value or values of Amax together with the molar absorptivity for each value of Amax. The UV spectrum of isoprene (2-methyl- l ,3-butadiene) is shown in Figure 15-27. This spectrum could be summarized as follows: A max

e = 20,000

= 222 nm

The value of Amax is read directly from the spectrum, but the molar absorptivity e must be calculated from the concentration of the solution and the path length of the cell. For an isoprene concentration of 4 X 1 0-5 M and a l -cm cell, the molar absorptivity is found by rearranging Beer's law (A = eel).

e=

1 .0

0.9

0.8



24 h

[

Diels-Alder product

]

-C02

---7)

final product

16

Aromatic Compounds

In 1 825, Michael Faraday isolated a pure compound of boiling point 80°C from the oily mixture that condensed from illuminating gas, the fuel burned in gaslights. Elemental analysis showed an unusually small hydrogen-to-carbon ratio of 1 : 1 , corresponding to an empirical formula of CH. Faraday named the new compound "bicarburet of hydrogen." Eilhard Mitscherlich synthesized the same compound in 1 834 by heating benzoic acid, isolated from gum benzoin, in the presence of lime. Like Faraday, Mitscherlich found that the empirical formula was CH. He also used a vapor-density measurement to determine a molecular weight of about 78, for a molecular formula of C6H6. Since the new compound was derived from gum benzoin, he named it benzin, now called benzene. Many other compounds discovered in the nineteenth century seemed to be related to benzene. These compounds also had low hydrogen-to-carbon ratios as well as pleas­ ant aromas, and they could be converted to benzene or related compounds. This group of compounds was called aromatic because of their pleasant odors. Other organic com­ pounds, without these properties, were called aliphatic, meaning "fatlike." As the un­ usual stability of aromatic compounds was investigated, the term aromatic came to be applied to compounds with this stability, regardless of their odors. The Kekule S tructure

In 1 866, Friedrich Kekule proposed a cyclic structure for benzene with three double bonds. Considering that multiple bonds had been proposed only recently ( 1 859), the cyclic structure with alternating single and double bonds was considered somewhat bizarre. The KekuIe structure has its shortcomings, however. For example, it predicts two different 1 ,2-dichlorobenzenes, but only one is known to exist. Kekule suggested (incor­ rectly) that a fast equilibrium interconverts the two isomers of 1 ,2-dichlorobenzene.

(X� I

Cl

cc�

# Cl 1,2-dichlorobenzene

T he Resonance Rep resentation

H----.... -,. ...

16- 1

I ntroduction: The D iscovery of Benzene

16-2

The Structure a n d Properties of Benzene

Cl

Cl

The resonance picture of benzene is a natural extension of Kekule's hypothesis. In a Kekule structure, the C - C single bonds would be longer than the double bonds. Spectroscopic methods have shown that the benzene

705

706

Chapter 16: Aromatic Compounds

ring is planar and all the bonds are the same length 0 .397 A). Because the ring is pla­ nar and the carbon nuclei are positioned at equal distances, the two Kekule structures differ only in the positioning of the pi electrons. B enzene is actually a resonance hybrid of the two Kekule structures. This repre­ sentation implies that the pi electrons are delocalized, with a bond order of 11 between adjacent carbon atoms. The carbon-carbon bond lengths in benzene are shorter than typical single-bond lengths, yet longer than typical double-bond lengths.

[0

1.397

all C-C bond lengths A

0]

=

bond order It combined representation

resonance representation

)

butadiene

The resonance-delocalized picture explains most of the structural properties of benzene and its derivatives-the benzenoid aromatic compounds. Because the pi bonds are delocalized over the ring, we often inscribe a circle in the hexagon rather than draw three localized double bonds. This representation helps us remember there are no localized single or double bonds, and it prevents us from trying to draw suppos­ edly different isomers that differ only in the placement of double bonds in the ring. We often use Kekule structures in drawing reaction mechanisms, however, to show the movement of individual pairs of electrons. P R O B L E M 16-1 Write Lewis structures for the Kekule representations of benzene. Show all the valence electrons.

Friedrich August Kekule von Stradoniz ( 1 829-1 896), pictured on a Belgian postage stamp.

Using this resonance picture, we can draw a more realistic representation of ben­ zene (Figure 1 6- 1 ). Benzene is a ring of six sp 2 -hybrid carbon atoms, each bonded to one hydrogen atom. All the carbon--carbon bonds are the same length, and all the bond angles are exactly 1 200• Each sp 2 carbon atom has an unhybridized p orbital perpendicular to the plane of the ring, and six electrons occupy this circle of p orbitals. At this point, we can define an aromatic compound to be a cyclic compound containing some number of conjugated double bonds and having an unusually large resonance energy. Using benzene as the example, we will consider how aromatic com­ pounds differ from aliphatic compounds. Then we will discuss why an aromatic struc­ ture confers extra stability and how we can predict aromaticity in some interesting and unusual compounds.

1.397 ?120 A

'}t!

H

2 Benzene is a flat ring of sp -hybrid carbon atoms with their unhybridized p orbitals all aligned and overlapping. The ring of p orbitals contains six electrons. The carbon-carbon bond lengths are all 1 .397 A, and all the bond angles are exactly 1 20°. � Figure 1 6- 1

1 6-2 The Structure and Properties of Benzene

The U nusual Reactions of Benzene

Benzene is actually much more stable than we would expect from the simple resonance-delocalized picture. Both the Kekule structure and the resonance-delocalized picture show that benzene is a cyclic conju­ gated triene. We might expect benzene to undergo the typical reactions of polyenes. In fact, its reactions are quite unusual. For example, an alkene decolorizes potassium per­ manganate by reacting to form a glycol (Section 8- 1 4). The purple permanganate color disappears, and a precipitate of manganese dioxide forms. When permanganate is added to benzene, however, no reaction occurs.

a::

C(

H

+

H

Long-term exposure

by a decrease in the number of red blood cells and an increase in the number of malfunctioning white blood cells.

ex:

Br2

0

Br2

-----7

DNA in the bone marrow, altering

the production of red and white blood cells.

MnO,

H

0:

CCl4

Br

no reaction

Addition of a catalyst such as ferric bromide to the mixture of bromine and benzene causes the bromine color to disappear slowly. HBr gas is evolved as a byproduct, but the expected addition of Br2 does not take place. Instead, the organic product results from substitution of a bromine atom for a hydrogen, and all three double bonds are retained. H

H

¢ H

�

H

H

H

Br2. FeBr3

CCl4

H

H

¢

The U nusual S tab ility of Benzene

H

�

H

B' H

+

HBrt

H

H

H

g �

�

Br Br

H H is not formed

B enzene's reluctance to undergo typical alkene reactions suggests that it must be unusually stable. By comparing molar heats of hydrogenation, we can get a quantitative idea of its stability. Benzene, cyclohexene, and the cyclohexadienes all hydrogenate to form cyclohexane. Figure 1 6-2 shows how the experimentally determined heats of hydrogenation are used to compute the reso­ nance energies of 1 ,3-cyclohexadiene and benzene, based on the following reasoning: 1. Hydrogenation of cyclohexene is exothermic by 1 20 kllmol (28.6 kcal/mol).

2 . Hydrogenation of 1 ,4-cyclohexadiene is exothermic by 240 kllmol (57.4 kcal/mol),

about twice the heat of hydrogenation of cyclohexene. The resonance energy of the isolated double bonds in 1 ,4-cyclohexadiene is about zero.

Metabolites of ben­

zene may react with proteins and

Most alkenes decolorize solutions of bromine in carbon tetrachloride (Section 8- 10). The red bromine color disappears as bromine adds across the double bond. When bromine is added to benzene, no reaction occurs, and the red bromine color remains. CCl4

benzene

can cause leukemia, characterized

no reaction

-----7

to

707

708

Chapter 1 6: Aromatic Compounds

o

energy

� Figure 16-2

The molar heats of hydrogenation and the relative energies of cyciohexene, l ,4-cyciohexadiene, 1 ,3-cyciohexadiene, and benzene. The dashed lines represent the energies that would be predicted if every double bond had the same energy as the double bond in cyciohexene.

-240

o

o

energy

j

0]

p"d;,ted)

------ . (- 359 predicted)

1 5 1 kJ resonance energy

8 kl resonance

------ . .;>

kllmol

energy

-208

-232 kllmol

- 1 20 kllmol

kllmol

3. Hydrogenation of 1 ,3-cyclohexadiene is exothermic by 232 kJ/mol (55.4 kcal/mol),

about 8 kJ ( 1 .8 kcal) less than twice the value for cyclohexene. A resonance energy of 8 kJ 0 .8 kcal) is typical for a conjugated diene. 4. Hydrogenation of benzene requires higher pressures of hydrogen and a more active catalyst. This hydrogenation is exothermic by 2 08 kJ/mol (49.8 kcal/mol), about 1 5 1 kJ (36 . 0 kcal) less than 3 times the value for cyclohexene. fl.HO

o

catalyst

high pressure

208 kllmo! 359 kJ/mol 15 1 kllmol

3 X cyclohexene resonance energy

The huge 1 5 1 kJ/mol (36 kcal/mol) resonance energy of benzene cannot be explained by conjugation effects alone. The heat of hydrogenation for benzene is actually smaller than that for 1 ,3-cyclohexadiene. The hydrogenation of the first double bond of benzene is endothermic, the first endothermic hydrogenation we have encountered. In practice, this reaction is difficult to stop after the addition of 1 mole of H 2 because the product ( l ,3-cyclohexadiene) hydrogenates more easily than benzene itself. Clearly, the benzene ring is exceptionally unreactive. catalyst )

o

benzene: 1,3-cyclohexadiene:

fl.Ho

LlH �ydrogen H Cr) Huracil adenine pyrimidine H guanine Determi ne which nichtrogen rings atoms of thesearebases are. aromatic. Predi c t whi basi c Docarbonyl any ofgroup thesetobasesformhavea phenol easilyicformed tvae.ut)omers that are aromatic? (Consider moving a proton from nitrogen to a deri v at i Consider the following compound, which has been synthesized and characterized: purine bases

��

N

H,

N

o

N

I

o

N

I

N

cytosine

"

H2N

H

purine

( J

NHo

N

N

N

N

N

"

" H

(a) (b) (c)

* 16-44

(a) (b)

Assumi ngthithsismolecul moleculee synthesi is entirelyzedconjugat ed, t-butyl do yousubsti expecttuents? it to beWhyaromat imake c, antithearomati c, ortutednonaromati c? and Why was wi t h three not unsubsti compound nstead?the nitrogen atom to be basic? Explain. DoAtstudyroom youit iexpect temperat ure, the protonAs the tespectrum shows only twoto -lsinlOgloC,ets oftheratilarger o sigThenal broadens smaller siandgnalseparat remainess unchanged at all temperatures. mperat u re i s l o wered iseparat nto twoe sinewnglesitsnglofets,areasone on eitExplai her sidne what of thethorieseginal chemi cialndishicatft.e Atabout- l tlhOeoC,bondithenspect rthumis molconsieculstseof. Howthreedoes dat a g i n your concl uansiooldn based one lathebeled "thymol data"agree wistthockroom your predishelctif.oAftn inerpartnoti(a)? Alowistudent found bot t l on the ciwingtha plD2e0.asantPropose odor, shea strobtuctauirenedforthtehy­fol­ n g mass, and spectra. The peak at di s appears on shaki n g mol, andand show show why how theyourresulti structnugreioins iconsi th thee. spectra. Propose a fragmentation to explain the peak at s relasttievntelywistabl (c) (d)

1:2.

NMR

1: 1: 1.

16-45

NMR

NMR

IR,

NMR

135,

100

80

g'" 60
'0 c::

.3 40 '"

20 o

--

I

I

I

I

I I

I

I

I

20

.III .I 30

I

40

MS

I

I thymol I

I

IIII

50

I

.d,.

60

I

70

IIII

I

I

I I

I

80

I

90

m/z

1 5

II

I

I

I

I

04.8

I

I

I

-I I

10

NMR

I

d. I

.,11 I

100 1 1 0 1 20

M+ 1150)

130

I

1 40

150

160

mlz

16:

746

2,5

10

8 6 I'

Chapter Aromatic Compounds

. �

'I AN If � it l T R

I' iI',

l.t- T 4 I I; T

iJ,W AN

I

II� 'I

2

mli!

180

I

,; "

�

+' j

-'-

'

,I T

T

H

10

*1 6-46

I

'i I 1

j

,

'

I,

T

T

9

I

, I

II, i

-, i I ,

, "

11

i

'

,

I

I

T

'I' ' ''I,i

I I

LHlilit ' Ifli HH .. I

100

I

'''ymol

7

8

' '

I[+ +,H

_

120

:i

JJ,'

"

,

!!

, '

.

"

Iii

5

6

ill

J II I

l200

1000

40

I ""

I I i iii

," "

i! "

15 (ppm)

'" , ,

1:, 1 i '

i' 1

"

"

,

" "

II

IIII ,:UJ '

II

4

3

i IT h

1

!1

I

I

' "

' ..l _

o

T

e

"

!

I

I

·

•

,

,

i

600

20

CDCIJ I

'

#tt

itt

800

60

,

-H+�

1

80

1

-++1 ! I r

i + ,H-/-H+--I-H+ I ! ,

1

I

' ,. ,

,

-"-_

lJl Jll ! i II ill! i!11 2000 1800 1600 1400 wavenumber (em-I)

2500

"

I I

I W

'rt'i

I,

ItH

L

ilil I'

ii Ll

140

,

I"

, ', '" � "

thymol

I

, ,

I

II

160

,

I

I

lUll 1.1 1 ill! 3000

3500

200

I

i

' I' 1,1

4000

wavelength (/Lm)

i, " ' t::±:�Ji

" '

\l

j' i

,

, i j j

I,

2

!

!I'M ," j i i "

,-;.

o

An unknown compound gia,vesShowthe thefollowing mass, IR,thatandgiNMR spect ra, Propose a atstructure,127andandshowinhowtheit is consi s tent wi t h t h e spect r fragmentations v e t h e promjnent peaks mass spectrum, 155

mlz

I

100

80

g 60 Q)

'" "0 c:

,

20

1---1'" I"

Iii

I I

I " -1, I t

----- -'- �l.-

.E 40 '"

I

I

I 1

1

,,'rtrh

i

I

�I--

-l, I

I�S

1127

I

I

MJ

I

170)

i

! i i

+-

I

I 1--

--

-- ---.-

,I ' ' ' ' ' ',I ' " ' ' I o ' ' ' ' ' ' 80 90 100 1 10 120 130 140 150 160 10 20 30 40 50 60 70

' 170 180

ll�

lo

! i

II' li!III!

I

II '

,

!

I

I

!

� I I' M il!'

60

I jill I Ii,! 111!illll II IIli i I !' 'I!II I! I, II I'

f I II � I, , E

, !A

20

,I, "

'I

I

o 11111 'I 4000

200

I

I "Ii I,

3000

160

180

iil'i 'II I

,',I "II!

o

I!

il I ! I!i'

I Ii

ill

�!!I !! n i i

Ii

I'

jl 'I '. i !. , ill 1600 1800 2000 wavenumber (em -I)

Ill',

2500

747

\I

mrrm fH+ H*lm � : * ; H� I+tttl�* H+� H- I+tI�'�H�H*thl+ II I,

"I

3500

wavelength (fLm) ? 5 5.5 'j I 1 1' 1, III, !I!

4.5 illlll;; II 1I11 Ii!! I, II "I" ,

4

1 � '11 Ii r ': I ,' � , I'I '

80

40

!

II"

3,5

Study Problems

140

,

LOO

120

1400

I!I

il! 1200

80

60

I �+ ':;". il�-+'+�4r'�-HI-t��- I-�++- 1 I I I ! t 000 600 800

40

Offset: 40 ppm

II L

J

, ! '

o

20

:

:

i

1: �'',

:: 9

10 *16-47

8

7

6

1

,

,

.

II

I

5 o (ppm)

i

4

3

o

2

sp2

Hexahel icfleneat. Neverthel (adjacent)ess,seemshexahel a pooricenecandihasdatbeene forsynthesi opticalzacted iandvity separat becauseedalinl toitsenanti carbonomers. atomsItsareopticalhybri dtisonandis enor­ pre­ sumabl y rot a mous: [0']0 3700°. Explain why hexahelicene is optically active, and speculate as to why the rotation is so large. =

hexahelicene 16-48

Four shown. Thesewithcompounds more quiInckleach y, or case, they react favorablreactie equivity.librium con­ stants,compounds than similaarer compounds less conjugatreacted systems. explaiwin ttheh more enhanced

o o

(a)

6 0

i, more b"inh"

(b)

Cl U o

ionizes more readily than

0UCI

748

Chapter 16: Aromatic Compounds 0H � OH rY dehydrates under much milder conditions than V V Umbelliferone (7-hydroxycoumarin) is a common plant product that is used in sunscreen lotions. HO � HO � O y O i s more aci d i c t h an Gv V umbelliferone Hops areon,added durithengmajor the brewi ngonesof beerin hops, to proviis used de biasttear bacteri flavoriongstatiagentc agent s callthated thcane alsurvi pha-acive dautocl s, or ahumul ones.retain its Humul one of humul vi n g and bacteriostatic effect. Is humulon aromatic? OH (c)

(d)

16-49

o

o

16-50

HO

OH

humulon

The proton NMR chemi cal shi(ofnts theof thecarbons hydrogens in tpyri dtriogen) ne areareshown. These areto 88.typi60.cal aromat iec oxichemidizcinalg shiagentfts, except t h at t h e ort h o protons bonded o ni deshi e l d ed sui t abl (such d) canprotons add an oxygenfromatom88.to60pyritod88.19. ine to The give metpyriadprotons ine N-oxiarede.shiTheftedeffect of thisfrom added87.oxygen at87.o4m0.iassThetoa peroxyaci shiparaft tprotons he artho 25 to are shi f ted from 87. 6 4 t o 87. 3 2. Expl a i n thi s curi o us effect , shi f ti n g some protons upfield and others downfield. 0N " H oxidation N " H I I H H Hpyridine N-oxide H pyridine A

upfield upfield,

({t �

87.64

downfield

87.25

({t 1

88.60

)

+

88.19

�

87.40

87.32

17

Reactions of Aromatic Compounds EPM

of anisole

W

ith an understanding of the properties that make a compound aromatic, we now consider the reactions of aromatic compounds. A large part of this chapter is devoted to electrophilic aromatic substitution, the most impor­ tant mechanism involved in the reactions of aromatic compounds. Many reactions of benzene and its derivatives are explained by minor variations of electrophilic aromat­ ic substitution. We will study several of these reactions and then consider how sub­ stituents on the ring influence its reactivity toward electrophilic aromatic substitution and the regiochemistry seen in the products. We will also study other reactions of aro­ matic compounds, including nucleophilic aromatic substitution, addition reactions, reactions of side chains, and special reactions of phenols.

Like an alkene, benzene has clouds of pi electrons above and below its sigma bond framework. Although benzene's pi electrons are in a stable aromatic system, they are available to attack a strong electrophile to give a carbocation. This resonance-stabi­ lized carbocation is called a sigma complex because the electrophile is joined to the benzene ring by a new sigma bond. +

E

H

----?

17-1

Electrophilic Aromatic Substitution

H

E

H

sigma complex

The sigma complex (also called an arenium ion) is not aromatic because the sp3-hybrid carbon atom interrupts the ring of p orbitals. This loss of aromaticity con­ tributes to the highly endothermic nature of this first step. The sigma complex regains aromaticity either by a reversal of the first step (returning to the reactants) or by loss of the proton on the tetrahedral carbon atom, leading to the substitution product. The overall reaction is the substitution of an electrophile (E+ ) for a proton (H+ ) on the aromatic ring: electrophilic aromatic substitution. This class of reactions includes

substituted

749

Chapter 17: Reactions of Aromatic Compounds a

7 50

substitutions by wide variety of electrophilic reagents. Because it enables us to intro­ duce functional groups directly onto the aromatic ring, electrophilic aromatic substitution is the most important method for synthesis of substituted aromatic compounds .

:�:.'-�-=- KEY MECHANISM 17-1

E le ctro philic Aromatic Substitution

•...

Step

1:

H

Step 2:

* * �

Attack on the electrophile forms the ig a complex.

H

H

sm

H

H

H

E+

*

H H

�

H

+

H

H E

H

* H H

H

+-

H

H E

sigma complex (arenium ion) u u od

H

* H H

+ -

H H

H E

Loss of a proton regai ns rom tic ity and gives the s bsti t tion pr uct.

a a H

H

H "

\\ +

H

H

H� E

+

----;.

base- H

EXAMPLE: Iodination of toluene

Preliminary step: Formation of the electrophile, 1+ (the iodine cation).

Step

1:

Attack on the electropbile forms the sigma complex.

Step 2: Deprotonation regains aromaticity and gives the substitution product.

Q+ \

H

+

I

CH3

\ Hi):

�

Q \ /;

I

CH3

(plus other isomers)

PROBLEM 17-1

Sttheepsi2gmaof thcompl e iodienx.atiWeon ofdidbenzene shows wate possi er actbiinligtyasofawatbaseerandactiremovi nnucl g a prot onlefromand not consi d er t h n g as a e ophi atwoultackid noccur g the icarbocati on, easd asin aannucleophi electrophileliandc addiaddedtion ttoo tanhe alcarbocati kene. Draw the react ion tthihats f wat e r react o n. Expl a i n why type of addition is rarely observed.

17-2

Halogenation of Benzene

twirotphih benzene, lic aromatandic tshuebsftoirtumtiatoin.onBrBrofooBrmimin+nateisiiotdisnefflffiocilulsonottws. AtsshuterffoigenerngcieLewintalyl elmechani es actcidrosphiuscmhlicasfotorFeBrreleeacc­3t ion,eshowever Brwicatt+ahl.yaBrzesweakened omitheneredonatactBr-Br a pabondir of, byelandefctoarrmoparnsingttioaalFeBrcposomplit3iev, xefowicharmthirngBrge aon2 tshtoneratongerreofacttelsheemucctbrroohmiphiliknleee aandatweakomsHBr. Atbas, tandaecktorbyergeneremovebenzeneatianprgftoohrteomncatsftrhaoelmyssittg.hmae sicompl gma complex. Breox,migidveiniognthferoarmoFeBrmatic;pracotducts as Fonnation of a stronger electrophile. Fe � r eB ElecHtropbilic attack and formation of the sigmaHcomplex. H H i/ :¢t BrH H H:¢cH . :: 3 H* H ,:/ H / , : H HHH HHH HHH HHH Loss of a proton gives the prHoducts. H e H� H :¢c ,:/ H..--/ HHH H H l H Fe For m at i o n of t h e s i g ma compl e x i s r a t e l i m i t i n g, and t h e t r a ns i t i o n s t a t e l e adi n g tios sittroccupi es thehhiergmhesic tbecaus energey ipoit fonrtmons athnonare enerogmaty diicagrcaarmboc(Faitgiourn.eThe secondThis sstteepp o ngl y endot isklrelgaimo!ned(10and.8 kcala mol/meolcul). e of HBr is evolved. The overis exotal hrereacmtiicobecaus n is exote harerommaticibycityBenzene etaciontivpreoasductalskenes(Sect, iowhin ch reForact rexampl apidly ewi, cyclth broohexene mine atreroacomts ttoempergiveattruarnse t-iols,gi2not-vdeibasaddirormocycl exothermic by about 121 kllmol (29 kcal/mo\). ohexane. This reaction is -121kcalkJ ) ex: + Br, the JoThes ofanalaroomatgousicaddistabitiloitny.ofThebromiaddinetitoonbenzene is not siesen under normbecaus al circeumsit retaquincesres. 17-2

Bromination o f Benzene

Halogenation of Benzene

MECHANISM 17-2 Step

Bromination of Benzene

1:

:

B'r -:Eir : � ••

Br3

••

(

)

t B..r - ;.. - F rJl '

Br2 . FeBr3 i ntermediate (a stronger electrophile than Br2)

Step 2:

�

Br

Br

-

1 '-----"' :Br-Br-FeBr .. \..: ..

�

+

+

1

#

sigma complex

Step 3:

r

�

F Bri

Br

+)

+

�

+

bromobenzene

1 7- 1).

45

Comparison with Alkenes

8-8).

I:1HO

=

endothermic

(-29

HBr

B r3

+

#

+ FeBr i

751

752

17:

trraantes-iltiomnitstnagte HBr -PeBr4 reactants intermediate I _�����/_n� l reaction c o r d i n a t e

Chapter Reactions of Aromatic Compounds

+1

1

� Figure 17-1

The energy diofagram for theshows that bromi n ati o n benzene tratheef-lirstimistep iands endotthe hsecond ermic andstep t i n g is rong y exothermic.

st l

� V

prodBruc+tsHBreBr -=---_V +P 3

______________________ _

--_

Thetution is exotherofmibrc,obutmineitforerquia hydrres oagenLewiatosmacgiidvescatanalayrsotmatoticonver c productt br.oTheminseubstotia­ stronger electrophile. H H H J¢


QH E

A H V E

-0

0 "'- / +N

+N

+N

6

-0

0 " / �

QH E

especially ullstable

0 "'- ,I' +N

A H VE

H 0 V E

Para attack

-0

0 "'- /

0 " / +N

�

¢H E

In the sigma complex for meta substitution, the carbon bonded to the nitro group does not share the positive charge of the ring. This is a more stable situation because the positive charges are farther apart. As a result, nitrobenzene reacts primarily at the meta position. We can summarize by saying that the nitro group is a deactivating group and that it is a meta-director (or meta-allower). The energy diagram in Figure 1 7-3 compares the energies of the transition states and intermediates leading to ortho, meta, and para substitution of nitrobenzene with

17 -7

N02

ortho, para

6=

&H H Gf E

1

�

E

Deactivating, Meta-Directing Substituents

NO?

¢E H

� Figure 17-3 e g r es

En r yNip tofilrobenzene with aisdeacti vvatiatedng group. deacti toward licposiaromati sdeactititvuateliioeoctrophi nn is strongest tioatn,thecbutortho oins occurs ath thenparahemetarepopositioontiosofn,Reacti e ze e lowe ubs

and

reaction coordinate

---�.

those for benzene. Notice that a higher activation energy is involved for substitution of nitrobenzene at any position, resulting in slower reaction rates than for benzene. Just as activating substituents are all ortho, para-directors, most deactivating substituents are meta-directors. In general, deactivating substituents are groups with a positive charge (or a partial positive charge) on the atom bonded to the aromatic ring. As we saw with the nitro group, this positively charged atom repels any positive charge on the adj acent carbon atom of the ring. Of the possible sigma complexes, only the one corresponding to meta substitution avoids putting a positive charge on this ring carbon. For example, the partial positive charge on a carbonyl carbon allows substitu­ tion primarily at the meta position: Ortho attack

8- 0� / CH3

+ char(unfavorabl ges adjacent e)

�E (+)VC+) + charge here in other

,-------."

acetophenone

Meta attack

resonance forms

E

765

+

charge hereformsin other resonance Thithe sposisigtmaivecompl exondoesthe notringplace charge carbon bearing the carbonyl group.

The following summary table lists some common substituents that are deactivat­ ing and meta-directing. Resonance forms are also given to show how a positive charge arises on the atom bonded to the aromatic ring.

t a t

t

at any

si i n .

act

but it s b n n .

r

Chapter 17: Reactions of Aromatic Compounds

766

«

S U M MARY

Deactivating, Meta-Directors

Group

sulfonic acid

ro

l

cyano

.. . .

-S-O-H . 0II .

0

-N '0::' _

nitro

II

Resonance Forms

+,/'

:

�

[-C =N :

�

-: 0 : +

1

-S-O-H

.)�] �

-N 0:

�

. 0II .

�

Example

-C=N:� J +

o J

J

..

-S-O-H -: 0I :

ni trobenzene

..

benzenesulfonic acid

benzonitrile

o

II

-C-R

ketone o r aldehyde o

II

-C-O-R ester

[ "0' II

: 0 :I

..

acetophenone

..

-C-O-R

-C-Q-R

+

-C=O-RJ : 0 :-

I

.. +

methyl benzoate

+

-NR3

quaternary ammonium

trimethylanilinium iodide

PROBLEM 17-11

Ingivane 2,aqueous solutionlcontai ntirating osodin ofumanibilicnarbonate, aniverylinestreact scondi quickltioyns,withowever, h bromineandto 4 , 6 -t r i b romoani i n e. Ni e requi r es r ong the yiWhat elds condi (mostltiyonsm-niaretroanil iforne) niaretratipoor.on, and what form of aniline is present under these used tiinons?why nitration of aniline is so sluggish and why it gives mostly meta substitution. (b) condi Expl a Alacetani thoughlideni(PhNHCOCH3) tration of anilinegoesis sloquickl w andy giandvesgimostl y metasy substi tution,tutiniotratin. Useon of v es mostl para substi resonance forms to explain this difference in reactivity. (a)

*(c)

17-8

Halogen Substituents: Deactivating, but Ortho, Para-Directing

The halobenzenes are exceptions to the general rules. Halogens are deactivating groups, yet they are ortho, para-directors. We can explain this unusual combination of properties by considering that 1. the halogens are strongly electronegative, withdrawing electron density from a carbon atom through the sigma bond (inductive withdrawal), and

2. the halogens have nonbonding electrons that can donate electron density through pi bonding (resonance donation).

17-8

Halogen Substi uents:

These inductive and resonance effects oppose each other. The carbon-halogen bond (shown at right) is strongly polarized, with the carbon atom at the positive end of the dipole. This polarization draws electron density away from the benzene ring, mak­ ing it less reactive toward electrophilic substitution. If an electrophile reacts at the ortho or para position, however, the positive charge of the sigma complex is shared by the carbon atom bearing the halogen. The nonbonding electrons of the halogen can further delocalize the charge onto the halo­ gen, giving a halonium ion structure. This resonance stabilization allows a halogen to be pi-donating, even though it is sigma-withdrawing. o rtho

attack

+

ge

ch a r

Para attack here i n other

bromonium ion (plus other structures)

.. (Br:

C;E

-----7

-

I -f->

C

I

-

x less electron-rich

Meta attack

: :8.[+

Br

VE

(+)

)+

767

Deactivating, but Ortho, Para-Directing

I

H

I

6 E+ �

(+)

bromonium ion (plus other structures)

I�

�

+

Br

OE

( ) �(+)

H + H

no bromonium ion

Reaction at the meta position gives a sigma complex whose positive charge is not delocalized onto the halogen-bearing carbon atom. Therefore, the meta intermedi­ ate is not stabilized by the halonium ion structure. The following reaction illustrates the preference for ortho and para substitution in the nitration of chlorobenzene.

(yNO' HN03 6 H2SO4 6lNO2 ¢N02 CI

CI

CI

+

chlorobenzene

CI

+

meta

altho

( 1 %)

(35%)

para

(64%)

Figure 1 7-4 shows the effect of the halogen atom graphically, with an energy diagram comparing energies of the transition states and intermediates for electrophilic attack on chlorobenzene and benzene. Higher energies are required for the reactions of chlorobenzene, especially for attack at the meta position.

PROBLEM 1 7- 1 2

Draw almeta, l theandresonance forms ofPoithent sioutgmawhycompl enxtermedi for nitaratite foron ofmetabromobenzene ats ltheess ortho, para posi t i o ns. the i substi t uti o n i stable than the other two. Predict thea mechani structuresmofwitheth product formed when addsyourtoprediI-bromocycl ohexene. (b) Propose resonance forms t o support c ti o n. Expl n how thic ris npredi on anaiaromati g. ction is in accord with the ortho, para-directing effect of bromine PROBLEM 1 7- 1 3

(a)

(c)

Hel

PROBL EM-SOLVING

H?nv

Remember which substituents are

activating and which are deactivating. Activators are ortho, para-directing, and deactivators are meta-directing, except for the halogens.

Chapter 17: Reactions of Aromatic Compounds

768

1

� Figure 17-4

Energytuents. profilThees wienergi th haleosgenof the substi ingteherrmediforachltesoandrobenzene transitiothnanstafortes are hibenzene. Thetutihiognhestat theenergy results from substi meta positsubsti ion; thetutienergi esslforightlortyhloowerand para o n are lization by the halbecause oniumofiostabi n structure. I

S U M MARY

reaction co rdinate u

Donors

-R alkyl

- NH2 - OH .

.

�)

- OR

..

- NHCOCH3

ortho, para-directing aryl (weak pi donor)

Halogens

-Br-I -F

-CI

Effects of Multiple Substituents on Electrophilic Aromatic Substitution

Carbonyls

- S 03H

0

II

-C-R

-C=N - N02 +

0

II

Other

- NR3

- C - OH 0

II

meta-directing

- C - OR

ACTIVATING

1 7-9

--_

Directing Effects of Su bstituents

Donors

7T

meta

DEACTIVATING

Two or more substituents exert a combined effect on the reactivity of an aromatic ring. If the groups reinforce each other, the result is easy to predict. For example, we can predict that all the xylenes (dimethylbenzenes) are activated toward electrophilic sub­ stitution because the two methyl groups are both activating. In the case of a nitroben­ zoic acid, both substituents are deactivating, so we predict that a nitrobenzoic acid is by electrophile.

deactivated toward attack an

ac-xtiyvlaetned m-nidtreoabcetinvzaoteicdacid m-nottoloubivcioacuisd

1 7-9

Effects of Multiple Substituents on Electrophilic Aromatic Substitution

The orientation of addition is easily predicted in many cases. For example, in m-xylene there are two positions ortho to one of the methyl groups and para to the other. Electrophilic substitution occurs primarily at these two equivalent positions. There may be some substitution at the position between the two methyl groups (ortho to both), but this position is sterically hindered, and it is less reactive than the other two activated positions. In p-nitrotoluene, the methyl group directs an electrophile to­ ward its ortho positions. The nitro group directs toward the same locations because they are its meta positions. each is arrha to one CH3,

para to

other � �

th e

CH3 6: O

��==�

__

PROBLEM 1 7 - 1 4

CH3

�CH' N02

ortho to both

C H3' s, but hindered

m-xylene

major product

NO' � N02(99%) (65%)

Q p-nitroN02toluene

major product

Predio-nict thetrotolmononi tration products of the follom-chl wingocompounds. u ene rotoluene c acid p-methoxybenzoi a-bromobenzoi c aci d m-cresol (m-methylphenol)

(a) (c) (e)

(b) (d)

When the directing effects of two or more substituents conflict, it is more difficult to predict where an electrophile will react. In many cases, mixtures result. For example, a-xylene is activated at all the positions, so it gives mixtures of substitution products.

&CH' H2SO4 �CH' 6cCH'N02 a-xylene N02(58%) (42%) HN03

)

+

When there is a conflict between an activating group and a deactivating group, the activating group u sually directs the substitution. We can make an impor­ tant generalization: Activating groups are usually stronger directors than deactivating groups.

-OR, NR2

In fact, it is helpful to separate substituents into three classes, from strongest to weakest.

1. Powerful ortho, para-directors that stabilize the sigma complexes through reso­ nance. Examples are - OH and groups. ,

2. Moderate ortho, para-directors, such as alkyl groups and halogens.

769

770

All meta-directors. -OR, - R2 -R, - -C-R, -S03H, - N02 -OH, tarhee lsiukIbself tytiw.tuIoentnstuhbsienfttiohtuleentoswitrsonngedig rreercacclt tanaiosn,inprcomitheedomistnrognngerelateecst.grrIofophibotuplehprtareodwaeomiinrdtnhdiaetffessaemerandentcdirlaersace,cttimiosntxhsteuitreiness­, csuobsmitnitgutsiounbsoctitcuuentrs or. Thetho andmethparoxyla togrthoeupmetis haoxylstrongergroup.dirStecetroicrethffanecttshpre eniventtro grmuchoup,sandub­ stitution at the crowded position ortho to both the methoxyl group and the nitro group. SH20SO3 4 ) J ;5S0JH O'N�S03H 02N ONi2 )' m-nitroanisole major products

Chapter 17: Reactions of Aromatic Compounds 3.

N

>

o

II

>

X

+

PROBLEM-SOLVING

HinZ;-

To predict products of compounds

with mUltiple substituents, look for the most strongly activating substituent(s).

SOLV ED PROBLEM 17-1

.

Predict the major product(s) of bromination of p-chloroacetanilide. CI -0��. -C-CH3 H The ( -NHCOCH3) iselaestctrrongons actis bonded ivating andto thediraromat ecting igroup because tdheegroup nitro­ genis a atamistoronger mdewigroup thdiirtsectnonbondi n g pai r of c ri n g. The ami heananthalekoxyl chlorigroup, ne atom,theandamidsubst iatuparttionicoccurs most lyactativthateinposig group, tions ortandhothetoreact the amiion dgie.voesLir ktsome e i s ul a rl y st r ong of the dibrominated product. H H C-CH, CH' B' + CI p-chlCIoroacetanilide o

II

SOLU T I ON

¢ I

�

0

II

I

0

II

PROBLEM 17 - 1 5

Predip-methyl ct the mononit ration products of the followim-ningtraromat ic compounds. a ni s oi e ochl o robenzene m-nitroanisole p-chlorophenol < O(- NH- � -CH, CHJ -� -NH -o- � -NH' (Cisonsiactidveratitnhg,e standructthueresothoferthisesedeactigroups.vating.) CH 3 One o-methylacetanilide

(a) (c) (,j

o

(b) (d) (fj

0

0

-10

17 The isdetermi two benzene ringswhijoichnphenyl ed by ria nsignigls more e bond.actiThe si(toer oflesssubsti tutivated), on foranda bi(2)phenyl i s n ed by (1) v at e d deacti ch posirectitionnong. that ring is most reactive, using the fact that a phenyl substituent is ortho,whiUsepara-di resonance fOnTISng. of a sigma complex to show why a phenyl substituent should be ortho, para-di r ecti Predict the mononitration products of the following compounds.

Friedel-Crafts Alkylation

771

PRO BLEM 17- 1 6

Biphenyl

(a)

(b)

W

(0) (0) (0) (o�H (o( (0) (0) (0) (0) (U)

biphenyl

(;v)

ring, first decide which ring is more a ctivated (or less deactivated). Then consider only that ring, and decide which position is most reactive.

N02

Carbocations are perhaps the most important electrophiles capable of substituting onto aromatic rings, because this substitution forms a new carbon-carbon bond. Reactions of carbocations with aromatic compounds were first studied in 1 877 by the French alkaloid chemist Charles Friedel and his American partner, James Crafts. In the presence of Lewis acid catalysts such as aluminum chloride ( AICI 3 ) or ferric chloride (FeCI 3 ) , alkyl halides were found to alkylate benzene to give alkylbenzenes. This useful reaction is called the Friedel-Crafts alkylation.

'"'-&' r:;;;:p

�

a

H

+

R-X

(X

CI,

=

Lewis acid (AlCI3, FeBr3, etc.)

Br, I)

� a

+

R

H-X

For example, aluminum chloride catalyzes the alkylation of benzene by t-butyl chlo­ ride. HCI gas is evolved.

0 benzene

+

CH3 I CH 3 - C - Cl I CH3

AlCl3

)

Qr

CH3 I C -CH,

o �H3

t-but(90%) ylbenzene

t-buty chloride I

+

HCI

This alkylation is a typical electrophilic aromatic substitution, with the t-butyl cation acting as the electrophile. The t-butyl cation is formed by reaction of t-butyl chloride with the catalyst, aluminum chloride. The t-butyl cation reacts with benzene to form a sigma complex. Loss of a proton gives the product, t-butylbenzene. The aluminum chloride catalyst is regenerated in the final step. Friedel-Crafts alkylations are used with a wide variety of primary, secondary, and tertiary alkyl halides. With secondary and tertiary halides, the reacting elec­ trophile is probably the carbocation. R-X (R is secondary or tertiary)

+

AICl3 (

)

reacting electrophile

HinZ;

for compounds with more than one

(v)

Friedel-Crafts alkylation

PROBLEM-SOLVING

When predicting substitution products

1 7- 1 0

T he Friedel-Crafts Al kylation

772

Chapter

17:

Reactions of Aromatic Compounds Friedel-Crafts Al kylatio n

MECHANISM 17-5

Friedel-Crafts alkylation is an electrophilic aromatic substitution in which an alkyl cation acts as the electrophile. EXAM PLE: Al kylation of benzene by the t-butyl cation.

Step 1:

Formation of a carbocation.

CH3 CI .� I 1 CH3 - Cl '--rC] ·· : + AI-Cl 1 CH3 Cl

t-butyl chloride

Step 2:

CH3 1 CH3 -C + 1 CH3

Cl 1 CI -Al=- Cl I Cl

+

t-butyl cation

Electrophilic attack forms a sigma complex.

sigma complex

Step 3:

Loss of a proton regenerates the aromatic ring and gives the alkylated product.

With primary alkyl halides, the free primary carbocation is too unstable. The ac­ tual electrophile is a complex of aluminum chloride with the alkyl halide. In this com­ plex, the carbon-halogen bond is weakened (as indicated by dashed lines) and there is considerable positive charge on the carbon atom. The mechanism for the aluminum chloride-catalyzed reaction of ethyl chloride with benzene is as follows:

CH3 -CH2- c]

O

�

+

AlCl3

8+ 8CH3 -CH2 -- -Cl- - -AICl3

CH3 8(2 8+ I H ' � � �CI ��� AICI, Cl -AICI3

H

-

CI

17- 1 0

The Friedel-Crafts Alkylation

PROBLEM 17 - 17

Propose productsohexane (if any)wiandth benzene mechanisms formethyl the folchllowioringdeAIwiCtlh3-catal yezed reactions: chl3-chlorocycI ani s ol oro-2,2-dimethylbutane with isopropylbenzene

(a) (c)

(b)

Friedel-Crafts Alkylation U sing Other Carbocation S ources We have seen several ways of generating carbocations, and most of these can be used for Friedel-Crafts alkylations. Two common methods are protonation of alkenes and treatment of alcohols with BF3. Alkenes are protonated by HF to give carbocations. Fluoride ion is a weak nucleophile and does not immediately attack the carbocation. If benzene (or an activat­ ed benzene derivative) is present, electrophilic substitution occurs. The protonation step follows Markovnikov ' s rule, forming the more stable carbocation, which alkylates the aromatic ring.

.. :F: +

HF

Alcohol s are another source of carbocations for Friedel-Crafts alkylations. Alcohols commonly form carbocations when treated with Lewis acids such as boron trifluoride (BF3 ) ' If benzene (or an activated benzene derivative) is present, substitu­ tion may occur.

H-O-BF [)-H

Formation o/ the cation

..

Electrophilic substitution on benzene

3

F

.. 1 : F� ' � B - OH I

H-F

F

H

H

sigma complex

The BF3 used in this reaction is consumed and not regenerated. A ful l equivalent of the Lewis acid is needed, so we say that the reaction is promoted by BF3 rather than catalyzed by BF3. PRO BLEM 17-1 8

and

For each reaction,cycIshowohexene the generati on of the e1ectrophilet-butylpredialccoholt the products. benzene HF + BF3 B 3 t-butylbenzene 2-methylpropene + HF 2-propanol toluenebenzene (a) (c)

+

+

+

(b) (d)

+

+

+

F

773

774

Chapter 17:

Reactions of Aromatic

Compounds

Although the Friedel-Crafts alkyl­ ation looks good in principle, it has three major limitations that severely restrict its use.

Limitations of the Friedel-Crafts Alkylation Limitation 1

Friedel-Crafts reactions work only with benzene, activated benzene derivatives, and halobenzenes. They fail with strongly deactivated systems such as nitrobenzene, benzenesulfonic acid, and phenyl ketones. In some cases, we can get around this limitation by adding the deactivating group or changing an activating group into a deactivating group after the Friedel-Crafts step. S O LV ED PRO B L EM 17-2

Devise a synthesis of p-nitro-t-butylbenzene from benzene. Tot-butylmakebenzene. p-nitro-t-but yolnbenzene, wecorrect woulproduct. d first use wea Friwereedelto-Craft s nireact ion to make Ni t r at i gi v es t h e make t robenzene first, the Friedel-Crafts reaction to add the t-butyl group would faiH l. 3h C(C rh ¥ o (pi", ortho) N02 (reaction fails) (deactivated) SOLUTION

If

Good

HN03

H2S04

)

Bad

PROBLEM-SOLVING

Hi-ltv

Friedel-Crafts reactions fail with strongly deactivated systems.

PROBLEM-SOLVING

Hi-ltv

Alkyl carbocations for Friedel-Crafts alkylations are prone to rearrangements.

(CH3hC- CI

--------'»

AlCl3

2 Like other carbocation reactions, the Friedel-Crafts alkylation is sus­ ceptible to carbocation rearrangements. As a result, only certain alkylbenzenes can be made using the Friedel-Crafts alkylation. t-Butylbenzene, isopropylbenzene, and ethylbenzene can be synthesized using the Friedel-Crafts alkylation because the corresponding cations are not prone to rearrangement. Consider what happens, how­ ever, when we try to make n-propylbenzene by the Friedel-Crafts alkylation.

Limitation

Ionization with rearrangement gives isopropyl cation

H 15 I� o + - - Cl --- AlCl3 CH 3 -C-CH 2 -\..,.I H

+

CH 3 - C -CH 3 + I H

-

AICl4

Reaction with benzene gives isopropylbenzene

+ CH 3 -C-CH 3 I H

PROBLEM-SOLVING

-AICl4

Hi-ltv

Friedel-Crafts a l kylations are prone to mUltiple alkylation.

3 Because alkyl groups are activating substituents, the product of the Friedel-Crafts alkylation is more reactive than the starting material. Multiple alkylations are hard to avoid. This limitation can be severe. If we need to make ethylbenzene, we might try adding some AICl3 to a mixture of I mole of ethyl chloride and 1 mole of ben­ zene. As some ethylbenzene is formed, however, it is activated, reacting even faster than benzene itself. The product is a mixture of some (ortho and para) diethylbenzenes, some triethylbenzenes, a small amount of ethylbenzene, and some leftover benzene.

Limitation

0 I mole

+

CH3CH2 -CI

AI Cl 3

------7

1 mole

QC

H;

CH2CH3

& +

CH?CH3 CH'CH'

+

6

17-11 The Friedel-Crafts Acylation

triethylbenzenes

775

CH;

+ +

benzene

The problem of overalkylation can be avoided by using a large excess of ben­ zene. For example, if 1 mole of ethyl chloride is used with 50 moles of benzene, the concentration of ethylbenzene is always low, and the electrophile is more likely to react with benzene than with ethylbenzene. Distillation separates the product from excess benzene. This is a common industrial approach, since a continuous distillation can recycle the unreacted benzene. In the laboratory, we must often alkylate aromatic compounds that are more expensive than benzene. Because we cannot afford to use a large excess of the starting material, a more selective method is needed. Fortunately, the Friedel-Crafts acylation, discussed in Section 1 7- 1 1 , introduces j ust one group without danger of polyalkyla­ tion or rearrangement. P ROBLEM 1 7- 1 9

Predi(excess) ct the products (if any)isobutyl of thechlfololriodwie n+g react i3ons. benzene AI C l (b) (excess) tolniturobenzene ene + I-butanol +oropropane BF + AICl3 (excess) + 2-chl (excess) benzene + 3,3-dimethyl- I-butene + HF

(a)

+

(c) (d)

3

P ROBLEM 1 7 - 2 0

Whi l produce the desi rnedeachproduct inForgoodthe reacti yieldo?nsYouthatmaywilassume tehatgood alu­ miyienldumchofreacti chltheodesi riodnserisedwiadded as a catal y st i case. not gi v product, predict the major products. n-butylbenzene benzene n-butyl bromide (b) ethylbenzene + t-butyl chloride p-ethyl-t-butylbenzene bromobenzene + ethyl chloride p-bromoethylbenzene p-bromoethylbenzene ethylbenzene bromine anisole + methyl iodide (3 moles) 2,4,6-trimethylanisole Showp-t-butyl how younitrobenzene would synthesize the(b)folp-tol lowinuenesul g aromatfoniiccderiacivdatives from benzene. p-chlorotoluene a

Desired Product

Reagents

(a)

+

(c)

(d)

+

(e)

P ROBLEM 1 7-2 1

(a)

(c)

.

acyl group is a carbonyl group with an alkyl group ch d groups are named systematically by dropping the final -e from the alkane name and adding the -oyl suffix . Historical names are often used for the formyl group, the acetyl group, and the propionyl group, however.

An

atta e Acyl

0

II R -C -

0

II H-C(formyl)

acyl

g

roup

methanoyl

0

II CH3 - C -

et(ahcetanoylyl)

0

II CH3 CH2-C (propionyl) propanoyl

0

-� benzoyl

1 7- 1 1

T he Friedel-Crafts Acylation

776

Chapter

17:

Reactions of Aromatic Compounds An acyl chloride is an acyl group bonded to a chlorine atom. Acyl chlorides are made by reaction of the corresponding carboxylic acids with thionyl chloride. There­ fore, acyl chlorides are also called acid chlorides. We consider acyl chlorides in more detail when we study acid derivatives in Chapter 2 1 . o

(an acid chloride) an acyl chloride

acetyl chloride

o

o

II R - C - OH a carboxylic acid

o

-Cl

benzoyl chloride

+

I R - C -CI

II CI - S - Cl

+

-�

o

I R-C -CI

thionyl chloride

HCl i

a n acyl chloride

In the presence of aluminum chloride, an acyl chloride reacts with benzene (or an activated benzene derivative) to give a phenyl ketone: an acylbenzene. The Friedel-Crafts acylation is analogous to the Friedel-Crafts alkylation, except that the reagent is an acyl chloride instead of an alkyl halide and the product is an acylben­ zene (a "phenone") instead of an alkylbenzene. Friedel-Crafts acylation

OJ

+

benzene

benzene

1 7- 1 1 A

0

II R-C-CI acyl halide

'-

R

+

HCl

an acylbenzene

0

+

Q

AlCl 3

----i>

II (

(a phenyl ketone)

Example

0

0

II CH3 -C-Cl

AICl 3

----i>

acetyl chloride

IQr

0

II C '-

CH,

(95%)

acetylbenzene (acetophenone)

+

HCl

Mechan ism of Acylation

The mechanism of Friedel-Crafts acylation (shown next) resembles that for alkylation, except that the carbonyl group helps to stabilize the cationic intermediate. The acyl halide forms a complex with aluminum chloride; loss of the tetrachloroaluminate ion CAICI4 ) gives a resonance-stabilized acylium ion. The acylium ion is a strong electrophile. It reacts with benzene or an activated benzene derivative to form an acylbenzene.

MECHANISM 17-6

--------,

Friedel-Crafts Acylation

Friedel-Crafts acylation is an electrophilic aromatic substitution with an acylium ion acting as the electrophile.

Step 1: Formation of an " 0"

acylium ion.

II .� R -C -Cl : AICl3 acyl chloride

+

(

)

" "0

+ II R - C l Q-AICI3 ".. _

complex

-AICI4

+

+

.

+

[R- C = O:

R - C = O :] acylium ion

Step 2:

17-11

Electrophilic attack forms a sigma complex.

The Friedel-Crafts Acylation

o

I

�C """"' R (+ )�; sigma complex

Step 3:

Loss of a proton regenerates the aromatic system. o

'R : (;f Hp

O- Na.

dim(oetrhyl sulfate )

CH30SOPCH3

(licoraicnetfhlaovleoring)

CH3I)

-

All the alcohol-like reactions shown involve breaking of the phenolic 0 - R bond. This is a common way for phenols to react. It is far more difficult to break the C - 0 bond of a phenol, however. Most alcohol reactions in which the C 0 bond breaks are not possible with phenols. For example, phenols do not undergo acid­ catalyzed elimination or SN 2 back-side attack. Phenols also undergo reactions that are not possible with aliphatic alcohols. Let's consider some reactions that are peculiar to phenols.

1 7- 1 5A

Oxidation of Phenols to Q u i nones

Phenols undergo oxidation, but they give different types of products from those seen with aliphatic alcohols. Chromic acid oxidation of a phenol gives a conjugated 1 ,4diketone called a quinone. In the presence of air, many phenols slowly autoxidize to dark mixtures containing quinones. o

OR

6l m-cresol

CH3

(l 2 - m e t hy l - I , 4 - b e n z o q u i n o n e o

CH'

Rydroquinone ( l ,4-benzenediol) is easily oxidized because it already has two oxygen atoms bonded to the ring. Even very weak oxidants like silver bromide (AgBr) can oxidize hydroquinone. Silver bromide is reduced to black metallic silver in a light­ sensitive reaction: Any grains of silver bromide that have been exposed to light ( AgBr* ) react faster than unexposed grains.

+ 9

Large quantities of acetaminophen can be toxic because the body trans­ forms it to a benzoquinoneimine (a nitrogen replaces one oxygen of benzoquinone). This highly reactive metabolite causes extensive liver damage and can be deadly.

o

OR

¢ ¢ aceta m i n o p h e n b e n z o q u i n o n e i m i n e [0]

----,">

HNCOCR3

NH

(lh,4y-dbreonqzueineodnieol) OR

+ + ¢ o

OR

2 AgBr*

(1 ,4-bqeuniznoqnueinone)

2 Ag -t

2 RBr

o

Black-and-white photography is based on this reaction. A film containing small grains of silver bromide is exposed by a focused image. Where light strikes the film, the grains are activated. The film is then treated with a hydroquinone solution (the de­ veloper) to reduce the activated silver bromide grains, leaving black silver deposits where the film was exposed to light. The result is a negative image, with dark areas where light struck the film.

17-15

Reactions of

Phenols

793

PROBLEM 17-39

The(see photo). bombardiThiersbeetl e defends itselfbybythesprayi ng a hot quiyzednoneoxidsolatiuotinonoffrom its abdomen sol u tion i s formed enzyme-catal hydroqui none by hydrogen peroxide. Write a balanced equation for this oxidation. Quinones occur widely in nature, where they serve as biological oxidation­ reduction reagents. The uin ne (C o Q) is because it seems ubiquitous (found everywhere) in oxygen-consuming organisms. Coenzyme q

o

coenzyme Q

also called ubiquinone

Q serves as an oxidizing agent within the mitochondria of cells. The following reac­ tion shows the reduction of coenzyme Q by NADH (the reduced form of nicotinamide adenine dinucleotide), which becomes oxidized to NAD +.

�

+

coenzyme Q oxidized form

1 7-15 8

0

H

H

N I su ar I

�

NH

+

'

H+

CH,o

yY

CHP

�

CH, +

R

+

�

coenzyme Q reduced form

reduced form

� l_ _J

N�

N I su ar I

OH

NADH

0

H

OH

NAD+

oxidized form

E lectro p h i l ic Aromatic Substitution of Phenols

Phenols are highly reactive substrates for electrophilic aromatic substitution because the nonbonding electrons of the hydroxyl group stabilize the sigma complex formed by attack at the ortho or para position (Section 1 7-6B). Therefore, the hydroxyl group is strongly activating and artho, para-directing. Phenols are excellent substrates for halogenation, nitration, sulfonation, and some Friedel-Crafts reactions. Because they are highly reactive, phenols are usually alkylated or acylated using relatively weak Friedel-Crafts catalysts (such as HF) to avoid overalkylation or overacylation. OH

6

HF

------7

+

Phenoxide ions, easily generated by treating a phenol with sodium hydrox­ ide, are even more reactive than phenols toward electrophilic aromatic substitu­ tion. Because they are negatively charged, phenoxide ions react with positively charged electrophiles to give neutral sigma complexes whose structures resemble quinones. Be"

phenoxide ion

s

Be"

B,

h

y Br

igma complex

Whene tmihreatxesehydroqui ned, the bombardi eH202 r beetl n one and wihydroqui th enzymes. Peroxi de oxianddizteshe n one t o qui n one, strongl theirritsolatinuygtiexothermi olinquitodboisprays licnreact g. from Theionhottheheat, tips of the insect's abdomen. B'

Chapter 17 : Reactions of Aromatic Compounds

794

Phenoxi are so distroonglxide,y acta weak ivatedeltehctatrtophiheylundergo electrophiatiolinc aromat ic sub­de stiointuitsioanndwieinitdustohnscarbon e . The carboxyl of phenoxi shown on p. rial synthesis of salicylic acid, which is then converted to aspirin, as 79 1 .

&) "0"

'

I

'

�

0II C-OH �-OH

�

:0: 0 &:,0 . - 0II

�

salicylic acid

PROBLEM 1 7 - 4 0

Predict the products formed when m-cresol (m-methylphenol) reacts with II acet y l chl o ri d e, CH3-C-Cl NaOH and t h en ethyl bromi d e bromiumnediincluCCI'omat4 ine thein H2S04 dark excessequibromi sodi two valentsne iofn CClt-but4yilnchltheorilidgehtand AICI3 °

(b)

(a) (c) (e)

(d) (f)

PROBLEM 1 7 - 4 1

er dieonophi Benzoqui noneeisnea good Diels-Al1,d3-cycl 1,3-butadi hexadilee. nePredict the productfurans of its reaction with (a)

(c)

(b)

PROBLEM 1 7 - 4 2

Phenol reacts withWhen three equi vproduct alents ofisbromi netoinbromi CC14n(ine wattheer,dark)a yeltologiwvsole aidproduct ofeculfora­r mul a C6H30Br3' thi s added of mol formul aa C6H20Br 4 preciiopnit(much ates outliofke thethatsolofutiaoquin. Thenone)IRaround spectrum1680of thecm yellPropose ow precistpirtuc­ate shows st r ong absorpt tures for the two products. -1.

t

1.

SUMMARY

a.

Reactions of Aromatic Compounds

(Section

Electrophilic aromatic suhstitution. Halogenation. 17-2)

Br HBr (1 � +

bromobenzene b.

Nitration

(Section

17-3)

Nitration foLLowed by redu.ction gives anilin.es.

o

II N-O + (1 � nitrobenzene

+

H20

17-15

c.

Sulfonation

(Section

Reactions of Phenols

795

o

17-4) (

H2S04

H30+, heat

I

S-OH

or�

)

benze nesulfo nic a cid

d.

Friedel-Crafts alkylation

(Section

17-10)

t-bu ty lbe nze ne

e.

Friedel-Crafts acylation

©

+

(Section

�

( 1 ) AlCl3

)

(Section

l8J

C-CH2CH3

+

HCI

+

o

17-11C) CO,HCI

i

A1Cl Cu Cl

or

)

I

C-H

benza ldehyde

(Sections through -R, - QR, - QH, Q -NR2 (amines, amides) -Br, -I -N02, -S03H, -NR3, 17-5

17-9)

-

Activating, ortho, para-directing:

Deactivating, ortho, para-directing:

Nucleophilic aromatic substitution

+

(Section

JO(X G G a halobe nzene

:-,

-Cl,

Deactivating, meta-allowing:

2.

rcy

° II

pro pio pheno ne

o

Substituent effects

°

CH3CH2- -Cl

f. Gatterman-Koch synthesis

g.

17-11)

17-12) +

(0 = strong withdrawing group)

Nuc:strong nu cleophile

I

-C=O, -C

N

Nuc

rAr + x� G G ( Continued)

Chapter 17: Reactions of Aromatic Compounds

796

Example

If is not a strong electron-withdrawing group, severe conditions are required, and a benzyne mechanism is involved. a. (Section l7-13 A) G

3.

2 NH � + NaCl ey 022,N4-dinitroaniliN02 ne

Cl + NaNH2 � NO 02,42-diNnitrey 2 ochlorobenzene

Addition reactions Chlorination

heat and pressure ) or light

benzene b.

Catalytic hydrogenation

(Section 17-13 B)

ClH�ClH ClM Cl H Cl benzene hexachloride (BHC) H H,CH3 a� CH2 C H3 1 ,2-diethylcyclohexane

Ru or Rh catalyst)

100°C, 1000 psi

c.

o-diethylbenzene (Section l7-13C)

(mixture of cis and trans)

Birch reduction

Na orLi

4.

a.

ethylbenzene l-ethyl-l,4-cyclohexadiene (converts acylbenzenes to alkylbenzenes, Section 17-1lB)

Side-chain reactions The Clemmensen reduction

b. c.

0

HCl

+

3

phenol

17-60

bisphenol A

acetone

Reaction Temperature

Isomer of the Product

Oo(

1000(

43% 4% 53%

13% 8% 79%

enesul m-p-to-ttooollluuuenesul enesulfffonionionicccaciaciaciddd Expl acint what the change in the when producttheratproduct ios whenmixtheturetefrom mperattheurereactis increased. Predi wil l happen at O°Canyis necessary heated to 100°C. Devise a synthesis of the following compound, starting from toluene andionusing reagents. (a) (b)

17-61

H C JV When antofhformul raceneaisC20Hadded resul to thets.reactTheioproton n of chlNMR orobenzene witofh theconcent ratedshows NaOHa atsingl350°C, anareaint2erestarounding Diel s-Alader adduct spect r um product e t of 83 and broad singlereact t of earead as12a diene. around 87. Propose a structure for the product, and explain why one of the aromatic rings of anthracene In Chapter 14,by wethe sawpartithatal reactAgentionOrange contains (2,o4robenzene ,5-trichlorophenoxy) acethydroxi ic acid,de,calfollelodwed2,4,5by-T.reactThisiocompound is­ synthesized of 1, 2 , 4 , 5 t etrachl with sodi u m n wi t h sodi um chlDraworoacet atreu,ctCIuresCH2C02Na. the st of thinesethecompounds, and wrusedite equat ions forwasthese2,react i-tetrachl ons. orodibenzodioxin (2,3,7,8-TCDD), One of the impuri t ies Agent Orange in Vietnam 3 , 7 , 8 often y called "dioxin." Propose a mechanism to show how 2,3,7 ,8 -TCDD is formed in the synthesis of 2,Show4,5-T.incorrectl synthesis.how the TCDD contamination might be eliminated, both after the first step and on completion of the Cl CI O Cl n OCH?COOH V CI,8-V � O � CI CI o Cl tetrachlorodibenzodioxin (TCDD) 3

*17-62

17-63

14

(a) (b)

(c)

2,4,5-T

2,3,7

804 "'17-64

Chapter 17: Reactions of Aromatic Compounds Phenol common is alisoon ofaciphthal d-baseic anhydri indicatdoer thatwithis2 colequiorlvealssentins aciof phenol d and red. in base. Phenolpphththhalaleein,in ias synt hesizednonprescri by the aciptido-ncatlaaxatlyzedive,react HO OH an

H2S04) heat

phthalic anhydride o

phenolphthalein

red dianion

Propose aa mechanism mechanism forfor thethe conversi synthesisonofofphenol phthal ein.ein to its red dianion in base. Propose phenol p ht h al Use resonance structphenol ures ptohthalshoweinthatdianion. the two phenolic oxygen atoms are equivalent (each with half a negative charge) in t h e red Because the2,S03H group canuenebecanaddedbe made to a benzene rieneng andusingremoved latioenr, andit isdesul sometfonatimesiocaln asledintermedi a blockingate group. Show how 6 -di b romotol from t o l u sul f onat steps in the synt h esis. A graduate studentAftrtieredobtto amake o-f1puzzluorophenyl ms agnesium bromide sheby addi ngedmagnesium ton byan using ether solas usoltiovnentofsome o-f1uo­ robromobenzene. ining i ng resul t wi t h t h i s reaction, repeat t h e react i o tfoletrahydrofuran contained aforsmalits lformation. amount of furan. From this reaction, she isolated a fair yield of the compound that lows. Proposethata mechanism (a) (b) (c)

*17-65

"'17-66

Mg

0 ) 0,0 o

*17-67

A common illalicciohol t syntishesiaddeds oftometliqhuiamphet aminae, invol vwedes anbyintseveral erestinpieces g variaoftiolnithofiumthemetal Birch. ThereductBirchion. reduct A soluitoionnusual of ly ephedr i ne i n d ammoni fol l o reduces the aromat icPropose ring (Secta mechanism, ion 17-13C),simbutilairntothtihs atcaseforittheeliBirch minatereduction, s the hydroxyl group ofisephedr inecourse to givofe the met h amphet a mine. to expl a i n t h unusual reaction. Li

17-68

ephedrine methamphetamine The antiandoxidants BRA and BHT are commonly used as food preservatives. Show how BHA and BHT can be made from phenol hydroquinone. OH OH � CCCH3)3 (CH3)3C yY CCCH3)3

YOCH3

17-69

YOCR3

BHA BHT When an,2equation -dibromo-3,for 5this-dinitreaction, robenzeneshowiis tnrgeattheedproduct with excess NaOH atGi50°C, only one oftotheaccount bromiforne atthoemsformat is replionaced.of Draw you expect. v e a mechanism your proposed product. I

18

Ketones and Aldehydes

Wetheywiarle stofucent dy compounds conttaoinorgani ing thce chemistry, biochemiCstry, andin bidetaiololgy.because r al i m port a nce Some of theCarbonyl common compounds types of carbonyl compounds ar e l i s t e d i n Tabl e In addi tplioansttiocs,thandeir usesdrugs.as Natreagent sy andoc­ solcurriventngscarbonyl , they are compounds constituentares iofneverywhere. fabri c s, fl a vori n gs, u ral l clnuthdee nextprotefewins,chapt carbohydrat ewis,l anddiscussnucltehice propert acids thieats make up al l pl a nt s and ani m al s . I e rs, we reactions chemi of simspltrye tcarbonyl compounds. Then, ins,Chapt eorsteins.and we apply tandhis carbonyl o carbohydrat e s, nucl e i c aci d and pr compounds areecarbonyl ketones carbon and aldatehydes. A hashas tonewTheo alalksikylylmpl(or(orestarylarylcarbonyl ) )groups bonded t o t h o m. An group and one hydrogen atom bonded to the carbonyl carbon atom. I II II C C R R' R H carbonyl group

(

0)

1 8- 1

1 8-1.

Ca rbonyl Compo u n ds

=

23

24,

ketone

aldehyde

co nde nsed structures:

Ketone: Aldehyde:

TABLE 18-1 Class

0

0

0

/'"

c

/ '"

/ '"

ketone

aldehyde

carbony l group

RCOR'

RCHO

Two alkyl groups bonded to a carbonyl group. One alkyl group and one hydrogen bonded to a carbonyl group. Some Common Classes of Carbonyl Compounds General Formula

0 0 0 II

ketones

R-C-R'

carboxy lic acids

R-C-OH

esters

R-C-O-R'

II

II

Class

General Formula

0 0 0 II

a ldehydes

R-C-H

a cid ch lor ides

R-C-Cl

amides

R-C-NH2

II

II

805

806

1 8-2

Chapter Ketones and Aldehydes Ketare osome nes anddiffaledrences, ehydeshowever, are similapartr i nistculructarluyre,i nandtheitrheyreacthaveionssimwiiltahr proxioperdiztiinegs.agent Theres andreasonswithwenucldisecussophishort les. Inly.most cases, aldehydes are more reactive than ketones, for Thecoplacarnarbonylsigmacarbonds bon atoriomentised abouthybridizedapartand. bonded to threedizedotherorbiatotmsal overl throughaps The unhybri with a isorbisimtaillaofr tooxygen to form a pidoubl bond.e bond, The doublexcepte bondthat betthewcareenbonylcarbondoublande oxygen an al k ene bond is shorter, stronger, and polarized. kJ/kcalmollmol ) ketone 0 bond kllkcalmol/moi) alkene bond e bondiveofthanthecarcarbbon,onylandgroupthe bondi has anlagrgeelectdirponsolearemoment becauseequaJoxygen ipars morticTheuleaelr,edoubl ctthreolnegat not shared ly. In e ss t i g ht l y hel d pi el e ct r ons ar e pul l e d mor e st r ongl y t o war d t h e oxygen and aldehydes rger dipizoleethmoment an mostofaltkhyle pihaleliedctesrons.and etatohers.m, giWevincang ketuseonesresonance forms tolasymbol is unequals thshming 18:

2 sp

1 200

Struct u re of the Carbo nyl Group

p

p

C=C

energy

length

C=

1 .23 A

745 ( 1 78

C=C

1 .34 A

6 11 (146

�R�. R

R� . R J

�

� C=O:.

+ C-O : � 00

Thechargefirssepar t resonance formcontis rmore inmofportthaentsecond becausestructit invol vsesevimore bondsby tandhe lalrgeess a t i o n. The i b ut i o u re i d enced di" "pole moments of the ketones and aldehydes shown here . 0 III III Compare with: III � �C� :0"1 major

minor

.,0"

H

�C

CI

H-C-H

CH3

H3C

P, = 2.70 acetaldehyde

CH3

H3C

H

P, = 2.90

P, = 1.90 chloromethane

/ � CH3

P, = 1.300 dimethyl ether

Thialdehydes: s polarizTheatioposin oftitvheleycarpolbonyl group contatorimbutactesstaso tanhe elreeactctriophivity lofe (Lewi ketoness aciandd), a ri z ed cmb on and the negati vely polarized oxygen acts as a nucleophile (Lewis base). nes arbecomes e derived "albykreplanone.acin"gIntheopen-chai final inn tketheoalnes,kanewenamenumberwiSystththeematilongestcThenameschai"alkofnane"thketatoinclname utdeshe posithe tcarionbonylof thecarcarbonbonylfromgroup the endby cla number. osest to tInhecyclcarbiconylketogroup, and we i n di c at e nes, the carbonyl carbon atom is assigned the number

1 8-3

acetone

-e

IUPAC Names

-one.

No menclat u re of Keto n es a n d Aldehyd es

1.

o

II CH3-C-CH2 -CH3 I

2

3

2-butanone butan-2-one

4

CH3 0 CH3 I II I CH3-CH-C-CH-CH3 I

2

3

4

5

2,4-d i methy 1- 3-pentanone 2,4-dimethylpentan-3-one

l -phenyl - l -propanone I-phenylpropan- l -one

18-3 Nomenclature of Ketones and Aldehydes

a

:0:

CH3-C-CH2-C-CH3 I 2 3 41 5 CH3

2-cyclohexenone cyclohex-2-en- l -one

4-h ydrox y-4-methy 1-2- pentanone 4-h ydrox y-4-methyl pentan-2-one

CH3 3-methylcyclopentanone

a

aH

II

I

ematwiicthnamesAnforaldalehyde dehydescarbonare isderiat vthede endby rofeplaachaicingn,thsoe ifitnials numberof the alIfktaneheSystname aldehydeis used.group is attached to large unit (usually a ring), the suffix BrI I I II -e

-al.

1.

a

carbaldehyde

CH

aH

a

3

CH3-CH -CH2-C-H 3 2 4 I

4-bromo-3-meth y Iheptanal

3-hydroxybutanal

o-

CH3-CH2-CH=CH-CHa 5 4 3 2 I 2-pentenal pent-2-enal

•

CHO

(X 3

CHO aH

2-hydroxycyclopentane- l-carbaldehyde

cyclohexanecarbaldehyde

wipretfih xanotherandfunctortheional groupgroup as itsis rootnamed. Theas ketafoneycarbonyl is desiboxylgnatic eacid dbys fre­the group group. Car quently contain ketone or aldehyde groups named as substituents. aldehyde

can also be named as a substituent on a molecule

-CHa

axa-,

a nn

�

a

II

�

CH3CH2-C-CH2-CHa 3-oxopentanal

a II C-H

l

a

I

CH3-C-CH2-CaaH

COOH

3-oxobutanoic acid

2-formylbenzoic acid

As winames th otherinstcleaadssesofofthcompounds, ketic IUPAC ones andnames. aldehydesKetoarene ofcommon ten callenames d by common ei r syst e mat areloforcatmioedns byare nami ngusithnegtwGreek o alkylletgroups bonded totthhethcarbonyl group. Subst i t u ent gi v en e rs, begi n ni n g wi e carbon the carbonyl group. Common Names

next ta

CH3 a

a

II

{3

CH3

II

I

I

CH3CH2-C-CH3

CH3CH2-CH -C-CH -CH2CH3

methyl ethyl ketone

di-sec-butyl ketone

a

I

CH3

I

Br-CH2-CH2-C-CH-CH3 "

II

CH3CH2CH2-CH -CH -CH2-C-H 7 6 5 4 3 2 I

ethanal

A ketone

a

f3-bromoethyl isopropyl ketone

y

{3

aCH3 a

"I

II

CH3-CH2-CH -C-CCCH3)3 I-butyl a-methoxypropyl ketone

807

808

Chapter 1 8: Ketones and Aldehydes Some ketones have historical common names. Dimethyl ketone is always called ketones are usually named as the acyl group followed by the

acetone, and alkyl phenyl suffix -phenone.

a

a

II

o

CH3-C-CH3 ac etone

b enzophenone

p ropiophenone

acetophenone

Common names of aldehydes are derived from the common names of carboxylic acids (Table 1 8-2). These names often reflect the Latin or Greek term for the original source of the acid or the aldehyde. Greek letters are used with common names of aldehydes to give the locations of substituents. The first letter (a) is given to the carbon atom next to the carbonyl group. Br

a

I

II

CH3-CH-CH2-C-H y

Common name:

{3

OCH30

"I

II

CH3-CH-C-H

{3 f3-bromobutyraldehyde

ll'-methoxypropionaldehyde

3-bromobutanal

2- methoxy propanal

IUPAC name:

a

Common Names of Aldehydes

TABLE 1 8-2

Carboxylic Acid

Derivation

0

Aldehyde 0

II

formica, "ants"

H-C-OH formic acid 0

II

H-C-H formaldehyde (methanal) 0

II

CH3-C-OH acetic acid

acetum, "sour"

0

II

CH3-C- H acetaldehyde (ethanal) 0

II

CH3-CHz-C-OH propionic acid

protos pion, "first fat"

II

CH3-CHz-C- H propionaldehyde (propanal)

0

II

CH3-CHz-CH2-C-OH butyric acid



mlz 72

mlz 44

+

28

-- -.:::::-===

loetsshylofene McLafferty rearrangement of butyraldehyde

enol

28

H l:

+ C - CI 'Y- R f3

C / "

1 00

CH

H

mlz 7 2

a

II

loss of

C

cleavage

McLaf ferty rearrangement

II R' - C -

__

0

l

90

I

-

I

,

-

I

.. / H :0

�, y

H

mlz 29

mlz 7 2

I

-

M+

I

1 -�

60

,

I

- 7 2-

1 R

R H // R o� C

II

C

t

�

I

C-

B

/11lz 44

+

/A

l O-H l: I

R' - C = C' l\

+

loss of alkene enol A AB McLafferty rearrangement of a general ketone or aldehyde Mechani s rearrangement hydrogensmmayof bethetMcLaffert ransferredyfirearrangement. rst, followed byThifragmentati on. may be concerted, as shown here, or the 0

... Figure 1 8-5

/ " / \

R'

\

815

y

Chapter 18: Ketones and Aldehydes Htf1Z; Why were no products from McLafferty rearrangement observed i n the spectrum of 2-bu­ tanone (Figure 18-3)7

816

PROB L E M 1 8 - 3

PROBLEM-SOLVING

The McLafferty rearrangement is

equ ivalent to a cleavage between the

a

a n d f3 carbon atoms, plus one mass

u n it for the H that is transferred.

PROB L E M 1 8 - 4

Use equati ons to show the fragmentation leading to each numbered peak in the mass spec­ trum of 2-octanone. 43 80 11 Cf2CH2CH2CH2CH20l-r3 H3C+- C� -.., �4 + � + �1 � + � 60 - -4 - -- '-- - --- -- -8-5 -- - 4-- --1---+--4---1 4 1 i 40 1---4---+_��--+-�.�--+_�� ;�--�--41--�--4_ �_ --8 12 � 13 20 1-_ 10 20 30 40 60 70 80 90 100 110 120 130 140 160 I--+--+---I-+----+---+-�-+- I I

.1

I

1

g

�

1

-+--+--IlI- Ih �.I,II II+-+-+- I-!---i----+·� I;.:-+: --,'+ i--+-+-- 1 , I I I LL��.I �-.LL.-L-�-.� I O L,�-.��rna��illLywillL�� � SO

PROBLEM-SOLVING

Htf1Z;

Conj u g ated carbonyl compou nds have cha racteristic

7T

--'>

the UV spectru m.

7T*

absorptions in

o

Base value:

II � 0-

210nm

H

An additional conj u g ated ( = C increases

Amax about 30

nm; a n

1 8- S E

I

I SO

mlz

U ltraviolet Spectra of Ketones and Aldehydes

The n � n * Transition

The strongest absorptions in the ultraviolet spectra of aldehydes and ketones are the ones resulting from � electronic transitions. As with alkenes, these absorptions are observable (A.max > 200 nm) only if the carbonyl double bond is conjugated with another double bond. The simplest conjugated car­ bonyl system is propenal, shown next. The � transition of propenal occurs at "-max of 2 1 0 nm (e = 1 1 ,000 ) . Alkyl substitution increases the value of "-max by about 1 0 nm per alkyl group. An additional conjugated double bond increases the value of "-max by about 30 nm. Notice the large values of the molar absorptivities (e > 5000) , as we also observed for the � transitions of conjugated dienes. 7T

7T

7T

7T

7T

7T

*

*

*

additi o n a l a l kyl g r o u p i n creases it

'P

a bout 10 n m .

A,nax

The

n

propenal

=

210

nm, f =

1 1 ,000

Ama,

o

three alkyl groups

three alkyl groups =

237

nm, f =

1 2,000

"'nax

=

244

nm, f =

� n * Tra nsition

1 2,500

An additional band of absorptions in the ultraviolet spec­ tra of ketones and aldehydes results from promoting one of the nonbonding electrons on oxygen to a anti bonding orbital. This transition involves a smaller amount of energy than the � transition because the promoted electron leaves a nonbond­ ing (n) orbital that is higher in energy than the bonding orbital (Figure 1 8-6). Because the n � transition requires less energy than a � transition, it gives a lower frequency (longer wavelength) absorption. The n � transitions of sim­ ple, unconjugated ketones and aldehydes give absorptions with values of "-max between 280 and 300 nm. Each double bond added in conjugation with the carbonyl group increases the value of "-max by about 30 nm. For example, the n � transition of ace­ tone occurs at "-max of 280 nm (e 1 5 ) . Figure 1 8-7 shows the UV spectrum of a ketone conjugated with one double bond, having "-max around 3 1 5 to 330 nm (e 1 1 0 ) . Figures 1 8-6 and 1 8-7 show that n � transitions have small molar absorp­ tivities, generally about 1 0 to 200. These absorptions are around 1 000 times weaker than � transitions because the n � transition corresponds to a "forbidden" electronic transition with a low probability of occurrence. The nonbonding orbitals 7T

*

7T

7T

*

7T

7T

*

7T

7T

7T

*

7T

*

*

=

7T

7T

7T

*

7T

*

*

=

18-5 Spectroscopy of Ketones and Aldehydes (+l !-) . V V -.,---- - " 'I C 0

f

1T

*

d

���� �

71' *

� � � � � � � �

�

�

�

� � � �

(+l !-) . V V " 'IC _ O

-,-�_or_bi_d:_e

l ph l attack

n uc e o

R'

" /

MaX b

C=N .

Mg salt

R

i ic

/

------i>

H+

R' R

" /

H / C=N..

H30+

R imine

of imine

R'

" /

C=o":.

ketone

a

Example ------'.>

ether

benzonitrile

+ NHt

phenylmagnesium bromide

o o ( ag s salt) benzophenone i mine m ne i u m

benzophenone (80%)

PROBL E M 1 8 - 9

ct the products of the followingCH3CH2-MgBr, reactions. then H)O+ a) benzyl CH)CH2CH2CH2-C==N (Predi um cyani de um bromide, then acidic hydrolysis productbromide cycJsodiopentyl magnesi +

(b) (c)

of (b) +

+

PROBLEM 1 8 - 1 0

Show howtionalthe reagent followisngthattransformati ons may be accomplished in good yield. You may use any(a) addi are needed. bromobenzene propioaphenone pentanoi c acid 3-hept none CH3CH2CN toluene benzyl3-heptanone cyc\opentyl ketone (c)

---?

---?

(b) (d)

---?

---?

1 8- 1 1 Synthesis of Aldehydes and Ketones from Acid Chlorides Because aldehydes are easily oxidized to acids, one might wonder whether acids are easily 18- 1 1 reduced back to aldehydes. Aldehydes tend to be more reactive than acids, however, and reducing agents that are strong enough to reduce acids also reduce aldehydes even faster. Synthesis

0 I R-C-OH

�-LJ aldehyde

LiA1H 4 (slow)

acid

(not isolable)

LiA1H 4

of Aldehydes a n d Ketones from Acid Chlorides

R-CH2-O -

alkoxide

(fast)

Acids can be reduced to aldehydes by first converting them to a functional group that is easier to reduce than an aldehyde: the acid chloride. Acid chlorides (acyl chlo­ rides) are reactive derivatives of carboxylic acids in which the acidic hydroxyl group is replaced by a chlorine atom. Acid chlorides are often synthesized by treatment of carboxylic acids with thionyl chloride, SOCI2. o

o

I R-C-OH

+

acid

o

II CI- S - Cl

I R-C-Cl

thionyl chloride

acid chloride

+

HCI

+

Strong reducing agents like LiAIH4 reduce acid chlorides all the way to primary alcohols . Lithium aluminum tri(t-butoxy)hydride is a milder reducing agent that reacts faster with acid chlorides than with aldehydes. Reduction of acid chlorides with lithium aluminum tri(t-butoxy)hydride gives good yields of aldehydes. o

II R-C-Cl

Li +

acid chloride

o

I R-C- H

- AlH(O-t-Bu)3

aldehyde

l i thium aluminum tri(t-butoxy)hydride

Example

0 CH 3 II I CH3CHCH2-C - OH

CH3 0 I I CH3CHCHz-C-CI

Li+ -AIH(O-f-B u)

isovaleroyl chloride

isovaleric acid

CH3 0 I I CH3CHCH2-C-H

3)

isovaleraldehyde (65%)

Synthesis of Ketones Grignard and organolithium reagents react with acid chlo­ rides much like hydride reagents: They add R- where a hydride reagent would add H-. As we saw in Section 1 0-9, Grignard and organolithium reagents add to acid chlorides to give ketones, but they add again to the ketones to give tertiary alcohols.

o

II R ' -C - Cl

RM"X b )

(fast)

acid chloride

ketone

RM"X )

(fast) b

O- +MgX I R'-C-R I R

alkoxide

To stop at the ketone stage, a weaker organometallic reagent is needed: one that reacts faster with acid chlorides than with ketones. A lithium dialkylcuprate (Gilman reagent) is such a reagent. o

R2CuLi

a lithium dialkylcuprate +

II R' -C- Cl

o

II R' -C-R

+

R-Cu

+

825

LiCI

ha r 18:

C pte

826

s and Ald hyd

K etone

e

es

The lithium dialkylcuprate is formed by the reaction of two equivalents of the corre­ sponding organolithium reagent (Section lO-8B) with cuprous iodide . R -Li

2

CuI

+

( 1) Li

(2) CuI

(� CULi

)

R2CuLi

r:OJ

Example

� Cl

-----'.>

°

II

C '--

+

LiI o

Cl

II

orc� 80%

PROBLEM 1 8 - 1 1

Predict the products of the following reactions: C --""" C1 o

U

II

lVJ

(a)

( I ) LiAlH4

(C) � Cl excess o

( 1)

o

I 1.

SUMMARY

)

)

(Section 11-2)

Oxidation of alcohols Secondary alcohols � ketones

o

OH I R - CH - R'

II R-C-R'

secondary alcohol

ketone

Primary alcohols � aldehydes

primary alcohol (Section 8-15B) R-CHPH

2.

MgCl

Syntheses of Ketones and Aldehydes

a.

b.

�

(�CuLi

�Cl

(d)

LiAlH(0-t-Bu) 3

)

Ozonolysis of alkenes

R H

"/

/

C=C"-

R'

CSH S NH+ Cr0 3 CnpCC) )

( 1 ) 03

(2 ) (CH3 )2 S )

R"/

C=O

o

II R-C-H

aldehyde

+

O=C

/ "-

R'

al(gikeneves aldehydes or ketones, dependinalgdonehydethe starting alkene)ketoneR" R"

H

)

18-11 3.

(Section � + G lA 0 RG alhydrogen, kyl or arylan; activating group, or halogen. (Section 17-1lC)

Friedel-Crafts acylation

Synthesis of Aldehydes and Ketones from Acid Chlorides

827

17- 1 1 )

-o-!

o

R - C - Cl

O

(+

-R

aryl ketone

=

=

oethO)

The Gatterman-Koch formylation

+ + GV G hydrogen, an activating group,� or halogen. (Section a. HCI

CO

)

O

-H

benzaldehyde derivative

=

4.

-o- !

o

AlCI3 , CuCI

Hydration of alkynes 9-9F) Catalyzed by acid and mercuric salts (Markovnikov orientation)

o

"

alkyne

methyl ketone

R - C = C -H

b.

enol (not isolated)

Hydroboration-oxidation (anti-Markovnikov orientation)

R - C - C -H

alkyne

5.

R - C - CH3

Alkylation of 1,3-dithianes

(l

2B

(Section

alkylation

H

II

)

R - CH 2 - C -H

aldehyde

18-8)

) (2) lO R - X

1,3-dithiane

o

enol (not isolated)

( 1 ) BuLi

SXS H

(1) Sia H

(l

SXS

thioacetal R

1

H

(1) alkylation BuLi

) (2) lO R'-X

(l

SXS

thioketal R

Hp+, HgCl2

Example

/

o

C

C

"

" H

aldehyde

(1) S S(1 X H 1,3-dithHiane

BuLi

(2) PhCH2Br

HP+, HgCl2

o

"

R

l

R'

S S(1 X HthioacetCH2Phal

R

/

" R'

ketone

(I) BuLi (2) BuBr

thioketal

l-phenyl-2-hexanone

( Continued)

828 6.

Chapter 1 8 : Ketones and Aldehydes

Synthesis of ketones using organolithium reagents with carboxylic acids (Section 1 8-9) a

II

2 R'

R - C - aH carboxylic acid

-

aLi

o

R-C-OLi

R - C -R'

I

Li

II

I

ketone

R' dianion

Example

a

a

a

II

C '-

OH

+

�

----7

2 CH3Li methyllithium

cyclohexane

dianion

ketone

Synthesis of ketones from nitriles (Section 18- 10) N-MgX +

II

�

R' -Mg-X

Mg salt

(or R' -Li)

� H30+

R-C-R'

o

II

R-C-R' ketone

of imine

Example ( 1 ) CH3CH2CH2-MgBr (2) H30 +

)

benzonitrile

8.

butyrophenone

Aldehyde synthesis by reduction of acid chlorides ( Section 1 8- 1 1 ) o

II

Li + -AIH(O-t-Bu)3

acid chloride

(or H2, Pd, BaS04' S)

R - C - Cl

Example

0

Ph

I

II

CH3- CH -CH2- C - Cl

Li + -AIH(O-t-Bu)3

3-phenylbutanoyl chloride

9.

o

II

R-C-H aldehyde

Ph

I

a

R' - C - Cl

II

+

R' - C - R

acid chloride

Example +

II

CH3- CH - CH2- C -H

Ketone synthesis from acid chlorides (Section 18- 1 1 )

II

0

3-phenylbutanal

a

ketone

(�

CULi

II

C '-

CHJ

cyclohexyl methyl

carboxylic acid

7.

a

1 8- 1 2

Reactions of Ketones and Aldehydes: Nucleophilic Addition 18- 1 2

Ketones and aldehydes undergo many reactions to give a wide variety of useful deriv­ atives. Their most common reaction is nucleophilic addition, addition of a nucle­ ophile and a proton across the C = O double bond. The reactivity of the carbonyl group arises from the electronegativity of the oxygen atom and the resulting polariza­2 tion of the carbon-oxygen double bond. The electrophilic carbonyl carbon atom is sp hybridized and flat, leaving it relatively unhindered and open to attack from either face of the double bond. As a nucleophile attacks the carbonyl group, the carbon atom changes hybridization from sp 2 to sp 3 . The electrons of the pi bond are forced out to the oxygen atom to form an alkoxide anion, which protonates to give the product of nucleophilic addition.

�

Nlic Nuc " " - H - Nlic) : -O. . / '\\C - O: .. R ,.,\\C R" � � R R "

nucleophilic attack

alkoxide

H

+

Reactions of Ketones a n d Ald ehydes: N uc l eop h i l ic Addition

Nuc : -

product

We have seen at least two examples of nucleophilic addition to ketones and alde­ hydes. A Grignard reagent (a strong nucleophile resembling a carbanion, R : -) attacks the electrophilic carbonyl carbon atom to give an alkoxide intermediate. Subsequent protonation gives an alcohol . 0-

CH3CH2

ethyl magnesi um bromide

CH

',J+

CH3

CH3

I

I .. + CH3CH?-C-0 - I .. :- MgBr

�C �.: 0'

8-.

CH3

CH3 CH 2 -C - O - H 1

CH3

CH3

2-methyl-2-butanol

alkoxide

acetone

Hydride reduction of a ketone or aldehyde is another example of nucleophilic addition, with hydride ion ( H : - ) serving as the nucleophile. Attack by hydride gives an alkoxide that protonates to form an alcohol. H

I

" 0"

II�

Na+ H-B-� /C" CH3 "'CH3 I H

acetone

----7

: 0 :I H-C-CH3 I CH3

�

CH3CHPH (solvent)

.

.

: O-H )

I

+

H -C-CH3 I CH3

CH3CH2O-

2-propanol

alkoxide

Weak nucleophiles, such as water and alcohols, can add to activated carbonyl groups under acidic conditions. A carbonyl group is a weak base, and it can become protonated in an acidic solution. A carbonyl group that is protonated (or bonded to some other electrophile) is strongly electrophilic, inviting attack by a weak nucleophile. ..

�

: O -H I /+ C "" 'R R activated carbonyl

Nuc : -

)

..

: O-H I R-C-Nuc I R

829

830

Chapter 1 8 : Ketones and Aldehydes The following reaction is the acid-catalyzed nucleophilic addition of water across the carbonyl group of acetone. This hydration of a ketone or aldehyde is discussed in Section 1 8- 14. /

H ,+ O :

H

CH '-"'" ""- H ""3 '.r ) C=O.. ( / CH3

+ protonated, acti vated acetone

acetone

CH3 .. I .. H -O-C-O-H .. .. I CH3

Hi-ltv

Please become fam i liar with these simple mechanisms. You will see many exa mples in the next few pages. Also, most of the important m u ltistep mechanisms in this chapter are com binations of these simple steps.

H30 +

acetone hydrate

attack by water

PROBLEM-SOLVING

+

In effect, the base-catalyzed addition to a carbonyl group results from nucleophilic attack of a strong nucleophile followed by protonation . Acid-catalyzed addition begins with protonation, followed by the attack of a weaker nucleophile . Many additions are reversible, with the position of the equilibrium depending on the relative stabilities of the reactants and products. In most cases, aldehydes are more reactive than ketones toward nucleophilic additions . They usually react more quickly than ketones, and the position of the equi­ librium usually lies more toward the products than with ketones. We explain aldehydes' enhanced reactivity by noticing that an aldehyde has only one electron-donating alkyl group, making the aldehyde carbonyl group slightly more electron-poor and electrophilic (an electronic effect). Also, an aldehyde has only one bulky alkyl group (compared with two in a ketone), leaving the carbonyl group more exposed toward nucleophilic attack. Especially with a bulky nucleophile, the product of attack on the aldehyde is less hindered than the product from the ketone (a steric effect). 0"- Nuc 'C/ / '" R R ketone less electrophi lic

11 1

)

alkoxide more crowded

o

Nuc:-

ij'C", R H

H - Nuc

)

aldehyde more electroph i I ic

0"- Nuc 'C/ " � 'H

product more crowded

H -Nuc

)

alkoxide less crowded

HO" Nuc ,C/ / '" R H product less crowded

PROBLEM 1 8 - 1 2 Show how you would accomplish the following synthetic conversions. You may use any additional reagents and solvents you need. o o

II

(a) Ph -CHO--7 Ph - C - Ph o

II

(c) Ph - C - Ph

OH --7

I

Ph - CH - Ph

II

(b) Ph-C - Ph

--7

Ph 3 C - OH

Reactions of Ketones and Aldehydes: Nucleophilic Addition Sodi um trimuch acetoxyborohydri dye,thanNa(CH3COOhBH, isuseda mitoldreduce reducinalgdehydes agent thatin threduces alencedehydes more qui c kl ketones. It can be e pres­ of ketones, such as in the following reaction: OHI II CH3-C-CH2-CH2 Draw a compl ete Lewi cetoxyborohydri Propose a mechani sm fors structure the reductiforosodi n ofuanm altrdiaehyde by sodium dtre.iacetoxyborohydride. 1 8- 1 2

831

PR O B L E M 1 8 - 1 3

o

Na(CH3COO)3B H CH3 COOH

)

(a) (b)

The following box summarizes the base-catalyzed and acid-catalyzed mecha­ nisms for nucleophilic addition, together with their reverse reactions.

e;.��7'�

N ucleoph i l ic Add itions to Ca rbonyl G ro u ps

KEY M E C H A N I S M 1 8- 1

Basic Conditions (strong n ucleophile) Step 1 :

A strong nucleophile adds to the carbonyl group to form an alkoxide. N uc: -�C /

Step 2:

A weak acid protonates the alkoxide to give the addition product. I

. .

Nuc - C - O: I ..

_ /*" H � uc

I I

" "

Nuc - C-O:-

0".' \..;:

I

) Nuc - C- O - H • •

I

..

EXAM PLE: Formation of a Cya nohyd rin (covered i n Section 1 8-1 5) Step 1:

A strong nucl eophil e adds to the carbonyl group to form an alkoxide .�)

Q' �

:C

.

N:

benzaldehyde

Step 2:

A weak acid protonates the alkoxide to give the addition product.

Q' I � �

:O- H I C -H \

C

+ N:

benzaldehyde cyanohydrin Reverse reaction:

Reverse reaction:

Deprotonation, followed by loss of the nucleophile. I

" ....,

Nuc-C -O-H

I

,Y":.

-----,)

Nuc-C -O: "-

"C - O·..

Nuc:/ -. � I I " PROBLEM: The cyanohydrin formation used in the example is reversible. Draw a mechanism for the reverse reaction. /,,::Nuc-

----'>

( Continued)

832

Chapter 1 8 : Ketones and Aldehydes

Acidic Conditions (weak nucleophile, activated carbonyl) Step 1 :

Protonation activates the carbonyl group toward nucleophilic attack.

Step 2:

A weak nucleophile adds to the activated (protonated) carbonyl group.

[

"

/

/ +

C=O..

H

I

�

Nuc-C - O -H I

Formation of a Hemiacetal (covered in Section 1 8- 1 8) Step 1: Protonation activates the carbonyl group toward nucleophilic attack. EXAM PLE:

·o·�

U

II C '-

H + , CH30H

H

·

'

benzaldehyde

Step 2:

A weak nucleophile adds to the activated (protonated) carbonyl group. Deprotonation of the product gives the hemiacetal. . . /H :0 I + C-H � \ I � :Q-CH3

U

a hemiacetal

Reverse reaction:

Loss of the weak nucleophile, followed by deprotonation.

Reverse reaction:

I Nuc -C � I - O-H PROBLEM :

" o' C= .' /

Nuc-H

The hemiacetal formation used in the example is reversible. Draw a mechanism for the reverse reaction.

18- 13 The Wittig Reactio n

We have seen carbonyl groups undergo addition by a variety of carbanion-like reagents, including Grignard reagents, organolithium reagents, and acetylide ions. In 1 954, Georg Wittig discovered a way of adding a phosphorus-stabilized carbanion to a ketone or aldehyde. The product is not an alcohol, however, because the intermediate undergoes elimination to an alkene. In effect, the Wittig reaction converts the car­ bonyl group of a ketone or an aldehyde into a new C = C double bond where no bond existed before. This reaction proved so useful that Wittig received the Nobel Prize in chemistry in 1 979 for this discovery.

1 8- 1 3

The Wittig Reaction

The Wittig reaction

R' R'

" C=O /

+

ketone or aldehyde

R,, _ Ph +/ : C-P -Ph " Ph / H

R'

� �

R'

" / C=C " /

R H

alkene

phosphorus ylide

The phosphorus-stabilized carbanion is an ylide (pronounced "iW-id")-a molecule that bears no overall charge but has a negatively charged carbon atom bonded to a positively charged heteroatom. Phosphorus ylides are prepared from triphenylphosphine and alkyl halides in a two-step process. The first step is nucle­ ophilic attack by triphenylphosphine on an unhindered (usually primary) alkyl halide. The product is an alkyltriphenylphosphonium salt. The phosphonium salt is treated with a strong base (usually butyllithium) to abstract a proton from the carbon atom bonded to phosphorus. 8+

2 -Li H �c5-CH I Ph ,, CH2CH2CH3 + 17 Ph- P- C- H butyllithium ) 1 Ph / R X­

H

Ph " 1 Ph- P : + H- C - X Ph / � RI U triphenylphosphine

alkyl halide

+

LiX

phosphonium salt phosphorus yJide

Examples

�

+ Ph3P-CH3

B u - Li

)

y lide

methyltriphenylphosphonium salt

�

+ Ph3P-CH2 -Ph

+ .. Ph3P-CH2

B u - Li

)

+ .. Ph3P-CH-Ph

benzyltriphenylphosphonium salt

The phosphorus ylide has two resonance forms: one with a double bond be­ tween carbon and phosphorus, and another with charges on carbon and phospho­ rus. The double-bonded resonance form requires ten electrons in the valence shell of phosphorus, using a d orbital. The pi bond between carbon and phosphorus is weak, and the charged structure is the major contributor. The carbon atom actual­ ly bears a partial negative charge, balanced by a corresponding positive charge on phosphorus. PROB L E M 1 8 - 1 4

hylphosphi is muchng most less expensi ve thylanidtres?iphenylphosphine. Why is trimethylphos­ phiTrimneetunsui table forne maki phosphorus Because of its carbanion character, the ylide carbon atom is strongly nucle­ ophilic. It attacks a carbonyl group to give a charge-separated intermediate called a betaine (pronounced "bay'-tuh-ene"). A betaine is an unusual compound because it contains a negatively charged oxygen and a positively charged phosphorus on adjacent carbon atoms. Phosphorus and oxygen form strong bonds, and the attraction of oppo­ site charges promotes the fast formation of a four-membered oxaphosphetane ring. (In

yJide

833

834

Chapter 1 8 : Ketones and Aldehydes some cases, the oxaphosphetane may be formed directly by a cycloaddition, rather than via a betaine.) The four-membered ring quickly collapses to give the alkene and triphenylphos­ phine oxide. Triphenylphosphine oxide is exceptionally stable, and the conversion of triphenylphosphine to triphenylphosphine oxide provides the driving force for the Wit­ tig reaction. The Wittig Reaction

M E C H A N I S M 1 8-2 Step 1:

The ylide attacks the carbonyl to form a betaine. +

H R' "' / . Ph3P-C:==------....- C ....0:.",. . / '" R R' +

\

ylide

�

.•

Ph3P : 0 : I I H-C-C-R' I I R R' a betaine

ketone or aldehyde

Step 2:

The betaine closes to a four-membered ring oxaphosphetane (first P-O bond formed). + )"\ . .

Ph3P : 0 :I I H-C-C-R' I I R R'

�

a betaine

Step 3:

Ph3P- 0: I I H-C - C-R' I I R R' oxaphosphetane

The ring collapses to the products (second P-O bond formed). Ph3P= O:: H R' / '" C=C / '" R R'

Ph3P -C0 : I� I H-C-C-R' I I R R' four-membered ring

triphenylphosphine oxide + alkene

The following examples show the formation of carbon-carbon double bonds using the Wittig reaction. Mixtures of cis and trans isomers often result when geomet­ ric isomerism is possible.

0°

Q H

C=O /

+

+

+

..

Ph3P-CH2

�

O

CH,

85%

+

Q

Ph3P-C : '"

H

�

Q

C �C

H

/

/H

O

1 8- 1 3 PROBLEM 1 8- 1 5

Like other(a betai strongne)nuclquiecophikly lcycl es, triizesphenyl phosphi ne attacksane andthat opens epoxitodanes. alThekeneiniandtial product to an oxaphosphet col l a pses triphenyl peach hosphistepne oxiin thede. reaction of trans-2,3-epoxybutane with triphenylphosphine to Show giShowve 2-butene. What is themigsthtereochemi e stry of the double bond in the product?

(a) (b)

how this reaction

b

used to convert cis-cyc]ooctene to tran.s-cyclooctene.

Plan n i n g a Wittig Synthesis The Wittig reaction is a valuable synthetic tool that converts a carbonyl group to a carbon-carbon double bond. A wide variety of alkenes may be synthesized by the Wittig reaction. To determine the necessary reagents, mentally divide the target molecule at the double bond and decide which of the two components should come from the carbonyl compound and which should come from the ylide. In general, the ylide should come from an unhindered alkyl halide. Triphenyl­ phosphine is a bulky reagent, reacting best with unhindered primary and methyl halides. It occasionally reacts with unhindered secondary halides, but these reactions are sluggish and often give poor yields. The following example and Solved Problem show the planning of some Wittig syntheses. Analysis

(preferred)

>

could come from

or

Synthesis

(2) BuLi

S O LV E D P R O B L E M 1 8 - 2

Show how you would use a Wittig reaction to synthesize I-phenyl- l,3-butadiene.

o

H / C=C H "/ / C=C H / "H H

I -pheny l- l ,3-butadiene

This mole bondeculcoul e hasdtbewoformed double ibonds thatof mitwgohtways. be formed bythWieset isynt g reactihesesons.wiThel probabl centraly doubl n ei t her Bot h of work, and both wil produce a mixture of cis and trans isomers. (Continued) SOLUTION

The Wittig Reaction

835

836

Chapter 18: Ketones and Aldehydes Analysis

could come from >

or

U You shoul d complete this solution by drawing out the syntheses indicated by thjs analysis (Probl em 18-16). H

PROBLEM-SOLVING

HiltZ;

P l a n a Wittig synthesis so that the less hi ndered end of the d o u b l e bond comes from the ylide. Remember that the yl ide is made by SN2 attack of

tri phenylphosphine on a n u n h i ndered a l ky l ha lide, fo l lowed by deprotonation.

CH =CH2

Outl ed in Solved Problem 18-2, beginning with aldehydes and alBothkylinofhale ttihhdeesees.syntsyntheseshesesindiofcat1-phenyl form ntheg thecentterrmjal doubl e bond.e bond. Show how you would synthesize thjs target- 1mol,3-buteculadieebyne formi nal doubl ze the following compounds. In each Show case,Ph-CH=C sthowart wiWitht anig alreact k( CH3ylihalohnsidmie andght bea ketusedone toro synthesi an aldehyde. Ph-C ( CH3 ) =CH2 � CH (c) Ph-CH=CH-CH=CH-Ph

(a)

(b)

PROBLEM 1 8 - 1 7

(b) (d)

Hyd ration of Ketones a n d Ald ehydes

H

PROBLEM 1 8 - 1 6

(a)

18- 14

/ C = PPh 3 + O = C "" /

U

2

In an aqueous solution, a ketone or an aldehyde is in equilibrium with its hydrate, a geminal diol. With most ketones, the equilibrium favors the unhydrated keto form of the carbonyl. R R OH '" '" / [hydrate] C K = (jfj / C=O + H2O � [ketone] [H2O] / '" R R OH

(a geminal dial)

keto form

hydrate

Example

° II

CH3-C-CH3 acetone

+

H2 O

�

HO OH \ / CH3-C-CH3

acetone hydrate

K =

0.002

Hydration occurs through the nucleophilic addition mechanism, with water (in acid) or hydroxide ion (in base) serving as the nucleophile. Aldehydes are more likely than ketones to form stable hydrates. The elec­ trophilic carbonyl group of a ketone is stabilized by its two electron-donating alkyl

Hydration of Ketones and Aldehydes

1 8- 1 4

M E CHAN I S M 1 8-3

837

Hydration of Ketones and Aldehydes

In acid

The acid-catalyzed hydration is a typical acid-catalyzed addition to the carbonyl group. Protonation, followed by addition of water, gives a protonated product. Deprotonation gives the hydrate. Step 1: Protonation. Step 2: Water adds. Step 3: Deprotonation. ..

..

: O-H I R-C-R . I : 0+ /-..) " H H

( )

: O-H I

( )

R-C-R I

: O- H

H2 O : �

In base

The base-catalyzed hydration i s a perfect example of base-catalyzed addition to a carbonyl group. The strong nucleophile adds, then protonation gives the hydrate. Step 1: Hydroxide adds. Step 2: Protonation. ..

..

: O : =------.... I

HO-C-R I R

H-O-H \.,( ..

OH )

I

+

HO-C-R I R

- OH

groups, but an aldehyde carbonyl has only one stabilizing alkyl group; its partial positive charge is not as well stabilized. Aldehydes are thus more electrophilic and less stable than ketones. Formaldehyde, with no electron-donating groups, is even less stable than other aldehydes.

R

ssjl Y' C ""'-

s-

R

R

o+� 1 Y' C """'--.

H

H

ssjl / C """'--.

usua l l y protonates the carbonyl to activate it toward attack by a weak

H

formaldehyde

two alkyl groups

less stabilization

relatively unstable

nucleo p h i l e .

These stability effects are apparent in the equilibrium constants for hydration of ketones and aldehydes. Ketones have values of Keg of about 1 0-4 to 1 0-2 . For most aldehydes, the equilibrium constant for hydration is close to 1 . Formaldehyde, with no alkyl groups bonded to the carbonyl carbon, has a hydration equilibrium constant of about 40. Strongly electron-withdrawing substituents on the alkyl group of a ketone or aldehyde also destabilize the carbonyl group and favor the hydrate. Chloral (trichloroacetaldehyde) has an electron-withdrawing trichloromethyl group that favors the hydrate. Chloral forms a stable, crystalline hydrate that became famous in the movies as knockout drops or a Mickey Finn. 0 HO OH \/ II ) CH3-CH2 -C-H + H2O CH3-CH2 -C-H K 0.7 (

I

H

=

propanal hydrate

/ C """'--.

H

formaldehyde

+

H2O

�

HO """'--. / OH /C H """'--. H formalin

nucl eoph i l e u s u a l ly adds d i rectly to the carbonyl group. In acidic

rarely present; an acid (or Lewis acid)

aldehyde

0

H?/tZ;

In basic conditions, a strong

conditions, strong nucleophi les are

ketone

propanal

PROBLEM-SOLVING

K

=

40

838

Chapter 1 8: Ketones and Aldehydes o

The body rapidly reduces chloral (trichloroacetaldehye)

to

II

trichloro­

HO OH \/ CI3C-C-H

K

Hp chloral hydrate chloral Propose mechani slmsyzedforhydration of chloral to form chloral hydrate. The aci d -cat a (b) The base-catalyzed hydration of acetone to form acetone hydrate. Rank the following compounds in order of increasing amount of hydrate present at equilibrium. CI3C-C-H

ethanol, which is responsible for the drug's sleep-inducing effect.

+

=

500

PROB L E M 1 8 - 1 8

PROBLEM-SOLVING HiltP Don't be surprised to see some 0 - H

(a)

stretch, from the hydrate, in the I R spectra o f many a l dehydes.

PROB L E M 1 8 - 1 9

1 8- 1 5 Formation of Cya n ohyd ri n s

M E C H A N I S M 1 8-4

Hydrogen cyanide (H -C - N) is a toxic, water-soluble liquid that boils at 26°C. Because it is mildly acidic, HCN is sometimes called hydrocyanic acid. pK 9.2 The conjugate base of hydrogen cyanide is the cyanide ion C: C = N: ). Cyanide ion is a strong base and a strong nucleophile. It attacks ketones and aldehydes to give addition products called cyanohydrins. The mechanism is a base-catalyzed nucleophilic addition: attack by cyanide ion on the carbonyl group, followed by protonation of the intermediate.

a

=

Formation of Cya notlydrins

� __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ -J __ __

Cyanohydrin formation is a perfect example of base-catalyzed addition to a carbonyl group. The strong nuc1eophile adds in fIrst step to give an alkoxide. Protonation gives the cyanohydrin. Step 2: Protonation gives the cyanohydrin. Step 1: Cyanide adds to the carbonyl. .. : O-H r>I I H-C=N - ) R -C-R' R-C-R' I I C=N C=N :

the

:O�

ketone or aldehyde

intermediate

cyanohydrin

EXAM PLE: Formation of benzaldehyde cy,,\nohydrin.

Step 1:

Cyanide adds to the carbonyl.

·VJ

a� benzaldehyde

:C N

Step 2 :

Protonation gives the cyanohydrin. : O -H I C-H � \ I � C N

a

+

benzaldehyde cyanohydrin

Cyanohydrins may be formed using liquid HCN with a catalytic amount of sodi­ um cyanide or potassium cyanide. HCN is highly toxic and volatile, however, and therefore dangerous to handle. Many procedures use a full equivalent of sodium or potassium cyanide (rather than HCN), dissolved in some other proton-donating solvent.

1 8- 1 5

Formation of Cyanohydrins

Cyanohydrin formation is reversible, and the equilibrium constant may or may not favor the cyanohydrin. These equilibrium constants follow the general reactivity trend of ketones and aldehydes: formaldehyde > other aldehydes > ketones Formaldehyde reacts quickly and quantitatively with HCN. Most other alde­ hydes have equilibrium constants that favor cyanohydrin formation. Reactions of HCN with ketones have equilibrium constants that may favor either the ketones or the cyanohydrins, depending on the structure. Ketones that are hindered by large alkyl groups react slowly with HCN and give poor yields of cyanohydrins. 0 HO ....... / CN II ----';> + HCN, KCN C """"' / C """"' CH3CH2 H CH3CH; H

propanal

propanal cyanohydrin ( 1 00%)

0 II / C ....... CH3CH2 CH3

+ HCN, KCN

2-butanone

0 II / C .......

HO ....... / CN / C ....... CH3CH2 CH3

2-butanone cyanohydrin (95%)

+

di-t-butyl ketQCH3)3 one

(CH3)3C

�

HCN, KCN

�

HO ....... / CN « / C ....... (CH3)3C qCH3)3

The mjesl aipmiedexture of HCN and secret dsehyde to iprevent othmjer l ipede anibenzal m al from eat n g i t . The stores mandeln) oinniatrireservoi le (benzalr. When dehyde cyanohydri attacked,aitreacti discharges mandelonitrile through o n chamber containinogn enzymes that catalnyzeto the conversi of the cyanohydri benzaldehyde and HCN.

5 %)

slow reaction, poor yields

The failure with bulky ketones is largely due to stelic effects. Cyanohydrin formation involves rehybridizing the sp 2 carbonyl carbon to sp 3 , narTowing the angle between the alkyl groups from about 1 20° to about 1 09.5°, increasing their steric interference. PRO B L E M 1 8-20

Propose a mechanism for each cyanohydrin synthesis just shown. Organic compounds containing the cyano group ( - C - N ) are called nitriles. A cyanohydrin is therefore an a-hydroxynitrile. Nitriles hydrolyze to carboxylic acids under acidic conditions (discussed in Section 2 1 -7D), so cyanohydrins hydrolyze to a-hydroxy acids. This is the most convenient method for making many a-hydroxy acids. OH 0 OH o "I a l II II R-C -C -OH R-C -CN R-C-H + HCN I I H H

aldehyde

cyanohydrin Showacetophenone how you would accompl ish thecyanohydri following syntheses. acetophenone I-hexanol hexanal cyanohydrin n C-H ('l U H PRO B L E M 1 8- 2 1

(a) (b)

o

II

�

�

a-hydroxy acid

839

Apheloria corrugata

840

Chapter 1 8 : Ketones and Aldehydes

1 8- 1 6 Formation of I m i nes

Under the proper conditions, either ammonia or a primary amine reacts with a ketone or an aldehyde to form an imine. Imines are nitrogen analogues of ketones and aldehydes, with a carbon-nitrogen double bond in place of the carbonyl group. Imines are commonly involved as synthetic intermediates, both in biosynthesis and in industrial synthesis. One of the best methods for making amines (both in organ­ isms and in the lab) involves making an imine, then reducing it to the amine (Section 1 9- 1 9) . Like amines, imines are basic; a substituted imine is also called a Schiff base. Imine formation is an example of a large class of reactions called condensations, reactions in which two (or more) organic compounds are joined, often with the loss of water or another small molecule. o

II C

/ '"

+

ketone or aldehyde

..

R - NH2 primary amine

[

OH I -C-

�

J

E )

R - -H

carbinolamine

'" / C

II R -N

+

HP

imine (Schiff base)

The mechanism of imine formation (Key Mechanism 1 8-5) begins with an acid­ catalyzed nucleophilic addition of the amine to the carbonyl group. Attack by the amine, followed by deprotonation of the nitrogen atom, gives an unstable intermediate called a carbinolamine. A carbinolamine converts to an imine by losing water and forming a double bond: a dehydration. This dehydration follows the same mechanism as the acid-catalyzed dehydration of an alcohol (Section 1 1- 1 0). Protonation of the hydroxyl group convelts it to a good leaving group, and it leaves as water. The resulting cation is stabilized by a resonance structure, with all octets filled and the positive charge on nitrogen. Loss of a proton gives the imine.

,.;��

•

KEY M EC H A N I S M 1 8- 5

This mechanism is more easily remembered by dividing it into two parts:' 1 . Acid-catalyzed addition of the amine to the carbonyl group. 2. Acid-catalyzed dehydration. First part: Acid-catalyzed addition of the amine to the carbonyl group. Step 1: Protonation of the carbonyl. Step 2: Addition of the amine.

:

..

: OH I -C-

..

R-NH2

Step 3:

�(==::z

1 + R-N-H----.. -.J -: I H20 : H

Deprotonation

..

O -H I

-c-

I

+

H3 0+

R - N-H

.

carbinolamine

'This mechanism takes place at slightly acidic pH. The amine can act as a strong nucleophile, so the first half of this mechanism (addition to the carbonyl) may be drawn as either acid-catalyzed or as base-catalyzed. The second half (dehydration) is acid-catalyzed, so the entire mechanism is shown here as acid-catalyzed to be consistent.

1 8 - 17 Condensations with Hydroxylamine and Hydrazines

843

same pathway but in reverse order. Therefore, the mechanism for hydrolysis of an imine is simply the reverse of the mechanism for its formation.

P R O B L E M 18- 25 Propose a mechanism for the hydrolysis of benzaldehyde methyl imine just shown.

Ketones and aldehydes also condense with other ammonia derivatives, such as hydroxyl­ amine and substituted hydrazines, to give imine derivatives. The equiljbrium constants for these reactions are usually more favorable than for reactions with simple amines. Hydroxylamine reacts with ketones and aldehydes to form oximes; hydrazine deriva­ tives react to form hydrazones; and semicarbazide reacts to form sernicarbazones. The mechanisms of these reactions are similar to the mechanism of imine formation.

(JY

H

"

O

(

�O

+

�

H+

oH

hydroxylamine

H H

benzaldehyde

"N NH2 / .. �

�

H+

H " .. N_ H_P----h' N -lL/ H

cyclohexanone

o

II

C / "

CH3

CH2CH3

2-butanone

I

+

phenyl-2-propanone oxime

hydrazine

+

�" .

H

phenyl-2-propanone H

" N ---@ill /

+

H dC

NiNH'1 �

+

1 8- 1 7 Condensations with Hyd roxylamine a n d Hyd razi nes

Hp

Hp

benzaldehyde hydrazone

I

phenylhydrazine

0N-!NH-Phl

+

cyclohexanone phenyl hydrazone

+ semicarbazide

2-butanone semicarbazone

These derivatives are useful both as starting materials for further reactions (see Section 19-19) and for characterization and identification of the original carbonyl compounds. ximes, semicarbazones, and phenylhydrazones are often solid com­ pounds with O characteristic melting points. Standard tables give the melting points of these derivatives for thousands of different ketones and aldehydes.

844

Chapter 1 8: Ketones and Aldehydes

Anhydrous hydrazine is a common rocket fuel. In most cases, liquid oxygen serves as the oxidizer. The fuel and oxidizer are sprayed into the

combustion chamber,

where

they react to generate heat and pressure, forcing the reaction prod­ ucts out through the rocket nozzle.

If an unknown compound forms one of these derivatives, the melting point can be compared with that in the table. If the compound's physical properties match those of a known compound and the melting point of its oxime, semicar­ bazide, or phenylhydrazone matches as well, we can be fairly certain of a correct identification. P R O B L E M 18- 26

2,4-Dinitrophenylhydrazine is frequently used for making derivatives of ketones and alde­ hydes because the products (2,4-dinitrophenylhydrazones, called 2,4-DNP derivatives) are even more likely than the phenylhydrazones to be solids with sharp melting points. Propose a mechanism for the reaction of acetone with 2,4-dinitrophenylhydrazine in a mildly acidic solution.

I

SU M MARY

Condensations of Amines with Ketones and Aldehydes

ZinZ-NH2

Reagent

Hir

-H

H2N

Product

� primary amine

-fQ[j hydroxylamine H2N -fNH2 l hydrazine H2N � phenylhydrazine

-R

H2N

-OH

-NHPh o

"

-NHCNH2

H2N

J

�

NH- -NH2

semicarbazide

PROBLEM-SOLVING

::C= N -tEJ an imine ::C=N -ill an imine (Schiff base) ::C= N -fQ[j an oxime ::c= N -fNH2 1 a hydrazone ::C= N � a phenylhydrazone

-tEJ ammonia

Hilti:;

Please learn these common

:: C�N

�

NH

J

-NH,

a semicarbazone

P R O B L E M 18- 27 Predict the products of the following reactions.

derivatives. You will see many

examples, espec ially in the laboratory.

1

(a)

�O U

+ HO-NH2

H+

�

o

(b)

ro

o "

(c) Ph-CH=CH-CHO + H2N-C-NH-NH2 o

\I

(d) Ph-C-Ph + Ph-NH-NH2

W �

I

18-16 Formation of Imines Second part: Step 4:

"

H

Acid-catalyzed dehydration.

:O:�

/

I I

(

-c-

R

"

N:

w

H

)

H H " / :0+

P I

-c-

R

carbinolamine

/ "

N:

/

Step 5:

Protonation of the -OHgroup.

�'

( )

R

H

protonated

c+

I

/ "

N:

'C/ / .... 6 Step 6:

Loss of H20. �

H

minor

II

R

"

N+

major

H

+

Deprotonation.

..

H)O: -

( )

"

R

intermediate (all octets filled)

/

/

c

II

N: + HO+ 3

imine

E XAMPLE : Formation of benzaldehyde methyl i m i ne First part: Step 1:

Acid-catalyzed addition of the amine to the carbonyl group.

Protonation of the carbonyl.

Step 2:

Addition of the amine.

Step 3:

Deprotonation . '-'::::

:0 /

1.0U benzaldehyde

Second part: Step 4:

Acid-catalyzed dehydration.

Protonation of the -OHgroup. Step 5: Loss of H20. H ",..-y H

'-'::::

1.0U

I

\

�

.

C -H

Step 6:

\

�

C- H

+

: -C H 3 H

Deprotonation.

.

: -C H 3 H

a carbinolamine

PROB L E M (a) What would happen if the reaction were made too acidic by the addition of too much acid? (b) What would happen if it were too basic?

The proper pH is crucial to imine formation. The second half of the mecha­ nism is acid-catalyzed, so the solution must be somewhat acidic. If the solution is too acidic, however, the amine becomes protonated and non-nucleophilic, inhibit­ ing the first step. Figure 18-8 shows that the rate of imine formation is fastest around pH

4.5 .

I

H

a carbinol amine

methylamine

0)

..

imine

841

842

Chapter

18:

Ketones and Aldehydes -- --.----

H R-N: � H+ H -

-

� Figure 18-8

H1+ R-N-H H

------- -�-.�--- ----

-

1

t

1

1

Although dehydration of the carbinolamine is acid-catalyzed, excess acid stops the first step by protonating the amine. Formation of the imine is fastest around pH 4.5.

nucleophilic

-------

---- ---.-

nonnucleophiLic

--_.._------

-------

-----------------

The following equations show some typical imine-forming reactions. In each case, that theof theC=O C Nnotice -R group imine.group of the ketone or aldehyde is replaced by the

NH er

=

NH3 a er NH' O (Y O ac�a CH3-NH2. ..

+

ammonia

H+

(

)

cyclohexanone

cyclopentanone

(

H+ )

.

(

H+ )

methylamine

benzaldehyde

PROBLEM-SOLVING

HinZ;

Imine formation is one of the important mechanisms in this chapter. It is more easily remembered as consisting of two simple mechanisms: 1. acid-catalyzed nucleophilic addition to the carbonyl, and 2. acid-catalyzed dehydration (as with an alcohol).

H2O

(YNlO H2O H....... C�N-CH' . H2 O O +

cyclopentanone phenyl imine

aniline

+

+

cyclohexanone imine

..

+

..

+

benzaldehyde methyl imine

P R O B L E M 18- 2 2

Propose mechanisms for the three imine-forming reactions just shown . P R O B L E M 18- 23

Depending on the reaction conditions, two different i mines of formula CgH9N might be formed by the reaction of benzaldehyde with methylamine. Explain, and give the structures of the two imines. PROB L E M 18- 24

Give the structures of the carbonyl compound and the amine used to form the following imines. NH (b)

II

C / ",

( Ny CH3

CH3

CH2CH3

(d) V

Imine formation is reversible, and mostofimmiinescroscopic can be hydrol yzed back(Section to the amine and the ketone or aldehyde. The principle reversibility states that the reverse reaction taking place under the same conditions should follow8-4A)the

18-18 Formation of Acetals P R O B L E M 18- 28

Show what amines and carbonyl compounds combine to give the following derivatives. II

o

(a) Ph-CH=N-NH-C-NH2

(b)

k DJ'

00

N-NHPh

NOH

(,)

Just asalcohols ketonesto and aldehydes Acetals react with waterofto theformmosthydrates, theyorganic also com­ react with form are some common pounds cottoncarbohydrate fabric, and aacetals woodenandshiptheir are allpolymers composedin of acetals.in theWeworld. coverTable thesesugar, common ChapterIn the 23. formation of an acetal, two molecules of alcohol add to the carbonyl group, and one molecule of water is eliminated. oII R'O """C / OR' + Hp C + 2 R'-OH " R/ ""H R / "" " H acetals.2

acetal

aldehyde o

II C R / """R' + 2 R"-OH

R"O """C/" OR" + H2O R/ ""AC) R' acetal (IUP ketal (common)

ketone

Although hydration is catalyzed bytheeither acid orofbase, acetal formation must be acid­ catalyzed. For example, consider reaction cyclohexanone with methanol, cat­ alyzed by p-toluenesulfonic acid. Overall reaction

o

6

+ 2 CHPH

o

-o-�

-OH

o (Ts-OH) p-toluenesulfonic acid

+ cyclohexanone dimethyl acetal

cyc1ohexanone

The addition mechanismto theforcarbonyl this reaction follows. Thecatalyst first protonates step is a typical acid­ catalyzed group. The acid the carbonyl group, and theof aalcohol (a weak nucleophile)charged attacksintermediate the protonated, activated car­ bonyl. Loss proton from the positively gives a The gets itsof name from thetheGreek prefixis halfwaymeaning "half." aHaving addedhemiacetal one molecule the alcohol, hemiacetal to becoming "full" hemi-,

hemiacetal.

'Acetals formed from ketones are often called ketals, although this term has been dropped from the IUPAC nomenclature.

1 8- 1 8 Formation of Acetals

845

846

Chapter Ketones and Aldehydes Like theandhydrates of ketones and aldehydes, most hemiacetals are too unstable toacetal. be The isolated purified. second haloff the of thehydroxyl mechanism converts thebyhemiacetal to thegives moreastable acetal. Protonation group, followed loss of water, reso­ nance-stabilized carbocation. Attack on the carbocation by met h anol, followed by loss of a proton, gives the acetal . 18:

:;�'�-=

•

KEY MECHANISM 18-6

Formation of Acetals

Like im ine formation, acetal formation is easily remembered by divid ing it into two simple processes: The first half is an acid-catalyzed addition of the alcohol to the carbonyl group. The second half begins like an acid-catalyzed dehydration, but then another alcohol adds. Acid-catalyzed addition of the alcohol to the carbonyl group. Alcohol adds. Protonation. Deprotonation. 1.

2.

First part: Step 1:

ketone

Step 2:

Step 3:

protonated (activated) ketone

hemiacetal

The second half begins like an acid-catalyzed dehydration Protonation. Loss of water.

Step 4:

Step 5:

hemiacetal

Step 6:

Second alcohol adds.

protonation, loss of water

resonance-stabilized carbocation

Deprotonation. HI H"""--"""':O-CH . .. I] CH'- -CH, Step 7:

o

.

attack by methanol

3

o-C:

CH'-

..

"

acetal

+

CH -O-H HI 3

P R O B L E M 18- 2 9

Propose a mechanism for the acid-catalyzed reaction of benzaldehyde with methanol to give acetal .

benzaldehyde dimethyl

Sinceishydration isonlycatalyzed byIneither acid or base, you might wonder why acetal formation catalyzed by acid. fact, the first step (formation of the hemiacetal) can beThebase-catalyzed, involving attack byofalkoxide ion and -OH protonation ofbythethealkox­ ide. second step requires replacement the hemiacetal group alco­so hol -OR" group. Hydroxide ion is a poor leaving group for the SN2 reaction, alkoxide cannot displace theprotonation -OH group. This-OH replacement occurs under acidic conditions, however, because of the group and loss of water gives a resonance-stabilized cation.

18-18 Formation

of Acetals

847

e?H I :OI � H00-R") -:O-R" R-C-R' R-C-R' I I � ·· :O-R" OR"

Attempted base-catalyzed acetal formation

poor leaving group

..

attack on ketone (or aldehyde)

hemiacetal

(no SN2 displacement)

Acetalof reactants formationandis products reversible,present and theat equi­ librium constant determines the proportions equi­ librium. For simple aldehydes, the equilibrium constants generally favor the acetal products. the acid-catalyzed reaction of acetaldehyde with ethanol gives a goodWith yieFor ldhindered ofexample, the acetal. aldehydes rather and with most ketones,Totheenhance equilibrium constants favor the carbonyl compounds than the acetals. these reactions, the alcohol isbyproduct often used as the bysolvent to assure a large excess. Thetoward water formed as a is removed distillation to force the equilibrium the right. Conversely, most acetals are hydrolyzed simply by shaking them with dilute acid inhyde.water. The large excess of water drives the equilibrium toward the ketone or al de­ The mechanism is simply the reverse of acetal formation. For example, cyclo­ hexanone dimethyl quantitatively hydrolyzed to cyclohexanone by brief treatment with diluteacetal aqueousis acid. Equilibrium of Acetal Formation

+ 2 CHPH P R O B L E M 18-30

Propose a mechanism for the acid-catalyzed hydrolysis of cyclohexanone dimethyl acetal.

Formation of more an acetal usingequilibrium a diol as theconstants, alcohol since givesthere a cyclic acetal. Cyclic acetal s often have favorable is a smaller entropy loss when two molecules (a ketone and a diol) condense than when three (a ketone andacetals; two molecules ofareancalled alcohol) condense. Ethylene glycol is oftenmolecules used to make cyclic its acetals (or thioacetalsDithiane (sulfur(Section acetals). 18-8) and its alkylated derivatives are examples of cyclic Cyclic Acetals

ethylene acetals

ketals).

ethylene

PROBLEM-SOLVING

HiltZ;

Acetal formation is one of the important mechanisms in this chapter. Remember it as a two-part process consisting of these two simple mechanisms: 1. Acid-catalyzed nucleophilic addition to the carbonyl group, and 2. SN1 by protonation and loss of the OH group, then attack by the alcohol.

848

Chapter

18:

Ketones and Aldehydes

HI HI

Fluocinolone acetonide is a steroid acetal used to treat skin conditions

/

solubility of the parent steroid,

H

I

allowing for a longer duration of action.

benzaldehyde

II

o H

H

formaldehyde

ethylene acetal

+ (l SH

H+

(

SH

)

propalle-I,3-dithiol

(l +

S�S

H 2O

dithiane

Sugars and othere, glucose carbohydrates most commonly exist asstable cyclicas acetals and hemiacetal s . For exampl i s a six-carbon sugar that i s most a hemiacetal. LactoseWeis adiscuss disaccharide of two sugar units) one acetal and one hemiacetal. the str (composed uctures of carbohydrates in detailthatinhasChapter Carbohyd rates

23.

I

CHO

OH

HO

H

H

H

/ "'C

+

benzaldehyde

ethylene glycol

0

fluocinolone acetonide

a9 N �N . :CNJ H

0N

4

?

2-methylpyridine

I

piperidine

.

pyrimidine

P R O B L E M 19- 1

I

5

4

N7 � .

\

19-3 Structure of Amines

8

purine

Determine which of the heterocyclic amines just shown are aromatic. Give the reasons for your conclusions.

1 9-2B

IU PAC Names

con­ longest forThealcohols.endiThe to that nes is similar for ami nomenclature IUPAC The alkane the in g n name. root the determines atoms carbon of chain tinuous the aminoandgroup of numbers, the position shows chain a onnumber to substitand is changed name the given are carbon the uents Other chain. the along prefix is used for each substituent on nitrogen. -e

-amine,

N-

2-butanamine

3-methyl- l - butanamine

CH3 CH3 I I CH3CH2CH

�

+

Ph - Cl

N2+ Cl -

� 02N

N2 I

&

chlorobenzene

benzenediazonium chloride

02N

Ar - X + Nz f

CI, Br, C = N

C =N

�

+ N2 1

p-nitrobenzonitrile

p-nitrobenzenediazonium chloride

(III) Replacement by fluoride or iodide

+ Ar - N = N: Cl-

�

+ Ar - N = N: BF4-

+ Ar - N = N: Cl -

�

Ar -I

HBF4

KI

+

N2 I

+

heat

---7

Ar -F

+

N2 I

+

BF3

KCl

Example

2-iodonaphthalene

(IV) Reduction to hydrogen

� NH2

Example

�

)

( 1 ) NaN02, HCI

CH3CH2

ethyl benzene

p-ethylaniline

(V) Diazo coupling +

Ar - N = N : + H - Ar' diazonium ion

(activated)

�

Ar -N = N - Ar' + H + an azo compound

Example

.. � -lQ\� N.. = N.. � �. N(CH3)z

o

-O-

11

o

methyl orange (an indicator)

+

HCl

1 9-19

Synthesis of Amines by Reductive Amination

Many methods are available for making amines. Most of these methods derive from the reactions of amines covered in the preceding sections. The most common amine syntheses start with ammonia or an amine and add another alkyl group. Such a process converts ammonia to a primary amine, or a primary amine to a secondary amine, or a secondary amine to a tertiary amine. :NH3

-----;.

-----;. R -NH 2

" .. N -H /

1 9- 1 9 Synthesis of Amines by Reductive Amination

1° amine

ammonia

-----;. -----;.

I ° or 2° amine

" .. N-R

/ 2° or 3° amine

i s the most general amine synthesis, capable of adding a primary or secondary alkyl group to an amine. Reductive amination is a two-step procedure. First we form an imine or oxime derivative of a ketone or aldehyde, and then reduce it to the amine. In effect, reductive amination adds one alkyl group to the nitrogen atom. The product can be a primary, secondary, or tertiary amine, depending on whether the starting amine had zero, one, or two alkyl groups. Reductive amination

o

II

" .. N-H

( I ) R - C - R'

(2) reduction

/

H " .. I ) N-C-R / I R'

Pri mary Ami nes Primary arnines result from condensation of hydroxylamine (zero alkyl groups) with a ketone or an aldehyde, followed by reduction of the oxime. Hy­ droxylamine is used in place of ammonia because most oximes are stable, easily iso­ lated compounds. The oxime is reduced using catalytic reduction, lithium aluminum hydride, or zinc and HC!.

0

� Examples

II R-C-R'

H zi'.J - OH

ketone or aldehyde

H+

N-OH II ) R-C-R'

reduction

NH? I ) R-CH-R' 1 ° amine

oxime

II CH3CH2CH2 -C -CH3

N-OH II CH3CH2CH2 -C -CH3

NH2 I CH3CH2CH2 -CH -CH3

2-pentanone

2-pentanone oxime

2-pentanamine

o

@-�

@-�

N-OH

o

-H

benzaldehyde

-H

benzaldehyde oxime

( I ) LiAIH4

) benzylamine

Seco n d a ry A m i n es Condensation of a primary amine with a ketone or alde­ hyde forms an N-substituted imine (a Schiff base). Reduction of the imine gives a secondary amine.

911

91 2

Chapter 1 9: Amines o

II R-C-R' ketone or aldehyde

1 ° amine R" - NH? -)

N-R" II R- C -R'

reduction

---�)

NHR" I R-CH-R' 2° amine

N-substituted imine

Example

o

II CH3-C-CH3

( I ) LiAlH4 (2) H20

acetone

NHPh I CH3- CH -CH3 phenylisopropylamine

(75 %)

Condensation of a secondary amine with a ketone or aldehyde gives an iminium salt. Iminium salts are frequently unstable, so they are rarely isolated. A reducing agent in the solution reduces the iminium salt to a tertiary amine. The reducing agent must reduce the iminium salt, but it must not reduce the carbonyl group of the ketone or aldehyde. Sodium triacetoxyborohydride [Na(CH3COOhBH or Na(AcO)3BH] is less reactive than sodium borohydride, and it does not reduce the carbonyl group. Sodium triacetoxyborohydride has largely replaced an older reagent, sodium cyanoborohydride (NaBH3CN), which is more toxic and not as effective.

Terti a ry A m i nes

2° amine

0

c::tr PROBLEM-SOLVING

HtnZ;

Reductive a m ination is the most useful a m i ne synthesis: It adds a

1 0 or 20 a l kyl group to nitrogen. Use a n a l dehyde to add a 1 0 group,

and a ketone to add a 20 group.

H I Y-N-H

R - NH-R

II R'-C-R"

H+

.

/

.

.

1

[Na(AcO)3BH to make tertiary ammes

]

R

" R

. LIAlH4

H H I I Y-N - C - R I .. R

----

hydroxylamine -- primary amine primary amine

secondary amine

secondary amine tertiary amine

R,- -R"

Na(CH3C00)3BH CH3COOH

iminium salt

ketone or aldehyde

R-N-R I ) R'-CH- R" 3° amine

Example

o

6

Na(CH3COOhBH

(85 %)

CH3COOH

cyclohexanone

Y-N=C .

[ g+ J R-N-R

iminium salt

N,N-dimethycyclohexylamine

SO L VED P R O B L EM 1 9 -5 Show how to synthesize the following amines fro m the indicated starting materials. (a) N-cyclopentylaniline fro m aniline

(b) N-ethylpyrrolidine from pyrrolidine

SO L U TIO N

(a)

This synthesis requires adding a cyclopentyl group to aniline (primary) to make a sec­ ondary amine. Cyclopentanone is the carbonyl compound.

I

H Ph-�-H aniline

+ 0

=(J � H+

cyclopentanone

(b) This synthesis requires adding

an

Ph -

N

=(J

� -(J H

Ph-

ethyl group to a secondary amine to make a tertiary

amine. The carbonyl compound is acetaldehyde. Formation of a tertiary amine by reductive

1 9-20 S ynthesis of Amin e s amination involves an iminium intennediate, which is reduced by triacetoxyborohydride).

C

..

O

,I'

N - H + CH3-C

pyrrolidine

" H

(

H+

)

acetaldehyde

�C

]

/

H

+

N =C

"

by Acylation-Reduction

913

Na(AcOhBH (sodium H

C·· ?I

N - -H

CH3

CH3

PROB LEM 1 9-26 Show how to synthesize the following amines from the indicated starting materials by reduc­ tive amination. (a) benzylmethylamine from benzaldehyde

I

: NH2 (b)

Ph -CH2-CH- CH3 I-phenyl-2-propanone

( ::!:: )-amphetamine

(c) N-benzylpiperidine from piperidine (d) N-cycIohexylaniline from cyclohexanone (e) cycIohexylamine fro m cyclohexanone (f)

C -o N

from compounds containing no more than five carbon atoms

The second general synthesis of amines is acylation-reduction. Like reductive ami­ nation, acylation-reduction adds one alkyl group to the nitrogen atom of the starting amine. Acylation of the starting amine by an acid chloride gives an amide, which is much less nucleophilic and unlikely to overacylate (Section 1 9- 1 3). Reduction of the amide by lithium aluminum hydride ( LiAIH4 ) gives the corresponding amine.

..

I

.. I

o

o

R - NH2 + Cl-C-R' amine

acid chloride

acylation pyridine or NaOH

)

R - NH -C-R'

reduction ( I ) LiAIH4

---�)

arrude

(2) H20

1 9-20 Synthesis of Amines by Acylation-Reduction

R- NH- CH2-R' alkylated arrune

Acylation-reduction converts ammonia to a primary amine, a primary amine to a secondary amine, or a secondary amine to a tertiary amine. These reactions are quite general, with one restriction: The added alkyl group is always 1 0 because the carbon bonded to nitrogen is derived from the carbonyl group of the amide, reduced to a methylene ( - CH2 - ) group. Primary amines

I

I

0

0

r:Jr

R - C - Cl acid chloride

Example CH3

I

I

+

�

NH3 ammonia

1 0 arrude

3-methylbutanoyl chloride

I

CH3

0

CH3- CH- CH2-C - CI

( 1 ) LiAlH4

(2) H2O

R-C- NH2

+

NH3

�

)

R- CH2- NH2 1 0 amine

I ..

0

CH3- CH-CH2-C- NH2

( I ) LiAIH4

)

3-methylbutanamide

3-methyl-I-butanamine

914

Chapter 19: Amines

Secondary amines

"

R - C - Cl acid chloride

I

o

o

+

R'- NH2 primary amine

�

Example

\

. •

R-C- NH - R ' N-substituted amide

( 1 ) LiAIH4 ) (2) H20

R - CH2- NH - R ' 20 amine

+

butanoyl chloride

ani l i ne

Tertiary amines

\I

0

0

r::ir Example

N-butylaniline

N-phenylbutanamide

R - C - Cl + R2NH secondary acid chloride amine

�

\I

R - C- NRz N, N-disubstituted amide

PROBLEM-SOLVING

HiltZ;

Like reductive a m i nation, acylation-red uction adds an a l kyl group to nitrogen. It i s more restrictive, though, because the g ro u p added i s always 1 °.

X-N-H

,

y

..

jRJ-Cl o

\I

X-N -C-R

,

Y

1

LiAIH4

(2) H2 O

) R - CH2- NRz 30

diethylamine

N,N-diethylbenzamide

amine

(CH3CH2hN ( 1 ) LiAIH4

+ benzoyl chloride

( I ) LiAIH4

(2) H20

""

©

)

benzyldiethylamine

SO L VED P R O B L E M 1 9 - 6

Show how to synthesize N-ethylpyrrolidine from pYITolidine using acylation-reduction. SO L U TIO N

This synthesis requires adding an ethyl group to pyrrolidine to make a tertiary amine. The acid chloride needed will be acetyl chloride (ethanoyl chloride). Reduction of the amide gives N-ethylpyrrolidine.

C· ·

,f'0 pyri d i ne

--�) N-H + CH3-C, " CI pyrrolidine acetyl chloride

C· · I

°

( I ) LiAIH4

N-C-CH3 -----''-i>l (2) H20

C

·· I

H

N-C - H , CH3

X - N - CH2- R

Compare this synthesis with Solved Problem 1 9-5(b) to show how reductive amination and acylation-reduction can accomplish the same result.

ammonia � primary amine primary amine � secondary amine secondary amine � tertiary amine

Show how to synthesize the following amines from the indicated starting materials by acyla­ tion-reduction. (a) N-propylpiperidine from piperidine (b) N-benzylaniline from aniline

�I

1 0 group added

P R O B L EM 1 9 - 2 7

1 9-2 1 Syntheses Limited to Primary Amines

Primary amines are the most common class of amines, and they are also used as start­ ing materials for synthesis of secondary and tertiary amines. Many methods have been developed for making primary amines, ranging from simple alkylation of ammonia to sophisticated multistep syntheses. We will consider some of the more common syntheses.

1 9-2 1 A

1 9-21 Syntheses Limited to Primary Amines

D i rect A l kylation and G a briel Synthesis

The SN2 reaction of amines with alkyl halides i s complicated by a tendency for over-alkylation to form a mixture of monoalkylated and polyalkylated products (Section 1 9- 1 2) . S i mple primary amines can be synthesized, however, by adding a halide or tosylate ( must be a good SN2 substrate) to a large excess of ammonia. Because there i s a large excess of ammonia present, the probability that a molecule of the halide will alkylate ammonia is much larger than the probability that it will overalkylate the amine product.

Example I-pentanamine

I -bromopentane

P R O B L EM 1 9 - 2 8 Show how you would use direct alkylation of ammon i a to synthesize I -heptanamine.

In

1 887, Siegmund Gabriel (at the University of Berlin) developed the

Gabriel

for making primary amines without danger of overalkylation. He used the phthalimide anion as a protected form of ammonia that cannot alkylate more than once. Phthalimide has one acidic N - H proton ( pKa 8.3) that is abstracted by potassium hydroxide to give the phthalimide anion.

amine synthesis

N -

�N - H

�

o

9·

phthalimide

resonance-stabilized phthalimide anion

The phthalimide anion is a strong nucleophile, displacing a halide or tosylate ion from a good SN2 substrate. Heating the N-alkyl phthalimide with hydrazine displaces the primary amine, giving the very stable hydrazide of phthalimide.

qj=

cgQN ' -

0 phthalimide anion

0

0

0 R-X

alkyl halide (usually n

)

cgQN -R

0 N-alkyl phthalimide

H?N - NH?heat

)

915

H

+

©¢ (

o

R -NH2

H

phthalimide hydrazide

primary amine

916

Chapter

1 9 : Amines

Example

©rJN-C 2C J:� H

(phthalimide anion)

H

H3

hydrazine

o

isopentyJamine (95 %)

N-isopentyJphthalimide

isopentyl bromide

PRO B L E M 1 9-29

Show how Gabriel syntheses might be used to prepare the following amines. (c) y-aminobutyric acid (b) I -hexanamine (a) benzylamine

1 9-21 B

Red uctio n of Azides and N itriles

Just as Gabriel used the anion of phthalimide to put the nitrogen atom into a primary amine, we can use other nucleophiles as well. We need a good nucleophile that can alkylate only once and that is easily converted to an amino group. Two good nucle­ ophnes for introducing a nitrogen atom are the azide ion and the cyanide ion. Azide ion introduces (after reduction) an - N H z group, and cyanide ion introduces a CH z N Hz group.

- -

Formation and Reduction of Azides Azide ion (-N3) is an excellent nucleophile that displaces leaving groups from unhindered primary and secondary alkyl halides and tosylates. The products are alkyl azides ( RN3 ), which have no tendency to react further. Azides are easily reduced to primary amines, either by LiAIH4 or by catalytic hydro­ genation. Alkyl azides can be explosive, so they are reduced without purification.

R-X

-

+

halide or tosylate (must be I ° or 2°)

Na+ :N=N=N: . 0

+

-

sodium azide

[R-N=N=N:

• •

+

�

+ R-N-N=N:] .

'

or

an alkyl azide

H2/Pd

)

R-NH2 1° amine

Examples

@-CH2CH2-N=N=N: I -bromo-2-pheny lethane

cyclohexyl bromide

2-phenylethyl azide

cyclohexyl azide

( I ) LiAIH4 (2) H20

( I ) LiAIH4 (2) H20

@-CHZCH2 -NH2 2-phenylethylamine (89 %)

cyclohexylamine (54%)

Azide ion also reacts with a variety of other electrophiles. The following example shows how an azide ion opens an epoxide, and the product can be reduced to an amino alcohol:

�: N=N=N: O C; o

..

H

epoxycyclohexane

Formation

and

Reduction

..

of

)

N rJL.� �OH if

Nitri les

H2, Pd

)

� :2 V�OH if

Like the azide ion, cyanide ion

C:C = N : ) is a good SN2 nucleophile; it displaces leaving groups from unhindered

1 9-2 1 Syntheses Limited to Primary Amines

917

primary and secondary alkyl halides and tosylates. The product i s a nitrile ( R - C = N ) , which has no tendency to react further. Nitriles are reduced to primary amines by lithium aluminum hydride or by catalytic hydrogenation. R-X

Example

+

halide or tosylate (must be 10 or 20)

-: C - N :

�

R -C == N : nitrile

CH3CH2CH2-C

or H2/catalyst

( I ) LiAIH4

N:

(2) H20

butanenitrile

R-CH2 -NH2 amme (one carbon added)

CH3CH2CH2-CH2 -NH2

)

I -butanamine (70%)

I-bromopropane

When the cyano ( - C - N ) group i s added and reduced, the resulting amine has an additional carbon atom. In effect, the cyanide substitution-reduction process is like adding - CH2 - NH2. The following synthesis makes 2-phenylethylamine, which we also made by the azide synthesis:

� H2

Ni

-

CH2CH2NH2

2-phenylethylamine

phenylacetonitrile benzyl bromide

Notice that the starting material in this case has one less carbon atom because the cyanide synthesis adds both a carbon and a nitrogen. We have seen (Section 18- 15) that cyanide ion adds to ketones and aldehydes to form cyanohydrins. Reduction of the - C = N group of the cyanohydrin provides a way to synthesize ,B-hydroxy amines.

HeN cyclopentanone

�OH U �CN

( 1 ) LiAIH4

cyclopentanone cyanohydrin

)

U

OH CH2NH2

l -(methylamino)cyclopentanol

PRO B L E M 1 9-30 Show how you would accomplish the following synthetic conversions. (a) benzyl bromide � benzylamine (b) I -bromo-2-phenylethane � 3-phenylpropanamine (c) pentanoic aci d � I -pentanamine (d) pentanoic acid � I - hexanamine (e) (R)-2-bromobutane � (S)-2-butanamine (0 (R)-2-bromobutane � (S)-2-methyl- l -butanamine (g) 2-hexanone � l -amino-2-methyl-2-hexanol

1 9-2 1 C

Redu ction of N itro Com p o u n d s

B oth aromatic and aliphatic nitro groups are easily reduced t o amino groups. The most common methods are catalytic hydrogenation and acidic reduction by an ac­ tive metal.

PROBLEM-SOLVING

Htnp

To convert an a l ky l ha l ide (or alcohol, via the tosylate) to an a m ine, form the azide and reduce. To convert it to an a m i n e with an additional carbon atom, form the nitrile and reduce. In either case, the a l kyl g ro u p must be suitable for SN2 displacement.

918

C hap ter 19: Amines H2!catalyst

R -N02

)

or

R -NH2

active metal and H + catalyst N i , Pd, o r Pt active metal Fe, Zn, or S n =

=

©r

Examples

NO'

H?, N i �

CH3

©r

NH2 CH3

o-toluidine

o-nitrotoluene

I

I-

(90 %)

:NH?

N02

CH3CH2CH2- CH-CH3

CH3CH2CH2- CH - CH3 2-n itropentane

2-aminopentane (85 %)

The most common reason for reducing aromatic nitro compounds is to make substituted anilines. Much of this chemistry was developed by the dye industry, which uses aniline derivatives for azo coupling reactions (Section 19- 18) to make aniline dyes. Nitration of an aromatic ring (by electrophilic aromatic substitution) gives a nitro compound, which is reduced to the aromatic amine. A r - NO2

Ar - H

reduction

)

For example, nitration followed by reduction is used in the synthesis of benzocaine (a topical anesthetic), shown below. Notice that the stable nitro group is retained through an oxidation and esterification. The final step reduces the nitro group to the relatively sensitive amine (which could not survive the oxidation step).

I

0

HN03 H2S04 nitration

�

C - OH

(I)

KMn04. - OH

(2) H +

oxidation

N02

)

¢I 0

¢ N02

CH3CH20H, H + (see Section

1 1 - 1 2)

)

OCH'CH3

esterification

¢I

N02

o

Zn, HCI CH3CH20H

OCH,CH3

)

reduction

:NH2 ' HC] benzocaine · HC]

PRO B L E M 1 9-3 1 Show how you would prepare the following aromatic amines by aromatic nitration, followed by reduction. You may use benzene and toluene as your aromatic starting materials. (a) aniline (b) p-bromoaniline (c) m-bromoaniline (d) m-aminobenzoic acid

Syntheses Limited to Primary Amines

1 9-2 1

1 9- 2 1 D

919

T h e Hofm a n n Rearra n gement o f A m ides

In the presence of a strong base, primary amides react with chlorine or bromine to form shortened carbon atom. This reaction, called the amines with loss of istheusedcarbonyl to synthesize primary alkyl and aryl am1l1es. Hofmann rearrangement,

The Hofinann rearrangement o

II . . + X2 + 4 NaOH R -C-NH2 primary amjde

(X2

=

R-NH2 + 2 NaX + Na2C03 + H20 2

Cl2 or 8r2)

amine

Although we have several other methods for making primary amines, most of them depend on SN2 displacements, which fail with 3° alkyl groups. The Hofmann rearrangement can produce primary amines with 1 °, 2°, or 3° alkyl groups, or aryl amines. The following examples suggest the wide variety of amines that are accessible by the Hofmann rearrangement: 0II ) CH3CH2CH2CH2CH2- NH2 CH3CH2CH2CH2CH2-C-NH2 Examples

C12. - O H H2O

hexanamjde

CH3 0 @o -C-NH2 I

T CH3

II

C12. -OH H2O

I -pentanamjne (90 %)

)

2-methyl-2-phenylpropanamide

02N

-@-

0 �-NH2

@-{

CH3 -NH2 CH3

2-phenyl-2-propanamine

)

8r2. -OH H2O

02N

-@-NH2 p-nitroaniline

p-nitrobenzamide

The mechanism of the Hofmann rearrangement is particularly interesting because it involves some intermediates that we have not encountered before. The first step is the replacement of one of the hydrogens on nitrogen by a halogen. This step is possible because the amide - H protons are slightly acidic, and a strong base deprotonates a small fraction of the amide molecules. The deprononated amide is a strong nucleophile, and it attacks bromine to form the N-bromo amide. N

MECHANISM 1 9-7 Part 1:

The Hofmann Rearrangement of Amides

Deprotonation of the amide and nucleophilic attack on bromine. rOH H II R-C-N: H o

f7

"­

primary amjde

'�' .. [R-C-N

(

H

�

:9:R-C=N\

deprotonated amide

l

H

o

Br / R-C-N: H II

"­

N-bromo amide

( Continued)

920

Chapter 19: Amine s

Part 2: Second deprotonation. The presence of a leaving group (bromine) allows the alkyl group to migrate and the leaving group to leave. o

f7 R -C-N: "

II

f-OH H

(

)

Br

N-bromo amide

[

R_

N-Br .. �

deprotonated

�

R-N=C=Q:

+ Br"

an isocyanate

Part 3: Isocyanates react rapid l y with water to give carbamic acids. Hydroxide ion attacks the carbonyl group of the isocyanate. .. 1:. R-N=C=O: � � OH isocyanate

�

:0:.. I

0 " _ II '

R-N=C-OH

�

R-t'{

'

J

�

HrO-H

H

I

0

II

) R-t'{-C-OH + - OH a carbamic acid

Part 4: Carbamic acids tend to lose CO2 spontaneously. Decarboxylation (loss of CO2) gives the amine. o

..

II

(\ �OH

R - NH-C-O-H

�

..

o

II �

R-Nt!,7 C-O -

R-f'F-H

+ O=C=O

PRO B L E M 1 9-32 Propose a mechanism for the following Hofmann rearrangement used i n the synthesis of phentermine, an appetite suppressant.

0 _,

PROBLEM-SOLVING

Htltl/

The Hofmann rearrangement

mechanism is long and com p l icated, but it can be broken down: 1 . Deprotonation and bromi nation to form the bromoa m ide, then another deprotonation.

2 . Rearrangement to a n isocyanate,

3.

4.

with bromide as the leaving group. Hydroxide attack on the carbonyl of the isocyanate. Decarboxylation of the carba m i c acid.

(Step 2, the actual rearrangement, may be easier to understand if you

compare it with the Curtius rearra ngement, in Problem

1 9-34.)

0 ,

�I �

CH3 0

CH2 - - -NH2

_

CH3

CH3

�I

CH2 - - NH2

CH3 phentermine

P R O B L E M 1 9-33 When (R)-2-methylbutanamide reacts with bromine in a strong aqueous solution of sodium hydroxide, the product is an optically active amine. Give the structure of the expected prod­ uct, and use your knowledge of the reaction mechanism to predict its stereochemistry. * P R O B L E M 1 9-34 The Curtius rearrangement accomplishes the same synthetic goal as the Hofmann rearrange­ ment, and it takes place by a similar mechanism. An acid chloride reacts with azide ion to give an acyl azide, which undergoes Curtius rearrangement when heated.

R

o

II

C / " Cl

IX . + lR

�

- W-N==N:

/.��

+

l J

R �t'{ � ==N acyl azide

R-t'{=C=Q: isocyanate

19- 2 1

Syntheses Limited to Primary Amines

921

The Curtius rearrangement takes place through a much shorter mechanism than the Hofmann rearrangement. Which step(s) of the Hofmann rearrangement resemble the Curtius rearrangement? (b) Bromide serves as the leaving group in the Hofmann rearrangement. What is the leaving group in the Curtius rearrangement? (c) Propose a mechanism for the following reaction:

(a)

I

S U M MARY

Synthesis of Amines

1 . Reductive amination

a. Primary amines

Example

(Section 19-19)

I

b. Secondary amines

1\

uN

R - C - R' ketone or aldehyde

I

o

CH3 - C- CH3 acetone

c. Tertiary amines

I

o

R'- C -R" ketone or aldehyde

-OH

1° amine R" - NH?H

+

Ph- N H2

)

I

:N-R"

R - C - R' N-substituted imine

(2) H20

I

:NHR"

I

reduction

( 1 ) LiAlH4

)

R - CH -R' 1 ° amine

cyclopentylamine

cyclopentanone oxime

o

I-

:NH?

reduction

R - C-R' oxime

R -C-R' ketone or aldehyde

cyclopentanone

Example

I

:N-OH

o

)

R - CH -R' 2° amine

I

N

CH3 - CH - CH3 phenylisopropylamine

+

2° amine R-N H - R

)

R - N-R

R'- C - R" iminium salt

HPh

Na(CH 3 COOhBH

)

N R- b - R

R,- H -R" 3° amine

Example

6 cyclohexanone

iminium salt

N,N-dimethylcyclohexylamine

( Continued)

922

Chapter 1 9 : A mines

2. Acylation-reduction of amines

(Section 1 9-20)

II

II

R - NH2 + R'- C - CI

acylation

----"---0»

reduction

R' -C -NH -R

( 1 ) LiAIH4 (2) HzO

amide (acylated amine)

acid chloride

amine

..

o

o

)

N R'- CH2- H-R alkylated amine

Example

N-phenylbutanamide

butanoyl chloride

ani line

( 1 ) LiAIH4

(2) HzO

3. Alkylation of ammonia (Section Example

N-butylaniline (2°)

1 9-21 A)

N R-CH2 - X + ex c e ss H3

�

(excess)

N R-CH2- H2 + HX

,

benzyl bromide 4. The Gabriel synthesis ofprimary amines

�N: o

(Section 1 9-21A)

K+

«" 0

----------� ) ' phtha l'lffil'de anton

N- R

heat

o

N-alkyl phthalimide

5. Reduction of azides (S ec ti o n "

benzylamine

o

R -X alky I halide

+

1 9-2I B) " -

R -N=N=N: alkyl azide

or

Hz/Pd

Example

)

( 1 ) LiAIH4

6. Reduction of nitriles

cyclohexyl azide

(Section 1 9-2IB) nitrile

N R - H2 1° amine

N R- H2 1° amine (2) H20

cyclohexyl bromide

)

H ©f N-CH2CH2CH2CH3

Hz/catalyst or LiAIH4

)

cyclohexyJamine

N R-CH2- H2 1° amine

Chapter 1 9 Glossary

Example

benzyl bromide

7. Reduction of nitro compounds

(Section 1 9-2 1C)

phenylacetonitlile H2/catalyst

active metal and H+

or

)

catalyst Ni, Pd, or Pt metal Fe, Zn, or Sn =

active

Example

,B-phenylethylamine N R - H2

=

nitrobenzene

8. The Hofmann rearrangement

Example

aniline

(Section 1 9-2 1 D)

1 0 amide

amine

II

o

CH3CH2CH2CH2CH2 - C-NH2 hexanarnide

9. Nucleophilic aromatic substitution

pentanarrune

(Section 1 7- 1 2)

R -NH2 + Ar-X ----'> R-NH-Ar + HX (The aromatic ring should be activated toward nucleophil.ic attack.)

�

O?N

Example

ethylamine

CH3CH2 -NH

2,4-dinitrofluorobenzene

N-ethyl-2,4-dinitroaniline

I

o

acylation Addition of an acyl group (R - C -), usually replacing a hydrogen atom. Acylation of an amine gives an amide. (p. 893)

I

II

o

o

R -NH2 + Cl - C - R ' acid chloride amine acetylation :

�

R -NH-C-R' + HCl amide

II

o

Acylation by an acetyl group (CH3 -C-).

N02

Chapter 1 9 G lossa ry

923

924

Chapter 1 9: Amines

acylation-reduction A method for synthesizing amines by acylating ammonia or an amine, then reducing the amide. (p. 9 1 3)

..

I

o

II

..

o

( 1 ) LiAIH4

) R - NH2 + R ' - C - CI � R - NH - C-R' alkylated amine amide acid chloride amine amine A derivative of ammonia with one or more alkyl or aryl groups bonded to the nitrogen atom. (p. 870) A primary amine: ( 1 0 amine) has one alkyl group bonded to nitrogen. A secondary amine: (2° amine) has two alkyl groups bonded to nitrogen. A tertiary amine: (30 amine) has three alkyl groups bonded to nitrogen. H R" H

I

R -N"- H primary amine

I

R - N"-R' secondary amine

I

R - N"- R' tertiary amine

amino group: The - NH2 group. If alkylated, it becomes an alkylamino group, -NHR or a dialkylamino group, - NR2. (p. 872) amine oxide An amine with a fourth bond to an oxygen atom. In the amine oxide, the nitrogen atom bears a positive charge, and the oxygen atom bears a negative charge. (p. 900)

0-

I

I

R-N±..... R '

R" an amine oxide

R-NH3+ X­ an ammonium salt

R

�

R- ±""'R X-

I

R a quaternary ammonium salt

ammonium salt (amine salt) A derivative of an amine with a positively charged nitrogen atom having four bonds. An amine is protonated by an acid to give an ammonium salt. (p. 880) A quaternary ammonium salt has a nitrogen atom bonded to four alkyl or aryl groups. (p. 87 1 ) azide A compound having the azido group, - N3. (p. 9 1 6)

[CH3CH2-f'F-N ==N:

j

N N CH3CH2- =N = :ethyl azide �

base-dissociation constant (Kb) A measure of the basicity of a compound such as an amine, defined as the equilibrium constant for the following reaction. The negative log 10 of Kb is given as pKb. (p. 877)

1 I

H

+ R-N-H + H

Cope elimination A variation of the Hofmann elimination, where a tertiary amine oxide elim­ inates to an alkene with a hydroxylamine serving as the leaving group. (p. 900) diazo coupling The use of a diazonium salt as an electrophile in electrophilic aromatic substi­ tution. (p. 907)

Ar- N ==N + H

diazonium ion

-@(activated)

Y

diazotization of an amine The reaction nium salt. (p. 902) exhaustive alkylation Treatment of an

�

Ar-N=N

-@-

an azo compound

Y + H+

of a primary amine with nitrous acid to form a diazo­

amine with an excess of an alkylating agent (often methyl iodide) to form the quaternary ammonium salt. (p. 892) R -NH2

+ excess CH3I ) R - N(CH3 h Iexhaustive methylation of a primary amine

Gabriel amine synthesis Synthesis of primary amines by alkylation of the potassium salt of phthalimide, followed by displacement of the amine by hydrazine. (p. 9 1 5)

Elimination of a quaternary ammonium hydroxide with an amine as the leaving group. The Hofmann elimination usually gives the least-substituted alkene. (p. 897)

Hofmann elimination

�

H

H

It I R-C-C-H

HO-

In

heat

�

H + N(CH3h

Hofmann rearrangement of amides (Hofmann degradation) Treatment of a primary amide with sodium hydroxide and bromine or chlorine gives a primary amine. (p. 9 1 9)

The compound H2NOH; or generically, an amine in which a hydroxyl group is one of the three substituents bonded to nitrogen. (p. 900)

hydroxylamine

R'

I

R -N-OH

nitrile A compound of formula R - C - N, containing the cyano group, -C = N. nitrogen inversion (pyramidal inversion) Inversion of configuration of a nitrogen

(p. 9 1 6) atom in which the lone pair moves from one face of the molecule to the other. The transition state is pla­ nar, with the lone pair in a p orbital. (p. 874) N-nitrosoamine (nitrosamine) An amine with a nitroso group ( - N 0) bonded to the amine nitrogen atom. The reaction of secondary amines with nitrous acid gives secondary N-nitrosoamines. (p. 903) phase-transfer catalyst A compound (such as a quaternary ammonium halide) that is soluble in both water and organic solvents and that helps reagents move between organic and aqueous phases. (p. 882) reductive amination The reduction of an imine or oxime derivative of a ketone or aldehyde. One of the most general methods for synthesis of amines. (p. 9 1 1 ) N N HR" - R" o =

II

II

reduction

I

) R- CH- R' R- C-R' 2° amine N-substituted imine + Sandmeyer reaction Replacement of the - N = N group in an arenediazonium salt by a cuprous salt; usually cuprous chloride, bromide, or cyanide. (p. 905)

R-C-R' ketone or aldehyde

CuX

------�)

(X

sulfonamide

=

Cl, Br, C=N)

An amide of a sulfonic acid. The nitrogen analogue of a sulfonate ester. (p. 895)

II

o

o

R - NH - S - R '

II o

I 1. 2.

3. 4.

Ar-X + N2 i

a sulfonamide

R - NH -

U--

� CH3 IIo�

a p-toluenesulfonamide (a tosylamide)

Essential Pro b l e m -Solving S k i l l s i n Chapter 1 9

Name amines, and draw the structures from their names.

Interpret the IR, NMR, and mass spectra of amines, and use the spectral information to determine the structures. Explain how the basicity of amines varies with hybridization and aromaticity. Contrast the physical properties of amines with those of their salts.

Chapter 1 9 Glossary

925

926

Chapter 19: A mi nes

Predict the products of reactions of amines with the following types of compounds; propose mechanisms where appropriate. (a) ketones and aldehydes (b) alkyl halides and tosylates (c) acid chlorides (d) sulfonyl chlorides (e) nitrous acid (t) oxidizing agents (g) arylamines with electrophiles 6. Give examples of the use of arenediazonium salts in diazo coupling reactions and in the synthesis of aryl chlorides, bromides, iodides, fluorides, and nitriles. 5.

7.

8.

9. 10.

Illustrate the uses and mechanisms of the Hofmann and Cope eliminations, and predict the major products.

Use your knowledge of the mechanisms of amine reactions to propose mechanisms and products of similar reactions you have never seen before.

Show how to synthesize amines from other amines, ketones and aldehydes, acid chlo­ rides, nitro compounds, alkyl halides, nitriles, and amides.

Use retrosynthetic analysis to propose effective single-step and multistep syntheses of compounds with amines as intermediates or products, protecting the amine as an amide if necessary.

Study Problems 19-35

19-36

Define each term and give an example. (a) acylation of an amine (b) (d) a 3° amine (e) (g) an aliphatic heterocyclic amine (h) (k) (j) a diazo coupling reaction (m) Gabriel synthesis of an amine (n) (p) an N-nitrosoamine (q) (s) a sulfonamide (t) For each compound, (1) classify the nitrogen-containing functional (2) provide an acceptable name. CH3 (a)

1

CH3-C-CH? -NH2

1

-

a 1 amine an aromatic heterocyclic amine a quaternary ammonium salt exhaustive methylation the Hofmann elimination reductive ami nation an azide

groups.

(b)

(c)

rb 0-

1 9-37

1+ CH,CH,

(g)

(c)

1+

g er g lQJ

H

,,N � N

\ I

cQJ N

Rank the amines in each set in order of increasing basicity. H NH2 N NH2

(a)

[Qj N

CH3

(,) CA3

O

/

H

(c)

°

H

(b)

a 2° amine

(0 a tertiary amine oxide

(i)

diazotization of an amine

(r) (u)

the Sandmeyer reaction a nitrile

(I) a sulfa drug (0) the Hofmann rearrangement

NO,

(d)

Q �C /

CH3 (h)

CI

O

/H

©J

"-

I-

CH3

CH2CH3 H N

o

H'

(e)

(f) 19-38

I �

N

I �

I �

I �

H2N

N

H3C

Which of the fol lowi ng compounds are capable of being resolved into enantiomers? (b) 2-methylpiperidine (c) I -methylpiperidi ne

(g)

1 ,2,2-trimethylaziridine

(:f'

N

0

(e)

N

(f)

CI-

/ � CH3 CH2CH3 (h)

\

H

(:f'

N

(c)

©+

N pyridine

�

CH3COOH acetic acid

©

] C-

+

(b)

CI/ � CH3 CH2CH3

\ CH3

0 N H pyrrol e

0

�

+ CH3COOH acetic acid

Q

�

H pyrrolidine

H H anilinium chloride pyridinium chloride piperidine Predict the organic products formed when the following amides are treated w ith alkaline bromine water.

I

o

o

(c)

Predict the products of the fol lowing reactions:

P:

(a) excess N H 3 (c)

(e)

(g)

( i)

+

Ph-CH 2CH zCH2Br

o CO cr + + a H20 Z

--->

--->

CH3NH2

+

(b) I -bromopentane

I

H

(f) �

11

H

( 1 ) NaN3

-'--'--=-­

(2) LiAIH4

heat

----->

(h)

©r

product from part (e)

�

m

(2) Ag20 (3) heat

N02 Zn, HCl

CH 2-C-C1 0

NH2

( I ) excess CH 3 1

(3) heat

He l

�

1I'''

(d) product from part (c)

( l ) excess CH 3 1 (2) Ag20

N oNO,

�\ �

o

(a) Ph-CH2CH2-C-NH2 19-41

0 N

Complete the following proposed acid-base reactions, and predict whether the reactants or products are favored.

( a)

19-40

CONH2

U U U -(-{ -(-{ -( fJ fJ cx:�

(a) N-ethyl-N-methyl an i l i ne (d)

19-39

CH2NH2

NH2

Study Problems

product from part(i )

)

LiAIH4

--�

927

928

Chapter 19: Amines

I

NCH3

LiAIH4

(k) CH3- (CH2)3-C-CH2CH3 (m) 2-butanone + diethylamine (0) 3-nitroaniline

19-42

Na(CH3COOhBH

(a)

( 1 ) HCI, NaNO?

-)

lQj

C=N

(b)

m-toluonitrile

(d)

19-43

CH3

lQj

CH'

4-fluoropyridine

CH3

CH,NH'

m-toluidine CH3 (c)

(e)

CH3

:rQr 0

NH'

lQj

(0

CH'

)

)

NH'

m-iodotoluene

m-methy I benzy lamine

OH

NaOCH2CH3

( l ) KCN, HCN ) (2) LiAIH4

lQj

(p) 2-butanone

lQj

LiAIH4

Ph -CH2-CH -CH3

'

lQj�--O H

02N m-cresol N-cyclopentyl-m-toluidine 3-methyl-4-nitroaniline The mass spectrum of t-butylamine follows. Use a diagram to show the cleavage that accounts for the base peak. Suggest why no molecular ion is visible in this spectrum. 1 00 , I I 80

-

40 E co

-

'"

g 60

co "0 "

20

o

19-44

(n)

Show how m-toluidine can be converted to the following compounds, using any necessary reagents. CH3

I

(I)

--'-"----'-"--'» )

(2) H3P02

CN

f---

--

.I I

r-

I

I

.11.

�

I� -0

--

I

I

I

i

I (CH3 hCNH2 I

I

I I 1 1I -1---1-I I

1 0 20 30 40 50 60 70 80 90 100 1 1 0 1 20 130 140 1 50 1 60 Using any necessary reagents, show how you would accomplish the following syntheses. (a)

(c)

(g)

0

�

C CH3

Qr

o

11

NH-C

N-H

� COOH

Q

h ( )

(d)

LQr

O

d

-----'>

(mosquito repellent)

NH'

�

�

d

NH'

19-45

Study Problems

The following drugs are synthesized using the methods in this chapter and in previous chapters. Devise a synthesis for each, starting with any compound containing no more than six carbon atoms. (a) Phenacetin, used with aspirin and caffeine in pain-relief medications. (b) Methamphetamine, once considered a safe diet pill, but now known to be addictive and destructive to brain tissue. (c) Dopamine, one of the neurotransmitters in the brain. Parkinson's disease is thought to result from a dopamine deficiency. CH3CH20

19-46

(b)

1 9-48

19-49

NH- -CH3

phenacetin

+ (CH3hNH

W

H

�

N

�Pt ,

Na(AcO)3BH H

--@HO

HO

+

)

methamphetamine

()

CH2CH2NH2

dopamine

N(CH3h

CO

H The two most general amine syntheses are the reduction of amides and the reductive ami nation of carbonyl compounds. Show how these techniques can be used to accomplish the following syntheses. (a) benzoic acid � benzylamine (b) benzaldehyde � benzylamine (c) pyrrolidine � N-ethylpyrrolidine (d) cyclohexanone � N-cyclohexylpyrrolidine. (e) HOOC - (CH2 h - COOH � 1 ,5-pentanediamine (cadaverine)

Several additional amine syntheses are effectively limited to making primary amines. The reduction of azides and nitro compounds and the Gabriel synthesis leave the carbon chain unchanged. Formation and reduction of a nitrile adds one carbon atom and the Hofmann rearrangement eliminates one carbon atom. Show how these amine syntheses can be used for the following conversions. (a) allyl bromide � allylamine (b) ethylbenzene � p-ethylaniline (c) 3-phenylheptanoic acid � 2-phenyl- l -hexanamine (d) I -bromo-3-phenylheptane � 3-phenyl - I -heptanamine (e) I -bromo-3-phenylheptane � 4-phenyl- l -octanamine (a) Guanidine (shown) is about as strong a base as hydroxide ion. Explain why guanidine is a much stronger base than

(b) * (c)

19-50

ero

--

o�

Propose mechanisms for the following reactions. (a)

19-47

929

most other amines. Show why p-nitroaniline is a much weaker base (3 pKb units weaker) than aniline. Explain why N,N,2,6-tetramethylaniline (shown) is a much stronger base than N,N-dimethylaniline.

N, N,2,6-tetramethylaniline

guanidine

N, N-dimethylaniline

Show how you would synthesize the following compounds starting with benzene, toluene, and alcohols containing no more than four carbon atoms as your organic starting materials. Assume that para is the major product (and separable from artho) in ortho, para mixtures. (b) N-methyl-l -butanamine (a) l -pentanamine (d) benzyl-n-propylamine (c) N-ethyl-N-propyl-2-butanamine (e)

(g)

@- -N=N

4-isobutylaniline

OH

(f)

3-propylaniline

930 19-51

Chapter 1 9: Amines

Using any necessary reagents, show how you would accomplish the following multistep syntheses. (a)

©��

CH2CH2CH2CH3

« )

19-52

19-53

©--@-

N o,

CH?CH3

� ©---�-@-

H,NH,

The alkaloid coniine has been isolated from hemlock and purified. Its molecular formula is C s H 1 7 N . Treatment of coniine with excess methyl iodide, followed by silver oxide and heating, gives the pure (S)-enantiomer of N,N­ dimethyloct-7-ene-4-amine. Propose a complete structure for coniine, and show how this reaction gives the observed product. A chemist is summoned to an abandoned waste-disposal site to determine the contents of a leaking, corroded barrel. The barrel reeks of an overpowering tishy odor. The chemist dons a respirator to approach the barrel and collect a sample, which she takes to her laboratory for analysis. The mass spectrum shows a molecular ion at mlz 1 0 1 , and the most abundant fragment is at mlz 86. The IR spec­ trum shows no absorptions above 3000 cm - I , many absorptions between 2800 and 3000 cm - I , no absorptions between 1 500 and 2800 cm-I , and a strong absorption at 1 200 cm- I . The proton NMR spectrum shows a triplet (J 7 Hz ) at 0 1 .0 and a quartet (J 7 Hz ) at 02.4, with integrals of 1 7 spaces and 1 1 spaces, respectively. (a) Show what structural information is implied by each spectrum, and propose a structure for the unknown toxic waste. (b) Current EPA regulations prohibit the disposal of liquid wastes because they tend to leak out of their containers. Propose an inexpensive method for converting this waste to a solid, relatively odorless form for reburial. (c) Suggest how the chemist might remove the fishy smell from her clothing. Pyrrole undergoes electrophilic aromatic substitution more readily than benzene, and mild reagents and conditions are sufficient. These reactions normally occur at the 2-position rather than the 3-position, as shown in the fol lowing example. =

=

* 19-54

4

3

502 Nl

I

H

pyrrole

+

oI

I

0

CH3- C - 0 -C-CH 3 acetic anhydride

4

3

sO Nl

I

H

�

C

II

CH3 + CH3-C-OH /

0

2-acetylpyrrole

(a) Propose a mechanism for the acetylation of pyrrole just shown. You may begin with pyrrole and the acylium ion, CH3 -C - 0 + . Be careful to draw all the resonance structures of the intermediate.

19-55

(b) Explain why pyrrole reacts more readily than benzene, and also why substitution occurs primarily at the 2-position rather than the 3-position. Section 1 7- 1 2 showed how nucleophilic aromatic substitution can give aryl amines if there is a strong electron-withdraw­ ing group ortho or para to the site of substitution. Consider the following example.

(a) Propose a mechanism for this reaction.

(b) We usually think of fluoride ion as a poor leaving group. Explain why this reaction readily displaces fluoride as the leaving group. (c) Explain why this reaction stops with the desired product, rather than reacting with another dinitrofluorobenzene.

93 1

Study Problems

The following spectra for A and B correspond to two structural isomers. The NMR singlet at 8 1 . 1 6 i n spectrum A dis­ appears when the sample i s shaken with D 2 0 . The singlet at 80.6 ppm i n the spectrum of B d isappears on shaking with D 2 0 . Propose structures for these i somers, and show how your structures correspond to the spectra. Show what cleav­ age is responsible for the base peak at m/z 4 4 in the mass spectrum of A and the prominent peak at m/z 58 in the mass spectrum of B.

19-56

1 00

80

..,

60

u " '" u "

40

� '"

A

I _�, -I44 �]--_+--_+--_4----J----L---�i --+_--4_--�I 1----+_--4_--

20

o 10

2.5 ,,,., ,,.ll ," 1 0 0 6� lli,,,,", i ll� 80

1

I

1.1I 11!111

I II I

ill

I�

1�

60 II ::�

40 20

II �

I

,

li t ,

,

I 'f; ;II

20

0 1 4000

200

TO mmTITTTCOTiTTTITTIT' !

ill LH

40

50

j.

iililliil

Ii illlI ,,:: 3500

I ,

I

I

1 00

1 10

,

I

---

"----

4

70

4.5

I,!

iliMY,

60

"i

I I I I I

I I', /"1 1

m/z

90

1 20

1 30

1 40

1 50

1 60

wavelength (/Lm) 5.5 I + I+++H+++'A-H-H H+lH-I-+-+-+--H-+-++-H-+--1-+--H---H�-!-- 1 I i , ! I-H-iH++++ Ii

5

; ,I I I lill

80

:I

" "

'j

! II I i I II!! I'll iii ll ! i ;1 1 !I d i j! I I I; I I +I+++H-H ! Ifl- I+!+H-,H-;H-I H� l , I *I!H� W.,'+t I --, l+I-t H I+ I+H-I + ) iH-H+ � 'H+I+lH-H -HI H il� l'� IW.'_H*H� -H++�++++ II+ # I

I-!+t+!!il-i+l-Htf+t lfi+i,' •

**H I�*fi+'jlw.i' +Lt �iJi1i+1 I

I

'I H ! HI+i,H j* l�ril� i H� I+H�,I+,t:t H�4+�HII I�H I W.l� I t-+ ,-t-1-1-++4-!f- l-+-+ /+-H I-+�-HHI I,li\¥+H- I�H+I\� K� �H�-HH-H+ H I+H� -

::iIi::tii

,

jtti :*[

i" +t ,l8 i:t it ll t llit i i i dI I , I I I: ' I! I I" ';: IIIi :i i , iq ,

'

1 60

1 80

Ii Ii I

'

I

Ii: ,i ::i

I :: :

' +++!+!++++! i +!+i+tt+++i "'+i+ ' "+i-!'tt +! +++!+H-i+tt.iJ--H-+-++-H--H-++-++-H-+-+-+-H i 1+l++i-!-++t++ l++ CH 1 1!1 ' II I , "

�IH�+ttt+1tij-it+iii+�i�+tttlHti+ilf+lltH_t!+t,tiH:HWf+�+tit+iti++:++ltit-HttH-tlHti++-Htit+tti+ttHtlH�-tH+til+tlHti++=-�j-t+�_t�-t=_�I-t+t+��=-If:i_tI:-t+-+��-1=j�-t-l-�+�-tj_�f-qH

1 ill1..L!.l:L!-ill ll.Lll.'':l.LL�!JJJ.w.-!: 1 " lli, W!.L!.llll�1J,LL;,wjJ,1I..II� L!:'", .L l. .l.. �l.!..l�� .Ll...L.L .L .Ll�!:-;!-...LJ....L.Ll...�

i l ll l lWJ.l.l l llli i' .LLlJ.ll!J !lL!.!J 3000 2500

l!!J l!lliifil ll. lll lll lli!..l .LLlJ.l l ll iw i tlJilll.wJil 'W I '

A

30

ill ' lll i f+l i + , ' I H+H+t+itit+itiiI ,;

I!! I

I

3.5

\H++t+H11 fi+,i+t1i+t i '* I ,fi+t iH ' I� !i ; I I+ lifH �f� I� *H+H+ ' 1 I II1 liil !L �I* -,,-* ITt " I! '" H+H+ I ' ITt II' I i I I i II Ii' I I

-I : 1

-,

I

-!-- ---I __

]=r-;� t+--��;;; � j t==F�-

I �++J.H+fIrr,.".,+H-l+tl-i+:�I:I�itt'lfi++t,:

. I

,I

I

140

2000

1 800

1 600

wavenumber (em - I ) 1 00

1 20

1 400

80

1 200

1 000

800

o

20

40

60

600

I

•

.

' .

...

-

••

••

3

2(

f ·· · . ·

H

10

9

8

7

6

5

8 (ppm)

4

3

: ..

.

• • •

2

rt.

-, • . -,-, � '. '--,-,-� ' ...:. r� o

932

Chapter 1 9: Amines 1 00 80 f. I! R�;O, : -:+ + - H-H+-I-l-H- H Ii! , I 69 ,:+HI '+H-l+ I+� \-�+m�-H-++-I--H--l-+- H-+-r+-H�H-1 ,'III!I II !II 1111111 iii' .. ; ) Ii r ?l ,I ii! I 2500 2000 1 800 1600 1400 1200 1 000 600 800 wavenumber (em - I )

, 1 48++fH+ HH+I HH H CH2

iliIiH!1i 3000

=

.... Figure 20-5 IR spectrum of 2-methyl-propenoic acid.

PRO B L E M 20-7

The IR spectrum of trans-2-octenoic acid is shown. Point out the spectral characteristics that allow you to teU that this is a carboxylic acid, and show which features lead you to conclude that the acid is unsaturated and conjugated.

1 1111 :'1

1 05 1

11111 !I

�

'HH1+HH1+ 1+H+III+mlli', I i++HH+1H+', I,', TR i++-HH+ III III! il II 60 : � ttt+rtn +i1 !++ l' lm +m mlH+ , � tHH\-rmHl lHil HH,H',H+ 1 , I I I ! !"+H H\­ 40 I : i i++H\-H+iH+H+ , � tHHHlrmH , H ,I m , 1,+H, IH, I+;1+ , � I-HmtH+Hi Hi,,l "IIHii 'f; iH,1mI H:t 20 I I " 1.,I1 ,111' **Ht!+lI+t1 Kt I IItt+ i III! :I+HI;+++ , �, I, 80

,

�

35 !I"I;;I"" I'

I I

4000

20-7B

��;;1�� I � ' iI�I�� I ,! = 3500

I

iii! iii. !In j" ii-i

I! IiI " Ii Ii. II, i:l Iii! ii.! iii i iii i ' , I I I '

!

j ii! 11!l l!' l :iii

i+H,

o

I'

4;5

wavelength (/Lm) 55 ; ' !il! Ii

i ]! i i i I;

,III iIi ! i

Ill!

' c ' \- I- lf"t-++-1 , ,m +l-m 'ffi+ l+fl\l� ' m"'+-++-+-t- H-�-J-H-++' ++-H+ �Il-H, +rrH'lT'N-d±i -�,,i '! +it H Ifr,+l i+1 1 1+ i+ , ++i*+++++++t+IrH+�t-++-+-+-r+-++�H- ++-HI-I- r++-I r t++i��1�t+ r +i-++Hf+' 1 :tt� , , �t+tI , ++ 1 1 '* : r

;ii iiIi iii!HfI fHI ,H1HHHHf+1+H++1++ I!1j! ++ 1 i+H�+H- Hi! tHlli iIH+tHH-!'H i' iH:++H# H+t+l-H+t Hi-t+ I-+-+- i-H-+--t-+-iH-++-+-i- I-++++- 1 II;i I l I I 'il , I I' ! iii; Uii Iii 11111 1

, i .Iii

,'Ii"i"

3000

iI

2500

I I'

I

,

II I"

2000 1 800 1 600 wavenumber (em- I )

1 400

N M R S pectroscopy

Carboxylic acid protons are the most deshielded protons we have encountered, absorb­ ing between 810 and 8 1 3 . Depending on the solvent and the concentration, this acid proton peak may be sharp or broad, but it is always unsplit because of proton exchange. The protons on the a carbon atom absorb between 82.0 and 82.5, in about the same position as the protons on a carbon atom alpha to a ketone or an aldehyde. The proton NMR spectrum of butanoic acid is shown in Figure 20-6. sextet (overlapping quartet of triplets)

o

/

II

H - O - C-CH / 2 - CH\2 - CH\3

I

811.2 singlet

82.4 triplet

8 1 .6

8 1 .0 triplet

H

0

I II -C-C-O-H "" 8 1 0 I H

� 82.0-82.5

-8 1 3

950

Chapter 20: Carboxylic Acids 200

I I '--iI-'-, �:III II' -'--'-+- � I'--i-'--c- ' '---4-'-- : '--i-'-11-+ H--'-+-'-I '--i"

a

'

.

I" I ' II: I :

:

j

i

,

II

o

20

40

60

80

100

1 20

1 40

1 60

1 80

l ��������+�1 H7�=d±±��=F� rH��I�� Ho-- c--c� 2 ,�c� �c�I��������� , 4. T Offset: 2.0 ppm I ' : II I I Ii I i l 'l 1 ' 1; i

"" ]'

,

I ' I

i

:

:: ;

9

to

1

,

i

6

7

8

I

'

,

I

!

I

i

4

5 o (ppm)

3

2

o

.A. Figure 20-6

Proton NMR spectrum of butanoic acid.

The carbon NMR chemical shifts of carboxylic acids resemble those of ketones and aldehydes. The carbonyl carbon atom absorbs around 1 80 ppm, and the a-carbon atom absorbs around 30 to 40 ppm. The chemical shifts of the carbon atoms in hexa­ noic acid are the following: o

II

112 12 12 12 13

HO -C -CH 181

34

-

CH

-

25

CH

-

31

CH

-

22

CH 14

(ppm)

P R O B L E M 2 0-8

(a) Determine the structure of the carboxylic acid whose proton NMR spectrum appears below. (b) Draw the NMR spectrum you would expect from the conesponding aldehyde whose oxidation would give this carboxylic acid. (c) Point out two distinctive differences in the spectra of the aldehyde and the acid. ,

i

: 1

i;

Offset: 2 . 1 ppm .

:

:

I I, I i i ,

I

.

i

I

,

.

i

, ,

i'

to

9

i

,

8

i

i

i

i '

i I I

i

iI

' I-+-

I I i

i

I

,

,

I i .:

7

6

5 o (ppm)

4

3

2

;

o

20-7C

20- 7 S pectroscopy of Carbox ylic Acids

Ultra vio let Spectroscopy

Saturated carboxylic acids have a weak n � 7T* transition that absorbs around 200 to 2 1 5 nm. This absorption cOlTesponds to the weak transition around 270 to 300 nm in the spectra of ketones and aldehydes. The molar absorptivity i s very smaIl (about 30 to 1 00), and the absorption often goes unnoticed. Conjugated aci ds show much stronger absorptions. One C = C double bond conjugated with the carboxyl group results in a spectrum with Amax still around 200 nm, but with molar absorptivity of about 1 0,000. A second conjugated double bond raises the value of Amax to about 250 nm, as illustrated by the following examples: o

II

CH2=CH- C - OH o

II

CH3 - CH =CH - CH =CH - C - OH

20-70

Amax

=

200 nm

8 =

1 0,000

Amax

=

254 nm

8 =

25 ,000

Mass S pectrometry

The molecular ion peak of a carboxylic acid is usually small because favorable modes of fragmentation are available. The most common fragmentation is loss of a molecule of an alkene (the McLafferty rearrangement, discussed in Section 1 8-5D). Another common fragmentation is loss of an alkyl radical to give a resonance­ stabilized cation with the positive charge delocalized over an allylic system and two oxygen atoms. + +

McLafferty rearrangement

loss of an alkyl group

(111/� is even)

resonance-stabilized cation (mlz is odd)

The mass spectrum of pentanoic acid is given in Figure 20-7. The base peak at mlz 60 corresponds to the fragment from loss of propene via the McLafferty rearrangement. The strong peak at mlz 73 corresponds to loss of an ethyl radical with rearrangement to give a resonance-stabilized cation. P R O B L E M 20-9

\/ C " C /\

Draw all four resonance forms of the fragment at mlz 73 in the mass spectrum of pentanoic acid.

951

952

Chapter 20: Carboxyl ic Ac ids 100

'"

60 g OJ

"'"

�

60

80 1---

I

-

40

20 ---- - - - -. . -

o

10

20

III

30

II�-I

--

I,

SO

40

-

_._

60

73 -- -

70

--

�

�---+-I

I

I �

I

I

----

CH3 CH2CH2CH2COOH

II I I I i J

I I I 1 i i l - 1 - --TJ02t---- r - --r- - -r-1 \

-

80

mlz

90

1 00

1 10

I

i

1 20

1 30

I

1 40

ISO

160

+

II1lz

� Figure 20-7

The mass spectrum of pentanoic acid shows a weak parent peak, a base peak from the McLafferty rearrangement, and another strong peak from loss of an ethyl radical.

neutral, not observed

60

C H3C H2' + neutral , not observed

P R O B L E M 2 0- 1 0

Use equations to explain the prominent peaks at mlz 74 and mlz 87 in the mass spectrum of 2-methylpentanoic acid. 1 00

'"

80

g 60 OJ

"'"

.E 40 OJ

20

o

I

I

+ 1�--1- -i---- ---J �-I+-

r------ -

I

10

20

30

40

I

I

1

I

I

SO

I

II

-------j-

60

!I

70

I

I

I

CH3CH2CH2CHCOOH

I,

I

-

I

741

80

I

I

CH3

I-

· r l-+t=�+=

'7--1--190 100

mlz

J JO

1 20 1 30 1 40 1 50 1 60

20-8

20-SA

Synthesis of Ca rboxy l i c Acids

We have already encountered three methods for preparing carboxylic acids: ( 1 ) oxidation of alcohols and aldehydes, (2) oxidative cleavage of alkenes and alkynes, and (3) severe side-chain oxidation of alkylbenzenes.

Review of Previous Syntheses

1. Primary alcohols and aldehydes are commonly oxidized to acids by chromic

acid (H 2Cr04, formed from Na 2 Cr207 and H 2 S04). Potassium permanganate is occasionally used, but the yields are often lower (Sections 1 1-2B and 1 8-20).

20-8 Synthesis of Carboxylic Acids a

a

R-CH2 - OH

II R- C-H

II R- C - OH

primary alcohol

aldehyde

carboxylic acid

Example

a

Ph - CH2-CH2-CH2- OH

NazCrz07, H2S04

II

)

Ph - CH2- CH2- C - OH

3-phenylpropanol

3-phenylpropanoic acid

2. Cold, dilute potassium pennanganate reacts with alkenes to give glycols. Warm,

concentrated permanganate solutions oxidize the glycols further, cleaving the central carbon-carbon bond. Depending on the substitution of the original double bond, ketones or acids may result (Section 8-15A).

concd. KMn04

r?? R

)

R'

J

H - - - R" HO

OH

R -COOH acid

+

/ R' O=C ""R" ketone

glycol (not isolated)

Examples concd. KMn04

o

coned. KMn04

)

C

COOH COOH

adipic acid

cyC\ohexene

With alkynes, either ozonolysis or a vigorous permanganate oxidation cleaves the triple bond to give carboxylic acids (Section 9- 1 0).

r:jj=

R- C == C - R'

coned. KMn04

alkyne

)

[ JJ ,] R

-R

(not isolated)

Example

3. Side chains of alkylbenzenes are oxidized to benzoic acid derivatives by

treatment with hot potassium permanganate or hot chromic acid. Because this oxidation requires severe reaction conditions, it is useful only for mak­ ing benzoic acid derivatives with no oxidizable functional groups. Oxidation-

R-COOH

+

HOOC- R'

carboxylic acids

953

954

Chapter 20: Carboxylic Aci ds

resistant functional groups such as may be present (Section 1 7- 1 4A).

W

RCalkYll

-CI,

- N02,

- H, -COOH WCOOH S03

Na2Cr2 07, H2 S04, heat or KMn04, H2 0, heat

and

)

Z

Z

an alkyl benzene (Z must be oxidation-resistant)

a benzoic acid

Example

CH3I aCH-CHl CI�

Na2Crp7' H2 S04 heat

) p-chlorobenzoic acid

p-chloroisopropylbenzene

20-S8

Ca rboxylation of Gri g n a rd Reagents

We have seen how Grignard reagents act as strong nucleophiles, adding to the car­ bonyl groups of ketones and aldehydes (Section 1 0-9). Similarly, Grignard reagents add to carbon dioxide to form magnesium salts of carboxylic acids. Addition of dilute acid protonates these magnesium salts to give carboxylic acids. This method is useful because it converts a halide functional group to a carboxylic acid functional group with an additional carbon atom. R-X (alkyl or aryl halide)

Mo �

�c '�i:' ) '

R -MgX Example

� ether bromocyclohexane

The vinegaroon (whip-tail scorpion) expels a defensive spray consisting of 84% acetic acid, 5% caprylic acid, and 11 % water. Caprylic acid acts as a wetting and spreading agent.

-M

R

ether

"

0 II .. -

R

g : + gX

o

0 II ..

"

-C- - M aMgB,

rY�cro) V 20-SC

gX

II C-O-MgB, a

'

R-C-OH

+

cyclohexanecarboxylic acid

Formation and Hydrolysis of N itri les

Another way to convert an alkyl halide (or tosylate) to a carboxylic acid with an additional carbon atom is to displace the halide with sodium cyanide. The prod­ uct is a nitrile with one additional carbon atom. Acidic or basic hydrolysis of the nitrile gives a carboxylic acid by a mechanism discussed in Chapter 2 1 . This

20-8 Synthesis of CarboxyEc Acids

955

method is limited to halides and tosylates that are good SN2 electrophiles: usually primary and unhindered.

R-CH2-C-N:

NaCN acetone

R-CH2-X R-CH2-C-N:

Example

or OH, Hp

QrCH,-

B,

NaCN acetone

benzyl bromide

phenylacetonitrile

phenylacetic acid

phenylacetonitrile

PROBLEM·SOLVING

Oxidation of alcohols does not change

P R O B L E M 2 0· 1 1

Show how you would synthesize the following carboxylic acids, using the indicated start­ ing materials. (b) trans -cyclodecene � decanedioic acid (a) 4-octyne � butanoic acid (d) 2-butanol � 2-methylbutanoic acid (c) bromobenzene � phenylacetic acid (0 allyl iodide � 3-butenoic acid (e) p-xylene � terephthalic acid

(

SUMMARY

Hinl/

the number of carbon atoms. Oxidative cleavages of alkenes and alkynes decrease the number of carbon atoms (except in cyclic cases). Carboxylation of Grignard reagents and formation and hydrolysis of nitriles increase the number of carbon atoms by one.

Syntheses of Carboxylic Acids

1. Oxidation of primary alcohols and aldehydes

R-CH2-OH

or KMn04

primary alcohol

(Section 11-2B and 18-20)

o II

)

2. Oxidation cleavage of alkenes and alkynes (Section

R R' C=C H R" R-C=C-R' "'-

/

/

alkene

alkyne

"'-

coned.

R-C-H aldehyde

R-C-OH

)

carboxylic acid

8-15A and 9-10)

KMn04

concd. KMn04 or (1) 03

or KMn04

o II

(2)�0

)

R' R-COOH O=C R" R-COOH HOOC-R' /

+

acid

+

"'-

ketone

carboxylic acids

956

Chapter 20: Carboxylic Acids

rgr

3. Oxidation ofalkylbenzenes (Section 1 7- 1 4A)

Na2Cr207, H2S04 or

KMn04, H20

Z

an alkyl benzene (Z must be oxidation-resistant)

4. Carboxylation of Grignard reagents (Section 20-8B)

R-X alkyl or aryl halide Example

Mg

--7

ether

O=C=O

-----'»

R -MgX

CH3 - CH - CH3 I CH2Br isobutyl bromide

a benzoic acid

o

II R - C - O- +MgX (1) CO2 + (2) H

Mg

--7

-----;.)

ether

5. Formation and hydrolysis of nitriles

CQOH

(Section 20-8C)

NaCN ) acetone Example

(1) NaCN, acetone + (2) H , H20

rgr

benzyl bromide

H+

----7

o

II R - C -OH acid

CH3 -CH -CH3 I CH2-COOH isovaleric acid

o

II CH2 - C - OH

phenylacetic acid

6. The haloform reaction (converts methyl ketones to acids and iodoform; Chapter 22) o

II R-C-CH3

X2

-----'»

x

- OH Cl, Br, I

=

o

Example

II Ph - C - CH3 acetophenone

o

II R- C - O-

+

o

(1) X2, - OH II -----7) Ph - C - OH + (2) H benzoic acid

7. Malonic ester synthesis (makes substituted acetic acids; Chapter 22)

COOEt I CH2

I

Example

(1) Na+ - OCH2CH3

(2) R-X

COOEt COOEt I CH2 I COOEt

(1) Na + - OCH2CH3 (2) CH3CH2CH2CH2Br

)

)

HCX3

COOEt I R - CH I COOEt

COOEt I n-B u - CH I COOEt

(1) -OH ) (2) H + , heat

(1) -OH ) (2) H + , heat

0 II R - CH2 - C-OH

0 II n-B u - CH2 - C - OH hexanoic acid

Reactions of Carboxylic Acids and Derivatives; Nucleophilic Acyl Substitution Ketreactones,ions ofaldaciehydes, andtecarboxyl icfromacidthose s all ofcontai nnesthande carbonyl group,Ketoyetnes andthe d s are qui di ff erent ket o al d ehydes. aldehydes y rreact byativnucles) more eophicommonl lic additioynreactto thbye carbonyl group; but car­ boxyl icwhere acidcommonl s (and thei deri v one nucleophile replaces another on the acyl (C 0) carbon atom. 20-9

957

20-9

nucleophilic acyl substi­

=

tution,

··0·

Nucleophilic acyl substitution

0

·· ·

II R-C-X Nuc:

I R-C-Nuc :X-

+

+

A cid derivatives

o II

Reactions of Ca rboxy l i c Acids and Derivatives; N u cleoph i l ic Acyl S u bstitution

o II

0II o o II II R-C-OH R-C-X R-C-O-C-R R-C-O-R' R-C-NH 2 Aci-OH d deriinvatithevesacididff, -Cl er in ithen thenataciured chloforithdee,nucl-OR' eophiinlethbonded to t-NH he acyl2 carbon: e est e r, and ne) in nthgethese amide.deriNuclvatievophi lic acylwil substi tutioexampl n is theemost common metlic acylhod for(orsubstianintteamiutirconverti es. We see many s of nucl e ophi on in thsismschapter andon ithen Chapter ("Carboxyl ic AcibutdweDericanvatigroup ves").thTheem speci f i c mechani depend reagents and condi t i o ns, generalUnder ly accordi ngcondi to whether theystrongtake nucl placeeophiunderle acicandiaddc or basi cecondi tions. group basi c t i o ns, a t o t h carbonyl to give a tetrahedral inytesirmedi ateest. Thierstointermedi ate thenate expel s tanhe acileavid nisgangroup. Thee base-catal y zed hydrol s of an t h e carboxyl sal t of exampl ofto thigivsemechani sm (Mechani s m aBoxte. The tetrahedral Hydroxidienitoermedi n addsatteo stabi the carbonyl group a t e t r ahedral i n termedi l i z es i t sel f by expel l i n g an al k oxi d e i o n. The al k oxi d e i o n qui c kl y react s wi t h the aci d (pKa to give an alcohol (pKa 16) and a carboxylate ion. carboxylic acid

o I

acyl halide

anhydride

ester

amide

21

20- 1 ).

=

5)

=

M ECHAN I S M 20- 1 Step 1:

Nucleoph i l i c Acyl S ubstitution i n the Basic Hydrolysis of a n Ester

Hydroxide rong anuclte. eophile) adds to the carbonyl group, forming a tetrahedralioinnt(stermedi · 0) I .. I .. R-C-OR' R-C-OR' . I . : OH C-:o�· An alkoxide ion leaves, regenerating the C=O double bond. ·0· :°5 R-C :O-R' R-C-OR' I :O-H : OH :

ester + -OH

Step 2:

I

..

u- . .

tetrahedral intermediate

0 :-

tetrahedral intermediate

I!

\

acid + alkoxide

( Continued)

958

C apt Carboxylic Acids h

er 20:

Step 3 : A fast, exothermic

'0-

proton

transfer drives the reaction

'0 '

.. I! :OR-C . . R' \ :O� H'"")

I! R-C

to completion. H- O -R'

':Q:-

carboxylate + alcohol

acid + alkoxide

Basic hydrolysis of ethyl benzoate. Eldeim. ination Addide.tion of al k oxi of hydroxi '0) II . . Ph :°5II Vf ." . Example:

Step 3:

Step 2:

Step 1 :

-C

Ph - C - QCH2CH3

l-: OH

I

.OCH2CH3 (

: OH

ester + -OH

)

Ph

Proton transfer.

' 0-

.. I! -C :OCH2CH3 .. \ :O� H'"")

tetrahedral intermediate

acid + alkoxide

carboxylate + alcohol

Nuclnucleophieophilic acylle is substi tutitooatn talackso thetakescarbonyl place ingroup. acid. TheUndercarbonyl acidic group conditmust ions, nobecome strong present protonated, actes aivtatetirnahedral g it toward nuclaete.ophiInlimost c acylcases,substithetutileoavin. nAtg tgroup ack bybe­a weak nucl e ophi l e gi v i n termedi comes protonatexampl ed before it leaciaves,d-catalso iyt zedleavesnuclaseophi a neutlircalacylmolsubsti ecule.tWeutionow cover the most useful e of an n: the Fi s cher esterification. 20-1 0 Conde nsation of Acids with Alcohols: The Fischer Este rification

The convert ssubsti carboxyl icn.aciTheds netandreacti alcoholon sisdireplrectlacement y to esteofrstbyhe anacidacid-catalgroup yzed nuclby tehophi l i c acyl t uti o e group of the alcohol. Fischer esterification

- OH

- OR

o II

R- C - OH acid Examples

a

o

+

R'- OH

II R- C - O - R'

alcohol

ester

o II

C� - C - O - CH�� COOH

� COOH phthalic acid

COOCH3 dimethyl phthalate

+

HzO

+

�O

Condensation of Acids with Alcohols: The Fischer Esterification The Fizedschernuclesteri filcatiic acyl on mechani smon.(seeThethecarbonyl Key Mechani sofm abox)carboxyl is anic aciacidd­ catal y e ophi substi t uti group ciently elgroup ectrophiandlicacttoivbeatesatittacked bynuclan aleophi cohollic. Theattack.acidLosscatalofysta protprotoon­ natgiis vnoteess thetsuffi he carbonyl t o ward hydrat e eofr from an esttheer. hydrate of the ester occms by the same mechanism as loss Loss of wat of water from thealhydrat et tofo leaaveketoasnewat(Sectier, formi on ng a resonance-stabi Protonation oflieizedthercationeon.ofLossthe hydroxyl groups l o ws i of a protTheonmechani from thsemsecond hydroxyl groupficgiativoesn twoul he estdeseem r. l ong and complicated if of t h e Fi s cher esteri tmechani ried to memori zeaciit, dbut-catalweycanzed understand itthbye albreaki ntgoitthedowncarbonyl into twando si(2)m­ placiyouedr-catal s ms: addi t i o n of c ohol understand thesehavimechani tic components, write the Fiyzedscherdehydrati esterification. oIfnyoumechani sm without ng tosmemori ze it. you can 20- 1 0

959

1 8- 14).

(1)

-;i>-:-'""- KEY M EC H A N I S M 20-2

Fischer Esterification

Part 1: A cid-catalyzed addition of the alcohol to the carbonyl group.

Prot o nat i o n act i v at e s The al c ohol adds. the carbonyl. l ) I

:" l

R -r H

R'- O - H

Part 2: Ac id-catalyzed dehydration.

Water leaves.

Prot ion prepares the onatgroup to leave. OH

Deprotonation completes the reaction.

�

:O-H

: O -H I .. R - C - OH ..

+ I .. R - C - OH .. + R'OH2

I

I

d- R' Ht � .. R'- O - H

[

: OR'

ester hydrate

Deprotonation completes the reaction.

P Acid-catalyzed formation of methyl benzoate from methanol and benzoic acid. Deprotonation completes the reaction. onation act. ivates Methanol adds. tProthe carbonyl 1 Ph Ph [Ph :O-H

I .� H+ R-C-O .. - H ( I

)

: OR'

(: O - H 'I .. / H ( ) R - C �.z. H I : OR'

EXAM PLE:

+ O-... -.J H

II

R -C

":OR' ..

+

" .. ( ) H20

Part 1: A c id-catalyzed addition of methanol to the carbonyl group.

.+ / H '0

II) . .

} ..

- C - OH

CH3 - q -H

:O -H

I

..

-C - OH .. 1+ 3 H-O . - CH f. � . CH 3- q -H

:O -H

I I :

..

- C - OH .. Q CH3

ester hydrate

(Continued)

Chapter 20: Carboxylic Acids

960

Part 2: Acid-catalyzed dehydration.

Water leaves.

Protonation prepares

the

OH

group to leave.

: O -H

I

e r n n e he r n p oto ati o

c ompl

tes

t

eacti o .

. '-----'- H+

Ph - C - O . -H (

I

D

.

)

: QCH3

QU ESTION : Why can ' t

PROBLEM-SOLVING

the

Fischer esterification take place P R O B L E M 20- 1 2

Htnp

methyl benzoate

protonated ester

under

basic catalysis?

(a) The Key Mechanism for Fischer Esterification omitted some important resonance forms of the intermediates shown in brackets. Complete the mechanism by drawing all the res­ onance forms of these two intermediates. (b) Propose a mechanism for the acid-catalyzed reaction of acetic acid with ethanol to give ethyl acetate. (c) The Principle of Microscopic Reversibility states that a forward reaction and a reverse reaction taking place under the same conditions (as in an equilibrium) must follow the same reaction pathway in microscopic detail. The reverse of the Fischer esterification is the acid-catalyzed hydrolysis of an ester. Propose a mechanism for the acid-catalyzed hydrolysis of ethyl benzoate, PhCOOCH2CH3.

The Fischer esterification mechanism

is a perfect example of acid-catalyzed nucleophilic acyl substitution. You should understand this mechanism well.

PRO B L E M 20- 1 3

Most of the Fischer esterification mechanism is identical with the mechanism of acetal for­ mation. The difference is in the final step, where a carbocation loses a proton to give the ester. Write mechanisms for the following reactions, with the comparable steps directly above and below each other. Explain why the final step of the esterification (proton loss) cannot occur in acetal formation, and show what happens instead. o

II Ph - C - H aldehyde

o

II

Ph - C - OH acid

W, CHPH

)

CHP

OCH}

\/

Ph -C-H

+

H2 O

+

H2O

acetal

0

II

Ph- C -OCH} ester

P R O B L E M 2 0- 1 4

A carboxylic acid has two oxygen atoms, each with two nonbonding pairs of electrons. (a) Draw the resonance forms of a carboxylic acid that is protonated on the hydroxyl oxygen atom. (b) Compare the resonance forms with those given previously for an acid protonated on the carbonyl oxygen atom. (c) Explain why the carbonyl oxygen atom of a carboxylic acid is more basic than the hydroxyl oxygen.

ationlaisrge.anForequiexampl librium,e, iandf moltypiecalofequi liicbrium constmixaedntswiforth esterimolfiFiecofsaticherethanol on areesteri,notthfiecvery acet aci d i s libacetriumic miacixdtuandre contai ns . Esterimolficeatieachon usiofnetghsecondary yl acetate andand wattertiearryandalcoholsmolgiveeseachequi of ethanol . even smaller equilibrium constants. 1

1

0.65

0.35

20- 1 0 Condensation of Acids with Alcohols: The Fischer Esterification

961

Equilibrium mixture

°

II

CH3- C - OH 0 . 3 5 mole

+

C H3CH20H

(

Keq

=

3.38

)

0.35 mole

°

II

C H3- C - OCH2CH3 + 0.65 mole

Esteri fbyicatiremovi on mayng beonedriofventhetoproducts. the rightForeitherexamplby usie, inngformi an ng ethyl esters, ex­the react a nt s or cessternati ethanolvely,iswatoftener mayusedbeto removed drive the eiequitherlibbyriudimstil faring asit outpossiorblbye toaddiwardngthea dehy­ ester. Aldrati g agent as magnesium sulfate or (dehydrated zeolite crys­ tals thatnBecause adsorbsuchofwater). inconveni eannceaciofd drichlvoinrigdethewiFithsancheralcesteri fforicatitheonlatoborat comploryetisyn­on, wethesimis gofhtesters. preferThethethereacti o n of ohol Fisgicherve good esterifyiiceatildsonofisproducts preferredandin inavoidustdry,thehowever, where theof techni q ues menti o ned expensi v e step converting the acid to its acid chloride. as

excess o f one of

molecular sieves

P R O B L E M 20- 1 5

Show how Fischer esterification might be used to form the following esters. In each case, suggest a method for driving reaction to completion. (a) methyl salicylate (b) methyl formate (bp 32°C) (c) ethyl benzoate

the

PROBLEM-SOLVING

Htl1P

I n equilibr ium reactions, look for ways to use an excess of a reagent or else to remove a product as it forms. Is it possible to use one of the reagents as

P R O B L E M 20- 1 6

a solvent? Can we distill off a product

The mechanism of the Fischer esterification was controversial until 1 938, when Irving Roberts and Harold Urey of Columbia University used isotopic labeling to follow the alcohol oxygen atom through the reaction. A catalytic amount of sulfuric acid was added to a mixture of 1 mole of acetic acid and 1 mole of special methanol containing the heavy 1 80 isotope of oxygen. After a short period, the acid was neutralized to stop the reaction, and the compo­ nents of the mixture were separated.

or drive off water?

CH3�O - H

(a) Propose a mechanism for this reaction. (b) Follow the labeled 1 80 atom through your mechanism, and show where it will be found

in the products.

(c) The 1 80 isotope is not radioactive. Suggest how you could experimentally determine the

amounts of 1 80 in the separated components of the mixture.

S O LV E D P R O B L E M 2 0 - 1

Ethyl orthoformate hydrolyzes easily in dilute acid to give formic acid and three equivalents of ethanol . Propose a mechanism for the hydrolysis of ethyl orthoformate. OCH2CH3

I I

H - C - OCH2CH3 OCH2CH3

SO LUTI ON

ethyl orthoform ate

formic acid

ethanol

Ethyl orthoformate resembles an acetal with an extra alkoxy group, so this mechanism should resemble the hydrolysis of an acetal (Section 1 8- 1 8). There are three equivalent basic sites: the three oxygen atoms. Protonation of one of these sites allows ethanol to leave, giving a resonance-stabilized cation. Attack by water gives an intermediate that resembles a hemiacetal with an extra alkoxy group. ( Continued)

Chapter 20: Carboxylic Acid

962

: OEt

I

s

: OEt : OEt . -0 H20 : I + f> H ...I .. ( ) H-C-OH H-C-O · · "-. .. .H I I : OEt : OEt

�

l

H+

H - C - OEt ( )

I

: OEt

..

Protonation and loss of a second ethoxyl group gives an intermediate that is simply a protonated ester.

: 6� I

..

H - C- OH (

I

: OEt

..

Et

H

(clI

·0·

II

· .. ) H - C - OH

H+

I

: OEt

.. H - C-O-Et

..

protonated ester

ethyl formate

Hydrolysis of ethyl formate follows the reverse path of the Fischer esterification. This part of the mechanism is left to you as an exercise. P RO B L E M 20- 1 7

20-1 was missing some important resonance forms of the intermediates shown in brack­ ets. Complete this mechanism by drawing all the resonance forms of these intermediates. Do your resonance forms help to explain why this reaction occurs under very mild conditions (water with a tiny trace of acid)? (b) Finish the solution for Solved Problem 20- 1 by providing a mechanism for the acid-catalyzed hydrolysis of ethyl formate. (a) The solution given for Solved Problem

20- 1 1 Este rification Using D i azometh a n e

Carboxyl s are hconvert ed tonlo theiy byproduct r methyl esters very sigas,mplyandby anyaddinexcess g an etdihaer­ solzomethane ution ofic alacidisaodzomet ane. The i s ni t r ogen s. Puriy fquanti icatiotnatiofvethein ester of the solvent. Yievaporat elds areenearl most usual cases.ly involves only evaporation II II CH2N2 R-C-O-CH3 N2 i (tr R-C-OH COOH COOCH3 N2 i 0 O Diazomet hhaneer solisuatiotoxins.cThe, explreactosivieonyeloflodiwazomet gas thathanediswisolthvescarboxyl in ethericandacidiss prob­ fairly safe t o use i n et ves trleansfer proton,witgih vniintrgogena methyl t. This diazonium salabltyisinanvolexcel nt metofhylthaetinacig dagent, gas asdiaazonileaviumngsalgroup. 0

Example

Diazomethane

is

often

used

to

esterify polar or reactive compounds for mass spectrometry analysis. For example, a urine test for cocaine might involve treating the sample with diazomethane to convert ben­ zoylecgonine,

the

major

urinary

metabolite of cocaine, to its volatile methyl ester for MS analysis.

M ECHAN I S M 20-3 Step 1:

acid

0

+

�

diazomethane CH2 N2

cyclobutanecarboxylic acid

+

methyl ester

+

methyl cyclobutanecarboxylate ( 1 00%)

Esterification Using D iazomethane

Proton transfer, forming a carboxylate ion and a methyldiazonium ion. ."0.

II . . R-C-O:

carboxylate ion

+

+

CH3-N=N:

methyldiazonium ion

20- 1 3

Reduction of Carboxylic Acids

Nucleophilic attack on the methyl group displaces nitrogen. '"0"" II .. II .. � CH3 � =N: R-C-9: R-C-O-CH diazomet hane is hazardous inThelargeyiequanti timethyl es, it is rarel y Larelsedexcel industrilent,­ alhowever, ly orBecause in solarge-scal e l a boratory reacti o ns. l d s of est e rs diazomet delicate carboxyl ic acihdanes. is often used for small-scale esterifications of valuable and Ami des canthebereacti synthesi zedcompldirectletioyn.fromThe carboxyl icd-base acids, react usingioheat taocarboxyl dlive officwataciedr andwithforce o n t o i n i t i a l aci n of es an ammoni umis carboxyl aophi te sallitc.,Theso thcarboxyl antestioonpsisata thipoors poielenc­t. tHeat rophiiannleg,amithiandsnsalethgiet vtoammoni u m i o n not nucl e e reacti o drives offandsteam formswelanlamiin tdhee. Thilaborats diroectry. synthesis is an importwelalntabove industri100°C al process, it ofteandn works Step 2:

" 0"

+

..

o II

R-C-OH R' -NH2 +

acid

o II

..

amine

+

-----i>

heat

R-C-O- H3N-R'

3

+

:

N=N

:t

20- 1 2 Co n d e n sation of Aci ds with A m i nes: D i rect Synthesis of Amides

o II

.. R-C-NH-R' Hp i +

amide

an ammonium carboxylate salt

963

Example

Qr

COOH

-----i>

heat

ethylamine

benzoic acid

PRO B L E M 20- 1 8

Show how CH 3

(a)

to sy t n

N-ethylbenzamide

ethylammonium benzoate

c o

h es i ze the following compounds, using appropriate a r b xy l ic acids and amines.

lOJ O

o

- -

II

C - N(CH?CH3)?

b ( )

N.N-diethyl-meta-toluamide (DEET insect repel lent)

Q'

o

II

NH - C - CH3

acetanilide

Lialcthoholi ums. alTheumialnumdehydehydriisdean(Liintermedi AlH4 orateLAH) reduces carboxyl iccannot acids tbeo priisolmaaryted in thi s reducti o n, but i t because it is reduced more easily than the original acid. o II R-CH2-OH R-C-OH ( I ) LiAIH4

Example

aci d



o \I R - C - O-R' ester

>

o II R - C-NH2 amide

2 1 - 1 5 Thioesters

[

ester '0'

c-o

good C - O overlap n:

: 0 :- ] [

I R - C - q+- R'

R -C- q - R' stronger overlap J II

..

� n:

====-===-- --- _. ---.,- -..

thioester

c- S

'0'

poor C - S overlap n:

: 0:-

J

I + R -C - ..� - R' R -C - � - R' weaker overlap J II

� n:

----

� F ig u r e 2 1 - 1 2

The resonance overlap in a thioester is not as effective as that in an ester.

The enhanced reactivity of thioesters results from two major differences. First, the resonance stabilization of a thioester is less than that of an ester. In the thioester, the second resonance form involves overlap between a 2p orbital on carbon and a 3p orbital on sulfur (Figure 2 1 - 1 2) . These orbitals are different sizes and are located at different distances from the nuclei. The overlap is weak and relatively ineffective, leaving the C - S bond of a thioester weaker than the C - 0 bond of an ester. The second difference is i n the leaving groups: An alkyl sulfide anion C: � - R) is a better leaving group than an alkoxide anion C:g- R) because the sulfide is less basic than an alkoxide, and the larger sulfur atom carries the negative charge spread over a larger volume of space. Sulfur is also more polarizable than oxy­ gen, allowing more bonding as the alkyl sulfide anion is leaving (Section 6- 1 1 A). Living systems need acylating reagents, but acid halides and anhydrides are too reactive for selective acylation. Also, they would hydrolyze under the aqueous condi­ tions found in living organisms. Thioesters are not so prone to hydrolysis, yet they are excellent selective acylating reagents. For these reasons, thioesters are common acylat­ ing agents in living systems. Many biochemical acylations involve transfer of acyl groups from thioesters of coenzyme A (CoA). Figure 2 1 - 1 3 shows the structure of - -----:-_ . ......:: =--=. -----,-===

+ NH3

�

N �N 00CH3 H 0 H 0 o I I I ' II I II I I CH 3 -C - S - CH2CH2N -C - CH2CH2N - C - CH - C - CH2 -0 - P - 0 - P - O- CH2 0 I II II I H H thioester OH CH 3 0 0 H

N, N!

~ o

'0'

� CH3 - C - S CoA + Nuc:� II)

..

acetyl coenzyme A

I

o = p - o-

CH3 - C I

.. ", s

..

� CoA

Nuc tetrahedral intermediate

H OH

I

coenzyme A (CoA)

:05 I

----i>

0 �

CH 3 - C

'

'

""- Nuc

acylated product

OH

+ :� � CoA

Coenzyme A (CoA) is a thiol whose thioesters serve as biochemical acyl transfer reagents. Acetyl CoA transfers an acetyl group to a nucleophile, with coenzyme A serving as the leaving group. ... F ig u re 2 1 -1 3

1027

10 28

Chapter 2 1 : Carboxylic Acid Derivatives acetyl coenzyme A, together with the mechanism for transfer of the acetyl group to a nucleophile. In effect, acetyl CoA serves as a water-stable equivalent of acetyl chloride (or acetic anhydride) in living systems.

2 1 -1 6

Carbonic acid ( H2C03 ) is formed when carbon dioxide dissolves in water.

Although carbonic acid itself is always in equilibrium with carbon dioxide and water, it has several important stable derivatives. Carbonate esters are diesters of carbonic acid, with two alkoxy groups replacing the hydroxyl groups of carbonic acid. Ureas are diamides of carbonic acid, with two nitrogen atoms bonded to the carbonyl group. The unsubstituted urea, simply called urea, is the waste product excreted by mammals from the metabolism of excess protein. Carbamate esters (urethanes) are the stable esters of the unstable carbamic acid, the monoamide of carbonic acid.

Esters and Amides of Carbonic Acid

O=C=O o II R-O- C - O - R a carbonate ester

o II R-NH-C - NH-R a

substitutecl urea

+

[

HzO

H-O

J

]

- O- H

carbonic acid (unstable)

o-

o II CH3CH2 -O-C - O - CH2CH3 diethyl carbonate

o II (CH3hN-C-N(CH3)2

urea

o

I

R-NH -C -O - R a carbamate or urethane

r

H2N - C - OEt carbamic acid (unstable)

r60

tetramethylurea

CH3-

II

�

O - - O-CH2 CH3

cyclohexyl ethyl carbonate

o II H2N -C-NH2

o

o

ethyl carbamate

o II C

l -naphthyl-N-methylcarbamate (Sevin® insecticide)

Most of these derivatives are synthesized by nucleophilic acyl substitution from phosgene, the acid chloride of carbonic acid. o II CI -C -C1 phosgene

o II CI - C- CI

o +

2 CH)CH2- OH

I

CH3CH2- O-C -O-CH2CH3 diethyl carbonate

o-NH2

)

o-y H

+

2 HCI

o

�

- - OCH,CH3

ethyl N-cyclohexyl carbamate

o II (CH3hN-C - N(CH3)2 tetramethylurea

+

2 HCI

2 1 - 1 6 Esters and A mides of Carbonic Acid

10 29

Another way of making urethanes is to treat an alcohol or a phenol with an isocyanate, which is an anhydride of a carbamic acid. Although the carbamic acid is

unstable, the urethane is stable. This is the way Sevin® insecticide is made. R- N = C = O an isocyanate

[

+

]

J

R-NH - OH a carbamic acid +

R- N = C = O an isocyanate

+

R-NH 2 an amine

CO2

o II

HO -R' alcohol

R-NH - C - O - R' a carbamate ester (urethane)

Example OH

CH3- N = C = O methyl isocyanate

00

+

Sevin® insecticide

I -naphthol

P R O B L E M 2 1 -40

Propose a mechanism for the reaction of methyl isocyanate with I -naphthol to give Sevin® insecticide.

m ine, which has a methyl carba­ led to the development of potent

For each heterocyclic compound, (i) Explain what type of acid derivative is present. (ii) Show what compounds would result from complete hydrolysis. (iii) Are any of the rings aromatic? Explain.

(d)

studies on the a l kaloid physostig­ mate portion. These studies also

P R O B L E M 2 1 -4 1

(a)

The development of Sevin™ and related insecticides resulted from

CX:ro 0 HNANH

(b)

(e)

\ /

P RO B L E M 2 1 -42

Cr°

(c)

H(J0H

(f)

� �

nerve gases such as SarinTM.

(:)=0 oO�NH

0

H3C

\

II

N-C

\

/

H

0

\-1

Biosynthesis of the pyrimidine bases used in DNA occurs via N-carbamoylaspartate, which is formed as follows: o

0

II

II

H2N-C-O-P-O0-

I

carbamoyl phosphate

+

H I H 2N-C-COOI CH2COOH

e nzy

aspartate

me ,

o

H I H2N-C-N-C-COOI I H CH2COOH II

N-carbamoylaspar1ate

N-Carbamoylaspartate cyciizes under the control of an enzyme, giving dihydroorotate, which is dehydrogenated to orotate, a direct precursor to the pyrimidine bases. (The structures of the pyrimidine bases are shown in Section 23-2 1 .) 0

o

HO-C " HN CH

II

2

2

�C " /,CHCOON O d-

I

I

H

I

N-carbamoylaspartate

enzyme )

(- HP)

HN /'

C

II

"

CH

2

-9'"C " /,CHCOOO N

I

H

I

I

dihydroorotate

orotate

physostigmine (CH3hCHO

", CH3 - P = 0

/

F

Sarin

1030

Chapter 2 1 : Carboxylic Acid Derivatives (a) What kind of compound is carbamoyl phosphate? Would you expect such a compound

to react with an amine to give an amide?

(b) What special kind of amide is N-carbamoylaspartate? (c) What kind of reaction cyciizes N-carbamoylaspartate to dihydroorotate? (d) Is orotate aromatic? Draw the structure of pyrimidine. Why is orotate called a "pyrimi­

dine base"? (Hint: Consider tautomers.)

Before the development of tough, resilient polyurethane wheels, street roller skates used steel wheels that stopped dead when they hit the smallest pebble or crack in the pavement. Rollerblades would not exist without polymer technology, both in the wheels and in the strong ABS plastic used for the uppers. The helmet is molded from Lexan® polycarbonate.

Polycarb onat es and Polyu reth anes The chemistry of carbonic acid derivatives is particularly important because two large classes of polymers are bonded by linkages containing these functional groups: the polycarbonates and the polyurethanes. Poly­ carbonates are polymers bonded by the carbonate ester linkage, and polyurethanes are polymers bonded by the carbamate ester linkage. Lexan® polycarbonate is a strong, clear polymer used in bulletproof windows and crash helmets. The diol used to make Lexan® is a phenol called bisphenol A, a common intermediate in polyester and polyurethane synthesis.

o

II

CI - C - Cl phosgene

+

Ho

-o-F,

phosgene fastest

0

>

o

II

C

>

II

C

99 66 N02 N02

bis(p-nitrophenyl) carbonate diphenyl carbonate

o

o

dimethyl carbonate slowest

(b) Another phosgene substitute is triphosgene, a stable solid that can be handled more easily than phosgene. Write a

mechanism to show how triphosgene reacts with methanol to form 3 equivalents of dimethyl carbonate. +

21-60

21-61

II

II

3

/ C .......

/ C .......

H3CO OCH3 dimethyl carbonate

Cl3CO OCCI3 triphosgene

+

6 HCl

In Section 2 1 - 1 6, we saw that Sevin® insecticide is made by the reaction of I-naphthol with methyl isocyanate. A Union Carbide plant in Bhopal, India, once used this process to make Sevin® for use as an agricultural insecticide. On December 3, 1 984, either by accident or by sabotage, a valve was opened that admitted water to a large tank of methyl isocyanate. The pressure and temperature within the tank rose dramatically, and pressure-relief valves opened to keep the tank from bursting. A large quantity of methyl isocyanate rushed out through the pressure-relief valves, and the vapors flowed with the breeze into populated areas, killing about 2500 people and injuring many more. (a) Write an equation for the reaction that took place in the tank. Explain why the pressure and temperature rose dramatically. (b) Propose a mechanism for the reaction you wrote in part (a). (c) Propose an alternative synthesis of Sevin® . Unfortunately, the best alternative synthesis uses phosgene, a gas that is even more toxic than methyl isocyanate. The structures of six useful polymers are shown, together with some of their best-known products. In each case, (i) Determine the kind of polymer (polyamide, polyester, etc.). (ii) Draw the structures of the monomers that would be released by complete hydrolysis. (iii) Suggest what monomers or stable derivatives of the monomers might be used to make these polymers.

(a) - o

J -oJ

�

{

Ir\

OCH'

CH,o

soft, sheer fabrics; synthetic silk

�l

(b) -NH- (CH2)s-C -NH- (CH2)s-C

�l

NH- (CH2)5- C

climbing ropes, violin strings

"

II

high-strength fabrics; bulletproof vests

�

NH - (CH2)s- C -

-NH-o-�{NH-o-�}NH-o-�crash helmets, bulletproof "glass"

(d)

l

-O J-o-� �::::O-

CH'O-

Study Problems o

/I

H (el -c-N

IOC

0 H II N-c

o

- -

II

H O-CH,CH, - O-C-N

CH3

IA-r

V

1 037

0 H II N -C CH3

skateboard wheels, foam mattresses

/I

fishing lines and nets The following two compounds, which exhibit powerful antibacterial activity, were isolated from the fungus Cephalosporium acremonium in 1 948 from the sea near a sewage outlet on the Sardinian coast.

21-62

o

):C:(

H2N" /H II / CH- (CH2)3 -C-N HOOC o

S

N

cephalosporin N

CH3

CH COO�

(a) Name the class of antibiotics represented by each compound. (b) After its structure had been determined, one of these compounds came to be known by a more appropriate name.

Determine which compound is poorly named, and replace the inappropriate part of the name. Methyl p-nitrobenzoate has been found to undergo saponification faster than methyl benzoate. (a) Consider the mechanism of saponification, and explain the reasons for this rate enhancement. (b) Would you expect methyl p-methoxybenzoate to undergo saponification faster or slower than methyl benzoate? A student has just added ammonia to hexanoic acid and has begun to heat the mixture when he is called away to the tele­ phone. After a long telephone conversation, he returns to find that the mixture has overheated and turned black. He distills the volatile components and recrystallizes the solid residue. Among the components he isolates are compounds A (a liq­ uid; molecular formula C6H I I N) and B (a solid; molecular formula C6H l3NO). The infrared spectrum of A shows a strong, sharp absorption at 2247 em-I . The infrared spectrum of B shows absorptions at 3390, 3200, and 1 665 em- I . Determine the structures of compounds A and B. A chemist was called to an abandoned aspirin factory to determine the contents of a badly corroded vat. Knowing that two salvage workers had become ill from breathing the fumes, she put on her breathing apparatus as soon as she noticed an overpowering odor like that of vinegar but much more pungent. She entered the building and took a sample of the con­ tents of the vat. The mass spectrum showed a molecular weight of 1 02, and the NMR spectrum showed only a singlet at 02. 1 5 . The IR spectrum, which appears here, left no doubt about the identity of the compound. Identify the compound, and suggest a method for its safe disposal.

*21-63

21-64

21-65

2.5 , "' 1 00 mrn"' , I,ICT , ,"' II III I I , . O J Ii ':' " II , : 80

wavelength (J.Lm)

'"

60

40

,' ! 11 11.1 1 ; I � 11 '1 1 ! , : � II !I' , M

'

11 11'

' T ' ',

"

: ! ,I : � In . E

II

I ,r;

, il I I I ' II

,' I i I i i ! i iH' o Ii l i l i , ; 3500 4000

20

45 :

HI f "

I I 1Ii 'I " I

'I I I I

" , ,j

1111 1 11I

i ! ! I II! , II I II I !I I

I II ' 1" Ii i

Ilil ll

III I I I

I, " , I \

I 1 '1

, 'il II I I

II

iii

1 11 1

'II i H i 'I PH !

11111

nd ]II"

II I

I

I

.

II

I'

' Ii IHi 2500

iii II 2000

II I

I ii , II ,I I

1 ,3

14

15 16

!

Ii

t.

Ii

II I I

i !

, I

1 800

12 ,

ji

m

1'1

!I

'

I 3000

I

J :5

1 ,f+f+++H+ 1 i i+ +++ ,:!+++++Hi-H-ii+di+i+AA+++i1 -H mi+H+1++i:H+ H+

I 1111" , I

II

1 600

wavenumber (cm - I )

1 400

1 200

1 000

800

600

1 038

Chapter 21: Carboxylic Acid Derivatives

An unknown compound gives a mass spectrum with a weak molecular ion at mlz 1 1 3 and a prominent ion at mlz 68. Its NMR and IR spectra are shown here. Determine the structure, and show how it is consistent with the observed absorp­ tions. Propose a favorable fragmentation to explain the prominent MS peak at mlz 68.

21-66

2.5 100

80

60

I� i;

40

20

a: I ' I, I i

"II

,

"

I II HI I I

I 11,1, , 1 1,1 i III .

jih

I

"

I! I

0 4000

3500

200

180

I

,

I

"

. i j

If l l! !, I 1 1 1 .11

IIlIil ifl

I I 'i! I ! I J! ' : I

Iii d I

I Ill,

' 1111

11 1 11 i1 III '

'

i i'

. ' ' 111 1

11" 1 11 1 1 '

,

I.

iii! !il! i i ' 1 1 ' i i l! II I iii ll!i I ,II II!! III

!!i, !

-

wavelength (p.m) 5 5 ,

45

3.5

I'

iiil ))!) "

j j' i il! ! .iIi , i i i ! .l i i i i ;

,

,I

J' i ' i" ,, 1 , ii'j i ii i l, ' . l l , I I

'

'Il! ! II i - ! H ! II :,j""UH !I I_ '

�

;! l�

:

i;

! 'l>

I

3

I!

I 01

I I i 'I III I I

I

I

.;

'

I

l UI

p

3000

I! ' I I

,I

I

II ' , I

,

I

III

2000

2500

"

Ii i

II

120

-

1 600

1

1 00

I

i)

I I

ii i

I I

!

I I , ,

I I

I

200

,

,

800

1 000

600

0

20

40

: : : -:

i

! : " I

:

rr

!

i I

,

: : :

i

9

10

*21-67

! "

7

8

,

a

6

I

i

i

!

i

,

I' 5

16

I

60

�

:

15

III

80

!

14

, i I

,

, I

i

1 400

!

I .

Ii

Ii ' !

13

12

! i l

:

i Ii i ii i i I I 1 1 ,I

II

0

1 '( ,

Ii II

I! I

II I

1 800

I II , I I ,

• Ij

;j

Ii

I

wavenumber (em 1 )

1 40

1 60

I

,

"

ii i

I

" "

!I

III, II I !

I ,II I I

, ,

i

_IIJ.! I j in I '! I

,I I,

,I ,

j l.!: ,:11

ill I i

ilI

"! ! d i" ii(H j

,

! I

I

III

1 .1 i ! · ,

'

1, 1 I I

II

,\

,

"

I

i

I,

:

,

4

i!

,

3

i o

2

(ppm)

An unknown compound of molecular formula CSH9NO gives the IR and NMR spectra shown here. The broad NMR peak at 07.55 disappears when the sample is shaken with D20. Propose a structure, and show how it is consistent with the absorptions in the spectra.

11

2.5 1 00

" ' '1

' !! I I : .i

i., I I, I i l, !

80

60 40

20

'II '''i · , ·' ",

: A !� N i� 'M

wavelength (p.m)

I 'i,ill,,!:, 11!

, II! .

i.lf; I "

"

, I i : : Ii j l!!N ! ! ! 'I ' ! i , I l l ' II :

, I I! H

I " "'88:'*:: :; *1: ++ffif8t1 11'" 11111 1 1, ,

IIIIiIt

o lllll ! 4000

II

1 8 CrT , i l H+ ll "

ttttf+tt+! t +tiH+ : !tt l +t ! it i: I III!

l i lli "

Ii i

I!ii

"

1 1 1 1 ' II !

lI I l Iff 1tlj 8 i I,

' ,, ,ittltt++,1+H'"' IfH+I, -If Iii

I' !! I '

II

Lilili WJ.l1 J.!.ll!lJ l.!J

3500

, I iHI ' i! ll1

HtHltttt ltlttltHfHtttHtttHi t l .ttltt ,t

II

3000

l f,

45 ,

1 11 1,'

, , }iii ,i!Nl l l i I i i

I I"

5 5 6 7 8 , ! II! I lill ! tIl 1 iHi ffitttHl* HiI i* ll!ii mtt 11 ++t1-+H+i ++ ++t-H++ I I+++i� i I ' tN-l+t+H+i++t-H++t+Hi+,!+ I!j 'I"!: ', . n l I I!,.w.Im

3 5 ;

t" t ' II ! ' ; I 'tt , : ::

"II I i' 'Iii !

I " : Ii i lil i i i : ' I iiil Ilii : ,!ll

I

i

'Ii' II:

, ' II I i i i I:

I I

' , ' II

1 1 1 1, I

II I ,

2500

i

l2 ,

13 ,

l4

15 16

I �

�U�I+�Ii '1+U�+�1+U�t�1+U ���ttU�ttt" �I �,�"' � i l-H+I i+I! H-i+H-H+I+l-H'I+tt\:-! -ltHIII+HI i- f;1H-itHi- lt HHtttl-H-IH-I-+-f\ci--l:::-I-H-!71"'-I-tH :H- II-I TI+!+I -+ I i H+i ++ ++t-H+t+H + +l+H-H-Y++++ \! -HI!+IIlf+f-I++-HItH-H--H-tH-I-H--'Hi--" ! 4 " 4-----li -HiI+-H iI l\' tt:' ttH ' :, !I' \ Iuttt t tt l H tI H� -1H-+ '-+ I ; +i t+-1++++-1-1 "'

I

_L .I ,i

,i

I

, I

i: :

,

i!

!

I I

II

+I-I IhttrHIi+i' Ht++I -H----j-I -H--H- I-H---t,-1I-1-H I I

mIT : ,ttHttttitfiitHtt-ttHtt++titftHtHttjit+llilH+t++I+++f++i++t--H+H+-i++-++f+'+'+-i+++-l itlH,+,1+, Htttl+,HtlttH'l+f-ttllttt/-Htt +Ut1+,Hi HH-lttI+H-H+I+HH-I++-++f++++H--H- I-++--H� 1 I

f-L+++-l-++++-l-++-lH-H+-l t i I mt iT tmtttlTI I ttHT l I mt I ttHtH-HtTIIH-H-+rr i ihlltH+++Il+t+tlthl+ i+t TI I

I

: !!', 'I

II I i ' ,I! I

I

I

I I,

2000

I' 1 800

-

1 600

wavenumber (em 1 )

1 , 1

I'

1 400

,

II !

I ,

II

1 200

1 000

i

800

I

I

600

1 039

Study Problems 200

1 80

"

i

I:: I ! . ... ' . ; �i ! : !I. ; : : i ' i , i

�

--

i!

!

10

21-68

1 40

1 60

80

60

-

'

,

,

,

�

9

I

; : i II ; !

'

:jl l

'

U :

(�

protonated carbonyl

Compare the base-catalyzed and acid-catalyzed mechanisms shown for keto-enol tautomerism. In base, the proton is removed from carbon, then replaced on oxygen. In acid, oxygen is protonated first, then carbon is deprotonated. Most proton-transfer mechanisms work this way. In base, the proton is removed from the old location, then replaced at the new location. In acid, protonation occurs at the new location, followed by deprotonation at the old location. In addition to its mechanistic i mportance, keto-enol tautomerism affects the stere­ ochemistry of ketones and aldehydes. A hydrogen atom on an a carbon may be lost and regained through keto-enol tautomerism; such a hydrogen is said to be enolizable. If an asymmetric carbon atom has an enolizable hydrogen atom, a trace of acid or base allows that carbon to invert its configuration, with the enol serving as the intermediate. A racemic mixture (or an equilibrium mixture of diastereomers) is the result. enolizable hydrogens

a

carbons

(R) configuration

enol (achiral)

(S) configuration

.

.

HO:

/' "C=C " /' enol form

+

H30+

1 044

Chapt er 2 2 : Condensati o n s and

H?nv

PROBLEM-SOLVING

Alpha Substitutions

of Carbonyl Compou nd s

PROBLEM 22-1 Phenylacetone can form two different enols. (a) Show the structures of these enols. (b) Predict which enol will be present in the larger concentration at equilibrium. (c) Propose mechanisms for the formation of the two enols in acid and in base.

In acid, proton transfers usua l ly occur by adding a proton i n the new position, then deprotonating the old position. In base, proton transfers usua l ly occur by deproto nating the o l d position, t h e n reprotonating a t t h e new position.

P R O B L E M 22-2 Show each step in the mechanism of the acid-catalyzed interconversion of (R)- and (S)-2-methyl­ cyclohexanone.

PROBLEM 22-3 When cis-2,4-dimethylcyclohexanone i s dissolved in aqueous ethanol containing a trace of NaOH, a mixture of cis and trans isomers results. Propose a mechanism for this isomerization.

Formation and Stability of Enolate Ions

22-2B

carbonyl group dramatically increases the acidity of the protons on the a carbon atom because most of the enol ate ion's negative charge resides on the electronegative oxygen atom. The pKa for removal of an a proton from a typical ketone or aldehyde is about 20, showing that a typical ketone or aldehyde is much more acidic than an alkane or an alkene (pKa > 40), or even an alkyne (pKa 25). Still, a ketone or aldehyde is less 1 5 .7) or an alcohol (pKa 16 to 1 8). When a simple ketone acidic than water (pKa or aldehyde is treated with hydroxide ion or an alkoxide ion, the equilibrium mixture contains only a small fraction of the deprotonated, enolate form. A

=

=

Of

�;-\

R - C - C - R'

I

+

-OR

H

ketone or aldehyde

=

0

II

:0:I

R' / R - C - C :'" H ··

·

R - C=C

minor

/

'"

major

R'

+

H

ROH

enolate ion

Example

+

CH3CHPH pKa

cyclohexanone pKa

=

19

ethoxide ion

=

15.9

cyclohexanone enolate

(equilibrium lies to the left)

Even though the equilibrium concentration of the enol ate ion may be small, it serves as a useful, reactive nucleophile. When an enolate reacts with an electrophile (other than a proton), the enol ate concentration decreases, and the equilibrium shifts to the right (Figure 22-1). Eventually, all the carbonyl compound reacts via a low con­ centration of the enolate ion. P R O B L E M 2 2 -4 Give the i mportant resonance forms for the enolate i o n of (b) cyclopentanone (a) acetone

(c) 2,4-pentanedione

22-2

o

y

�

II

\ � '- E+

+-OH

E+

o

R-C - CH2 - R'

.... Figure 22-1

Reaction of the enolate ion with an electrophjle removes it from the equilibrium, shifting the equilibrium to the right.

II

R - C - CH - R'

I E

Sometimes this equilibrium mixture of enol ate and base won't work, usually be­ cause the base (hydroxide or alkoxide) reacts with the electrophile faster than the eno­ late does. In these cases, we need a base that reacts completely to convert the carbonyl compound to its enolate before adding the electrophile. Although sodium hydroxide and alkoxides are not sufficiently basic, powerful bases are available to convert a car­ bonyl compound completely to its enolate. The most effective and useful base for this purpose is lithium diisopropylamide (LDA), the lithium salt of diisopropylamine. LDA is made by using an alkyllithium reagent to deprotonate diisopropylarnine. CH3

CH3

I

I

CH3 - C � .. ......--N - H CH3 -

y

H

+

C 4 H 9Li

CH3 - C!f .. 'N : -

+

Li+

......--

CH3 - CH

n-butyllithium

I

CH3

CH3

diisopropylamjne

lithium diisopropylamide (LDA)

Diisopropylamine has a pKa of about 36, showing that it is much less acidic than a typical ketone or aldehyde. B y virtue of its two isopropyl groups, LDA is a bulky reagent. It does not easily attack a carbon atom or add to a carbonyl group. Thus, it is a powerful base, but not a strong nucleophile. When LDA reacts with a ketone, it abstracts the Il' proton to form the lithium salt of the enolate. We will see that these lithium enolate salts are very useful in synthesis. 0

II

0-

H

I

R-C-C-

I

+

(i-C3H7hN-

Li+

�

Li+

R - C=C

"""-......

+

r

(equilibrium lies to the right) Example

cyclohexanone (pK, 19) =

n

O - Li+

0

a:

=

(i-C3H7hN - H

diisop opylami e (pK" 36)

i a of ola

lith u m s l t en te

LDA

ketone == 20)

(pK"

I

+

(i-C3H7hN-

LDA

Li +

�

0

H

lithium enol ate of cyclohexanone (100%)

1 045

ion

enolate eacts with

R - C�CH - R' +H20

Enols and Enolate Ions

+

(i-C3H7)2N - H

(pKa

=

36)

Ch apter 22: Condensation s and Alpha Substitutions of Carbonyl Compounds

1 046

Base-Promoted Cl' Halogenation

22-3

22-3A

Alpha Ha loge nation of Ketones

When a ketone is treated with a halogen and base, an a-halogenation reaction occurs.

o

H I

II

-C-C-

I ketone

o X +

-OH

+

I

I I

-C-C-

X2

+

X-

+

HP

a-haloketone

Example

o

o

lfc

a:

cyclohexanone

l

+

CI-

+

Hp

2-chlorocyclohexanone

The base-promoted halogenation takes place by a nucleophilic attack of an enolate ion on the electrophilic halogen molecule. The products are the halogenated ketone and a halide ion.

MECHANISM 22-6 Step 1:

B ase-Promoted Halogenation

------

Deprotonation of the a carbon forms the enolate ion.

� �)\-OH

- C- C -

I

Step 2:

"0II

�

The enolate ion attacks the electrophilic halogen. "y

-C- C

+

enolate ion

+

x�x

"-

HP

EXA M PLE: Base-promoted bromination of cycl ohex anone.

Br--cJ3r

)

o X )

II

I

- C-C-

00''

I

Br H

+

X-

+

enolate ion

This reaction is called base-promoted, rather than base-catalyzed, because a full equivalent of the base is consumed in the reaction.

PROBLEM-SOLVING

Hint::

In drawing mechanisms, you can show e ither resonance form of a n enol ate attacking the electro p h i le. It is often more intuitive to show the carbanion form attacking.

S O LV E D P R O B L E M 2 2 - 1 Propose a mechanism for the reaction of 3 -pentanone with sodium hydroxide and bromjne to give 2-bromo-3 -pentanone.

SO LUTION I n the presence of sodium hydroxide,

a small amount of 3 -pentanone is present as its enolate. � enolale

22 -3 Alpha Halogenation of Ketones The enolate reacts with bromine to give the observed product.

O�

CH3

I

C-C-Br I CH3CIf-? H

+ Br-

a-haloketone

PROBLEM 22-5 Propose a mechanism for the example showing the formation of 2-chlorocyclohexanone above.

M u ltiple Halogenation In many cases, base-promoted halogenation does not stop with replacement of just one hydrogen. The product (the a-haloketone) is more reactive toward further halogenation than is the starting material, because the electron­ withdrawing halogen stabilizes the enolate ion, enhancing its formation.

o X

-C-CHI II

I

-OH

+

+

[:o�C-C:

:(\� yX C=C

�X "

/

"

/

]

(enol ate stabilized by X )

For example, bromination of 3-pentanone gives mostly 2,2-dibromo-3-pentanone. After one hydrogen is replaced by bromine, the enolate ion is stabilized by both the carbonyl group and the bromine atom. A second bromination takes place faster than the first. Notice that the second substitution takes place at the same carbon atom as the first, because that carbon atom bears the enol ate-stabilizing halogen.

O H� II I) CH CH -C-C-CH I

-OH

(----7

3

2

3

Br

monobrominated ketone

O� CH CH -�-C�CH 3

2

I Br

3

B r �r

o

Br I

CH CH -C-C-CH I

)

II

2

3

Br

stabilized by Br

second brornination

Because of this tendency for multiple halogenation, base-promoted halogenation is rarely used for the preparation of monohalo ketones. The acid-catalyzed procedure (discussed in Section 22-3C) is preferred. PROBLEM 22-6 Propose a mechanism to show how acetophenone undergoes base-promoted chlorination to give trichloroacetophenone.

22-38

The Haloform Reaction

With most ketones, base-promoted halogenation continues until the a-carbon atom is completely halogenated. Methyl ketones have three a protons on the methyl carbon, and they undergo halogenation three times to give trihalomethyl ketones.

o

R-C-CH3 II

methyl ketone

+

3 X2

+

3

-OH

o

II R-C-CX3 trihalomethyl ketone

+

3

X-

+

3

Hp

3

1 047

1 048

Chapter 22: Condensations and Alpha Substitutions of Carbonyl Compounds

With three electron-withdrawing halogen atoms, the trihalomethyl group can serve as a reluctant leaving group for nucleophilic acyl substitution. The trihalomethyl ketone reacts with hydroxide ion to give a tetrahedral intermediate that expels the tlihalomethyl anion CCX3 ) leaving a carboxylic acid. A fast proton exchange gives a carboxylate ion and a haloform (chloroform, CHCI3; bromoform, CHBr3; or iodoform, CHI3). The overall reaction is called the haloform reaction. '

MECHANISM 22-7

Final Steps of th e Haloform Reac1iion

--------�

The conclusion of the h aloform reaction is a nucleophilic acyl substitution, with hydroxide as

the nucleophile and -CX3

as the

leaving group.

Step 1:

Step 2:

Hydroxide adds to the carbonyl group. "0"

-CX3 leaves.

:?5

II�

.. R-C ::- CX, /-:OH .. �

Step 3:

Fast proton transfer from the acid.

..

�O:

", 0 : �

) :CX3 :O - H ...-/

R-C-CX3

R-C ""

I '-.::;

:OH

R-C.......

�

...... 0 : -

+

a car boxy late ion

nu cleoph ilic a cy l su bstitu tion

H CX3 a ha lo fo rm

The overall haloform reaction is summarized next. A methyl ketone reacts with a halogen under strongly basic conditions to give a carboxylate ion and a haloform.

o

II R-C-CH3

ex ce ssX2' -OH

)

a methyl ketone

o

[R-LcxJ

II R -C-O-

HCX}

+

a car boxylate

a tr ih alo methyl ketone (not isolated)

a halo for m

Example

o

II CH3CH2 -C-CH 3

ex ce ssBr2

2- butanone

-OR

)

o

II CH3CH2-C-CBr3

o

II CH3CH2-C-O-

+

prop ionate

HCBr3 bro mo for m

When the halogen is iodine, the haloform product (iodoform) is a solid that separates out as a yellow precipitate. This iodoform test identifies methyl ketones, which halo­ genate three times, then lose -C13 to give iodoform.

o II

Ph -C-CH 3 acetophenone

ex ce ss12 -OR

0

)

II

Ph-C-CI 3 a,a,a-tr iiodoacetophenone

0 �

-OR

II

Ph -C-Obenzoate

+

HCl3 t

iodo for m

Iodine is an oxidizing agent, and an alcohol can give a positive iodoform test if it oxidizes to a methyl ketone. The iodoform reaction can convert such an alcohol to a carboxylic acid with one less carbon atom.

22-3 Alpha Halogenation of Ketones o

-

II R- C - CH3

+

2 HI

excess

-OH

o

II R- C-O-

-)

I?

+

HCI3 �

(one less carbon)

Example

OH I CH3(CH)3- CH - CH3

I CH3(C�)3 - C - CH3

CH3(CH2)3 - C - O-

2-hexanol

2-hexanone

pentanoate

o

o

II

PROBLEM 22-7 Propose a mechanism for the reaction of methyl cyclohexyl ketone with excess bromine in the presence of sodium hydroxide.

P R O B LE M 22-8 Predict the products of the following reactions. (a) methyl cyclopentyl ketone + excess el2 + excess NaOH (b) l -cyclopentylethanol + excess I2 + excess NaOH (c) propiophenone + excess B r2 + excess NaOH

P R O BlE M 2 2 - 9 Which compounds will give a positive iodoform test? (a) I -phenylethanol (b) 2-pentanone (c) 2-pentanol (d) 3 -pentanone (e) acetone (0 isopropyl alcohol

22-3C

Acid-Catalyzed Alpha Halogenation

Ketones also undergo acid-catalyzed a halogenation. One of the most effective pro­ cedures is to dissolve the ketone in acetic acid, which serves as both the solvent and the acid catalyst. In contrast with basic halogenation, acidic halogenation can selectively replace j ust one hydrogen or more than one, depending on the amount of the halogen added.

or acetophenone

0

II C - CR,Bf HBr

a -bromoacetophenone (70%)

or acetophenone

+

0

II C-CRCl, +

2 HCl

a,a -dichloroacetophenone

The mechanism of acid-catalyzed halogenation involves attack of the enol form on the electrophilic halogen molecule. Loss of a proton gives the a-haloketone and the hydrogen halide.

+

HCI3 �

1 049

1 05 0

Chapter 2 2 : Condensations and Alpha Substitutions o f Carbonyl Compounds

MECHANISM 22-8

Acid-Catalyzed Alpha Halog enation

--------'

Acid-catalyzed halogenation results when the enol form of the carbonyl compound serves as a nuc1eophile to attack the halogen (a strong electrophile). Deprotonation gives the a-haloketone.

Step 1: The enol attacks the halogen.

H-O'Q

C

� x

I

II a (3 -c-c=cI

I

ketone

(fD

0

3

2-cyclohexenone

cyclohexanone

a,p-unsaturated

aD

----i>

The Hell- Volhard-Zelinsky (HVZ) reaction replaces a hydrogen atom with a bromine atom on the 0' carbon of a carboxylic acid. The carboxylic acid is treated with bromine and phosphorous tribromide, followed by water to hydrolyze the intermediate a-bromo acyl bromide.

The HVZ reaction

Br2/PBr3

)

Br 0 II I R - CH - C - Br a-bromo acyl bromide

Br 0 I II R - CH - C - OH

Br 0 I II R - CH - C - Br

+

HBr

+

HBr

a-bromoacid

a-bromo acyl bromide

Example

o

II CH3CH2CH2- C - OH

Br 0 I I CH3CH2CH - C - Br

butanoic acid

2-bromobutanoyl bromide

Br 0 II I CH3CH2CH - C -B r

Hp

----'>

Br 0 I II CH3CH2CH - C - OH 2-bromobutanoic acid

2-bromobutanoyl bromide

The mechanism is similar to other acid-catalyzed 0' halogenations; the enol form of the acyl bromide serves as a nucleophilic intermediate. The first step is formation of acyl bromide, which enolizes more easily than does the acid. H 0 I / R-C-C " I OH H

H I ,;P R - C-C " I Br H

acid

acyl bromide keto form

R H

" /

C=C

/ "

O-H Br

enol form

22-4 a

Bro m i nation of Aci ds: Th e HVZ Reacti o n

1052

Alpha Substitutions of Carbonyl Compounds

Chapter 22: Condensations and

The enol is nucleophilic, attacking bromine to give the a-brominated acyl bromide. R

Br-Br

"

�

C

H

v

/� y ;0..:; H1

..

9-0 - H "

Br-

)

(

Br - C - C

Br

I

"

H

R

"0'

I

/ ' Br - C - C " I Br H

�

Br

+

HBr

a-bromo acyl bromide

enol

I f a derivative of the a-bromoacid is desired, the a-bromo acyl bromide serves as an activated intermediate (similar to an acid chloride) for the synthesi s of an ester, amide, or other derivative. If the a-bromoacid itself is needed, a water hydrolysis com­ pletes the synthesis. P R O B L E M 22- 1 2 Show the products of the reactions of these carboxylic acids with PBr3/ Br2 followed by hydrolysis. (d) oxalic acid (c) succinic acid (a) propanoic acid (b) benzoic acid

22-5

We have seen many reactions where nucleophiles attack unhindered alkyl halides and tosylates by the SN2 mechanism. An enolate ion can serve as the nucleophile, becom­ ing alkylated in the process. Because the enol ate has two nucleophilic sites (the oxy­ gen and the a carbon), it can react at either of these sites. The reaction usually takes place primarily at the a carbon, forming a new C - C bond. In effect, this is another type of a substitution, with an alkyl group substituting for an a hydrogen.

Alkylati on of E n o l ate Io ns

a

+

I

..

1

:0:

- C=C

/

"

/

II

+

- C - C - CH - R

R - CH2\...; -X

" 2 C-alkylation product (more common)

x-

O - CH - R

+

I

- C=C

R -CH 2\...; -X

/

" O-alkylation product (less common)

+

X-

Typical bases such as sodium hydroxide or an alkoxide ion cannot be used to form enolates for alkylation because at equilibrium a l arge quantity of the hydroxide or alkoxide base is still present. These strongly nucleophilic bases give side reactions with the alkyl halide or tosylate. Lithium diisopropylamide (LDA) avoids these side reactions. Because it is a much stronger base, LDA converts the ketone entirely to its enolate. All the LDA is consumed in forming the enolate, leaving the enolate to react without interference from the LDA. Also, LDA is a very bulky base and thus a poor nucleophile, so it generally does not react with the alkyl halide or tosylate. a

R'

II

R - C - CH - R '

+

(i-Pr)zN -

enolizable ketone a

"

� :9: r L i+

I

LOA

+Li

R'

1-

R - C-C - R '

+

�

R"- CH -X

2

unhindered halide

.�. ��

l

R - C=C - R' � R - C - C - R ' a

Li+" � enolate

�

Li+

I

enolate

R'

I

R-C - C - R'

I

CH 2 - R"

alkylated

+

LiX

+

(i-Pr)zN -H

diisopropylamine

22-6

Example

o

CH3 I Ph-C-CH-CH3

o

I

Formation and Alkylation of Enamines

CH3 I Ph-C-C-CH I CH -Ph 2 Direct alkylation of enolates (using LDA) gives the best yields when only one I

( I) LDA

3

kind of a hydrogen can be replaced by an alkyl group. If there are two different kinds of a protons that may be abstracted to give enolates, mixtures of products alkylated at the different a carbons may result. Aldehydes are not suitable for direct alkylation because they undergo side reactions when treated with LDA. PROBLE M 22- 1 3 Pr edic t the majo r prod uc ts o f the fol l o win g reac tio ns . ( l ) LDA

(a)

ac eto n e

(b)

&=:

o

( l ) LDA

(c)

06:)

A milder alternative to direct alkylation of enolate ions is the formation and alkylation of an enamine derivative. An enamine (a vinyl amine) is the nitrogen analogue of an enol. The resonance picture of an enamine shows that it has some carbanion character.

R I R-N: \ C=C/

R 1 R-N+ / \ C-c:-

ma jo r

minor

/

\

/

22-6 Fo rmati on and A l kylation of E n a m i nes

\

(EPM)

The electrostatic potential map of a simple enarnine shows a high negative electrostatic potential (red) near the a carbon atom of the double bond. This is the nucleophilic carbon atom of the enamine. n ucl eophilic c ar bon

pyrrol id in e en amin e o f cyc Io h ex anon e

el ec tros tatic po ten ti al map

An enamine is a stronger nucleophile than an enol, but still quite selective in its alkylation reactions. The nucleophilic carbon atom attacks an electrophile to give a resonance-stabilized cationic intermediate (an irninium ion).

R'"RI Ny � E+ /C=C'" an en amin e

el ectrop hil e

R ",IR N� 1 /+C-C-E 1

R",IR N+� 1 /C-C-E 1

minor

ma jo r

1053

1054

Chapter 22: Condensations and Alpha Substitutions of Carbonyl Compounds

An enamine results from the reaction of a ketone or aldehyde with a secondary amine. Recall that a ketone or aldehyde reacts with a primary amine (Section 18- 16) to form a carbinol amine, which dehydrates to give the C = N double bond of an imine. But a carbinolamine from a secondary amine does not form a C=N double bond because there is no proton on nitrogen to eliminate. A proton is lost from the CI' carbon, forming the C = C double bond of an enamine.

0

/

R

II C

+

"

R

N:/ "I H

(

" / o� .. H+

R R " :N/ C

)

/

(

)

H R R " .+ :N/ ·0/ J" H C:" C/. / "

"

�

2° carbino la mine

2° a mine

R"

R

(

)

l' /R

R

" ;( II � c " / " / no proton on N :N I c+

1

�

:N/ I .. H2 VO.:') C+ / H _____ " C / re mo ves a proton " /

�

Example

QC H

cyc lo hexanone

pyrroli dine

pyrroli dine ena mi ne o f cyclohexanone

P R O B L E M 2 2- 1 4 Propose a mechanism for the acid-catalyzed reaction of cyclohexanone with pyrrolidine.

Enamines displace halides from reactive alkyl halides, giving alkylated iminium salts. The i minium ions are unreactive toward further alkylation or acylation. The following example shows benzyl bromide reacting with the pyrrolidine enamine of cyclohexanone.

ena mine

ben zyl bro mi de

alkylate d i miniu m sa lt

alkylate d ketone

The alkylated iminium salt hydrolyzes to the alkylated ketone. The mechanism of this hydrolysis is similar to the mechanism for acid-catalyzed hydrolysis of an lIrune (Section 1 8-16).

22-6 Formation

and Alkylation of Enamines

PROBLEM 2 2 - 1 5 Without looking back, propose a mechanism for the hydrolysis of t hi s iminium salt to the alkyl ­ ated ketone. The first step is attack by water, followed by loss of a proton to give a carbinol­ amine. Protonation on nitrogen allows pyrrolidine to leave, giving the protonated ketone.

We can summarize the overall enamine alkylation process: 1. convert the ketone to an enamine

2. alkylate

with a reactive alkyl (or a cyl ) h a l i de

3. hydrolyze the irninium salt.

OveraLL reaction

R"

0 II

R2NH, H+

H C / " / C

rff='

C

R,,+ R / N II C / " / C

en amine

i mini u m s alt

/ N'

R

I) E+ C �-7 � / /

)

I"

I

1"-

E

H 0+ �

0

II C / " / C

1"-

E

The en amine alkylation procedure is sometimes called the Stork reaction, after its inventor, Gilbert Stork of Columbia University. The Stork reaction is often the best method for alkylating or acylating ketones, using a variety of reactive alkyl and acyl halides. Some halides that react well with enamines to give alkylated and acylated ketone derivatives are the following:

o II

" I / C=C - CH 2-X benzy l halides

R-C -Cl

allylic hal i des

met hyl halides

acy l halides

The following sequence shows the acylation of an enamine to synthesize a f3-diketone. The initial acylation gives an acyl iminium salt, which hydrolyzes to the f3-diketone product. As we will see i n Section 22- 15, f3-dicarbonyl compounds are easily alkylated, and they serve as useful intermediates in the synthesis of more com­ plicated molecules.

H

en amine

(

acyl c hloride

)

ClL�JQ"""" ' - �C-CH' o I CII �C-CH' �H 'CH o

3

inter medi ate

PROBLEM 2 2 - 1 6

-7

N

cyclohexanone + aniline

acyl i miniu m s alt

(d) cyclohexanone + piperidine

PROBLEM 2 2 - 1 7 Show how you would accomplish each conversion using an enamine synthesis. (a) cyclopentanone � 2-allylcyclopentanone

(b) 3 -pentanone

�

I I -C-CH

2-methyl-l-phenyl-3 -pentanone a

(c)

acetophenone

Ph

a

2 - C - Ph

a

a

H 0+ �

Give the expected products of the following acid-catalyzed reactions. (a) acetophenone + methylamine (b) acetophenone + dimethylamine

(c)

a

f3-diketone

1055

1056

Chapter 22: Condensations and Alpha Substitutions of Carbonyl Compounds

22-7

Condensations are some of the most important enolate reactions of carbonyl com­ pounds. Condensations combine two or more molecules, often with the loss of a small molecule such as water or an alcohol. Under basic conditions, the aldol condensation involves the nucleophilic addition of an enolate ion to another carbonyl group. The product, a ,B-hydroxy ketone or aldehyde, is called an aldol because it contains both an aldehyde group and the hydroxy group of an alcohol. The aldol product may dehy­ drate to an a,,B-unsaturated carbonyl compound.

Th e Aldol Condensati on of Ketones a n d Ald ehydes

The aldol condensation

I I a

o

OH III R-C -CH2 -R' I R - C-CH- R ' II

R - C - CH2 - R ' R - C - CH2 -R'

o

aldol products, most commonly in the metabolism of carbohydrates or sugars. In contrast to the chemical reaction, aldolases generate just one product

stereospecifically.

Hence,

they are sometimes used in organic synthesis for key transformations.

��):-�

KEY MECHANISM 22-9

Il

I a

heat

R-C-C-R'

o

o

aldol product

ketone or aldehyde

Aldolases are enzymes that form

i

R- - CH2- R'

22-7 A

1: A base removes an

a

Base-Catalyzed Aldol Condensations

Under basic conditions, the aldol condensation occurs by a nucleophilic addition of the enolate ion (a strong nucleophile) to a carbonyl group. Protonation gives the aldol product. Note that the carbonyl group serves as the electrophile that is attacked by the nucleophilic enolate ion. From the electrophile's viewpoint, the reaction is a nucle­ ophilic addition across the carbonyl double bond. From the viewpoint of the enolate ion, the reaction is an alpha substitution: The other carbonyl compound replaces an alpha hydrogen.

Base-Catalyzed Aldol Condensation

[00'0

proton to form an enolate ion.

;;/ "\. C-C '" /

-:O� / enolate ion

Step 2:

H 2O

a,{3-unsaturated ketone or aldehyde

The base-catalyzed aldol involves the nucleophilic addition of an enolate ion to a carbonyl group. Step

+

The enolate ion adds to the carbonyl group.

enolate

carbonyl

alkoxide

J

/ + Hp C=C

'"

22-7 The Aldol Condensation of Ketones and Aldehydes Step 3:

Protonation of the alkoxide gives the aldol product.

"6 '\.

'O �

_tL

I

"

I c-c/

..

I

:O-H 1 /3 � '0' -cR-O"'l.H =:2: '\./c-c� + RO-

�( ===

�

I

alkoxide

EXAM PLE:

1057

aldol product

Aldol condensation of acetaldehyde. The enol ate ion of acetaldehyde attacks the carbonyl group of another acetaldehyde molecule. Protonation gives the aldol product.

Step

1: A base removes an a proton to form an enol ate ion.

acetaldehyde

Step 2:

base

enol ate of acetaldehyde

I II

The enolate ion adds to the carbonyl group.

:0:H-C-CH I H-C-C-H o H enolate

Step 3:

acetaldehyde

Protonation of the alkoxide gives the aldol product.

II I �

..

'0'-

H�{ H3 H-C-C-H o H

�

H-O-H

)

I

3

:O-H 1/3 H-C-CH 3 H-C-C � H + -OH II I o H aldol product

(50%)

The aldol condensation is reversible, establishing an equilibrium between re­ actants and products. For acetaldehyde, conversion to the aldol product is about 50%. Ketones also undergo aldol condensation, but equilibrium concentrations of the products are generally small. Aldol condensations are sometimes accomplished by clever experimental methods. For example, Figure 22-2 shows how a good yield of the acetone aldol product ("diacetone alcohol") is obtained, even though the equi­ librium concentration of the product is only about 1 %. Acetone is boiled so it con­ denses into a chamber containing an insoluble basic catalyst. The reaction can take

1058

Chapter 22: Condensations and Alpha S ubstitutions of Carbonyl Compounds

CH3 I CH3-C-CH - C-OH o

_ o ut

I�

a

2

__

I CH3

2 acetone

4-hydroxy-4-methyl-2-pentanone

(99%)

"diacetone alcohol"

cold water in

( 1 %) "'--- Ba(OH) catalyst 2

� Figure 22-2 Driving an aldol condensation to completion. The aldol condensation of acetone gives only 1 % product at equilibrium, yet a clever technique gives a good yield. Acetone refluxes onto a basic catalyst such as Ba ( OHh The nonvolatile diacetone alcohol does not reflux, so its equiliblium concentration gradually increases until all the acetone is conveI1ed to diacetone alcohol.

acetone vapor

acetone mixture

place only in the catalyst chamber. When the solution returns to the boiling flask, it contains about 1 % diacetone alcohol. Diacetone alcohol is less volatile than acetone, remaining in the boiling flask while acetone boils and condenses (refluxes) in con­ tact with the catalyst. After several hours, nearly all the acetone is converted to diacetone alcohol. SOLVED PROBLEM 22-3 Propose a mechanism for the base-catalyzed aldol condensation of acetone (Figure 22-2).

SOLUTION The first step is formation of the enolate to serve as a nucleophile.

:0· H

II I�� . + -:OH .. I

CH3-C-C-H H

acetone

enolate ion

The second step is a nucleophilic attack by the enolate on another molecule of acetone. Protonation gives the aldol product.

..

:O�

CJ::b

'.

H ...c C-C:"'-------.- C 0 /' \..:;. H3 C/ H CH3

.. :O� H3C/

H

I

CH3

I

�n

./

C-C-C-O:

I

H

I

CH

_

k.1.- 0 -H

)

3

PROBLEM 22-18 Propose a mechanism for the aldol condensation of cyclohexanone. Do you expect the equilibrium to favor the reactant or the product?

PROBLEM 22-19 Give the expected products for the aJdol condensations of (a) propanal (b) phenylacetaldehyde

22-7 The Aldol Condensation of Ketones and Aldehydes

1059

PROBLEM 22-20 A student wanted to dry some diacetone alcohol and allowed it to stand over anhydrous potassium carbonate for a week. At the end of the week, the sample was found to contain nearly pure acetone. Propose a mechanism for the reaction that took place.

22-78

Acid-Catalyzed A l d o l Condensations

Aldol condensations also take place under acidic conditions. The enol serves a s a weak nucleophile to attack an activated (protonated) carbonyl group. As an example, consider the acid-catalyzed aldol condensation of acetaldehyde. The first step is for­ mation of the enol by the acid-catalyzed keto-enol tautomerism, as discussed earlier. The enol attacks the protonated carbonyl of another acetaldehyde molecule. Loss of the enol proton gives the aldol product.

MECHANISM 22-10

Acid-Catalyzed Aldol Condensation

The ac id- catalyzed aldol invo lves

nuc leophi li c addition of an enol to a protonated carbonyl group.

Step 1:

protonation on 0 followed by deprotonation on C.

Formation of the enol,

by

"�'T'\ +

R'/

I

C - C -H

.

H 3 O+ H

H

keto form

Step 2:

I II

/HH

'0+

..

H O: H "" / =C C "" / H H

/ C-C-H H

enol form

protonated carbonyl

Addition of the enol to the protonated carbonyl. .

.

.

I

I

H-C-CH3

I I

H"-. +C-C-H H-O f .. H attack by enol

Step 3:

.

:O-H

:O- H

I

H-C-CH3

H"" I C-C-H -7"' + H- O .. H

�

resonance-stabilized intermediate

Deprotonation to give the aldol product.

.

I

.

..

:O-H

:O-H

I

I

lose W �=::z)

H""

-7'" :0 ..

resonance-stabilized intermediate

PROBLEM 22-21 Propose a complete mechanism for the acid-catalyzed aldol condensation of acetone.

H-C-CH3 C-C-H

I

H

aldol product

+

+

R OH2

+

1060

Chapter 22: Condensations and Alpha Substitutions of Carbonyl Compounds

22-8 Dehyd rati on of Aldol Prod u cts

Heating a basic or acidic mixture of an aldol product leads to dehydration of the alco­ hol functional group. The product is a conjugated a,j3-unsaturated aldehyde or ketone. Thus, an aldol condensation, followed by dehydration, forms a new carbon-carbon double bond. Before the Wittig reaction was discovered (Section 1 8- 1 3), the aldol with dehydration was probably the best method for joining two molecules with a double bond. It is still often the cheapest and easiest method.

I? HI CH -CLCH HC 1 '-... C -C� 3

3

O

�

3

I�

heat

H

4-methyl-3-penten-2-one (mesityl oxide)

diacetone alcohol

Under acidic conditions, dehydration follows a mechanism similar to those of other acid-catalyzed alcohol dehydrations (Section 1 1 - 1 0). We have not previously seen a base-catalyzed dehydration, however. Base-catalyzed dehydration depends on the acidity of the a proton of the aldol product. Abstraction of an a proton gives an enolate that can expel hydroxide ion to give a more stable product. Hydroxide is not a good leaving group in an E2 elimination, but it can serve as a leaving group in a strongly exothermic step like this one, which stabilizes a negatively charged i ntermediate. The following mechanism shows the base-catalyzed dehydration of 3-hydroxybutanal.

.�,...,.. KEY MECHANISM 22-1 1

B ase-Catalyzed Dehydration of an Aldol

Unlike most alcohols, aldols undergo dehydration in base. Abstraction of an gives an enolate that can expel hydroxide ion to give a conjugated product. Step

1: Formation of the enolate ion.

1

: O -H

1 H-C-CH

..

3

H '-... � : 9H � C-C----- () H � O H

1

removal of Q' proton

Step 2:

H-C-CH3

I

H .. 1 _ '-. §C-C:

0 :7"

+

H20

H

resonance-stabilized enolate

Elimination of hydroxide.

I)

proton

..

..

: O -H

I

a

..

: O -H

H-C-CH H '-... �C _ C� O

1 1 H

3

resonance stabilized enolate

conjugated system

Even when the aldol equilibrium is unfavorable for formation of a j3-hydroxy ketone or aldehyde, the dehydration product may be obtained i n good yield by heat­ i ng the reaction mixture. Dehydration is usually exothermic because it leads to a con­ jugated system. In effect, the exothermic dehydration drives the aldol equilibrium to the right.

22-9 Crossed Aldol Condensations

PROBLEM 22-22 Propose a mechanism for the dehydration of diacetone alcohol to mesityl oxide (b) in base (a) in acid

PROBLE M 22-23

.

When propionaldehyde is warmed with sodium hydroxide, one of the products is 2-methyl2-pen tena1 Propose a mechanism for t h i s reaction.

PROBLEM 22-24 Predict the products of aldol condensation, followed by dehydration, of the following ketones and aldehydes. (a) butyraldehyde (b) acetophenone (c) cyclohexanone

When the enolate of one aldehyde (or ketone) adds to the carbonyl group of another, the result is called a crossed aldol condensation. The compounds used in the reaction must be selected carefully, or a mixture of several products will be formed. Consider the aldol condensation between acetaldehyde (ethanal) and propanal shown below. Either of these reagents can form an enolate ion. Attack by the enolate of ethanal on prop anal gives a product different from the one formed by attack of the eno­ late of propanal on ethanal. Also, self-condensations of ethanal and propanal continue to take place. Depending on the reaction conditions, various proportions of the four possible products result. Enolate of ethanal adds t o propanal

Crossed Aldol Co ndensations

Enolate of propanal adds to ethanal

o

CH3CH2

22-9

OH

"(C - H II }

( )

OH

I

I

CH 3 - C - H

CH2-CHO

CH3 -CH-CHO

CH3CH2 - C - H

I

- : CH -CHO 2

Self-condensation of ethanal

I

Self-condensation of propanal

OH

OH

I

I

CH 3 - C - H

CH3CH2 -C-H I CH3 - CH - CHO

I

CH2- CHO

A crossed aldol condensation can be effective if it is planned so that only one of the reactants can form an enolate ion and so that the other compound is more likely to react with the enol ate. If only one of the reactants has an Q' hydrogen, only one enol ate will be present in the solution. If the other reactant is present in excess or contains a particularly electrophilic carbonyl group, it is more likely to be attacked by the enolate ion. The following two reactions are successful crossed aldol condensations. The aldol products may or may not undergo dehydration, depending on the reaction condi­ tions and the structure of the products. CH3

I �O CH - C - C " 3 I H CH3

excess, no a protons

+

�o

CH 3 - C " a

protons

H

CH3 0H

I I CH3 -C-C-H I CH3

I

�

CH2 - C - H

aldol

CH3 I /H CH3 -C-C � �O C-C" I CH3 H I H dehydrated (75%)

106 1

1062

Chapter

22:

Condensations and Alpha Substitutions of Carbonyl Compounds

excess, no ex protons

ex

{) � � �l

H

�o CH-C " H I CH3

protons

�O��� H

�O C-C" H I CH3

dehydrated (80%)

aldol To carry out these reactions, slowly add the compound with a protons to a basic solution of the compound with no a protons. This way, the enol ate ion i s formed in the presence of a l arge excess of the other component, and the desired reaction i s favored .

P R O B L E M - S O LV I N G S T R AT E G Y

Proposi n g Reaction M ec h a n isms

The general principles for proposing reaction mechanisms, first introduced in Chapter 4 and summarized in Appendix 4, are applied here to a crossed aldol condensation. This example emphasizes a base-catalyzed reaction involving strong nucleophiles. In drawing mecha­ nisms, be careful to draw all the bonds and substituents of each carbon atom involved. Show each step separately, and draw curved arrows to show the movement of electrons from the nucleophile to the electrophile. problem is to propose a mechanism for the base-catalyzed reaction of methylcyclo­ hexanone with benzaldehyde:

Our

o

o

First, we must determine the type of mechanism. Sodium ethoxide, a strong base and a strong nucleophile, implies the reaction involves strong nucleophiles as intermediates. We expect to see strong nucleophiles and anionic intermediates (possibly stabilized car­ banions), but no strong electrophiles or strong acids, and certainly no carbocations or free radicals. 1 . Consider the carbon skeletons of the reactants and products, and decide which

carbon atoms in the products are likely derived from which carbon atoms in the reactants. Because one of the rings is aromatic, it is clear which ring in the product is derived from which ring in the reactants. The carbon atom that bridges the two rings in the products must be derived from the carbonyl group of benzaldehyde. The two a protons from methylcyclohexanone and the carbonyl oxygen are lost as water.

2. Consider whether any of the reactants is a strong enough nucleophile to react

without being activated. If not, consider how one of the reactants might be con­ verted to a strong nucleophile by deprotonation of an acidic site or by attack on an electrophilic site. Neither of these reactants is a strong enough nucleophile to attack the other. If ethoxide removes an a proton from methylcyclohexanone, however, a strongly nucleophilic enol ate ion results.

22-9 Crossed Aldol Condensations

3. Consider how an electrophilic site on another reactant (or, in a cyclization, another

part of the same molecule) can undergo attack by the strong nucleophile to form a bond needed in the product. Draw the product of this bond formation. Attack at the electrophilic carbonyl group of benzaldehyde, followed by protonation, gives a ,B-hydroxy ketone (an aldol).

aldol 4. Consider how the product of nucleophilic attack might be converted to the final

product (if it has the right carbon skeleton) or reactivated to form another bond needed in the product. The ,B-hydroxy ketone must be dehydrated to give the final product. Under these basic conditions, the usual alcohol dehydration mechanism (protonation of hydroxyl, followed by loss of water) cannot occur. Removal of another proton gives an enolate ion that can lose hydroxide in a strongly exothermic step to give the final product.

+ aldol

enolate

dehydrated

5. Draw out all the steps using curved arrows to show the movement of electrons. Be

careful to show only one step at a time. The complete mechanism is given by combining the equations shown above. We suggest you write out the mechanism as a review of the steps involved. As further practice in proposing mechanisms for base-catalyzed reactions, do Problems 22-25 and 22-26B, using the steps just shown.

PROBLEM 22- 2 5 Propose mechanisms for the following base-catalyzed condensations. (a) 2,2-dimethylpropanal with acetaldehyde (b) benzaldehyde with propionaldehyde

PROBLEM 22- 26 When acetone is treated with excess benzaldehyde in the presence of base, the crossed condensation adds two equivalents of benzaldehyde and expels two equivalents of water. Propose a structure for the condensation product of acetone with two molecules of benzaldehyde.

1063

1064

Chapter 22: Condensations and Alpha S ubstitutions of Carbonyl Compounds

PROBLEM-SOLVING

HiltZ;

PROBLEM 22-27 In the problem-solving feature above, methyIcycIohexanone was seen to react at its unsub­ stituted IX carbon. Try to write a mechanism for the same reaction at the methyl-substituted carbon atom, and explain why this regiochemistry is not observed.

Practice predicting the structures of a ld o l products (before and after dehydration) a n d drawing the mechanisms. These reactions are among the most i m portant in this chapter.

PROBLEM 22-28 Predict the major products of the following base-catalyzed aldol condensations with dehydration. (a) benzophenone ( PhCOPh ) + propionaldehyde (b) 2, 2-dimethylpropanal + acetophenone

PROBLEM 22-29 Cinnamaldehyde is used as a flavoring agent in cinnamon candies. Show how cinnamalde­ hyde is synthesized by a crossed aldol condensation followed by dehydration.

cinnamaldehyde:

22- 1 0

Intramolecular aldol reactions of diketones are often useful for making five- and six­ membered rings. Aldol cyclizations of rings larger than six and smaller than five are less common because larger and smaller rings are less favored by their energy and entropy. The following reactions show how a 1 ,4-diketone can condense and dehydrate to give a cyclopentenone and how a 1 ,5-diketone gives a cyclohexenone.

Aldol Cyc l i zations

q "

)

0)

H

-OH

(

R'

)

u :,

0

----7

-OH

OH

enol ate of a I ,4-diketone

oe:.

+

H2 O

+

H2 O

a cyc1opentenone

aldol product

Example

0

0 0

-OH

(

§

)

----7

-OH

C H3

CH3

OH

cis-8-undecene-2,5-dione

aldol product

cis-jasmone (a perfume)

(90%)

0

a: , 0)

enolate of a 1 ,5-diketone

0

-OH

(

)

a:, OH

aldol product

0

----7

-OH

6:,

a cyc1ohexenone

+

H2 O

22- 1 1 Planning Syntheses Using Aldol Condensations

Example

o

1065

o

Q'

+

aCH'

CH3

0CH 3

OH

2,6-heptanedione (a 1 ,5-diketone)

aldol product

3-methylcyclohex-2-enone

The following example shows how the carbonyl group of the product may be outside the ring in some cases.

0

0 CH'

-OH

(

Q CH3 2,7-octanedione

)

0

d:CH3 CH3

OH

aldol product

-OH

-------'.>

otCH3

+ CH3

H 2O

l -acetyl-2-methylcyclopentene

PROBLEM 22-30 Show how 2,7-octanedione might cyclize to a cycloheptenone. Explain why ring closure to the cycloheptenone is not observed.

PROBLEM 22-31 When 1 ,6-cyclodecanedione is treated with sodium carbonate, the product gives a UV spec­ trum similar to that of l -acetyl-2-methylcycIopentene. Propose a structure for the product, and give a mechanism for its formation.

6 o

1 ,6 - cy clodecan edio ne

As long as we remember their limitations, aldol condensations can serve as useful syn­ thetic reactions for making a variety of organic compounds. In particular, aldol con­ densations (with dehydration) form new carbon-carbon double bonds. We can use some general principles to decide whether a compound might be an aldol product and which reagents to use as starting materials. Aldol condensations produce ,B -hydroxy aldehydes and ketones (aldols) and a,,B-unsaturated aldehydes and ketones. If a target molecule has one of these func­ tionalities, an aldol should be considered. To determine the starting materials, divide the structure at the a,,B bond. In the case of the dehydrated product, the a,,B bond is the double bond. Figure 22-3 shows the division of some aldol products into their starting materials.

22- 1 1 Pl a n n i n g Synth eses U s i ng Aldol Condensations

1066

Chapter 22: Condensations and Alpha Substitutions of Carbonyl Compounds

OI H II CH -CH -CpI taCH-C-H H I 0

3

2

came from

;

CH3

break at the a, (3 bond

0 01 o TP taTH-C1 -0 0 OH

H

break at the

� Fi gure 22-3

Aldol products are (3-hydroxyl aldehydes and ketones, or a,(3-unsaturated aldehydes and ketones. An aldol product is dissected into its starting materials by mentally breaking the a, (3 bond.

CH,� - CH

a,

;

o- T�C-C-H H

came fro m

;

propanal

+

CH3

0-(0 CH3

propiophenone

+

CH3 - C

came from

- �'" .�----.

;

- -

�O T o- H benzaldehyde

O �IO

acetophenone

acetophenone

CH3 - CH2 O

break at the double bond

I -

CH3

�O TH2-C1 -0 0 T o- H benzaldehyde

CH3 break at the doubl e bond

II

II

CH,) - C - H

+

0

came from

(3 bond

\

0

propanal

CH3

HI � O o- T \C-C IO

2-C�"-OH

CH3 - CH2

I

II

CH2- C - H

+

-�

0

butanal

,-,- - -

P R O B L E M 22 - 3 2 Show how each compound can be dissected into reagents joined by an aldol condensation, then decide whether the necessary aldol condensation is feasible.

(a)

OH CHFH2CH2-CH-rH-CHO CH2CH2CH3 I

(d)

P R O B L E M 22 - 3 3

OHI CH3I CH2I CH3

0

II

(b) Ph-C-CH - C -Ph

C-CH, aOH o

II

The following compound results from base-catalyzed aldol cyc1ization of a 2-substituted cyclohexanone. (a) Show the diketone that would cyclize to give this product. (b) Propose a mechanism for the cyclization.

22- 1 2 The Claisen Ester Condensation The

a

hydrogens of esters are weakly acidic, and they can be deprotonated to give eno­

late ions. Esters are less acidic than ketones and aldehydes because the ester carbonyl

group is stabilized by resonance with the other oxygen atom. This resonance makes the carbonyl group less capable of stabilizing the negative charge of an enol ate ion.

[ I .

: 0 :-

0

I

R-C-q-R'

A typical pKa for a n

a

+

R-C=q-R'

proton o f a n ester i s about

]

22- 1 2 The Claisen Este r Condensation

24, compared with a pKa o f about 2 0

[ I ;:

for a ketone o r aldehyde. Even so, strong bases d o deprotonate esters.

6

'

acetone (pK" = 20)

I

+

CH 3 0 :-

IL

o CH 3 - O - C - CH 3

methyl acetate (pK" 24) =

+

�

'

V

'

CH 3 -

enolate of acetone

;:

�

CH3 - O - C - CH2

CH 3 Q :-

]

' 0 '-

'

CH3 - C - CH 2

••

1067

�

= CH2

:

1

:-

+ CH30H (pKa

]

CH 3 - O - C = CH2

enolate of methyl acetate

=

1 6)

+ CHpH (pKa

=

1 6)

Ester enolates are strong nucleophiles, and they undergo a wide range of interest­

ing and useful reactions. Most of these reactions are related to the Claisen condensation,

the most important of all ester condensations.

The Claisen condensation results when an ester molecule undergoes nucle­

ophilic acyl substitution by an enol ate. The i ntermediate has an alkoxy ( - OR ) group that acts as a leaving group, leaving a f3 -keto ester. The overall reaction combines two

ester molecules to give a f3-keto ester.

. �V:-_' KEY MECHANISM 22-1 2

The Claisen Ester Condensation

The Claisen condensation is a nucleophilic acyl substitution on an ester, in which the attacking nucleophile is an enolate ion.

Step 1: Formation of the enolate ion.

II

H

�

b

R'O - C - C - R

.Q.

-OR'

+

E�

R'OH

H

ester enolate ion

Step 2: Addition of the enolate to give a tetrahedral intermediate.

"a-

IIJ R -CH2 -(._ C -OR'

I o

R ' O - C - CH - R a

ester enolate

I '-.- . I .

:0 5

_

R - CH 2 - C - OR '

I

R'O-C-CH-R

o

tetrahedral intermediate ( Continued)

1068

Chapter 22: Condensations and Alpha Substitutions of Carbonyl Compounds

Step 3: Elimination of the alkoxide leaving group.

I '-..

) I .. R-CH -C -OR' 2 .. R'O-C-CH-R I o :0

0

· ·

·

R-CH -cf -:OR' 2 I R'O-C-CH-R I " o

a /3-keto ester

tetrahedral intermediate

Notice that one molecule of the ester (deprotonated, reacting as the enolate) serves as

the nucleophile to attack another molecule of the ester, which serveS as the acylating reagent in this nucleophilic acyl substitution.

The j3-keto ester products of Claisen condensations are more acidic than simple

ketones, aldehydes, and esters because the negative charge of the enolate is delocal­ ized over both carbonyl groups. j3-Keto esters have pKa values around

1 1,

showing

they are stronger acids than water. In strong base such as ethoxide ion or hydroxide ion, the j3-keto ester is rapidly and completely deprotonated.

O :­ � R-CH -C 2 R'O-C-CI I "R

0

·

.,7'0

·

� R'-OH + R-CH -C R-CH2 -C 2 I I R'O-C-C-./ H) (pKa 16- 18) R'O-C-C:I "R I I .. o R -:OR' .,7'

=

(pKa

=

I I)

.

:0:

.

a J3-keto ester

0

R-CH -C.,7' 2 R'O-C=CI I "R : 0 :-

-

resonance-stabil ized enolate ion

.p .

Deprotonation of the j3-keto ester provides a driving force for the Claisen con­

densation. The deprotonation is strongly exothermic, making the overall reaction

exothermic and driving the reaction to completion. Because the base is consumed in

the deprotonation step, a full equivalent of base must be used, and the Claisen conden­ sation is said to be base-promoted rather than base-catalyzed. After the reaction is

complete, addition of dilute acid converts the enolate back to the j3-keto ester.

The following example shows the self-condensation of ethyl acetate to give ethyl

acetoacetate (ethyl 3-oxobutanoate). Ethoxide is used as the base to avoid transesteri­

fication or hydrolysis of the ethyl ester (see Problem 22-34). The initial product is the

enolate of ethyl acetoacetate, which is reprotonated in the final step. o

o

2

o

I

I

CH3-C-OCH2CH3

sodium ethoxide

ethyl acetate

CH3 -C 0 I I Na+ -:CH-C-OCH2CH3 keto ester enol ate

H

0+

�

I

CH J -C J3 0 . a CH2-C-OCHFH3 I II ethyl acetoacetate (75%)

S O LV E D P R O B L E M 22-4 Propose a mechanism for the self-condensation of ethyl acetate to give ethyl acetoacetate.

S O LUTION

The first step i s formation of the ester enolate. The equi librium for this step lies far to the left; ethoxide deprotonates only a small fraction of the ester. H

I

0

· ·

·

II

CH2 - C - OCH2CH3 +

(pKa

=

24)

- : qCH2CH3

(�

enolate

(pK"

=

1 6)

22- 1 2 The Claisen Ester Condensation

1069

The enol ate ion attacks another molecule of the ester; expulsion of ethoxide ion gives ethyl acetoacetate. '

'0'

'

II)

E

CH3- C - OCH2CH3

\

)

:0 I)

..

CH3 - C I OCI-I,C H, ..

I� �

�

- : CH1- C - OC H2CH3 nucleophilic attack

-

E

)

"

CH1 - C -OCH1CH3 expulsion of ethoxide

ethyl acetoacetate

In the presence of ethoxide ion, ethyl acetoacetate is deprotonated to give its enolate. This exothermic deprotonation helps to drive the reaction to completion. O

H

II

�

I)

II

0

..

CH3 -C-CH-C-OCH2CH3 + - : qCH2CH) (pKa

o

=

�

II

0

;;

II

CH)-C-CH-C-OCH1CH3

g

enol ate

1 1)

0

o

0

+

H- CHzCH) (pK" 16) =

When the reaction is complete, the enolate ion is reprotonated to give ethyl acetoacetate.

II

::

II

H

II

CH3 -C-CH -C-OCH1CH3

I

II

CH,-C-CH -C-OCH1CH 3

enolate

ethyl acetoacetate

PRO B L E M 22-34 Ethoxide is used as the base i n the condensation of ethyl acetate to avoid some unwanted side reactions. Show what side reactions would occur if the following bases were used. (a) sodium methoxide (b) sodium hydroxide

PROBLEM 22-35

Esters with only one lX hydrogen generally give poor yields in the Claisen condensation. Pro­ pose a mechanism for the Claisen condensation of ethyl isobutyrate, and explain why a poor yield is obtained.

Predict the products of self-condensation of the following esters. (a) methyl propanoate + NaOCH 3 (b) ethyl phenylacetate + NaOCH 2CH 3

U

reactions to generate many useful natural products, such as the an­ tibiotic erythromycin (page

o II

CH 2-C-OCH3

+

NaOCH3

S O LV E D P R O B L E M 2 2 - 5

o

0

Show what ester would undergo Claisen condensation to give the following f3-keto ester.

II

II

Ph-CH 2-CH 2-C-CH-C-OCH 3

I

CH2-Ph

SOLUTION

First, break the structure apart at the lX, f3 bond (lX, f3 to the ester carbonyl). This is the bond formed in the Claisen condensation.

Ph-CH 2 -CH 2

o

-�- I I (3

a

o II

-CH-C-OCH 3

I

CH2-Ph

( Continued)

1021).

These enzymes u s e thioesters in­ stead of the oxygen esters.

PROBLEM 22-36

(c)

Enzymes called p olyketide synthe­ ses catalyze a series of Claisen-type

1070

Chapter 22: Condensations and Alpha Substitutions of Carbonyl Compounds Next, replace the a proton that was lost, and replace the alkoxy group that was lost from the carbonyl. Two molecules of methyl 3-phenylpropionate result. o

o

II

II

H - C H - C - OCH3

Ph -CH2 -CH2 - C - OCH3

I

CH2-Ph

Now draw out the reaction. Sodium methoxide is used as the base because the reactants are methyl esters. 0

o

II

II

Ph-CH 2 -CH 2 - C - C H -C - OCK.}

I

CH2 -Ph PROBLEM-SOLVING

Htnv

P R O B L E M 22-3 7 Propose a mechanism for the self-condensation of methyl 3-phenylpropionate catalyzed by sodium methoxide.

The Cla isen condensation occurs by a n ucleoph i l ic acyl substitution, with d ifferent forms of the ester acti ng as both the nucleo p h i l e (the enol ate) and the el ectrophile (the ester carbonyl).

P R O B L E M 22-38 Show what esters would undergo Claisen condensation to give the following .a -keto esters.

(a) CH3CH2CH2-C

I

�O

(b)

0

II

Ph -CH?-C

(e)

II

r

I

0

II

Ph-CH - C - OCH3

CH3CH2 -CH-C-OCH2CH3 o

-

�O

0

II

(CH 3 hCHCH2 - C - H-C-OEt CH(CH3)2

An internal Claisen condensation of a diester forms a ring. Such an internal Claisen

22-1 3 The Dieckm a n n Con densation: A Claisen Cyclization

cyclization is called a Dieckmann condensation or a Dieckmann cyclization. Five­

and six-membered rings are easily formed by Dieckmann condensations. Rings smaller

than five carbons or larger than six carbons are rarely formed by this method.

The following examples of the Dieckmann condensation show that a 1 ,6-diester

gives a five-membered ring, and a I ,7-diester gives a six-membered ring.

o II C

/ "-

CH2 a

C II

OCH2CH3

�

C - OCH2 CH3

o

diethyl adipate (a 1 ,6-diester)

�

a

o

I

C

'

OCH,CH,

o

cyclic .a-keto ester

(80%)

o II

C

C a / "CH2

OCH3

C - OCH3

-OCH

3)

II

o

dimethyl pimelate (a l ,7-diester)

cyclic .a-keto ester

22- 1 4

Crossed Claisen Condensations

107 1

PRO B L E M 22-39

Propose mechanisms for the two Dieckmann condensations just shown. PRO B L E M 2 2 -40

Some (but not all) of the following keto esters can be formed by Dieckmann condensations. Determine which ones are possible, and draw the starting diesters. II [y C-OCH2CH 3 -a

CH=C / ......

� - Ph

(Hint: aldol)

CR3

Show how you would use the malonic ester synthesis to make the following compounds.

il

a

O

c -aH

Show how you would use the acetoacetic ester synthesis to make the following compounds.

0-11

a

(b)

C - CR3

a

(el

� V

Br

Ph -CH=CH -CH3

a

(c) 22-71

�

CH3

CH3

(e)

(b)

CH3

CH3

22-70

)

CR3

(Hint: Consider using 2,6-heptanedione as an intermediate.)

1095

Study Problems *22-72

6-0

The following compounds can be synthesized by aldol condensations, followed by further reactions. (In each case, work backward from the target molecule to an aldol product, and show what compounds are needed for the condensation.)

o

(b)

22-73

�

C

CH 3

II

C-Ph

(e)

C-OCH3

II

� NOz

o

Propose mechanisms for the following reactions.

(a)

(10+

o II

(b)

PhCHO

Ph-C-OCH3

o (1) MVK

) (2) Hp+

22-74

*22-75

Write equations showing the expected products of the following enarnine alkylation and acylation reactions. Then give the final products expected after hydrolysis of the iminium salts. (a) pyrrolidine enarnine of 3-pentanone + allyl chloride (b) pyrrolidine enarnine of acetophenone + butanoyl chloride (c) piperidine enamine of cyclopentanone + methyl iodide (d) piperidine enarnine of cyclopentanone + methyl vinyl ketone Show how you would accomplish the following multistep conversions. You may use any additional reagents you need.

(a) dimethyl adipate and allyl bromide

(e)

(b) CH3

*22-76

o

o

HO

6

o OEt

Many of the condensations we have studied are reversible. The reverse reactions are often given the prefix retro-, the Latin word meaning "backward." Propose mechanisms to account for the following reactions.

(a)

� U

CH3

(b)

6b

(retro-aldol)

OH

N

,-

,

6

(retro-Michael)

c¢;H' 6 �

CH3 OH (relro-aldol and further condensation)

OH

(e)

0

o

o

+

H2C=CH-CN

1096 22-77

Chapter 22: Condensations and Alpha S ubstitutions of Carbonyl Compounds Show how you would use the Robinson annulation to synthesize the following compounds. (')

22-78

� UJ

o

Propose a mechanism for the following reaction. Show the structure of the compound that results from hydrolysis and decarboxylation of the product.

hydrolysis, decarboxylation benzaldehyde

22-79

malonic ester

A reaction involved in the metabolism of sugars is the splitting of fructose 1,6-diphosphate to give glyceraldehyde 3 phosphate and dihydroxyacetone phosphate. In the living system, this retro-aldol is catalyzed by an enzyme called aldolase; however, it can also be catalyzed by a mild base. Propose a mechanism for the base-catalyzed reaction. o

o

II C H -O-p-OI I 0C=O

2

II

CH 2 -O-P-O-

I

I

C=O

I

0-

I HO-C - H I H-C-OH

(

I

aldolase or -OH

CH20H dihydroxyacetone phosphate

)

\/

o

C I 0 H -C -OH II I CH2 -O-P-OI 0glyceraldehyde 3-phosphate

0 H -C-OH II I CH2 -O-P-OI 0fructose 1,6-diphosphate 22-80

H

Biochemists studying the structure of collagen (a fibrous protein in connective tissue) found cross-links containing a,,B-unsaturated aldehydes between protein chains. Show the structures of the side chains that react to form these cross­ links, and propose a mechanism for their formation in a weakly acidic solution. �

�

H-N rno N-H 1 1 1 H-C-CH2 -CH2 -CH2-CH= C-CH2 -CH2 -C-H 1 1 O=C C=O �

�

protein chain *22-81

protein chain

Show reaction sequences (not detailed mechanisms) that explain these transformations: (a) CH20

+

2 CH2(COOEt)2

(1) NaOEt ) (2) W

EtOOC � COOEt

1

EtOOC

(I) NaOEt

)

+ 2

o

0

� OEt

(1) NaOEt ) (2) W

1

COOEt

23

Carbohydrates and Nucleic Acids

Carbohydrates are the most abundant organic compounds in nature. Nearly all plants and animals synthesize and metabolize carbohydrates, using them to store energy and deliver it to their cells. Plants synthesize carbohydrates through photo­ synthesis, a complex series of reactions that use sunlight as the energy source to convert carbon dioxide and water into glucose and oxygen. Many molecules of glu­ cose can be linked together to form either starch for energy storage or cellulose to support the plant. 6 CO2 + 6 H20

light

)

6 O2 + C6HI206 glucose

�

23- 1 Int roduct i on

starch, cellulose + H20

Most living organisms oxidize glucose to carbon dioxide and water to provide the energy needed by their cells. Plants can retrieve the glucose units from starch when needed. In effect, starch is a plant's storage unit for solar energy for later use. Animals can also store glucose energy by linking many molecules together to form glycogen, another form of starch. Cellulose makes up the cell walls of plants and forms their structural framework. Cellulose is the major component of wood, a strong yet supple material that supports the great weight of the oak, yet allows the willow to bend with the wind. Almost every aspect of human life involves carbohydrates in one form or another. Like other animals, we use the energy content of carbohydrates in our food to produce and store energy in our cells. Clothing is made from cotton and linen, two forms of cellulose. Other fabrics are made by manipulating cellulose to convert it to the semi­ synthetic fibers rayon and cellulose acetate. In the form of wood, we use cellulose to construct our houses and as a fuel to heat them. Even this page is made from cellulose fibers. Carbohydrate chemistry is one of the most interesting areas of organic chem­ istry. Many chemists are employed by companies that use carbohydrates to make foods, building materials, and other consumer products. All biologists need to understand carbohydrates, which play pivotal roles throughout the plant and animal kingdoms. At first glance, the structures and reactions of carbohydrates may seem complicated. We will learn how these structures and reactions are consistent and predictable, however, and we can study carbohydrates as easily as we study the simplest organic compounds.

1097

1098

Chapter 23: Carbohydrates and Nucleic Acids

23 -2 Classification of Carbo hyd rates

PROBLEM-SOLVING

H?nv

The Fischer projection rep resents each asym metric carbon atom by a cross, with the horizontal bonds p rojecting

The term carbohydrate arose because most sugars have molecular formulas Cn (H20 ) m, suggesting that carbon atoms are combined in some way with water. In fact, the empirical formula of most simple sugars is C(H20) . Chemists named these compounds "hydrates of carbon" or "carbohydrates" because of these molecular for­ mulas. Our modern definition of carbohydrates includes polyhydroxyaldehydes, poly­ hydroxyketones, and compounds that are easily hydrolyzed to them. Monosaccharides, or simple sugars, are carbohydrates that cannot be hydrolyzed to simpler compounds. Figure 23-1 shows Fischer projections of the monosaccharides glucose andfructose. Glucose is a polyhydroxyaldehyde, and fructose is a polyhydroxy­ ketone. Polyhydroxyaldehydes are called aldoses (ald- is for aldehyde and -ose is the SUffIX for a sugar), and polyhydroxyketones are called ketoses (ket- for ketone, and -ose for sugar). We have used Fischer projections to draw the structures of glucose and fructose because Fischer projections conveniently show the stereochemistry at all the asymmet­ ric carbon atoms. The Fischer projection was originally developed by Emil Fischer, a carbohydrate chemist who received the Nobel Prize for his proof of the structure of glucose. Fischer developed this shorthand notation for drawing and comparing sugar structures quickly and easily. We will use Fischer projections extensively in our work with carbohydrates, so you may want to review them (Section 5-10) and make models of the structures in Figure 23-1 to review the stereochemistry implied by these struc­ tures. In aldoses, the aldehyde carbon is the most highly oxidized (and numbered 1 in the IUPAC name), so it is always at the top of the Fischer projection. In ketoses, the carbonyl group is usually the second carbon from the top.

toward the viewer and the vertical bonds projecting away. The carbon chain is arranged along the vertical bonds, with the most oxidized end

(or carbon #1 in the IUPAC name) at

PROBLE M 23-1

Draw the mirror images of glucose and fructose. Are glucose and fructose chiral? Do you expect them to be optically active?

the top.

CHO =*

H-C-OH

A disaccharide is a sugar that can be hydrolyzed to two monosaccharides. For example, sucrose ("table sugar") is a disaccharide that can be hydrolyzed to one mol­ ecule of glucose and one molecule of fructose. 1 sucrose

For more than one asymmetric carbon atom, the Fischer projection represents a tota l l y eclipsed conformation. This is not the most stable conformation, but it's usually the most sym m etric conformation, which is most helpful for comparing stereochem istry.

� Figure 23-1

heat

1 glucose

+

1 fructose

Both monosaccharides and disaccharides are highly soluble in water, and most have the characteristic sweet taste we associate with sugars. Polysaccharides are carbOhydrates that can be hydrolyzed to many monosaccha­ ride units. Polysaccharides are naturally occurring polymers (biopolymers) of carbohy­ drates. They include starch and cellulose, both biopolymers of glucose. Starch is a polysaccharide whose carbohydrate units are easily added to store energy or removed to

CHO

Fischer projections of sugars. Glucose and fructose are monosaccharides. Glucose is an aldose (a sugar with an aldehyde group), and fructose is a ketose (a sugar with a ketone group). Carbohydrate structures are commonly drawn using Fischer projections.

H30+

H-C-OH I HO-C-H I H-C-OH

CHO H or

I

H-C-OH CH20H glucose

HO

OH H

CHpH

CHpH

\

I

c=o

c=o -

H

OH

HO-C-H I H-C-OH

H

OH

H-C-OH

CH20H

CH20H

or

HO

I

H

H

OH

H

OH CH20H

fructose

23-3 Monosaccharides

1099

provide energy to cells. The polysaccharide cellulose is a major structural component of plants. Hydrolysis of either starch or cellulose gives many molecules of glucose. starch cellulose

+ H30 heat

+ H30 heat )

over 1000 glucose molecules over 1 000 glucose molecules

To understand the chemistry of these more complex carbohydrates, we must first learn the principles of carbohydrate structure and reactions, using the simplest mono­ saccharides as examples. Then we will apply these principles to more complex disac­ charides and polysaccharides. The chemistry of carbohydrates applies the chemistry of alcohols, aldehydes, and ketones to these polyfunctional compounds. In general, the chemistry of biomolecules can be predicted by applying the chemistry of simple organ­ ic molecules with similar functional groups. 23-3A

23-3

Classification of Monosaccharides

Most sugars have their own specific common names, such as glucose, fructose, galactose, and mannose. These names are not systematic, although there are simple ways to remember the common structures. We simplify the study of monosaccharides by grouping similar structures together. Three criteria guide the classification of monosaccharides: 1. Whether the sugar contains a ketone or an aldehyde group

2. The number of carbon atoms in the carbon chain

3. The stereochemical configuration of the asymmetric carbon atom farthest from

the carbonyl group

As we have seen, sugars with aldehyde groups are called aldoses, and those with ketone groups are called ketoses. The number of carbon atoms in the sugar generally ranges from three to seven, designated by the terms triose (three carbons), tetrose (four carbons), pentose (five carbons), hexose (six carbons), and heptose (seven carbons). Terms describing sugars often reflect these first two criteria. For example, glucose has an aldehyde and contains six carbon atoms, so it is an aldohexose. Fructose also con­ tains six carbon atoms, but it is a ketone, so it is called a ketohexose. Most ketoses have the ketone on C2, the second carbon atom of the chain. The most common natu­ rally occurring sugars are aldohexoses and aldopentoses. I CHO

I CHPH

2CHOH

2C=O

3CHOH

3CHOH

ICHO

ICHPH

4CHOH

4CHOH

2CHOH

2C=O

sCHOH

5CHOH

3CHOH

3CHOH

6CHPH

6CHPH

4CHPH

4CHPH

an aldohexose

a ketohexose

an aldotetrose

a ketotetrose

I I

I

I

I

PROBLE M 23-2

I

I I I

I

I I

I

I

I I

(a) How many asymmetric carbon atoms are there in an aldotetrose? Draw all the aldotet­

rose stereoisomers. (b) How many asymmetric carbons are there in a ketotetrose? Draw all the ketotetrose stereoisomers. (c) How many asymmetric carbons and stereoisomers are there for an aldohexose? For a ketohexose? (continued)

Monosaccha rides

1 100

Chapter 23: Carbohydrates and Nucleic Acids PROBLEM 23-3

(a) There is only one ketotriose, called dihydroxyacetone. Draw its structure. There is only one aldotriose, called glyceraldehyde. Draw the two enantiomers of glyceraldehyde.

(b)

23-38

The D and L Configurations of Sugars

Around 1 880-1900, carbohydrate chemists made great strides in determining the struc­ tures of natural and synthetic sugars. They found ways to build larger sugars out of smaller ones, adding a carbon atom to convert a tetrose to a pentose and a pentose to a hexose. The opposite conversion, removing one carbon atom at a time (called a degradation), was also developed. A degradation could convert a hexose to a pentose, a pentose to a tetrose, and a tetrose to a triose. There is only one aldotriose, glyceraldehyde. These chemists noticed they could start with any of the naturally occurring sugars, and degradation to glyceraldehyde always gave the dextrorotatory ( + ) enan­ tiomer of glyceraldehyde. Some synthetic sugars, on the other hand, degraded to the levorotatory ( - ) enantiomer of glyceraldehyde. Carbohydrate chemists started using the Fischer-Rosanoff convention, which uses a D to designate the sugars that degrade to (+) glyceraldehyde and an L for those that degrade to ( - ) glyceralde­ hyde. Although these chemists did not know the absolute configurations of any of these sugars, the D and L relative configurations were useful to distinguish the natu­ rally occurring D sugars from their unnatural L enantiomers. We now know the absolute configurations of (+) and ( - ) glyceraldehyde. These structures serve as the configurational standards for all monosaccharides.

CHO IH-C-OHI CH20H (+ )-glyceraldehyde

CHO IHO-C-HI CH20H

( - )-glyceraldehyde L series of sugars

D series of sugars

Figure 23-2 shows that degradation (covered in Section 23-14) removes the aldehyde carbon atom, and it is the bottom asymmetric carbon in the Fischer projec­ tion (the asymmetric carbon farthest removed from the carbonyl group) that deter­ mines which enantiomer of glyceraldehyde is formed by successive degradations.

CHO

C02t

H-C-OH

CHO

I I

HO-C-H I

H-C-OH

I HD- (+)

i

-OH 1

CH20H

-glucose

C02t

I

degrade )

HO-C-H I

H-C-OH

i

I H-

-OH 1

CH20H

D-

(-) - arabinose

degrade

)

CHO

co2t

H-C-OH

CHO

I

i

I H-

-OH 1

degrade )

CHpH

D-

(-) -erythrose

i

I H-

-OH I

CH20H

D-

(+) -glyceraldehyde

... Figure 23-2

Degradation to glyceraldehyde. Degradation of an aldose removes the aldehyde carbon atom to give a smaller sugar. Sugars of the 0 series give ( + )-glyceraldehyde on degradation to the triose. Therefore, the OH group of the bottom asymmetric carbon atom of the 0 sugars must be on the right in the Fischer projection.

23-3 Monosaccharides

We now know that the ( + ) enantiomer of glyceraldehyde has its OH group on the right in the Fischer projection, as shown in Figure 23-2. Therefore, sugars of the D series have the OH group of the bottom asymmetric carbon on the right in the Fischer projec­ tion. Sugars of the L series, in contrast, have the OH group of the bottom asymmetric carbon on the left. In the following examples, notice that the D or L configuration is determined by the bottom asymmetric carbon, and the enantiomer of a D sugar is always an L sugar. CH10H CHO

I

C=O

CHO

CH10H H

O=C

�~ ~~ H

HO

OH

H

H

CHpH

CHpH

o-threose

OH

HO

o-ribulose

OH

H

HO H

H

HO

OH

H

CH20H

CH20H

L-threose

CHO

CHO

I -

CH20H

CHpH o-xylose

L-ribulose

L-xylose

As mentioned earlier, most n aturally occurring sugars have the D configuration, and most members of the D family of aldoses (up through six carbon atoms) are found in nature. Figure 23-3 shows the D family of aldoses. Notice that the D or L configu­ ration does not tell us which way a sugar rotates the plane of polarized light. This must be determined by experiment. Some D sugars have (+) rotations, and others have ( ) rotations.

CHO IH*CH2OH0HI

-

jCHO

H+OH l � CH20H I CHO CHO HH OH [H OH 1 H CHpHOH � [CHO CHO CHO CHO HH OHOH HOH HOH HOH HOH HOHO HH H OH H OH H OH H OH CH20H CH20H CHpH CHpH 0-(-)-erythrose

0-(-)-arabinose

0-(+)-allose � Figure

The

0-(+)-altrose

23-3

ICHO

HO+H l � CHpH CHO CHO H ["0 1 ["0� H 1 H CHpHOH H CHpHOH ~ CHO CHO CHO CHO HH OHOH HOH HOH HOH HOH HOHO HH HO H HO H HO H HO H CH20H CHpH CHpH CHpH

0-(+)-glyceraldehyde

0-(+)-glucose

o-(+)-mannose

0-(-)-threose

0-(-)-lyxose

0-(+)-xylose

0-(-)-gulose

0-(-)-idose

0-(+)-galactose

D family of aldoses. All these sugars occur naturally except for threose, lyxose, allose, and gulose.

0-(+)-talose

1 101

1102

Chapter 23: Carbohydrates and Nucleic Acids

On paper, the family tree of D aldoses (Figure 23-3) can be generated by starting with D-( + )-glyceraldehyde and adding another carbon at the top to generate two aldotetroses: erythrose with the OH group of the new asymmetric carbon on the right, and threose with the new OH group on the left. Adding another carbon to these aldotetroses gives four aldopentoses, and adding a sixth carbon gives eight aldohexoses.* In Section 23- 1 5, we describe the Kiliani-Fischer synthesis, which actually adds a carbon atom and generates the pairs of elongated sugars just as we have drawn them in this family tree. At the time the D and L system of relative configurations was introduced, chemists could not determine the absolute configurations of chiral compounds. They decided to draw the D series with the glyceraldehyde OH group on the right, and the L series with it on the left. This guess later proved to be correct, so it was not necessary to revise all the old structures. PROBLEM-SOLVING

HritZ:-

Most naturally occurring sugars are of the

D

series, with the OH group of the

bottom asymmetric carbon on the right in the Fischer projectio n.

23-4 Eryt h ro and Th reo Diastereome rs

PROBLEM 23-4

Draw and name the enantiomers of the sugars shown in Figure 23-2. Give the relative config­ uration (0 or L) and the sign of the rotation in each case. PROBLEM 23-5

Which configuration (R or S) does the bottom asymmetric carbon have for the 0 series of sugars? Which configuration for the L series?

Erythrose is the aldotetrose with the OH groups of its two asymmetric carbons situat­ ed on the same side of the Fischer projection, and threose is the diastereomer with the OH groups on opposite sites of the Fischer projection. These names have evolved into a shorthand way of naming diastereomers with two adjacent asymmetric carbon atoms. A diastereomer is called erythro if its Fischer projection shows similar groups on the same side of the molecule. It is called threo if similar groups are on opposite sides of the Fischer projection. For example, syn hydroxylation of trans-crotonic acid gives two enantiomers of the threo diastereomer of 2,3-dihydroxybutanoic acid. The same reaction with cis-crotonic acid gives the erythro diastereomer of the product. """

CH3 H

/

C=C

/

H

"-

COOH

H HO

trans-crotonic acid

+�: t

+ t

COOH

HO

H

H

CH3

H OH

CH3

(2R,3S)

(2S,3R) threo-2,3-dihydroxybutanoic acid

+ t

COOH

"""

CH3 H

/

C=C

/

"-

COOH H

cis-crotonic acid

H H

OH

HO

OH

HO

CH3

+� t

H

H

CH3

(2R,3R) (2S,3S) erythro-2,3-dihydroxybutanoic acid

"'Drawn in this order, the names of the four aldopentoses (ribose, arabinose, xylose, and Iyxose) are remembered by the mnemonic "Ribs are extra lean." The mnemonic for the eight aldohexoses (allose, altrose, glucose, mannose, gulose, idose, galactose, and talose) is "All ailluists gladly make gum in gallon tanks."

- --�--

i i f f �

�

2 H3

H H

23-5 Epimers

��M�__

2 H'

,

B'

Br

H

CH3

Br

CH3

H

Cl

H

OH

H

ci

HO

H

CH3

erythro threo 2,3 - dibromopentane

CH3

+ f �+i: -:+�: CH3

COOH

H

H

H

Br

HO

CH3

(±) or (d,!)

COOH

meso

Elythro and threo nomenclature. The terms eJythro and threo are used with dissymmetric molecules whose ends are different. The erythro diastereomer is the one with similar groups on the same side of the Fischer projection, and the threo diastereomer has similar groups on opposite sides of the Fischer projection. The terms meso and (±) [or (d,l)] are preferred with symmetric molecules.

COOH

B'

2,3 -dibromobutane

.... Figure 23-4

erythro threo 3-chloro-2-butanol

CH3

meso

OH H

COOH

(±) or (d,l)

tartaric acid

-- �-

The terms erythro and threo are generally used only with molecules that do not have symmetric ends. In symmetric molecules such as 2,3-dibromobutane and tartaric acid, the terms meso and Cd,l) are preferred because these terms indicate the diastere­ orner and tell whether or not it has an enantiorner. Figure 23-4 shows the proper use of the terms erythro and threo for dissymmetric molecules, as well as the terms meso and Cd,l) for symmetric molecules. PROBLEM 23-6

Draw Fischer projections for the enantiomers of threo-l ,2,3-hexanetriol.

HOCH2-CH(OH)-CH(OH)-CH2CH2CH3 PROBLEM 23-7

The bronchodilator ephedrine is erythro-2-(methylamino)- 1 -phenylpropan- l -ol. The decon­ gestant pseudoephedrine is threo-2-(methylamino)- l -phenylpropan- l -ol. (a) Draw the four stereoisomers of 2-(methylamino)- 1 -phenylpropan-I-ol, first as Fischer projections, and then as three-dimensional representations (dotted lines and wedges). (b) Label ephedrine and pseudoephedrine. What is the relationship between them? (c) Label the D and L isomers of ephedrine and pseudoephedrine using the Fischer-Rosanoff convention. (d) Both ephedrine and pseudoephedrine are commonly used as racemic mixtures. Ephedrine is also available as the pure levorotatory ( ) isomer (Biophedrine® ), and pseudoephedrine is also available as the more active ( + ) isomer (Sudafed®). Can you label the ( ) isomer of ephedrine and the ( +) isomer of pseudoephedrine?

-

1103

-

Many common sugars are closely related, differing only by the stereochemistry at a single carbon atom. For example, glucose and mannose differ only at C2, the first asymmetric carbon atom. Sugars that differ only by the stereochemistry at a single carbon are called epimers, and the carbon atom where they differ is generally stated. If the number of a carbon atom is not specified, it is assumed to be C2. Therefore, glucose and mannose are "C2 epimers" or simply "epimers." The C4 epimer of glucose is galactose, and the C2 epimer of erythrose is threose. These relationships are shown in Figure 23-5.

23- 5 Eplme rs •

1104

Chapter 23: Carbohydrates and Nucleic Acids

iCHO

iCHO

iCHO

C2 epimers

H HO H H 4 OH HS OH 6CH2OH

2

3

H

H

OH 6CH2OH

OH 6CH20H 5

o-galactose

o-glucose

o-mannose

OH H

iCHO H

iCHO

C2 ep ime rs

H

OH 4CHPH 3

OH 4CH20H 3

o-threose

o-erythrose

.. F i g u re 23-5

Epimers. Epimers are sugars that differ only by the stereochemistry at a single carbon atom. If the number of the carbon atom is not specified, it is assumed to be C2.

PROBLEM 23-8 (a) Draw o-allose, the C3 epimer of glucose.

(b)

(c) Draw D-idose, the C3 epimer of D-talose. Now compare your answers with Figure 23-3.

Draw D-talose, the C2 epimer of D-galactose.

(d) Draw the C4 "epimer" of D-xylose. Notice that this "epimer" is actually an L-series sugar, and we have seen its enantiomer. Give the correct name for this L-series sugar.

23-6 Cyc l i c Structures of Monosacch a rid es

Cyclic Hemiacetals In Chapter 1 8, we saw that an aldehyde reacts with one molecule of an alcohol to give a hemiacetal, and with a second molecule of the alcohol to give an acetal. The hemiacetal is not as stable as the acetal, and most hemiacetals decompose spontaneously to the aldehyde and the alcohol. Therefore, hemiacetals are rarely isolated. If the aldehyde group and the hydroxyl group are part of the same molecule, a cyclic hemiacetal results. Cyclic hemiacetals are particularly stable if they result in five- or six-membered rings. In fact, five- and six-membered cyclic hemiacetals are often more stable than their open-chain forms.

MECH ANISM 23-1 Step 1:

Formation of a Cyclic Hemiacetal

Protonation of the carbonyl.

H H C.o : II · :0:

Step 2:

The OH group adds as a nucleophile.

H 0+ I C-H ··/ I :O-H

Cy Step 3:

C

�

�

8-hydroxyaldehyde

Deprotonation gives a cycl i c hemiacetal.

H 0+ H20: ( ) I ·C-H · -j' �.. I :O-H

C

derived from the OH group

+

H3 0+

derived from the CHO group

1105

23-6 Cyclic S tructures of Monosaccharides I

�

CHO 2 H OH HO H H

3

4 5

6

H OH OH

CHZOH

H

6

Fischer projection

HO

on right side

� 6

l

CH20H .___H __ H 0: .. ,-:":/: 0+ H 4 _____ OH H HO 3 2

;

';1

CH20H

O OH

4

H

H

2 H OH

3

HO

OH

C6 rotated up

H OH

Haworth projection

Cl is the only carbon atom bonded to 2 oxygens H

chair conformation (all substituents equatorial)

chair conformation (OH on CI axial)

.... Figure 23-6

Glucose exists almost entirely as its cyclic h emiacetal form.

The Cyclic Hemiacetal Form of G lucose

Aldoses contain an aldehyde group and several hydroxyl groups. The solid, crystalline form of an aldose is normally a cyclic hemi­ acetal. In solution, the aldose exists as an equilibrium mixture of the cyclic hemiacetal and the open-chain form. For most sugars, the equilibrium favors the cyclic hemiacetal. Aldohexoses such as glucose can form cyclic hemiacetals containing either five­ membered or six-membered rings. For most common aldohexoses, the equilibrium favors six-membered rings with a hemiacetal linkage between the aldehyde carbon and the hy­ droxyl group on C5. Figure 23-6 shows formation of the cyclic hemiacetal of glucose. No­ tice that the hemiacetal has a new asymmetric carbon atom at Cl. Figure 23-6 shows the Cl hydroxyl group up, but another possible stereoisomer would have this hydroxyl group directed down. We discuss the stereochemistry at Cl in more detail in Section 23-7. The cyclic structure is often drawn initially in the Haworth projection, which depicts the ring as being flat (of course, it is not). The Haworth projection is widely used in biology texts, but most chemists prefer to use the more realistic chair confor­ mation. Figure 23-6 shows the cyclic form of glucose both as a Haworth projection and as a chair conformation. Draw i n g Cycl ic Monosaccharides Cyclic hemiacetal structures may seem com­ plicated at first glance, but they can be drawn and recognized by following the process illustrated in Figure 23-6.

1. Mentally lay the Fischer projection over on its right side. The groups that were on the right in the Fischer projection are down in the cyclic structure, and the groups that were on the left are up.

2. C5 and C6 curl back away from you. The C4 - C5 bond must be rotated so that the CS hydroxyl group can form a part of the ring. For a sugar of the D series, this rota­ tion puts the terminal -CHzOH (C6 in glucose) upward.

3. Close the ring, and draw the result. Always draw the Haworth projection or chair conformation with the oxygen at the back, right-hand corner, with Cl at the far right. Cl is easily identified because it is the hemiacetal carbon-the only car­ bon bonded to two oxygens. The hydroxyl group on Cl can be either up or down, as discussed in Section 23-7. Sometimes the ambiguous stereochemistry is symbolized by a wavy line.

I

1106

Chapter 23: Carbohydrates and Nucleic Acids

Chair conformations can be drawn by recognizing the differences between the sugar in question and glucose. The following procedure is useful for drawing D-aldohexoses.

1. Draw the chair conformation puckered, as shown in Figure 23-6. The hemiacetal

carbon (C l ) is the footrest.

2. Glucose has its substituents on alternating sides of the ring. In drawing the chair

conformation, just put all the ring substituents in equatorial positions. (In the Haworth projection, the - OH on C4 is opposite the - CH 20H on C5, and the - OH on C3 is opposite that on C4.) 3. To draw or recognize other common sugars, notice how they differ from glucose and make the appropriate changes. PROBLEM-SOLVING

HinZ;

Learn to draw g l ucose, both in the Fischer projection and in the chair conformation (a l l su bstituents

equatorial). Draw other pyranoses by noticing the differences from g l ucose and changing the g l u cose structure as

((2: man nose; (3: al lose; and (4: galactose). To recognize other needed. Remem ber the epimers of

g l ucose

sugars, look for axial substituents where they differ from g l u cose.

SOLV E D PROBLEM 23- 1

Draw the cyclic hemiacetal forms of D-mannose and D-galactose both as chair conforma­ tions and as Haworth projections. Mannose is the C2 epimer of glucose, and galactose is the C4 epimer of glucose. S OL U T I O N

The chair conformations are easier t o draw, s o w e will d o them first. Draw the rings and number the carbon atoms, starting with the hemiacetal carbon. Mannose is the C2 epimer of glucose, so the substituent on C2 is axial, while all the others are equatorial as in glucose. Galactose is the C4 epimer of glucose, so its substituent on C4 is axial. H

OH C4 H OH H

PROBLEM-SOLVING

HinZ;

Groups on the right in the Fischer

projection are down in the usual cyclic structure, and groups that were on the left in the Fischer projection are up.

OH

H

C2 D-mannose

H D-galactose

H

The simplest way to draw Haworth structures for these two sugars is to draw the chair conformations and then draw the flat rings with the same substituents in the up and down posi­ tions. For practice, however, let's lay down the Fischer projection for galactose. You should fol­ low along with your molecular models. l . Lay down the Fischer projection: right � down and left � up. I CHO H

H

OH

HO

H

HO

H OH

H

6 CHoOH 0 � IC "

HO

H

6

H

OH

CH20H D-galactose 2. Rotate the C4 - C5 bond to put the C5 - OH - CI-I20H goes up.) H� 6 ) CHPH � OH �O I c" OH H H _

4

H

H

OH

�

HO H

111

place. (For a D sugar, the

6 CHPH OH H OH

H

H

OH

O C� ' ' 'H

23-6 Cyclic Structures of Monosaccharides 3. Close the ring, and draw the final hemiacetal. The hydroxyl group 0 11 C l can be either up or down, as discussed in Section 23-7. Sometimes this ambiguous stereochemistry is symbolized by a wavy line.

"'

H

� �>

H

H O '

equivocal

OH

H

'

H

OH

H

H

PRO B LEM 23- 9

Draw the Haworth projection for the cyclic structure of D-mannose by laying down the Fischer projection. PRO B LEM 23- 1 0

Allose is the C3 epimer of glucose. Draw the cyclic hemiacetal form of D-allose, first in the chair conformation and then in the Haworth projection.

Not all sugars exist as six-membered rings in their hemiacetal forms. Many aldopentoses and keto­ hexoses form five-membered rings. The five-membered ring of fructose is shown in Figure 23-7. Five-membered rings are not puckered as much as six-membered rings, so they are usually depicted as flat Haworth projections. The five-membered ring is customarily drawn with the ring oxygen in back and the hemiacetal carbon (the one bonded to two oxygens) on the right. The - CH20H at the back left (C6) is in the up position for D-series ketohexoses.

The Five - M embered Cyclic H e m iacetal Form of Fructose

F u ra nose Names Cyclic structures of monosaccharides are named according to their five- or six-membered rings. A six-membered cyclic hemiac­ etal is called a pyranose, derived from the name of the six-membered cyclic ether pyran. A five-membered cyclic hemiacetal is called a furanose, derived from the name of the five-membered cyclic ether Juran. The ring is still numbered as it is in the sugar, not beginning with the heteroatom as it would be in the heterocyclic nomencla­ ture. These structural names are incorporated into the systematic names of sugars. Pyranose and

l CHPH

I

2C = O HO

3

H

H

4

OH

H

5

OH

6CH20H

6 OH HOCH2 0 �2 / HO C'l' 5 H

H4 CH20H 3 OH H cyclic form

D-fructose • Figure 23-7

Fructose forms a five-membered cyclic hemiacetal. Five-membered rings are usually represented as flat Haworth structures.

1107

1 1 08

� 3

Chapter 23: Carbohydrates and Nucleic Acids

4 0 5 31 6 1 2

HO 2 OH S H H I HO 0 6 CH2OH

pyran

a pyranose

o

H OH

4

�

6 HOCHz 0 _ OH 5 H HO �HzOH H OH H

H

4 3

o o

D-fructofuranose

a furanose

furan

D-glucopyranose PROBLEM 23-1 1

Talose is the C4 epimer of mannose. Draw the chair conformation of o-talopyranose. PROBLEM 23-1 2 (a) Figure 23-3 shows that the degradation of o-glucose gives o-arabinose, an aldopentose.

Arabinose is most stable in its furanose form. Draw D-arabinofuranose. (b) Ribose, the C2 epirner of arabinose, is most stable in its furanose form. Draw o-ribofuranose.

PROBLEM 23-1 3

The carbonyl group in o-galactose may be isomerized from C l to C2 by brief treatment with dilute base (by the enediol rearrangement, Section 23-8). The product is the C4 epimer of fructose. Draw the furanose structure of the product.

23-7 Anome rs of Monosaccharides; Mutarotati on

W hen a pyranose or furanose ring closes, the hemiacetal carbon atom is converted from a flat carbonyl group to an asymmetric carbon. Depending on which face of the (protonated) carbonyl group is attacked, the hemiacetal -OR group can be directed either up or down. These two orientations of the hemiacetal -OR group give diastereomeric products called anomers. Figure 23-8 shows the anomers of glucose. The hemiacetal carbon atom is called the anomeric carbon, easily identified as the only carbon atom bonded to two oxygens. Its -OR group is called the anomeric hydroxyl group. Notice in Figure 23-8 that the anomer with the anomeric -OR group down (axial) is called the a (alpha) anomer, and the one with the anomeric -OR group up (equatorial) is called the {3 (beta) anomer. We can draw the a and {3 anomers of most aldohexoses by remembering that the {3 form of glucose ({3-o-glucopyranose) has all its substituents in equatorial positions. To draw an a anomer, simply move the anomelic -OR group to the axial position. Another way to remember the anomers is to notice that the a anomer has its anomeric hydroxyl group trans to the terminal - CR20R group, but it is cis in the {3 anomer. This rule works for all sugars, from both the 0 and L series, as well as for furanoses. Figure 23-9 shows the two anomers of fructose, whose anomeric carbon is C2. The a anomer has the anomeric -OR group down, trans to the terminal -CR20R group, while the {3 anomer has it up, cis to the terminal - CR20H.

anomeric carbon H HO

6

HO

HO �O

c Y'

OH

l "-..... H

u-D-glucopyranose ... Figure 23-8

open-chain form

!3-D-glucopyranose

The anomers of glucose. The hydroxyl group on the anomeric (hemiacetal) carbon is down (axial) in the a anomer and up (equatorial) in the f3 anomer. The f3 anomer of glucose has all its substituents in equatorial positions.

23-7 Anomers of Monosaccharides; Mutarotation --, ----- " '�,�..

ciS =

�

HOCH2 0 6

5

,---'-----,

OH 2

H

H trans = a

a -D- fructofuranose

� -D-fructofuranose

... F i g u re 23-9

0' anomer of fructose has the anomeric - OH group down, trans to the terminal - CH20H group. The {3 anomer has the anomeric hydroxyl group up, cis to the terminal - CH20H.

The

PROBLEM 23-1 4

Draw the following monosaccharides, using chair conformations for the pyranoses and Ha­ worth projections for the furanoses. (a) O'-o-mannopyranose (C2 epimer of glucose) (b) {3-o-galactopyranose (C4 epimer of glucose) (c) {3-o-allopyranose (C3 epimer of glucose) (d) O'-o-arabinofuranose (e) {3-o-ribofuranose (C2 epimer of arabinose) Properties of Anomers: M utarotation Because anomers are diastereomers, they generally have different properties. For example, a-D-glucopyranose has a melt­ ing point of 146°C and a specific rotation of + 1 1 2.2°, while ,B-o-glucopyranose has a melting point of I S0°C and a specific rotation of + 1 8.7°. When glucose is crystallized from water at room temperature, pure crystalline a-o-glucopyranose results. If glu­ cose is crystallized from water by letting the water evaporate at a temperature above 98°C, crystals of pure ,B-D-glucopyranose are formed (Figure 23- 1 0). --- -

-,..

H

H

HO

H

HO

HO

H

OH H

a anomer crystallize

t

open-chain form

� anomer

equilibriulll in solution

below 9SoC

crystallize

t

above 9SoC

H

H HO H

�

HO equilibrium mixture of a and [aj +52.6°

�

OH

=

H

H pure a anomer mp 146°C, [aj + 1 1 2.2° =

pure � anomer Illp 1 50°C, [aj = + 1 S.7°

... Fig ure 23-1 0

An aqueous solution of o-glucose contains an equilibrium mixture of O'-o-glucopyranose, {3-o-glucopyranose, and the intermediate open-chain form. Crystallization below 98°C gives the 0' anomer, and crystallization above 98°C gives the {3 anomer.

1109

1110

Chapter 2 3 : Carbohydrates and Nucleic Acids

In each of these cases, all the glucose in the solution crystallizes as the favored anomer. In the solution, the two anomers are in equilibrium through a small amount of the open-chain form, and this equilibrium continues to supply more of the anomer that is crystallizing out of solution. When one of the pure glucose anomers dissolves in water, an interesting change in the specific rotation is observed. When the a anomer dissolves, its specific rotation gradually decreases from an initial value of + 1 1 2.2° to +52.6°. When the pure f3 anomer dissolves, its specific rotation gradually increases from + 1 8.7° to the same value of +52.6°. This change ("mutation") in the specific rotation is called muta­ rotation. Mutarotation occurs because the two anomers interconvert in solution. When either of the pure anomers dissolves in water, its rotation gradually changes to an intermediate rotation that results from equilibrium concentrations of the anomers. The specific rotation of glucose is usually Listed as + 52.6°, the value for the equilibrium mixture of anomers. The positive sign of this rotation is the source of the name dex­ trose, an old common name for glucose. SOLV E D PROBLEM 23-2

Calculate how much of the a anomer and how much of the f3 anomer are present in an equi­ librium mixture with a specific rotation of +S2.6°. SOLU T IO N

If the fraction of glucose present as the a anomer ( [a] + 1 1 2.2° ) is a, the fraction present as the f3 anomer ( [a] + 1 8.7°) is b, and the rotation of the mixture is +S2.6°, we have =

=

There is very little of the open-chain form present, so the fraction present as the a anomer (a) plus the fraction present as the f3 anomer (b) should account for all the glucose: a + b = l Substituting ( 1

-

or

b= l - a

a) for b in the first equation, we have

Solving this equation for a, we have a 0.36, or 36%. Thus, b must be ( 1 - 0.36) or 64%. The amounts of the two anomers present at equilibrium are =

a anomer, 36%

=

0.64,

f3 anomer, 64%

When we remember that the anomelic hydroxyl group is axial in the a anomer and equatorial in the f3 anomer, it is reasonable that tlle more stable f3 anomer should predominate. P ROBLEM 23-1 5

Like glucose, galactose shows mutarotation when it dissolves in water. The specific rotation of a-D-galactopyranose is + I S0.7°, and that of the f3 anomer is +S2.8°. When either of the pure anomers dissolves in water, the specific rotation gradually changes to + 80.2°. Deter­ mine the percentages of the two anomers present at equilibrium.

23-8 Reactions of Monosaccha rides: Side React i ons in Base

Sugars are multifunctional compounds that can undergo reactions typical of any of their functional groups. Most sugars exist as cyclic hemiacetals, yet in solution they are in equilibrium with their open-chain aldehyde or ketone forms. As a result, sugars undergo most of the usual reactions of ketones, aldehydes, and alcohols. Reagents commonly used with monofunctional compounds often give unwanted side reactions with sugars, however. Carbohydrate chemists have developed reactions that work well with sugars while avoiding the undesired side reactions. As we learn about the unique reactions of simple sugars, we will often draw them as their open-chain forms because it is often the small equilibrium amount of the open-chain form that reacts.

23-8 Reactions of Monosaccharides: Side Reactions in B ase

1111

One of the most important aspects of sugar chemistry is the inability, in most cases, to use basic reagents because they cause unwanted side reactions. Two common base-catalyzed side reactions are epimerization and enediol rearrangement. Under basic conditions, the proton alpha to the aldehyde (or ketone) carbonyl group is reversibly removed (shown in the following Mechanism Box 23-2). In the resulting eno­ late ion, C2 is no longer asymmetric, and i ts stereochemistry is lost. R eprotonati on can occur on either face of the enolate, giving either configuration. The result is an equiliblium mixture of the original sugar and its C2 epimer. Because a mixture of epimers results, this stereochemical change is called epimerization. The mechanism involves rapid base­ catalyzed equilibration of glucose to a mixture of glucose and its C2 epimer, mannose. Epimerization and the Enediol Rea rrangement

M EC H A N I S M 23-2 Step

1:

Base-Catalyzed Epimerization of Glucose

Step 2: Reprotonation on the other face .

Abstraction of the a proton. "'- �

H

H

0:

C I -:C-OH

H

HO H

OH

H

OH

-OH

----7 �

H

HO



I

c=o H

HO H

OH

H

OH

CH20H D-fructose

CH20H enolate

CH20H enediol

I

H - C - OH

Under strongly basic conditions, the combination of enediol rearrangements and epimerization leads to a complex mixture of sugars. Except when using protected sugars, most chemists doing sugar chemistry employ neutral or acidic reagents to avoid these annoying side reactions. PROBLEM 23-1 7

Show how C3 of fructose can epimerize under basic conditions. PROBLEM 23- 1 8

Show how another enediol rearrangement can move the carbonyl group from C2 i n fruc­ tose to C3.

Like other aldehydes and ketones, aldoses and ketoses can be reduced to the corre­ sponding polyalcohols, called sugar alcohols or alditols. The most common reagents are sodium borohydride or catalytic hydrogenation using a nickel catalyst. Alditols are named by adding the suffix -itol to the root name of the sugar. The following equation shows the reduction of glucose to glucitol, sometimes called sorbitol.

23-9 Reduct i on of Monosacc h a rides

H

'"

H

H HO OH H �-o-glucopyranose

(--0.

HO

C=O

CHpH

OH H

H

OH

H

OH CH20H

open-chain aldehyde

H ----;.

H2, Ni

HO

OH H

H

OH

H

OH CH20H

o-glucitol (o-sorbitol) an alditol

Reduction of a ketose creates a new asymmetric carbon atom, formed in either of two configurations, resulting in two epimers. Figure 23- 1 1 shows how the reduction of fructose gives a mixture of glucitol and mannitol. Sugar alcohols are widely used in industry, primarily as food additives and sugar substitutes. Glucitol has the common name sorbitol because it was first isolated from

23- 1 0 Oxid ation of M onosaccharides ; Reduci ng S ugars

1 1 13

- ---

�

HOCH2 0 H

CH?OH C=O

CHpH

HO

H

CH20H

I -

HO

H

H NaBH4

)

HO

CH20H

OH

HO

H

H

HO

H

OH

H

OH

H

OH

R

OR

a-o-fructofuranose

H

OH

H

OH

H

OH

OH

H

CH20H open-chain ketone

CH20H

CH20H

o-glucitol + o-mannitol a mixture of alditols

.... Figure 23-1 1

Reduction of fructose creates a new asymmetric carbon atom, which can have either configuration. The products are a mixture of glucitol and mannitol.

the benies of the mountain ash, Sorbus aucuparia. Industrially, sorbitol is made by catalytic hydrogenation of glucose. Sorbitol is used as a sugar substitute, a moistening agent, and a starting material for making vitamin C. Mannitol was first isolated from plant exudates known as mannas (of Biblical fame), the origin of the names mannose and mannitol. Mannitol is derived commercially from seaweed, or it can be made by catalytic hydrogenation of mannose. Galactitol (dulcitol) also can be obtained from many plants, or it can be made by catalytic hydrogenation of galactose. PROBLEM 23 - 1 9

When o-glucose is treated with sodium borohydride, optically active glucitol results. When optically active o-galactose is reduced, however, the product is optically inactive. Explain this loss of optical activity. PROBLEM 23-20

Emil Fischer synthesized L-gulose, an unusual aldohexose that reduces to give o-glucitol. Suggest a structure for this L sugar, and show how L-gulose gives the same alditol as o-glu­ cose. (Hint: o-Glucitol has - CH 2 0H groups at both ends. Either of these primary alcohol groups might have come from reduction of an aldehyde.)

Monosaccharides are oxidized by a variety of reagents. The aldehyde group of an aldose oxidizes easily. Some reagents also selectively oxidize the terminal - CH20H group at the far end of the molecule. Oxidation is used to identify the functional groups of a sugar, to help to determine its stereochemistry, and as part of a synthesis to convert one sugar into another. Bromine water oxidizes the aldehyde group of an aldose to a carboxylic acid. Bromine water is used for this oxidation because it does not oxidize the alcohol groups of the sugar and it does not oxidize ketoses. Also, bromine water is acidic and does not cause epimerization or rearrangement of the carbonyl group. Because bromine water oxidizes aldoses but not ketoses, it serves as a useful test to distinguish aldoses from ketoses. The product of bromine water oxidation is an aldonic acid (older term: glyconic acid). For example, bromine water oxidizes glucose to gluconic acid.

Brom i n e Water

23- 1 0 Oxidat i on of Monosacc h a rides; Reducing Sug ars

C

1 1 14

h apter 23:

C

arboh y d rates and Nucleic Acids Example

CHO H OH HO H H OH H OH CHzOH

acid

aldehyde

I O �(H I (9HOH)n CH20H

Brz z

------0> HO

I O � OH I ( (9HOH)n CHzOH aldonic acid (glyconic acid)

aldose

Brz z

------0> HO

COOH H OH HO H H OH H OH CHzOH gluconic acid

glucose

PROBLEM 23-21

Draw and name the products of bromine water oxidation of (c) D-fructose (b) D-galactose (a) D-mannose Nitric Acid Nitric acid is a stronger oxidizing agent than bromine water, oxidizing both the aldehyde group and the telminal group of an aldose to carboxylic acid groups. The resulting dicarboxylic acid is called an aldaric acid (older terms: glycaric acid or saccharic acid). For example, nitric acid oxidizes glucose to glucaric acid.

-CH20H

Example

aldehyde

alcohol

ITHO I

(CHOH)n ItH20H I

HN03 -------7

CHO H OH HO H H OH H OH CH20H

aCid lTooHI (CHOH)n ItooHI acid

aldaric acid (glycaric acid)

aldose

HN03 -------7

glucaric acid

glucose

Diabetics must monitor their blood gl ucose levels several times a day. A common diagnostic kit uses the enzymatic oxidation of g lucose to determ ine its concentration. A paper strip is treated with gl ucose oxidase. which specifically cat­ alyzes oxidation of the aldehyde group by oxygen in the a i r. The prod ucts are gl uconic acid and hydrogen peroxide. Peroxide oxi­ dizes a dye on the same strip of paper to produce a color change.

R

C-H -6 II

aldehyde

+

COOH H OH HO H H OH H OH COOH

PROBLEM 23-2 2

Draw and name the products of nitric acid oxidation of (a) D-mannose (b) D-galactose PROBLEM 23-23

Two sugars, A and B, are known to be glucose and galactose, but it is not certain which one is which. On treatment with nitric acid, A gives an optically inactive aldaric acid, while B gives an optically active aldaric acid. Which sugar is glucose, and which is galactose? Tollens test detects aldehydes, which react with Tollens reagent to give carboxylate ions and metallic silver, often in the form of a silver mirror on the inside of the container.

Tollens Test

( H -OH

+ 2 Ag N 3h

Tollens reagent

+

-OH

+ 2 A gJ oxidized acid anion

+

4 N H3

+

reduced silver mirror

In its open-chain form, an aldose has an aldehyde group, which reacts with Tollens reagent to give an aldonic acid and a silver mirror. This oxidation is not a good syn­ thesis of the aldonic acid, however, because Tollens reagent is strongly basic and

23- 1 1 Nonreducing Sugars: Formation of Glycosides

1 1 15

promotes epimerization and enediol rearrangements. Sugars that reduce Tollens reagent to give a silver mirror are called reducing sugars. aldehyde

H

H HO OH H

OH

H

H

A g( NH3)� OH

H

OH

(Tollens reagent)

H

OH

HO

�

acid

{3-o-glucose

-

)

OH

HO

CH20H

H

H

OH

H

OH

Agt

+

CHzOH

open-chain form

gluconic acid (+ side products)

Tollens test cannot distinguish between aldoses and ketoses because the basic Tollens reagent promotes enediol rearrangements. Under basic conditions, the open-chain form of a ketose can isomerize to an aldose, which reacts to give a pos­ itive Tollens test.

I

C=O

I

H ....... �O C

H ....... / OH C

CH20H

(

-OH

II

)

(

C - OH

I

R a ketose

R enediol intermediate

-OH

)

I

O � /0 C

H - C - OH

Ag(NH3 )! -OH

I

I

R an aldose

HO H

CH20H acetal 0,.----,1--, H

+ Agt

R positive Tollens test

What good is Tollens test if it doesn' t distinguish between aldoses and ketoses? The an­ swer lies in the fact that Tollens reagent must react with the open-chain form of the sugar, which has a free aldehyde or ketone. If the cyclic form cannot open to the free carbonyl compound, the sugar does not react with Tollens reagent. Hemiacetals are eas­ ily opened, but an acetal is stable under neutral or basic conditions (Section 1 8- 1 8). If the carbonyl group is in the form of a cyclic acetal, the cyclic form cannot open to the free carbonyl compound, and the sugar gives a negative Tollens test (Figure 23- 1 2).

H

I

H -C - OH

NH:

23- 1 1 N onreducing Sug a rs: Formation of G l ycosi des

- - -----

no reaction

OH H a glycoside Examples of nonreducing sugars

HO

H

acetal

1

.... Figu re 23-1 2

H methyl �-D-glucopyranoside (or methyl �-D-glucoside)

� acetal ethyl cx.-D-fructofuranoside (or ethyl cx.-D-fructoside)

Glycosides. Sugars that are full acetals are stable to Tollens reagent and are nonreducing sugars. Such sugars are called glycosides.

1 1 16

Chapter 2 3 : Carbohydrates and Nucleic Acids Sugars in the form of acetals are called glycosides, and their names end in the -oside suffix. For example, a glycoside of glucose would be a glucoside, and if it were a six-membered ring, it would be a glucopyranoside. Similarly, a glycoside of ribose would be a riboside, and if it were a five-membered ring, it would be a ribofuranoside. In general, a sugar whose name ends with the suffix -ose is a reducing sugar, and one whose name ends with -oside is nonreducing. Because they exist as stable acetals rather than hemiacetals, glycosides cannot spontaneously open to their open-chain forms, and they do not mutarotate. They are locked in a particular anomeric form. We can summarize by saying that Tollens test distinguishes between reducing sug­ ars and nonreducing sugars: Reducing sugars (aldoses and ketoses) are hemiacetals, and they mutarotate. Nonreducing sugars (glycosides) are acetals, and they do not mutarotate. PROBLEM 23-24

Which of the following are reducing sugars? Comment on the common name table sugar. (b) f3-L-idopyranose (an aldohexose) (a) methyl a-o-galactopyranoside (c) a-D-allopyranose (d) ethyl f3-D-ribofuranoside (e)

(0

H

sucrose

for

H HO H

k: Y By \ H

HOC

°

O

T

OH

OH H sucrose

CH20H

PROBLEM 23-25

Draw the structures of the compounds named in Problem 23-24 parts (a), (c), and (d). Allose is the C3 epimer of glucose, and ribose is the C2 epimer of arabinose. Formation of G lycosides Recall that aldehydes and ketones are converted to acetals by treatment with an alcohol and a trace of acid catalyst (Section 1 8- 1 8). These condi­ tions also convert aldoses and ketoses to the acetals we call glycosides. Regardless of the anomer used as the starting material, both anomers of the glycoside are formed (as an equilibrium mixture) under these acidic conditions. The more stable anomer predomi­ nates. For example, the acid-catalyzed reaction of glucose with methanol gives a mixture of methyl glucosides.

H

H

HO H

H +

Hp, W

f3 glycosidic bond

HO

H O'-o-glucopyranose (either 0' or f3)

0'

glycosidic bond aglycone methyl O'-D-glucopyranoside

aglycone methyl f3-o-glucopyranoside

Like other acetals, glycosides are stable to basic conditions, but they hydrolyze in aqueous acid to a free sugar and an alcohol. Glycosides are stable with basic reagents and in basic solutions.

23- 1 2 Ether and Ester Formation

1 117

aglycone

H

H

HO

HO

aglycone ethyl <X -D-glucopyranoside

cytidine, a nucleoside (Section 23-2 1 )

salicin, from willow bark aglycone

H HO

aglycone

RO

C N I O - CH - Ph

H H

amygdalin a component of laetrile, a controversial cancer drug

a glycoprotein N-glycoside (showing the linkage from carbohydrate to protein)

... Figure 23-1 3

Aglycones. The group bonded to the anomeric carbon of a glycoside is called an aglycone. Some aglycones are bonded through an oxygen atom (a true acetal), and others are bonded through other atoms such as nitrogen.

An aglycone is the group bonded to the anomeric carbon atom of a glycoside. For example, methanol is the aglycone in a methyl glycoside. Many aglycones are bonded by an oxygen atom, but others are bonded through a nitrogen atom or some other heteroatom. Figure 23- 1 3 shows the structures of some glycosides with interesting aglycones. Disaccharides and polysaccharides are glycosides in which the alcohol forming the glycoside bond is an OR group of another monosaccharide. We will consider disaccharides and polysaccharides in Sections 23- 1 8 and 23- 19. -

PROBLEM 23- 26

The mechanism of glycoside formation is the same as the second part of the mechanism for acetal formation. Propose a mechanism for the formation of methyl f3 -D-glucopyranoside. PROB LEM 23-27

Show the products that result from hydrolysis of amygdalin in dilute acid. Can you suggest why amygdalin might be toxic to tumor (and possibly other) cells? PROBLEM 23- 28

Treatment of either anomer of fructose with excess ethanol in the presence of a trace of HCI gives a mixture of the a and f3 anomers of ethyl-D-fructofuranoside. Draw the starting materi­ als, reagents, and products for this reaction. Circle the aglycone in each product.

Because they contain several hydroxyl groups, sugars are very soluble in water and rather insoluble in organic solvents. Sugars are difficult to recrystallize from water because they often form supersaturated syrups like honey and molasses. If the hydroxyl groups are alkylated to form ethers, sugars behave like simpler organic compounds. The ethers are soluble in organic solvents, and they are more easily purified by recrystalliza­ tion and simple chromatographic methods.

Many diabetics have long-standing elevated blood gl ucose levels. In the open-chain form, glucose con­ denses with the amino groups of proteins. Th is glycosylation of pro­ teins may cause some of the chron­ ic effects of diabetes.

23- 1 2 Eth e r and Ester Format i on

1118

Chapter 23: Carbohydrates and Nucleic Acids

Example

ether

polarized CH} I

sugar hydroxyl group

H

excess

HO H H

methyI 2,3,4,6-tetra-O-methyl-cx-D-glucopyranoside

cx-D-glucopyranose .A.

Figure 23-14

Formation of methyl ethers. Treatment of an aldose or a ketose with methyl iodide and silver oxide gives the totally methylated ether. If the conditions are carefully controlled, the stereochemistry at the anomeric carbon is usually preserved.

Treating a sugar with methyl iodide and silver oxide converts its hydroxyl groups to methyl ethers. Silver oxide polarizes the H 3 C - I bond, making the methyl carbon strongly electrophilic. Attack by the carbohydrate - OH group, followed by deprotonation, gives the ether. Figure 23- 1 4 shows that the anomeric hydroxyl group is also converted to an ether. If the conditions are carefully controlled, the hemiacetal C - 0 bond is not broken, and the configuration at the anomeric carbon is preserved. The Williamson ether synthesis is the most common method for forming simple ethers, but it involves a strongly basic alkoxide ion. Under these basic conditions, a sim­ ple sugar would isomerize and decompose. A modified Williamson method may be used if the sugar is first converted to a glycoside (by treatment with an alcohol and an acid cat­ alyst). The glycoside is an acetal, stable to base. Treatment of a glycoside with sodium hydroxide and methyl iodide or dimethyl sulfate gives the methylated carbohydrate. H

H NaOH o

H

II

CH3- O -S-O-CH3

II

o

methyl a-D-glucopyranoside (stable to base) Oxidation of glucose at (-6 pro­ duces glucuronic acid. This sugar is attached to many drugs to make a glucuronide derivative, a water­ soluble metabolite that is readily excreted.

(dimethyl sulfate)

H methyl 2,3,4,6-tetra-O-methyl-a-D-glucopyranoside

PROB LEM 2 3 - 2 9

Propose a mechanism for methylation of any one of the hydroxyl groups of methyl a-D-glucopy­ ranoside, using NaOH and dimethyl sulfate. PROBLEM 23-30

Draw the expected product of the reaction of the following sugars with methyl iodide and silver oxide. Ca) a-D-fructofuranose (b) ,B -D-galactopyranose

H HO OH glucuronic acid

Ester Formation Another way to convert sugars to easily handled derivatives is to acyl ate the hydroxyl groups to form esters. Sugar esters are readily crystallized and purified, and they dissolve in common organic solvents. Treatment with acetic anhy­ dride and pyridine (as a mild basic catalyst) converts sugar hydroxyl groups to acetate

23- 1 3 Reactions with Phenylhydrazi ne : Osazone Formation :0:

: 0:

I I .)

II

: ?j

CH 3 - C - O - C - CH 3

�

R

sugar

HO

'-.;

�� ��C o -

I

:0+

./

"-

.....::; 0 CH 3 _ c oY

.

I

R -O:

H

H

o +

II

HO - C - CH 3

acetate ester o

H20H

OH

�.

CH 3 - C - O - C - CH 3

R- O-H

Example

..

1 1 19

excess

o

II

II

CH 3 - C - 0 - CH2

0 - C - C H3 o

II

pyridine

H

0

� -o-fructofuranose

"

H C - CH 3

CH20 - C - CH 3

II

o

penta-O-acety I-� -0- fructofuranoside ... Figure 23-1 5

Formation of acetate esters. Acetic anhydride and pyridine convert all the hydroxyl groups of a sugar to acetate esters. The stereochemistry at the anomeric carbon is usually preserved.

C

esters, as shown in Figure 23- 1 5 . This reaction acetylates all the hydroxyl groups, including that of the hemiacetal on the anomeric carbon. The anomeric ° bond is not broken in the acylation, and the stereochemistry of the anomeric carbon atom is usually preserved. If we start with a pure a anomer or a pure {3 anomer, the product is the corresponding anomer of the acetate. -

P R OBLEM 2 3-31

Predict the products formed when the following sugars react with acetic anhydride and pyridine. (b) !3-D-ribofuranose (a) a-D-glucopyranose

Before spectroscopy, one of the best ways to identify ketones and aldehydes was conversi on to crystalline hydrazones, especially phenylhydrazones and 2,4dinitrophenylhydrazones (Section 1 8- 1 7). In his exploratory work on sugar struc­ tures, Emil Fischer often made and used phenylhydrazone derivatives. In fact, his constant use of phenylhydrazine ultimately led to Fischer's death in 1 9 1 9 from chronic phenylhydrazine poisoning. R' R ketone or aldehyde

�C=N-NH-

H

H free hemiacetal

pentamethyl derivative

OCH3 H

CHP H

4

OCH3

H

5

OH

open-chain form 2,3,4,6,-O-tetramethyl-D-glucose PROBLEM 23-40 (a) Show the product that results when fructose is treated with an excess of methyl iodide

and silver oxide. Show what happens when the product of part (a) is hydrolyzed using dilute acid. (c) Show what the results of parts (a) and (b) imply about the hemiacetal structure of fructose.

(b)

Periodic Acid Cleavage of Carbohyd rates Another method used to determine the size of carbohydrate rings is cleavage by periodic acid. Recall that periodic acid cleaves vicinal diols to give two carbonyl compounds, either ketones or aldehydes, depending on the substitution of the reactant (Section I I - l I B ) .

R

I

H

I

R - C -C - R '

I

I

OH OH

R +

HI04

periodic acid

----7

R

"/

/

H

C=O

+

O=C

"-

R'

+

HI03

+

Hp

ketones and aldehydes

vicinal diol

Because ether and acetal groups are not affected, periodic acid cleavage of a glycoside can help to determine the size of the ring. For example, periodic acid oxidation of methyI f3-D-glucopyranoside gives the following products. The structure of the fragment con­ taining C4, C5, and C6 implies that the original glycoside was a six-membered ring bonded through the C5 oxygen atom. H 6 CH OH 2

4

4

H

HO H

methyl j3-D-glucopyranoside

H

t

C HO

OH

6 CH 0H 2

D-glyceraldehyde

I CHO

+ CH30H 21 CHO o

3 11

H-C-OH

On the other hand, if glucose were a furanose (five-membered ring), the periodic acid cleavage would give an entirely different set of products. Because glucose actually exists as a pyranose (six-membered ring), these products are not observed.

23- 1 8 Disaccharides

�

1129

6

HO-CH2

+

OCH3

HO -F 4

H

O

0H 3

H

'

' CHO

I 2 CHO

HI04

------'.>

6

2 H

H

+ CH3 0H

H2C = O

OH

methyl ,B-D-glucofuranoside

(not observed)

PROBLEM 23-41

(a) Draw the reaction of methyl j3 -D-fructofuranoside with periodic acid, and predict

the products.

(b) Draw the structure of a hypothetical methyl j3 -D-fructopyranoside, and predict the

products from periodic acid oxidation. (c) The reaction of methyl j3-D-glucopyranoside with periodic acid (shown above) gives

only the 0-( + ) enantiomer of glyceraldehyde (among other products). If you oxidized an aldohexose glycoside with periodic acid and one of the products was the L- ( - ) enantiomer of glyceraldehyde, what would that tell you about the sugar?

As we have seen, the anomeric carbon of a sugar can react with the hydroxyl group of an alcohol to give an acetal called a glycoside. If the hydroxyl group is part of another sugar molecule, then the glycoside product is a disaccharide, a sugar composed of two monosaccharide units (Figure 23- 1 6).

PROBLEM-SOLVING

bon atoms that bear hydroxyl groups.

23- 1 8 Disaccha ri des

H

H HO

R - OH

OH H

sugar 1

a glycoside

glycosidic bond

OH

sugar 2

OH

a disaccharide .... Figure 23-16

Hi-ltv

Cleavage occurs only between two car­

Disaccharides. A sugar reacts with an alcohol to give an acetal called a glycoside. When the alcohol is part of another sugar, the product is a disaccharide.

1 1 30

Chapter 23: Carbohydrates and Nucleic Acids

In principle, the anomeric carbon can react with any of the hydroxyl groups of another sugar to form a disaccharide. In naturally occUlTing disaccharides, however, there are three common glycosidic bonding arrangements. 1. A 1 ,4' link. The anomeric carbon is bonded to the oxygen atom on C4 of the second

sugar. The prime symbol ( ' ) in 1 ,4' indicates that C4 is on the second sugar.

2. A

] ,6' link. The anomeric carbon is bonded to the oxygen atom on C6 of the second sugar. 3. A I , l ' link. The anomeric carbon of the first sugar is bonded through an oxygen atom to the anomeric carbon of the second sugar.

We will consider some naturally occurring disaccharides with these common glycosidic linkages. 23-1 8A

The 1 ,4 ' L i n ka g e : C e l l o b i ose, M a ltose, a n d Lactose

The most common glycosidic linkage is the 1 ,4' link. The anomeric carbon of one sugar is bonded to the oxygen atom on C4 of the second ring. Cellobiose: A 13 - 1 ,4 ' G l ucosidic Lin kage Cellobiose, the disaccharide obtained by partial hydrolysis of cellulose, contains a 1 ,4' linkage. In cellobiose, the anomeric carbon of one glucose unit is linked through an equatorial ( [3 ) carbon-oxygen bond to C4 of another glucose unit. This [3 - 1 ,4' linkage from a glucose acetal is called a [3- 1 ,4' glucosidic linkage.

Cellobiose, 4-0-([3-D-glucopyranosyl)-[3-D-glucopyranose or 4-0-([3- D-glucopyranosyl)-D-glucopyranose f3-glucosidic linkage

f3 -glucosidic linkage

H

HO

HO OH

H

H

H

H

H

Two alternative ways of drawing and naming cellobiose

The complete name for cellobiose, 4-0-([3-o-glucopyranosyJ)-[3-o-glucopyra­ nose, gives its structure. This name says that a [3-o-glucopyranose ring (the right-hand ring) is substituted in its 4-position by an oxygen attached to a ([3-o-glucopyranosyl) ring, drawn on the left. The name in parentheses says the substituent is a [3-glucose, and the -syl ending indicates that this ring is a glycoside. The left ring with the -syl ending is an acetal and cannot mutarotate, while the right ring with the -ose ending is a hemiacetal and can mutarotate. Because cellobiose has a glucose unit in the hemiacetal form (and therefore is in equilibrium with its open-chain aldehyde form), it is a reducing sugar. Once again, the -ose ending indicates a mutarotating, reducing sugar. Mutarotating sugars are often shown with a wavy line to the free anomeric hydrox­ yl group, signifying that they can exist as an equilibrium mixture of the two anomers. Their names are often given without specifying the stereochemistry of this mutarotating hydroxyl group, as in 4-0-([3-o-glucopyranosyl)-D-glucopyranose. Maltose is a disaccharide formed when starch is treated with sprouted barley, called malt. This malting process is the first step in brewing beer, converting polysaccharides to disaccharides and monosac­ charides that ferment more easily. Like cellobiose, maltose contains a 1 ,4' glycosidic linkage between two glucose units. The difference in maltose is that the stereochem­ istry of the glucosidic linkage is Q' rather than [3 .

M a ltose: An a- 1 ,4 '-G l u cosidic Li n kage

23- 1 8 D isaccharides

Maltose, 4-0-(a-D-glucopyranosyl)-D -glucopyranose

H HO

Q'

- 1 ,4' glucosidic linkage

I'

H

OH

H

Like cellobiose, maltose has a free hemiacetal ring (on the right). This hemiacetal is in equilibrium with its open-chain fonn, and it mutarotates and can exist in either the a or [3 anomeric form. Because maltose exists in equilibrium with an open-chain alde­ hyde, it reduces Tollens reagent, and maltose is a reducing sugar. PROBLE M 23-42

Draw the structure of the individual mutarotating Q' and f3 anomers of maltose. PROB L E M 2 3-43

Give an equation to show the reduction of Tollens reagent by maltose. Lactose: A 13 -1,4 ' Galactosidic Li n kage Lactose is similar to cellobiose, except that the glycoside (the left ring) in lactose is galactose rather than glucose. Lactose is composed of one galactose unit and one glucose unit. The two rings are linked by a f3-glycosidic bond of the galactose acetal to the 4-position on the glucose ring: a [3 - 1 ,4' galactosidic linkage.

Lactose, 4-0-([3-D-galactopyranosy/)-D -glucopyranose

H {3 -galactosidic linkage Lactose occurs naturally in the milk of mammals, including cows and humans. Hydrolysis of lactose requires a j3-galactosidase enzyme (sometimes called lactase). Some humans synthesize a j3-galactosidase, but others do not. This enzyme is pres­ ent in the digestive fluids of normal infants to hydrolyze their mother's milk. Once the child stops drinking milk, production of the enzyme gradually stops. In most parts of the world, people do not use milk products after early childhood, and the adult population can no longer digest lactose. Consumption of milk or milk products can cause digestive discomfort in lactose-intolerant people who lack the [3-galactosidase enzyme. Lactose-intolerant infants must drink soybean milk or an­ other l actose-free formula. PROBLE M 23-44

Does lactose mutarotate? Is it a reducing sugar? Explain. Draw the two anomeric forms of lactose.

1131

1 1 32

Chapter 23: Carbohydrates and Nucleic Acids 23-1 8 8

The 1 ,6 ' Linkage: Gentiobiose

In addition to the common 1 ,4' glycosidic linkage, the 1 ,6' linkage is also found in naturally occurring carbohydrates. In a 1 ,6' linkage, the anomeric carbon of one sugar is linked to the oxygen of the terminal carbon (C6) of another. This linkage gives a dif­ ferent sort of stereochemical arrangement, because the hydroxyl group on C6 is one carbon atom removed from the ring. Gentiobiose is a sugar with two glucose units j oi ne d by a 13- 1 ,6' glucosidic linkage. Gentiobiose, 6-0-(f3-0-glucopyranosyl)-0 -glucopyranose i3 -glucosidic l inkage

H

H Although the 1 ,6' linkage is rare in disaccharides, it is commonly found as a branch point in polysaccharides. For example, branching in amylopectin (insoluble starch) occurs at 1 ,6' linkages, as discussed in Section 23- 1 9B . PROBLEM 23-45 Is gentiobiose a reducing sugar? Does it mutarotate? Explain your reasoning.

23-1 8C

Linkage of Two Anomeric Ca rbons: Sucrose

Some sugars are joined by a direct glycosidic linkage between their anomeric carbon atoms: a I ) ' linkage. Sucrose (common table sugar), for example, is composed of one glucose unit and one fructose unit bonded by an oxygen atom linking their anomeric carbon atoms. (Because fructose is a ketose and its anomeric carbon is C2, this is actually a 1 ,2' linkage.) Notice that the linkage is in the a position with respect to the glucose ring and in the f3 position with respect to the fructose ring. Sucrose, a-o-glucopyranosyl-f3-ofructofuranoside (or f3-ofructofuranosyl-a-o-glucopyranoside)

HO a-glycosidic l inkage on glucose

o i3-glycosidic linkage on fructose Sucrose (a nonreducing sugar) is not as easily oxidized as a reducing sugar, so it is much more useful for preserving foods such as jams and jellies. A reducing sugar like glucose would oxidize and spoil.

OH

H

Both monosaccharide units in sucrose are present as acetals, or glycosides. Neither ring is in equilibrium with its open-chain aldehyde or ketone form, so sucrose

23- 1 8 Disaccharides

does not reduce Tollens reagent and it cannot mutarotate. Because both units are gly­ cosides, the systematic name for sucrose can list either of the two glycosides as being a substituent on the other. Both systematic names end in the -aside suffix, indicating a nonmutarotating, nonreducing sugar. Like many other common names, sucrose ends in the -ose ending even though it is a nonreducing sugar. Common names are not reli­ able indicators of the properties of sugars. Sucrose is hydrolyzed by enzymes called invertases, found in honeybees and yeasts, that specifically hydrolyze the [3-D-fructofuranoside linkage. The resulting mixture of glucose and fructose is called invert sugar because hydrolysis converts the positive rotation [+66.5°] of sucrose to a negative rotation that is the average of glu­ cose [ + 52. 7°] and fructose [ -92.4 0]. The most common form of invert sugar is honey, a supersaturated mixture of glucose and fructose hydrolyzed from sucrose by the invertase enzyme of honeybees. Glucose and fructose were once called dextrose and levulose, respectively, according to their opposite signs of rotation. SOLV E D PROBLEM 23-3

An unknown carbohydrate of formula C ' 2H220 1 1 reacts with Tollens reagent to form a silver mirror. An a-glycosidase has no effect on the carbohydrate, but a ,B-galactosidase hydrolyzes it to D-galactose and D-mannose. When the carbohydrate is methylated (using methyl iodide and silver oxide) and then hydrolyzed with dilute HCl, the products are 2,3,4,6-tetra-O-methylgalactose and 2,3,4-tri-O-methylmannose. Propose a structure for this unknown carbohydrate. SOL U T ION

The formula shows this is a disaccharide composed of two hexoses. Hydrolysis gives o-galactose and D-mannose, identifying the two hexoses. Hydrolysis requires a ,B-galactosidase, showing that galactose and mannose are linked by a ,B-galactosyl l inkage. Since the original carbohydrate is a reducing sugar, one of the hexoses must be in a free hemiacetal form. Galactose is present as a glycoside; thus mannose must be present in its hemiacetal form. The unknown carbohydrate must be a ( ,B-galactosyl)-mannose. The methylation/hydrolysis procedure shows the point of attachment of the glycosidic bond to mannose and also confirms the size of the six-membered rings. In galactose, all the hydroxyl groups are methylated except C l and C5. C l is the anomeric carbon, and the C5 oxy­ gen is used to form the hemiacetal of the pyranose ring. In mannose, all the hydroxyl groups are methylated except C l , C5, and C6. The C5 oxygen is used to form the pyranose ring (the C6 oxygen would form a less stable seven-membered ring); therefore, the oxygen on C6 must be involved in the glycosidic linkage. The structure and systematic name are shown here.

H 6-0-(,B-D-galactopyranosyl)-D-mannopyranose PROBLEM 23-46

Trehalose is a nonreducing disaccharide ( C 1 2H220 1 1 ) isolated from the poisonous mush­ room Amanita muscaria. Treatment with an a-glucosidase converts trehalose to two mole­ cules of glucose, but no reaction occurs when trehalose is treated with a ,B-glucosidase. When trehalose is methylated by dimethyl sulfate in mild base and then hydrolyzed, the only product is 2,3,4,6-tetra-O-methylglucose. Propose a complete structure and systematic name for trehalose.

1 1 33

1134

Chapter 23: Carbo hy drate s and Nuc le i c Ac ids PROBLEM 23-47

Raffinose is a trisaccharide ( C l sH32016) isolated from cottonseed meal . Raffinose does not reduce Tollens reagent, and it does not mutarotate. Complete hydrolysis of raffinose gives o-glucose, o-fructose, and o-galactose. When raffinose is treated with invertase, the products are o-fructose and a reducing disaccharide called melibiose. Raffinose is unaffected by treat­ ment with a f3-galactosidase, but an a-galactosidase hydrolyzes it to o-galactose and sucrose. When raffinose is treated with dimethyl sulfate and base followed by hydrolysis, the products are 2,3,4-tri-O-methylglucose, 1 ,3,4,6-tetra-O-methylfructose, and 2,3,4,6-tetra-O-methyl­ galactose. Determine the complete structures of raffinose and melibiose, and give a system­ atic name for melibiose.

23- 1 9 Po lysaccharides

are carbohydrates that contain many monosaccharide units joined by glycosidic bonds. They are one class of biopolymers, or naturally occurring polymers. Smaller polysaccharides, containing about three to ten monosaccharide units, are sometimes called oligosaccharides. Most polysaccharides have hundreds or thou­ sands of simple sugar units linked together into long polymer chains. Except for units at the ends of chains, all the anomeric carbon atoms of polysaccharides are invol ved in acetal glycosidic links. Therefore, polysaccharides give no noticeable reaction with Tollens reagent, and they do not mutarotate. Polysaccharides

2 3 -19A

Cel l u l ose

a polymer of o-glucose, is the most abundant organic material. Cellulose is synthesized by plants as a structural material to support the weight of the plant. Long cellulose molecules, called microfibrils, are held in bundles by hydrogen bonding between the many - OH groups of the glucose rings. About 50% of dry wood and about 90% of cotton fiber is cellulose. Cellulose is composed of o-glucose units linked by ,8- 1 ,4' glycosidic bonds. This bonding arrangement (like that in cellobiose) is rather rigid and very stable, giving cellulose desirable properties for a structural material. Figure 23- 1 7 shows a partial structure of cellulose. Humans and other mammals lack the ,8-glucosidase enzyme needed to hydrolyze cellulose, so they cannot use it directly for food. Several groups of bacteria and protozoa can hydrolyze cellulose, however. Termites and ruminants maintain colonies of these bacteria in their digestive tracts. When a cow eats hay, these bacteria convert about 20% to 30% of the cellulose to digestible carbohydrates. Rayon is a fiber made from cellulose that has been converted to a soluble deriv­ ative, and then regenerated. In the common viscose process, wood pulp is treated with carbon disulfide and sodium hydroxide to convert the free hydroxyl groups to xan­ thates, which are soluble in water. The viscous solution (called viscose) is forced through a spinneret into an aqueous sodium bisulfate solution, where a fiber of insolu­ ble cellulose is regenerated. Alternatively, the viscose solution can be extruded in sheets to give cellophane film. Rayon and cotton are both cellulose, yet rayon thread

Cellulose,

The acoustic properties of cellulose have never been surpassed by other substances. Here, a luthier carves maple for use in a violin.

H

H

� F i g u re 23-17

Partial structure of cellulose. Cellulose is a 13- 1 ,4 ' polymer of o-glucose, systematicall y named poly( 1 ,4' -O-f3-o-glucopyranoside).

H o

f3-glucosidic l inkage - -=---=--- .:-: ::--====== =======

/"\

23- 1 9 Polysaccharides

1 1 35

can be much stronger because it consists of long, continuously extruded fibers, rather than short cotton fibers spun together.

ROH

+

NaOH

+ extruded into solution

+

Hp

xanthate derivatives (viscose)

cellulose

NaHS04

ROH

+

CS 2

rayon (regenerated cellulose)

PROBLEM 23-48

Cellulose is converted to cellulose acetate by treatment with acetic anhydride and pyridine. Cellulose acetate is soluble in common organic solvents, and it is easily dissolved and spun into fibers. Show the structure of cellulose acetate.

23 - 1 98

Starches: Amylose, Amylopecti n, a n d G l ycogen

Plants use starch granules for storing energy. When the granules are dried and ground up, different types of starches can be separated by mixing them with hot water. About 20% of the starch is water-soluble amylose, and the remaining 80% is water-insoluble amylopectin. When starch is treated with dilute acid or appropriate enzymes, it is pro­ gressively hydrolyzed to maltose and then to glucose. Like cellulose, amylose is a linear polymer of glucose with 1 ,4' glycosidic linkages. The difference is in the stereochemistry of the linkage. Amylose has a- I ,4 ' links, while cellulose has ,8 - 1 ,4' links. A partial structure of amylose is shown in Figme 23- 1 8. The subtle stereochemical difference between cellulose and amylose results in some striking physical and chemical differences. The a linkage in amylose kinks the polymer chain into a helical structure. This kinking increases hydrogen bonding with water and lends additional solubility. As a result, amylose is soluble in water, and cellu­ lose is not. Cellulose is stiff and sturdy, but amylose is not. Unlike cellulose, amylose is an excellent food source. The a- l ,4' glucosidic linkage is easily hydrolyzed by an a-glucosidase enzyme, present in all animals. The helical structure of amylose also serves as the basis for an interesting and use­ ful reaction. The inside of the helix is just the right size and polarity to accept an iodine Amylose

� Fig u re 23-18

Partial structure of amylose. Amylose is an a- l ,4' polymer of glucose, systematically named poly ( 1 ,4' -O-a-D-glucopyranoside). Amylose differs from cellulose only in the stereochemistry of the glycosidic linkage.

1 1 36

Chapter 23: C arb ohydrates and Nucleic Acids

� Figure 23-1 9

The starch-iodine complex of amylose. The amylose helix forms a blue charge-transfer complex with molecular iodine.

(12) molecule. When iodine is lodged within this helix, a deep blue starch-iodine com­ plex results (Figure 23- 19). This is the basis of the starch-iodide test for oxidizers. The material to be tested is added to an aqueous solution of amylose and potassium iodide. If the material is an oxidizer, some of the iodide (1-) is oxidized to iodine (12)' which forms the blue complex with amylose. Amylopectin, the insoluble fraction of starch, is also primarily an a- 1 ,4' polymer of glucose. The difference between amylose and amylopectin lies in the branched nature of amylopectin, with a branch point about every 20 to 30 glucose units. Another chain starts at each branch point, connected to the main chain by an a- 1 ,6' glycosidic linkage. A partial structure of amylopectin, including one branch point, is shown in Figure 23-20.

Amylopectin

low-carbohydrate diets restrict the intake of carbohydrates, sometimes resulting in rapid weight loss. The weight is lost because glycogen and fatty acids are burned to maintain blood gl ucose levels.

G l ycogen Glycogen is the carbohydrate that animals use to store glucose for read­ ily available energy. A large amount of glycogen is stored in the muscles themselves, ready for immediate hydrolysis and metabolism. Additional glycogen is stored in the

H H

}

a- l ,6' glUCOSldic Imkage branch point

H H

H

H ... F i g u re 23-20

Partial structure of amylopectin. Amylopectin is a branched a- I ,4 ' polymer of glucose. At the branch points, there is a single a- I ,6' linkage that provides the attachment point for another chain. Glycogen has a similar structure, except that its branching is more extensive.

23-20 Nucleic Acids: Introduction

1 1 37

liver, where it can be hydrolyzed to glucose for secretion into the bloodstream, provid­ ing an athlete with a "second wind." The structure of glycogen is similar to that of amylopectin, but with more exten­ sive branching. The highly branched structure of glycogen leaves many end groups available for quick hydrolysis to provide glucose needed for metabolism. 23- 1 9C

Ch iti n: A Polymer of N-Acety l g l ucosa m i n e

Chitin (pronounced ki' -t' n, rhymes with Titan) forms the exoskeletons of insects. In crustaceans, chitin forms a matrix that binds calcium carbonate crystals into the exoskeleton. Chitin is different from the other carbohydrates we have studied. It is a polymer of N-acetylglucosamine, an amino sugar (actually an amide) that is common in living organisms. In N-acetylglucosamine, the hydroxyl group on C2 of glucose is replaced by an amino group (forming glucosamine), and that amino group is acetylated.

N-A ce tylg lucosamine,

or 2 -acetamido-2 -deoxy-D-glucose

This cicada is shedding its nymphal exoskeleton. Chitin lends strength and rigidity to the exoskeletons of insects, but i t cannot grow and change with the insect.

Chitin is bonded like cellulose, except using N-acetylglucosamine instead of glu­ cose. Like other amides, N-acetylglucosamine forms exceptionally strong hydrogen bonds between the amide carbonyl groups and N H protons. The glycosidic bonds are .B-1,4' links, giving chi tin structural rigidity, strength, and stabili ty that exceed even that of cellulose. Unfortunately, this strong, rigid polymer cannot easily expand, so it must be shed periodically by molting as the animal grows. -

Chitin synthase inhibitors are used commercially to kill termite colonies. These

compounds

prevent

trapping the termites in an old exoskeleton that cannot grow with them.

Chitin, or poly (1,4' - O-.B-2- acetamido-2-deoXY-D-glucopyranoside), a .B- l,4-linked polymer ofN-acetylglucosamine

o

Nucleic acids are substituted polymers of the aldopentose ribose that carry an organism's genetic information. A tiny amount of DNA in a fertilized egg cell deter­ mines the physical characteristics of the fully developed animal. The difference between a frog and a human being is encoded in a relatively small part of this DNA. Each cell carries a complete set of genetic instructions that determine the type of cell, what its function will be, when it will grow and divide, and how it will synthesize all

the

formation of a new exoskeleton,

23-20 Nucleic Acids: Introduction

1 1 38

Chapter 23: C arbohydrates and Nucleic Acids

HIV (the AIDS virus) is shown here attacking a T-4 lymphocyte. HIV is an RNA virus whose genetic material must be translated to DNA before inserting itself into the host cell's DNA. Several of the anti-AIDS drugs are directed toward stopping this reverse transcription of RNA to DNA. (Magnification lOOOX)

the proteins, enzymes, carbohydrates, and other substances the cell and the organism need to survive. The two major classes of nucleic acids are ribonucleic acids (RNA) and deoxy­ ribonucleic acids (DNA) . In a typical cell, DNA is found primarily in the nucleus, where it carries the permanent genetic code. The molecules of DNA are huge, with molecular weights up to 50 billion. When the cell divides, DNA replicates to form two copies for the daughter cells. DNA i s relatively stable, prov id i n g a medium for trans­ mission of genetic information from one generation to the next. RNA molecules are typically much smaller than DNA, and they are more eas­ ily hydrolyzed and broken down. RNA commonly serves as a working copy of the nuclear DNA being decoded. Nuclear DNA directs the synthesis of messenger RNA, which leaves the nucleus to serve as a template for the construction of protein mole­ cules in the ribosomes. The messenger RNA is then enzymatically cleaved to its component parts, which become available for assembly into new RNA molecules to direct other syntheses. The backbone of a nucleic acid is a polymer of ribofuranoside rings (five-mem­ bered rings of the sugar ribose) linked by phosphate ester groups. Each ribose unit car­ ries a heterocyclic base that provides part of the information needed to specify a particular amino acid in protein synthesis. Figure 23-2 1 shows the ribose-phosphate backbone of RNA. DNA and RNA each contain four monomers, called nucleotides, that differ in the structure of the bases bonded to the ribose units. Yet this deceptively simple structure encodes complex information just as the 0 and 1 bits used by a computer encode com­ plex programs. First we consider the structure of individual nucleotides, then the bond­ ing of these monomers into single-stranded nucleic acids, and finally the base pairing that binds two strands into the double helix of nuclear DNA. ====-====����==�===----------��=�-

15

I

ro 0""'"

',"'110

CI H2 H H

o HO 10 t o-I o

CI H 20,'---- _.:.J H H I H H HO

� Figure 23-21 A short segment of the RNA polymer. Nucleic acids are assembled on a backbone made up of ribofuranoside units linked by phosphate esters.

o 10 to-I o

I CH20,'--- _"'-' -I H HO

oT 10 to-I o

Symbolically,

o 10 t o-I o o 10 t o-I o

23-21 Ribo n ucleosides

Ribonucleosides are components of RNA based on glycosides of the furanose form of o-ribose. We have seen (Section 23-11) that a glycoside may have an aglycone (the substituent on the anomeric carbon) bonded by a nitrogen atom. A ribonucleo­ side is a ,B-o-ribofuranoside (a ,B-glucosidase of o-ribofuranose) whose aglycone is a heterocyclic nitrogen base. The following structures show the open-chain and furanose forms of ribose, and a ribonucleoside with a generic base bonded through a nitrogen atom.

�

CHO H

OH

H

OH

H

OH

base

23-21 Ribonucleosides and Ribonucleotides

Uric acid is one of the principal end products

of

purine

metabolism.

Gout is caused by elevated levels uric acid crystals to precipitate in the joints.

/3-o-ribofuranose

a ribonucleoside

o-ribose

The four bases commonly found in RNA are divided into two classes: The monocyclic compounds cytosine and uracil are called pyrimidine bases because they resemble substituted pyrimidines, and the bicyclic compounds adenine and guanine are cal led purine bases because they resemble the bicyclic heterocycle purine (Section 19-3).

�

C; N

N

pyrimidine

H

Q: 0

0

N

�

NH?

H

�;

0

1

N

H

H

H

cytosine (C) uracil (U) pyrimidine bases

adenine (A)

puri ne bases

I

0

N

N

uric acid

/H

(£;

A NH2

N

/

H

guanine (G)

purine

When bonded to ribose through the circled nitrogen atoms, the four heterocyclic bases make up the four ribonucleosides cytidine, uridine, adenosine, and guanosine (Figure 23-22). Notice that the two ring systems (the base and the sugar) are num­ bered separately, and the carbons of the sugar are given primed numbers. For exam­ ple, the 3' carbon of cytidine is C3 of the ribose ring.

NH?

5 �] ' I 0 5 2 0 JN � HO-C 4' H H H 2' H OH OH

f �

I'

3'

cytidine (C)

H

51 N]/ ., 2 0 6 N� I 0 HO�C 4' H H I' H 2' H OH OH o

�C 3'

uridine (U)

1139

of uric acid in the body, causing

HO-C H2 0 OH H H H H OH OH

CHpH

and R ibonucleo tid es

H 6N/ �t · . ,i;�; N 5' � HO�CH2 9N HO-C 2 NH2 �� H H o

4,

0V

I'

4

��� OH OH

adenosine (A)

... Figure 23-22 The four common ribonucleosides are cytidine, uridine, adenosine, and guanosine.

� -

4'

H

8� I � hi 9 4 �

2' H OH OH 0

3'

I'

g u anos i ne (G)

1 1 40

Chapter 23: Carbohydrates and Nucleic Acids PROBLE M 2 3 -49

Cytosine, uracil, and guanine have tautomeric forms with phenolic hydroxyl groups. Draw these tautomeric forms. PROBLE M 23-50

(a) An aliphatic aminoglycoside is relatively stable to base, but it is quickly hydrolyzed by dilute acid. Propose a mechanism for the acid-catalyzed hydrolysis.

� R

HO-C H 2 0

'"

R

N

H

H

OH

H OH

H

/

+

R 2NH 2 + sugar

an aliphatic riboside

(b) Ribonucleosides are not so easily hydrolyzed, requiring relatively strong acid. Using

your mechanism for part (a), show why cytidine and adenosine (for example) are not so readily hydrolyzed. Explain why this stability is important for living organisms.

Ribonucleic acid consists of ribonucleosides bonded together into a polymer. This polymer cannot be bonded by glycosidic linkages like those of other polysaccharides because the glycosidic bonds are already used to attach the het­ erocyclic bases. Instead, the ribonucleoside units are linked by phosphate esters. The 5'-hydroxyl group of each ribofuranoside is esterified to phosphoric acid. A ribonu­ cleoside that is phosphorylated at its 5' carbon is called a ribonucleotide ("tied" to phosphate). The four common ribonucleotides, shown in Figure 23-23, are simply phosphorylated versions of the four common ribonucleosides. The phosphate groups of these ribonucleotides can exist in any of three ioniza­ tion states, depending on the pH of the solution. At the nearly neutral pH of most organisms (pH 7.4), there is one proton on the phosphate group. By convention, however, these groups are usually written completely ionized.

Ribonucleotides

=

II HO - P - O - ribose I OH

I -0- P - O -ribose I OH

in acid

nearly neutral

o

NH2

o

II

�eA N

5'

-O-P-O-C ?- 0

I

0-

H

H

OH

N

O

H

H OH

cytidine monophosphate, CMP (cytidylic acid)

o

II

t �

5'

N

-O-P-O-C ?- 0

I

0-

H

H

OH

H N/

Ao

H

H OH

lIridine monophosphate, UMP (liridylic acid)

o

0

II

�

5'

N

-O-P-O-C ? 0

1

0-

H

-

H

OH

�N

H

o

I

-0- P- O -ribose

I

0in base (usually written)

�� JL .. H

NJ

H OH

adenosine monophosphate, AMP (adenylic acid)

o

II

�

t

5'

-o-p-o-c 2 o

I

0-

H

H

OH

N

H

-{l JL ..� o

N

/H w NH2

H OH

guanosine monophosphate, OMP (guanidylic acid)

.A. Figure 23-23 Four common ribonllcleotides. These are ribonucleosides esterified by phosphoric acid at their 5'-position, the -CH 2 0H at the end of the ribose chain .

23-23 Deoxyribose and the Structure of Deoxyribonucleic Acid

l r 1 o

0-

o

5, 1 CH2

(3' -h drox l� 4'

Y

Y

l r 1 o

o

H

2' H OH OH

H

3'

� o 5'1 CH2

0-

5,1 C H2 0 'l--I-.!J

H

(5'-phosphater lrl

3'

t

o

I

OH + HzO

o� -o- I o

5,1 CH2 0 '-'

I_=-i

___

Now that we recognize the individual ribonucleotides, we can consider how these units are bonded into the RNA polymer. Each nucleotide has a phosphate group on its 5' car­ bon (the end carbon of ribose) and a hydroxyl group on the 3 ' carbon. Two nucleotides are joined by a phosphate ester linkage between the 5 ' -phosphate group of one nucleotide and the 3 ' -phosphate group of another (Figure 23-24). The RNA polymer consists of many nucleotide units bonded this way, with a phosphate ester linking the 5' end of one nucleoside to the 3 ' end of another. A mole­ cule of RNA always has two ends (unless it is in the form of a large ring). One end has a free 3' group, and the other end has a free 5' group. We refer to the ends as the 3' end and the 5' end, and we refer to directions of replication as the 3 ' � 5' direction and the 5' � 3 ' direction. Figures 23-2 1 and 23-24 show short segments of RNA with the 3 ' end and the 5' end labeled. All our descriptions of ribonucleosides, ribonucleotides, and ribonucleic acid also apply to the components of DNA. The principal difference between RNA and DNA is the presence of D-2-deoxyribose as the sugar in DNA instead of the D-ribose found in RNA. The prefix deoxy- means that an oxygen atom is missing, and the number 2 means it is missing from C2. base

ICHO

®

no OH

H

3

OH

H

�

OH

sCH20H D-2-deoxyribose

f3-D-2-deox yri bofuranose

1 1 41

a deoxyribonucleoside

� Figure 23-24 Phosphate l inkage of nucleotides in RNA. Two nucleotides are joined by a phosphate linkage between the 5'-phosphate group of one and the 3'-hydroxyl group of the other.

23-22 The Structure of Ribonucleic Acid

23-23 Deoxyribose and the Structure of Deoxyribonucleic Acid

1142

Chapter 23: Carbohydrates and Nucleic Acids Another key difference between RNA and DNA is the presence of thymine in DNA instead of the uracil in RNA. Thymine is simply uracil with an additional methyl group. The four common bases of DNA are cytosine, thymine, adenine, and guanine.

�� �o

NH?

H

H

cytosine (C) thymine (T) pyrimidine bases

� H

adenine (A)

guanine (G) puri ne bases

These four bases are incorporated into deoxyribonucleosides and deoxyribo­ nucleotides similar to the bases in ribonucleosides and ribonucleotides. The following structures show the common nucleosides that make up DNA. The corre­ sponding nucleotides are simply the same structures with phosphate groups at the 5' positions. The structure of the DNA polymer is similar to that of RNA, except there are no hydroxyl groups on the 2' carbon atoms of the ribose rings. The alternating deoxyri­ bose rings and phosphates act as the backbone, while the bases attached to the de­ oxyribose units carry the genetic information. The sequence of nucleotides is called the primary structure of the DNA strand.

o'{

Four common deoxyribonucleosides that make up DNA

��io

HOC 2 0 H H H H OH H

deoxycytidine

H3C

/H

�� I

N HOC 2 0 H H H H OH H

N

deoxythymidine

23-23A

NH2

O

o

�(=6 �c,

'?-,,�

� L �e,

�

�

.�

In the amino acid analyzer, the components of the hydrolysate are dissolved in an aqueous buffer solution and separated by passing them down an ion-exchange column. The solution emerging from the column is mixed with ninhydrin, which reacts with amino acids to give the purple ninhydrin color. The absorption of light is recorded and printed out as a function of time. The time required for each amino acid to pass through the column (its retention time) depends on how strongly that amino acid interacts with the ion-exchange resin. The retention time of each amino acid is known from standardization with pure amino acids. The amino acids present in the sample are identified by comparing their reten­ tion times with the known values. The area under each peak is nearly proportional to the amount of the amino acid producing that peak, so we can determine the relative amounts of amino acids present. Figure 24- 13 shows a standard trace of an equimolar mixture of amino acids, fol­ lowed by the trace produced by the hydrolysate from human bradykinin (Arg-Pro-Pro­ Gly-Phe-Ser-Pro-Phe-Arg). Seq u e n c i n g the Pepti de: Te r m i n a l Res i d ue Ana lys i s The amino acid ana­ lyzer determines the amino acids present in a peptide, but it does not reveal their sequence: the order in which they are linked together. The peptide sequence is destroyed in the hydrolysis step. To determine the amino acid sequence, we must cleave j ust one amino acid from the chain and leave the rest of the chain intact. The cleaved amino acid can be separated and identified, and the process can be repeated on the rest of the chain. The amino acid may be cleaved from either end of the peptide (either the N terminus or the C terminus), and we will consider one method used for each end. This general method for peptide sequencing is called

terminal residue analysis.

24-9C

Seq uencing from the N Term i n us: The E d m a n Degradatio n

The most efficient method for sequencing peptides is the Edman degradation. A pep­ tide is treated with phenyl isothiocyanate, followed by acid hydrolysis. The products are the shortened peptide chain and a heterocyclic derivative of the N-terminal amino acid called a phenylthiohydantoin. This reaction takes place in three stages. First, the free amino group of the N-ter­ minal amino acid reacts with phenylisothiocyanate to form a phenylthiourea. Second,

24-9 Peptide Structure Determination

1 1 77

the phenylthiourea cyclizes to a thiazolinone and expels the shortened peptide chain. Third, the thiazolinone isomerizes to the more stable phenylthiohydantoin. Step J : Nucleophilic attack by the free amino group on phenyl isothiocyanate, followed by a proton transfer, gives a phenylthiourea.

..

..

.. S: Ph - NH -C .p I HN-CH -C-NH-fpeptide II .. I

Ph -N=C= S :

t V

Hil -CH-C-NH -fpeptide

R'I

0I

I

R'

0

I

a phenylthiourea

Step 2: Treatment with Hel induces cyclization to a thiazolinone and expulsion of the shortened peptide chain.

� �

NHPh

NHPh I 9C , . + HN S: \

r

H-C-C-NH-fpeptide

I I R' OH

I

� �

I

:N \

.,C .... . /S.:

+ H --+peptide H - C -C , N

I R'I : 0\

-

2

" "H

H 2O , )

protonated phenylthiourea

Step 3: In acid, the thiazolinone isomerizes to the more stable phenylthiohydantoin.

AS: )-{0 Rl :N

thiazolinone

II

S

NHPh HC l

C / ......

HN N- Ph \ I H-C-C �0 Rl

I

a phenylthiohydantoin

The phenylthiohydantoin derivative is identified by chromatography, by compar­ ing it with phenylthiohydantoin derivatives of the standard amino acids. This gives the identity of the original N-terminal amino acid. The rest of the peptide is cleaved intact, and further Edman degradations are used to identify additional amino acids in the chain. This process is well suited to automation, and several types of automatic sequencers have been developed. Figure 24- 1 4 shows the first two steps in the sequencing of bovine oxytocin. Be­ fore sequencing, the oxytocin sample is treated with peroxyformic acid to convert the disulfide bridge to cysteic acid residues. In theory, Edman degradations could sequence a peptide of any length. In practice, however, the repeated cycles of degradation cause some internal hydrolysis of the peptide, with loss of sample and accumulation of byproducts. After about 30 cycles of degrada­ tion, nuther accurate analysis becomes impossible. A small peptide such as bradykinin can be completely determined by Edman degradation, but larger proteins must be broken into smaller fragments (Section 24-9E) before they can be completely sequenced. PROBL EM 24- 2 1

Draw the structure o f the phenylthiohydantoin derivatives of (a) alanine (b) valine (c) lysine (d) proline PRO B L EM 24 - 2 2

Show the third and fourth steps i n the sequencing of bovi ne oxytocin. Use Figure 24- 1 4 a s a guide.

I

� �

NHPh I q-C , . :N S: r \ H - C -C � I

Rl

""0

a thiazolinone

+

.. �

H,N

-

peptide

+

HP +

1 1 78

Chapter 24: Amino Acids, Peptides, and Proteins

Step

o

J: Cleavage and determ inatio n of the N-terminal amino acid

S

II

..

H2N - C H - C - NH - Tyr - I le - Gln -ipeptide[

I

CH?

I -

( \ ) Ph - N = C = S (2)

H30+

S03H cysteic acid

o

II

,./ C """ HN N - Ph /

\

+

.. H2N - Tyr - I le - Gln -ipeptide[

CH- C� I 0 CH2S03H cysteic acid phenylthiohydantoin

Step 2 : Cleavage and determination of the second amino acid (the new N-terminal amino acid)

..

S

II

H2N - CH - C - NH- Ile - Gln -ipeptide[

I

9

( \ ) Ph - N = C = S

(2)

H30+

II

¢-c�o

,./ C """ HN N - Ph \

/

+

.. H2N - Ile - Gln -ipeptide[

OH tyrosine p henylthi ohydantoi n

OH

... Figure 24-14 The first two steps in sequencing bovine oxytocin. Each Edman degradation cleaves the N-terminal amino acid and forms its phenylthiohydantoin derivative. The shortened peptide is available for the next step.

PROB LEM 24- 2 3

The Sanger method for N-terminus determination i s a less common alternative to the Edman degradation. In the Sanger method, the peptide is treated with the Sanger reagent, 2,4-din i trofluorobenzene, and then hydrolyzed by reaction with 6 M aqueous HC!. The N-terminal amino acid is recovered as its 2,4-dinitrophenyl derivative and identified. The Sanger method

O,N

-Q-

F

+

N02

peptide

2,4-dinitrofluorobenzene (Sanger reagent)

6 M HCl, heat

02N

-o�

9

NH - H - COOH

N02

+

)

amino acids

Rl

2,4-din itrophenyl derivative

(a) Propose a mechanism for the reaction of the N terminus of the peptide with 2,4-dinitro­ fluorobenzene. (b) Explain why the Edman degradation is usually preferred over the Sanger method.

24-9 24-9D

Peptide Structure Determination

1 1 79

C-Te r m i n a l Res i d u e Ana lysis

There is no efficient method for sequencing several amino acids of a peptide starting from the C terminus. In many cases, however, the C-terminal amino acid can be iden­ tified using the enzyme carboxypeptidase, which cleaves the C-terminal peptide bond. The products are the free C-terminal amino acid and a shortened peptide. Fur­ ther reaction cleaves the second amino acid that has now become the new C terminus of the shortened peptide. Eventually, the entire peptide is hydrolyzed to its i ndividual amino acids. o

0

II II Ipeptidet-NH-CH -C -NH-CH-C-OH I I R/- 1

R"

carboxypeptidase

)

T h e selective enzymatic cleavage of proteins is critical to many bio­ logical processes. For example. the clotting of blood depends on the enzyme thrombin cleaving fibrino­ gen at specific points to produce fibrin. the protein that forms a clot.

o

0

II II I peptidet- NH-CH- C-OH + H?N -CH -C-OH . I I R/ - 1

R"

�

free amino acid

(further cleavage)

A peptide is i ncubated with the carboxypeptidase enzyme, and the appearance of free amino acids is monitored. In theory, the amino acid whose concentration increases first should be the C terminus, and the next amino acid to appear should be the second residue from the end. In practice, different amino acids are cleaved at different rates, making it difficult to determine amino acids past the C terminus and occasionally the second residue in the chain. 24-9E

B reaki n g the Pept i d e i nto S h o rter Chains: Partial Hydrolysis

Before a large protein can be sequenced, it must be broken into smaller chains, not longer than about 30 amino acids. Each of these shortened chains is sequenced, and then the entire structure of the protein is deduced by fitting the short chains together like pieces of a j igsaw puzzle. Partial cleavage can be accomplished either by using dilute acid with a short reac­ tion time or by using enzymes, such as trypsin and chymotlypsin, that break bonds between specific amino acids. The acid-catalyzed cleavage is not very selective, leading to a mixture of short fragments resulting from cleavage at various positions. Enzymes are more selective, giving cleavage at predictable points in the chain.

TRYPSIN: Cleaves the chain at the carboxyl groups of the basic amino acids lysine and arginine. CHYMOTRYPSIN: C leaves the chain at the carboxyl groups of the aromatic amino acids phenylalanine, tyrosine, and tryptophan. Let ' s use oxytocin (Figure 24-9) as an example to illustrate the use of partial hydrolysis. Oxytocin coul d be sequenced directly by C-terminal analysis and a series of Edman degradations, but it provides a simple example of how a structure can be pieced together from fragments. Acid-catalyzed partial hydrolysis of oxytocin (after cleavage of the disulfide bridge) gives a mixture that includes the following peptides: I1e-Gln-Asn-Cys

Gln-Asn-Cys-Pro

Pro-Leu-Gly · NH2

Cys-Pro-Leu-Gly

Cys-Tyr-I1e-Gln-Asn

When we match the overlapping regions of these fragments, the complete sequence of oxytocin appears: Ile-G In-Asn-Cys Gln-Asn-Cys-Pro Cys-Tyr-lle-Gln-Asn Cys-Pro-Leu-Gly Pro-Leu-Gly . NH2 Complete structure

Proteolytic zymes

(protein-cleaving) have

en­

applications in

consumer products.

For example.

papain (from papaya extract) serves as a meat tenderizer. It cleaves the fibrous

Cys-Tyr-Ile-Gln-Asn-Cys-Pro-Leu-Gly · NH2

also

proteins

at several

loca­

tions. making the meat less tough.

1 1 80

C h apter 24: Amino Ac ids , Peptides, and Proteins

The two Cys residues in oxytocin may be involved i n disulfide bridges, either linking two of these peptide units or forming a ring. By measuring the molecular weight of oxytocin, we can show that it contains j ust one of these peptide units; therefore, the Cys residues must link the molecule in a ring. PROBLEM 24 - 24

Show where trypsin and chymotrypsin would c leave the following pepti de

.

Tyr-Ile-Gln-Arg-Leu-Gly-Phe-Lys-Asn-Trp-Phe-Gly-Ala-Lys-Gl y-Gln-Gln . NH2 PROBLEM 24-25

After treatment with peroxyformic acid, the peptide hormone vasopressin is partially hy­ drolyzed. The following fragments are recovered. Propose a structure for vasopressin.

24- 1 0 Solution-Phase Peptide Synthesis

24- 1 0A

Phe-Gln-Asn

Pro-Arg-Gly ' NH2

Asn-Cys-Pro-Arg

Tyr-Phe-Gln-Asn

Cys-Tyr-Phe

I ntrod uction

Total synthesis of peptides is rarely an economical method for their commercial pro­ duction. Important peptides are usually derived from biological sources. For example, insulin for diabetics was originally taken from pork pancreas. Now, recombinant DNA techniques are improving the quality and availability of peptide pharmaceuticals. It is possible to extract the piece of DNA that contains the code for a particular protein, insert it into a bacterium, and induce the bacterium to produce the protein. Strains of Escherichia coli have been developed to produce human insulin that avoids dangerous reactions in people who are allergic to pork products. Laboratory peptide synthesis is still an important area of chemistry, however. When the structure of a new peptide is determined, a synthesis is usually attempted. The purpose of the synthesis is twofold: If the synthetic material is the same as the natural material, it proves that the proposed structure is correct; and the synthesis provides a larger amount of the material for further biological testing. Also, synthetic peptides can be made with altered amino acid sequences to compare their biological activity with the natural pep­ tides. These comparisons can point out the critical areas of the peptide, which may sug­ gest causes and treatments for genetic diseases involving similar abnormal peptides. Peptide synthesis requires the formation of amide bonds between the proper amino acids in the proper sequence. With simple acids and amines, we would form an amide bond simply by converting the acid to an activated derivative (such as an acyl halide or anhydride) and adding the amine. o

II R - C -X

o

+

II R - C - NH - R '

+

H-X

(X i s a good leaving group, preferably electron-withdrawing)

Amide formation is not s o easy with amino acids, however. Each amino acid has both an amino group and a carboxyl group. If we activate the carboxyl group, it reacts with its own amino group. If we mix some amino acids and add a reagent to make them couple, they form every conceivable sequence. Also, some amino acids have side chains that might i nterfere with peptide formation. For example, glutamic acid has an extra carboxyl group, and lysine has an extra amino group. As a result, peptide synthe­ sis always involves both activating reagents to form the correct peptide bonds and pro­ tecting groups to block formation of incorrect bonds. Chemists have developed many ways of synthesizing peptides, falling into two maj or groups. The classical solution-phase method involves adding reagents to solutions of growing peptide chains and purifying the products as needed. The

24- 1 0

Solution-Phase Peptide Synthesis

1 1 81

i nvolves adding reagents to growing peptide chains bonded to solid polymer particles. The solid-phase method has an advantage because spent reagents and impurities are easily washed from the solid particles holding the pep­ tide. Many different reagents and procedures are available for each of these meth­ ods, but we will consider only one set of reagents for the solution-phase method and one set for the solid-phase method. The general principles are the same regardless of the specific reagents. solid-phase method

24- 1 08

o

0

Solution-Phase M ethod

0

Consider the structure of alanylvalylphenylalanine, a simple tripeptide:

II

I

II

H2 N- CH - C - NH - CH - C - NH - CH - C - OH

I

CH3

alanyl

I

CH(CH )2 3 valyl

Ala-Val-Phe

I

CH2Ph

phenylalanine

Solution-phase peptide synthesis begins at the N terminus and ends at the C terminus, or left to right as we draw the peptide. The first major step is to couple the carboxyl group of alanine to the amino group of valine. This cannot be done simply by activating the carboxyl group of alanine and adding valine. If we activated the car­ boxyl group of alanine, it would react with another molecule of alanine. To prevent side reactions, the amino group of alanine must be protected to make it nonnucleophilic. Treating the free amino acid with benzyl chloroformate (also called benzyloxycarbonyl chloride) forms a urethane, or carbamate ester, that is easily removed at the end of the synthesis. This protecting group has been used for many years, and it has acquired several names. It is called the benzyloxycarbonyl group, the carbobenzoxy group (Cbz), or simply the Z group (abbreviated Z). Preliminary step: Protect the amino group with Z.

o H,N-CH-C-OH CH3 -

benzyl chloroformate Z-Cl

..

II

I

alanine Ala

Io-CH,-OJ�NH-CH,THJ-OH HCl Z group

+

benzyloxycarbonyl alanine Z-Ala

The amino group in Z-Ala is protected as the nonnucleophilic amide half of a carbamate ester. The carboxyl group can be activated without reacting with the pro­ tected amino group. In the solution-phase synthesis, the carboxyl group is activated by treatment with ethyl chloroformate. The product is a mixed anhydride of the amino acid and carbonic acid. It is strongly activated toward nucleophilic attack. Step

1:

Activate the carboxyl group with ethyl chloroformate.

o I

Z -NHCH- C - OH I CH3 protected alanine

o 0

anhydride of carbonic acid

I

+

ethyl chloroformate

I

Z- NHCH - C - O- C -OCH?CH - 3 I CH, mixed anhydride

+

HCI

1 1 82

Chapter 24: Amino Acids, Peptides, and Proteins When the second amino acid (valine) i s added to the protected, activated alanine, the nucleophilic amino group of valine attacks the activated carbonyl of alanine, dis­ placing the anhydride and forming a peptide bond. (Some procedures use an ester of the new amino acid to avoid competing reactions fro m its carboxylate group.)

Step

2: Form an amide bond to couple the next amino acid.

0

o

I

II

Z- NHCH -C - O -C - OCH?CH - 3 + I

o

..

II

H 2N - CH - C - O H

I

CH(CH3) 2

C H3

valine

protected, activated alanine

0

o

II

II

+ C021

Z - NHCH - C - N H - CH -C - OH

+

I

I

CH3

CH(CH3 )2

CH3CHPH

Z-Ala-Val

PROB LEM 24- 2 6

Give complete mechanisms for the formation of Z-Ala, its activation b y ethyl chloroformate, and the coupling with valine.

At this point, we have the N-protected dipeptide Z-Ala-Val. Phenylalanine must be added to the C terminus to complete the Ala-Val-Phe tripeptide. Activation of the valine carboxyl group, followed by addition of phenylalanine, gives the protected tripeptide. Step

1: Activate the carboxyl group with ethyl chloroformate.

o

0

II

II

Z-NHCHCNHCH -C

I

CH3

ala

Step

1'

Z-AI, -NH H -

�J �

CH(CH3)2

val

I

I

---t::Qill

0

I

O

0

+ Cl -C-OEt

--->

I

II

Z -NHCHCNHCH - C

I

CH(CH3)2

CH3

I

val

I

O -C-OEt

CH(CH3h

ala

val

J

0

I

+ HCl

2: Form an amide bond to couple the next amino acid.

1'

�

o - - OEt + H'N - H - -OH

�

1'

Z -AI, -NH H -

CH2-Ph

phenylalanine

�

1'

�

NH - H - -OH

H3C-CH-CH3

CH2 - Ph

Z-Ala-Val-Phe

To make a larger peptide, repeat these two steps for the addition of each amino acid residue: 1. Activate the C terminus of the growing peptide by reaction with ethyl chloroformate.

2. Couple the next amino acid.

The final step in the solution-phase synthesis is to deprotect the N terminus of the completed peptide. The N-terminal amide bond must be cleaved without breaking any of the peptide bonds in the product. Fortunately, the benzyloxycarbonyl group is partly an amide and partly a benzyl ester, and hydrogenolysis of the benzyl ester takes place under mild conditions that do not cleave the peptide bonds. This mild cleavage is the reason for using the benzyloxycarbonyl group (as opposed to some other acyl group) to protect the N terminus.

.

24- 1 1 S o lid-Phase Peptide Synthesis

Final step: Remove the protecting group.

-

o

�

0

?�

CH2 -0- -NH H - Val -Phe Z-Ala-Val-Phe

CH3

?

o

.. II H 2N HC- Val-Phe Ala-Val-Phe

C0 21

o-

PROBLEM 24- 2 7

Show how you would synthesize Ala-Val-Phe-Gly-Leu starting with Z-Ala-Val-Phe. PROBLEM 24- 2 8

Show how the solution-phase synthesis would b e used t o synthesize Ile-Gly-Asn.

The solution-phase method works well for small peptides, and many peptides have been synthesized by this process. A large number of chemical reactions and purifications are required even for a small peptide, however. Although the individual yields are excellent, with a large peptide, the overall yield becomes so small as to be unusable, and several months (or years) are required to complete so many steps. The large amounts of time required and the low overall yields are due largely to the purification steps. For larger peptides and proteins, solid-phase peptide synthesis is usually preferred.

In 1 962, Robert Bruce Merrifield of Rockefeller University developed a method for synthesizing peptides without having to purify the intermediates. He did this by at­ taching the growing peptide chains to solid polystyrene beads. After each amino acid is added, the excess reagents are washed away by rinsing the beads with solvent. This ingenious method lends itself to automation, and Merrifield built a machine that can add several amino acid units while running unattended. Using this machine, Merrifield synthesized ribonuclease ( 1 24 amino acids) in just six weeks, obtaining an overall yield of 1 7%. Merrifield ' s work in solid-phase peptide synthesis won the Nobel Prize in chemistry in 1 984. The I n d iv i d u a l Reactions

There are three important reactions to consider before we use solid-phase peptide syn­ thesis. We must first learn how an amino acid is attached to the solid support, how and why the amino group is protected, and how peptide bonds are formed. The greatest difference between solution-phase and solid-phase peptide synthesis is that solid-phase synthesis is done in the opposite direction: starting with the C terminus and going toward the N termi­ nus, right to left as we write the peptide. The first step is to attach the last amino acid (the C terminus) to the solid support. The solid support is a special polystyrene bead in which some of the aromatic rings have chloromethyl groups. This polymer, often called the Merrifield resin, is made by copolymerizing styrene with a few percent of p-(chloromethyl)styrene.

Attaching the Peptide to the Sol id Support

+

CH3

+

24- 1 1 A

1 1 83

CH3

PROBLEM-SOLVING

Htnp

Remember that classica l (solution­ phase) peptide synthesis: 1 . Goes N C. Protect N terminus (Z group) fi rst, deprotect last. 2. Couple each a m ino acid by activating the C terminus (ethyl chloroformate), then adding new a mino acid. --'>

24- 1 1 Solid -Phase Peptide Synthesis

1 1 84

Chapter 24: Amino Acids, Peptides, and Proteins

Formation of the Merrifield resin

CI

CI

Q

H

/

C=C

Q

+

/ "'-

H H

styrene

H

/

�

C=C

/

"'-

H

Q

Q

Q

+ CH-CH2 - CH - CH2 - CH -CH2 +

H

p-(chloromethy l )styrene

polymer

P abbreviation

Like other benzyl halides, the chI oromethyl groups on the polymer are reactive toward SN2 attack. The carboxyl group of an N-protected amino acid displaces chlo­ ride, giving an amino acid ester of the polymer. In effect, the polymer serves as the alcohol part of an ester protecting group for the carboxyl end of the C-terminal amino acid.

protecting group

o

Attachment of the C-terminal amino acid

H

H

I

R

o

..

-

II .. protecting r- NH -CH - C - O -CH? group

\/ .. :-NH - CH -C - O CI .. :� C � II

I R

o p

o p

Once the C-terminal amino acid i s fixed to the polymer, the chain i s built on the amino group of this amino acid. At the completion of the synthesis, the ester bond to the polymer is cleaved by anhydrous HF. Because this i s an ester bond, it i s more eas­ ily cleaved than the amide bonds of the peptide.

o

Cleavage of the finished peptide

II

I peptider C -O-CH2

o

�

HF

I pePtider

o�

- OH

+

0 p

p

Use of the t-Butyloxyca rbonyl (Boc) Protecting G roup The benzyloxy­ carbonyl group (the Z group) cannot be used with the solid-phase process because the Z group is removed by hydrogenolysis in contact with a solid catalyst. A polymer-bound peptide cannot achieve the inti mate contact with a solid catalyst required for hydrogenolysis. The N-protecting group used in the Merrifield proce­ dure is the t-butyloxycarbonyl group, abbreviated Boc or t-Boc. The Boc group i s similar t o the Z group, except that it has a t-butyl group i n place o f the benzyl group. Like other t-butyl esters, the Boc protecting group is easily removed under acidic conditions. The acid chloride of the Boc group is unstable, so we use the anhydride, di-t­ butyldicarbonate, to attach the group to the amino acid.

1 1 85

24- 1 1 Solid-Phase Peptide Synthesis

Protection of the amino group as its Boc derivati ve 0

CH3

0

CH 3

CH3

0

CH3

II II I I CH3 - C - O - C - O -C - O - C -CH .,) I I

+

CH3

-

H?N - CH- COOH

I

�

CH3

R

di-t-butyldicarbonate

I I CH 3 -C - O - C I-- NH - CH - COOH I I

amino acid

R

Boc-amino acid CH3

I

CH3- C -OH

I

CH3

The Boc group is easily cleaved by brief treatment with trifluoroacetic acid Loss of a relatively stable t-butyl cation from the protonated ester gives an unstable carbamic acid. Decarboxylation of the carbamic acid gives the deprotected amino group of the amino acid. Loss of a proton from the t-butyl cation gives isobutylene. (TFA), CF3COOH .

CH3

"0"

I

'

I

I (II CH 3- C - O - C - NH - C H - COOH I I \2:.) .•

CH3-C - O - C - NH - CH - COOH

I

CH3

I CH3-C+ I

CH3

I

Boc-amino acid

CH3

+

l

'OL H

CH3

'

. : O-H

::o �

b .

CH3

R

- NH -

r

]

a carbamic acid

R

protonated

CH2 =C

- COOH

free amino acid

�

/ CH,0

�

+

C0 2i

CH3

isobutylene

People who synthesize peptides generally do not make their own Boc-protected amino acids. Because they use all their amino acids in protected form, they buy and use commercially available B oc-amino acids. Use of DCC as a Peptide Cou p l ing Agent The final reaction needed for the Merrifield procedure is the peptide bond-forming condensation. When a mixture of an amine and an acid is treated with N, N ' -dicyclohexylcarbodiimide (abbreviated DCC), the amine and the acid couple to form an amide. The molecule of water lost in this condensation converts DCC to N, N ' -dicyclohexyl urea (DCU) .

o

II R - C - NH -R' acid

amine

N,N' -dicyclohexylcarbodiimide (DCC)

amide

+ o-�-�-�-o

The mechanism for DCC coupling is not as complicated as it may seem. The carboxylate ion adds to the strongly electrophilic carbon of the diimide, giv­ ing an activated acyl derivative of the acid. This activated derivative reacts readily with the amine to give the amide. I n the final step, DCU serves as an excellent leaving group.

H

0

H

N,N' -dicyclohexyl urea (OCU)

1 1 86

Chapter 24: Amino Acids, Peptides, and Protei n s

Formation of a n activated acyl derivative

o-

� = C 0�

)

o II .. R -C - O :

-o

activated

Coupling with the amine and loss of DCU

o

N II R-C-O-C

II ) �

R'- NH,

D

""'NH

6

---7

,6j I

e

'N o II II R-C- O-C "",

_

r"

N II

D

R - C - O -C 1+ " ""' NH R -N ' Il,

c5

---7

6

NH I R - C-NHR' + o =c o II

amide

PROBLEM 24- 29

C

(5

Propose a mechanism for the coupling of acetic acid and aniline using Dee as a coupling agent.

Now we consider an example to illustrate how these procedures are combined in the Merrifield solid-phase peptide synthesis. 24- 1 1 B

An Exa m p l e of Sol id-Phase Peptide Synthesis

For easy comparison of the solution-phase and solid-phase methods, we will con­ sider the synthesis of the same tripeptide we made using the solution-phase method. Ala-Val-Phe The solid-phase synthesis is carried out in the direction opposite that of the solu­ tion-phase synthesis. The first step is attachment of the N-protected C-terminal amino acid (Boc-phenylalanine) to the polymer.

d

o

24-11

3C-O-� NH-CH-�-O-

Me

Boc

I

Ph-CH2

+

CH2-CI

Boc-Phe

d

1187

C-O-� NH-CH-�-O-CH? ' I Boc Ph-CH2

Me3

�

o �

o

Solid-Phase Peptide Synthesis

�O

/

P

Boc-Phe

----®

P

Trifluoroacetic acid (TFA) cleaves the Boc protecting group of phenylalanine so that its amino group can be coupled with the next amino acid.

d C-O-�

Me3

Boc

o

CF3COOH (TFA)

NH-CH-�-O-CHo ' I

Boc-Phe

H3N-CH-C-0-CHo

--0

II

I

j

�

Ph-CH?

D

Ph-CH2

o

+

Phe

-

--®

+

-

0

/CH3 CH?=C "CH3

�NH-CH-C-OI (CH,)2CH II

+

�

o

DCC

H3N- H-C-0 CH2 T Ph-CH2 II

+

Boc-Val

Phe

0

--0

�

CO?1' -I

P

The second amino acid (valine) is added in its N-protected Boc form so that it cannot couple with itself. Addition of Dee couples the valine carboxyl group with the free -NH2 group of phenylalanine.

o

+

�

o

0 II � NH-TH-C-NH-TH-C-0 CH2 Ph-CH l""'T(CH,),CH II

0

,

--0

P

Boc-val-Phe

+

DCU

P

To couple the final amino acid (alanine), the chain is first deprotected by treat­ ment with trifluoroacetic acid. Then the N-protected Boc-alanine and Dee are added.

o

Step 1: Deprotection

0

� � NH-CH-C-NH-CH-C-O-CH I I (CH3)2CH Ph-CH II

II

h LOJ

2

Boc-Val-Phe

--®

CF3COOH (TFA)

0 0 II II ) H N-CH-C-NH-CH-C-O-CH I Ph-CH2 (CH3)2CH

3 1 +

h LOJ

--®

+

3

P

Val-Phe

P

+

C K' I CH -C=CH2 CO2

Step 2: Coupling

o

o

�

o

� NH-?H - C - O0 0 0 II \I II II II CH3 ' � NH-CH-C-NH-CH-C-NH-CH-C-O-CH? H3N-CH-C-NH-CH-C-0-CH2 DCC I I I I 1 · CH3 (CH3hCH Ph - CH2 (CH3hCH Ph-CH2 +

----0

Val-Phe

0 P

�

II

--®

Boc-Ala-Val-Phe

0 P

+

DCU

1188

Chapter 24: Amino Acids, Peptides, and Proteins

0

0

If we were making a longer peptide, the addition of each subsequent amino acid would require the repetition of two steps: 1. Use trifluoroacetic acid to deprotect the amino group at the end of the growing chain. 2. Add the next Boc-amino acid, using DeC as a coupling agent. Once the peptide is completed, the final Boc protecting group must be removed, and the peptide must be cleaved from the polymer. Anhydrous HF cleaves the ester linkage that bonds the peptide to the polymer, and it also removes the Boc protecting group. In our example, the following reaction occurs: 0

II II II � NH-CH-C-NH-CH-C-NH-CH-C-O-CH2

I

I

I

CH3

(CH3hCH

Ph-CH 2

0

-®

p

Boc-Ala-Val-Phe

PROBLEM-SOLVING

HF

�

HiltZ;

Remember that solid-phase peptide synthesis:

1. Goes C � N. Attach the

Ala-Val-Phe

Boc-protected C terminus to the bead first.

2. Couple each amino acid by

removing (TFA) the Boc group

PRO B LE M 2 4 - 3 0

from the N terminus, then add the

Show how you would synthesize Leu-Gly-Ala-Val-Phe starting with Boc-Ala-Val-Phe-®

with DCC.

PRO B LE M 2 4- 3 1

next Boc-protected amino acid

3. Cleave (HF) the finished peptide from the bead.

24- 1 2 Classification of Proteins

Show how solid-phase peptide synthesis would be used to make Ue-Gly-Asn.

Proteins may be classified according to their chemical composition, their shape, or their function. Protein composition and function are treated in detail in a biochemistry course. For now, we briefly survey the types of proteins and their general classifications. Proteins are grouped into simple and conjugated proteins according to their chemical composition. Simple proteins are those that hydrolyze to give only amino acids. All the protein structures we have considered so far are simple proteins. Exam­ ples are insulin, ribonuclease, oxytocin, and bradykinin. Conjugated proteins are bonded to a nonprotein prosthetic group such as a sugar, a nucleic acid, a lipid, or some other group. Table 24-3 lists some examples of conjugated proteins. Proteins are classified as .fibrous or globular depending on whether they form long filaments or coil up on themselves. Fibrous proteins are stringy, tough, and usu­ ally insoluble in water. They function primarily as structural parts of the organism. Examples of fibrous proteins are a-keratin in hooves and fingernails, and collagen in TABLE 24-3

Classes of Conjugated P rotei ns

Class

Prosthetic Group

Examples

glycoproteins nucleoproteins lipoproteins metalloproteins

carbohydrates nucleic acids fats, cholesterol a complexed metal

y-globulin, interferon ribosomes, viruses high-density lipoprotein hemoglobin, cytochromes

24-13

Levels of Protein Structure

1189

tendons. Globular proteins are folded into roughly spherical shapes. They usually function as enzymes, hormones, or transport proteins. Enzymes are protein-contain­ ing biological catalysts; an example is ribonuclease, which cleaves RNA. Hormones help to regulate processes in the body; an example is insulin, which regulates glucose levels in the blood and its uptake by cells. Transport proteins bind to specific mole­ cules and transport them through cell membrane or in the blood; an example is hemo­ globin, which transports oxygen in the blood from the lungs to the tissues. 24-13A

24- 1 3

Primary Structure

Up to now, we have discussed the primary structure of proteins. The primary struc­ Levels of Protein ture is the covalently bonded structure of the molecule. This definition includes the sequence of amino acids, together with any disulfide bridges. All the properties of the Structure protein are determined, directly or indirectly, by the primary structure. Any folding, hydrogen bonding, or catalytic activity depends on the proper primary structure. 24-138

Secondary Structure

Although we often think of peptide chains as linear structures, they tend to form order­ ly hydrogen-bonded arrangements. In particular, the carbonyl oxygen atoms form hydrogen bonds with the amide ( N -H ) hydrogens. This tendency leads to orderly patterns of hydrogen bonding: the helix and the pleated sheet. These hydrogen­ bonded arrangements, if present, are called the secondary structure of the protein. When a peptide chain winds into a helical coil, each carbonyl oxygen can hydro­ gen-bond with an N -H hydrogen on the next turn of the coil. Many proteins wind into an a helix (a helix that looks like the thread on a right-handed screw) with the side chains positioned on the outside of the helix. For example, the fibrous protein ker­ atin is arranged in the a-helical structure, and most globular proteins contain segments of a helix. Figure 24-15 shows the a-helical arrangement. Segments of peptides can also form orderly arrangements of hydrogen bonds by lining up side-by-side. In this arrangement, each carbonyl group on one chain forms a hydrogen bond with an N -H hydrogen on an adjacent chain. This arrangement may involve many peptide molecules lined up side-by-side, resulting in a two-dimensional sheet. The bond angles between amino acid units are such that the sheet is pleated (creased), with the amino acid side chains arranged on alternating sides of the sheet. Silk fibroin, the principal fibrous protein in the silks of insects and arachnids, has a pleated sheet secondary structure. Figure 24-16 shows the pleated sheet structure. a

a

Spider web is composed mostly of fibroin, a protein with pleated-sheet secondary structure. The pleated­ sheet arrangement allows for multiple hydrogen bonds between molecules, conferring great strength.

C

N

o

R

= = = =

gray blue red green

... Figure 24-15

The IX helical arrangement. The peptide chain curls into a helix so that each peptide carbonyl group is hydrogen-bonded to an N - H hydrogen on the next tum of the helix. Side chains are symbolized by green atoms in the space-filling structure.

1190

Chapter

24:

A m ino Acids, Peptides, and Proteins o

,R 0 H ,-,-" II I " ./N" ./Cl-!" ./c" N C CH I II /' H 0 R

,-,-" II /CH" /c" . . N I H 0 H I II N " ./C " N./ CH I /

-� ,,,

Micrograph of normal human brain tissue. The nuclei of neurons appear as dark spots.

Brain tissue of a patient infected with vClD. Note the formation of (white) vacuole spaces and (dark, irregular) plaques of prion protein. (Magnification 200X)

24-14A

Reversible and Irreversible Denaturation

24-14B

Prion Diseases

The cooking of egg white is an example of protein denaturation by high temperature. Egg white contains soluble globular proteins called albumins. When egg white is heat­ ed, the albumins unfold and coagulate to produce a solid rubbery mass. Different pro­ teins have different abilities to resist the denaturing effect of heat. Egg albumin is quite sensitive to heat, but bacteria that live in geothermal hot springs have developed pro� teins that retain their activity in boiling water. When a protein is subjected to an acidic pH, some of the side-chain carboxyl groups become protonated and lose their ionic charge. Conformational changes result, leading to denaturation. In a basic solution, mmno groups become deprotonated, simi­ larly losing their ionic charge, causing conformational changes and denaturation. Milk turns sour because of the bacterial conversion of carbohydrates to lactic acid. When the pH becomes strongly acidic, soluble proteins in milk are denatured and precipitate. This process is called curdling. Some proteins are more resistant to acidic and basic conditions than others. For example, most digestive enzymes such as amylase and trypsin remain active under acidic conditions in the stomach, even at a pH of about 1. In many cases, denaturation is irreversible. When cooked egg white is cooled, it does not become uncooked. Curdled milk does not uncurdle when it is neutralized. Denaturation may be reversible, however, if the protein has undergone only mild denaturing conditions. For example, a protein can be salted out of solution by a high salt concentration, which denatures and precipitates the protein. When the precipitated protein is redissolved in a solution with a lower salt concentration, it usually regains its activity together with its natural conformation.

Up through 1980, people thought that all infectious diseases were caused by microbes of some sort. They knew about diseases caused by viruses, bacteria, protozoa, and fungi. There were some strange diseases, however, for which no one had isolated and cultured the pathogen. Creutzieldt-lakob Disease (CJD) in humans, scrapie in sheep, and transmissible encephalopathy in mink (TME) all involved a slow, gradual loss of mental function and eventual death. The brains of the victims all showed unusual plaques of amyloid protein surrounded by spongelike tissue. Workers studying these diseases thought there was an infectious agent involved (as opposed to genetic or environmental causes) because they knew that scrapie and TME could be spread by feeding healthy animals the ground-up remains of sick animals. They had also studied kuru, a disease much like CJD among tribes where family members showed their respect for the dead by eating their brains. These diseases were generally attributed to "slow viruses" that were yet to be isolated. In the 1980s, neurologist Stanley B. Prusiner (of the University of California at San Francisco) made a homogenate of scrapie-infected sheep brains and systemati­ cally separated out all the cell fragments, bacteria, and viruses, and found that the remaining material was still infectious. He separated out the proteins and found a protein fraction that was still infectious. He suggested that scrapie (and presumably similar diseases) is caused by a protein infectious agent that he called prion' protein. This conclusion contradicted the established principle that contagious diseases require a living pathogen. Many skeptical workers repeated Prusiner's work in hopes of finding viral contaminants in the infectious fractions, and most of them finally came to the same conclusion. Prusiner received the 1 998 Nobel Prize in medicine for this work. Since Prusiner's work, prion diseases have become more important because of their threat to humans. Beginning in 1996, some cows in the United Kingdom

Chapter 24 Glossary developed "mad cow disease" and would threaten other animals, wave their heads, fall down, and eventually die. The disease, called bovine spongiform encephalopa­ thy, or BSE, was probably transmitted to cattle by feeding them the remains of scrapie-infected sheep. The most frightening aspect of the BSE outbreak was that people could contract a fatal disease, called new-variant Creutzfeldt-lakob Disease (vCJD) from eating the infected meat. Since that time, a similar disease, called chronic wasting disease, or CWD, has been found in wild deer and elk in the Rocky Mountains. All of these (presumed) prion diseases are now classified as trans­ missible spongiform encephalopathies, or TSEs. The most widely accepted theory of prion diseases suggests that the infectious prion protein has the same primary structure as a normal protein found in nerve cells, but it differs in its tertiary structure. In effect, it is a misfolded, denatured ver­ sion of a normal protein that polymerizes to form the amyloid protein plaques seen in the brains of infected animals. When an animal ingests infected food, the polymer­ ized protein resists digestion. Because it is simply a misfolded version of a normal protein, the infectious prion does not provoke the host's immune system to attack the pathogen. When the abnormal prion interacts with normal version of the protein on the membranes of nerve cells, the abnormal protein somehow induces the normal mole­ cules to change their shape. This is the part of the process we know the least about. (We might think of it like crystallization, in which a seed crystal induces other molecules to crystallize in the same conformation and crystal form.) These newly misfolded protein molecules then induce more molecules to change shape. The polymerized abnormal protein cannot be broken down by the usual protease enzymes, so it builds up in the brain and causes the plaques and spongy tissue asso­ ciated with TSEs. We once thought that a protein with the correct primary structure, placed in the right physiological solution, would naturally fold into the correct tertiary structure and stay that way. We were wrong. We now know that protein folding is a carefully controlled process in which enzymes and chaperone proteins promote correct folding as the protein is being synthesized. Prion diseases have taught us that there are many factors that cause proteins to fold into natural or unnatural conformations, and that the folding of the protein can have major effects on its biological properties within an organism.

active site The region of an enzyme that binds the substrate and catalyzes the reaction. (p. 1191) amino acid Literally, any molecule containing both an amino group ( -NH2) and a carboxyl group (-COOH). The term usually means an a-amino acid, with the amino group on the carbon atom next to the carboxyl group. (p. 1154) biomimetic synthesis A laboratory synthesis that is patterned after a biological synthesis. For

example, the synthesis of amino acids by reductive amination resembles the biosynthesis of glutamic acid. (p. 1162) complete proteins Proteins that provide all the essential amino acids in about the right proportions for human nutrition. Examples include those in meat, fish, milk, and eggs. Incomplete proteins are severely deficient in one or more of the essential amino acids. Most plant proteins are incomplete. (p. 1157) conjugated protein A protein that contains a nonprotein prosthetic group such as a sugar, nucleic acid, lipid, or metal ion. (p. 1188) C terminus (C-terminal end) The end of the peptide chain with a free or derivatized carboxyl group. As the peptide is written, the C terminus is usually on the right. The amino group of the C-termjnal amjno acid links it to the rest of the peptide. (p. 1172) denaturation An unnatural alteration of the conformation or the ionic state of a protein. Denaturation generally results in precipitation of the protein and loss of its biological activity. Denaturation may be reversible, as in salting out a protein, or irreversible, as in coobng egg white. (p. 1191)

Chapter 24 Glossary

1193

1 194

Chapter

24:

Amino Acids, Peptides, and Proteins dipolar ion (zwitterion) A structure with an overall charge of zero but having a positively

charged substituent and a negatively charged substituent. Most amino acids exist in dipolar ionic forms. (p. 1158) °

o

II

HN-CH-C-O-

HN-CH-C-OH

I

2

II

+

3

R

I

R

dipolar ion, or zwitterion (major component)

uncharged structure (minor component)

disulfide linkage (disulfide bridge) A bond between two cysteine residues formed by mild oxidation of their thiol groups to a disulfide. (p. 1172) Edman degradation A method for removing and identifying the N-terminal amino acid from a peptide without destroying the rest of the peptide chain. The peptide is treated with phenyliso­ thiocyanate, followed by a mild acid hydrolysis to convert the N-terminal amino acid to its phenylthiohydantoin derivative. The Edman degradation can be used repeatedly to determine the sequence of many residues beginning at the N terminus. (p. 1176) electrophoresis A procedure for separating charged molecules by their migration in a strong electric field. The direction and rate of migration are governed largely by the average charge on the molecules. (p. 1160) enzymatic resolution The use of enzymes to separate enantiomers. For example, the enan­ tiomers of an amino acid can be acylated and then treated with hog kidney acylase. The enzyme hydrolyzes the acyl group from the natural L-amino acid, but it does not react with the D-amino acid. The resulting mixture of the free L-amino acid and the acylated D-amino acid is easily separated. (p. 1167) enzyme A protein-containing biological catalyst. Many enzymes also include prosthetic groups, nonprotein constituents that are essential to the enzyme's catalytic activity. (p. 1189) essential amino acids Ten standard amino acids that are not biosynthesized by humans and must be provided in the diet. (p. 1157) fibrous proteins A class of proteins that are stringy, tough, threadlike, and usually insoluble in water. (p. 1188) globular proteins A class of proteins that are relatively spherical in shape. Globular proteins gen­ erally have lower molecular weights and are more soluble in water than fibrous proteins. (p. 1189) ll' helix A helical peptide conformation in which the carbonyl groups on one turn of the helix are hydrogen-bonded to N - H hydrogens on the next turn. Extensive hydrogen bonding makes this helical arrangement quite stable. (p. 1189) hydrogenolysis Cleavage of a bond by the addition of hydrogen. For example, catalytic hydrogenolysis cleaves benzyl esters. (p. 1168) o

�

R- -O-CH2

-- (d) (D,L)-proline ----2-og k -:id ne ": a:' cyy ] a s e, H -2O ( ) h-

({) product from part (e)

Hp (g) 4-methylpentanoic acid

+

Br2/PBr3

�

H30

+

)

(h) product from part ( g) + excess NH 3

�

H2, Pd

)

Study Problems 24-35

1197

Show how you would synthesize any of the standard amino acids from each starting material. You may use any necessary reagents.

(b)

CH3-CH -CH2-COOH

I

CH2CH3

24-36

24-37

24-38 24-39 24-40

24-41

Show how you would convert alanine to the following derivatives. Show the structure of the product in each case. (a) alanine isopropyl ester (b) N-benzoylalanine (c) N-benzyloxycarbonyl alanine (d) t-butyloxycarbonyl alanine Suggest a method for the synthesis of the unnatural lactic acid.

D

enantiomer of alanine from the readily available L enantiomer of

lactic acid

Show how you would use the Gabriel-malonic ester synthesis to make histidine. What stereochemistry would you expect in your synthetic product? Show how you would use the Strecker synthesis to make tryptophan. What stereochemistry would you expect in your synthetic product? Write the complete structures for the following peptides. Tell whether each peptide is acidic, basic, or neutral. (a) methionylthreonine (b) threonylmethionine (c) arginylaspartyllysine (d) Glu-Cys-Gln The following structure is drawn in an unconventional manner.

0

CH

I

0

II

II

3

CH 3CH2-CH-CH-NH-C-CH-CH2CH2-C-NH 2

I

CONH2

24-42

24-43

I

NH-CO-CH 2NH2

(b) Label the peptide bonds. (a) Label the N terminus and the C terminus. (c) Identify and label each amino acid present. (d) Give the full name and the abbreviated name. Aspartame (Nutrasweet®) is a remarkably sweet-tasting dipeptide ester. Complete hydrolysis of aspartame gives phenyl­ alanine, aspartic acid, and methanol. Mild incubation with carboxypeptidase has no effect on aspartame. Treatment of aspartame with phenyl isothiocyanate, followed by mild hydrolysis, gives the phenylthiohydantoin of aspartic acid. Propose a structure for aspartame.

A molecular weight determination has shown that an unknown peptide is a pentapeptide, and an amino acid analysis shows that it contains the following residues: one Gly, two Ala, one Met, one Phe. Treatment of the original pentapeptide with carboxypeptidase gives alanine as the first free amino acid released. Sequential treatment of the pentapeptide with phenyl isothiocyanate followed by mild hydrolysis gives the following derivatives: first time

second time

third time

S

S

Ph-N

A NH

H

O 24-44 24-45 24-46

" CH?Ph H

Ph-N

A NH

o}--l

Propose a structure for the unknown pentapeptide. Show the steps and intermediates in the synthesis of Leu-Ala-Phe (b) by the solid-phase process. (a) by the solution-phase process. Using classical solution-phase techniques, show how you would synthesize Ala-Val and then combine it with I1e-Leu-Phe to give I1e-Leu-Phe-Ala-Val. Peptides often have functional groups other than free amino groups at the N terminus and other than carboxyl groups at the C terminus. (a) A tetrapeptide is hydrolyzed by heating with 6 M H Cl, and the hydrolysate is found to contain Ala, Phe, Val, and Glu. When the hydrolysate is neutralized, the odor of ammonia is detected. Explain where this ammonia might have been incorporated in the original peptide.

1198

Chapter 24: Amino Acids, Peptides, and Proteins (b) The tripeptide thyrotropic hormone releasing factor (TRF) has the full name pyroglutamylhistidylprolinamide. The

structure appears here. Explain the functional groups at the N terminus and at the C terminus. H2 H2 0 0 C-C -H2C" \ / "CH2 II H II CH-C-N-C-C-N I I / H ..--C H2 C I � ---N CH2 / o \

�

r(

H

-=

N

VN.

......

H

H2 N-C,\0

(c) On acidic hydrolysis, an unknown pentapeptide gives glycine, alanine, valine, leucine, and isoleucine. No odor of 24-47

ammonia is detected when the hydrolysate is neutralized. Reaction with phenyl isothiocyanate followed by ITuid hydrol­ ysis gives no phenylthiohydantoin derivative. Incubation with carboxypeptidase has no effect. Explain these findings. Lipoic acid is often found near the active sites of enzymes, usually bound to the peptide by a long, flexible amide linkage with a lysine residue.

�

NH

o

� COOH \

I

\

S-S

I

II

�C " I

N

�

I

S-S

H bound to lysine residue

lipoic acid

H

T C=O �

(a) Is lipoic acid a mild oxidizing agent or a mild reducing agent? Draw it in both its oxidized and reduced forms. (b) Show how lipoic acid might react with two Cys residues to form a disulfide bridge. (c) Give a balanced equation for the hypothetical oxidation or reduction, as you predicted in part (a), of an aldehyde by

lipoic acid. o

II

+

R-C-H

�

COOH

S-S

24-48

24-49 24-50

24-51

Histidine is an important catalytic residue found at the active sites of many enzymes. In many cases, histidine appears to remove protons or to transfer protons from one location to another. (a) Show which nitrogen atom of the histidine heterocycle is basic and which is not. (b) Use resonance forms to show why the protonated form of histidine is a particularly stable cation. (c) Show the structure that results when histidine accepts a proton on the basic nitrogen of the heterocycle and then is deprotonated on the other heterocyclic nitrogen. Explain how histidine might function as a pipeline to transfer pro­ tons between sites within an enzyme and its substrate. Metabolism of arginine produces urea and the rare amino acid ornithine. Ornithine has an isoelectric point close to 10. Propose a structure for ornithine. Glutathione (GSH) is a tripeptide that serves as a mild reducing agent to detoxify peroxides and maintain the cysteine residues of hemoglobin and other red blood cell proteins in the reduced state. Complete hydrolysis of glutathione gives Gly, Glu, and Cys. Treatment of glutathione with carboxypeptidase gives glycine as the first free amino acid released. Treatment of glutathione with phenyl isothiocyanate gives the phenylthiohydantoin of glutamic acid. (a) Propose a structure for glutathione consistent with this information. (b) Oxidation of glutathione forms glutathione disulfide (GSSG). Propose a structure for glutathione disulfide, and write a balanced equation for the reaction of glutathione with hydrogen peroxide. Complete hydrolysis of an unknown basic decapeptide gives Gly, Ala, Leu, Ile, Phe, Tyr, Glu, Arg, Lys, and Ser. Terminal residue analysis shows that the N terminus is Ala and the C terminus is IIe. Incubation of the decapeptide with chy­ motrypsin gives two tripeptides, A and B, and a tetrapeptide, C. Amino acid analysis shows that peptide A contains Gly, Glu, Tyr, and NH3; peptide B contains Ala, Phe, and Lys; and peptide C contains Leu, lie, Ser, and Arg. Terminal residue analysis gives the following results. A

B C

N term i n us

C term i n u s

Glu Ala Arg

Tyr Phe IIe

Study Problems

24-52

Incubation of the decapeptide with trypsin gives a dipeptide D, a pentapeptide E, and a tripeptide F. Terminal residue analy­ sis of F shows that the N terminus is Ser, and the C terminus is Ile. Propose a structure for the decapeptide and for fragments A through F. There are many methods for activating a carboxylic acid in preparation for coupling with an amine. The following method converts the acid to an N-hydroxysuccinimide (NHS) ester.

R-\

o +

F3C

OH

:l

� y

O-N

� -\ R y O

tN

E �

O

O-N

+

0

o

(a) Explain why an NBS ester is much more reactive than a simple alkyl ester. (b) Propose a mechanism for the reaction shown.

24-53

1199

R

(e) Propose a mechanism for the reaction of the NBS ester with an amine, NH2 . Sometimes chemists need the unnatural D enantiomer of an amino acid, often as part of a drug or an insecticide. Most L-amino acids are isolated from proteins, but the D amino acids are rarely found in natural proteins. D-amino acids can be synthesized from the con'esponding L amino acids. The following synthetic scheme is one of the possible methods. COOH

H

L

�"R

H configuration

NaNG3 He]

)

intermediate 1

NaN3

-

J:

H

)

intermediate 2

� Pd

(a) Draw the structures of intermediates 1 and 2 in this scheme. (b) How do we know that the product is entirely the unnatural D configuration?

R NH2 H o configuration

25

Li pids

25-1 Introd uction

2 5-2 Waxes

1 200

What do the following have in common? An athlete is disqualified from the Olympics for illegal use of anabolic steroids. You spray a bread pan with canola oil to keep the bread from sticking. Your mother is rushed to surgery to remove a gallbladder packed with cholesterol. You wax your shiny new car with carnauba wax. Your father is treat­ ed with a prostaglandin to lower his blood pressure. An artist uses turpentine to thin her brushes after painting the brilliant autumn colors. All these actions involve the use, misuse, or manipulation of lipids. Steroids, prostaglandins, fats, oil, waxes, terpenes, and even the colorful carotenes in the falling leaves are all lipids. In our study of organic chemistry, we have usually classified com­ pounds according to their functional groups. Lipids, however, are classified by their solubility: Lipids are substances that can be extracted from cells and tissues by non­ polar organic solvents. Lipids include many types of compounds containing a wide variety of function­ al groups. You could easily prepare a solution of lipids by grinding a T-bone steak in a blender and then extracting the puree with chloroform or diethyl ether. The resulting solution of lipids would contain a multitude of compounds, many with complex struc­ tures. To facilitate the study of lipids, chemists have divided this large family into two major classes: complex lipids and simple lipids. Complex lipids are those that are easily hydrolyzed to simpler constituents. Most complex lipids are esters of long-chain carboxylic acids called fatty acids. The two major groups of fatty acid esters are waxes and glycerides. Waxes are esters of long-chain alcohols, and glycerides are esters of glycerol. Simple lipids are those that are not easily hydrolyzed by aqueous acid or base. This term often seems inappropriate, because many so-called "simple" lipids are quite complex molecules. We will consider three important groups of simple lipids: steroids, prostaglandins, and terpenes. Figure 25- 1 shows some examples of complex and simple lipids. Waxes are esters of long-chain fatty acids with long-chain alcohols. They occur widely in

nature and serve a number of purposes in plants and animals. Spermaceti (Figure 25- 1 ), found in the head of the sperm whale, probably helps to regulate the animal's buoyancy for deep diving. It may also serve to amplify high-frequency sounds for locating prey. Beeswax is a mixture of waxes, hydrocarbons, and alcohols that bees use to form their honeycomb. Carnauba wax is a mixture of waxes of very high molecular weights. The camauba plant secretes this waxy material to coat its leaves to prevent excessive loss of water by evaporation. Waxes are also found in the protective coatings of insects' exoskeletons, mammals' fur, and birds' feathers. In contrast to these waxes, the "paraffin

25-3 Examples of complex lipids

Triglycerides

1 20 1

Examples of simple lipids

o

II

CH2- O - C - (CH2)1 6CH3

I I

� �

CH - O - C - (CH2)1 6CH3 CH2- O - C - (CH2) 1 6CH3 tristearin, a fat

HO

cholesterol, a steroid .... Figure 25-1

o

II

CH3(CH2)lS - O - C - (CH2 ) 14CH3 spermaceti (cetyl palmitate), a wax

<x-pinene, a terpene

Examples of lipids. Complex lipids contain ester functional groups that can be hydrolyzed to acids and alcohols. Simple lipids are not easily hydrolyzed.

wax" used to seal preserves is not a true wax; rather, it is a mixture of high-molecular­ weight alkanes. o

II

CH3 (CH2)29 - 0-C - (CH2)24C H3 a component of beeswax

o

II

CH3(CH2) 33 - 0 - C - (CH2)26CH3 a component of camauba wax

For many years, natural waxes were used in making cosmetics, adhesives, varnishes, and waterproofing materials. Synthetic materials have now replaced natural waxes for most of these uses. Glycerides are simply fatty acid esters of the trial glycerol. The most common glyc­ 2 5-3 erides are triglycerides (triacylglycerols), in which all three of the glycerol - OH groups have been esterified by fatty acids. For example, tristearin (Figure 25-1) is a Trig lycerides component of beef fat in which all three - OH groups of glycerol are esterified by stearic acid, CH 3 ( CH 2 ) 16COOH. Triglycerides are commonly called fats if they are solid at room temperature and oils if they are liquid at room temperature. Most triglycerides derived from m ammals are fats, such as beef fat or lard. Although these fats are solid at room temperature, the warm body temperature of the living animal keeps them somewhat fluid, allowing for body move­ ment. In plants and cold-blooded animals, triglycerides are generally oils, such as com oil, peanut oil, or fish oil. A fish requires liquid oils rather than solid fats because it would have difficulty moving if its triglycerides solidified whenever it swam in a cold stream. Fats and oils are commonly used for long-term energy storage in plants and ani­ mals. Fat is a more efficient source of long-term energy than carbohydrates because me­ tabolism of a gram of fat releases over twice as much energy as a gram of sugar or starch. An average 70-kg adult male stores about 4000 kJ (about 1 000 kcal) of readily available energy as glycogen (0.2 kg), and about 600,000 kJ (about 140,000 kcal) of long-term en­ ergy as fat (15 kg): enough to supply his resting metabolic needs for nearly three months! The fatty acids of common triglycerides are long, unbranched carboxylic acids with about 1 2 to 20 carbon atoms. Most fatty acids contain even numbers of carbon atoms be­ cause they are derived from two-carbon acetic acid units. Some of the common fatty acids have saturated carbon chains, while others have one or more carbon-carbon double bonds. Plant leaves often have a wax coating to prevent excessive loss of water. Table 25-1 shows the structures of some common fatty acids derived from fats and oils.

1 202

Chapter 25: Lipids TABLE 25-1

Structures and Melting Points of Some Common Fatty Acids

Name

�COOH

Satu.rated acids

lauric acid

M elti ng Point (DC)

Structure

Carbons

12

44

COOH

myristic acid palmitic acid

59

14 COOH

16

64 COOH

stearic acid

70

18 COOH

arachidic acid

76

20

Unsaturated acids

oleic acid

18

COOH

linoleic acid

18

COOH

linolenic acid

18

COOH

eleostearic acid

18

COOH

arachidonic acid

20

COOH

4 -5 -11 49 - 49

PRO B L E M 2 5 - 1

Trimyristin, a solid fat present in nutmeg, is hydrolyzed to give one equivalent of glycerol and three equivalents of myristic acid. Give the structure of trimyristin.

Table 25- 1 shows that saturated fatty acids have melting points that increase gradually with their molecular weights. The presence of a cis double bond lowers the melting point, however. Notice that the 1 8-carbon saturated acid (stearic acid) has a melting point of 70° C, while the 1 8-carbon acid with a cis double bond (oleic acid) has a melting point of 4 C. This lowering of the melting point results from the unsaturated acid's "kink" at the position of the double bond (Figure 25-2). Kinked molecules cannot pack as tightly together in a solid as the uniform zigzag chains of a saturated acid. A second double bond lowers the melting point further (linoleic acid, mp - 5°C) , and a third double bond lowers it still further (linolenic acid, mp - 1 1 C) . The trans dou­ ble bonds in eleostearic acid (mp 49° C) have a smaller effect on the melting point than the cis double bonds of linolenic acid. The geometry of a trans double bond is similar to °

°

COOH

stearic acid, mp 70DC

... Figure 25-2

oleic acid, mp 4°C

Comparison of stearic acid and oleic acid. The cis double bond in oleic acid lowers the melting point by 66° C.

25-3

TrigJ ycerides

1 203

o

II �

C H? - O - C

,,

�� �

CH - O - C

CH 2 - O - C

tristeari n,

�

mp noe

� Figure 25-3 Unsaturated triglycerides have lower melting points because their unsaturated fatty acids do not pack as well in a solid lattice.

the zigzag conformation of a saturated acid, so it does not kink the chain as much as a cis double bond. The melting points of fats and oils also depend on the degree of unsaturation (especially cis double bonds) in their fatty acids. A triglyceride derived from saturated fatty acids has a higher melting point because it packs more easily into a solid lattice than a triglyceride derived from kinked, unsaturated fatty acids. Figure 25-3 shows typical conformations of triglycerides containing saturated and unsaturated fatty acids. Tristearin (mp 72° C) is a saturated fat that packs well in a solid lattice. Triolein (mp -4°C) has the same number of carbon atoms as tristearin, but triolein has three cis double bonds, whose kinked conformations prevent optimum packing in the solid. Most saturated triglycerides are fats because they are solid at room temperature. Most triglycerides with several unsaturations are oils because they are liquid at room temperature. The term polyunsaturated simply means there are several double bonds in the fatty acids of the triglyceride. Most naturally occurring fats and oils are mixtures of triglycerides containing a variety of saturated and unsaturated fatty acids. In general, oils from plants and cold­ blooded animals contain more unsaturations than fats from warm-blooded animals. Table 25-2 gives the approximate composition of the fatty acids obtained from hydrol­ ysis of some common fats and oils. TABLE 2 5-2

Fatty Acid Composition of Some Fats and O i ls. Percent by Weight Satu rated Fatty Acids

Sou rce

La u ric

beef fat

0 0 I 0 0 0 0.2 0 0

lard human fat herring oi l corn oil olive oi l soybean oil canol a oil linseed oil

Myristic

6

I 3 5 I 0. 1 0. 1 0 0.2

U nsaturated Fatty Acids

Pa l m itic

Stearic

Oleic

Linol eic

Linolenic

27 24 27 14 10 7 10 2 7

14 9 8 3 3 2 2 7

49 47 48 0 50 84 29 54 20

2 10 10 0 34 5 51 30 20

0 0 0 30" 0 2 7 7 52

"Contains large amounts o f even more highly unsaturated fatty acids.

1 204

Chapter 25: Lipids

For many years, lard (a soft, white solid obtained by rendering pig fat) was com­ monly used for cooking and baking. Although vegetable oil could be produced more cheaply and in greater quantities, consumers were unwilling to use vegetable oils because they were accustomed to using white, creamy lard. Then vegetable oils were treated with hydrogen gas and a nickel catalyst, reducing some of the double bonds to give a creamy, white vegetable shortening that resembles lard. This "partially hydro­ genated vegetable oil" largely replaced lard for cooking and baking. Margarine is a similar material flavored with butyraldehyde to give it a taste like that of butter. More recently, consumers have learned that "polyunsaturated" vegetable oils are more easily digested, prompting a switch to natural vegetable oils. Some consumers are also concerned with the presence of unnatural trans fatty acids i n "partially hydro­ genated vegetable oil." During the hydrogenation process, the catalyst lowers the acti­ vation energy of both the forward (hydrogenation) and reverse (dehydrogenation) processes. The naturally occurring cis double bonds in vegetable oils can hydrogenate and the products can dehydrogenate. The double bonds end up in random positions, with either cis or trans stereochemistry. The white, creamy product has fewer double bonds overall, but some of the remaining double bonds may be in positions or stereo­ chemical configurations that never occur in nature. P R O B L E M 2 5-2

Give an equation for the complete hydrogenation of triolein using an excess of hydrogen. What is the name of the product, and what are the melting points of the starting material and the product?

Saponification is the base-promoted hydrolysis of the ester linkages in fats and oils

2 5-4

(review Section 2 1-7B). One of the products is soap, and the word sapon ification is derived from the Latin word sapon is, meaning "soap." Saponification was discovered (before 500 B.C.), when it was found that a curdy material resulted when animal fat was heated with wood ashes. Alkaline substances i n the ashes promote hydrolysis of the ester linkages of the fat. Soap is currently made by boiling animal fat or vegetable oil with a solution of sodium hydroxide. The following reaction shows formation of soap from tristearin, a component of beef fat.

Saponification of Fats and Oils; Soaps and Deterg ents

o

CH? - OH

II

I -

CH2 -O- C- (CH2) J 6CH3

I I

�

CH-O- C-(CH2) J 6CH3

�

CH2 -O-C- (CH2) J 6CH3 tristearin, a fat

CH- OH

I

+

CH2-OH glycerol

3 NaOH

o

+

II

3 Na+ - O-C-(CH2 ) J 6CH3 sodium stearate, a soap

Chemically, a soap is the sodium or potassium salt of a fatty acid. The negatively charged carboxylate group is hydrophilic ("attracted to water"), and the long hydrocarbon chain is hydrophobic ("repelled by water") and lipophilic ("attracted to oils"). The electrostatic potential map of the stearate ion is shown in Figure 25-4. Notice the high electron density (red) around the negatively charged carboxylate end of the mol­ ecule. The carboxylate oxygen atoms share the negative charge and participate in strong hydrogen bonding with water molecules. The rest of the molecule (green) is a hydrocar­ bon chain that cannot participate in hydrogen bonding with water.

25-4 Saponification of Fats and Oils; Soaps and Detergents

1 205

Ionic head

Water

boundary of water into structure

Hyd rocarbon tail

.... Figure 25-4

Aggregation of soap in micelles. The electrostatic potential map of a soap molecule shows high electron density in the negatively charged head and medium electron density (green) in the hydrocarbon tail. In water, soap forms a cloudy solution of micelles, with the hydrophilic heads in contact with water and the hydrophobic tails clustered in the interior. The Na+ ions (not shown) are dissolved in the water surrounding the micelle. ---

_ ._---

In water, soap forms a cloudy solution of micelles: clusters of about 1 00 to 200 soap molecules with their polar "heads" (the carboxylate groups) on the surface of the cluster and their hydrophobic "tails" (the hydrocarbon chains) enclosed within. The micelle (Figure 25-4) is an energetically stable particle because the hydrophilic groups are hydrogen-bonded to the surrounding water, while the hydrophobic groups are shielded within the interior of the micelle, interacting with other hydrophobic groups. Soaps are useful cleaning agents because of the different affinities of a soap mol­ ecule's two ends. Greasy dirt is not easily removed by pure water because grease is hy­ drophobic and insoluble in water. The long hydrocarbon chain of a soap molecule dissolves in the grease, with the hydrophilic head at the sllIface of the grease droplet. Once the sllIface of the grease droplet is covered by many soap molecules, a micelle can fonn around it with a tiny grease droplet at its center. This grease droplet is easily suspended in water because it is covered by the hydrophilic carboxylate groups of the soap (Figure 25-5). The resulting mixture of two insoluble phases (grease and water), with one phase dispersed throughout the other in small droplets, is called an emulsion. We say the grease has been emulsified by the soapy solution. When the wash water is rinsed away, the grease goes with it. The usefulness of soaps is limited by their tendency to precipitate out of solution in hard water. Hard water is water that is acidic or that contains ions of calcium, magnesium, or iron. In acidic water (such as the "acid rain" of environmental concern), soap molecules are protonated to the free fatty acids. Without the ionized carboxylate group, the fatty acid floats to the top as a greasy "acid scum" precipitate.

1 206

Chapter 25: Lipids - --------- ----::..-=--_.::..-_:.=:.= --- ,-:..�--=---,==-- --

Wafer

Water

� Figure 25-5 Emulsification of grease. In a soapy solution, grease is emulsified by forming micelles coated by the hydrophilic carboxylate groups of the soap.

o

o

II

CH3(CH2)II - C - O- +Na

+

H+

II

CH3(CHJII - C- OH �

+

Na+

acid scum

a soap

Many areas have household water containing calcium, magnesium, and iron ions. Al­ though these mineral-rich waters can be healthful for drinking, the ions react with soaps to form insoluble salts called hard-water scum. The following equation shows the reaction of a soap with calcium, common in areas where water comes in contact with limestone rocks. o

o

II

2 CH3(CH2)II - C - O- +Na + a soap

Ca+

2

II

[CH3(CH2)n - C -O]2Ca �

+

2 Na+

hard-water scum

P RO B L E M 25-3

Give equations to show the reactions of sodium stearate with + 2+ (b) Mg 2+ (c) Fe 3 (a) Ca P R O B L E M 2 5 -4

Several commercial laundry soaps contain water-softening agents, usually sodium carbonate ( Na2C03 ) or sodium phosphate (Na3 P04 or Na2HP04). Explain how these water-softening agents allow soaps to be used in water that is hard by virtue of its (b) dissolved Ca2+, Mg 2+, and Fe 3+ salts (a) low pH

Soaps precipitate in hard water because of the chemical properties of the carboxylic acid group. Synthetic detergents avoid precipitation by using other functional groups in place of carboxylic acid salts. Sodium salts of sulfonic acids are the most widely used class of synthetic detergents. Sulfonic acids are more acidic than carboxylic acids, so their salts are not protonated, even in strongly acidic wash water. Calcium, magnesium, and iron salts of sulfonic acids are soluble in water, so sulfonate salts can be used in hard water without forming a scum (Figure 25-6). Like soaps, synthetic detergents combine hydrophilic and hydrophobic regions in the same molecule. Hydrophobic regions are generally alkyl groups or

25-4 S aponification of Fats and Oils; S oaps and Detergents

1 207

A n alkylbenzenesulfonate detergent

o

II

S - O- +

II

o

Na+

{ �:2:} Mg- + Fe3+

_

no precipitate

Examples of other types of detergents

o� -

CH3 CH2 -

� L (CH?) 1 5 - CH3 1

CH3

-

CI-

Nonoxynol ® , Ortho Pharmaceuticals

benzylcetyldimethylammonium chloride (benzalkonium chloride)

o

II

O - S - O-

II

sodium dodecyl sulfate (sodium lauryl sulfate)

Na +

o

N-Iauroyl-N-methylglycine, sodium salt Gardol® , Colgate-Palmolive Co.

.. Figure 25-6

Synthetic detergents may have anionic, cationic, or nonionic hydrophilic functional groups. Of these detergents, only Gardol® is a carboxylate salt and forms a precipitate in hard water.

aromatic rings. Hydrophilic regions may contain anionic groups, cationic groups, or nonionic groups containing several oxygen atoms or other hydrogen-bonding atoms. Figure 25-6 shows the electrostatic potential map of a sulfonate detergent, with red (electron-rich) regions around the hydrophilic sulfonate group. Figure 25-6 also shows more examples of anionic, cationic, and nonionic detergents. PRO B L E M 2 5-5

Draw a diagram, similar to Figure 25-5, of an oil droplet emulsified by the alkylbenzenesul­ fonate detergent shown in Figure 25-6. PRO B L E M 25-6

Point out the hydrophilic and hydrophobic regions i n the structures of benzalkonium chlo­ ride, Nonoxynol® , and Gardol® (Figure 25-6). PROBLEM 25-7

The synthesis of the alkylbenzenesulfonate detergent shown i n Figure 25-6 begins with the partial polymerization of propylene to give a pentamer. acidic catalyst

) a pentamer

Show how Friedel-Crafts reactions can convert this pentamer to the final synthetic detergent.

C h apter 25: Lipids

1 208

Phospholipids are lipids that contain groups derived from phosphoric acid. The most common phospholipids are phosphoglycerides, which are closely related to common

25-5 Ph osp h olipid s

fats and oils. A phospho glyceride generally has a phosphoric acid group in place of one of the fatty acids of a triglyceride. The simplest class of phosphoglycerides are phosphatidic acids, which consist of glycerol esterified by two fatty acids and one phosphoric acid group. Although it is often drawn in its acid form, a phosphatidic acid is actually deprotonated at neutral pH.

o

nonpolar. hydrocarbon

II

tails

CH2 -O-C -(CH2 )Il CH3

I I

� �

CH-0-C-(CH2)I/ICH3

CH2 - O-P-OH

I

--- polar head

OH ionized form

a phosphatidic acid

schematic representation

P R O B L E M 2 5 -8

Dipal mitoyl phosphatidyl cho l i n e is one component of h u m a n lu n g su�

Draw the i mportant resonance forms for a phosphatidic acid that has lost

factant, which coats the inner sur­

(a) one proton

faces of the lung membranes and

(b) two protons

p revents them from clinging to­ gether a nd col l a ps i n g . Premature

Many phospholipids contain an additional alcohol esterified to the phosphoric acid group. Cephalins are esters of ethanolamine, and lecithins are esters of choline. Both cephalins and lecithins are widely found in plant and animal tissues.

infants often produce little or no lung surfactant, which can lead to severe difficulty in breathing.

I

o

HO-CH2CH2-N(CH3 )3

ethanolamine

choline

II �

CH2 -O- C

�

CH- O-C

I

o

+

HO- CH2CH2 -NH2

�r

nonpolar

�

+

CH2 -0 - -0-CH2CH2 -NH3 0-

polar

II �

I

CH2 - O-C

o

�

nonpolar

r- -��

CH? -O-P - O - CH?CH?- -N(CH3 )3 _

I

-

polar

0-

a cephal in.

a lecithin,

or phosphatidyl ethanolamine

or phosphatidyl choline

Like phosphatidic acids, lecithins and cephalins contain a polar "head" and two long, nonpolar hydrocarbon "tails." This soap-like structure gives phospholipids some inter­ esting properties. Like soaps, they form micelles and other aggregations with their polar heads on the outside and their nonpolar tails protected on the inside. Another stable form of aggregation is a lipid bilayer, which forms animal cell membranes (Figure 25-7). In a lipid bilayer, the hydrophilic heads coat the two

25-6 Steroids

1 209

0-

hydrophilic surface

I

o= p - o-

exposed to water

en larged

I

�

� CH? I CH I

o

hydrophobic interior

hydrophi l i c surface � Figure 25-7

-

I C=O I CH? I CH? I CH? I I

CH? I -

0

I C=O I CH? 1 CH? 1 CH? I I

Structure of a lipid bila yer membrane. Phosphoglycerides can aggregate into a bilayer membrane with their polar heads exposed to the aqueous solution and the hydrocarbon tails protected within. This lipid bilayer is an important part of the cell membrane.

surfaces of a membrane, and the hydrophobic tails are protected within. Cell mem­ branes contain phosphoglycerides oriented in a lipid bilayer, forming a barrier that restricts the flow of water and dissolved substances. Steroids are complicated polycyclic molecules found in all plants and animals. They are classified as simple lipids because they do not undergo hydrolysis like fats, oils, and waxes do. Steroids encompass a wide variety of compounds, including hormones, emul­ sifiers, and components of membranes. Steroids are compounds whose structures are based on the tetracyclic androstane ring system, shown here. The four rings are designat­ ed A, B, C, and D, beginning with the ring at lower left, and the carbon atoms are num­ bered beginning with the A ring and ending with the two "angular" (axial) methyl groups. 12

18 17 16

H androstane

We have seen (Section 3-16B) that fused ring systems such as androstane can have either trans or cis stereochemistry at each ring junction. A simple example is the geometric isomerism of trans- and cis-decalin shown in Figure 25-8. If you make models of these isomers, you will find that the trans isomer is quite rigid and flat (aside from the ring puckering) . In contrast, the cis isomer is relatively flexible, with the two rings situated at a sharp angle to each other. Each of the three ring junctions is trans in the androstane structure shown above. Most steroids have this all-trans structure, which results in a stiff, nearly flat molecule with the two axial methyl groups perpendicular to the plane. In some steroids, the junction between rings A and B is cis, requiring the A ring to fold down below the rest of the ring system. Figure 25-9 shows the androstane ring system with both trans and cis A-B ling junctions. The B-C and C-D ring junctions are nearly always trans in natural steroids.

2 5-6 Steroid s

1210

Chapter 25 : Lipids ---- - ======-=----

- - --

H -=====(one hydrogen up)

�� H

� Figure 25-8

:::( one hydrogen down)

Cis-trans isomers of decahn. In trans­ decalin, the two bonds to the second ring are trans to one another, and the hydrogens on the junction are also trans. In cis-decalin, the bonds to the second ring are cis, and the junction hydrogens are also cis.

Most steroids have an oxygen functional group ( = 0 or OR) at C3 and some kind of side chain or other functional group at C17. Many also have a double bond from C5 to either C4 or C6. The structures of androsterone and cholesterol serve as examples. Androsterone, a male sex hormone, is based on the simple androstane ring system. Cho­ lesterol is a common biological intermediate and is believed to be the biosynthetic pre­ cursor to other steroids. It has a side chain at C 1 7 and a double bond between C5 and C6. -

R

RO androsterone

R cholesterol

The principal sex hormones have been characterized and studied extensively. Testosterone is the most potent of the natural male sex hormones, and estradiol is the most

HO

a trans A-B steroid

H HO

a cis A-B steroid H

... Figure 25-9

Common steroids may have either a cis or a trans A-B ring j unction. The other ring junctions are normally trans.

121 1

25-6 Steroids

potent natural female hormone. Notice that the female sex hormone differs from the male hormone by its aromatic A ring. For the A ring to be aromatic, the C 1 9 methyl group must be lost. In mammals, testosterone is converted to estradiol in the female's ovaries, where enzymes remove C 1 9 and two hydrogen atoms to give the aromatic A ring.

H HO

H

estradiol

testosterone

These gallstones, shown here within the gallbladder, are composed mostly of cholesterol.

PRO B L E M 2 5-9

How would you use a simple extraction to separate a mixture of testosterone and estradiol?

When steroid hormones were first isolated, people believed that no synthetic hormone could rival the astonishing potency of natural steroids. In the past 20 years, however, many synthetic steroids have been developed. Some of these synthetic hormones are hundreds or thousands of times more potent than natural steroids. One example is ethynyl estradiol, a synthetic female hormone that is more potent than estradiol. Ethynyl estradiol is a common i ngredient in oral contraceptives.

The biological activities of steroid hormones result from i nteractions with specific receptors. For exam­ ple. estradiol i nteracts with the es­ trogen

receptor.

Some

i ndustrial

chemicals. such as DDT and poly­ chlorinated biphenyls. may possi­ bly i nteract with these receptors and

cause

hormonal

effects

on

peop l e and wildl ife.

HO

ethynyl estradiol

Some of the most important physiological steroids are the adrenocortical hor­ mones, synthesized by the adrenal cortex. Most of these hormones have either a car­ bonyl group or a hydroxyl group at C l l of the steroid skeleton. The principal adrenocortical hormone is cortisol, used for the treatment of i nflammatory diseases of the skin (psoriasis), the j oints (rheumatoid arthritis), and the lungs (asthma). Figure 25-10 compares the structure of natural cortisol with two synthetic

CHpH

CHpH

\

\

CH3 c = o I ' ' ' ' OH CH3 " H

CH3 C = O I ' ' ' ' OH

I

o

o

o

H cortisol ..... Figure 25-10

,

F

f1uocinolone acetonide

beclomethasone

Cortisol is the major natural hormone of the adrenal cortex, Fluocinolone acetonide is more potent for treating skin inflammation, and beclomethasone is more potent for treating asthma,

1212

Chapter 2S : Lipids corticoids: fluocinolone acetonide, a fluorinated synthetic hormone that is more potent than cortisol for treating skin inflammation; and beclomethasone, a chlorinat­ ed synthetic hormone that is more potent than cortisol for treating asthma. PRO BLEM 25- 1 0

Draw each molecule in a stable chair conformation, and tell whether each red group i s axial or equatorial.

o

(c)

(d) HO

H

H H digitoxigenin, a cardiac stimulant

androsterone

2 5-7 Prostagland ins

Prostaglandins are fatty acid derivatives that are even more powerful biochemical

regulators than steroids. They are called prostaglandins because they were first isolat­ ed from secretions of the prostate gland. They were later found to be present in all body tissues and fluids, usually in minute quantities. Prostaglandins affect many body systems, including the nervous system, smooth muscle, blood, and the reproductive system. They play important roles in regulating such diverse functions as blood pressure, blood clotting, the allergic inflammatory response, activity of the digestive system, and the onset of labor. Prostaglandins have a cyclopentane ring with two long side chains trans to each other, with one side chain ending in a carboxylic acid. Most prostaglandins have 20 carbon atoms, numbered as follows: H 10 I I

12

H

I

7

13

14

15

16

17

18

COOH CH3 20

19

Many prostaglandins have hydroxyl groups on C I I and C I S , and a trans double bond between C13 and C I 4. They also have a carbonyl group or a hydroxyl group on C9. If there is a carbonyl group at C9, the prostaglandin is a member of the E series. If there is a hydroxyl group at C9, it is a member of the F series, and the symbol a means the

25-8 Terpenes

�

OH

2 °2

COOH

1213

reduction

enzyme

arachidonic acid

PGF2a

� Figure 25-1 1

Biosynthesis of prostaglandins begins with an enzyme-catalyzed oxidative cyclization of arachidonic acid.

hydroxyl group is directed down. Many prostaglandins have a cis double bond between C5 and C6. The number of double bonds is also given in the name, as shown here for two common prostaglandi ns.

Despite th e i r range of activities, many

naturally

occurring

prosta­

glandins would make poor drugs because they are rapidly converted to i nactive products. Synthetic mod­ ifications can prolong their activity.

COOH ''' HO '"' H H II

13

H



H

H H

" /

C=C

/ "

H

I / R - C- C+ " I

H

ethylene

growi ng chain

R+

"



CH3 (good) CH 3

tertiary carbocation

Poor monomers for cationic polymerization

H

( "ood)

benzylic carbocation H

growing chain

H

H (poor) H

primary carbocation H

CN ----i>

H

acry lonitri Ie

I / R - C- C+ " I

H

CN (poor) H

destabilized carbocation

A major difference between cationic and free-radical polymerization is that the cationic process needs a monomer that forms a relatively stable carbocation when it reacts with the cationic end of the growing chain. Some monomers form more stable intermediates than others. For example, styrene and isobutylene undergo cationic polymerization easily, while ethylene and acrylonitrile do not polymerize well under these conditions. Figure 26-2 compares the intermediates involved in these cationic polymerizations. PROBLEM 26-4

The mechanism given for cationic polymerization of isobutylene (Mechanism 26-2) shows that all the monomer molecules add with the same orientation, giving a polymer with methyl groups on alternate carbon atoms of the chain. Explain why no isobutylene molecules add with the opposite orientation. PRO B L E M 26-5

Suggest which of the following monomers might polymerize well on treatment with BF3. (b) propylene (c) methyl a-cyanoacrylate (a) vinyl chloride PROBLEM 26-6

Chain branching occurs i n cationic polymerization much as it does in free-radical polymer­ ization. Propose a mechanism to show how branching occurs in the cationic polymerization of styrene. Suggest why isobutylene might be a better monomer for cationic polymerization than styrene.

... Figure 26-2 Cationic polymerization requires relatively stable carbocation intermediates.

1 228

Chapter 26: Synthetic Polymers 26-2C

A n i o n i c Polymerization

Anionic polymerization occurs through carbanion intermediates. Effective anionic

polymerization requires a monomer that gives a stabilized carbanion when it reacts with the anionic end of the growing chain. A good monomer for anionic polymerization should contain at least one strong electron-withdrawing group such as a carbonyl group, a cyano group, or a nitro group. The following reaction shows the chain-lengthening step i n the polymerization of methyl acrylate. Notice that the chain-growth step of an anionic polymerization is simply a conjugate addition to a Michael acceptor (Section 22- 1 8). Chain-growth step in anionic polymerization

growing chain

polymer

stabilized anion

methyl acrylate PROBLEM 26-7

Draw the important resonance forms of the stabilized anion formed i n the anionic polymer­ ization of methyl acrylate.

Anionic polymerization is usually initiated by a strong carbanion-like reagent such as an organolithium or Grignard reagent. Conjugate addition of the initiator to a monomer molecule starts the growth of the chain. Under the polymerization condi­ tions, there is no good proton source available, and many monomer units react before the carbanion is protonated. The following reaction shows a butyllithium-initiated anionic polymerization of acrylonitrile to give Orlon® . M EC H A N I S M 26-3 Initiation step:

A n i o m i c Polymeriza:l!ion

The initiator adds to the monomer to form an anion. 5-

Bu

H

�C 5+

H

_ _ _

{� I

H

_C

::

CN I / Bu- C-C : "H I H

H

acrylonitrile

butyllithium Propagation step:

\'viC

H

CN

{

Li+

stabilized anion

Another molecule of monomer adds to the chain.

/ � C=C v "H H/ "H

growing chain

CN

_ _ _

acrylonitrile

� _ CN � _ H� _

H

I

H

I

H

I

C

H

� CN "H

elongated chain

- - -

f; �l H

-

H

polymer

11

---

P R O B L E M 2 6-8

Methyl a-cyanoacrylate (Super Glue) is easily polymerized, even by weak bases. Draw a mechanism for its base-catalyzed polymerization, and explain why this polymerization goes so quickly and easily.

H" / COOCH3 /C = C " H CN

methyl a-cyanoacrylate

26-3 Stereochemjstry of Polymers PROBLEM 26-9

Chain branching i s not as common with anionic polymerization as it i s with free-radical polymerization and cationic polymerization. (a) Propose a mechanjsm for chain branching in the polymerization of acrylonitrile. (b) Compare the relative stab i lities of the intermediates in this mechanism with those you drew for chain branching i n the cationic polymerization of styrene (Problem 26-6). Explain why chain branching is less common in this anionic polymerization.

Chain-growth polymerization of alkenes usually gives a head-to-tail bonding arrange­ ment, with any substituent(s) appearing on alternate carbons of the polymer chain. This bonding arrangement i s shown here for a generic polyalkene. Although the poly­ mer backbone is joined by single bonds (and can undergo conformational changes), it is shown in the most stable all-anti conformation. polymerize )

[�] H R H R H R H R

The stereocherll i stry of the side groups (R) i n the polymer has a major effect on the polymer's properties. The polymer has many chirality centers, raising the possibility of millions of stereoisomers. Polymers are grouped into three classes, according to their predominant stereocherlli s try. If the side groups are generally on the same side of the polymer backbone, the polymer is called isotactic (Greek, iso, meaning "same," and tactic, meaning "order") . If the side groups generally alternate from one side to the other, the polymer is called syndiotactic (Greek, meaning "alternating order") . If the side groups occur randomly on either side of the polymer backbone, the polymer is called atactic (Greek, meaning "no order"). In most cases, isotactic and syndiotactic polymers have enhanced strength, clarity, and thermal properties over the atactic form of the polymer. Figure 26-3 shows these three types of polymers. ===================--=-======. --

An isotactic polymer (side g roups on the sam.e side of the backbone)

H R

H R H R

H R

- ;-�R H� �R J =

H �R

H

A syndiotactic polym.er (side groups on alternating sides of the backbone)

H R

1"-'--. R H H R

///-.

R H H R

An atactic polymer (side groups on random sides of the backbone)

H R

,�.

H R

1"-'--. F. R H H R

A Figure 26-3 Three stereochemical types of addition polymers.

26-3 Stereoch em istry of Polymers

1 229

1 230

Chapter 26: Synthetic Polymers P R O B L E M 26- 1 0

Draw the structures of isotactic poly(acrylonitrile) and syndiotactic polystyrene.

26-4 Stereoch em ical Control of Polymerization; Zieg l er-Natta Catalysts

For any particular polymer, the three stereochemical forms have distinct properties. In most cases, the stereoregular isotactic and syndiotactic polymers are stronger and stiffer because of their greater crystallinity (a regular packing arrangement). The con­ ditions used for polymerization often control the stereochemistry of the polymer. Anionic polymerizations are the most stereoselective; they usually give isotactic or syndiotactic polymers, depending on the nature of the side group. Cationic polymer­ izations are often stereoselective, depending on the catalysts and conditions used. Free-radical polymerization is nearly random, resulting in branched, atactic polymers. In 1 95 3 , Karl Ziegler and Giulio Natta discovered that aluminum-titanium initiators catalyze the polymerization of alkenes, with two major advantages over other catalysts: 1. The polymerization is completely stereoselective. Either the isotactic form or the

syndiotactic form may be made, by selecting the proper Ziegler-Natta catalyst.

2. Because the intermediates are stabilized by the catalyst, very little hydrogen

abstraction occurs. The resulting polymers are linear with almost no branching.

A Ziegler-Natta catalyst is an organometallic complex, often containing titani­ um and aluminum. A typical catalyst is formed by adding a solution of TiCl4 (titanium tetrachloride) to a solution of ( CH3CH2 ) 3Al (triethyl aluminum). This mixture is then "aged" by heating it for about an hour. The precise structure of the active catalyst is not known, but the titanium atom appears to form a complex with both the growing polymer chain and a molecule of monomer. The monomer attaches to the end of the chain (which remains complexed to the catalyst), leaving the titanium atom with a free site for complexation to the next molecule of monomer. With a Ziegler-Natta catalyst, a high-density polyethylene (or linear polyethylene) can be produced with almost no chain branching and with much greater strength than common low-density polyethylene. Many other polymers are produced with improved properties using Ziegler-Natta catalysts. In 1 963, Ziegler and Natta received the Nobel Prize for their work, which had revolutionized the polymer industry in only ten years.

26-5 Natural and Synth etic Rubbers

Natural rubber is isolated from a white fluid, called latex, that exudes from cuts in the bark of Hevea brasiliensis, the South American rubber tree. Many other plants secrete this polymer, as well. The name rubber was first used by Joseph Priestly, who used the crude material to "rub out" errors in his pencil writing. Natural rubber is soft and sticky. An enterprising Scotsman named Charles Macintosh found that rubber makes a good waterproof coating for raincoats. Natural rubber i s not strong or elastic, howev­ er, so its uses were limited to waterproofing cloth and other strong materials. Like many other plant products, natural rubber is a terpene composed of isoprene units (Section 25-8). If we imagine lining up many mol­ ecules of isoprene in the s-cis conformation, and moving pairs of electrons as shown in the following figure, we would produce a structure similar to natural rubber. This poly­ mer results from l ,4-addition to each isoprene molecule, with all the double bonds in the cis configuration. Another name for natural rubber is cis- l ,4-polyisoprene. Structure of Natural Ru bber

Imaginary polymerization of isoprene units

26-5 Natural and Synthetic Rubbers

1231

Natural rubber

The cis double bonds in natural rubber force it to assume a kinky conformation that may be stretched and still return to its shorter, kinked structure when released. Unfortunately, when we pull on a mass of n atural rubber, the chains slide by each other and the material pulls apart. This is why natural rubber is not suitable for uses requiring strength or durability. Vulcanization: Cross-Linking of Rubber In 1 839, Charles Goodyear acciden­ tally dropped a mixture of natural rubber and sulfur onto a hot stove. He was surprised to find that the rubber had become strong and elastic. This discovery led to the process that Goodyear called vulcanization, after the Roman god of fire and the volcano. Vul­ canized rubber has much greater toughness and elasticity than natural rubber. It with­ stands relatively high temperatures without softening, and it remains elastic and flexible when cold. Vulcanization also allows the casting of complicated shapes such as rubber tires. Natural rubber is putty-like, and it is easily mixed with sulfur, formed around the tire cord, and placed into a mold. The mold is closed and heated, and the gooey mass of string and rubber is vulcanized into a strong, elastic tire carcass. On a molecular level, vulcanization causes cross-linking of the cis- l ,4-poly­ isoprene chains through disulfide ( - S - S - ) bonds, similar to the cysteine bridges that link peptides (Section 24-8C). In vulcanized rubber, the polymer chains are linked together, so they can n o longer slip past each other. When the material is stressed, the chain s stretch, but cross-linking prevents tearing. When the stress is released, the chains return to their shortened, kinky conformations as the rubber snaps back. Figure 26-4 shows the structure of rubber before and after vulcanization . Rubber can be prepared with a wide range of physical properties by controlling the amount of sulfur used in vulcanization. Low-sulfur rubber, made with about 1 to 3% sulfur, is soft and stretchy. It is good for rubber bands and inner tubes. Medium­ sulfur rubber (about 3 to 1 0% sulfur) is somewhat harder, but still flexible, making good tires. High-sulfur rubber (20 to 30% sulfur) is called hard rubber and was once used as a hard synthetic plastic.

s ----;> heat

White latex drips out of cuts in the bark of a rubber tree in a Malaysian rubber plantation.

.... Figure 26-4

Vulcanization of rubber introduces disulfide cross-links between the polyisoprene chains. Cross-linking forms a stronger, elastic material that does not pull apart when it is stretched. --- -.---==---...:::---:-�---�-- -=======

1232

Chapter 26: Sy n theti c

Polymers P ROBL EM 26-11

Draw the structure of gutta-percha, a natural rubber with all its double bonds in the trans configuration. (b) Suggest why gutta-percha is not very elastic, even after it is vulcanized.

(a)

Synthetic Rubber There are many different formulations for synthetic rubbers, but the simplest is a polymer of 1 ,3-butadiene. Specialized Ziegler-Natta catalysts can produce 1 ,3-butadiene polymers where 1 ,4-addition has occurred on each butadiene unit and the remaining double bonds are all cis. This polymer has properties similar to those of natural rubber, and it can be vulcanized in the same way.

����� l,4-polymerization of 1,3-butadiene Wallace Carothers, the inventor of nylon, stretches a piece of synthetic rubber in his laboratory at the DuPont company.

26-6

Copolymers of Two or M o re Monomers

cis-! ,4-polybutadiene

All the polymers we have discussed are homopolymers, polymers made up of identi­ cal monomer units. Many polymeric materials are copolymers, made by polymeriz­ ing two or more different monomers together. In many cases, monomers are chosen so that they add selectively in an alternating manner. For example, when a mixture of vinyl chloride and vinylidene chloride ( l ,l-dichloroethylene) is induced to polymer­ ize, the growing chain preferentially adds the monomer that is not at the end of the chain. This selective reaction gives the alternating copolymer Saran®, used as a film for wrapping food. Overall reaction H H

"-

/

C=C

/

"-

CI

+

---

H

vinyl chloride

vinylidene chloride

f

CH 2

l

�� -� --I

CH2

H

Saran®

I

CII/

Three or more monomers may combine to give polymers with desired proper­ ties. For example, acrylonitrile, butadiene, and styrene are polymerized to give ABS plastic, a strong, tough, and resilient material used for bumpers, crash helmets, and other articles that must withstand heavy impacts. P ROBL E M 26-1 2

Isobutylene and isoprene copolymerize to give "butyl rubber." Draw the structure of the repeating unit in butyl rubber, assuming that the two monomers alternate.

26-7

Condensation Polymers

Condensation polymers result from formation of ester or amide linkages between difunc­ tional molecules. The reaction is called step-growth polymerization. Any two monomer molecules may react to form a dimer, dimers may condense to give tetramers, and so on. Each condensation is an individual step in the growth of the polymer, and there is no chain reaction. Many kinds of condensation polymers are known. We discuss the four most common types: polyamides, polyesters, polycarbonates, and polyurethanes.

26-7 C ondensation Polymers

26-7A

1233

Polyamides: Nylon

When Wallace Carothers of DuPont discovered nylon in 1 935, he opened the door to a new age of fibers and textiles. At that time, thread used for clothing was made of spun animal and plant fibers. These fibers were held together by friction or sizing, but they were weak and subject to rotting and unraveling. Silk (a protein) was the strongest fiber known at the time, and Carothers reasoned that a polymer bonded by amide link­ ages might approach the strength of silk. Nylon proved to be a completely new type of fiber, with remarkable strength and durability. It can be melted and extruded into a strong, continuous fiber, and it cannot rot. Thread spun from continuous nylon fibers is so much stronger than natural materials that it can be made much thinner. Availability of this strong, thin thread made possible stronger ropes, sheer fabrics, and nearly invisible women's stockings that came to be called "nylons." Nylon is the common name for polyamides. Polyamides are generally made from reactions of diacids with diamines. The most common polyamide is called nylon 6,6 because it is made by reaction of a six-carbon diacid (adipic acid) with a six-carbon diamine. The six-carbon diamine, systematically named 1,6-hexanediamine, is com­ monly called hexamethylene diamine. When adipic acid is mixed with hexamethylene diarnine, a proton-transfer reaction gives a white solid called nylon salt. When nylon salt is heated to 2500 C, water is driven off as a gas, and molten nylon results. Molten nylon is cast into a solid shape or extruded through a spinneret to produce a fiber. o

o

o

II

Scanning electron micrograph of the material in a nylon stocking. Sheer stockings require long, continuous fibers of small diameter and enormous strength. (Magnification lS0X.)

o

II

II

HO-C-(CH2)4-C -OH

+

+

adipic acid

(CH,l,

+

H3N - (CH2)6- NH3

hexamethylene diamine

---!- -!t

II

-O - C - (CH2)4- C - O-

nylon salt

!

NH - (CH,l, -NH - -(CH,l, -

�_-H'O !1

NH - (CH,l. - NH

--­

poly(hexamethylene adipamide), called nylon 6,6

Nylon can also be made from a single monomer having an amino group at one end and an acid at the other. This reaction is similar to the polymerization of a-amino acids to give proteins. Nylon 6 is a polymer of this type, made from a six-carbon amino acid: 6-aminohexanoic acid (e-aminocaproic acid). This synthesis starts with e-caprolactam. When caprolactam is heated with a trace of water, some of it is hydrolyzed to the free amino acid. Continued heating gives condensation and poly­ merization to molten nylon 6. Nylon 6 (also called Perlon®) is used for making strong, flexible fibers for ropes and tire cord.

6

H

-!-

II

H3N- (CH2)s- C - O-

B-caprolactam

- -- NH-(CIl,),

o

+

�-H'O -!± -!--B-aminocaproic acid

!t

NH-(CH,\ -

NH-(CH,l,

/I

NH - (CH,l,

poly(6-aminohexanoic acid), called nylon 6 or Perlon®

Kevlar® body armor works by "catching" a bullet in a multilayer web of woven fabrics. Kevlar's extraordinary strength resists tearing and cutting, allowing the stressed fibers to absorb and disperse the impact to other fibers in the fabric.

1234

Chapte r 26: Sy nt hetic

Polymers P ROBL E M 2 6-1 3

Nomex®, a strong fire-resistant fabric, is a polyamide made from meta-phthalic acid and meta-diaminobenzene. Draw the structure of Nomex®. (b) Kevlar®, made from terephthalic acid (para-phthalic acid) and para-diaminobenzene, is used in making tire cord and bulletproof vests. Draw the structure of Kevlar®. (a)

26-7B

Test inflation of an Echo satellite in a blimp hangar at Weeksville, NC on August 5, 1965 .

Polyesters

The introduction of polyester fibers has brought about major changes in the way we care for our clothing. Nearly all modern permanent-press fabrics owe their wrinkle­ free behavior to polyester, often blended with other fibers. These polyester blends have reduced or eliminated the need for starching and ironing clothes to achieve a wrinkle-free surface that holds its shape. The most common polyester is Dacron®, the polymer of terephthalic acid (paraphthalic acid or benzene-l ,4-dicarboxylic acid) with ethylene glycol. In principle, this polymer might be made by mixing the diacid with the glycol and heating the mixture to drive off water. In practice, however, a better product is obtained using a transester­ ification process (Section 2 1 -5). The dimethyl ester of terephthalic acid is heated to about 1 500 with ethylene glycol. Methanol is evolved as a gas, driving the reaction to completion. The molten product is spun into Dacron® fiber or cast into Mylar® film.

C 0CHP-C�C-OCH3 ·J-o-�fO-CH'CH'-oJ-o- qo-CH,CH, 0 11

o

11-

dimethyl terephthalate

heat, loss of CH30H ) -OCH3 catalyst

ethylene glycol

-0-·

poly(ethylene terephthalate) or PET, also called Dacron® polyester or Mylar® film

Dacron ® fiber is used to make fabric and tire cord, and Mylar® film is used to ® make magnetic recording tape. Mylar film is strong, flexible, and resistant to ultravi­ ® olet degradation. Aluminized Mylar was used to make the Echo satellites, huge bal­ loons that were put into orbit around the Earth as giant reflectors in the early 1 960s. Poly(ethylene terephthalate) is also blow-molded to make plastic soft-drink bottles that are sold by the billions each year. P ROBLE M 2 6-1 4

A polyester of €-caprolactone is

used to make bio-absorbable mate­ rials for use in the body. For exam­ ple,

suture

thread

for

surgical

stitches can be made from poly (€-caprolactone) .

Kodel® polyester is formed by transesterification of dimethyl terephthalate with 1,4-di­ (hydroxymethyl)cyclohexane. Draw the structure of Kodel®. P ROBLE M 2 6 -15

Glyptal® resin makes a strong, solid polymer matrix for electronic parts. Glyptal® is made from terephthalic acid and glycerol. Draw the structure of Glyptal®, and explain its remark­ able strength and rigidity. 26-7C

Polycarbonates

A carbonate ester is simply an ester of carbonic acid. Carbonic acid itself exists in equi­ librium with carbon dioxide and water, but its esters are quite stable (Section 2 1 - 1 6).

HO-C-OH o

II

carbonic acid

R-O-C-O-R' o

II

a carbonate ester

26-7 Condensation Po lym ers

Carbonic acid is a diacid; with suitable diols, it can form polyesters. For example, when phosgene (the acid chloride of carbonic acid) reacts with a diol, the product is a poly(carbonate ester). The following equation shows the synthesis of Lexan® polycar­ bonate: a strong, clear, and colorless material that is used for bulletproof windows and crash helmets. The diol used to make Lexan ® is a phenol called bisphenol A, a common intermediate in polyester and polyurethane synthesis. o

II

CI-C-Cl

phosgene

+

CH3 H°-o-?VOH CH3

Polycarbonate

is

a

1235

tough,

clear

material that withstands repeated sterilization.

These

properties

account for its wide use in medical devices such as blood filters, surgi­ cal instruments, and intravenous line components.

bisphenol A

Lexan® polycarbonate P ROBL E M 2 6 -1 6

Propose a mechanism for the reaction of bisphenol A with phosgene. P ROBL E M 2 6 -1 7

Bisphenol A is made on a large scale by a condensation of phenol with acetone. Suggest an appropriate catalyst, and propose a mechanism for this reaction. (Hint: This is a condensation because three molecules are joined with loss of water. The mechanism belongs to another class of reactions, though.)

26-70

Polyurethanes

(R-NH-COOH),

A urethane (Section 2 1 - 1 6) is an ester of a carbamic acid a half­ amide of carbonic acid. Carbamic acids themselves are unstable, quickly decomposing to amines and Their esters (urethanes) are quite stable, however.

CO2. R-NH-C-OH o

II

a carbamic acid

�

amine

ments. "Plastic condoms" made of

II

R-NH-C-O-R'

+

a urethane or carbamate ester

Because carbamic acids are unstable, normal esterification procedures cannot be used to form urethanes. Urethanes are most commonly made by treating an isocyanate with an alcohol or a phenol. The reaction is highly exothermic, and it gives a quantitative yield of a carbamate ester.

R-N=C=O isocyanate

+

Example

-N�C�O HO-CH,CH, +

phenyl isocyanate

ethanol

HO-R' alcohol

deteriorate over time or in contact with oil-based lubricants or oint­

o

R-NH2 CO2

Latex condoms cause severe aller­ gies for some people, and they

polyurethane cause fewer allergic reactions. These polyurethane ma­ terials can be used with oil- or water-based

lubricants,

and

deteriorate less in storage.

o

II

R-NH-C-O-R' carbamate ester (urethane) o

-NH-�-O-CH,CH' ethyl-N-phenylcarbamate

they

1236

Chapter 26: Synthetic Polymers P ROBLEM 2 6-1 8

Propose a mechanism for the reaction of phenyl isocyanate with ethanol. A polyurethane results when a diol reacts with a diisocyanate, a compound with two isocyanate groups. The compound shown next, commonly called toluene diiso­ cyanate, is frequently used for making polyurethanes. When ethylene glycol or anoth­ er diol is added to toluene diisocyanate, a rapid condensation gives the polyurethane. Low-boiling liquids such as butane are often added to the reaction mixture. Heat evolved by the polymerization vaporizes the volatile liquid, producing bubbles that convert the viscous polymer to a frothy mass of polyurethane foam.

O�C�N

�

N�C�O + CH3

HO -CH2CH2-OH

ethylene glycol

toluene diisocyanate

o

II

H O- CH'CH'-O - C - N

�

a polyurethane

0

II H N-C CH3

7 11

P ROBL E M 2 6-1 9

Explain why the addition of a small amount of glycerol to the polymerization mixture gives a stiffer urethane foam. P ROBLE M 2 6- 2 0

Give the structure of the polyurethane formed by the reaction of toluene diisocyanate with bisphenol A.

26-8

Polymer Struct u re a n d Properties

Although polymers are very large molecules, w e can explain their chemical and physi­ cal properties in terms of what we already know about smaller molecules. For example, when you spill a base on your polyester slacks, the fabric is weakened because the base hydrolyzes some of the ester linkages. The physical properties of polymers can also be explained using concepts we have already encountered. Although polymers do not crys­ tallize or melt quite like smaller molecules, we can detect crystalline regions in a poly­ mer, and we can measure, the temperature at which these crystallites melt. In this section, we consider briefly some of the important aspects of polymer crystallinity and thermal behavior.

26-SA

Polymer Crystallinity

Polymers rarely form the large crystals characteristic of other organic compounds, but many do form microscopic crystalline regions called crystallites. A highly regular polymer that packs well into a crystal lattice will be highly crystalline, and it will gen­ erally be denser, stronger, and more rigid than a similar polymer with a lower degree of crystallinity. Figure 26-5 shows how the polymer chains are arranged in parallel lines in crystalline areas within a polymer. Polyethylene provides an example of how crystallinity affects a polymer's physical properties. Free-radical polymerization gives a highly branched, low-density polyethylene that forms very small crystallites because the random chain branching

26-8

Polymer Structure and Properties

1237

-------"-----------

.... Figure 26-5

------

Crystallites are areas of crystalline structure within the large mass of a solid polymer.

destroys the regularity of the crystallites. An unbranched, high-density polyethylene is made using a Ziegler-Natta catalyst. The linear structure of the high-density mate­ rial packs more easily into a crystal lattice, so it forms larger and stronger crystallites. We say that high-density polyethylene has a higher degree of crystallinity, and it is therefore denser, stronger, and more rigid than low-density polyethylene. Stereochemistry also affects the crystallinity of a polymer. Stereoregular isotac­ tic and syndiotactic polymers are generally more crystalline than atactic polymers. B y careful choice o f catalysts, we can make a linear polymer with either isotactic o r syn­ diotactic stereochemistry.

26-88

Thermal Properties

At low temperatures, long-chain polymers are glasses. They are solid and unyielding, and a strong impact causes them to fracture. As the temperature is raised, the polymer goes through a glass transition temperature, abbreviated Tg. Above Tg, a highly crys­ talline polymer becomes flexible and moldable. We say it is a thermoplastic because application of heat makes it plastic (moldable). As the temperature is raised further, the polymer reaches the crystalline melting temperature, abbreviated Tm. At this temper­ ature, crystallites melt and the individual molecules can slide past one another. Above Tm, the polymer is a viscous liquid and can be extruded through spinnerets to form fibers. The fibers are immediately cooled in water to form crystallites and then stretched (drawn) to orient the crystallites along the fiber, increasing its strength. Long-chain polymers with low crystallinity (called amorphous polymers) become rubbery when heated above the glass transition temperature. Further heating causes them to grow gummier and less solid until they become viscous liquids without definite melting points. Figure 26-6 compares the thermal properties of crystalline and amorphous long-chain polymers.

Drugs are sometimes enclosed in water-soluble polymers to control

the rate of drug release. The poly­ mer breaks down in the body over

time at a predictable rate and grad­ ually releases the drug.

------ --

t �

�

.... ..,

liquid

f------i

T,II

(sharp)

thermoplastic

Ef---.,.,��---ITg 2:l glass

crystalline polymer

t

.., .... ::J

� ....

liquid gummy

.., 0.

E 2:l

broad

rubber.y

gla ss

Tg

amorphous polymer

T,II .... Fig u re 26- 6

Crystalline and amorphous long­ chain polymers show different physical properties when they are heated.

1238

Chapter

26:

Synthetic Polymers These phase transitions apply only to long-chain polymers. Cross-linked poly­ mers are more likely to stay rubbery, and they may not melt until the temperature is so high that the polymer begins to decompose.

26-8C

Plasticizers

In many cases, a polymer has desirable properties for a particular use, but it is too brit­ tle-either because its glass transition temperature (Tg) is above room temperature or because the polymer is too highly crystalline. In such cases, addition of a plasticizer often makes the polymer more flexible. A plasticizer is a nonvolatile liquid that dis­ solves in the polymer, lowering the attractions between the polymer chains and allow­ ing them to slide by one another. The overall effect of the plasticizer is to reduce the crystallinity of the polymer and lower its glass transition temperature (Tg). A common example of a plasticized polymer is poly(vinyl chloride). The com­ mon atactic form has a Tu of about 80° C, well above room temperature. Without a plasticizer, "vinyl" is stiff and brittle. Dibutyl phthalate (see the structure below) is added to the polymer to lower its glass transition temperature to about 0° C. This plas­ ticized material is the flexible, somewhat stretchy film we think of as vinyl raincoats, shoes, and even inflatable boats. Dibutyl phthalate is slightly volatile, however, and it gradually evaporates. The soft, plasticized vinyl gradually loses its plasticizer and becomes hard and brittle.

Chapter 26 Glossary

addition polymer (chain-growth polymer) A polymer that results from the rapid addition of one molecule at a time to a growing polymer chain, usually with a reactive intermediate (cation, radical, or anion) at the growing end of the chain. (p. 1222) amorphous polymer A long-chain polymer with low crystallinity. (p. 1237) anionic polymerization The process of forming an addition polymer by chain-growth polymerization involving an anion at the end of the growing chain. (p. 1228) atactic polymer A polymer with the side groups on random sides of the polymer backbone. (p. 1229) cationic polymerization The process of forming an addition polymer by chain-growth polymerization involving a cation at the end of the growing chain. (p. 1226) chain-growth polymer See addition polymer. (p. 1223) condensation polymer (step-growth polymer) A polymer that results from condensation (bond formation with loss of a small molecule) between the monomers. In a condensation polymerization, any two molecules can condense, not necessarily at the end of a growing chain. (p. 1232) copolymer A polymer made from two or more different monomers. (p. 1232) crystalline melting temperature (Tm) The temperature at which melting of the crystal­ lites in a highly crystalline polymer occurs. Above Tm, the polymer is a viscous liquid. (p. 1237) crystallinity The relative amount of the polymer that is included in crystallites, and the rela­ tive sizes of the crystallites. (p. 1236) crystallites Microscopic crystalline regions found within a solid polymer below the crystalline melting temperature. (p. 1236) free-radical polymerization The process of forming an addition polymer by chain-growth polymerization involving a free radical at the end of the growing chain. (p. 1223)

Chapter 26 Glossary glass transition temperature (Tg) The temperature above which a polymer becomes rubbery or flexible. (p. 1237) homopolymer A polymer made from identical monomer units. (p. 1232) isotactic polymer A polymer with all the side groups on the same side of the polymer back­ bone. (p. 1229) monomer One of the small molecules that bond together to form a polymer. (p. 1222) nylon The common name for polyamides. (p. 1233) plasticizer A nonvolatile liquid that is added to a polymer to make it more flexible and less brittle below its glass transition temperature. In effect, a plasticizer reduces the crystallinity of a polymer and lowers To. (p. 1238) polyamide (nylon) A polymer whose repeating monomer units are bonded by amide linkages, much like the peptide linkages in protein. (p. 1233) polycarbonate A polymer whose repeating monomer units are bonded by carbonate ester link­ ages. (p. 1 234) polyester A polymer whose repeating monomer units are bonded by carboxylate ester link­ ages. (p. 1234) polymer A large molecule composed of many smaller units (monomers) bonded together. (p. 1222) polymerization The process of linking monomer molecules into a polymer. (p. 1222) polyurethane A polymer whose repeating monomer units are bonded by urethane (carbamate ester) linkages. (p. 1 235) rubber A natural polymer isolated from the latex that exudes from cuts in the bark of the South American rubber tree. Alternatively, synthetic polymers with rubberlike properties are called synthetic rubber. (p. 1230) step-growth polymer See condensation polymer. (p. 1232) syndiotactic polymer A polymer with the side groups on alternating sides of the polymer backbone. (p. 1229) thermoplastic A polymer that becomes moldable at high temperature. (p. 1237) vulcanization Heating of natural or synthetic rubber with sulfur to form disulfide cross-links. Cross-linking adds durability and elasticity to rubber. (p. 1231) Ziegler-Natta catalyst Any one of a group of addition polymerization catalysts involving titanium-aluminum complexes. Ziegler-Natta catalysts produce stereoregular (either isotactic or syndiotactic) polymers in most cases. (p. 1 230)

(I

Essential Problem-Solving Skills in Chapter 26

1.

Given the structure of a polymer, determine whether it is an addition or condensation polymer, and determine the structure of the monomer(s).

2.

Given the structure of one or more monomers, predict whether polymerization will occur to give an addition polymer or a condensation polymer, and give the general structure of the polymer chain.

3.

Use mechanisms to explain how a monomer polymerizes under acidic, basic, or free­ radical conditions. For addition polymerization, consider whether the reactive end of the growing chain is more stable as a cation (acidic conditions), anion (basic condi­ tions), or free radical (radical initiator). For condensation polymerization, consider the mechanism of the step-growth reaction.

4.

Predict the general characteristics (strength, elasticity, crystallinity, chemical reactivity) of a polymer based on its structure, and explain how its physical characteristics change as it is heated past Tg and Tm.

5.

Explain how chain branching, cross-linking, and plasticizers affect the properties of polymers.

6.

Compare the stereochemistry of isotactic, syndiotactic, and atactic polymers. Explain how the stereochemistry can be controlled during polymerization and how it affects the physical properties of the polymer.

1239

1240

Chapter 26: Synthetic Polymers

Study Problems 26-21

26-22

26-23

Define each term and give an example. (c) copolymer (b) condensation polymer (a) addition polymer (e) isotactic polymer (I) syndiotactic polymer (d) atactic polymer (i) anionic polymerization (h) cationic polymerization (g) free-radical polymerization (k) amorphous polymer (I) monomer (j) crystalline polymer (0) Ziegler-Natta catalyst (n) vulcanization (m) plasticizer (r) polyester (q) polyamide (p) glass transition temperature (t) polyurethane (u) polycarbonate (s) crystalline melting temperature Polyisobutylene is one of the components of butyl rubber used for making inner tubes. (a) Give the structure of polyisobutylene. (b) Is this an addition polymer or a condensation polymer? (c) What conditions (cationic, anionic, free-radical) would be most appropriate for polymerization of isobutylene? Explain your answer. Poly(trimethylene carbamate) is used in high-quality synthetic leather. It has the following structure.

What type of polymer is poly(trimethylene carbamate)? Is this an addition polymer or a condensation polymer? Draw the products that would be formed if the polymer were completely hydrolyzed under acidic or basic conditions. Poly(butylene terephthalate) is a hydrophobic plastic material widely used in automotive ignition systems.

(a) (b) (c) 26-24

�CH'CH'CH'CH'-O-�-o-�-O* poly(butylene terephthalate) What type of polymer is poly(butylene terephthalate)? Is this an addition polymer or a condensation polymer? Suggest what monomers might be used to synthesize this polymer and how the polymerization might be accomplished. Urylon fibers are used in premium fishing nets because the polymer is relatively stable to UV light and aqueous acid and base. The structure of Urylon is (a) (b) (c)

26-25

What functional group is contained in the Urylon structure? Is Urylon an addition polymer or a condensation polymer? Draw the products that would be formed if the polymer were completely hydrolyzed under acidic or basic conditions. Polyethylene glycol, or Carbowax® [( -O-CH2-CH2- )/1]' is widely used as a binder, thickening agent, and packag­ ing additive for foods. (a) What type of polymer is polyethylene glycol? (We have not seen this type of polymer before.) (b) The systematic name for polyethylene glycol is poly(ethylene oxide). What monomer would you use to make polyethylene glycol? (c) What conditions (free-radical initiator, acid catalyst, basic catalyst, etc.) would you consider using in this polymerization? (d) Propose a polymerization mechanism as far as the tetramer. Polychloroprene, commonly known as neoprene, is widely used in rubber parts that must withstand exposure to gasoline or other solvents. (a)

(b) (c)

26-26

26-27

(a) (b)

What type of polymer is polychloroprene? What monomer is used to make this synthetic rubber?

Study Problems 26-28

26-29

26-30

26-31

26-32

"'26-33

*26-34

26-35

124 1

Polyoxymethylene (polyformaldehyde) is the tough, self-lubricating Delrin® plastic used in gear wheels. (a) Give the structure of polyformaldehyde. (b) Formaldehyde is polymerized using an acidic catalyst. Using H+ as a catalyst, propose a mechanism for the polymerization as far as the trimer. (c) Is Delrin® an addition polymer or a condensation polymer? Acetylene can be polymerized using a Ziegler-Natta catalyst. The cis or trans stereochemistry of the products can be con­ trolled by careful selection and preparation of the catalyst. The resulting polyacetylene is an electrical semiconductor with a metallic appearance. cis-Poly acetylene has a copper color, and trans-polyacetylene is silver. (a) Draw the structures of cis- and trans-polyacetylene. (b) Use your structures to show why these polymers conduct electricity. (c) It is possible to prepare polyacetylene films whose electrical conductivity is anisotropic. That is, the conductivity is higher in some directions than in others. Explain how this unusual behavior is possible. Use chemical equations to show how the following accidents cause injury to the clothing involved (not to mention the skin under the clothing!). (a) An industrial chemist spills aqueous H2S04 on her nylon stockings but fails to wash it off immediately. (b) An organic laboratory student spills aqueous NaOH on his polyester slacks. Poly(vinyl alcohol), a hydrophilic polymer used in aqueous adhesives, is made by polymerizing vinyl acetate and then hydrolyzing the ester linkages. (a) Give the structures of poly(vinyl acetate) and poly(vinyl alcohol). (b) Vinyl acetate is an ester. Is poly(vinyl acetate) therefore a polyester? Explain. (c) We have seen that basic hydrolysis destroys the Dacron® polymer. Poly(vinyl acetate) is converted to poly(vinyl alcohol) by a basic hydrolysis of the ester groups. Why doesn't the hydrolysis destroy the poly(vinyl alcohol) polymer? (d) Why is poly(vinyl alcohol) made by this circuitous route? Why not just polymerize vinyl alcohol? In reference to cloth or fiber, the term acetate usually means cellulose acetate, a semisynthetic polymer made by treating cellulose with acetic anhydride. Cellulose acetate is spun into yarn by dissolving it in acetone or methylene chloride and forcing the solution through spinnerets into warm air, where the solvent evaporates. (a) Draw the structure of cellulose acetate. (b) Explain why cellulose acetate is soluble in organic solvents, even though cellulose is not. (c) (A true story) An organic chemistry student wore a long-sleeved acetate blouse to the laboratory. She was rinsing a warm separatory funnel with acetone when the pressure rose and blew out the stopper. Her right arm was drenched with acetone, but she was unconcerned because acetone is not very toxic. About ten minutes later, the right arm of the student's blouse disintegrated into a pile of white fluff, leaving her with a ragged short sleeve and the tatters of a cuff remaining around her wrist. Explain how a substance as innocuous as acetone ruined the student's blouse. (d) Predict what usually happens when students wear polyvinyl chloride shoes to the organic laboratory. One of the earliest commercial plastics was Bakelite®, formed by the reaction of phenol with a little more than one equiva­ lent of formaldehyde under acidic or basic conditions. Baeyer first discovered this reaction in 1 872, and practical methods for casting and molding Bakelite® were developed around 1909. Phenol-formaldehyde plastics and resins (also called phenolics) are highly cross-linked because each phenol ring has three sites (two ortho and one para) that can be linked by condensation with formaldehyde. Suggest a general structure for a phenol-formaldehyde resin, and propose a mechanism for its formation under acidic conditions. (Hint: Condensation of phenol with formaldehyde resembles the condensation of phenol with acetone, used in Problem 26-17, to make bisphenol A.) Plywood and particle board are often glued with cheap, waterproof urea-formaldehyde resins. Two to three moles of formaldehyde are mixed with one mole of urea and a little ammonia as a basic catalyst. The reaction is allowed to proceed until the mixture becomes syrupy, then it is applied to the wood surface. The wood surfaces are held together under heat and pressure, while polymerization continues and cross-linking takes place. Propose a mechanism for the base-catalyzed condensation of urea with formaldehyde to give a linear polymer, then show how further condensation leads to cross­ linking. (Hint: The carbonyl group lends acidity to the N -H protons of urea. A fust condensation with formaldehyde leads to an imine, which is weakly electrophilic and reacts with another deprotonated urea.) The polyester named Lactomer® is an alternating copolymer of lactic acid and glycolic acid. Lactomer is used for absorbable suture material because stitches of Lactomer hydrolyze slowly over a two-week period and do not have to be removed. The hydrolysis products, lactic acid and glycolic acid, are nonnal metabolites and do not provoke an inflamma­ tory response. Draw the structure of the Lactomer polymer.

HO

�

HO

o

glycolic acid

OH

o

� CH3

lactic acid

OH

1242 26-36

26-37

Chapter 26: Synthetic Polymers Compare the molecular structures of cotton and polypropylene, the two major components of thermal underwear. One of these gets wet easily and holds the water in contact with the skin. The other one does not get wet, but wicks the water away from the skin and feels relatively dry to the touch. Explain the difference in how these two fabrics respond to moisture. For each polymer shown below, (i) draw the monomer or monomers that were needed to make the polymer. (ii) explain whether the polymer is an addition polymer or condensation polymer. (iii) suggest what reagents and conditions one might use to synthesize the polymer. Cl

(a)

Cl

Cl

� Cl Cl Cl

""

'/

(b)

o

(d) *26-38

'

/

�� 0

�� ��

��

o

T(

'

0

The strongly hydrophilic polymer shown below is used in soft contact lenses. (a) Suggest how you might synthesize this polymer from methacrylic acid and any other reagents you need. (b) What is it about this polymer that makes it strongly hydrophilic? Explain why it is so important that the plastic in soft contact lenses be hydrophilic.

O�

0

O�

0

O�

0

O�

0

O�

0

O�

0

( ( ( ( ( (

OR

OR

OR

OH

OH

OH

Appendices

ee:

1A

NMR: Proton Chemical Shifts 1244

1B

NMR: Spin-Spin Coupling Constants 1246

1C

NMR: 13C Chemical Shifts in Organic

2A 2B 3

Compounds 1247 IR: Characteristic Infrared Group Frequencies 1248 IR: Characteristic Infrared Absorptions of Functional Groups 1251 UV: The Woodward-Fieser Rules for Predicting UV-Visible Spectra 1253

4A

Methods and Suggestions for Proposing

4B

Suggestions for Developing Multistep

5

Mechanisms 1257

Syntheses 1260 pKa Values for Representative Compounds 1261

1243

1244

Appendices

APPENDIX 1 A NMR: Proton Chemical Shifts o

Structural type

14

13

12

11

9

10

Value and range"

7

8

6

-' -§

4

TMS, O.OOO . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... . . - CH2 - , cyclopropane

. . ..

CH4

.... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... ....

....

.. . .

. . . .

. . . .

. .. .

. ...

.... ....

. . . .

. . . .

.. . .

. . . .

....

. ...

....

....

....

....

....

....

....

. . ..

....

. .. .

....

....

....

. . . .

. . . .

.. . .

3

.

. ...

. .... . +. . . . .

. . . .

ROH, monomer, very dilute solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . .. . . . . . . . . . . . . . . . . . . . . . ... . . . . . . .. . . .

I

C H3 - C - (saturated)

. . . .

.

.. . . . . .. . .

....

....

....

....

I R2NHb , 0.1-0.9 mole fraction in an inert solvent CH3 -

rI - rI - X (X

=

.

.

....

....

..

.

.

. . . .

. . . .

. . . .

. . . . ....

...

....

. . . .

. . . .

. . . .

. . . .

..

.

.

. . . . . . . .

. . . .

. .

.... . . . .

. . . .

. . . .

....

..

. . . .

....

....

. . . .

. . . .

....

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. . . .

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. . . .

. . . .

....

....

. . . .

....

.

....

.

..

....

.

..

....

. .

.

. ...

.

+.

.

. . . .

. . . .

..

. . . .

. . . .

....

I I I

. . . .

. . . .

...

CI, Br, I, OH, OR, C=O, N)

o

.

:::::;� . ...

. C'J ..

..:::;j:c ....... ....

..

;]� ... ...

-CH2- (saturated) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... . . . . . . . . . . . � .... ... RSHb b ::::::;� .... ... RNH2 , 0.1-0.9 mole fraction in an inert solvent. . . . ....

-

I

T

CH3

. . . .

. ..

....

.... . . . .

��� : 'rurnt

r

/" C=C, /"

.. . .

..

- F, CI,

X (X

CH3,

)

. . . .

. . . .

. . . .

. . . .

....

. . . .

.

. . . .

. .. .

. . .

....

.

...

....

....

. . . .

....

. . ..

.

.. .

....

.

.. .

I

..

....

CH3 - S -

.

. . .

....

.

.

. .

....

.

.

. .

....

. . . .

. . ..

....

....

. . . .

....

....

..

....

....

. . . .

. . . .

....

...

.

....

. . . .

.. . . .. . .

Br, I, OH, OR, OAr, N .

...

. .. . .

. . . .

. . . .

....

....

. . . .

.

....

....

....

..

....

. .

/" C H3 - N, . . . . . . . . . . . . . . . . . . . . . . .

.

. . . .

..

.

...

....

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····

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r

.

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.. . .

....

. . . .

... . . . . .

.

.. .

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....

....

....

.

. ...

....

. . . .

. . . .

....

....

. . . .

. . . .

. . . .

....

. . . . . . . .

....

. . . .

....

. . . .

...

....

....

. . . .

....

C - , conjugated . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . .... .. . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . =

F, CI, Br, I, 0)

. . . .

.

.

..

. . . .

. . . .

..

..

....

....

. . . .

. . . .

....

....

....

....

....

.

... . . . .

. . . .

. . . .

. . . .

. . . .

. . .. . . . .

. . . .

....

. . . .

. .. .

. . . .

14

....

....

13

....

....

12

. . ..

....

....

....

....

....

.. ..

. ...

..

.

.

.

..

....

....

I;:;:q·. ....�:=I···· .... .... ...

. . . . ..

....

10

9

8

7

6

5

4

a

. .

.

.

.

.

....

.. ... ..... ... .... . .. .... . . . ... .. . .... . .. .... .... .... .���. ..... .... .... .... 11

.

.

.. .... .... ........ .... .... ........ .... .... .... ............ .... .... .... .... ........ .... .... .... ............ ........ ....,..� .... .... .... .... ... CH3-O. . . .. ��...... ......... . . . .

... .

.

ArSHb

. . . .

.

.

H-C

I

.

_ .... .... ...

C - , nonconjugated . . . . . . . . . . . . . . . . . . . . . . .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . .

I

.

.

.

. .

.

i-� .... ...

H-C

H - C - X (X

....

. .... .... ........ .... .... .... .... .... .... .... .... .... .... .... .... .... ... � .... .... .... .. �. ............... -� .... .... .... ... . . .. .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... -� .... .... .... ...

CH3-C=0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . CH3Ar

. . . .

.

3

2

.

o

Normally, absorptions for the functional groups indicated will be found within the range shown in black. Occasionally, a functional group will absorb outside this range. Approximate limits are indicated by extended outlines. b

Absorption positions of these groups are concentration-dependent and are shifted to lower 8 values in more dilute solutions.

1245

Appendices APPENDIX 1 A NMR: Proton Chemical Shifts o

Structural type 14

L3

12

9

10

IL

H H

� , nonconjugated . . .. . . . . . . . . .

" . /' /, C=C " , acyclIc, nonconJ ugated .

. . .

. . . . . . . . . . . . . . . . .� . . .

. . . .

. . . .

....

.

. . .

. . . .

. . . .

. . . .

. . . .

. . . .

� , conjugated . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

ArOHb , polymeric association H" /' . ,C=C " , conJugated /

. . . .

. . . .

.

.. .

. . . .

. . . .

. . . .

.. . .

....

....

. . . .

....

. . . .

. . . .

....

. . . .

.

.

. .

....

. . . .

. . . .

. . . .

. . . .

....

ArH, benzenoid

.

. . . .

. . . .

. . .

.

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . .

. .

.

. . .

.

. . . .

. . .

. . . .

. . . .

. . . .

. .

. . . .

. . . .

. . . .

. . . .

. . . .

.

. . .

. . . .

....

. . . .

. .. .

. . . .

. . . .

. . . .

. . .

.

. .

. . . .

. .

. . . .

. . . .

.

. . . .

. . . .

. .

.

.

. . .

....

. . . .

. . . .

. . . .

.

.

. . . .

. . . .

. . . .

. . . .

. . . .

..

.

ArH, nonbenzenoid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . + + RNH3 , R2NH2 + , and R3NH , (trifluoroacetic acid solution)

. . .

.

-;:/0 H-C " . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . N -;:/0 H-C, ·· ···· ···· ···· ···· ···· ···· ···· O_

. . . .

. . . .

. . . .

. . . .

. . . .

.

..

. . . .

. . . .

. .

.

....

.

. .

/'

C=N "

. . .

OH b

RCHO, aliphatic,

. . . .

ex,

. . . .

. . . .

....

. . . .

. . . .

. . . .

. . . .

{3-unsaturated . . . . . . . .. ..

. . . .

. . . .

. . . .

. . . .

. . . .

. .

. .

. ...

. . . .

.

. .

.

.

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

.

14

. . . .

. .

.

. . . .

. . . .

......_..........._1 ....

. . . .

. . .

. . . .

. . . .

.... .... .... ..

. . . .

.

.

. .

. .

..::::I::: c:::. t: ·::· .. ::J .c::.p.... .... .... .... .... .... .... .... .... .... .. . .

.

. .

.

t:=)I•.=J ... .... .... .... .... .... .... .... .... .... .... ..

. .

.

. .

.

. . . .

.

. .

....

....

. . . .

. . .

. . . .

. . . .

. . . .

. .

. .

•

....

•....

. . .

. . .

. .

.

.

. . . . . . . .

. . . .

.

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . . . . . .

. . . . . . . .

....

. . .

.

. .. .

. . . .

. . . .

. . ..

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . . . . . .

. . . .

. . . .

. . . .

..

. . . .

. . . .

. . . .

. . .

. . .

. . . .

. . . . . . . .

.

....

. . .

.

. . . .

. . . .

. . . .

. . . .

. .

.

. . . .

. . . . . . . .

_� . . . . .......

. . . .

.

. .

•.

.

....

....

....

..

.

. . .

. .

. .

. . . .

1-...

. . . .

. . . .

. . . .

. . . .

.��....� . q :j:z;:.... .... .... .... .... 13

. . . .

. .

.

...

. . . .

RC02H, dimer, in nonpolar solvents

I.:::::.� .... .... .... .... .... .... .... .... .... ..

. . . .

. . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .� . . . . . . . . . .

.

ArOH, i ntermolecularly bonded . . . . . . . . . . . . . . . . . . . - S03H

. .

. . . .

· .... ·E.... .... ....

RCHO, aliphatic . . . ArCHO

....

.

o

. .

�....,............� .... .... .... .... . . . r::·=t · ·c: ·,= ::: · :::JI....-t=1 . .:;:t I ....� � ... -.t:. $ : . ti;:q . =:j ... .... . . . .

. . . .

+ ArNH3 , ArRNH2 + , and ArR2NH + , (trifluoroacetic acid solution) "

...

2

3

.1 .

.... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .::::�I- . . . . . . . . . . . . . . . . . . . . . . ..

. .

. .

. . . .

:::C = C � , acyclic, conjugated. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .... . . .. . .. . . . .. . . . .

I -;:/0 H - N-O "

4

5

6

.

.

:::C=C � , cyclic, nonconjugated

CH2=C

H

.

7

8

ROHb, 0 . 1 - 0.9, mole fraction in an i nert solvent . . .. .... . .. . . . . . . . . . . . . . ... . . . CH2 = C

Value and range"

L2

II

= . . . .

LO

9

8

.

7

6

5

4

3

2

o

" Normally, absorptions for the functionaL groups indicated will be found within the range shown in black. Occasionally, a functional group will absorb outside this range. Approximate limits are indicated by extended outlines. b

Absorption positions of these groups are concentration-dependent and are shifted to Lower () values in more diLute solutions.

1246

Appendices APPENDIX 1 B N M R: Spin-Spin Coupling Constants J, Hz

Type " /

H

C

/ "

12-15 H

/

"-

/ C H - C IZ

2-9

with free rotation

-7

1 I

1 I

I " I

-C -(-C -)-C1-1

C I-1 3 C H3

/CH -X

n

I-I - C - C - I-I

x

"- C / =C

1-1,,- C /

I

y

/H

/ 1-1 = C ,,-

/ / C = C ,,-

=

"-c /

/

I-I

ax ial, e

=

I-I

equatorial

0.5 - 2.5

"-

,,- I

/C - I-I / C = C ,,-

H ,,-

6.5 -7.5

5.5 - 7.0

a,a 5 - 1 0 a,e 2- 4 e,e 2 - 4

0.5 - 3

"- H

1-1 ,,-

a

1-1

"- C = C /

4- 1 0

"- C = C I-I - C H / =C

,,-

I

/H "- C / = C ,,/C- H 1

-0

1 -1

C I-I3 - CI-I2 - X

J, H z

Type

-0 / "-

9-13

"/ CH - C = C - I-I

2-3

/H "- C I-j - C� /

1 -3

a

/ 1-1 "- C / = C ,,- / H C

6-8

II

7- 12

13-1 8

a

0

6-9 1 -3 0-1

Appendices

I�II'I -

.!.,:.I:.I:I ·I'�'" " C ° /' " C ° /' " C=O /' " C=O /' " /' C=S " ,C=N / "

-C =N " " ,C=N " / /' " C c /' " /' " C C /' " /' " C C /' " -C =C /' " C C /' " " C-O /' " /' " ,C-N " / " C-S /' " "

c

Ketone

.

.

W' IimiW.I . . !lll!I.!I.' !I

....

'1'1

Aldehyde

I•

Acid Ester, amide Thioketone Azomethine

•

•

•

••

Nitrile

.-

.-

Heteroaromatic Alkene Aromatic Heteroaromatic

•••

Alkyne (C Quaternary)

•

•

•-

-•

Halogen / C (C Tertiary)

/' " CH /' " " ,CH- O " / / " CH-N /' " " ,CH- S " / " CH Halogen /' /' (C Secondary) CH2 C " CH2 0 " /' - CH2 - N " ,CH2- S / " CH2

Halogen

H3C

C

/ "

1 -. I -•

••

I

Cy pro

••

.-•

(C Primary)

1-.

H3C - 0

" /' H3C N " H3C - S " H3C

;��I

I-

Halogen

Resonances of common solvents ppm (TMS)

CH, coI cs\

\

c,loJH

c

1\ IJ C1H3)2 �6jJ,g: OH Cl4 1 CH l ,� CH,PHD1'y1S0 •

I,

oxal e

220 2 1 0 200 1 90 1 80 l 70 1 60 1 50 1 40 1 30 1 20 J 10 1 00 90 80 70 60 50 40 30 20

"'Relative to internal tetramethylsilane. Copyright 1 998 by Bruker Analytik GmbH. Used by permission.

:J

I

0

10

0

1247

APPENDIX 2A IR: Characteristic I nfra red Group Frequencies (s 4000 em - I

3000

3500

2500

l

=

strong, m

medium, w

1600

1800

2000

=

=

weak; overtone bands are marked 2v) 1200

1400

1000

800

600

400

. . . . 'Sf---+-i " ?-pr�pyl . . . !--+-i' . . . . . . . . . . . f-::-IM M I ·· MM···· .... ·r-rw · .... .... .. . I isopropyl .... .... .. .. , , .. .. ·'� . . . . . . . . . . . rwf---< .... .... .... tertiary butyl .... Sr " Srw M· · M .1 1 1 1 . .... .... . vmyl -CH=CH, . . .... � .. . . ' .. . (conj .). .... 'M'eg- " ALKENE I W " rsrs " S � . .(cdnj.). rs : .. . 81'--< . . . . ';:c �c -;; (t�ansj� . . . > . ' rs M . . ) . S "� sr � � ���r H C =C H : . .. � . � I : . . 'r-M ls :.' . W s M� M ' ' ;;r S · EC � = rn ' , - , . . (colnj .). . . . . 's " 'T . .. . . ... . 'Mf >-:-:- . " IC�CIH ALKYNE - c ==cI, 1 �� 'w" . . . . .. . . SfJS:'' ' .... � lUI monosubstituted benzene . . . MTM" 'Mr ... . . . . . . . . . . . 'W-i[w >;-;;t-;-- . . . . . . . . AROMATIC I ' I-rw :� .... . . . . di � U b� �'t� tel d :: I-rw­ . · · . . · · I-rw r-I- . . �� :J, l. I l ¢ par� ... .... .... . . .. r-r--- . . .. M · . . 0:: vicinal trisubstituted . ... 1 1 .1 '1' N . . . . unsymmetncal . r-r� I. ...lUI.- sYimetncal .... r-r--- . . . . I . . 00 ... a-naphthalenes . . .. f=M . . . . . 00" /3-n�phthalehes . . ........, ·r-f-< · . . · . .... .. . . .. . r-- . . ...f--< 'M... .... . . . . . . ·,...-:-: ...., M+--I pl ethfrs.. al . h ati� CH . . .q-f21 � I 2 " i " " " " " " " ETHERS I .. aromatic ethers . . .. . .. 0-0 CH, . .. .. . (m-h,gh) s M ALCO�OL . .. r-:::-:-1----1 . . . .... . . . . . . (unbondmg (bolnde,d) . . . . . .... .... .... .... . . .. primII aryllalc;hols . . . . RCL,-ll oH I Ind,�g, lowers) I ' S (free) _ M S1 .... . sec7ndary (�nbol . . . . f-2C� -pH . . . . loters( .... �f-r" " " (broad . . .... .... .... .... . M M S1 ( haIrp) f-I lary.[ . . . . . .. .R ,C I OH, . . .. .. . M . . ... . . ...r--s--l . �nbolnd, �g lolvers! l ..h.. ...h) .. ((rbOndmg S l �=-:�'-:+-f-::-S-+I-+ l .... .... . . . . . .... .... .... .... .... ... . lert' aromatic . . . . . . . . . . . . o-OH .. (m lorersl l j M M M S r-I (broad) (broad) (abrent In monorer) . ... .... .... .... .... ..carboxtic lacidf " · . . . . COi)H :'; o +-s-l . . . . . . . . 'M-s . . ed carbixyll(salis, 'j'tterlOni' ell ) .. . . · C ( 1 M IOn · 1 ,!o ,) ' " '-s t I rs 1 1 1 � I I · CH;' C , CH,-(C=O) ,

:1

'

.

I

.

.

·

. . . .

. . . .

'

l I �

M

..

r-r---

·

.

::��

.

.

I

JT '"

.

w

.

I

M

.... ...

.. ..

.

�

.

..

4000 em- I 2.50 /1.111

3500

2.75

3000

3.00

1

3.25

2500

3.50

3. 75

4.00

2000

4.5

1

5.0

[

"

5.5

'

.

' -

. . . �

>---+--I w

.

"

..

..

F� '

.

.

J T ' T w

.... .... ....

' '

'

..

...

.

I . j&

. .

w

. .

..w ttl' '

.

w

1600

.

6.0

1400

6.5

1

7.0

1200

7.5

Courtesy of N. B. Colthup, Stanford Research Laboratories, American Cyanamid Company, and the editor of the Journal of the

8.0

Optical Society.

I+--


+------>

t----+

+---



( CH 3 h C ' > (CH3hCH ' > CH3CH2 ' > CH3 · · =

=

=

=

=

=

=

CHAPTE R 5

(d) 2-bromopropane; (e) 2-iodo-2-methylbutane; (f) 2-bromo2-methylbutane. 6.27. (a) ( CH 3 hC(OCOCH3 )CH2CH3, flrst order; (b) l -methoxy-2-methylpropane, second order; (c) I -ethoxy- l -methyl­ cyclohexane, first order; (d) methoxycyclohexane, fust order; (e) ethoxycyclohexane, second order. 6.36. 3-methyl- l -butene by E2 (minor); 2-methy[-2-butene by E2 (major); and 2-ethoxy-3-methy[bu­ tane (trace) by SN2. 6.43. (a) 2-bromo-2-methylpentane; (b) [ chloro­ I-methylcyclohexane; (c) 1 , [ -dichloro-3-fluorocycloheptane, (d) 4-(2-bromoethyl)-3-(fluoromethy[)-2-methylheptane; (e) 4,4-dichloro-5-cyclopropy[- [ -iodoheptane; (f) cis- I ,2-dichloro­ I -methykyclohexane. 6.44. (a) l -chlorobutane; (b) l -iodobutane; (c) 4-chloro-2,2-dimethylpentane; (d) l -bromo-2,2-dimethylpentane; (e) chloromethylcyclohexane; (f) 2-methyl- l -bromopropane. 6.45. (a) t-butyl chloride; (b) 2-chlorobutane; (c) bromocyclohexane; (d) iodocy­ clohexane; (e) t-butyl bromide; (f) 3-bromocyclohexene. 6.48. (a) rate doubles; (b) rate multiplied by six; (c) rate increases. 6.55. (a) (R)2-cyanobutane (inversion); (b) (2S,3R)-3-methyl-2-pentano[ (inversion); (c) racemic mixture of 3-ethoxy-2,3-dimethy[pentanes (racemization). 6.56. (a) diethyl ether; (b) PhCH2CH2CN; (c) PhSCH2CH3; (d) I -do­ decyne; (e) N-methy[pyridinium iodide; (f) (CH3hCCH2CH2NH2; (g) tetrahydrofuran; (h) cis-4-methylcyclohexanol. 6.58. (a) o.p. e.e. 15.58/ 1 5 .90 98% (99%(S) and 1 %(R); (b) The e.e. of (S) decreases twice as fast as radioactive iodide substitutes, thus gives the (R) enantiomer; implies the SN2 mechanism. 6.64. NBS provides [ow conc. Br2 for free-radical bromination. Abstraction of one of the CH2 hydrogens gives a resonance-stabilized free radical; product PhCHBrCH3· =

5.1. chiIal: corkscrew, desk, screw-cap bottle, rifle, knot. 5.2. (b), (d), (e), and (f) are chiral. 5.3. (a) achira[, no C*; (b) achira[, no C*; (c) chira[, one C*; (d) achiral, no C*; (e) achiral, no C*; (f) chiral, one C*; (g) achi­ rat, two C*; (h) chiral, two C*; (i) achira[, no C*; (j) chiral, one C*; (k) chiral, two C* 5.5. (a) mirror, achiral; (b) mirror, achiral; (c) chiral, no mirror; (d) chiral, no mirror; (e) chiral, no mirror; (f) mirror, achiral; (g) mirror, achiral; (h) chiral, no mirror. 5.6. (a) (R); (b) (S); (c) (R); (d) (S) , (S); (e) (R), (S); (f) (R), (S) ; (g) (R), (R); (h) (R); (i) (S). 5.8. + 8.7°. 5.10. Dilute the sample. If clockwise, will make less clock­ wise, and vice-versa. 5.12. e.e. 33.3%. Specific rotation 33.3% of + 1 3 .5° +4.5°. 5.15. (a), (b), (f), and (h) are chiral; only (h) has chiral carbons. 5.16. (a) enantiomer, enantiomer, same; (b) same, enantiomer, enantiomer; (c) enantiomer, same, same. 5.18. (a), (d), and (f) are chiral. The others have internal mirror planes. 5.19. (from 5-1 7) (a) (R); (b) none; (c) none; (d) (2R), (3R); (e) (2S), (3R); (f) (2R), (3R); (new ones) (g) (R); (h) (S); (i) (S). 5.20. (a) enantiomers; (b) diastereomers; (c) diastereomers; (d) constitutional isomers; (e) enantiomers; (f) diastere­ omers; (g) enantiomers; (h) same compound; (i) diastereomers. 5.23. (a), (b), and (d) are pairs of diastereomers and could theoretically be separated by their physical properties. 5.30. (a) same compound; (b) enantiomers; (c) enantiomers; (d) enantiomers; (e) enantiomers; (f) diastereomers; (g) enantiomers; (h) same compound. 5.34. (b) ( - ) 1 5.90°; (c) 7 .95°/ 1 5.90° 50%e.e. Composition is 75% (R) and 25% (S). =

=

=

=

CHAPTE R 6 6.1. (a) vinyl halide; (b) alkyl halide; (c) aryl halide; (d) alkyl halide; (e) vinyl halide; (f) aryl halide. 6.5. The C - CI bond has consider­ ably more charge separation (0.23 e) than the C- I bond (0. [ 6 e). 6.7. Water is denser than hexane, so water forms the lower layer. Chloroform is denser than water, so chloroform forms the lower layer. 6.11. (a) substitution; (b) elimination; (c) elimination, also a reduction. 6.13. 0.02 mol/L per second. 6.14. (a) ( CH 3 ) 3COCH2CH3; (b) HC = CCH2CH2CH2CH3; (c) (CH3 hCHCH2NH2; (d) CH3CH2C = N; (e) l -iodopentane; (f) l -fluoropentane. 6.16. (a) (CH3CH2)2NH, less hindered; (b) (CH3hS, S more polarizable; (c) Plh, P more polarizable; (d) CH3S-, neg. charged; (e) ( CH3hN, N less electronegative; (f) CH 3S-, neg. charge, more polarizable; (g) CH3CH2CH20-, Jess hindered; (h) r-, more polarizable. 6.18. methyl iodide > methyl chloride > ethyl chloride > isopropyl bromide » neopentyl bromide, t-butyl iodide. 6.19. (a) 2-methyl- I -iodopropane; (b) cyclohexyl bromide; (c) isopropyl bro­ mide; (d) 2-chlorobutane; (e) isopropyl iodide. 6.23. (a) 2-bromo­ propane; (b) 2-bromo-2-methylbutane; (c) allyl bromide;

=

=

CHAPTER 7 7.4. (a) one; (b) one; (c) three; (d) four; (e) five. 7.5. (a) 4-methyl- l ­ pentene; (b) 2-ethyl- l -hexene; (c) 1 ,4-pentadiene; (d) 1 ,2,4-pentatriene; (e) 2,5-dimethy[- 1 ,3-cyclopentadiene; (f) 4-vinylcyclohexene; (g) allyl­ benzene or 3-phenylpropene; (h) trans-3,4-dimethylcyclopentene; (i) 7-methylene- I ,3,5-cycloheptatriene. 7.6. ( 1 ) (a), (c) and (d) show geometric isomerism. 7.7. (a) 2,3-dimethyl-2-pentene; (b) 3-ethyl1 ,4-hexadiene; (c) I -methylcyclopentene; (d) give positions of double bonds; (e) specify cis or trans; (f) (E) or (2), not cis. 7.9. 2,3-dimethyl2-butene is more stable by 6.0 kJ/mole. 7.11. (a) stable; (b) unstable; (c) stable; (d) stable; (e) unstable (maybe stable cold); (f) stable; (g) unstable; (h) stable (i) unstable (maybe stable cold) 7.12. (a) cis1 ,2-dibromoethene; (b) cis (trans has zero dipole moment), (c) 1 ,2-dichlorocyclohexene. 7.17. There is no hydrogen trans to the bromide leaving group. 7.23. In the first example the bromines are axial; in the second, equatorial. 7.26. (a) I1G > 0, disfavored. (b) I1G < 0, favored. 7.27. (a) strong bases and nucleophiles; (b) strong acids and electrophiles; (c) free-radical chain reaction; (d) strong acids and electrophiles. 7.32. (a) 2-ethyl - I -pentene; (b) 3-ethyl-2-pentene; (c) (3E,6E)-1 ,3,6-octatriene; (d) (E)-4-ethyl-3heptene; (e) l -cyclohexyl- I ,3-cyclohexadiene; (f) (3Z,5E)-6-chloro-3(chloromethyl)- 1 ,3,5-octatriene. 7.36. (b), (c), (e) and (f) show geometric isomerism. 7.38. (a) cyclopentene; (b) 2-methyl-2-butene (major) and 2-methyl- l -butene (minor); (c) I -methylcyclohexene (major) and methylenecyclohexane (minor); (d) 1 methylcyclohexene (minor) and methylenecyclohexane (major). 7.42. (a) a I -halobutane; (b) a t-butyl halide; (c) a 3-halopentane; (d) a halomethylcyclohexane; (e) a 4-halocyclohexane (preferably cis). 7.44. (a) 2-pentene; (b) I -methylcyclopentene; (c) I -methylcyclohexene; (d) 2-methyl-2butene (rean·angement). 7.56. E l with rearrangement by an alkyl shift. The Zaitsev product violates Bredt's rule.

CHAPTER 8 8.1. (a) 2-bromopropane; (b) 2-chloro-2-methylpropane; (c) l -iodo- [ ­ methylcyclohexane; (d) mixture o f cis and trans l -bromo-3-methyl- and l -bromo-4-methylcyclohexane. 8.3. (a) I -bromo-2-methylpropane; (b) l -bromo-2-methylcyclopentane; (c) 2-bromo- I -phenylpropane.

Answers to Selected Problems 8.6. (a) l - methylcyclopentanol; (b) 2-phenyl-2-propanol; (c) l -phenylcy­ clohexanol . 8.10. (b) I -propanol; (d) 2-methyl-3-pentanol; (f) trans2-methylcyclohexanol. 8.13. (a) trans-2-methylcycloheptanol; (b) mostly 4,4-dimethyl-2-pentanol; (c) - OH exo on the less substituted carbon. 8.16. The carbocation can be attacked from either face. 8.22. (a) CI2/H20; (b) KOH/heat, then CI2/H20; (c) H2S04/heat, then CI2/H20. 8.28. (a) CH212 + Zn( C u ) ; (b) CH2Br2, NaOH, H20, PTC; (c) dehydrate ( H2S04 ) , then CHCI3, NaOH/H20, PTe. 8.34. (a) cis­ cyclohexane-l ,2-diol; (b) lrans-cyclohexane- l ,2-diol; (c) and (f) (R,S)2,3-pentanediol ( + enantiomer ) ; (d) and (e) (R, R )-2,3-pentanediol ( + enantiomer) . 8.35. (a) Os04/H202; (b) CH3C03H/H30+, (c) CH3C03H/H30+; (d) Os04/H202' 8.54. H2C = CH ­ COOCH2CH3. 8.55. (a) 2-methylpropene (3° cation); (b) l -methyl­ cyclohexene (3° cation); (c) l ,3-butadiene (resonance-stabilized cation). 8.59. (a) l -methylcyclohexene, RC03H/H30+; (b) cyclooctene, Os04/H202; (c) trans-cyclodecene, B r2; (d) cyclohexene, CI2/H20. 8.62. CH3(CH2) 1 2CH = CH ( CH2hCH3, cis or trans unknown.

CHAPTER 9 9.3. decomposition to its elements, C and H2. 9.4. Treat the mixture with NaNH2 to remove the I -hexyne. 9.5. (a) N a+ -C = CH and NH3; (b) Li+-C = CH and CH4; (c) no reaction; (d) no reaction; (e) acetylene + NaOCH3; (f) acetylene + NaOH; (g) no reaction; (h) no reaction; (i) NH3 + NaOCH3. 9.7. (a) NaNH2; butyl halide; (b) NaNH2; propyl halide; NaNH2; methyl halide. (c) NaNH2; ethyl halide; repeat; (d) SN2 on sec-butyl halide is unfavorable; (e) NaNH2; isobutyl halide (low yield); NaNH2; methyl halide; (f) NaNH2 added for second substitution on 1 ,8-dibromooctane might attack the halide. 9.8. (a) sodium acetylide + formaldehyde; (b) sodium acetylide + PhCOCH3; (c) sodium acetyl ide + CH3I, then NaNH2, then CH3CH2CH2CHO; (d) sodium acetylide + CH3I, then NaNH2, then CH3CH2COCH3. 9.10. About 1 :7 5 . 9.13. (a) H2, Lindlar; (b) Na, NH3; (c), (d) Add halogen, dehydrohalogenate to the alkyne, reduce as needed. 9.16. (a) CH3CCl2CH2CsH I I and CH3CH2CCI2CsH I I ; (b) Lone pairs on C I help stabilize the carbocation. 9.18. (a) C12; (b) HBr, peroxides; (c) HBr, no peroxides; (d) excess Br2; (e) reduce to l -hexene, add HBr; (f) excess HBr. 9.20. (a) The two ends of the triple bond are equivalent. (b) The two ends of the triple bond are not equiva­ lent, yet not sufficiently different for good selectivity. 9.21. (a) 2-hexa­ none; hexanal; (b) mixtures of 2-hexanone and 3-hexanone; (c) 3-hexanone for both; (d) cyclodecanone for both. 9.24. (a) CH3C = C (CH2)4C = CCH3 9.28. (a) ethylmethylacetylene; (b) phenylacetylene; (c) sec-butyl-n-propylacetylene; (d) sec-butyl-t­ butylacetylene. 9.38. I ,3-cyclohexadiene with ( HC = C - CH = CH - ) at the I position (cis or trans).

CHAPTER 1 0 10.1. (a) 2-phenyl-2-propanol; (b) 5-bromo-2-heptanol; (c) 4-methyl3-cyclohexen- I -ol; (d) trans-2-methylcyclohexanol; (e) (E)-2chloro3-methyl-2-penten- l -ol; (f) (2R,3S)-2-bromo-3-hexanol. 10.4. (a) 8,8-dimethyl-2,7-nonanediol; (b) I ,8-octanediol; (c) cis-2-cyclohexene- l ,4diol; (d) 3-cyclopentyl-2,4-heptanediol; (e) trans- l ,3-cyclobutanediol. 10.5. (a) cyclohexanol; more compact; (b) 4-methylphenol; more compact, stronger H-bonds; (c) 3-ethyl-3-hexanol; more spherical; (d) cyclooctaneI ,4-diol; more OH groups per carbon; (e) enantiomers; equal solubility. 10.7. (a) methanol ; less substituted; (b) 2-chloro- I -propanol; chlorine clos­ er to the OH group; (c) 2,2-dichloroethanol; two chlorines to stabilize the alkoxide; (d) 2,2-difluoro- l -propanol; F is more electronegative than CI, stabilizing the alkoxide. 10.9. The anions of 2-nitrophenol and 4-nitro­ phenol (but not 3-nitrophenol) are stabilized by resonance with the nitro group. 10.10. (a) The phenol (left) is deprotonated by sodium hydroxide; it dissolves. (b) In a separatory funnel, the alcohol (right) will go into an ether layer and the phenolic compound will go into an aqueous sodium hydroxide layer. 10.11. (b), (f), (g), (h). 10.15. (a) 3 ways: (i) CH3CH2MgBr + PhCOCH2CH2CH3; (ii) PhMgBr +

A3

CH3CH2COCH2CH2CH3; (iii) CH3CH2CH2MgBr + PhCOCH2CH3; (b) PhMgBr + PhCOPh. (c) CH3Mg1 + cyclohexanone. (d) CH3CH2CH2MgBr + dicyclohexyl ketone. 10.17. (a) 2 PhMgBr + PhCOCI; (b) 2 CH3CH2MgBr + (CH3 hCHCOCI; (c) 2 c-HxMgBr + PhCOCI. 10.19. (a) PhMgBr + ethylene oxide; (b) (CH3hCHCH2MgBr + ethylene oxide; (c) 2-methylcyclohexyl­ magnesium bromide + ethylene oxide. 10.23. (a) Grignard removes NH proton; (b) Grignard attacks ester; (c) Water will destroy Grignard; (d) Grignard removes OH proton. 10.26. (a) heptanoic acid + LiAIH4; or heptaldehyde + NaBH4; (b) 2-heptanone + NaBH4; (c) 2-methyl-3-hexanone + NaBH4; (d) ketoester + NaBH4. 10.34. (a) I -hexanol, larger surface area; (b) 2-hexanol, hydrogen-bonded; (c) I ,5-hexanediol, two OH groups; (d) 2-hexanol. 10.38. (a) cyclohexyl­ methanol; (b) 2-cyclopentyl-2-pentanol; (c) 2-methyl- l -phenyl- I -propanol; (d) methane + 3-hydroxycyclohexanone; (e) cyclopentylmethanol; (f) triphenylmethanol; (g) l -cyclopentyl-l , l-diphenylmethanol; (h) 5-phenyl-5-nonanol; (i) reduction of just the ketone, but not the ester; U) 3-(2-hydroxyethyl)cyclohexanol from reduction of ketone and ester; (k) the tertiary alcohol from Markovnikov orientation of addition of H - OH; (I) the secondary alcohol from anti-Markovnikov orientation of addition of H - OH ; (m) (2S,3S)-2,3-hexanediol ( + enantiomer ) ; (n) (2S,3R)-2,3hexanediol ( + enantiomer) ; (0) 1 ,4-heptadiene. 10.39. (a) EtMgBr; (b) Grignard with formaldehyde; (c) c-HxMgBr; (d) cyclohexylmagnesium bromide with ethylene oxide; (e) PhMgBr with formaldehyde; (f) 2 CH3MgI; (g) cyclopentylmagnesium bromide; (h) CH3C = CMgX.

CHAPTER 1 1 11.1. (a) oxidation, oxidation; (b) oxidation, oxidation, reduction, oxida­ tion; (c) neither (C2 is oxidation, C3 reduction); (d) reduction; (e) neither; (f) oxidation; (g) neither; (h) neither; (i) oxidation; (j) oxidation then neither (k) reduction then oxidation, no net change. 11.6. (a) PCC; (b) chromic acid; (c) chromic acid or .Tones reagent; (d) PCC; (e) chromic acid; (f) dehydrate, hydroborate, oxidize (chromic acid or Jones reagent). 11.7. An alcoholic has more alcohol dehydrogenase. More ethanol is needed to tie up this larger amount of enzyme. 1 1.8. CH3COCHO (pyruvaldehyde) and CH3COCOOH (pyruvic acid). 1 1.10. Treat the tosylate with (a) bromide; (b) ammonia; (c) ethoxide; (d) cyanide. 11.14. (a) chromic acid or Lucas reagent; (b) chromic acid or Lucas reagent; (c) Lucas reagent only; (d) Lucas reagent only; allyl alcohol forms a resonance-stabilized carbocation. (e) Lucas reagent only. 11.19. (a) thionyl chloride (retention); (b) tosylate (retention), then SN2 using chloride ion (inversion). 11.20. resonance-delocalized cation, positive charge spread over two carbons. 11.22. (a) 2-methyl-2-butene ( + 2-methyl-l -butene) ; (b) 2-pentene ( + l -pentene) ; (c) 2-pentene ( + I -pentene) ; (d) c-Hx= C ( C H 3 ) z ( + l -isopropylcyclohexene) ; (e) I -methylcyclohexene ( +3-methylcyclohexene) . 11.25. Using R - OH and R ' - OH will form R - O - R, R ' - O - R', and R - 0 - R ' . 1 1.31. (a) CH3CH2CH2COCI + I -propanol; (b) CH3CH2COCI + l -butanol; (c) (CH3 hCHCOCI + p-methylphenol ; (d) PhCOCI + cyclopropanol. 1 1 .33. An acidic solu­ tion (to protonate the alcohol) would protonate methoxide ion. 11.34. (a) the alkoxide of cyclohexanol and an ethyl halide or tosylate; (b) dehy­ dration of cyclohexanol. 11.42. (a) Na, then ethyl bromide; (b) NaOH, then PCC to aldehyde; Grignard, then dehydrate; (c) Mg in ether, then CH3CH2CH2CHO, then oxidize; (d) PCC, then EtMgBr. 11.45. Use CH3S02Cl. 1 1.46. (a) thionyl chloride; (b) make tosylate, displace with bromide; (c) make tosylate, displace with hydroxide. 1 1.52. Compound A is 2-butanol. 1 1.59. X is l -buten-4-01; Y is tetrahy­ drofuran (5-membered cyclic ether).

CHAPTER 1 2 12.3. (a) alkene; (b) alkane; (c) terminal alkyne. 12.4. (a) amine (second­ ary); (b) acid; (c) alcohol. 12.5. (a) conjugated ketone; (b) ester; (c) primary amide. 12.6. (a) 3070 C - H; 1 642 C = C alkene; (b) 27 1 2, 28 1 4 - CHO; 1 69 1 carbonyl-aldehyde; (c) over-inflated C - H

A4

Answers to Selected Problems

region - COOH; 1 703 carbonyl (maybe conjugated); 1 650 C = C (maybe conjugated)-conjugated acid; (d) 1 742 ester (or strained ketone)­ eSler. 12.7. (a) bromine ( C6H5Br); (b) iodine ( C2H5I ) ; (c) chlorine ( C4H7C I ) ; (d) nitrogen (C7H 1 7 N ) . 12.8. the isobutyl cation, (CH3hCHCH/ 12.11. 1 26: loss of water; I l l : allylic cleavage; 87: cleavage next to alcohol. 12.14. (a) about 1 660 and 1 7 1 0 ; the carbonyl is much stronger; (b) about 1 660 for both; the ether is much stronger; (c) about 1 660 for both; the imine is much stronger; (d) about 1 660 for both; the terminal alkene is stronger. 12.16. (a) CH2 = C (CH3 )COOH; (b) (CH3hCHCOCH3; (c) PhCH2C = N ; (d) PhNHCH2CH3. 12.17. (a) 86, 7 1 , 43; (b) 98, 69; (c) 84, 87, 45. 12.20 (a) I -bromobutane. 12.23 (c) I -octyne.

CHAPTER 1 3 13.1 . (a) 82. 1 7 ; (b) 0. 1 53 gauss; (c) 82. 1 7; (d) 1 30 Hz. 13.3. (a) three; (b) two; (c) three; (d) five. 13.6. (a) 2-methyl-3-butyn-2-01; (b) p-dimethoxybenzene; (c) 1 ,2-dibromo-2-methylpropane. 13.10. trans CHC1 = CHCN. 13.11. (a) l -chloropropane; (b) methyl p-methylbenzoate, C H3C6H4COOCH3. 13.14. (a) H", 89.7 (doublet); Hb, 86.6 (multiplet); H", 87.4 (doublet); (b) Jab 8 Hz, Jbc 1 8 Hz (approx). 13.18. (a) Five; the two hydrogens on C3 are diastereotopic. (b) Six; all the CH2 groups have diastereotopic hydrogens. (c) Six; three on the Ph, and the CH2 hydrogens are diastereotopic. (d) Three; The hydrogens cis and trans to the CI are diastereotopic. 13.21. (a) butane1 ,3-diol; (b) H2NCH2CH20H. 13.24. (a) (CH 3 hCHCOOH; (b) PhCH2CH2CHO; (c) CH3COCOCH2CH3 (d) C H 2 = CHCH( OH)CH3; (e) CH3CH2C (OH) (CH3)CH( CH3h 13.29. (a) allyl alcohol, H2C = CHCH20H. 13.30. (a) 4-hydroxybutanoic acid lactone (cyclic ester). 13.31. (a) cyclohexene. 13.32. isobutyl bromide. 13.36. (a) isopropyl alcohol. 13.38. (a) PhCH2CH20COCH3. 13.42. 1 , 1 ,2trichloropropane. 13.45. A is 2-methyl-2-butene (Zaitsev product); B is 2-methyl- l -butene. 13.47. PhCH2CN. =

=

CHAPTER 1 4 1 4.2. ( CH 3C H 2 hOrAICI3. 14.4. (a) cyclopropyl methyl ether; methoxycyclopropane; (b) ethyl isopropyl ether; 2-ethoxypropane; (c) 2-chloroethyl methyl ether; l -chloro-2-methoxyethane; (d) 2-ethoxy-2,3-dimethylpentane; (e) sec-butyl I-butyl ether; 2-1butoxybutane; (f) Irans-2-methoxycyclohexanol (no common name). 14.6. (a) dihydropyran; (b) 2-chloro- 1 ,4-dioxane; (c) 3-isopropyl­ pyran; (d) trans-2,3-diethyloxirane or trans-3,4-epoxyhexane; (e) 3-bromo-2-ethoxyfuran; (f) 3-bromo-2,2-dimethyloxetane. 14. 1 1. Intermolecular dehydration of a mixture of methanol and ethanol would produce a mixture of diethyl ether, dimethyl ether, and ethyl methyl ether. 14.13. I ntermolecular dehydration might work for (a). Use the Williamson for (b). Alkoxymercuration is best for (c). 1 4. 1 5. (a) bromocyclohexane and ethyl bromide; (b) 1 ,5-diiodopen­ tane; (c) phenol and methyl bromide; (e) phenol, ethyl bromide, and 1 ,4-dibromo-2-methylbutane. 14.22. Epoxidiation of ethylene gives ethylene oxide, and catalytic hydration of ethylene gives ethanol. Acid-catalyzed opening of the epoxide in ethanol gives cellosolve. 14.26. (a) CH3CH20CH2CH20- Na+; (b) H2NCH2CH20- Na+; (c) Ph - SCH2CH20- Na+; (d) PhNHCH2CH20H; (e) N = C - CH2CH20- Na+; (f) N3CH2CH20- Na+ 14.27. (a) 2-methyl- l ,2-propanediol, \ 80 at the C2 hydroxyl group; (b) 2-methyl- I ,2-propanediol, \ 80 at the C I hydroxyl group; (c), (d) same products, (S,S) and (R,R). 14.28. (a) (CH3 hCH ­ CH2CH 2 - OH ; (b) CH3CH2C(CH 3 h - OH; (c) I cyclopentyl­ I -butanol. 14.34. (a) The old ether had autoxidized to form perox­ ides. On distillation, the peroxides were heated and concentrated, and they detonated. (b) Discard the old ether or treat it to reduce the per­ oxides. 14.38. (a) epoxide + PhMgBr; (b) epoxide + NaOCH3 in methanol; (c) epoxide + methanol, H+ 14.42. Sodium then ethyl iodide gives retention of configuration. Tosylation gives retention,

then the Williamson gives inversion. Second product ( + ) 1 5 .6°. 14.46. (CH30CH2CH2hO 14.47. phenyloxirane.

CHAPTER 1 5

15.1. (a) 2,4-hexadiene < 1 ,3-hexadiene < 1 ,4-hexadiene < 1 ,5-hexadiene < 1 ,2-hexadiene < 1 ,3 ,5-hexatriene; (b) third < fifth < first < fourth < second. 1 5.6. 3-ethoxy- l -methy1cy­ clopentene and 3-ethoxy-3-methylcyclopentene. 15.8. (a) A is 3,4dibromo- I -butene; B is 1 ,4-dibromo-2-butene; (c) Hint: A is the kinetic product, B is the thermodynamic product; (d) Isomerization to an equilibrium mixture. 1 0% A and 90% B. 15.9. (a) 1 -(bromomethyl) cyclohexene and 2-bromo- l - methylenecyclohexane. 15.11. (a) 3-bro­ mocyclopentene; (b) (cis and trans) 4-bromo-2-pentene. (c) PhCH2Br 15.12. Both generate the same allylic carbanion. 15.13. In this reaction, alkyllithiums or Grignard reagents can be used interchangeably. (a) allyl bromide + n-butyllithium; (b) isopropyllithium + l -bromo-2-butene; (c) 1 ,4-dibromo-2-butene + two equivalents of propyllithium. 15.20. (b) [4 + 2] cycloaddition of one butadiene with just one of the double bonds of another butadiene. 15.21. 800. 15.22. (a) 353 nm; (b) 3 1 3 nm; (c) 232 nm; (d) 273 nm; (e) 237 nm. 15.24. (a) isolated; (b) conjugated; (c) cumulated; (d) conjugated; (e) conjugated; (f) cumu­ lated and conjugated. 15.25. (a) allylcyclohexane; (b) 3-chlorocyclopen­ tene; (c) 3-bromo-2-methylpropene; (d) 3-bromo- l -pentene and l -bromo-2-pentene; (e) 4-bromo-2-buten-l -ol and I -bromo-3-buten-2-01; (f) 5,6-dibromo- I ,3-hexadiene, 1 ,6-dibromo-2,4-hexadiene, and 3,6-dibromo- l , 4-hexadiene (minor); (g) 1-(methoxymethyl)-2-methylcy­ clopentene and I -methoxy- l -methyl-2-methylenecyclopentane; (h), (i), and U) Diels-Alder adducts. 15.26. (a) allyl bromide + isobutyl Grignard; (b) l -bromo-3-methyl-2-butene + CH3CH2C(CH3hMgBr; (c) cyclopentyl-MgBr + I -bromo-2-pentene. 15.28. (a) 1 9,000; (b) second structure. 15.29. 3-bromo- l -hexene, and (cis and trans)­ l -bromo-2-hexene. 15.32. (a) The product isomelized, 1 630 suggests conjugated; (b) 2-propyl-1 ,3-cyclohexadiene.

CHAPTER 1 6 16.2. (a) + 3 1 .8 kJ/mole; (b) - 88.6 kJ/mole; (c) - 1 1 2.0 kJ/mole. 16.5. Two of the eight pi electrons are unpaired in two non-bonding orbitals, an unstable configuration. 16.7. (a) nonaromatic (internal H's prevent planarity); (b) nonaromatic (one ring atom has no p orbital); (c) aromatic, [ J 4]annulene; (d) aromatic (in the outer system). 16.8. Azulene is aromatic, but the other two are antiaromatic. 16.10. The cation (cyclopropenium ion) is aromatic; the anion is anti aromatic. 16.12. (a) antiaromatic if planar; (b) aromatic if planar; (c) aromatic if planar; (d) anti aromatic if planar; (e) nonaromatic; (f) aromatic if planar. 16.14. cyclopropenium fluoroborate. 16.18. (a) aromatic; (b) aromatic; (c) nonaromatic; (d) aromatic; (e) aromatic; (f) nonaromatic; (g) aro­ matic. 16.22. (a) fluorobenzene; (b) 4-phenyl- l -butyne; (c) 3-methyl­ phenol or m-cresol; (d) o-nitrostyrene; (e) p-bromobenzoic acid; (f) isopropyl phenyl ether; (g) 3,4-dinitrophenol; (h) benzyl ethyl ether. 16.25. 3-phenyl-2-propen- l -ol 16.28. (a) o-dichlorobenzene; (b) p-nit:roanisole; (c) 2,3-dibromobenzoic acid; (d) 2,7-dimethoxynaph­ thalene; (e) m-chlorobenzoic acid; (f) 2,4,6-trichlorophenol; (g) 2-sec­ butyl benzaldehyde; (h) cyclopropenium fluoroborate. 16.30. The second is deprotonated to an aromatic cyclopentadienyl anion. 16.31. (d), (e) The fourth structure, with two three-membered rings, was considered the most likely and was called Ladenburg benzene. 16.37. (a) three; (b) one; (c) mela-dibromobenzene. 16.38. (a) a-chloroacetophenone; (b) 4-bromo- I -ethylbenzene. 16.45. 2-iso­ propy 1-5-methyIphenol.

CHAPTER 1 7 17.4. The sigma complex for p-xylene has the + charge on two 2° carbons and one 3° carbon, compared with three 2° carbons in benzene.

Answers to Selected Problems 17.10. Bromine adds to the alkene but substitutes on the aryl ether, evolving gaseous HBr. 17.11. Strong acid is used for nitration, and the amino group of aniline is protonated to a deactivating - NH3 + group. 17.13. I -Bromo- l -chlorocyclohexane; the intermediate cation is stabi­ lized by a bromonium ion resonance form. 17.14. (a) 2,4- and 2,6dinitrotoluene; (b) 3-chloro-4-nitrotoluene and 5-chloro-2-nitrotoluene; (c) 3- and 5-nitro-2-bromobenzoic acid; (d) 4-methoxy-3-nitrobenzoic acid; (e) 5-methyl-2-nitrophenol and 3-methyl-4-nitrophenol. 17.17. (a) phenylcyclohexane; (b) 0- and p-methylanisole, with overalkylation products; (c) l -isopropyl-4-( l , I ,2-trimethylpropyl)benzene. 17.18. (a) phenylcyclohexane; (b) t-butylbenzene; (c) p-di-t-butylbenzene; (d) 0and p-isopropyl toluene. 17.19. (a) t-butylbenzene; (b) 2- and 4-sec­ butyltoluene; (c) no reaction; (d) ( l , 1 ,2-trimethylpropyl)benzene. 17.20. (a) sec-butylbenzene and others; (b) OK; (c) +disub, trisub; (d) OK (some orrho); (e) OK. 17.22. (a) (CH3 hCHCH2COCI, benzene, AICI3; (b) ( CH 3hCCOCI, benzene, AICI3; (c) PhCOCI, benzene, AICI 3; (d) CO/HCI, AICI3/CuCI, anisole; (e) Clemmensen on (b); (f) CH3 (CH2hCOCI, benzene, AICI3 then Clemmensen. 17.23. Fluoride leaves in a fast exothermic step; the C - F bond is only slight­ ly weakened in the reactant-like transition state (Hammond postulate). 17.25. (a) 2,4-dinitroanisole; (b) 2,4- and 3,5-dimethylphenol; (c) N-methyl-4-nitroaniline; (d) 2,4-dinitrophenylhydrazine. 17.29. (a) (trichloromethyl) hexachlorocyclohexane; (b) l -methyl- I ,4-cyclohexa­ diene; (c) cis and trans- I ,2- dimethylcyclohexane; (d) 1 ,4-dimethyl1 ,4-cyclohexadiene. 17.30. (a) benzoic acid; (b) terephthalic acid (benzene- I ,4-dicarboxylic acid); (c) phthalic acid (benzene- l ,2-dicar­ boxylic acid). 17.33. 60% beta, 40% alpha; reactivity ratio 1 .9 to l . 17.36. (a) I -bromo- I -phenylpropane. 17.38. (a) HBr, then Grignard with ethylene oxide; (b) CH3COCI and AICI3, then Clemmensen, Br2 and light, then -OCH3; (c) nitrate, then Br2 and light, then NaCN. 17.40. (a) 3-ethoxytoluene; (b) m-tolyl acetate; (c) 2,4,6-tribromo3-methy Iphenol; (d) 2,4,6-tribromo-3-(tribromomethyl)phenol; (e) 2-methyl- 1 ,4-benzoquinone; (f) 2,4-di-t-butyl-3-methyl-phenol. 17.50. indanone 17.55. The yellow species is the triphenylmethyl cation. 17.60. kinetic control at 0°, thermodynamic control at 1 00°. 17.61. Brominate, then Grignard with 2-butanone. =

CHAPTER 1 8 18.1. (a) 5-hydroxy-3-hexanone; ethyl {3-hydroxypropyl ketone; (b) 3-phenylbutanal: {3-phenylbutyraldehyde; (c) trans-2-methoxycy­ clohexanecarbaldehyde; (d) 6,6-dimethyl-2,4-cyclohexadienone. 18.2. (a) 2-phenylpropanal; (b) acetophenone. 18.3. No y-hydrogens. 18.5. (a)