Schaum's Outline of Theory and Problems of Organic Chemistry

SCHAUM’S OUTLINE OF

THEORY AND PROBLEMS OF

ORGANIC CHEMISTRY Second Edition

HERBERT MEISLICH, Ph.D. Professor Emeritus of Chemistry City College of CUNY

HOWARD NECHAMKIN, Ed.D. Professor Emeritus of Chemistry Trenton State College and

JACOB SHAREFKIN, Ph.D. Professor Emeritus of Chemistry Brooklyn College of CUNY

SCHAUM’S OUTLINE SERIES McGRAW-HILL

New York St. Louis San Francisco Auckland Bogoti Caracas Lisbon London Madrid Mexico City Milan Montreal New Delhi San Juan Singapore Sydney Tokyo Toronto

HERBERT MEISLICH holds a B.A. degree from Brooklyn College and an M.A. and Ph.D. from Columbia University. He is a professor emeritus from the City College of CUNY, where he taught Organic and General Chemistry for forty years at both the undergraduate and doctoral levels. He received the Outstanding Teacher award in 1985, and has coauthored eight textbooks, three laboratory manuals in General and Organic Chemistry and 15 papers on his research interests. HOWARD NECHAMKIN is Professor Emeritus of Chemistry at Trenton State College; for 11 years of his tenure he served as Department Chairman. His Bachelor’s degree is from Brooklyn College, his Master’s from the Polytechnic Institute of Brooklyn and his Doctorate in Science Education from New York University. He is the author or coauthor of 53 papers and 6 books in the areas of inorganic, analytical, and environmental chemistry. JACOB SHAREFKIN is Professor Emeritus of Chemistry at Brooklyn College. After receiving a B.S. from City College of New York, he was awarded an M.A. from Columbia University and a Ph.D. from New York University. His publications and research interest in Qualitative Organic Analysis and organic boron and iodine compounds have been supported by grants from the American Chemical Society, for whom he has also designed national examinations in Organic Chemistry.

Schaum’s Outline of Theory and Problems of ORGANIC CHEMISTRY Copyright 0 1991,1977 by The McGraw-Hill Companies Inc. All rights reserved. Printed in the United States of America. Except as permitted under the Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a data base or retrieval system, without the prior written permission of the publisher. 8 9 10 1 1 12 13 14 15 16 17 18 19 20 PRS PRS 9 8 7

ISBN 0-07-041458-0 Sponsoring Editor: David Beckwith Production Supervisor: Fred Schulte Editing Supervisors: Patty Andrews, Meg Tobin, Maureen Walker

Library of Congress Cataloging-in-PublicationData

Meislich. Herbert. Schaum’s outline of theory and problems of organic chemistry / Herbert Meislich, Howard Nechamkin, and Jacob Sharefkin. - - 2nd ed. cm. - - (Schaum’s outline series) p. Includes index. ISBN 0-07-041458-0 1. Chemistry, Organic- -Problems, exercises, etc. 2. Chemistry, Organic- -Outlines, syllabi, etc. I . Nechamkin, Howard, 1 1 . Sharefkin. Jacob. 111. Title. QD257.M44 1991 547- -dc20 90-6383 CI P

McGraw -Hill A Division of The McGmwHiU Companies

gz

To Amy Nechamkin, Belle D. Sharefkin, and John B. Sharefkin

This page intentionally left blank

The beginning student in Organic Chemistry is often overwhelmed by facts, concepts, and new language. Each year, textbooks of Organic Chemistry grow in quantity of subject matter and in level of sophistication. This Schaum’s Outline was undertaken to give a clear view of first-year Organic Chemistry through the careful detailed solution of illustrative problems. Such problems make up over 80% of the book, the remainder being a concise presentation of the material. Our goal is for students to learn by thinking and solving problems rather than by merely being told. This book can be used in support of a standard text, as a supplement to a good set of lecture notes, as a review for taking professional examinations, and as a vehicle for self-instruction. The second edition has been reorganized by combining chapters to emphasize the similarities of functional groups and reaction types as well as the differences. Thus, polynuclear hydrocarbons are combined with benzene and aromaticity. Nucleophilic aromatic displacement is merged with aromatic substitution. Sulfonic acids are in the same chapter with carboxylic acids and their derivatives, and carbanion condensations are in a separate new chapter. Sulfur compounds are discussed with their oxygen analogs. This edition has also been brought up to date by including solvent effects, CMR spectroscopy, an elaboration of polymer chemistry, and newer concepts of stereochemistry, among other material. Our thanks and appreciation to Mr. David Beckwith and Maureen Walker for their careful and thorough proofreading. HERBERT MEISLICH HOWARD NECHAMKIN JACOBSHAREFKIN

V

This page intentionally left blank

Contents Chapter

Chapter

Chapter

Chapter

1

2

3

4

STRUCTURE AND PROPERTIES. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Carbon Compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Lewis Structural Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Types of Bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Functional Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Formal Charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5

1 2 5 6 6

BONDING AND MOLECULAR STRUCTURE . . . . . . . . . . . . . . . . . . . . 2.1 Atomic Orbitals., . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Covalent Bond Formation-Molecular Orbital (MO) Method . . . . . . . . . . . 2.3 Hybridization of Atomic Orbitals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Electronegativity and Polarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Oxidation Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Intermolecular Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 Solvents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8 Resonance and Delocalized 7~ Electrons . . . . . . . . . . . . . . . . . . . . . . . . . . .

12

CHEMICAL REACTIVITY AND ORGANIC REACTIONS . . . . . . . . . . 3.1 Reaction Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Carbon-Containing Intermediates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Types of Organic Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Electrophilic and Nucleophilic Reagents . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Bond-Dissociation Energies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 Chemical Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.8 Rates of Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.9 Transition-State Theory and Enthalpy Diagrams . . . . . . . . . . . . . . . . . . . . . 3.10 Bronsted Acids and Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.11 Basicity (Acidity) and Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.12 Lewis Acids and Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

30

ALKANES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

49

12 13 16 19 20 20 21 22

30 30 32 33 34 35 36 37 38 40 41 43

Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Nomenclature of Alkanes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Preparation of Alkanes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chemical Properties of Alkanes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Summary of Alkane Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

49 54 55 57 61

STEREOCHEMISTRY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Stereoisomerism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Optical Isomerism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

68

4.1 4.2 4.3 4.4 4.5

Chapter

1

vii

68 69

...

CONTENTS

Vlll

5.3 Relative and Absolute Configuration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Molecules with More Than One Chiral Center . . . . . . . . . . . . . . . . . . . . . . 5.5 Synthesis and Optical Activity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

71 76 78

ALKENES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Nomenclature and Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Geometric (cis-tram) Isomerism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Preparation of Alkenes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Chemical Properties of Alkenes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 Substitution Reactions at the Allylic Position . . . . . . . . . . . . . . . . . . . . . . . 6.6 Summary of Alkene Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

87

5.4

Chapter

Chapter

Chapter

6

7

8

ALKYL HALIDES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Synthesis of RX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Chemical Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 Summary of Alkyl Halide Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

119

ALKYNES AND DIENES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

142

Chapter

9

10

119 120 121 132

Alkynes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chemical Properties of Acetylenes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Alkadienes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . MO Theory and Delocalized T Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . Addition Reactions of Conjugated Dienes . . . . . . . . . . . . . . . . . . . . . . . . . . Polymerization of Dienes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Cycloaddition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Summary of Alkyne Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Summary of Diene Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

142 145 148 149 152 155 156 157 157

CYCLIC HYDROCARBONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Nomenclature and Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Geometric Isomerism and Chirality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Conformations of Cycloalkanes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4 Synthesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5 Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.6 MO Theory of Pericyclic Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.7 Terpenes and the Isoprene Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

166

8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9

Chapter

87 88 91 96 107 108

BENZENE AND POLYNUCLEAR AROMATIC COMPOUNDS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Aromaticity and Huckel's Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10.3 Antiaromaticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4 Polynuclear Aromatic Compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

166 167 170 179 181 184 188

198 198 202 203 206

CONTENTS

10.5 Nomenclature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6 Chemical Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.7 Synthesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Chapter

11

12

13

207 209 212

.

AROMATIC SUBSTITUTION ARENES . . . . . . . . . . . . . . . . . . . . . . . . 11.1 11.2 11.3 11.4 11.5

Chapter

ix

215’ Aromatic Substitution by Electrophiles (Lewis Acids. E’ or E) . . . . . . . 215 Electrophilic Substitutions in Syntheses of Benzene Derivatives . . . . . . . . 224 Nucleophilic Aromatic Substitutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 Arenes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228 Summary of Arene and Aryl Halide Chemistry . . . . . . . . . . . . . . . . . . . . . 234

SPECTROSCOPY AND STRUCTURE . . . . . . . . . . . . . . . . . . . . . . . . . . 12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2 Ultraviolet and Visible Spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3 Infrared Spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.4 Nuclear Magnetic Resonance (Proton. PMR) . . . . . . . . . . . . . . . . . . . . . . . 12.5 ‘ T N M R ( C M R ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.6 Mass Spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

242 242 243 244 248 257 259

Chapter

ALCOHOLS AND THIOLS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A . Alcohols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.1 Nomenclature and H-Bonding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2 Preparation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.3 Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.4 Summary of Alcohol Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B. Thiols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.5 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.6 Summary of Thiol Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

269 269 269 270 274 279 280 280 281

Chapter

14

ETHERS. EPOXIDES. GLYCOLS. AND THIOETHERS. . . . . . . . . . . . 292 A . Ethers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292 292 14.1 Introduction and Nomenclature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2 Preparation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292 14.3 Chemical Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296 298 14.4 Cyclic Ethers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300 14.5 Summary of Ether Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B. Epoxides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300 14.6 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300 14.7 Synthesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300 301 14.8 Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303 14.9 Summary of Epoxide Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . C . Glycols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304 304 14.10 Preparation of 1.2.Glycols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.11 Unique Reactions of Glycols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304 14.12Summary of Glycol Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306

CONTENTS

X

D. Thioethers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

14.13 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.14 Preparation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.15 Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Chapter

15

CARBONYL COMPOUNDS: ALDEHYDES AND KETONES . . . . . . . . 315 15.1 15.2 15.3 15.4 15.5 15.6 15.7 15.8 15.9

Introduction and Nomenclature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Preparation . . . ......................................... Oxidation and Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Addition Reactions of Nucleophiles to ‘C=O . . . . . . . . . . . . . . . . . . . . . Addition of Alcohols: Acetal and Ketahormation . . . . . . . . . . . . . . . . . . Attack by Ylides; Wittig Reaction . . . . ........................ Miscellaneous Reactions ........................ Summary of Aldehyde Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Summary of Ketone Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ~

Chapter

16

17

~~

Introduction and Nomenclature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Preparation of Carboxylic Acids . . ............................ Reactions of Carboxylic Acids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Summary of Carboxylic Acid Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . Polyfunctional Carboxylic Acids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ....... Transacylation; Interconversion of Acid Derivatives . . . . . More Chemistry of Acid Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Summary of Carboxylic Acid Derivative Chemistry . . . . . . . . . . . . . . . . . Analytical Detection of Acids and Derivatives . . . . . . . . . . . . . . . . . . . . . Carbonic Acid Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Summary of Carbonic-Acid-Derivative Chemistry ....................... Synthetic Condensation Polymers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Derivatives of Sulfonic Acids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

CARBANION-ENOLATES AND ENOLS . . . . . . . . . . . . . . . . . . . . . . . . . 17.1 Acidity of H’s a to C=O; Tautomerism . . . . . . . . . . . . . . . . . . . . . . . . . .

17.2 Alkylation of Simple Carbanion-Enolates . . . . . . . . . . . . . . . . . . . . . . . . . 17.3 Alkylation of Stable Carbanion-Enolates . . . . . . . . . . . . . . . . . . . . . . . . . . 17.4 Nucleophilic Addition to Conjugated Carbonyl Compounds: Michael 3,4-Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.5 Condensations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Chapter

18

315 317 323 325 330 332 334 336 337

CARBOXYLIC ACIDS AND THEIR DERIVATIVES . . . . . . . . . . . . . . . 344 16.1 16.2 16.3 16.4 16.5 16.6 16.7 16.8 16.9 16.10 16.11 16.12 16.13

Chapter

307 307 307 307

344 347 348 354 354 358 360 367 367 369 370 371 372

384 384 389 391 396 397

AMINES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

412

Nomenclature and Physical Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . Preparation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chemical Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Reactions of Quaternary Ammonium Salts . . . . . . . . . . . . . . . . . . . . . . . . Ring Reactions of Aromatic Amines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Spectral Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Reactions of Aryl Diazonium Salts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.8 Summary of Amine Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

412 414 418 424 425 426 426 429

18.1 18.2 18.3 18.4 18.5 18.6 18.7

CONTENTS

Chapter

19

PHENOLIC COMPOUNDS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

440

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Preparation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chemical Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Analytical Detection of Phenols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Summary of Phenolic Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Summary of Phenolic Ethers and Esters . . . . . . . . . . . . . . . . . . . . . . . .

440 441 443 450 451 451

19.1 19.2 19.3 19.4 19.5 19.6

Chapter

Chapter

20

21

AROMATIC HETEROCYCLIC COMPOUNDS . . . . . . . . . . . . . . . . . . . 457 20.1 Five-Membered Aromatic Heterocycles with One Heteroatom . . . . . . . . . 457

20.2 Six.Membered. Heterocycles with One Heteroatom . . . . . . . . . . . . . . . . . 20.3 Compounds with Two Heteroatoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.4 Condensed Ring Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

463 467 468

AMINO ACIDS. PEPTIDES. PROTEINS . . . . . . . . . . . . . . . . . . . . . . . . .

474

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Preparation of a-Amino Acids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chemical Properties of Amino Acids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Peptides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Proteins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Summary of Amino Acid Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

474 476 478 481 486 488

CARBOHYDRATES AND NUCLEIC ACIDS . . . . . . . . . . . . . . . . . . . . .

494

21.1 21.2 21.3 21.4 21.5 21.6

Chapter

22

xi

22.1 22.2 22.3 22.4 22.5 22.6 22.7

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chemical Properties of Monosaccharides . . . . . . . . . . . . . . . . . . . . . . . . . . Evidence for Hemiacetal Formations as Exemplified with Glucose . . . . . Stereochemistry and Structure Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Disaccharides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Polysaccharides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Nucleic Acids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

INDEX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

494 495 498 500 507 510 511

517

This page intentionally left blank

Chapter 1 Structure and Properties 1.1 CARBON COMPOUNDS

Organic chemistry is the study of carbon (C) compounds, all of which have covalent bonds. Carbon atoms can bond to each other to form open-chain compounds, Fig. l-l(o), or cyclic (ring) compounds, Fig. 1-l(c). Both types can also have branches of C atoms, Fig. 1-l(b) and (d). Saturated compounds have C’s bonded to each other by single bonds, C-C; unsaturated compounds have C’s joined by multiple bonds. Examples with double bonds and triple bonds are shown in Fig. 1-l(e). Cyclic compounds having at least one atom in the ring other than C (a heteroatom) are called heterocyclics, Fig. 1-l(f). The heteroatoms are usually oxygen (0),nitrogen (N), or sulfur (S). H H H H H

I I

l I

H-C-C-C-C-H

l I

1 I

H H H

I I

l l

H H ‘C/ H q \C,H \ / H H

l l

H-C-C-C-H

‘

H C H

H H H H

/I

H H H

H

‘c’

H

‘c’

‘H

/ \C/H

H\

F-

\

H

H

n- Butane

lsobutane

un branched, open-chain

open -chain

Cyclopropane

Met hylcyclopropane

unbranched. cyclic

branched. cyclic

(a)

(b)

(c)

(d)

branched.

H H

I

H, ,C=C,

H

HH

H

I

H-F sp2 > sp

s

Problem 2.11 What effect does hybridization have on the stability of bonds?

4

Hybrid orbitals can ( a ) overlap better and ( b ) provide greater bond angles, thereby minimizing the repulsion between pairs of electrons and making for greater stability.

By use of the generalization that each unshared and o-bonded pair of electrons needs a hybrid orbital, but 7~ bonds do not, the number of hybrid orbitals (HON)needed by C or any other central atom can be obtained as HON = (number of

(J

bonds) + (number of unshared pairs of electrons)

CHAP. 21

19

BONDING AND MOLECULAR STRUCTURE

The hybridized state of the atom can then be predicted from Table 2-3. If more than four HO's are needed, d orbitals are hybridized with the s and the three p's. If five HO's are needed, as in PCI,, one d orbital is included to give trigonal-bipyramidal sp3d HO's, Fig. 2-9(a). For six HO's, as in SF,, two d orbitals are included to give octahedral sp3d2HO's, Fig. 2-9(b). The above method can also be used for the multicovalent elements in the second and, with few exceptions, in the higher periods of the periodic table. Table 2-3

Predicted Hybrid State

sp sp ' d sp'd

I

6

%3 120"

90"

(a) sp& HO's

( 6 ) s p s d 2 HO's Fig. 2-9

Problem 2.12 Use the HON method to determine the hybridized state of the underlined elements: (a) CHCl,

Number of

U

(6) H 2 C e H , Bonds

+

(c) -4

Number of Unshared Electron Pairs

4

3 2 2 1

3

2.4

(d) H e N :

0 0 0 0 1 1

=

( e ) H&+

4

HON

Hybrid State

4

SP3

3 2 2 2

SP

4

sps

SP2 SP

SP

ELECTRONEGATIVITY AND POLARITY

The relative tendency of a bonded atom in a molecule to attract electrons is expressed by the term electronegativity. The higher the electronegativity, the more effectively does the atom attract and hold electrons. A bond formed by atoms of dissimilar electronegativities is called polar. A nonpolar covalent bond exists between atoms having a very small or zero difference in electronegativity. A few relative electronegativities are F (4.0) > O(3.5) > C1, N (3.0) > Br (2.8) > S , C, I ( 2 . 5 ) > H (2.1)

20

B O N D I N G A N D M O L E C U L A R STRUCTURE

[CHAP. 2

The more electronegative element of a covalent bond is relatively negative in charge, while the less electronegatirie eienient is relutiveiy positive. The symbols 6 + and S - represent partial charges (bond polarity). These partial charges should not be confused with ionic charges. Polar bonds are indicated by ++; the head points toward the more electronegative atom. The vector sum o f all individual bond moments gives the net dipole moment of the molecule. Problem 2.73 What d o the molecular dipole moments g the shapes o f thcse molecules'?

=0

for CO2 and

,U

= 1.84

D for H,O tell you about

4

In CO,

0 is more electronegative than C, and each C-0 bond is polar as shown. A zero dipole moment indicates a symmetrical distribution o f 6 - charges about the 6 + carbon. The geometry must be linear; in this way, in div id ua I bond momen t s ca nce I : c--fct---,

o=c=o

H , O also has polar bonds. However, since there is a net dipole moment, the individual bond moments do not cancel. and the molecule must have a bent shape:

pqH H

f rmrltunt moment

2.5 OXIDATION NUMBER

The oxidation number (ON) is a value assigned to an atom based on relative electronegativities. It equals the group number minus the number of assigned electrons, when the bonding electrons are assigned to the more electronegative atom. The sum of all ( 0 N ) ' s equals the charge on the species. Problem 2.74 Determine the oxidation number of each C, (ON),., in: ( a ) CH,. (6) C H , O H , (c) CH,NH,, 4 ( d ) H,C=CH,. Use the data ( O N ) , = -3; (ON),, = 1; (ON),, = -2. All examples are molecules; therefore the sum of all ( O N ) values is 0. (U)

(ON), + 4(ON),, = 0;

(ON),

+ (4 x

1) = O ;

(ON),. = -4

( 6 ) ( O N ) , + (ON),) + 4(ON),, = 0;

(ON),. + (-2)

(c) ( O N ) , + ( O N ) , + 5(ON),,

(ON),

= 0;

+ (-3)

+ 4 =0; +5 =O;

(ON),. = -2

(ON),. = -2

( d ) Since both C atoms are equivalent,

2(ON),

2.6

+ 4(ON),,

= 0;

2(ON), + 4 = 0 ;

(ON),.= - 2

INTERMOLECULAR FORCES

(a) Dipole-dipole interaction results from the attraction of the 6 + end of one polar molecule for the 6- end of another polar molecule. (b) Hydrogen-bond. X-H and :Y may be bridged X-H---:Y if X and Y are small, highly electronegative atoms such as F, 0, and N. H-bonds also occur intramolecularly. (c) London (van der Waals) forces. Electrons of a nonpolar molecule may momentarily cause an imbalance of charge distribution in neighboring molecules, thereby inducing a temporary dipole moment. Although constantly changing, these induced dipoles result in a weak net attractive force.

CHAP. 21

BONDING AND MOLECULAR STRUCTURE

21

The greater the molecular weight of the molecule, the greater the number of electrons and the greater these forces. The order of attraction is H-bond S dipole-dipole > London forces Problem 2.15 Account for the following progressions in boiling point. (a) CH,, -161.5"C; Cl,, -34°C; 4 CH,CI, -24 "C. ( b ) CH,CH,OH, 78 "C; CH,CH,F, 46 "C; CH,CH,CH,, -42 "C.

The greater the intermolecular force, the higher the boiling point. Polarity and molecular weight must be considered. ( a ) Only CH,CI is polar, and it has the highest boiling point. CH, has a lower molecular weight (16 g/mole) than has Cl, (71 g/mole) and therefore has the lowest boiling point. (6) Only CH,CH,OH has H-bonding, which is a stronger force of intermolecular attraction than the dipole-dipole attraction of CH,CH,F. CH,CH,CH, has only London forces, the weakest attraction of all.

Problem 2.16 Account for the following progression in boiling point: CH,CI, -24 "C; CH,Br, 5 "C; CH,I, 43 "C. 4

The order of polarity is CH,CI>CH,Br>CH,I. The order of molecular weight is CH,I>CH,Br> CH,CI. The two trends oppose each other in affecting the boiling point. The order of molecular weight predominates here. Problem 2.17 The boiling points of n-pentane and its isomer neopentane are 36.2 "C and 9.5 " C , respectively. Account for this difference. (See Problem 1.4 for the structural formulas.) 4

These isomers are both nonpolar. Therefore, another factor, the shape of the molecule, influences the boiling point. The shape of n-pentane is rodlike, whereas that of neopentane is spherelike. Rods can touch along their entire length; spheres touch only at a point. The more contact between molecules, the greater the London forces. Thus, the boiling point of n-pentane is higher. Problem 2.18 Ethyl alcohol, C,H,OH, boils at 78.3"C. Its isomer, dimethyl ether, CH,OCH,, boils at -24 "C. Explain. 4

Both are polar molecules and have dipole-dipole attraction. However. in ethyl alcohol, H-bonding can exist:

H

I

H,CCH,O----H-OCH,CH, In CH,OCH, the H's are on C and so cannot H-bond.

2.7 SOLVENTS

The oppositely charged ions of salts are strongly attracted by electrostatic forces, thereby accounting for the high melting and boiling points of salts. These forces of attraction must be overcome in order for salts to dissolve in a solvent. Nonpolar solvents have a zero or very small dipole moment. Protic solvents are highly polar molecules having an H that can form an H-bond. Aprotic solvents are highly polar molecules that do not have an H that can form an H-bond. Problem 2.19 Classify the following solvents: (a) (CH,),S=O, tetrachloride; (c) C,H,, benzene;

dimethyl sulfoxide; ( b ) CCI,, carbon

22

BONDING AND MOLECULAR STRUCTURE

[CHAP. 2

F

(d) H N(CH,)2 Dimethylformamide 0

(e) CH,OH, methanol; ( f ) liquid NH,.

4

Nonpofar: (b) Because of the symmetrical tetrahedral molecular shape, the individual C-CI bond moments cancel. (c) With few exceptions, hydrocarbons are nonpolar. Protic: (e) and ( f ) . Aprotic: ( a ) and (d). The C--0and C=O groups are strongly polar and the H's attached to C do not typically form H-bonds. Problem 2.20 Mineral oil, a mixture of high-molecular-weight hydrocarbons, dissolves in n-hexane but not in 4 water or ethyl alcohol, CH,CH,OH. Explain. Attractive forces between nonpolar molecules such as mineral oil and n-hexane are very weak. Therefore, such molecules can mutually mix and solution is easy. The attractive forces between polar H,O or C,H,OH molecules are strong H-bonds. Most nonpolar molecules cannot overcome these H-bonds and therefore do not dissolve in such polar protic solvents. Problem 2.21 Explain why CH,CH,OH is much more soluble in water than is CH,(CH,),CH,OH.

4

The OH portion of an alcohol molecule tends to interact with water-it is hydrophilic. The hydrocarbon portion does not interact, rather it is repelled- it is hydrophobic. The larger the hydrophobic portion, the less soluble in water is the molecule. Problem 2.22 Explain why NaCl dissolves in water but not in hexane.

4

Water, a protic solvent, helps separate the strongly attracting ions of the solid salt by solvation. Several water molecules surround each positive ion (Na') by an ion-dipole attraction. The 0 atoms, which are the negative ends of the molecular dipole, are attracted to the cation. H,O typically forms an H-bond with the H-bond negative ion (in this case CIF). ion-dipole attraction

artracrion

61

6*

CI----H

H

6 -/ Na'---0

\6+

H

and

\a-

0

6+/

H

Problem 2.23 Compare the ways in which NaCl dissolves in water and in dimethyl sulfoxide.

4

The way in which NaCI, a typical salt, dissolves in water, a typical protic solvent, was discussed in Problem 2.22. Dimethyl sulfoxide also solvates positive ions by an ion-dipole attraction; the 0 of the C-0 group is attracted to the cation. However, since this is an aprotic solvent, there is no way for an H-bond to be formed and the negative ions are not solvated when salts dissolve in aprotic solvents. The S, the positive pole, is surrounded by the methyl groups and cannot get close enough to solvate the anion.

The bare negative ions discussed in Problem 2.23 have a greatly enhanced reactivity. The small amounts of salts that dissolve in nonpolar or weakly polar solvents exist mainly as ion-pairs or ion-clusters, where the oppositely charged ions are close to each other and move about as units. Tight ion-pairs have no solvent molecules between the ions; loose ion-pairs are separated by a small number of solvent molecules. 2.8

RESONANCE AND DELOCALIZED n ELECTRONS

Resonance theory describes species for which a single Lewis electron structure cannot be written. As an example, consider dinitrogen oxide, N,O:

23

BONDING AND MOLECULAR STRUCTURE

CHAP. 21

-

..

-

t

:N=N=O:

Calculated bond length, nm Observed bond length, nm

:N-N-Q:

+

.. -

resonance

0.120

0.115

0.110

0.147

0.112

0.119

0.112

0.119

(with formal charges)

A comparison of the calculated and observed bond lengths shows that neither structure is correct. Nevertheless, these contributing (resonance) structures tell us that the actual resonance hybrid has some double-bond character between N and 0, and some triple-bond character between N and N . This state of affairs is described by the non-Lewis structure 6-

4;

+

:N=NcO:

in which broken lines stand for the partial bonds in which there are delocalized p electrons in an extended ?T bond created from overlap of p orbitals on each atom. See also the orbital diagram, Fig. denotes resonance, not equilibrium. 2-10. The symbol

Fig. 2-10

The energy of the hybrid, E,, is always less than the calculated energy of any hypothetical contributing structure, E,. The difference between these energies is the resonance (delocalization) energy, E,:

E, = E,

-

E,,

The more nearly equal in energy the contributing structures, the greater the resonance energy and the less the hybrid looks like any of the contributing structures. When contributing structures have dissimilar energies, the hybrid looks most like the lowest-energy structure. Contributing structures ( a ) differ only in positions of electrons (atomic nuclei must have the same positions) and (6) must have the same number of paired electrons. Relative energies of contributing structures are assessed by the following rules.

1. Structures with the greatest number of covalent bonds are most stable. However, for second-period elements (C, 0, N) the octet rule must be observed. 2. With few exceptions, structures with the least number or amount of formal charges are most stable. 3. If all structures have formal charge, the most stable (lowest energy) one has - on the more electronegative atom and + on the more electropositive atom. 4. Structures with like formal charges on adjacent atoms have very high energies. 5 . Resonance structures with electron-deficient, positively charged atoms have very high energy, and are usually ignored. Problem 2.24 Write contributing structures, showing formal charges when necessary, for ( a ) ozone, 0,; ( 6 ) CO,; (c) hydrazoic acid, HN,; ( d ) isocyanic acid, HNCO. Indicate the most and least stable structures and give 4 reasons for your choices. Give the structure of the hybrid.

[CHAP. 2

BONDING AND MOLECULAR STRUCTURE

24

:o=o-o: It/ + b :

t--*

:o-O=$: 0

(equal-energy structures).

9.-

2

The hybrid is :O-0-0:.

4;

( 1 ) is most stable; it has no formal charge. (2) and (3) have equal energy and are least stable because they have formal charges. In addition, in both (2) and ( 3 ) , one 0, an electronegative element, bears a + formal charge. Since (1) is so much more stable than ( 2 ) and (3), the hybrid is :O=C=O:, which is just ( 1 ) .

(1) And (2) have about the same energy and are the most stable, since they have the least amount of formal charge. (3) has very high energy since it has + charge on adjacent atoms and, in terms of absolute value, a total formal charge of 4. (4) has a very high energy because the N bonded to H has only six electrons. The hybrid, composed of (1) and (2), is

(1) has no formal charge and is most stable. (2) is least stable since the - charge is on N rather than on the

more electronegative 0 as in (3). The hybrid is H-N=C-+: contributing structure.

(the same as ( I ) ) , the most stable

Problem 2.25 ( a ) Write contributing structures and the delocalized structure for (i) NOT and (ii) NO;. ( b ) Use p AO’s to draw a structure showing the delocalization of the p electrons in an extended T bond for (i) and 4 (ii). (c) Compare the stability of the hybrids of each.

The - is delocalized over both 0’s so that each can be assumed to have a the same bond length.

-

f charge. Each N - 4 bond has

The - charges are delocalized over three 0’s so that each has a - $ charge. (6) See Fig. 2-11. (c) We can use resonance theory to compare the stability of these two ions because they differ in only one feature-the number of 0 ’ s on each N, which is related to the oxidation numbers of the N’s. We could not, for example, compare NOT and HSO;, since they differ in more than one way; N and S are in different groups and periods of the periodic table. NO, is more stable than NOT since the charge on NOS is delocalized (dispersed) over a greater number of 0 ’ s and since NO, has a more extended T bond system. Problem 2.26 Indicate which one of each of the following pairs of resonance structures is the less stable and is an unlikely contributing structure. Give reasons in each case.

25

BONDING A N D MOLECULAR STRUCTURE

CHAP. 21

P

Y

f+l

Fig. 2-11

I

I1

(c) H,C=CH-CH,

+

.c-+

V (e)

H,c-CI: IX

-

H,C-CH-CH,

-

-

*

VI ++

H,C=~I:

4

X

I has fewer covalent bonds, more formal charge and an electron-deficient N. (6) IV has + on the more electronegative 0. (c) VI has similar - charges on adjacent C's, fewer covalent bonds, more formal charge and an electrondeficient C. ( d ) VII has fewer covalent bonds and a + on the more electronegative N , which is also electron-deficient. (e) C in X has 10 electrons; this is not possible with the elements of the second period. (a)

Supplementary Problems Problem 2.27 Distinguish between an AO, an HO, an MO and a localized MO.

4

An A 0 is a region of space in an atom in which an electron may exist. An HO is mathematically fabricated from some number of AO's to explain equivalency of bonds. An M O is a region of space about the entire molecule capable of accommodating electrons. A localized MO is a region of space between a pair of bonded atoms in which the bonding electrons are assumed to be present. Problem 2.28 Show the orbital population of electrons for unbonded N in (a) ground state, (6) sp3, (c) sp'. 4 and (d) sp hybrid states.

BONDING AND MOLECULAR STRUCTURE

26

(a)

t.l

Is

fi 2s

r r2P

t

(4 TJ.

tL t

t

- --

2sp‘

[CHAP. 2

1 2p

ts- t -t t ( d ) 7s. -2sp -’ 2SP 2P Note that since the energy difference between hybrid and p orbitals is so small, Hund’s rule prevails over the Aufbau principle. (6)

tS. t.l -t t t

Problem 2.29 ( a ) NO; is linear, ( 6 ) NO, is bent. Explain in terms of the hybrid orbitals used by N.

4

(a) NO;, :&N=6:. N has two a bonds, no unshared pairs of electrons and therefore needs two hybrid orbitals. N uses s p hybrid orbitals and the U bonds are linear. The geometry is controlled by the arrangement of the sigma bonds. ( 6 ) NO,, : b N : O F . N has two U bonds, one unshared pair of electrons and, therefore, needs three hybrid orbitals. N uses sp‘ hybrid HO’s, and the bond angle is about 120”. Problem 2.30 Draw an orbital representation for the cyanide ion. :GENT.

4

See Fig. 2-12. The C and N each have one U bond and one unshared pair of electrons, and therefore each needs two s p hybrid HO’s. On each atom one s p hybrid orbital forms a (r bond while the other has the unshared pair. Each atom has a p , A 0 and a p , AO. The two p , orbitals overlap to form a n, bond in the xy-plane; the two p , orbitals overlap to form a T , bond in the xz-plane. Thus, two T bonds at right angles to each other and a U bond exist between the C and N atoms.

SP

only

Fig. 2-12

Problem 2.31 (a) Which of the following molecules possess polar bonds: F,, HF, BrCl, CH,, CHCI,, CH,OH? ( b ) Which are polar molecules? 4 (a)

HF, BrCl, CH,, CHCI,, CH,OH.

( b ) HF, BrCl, CHCI,, CH,OH. The symmetrical individual bond moments in CH, cancel.

Problem 2.32 Considering the difference in electronegativity between 0 and S, would H’O or H,S exhibit greater (a) dipole-dipole attraction, (b) H-bonding? 4 (a)

H,O,

(6) H,O.

Problem 2.33 Nitrogen trifluoride (NF,) and ammonia (NH,) have an electron pair at the fourth corner of a tetrahedron and have similar electronegativity differences between the elements (1.0 for N and F and 0.9 for N and H). Explain the larger dipole moment of ammonia (1.46 D) as compared with that of NF, (0.24 D). 4

CHAP. 21

27

BONDING AND MOLECULAR STRUCTURE

Net dipole moment is very small; actual direction of moment is not known.

(It

.. ..

:F

yjp 1

:F..

H

Netmoment

H H

The dipoles in the three N-F bonds are toward F, see Fig. 2-13(a), and oppose and tend to cancel the effect of the unshared electron pair on N. In NH,, the moments for the three N-H bonds are toward N , see Fig. 2-13(6), and add to the effect of the electron pair. Problem 2.34 Explain why the dipole moment of CH2C1, is greater than that of CHCl,.

4

The resultant moment of the H and the encircled CI in CHCI, opposes the resultant moment of the other two Cl's, as shown in Fig. 2-14(b). The actual values for CH,Cl2 and CHCI, are 1.6 and l.OD, respectively.

Net moment ( a1

Net moment

Fig. 2-14

( b1

Problem 2.35 When 50 mL of C,H,OH and 50 mL of H,O are mixed, the volume of the solution is less than 100 mL. Explain. 4

The H-bonds between H,O and C,H,OH molecules are stronger than the H-bond between the like molecules of the individual compounds. The H,O and C,H,OH molecules are closer when mixed, and the volume shrinks. Problem 2.36 NH; salts are much more soluble in water than are the corresponding Na' salts. Explain.

Na' is solvated merely by an ion-dipole interaction. NHZ is solvated by H-bonding

H

which is a stronger attractive force. Problem 2.37 The F - of dissolved NaF is more reactive in dimethyl sulfoxide,

P

CH, CH, and in acetonitrile, CH,C=N, than in CH,OH. Explain.

4

28

[CHAP. 2

BONDING AND MOLECULAR STRUCTURE

H-bonding prevails in CH,OH (a protic solvent), CH,OH---F-, thereby decreasing the reactivity of F-. CH,SOCH, and CH,CN are aprotic solvents; their C-H H’s do not H-bond. Problem 2.38 Find the oxidation number of the C in ( a ) CH,CI, ( 6 ) CH,CI,, (c) H,CO, (d) HCOOH, and 4 (e) CO,, if (ON),., = - 1 . From Section 2.5:

Problem 2.39 Give a True or False answer to each question and justify your answer. ( a ) Since in polyatomic anions XYL,-(such as SO:- and B F I ) , the central atom X is usually less electronegative than the peripheral atom Y,it tends to acquire a positive oxidation number. (6) Oxidation numbers tend to be smaller values than formal charges. (c) A bond between dissimilar atoms always leads to nonzero oxidation numbers. ( d ) Fluorine 4 never has a positive oxidation number. True. The bonding electrons will be alloted to the more electronegative peripheral atoms, leaving the central atoms with a positive oxidation number. False. In determining formal charges an electron of each shared pair is assigned to each bonded atom, In determinating oxidation numbers pairs of electrons are involved, and more electrons are moved to or away from an atom. Hence larger oxidation numbers result. False. The oxidation numbers will be zero if the dissimilar atoms have the same electronegativity, as in PH,. True. F is the most electronegative element; in F, it has a zero oxidation number. Which of the following transformations of organic compounds are oxidations, which are Problem 2.40 reductions, and which are neither?

( a ) H,C=CH,-CH,CH,OH

( c ) CH,CHO-CH,COOH

(6) CH,CH20H-CH,CH=0

(d) H2C=CH,-CH,CH,CI

(e) HC%CH-H,C=CH, 4

To answer the question determine the average oxidation numbers (ON) of the C atoms in reactant and in product. An increase (more positive or less negative) in ON signals an oxidation; a decrease (more negative or less positive) signals a reduction; no change means neither. ( a ) and (d) are neither, because (ON),. is invariant at -2. (b) and (c) are oxidations, the respective changes being from -2 to -1 and from -1 to 0. (e) is a reduction, the change being from - 1 to -2. Problem 2.41 Irradiation with ultraviolet (uv) light permits rotation about a bonding and antibonding MO’s.

7r

bond. Explain in terms of 4

Two p AO’s overlap to form two pi MO’s, 7r (bonding) and 7r* (antibonding). The two electrons in the original p AO’s fill only the 7r MO (ground state). A photon of uv causes excitation of one electron from 7~ to 7r* (excited state).

t J. -

-

7r

7r*

(ground state)-

uv

.1 t 7 7r

(excited state)

(Initially the excited electron does not change its spin.) The bonding effects of the two electrons cancel. There is now only a sigma bond between the bonded atoms, and rotation about the bond can occur. Problem 2.42 Write the contributing resonance structures and the delocalized hybrid for ( a ) BCI,, (b) H,CN, (diazomethane). 4

CHAP. 21

29

BONDING AND MOLECULAR STRUCTURE

( a ) Boron has six electrons in its outer shell in BCI, and can accommodate eight electrons by having a B-CI bond assume some double-bond character.

Problem 2.43 Arrange the contributing structures for (a) vinyl chloride, H,C==CHCI, and (6) formic acid, HCOOH, in order of increasing importance (increasing stability) by assigning numbers starting with 1 for most important and stable. 4

H,CkHKCl:

( a)

-

H,C--CH=el:

I

c-*

H,&-CH=CI; 111

I1

I is most stable because it has no formal charge. 111 is least stable since it has an electron-deficient C. In 111, Cl uses an empty 3d orbital to accommodate a fifth pair of electrons. Fluorine could not do this. The order of stability is :O:

II

H--C-+-H

( b)

-

V

I ( 1 ) > II(2) > III(3)

:o; I +

H-c=~-H

:or c--*

+

:():

I

H-4-0-H

c--,

VII

VI

H-C-Q-H

..

VIII

V and VI have the greater number of covalent bonds and are more stable than either VII or VIII. V has no formal charge and is more stable than VI. VIII is less stable than VII since VIII's electron deficiency is on 0, which is a more electronegative atom than the electron-deficient C of VII. The order of stability is

V( 1 ) > VI(2) > VII(3) > VIII(4) Problem 2.44

What is the difference between isomers and contributing resonance structures?

4

Isomers are real compounds that differ in the arrangement of their atoms. Contributing structures have the same arrangement of atoms; they differ only in the distribution of their electrons. These imaginary structures are written to give some indication of the electronic structure of certain species for which a typical Lewis structure cannot be written. Problem 2.45 Use the HON method to determine the hybridized state of the underlined elements: (a) H - m H

( 6 ) H&=O

Number of

U

Bonds

(c) H m - -

+

( d ) &ICI:-

(e) PF,

Number of Unshared Electron Pairs

=

(f) HON

CH,QCH, Hybrid State SP

sp SP

4

sp'd' sp ' d sp

4

Chapter 3 Chemical Reactivity and Organic Reactions 3.1

REACTION MECHANISM

The way in which a reaction occurs is called a mechanism. A reaction may occur in one step or, more often, by a sequence of several steps. For example, A + B-* X + Y may proceed in two steps: (1) A - I + X

followed by

(2) B + I + Y

Substances such as I, formed in intermediate steps and consumed in later steps, are called intermediates. Sometimes the same reactants can give different products via different mechanisms. 3.2 CARBON-CONTAINING INTERMEDIATES Carbon-containing intermediates often arise from two types of bond cleavage:

Heterolytic (polar) cleavage. Both electrons go with one group, e.g. A:B-A'

+ :B-

or

A:-

+ B'

Homolytic (radical) cleavage. Each separating group takes one electron, e.g.

A:B----* A. + .B 1. Carbocations are positively charged ions containing a carbon atom having only six electrons in three bonds:

2. Carbanions are negatively charged ions containing a carbon atom with three bonds and an unshared pair of electrons: - .

I

3. Radicals (or free radicals) are species with at least one unpaired electron. This is a broad category in which carbon radicals,

I

are just one example. 4. Carbenes are neutral species having a carbon atom with two bonds and two electrons. There are two kinds: singlet

15.

-C-

in which the two electrons have opposite spins and are paired in one orbital, and triplet

t t

-C-

in which the two electrons have the same spin and are in different orbitals. 30

CHAP. 31

31

CHEMICAL REACTIVITY AND ORGANIC REACTIONS

Problem 3.1 Determine the hybrid orbital number (HON) of the five C-containing intermediates tabulated below and give the hybrid state of the C atom. Unpaired electrons do nor require an HO and should not be counted in your determination. 4 Number of Bonds

U

+

Number of Unshared Electron Pairs

( a ) carbocation (b) carbanion (c) radical (d) singlet carbene (e) triplet carbene

=

HONl Hybrid State

sp sp .’ sp sp

SP

Recall that the two unshared electrons of the triplet carbene are not paired and, hence, are not counted; they are in different orbitals. Problem 3.2 Give three-dimensional representations for the orbitals used by the C’s of the five carbon 4 intermediates of Problem 3.1. Place all unshared electrons in the appropriate orbitals. A carbocation has three trigonal planar sp2 HO’s to form three 0 bonds, and an empty P A 0 perpendicular to the plane of the U bonds. See Fig. 3-l(a). (b) A carbanion has four tetrahedral sp3 HO’s; three form three 0 bonds and one has the unshared pair of electrons. See Fig. 3-l(b). ( c ) A radical has thesame ‘orbitals as the carbocation. The difference lies in the presence of the odd electron in the p orbital of the radical. See Fig. 3-l(c). (d) A singlet carbene has three sp2 HO’s; two form two U bonds and the third holds the unshared pair of electrons. It also has an empty p A 0 perpendicular to the plane of the sp’ HO’s. See Fig. 3-l(d). (e) A triplet carbene uses two sp HO’s to form two linear U bonds. Each of the two unhybridized p AO’s has one electron. See Fig. 3-l(e). (a)

P,

P,

Fig. 3-1

32

CHEMICAL REACTIVITY A N D O R G A N I C REACTIONS

Problem 3.3 Write formulas for the species resulting from the of the C-C bond in ethane. C,H,,, and classify these species. H ,C:CH

((1)

[CHAP. 3

homolytic cleavage. ( b ) heterolytic cleavage 4

H ,C*+ *CH,

+

tth'rnc

hleth>l radical*,

H,C:CH,-H,C' Et hanc

+ :CH,

('arbocLition ('arbanion

3.3 TYPES OF ORGANIC REACTIONS 1. Displacement (substitution). An atom o r group of atoms in a molecule or ion is replaced by another atom or group. 2. Addition. Two molecules combine t o yield a single molecule. Addition frequently occurs at a double or triple bond and sometimes at small-size rings. 3. Elimination. This is the reverse of addition. Two atoms or groups are removed from a molecule. If the atoms o r groups are taken from adjacent atoms (p-elimination), a multiple bond is formed; if they are taken from other than adjacent atoms, a ring results. Removal of two atoms or groups from the same atom (a-elimination) produces a carbene. 4. Rearrangement. Bonds in the molecule are scrambled, converting it to its isomer. 5 . Oxidation-reduction (redox), These reactions involve transfer o f electrons or change in oxidation number. A decrease in the number o f H atoms bonded to C and an increase in the number o f bonds to other atoms such as C, 0, N , CI, Br, F. and S signals oxidation. Problem 3.4 The following represents the steps in the mechanism for chlorination of methane: Initiation Step

Propagation Steps

(1)

I I

:CI:CI: + energy

__*

:CI-+ .CI: Chlorine radicals

H (2) H:C:H H

H

+ .CI: --, H:CI:+ H:C. H

Methyl radical

H

H

The propagation \teps constitute the overall reaction. ( U ) Write the equation for the overall reaction. (b) What are the intermediate\ in the overall reaction'! (c) Which reactions are homolytic? (d) Which is a displacement reaction'? ( U ) I n which reaction is addition taking place? (f')The collision of which species would lead to side 4 products'? Add Steps 2 and 3: CH, + CI,-+CH,CI + HCI. The intermediates formed and then consumed are the H,C. and radicals. Each step is homolytic. In Steps 1 and 3, Cl, cleaves; in Step 2, CH, cleaves. Step (3) involves the displacement of one .cl: of CI, by a C H , group. I n Step 2, .Cl: displaces a C H , group from an H. None. H ,C- + .CH, -+ H ,CCH (ethane) Problem 3.5

ldcntify each of the following as ( 1 ) carbocations, (2) carhanions, (3) radicals o r (4) carbenes:

4

CHAP. 31

CHEMICAL REACTIVITY AND ORGANIC REACTIONS

33

Problem 3.6 Classify the following as substitution, addition, elimination, rearrangement, or redox reactions. (A reaction may have more than one designation.)

+ Br,-CH,BrCH,Br C,H,OH + HCI-C,H,CI + H,O CH ,CH=CHCH , + ZnCI, CH ,CHCICHCICH ,+ Zn CH,=CH,

NH:(CNO)--H,NCNH,

-

-

II

0

CH,CH,CH,CH,

-

+ 2Mn03 + O H - A 3CH,COO- + 2Mn0, + 2H,O + OH :CCI, + H,O + CI

3CH,CHO HCCl,

(CH ,),CH

BrCH,CH,CH,Br

+ Zn-,

H,C--&H,

(A means heat.)

+ ZnBr,

Addition and redox. In this reaction the two Br’s add to the two double-bonded C atoms (1.2-addition). The oxidation number (ON) for C has changed from 4 - 2(2) - 2 = -2 to 4 - 2(2) - 1 = - 1; (ON) for Br has changed from 7 - 7 = 0 to 7 - 8 = -1. Substitution of a CI for an OH. Elimination and redox. Zn removes two Cl atoms from adjoining C atoms to form a double bond and ZnCI, (a p-elimination). The organic compound is reduced and Zn is oxidized. Rearrangement (isomerization). Rearrangement (isomerization). Addition and redox. The Br’s add to two C atoms of the ring. These C’s are oxidized and the Br’s are reduced. Redox. CH,CHO is oxidized and MnO, is reduced. Elimination. An H’ and Cl- are removed from the same carbon (a-elimination). Elimination. The two Br’s are removed from nonadjacent C’s giving a ring. Redox [see (c)].

ELECTROPHILIC AND NUCLEOPHILIC REAGENTS Reactions generally occur at the reactive sites of molecules and ions. These sites fall mainly into two categories. One category has a high electron density because the site ( a ) has an unshared pair of electrons or (6) is the S- end of a polar bond or (c) has C=C n electrons. Such electron-rich sites are nucleophilic and the species possessing such sites are called nucleophiles or electron-donors. The second category ( a ) is capable of acquiring more electrons or (b) is the S + end of a polar bond. These electron-deficient sites are electrophilic and the species possessing such sites are called electrophiles or electron-acceptors. Many reactions occur by coordinate covalent bond formation between a nucleophilic and an electrophilic site.

Nu:

+ E----*

Nu:E

Problem 3.7 Classify the following species as being (1) nucleophiles or ( 2 ) electrophiles and give the reason for your classification: ( a ) HO:, (b) :C-=N:-, (c) : B r + , ( d ) BF,, (e) H 2 0 : . ( f ) AICI,, ( g ) :NH,, ( h ) H,Cf 4 (a carbanion), (i) SiF,, ( j ) Ag’, (k) H,C+ (a carbocation), ( I ) H,C: (a carbene), ( m ) ::I:. (1) ( a ) , ( b ) , (e), (g), ( h ) , and ( m ) . They all have unshared pairs of electrons. All anions are potential nucleophiles.

34

CHEMICAL REACTIVITY AND ORGANIC REACTIONS

[CHAP. 3

(2) (d) and ( f ) are molecules whose central atoms (B and AI) have only six electrons rather than the more desirable octet; they are electron-deficient. (c), ( j ) and (k)have positive charges and therefore are electron-deficient. Most cations are potential electrophiles. The Si in (i) can acquire more than eight electrons by utilizing its d orbitals, e.g. SiF, (an electrophile) + 2:F:

-

SiFi-

Although the C in (I) has an unshared pair of electrons, (I) is electrophilic because the C has only six electrons. Problem 3.8 Why is the reaction CH,Br

+ OH- --* CH,OH + Br a nucleophilic displacement?

The :OH- has unshared electrons and is a nucleophile. Because of the polar nature of the C-Br \ii+

-C-Br /

4

bond,

6

C acts as an electrophilic site. The displacement of Br- by OH- is initiated by the nucleophile HO: Problem 3.9 Indicate whether reactant (1) or (2) is the nucleophile or electrophile in the following reactions:

H,C=CH2 (1) + Br, (2)+ BrCH,-CH,Br CH,NH; (1) + CH,COO- (2)--*CH,NH2 + CH,COOH C H , C X l : (1) + AlCl, (2)+ CH, +AICL II 0 0

ii'

CH,CH--V (1) + :SO,H- (2)+ CH,--CHSO,H

4 4 1 I

0-

4

Nucleophile

THERMODYNAMICS thermodynamics and the rate of a reaction determine whether the reaction proceeds. The thermodynamics of a system is described in terms of several important functions: (1) A E , the change in energy, equals q,, the heat transferred to or from a system at constant volume: A E = q,. ( 2 ) A H , the change in enthalpy, equals q p ,the heat transferred to or from a system at constant pressure: A H = qp. Since most organic reactions are performed at atmospheric pressure in open vessels, A H is used more often than is A E . For reactions involving only liquids or solids: A E = A H . A H of a chemical reaction is the difference in the enthalpies of the products, H , , and the reactants, H R :

AH=Hp

- HR

If the bonds in the products are more stable than the bonds in the reactants, energy is released, and A H is negative. The reaction is exothermic. (3) AS is the change in entropy. Entropy is a measure of randomness. The more the randomness, the greater is S ; the greater the order, the smaller is S. For a reaction, A S = s, - SR

CHEMICAL REACTIVITY AND ORGANIC REACTIONS

CHAP. 31

(4)

AG = G,

-

35

G, is the change in free energy. At constant temperature, AG

= AH

- 7' AS ( T = absolute temperature)

For a reaction to be spontaneous, AG must be negative. Problem 3.10 State whether the following reactions have a positive or negative AS and give a reason for your choice.

H2 + H2C=CH2

-

H,CCH,

-

H,C-CH=CH2 CH,COO-(aq)

+ H,O'(aq)

CH,COOH

+ H20

4

Negative. Two molecules are changing into one molecule and there is more order (less randomness) in the product (S, < SR). (6) Positive. The rigid ring opens to give compound having free rotation about the C - C single bond. There is now more randomness (S, > SR). (c) Positive. The ions are solvated by more H,O molecules than is CH,COOH. When ions form molecules, many of these H,O molecules are set free and therefore have more randomness (S,> S,). (a)

Problem 3.11 Predict the most stable state of H,O (steam, liquid, or ice) in terms of ( a ) enthalpy, (6) 4 entropy, and (c) free energy. ( a ) Gas-, Liquid- Solid are exothermic processes and, therefore, ice has the least enthalpy. For this reason,

ice should be most stable. (6) Solid-, Liquid-, Gas shows increasing randomness and therefore increasing entropy. For this reason, steam should be most stable. (c) Here the trends to lowest enthalpy and highest entropy are in opposition; neither can be used independently to predict the favored state. Only G, which gives the balance between Hand S, can be used. The state with lowest G or the reaction with the most negative AG is favored. For H,O, this is the liquid state, a fact which cannot be predicted until a calculation is made using the equation G = H - TS.

3.6

BOND-DISSOCIATION ENERGIES

The bond-dissociation energy, AH, is the energy needed for the endothermic homolysis of a covalent bond A:B- A. + -B; AH is positive. Bond formation,the reverse of this reaction, is exothermic and the AH values are negative. The more positive the AH value, the stronger is the bond. The AH of reaction is the sum of all the (positive) AH values for bond cleavages plus the sum of all the (negative) A H values for bond formations. Problem 3.12 Calculate AH for the reaction CH, + Cl,-, CH,Cl + HCl. The bond-dissociation energies, in kJ/mol, are 427 for C-H, 243 for C l - C l , 339 for C - C l , and 431 for H - C l . 4

The values are shown under the bonds involved:

+ Cl--Cl+ 427 + 243 +

-

H,C-H

cleavages (endothermic)

-

H,C--Cl+ H-Cl (-339) + ( 4 3 1 )

=-loo

formations (exothermic)

The reaction is exothermic, with AH = - 100 kJ/mol.

36

CHEMICAL REACTIVITY AND ORGANIC REACTIONS

8

[CHAP. 3

Problem 3.13 Compare the strengths of bonds between similar atoms having: ( a ) single bonds between atoms with and without unshared electron pairs; (6) triple, double, and single bonds. 4 ( a ) Bonds are weaker between atoms with unshared electron pairs because of interelectron repulsion. (6) Overlapping of p orbitals strengthens bonds, and bond energies are greatest for triple and smallest for single bonds.

3.7 CHEMICAL EQUILJBRIUM Every chemical reaction can proceed in either direction, dA + eB Z f X + gY, even if it goes in one direction to a microscopic extent. A state of equilibrium is reached when the concentrations of A , B , X , and Y no longer change even though the reverse and forward reactions are taking place. Every reversible reaction has an equilibrium expression in which K,, the equilibrium constant, is defined in terms of molar concentrations (mol/L) as indicated by the square brackets: dA + eB= f X Ke

=

+ gY

Products favored; K , is large Reactants favored; K, is small

[X]f[Y]x [A]'[B]'

K , varies only with temperature. The AG of a reaction is related to K , by AG where R is the molar gas constant (R

=

-2.303 RT log K,

= 8.314 J

mol-' K - ' ) and T is the absolute temperature (in K).

Problem 3.14 Given the reversible reaction

C,H,OH

+ CH,COOH eCH,COOC,H, + H,O

what changes could you make to increase the yield of CH,COOC2H,?

4

The equilibrium must be shifted to the right, the side of the equilibrium where CH,COOC2H, exists. This is achieved by any combination of the following: adding C,H,OH, adding CH,COOH, removing H,O, removing CH ,COOC,H,. Problem 3.15 Summarize the relationships between the signs of A H , T A S , and A G , and the magnitude of K,, and state whether a reaction proceeds to the right or to the left for the reaction equation as written. 4

See Table 3-1 Table 3-1

AH

-

TAS

=

AG

Reaction Direction

K,

-

+

-

Forward + right

>1

+

-

+

Reverse --* left

+60 kJ/mol

Depends on conditions

?

CHAP. 31

CHEMICAL REACTIVITY A N D ORGANIC REACTIONS

37

Problem 3.16 At 25°C the following reactions have the indicated K , values:

CH,CH,OH Ethanol

H2C-OH

+ CH,COOH

CH,COOCH,CH,

Acetic acid

Ethyl acetate

+ H20

K , = 4.0

Water

Hg-0 a

lactone

Since the changes in bonding in these reactions are similar, both reactions have about the same A H . Use 4 thermodynamic functions to explain the large difference in the magnitude of K , . A larger K,, means a more negative AG. Since S H is about the same for both reactions, a more negative AG means that AS for this reaction is more positive. A more positive AS (greater randomness) is expected in reaction (6) because one molecule is converted into two molecules, whereas in reaction ( a ) two molecules are changed into two other molecules. When two reacting sites, such as O H and C O O H , are in the same molecule, the reaction is intramolecular. When reaction sites are in different molecules. as in ( a ) , the reaction is intermolecular. Intramolecular reactions often have a more positive AS than similar intermolecular reactions.

3.8 RATES OF REACTIONS The rate of a reaction is how fast reactants disappear or products appear. For the general reaction + gD, the rate is given by a rate equation

dA + eB-.fC

Rate = k[A].'[B]' where k is the rate constant at the given temperature, T, and [ A ] and [ B ]are molar concentrations. Exponents x and y may be integers, fractions, or zero; their sum defines the order of the reaction. The values of x and y are found experimentally, and they may differ from the stoichiometric coefficients d and e. Experimental conditions, other than concentrations, that affect rates of reactions are: Temperature. A rough rule is that the value of k doubles for every rise in temperature of 10°C. Particle size. Increasing the surface area of solids by pulverization increases the reaction rate. Catalysts and inhibitors. A catalyst is a substance that increases the rate of a reaction but is recovered unchanged at the end of the reaction. Inhibitors decrease the rate.

At a given set of conditions, the factors that determine the rate of a given reaction are: 1 . Number of collisions per unit time. The greater the chances for molecular collision, the faster the reaction. Probability of collision is related to the number of molecules of each reactant and is proportional to the molar concentrations. 2. Enthalpy of activation (activation energy, EpCt) ( A H ' ) . Reaction may take place only when colliding molecules have some enthalpy content, AH', in excess of the average. The smaller the value of AH', the more successful will be the collisions and the faster the reaction. (AH' = E,,, at constant pressure .) 3. Entropy of activation ( A S ' ) , also called the probability factor. Not all collisions between molecules possessing the requisite AH' result in reaction. Often collisions between molecules must also occur in a certain orientation, reflected by the value of AS'. The more organized or less random the required orientation of the colliding molecules, the lower the entropy of activation and the slower the reaction.

CHEMICAL REACTIVITY AND ORGANIC REACTIONS

38

[CHAP. 3

Problem 3.17 Predict the effect on the rate of a reaction if a change in the solvent causes: ( a ) an increase in AHS and a decrease in ASS,(6) a decrease in AH' and an increase in ASS,(c) an increase in AHSand in AS*, (d) a decrease in AHS and in AS'. 4 ( a ) Decrease in rate. (6) Increase in rate. (c) The increase in AHS tends to decrease the rate but the increase in ASS tends to increase the rate. The combined effect is unpredictable. (d) The trends here are opposite to those in part (c); the effect is also unpredictable. In many cases the change in AH' is more important than the change in ASs in affecting the rate of reaction.

3.9 TRANSITION-STATE THEORY AND ENTHALPY DIAGRAMS When reactants have collided with sufficient enthalpy of activation (AHt) and with the proper orientation, they pass through a hypothetical transition state in which some bonds are breaking while others may be forming. The relationship of the transition state (TS) to the reactants (R) and products (P) is shown by the enthalpy (energy) diagram, Fig. 3-2, for a one-step exothermic reaction A + B 4 C + D. At equilibrium, formation of molecules with lower enthalpy is favored, i.e., C + D. However, this applies only if A H of reaction predominates over T AS of reaction in determining the equilibrium state. The reaction rate is actually related to the free energy of activation, AC', where AG' = AH' - T AS'.

T

Progress of Reaction (Reaction Coordinate) Fig. 3-2

In multistep reactions, each step has its own transition state. The step with the highest-enthalpy transition state is the slowest step and determines the overall reaction rate. The number of species colliding in a step is called the molecularity (of that step). If only one species breaks down, the reaction is unimolecular. If two species collide and react, the reaction is bimolecular. Rarely do three species collide (termolecular) at the same instant. The rate equation gives molecules and ions and the number of each (from the exponents) involved in the slow step and in any preceding fast step@). Intermediates do not appear in rate equations, although products occasionally do appear. Problem 3.18 Draw an enthalpy diagram for a one-step endothermic reaction. Indicate AH of reaction and AH*. 4 See Fig. 3-3. Problem 3.19 Draw a reaction-enthalpy diagram for an exothermic two-step reaction in which ( a ) the first step 4 is slow, (6) the second step is slow. Why do these reactions occur?

T

39

CHEMICAL REACTIVITY AND ORGANIC REACTIONS

CHAP. 3)

TS

I

Progress of Reaction

> React ion Progress (a)

Fig. 3-4

Fig. 3-3

See Fig. 3-4, in which R = reactants, I = intermediates, P = products, TS, = transition state of first step, TS, = transition state of second step. Because the reactions are exothermic, H , < H , . Problem 3.20 In Problem 3.19(6), the first step is not only fast but is also reversible. Explain.

4

The A H S for I to revert to reactants R is less than that for I to form the products P. Therefore, most 1’s re-form R, so that the first step is fast and reversible. A few 1’s have enough enthalpy to go through the higher-enthalpy TS, and form the products. The A H S for P to revert to I is prohibitively high; hence the products accumulate, and the second step is at best insignificantly reversible. Problem 3.21 Catalysts generally speed up reactions by lowering A H S . Explain how this occurs in terms of 4 ground-state and transition-state enthalpies ( H , and H T s ) .

A H S can be decreased by ( a ) lowering H T s , (6) raising H , or (c) both of these. Problem 3.22 The reaction A + B + C

+ D has ( a ) rate mechanisms consistent wih these rate expressions.

or (6) rate

= k[A][B],

= k[A].

Offer possible

4

( a ) Molecules A and B must collide in a bimolecular rate-controlling step. Since the balanced chemical

equation calls for reaction of one A molecule with one B molecule, the reaction must have a single (concerted) step. ( 6 ) The rate-determining step is unimolecular and involves only an A molecule. There can be no prior fast steps. Molecule B reacts in the second step, which is fast. A possible two-step mechanism is: Step 1: Step 2:

C + I ( I = intermediate) B+1 4 D

A+

Adding the two steps gives the balanced chemical equation: A + B-, C + D. Problem 3.23 For the reaction 2A + 2B-, C + D, rate = k[Al2[B). Give a mechanism using only unimolecular or bimolecular steps. 4

One B molecule and two A molecules are needed to give the species for the slow step. The three molecules do not collide simultaneously since we are disregarding the very rare termolecular steps. There must then be some number of prior fast steps to furnish at least one intermediate needed for the slow step. The second B molecule which appears in the reaction equation must be consumed in a fast step following the slow step.

-

Mechanism 1

A+B

fast

AB + AA,B

+ B-c

AB (intermediate)

slow

fast

A,B

(intermediate)

+D

A +A

A,

-

+ B-

A,B

Mechanism 2

fast

slow

+ B-C

fast

(intermediate)

A,

A,B

(intermediate)

+D

CHEMICAL REACTIVITY AND ORGANIC REACTIONS

40

Problem 3.24 For the reaction A + 2B-C determining step is unimolecular.

+ D,

rate

=

[CHAP. 3

k[A][B]'. Offer a mechanism in which the rate4

The slow step needs an intermediate formed from one A molecule and two B molecules. Since the rate expression involves the same kinds and numbers o f molecules as does the chemical equation, there are no fast steps following the slow step. Mechanism 2

Mechanism I ld*1

A+B-AB

B+B-B,

AB + B-AB,

B2 + A-

\IOU

AB,-C+D

A,B-C

fd\t

fast

A,B

+D

clow

Notice that often the rate expression is insufficient to allow the suggestion of an unequivocal mechanism. More experimental information is often needed.

BRONSTED ACIDS AND BASES

3.10

In the Bronsted definition, an acid donates a proton and a base accepts a proton. The strengths of acids and bases are measured by the extent to which they lose or gain protons, respectively. In these reactions acids are converted to their conjugate bases and bases to their conjugate acids. Acid-base reactions go in the direction of forming the weaker acid and the weaker base. Problem 3.25 Show the conjugate acids and bases in the reaction of H,O with gaseous ( a ) HCI, ( b ) :NH,.

4

H,O, behaving as a Bronsted base, accepts a proton from HCI, the Bronsted acid. They are converted to the conjugate acid H,O + and the conjugate base Cl -, respectively. HCI + H,O,-H,O'

+ CK

acid ba-id? base I (stronger) (stronger) (weaker) (weaker)

The conjugate acid-base pairs have the same subscript and are bracketed together. This reaction goes almost to completion because HCI is a good proton donor and hence a strong acid. H,O is amphoteric and can also act as an acid by donating a proton to :NH,. H,O is converted to its conjugate base, OH-., and :NHl to its conjugate acid. NHf. H : O H + N H , e N H i +:OHagd I

ba-di,

base,

:NH, is a poor proton acceptor (a weak base); the arrows are written to show that the equilibrium lies mainly to the left. To be called an acid, the species must be more acidic than water and be able to donate a proton to water. Some compounds, such as alcohols, are not acidic toward water, but have an H which is acidic enough to react with very strong bases or with Na. Problem 3.26 Write an equation for the reaction of ethanol with ( a ) NH;, a very strong base; ( b ) Na.

(4 (b)

CH,CH,OH

+ :NHi -CH,CH20:-

2CH,CH20H + 2Na-2CH,CH20:-

4

+ :NH, + 2Na' + H,

Relative quantitative strengths of acids and bases are given either by their ionization constants, K , and K,, o r by their pK, and p K , values as defined by:

41

C H EM I CA L RE ACT1V ITY AN D 0RG A N I C REACT10N S

CHAP. 31

PK,, = -log K,,

P K , = - 1% K',

The stronger an acid or base, the larger its ionization constant and the smaller its pK value. The strengths of bases can be evaluated from those of their conjugate acids; the strengths of acids can be evaluated from those of their conjugate bases. The strongest acids have the weakest conjugate bases, and the strongest bases have the weakest conjugate acids. This follows from the relationships K,,

= (K[,)(K,,)= 1Wl4

PK,,+ PK,, = P K ,

=

14

in which K w , the ion-product of water = (H,O'][OH-1.

3.1 1 BASICITY (ACIDITY) AND STRUCTURE

The basicity of a species depends on the reactivity of the atom with the unshared pair of electrons, this atom being the basic site for accepting the H'. The more spread-out (dispersed, delocalized) is the electron density on the basic site, the less basic is the species. The acidity of a species can be determined from the basicity of its conjugate base.

DELOCALIZATION RULES (1) For bases of binary acids (H,,X) of elements in the same Group, the larger the basic site X , the more spread-out is the charge. Compared bases must have the same charge. (2) For like-charged bases of binary acids in the same period, the more unshared pairs of electrons the basic site has, the more delocalized is the charge. (3) The more s character in the orbital with the unshared pair of electrons, the more delocalized the electronic charge. (4) Extended p - p T bonding between the basic site and an adjacent T system (resonance) delocalizes the electronic charge. ( 5 ) Extended p - d T bonding with adjacent atoms that are able to acquire more than an octet of electrons delocalizes the electronic charge. (6) Delocalization can occur via the inductive effect, whereby an electronegative atom transmits its electron-withdrawing effect through a chain of 0 bonds. Electropositive groups are electrondonating and localize more electron density on the basic site. are

With reference to (4) and ( 5 ) . some common

I

-C=O carbonyl

&N nitrile

-N=O

I

0

nitro

7~

systems that participate in extended 0

I -s=o I

7~

bonding

C1, > Br,.

2

or A

2 RX + HX

I, does not react; F, destroys the molecule)

Chlorination (and bromination) of alkanes such as methane, CH,, has a radical-chain mechanism, as follows: INITIATION STEP C I : C I A 2C1. or A

A H = +243 kJ/mol

The required enthalpy comes from ultraviolet (uv) light or heat. PROPAGATION STEPS

(i) H,C:H + C1.(ii) H,C. + C1:Cl-

H,C. + H:CI H,C:Cl + C1.

AH = -4 kJ/mol (rate-determining) AH = -96 kJ/mol

58

[CHAP. 4

ALKANES

The sum of the two propagation steps is the overall reaction,

CH,

+ Cl,--+CH,CI + HCI

AH = -100 kJ/mol

In propagation steps, the same free-radical intermediates, here CI. and H,C., are being formed and consumed. Chains terminate on those rare occasions when two free-radical intermediates form a covalent bond:

Inhibitors stop chain propagation by reacting with free-radical intermediates, e.g.

The inhibitor-here 0,-must be consumed before chlorination can occur. In more complex alkanes, the abstraction of each different kind of H atom gives a different isomeric product. Three factors determine the relative yields of the isomeric product. (1) Probability factor. This factor is based on the number of each kind of H atom in the molecule. For example, in CH,CH,CH,CH, there are six equivalent 1" H's and four equivalent 2" H's. The odds on abstracting a 1" H are thus 6 to 4, or 3 to 2. (2) Reactivity of H. The order of reactivity of H is 3" > 2" > 1". (3) Reactivity of X-. The more reactive CI. is less selective and more influenced by the probability factor. The less reactive Br. is more selective and less influenced by the probability factor. As summarized by the reactivity-selectivityprinciple: If the attacking species is more reactive, it will be less selective, and the yields will be closer to those expected from the probability factor. Problem 4.19 ( a ) List the monobromo derivatives of (i) CH,CH2CH,CH, and (ii) (CH,),CHCH,. (6) Predict the predominant isomer in each case. The order of reactivity of H for bromination is

3"(16OO) > 2"(82) > 1"( 1 )

4

H's, and there are two possible isomers for each compound: (i) CH,CH,CH,CH,Br and CH,CHBrCH,CH,; (ii) (CH,),CHCH,Br and (CH,),CBrCH,. (6) In bromination, in general, the difference in reactivity completely overshadows the probability effect in determining product yields. (i) CH,CHBrCH,CH, is formed by replacing a 2" H; (ii) (CH,),CBrCH, is formed by replacing the 3" H and predominates. ( a ) There are two kinds of

Problem 4.20

Using the bond dissociation energies for X,,

show that the initiation step for halogenation of alkanes,

x, is not rate-determining.

+2x.

CHAP. 41

ALKANES

59

The enthalpy AH' (Section 3.8) is seldom related to A H of the reaction. In this reaction, however, A H S and A H are identical. In simple homolytic dissociations of this type, the free radicals formed have the same enthalpy as does the transition state. On this basis alone, iodine, having the smallest A H and AH', should react fastest. Similarly, chlorine, with the largest A H and AH', should react slowest. But the actual order of reaction rates is F, > CI, > Br, > I, Therefore, the initiation step is not rate-determining; the rate is determined by the first propagation step, H-abstraction.

Problem 4.21 Draw the reactants, transition state, and products for

Br.

+ CH,

+ CH,+HBr

4

In the transition state, Br is losing radical character while C is becoming a radical; both atoms have partial radical character as indicated by 6 . . The C atom undergoes a change in hybridization as indicated: REACTANTS

H

\

.C-H

He* 1

+

Br.

H

-

TRANSITION STATE 1

-

C is becoming s p * (trigonal planar)

C is sp' (tetrahedral)

PRODUCTS

H

I

.C.

+

H-Br

H" H ' C is s p 2 (trigonal planar)

Problem 4.22 Bromination of methane, like chlorination, is exothermic, but it proceeds at a slower rate under the same conditions. Explain in terms of the factors that affect the rate, assuming that the rate-controlling step is

X.

HX + *CH,

+ CH,-

4

Given the same concentration of CH, and C1. or Bra, the frequency of collisions should be the same. Because of the similarity of the two reactions, A S s for each is about the same. The difference must be due to the AH*, which is less (17 kJ/mol) for C1. than for Br- (75 kJ/mol). Problem 4.23 2-Methylbutane has l", 2", and 3" H's as indicated: 3"

2"

I"

(cH, ),CHCH,CH,

( a ) Use enthalpy-reaction progress diagrams for the abstraction of each kind of hydrogen by .X.( b ) Summarize the relationships of relative (i) stabilities of transition states, (ii) A H s values, (iii) stabilities of alkyl radicals, and (iv) rates of H-abstraction. 4 ( a ) See Fig. 4-7. (6) (i) 3" > 2"> 1" since the enthalpy of the TS,, is the greatest and the enthalpy of the TS,. is the smallest. (ii) A H : . < A H i . < A H : , . (iii) 3">2">1". (iv) 3">2">1".

Problem 4.24 List and compare the differences in the properties of the transition states during chlorination 4 and bromination that account for the different reactivities for l", 2", and 3"H's.

ALKANES

60

t

[CHAP. 4

TS,.

TS,.

6.

6.

(CH,),C---H --X

+ HX

Reaction Progress Fig. 4-7

The differences may be summarized as follows:

1. Time of formation of transition state 2. Amount of breaking of C-H bond 3. Free-radical character ( 6 . ) of carbon 4. Transition state more closely resembles

Chlorination Earlier in reaction

Bromination Later in reaction

Less, H,C - - - H - - - - - C1

More, H,C- - - - H- - - Br

Less

More

Reactants

Products

These show that the greater selectivity in bromination is attributable to the greater free-radical character of carbon. With greater radical character, the differences in stability between l", 2", and 3" radicals become more important, and reactivity of H (3" > 2" > 1") also becomes more significant.

Problem 4.25 Account for the order of stability of radicals: 3" > 2" > 1".

4

The carbon atom with the odd electron is an electron-deficient site. Alkyl groups attached to such a C have an electron-releasing inductive effect which attenuates the electron deficiency, thereby stabilizing the radical.

4.

Isomerization

AICIj. HCI

CH3

I

CH3CH2CHzCH3 I CH3-CH-CH3

CHAP. 41

4.5

61

ALKANES

SUMMARY OF ALKANE CHEMISTRY

-

PREPARATION 1. C Skeleton Preserved (a) Direct Replacement of X by H

CH ,CH,CH,CH,X

PROPERTIES 1. Thermal Dehydrogenation

H2C=CH-CH=CH2

+ 2H,

2. Combustion

or

+ O,-CO,

CH,CH,CHXCH3

3. Halogenation (X = C1, Br)

(6) From Organometal

+ X,+ h

(M = Li, MgX) CH,CH,CH,CH,M* or CH ,CH,CHMCH

+ H,O

H,O --+

11

CH,CH,CHXCH, (major) + CH,CH,CH,CH,X

CH,CH2CH2CH, Butane

,

(c) Hydrogenation of Alkenes

CH,CH,CH=CH, or

CH,CH=CHCH, 4. Isomerization

2. Coupling of RX

(CH,),CH

(a) With Cuprate (Corey-House)

CH,CH,X-

1 LI

2 Cul

(CH,CH,),CuLi

( b ) With Na (Wurtz)-poor 2 CH,CH,X

+ C,H5X-

yield Na

*From RX with Li or Mg. ' Also from corresponding alkynes.

Supplementary Problems Problem 4.26 Assign numbers, ranging from (1) for LOWEST to (3) for HIGHEST. to the boiling points of the following hexane isomers: 2,2-dimethylbutane, 3-methylpentane, and n- hexane. Do not consult any tables for data. 4 Hexane has the longest chain. the greatest intermolecular attraction. and, therefore, the highest boiling point, (3). 2,2-Dimethylbutane is ( I ), since, with the most spherical shape, it has the smallest intermolecular contact and attraction. This leaves 3-methylpentane as (2).

62

ALKANES

[CHAP. 4

Problem 4.27 Write structural formulas for the five isomeric hexanes and name them by the IUPAC system.

4

The isomer with the longest chain is hexane, CH,CH,CH,CH,CH,CH,. If we use a five-carbon chain, a CH, may be placed on either C’ o r CJ to produce 2-methylpentane, or on C’ to give another isomer, 3-methylpentane.

CH3

CH3 ’

1

1

4

7

dHICH2CHCH2CHI

bH$HtHztH,6H, 2-Methylpentane

3-Methylpentane

With a four-carbon chain either a CH,CH, o r two CH,’s must be added as branches to give a total of 6 C’s. Placing CH,CH, anywhere on the chain is ruled out because it lengthens the chain. Two CH,’s are added, but only to central C’s to avoid extending the chain. If both CH,’s are introduced on the same C, the isomer is 2J-dimethylbutane. Placing one CH, on each of the central C’s gives the remaining isomer, 2,3-dimethylbutane.

CH, CH,

CH,

I

I I

CH,-C-CH,CH3

I I

CH,-C-C--CH, H

bH3

H

2,3-Dimethylbutane

2.2-Dimethylbutane

Problem 4.28 Write the structural formulas for (a) 3.4-dichloro-2,5-dimethylhexane; (b) 5-( 1.24 dimethylpropyl)-6-methyldodecane.(Complex branch groups are usually enclosed in parentheses.)

CH, CI

I

I

Cl

I

CH,

I

CH3-CH-CH-CH-CH-CH, (6) The group in parentheses is bonded to the fifth C. It is a propyl group with CH,’s on its first and second C’s (denoted I ’ and 2 ’ ) counting from the attached C.

( a ) 2,2,3-Trimethylpentane. (6) The longest chain has ten C’s, and there is an isobutyl group on the fifth C. The name is 5-(2-methylpropy1)decane o r 5-isobutyldecane.

Problem 4.30 Write the structural formulas and give the IUPAC names for all monochloro derivatives of ( a ) isopentane, (CH,),CHCH,CH,; (6) 2,2,4-trimethylpentane, (CH,),CCH,CH(CH,),. ( a ) Since there are four kinds of equivalent H’s,

4

CHAP. 41

63

ALKANES

there are four isomers: CH,

CH3

I CIC H,CC H,C H I

I

CH,-C-CH,CH,

I

1-Chloro-2-methyl-

I

C H,-C-CH2CH2CI

I

I

H CI

2-Chloro-2-methylbutane

butane

I

CH,-C-CHCH,

CI

H

CH,

CH3

I

H

2-C hloro-3-methylbutane

1 -Chloro-3-methyl-

butane

(6) There are four isomers because there are four kinds of H‘s: (CH,),CCH2CH(Ch,),. CH,

CH, H

CH,

I

CICH,-C-CH,-C-CH,

I

CH,

I I

I

l

I

l

CH,-C-C-C-CH,

CH,

l

l

CH, C1

H

1 -Chloro-2,2.4-

CH,

I

CH,-C-CH,-C-CH~

I

H

CH,

3-ChIoro-2,2,4trimethylpentane

trirnethylpentane

CH,

CH,

CH,

H

CH,

I I

CI

2-ChIoro-2,4,4trimethylpentane

I I CH,-C-CH,-C-CH,CI I I

1 -Chloro-2,4,4trirnethylpentane

Problem 4.31 Derive the structural formulas and give the IUPAC names for all dibromo derivatives of propane. 4

The two Br’s are placed first on the same C and then on different C’s. Br

Br

I

Br-CHCH,CH,

I I

BrCH,CHBrCH,

BrCH,CH,CH,Br

2.2-Dibromopropane

1,2-Dibromopropane

1,3-Dibrornopropane

CH,CCH,

Br 1 . 1 -Dibrornopropane

Problem 4.32 Give topological structural formulas for (a) propane, ( 6 ) butane, (c) isobutane, (d) 2,2dimethylpropane, (e) 2,3-dimethylbutane, ( f ) 3-ethylpentane, ( g ) l-chloro-3-methylbutane,( h ) 2,3-dichloro2-methylpentane, ( i ) 2-chloro-2,4,4-trimethylpentane. 4

In this method one writes only the C - - C bonds and all functional groups bonded to C. The approximate bond angles are used.

Problem 4.33 Synthesize ( a ) 2-methylpentane from CH,CH=CH--CH(CH,),, ( 6 ) isobutane from isobutyl chloride, (c) 2-methyl-2-deuterobutane from 2-chloro-2-methylbutane. Show all steps. 4 ( a ) The alkane and the starting compound have the same carbon skeleton.

CH3

CH,

I

CH,-CH=CH-CH-CH,

H p t __*

I

CH,-CH,-CH,--CH-CH~

ALKANES

64

[CHAP. 4

( 6 ) The alkyl chloride and alkane have the same carbon skeleton. CH,

CH,

I

I

CH3CHCH2-CI

CH3CHCH,-H

( c ) Deuterium can be bonded to C by the reaction of D,O with a Grignard reagent.

-

CH3

CH3

I dryyfhcr+ CH,CH,C-MgCI I

I CH3CH2C-CI I CH3

DZO

CH,

2-Chloro-2-methylbutane

Grignard reagent

CH3

I I

CH,CH,C-D CH3

2-Deutero-2-methylbutane

Problem 4.34 RC1 is treated with Li in ether solution to form RLi. RLi reacts with H,O to form isopentane. Using the Corey-House method, RCI is coupled to form 2,7-dimethyloctane. What is the structure of RCI? 4 To determine the structure of a compound from its reactions, the structures of the products are first considered and their formation is then deduced from the reactions. The coupling product must be a symmetrical molecule whose carbon-to-carbon bond was formed between C4 and C 5 of 2,7-dimethyloctane. The only RCI which will give this product is isopentyl chloride:

( CH3

CH,CHCH,CH,

)

CH, ,CuLi

I

+ CICH,CH,CHCH,

Lithium dusopentylcuprate

-

This alkyl halide will also yield isopentane. CH,

CH,

I

formed bond

IT

*

CH,

I

CH,CHCH,CH,-CH,CH,CHCH, 2,7-Dimethyloctane

CH,

I

CH,CHCH2CH2Li

2 CH,AHCH,CH, (Isopentane)

Problem 4.35 Give steps for the following syntheses: ( a ) propane to (CH,),CHCH(CH,),, 2-methylpentane, ( c ) "CH,CI to 'JCH,'JCH2'4CH,'4CH,.

(6) propane to 4

( a ) The symmetrical molecule is prepared by coupling an isopropyl halide. Bromination of propane is

preferred over chlorination because the ratio of isopropyl to n-propyl halide is 96%/4% in bromination and only 56%/44% in chlorination. CH3CH2CH,

Br (127 " C )

(CH,),CHBr

CH,CH,CH,CI

I

l.,!& 3

CH,CH,CH,

CH, CH,

I

CH3CH-CHCH,

CH3CHBrCH3

+ (CH,),CHCI

(sepurate the mixture)

Problem 4.36 Synthesize the following deuterated compounds: ( a ) CH,CH,D, (6) CH,DCH,D. ( a ) CH,CH,Br=

(6) H,C=CH,

Mg

+ D,-

CH,CH,MgBrPt

H2CDCH,D

CH,CH,D

4

CHAP. 41

65

ALKANES

Problem 4.37 In the dark at 150 OC, tetraethyl lead, Pb(C,H,),, catalyzes the chlorination of CH,. Explain in terms of the mechanism. 4

Pb(C,H,), readily undergoes thermal homolysis of the P b - C bond. Pb(C,H,), The CH,CH;

then generates the C1- that initiates the propagation steps. CH,CH2*+ C1:CI-

CH,CH,CI

CH, H,C* + Cl:Cl+

Problem 4.38

*Pb*+ 4CH3CH,*

+

+ C1.

1

H,C* H C 1 (propagation steps) H,CCI + CI-

"*-

+

Hydrocarbons are monochlorinated with tert-butyl hypochlorite, t-BuOC1. t-BuOCI + RH-

RCI + t-BuOH

Write the propagating steps for this reaction if the initiating step is t-BuO*+ CI.

I-BuOCl-

4

The propagating steps must give the products and also form chain-carrying free radicals. The formation of t-BuOH suggests H-abstraction from RH by t-BuO-, not by CI.. The steps are: RH + t-BuO*---* R. + t-BuOH RCI + t-BuO. R*+ t-BuOCl+

R. and t-BuO. are the chain-carrying radicals. Problem 4.39 Calculate the heat of combustion of methane at 25°C. The bond energies for C-H, C=O, and O - H are respectively 413.0, 498.3, 803.3, and 462.8 kJ/mol.

-

O=O, 4

First, write the balanced equation for the reaction. CH,

+ 20,

CO,

+ 2H,O

The energies for the bonds broken are calculated. These are endothermic processes and A H is positive. C + 4H

CH,-

2 0 , 4 4 0

AH

= 4( +413.0) =

+ 1652.0kJ/mol

AH

= 2( +498.3) =

+996.6 kJ/mol

Next, the energies are calculated for the bonds formed. Bond formation is exothermic, so the A H values are made negative. C + 20-+0=C=O 4H + 2 0 +

2 H U H

A H = 2(-803.3)

=

-1606.6 kJ/mol

A H = 4(-462.8)

= - 1851.2 kJ/mol

The enthalpy for the reaction is the sum of these values:

+ 1652.0+ 996.6 - 1606.6 - 1851.2 = -809.2

kJ/ mol (Reaction is exothermic)

Problem 4.40 ( a ) Deduce structural formulas and give IUPAC names for the nine isomers of C,H,,. (6) Why 4 is 2-ethylpentane not among the nine? (a)

seven-C chain

1. CH,CH,CH,CH,CH,CH,CH,

Heptane

CH, six-C chain

I

2. CH,CHCH,CH,CH,CH,

2-Methyl hexane

CH

1 ,

3. CH,CH,CHCH,CH,CH,

3-Methylhexane

ALKANES

66

[CHAP. 4

CH,

I

five-C chain

2,3-Dimet h ylpent ane

4. CH,CHCHCH,CH,

I

CH, CH,

CH,

I

I

2,4-Dimethylpentane

5. CH,CHCH,CHCH, CH 6.

4' 4'

CH, CH,CH,CH,

2,2-Dimet hylpentane

CH,

CH

7. CH,CH, CH,CH,

3,3-Dimet hylpen tane

CH.3

CH

:{.

8. CH,CH,CHCH,CH,

3-Ethylpentane

CH, CH,

I I

four-C chain

I

9. CH,C---CHCH,

2,2,3-Trimethylbutane

CH, (6) Because the longest chain has six C's, and it is 3-methylhexane. Problem 4.41 Singlet methylene, :CH, (Section 3.2), may be generated from diazomethane, CH2N2;the other product is N,. It can insert between C-H bonds of alkanes: :CH,

I

+ --C-H+--C CH,CH, > CH, > H , while that for I11 is C1 > (CH,),CH > CH, > H. The con figurations a re :

Problem 5.14 Write structural formulas for the following, indicating enantiomers, if any. and their configura4 tions: ( a ) 3-methyl-3-pentanol. ( b ) 2,2-dimethyl-3-bromohexane. ( c ) 3-phenyl-3-chloro- I-propene.

The alcohol

CH~CH*- ~ - - c H ~ c H ,

I

OH

has no atom bonded to four different groups and no enantiomers; it is achiral.

CHAP. 51

75

STEREOCHEMISTRY

*

(6) (H,C),CCHBr(CH,),CH, has one chiral C (shown with an asterisk).

Relative configuration is the experimentally determined relationship between the configuration of a given chiral molecule and the arbitrarily assigned configuration of a reference chiral molecule. The D,L system (1906) uses glyceraldehyde, HOCH,C*H(OH)CHO, as the reference molecule. The (+)and (-)-rotating enantiomers were arbitrarily assigned the configurations shown below in Fischer projections, and were designated as D and L, respectively.

H+OH CHO

CHzOH

OHC HO+H HOH2C

Relative configurations about chiral C's of other compounds were then established by synthesis of these compounds from glyceraldehyde. It is easy to see that D-(+)-glyceraldehyde has the R configuration and the L-(-) isomer is S. This arbitrary assignment was shown to be correct by Bijvoet (1951), using X-ray diffraction. Consequently, we can now say that R and S denote absolute configuration, which is the actual spatial arrangement about a chiral center. Problem 5.15 D-( +)-Glyceraldehyde is oxidized to (-)-glyceric acid, HOCH,CH(OH)COOH. Give the designation of the acid.

D,L

- r0;:

Oxidation of the D-( +)-aldehyde does not affect any of the bonds to the chiral C. The acid has the same configuration even though the sign of rotation is changed. ~~

I

CHO OH CH,OH

D-(+)-Glyceraldehyde

CH,OH

D-(-)-Glyceric acid

Problem 5.16 Does configuration change in the following reactions? Designate products

H

(D,L)

and (R,S).

4 D

76

STEREOCHEMISTRY

[CHAP. 5

4

No bond to the chiral C is broken and the configuration is unchanged. Therefore, the configuration of both reactant and product is D . The product is R: S converts to R because there is a priority change. Thus, a change from R to S does not necessarily signal an inversion of configuration; it does only if the order of priority is unchanged. A bond to the chiral C is broken when I - displaces Cl . An inversion of configuration has taken place and there is a change from D to L . The product is R. This change from S to R also shows inversion, because there is no change in priorities. Inversion has occurred and there is a change from D to L . The product is R. Even though an inversion of configuration occurred, the reactant and product are both R. This is so because there is a change in the order of priority. The displaced Cl has priority (1) but the incoming CN- has priority (2).

In general. the D , L convention signals a configuration change, if any, as follows: D--* D or L-+ L means no change (retention), and D - L or L + D means change (inversion). The RS convention can also be used, but it requires working out the priority sequences. Problem 5.17 Draw three-dimensional and Fischer projections, using superscripts to show priorities, for (a) (S)-2-bromopropanaldehyde, (6) (R)-3-iodo-2-chloropropanol. 4

5.4 MOLECULES WITH MORE THAN ONE CHIRAL CENTER

With n dissimilar chiral atoms the number of stereoisomers is 2" and the number of racemic forms is 2''-', as illustrated below for 2-chloro-3-bromobutane (n = 2). The R,S configuration is shown next to the C's.

cl-7

(R) R r (R)

H

CH3 I

Br

CH, 11

CH3 111

H

(R) (S)

CH3 IV

If n = 2 and the two chiral atoms are identical in that each holds the same four different groups, there are only 3 stereoisomers, as illustrated for 2,3-dichlorobutane.

77

STEREOCHEMISTRY

CHAP. 51

Mirror

c1.F

(R)

H

(R)

c1

CH3 V

H F I (S) Cl H (s)

H

H CH3

CH3 VI

(s)

Plane

CH,

VII

VIII meso

2( S)-3(R)Dichlorobutane

2(R)-3(S)Dichlorobutane

Structures VII and VIII are identical because rotating either one 180" in the plane of the paper makes is superposable with the other one. VII possesses a symmetry plane and is achiral. Achiral stereoisomers which have chiral centers are called meso. The meso structure is a diastereomer of either of the enantiomers. The meso structure with two chiral sites always has the (RS) configuration. Problem 5.18 For the stereoisomers of 3-iodo-2-butano1, ( a ) assign R and S configurations to C-' and C7.(6) Indicate which are (i) enantiomers and (ii) diastereomers. ( c ) Will rotation about the C - C bond alter the configurations? 4

I H

I1

I11

CH,

CH3

IV I

I

I

I

I

H

OH

OH

(i) I and IV, I1 and 111. (ii) I and IV diastereomeric with I1 and 111; I1 and I11 diastereomeric with I and IV. No. Problem 5.19 Compare physical and chemical properties of ( a ) enantiomers, (6) an enantiomer and its 4 racemic form, and ( c ) diastereomers. With the exception of rotation of plane-polarized light, enantiomers have identical physical properties, e.g., boiling point, melting point, solubility. Their chemical properties are the same toward achiral reagents, solvents, and conditions. Towards chiral reagents, solvents, and catalysts, enantiomers react at different rates. The transition states produced from the chiral reactant and the individual enantiomers are diastereomeric and hence have different energies; the A H S values are different, as are the rates of reaction. Enantiomers are optically active; the racemic form is optically inactive. Other physical properties of an enantiomer and its racemic form may differ depending on the racemic form. The chemical properties are the same towards achiraf reagents, but chiral reagents react at different rates. Diastereomers have different physical properties, and have different chemical properties with both achiral and chiral reagents. The rates are different and the products may be different.

STEREOCHEMISTRY

78

[CHAP. 5

Problem 5.20 How can differences in the solubilities of diastereomers be used to resolve a racemic form into individual enantiomers? 4 The reaction of a racemic form with a chiral reagent, for example, a racemic ( 5 ) acid with a (-) base, yields two diastereomeric salts (+ )( -) and (-)( -) with different solubilities. These salts can be separated by fractional crystallization, and then each salt is treated with a strong acid (HCI) which liberates the enantiomeric organic acid. This is shown schematically:

Racemic Form (2)RCOOH

+

Chiral Base (-)B

Diastereomeric Salts (+)RCOO-(-)BH' + (-)RCOO-(-)BH'

---+

HCI

BH'CI-

sepu ru t ed

5.

+ (+)RCOOH

5. HC1 (-)RCOOH + BH'CI'

separated enant iomeric acids

The most frequently used chiral bases are the naturally occurring, optically active alkaloids, such as strychnine, brucine, and quinine. Similarly, racemic organic bases are resolved with naturally occurring, optically active, organic acids, such as tartaric acid.

5.5

SYNTHESIS AND OPTICAL ACTIVITY

1. Optically inactive reactants with achiral catalysts or solvents yield optically inactive products. With a chiral catalyst, e.g., an enzyme, any chiral product will be optically active. 2. A second chiral center generated in a chiral compound may not have an equal chance for R and S configurations; a 50 : 50 mixture of diastereomers is not usually obtained. 3. Replacement of a group or atom on a chiral center can occur with retention or inversion of configuration or with a mixture of the two (complete or partial racemization), depending on the mechanism of the reaction. Problem 5.21 ( a ) What is the stereochemistry of CH,CHC1CH2CH, isolated from the chlorination of butane? ( b ) How does the mechanism of chlorination of RH account for the formation of the product in (a)? 4 Since all reactants are achiral and optically inactive, the product is optically inactive. Since the product has a chiral center, C 2 , it must be a racemic form. See Fig. 5-5. The first propagation step produces CH,CHCH,CH,. The C of this radical uses sp2 hybridized orbitals and its three bonds are coplanar. The p orbital with the odd electron is at right angles to this plane, which is said to have two faces. Approach of a reactant to one or the other face makes contact with one or the other lobe of the p AO. In the second propagation step, the p A 0 of the sec-butyl radical can add a CI atom from CI, on either face to give rise to one or the other of a pair of enantiomers. For this reason the faces are said to be enantiotopic faces. Since the probability of attack is identical on either face, chlorination gives equal amounts of enantiomers and hence a racemic form. CI, (on either 4de of p orbital)

-

C,H,

H n - Bu tane

CI. + C,H,

HCI +

CI + CI

H

sp' sec-Butyl

free radical

(R)-2-Chlorobutane Fig. 5-5

C,H5

H (S)-2-Chlorobutane

CHAP. 51

79

STEREOCHEMISTRY

Problem 5.22 ( a ) What two products are obtained when C.’ of (R)-2-chlorobutane is chlorinated? (6) Are these diastereomers formed in equal amounts? (c) In terms of mechanism account for the fact that

*

is obtained when (R)-CICH,-CH(CH,)CH,CH, is chlorinated. CI

CI H

H

H CI (RI

4

CI CI

(RR) optically active en a n t iome r

In the products C’ retains the R configuration since none of its bonds were broken and there was no change in priority. The configuration at C3,the newly created chiral center, can be either R or S. As a result, two diastereomers are formed, the optically active RR enantiomer and the optically inactive RS meso compound. No. The numbers of molecules with S and R configurations at Cz are not equal. This is so because the presence of the C’-stereocenter causes an unequal likelihood of attack at the faces of C3.Faces which give rise to diastereomers when attacked by a fourth ligand are diastereotopic faces. Removal of the H from the chiral C leaves the achiral free radical ClCH,C(CH,)CH,CH,. Like the radical in Problem 5.21, it reacts with C1, to give a racemic form.

The C3 of 2-chlorobutane7 of the general type RCH,R’, which becomes chiral when one of its H’s is replaced by another ligand, is said to be prochiral. Problem 5.23 Answer True or False to each of the following statements and explain your choice. ( a ) There are two broad classes of stereoisomers. (6) Achiral molecules cannot possess chiral centers. (c) A reaction catalyzed by an enzyme always gives an optically active product. (d) Racemization of an enantiomer must result in the breaking of at least one bond to the chiral center. (e) An attempted resolution can distinguish a racemate 4 from a meso compound.

( a ) True: Enantiomers and diastereomers. (6) False: Meso compounds are achiral, yet they possess chiral centers. ( c ) False: The product could be achiral. (d) True: Only by breaking a bond could the configuration be changed. (e) True: A racemate can be resolved but a meso compound cannot be, because it does not consist of enantiomers. Problem 5.24 In Fig. 4-4 give the stereochemical relationship between ( a ) the two gauche conformers; (6) the 4 anti and either gauche conformer.

( a ) They are stereomers, because they have the same structural formulas but different spatial arrangements. However, since they readily interconvert by rotation about a U bond, they are not typical, isolatable, configurational stereomers; rather they are conformational stereomers. The two gauche forms are nonsuperimposable mirror images; they are conformational enantiomers. (6) They are conformational diastereomers, because they are stereomers but not mirror images.

Configurational stereomers differ from conformational stereomers in that they are interconverted only by breaking and making chemical bonds. The energy needed for such changes is of the order of 200-600 kJ/mol, which is large enough to permit their isolation, and is much larger than the energy required for interconversion of conformers.

STEREOCHEMISTRY

80

[CHAP. 5

Supplementary Problems Problem 5.25 (a) What is the necessary and sufficient condition for the existence of enantiomers? (6) What is the necessary and sufficient condition for measurement of optical activity? (c) Are all substances with chiral atoms optically active and resolvable? (d) Are enantiomers possible in molecules that do not have chiral carbon atoms'? (e) Can a prochiral carbon ever be primary or tertiary? ( f ) Can conformational enantiomers ever be resolved'? 4 ( a ) Chirality in molecules having nonsuperimposable mirror images. (6) An excess of one enantiomer and a specific rotation large enough to be measured. (c) No. Racemic forms are not optically active but are resolvable. Meso compounds are inactive and not resolvable. (d) Yes. The presence of a chiral atom is a sufficient but not necessary condition for enantiomerism. For example, properly disubstituted allenes have no plane or center of symmetry and are chiral molecules even though they have no chiral C's:

Mirror

*

a chiral allene

(nonsuperimposable enantiomers)

(e) No. Replacing one H of U 1" CH, group by an X group would leave an achiral --CH,X group. In order for a 3" CH group to be chiral when the H is replaced by X, it would already have to be chiral when bonded to the H. ( f ) Yes. There are molecules which have a large enthalpy of activation for rotating about a (+ bond because of severe steric hindrance. Examples are properly substituted biphenyls, e.g. 2,2'-dibromo-6,6'-dinitrobiphenyl, Figs. 5-6. The four bulky substituents prevent the two flat rings from being in the same plane, a requirement for free rotation.

+

Mirror

-

NO,

-

Br

Br NO,

Fig. 5-6

Problem 5.26 Select the chiral atoms in each of the following compounds: CH.

4

.J

CI

CH3

H

I

[CH,CHCH,NCH,CH,CH,] ' CI

I

OH

OH Cholesterol (0)

I

CH,CH,

0

CHAP. 5 )

81

STEREOCHEMISTRY

(a) There are eight chiral C’s: three C’s attached to a CH, group, four C’s attached to lone H’s, and one C attached to OH. ( 6 ) Since N is bonded to four different groups, it is a chiral center, as is the C bonded to the OH. ( c ) There are no chiral atoms in this molecule. The two sides of the ring -CH2CH2- joining the C’s bonded to Cl’s are the same. Hence, neither of the C’s bonded to a CI is chiral. Problem 5.27 Draw examples of (a) a meso alkane having the molecular formula C,H,, and (6) the simplest alkane with a chiral quaternary C. Name each compound. 4

(4 3.4-Dimethylhexane

(6) The chiral C in this alkane must be attached to the four simplest alkyl groups. These are CH,-, CH,CH,-, CH,CH, CH,-, and (CH,),CH-, and the compound is

CH,CH,CH,

qTH3 CHCH,

CH,CH,

2,3-Dimet hyl-3-et h ylhexane

Problem 5.28 Relative configurations of chiral atoms are sometimes established by using reactions in which there is no change in configuration because no bonds to the chiral atom are broken. Which of the following 4 reactions can be used to establish relative configurations?

(S)-CH,CHClCH,CH,

(4

+ Na’OCH,

CH,CH(OCH,)CH,CH,

+ Na’Cl-

+ CH,Br+CH,CH,

(6)

CH3

(4 (4 (4

-+

--

CH3

I + PC1, CH,CH,CClCH,CI + POCI, + HCI (CH,),C(OH)CH(CN)CH, + Na’Br(S)-(CH,),C(OH)CHBrCH, + Na’CNCH,CH,CHCH, + ;H2 (R)-CH,CH,CHCH, + Na I I 1

(R)-CH,CH,COHCH,CI

OH

-

0-Na’

(6) and (e). The others involve breaking bonds to the chiral C. Problem 5.29 Account for the disappearance of optical activity observed when (R)-2-butanol is allowed to 4 stand in aqueous H,SO, and when (S)-2-iodooctane is treated with aqueous KI solution.

Optically active compounds become inactive if they lose their chirality because the chiral center no longer has four different groups, or if they undergo racemization. In the two reactions cited, C remains chiral and it must be concluded that in both reactions racemization occurs. Problem 5.30 For the following compounds, draw projection formulas for all stereoisomers and point out their R,S specifications, optical activity (where present), and meso compounds: (a) 1,2,3,4-tetrahydroxybutane, (6) 1-chloro-2,3-dibromobu tane , (c) 2,4-diiodopentane, (d) 2,3,4-tribromohexane , (e) 2,3,4-tribromopentane .

*

*

4

(a) HOCH,CHOHCHOHCH,OH, with two similar chiral C’s, has one meso form and two optically active enantiomers.

STEREOCHEMISTRY

82

HF H3

*

OH

[CHAP. 5

‘CH,OH

‘CH,OH

H;$H

H& -H HO

OH

‘CH,OH

4CH,0H

4CH,0H

(2S.3R) meso

(2S.X)

(2R,3R)

,

1

Racemic form

*

(6) CICH,CHBrCHBrCH, has two different chiral C’s. There are four ( Z 2 ) optically active enantiomers. CH,CI

CH,CI

:+E

H+:; CH, (2S.3R)

CH,CI

CH,CI

H:* Br CH3

CH,

CH3

(2R.X)

(2S.3S)

(2R.3R)

threo enantiomers

erythro enantiomers

The two sets of diastereomers are differentiated by the prefix eryrhro for the set in which at least two identical o r similar substituents on chiral C’s are eclipsed. The other set is called threo.

(c) CH3?!HICH,6HICH, has two similar chiral C’s, C‘ and C‘, separated by a CH, group. There are two enantiomers comprising a ( 2 )pair and one meso compound.

meso

*

*

*

Brr

Racemate

( d ) With three different chiral C’s in CH,CHBrCHBrCHBrCH,CH,, there are eight (2’) enantiomers and four racemic forms.

l3;3

H t‘CH, ;;

H H

H” H4

Br

C,H, I (2S,3S,4R) Racemate

Hf; H Br

(2S,3S,4S) \

C2Hs

C,H,

111

IV

(2R,3R,4S)

(2S,3R,4R)

(2R,3S,4S)

Br3r

Racemate

(2R,3R,4R) 2

$

Racemate

Br H*

Br

Br

C2Hs

*

H

I1

v1

V

Br

C2Hs

Br

C,H,

H

Br

C2Hs

.

C,H,

VII

Vill

(2S,3R,4S)

(2R.3S.JR) Y

Race mate

J

CHAP. 51 I

1

83

STEREOCHEMISTRY

*

3

* J

5

(e) CH,CHBrCHBrCHBrCH, has two similar chiral atoms (C' and C'). There are two enantiomers in which the configuratiosn of C' and C' are the same, RR or SS. When C' and C' have different configurations, one R and one S, C7becomes a stereocenter and there are two meso forms.

Problem 5.31 The specific rotation of (R)-(-)-2-bromooctane is -36". What is the percentage composition 4 of a mixture of enantiomers of 2-bromooctane whose rotation is +18"?

Let x

= mole

fraction of R, 1 - x

= mole

x(-336')

fraction of S.

+ (1 - x)(36") = 18"

or

x=

The mixture has 25% R and 75% S; it is 50% racemic and 50% S. Problem 5.32 (S)-2-Chlorobutane yields a mixture of 71 % meso and 29% 2(S)-3( S)-2,3-dichlorobutane, while (R)-2-chlorobutane similarly gives a mixture of 71% meso and 29% 2(R)-3(R)-2,3-dichlorobutane on chlorination. 4 Account for these products by conformational analysis.

See Fig. 5-7. In the first step the radicals CH,~HCI--kH--CH, (111 through VI) are formed on an atom adjacent to a chiral C. The formation of the free radical does not change the configuration of the chiral center, which is S in one case and R in the other. A second chiral center, which may be R or S, is generated after step two. I

I

I I

I

H

I

I

/ 1

(I)

\

1

1

(VII)

(VIII)

(2R.3R)

(2R.X) (71% meso)

(2Wo enantiomer)

!

Fig. 5-7

I I

I

1

(IX)

(X)

(ZS.3R) (71% meso)

(7S.3S)

(2% enantiomer)

[CHAP. 5

STEREOCHEMISTRY

84

(R)-2-Chlorobutane ( I ) forms free radicals (111 and IV) which are conformational diastereomers with different stabilities and populations. This is also true of 1's enantiomer 11, (S)-2-chlorobutane, which gives free radicals V and VI. The more stable free-radical conformers are IV and V because their CH,'s are anti-like. The transition states for CI abstractions arising from conformations IV and V have lower AH' values than the diastereomeric transition states from the more crowded gauche-like conformations, 111 and VI. The major product is therefore the meso compound, VIII = I X . Problem 5.33 Predict the yields of stereoisomeric products, and the optical activity of the mixture of products, formed from chlorination of a racemic mixture of 2-chlorobutane to give 2,3-dichlorobutane. 4

The (S)-2-chlorobutane comprising 50% of the racemic mixture gives 35.5% of the rneso (SR) product and

14.5% of the RR enantiomer. The R enantiorner gives 35.5% rneso and 14.5% RR products. The total yield of meso product is 71% and the combination of 14.5% RR and 14.5% SS gives 29% racemic product. The total reaction mixture is optically inactive. This result confirms the generalization that optically inactive starting

materials, reagents and solvents always lead to optically inactive products.

Problem 5.34 For the following reactions give the number of stereoisomers that are isolated, their R,S configurations and their optical activities. Use Fischer projections. (4

(6)

meso- HOCH ,CHOHCHOHCH,OH ( R)-CICH ,CH( CH

,)CH2CH,

o x 1J.I I I t 1 11

CUprdlL.

0

I1

t12 NI

t)-CH,-C--CHOH--CH,-CH

~UC(

(d)

HOCH ,CHOHCHOHCOOH

CH 3CH,CH( CH ,)CH ,CH,CH( CH, )CH,CH ,

\ 1.1

(S)-CH,CH,

-EH2 c=c /c=c\

\CH3

\H

H

H H

\c=c

H

H 3c /

\H

\H

H'

rrurts, rruns -2A-Hexadiene or

cis, truns-2.4-Hexadiene or (2,E )-ZP-Hexadiene

cis. cis -2.4-Hexadiene or (2,Z )-2,4-He xadiene

'

\c=c

( E , E )-2,CHexadiene

Problem 6.6 Write structural formulas for ( a ) (E)-2-methyl-3-hexene ( t r a m ) , (b) (S)-3-chloro-l-pentene, ( c ) (R) ,( 2)-2-chlor0-3-heptene (cis). 4

Problem 6.7 Supply structural formulas and systematic names for all pentene isomers including stereoisomers. 4 These are derived by first writing the carbon skeletons of all pentane isomers and then introducing one double bond. A double bond can be placed in n-pentane in two ways to give two isomers.

H2C=CHCH2CH,CH,

CH,CH=CHCH,CH, 2-Pentene

I-Pentene

2-Pentene has geometric isomers:

cis- or (2)-?-Pentene

rrans- or (E)-2-Pentene

From isopentane we get three isomeric alkenes but none has geometric isomers.

CH3

CH, H2C=CCH,CH3 I

CH,C=CHCH, I

?-Methyl- I-butene

2- Methyl-2-butene

CH,

I

CH,CHCH=CH, 3-Methyl- I -butene

N o alkene can be formed from neopentane

because a double bond would give the central C five bonds. None of the isomeric pentenes are chiral. Atoms with multiple bonds cannot be chiral centers. Problem 6.8 How do boiling points and solubilities of alkenes compare with those of corresponding alkanes?

4

Alkanes and alkenes are nonpolar compounds whose corresponding structures have almost identical molecular weights. Boiling points of alkenes are close to those of alkanes and similarly have 20" increments per

91

ALKENES

CHAP. 61

C atom. Both are soluble in nonpolar solvents and insoluble in water, except that lower-molecular-weight alkenes are slightly more water-soluble because of attraction between the 'TT bond and H,O. Problem 6.9 Show the directions of individual bond dipoles and net dipole of the molecule for ( a ) 4 1,I -dichloroethylene, (6) cis- and trans-l,2-dichloroethylene.

The individual dipoles are shown by the arrows on the bonds between C and C1. The net dipole for the molecule is represented by an arrow that bisects the angle between the two Cl's. C-H dipoles are insignificant and are disregarded.

cis

In the truns isomer the C - 4 a zero dipole moment.

trans

moments are equal but in opposite directions; they cancel and the tram isomer has

Problem 6.10 How can heats of combustion be used to compare the differences in stability of the geometric isomers of alkenes? 4

The thermodynamic stability of isomeric hydrocarbons is determined by burning them to CO, and H,O and comparing the heat evolved per mole ( - A H combustion). The more stable isomer has the smaller ( - A H ) value. Truns alkenes have the smaller values and hence are more stable than the cis isomers. This is supported by the exothermic ( A H negative) conversion of cis to tram isomers by ultraviolet light and some chemical reagents. The cis isomer has higher energy because there is greater repulsion between its alkyl groups on the same side of the double bond than between an alkyl group and an H in the trans isomer. These repulsions are greater with larger alkyl groups, which produce larger energy differences between geometric isomers. Problem 6.11 Suggest a mechanism to account for interconversion of cis and tram isomers by electromagnetic 4 radiation of wavelength less than 300nm.

See Problem 2.41. The excited molecule has the indicated major conformation. Return of the excited electron from this conformation to the ground-state 'TT bond results in formation of a mixture of geometric isomers.

excited molecule

6.3 PREPARATION OF ALKENES 1.

1,2-Eliminations

Also called p-eliminations, these constitute the principal laboratory method whereby two atoms or groups are removed from adjacent bonded C's. (a)

Dehydrohalogenation

H X I I B; + -C-C-

I

I

+ -C=C-

I

I

+

B:H

+

X-

KOH in ethanol is most often used as the source of the base, BT, which then is mainly C,H,O-.

92

ALKENES

(6) Dehydration

H OH

I I

-C-C(c)

[CHAP. 6

Dehalogenation

I I

acid

--C=C-

I

I

+

HzO

+

MgBr,(or ZnBr,)

Br Br

Mg(or Zn)

I I + -C-CI I

+

-C=C-

I

I

(d) De hydrogenation H H

I I

-C-C--

I I

Pt or Pd

-C=C-

I

I

+ H2

(mainly a special industrial process)

Cracking (Section 4.4) of petroleum hydrocarbons is the source of commercial alkenes. In dehydration and dehydrohalogenation the preferential order for removal of an H is 3”> 2” > 1” (Saytzeff rule). We can say “the poor get poorer.” This order obtains because the more R’s on the C = C group. the more stable is the alkene. The stability of alkenes in decreasing order of substitution by R is R,C=CR, > R,C=CRH > R,C=CH,, 2.

RCH=CHR > RCH=CH, > CH,=CH,

Partial Reduction of Alkynes

R

\c=c

cis-alkene

H

RC-CR an alkyne

/R

R\

,c=c

H

/H

?runs-a1kene

\R

These reactions, which each give only one of two possible stereomers, are called stereoselective. In this case they are more specifically called diastereoselective, because the stereomers are diastereomers. Problem 6.12 (a) How does the greater enthalpy of cis- vis a vis tram-2-butene affect the ratio of isomers formed during the dehydrohalogenation of 2-chlorobutane? (6) How does replacing the CH, groups of 2-chlorobutane by t-butyl groups to give CH,C(CH,),CH,CHCIC(CH,),CH, alter the distribution of the alkene geometric isomers? 4 ( a ) The transition states for the formation of the geometric isomers reflect the relative stabilities of the

isomers. The greater repulsion between the nearby CH,’s in the cis-like transition state (TS), causes this TS to have a higher enthalpy of activation ( A H ’ ) than the rrans-like TS. Consequently, the tram isomer predominates. (6) The repulsion of the bulkier r-butyl groups causes a substantial increase in the A H ’ of the cis-like TS, and the tram isomer is practically the only product.

Problem 6.13 Give the structural formulas for the alkenes formed on dehydrobromination of the following alkyl bromides and underline the principal product in each reaction: ( a ) 1-bromobutane, (6) 2-bromobutane, (c) 3-bromopentane, (d) 2-bromo-2-methylpentane, (e) 3-bromo-2-methylpentane, ( f ) 3-bromo-2,3-dimethylpen tane . 4

CHAP. 61

93

ALKENES

-

The Br is removed with an H from an adjacent C.

H2C-CHCH,CH,

( a1

$r

I, I I

-

4

H,C=CHCH,CH,

H Br H2 (c1

(only 1 adjacent H; only 1 product)

A 1

H2C-CHCHCH,

(61

H2C=CHCH2CH,

CH,CHCHCHCH,

I, I I

H Br HI

+ cis- and truns-CH,CH=CHCH, (-H2)

(-HI)

cis- and trans-CH,CH=CHCH,CH,

di-R-substituted

(adjacent H's are equivalent)

CH,

CH3

I

I

+ H2C=C-CH2CH2CH3

(d1

and CH,C=CHCH,CH,

(-H') (CH,),C-CHCHCH,

(e)

I, I I H Br H2

* (CH,),C=CHCH,CH, (-H'

)

(-H2)

+ cis- and trans-(CH,),CHCH-CHCH,

tri-R-substituted

(-H2)

H,CH~

di-R-substituted

CH,

I CH,CH-C-C(CH,), I I I H' Br H3

I

__+

cis - and trans-CH,CH=CCH(CH,), (-HI) tri-R-alkene CH,

I + C H ,C H,C=C

CH2

(C H 3,

II + C H ,C H2-C-C

(-H3)tetra-R-alkene

(

H(C H3)2

- H 2 ) di-R-alkene

Problem 6.14 ( a ) Suggest a mechanism for the dehydration of CH,CHOHCH, that proceeds through a carbocation intermediate. Assign a catalytic role to the acid and keep in mind that the 0 in ROH is a basic site like the 0 in H,O. (b) Select the slow rate-determining step and justify your choice. (c) Use transition states to 4 explain the order of reactivity of ROH: 3" > 2" > 1".

H

H (U)

Step 1

I

CH,CHCH,

I

+

H,SO,

H:~)H

:OH base

I

e CH,CHCH, + HSO; I fast

+

acid,

acid,

basez

an onium ion

Step 2

CH,C-CH,

CH,eHCH,

I

H,b'

+

H20

H lsopropyl cation

+

5CH,CH=CH,

Step 3

+

H,SO,

very Strong

acid I

base,

base I

acid

,

Instead of HSO;, a molecule of alcohol could act as the base in Step 3 to give ROH;. ( 6 ) Carbocation formation, Step 2, is the slow step, because it is a heterolysis leading t o a very high-energy carbocation possessing an electron-deficient C. ( c ) The order of reactivity of the alcohols reflects the order of stability of the incipient carbocation (3" > 2" > 1") in the TS of Step 2, the rate-determining step. See Fig. 6-2.

ALKENES

94

[CHAP. 6

REH~ - - OH^ 8+

8+

6+

R,C---OH,

+

R3

R ,CH~H,

~

0

~

2

Reaction Prope ss Fig. 6-2

Problem 6.15 Account for the fact that dehydration of: (a) CH,CH,CH,CH,OH yields mainly CH,CH==CHCH, rather than CH,CH,CH=CH,, (b) (CH,),CCHOHCH, yields mainly (CH,),C=C(CH,),.

4

( a ) The carbocation (R') formed in a reaction like Step 2 of Problem 6.14(a) is 1" and rearranges to a more stable 2" R,CH' by a hydride shift (indicated as -H:; the H migrates with its bonding pair of electrons).

H

H

I +ROH 2C H , C H 2 C + H 2 T +

I" RCH; (n-Butyl cation)

H

&

2" RICH+ (3,3-Dimethyl2-butyl cation)

+ RGH,

2" RICH+ (sec-Butyl cation)

(b) The 2" R,CH+ formed undergoes a methide shift (-:CH,)

(CH,), -CHCH3

CH,CH=CHCH,

-.CHj

+

(CH,),C 1") has a lower A N S for the transition state and forms more rapidly (Fig. 6-4). Markovnikov additions are called regioselective since they give mainly one of several possible structural isomers. Problem 6.24 Give the structural formula of the major organic product formed from the reaction of CH,CH=CH, with: ( a ) Br,, (6) HI, (c) BrOH, (d) H,O in acid, (e) cold H,SO,, ( f ) BH, from B,H,, (g) peroxyformic acid ( H , 0 2 and HCOOH). 4

CHAP, 61

99

ALKENES

I

(CH3)2C2= C'H2 + H' Reaction Progress Fig. 6-4

The positive +

(a+)

+

part of the addendum is an electrophile (E') which forms CH,CHCH,E rather than

CH,CHECH,. The Nu: part then forms a bond with the carbocation. The E' is in a box; the Nu:- is encircled.

(6) CH3-CH-CH2

6h

2CH CHICH

A (CH3CH,CH3,B (Anti-Markovnikov orientation; with nonbulky alkyl groups, all H's of BH, add to form a trialkylborane)

Problem 6.25 Account for the anti-Markovnikov orientation in Problem 6.24( f ) .

4

The electron-deficient B of BH,, as an electrophilic site, reacts with the ?r electrons of C=C, as the nucleophilic site. In typical fashion, the bond is formed with the C having the greater number of H's, in this case the terminal C. As this bond forms, one of the H's of BH, begins to break away from the B as it forms a bond to the other doubly bonded C atom giving a four-center transition state shown in the equation. The product from this step, CH,CH,CH,BH, (n-propyl borane), reacts stepwise in a similar fashion with two more molecules of propene, eventually to give (CH,CH,CH,),B.

CH,CH=CH,

+ BH, --+

U +

CH,q+H,+ 1 1 H-4H,

CH,CH2CH2BH,

-

2 CH2=CHCH3

(CH,CH,CH,),B

This reaction is a stereoselective and regioselective syn addition.

Problem 6.26 ( a ) What principle is used to relate the mechanisms for dehydration of alcohols and hydration of alkenes? (6) What conditions favor dehydration rather than hydration reactions? 4 ( a ) The principle of microscopic reversibility states that every reaction is reversible, even if only to a

microscopic extent. Furthermore, the reverse process proceeds through the same intermediates and transition states, but in the opposite order.

RCH2CH20H

H+

RCH=CH,

+H20

ALKENES

100

[CHAP. 6

(6) Low H,O concentration and high temperature favor alkene formation by dehydration, because the volatile alkene distills out of the reaction mixture and shifts the equilibrium. Hydration of alkenes occurs at low temperature and with dilute acid which provides a high concentration of H,O as reactant. Problem 6.27 Why are dry gaseous hydrogen halides (HX) acids and not their aqueous solutions used to 4 prepare alkyl halides from alkenes?

Dry hydrogen halides are stronger acids and better electrophiles than the H 3 0 + formed in their water solutions. Furthermore, H,O is a nucleophile that can react with R' to give an alcohol. Problem 6.28 Isobutylene gas dissolves in 63% H,SO, to form a deliquescent white solid. If the H,SO, solution is diluted with H,O and heated, the organic compound obtained is a liquid boiling at 83°C. Explain.

CH3-C CH, -I

I I

+ H:OS03H

CH,-C-OSO,H

I

H O

A CH,-C-OH

I

+ H,SO,

CH3

CH3

Isobutylene

4

CH,

CH,

terl-Butyl hydrogen sulfate (reactive ester of H2S04) (white solid)

red-Butyl alcohol (b.p. 83 "C)

Problem 6.29 Arrange the following alkenes in order of increasing -reactivity on addition of hydrohalogen acids: (a) H,C=CH,, (6) (CH,),C=CH,, ( c ) CH,CH=CHCH,. 4

The relative reactivities are directly relattd to the stabilities of the intermediate R"s. Isobutylene, (b), is most reactive beca+use it forms the 3" (CH,),CCH,. The ne+xt-most reactive compound is 2-butene, (c), which forms the 2" CH,CHCH,CH,. Ethylene forms the 1" CH,CH, and is least reactive. The order of increasing reactivity is: ( a ) < (c) < ( b ) . Problem 6.30 The addition of HBr to some alkenes gives a mixture of the expected alkyl bromide and an isomer formed by rearrangement. Outline the mechanism of formation and structures of products from the 4 reaction of HBr with ( a ) 3-methyl-l-butene, (b) 3,3-dimethyl-l-butene.

No matter how formed, an R' can undergo H: or :CH, (or other alkyl) shifts to form a more stable R".

7 w

CH3

CH3

I

CH3CHBrcHCH3

2-Bromo-3-methylbutane

H2C-LHCHCH3 -% CH3-yH-C-CH3

I

CH3

I CH,CH,-:-CH,

H 2"

Br-

3"

CH3

CH,

CH,

I

H2C=-CH--C--CH,

H* I__*

I

CH3-:H-C-CH3

I

CH3

&H3 2"

lesr srahle

/

I

CH3

I 1

CH3CH2-C-CH3

Br

2-Bromo-2-methylbutane

CH,-CH-k-CH,

I

I

Br CH3 3-Bromo-2,2-dimethylbutane

k3

CH3

I

+

CH3-CH-C

I

CH, Br

I

I

-CH3 Br-_ CH,CH-C-CH,

CH3 3" more srable

I

CH, 2-Bromo-2,3-dimethylbutane

CHAP. 61

101

ALKENES

Problem 6.31 Compare and explain the relative rates of addition to alkenes (reactivities) of HCI, HBr and HI.

4

The relative reactivity depends on the ability of HX to donate an H ' (acidity) to form an R ' in the rate-controlling first step. The acidity and reactivity order is HI > HBr > HCI.

Problem 6.32 (a) What does each of the following observations tell you about the mechanism of the addition of Br, to an alkene? (i) In the presence of a C1- salt, in addition to the vic-dibromide, some vicbromochloroalkane is isolated but no dichloride is obtained. (ii) With cis-2-butene only rac-2,3-dibromobutane is formed. (iii) With tram-2-butene only rneso-2,3-dibromobutane is produced. (b) Give a mechanism 4 compat ible with these observations. (a) (i) Br, adds in two steps. If Br, added in one step, no bromochloroalkane would be formed. Furthermore, the first step must be the addition of an electrophile ( the Br ' part of Br, ) followed by addition of a

nucleophile, which could now be Br- or CI-. This explains why the products must contain at least one Br. (ii) One Br adds from above the plane of the double bond, the second Br adds from below. This is an anti (trans) addition. Since a Br' can add from above to either C, a racemic form results.

Br cis rucemic

(iii) This substantiates the anti addition.

trans

meso

The reaction is also stereospecific because different stereoisomers give stereochemically different products, e.g. cis + racemic and \runs + meso. Because of this stereospecificity , the intermediate cannot be the free carbocation CH,CHBrCHCH,. The same carbocation would arise from either cis-or trans-2-butene, and the product distribution from both reactants would be identical. ( 6 ) The open carbocation is replaced by a cyclic bridged ion having Br' partially bonded to each C (bromonium ion). In this way the stereochemical differences of the starting materials are retained in the intermediate. In the second step, the nucleophile attacks the side opposite the bridging group to yield the anti addition product.

(2S,3R)-2,3-Dibromobutane (meso )

H

H

H trans

(2R,3S)-2,3-Dibromobutane (meso)

a bromonium ion

Br, does not break up into Br' and Br-. More likely, the other as an anion (Fig. 6-5).

T

electrons attack one of the Br's. displacing the

102

ALKENES

[CHAP. 6

..

Bromonium ion + :BJ :

-

Fig. 6-5

Problem 6.33 Alkenes react with aqueous C1, or Br, to yield vic-halohydrins, -CXCOH. Give a mechanism 4 for this reaction that also explains how Brz and (CH,),C=CH2 give (CH,),C(OH)CH,Br. The reaction proceeds through a bromonium ion [Problem 6.32(6)] which reacts with the nucleophilic H,O to give

I+ I

-CBrCOH,

I

This protonated halohydrin then loses H' to the solvent, giving the halohydrin. The partial bonds between the C's and Br engender S+ charges on the C's. Since the bromonium ion of 2-methylpropene has more partial positive charge on the 3" carbon than on the 1" carbon, H,O binds to the 3°C to give the observed product. In general, X appears on the C with the greater number of H's. The addition, like that of Br,, is anti because H,O binds to the C from the side away from the side where the Br is positioned. Problem 6.34 ( U ) Describe the stereochemistry of glycol formation with peroxyformic acid (HC0,H) if cis-2-butene gives a racemic glycol and rruns-2-butene gives the meso form. (6) Give a mechanism for cis. 4 The reaction is a stereospecific anti addition similar to that of addition of Br,.

OH

racemic

I

HK..

.

I

OH racemic

Problem 6.35 Describe the stereochemistry of glycol formation with cold alkaline aqueous KMnO, if cis-2-butene gives the meso glycol and truns-2-butene gives the racemate. 4

ALKENES

CHAP. 61

103

The reaction is a stereospecific syn (cis) addition since both OH groups bond from the same side.

cis-2-Butene

(R,S)-meso-Glycol

1! __*

Cyclic intermediate from topside attack

tram-2-Butene

from underside attack

roc-Glycol

DIMERIZATION AND POLYMERIZATION Under proper conditions a carbocation (R’), formed by adding an electrophile such as H’ or BF, to an alkene, may add to the C=C bond of another alkene molecule to give a new dimeric R“; here, R’ acts as an electrophile and the 7r bond of C=C acts as a nucleophilic site. R“ may then lose an H’ to give an alkene dimer. Problem 6.36 ( a ) Suggest a mechanism for the dimerization of isobutylene, (CH,),C=CH,. (6) Why does (CH,),C’ add to the “tail” carbon rather than to the “head” carbon? (c) Why are the Bronsted acids H,SO, 4 and HF typically used as catalysts, rather than HCl, HBr, or HI?

n

Step 2 Me,C’

+ H,C=C(CH r \

CH, ) +Me,C-CHC 3 2

“tail” “head”

electrophile

Step 3 R‘+ - H’

R”

-

I

H

I

+TH H’

nucleophile

3” dimeric R’’

Me,C--CH==C(CH,),

(major, Saytzeff product) (minor, non-Saytzeff product)

- H’+ ---+ Me,C--CH,C(CH,)=CH,

( 6 ) Step 2 is a Markovnikov addition. Attack at the “tail” gives the 3” R“; attack at the “head” would give the much less stable, 1” carbocation ‘CH,C(CH,),CMe,. ( c ) The catalytic acid must have a weakly nucleophilic conjugate base to avoid addition of HX 60 the C=C. The conjugate bases of HCl, HBr, and H1 (Cl-, Br-, and I - ) are good nucleophiles that tind to R’.

104

[CHAP. 6

ALKENES

The newly formed R’’ may also add to another alkene molecule to give a trimer. The process whereby simple molecules, or monomers, are merged can continue, eventually giving high-molecularweight molecules called polymers. This reaction of alkenes is called chain-growth (addition) polymerization. The repeating unit in the polymer is called the mer. If a mixture of at least two different monomers polymerizes, there is obtained a copolymer. Problem 6.37 Write the structural formula for ( a ) the major trimeric alkene formed from (CH,),C=CH,, labeling the mer; ( b ) the dimeric alkene from CH,CH=CH,. [Indicate the dimeric R’.] 4

mer The individual combining units are boxed. (b)

(CH,),CHCH=CHCH,

; [( CH, ),CHCH,CHCH,]

S TEREOCHEMISTR Y OF POL YMERIZATION

The polymerization of propylene gives stereochemically different polypropylenes having different physical properties. H

-c-

or

H *.,

*IR

4-

The C’s of the mers are chiral, giving millions of stereoisomers, which are grouped into three classes depending on the arrangement of the branching Me (R) groups relative to the long “backbone” chain of the polymer (Fig. 6-6). CH,H CH,H CH,H CH,H CH,H CH,H

Isotactic (Me or R all on same side)

CH,H HCH, CH,H HCH, CH,H H C H , CH,H

Syndiotactic (Me or R on alternate sides)

CH,H CH,H

HCH, CH,H H C H , H C H ,

Atactic (Me or R randomly distributed)

Fk.6-6

CHAP. 6)

105

ALKENES

Problem 6.38 Suggest a mechanism for alkane addition where the key step is an intermolecular hydride (H:) transfer. 4 See Steps 1 and 2 in Problem 6.36 for formation of the dimeric R'. CH 3 CH (CH,),CCH,L

I

4'

~-@C(CH,),-+(CH,),CCH,

@+kCH,),

CH, CH 3 dimeric R' This intermolecular H: transfer forms the (CH,),C' ion which adds to another molecule of (CH,),C=CH, to continue the chain. A 3"H usually transfers to leave a 3" R'. FREE -RADICAL ADDITIONS

-

-

RCHSH,

+ HBr

RCHSH,

+ HSH

RCHSH,

+ HCCl,

RCHSH,

+ BrCCI,

RCH,CH,Br

ROOR

(anti-Markovnikov; not with HF, HCI or HI)

RCH,CH,SH

ROOR

(anti-Markovnikov)

RCHCH,[ml

-Lh ROOR

(anti-Markovnikov)

1-1

RCHmCH,

Problem 6.39 Suggest a chain-propagating free-radical mechanism for addition of HBr in which Br. attacks the alkene to form the more stable carbon radical. 4 Initiation Steps R-0-U-RRO. + HBr-

heat

2R-0

a

bond is weak)

(*

Br. + R U H

Propagation Steps For Chain Reaction CH,CHBrCH, +ICH,CH==CH,I+

Br---+ CH,CHCH2Br

( I " radical)

CH,CHCH,Br

(2" radical)

+ HBr-CH,CH,CH,Br

+ Br.

The Br. generated in the second propagation step continues the chain.

CARBENE ADDITION

1

H H

WN2

Diazomethane

CLEAVAGE REACTIONS

Ozonolysis

Ozonide

Carbonyl compounds

ALKENES

106

[CHAP. 6

Problem 6.40 Give the products formed on ozonolysis of ( a ) H,C=CHCH,CH,, ( b ) CH,CH=CHCH,, (c) (CH,),C==CHCH,CH,, (d) cyclobutene, (e) H,C=CHCH,CH=CHCH,. 4

To get the correct answers, erase the double bond and attach a =O to each of the formerly double-bonded The total numbers of C’s in the carbonyl products and in the alkene reactant must be equal.

H,C=O + O=CHCH,CH,. CH,CH=O; the alkene is symmetrical and only one carbonyl compound is formed. (CH,),C=O + O=CHCH,CH,. O=CHCH,CH,CH-V; a cycloalkene gives only a dicarbonyl compound. H , C = O + O = C H C H , C H = O + O=CHCH,. Noncyclic polyenes give a mixture of monocarbonyl compounds formed from the terminal C’s and dicarbonyl compounds from the internal doubly bonded C’S. Problem 6.41 Deduce the structures of the following alkenes. 0

II

An alkene CIOHZO o n ozonolysis yields only CH,--C--CH,CH,CH,.

0

H

I

II

An alkene C,H,, on ozonolysis gives (CH,),CC--V and C H , - C - C H , C H , . A compound C,H,, adds one mole of H, and forms on ozonolysis the dialdehyde

CH3

CH3

I

I

O=CHCHCH,CH,CHCH=O A compound C X H l 2adds two moles of H, and undergoes ozonolysis to give two moles of the dialdehyde O=CHCH2CH,CH=0. 4 The formation of only one carbonyl compound indicates that the alkene is symmetrical about the double bond. Write the structure of the ketone twice so that the C = O groups face each other. Replacement of the two 0 ’ s by a double bond gives the alkene structure.

CH3

CH3

I

CH,CH,CH,C=O

+ O=CCH2CH2CH,

CH3

I CH,-C-CH=O I

H3C CH,

I

I

CH3

CH3

I

CH,CH2CH2C=CCH2CH2CH3

t-

CHj

I I +CH3-C-C=C-CH2CH, I 1

I

+ 0-C-CH2CH3

cis or tram

CH,H

CH3

C,H,, has four fewer H’s than the corresponding alkane, CxH,,. There are two degrees of unsaturation. One of these is accounted for by a C = C because the alkene adds 1 mole of H,. The second degree of unsaturation is a ring structure. The compound is a cycloalkene whose structure is found by writing the two terminal carbonyl groups facing each other.

CH,-CH

CH=O O=CH /

‘cH-cH3 ‘C H*-cH /

’

2

01 HZO(Zn)

CH3-C

yCH=T ,CH-CHI \CH,-CH~

(d) The difference of six H’s between C,H,, and the alkane C,H,, shows three degrees of unsaturation. The 2mol of H, absorbed indicates two double bonds. The third degree of unsaturation is a ring structure. When two molecules of product are written with the pairs of C = O groups facing each other, the compound is seen to be a cyclic diene.

CHAP. 61

107

ALKENES

CH=O /

7

0-CH

I

HX\

CH=O

\

CH2

H*C

O=CH

I

~H=CH R / '&i2 H2C

o3

1.

'2.

17

HZO(Zn)

,CH2

H2C

14

5

'~H=CH

/CH2

13-Cyclooctadiene

6.5 SUBSTITUTION REACTIONS AT THE ALLYLIC POSITION

-

Allylic carbons are the ones bonded to the doubly bonded C's; the H's attached to them are called allylic H's.

+ H,C=CHCH, Br, + H,C=CHCH, C1,

high temperature

H,C=CHCH,CI

low concentration

H,C=CHCH,Br

of Br,

+ HCI + HBr

The low concentration of Br, comes from N - bromosuccinimide (NBS).

NBS

S02C12 + H2C=CHCH3

product of bromina tion

Succinimide

H2C=CHCH2C1

Sulfuryl chloride

+ HCl + SO2

These halogenations are like free-radical substitutions of alkanes (see Section 4.4). The order of reactivity of H-abstraction is allyl > 3" > 2" > 1" > vinyl Problem 6.42 Use the concepts of ( a ) resonance and ( b ) extended account for the extraordinary stability of the allyl-type radical.

7~

orbital overlap (delocalization) to

-LA-&

4

( a ) Two equivalent resonance structures can be written:

therefore the allyl-type radical has considerable resonance energy (Section 2.7) and is relatively stable. (6) The three C's in the allyl unit are sp2-hybridized and each has a p orbital lying in a common plane (Fig. 6-7). These three p orbitals overlap forming an extended 7r system, thereby delocalizing the odd electron. Such delocalization stabilizes the allyl-type free radical.

Problem 6.43 Designate the type of each set of H's in CH,CH=CHCH,CH,--CH(CH,), (e.g. 3", allylic, etc.) and show their relative reactivity toward a Br. atom, using (1) for the most reactive, (2) for the next, etc. Labeling the H's as

we have: ( a ) l", allylic (2); (6) vinylic (6); (c) 2", allylic ( 1 ) ; (d) 2" (4); ( e ) 3" (3); ( f ) 1" ( 5 ) .

4

[CHAP. 6

ALKENES

108

-7-7 0

....

AlJ

-c-c\

--U---

Fig. 6.7

6.6

SUMMARY OF ALKENE CHEMISTRY

PROPERTES

PREPARA TlON 1.

Dehydrohalogenation of RX

RCHXCH,, RCH,CH,X 2.

1. Addition Reactions

+ alc. KOH

(a) Hydrogenation

Dehydration of ROH

RCHOHCH,, RCH,CH,OH

+ H,SO,

,Homogeneous: H,

3. Dehalogenation of vic - Dihalide

4.

+ Zn A

I

Rb(PP-

RCH,CH, 2 Chemical: (BH,), CH,CO,H

heat

RCHXCH,X

-

Heterogeneous: H,/Pt

1I

///

( b ) Polar Mechanism

Dehydrogenation of Alkanes

X2+ RCHXCH,X (X = C1, Br) HX-, RCHXCH, + X,, H 2 0 RCH( OH)CH,X + H,O- H RCH(OH)CH,

RCH,CH,, heat, Pt/Pd '

+ H,SO,-+

-+

RCH(OSO,H)CH, + BH,, H,O, + NaOH- RCH,CH,OH + dil. cold KMnO,-+ RCH(OH)CH,OH + hot KMnO, RCOOH + CO, + R'C0,H- - H + R C H - C H ,

5.

--j

0 '' + RCO,H, H,O++ RCH(OH)CH,OH + HF, HCMe, 0" RCH,CH,CMe, + 0,, Zn, H,O+ RCH-0 + CH2=0 __.*

(c)

Free-Radical Mechanism

+ HBr + RCH,CH,Br

+ CHCl, + RCH,CH,CCl, + H' or BF,-,polymer

2.

Allylic Substitution Reactions

R--CH,--CH=CH,

+ X, A or uv

R--CHX-CH==CH,

109

ALKENES

CHAP. 61

Supplementary Problems Problem 6.44 Write structures for the following: 2,3-dimethyl-2-pentene (6) 4-chloro-2,4-dimeth yl-2-pen tene (c) ally1 bromide (d) 2,3-dimethylcyclohexene (e) 3-isopropyl-1- hexene ( f ) 3-isopropyl-2,6-dimethyl-3-heptene (a)

CH

I '

H,C--C----C--CH,--CH,

(U)

I

CH' CH

CH

I '

I '

(6) H,C--C=CH--C-CH, A1

Problem 6.45 (a) Give structural formulas and systematic names for all alkenes with the molecular formula C,H,, that exist as stereomers. (b) Which stereomer has the lowest heat of combustion (is the most stable), and 4 which is the least stable? (a) For a molecule to possess stereomers, it must be chiral, exhibit geometric isomerism, or both. There are five such constitutional isomers of C,H,,. Since double-bonded C's cannot be chiral, one of the remaining four C's must be chiral. This means that a sec-butyl group must be attached to the C=C group. and the enantiomers are:

H

,

H C=CH -C-CH~

CH

(R)-3-Methyl- I-pentene

(S)-3-Methyl-I-pentene

A molecule with a terminal = C H , cannot have geometric isomers. The double bond must be internal and is first placed between C' and C3to give three more sets of stereomers: H H \ / C=C H,C

/

\

H \

c=c

CH,CH,CH,

( 2 )-2-Hexene

CH,CH,CH, /

H,C

/

\

H

\

/ C=C

H

H

\

c=c

CH(CH,),

/

H

(E)-2-Hexene

(E)-3-Methyl-2-pentene

(2)-4-Methyl-2-pentene

(E)-4-Methyl-2-pentene

( 2 )-3-Meth yl-2-pen tene

110

[CHAP. 6

ALKENES

The double bond is now placed between C-i and CJ to give the isomers: CH,CH,

‘c=c

/

CH,CH H YC=C< H CH,CH,

CH,CH,

H ‘

H’

(Z ) -3-Hexene

(E)-3-Hexene

(6) The 3-methyl-2-pentenes each have three R groups on the C = C group and are more stable than one or the other isomers, which have only two R’s on C=C. The E isomer is the more stable because it has the smaller CH,’s cis to each other, whereas the 2 isomer has the larger CH,CH, group cis to a CH, group. 3-Methyl-1-pentene is the least stable, because it has the fewest R groups, only one, on the C=C group. Among the remaining isomers, the truns are more stable than the cis because the R groups are on opposite sides of the C=C and are thus more separated. Problem 6.46 Write structural formulas for the organic compounds designated by a ? and show the stereochemistry where requested. CH3

I

(U)

CH,-C-CH,CH,CH, I

?(major) + ?(minor)

+ alcoholic KOH-

Br H

\

,c=c\ /H

CH;

+

Br,

CH3

? (alkene) + ? (reagent)

(4

\ CH, / CH2, CH-C \

(Stereochemistry)

CH,

CH,

1

CH,-C-CH,-C-CH,

I

H

I

I

OH

+ CHBr, + peroxide +?

HOBr

-

’? (Stereochemistry)

CH3

CH3

cf)

+

+?

-

CH,CH,CH,CH2CH=CH, ,CH,CH,

?

-+

I

CH,CH,CHCH,OH

+ H,SO,,

major: has 3 R‘s on the C 4 .

heat --+?(major) + ?(minor)

minor: has only 2 R’s.

(6) Anti addition to a cis diastereomer gives a racemic mixture, Br H H Br CH,+CH, + CH,--$--$-CH, Br H H Br (c) The 3” alcohol is formed by acid-catalyzed hydration of CH, CH,

I

I CH,-C-CH=C-CH, I H

or

CH,

CH,

I I CH,-C-CH2-C=CH, I H

111

ALKENES

CHAP. 61

The reagent is dilute aq. H,SO,.

(4

CH,CH,CH,CH,CH,CH,CBr,.

(e)

In this polar Markovnikov addition, the positive Br adds to the C having the H. The addition is anti, so the Br will be tram to OH but cis to CH,. The product is racemic:

I

H

HO

H.

The formation of products is shown:

(f)

H CH,CH,CHCHDH CH3 I

H+ ---+

I

CH3CH2CHCH21)H CH3 I

-H20

OH

i"+ /

CH,CH,CHCH; (I")

I

;

CH3CH2C=CH2

I

- H:

CH,C H=C(CH,), _-Ht CH$H,-C(CH,), (3")

(major)

Problem 6.47 Draw an enthalpy-reaction progress diagram for addition of Br, to an alkene. See Fig. 6-8.

A

-xa d

5 C w

/---

+ Br,

Reaction Progress Fig. 6-8

Problem 6.48 Write the initiation and the propagation steps for a free-radical-catalyzed (RO-) addition of CH,CH=O to I-hexene to form methyl n-hexyl ketone,

0

II

CH,CH,CH,CH,CH,CH,-C-CH, The initiation step is

0

II

CH,-C:H

0

+ RO.

4

I + RO:H

CH,--C.

4

112

ALKENES

[CHAP. 6

and the propagation steps are

0 n-C,H,-CH=CH,

0

I

+ C-CH,

+ H-C-CH,

n-C,H,-CH-CH,-C-CH,

n-C,H,-CH-CH,-C-CH,

__*

__*

II

n-C4H9--CH2CH2-

e e

(adds to give 2" R.)

-CH,

+-

-CH,

(regenerates)

Problem 6.49 Suggest a radical mechanism to account for the interconversion of cis and tram isomers by heating with I,. 4

I, has a low bond dissociation energy (151 kJ/mol) and forms 21- on heating. 1- adds to the C==C to form a carbon radical which rotates about its sigma bond and assumes a different conformation. However, the C-I bond is also weak (235 kJ/mol) and the radical loses 1. under these conditions. The double bond is reformed and the two conformations produce a mixture of cis and tram isomers.

'

'IJ

'

+"

tram

-\

cis

Problem 6.50 Write structures for the products of the following polar addition reactions: (a) (CH,),C=CHCH, (c)

(cH,),N--cH=cH,

+ I-Cl-?

+ HSCH,+?

( b ) (CH,),C=CH,

+ HI-?

( d ) H,C=CHCF, 6 +

8-

(CH,),C@CHmCH,. I is less electronegative than Cl in I -Cl (CH 3 ) q

+ HCI-

?

4

and adds to the C with more H's.

H2

CH ,S 8 i

6-

H is less electronegative than S; H-SCH,. (CH, ),N-cH~--cH,--I

+

+

The + +c+harge on N destabilizes an adjacent + charge. (CH,),NCH,CH, is more stable than (CH,),NCHCH,. Addition is anti-Markovnikov. CICH,CH2CF,. The strong electron-attracting C+F3group destabilizes an adjacent + charge so that CH2CH,CF, is the intermediate rather than CH,CHCF,. Problem 6.51 Explain the following observations: (a) Br, and propene in C,H,OH gives not only BrCH,CHBrCH, but also BrCH,CH(OC,H,)CH,. (6) Isobutylene is more reactive than 1-butene towards peroxide-catalyzed addition of CCI,. (c) The presence of A g t salts enhances the solubility of alkenes in H,O. 4

(a) The intermediate bromonium ion reacts with both Br- and C,H,OH as nucleophiles to give the two products. (6) The more stable the intermediate free radical, the more reactive the alkene. H,C=CHCH,CH, adds -CCI, to give the less stable 2" radical CI,CCH,CHCH2CH,, whereas H,C==C(CH,), reacts to give the more stable 3" radical CI,CCH,C(CH,)2. (c) Ag' coordinates with the alkene by p-d 7r bonding to give an ion similar to a bromonium ion, but more stable:

CHAP. 61

113

ALKENES

Problem 6.52 Supply the structural formulas of the alkenes and the reagents which react to form: (a) 4 (CH,),CI, (6) CH,CHBr,, ( c ) BrCH,CHClCH,, (d) BrCH,CHOHCH,Cl.

CH

1 ,

+ HI + HBr + peroxide H,C==CHCH, + BrCl

( 6 ) H,C=CHBr + HBr (d) BrCH,--CH=CH, + HOCl

(a) CH,--C=CH,

( c ) H,C=CCI--CH,

or

or H,C=CH--CH,CI

+ HOBr

Problem 6.53 Outline the steps needed for the following syntheses in reasonable yield. Inorganic reagents and solvents may also be used. (a) 1-Chloropentane to 1,2-dichloropentane. (b) 1-Chloropentane to 2-chloropentane. ( c ) 1-Chloropentane to 1-bromopentane. (d) 1-Bromobutane to 1,2-dihydroxybutane. ( e ) Isobutyl chloride to CH, CH,

I I

I I

CH,-C-CH,-C-CH,

4

I

CH3

Syntheses are best done by working backwards, keeping in mind your starting material.

-

(a) The desired product is a vic-dichloride made by adding C1, to the appropriate alkene, which in turn is made by dehydrochlorinating the starting material. ClCH,CH,CH,CH,CH,

KOH- H,C==CHCH,CH,CH, alc

CI 7

ClCH,CHCICH,CH,CH,

(b) To get a pure product add HCI to 1-pentene as made in part ( a ) .

+ HCI+

H,C=CHCH,CH,CH,

H,CCHCICH,CH,CH,

( c ) An anti-Markovnikov addition of HBr to 1-pentene [part (a)]. H,C=CHCH,CH,CH,

+ HBr-

peroxide

BrCH,CH,CH,CH,CH,

( d ) Glycols are made by mild oxidation of alkenes. alc KMn04 BrCH, CH, CH, CH, KOH- H, M H C H , CH, RTHOCH, CHOHCH, CH, (e) The product has twice as many C's as does the starting material. The skeleton of C's in the product corresponds to that of the dimer of (CH,),C=CH,. CH3 (CH3)2CHCH2Cl

:!i+ (CH&C=CH,

"2'O4

I .(CH,),CCH=C(CH,),+ (CH,),CCH2C=CH2

1

!

(CH,),CCH,CI(CH,), Problem 6.54 Show how propene can be converted to (a) 1,s-hexadiene, (6) 1-bromopropene, ( c ) 4-methyl-lpentene. 4

I

Br CH,CH=CH,

Cl*. so0 "C

I

CH3

MgBr

+

* CICH,CH=CH,

CuBr p

,

I

CH,-CH-CH,CH=CH,

114

ALKENES

[CHAP. 6

Problem 6.55 Dehydration of 3,3-dimethyl-2-butanol, (CH,),CCHOHCH,, yields two alkenes, neither of 4 which is (CH,),CCH=CH,. What are their structures? +

The initially formed 2" (CH,),CCHCH, undergoes a :CH, shift to give H

(C€-13)2k-&CH,)2 (3" carbocation)

which forms

CH,

CH,

I

I

and H,C=C-CH-CH,

I

CH3 J

2

3

4

Problem 6.56 (a) Br, is added to (S)-H,C=CHCHBrCH,. Give Fischer projections and R,S designations for 4 the products. Are the products optically active? (6) Repeat ( a ) with HBr. ( a ) C' becomes chiral and the configuration of C-' is unchanged. There are two optically active diastereomers of 1,2,3-tribromobutane. It is best to draw formulas with H's on vertical lines.

Br H2k=EH+dH, H (S)

Br Br

__*

BrCH,

H

Br

W t H ,+ B r & - i 2 - ~ ~ H ,

H H (2R,3S) optically active

r H (2S.3S) optically active

(6) There are two diastereomers of 2,3-dibromobutane:

Problem 6.57 Polypropylene can be synthesized by the acid-catalyzed polymerization of propylene. ( a ) Show 4 the first three steps. ( b ) Indicate the repeating unit (mer).

-

+=PCH,

(U)

CH,CH=CH, -% (CH,),C+

I

H monomer

monomeric carbocation

H H

-

I l&&-&Hj (CH,),C-C--C+ I l l H H CH, dimeric carbocat ion

H

I (CH,),CH-CH2-C-CH2-C+ I

H

I

CH3 trimeric carbocation

H -CH2-C-

I

I

CH,

Problem 6.58 List the five kinds of reactions of carbocations and give an example of each. ( a ) They combine with nucleophiles.

4

CHAP. 61

ALKENES

115

( b ) As strong acids they lose a vicinal H [deprotonate), to give an alkene.

CH

I '

CH,--C--CH, +

-H4

CH

l 3

CH,--C=CH,

(c) They rearrange to give a more stable carbocation.

CH

I

CH

I '

'+

3"

1"

(d) They add to an alkene to give a carbocation with higher molecular weight.

I

C H , - C+ - C H ,

I ,

I

+ H,C=C-CH,---.+

electrophile

CH

CH

CH'

CH'

1 '

nucleophile

( e ) They may remove :H (a hydride transfer) from a tertiary position in an alkane.

CH

CH

1 ,

+ CH,-C'

Problem 6.59 From propene, prepare ( a ) CH,CHDCH,D, ( 6 ) CH'CHDCH,, (c) CH,CH,CH,D. (a)

Add D, in the presence of Pd.

(6) CH,CH==CH,

(4

+ HCl-

4

CH,CHClCH,

CH,CH=CH,

5 CH,CHMgCICH,

87Hg

CH,CHDCH,

or propene + B,D,-

(CH,CH,CH,),B-

(CH,CHDCH,),B-

C H COOD

C H COOH

product

CH,CH,CH,D

Problem 6.60 Ethylene is alkylated with isobutane in the presence of acid (HF) to give chiefly (CH,),CHCH(CH,), , not (CH,),CCH,CH,. Account for the product. 4

H,C=CH2

- HF

H,C==C(CH,),

+ q c (CH 3 l 3

H,CCH,

H'CCH,

+

+ C(CH,),

+

Problem 6.61 Give 4 simple chemical tests to distinguish an alkene from an alkane.

116

[CHAP. 6

ALKENES

A positive simple chemical test is indicated by one or more detectable events, such as a change in color, formation of a precipitate, evolution of a gas, uptake of a gas, evolution of heat.

' '+ \C=C

/

-A--&I I

5

Br,

(red)

(loss of color)

Br Br

(colorless)

(loss of color and formation of precipitate) (colorless)

\ / C-C + H,SO,(conc.) / \ \C=C

/

/

\

+ H,'*

-A-(!.?,I 1

+

+ heat

(solubility in sulfuric acid)

H OS0,H

-k-k-I I

(uptake of a gas)

H H

Alkanes give none of these tests. Problem 6.62 Give the configuration, stereochemical designation and R,S specification for the indicated tetrahydroxy products.

H H H H

I I I I I I

dil. cold

CH,-C-C=C-C-CH, OH (S)

OH

cis, meso

H H

I I

I

(RI

q

I

I

dil. cold

H OH

(U)

?+?

(6)

+?

(c)

?+?

(d)

?

(e)

?+?

cf)

?

KMno4,0H-+

I

+?

I.

H

CH,-C-C=C-C-CH, OH (9

?

KMnO,.OH-

I.

trans

(optically active)

OHH H H

I I l CH,-C-C=C-C--CH, I

H (RI

l

I

OH

cis (optically active)

(a) syn Addition of encircled OH'S:

dil. cold

KMn04. OH-

*

I

CHAP. 61

117

ALKENES

( 6 ) anti Addition of encircled OH’S:

racemic form

( c ) syn Addition; same products as part (b). (d) anti Addition; same products as part ( a ) . ( e ) syn Addition:

CH3,

w

C

, = C

H OH

H

3

~ OH

w C € H OH

?

,

One optically active stereoisomer is formed.

(f) anti Addition:

(RI (RI (RI (RI Two optically active diasteriomers. Problem 6.63 Describe ( a ) radical-induced, and ( b ) anion-induced, polymerization of alkenes. (c) What kind 4 of alkenes undergo anion-induced polymerization?

-

( a ) See Problem 6.39 for formation of a free-radical initiator, RO-, which adds according to the Markovnikov rule. a = c H c H ,

RO. + H,C--LHCH,

H,C--C*

I\

/

RO

H CH ,

H,C--C--CH,C.

I\ I\ R b H CH, H CH,

-

The polymerization can terminate when the free-radical terminal C of a long chain forms a bond with the terminal C of another long chain (combination), RC- + C R ’ RC-CR’. Termination may also occur when the terminal free-radical C’s of two long chains disproportionate, in a sort of auto-redox reaction. One C picks off an H from the C of the other chain, to give an alkane at one chain end and an alkene group at the other chain end:

RCHC. I + *CCHR’---* I RCHCH I + ,C&R’ \

I I

II

I I

I

(6) Typical anions are carbanions, R: -, generated from lithium or Grignard organometallics.

-

RCH,C:- __* RCH,C-CH2C: /\ /\ I\ X CH, H,C X H,C X

These types of polymerizations also have stereochemical consequences.

(c)

[CHAP. 6

ALKENES

118

Since alkenes do not readily undergo anionic additions, the alkene must have a functional group X (such as - - C N or O=COR') on the C=C that can stabilize the negative charge: I

Problem 6.64 Alkenes undergo oxidative cleavage with acidic KMnO, and as a result each C of the C=C ends up in a molecule in its highest oxidation state. Give the products resulting from the oxidative cleavage of ( a ) 4 H,C==CHCH,CH,, ( b ) (E)- or (2)-CH,CH=CHCH,, (c) (CH,),C=CHCH,CH,.

CH,CH=CHCH,

(6) CH3 (c)

-

H,C==CHCH,CH,

(4

I

CH,--C=CHCH,CH,CH,

KMn04

KMn04

+ HOOCCH,CH, CH,COOH + HOOCCH, CO,

(H,C== gives CO,) (RCH= gives RCOOH)

CH.3

KMnO.,

I

CH,-C=O

+ HOOCCH,CH,CH,

(R,C= gives R,C=O)

Chapter 7 Alkyl Halides 7.1 INTRODUCTION Alkyl halides have the general formula RX, where R is an alkyl or substituted alkyl group and X is any halogen atom (F, C1, Br, or I). Problem 7.1 Write structural formulas and IUPAC names for all isomers of: ( a ) C,H,,Br. Classify the isomers as to whether they are tertiary (3"), secondary (2"), or primary (1"). (6) C,H,CI,. Classify the isomers which are gem-dichlorides and uic-dichlorides. 4 Take each isomer of the parent hydrocarbon and replace one of each type of equivalent H by X. The correct IUPAC name is written to avoid duplication. ( a ) The parent hydrocarbons are the isomeric pentanes. From pentane, CH3CH,CH,CH,CH3, we get three monobromo products, shown with their classification. BrCH,CH,CH,CH,CH,

CH,CHBrCH,CH,CH,

I-Bromopentane (1")

CH,CH,CHBrCH,CH,

2-Bromopentane (2")

3-Bromopentane (2")

Classification is based on the structural features: RCH,Br is l", R,CHBr is 2", and R,CBr is 3". From isopentane, (CH,),CHCH,CH,, we get four isomers. CH3

CH3

CH3

I

I

I

BrCH,CHCH,CH,

CH,C BrCH,C H,

C H,C HC H BrC H

I-Bromo-2-methylbutane (1")

2-Bromo-2-methylbutane (3")

2-Bromo-3-methylbutane (2")

CH,

I

,

C H ,C HC H,C H,B r 1 -Bromo-3-methyl-

butane ( 1 ")

CH,Br

I

CH,CHCH,CH, is also 1-bromo-2-methylbutane;the two CH,'s on Cz are equivalent. Neopentane has 12 equivalent H's and has only one monobromo substitution product, (CH,),CCH,Br (l"), l-bromo-2,2-dimethylpropane. (b) For the dichlorobutanes the two Cl's are first placed on one C of the straight chain. These are geminal or gem-dichlorides. CH,CCI,CH,CH,

CI,CHCH,CH,CH,

2 ,2-Dichlorobutane

1 .I-Dichlorobutane

Then the Cl's are placed on different C's. The isomers with the Cl's on adjacent C's are vicinal or uic-dichlorides. CI

Cl

I

I

CICH,CHCH,CH,

ClCH ,C H,CHC H,

1.2-Dichlorobutane ( u i c )

1,3-Dichlorobutane

CICH,C H,CH,C H,CI

C H,C HClC HC IC H,

1 ,CDic hloro-

2.3-Dichlorobutane ( u i c )

butane

From isobutane we get CH,

I

C12CH-CH-CH3 1.1 -Dichloro-2-methyl-

propane (gem )

CH3

CH,

I CICH2-CCI-CH3

CICH2-CH-CH2CI

1,2-Dichloro-2-methylpropane (uic )

1,3-Dichloro-2-methylpropane

119

I

ALKYL HALIDES

120

[CHAP. 7

Problem 7.2 Give the structural formula and IUPAC name for ( a ) isobutyl chloride, (b) r-amyl bromide (amylz pentyl). 4 (a)

CH,,

,

CH3

I I

C H-CH2CI

( b ) CH3-C-CH2CH3

CH3

Br

1 -Chloro-2-methylpropane

2-Bromo-2-methylbutane

Problem 7.3 Compare and account for differences in the ( a ) dipole moment, (6) boiling point, (c) density and 4 (d) solubility in water of an alkyl halide RX and its parent alkane RH. ( a ) RX has a larger dipole moment because the C-X bond is polar. (6) RX has a higher boiling point since it has a larger molecular weight and also is more polar. (c) RX is more dense since it has a heavy X atom; the order of decreasing density is RI > RBr > RCl > RF. (d) RX, like RH, is insoluble in H,O, but RX is somewhat more soluble because some H-bonding can occur: 8+

8-

&+

8-

R-X :-- -H-OH

This effect is greatest for RF.

7.2 SYNTHESIS OF RX 1. Halogenation of alkanes with C1, or Br, (Section 4.4). 2. From alcohols (ROH) with HX or PX, (X= I, Br, Cl); SOC1, 3. Addition of HX to alkenes (Section 6.4). 4. X, ( X = Br, C1) + alkenes give vic-dihalides (Section 6.4). 5 . RX + X’ - +RX’ + X- (halogen exchange).

(major method).

Problem 7.4 Give the products of the following reactions:

(a)

+ HI(c) CH3CH,0H + PI,(P + I,)+ (e) H,C=CH, + Br2+ (g) CH,CH,CH,Br + ICH,CH,CH,I + H,O

(c)

CH,CH,I

( a ) CH,CH,CH,OH

-

+ H,PHO, (phosphorous acid)

(e) H,CBrCH,Br

( f ) CH,CH,CI(CH,),

(d)

+ NaBr + H,SO,(CH,),CHCH,OH + SOCI, +

(f)

CH,CH=C(CH,),

(b) n-C,H,OH

+ HI+ 4

+ NaHSO, + H,O (CH,),CHCH,CI + HCl(g) + SO,(g)

(6) n-C,H,Br (d)

( 8 ) CH,CH,CH,I

+ Br-

Problem 7.5 Which of the following chlorides can be made in good yield by light-catalyzed monochlorination of the corresponding hydrocarbon?

To get good yields, all the reactive H’s of the parent hydrocarbon must be equivalent. This is true for

121

ALKYL HALIDES

CHAP. 7)

Although the H's of (f),propene, are not equivalent, the three allylic H's of CH, are much more reactive than the inert vinylic H's on the double-bonded C's. The precursors for (b) and (d), which are CH,CH,CH,CH, and (CH,),CH, respectively, both have more than one type of equivalent H and would give mixtures. Problem 7.6 Prepare

(c)

(a) CH,CHBrCH,

( b ) CH,CH,CH,CH,I

CH3

CH,

I CH,-C-CH, I

I

( d ) CICH,C=C-CH,

CI

CH,CHCH,

(e)

I

I t

CI CI

CH,

from a hydrocarbon or an alcohol. Two ways:

4

CH3CH=CH, + H B r - m l a C H , C H O H C H ,

Two ways: CH,CH,CH=CH,

+ HBr-

peroxide

I-

CH3CH2CH,CH,Br-

acetone

ICH,CH,CH ,CH,

P'3

(P+Iz)

CH,CH,CH,CH,OH

HI does not undergo an anti-Markovnikov radical addition. Two ways: CH3

I

+ HCI

CH,--C=CH,

I

I

Cl

CH3

I CH,-C=C-CH, I

CH,

I

+ CI,

CH, CH,CH=CH,

CICH,-C=C-CH,

1

+ CI, -+

+ HCI

CH3 CH,CHCH,

I t

CI CI

CHEMICAL PROPERTIES Alkyl halides react mainly by heterolysis of the polar C6:X6-

bond.

NUCLEOPHILIC DISPLACEMENT or

"B

+?R X

Nu:/ + Substrate

Nucleophile

-+

--

R:Nu

or

[R:Nu]' Product

+ :x+

Leaving Group

The weaker the Bronsted basicity of X-, the better leaving group is X - and the more reactive is RX. Since the order of basicities of the halide ions X- is I- < Br- < C1- < F-, the order of reactivity of RX is RI > RBr > RCI > RF. The equilibrium of nucleophilic displacements favors the side with the weaker Bronsted base; the stronger Bronsted base displaces the weaker Bronsted base. The rate of the displacement reaction on the C of a given substrate depends on the nucleophilicity of the attacking base. Basicity and nucleophilicity differ as shown:

ALKYL HALIDES

122

basicity BTH&

w *I

B-H

nucleophilicity

[CHAP. 7

+ :x

I

+ :X -

B-C-

I

Problem 7.7 What generalizations about the relationship of basicity and nucleophilicity can be made from the following relative rates of nucleophilic displacements: ( a ) O H - + H H , O and (c)

(b) H,C:- > :OH- > :F:-

NH,+NH,

:I: - > :Br: - >:Cl: - > :F:-

( d ) CH,O- > O H - - > CH,COO-

4

Bases are better nucleophiles than their conjugate acids. In going from left to right in the Periodic Table, basicity and nucleophilicy are directly related-they both decrease. (c) In going down a Group in the Periodic Table they are inversely related, in that nucleophilicity increases and basicity decreases. (d) When the nucleophilic and basic sites are the same atom (here an 0),nucleophilicity parallels basicity. (a)

(b)

The order in Problem 7 . 7 ( c ) may occur because the valence electrons of a larger atom could be more available for bonding with the C, being further away from the nucleus and less firmly held. Alternatively, the greater ease of distortion of the valence shell (induced polarity) makes easier the approach of the larger atom to the C atom. This property is called polarizability. The larger, more polarizable species (e.g. I, Br, S, and P) exhibit enhanced nucleophilicity; they are called soft bases. The smaller, more weakly polarizable bases (e.g. N, 0, and F) have diminished nucleophilicity; they are called hard bases. Problem 7.8 Explain why the order of reactivity of Problem 7.7(c) is observed in nonpolar, weakly polar 4 aprotic, and polar protic solvents, but is reversed in polar aprotic solvents. In nonpolar and weakly polar aprotic solvents, the salts of : N u - are present as ion-pairs (or ion-clusters) in which the nearby cations diminish the reactivity of the anion. Since, with a given cation, ion-pairing is strongest with the smallest ion, F-, and weakest with the largest ion, I-, the reactivity of X- decreases as the size of the anion decreases. In polar protic solvents, hydrogen-bonding, which also lessens the reactivity of X-, is weakest with the largest ion, again making the largest ion more reactive. Polar aprotic solvents solvate only the cations, leaving free, unencumbered anions. The reactivities of all anions are enhanced, but the effect is more pronounced the smaller the anion. Hence, the order of Problem 7 . 7 ( c ) is reversed. Problem 7.9 Write equations for the reaction of RCH,X with (a)

:c

( b ) :OH-

( c ) :OR‘-

( d ) R’-

Y0

( e ) RC,

U)

H3N:

10and classify the functional group in each product.

(6)

+ RCH,X -:OH + RCH,X

(4

-:OR’+ RCH,X

(4

-R’+ RCH,X-

(0)

(4 (f1 (g )

:I-

__*

_I*

-:CN

+ RCH,X

+ :XF

RCH,OH + :X7

+ :XF RCH,NH,’ + :@ RCH,CN + :XF

__*

__*

__*

+

+ :XT

:*

RCH,OR’

RCH,R’

7 0 0 C R ’ + RCH,X :NH3 + RCH,X

RCH,I

RCH,OOCR’

Iodide Alcohol Ether Alkane (coupling) Ester Ammonium salt Nitrile (or Cyanide)

(g)

CN-

CHAP. 71

123

ALKYL HALIDES

Problem 7.10 Compare the effectiveness of acetate (CH,COO- ), phenoxide (C,H,O- ) and benzenesulfonate (C,H,SO,) anions as leaving groups if the acid strengths of their conjugate acids are given by the pK, 4 values 4.5, 10.0, and 2.6, respectively.

The best leaving group is the weakest base, C,H,SOJ; the poorest is C,H50-, which is the strongest base.

Sulfonates are excellent leaving groups-much better than the halides. O n e of the best leaving groups (10' times better than B r - ) is CF,SOi, called triflate. 1.

s N 1 and s N 2 Mechanisms

The two major mechanisms of nucleophilic displacement are outlined in Table 7-1. Table 7-1

TWO: ( 1 ) R : X ~ R +

Steps

+:x-

(2) R + + : N U H ~ ~ S ' - R N U H + = k[RX]

(1st-order)

RNu + :XR:X + :Nu-R:X + :Nu+ RNu':XR:X' + Nu- __* RNu + X: R:X' + Nu+ RNu' + X:

One: or or or

carbocation

I

= k[RX][:Nu-] s-

TS of slow step

I

Stereochemistry ~

Unimolecular

Bimolecular

Inversion and racemization

Inversion (backside attack)

3"> 2" > 1" > CH, Stability of R' RI > RBr > RCI > RF Rate increase in polar solvents

CH, > 1" > 2"> 3" Steric hindrance in R group RI > RBr > RCl > RF With Nu- there is a large rate increase in polar aprotic solvents

~~

Reactivity Structure of R Determining factor Nature of X Solvent effect on fate Effect of nucleophile

R' reacts with nucleophilic solvents rather than with :Nu(solvolysis), except when R' is relatively stable

Cat a1ysis

Lewis acid, e.g. Ag', AlCI,, ZnCI,

Competition, reaction

Elimination, rearrangement

I

Rate depends on nucleophilicity

I - > Br- > Cl-; RS:- > R0:Equilibrium lies towards weaker Bronsted base (1) Aprotic polar solvent (2) Phase-transfer

Problem 7.11 Give the 3 steps for the mechanism of the SN1 hydrolysis of

(I ) Me,C:Br: (3) Me,COH;

1 s-

:Nu---C---X (with :Nu-)

Molecularity ~

*'*.

(2d-order)

--+

Me,C'

+ :Br:-

+ H,O--tMe,COH

(2) Me,C'+ :OH,--tMe,C:OH;

+ H,O+

Elimination, especially with 3" RX in strong Bronsted base 3O

Rx, Me,CBr.

ALKYL HALIDES

124

[CHAP. 7

Problem 7.12 Give examples of the four charge-types of SN2 reactions, as shown in the first line of Table 7-1. 4 CH,CH,Br + :OH- +CH,CH20H + :Bi:

-

+ :NH,-CH,CH,NH; + :Brr (cH,),S:CH, + :OHCH, OH + (cH,), S:

CH,CH,Br +

*.-

-

a sulfonium cation

(CH,),&CH,

(leaving group)

Dimethyl sulfide

+ :N(CH,),+CH,N(CH,)f

+ (CH3)2S:

Problem 7.13 ( a ) Give an orbital representation for an s N 2 reaction with (S)-RCHDX and :Nu-,if in the transition state the C on which displacement occurs uses sp2 hybrid orbitals. ( 6 ) How does this representation explain (i) inversion, (ii) the order of reactivity 3" > 2" > l"? 4 ( a ) See Fig. 7-1. (6) (i) The reaction is initiated by the nucleophile beginning to overlap with the tail of the sp3 hybrid orbital holding X. In order for the tail to become the head, the configuration must change; inversion occurs. (ii) As H's on the attacked C are replaced by R's, the TS becomes more crowded and has a higher enthalpy. With a 3" RX, there is a higher A H S and a lower rate.

R

sp

l*

R

\

hybridized

S Configuration

sp hybridized withp orbital

TS Fig. 7-1

R

'

sp hybridized

R Configuration (Nu and X of same priority)

Problem 7.14 (a) Give a representation of an S,1 TS which assigns a role to the nucleophilic protic solvent molecules (HS:) needed to solvate the ion. (6) In view of this representation, explain why (i) the reaction is first-order; (ii) R' reacts with solvent rather than with stronger nucleophiles that may be present; (iii) catalysis by Ag' takes place; (iv) the more stable the R', the less inversion and the more racemization occurs. 4

Solvent-assisted S, 1 TS

(6) (i) Although the solvent HS: appears in the TS, solvents do not appear in the rate expression. (ii) HS: is already partially bonded, via solvation, with the incipient R'. (iii) Ag' has a stronger affinity for X- than has a solvent molecule; the dissociation of X - is accelerated. (iv) The HS: molecule solvating an unstable R' is more apt to form a bond, causing inversion. When R' is stable, the TS gives an intermediate that reacts with another HS: molecule to give a symmetrically solvated cation,

which collapses to a racemic product:

. .

CHAP. 71

ALKYL HALIDES

125

The more stable is R', the more selective it is and the more it can react with the nucleophilic anion Nu-. Problem 7.15 Give differences between SN1 and sN2 transition states.

4

1. In the SN1 TS there is considerable positive charge on C; there is much weaker bonding between the attacking group and leaving groups with C. There is little or no charge on C in the SN2TS. 2. The SN1 TS is approached by separation of the leaving group; the SN2TS by attack of :Nu- or :Nu. 3. The A H * of the SN1 TS (and the rate of the reaction) depends on the stability of the incipient R'. When R' is more stable, AH' is lower and the rate is greater. The A H * of the SN2 TS depends on the steric effects. When there are more R's on the attacked C or when the attacking :Nu- is bulkier, A H * is greater and the rate is less. Problem 7.16 How can the stability of an intermediate R' in an SN1 reaction be assessed from its enthalpy-reaction diagram? 4

The intermediate R' is a trough between two transition-state peaks. More stable R"s have deeper troughs and differ less in energy from the reactants and products. Problem 7.17 Give the solvolysis products for the reaction of (CH,),CCl with ( a ) CH,OH, (6) CH,COOH. 4

0

II

(6) (CH,),COC-CH, t-Butyl acetate.

( a ) (CH,),COCH, Methyl t-butyl ether.

Problem 7.18 ( a ) Formulate (CH,),COH + HCl+ (CH,),CCl + H,O as an s N 1 reaction. (6) Formulate the reaction CH,OH + HI+ CHJ + H,O as an s N 2 reaction. 4 Step 1

(CH,),COH base

+ HCI e(cH,),cOH,+ fast

acid,

acid I

CI-

base,

oxonium ion

+

Step 2

(CH,),COH,

Step 3

(CH,),C'

SteD 1

CH,OH

-

slow

+ C1-

(CH,),C'

+ H,O

(CH,),CCl

+

+ H I e C H , O H , + Ifast

Problem 7.19 ROH does not react with NaBr, but adding H,SO, forms RBr. Explain.

4 +

Br-, an extremely weak Bronsted base, cannot displace the strong base OH-. In acid, ROH, is first formed. Now, Br- displaces H,O, which is a very weak base and a good leaving group. Problem 7.20 Optically pure (S)-( +)-CH,CHBr-n-C,H,, has [ a ] g= +36.0". A partially racemized sample having a specific rotation of +30" is reacted with dilute NaOH to form (R)-(-)-CH,CH(OH)-n-C,H,, ([a]: = -5.97'), whose specific rotation is -10.3" when optically pure. ( a ) Write an equation for the reaction using projection formulas. (6) Calculate the percent optical purity of reactant and product. ( c ) Calculate percentages of racemization and inversion. ( d ) Calculate percentages of frontside and backside attack. (e) Draw a conclusion concerning the reactions of 2" alkyl halides. (f) What change in conditions would increase inversion? 4

126

ALKYL HALIDES

[CHAP. 7

(6) The percentage of optically active enantiomer (optical purity) is calculated by dividing the observed specific rotation by that of pure enantiomer and multiplying the quotient by 100%. The optical purities are: - 5.97" Bromide = (100%) = 83% Alcohol = -(100%) = 58% + 30" 36" - 10.3" ( c ) The percentage of inversion is calculated by dividing the percentage of optically active alcohol of opposite configuration by that of reacting bromide. The percentage of racemization is the difference between this percentage and 100%. +

58%

Percentage inversion = -(100%) = 70% 83% Percentage racemization = 100% - 70% = 30%

( d ) Inversion involves only backside attack, while racemization results from equal backside and frontside attack. The percentage of backside reaction is the sum of the inversion and one-half of the racemization; the percentage of frontside attack is the remaining half of the percentage of racemization. Percentage backside reaction = 70%

+ $ (30%) = 85%

Percentage frontside reaction = f (30%) = 15% The large percentage of inversion indicates chiefly SN2 reaction, while the smaller percentage of racemization indicates some S,1 pathway. This duality of reaction mechanism is typical of 2" alkyl halides. (f) The S,2 rate is increased by raising the concentration of the nucleophile-in this case, OH . (e)

Problem 7.21 Account for the following stereochemical results:

I 2% inversion Ho-c-c6H5 HBr 98% racemization I +

H

CH3 CH30-c-c6H5

II

+

27% inversion HBr 73% racemization

4

H fit

H,O is more nucleophilic and polar than CH,OH. It is better able to react to give HS:--- R---:SH (see Problem 7.14), leading to racemization. Problem 7.22 NH, reacts with RCH,X to form an ammonium salt, RCHzNH f X - . Show the transition state, 4 indicating the partial charges.

N gains S + as it begins to form a bond. Problem 7.23 Explain why neopentyl chloride, (CH,),CCH,CI, a 1" RCI, does not participate in typical SN2 react ions. 4

The bulky (CH,),C group sterically hinders backside attack by a nucleophile. Problem 7.24 H,C=CHCH,CI is solvolyzed faster than (CH,),CHCI. Explain.

4

127

ALKYL HALIDES

CHAP. 71

Solvolyses go by an SN1 mechanism. Relative rates of different reactants in S,1 reactions depend on the stabilities of intermediate carbocations. H,C=CHCH,CI is more reactive because

[H,C= CH--CH,]+ is more stable than (CH,),CH'. radical.)

See Problem 6.42 for a corresponding explanation of stability of an ally1

Problem 7.25 In terms of ( a ) the inductive effect and (6) steric factors, account for the decreasing stability of R': Compared to H, R has an electron-releasing inductive effect. Replacing H's on the positive C by CH,'s disperses the positive charge and thereby stabilizes R'. Steric acceleration also contributes to this order of R' stability. Some steric strain of the three Me's in Me,C-Br separated by a 109" angle ( s p , ) is relieved upon going to a 120" separation in R' with a C using sp2 hybrid orbitals. Carbocations have been prepared as long-lived species by the reaction RF + SbF, + R' + SbF,. SbF,, a covalent liquid, is called a superacid because it is a stronger Lewis acid than R'. Problem 7.26 The following are relative rates observed for the formation of alcohols from the listed alkyl halides in 80% water and 20% ethanol at 25°C. Account for the minimum value for (CH,),CHBr. i

Compound Relative rate

CH,Br

CH,CH,Br

(CH,),CHBr

(CH,),CBr

2140

171

4.99

1010

4

The first three halides react mainly by the sN2 pathway. Reactivity decreases as CH,'s replace H's on'the attacked C, because of increased steric hindrance. A change to the S,1 mechanism accounts for the sharp rise in the reactivity of (CH3),CBr. Problem 7.27 How does the S,1 mechanism for the hydrolysis of 2-bromo-3-methylbutane, a 2" alkyl bromide, to give exclusively the 3" alcohol 2-methyl-2-butano1, establish a carbocation intermediate? 4 The initial slow step is dissociation to the 2" 1,2-dimethylpropyl cation. A hydride shift yields the more stable 3" carbocation, which reacts with H,O to form the 3" alcohol. , r B r -

+

+

I CH

l 3

Br 2o

I ,

C H , - C H - C - C H , ~C H , - C H - C - C H , I + fast I I k 6H k 3"

Problem 7.28 RBr reacts with AgNO, to give RNO, and RONO. Explain. The nitrite ion,

CH

[":]-

+H

~ O

4

\b:

has two different nucleophilic sites: the N and either 0. Reaction with the unshared pair on N gives RNO,, while RONO is formed by reaction at 0. [Anions with two nucleophilic sites are called ambident anions.]

128

ALKYL HALIDES

[CHAP. 7

2. Role of the Solvent Polar solvents stabilize and lower the enthalpies of charged reactants and charged transition states. The more diffuse the charge on the species, the less effective the stabilization by the polar solvent . Problem 7.29 In terms of transition-state theory, account for the following solvent effects: (a) The rate of solvolysis of a 3" RX increases as the polarity of the protic nucleophilic solvent (:SH) increases, e.g.

H,O > HCOOH > CH,OH > CH,COOH

(6) The rate of the S,2 reaction :Nu- + RX-+RNu + :X- decreases slightly as the polarity of protic solvent increases. (c) The rate of the S,2 reaction :Nu + RX+ RNu' + :X- increases as the polarity of the solvent increases. (d) The rate of reaction in (6) is greatly increased in a polar aprotic solvent. (e) The rate of the reaction in (6) is less in nonpolar solvents than in aprotic polar solvents. 4 See Table 7-2. Table 7-2 Ground State (GS) (a)

6+

I

6-

Nu - - -R- - -X

1

RX + Nu

(c)

None in GS charge in TS

6-

HS---R---X---HS:

RX + HS: RX +Nu-

(b)

Relative Charge

TS

1

6-

Nu- - -R-- -X I

(d)

Sameas (6)

Full in GS diffuse in TS None in GS charge in TS

-~~

I I

1

Effect of Solvent Change

AHt

I I

Rate

I

A lower H of TS A lower H of GS a less lower H of TS

I I

Decreases Increases

I

Increases Decreases

A lower H of TS

Decreases

Increases

A big rise in H of GS* A rise in H of TS

Decreases

Increases

~~

* Aprotic solvents do not solvate anions.

'. ( e ) In nonpolar solvents, N u - is less reactive because it is ion-paired with its countercation, M S,2 reactions with Nu-'s are typical of reactions between water-soluble salts and organic substrates that are soluble only in nonpolar solvents. These incompatible reactants can be made to mix by adding small amounts of phase-transfer catalysts, such as quarternary ammonium salts, Q'A-. Q' has a water-soluble ionic part and nonpolar R groups that tend to be soluble in the nonpolar solvent. Hence, Q' shuttles between the two immiscible solvents while transporting Nu-, as Q'Nu-, into the nonpolar solvent; there Nu- quickly reacts with the organic substrate. 0' then moves back to the water while it transports the leaving group, X-, as Q'X-. Since the positive charge on the N of Q' is surrounded by the R groups, ion-pairing between Nu- and Q' is loose, and N u - is quite free and very reactive. Problem 7.30 Write equations for and explain the use of tetrabutyl ammonium chloride, Bu,N'Cl-, facilitate the reaction between 1-heptyl chloride and cyanide ion.

Bu,N'Cln-C,H,,CI

+ Na'CN--Bu,N'CN-

+ Bu,N'CN-+

n-C,H,,CN

+ Na'C1+ Bu,N'Cl-

to

4

(in water) (in nonpolar phase)

The phase-transfer catalyst, Bu,N'Cl , reacts with CN- to form a quaternary cyanide salt that is slightly soluble in the organic phase because of the bulky, nonpolar, butyl groups. Reaction with CN- to form the nitrile is rapid because it is not solvated or ion-paired in the organic phase; it is a free, strong nucleophile. The

CHAP. 71

129

ALKYL HALIDES

phase-transfer catalyst, regenerated in the organic phase, returns to the aqueous phase, and the chain process is propagated.

ELIMINATION REACTIONS In a 6-elimination (dehydrohalogenation) reaction a halogen and a hydrogen atom are removed from adjacent carbon atoms to form a double bond between the two C's. The reagent commonly used to remove HX is the strong base KOH in ethanol (cf. Section 6.2). H H

H

le H-C-C-H

alc.KOH

____*

I I H X

I

H

H-C-C-H

ethyl halide

I

+ HX (as KX and H 2 0 )

Ethylene

n-Propyl bromide CH,CH,CH,Br-

-

alc. KOH

sec-Butyl chloride CH,CHCICH,CH,

CH,CH=CH,

alc. KOH

Propene

CH,CH=CHCH,

2-Butene (mainly truns)

1. El and E2 Mechanisms The two major mechanisms for p-eliminations involving the removal of an H and an adjacent functional group are the E2 and El. Their features are compared and summarized in Table 7-3. A third mechanism, Elcb,is occasionally observed. Table 7-4 compares E2 and SN2. Problem 7.31 ( a ) Why do alkyl halides rarely undergo the E l reaction? (6) How can the E l mechanism be promoted? 4

Rx reacts by the E l mechanism only when the base is weak and has a very low concentration; as the base gets stronger and more concentrated, the E2 mechanism begins to prevail. On the other hand, if the base is too weak or too dilute, either the R' reacts with the nucleophilic solvent to give the S,1 product or, in nonpolar solvents, RX fails to react. ( b ) By the use of catalysts, such as Ag', which help pull away the leaving group X-. (a)

2.

H-D Isotope Effect

The C-H bond is broken at a faster rate than is the stronger C-D bond. The ratio of the rate constants, k,lk,, measures this H-D isotope effect. The observation of an isotope effect indicates that C-H bond-breaking occurs in the rate-controlling step. Problem 7.32 Why does CH,CH,I undergo loss of HI with strong base faster than CD,CH,I loses DI?

4

These are both E2 reactions in which the C-H or C-D bonds are broken in the rate-controlling step. Therefore, the H-D isotope effect accounts for the faster rate of reaction of CH,CH,I. Problem 7.33 Explain the fact that whereas 2-bromopentane undergoes dehydrohalogenation with C,H,O-K' to give mainly 2-pentene (the Sayzteff product), with Me,CO-K' it gives mainly 1-pentene (the 4 anti-Sayzteff, Hofmanrr, product).

Since Me,CO- is a bulky base, its attack is more sterically hindered at the 2"H than at the 1"H. With Me,CO -,

, ,

CH ,==CHCH,CH CH major product

~

ICY \

Ii'Br /I

indercrl

+-

-~

2

m o r e hrndcrcd

0

CH,CH--LHCH,CH, ~

minor product

ALKYL HALIDES

130

[CHAP. 7

Table 7-3 El Steps

Two: (1)

H-&-&--L

I I

E2

H-L-&+

I

I

+ L-

R' intermediate

6-

Transition states

Indicates El Kinetics

Indicates E2

First-order Rate = k[RL] Ionization determines rate Unimolecular

Second-order Rate = k[RLJ[:B-] Bimolecular

Stereochemistry

Nonstereospecific

anti Elimination (syn when anti impossible)

Reactivity order factor

3" > 2" > 1" RX Stability of R'

3" > 2" > 1" RX Stability of alkenes (Saytzeff rule)

Re arrangements

Common

None

Deuterium isotope effect

None

0bse rved

Regioselect ivi t y

Saytzeff

SN2 (See Problem 7.33)

Alkyl group

3" > 2" > 1"

3" > 2" > 1"

Loss of H

No effect

Increased acidity

Base Strength Concentration

Weak Low

Strong High

Leaving group

Weak base I - > Br- > C1- > F-

Weak base I - > Br- > C1- > F-

~~~

~

~~

~

~~

Competing reaction Favors E l

Solvent

Favors E2

~~

Catalysis

~

~

~

Phase-transfer Polar protic

Polar aprotic

Problem 7.34 Assuming that anti elimination is favored, illustrate the stereospecificity of the E2 dehydrohalogenation by predicting the products formed from (a) meso- and ( b ) either of the enantiomers of 4 2,3-dibromobutane. Use the wedge-sawhorse and Newman projections.

ALKYL HALIDES

CHAP. 7)

131

Table 7-4

Structure of R Reagent

Favors E2

Favors

3" > 2" > 1"

1" > 2" > 3"

Strong bulky Bronsted base (e.g., Me,CO-)

sN2

Strong nucleophile

Temperature

High

LOW

Low-polarity solvent

Yes

No

I > Br > C1> F

I > Br > C l > F

Favors E2

Favors SN1

3" > 2" > 1"

3" % 2" > 1"

Structure of L

Structure of R Base Strength Concentration

I

Structure of L

&

Strong High

Very weak Low

I

I>Br>Cl%F

I>Br>Cl%F

cis - or (E)-2-Bromo-2-butene

H

mesa

__* anti

--HRr

*FMe Br

111

111

Br

Me

Me

H enantiorner (RR)

tram- or (2)-2-Brorno-2-butene

Problem 7.35 Dehalogenation of vic-dihalides with active metals (Mg or Zn) is also an anti elimination. Predict the products from ( a ) meso- and (6) either enantiomer of 2,3-bromobutane. 4

[CHAP. 7

ALKYL HALIDES

132

HxMe

unri eliminnkion

Mf3 +

+

_____)

Me

tram-2-Butene

meso

< 1 'H

elimination

aH

Me.

unki

Mg + Me&H

H

+

~

Me'

enantiomer (SS)

~

MgBr,

'H

ck-2-Butene

Problem 7.36 Account for the percentages of the products, (i) (CH,),CHOC,H, and (ii) CH,CH=CH,, of the reaction of CH,CHBrCH, with 4 (b) C2H,0H+3% (ii) +97% (i) ( a ) C,H,ONa/C,H,OH-,79% (ii) + 21% (i) (a) C,H,O- is a strong base and E2 predominates. (6) C2H,0H is weakly basic but nucleophilic, and SN1 is favored.

Problem 7.37 Account for the following observations: (a) In a polar solvent such as water the SN1 and E l reactions of a 3" RX have the same rate. (b) (CH,),CI+H,O-,(CH,),COH+HI but (CH,),CI+ 4 OH- --.* ( C H , ) , M H , + H,O + I-. (a) The rate-controlling step for both E l and S,1 reactions is the same:

and therefore the rates are the same. (6) In a nucleophilic solvent in the absence of a strong base, a 3" RX undergoes an S,1 solvolysis. In the presence of a strong base (OH-) a 3" RX undergoes mainly E2 elimination.

7.4

SUMMARY OF ALKYL HALIDE CHEMISTRY PREPA RA TION

PROPERTIES

1. Direct Halogenation

1.

/,+ 2.

Replacement of OH in ROH

RCH2CH20H + NaX +

+ H2S04

/+ soc12------+ RCH2CH2X A

-

Nucleophilic Substitution + OH- +RCHzCHzOH Alcohol SHRCHzCH2SH Mercaptan + CN- +RCH2CH2CN Nitrile + OR- --+ RCHzCHzOR Ether + SR- -+ RCH2CH2SR +Thioether + NH3 --+ RCH2CH2NH3X- Amine salt + NRJ--+ RCH,CH,NR;'X- Quarternary ammonium salt ( Q 'A- )

R'COO-

+ px,

+ LiAIH4 3. Addition of HX to Alkene

RCH=CH2

+ HX

RCH=CH2

+ HBr

-

__*

-

RCH2CH200CR' Ester

RCH2CH3

H+

__*

peroxide

RCH2CH2Br

4.

Formation of Organometallics

+ Mg-

RCH,CH,MgX Grignard reagent Ornanolithium reagent

+ Li --+RCH,CH,Li

133

ALKYL HALIDES

CHAP. 71

Supplementary Problems Problem 7.38

Write the structure of the only tertiary halide having the formula C,H,,Br.

rl

In order for the halide to be tertiary, Br must be attached to a C that is attached to three other C’s; i.e., to no H’s. This gives the skeletal arrangement C

I I

C-C-C Br involving four C’s. A fifth C must be added, which must be attached to one of the C’s on the central C, giving: C

I I

C-C-C-C

Br and with the H’s. CH3

I I

CH,-C-CH,-CH Br

2-Bromo-2-methylbutane

Problem 7.39 On substitution of one H by a C1 in the isomers of C5HI2,( a ) which isomer gives only a primary 4 halide? (b) Which isomers give secondary halides? (c) Which isomer gives a tertiary halide? ( a ) 2,2-Dimethylpropane. (b) CH,CH,CH,CH,CH, gives 2-chloropentane and 3-chloropentane. 2-Methylbutane gives 2-chloro-3-methylbutane. (c) 2-Methylbutane gives 2-chloro-2-methylbutane.

Problem 7.40 Complete the following table: Substance

1. 2.

CH,CH,

Treated with

Yields

(4

CH,CH,Cl CH, HCH,

SOCI,

(6)

’iCI

CH,CHCH,Br

I

Br 4.

HBr, peroxides

CH,CH==CH,

( a ) Cl,, UV

(b) CH,CHCH,

(c)

CH,CH==CH,

(4

4

(d) CH,CH,CH,Br

AH

--

Problem 7.41 Give the organic product in the following substitution reactions. The solvent is given above the arrow. (a) HS- + CH,CH,CHBrCH,

+ (CH,),CBr HCOOH CH,CH,Br + AgCN CH,CHBrCH, + CH,NH, CH,CHBrCH, + (CH3)2S:

(6) I(c)

(d) (e)

CH OH

2 products

-

ALKYL HALIDES

[ j: 1'-

CH,CH,Br

+ :P(C,H,),

CH,CH,Br

+

(thiosuffate ion)

S-S-0

CH ,CH ,C H SHCH

,

[CHAP. 7

-

(a mercaptan).

0

II

(CH,),COCH; 3" RX undergoes S, 1 solvolysis. CH,CH,CN

i

+ CH,CH,NC;

[I+

(CH,),CHNCH,

: m N :- is an ambident anion (Problem 7.28).

Br-; an ammonium salt.

[(CH,)2CHS(CH,)2]+ Br-; a sulfonium salt. Br-; a phosphonium salt. [CH3CH2P(C,H5),]+ 0

I I

CH,CH,-S-S-0-;

S is a more nucleophilic site than is 0.

0 Problem 7.42 Account for the observation that catalytic amounts of KI enhance the rate of reaction of 4 RCH,CI with OH- to give the alcohol RCH,OH. Since I - is a better nucleophile than OH-, it reacts rapidly with RCH,CI to give RCH,I. But since I - is a much better leaving group than Cl-, RCHzI reacts faster with OH- than does RCH,CI. Only a catalytic amount of I - is needed, because the regenerated I - is recycled in the reaction. RCHzCl + OH-i

slower

RCH,OH

.p

+ C1'

Problem 7.43 Account for the following products from reaction of CH,CHOHC(CH,), with HBr: (a) CH,CHBrC( CH,), , (6) H,CF---CHC(CH, ), , ( c ) (CH,),CHCHBr(CH,), , ( d ) (CH,),CH--LH( CH,), , ( e ) (CH,),CHC=CH, 4

b3

For this 2" alcohol, oxonium-ion formation is followed by loss of H,O to give a 2" carbocation that rearranges to a 3" carbocation. Both carbocations react by two pathways: they form bonds to Br- to give alkyl bromides or they lose H' to yield alkenes. CH,CHOHC(CH,),

HBr

-HO CH,CHC(CH,), --LCH,CHC(CH,), I 2" carbocation OH-. :CH, +

J-

(CH, 1, C H b h 1 2

CH,CHBrC(CH,), -H

(4

1 H2C=CHC(CH,)2 (6)

CHAP, 71

ALKYL HALIDES

-

135

Problem 7.44 Show steps for the following conversions: (a)

BrCH,CH,CH,CH,

(6) CH,CHBrCH,CH,

CH,CHBrCH,CH,

+CH,CICHClCH,CI

(c) CH,CH,CH,

-

BrCH,CH,CH,CH, 4

To do syntheses, it is best to work backward while keeping in mind your starting material. As you do this, keepasking what is needed to make what you want. Always try to use the fewest steps. ( a ) The precursors of the possible product are the corresponding alcohol, 1-butene, and 2-butene. The alcohol is a poor choice, because it would have to be made from either of the alkenes and an extra step would be needed. 2-Butene cannot be made directly from the 1" halide, but 1-butene can.

alc. K O H

BrCH,CH,CH,CH,

CH,=CHCH,CH,

CH,CHBrCH,CH,

-

(6) To ensure getting 1-butene, the needed precursor, use a bulky base for the dehydrohalogenation of the starting material. CH,CHBrCH,CH,

Me

CO-K+

HBr

CH,==CHCH,CH,

peroxides

BrCH,CH,CH,CH,

(c) The precursor for vic dichlorides is the corresponding alkene; in this case, H,C=CHCH,CI, which is made by allylic chlorination of propene. Although free-radical chlorination of propane gives a mixture of isomeric propyl chorides, the mixture can be dehydrohalogenated to the same alkene, making this particular initial chlorination a useful reaction.

C H , C H , C H , G CICH,CH,CH,

+

CH,CHCICH,

alc.

CH,=CHCH,

2CH,=CHCH,CI-% uv

ClCH,CHClCH,CI

Problem 7.45 Indicate the products of the following reactions and point out the mechanism as SN1, SN2,El

or E2.

( a ) CH,CH,CH,Br + LiAIH, (source of :H-) (6) (CH,),CBr + C,H,OH, heat at 60°C ( c ) CH,CH==CHCl+ NaNH,

(d) BrCH,CH,Br + Mg (ether) (e) BrCH,CH,CH,Br + Mg (ether) ( f ) CH,CHBrCH, + NaOCH, in CH,OH

( a ) CH,CH,CH,; an SN2reaction, :H- of AlHa replaces Br-.

(6)

(4 (4

+ (CH3),C+]*

CH CH OH

(CHJ,COCH,CH,

3"RX

major; S,1

CH,CH=CHCl + NaNH,CH,C%CH Vinyl halides are quite inert toward sN2 reactions. BrCH,CH,Br

+ Mg-

very minor; El

+ NH, + NaCl

H,C=CH,

(E2)

+ MgBr,

This is an E2 type of p-elimination via an alkyl magnesium iodide. Mg + BrCH,CH,BrBrMg-:CH,-CH,-Br+ + Q P MgBr, (e)

CH3

+ CH,C=CH, I

+ H,C=CH,

This reaction resembles that in ( d ) and is an internal S,2 reaction.

( f ) This 2" RBr undergoes both E2 and S,2 reactions to form propylene and isopropyl methyl ether.

CH,CHBrCH,

+ &aOCH,(CH,OH)-

CH,CH=CH,

+ CH,CHCH, I

OCH,

4

136

ALKYL HALIDES

[CHAP. 7

Problem 7.46 Give structures of the organic products of the following reactions and account for their formation: C H OH

CICH,CH2CH,CH,Bra,S;t~~CN(6) CH,CHBrCH, + NaI- a c e t o n e (c) CICH,CH=CH, + NaI(a)

4

( a ) CICH,CH,CH,CH,CN.

Br- is a better leaving group than Cl-. (6) CH,CHICH,. Equilibrium is shifted to the right because NaI is soluble in acetone, while NaBr is not and precipitates. (c) ICH,CH==CH,. NaI is soluble in acetone and NaCl is insoluble.

Problem 7.47 As a rule 1" RX is the least reactive in S,1 solvolysis reactions. Account for the high reactivity 4 observed in S, 1 displacements with CICH,OCH,CH, in ethanol.

The rapid S,l reaction is attributed to the stability of a C' adjacent to -&-. The empty p orbital on the C' can overlap with the p orbital on 0, thereby delocalizing the + charge. The positive ion then reacts with the nucleophilic alcohol, giving an ether. CH3CH2-O-CH2CI

+Cl-

+

CH3CH2-0-cH,'

$

+C2H50H

+

CH3CH,-O=CH2

-H+ CH3CH2-O-CH20C2H,

Problem 7.48 Account for the following observations when (S)-CH,CH,CH,CHID is heated in acetone solution with Naf: (a) The enantiomer is racemized. (6) If radioactive *I- is present in excess, the rate of racemization is twice the rate at which the radioactive * I - is incorporated into the compound. 4 (a) Since enantiomers have identical energy, reaction proceeds in both directions until a racemic equilibrium mixture is formed.

D

*I-

+ CH3CH2CH,+I

inversion inversion

*

D I+CH,CH,CH,

H

H

(R)

6)

+ I-

( s N 2 reaction)

(6) Each radioactive *I- incorporated into the compound forms one molecule of enantiomer. Now one unreacted molecule and one molecule of its enantiomer, resulting from reaction with *I-, form a racemic modification. Since two molecules are racemized when one *I- reacts, the rate of racemization will be twice that at which *I- reacts. Problem 7.49 Indicate the effect on the rate of S,1 and s N 2 reactions of the following: ( a ) Doubling the concentration of substrate (RL) or Nu-. (6) Using a mixture of ethanol and H,O or only acetone as solvent. (c) Increasing the number of R groups on the C bonded to the leaving group, L. (d) Using a strong Nu-. 4

Doubling either [RL] or [Nu-] doubles the rate of the s N 2 reaction. For S,1 reactions the rate is doubled only by doubling [RL] and is not affected by any change in [Nu-]. (6) A mixture of ethanol and H,O has a high dielectric constant and therefore enhances the rate of S,1 reactions. This usually has little effect on S,2 reactions. Acetone has a low dielectric constant and is aprotic and favors S,2 reactions. (c) Increasing the number of R's on the reaction site enhances s N 1 reactivity through electron release and stabilization of R'. The effect is opposite in S,2 reactions because bulky R's sterically hinder formation of, and raise AHt for, the transition state. (d) Strong nucleophiles favor S,2 reactions and do not affect S,1 reactions. (a)

Problem 7.50 List the following alkyl bromides in order of decreasing reactivity in the indicated reactions,

ALKYL HALIDES

CHAP. 71

137

CH3

I

CH,CH2CH2CH2CH2Br

CH3C-CH,CH,

8,

CH,CH2CHCH2CH,

8,

s N 1 reactivity, ( b ) s N 2reactivity, (c) reactivity with alcoholic AgNO,.

4

Reactivity for the S, 1 mechanism is 3"(I) > 2"(III) > l"(11). The reverse reactivity order for SN2 reactions gives 1"(11) > 2"(111) > 3"(I). Ag' catalyzes S, 1 reactions and the reactivities are 3"(I) > 2"( 111) > 1"( 11). Problem 7.51 Potassium rert-butoxide, K'OCMe,, is used as a base in E2 reactions. (a) How does it compare in effectiveness with ethylamine, CH,CH,NH,? (6) Compare its effectiveness in the solvents tert-butyl alcohol and dimethylsulfoxide (DMSO) (c) Give the major alkene product when it reacts with (CH,),CClCH,CH,. 4

( a ) K'OCMe, is more effective because it is more basic. Its larger size also precludes SN2 reactions. (6) Its reactivity is greater in aprotic DMSO because its basic anion is not solvated. Me,COH reduces the effectiveness of Me,CO- by H-bonding. (c) Me,CO- is a bulky base and gives the anti-Saytzeff (Hofmann) product CH2=C(CH,)CH2CH,.

Problem 7.52 Indicate and explain why the following may or may not be used as synthetic reactions.

(4

H 2 0 + CH,CHCICH,

CH,CH=CH,

4

(a) A 3"RX does not undergo S,2 displacement. Instead, strong bases induce E2 elimination to give CH,==C(CH,),. (6) H,O is too weakly basic to effect E2 elimination. The E l reaction does not occur because AH' for the transition state leading to an incipient R,CH+ is very large and the rate is extremely low. Problem 7.53 Give structures of all alkenes formed and underline the major product expected from E2 elimination of: (a) 1-chloropentane, (6) 2-chloropentane. 4 (4

CH,CH,CH,CH,CH,Cl+

CH,CH,CH,CH==CH,

A 1" alkyl halide, therefore one alkene.

(6)

CH,CH2CH2CH-CH3

I c1

CH3CH2CH,CH=CH2

+ CH3CH,CH=CHCH3

A 2" alkyl halide flanked by two R's; therefore two alkenes are formed. The more substituted alkene is the major product because of its greater stability. Problem 7.54 How is conformational analysis used to explain the 6 : 1 ratio of trans- to cis-2-butene formed on dehydrochlorination of 2-chlorobutane? 4 For either enantiomer there are two conformers in which the H and Cl eliminated are anti to each other.

ALKYL HALIDES

138

[CHAP. 7

-H$Hl' $2:; H

H

trans-2-Butene

4

H

H

CHI

a

L B:---H

H

H...

--I

cis-2-Butene

Conformer I has a less crowded, lower-enthalpy transition state than conformer 11. Its A H S is less and reaction rate greater; this accounts for the greater amount of trans isomer obtained from conformer I and the smaller amount of cis isomer from conformer 11. Problem 7.55 State whether each of the following R"s is stabilized or destabilized by the attached atom or group: +

( 4 F3C-C-

(b)

I

:F3C+

H2N 3" > 2" > 1". (b) The 1,4-orientation is similar to ionic addition because of the relative stabilities of the two products.

8.6 POLYMERIZATION OF DIENES

ELECTROPHILIC CATALYSIS

I

2

3

4

5

6

?

ECH,-CH=CHCH2CH2CH=CH:H2

+

MCHFCHCHPCH~)

(E[-CH2CH=CHCHt-],+2)+ mer of polymer

156

ALKYNES AND DIENES

[CHAP. 8

NUCLEOPHILIC OR ANIONIC POLYMERIZATION

CH3 CH3 I I Nu:CH2C=CH-CH2:CH2-C=CH-CH2

CH3

n(CHrL--CH-CH~ n +2

dirneric anion

mer of polymer

The reaction is stereospecific in yielding a polymer with an all-cis configuration. Conjugated dienes undergo nucleophilic attack more easily than simple alkenes because they form more stable allyl carbanions,

Like the allyl cation, the allylic anion is stabilized by charge delocalization through extended bonding.

7~

RADICAL POLYMERIZATION

monorner

dimeric free radical

monomeric free radical

Conjugated diene polymers are modified and improved by copolymerizing them with other unsaturated compounds, such as acrylonitrile, H,C=CH-C%N.

8.7 CYCLOADDITION A useful synthetic reaction is cycloaddition of an alkene, called a dienophile, to a conjugated diene by 1,4-addition. //CH2 CH

I

CH \CH2 1,3-Butadiene

+

CH2

II

CH2 Ethylene

F H 2 \

2 0 0 ~

___,

CH

II

C?

CH,

I

,CH2 CH* Cyclohexene

Diels-Alder reaction

CHAP. 81

157

ALKYNES AND DIENES

8.8 SUMMARY OF ALKYNE CHEMISTRY

PREPARA TION

PROPERTIES

1. Industrial

1. Addition Reactions

2CH4(- 3H2)-

0 2

+ Cu(NH,),Cl--+ H,C=CH--C-=CH

H-H

H,/Pt(Pb)d RCHSH,

/+

+ Na, NH,-RCH=CH, + HX- RCX=CH,-

--

+HX

(~yn) (anti) RCX,CH,

+ HOH(Hg++,H’)+

2. Laboratory

R-CO-CH, RCX=CHX++xz RCX,--CHX,

Triple-Bond Formation

+ R;BH

De hydrohalogen at ion RCHXCH,X or RCH,CHX,

RCHdHBR;

Alkylation of Acetylene

,

HC-=CH + NaNH, --+H-Na + RMgX- HC-3CMgX H-H

RCH CH=O

-

2. Replacement of Acidic Hydrogen -

+ NaNH,+ Ag(NH,): + R’MgX+

R m N a R m A g R m M g X + R’H

8.9 SUMMARY OF DIENE CHEMISTRY

PREPARA TION

PROPERTIES

1. Dehydration of Diols

1. Hydrogen Addition

HOCH2--CH2--CH2--CH2-OH 2.

\

Dehydrogenation (industrial)

r+ 2.

Alkanes: CH,CH,CH,CH, - 2H2 Alkenes: H,C=CHCH,CH, - H, H,C=CHCH=CH,

3. Dehydrohalogenation

,

Dihalides: CH CHXCH CH ,X Ally1 halides: H,C=CHCHXCH,

2H2Butane Polar Additions + CI, -+ 1,2- + 1,4-Dichlorobutenes + 2C1, 1,2,3,4-Tetrachlorobutane + HI +3-Iodo- + 1-Iodobutenes

/

-

+ --(CH2CH4H - O H > - O C O C H , . The -0-,with a full negative charge, is best able to donate electrons, thereby giving the very stable uncharged intermediate

s+

s-

In -OCOCH, the C of the C - 0 group has + charge and makes demands on the for electron density, thereby diminishing the ability of this U to donate electrons to the benzenonium ion. (b) The order is -NH, >-NHCOCH, for the same reason that OH is a better activator than --OCOCH,.

Table 11-2 extends the results of Problem 11.10. Table 11-2

rn-Directors

op-Directors weakly deactivating

activating

-F:

-R

C6Hs-

I

-C=L-

--cl:

--er:

-I:

deactivating

..

-N=o:

..

-NRf -NO2

-6.5 "C.

SYNTHESES Friedel-Crafts alkylations and acylations, followed by reduction of the c=O group, are most frequently used to synthesize arenes. Coupling reactions can also be employed. Ar,CuLi

+ RX-

ArR

1"

R,CuLi

or

+ ArX-

ArR

Problem 11.33 Explain the following observations about the Friedel- Crafts alkylation reaction. (a) In monoalkylating C,H6 with RX in AIX, an excess of C6H6is used. ( b ) The alkylation of PhOH and PhNH, gives poor yields. (c) Ph-Ph cannot be prepared by the reaction AICll

PhH + PhCl+> (d) At 0°C

PhH + 3CH,CI

AIC13

+ HCl

Ph-Ph

1,2,4-trimethyIbenzene

but at 100 "C one gets 1,3,5-trimethyIbenzene (mesitylene). (e) The reaction PhH

+ CH,CH,CH,Cl-

AICIl

PhCH,CH2CH,

+ HCI

gives poor yield, whereas PhH

+CH,CHClCH,z

PhCH(CH,),

+ HCl

gives very good yield.

4

( a ) The monoalkylated product, C6H,R, which is more reactive than C6H6 itself since R is an activating group, will react to give C6H,R, and some C6H,R3. To prevent polyalkylation an excess of C,H, is used to increase the chance for collision between R' and C6H6and to minimize collision between between R' and C6H,R. ( b ) OH and NH, groups react with and inactivate the catalyst. PhCl + AICI, +-B

(4

Ph +

+ AlCl;

Ph' has a very high enthalpy and doesn't form. (d) The alkylation reaction is reversible and therefore gives the kinetic-controlled product at 0°C and the thermodynamic-controlled product at 100 "C. (e) The R' intermediates, especially the 1" RCH; can undergo rearrangements. With CH,CH,CH,CI we get

CH,CH,CH;

--H\ CH,&HCH,

and the major product is PhCH(CH,),.

-

-

Problem 11.34 Prepare PhCH,CH,CH, from PhH and any open-chain compound.

-

PhH + CICH,CH==CH, or

PhH + CICOCH,CH,

AICI3

AICI3

H21 PI

PhCH,CH,CH,

Z n l H g . HCI

PhCH,CH,CH,

PhCH,CH=CH,

PhCOCH,CH,

4

(Clemmen5en reduction)

CH,CH,CH,Cl cannot be used because the intermediate 1" CH,CH,CHi rearranges to the more stable 2" (CH,),CH+, to give C,H,CH(CH,), as the major product. The latter method is for synthesizing PhCH2R.

230

AROMATIC SUBSTITUTION. ARENES

[CHAP. 11

Problem 11.35 Give the structural formula and the name for the major alkylation product: ( a ) C,H,

+ (CH,),CHCH,Cl-

( c ) C,H,

+ CH,CH,CH,CH,Cl=

AIC13

+ (CH,),CCH,OH-

(b) C,H,CH,

AIClj

(d) m-xylene + (CH,),CCl=

CH3

BF,

AIClj

4

CH3

I

I

3 CH,:CH,

CH, CH, x6H5

lsobutyl cation (1")

lsobutyl chloride

Neopentyl alcohol

Neopentyl cation (1")

1

rCH,CH,CH,CH;] ( c ) CH,CH,CH,CH,CI

-%

1-

H'

CH3CH,yHCH,

rerr-Butyl cation (3")

1

rerr- Pe nt y1 cation (3")

p-tert-Pentyltoluene

% 34% PhCH,CH,CH,CH, + 66% C,H,-CHCH,CH,

I

n-Butylbenzene

n-Butyl chloride

H 3c

fert-Butylbenzene

CH3

s-Butylbenzene

Thermodynamic product; it has less steric strain and is more stable than the kinetic-controlled isomer having a bulky t-butyl group orfho to a CH,.

Chemistry; Reactivity of the Benzylic (Ph The chemistries of the benzylic and allylic positions are very similar. Intermediate carbocations, free radicals and carbanions formed at these positions are stabilized by delocalization with the adjacent T system, the benzene ring in the case of the benzylic position. Another aspect of arene chemistry is the enhanced stability of unsaturated arenes having double bonds conjugated with the benzene ring. This property is akin to the stability of conjugated di- and polyenes. Problem 11.36 PhCH, reacts with Br, and Fe to give a mixture of three monobromo products. With Br, in light, only one compound, a fourth monobromo isomer, is isolated. What are the four products? Explain the 4 formation of the light-catalyzed product. With Fe, the products are 0-,p-, and some rn-BrC,H,CH,. In light the product is benzyl bromide, PhCH,Br. Like allylic halogenation (Section 6 . 5 ) , the latter reaction is a free-radical substitution: (1) Br2;2Br.

(2) Bra

(3) PhCH,

+ Br2-

+ PhCH, --+PhCH, + HBr PhCH,Br

+ Br.

Steps (2) and (3) are the propagating steps. Problem 11.37 Account for the following: PhCH,CH(CH,), with little or no PhCH,CBr(CH,), formed.

+ Br2&

PhCHBrCH(CH,), 4

CHAP. 111

231

AROMATIC SUBSTITUTION. ARENES

The H’s on the C attached to the ring (the benzylic H’s), although they are in this case 2”, are nevertheless more reactive toward Br- than are ordinary 3” H’s. Like a C=C group in the allylic system, the Ph group can stabilize the free radical by electron donation through extended p orbital overlap.

Problem 11.38 Irradiation of an equimolar mixture of cyclohexane, toluene and Br, in CCI, gives almost 4 exclusively benzyl bromide. A similar reaction with C1, gives mainly cyclohexyl chloride. Explain.

This is a competitive reaction that compares the reactivities of toluene and cyclohexane. Br. is less reactive and more selective than Cl., and reactivity differences of the respective H’s determine the product (benzyl > 2”). With the more reactive and less selective Cl., the statistical advantage of cyclohexane (12H’s) over toluene (three H’s) controls the product ratio. Problem 11.39 Which is more reactive to radical halogenation, PhCH, or p-xylene? Explain.

4

p-Xylene reactivity depends on the rate of formation of the benzyl-type radical. Electron-releasing groups such as CH, stabilize the transition state, producing the benzyl radical on the other CH, and thereby lowering the AH’ and increasing the reaction rate. Problem 11.40 Outline a synthesis of 2,3-dimethyl-2,3-diphenylbutane from benzene, propylene, and any needed inorganic reagents. 4

The symmetry of this hydrocarbon makes possible a self-coupling reaction with 2-bromo-2-phenylpropane. C6HS

C6HS

C6H6 + CH,CH=CH,

Br

I.

Li

CH3-L-L--CH, LH3

-

LH,

Problem 11.41 Give all possible products of the following reactions and underline the major product. ( a ) PhCH,CHOHCH(CH,),

(c) PhCH=CHCH,

+ HBr+

“ZS04

(e) P h C H - L H C H d H ,

(6) PhCH,CHBrCH(CI-I,),; (d) PhCH=CHCH,

+ Br,

(equimolar amounts)-

+ HBr-

alc

peroxide

4

PhCH,CH=C(CH,), + PhCH=CHCH(CH,),. The major product has the C=C conjugated with the benzene ring and therefore, even though it is a disubstituted alkene, it is more stable than the minor product, which is a trisubstituted nonconjugated alkene. Same as part ( a ) and for the same reason. PhCH,CHBrCH, + PhCHBrCH,CH,. H’ adds to C = C to give the more stable tenzyl-type PhCHCH,CH,. Reaction with Br- gives the major product. The benzyl-type cation PhCHR (like CH,=CHCH,’) can be stabilized by delocalizing the + to the op-positions of the ring:

PhCH,CHBrCH, + PhCHBrCH,CH,. Br. adds to give the more stable benzyl-type PhCHCHBrCH, rather than PhCHBrCHCH,. We have already discussed the stability of benzylic free radicals (Problem 11.37).

232

AROMATIC SUBSTITUTION. ARENES

[CHAP. 11

( e ) PhCHBrCHBrCH=CH, + PhCHSHCHBrCHBr + PhCHBrCH=CHCH,Br. The major product is the conjugated alkene, which is more stable than the other two products [see part ( a ) ] .

Problem 11.42 Explain the following observations. ( a ) A yellow color is obtained when Ph,COH (trityl alcohol) is reacted with concentrated H,SO,, or when Ph,CCI is treated with AICI,. On adding H,O, the color disappears and a white solid is formed. (6) Ph,CCI is prepared by the Friedel-Crafts reaction of benzene and CCI,. It does not react with more benzene to form Ph,C. (c) A deep-red solution appears when Ph,CH is added to a solution of NaNH, in liquid NH,. The color disappears on adding water. (d) A red color appears when Ph,CCI reacts with Zn in C,H,. 0, decolorizes the solution. 4

The yellow color is attributed to the stable Ph,Cf, whose rings. Ph,COH

+ H2S0,+Ph,Cf

is delocalized to the oppositions of the 3

+ H,O+ + HSO; + AlCI;

Ph,CCl

+ AlCI,+

Ph,C'

Ph,Cf

+ 2H,O--

Ph,COH

kwis

Lewis base

white solid

acid

+

+ H,O+

With AICI,, Ph,CCI forms a salt, Ph,C'AICl;, whose carbocation is too stable to react with benzene. Ph,C' may also be too sterically hindered to react further. The strong base :NHZ removes H' from Ph,CH to form the stable, deep red-purple carbanion Ph,CT, which is then decolorized on accepting H' from the feeble acid H,O.

n + :NH,

Ph,CH acid

,

Ph,Ci base,

**

__*

basez

+ H,Oacid,

H:NH, acid,

Ph,CH acid,

+ Ph,Ci base, (deep red)

+ OHbase,

The Ph,Ci is stabilized because the - can be delocalized to the oppositions of the three rings (as in the corresponding carbocation and free radicals). C1. is removed from Ph,CCI by Zn to give the colored radical Ph,C-, which decolorizes as it forms the peroxide in the presence of 0,. 2Ph,CCI

+ Zn-

2Ph,C- + ZnCl,

2Ph,C* + .c)-o*+ Ph,C:O:O:CPh, Problem 11.43 In the following hydrocarbons the alkyl H's are designated by the Greek letters a,p, y , etc. Assign each letter an Arabic number, beginning with 1 for LEAST, in order of increasing ease of abstraction by a Br..

4

See Table 1 1-3.

AROMATIC SUBSTITUTION. ARENES

CHAP. 111

233

Table 11-3

B

a

I

I

6

Y

€

(a)

1 (1")

2 (2")

4 (2", benzylic)

5 (2", dibenzylic)

3 ( I", benzylic)

(W

2 (1")

3 (2")

4 (allylic)

1 (vinylic)

5 (allylic, benzylic)

(c)

3 (OP to other CH,'s)

1 ( m o to two other CH,'s; more hindered)

2 (less sterically hindered than p ; mp to two

Problem 11.44 Use + and - signs for positive and negative tests in tabulating rapid chemical reactions that ;an be used to distinguish among the following compounds: ( a ) chlorobenzene, benzyl chloride and cyclohexyl 4 ;hloride; ( b ) ethylbenzene, styrene, and phenylacetylene. See Tables ll-4(a) and 11-4(b).

Table 11-4(a) Chlorobenzene

Benzyl chloride

Cyclohexyl chloride

Ring sulfonation is exothermic

+

+

-

Alc. AgNO, (forms AgCI, a white precipitate)

-

+ (very fast)*

+ (much slower)*

Reaction

+

*Ag' induces an S, 1 reaction; PhCH, > C,H:, .

Table ll-4(6) Reactions

Ethylbenzene

Styrene

Phenylacetylene

Br, in CCI, (is decolorized)

-

+

+

Ag(NH,),' (forms a precipitate)

-

-

+

234

[CHAP. 11

AROMATIC SUBSTITUTION. ARENES

11.5 SUMMARY OF ARENE AND ARYL HALIDE CHEMISTRY PREPARATlOU A.

PROPERTIES

ALKYLBENZENES 1. Oxidation

1. Direct Alkylation

ArH + CH,CH ArH + H,C=C ArH + CH,CH

+ CrO,CI, + KMnO, 2.

2. Coupling

ArCHC , H-,--

Ar,CuLi + RX (1") R,CuLi + ArX .-c-cC

fl -

ArA-CH, Ar-COOH

heat

Reduction +

3H2w

3. Side-Chain

+ C 1 , L ArCHCICH,

ArCH=CH2 + H, ArH + CH,COCI(AICI,)-* A r X - C H ,

-

(6) Ring (Electrophilic)

+ X,

Fe

op-C,H,C,H,X

H2S04

+ HN0,op-C,H,C,H,NO, + H,S,O,---* op-C2H,C,H,S0,H + RX- AICI, op-C,H,C,H,R

\-

+ RCOCl-C,H,C,H,COR AIClj

4.

Disproportionation

+

C6H6

B. ALKENYLBENZENES

1. Reduction

1. Dehydrogenation 2. Dehydration

ArCHOHCH, or ArCH,CH,OH

/--

+ H,

ArCH=CH2 ArCHXCH, or ArCH,CH2X + alc. KOH

-

ArCHOHCH,OH ArCOOH

x

1. Nucleophilic Displacements

+ NaOH ArOH + NaCl + NaNH, # ArNH, + HCI + NaCN--+ ArCN + NaCl

Fe HIO,/HNO, Ag'

2. Indirect Halogenation by Diazonium

ArH-

KMn0,-

pe

1. Direct Halogenation +

+ hot

+ X, ArCHXCH,X + H,O-ArCHOHCH, + HBr- r o tde ArCHBrCH, + HBrArCH,CH,Br

+ Zn

C. ARY LHALIDES

ArH xz ArH + I,

+ cold KMnO,

3. Other Addition Reactions

Dehalogenation

ArCHXCH,X

-

ArCH,CH,

2. Oxidation

\ \ // --

3. Dehydrohalogenation 4.

-

+ C,H4(C2HS)2

2. Metals

ArNH, __* Ar Nl+ X-

+ Li --+ArLi + Cu+

(see Chapter 18)

3.

CHzO

Ar-Ar

ArCH,OH

+ CuX,

Ullmann Reaction

Electrophilic Substitution

+ HN03/H2S04*

op-XC,H,NO,

AROMATIC SUBSTITUTION. ARENES

CHAP. 111

235

Supplementary Problems Problem 11.45 Write structural formulas for the principal monosubstitution products of the indicated reactions from the following monosubstituted benzenes. For each write an S or F to show whether reaction is SLOWER or FASTER than with benzene. ( a ) Monobromination, C,H,CF,. (6) Mononitration, C,H,COOCH,. (c) Monochlorination, C,H,OCH,. (d) Monosulfonation, C,H,I. (e) Mononitration, C,H,C,H,. (f)Monochlorination, C,H,CN. ( 8 ) Mononitration, C,H,NHCOCH,. (h) Monosulfonation, C,H,CH(CH,)CH,CH,. 4

Problem 11.46 Which xylene is most easily sulfonated?

4

m-Xylene is most reactive and sulfonates at C4 because its CH,’s reinforce each other [Rule 1; Problem 11.17(a)]. Problem 11.47 Write structures for the principal mononitration products of ( a ) o-cresol (o-methylphenol), (6) p-CH,CONHC,H,SO,H, (c) m-cyanotoluene (m-toluonitrile). 4

Problem 11.48 Assign numbers from 1 for monobromination of the following groups. (a)

(I) PhNH,,

(6) (I) PhCH,,

(11) PhNHlCI-,

LEAST

to 5 for

(111) PhNHCOCH,,

MOST

to designate relative reactivity to ring

(IV) PhCl, (V) PhCOCH,.

(11) PhCOOH, (111) PhH, (IV) PhBr, (V) PhNO,.

(c) (I) p-xylene, (11) p-C6H4(COOH),, (111) PhMe, (IV) p-CH,C6H,COOH, See Table 11-5. Table 11-5

(V) m-xylene.

4

236

AROMATIC SUBSTITUTION. ARENES

[CHAP. 11

Problem 11.49 Why are the negative poles of the dipole moments of C,H,OCH, and C,H,NH, in the ring 4 away from the 0 and N , even though these atoms are more electronegative than C?

Delocalization of the electron densities on 0 and N toward the ring submerges the electron-withdrawing inductive effect of these atoms. Problem 11.50 Explain why the oxidation of PhCH, to PhCOOH by KMnO, or K,Cr,O, and H' goes in 4 poor yield, whereas the same oxidation of p-O,NC,H,CH, to p-O,NC,H,COOH goes in good yield.

The oxidant is seeking electrons and is therefore an electrophile. As a side reaction, the oxidant can attack and destroy the ring. The NO2 deactivates the ring toward electrophilic attack, thereby stabilizing it toward degradative oxidation. Problem 11.51 Use PhH, PhMe, and any aliphatic or inorganic reagents to prepare the following compounds in reasonable yields: ( a ) rn-bromobenzenesulfonic acid, (6) 3 -nitro-4-bromobenzoic acid, (c) 3,4-dibromonitrobenzene , (d) 2,6-dibromo-4-nitrotoluene. 4

Br

(m-director is added first)

Nitration of p-BrC,H,CH, would have given about a 50-50 mixture of two products; 2-nitro-4bromotoluene would be unwanted. When oxidation precedes nitration, an excellent yield of the desired product is obtained.

Nitration followed by dibromination would give as the major product 2,5-dibromonitrobenzene (see Rule 2, preceding Problem 11.17).

Problem 11.52 Available organic starting materials for synthesizing the pairs of compounds below are PhH, PhMe, and PhNO,. Select the compound in each pair whose synthesis is more expensive and explain your choice. ( a ) p-C,H,(NO,), and rn-C,H,(NO,), , ( 6 ) rn-nitrotoluene and p-nitrotoluene, ( c ) 2,4,6-trinitrotoluene (TNT) and 1,3,5-trinitrobenzene7 (d) rn-C,H,CI, and p-C,H,Cl,. 4

( a ) p-C,H,(NO,),. NO, is rnetu-directing. The rnetu isomer is easily made directly by nitrating PhNO,. The paru isomer cannot be prepared directly and requires a more circuitous, expensive route. (6) rn-Nitrotoluene. Nitration of PhMe gives the pura isomer. PhNO, is too deactivated to undergo metu alkylation. (c) 1,3,5Trinitrobenzene. Two NO,'s deactivate the ring so that insertion of a third NO, is very difficult. This extreme deactivation is countered somewhat by Me. Therefore, PhMe can be more easily trinitrated than PhH. (d) rn-C,H,C12. C1 is op-directing. Problem 11.53 Give the product(s) from ( a ) mononitration and ( 6 ) dinitration of naphthalene.

4

CHAP. 111

237

AROMATIC SUBSTITUTION. ARENES

(6) Ring without NO, is deactivated less and undergoes cy -substitution:

( a ) a-Substitution; a-C,,H,NO,.

13-Dinitronaphthalene

1,&Dinitronaphthalene

Problem 11.54 Name and account for the products from the reaction of 2-ethylnaphthalene at

(U)

+ 96% H,SO,

40"C, (6) 140°C.

( a ) 2-Ethyl-1-naphthalenesulfonicacid. The activating group directs substitution to the reactive ortho position. ( 6 ) 2-Ethyl-6-naphthalenesulfonicacid. Higher-temperature p -sulfonation in the less hindered ring gives the thermodynamic product.

Problem 11.55 Supply structures for organic compounds (A) through (0).

PhBr + Mg

--

Ph-GSH

(4

Et 0

(F)

KOH

ArCH7CI

+ CH,MgX-(I)-(J)-(K)

PhCHBrCH,CH,

(A)

H C=CHCH2Br

(E)

(B) P h C H S H C H ,

(G)

N BS

(H)

LI. NH3

(C) PhCHOHCHOHCH,

(D) PhCOOH + CH,COOH ( 6 ) (E) PhMgBr

(F) PhCH,CH=CH,

(G) P h C H K H C H ,

(H) PhCH=CHCH2Br

(conjugated alkene is more stable)

Ph

(4

(I)

p-CH3C6H4

(4

(J)

PhC-CMgX (+CH,)

(L)

\

PhC=CCH,Ar Ph

/

(cis)

/'="\,

H

(M)

\ (K) ,C=C H

/

H

\cH,A~

p-CH C H CH-CH,Ph

'1

Br

+

', which is p-CH,C,H,CHCH,Ph In (M), H'+adds to give more stable R CH,C,H,CH,CHPh, because of electron release by p-CH,. p-BrC H

$9

6 4 \

(N)

/c=c\

H

/

H

CH3

( t run s )

(0)p-BrC,H,-cH,-CHCH,

8,

In (0),Br. adds to give more stable R., p-BrC,H,CHCHBrCH,, which is benzylic.

rather than p -

[CHAP. 11

AROMATIC SUBSTITUTION. ARENES

238

Problem 11.56 Assign numbers from 1 for LEAST to 3 for MOST to the Roman numerals for the indicated compounds to show their relative reactivities in the designated reactions. (a)

HBr addition to (I) P h C H S H , , (11) p - C H 3 C 6 H , C H d H 2 , (111) p-O,NC,H,CH=CH,.

(6) Dehydration of (I) p-O,NC,H,CHOHCH,, (11) p-H,NC,H,CHOHCH,,

(c)

CH3 I Dehydration of ( I ) P h - C - C H , C H , ,

CH, I (11) P h - C H - C H , O H ,

I

(111) C,H5CHOHCH,.

(111) PhCHOHCH,CH,CH,.

OH (d) Solvolysis of ( I ) C,HSCH,Cl, (11) /I-O,NC,H,CH,Cl,

(111) p-CH,OC,H,CH,Cl.

4

See Table 11-6. Table 11-6

I11

I1

I

3 (p-Me stabilizes the benzylic ) 'R

1 (p-NO, destabilizes the benzylic ) 'R

(b)

1 (p-NO, destabilizes benzylic ) ' R

3 (p-NH, stabilizes the benzylic ) ' R

2

(c)

3 (get 3" benzylic R')

I (get I", non-benzylic ) 'R

2 (get 2" benzylic ) 'R

1 (p-NO, destabilizes benzylic ) 'R

3 (p-CH,O stabilizes benzylic R')

I I* (d)

Problem 11.57 Show the syntheses of the following compounds from benzene, toluene and any inorganic reagents or aliphatic compounds having up to three C's:

CH3 (a)

p-BrC,H,CH,Cl

( b ) p-BriCbH,CH=CH,

(c) Ph-A-CH,

AH ( d ) p-0,NC,H4CH2Ph

C,H,CH,

(a)

C,H,-

(6)

C7H5CI AIC13

4

-

Er,. Fe

Ph,CHCH,

(e)

CI

p-BrC,H,CH,

uv

p-BrC,H,CH,Cl

Br2

CI2

C , H , C H 2 C H , ~p-BrC,H,CH,CH,:

p-BrC,H,CHClCH,

alc.

p-BrC,H,CH=CH,

The C = C must be introduced by a base-induced reaction. If acid is used, such as in the dehydration of an alcohol, the product would undergo polymerization.

(4 (4

PhH

+ (CH,),CHCl

5 Ph-CH(CH,),

--% Ph-C(CH,), Br I

HN03 C6H5CH3

CIZ

c:$b% Ph-C(CH,), AH

C6H6

p-02NC6H4CH3 7 p - 0 2 N C 6 H 4 C H 2 C 1 ~p-02NC6H4CH2CfjH5

CHAP. 111

239

AROMATIC SUBSTITUTION. ARENES

The deactivated C,H,NO, cannot be alkylated with PhCH,CI. PhH-

(4

C H C1 AICI:,

AIc.

PhCH,CH,

P h C H B r C H , G PhCH=CH,

Ph,CHCH,

Problem 11.58 Deduce the structural formulas of the following arenes. ( a ) (i) Compound A (C,6H16) decolorizes both Br, in CCl, and cold aqueous KMnO,. It adds an equimolar amount of H,. Oxidation with hot KMnO, gives a dicarboxylic acid, C,H,(COOH),, having only one monobromo substitution product. (ii) What structural feature is uncertain? (6) Arene B (C,oH14)has five possible monobromo derivatives (C,,,H1.7Br). Vigorous oxidation of B yields an acidic compound, C 8 H 6 0 4 ,having only one mononitro substitution product, C,H,O,NO,. 4 since it adds one H,. The other 8 degrees of unsaturation mean the presence of two benzene rings. Since oxidative cleavage gives a dicarboxylic acid, C,H,(COOH),, each benzene ring must be disubstituted. Since C,H,(COOH), has only one monobromo derivative, the COOH’s must be para to each other.

( a ) (i) Compound A has one C=C

Br

Br

(A)

\.

Br

(ii) It may be cis or tram. : 6 ) C,H,O, must be a dicarboxylic acid, C,H,(COOH),, and, as in part (a), has para COOH’s. Compound B must therefore be a p-dialkylbenzene.

@

CH,Br

/TCH3 H3C

6

B

4 6

r

F\HCH3 H3C

H3C

F\HCH3

CH H3C/ \CH2Br

monobromo substitution products

H,C

CH / \ CH3 (B)

COOH

COOH

COOH

COOH

CBr /\ H3C CH3

240

AROMATIC SUBSTITUTION. ARENES

[CHAP. 11

Problem 11.59 Outline practical laboratory syntheses from benzene or toluene and any needed inorganic reagents of: ( a ) p-chlorobenzal chloride, (6) 2,4-dinitroaniline1 ( c ) rn-chlorobenzotrichloride, (d) 2,5dibromonitrobenzene . 4

CCI,

Br

Br

Problem 11.60 Assign numbers from 1 for LEAST to 3 for MOST to show the relative reactivities of the compounds with the indicated reagents: (a) C,H,CH,CH,Br (I), C,H,CHBrCH, (11) and C,H,CH==CHBr (111) with alcoholic AgNO,; (b) CH,CH,CI ( I ) , C,H,CH,Cl (11) and C,H,Cl (111) with KCN; ( c ) M nitrochlorobenzene (I), 2,4-dinitrochlorobenzene(11) and p-nitrochlorobenzene (111) with sodium methoxide. See Table 11-7

I

I1

I11

( a ) An s N 1 reaction

2 (primary)

3 (benzylic)

1 (vinylic)

(b) An s N 2 reaction

2 (primary)

3 (benzylic)

1 (aryl)

(c) An aromatic nucleophilic displacement

1 (meta)

3 ( 0 and p )

2 (para)

Problem 11.61 Which C1 in 1,2,4-trichlorobenzene reacts with -OCH,COO- to form the herbicide “2,4-D”? 4 Give the structure of “2,4-D.” Cl’s are electron-withdrawing and activate the ring to nucleophilic attack. The Cl at C’ is displaced because it is ortho and para to the other Cl’s.

+

-OCH,COO-

-

0C H 2C00-

Qcl Cl “2.4-D”

+

CI-

CHAP. 11)

241

AROMATIC SUBSTITUTION. ARENES

Problem 11.62 Explain these observations: ( a ) p-Nitrobenzenesulfonic acid is formed from the reaction of p-nitrochlorobenzene with NaHSO, , but benzenesulfonic acid cannot be formed from chlorobenzene by this reaction. ( 6 ) 2,4,6-Trinitroanisole with NaOC,H, gives the same product as 2,4,6-trinitrophenetole with 4 NaOCH,. ( a ) Nucleophilic aromatic substitution occurs with p-nitrochlorobenzene, but not with chlorobenzene, because

NO, stabilizes the carbanion [Problem 11.24(a)]. . ? N O 1

+ Na':SO,H-

+02N

(O)S4H

+ NaCl

(6) The product is a sodium salt formed by addition of alkoxide.

Problem 11.63 After o-DC,H,F is reacted for a short time with NaNHJNH,, the recovered starting material contains some C,H,F. Under the same conditions o-DC,H,Br leaves no C,H,Br. Explain. 4

F- is a much poorer leaving group than is Br-.

&* QF

2

F

F

&D

w,

Br

Br

+ NH,

Chapter 12 Spectroscopy and Structure 12.1 INTRODUCTION Spectral properties are used to determine the structure of molecules and ions. Of special importance are ultraviolet (uv), infrared (ir), nuclear magnetic resonance (nmr), and mass spectra (ms). Free radicals are studied by electron spin resonance (esr). The various types of molecular energy, such as electronic, vibrational, and nuclear spin, are quantized. That is, only certain energy states are permitted. The molecule can be raised from its lowest energy state (ground state) to a higher energy state (excited state) by a photon (quantum of energy) of electromagnetic radiation of the correct wavelength.

ultraviolet (uv) 8 ' Far Near ultraviolet (uv)

5

?! w

8

Electronic Electronic Electronic Molecular vibration Spin (electronic

Visible Infrared (ir) Radio

100-200 nm 200-350 nm 350-800 nm 1-300 p m lm

Wavelengths ( A ) for ultraviolet spectra are expressed in nanometers (1 nm = 10-' m); for the infrared, micrometers are used ( 1 p m = 10-6 m). Frequencies (v) in the infrared are often specified by the wave number V , where V = 1/A; a common unit for V is the reciprocal centimeter (1 cm-' = 100m-l). The basic SI units for frequency and energy are the hertz (Hz) and the joule (J), respectively. Problem 12.1 ( a ) Calculate the frequencies of violet and red light if their wavelengths are 400 and 750nm, respectively. (6) Calculate and compare the energies of their photons. 4

The wavelengths are substituted into the equation v = c/A, where c = speed of light = 3.0 X 10' m / s . Thus

Red:

V =

3.0 x 10' m/s = 4 . 0 1014s--1 ~ =400THz 750 x 10-9 m

where 1 THz = 10'' Hz = 10" s-l. Violet light has the shorter wavelength and higher frequency. The frequencies from part (a) are substituted into the equation E = hv, where h = 6.624 x 10-34J - s (Planck's constant). Thus Violet:

E = (6.624 x 1 0 - J~. ~s)(7.5 x 1014s - I )

Red:

E = (6.624 x 10-34J * s)(4.0 X 1014S - ' ) = 2.7 x 10-" J

= 5.0 x

10-

J

Photons of violet light have more energy than those of red light. emblem 12.2 Express 10 micrometers ( a ) in centimeters, (6) in angstroms (1 A = 10-"' m), (c) in nanometers, (d) as a wave number. 4

(4

10 p m = (10 x 10-6 m)(

242

100 cm r ) 10-3 cm =

243

SPECTROSCOPY AND STRUCTURE

CHAP. 12)

1A 10- m = 105 A 10 p m = (10 x 10-6 m)( T) 10 p m = (10 x 10-6 m)(

109nm 7) = 104 nm

In a typical spectrophotometer, a dissolved compound is exposed to electromagnetic radiation with a continuous spread in wavelength. The radiation passing through or absorbed is recorded on a chart against the wavelength or wave number. Absorption peaks are plotted as minima in infrared, and usually as maxima in ultraviolet spectroscopy. At a given wavelength, absorption follows an exponential law of the form A=ECl

where

A = absorbance = -log,, (fraction of incident radiation transmitted) E = molar extinction coefficient, cm */mol C = concentration of solution, moI/L( = moI/cm3) l = thickness of solution presented to radiation, cm

The wavelength of maximum absorption, A,,, , and the corresponding of a compound. Units are normally omitted from specifications of E .

are identifying properties

12.2 ULTRAVIOLET AND VISIBLE SPECTROSCOPY

Ultraviolet and visible light cause an electron to be excited from a lower-energy HOMO to a high-energy LUMO. There are three kinds of electrons: those in U bonds, those in T bonds, and unshared electrons, which are designated by the letter n for nonbonding. These are illustrated in formaldehyde:

H

On absorbing energy, any of these electrons can enter excited states, which are either antibonding U * or T * . All molecules have CT and U * orbitals, but only those with n orbitals have 7r* orbitals. Only the n - , T * , n + , ~ * and , more rarely the n - , U * excitations occur in the near ultraviolet and visible regions, which are the available regions for ordinary spectrophotometers. Species which absorb in the visible region are colored, and black is observed when all visible light is absorbed. Problem 12.3 The relative energy for various electronic states (MO's) is: IT*

n I

(nonbonding)

U

List the three electronic transitions detectable by uv spectrophotometers in order of increasing A E.

4

n + n* < T + n* < n + a * Problem 12.4 List all the electronic transitions possible for (a) CH,, (6) CH,CI, (c) H,C==O. n-

( a ) U + U * .(6) u+ n* and n + n * .

U * and

n-,

U*

(there are no n or

T*

MO's). ( c ) U--+U * ,U +

4

7r*, T--+U * ,n--, U * ,

244

SPECTROSCOPY AND STRUCTURE

Problem 12.5 The uv spectrum of acetone shows two peaks of A,, = 280nm, E,,, = 100. (U) Identify the electronic transition for each. (6) Which is more intense?

[CHAP. 12

= 15

and A,,

= 190,

4

E,,,

( a ) The longer wavelength (280 nm) is associated with the smaller-energy (n+ n * ) transition. n-+ n* occurs

at 190 nm. ( 6 ) n+ n* has the larger

E,,,

and is the more intense peak.

Problem 12.6 Draw conclusions about the relationship of A,,, to the structure of the absorbing molecule from the following A,, values (in nm): ethylene (170), 1,3-butadiene (217), 2,3-dimethyl-l,3-butadiene(226), 1,3-~yclohexadiene(256), and 1,3,5-hexatriene (274). 4

1. Conjugation of n bonds causes molecules to absorb at longer wavelengths. 2. As the number of conjugated T bonds increases, A,, increases. 3. Cyclic polyenes absorb at higher wavelengths than do acyclic polyenes. 4, Substitution of alkyl groups on C=C causes a shift to longer wavelength (red shift). Problem 12.7 Account for the following variations in A,,,(nm)

of CH,X: X = C1( 173), Br(204), and I(258).

4

The transition must be n - + U * [Problem 12.4(6)]. On going from C1 to Br to I the n electrons ( a ) are found in higher principal energy levels (the principal quantum numbers are 3, 4, 5 , respectively), (6) are further away from the attractive force of the nucleus, and (c) are more easily excited. Hence absorption occurs at progressively higher A,, since less energy is required. Problem 12.8 Use MO theory (see Fig. 8-3) to explain why A,, (See Problem 12.6 for data.)

for 1,3-butadiene is higher than for ethylene. 4

For 1,3-butadiene, the excitation is from n,, the highest occupied MO (HOMO), to n:, the lowest unoccupied MO*(LUMO). The AE for this transition is less than the AE for n-+ n* for ethylene. Therefore A,, for 1,3-butadiene is greater than A,, for ethylene. Problem 12.9 Identify the two geometric isomers of stilbene, C,H,CH=CHC,H,, 294 nm and 278 nm.

from their A,,

values,

4

The higher-energy cis isomer has the shorter wavelength. Steric strain prevents full coplanarity of the cis phenyl groups, and the conjugative effect is attenuated. Problem 12.10 Each of three C,H,, isomers absorbs 2 mol of H, to form n-hexane. CO, and RCOOH are not formed on ozonolysis from any of the compounds. Deduce possible structures for the 3 compounds if their uv absorption maxima are 176, 211, and 216 nm. 4

The isomers are hexadienes. An allene or alkyne is not possible since CO, and RCOOH are not ozonolysis products. Absorption at 175 nm is attributed to the isolated 1,4-hexadiene. The conjugated cis- and truns-1,3hexadienes absorb at 21 1 and 216 nm, respectively. Problem 12.11 The complementary color pairs are: violet-yellow, blue-orange, and green-red. Given a red, 4 an orange and a yellow polyene, which is most and which is least conjugated?

The orange polyene absorbs blue, the red absorbs green and the yellow absorbs violet. The most conjugated polyene, in this case the red one, absorbs the color of longest wavelength, in this case green. Violet has the shortest wavelength and therefore the yellow polyene is the least conjugated.

12.3 INFRARED SPECTROSCOPY

Infrared radiation causes excitation of the quantized molecular vibration states. Atoms in a diatomic molecule, e.g. H-H and HC==C\

7-12

cis

/H =ch

0-3

Phenyl H’s

12.5

ortho

6-9

meta para

0- I

1-3

13C NMR (CMR)

Although I2C is not nmr-active, the I3C isotope, which has a natural abundance of about 1%, is nmr-active because it has an odd number of neutrons. Modern techniques of Fourier transform spectroscopy are capable of detecting signals of this isotope in concentrated solution. Since H and I3C absorb at different frequencies, proton signals do not appear in the I3C spectra. However, spin-spin coupling between attached H’s and a 13C (proton-coupling) is observed. The same (n + 1)-rule that is used for H’s on adjacent C’s is used to analyze the 13 C-H coupling pattern. Thus, the I’C of a I3CH, group gives rise to a quartet. To avoid this coupling, a proton-decoupled spectrum can be taken, in which each different I3C nucleus appears as a sharp singlet. Both types of spectra are useful. The decoupled spectrum permits the counting of different I3C’sin the molecule; the coupled spectrum allows the determination of the number of H’s attached to each ”C. Figure 12-9 shows ( a ) the proton-coupled and (6) the proton-decoupled spectrum of dichloroacetic acid, CHC1,COOH. Each I3C has a characteristic chemical shift 6, as listed in Table 12-5. Note that the 13Cof 1 3 C 4 is the most downfield, with aromatic 13C’ssomewhat more upfield. Integration of the peak areas affords the relative number of 13C’seach peak represents, as shown in Fig. 12-10 for p-BrC,H,COCH,. Problem 12.34 Why is spin-spin coupling between adjacent l3Cs not observed?

4

Since the natural abundance of this isotope is so low, the chance of finding two 13C’snext to each other is practically nil. However, if a compound were synthesized with only 13C’s,then coupling would be observed. Problem 12.35 How many peaks would be evidenced in the decoupled spectrum of ( a ) methylcyclohexane? (6) cyclohexene? (c) 1-methylcyclohexene? 4

33 5

(a)

Five peaks

0

11

2

(6) Three peaks

5

(c)

Seven peaks

258

[CHAP. 12

SPECTROSCOPY AND STRUCTURE

t

C2

x

.C1 U

2

U I

200

180

160

140

120 100 80 60 e--- Chemical Shift, 6 ppm ( a ) Proton-coupled

40

20

0

C2

t

x

.CI U

2

U

TMS

c (

2 200

180

160

140

120

80

100

I--- Chemical

1

I

1

I

60

40

20

0

Shift, 6 ppm

(6) Proton-decoupled

Fig. 12-9

Table 12-5

259

SPECTROSCOPY AND STRUCTURE

CHAP. 121

200

I

I

180

160

I

140

I

I

I

I

120 100 80 60 Chemical Shift, 6 ppm

1

1

1

40

20

0

Fig. 12-10

Problem 12.36 Sketch ( a ) the coupled and ( b ) the decoupled '3C-spectrumof

y . 3

The

l3C__o

1 p

2

p

3

group is electron-withdrawing and causes a downfield shift of other attached I3C atoms.

4

See Fig. 12-11.

C2

1

210

I

I

I

1

I

1

1

I

I

1

1

200

180

160

140

120

100

80

60

40

20

0

Fig. 12-11

12.6 MASS SPECTROSCOPY

When exposed to sufficient energy, a molecule may lose an electron to form a cation-radical, which then may undergo fragmentation of bonds. These processes make mass spectroscopy (ms)a useful tool for structure proof. Very small concentrations of the parent molecules (RS), in the vapor state, are ionized by a beam of energetic electrons ( e - ) ,

260

SPECTROSCOPY AND STRUCTURE

R:S+e-+2eParent molecule

+R*S'----*R*+S' Parent cation-radical

or

[CHAP. 12

R' + S .

Fragment ions

and a number of parent cation-radicals (RS') may then fragment to give other cations and neutral species. Fragment ions can undergo further bond-cleavage to give even smaller cations and neutral species. In a mass spectrogram, sharp peaks appear at the values of rnle (the mass-to-charge ratio) for the various cations. The relative heights (intensities) of the peaks represent relative abundances of the cations. Most cations have a charge of + 1 and therefore most peaks record the masses of the cations. Fragmentation tends to give the more stable cations. The most abundant cations are the most stable ones. The standard against which all peak intensities in a given mass spectrogram are measured is the most intense peak, called the base peak, which is arbitrarily assigned the value 100. If few parent molecules fragment-not a typical situation-the parent cation will furnish the base peak. Unless all the parent ions fragment, which rarely happens, the largest observed m l e value is the molecular weight of the parent (RS) molecule. This generalization overlooks the presence of naturally occurring isotopes in the parent. Thus, the chances of finding a I3C atom in an organic molecule are 1.11%; the chances of finding two are negligible. Therefore, the instrument detects a small peak at mRS+ 1, owing to the 13 C-containing parent. The chances of finding 'H in a molecule are negligible. The masses and possible structures of fragment cations, especially the more stable ones, are clues to the structure of the original molecule. However, rearrangement of cations complicates the interpretation. Mass spectra, like other spectra, are unique properties used t o identify known and unknown compounds. Problem 12.37 ( a ) What molecular formulas containing only C and H can be assigned to a cation with rnle equal to (i) 43, (ii) 65, (iii) 91? (Assume that e = + l . ) (6) What combination of C, H and N can account for an mle of (i) 43, (ii) 57? (Assume that e = + l . ) 4 Divide by 12 to get the number of C's; the remainder of the weight is due to H's. (i) C , H f , (ii) C,H;, (iii) C,H:. (b) ( i ) If one N is present, subtracting 14 leaves a mass of 29, which means 2 C's (mass of 24) are present. Therefore the formula is C,H,N+. If 2 N's are present, it is CH,Nl. (ii) CH,Nf, C2HSNI or C,H,N+

(a)

Problem 12.38 ( a ) Do parent (molecular) ions, RS', of hydrocarbons ever have odd rnle values? (6) If an RS' contains only C, H and 0, may its mle value be either odd or even? (c) If an RS' contains only C, H and N, may its mle value be either odd or even? ( d ) Why cannot an ion, rnle = 31, be C,H:? What might it be?

4

No. Hydrocarbons, and their parent ions, must have an even number of H's: CnHZn,_?, C n H z nC,H2n-2, , C n H Z n - 6etc. , Since the atomic weight of C is even (12), the mle values must be even. The presence of 0 in a formula does not change the ratio of C to H. Since the mass of 0 is even (16), the mass of RS' with C, H and 0 must be even. The presence of each N (rn = 14) requires an additional H (CnHZn+,N,C n H 2 n t l N ,C,H,,-,N). Therefore, if the number of N's is odd, an odd number of H's and an odd rnle value result. An even number of N's requires an even number of H's and an even mle value. These statements apply only to parent ions, nor to fragment ions. The largest number of H's for two C's is six (C,H,). Some possibilities are CH,O' and CH,N+. Problem 12.39 Which electron is most likely to be lost in the ionization of the following compounds? Write an 4 electronic structure for RS'. ( a ) CH,, (6) CH,CH,, ( c ) H,C=CH,, ( d ) CH,CI, (e) H , C = O . The electron in the highest-energy M O is most likely to be lost. The relative energies for electrons are: n > T > a (Problem 12.3). In a compound with no n or n electrons, the electron most likely comes from the highest-energy a bond.

CHAP. 121

H ( a ) H:C;H

(+

H

H H (b) H:C;C:H H H (c)

261

SPECTROSCOPY AND STRUCTURE

H,CLCH,

is the charge due to the missing electron)

(the C-C

bond is weaker than the C-H

(electron comes from the

7~

bond)

bond)

( d ) H3C:Cl? (an n electron is lost) (e)

H,C=O'

(an n electron is lost)

Problem 12.40 Write equations involving the electron-dot formulas for each fragmentation used to explain the following. ( a ) Isobutane, a typical branched-chain alkane, has a lower-intensity RS' peak than does n-butane, a typical unbranched alkane. (6) All 1" alcohols, RCH,CH,OH, have a prominent fragment cation at mle = 31. ( c ) All C,H,CH,R-type hydrocarbons have a prominent fragment cation at mle = 91. ( d ) Alkenes of the type H,C=CHCH,R have a prominent fragment cation at rnle = 41. ( e ) Aldehydes,

H

I

R-C=O show intense peaks at (rnle),, - 1 and mle = 29.

4

(a) Cleavage of C--C is more likely than cleavage of (the stronger) C-H. isobutane, H H

I

(H,C),C?CH,

-

I

(H,C),C+

Fragmentation of RS' for

+ CH,

gives a 2" R', which is more stable than the 1" R' from n-butane,

H H

H

I I

I + *CH,CH, IH

H,CC?CCH,

+H,CC+

I t

H H

Hence RS' of isobutane undergoes fragmentation more readily than does RS' of n-butane, and fewer RS' fragments of isobutane survive. Consequently, isobutane, typical of branched-chain alkanes, has a low-intensity RS' peak compared with n-butane.

H ,! R-&~:kH,OH

-

H

R-

I :b+*HI H'

A -€'next to an 0 is stabilized by extended generally undergoes cleavage of the bond

H

I I

C,H,: C?R

H

-

R.

*

T

+ CH,OH +

c--*

rnle

\a

H,C=()H

= 31

bonding (resonance). The RS' species of alcohols

*t 0

C E-OH

+ C,H,:CHr

__*

stable benzyl R'

. ,; a-.-

0:

H,CLCH-C;'R

a more stable aromatic cycloheptatrienyl cation, m l e = 91

+

H,C=CH-CH,

*\\

stable allylic

cation, m l e = 41

+ -R

SPECTROSCOPY AND STRUCTURE

262

[CHAP. 12

Problem 12.41 Why is less than 1mg of parent compound used for mass spectral analysis?

4

A relatively small number of molecules are taken to prevent collision and reaction between fragments. Combination of fragments might lead to ions with larger masses than RS', making it impossible to determine the molecular weight. The fragmentation pattern would also become confusing.

Problem 12.42 Suggest structures for three of the peaks (86, 43, and 42) in the mass spectrum of n-hexane.

4

The R S ' of n-hexane, CH,CH,CH,CH,CH,CH,, has a mass = 86. A split in half would give a fragment CH,CH,CH, of mass 43. The 42 fragment may be made up of three CH,'s coming from the middle of the chain: .CH,CH,CH,'. Problem 12.43 Give the structure of a compound, C,,H,,O, whose mass spectrum shows r n l e values of 15,

43, 57, 91, 105, and 148.

4

The value 15 suggests a 'CH,. Because 43 - 15 = 28, the mass of a c=O group, the value of 43 could mean an acetyl, CH,CO, group in the compound. The highest value, 148, gives the molecular weight. Cleaving an acetyl group (rnle = 43) from 148 gives 105, which is an observed peak. Next below 105 is 91, a difference of 14; this suggests a CH, attached to CH,CO. So far we have CH,COCH, adding up to 57, leaving 148; 57 = 91 to be accounted for. This peak is likely to be C,H:, whose precursor is the stable benzyl cation, C,H,CH,. The structure is CH, ---CH,---CH,-C,H,.

3 0

Problem 12.44 How could mass spectroscopy distinguish among the three deuterated forms of ethyl methyl ketone?

DCH,CH,COCH,

(1)

(2) CH,CH,COCH,D

(3) CH,CHDCOCH,

4

The expected peaks for each compound are shown in Table 12-6; each has a different combination of peaks.

b

m/e

15 16

29 30

43 44

DCH,CH,COCH, CHf DCH;

-

DCH,CH,' CH,CO'

-

CH,CH2COCH,D CHf DCH; CH,CH,'

-

DCH,CO'

CH,CHDCOCH, CHf

-

CH,CHD+ CH,CO+

-

Problem 12.45 A prominent peak in the mass spectrum of 2-methyl-2-phenylbutaneis at r n l e = 119. To what fragment cation is this peak due? Show how cleavage occurs to form this ion. 4

SPECTROSCOPY AND STRUCTURE

CHAP. 121

263

The RS is CH3 CH

' 5-1

+

NI

C-CHZCH, n

CH3

Cleavage between C" and C' gives C,H,C(CH,), , m / e = 119.

Supplementary Problems Problem 12.46 Match the type of spectrometer with the kind of information which it can provide the chemist.

1. 2. 3. 4.

Mass Infrared Ultraviolet Nuclear magnetic resonance

A. functional groups B. molecular weights C. proton environment D. conjugation

4

l B , 2 A , 3D, 4C. Problem 12.47 Why do colored organic compounds, such as @-carotene(the orange pigment isolated from carrots), have extended conjugation? 4

The more effective electron delocalization in molecules with extended conjugation narrows the energy gap between the HOMO and the LUMO. Thus, visible radiation of lower frequency (longer wavelength) is absorbed in the HOMO-, LUMO electron transition. Problem 12.48 ( a ) Account for the fact that benzene absorbs at 254nm in the uv, and phenol, C,H,OH, 4 absorbs at 280nm. ( b ) Where would one expect 1,3,5-hexatriene to absorb, relative to benzene? ( a ) The p orbital on 0, housing a pair of n electrons, overlaps with the cyclic v system of benzene, thereby

extending the electron delocalization and decreasing the energy for the HOMO +LUMO transition. Consequently, phenol absorbs at longer wavelengths. (b) The triene absorbs at a longer A,,, (275 nm). Since the cyclic 7r system of benzene has a lower energy than the linear T system of the triene, benzene absorbs radiation of shorter wavelength. Problem 12.49 Which peaks in the ir spectra distinguish cyclohexane from cyclohexene?

4

One of the C-H stretches in cyclohexene is above 3000cm-' (C,,z-H); in cyclohexane the C-H itretches are below 3000cm-' (CsP~--H). Cyclohexene has a C==C stretch at about 1650cm-'. Problem 12.50 The mass spectrum of a compound containing C, H, 0, and N gives a maximum mle of 121. [ts ir spectrum shows peaks at 700, 750, 1520, 1685, and 3100cm-', and a twin peak at 3440cm-'. What is a Feasonable structure for the compound? 4

The molecular weight is 121. Since the mass is odd, there must be an odd number of N's [Problem

12.38(c)]. The ir data indicate the following groups to be present:

1520 cm-' : aromatic ring (1450-1600 cm-' range) I 1685 cm-': C=O stretch of amide structure -CO-N (1630-1690 cm-' range) 3100 cm-I: aromatic C-H bond (3000-3100 cm-' range) 3440 cm- ' : -N-H in amine or amide (3300-3500 cm- ' range) 700, 750 cm-' : monosubstituted phenyl

264

SPECTROSCOPY AND STRUCTURE

A twin peak due to symmetric and antisymmetric N-H together we find that the compound is benzamide,

[CHAP. 12

stretches means an NH, group. By putting the pieces

0

II

C,H,-C-NH, Problem 12.51 The ir spectrum of methyl salicylate, o-HOC,H,COOCH,, has peaks at 3300, 1700, 3050, 1540, 1590, and 2990cm-'. Correlate these peaks with the following structures: ( a ) CH,, ( b ) C=O, (c) OH 4 group on the ring, and ( d ) aromatic ring. ( a ) 2990cm-'; ( b ) 1700cm-'; (c)

3300cm-'; ( d ) 3050, 1540, 1590cm-'.

Problem 12.52 Calculate E, for a compound whose maximum absorbance is A,,, = 1.2. The cell length ( 8 ) is 1.0cm and the concentration is O.O76g/L. The mass spectrum of the compound has the largest m / e value at 100. 4

The molecular weight is 100 g/mol; therefore, c = 7.6 x lO-, mol/L and

Emax =

-= Amax

CI

1.2 (7.6 x 10-,)( 1.0)

=

1600

Problem 12.53 Methanol is a good solvent for uv but not for ir determinations. Why?

4

Methanol absorbs in the uv in 183 nm, which is below 190 nm, the cutoff for most spectrophotometers, and therefore it doesn't interfere. Its ir spectrum has bands in most regions and therefore it cannot be used. Solvents such as CCI, and CS, have few interfering bands and are preferred for ir determinations. Problem 12.54 How can nmr distinguish between p-xylene and ethylbenzene?

4

p-CH;C6HtCHj has two kinds of H's, and its nmr spectrum has two singlets. C,H:CHfCHi has three kinds of H's, and its nmr spectrum has a singlet for H", a quartet for Hh, and a triplet for H'. Problem 12.55 A compound, C,H,O, contains a C=O compound is an aldehyde or a ketone?

group. How could nmr establish whether this

4

If an aldehyde, the compound is CH,CH,CHO, with three multiplet peaks and a downfield signal for

H

-c=o (6 = 9-10ppm). If a ketone, it is (CH,),C=O,

with one singlet.

Problem 12.56 Describe the expected appearance of the nmr spectrum of n-propylbenzene,

0

CH;-

CH,~-CH;

4

(All chemical shifts are given as 6 in ppm). At about 7.2 there is a signal due to the aromatic ring H's. At 2.3-3 there is a triplet for the benzylic H". If the coupling effects of H" and H' with Hb were similar, the multiplet for Hb would be a sextet ( 5 + 1). If the effects were different, Hb would be split into a triplet by an H", and each peak of the triplet would be further split into a quartet. Some of these peaks might coincide. At best, Hb is a complex multiplet at about 1.3. The H' signal is a triplet at about 0.9. The relative peak intensities will be: (ring H) : H" : Hb : H' = 5 : 2 : 2 : 3.

CHAP. 12)

SPECTROSCOPY AND STRUCTURE

265

Problem 12.57 The nmr spectrum of a compound, C,H,ClF,, has two noncoupled triplets. Triplet A is 1.5 4 times the intensity of the more downfield triplet B . Suggest a structure.

Since A is 1.5 times as intense as B and the molecular formula has five H's, A must come from CH, and B from CH,. A and B are not coupling each other; if they did, we would find a triplet and a quartet. A and B must each be coupled to two F's. B is more downfield than A , indicating a --CH,Cl group. The compound is ClCfi,CF,Cfi, See Problem 12.29 for H / F coupling. Problem 12.58 The nmr spectrum of a dichloropropane shows a quintuplet and, downfield, a triplet of about 4 twice the intensity. Is the isomer 1,l-, 1,2-, 1,3-, or 2,2-dichloropropane?

We would expect the following signals: 1,l-Dichloropropane, Cl,CH"CH,bCHi: a triplet (Hr), a complex multiplet more downfield (Hb), and a triplet still more downfield (H"). 1,2-Dichloropropane7ClCH;CH*CH;: a doublet (Hr), another doublet more downfield (H"), a complex

I

Cl multiplet most downfield (Hb). 1,3-Dichloropropane, C1CH;CHfCH;CI: a quintuplet (Hh), and, downfield, a triplet (H") C1

I

2,2-Dichloropropane, CHfl--C-CHfl: a singlet (H").

I

c1 The compound is 1,3-dichloropropane. Problem 12.59 Consider the coupled I3C nmr spectra of ( a ) 1,3,5-trimethylbenzene9( 6 ) 1,2,3-trimethylbenzene, ( c ) n-propylbenzene. (i) Write each structure and label the different kinds of C's by numerals I, 2, . . . . (ii) Show the splitting for each peak by the letters s for singlet, d for doublet, t for triplet, and q for quartet. 4

Problem 12.60 Indicate whether the following statements are true or false and give a reason in each case. ( a ) The ir spectra are identical for the enantiomers

Tir

B r T i H , and CH,

H

H

(6) The nmr spectra of the compounds in ( a ) are also identical. ( c ) The ir spectrum of 1-hexene has more peaks than the uv spectrum. (d) Compared to CH,CH2CH0, the n + 7r* for H,C=CHCHO has shifted to a shorter wavelength (blue shift). 4

266

SPECTROSCOPY AND STRUCTURE

(CHAP. 12

9

A

-

1

8

1

7

1

1

I

J

6

4

1

1

I

I

3

2

1

0

1 3

I 2

1 I

0

12

1 8

1 7

1

1

6

5

1 4

1

4 PPm

I 7

I

6

I

I 5

4 8 9

PPm

Fig. 12-12

I

3

I

2

I

I

I

0

CHAP, 121

267

SPECTROSCOPY AND STRUCTURE

(a) and ( b )

True. The compounds are enantiomers which have identical vibrational modes and proton resonances. ( c ) True. The ir spectrum has peaks for stretching and bending of all bonds, while the uv spectrum has only one peak for excitation of a ?r electron. ( d ) False. The shift is to longer wavelength (red shift), since

H,C=CHC=O

I

H is a conjugated system.

Problem 12.61 Assign the nmr spectra shown in Fig. 12-12 to the appropriate monochlorination products of 2,4-dimethylpentane (C,H,,CI) and justify your assignment. Note the integration assignments drawn in the spectra. 4 The three possible structures are:

CH3

CH3

CH3

CICH,-(!H-CH2-AH--CH3

CH3J

-CH2-CH-CH3 4 I

2IH33

I

CH3

CH3-CH--CH-&H-CH, I Cl

5

c1 1-Chloro-2,4-

2-Chloro-2.4-

dimethylpentane (I)

CH3

I

3-Chloro-2.4-

dimethylpentane (11)

dimethylpentane (111)

The best clue is the most downfield signal arising from the H's closest to CI. In spectrum ( a ) the signal with the highest 6 value is a doublet, integrating for two H's, that corresponds only to structure 1 (ClCH,-). This is confirmed by the nine H's of the 3 CH,'s that are most upfield and the four 2" and 3" H's with signals between these. In spectrum ( b ) the most downfield signal is a triplet, for one H, which arises from the

H C1 H

I l l -c-c-cI I!

grouping in 111. In addition, the most upfield signal is a doublet, integrating for 12 H's, which is produced by the H's of the 4 CH,'s split by the 3" H. This leaves I1 for spectrum ( c ) . The most downfield group of irregular signals, integrating for three H's, comes from the two 2" and one 3" H on C3 and C4, respectively. The most upfield doublet, integrating for six H's, arises from the two equivalent CH,'s on C4 split by the C4 3" H. The two CH,'s on C2 give rise (six H's) to the singlet of median S value. 0

I

I

1 2

I

3

I

4

I

J

1

6

I

7

I

8

I

9

1 10

Wavelength, pm Fig. 12-13. Infrared Spectrum

I I1

1 12

1

13

I

I4

1

IS

268

SPECTROSCOPY AND STRUCTURE

[CHAP. 12

Problem 12.62 Deduce structures for the compound whose spectral data are presented in Fig. 12-13, Table 12-7, and Fig. 12-14. Assume an 0 is present in the molecule. There was no uv absorption above 180nm. 4

1

m/e

Relative intensity , o/c of base peak

1 1 ::i 2:

Table 12-7. Mass Spectrum

2:

31

39

41

42

43

44

45

59

87

102

4

11

17

6

61

4

100

11

21

0.63

Fig. 12-14. nmr Spectrum (CDCI,) From the mass spectrum, we know the RS’ has molecular weight 102. The C, H portion of the compound has a mass of 102 - 16 (for 0)= 86. Dividing by 12 gives 6 C’s, leaving a remainder of 14 for 14 H’s. The molecular formula is C,H,,O. (Had we chosen 7 C’s we would have had C,H,O, which is an impossible formula.) The compound has no degrees of unsaturation. This fact is consistent with (but not proven by) the absence of uv absorption above 180 nm. Note also the absence of C-H stretch above 3000 cm-’ (H--C,,* or H-C,,,); the 0 must be present as C U H (an alcohol) or as C-0-C (an ether). The absence of a peak in the ir at 3300-3600cm-’ precludes the presence of an 0-H group. The strong band at about 1110cm-’ represents the C-0 stretch. The structure of the alkyl groups R U R of the ether is best revealed by the nmr spectrum. The downfield septet (see blown-up signal) and the upfield doublet integrate 1 : 6. This arrangement is typical for a CH(CH,), grouping [see Problem 12.26(b)]. Both R groups are isopropyl, since no other signals are present. The compound is (CH,),CHOCH(CH,),. The significant peaks in the m$ss spectrum that are cynsistent with our assignment are: m l e = 102 - 15 (CH,)=87, which is (CH,),CHOCHCH,; 43 is (CH,),CH; 41 would be the ally1 cation H,C----CHCH,’ formed from fragmentation of (CH,),CH. The most abundant peak, mle = 45, probably comes from rearrangement of fragment ions. It could have the formula C,H,O. The presence of this peak convinces us that 0 was indeed present in the compound. The peak cannot be from a fragment with only C and H; C,H; is impossible.

Chapter 13 Alcohols and Thiols A.

ALCOHOLS

13.1 NOMENCLATURE AND H-BONDING

ROH is an alcohol and ArOH is a phenol (Chapter 19). Some alcohols have common names, usually made up of the name of the alkyl group attached to the OH and the word "alcohol"; e.g., ethyl dcohol, C2H,0H. More generally the IUPAC method is used, in which the suffix -01 replaces the -e D f the alkane to indicate the OH. The longest chain with the OH group is used as the parent, and the C bonded to the OH is called the carbinol carbon. Problem 13.1 Give a common name for each of the following alcohols and classify them as 1". 2", or 3":

CH3

( a ) CH,CH,CH,OH

( b ) CH,CH,CHCH,

( c ) CH,CHCH,OH

I

AH CH3 (e)

I

CH,-CH-OH

I I

( d ) CH3-C--OH

CH3

CH3

CH3

cf) CH,-~-CH,OH

(g)

1

C,H,CH,OH

4

CH3 ( a ) n-propyl alcohol, 1"; (6) sec-butyl alcohol, 2"; (c) isobutyl alcohol, 1"; (d) r-butyl alcohol, 3"; (e) isopropyl alcohol, 2"; (f) neopentyl alcohol, 1"; ( g ) benzyl alcohol, 1".

Problem 13.2 Name the following alcohols by the IUPAC method.

CH,CH,CH, (U)

(6)

I

CH,CH,-C-OH I CH,CH3

H

( c ) Cl-CHCH,OH

I

c1

PH

( e ) a B OH r

OH I

H,C=CH&HCH,

( a ) 3-Ethyl-3-hexanol

(c) 2,2-Dichloroethanol

(6) 3-Buten-2-01

(d) 3-Phenyl-3-hexanol

{ote that in IUPAC the OH is given a lower number than C=C

H

(e) cis-2-Bromocyclohexanol or C1.

'roblem 13.3 Supply structures for alcohols having the IUPAC names ( a ) 2-propen-1-01, ( b ) 1-ethylcyclo4 ropanol, (c) 3-phenyl-2-butanol.

LT)

H,C=CHCH,OH Ally1 alcohol

'CH,CH,

269

ALCOHOLS AND THIOLS

270

[CHAP. 13

Problem 13.4 Explain why ( a ) propanol boils at a higher temperature than the corresponding hydrocarbon; ( 6 ) propanol, unlike propane or butane, is soluble in H,O; (c) n-hexanol is not soluble in H,O; (d) dimethyl ether (CH,OCH,) and ethyl alcohol (CH,CH20H) have the same molecular weight, yet dimethyl ether has a lower boiling point (-24°C) than ethyl alcohol (78°C). 4 ( a ) Propanol can H-bond intermolecularly. C , H , T - - - H - ( ( There is also a less important dipole-dipole interaction. H C,H, (6) Propanol can H-bond with H,O: C,H,T---H-?

H

H

(c) As the R group becomes larger, ROH resembles the hydrocarbon more closely. There is little H-bonding between H 2 0 and n-hexanol. When the ratio of C to OH is more than 4, alcohols have little solubility in water. (d) The ether CH,OCH, has no H on 0 and cannot H-bond; only the weaker dipole-dipole interaction exists.

Problem 13.5 The ir spectra of tram- and cis-1,2-cyclopentanediolshow a broad band in the region 3450-3570cm-'. On dilution with CCI,, this band of the cis isomer remains unchanged, but the band of the tram isomer shifts to a higher frequency and becomes sharper. Account for this difference in behavior. 4 The OH'S of the cis isomer participate in intramofecufarH-bonding, Fig. 13-1(u), which is not affected by dilution. In the tram isomer, the H-bonding is intermolecular, Fig. 13-1(6), and dilution breaks these bonds, causing disappearance of the broad band and its replacement by a sharp OH band at higher frequency. H

H H

cis

H- 0' Intermolecular H-bond

Intramolecular H-bond

(4

(4

Fig. 13-1

13.2 PREPARATION 1. RX + OH- --!% ROH + X- ( s N 2 or s N 1 Displacement, Table 7-1) 2. Hydration of Alkenes [see Problem 6.24(d)]

3. Hydroboration-Oxidationof Alkenes [see Problem 6.24( f )] Treatment of the alkylboranes with H,O, in OH- replaces -BRCH=CH2

+ (BH3)2

-

I

(RCHCH&-B

b

with OH.

: :RCHCH,

trialky lborant

The net addition of H - O H to alkenes is cis, anti-Markovnikov, and free from rearrangement.

27 1

ALCOHOLS AND THIOLS

CHAP. 131

4. Oxymercuration-Demercuration of Alkenes

not isolated

The net addition of H-OH

is Markovnikov and is free from rearrangement.

Problem 13.6 Give structures and IUPAC names of the alcohols formed from (CH,),CHCH=CH, reaction with ( a ) dilute H,SO,; (6) B,H,, then H,O,, OH-; ( c ) Hg(OCOCH,),, H,O, then NaBH,.

by

4

( a ) The expected product is 3-methyl-2-butano1, (CH,),CHCHOHCH,,

from a Markovnikov addition of H,O. However, the major product is likely to be 2-methyl-2-butano1, (CH,),COHCH,CH,, formed by rearrangement of the intermediate R'. +H+

(CH,),CHCH=CH,

*

(CH,),CH;HCH,

3(CH,),COHCH,CH,

(CH,),;CH,CH,

(2")

(3")

(6) Anti-Markovnikov HOH addition forms (CH, ),CHCH,CH,OH, 3-methyl-1-butanol. (c) Markovnikov HOH addition with no rearrangement gives (CH,),CHCHOHCH,, 3-methyl-2-butanol. Problem 13.7 Give the structure and IUPAC name of the product formed on hydroboration-oxidation of 1-meth ylcyclohexene. 4

H and OH add cis and therefore CH, and OH are tram.

0'"' CH,

'

_I___* b"2'

2.

H

OH-. H2O2

*.

H

trans-2-Methylcyclohexanol

H

5. Carbonyl Compounds and Grignard Reagents (Increase in Number of Carbons) Grignard reagents, RMgX and ArMgX, are reacted with aldehydes or ketones and the intermediate products hydrolyzed to alcohols.

-

H

[RJMgX

+ H-C-01

~-CH,-O(MgX)'

HO

2 \j-C H 2 -O H

(1" alcohol)

Formaldehyde

H mMgX

I

+ R'-C=O

H

H

I -

--+ R'-C-O(MgX)'

I

Hi0 __*

I

R'-C-OH

I

!E!

aldehyde

R" @MgX

I

+ R'-C=O ketone

R" --+

I

-

R'-C-O(MgX)'

I

rn

(2" alcohol)

alkoxide salts

R" "20

I

R'-C-OH

+

I

M

(3" alcohol)

ALCOHOLS AND THIOLS

272

[CHAP. 13

The boxed group in the alcohol comes from the Grignard; the remainder comes from the carbonyl compound.

R + Mg(X)(OH) electrophilic site

an alkoxide: a strong conjugate base of an alcohol

nucleoPhile

alcohol

ArMgX is best made from ArBr in ether or from ArCl in tetrahydrofuran (Arc1 is not reactive in ether). ether ArBr + MgArMgBr THF

ArCl + Mg-

ArMgCl

Problem 13.8 Give 3 combinations of RMgX and a carbonyl compound that could be used to prepare. CH,

I I

4

C,H,CH,C-CH,CH,

Off This 3” alcohol is made from RMgX and a ketone, R’COR”. The possibilities are:

0

CH, (I) m / - L - C H 2 C H 3

I

from

p ] M g C I

1I

and CH,CCH,CH,

OH

0

CH3

I

(2) C , H , C H , - - C - - ( C H ~ ~

from

1-MgBMgBr

II

and C,H,CH,CCH,

bH

P

(3) C,H,CH,-C--CH,CH,

I

0

from

MgI

II

and C,H,CH,CCH,CH,

OH

The best combination is usually the one in which the two reactants share the C content as equally as possible. In this case, (1) is best. Problem 13.9 Give four limitations of the Grignard reaction.

4

(1) The halide cannot possess a functional group with an acidic H , such as OH, COOH, NH, SH, or C-=C-H, because then the carbanion of the Grignard group would remove the acidic H and be reduced. For example: [HOCH,CH,MgBr]+ (BrMg)’ -OCH,CH,H HOCH,CH,Br + Mgunstable

(2) If the halide also has a C==O (or C=N--, B N , N 4 ,s----O, M H ) group, it reacts inter- or intramolecularly with itself. (3) The reactant cannot be a vic-dihalide, because it would undergo dehalogenation: BrCH,CH,Br + MgH,C==CH, + MgBr, (4) A ketone with two bulky R groups, e.g., ROH( 1") > ROH(2") > ROH(3") > R m H 9 RCH,

CHAP. 131

275

ALCOHOLS AND THIOLS

3. 1" and 2" alcohols have at least one H on the carbinol C and are oxidized to carbonyl compounds. They also lose H, in the presence of Cu (300°C) to give carbonyl compounds. 4. Formation of alkyl halides (Problem 7.4(a)-(d)). 5 . Intramolecular dehydration to alkenes [Problems 6.14 through 6.171. 6. Intermolecular dehydration 1to ethers. 2 R O H x ROR + H 2 0 H2S04

7. Ester formation. ROH

+ R'COH II

0

H2S04

R'COR

I1

+ H20

0 ester

8. Inorganic ester formation. With cold conc. H 2 S 0 4 , sulfate esters are formed. ROH + H,S04-

cold

H,O

+ ROS0,H-

ROH

ROS0,OR

an alkyl acid sulfate

a dialkyl sulfate

Similarly, alkyl phosphates are formed with H 3 P 0 4 , phosphoric acid. Problem 13.15 Supply equations for the formation from phosphoric acid of ( a ) alkyl phosphate esters by reactions with an alcohol, in which each acidic H is replaced and H,O is eliminated; (6) phosphoric anhydrides on heating to eliminate H,O. 4 / OH +ROH / O R +ROH O==P-OH-O=P-OH-O=P-OR-O=P--OR -HzO -HzO OH 'OH

'

Phosphoric acid

an alkyl dihydrogen phosphate

0

OH

'OH

II

0

II

+H3P0,

I

I

-HzO

OH

/OR

+ROH

'OR

-HzO

a dialkyl hydrogen phosphate

HO-P--O-P--OHOH

/OR

a trialkyl phosphate

0

0

II

II I

HO-PUP-O-P--OH

I

I

OH

OH

Diphosphoric acid

Phosphoric acid

0

II

OH

OH

Triphosphoric acid

Problem 13.16 Alkyl esters of di- and triphosphoric acid are important in biochemistry because they are stable in the aqueous medium of living cells and are hydrolyzed by enzymes to supply the energy needed for muscle :ontraction and other processes. ( a ) Give structural formulas for these esters. (6) Write equations for the hydrolysis reactions that are energy-liberating. 4

0

0

II II RO-P-O-P-OH I

OH

I

OH

an alkyl trihydrogen diphosphate

I

OH

I

OH

I

OH

an alkyl tetrahydrogen triphosphate

ALCOHOLS AND THIOLS

276

[CHAP. 13

( 6 ) Hydrolysis occurs at the ester(reaction I) or the anhydride bonds(reactions I1 and 111).

0

P /

0

11 II H 2 0+ R--O-P--O-PUP--OHI I I OH OH OH

\

I[

5;

RO-P4H OH I

I

OH

7 7

+HO-PUP-OH I I OH

I

OH

OH

I

OH

9. Alkyl sulfonates (esters of sulfonic acids), R S02RIAr) 0

RS-CI

8

+ HOR’ --+

1I 11

R S - O R ’ + HCI 0

a sulfonyl chloride

a sulfonate ester

Problem 13.17 How does sulfonate ester formation from sulfonyl chloride resemble nucleophilic displacements of alkyl halides? 4 The alcohol acts as a nucleophile and halide ion is displaced.

0 RS’-

11 /?

S-CI 0 II

-a 0

pyrldinc

11

R-S-ij-R’

+ (pyridine

H)’ C1-

Sulfonyl chlorides are prepared from the sulfonic acid or salt with PCl,.

Aromatic sulfonyl chlorides are also formed by ring chlorosulfonation with chlorosulfonic acid, HOS0,CI. ArH

+ HOS0,CI-

ArS0,CI

+ H,O

a sulfonyl chloride

Problem 13.18 ( a ) Why are sulfonate esters such as p-toluenesulfonyl (tosyl) esters widely used in S,2 reactions? ( 6 ) How is this reaction used for indirect displacement of --OH from ROH? 4 (a)

The p-toluenesulfonate anion, p-CH,C,H,SO;, like I -.

is a very weak base and is therefore a good leaving group,

(6) The alcohol is converted to a tosyl ester, which is then reacted with a nucleophile.

CHAP. 131

277

ALCOHOLS AND THIOLS

10. Oxidation. Alcohols with at least one H on the carbinol carbon (1" and 2") are oxidized to carbonyl compounds. R R RCH,OH1"

R C H S

\CHOH-+

and R

/

2"

>c=o

R

an aldehyde

a ketone

Aldehydes are further oxidized to carboxylic acids, RCOOH. To get aldehydes, milder reagents, such as the Jones (diluted chromic acid in acetone) or Collins reagent (a complex of CrO, with 2mol of pyridine), are used. Problem 13.19 Give the products for the reaction of isopropanol with conc. H,SO, at ( a ) O"C, (6) room 4 temperature, (c) 130 "C, (d) 180 "C. (a)

(CH,),CH&H,

+ HSO,

oxonium ion

(6) (CH,),CHOSO,H Isopropyl hydrogen sulfate ( c ) (CH,),CHOCH(CH,), Diisopropyl ether (d) CH,CH=CH, (high temperatures favor intramolecular dehydration) Problem 13.20 Write equations to show why alcohols cannot be used as solvents with Grignard reagents or with LiAlH,. 4

--

Strongly basic R i and HT react with weakly acidic alcohols.

n+ CH,CH,MgCl +

CH,OH

4CH,OH

+ LiAlH,

CH,CH, 4H,

+ (CH,O)-(MgCI)'

+ LiAI(OCH,),

Problem 13.21 Why may Na metal be used to remove the last traces of H,O from benzene but not from ethanol? 4

Ethanol is sufficiently acidic to react with Na, although not as vigorously as does H,O. 2C,H,OH

+ 2Na-+2C,HsO-Na' + H,

Problem 13.22 Give the main products of reaction of 1-propanol with ( a ) alkaline aq. KMnO, solution during 4 distillation; (6) hot Cu shavings; (c) CH,COOH, H'. (a) CH,CH,CHO.

Since aldehydes are oxidized further under these conditions, CH,CH,COOH is also obtained. Most of the aldehyde is removed before it can be oxidized. (6) CH,CH,CHO. The aldehyde can't be oxidized further. (c) CH,C--OCH,CH,CH,, an ester.

a

Problem 13.23 Explain the relative acidity of liquid l", 2", and 3" alcohols.

4

The order of decreasing acidity of alcohols, CH,OH > 1" > 2" > 3", is attributed to electron-releasing R's. These intensify the charge on the conjugate base, RO-, and destabilize this ion, making the acid weaker. Problem 13.24 Give simple chemical tests to distinguish ( a ) 1-pentanol and n-hexane; (6) n-butanol and t-butanol; ( c ) 1-butanol and 2-buten-1-01; (d) 1-hexanol and 1-bromohexane. 4

278

ALCOHOLS AND THIOLS

[CHAP. 13

( a ) Alcohols such as 1-pentanol dissolve in cold H,SO, [see Problem 13.19(a)]. Alkanes such as n-hexane are insoluble. (6) Unlike t-butanol (a 3" alcohol), n-butanol (a 1" alcohol) can be oxidized under mild conditions. The analytical reagent is chromic anhydride in H,SO,. A positive test is signaled when this orange-red solution turns a deep green because of the presence of Cr". (c) 2-Buten-1-01decolorizes Br, in CCI, solution; 1-butanol does not. (d) 1-Hexanol reduces orange-red CrO, to green Cr3+; alkyl halides such as 1-bromohexane do not. The halide on warming with AgNO,(EtOH) gives AgBr.

Problem 13.25 Write balanced ionic equations for the following redox reaction: CH,CHOHCH,

+ K,Cr,O, + H,SO,-CH,--CO--CH,

+ Cr,(SO,), + H 2 0 + K,SO,

hear

4

Write partial equations for the oxidation and the reduction. Then: (1) Balance charges by adding H' in acid solutions or OH- in basic solutions. (2) Balance the number of 0 ' s by adding H,O's to one side. (3) Balance the number of H's by adding H's to one side. The number added is the number of equivalents of oxidant or

-

OXIDATION (CH,),CHOH

Cr,O;-

(CH,),C=O

In acid balance charges with H ': (no change)

Cr,0,2-

+ 8H'

Balance 0 with H 2 0 : (no change)

Cr,Of-

+ 8H'

Balance H: (CH,),CHOH

-

(CH,),C=O

Balance equivalents:

Add:

--

REDUCTION

+ 2H

Cr,O:

2Cr3+

+X r 3 +

__*

Xr3'

+ 7H20

+ 8H' + 6H +2Cr3++ 7H20

3(CH3),CHOH-3(CH,),C=O + 6H Cr,O:- + 8 H t + 6H-2Cr3' + 7H20

3(CH,),CHOH

+ Cr,O:- + 8H'

-

3(CH3),C=0

+ 2Cr3' + 7H,O

Problem 13.26 How can the difference in reactivity of l", 2", and 3" alcohols with HCI be used to distinguish 4 among these kinds of alcohols, assuming the alcohols have six or less C's? The Lucas test uses conc. HCI and ZnCI, (to increase the acidity of the acid). 3" ROH + HCI+

soluble, by assumption

3' RCI + H,O insoluble liquid

The above reaction is immediate; a 2" ROH reacts within 5 min; a 1" ROH does not react at all at room temperature. Problem 13.27 In CCI, as solvent, the nmr spectrum of CH,OH shows two singlets. In (CH,),SO, there is a 4 doublet and a quartet. Explain in terms of the "slowness" of nmr detection. In CCI,, CH,OH H-bonds intermolecularly, leading to a rapid interchange of the H of 0 - H . The instrument senses an average situation and therefore there is no coupling between CH, and OH protons. In (CH,),SO, H-bonding is with solvent, and the H stays on the 0 of OH. Now coupling occurs. This technique can be used to distinguish among RCH20H, R2CHOH and R,COH, whose signals for H of OH are a triplet, a doublet and a singlet, respectively. Problem 13.28 H of OH has a chemical shift that varies with the extent of H-bonding. Why can we detect an 4 nmr signal from H of ROH by shaking the sample with D 2 0 and then retaking the spectrum? Through H-bonding the exchange reaction ROH + D,O+ ROD + DOH occurs. D-signals are not detected by a proton nmr spectrometer. If the signal in question disappears, it came from H of ROH. A new signal appears at about S =5ppm, due to DOH.

CHAP. 13)

279

ALCOHOLS AND THIOLS

13.4 SUMMARY OF ALCOHOL CHEMISTRY

PROP€RTES

PREPARATION 1. Hydration of Alkenes

1. Basicity

+ HX+

(a) Markovnikov RCH=CH2 + H 3 0 '

Acidity

2.

\

+ Na-

Oxymercuration-demercuration: RCHOHC

+ Hg(OAc),/H,O

RCHdH,

Hydrolysis

-P

2" (sN2): RCHXCH,

+ R'COOH/H'

(3")

+ conc. H,SO,=

( b ) Esters R'COOCH,CH,R

+ H,O'

Reduction of

- ;I:i

RCOOH + LiAlH, (b) Esters

R'COOCH,CH,R

+

, , RCH,CH,OH ~

+ LiAlH,

(c) Carbonyl Compounds Aldehydes (to 1" ROH):

ll

+ NaBH,

Formaldehyde

+ H,C=O

(b) Other Aldehydes

RLi + R'CH-0-

RCH(0H)R' (2" ROH)

(c) Ketones RMgX + R'C0R'-

(b) Ether Formation (Intermolecular)

\ 5.

+ H,SO,=

(RCH,CH,),O

Oxidation

\

-

(a) Aldehydes (from 1" ROH)

+ pyridine

*

CrO,

RCH,CHO

(b) Ketones (from 2" ROH)

Organometallic Reactions

RCH,MgX

(RCH,CH,O),SO,

(a) Alkene Formation (Intramolecular) H,SO,- heat RCH=CH,

(1")-

Ketones (to 2" ROH): RCOR' + NaBH, \-

(8)

R'COOCH,CH,R

c=O Group

(a) Carboxylic Acids

4.

__*

(c) Inorganic Esters (e.g., Sulfates)

RCH,CH,X

RCH,CHO

RCH,CH,X RCH,CH,Cl RCH,CH,Br RCH,CH,F

(b) Esterification

+ OH-

3" (sN1): R3CX + H,O-R,COH

3.

-

+ HX+ SOC1,+ PBr, + SF,

RCH=CH2 + 1. B,H,, 2. HO,

(a) Alkyl Halides, R

Alkoxide salt

(a) Formation of Alkyl Halides, RX (X = CI, Br, I

H ydroboration-Oxidation

1" (sN2):

RCH,CH,O- Na'

3. Displacements

(b) Anti-Markovnikov

2.

RCH2CH,0H; X- Onium salt

RR'R'COH (3" ROH)

+ KMnO,/H'

RCOR'

280

ALCOHOLS AND THIOLS

B.

[CHAP. 13

THIOLS

13.5 GENERAL

Also called mercaptans, these compounds, with the generic formula RSH, are sulfur analogs of alcohols, just as H2S is of H 2 0 . The 0 atom of the OH has been replaced by an S atom, and the -SH group is denominated sulfhydryl. A physical characteristic is their odor; 2-butene-1-thiol, CH,CH=CHCH,SH, and 3-methyl-lbutanethiol, CH,CH(CH,)CH,CH,SH, contribute to skunk smell. Thiols are biochemically important because they are oxidized with mild reagents to disulfides which are found in insulin and proteins.

=R-S-S-R Br2

2 R-S-H a thiol

Zn. H

a disulfide

Problem 13.29 How are thiols prepared in good yield?

4

The preparation of thiols by S,2 attack of nucleophilic HS- on an alkyl halide gives poor yields because the mercaptan loses a proton to form an anion, RS-, which reacts with a second molecule of alkyl halide to form a t hioether .

HS;+ R-X:

-:x:

-

+ RSIH-

-H+

-

RS:

t

~

R-9

--+

R-:+R

+ 1..x1

-

thioether

Dialkylation is minimized by using an excess of :SH-, and avoided by using thiourea to form an alkylisothiourea salt that is then hydrolyzed.

Thiourea

Urea

Problem 13.30 Show steps in the synthesis of ethyl ethanesulfonate, CH,CH2S0,0CH,CH,, from CH,CH,Br and any inorganic reagents. 4

The alkyl sulfonic acid is made by oxidizing the thiol, which in turn comes from the halide. excess

CH,CH,SH-

KMn04

CH,CH,SO,H A L ?

CH,CH,OH

+ CH,CH,SO,CI-

pyridine U

bust

CH,CH,SO,OCH,CH,

Problem 13.31 Why are mercaptans ( a ) more acidic ( K , = 10-l’) than alcohols ( K O= l0-l’) and (b) more 4 nucleophilic than alcohols? ( a ) There are more and stronger H-bonds in alcohols, thus producing an acid-weakening effect. Also, in the conjugate bases RS- and RO- the charge is more dispersed over the larger S, thereby making RS- the weaker base and RSH the stronger acid (Section 3.11). ( b ) The larger S is more easily polarized than the smaller 0 and therefore is more nucleophilic. For example, RSH participates more rapidly in S,2 reactions than ROH. Recall that among the halide anions, nucleophilicity also increases as size increases: F- < C1- < Br- < I - .

Problem 13.32 Offer mechanisms for

CH,CH=CH,

&CH,CH(SH)CH,

5Lc H , c H , c H , s H

Acid-catalyzed addition has an ionic mechanism (Markovnikov):

CH,CH=CH,

+ H + --+ CH,;HCH, Z CH, H‘ -

4

CHAP. 131

281

ALCOHOLS AND THIOLS

The peroxide- or light-catalyzed reaction has a free-radical mechanism (anti-Markovnikov):

H - S - H ~ HS. + . H

CH,CHSH,

+ *SH+

CH,CHCH2SH

C H , C H C H , S H x CH,CH,CH,SH

+ -SH

13.6 SUMMARY OF THIOL CHEMISTRY

PREPARA TlON

PROPERTIES

1. Alkyl Halide

1. Acidic Hydrogen

+ OH----,

RX + KSH or (H,N),C=S

.

2. Olefin

R'CH-CH, 3. Disulfide

R-S-S-R

/

Zn. H 2 S 0 4

+ RS-

( K , = 10-l')

2. Sulfur Atom as Nucleophile

+ H,S

+ R'--COCl-

+ R'--CH=O+ O H - ; R'X-

R'--COSR (thioester) R'--CH(SR,) (thioacetal) R-S-R' (thioether)

-

3. Oxidation

4. Aryl Sulfonyl Chloride

ArS0,Cl

H,O

+ Pb2+--+Pb( SR),

+ KMnO, --+R-SO,H (sulfonic acid) + I, R-S-S-R (disulfide)

ArSH

Supplementary Problems Problem 13.33 Give the IUPAC name for each of the following alcohols. Which are l", 2", and 3"? (U)

CH,CH3

I

(d) CH3C-CH2CH, AH

(e)

Cl-CHCH,CH=CHCH,OH

I

CH3-C-CH, L6H,- In-C 1

( a ) 2-Methyl- 1-butanol , 1". (6) 2-Methyl-3-phenyl-1-propanol, 1". (c) 1-Methyl-1-cyclopentanol, 3". (d) 3-Methyl-3-.pentanol, 3". (e) 5-Chloro-6-methyl-6-(3-chlorophenyl)-2-hepten-l-o1. (The longest chain with OH has seven C's and the prefix is hept-. Numbering begins at the end of the chain with OH; therefore, -1-01. The aromatic ring substituent has C1 at the 3 position counting from the point of attachment, and is put in parentheses to show that the entire ring is attached to the chain at C". C1 on the chain is at C5.)1".

Problem 13.34 Write condensed structural formulas and give IUPAC names for ( a ) vinylcarbinol, (6) 4 diphenylcarbinol, (c) dimethylethylcarbinol? (d) benzylcarbinol.

ALCOHOLS AND THIOLS

282

(a)

[CHAP. 13

H,C=CHCH:OH 2-Propen-1-01 (Ally1 alcohol)

( 6 ) (C,H,),CH--OH (c) (CH,),C--OH

Diphenylmethanol (Benzhydrol) 2-Methyl-2-butanol

LB H 2 W (d) C,H ,-CH,eH,OH 2-Phenylethanol ( /3 -Phenylethanol)

Problem 13.35 The four isomeric C,H,OH alcohols are (i) (CH,),COH

(ii) CH,CH,CH,CH,OH

(iii) (CH,),CHCH,OH

(iv) CH,CH(OH)CH,CH, Synthesize each, using a different reaction from among a reduction, an S,2 displacement, a hydration, and a 4 Grignard reaction. Synthesis of the 3" isomer, (i), has restrictions: the S,2 displacement of a 3" halide cannot be used because elimination would occur; nor is there any starting material that can be reduced to a 3" alcohol. Either of the two remaining methods can be used; arbitrarily the Grignard is chosen. CH,Mgl

+ (CH,),C=O-----+(CH,),COH after

hydrolvsrs

-

The S,2 displacement o n the corresponding RX is best for 1" alcohols such as (ii) and (iii); let us choose (ii) for this synthesis.

+ OH-

CH,CH,CH,CH,CI

-cI-

CH,CH,CH,CH,OH

The 1" alcohol, (iii), and the 2" alcohol, (iv) can be made by either of the two remaining syntheses. However, the one-step hydration with H,O to give (iv) is shorter than the two-step hydroboration-oxidation to give (iii). CH,CH,CH=CH,

+ H,O+ 2CH,CH(OH)CH,CH,

-

Finally, (iii) is made by reducing the corresponding RCH=O or RCOOH. (CH ,) ,CH ,COOH

I

LiAIHJ

2

Ht

(CH ,) ,CH ,CH ,OH

Problem 13.36 Identify the products from reaction of (CH,),C=O with: ( a ) NaBD, and H,O; ( 6 ) NaBD, and CH,OD. 4 ( a ) (CH,),CDOH; (6) (CH,),CDOD. In both reactions nucleophilic :D- first attacks and reduces the carbonyl to form a basic alkoxide salt:

I-

44 + D -

Problem 13.37 Prepare ethyl p-chlorophenylcarbinol by a Grignard reaction.

4

Prepare this 2" alcohol, p-CIC,H,CHOHCH,CH,, from RCHO and R'MgX. Since the groups on the carbinol C are different, there are two combinations possible: H (l)

1

p-C1C6H4C=0

, 1

CH,CH2MgCI H,U+

p-ClC,H,CHOHCH,CH, (ring Cl has not interfered)

(2) p-CICf34MgBr

I

CHICHZCHO H,Ot

+

p-CIC,H,CHOHCH2CH,

Br is more reactive than Cl when making a Grignard of p-CIC,H,Br.

CHAP. 131

283

ALCOHOLS AND THIOLS

Problem 13.38 Give the hydroboration-oxidation product from ( a ) cyclohexene, (b) ~&-2-phenyl-2-butene, (c) tram-2-phenyl-2-butene. 4 Addition of H,O is cis anti-Markovnikov. See Fig. 13-2. In Fig. 13-2(c) the second pair of conformations show the eclipsing of the H's with each other and of the Me's with each other. The more stable staggered conformations are not shown,

H OH Me

H

H

0

I

Me

Enantiomers

I

"'...p&+ ,

Ph staggered

eclipsed

\

7

#

Conformers

n

Q H Me 0

.**

Ph

Me

*:

Me

Ph

Me HO

'Enantiomers

Me

c o d bmer s

Fig. 13-2

Problem 13.39 Write a structural formula for a Grignard reagent and a carbonyl compound which react to give each of the following alcohols after hydrolysis: ( a ) 2-butanol, (6) benzyl alcohol, (c) 2,4-dimethyl-3pentanol, (d) t-butanol. 4 The carbinol C and all but one of its R or Ar groups come from the carbonyl compound. The substituent from the Grignard reagent is written in a box below.

284

ALCOHOLS AND THIOLS

[CHAP. 13

( a ) Two possible combinations of reactants are:

H

OH

I 7 CH,CH,C-m I

I

I

CH,CH,C=O

W p B r

A

H

OH

CH,C=O I

I

r w ] M s B r

2

H3O'

CH,-C-[-I

I I

H

I CH,] m h l p B r

I

CH,-C-CH,

II

I

CH,-C-CH,

I

OH

0

Problem 13.40 The following Grignard reagents and aldehydes or ketones are reacted and the products hydrolyzed. What alcohol is produced in each case'? ( a ) Benzaldehyde (C,H,CH=O) and C,H,MgBr. (b) Acetaldehyde and phenyl magnesium bromide. (c) Acetone and benzyl magnesium bromide. ( d ) Formaldehyde and cyclohexyl magnesium bromide. (e) Acetophenone (C,H,CCH 3 ) and ethyl magnesium bromide. 4

II

C6HTCH=0

+ C,H,MgBr

8

---

H

I

C,H,T-OH

1-Phenyl-1-propanol

C,H, H CH,CH=O

I

+ C,H,MgBr-C,H,

1-Phenyl- 1-ethano1

CH,

+ C,H,CH,MgBr-

CH,--C--CH,

II

C,H,CH,

I

-F""

l-Phenyl-2-methyl-2-propanol

OH

0 H,C=O

C,H,-C-CH,

b

+ C,H

I

,MgBr --+C6HI ,CH,OH

+ C,H,MgBr

I

__*

Cyclohexylcarbinol

2-Phenyl-2-butanol

CHAP. 131

285

ALCOHOLS AND THIOLS

Problem 13.41 Give the mechanism in each case:

CH, ( a) CH,CH,CH,CH,CH,OH

aCH,CH,CH,CH2CH2Cl

I

3 CH,CCH,CH,

( 6 ) CH,

AI CH3 ( c ) CH,-&-CH2CH3

-!% C H , - ~ ~ C H , C H ,

AH

&I

Why did rearrangement occur only in (b)?

4

( a ) The mechanism is S,2 since we are substituting C1 for H,O from 1" ROH,' (6) The mechanism is S,1.

CH,

I CH,-CH-CH-CH, AH

CH3

I CH3-C-CH-CH3 I +

CH,

I

3 CH,-C-CH,-CH,

HJ

2" R' (less stable)

CI

~

CH3

I

CH,-C-CH,-CH,

AI

3" R' (more stable)

+

( c ) S,1 mechanism. The stable 3" (CH,),CCH,CH, reacts with CI- with no rearrangement. Problem 13.42 Why does dehydration of 1-phenyl-2-propanolin acid form 1-phenyl-1-propene rather than 4 1-phenyl-2-propene?

1-Phenyl-1-propene, PhCH=CHCH,, is a more highly substituted alkene and therefore more stable than l-phenyl-2-propene, PhCH,CH=CH2. Even more important, it is more stable because the double bond is conjugated with the ring. Problem 13.43 Write the structural formulas for the alcohols formed by oxymercuration-demercuration from ( a ) 1-heptene, (6) 1-methylcyclohexene, (c) 3,3-dimethyl-l-butene. 4

The net addition of H,O is Markovnikov. CH,( CH, ),CHOHCH,

(CH,),CCHOHCH,

2-Heptanol

3,3-Dimethyl-2-butanol (no rearrangement occurs)

Problem 13.44 List the alcohols and acids that compose the inorganic esters ( a ) (CH,),COCl and (b) 4 CH,CH,ONO,. Name the esters.

Conceptually hydrolyze the 0 to the heteroatom bond while adding an H to the 0 and an OH to the heteroatom. ( a ) (CH,),COH and HOCl, t-butyl hypochlorite. (6) CH,CH,OH and HONO,, ethyl nitrate. Tert-butyl hypochlorite is used to chlorinate hydrocarbons by free-radical chain mechanisms. Problem 13.45 Starting with isopropyl alcohol as the only available organic compound, prepare 2,3-dimethyl2-butanol. 4

This 3" alcohol, (CH,),COHCH(CH,),,

is prepared from a Grignard reagent and a ketone.

ALCOHOLS A N D THIOLS

286

[CHAP. 13

CH, CH,

I

t

CH,-C4H-CH3=

C1

AMgBr -+ CH3CHCH3

CH, CH,

CH,CHCH3

I

I I

1

I

CH,C----CH-CH,

MgBr

Br

+ Mg” + Br-

OH Problem 13.46 Use formaldehyde and 2-butanol to prepare 1,2-dibrom0-2-methylbutane. The product, BrCH,CBr(CH,)CH,CH,,

4

is a vic-dibromide formed from the alkene 2-methyl-l-butene,

CH3 CH,CHL=CH,

which must be made from

CH3

I

CH,CH,CHCH,OH which in turn is prepared as follows:

CH,-CH,-CH-CH,

CH3CH2CHCH,

AH

ez*CH,CH,CHCH,

HCHO

MgBr I

Ar

CH,CH,CHCH, h,oH

2-Methyl- 1-butanol

However, dehydration of 2-methyl-1-butanol with H,SO, forms 2-methyl-2-butene by carbocation rearrangement.

CH,CH,CHCH,OH

:Hy

CH,CH,CHCH:a

LH3

CH3CH26-CH3

LH3

2CH,CH=C-CH,

LH,

LH3 2-Meth yt-2-butene

The needed alkene is prepared by dehydrohalogenation and then brominated.

Problem 13.47 Alcohols such as Ph,CHCH,OH rearrange on treatment with acid; they can be dehydrated by heating their methyl xanthates (Tschugaev reaction). The pyrolysis proceeds by a cyclic transition state. Outline the steps using Ph,CHCH,OH. 4

S

I

Ph,CHCH,OH 5 Ph,CHCH,O-Na’

Ph,CHCH,-0-C-S-Na’ a xanthate Ph,C

CH,-m,

IT

HLsf-scH3 an s-methylxanthate

The elimination is cis.

CH I

4

Ph,C=CH,

+ &C-0

+ HSCH,

CHAP. 131

287

ALCOHOLS AND THIOLS

F R I N:y*

Problem 13.48 Methyl ketones gives the haloform test: CHI,

+ RCOO-Na'

-

I

a methyl ketone

yellow precipitate

1, can also oxidize 1" and 2" alcohols to carbonyl compounds. Which butyl alcohols give a positive haloform test?

4

Alcohols with the

groups are oxidized to

cH31i-R

and give a positive test. The only butyl alcohol giving a positive test is

H

t:

CH, -CH2CH3 AH Problem 13.49 A compound, C,H,,O, is oxidized under vigorous conditions to benzoic acid. It reacts with CrO, and gives a positive iodoform test (Problem 13.48). Is this compound chiral? 4 Since benzoic acid is the product of oxidation, the compound is a monosubstituted benzene, C,H,G. Subtracting C,H, from C,H,,O gives C,H,O as the formula for a saturated side chain. A positive CrO, test means a 1" or 2" OH. Possible structures are:

CH

'-1

H H H C-C-CH H H OH

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

H

OH

I

H

H H H C,H,C-c-C-oH H H H

H

c6H5-

kH3

111

I1

YCH20H IV

Only I1 has the - C H( OH) C H, needed for a positive iodoform test. 11 is chiral.

Problem 13.50 Suggest a possible industrial preparation for ( a ) t-butyl alcohol, ( b ) ally1 alcohol, (c) glycerol (HOCH,CHOHCH,OH). 4 CH3 CH3 CH3 ( a)

I

';;rt bH2 + C H 3 - -IC = C H 2 7

CHJ-CH-CH, Isobutane

(b)

Propene

CICH2-CH=CH, Allyl chloride

AH

Isobut ylene CI

CH,-CH=CH,

ClCH2-CH=CH2

HOCH,CH=CH,

Allyl chloride

+

\

Allyl alcohol

CH2-C H-C H2

I

CI

I

I

CH3-C-CH3

I

OH C1

ALCOHOLS AND-TH!OLS

288

[CHAP. 13

Problem 13.51 Assign numbers, from 1 for the lowest to 5 for the highest, to indicate relative reactivity with HBr in forming benzyl bromides from the following benzyl alcohols: (a) p-CI--C,H,--CH,OH, (b) (C,H,),CHOH. (c) P-@,N--C,H,--CH,OH, ( d ) (C,H,);COH, ( e ) C,HsCH,OH. 4 Differences in reaction rates depend on the relative abilities of the protonated alcohols to lose H,O to form The stability of R' affects ihe A H S for forming the incipient R' in the transition state and determines the overall rate. Electron-attracting groups such as NO, and CI in tne par@ position destabilize R' by intensifying the positive charge. NO, of (c) is more effective s k e it destabilizes by both resonance and induction, while CI of (a) destabilizes only by induction. The more C,H,'s on the benzyl C, the more stable is R'.

I?'.

(0)

2

!cl

(b) 4

(4

1

(4 3

5

Problem 13.52 Supply structural formulas and stereochemical designations foi the organic compounds (A) through (H). H O ( a ) (A) + conc. H , S O , ~ C H , - C H C H , C H , ~ (B) bS03H H O

( b ) (R)-CH3CHOHCH2CH=CH2

(C) + (D)

+ KEr --+ (E) + (F) + ( G )

(c) (R)-CH3CH2-CH-CH=CH2

I OH

4

(A) is cis- or tram-CH,CH=CHCH, or H,C=CHCH,CH,; (B) is rac-CH,CHOHCH,CH,. The intermediate CH,C+HCH,CH, can be attacked from either side by HSOd to give afi optically inactive racemic hydrogen sulfate ester which is hydrolyzed to rac-2-butanol. (C) is (R,R)-CH,-T

cH 2 T C

H

3

$7

(D)is (R,S)-CH,

;

OH

CH2-C-CH,

H

H

m eso

optically active

Hydration follows Markovnikov's rule, and a new similar chiral C is formed. The original chiral C remains R, but the new chiral C may be R or S. The two diastereomers are not formed in equal amounts [see Problem 5.22(b)]. (E) is rac-CH,CH,CHBrCH=CH,; (F) is tram- and (G) cis-CH,CH2CH=CHCH,Br. The intermediate R' in this S, 1 reaction is a resonance-stabilized (charge-delocalized) allylic cation +

t

[CH,CH,CHCH=CH,

c---*

CH,CH,CH=CHCH,]

6+

or CH,CH,CH=CH-CH,

6+

+

~-

In the second step, Br- attacks either 05 the positively charged C's to give one of the three products. Since R' is flat the chiral C in (E) can be R or S; (E) is racernic. (F) is the major product because it is most stable (tram and disubstituted).

Problem 13.53 -How does the Lewis theory of acids and bases explain the functions of (a) ZnCl, in the Lucas 4 reagent? (b) ether as a solvent in the Grignard reagent? ZnC1, Lewis acid

+ 2 HCI

__*

ROH

H2ZnCl, --+ ROH;

(a stronger acid than HCI)

RCI

+ H20

R'MgX acts as a Lewis acid because Mg can coordinate with one unshared electron pair of each 0 of two ether molecules to form an addition compound,

R'

R I !! :OtMg:Q: R A R that is soluble in ether.

CHAP. 131

289

ALCOHOLS AND THIOLS

Problem 13.54 Draw Newman projections of the conformers of the following substituted ethanols and predict 4 their relative populations: ( a ) FCH,CH,OH, ( 6 ) H,NCH,CH,OH, (c) BrCH,CH,OH.

If the substituents F, H2N, and Br are designated by Z, the conformers may be generalized as anti or gauche.

gauche

anti

For ( a ) and (b) the gauche is the more stable conformer and has a greater population because of H-bonding with F and N. The anti conformer is more stable in ( c ) because there is no H-bonding with Br and dipole-dipole repulsion causes OH and Br to lie as far from each other as possible. Problem 13.55 Deduce the structure of a compound, C,H,,O, which gives the following nmr data: S = 0.8 (doublet, six H's), 6 = 1.7 (complex multiplet, one H), 6 = 3.2 (doublet, two H's) and S = 4.2 (singlet, one H; 4 disappears after shaking sample with D 2 0 ) .

The singlet at S = 4.2 which disappears after shaking with D 2 0 is for OH (Problem 13.28). The compound must be one of the four butyl alcohols. Only isobutyl alcohol, (CH,),CHCH,OH, has six equivalent H's (two CH,'s), accounting for the six-H doublet at S = 0.8, the one-H multiplet at S = 1.7, and the two-H doublet which is further downfield at S = 3.2 because of the electron-attracting 0. Problem 13.56 The attempt to remove water from ethanol by fractional distillation gives 95% ethanol, an

azeotrope that boils at a constant temperature of 78.15 "C. It has a lower boiling point than either water (100 "C)

or ethanol (78.3"C). A liquid mixture is an azeotrope if it gives a vapor of the same composition. How does 4 boiling 95% ethanol with Mg remove the remaining H 2 0 ? Mg

+ H,O

-

H,

+ Mg(OH), insoluble

The dry ethanol, called absolute, is now distilled from the insoluble Mg(OH),. Problem 13.57 Explain why the most prominent (base) peak of 1-propanol is at mle = 31, while that of allyl alcohol is at m / e = 57. 4 CH,CH,--CH,OH + cleaves mainly into +

CH,CH2- + 6 H 2 0 Hc-----,H2C=0H

( m / e = 31) rather than CH,CH,CHOH

+ H., because C - C is weaker

than C-H.

In allyl alcohol,

H

I

CH,=CHC-OH

I

H the C-H

bond cleaves to give

CH,=CHc-OH

I

H ( m l e = 57). This cation is stabilized by both the CH2=CH and the 0.

I

Problem 13.58 Inorganic acids such as H 2 S 0 , , H,PO,, and HOCl (hypochlorous acid) form esters. Write structural formulas for ( a ) dimethyl sulfate, ( 6 ) tribenzyl phosphate, (c) diphenyl hydrogen phosphate, (d) t-butyl nitrite, ( e ) lauryl hydrogen sulfate (lauryl alcohol is n-C,,H,,CH,OH), ( f ) sodium lauryl sulfate. 4

ALCOHOLS AND THIOLS

290

[CHAP. 13

Replacing the H of the OH of an acid gives an ester.

0

0

(U)

I CH,O-S-WH,

I

(6) (PhCH,O),P-O

A

(c) (PhO),P-OH

( d ) (CH,),CONO

0 (e)

I I 0

n-C, ,H,,CH,OS-OH

Cf)

n-C, ,H,,CH,OSO,O-Na'

(a detergent)

Problem 13.59 Dialkyl sulfates are good alkylating agents. ( a ) Write an equation for the reaction of dimethyl sulfate with C,H,OH in NaOH. (6) Explain why dimethyl sulfate is a good methylating agent. 4

( b ) Methyl sulfate anion (CH,OSO,) is the conjugate base of a strong acid, methyl sulfuric acid (CH,OSO,H). It is a very weak base and a good leaving group.

Problem 13.60 Explain why trialkyl phosphates are readily hydrolyzed with OH- to dialkyl phosphate salts, 4 whereas dialkyl hydrogen phosphates and alkyl dihydrogen phosphates resist alkaline hydrolysis.

n

HO-

OR

I + RO=P=O+HOR I

+

OR

dialkyl phosphate

-

HO-

1

OR

+ HO=P=O+ I

H20 +

OR

OH n~

2HO-

+ HO=P--C\---* bR

0

4

2H20 + RO-P-O

\

The hydrogen phosphates are moderately strong acids and react with bases to form anions (conjugate bases). Repulsion between negatively charged species prevents further reaction between these anions and OH-. Problem 13.61 The ir spectrum of RSH shows a weak -S-H stretching band at about 2600cm-' that does not shift significantly with concentration or nature, of solvent. Explain the difference in behavior of S - H and 0-H bonds. 4

The S-H bond is weaker than the O - H bond and therefore absorbs at a lower frequency. There is little or no H-bonding in S - H and, unlike the 0-H bond, there is little shifting of absorption frequency on dilution. Problem 13.62 Write structural formulas with stereochemical designations for the reaction products I and 11.

CHAP. 131

ALCOHOLS AND THIOLS

H

I

I = (S)-TsO--C--CH,

I

(no change in configuration)

H I1

I I

(R)-CH,-C-SH

(S,2 inversion)

29 1

Chapter 14 Ethers, Epoxides, Glycols, and Thioethers A.

ETHERS

14.1 INTRODUCTION AND NOMENCLATURE

Simple (symmetrical) ethers have the general formula R--O-R or Ar--O--Ar; mixed (unsymmetrior Ar-0-Ar' or Ar-0-R. The derived system names R and Ar as cal) ethers are R-0-R' separate words and adds the word "ether." In the IUPAC system, ethers (ROR) are named as alkoxy- (RO-) substituted alkanes. Cyclic ethers have at least one 0 in a ring. Problem 14.1 Give a derived and IUPAC name for the following ethers: (a) CH,OCH,CH,CH,CH,, (CH ,) ,CHOCH( CH ,)CH ,CH ,,(c) C,H,OCH,CH, , (d) p-NO,C,H,OCH , (e) CH,OCH ,CH,OCH, .

,

(6) 4

(a) Methyl n-butyl ether, 1-methoxybutane. (6) sec-Butyl isopropyl ether, 2-isopropoxybutane. (Select the longest chain of C's as the alkane root.) (c) Ethyl phenyl ether, ethoxybenzene (commonly called phenetole). (d) Methyl p- nitrophenyl ether, 4-nitromethoxybenzene (or p-nitroanisole). (e) 1,2-dimethoxyethane. Problem 14.2 Account for the following. (a) Ethers have significant dipole moments (-1.18 D). (6) Ethers have lower boiling points than isomeric alcohols. (c) The water solubilities of isomeric ethers and alcohols are comparable. 4 ( a ) The C - 0 - C bond angle is about 110" and the dipole moments of the two C-0 bonds do not cancel. (6) The absence of OH in ethers precludes H-bonding and therefore there is no strong intermolecular force of attraction between ether molecules as there is between alcohol molecules. The weak polarity of ethers has no appreciable effect. (c) The 0 of ethers is able to undergo H-bonding with H of H,O.

14.2 PREPARATION

SIMPLE ETHERS 1.

Intermolecular Dehydration of Alcohols (see Section 13.3)

2. 2" Alkyl Halides with Silver Oxide 2(CH,),CHCI

+ Ag,O-

(CH,),CHOCH(CH,),

+ 2AgCI

MIXED ETHERS 1.

Williamson Synthesis

R'OR a1koxide

+ Na' + X-

X = C1, Br, I, OS02R, OS02Ar

1"

Problem 14.3 Specify and account for your choice of an alkoxide and an alkyl halide to prepare the following 4 ethers by the Williamson reaction: (a) C,H,OC(CH,),, (6) (CH,),CHOCH,CH=CH,. (a) C,H,X

+ Na'

-OC(CH,),

(6) (CH,),CHO - N a t

292

+ XCH,CH=CH,

CHAP. 141

293

ETHERS, EPOXIDES, GLYCOLS, AND THIOETHERS

The 2" and 3" alkyl halides readily undergo E2 eliminations with strongly basic alkoxides to form alkenes. Hence, to prepare mixed ethers such as ( a ) and (6) the 1" alkyl groups should come from RX and the 2" and 3" alkyl groups should come from the alkoxide. Problem 14.4 Show how dimethyl sulfate, MeOSO,OMe, is used in place of alkyl halides in the Williamson 4 syntheses of methyl ethers.

Alkyl sulfates are conjugate bases of the very strongly acidic alkyl sulfuric acids and are very good leaving groups. Dimethyl sulfate is less expensive than CH,I, the only liquid methyl halide at room temperature. Liquids are easier to use than gases in laboratory syntheses.

CF

: R Q : T + 'Me:OSO,OMe

ROMe + MeOSO,

Dimethyl sulfate (very toxic)

2.

Alkoxymercuration-demercuration (Section 13.2 for ROH preparation)

Whereas mercuration-demercuration of alkenes in the presence of water gives alcohols, in alcohol solvents (free from H,O) ethers result. These reactions in the presence of nucleophilic solvents such as water and alcohols are examples of solvomercuration. The mercuric salts usually used are the acetate, Hg(OAc), (-OAc is an abbreviation for - O C C H , ) or the trifluoracetate,

-

Hg(OCOCF, 1,. RCH=CH2

+ R'OH + Hg(OC-CF,), II

tl

RCHCH2

I I

NaBH,

RCHCH,

I I

R'O HgOCOCF, R'O H

0

Problem 14.5 Suggest a mechanism consistent with the following observations for the solvomercuration of RCH=CH, with R'OH in the presence of Hg(OAc),, leading to the formation of RCH(OR')CH,Hg(OAc): (i) no rearrangement, ( i i ) Markovnikov addition, (iii) anti addition, and (iv) reaction with nucleophilic solvents. 4

Absence of rearrangement excludes a carbocation intermediate. Stereoselective anti addition is reminiscent of a bromonium-ion-type intermediate, in this case a three-membered ring mercurinium ion. The anti regioselective Markovnikov addition requires an S, 1-type backside attack by the nucleophilic solvent on the more substituted C of the ring-the C bearing more of the partial + charge (6 '). Frontside attack is blocked by the large HgOAc group. OR' OAc RCH=CH,

/

+ Hg-

\

OAC

I

-HOR'

RCH RCH-0

7 R2C=0

Account for this order in terms of steric and electronic factors.

> RC Y

A-

4

A change from a trigonal sp2 to a tetrahedral sp3 C in the transition state is accompanied by crowding of the four groups on C. Crowding and destabilization of the transition state is in the order

CH,=OcHCH*-H

CHAP. 15)

331

CARBONYL COMPOUNDS: ALDEHYDES AND KETONES

Step 2 Alkylation of W H :

Step 3 Unmasking of C=O group:

L;

CHCH,mCH,=

OCHCH,-CH,

+ HO(CH,),OH

Problem 15.37 In acid, most aldehydes form nonisolable hydrates (gem-diols). Two exceptions are the stable chloral hydrate, CI,CCH( OH), , and ninhydrin,

( a ) Given the bond energies 749, 464 and 360kJ/mol for

c----O, 0-H and C 4 , respectively, show why the 4 equilibrium typically lies toward the carbonyl compound. (6) Account for the exceptions. Calculating AH for

we obtain (749+ 2(464)]+[2(-360)+ 2(-464)]=AH (C-0) (O-H) cleavages

(C-0)

(O-H)

formations

cndothcnnic

exothcnnic

or AH = +29 kJ/mol. Hydrate formation is endothermic and not favored. The carbonyl side is also favored by entropy because two molecules, 'C=O

/

and H,O

are more random than 1 gem-diol molecule. Strong electron-withdrawinggroups on an a C destabilize an adjacent carbonyl group because of repulsion of adjacent + charges. Hydrate formation overcomes the forces 'of repulsion.

c1

+a+

a+

C1

H

Cl+-y-(f=O

npulsion fmm adjacent + chagcs

a-

Cl

OH

+ H,O ~ Z C L A - A - ( ) H I

(!I H Chloral hydrate less npulsion

Hydration of the middle carbonyl group of ninhydrin removes both pairs of repulsions.

'robkm 15.38 Write structures for the cyclic ketals or acetals prepared from ( a ) butanal 3) cyclohexanone + ethylene glycol.

+ 1,3-propanediol, 4

CARBONYL COMPOUNDS: ALDEHYDES AND KETONES

332

I

Problem 15.39 Convert --C=O

[CHAP. 15

to --CH2- via formation of an intermediate thioacetal or thioketal.

4

a thioketal

Problem 15.40 Show steps in the synthesis of cyclooctyne, the smallest ring with a triple bond, from 4 C2H ,OOC( CH ),COOC2H ? .

The IJ-diester is converted to an eight-membered ring acyloin, which is then changed to the alkyne.

0 0 C,H,-O-C-(CH~,-C-OC,H, It II

-%

0;". zz: 0a OH Cyclooctanone

an acyloin

1,l-Dichlorocyclooctane

Cyclooctyne (major)

1,2-CycIooctadiene (minor), an allene

15.6 ATTACK BY YLIDES; WlTTlG REACTION A carbanion C can form a p-d 7r bond (Section 3.11) with an adjacent P or S. The resulting charge delocalization is especially effective if P or S, furnishing the empty d orbital, also has a + charge. Carbanions with these characteristics are called ylides, e.g.

I+ I

-p-c---

..-

I

I

=c--

I

..-

-s-c-..+ I

I

c--*

-s=cI

I

The Wittig reaction uses P ylides to change 0 of the carbonyl group to

The carbanion portion of the ylide replaces the 0.

The ylide is prepared in two steps from RX.

n

Ph3P: + R C H 2 y X -% [Ph3kH,R] Xa phosphine

C,H+i+

Ph,;CHR

+ C,Hlo + LiX-

CHAP. 151

CARBONYL COMPOUNDS: ALDEHYDES AND KETONES

333

Sulfur ylides react with aldehydes and ketones to form epoxides (oxiranes):

H

H

The sulfur ylide is formed from the sulfonium salt,

with a strong base, such as sodium dimethyloxosulfonium methylide,

[ 7.1 CH3S-CH2

Na’

Problem 15.41 Which alkenes are formed from the following ylide-carbonyl compound pairs? ( a ) 2-butanone and CH,CH,CH,CH=P(C,H,),, (6) acetophenone and (C,H,),P=CH,, (c) benzaldehyde and C,H,--CH=P(C,H,), , (d) cyclohexanone and (C,H,),P=C(CH,),. (Disregard stereochemistry.) 4

The boxed portions below come from the ylide.

CH3

CH3

I

I

(a) CH,CH2C=ICHCH,CH2CH3]

( b ) C,H,-C=m\ n

Problem 15.42 Give structures of the ylide and carbonyl compound needed to prepare:

H +-

Ph,PCHCH3

I

+ C,H,C=O

H or

I

+ CH3C=0

Ph&C,H,

The cis-tram geometry of the alkene is influenced by the nature of the substituents, solvent and dissolved salts. Polar protic or aprotic solvents favor the cis isomer.

(61 (c1

0

+ Ph3iCH,

+ Ph,iCHC,H,

or

0

or

iPh,

+ 0-CH,

CH,CH,’CH(CH,), +!Ph,

(d)

(CH,),C=O

+

-

+ Ph,S-C(CHJ,

+ C,H,CHO

CARBONYL COMPOUNDS: ALDEHYDES AND KETONES

334

15.7 MISCELLANEOUS REACTIONS 1. Conversion to Dihalides

0

0

+

PClJ

[CHAP. 15

- m;

+ POC19

2. Reformatsky Reaction

Ketones or aldehydes can be reacted to form P-hydroxyesters.

R R'

'C-0

+ Zn + Brk-COOEt I

R 1. dryether 2. H1O

\

via

OH a p - hydroxyester

R' and R may also be H or Ar. Problem 15.43 Use the Reformatsky reaction to prepare (a ) (CH,)2C(OH)CH2COOC2H,

(b) PhC(OH)CHCOOC,H,

&H, LH,

( c ) PhC=CCOOH

4

LH, LH,

The structure in the box comes from the carbonyl compound (acceptor); the structure in the oval comes from the a - bromoester (carbanion source).

@ z C a

- EH, CH,-&

+ Zn + BrCH,COOC,H,

( b)

( c1

Product from ( b) .* PhC-C-COOH $

I

I

CH, CH,

3. Reactions of the Aldehydic H The chemistry of the aldehydic H,except for oxidation to OH, is meager. The C-H bond can be homolytically cleaved by participation of a free radical.

CARBONYL COMPOUNDS: ALDEHYDES AND KETONES

CHAP. 151

335

Problem 15.44 Propanal reacts with 1-butene in the presence of uv or free-radical initiators (peroxides, sources of RO.) to give CH,CH,COCH,CH,CH,CH,. Give steps for a likely mechanism. 4

7

The net reaction is in addition of H

I

RO. 2 CH,CH,C=O

Step 1

CH,CH,C=O

Step 2

r ? n CH,CH,C- F + H,CII-CHCH,CH,

Step 3

CH,CH,COCH,CHCH,CH,

II

-+

CH,CH,C-CH,CHCH,CH,

r-3+ CH,CH,

8

-0

-

CH,CH,COCH,CH,CH,CH,

+ CH3CH2C=0

Step 1 is the initiation step. Steps 2 and 3 propagate the chain.

Supplementary Problems Problem 15.45 ( a ) What properties identify a carbonyl group of aldehydes and ketones? (b) How can 4 aldehydes and ketones be distinguished? ( a ) A carbonyl group (1) forms derivatives with substituted ammonia compounds such as H,NOH, (2) forms sodium bisulfite adduct with NaHSO,, (3) shows strong ir absorption at 1690-1760cm-' (c=O stretching frequency), (4) shows weak n-v* absorption in uv at 289nm. ( 6 ) The H - C bond in RCHO has a unique ir absorption at 2720 cm-'. In nmr the H of CHO has a very downfield peak at S = 9-10 ppm. RCHO gives a positive Tollens' test.

Problem 15.46 What are the similarities and differences between C=O and C = C bonds?

4

Both undergo addition reactions. They differ in that the C of C=O is more electrophilic than a C of C=C, because 0 is more electronegative than C. Consequently, the C of C=O reacts with nucleophiles. The C = C is nucleophilic and adds mainly electrophiles. Problem 15.47 Give another acceptable name for each of the following: ( a ) dimethyl ketone, (b) l-phenyl-24 butanone, (c) ethyl isopropyl ketone, (d) dibenzyl ketone, (e) vinyl ethyl ketone. ( a ) acetone or propanone, (b) benzyl ethyl ketone, (c) 2-methyl-3-pentanone, (d) 1,3-diphenyl-2-propanone, (e) 1-penten-3-one.

Problem 15.48 Identify the substances (I) through (V).

(I) + H,

Pd (BsSO

CH,

I

(CH,),CH-CHO

+ NaOI

CH,-C-C-CH, &H3

)

! ! I

-

(11) + (111) (continued on pg. 338)

336

[CHAP. 15

CARBONYL COMPOUNDS: ALDEHYDES AND KETONES

15.8 SUMMARY OF ALDEHYDE CHEMISTRY

PREPARATION

PROPERTIES

1. Aliphatic Aldehydes

1. Aliphatic Aldehydes

-

(a) Oxidation

( a ) Carbonyl Addition

1" Alcohols:

+ H, Pd CH,CH,CH,OH + HCN+ CH,CH,CH(OH)CN + R M g X - - + CH,CH,CHR(OMgX) + 2ROH (HX)+ CH,CH,CH(OR),

CH,CH,CH,OH 1" Alkyl Halides: DMSO

CH,CH,CH21 Vinyl boranes: CH,CH=CH,

base

3(CH,CH,CH,),B

( b ) Hydrolysis NaOH

CH,CH,CHX,

+ Ar,P=CR-

1

CH,CH,CH=CR

-

( 6 ) Carbonyl Oxygen Replacement

+ PCl,

CH,CH,CHCl,

(c) Reduction CH ,CH,COCI

CH,CH,COOH

Pd I B a S 0 4 / H,

( d ) Free Radical ( R L O ) + Alkene

CH ,CH,COOR LIAIH(~-BuO),

CH,CH,CN

+ Ag

(with H,C=CH,)

2.

2. Aromatic Aldehydes

Aromatic Aldehydes

(a) Reduction

(a) Oxidation ArCH,OH + CrO, (Ac,O) ArCH,X + DMSO (NaOH)

/

\

/

\\ \

( b ) Hydrolysis x2

ArCH, --+ArCHX,

\\

NaOH

+Ar-CH-0

/As

+ H,/Pd-

ArCH,OH

( b ) Carbonyl Addition with aliphatic aldehydes, with HCN, H,NOH, RMgX, etc.

(c) Oxidation ArCOX + H, or ArCOX

( d ) Formylation

RC,H5

'I I

+ LiAIH(t-BuO),

+ CO + HCl (AlC1,)

ArCOOH

\+

( d ) Canniuaro Reaction

\

NaOH-

ArCH,OH

+ ArCOOH

1

337

CARBONYL COMPOUNDS: ALDEHYDES AND KETONES

CHAP. 15)

SUMMARY OF KETONE CHEMISTRY PREPARAnoN

PROPERTIES

Aliphatic and Aromatic Ketones

1. Additions to Carbonyl Group

C,H,CHOHCH, ArCHOHCH,

(a) Oxidation

ArCHOHCH,

CN-

C,H,CHOH(CN)CH, ArCHOH(CN)CH, a cyanohydrin

C,H,CR(OH)CH,

0

Ar-H ( c ) Acid-Derivative Reduction (C H , ) ~ L I C ~

ArCOCl H,C=CH ArH

9

C,H,CCH or ArCCH,

, +

( C,H,C(=NOH)CH, ArC(=NOH)CH,

H2NoH

an oxime

EtC( OH)(SO,Na)CH, + Name, + ArC(OH)(SO,Na)CH,

2

{ Arc( OR),CH, EtC(OR),CH,

( d ) From 1,3=Dithianes

Reduction to Methylene

RX + R’X (two steps) 2-Ar-1,3-dithiane + RX

\\ \\

3. Oxidation

Alicyclic Ketones (a) Oxidation

0

-OH?

CH3-

Cr207

CH,-

BaO heat

-

(CH,),C=O

-;F -,so

7’

(c) Acyloin Reaction E t O O C ( C H , ) , , C O O E t ~(CH,),C=O

‘ 4 H o H

+ NaOX-CHX,

+

C,H,COOH ArCOOH

(cyclic) 4.

+ CH,COOH

Replacement Reactions

(a) Formation of gem-Dichlorides

(d ) From Cycyloalkenes

Alkylborane Oxidation:

C, H, COONa

( 6 ) Strong Oxidants

( 6 ) Decarboxylation of Dicarboxylic Acids

HOOC(CH,),COOH-

(a) Haloform for Methyl Ketones

(6) Wittig Reaction +

ArJP=CR2

----*

338

CARBONYL COMPOUNDS: ALDEHYDES AND KETONES

(IV) + H 2 0

HaSO H S O ,

CH3CH2-C-CH3

II

CH2CH,

1('

"2.304 -H20

[CHAP. 15

0

' C H 3 C H 2 - 1 IK i - c H 2 c H 3

CH3 (a)

(I)

(CH,),CHC=O

I

( b ) CH3-C-COO-Na'

cI1

(II), CHI, (111)

AH3

( c ) H-C-C-CH,CH,

(IV)

or CH,-C-C-CH,

( d ) (CH,CH2),C-C(CH2CH,), AH

(V)

AH

Problem 15.49 By rapid test-tube reactions distinguish between ( a ) pentanal and diethyl ketone. (b) diethyl ketone and methyl n-propyl ketone, ( c ) pentanal and 2,2-dimethylpropanal, ( d ) 2-pentanol and 2-pentanone. 4

( a ) Pentanal, an aldehyde, gives a positive Tollens' test (Ag mirror). (b) Only the methyl ketone gives CHI, (yellow precipitate) on treatment with NaOI (iodoform test). ( c ) Unlike pentanal, 2,2-dimethylpropanal has no a H and so does not undergo an aldol condensation. Pentanal in base gives a colored solution. ( d ) Only the ketone 2-pentanone gives a solid oxime with H,NOH. Additionally, 2-pentanol is oxidized by CrO, (color change is from orange-red to green). Both give a positive iodoform test.

Problem 15.50 Show steps in the following syntheses: ( a ) acetyl chloride to acetal (CH,CH(OC,H,),), ( b ) ally1 chloride to acrolein (propenal), ( c ) ethanol to 2-butene. 4

(4

(4

H,C=CHCH,Cl CH,CH,OH

CH,CH,Br

CH SOCH

1.

&CH,CHO

Ph3P

H,C=CHCHO CH,CH=PPh,

Problem 15.51 Use benzene and any aliphatic and inorganic compounds to prepare ( a ) 1,l-diphenylethanol, (6) 4,4-diphenyl-3-hexanone. 4 (a) The desired 3" alcohol is made by reaction of a Grignard with a ketone by two possible combinations:

(C,H,),CO

+ CH,MgBr

or C,H,COCH,

+ C,H,MgBr

Since it is easier to make C,H,COCH, than (C,H,),CO from C,H,, the latter pair is used. Benzene is used to prepare both intermediate products.

CARBONYL COMPOUNDS: ALDEHYDES AND KETONES

CHAP. 151

339

4”

(6) The 4” C of CH,CH,COCPh,CH,CH, is adjacent to C = O , and this suggests a pinacol rearrangement of CH,CH2CPh-CPhCH2CH3 AH

AH

which is made from CH,CH,COPh as follows: C,H,

+ CH,CH,COCI

Ph

I

aPhCOCH,CH,

CH,CH,-i;i;CH,CH,

Ph

I

product

Problem 15.52 Use butyl alcohols and any inorganic materials to prepare 2-methyl-4-heptanone.

4

The indicated bond

1 CH3CH,CH2C-CH,CH(CH3)2 II

is formed from 2 four-carbon compounds by a Girgnard reaction. CH3

I

CH3CH2CH2CH,0H CH,-CHCH,Br n-Butyl alcohol

I

Cu.heat

CH,CH2CH2CH=0

lm

CH3

I aCH3-CH-CH20H Isobutyl alcohol

*

CH3

I

+ CH,-CHCH,MgBr

2.

H20.H’

CH3CH2CH2

KMd,

product

Problem 15.53 Compound (A), C,H,Cl,, is hydrolyzed to compound (B), C,H,O, which gives an oxime and 4 a negative Tollens’ test. What is the structure of (A)? (B) is a carbonyl compound because it forms an oxime, and therefore (A) is a gem-dihalide.

I

-CC12 Since (B) does not reduce Tollens’ reagent, (B) is not an aldehyde, but must be a ketone. The only ketone with

4 C’s is CH,COCH,CH,, and (A) is CH,CCl,CH,CH,.

Problem 15.54 Compound (A), CSH,OO,forms a phenylhydrazone, gives negative Tollens’ and iodoform tests and is reduced to pentane. What is the compound? 4

Phenylhydrazone formation indicates a carbonyl Compound. Since the negative Tollens’ test rules out an aldehyde, (A) must be a ketone. A negative iodoform test rules out the CH,C----O group, and the reduction product, pentane, establishes the C’s to be in a continuous chain. The compound is CH3CH,COCH,CH,. Problem 15.55 A compound (C,H,O,) is reduced to pentane. With H,NOH it forms a dioxime and also gives positive iodoform and Tollens’ tests. Deduce its structure. 4

Reduction to pentane indicates 5 C’s in a continuous chain. The dioxime shows two carbonyl groups. The positive CHI, test points to

340

CARBONYL COMPOUNDS: ALDEHYDES AND KETONES

[CHAP. 15

while the positive Tollens' test establishes a --CH=O. The compound is CH3-C-CHzCHz-CHO

8

Problem 15.56 The Grignard reagent of RBr(1) with CH,CH,CHO gives a 2" alcohol (11), which is converted to R'Br (III), whose Grignard reagent is hydrolyzed to an alkane (IV). (IV) is also produced by coupling (I). What are the compounds (I), (11), (III), and (IV)? 4

Since CH,CH,CHO reacts with the Grignard of (I) to give (11) after hydrolysis, (11) must be an alkyl ethyl carbinol H H RMgBr

-

I

CH~CHZCHO

HOH

bMgBr

I

cH3CHzY-R6H

Grignard of (I)

(11)

The conversion of (11) to (IV) is H

H

H

I

___* HOH

CH,CH,CH,-R

MgBr (IV) must be symmetrical, since it is formed by coupling (I). R is therefore --CH,CH,CH,. (I) is CH,CH,CH,Br. (IV) is n-hexane. (11) is CH,CH,CH(OH)CH,CH,CH,. (111) is CH,CH,CHBrCH,CH,CH,. Problem 15.57 Carry out the following synthesis: 4

Common oxidants cannot be used, because they oxidize the C-C bond. Since the starting compound is a methyl ketone, its reaction with NaOI removes CH, and converts.

0

II

-C-CH,

to --COO-, which is then acidified. Problem 15.58 Translate the following description into a chemical equation: Friedel-Crafts acylation of resorcinol( 1,3-dihydroxybenzene) with CH,( CH,),COCl produces a compound which on Clemmensen reduc4 tion yields the important antiseptic, hexylresorcinol.

Problem 15.59 Treatment of benzaldehyde with HCN produces a mixture of two isomers that cannot be separated by very careful fractional distillation. Explain. 4

Formation of benzaldehyde cyanohydrin creates a chiral C and produces a racemic mixture, which cannot be separated by fractional distillation.

CHAP. 151

34 1

CARBONYL COMPOUNDS: ALDEHYDES AND KETONES

H

H

I C,H,-C=O

I*

+ HCN

C,H,-C-OH

AN Problem 15.60 Explain the following reaction: C H C-C=O

7O1 HI

+ NaOH

Phen ylgl yoxal

-

C H CHC-0-Na'

4

II

,I

HO 0

sodium salt of mandelic acid

This is an internal crossed-Cannizzaro reaction: the keto group is reduced and the --CHO is oxidized. Problem 15.61 Prepare 1-phenyl-1-( p-bromopheny1)-1-propanol from benzoic acid, bromobenzene and ethanol. 4

The compound is a 3" alcohol, conveniently made from a ketone and a Grignard reagent as shown.

+

C,H,OH

C,H,Br

=Zr* C,H,MgBr

Problem 15.62 Convert cinnamaldehyde, C,H,-€H=CH--CH=O, propane, C,H,CHBrCHBrCH,CI.

to l-phenyl-l,2-dibromo-3-chloro4

We must add Br, to c=C and convert --CHO to -CH,CI. Since Br, oxidizes --CHO to - C O O H , --CHO must be converted to CH,Cl before adding Br,. 1 NaBH4

C,H,-CH=CH-CH-*~

PCl,

C,H,-CH=CH-CH,OH-----*

C,H,--CH=CH-CH,CI

2C,H,--CH--CH-CH,Cl Ar

$r

Problem 15.63 Compounds "labeled" at various positions by isotopes such as 14C (radioactive), D (deuterium) and " 0 are used in studying reaction mechanisms. Suggest a possible synthesis of each of the labeled compounds below, using I4CH,OH as the source of I4C, D,O as the source of D, and H,"O as the source of " 0 . Once a 14C-labeled compound is made, it can be used in ensuing syntheses. Use any other unlabeled compounds. ( a ) CH,I4CH,OH, (6) I4CH,CH,OH, (c) 14CH,CH,CH0, (d) C,H,14CH0, (e) 14CH,CHDOH, ( f ) CH,CHI80. 4 ( a ) The 1" alcohol with a labeled carbinol C suggests a Grignard reaction with 14

I . CHjMgBr

cu

C H , 0 H = H , ' 4 C - - 0 ~ (6) Now the Grignard reagent is labeled instead of H,CO. I4CH3OH--!%I4CH3Br

zo*

'*CH,MgBr

H,'4C=0.

CH,'~CH,OH

&I4CH3CH,OH I.

CHO

CARBONYL COMPOUNDS: ALDEHYDES AND KETONES

342

(c 1

-

14CH,MgBr [see (611 + H2C-CH2 \/

2.

n,o+

''CH,CH2CH20H

[CHAP. 15

314CH,CH2CH0

0

(e)

D on carbinol C is best introduced by reduction of a --CHO group with a D-labeled reductant. I4CH,CH,OH from (6) 5 heat I4CH,CHO then DJPt or LiAID, --+ ''CH,CHDOD---,

4 0

D of OD is easily exchanged with excess H,O. ( f ) Add CH,CHO to excess Hz'*Owith a trace of HC1. H,

H+

+ CH,CHO

CH,CHI80H

l80

OH hydrate

1

CH,CHI80

14CH,CHDOH

+ H20

The unstable half-labeled hydrate can lose H,O to give CH,CH'*O. Problem 15.64 Isopropyl chloride is treated with triphenylphosphine (Ph,P) and then with NaOEt. CH,CHO is added to the reaction product to give a compound, C,H,,. When C,H,, is treated with diborane and then 4 CrO,, a ketone is obtained. Give the structural formula for C,H,, and the name of the ketone. The series of reactions is:

-

Formation of Ylide (CH,),CHCI

+ Ph,P+

+

[(CH,),CH-PPh,]CI-

NaOEt

(CH,),C=PPh,

+ EtOH + Na'CI-

Wittig Reaction H (CH,),C=PPh,

+CH,d=O-

H (CH,),

Anti-Markovnikov Hydroboration-Oxidation I.

BHJ

(CH

3-Methyl-2-butanone

Problem 15.65 Trace the oxidation of 2-propanol to acetone by infrared spectroscopy. Observe the disappearance of the 0-H stretching band at about 1720 cm-I.

4

stretching band at 3600cm-* and the appearance of the C=O

Problem 15.66 Deduce the structure of a compound, C,H,O, with the following spectral data: (a) Electronic absorption at Amax = 213 nm, erns, = 7100 and A,,, = 320 nm, em,, = 27. (6) Infrared bands, among others, at 3000, 2900, 1675 (most intense) and 1602 cm-'. ( c ) Nmr singlet at S = 2.1 ppm (3 H's), three multiplets each integrating for 1 H at S = 5.0-6.0 ppm. 4 The formula C,H,O indicates two degrees of unsaturation and may represent an alkyne or some combination of two rings, C=C and C=O groups. (a)

A,, at 213 nm comes from the T + T * transition. It is more intense than the A,,, at 320 nm from the n-* T * transition. Both peaks are shifted to higher wavelengths than normal (190 and 280 nm, respectively), thus indicating an a,P-unsaturated carbonyl compound. The 2 degrees of unsaturation are a C-C and a C 4 .

CHAP. 151

343

CARBONYL COMPOUNDS: ALDEHYDES AND KETONES

( b ) The given peaks and their bonds are 3000cm-', sp2 C-H; 2900cm-', sp3 C-H; 1675cm-', c----O (probably conjugated to C=C); 1602cm-', C=C. All are stretching vibrations. Absence of a band at 2720cm-' means no aldehyde H. The compound is probably a ketone. (c) The singlet at 6 = 2.1 ppm is from a

0

II

-C-CH, There are also three nonequivalent vinylic H's (6 = 5.0-6.0 ppm) which intercouple. The compound is

H'

'c=c

/

Hd

/ \

Hh C-CH;

I

0

shown with nonequivalent H's. Problem 15.67 A compound, C,Hl,O, has a strong ir band at about 1700cm-I. The nmr shows no peak at 6 = 9-10 ppm. The mass spectrum shows the base peak (most intense) at mle = 57 and nothing at m / e = 43 or m / e = 71. What is the compound? 4

The strong ir band at 1700cm-' indicates a C=O, accounting for the one degree of unsaturation. The absence of a signal at 6 = 9-10 ppm means no 0

1I

-C-H proton. The compound is a ketone, not an aldehyde. Nmr is the best way to differentiate between a ketone and an aldehyde. Carbonyl compounds undergo fragmentation to give stable acylium ions: R-C=O:

I

.+

R' (P')

-

R-C=O; + .R' (or R~-c--o~ + .RI an acyliurn ion

The possible ketones are CH,CH,CCH,CH, 0 II (A)

a

CH,CCH,CH,CH,

CH,CCH(CH,),

b

(B) (C) Compounds (B) and (C) would both give some CH,C=O:' (mle = 43) and C , H , m : ' ( m l e = 71). These peaks were absent; therefore (A), which fragments to CH,CH,(+O:' ( m l e = 57), is the compound.

-

Problem 15.68 Give the steps in the preparation of DDT, (p-ClC,H,),CHCCl,, acetaldehyde) and chlorobenzene in the presence of H,SO,.

Cl3CCHO-

"2'O4

+

Cl,CCH(OH)-

PhCl

p-Cl--C,H,CHOHCCl~

"ZS04

-

from choral (trichloro+

p-CI--C,H,CHCCl,

PhCl

DDT

4

Chapter 16 Carboxylic Acids and Their Derivatives 16.1 INTRODUCTION AND NOMENCLATURE

Carboxylic acids (RCOOH or ArCOOH) have the carboxyl group, ----€--OH, which is an acyl II group, R-C-, bonded to OH. 0

II

Common names, such as formic (ant) and butyric (butter) acids, are based on the natural source of the acid. The positions of substituent groups are shown by Greek letters LY,p, y, S, etc. Some have names derived from acetic acid, e.g., (CH,),CCOOH and C,H,CH,COOH, are trimethylacetic acid and phenylacetic acid, respectively. Occasionally they are named as carboxylic acids, e.g.

(coo. is cyclohexanecarboxylic acid. For IUPAC names replace the -e of the corresponding alkane with -oic acid: thus, CH,CH,COOH is propanoic acid. The C’s are numbered; the C of COOH is numbered 1. C,HSCOOH is benzoic acid. Dicarboxylic acids contain two COOH groups and are named by adding the suffix -dioic and the word acid to the name of the longest chain with the two COOH’s. Problem 16.1 Give a derived and IUPAC name for the following carboxylic acids. Note the common names. (a) CH,( CH,),COOH (caproic acid); (6) (CH ,),CCOOH (pivalic acid); (c) (CH,),CHCH,CH,COOH (y-methylvaleric acid); (d) C,H,CH,CH,COOH (P-phenylpropionic acid); (e) (CH,),C(OH)COOH (a4 hydroxyisobutyric acid); ( f ) HOOC(CH,)2COOH (succinic acid) (no derived name).

To get the IUPAC name, find the longest chain of C’s including the C from COOH, as shown below by a horizontal line. To get the derived name, find and name the groups attached to the a C:

(a 1

m

O

O

H

]

n-Butylacetic acid, Hexanoic acid

(6 C’s in longest chain)

7H3 €Hd€€OOH

I Trimethylacetic acid, 2,2-Dimethylpropanoic acid

Isobutylacetic acid, CMethylpentanoic acid

Benzylacetic acid, 3-Phenylpropanoic acid 344

CHAP. 161

(e 1

(f)

CARBOXYLIC ACIDS AND THEIR DERIVATIVES

345

-3

Dimethylhydroxyacetic acid, 2-Hydroxy-2-methylpropanoic acid O H -

Butanedioic acid

Problem 16.2 Name the following aromatic carboxylic acids.

(a) p-nitrobenzoic acid; (6) 3,5-dibromobenzoic acid; (c) rn-formylbenzoic acid (COOH takes priority over CHO, wherefore this compound is named as an acid, not as an aldehyde); (d) o-methylbenzoic acid, but more commonly called o-toluic acid (from toluene).

Problem 16.3 Account for the following physical properties of carboxylic acids. (a) Only RCOOH’s with five or fewer C’s are soluble in water, but many with six or more C’s dissolve in alcohols. (6) Acetic acid in the vapor state has a molecular weight of 120 not 60. ( c ) Their boiling and melting points are higher than those of corresponding alcohols. 4 (a) RCOOH dissolves because the H of COOH can H-bond with H,O. The R portion is nonpolar and hydrophobic; this effect predominates as R gets large (over five C’s). Alcohols are less polar than water and are less antagonistic toward the less polar carboxylic acids of higher C content. (6) CH,COOH typically undergoes dimeric intermolecular H-bonding.

CH3-C

//O---Ho//\C--CH, \OH---O

(c) The intermolecular forces are greater for carboxylic acids. Problem 16.4 Write resonance structures for the COOH group and show how these and orbital hybridization account for: (a) polarity and dipole moments (1.7-1.9D) of carboxylic acids; (b) their low reactivity toward 4 nucleophilic additions, as compared to carbonyl compounds. (a) The C of COOH uses sp‘ HO‘s to form the three coplanar U bonds. A p A 0 of the 0 of the OH, accommodating a pair of electrons, overlaps with the 72 bond of C=O. In this extended T system there is negative charge on the lone 0 and positive charge on the other 0; the charge separation results in greater polarity and dipole moments.

(6) The electron deficiency of the C of C=O, observed in carbonyl compounds, is greatly diminished in the C of --COOH by the attached OH group. 0

II

Carboxylic acid derivatives (or acyl derivatives), RC-G, have the OH replaced by another electronegative functional group, G, such that it can be hydrolyzed back to the acid: R(Arv-4

II

0

+ H,O+

a carboxylic acid derivative

R(Ar)--C-OH

It

0

a carboxylic acid

+ HG

CARBOXYLIC ACIDS AND THEIR DERIVATIVES

346

General Formula

I

Type

Table 16-1

I

II

0

Ester

II

CH,C-OCH,CH,

II

0

0 R-C--R'

I

Acid anhydride CH,C-OCCH,

I

0

0 1

R-C-N0

A

I

II

0

Name

Example

R-C-CI*

RX-OR'

[CHAP. 16

Change

Acetyl chloride or Ethanoyl chloride

-ic acid to -yl chloride

Ethyl acetate or Ethyl ethanoate

Cite alkyl group attached to 0; then change -ic acid to -ate

Acetic anhydride

acid to anhydride

Acetamide or Et hanamide N-Methylpropanamide

-ic or -oic acid to -amide or -carboxylic acid to carboxamide

0

I

Amide

CH,C-N-H

II I

O H CH,CH ?CNHCH

II

0 ~~~~~

R-4-N-C-R'

Imide

CH,CO-NH-OCCH,CH,

Acetyl propiony 1 imide

-ic or -oic acid to -imide

RBN+

Nitrile

CH,C%N

Acetonitrile or Ethanenitrile

-ic or -oic acid to nitrile or add -nitrile to alkane name

*Some acid bromides are known. Although nitriles have no acyl group, they are grouped with acid derivatives because they are readily hydrolyzed to RCOOH.

CHAP. 161

CARBOXYLIC ACIDS AND THEIR DERIVATIVES

347

The common ones are given in Table 16-1, with conventions of nomenclature that involve changes of the name of the corresponding carboxylic acid; the G group is shown in bold type. Problem 16.5 Name the following acid derivatives:

c

(a) C,H,COCl

(4 CH,CH,CH,C-OCH,CH, It

(6) CH,CH, ,OCCH,CH,

II

0 0

(4

(4 C,HSCH,CNH,

It

It

It

( 3 3 - Z0 H 2

0

0

(g) HCN(CH,),

(f1

C,HsC--OC,Hs

4

( h ) C,H,COCCH,

It It

0

0 0

(a) Benzoyl chloride. (6) Propionic (or propanoic) anhydride. ( c ) Ethyl butyrate (or butanoate). ( d ) Phenylacetamide. (e) Phenyl benzoate. ( f ) Cyclohexanecarboxamide. ( g) N,N-Dimethylformamide. (h) Acetic benzoic anhydride. Problem 16.6 Give structural formulas for the following acid derivatives: ( a ) propionitrile, (6) isopropyl-24 fluorobutanoate, (c) 3-phenylhexanoyl chloride, (d) 3-chloropropyl benzoate. ( a ) CH,CH,C%N

(6) CH,CH,CHFCOOCH(CH,),

(c)

CH,CH,CH,

(d) C,H,COOCH,CH,CH2C1

HCH,COCl 'iC,H,

16.2 PREPARATION OF CARBOXYLIC ACIDS 1. Oxidation of 1" Alcohols, Aldehydes, and Arenes

R--CH,OH-

ArCHR,

2.

-

KMnO,

0-

RCHO-

KMn04 H+

KMn0,

RCOOH

ArCOOH

Oxidative Cleavage of Alkenes and Alkynes "NO,

HOOCCH,CH,CH,CH,COOH Adipic acid

Cyclohexene

a dicarboxylic acid

Problem 16.7 Account for the fact that on oxidative cleavage all substituted alkynes give carboxylic acids, whereas some alkenes give ketones. 4 Imagine that the net effect of the oxidation is replacement of each bond and H on the multiple-bonded C by OH. Intermediates with several OH'S on C are unstable and lose H,O, leaving c=O. [R,C(OH),J~R,C=O

and

[RC(OH),]-

RCOOH

Thus 101

R-R

__*

[RC(OH),] -HZO\ RCOOH

Alkenes with the structural unit R2C=C- would give [R,C(OH),]-

I

-H20

R2C=0

Multiple-bonded C's bonded only to H's would form [C(OH),), which loses two molecules of H 2 0 to give CO,.

CARBOXYLIC ACIDS AND THEIR DERIVATIVES

348

[CHAP. 16

3. Grignard Reagent and CO, I

R-MgX

+ R-C=O

+>C=%

H20

-O( I MgX)'

4.

bH

a carboxylate salt

-

Hydrolysis of Acid Derivatives and Nitriles

R4-G

II

+ H,O+

0

5.

+ Mg2++ 2X-

R-C=O

+ HG

R--C-OH

II

(or H,O+

+ Cl-, for G = Cl)

0

Haloform Reaction of Methyl Ketones

Although this reaction is used mainly for structure elucidation, it has synthetic utility when methyl ketones are readily prepared and halides are not. Problem 16.8 ( U ) Why isn't 2-naphthoic acid made from 2-chloronaphthalene? (6) How is 2-naphthoic acid 4 prepared in a haloform reaction? (a) Since most electrophilic substitutions of naphthalene, including halogenation, occur at the 1-position, 2-chloronaphthalene is not a readily accessible starting material. (6) Acetylation with CH,COCI in the presence of the solvent, nitrobenzene, occurs at the 2-position, which allows for the following synthesis: 0

II

2-Naphthoic acid

',-Acetonaphthalene (Methyl p-naphthyl ketone)

( p-Naphthoic acid)

Problem 16.9 Prepare the following acids from alkyl halides or dihalides of fewer C's. ( a ) C,H,CH,COOH, (6) (CH,) ,CCOOH, (c) HOCH,CH,CH,COOH, ( d ) HOOCCHICH,COOH (succinic acid). 4

Replace COOH by X to find the needed alkyl halide. The two methods for RX-*RCOOH are: I

COz

RX( 1"' 2"' or 3") Me\ R M g X - m l =

RCN ERX( 1")

2 H 3 0*

Either method can be used starting with C,H,CH,Br (a l o R X ) . With the 3"(CH,),C--Br, CN cannut be used because elimination rather than substitution would occur. HOCH2CH,CH,Br has an acidic H (0-H); hence the Grignard reaction can't be used. BrCH,CH,Br undergoes dehalogenation with Mg to form an alkene; therefore use the nitrile method. BrCH,CH,Br".,

N=tCH,CH2C+N---,

H'

product

16.3 REACTIONS OF CARBOXYLIC ACIDS

H OF COOH IS ACIDIC RCOOH acid,

+ H,O f- RCOO- + H,O' base

base,

acid,

pK,

=z

5 (Section 3.10)

CARBOXYLIC ACIDS AND THEIR DERIVATIVES

CHAP. 161

349

RCOOH forms carboxylate salts with bases; when R is large, these salts are called soaps. RCOOH + KOH2RCOOH + Na,CO,+

RCOO-K’ 2RCOO-Na’

+ H,O + H,O + CO,

Problem 16.10 Use the concept of charge delocalization by extended n- bonding (resonance) to explain why

( a ) RCOOH (pK, = 5 ) is more acidic than ROH ( p K , = 15), and ( b ) peroxy acids, RCOOH, are much weaker

than RCOOH.

8

4

( a ) It is usually best to account for relative strengths of acids in terms of relative stabilities of their conjugate

bases. The weaker (more stable) base has the stronger acid. Since the electron density in RCOO- is dispersed to both 0’s.

RCOO- is more stable and a weaker base than RO-, whose charge is localized on only one 0. ( b ) There is no way to delocalize the negative charge of the anion, RCOO-, to the C = O group, as can be done for RCO-.

b

b

Problem 16.11 Use the inductive effect (Section 3.11) to account for the following differences in acidity. ( a ) ClCH,COOH > CH,COOH, (6) FCH,COOH > ClCH,COOH, ( c ) CICH,COOH > ClCH,CH,COOH, (d) Me,CCH,COOH > Me,SiCH,COOH, (e) C1,CHCOOH > ClCH,COOH. 4 The influence of the inductive effect on acidity is best understood in terms of the conjugate base, RCOO-, can be summarized as follows:

0 Electron-withdrawing groups stabilize RCOO and strengthen the acid.

0 Electron-donating groups destabilize RCOO and weaken the acid.

Like all halogens, C1 is electronegative, electron-withdrawing, and acid-strengthening. Since, F is more electronegative than C1, it is a better EWG and a better acid-strengthener. Inductive effects diminish as the number of C’s between C1 and the 0’s increases. ClCH,COO- is a weaker base than ClCH,CH,COO-, and so CICH,COOH is the stronger acid. Si is electropositive and is an acid-weakening EDG. Two Cl’s are more electron-withdrawing than one C1. C1,CHCOO- is the weaker base and C1,CHCOOH is the stronger acid.

Resonance and induction, which affect AH, can be used with certainty to explain differences in acidities only when the K , values are different by a factor of at least 10. Smaller differences may be accounted for by solvation, which affects AS. Problem 16.12 Compare and account for the relative acid strengths of the following benzoic acids: C,H,COOH, p-CH,OC,H,COOH, and p-NO,C,H,COOH. 4 0 +/

is an acid-strengthening EWG by extended T bonding and induction; p-NO,C,H,COOH is the ‘0strongest acid. CH,O- is an EDG by extended 7r bonding, an effect that overshadows its electron-withdrawal by induction; p-CH,OC,H,COOH is the weakest acid.

-N

350

CARBOXYLIC ACIDS AND THEIR DERIVATIVES

[CHAP. 16

Acid-strengthening ring substituents also retard electrophilic, and enhance nucleophilic, aromatic substitution. Conversely, the acid-weakening groups accelerate electrophilic, and retard nucleophilic, aromatic substitution. Pl0b)cHn 16.13 Explain why highly branched carboxylic acids such as

are less acidic than unbranched acids.

4

The --CO, group of the branched acid is shielded from solvent molecules and cannot be stabilized by solvation as effectively as can acetate anion. Problem 16.14 Although p - hydroxybenzoic acid is less acidic than benzoic acid, salicyclic (o-hydroxybenzoic) 4 acid (K,, = 105 x 10-5) is 15 times more acidic than benzoic acid. Explain. The enhanced acidity is partly due to very effective H-bonding in the conjugate base, which decreases its basic strength. 0y O - H

-H+ ___,

Problem 16.15 The K, for fumaric acid (trans-butenedioic acid) is greater than that for maleic acid, the c i ~ isomer. Explain by H-bonding. 4 Both dicarboxylic acids have two ionizable H's. The concern is with the second ionization step.

Maleate monoanion (H-bond)

Fumarate monoanion (no H-bond)

Since the second ionizable H of maleate participates in H-bonding, more energy is needed to remove this H because the H-bond must be broken. The maleate monoanion is therefore the weaker acid. In general, H-bonding involving the acidic H has an acid-weakening effect; H-bonding in the conjugate base has an acid-strengthening effect.

NUCLEOPHIUCITY OF CARBOXYLATES

RCOO- acts as a nucleophile in S,2 reactions with RX to give esters. R C - O q ' X

1I

0 a carboxylate

___*

RC-OR' + X -

1I

0 an ester

35 1

CARBOXYLIC ACIDS AND THEIR DERIVATIVES

CHAP. 161

FORMATION OF ACID DERNATNES (OH-

G)

1 . Acyl Chloride (RCOCl) Formation, OH --+ C1

+ PCI,--+3 R--COCI + H,PO,(t) R - C O O H + PCl, --+ R - C O C I + HCl( g) + POCl,( e ) R - C O O H + SOC1,R-COCl + HCl(g) + SO,(g) 3 R-COOH

Thionyl chloride

Reaction with SOCl, is particularly useful because the two gaseous products SO, and HCI are readily separated from RCOCl. 2.

Ester (RCOOR') Formation, OH --+ OR' R-COOH

+ R'OH-

"2SO4

reflux

R-COOR'

+H20

(In dilute acid the reaction reverses.) Problem 16.16 Give the mechanism for the acid-catalyzed esterification of RCOOH with R'OH.

4

In typical fashion, the 0 of C=O is protonated, which increases the electrophilicity of the C of C=O and renders it more easily attacked in the slow step by the weakly nucleophilic R'OH.

1

+

HO-R'

\ RLOH RC-OH+ R'Oti

c-)

I

rlon

HO

RC-OR'

II

I HO

+ H,O+

0

tetrahedral intermediate

The tetrahedral intermediate undergoes a sequence of fast deprotonations and protonations, the end result being the loss of H' and H,O and the formation of the ester.

3. Amide (RCONH,) Formation, OH --+NH, R-COOH

+ NH,-

acid

ammonium salt

RCOOH + R'NH,-

RCOO- R'NH;

1" amine

RCOOH + R'RNH-

heat

R-COO-NH:

RCOO- R'RNHI-

heat

R-CONH, RCONHR'

heat

2" amine

+ H,O

1" amide

+ H,O

2" amide

RCONHR'R"

+ H,O

3" amide

Problem 16.17 Use ethanol to prepare CH,COOC,H,, an important commercial solvent.

4

CH3COOC2H,, ethyl acetate, is the ester of CH,CH,OH and CH,COOH. CH,CH20H is oxidized to CH,COOH. CH,CH,OH-

MnO-/H+

CH,COOH

Ethanol and acetic acid are then refluxed with concentrated H,S04. C 2 H , 0 H + CH,COOH

HzSOJ

heat

CH,COOC2H, + H,O

With added benzene and a trace of acid, the reversible reaction is driven to completion by distilling off H,O as an azeotrope (Problem 13.56).

CARBOXYLIC ACIDS AND THEIR DERIVATIVES

352

[CHAP. 16

REDUCTION OF C=O OF COOH (RCOOH-, RCH,OH)

Acids are best reduced to alcohols by LiAIH,. RCOOH HALOGENATION OF

LY

2 RCH,OH r I . LIAIHJ(ether)

H's. HELL-VOLHARD-ZELINSKY ( H V Z ) REACTION

One or more a H's are replaced by C1 or Br by treating the acid with C1, or Br,, using phosphorus as catalyst. RCHzCOOH

x IP

RCXzCOOH (X = Cl, Br)

RCHCOOH

I

X a -Halogenated acids react like active alkyl halides and are convenient starting materials for preparing other a -substituted acids by nucleophilic displacement of halide anion.

-

x- + R-CH-COOH I

OH

-H+.X-

R-CH-COO

I

NH,'

an a-hydroxy acid

an a-amino acid

REACTION OF COOH. DECARBOXYLATION 1.

Arylcarboxylic Acids

ArCOOH2.

soda lime

ArH (in poor yields) + CO,

0-Keto Acids and 0-Dicarboxylic Acids

1

I

-CO,

0

0

Decarboxylation proceeds readily and in good yields when the C that is p to COOH is a C=O. RC--CH2COOH+

-cn. -

I1

0

-

I

RC-CH,

II

I

O H

a 6-ketocarboxylic acid

I

a ketone

-CO,

I

H O C - C - C O O H ~H--C--COOH

II

0

I

a 6 -dicarboxylic (rnalonic) acid

I

a carhoxylic acid

Problem 16.18 ( a ) Suggest a mechanism for a ready decarboxylation of malonic acid, HOOCCH,COOH, that proceeds through an activated, intramolecular H-bonded, intermediate complex. (6) Give the decarboxylation products of (i) oxalic acid, HOOC-COOH, and (ii) pyruvic acid, CH,COCOOH. 4

353

CARBOXYLIC ACIDS AND THEIR DERIVATIVES

CHAP. 161

end

complex

Acetic acid

2

0 (6) (i)

II

-CO,

H-HM-H

H0-C

H

(ii) CH,--C

-

Formic acid

0

II

CH,-C-H

Acetaldehydc

Problem 16.19 Prepare n-hexyl chloride from n-butylmalonic ester.

n-Butylmalonic ester is hydrolyzed with base and decarboxylated to hexanoic acid. I

n-C,H,CH(COOC,H,),

CIH

n-C,H,CH,COOH

This acid and n-hexyl chloride have the same number of C's. 1. I A I H 4

n-C,H,CH,COOH--------*

3.

n-C,H, ,CH,OH-

SOCI,

n-C,H,,Cl

Conversion of RCOOH to RBr The Hunsdiecker reaction treats heavy-metal (e.g. , Ag' ) carboxylate salts with Br,.

RCOO- Ag+ + Br2-

RBr + CO,

+ AgBr

Problem 16.20 Suggest a typical free-radical mechanism for the Hunsdiecker reaction which requires the initial formation of an acyl hypobromite, R C - O B r , from the Ag' salt and Br,. 4

RC-OBr+

Propagation

(1) RC--O.---+R*+CO,

8

8

RC-0-

8+

Initiation

8

-0Br-

Br-

R-Br

+ RC-0-

b

ELECTROPHILIC SUBSTITUTIONS During aromatic substitution with ArCOOH or ArCOG, the electron-attracting --COOH or -COG group is rneta-directing and deactivating. Problem 16.21 ( a ) Account for the fact that ArCOOH, with a strongly activating substituent ortho or para to COOH, loses CO, during attempted electrophilic substitution. (6) Write the equation for the reaction of p-aminobenzoic acid and Br,. 4

( a ) When the electrophile attacks the ring C bonded to COOH, the intermediate phenonium ion first loses an H' from COOH and then loses CO,, which is a very good leaving group:

CARBOXYLIC ACIDS AND THEIR DERIVATIVES

354

COOH

Br

p-Aminobenzoic acid

16.4

1.3.5-Tribromoaniline

SUMMARY OF CARBOXYLIC ACID CHEMISTRY PREPARATION

--

PROPERTIES

1. Acidic Hydrogen of the -COOH

1. Oxidation

RCH2CH20H+ Cr20:- or MnO,, H’ RCH&H=O

[CHAP. 16

+ HOH

+ Ag(NH3);

RCH,CH=CH,

+ MnO,,

2.

H30++ RCH2COO-

Hydroxyl Group of the -COOH

+ PCI,, SOCI, RCH,COCl + R’OH -!% RCH,COOR’ + NH3 +RCH2COO-NHi &*

H’

2. Grlgnard

RC H2C ON H2 3.

Group of the -COOH

Hydrolysis

+ 1.

RCH,-CONH, + HOH RCHZCOOR’ + HOH (RCHIC0)20 + HOH RCH,COX + HOH RCHzCCl, + 3HOH

LiAIH,,,

2. HOH

__*

RCH2CH20H

+ heat 7 RCH2-CO-CH2R MnO

+ X2 5 RCHXCOOH 5.

-COOH

RCX2COOH

Group

+ N a O H T RCH, + Ag’, Br, RCH,Br --t

16.5 POLYFUNCTIONAL CARBOXYLIC ACIDS

DICARBOXYLJC ACIDS, HOOC(CH,), COOH; CYCLIC ANHYDRIDES The chemistry of dicarboxylic acids depends on the value of n. See Problem 16.18 for decarboxylations of oxalic acid (n = 0) and malonic acid (n = 1). When n = 2 or 3, the diacid forms cyclic anhydrides when heated. When n exceeds 3, acyclic anhydrides, often polymers, are formed. Problem 16.22 Compare the products formed on heating the following dicarboxylic acids: ( a ) succinic acid, ( b ) glutaric acid ( 1,5-pentanedioic acid), (c) longer-chain HOOC( CH,),COOH. 4 ( a ) Intramolecular dehydration and ring formation:

CH2-COOH CH2-COOH I n=2

hc.l

-H20

CHZ--C=O

I /

’ CH,-C=O‘0

Succinic anhydride

CHAP. 161

355

CARBOXYLIC ACIDS A N D THEIR DERIVATIVES

( b ) Intramolecular dehydration and ring formation:

CH2-COOH / CH2 \CH2-COOH n=3 (c)

heat -H*O’

CH2-C-0 / \ 0 Glutaric anhydride CH2 / ‘CH,--C=O

Longer-chain a,o-dicarboxylic acids usually undergo intermolecular dehydration on heating to form long-chain polymeric anhydrides (see Section 16.8). HOOC-(CH-J,-COOH n>3

-ac;

Problem 16.23 Show steps in the following syntheses, using any needed inorganic reagents:

(a) o-Xylene

C O\

(6)

0

II

__*

Tctrahydrof w a n

0

0

II

CIC(CH2),CCI 4

Adipoyl dichloride

acH3 ri,: acooH Phthalic anhydride

R0

CH3

oxidation

COOH

Phthalic acid

a c > O +H,O C O\

(6) The chain is increased from 4 to 6 C’s by forming a dinitrile.

) (

) ( NC

CN

HOOC

COOH

aCIC(CH2)&CI

II

0

HYDROXYACIDS; LACTONES Reactions of hydroxycarboxylic acids, HO(CH,).COOH, also depend on value of n. In acid solutions, y-hydroxycarboxylic acid (n = 3) and 6-hydroxycarboxylic acid (n = 4) form cyclic esters (lactones) with, respectively, five-membered and six-membered rings. Problem 16.24 Write the structure for the lactone formed on heating in the presence of acid ( a ) y 4 hydroxybutyric acid, ( 6 ) 8-hydroxyvaleric acid.

Since an OH and a COOH are present in each compound, intramolecular dehydration gives lactones with 5and 6-membe red rings, respective I y . ,CHZ-CHz H2c/CH2-cH2 \c=o -&b H’C.\O/ ‘c=o ( a) \OH

I

OH

y- Hydroxybutyric acid

y - Butyrolactone

(4-Butanolide)

CARBOXYLIC ACIDS AND THEIR DERIVATIVES

356

[CHAP. 16

6 -Valerolactone (5-Pentanolide)

6-Hydroxyvaleric acid

Problem 16.25 p-Hydroxyacids ( n = 1) readily undergo dehydration but do not yield a lactone. Give the 4 structure of the product and account for its formation.

RCH=CHy=O

8 x ; O

OH A double bond forms because the product is a stable, conjugated a,p-unsaturated acid [Problem 16.28(b)]. Problem 16.26 ( a ) When heated, 2 mol of an a-hydroxyacid ( n = 0) lose 2 mol of H,O to give a cyclic diester (a lactide). Give the structural formulas for two diastereorners obtained from lactic acid, CH,CHOHCOOH, and select the diastereomer which is not resolvable. (6) Synthesize lactic acid from CH,CHO. 4 0

unresol vable center of symmetry

/

0

CH,CHO

CH,

H3O+

CH,CH(OH)COOH

+ NHf

HALOCARBOXYLIC ACIDS (Amino Acids in Chapter 21 .) With the exception of the a-acid, haloacids behave like hydroxyacids. Heating an a-halogenated acid with alcoholic KOH leads to a,p-unsaturated acids when there is a p H in the molecule.

R-CH,CH-COOH

h

alc. KOH

R-CH-CH-COO-K’

-% R-CH-CH-COOH

a,&unsaturated acid salt

Problem 16.27 Prepare malonic acid (propanedioic acid, HOOC--CH,-€OOH)

from CH,COOH.

4

CH,COOH is first converted to CICH,COOH. The acid is changed to its salt to prevent formation of the very poisonous HCN when replacing the Cl by CN. The C=N group is then carefully hydrolyzed with acid instead of base, to prevent decarboxylation.

CH,COOH c12/p_ C1-CH,COOH Chloroacetic acid

C1-CH2CO06a Sodium chloroacetate

N=C-CH,COONa

+

Sodium cyanoacetate

HO+

3 HOOC-CH,-COOH

+ NHf

CHAP. 161

357

CARBOXYLIC ACIDS AND THEIR DERIVATIVES

Problem 16.28 What is the product when each of the following compounds is reacted with aqueous base (NaOH)? ( a ) 2-bromobutanoic acid, ( b ) 3-bromobutanoic acid, (c) 4-bromobutanoic acid, (d) 5-bromo4 pentanoic acid. (In (a) and (6) the initially formed salt is acidified.) ( a ) a-Hydroxyacid by S,2 substitution.

FOH

FBr

CH3CH2 H-COOH

CH3CH2 H-COOH

2-Bromobutanoic acid

2-Hydroxybutanoic acid

( 6 ) Dehydrohalogenation to an a,P-unsaturated acid. The driving force for this easy reaction is the formation of a conjugated system. OH I.

OH-

I

CH,CH=CH-C=O 2-Butenoic acid

Elimination is a typical reaction of 6-substituted carboxylic acid.

F

R HCH,COOH 5 RCH-CHCOOH

(c)

(Y = C1, Br, I, OH, NHJ

y-Haloacids undergo intramolecular SN2-typedisplacement of X- initiated by the nucleophilic carboxylate anion, to yield y-lactones. H2C-0 BrCH,CH,CH,COOH

Na'

0

/

__*

H,C,

\

,C=O

C H*

+ NaBr

y-Butyrolactone

( d ) Similar to part (c) to give a 6-lactone.

Br(CH,),C-0-

It 0 -

+

6-Valerolactone

Problem 16.29 Explain the fact that (R)-CH,CHBrCOO-Na' CH,CHOHCOO -Na '.

reacts with NaOH to give (R)4

On displacing Br by OH, the order of priorities (Section 5.3) is unchanged. Therefore, since both configurations are R, there was retention of configuration rather than the inversion expected from SN2reactions. It is believed that the --COO- group participates in first displacing the Br- with inversion to give an unstable a-lactone. Then O H - attacks the a-lactone with a second inversion to give the product. Two inversions add up to retention. Intervention by an adjacent group is called neighbring-group participation.

unstable a-lactone (S)

358

CARBOXYLIC ACIDS AND THEIR DERIVATIVES

[CHAP. 16

Intramolecular nucleophilic displacements, such as those in lactone formation, have faster reaction rates than intermolecular S,2 reactions because the latter require two species to collide. The neighboring participant is said to furnish anchimeric assistance. 16.6 TRANSACYLATION; INTERCONVERSION OF ACID DERIVATIVES

Transacylation is the transfer of the acyl group from one G group to another, resulting in the formation of various acid derivatives. Figure 16-1 summarizes the transacylation reactions. Notice that the more reactive derivatives are convertible to the less reactive ones. Because acetic anhydride reacts less violently, it is used instead of the more reactive acetyl chloride to make derivatives of acetic acid. In aqueous acid, the four kinds of carboxylic acid derivatives in the figure are hydrolyzed to RCOOH; in base, to RCOO-. R'C--O--CR'

II

II

0 RCOOH

Socl2

p/

II

II

0

1 t R'COOH (1)

\

R C - O H cRC---cl 0

0

R'OH

1. Transanhydride formation-an exchange of anhydrides. 2. Transesterification-an exchange of esters. 3. Also for H,NR' -P RCONHR' and HNR'R"+ RCONR'R"

Fig. 16-1

Problem 16.30 ( a ) Give the mechanism for the reaction of RCOCl with Nu:-. (6) Compare, and explain the difference in the reactivities of RCl and RCOCl with Nu:-. (c) What is the essential difference between nucleophilic attack on C==O of a ketone or aldehyde and of an acid derivative? 4 (a)

Nucleophilic substitutions of RCOG, such as RCOCl, occur in two steps. The first step (addition) resembles nucleophilic addition to ketones and aldehydes (Section 15.4) and the second step (elimination) is loss of G. in this case C1 as CI-. ADD IT10N

ELIMINATION b

React ant (trigonal sp')

Transition State ( C becoming sp')

Intermediate (C is tetrahedral sp', on electronegative 0)

Transition State ( C becoming s p 2 )

Product (trigonal s p 2 )

CHAP. 161

359

CARBOXYLIC ACIDS AND THEIR DERIVATIVES

Alkyl halides are much less reactive than acyl halides in nucleophilic substitution because nucleophilic attack on the tetrahedral C of RX involves a badly crowded transition state. Also, a U bond must be partly broken to permit the attachment of the nucleophile. Nucleophilic attack on C=O of RCOCI involves a relatively unhindered transition state leading to a tetrahedral intermediate. Most transacylations proceed by this mechanism, which is an example of addition-elimination reminiscent of aromatic nucleophilic substitution (Section 11.3). The addition step leading to the tetrahedral intermediate is the same in each case. However, the intermediates from the carbonyl compounds would have to eliminate an R:- (from a ketone) or an H:(from an aldehyde) to restore the C 4 . These are very strong bases and are not eliminated. Instead the intermediate accepts an H’ to give the adduct. The intermediate from RCOG can eliminate G:-as the leaving group. Problem 16.31 Account for the relative reactivities of RCOG with nucleophiles: RCOX > ( R C 0 , ) , 0 > 4 RCOOR’ > RCONH,. ,

Often this order of reactivity is related to the order of “leavability” of G-, which is the reverse of its order of basicity; CI- is the best leaving group and the weakest base.

increasing o r d e r of hasccirv

Although this rationale gives the correct answer, it overlooks an important feature of the mechanism-the step eliminating G - cannot influence the reaction rate because it is faster than the addition step. The first step, the slow nucleophilic addition, determines the reaction rate, and here resonance stabilization of the 4 - G group II is important: R-COOCH,

C l e C O O C H ,

111

IV

4

See Table 16-2. Table 16-2

Ranks I

I1

I11

IV

Rate-Determining Factors

2

4

1

3

Steric effects (branching on alcohol portion)

4

2

3

1

Steric factor (branching on acid portion)

4

1

2

3

Electron-attracting groups disperse, developing negative charge in transition state and increasing reactivity

Problem 16.42 Acid-catalyzed hydrolysis with H2l80of an ester of an optically active 3" alcohol, RCOOC*R'R"R, yields the partially racemic alcohol containing '*O, R'RR"'C'*OH. Similar hydrolyses of esters of 2" chiral alcohols, RCOOC*HR'R", produce no change in the optical activity of the alcohol, and "0is found in RC''0,H. Explain these observations. 4 Hydrolyses of esters of most 2" and 1" alcohols occur by cleavage of the 0-acyl

bond,

0 R-!tO-EHR'R" Since no bond to C* is broken, no racemization occurs. However, with 3" alcohols, there is an S,1 0-alkyl cleavage,

0 R-!!-O-/-CR'R"R"

364

CARBOXYLIC ACIDS AND THEIR DERIVATIVES

[CHAP. 16

producing RCOOH and a 3" carbocation, ' C R ' R R , which reacts with the solvent (H,"O) R'R"R"C''0H. This alcohol is partially racemized because ' CR'RR"' is partially racemized.

+

+ RR"R"C+

RCOOH

to form

H2"0

R'R"R"C'80H

+ H+

Problem 16.43 Write structures of the organic products for the following reactions:

(R)-CH,COOCH(CH,)CH,CH, C,H,COOC,H, + NH,C,H,COOC,H,

+ H,O-

NdOti

( 6 ) (R)-CH,COOCH(CH,)CH,CH, ( d ) C,H,COOC,H,

-

+ LiAIH,

0

+ n-C,H,OH-

+ H,O+

__*

H*

0

II II cf) C,H,C-0-0-C-C,H,

+ CH,O-Na'

CH,COO-Na' + (R)-HOCH(CH,)CH,CH,. The alcohol is 2"; we find 0-acyl in configuration of alcohol. RCOO-Na' forms in basic solution. CH,COOH + (R)-HOCH(CH,)CH,CH,. Again 0-acyl cleavage occurs. C,H,CONH, + C,H,OH. C,H, COO-n-C, H, + C, H OH. Acid-catalyzed transesterification. C,H,CH,OH + C 2 H , 0 H .

4

4

cleavage and no change

0

It

C6HsC-o-o-Na'

4-

C6H5COOCH3

Sodium peroxybenzoate

Problem 16.44 Supply a structural formula for the alcohol formed from (a) CH,CH,COOCH,

+ ZC,H,MgBr

( 6 ) C,H,CH,COOCH,

+ 2C,H,MgBr

4

Two R or Ar groups bonded to carbinol C come from the Grignard reagent, while the carbonyl C becomes the carbinol C.

Problem 16.45 Methyl esters can be prepared on a small scale from RCOOH and CH,N,. Suggest a mechanism involving S,2 displacement of N,. 4

Diazomethane is a resonance hybrid,

H

I

H

I

+

HC-N=N: and we have

c----,

H

HC-N-N: +

1 .

or

1

+

HC-N='N:

6-

6-

H

RCOOH

-6

acid,

I + + HC --N=?J: -

base:

6-

+RCOO-

?+ + CH,:N-N:

U

base,

acidz

__*

RCOOCH,

+ :N-N:

CHAP. 161

365

CARBOXYLIC ACIDS AND THEIR DERIVATIVES

Problem 16.46

Prepare CH3CH2CH2COOCH2CH2CH2CH3 from n-C4H90H.

4

The acid portion of the ester has the same number of C’s ( 4 ) as does the starting material, the alcohol. Therefore, oxidize the alcohol. The ester is best formed from the acid chloride: . n-C,H,OH

KMnOJ

CH,(CH,),COOH

S

OH

OH

R-T+-jlH

+R-Y-OH,

I

I

0

II

+ __*

R-C--"O-R'

+ H30+

I8O

I80H

I R'

I

R'

Problem 16.87 When reactive alcohols such as CH,=CHCH2"OH found. Explain.

are esterified in acid, some H,IXO is 4

Protonated ally1 alcohol undergoes an S, 1-type substitution reaction.

CH,=CHCH, "OH

2CH,=CHCH,-'80H

+

I H

S I

-% H,I80

+ CH,=CH;H,

0 t

CH,t

CH,=CHCvy

OH

CH,C-OCH,CH=CH,

--+

b

Problem 16.88 Show how phosgene,

CI-c-CI

b

is used to prepare ( a ) urea, (b) methyl carbonate, (c) ethyl chlorocarbonate, (d) ethyl N-ethylcarbamate (a urethane), (e) ethyl isocyanate (C2H,-N=C=O). 4

COCI,

NH,CONH,

2CH OH

COCI, A COCI,

-c2H50-8-c' C H O H ( I mole)

H

CXI,

C H NH

C,H5N-l-CI I

C H OH

H

C2Hs&-C-OC2H5

0 H

Problem 16.89 What carboxylic acid (A) has a neutralization equivalent (NE) of 52 and decomposes on heating to yield CO, and a carboxylic acid (B) with an NE of 60? 4

The NE o f a carboxylic acid is its equivalent weight (molecular weight divided by the number of COOH groups). Since CO, is lost on heating, there are at least 2 COOH's in acid (A), and loss of CO, can produce a

CHAP. 161

CARBOXYLIC ACIDS AND THEIR DERIVATIVES

38 1

monocarboxylic acid, (B). Since one mole of COOH weighs 45 g, the rest of (B) weighs 60 g - 45 g = 15 g, which is a CH, group. (B) is CH,COOH, and (A) is malonic acid (HOOC--CH,-COOH), whose NE is 104 g 2 eq

= 52 gleq

Problem 16.90 An acyclic compound, C 6 H I 2 O 2has , strong ir bands at 1740, 1250, and 1060cm-', and no bands at frequencies greater than 2950cm-I. The nmr spectrum has two singlets at 6 = 3.4 ( 1 H ) and 6 = 1.0 4 ( 3 H ) . What is the compound?

The one degree of unsaturation is due to a carbonyl group, indicated by the ir band at 1740cm-I. The lack of bands above 2950cm-I shows the absence of an O H group. Hence the compound is not an alcohol or a carboxylic acid. That the compound is probably an ester is revealed by the ir bands at 1250 and 1060cm-' ( C - 0 stretch). Two singlets in the nmr means two kinds of H's. The integration of 1 : 3 means the 12 H's are in the ratio of 3 : 9. The signal at 6 = 3.4 indicates a C H , 4 group. The nine equivalent protons at S = 1.0 are present in three CH,'s not attached to an electron-withdrawing group. A r-butyl group, (CH,),C--, fits these requirements. The compound is (CH,),CCOOCH,, methyl trimethylacetate (methyl pivalate). Problem 16.91 Predict the base (most prominent) peak in the mass spectrum of the compound in Problem 16.90. 4

Parent ions of esters resemble those of other acid derivatives and carboxylic acids in that they cleave into an acylium ion.

(CH,),CC-OCH,

[

A

I

(CH,),CC=Ot

+ OCH,

or CH,C-

+ &COCH,

The base peak should be m l e = 8 5 or 59. Problem 16.92 Prepare 0-bromotoluene from toluene without using direct bromination (which gives a large amount of the unwanted p-isomer). 4

Use -SO,H

as a blocking group and then remove it by desulfonation.

Toluene

p-Toluenesulfonic acid

Problem 16.93 Identify the compound C,H,,O,SCI(A), and also compounds (B), (C), and ( D ) , in the reactions: AgCl(s)

4

T : -1-

water soluble (B)

Br C H 3 0 C H 3 c-- CHQCH, CH3 (D) 1 -Bromo-2,4,6trimethylbenzene

CH1 (C) Mesitylene

-

H30+

C,H,, (C)

Br2

Fe

%

one monobromo derivative (D) 4

S03H SO*Cl C H , ~ C H t , CH3QCH, CH3

(B) 2.4.6-Trimethylbenzenesulfonic acid

CHi (A) 2,4,6-Trimethyl-

benzenesulfonyl chloride

382

CARBOXYLIC ACIDS AND THEIR DERIVATIVES

[CHAP. 16

Problem 16.94 ( U ) (S)-sec-C,H,OH, which has an optical rotation of +13.8', is reacted with tosyl chloride (see Problem 16.60) and the product is saponified. (b) Another sample of this alcohol is treated with benzoyl chloride and the product is also hydrolyzed with base. What is the rotation of the sec-C,H,OH from each reaction? 4

(b)

A

A

A

(S)(+13.8")

(S)(Tosyl ester)

(R)(- 13.8")

+ 13.8'. Reaction with benzoyl chloride causes no change in configuration about the chiral C of the alcohol. Hydrolysis of PhCOOR occurs by attack at the carbonyl group, with retention of alcohol configuration.

Problem 16.95 Reaction of the (R)-tosyl ester of a-phenylethyl alcohol (C,H,CHMeOTs) with CH,COOH 4 yields a mixture of ( S ) - and ruc-acetate. Explain. Inversion and racemization products show operation of an SN! reaction. The SN1 pathway is possible because the alcohol forms the stable 2" benzylic carbocation, C,H,CHCH,. Problem 16.96 Write structural formulas for the organic compounds (1) through (XI) and show the stereochemistry of (X) and (XI). (Write Ts for the tosyl group.)

ArS0,Cl

+ CH,OH

__+

(1)-

H

I*O+

(11) + (111)

C6H,S02Cl + n-C4H90H +(VII) ______* (VIII) + (IX) PhCH2M.Br

t,,

C6H13

TsCI

+ (S)-HO-

(I) ArS020CH3

-H

+(X)

(11) ArSO,H

CH COO-

3(XI)

4

(111) CH,"OH

Problem 16.97 Why are sulfonic acid derivatives less reactive toward nucleophilic substitutions than are the corresponding acyl derivatives? 4

Attack by Nu: - on the trigonal acyi C leads to a stable, uncrowded, tetrahedral intermediate (and transition state), with an octet of e-'s on C. The sulfonyl S is already tetrahedral, and attack by Nu:- gives a less stable, more crowded, intermediate (and transition state), with a pentavalent S having ten e-'s, as shown:

CHAP. 161

383

CARBOXYLIC ACIDS AND THEIR DERIVATIVES

-[

0 Ar-k-G

A

+ Nu:

N 5 f ]

pentavalent intermediate

Problem 16.98 ( a ) Show the mers of the following condensation polymers, formed from the two indicated (tough and optically monomers: (i) Polycarbonate (Lexan), phosgene, CI2C=O + (HOC,H,),C(CH,), clear); ( i i ) Kodel, dimethyl terephthalate

+ HOCH2--

0

-CH,OH

(strong fibers); (iii) Aramid, (terephthalic acid) + p-H,NC,H,NH, (tire cords). (6) What small molecules are split off during growth of the polymer chain? (c) What type of functional group links the mer units? 4 -CH20-€-

b

( 6 ) HCI

(c) ester

(iii)

Problem 16.99 Which would have the higher A,,, C,H,CH=CHCOOH? Why? Cinnamic acid, because it has a more extended

7r

value in the uv, benzoic acid or cinnamic acid, 4

system.

Problem 16.100 Tell how propanal and propanoic acid can be distinguished by their

spectra.

(a) ir, (6) nmr, (c) cmr

4

Propanal has a medium-sharp stretching band at 2720cm-' due to the aldehydic H. Propanoic acid has a very broad 0-H band at 2500-3300 cm- I. The C=O bands for both are in the same general region, about 1700cm-'. Propanal gives a triplet near 6 = 9.7 ppm (aldehydic H), a triplet near 0.9 (CH,), and a multiplet near 2.5 (CH,). Propanoic acid has a singlet in the range 10-13 (acidic H), a triplet near 0.9 (CH,) and a quartet near 2.5 (CH,). Carbonyl C's are sp'-hybridized and attached to electronegative 0 ' s and hence suffer little shielding. They absorb farther downfield than any other kind of C: 190-220 for carbonyl compounds (propanal) and somewhat less downfield, 150-185, for carboxylic acids (propanoic acid).

Chapter 17 Carbanion-Enolates and Enols 17.1 ACIDITY OF H’s a TO C 4 ; TAUTOMERISM

GENERAL Three base-catalyzed reactions of such a H’s are shown in Fig. 17-1. Since the reactions have the same rate expression, they have the same rate-determining step: the removal of an cy H to form a stabilized carbanion-enolate anion. The name of this anion indicates that the resonance hybrid has negative charge on C (carbanion) and on 0 (enolate).

D

I I

CH 3CH2CCHO

H”

H

*I I

OH-

CH,CH,-CCHO

rncemization

B

I I

CH3CH2CCH0

rate = k[RCHO][Base]

CH3

racemic

optically active

X

I

C H 3CH2CCHO

I

CH3

racemic

CH3

I

CH3CH2C=C+

lA

CH3

-I

or CH3CH2-C==C=0

H acid,

base

’

I H

+ H:B

I

base,

acid,

stable carbanion-enofate ion

Stabilization of the anion by charge delocalization causes the a H of carbonyl compounds to be more acidic than H’s of alkanes. Problem 17.1 17-1.

Show how the stable carbanion-enolate anion reacts to give the three products shown in Fig.

4

384

CARBANION-ENOLATES AND ENOLS

CHAP. 171

385

Table 17-1 Other Stable Carbanions

Reason for Stability

p-p

IT

bond

p-p

IT

bond

p-d

7r

bond

~

c1

cI1

c1 sp hybrid

CH,C=CC

H

H

or

H

)QH

H

H H

w: H

-

H2E-Q-N02

H 2 c = p 4 H * c Q N 0 *

NO,

-

RSXH-SR

Me2G*H,

Ph,PR$

+ Na'X-

For dialkylacetic acids, the second H of the a C is similarly replaced with another R or a different R' group. Et00C-CH-COOEt

I

R

NiOEt

--

[EtOOC-CR-COOEtINa'

RX

R

I I

Et00C-C-COOEt

R

+ :X-

392

CARBANION-ENOLATES AND ENOLS

[CHAP. 17

of the substituted malonic ester gives the malonic acid, which undergoes decarboxylation (loss of CO,) to form a substituted acetic acid. Step 3 Hydrolysis

Et00C-CH-COOEt

k

,. OH-’

hydrolysis

[ n

HOOCCHCOOH

-[ 2. H,O+

R’ Et00C-C-COOEt I

hydrolysis

I

R

R’

1

HOOC-C-COOH

a

RCH2COOH

]a

R‘-CH-COOH

I

R

For a general carboxylic acid, the parts are assembled as follows:

m 1

from alkyl halides

TRX and R’X

from malonic ester

where H* replaced COOEt.

Problem 17.13 Use malonic ester to prepare ( a ) 2-ethylbutanoic acid, ( b ) 3-methylbutanoic acid, (c) 4 2-methylbutanoic acid, ( d ) trimethylacetic acid. The alkyl groups attached to the a C are introduced by the alkyl halides. (a)

In 2-ethylbutanoic acid

CH,CH H C O O ~ C - from malonic ester r ( 2 I

from RX’s-

the circled R’s are both CH,CH,. Therefore each a H is replaced sequentially by an ethyl group (R= R’ = Et), using CH,CH,Br.

CH2(COOC,H,),

+ Na’OC,H;

-

+C2HjBr

C2H,0H + [CH(C0OC2H,),1 y

:cz,:*

~~CH(COOC,H,),

(6) Only (CH,),CH-Br

CH2(C00C2H5)2

aq;z-*

~~IC(COOC,H,),

is needed for a single alkylation (see product below). I. 2.

NaOEt i-PrBr

’

CH,-C H -C H(COOC2H,),

-CH(COO-), (c)

[C,H,le(COOC2HS),Na’

H Oi

CH,-CH

aqiz-’

-CH(COOH),

‘ i CH,-CH

-CH,COOH

To obtain

Alkylate first with CH,CH,Br and then with CH,I. The larger R is introduced first to minimize steric hindrance in the second alkylation step.

CHAP. 171

393

CARBANION-ENOLATES AND ENOLS

CH,(COOEt),

m]CH(COOEt),

(d) Trialkylacetic acids cannot be prepared from malonic ester. The product prepared from malonic ester must have at least one Q H, which replaces the lost COOH. Problem 17.14

Prepare an alkyl succinic acid: @-CH-COOH

4

Alkylsuccinic acids are disubstituted acetic acids as shown above. Alkylate with RX and then with BrCH,COOC,H, to give

R-C(COOC,H,), AH,COOC,H,

-

Hydrolysis and acidification of this ester yields a tricarboxylic acid which is heated and loses CO, from one of the gem-COOH’s to form an alkylsuccinic acid.

R-C(COOC,H,),

R-C(COOH),

AH,COOC2H5 Problem 17.15 Use CH,CH,CH,OH valeramide, CH,CH,CH,CH,CONH, .

heat

+

R-CH-COOH

CH,COOH I

and H,C(COOC,H,),

&H,-COOH as the only organic reagents to synthesize 4

Since valeric acid is n-propylacetic acid,

e C H , C O O H CH,CH,CH,Br is used to alkylate malonic ester.

Problem 17.16 Synthesize cyclopentanecarboxylic acid from malonic ester.

H

>C(COOR), H’

CH,CH, \ I C(COOR),/ CH, I H CH,

I

Br

L

-

NaOEl ---+

4

CH,CH,Br

C( COOR), ______+

H’

NaObt

CH2CH2, 1 CH c ( c 0 0 R ) 2 j I ’/&a+ CH,

$‘

CH2CH2~C(COOR),-I

I

CH,CH,

’

NaOH

2 H30i

CARBANION-ENOLATES AND ENOLS

394

ACETOACETIC ESTER

(ME)

[CHAP. 17

SYNTHESIS

1. Carbanion Formation Acetoacetic ester is acidic (pK, = 10.2) and forms a resonance-stabilized carbanion whose negative charge is delocalized over one C and two 0 ’ s .

O

H

It I I

-

0

II

+

+ NaOC2H5# C2H50H+

CH3-C-C-C-OC2H5 H

2.

Alkylation As with malonic ester (Problem 17.13)’ either one or two R’s can be introduced in acetoacetic

-

ester.

R

I

-

ia(CH3COCHCOOC2H5)5 CH3COCHCOOC2H5

W2HS

R

R

I

I I R’

CH3COCCOOC2H5 3 CH3COCCOOC2H5

3. Hydrolysis and Decarboxylation

Dilute acid or base hydrolyzes the COOC,H, group and forms acetoacetic acids, which decarboxylate to methyl ketones.

O

H

O H

0

II I II CH3-C-C-C-OC2H5 I

3 C2H5OH + H O+

4

R

II I I

CH3-C-C-H

+ CO,

R

This sequence of steps can be used to synthesize methyl ketones. For a general methyl ketone, the parts are assembled as follows:

‘

from the ester

from

alkyl halides

The H* replaced COOEt. Problem 17.17 Prepare 3-methyl-2-pentanone from acetoacetic ester. In the product

H

4

CHAP. 17)

CARBANION-ENOLATES AND ENOLS

395

the - C H , and -CH,CH, attached t o m a r e introduced by alkylation of the carbanion of acetoacetic ester with appropriate alkyl halides; in this case use BrCH,CH, and then CH,I. CH,CH, H

-

n -

+

I

NaOEt

CH3CH2Br

CH,COEHCOOC,H,

CH3 I CH,CO COOC,H,-

1. hydrolysis

'iCH,CH,

I

CH,COCHCOOC,H,

1 N~OEI

2. CH3I

TH3

CH,COCH

I

CH,CH,

Methylethylacetoacetic ester

3-Methyl-2-pentanone

Problem 17.18 Prepare 2,5-hexanedione (a y-diketone) from acetoacetic ester and any alkylating reagent, 4

Monobromoacetone is the alkylating agent. 0

0 CH,-CO-CH,

0

II

CH,-C-CH,-COOEt-

NaOEt

II Na' CH,-C-‘C fi HC , H,

0

I

-H20

C-C-CH,CH,

bH

H3 0

-

H

8

kH3

CH,CH,

I

CH3CH,C=C-CCH,CH3 AH3

I !

0

( e ) No aldol condensation, because C,H,CHO has no a H. Aldol condensations require dilute NaOH at room temperatures. With concentrated NaOH at higher temperatures, C,H,CHO undergoes the Cannizzaro reaction (Problem 15.20). Problem 17.26 Some condensations have little synthetic value because they give mixtures of P-hydroxycarbonyl compounds, Illustrate with ( a ) a mixture of two aldehydes, each with an a H; (b) an unsymmetrical 4 ketone having an H on each a C . ( a ) There are four possible products. Each aldehyde reacts with itself to give two products. If one aldehyde

reacts as carbanion and the other as acceptor, and vice versa, there are two more products. Acceptor Carbanion Source

H

‘ RCH2{Fi-C=0

1

H

R ’ C H , i F ! - - - C R’ =I O HI

I

+ R’CH,C-CbH

H __*

1

I

H

I

A H k

H

Mixed aldol

H

+RCH,C-C-C=O

H

0

Selfaldol

1

H

H

H

H

d’I

H

H

bH

IL

H

H

I I RCH,C-C-C=O

H C=O

A

( 6 ) Two products are possible. There is only one acceptor, but two different carbanions can be formed-one

from each a C.

Acceptor

Carbanion Source

RICH, H R C H , { F ~ - ( ( - IC H , R ~ 0

0

-

R‘CH, H RCH,C-C-C-CH2R’ I I b H k

8

CARBANION-ENOLATES AND ENOLS

400

-

RCH, H R C H 2 - { F C - C - C IH 2 R

It

I

R‘

0

[CHAP. 17

R’CH, H RCH,C-C-C-CH,R I I

I

II

I

OH R

0

0

Problem 17.27 Mixed aldol condensations are useful if ( a ) one of the two aldehydes has no a H, (b) a symmetrical ketone reacts with RCHO. Explain and illustrate. 4 ( a ) The aldehyde with n o a H, e.g., H 2 C 0 and C,H,CHO, is only a carbanion acceptor, so that only two

products are possible.

Acceptor

Carbanion Source

-

H

H

R ~ - - H - ~ ~ , ~ ~ ~ - ~ - - -R’CH,-C-C-C-H - H I

Selfaldol

H

I

(!!)

bH

For reasonably good yields of mixed aldol product, the aldehyde with the a H should be added slowly to a large amount of the one with no a H. ( b ) Ketones are poor carbanion acceptors but are carbanion sources. With symmetrical ketones and an RCHO having an a H, two products can be formed: (1) the self-aldol of RCHO and (2) the mixed aldol. If RCHO has no a H, only the mixed aldol results. As in part ( a ) , the correct sequence of addition can give a good yield of the mixed aldol products. Problem 17.28 Give the product for the reaction of C,HSCH=O in dilute base with ( a ) CH,CHO, (6) CH ,COCH, . 4

H (a)

I

(!)I

C6H5C-CH2C-H bH

H PhCH-CHCHO

(b)

+

PhCH=CHCCH,

II

0 tram

tram

Problem 17.29 Show how the following compounds are made from CH,CH,CHO. Do not repeat the synthesis of any compound needed in ensuing syntheses. (a)

CH ,CH,CH==C( CH,)CHO

(d) CH,CH,CH,CH(CH,)CH,OH

(6) CH ,CH,CH,CH(CH,)CHO (e)

CH,CH,CH,CH(CH,),

(c) CH,CH,CH=C( CH,)CH,OH

(f)

CH,CH,CHCH(CH,)COOH

I

OH

4

Each product has six C’s, which is twice the number of C’s in CH,CH,CHO. This suggests an aldol condensation as the first step.

(4

CH,CH,CHO OH-\ CH,CH,CHCHCHO

: :A

CH,CH,CH=C(CH,)CHO

(A)

HA LH, (6) - C H O can be protected by acetal formation to prevent its reduction when reducing C=C of (A).

CHAP. 171

(A)

C H OH

CARBANION-ENOLATES AND ENOLS

CH,CH,CH=C( CH,)CH( OCH,),

2CH,CH,CH,CH(

401

~,-d HCI

CH,)CH( OCH,),

CH,CH,CH,CH( CH,)CHO

-

(There are specific catalysts that permit reduction only of C=C.) ( c ) CHO is selectively reduced by NaBH,. (A)

(4 (4

NaBH4 H2/ Pt

(A)(A)T

HZNNH2. O H -

CH,CH,CH==C(CH,)CH,OH

CH,CH,CH,CH( CH,)CH,OH

CH,CH,CH=C(CH,),

CH,CH,CH,CH(CH,),

(f) Tollens' reagent, Ag(NH,),', is a specific oxidant for CHO-COOH. 1 Ag(NHj)t

CH,CH,CH( OH)CH( CH,)CHO 7 CH,CH,CH( OH)CH( CH,)COOH r

P

Problem 17.30 Crotonaldehyde (CH,CH R,N (3"). ( b ) Water solubility depends on H-bonding between the amine and H,O. Either the H of H,O bonds with the N of the amine or the H on N bonds with the 0 of H,O: -N

I - - - H-0-H I

or

-N-HI

- - - OH,

All three kinds of amines exhibit the first type of H-bonding, which is thus a constant factor. The more H's on N, the more extensive is the second kind of H-bond and the more soluble is the amine. Thus the order of water solubility is RNH, (1") > R,NH (2") > R,N (3"). Problem 18.4 Does n-propylamine or 1-propanol have the higher boiling point?

4

Since N is less electronegative than 0 (3.1 < 3 . 3 , amines form weaker H-bonds than do alcohols of similar molecular weights. The less effective intermolecular attraction causes the amine to have a lower boiling point than the alcohol (49 "C < 97 "C).

AMINES

414

[CHAP. I8

18.2 PREPARATION

BY NUCLEOPHIUC DISPLACEMENTS 1. Alkylation of NH,, RNH, and R,NH with RX RX + NH, __* RNH 3 X -

Step 1

(an sN2 reaction)

an ammonium salt

RNHiX-

WP2 Di-, tri-, and tetraalkylation:

+ NH,+RNH,

RX

RX

RNH,-R,NH-R,N-R,N+X-HX -HX 2"

1"

+ NHaXRX

3"

-HX

4"

ammonium salt

When RX = MeI, the sequence is called exhaustive methylation. Problem 18.5 ( a ) What complication arises in the synthesis of a 1" amine from the reaction of RX and NH,? (6) How is this complication avoided? ( c ) Which halides cannot be used in this synthesis? 4 ( a ) Overalkylation: di- and tri-alkylation can occur since both RNH, and R,NH, once formed, can react with more RX. (6) Use a large excess of NH, to increase the probability of RX colliding with NH, to give RNH,, rather than with RNH, to give R,NH. ( c ) 3" RX undergoes elimination rather than SN2displacement. Aryl halides do not undergo S,2 reactions. Some ArX compounds with properly substituted rings undergo nucleophilic aromatic substitution (Section 11.3) with amines.

Problem 18.6 Describe an efficient industrial preparation of allylamine, H,G=CHCH,NH,, from propene, 4 chlorine, and ammonia.

Propene is readily monochlorinated to ally1 chloride, and the reactive C1 undergoes nucleophilic substitution with NH,. H,C=CH-CH,

C'2

H , C = C H - C H , C l s H,C=CH-CH,NH,

2. Alkylation of Imides; Gabriel synthesis of 1" Amines

COO-K'

+ RNHZ COO- K' Phthalimide

Phthalimide anion

Phthalhydrazide

See Problem 16.50 for the acidity of imides.

CHAP. 181

415

AMINES

REDUCTION OF N-CONTAINING COMPOUNDS 1. Nitro Compounds I . Zn, HCI; 2. OHor H2/Pt

C6HSN02

N itrobenzene

NO,

'

C6HSNH2

Aniline

NO,

m-Dinitrobenzene

rn-Nitroaniline

-

LAlH

RCN 4 RCH2NH2

2. Nitriles

RC-NR2

3. Amides

I

0

UAIH,

RCH2NR2 (R

=

H, alkyl, aryl)

NOH

0

NH,

4. Oximes Cyclohex ylamine

Oxime

RX + N;

5. Azides

-X-R

LiAlH o r H,IPI

N

or Na. EIOH

, RNH, ~

alkyl azide

a i d e ion

1" amine

This method is superior to the reaction of NH, and RX for the preparation of RNH, because no polyalkylation occurs. However, alkyl azides are explosive and must be carefully handled. Rather than being isolated, they should be kept in solution and used as soon as made. 6. Reductive Amination of Carbonyl Compounds

CH,CH=O-

NH

H /NI

H20

[CH,CH=NH]-

CH,CH,NH2

(NH,

an imine

-

1" amine)

H /Ni (1" amine 2"amine) + CH,CH,NH, 4 CH,CH,CH,NHCH,CH, + -OH (2"amine- 3" amine) R,C=O + R;NH- H 2 / N i [R,C=NR;]R,CH-NR; RNH, + 2 H,C=O + 2HCOOHRN(CH,), + 2 H 2 0 + 2C0, (Dimethylation of 1" amine)

CH,CH,CHO

-P

HOFMANN DEGRADATION OF AMIDES

RCONH,

+ Br, + 4KOH+

RNH,

Mechanism: Step 1

R-i-NH,

+ K,CO, + 2KBr + 2H,O

OBr-

(Brz. OH3

0

R-f-N-Br

I

O H N-bromoamide

(The amine has one less C than the amide.)

+ OH-

416

AMINES

H Step 2

R-C-N-BrI

II

+ OH-

0

-

[CHAP. 18

-

[R-t-kBr

CO:

+ H20 ..

-

N-bromoamide anion

Step 3 (electrondeficient N

Step 4

20H-

Step 5

-

+ R-N=C=O

“20

R-NH2

Problem 18.7 Synthesize 2-phenylethanamine, PhCH,CH,NH,, reduction method.

PhCH=CH,

tlBr

peroxide

PhCH,CH,Br-

NaN,

+ CO:-

(hydrolysis)

1” amine

from styrene, P h C H d H , , by the azide

PhCH,CH,N,

4

Na

PhCH,CH,NH,

Problem 18.8 Prepare ethylamine by ( a ) Gabriel synthesis, (b) alkyl halide amination, ( c ) nitrile reduction, (d) reductive amination, (e) Hofmann degradation. 4

0

0

II

(a)

+ C,H,NH,

a ) N K’:

0

0

(4)

COOH

II

II

CH,CH,CONH,

CH,CH,NH,

Problem 18.9 Prepare p-toluidine from toluene.

4

CH3 I

This is the best way of substituting an NH, on a phenyl ring.

-

Problem 18.10 Prepare CH,NHCH,CH, from compounds with one or two C’s.

CH,CHO

+ CH,NH,

ti?) N I

CH,CH,NHCH,

CHAP. 18)

417

AMINES

-

Problem 18.11 Prepare C,H,N(CH,),

from C,H,NH,.

2CH O H

C,H,NH,

4

C,H,N(CH,),

(Industrial method)

With CHJ, C,H,N(CH,)fI- is a by-product. H,CO and HCOOH cannot be used because the intermediate electrophile, H,COH, can attack the activated ring to give p-H,NC,H,CH,OH. Problem 18.12 Synthesize the following compounds from n-C ,H,,COOH and inorganic reagents:

(4 (d)

C,,H,,NH, NH2

I

1 2H2S--CH--C

I3 H 2 7

(4 Cl 2H25NH2

(b)

C13H27NH2

(e)

C12H2SNHC13H27

(f)

(C13H27)2NH

Do not repeat preparation of any needed compound. First, note the change, if any, in carbon content.

-

(a) Chain length is increased by one C by reducing RCH,CN (R = n-C,,H,,), prepared from RCH,Br and CN-. n-C,,H,,COOH-

LiAIHJ

PBr

n - C I 2 H , , C H , O H ~n-C,,H,,CH,Br+

KCN

n-C,,C2,CN

LiAIHJ

n-C,,H,,NH,

(6) The chain length is unchanged. SOCI?

~-CI,H2,COOH-

n-C,,H,,COCI+

NH,

LiAIH,

~z-C,,H,,CONH,-----*

n-Cl,H2,NH2

( c ) Chain length is decreased by one C; use the Hofmann degradation. NaOBr

~-C,,H,,CONH, ___* n-C,,H2,NH2

I-'

(d) The C content is doubled. The 1" amine is made from the corresponding ketone.

n-C,,H&OCI

n-C 1 2H,5CH 2 [see part ( a ) ]

I

LI

[see p a t

+

(n-C1 3

27 2

@)I

cuLi

n-C12Ha

"-C12H,Z

'c=o

n-C 13H27

;ZIb

n-C13H27

H ,

A

NH2

-

(e) Secondary amines may be prepared by reductive amination of an aldehyde (from RCOCl reduction), using a 1" amine. H , , Pd / B a S O J n - C 1 H 75N H n-C ,sCOCI n-C H2,CH0 n-C ,H2,NHC ,H,, H2/Ni 7

7

(f) The acyl halide C,,H,,COCI from (e) is reacted with the amine C,,H,,NH, from (b), to form an amide 0

that is then reduced.

C12H25COCl'

II

H2Nc13H27+

LIAIH,

c12H,,--C-NHc,,H,7-----+(C,,H,,),NH

Problem 18.13 There is no change in the configuration of the chiral C in sec-butylamine formed from the Hofmann degradation of (S)-2-methylbutanamide. Explain. 4

:R migrates with its electron pair to the electron-deficient :N, and configuration is retained because C--C is being broken at the same time that C-N is being formed in the transition state.

Intermediate

Transition State

AMINES

418

[CHAP. 18

Problem 18.14 A rearrangement of R, from C to an electron-deficient N, occurs in the following reactions. The substrates and conditions are given. Indicate how the intermediate is formed and give the structure of each product. (a)

Cwtius, RC-N,

II

(an acylazide) with heat

(6) Lessen, RC-NHOH

Schmidt, RCOOH

+ HN,

with H,SO,

(a hydroxamic acid) with base

8

(c)

or

Beckmann, R-C--R'

ll

with strong acid

4

NOH

Intermediate

R-C-N-OH

,,OH-H+ + H,O + 2.

+O=C=N-R

H20 OH-)

RNH,

+ C@-

-0H-

Intermediate (c)

R-f-R' NOH

R-fi-R'

+H,O

:N?H,

+ R-

$-R,O=C-R'

[\:i+ ] -R'

__*

+ H20

R-N:

Intermediate

I

R-N-H

The group tram to the OH (R) migrates as H,O leaves. 18.3 CHEMICAL PROPERTIES

STEREOCHEMISTRY

Amines with three different substituents and an unshared pair of electrons have enantiomers. However, in most cases an :NRR'R"-type amine cannot be resolved. The amine undergoes a very rapid nitrogen inversion similar to that for a C undergoing an S,2 reaction (Fig. 18-2).

transition state

Fig. 18-2

Simple carbanions, - - -):

-, behave similarly. chiral, (ii) resolvable? Give reasons in each case.

C6HSN(CH3)(C2H5) CH3

I

CH3CH2CH-N(C H3)(C2H,)

4

CHAP. 181

419

AMINES

(a) Not chiral and not resolvable, because N has two identical groups (C,H,). (6) Chiral, but the low energy barrier (25 kJ/mol) to inversion of configuration prevents resolution of its enantiomers. ( c ) Chiral and resolvable. N has four different substituents. The absence of an unshared pair of electrons on N prevents inversion as in part (6). (d) Chiral and resolvable. An asymmetric C is present.

CH,CH:bHN(CH,)(C,H,) Problem 18.16 Explain why 1,2,2-trimethylaziridine has isolable enantiomers.

4

This 3" amine is chiral because N has three different ligands and an unshared pair of e-'s. Unlike typical amines this molecule cannot undergo nitrogen inversion. The 3-membered ring requires ring angles of approximately 60" and restrains the N atom from attaining bond angles of 120°, as needed for the inversion transition state.

BASICITY AND SALT FORMATION [see Problem 3.25(b)] The lone pair of electrons on the N atom of amines accounts for their base strength and nucleophilicity. They abstract protons from water, react with Lewis acids, and attack electrophilic sites such as carbonyl carbon. Problem 18.17 ( a ) Why does aqueous CH,NH, turn litmus blue? (6) Why does C,H,NH, dissolve in aqueous HCl? 4 ( a ) Methylamine (pK, = 3.36) is a weak base, but a stronger base than water.

CH,NH, base,

+ H,O acid,

-

CH,NHl

+ OH-

acid,

base

( 6 ) A water-soluble salt forms.

C,H,NH,

+ H,O+ + Cl-+C,H,NHfCl-

+ H,O

Anilinium chloride

Problem 18.18 Account for the following observations. ( a ) In the gas phase the order of increasing basicity is NH, < CH,NH, < (CH,),NH < (CH,),N. (6) In water the order is NH, < CH,NH, = (CH,),N < (CH3)ZNH. 4 ( a ) Me, a typical alkyl group, has an electron-donating inductive effect and is base-strengthening. Basicity

increases with the increase in the number of Me groups. (6) Water draws the acid-base equilibrium to the right by H-bonding more with the ammonium cation than with the free base. The greater the number of H's in the ammonium cation, the greater the shift to the right and the increase in basicity. This effect should make NH, the strongest base and Me,N the weakest base. However, opposing the shift is the enhanced inductive effect of the increasing number of Me groups. The inductive effect predominates for MeNH, and Me,NH. In Me,N, the loss of an H more than cancels out the additional inductive effect of the third Me. Consequently, the 3" amine is less basic in water than the 2" amine.

It is not unusual that relative basicities in gas phase and water differ because of the solvation effects in H,O.

420

[CHAP. 18

AMINES

PrObl8m 18.19 (a) Assign numbers from 1 for LEAST to 4 for MOST to indicate relative base strengths of (i) C,H,NH,, (ii) (C,H,),NH, (iii) (C,H,),N, (iv) NH,. (6) Explain. 4 ( a ) (i) 3, (ii) 2, (iii) 1, (iv) 4, (b) C,H,NH2, an aromatic amine, is much less basic than NH, because the electron density on N is delocalized to the ring, mainly to the ortho and para positions [Fig. 18-3(a)]. With more phenyls bonded to N there is more delocalization and weaker basicity.

/ - -

N+

- \

....

\

Fig. 18-3 Problem 18.20 Compare the reactions of (CH,),N and (C,H,),N with BF,.

We have:

F

(CH,),N: Lewis base

F

I+ 8 : F +(CH,),N:B:F I f: F I

+

Lewis acid

(C,H,),N does not react, because the electron pair on N needed to bond to B is delocalized to 3 benzene rings. Problem 18.21 Assign numbers from 1 for following:

I ( a ) CH,NH-Na'

LEAST

I1

to 4 for MOST to indicate relative base strengths of the

111 (i-C,H,),N

C2H5NH2

IV CH,CONH2 4

I 4

I1 3

I11 2

IV 1

CH,NH, is a stronger base than NH, because the electron-donating effect of the Me group increases the electron density on N , the basic site. The three bulky i-propyl groups on N cause steric strain, but with an unshared electron pair on N , this strain is partially relieved by increasing the normal C-N--C bond angle (109") to about 112". If the unshared electron pair forms a bond to H, as in R,NH+, relief of strain by angle expansion is prevented. With bulky R groups, 3" amines therefore resist forming a fourth bond and suffer a decrease in basicity. Acyl R - C = O groups are strongly electron-withdrawing and base-weakening because electron density from N can be delocalized to 0 of the carbonyl group by extended T bonding [Fig. 18-3(6)]. I 3

I1 1

I11

IV

2

4

The strongly electron-attracting NO, group decreases electron density on N. It thus also decreases base strength by an inductive effect in the meta position, and to a greater extent, by both extended 7r bonding and inductive effects, in ortho and para positions.

CHAP. 181

42 1

AMINES

Since OCH, is electron-donating through extended 7r bonding, it increases electron density on N and the base strength of the amine because the ring accepts less electron density from N. Problem 18.22 Account for the following order of decreasing basicity:

RN H~> R N = C H R ~> R ~ N :

4

The hybrid atomic orbitals used by N to accommodate the lone pair of electrons in the above compounds are RNH,(sp3), RN=CHR'(sp'), RC%N:(sp). The nitrile (RCN)N has the most s character and is the least basic. The 1" amine has the least s character and is the most basic [Problems 8.3 and 8.5(b)].

REACTION WITH NITROUS ACID, HONO 1. Primary Amines (Diazonium Ion Formation) ( a ) Aromatic (ArNH,).

C,H, NH, ( 6 ) Aliphatic (RNH,). HONO

CH3CHZCHzNH2 ~

HONO

HCI, 5 "C

+

Benzenediazonium chloride

[C,H,-N=N:]Cl-

~ [CH3CH2CH2NJ+CIl r (unstable)

CH3CH=CH2

CH3CHCH3 CH3CHCH3

I

I c1

OH 1-Propanol

1-Chloropropane

Propene

2-Propanol

2-Chloropropane

This reaction of RNH, has no synthetic utility, bu11 the appearance of N, gas signals the presence of NH,. Problem 18.23 ( a ) Outline the steps in the formation of a diazonium cation from nitrous acid and a 1" amine. (b) Why are diazonium ions of aromatic amines more stable than those of aliphatic amines? 4 ( a ) Initially a nitrosonium cation is formed from HNO, in acid solution.

H Nitrosonium cation

Nucleophilic attack by the electron pair of the amine on the nitrosonium ion produces an N-nitrosamine. It undergoes a series of proton transfers that produce an OH group on the second N-N bond. H

H

H * *

I

k - *

.. ..

H

I

-H+

k R-N=N-OH-L

+H20

N-Nitrosamine + H Of

a diazoic acid

R-N=N:~+-H-

-H 0

+

R-NEN: Diazonium cation

AMINES

422

[CHAP. 18

(6) Aryldiazonium cations are stabilized by resonance involving the release of electrons from the ortho and puru positions of the ring.

2. Secondary Amines (Nitrosation) Ar(R)NH (or R2NH) + HONO-

Ar(R)N-NO

+ H,O

(or R,N-NO)

an N-nitrosamine (insoluble in acid)

N-Nitrosamines are cancer-causing agents (carcinogens). 3. Tertiary Amines

No useful reaction except for N ,N-dialkyl arylamines.

Problem 18.24 Prepare p-H,NC,H,N( CH,), from C,H,N( CH,), N(CH3)2

HONO

.

4

product

~

-H20

ON

With HN03,polynitration would occur since NMe, is very activating. NO' is less electrophilic than NO;.

REACTIONS WITH CARBOXYUC ACID DERWATWES; TRANSACYLATIONS 2

(R'COhO

R'CONHR

+ HCl

R'CONHR

+ R'COOH

REACTIONS WITH OTHER EUCTROPHILIC REAGENTS R'CH4

C1-

F

-C1

-H 0 + RNH, 2 R'CH-NR

+ RNH,

-

2HC1 + RNH-

e

(Schiff base, imine or azomethine)

-NHR

-

(symmetrical disubstituted urea)

0

II

R'NH-C-NHR

an isocyanate

(unsymmetrical disubstituted urea)

S

R'-N=C--C an isothiocyanate

+ HINR

II

__*

R'NH-C-NHR

(a thiourea)

423

AMINES

CHAP. 181

NUCLEOPHILIC DISPLACEMENTS 1.

Carbylamine Reactions of 1" Amines RNH2 + CHC13 + 3KOH

+ 3KC1+ 3H20

R-%:

__*

an isocyanide (foul smelling)

1: ] [

Nucleophilic RNH, attacks electrophilic intermediate [:CC12] [Problem 7.56(c)].

H

R-N+ m-O,NC,H,OH > C,H,OH

(b)

m-CIC6H40H> p-CIC,H40H > C,H,OH

4

( a ) The -NO,

is electron-withdrawing and acid-strengthening. Its resonance effect, which occurs only from para and ortho positions, predominates over its inductive effect, which occurs also from the rneta position. Other substituents in this cateogry are >C=O

-CN

X 0 0 R

-SO,R

( b ) C1 is electron-withdrawing by induction. This effect diminishes with increasing distance between C1 and OH. The meta is closer than the para po,sition and rn-C1 is more acid-strengthening than p-C1. Other substituents in this category are F, Br, I, NR,. Problem 19.14 Assign numbers from 1 for LEAST to 4 for MOST to indicate the relative acid strengths in the following groups: ( a ) phenol, rn-chlorophenol, rn-nitrophenol, rn-cresol; ( 6 ) phenol, benzoic acid, p-nitrophenol, carbonic acid; (c) phenol, p-chlorophenol, p-nitrophenol, p-cresol; (d) phenol, o-nitrophenol, rnnitrophenol, p-nitrophenol; (e) phenol, p-chlorophenol, 2,4,6-trichlorophenol, 2,4-dichlorophenol; ( f ) phenol, 4 benzyl alcohol, benzenesulfonic acid, benzoic acid. (a) 2, 3, 4, 1. Because

has

+ on N, it has a greater electron-withdrawing

inductive effect than has C1.

PHENOLIC COMPOUNDS

CHAP. 191

445

( b ) 1, 4, 2, 3. ( c ) 2, 3, 4, 1. The resonance effect of p-NO, exceeds the inductive effect of p-CI. p-CH, is electron-releasing. (d) 1, 3, 2, 4. Intramolecular H-bonding makes the o isomer weaker than the p-isomer. (e) 1, 2, 4, 3. Increasing the number of electron-attracting groups increases the acidity. ( f ) 2, 1, 4, 3.

FORMATION OF ETHERS 1.

Williamson Synthesis

+ X-

Ar-0-R'

A r O H Z ArOAr-MH,

+ CH,OSOJ

2. Aromatic Nucleophilic Substitution

2,CDinitrophenetole (2,CDinitrophenyl ethyl ether)

Problem 19.15 Why does cleavage of aryl ethers, ArOR, with HI yield only ArOH and RI?

4

An S,2 displacement on the onium ion of the ether occurs only on the C of the R group, not on the C of the Ar group. ArOR

+ HI-

ArORT-

A

+ArOH

+ RI

FORMATION OF ESTERS

Phenyl esters (RCOOAr) are not formed directly from RCOOH. Instead, acid chlorides or, anhydrides are reacted with ArOH in the presence of strong base. (CH,CO),O

+ C,H,OH + NaOH+

+ CH,COO-

CH,COOC,H,

Na'

+ H,O

Phenyl acetate

C,H,COCl

+ C,H,OH + NaOH-

C,H,COOC,H,

+ Na'

C1-

+ H,O

Phenyl benzoate

OH- converts ArOH to the more nucleophilic ArO- and also neutralizes the acids formed. Problem 19.16 Phenyl acetate undergoes the Fries rearrangement with AICI, to form ortho- and puruhydroxyacetophenone. The ortho isomer is separated from the mixture by its volatility with steam.

0

C-CH3

II

0 o-Hydroxyacetophenone

p-Hydroxyacetophenone

446

PHENOLIC COMPOUNDS

[CHAP. 19

( a ) Account for the volatility in steam of the ortho but not the puru isomer. ( 6 ) Why does the para isomer predominate at low, and the ortho at higher, temperatures? ( c ) Apply this reaction to the synthesis of the antiseptic 4-n-hexylresorcinol, using resorcinol, aliphatic compounds and any needed inorganic reagents. 4

The ortho isomer has a higher vapor pressure because of chelation, O-H---O----C (see Problem 19.5). In the para isomer there is intermolecular H-bonding with H,O. The para isomer (rate-controlled product) is the exclusive product at 25 "C because it has a lower AHSand is formed more rapidly. Its formation is reversible, unlike that of the ortho isomer which is stabilized by chelation. Although it has a higher A H S , the ortho isomer (equilibrium-controlled product) is the chief product at 165°C because it is more stable. Two activating OH groups in meta positions reinforce each other in electrophilic substitution and permit Friedel-Crafts reactions of resorcinol directly with RCOOH and ZnC!, .

Problem 19.17 Alkylation of PhO- with an active alky! halide such as CH,=CHCH,CI gives, in addition to phenyl ally1 ether, some o-allylphenol. Explain. 4 PhO- is an ambident anion with negative charge on both 0 and the ortho C atoms of the ring. Attack by 0 gives the ether; attack by the ortho carbanion gives o-allylphenol.

DISPLACEMENT OF OH GROUP Phenols resemble aryl halides in that the functional group resists displacement. Unlike ROH, phenols do not react with HX,SOCl,, or phosphorus halides. Phenols are reduced to hydrocarbons, but the reaction is used for structure proof and not for synthesis. ArOH

+ Zn A ArH + ZnO

(poor yields)

BUCHERER REACTION FOR INTERCONVERTING P-NH, AND @-OHNAPHTHALENES

&Naphthol

B-Naphthylamine

The ready availability of P-naphthol [see Problem 19.8(d)] makes this a good method for synthesizing P-naphthylamine. REACTIONS OF THE BENZENE RING 1. Hydrogenation

Cyclohexanol

447

PHENOLIC COMPOUNDS

CHAP. 191

2. Oxidation to Quinones

0

0 (KS03)2N0

,

(Fremy’s salt)

0 Phenol

Quinone

3. Electrophilic Substitution The - O H and even more so the 4-(phenoxide) are strongly activating and op-directing.

Phenol

Phenoxide anion

oxonium cations

fairly stable unsaturated ketones

Special mild conditions are needed to achieve electrophilic monosubstitution in phenols because their high reactivity favors both polysubstitution and oxidation. ( a ) Halogenation

0 OH

BrZ.H20+

OH BQB~

(goes via A ~ ono -; monobromophenols)

Br Monobromination is achieved with nonpolar solvents such as CS, to decrease the electrophilicity of Br, and also to minimize phenol ionization.

o-Bromophenol

(6) Nitrosation.

p-Bromophcnol (major)

6 6 -0 0

tautornerire

HONO+ 73T

N-0

p-Nitroso-

phenol

NOH Quinone monoxime

448

PHENOLIC COMPOUNDS

[CHAP. 19

Nitration. Low yields of p-nitrophenol are obtained from direct nitration of PhOH because of ring oxidation. Sulfonation.

o-Phenolsulfonic acid (rate-controlled)

p-Phenolsulfonic acid (equilibrium-controlled)

Diazonium salt coupling to form azophenols (Section 18.7). Ring alkylation.

RX and AICI, give poor yields because AICI, coordinates with 0. Ring acylation. Phenolic ketones are best prepared by the Fries rearrangement (Problem 19.16). Kolbe synthesis of phenolic carboxylic acids. C,H,O

Na'

+ O=C=O-

12s "C h atm

o-HOC,H,COO- Na'-

H 0 '

o-HOC,H,COOH Salicylic acid

Sodium salicylate

Reimer-Tiemann synthesis of phenolic aldehydes.

Condensations with carbonyl compounds; phenol-formaldehyde resins. Acid or base catalyzes electrophilic substitution of carbonyl compounds in ortho and puru positions of phenols to form phenol alcohols (Lederer-Manasse reaction).

RCH-0 1.

OH-(OnPhOH)

2.

H+

I

R-C-H

I

OH base catalysis

acid catalysis

Rearrangements from 0 to ring. (1) Fries rearrangement of phenolic esters to phenolic ketones (Problem 19.16). (2) Claisen rearrangement. The reaction is intrumofecufurand has a cyclic mechanism.

CHAP. 191

PHENOLIC COMPOUNDS

449

(3) Alkyl phenyl ethers.

0-H

Phenetole

o-Ethylphenol

0-H

p-Ethylphenol

Problem 19.18 Outline a mechanism for the ( a ) Kolbe reaction, (b) Reimer-Tiemann reaction. (a)

4

Phenoxide carbanion adds at the electrophilic carbon of CO,.

The conjugated ketonic diene tautomerizes to reform the more stable benzenoid ring.

(b) The electrophile is the carbene :CCl,

.n cl :OH-+ H:C:Cl__* H:OH +

c1

Cl

a b e n d chloride

Problem 19.19 Use phenol and any inorganic or aliphatic reagents to synthesize ( a ) aspirin (acetylsalicyclic acid), (b) oil of wintergreen (methyl salicylate). Do not repeat the synthesis of any compound. 4 OH

OH

OCOCHa

PHENOLIC COMPOUNDS

450

[CHAP. 19

Problem 19.20 Predict the product of the Claisen rearrangement of (a) a l l ~ l - 3 - ' ~ phenyl C ether, (6) 2,6-dimethylphenyl allyl-3-I4Cether. 4 (a) In this concerted intramolecular rearrangement, the ends of the allyl system interchange so that the y C is bonded to the ortho C.

'

0

1

OH

( 6 ) When the ortho position is blocked, the allyl group migrates to the para position by two consecutive rearrangements and the '"C is in the y position of the product.

P M ~ 19.21 I Use phenol and any aliphatic and inorganic compounds to prepare: ( a ) o-chlorornethylphenol, 4 (6) o-methoxybenzyl alcohol, ( c ) p-hydroxybenzaldehyde.

( 6 ) From (a),

o-HOC,H,CH,OH

NaOH

( C H J)2SO4

o-CH,OC,H,CH,OH

Only the phenolic OH is converted to its conjugate base with NaOH.

(4

C,H,OH

+ HCN-

ZnClZ HCI

p-HOC,H,CHO

19.4 ANALYTICAL DETECTION OF PHENOLS

Phenols are soluble in NaOH but not in NaHCO,. With Fe3' they produce complexes whose characteristic colors are green, red, blue, and purple. Infrared stretching bands of phenols are 3200-3600cm-' for the 0-H (like alcohols), but 1230 cm-' for the C-0 (alcohols: 1050-1150 cm-I). Nmr absorption of OH depends on H-bonding and the range is S = 4-12 ppm.

CHAP. 191

45 1

PHENOLIC COMPOUNDS

19.5 SUMMARY OF PHENOLIC CHEMISTRY PREPARATION

PROPERTIES 1. OH Group Reactions

1. Industrial

(a) H of OH

(b) Raschig

Ar-H

+ Fe3'

(c) NaOH Fusion ArS0,Na

-

+ NaOH+

+ HCl + 02+ NaOH+

ArO-Na'

ArO-Fe2'

+H20

(acidity)

(colored)

(d) Cumene Peroxide

+ RCOX-

base

ArO-COR

2. Benzene Ring Reactions (a) Reduction

+ 3H2/Ni-C,H,,0H

-

M 6 H 4 4 p-Quinone

]

(c) Electrophilic Substitution

+ Br2(H,0 sol.)-

2,4,6-Br,C6H20H p-BrC6H40H + H2S04+p - H M 6 H 4 - S o 3 H + dil. HN0,- 0-and p-0,NC,H40H

+ Br2(CS,)

2. Laboratory (a) Diazonium

5 "C

( b ) Sulfonic Acid Salt

+ H,C=O, + RCHO-

+ NaOH

ArSOiNa'

,-N=O,

p-O=C,H4=NOH

+ Ar'NlX-- p-HO-Ar-N-N-Ar' + ROH/H2S04+ p - H W b H 4 - R + CH,=O0- and p -Ho -c6 H,CH2 0H

HCl0- and p-HOC6H,CH2Cl RCH(C6H40H-p)2

--

19.6 SUMMARY OF PHENOLIC ETHERS AND ESTERS

ArO-Na'

+ X-CH,CH=CH,

ArO-Na'

+ X-CH2CH-CH3 0

ArO-Na'

II + X-C-CH3

heat

+ArO-CH,CH=CH, __*

-

ArO-CH2CH(CH3)2

ArO-

P

-R

-

+ AlC13

o-HOC,H4CH2CH=CH2 (A-Ph)

AlClj

p-HO-c6H4C(CH3)3 (Ar=Ph)

p- or o-HO-C6H,

P

-R

(Ar=Ph)

PHENOLIC COMPOUNDS

452

[CHAP. 19

Supplementary Problems Problem 19.22 Name the following compounds:

OH

OH

( a ) rn-ethoxytoluene, (6) methylhydroquinone, (c) p-allylphenol.

Problem 19.23 Write the structure for ( a ) phenoxyacetic acid, (6) phenyl acetate, (c) 2-hydroxy-3-phenylben4 zoic acid, (d) p-phenoxyanisole.

(a) G O C H , C O O H

(6) O O - I - C H 3

0

HOOC,

I

OH

Ether layer (upper)

Aq.

NaOH layer (lower)

I

PhO-Na'. PhCOO-Na'

PhCHIOH

Solution

r PhCOO-Na'

Precipitate

I

PhOH

(add H,O+to regenerate acid) Fig. 19-1

Problem 19.25 What product is formed when p-cresol is reacted with ( a ) (CH,CO),O, ( b ) PhCH,Br and 4 base, ( c ) aqueous NaOH, (d) aqueous NaHCO,, (e) bromine water? ( a ) p-CH,C,H,OCOCH,, p-cresyl acetate; (6) p-CH,C,H,OCH,Ph, p-tolyl benzyl ether; (c) pCH,C,H,O- Na +, sodium p-cresoxide; (d) no reaction; (e) 2,6-dibromo-4-methylphenol.

Problem 19.26 Use simple test tube reactions to distinguish ( a ) p-cresol from p-xylene, (6) salicylic acid from 4 aspirin (acetylsalicylic acid). ( a ) Aqueous NaOH dissolves the cresol. (6) Salicyclic acid is a phenol which gives a color (purple in this case) with FeCI,.

CHAP. 191

453

PHENOLIC COMPOUNDS

Problem 19.27 Identify compounds (A) through (D).

(CPSF4 OH-

' (A)

Zn*HCI

--

(B)

N a N 4 . HCl

c 6 H F

( 0

(D)

4

OH (A)

(B)

p-C,H,0C,H,N02

p-C,H,OC,H,NH,

( p-phenetidine)

(C)

p-C,H,OC,H,N

C1-

Problem 19.28 Prepare ( a ) 2-bromo-4-hydroxytoluene from toluene, (6) 2-hydroxy-5-methylbenzaldehyde 4 from p-toluidine, (c) rn-methoxyaniline from benzenesulfonic acid.

(a)

6 6 6 6 6 6 0 z* 2 +6

z i ,

B

NO2

NO2 % ,:N

(6

B

r

1.

N.OH

hue with NaOH

,

NO2

r

~

~

G0 B,

1.

r

OH

NiCICHCI3,NmOH

G C H O

2. H,O+

OH

0

S03H

2. 3.

B

OH

N;CI-

O N O 2

G

* " N ;

NH2

"0.

NH2

S03H

G

r

":%?,

H+

0 OCH3

OCH3

Zn.HCI

+

NO2

NH2

Problem 19.29 ' Use the Bucherer reaction to prepare 2-(N-methyl)- and 2-(N-pheny1)naphthylamines.

4

CH,NH, and C,H5NH, replace NH,.

Problem 19.30

(U)

Give products, where formed, for reaction of PhNlCl- with (i) a-naphthol, (ii)

p -naphthol, (iii) 4-methyl-1-naphthol, (iv) 1-methyl-2-naphthol. (b) How can these products be used to make

m~ 00 00

the corresponding aminonaphthols?

4

p ) to OH; (iii) ortho ( p ) to OH, since para ( a )position is blocked. (iv) No reaction. An activating @-substituentcannot activate other @ positions.

( a ) Structural formulas are given below. Coupling occurs (i) paru ( a )to OH; (ii) ortho (a, not

@

00

N-NPh

N-NPh (i) 4-PhenylazoI-naphthol

@N=NPh

H &

00

CH3 (ii)

1-Phenylazo2-naphthol

(iii) 4-Methyl-2-phenylazo1-naphthol

(iv)

no reaction

PHENOLIC COMPOUNDS

454

[CHAP. 19

(6) Reduction of the azo compounds with LiAlH, or Na,S,O, or Sn, HCl yields the amines.

NH2 4-Amho- 1-naphthol from (i)

CH, 1-Amino-2-naphthol from ( i i )

2-Amino4mcthyl- 1-naphthol from (iii)

Problem 19.31 From readily available phenolic compounds, synthesize the antioxidant food preservatives ( a ) BHA (tert-butylatedhydroxyanisole, a mixture of 2- and 3-tert-butyl-4-methoxyphenol; (6) BHT (tert-butylated hydroxytoluene). 4 ( a ) p-CH,OC,H,OH + (CH,),C==CH, gives the mixture. This Friedel-Craft monoalkylation occurs both orfho to the OCH, and to the OH group. (6) p-cresol, p-HOC,H,CH,, is tert-butylated with (CH,),CH=CH,.

Problem 19.32 The acid-base indicator phenolphthalein is made by using anhydrous ZnCl ,to condense 2 mol of phenol and one mol of phthalic anhydride by eliminating 1 mol of H,O. What is its foiSmula? 4

OH-

H+

0

0 Phenolphthalein (colorless)

(red)

(colorless)

Problem 19.33 From phenol prepare ( a ) p-benzoquinone, (b) p-benzoquinone dioxime, (c) quinhydrone (a 4 1 : 1 complex of p-benzoquinone and hydroquinone). OH

0

NO

tautomers

NOH

a charge-transfer complex

CHAP. 191

PHENOLIC COMPOUNDS

455

Problem 19.34 HCI adds to p-benzoquinone by a 1,4-addition. Show the steps and predict the structure of the 4 phenolic product.

6ijCl4 0

tautornerire

OH 2-Chloro- 1 ,Mihydroxybenzene

Problem 19.35 Write a structural formula for the product of the Diels-Alder reaction of p-benzoquinone with: ( a ) Butadiene

(J

(b)

(c)

r

4

1.1 '-Bicyclohexadienyl

1,3-CycIohexadiene

OH

- - 0J

OH

OH

V

taulomers

Problem 19.36 The ir OH stretching bands for the three isomeric nitrophenols in KBr pellets and in dilute 4 CCI, solution are identical for the ortho but different for mefa and para isomers. Explain.

In KBr (solid state) the OH for all three isomers is H-bonded. In CCI,, the H-bonds of mefa and para isomers, which are intermolecular, are broken. Their ir OH absorption bands shift to higher frequencies (3325-3520 cm-'). There is no change in the absorption of the ortho isomer (3200cm-'), since intramolecular H-bonds are not broken upon dilution by solvent.

-

Problem 19.37 Show all major products for the following reactions: (U)

C H 3 e O H + CHCI,

+ NaOH

(c) Br O O C C H ,

II

+

70

-c

(6) C,H,CH,OOH

+ acid

4

CzHS 0

0

(a) Reimer-Tiemann reaction;

CH,OCE:' ( b ) This is a 1.2-shift of phenyl to an electron-deficient 0.

C,H,CH,OOH

+ HX

n

c6H5cH20-p?+H

H

-HO

-t

2CH,OC,H,

3

H, W,H,

[6,

1

__*

H&=O

+ C,H,OH

a hemiacetal

( c ) Intramolecular rearrangement gives two products, (A) and (B), and intermolecular rearrangement gives two more products, (C) and (D). These are Fries rearrangements.

PHENOLIC COMPOUNDS

456

(*) G C O C H ,

Br

(c) @OC*H5

[CHAP. 19

(D) G C O C H ,

&0C2H5

C2H5

C2H5

Br

Problem 19.38 ( a ) Compound (A), C,H,O, is insoluble in aqueous NaHCO, but dissolves in NaOH. When treated with bromine water, (A) rapidly forms compound (B), C,H,OBr,. Give structures for (A) and (B). (6) What would (A) have to be if it did not dissolve in NaOH? 4 ( a ) With four degrees of unsaturation, (A) has a benzene ring. From its solubility, (A) must be a phenol with a methyl substituent to account for the seventh C. Since a tribromo compound is formed, (A) must be (6) (A) could not be a phenol. It would have to be an ether, rn-cresol and (B) is 2,4,6-tribromo-3-methylphenol. C,H,OCH, (anisole).

Problem 19.39 A compound (C,,H,,O) dissolves in NaOH but not in NaHCO, solution. It reacts with Br, in water to give C,,,H,,Br,O. The ir spectrum shows a broad band at 3250cm-I and a strong peak at 830cm-’. The proton nmr shows: a singlet at S = 1.3 (9 H), a singlet at S = 4.9 (1 H), a multiplet at 6 = 7.0 (4 H). Give the structure of the compound. 4

The level of acidity, the facile reaction with Br,, the broad band at 3250 cm-’, and the singlet at S = 4.9 suggest a phenolic compound. The multiplet at S = 7.0, the dibromination (not tribromination), and the strong band at 830cm-’ indicate a p-substituted phenol. The singlet at S = 1.3 is typical of a t-butyl group. The compound is p-HOC,H,C(CH,),. Problem 19.40 Dioxin (2,3,7,8-tetrachlordibenzo-p-dioxin),

a very toxic compound, is a by-product of the manufacture of 2,4,5-trichlorophenoI by treating 1,2,4,54 tetrachlorobenzene with NaOH. Suggest a mechanism for the formation of dioxin. NaOH converts 2,4,5-trichlorophenol to its phenoxide anion, which undergoes aromatic nucleophilic displacement with unreacted 1,2,4,5-tetrachIorobenzene.

This product undergoes another aromatic nucleophilic displacement with OH -, to give a phenoxide thaf reacts by an intramolecular displacement to give dioxin.

Chapter 20 Aromatic Heterocyclic Compounds The chemistry of saturated heterocyclic compounds is characteristic of their functional group. For example, nitrogen compounds are amines, oxygen compounds are ethers, sulfur compounds are sulfides. Differences in chemical reactivity are observed for three-membered rings, e.g., epoxides, whose enhanced reactivity is driven by the relief of their severe ring strain. This chapter discusses heterocycles that are aromatic and have unique chemical properties. 20.1 FIVE-MEMIBERED AROMATIC HETEROCYCLES WITH ONE HETEROATOM

NOMENCLATURE; AROMA TICITY The ring index system combines ( 1 ) the prefix oxa- for 0, aza- for N or thia- for S; and (2) a stem for ring size and saturation or unsaturation. These are summarized in Table 20-1. Table 20-1. Ring Index Heterocyclic Nomenclature No N

Ring Size

I

Stem Saturated

Unsaturated

irane etane olane

irene

epane ocane

epine ocin

ete ole

ane

ine

The three most common five-membered aromatic rings are furan, with an 0 atom; pyrrole, with an N atom; and thiophene, with an S atom. Problem 20.1 Name the following compounds, using (i) numbers and (ii) Greek letters. ( a1

HC -CH HC,II ,C-CH3 I1

HC-CH (6 ) H,C-C, II ,C-CH, 11

0

S (c)

HC --C--CHJ HIC-C,

I1

II

(d)

,CH 0

HC-CH Br-C,

I1

11 ,C-COOH

4

N

( a ) 2-methylthiophene (2-methylthiole) or a-methylthiophene, ( b ) 2,5-dimethylfuran (2,s-dimethyloxole) or a , a '-dimethylfuran, (c) 2,4-dimethylfuran or a , /3 '-dimethylfuran (2,4-dimethyloxole), (d) 1-ethyl-5-bromoacid). 2-pyrrolecarboxylic acid or N-ethyl-a - bromo-a '-pyrrolecarboxylic acid (N-ethyl-5-bromazole-2-carboxylic

457

AROMATIC HETEROCYCLIC COMPOUNDS

458

[CHAP. 20

Problem 20.2 Write structures for ( a ) 2-benzoylthiophene, (6) 3-furansulfonic acid, ( c ) a,p ’-dichloropyrrole. 4 SO,H

i

CI

\

H Problem 20.3 Account for the aromaticity of furan, pyrrole and thiophene which are planar molecules with 4 bond angles of 120”.

See Fig. 20-1. The four C’s and the heteroatom Z use sp2-hybridized atomic orbitals to form the U bonds. When Z is 0 or S, one of the unshared pairs of e-’s is in an sp2 HO. Each C has a p orbital with one electron and the heteroatom 2 has a p orbital with two electrons. These five p orbitals are parallel to each other and overlap side-by-side to give a cyclic 7~ system with six p electrons. These compounds are aromatic because six electrons fit Huckel’s 4n + 2 rule, which is extended to include heteroatoms. It is noteworthy that typically these heteroatoms would use s p 3 HO’s for bonding. The exceptional sp2 HO’s lead to a p A 0 for the cyclic aromatic w system. Problem 20.4 (toward 0).

Account for the following dipole moments: furan, 0.7 D (away from 0);tetrahydrofuran, 1.7 D 4

In tetrahydrofuran the greater electronegativity of 0 directs the moment of the C - 0 bond toward 0. In furan, delocalization of an electron pair from 0 makes the ring C’s negative and 0 positive; the moment is away from 0. See Fig. 20-1.

Tetrahydrofuran

Furan

s p 2-s

U

bond

Fig. 20-1

PREPARATION Problem 20.5 Pyrroles, furans, and thiophenes are made by heating 1,4-dicarbonyl compounds with (NH,),CO,, P40,0,and P2S,, respectively. Which dicarbonyl compound is used to prepare ( a ) 3,4-dimethylfuran. (6) 2,5-dimethylthiophene. ( c ) 2,3-dimethylpyrrole? 4

The carbonyl C’s become the a C’s in the heterocyclic compound.

459

AROMATIC HETEROCYCLIC COMPOUNDS

H H

I I

H3c-7-7-cH3 H-C C-H

-

~

H3C-E-y-CH3 H- ,..,C-H

___* p4°10

-"l0

II II

8

0 0 2,3-Dimethylbutanedial

a dienediol

Ace tonylacetone

H2C-cHCH 3 H-C I C-CH3 1

3,CDimethylf uran

2,5-Dimethylthiophene

(NH&CO3

d NL

____*

H

3

I

H 3-Methyl4oxopentanal

2,3-Dimeth ylpyrrole

Problem 20.6 Prepare pyrrole from succinic anhydride.

4

Zn

N

N

H

H

Succinimide

23-Dihydrox ypyrrole

I

Succinic anhydride

I

Problem 20.7 Identify compounds (A) through (D).

CH,COOC,H,

Pyrrole

+ (A) 5C,H,COCH,COOC,H,

N + ' a

(B)

dil. NmOH

(C)

(D)

4

(A) PhCOOEt

Problem 20.8 Use any necessary inorganic reagents to prepare (a) a,a'-dimethylthiophene from 4 CH,COOC,H,, ( 6 ) phenyl 2-thienyl ketone from PhCOOH and OHCCH,CH,CHO.

a ) 2CH3COOC,H,

CH,COCH,COOC,H,

NaOE1 ,2 b

CH3CO-CH-CHCOCH3

I

C,H,OOC

t

COOC,H,

AROMATIC HETEROCYCLIC COMPOUNDS

460

PhCOOH 5 PhCOCI

[CHAP. 20

1

CHEMICAL PROPERTIES Problem 20.9 (a) In terms of relative stability of the intermediate, explain why an electrophile (E') attacks the (Y rather than the p position of pyrrole, furan, and thiophene. (6) Why are these heterocyclics more reactive 4 than C,H, to E'-attack?

( a ) The transition state and the intermediate R' formed by a-attack is a hybrid of three resonance structures which possess less energy; the intermediate from p-attack is less stable and has more energy because it is a hybrid of only two resonance structures. I and I1 are also more stable allylic carbocations; V is not allylic.

major product

c

111

I

I1

V

4

IV I

less stable

more stable

very minor product

( b ) This is ascribed to resonance structure 111, in which Z has + charge and in which all ring atoms have an octet of electrons. These heterocyclics are as reactive as PhOH and PhNH,. Problem 20.10 Explain why pyrrole is not basic.

4

The unshared pair of electrons on N is delocalized in an "aromatic sextet.,' Adding an acid to N could prevent delocalization and destroy the aromaticity. Problem 20.11 Give the type of reaction and the structures and names of the products obtained from: (a) furfural,

O 0C H O a -Furancarboxaldehyde

and concentrated aq. KOH; (6) furan with (i) CH,CO-ONO, (acetyl nitrate), (ii) (CH,CO),O and BF, and then H,O; (c) pyrrole with (i) SO, and pyridine, (ii) CHCI, and KOH, (iii) PhN;Cl-, (iv) Br, and C,H,OH; ( d ) thiophene and ( i ) H,SO,, ( i i ) (CH,CO),O and CH,COONO,, (iii) Br, in benzene. 4 (a)

Cannizzaro reaction;

Potassium furoate

(6)

(i) Nitration; 2-nitrofuran,

Furfuryl alcohol

CHAP. 201

AROMATIC HETEROCYCLIC COMPOUNDS

46 1

(ii) Acetylation; 2-acetylfuran,

(c)

(i) Sulfonation; 2-pyrrolesulfonic acid, O S O , H

N

(H,SO, alone destroys ring).

H

(ii) Reimer-Tiemann formylation ; 2-pyrrolecarboxaldeh yde (2-formyl pyrrole) ,

(iii) Coupling; 2-phenylazopyrrole,

O N = N P h

N H (iv) Bromination; 2,3,4,5tetrabromopyrrole. (d) (i) Sulfonation; thiophene-2-sulfonic acid,

0 S 0 3 H (ii ) Nitration ; 2-nit rot hiop hene ,

(iii) Bromination; 2,5-dibromothiophene. (Thiophene is less reactive than pyrrole and furan.) Problem 20.12 Write structures for the mononitration products of the following compounds and explain their formation: ( a ) 3-nitropyrrole, (6) 3-methoxythiophene, (c) 2-acetylthiophene, (d) 5-methyl-2-methoxy4 thiophene, (e) 5-methylfuran-2-carboxylic acid. ( a ) Nitration at C" to form 2,4-dinitropyrrole. After nitration C5(i) becomes

at C' (ii) would form an intermediate with a

C2,and C" becomes C'. Nitration

+ on C3, which has the electron-attracting -NO,

1

I1

IdocH3

( d ) H,C

2-Ni tro-3-methox ythiophene (Q and ortho to OCH,)

2-Acet yl-5-nitro-

thiophene

(a-attack)

2-Methoxy-3-nitro-5-methylthiophene (ortho to OCH,,a stronger activating group than CH,)

group.

AROMATIC HETEROCYCLIC COMPOUNDS

462

[CHAP. 20

Attack of NO; at C2, followed by elimination of CO, and H'. Problem 20.13 Name the products formed when ( a ) furan, (6) pyrrole are catalytically hydrogenated. (a)

CO)

( tetrahydrofuran, oxacyclopentane)

(6)

(, )

(pyrrolidine, azacyclopentane)

4

N

H Problem 20.14 Give the Diels-Alder product for the reaction of furan and maleic anhydride.

4

Furan is the least aromatic of the five-membered ring heterocyclics and acts as a diene toward strong dienophiles.

Furan

p-0

Maleic anhydride

0

Problem 20.15 Give the products of reaction of pyrrole with (a) I, in aqueous KI; (6) CH,CN followed by hydrolysis; ( c ) CH,MgI. (a)

2,3,4,5-Tetraiodopyrrole

A, (stronger

acid)

+ HCI,

4

(6) a-Acetylpyrrole [see Problem 19.21(c))

B, (stronger base)

B, (weaker base)

A, (weaker acid)

Problem 20.16 Give the sequence of steps for the formation of the open-chain products from reaction of ( a ) 2,5-diethylfuran with acid (H,O' ); ( b ) pyrrole with hydroxylamine, NH,OH+ Cl-. 4 ( a ) This is the reverse of the formation of furans from 1, 4-dicarbonyl compounds [Problem 20.5(a)]. Here the

product is the diketone formed after tautomerization of the intermediate dienediol.

463

AROMATIC HETEROCYCLIC COMPOUNDS

CHAP. 201

00

H,C - CH,

H,C--CH,

(6)

I

I

HC CH

\ /

OH OH

N

H

I

t

HC CH

It

HON

II

NOH

Butanedial dioxime

20.2 SIX-MEMBERED HETEROCYCLES WITH ONE HETEROATOM

The most important example of this category is pyridine (azabenzene), C,H,N: . Problem 20.17 Write the structural formulas and give the names of the isomeric methylpyridines.

4

There are three isomers. CH3

2- or a-Methylpyridine (a-Picoline)

3- or 8-Methylpyridine (B-Picoline)

4- or y-Methylpyridine (y-Picoline)

Problem 20.18 (a) Account for the aromaticity of pyridine, a planar structure with 120" bond angles. (6) IS pyridine basic? Explain. (c) Explain why piperidine (azacyclohexane) is more basic than pyridine. ( d ) Write the 4 equation for the reaction of pyridine with HCI.

( a ) Pyridine (azabenzene) is the nitrogen analog of benzene, and both have the same orbird picture (Figs. 10-1 and 10-2). The three double bonds furnish six p electrons for the delocalized T system, in accordance with Huckel's rule. (6) Yes. Unlike pyrrole, pyridine does not need the unshared pair of electrons on N for its aromatic sextet. The pair of electrons is available for bonding to acids.

y Piper idine

s p hybridized (less s character)

& %s p

hybridized (more s character)

P yridine

The less s character in the orbital holding the unshared pair of electrons, the more basic the site.

Pyridinum chloride a salt

Problem 20.19 Explain why pyridine (a) undergoes electrophilic substitution at the /3 position, and (6) is less 4 reactive than benzene. (a) The R"s formed by attack of E' at the a or y positions of pyridine have resonance structures (I, IV) with

a positive charge on N having a sextet of electrons. These are high-energy structures.

AROMATIC HETEROCYCLIC COMPOUNDS

464

I

a-Attack

I

111

I1

I

[CHAP. 20

y- Attack

IV

V

VI

With p-attack, the + charge in the intermediate is distributed only to C's. A + on C with six electrons is not as unstable as a + on N with six electrons, since N is more electronegative than C. P-Electrophilic substitution gives the more stable intermediate. 0-Attack

(6) N withdraws electrons by induction and destabilizes the R' intermediates formed from pyridine. Also, the N atom reacts with electrophiles to form a pyridinium cation, whose + charge decreases reactivity. Problem 20.20 How do the ' H nmr spectra of pyridine and benzene differ?

4

These are both aromatic compounds, and their ring-H signals are decidedly downfield. As all the H's of benzene are alike, one signal is observed. Pyridine gives three signals (not counting spin-spin coupling): 6-8.5 (two C2 H's), S = 7.06 (two C-?H's) and S = 7.46 (lone C' H). Notice that the C' H-signal is most downfield because the N is electron-withdrawing and less shielding. Problem 20.21 Give structural formulas and the names of the products formed when pyridine reacts with ( a ) 4 Br, at 300°C; ( 6 ) KNO,, H,SO, at 300°C and then KOH; (c) H,SO, at 350°C; (d) CH,COCI/AICI,.

(a)

aBr BrmBr aNo2 (6)

N

N

N

3-Bromo- and 33-Dibromopyridine

N+

I

(KOH neutralizes the formed pyridinum salt)

3-Nitropyridine

( d ) N o reaction. Pyridine is too unreactive to undergo FriedelCrafts acylations.

H 3-P yridinesulfonic acid (a dipolar ion)

Problem 20.22 Compare, and explain the difference between, pyridine and pyrrole with respect to reactivity 4 toward electrophilic substitution.

Pyrrole is more reactive than pyridine because its intermediate is more stable. For both compounds the intermediate has a + on N . However, the pyrrole intermediate is relatively stable because every atom has a complete octet, while the pyridine intermediate is very unstable because N has only six electrons. Problem 20.23 If oxidation of an aromatic ring is an electrophilic attack, predict the product formed when a-phenylpyridine is oxidized. 4

The pyridine ring is less reactive, and the benzene ring is oxidized to a-picolinic acid (a-NC,H,COOH).

CHAP. 201

AROMATIC HETEROCYCLIC COMPOUNDS

465

Problem 20.24 Predict and account for the product obtained and conditions used in nitration of 2aminopyridine. 4

The product is 2-amino-5-nitropyridine because substitution occurs preferentially at the sterically less hindered p position para to NH,. The conditions are milder than those for nitration of pyridine, because NH, is activating. Problem 20.25 Explain why (a) pyridine and NaNH, give a -aminopyridine, (6) 4-chloropyridine and NaOMe 4 give 4-methoxypyridine, ( c ) 3-chloropyridine and NaOMe give no reaction.

Electron-attracting N facilitates attack by strong nucleophiles in a and y positions. The intermediate is a carbanion stabilized by delocalization of - to the electronegative N. The intermediate carbanion readily reverts to a stable aromatic ring by ejecting an H:- in (a) or a :Cl:- in (6).

(c)

P-Nucleophilic attack does not give an intermediate with - on N.

Problem 20.26 Give the product(s) formed when pyridine reacts with (a) BMe,, (6) H,SO,, (c) t-BuBr,

Peroxybenzoic acid

4

+

Pyridine is a typical 3" amine. (a) C,H,N-BMe,; (6) (C,H,NH):SO:-, pyridinium sulfate; ( c ) [C,H,N'-Et]I-, N-ethylpyridinium iodide; (d) C,H,NH'Br- + Me,C----CHMe (the 3" halide undergoes elimination rather than S,2 displacement); (e)

0

Pyridine-N-oxide

N

I

0Problem 20.27 Account for the following orders of reactivity:

(a) Toward H,O+ :

2,6-dimethylpyridine (2,6-lutidine) > pyridine

(b) Toward the Lewis acid BMe,:

pyridine > 2,6-lutidine

4

( a ) Alkyl groups are electron-donating by induction and are base-strengthening. (6) BMe, is bulkier than an H,O+. The Me's at C 2 and C6 flanking the N sterically inhibit the approach of BMe,, causing 2,6-lutidine to be less reactive than pyridine. This is an example of F-strain (Front strain).

AROMATIC HETEROCYCLIC COMPOUNDS

466

[CHAP. 20

Problem 20.28 Compare, and account for the products formed from nitration of pyridine and of pyridine N-oxide. 4 Pyridine is nitrated in the p position only under vigorous conditions [see Problem 20.21(6)]. However, its N-oxide is nitrated readily in the y position; the intermediate formed by such attack is very stable because all atoms have an octet of electrons. The N-oxide behaves like phenoxide ion, C,H,O- [Problem l9.l8(a)].

Problem 20.29 Pyridine N-oxide is converted to pyridine by PCI, or by zinc and acid. Use this reaction for the 4 synthesis of 4-bromopyridine from pyridine.

0Pyridine

Pyridine N-oxide

04-Bromopyridine N-oxide

4-Bromopyridine

Problem 20.30 Account for the fact that the CH,’s of a - and y-picolines (methylpyridines) are more acidic than the CH of toluene. 4 They react with strong bases to form resonance-stabilized anions with - on N.

anion (base I )

u-Picoline (acid, )

(base:)

(acid:)

anion (base )

Problem 20.31 Give the products formed when y-picoline reacts with C,H,Li and then with (a) CO, then 4 H,O’ ; ( h ) C,H,CHO then H,O‘.

AROMATIC HETEROCYCLIC COMPOUNDS

CHAP. 201

467

CH,COOH I

6

/

4-Pyridylacetic acid

N

4-Stilbazole

Problem 20.32 From picolines prepare ( a ) the vitamin niacin (3-pyridinecarboxylic acid), ( b ) the anti4 tuberculosis drug isoniazide (4-pyridinecarboxylic acid hydrazide) .

6 N

3-Picoline

Niacin

0

COOH

KMnO4,

2. I.

H soclz ~ N N H I GHNH2 ~

N

QPicoline

N Isoniazide

4-P yridine-

carboxylic acid

20.3 COMPOUNDS WITH TWO HETEROATOMS

We use the oxa-aza-thiu system to name these compounds; the suffixes -ole and mine indicate five- and six-membered rings, respectively. When there is more than one kind of ring heteroatom, the atom of higher atomic number receives the lower number. Problem 20.33 Name the following compounds:

(4

r y S

(4

0

(4

l 0g

I

H ( a ) 1,3-diazine (pyrimidine); ( b ) 1,3-thiazole; ( c ) 1,4-diazine (pyrazine); ( d ) 1,2-0xazole; (e) imidazole.

H3cc10

Three pyrimidines are among the constituents of nucleic acids:

60, NH,

Cytosine

OH

CLOH Uracil

OH

Thymine

468

AROMATIC HETEROCYCLIC COMPOUNDS

[CHAP. 20

Problem 20.34 Write the tautomeric structures of these pyrimidines.

Cytosine

Uracil

4

Thy mine

Problem 20.35 ( a ) What makes imidazole (Prob. 2 0 . 3 7 ~ )aromatic? (b) Explain why midazole, unlike pyrrole, 4 is basic. Which N is the basic site? ( a ) Imidazole is aromatic because it has a sextet of electrons from the two doubl : bonds and the electron pair on the N bonded to H. (b) The proton acceptor is the N not bonded to H, because its lone pair is not part of the aromatic sextet.

20.4

CONDENSED RING SYSTEMS

Many biologically important heterocyclic compounds have fused (condensed) ring systems. In particular, the purines adenine and guanine are found in DNA (with cytosine, 5-methylcytosine, and thymine) and also in RNA (with cytosine and uracil).

Adenine

Guanine

Quinoline

Isoquinoline

Indole

QUINOLINE ( I-AZANAPHTHALE") Problem 20.36 Which dicarboxylic acid is formed on oxidation of quinoline? The pyridine ring is more stable than the benzene ring [Problem 20.19(6)]. COOH

N Quinoline

COOH

Quinolinic acid

Problem 20.37 Quinoline is prepared by the Skraup reaction of aniline, glycerol and nitrobenzene. Suggest a mechanism involving Michael addition of aniline to an a,P -unsaturated aldehyde, ring closure, and then 4 dehydration and oxidation. The steps in the reaction are: ( 1) Dehydration of glycerol to acrolein (propenal).

H,COHCHOHCH,OHGlycerol

H2S04

H,C=CHCHO Acrolein

+ 2H,O

CHAP. 20)

469

AROMATIC HETEROCYCLIC COMPOUNDS

(2) Michael-type addition (Section 17.4) c

L

J

( 3 ) Ring closure by attack of the electrophilic carbonyl C on the aromatic ring ortho to the electronreleasing -NH. The 2" alcohol formed is dehydrated to 1.2-dihydroquinoline by the strong acid.

1,2-Dihydroquinoline

Quinoline

(4) PhNO, oxidizes the dihydroquinoline to the aromatic compound quinoline. PhNOz is reduced to PhNH,, which then reacts with more acrolein. This often violent reaction is moderated by added FeSO,. Problem 20.38 Give the expected products of (a) monobromination of quinoline, (6) catalytic reduction of 4 quinoline with 2 mol of H,. (a) The more reactive phenyl ring undergoes E +-attack in the

(Y

positions, giving

Br A

Br

I

I

8-Bromoquinoline

5-Bromoquinoline

(b) The more electron-deficient pyridine ring is more easily reduced.

I ,2,3,4Tetrahydroquinoline

Problem 20.39 Give structures for the products of reaction of quinoline with ( a ) HNO,, H,SO,; (b) NaNH,; (c) PhLi. 4 (a) 5-nitro- and 8-nitroquinoline; (b) 2-amino- and 4-aminoquinoline (like pyridine, quinoline undergoes nucleophilic substitution in the 2 and 4 positions); (c) 2-phenylquinoline. Problem 20.40 Outline a mechanism for the Bischler-Napieralski synthesis of 1-methylisoquinoline from N-acetylphenylethylamine by reaction with strong acid and P,O,, and then oxidation of the dihydroisoquinoline intermediate. 4

470

AROMATIC HETEROCYCLIC COMPOUNDS

[CHAP. 20

The mechanism is similar to that of the Skraup synthesis (Problem 20.37) in that carbonyl 0 is protonated and the electrophilic C attacks the benzene ring in cyclization, to be followed by dehydration and oxidation.

Supplementary Problems Problem 20.41 Supply systematic names for:

( a ) 4-phenyl-l,2-oxazole, ( 6 ) 3-methyl-5-bromo-l,2,4-triazine-6-carboxylic acid, (c) 2,4-dimethyl-I ,3thiazole, (d) 1,2,3,4-thiatriazoIe, (e) 2,3-benzazole (indole).

Problem 20.42 Name the following compounds systematically: (a)

11) N

H

(6)

c- 0 0 (d)

(c)

S

0

(e)

(sB S

0

U)

CJ 4

N

( a ) azole (pyrrole), (b) 1,3-thiazole, (c) 2H-oxirine, ( d ) 4H-oxirine (pyran), (e) 1,4dithiazine, ( f )

I ,3-diazine (pyrimidine). 2H- and 4H- are used in (c) and (d) to differentiate the position of the saturated sp3 atom, Common names are given in parentheses.

Problem 20.43 Write structures for ( a ) oxirane, (6) 1,2-oxazole, (c) 1,4-diazine (pyrazine), (d) 1-thia-4-oxa4 6-azocine, (e) 3H-1,2,4-triazole, ( f ) azepane.

N ( a ) H2C-CH2

0 ''

N

H

Problem 20.44 How many thiophenyl-thiophenes (bithienyls) are possible?

2,2'-Bithienyl

2,3'-Bithienyl

3,3'-Bithienyl

4

CHAP. 201

47 1

AROMATIC HETEROCYCLIC COMPOUNDS

Problem 20.45 Which methylquinoline is vigorously oxidized to a tricarboxylic acid that is then dehydrated to 4 a mixture of two carboxylic acid anhydrides?

If the CH, is on the benzene ring, oxidation gives only 2,3-pyridinedicarboxylic acid, regardless of the position of the CH,. If the CH, is on the pyridine ring, there are three possible tricarboxylic acids:

W

C

I

H

WCH3

3

HoococooH I oxid.

oxid.

oxid.

N

N

HOOC

H HOOC o o C ONC O O H

HOOC

- H20

0

0

0

II

II

F D C O O H

4‘

O\

‘

II

N

H

II

0

0

HOOC

N

N

0

Two anhydrides are obtained only from 4-methylquinoline.

Problem 20.46 Identify the compounds represented by Roman numerals.

Quinoline

P -CsHs OrH

Pyrrole (PyH) Furan (FuH)

HNOj

I

+ p-HO3SC6H4N2X-

+I-

(CH,CO),O

(CzHs)zO. BF3

I

PCI

11 A 111

NaOl

Sn HCI

I1 + I11

f u m . H SO

I1A I11

I

I1

Quinoline N-oxide

4-Nitroquinoline N-oxide

4-Nitroquinoline

2-PyN=NC,H4SO3 H-p

2-PyNH2

p-NH,C,H,SO,

2-FuCOCH3

2-FuCOO-Na’

5-H03S-2-FuCOOH

4

I11

Problem 20.47 Prepare ( a ) 3-aminopyridine from P-picoline, (b) 4-aminopyridine from pyridine, ( c ) 8hydroxyquinoline from quinoline, (d) 5-nitro-2-furoic acid from furfural, (e) 2-pyridylacetic acid from pyridine. 4

AROMATIC HETEROCYCLIC COMPOUNDS

472

0Ouinoline

_*,

H2S04

[CHAP. 20

0-

Ouinoline-8-sulfonic acid

OH-

I.

-

.. _ _ _

U

-

(COOH stabilizes the ring towards acid cleavage of the ether bond.)

(4

Pyridine

NHj

2-Aminopyridine

-

2-Bromopyridine + [CH(COOC,H,),l

Na'

__*

nucleophilic nucleophilic displacement

N

I

N a N 0 2 . H ' . 5 "C Cuer

2-Bromopyridine

:::

CH(COOC,H,),

'.

0 CHZCOOH

A. -cq

N N LV

Problem 20.48 ( a ) Explain why pyran [Problem 20.42(d)] is not aromatic. (6) What structural change would theoretically make it aromatic? 4 ( a ) There are six electrons available: four from the two bonds and two from the 0 atom. However, C4 is sp'-hybridized and has no p orbital available for cyclic p orbital overlap. (6) Convert C4 to a carbocation. C4 would now be sp2-hybridized and would have an empty p orbital for cyclic overlap.

Problem 20.49 How can pyridine and piperidine be distinguished by infrared spectroscopy?

4

Piperidine has an N-H bond absorbing at 3500 cm-' and H - C ( s p 3 ) stretch below 3000 cm-'. Pyridine has no N-H; has H--C(sp2) stretch above 3000cm-'; C = C and C=N stretches near 1600 and lSOOcrn-', respectively; aromatic ring vibrations near 1200 and 1050 cm - ; and C-H deformations at 750 cm - The peak at 750 varies with substitution in the pyridine ring. Problem 20.50 How can nmr spectroscopy distinguish among aniline, pyridine and piperidine?

4

The NH, of aniline is electron-donating and shields the aromatic H's; their chemical shift is 8 = 6.5-7.0 (for benzene the chemical shift is S = 7.1). The N of pyridine is electron-withdrawing and weakly shields the aromatic H's (6 = 7.5-8.0). Piperidine is not aromatic and has no signals in these regions. Problem 20.51 From pyridine (PyH), 2-picoline (2-PyMe), and any reagent without the pyridine ring prepare

( a ) 2 -acetylpyridine, ( 6 ) 2-vinylpyridine (c) 2 -cyclopropylpyridine (d) 2 -PyCH,CH,CH,COOH ' (e) 2-

PyC( Me)==CHCH,, ( f ) 2-pyridinecarboxaldebyde. Any synthesized compound can be used in ensuing steps.

4

-

( a ) By a crossed-Claisen condensation

2-Py Me

KMnOI

-

EtOH

2-PyCOOH 7 2-PyCOOEt

2-PyCOCH2COOEt

(6)

(4

OH -

heat

2-PyCOCH3-

C H COOEt

NaBHI

-

-

2[2-PyCOCH2COOH]

2-PyCOCH2COO-

-

2-PyCHOHCH3

2-PyCH=CH2+ CH,N,

uv

PP,"

2-PyCH=CH2

product

- CO2

2-PyCOCH3

CHAP. 201

AROMATIC HETEROCYCLIC COMPOUNDS

( d ) By a Michael addition:

-

2-PyCH=CH2 + (EtOOC),CH Na'

473

--+ 2-PyCHCH2CH(C0OEt),Na'

This a-pyridylcarbanion is stabilized by charge delocalization to the ring N (Problem 20.30). Refluxing the salt in HCl causes decarboxylation and gives the pyridinium salt of the product, which is then neutralized with OH-. ( e ) Use the Wittig synthesis: 2-PyCOMe + Ph,P=CHCH, +products (?runs and cis).

(f)

1. 0

2-PyCH=CH2 3 2-PyCHO 2. Zn, HOAc

Problem 20.52 Suggest a mechanism for the formation of 2-pyridone from the reaction of pyridine with KOH.

4

As with the strongly basic :NHS [Problem 20.25(u)], nucleophilic attack by :OH- at Cz, followed by :H--elimination, yields an enolic intermediate that tautomerizes to 2-pyridone.

Problem 20.53 ( a ) Account for the aromaticity of a- and y-pyridones. ( b ) Explain why a-pyridone (Problem 20.52) predominates over a -pyridinol, especially in the solid state. (The same is true for the y-tautomers.) (c) How can ir spectroscopy show which tautomer -predominates? 4 (a)

See Fig. 20-2 for the cyclic aromatic extended .rr-bonding with six electrons. The carbonyl carbon furnishes an empty p AO, N furnishes a p A 0 with a pair of electrons, and each doubly bonded C furnishes a p A 0 with one electron.

Fig. 20-2

( b ) The N-H forms a strong intermolecular H-bond with 0 of C=O. This bond, repeated throughout the crystalline solid, links molecules in endless helices. ( c ) The ir spectra of the solid (and solution) show a strong C=O stretzhing band.

Chapter 21 Amino Acids, Peptides, Proteins 21.1

INTRODUCTION

This chapter deals with the very important a-amino acids, the building blocks of the proteins that are necessary for the function and structure of living cells. Enzymes, the highly specific biochemical catalysts are proteins. a-Amino acids are dipolar ions (zwitterions), RCH(N+H,)COO-, as is indicated by their crystallinity, high melting point, and solubility in water rather than in nonpolar solvents. The standard (naturally occurring) amino acids are listed in Table 21-1; those marked with an asterisk are essential amino acids that cannot be synthesized in the body and so must be in the diet. They have 1' NH,'s except for proline and hydroxyproline (2'). They have different R groups. Table 21-1. Natural a-Amino Acids

Name

Formula

Symbol

MonoaminomonocarboxyIic Glycine Alanine Valine* Leucine* Isoleucine* Serine Threonine*

I

+

H,NCH,COOH,hCH(CH,)COOH,N+ C H( i-Pr)COOH,N+ CH(i-Bu)COOH,NC H( s-Bu)C 00+ H,NCH(CH,OH)COO+ H,NCH(CHOHCH,)COO-

GlY Ala Val LeU

Ileu Ser Thr

Monoaminodicarboxylic and Amide Derivatives Aspartic acid Asparagine Glutamic acid Glutamine

ASP ASP(NH3 Glu Glu(N H,)

+

HOOCCH,CH( N+H,)COOH,NCOCH,CH(NH,)COO+ HOOC( C H2),C H(N H ,)COOH,NCOCH,CH,CH(kH,)COO-

Diaminomonocarbox y lic ~

+

~~

H ,N( C H,),C H( N H,)COO-+ H,NCH, HCH,CH,CH(NH,)COO-

Lysine* H ydrox ylysine

'iOH

H2N\

+..yC-=N

Arginine*

H (CH 2)3CH(N HJCOO-

H2N

C y steine Cystine Met hionhe*

CySH cysscy Met

+

H,NCH(CH,SH)COO-

-OOCCH(~H,)CH,S-SCH,CH(hH,)COOC H ,SC H ,C H,C H

474

(6H ,)COO-

475

AMINO ACIDS, PEPTIDES, PROTEINS

CHAP. 211

I

Table 21-1. Natural a-Amino Acids (Continued) Name

Phenylalanine* Tyrosine

Formula

Symbol

Phe TYr

+

PhCH,CH(NH,)CO?p-HOC,H,CH,CH( NH3)COO-

Histidine* ~

,

Proline

Hydrox yproline

H o br3 C O O -

H'

'H

&

C H2C HCOO-

Tryptophane*

+

L

3

H

Except for glycine, they are chiral molecules with the configuration at the a C.

L

or S (or R, for cysteine and cystine)

Problem 21.1 (a) Find the absolute configuration of the following L-amino acids:

coo-

coo-

coo-

CH,SH

CH,OH

CH,CH,SCH,

(i) L-Cysteine

(ii) L-Serine

(iii) L-Methionine

( 6 ) Why is L-cysteine (along with L-cystine) unique among the twenty a-amino acids in having the R 4 configuration? (a) Redrawing the Fischer formulas with the lowest-priority ligand (H) away from the viewer and assigning priorities, we obtain:

AMINO ACIDS, PEPTIDES, PROTEINS

476

/

\H

[CHAP. 21

(iii) S

(ii) S

(i) R

( b ) The CH,SH group of cysteine is one of two R's (the other is part of cystine) that outrank COO- because S has a higher atomic weight than 0.

21.2

PREPARATION OF a-AMlNO ACIDS

LABORATORY AND INDUSTRIAL METHODS Problem 21.2 Synthesize leucine, (CH,),CHCH,CH(NH,)COO , by ( a ) Hell-Volhard-Zelinsky reaction (Section 16.3) followed by ammonolysis, (6) Gabriel synthesis, (c) phthalimidomalonic ester synthesis, ( d ) reductive amination of a keto acid, (e) Strecker synthesis (addition of NH,, HCN to RCH=O), ( f ) 4 acet yla m i nomalonic ester . Hr2 P

(CH,),CHCH,CH,COOH-

excess

( C H , ) 2 C H - - € H 2 C H B r C O O H r Leu

QQt;; Q&pT 0

Br

C

cII

COOEt

COOEt HjO+

CH2 I

II

11

CH(C H A

-L&u+Q

COOH COOH

I

CH(CHh Use ester rather than acid to prevent conversion of anion to phthalimide. 0

C H , ( C O O E t ) , z BrCH(COOEt),

-

0

II

1. 2,

PhthN--CH(COOEt)2

COOEt -CH,CHMe,

0 (PhthN -)

(use below)

NaOH HjOf

1. 2.

Ns+OEtBcH2CHMe2

Leu

3. hept

'

+ Phthalicacid + CO2 + 2EtOH

AOOEt

R

NH

,

ME~CHCH~CCOOHT

H

I

Me,CHCH,C=O

LEU

F

H

+ NHf

0'

Me,CHCH, HCN A Leu

NH2 In the presence of NH, an aminonitrile is formed instead of a cyanohydrin.

( f ) H,C(COOEt),

HON=C(COOEt), an oximino

ester

1. 2,

H2IPI

Ac.O

+

AcNHCH(COOEt),

NaOEt BcH2CHM9+

Acet y laminomalonic

ester

COOEt

1.

2. 3,

NaOH HJO"

b

Leu

+ CO, + 2EtOH

AMINO ACIDS, PEPTIDES, PROTEINS

CHAP. 211

477

Problem 21.3 Write structural formulas for compounds (A) through (C) and classify the reactions. 4 2

'CH2COOC2H, (A) 'L=O I

(crossed-Claisen condensation)

*COOC2H, 1

3

4

(B) CH,-C-COOH

I

0 (C) CH,-CH-COO-

I

(ester hydrolysis and decarboxylation of a P-keto acid; see Section 17.3) (reductive amination)

+NH3 Problem 21.4 Outline the preparation of ( a ) methionine from acrolein by the Strecker synthesis [Problem 4 21.2(e)], (6) glutamic acid by the phthalimidomalonic ester synthesis [Problem 21.2(c)].

H NHJ Methionine

Phthalimidomalonic ester

oddition

COOEt

I

CH,CHCOOEt % H3A-

phthN-lOOEt

Glutamic acid

BIOSYNTHESIS: TRANSAMINATION In the transamination reaction, in the presence of transaminase enzymes, (S)-glutamic acid reacts with the a-keto acid analog of the desired a-amino acid to give the desired (S)-amino acid and the keto analog of glutamic acid. HOOCCH,CH,CHCOO-

I NH ;

+ HOOCCH,CCOO-

Glutamic acid

b

transaminase

Oxaloacetic acid

HOOCCH,CH,CCOO-

b

+ HOOCCH,CHCOOI

a-Ketoglutaric acid

NH_:

Aspartic acid

Problem 21.5 (a) What is the relationship between reactants and products in the transamination reaction? (6) Which ketoacid is needed to give (i) alanine? (ii) leucine? (iii) serine? (iv) glutamine? (c) Which amino acids cannot be made by transamination? 4 (0)

Glutamic acid + Oxalacetic acid+ oxidant

reductant

a-Ketoglutaric acid + Aspartic acid reductant

( b ) (i) CH,COCOOH; (ii) (CH,),CHCH,COCOOH; (iii) HOCH,COCOOH;

(iv) H,NCOCH,CH,COCOOH.

oxidantz

AMINO ACIDS, PEPTIDES, PROTEINS

478

(c)

[CHAP. 21

Proline and hydroxyproline; they are 2' amines, and only 1' amines can be made this way.

21.3 CHEMICAL PROPERTIES OF AMINO ACIDS

ACID-BASE (AMPHOTERIC1 PROPERTIES Problem 21.6 Write equilibria showing the amphoteric behavior of an H 2 0 solution of glycine. All monoaminomonocarboxylic acids are slightly acidic. 4 H,O+

+ H,NCH,COO-

-

K,] = 1 0 -2

KOZ=16 X 1 0 - I o

Anion (acts as RNH2)

Ampholyte

H,kCH,COOH

+ OH-

Cation (acts as RCOOH)

Since the solution is acidic, the reaction to the left is favored and the anion predominates. Problem 21.7 ( a ) How does one repress the ionization of an a-amino acid? (6) The pH at which [anion] = [cation] is called the isoelectric point. In an electrolysis experiment, do a-amino acids migrate at the isoelectric point? 4 ( a ) Since the a-amino acid solution is slightly acidic, the anion predominates, and acid is added to repress the ionization. (6) No.

Problem 21.8 Why is H,NCH,COOH (pK,, = 2.4) more acidic than RCH,COOH (pK, t

a -H,N

= 4-5)?

4

increases acidity because of its electron-withdrawing inductive effect.

Problem 21.9 ( a ) Why does sulfanilic (p-aminobenzenesulfonic) acid exist as a dipolar ion, while p 4 aminobenzoic acid does not? (6) Why is sulfanilic acid soluble in bases but not in acids? ( a ) -SO,H is strongly acidic and donates H' to the weakly basic arylamino grpup. ArCOOHjs not acidic enough to transfer H' to the arylamino group. (6) In dipolar sulfanilic acid, p-H,NC,H,SOJ, H,N- is acidic enough to transfer H' to bases to give the soluble anion, p-H,NC,H,SO;, -SO; is too feebly basic and cannot accept H' from acids.

Problem 21.10 Predict whether the isoelectric points for the following a-amino acids are considerably acidic, slightly acidic, or basic; ( a ) alanine, (6) lysine, (c) aspartic acid, (d) cystine, (e) tyrosine. (See Table 21-1 and Problem 21.6.) 4 (a) Alanine is a monoaminomonocarboxylic acid. Its isoelectric point is slightly acidic (pH = 6.0). (6) Lysine is a diaminomonocarboxylic acid.

H,N(CH,),CH(NH2)COO-

+ H 2 0=H,N(CH,),CH(NH,)COO-

+ OH-

Additional base is needed to repress this ionization. The isoelectric point of lysine is basic (pH = 9.6). ( c ) Aspartic acid is a monoaminodicarboxylic acid. HOOCCH,CH(~H,)COO- + H,O-

-OOCCH,CH(hH,)COO-

+ H,O+

A more strongly acidic solution is needed to repress this ionization. Aspartic acid's isoelectric point is strongly acidic (pH = 2.7). ( d ) Cystine is a diaminodicarboxylic acid and behaves like a monoaminomonocarboxylic acid. The isoelectric point is slightly acid (pH = 4.6). (e) Tyrosine is a monoaminomonocarboxylic acid containing a phenolic OH, which however is too weakly acidic to ionize to any significant extent. The isoelectric point is slightly acidic (pH = 5.6).

CHAP. 211

479

AMINO ACIDS, PEPTIDES, PROTEINS

Problem 21.11 Why is the isoelectric point of arginine (pH = 10.7) greater than that of lysine (pH = 9.6)? 4

The second basic site in arginine is an R-substituted guanidino group. NH

II

H,N-C-NHR This group is more basic than the charge-delocalized cation:

E

NH, of lysine because, on accepting H', the guanidino group gives a

Problem 21.12 How can lysine be separated from glycine?

4

The isoelectric points are different: pH = 9.6 for lysine and pH = 5.97 for glycine. An aqueous solution of the mixture is placed between two electrodes, the pH adjusted to either 5.97 or 9.6, and an electric current applied. At pH 5.97, glycine doesn't migrate, while cationic lysine migrates to the cathode, where it is collected. At pH 9.6, lysine doesn't migrate, while anionic glycine migrates to the anode. Problem 21.13 Give the products of the reaction of glycine with ( a ) aqueous KOH, (b) aqueous HCI. ( a ) H,NCH,COO

- K ',

4

(6) CI- H,NCH,COOH.

REACTIONS OF FUNCTIONAL GROUPS Problem 21.14 Show that amino acids have properties typical both of - C O O H and -NH,, by giving the products formed when glycine reacts with ( a ) CH,COCI; (6) NaNO, + HCI(HON0); (c) EtOH + HCI; (d) PhCOCl+ NaOH; (e) Ba(OH),, heat. 4 ( a ) CH,CONHCH,COOH (6) HOCH,COOH + N, (c) C1- H, NCH,COOE t

(d) PhCONHCH,COO - Na +

(e) CH,NH,

+ BaCO, + H,O

(decarboxylation)

Problem 21.15 On being heated, amino acids behave like hydroxyacids (Problems 16.24 an$ 16.25). giving N-analog products. Write the structural formulas for the products formed on heating ( a ) o-RCHNH,COO-, (6)

p-RCHhH,CH,COO-, (c) yH,N(CH,),COO-, (d) 6-H3N(CH,),C00-. (e) E-H~N(CH,),COO-. R O

O R

an a-amino acid

a diketopiperazine

(b) RCH=CHCOO-NH: (c)

d ; H

.heat*

6: N H

+ H,O

H a y-amino acid

a y-lactam (a cyclic amide)

4

[CHAP. 21

AMINO ACIDS, PEPTIDES, PROTEINS

480

mer of Nylon 6

Intramolecular cyclization would give a 7-membered ring, which is formed with difficulty. Hence the more facile intermolecular reaction occurs. Since the substrate is bifunctional, polymerization occurs (see Section 16.12). Problem 21.16 When the methyl ester of rac-alanine is heated, two diastereomeric dimethyldiketopiperazines [Problem 21.15(a)j are obtained. One of them cannot be resolved. Give structures for the two products, and account for their stereochemistry. 4

The cis is resolvable.

CH3

CH3

cis ;racemic

trans ;meso

Problem 21.17 Threonine has two chiral C's. Write Fischer projections for its stereoisomers.

cooH3k&-I H OH \

coo-

cooH$H3 HO

CH3

CH3

H3i$ HO

coo-

OH

CH3

,

,3H3

4

CH3

,

racemate 2 (erythro)

racemate 1 (threo)

Problem 21.18 Write structural formulas for the products of the reactions of: (a) alanine + carbobenzoxy 4 chloride (C,H,CH,OCOCI); (6) tyrosine + aq. Br,, and the product reacted with (CH,),SO, + NaOH.

O H H

I I

I

C,HsCH,OC-N--C-COOH

'

CH30bCH2-CH-C Br

NH2

No \O-Na+

Problem 21.19 Can the acid chloride of an a-amino acid be made by adding SOCI, to the acid?

4

48 1

AMINO ACIDS, PEPTIDES, PROTEINS

CHAP. 211

No. Two or more molecules of amino acid chloride react to give peptides. H,NCHRCOCI

+ H,NCHRCOCI+

H,NCHRCONHCHRCOCI

ANALYTICAL DETECTION Amino acids are separated by chromatography and detected with ninhydrin; they give an extensively charge-delocalized, purple anion. 0

03=N-b 0-

0

+ NH;

CHRCOO----,

II

II

0

II

0

Ninhydrin hydrate

+ R c H o + CO, + H +

0 purple

Problem 21.20 Many a-amino acids react with HONO to give a quantitative volume of N2 (Van Slyke method of analysis). Which amino acids in Table 21-1 cannot be analyzed by this method? 4

The amino acid must have an -NH,. HONO.

Proline and hydroxyproline are 2" amines and do not evolve Nzwith

Problem 21.21 The ir spectra of amino acids have bands near 1400 and 1600cm-' that correspond to the COO- group. Why does a strong peak appear at 1720cm-' in highly acidic solution? 4

The 1400- and 1600-cm- I peaks are caused by the symmetrical and antisymmetrical C - 0 stretching in COO-. Acid converts COO- to COOH, whose C=O produces the 1720-cm-' peak.

21.4

PEPTIDES

INTRODUCTION Problem 21.22 Peptides are polyamides, like the product in Problem 21.15(e). Draw the structural formula for (a) a dipeptide, (h) a polypeptide unit of alanine (indicate the mer). 4 N-terminal acid

C-terminal acid

+

1

,

H3N-(fH-E-Y-TH-COOMe H Me H

I

H

H

I

I

H

I

a polypeptide

Problem 21.23

Name the tripeptide, Ala.Gly.Va1.

4

The full name of the C-terminal amino acid is used. For all other amino acids, the suffix -yl replaces - h e of the name. The name is alanylglycylvaline. Problem 21.24 Draw the structural formula for the pentapeptide glycylglycylalanylphenylalanylleucine (Gly .Gly .Ala. Phe. Leu). 4

482

AMINO ACIDS, PEPTIDES, PROTEINS

[CHAP. 21

+

The sequence starts at the left end with the amino acid having the free a NH, and ends on the right with the amino acid having the free COO-. GlY /

+

L

' -

GlY

Ala

Phe

A

A

f

H3NCH2~--~-CH2~-~-

I

LeU

II

IH3

A

I I LH2

H CH,CH(CH,),

hh Problem 21.25 ( a ) The artificial sweetener aspartame is a synthetic dipeptide, Asp.Phe. How many stereo4 men can aspartame have? (b) Which isomer would result if naturally occurring amino acids were used?

( a ) Since each amino acid component has one stereocenter. aspartame has two different stereocenters. There are four stereomers: SS and RR to give one racemic form, and SR and RS to give another. (6) The naturally occurring amino acids are both S, and their combination would give only the SS stereomer.

STRUCTURE DETERMINATION 1.

Total Hydrolysis to Amino Acids

Peptides (and proteins) are hydrolyzed with acid in amino acid analyzers to give the kind and number of each amino acid constituent. 2. Partial Hydrolysis to Smaller Polypeptide Chains Large peptides are partially hydrolyzed with enzymes (trypsin and chymotropsin) or acid to mixtures of di- and tripeptides. From the composition of these small units we establish the sequence of amino acids in the large polypeptide. Problem 21.26 What products are expected from ( a ) complete and (6) partial hydrolysis of the tetrapeptide 4 Ala .Glu .G1y .Leu? ( a ) The amino acids Ala, Glu, Gly, and Leu. (6) The dipeptides Ala.Glu, Glu.Gly, Gly.Leu and the tripeptides Ala. Glu.Gly and Glu .Gly .Leu.

Problem 21.27 What is the sequence of amino acids in the following peptides? ( a ) a tripeptide Gly,Leu,Asp (commas indicate an unknown sequence) that is partially hydrolyzed to the dipeptides Gly.Leu and Asp.Gly; (6) a heptapeptide Gly,Ser,His, ,Ala, ,Asp that is hydrolyzed to the tripeptides Gly.Ser.Asp, His. Ala.Gly, and 4 Asp. His. Ala.

S i y e Gly is bonded to Leu in one dipeptide and to Asp in the other, Gly must be in the middle. The free -NH, is in aspartic acid in Asp.Gly and the free --COO- is in leucine in Gly.Leu. The sequence is Asp.Gly. Leu. The N-terminal acid in the heptapeptide must be the first amino acid in one of the three tripeptides. The possibilities are Gly,His, or Asp. However, Gly and Asp, which are one of a kind in the heptapeptide, are also found in the third position of different tripeptides. Hence, Gly and Asp cannot be first in the original compound. In the heptapeptide, there are two His units, one of which must be the first amino acid. His is first in His.Ala.Gly and this unit comes from the first three amino acids of the heptapeptide. The C-terminal amino acid must be the last amino acid in one of the three tripeptides. The possibilities are Asp, Gly, or Ala. Since the one-of-a-kind Gly and Asp are multi-positioned in the tripeptides, they cannot be terminal in the heptapeptide. Ala must be the end amino acid in the heptapeptide and the tripeptide Asp.His.Ala comes from the last 3 amino acids. The seventh and middle amino acid is Ser. The tripeptides are arranged in order, and cross lines remove duplicating amino acids. His.Ala.Gly

+ @$.Ser.Mp + Asp.His.Ala = His.Ala.Gly.Ser.Asp.His.Ala

CHAP. 211

483

AMINO ACIDS, PEPTIDES, PROTEINS

SEQUENCING

By lopping off one terminal amino acid at a time from either the free amino end (N-terminal) or the free COOH end (C-terminal), sequencing techniques give the linkage order of the amino acids in the peptide. 1. Sanger method

The N-terminal acid is detected with 2,4-dinitrofluorobenzene

H

H

NO*

peptide

DNFB

(DNFB).

N-(2,4-dinitrophenyl) peptide (N-DNPpeptide)

R

O

H I 11 w-CH-C-OH

(This step is an aromatic nucleophilic displacement.)

+

+

R'

I

H3N-C-COO-

I

H N-( 2,4-dinitrophenyl)amino acid (identified)

unlabeled amino acid

2. Edman Method

The initial reagent for detecting the N-terminal group is an aryl isothiocyanate. R O H R ' O H R " Ar-N-C-S

-!.

I

+ H-N-

11 I

I

I1 I

-C-N-~-C-N-~-COOH H

aryl isothiocyanate

I

*

H

peptide

H

S H

Ar-N-C-yI I1

I

R 0 H R ' O H R"

I -CtN-C-C-N-C-COOH II I I II I I I I H

dry HCI

H

Thiourea-labcled peptide

S

II

R' 0 H

C

Ar-N' 'h-H 0-C-C-R I I

H

Phenylthiohydantoinlabeled amino acid (identified)

+ H3A-(!f

R"

-c-N-rcO " I

H

I

H

peptide less H,N terminal amino acid (can repeat degradation)

AMINO ACIDS, PEPTIDES, PROTEINS

484

[CHAP. 21

3. Carboxypeptidase enzyme is used to detect the C-terminal amino acid.

R 0 H R 0 H R' +

I

I I

I

I I

H3N-C-C-N-C-C-/-N-C-COO-

I

I

H

H

I I

carboxy-

peptidase b

R O H R

H

H

H

R"

+ I + H3N-C-COOI H

+ I I I I I H,N-C-C-N-C-COOI I

Problem 21.28 Determine the amino acid sequence of a peptide which: (i) on complete hydrolysis, separation of the amino acids, and determination of the amounts, shows the composition Le u z, A rg ,Cy SH ,G lu ,I le u ,Val ,Tyr ,Phe (ii) in sequencing studies reacts with DNBF to produce N-DNP tyrosine on hydrolysis, and yields valine with carboxypeptidase; (iii) undergoes partial hydrolysis to the tripeptides Leu.Val.Va1 + Leu.Arg.CySH

+ Tyr.Ileu.Phe + Phe.Glu.Leu + Arg.CySH.Leu

4

(i) There are 10 amino acids in the chain, but since there are 2 mol each of Leu and Val, only 8 are distinct. (ii) Sequencing establishes tyrosine as the first ( l ) , and one of the valines as the last ( l O ) , amino acid in the chain. (iii) From partial hydrolysis, the tripeptides establish the first three amino acids as Tyr(l), Ileu(2), Phe(3); and the last three as Leu(8), Val(9), and Val( 10). The remaining amino acids (4) through (7) are placed from the other tripeptides. There is only one Phe(3) and thus the tripeptide Phe(3).Glu(4).Leu(S) gives the fourth and fifth acids. There are two Leu molecules and one has been identified as Leu(8), so that the second Leu, which is (5) in the chain, is in the tripeptide Leu(5).Arg(6).CySH(7). Thus amino acids (6) and (7) are established. Placing the tripeptides in correct position and eliminating common acids, we have:

The position of CySH(7) is confirmed from the tripeptide Arg(6).CySH(7).Leu(8). The sequence in the peptide is Tyr.Ileu.Phe.Glu.Leu.Arg.CySH.Leu.Va1 .Val.

SYNTHESIS The problem in combining different amino acids in a specific sequence to synthesize polypeptides is to prevent reaction between the COOH (or COX) and NH2 of the same amino acid (Problem 21.19). Blocking the N-Terminal Amino Group

1.

This reaction is blocked by attaching a group (B) to N of NH,. After the desired peptide is constructed, the blocking group is removed without destroying the peptide linkages. The general scheme is: +

H,NCHRCOO-

block NH,

(BPNHCHRCOOH

activate COOH

(Bb-NHCHRCOX

H~~HR'COO-

0 NHCHR'COOH

II

H3NCHRCNHCHR'COO+

485

AMINO ACIDS, PEPTIDES, PROTEINS

CHAP. 21)

Table 21-2

Blocking

(B) Group PhCH,OC=O

I

Unblocking

Reagent

Name

Reagent

PhCH20COCl

Carbobenzoxy chloride

HJPt or cold HBr, HOAc

di-t-Butyl di-carbonate

CF,COOH

(2)

t-BuOC=O

,

(f-Buo% )

I

(Boc)

2°

c

Products PhCH, + CO, + peptide PhCHzBr+ CO, + peptide (CH3),C=CHZ + peptide

+ CO,

2. Activating the COOH The --COOH is activated to form a peptide bond with an N-terminal NH2 by reacting the N-blocked amino acid with ethyl chloroformate, CIC-OEt

8

A mixed anhydride of an alkyl carbonic acid is formed. This technique minimizes racemization of the chiral (Y C. Unactivated COOH groups form peptide bonds directly with N-terminal amino groups in the presence of

(N ,N-dicyclohexylcarbodimideor

cyclohexyl-N=C=N-cyclohexyl

DCC)

a powerful dehydrating agent that forms the urea derivative H--c y cloh e x y I

cy clohex y l-N H-C-N

tl

Problem 21.29 Prepare Gly.Val by the ( a ) carbobenzoxy and (6) Boc methods. Use symbols Z and Boc as in 4 Table 21-2, and X= - 0 C O O E t . 0 0 (U)

Z-CI

+ H,N+ CH,COO-

-

II

ClCOEt

Z-NHCH~COO- -Z-NHCH~C--%

II

H,I Pt

- PhCH3. -CO?

H,N'CH,CONHCH( i-Pr)COO-

(6) Boc-0-Boc+ H,N' CH,COO--Boc-NHCH,COO-. fluoroacetic acid, CF,COOH, is used instead of HJPt.

3.

k i 3 N < . t i ( i .Pr ) C ' O O (va,,ne)

*

N-protected, COOH activated

N-protected

0

2-NHCH,CNHCH(i-Pr)COOH-

II

(product)

Repeat as in ( a ) , changing Z to Boc. Tri-

Merrifield Solid-Phase Synthesis

To expedite peptide syntheses of up to 124 amino acid units, beads of solid polystyrene-with chloromethyl groups, - C H , C I , para to about every hundredth phenyl group-are used in an automated process. The sequence of steps is: (1) attach C-terminal Boc-protected amino acid by

AMINO ACIDS, PEPTIDES, PROTEINS

486

[CHAP. 21

replacing C1 of the CH,C1 group, and wash; (2) remove Boc and wash; (3) add Boc-protected amino acid with the aid of DCC, and wash. Steps (2) and (3) are repeated as many times as is necessary to make the desired peptide. In the final step the completed peptide is detached by hydrolysis of the ester bond that holds the peptide to the solid surface. The average yield for each step exceeds 99%.

Problem 21.30 Describe the Merrifield solid-phase synthesis when carried out in the reverse order of the classical Sanger and Edman methods. 4

A Boc-protected amino acid is first attached to the resin and then the Boc is removed to liberate a free amino group. This is reacted with another Boc-protected amino acid. To synthesize, e.g., Ala.Gly, the Gly is first attached to the resin.

Problem 21.31 Which a-amino acid can act to cross-link peptide chains?

4

Cysteine. Cysteine linkages also occur between distant parts of the same chain.

-

Cross-linking

21.5

s

PROTEINS

INTRODUCTION

,

Proteins with molecular weights in the millions are the major constituents of all living cells. Simple proteins are hydrolyzed only to amino acids. Coqjugated proteins are hydrolyzed to amino acids and nonpeptide substances known as prosthetic groups. These prosthetic groups include nucleic acids of nucleoproteins, carbohydrates of glycoproteins, pigments (such as hemin and chlorophyll) of chromoproteins, and fats or lipids of lipoproteins. PROPERTIES 1.

Amphoteric Properties. Isoelectric Points and Electrophoresis

Proteins have different isoelectric points, and in an electrochemical cell they migrate to one of the electrodes (depending on their charge, size, and shape) at different speeds. This difference in behavior is used in electrophoresis for the separation and analysis of protein mixtures.

CHAP. 211

AMINO ACIDS, PEPTIDES, PROTEINS

487

2. Hydrolysis Proteins are hydrolyzed to a-amino acids by heating with aqueous strong acids or, at room temperature, by digestive enzymes such as trypsin, chymotrypsin, or pepsin. 3.

Denaturation

Heat, strong acids or bases, ethanol, or heavy-metal ions irreversibly alter the secondary structure of proteins (see below). This process, known as denaturation, is exemplified by the heat-induced coagulation and hardening of egg white (albumin). Denaturation destroys the physiological activity of proteins.

STRUCTURE 1. The primary structure consists of the number and sequence of the constituent amino acids. 2. The secondary structure arises from different conformations of the protein chains; these conformations are best determined by X-ray analysis. There are three types. ( a ) The a-helix is a mainly right-handed coiled arrangement maintained by H-bonds between an N-H and O=C that are four peptide bonds apart. (b) The pleated sheet has chains lying side by side and linked through N - H - - - - O K H-bonds. The a C’s rotate slightly out of the plane of the merger (to minimize repulsions between their bulky R groups), which gives rise to the “pleats.” (c) Random structures have no repeating geometric pattern. However, there are structural constraints on the randomness leading to an ordered random orientation. 3. Tertiary structure is determined by any folding of the chains. There are two types. ( a ) Fibrous proteins are water-insoluble, elongated, threadlike helixes (occasionally pleated sheets) made up of chains which are bundled together intermolecularly through N-H - - - O=C H-bonding. They include fibroin (found in silk), keratin (in hair, skin, feathers, etc.), and myosin (in muscle tissue). (b) Globular proteins, or globulins, are folded into compact spheroid shapes such that hydrophilic R groups point outward toward the water solvent and the hydrophobic (lipophilic) R groups turn inward. As a result, globulins can dissolve in or easily emulsify with water. The shape is maintained by intramolecular H-bonding. The secondary structure is a combination of random (always present), helical, and pleated structures. Globulins include all enzymes, antibodies, albumin of eggs, hemoglobin, and many hormones such an insulin. 4. Quaternary structure exists when two or more polypeptide chains are linked only by weak forces of attraction among R groups at the surface of the chains. Such proteins are called oligomers (dimers, trimers, and so on).

Problem 21.32 The position of an amino acid along the chain and the nature of its R group critically influence protein function. This is revealed by a study of sickle-cell anemia, a hereditary disease affecting red blood cells (erythrocytes). Normal hemoglobin, part of disk-shaped erythrocytes, has a glutamic acid residue at position 6, whereas abnormal hemoglobin, part of crescent-shaped (“sickle”) cells, has a valine residue. Explain how this single error, (there are 146 amino acid units) causes sickling. 4 Glutamic acid has an R group with a polar ionic end (--COO-) that extends outward toward the water solvent, giving the normal disklike shape. Valine has a lipophilic R group, i-propyl (isopropyl), which pulls the portion of protein where it occurs inward away from the water, causing sickling.

488

[CHAP. 21

AMINO ACIDS, PEPTIDES, PROTEINS

21.6 SUMMARY OF AMINO ACID CHEMISTRY

PREPARATION

PROPERTIES

1. a-Amino Acids

1. a-Amino Acids

(a) Ammonolysis

(a) Amino Group

+ NaOH -+ H , N - C H R - C O O -

X

I

OH

+ CHR-COOH

Halo Acids: NH,

+ HNO,+

I

CHR-COO-

Gabriel: Qf>N--CHR--COOH

+ R'-C(t-X+ + CH,--V+ Ba(OH),-

R

Malonic Acid: NH,

Na'

I + X-C-COOH

HI

R'--CONCHR--COOH,C=N--CHR--COOH,N--CH,R + BaCO,

AOOH COOH

I

HN--CHR--C----O

+ NH, + H,

Keto Acids: R&O

I

W-CHR-NH

\

( b ) Alkylation

( b ) Carboxyl Group

\+

/

Acetylaminomalonic Ester:

R\/C \ CH3%-N COOR'

I

H +--+ H,&--CHR--COOH + R'OH+ H,~--CHR--COOR' + HOH 0

+'

SOCl,+

II

H,k-CHR--C--CI

Azlactone Synthesis H 0

I

II

R (c) Aminonitrile RCH-0 2.

+ NH,CN

-

@-AminoAcids

NH,

I

RCH-CN

2.

/

(a) Halo Acids, Gabriel, etc. X-CH,CH,COOH + NH, ( b ) Unsaturated Acids H , C= C-C O O H 3.

k

I :N-"\

N=C;

+

H,N-CH,CH,COO-

/

p-Amino Acids

+ HNO,

T H , C - - C - - IC O O H

k

+ NH,'

y- and &Amino Acids

(a) Halo Acids X--CH,CH,CH,--COOH

H,C=C--COOH

-

+ NH3

3.

k

+ NH,

7- and &Amino Acids

H3N--CH2CH2CH2-COO--

heat

NH

+ N,

CHAP. 21)

AMINO ACIDS, PEPTIDES, PROTEINS

489

Supplementary Problems Problem 21.33 (a) How many dipeptides can be synthesized from the amino acids in Table 21-I? (6) How many dipeptides in (a)have the empirical formula (i)X2,(ii)X,Y? ( c ) How many dipeptides are there with (i) two (ii) one essential amino(s)? (6)Give the number of possible proteins with 100 amino acids. (a) There are 23 choices for X, followed by 23 choices for Y. Thus there are 23 x 23 = 529 possible dipeptides. (6) (i) There are 23X,'s. (ii) For X, Y, there are 23 x 22 = 506. (c) There are 10 essential amino acids X* and 13 non-essential ones, (i) lOX: (ii) 10 x 13 = 130 when the H,N terminal amino acid is essential (X?Y) and 13 x 10 = 130 when the COOH terminal amino acid is essential (Y.X*). The total =260. (6)23100. Problem 21.34 Name the tetrapeptide

0 0 H ,N-CH-C-N+H--C-N+H-C-NH-CH~-COO 11 H F H

0

II

-

4

Val.Phe. Ala.Gly, or valylphenyalanylglycine. Problem 21.35 Count the possible amino acid sequences for a tetrapeptide having the empirical formula 4 Gly, ,Ala,. (The number of sequences equals the number of ways of reserving two out of four positions €or glycine, which is 4! G.G.AA, G.A.G.A., G.A.A.G, A.A.G.G, A.G.A.G, A.G.G.A Problem 21.36 Write the structural formula for tyrosylglycylalanine.

4

+

-

Problem 21.37 Identify the compounds (I), (11), (111). (I)

+ H,NCHMeCOOCH,CONHCHMeCOO(11) + HONO+ N, + HOCH,CH(n-Pr)COOH (111) + NH, + H2-MeCH(NH3)COONaOH

Na' 4

(I) ( C H , C 0 ) 2 0 ; (11) H,NCH,CH(n-Pr)COO- ; (111) MeCOCOOH. Problem 21.38 Identify the compounds ( I ) through (IV).

aP

\ N - K +5 (11) /

C

II

0

hydrolysis

+

4

PhCOOH

(111)

AMINO ACIDS, PEPTIDES, PROTEINS

490

[CHAP. 21

(I) ethyl bromoacetate (or any haloacetate); (11) PhthN--CH,-COOC2H, [Problem 21.2(c)]; (111) PhCOCl; (IV) hippuric acid, PhCONHCH,COOH. Problem 21.39

synthesis.

Synthesize valine ( a ) from isobutyl alcohol by a Strecker reaction, (6) by malonic ester

4

I

(a 1

CH, NH2

CH3

CH,

I

NHlCl NaCN

* CH$HCH=O

CHJCHCHZOH

+

I

CH, NH,

I

I

H O+

I

* CH,CH-CHCN 3 CH,CH-CHCOOValine

CH, CH,CHBr

(b)

CH, H

CH3

I

I

&a:CH(COOR)2 +

CH3CHCH(C00R)2

I 2.

3.

NaOH.hea1 HCI Br2/p

4

NH3(xs)

I

t

+ CH3cH--F-c00-+NH, Valine

Problem 21.40 Starting with acetamidomalonic acid, prepare phenylalanine.

COOEt

I

HCNHCOCH,

Loon

::

4

-

COOEt

I

* PhCH2-C-NHCOCH,

'NH,

I

PhCH,-CH-COO-

AOOEt

Problem 21.41 Use p-methoxytoluene and any other needed reagents to prepare tyrosine.

0'' :: '

CH30

C H 3 0 ~ c H 2 B r If,

CH(COOEI)~,

CH,O

CH,-CH(COOEt),

4

i, heal

3

'

$JH3 0cH2cH2c00H CH3O

CH2dHc00-

I.

Br2/P

2.

N H J (excess)*

CH3O NH3BrCH2CHCOOH I

HO

conc. -CHjBr HBr. A

~

-0

NH3

I

CH2CHCOO-

H Tyrosine

p-CH,C,H,OH itself cannot be used because the ring would be brominated and the acidic H of OH would convert basic malonate carbanion to malonic ester. t

Problem 21.42 Calculate the isoelectric point of alanine, given for H,NCHMeCOOH pK,, = 2.3 and pK,, = 9.7. 4

Isoelectric point = 4 (2.3 + 9.7) = 6.0. Problem 21.43 What occurs when an electric current is passed through an aqueous solution, buffered at pH = 6.0, of alanine (6.0), glutamic acid (3.2), and arginine (10.7)? The isoelectric points are shown in parentheses. 4

Glutamic acid is present as an anion and migrates to the anode. Arginine is a cation and migrates to the cathode. Alanine is a dipolar ion and remains uniformly distributed in solution.

CHAP. 211

491

AMINO ACIDS, PEPTIDES, PROTEINS

Problem 21.44 Supply structural formulas for principal organic compounds (I) through (XIV).

02N0N:2

Ph-N=C=S

-

loo-

+ H3N+

H-CH,--CH,-S-CH, +

+ (CH3),CHCH-NH3

(I)

(111)

1(1;)

Loo-

F

2 Ph- COCl+ -0OC H ( C H , ) , N H , ’ Z (IV) NH2

+

NH3

I

C&CH2-cH-cOOHooCOO-

+ (CH3CO)20OH-\ (V)

+ C,H,OH

(VI)

H/”\H

[

=(XI) -

I -

-0OC-CHCHzCOO-

H+

H+

OH-

OH-

(:;I)] (XIII)

02N@H-kH-CH2-CH2SCH3

loo-

PhCO-NH-

H(CHJ4-NH-COPh

HO ON C O O C 2 H , (VI) H’

‘H

H+

OH-

(I)

(IV)

(XIV)

AMINO ACIDS, PEPTIDES, PROTEINS

492

1

(VII) -OOC--SH(CH,),NH,

+kH3

U)

-

+

+

-OOC-CH(CH2),NH3

H0OC-CH(CH2),NH3

I

L

‘NH,

I

NH,

[CHAP. 21

+

(1x1

+

11 (XI)

Problem 21.45

Explain the use of the enzyme deacylase in the resolution of racemic amino acids.

4

The amino group of the racemic amino acid is acetylated with acetic anhydride to give racemic N-acetylamino acid. The deacylase hydrolyzes only the L enantiomer, which is thereby separated from the D enantiomer. The latter is then chemically hydrolyzed to give the free D amino acid. Problem 21.46 Use DNFB

(see Section 21.4, “Sequencing”) to distinguish between Lys.Gly and Gly.Lys.

DNFB

+ H,F;(cH?),CH(NH,)CONHCH,COOLys.Gly

--+

HN(CH,),CHCONHCH,COOH=

I

DNP

I

DNPNH

HN(CH3,CHCOOH

I

The a and

E

NH2’s are bonded to 0 H

DNFB

1I I

COO-

I

DNP

H

DNP

.

I

DNPNH

+ H,NCH,COOGlycine

COOH

I

+

+ H,NCH2C-N-~(CH2),NH3

4

__*

H O+

NH --L

)

DNP

DNP

Gly.Lys

HNCH,COOH

I DNP

+ HOOC-CH(CH,),NH +kH,

I

DNP

After hydrolysis of the DNP-derivative of Gly.Lys, both amino acids have a D N P attached. However, only the amino acid with DNP attached to the a NH, (in this case, glycine) is the NH, terminal amino acid. When a diaminomonocarboxylic acid is part of a peptide chain, but not in an NH, terminal position, it will be derivatized by D N P but not at the a NH,. Problem 21.47 Among the hydrolysis products of the pentapeptide Gly,,Ala,Phe are Ala.Gly and Gly.Ala. 4 Give two possible structures if this substance gives no N, with HONO (Van Slyke method).

Since N, is not obtained, there is no free NH,. The compound must be a cyclic peptide. It has the partial sequence Gly.Ala.Gly. The remaining Gly and Phe units can be placed in two sequences, giving

Gly .Ala.GIy Gly.Ala.Gly . or . Gly.Phe Phe.Gly Problem 21.48 Indicate the products obtained after a second Edman degradation of Ala.Phe.Gly.Val.

4

CHAP. 211

493

AMINO ACIDS, PEPTIDES, PROTEINS

S

II

C

Gly.Val

and

Ar-N

,./ \ NH

I

I

O=C--CHPhe

-

-

Problem 21.49

Prepare glycylalanyltyrosine from the free amino acids by the carbobenzoxy method.

C,H,CH,OCOCI

+ H,NCH,COO-

+

4

alanine

C,H,CH,OCONHCH,COOH~

-GlY

GiY

C,H,CH~OCONHCH,CONHCH(CH,)COOHGlY

Ala

C,H,CH,OCONHCH,CONHCH(CH,)CONH . . . COOH[Dcc-see

page 485-forms

H,i

I

\

GlY

Ala

a peptide from the acid itself.]

Pd

C,H,CH,

+ CO, + Gly.Ala.Tyr

TYr

Use Boc (Table 21-2) as a protecting group to synthesize the tripeptide Ala.Gly.Va1. (Let 4 H CH H CH CH,

Problem 21.50

X- - 0 C O O E t .)

+

I

H,N--C--COO-

A

H

CH,

H3N+--CH2--C00-

Eoc-0-Eoc ______*

HI

I k H O

+

(1-Pr)

I I

H,N-C--COO-

H

CH

H

k Problem 21.51 Account for the stereochemical specificity of enzymes with chiral substrates.

4

Since enzymes are proteins made up of optically active amino acids, enzymes are themselves optically active and therefore react with only one enantiomer of a chiral substrate. Problem 21.52 Explain how the following denature proteins; ( a ) Pb2+and Ag', (6) EtOH, (c) urea, and (d) heat, 4 ( a ) These heavy-metal cations form insoluble salts with the free COO- groups. (6) EtOH interferes with H-bonding by offering its own competing H. (c) Urea is a good H-bond acceptor and interferes with H-bonding in the protein. (d) Heat causes more random conformations in the protein and decreases the conformational population needed for helix formation.

Chapter 22 Carbohydrates and Nucleic Acids 22.1

INTRODUCTION

Carbohydrates (saccharides) are aliphatic polyhydroxyaldehydes (aldoses), polyhydroxyketones (ketoses), or compounds that can be hydrolyzed to them. The suffix -ose denotes this class of compounds. The monosaccharide D-( +)-glucose, an aldohexose, is formed by plants in photosynthesis and is converted to the plysaccharides cellulose and starch. Simple saccharides are called sugars. Polysaccharides are hydrolyzable to monosaccharides; e.g., a mol of trisaccharide gives 3 mol of monosaccharides. Problem 22.1 Classify the following monosaccharides: (a)

( 6 ) HO-CH2-

g-l-CH2-OH

( c ) HO-CH2-

H6H

-

6

H-

6,

H-

H-

6H 6H

H-CH2-OH

6,

H-CH-0 4

6H

The number of C's in a chain is indicated by the prefix -di-, -tri-, etc. (a) ketotriose, (6) aldohexose, (c) ketopen tose . Problem 22.2 ( a ) Write a structural formula for (i) an aldotriose, (ii) a ketohexose, (iii) a deoxypentose. ( b ) Write molecular formulas for (i) a hexose trisaccharide, (ii) a pentose polysaccharide. (c) What is the general formula for most carbohydrates? 4 ( a ) (i)

HO-CH,

(ii)

H--CH=O

(iii)* 0 OH OH OH

OH

TH=O CH2 I HCOH

I

* "Deoxy" means H in place of OH.

HCOH

I

CH20H

(6) Because n moles of C,H,,O, (monosaccharide) = 1 rnol of polysaccharide + (n - I ) rnol of H 2 0

we have: (9 n (c)

= 3,

C,,H,,O,,;

(ii) (C,H,O,),

C,(H,O),. This formula accounts for the name carbohydrate.

Problem 22.3 Deduce the gross structure of glucose from the following data. ( a ) The elementary composition is C=40.0%, H = 6 . 7 % , 0=53.3%. (6) A solution of 18g in 100g of H,O freezes at -1.86"C. (c) Mild oxidation with Br2/H,0, Ag(NH,)i (Tollens' reagent) or Cu" (tartrate) complex (Fehling's test) gives a carboxylic acid with the same number of C's. (4 This acid, an aldonic acid, reacts with HI and P to give n-hexanoic acid. (e) One mole of glucose reacts with 5 rnol of Ac,O to form an ester. (#) Vigorous oxidation of glucose with HNO, gives a dicarboxylic acid (an aldaric acid) with the same number of C's. (g) Reduction of glucose with H,/ Ni or Na/ Hg gives a product that reacts with 6 mol of Ac,O. 4

494

CARBOHYDRATES AND NUCLEIC ACIDS

CHAP. 221

495

(a) The molar amounts of C, H, and 0 in 1OOg of glucose would be

53.3

6.7 - = 6.7 1

4012.0= 3 * 3 3

-= 3.33 16

Thus, C : H : 0 = 1 : 2 : 1, and the empirical formula is CH,O (empirical mol. wt. = 30). (6) 18 g in 100 g of H,O is equivalent to 180 g in 1 kg of H 2 0 . Since Afp(freezing point) = 1.86, this solution is one-molal (Afp = K,,rn), and 180g is the gram molecular weight. The molecular formula is (CH,O),, where 180(molecularweight) =6 30(empirical weight)

n=

Thus glucose is C,H,,O,. (c) Glucose has a HC=O group, which undergoes mild oxidation to COOH. ( d ) The six C’s of glucose are in a chain sequence (there are no branching C’s). (e) Glucose has five OH’s that are acetylated to a pentaacetate. ( f )One of the OH’s is l”, since it was oxidized to a COOH along with CHO. (g) CHO is reduced to CH,OH to give a hexahydroxy compound (an alditol). The gross structure of glucose is HOCH,CHOHCHOHCHOHCHOHCHO.

---

Problem 22.4 ( a ) The monosaccharide

Fructose

HCN

D-(

-)-fructose undergoes the reaction sequence

H30’

HI/P

CH,CH,CH,CH,CHCH,COOH

Classify fructose. (6) Which carboxylic acid is formed when glucose is submitted to the same sequence?

4

( a ) Fructose is a hexose since the final product has seven C’s, one of which came from HCN addition to a

M.The precursor of COOH is CN, which added to C=O. COOH is attached to the C that was originally the C of -0. Fructose is a 2-ketohexose. The sequence is:

-

=O

HO-

HCN

H20H 2-ketohexose

cyanoh ydrin

hydroxy acid

2-Methylhexanoic acid

(6) By this sequence, glucose gives heptanoic acid. Problem 22.5 ( a ) How many chiral C’s are there in a typical (i) aldohexose (C,H,,O,), (ii) 2-ketohexose? (6) How many stereomers should an aldohexose have? 4 ( a ) (i) Four:

6

5

4

3

2

6

I

5

4

(6) There are four different chiral C’s and 2, = 16 stereomers.

22.2 CHEMICAL PROPERTIES OF MONOSACCHARIDES 1.

Oxidation [see Problem 22.3(c) and ( f ) ]

Problem 22.6

I

6

What are the products of reaction of HOCH,(CHOH),CHO (I) and

0 2

II

HO~H,(CHOH),-C-&H,OH with ( a ) Br,

+ H,O?

(6) HNO,? (c) HIO,?

3

2

1

(ii) 3: HOCH,$HOH$HOH$HOHCOCH,OH

HoCH,$HOH$HOH$HOH$HOHCHO

(11)

496

[CHAP. 22

CARBOHYDRATES AND NUCLEIC ACIDS

(a)

( b) (c)

I

I

I

I

HOCH,(CHOH),COOH

N o reaction (ketones are not oxidized)

0 II HOOC(CHOH),-C-COOH

HOOC(CHOH),COOH tH,=O

I1

l6

+ 4HCOOH + HdOOH

(1" OH'S

react)

+ 3HCOOH + 60, + dHz=O

CH,=O

2. Reaction in Base Problem 22.7 ( a ) Explain how in basic solution an equilibrium is established between an aldose, its Cz epimer (a diastereomer with a different configuration at one chiral C) and a 2-ketose. (b) Will fructose give a positive Fehling's test which is done in a basic solution? 4 (a)

In basic solutions, aldoses and ketoses tautomerize to a common intermediate enediol, and the following equilibria are established: achiral, sp'

H-C-0 H I~

chiral, s p 3

U

1 .

C

~

O

I

H

H-C-OH

I C-o

~

I

I

H-C-OH I

H-~--oH

I

H-&OH

I

enediol

aldose

ketose

When the aldose is reformed from the enediol, H' can attack the now achiral C2 from either side of the double bond to give C' epimers. chiral, sp'

H-A-OH

I

I

enediol

I C 2 epimeric aldoses

(6) Since fructose isomerizes to an aldohexose (glucose) in base, it gives a positive test: Cupric tartrate (blue solution)

C ~ , O(red solid)

3. Reduction to Alditols [Problem 22.3(g)J Problem 22.8 Glucose is reduced to a single alditol; fructose is reduced to two epimers, one of which is identical to the alditol from glucose. Explain in terms of configurations. 4

In glucose no new chiral C is formed on going from CHO to CH,OH. In fructose the carbonyl C (C') becomes chiral; there are two configurations and we get epimers.

CH,OH

>c=o R'

---

CH,OH

~--foH

+H

CH,OH o k H

The chiral C-', C', and C' [see Problem 22.5(a)] of glucose and fructose have the same configuration. The identical alditols also have the same configuration at C'.

CHAP. 221

CARBOHYDRATES AND NUCLEIC ACIDS

497

4. Kiliani-Fischer Step-up Method (Addition of One Carbon)

CN

H-

COOH

&-OH),

-!%(H-t-oH),

-OH

H-

H-C-0

H-

-OH (!H20H

I

LH20H

LH20H

HCN. NaCN

COOH

CN

I

HO-

-H

I (H-C-OH)3

CH2OH

I

H O*

A

H-A-OH

I

I H-

I

CH2OH aldohexose

H-C=O

CH20H

cyanohydrins

glyconic acids

I

CHzOH I

CH20H

glyconolactones

aldoheptoses

diastereomeric and epimeric

Problem 22.9 Why does the Kiliani-Fischer method give unequal amounts of diastereomeric products?

4

The presence of stereocenters in sugars causes their C = O groups to have diastereotopic faces (Section 5.4) that react at different rates, resulting in unequal amounts of diastereomers.

5. Degradation (Loss of One Carbon) (a)

Ruff degradation:

glyconic acid

aldohexose

aldopentose

( b ) Wohl degradation: Problem 22.10 Outline the steps in the Wohl degradation, which employs a dehydration of an aldose oxime and is thus a reversal of the Kiliani-Fischer step-up method. 4 HC=O

I

HC=NOH

I

HCN is removed from the cyanohydrin by Ag'.

C=N

I

498

6.

[CHAP. 22

CARBOHYDRATES AND NUCLEIC ACIDS

Reaction with Phenylhydrazine. Osazone Formation

CH-0

'

I

I

CHoH

H-C-OH

PhNHNH

I

A

(CHOH)3 AH20H

H-C=N

NHPh7

I

C-NNHPh

PhNHNH

PhN H N H2

d

-PhNH2. - N H J *

(H- A-OH), I 6H20H

H-C-NNHPh

(H-k-OH),

J

LH20H

phen ylh ydrazone

aldohexose

osazone

Problem 22.11 The C'-epimeric aldohexoses glucose and mannose give the same osazone as fructose. Write 4 equations and explain the configurational significance.

Since mannose and glucose are C' epimers, they are identical at the C3,C', C-', and C" portions of the molecule, which are unaltered during osazone formation. The chiral C 2 , which differs in each hexose, loses chirality in the osazone and becomes identical for both mannose and glucose. 2-Ketohexoses give osazones in which the C ' CH,OH is oxidized. Since C ', CJ, C.', and C" of fructose and glucose are identical (Problem 22.8), fructose also gives the same osazone. Identical portions are in the boxes below.

k, CHO

H-

H-C-NNHPh

CHO

-OH

Or

3PhNHNH2

or

H o - l ,

Glucose

=NNHPh

same osazone

Mannose C' epimers

: :1

Problem 22.12 Osazones are converted by PhCHO to 1,2-dicarbonyl compounds called osones. Use this 4 reaction to change glucose to fructose.

H-C=O

H-C-NNHPh

I

3PhNHNH2*

(H-

k="Hph -OH),

H-C=O 1-2 PhCH-NNHPh) phCHO

H20H Glucose

-CHO

22.3

1.

osazone

*

Zn.HAc,

(H-

-OH),

(H-

-OH),

H20H osone

Fructose

rather than C=O of the osone is reduced.

EVIDENCE FOR HEMIACETAL FORMATION AS EXEMPLIFIED WITH GLUCOSE

Some Negative Aldehyde Tests

Glucose does not form a bisulfite adduct or give a magenta color with the Schiff reagent. 2.

Mutarotation

Naturally occurring (+)-glucose is obtained in two forms: mp = 146 "C, [a],= +112" and mp = 150 O C , [a],= + 19". The specific rotation of each changes (mutarotates) in water, and both reach a constant value of +52.7".

499

CARBOHYDRATES AND NUCLEIC ACIDS

CHAP. 221

3. Acetal (Glycoside) Formation Unlike typical aldehydes, 1 mol of glucose reacts with only 1 mol of ROH in dry HCI to give two isomeric acetals (alkyl glycosides). See Problem 15.35. To explain the above results, glucose is assumed to exist as a cyclic hemiacetal in equilibrium with a small amount of the open-chain aldehyde. In the hemiacetal, C' is chiral and two diastereomers (anomers) are possible; they are called a- and p-glucosides. The C of C=O is always the anomeric position. Cyclic hemiacetals form when an OH and a CHO group in the same molecule can form a five- or six-membered ring that is more stable than the open chain. 1

-I

- H-I"-oHb c-

I

aGlucose: [aID=

CHJOH/dVHCI

+112",

C' OH on right side

I CHzOH

Hk

- 0 ~H

I

0

Methyl a-glucoside

8

CHjOHldry HCI

-CH)O-

F

Glucose (open-chain)

-C5

I

@-Glucose:[ a ] , = + 1 9 , C' OH on left side

i i I

Methyl 6-glucoside

Hemiacetal Anomers

Problem 22.13 Why do aldoses give positive Fehling and osazone reactions but negative Schiff and bisulfite tests? 4 These reactions are typical of the -CHO group. The Schiff and bisulfite reactions are reversible, and the equilibrium favors unreacted hemiacetal. Since osazone and Fehling reactions are irreversible, equilibrium is shifted to restore the low concentration (0.02%) of open-chain aldehyde as soon as some reaction occurs, and eventually all the aldose reacts. Problem 22.14 Glycosides do not react with either Fehling's or Tollens' reagents and do not mutarotate. Explain. 4 Glycosides are acetals. They are stable in the basic Fehling's and Tollens' solutions and in aqueous solutions used in mutarotation. Glycosides are nonreducing. Problem 22.15 Calculate the percentages of a- and /3-glucose present in the equilibrium mixture whose rotation is +52.7". 4

Let a and b be the mole fractions of the a- and p-anomers. Solving the simultaneous equations a+b=l

112"a + 19'6 = 52.7" gives

ax

100% = 36.2% and b x 100% = 63.8%.

500

CARBOHYDRATES AND NUCLEIC ACIDS

[CHAP. 22

Problem 22.16 Acetylation of glucose produces two isomeric pentaacetates that do not react with either phenylhydrazine or Tollens' solution. Explain. 4

The C' OH of the hemiacetal is acetylated instead of the OH that forms the ring. Since equilibrium with the open-chain aldehyde is prevented, these reactions are negative.

22.4

STEREOCHEMISTRY AND STRUCTURE PROOF

Problem 22.17 ( +)-Glucose, (+)-mannose and (+)-galactose are among the hexoses isolated after three Kiliani-Fischer step-ups, beginning with D-(+)-glyceraldehyde. Do these hexoses belong in the D or L families? 4

During the step-ups the D configuration of the chiral C' of glyceraldehyde (glycerose) does not change. This C becomes C' of (+)-glucose, (+)-mannose and (+)-galactose. Configuration of C s determines the family, and these monosaccharides are D-sugars. See Fig. 22-1. Notation A = CH-0 (aldehyde)

0

=

CH,OH

=

C=O (ketone)

- = OH