Textbook Question
Optically active 2-bromobutane undergoes racemization on treatment with a solution of KBr. Propose a mechanism for this racemization.
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Ch.6 - Alkyl Halides; Nucleophilic Substitution
Wade9th EditionOrganic ChemistryISBN: 9780135213728
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Table of contents
- Ch.1 - Structure and Bonding107
- Ch. 2 - Acids and Bases; Functional Groups115
- Ch.3 - Structure and Stereochemistry of Alkanes100
- Ch.4 - The Study of Chemical Reactions99
- Ch.5 - Stereochemistry93
- Ch.6 - Alkyl Halides; Nucleophilic Substitution118
- Ch. 7 - Structure and Synthesis of Alkenes; Elimination158
- Ch.8 - Reactions of Alkenes171
- Ch.9 - Alkynes91
- Ch.10 - Structure and Synthesis of Alcohols143
- Ch.11 - Reactions of Alcohols104
- Ch. 12 - Infrared Spectroscopy and Mass Spectrometry22
- Ch. 13 - Nuclear Magnetic Resonance Spectroscopy45
- Ch. 14 - Ethers, Epoxides, and Thioethers72
- Ch. 15 - Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy89
- Ch. 16 - Aromatic Compounds74
- Ch. 17 - Reactions of Aromatic Compounds126
- Ch. 18 - Ketones and Aldehydes193
- Ch. 19 - Amines129
- Ch. 20 - Carboxylic Acids77
- Ch. 21 - Carboxylic Acid Derivatives141
- Ch. 22 - Condensations and Alpha Substitutions of Carbonyl Compounds175
- Ch. 23 - Carbohydrates and Nucleic Acids93
- Ch. 24 - Amino Acids, Peptides, and Proteins54
Chapter 6, Problem 49c
Optically active butan-2-ol racemizes in dilute acid. Propose a mechanism for this racemization.
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Identify the structure of butan-2-ol: It is a chiral alcohol with the formula CH₃-CH(OH)-CH₂-CH₃. The chirality arises from the carbon atom bonded to the hydroxyl group (-OH), a methyl group (-CH₃), an ethyl group (-CH₂CH₃), and a hydrogen atom.
Understand the concept of racemization: Racemization occurs when an optically active compound (one enantiomer) converts into a 1:1 mixture of both enantiomers, resulting in a loss of optical activity. This typically involves the formation of an achiral intermediate or transition state.
Propose the first step of the mechanism: In the presence of dilute acid, the hydroxyl group (-OH) of butan-2-ol is protonated by H⁺, forming a good leaving group, water (H₂O). This step can be represented as: CH₃-CH(OH)-CH₂-CH₃ + H⁺ → CH₃-CH(OH₂⁺)-CH₂-CH₃.
Describe the formation of the carbocation intermediate: The protonated alcohol (CH₃-CH(OH₂⁺)-CH₂-CH₃) loses water, resulting in the formation of a planar carbocation intermediate (CH₃-C⁺H-CH₂-CH₃). This intermediate is achiral because the positively charged carbon is sp² hybridized and planar, allowing for attack from either side.
Explain the final step leading to racemization: Water (H₂O) or another nucleophile can attack the planar carbocation from either side, leading to the formation of both (R)- and (S)-butan-2-ol in equal amounts. This results in a racemic mixture, which is optically inactive.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Optical Activity
Optical activity refers to the ability of a chiral compound to rotate the plane of polarized light. Chiral molecules have non-superimposable mirror images, known as enantiomers, which can rotate light in opposite directions. Butan-2-ol has a chiral center, making it optically active, and its racemization involves the interconversion of these enantiomers.
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Mutorotation and Optical Activity
Racemization
Racemization is the process by which an optically active compound converts into a racemic mixture, containing equal amounts of both enantiomers. In the case of butan-2-ol, racemization occurs when the chiral center is temporarily converted to a planar carbocation intermediate, allowing for the reformation of the chiral center in either configuration.
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Acid-Catalyzed Mechanism
An acid-catalyzed mechanism involves the use of an acid to facilitate a chemical reaction. In the racemization of butan-2-ol, dilute acid protonates the hydroxyl group, converting it into a better leaving group. This leads to the formation of a carbocation, which can then be attacked by a nucleophile from either side, resulting in the formation of both enantiomers.
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Related Practice
Textbook Question
Give a mechanism to explain the two products formed in the following reaction.
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Textbook Question
Using 1,2-dimethylcyclohexene as your starting material, show how you would synthesize the following compounds. (Once you have shown how to synthesize a compound, you may use it as the starting material in any later parts of this problem.) If a chiral product is shown, assume that it is part of a racemic mixture.
(f)
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Textbook Question
In contrast, optically active butan-2-ol does not racemize on treatment with a solution of KOH. Explain why a reaction like that in part (a) does not occur.
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Textbook Question
Because the SN1 reaction goes through a flat carbocation, we might expect an optically active starting material to give a completely racemized product. In most cases, however, SN1 reactions actually give more of the inversion product. In general, as the stability of the carbocation increases, the excess inversion product decreases. Extremely stable carbocations give completely racemic products. Explain these observations.
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Textbook Question
a. Optically active 2-bromobutane undergoes racemization on treatment with a solution of KBr. Give a mechanism for this racemization.
b. In contrast, optically active butan-2-ol does not racemize on treatment with a solution of KOH. Explain why a reaction like that in part (a) does not occur.
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