Ch.6 - Alkyl Halides; Nucleophilic Substitution

Wade9th EditionOrganic ChemistryISBN: 9780135213728Not the one you use?Change textbook

Wade 9th Edition

Ch.6 - Alkyl Halides; Nucleophilic Substitution

Problem 49b

Chapter 6, Problem 49b

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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Understand the concept of racemization: Racemization occurs when a chiral molecule converts into a racemic mixture (equal amounts of enantiomers), typically through a reaction that involves the loss of chirality. This often requires the formation of an intermediate that is achiral or planar, allowing for free rotation or attack from either side.

Analyze the structure of butan-2-ol: Butan-2-ol is a secondary alcohol with a chiral center at the second carbon. Its chirality arises from the four different groups attached to this carbon.

Consider the role of KOH: KOH is a strong base, and in the presence of an alcohol, it can deprotonate the hydroxyl group to form an alkoxide ion. However, this does not lead to the loss of chirality because the alkoxide ion retains the tetrahedral geometry around the chiral carbon.

Compare with part (a): In part (a), racemization likely occurred due to the formation of a planar intermediate, such as a carbocation, which allows for attack from either side. However, in the case of butan-2-ol, the reaction with KOH does not involve the formation of a carbocation or any planar intermediate.

Conclude why racemization does not occur: Since the reaction with KOH does not disrupt the tetrahedral geometry of the chiral center in butan-2-ol, the molecule retains its chirality, and no racemization takes place.

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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. This property arises from the presence of a chiral center, which is typically a carbon atom bonded to four different substituents. In the case of butan-2-ol, the presence of a chiral center makes it optically active, meaning it can exist in two enantiomeric forms that rotate light in opposite directions.

Racemization

Racemization is the process by which an optically active compound converts into a racemic mixture, containing equal amounts of both enantiomers. This typically occurs when the chiral center can be temporarily converted into a prochiral state, allowing for the formation of both enantiomers. In the case of butan-2-ol with KOH, the conditions do not favor racemization, as the reaction does not lead to the necessary transition state for this process.

Base-Induced Elimination Reactions

Base-induced elimination reactions, such as those involving KOH, typically lead to the formation of alkenes through the removal of a leaving group and a hydrogen atom. In butan-2-ol, the presence of a strong base like KOH can promote elimination rather than substitution, resulting in the formation of an alkene without racemization. This is because the reaction pathway does not involve the formation of a planar intermediate that would allow for the reformation of both enantiomers.

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Related Practice

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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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

A solution of pure (S)-2-iodobutane ([α] = +15.90°) in acetone is allowed to react with radioactive iodide, ¹³¹I^–, until 1.0% of the iodobutane contains radioactive iodine. The specific rotation of this recovered iodobutane is found to be +15.58°.

b. What does this result suggest about the mechanism of the reaction of 2-iodobutane with iodide ion?

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Textbook Question

Optically active butan-2-ol racemizes in dilute acid. Propose a mechanism for this racemization.

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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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