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Chapter 11: Q11-71E (page 350)

The reaction of HBr with (R)-3-methyl-3-hexanol leads to racemic 3-Bromo-3-methyl hexane. Explain.

Short Answer

Expert verified

Formation of carbocation

Formation of the desired product

Step by step solution

01

Tertiary alcohol

Whenever a tertiary alcohol is treated with a strong acid, a carbocation intermediate is formed.

Formation of carbocation

This carbocation is planar due to the empty p-orbital of the carbocation carbon. Moreover, the tetrahedral chirality that was present in the starting material was lost.

02

Explanation

The chiral starting material goes through an achiral intermediate, such as a carbocation, and the stereochemical information is destroyed. Hence, treatment of 3 -methyl -3-hexanol with HBr forms a carbocation intermediate because 3-methyl-3-hexanol is a tertiary alcohol. This intermediate loses all stereochemical information so the product is racemized.

Formation of the desired product

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Most popular questions from this chapter

Question: There are eight diastereomers of 1,2,3,4,5,6-hexachlorocyclohexane. Draw each in its more stable chair conformation. One isomer loses HCl in an E2 reaction nearly 1000 times more slowly than the others. Which isomer reacts so slowly, and why?

(S)-2-Butanol slowly racemizes on standing in dilute sulfuric acid. Explain.

Question:What products would you expect from the reaction of 1-bromopropane with each of the following?

(a) NaNH2

Question: Order each of the following sets of compounds with respect to reactivity:

b. (CH3)3 CCI (CH3)3 CBr (CH3)3 COH

How might you prepare each of the following molecules using a nucleophilic substitution reaction at some step?

(a)
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