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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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