Q11-71E - Organic Chemistry (Mcmurry)

Question

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

Step-by-Step Solution

Answer

Formation of carbocation

Formation of the desired product

Step 1: 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.

Step 2: 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