Question

Butyl bromide (2-brom-2-methylpropane) can be prepared by simply taking t-butyl alcohol and shaking it with an aqueous solution of HBr at room temperature. The reaction is much faster than with n-butyl alcohol and is essentially 100% complete within a few minutes. Give a mechanism for this reaction. Note: it is not the same mechanism as for the lab preparation of 1-bromobutane. What is this reaction called?

Answer 1

Answer:

This is an example of $S_{N}1$(substitution nucleophilic bimolecular) reaction

Explanation:

Conversion of t-butyl alcohol to t-butyl bromide proceeds through  $S_{N}1$(substitution nucleophilic bimolecular)mechanism.

In the first step, -OH group in t-butyl alcohol gets protonated.

In the second step, stable tertiary carbocation (t-butyl cation) is produced by removal of $H_{2}O$ .

In the third step, bromide ion attacks to t-butyl cation and produces t-butyl bromide.

Removal of $H_{2}O$  from n-butyl alcohol will produce an unstable primary carbocation (n-butyl cation). Hence, to produce this unstable carboction, large amount of activation energy is required. Therefore, n-butyl alcohol gives much slower reaction with HBr.

Reaction mechanism has been shown below.