Chapter 9, Problem 9.55P

Interpretation Introduction

Interpretation: An explanation for the given observation that when $\text{3-methylbutan-2-ol}$ is treated with $\text{HBr}$, a single alkyl bromide is isolated, resulting from a $1\text{, 2-shift}$ and when $\text{2-methylpropan-1-ol}$ is treated with $\text{HBr}$, no rearrangement occurs to form an alkyl bromide is to be stated.

Concept introduction: Hydrogen bromide is useful to convert all alcohols $\left(1\text{°},2\text{°},\text{and}3\text{°}\right)$ into alkyl bromides. The reaction proceeds through ${\text{S}}_{\text{N}}\text{1}$ pathway for $\text{2}\text{°}\text{and}\text{3}\text{°}$ alcohols, and ${\text{S}}_{\text{N}}2$ pathway for $\text{1}\text{°}$ alcohols. In ${\text{S}}_{\text{N}}\text{1}$ pathway (two steps), the rearrangement of carbocation occurs to form stable carbocation, through $1\text{, 2-shift}$ of hydrogen or alkyl group. However, ${\text{S}}_{\text{N}}2$ pathway (single step) does not involve carbocation rearrangement.