Chapter 7.6, Problem 7.10E

Interpretation Introduction

Interpretation: The reason behind $30\text{ \% }\left({\text{CH}}_{\text{3}}\right){\text{COCH}}_{\text{2}}{\text{CH}}_{\text{3}}$ , $60\text{ \% }\left({\text{CH}}_{\text{3}}\right)\text{COH}$ and $10\text{ \% }\left({\text{CH}}_{\text{3}}\right){\text{C=CH}}_{\text{2}}$ formed upon reaction of $\text{2-bromo-2-methylpropane}$ in aqueous ethanol should be explained.

Concept introduction: Haloalkane solvolysis with ethanol, methanol or water is a typical example for unimolecuar substitution. It proceeds via two-step mechanism. The first slow step that determines rate is the removal of leaving group from the substrate haloalkane and generates a carbocation. Since the rate is only governed by substrate alone and no other nucleophile or solvent it is termed as unimolecuar substitution. The final step is attack of nucleophile on carbocation generated and formation of racemic products.

Tertiary or secondary halides undergo fastest unimolecuar substitution as they can readily form the tertiary carbocation followed by secondary and least reactive are primary.

Carbocations generated in solvolysis have two fates; they can be either trapped by nucleophiles to give substitution product or may deprotonated to yield small amount of alkene.