Chapter 9, Problem 9.22P

Interpretation Introduction

Interpretation:

Whether the substrates X, Y, and Z will undergo an ${\text{S}}_{\text{N}}2$ reaction fastest, ${\text{S}}_{\text{N}}\text{1}$ reaction fastest, or an $\text{E1}$ reaction fastest, is to be identified.

Concept introduction:

The ${\text{S}}_{\text{N}}\text{1}$ and $\text{E1}$ reactions proceed with the formation of a carbocation intermediate in the rate determining step, thus, the stability of the carbocation formed increases the rate of the reaction. The electron donating groups (alkyl substituents) adjacent to the carbocation increases its stability. The order of stability of carbocation is ${\text{CH}}_{\text{3}}\text{}{\text{< 1}}^{\text{o}}{\text{ < 2}}^{\text{o}}{\text{ < 3}}^{\text{o}}$. The rate of ${\text{S}}_{\text{N}}\text{1}$ and $\text{E1}$ reactions is fastest when the leaving group is bonded to a tertiary carbon.

In ${\text{S}}_{\text{N}}2$ reaction, the nucleophile attacks the carbon where the leaving group is attached and forces off the leaving group. The alkyl group surrounding the carbon atom bonded to the leaving group increases steric hindrance, and so, it becomes more difficult for nucleophile to attack the carbon. Thus, the rate of ${\text{S}}_{\text{N}}2$ reaction decreases as the number of alkyl substituents attached to the carbon increase. The rate of ${\text{S}}_{\text{N}}2$ reaction is the fastest when the leaving group is bonded to a methyl or a primary carbon.