Chapter 9, Problem 9.25P

Interpretation Introduction

Interpretation:

It is to be determined in which of the two solvents, dimethyl sulfoxide or ethanol, the given reaction will proceed faster via an ${\text{S}}_{\text{N}}\text{1}$ mechanism. It is also to be determined in which of the two solvents the reaction proceeds faster via an ${\text{S}}_{\text{N}}\text{2}$ mechanism.

Concept introduction:

An ${\text{S}}_{\text{N}}\text{1}$ mechanism is a two-step mechanism. In the first rate determining step, the leaving group breaks off with its bond pair. The result is a carbocation with the carbon that was bonded to the leaving group carrying a full positive charge. The leaving group is in most cases a negatively charged species as it takes away the electron pair from its bond with the carbon. Being charged species, both the carbocation and the leaving group can be stabilized by protic solvents. The high partial positive charge on the hydrogen atom in a protic solvent stabilizes the negatively charged leaving group by strongly solvating it. A protic solvent will also solvate the incoming nucleophile, but as it is not involved in the rate determining step, this does not affect the reaction rate.

An ${\text{S}}_{\text{N}}\text{2}$ mechanism is a single-step mechanism, in which both the substrate and the nucleophile are involved. Therefore, any effect that hampers the approach of the nucleophile at the site of reaction will tend to slow down the reaction. Since the nucleophile must start forming a bond with the reaction center before the leaving group breaks off, a strong, often negatively charged nucleophile is required for an ${\text{S}}_{\text{N}}\text{2}$ reaction. A protic solvent will strongly solvate a negatively charged nucleophile, preventing its close approach to the reaction center. Therefore, a protic solvent considerably reduces the rate of an ${\text{S}}_{\text{N}}\text{2}$ reaction. An aprotic solvent interacts much less strongly with the negative charge of the nucleophile. Therefore, an aprotic solvent is preferred for an ${\text{S}}_{\text{N}}\text{2}$ reaction.