Chapter 10, Problem 10.4P

Interpretation Introduction

Interpretation:

The product of the given reaction is to be predicted, and the complete and detailed mechanism with correct stereochemistry is to be drawn for the same reaction.

Concept introduction:

The alcoholic ($\text{-OH}$) group of a substrate is replaced by a Br atom by using ${\text{PBr}}_{\text{3}}$ via back to back ${\text{S}}_{\text{N}}\text{2}$ reactions.

In the first step of the reaction, the oxygen atom of an alcohol is electron rich and acts as a nucleophile whereas the phosphorous atom in ${\text{PBr}}_{\text{3}}$ is electron deficient and acts as the substrate. This reaction generates a bromide ion, which is a good leaving group. In step 2, the bromide ion generated in the previous step acts as a nucleophile, and the phosphorous containing species acts as the substrate. The leaving group is ${\text{HOPBr}}_{\text{2}}{\text{or HOPCl}}_{\text{2}}$. Thus, overall two back-to-back ${\text{S}}_{\text{N}}\text{2}$ reactions occur, which results in the inversion of configuration at the chiral carbon atom.