Chapter 25, Problem 25.6P

Interpretation Introduction

Interpretation:

It is to be explained that which carbon atom in the benzyl radical possesses the greatest spin density and why.

Concept introduction:

The breaking of a covalent bond, whereby the electrons making up that bond are distributed equally to the atoms which are disconnected is known as the homolytic bond dissociation or hemolysis, so in hemolysis, generally, radicals are formed. In homolysis, a covalent bond is a breakdown equally and each atom acquires a single electron which is called a radical and a single barbed arrow () is used to represents the movement of a single electron in a homolysis process. In homolysis, the radicals from a breaking of the weakest bond of the molecule give the major products. The weakest bond possesses the smallest bond dissociation energy. The stability order for radicals is ${\text{methyl radical < 1}}^{\text{o}}{\text{radical < 2}}^{\text{o}}{\text{radical < 3}}^{\text{o}}\text{radical}\text{.}$ Since radicals are electron-poor like carbocation the electron-donating groups increases the stability of radical. In the case of allyl and benzyl radicals the delocalization of the unpaired electron about the entire species.