Question

How would you prepare 1-phenyl-2-butanone, ${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{5}}{\mathbf{CH}}_{\mathbf{2}}{\mathbf{COCH}}_{\mathbf{2}}{\mathbf{CH}}_{\mathbf{3}}$, from

benzyl bromide,${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{5}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{Br}$? More than one step is required.

Short answer

The preparation of 1-phenyl-2-butanone from benzyl bromide is shown in two step as

follows:

Step-by-step solution

Conversion of benzyl bromide into benzylnitrile.

The given starting material, benzyl bromide is attacked by a nucleophile

cyanide,${\mathrm{CN}}^{-}$via a second order nucleophilic substitution pathway. This leads

to the formation of benzylnitrile as shown below.

Conversion of benzylnitrile into 1-phenyl-2-butanone.

Benzylnitrile obtained in the first step is reacted with ethylmagnesium bromide gives

the imine salt which is further acidified with aqueous acid to produce 1-phenyl-2-

butanone as the final product.