Question

Propose a mechanism for the reaction of cyclohexyl methyl ketone with excess bromine in the presence of sodium hydroxide.

Short answer

Haloform reaction is a nucleophilic acyl substitution, in which hydroxide acts as the nucleophile and${\text{-CX}}_{\text{3}}$ acts as the leaving group.

The reaction proceeds through halogenation of the methyl carbon followed by elimination of the ${\text{-CX}}_{\text{3}}$group to give a carboxylic acid having one carbon atom lesser than the starting compound.

Step-by-step solution

Haloform Reaction

Haloform reaction is a nucleophilic acyl substitution, in which hydroxide acts as the nucleophile and${\text{-CX}}_{\text{3}}$ acts as the leaving group.

The reaction proceeds through halogenation of the methyl carbon followed by elimination of the ${\text{-CX}}_{\text{3}}$group to give a carboxylic acid having one carbon atom lesser than the starting compound.

Mechanism of haloform reaction of cyclohexyl methyl ketone

In the Haloform reaction of cyclohexyl methyl ketone, there is an addition of hydroxide to the carbonyl group of a cyclohexyl methyl ketone. Cyclohexyl methyl ketone has three protons on the methyl carbon, and they undergo halogenation three times to give 2,2,2-tribromo cyclohexyl ethanone. Then the leaving group ${\text{-CBr}}_{\text{3}}$leaves. In the last, there is a fast proton transfer from the acid in the last step to form the product bromoform.

Formation of bromoform