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

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.

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