In the mechanism of the haloform reaction of a methyl ketone () with a halogen () and aqueous alkali — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

The haloform reaction proceeds via exhaustive halogenation of the methyl group to form a trihalomethyl ketone ($R-CO-CX_3$). Attack by $OH^-$ on the carbonyl carbon forms a tetrahedral intermediate. When the carbonyl reform, the $C-C$ bond cleaves, and the trihalomethyl carbanion ($CX_3^-$) acts as a relatively stable leaving group due to the strong -I effect of three halogens. It rapidly abstracts a proton to form the haloform ($CHX_3$).