In the mechanism of the haloform reaction of a methyl ketone () with a halogen () and aqueous alkali (), what is the identity of the leaving group during the critical carbon-carbon bond cleavage step?

Question

In the mechanism of the haloform reaction of a methyl ketone ($R-CO-CH_3$) with a halogen ($X_2$) and aqueous alkali ($NaOH$), what is the identity of the leaving group during the critical carbon-carbon bond cleavage step?

Prashant Kotian · Accepted Answer

Correct answer: B. The haloform reaction proceeds via exhaustive halogenation of the methyl group to form a trihalomethyl ketone (). Attack by on the carbonyl carbon forms a tetrahedral intermediate. When the carbonyl reform, the bond cleaves, and the trihalomethyl carbanion () 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 ().

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