The haloform reaction: conversion of methyl ketones to carboxylic acids
Description: Addition of a dihalogen such as iodine, bromine or chlorine to a methyl ketone in the presence of base results in a carboxylic acid and a haloform (such as iodoform, pictured)
Notes: Methyl ketones work best for this reaction. More sterically bulky ketones are less effective.
Notes: Note that where there is a choice between two alkyl groups, the less bulky one (CH3) disappears, as in a third example.
Mechanism: Removal of a proton from the methyl ketone by hydroxide ion (Step 1, arrows A and B) forms an enolate, which attacks iodine (Step 2, arrows C and D). After a second deprotonation (Step 3, arrows E and F) and a second attack of iodine (Step 4, arrows G and H), and finally a third deprotonation (Step 5, arrows I and J) and a third iodination (Step 6, arrows K and L), attack of hydroxide ion on the carbonyl carbon (Step 7, arrows M and N) followed by elimination of I3C(–) (Step 8, arrows O and P) results in the carboxylic acid.
Notes: Note that I3C(–) is an excellent leaving group due to the presence of the three iodine atoms.