The haloform reaction: conversion of methyl ketones to carboxylic acids

by James

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.


{ 3 comments… read them below or add one }

Petr Menzel

Hi, in the Description I can read “iodoform” but in the picture of reaction I can see “HCl3”. Thx, P.


Petr Menzel

Sorry… it is not “H CI3 = H CL3” but “H C I3” :-)


James Briscoe

So I understand that the haloform reaction when using Iodine provides a nifty way to identify methyl ketones because the LG ‘CI3 precipates as a yellow solid in the form of Iodoform; I also understand that secondary methyl alcohols are oxidized by I2 in a radical-led oxidation rxn to a methyl ketone, which obviously goes through a subsequent reaction with I2. However, why do primary alcohols such as ethanol or propanol not react with I2? Would methanol react? My book does not expand anymore than just that I2 will react with methyl ketones and secondary methyl alcohols.



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