Haloform reaction (iodoform test): practice and mechanism
Methyl ketones are iodinated three times at the CH₃ group under basic conditions. CI₃⁻ then acts as a leaving group: iodoform (yellow precipitate) and a carboxylic acid with one carbon fewer form.
Example
In base the methyl group of a methyl ketone is iodinated three times. The CI₃ group is then a good leaving group: hydroxide attacks the carbonyl carbon and iodoform (CHI₃, yellow precipitate) is released. The carboxylic acid with one carbon fewer remains.
Mechanism step by step
- 1
NaOH forms the enolate at the CH₃ group and it attacks I₂. Each I makes the remaining H more acidic, so three iodinations follow quickly: –CO–CI₃.
- 2
Hydroxide attacks the carbonyl carbon; CI₃⁻ is stabilised by the three I and leaves.
- 3
CI₃⁻ takes a proton: iodoform (CHI₃, yellow) precipitates. The carboxylate remains, the acid after acidification.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
Often confused with: α-Bromination of ketones (acidic)
Frequently asked questions
What does the iodoform test show?
A CH₃–CO group (methyl ketone, acetaldehyde) or a CH₃–CH(OH) group, which is first oxidised to the methyl ketone (e.g. ethanol).
Why does it stop at one halogen under acid?
Under acid halogenation goes via the enol. The halogen makes the C=O group less basic, so the next enolisation is slower. Under base it is the reverse.