Haloform (iodoform) reaction template

Enolate formation and iodination (three times), then hydroxide adds and expels CI₃⁻; proton transfer gives benzoate and yellow iodoform.

Mechanism step by step

A methyl ketone is triiodinated in base, then cleaved to a carboxylate and CHI₃

1. Enolate – Hydroxide removes an α-H
2. Iodination – Repeated twice more, each time faster (I makes the C–H more acidic)
3. Hydroxide adds – Attack on the triiodomethyl ketone
4. C–C cleavage – CI₃⁻ is a good leaving group
5. Proton transfer – Iodoform precipitates as a yellow solid

Key points

• Iodoform test: a yellow precipitate shows CH₃C=O or CH₃CH(OH) (ethanol gives it too).
• Each iodine makes the next α-H more acidic, so the CH₃ is fully iodinated.
• With Cl₂ or Br₂ the same reaction gives chloroform or bromoform.
• A route from methyl ketones to carboxylic acids with one carbon fewer.

Drawn with 144 atoms, 16 arrows, 7 shapes – every one editable in MolDraw.

Sources

• Clayden, Greeves & Warren, Organic Chemistry, 2nd ed., OUP 2012 Textbook