α-Halogenation of a ketone (acid) template

Propanone with Br₂ in ethanoic acid: O-protonation, slow loss of an α-H to the enol, attack of the enol on Br₂ and loss of H⁺ give bromopropanone.

Mechanism step by step

The slow step is enol formation; the enol then attacks the halogen

1. Protonation – Acid activates the carbonyl
2. Enol (slow) – Water removes an α-H
3. Enol attacks Br₂ – The C=C is the nucleophile
4. Deprotonation – Bromopropanone and HBr

Key points

• Rate = k[ketone][H⁺], independent of [Br₂]: Cl₂, Br₂ and I₂ react at the same rate (the iodination of propanone experiment).
• In acid it stops after one halogen: the C–Br makes the next enol slower to form.
• In base every α-H is replaced; methyl ketones go on to the haloform reaction.
• HBr is a product, so the reaction speeds up as it goes (autocatalysis).

Drawn with 59 atoms, 15 arrows, 8 shapes – every one editable in MolDraw.

Sources

• Clayden, Greeves & Warren, Organic Chemistry, 2nd ed., OUP 2012, ch. 20 Textbook
• Smith, March's Advanced Organic Chemistry, 8th ed., Wiley 2020, ch. 12 Textbook
• Lapworth, J. Chem. Soc., Trans. 1904, 85, 30 Original paper