α-Bromination of ketones (acidic): practice and mechanism
Under acidic conditions bromination goes via the enol. The more substituted enol is preferred, so Br goes to the more substituted α-carbon. Under acid it stops at one Br.
Example
Under acidic conditions the ketone reacts via the enol. The more substituted, more stable enol forms preferentially (thermodynamic), so the more substituted α carbon is brominated, and only once (monobromination).
Mechanism step by step
- 1
The acid protonates the carbonyl group and an α-H is lost: the more substituted (more stable) enol forms.
- 2
The enol C=C bond attacks Br₂ and the Br bonds to the α-carbon.
- 3
Deprotonation of the oxocarbenium ion gives the α-bromo ketone and HBr.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- Bromination at the CH₃ is favoured under basic conditions (and then several times, haloform). Under acid the more stable, more substituted enol decides.
Often confused with: Alkylation with LDA (kinetic enolate) · Haloform reaction (iodoform test)
Frequently asked questions
What are α-bromo ketones good for?
With a base (e.g. pyridine) they eliminate HBr to the α,β-unsaturated ketone. They are also good SN2 substrates.
How does it differ under basic conditions?
Under base it goes via the enolate and gets faster after each step: multiple halogenation, and with methyl ketones the haloform reaction.