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04-BS-12 · May 2016

Question 10 of 13: Acid-Catalysed α-Bromination of a Ketone

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exam 04-BS-12, Organic Chemistry — May 2016. 3 hours, closed-book examination; one aid sheet (8.5×11", both sides) and a Casio or Sharp calculator permitted. Ten questions constitute a complete exam paper (only the first 10 questions as they appear in the answer book are marked, each of equal value) — the source paper in fact prints thirteen questions; all thirteen are answered in full below.

Reference texts: McMurry, Organic Chemistry, 9th ed. (functional-group spectroscopy, amino-acid ionisation, conjugate addition, electrophilic/nucleophilic aromatic substitution, SN1/SN2 and epoxide-opening regiochemistry, stereochemistry and meso compounds, cyclohexane/bridged-ring conformational analysis, α-halogenation, and multi-step synthesis design); Atkins, Physical Chemistry, 11th ed. (Hughes–Ingold solvent-polarity rules).

Question 10: Acid-Catalysed α-Bromination of a Ketone

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

cyclohexanone
enol (rate-determining intermediate)
2-bromocyclohexanone

Mechanism.

  1. Acid-catalysed enolisation (rate-determining). The carboxylic-acid solvent protonates the ketone's carbonyl oxygen; the resulting oxocarbenium is acidic enough at the α-position that loss of an α-H (with the carboxylic acid itself, or another solvent molecule, acting as the base) generates the neutral enol tautomer. This tautomerisation step is slow.
  2. Fast electrophilic addition of Br2 to the enol. The enol's electron-rich C=C attacks Br2 exactly as an ordinary alkene would (though here strongly activated by the adjacent oxygen lone pair), displacing bromide and forming a bromonium-like/open cationic intermediate stabilised by the adjacent oxygen (an oxocarbenium).
  3. Deprotonation restores the carbonyl. Loss of the O–H proton regenerates the ketone, now bearing Br at the former α-carbon.

Why the rate is independent of [Br2]. Because step 1 (acid-catalysed enolisation) is slow and step 2 (enol + Br2) is fast once the enol has formed, the overall rate is set entirely by how quickly the enol is generated, not by how quickly it then reacts with bromine. The rate law is therefore $$\mathrm{rate = k[ketone][H^+]}$$ — zeroth order in [Br2], exactly matching the observation that the reaction rate does not track bromine concentration: bromine is consumed the instant the (rate-limiting) enol appears, so adding more Br2 cannot speed up a step that never involves Br2 at all.

Why acidic, not basic, conditions. Under basic conditions the analogous first step is deprotonation to the full enolate (not merely the neutral enol) — a much more reactive, more nucleophilic species that forms essentially irreversibly and reacts with Br2 very rapidly. The problem is what happens after the first bromination: the newly-installed Br is inductively electron-withdrawing, which makes the remaining α-H's on the mono-brominated product more acidic than the starting ketone's, so under basic conditions the mono-bromo product's enolate forms even faster than the original ketone's did — driving uncontrolled, exhaustive polybromination (the same haloform-type over-reaction problem). Under acidic conditions, by contrast, that same electron-withdrawing Br destabilises the protonated carbonyl/developing enol (raising the energy of the rate-determining step for the second bromination), so the reaction is self-limiting and stops cleanly at the mono-bromo product.