In an acid-base reaction of $\ce{HO-}$ with 2-methylcyclopent-2-en-1-one, which is the most acidic proton?


I had assumed it was one of them coming off C-5 according to IUPAC numbering (the sp3 carbon α to the carbonyl group). However, the answer book claims it to be C-4 (γ when going in the direction of the double bond).

I assumed C-5, since it seemed it would distribute the charge due to resonance more with the $\ce{O}$ atom. What am I doing wrong?


1 Answer 1


There are two possible enolates that can be formed from your compound: the α deprotonation (left equilibrium arrow in the scheme below) and the γ deprotonation. Both protons have roughly the same $\mathrm{p}K_\mathrm{a}$ values, so deciding which one is more acidic is non-trivial.

Enolates of 2-methylcyclopentenone

However, experimental procedures give us a hint. Usually, the α deprotonation is achieved with strong bases at low temperatures — kinetic conditions. Typically conditions such as LDA in THF at $-78~\mathrm{^\circ C}$. This leads to the conclusion, that the α deprotonation is kinetically favoured. This is more evident in open chain compounds and especially if the α carbon is a methyl group.

The γ deprotonation is generally achieved under more thermodynamic conditions: Weaker bases, warmer temperatures. I didn’t do it myself, so I cannot state conditions off the top of my head, but I would guess conditions such as $\ce{$t$BuOK}$ and $0~\mathrm{^\circ C}$. Therefore, this system must be the thermodynamically favoured one and the γ proton must be slightly more acidic. (But only a tad, or the kinetic product would not be so easily accessable.)

I cannot provide a proper explanation, but it would seem to me like the non-branched π system achieved by γ deprotonation is more stable than the branched one following α deprotonation.


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