Consider the base catalysed aldol condensation of methyl ethyl ketone. The first step involves the abstraction of a proton by the base. However I am having trouble deciding whether the proton will be abstracted from the $\ce{-CH2-{}}$ group or the $\ce{-CH3}$ group.

I think it should be from the latter as the $\ce{CH^-}$ ion so formed in the former experiences undesired $+I$ effect by the two groups neighbouring it; but I'm not sure.


1 Answer 1


There have been tons of articles, review articles and even books written on the subject. The tl;dr version is that it depends on the conditions you use.

In general, the less-hindered enolisation (in your case: enolising towards the methyl) will be more rapid. This leads to the possibility of kinetic control to selectively generate the terminal enolate by very strong bases and low temperatures such as LDA or even BuLi if you’re daring (both at $\pu{-78^\circ C}$).

On the other hand, the more substituted enolate is more stable thermodynamically as you noted. Thus, if you use a weak base that binds and deprotonates reversibly and if you use relatively high temperatures or longer reaction times, the more stable enolate will be formed. An example for these conditions would be an alcoholate base at room temperature.

Unfortunately, I can give you no simple rules. You will have to check the literature or your lecture notes to see which combinations of reagents give which selectivity.

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    $\begingroup$ Why make LDA from nBuLi when you can just use nBuLi, amirite... Anyway, I don't think OP said that the more substituted enolate is more stable; they were implying that the more substituted enolate was less stable, because the carbanion resonance form would be a secondary carbanion, vs a primary carbanion when abstracting a proton from the methyl group. $\endgroup$ Commented Nov 4, 2017 at 13:38
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    $\begingroup$ @orthocresol Now that I reread OP’s statement, yes they did say that. However, since the charge is on the oxygen in the canonical structure, I’ll just stick with it ;) $\endgroup$
    – Jan
    Commented Nov 5, 2017 at 12:42

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