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Intramolecular aldol condensation of hexane-2,4-dione

In the intramolecular aldol condensation of hexane-2,4-dione (above), why does the base abstract one of the $\ce{-CH3}$ protons instead of one of the $\ce{-CH2}$ protons? Both sets of protons are α to a carbonyl group and should therefore be acidic.

I'm confused because I've seen similar reactions where the proton is taken from the $\ce{-CH2}$ group.

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Here, both the α-$\ce{CH2}$ and α-$\ce{CH3}$ groups have very similar acidities since they are both α to only one carbonyl group. [This is different compared to, for example, a 1,3-dicarbonyl such as acetylacetone, where one $\ce{CH2}$ group is α to two carbonyl groups and is therefore significantly more acidic.] In our substrate, the two protons differ very slightly in acidities, and in general, whether one deprotonates at the more or less substituted carbon can be controlled by reaction conditions.

The mechanism you are drawing is the first step of what is known as an aldol condensation. Because all the steps in this reaction are easily reversible, simple aldol condensations with $\ce{OH-}$ are typically under thermodynamic control, meaning that the most stable product is preferentially formed.

Therefore, let's consider what would happen if we deprotonated the $\ce{-CH2}$ group. The deprotonation itself is definitely a possibility, but what happens after that is quite implausible:

Formation of thermodynamically disfavoured product

Once you form the enolate, the next step in an aldol condensation is the enolate attacking the other carbonyl group. In this case, it would lead to a three-membered ring, where there is a lot of angle strain and the substituents eclipse each other. The dehydration to form the α,β-unsaturated ketone is even less likely, because the cyclopropene ring thus formed would possess incredible angle strain.

On the other hand, deprotonation of the terminal $\ce{-CH3}$ group eventually leads to the formation of a five-membered ring after nucleophilic attack on the other carbonyl group. This is much more stable, and you can eliminate water via an E1cb-type mechanism to get a cyclopentenone.

Formation of thermodynamically favoured product

So, you could certainly deprotonate the $\ce{-CH2}$ position. It is just that, after it gets deprotonated, nothing very good can come out of it. This deprotonation is said to be unproductive.

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  • $\begingroup$ Plus, isn't a primary carbanion more stable than a secondary carbanion, in general? It might also be worth mentioning that 1,3-dicarbonyls, in contrast to 1,4-dicarbonyls, would behave quite differently. The acidity of the hydrogens boudn to the "di"-alpha carbon is much higher than the acidity of the alpha carbons in 1,4-dicarbonyls. $\endgroup$ – Curt F. Oct 8 '16 at 5:53
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    $\begingroup$ @CurtF A primary carbanion would be more stable, but in this case, the carbanion is not the predominant resonance form - the enolate is. The way I learnt it was that more substituted double bond = more stable, so deprotonation of the inner position is more thermodynamically favourable, as stated in the other answer. i.stack.imgur.com/i5ptd.png $\endgroup$ – orthocresol Oct 8 '16 at 8:18
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You can actually take either. The other two $\alpha$ hydrogens are more acidic thermodynamically, but if you used a hindered (kinetic) base at a low temperature, you can predominatly de-protonate at the less hindered methyl (as in your diagram).

So, the answer is, it does sometimes, but it depends on the base and reaction conditions. This is why organic chemistry is interesting.

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