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I know the mechanism of aldol condensation and general method to determine products in crossed aldol reactions, or reactions between dissimilar molecules. But the following statement in my class notes baffles me:

In presence of a base (alkaline medium) the $\alpha$ hydrogen of the lower aldehyde is more acidic and hence is removed in the first step, while in presence of an acid, the $\alpha$ hydrogen of the higher aldehyde is more acidic and hence migrates.

According to this, on reaction of propanal with ethanal, basic medium yields 3-hydroxy pentanal while acidic medium yields 3-hydroxy 2-methyl butanal.

Why is this statement true (if it is)? And can aldol condensation proceed in an acidic medium, since the presence of the base is necessary for the removal of the slightly acidic $\alpha$-hydrogen to start the reaction?

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    $\begingroup$ I think the issue is much more complicated. For one thing, the particular reaction conditions are relevant (i.e., thermodynamic vs. kinetic control). I'm dubious as to whether the premise that, under basic conditions, the lower aldehyde will form the enolate is accurate, unless conditions are selected for kinetic control (i.e., use of low temperatures and strong, sterically hindered bases). The enolate with the more substituted double-bond is more thermodynamically stable due to hyperconjugation. On the other hand, the smaller enolate may be better stabilized by solvation. $\endgroup$ – Greg E. Dec 23 '13 at 4:37
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I believe your conception of the need for deprotonation under acidic conditions is misconceived. Whenever you are under acidic conditions, your mechanisms must necessarily start with protonation.

The key concept here is the enol tautomerization. The enol is more favored than for propanal as compared to ethanal because it results in more substituted alkene. Thus there is likely a higher concentration of this nucleophile in the reaction mixture. The mechanism of tautomerization is difficult to elucidate, but presumably starts with protonation of the carbonyl in acidic conditions. Then, likely another carbonyl deprotonates the the recently protonated carbonyl at its alpha position, forming the enol. This transformation has produced a nucleophile (enol -- 1-propene-1-ol ) and also a strong electrophile (a protonated carbonyl -- protonated ethanal). The reaction occurs, leading to a protonated aldol product (proton on the carbonyl) which then is deprotonated by the next carbonyl and the process continues.

As far as the base catalyzed case, ethanal is more acidic because it does not have the extra methyl group donating electron density disfavoring formation of a negative charge. Hence ethanal becomes the nucleophile and propanal becomes the electrophile.

In each case, these are only major products. I don't expect the energy differences are large enough to give great selectivity.

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