# Does aldol condensation give different products in different conditions?

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?

• 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. – Greg E. Dec 23 '13 at 4:37