[leelee]

In an aldol rxn, if it is a ketone, which side would have a base abstracted and form the enolate ion?

I think it is the side that is more substituted because it is able to distribute the negative charge better.

Eg,
in the picture, OH- would remove the H from -CHR2.

Is this correct?

[Albert]

Yes, you are right. But, unfortunately, as you can see below, it is almost impossible to get only one product from these reagents.

[movies]

You can get decent selectivity for the more substituted side if you use a thermodynamic enolate method like t-BuOH/t-BuOK, as Albert said.  You can get selectivity at the less hindered site by treating with a kinetic base such lithium diisopropylamide.  In the kinetic case, the base removes the first proton it sees, which would more likely be one of the three on the methyl group, and sterically the less hindered site is preferred.  You also run these kinetic enolate reactions in such a way that you prevent isomerization of the enolate to the more thermodynamically favored, more substituted enolate.