# Deciding which α-Η gets abstracted in aldol condensation

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.

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 $\pu{-78^\circ C}$).