# Reduction of carbonyl groups

In such reactions as in the pic, how can we know which carbonyl group will undergo a reaction to give an alcohol. Also, can both of the sites undergo the reaction?

• R-O-R' is an ether And R-OO-R' is an ester. Am I right? – Hani Sep 29 '16 at 15:49

Sodium borohydride is a weaker reducing agent than, say, lithium aluminium hydride, so, although it will react with ketones and aldehydes, it is not strong enough to react with carboxylic acids or esters. This generalization is quite broadly applicable, so you already know your borohydride will reduce the ketone carbonyl on that molecule. $\ce{NaBH4}$ is a weaker reducing agent than $\ce{LiAlH4}$ because boron is less electropositive than aluminium. Boron has a Pauling electronegativity of 2.04 while aluminium's pauling electronegativity is only 1.61, therefore $\ce{NaBH4}$'s hydrides are significantly less nucleophilic and they won't be able to react with a weaker electrophile.
The ketone has two resonance forms, while the ester has three, therefore the partial positive charge on the carbon is more delocalized on the ester, making the ketone more electrophilic. The weak $\ce{H-}$ nucleophile provided by $\ce{NaBH4}$ will only be able to react with the stronger electrophile, the ketone.