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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?

enter image description here

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

To answer your question directly, you have to think about which carbonyl is more electrophilic. Generally, resonance effects are more important than inductive effects when considering something's nucleophilicity/electrophilicity.

resonance forms

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.

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  • $\begingroup$ What if it were two ketones? $\endgroup$ – Hani Sep 30 '16 at 0:09
  • $\begingroup$ No reason it couldn't reduce them both. $\endgroup$ – gannex Sep 30 '16 at 0:13
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The key difference between the carbonyl groups is that one is a ketone and the other is an ester. Esters are not generally able to be reduced by sodium borohydride, while ketones can be.

The key to understanding the different reactivity is being able to understand the different functional groups.

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