LiAlH4 vs. NaBH4 (concept question)

[pineappletree]

I know that LiAlH4 is capable of reducing both carbonyls with leaving groups and aldehyde/ketones, while NaBH4 is only capable of reducing aldehydes/Ketones.

I also happen to know that LiAlH4 is stronger that NaBH4 here, which can also be deduced from the above reactions.




However, I was wondering why carbonyls with leaving groups are more difficult to reduce than aldehydes/ketones...
Also, if both reducing agents can be used for aldehydes/ketones, why does NaBH4 even exist? (perhaps for selective reduction?)

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[Rayhaan]

Yes, NaBH4 is for selective reduction. If you had a molecule that contains a carboxylic section AND a ketone; maybe you only want to reduce the ketone portion and not the acid, so use NaBH4 since it can't reduce the acid.

 

[orgohacks]

I know that LiAlH4 is capable of reducing both carbonyls with leaving groups and aldehyde/ketones, while NaBH4 is only capable of reducing aldehydes/Ketones.

I also happen to know that LiAlH4 is stronger that NaBH4 here, which can also be deduced from the above reactions.




However, I was wondering why carbonyls with leaving groups are more difficult to reduce than aldehydes/ketones...
Also, if both reducing agents can be used for aldehydes/ketones, why does NaBH4 even exist? (perhaps for selective reduction?)

1) Carbonyls w/ leaving groups are more difficult to reduce than aldehydes/ketones because Pi-donation from the lone pair into the carbonyl carbon reduces the electrophilicity. The stronger the pi donor, the harder it will be to reduce: for more info see point #3 in this post - http://is.gd/cOtK5

2) You are right about selective reduction. LiAlH4 is a real sledgehammer of a reagent. It will reduce ketones as well as esters. In molecules with multiple functional groups (which is often the case in complex molecule synthesis) NaBH4 is useful for selective reduction of ketones/aldehydes.

hope this helps - james