Question

Expt 55 Micro: Borohydride Reduction of a Ketone: Hydrobenzoin from Benzil

3. Sodium borohydride can be used in ethanol solutions to rapidly reduce aldehydes and ketones at room temperature. Even though lithium aluminum hydride (LiAlH₄) is much more reactive than sodium borohydride (NaBH₄) it can’t be used to reduce aldehydes and ketones in ethanol solutions. Explain why not? NOTE: The answer is not the quote from the lab book “Unlike lithium aluminum hydride, sodium borohydride is insoluble in ether and soluble in methanol and ethanol”.

Answer 1

NaBH4 is a “weak reducing agent” and can only take aldehydes and ketones to alcohols easily.

NaBH4 can handle esters, but it is very slow at converting them and thus not preferable.

We then learn that if you want to turn esters and carboxylic acids into alcohols, we better use LiAlH4 because it is a strong reducing agent.

LiAlH4 can convert aldehydes, ketones, esters, and carboxylic acids all to alcohols in the blink of an eye.

LiAlH4 stronger than NaBH4.

This can be explained by their electronegativities.

The atom holding on to the hydrides should be examined here. In the case of NaBH4 we have boron with an electronegativity of around 2.0 where as with LiAlH4 we have aluminum with an electronegativity around 1.6.

The more electronegative an atom is, the less likely it will be to share or give up electrons.

A hydride (H-) is just that, extra electrons (one to be exact) on a hydrogen. So if the hydride is bound to boron, it is less likely to be released for organic transformation when compared to aluminum (because boron will hold on to electrons, and thus the hydride, better).

An argument can also be made that lithium will better stabilize the negative oxygen intermediate that results from the hydride attacking the carbonyl group.

So why use NaBH4 at all? Because it is stable compared to LiAlH4, it can be used in aqueous solutions and measured in open air where as LiAlH4 will react violently to moisture and water and ethanol.