Question

What would be the difference in reactivity if ethyl 3-methylbenzoate was treated in a reduction with NaBH4 vs LiAlH4? I know that NaBH4 is only powerful enough to reduce ketones and aldehydes, and ethyl 3-methylbenzoate has a ketone group, but I am confused as to how different LAH would be?

Answer 1

Ethyl 3-methylbenzoate has a ester group, not a ketone group. In LiAlH₄ the hydride source is AlH₄^-, and in NaBH₄ the hydride source is BH₄^-, the hydride source AlH₄^- is a stronger hydride source than BH₄^-.. This is beacuse aluminium is less electronegative than boron, and thus the electron cloud is shifted more towards the hydrogen in AlH₄^- compare to BH₄^-. This is the reason aluminium hydride is a better hydride donor.

Anoter reason is from the comparision of metals (Lithium and sodium). Lithium is more electropositive than sodium and thus is a better lewis acid, and is therefore able to activate the carbonyl more effectively compared to sodium.

The carbonyl carbon of esters have a lone pair which is in resonance with the double bond of carbonyl group. NaBH₄ being a weaker reducing agent cannot reduce any of these. Thus esters are mostly reduced by LiAlH₄ and not by NaBH₄.

The detailed mechanism of the reduction of esters by LiAlH₄ is given below:

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