Why do BH3 and NaBH4 have different selectivities?

Question

I've always known borane ($\\ce{BH3}$), as a reagent for alkene hydroboration. Recently in class we talked about its use in the reduction of carboxylic acids to alcohols as well. Now obviously $\\ce{BH3}$ is an electrophile and I don't think it reacts by donating hydrides like sodium borohydride ($\\ce{NaBH4}$) does - in any case it's not charged like $\\ce{BH4-}$. Also $\\ce{BH3}$ should be a pretty weak hydride donor, considering the covalent nature of the bonds and lack of any charge. It can reduce carboxylic acids, however, which $\\ce{NaBH4}$ can't do! Where's the difference and what's the reason for this contradiction? And what prevents borane from reducing acid chlorides (my instructor told me this)? This is something $\\ce{NaBH4}$ is successful at.

Answer 1

BH3 becomes a source of "H minus" after the boron complexes to a carbonyl oxygen. It would seem acid chlorides are not reduced because the carbonyl oxygen is very electron poor due to electron withdrawing chlorine, which is bad at pi donation due to its large size, and also very electronegative. LiAlH(OtBu)3 is good for reducing acid chlorides. H2/Lindlar's catalyst will also reduce it to an aldehyde.

Based ONLY on electron richness, you would think BH3 would be best at reducing amides, and equally OK at reducing carboxylic acids and esters, and the worst at reducing acid chlorides. But it really does reduce carboxylic acids the best, among all the carbonyl electrophiles.

An emeritus professor/blogger wrote about this issue: http://www.ch.imperial.ac.uk/rzepa/blog/?p=5114 . When interpreting the graphic, keep in mind "R" can be equal to hydrogen. He says the rate determining step is TS1. You don't need to understand the computations to understand his final conclusion.

I doubt this mechanism has been on any organic chem exam in recorded human history. Computations like this do not exactly give precise answers, so the author can only speculate.