How to determine which compound will get oxidised and which will be reduced in a cross cannizzaro reaction?

Question

How to determine which compound will get oxidised and which will be reduced in a cross cannizzaro reaction?

As a general rule of thumb, I have always been told that Formaldehyde is always oxidised, and the other is reduced.

But no general principle was ever given to me. What factors (eg electronic, steric, etc) affect this ??

There is this question already here, but the both the answers seems to be contradictory. @Jan says , the one with more hydrate forming tendancy oxidised , whereas @RE60K says less Nucleophilic aldehyde oxidises (which I believe will have a lesser tendancy to form hydrates)

Answer 1

It correlates very strongly with ability to form a hydrate as @Jan's answer says, which is why formaldehyde is usually the sacrificial oxidised component. In general electron withdrawn aldehydes are more easily oxidised, electron rich aldehydes reduced.

This paper from J. Org. Chem. here compared several pairs of aldehydes and the conclusions were:

m-nitrobenzaldehyde is (relatively) extensively oxidised as contrasted to p-anisaldehyde, which is much less extensively oxidised - i.e., much more extensively reduced. On the bases of this and prior investigations, the five aldehydes studied can be arranged (in order of diminishing susceptibility to oxidation): m-nitrobenzaldehyde, furfural, p-bromobenzaldehyde, benzaldehyde, and p-anisaldehyde. With reference to benzaldehyde and the substituted benzaldehydes, this order correlates with expected combined resonance and inductive influences of the substituent groups on the formyl group, as these should contribute to ease of a) hydroxide ion attack and b) hydranion transfer

Answer 2

It is a kinetically controlled reaction so OH- will attack that carbon which is having more Electrophilic character(more partaol positive charge) to yield the major product