In my previous post I speculated why bis(trifluoromethyl) ketone tends to fully form a hydrate when dissolved in water, but acetone does not. Here I turn to asking why formaldehyde is also 80% converted to methanediol in water? Could it be that again, the diol is somehow preferentially stabilised compared to the carbonyl precursor and if so, why?
The lowest energy geometry is shown above. Conspicuously, it does not form an intramolecular O-H…O hydrogen bond, but adopts a C2-symmetric form. NBO analysis for this geometry reveals two interactions larger than the rest. The first, shown below, involves overlap of an oxygen lone pair (Lp) donor orbital with a C-H acceptor (purple+blue, orange-red), and this is worth E(2) 6.1 kcal/mol (there are two of these). Unfortunately, the analogous NBO interaction in acetone itself originating from a C-Me bond as acceptor is 6.3 kcal/mol and so this interaction does not differentiate between the two.
The larger NBO interaction of E(2) = 16.9 kcal/mol arises from the same donor orbital interacting with the C-O acceptor (the presence of the more electronegative oxygen accounts for it being the better acceptor). In acetone however, this too has the high value of 16.8 kcal/mol. Another possible interaction might be from a H-C donor to a C-O acceptor. But as you can see below, the positive overlap (red+orange) is matched by the negative overlap (orange+blue) and this interaction turns out to be insignificant. We have to seek elsewhere for differentiation between formaldehyde and acetone. To do this, I have added four explicit water molecules as solvent, and looked at the free energies of diol formation from the carbonyl (wB97XD/6-311G(d,p)/scrf=water).
The water molecules combine with the methanediol to form an elegant lattice of hydrogen bonds, involving two rings of three oxygens and one ring of four oxygens. This compact motif is less stable for propanediol, which instead prefers a structure forming fewer hydrogen bonds, largely because of the presence of the hydrophobic methyl groups. The result is that the free energy of hydration of formaldehyde to the diol, assisted by hydrogen bonds formed to four water molecules, is exothermic at -1.2 kcal/mol, whereas that for acetone is endothermic at +7.5 kcal/mol.
As with most things water, a proper stochastic exploration of all the possible configurations of the hydrogen bonds is necessary for a definitive explanation. But it does seem that a probable theory for why formaldehyde readily forms a diol whereas acetone does not lies not so much in stereoelectronic donor-acceptor interactions but in the hydrogen bonds set up in the solvated diol.