Why does formaldehyde exist primarily as the gem-diol in aqueous solution?

Question

From Wikipedia,

Methanediol is the product of the hydration of formaldehyde $\ce{H2C=O}$, and predominates in water solution: the equilibrium constant being about $10^3$ and in a 5% by weight solution of formaldehyde in water, 80% is in the methanediol form.

As far as I know, carbonyl compounds usually don't exist in their gem-diol form, unless there's a stabilizing effect (like H-bonding, release of angle strain, inductive effect etc.), which we observe in ninhydrin, chloral, hexafluoroacetone, etc. (relevant image) But such effects don't seem to be the case for formaldehyde.

My question: What is the stabilizing factor in the case of hydration of formaldehyde?

Answer 1

In acetone the carbonyl form is more stable than the gem-diol, whereas with formaldehyde the situation is reversed and the gem-diol form is preferred. Overall, the equilibrium shifts by 6 powers of 10 between these two compounds.

It takes $\pu{\approx1.4 kcal/mol}$ to shift an equilibrium by 1 power of 10. Therefore in the acetone-formaldehyde comparison there is a stabilization change of $\pu{\approx8.4 kcal/mol}$.

It is difficult to see how changing the substituents in the gem-diol could have such a large effect, after all the gem-diol carbon is a saturated $\mathrm{sp^3}$ carbon, no p-orbitals are avaliable for resonance interaction. Considering steric effects as proposed in the other answer, methyl groups and hydroxyl groups are not particularly large, nor is the gem-diol carbon especially congested. Hence it is difficult to see how steric factors could provide the necessary $\pu{8 kcal/mol}$ change in stabilization. For reference, Cis-di-t-butylethylene only has $\pu{9.3 kcal/mol}$ of strain energy.⁽¹⁾

On the other hand we might expect the substituents on the carbonyl carbon (an unsaturated $\mathrm{sp^2}$ carbon with a p-orbital available for interaction) to have a significant effect on stability. Just as alkyl substituents stabilize a carbon-carbon double bond through both an electron-releasing inductive effect ⁽²⁾ and a hyperconjugative effect, we would expect the same effects to operate with a carbon-oxygen double bond. In fact, we might anticipate that the alkyl stabilization effect will be even greater with the carbonyl double bond since the carbonyl carbon is more electron deficient than the carbon in a carbon-carbon double bond.

If we compare the heats of combustion for the isomers acetone ($\pu{-427.9 kcal/mol}$) and propanal ($\pu{-434.1 kcal/mol}$), we see that attaching a second alkyl group to the carbonyl carbon stabilizes the system by $\pu{6.2 kcal/mol}$.⁽³⁾ Of course, adding the "first" alkyl group will also provide significant stabilization. Hence, it is clear that adding two alkyl groups to formaldehyde will provide enough electronic stabilization (inductive and hyperconjugative) to shift the equilibrium by $10^6$ from the enol to the keto form.

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1. see here

2. the alkyl group has an $\mathrm{sp^3}$ carbon while the double bond carbon is $\mathrm{sp^2}$ hybridized; since the latter is more electronegative, electron density will flow inductively (through sigma bonds) from the alkyl group to the $\mathrm{sp^2}$ carbon.

3. data from K. Peter C. Vollhardt, Neil E. Schore, "Organic Chemistry, Fourth Edition: Structure and Function" p. 125; see here

Answer 2

The reason is mainly a lack of steric hindrance due to the hydrogen atoms. The substituents on the carbonyl in formaldehyde - hydrogen, offers negligibly small steric hindrance in shifting from a $\ce{sp^2}$ hybridization on the carbonyl to a $\ce{sp^3}$ hybridization of the gem-diol, the $\ce{C-H-C}$ angle decreasing from 120 degrees to ~109.5 degrees.

The other examples you mentioned are because of electronic effects (chloral hydrate, hexafluoroacetone), hydrogen bonding (ninhydrin) or angle strain release (cyclopropanone). Read the relevant portion from Clayden/Warren's textbook for a sound comprehension!