organic chemistry - Is the compound CH2(OH)(ONa) possible?

I was just pondering about the reaction mechanism of Cannizzaro reaction.

$\ce{H2C^{+}-O- + OH- -> H2C(OH)O-}$

$\ce{HCHO + H2C(OH)O- -> H3CO- + HCOOH}$

$\ce{H3CO- + HCOOH -> HCOO- + H3COH}$

I wondered why in 2nd step instead of receiving $\ce{H-}$ from $\ce{H2C(OH)O-}$; $\ce{H2C^{+}-O-}$ wouldn't react with NaOH to form $\ce{H2C(OH)ONa}$.

I don't think this compound will be able to exist. But don't know why! Everything seems right about it.

Answer

Of course, you shouldn’t start from $\ce{H2C+-O-}$ unless you note that it is a mesomeric form of $\ce{H2C=O}$ — the more relevant mesmeric structure.

The mechanism as you wrote it and as it is typically written is given without any counterion, i.e. using just $\ce{OH-}$ and not $\ce{NaOH}$. You could instead write the mechanism using a counterion for $\ce{OH-}$ — $\ce{Na+}$, which would result in the first step being:

$$\ce{H2C=O + NaOH -> H2C(ONa)(OH)}\tag{1}$$

This is essentially the compound you are suggesting. Hence it exists, but it is a minor constituent of the overall solution. It’s not part of the second step, it is the first step. For the second step to happen, you need another molecule of $\ce{H2C=O}$, which has not yet reacted to the geminal diolate $\ce{H2C(ONa)(OH)}$.

See this scheme for the transition state of the Cannizzaro reaction and where $\ce{H2C(ONa)(OH)}$ comes into play: