How does sodium react with methanol?

Question

Strong acids of $\ce{AH}$ type readily dissociate to its constituent ions (considering the solvent is water): $\ce{A-}$ and $\ce{H+}$. But the mechanism of the reaction between sodium metal and methanol $(\ce{Na + MeOH})$ as taught by my teacher appears to contradict with this statement:

$$ \begin{align} \ce{Na &-> Na+ + e-} \tag{R1} \\ \ce{MeOH &-> MeO- + H+} \tag{R2} \\ \hline \ce{Na + MeOH &-> Na+MeO- + H^.} \tag{R3} \end{align} $$

$$\ce{2 H^. -> H2} \tag{R4}$$

$\mathrm{p}K_\mathrm{a}(\ce{MeOH}) = 15.5$ according to Wikipedia — Methanol and reference therein [1], which is an indication that methanol is a weak acid. But then how the hydrogen is evolved?

Secondly, how does sodium gives its electron freely in the water? I can't believe free electron could exist in solution. Does electron solvation take place?

Reference

1. Ballinger, P.; Long, F. A. Acid Ionization Constants of Alcohols. II. Acidities of Some Substituted Methanols and Related Compounds. J. Am. Chem. Soc. 1960, 82 (4), 795–798. DOI: 10.1021/ja01489a008.

Answer 1

Caution

Sodium ($\ce{Na}$) reacts explosively with water ($\ce{H2O}$); I would recommend carrying out the experiment in absence water under inert gas atmosphere (to avoid moisture). Moreover, even the reaction of $\ce{Na}$ with methanol ($\ce{MeOH}$) is highly exothermic, and ignition is likely. Remember, alcohols and evolved dihydrogen ($\ce{H2}$) form good fuels. Thus, the reaction would have to be carried out at low temperatures or using very small amount of $\ce{Na}$ in inert atmosphere.

The Reaction

$\ce{MeOH}$ is acidic enough to react (exothermically) with Direct $\ce{Na}$. Reaction of $\ce{Na}$ and $\ce{MeOH}$, and subsequent vaporization of excess $\ce{MeOH}$ is a common route to the preparation of pure sodium methoxide $\ce{MeO-Na+}$ crystals.

$$ \begin{multline} \ce{2Na(s) + 2ROH_\text{excess}(l) -> 2RO- Na+ (ROH) + H2 ^}\\ \ce{->[vacuum distillation][] 2RO- Na+ + ROH(l)} \end{multline} $$

In the article Synthesis and Characterization of Sodium Alkoxides$^\text{1}$ by Chandran K., Nithya K., Sankaran, K., and Gopalan K. (2006), the authors used $\pu{100 mL}$ of alcohol ($\ce{ROH}$) and reacted it with $\pu{500 mg}$~$\pu{1000 mg}$ of $\ce{Na}$ to produce corresponding sodium alkoxides ($\ce{RO- Na+}$). Upon subsequent vacuum distillation, pure $\ce{RO- Na+ (s)}$ could be isolated.

Solvation of Electrons

There is no possibility of solvation of (highly basic) electrons ($\ce{e-}$) in acidic $\ce{ROH}$ (except at very high pressures, order of 10-100 $\pu{Mbar}$, as pointed out in the comments).

$$ \ce{2ROH + 2e- -> 2RO- + H2} $$

This would be an immediate reaction. However, solvation of electrons is possible; for example, in when $\ce{Na}$ is dissolved in (much less acidic) ammonia ($\ce{NH3}$), the characteristic blue color is obtained due to solvated electrons (video demonstration).

Reference

1. Chandran K., Nithya K., Sankaran, K., and Gopalan K. (2006). Synthesis and characterization of sodium alkoxides. Bull Mater Sci 29, 173–179. 10.1007/BF02704612