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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.
andselisk
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Chesx
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    Same reason it reacts with water, even if difference in speed is huge. – Mithoron May 24 '23 at 18:33
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    Ultrafast submilisecond camera can catch the blue colour of hydrated electrons in water. Solution of sodium in liquid ammonia keeps the dark color of solvated electrons relatively long, as ammonia is much weaker acid then water. Methanol is much closer to water than to ammonia, being somewhat weaker acid than water. – Poutnik May 24 '23 at 18:36
  • related https://chemistry.stackexchange.com/questions/41669/what-is-the-reason-for-the-blue-color-of-an-ammoniated-electron – Mithoron May 24 '23 at 18:36
  • @Mithoron If I already knew that reason why would I ask here? And what do you mean by "It"? – Chesx May 24 '23 at 18:39
  • @Poutnik but how would that explain that reaction takes place since methanol is still weak acid. – Chesx May 24 '23 at 18:41
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    Water, methanol and liquid ammonia are all weak acids (ammonia many orders weaker than both others) and sodium reacts with weak acids. More exactly, solvated electrons react with weak acids. – Poutnik May 24 '23 at 18:43
  • It's not an acid base, but redox reaction. – Mithoron May 24 '23 at 18:53
  • @Poutnik could you please elaborate mechanism? – Chesx May 24 '23 at 18:55
  • @Mithoron Yes But I never said it is acid base. I meant to say that the hydrogen dissociated from MeOH is in very low amount then how does reaction takes place since Na is in higher amount. – Chesx May 24 '23 at 18:59
  • $\ce{e- + H^{(+)}-A^{(-)} <=> H^{(-)}-A^{(-)} -> H^{.} + A-}$ – Poutnik May 24 '23 at 19:55

1 Answers1

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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
ananta
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