0
$\begingroup$

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.
$\endgroup$
10
  • 1
    $\begingroup$ Same reason it reacts with water, even if difference in speed is huge. $\endgroup$
    – Mithoron
    May 24, 2023 at 18:33
  • 2
    $\begingroup$ 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. $\endgroup$
    – Poutnik
    May 24, 2023 at 18:36
  • $\begingroup$ related chemistry.stackexchange.com/questions/41669/… $\endgroup$
    – Mithoron
    May 24, 2023 at 18:36
  • $\begingroup$ @Mithoron If I already knew that reason why would I ask here? And what do you mean by "It"? $\endgroup$
    – Chesx
    May 24, 2023 at 18:39
  • 1
    $\begingroup$ 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. $\endgroup$
    – Poutnik
    May 24, 2023 at 18:43

1 Answer 1

-3
$\begingroup$

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
$\endgroup$
6
  • 3
    $\begingroup$ "People saying it is not possible should not stand in path of people doing it." (Chinese proverb). :-) // If it is possible to catch hydrated electrons in water, it would be surely possible doing so for methanol. Search for "hydrated electrons", namely work of prof Pavel Jungwirth. $\endgroup$
    – Poutnik
    May 25, 2023 at 4:37
  • 1
    $\begingroup$ Closely related Nature - Spectroscopic evidence for a gold-coloured metallic water solution $\endgroup$
    – Poutnik
    May 25, 2023 at 5:45
  • 1
    $\begingroup$ One of multiple his papers: ACS - Unraveling the Complex Nature of the Hydrated Electron $\endgroup$
    – Poutnik
    May 25, 2023 at 5:51
  • 4
    $\begingroup$ This isn't acid base reaction. You said MeOH is acidic enough to react but that doesn't explain mechanism. According to me it involves Single electron transfer mechanism. $\endgroup$
    – Leibniz-Z
    May 25, 2023 at 15:38
  • 4
    $\begingroup$ Also you're too cautious in first section - reacting Na with methanol is a routine way to get rid of it. Fume hood is more than enough. Another issue is I don't get why you call it vacuum distillation, if it's the precipitate that gets purified. $\endgroup$
    – Mithoron
    May 26, 2023 at 0:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.