Prepic
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Sodium methoxide from methanol, NaOH and Aluminium
Hey all,
I was wandering if it might be possible to make sodium methoxide by reacting sodium hydroxide, methanol and aluminium together.
Ideally, the water formed from the equilibrium reaction between sodium hydroxide and methanol will react with excess sodium hydroxide and aluminium to
form sodium aluminate thereby shifting the equilibrium to favour sodium methoxide.
I already know that sodium methoxide can be made just with NaOH and methanol and that molecular sieves are a viable option too however I'd argue
aluminium is more accessible and should irreversibly consume any water formed.
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njl
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I think the reaction forming sodium aluminate would be much faster than the reaction forming sodium methoxide. Also the aluminum oxidation is
extremely exothermic, and although it would be slower in methanol than in water, the boiling point of methanol is so low that I can't see this being
easy to control.
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clearly_not_atara
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Even easier would be the (as yet hypothetical) reaction between Na2CO3 and solutions of Mg(OMe)2 made by the dissolution of Mg in MeOH. Sodium
carbonate is poorly soluble, but MgCO3 should be completely insoluble.
[Edited on 04-20-1969 by clearly_not_atara]
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njl
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What about quicklime (CaO) and methanol to give a mixed alkoxide/hydroxide that could maybe disproportionate into insoluble Ca(OH)2 and Ca(OMe)2?
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AJKOER
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A review of the chemistry is interesting. Here Al or Mg ribbons with NaOH can also create surface absorbed hydrogen atoms (so-called hydrogen
embrittlement).
The formed •H atoms are apparently active on the metal surface even after removal from the NaOH/Metal reaction chamber and have been actually
suggested for employment in the commercial leaching of ores. Source: my prior citation of a hydrometallurgy reference, to quote:
"One may assume that the •H radical functional behaves (per its seemingly reversible formation reaction: e- + H+ = •H ) as apparently a (e-,H+)
pair acting on ions. For an example from 'Hydrometallurgy 2008: Proceedings of the Sixth International Symposium', p. 818, a commercial reductive
leaching equation, to quote:
" PbS + 2 •H = Pb + H2S (5) " (see https://books.google.com/books?id=1etfSdk55SYC&pg=PA818&... )"
In the current context, possible reactions:
CH3OH (aq) <=> H+ + CH3O-
as methanol is more acidic than water. Further conceivable interactions with the Hydrogen atom radical (as a source of e- and H+):
CH3OH (aq) + e- (aq) <=> •H + CH3O-
•H + •H -> H2 (g)
In the presence of Sodium ions:
Na+ + CH3O- -> NaCH3O
With Mg metal, for example:
Mg -> Mg(2+) + 2 e-
H+ + e- = •H
This review may assist.
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njl
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That's very interesting. In what way does the hydrogen radical participate in the reaction though? Other than escaping as H2 gas.
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Prepic
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I decided to try out my reaction.
To a 250ml flat bottom flask I added 22.0 g of NaOH and 100 ml of methanol along with a stir bar. The flask warmed up significantly but didnt truly
reach a reflux. After about 30 min, majority of the sodium hydroxide dissolve and I began adding aluminum.
First I added rolled aliminium foil, ~0.25g and some reaction was noticed albeit slow. I then tried adding portions of aluminium powder, not too
vigorous and I successfully added all the aluminum powder over the course of about 20 min.
Feeling confident there wont be any runaway reaction, I am allowing the mixture to reflux gently.
Is there a easy method to test for the concentration of sodium methoxide which will differentiate from dissolved sodium hydroxide in methanol?
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unionised
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Quote: Originally posted by njl  | | That's very interesting. In what way does the hydrogen radical participate in the reaction though? Other than escaping as H2 gas.
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None at all.
AJKOER is noted for posting stuff about free radicals, whether his stuff is relevant (or even correct) or not.
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