Question bout alkoxides

[big314mp]

This is purely for my own curiosities sake...

 

If one were to take an ionic solution in an alcohol, and then run current through it, would the product be the alkoxide in solution?

 

Something along the lines of sodium chlorode in methanol with a few volts using graphite electrodes. Would this produce hydrogen at one electrode (from the alcohol) and chlorine at the other electrode (from the salt) leaving sodium methoxide in solution?

 

What if the anion was something more difficult to oxidize, such as sulphate?

 

Thanks!

[Theophrastus]

I think I have realised a minor flaw in your process. It certainly seems rational that if alkali metals react with water, to isolate them, would require the use of a nonreactive, nonpolar solution, however, in the case of solutes and solvents, like dissolves like, and as such, the ionic solute, which would maximize the electrolytic properties of a solution, would be unable to dissolve,in the solution, and thus, the electrolysis would be deemed ineffective. Sorry to spoil the fun. (though let me know if I'm wrong; alkali metals, are rather fun!) Best wishes!

Edited March 20, 2009 by Theophrastus
addition of content

[UC]

This is purely for my own curiosities sake...

 

If one were to take an ionic solution in an alcohol, and then run current through it, would the product be the alkoxide in solution?

 

Something along the lines of sodium chlorode in methanol with a few volts using graphite electrodes. Would this produce hydrogen at one electrode (from the alcohol) and chlorine at the other electrode (from the salt) leaving sodium methoxide in solution?

 

What if the anion was something more difficult to oxidize, such as sulphate?

 

Thanks!

 

Well, sodium chloride doesn't dissolve in alcohol, but that's besides the point. The easiest way to do this is by azeotropic drying or using molecular sieves. An alkali metal hydroxide dissolved in alcohol exists in the following equilibrium: [ce] KOH + MeOH <-> KOMe + H2O [/ce]. If the alcohol is truly dry, the equilibrium will lie to the right unless the concentration of KOH is very high (and therefore the concentration of water would be high). If you can remove the water from the reaction mix, then Le Chatlier's Principle states that the equilibrium will push to the right. Either you use molecular sieves to adsorb the water or you add a solvent that has an azeotrope with water, but not MeOH, which will boil off below the boiling point of methanol. simply distill off the azeotrope to drive the proportion of methoxide up.

 

Also a problem with the original idea is that alcohols are not nearly as polar as water and as such, even ionic compounds that dissolve in them don't disassociate into ions nearly as well. These solutions are in most cases not good conductors.

Edited March 21, 2009 by UC