I'm having trouble understanding why dissolving metal reductions, like (for example) sodium with alkenes in liquid ammonia, sodium with aldehydes/ketones in ethanol, or the Bouveault–Blanc reduction, reduce the substrate instead of the solvent. If you put sodium in ethanol (or any protic solvent), the ethanol is reduced and hydrogen gas is released. I can't see how the presence of a ketone (for example) will create a more favorable reaction.

For once, it doesn't seem thermodynamically favorable - reduction of the ethanol will release hydrogen gas, largely increasing the entropy, while reduction of the ketone produces no gas. Even if it is thermodynamically favorable, however, reduction of the ethanol seems much more favorable kinetically, since there are much more solvent molecules than substrate molecules. What is going on here?

  • $\begingroup$ Looking at some typical reaction conditions (e.g. organic-chemistry.org/namedreactions/…) I see that the proton source (e.g. ethanol, isopropanol) is added in stoichiometric amounts and is not employed as the solvent. It would strike me as very odd if the sodium would not just be quenched if it is dissolved in ethanol. In liquid ammonia, dissolved electrons are formed, just have a look at the wikipedia article of the Birch reduction. $\endgroup$ Oct 15, 2016 at 18:19

1 Answer 1


There are plenty of examples where ethanol is present at over 10ml/g for this reaction. The key thing is that this is transfer of a single electron from Na initially. Where would this go onto ethanol? Clearly, it is faster to transfer an electron to an unsaturated compound i. e. The carbonyl compound. Once this occurs, protons are transferred. The movement of electrons and protons is fast compared to getting the two requisite H's together to form hydrogen and drive the alternative reaction.


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