According to my teacher, Na in ethanol reduces carbonyl compounds (e.g. aldehydes and ketones) to alcohols. However, I can't see where the hydrogen needed for reduction comes from. How does the reduction take place? (or is my teacher wrong?)

  • $\begingroup$ Na provides a single electron in one mechanism. and the use of ethanol is as of a proton donor. The intermediate of Blanc reduction is an aldehyde which is obtained after repeating the above step twice $\endgroup$ – Dr.Martin Yin Jun 5 at 12:36

Sodium is a powerful reducing agent. It wants to donate an electron to things that are electron deficient.

The mechanism of the reduction of aldehydes and ketones by sodium would be similar to that of the Bouveault-Blanc reduction. Ethanol is the source of the protons (at least early in the reaction). The dots in the reactions below signify radicals (unpaired electrons), which are represented to the best mchem and MathJax's ability (which is not great).

Step 1 - one electron transfer from sodium to carbonyl group:

$$\ce{Na. + RCH=O->Na+ + RCH.-O^-}$$

Step 2 - abstraction of a proton from ethanol:

$$\ce{CH3CH2OH + RCH.-O^- <=> CH3CH2O- + RCH.-OH}$$ This step is an equilibrium step, and is driven toward products by the second electron transfer (which consumes the $\ce{RCH.-OH}$.

Step 3 - one electron transfer from a second sodium atom to the organic radical:

$$\ce{Na. +RCH.-OH -> Na+ + RCH:^{-}{}-OH}$$

Step 4 - second proton transfer from ethanol:

$$\ce{CH3CH2OH + RCH:^{-}{}-OH -> CH3CH2O- + RCH2OH}$$

This reaction, while technically an acid-base equilibrium, essentially goes to completion because of the vast disparity in acidity of $\ce{C-H}$ and $\ce{O-H}$.

The net reaction is:

$$\ce{2Na + RCH=O + 2CH3CH2OH -> RCH2OH + 2CH3CH2ONa}$$ $%edit$

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    $\begingroup$ Can sodium-ethanol reduce amides, acyl chlorides, acid anhydrides and esters ? $\endgroup$ – user39812 Jan 12 '17 at 11:27
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    $\begingroup$ @anonymous - feel free to ask a new question. $\endgroup$ – Ben Norris Jan 13 '17 at 3:05
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    $\begingroup$ This scheme looks plausible, except that I'd expect $\ce{RCH:^{-}{}-OH}$ to immediately isomerize to $\ce{ RCH2O-}$ which will then be in equilibrium with the ethoxide as pKas are not vastly different anymore. $\endgroup$ – Matyas Koszik Oct 8 '18 at 14:14
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    $\begingroup$ @BenNorris I've also read in my book that Na+Alcohol is a strong base and is used for dehydrohalogenation. Is is true that the above combination acts as both reducing agent and a strong base? I'm confused since a strong base must accept a proton instead. $\endgroup$ – Dante Dec 13 '18 at 4:02
  • $\begingroup$ Thanks for the explanation sir, but isn't there a major possibility of formation of vicinal diols in these case?. $\endgroup$ – Chirag Lathi Feb 9 at 8:45

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