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?)
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$\begingroup$ 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 YinCommented Jun 5, 2020 at 12:36
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1 Answer
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Metallic sodium $(\ce{Na})$ 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):
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:
$$\ce{2Na + RCH=O + 2CH3CH2OH -> RCH2OH + 2CH3CH2ONa}$$ $%edit$
answered Oct 16, 2013 at 12:02
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2$\begingroup$ Can sodium-ethanol reduce amides, acyl chlorides, acid anhydrides and esters ? $\endgroup$– user39812Commented Jan 12, 2017 at 11:27
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1$\begingroup$ @anonymous - feel free to ask a new question. $\endgroup$ Commented Jan 13, 2017 at 3:05
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2$\begingroup$ 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. $\endgroup$ Commented Oct 8, 2018 at 14:14
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1$\begingroup$ @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. $\endgroup$– DanteCommented Dec 13, 2018 at 4:02
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$\begingroup$ Thanks for the explanation sir, but isn't there a major possibility of formation of vicinal diols in these case?. $\endgroup$ Commented Feb 9, 2020 at 8:45