In Clemmensen reduction, $\ce{Zn-Hg}$/ $\ce{ conc. HCl}$ reduces the carbonyl $\ce{>C=O}$ to $\ce{>CH2}$, by the following mechanism(?): Clemmensen Reduction Mechanism

What happens to the carbonyl group if we replace Zinc with any other metal such as Al or Na?

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Will this reaction take place if X is $\ce{Al-Hg / conc. HCl}$ ?

  • $\begingroup$ A more interesting reaction to explore is first react a sheet of Al in HCl. Extract the Al sheet which now holds (for a limited time) H chemisorbed hydrogen. The latter .H surface radical acts per the reaction: H+ + e- = .H, as working with an e- followed by H+. For example, with PbS + 2 .H -> Pb + H2S. That is, Pb2+ + 2 e- = Pb and S(2-) + 2H+ = H2S. Depending on the organic exposed to the Al sheet, expect a variation of the inorganic chemistry which may (or may not) display particular selectivity to produce one or more products. The example I cited comes from application in Hydrometallurgy. $\endgroup$
    – AJKOER
    Commented Sep 19, 2020 at 17:31
  • $\begingroup$ Alluded to source reference: 'Hydrometallurgy 2008: Proceedings of the Sixth International Symposium', p. 818, a commercial reductive leaching equation, to quote: " PbS + 2 •H = Pb + H2S (5) " (see books.google.com/… ). $\endgroup$
    – AJKOER
    Commented Sep 19, 2020 at 17:41

1 Answer 1


In comparision to Clemmensen reduction, both $\ce{Na-Hg}$ and $\ce{Al-Hg}$ cannot be used as direct substitutes for $\ce{Zn-Hg}$.

Reaction with $\ce{Na-Hg}$ amalgam:

$\ce{Na-Hg}$ amalgam is also a good reducing agent and is capable of reducing ketones/aldehydes to alcohols1. However, Do not use conc. $\ce{HCl}$ with $\ce{Na-Hg}$ amalgam, as there would be a vigorous reaction between the acid and the sodium in the amalgam. Instead, use an organic solvent such as absolute alcohol.

The mechanism for this reduction is a bit different than standard Clemmensen reduction:

\begin{align} \ce{2(C6H5)2CO +2Na &->2(C6H5)2C-ONa} \\ \ce{2(C6H5)2C-ONa&<=>(C6H5)2C(ONa)-C(C6H5)2-ONa\tag{1}} \\ \ce{2(C6H5)2C-ONa + 2C2H5OH &-> 2(C6H5)2C-OH + 2C2H5ONa} \tag{2a} \\ \ce{2(C6H5)2CO &-> (C6H5)2CO +(C6H5)2CHOH \tag{2b}} \end{align}

Reaction with $\ce{Al-Hg}$ amalgam:

Aluminium amalgam, like Sodium amalgam, vigorously reacts with water, and you cannot use concentrated $\ce{HCl}$ with it. With ethanol as a solvent, it is a mild reducing agent, can reduce cycloalkanones and aldehydes to alcohols. Acyclic ketones remain almost inert2

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In some cases, it behaves similarily to $\ce{Mg-Hg}$ amalgam and promotes dimerization of ketones to form pinacols. Thus, your reaction would not take place, and instead pinacol would be the major product.

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This question and this one cover some other reduction uses of $\ce{Al-Hg}$, such as reduction of imines to amines.


  1. Bachmann, W. E. “The Mechanism of Reduction by Sodium Amalgam and Alcohol. I. The Reduction of Aromatic Ketones to Hydrols.” Journal of the American Chemical Society, vol. 55, no. 2, Feb. 1933, pp. 770–74. doi:10.1021/ja01329a051.
  2. Troyansky, Emmanuil I., and Meghan Baker. “Aluminum Amalgam.” Encyclopedia of Reagents for Organic Synthesis, John Wiley & Sons, Ltd, 2016, pp. 1–6. doi:10.1002/047084289X.ra076.pub2.

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