I've read that sodium metal is used to form ethynides but for forming propynide, they have used $\ce{NaNH2}$. Is there some reason for not using sodium metal for propynides as well?

Source: Chemistry NCERT class 11 part-II page 394


2 Answers 2


Sodium metal can indeed be used to deprotonate propyne, although as we will see sodium amide would be more selective.

Addison et al. [1] report the reactions of melted sodium metal with both ethyne and propyne. In both cases a mixture of products is obtained; both displacement of the acetylenic hydrogen to form an acetylide salt and reduction of the alkyne to an alkene occur. This is in contrast with sodium amide, which has no reducing power and thus the acetylide salt would be obtained more selectively with the amide.

One difference between the two hydrocarbon substrates is the acetylide salt that is obtained. With propyne the 1:1 salt is found ($\ce{Na^+C3H3^-}$), as the methyl group is not deprotonated. With ethyne, however, the deprotonation goes all the way to the di-anion, forming $\ce{(Na^+)2C2^{2-}}$ as the salt product. Sodium amide would be expected to accomplish only one deprotonation with either ethyne or propyne.


1. C. C. Addison, M. R. Hobdell and R. J. Pulham, "The reactions of acetylene and propyne at the surface of liquid sodium ", Journal of the Chemical Society A (1971), 1704-8. https://doi.org/10.1039/J19710001704

  • $\begingroup$ Why would sodium amide deprotonate only from one side? $\endgroup$
    – TRC
    Aug 12, 2021 at 3:56
  • 2
    $\begingroup$ Would not be a strong enough base to remove the second proton from ethyne. Sodium metal can reduce both protons. $\endgroup$ Aug 12, 2021 at 5:14

I think the reason for using NaNH2 for propyne and Na(with suitable solvent) for ethyne has to do something with the acidity of ethyne and propyne.

While any terminal alkyne is generally very acidic,ethyne is more acidic than propyne.In propyne,there is a methyl group connected to the sp hybridized carbon leading to a slight reduction in the acidic nature of the terminal hydrogen(+I effect of methyl group).Hence it needs a better base like NaNH2.


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