The hydrocarbon which can react with sodium in liquid ammonia is

(1) $\ce{CH3CH2CH2C#CCH2CH2CH3}$
(2) $\ce{CH3CH2C#CH}$
(3) $\ce{CH3CH=CHCH3}$
(4) $\ce{CH3CH2C#CCH2CH3}$


Solve it as a case of terminal alkynes, otherwise all alkynes react with sodium in liquid ammonia.


I know sodium in liquid ammonia is used to produce trans-alkenes from alkynes. But from the options, it seems all of them react with sodium in liquid ammonia.

The solution doesn't make any sense to me. If all of the alkynes given react with sodium in liquid ammonia, then why the question? And what difference does it make if the alkyne is terminal?

  • $\begingroup$ Could the answer actually be all of them? Sometimes they do that on multiple choice questions. In any case the second choice should have an additional carbon between the triple bond and the methyl group on the right. They apparently mistyped. $\endgroup$ – Oscar Lanzi Apr 6 '18 at 9:24
  • $\begingroup$ The terminal alkyne will not give the birch reduction as they will just form the aklyne anion. Maybe this was something the textbook was referring to. $\endgroup$ – user2652780 Aug 10 '18 at 4:55
  • $\begingroup$ A Birch reduction uses alcohol as a proton source. No mention of alcohol here. Choice (2) was probably meant to be 1-butyne. $\endgroup$ – user55119 Aug 9 '20 at 16:35

Although this is more of a case of communication rather than actual chemistry, I'll explain exactly what the author meant.

They basically want you to ignore the Birch reduction reaction. They want you to find the compound that reacts with sodium in liquid ammonia, but not via Birch reduction.

I however, do stand by that the question is very poorly framed. It is definitely ambiguous, and relies on strange, unrealistic assumptions.


The terminal alkyne has acidic $\ce{H}$ and therefore will give acid base reaction with $\ce{NaNH3}$ and go $\ce{NH4+}$ and $\ce{RC#C- Na+}.$


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