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Firstly, I would like to ask why the alkyne reacts at all with the solvated electron? Surely the alkyne group is extremely nucleophilic (more so than the alkene). This, to me, seems to suggest that an alkene would be more likely to react in this manner than the alkyne. However, this is clearly not the case; why?

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As shown in your diagram, when the initial electron is transferred to the alkyne a radical anion is formed; the transferred electron pairs with an electron from the alkyne to form an anion at one end of the molecule. These two electrons are in an $\ce{sp^2}$ orbital. If you try to repeat this process with an alkene, you wind up with the paired electrons in an $\ce{sp^3}$ orbital. An $\ce{sp^2}$ is lower in energy than an $\ce{sp^3}$ orbital since it contains more s character and an s orbital is lower in energy than a p orbital. The electron solvated in ammonia has enough energy to transfer into a lower energy $\ce{sp^2}$ orbital, but not enough energy to transfer into a higher energy $\ce{sp^3}$ orbital.

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  • $\begingroup$ But we can do Birch reduction of Benzene, wouldn't in that case too, the electrons pair in sp3 only? $\endgroup$ Oct 29, 2022 at 13:13
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    $\begingroup$ @GurjotSingh But in the case of benzene the intermediate radical anion is further stabilized by conjugation. Generally simple alkenes won't undergo Birch reduction, but conjugated systems will. 1,3-Cyclohexadiene is reduced to cyclohexene, but 1,4-cyclohexadiene is not. $\endgroup$
    – ron
    Oct 30, 2022 at 15:14

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