Reactions of alkynes with sodium/ammonia vs sodium amide

What happens when alkynes react with sodium in ammonia ($\ce{Na/NH3}$), or when they react with sodium amide ($\ce{NaNH2}$)?


Are these two sets of conditions different?




Sodium in ammonia ($\ce{Na/NH3}$) and sodium amide ($\ce{NaNH2}$) are very different sets of conditions, despite superficially looking similar!




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Sodium metal itself, $\ce{Na}$, is a one-electron reducing agent – it is oxidised to $\ce{Na+}$ in the process. When it reduces neutral hydrocarbons, negatively charged species are formed, which pick up a proton from the ammonia solvent. Overall, $\mathbf{Na/NH\_3}$ is a way of **reducing** some organic compounds (including alkynes).


Internal alkynes $\ce{R-C#C-R}$ can be reduced to *trans*-alkenes using $\ce{Na/NH3}$. Specifically, solvated electrons are created which add to the triple bond as shown. Ammonia acts as a proton source to protonate the carbanions formed. The vinyl anion intermediate **4** can interconvert between a *cis* and *trans* geometry, and the *trans* geometry is preferred as this minimises steric repulsion between $\ce{R^1}$ and $\ce{R^2}$. When this anion is protonated, the *trans*-alkene is formed. This is a nice synthetic procedure for the preparation of *trans*-alkenes, complementary to Lindlar hydrogenation which produces *cis*-alkenes from alkynes.


[![Mechanism of Na/NH3 reduction of alkynes](https://i.stack.imgur.com/3MWvk.png)](https://i.stack.imgur.com/3MWvk.png)




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On the other hand, in sodium amide $\ce{NaNH2}$, sodium is already in the +1 oxidation state and is no longer a reducing agent. However, the amide ion $\ce{NH2-}$ is a very strong base (the $\mathrm pK\_\mathrm a$ of ammonia, $\ce{NH3}$, is 38). So $\mathbf{NaNH\_2}$ is a way of **deprotonating** some organic compounds (including terminal alkynes).


The proton attached to the terminal carbon in a terminal alkyne is [acidic enough](https://chemistry.stackexchange.com/q/32341/16683) ($\mathrm pK\_\mathrm a \sim 25$) to react with sodium amide, forming the corresponding carbanion **7**. Such carbanions are good nucleophiles and can be used in many synthetic procedures involving nucleophiles, such as addition to a carbonyl compound, as illustrated here.


[![Deprotonation of terminal alkyne with NaNH2](https://i.stack.imgur.com/HqReP.png)](https://i.stack.imgur.com/HqReP.png)


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In the first case hydrogenation occurs(anti addition) whereas in the second case, acid base neutralization occurs. So they aren't identical.


According to [Wikipedia](https://en.wikipedia.org/wiki/Sodium_amide#Preparation_and_structure),



> 
> $$\ce{2 Na + 2 NH3 -> 2 NaNH2 + H2}$$
> 
> 
> 


This shouldn't occur "usually" in reactions, unless it is specified at very low temperature.


However, using FeCl3 as a catalyst one can produce sodamide and hydrogen from sodium metal and ammonia in appreciable amounts.


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