# 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!

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.

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 ($$\mathrm pK_\mathrm a \sim 25$$) 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.

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,

Sodium amide can be prepared by the reaction of sodium with ammonia gas. The reaction is fastest at the boiling point of the ammonia, $$\pu{−33 °C}$$. An electride, $$\ce{[Na(NH3)6]+e-},$$ is formed as a reaction intermediate.

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

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.