Terminal alkynes `(RC-=CH)` are not reduced by alkali metals (e.g., Na, K, or Li) in liq. `NH_3`, but reduction takes place when `(NH_4)_2SO_4` is added in the reaction mixture. Explain why.

Explanation

`R-C-=CH` is initially converted to `R-C-=C^(overset(..)ө)` by `NaNH_2`. The negative charge on the species does not allow the donation of `bare` by `Na`, the first step in a dissolving metal resolution.
When `(NH_4)_2SO_4` is present, alkynide ion is protonated `(R-C-=C^(ө)overset(Noverset(o+)H_4)rarrR-C-=CH+NH_3`), and the neutral species will undergo a dissolving metal reduction to an alkene.