ohh my god...stuck again..

Because the H come from the more electron withdrawing C, which in this case, is on the right of the triple bond

alkyne has acidic H

H₃CH <<<< acidic

Alkyl groups are electron donating...or, proton withdrawing. In this problem, a proton is being given to NH2, so u want proton donating. CH3 is the alkyl group here, so it will keep in its Hydrogens. So comparatively, its easier to pull H from the C on the right.

carbon with sp hybridisation is more electronegative since s- character is 50%.
hence hydrogen attached will be more acidic and NH₂^- will abstract hydrogen from the triple bond carbon....

So an acid base reaction culminates.

CH₃C≡CH + NH₂^- ---->NH₃⁺ + CH₃C≡C^-

This alkynyl anion can be used as a nucleophile in preparation of alkynes by nucleophilic substitution reactions. The acid base rn you have given here is the typical reaction of acetylene with sodamide in liquid ammonia.