Chemistry - GOC Question with Solution | TestHub

Removing ${\mathrm{H}}_1$ gives the acetylide :

Removing ${\mathrm{H}}_2$ gives the amide anion :

Factor (4) orbital : the acetylide holds its lone pair in an sp orbital (50% s-character), which holds the charge close to the nucleus and stabilises it, while the amide anion holds its lone pair in an sp3 orbital (25% s-character). More s-character means a more stable conjugate base (terminal alkyne $\mathrm{p}{K}_{\mathrm{a}}$ ~ 25 vs amine N-H $\mathrm{p}{K}_{\mathrm{a}}$ ~ 38), so the C-H proton ${\mathrm{H}}_1$ is more acidic.