Chemistry - GOC Question with Solution | TestHub

Removing ${\mathrm{H}}_1$ puts the negative charge on the amino nitrogen :

Removing ${\mathrm{H}}_2$ puts the negative charge on the amide nitrogen :

In the first conjugate base the charge is localised on nitrogen. In the second, the charge is delocalised by resonance over the amide nitrogen and the carbonyl oxygen. A delocalised charge is more stable, so the second conjugate base wins and the amide N-H proton ${\mathrm{H}}_2$ is more acidic.