Chemistry - GOC Question with Solution | TestHub

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

Removing ${\mathrm{H}}_2$ gives the carboxylate :

Both anions are resonance-stabilised, so the tie is broken by the atom factor : sulphur lies below oxygen and is larger, so it disperses the negative charge better. The dithiocarboxylate is more stable, so the S-H proton ${\mathrm{H}}_1$ is more acidic.