Chemistry - GOC Question with Solution | TestHub

Removing ${\mathrm{H}}_1$ gives the secondary alkoxide :

Removing ${\mathrm{H}}_2$ gives the tertiary alkoxide :

Both charges are on oxygen with no resonance, so compare induction : the carbon bearing the second alkoxide carries two electron-donating methyl groups, which destabilise the negative charge. The less substituted alkoxide is more stable, so the secondary O-H proton ${\mathrm{H}}_1$ is more acidic.