Chemistry - GOC Question with Solution | TestHub

Removing ${\mathrm{H}}_1$ gives the imine anion :

Removing ${\mathrm{H}}_2$ gives the ring carbanion :

The imine anion has its lone pair in an sp2 orbital lying in the plane of the ring, orthogonal to the π system, so the charge stays localised. The ring carbanion sits next to the $\mathrm{C}=\mathrm{N}$ π bond and delocalises onto the exocyclic nitrogen, like an aza-enolate. Delocalisation beats the atom comparison here, so ${\mathrm{H}}_2$ is more acidic.