Chemistry - STRUCTURAL ISOMERISM Question with Solution | TestHub

In reaction (1), 1H-pyridin-2-one goes to 2-hydroxypyridine : the keto (lactam) form is the stable one - the amide-like $\mathrm{C}=\mathrm{O}$ with N-H beats the aromatic hydroxypyridine - so the equilibrium lies backward (toward the keto form). In reaction (2), cyclohexa-2,4-dien-1-one goes to phenol : the enol is aromatic, so the equilibrium lies overwhelmingly toward phenol - forward. Hence backward, forward.