Which among the following molecules (I, II, III) can exhibit tautomerism?

Solution & Explanation

\textbf{Answer:} (A) III only. \textbf{Solution:} Keto-enol tautomerism requires an $\alpha$-hydrogen that can shift to give an enol $\mathrm{C}=\mathrm{C}-\mathrm{OH}$. In I, enolisation would create a double bond at a bridgehead carbon, an anti-Bredt alkene forbidden by Bredt's rule, so I cannot tautomerise. In II the $\alpha$-carbon bears two phenyl groups and has no $\alpha$-hydrogen, so no enol can form. Only III (norbornanone-type) has a removable $\alpha$-hydrogen on a non-bridgehead carbon, so only III tautomerises.