Which among the given bicyclic ketone molecules (I), (II) and (III) can exhibit tautomerism?

Solution & Explanation

\textbf{Answer:} (a) III only \textbf{Solution:} Keto-enol tautomerism requires an alpha-hydrogen AND that the resulting enol $\mathrm{C}=\mathrm{C}$ does not place a double bond at a bridgehead (Bredt's rule). In (I) the enol would put a $\mathrm{C}=\mathrm{C}$ at the bridgehead (an anti-Bredt, unstable bridgehead alkene), so no tautomerism. In (II) the alpha-carbon next to the $\mathrm{C}=\mathrm{O}$ bears two phenyl groups and has no alpha-H to enolise. Only (III) has a normal alpha-$\mathrm{CH}$ that can enolise without violating Bredt's rule. Hence only III shows tautomerism. Hence (a).