Question

Product(s) formed in the given reaction sequence is(are) 

Hint:

Ozonolysis of a ring alkene \(⇒\) 1,6-dicarbonyl; under base, intramolecular aldol can “stitch” an aromatic ring. Keto–enol tautomerization of a conjugated diketone generated in such annulations often favors the {phenol} (naphthol) form. Pd/C commonly serves to aromatize dihydroarenes; it doesn’t remove phenolic OH in the absence of specific reagents.

Answer 1

Step 1: Ozonolysis outcome.
Cleavage of the endocyclic C=C by O\(_3\) gives a 1,6-dicarbonyl (dialdehyde/aldehyde–ketone) embedded in the bicyclic framework. The tertiary alcohol is retained under these conditions (no oxidative removal of –OH).
Step 2: Base-promoted intramolecular aldol and aromatization.
In NaOH, the 1,6-dicarbonyl undergoes intramolecular aldol cyclization. Two regioisomeric closures are possible (attack of either carbonyl), each followed by dehydration and keto–enol tautomerization to the phenolic form. This furnishes two alkyl-substituted naphthols (positional isomers).
Step 3: Final Pd/C step.
Pd/C here acts as a dehydrogenation/aromatization catalyst to ensure full aromaticity of the newly formed ring; it does not remove the phenolic OH under these conditions. Hence, the products remain as naphthols.
Conclusion: The reaction delivers two positional naphthol isomers, corresponding to options (A) and (B); options without –OH (e.g., (C)) are not consistent with the sequence.