See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify Compound A: Cyclohexanol treated with PCC (pyridinium chlorochromate) oxidizes the secondary alcohol to a ketone without over-oxidation. Therefore, compound (A) is cyclohexanone. Step 2 - Aldol Condensation of Cyclohexanone with dilute NaOH and heat: When cyclohexanone is treated with dilute NaOH and heated, it undergoes an intermolecular aldol condensation. The base deprotonates the alpha carbon of one cyclohexanone molecule to form an enolate, which attacks the carbonyl carbon of a second cyclohexanone molecule. This gives a beta-hydroxy ketone (aldol product) which, upon heating, undergoes dehydration (aldol condensation) to yield an alpha,beta-unsaturated ketone. Step 3 - Structure of the product: The condensation of two cyclohexanone molecules produces 2-(cyclohexylidene)cyclohexan-1-one. In this product, one cyclohexanone ring retains its carbonyl group at C1, and at the alpha position (C2) an exocyclic double bond connects to the second cyclohexane ring (which lost its carbonyl oxygen as water). This is the cross-conjugated enone structure shown in option (b). Step 4 - Why other options fail: - Option (a) shows a cycloheptylidene group, which would require a 7-membered ring - incorrect ring size. - Option (c) shows a fused bicyclic system (naphthalene-type skeleton) which would require an intramolecular reaction or different connectivity - not the product of simple intermolecular aldol condensation of cyclohexanone. - Option (d) shows a fully saturated decalin-type ketone with no double bond, which cannot result from an aldol condensation (which always gives an alpha,beta-unsaturated carbonyl upon dehydration under heating conditions). Therefore, the correct answer is B.