See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: The starting material is o-phenylenediacetaldehyde (1,2-bis(2-oxoethyl)benzene), which has two -CH2CHO groups on adjacent (ortho) positions of a benzene ring. Under basic conditions (NaOH, ethanol/water, heat), an intramolecular aldol condensation can occur. Step 1 – Identify the reactive groups: Both -CH2CHO groups are aldehydes with alpha-CH2 protons. NaOH deprotonates one alpha-carbon to form an enolate. Step 2 – Intramolecular aldol condensation: The enolate of one -CH2CHO attacks the carbonyl carbon of the other -CH2CHO intramolecularly. This forms a five-membered ring (closing through the two CH2 groups and the two carbons of the ortho positions on benzene), giving a beta-hydroxy aldehyde intermediate. Step 3 – Dehydration: Under heating with base, the beta-hydroxy aldehyde undergoes elimination of water (H2O is shown as a byproduct in the equation) to give an alpha,beta-unsaturated aldehyde (conjugated enal). Step 4 – Product identification: The intramolecular aldol condensation followed by dehydration of the ortho-disubstituted compound forms a five-membered ring fused to benzene (indene skeleton) with an exocyclic CH=O group. The resulting product is 1H-indene-1-carbaldehyde (indene ring with CHO substituent), which corresponds to option (b). Why other options fail: - (a) A four-membered ring product would require a very strained cyclobutane formation and is not favored; also no dehydration would be needed. - (c) A naphthalenediol would require an entirely different reaction pathway (not an aldol) and would not produce water as byproduct in this manner. - (d) The diketone of a tetralin framework would require oxidation, not an aldol condensation/dehydration. The formation of the five-membered ring via intramolecular aldol condensation with loss of water is well-established for 1,2-bis(oxoethyl)benzene systems, giving the indene-carbaldehyde product. Therefore, the correct answer is B.