See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material and reaction: The starting material is 2,2'-diformylbiphenyl, which has an aldehyde (CHO) group at each ortho position of the two phenyl rings connected by a C-C single bond. It is treated with concentrated KOH under heat (Cannizzaro conditions, since these are aromatic aldehydes with no alpha-H). Step 2 - Reaction with Conc. KOH / Delta (Cannizzaro reaction): Because 2,2'-diformylbiphenyl has no alpha-hydrogen atoms, it undergoes an intramolecular Cannizzaro reaction under concentrated KOH and heat. In the Cannizzaro reaction, one aldehyde is oxidized to a carboxylate (COO-K+) and the other is reduced to an alcohol (CH2OH). Since both CHO groups are on the same molecule (intramolecular), one CHO becomes COO-K+ (carboxylate salt) and the other becomes CH2OH. Product (A) is therefore the potassium salt of 2-(hydroxymethyl)biphenyl-2'-carboxylic acid, i.e., the biphenyl compound bearing -COOK on one ortho position and -CH2OH on the other ortho position. Step 3 - Reaction of (A) with H(+) / Delta (acid-catalyzed lactonization): Upon treatment with acid (H+) and heat, the potassium carboxylate is converted to the free carboxylic acid (-COOH), and then under heating the carboxylic acid and the adjacent (intramolecular) hydroxymethyl group (-CH2OH) undergo esterification to form a cyclic ester (lactone). The -COOH and -CH2OH on the two ortho positions of biphenyl are in close proximity, so they readily cyclize to form a six-membered lactone ring (the ring contains: Ar-C(=O)-O-CH2-Ar, a six-membered ring bridging across the biphenyl 2,2' positions). This is a dibenzofuranone-type or more precisely a 6H-dibenzo[b,d]pyran-6-one precursor - actually a six-membered lactone: the bridging ring is -C(=O)-O-CH2- connecting the two ortho carbons of biphenyl, giving a structure that matches option (a). Step 4 - Why other options fail: Option (b) is the free acid/alcohol (the intermediate before lactonization, but without the K, and before cyclization) - this is essentially the acid form of (A) before cyclization, not the final cyclic product. Option (c) shows a lactone fused in a different ring system (appears to be a naphthalene-based or direct aryl lactone without the CH2 spacer), which would require direct esterification between COOH and an ArOH, but there is no phenolic OH in this reaction. Option (d) is essentially compound (A) itself (the potassium salt form), not the cyclic product (B). Step 5 - Conclusion: The cyclic product (B) is the six-membered lactone formed by intramolecular esterification between the carboxylic acid and hydroxymethyl groups at the 2,2' positions of biphenyl, which corresponds to structure (a). Therefore, the correct answer is A.