See image — Biomolecules Chemistry Question

💡 Solution & Explanation

Concept: Lobry de Bruyn–Van Ekenstein (LdBVE) rearrangement. When D-glucose is treated with dilute base (NaOH or similar), it undergoes base-catalyzed isomerization via an enediol intermediate. Step 1: Identify the reaction. The double arrow with 'HO' (hydroxide/base) above it indicates the Lobry de Bruyn–Van Ekenstein rearrangement, which is a base-catalyzed epimerization/isomerization of aldoses and ketoses. Step 2: Mechanism. Under basic conditions, D-glucose forms a 1,2-enediol intermediate. This enediol can re-close in two ways: - Reprotonation at C-1 from the re face gives D-mannose (C-2 epimer of D-glucose) - Reprotonation at C-2 to give a ketose yields D-fructose (a 2-ketose with the same carbon chain) Step 3: Products. The LdBVE rearrangement of D-glucose yields a mixture of D-glucose (starting material), D-mannose (C-2 epimer), and D-fructose (2-ketose isomer). Thus A = D-mannose and B = D-fructose. Step 4: Why other options fail: - (a) D-mannitol is a sugar alcohol (reduction product), not formed by base treatment alone. - (c) D-Allose and D-Altrose are C-3 and C-4 epimers; the enediol intermediate from C-1/C-2 does not produce these. - (d) D-Glucose and D-Idose would require isomerization at C-3 or higher, which is not the primary LdBVE product. Therefore, the correct answer is B.