See image — Biomolecules Chemistry Question

💡 Solution & Explanation

Concept: The Lobry de Bruyn–Alberda van Ekenstein (LdB–AvE) rearrangement occurs when an aldose or ketose is treated with dilute base (HO⁻). Under basic conditions, the carbonyl sugar undergoes enolization at C1–C2 to form a common enediol intermediate. This enediol can then collapse to give interconverted sugars at C1 and C2. Step 1: D-Mannose treated with HO⁻ gives the C1–C2 enediol intermediate, which can re-close to give D-Glucose (epimer at C2) or D-Fructose (2-ketose). This is the well-known interconversion: D-Mannose ⇌ enediol ⇌ D-Glucose ⇌ enediol ⇌ D-Fructose. Step 2: D-Glucose treated with HO⁻ again forms the same C1–C2 enediol. This enediol intermediate can collapse to regenerate D-Glucose, form D-Mannose (C2 epimer), or form D-Fructose (ketose at C2). Step 3: Product (A) is therefore D-Fructose, the 2-ketose that arises from the enediol intermediate shared by D-Glucose and D-Mannose under basic conditions. This is a classic result of the LdB–AvE rearrangement. Why other options fail: - (a) D-Glucose: This is the starting material in the second equilibrium, not a new product (A). - (c) D-Talose: D-Talose is the C2 epimer of D-Galactose, not related by a simple C1–C2 enediol from D-Glucose. - (d) D-Idose: D-Idose would require epimerization at a different carbon, not produced by the C1–C2 enediol rearrangement. Therefore, the correct answer is B.