See image — Biomolecules Chemistry Question

💡 Solution & Explanation

Concept: D-glucose is an aldose (aldehyde sugar) while D-fructose is a ketose (ketone sugar). We need a reagent that can distinguish between an aldehyde sugar and a ketone sugar. Step 1: Evaluate options (a), (b), and (c). - Fehling solution (option a): Both D-glucose and D-fructose give a positive Fehling test. Fructose, despite being a ketose, is oxidized by Fehling solution under alkaline conditions because it undergoes tautomerization to an aldose (enediol rearrangement) in alkaline medium. Therefore, Fehling solution cannot differentiate them. - Tollens reagent (option b): Similarly, Tollens reagent is used under alkaline conditions. D-fructose also gives a positive Tollens test due to the same enediol tautomerization in alkaline medium. Therefore, Tollens reagent cannot differentiate them. - Benedict test (option c): Benedict reagent also works under mildly alkaline conditions. Both glucose and fructose give a positive Benedict test for the same reason as above. Therefore, Benedict test cannot differentiate them. Step 2: Evaluate option (d) — Br2/H2O. - Bromine water (Br2/H2O) is a mild oxidizing agent that works under acidic/neutral conditions. Under these conditions, tautomerization of fructose does NOT occur. - D-glucose (aldose) is readily oxidized by Br2/H2O because the aldehyde group is directly oxidized to a carboxylic acid (gluconic acid), decolorizing the bromine water. - D-fructose (ketose) does NOT react with Br2/H2O under neutral/acidic conditions because ketones are not oxidized by bromine water in the absence of base. The bromine water remains colored (orange-yellow). - Therefore, Br2/H2O can selectively differentiate D-glucose (decolorizes Br2 water) from D-fructose (does not decolorize Br2 water). Why other options fail: All of Fehling solution, Tollens reagent, and Benedict test operate under alkaline conditions, where fructose undergoes enediol tautomerization and behaves as a reducing sugar similar to glucose, making differentiation impossible. Therefore, the correct answer is D.