See image — Biomolecules Chemistry Question

💡 Solution & Explanation

Concept: Glucose is an aldose (aldehyde sugar) while fructose is a ketose (ketone sugar). A reagent that distinguishes between them must react with one but not the other. Step 1 - Bromine water (reagent I): Bromine water is a mild oxidizing agent that specifically oxidizes aldehydes but does NOT oxidize ketones under mild conditions. Glucose (aldehyde group) decolorizes bromine water, whereas fructose (ketone group) does not decolorize it. Therefore, bromine water CAN distinguish glucose from fructose. Step 2 - Tollen's reagent (reagent II): Tollen's reagent oxidizes both aldehydes and alpha-hydroxy ketones. Fructose, despite being a ketose, gives a positive Tollen's test because it undergoes tautomerization (through enediol intermediate) in the basic conditions of Tollen's reagent, converting to an aldehyde form. Both glucose and fructose give a positive silver mirror test. Therefore, Tollen's reagent CANNOT distinguish between glucose and fructose. Step 3 - Schiff's reagent (reagent III): Schiff's reagent tests for free aldehydes. However, under the acidic conditions of Schiff's test, fructose can also tautomerize or enolize and give a positive result. In practice, both glucose and fructose give a positive Schiff's test, so it cannot reliably distinguish them. Therefore, Schiff's reagent CANNOT distinguish between glucose and fructose. Conclusion: Only bromine water (reagent I) can selectively distinguish glucose (aldehyde) from fructose (ketone) because it oxidizes aldehydes but not ketones under mild conditions, and the basic/alkaline conditions required by Tollen's and Schiff's reagents cause fructose to tautomerize and also give positive results. Therefore, the correct answer is C.