Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution?
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The Correct Option is D
Solution and Explanation
A sugar behaves as reducing when the anomeric carbon has a free OH, so that the ring can open to the free carbonyl form (and, in base, isomerise through an enediol).
The glycoside (2) is a ketal, and ethers such as (3) are stable to base, so they stay locked and are non-reducing.
Compound (4) has an ester (\(\mathrm{OCOCH_3}\)) at the anomeric carbon. In aqueous KOH the ester is hydrolysed: \(\mathrm{R-OCOCH_3 + OH^- \rightarrow R-OH + CH_3COO^-}\), liberating the free anomeric hemiketal OH. The ring can then open to give free fructose, which is a reducing sugar (it gives the Tollens'/Fehling's test through the enediol).
Compounds (1) and (3) have no hydrolysable group at the anomeric carbon.
Answer: (4) the anomeric acetate
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