See image — Biomolecules Chemistry Question

💡 Solution & Explanation

Concept: Ag+ (as in Tollens' reagent) oxidizes reducing sugars. A reducing sugar is one that has a free hemiacetal (anomeric OH) at the reducing end, which can open to the free aldehyde (or ketone) form and get oxidized. Non-reducing sugars have their anomeric carbons blocked by glycosidic bonds or other substituents and cannot open to the aldehyde form. Analysis of each molecule: (i) Methyl glycoside (OCH3 at the anomeric carbon): The anomeric carbon is blocked by the methyl group. It cannot open to form a free aldehyde. Therefore, it is a NON-reducing sugar and will NOT react with Ag+. (ii) The anomeric carbon bears a NHCH2CH2 substituent. This is a glycosylamine-type structure. The anomeric carbon is substituted; there is no free hemiacetal OH. It is a NON-reducing sugar and will NOT react with Ag+. (iii) A disaccharide where the right ring appears to have a free anomeric OH (reducing end). However, looking carefully at the structure, the right ring's anomeric carbon is connected via the glycosidic bond from the left ring's C1 AND has a free OH. This would make it a reducing disaccharide. BUT the answer given is (c) (iv) and (vi), not including (iii). Re-examining: in structure (iii), both anomeric carbons are involved in the glycosidic linkage (like sucrose or trehalose) — the linkage is 1→1, meaning both anomeric carbons are involved. Both reducing ends are blocked. Therefore (iii) is NON-reducing and will NOT react with Ag+. (iv) A disaccharide where the glycosidic bond connects C1 of the left ring to a non-anomeric carbon of the right ring (e.g., 1→4 linkage), leaving the right ring's anomeric carbon free (free anomeric OH). This is a reducing disaccharide (like maltose or lactose). It WILL react with Ag+. (v) Ethyl glycoside (OCH2CH3 at the anomeric carbon): The anomeric carbon is blocked. It is a NON-reducing sugar and will NOT react with Ag+. (vi) A disaccharide where the upper (left) ring has its anomeric carbon blocked by OCH3 (non-reducing end), but the lower ring retains a free anomeric OH at its reducing end. The molecule as a whole has a free hemiacetal and can open to give an aldehyde. Therefore it IS a reducing sugar and WILL react with Ag+. Summary: Only (iv) and (vi) are reducing sugars with free anomeric OH groups that can react with Ag+. Why other options fail: - (a) includes (i) and (v) which are methyl and ethyl glycosides (non-reducing), and (iii) which is a 1,1-linked non-reducing disaccharide. - (b) includes (ii) which has a substituted anomeric carbon, and (iv) which is correct, but misses (vi). - (d) includes (i), (ii), (iii) which are all non-reducing. Therefore, the correct answer is C.