See image — Biomolecules Chemistry Question

💡 Solution & Explanation

Step 1: Count the total carbon atoms. The furanose ring contains 4 carbons plus the exocyclic CH2OH group gives 5 carbons total. This makes it a pentose (5-carbon sugar), eliminating hexose (4) and hexulose (3). Step 2: Identify the carbonyl type (aldose vs ketose). In the cyclic (Haworth) structure shown, the ring oxygen bridges between C1 and C4 of the open-chain form. The structure is a fructose-type sugar where the anomeric carbon (C2 in the open chain) bears the ring oxygen and had the ketone group before ring closure, making it a ketose (6) — specifically a 2-ketose. This rules out aldose (5) and confirms ketose (6). Step 3: Identify ring size. The ring is five-membered (four carbons + one oxygen), which is a furanose (8), not a pyranose (7-membered ring with 5 carbons + 1 oxygen). Step 4: Combine the terms. The sugar is a 5-carbon ketose in furanose form: Pentulose (2) + Ketose (6) + Furanose (8) = option (a) 2, 6, 8. Why other options fail: - (b) 2, 6, 7: pyranose is wrong because the ring is 5-membered, not 6-membered. - (c) 1, 5, 8: pentose and aldose are wrong because the carbonyl was at C2 (ketone), not C1 (aldehyde). - (d) is wrong because option (a) correctly identifies it. Therefore, the correct answer is A.