See image — Biomolecules Chemistry Question

💡 Solution & Explanation

Concept: Mutarotation is the process by which the anomeric carbon of a cyclic sugar interconverts between alpha and beta forms through ring opening and reclosure in aqueous solution. For mutarotation to occur, the anomeric carbon must bear a free hydroxyl group (hemiacetal or hemiketal). If the anomeric -OH is replaced by an -OR group (i.e., it is a glycoside / acetal), the ring cannot open under neutral aqueous conditions, and mutarotation does not occur. Step 1 - Identify the anomeric carbon in each structure and check whether it carries a free -OH or an -OCH3 (methoxy) group. Structure (I): The anomeric carbon has a free -OH group (shown on wedge at the ring junction). This is a free hemiacetal. It CAN undergo mutarotation. Structure (II): The anomeric carbon has -OCH3 (shown as OCH3 at the top-right of the furanose ring). This is a glycoside (full acetal). It CANNOT undergo mutarotation. Structure (III): The anomeric carbon has a free -OH group (shown on wedge). Even though the primary hydroxyl is methylated (CH2OCH3), the anomeric position is a free hemiacetal. It CAN undergo mutarotation. Structure (IV): The anomeric carbon has -OCH3 (shown as OCH3 at the top-right of the pyranose ring). This is a glycoside (full acetal). It CANNOT undergo mutarotation. Step 2 - Conclusion: Structures II and IV have their anomeric -OH blocked as methyl glycosides, so they will NOT undergo mutarotation. Structures I and III retain a free anomeric -OH and will undergo mutarotation. Why other options fail: - (a) II only: misses IV, which is also a glycoside. - (b) I, III and IV only: incorrectly includes I and III (which have free anomeric -OH) and misidentifies IV. - (d) I and III only: incorrectly identifies I and III as non-mutarotating; they actually have free anomeric -OH and do mutarotate. Therefore, the correct answer is C.