See image — Biomolecules Chemistry Question

💡 Solution & Explanation

Step 1 - Concept review: An acetal carbon (in a sugar context, a glycosidic carbon) is bonded to two -OR groups (two oxygens from different sources, no free -OH). A hemiacetal carbon is bonded to one -OR and one free -OH (the anomeric -OH that is not involved in a glycosidic bond). Step 2 - Identify the glycosidic linkage (acetal center): Isomaltose is a disaccharide with an alpha-1,6-glycosidic bond. The C1 of the left (non-reducing) glucose ring is connected via an oxygen to the C6 of the right glucose ring AND is also part of the ring oxygen. C1 of the left ring is therefore bonded to two oxygens (ring oxygen and glycosidic oxygen) with no free -OH. This makes C1 of the left ring a full acetal. Count of acetals = 1. Step 3 - Identify the hemiacetal center: The right (reducing-end) glucose ring retains a free anomeric -OH at C1 (the reducing end is not involved in the glycosidic bond). C1 of the right ring is bonded to one ring oxygen and one free -OH, making it a hemiacetal. Count of hemiacetals = 1. Step 4 - The structure shown confirms this: the left ring C1 has the O with lone pairs (indicating the glycosidic oxygen, no free OH — acetal), while the right ring C1 has an -OH group drawn (hemiacetal). The question image also shows an H-O on the left ring top, but that is the ring oxygen bridge, not a free anomeric OH. Step 5 - Evaluating options: (a) 0 acetals, 0 hemiacetals — incorrect, ignores both centers. (b) 1 acetal, 0 hemiacetals — incorrect, ignores the reducing-end hemiacetal. (c) 0 acetals, 1 hemiacetal — incorrect, ignores the glycosidic acetal. (d) 1 acetal, 1 hemiacetal — correct. Therefore, the correct answer is D.