See image — Biomolecules Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: L-glucose is the mirror image (enantiomer) of D-glucose. In Fischer projection, D-glucose has the following configuration (from C2 to C5): C2: OH on right; C3: OH on left; C4: OH on right; C5: OH on right. The designation L vs D is determined by the configuration at the highest-numbered chiral center (C5 for glucose): D-glucose has OH on the right at C5, so L-glucose has OH on the left at C5. Step 2 - L-glucose structure: L-glucose is the complete mirror image of D-glucose, so every OH that is on the right in D-glucose moves to the left, and every OH on the left moves to the right. Thus L-glucose Fischer projection is: C2: OH on left (HO-H); C3: OH on right (H-OH); C4: OH on left (HO-H); C5: OH on left (HO-H). Step 3 - Matching to options: Option (a) shows: C2: HO on left (OH left), H on right; C3: H on left, OH on right; C4: HO on left, H on right; C5: HO on left, H on right. This gives OH at C2: left, C3: right, C4: left, C5: left — which exactly matches the mirror image of D-glucose (D-glucose: C2 right, C3 left, C4 right, C5 right). This is L-glucose. Step 4 - Why other options fail: Option (b) has C2: OH right, C3: OH left, C4: OH right, C5: OH right — this is D-glucose itself. Option (c) has C2: OH right, C3: OH left, C4: OH right, C5: OH left — this is a different diastereomer (L-idose or similar), not L-glucose. Option (d) is eliminated since option (a) is correct. Therefore, the correct answer is A.