See image — Biomolecules Chemistry Question

💡 Solution & Explanation

Concept: When an aldopentose is oxidized (e.g., with HNO3), both the aldehyde (CHO) and the primary alcohol (CH2OH) are converted to carboxylic acid (COOH) groups, yielding an aldaric acid (saccharic acid). The resulting dibasic acid is optically active only if it lacks an internal plane of symmetry (i.e., it is not a meso compound). Additionally, since we want an L-sugar, the configuration at the highest-numbered chiral center (C4 for a pentose) must have the OH on the left in the Fischer projection. Step 1 - Identify L-sugars: In pentoses, the L-configuration is defined by the OH group at C4 being on the LEFT in the Fischer projection. - (a): C4 has HO on left → L-sugar ✓ - (b): C4 has H on left, OH on right → D-sugar ✗ - (c): C4 has H on left, OH on right → D-sugar ✗ - (d): C4 has HO on left → L-sugar ✓ Step 2 - Among L-sugars (a) and (d), determine which gives an optically active aldaric acid upon oxidation: Option (a): Configuration is C2(H/OH), C3(HO/H), C4(HO/H) — reading OH positions: C2-OH right, C3-OH left, C4-OH left. This corresponds to L-Arabinose (or a related sugar). The aldaric acid obtained from L-Arabinose is arabinaic acid. Check for symmetry: the acid has COOH at both ends. The internal carbons are C2(OH-right), C3(OH-left), C4(OH-left). Since C2 and C4 are not mirror images of each other relative to C3, the molecule has no internal plane of symmetry → optically active ✓ Option (d): Configuration is C2(HO/H), C3(H/OH), C4(HO/H) — OH positions: C2-OH left, C3-OH right, C4-OH left. The aldaric acid from this sugar: COOH at top and bottom, C2(OH-left), C3(OH-right), C4(OH-left). This molecule has a plane of symmetry through C3 because C2 and C4 are mirror images → meso compound → optically inactive ✗ Step 3 - Conclusion: Only option (a), an L-sugar with the configuration giving a non-meso aldaric acid upon oxidation, satisfies both conditions (L-sugar AND optically active dibasic acid). Therefore, the correct answer is A.