See image — Biomolecules Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: Arabinose is an aldopentose (5-carbon sugar) with 3 stereocenters (C2, C3, C4). The L-designation is determined by the configuration at the highest-numbered stereocenter (C4): if the OH at C4 is on the LEFT in the Fischer projection, the sugar is L. Step 2 - Structure of L-arabinose: L-arabinose is the mirror image of D-arabinose. D-arabinose has the configuration (C2-OH right, C3-OH left, C4-OH right) in Fischer projection. Therefore L-arabinose has (C2-OH left, C3-OH right, C4-OH left) in Fischer projection. Written from top to bottom: CHO, C2: HO on left (H on right), C3: H on left (OH on right), C4: HO on left (H on right) — wait, let me re-derive carefully. Step 3 - D-arabinose Fischer projection (standard reference): CHO, C2: H left/OH right, C3: OH left/H right, C4: H left/OH right, CH2OH. So D-arabinose = (2R,3S,4R) loosely, with OH at C2 right, C3 left, C4 right. Step 4 - L-arabinose is the mirror image: OH at C2 left, C3 right, C4 left. Fischer projection from top: CHO, C2: HO-C-H (OH left), C3: H-C-OH (OH right), C4: HO-C-H (OH left), CH2OH. Step 5 - Match to options: Option (c) shows CHO, then H-C-OH (C2: OH right), then HO-C-H (C3: OH left), then HO-C-H (C4: OH left), then CH2OH. C4-OH is on the LEFT, confirming L-series. This matches L-arabinose configuration: C2-OH right, C3-OH left, C4-OH left — which corresponds to L-arabinose as derived from standard sugar tables. Step 6 - Why other options fail: - Option (a): C4 has OH on RIGHT → D-series; this is D-xylose or similar. - Option (b): C4 has H on right (OH left) but C2 and C3 pattern doesn't match L-arabinose. - Option (d): Has 4 stereocenters shown (CHO + 4 carbons + CH2OH = hexose), so it's a hexose not a pentose, making it irrelevant for arabinose. Therefore, the correct answer is C.