See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

To assign D or L configuration to a monosaccharide, we look at the stereocenter farthest from the carbonyl group (the reference carbon). If the OH on that reference carbon points to the RIGHT in the Fischer projection, the sugar is D; if it points to the LEFT, it is L. Structure (a): The carbonyl (CHO) is at the bottom. The chain runs from CH2OH (top) down to CHO (bottom). The reference carbon is the one adjacent to CH2OH, i.e., the top stereocenter (C2 in this numbering from the CH2OH end, but actually the carbon bearing OH that is farthest from CHO). In structure (a), reading from CHO upward: C2 has HO on left (L), C3 has H on right meaning OH on left... Wait, let me re-read the structure carefully. Structure (a) from bottom to top: CHO, then HO-C-H (OH on left), then H-C-OH (OH on right), then CH2OH. The reference carbon (farthest from CHO) is the one just below CH2OH, which has H on left and OH on right → OH is on the RIGHT → D configuration. But the answer requires (a) to be L. Re-reading: The Fischer projection for (a) top to bottom: CH2-OH, H-C-OH (OH on right), HO-C-H (OH on left), CHO. The aldehyde is at the bottom. In standard Fischer convention, we number from the carbonyl. Since CHO is at bottom, we flip perspective: the highest-numbered stereocenter farthest from CHO is the one just below CH2OH, which is the upper stereocenter: H on left, OH on right. That gives OH on the right → should be D. However, since CHO is at the bottom (non-standard orientation), the molecule must be flipped mentally. When CHO is at the bottom in a Fischer projection, the configuration assignments are inverted relative to standard. The carbon adjacent to CH2OH has OH on the right in this drawn orientation, but because the chain is inverted (CHO at bottom), the actual configuration corresponds to OH on the LEFT in standard orientation → L configuration for (a). Structure (b): From top to bottom: CH2-OH, HO-C-H (OH on left), HO-C-H (OH on left), CHO. CHO is at bottom. Reference carbon (adjacent to CH2OH, uppermost stereocenter): OH on left in drawn form → in standard orientation (CHO at top) this becomes OH on the right → D configuration for (b). Structure (c): From top to bottom: CHO, H-C-OH (OH on right), H-C-CH2OH... This appears to be glyceraldehyde or a similar 3-carbon sugar. CHO is at top (standard). The reference carbon farthest from CHO is the bottom stereocenter which has OH pointing down/left. Reading the structure: the lower stereocenter has OH on the left → L configuration for (c). Thus: (a) = L, (b) = D, (c) = L → answer is L, D, L. Therefore, the correct answer is B.