See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the left structure. The left Fischer projection has CHO at top and CH2OH at bottom. Reading substituents at each stereocenter from C2 to C5 (top to bottom): C2: H(left), OH(right) C3: H(left), OH(right) C4: HO(left), H(right) C5: H(left), OH(right) This corresponds to D-glucose (configurations: R,S,R,R at C2–C5 using standard numbering, or simply the well-known glucose pattern: OH right, OH right, OH left, OH right from C2 to C5). Step 2 - Analyze the right structure. The right Fischer projection has CH2OH at top and CHO at bottom. In a Fischer projection, the chain can be written either direction; conventionally CHO is at the top. To compare, we must flip the right structure so CHO is at top. Flipping a Fischer projection end-for-end (rotating 180° in the plane) inverts ALL stereocenters if done improperly, but rotating the entire projection 180° in the plane of the paper produces the mirror image unless we account for the even/odd inversion rule. Specifically, rotating a Fischer projection 180° in the plane gives the same compound (an even number of inversions for a chain with an even number of carbons = 4 stereocenters → 4 inversions → same compound... wait, let us use the correct method). Correct method: Rotating a Fischer projection 180° in the plane of the paper gives the same stereoisomer (equivalent representation). When the right structure is rotated 180°, CHO moves to the top and CH2OH moves to the bottom, and all left/right assignments are preserved (both ends swap, each center's left/right swap twice effectively). After rotating the right structure 180°: New top: CHO New C2 (originally C5 before rotation): HO→H becomes H on left... Let us be careful. Original right structure top-to-bottom: CH2OH C(top-1): HO(left), H(right) C(top-2): H(left), OH(right) C(top-3): HO(left), H(right) C(top-4): HO(left), H(right) CHO After 180° rotation in-plane, reading top-to-bottom (now CHO on top): CHO New-C2 (was old C5=top-4): left/right swap → H(left), OH(right)... Actually, upon 180° rotation in the plane of a Fischer projection, left and right substituents at each center are INVERTED (left becomes right and vice versa) AND the top-to-bottom order reverses. Since both effects occur simultaneously, the net result for each center is: the substituent that was on the left is now on the right and vice versa, but the center is read in the new order. Let us label the right structure centers 1–4 from top (below CH2OH) to bottom (above CHO): Center 1: HO(left), H(right) Center 2: H(left), OH(right) Center 3: HO(left), H(right) Center 4: HO(left), H(right) After 180° in-plane rotation, the new order from top (below CHO) to bottom (above CH2OH) reads the old centers in reverse (4,3,2,1) with left/right swapped: New C2 (old center 4): H(left), OH(right) New C3 (old center 3): H(left), OH(right) New C4 (old center 2): HO(left), H(right) New C5 (old center 1): H(left), OH(right) Step 3 - Compare. Left structure: C2=H/OH, C3=H/OH, C4=HO/H, C5=H/OH Right structure (after rotation): C2=H/OH, C3=H/OH, C4=HO/H, C5=H/OH All four stereocenters match exactly. The two structures are identical. Step 4 - Why other options fail. (a) Enantiomers: would require mirror-image configurations at all centers — not the case here. (b) Diastereomers: would require same molecular formula but different configurations at one or more (but not all) centers — not applicable since they are identical. (d) Structural isomers: would require different connectivity — both are aldohexoses with the same connectivity, so not applicable. Therefore, the correct answer is C.