See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

The compound is a Fischer projection of a 4-carbon aldehyde acid (2,3-dihydroxybutanedioic acid analogue but with CHO at bottom): COOH at top, CHO at bottom, C2 has H on left and OH on right, C3 has OH on left and H on right. Step 1: Assign priorities at C2. C2 is bonded to: COOH (C1, top), C3 (bottom chain containing OH, H, CHO), OH (right), H (left). Priority order: OH (O directly attached) > COOH (O,O attached carbon) > C3 chain (O attached via next carbon) > H. Priorities: 1=OH, 2=COOH, 3=C3 chain, 4=H. Step 2: In a Fischer projection, horizontal bonds point toward the viewer and vertical bonds point away. At C2, H is on the left (toward viewer). Since the lowest priority (H) is on a horizontal bond (toward viewer), we must determine the arrangement of 1,2,3 and then invert. Arrangement of groups 1(OH-right), 2(COOH-top/away), 3(C3-bottom/away) as seen with H toward viewer: - OH is to the right, COOH is up (going away), C3 chain is down (going away). - Reading 1→2→3: from right(OH) to top(COOH) to bottom(C3) = this traces counterclockwise (right → up → down = CCW) = S arrangement as seen from front. - Since H (priority 4) is toward us (horizontal), we invert: S → R. - C2 configuration = R. Step 3: Assign priorities at C3. C3 is bonded to: C2 chain (top, containing OH, H, COOH), CHO (bottom, C with =O, H), OH (left), H (right). Priority order: 1=OH (O directly attached, left), 2=CHO (C doubly bonded to O, counts as C bonded to O,O,H phantom), 3=C2 chain (C bonded to O,C,H), 4=H (right). Priorities: 1=OH, 2=CHO, 3=C2 chain, 4=H. Step 4: At C3, H is on the right (horizontal, toward viewer). Arrangement of 1(OH-left), 2(CHO-bottom/away), 3(C2-top/away): - OH is left, CHO is down (away), C2 is up (away). - Reading 1→2→3: left(OH) → down(CHO) → up(C2) = this is clockwise = R as seen from front. - Since H is toward us (horizontal), invert: R → S. - C3 configuration = S? Re-examining C3: Let me recheck the direction 1→2→3 with OH on left, CHO on bottom, C2 on top. Going OH(left)→CHO(bottom)→C2(top): left to bottom to top traces: left→down is quarter turn clockwise, then bottom→top goes up through right side = overall the rotation left→bottom→top is counterclockwise (CCW) = S before inversion. Inverting because H is on horizontal (toward viewer): S → R. C3 = R. So C2=R, C3=R, matching answer (a) R, R. Verification: The other options S,S / R,S / S,R do not match the correct CIP assignments derived above. Therefore, the correct answer is A.