See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify meso-2,3-butanediol structure: meso-2,3-butanediol has the configuration (2R,3S) or equivalently (2S,3R) - the two stereocenters have opposite configurations, making the molecule achiral due to an internal mirror plane. Each carbon bears: Me, OH, H, and the other chiral carbon. Step 2 - Determine configurations in each Newman projection by assigning priorities to substituents on each carbon: For C2 (front) and C3 (back): Priority order is OH > CH(OH)CH3 > CH3 > H. Step 3 - Analyze projection P: Front carbon (C2): Me (top), H (left), OH (right). Rotating to assign: OH, then the carbon chain, then Me, with H pointing back - this gives one configuration. Back carbon (C3): HO (left), H (right), Me (bottom). In P, the front carbon has OH and Me and H arranged so that one center is R and the other is S - consistent with meso. P represents an antiperiplanar (anti) conformation of meso-2,3-butanediol where OH groups are anti to each other and Me groups are anti to each other. The front C: Me up, H left, OH right. Back C: HO left, H right, Me down - OH groups are anti (180°), Me groups are anti. This is a valid anti conformation of the meso compound. Step 4 - Analyze projection R: Front carbon: Me (top), HO (left), Me (right) - wait, this shows two Me groups on front carbon which cannot be correct for 2,3-butanediol. Re-reading: Front carbon has Me (top), HO (upper-left), Me (upper-right) - but each carbon only has one Me. Re-examining: R shows front carbon with Me top, HO left, and back carbon Me upper-right (this is a back substituent). Correctly reading R: front carbon has Me (top), HO (left), and the circle separates front from back. Back carbon has Me (upper right extending back), HO (lower left), H (lower right), H (bottom). So front: Me, HO, H (implicit) and back: Me, HO, H. This represents a gauche conformation where both OH groups are on the same side - a gauche conformation of meso compound where OHs are gauche. Assigning configurations: front C2 has OH, Me, H - R or S; back C3 has OH, Me, H with opposite configuration = meso confirmed. Step 5 - Analyze Q: Front carbon Me top, H left, OH right; Back carbon H left, OH right, Me bottom. Here the OHs are eclipsed or gauche, and configurations: if front is R then back should be S for meso. Checking Q - OHs are syn (both on right side, 0° or 60° apart). This could represent the (2R,3R) or (2S,2S) racemate conformation, not meso. Step 6 - Analyze S: Front carbon Me top, Me left, H right; Back carbon H left, OH right, OH bottom. Having Me on both positions of front seems off - re-reading: front has Me (top) and Me (upper-left) which would mean two methyls on C2, impossible. This likely represents a different arrangement. S appears to represent the racemic diastereomer conformation. Step 7 - Conclusion: P represents the anti conformation of meso-2,3-butanediol (OHs anti, Mes anti, which in meso means the two chiral centers are R and S). R represents a gauche conformation of the same meso compound (different rotation around C2-C3 bond). Both P and R are Newman projections of meso-2,3-butanediol in different conformations. Q and S represent conformations of the racemic (dl) 2,3-butanediol. Why other options fail: - (a) P, Q: Q is a conformation of dl-2,3-butanediol, not meso - (c) R, S: S is a conformation of dl-2,3-butanediol, not meso - (d) Q, S: Both are dl-2,3-butanediol conformations Therefore, the correct answer is B.