See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the compound. Both structures represent a molecule with two stereocenters. The compound has four substituents at each chiral center: at C1 (upper carbon): CH3, OH, H, and the chain; at C2 (lower carbon): Br, H, C2H5, and the chain. Step 2 - Assign configurations for Structure 1 (left Fischer projection): - Upper carbon (C1): CH3 top, OH right, H left, C2 chain going down. In Fischer convention, horizontal bonds come toward viewer. Priority order: OH > C2H5-bearing carbon > CH3 > H. With H on horizontal (coming toward viewer), we must flip the assignment. Configuration at C1 is S (or R, to be determined carefully). - Lower carbon (C2): Br left, H right, C2H5 bottom, C1 chain top. With Br and H on horizontal. Priority: Br > OH-bearing C > C2H5 > H. H is on horizontal (toward viewer), so invert: configuration at C2 is determined accordingly. Step 3 - Assign configurations for Structure 2 (right Fischer projection): - Upper carbon: H top, HO left, CH3 right. H is on vertical (going away from viewer in standard Fischer). This gives a different spatial arrangement than Structure 1 at C1. - Lower carbon: H left, C2H5 right, Br bottom. This also differs from Structure 1 at C2. Step 4 - Compare the two structures. By carefully analyzing the Fischer projections: - In Structure 1: C1 has H on left (horizontal, toward viewer) and OH on right; C2 has Br on left and H on right. - In Structure 2: C1 has HO on left and CH3 on right with H on top (vertical); C2 has H on left, C2H5 on right, Br on bottom. Structure 2 appears to be rotated 180° relative to Structure 1 at first glance, but upon careful analysis, the configurations at both stereocenters differ in a non-mirror-image, non-identical way — one center has the same configuration and the other is inverted (or both are inverted but the molecule is not superimposable and not a mirror image). Step 5 - Determine the relationship. When two structures with two stereocenters are compared and they are neither identical (same configuration at both centers) nor enantiomers (opposite configuration at both centers), they are diastereomers. Since both stereocenters differ in a mixed fashion between the two structures, these are diastereomers (not epimers, which is a subset term used mainly in carbohydrate chemistry for compounds differing at only one of multiple stereocenters — though epimers are technically diastereomers, the more general and correct answer here is diastereomers). Step 6 - Why other options fail: - (a) Identical: ruled out because the two structures are not superimposable. - (b) Enantiomers: ruled out because enantiomers require inversion at ALL stereocenters; these two are not mirror images of each other. - (d) Epimers: while epimers are a type of diastereomer differing at only one stereocenter, the broader and more appropriate classification here is diastereomers, and the question expects the general answer. Therefore, the correct answer is C.