See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Step 1 – Pair I (Newman projection vs. wedge-dash of the same compound): The Newman projection shows: front carbon – F (top), Br (upper-left), H (upper-right); back carbon – H (lower-left), CH3 (lower-right), H (bottom). The wedge-dash shows a single stereocenter with H (wedge), CH2Br... wait – re-examining: the right structure of pair I has H3C, H, CH2Br, and F on one carbon, making it a single chiral center. The Newman also encodes one chiral center (the front carbon bears F, Br, H; but actually only one carbon is chiral in a Newman if the back carbon has two H's). Assign R/S to both: In the Newman, front carbon has F, Br, H – priority F>Br>H, with the carbon chain going back. In the wedge-dash, the same four groups are present. Careful comparison of the spatial arrangement reveals they are non-superimposable mirror images → enantiomers. Step 2 – Pair II (two stereocenters each): Each molecule has two stereocenters bearing Cl, H, CH3, CH2CH3 (i.e., 2,3-dichloropentane framework or similar). Assign R/S at each stereocenter for both structures. For the left structure, assign configurations at C1 and C2; for the right structure, assign configurations at C1 and C2. The analysis shows that the configurations at the two centers are (R,S) in one and (S,R)... or more carefully, one center is inverted and the other is not between the two structures. When one but not both stereocenters are inverted, the compounds are diastereomers (not mirror images of each other overall). Therefore Pair II = diastereomers. Step 3 – Pair III (two cross/sawhorse structures of 2,3-dibromobutane): Left structure: top carbon – CH3 (up), H (left), Br (right); bottom carbon – CH3 (left), H (right), Br (down). This corresponds to one configuration at each carbon. Right structure: top carbon – CH3 (up), Br (left), H (right); bottom carbon – H (left), CH3 (right), Br (down). By rotating/flipping one structure, we can check if they superimpose. Rotating the right structure 180° about the C–C bond axis maps it onto the left structure exactly. Hence they are identical (same compound). Summary: I = enantiomers, II = diastereomers, III = identical → matches option (c). Why other options fail: (a) says III = enantiomers, but III structures are identical after rotation. (b) says I = identical, but they are non-superimposable mirror images. (d) says II = identical, but the two stereocenters are not both inverted, giving diastereomers. Therefore, the correct answer is C.