See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Step 1 – Identify compound (X). The structure shown is a carbon bearing four different groups arranged in a specific spatial configuration: H_a (wedge), H_b (dash), Br (in-plane right), and CH3 (in-plane below), with H in-plane left. The full molecule is CH3CHBrH2 type — actually it is BrCH(CH3) with two hydrogens labeled H_a and H_b at the same carbon or an adjacent carbon. Reading more carefully: the central carbon has H (left, in-plane), Br (right, in-plane), CH3 (below, in-plane), H_a (top wedge) and H_b (top dash) — this suggests the top carbon bears H_a and H_b, making it a CH2 group adjacent to the CHBr(CH3) center. So X is CH2(Ha)(Hb)-CHBr-CH3 type, i.e., the molecule has at least one stereocenter at the CHBr carbon. Step 2 – Understand the substitution operation. H_b is replaced with D (deuterium) and H_a is replaced with H (no change for H_a, it stays H). This means in compound Y, the CH2 group becomes CHD, introducing a new stereocenter (since C with H, D, and the rest of the chain becomes a stereocenter due to H/D isotopic substitution). The CHBr center remains unchanged. Step 3 – Determine the stereochemical relationship. Compound X has one stereocenter (the CHBr carbon). Compound Y has two stereocenters: the original CHBr carbon and the new CHD carbon (isotopically differentiated). Both X and Y have the same molecular connectivity (constitutional structure is essentially the same, differing only in H vs D at one position — isotopically). Since X has one stereocenter and Y has two stereocenters (same connectivity, different number of stereocenters due to isotopic substitution creating a new chiral center), they are stereoisomers but not mirror images of each other. They have the same connectivity but differ in spatial arrangement at two centers (one center is present in both; the CHD center is new in Y). Since they are stereoisomers that are not mirror images of each other (X lacks the CHD stereocenter entirely while Y has it), they are classified as diastereomers. Step 4 – Why other options fail. (a) Enantiomers: enantiomers are non-superimposable mirror images with identical connectivity and same stereocenters — X and Y differ in the number of stereocenters, so they are not enantiomers. (c) E and Z isomers require a double bond with restricted rotation — not applicable here. (d) Constitutional isomers have different connectivity — X and Y have the same connectivity (H and D are isotopes of hydrogen, same element for connectivity purposes, but the compounds are still considered the same constitution). Since they are stereoisomers (not constitutional isomers) that are not enantiomers, they must be diastereomers. Therefore, the correct answer is B.