See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Concept: A compound has a plane of symmetry (is a meso compound) when it contains stereocenters but an internal mirror plane makes it achiral overall. For a molecule with two stereocenters bearing identical substituents, the meso form has one R and one S configuration at the two centers — specifically the (R,S) or equivalently (S,R) diastereomer. Step 1: Define the groups. R-group = -CH(Cl)(CH3) and S-group = -CH(Cl)(Br). These two groups are different from each other (R has CH3, S has Br), so they are NOT identical substituents on the central carbon. Step 2: For a meso compound to exist, the molecule must have at least two stereocenters with identical substituents so that an internal mirror plane can relate them. Meso compounds arise when the same substituents appear in opposite configurations, canceling chirality. Step 3: Evaluate option (a): all four positions are R-configured groups. This gives a molecule where all stereocenters are R — no internal mirror plane exists because all centers have the same configuration. This is a chiral molecule (or if identical groups, a specific enantiomer). No plane of symmetry. Step 4: Evaluate option (b): all four positions are S-configured groups. Similarly, all stereocenters are S — no internal mirror plane. This is the enantiomer of (a). No plane of symmetry. Step 5: Evaluate option (c): the central carbon has R on top and bottom, S on left and right. However, since R-group (-CH(Cl)(CH3)) and S-group (-CH(Cl)(Br)) are structurally different substituents, simply having opposite configurations at those positions does not automatically create a plane of symmetry. A plane of symmetry requires that the two halves of the molecule be mirror images. Because R-group ≠ S-group (different substituents), no internal mirror plane exists. Step 6: Since none of the structures in (a), (b), or (c) possess a plane of symmetry — (a) and (b) because all centers have the same configuration, and (c) because the two different types of substituents (R-group with CH3 vs S-group with Br) cannot be related by a mirror plane — the answer is (d) None of these. Therefore, the correct answer is D.