See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Concept: Optical activity of a mixture depends on the net rotation produced by all optically active species present. A racemic mixture (equal amounts of enantiomers) is optically inactive because the rotations cancel exactly. A meso compound is also optically inactive due to an internal plane of symmetry. Step 1: Analyze the given information. The aqueous solution of A and B (which are stereoisomers of each other) shows optical activity, meaning there is a net non-zero optical rotation. Step 2: Evaluate option (b) - A and B are enantiomers. If A and B are enantiomers present in unequal amounts, the solution would still show optical activity (net rotation ≠ 0). So this is possible, provided they are not in equal amounts. However, consider option (b) alone: if A and B are enantiomers, there is a specific scenario where optical activity could still be observed if they are unequal. So option (b) is not necessarily impossible on its own. Step 3: Evaluate option (a) - A has two chiral centers, B does not have any because it has a symmetry plane (i.e., B is a meso compound). If B is meso, it is optically inactive. A has two chiral centers and is chiral, so A is optically active. The mixture would show optical activity due to A alone. This is entirely consistent with the observation. So option (a) can be correct. Step 4: Re-evaluate option (b) more carefully. If A and B are enantiomers, they rotate plane-polarized light by equal magnitude but opposite directions. For the solution to show optical activity, they must be present in unequal amounts (option d would then also be true). But if A and B are enantiomers present in equal amounts, the solution would be a racemic mixture and optically inactive — contradicting the observation. The question asks which possibility CANNOT be correct. Option (b) states only that A and B are enantiomers, without specifying equal amounts. But stereoisomers that are enantiomers: if the solution shows optical activity, they simply cannot be present as a racemic (equal) mixture. Being enantiomers alone does not prevent optical activity if amounts differ. Step 5: The key reasoning for option (a) being the answer that CANNOT be correct: If A has two chiral centers and B is a meso compound (has a symmetry plane), then A and B are diastereomers (they are stereoisomers but not mirror images). B being meso means B is achiral and optically inactive. A being chiral means A contributes optical rotation. The mixture shows optical activity — this is perfectly consistent. So option (a) CAN be correct. Step 6: Re-examine: The question asks which CANNOT be correct. Option (b): A and B are enantiomers — if they are enantiomers and the solution is optically active, they must be in unequal amounts. This is possible. Option (b) can be correct (with unequal amounts). Step 7: For option (a): if B has a symmetry plane (meso), B is not chiral and has no chiral centers in the effective sense — but wait, the statement says B does not have any chiral centers because it has a symmetry plane. A meso compound DOES have chiral centers (stereocenters) but the internal symmetry renders it achiral. So saying B 'does not have any [chiral centers] because it has a symmetry plane' is chemically incorrect — a meso compound has stereocenters but is achiral due to internal symmetry. The statement in option (a) incorrectly equates having a symmetry plane with having no chiral centers. This is a factually incorrect statement about stereochemistry, meaning option (a) as stated cannot be correct because if B truly had no chiral centers at all, it would not be a stereoisomer of A (which has two chiral centers). For B to be a stereoisomer of A, B must have the same connectivity and same number of stereocenters — a meso compound has stereocenters but is achiral. The description in (a) is self-contradictory: B cannot be a stereoisomer of A (with two chiral centers) while having no chiral centers at all. Therefore, option (a) cannot be correct because if B truly has no chiral centers, it cannot be a stereoisomer of A which has two chiral centers. A meso compound has chiral centers but is made achiral by the symmetry plane — so describing it as having no chiral centers is wrong and internally inconsistent with being a stereoisomer of A. Therefore, the correct answer is A.