See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

To determine which beaker's mixture rotates plane-polarized light, we must assess whether each mixture is optically active (net rotation ≠ 0). Concept: A mixture rotates plane-polarized light if it has a net optical rotation. A racemic mixture (equal amounts of enantiomers) gives zero net rotation. A mixture of diastereomers, or an unequal mixture of enantiomers, or a single chiral compound, will rotate plane-polarized light. A meso compound (internally compensated) does not rotate light. A mixture of a chiral compound with an achiral compound or with its diastereomer generally does rotate light. Beaker (a): The first compound is a trans-decalin-like structure with HO and CH3 substituents — it is a chiral molecule (no internal plane of symmetry). The second compound is a substituted cyclohexane (4-methylcyclohexanol with an extra methyl). These two structures are NOT enantiomers of each other — they are structurally different (one is bicyclic, one is monocyclic). Since they are not enantiomers, there is no racemic cancellation; each compound has its own optical rotation, and the mixture will show net rotation. Therefore beaker (a) rotates plane-polarized light. Beaker (b): The two structures shown are a pair of diastereomers (they differ in configuration at one or both stereocenters but are not mirror images of each other — the Cl stereocenter is inverted but the OH stereocenter appears the same, or vice versa). Diastereomers have different physical properties and different specific rotations. They are not enantiomers, so their rotations do not cancel. The mixture of two diastereomers will generally rotate plane-polarized light. Therefore beaker (b) rotates plane-polarized light. Beaker (c): The first structure is (2R,3S)- or (2S,3R)-2,3-butanediol drawn in Fischer projection — this could be the meso form (if H and OH are on opposite sides at C2 and C3) or a chiral form. Looking carefully: C2 has H left, OH right; C3 has H left, OH right — this is the (R,R) or (S,S) form (both OH on same side), making it a chiral molecule. The second structure has C2 with H left, OH right, and C3 with H left, H right — this means C3 is not a stereocenter (it has two H's), so this is simply (R)- or (S)-1,2-propanediol extended, i.e., butan-2-ol type — actually it is 1,3-butanediol or similar achiral/chiral molecule. The key point: these two structures are not enantiomers of each other (they are constitutional isomers or at least not mirror images), so there is no cancellation. The mixture contains at least one chiral, optically active component, giving net rotation. Therefore beaker (c) rotates plane-polarized light. Since all three beakers (a), (b), and (c) contain mixtures that rotate plane-polarized light, the answer is all of these. Why other options fail: Options (a), (b), and (c) alone are each correct individually, but the question asks which beaker's mixture rotates plane-polarized light — since all three do, option (d) 'All of these' is the most complete and correct answer. Therefore, the correct answer is D.