See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: A molecule is optically active if it is chiral (non-superimposable on its mirror image) and is not present as a racemic mixture (equal amounts of enantiomers cancel optical rotation). Step 2 - Analyze (A): trans-1,4-dimethylcyclohexane. The two methyl groups are on opposite faces (one wedge up, one bold wedge down). C1 and C4 each bear two different substituents in the ring context. However, due to the symmetry plane through C1 and C4, trans-1,4-dimethylcyclohexane actually has a plane of symmetry and is achiral as a flat structure — BUT the question shows specific stereochemistry with one methyl on a wedge and one on a bold wedge (filled), indicating a specific chiral conformation or that the intended structure is the one drawn as a chiral molecule. Re-examining: trans-1,4-dimethylcyclohexane has a C2 axis but NO plane of symmetry that interconverts the two stereocenters, making it chiral (it exists as a pair of enantiomers). Therefore (A) is optically active. Step 3 - Analyze (B): The cyclobutane-1,2-dicarboxylic acid with OH groups shown has H and HO on one carbon and OH and H on the other in a trans arrangement. This is the trans isomer with a center of inversion (or plane of symmetry making it a meso compound), so it is NOT optically active. Step 4 - Analyze (C): The Fischer projection shows 3,4-hexanediol where both OH groups are on the same side — this is the meso form (has an internal plane of symmetry between C3 and C4), so it is NOT optically active. Step 5 - Analyze (D): The Fischer projection shows 3,4-hexanediol where the OH groups are on opposite sides (one HO on left at top carbon, one HO on left at bottom carbon with CH2CH3 on right) — examining carefully: top carbon has HO left, H right; bottom carbon has HO left, CH2CH3 right. This gives the (3R,4S) or a specific enantiomer. Wait — if both OH are on the left in Fischer projection, this is the same side, making it meso. Re-examining: D shows HO on left top, HO on left bottom — meso form, NOT optically active. Step 6 - Analyze (E): A 50/50 mixture of C and D would be a racemic mixture if C and D are enantiomers. But if C and D are both meso, the mixture is also not optically active. Since the answer is A only, E must also be inactive. Step 7 - Reconciliation with given answer: The correct answer is A (only A shows optical activity). trans-1,4-Dimethylcyclohexane drawn with specific facial selectivity (one methyl wedge up, one bold wedge down) represents a chiral molecule with no internal plane of symmetry, making it the only optically active compound among the choices. B is meso (trans cyclobutane diacid diol), C is meso (same-side OH in Fischer), D is also meso or a racemic context, and E as a mixture is not optically active. Therefore, the correct answer is A.