See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the structure. The molecule is a 1,3,5-trisubstituted cyclohexane trioxime: a cyclohexane ring bearing three R groups (all on wedge bonds, i.e., all on the same face) at alternating carbons (C1, C3, C5) and three C=N–OH (oxime) groups at the remaining alternating carbons (C2, C4, C6). The three lone pairs drawn on the nitrogen atoms emphasize their stereochemical significance. Step 2 – Check for a C3 axis. Because the three R groups are all equivalent and disposed at 120° intervals around the ring (C1, C3, C5), and the three oxime groups are likewise disposed at 120° intervals (C2, C4, C6), rotation by 120° about the axis perpendicular to the mean plane of the ring maps the molecule onto itself. Therefore the molecule possesses a C3 axis of symmetry. Option (b) is correct. Step 3 – Check for chirality. A C3 axis is a proper rotation axis, not an improper axis (Sn). The presence of a C3 axis alone does not make a molecule achiral. For a molecule to be achiral it must possess an improper symmetry element (a mirror plane σ, an inversion center i, or an Sn axis). This molecule has no mirror plane (the three R groups are all on the same face, so the top and bottom faces are different), no inversion center, and no S6 or other improper axis. Therefore the molecule is chiral. Option (a) is correct. Step 4 – Check optical activity. A chiral molecule that is not a racemic mixture is optically active (it rotates plane-polarized light). Since the structure drawn represents a single stereoisomer (all R groups on the same face, all oxime geometries defined), the compound is optically active. Option (c) is correct. Step 5 – Conclusion. Since options (a), (b), and (c) are all correct, option (d) 'All of these' is the answer. Therefore, the correct answer is D.