See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

The question asks which molecules possess both an axis of symmetry (Cn, n≥2) and a coaxial plane of symmetry (a plane that contains the symmetry axis, i.e., a vertical or dihedral mirror plane). Concept: A coaxial plane of symmetry is a mirror plane that contains (is coincident with) the principal rotation axis. This means the plane includes the Cn axis. (a) Formaldehyde (HCHO): It is a planar molecule with C2v symmetry. The principal C2 axis passes through O, C, and bisects the H-C-H angle. There is a plane of symmetry that contains this C2 axis (the molecular plane itself, and also a plane perpendicular to the molecular plane containing C2). So it has a C2 axis and coaxial planes of symmetry. YES. (b) 1,3-Dichlorobenzene: It has C2v symmetry. The C2 axis passes through the carbon between the two Cl-bearing carbons (C2 position) and through C5. The molecular plane is a coaxial plane, and there is another mirror plane (perpendicular to the ring, containing the C2 axis) that bisects the molecule. So it has a C2 axis and a coaxial plane of symmetry. YES. (c) CH2Cl2 (dichloromethane): It has C2v symmetry. The C2 axis bisects the H-C-H angle and the Cl-C-Cl angle simultaneously. Two mirror planes each contain this C2 axis (one through H-C-H and one through Cl-C-Cl). So it has a C2 axis and coaxial planes of symmetry. YES. (d) Triphenylene (or the shown tricyclic aromatic): The structure shown appears to be triphenylene with D3h symmetry, which has a C3 axis and multiple coaxial planes (vertical mirror planes containing the C3 axis). Alternatively if it is phenanthrene, it has C2v symmetry with a C2 axis and coaxial mirror plane. Either way, it has both an axis of symmetry and a coaxial plane of symmetry. YES. Since all four individual molecules (a), (b), (c), and (d) each possess at least one axis of symmetry and at least one coaxial (axis-containing) plane of symmetry, the answer is (e) All of these. Therefore, the correct answer is E.