See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Concept: An axis of symmetry (Cn axis) is a rotational axis such that rotation by 360°/n produces an indistinguishable structure. We identify the order n for each compound. (1) Cyclobutane with alternating Me/Cl substituents (all up on one diagonal, all down on the other): Rotating 180° about an axis perpendicular to the ring (or through the ring midpoint) interconverts equivalent positions, giving a C2 axis. No C4 because the substituents alternate in configuration; no C3 possible for a 4-membered ring with this pattern. (2) All-equatorial or alternating cyclohexane with six ethyl groups in alternating up/down pattern: The molecule has C3 rotational symmetry. Rotating 120° about an axis through the center perpendicular to the mean plane maps carbon 1→3→5 and 2→4→6, and since all substituents are identical (Et), the molecule looks the same. Hence C3 axis. (3) The macrocyclic triester-triamide has three identical repeating units arranged in a ring. A 120° rotation maps one unit onto the next, giving a C3 axis. (4) The norbornane (bicyclo[2.2.1]heptane) framework with three Cl atoms: The three Cl groups are placed symmetrically at three equivalent positions related by 120° rotation. A C3 axis passes through the center of the molecule along the C1–C4 bridgehead axis (actually through the molecule in a direction that relates the three bridge positions). Hence C3 axis. (5) The adamantane-like framework bearing three Br and three Cl in alternating positions: Rotation by 120° about the C3 axis (through the two bridgeheads of the adamantane core) maps Br→Br→Br and Cl→Cl→Cl, giving C3 symmetry. (6) 1,3,5-tri-tert-butyl-1,3,5-triazinane: The three N–CMe3 groups are equivalent, related by 120° rotation about the C3 axis perpendicular to the ring plane. Hence C3 axis. (7) The triketone with three identical aryl-carbonyl arms arranged around a central N atom: Three-fold C3 rotation maps arm 1→2→3. Additionally, a C2 axis exists through each arm and the center (or through the N and the midpoint of the opposite C=O). Both C3 and C2 axes are present. (8) Triglycidyl isocyanurate (triazine-trione with three epoxide groups): The three N–CH2–epoxide arms are identical and related by 120° rotation, giving a C3 axis. (9) The bicyclic compound with three Cl and three H at alternating bridge positions: A C3 axis passes through the center of the molecule, and 120° rotation maps (Cl,H) at one bridge to equivalent (Cl,H) at the next bridge. (10) Norbornone (bicyclo[2.2.1]heptan-2-one): The C=O group sits on one bridge. A C2 axis passes through the carbonyl carbon and the midpoint of the C1–C4 axis (or through C7), making the two halves mirror images related by 180° rotation. Hence C2 axis. (11) 7,7-dichloronorbornane: The gem-dichloro group at C7 and the bicyclic framework give a C2 axis passing through C7 and the midpoint of C1–C4 (the two bridgeheads), because the two halves of the molecule (the two CH2–CH2 bridges) are equivalent. (12) Norbornene (bicyclo[2.2.1]hept-2-ene): The double bond and the C7 methylene bridge define a plane of symmetry and a C2 axis through C7 and the midpoint of the C2=C3 double bond. (13) The bicyclic diketone with two C=O groups symmetrically disposed: A C2 axis bisects the molecule, mapping one carbonyl onto the other. Therefore, the correct answer is (1) C2-axis; (2) C3-axis; (3) C3-axis; (4) C3-axis; (5) C3-axis; (6) C3-axis; (7) C3-axis & C2-axis; (8) C3-axis; (9) C3-axis; (10) C2-axis; (11) C2-axis; (12) C2-axis; (13) C2-axis.