See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Concept: A molecule possesses a plane of symmetry (sigma) if there is a mirror plane that bisects it into two mirror-image halves, and a centre of symmetry (inversion centre, i) if every atom has an equivalent atom at an equal distance on the opposite side through a central point. A molecule that is achiral can have both elements of symmetry simultaneously. Analysis of each option: Option (a): 1,2-disubstituted cyclobutane with methyl (wedge, up) at C1 and Cl (dash, down) at C2. These are different substituents on adjacent carbons with opposite configurations. This molecule is chiral (no plane of symmetry, no centre of symmetry). Option (b): 1,2-dichlorocyclobutane (trans configuration) - C1 has Cl on dash and C3 has Cl on wedge (or trans-1,2). For trans-1,2-dichlorocyclobutane: the two Cl groups are on opposite faces. This molecule has a centre of symmetry (the two Cl atoms are related by inversion through the ring centre) but the plane of symmetry analysis: a plane through C1 and C3 (or C2 and C4) does not make it symmetric because the Cl groups are on opposite faces. Actually trans-1,2-dichlorocyclobutane has a centre of inversion but no plane of symmetry - it is a chiral molecule with Ci symmetry only. Option (c): 1,3-dimethylcyclobutane with both CH3 groups on wedge (cis-1,3-dimethylcyclobutane). Both methyls point up on the same face. This molecule has a plane of symmetry (the plane bisecting the ring through C1 and C3 is a mirror plane) but does NOT have a centre of symmetry because there is no inversion centre relating the two CH3 groups (they are both up, so inversion would place them both down, which is the same molecule - wait, actually cis-1,3 does have a plane of symmetry through C2 and C4 axis). Cis-1,3-dimethylcyclobutane has a plane of symmetry but no centre of symmetry. Option (d): 1,3-dimethylcyclobutane with CH3 on wedge at top-right carbon (C1) and CH3 on dash at bottom-left carbon (C3) - this is trans-1,3-dimethylcyclobutane. In this molecule, C1 has CH3 up and C3 has CH3 down. A plane of symmetry exists: the plane containing C2 and C4 (perpendicular to the C1-C3 axis, bisecting the ring) maps C1 onto C3 and reflects the up-CH3 to a down position matching C3, so this IS a plane of symmetry. Additionally, the centre of the ring serves as a centre of symmetry: inverting through the centre maps C1 (CH3 up) to C3 (CH3 down) and vice versa, satisfying the inversion centre requirement. Thus trans-1,3-dimethylcyclobutane possesses BOTH a plane of symmetry and a centre of symmetry, making it the correct answer. Therefore, the correct answer is D.