See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Concept: Optical activity (rotation of plane-polarized light) requires chirality. In biphenyls, chirality can arise from restricted rotation about the central C–C bond (atropisomerism) when the two rings bear substituents that prevent free rotation AND the molecule lacks an internal plane of symmetry. This is called axial chirality. Step 1 – Option (a): Unsubstituted biphenyl. The two phenyl rings are unsubstituted and can rotate freely about the central bond. There is no restricted rotation, and even if rotation were restricted, the molecule would have a plane of symmetry. Not optically active. Step 2 – Option (d): Two cyclobutane rings joined by a C=C double bond (bicyclobutylidene). This molecule has a plane of symmetry through the double bond and is achiral. Not optically active. Step 3 – Option (c): 2,2',6,6'-Tetramethylbiphenyl. Each ring bears methyl groups at BOTH ortho positions (2,6 and 2',6'). The steric bulk of four methyl groups forces the rings to be perpendicular and restricts rotation. However, because each ring has identical substituents at both ortho positions (the substitution pattern on each ring is symmetric: methyl at 2 and 6), each ring individually has a plane of symmetry. The overall molecule has a C2 axis and a plane of symmetry (the molecule is its own mirror image — it is a meso-like situation for axial chirality). Both rings have the same substitution pattern and there is no non-superimposable mirror image; it is achiral. Not optically active. Step 4 – Option (b): 2,2'-Dimethylbiphenyl (one methyl on each ring at one ortho position only). Each ring bears a methyl at one ortho position only. The large ortho substituents (methyl groups) create sufficient steric hindrance to restrict rotation about the central C–C bond (atropisomerism). The two rings are now locked in a non-planar conformation. With one methyl per ring at the 2 and 2' positions, each ring has an asymmetric substitution pattern (one ortho position has methyl, the other does not). This creates a non-superimposable mirror image (the molecule lacks both a plane of symmetry and a center of inversion when the rings are locked). However, the key question is whether rotation is restricted enough at room temperature. With two ortho methyl groups (one per ring), the rotational barrier is sufficient to prevent interconversion of atropisomers at room temperature, so the compound can be resolved into enantiomers and will rotate plane-polarized light. Why (b) and not (c): In (c), even though rotation is restricted, the molecule is achiral due to the symmetrical substitution (2,6,2',6'-tetramethyl gives a molecule with an internal mirror plane). In (b), the 2,2'-disubstitution gives genuine axial chirality with no compensating symmetry element, making it optically active. Therefore, the correct answer is B.