See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: A compound is optically inactive if it is achiral. For cyclic compounds with two stereocenters, optical inactivity can arise either because the molecule is a meso compound (has an internal plane of symmetry despite having stereocenters) or because it is a racemic mixture. A meso compound is achiral due to an internal mirror plane. Step 2 - Analyze option (c): The structure shown is cis-1,3-dimethylcyclopentane. In this molecule, both methyl groups are on the same face of the ring (both on wedge bonds pointing up), and both H atoms are on the opposite face (both on dash bonds pointing down). Step 3 - Symmetry check for (c): In cis-1,3-dimethylcyclopentane, the two substituents are identical (both CH3) and both are on the same face. There is a plane of symmetry passing through C2 (the carbon between the two methyl-bearing carbons) and bisecting the ring perpendicular to the C1-C3 axis. This mirror plane makes C1 and C3 mirror images of each other within the molecule. Because the molecule has an internal plane of symmetry, it is a meso compound and therefore optically inactive. Step 4 - Why other options are optically active: (a) trans-1,3-cyclopentanediol: the two OH groups are on opposite faces; no internal plane of symmetry; chiral (optically active). (b) cis-1,2-cyclopentanediol: both OH on same face at adjacent carbons; this is chiral (no internal symmetry plane for 1,2-substitution pattern); optically active. (d) trans-3-methylcyclopentan-1-ol: two different substituents (OH and CH3) at non-equivalent positions with trans relationship; no plane of symmetry; chiral and optically active. (e) trans-1,3-dimethylcyclopentane: the two methyl groups are on opposite faces of the ring; C1 and C3 are stereocenters with opposite configurations; no internal plane of symmetry; this is a chiral molecule and optically active. Step 5 - Conclusion: Only option (c), cis-1,3-dimethylcyclopentane, is a meso compound with an internal plane of symmetry, making it optically inactive. Therefore, the correct answer is C.