See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Solution **Step 1: Identify compound A ($C_6H_{12}$)** The problem states A is optically active with formula $C_6H_{12}$. From option (a), A is shown as **1,2-dimethylcyclobutene** (a cyclobutene ring with two methyl substituents). This structure has a chiral center, making it optically active. ✓ **Step 2: Determine the hydrogenation product B ($C_6H_{14}$)** Catalytic hydrogenation of the $C=C$ double bond in the cyclobutene ring converts it to a cyclobutane ring: $$\text{A: Cyclobutene with } C=C \rightarrow^{H_2, \text{catalyst}} \text{B: Cyclobutane (saturated)}$$ The double bond reduction adds 2 hydrogens: $C_6H_{12} + H_2 \rightarrow C_6H_{14}$ ✓ **Step 3: Verify B is optically inactive** After hydrogenation, compound B becomes **1,2-dimethylcyclobutane**. This saturated cyclic structure has a **plane of symmetry** passing through the ring, making it a meso compound. Therefore, B is optically inactive despite having chiral centers. ✓ **Step 4: Eliminate other options** - Options (B), (C), (D): These contain open-chain alkenes or alkanes that would either remain chiral after hydrogenation or don't properly account for the optical activity change. - Option (a) uniquely explains why A is optically active and why its hydrogenation product B becomes optically inactive. **Answer: (a)** — A is 1,2-dimethylcyclobutene and B is 1,2-dimethylcyclobutane.