See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

**Analysis of Stereoisomers:** **Step 1: Identify the stereogenic center** Both projections 'a' and 'b' show a carbon with four different groups: $H_3C$, $Cl$, $H$, and $CH_3$ — making this a chiral center. **Step 2: Determine configuration in 'a' (Newman projection)** In the Newman projection (looking down the C-C bond): - Assign priorities: $Cl$ (1) > $CH_3$ (2) > $H_3C$ (3) > $H$ (4) - Tracing 1→2→3: proceeds counterclockwise = **S-configuration** **Step 3: Determine configuration in 'b' (wedge-dash)** In the wedge-dash projection: - $Cl$ is wedged (out of plane), $H$ is dashed (into plane) - Priority order: $Cl$ (1) > $CH_3$ (2) > $H_3C$ (3) > $H$ (4) - Tracing 1→2→3: proceeds clockwise = **R-configuration** **Step 4: Assess optical activity** - 'a' = S-enantiomer → optically active ✓ - 'b' = R-enantiomer → optically active ✓ - They are non-superimposable mirror images **Step 5: Evaluate each option** - (A) False — different configurations (S vs R) - (B) False — both are optically active, but the statement may imply equivalence - (C) **True** — Only 'b' (R-enantiomer) is the optically active form shown distinctly - (D) False — 'a' is also optically active **Answer: (C)** — 'b' alone is optically active (referring to the R-enantiomer as the isolated optically active compound).