See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Solution: Optical Rotation of Stereoisomers **Step 1: Identify chiral centers** All four compounds have a chiral carbon (C bonded to 4 different groups). Only compounds I, II, and III are actual chiral molecules. **Step 2: Assign R/S configuration** **(I)** Priority: $Cl(1) > Et(2) > Me(3) > H(4)$ Arrangement: counterclockwise → **S-configuration** **(II)** Priority: $Cl(1) > Et(2) > Me(3) > H(4)$ Arrangement: clockwise → **R-configuration** **(III)** Priority: $Cl(1) > Et(2) > Me(3) > H(4)$ Arrangement: clockwise → **R-configuration** **(IV)** This is **achiral** (two H atoms attached to chiral center) → **optically inactive** (rotation = 0°) **Step 3: Identify enantiomers** - Compounds I and II are **enantiomers** (S and R configurations, same molecular formula) - Enantiomers have **equal but opposite optical rotations**: $\alpha_I = -\alpha_{II}$ Compound III (R) is a **diastereomer** of II (not an enantiomer) and has different rotation. **Answer: (A) I, IV** Compounds I and IV have the **same optical rotation of 0°** because: - **I rotates light** by some angle (S-enantiomer) - **IV is achiral** and doesn't rotate light (rotation = 0°) Wait—this needs clarification: If the question asks for same non-zero rotation, none work. But if it includes zero rotation, **(A) is correct** as IV is optically inactive. Alternatively, **(B) I, II** would show equal-but-opposite rotations. Check if "same rotation" means equal magnitude or identical value in the problem context.