See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Solution: Isomers from Reduction of CFCl.CBrI **Step 1: Identify the starting material structure** $CFCl.CBrI$ represents: $F-C(Cl)-C(Br)-I$ This molecule has **two chiral centers** (both carbon atoms bear four different groups). **Step 2: Determine the stereoisomers before reduction** With 2 chiral centers, maximum stereoisomers = $2^2 = 4$ (E,Z + E,Z isomers are noted) **Step 3: Analyze the reduction reaction** $H_2/Ni$ is a **heterogeneous catalytic reduction** that adds $H_2$ across the double bond: - C-F bond → C-H (replacing fluorine) - C-Br bond → C-H (replacing bromine) - C-I bond → C-H (replacing iodine) The product is: $H_3C-CH_3$ (ethane) **Step 4: Account for stereochemistry during reduction** The reduction occurs with **syn addition** (both hydrogens add from the same face). Depending on which face of the double bond is attacked: - Attack from one face → one stereoisomeric product - Attack from opposite face → enantiomeric product However, the resulting ethane itself has **no chiral centers**, but the reduction of the starting stereoisomers occurs with stereoselectivity based on: - Steric approach control - The E/Z geometry of the intermediate **Step 5: Count distinct isomeric products** Considering: - Two possible geometries of approach (above/below plane) - Conformational/stereochemical outcomes from 2 different starting E/Z isomers - Restricted rotation effects during reduction This yields **4 distinct isomeric products** maximum. **Answer: (C) 4**