See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Finding Isomers of $C_2BrClFI$ **Step 1: Identify the molecular structure** The compound has 2 carbons with 4 different substituents: Br, Cl, F, and I. **Step 2: Determine the carbon skeleton** With only 2 carbons, the structure is: $C - C$ Each carbon can bear substituents from {Br, Cl, F, I, H}. **Step 3: Analyze bonding possibilities** - **Case 1:** Both halogens on the same carbon: $CHX_2-CH_3$ where X represents the halogen positions - **Case 2:** Halogens distributed on different carbons: $CHX-CH_2Y$ type arrangements **Step 4: Enumerate distinct isomers** With Br, Cl, F, and I all different, we can have: 1. $CHBrClF-CH_2I$ (Br, Cl, F on C1; I on C2) 2. $CHBrClI-CH_2F$ (Br, Cl, I on C1; F on C2) 3. $CHBrFI-CH_2Cl$ (Br, F, I on C1; Cl on C2) 4. $CHClFI-CH_2Br$ (Cl, F, I on C1; Br on C2) 5. $CBr(Cl)(F)(I)-CH_3$ (all four halogens on one carbon) 6. $CHBrCl-CHFI$ (Br, Cl on one C; F, I on other C) — exists as cis/trans isomers, but counted as one structural isomer Accounting for all possible distributions of the four different halogens across the two-carbon chain yields **6 distinct structural isomers**. **Answer: (D) 6**