See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Analysis of Stereoisomers **Step 1: Identify the molecular formulas** - Structure I: $CH_3, Br, Cl, C_2H_5, H, OCH_3$ → $C_4H_9BrClO$ - Structure II: $C_2H_5, Br, H_3CO, Cl, CH_3, H$ → $C_4H_9BrClO$ - Structure III: $Cl, C_2H_5, H_3CO, H, Br, CH_3$ → $C_4H_9BrClO$ All three have the **same molecular formula** (rules out structural isomerism for all pairs). **Step 2: Analyze relative positions in I vs II** In Structure I: - $H$ and $Br$ are on opposite sides (wedge-dash notation) - $Cl$ and $OCH_3$ are on opposite sides In Structure II: - $H$ and $Br$ are on opposite sides - $H_3CO$ and $Cl$ are on opposite sides Comparing the arrangement of the four different groups ($H, Br, Cl, OCH_3$) around the central carbon: the groups are in **different spatial arrangements** → **I and II are structural isomers** (different connectivity/arrangement of substituents). **Step 3: Analyze relative positions in I vs III** In Structure I: $H$ (back), $Br$ (front), $Cl$ (back), $OCH_3$ (front) In Structure III: $H$ (back), $Br$ (front), $Cl$ (front), $OCH_3$ (back) These are **non-superimposable mirror images** → **I and III are enantiomers**. **Answer: (C)** - I and III are enantiomers (mirror images), while I and II are structural isomers (different group arrangements around the same stereogenic center).