See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Optically Active Isomers of Monochloroisopentanes **Step 1: Identify the parent structure** Isopentane (2-methylbutane) has the structure: $CH_3-CH(CH_3)-CH_2-CH_3$ **Step 2: Locate possible chlorination sites** Chlorine can substitute at: - C1 (terminal CH₃) - C2 (tertiary carbon with methyl branch) - C3 (secondary CH₂) - C4 (terminal CH₃) **Step 3: Identify chiral centers** For optical activity, we need a carbon with **four different groups** (stereogenic center): 1. **1-chloro-3-methylbutane**: $CH_3-CHCl-CH(CH_3)-CH_3$ - C2 is chiral (bonded to: Cl, H, CH₃, and $CH(CH_3)CH_3$) → **1 enantiomeric pair** 2. **2-chloro-3-methylbutane**: $CH_3-CCl(CH_3)-CH(CH_3)-CH_3$ - No chiral center (C2 has two CH₃ groups) 3. **3-chloro-2-methylbutane**: $CH_3-CH(CH_3)-CHCl-CH_3$ - C3 is chiral (bonded to: Cl, H, CH₃, and $CH(CH_3)CH_3$) → **1 enantiomeric pair** 4. **1-chloro-2-methylbutane**: $CH_3-CH(CH_3)-CH_2-CH_2Cl$ - No chiral center **Step 4: Count total optically active isomers** - Compound 1: 2 enantiomers (R and S forms) - Compound 3: 2 enantiomers (R and S forms) **Total = 4 optically active isomers** **Answer: (D) 4**