See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Solution: Optically Active Monochloroisopentane Isomers **Step 1: Identify the structure of isopentane** Isopentane has the structure: $(CH_3)_2CHCH_2CH_3$ or 2-methylbutane **Step 2: Determine possible monochlorination sites** Chlorine can substitute at four different carbon environments: - Primary carbons (two equivalent $-CH_3$ on C2) - Secondary carbon (C2 – the branched carbon) - Primary carbon (C4) - Secondary carbon (C3) **Step 3: Analyze each monochloroisomer for optical activity** For a molecule to be optically active, it must have a chiral center (asymmetric carbon with 4 different groups): 1. **Chlorination at C1 or C5** (terminal primary carbons): Produces $CH_2ClCH(CH_3)CH_2CH_3$ and isomers – **no chiral center** (C2 still has two identical $CH_3$ groups) 2. **Chlorination at C2** (the branching point): Produces $CH_3CHClC(CH_3)_2$ – **no chiral center** (has two identical $CH_3$ groups) 3. **Chlorination at C3** (secondary carbon next to branched carbon): Produces $(CH_3)_2CHCH_2CHCl_2$ – **no chiral center** (C3 has two Cl atoms? No—it's $CH_3CHCl...$). Actually: **C3 becomes chiral** with groups: $-Cl$, $-H$, $-CH_2CH_3$, $-CH(CH_3)_2$ → **ONE optically active isomer** ✓ **Step 4: Conclusion** Only **1-chloro-3-methylbutane** (chlorination at C3) produces an optically active compound with one stereogenic center. **Answer: (D) one**