See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Identifying Optically Active Compounds **Key Principle:** A compound is optically active if it contains a **chiral center** (stereogenic center) — a carbon atom bonded to four different groups. **Analysis of each option:** **(1) n-propanol:** $CH_3CH_2CH_2OH$ - All carbons have at least two identical groups - No chiral center → optically inactive **(2) 2-chlorobutane:** $CH_3CHClCH_2CH_3$ - C2 is bonded to: $Cl$, $H$, $CH_3$, and $CH_2CH_3$ (four different groups) - **Contains one chiral center** → optically active ✓ **(3) n-butanol:** $CH_3CH_2CH_2CH_2OH$ - All carbons lack four different groups - No chiral center → optically inactive **(4) 4-hydroxyheptane:** $CH_3CH_2CH_2CH(OH)CH_2CH_2CH_3$ - C4 is bonded to: $OH$, $H$, $CH_2CH_2CH_3$, and $CH_2CH_2CH_3$ (two identical alkyl chains) - No chiral center → optically inactive **Answer: (2) 2-chlorobutane** is optically active because it possesses one chiral center at the C2 atom.