See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Optical Activity Analysis **Optical activity requires a chiral center** — a carbon atom bonded to four different groups. **Examining each option:** **(1) $C_6H_5-C(=O)-OH$** - The carbonyl carbon has a double bond to oxygen - Not a chiral center (sp² hybridized) - **Not optically active** **(2) $CH_3-CH(OH)-C_2H_5$** - The $CH$ carbon bonds to: $CH_3$, $OH$, $C_2H_5$, and $H$ - Four different groups ✓ - This IS a chiral center - **Optically active** (but not option 3) **(3) $C_6H_5-CH(OH)-H$** - The $CH$ carbon bonds to: $C_6H_5$, $OH$, $H$, and another $H$ - **Two hydrogen atoms** attached to the same carbon - Not all four groups are different - **Not optically active** **(4) $C_6H_5-CH(CH_3)-$** - The $CH$ carbon bonds to: $C_6H_5$, $CH_3$, $H$, and another $H$ - **Two hydrogen atoms** attached to the same carbon - Not a true chiral center - **Not optically active** **Answer: Option (2)** is the optically active molecule as it contains a chiral center with four distinct groups ($C_6H_5$, $OH$, $C_2H_5$, $H$). *Note: If option (2) is labeled as option C in your question, this is the correct answer.*