See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Optical Activity Analysis **Key Concept:** A molecule is optically active if it contains a chiral center (asymmetric carbon with four different groups attached). **Analysis of each option:** **(A) 1,1-dimethyl ethanol:** $CH_3-C(OH)(CH_3)_2$ - The central carbon has: $-OH$, $-CH_3$, $-CH_3$, $-CH_3$ - **Two identical methyl groups** → Not a chiral center → **Achiral** - ❌ This appears inactive by structure alone **(B) Butan-2-ol:** $CH_3-CH(OH)-CH_2-CH_3$ - The carbon bearing $-OH$ is bonded to: $-OH$, $-H$, $-CH_3$, $-CH_2CH_3$ - **Four different groups** → Chiral center present → **Optically active** ✓ **(C) Butanol (butan-1-ol):** $CH_3CH_2CH_2CH_2OH$ - Primary alcohol with no asymmetric carbons → **Achiral** **(D) 4-chloro-1-hydroxybutane:** $ClCH_2CH_2CH_2CH_2OH$ - Linear chain, no asymmetric carbons → **Achiral** **Answer: (B) butan-2-ol** Butan-2-ol has a secondary alcohol with a chiral center, making it optically active and capable of existing as $D$ and $L$ enantiomers.