Which of the following is an optically active compound?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → pentan-2-ol
A compound is optically active when its molecule is chiral; the usual sign is a carbon atom carrying four different groups (an asymmetric carbon). Checking the marked carbon of each structure:
Option (2) \( \ce{CH3-CH(OH)-CH2CH2CH3} \) (pentan-2-ol): C-2 carries H, OH, \( \ce{CH3} \) and \( \ce{C3H7} \) — four different groups, so the molecule is chiral and optically active.
Why the other options are incorrect
Option (1) pentan-3-ol — Incorrect. \( \ce{CH3CH2-CH(OH)-CH2CH3} \): C-3 carries two identical ethyl groups, so there is no asymmetric carbon and the molecule is achiral.
Option (3) 4-methylheptane — Incorrect. The marked carbon carries H, \( \ce{CH3} \) and two identical propyl groups, so it is not chiral.
Option (4) — Incorrect. The structure as printed shows no carbon with four different groups, so it does not represent an optically active compound.
NCERT (2026-27 edition):
Molecular Asymmetry, Chirality and Enantiomers – …
