Question

A compound shows optical isomerism and has molecular formula \( \text{C}_4\text{H}_{10}\text{O} \). Which of the following compounds can exhibit chirality?

• A. Butan-1-ol
• B. Butan-2-ol
• C. 2-Methylpropan-2-ol
• D. Methoxypropane

Hint:

A carbon atom becomes chiral only if all four attached groups are different. Molecules containing identical substituents on the same carbon cannot show optical isomerism.

Answer 1

The Correct Option is B

Concept:
Optical isomerism arises due to the presence of a chiral carbon atom. A carbon atom is said to be chiral when it is attached to four different groups. Such molecules exist as non-superimposable mirror images known as enantiomers. To determine chirality:
• Identify tetrahedral carbon atoms.
• Check whether all four attached groups are different.
• If yes, the compound is optically active.

Step 1: Examine Butan-1-ol.
Structure: \[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \] No carbon atom is attached to four different groups. Hence, it is not chiral.

Step 2: Examine Butan-2-ol.
Structure: \[ \text{CH}_3\text{CH(OH)CH}_2\text{CH}_3 \] The second carbon atom is attached to: \[ \text{CH}_3,\quad \text{OH},\quad \text{H},\quad \text{CH}_2\text{CH}_3 \] All four groups are different. Therefore, this carbon is chiral. Hence, Butan-2-ol exhibits optical isomerism.

Step 3: Examine remaining compounds.
For 2-Methylpropan-2-ol: \[ (\text{CH}_3)_3\text{COH} \] The central carbon has three identical methyl groups. Hence, it is achiral. For Methoxypropane: \[ \text{CH}_3\text{OCH}_2\text{CH}_2\text{CH}_3 \] No chiral center exists. Therefore, the correct answer is: \[ \boxed{\text{Butan-2-ol}} \]