A compound having molecular formula $\mathrm{C}_4 \mathrm{H}_{11} \mathrm{N}$, reacts with $\mathrm{CHCl}_3$ in alcoholic $\mathrm{KOH}$, on heating, to form a compound with a foul smell. Identify the optically active isomer of the compound which also shows the above reaction.

The reaction of a compound with $\mathrm{CHCl}_3$ and alcoholic $\mathrm{KOH}$ to produce a foul-smelling substance (isocyanide) is the carbylamine test. This test is specific to primary amines ($R-NH_2$).

The molecular formula $\mathrm{C}_4 \mathrm{H}_{11} \mathrm{N}$ corresponds to a saturated primary amine. The possible primary amine isomers are:
1. Butan-1-amine: $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NH}_2$
2. Butan-2-amine: $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}(\mathrm{NH}_2)-\mathrm{CH}_3$
3. 2-Methylpropan-1-amine: $(\mathrm{CH}_3)_2\mathrm{CH}-\mathrm{CH}_2-\mathrm{NH}_2$
4. 2-Methylpropan-2-amine: $(\mathrm{CH}_3)_3\mathrm{C}-\mathrm{NH}_2$

Among these, we check for optical activity. A molecule is optically active if it contains a chiral center (a carbon atom bonded to four different groups).

In Butan-2-amine, the carbon at position 2 is bonded to a methyl group, an ethyl group, a hydrogen atom, and an amino group. Since all four groups are distinct, this carbon is chiral. Therefore, Butan-2-amine is optically active.

Answer: Butan-2-amine