An organic compound [A] (C$$_4$$H$$_{11}$$N), shows optical activity and gives N$$_2$$ gas on treatment with HNO$$_2$$. The compound [A] reacts with PhSO$$_2$$Cl producing a compound which is soluble in KOH. The structure of A is:

Compound A has molecular formula C$$_4$$H$$_{11}$$N. The given clues are:

1. Optical activity: A has a chiral center.

2. Gives N$$_2$$ with HNO$$_2$$: A is a primary amine ($$-NH_2$$).

3. Reacts with PhSO$$_2$$Cl to give a product soluble in KOH: This is the Hinsberg test result for a primary amine (the sulfonamide product has an acidic N-H that dissolves in KOH).

So A is an optically active primary amine with formula C$$_4$$H$$_{11}$$N, which is C$$_4$$H$$_9$$NH$$_2$$.

Among the isomers of C$$_4$$H$$_9$$NH$$_2$$:

- n-Butylamine: no chiral center

- Isobutylamine: no chiral center

- sec-Butylamine (2-aminobutane): CH$$_3$$CH(NH$$_2$$)CH$$_2$$CH$$_3$$ — has a chiral center at C-2 ✔

- tert-Butylamine: no chiral center

Therefore, A is sec-Butylamine (2-aminobutane).