Given below are four compounds :
(a) n-propyl choride
(b) iso-propyl chloride
(c) sec-butyl chloride
(d) neo-pentyl chloride
Percentage of carbon in the one which exhibits optical isomerism is:

We need to identify which compound exhibits optical isomerism and find its carbon percentage.

(a) n-propyl chloride: $$CH_3CH_2CH_2Cl$$ — no chiral center.

(b) iso-propyl chloride: $$(CH_3)_2CHCl$$ — no chiral center (the carbon bonded to Cl has two identical methyl groups).

(c) sec-butyl chloride: $$CH_3CH(Cl)CH_2CH_3$$ — has a chiral center at C-2 (bonded to H, Cl, CH₃, C₂H₅ — all different). Exhibits optical isomerism.

(d) neo-pentyl chloride: $$(CH_3)_3CCH_2Cl$$ — no chiral center.

sec-butyl chloride is $$C_4H_9Cl$$. Molecular formula: $$C_4H_9Cl$$.

Molar mass = 4(12) + 9(1) + 35.5 = 48 + 9 + 35.5 = 92.5 g/mol

% Carbon = $$\frac{48}{92.5} \times 100 = 51.89\% \approx 52\%$$

The answer is Option 2: 52.