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

### Core Logic Among the given compounds, sec-butyl chloride (2-chlorobutane) is optically active and contains a chiral center. Molecular formula of 2-chlorobutane rightarrow textC_4textH_9textCl$\rightarrow \text{C}_4\text{H}_9\text{Cl}$. Molar mass = 4(12) + 9(1) + 35.5 = 48 + 9 + 35.5 = 92.5 text g/mol$= 4(12) + 9(1) + 35.5 = 48 + 9 + 35.5 = 92.5 \text{ g/mol}$. ### Step 1: Calculating Percentage of Carbon text\% of C = frac4892.5 times 100 = 51.89\% approx 52\%$$\text{\% of C} = \frac{48}{92.5} \times 100 = 51.89\% \approx 52\%$$ ### Pattern Recognition Sees: optical isomerism identification combined with elemental percentage composition calculation. Trap: Selecting the wrong halogen derivative for optical activity. ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes