Question

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 :

• A. 56
• B. 40
• C. 46
• D. 52

Hint:

For a carbon to be chiral, all four groups attached to it must be distinct. sec-butyl is the smallest alkyl group that can form a simple chiral halide.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:  
Optical isomerism occurs in molecules containing at least one chiral center (a carbon atom bonded to four different groups). 
We must identify the chiral compound among the options and calculate its percentage of carbon by mass. 

Step 2: Key Formula or Approach: 
The mass percentage of carbon is calculated as: 
\[ %C = \frac{\text{Mass of Carbon atoms}}{\text{Molar mass of the compound}} \times 100 \] 

Step 3: Detailed Explanation: 
Let's analyze each compound : 
(a) n-propyl chloride: \( CH_3CH_2CH_2Cl \). No chiral carbon. 
(b) iso-propyl chloride: \( (CH_3)_2CHCl \). No chiral carbon. 
(c) sec-butyl chloride: \(CH_3C^*H(Cl)CH_2CH_3\). The second carbon is bonded to -H, -Cl, -\( CH_3 \), and -\( C_2H_5 \). It is a chiral center. Thus, this compound exhibits optical isomerism. 
(d) neo-pentyl chloride: \( (CH_3)_3CCH_2Cl \). No chiral carbon. 
Now, calculate the percentage of carbon for sec-butyl chloride (\( C_4H_9Cl \)) : 
Mass of Carbon = \( 4 \times 12 = 48 \text{ g/mol} \) 
Mass of Hydrogen = \( 9 \times 1 = 9 \text{ g/mol} \) 
Mass of Chlorine = \( 35.5 \text{ g/mol} \) 
Molar mass = \( 48 + 9 + 35.5 = 92.5 \text{ g/mol} \) 
\[ %C = \frac{48}{92.5} \times 100 \approx 51.89% \] 
Rounding to the nearest integer, we get 52%. 

Step 4: Final Answer: 
The percentage of carbon in sec-butyl chloride is 52.