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

To determine the compound that exhibits optical isomerism and identify its percentage of carbon, let us examine the structures: 

1. **n-Propyl chloride (CH₃CH₂CH₂Cl):** This compound is a straight-chain alkyl halide and does not have a chiral center, hence does not exhibit optical isomerism.
2. **iso-Propyl chloride ((CH₃)₂CHCl):** This compound has no chiral center, thus it does not show optical isomerism.
3. **sec-Butyl chloride (CH₃CH(Cl)CH₂CH₃):** This compound has a chiral center at the second carbon (C-2), hence it can exhibit optical isomerism.
4. **neo-Pentyl chloride (C(CH₃)₃CH₂Cl):** This compound does not have a chiral center due to the symmetric branching of methyl groups.

From the above analysis, only sec-butyl chloride can exhibit optical isomerism due to the presence of a chiral center at the C-2 position.

Now, we calculate the percentage of carbon in sec-butyl chloride:

1. Molecular formula of sec-butyl chloride is C₄H₉Cl.
2. Molecular weight calculation:
   • Carbon (C): 12 g/mol, so 4 carbons contribute 4 × 12 = 48 g/mol.
   • Hydrogen (H): 1 g/mol, so 9 hydrogens contribute 9 × 1 = 9 g/mol.
   • Chlorine (Cl): 35.5 g/mol, contributes 35.5 g/mol.
3. Total molecular weight = 48 + 9 + 35.5 = 92.5 g/mol.
4. Percentage of carbon = \(\frac{48}{92.5} \times 100 \approx 51.89\%\).

Upon rounding, the percentage of carbon in sec-butyl chloride is approximately 52%. Thus, the correct answer is 52.