If relative rates of substitution of `1^(@)` and `2^(@)` H are in the ratio 1 : 3.8, show that in the presence of light at 298 K, the chlorination of n-butane gives a mixture of 72% 2-chlorobutane and 28% 1-chlorobutane.

Explanation

To solve the problem, we need to show how the chlorination of n-butane leads to a mixture of 72% 2-chlorobutane and 28% 1-chlorobutane when the relative rates of substitution of primary and secondary hydrogens are in the ratio of 1:3.8.

### Step-by-Step Solution:

1. **Identify the Structure of n-Butane**:
- The structure of n-butane (C₄H₁₀) can be represented as:
\[
CH₃-CH₂-CH₂-CH₃
...