Statement-I: Free radical chlorination of n-butane gives `72%` of 2-chlorobutane and `28%` of 1-chlorobutane though it has six primary and four secondary hydrogens.
Because Statement-II: A secondary hydrogen is abstracted more easily than the primary hydrogen.

Options

A. Statement-I is True, Statement-II is True, Statement-II is a correct explanation for Statement-I
B. Statement-I is True, Statement-II is True, Statement-II is NOT a correct explanation for Statement-I
C. Statement-I is True, Statement-II is False.
D. Statement-I is False, Statement-II is True.

Answer

To analyze the given statements regarding the free radical chlorination of n-butane, we will break down the reasoning step by step.

### Step 1: Understand the Structure of n-Butane
n-Butane has the molecular formula C4H10 and can be represented as:
\[ \text{CH}_3 - \text{CH}_2 - \text{CH}_2 - \text{CH}_3 \]
In this structure, we identify the types of hydrogens:
- **Primary hydrogens**: These are attached to the terminal carbon atoms (CH3 groups). There are a total of 6 primary hydrogens (3 from each CH3).
- **Secondary hydrogens**: These are attached to the middle carbon atoms (CH2 groups). There are a total of 4 secondary hydrogens (2 from each CH2).

### Step 2: Free Radical Mechanism of Chlorination
When n-butane undergoes chlorination in the presence of chlorine (Cl2), it follows a free radical mechanism. This involves:
1. **Initiation**: Formation of chlorine radicals.
2. **Propagation**: Chlorine radicals abstract hydrogen atoms from n-butane, forming alkyl radicals.
3. **Termination**: Combination of radicals to form products.

### Step 3: Formation of Radicals
During the chlorination process, two types of radicals can be formed:
- **Primary free radicals**: Formed by the abstraction of primary hydrogens.
- **Secondary free radicals**: Formed by the abstraction of secondary hydrogens.

### Step 4: Stability of Radicals
The stability of free radicals follows this order:
- Tertiary > Secondary > Primary
Since secondary radicals are more stable than primary radicals, the formation of secondary radicals will be favored.

### Step 5: Product Distribution
Due to the higher stability of secondary free radicals, the chlorination of n-butane will yield:
- **2-Chlorobutane** (from secondary radical): 72%
- **1-Chlorobutane** (from primary radical): 28%

This explains why we have a higher percentage of 2-chlorobutane despite having more primary hydrogens.

### Step 6: Evaluate the Statements
- **Statement I**: Free radical chlorination of n-butane gives 72% of 2-chlorobutane and 28% of 1-chlorobutane. This is true based on the analysis above.
- **Statement II**: A secondary hydrogen is abstracted more easily than a primary hydrogen. This is also true, as the formation of a more stable secondary radical is favored.

### Conclusion
Both statements are true, and Statement II correctly explains Statement I.

### Final Answer
- Statement I is true.
- Statement II is true.
- Statement II is the correct explanation of Statement I.

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