In which of the following pairs the bromination of first member is easier than the second member ?

Options

A. Isobutane, n-Butane
B. n-Butane, Isobutane
C. Methane, Ethane
D. None of these

Answer

To determine which of the following pairs has the first member undergoing bromination more easily than the second member, we need to analyze the stability of the free radicals formed during the bromination process of each alkane. The stability of free radicals is influenced by the number of alkyl groups attached to the carbon atom bearing the unpaired electron. More alkyl groups lead to greater stability due to hyperconjugation and inductive effects.

Let's analyze the pairs given in the question:

1. **Isobutane (C4H10) vs. n-Butane (C4H10)**:
- **Isobutane**: The structure is (CH3)3C- (tertiary carbon). When brominated, it forms a tertiary free radical which is highly stable due to the three alkyl groups surrounding the carbon with the unpaired electron.
- **n-Butane**: The structure is CH3-CH2-CH2-CH2- (primary carbon). The free radical formed here is a primary free radical, which is less stable than a tertiary free radical.
- **Conclusion**: Isobutane undergoes bromination more easily than n-butane.

2. **n-Butane (C4H10) vs. Isobutane (C4H10)**:
- This is the same pair as the first one but in reverse order. The conclusion remains the same: n-butane is less reactive than isobutane.

3. **Methane (CH4) vs. Ethane (C2H6)**:
- **Methane**: Forms a primary free radical (CH3•).
- **Ethane**: Also forms a primary free radical (CH3-CH2•).
- **Conclusion**: Both have similar stability, but ethane has a slightly higher chance of bromination due to having more hydrogen atoms available for reaction.

4. **None of these**: This option suggests that none of the pairs fit the criteria.

Based on the analysis, the only pair where the bromination of the first member is easier than the second member is **Isobutane vs. n-Butane**.

### Final Answer:
The correct option is **Isobutane (C4H10) and n-Butane (C4H10)**.