Arrange the following compounds in the increasing order of homolytic (C--C) bond dissociation energy.
I. Propane II. Ethane
III. 2,2-Dimethyl propane IV. 2-Methyl propane

Options

A. (a) `(III) lt (IV) lt (II) lt (I)`
B. (b) `(II) lt (I) lt (IV) lt (III)`
C. (c) `(III) lt (IV) lt (I) lt (II)`
D. (d) `(I) lt (III) lt (II) lt (IV)`

Answer

To arrange the compounds in the increasing order of homolytic (C--C) bond dissociation energy, we need to analyze the stability of the free radicals formed when the C--C bonds are cleaved. The more stable the free radical, the lower the bond dissociation energy, and vice versa.

### Step-by-Step Solution:

1. **Identify the Compounds**:
- I. Propane (C3H8) - CH3-CH2-CH3
- II. Ethane (C2H6) - CH3-CH3
- III. 2,2-Dimethylpropane (C5H12) - (CH3)4C
- IV. 2-Methylpropane (C5H12) - CH3-CH(CH3)-CH3

2. **Determine the Type of Free Radicals Formed**:
- **Ethane (II)**: When the C–C bond breaks, it forms two methyl radicals (CH3•). This is a primary radical.
- **Propane (I)**: When the C–C bond breaks, it forms one ethyl radical (CH3-CH2•) and one methyl radical (CH3•). The ethyl radical is a secondary radical.
- **2-Methylpropane (IV)**: When the C–C bond breaks, it forms a secondary radical (CH3-CH•-CH3) and a methyl radical (CH3•).
- **2,2-Dimethylpropane (III)**: When the C–C bond breaks, it forms a tertiary radical (C(CH3)2•) and a methyl radical (CH3•).

3. **Stability of Free Radicals**:
- The stability order of free radicals is: tertiary (3°) > secondary (2°) > primary (1°) > methyl (0°).
- Thus, the stability of the radicals formed from each compound is:
- Ethane (II): Primary radical (least stable)
- Propane (I): Secondary radical (more stable)
- 2-Methylpropane (IV): Secondary radical (more stable)
- 2,2-Dimethylpropane (III): Tertiary radical (most stable)

4. **Order of Homolytic Bond Dissociation Energy**:
- Since lower stability of the radical corresponds to higher bond dissociation energy, we can arrange the compounds in increasing order of bond dissociation energy:
- Ethane (II) > Propane (I) > 2-Methylpropane (IV) > 2,2-Dimethylpropane (III)

5. **Final Arrangement**:
- Therefore, the increasing order of homolytic C–C bond dissociation energy is:
- II < I < IV < III

### Conclusion:
The final answer is:
**II (Ethane) < I (Propane) < IV (2-Methylpropane) < III (2,2-Dimethylpropane)**