Arrange the following in correct order of reactivity towards `Cl_(2)//hupsilon`
(A) `CH_(4)` (B) `CH_(3)CH_(3)`
(C ) `CH_(3)CH_(2)CH_(3)`
(D) `CH_(3)-overset(CH_(3))overset(|)(CH)-CH_(3)`

Options

A. `A gt B gt C gt D`
B. `D gt C gt B gt A`
C. `B gt C gt A gt D`
D. `C gt B gt D gt A`

Answer

To determine the correct order of reactivity of the given hydrocarbons towards chlorination using `Cl2` in the presence of light (denoted as `Cl2//hupsilon`), we need to consider the stability of the free radicals that will be formed during the reaction. The mechanism involved is a free radical substitution mechanism.

### Step-by-Step Solution:

1. **Identify the Compounds**:
- (A) `CH4` (Methane)
- (B) `CH3CH3` (Ethane)
- (C) `CH3CH2CH3` (Propane)
- (D) `CH3-CH(CH3)-CH3` (Isobutane)

2. **Understand Free Radical Stability**:
- The stability of free radicals increases with the number of alkyl groups attached to the carbon bearing the unpaired electron. This is due to hyperconjugation and the electron-donating effect of alkyl groups.

3. **Analyze Each Compound**:
- **(A) Methane (`CH4`)**: Forms a primary free radical, which is the least stable.
- **(B) Ethane (`CH3CH3`)**: Also forms a primary free radical, slightly more stable than methane due to the presence of two methyl groups.
- **(C) Propane (`CH3CH2CH3`)**: Forms a secondary free radical, which is more stable than primary radicals.
- **(D) Isobutane (`CH3-CH(CH3)-CH3`)**: Forms a tertiary free radical, which is the most stable due to the presence of three methyl groups.

4. **Rank the Stability of Free Radicals**:
- **Most Stable**: (D) Isobutane (tertiary radical)
- **Next**: (C) Propane (secondary radical)
- **Then**: (B) Ethane (primary radical)
- **Least Stable**: (A) Methane (primary radical)

5. **Order of Reactivity**:
- The order of reactivity towards `Cl2//hupsilon` will be the same as the order of stability of the free radicals formed:
- (D) > (C) > (B) > (A)

### Final Order of Reactivity:
- **(D) > (C) > (B) > (A)**