When a mixture of 2-merthyl propane and `C C1_(4)` is reacted at 403-413 in the presence of a radical intiator, t-butyl peroxide, 2-chloro-2-methyl propane and chloroform `(CHC1_(3)` are formed. Give the mechanism of the reaction.

Explanation

To understand the mechanism of the reaction between 2-methylpropane and carbon tetrachloride (CCl4) in the presence of a radical initiator, we can break it down into several steps: initiation, propagation, and termination.

### Step 1: Chain Initiation
1. **Formation of Radicals**: The reaction begins with the thermal decomposition of the radical initiator, t-butyl peroxide (C4H10O2). When heated (or exposed to light), t-butyl peroxide undergoes homolytic cleavage, breaking the O-O bond to form two t-butyl radicals:
\[ \text{(CH}_3\text{)}_3\text{COO-} \xrightarrow{\text{heat/light}} \text{(CH}_3\text{)}_3\text{C}^\cdot + \text{(CH}_3\text{)}_3\text{C}^\cdot \]

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