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The chlorination of an alkane proceeds through the formation of free radicals. The chlorine radical (Cl·) can attack either side, leading to the production of a racemic mixture. Which of the following represents the initiation step of this reaction?

1. Cl2, when exposed to light (hv), splits into two chlorine radicals (Cl·).
2. When butane reacts with chlorine under light, it forms a secondary radical intermediate.
3. A secondary radical intermediate reacts further to form a chlorinated product.
4. A secondary radical intermediate rearranges to form another isomeric radical.

Correct answer: Cl2, when exposed to light (hv), splits into two chlorine radicals (Cl·).

Solution

The chlorination of alkanes proceeds via a free radical mechanism initiated by the homolytic cleavage of Cl2 under UV light, producing two chlorine radicals (Cl·). This is represented by option A.

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