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QUESTION 1 If n-butane is chlorinated as shown below, approximately what percentage of the product mixture will be 1-chlorobu

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Answer #1

In the presence of sunlight and chlorine, the n-butane undergoes halogenation in the 2 positions and also in the 1 positions. The scheme is shown below:

CH3CH2CH2CH3+ Cl2 (Sunlight) ---------------> CH3CH2CH2CH2Cl (25%) + CH3CH2CHClCH3 (75%)

This reaction undergoes via a radical mechanism. The 2 degree radical is more stable than 1 degree radical. Hence the 2-chloro butane is the major product.

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