I read about the following reaction in my book:

$$\ce{CH3-CH=CH2->[Cl2][\pu{773 K}] (Cl)CH2-CH=CH2}$$

It surprises me that the $\ce{Cl}$ doesn't attack the $\pi$ bond, rather it attacks the $\sigma$ bond.

It seems to me that the reaction follows free radical mechanism, for the high temperature of the reaction system. That's why, the $\ce{Cl}$ attacks the $\sigma$ bond.

But, I am not sure, why won't free radical $\ce{Cl}$ attack the $\pi$ bond.


1 Answer 1



It's not that the chlorine radical doesn't attack the $\pi$-bond, it just doesn't do it as fast. Alkyl radicals are much less stable than allyl radicals, so hydrogen abstraction will occur much faster than radical addition ($k_1 >> k_2$).

  • 6
    $\begingroup$ Isn't the question rather why the chlorine abstracts the hydrogen in the first place instead of directly adding to the double bond? $\endgroup$ Jun 12, 2017 at 5:45

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