In Free Radical halogenation using chlorine and sunlight, why does the chlorine-free radical get attached to the allylic carbon?

Will it be correct to say that it follows the anti-Markovnikov rule?

But if so, there is no $\ce{H2O2}$ present...

The substrate can be considered as but-2-ene.

  • $\begingroup$ What's the substrate for free radical halogenation? $\endgroup$
    – Zenix
    Commented May 8, 2020 at 15:16
  • $\begingroup$ @zenix but-2-ene $\endgroup$
    – Devaesh
    Commented May 10, 2020 at 7:55
  • 1
    $\begingroup$ This is not $\ce{HX}$ addition to an alkene. Therefore, Markovnikov rule does not apply here. You should change your question accordingly. $\endgroup$ Commented Sep 18, 2020 at 17:38
  • 2
    $\begingroup$ More specifically, this is not HBr addition in the presence of peroxide (Kharasch reaction). HCl and HI do not work in the Kharasch reaction. $\endgroup$
    – user55119
    Commented Sep 18, 2020 at 18:06

1 Answer 1


H. C. Brown has suggested that the halogen atom does add but, at high temperatures, is expelled before the second step of free radical addition can occur$^1$.

enter image description here

Addition of halogen atom gives this radical, which falls apart to regenerate the starting material if the temperature is high or if it does not soon encounter a halogen molecule to complete the addition. Hence, low concentration of halogen can be used with high temperature to favour substitution over addition$^1$.

The allylic radical, on the other hand, once formed, has little option but to wait for a halogen molecule, however low the halogen concentration$^1$.

enter image description here

For the formation of minor product you may look into this question.

To follow the Markovnikov's or anti-Markovnikov's rule, there should be difference of electronegativity between the two atoms. Therefore, I do not understand what you are trying to ask.


  1. Organic Chemistry- Morrison and Boyd (6E); Page 389-390
  • $\begingroup$ The first step of the allylic mechanism should have the half arrow going to the chlorine radical. $\endgroup$
    – Zhe
    Commented Sep 18, 2020 at 17:33
  • $\begingroup$ I would appreciate a reference to H. C. Brown's suggestion. $\endgroup$
    – user55119
    Commented Sep 18, 2020 at 18:15
  • $\begingroup$ @Zhe It's done now. Can you please verify the linked question? $\endgroup$
    – Apurvium
    Commented Sep 19, 2020 at 2:28
  • $\begingroup$ @Apurvium Sorry, but I don't see a fix, and the linked question has the same problem. The product from step 1 does not correspond with the arrows (the product based on arrows are the allylic radical, hydrogen radical, and chlorine radical, with no new bond formed between chlorine and hydrogen). $\endgroup$
    – Zhe
    Commented Sep 19, 2020 at 10:58

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