Let's assume we have propene and it's undergoing two different reactions: a Kharasch reaction and allylic substitution reaction.

In both reaction bromine radicals are formed, and the attacking particle is definitely bromine radical, but in one case it goes to the double bond and addition happens, in the other - it breaks C-H bond, making an allylic radical.

Why in these two similar reaction, where the most "powerful" piece is the Br radical, such different products are formed?

My guess is that even if an allylic radical is formed in the case of using HBr and peroxide, the first molecule it meets is HBr, which makes H add to the radical and frees the Br radical again, so no effective interaction happens in summary. And in the case of allylic bromination the Br radical cleaves the C-H bond, since it is like less stable in the allylic position, and the next molecule this radical meets is either NBS or Br2, depending on the reagent.

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    $\begingroup$ To form propylbromide in the above case, a hydrogen source is needed, while to form allylbromide in the bottom case, no hydrogen source is added. This should control, which of the two reactions is observed. $\endgroup$ Oct 14, 2017 at 9:33
  • $\begingroup$ @deusexmachina agree. Although Kharasch reaction sometimes involves using of H2O2 itself. Which is able to produce H radicals and OOH radicals. $\endgroup$
    – wolphram
    Oct 14, 2017 at 15:21
  • $\begingroup$ This answer will help. $\endgroup$
    – Apurvium
    Sep 19, 2020 at 7:20


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