From what i have learnt, kharasch Peroxide effect is only applicable for HBr. The reason for this is for the other halogens, HCl and HI the propagation steps in the radical mechanism are endothermic. Now, can't this problem be solved by heating? Theoretically, then, Peroxide effect could be shown by HCl and HI in presence of both heat and sunlight ( required for radical mechanism)

  • $\begingroup$ There are two propagation steps, only with $\ce{HBr}$ are they both exothermic. Take a look at the numbers in this earlier answer. $\endgroup$
    – ron
    Jan 28 at 14:09

1 Answer 1


If the reaction is endothermic, then it cannot be proceeded by just applying heat.

$\ce{HCl}$ is a very stable acid with $\ce{H-Cl}$ much stronger than $\ce{H-Br}$ bond and hence not broken asymmetrically by the free radical generated by peroxide. Moreover, chlorination is not that selective when compared to bromination. In this case, non radical $\ce{HCl}$ are forming Markovnikov products before the radicals come to form the anti-Markovnikov products.

With $\ce{HI}$, the first propagation step is endothermic, because $\ce{H-I}$ bond is weak and it undergo homolysis readily to form free iodine radical which in turn combines to form iodide rather than getting added to the ethylenic bond.


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