I am reading Organic Chemistry from Morrison & Boyd and there is this problem in it:

Problem 6.8 From the addition of $\ce{CCl4}$ to alkenes (in presence of peroxides), $\ce{RCH2=CH2}$, there is obtained not only $\ce{RCHClCH2CCl3}$ but also $\ce{RCHClCH2CH(R)CH2CCl3}.$ Using only the kind of reactions you have already encountered, suggest a mechanism for formation of this second product.

I know that the addition of CCl4 to alkenes in presence of peroxides is a free radical reaction. So what I have thought is, while the reaction is going on, there are the following species present: RCH=CH2 (reactant), RC^HCH2CCl3 (^ represents the electron on the free radical) and the product $\ce{RCHClCH2CCl3}$.

When the free radical reacts with $\ce{RCH=CH2}$ it gives the main product $\ce{RCHClCH2CCl3}$ and $\ce{.CCl3}$. So probably the second product is formed by reaction of $\ce{.CH(R)CH2CCl3}$ with $\ce{RCHClCH2CCl3}$?

$$\ce{RCHClCH2-CCl3 + .CH(R)CH2CCl3 ->[???] RCHClCH2CH(R)CH2CCl3 + .CCl3}$$

However this reaction I have never encountered while the question explicitly states that I have to use only the kind of reactions I have already encountered. I don't think anyway that this reaction is possible.

There must be something big that I am missing here ... Can someone give me a little hint?

  • $\begingroup$ Is the chain mechanism not given directly above the question on the same page? Using that you should be able to rationalise where an issue of selectivity might arise to give the various product. $\endgroup$
    – NotEvans.
    Jul 24, 2016 at 11:47
  • $\begingroup$ @NotNicolaou Yes, it is given just above that and I have read it but still I don't get it. $\endgroup$
    – Kartik
    Jul 24, 2016 at 11:53
  • $\begingroup$ google://radical+polymerization $\endgroup$
    – permeakra
    Jul 28, 2016 at 6:30
  • $\begingroup$ @permeakra I've searched google but nowhere is this product of the reaction mentioned. If you know the answer why don't you post it? $\endgroup$
    – Kartik
    Jul 28, 2016 at 10:38
  • 3
    $\begingroup$ @Kartik hint: alkene does not care much, which radical will attack it, as long as it is active enough. Alkyl radicals in general are quite active. $\endgroup$
    – permeakra
    Jul 28, 2016 at 14:06

1 Answer 1


The propagation series starts with the addition of the trichloromethyl radical to the alkene to give the more stable secondary radical. This radical can abstract a chloride from carbon tetrachloride to give the expected product. Alternatively, this radical can add to the alkene to give an a new carbon-centered radical with an extended chain. Abstraction of chloride gives the 'unexpected' product.


The key realization here, as permeakra suggested in the comments, is that any radical in solution can theoretically add to the alkene. This means that the penultimate radical intermediate could add to the alkene as well extending the chain. This is the basis of a radical polymerization.

  • $\begingroup$ Thank you. I had found the answer myself a few days ago but I didnt have internet at that time so I couldnt reply. $\endgroup$
    – Kartik
    Aug 2, 2016 at 12:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.