# Reaction between RCH2=CH2 and CCl4 — Why does it form RCHClCH2CH(R)CH2CCl3 along with RCHClCH2CCl3?

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?

• 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. – NotEvans. Jul 24 '16 at 11:47
• @NotNicolaou Yes, it is given just above that and I have read it but still I don't get it. – Kartik Jul 24 '16 at 11:53