This reaction always occurs in the following, regiospecific way:

$$\ce{CCl4 + R-CH=CH2 ->[$h\nu$] R-CHCl-CH2-CCl3}$$

The same should apparently also happen when there is a carbon that has two substituents (e.g. methyl groups) and yield only one major product:

$$\ce{CCl4 + (CH3)2C=CH2 ->[$h\nu$] (CH3)2CCl-CH2-CCl3}$$

I believe the initiation step goes like this:

$$\ce{CCl4 ->[$h\nu$] CCl3^. + Cl^.}$$

Can't both of these radicals, that form in the initiation step, react with the double bond and form a secondary or tertiary radical carbon? Why does only $\ce{CCl3^.}$ apparently attack the double bond?

  • $\begingroup$ Is there any other information missing from this problem or in the wording of it? I would have expected both of the radicals to react significantly, although I would expect the CCl3 radical to be a less stable intermediate than the Cl radical. $\endgroup$ – airhuff Jan 25 '17 at 19:54
  • $\begingroup$ Is it not just a steric effect making the $\ce{CCl3\cdot}$ react more slowly and so the $\ce{Cl\cdot}$ wins out at the more crowded site? $\endgroup$ – porphyrin Jan 26 '17 at 14:53

While this reaction is highly regioselective, it is not 100% regioselective. Consider the nearly identical reactions:

$$\begin{align} \ce{CCl4 + H2C=CH-(CH2)5CH3 ->[(PhCO2)2][90\text{-}105ºC] Cl3C-CH2-CHCl-(CH2)5CH3} \hspace{.63cm} \text{75%}^{[1]}\\ \ce{CCl4 + H2C=CH-COH-(CH3)2 ->[(PhCO2)2][80ºC] Cl3C-CH2-CHCl-CH(OH)-(CH2)5CH3} \hspace{.63cm} \text{70%}^{[2]}\\ \end{align} $$

Granted these reactions use radical initiators instead of light, but that wouldn't make the difference between 100% and 70-75% regioselectivity. The reality is that at the start of the reaction, $\ce{.CCl3}$ and $\ce{.Cl}$ are both competing to attack the alkene, and $\ce{.Cl}$ does in fact attack faster. Looking at the activation energies for the abstraction of hydrogen atoms by different radicals, we gain insight into their relative stability:

$$ \small \begin{array}{lcc} \hline \text{Radical} & \ce{CH3-H} & \ce{CH3CH2-H} & \ce{(CH3)2CH-H} & \ce{(CH3)C-H}^{[3]} \\ \hline \ce{.F} & \text{1-1.5} & \text{<1} & & \text{<1} \\ \ce{.Cl} & \text{3.4} & \text{1.1} \\ \ce{.Br} & \text{17.5} & \text{13.0} & \text{9.5} & \text{6.9} \\ \ce{.CH3} & \text{14.0} & \text{11.6} & \text{9.6} & \text{8.1} \\ \ce{.CF3} & \text{10.9} & \text{8.0} & \text{6.5} & \text{4.9} \\ \ce{.CCl3} & \text{17.9} & \text{14.2} & \text{10.6} & \text{7.7*} \\ \hline _{^* \text{units are } \mathrm{kcal\ mol^{-1}}} \end{array} $$

As this reaction continues, however, chlorine atoms are abstracted from $\ce{CCl4}$, creating the product along with $\ce{.CCl3}$ radicals. Abstraction of $\ce{.CCl3}$ is highly disfavored due to steric effects. Thus, the radical chain reaction is as follows:

$$\ce{CCl4 ->[$h\nu$] .CCl3 + Cl.}$$

$$\ce{.CCl3 + CH2=CH-R -> Cl3C-CH2-\stackrel{\displaystyle .}{\ce{C}}H-R}$$

$$\ce{Cl3C-CH2-\stackrel{\displaystyle .}{\ce{C}}H-R + CCl4 -> Cl3C-CH2-CHCl-R + .CCl3}$$

The less than ideal yields of the radical addition of $\ce{CCl4}$ can be attributed to the strength of the $\ce{C-Cl}$ bond (higher activation energies correspond to slower rates of reactions). Because $\ce{CCl4}$ is in competition with the polymerization of the $\ce{.CCl3}$-alkene adduct, $\ce{BrCCl3}$ and $\ce{CBr4}$ are generally preferred for their higher selectivity, owed to the lower strength of the $\ce{C-Br}$ bond. One such example is shown below.

$$\ce{BrCCl3 + H2C=C(C2H5)2 ->[$h \nu$] Cl3C-CH2-CBr(C2H5)2} \hspace{1cm} {\text{91%}^{[4]}}$$

$^{[1]}$M. S. Kharasch, E. W. Jensen, and W. H. Urry, J. Am. Chem. Soc., 69, 1100 (1947).
$^{[2]}$P. D. Klemmensen, H. Kolind-Andersen, H. B. Madsen, and A. Svendsen, J. Org. Chem., 44, 416 (1979).
$^{[3]}$F. A. Carey, R. J. Sundberg, Advanced Organic Chemistry Part A: Structure and Mechanisms, Springer Science, New York, 2007, pg. 1034.
$^{[4]}$M. S. Kharasch and M. Sage, J. Org. Chem., 14, 537 (1949).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy