Why does the light catalyzed radical addition of tetrachloromethane to an alkene produce only a single regioisomer?
While this reaction is highly regioselective, it is not 100% regioselective. Consider the nearly identical reactions:

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

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:

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

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:

$$\ce{CCl4 -&gt;[$h\nu$] .CCl3 + Cl.}$$

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

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

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.

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




$^{[1]}$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).