The reaction of benzene with $\ce{Cl2}$ in presence of sunlight breaks the double bonds and 6 chlorine atoms attach to the ring.
However, the same reaction done with toluene does not lead to removal of the double bonds. Instead, the $\ce{H}$ atom of $\ce{CH3}$ is substituted.
Can somebody please explain why does this happen. If more amount of $\ce{Cl2}$ is taken in sunlight and the reaction is carried out for long time, will the double bonds break?
Both halogenations can be, according to procedures on SciFinder, performed at ambient temperature, simple sunlight and in reasonable reaction times.
The first step in both transformations is the formation of chlorine radicals by photoinduced homolytic cleavage of the $\ce{Cl-Cl}$ bond (equation $(1)$).
$$\ce{Cl-Cl ->[$h \nu$] 2 Cl^.}\tag{1}$$
These chlorine radicals want to react with something to lose their radical configuration. In the case of toluene, the kinetically favoured attack is that generating the stabilised benzyl radical as shown in equation $(2)$.
$$\ce{Ph-CH3 + Cl^. -> Ph-CH2^. + HCl}\tag{2}$$
This benzyl radical can then react with another chlorine molecule forming benzyl chloride (equation$(3)$), and so on in two further steps to form trichloromethylbenzene (equations not shown but analogous to $(2)$ and $(3)$).
$$\ce{Ph-CH2^. + Cl2 -> Ph-CH2-Cl + Cl^.}\tag{3}$$
Once we have reached the fully chlorinated methyl group, we can again imagine a chlorine radical looking for a reaction partner. Again, we should ask us which the favoured attack is — the one leading to the most stable products. And rather than attacking the aromatic ring, the benzylic carbon can again be attacked as shown in equation $(4)$:
$$\ce{Ph-CCl3 + Cl^. -> Ph-CCl2^. + Cl2}\tag{4}$$
Thus, we are ‘destroying’ our product because this attack is still better (leads to more stable products) than attacking the aromatic system.
Why does benzene react in the way it does? Well, equation $(1)$ is identical. However, there is no benzylic group; there are only two possibilities how a radical can react with benzene:
The second reaction is clearly favoured over the first, since while it does break the aromatic system, it still generates a p-orbital centred pentadienyl-type radical which is much more stabilised than a radical in an $\mathrm{sp^2}$ orbital.
However, toluene or its reaction products will hardly ever go this route since a benzyl radical is still much more stabilised than a pentadienyl-type one. The minute quantities of toluene products that do chlorinate this way will be fully chlorinated, but their yields will be very insignificant.
