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?

  • $\begingroup$ the reactions are discussed here for instance... chemguide.co.uk/organicprops/arenes/halogenation.html $\endgroup$ – MaxW Sep 14 '16 at 15:09
  • $\begingroup$ @MaxW I went through the link but I wanted to know will the double bonds be broken once all H atoms from CH3 are substituted? $\endgroup$ – Amritansh Singhal Sep 14 '16 at 15:16
  • $\begingroup$ Sorry, I was just trying to point out a source where the reaction conditions were discussed in more detail. I agree that the source i referenced doesn't have an answer. $\endgroup$ – MaxW Sep 14 '16 at 15:19
  • $\begingroup$ @MaxW It's ok. If you have any explanation, do tell $\endgroup$ – Amritansh Singhal Sep 14 '16 at 15:22

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:

  • abstract a hydrogen radical forming a phenyl ($\mathrm{sp^2}$-hybridised) radical
  • attack the aromatic ring generating a 6-chlorocyclohexa-1,3-diene-5-yl radical

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.

  • $\begingroup$ I very well understand the reaction you mentioned for toluene. But, in case of benzene, if the reaction intermediates have quite less stability , why do they react after all? (Though I understood why the aromaticity is lost if the reaction takes place). And yes, thanks for the answer. $\endgroup$ – Amritansh Singhal Sep 14 '16 at 16:31
  • 2
    $\begingroup$ @AmritanshSinghal Because the chlorine radical is a very reactive species, so it wants to react. It cannot react to form the more stable radicals, so it just chooses the least instable one. And once a single double bond of benzene is broken, it will continue to chlorinate all of that molecule’s double bonds since the stabilising aromaticity is gone. $\endgroup$ – Jan Sep 14 '16 at 16:35

Your Answer

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

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