When talking about the example of the alkylation of toluene by chloromethane in the presence of $\ce{AlCl3}$, Hepworth, Waring and Waring (2002) mentioned that:

At room temperature, a mixture of 1,2-dimethylbenzene and 1,4-dimethylbenzene results, but at 80 °C the product is mainly 1,3-dimethylbenzene. In fact, heating either of the 1,2- or 1,4-isomers in the presence of $\ce {AlCl_3}$/$\ce {HCl}$ results in rearrangement to the more stable 1,3-dimethylbenzene.

This leads me to question: Why would the 1,3-isomer be more thermodynamically stable than the 1,2- and 1,4-isomers? Clearly, there must be an activation energy barrier to overcome as heat is required for the conversion. The methyl group being an ortho/para-director would lead to the formation of 1,2- and 1,4- isomers. Thus, wouldn't these isomers also be the more thermodynamically stable isomers?


Hepworth, J. D., Waring, D. R., & Waring, M. J. (2002). Aromatic Chemistry. United Kingdom: The Royal Society of Chemistry.

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    $\begingroup$ Methyl group being o/p directing is a kinetic property based on the energy of a transition state (or the energy of an intermediate by the Hammond postulate), whereas thermodynamic stability has to be based on the energy of the product. $\endgroup$ – orthocresol Jan 30 '18 at 11:42
  • $\begingroup$ All of these Friedel-Crafts reactions are reversible. Hence, if the reaction is run for a long enough time, the thermodynamic product (m-xylene) will predominate over the kinetic products (o- and p-xylene). It is curious though, that m-xylene appears to be slightly more stable that p-xylene. Why that is, is a good question. $\endgroup$ – ron Jan 31 '18 at 1:49

The thermodynamic stability of m-xylene over o-xylene or p-xylene can be deduced by considering the hyperconjugative effect. More specifically, resonance forms with a positive charge on methyl-substituted carbons are more important than resonance forms with a negative charge on them, as the methyl group stabilises the positively charged carbon by hyperconjugation.

In the o- or p- isomers, the hyperconjugation of one methyl group produces negative charge at ortho and para position where another methyl group resides. This is highly destabilising. But, in the m-isomer, hyperconjugation does not destabilise the structure and both methyl groups' hyperconjugation effects act in a cooperative way with each other.

Resonance structures of meta-xylene

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    $\begingroup$ @SoumikDas This may seem very basic, but I find the representation of hyperconjugation in your diagram a little weird. Traditionally, when we make hyperconjugation structures in alkenes, don't we break the C-H bond and shift the π bond one place to get a sort of allylic carbanion with the release of a proton? $\endgroup$ – Yusuf Hasan Nov 4 '18 at 15:54
  • $\begingroup$ @YusufHasan This is just one step away from breaking the C–H bond; all you need to do would be to detach that H+ from the CH3 group and make a double bond between the CH2 and the adjacent positive charge: $$\ce{H-CH2-C+ <-> H+$\cdots$CH2=C}$$ It doesn't affect the subsequent analysis that the negative charge is delocalised onto C-2 and C-4 of the benzene ring. $\endgroup$ – orthocresol Jun 8 '20 at 7:17
  • $\begingroup$ Can anyone please explain how does methyl groups' hyperconjugation works? $\endgroup$ – Divyansh Verma Apr 9 at 2:54

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