# Thermodynamic stability of meta-xylene over ortho- and para-isomers

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?

Reference

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

• 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. – orthocresol Jan 30 '18 at 11:42
• 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. – ron Jan 31 '18 at 1:49