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

Question

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.

Answer 1

The 1,3 isomer must be more stable due to steric effects. I think you are mixing up kinetic and thermodynamic effects. The O,P products form faster because the reactive intermediate, the non-aromatic carbocation, is more stable. Formation of the carbocation intermediate is the rate limiting step.

But once the product forms, the molecules can't "remember" the carbocation intermediate. They only know how stable they are as final products. If the reaction is reversible, such as with higher temperatures, eventually an equilibrium is reached where the 1,3 product predominates.

I have never actually seen this discussed in an introductory class. Normally, they just want you to draw the O/P products. Other reactions are used to illustrate kinetic vs thermodynamic effects, such as addition to conjugated dienes.