Reaction of toluene at high temperatures gives us (o-m)toluene; whereas at normal conditions of electrophilic attack, it gives us (o-p) directing.

Why does this happen at high temp although as methyl group is activating and o-p directing?

  • 5
    $\begingroup$ I could imagine that at high temperatures the higher activation energy for reacting at the meta position could be overcome by the higher number of meta positions (2) available for the reaction compared to the para position (only 1). But that is just a guess... $\endgroup$
    – Philipp
    Jul 22, 2013 at 17:21
  • $\begingroup$ What specific reaction are you talking about?' $\endgroup$
    – permeakra
    Jul 23, 2013 at 7:43
  • $\begingroup$ Alkylation at high temp of toluene $\endgroup$ Jul 23, 2013 at 8:05

1 Answer 1


Assuming you're talking about Friedel-Crafts alkylation of toluene, this can be explained by the reaction being reversible and thermodynamic control taking precedence at sufficiently high temperatures. The directing ability of substituents in electrophilic aromatic substitution reactions is a function of the reaction being under kinetic control at lower temperatures. According to Fleming's book on MO theory, a single electron-donating substituent on benzene conjugated to the ring's $\pi$ system disrupts the normal degeneracy of the $\psi_2$ and $\psi_3$ molecular orbitals of unsubstituted benzene. Comparing the intermediates of an EDG-substituted benzene undergoing electrophilic attack, the HOMO of the $\pi$ system, $\psi_3$, is higher in energy when attack occurs at the meta- position than when it occurs at the ortho- or para- positions. Using the concept of resonance, this can be illustrated in an abstract sense by considering the resonance structures of the intermediates resulting from attack at the various positions; an additional resonance structure exists for ortho- and para- attack if the substituent is conjugated to the $\pi$ electron system of the ring.

With alkyl groups, the electrons are donated by hyperconjugation, which is a weaker effect (and not apparent from resonance structures), but nonetheless has a similar impact on regioselectivity and the $\pi$ molecular orbitals. Additionally, the coefficients on the atoms of the HOMO of the $\pi$ system are lowest at the meta positions when an electron donating group is present (including alkyl groups), indicating electron density is higher at the ortho- and para- positions.

(I should clarify, of course, that resonance is not an actual physical process/phenomenon, merely an abstraction which provides a qualitative picture of the extent of electron delocalization and which often correlates well with the actual distribution of electron density in the molecular orbitals.)

Ultimately, at low temperatures, when the reaction is kinetically controlled, the favored pathways are the ones in which the transition states and intermediates are stabilized, leading to ortho-/para- substitution.

At higher temperatures, the preferred product is the thermodynamic one, and the advantage of meta- substitution is that there is less crowding and reduced steric hindrance in, e.g., a 1,3,5-trisubstituted benzene by comparison to a 1,2,4-trisubstituted benzene. Additionally, Friedel-Crafts is actually reversible at high temperatures. Even if the kinetic product forms initially, then it can decompose by dealkylation in the presence of Lewis-Acid catalysts and acid at high temperatures, at which point equilibrium can be established with the thermodynamic product.

According to March, experimental evidence suggests that methyl groups will mainly migrate intramolecularly in the isomerization reaction to give the thermodynamically preferred meta-substituted benzenes, while both intramolecular rearrangements and complete dealkylation with subsequent intermolecular reactions are possible with other types of alkyl groups.

  • $\begingroup$ Why is meta substituted benzene thermodynamically preferred product in reactions like friedel crafts and sulphonation nitration? $\endgroup$ Apr 16, 2022 at 5:58

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