Why methyl group is 2,4-directing? I am an A level student and this isn't required at this level. I am looking for a simple explanation which a high school student can understand.

  • $\begingroup$ Do you understand the rationale for other directing groups? $\endgroup$
    – jerepierre
    Commented Nov 14, 2014 at 16:30
  • $\begingroup$ To solve this problem, draw the intermediates of reaction. It will be obvious at that point-- or otherwise for the basis for a useful discussion. $\endgroup$
    – Lighthart
    Commented Nov 14, 2014 at 23:23

1 Answer 1


Why methyl group is 2,4-directing?

When toluene undergoes electrophilic aromatic substitution the products are primarily the ortho and para isomers; usually only a small amount of the meta isomer is produced. In order to explain this observation there are two effects to consider, inductive and resonance effects.

Inductively the methyl group releases electron density into the benzene ring. This is because the methyl group, being $\ce{sp^3}$ hybridized, is less electronegative than the $\ce{sp^2}$ hybridized aromatic carbon. The carbon with more s-character is more electronegative, electron density likes to flow from low to high s-character orbitals since the more s-character in an orbital, the lower its energy. If inductive arguments were most important we would expect the ortho isomer to predominate followed by the meta and para isomers in that order. This is not observed, the meta isomer is only a minor isomer.

Resonance arguments are based on the concept of hyperconjugation involving the $\ce{C-H}$ bonds in the methyl group. The following figure shows some of the possible hyperconjugative resonance structures for toluene.

enter image description here

Additional resonance structures can be drawn using the other methyl $\ce{C-H}$ bonds as well. As the resonance structures show, electron density is increased at the ortho and para positions, not the meta. This hyperconjugative resonance effect makes the ortho and para carbons more susceptible to electrophilic attack.

Hyperconjugation explains how alkyl groups can be electron releasing in a resonance sense. Hyperconjugation explains the mild activating effect of an alkyl group in electrophilic aromatic substitution and its ortho-para directing effect.

  • $\begingroup$ I find this argument is a lot more convincing when applied to the EAS intermediate, not the putative hyperconjugated-dissociated zwitterion. $\endgroup$
    – Lighthart
    Commented Nov 14, 2014 at 23:22
  • 2
    $\begingroup$ @Lighthart Why should the argument be more or less convincing if we apply it (hyperconjugation) from the starting material or the intermediate side? Hyperconjugation is applied to other neutralmolecule phenomenon like the rotational barrier in ethane and why more highly substituted alkenes are more stable. $\endgroup$
    – ron
    Commented Nov 14, 2014 at 23:44
  • 2
    $\begingroup$ Hyperconjugation is too quantum mechanicky for mere mortals to grasp easily. I have found students grasp the energy of the resulting cation intermediate more quickly. So the people I am convincing more easily are novices. Your explanation is quite good, but very sophisticated. $\endgroup$
    – Lighthart
    Commented Nov 17, 2014 at 19:52
  • $\begingroup$ @ron But I learnt that hyperconjugation occurs when there is partial overlap of C-H sigma bond and adjacent empty-p orbital of carbon. However your structures suggests to contradicts this as the C-H bond is fully broken, C=C bond is completely formed and a unit positive charge is involved in the resonance? $\endgroup$ Commented Aug 1, 2022 at 13:30
  • 1
    $\begingroup$ @An_Elephant What you say is correct, but what I've drawn is just a resonance structure. It is an extreme form with charge separation, but it is meant to indicate that the $\ce{C-H}$ bond donates electron density into the ring (e.g. hyperconjugation). $\endgroup$
    – ron
    Commented Aug 2, 2022 at 14:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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