In electrophilic aromatic substitution:

I have just learnt a 'rule' about benzene with more than 1 substituents attached (excluding the effect on steric hindrance):

  1. if a strong activating group (EDG) and a strong deactivating group (EWG) are attached, activating group wins hence the electrophile adds on ortho & para position relative to the activating group.

  2. if a weak activating group (EDG) and a strong deactivating group (EWG) are attached, deactivating group wins hence the electrophile adds on meta position relative to the deactivating group.

I understand rule 2 (strong must dominate the weak), but I do not understand what is the principle behind rule 1:


e.g. methoxide group wins over nitro group, although they are both strong EDG & EWG. I understand if the methoxide group is replaced by a methyl group, nitro group wins as $\ce{R-CH3}$ only has a much weaker inductive effect (compared to strong resonance effect). However, both methoxide and nitro groups use resonance to exert their respective directing effect, why activating groups always win?

  • $\begingroup$ Is there any information on the positions of EDG and EWG in the first case? As in, was it referenced using an example? $\endgroup$ – Safdar Aug 6 '20 at 13:15
  • $\begingroup$ @Safdar Do you mean the rule 1? The molecule shown in the photo (with methoxide & nitro) is the example I just came across $\endgroup$ – 234ff Aug 6 '20 at 13:18
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    $\begingroup$ Came across this on the related sidebar. Does this answer your question? Electrophilic substitution on disubstituted benzene rings $\endgroup$ – Safdar Aug 6 '20 at 13:21
  • $\begingroup$ @Safdar Thanks, I have read this post before posting this question. The solver tells us that a strong activating group has lone pair electrons for charge stabilization, hence it trumps over the deactivating group (the most-stable extra resonance structure). However, I don't think this is the sole factor, as the solver also says it maybe overall deactivating. $\endgroup$ – 234ff Aug 6 '20 at 13:31
  • $\begingroup$ @Safdar I began writing the answer but then got mixed. The neutralization of charge accumulation is observed at the same location when the substituents are meta. The only reason I can think of is simultaneous resonace structures cannot be constructed. If electron density increases at a point ortho or para then that same charge accumulation cannot be again decreased by the EWG at the meta position, by drawing resonace structures. Thus the M- effect is lost. And only I- effect occurs wrt the EWG. Does this seem somewhat correct? $\endgroup$ – Aditya Roychowdhury Aug 6 '20 at 15:58

I believe that if we ignore the inductive effects of the substituents for the time being, then the electron-donating group(which donates electrons via resonance effect) would manage to activate a few positions in one or more resonating structures, and the electron-withdrawing group can only take up this delocalized pair of $\pi$ electrons when it is present at exactly an ortho or para position where the $\ce{+M}$ effect operates, and the negative charge lies on the carbon alpha to the electron-withdrawing group.

It can be seen here that the negative charge in the resonance structures of $2,4$-dinitroaniline can be taken up by the nitro group only when the negative charge becomes alpha to it:

Resonance in 2,4-dinitroaniline

So the only way an electron-withdrawing group can "win" is if it placed against a very poor electron-donating group.

First, in the case of $\ce{-CH3}$ or halogens (who neither use $\pi$ conjugation for electron donation nor have an electron-donating nature), the electron density donated to the ring is unable to emerge in the intermediates without being delocalised by the EWG, or second when the EWG is present exactly at positions where the negative charge reaches by $\pi$ conjugation(in case of "strong" electron donators).

If the group donating via the $\pi$ bonds is even a relatively better donating group, then unless all the ortho and para positions are having electrons being withdrawn by the presence of electron withdrawing groups, there will always be resonance structures donated having some partial activation at ortho or para positions.

Hence, the electron-donating group "wins" in such situations unless there is no steric hindrance during the substitution taking place.

  • $\begingroup$ Thanks. In the presence of strong EDG & EWG, from resonance structures, the EDG donates electron density to ortho & para positions (those positions become more nucleophilic = activation) unless EWG is bonded to ortho/para, which withdraws the delocalized electrons. Therefore, in my case of meta-position nitro group, EDG trumps? $\endgroup$ – 234ff Aug 7 '20 at 2:54
  • $\begingroup$ And in the case of weak EDG & strong EWG, since the EDG only has an inductive effect only (no donation of electrons by resonance), EWG meta-directing wins due to the partial positive charge (less nucleophilic) from the resonance structures (resonance effect > inductive effect)? $\endgroup$ – 234ff Aug 7 '20 at 3:00
  • $\begingroup$ Yep, pretty much what you have said sums it up @234ff $\endgroup$ – Yusuf Hasan Aug 7 '20 at 8:00
  • $\begingroup$ OK thank you for the explanation $\endgroup$ – 234ff Aug 7 '20 at 8:01

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