A molecule of phenol is much more inclined to under go an electrophilic substitution reaction than a molecule of benzene because the $\ce{-OH}$ group is highly reaction favoring. From what I understand, this is because when the $\ce{-OH}$ group is attached, it contributes a resonance structure involving a double bond between the oxygen and the ring, which stabilizes the intermediate arenium ion.

However, from what I also understand, oxygen is also a highly electronegative atom and therefore inductively draws electron density away from the benzene ring. Despite these two competing effects (electron donating via resonance and electron withdrawing via induction), the overall effect of the $\ce{-OH}$ group is electron donating. This also applies to $\ce{-NH2}$ group.

Am I right in concluding that resonance effects are stronger than inductive effects, if not in general, then at least for electrophilic substitution of substituted aryls?

  • $\begingroup$ Considering benzoic acid, I would say no because the carbon of $-COOH$ can freely rotate but still be in its plane, but still we say that it is a meta directing group. When I do get the time I will post my answer. $\endgroup$
    – Rohinb97
    Commented Feb 4, 2015 at 13:17

3 Answers 3


Often, but not always, mesomeric displacement leads to a shift in prevailing over the inductive effect.

Pi-electrons are at the peripheral orbitals. Association with the nucleus is less strong than at the sigma-electrons. The ionization potential of pi-electrons is smaller and chemical bond is more polarizable. Therefore, the dipole moment associated with the mesomeric effect can prevail over the dipole moment associated with the inductive effect.

However, there are groups in which the inductive effect prevails, such as halogens.

Also, you can read about hyperconjugation. Hyperconjugation can have influence to the mesomeric and inductive effects too.

Source: Reutov O.A., Kurts A.L., Organic Chemistry, vol. 1 of 4, ch. 2.2.2, MSU

  • $\begingroup$ thank you for the informative answer. However, my prof does not believe in the hyperconjugative effect and only considers the inductive effect - which often seems to have the same effect as the hyperconjugative effect. Also can't the hyperconjugative effect be lumped under resonance/mesomeric effects? $\endgroup$
    – Dissenter
    Commented Feb 4, 2015 at 18:07
  • 1
    $\begingroup$ @Dissenter This effect (hyperconjugation or sigma,pi-conjugation) - something between a mesomeric and inductive effects. It doesn't have a strong influence in bond C-H, but bond between carbon and metals we can't describe without it. Mesomeric effect of halogens is very different from all other substituents $\endgroup$ Commented Feb 4, 2015 at 18:19

Am I right in concluding that resonance effects are stronger than inductive effects, if not in general, then at least for electrophilic substitution of substituted aryls?

No, this is not a general phenomenon.

Consider the case of electrophilic aromatic substitution of fluorobenzene. Fluorobenzene undergoes electrophilic nitration roughly 10 times slower than benzene, yet it strongly directs ortho and para. Nitration of fluorobenzene produces primarily ortho (13%) and para (86%) product, only a trace of meta product is observed.

The fact that that substitution occurs primarily at the ortho and para positions signals that the fluorine has a +R effect. The fluorine is donating electrons through resonance into the benzene ring - but only to the ortho and para positions; resonance structures cannot be drawn that donate electron density to the meta position (this applies to both the ground state and transition state resonance structures).

Fluorine is much more electronegative than hydrogen, so inductively fluorine will remove electron density from the ring (-I effect).

Since the reaction rate for electrophilic nitration is decelerated in fluorobenzene compared to benzene, this indicates that there is less stabilization (less electron density to stabilize the positive charge) of the transition state in the fluorobenzene reaction than in the benzene raection. If the fluorine substituent is donating electron density through resonance, but overall the electronic stabilization has decreased, then the -I effect must overwhelm the +R effect in this case.

There is a continuous rebalancing of inductive and resonance effects as we move through the series benzene, toluene, dimethyl aniline, anisole and fluorobenzene. With dimethyl aniline and anisole, the resonance effects far outweigh the inductive effects, but by the time we reach fluorine the inductive effect outweighs the resonance effect.

  • $\begingroup$ what about cases other than halogenated benzenes? I believe that the general trend of resonance > inductive effect holds for these cases (at least according to my prof). $\endgroup$
    – Dissenter
    Commented Feb 3, 2015 at 18:30
  • 1
    $\begingroup$ Yes, the general trend holds for the common groups like alkyl, OR and NR2. But obscure groups like -SR (maybe when R=CF3), pentafluorophenyl, etc. may also violate the "general" trend of resonance effects outweighing inductive effects. $\endgroup$
    – ron
    Commented Feb 3, 2015 at 20:22
  • $\begingroup$ "resonance structures cannot be drawn that donate electron density to the meta position" ... how do resonance structures correlate with reality? Why can't more e- density be delocalized to the meta position? Aren't all the p-orbitals connected anyway? $\endgroup$
    – Dissenter
    Commented Feb 5, 2015 at 7:39
  • 2
    $\begingroup$ @Dissenter 1) Resonance structures correlate pretty well with reality, that's why they still find use. 2) All p-orbitals are connected, but some have very small coefficients in a given molecular orbital, effectively disconnecting them. Perhaps you can run an MO calculation on toluene at school, I suspect some orbital(s) close to the HOMO would have smaller or oppositely signed coefficients for the meta carbons. $\endgroup$
    – ron
    Commented Feb 5, 2015 at 15:01
  • $\begingroup$ What's the physical reasoning behind small coefficients? $\endgroup$
    – Dissenter
    Commented Feb 5, 2015 at 15:16

(I thought the previous answers were incomplete, mainly because previous answers have talked about halo arenes, particularly about fluorobenzene)

Your question reminds me of a doubt I had once. Consider the two structures shown below:enter image description here

Can I say these two are geometrical isomers? It sounds a pretty dumb question, but here is the thing. One could argue that whenever we show that the $-COOH$ group is meta directing, we draw resonating structures that have a double bond between the carbons of the $-COOH$ and the carbon of benzene to which the group is joined to. Therefore one could say that the rotation of the carbon is restricted and hence are geometrical isomers.

But the thing is, that the rotation of the carbon in $-COOH$ is actually free(not exactly free, since it has to be in the same plane as the rest of the benzene, since it has $sp^2$ hybridization which confines it in a plane) and the $\pi e^-$ of the $-COOH$ group remains with it and doesn't get localized in the benzene ring. But this leads to a very natural question here- then how can we explain that $-COOH$ is meta directing? By inductive effect. Or at least that is what my teacher said.

-R (or -M) is shown by a group if a more electronegative element is attached by a double bond directly then it decreases electron density at ortho and para positions wrt meta and hence the group is deactivating. Considering the case of benzoic acid, electronegativity of oxygen plays a vital role , mainly because it attracts the $\pi e^-$s of the ring and decreases electron density over the ring. Although we can say that resonance does take place (which doesn't contribute much to explain why $-COOH$ is meta directing, as explained above), but the more contributing reason here is that oxygen pulls the electron density towards itself making the ring deactivated. So therefore in this case inductive effect plays a vital role here, even more important than resonance.

It could also be explained as to why $-COO^-$ is activating by the above. Since a -ve charge over a very electronegative element (oxygen) makes it very stable and kinda "calms it down" from attracting more electrons, therefore oxygen won't attract the electrons of benzene so much therefore relaxing the electron density of benzene a little bit and hence make it ortho and para directing.

Hence, as we can clearly see the electronegativity of oxygen here plays a vital role in making $-COOH$ meta directing and $-COO^-$ ortho and para directing. This is the best example to disprove your conclusion because here the overall effect by the oxygen is attracting electrons demonstrates a case where inductive effect is more dominant than resonance.

  • 1
    $\begingroup$ I fail to see how these two molecules you draw are different and how they refer to the point you are making, especially when you consider that the carboxylic group to be freely rotatable (It's about 7 kcal/mol, so this is an acceptable assumption). However, in its minimum conformation, the $\pi$ system is of course delocalised over the whole molecule, withdrawing more electron density from the ring. $\endgroup$ Commented Feb 9, 2015 at 13:07
  • $\begingroup$ ^ That is my exact point @Martin. The carboxylic group is freely rotatable. When we draw the resonant structures of the system and show the delocalisation of the $\pi e^-$ of the carboxylic group over the ring, there is a $\pi$ bond b/w the carbon and benzene, and one could say that the structures I drew are geometrical isomers. But they aren't. There is no difference between the structures I drew. I explained it in my answer. This was a doubt I had before. $\endgroup$
    – Rohinb97
    Commented Feb 9, 2015 at 13:14
  • $\begingroup$ But how does that explain that the carboxylic group is meta directing? If it was only the inductive effect, then clearly the COOH must be drawing a lot of electron density from the $\sigma$ system. Hence the para position would be least affected by this group. $\endgroup$ Commented Feb 9, 2015 at 13:27
  • $\begingroup$ @Martin I have also addressed this in my answer. To some extent, the $\pi$ electron of the carboxylic group does go in the ring by resonance. It is just that the more dominant reason for the $-COOH$ group to be meta directing is the pulling of $e^-$s by the oxygens. I do not deny to the fact that the $\pi$ electron of the carboxylic group doesn't get delocalised. $\endgroup$
    – Rohinb97
    Commented Feb 10, 2015 at 8:10
  • $\begingroup$ And who down voted this? What is wrong here? I'm open to your suggestions. $\endgroup$
    – Rohinb97
    Commented Feb 10, 2015 at 8:12

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