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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?

M.A.R.
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Tyler
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  • 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. – Rohinb97 Feb 04 '15 at 13:17

3 Answers3

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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

Alexander SHevyakov
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  • 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? – Dissenter Feb 04 '15 at 18:07
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    @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 – Alexander SHevyakov Feb 04 '15 at 18:19
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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.

ron
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  • 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). – Dissenter Feb 03 '15 at 18:30
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    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. – ron Feb 03 '15 at 20:22
  • "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? – Dissenter Feb 05 '15 at 07:39
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    @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. – ron Feb 05 '15 at 15:01
  • What's the physical reasoning behind small coefficients? – Dissenter Feb 05 '15 at 15:16
  • @Dissenter The wave function is telling us that there is not much charge (or opposite sign charge, depending upon the sign of the coefficient) at that position. – ron Feb 05 '15 at 16:02
  • How can we rationalise the relative lack of electron density at the meta position without resonance structures and save functions – Dissenter Feb 05 '15 at 16:30
  • @Dissenter You can't. Chemists brominated toluene and found that the o- and p-positions were substituted. They repeated the experiment with nitrobenzene and found that the meta isomer predominated. They realized that inductive effects alone could not explain their observations, so they looked further and applied\developed resonance theory. – ron Feb 05 '15 at 16:35
  • Who downvoted this answer? I gave this answer the bounty because I thought it was very through and well-written; if the downvoter has any comments/suggestions I'd be open to hear them (and I'm sure ron would be too). – Dissenter Feb 10 '15 at 00:01
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(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.

Rohinb97
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    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. – Martin - マーチン Feb 09 '15 at 13:07
  • ^ 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. – Rohinb97 Feb 09 '15 at 13:14
  • 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. – Martin - マーチン Feb 09 '15 at 13:27
  • @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. – Rohinb97 Feb 10 '15 at 08:10
  • And who down voted this? What is wrong here? I'm open to your suggestions. – Rohinb97 Feb 10 '15 at 08:12
  • Well, I did not downvote, but It is really hard to follow your train of thought. I still do not understand what your point is. And don't be offended by downvotes, I get a couple of them every now and again. – Martin - マーチン Feb 10 '15 at 08:29
  • Did you at least get my point @Martin ? Then you might help and edit the question. – Rohinb97 Feb 10 '15 at 08:42
  • I downvoted for two reasons: (1) The compounds shown do not have ortho, meta, and para positions. They only have four positions. (2) Resonance DOES account for why a carbonyl is a meta director. Both resonance and inductive effect play a role. The question is looking for instances where resonance and inductive effect are competing. The carboxylate analog could be an effective example, but that hasn't been demonstrated in this answer. If that can be explained more clearly, and the irrelevant part about the diacid removed, it could be a useful answer. – jerepierre Feb 10 '15 at 20:45
  • I guess you didn't get my answer. I had a doubt in which I learned that the rotation of the carbon in carboxylic group is actually free and not restricted because the pi electron of the carbonyl group DOESNT delocalised in the benzene ring. This is a hard fact to digest and that is why I used my doubt as an example to prove my point. Moreover, when I finished explaining my doubt and why was I wrong, then I was talking about benzoic acid (I was referring to ortho, meta and para positions in benzoic acid ((1))). – Rohinb97 Feb 11 '15 at 04:34
  • Although I agree to the fact that resonance does contribute to carbonyl being meta directing, but the more DOMINANT factor for it being deactivating and $-COO^-$ being o-p-directing is inductive effect and not resonance. Here the effects aren't competing with each other, but the inductive effect is more dominating here. It wasn't necessary to give only an answer which showed inductive and resonance effects competing with each other. I do hope that now I'm clear enough @jerepierre ? – Rohinb97 Feb 11 '15 at 04:41
  • @Rohinb97 The digression about the diacids greatly confuses your answer. I don't argue that the carbonyl pi bond will not donate electrons to the aromatic ring. The relevant resonance structure (to explain the carbonyl as a meta-director) is the ring donating to the carbonyl. Just because there is free rotation doesn't mean that the resonance contribution is zero. In any case, what is the evidence that inductive effect is dominant over resonance? The question poses groups with opposite effects. We can look at the product ratios to determine which effect is dominant. – jerepierre Feb 11 '15 at 16:14
  • I'm sorry if I wasn't clear enough. But I have edited my answer and I hope you do get my point. And to answer the above, the statement that the ring donates the electrons is technically wrong, it's the carbonyl group which attracts it, but anyways, my point is that the inductive effect is more dominant than resonance effect because the pi electron of the carbonyl group doesn't get delocalised in the ring (because if it had, then the above structures drawn would've been geometrical isomers) and hence resonance takes place in a small amount. But still we say that $-COOH$ is meta directing. Why? – Rohinb97 Feb 11 '15 at 21:04
  • Because the oxygens attract the electron density of the ring and therefore makes it deactivating. Since resonance does take place(very little), it increases electron density at meta wrt ortho and para and hence carbonyl is meta directing. The above can also be supported by the fact that $-COO^-$ is an activating group. A -ve charge over an electro-ve element makes it more stable and decreases it's electro-vity to some extent. Hence the major effect responsible for carboxylic being meta directing is nullified, and hence is ortho and para directing. – Rohinb97 Feb 11 '15 at 21:10
  • I've written my whole answer again in these comments now @jerepierre. I do hope you get the point I've been trying to make. – Rohinb97 Feb 11 '15 at 21:12
  • @Rohinb97 As stated in the question, the hydroxyl group is inductively withdrawing and resonance donating. Based on the product distribution of an EAS reaction of phenol, we can see that resonance dominates over the inductive effect. You assert that the directing effect of -COOH is primarily due to inductive effect because of free rotation which disrupts resonance to the ring. However, the C-O bond in phenol is freely rotating as well, yet the resonance effect still dominates. This tells me that free rotation alone is not enough to conclude that inductive effect can dominate over resonance. – jerepierre Feb 11 '15 at 21:54
  • @jerepierre I guess you are right. Didn't think this. Thanks. But it is a fact that resonance doesn't take place (so much) in benzoic acid. The pi electron is tightly held up by the two oxygens and is in resonance with the lone pairs of the next oxygen. The pi electron of the carboxylic group doesn't delocalise in the ring. But still it is meta directing. Inductive effect is more dominant in this case. – Rohinb97 Feb 12 '15 at 14:13