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Mechanism of benzene nucleophilicity activation

  1. Apr 30, 2014 #1
    I have some questions how various substituents are transfering benzene conjugated (π) electrons (i.e. activating/deactivating). As usual, no chemist provides an explanation.

    1)As I see, the first important thing when forming a carbocation during electrophilic aromatic substitution is to find out the part of the conjugated positive charge on the carbon that is binding the substituent. This seems clear for me, as this answers the resonance isomers.

    2)As I see, the transfer of electron density just depends on the carbon positive charge. But there I'm not sure anymore. In benzene the carbon hybridization is to be (sp2), besides a substituent it could be maybe slightly crumpled. Thus, when (any part of) positive charge on the carbon, its hybridization shouldn't change, right?
    It seemed to me probable that the electron density transfer (during conjugation, hyperconjugation) occurs because of different degree of orbital overlap because of different charge on an atom...

    3)The second factor is the inductive effect. Wikipedia says a donation of electron density into conjugated system is via inductive effect. Does this mean a hyperconjugation? I have no idea how inductive effect could affect the electron density without hyperconjugation (e.g. with deactivating trifluorinemethyl R-CF3 the transfer from the benzene is greater with greater carbon positive charge).

    4)Also, when (sp2) substituent (for example R-NH2), the hyperconjugation shouldn't occur. Is it true?
  2. jcsd
  3. May 1, 2014 #2


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    I don't quite understand what you are asking in question 2.
    Concerning three, I got to the conclusion that the inductive effect is an empirical relation which may have different physical explanations in different cases.
    To 4, take in mind that an amine has a lone pair that can participate mesomerically.
  4. May 1, 2014 #3
    My question in two:
    In meta/(ortho,para) position there is different electron density transfer (i.e. different carbocation stability as a result). There is also different part of the positive charge on the binding carbon. The only explanation, why the ED transfer should differ in these cases, seemed to me due to different (binding carbon <-> substituent) orbital overlap due to different positive charge on the carbon. However, the orbitals in carbon cation (sp2) are not different from those in carbon atom (sp2) in aromatic ring. Or do they vary?

    There are the products of nitration:

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    Last edited: May 1, 2014
  5. May 1, 2014 #4
    I suppose the way these things are taught may have changed since I was at university, but I'll tell you what I remember from back then, which to me seemed a bit simpler to apply in real situations.

    1) Reactivity. For a benzene ring to be reactive toward electrophiles, you want it to be electron-rich. So any substituent that 'gives' electrons to the ring will favour aromatic electrophilic substitution (SEAr), and viceversa. I suppose you could say that you need to look at how the substituent affects the 'electron density' of the ring.

    This is what I interpreted as inductive effect. My understanding was that electronegativity played a major role. A substituent that wants to keep electrons close to itself will distort the electron cloud so that the ring becomes electron-poorer, and viceversa. So F deactivates the ring more than Cl, Cl more than Br, Br more than I.
    For more complex substituents, the story may be more complicated. e.g. OMe has an electronegative oxygen atom bound to the ring, but it also has two 'free' electron pairs and a methyl group that pushes further electrons into it, and overall it's found to be an activating substituent for SEAr.
    A substituent like CF3 is deactivating because with those 3 fluorine atoms, overall it 'prefers' to pull electrons to itself than to give them to the ring.
    I may be wrong, but I don't think this requires hyperconjugation. I think a substituent can affect the electron density of the ring just on the basis of which atomic nuclei constitute it, without the need for formally charged structures to get involved.

    2) Regiochemistry. The preferred position of substitution is determined by the relative stability of the reaction intermediates, i.e. the species where the electrophile and the benzene ring are linked but the proton hasn't left yet. An intermediate is more stable when you can draw more canonical resonance structures for it, because that means the positive charge can be spread over a larger area. Apparently that's because a high density of charge increases the energy of the molecule, i.e. it decreases its stability.
    Based on the energy of each intermediate, the distribution of kinetic energy of the molecules in the reaction mixture (which depends on the temperature), the presence of solvents and catalysts that may change the picture by stabilising different species, etc..., different proportions of each intermediate will form, resulting in different proportions of each final product (but bear in mind that SEAr is reversible in some cases - e.g. sulphonation).

    This was the basis of what I understood as mesomeric effect.
    An OMe (or NR2...) substituent will favour SEAr in ortho- and para- (over meta-) because it's only by putting the electrophile in those positions that you can use the electron pair of the OMe group to further stabilise the positive charge (i.e. you can draw one additional resonance structure). It's like putting a positive charge next to a negative one, it's a state of low potential energy.
    A substituent like NO2 does not 'activate' the meta position, mind, it only deactivates it less than the ortho- and para- positions, because in the latter cases you end up with resonance structures where the positive charge is at the carbon bearing the NO2 group. This is bad for the stability of the intermediate, because it's like putting two positive charges close to each other, it's a state of high potential energy.

    I'm not sure why you refer to hybridisation, again I may be wrong, but I thought that was essentially an automatic aspect of the process, which only depends on how many bonds there are to other atoms or electron pairs.

    Another thing that they used to teach is that, unlike the mesomeric effect that can 'reach' the whole ring, the inductive effect acts mostly at closer range. That means an electronegative atom will deactivate the ortho- positions more than the para-.

    Finally, note that it's become very frequent to use methods different from SEAr to attach substituents onto CH positions of aromatic rings, especially in heterocyclic chemistry and especially when the electrophile would be too weak to react with the aromatic ring as such, or when the activation of the electrophile would require too harsh conditions.
    Typically, metalation (abstraction of a proton by a very basic species like BuLi, iPrMgCl, LiN(iPr)2...) followed by addition of an electrophile (I2, RCHO, CO2...); and transition-metal catalysed 'CH activation', where a certain CH group in the aromatic ring reacts with an electrophile thanks to the presence of, say, a Pd complex.
    These are very good methods, they allow functionalisation under mild conditions and often with regioselectivity that is complementary to SEAr.

    I hope this helps. Maybe I'm not answering your questions directly, but I'm trying to give you a point of view that so far has proven sufficiently effective - at least in passing exams and job interviews ;O) .
    Last edited: May 1, 2014
  6. May 1, 2014 #5
  7. May 2, 2014 #6


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    I'd rather argued that the regiochemistry is different due to hyperconjugation. Evidently a resonance structure with a H+ on the CH3 group is much more favourable than a structure with an F+ in CF3.
  8. May 7, 2014 #7
    That makes me sense only half.

    With ortho/para substituent, I can draw a resonance isomer with three double bonds (with positive charge on CH3 hydrogens). Let's suppose that isomer is very stable, also its relative rate in the molecule is large, i.e. the molecule is stable. Other isomers posses only two double bonds, also they are less stable, with smaller relative rate in the molecule. The molecule is thus stable.
    With meta substituent, I can draw resonance isomers with at most two double bonds. It is the largest relative rate in the molecule, also the molecule is less stable. OK

    With ortho/para substituent, resonance isomer with three double bonds is unstable (small shielding of fluorines <-> large work on removing electrons from fluorines...) = small relative rate. The isomers with the largest relative rate have two double bonds.
    But, with meta substituent, there are also isomers with the largest relative rate having two double bonds, as well. It is not apparent for me that molecule should be significantly more stable in this case.

    Maybe in certain quantity (isomer with three double bonds has not so high energy), the ortho/para molecule could truly be significantly more unstable...

    What's your opinion?
    Last edited: May 7, 2014
  9. May 7, 2014 #8


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    But you can draw structures with F- and a positive charge on the ring when CF3 in ortho or para. Not a good starting point for a nucleophilic attack.
  10. May 7, 2014 #9
    I don't know what do you mean. I can do that also in meta position. The problem is I'm not already sure about resonance isomer's stability, as there are plenty similar of them. Though, the whole problem must be expressible by r. isomers if someone is aware of their energy.

    Anyway, one can imagine substituted aromatic carbocation. There is unevenly distributed positive charge in delocalized orbital. Larger part of it possess the carbon across from a substituent. It is also expressed by the resonance isomers (stable one of them is):

    If an activating substituent connected to:
    1) carbon with larger positive charge, more degree of electron density is transferred to the aromatic ring, probably making the molecule more stable (more electron density on the more electronegative atoms).
    2) carbon with smaller positive charge, lesser degree of electron density is transferred to the aromatic ring making the molecule less stable.

    If a deactivating substituent connected to:
    1) carbon with larger positive charge, lesser degree of electron density is transferred from the aromatic ring, probably making the molecule less stable (less electron density on the more electronegative atoms).
    2) carbon with smaller positive charge, more degree of electron density is transferred from the aromatic ring making the molecule more stable.

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    Last edited: May 7, 2014
  11. May 10, 2014 #10


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    If you have problems to work out the answer using resonance structures, you could also use alternative formulations from MO theory, like Fukui's frontier orbital theory.
  12. May 11, 2014 #11
    Can you help me with those resonance isomers? What is the stable deactivated r. isomer with meta substituent, that doesn't exist with ortho/para substituent?

    I found the stable activated r. isomer with ortho/para substituent:
    It truly cannot be done with meta substituent.

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  13. May 11, 2014 #12


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    I can only think of something like sub+ (no bond between sub and the phenyl ring), ##CF_3^-=## and a positive charge in the ring.
  14. May 12, 2014 #13
    Sorry to intervene again in this discussion.
    I hope I'm not missing the point, but the way they taught me this was the following.
    I attached a pdf with the resonance structures for an ortho and meta attack on toluene and trifluoromethylbenzene - I tried to make a jpg of the scheme, but apparently the system doesn't accept it.

    View attachment SEAr.pdf

    In an ortho attack on Ph-CH3 there are 3 resonance structures. One of these has the positive charge next to CH3 (A in the scheme). This can be seen as an extra stabilisation, because CH3 is a group that can donate electron density, so it 'doesn't mind' sitting next to a positive charge. You can use the hyperconjugation argument if you like, and add one more resonance structure (although, 5 bonds on a carbon atom...?), or sophisticated proofs from recent / advanced theories, but anyway, it's an experimental fact.

    In an ortho attack on Ph-CF3 there are 3 resonance structures. One of these has the positive charge next to CF3 (A' in the scheme). This can be seen as a destabilisation, because a very electronegative group doesn't want to sit next to a positive charge. Again, other people will prefer to justify this differently, but it's still an experimental fact.

    In a meta attack on Ph-CH3 there are still 3 resonance structures. BUT none of these has the extra stabilisation resulting from the positive charge sitting next to CH3 as in the ortho attack.
    This is why electrophiles substitute toluene in ortho more than in meta.

    In a meta attack on Ph-CF3 there are still 3 resonance structures. BUT none of these has the destabilisation resulting from the positive charge sitting next to CF3 as in the ortho attack.
    This is why electrophiles substitute trilfuoromethylbenzene in meta more than in ortho.

    When I was at university I considered this a sufficient explanation of the observed regioselectivities in SEAr reactions. I may be wrong, but it still sounds OK to me.
  15. May 12, 2014 #14
    Yeah, that is for me the best explanation so far. That's what I published above.
    CH3, CF3 probably attach their bonding electrons to the aromatic ring by hyperconjugation. It explains their interaction with delocalized orbital.
    Still, I have no clue to find out the electron density distribution in delocalized orbital (as a degree of number of electrons, atoms electronegativity, orbital overlap etc...). That would clarify the entire problem. At present, there is only a very abstract explanation that more deactivating substituent is more stable attached to a carbon with bigger electron density. I could argue that (CH3, CF3) interact with the whole delocalized orbital, so why should the binding carbon charge affect the transferred electron density if carbon hybridisation remain unchanged?
    Or, binding carbon charge does not affect transferred electron density via hyperconjugation, just via the inductive effect?
    Last edited: May 12, 2014
  16. May 12, 2014 #15
    I have pretty similar question now (hope I can post it here):
    Deactivated benzoic acid possess greater acidity, that seems OK.
    Acidity with ortho substituent is greater. My question:
    Is it only due to the inductive effect, which takes more effect if closer to the hydroxyl? Or does the degree of (NO2) conjugation also differs in those isomers?

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    Last edited: May 12, 2014
  17. May 14, 2014 #16
    I admit I have considerable trouble understanding your questions.
    You initially seemed to be after a practical explanation of regioselectivity in SEAr reactions, then you seemed to be more interested in a theoretical understanding of the electron density distribution. In either case, you basically rejected most 'rules' or justifications people provided to you, because you were never satisfied on the theoretical level, or weren't convinced enough, or found the explanations too vague, etc.

    In practice there are two things you need to consider.

    There are many applications that calculate the electron density in any molecule you want. We often use them at work.
    So if your problem is to make an electron density map for a molecule, *this* specific problem has long been solved.
    I believe these applications are based on sound principles from physical chemistry, so the results are probably correct.

    We are sure that electron density is non-homogeneous in molecules like toluene and PhCF3, and that 'asymmetry' is obviously due to where the substituent is placed.
    You don't even need to believe the theory or the software that makes the ED map: we have direct experimental evidence of it.
    You could say that the results of SEAr reactions are indirect evidence. Fair enough.
    Not proton NMR though: that's almost as good as looking at the molecule itself. And it is an experimental fact that some hydrogens in toluene or PhCF3 are more shielded than others, compared to benzene where all CH's are equivalent.
    If this is no proof that the ED is different and that the CH3 group is responsible for it it, I don't know what is.

    People may not be able to provide theoretical explanations of why a certain group causes this ED distribution that sound sufficiently convincing to you, but it remains a fact that these things exist and are observed every day in laboratories, and if you want to do science you'll sometimes have to accept evidence even in the absence of a completely satisfactory theory.

    Chemistry is still an experimental science. So the advice I can give you is: use as much theory as you need to, but no more that that.
    In science you can bring down a theory on the basis of experimental evidence, but you can never reject experimental evidence solely on the basis of a theory.
  18. May 14, 2014 #17
    Hey, don't misunderstand my statements! I'm extremely sure there exists an explanation for the problem I'm dealing with. I'm also pretty sure that e.g. CH3 group is responsible for electron density transfer, as you write I'm denying.

    That means IN MY MIND/IN THIS THREAD there is a very abstract explanation... Sorry for my poor translation. No further comments on your post.

    On the other hand, I don't see why is my question obscure for you. I would be also very interested for your opinion to the deactivated benzoic acid (if you have no problems understanding the question). Because that is half of the problem I'm solving right now.
    Last edited: May 15, 2014
  19. May 15, 2014 #18


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    I also tried to point out to you which resonance structures are relevant in the case of e.g. CF3. I am not going to try to draw them. I think this is enough info for you to work it out yourself.
  20. May 15, 2014 #19
    And by abstract explanation, I'm referring to that the verdict is true, nevertheless it doesn't explicitly provide the primary cause explanation:

    By primary causes explanation I mean e.g. to calculate binding energy of substituent with carbon with various positive charge and compare the values, then draw a conclusion. Maybe I'm sick, but this verdict is not sufficiently probative for me, without having it postulated.

    And I know, the recommendation for me is GO TO QM CALCULATIONS!. I was just merely hoping somebody QM stuff knowledgeable is there.
    Last edited: May 15, 2014

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