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According to my course of organic chemistry, para-nitro-aminobenzene has to break his aromaticity to delocalize the electrons of the amino-group in the nitro-group. I don't really see this, when I draw the resonance structures.

resonance structures

Martin - マーチン
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Simon Ravelingien
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    @Simon Ravelingien - why does your diagram use equilibrium arrows to show resonance contributors? – Dissenter Dec 26 '14 at 17:34
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    As far as I am aware the aromaticity is not broken – bon Dec 26 '14 at 17:46
  • You need three conjugated pi bonds in the ring to conserve aromaticity in this system. The molecule's "true" structure is a weighted sum of the resonance structures; you therefore would expect a partial loss of aromaticity compared to benzene. – J. LS Dec 26 '14 at 18:15
  • What do you mean by loss of aromaticity? – Dissenter Dec 27 '14 at 12:00
  • A broken conjugated system or 4 instead of 6 electrons in the conjugated pi bonds – Simon Ravelingien Dec 27 '14 at 20:05
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    I agree with Dissenter, the convention is to use a double headed arrow for resonance structures. It helps people be aware of the difference between resonance structures and different species in equilibrium. – iad22agp Dec 27 '14 at 23:23
  • @Bon Aside equilibrium arrows,the intermediate 5 structures do not seem aromatic. – Chakravarthy Kalyan Jun 08 '19 at 05:01
  • *diImine quinone would undergo a hydride-shift, not simply "quinone would undergo a hydride shift." – Zach Earle Apr 03 '20 at 18:56
  • One could argue the delocalization and resonance stabilization of the pi electrons is extended beyond just the ring. Aromaticity is not really broken, but superseded. – Oscar Lanzi Apr 03 '20 at 23:51

3 Answers3

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Your teacher is correct, some of the resonance structures in your figure are not aromatic. Still though, as resonance structures they contribute to the overall description of the molecule, just less than the structures that are aromatic. The resonance structures at the right and left end of your figure are aromatic. The fourth resonance structure (the one in the middle) is referred to as being a "quinone" rather than an aromatic.

o- and p-benzoquinone are examples of quinones; they are not aromatic compounds (ref, note the sentence "Quinones are conjugated but not aromatic").

enter image description here

Likewise, their carbon analogues, o- and p-xylylene are also non-aromatic quinoid-like compounds.

For a compound to be aromatic it must be

  • cyclic
  • contain 4n+2 pi electrons
  • have a continuous loop of p-orbitals
  • and you must be able to draw double bonds - internal to the ring - from each ring carbon atom.

It is this last point that serves to differentiate quinones and quinone-like molecules from aromatic compounds.

Finally, aromatic compounds need not be planar, deviations in excess of 15° from planarity can be tolerated and the molecule can remain aromatic (see, for example, [6]-paracyclophane)

enter image description here

ron
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    Can benzoquinone be called anti-aromatic since it has 4n pi electrons? – ChemExchange Mar 28 '15 at 04:34
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    @ChemExchange No, benzoquinone is not anti-aromatic, it is not a simple cyclic array of 4 pi electrons. In fact, anti-aromatic compounds distort to some other geometry so they are no longer anti-aromatic and hence, more stable. – ron Mar 28 '15 at 13:46
  • @ChemExchange ron is right, but your point is valid - you can indeed write some anti-aromatic mesomeric structures for quinones, but as their are destabilizing, they aren't major contributors - in tropone it's the other way round. – Mithoron Mar 28 '15 at 16:00
  • @Mithoron But all resonance structures have 4 pi electrons so all have destabilizing effect. – ChemExchange Mar 29 '15 at 17:42
  • @ron So Huckel's rule fails here? – ChemExchange Mar 29 '15 at 17:43
  • @ChemExchange What do you mean by "here"? – ron Mar 29 '15 at 17:44
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    @ron Here in the case of benzoquinone. It has 4 pi lectrons inside the ring. – ChemExchange Mar 29 '15 at 17:46
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    @ChemExchange Huckel's rule applies to cyclic systems where all of the p-orbitals and electrons are contained within the ring. With quinones we have some orbitals and electrons external to the ring, so Huckel's rule doesn't really apply in such cases.. – ron Mar 29 '15 at 17:52
  • @ron So the 4 bulleted points in your answer are not Huckel's rule but a modification of it? – ChemExchange Mar 29 '15 at 17:54
  • @ChemExchange I never mentioned Huckel's rule in my answer. The 4 bullets represent a few points, likely incomplete, on what I think is needed for aromaticity. – ron Mar 29 '15 at 18:17
  • @ron Yes, you never mentioned it. I meant whether they were a modification or not. Anyway, I got the answer to it. Thank you. – ChemExchange Mar 29 '15 at 18:20
  • Let us continue this discussion in chat. – ChemExchange Mar 29 '15 at 18:21
  • @ChemExchange I need to go somewhere now, I'll see if you're around later. – ron Mar 29 '15 at 18:22
  • Alright. I will leave my message in that chat. – ChemExchange Mar 29 '15 at 18:25
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    I believe Quinones are aromatic according to the goldbook definition of aromaticity. I cannot find evidence for the statement that they would not be other than the wikipedia page, and it has been disputed there unfortunately without citation. – Martin - マーチン Mar 31 '15 at 09:57
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    @ChemExchange related http://chemistry.stackexchange.com/questions/15272/aromaticity-of-cyclic-ketone, http://chemistry.stackexchange.com/questions/27360/resonance-of-aromatic-compounds – Mithoron Apr 08 '15 at 22:34
  • @Martin Why? Can you contribute your reasoning here? – ChemExchange Apr 10 '15 at 15:08
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    @ChemExchange I might post an answer in a while, but I am not going to participate in an extensive chat about it. Sorry I am too busy for that. In the meantime you can have a look at this answer of mine. Aromaticity is not a very well defined concept and I also believe that it is hard to tell apart from overall resonance stabilisation. Hückel's rules are nice guidelines, but they are not the definition of aromaticity. This is always taught wrong. They also do not explain Y-aromaticity and sigma aromaticity... – Martin - マーチン Apr 14 '15 at 11:47
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Aromaticity is not a very well defined concept (there are no strict rules) and I also believe that it is hard to tell apart from overall resonance stabilisation. The IUPAC goldbook states for aromatic:

aromatic

  1. In the traditional sense, 'having a chemistry typified by benzene'.
  2. A cyclically conjugated molecular entity with a stability (due to delocalization ) significantly greater than that of a hypothetical localized structure (e.g. Kekulé structure ) is said to possess aromatic character. If the structure is of higher energy (less stable) than such a hypothetical classical structure, the molecular entity is 'antiaromatic'. The most widely used method for determining aromaticity is the observation of diatropicity in the 1H NMR spectrum.
    See also: Hückel (4n + 2) rule, Möbius aromaticity
  3. The terms aromatic and antiaromatic have been extended to describe the stabilization or destabilization of transition states of pericyclic reactions The hypothetical reference structure is here less clearly defined, and use of the term is based on application of the Hückel (4n + 2) rule and on consideration of the topology of orbital overlap in the transition state. Reactions of molecules in the ground state involving antiaromatic transition states proceed, if at all, much less easily than those involving aromatic transition states.

Hückel's set of rules are often used as nice guidelines, but they are not the definition of aromaticity. In fact they are very, very rigid and strict and most of the compounds you'd usually consider being aromatic in this context, do not comply with these rules. The IUPAC goldbook defines:

Hückel (4n + 2) rule
Monocyclic planar (or almost planar) systems of trigonally (or sometimes digonally) hybridized atoms that contain (4n + 2) π-electrons (where n is a non-negative integer) will exhibit aromatic character. The rule is generally limited to n = 0–5. This rule is derived from the Hückel MO calculation on planar monocyclic conjugated hydrocarbons (CH)m where m is an integer equal to or greater than 3 according to which (4n + 2) π-electrons are contained in a closed-shell system.

In this framework only the first and the last structure obey Hückel's rule. The proton and carbon NMR clearly proof that p-nitroanniline is an aromatic compound, the spectrum is very close to the data of benzene.

Resonance structures are just a crutch to explain a very complicated and complex electronic structure in simple and understandable terms, i.e. Lewis structures. In highly delocalised systems a superposition of all resonance structures has to be considered to paint a somewhat conclusive picture of the bonding situation. There are contributions of different structures and not all of them may be equal in the overall description, however, they do not change the fact, that the experimental property is observed. This is again a case, where you can see the limitations of the resonance concept.

Martin - マーチン
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All the resonating structures follow Huckel's Rule and so they are not losing their aromaticity.

  • Planar Cyclic Ring
  • 4n+2 pi-electrons (the exocyclic pi electrons are not considered)1,2.
  • All atoms of the ring in conjugation

A classic example of an aromatic cabanion is cyclopentadiene ion.

The negative charge or the lone pair is in resonance and is in an unhybridised orbital. Hence it gives $sp^2$ and not $sp^3$. So its planar.

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ChemExchange
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  • Are the structures with a negative charge at a carbon atom really planar? I would expect tetrahedral bonding angles for those. – Curt F. Mar 27 '15 at 18:12
  • @CurtF. the phenyl ring is a hexagon (or near enough) so it cant have tetrahedral angles – bon Mar 27 '15 at 19:17
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    Can you really say that any of the resonance structures in the middle have 6 pi electrons in the ring? Each (ring) double bond counts for two electrons, the lone pair counts for two electrons so we're up to 6 electrons right there. The exocyclic pi bond counts for zero? One? Two? – jerepierre Mar 27 '15 at 19:52
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    @CurtF. asked, "Are the structures with a negative charge at a carbon atom really planar?" Yes. The cardinal rule for resonance structures is that nuclear positions cannot change. So since the aromatic ring is planar, so must it be in the resonance structures. You're right that carbon with a negative charge would prefer to be non-planar. Since it can't deform in the resonance structures, that tells us that those resonance structures contribute less to the "real" structure. – ron Mar 27 '15 at 19:52
  • @CurtF. A classic example of an aromatic cabanion is cyclopentadiene – ChemExchange Mar 28 '15 at 04:28
  • The negative charge or the lone pair is in resonance and is in an unhybridised orbital. Hence it gives sp2 and not sp3. @ron Am I correct in saying this? – ChemExchange Mar 28 '15 at 04:30
  • @ChemExchange Yes, the carbon is sp2 hybridized. – ron Mar 28 '15 at 13:50
  • @ChemExchange - thanks for the extra info. I was wrong to downvote you, but I can't reverse it unless you make some kind of edit to your initial answer. It would probably be great to add in the extra info about cyclopentadiene etc into your answer. But if you don't have time, just add in a whitespace or something and I'll reverse my vote. – Curt F. Mar 29 '15 at 02:57
  • @jerepierre I don't understand why the exocyclic bonds have to be included in the 6 pi electrons. – ChemExchange Mar 29 '15 at 17:52
  • @ChemExchange My point is just that depending on your decision there, the aromaticity changes: zero=aromatic, one=non-aromatic, two=anti-aromatic. Without addressing the electron count, you can't say it follows 4n+2. – jerepierre Mar 30 '15 at 13:27
  • @jerepierre True. I need to mention that in my answer too. – ChemExchange Apr 10 '15 at 15:16