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I've learned that if any one ring of a structure like the below is aromatic, the whole compound is aromatic.


But one of my friends suggested that such compounds cannot be aromatic and gave the following resonance structures as the reason:

resonance delocalisation in 1,2-dihydronaphthalene

I feel that the non-aromatic/anti-aromatic resonance structures wouldn't make significant contribution. Is my reasoning correct?


marked as duplicate by Todd Minehardt, jerepierre, airhuff, pentavalentcarbon, paracetamol Jun 6 '17 at 6:26

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    $\begingroup$ Antiaromatic species are unstable. No molecule would attempt to take up an antiaromatic contributor. $\endgroup$ – Pritt Balagopal Jun 5 '17 at 4:36
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    $\begingroup$ I edited your post to make the picture more readable. I used Web Chemdoodle for this purpose. See my guide on how to format chemical compounds and mechanisms. I used Microsoft Paint to write "Aromatic" and "2e" and stuff. $\endgroup$ – Pritt Balagopal Jun 5 '17 at 5:05
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    $\begingroup$ I think I address this in the answer to the very similar (possibly duplicate) question: Is 1,2-dihydronaphthalene aromatic? $\endgroup$ – Martin - マーチン Jun 5 '17 at 9:38

Aromaticity is a quite difficult concept but in this case we can answer the question by simply looking at similar compounds and experimental data.

What you got here is a styrol derivative, the pi-system (which we need to look at if we are talking about aromaticity) is the same. So if this compound is not aromatic then styrol shouldn't be either.

But styrol is in fact aromatic (or better: the benzene ring in styrol is). It reacts like an aromatic compound should, it got the right shift in NMR, the bond lengths are in accordance with aromaticity and so is the UV/VIS spectrum.

So a conjugated double bond clearly doesn't make a benzene (significantly) less aromatic.


According to Huckel Rule the compound is aromatic only if the conjugation is spreaded all over the structure and the compound is flat or nearly so, so that the sideways overlapping of pi orbitals can take place for conjugation and there should be conjugation of (4n+2)pi electrons on the whole compound and as can be seen from your provided structure the second(right hand sided) ring does have conjugation on the whole structure.

  • $\begingroup$ So, is the compound aromatic or not? $\endgroup$ – user38977 Jun 5 '17 at 5:38
  • $\begingroup$ I don't think so! $\endgroup$ – Vaibhav Dixit Jun 5 '17 at 5:41
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    $\begingroup$ The way you describe the rule implies that substituted benzenes would not be aromatic as electrons are not delocalised over the substituents. That's just wrong. This compound contains an aromatic ring. $\endgroup$ – matt_black Jun 5 '17 at 8:19
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    $\begingroup$ This answer is plain wrong. $\endgroup$ – Pritt Balagopal Jun 5 '17 at 9:50
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    $\begingroup$ @VaibhavDixit if a chemist says that a compound is aromatic it means that there's at least one aromatic structure in it. $\endgroup$ – DSVA Jun 5 '17 at 16:51

For a compound to be aromatic or anti aromatic there should be conjugation throughout the peripheral , this compound lacks conjugation throughout the peripheral so it falls under non-aromatic category.

  • $\begingroup$ Huckel baba fails many time for prediction of aromaticity of a compound, and the chart for checking if the compound is aromatic or not goes like this: 1- we check if compound is cyclic and presence of conjugation , 2- planarity of molecule,3- 4n+2 pi electrons .There are many examples in which some of double which can conjugate with the ring doesn't conjugate and resides localized on two carbons just to maintain aromaticity like Pyrene ,in this exmple also the double bond should be localised on those two cabon atoms and not participating in resonance and then only it only it can be aromatic $\endgroup$ – Kushal Tripathi Jun 6 '17 at 4:42
  • $\begingroup$ @KushalTripathi I don't know which "chart" you are speaking of but you should only count the pi electrons in the conjugated ring system, not every pi-system in the molecule and also not pi-systems conjugated to the cyclic one. If we do that here we got a 6-membered ring with presence of conjugation and 6 pi-electrons, which fulfills the hückel rule. And for the second part: I don't nderstand what you mean at all, that's wild talks. And also Pyrene is completly aromatic. $\endgroup$ – DSVA Jun 6 '17 at 6:53
  • $\begingroup$ @DSVA yes , Pyrene is completely aromatic . I was relating this molecule to the Pyrene because in pyrene also the total no of pi electrons are 4n and not 4n+2 . But the pi bond at the center of pyrene is not considered to be conjugating with other pi bonds of molecule, so we do not count that pi electron and that's why we say it has 4n+2 pi electrons, simply in this example double bond in the non benzenoid ring should not be disturbing the conjugation of the benzene ring and electrons of benzene are only counted. It is an aromatic compound ,my answer is wrong and I was explaining this only $\endgroup$ – Kushal Tripathi Jun 6 '17 at 7:39
  • $\begingroup$ @KushalTripathi the hückel rule doesn't work for fused polycyclic compounds, in fact it does only work correcly for monocyclic systems. The middle bond in Pyrene is in conjugation with the others... $\endgroup$ – DSVA Jun 6 '17 at 9:11
  • $\begingroup$ @DVSA It may be wrong whether the middle pi bond conjugate with rest of the double bonds, but to the best of my knowledge that bond is not considered as pi bond and this is why we say that no. of non pi electrons in pyrene is 2. You can confirm this from the fact that reaction of pyrene with Br2/CCl4 gives the addition reaction at the middle pi bond. $\endgroup$ – Kushal Tripathi Jun 7 '17 at 5:12