Which of the following compounds is aromatic?

From the 4n+2 rule, it must be either option 2 or 4. The answer given in my book is option 4. Why? I think It might be due to one being planar and the other not, but to me both seem to be planar. Is there a way to check if one is planar or not?


2 Answers 2


Number 2 is not aromatic because transannular $\ce{C-H}$ interactions cause the molecule to be non-planar although it satisfies $4n+2$. This problem was solved by the synthesis of number 4, which, while a little less than planar but rigid, displays aromatic properties (NMR, etc., Ref 1).


  1. E. Vogel, H. D. Roth, "The Cyclodecapentaene System," Angew. Chem. Int'l. Ed. 1964, 3(3), 228-229 (DOI: https://doi.org/10.1002/anie.196402282).
  • 2
    $\begingroup$ Yeah, there was also analogous ether IIRC. $\endgroup$
    – Mithoron
    Dec 13, 2017 at 0:58

If you have a molecular model kit, or a computer simulation that plays a similar role, use it to build both molecules. You discover that neither (2) nor (4) is entirely planar: (4) is not planar because the internal methylene group is out of plane (the only way to get realistic bond angles at the central carbon), and you find (2) is not planar because of the internal hydrogen atoms pushing each other out of the way and distorting the ring accordingly. But (4) comes much closer to being planar. Thus (4) is more likely to show at least some of the conjugation which, with an optimal $4n+2$ electron count, makes the ring aromatic.


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