Is cyclohepta-1,3,5-triene aromatic?
I think it should not be, because even though the π-system possesses six π-electrons, the conjugation is not complete around the ring.
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$\begingroup$ Related: chemistry.stackexchange.com/questions/28342/are-these-aromatic/… $\endgroup$– Klaus-Dieter WarzechaCommented Jul 16, 2016 at 6:07
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3 Answers
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You are correct. The conjugation does not extend all the way around the ring, so the molecule is not aromatic. In addition, the single methylene group ($\ce{CH2}$) sticks out of the plane of the rest of the molecule.
(On the other hand, removing a hydride ion $\ce{H-}$ from the $\ce{CH2}$ group leaves the tropylium ion $\ce{C7H7+}$, which is planar and aromatic, with resonance structures putting the positive charge on any of the seven carbons.)
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6$\begingroup$ If the world was black and white, you'd be correct. Unfortunately, aromaticity is among those concepts which have a large grey area: en.wikipedia.org/wiki/Homoaromaticity $\endgroup$– Martin - マーチン ♦Commented Jul 14, 2017 at 5:23
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Actually the comments above are not entirely correct. This molecule does have some aromatic character.
These systems are called Homoaromatic. We discussed them at length in my Physical Chemistry course several years ago. Wikipedia has a page on this and here's a recent primary source as well: Williams et al., J. Am. Chem. Soc. 2012, 134, 40, 16742–16752
The CH2 rotates out of the plane of the other carbon atoms which brings the adjacent p orbitals close enough together to have some pi-pi overlap. This gives a slightly stabalized 4n+2 homoaromatic system. The stabalization seems to be about 20-50% that of benzene in the case of this molecule, according to the article above.
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To define an aromatic compound, there are four basic rules:
- P(lanarity)
- D(elocalization)
- H(ückel's rule): $4n + 2$ π-electrons
- C(onjugation)
In the above compound, if any one of the carbon is $\mathrm{sp^3}$ hybridised, then it is not planar. In the above compound, there is an $\mathrm{sp^3}$ carbon, so it is not aromatic.
