I am learning to determine whether a monocyclic conjugated alkene is aromatic or anti-aromatic or not.

question 1:

cyclobutadiene: according to Huckel's rule, 4 * 1 = 4, it should be antiaromatic, like what is said here. However, Wikipedia says the 2 pi bonds have a 'more double bond' character, making it a rectangular shape (electrons not delocalized). Therefore, it is more like a conjugated diene, hence non-aromatic. What is the truth?

question 2:

planar / non-planar: for [18]annulene, it is aromatic (4 * 4 + 2 = 18). From Wiki, ''it is large enough to comfortably accommodate six hydrogen atoms in its interior, allowing it to adopt a planar shape, thus satisfying Hückel's rule.'' So in this molecule, there is an extremely small steric effect. So I am wondering, is size of the inner space also a factor in determining whether a cyclic molecule is aromatic or not? If the inner lumen of the molecule is larger, it can follow Huckel's rule? Like Cyclodecapentaene, as the 2 hydrogens cannot be in planar (not comfortably accommodate) (figure 3 in Wikipedia). It is non-aromatic rather than aromatic (do not follow Huckel's rule).

question 3: cyclooctatetraene is non-aromatic rather than antiaromatic (4 * 2 = 8) since it avoids by changing into a 'tub' shape (avoiding being the unstable planar conformation). I am wondering if other supposedly antiaromatic molecules can change their conformation to be non-aromatic (become more stable) as cyclooctatetraene does? Are there any criteria for changing the shape, or cyclooctatetraene is the only example? If yes, why? If no, what monocyclic conjugated alkene molecule is actually antiaromatic then?

Thank you very much.

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    $\begingroup$ In all three questions, you are probably missing the following: for coaxial pi orbitals to form, C-C bonds must be at certain angle, ideally 60 degrees, as in benzene. Cyclobutadiene cannot do that. Long cycles in some conformations can, but not all of them can also become flat so that these pi orbitals can effectively overlap. $\endgroup$ – sa7 Aug 5 '20 at 11:55
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    $\begingroup$ The orbitals in cyclobutadiene still overlap in the rectangular shape. It would be even more antiaromatic if you somehow forced it to be quadratic. To become nonaromatic, it would have to adopt a twisted shape, but double bonds (i.e. sp2 atoms) want to be planar, so that is also not possible. $\endgroup$ – Karl Aug 5 '20 at 12:32
  • $\begingroup$ @Karl I agree that the conformation of cyclobutadiene is impossible to be changed. However, for larger cyclic molecules which supposedly to be antiaromatic (e.g. cyclooctatetraene), in general, can they all twist into a non-planar shape to avoid being antiaromatic (more unstable)? Thanks $\endgroup$ – 234ff Aug 5 '20 at 14:06
  • $\begingroup$ @sa7 OK so for larger cycles, apart from using Huckel's rule, I need to check the C-C bond angles to see whether coaxial pi orbitals can form, hence to check whether it can be planar (flat), and eventually conclude whether it is aromatic or not? Thanks $\endgroup$ – 234ff Aug 5 '20 at 14:14
  • $\begingroup$ I wonder why you apparently don't get that asking about there different compounds is two questions too many... $\endgroup$ – Mithoron Aug 5 '20 at 15:31

Steric effects play an important role in determining the planarity, aromaticity of a molecule. Cyclobutadiene is sterically unstable(90 degrees angle between bonds), hence having high energy, and thus anti-aromatic.

A lot of compounds change their conformations for reducing their overall energy. Check for the conformations of glucose, for instance. A compound will always prefer its low energy conformation, which occurs when Cyclooctatetraene is non-planar depicted in the right hand perspective drawing



This compound looks like the big brother of benzene, which may incorrectly induce the notion of it being aromatic, or, by Huckel rule, anti-aromatic. But the inner angle of a regular octane shape is 135 degress, while a much stable angle nearby it is 120 degrees(an sp2 config.), which only occurs in the tub config. of Cyclooctatetraen. 120 degrees is not achieved, but ~126 degrees is achieved.

Frozen (at 129 K) cyclooctatetraene's structure was characterized by single crystal diffraction analysis by Claus and Rüger. You find the corresponding interactive model here.

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    $\begingroup$ Cyclobutadiene distorts to a rectangular geometry, so it's not exactly a perfectly antiaromatic square molecule. Also, I feel like "high energy" isn't a sufficient condition for being antiaromatic, much like how "low energy" doesn't quite mean aromatic. $\endgroup$ – orthocresol Aug 5 '20 at 12:58
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    $\begingroup$ I think the usual understanding of the words aromatic and antiaromatic, when used in conjunction with Huckel's rule, is something along the lines of the Dewar resonance energies, cf. first part of my answer chemistry.stackexchange.com/a/82690/16683. Colloquially one would say that benzene is aromatic because its π system is more stable than hexatriene. $\endgroup$ – orthocresol Aug 5 '20 at 13:02
  • $\begingroup$ It is rectangular, which means it is sterically hindered(90 degrees angle), while a double bonded molecule should have angle nearby 120 degrees. It is very sterically hindered. Thus it will have a high energy: the sense that it will be 'unstable'. Stability is a criteria or a result of aromaticity. Benzene is less reactive in many substitution reactions. $\endgroup$ – losnihciL Aug 5 '20 at 13:17
  • $\begingroup$ @losnihciL A double bond primarily needs to be planar. If you twist the two sides of a double bond out of plane, it becomes antibinding. There is no sterical hindrance in cyclobutadiene, the problem is electronic, not steric. $\endgroup$ – Karl Aug 5 '20 at 13:52
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    $\begingroup$ I recommend you learn chemistry from appropriate sources, not merriam webster. And yes, an sp2 hybridised carbon atom with a 90° bond angle is very unhappy, things being the way they are. This is however NOT called "steric hindrance". Total misnomer. $\endgroup$ – Karl Aug 5 '20 at 21:32

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