I know that cyclooctatetraene is non-planar because of

  1. Angle strain and
  2. Antiaromaticity if it was planar.

But what about cycloocta-1,3,5 triene or cycloocta-1,3 diene. It doesn't risk being antiaromatic, but there still is angle strain. Would it lose its planarity and hence inhibit resonance?

In general, when is the angle strain enough, that releasing the angle strain but also inhibiting resonance at the same time lead to a net increase in stability?

  • 2
    $\begingroup$ Guess those would be somewhat non-planar, but not enough so to break the conjugation completely. Moreover, I expect the conjugated part to be more planar than the rest. $\endgroup$ – Ivan Neretin Mar 23 '17 at 5:24
  • $\begingroup$ That's what I'm guessing too, but I posted this question so that someone could confirm/debunk this thought. $\endgroup$ – xasthor Mar 23 '17 at 5:35
  • $\begingroup$ I think the more dominating factor for the non-planarity of cyclooctatetraene is antiaromaticity, rather than angle strain. So in case of cycloocta-1,3,5 triene and cycloocta-1,3 diene though angle strain may be a destabilizing factor but the other one i.e. antiaromaticity is absent, on the other hand they have a conjugated system of 6 and 4 carbons respectively which gives them stability to some extent. So I agree to Ivan Netetin's comment that the conjugated part will be more planar than the rest. $\endgroup$ – jyoti proy Mar 23 '17 at 5:42
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    $\begingroup$ Closely related, but not a duplicate: Why is cyclooctatetraene non planar but the cyclooctatetraenide anion planar? $\endgroup$ – hBy2Py Mar 23 '17 at 17:05

Short answer: there's some twisting of the conjugated structure, particularly in the diene

I ran initial calculations for these using B3LYP density functional theory. If anything this method has a slight tendency for extra delocalization.

I'll follow up later with other methods, but I suspect the general answer will be the same.


Honestly, I expected the delocalization to be retained in the central conjugated section (albeit not perfectly flat) with some twisting on the outer double bonds.

Indeed, the dihedral angle around that central C=C is ~6.9°, but it's not that different around the other CC=CC bonds (~3° and 5°, respectively).

The whole C=CC=CC=C is fairly co-planar, with the remaining two carbons profoundly out of plane, likely relieving ring strain.

It's clear there's some delocalization because the "single" bonds between the double bonds have bond lengths ~1.46Å, smaller than usual.

Thus, in the triene, there's slight non-planarity to the conjugated section, but it's fairly delocalized.

enter image description here


Interestingly, the result is a bit different for the diene. While the C=C bonds are planar (i.e., dihedral angle ~2°) the middle C-C bond is strongly non-planar (i.e., dihedral angle ~38.4°). Moreover, it's 1.475Å in length, longer than the comparable bonds in the triene.

Here, I think the delocalization of two double bonds isn't enough to balance the ring strain.

enter image description here

  • 1
    $\begingroup$ With more saturated carbon atoms and their hydrogen atoms sticking out above and below the "plane" of the ring, the diene could be more affected by conformal strain than the triene. That disfavors planar bond geometry. $\endgroup$ – Oscar Lanzi Mar 23 '17 at 18:34
  • $\begingroup$ Agreed. There's several things going on that make the diene different than the triene. $\endgroup$ – Geoff Hutchison Mar 23 '17 at 18:43

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