Is cycloocta-1,3 diene planar? I know that cyclooctatetraene is non-planar because of 


Angle strain and 
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?
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.

Cycloocta-1,3,5-triene

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.



Cycloocta-1,3-diene

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.