[ssssss]

Any open chain conjugated system is taken to be planar.[like butadiene CH2=CH-CH=CH2].Now it is not neccesary for a sp2 hybridized C atom to be in plane with other sp2 hybridized C atom.Then why do we consider a open chain conjugated system as planar?



[hint-C8H8[cyclooctatraene] has each atom sp2 hybridized but still does not show aromatic character]

[Yggdrasil]

Any open chain conjugated system is taken to be planar.[like butadiene CH2=CH-CH=CH2].Now it is not neccesary for a sp2 hybridized C atom to be in plane with other sp2 hybridized C atom.Then why do we consider a open chain conjugated system as planar?



[hint-C8H8[cyclooctatraene] has each atom sp2 hybridized but still does not show aromatic character]

For conjugation to occur, the p orbitals of the adjacent sp²-hybridized carbons must be parallel.  Since the p orbital is perpendicular to the plane formed by the sp² orbitals, making adjacent p orbitals parallel makes adacent sp² orbitals be in the same plane.

Cyclooctatetraene is not conjugated because if conjugation were to occur, it would form a molecule with two unpaired electrons in non-bonding orbitals, a highly unstable configuration.  Such molecules are known as anti-aromatic, because of their great instability which results from conjugation.  In order to prevent such conjugation from forming a highly unstable anti-aromatic ring, cyclooctatetraene "puckers" so that it is not a planar molecule and a conjugated system cannot form.

[Donaldson Tan]

the planar configuration would be the least energetic because it facilitates electron delocalisation. hence, it would be the most thermodynamically favoured configuration.

[ssssss]

Yes i agree with you geodome.Your reason is somewhat incomplete Ygg.Here i am posting the brief explanation.

No need to say what Ygg has already mentioned so i am completing his answer.It has to do with the angular and torrisional theory of rings.

Suppose we take cyclooctaraene as a planar molecule and assume it to be planar.Now it will provide a more stable structure to it by lowering its energy state due to delocalized molecular orbital formed by constructive overlaping of 8 p orbitals.
But on the other hand it will become highly unstable due to angular and torrisional strain in it.[Remember if we take a conjugated system of cyclooctatraene its bond angle should be 135 degree mathematically but considering sp2 hybrid C the bond angle should be 120 degree.hence it produces enormous angular strain].So the stabilizing energy it has attained due to conjugation is far less to compensate with the instability that has occured due to Ring strain theory.

Hence the cyclooctatraene in reality gets twisted and forms some kind of chair[not exactly chair].Considering all these points E Huckel formulated a formulae for aromoticity Known as 4n+2 pie rule which can be applied to rings along with some other eligibilities.

Now if we sonsider an open chain molecule we dont have a closed ring strain factor which make them unstable so we can consider them planar due to the reason the geodome has already mentioned.

[Yggdrasil]

Suppose we take cyclooctaraene as a planar molecule and assume it to be planar.Now it will provide a more stable structure to it by lowering its energy state due to delocalized molecular orbital formed by constructive overlaping of 8 p orbitals.

Cyclooctatetraene will not exhibit added stability from pi-electron delocalization.  The exact opposite will happen; cyclooctatetraene will become more unstable when pi-electron delocalization occurs.

You mentioned Huckle's 4n+2 rule earlier.  Huckle also predicted the existence of anti-aromatic compounds which would exhibit decreased stability, and these compounds can be identified by the 4n rule.  Planar cyclooctatetraene, with its 8 (n=2) electrons, fits Huckle's rule for anti-aromaticity.  If you consider the molecular orbital configuration of cyclooctatetraene, you will have three bonding orbitals, three anti-bonding orbitals, and two nonbonding orbitals.  Six electrons will fill the bonding orbitals, while the other two will be left in the two non-bonding orbitals.  This configuration results in a highly unstable diradical species.  Since formation of this diradical is so unfavored, cyclooctatetraene will rarely be found in a planar form.  These electronic considerations are the primary reason why cyclooctatetraene is not a planar molecule.  Sterics, while they do play some part, are only a minor factor.

Here's a link on anti-aromaticity:
http://en.wikipedia.org/wiki/Anti-aromatic

[ssssss]

Yeah i know rule for anti aromatic compound.

[Winga]

So, will cyclooctatetraene prefer planar conformation if we either add or remove a pair of e- of the pi system.

[movies]

It should.  I think that those variants have been made.

[Mitch]

Yes they have. I briefly reviewed such systems over a year ago.

[FeLiXe]

just to complete what everyone's already said: Hückel's theory only talks about p-orbitals. p-orbitals don't care about binding angles though. parallel pi-overlapping stays the same. Only the sigma-bond is affected.

It is coincidence that the aromatic benzene (with stable pi-electron energies) also has the ideal angles for the sigma bonds.