How would you know if a cyclic organic molecule is planar?
To be planar, the bond angles in the planar ring have to be close to those predicted from VSEPR theory or there will be too much bond strain and the stucture will be distorted from planarity. To be aromatic, there have to be (4n+2) pi-electrons in a planar delocalised cyclic system (where n is a positive integer). In benzene (which can be written in its Kekulé form of three single and three double bonds, cyclohexatriene) there are 6 pi-electrons (i.e 4n+2 where n=1) so it is aromatic and planar with all bond angles being 120° which is the ‘ideal’ value for an sp²-hybridised carbon. In cyclooctatetraene (which has four single and four double bonds) there are 8 pi-electrons (i.e 4n+2 where n=1.5 and not an integer) so it is non-aromatic and non-planar. In cyclodecapentaene (which has five single and five double bonds) there are 10 pi-electrons (i.e 4n+2 where n=2) so it could be aromatic but the bond angles would have to be 144° (very different from 120°) so the strain would be too great
