How to Determine the Aromaticity of a Ring System

• It must be a ring.

• It must be flat (planar).

• It must have in each atom of the ring a p orbital that's orthogonal to the plane of the ring. In other words, no atom in the ring can be sp³ hybridized.

• It must have a Hückel number of pi electrons, following the 4n + 2 rule.

The first condition is that only rings can be aromatic. Acyclic systems cannot be aromatic.

The second condition involves the shape of the ring. Ringed systems can be flat or three-dimensional. Most conjugated ring systems are flat in order to maximize the overlap between the p orbitals. But some exceptions exist. Cyclodecapentaene, for example, shown in the figure, is puckered because two hydrogens in the ring are pushed into each other's space, and the ring is forced out of planarity so that the hydrogens can occupy separate spaces. Because of its nonplanar structure, cyclodecapentaene is non-aromatic. Removing the two hydrogens and replacing them with a carbon bond to make naphthalene erases this problem; naphthalene is planar (and aromatic).

Cyclooctatetraene (shown in the next figure) is another example of a molecule that has alternating double bonds around the ring but isn't planar. Instead, this ring system adopts a puckered tub-shaped conformation. Why? Because if it remained flat, it would become anti-aromatic as a result of having a non-Hückel number of pi electrons (8)! Because anti-aromatic systems are so unstable, this compound puckers and sacrifices the p orbital overlap so that it won't have to endure the instability associated with being anti-aromatic.

The third condition involves the orbitals. Aromatic systems must have an unbroken ring of p orbitals, so any ring that contains an sp³ hybridized carbon will not be aromatic. Cycloheptatriene, shown in the next figure, is non-aromatic because one of the ring carbons is sp³ hybridized. However, carbocations (positively charged carbons) are sp² hybridized (and have an empty p orbital), so the cycloheptatriene cation has an unbroken ring of p orbitals and is an aromatic compound.

Finally, to be aromatic, a ring must have a Hückel number of electrons (4n + 2). The table gives solutions to these equations for low values of n.