Ch 11 : Aromaticity

Chapter 11 : Arenes and Aromaticity

Aromaticity Criteria

Qu: How can we recognise systems that are aromatic ?
Ans: By applying the criteria for aromaticity outlined below.

Based on the properties of aromatic compounds, there are FOUR criteria about the π system that need to be met inorder for the "special" aromatic stabilisation to be observed:

Conjugated : there needs to one "p" orbital from each atom in the ring, so each atom must be either sp² or sp hybridised.
Cyclic : linear systems are not aromatic, all atoms in the ring must be involved in the π system (i.e. no sp³ atoms)
Planar : if the ring is planar flat then this means there is good overlap / interaction between the "p" orbitals....not always easy to consider.
The Huckel Rule..... 4n+2 π electrons in the cyclic conjugated π system (n = 0, 1, 2, 3 etc.) This is equivalent to an odd number of π-electrons pairs).

Study Tips:
One way to think about the first 3 criteria is to compare the p orbitals of the system to "fence posts" surrounding a field. The posts need to be regularly spaced (i.e. no gaps), all around the field with all the posts standing upright in order to have an effective fence.

The Huckel rule....just think of "n" as an integer 0, 1, 2, 3... etc. it's just a number to feed into the simple formula to generate a set of numbers that correspond to the numbers π electrons in aromatic systems : Those numbers are 2, 6, 10, 14 etc. π electrons. So, if a cyclic, conjugated, planar π system contains one of these numbers of π electrons, then it's an aromatic system. For example, benzene is a 6 π electron system. "n" does not really relate to a simple structural property in the molecule.

In order for a compound to be aromatic, all FOUR of these criteria must be met.

The most important and well known aromatic system is benzene. It consists of a set of 6 sp² C arranged in a ring. This creates a set of 6 p orbitals, no gaps in the π system so we have a cyclic, conjugated π system. The planarity of the ring means that the interaction of the adjacent p orbitals is optimal. The 3 C=C in benzene mean that we have 3 pairs of π electrons = 6 π electrons = a 4n+2 number where n=1.

Benzene

To aid in counting the electrons the following factors may help:

Each atom in the conjugated, cyclic system can only contribute a maximum of one pair of electrons (i.e. one p orbital)
"Refining" your "predicted" hybridisation for heteroatoms (esp. O, N andS) will allow lone pairs to be part of the πsystem.
Carbocations and boron atoms (C⁺ & B) are sp² hybridised and have empty "p" orbitals. This allows them to be part of the conjugated π system but they contribute NO electrons.