Question

What are the necessary conditions for any system to be aromatic?

Answer 1

Hint:
We know that the conditions for any compound to be aromatic are also known as Huckel’s rules. Some examples of aromatic compounds are benzene, pyrrole, Tropylium, etc. Some examples of non-aromatic compounds are tetrahydrofuran, cyclohexene etc.

Complete step by step solution
There are basically three conditions for any system to be aromatic, let’s discuss them.

1. The conditions for a molecule to be Aromatic-Cyclic Rule: -First and most important rule is that molecule must be cyclic in nature. Cyclic structure means that the molecule must be closed. For example, cyclohexane is a type closed, benzene is also a closed structure.

2. The atom present in the ring must be conjugated; Cyclic is not only a condition for aromaticity. For example, in case of cyclohexane which is cyclic structure but it is not aromatic. It is due to the absence of double bonds. Due to the presence of two double bonds alternate to each other bond conjugation will take place.

3. The molecule must have $\left[ {4n + 2} \right]pi$electrons where n is the number of pi-bonds. The famous example is the benzene and cyclooctatetraene. Benzene has \[6{\rm{ }}pi\] electrons while cyclooctatetraene has $8\,pi$ electrons. Due to which benzene is an aromatic compound while cyclooctatetraene is not an aromatic compound.

So, for $n = 2$ $n = 0$,we have two pi electrons.
For $n = 1$,we have six electrons.
For $n = 2$we have ten electrons.

Hence, we can say that if a compound is cyclic, conjugated and has $\left[ {4n + 2} \right]pi$ electrons are known as an Aromatic compound.

Note:
To explain the extra stability of benzene type molecules, a special term is needed to know that is called Aromaticity. On seeing the energy level diagrams for the rings, we will find that there exists always a single low-energy bonding orbital and then pairs of degenerate orbitals. This is the basis of Huckel rule.