Question

How would you differentiate between aromatic, anti-aromatic and non-aromatic compounds?

Answer 1

Hint: Use Huckel’s formula for aromatic and antiaromatic compounds. These systems have a conjugation so every carbon should be $S{P}^2$ hybridised including all cations and anions if present in the system. Simply if you are seeing any benzene ring in the structure termed it as aromatic and if it is not present then the compound should be among anti-aromatic and non-aromatic.

Complete step by step answer:
For giving organic compounds a word of stability, that they are stable or unstable we use huckel orbital theory. Using huckel’s theory we can easily tell by seeing whether a system is stable or not. What stability means in chemistry, is that a particular system if it is stable must have a lower amount of energy and a system which is less stable has a greater amount of energy than the preceding one.

To understand this phenomenon of stability, a scientist named Huckel gives a theory according to which firstly, we must have a conjugated system. It means this theory is applicable to mostly structures which are conjugated, having double bonds at the alternative position.

There are many tests by which we can estimate that a system given to us is aromatic or non aromatic. For aromatic structure, the system must be conjugated first, having $s{p}^2$ hybridized carbon atoms in it, or we can say it contains a benzene ring in their system. Number of pi- electrons should be equal to $(4n+2)e^{-}$ where $n=0,1,2,3......$. There are many systems that satisfy this statement, like for example if we consider benzene- it has $6$pi electrons so when we put the term $(4n+2)e^{-}=6$ we get $n=1$ from there. It means that benzene satisfies huckel theory and it is an aromatic system. More examples we have like naphthalene, pyridine etc.

For anti-aromatic compounds they don't have a benzene ring in their structure but the pi electrons should be equal to $(4n)pi{e}^{-}$ so easily a trick can be made from here. That those systems which have double bonds as a multiple of $4$ are antiaromatic in nature. In terms of energy these anti-aromatic compounds are having energy in between the aromatic and non-aromatic system.

The criteria of non-aromatic is somehow very much different from the above systems. They do not have a conjugated system, so if we place a $s{p}^3$ hybridized carbon atom in any of the system aromatic or antiaromatic, they easily get converted into a non-aromatic system. As they don’t have a conjugated system, they will not show resonating structure and have very high energy as compared to aromatic compounds. We can take examples of non-aromatic systems such as alkanes, alkenes etc.

Note: When you are counting pi electrons for putting in the formula of aromatic and antiaromatic compound keep in mind that a positive charge $(+)$ of any cation is not counted in that but we can count negative charge $(-)$of anion as $1$. Always check that the system given to you has all $s{p}^2$ hybridized carbon atoms or not, if not it will be a non-aromatic system.