## Introduction to Huckel’s rule

Huckel formulated a rule to identify aromatic compounds that is known as **Huckel’s rule**. The conditions of aromaticity are as follows:

a) Aromatic compounds should be cyclic.

b) The compound should have a planar structure.

c) There should be delocalized pi electrons in the compound.

d) There should be (4n + 2) pi electrons in the compound. The value of n =0, 1, 2, 3…

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The rule d) given here is the Huckel’s rule.

Any compound that does not follow one or more than one rules as stated above, are not aromatic – they are non-aromatic.

Substituting n=0, 1, 2, 3 … in (4n+2), we get 2, 6, 14 … respectively. So, in short any planar cyclic system should contain 2, 6, 14 … pi electrons, with at least one delocalized pi electron.

## Application of Huckel’s Rule

Let us try a few examples in finding whether the compound is aromatic or non-aromatic.

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**Solutions of problems on huckel’s rule:**

**
a) ** There are 3 double bonds. So, we have 3 pairs of pi electrons there. We can also see 2 other lone pairs of electrons there. So, totally, we have

(3×3) + (2×2) pi electrons

9 + 4 pi electrons

13 electrons

It is not aromatic because it is not in accordance with the Huckel’s rule.

**b) ** (2×1) electrons from the lone pair present + (3×2) electrons from the 3 double bonds present

2+6 pi electrons

8 electrons

The compound is non aromatic because it does not follow the Huckel’s rule.

**c) ** There are 3 double bonds which implies there are

(2×3) pi electrons

6 pi electrons

This is an aromatic compound as it is accordance with the Huckel’s rule.

**
d) ** (3×2)+ (2×1) pi electrons

6+2 = 8 pi electrons

This is a non- aromatic compound.

**
e) ** There are (2×2) pi electrons

= 4 pi electrons

This a non-aromatic compound

**
f) ** (3×2) + (3×2)

= 6+6

= 12 pi electrons

This is a non-aromatic compound.

## Some Practice Questions on Huckel’s Rule

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