Question
How does Huckel rule help in determining the aromaticity of pyridine?
✓
Answer
$i.$ Pyridine has three double bonds and $6 \pi$ electrons.
$ii.$ The six $p$ orbitals containing six electrons form delocalized $\pi$ molecular orbital.
$iii.$ The unused $sp^2$ hybrid orbital of nitrogen containing two non$-$bonding electrons is as it is.
$iv.$ According to Huckel rule, this compound is aromatic if, $4n + 2 =$ Number of $\pi$ electrons
$4n + 2 = 6,$
$\therefore 4n = 6 – 2 = 4$
$n = 4/4 = 1,$ here $‘n \ ’$ comes out to be an integer. Hence, pyridine is aromatic.
$ii.$ The six $p$ orbitals containing six electrons form delocalized $\pi$ molecular orbital.
$iii.$ The unused $sp^2$ hybrid orbital of nitrogen containing two non$-$bonding electrons is as it is.
$iv.$ According to Huckel rule, this compound is aromatic if, $4n + 2 =$ Number of $\pi$ electrons
$4n + 2 = 6,$
$\therefore 4n = 6 – 2 = 4$
$n = 4/4 = 1,$ here $‘n \ ’$ comes out to be an integer. Hence, pyridine is aromatic.
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