Question

Four structures are given in options (A) to (D). Examine them and select the aromatic structures.

• A.
• B.
• C.
• D.

Answer 1

Answer: (B), (D)

Huckel’s rule of aromaticity:

Aromaticity condition:

A). Cyclic and planar compound ( involving $s{p}^2$ hybridisation)

B). Complete delocalisation of $\pi$ electrons in the ring.

C). Presence of $(4n+2)\pi$ electrons in the ring. Where $n=0,1,2,3\dots$

$\bullet$ If presence of $4n\pi$ electrons in the ring, along with satisfaction of condition A and B, then the compound is said to be
anti-aromatic.

A)Analysing the given options


$\bullet$ This molecule is planar ($s{p}^2$ hybridised carbons) and cyclic.

$\bullet$ Number of delocalized $\pi$ electrons = $2$,
So, $(4n+2)\pi$ electrons= $2$
$=>n=1$

$\bullet$ Therefore, the ring is aromatic in nature.

Option B)
It is a tub shaped molecule. Therefore, this molecule is non-planar.

So, it doesn’t satisfy the first condition for huckel’s rule of aromaticity.

Hence, it is not an aromatic compound

Cyclooctatetraene escapes from
anti-aromaticity by twisting itself into a
tub-shaped structure.

Option C)

$\bullet$ The compound is planar (all carbon has 𝑠𝑝2 hybridisation) and cyclic.

$\bullet$ Complete delocalisation of $\pi$ electrons in the ring.

$\bullet$ Number of $\pi$ electrons =$6$,

$4n+2=6$

$=>n=1$ (Considering each ring as a
separate entity).

Therefore, the ring is aromatic in nature.

Option D)

This molecule is planar ($s{p}^2$ hybridised carbon) and cyclic.

Number of π electrons involving in
delocalisation = $4$ ,

$4n=4$

$=>n=1$

Hence, it follows the 4n π electrons rule, so the molecule is anti-aromatic.