
Azulene is antiaromatic.
Answer
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Hint: There are some necessary conditions for the compounds to be aromatic. The compound should have a single cyclic cloud of delocalized $\pi -electrons$, the compound must be planar, and it should contain ($4n+2$)$\pi -electrons$ where n = 1, 2, 3…....etc.
Complete step by step answer:
Based on the molecular orbital theory, Hukel has predicted that electrons in cyclic conjugated polyenes (cyclic polyenes having alternate single and double bonds) containing ($4n+2$)$\pi -electrons$ where n = 1, 2, 3, 4, ….etc are completely delocalized. This makes the compound stable and such compounds are called aromatic compounds.
So the necessary and sufficient conditions for the compound to aromatic are that it should have a single cyclic cloud of delocalized $\pi -electrons$ above and below the plane of the molecule, the compound must be planar, and the compound must contain ($4n+2$)$\pi -electrons$ where n = 1, 2, 3, …….etc.
The molecule that doesn't satisfy any one or more of the above conditions then is non-aromatic.
And if the compound is having ($4n$)$\pi -electrons$then the compound is antiaromatic.
So, in Azulene there are 5 double bonds which means that it has 10 $\pi -electrons$ as shown below:
So, when we put n as 2 in the equation of aromatic compound, we get:
\[(4n+2)\text{ }\pi -electrons\]
\[(4(2)+2)\text{ }\pi -electrons\]
$10\,\pi -electrons$
So, the Azulene compound is aromatic, not antiaromatic and it will give all the reactions of an aromatic compound. Thus the statement is false.
Note: The $10\,\pi -electrons$ in Azulene is contributed by two aromatic ions, by tropylium cation and the cyclopentadienyl anion. Azulene gives Friedel-Crafts-like reactions and both its peripheral bonds have the same length.
Complete step by step answer:
Based on the molecular orbital theory, Hukel has predicted that electrons in cyclic conjugated polyenes (cyclic polyenes having alternate single and double bonds) containing ($4n+2$)$\pi -electrons$ where n = 1, 2, 3, 4, ….etc are completely delocalized. This makes the compound stable and such compounds are called aromatic compounds.
So the necessary and sufficient conditions for the compound to aromatic are that it should have a single cyclic cloud of delocalized $\pi -electrons$ above and below the plane of the molecule, the compound must be planar, and the compound must contain ($4n+2$)$\pi -electrons$ where n = 1, 2, 3, …….etc.
The molecule that doesn't satisfy any one or more of the above conditions then is non-aromatic.
And if the compound is having ($4n$)$\pi -electrons$then the compound is antiaromatic.
So, in Azulene there are 5 double bonds which means that it has 10 $\pi -electrons$ as shown below:
So, when we put n as 2 in the equation of aromatic compound, we get:
\[(4n+2)\text{ }\pi -electrons\]
\[(4(2)+2)\text{ }\pi -electrons\]
$10\,\pi -electrons$
So, the Azulene compound is aromatic, not antiaromatic and it will give all the reactions of an aromatic compound. Thus the statement is false.
Note: The $10\,\pi -electrons$ in Azulene is contributed by two aromatic ions, by tropylium cation and the cyclopentadienyl anion. Azulene gives Friedel-Crafts-like reactions and both its peripheral bonds have the same length.
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