1. Employing the Huckel rule, clearly show how both molecules A and B are both aromatic. A B Huckel rule states that for a molecule to be aromatic the number of electrons must be equal to 4n + 2, where n = 0,1,2,3... etc. For compound A there are 6 ? electrons from the three double bonds. The lone on the nitrogen does not partake in the resonance. solutionspile.com

Question

1. Employing the Huckel rule, clearly show how both molecules A and B are both aromatic. A B Huckel rule states that for a molecule to be aromatic the number of electrons must be equal to 4n + 2, where n = 0,1,2,3... etc. For compound A there are 6 ? electrons from the three double bonds. The lone on the nitrogen does not partake in the resonance. Therefore, there are 6 = 4n+ 2 ? electrons, and n = 1. For compound B there are 4 r electrons from the double bonds. The lone pair on the nitrogen atom participates in the resonance. Therefore, there are 6 = 4n+ 2 ? electrons in total, and n = 1. Both compounds A and B are both aromatic because: In A, the lone pair is not part of the system making n = 1 for pyridine (A). In B, the lone pair participates in the system making n = 1 for pyrrole (B) as well.

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Employing the Huckel rule, clearly show how both molecules

A

and

B

are both aromatic. B Huckel rule states that for a molecule to be aromatic the number of

?

electrons must be equal to

4 n + 2

, where

n = 0, 1, 2, 3 \dots

etc. For compound

A

there are

6 ?

electrons from the three double bonds. The lone on the nitrogen does not partake in the resonance. Therefore, there are

6 = 4 n + 2 ?

electrons, and

n = 1

. For compound B there are

4 ?

electrons from the double bonds. The lone pair on the nitrogen atom participates in the resonance. Therefore, there are

6 = 4 n + 2 ?

electrons in total, and

n =

1. Both compounds A and B are both aromatic because: In A, the lone pair is not part of the

?

system making

n = 1

for pyridine (A). In

B

, the lone pair participates in the

?

system making

n = 1

for pyrrole (B) as well.

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