Aromatic N-containing rings

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

nfg05

Full Member
10+ Year Member
15+ Year Member
Joined
May 20, 2008
Messages
159
Reaction score
2
I'm having some trouble deciding whether lone pairs on N atoms involved in a six-membered ring contribute to the pi electron system or not, and subsequently whether or not these N-containing rings are aromatic. Anyone have a good system?
 
I'm having some trouble deciding whether lone pairs on N atoms involved in a six-membered ring contribute to the pi electron system or not, and subsequently whether or not these N-containing rings are aromatic.


Yes they do, and yes they are.
 
Count the number of pi electrons. If the lone pair would satisfy Huckel's rule (4n+2) and make, say, 6 pi electrons, then it does contribute to the aromaticity of the N ring. However, if there already exists 6 pi electrons without the N lone pair, then the electrons do not contribute to the aromaticity, and the N can be used as a Lewis base (free electron pair).
 
Count the number of pi electrons. If the lone pair would satisfy Huckel's rule (4n+2) and make, say, 6 pi electrons, then it does contribute to the aromaticity of the N ring. However, if there already exists 6 pi electrons without the N lone pair, then the electrons do not contribute to the aromaticity, and the N can be used as a Lewis base (free electron pair).

this isn't quite correct though. you can't just choose to count the lone pair if it helps you satisfy Huckel's rule, or to not count it if it doesn't help. (look at cyclobutadiene, and replace one of the carbons with a nitrogen. it's still antiaromatic, even though you could count the lone pair and say it's aromatic)

Basically, you count the lone pair if it extends perpendicular to the plane of the ring. This way it can participate in the resonance delocalization of the rest of the pi electrons.

Soooo, the way I teach it is this:
When deciding whether or not to count heteroatom lone pairs as pi electrons:

Case I: The heteroatom has one lone pair:

1. Is the heteroatom double bonded to any other atom? If so, then don't count the lone pair as pi electrons.

2. If not, count the lone pair as two pi electrons.

Case II: The heteroatom has two lone pairs (this generally applies to oxygen and sulfur):

1. Count only ONE of the two lone pairs as two pi electrons.


This way applies very broadly. Like I said, it's all because the lone pair extends perpendicular to the plane of the ring. When you have only one lone pair, if there heteroatom is double bonded to another atom, then if you draw out the geometry, you find that the only place you can put the lone pair is IN the plane of the ring, so it can't participate in the fully conjugated system. Instead, the two electrons in the double bond are perpendicular to the plane and are counted. If there is no double bond, then when you draw out the geometry you are forced to put the lone pair perpendicular to the ring so it does count.

If there are two lone pairs, obviously they both can't be perpendicular, so one is perpendicular, the other is parallel, and so you only count the perpendicular one.
 
yeah sorry i forgot to add a corollary to my explanation. You have to make sure that each atom in the ring is sp2 hybridized. You're right that the cyclobutadiene with a N wouldn't work because if the N lone pair donated to the ring to satisfy Huckel's rule, the N would no longer be sp2 hybridized. Anyways, u provided an excellent explanation
 
Top