Aromatic? Destroyer #100 2014

[TheEpicFruitCake]

How is answer choice C not follow the 4n+2 rule? The solution says it has 4 pi electrons, but I don't see how. The lone pair shouldn't be contributing to resonance like answer choice A right? Either way if it did contribute the carbon would become sp hybridized with only 2 RHEDs which I have never seen in my ochem class.
Thx!

Alright I pulled out my old exam and I have the same compound as in answer choice C except with a hydrogen (and positive charge) instead of the lone pair. This compound was considered aromatic according to the answer key. The lone pair should act the same as the hydrogen in this situation shouldn't it? Making the carbon sp2 and not contributing to pi electrons

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[nickh]

Chad gave a simplified version of the 4n-2 rule. So for an aromatic compound, you want to have an odd number of double bonds (and 1 lone pair = 1 double bond). For C, it has 1 double bone + 1 lone pair = 2 double bones, which is an even number, making it anti aromatic.

For A, you have 2 double bonds, and 1 lone pair = 3 double bonds, which is an odd number, making it aromatic. Hope this helps

 

[nickh]

I find it is actually easier to count how many double bonds are there, as long as you know you know why you want an odd number (because represents the 4n-2 rule), you are good

 

[TheEpicFruitCake]

I know A is correct for sure. My question is how C is anti-aromatic. The lone pairs in C don't contribute to pi electrons, else that carbon would be sp, which I don't see as working. The electrons should only contribute if it makes an sp2 carbon I thought. I guess I've never seen an aromatic compound with an sp carbon.

Just to be sure then, sp hybridized carbons are stable in aromatics? I've just never seen it ever

 

[eugpeng]

The lone pairs are actually contributing so you end up with 2 double bonds, making it anti aromatic

 

[TheEpicFruitCake]

Thanks guys, usually a carbon with a lone pair also has a negative charge (implying a hydrogen bond as well) so I just got I confused. If it was negatively charged then definitely the lone pairs contribute. I will accept it lol

Thanks again!