aromacity on cyclopropene

[d wannabe]

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For A
I think it has 2 pi electron --> Aromatic

For B
It has 4 pi electron --> Anti aromatic

For C
I'm not sure about this one. Does it have 4 pi electron making antiaromatic? Or 2 pi electron making nonaromatic?

 

[harrygt]

View attachment 10586

For A
I think it has 2 pi electron --> Aromatic

For B
It has 4 pi electron --> Anti aromatic

For C
I'm not sure about this one. Does it have 4 pi electron making antiaromatic? Or 2 pi electron making nonaromatic?

Regarding C, the carbon at the top has no formal charge, therefore it is a CARBENE. THe electron pair is in an sp2 orbital, therefore you can't count it as Pi electrons =====> Compound C is not aromatic.

 

[d wannabe]

Regarding C, the carbon at the top has no formal charge, therefore it is a CARBENE. THe electron pair is in an sp2 orbital, therefore you can't count it as Pi electrons =====> Compound C is not aromatic.

So if it has 2 pi electron, then wouldn't that be aromatic?
4n+2 = 2 --> n=0?

I just found this structre in Org Destroyer #27.
The explanation says that this is Anti aromatic.
Which means it has 4 pi electrons.. but doesn't make sense.

Anyone?

 

[WayneSU]

it is anti......think of resonance

 

[d wannabe]

it is anti......think of resonance

Hmm
can someone explain little bit more?

 

[harrygt]

I see. What you say makes sense. I'm gonna see my Ochem prof. next week. I will definitely ask him this question.

 

[y0ssarian]

Tough question. I believe it is aromatic, but it's not your standard aromatic molecule. First, you start out with two electrons from the double bond (the easy part). To obtain a fully conjugated system, you need the diradical carbon to adopt an sp2 conformation. With the normal rules, this would put one of the radical electrons into an sp2 orbital (this is the orbital that doesn't interact with the conjugated pi system), and put the other electron into the participating p orbital, making for three total electrons in the system. However, because aromaticity is a favorable state for cyclic molecules, both the radical electrons will be pushed into an sp2 orbital, leaving the p orbital of the diradical carbon empty. This allows the molecule to be fully conjugated, and be aromatic. I hope that helps.

 

[license2kill]

According to Huckel's rule, the 4n+2 rule only applies to a planar structure thus sp2 system. In A, the + charge indicates that there is a bond with one hydrogen which makes it an sp2 orbital (n=0). In the second case, the - charge indicates that there is one bond with hydrogen as well, thus the lone pair is not used in 4n+2 rule because it is sp3 orbital. In the third case, the "no charge" indicates that there are no bonds associated with hydrogen on that specific carbon thus making it a sp2 orbital thus lone pair is counted in 4n+2 rule (n=1/2) making it antiaromatic.

This would be my take, feel free to correct if my thinking is wrong. 🙂