Can you explain this picture?

[YoonS]

Advertisement - Members don't see this ad

Okay I was supposed to find anti-aromatic compound, which I got right,

but I got the answer right by doing elimination.

I understand that in order to be anti-aromatic, it has to be planar

and also has hucker number of 4,8,12, and so on.

But this picure it seemed like there is one radical on each of two carbons.

Is it right? or is it just two electrons in one carbon?

Thanks

 

[addis4ever]

according to chad, when u have electrons on the outside of the ring, u see them as lone pairs and u count them. so in this case u have 4 e- , which doesn't follow the 4n... rule so it's anti-aromatic.

 

[yappy]

Not True. Identifying pi electrons in a conjugated system doesnt have to do with what side of the ring the electrons appear on.

My understanding is that a lone pair electrons count as apart of the pi ssytem when they are attached to an atom with a non-hybridized/open P orbital (such as sp2). Therefore, it doesnt matter where you draw electrons. They may be inside or outside a ring and they still would or would not count as pi.

4n+2. aromatic
4n antiaromatic

That structure has one double bond and what looks like two radical electrons. Because the carbons are sp2 that have the radicals I'm going to go out on a limb and say that they participate in the pi system. Therefore, we have four electrons in the pi system consistant with antiaromatic.

Do you guys agree?

according to chad, when u have electrons on the outside of the ring, u see them as lone pairs and u count them. so in this case u have 4 e- , which doesn't follow the 4n... rule so it's anti-aromatic.

 

[invictusx]

Not True. Identifying pi electrons in a conjugated system doesnt have to do with what side of the ring the electrons appear on.

My understanding is that a lone pair electrons count as apart of the pi ssytem when they are attached to an atom with a non-hybridized/open P orbital (such as sp2). Therefore, it doesnt matter where you draw electrons. They may be inside or outside a ring and they still would or would not count as pi.

4n+2. aromatic
4n antiaromatic

That structure has one double bond and what looks like two radical electrons. Because the carbons are sp2 that have the radicals I'm going to go out on a limb and say that they participate in the pi system. Therefore, we have four electrons in the pi system consistant with antiaromatic.

Do you guys agree?

Agree.
Everything about aromaticity applies to both antiaromatics and aromatics except that for anti-aromatics, it's 4n.

 

[mmoosavi41]

i got this same question on topscore test 1 yesterday. i was confused too but watched chad's vids and it cleared it up for me.

 

[yappy]

Was chads video consistant with what you're seeing here? Did he talk about radicals?

i got this same question on topscore test 1 yesterday. i was confused too but watched chad's vids and it cleared it up for me.