[kelaklub]

Due to all the help I received on this forum, I was able to answer most of the aromatic questions on my sample tests correctly. This one however has me a little stumped. It may be a typo, I don't know. The answer is stated as choice C.

But I thought that choice D also represents an Aromatic molecule, because

• It is cyclic
• All the carbons are sp² hybridized.
• It satisfies Huckels rule, because I don't think the lone pairs of electrons will take part in the resonance because that would not make all the carbons sp2 hybridized.
  

Are these valid reasons? Thanks.

[Winga]

As you said all the carbons are sp2 hybridrized, the lone pairs of that carbon can also be delocalized into the pi system (remember, there is a hydrogen attached to that carbon). Now the no. of e- in the pi system is 4 which cannot fulfill 4n + 2 rule. If the lone pair doesn't delocalize to pi system, this carbon will become sp3 hybridrized that blocks the delocalization of e- in pi system.

[Dan]

I agree with kelaklub. If it's a singlet carbene, I would answer with E (ie. C and D).

The reason I would expect D to be aromatic in it's singlet state (that's with a lone pair on the C rather than a diradical) is because, when sp2 hybridised, you can put the lone pair in an sp2 orbital, leaving an empty p orbital => 2 electrons in the pi system => aromatic.

I don't think the lone pairs of electrons will take part in the resonance because that would not make all the carbons sp2 hybridized

I don't agree with that argument.
The only way anything on that c will interact with the pi system (take part in resonance) is if it is in a p orbital - so in the hypothetical reverse situation to what I described above, the lone pair is in a p orbital, and there is an empty sp2 hybrid orbital. All the carbons would still be still sp2! - but it is not aromatic (and so the cofiguratiion would be unstable) because the number of electrons is 4.

[Dan]

(remember, there is a hydrogen attached to that carbon).

I don't think there is... I've never seen a trivalent carbene

[Yggdrasil]

The problem probably meant to put a negative formal charge on the carbon atom with the lone pair making it a standard carbanion.  In this case, because there is a hydrogen and two C-C bonds occupying the sp<sup>2[/sub] orbitals, the lone pair must be in the p-orbital creating a system with 4n electrons which is antiaromatic by Huckle's criteria.

However, as the problem is written with no formal charge on the carbon, I agree with Dan's answer.  But I really doubt someone would mean to put a carbene into a problem like this.</sup>

[AWK]

Due to all the help I received on this forum, I was able to answer most of the aromatic questions on my sample tests correctly. This one however has me a little stumped. It may be a typo, I don't know. The answer is stated as choice C.

But I thought that choice D also represents an Aromatic molecule, because

• It is cyclic
• All the carbons are sp² hybridized.
• It satisfies Huckels rule, because I don't think the lone pairs of electrons will take part in the resonance because that would not make all the carbons sp2 hybridized.
  

Are these valid reasons? Thanks.

Only pyridine is aromatic (C)

[Dan]

Only pyridine is aromatic (C)

What is your justification for D not being aromatic?

[Organishe]

I think that the authors of this question made a silly mistake, in that they did not include the formal charge. Let's assume for now that it is indeed a carbanion, and not a carbene. To have p orbitals available for resonance at every carbon, the carbanion must be sp² hybridized. At the same time, doing this does not satisfy Hückel's rule.  If we leave the carbon sp³ hyrbridized, then it is not a fully conjugated system.  So, assuming it is indeed a carbanion, it is certainly not aromatic.

Taking the case of the carbene, well I don't know enough about carbenes to answer it, honestly.  But from the context of the question, I highly doubt the authors would intend for you to consider a carbene.

So, my answer: C

[AWK]

Only pyridine is aromatic (C)

What is your justification for D not being aromatic?

As Winga after Huckel said - 4n+2 rule

[Dan]

But look at the structure of a singlet carbene, http://wngr343-pc1.science.oregonstate.edu/gablek/CH630/carbenorbs.gif
The empty p orbital (LUMO) overlaps with the pi system, giving 2 electrons in the pi system, ie n=0 for Huckel's rule => aromatic. The HOMO is orthogonal to the pi system.

If it was a carbanion, then it wouldn't be aromatic, but it isn't, it's a carbene.

[Winga]

Will the structure D achieve maximum multipilicity?

[barcrphd]

i think there is nothing too much in this question which worth such a lengthy debate. as suggested above, the system D doesnt obey the 4n+2 rule. hence it is not aromatic.

[Dan]

the system D doesnt obey the 4n+2 rule. hence it is not aromatic.

My point is that D will obey the 4n + 2 rule in the singlet state (where n = 0). The LUMO of the singlet state is a p orbital, see link above.

Sure, the triplet state will not be aromatic, but I expect the singlet state is lower in energy than the triplet due to the aromatic stabilisation of the singlet state => the ground state the carbene D is the singlet state which is aromatic.

I have drawn a pretty picture http://users.ox.ac.uk/~some1599/Images/SingletCarbeneD.jpg

If  there is a formal negative charge then, I agree, it will not be aromatic.

[Winga]

Can the structure D undergo aromatic reactions? (e.g. electrophilic aromatic substitution)

[Dan]

Well, I have never made it so I dont know, but I would have thought that it's reactivity resembles that of a carbene more than benzene. An example would be (rapid) insertion to C-H bonds.
I don't think it would be stable.
If the point you are getting at is "If it doesn't react like an aromatic compound, is it really aromatic?", then I can see exactly where you are coming from! Technically it is aromatic, in the sense that it obeys the 4n+2 rule, but in practice I would be suprised if this label is meaningful.