[timeless]

i'm studying for the acs test i have on monday and came across this question:

which structures would be classified as aromatic:


A) I, II, III
B) II and III only
C) I and II only
D) I and III only

the answer is given to be A, I, II, and III.
but i thought the qualifications to be aromatic are:
1. completely conjugated cyclic compound (which they are)
2. follows huckels rule (which they do)
3. flat molecule
4. NO NONBONDING ORBITALS
that means that III should not be in the answers because if u draw it with one corner pointing down, and do MO theory, there are bonding sites on the zero of energy.

so... ?

[Yggdrasil]

If you move the plus sign to the other carbon with no double bonds on III, it be aromatic.  Although it does have two nonbonding orbitals, it just has 6 pi electrons, so none of the electrons will occupy the nonbonding orbitals.  If the molecule had two extra electrons, it would be antiaromatic since the electrons would each occupy a nonbonding orbital, creating a diradical species, which is highly unfavorable.

[timeless]

ooo ok. so it doesn't matter that there are bonding SITES on the zero of energy, so long as there are no ELECTRONS in the bonding sites. if there was, then it would be NONaromatic...

yah?

[Yggdrasil]

Yup, that is correct.

[timeless]

hrm... that's really weird because i have it written in my notes many times that it's not aromatic if there's bonding SITES on the zero of energy. i've also tried to look aromaticity online and there aren't anything that says that that is one of the rules. just that everything has to be flat, p orbitals have to be parallel, huckel's rule and obviously cyclic. i'll ask my professor tmrw before i take the acs test. thanks

[Yggdrasil]

Yeah, I've never heard of a rule preventing there from being empty nonbonding orbitals.  That's certainly not one of the rules listed in my organic chem text (Brown and Foote).

Anyway, good luck on your test.

[kelaklub]

Just out of curiousity, but how is the third structure conjugated when conjugation is defined as double bonds being seperated by one single bond. The leftmost double bond has more than one single bond seperating it on both sides from the next double bond.

[Yggdrasil]

The left-most double bond is adjacent to two carbocations (note the structure is drawn incorrectly since the second formal charge should be on the carbon with no double bonds and not on one of the double bonds).  Carbocations have empty p-orbitals which are capable of forming a conjugated pi system.  So conjugation doesn't necessarily require double bonds being separated by one single bond; all it requires is adjacent atoms with parallel p-orbitals.