Why is this not "NON- aromatic"

[rippinitez]

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I'm trying to figure out why this molecule is not considered "non-aromatic".

 

[Blarelli]

It follows Huckel's rule (4n+2), each carbon has P-orbital electrons, and I have no proof it isn't planar.

 

[CANgnome]

anything above a 7 carbon ring is non-planar

 

[lex1489]

I was never really sure about the 'planar' part of the rule. Benzene is non-planar.. Chair, boat, twisted boat

 

[Huskk]

It follows Huckel's rule (4n+2), each carbon has P-orbital electrons, and I have no proof it isn't planar.

Doesn't it have to be non-hybridized p orbitals? As far as I can tell those electrons are sitting in sp3 orbitals. I may be having a dumb moment though.

 

[LuckMC11]

1. must have a ring
2. every atom must have a p orbital. otherwise --> nonaromatic
3. ring must be flat. otherwise --> nonaromatic
4. 4n+2. if 4n=antiaromatic

If a 4n structure can conform to a nonplanar configuration, it is nonaromatic. I believe the rule of thumb is that any ring with over 8 atoms can conform to nonplanar.

here

 

[rippinitez]

So the consensus is that it is aromatic?

 

[Blarelli]

anything above a 7 carbon ring is non-planar

Not true. Look at 18-annulene. It's planar.

I was never really sure about the 'planar' part of the rule. Benzene is non-planar.. Chair, boat, twisted boat

Also not true. Cyclohexane follows the chair, boat, twisted conformations. Benzene is planar.

 

[wantVCUdental]

it's not aromatic, can you show me how these electrons are delocalized? YOU CAN'T because the two negative charges are not part of the system... you can't bond 10 electrons on a carbon!

this Q has nothing to do w/ planar geometry

 

[lex1489]

Not true. Look at 18-annulene. It's planar.

Also not true. Cyclohexane follows the chair, boat, twisted conformations. Benzene is planar.

Touche`. I always do that 😳

 

[rippinitez]

The question asks "which is not aromatic". The figure I attached in the first post was not the correct answer. The correct answer was this:

 

[rippinitez]

Hopefully what I'm asking makes sense.

 

[rippinitez]

I'm just trying to figure out why the structure from post 1 is not considered "non aromatic".

 

[iamgosu]

The question asks "which is not aromatic". The figure I attached in the first post was not the correct answer. The correct answer was this:

this one's easy
4n+2 = aromatic.

this has 8 pi electrons (2 from each double-bond)
if n=1, 6pi e-
if n=2, 10pi e-

 

[UCB05]

it's not aromatic, can you show me how these electrons are delocalized? YOU CAN'T because the two negative charges are not part of the system... you can't bond 10 electrons on a carbon!

this Q has nothing to do w/ planar geometry

the two pairs can sit in pi orbitals. It has 4n+2 = 10 electrons sitting in pi orbitals that become delocalized into the ring. It's the same concept that explains why cyclopentadiene anion is aromatic. The concensus on a google search on "cyclooctatetraene dianion aromaticity" seems to indicate it is aromatic as well (there's a paper whose abstract concludes that the ring may not be aromatic because it is not planar).

I personally would've guess it was nonaromatic though. It's got a lot of carbons in that thar ring.

 

[chromosome21]

guys, the question asks why the structure is "not" "non-aromatic". that means it can be aromatic.
1. it's a ring (meets the requirement for a molecule to be aromatic)
2. conjugative pi bonds. or every atom must have a p orbital. This molecule does have all the atoms that have p orbital.
3. planar ring .. umm.. not sure about this one for this molecule, but let's just consider the similar kind of a molecule is planar, so this one might also be planar.
4. 4(2)+2 = 10, and the molecule does have 10 pi electrons. meets the huckle rule : 4n+2 requirement. [antiaromatic = 4n]

woh this molecule IS definitely NOT non-aromatic. This might be an aromatic compound. Now why did i count the four non bonded electrons? well, they seem to be in the P orbital of the atoms like pyrrole. Don't forget the hydrogens are not shown, and all the atoms of the molecule have one hydrogen attached to them. These two carbons seem to act like pyrrole, that's why I counted those four non-bonded electrons.

the molecule kinda looks like if a benzene's one hydrogen is pulled out, and one bond has not been drawn. draw such molecule and you'd find a similar looking compound.

Not sure if my explanation makes sense. I could be wrong!

 

[UCB05]

guys, the question asks why the structure is "not" "non-aromatic". that means it can be aromatic.
1. it's a ring (meets the requirement for a molecule to be aromatic)
2. conjugative pi bonds. or every atom must have a p orbital. This molecule does have all the atoms that have p orbital.
3. planar ring .. umm.. not sure about this one for this molecule, but let's just consider the similar kind of a molecule is planar, so this one might also be planar.
4. 4(2)+2 = 10, and the molecule does have 10 pi electrons. meets the huckle rule : 4n+2 requirement. [antiaromatic = 4n]

woh this molecule IS definitely NOT non-aromatic. This might be an aromatic compound. Now why did i count the four non bonded electrons? well, they seem to be in the P orbital of the atoms like pyrrole. Don't forget the hydrogens are not shown, and all the atoms of the molecule have one hydrogen attached to them. These two carbons seem to act like pyrrole, that's why I counted those four non-bonded electrons.

the molecule kinda looks like if a benzene's one hydrogen is pulled out, and one bond has not been drawn. draw such molecule and you'd find a similar looking compound.

Not sure if my explanation makes sense. I could be wrong!

exactly. Really the only thing for me is determining whether or not that thing is planar. I don't know if the question says so, so I could've swung either way. In normal testing conditions, between this and another choice, I would hope the other (right) choice was blatantly nonaromatic.

 

[chromosome21]

The question asks "which is not aromatic". The figure I attached in the first post was not the correct answer. The correct answer was this:

the second molecule picture is definitely not aromatic. it got 8 pi electrons, and does not meet the requirement of having 4n+2 pi electrons. This one MUST not be the aromatic. (it's actually an anti-aromatic). So this one should be picked as an answer cause it's not "not non-aromatic" (or not "aromatic"). Hope this makes sense?

 

[UCB05]

this molecule is definitely not aromatic. it got 8 pi electrons, and does not meet the requirement of having 4n+2 pi electrons. This one MUST not be the aromatic. (it's actually an anti-aromatic). So this one should be picked as an answer cause it's not "not non-aromatic" (or not "aromatic"). Hope this makes sense?

Ohay, didn't see that post in the middle of the thread. That's a definite answer.

 

[chromosome21]

Ohay, didn't see that post in the middle of the thread. That's a definite answer.

yep yep. I didn't see that at first, but then read the whole post, and that was the answer right there!

 

[dhk5]

i saw this on achiever too but i'm pretty sure it's wrong....b/c planar means every atom is sp2 hybridized and having two electrons there makes that an sp3. for example a carbanion CH4 with 2e- is sp3 and CH3+ missing 2e- is sp2 (like in this picture)

therefore, in your first image is not aromatic, even though it follows 4n+2. but in your second image that you provided, anti-aromatic because it follows 4n which means that it will avoid being planar and conjugated system because it is energetically unfavorable.

on a side note, not sure if you noticed, but in another achiever test, they add HNO3/H2SO4 to aniline, but you cannot carry out a nitration of that because it is an oxidizing agent and you would have to protect the amino group of benzene...at least that's what it says in my orgo book

 

[UCB05]

i saw this on achiever too but i'm pretty sure it's wrong....b/c planar means every atom is sp2 hybridized and having two electrons there makes that an sp3. for example a carbanion CH4 with 2e- is sp3 and CH3+ missing 2e- is sp2 (like in this picture)

therefore, in your first image is not aromatic, even though it follows 4n+2. but in your second image that you provided, anti-aromatic because it follows 4n which means that it will avoid being planar and conjugated system because it is energetically unfavorable.

Not necessarily. The anion does not exist solely as a set of double bonds and a lone pair on a carbon. The charge is delocalized and the pi electrons exist as a cloud distributed in the ring. Such is the case with cyclopentadiene anion, which is an aromatic carbanion ring. In these instances, the lone pair can be considered as occupying a pi orbital rather than a molecular sp3 orbital as the molecule adopts a planar shape, resulting in aromaticity. The properties of methane carbocation/carbanion molecular geometry are not relevant to this question, as it does not involve any ring structure or resonance charge delocalization.

 

[wantVCUdental]

Not necessarily. The anion does not exist solely as a set of double bonds and a lone pair on a carbon. The charge is delocalized and the pi electrons exist as a cloud distributed in the ring. Such is the case with cyclopentadiene anion, which is an aromatic carbanion ring. In these instances, the lone pair can be considered as occupying a pi orbital rather than a molecular sp3 orbital as the molecule adopts a planar shape, resulting in aromaticity. The properties of methane carbocation/carbanion molecular geometry are not relevant to this question, as it does not involve any ring structure or resonance charge delocalization.

I stand by my answer this is not aromatic, and if you work out all the resonance forms, you will notice that in each resonance system you can only delocalize 1 of the lone pairs, therefore the whole electron system is NOT delocalized and therefore there is no resonance

 

[UCB05]

I stand by my answer this is not aromatic, and if you work out all the resonance forms, you will notice that in each resonance system you can only delocalize 1 of the lone pairs, therefore the whole electron system is NOT delocalized and therefore there is no resonance

resonance is not the reason for aromaticity per se. It's just how we rationalize the phenomena diagrammatically . In the op's molecule, the lone pairs don't bounce around in different resonance forms. Rather, all the electrons exist in the extended pi cloud. Whether or not you can draw the lone pairs moving simultaneously from one drawing to the next does not change the fact that there is a continuous pi system in the dianion molecule.

 

[americanpierg]

I stand by my answer this is not aromatic, and if you work out all the resonance forms, you will notice that in each resonance system you can only delocalize 1 of the lone pairs, therefore the whole electron system is NOT delocalized and therefore there is no resonance

You can delocalize both... all it means to delocalize the electrons is that they can be moved from the original atom they were with to a neighboring atom. You can create a new resonance form just by shifting the double bonds and lone pairs around the molecule (in reality youre just moving a single electron around the ring and new double bonds and lone pairs are created). Remember that no single resonance form dominates, they all exist simultaneously. So if you rotate all resonance forms fast enough it all blurs together evenly... and this is what you get when you think "delocalized."

This is much easier if you draw things out like you learned when you first started learning about organic chemistry. Draw out the molecule and draw out the orbital on each atom of the ring. If you draw the orbitals in a up-down orientation, single electrons will reside in each carbon that shares a double bond. the electron pair will be present in the two orbitals above the remaining two carbons. These electron pairs are part of the conjugated system, ie they are part of the ring and can therefore delocalize around the ring. The lone pairs would NOT delocalize if they are not part of the ring structure (ie the lone pairs of pyridine), but in this case they are part of the ring.


http://www.chemtube3d.com/model/aromatic_heterocycles/pyrrole/pyrrole-stucture.jpg
Click there for the oribital structure of pyrolle to give you a picture of these orbitals.

Also, whether a molecule is planar or not is also another requirement for aromaticity so this question does ask about planar geometry even if it didnt specifically say it since its asking about aromaticity

 

[chromosome21]

i saw this on achiever too but i'm pretty sure it's wrong....b/c planar means every atom is sp2 hybridized and having two electrons there makes that an sp3. for example a carbanion CH4 with 2e- is sp3 and CH3+ missing 2e- is sp2 (like in this picture)

therefore, in your first image is not aromatic, even though it follows 4n+2. but in your second image that you provided, anti-aromatic because it follows 4n which means that it will avoid being planar and conjugated system because it is energetically unfavorable.

on a side note, not sure if you noticed, but in another achiever test, they add HNO3/H2SO4 to aniline, but you cannot carry out a nitration of that because it is an oxidizing agent and you would have to protect the amino group of benzene...at least that's what it says in my orgo book

This diagram does NOT have conjugated pi system where the electrons can localize. It has to be an SP3 configuration. However, if you think of the N atom in pyrrole, it should also be SP3 configuration based on what your drawing suggests. But it's not. It acts as trigonal planer, and the dumbbell shaped p orbital sticks out up and down and the nonbonded electrons then participate in the delocalization of conjugated electrons. So I don't support your theory.

Could you plz upload the achiever question you were talking about in the second paragraph? Thx!

 

[chromosome21]

You can delocalize both... all it means to delocalize the electrons is that they can be moved from the original atom they were with to a neighboring atom. You can create a new resonance form just by shifting the double bonds and lone pairs around the molecule. Remember that no single resonance form dominates, they all exist simultaneously. So if you rotate all resonance forms fast enough it all blurs together evenly... and this is what gets when you think "delocalized."

This is much easier if you draw things out like you learned when you first started learning about organic chemistry. Draw out the molecule and draw out the orbital on each atom of the ring. If you draw the orbitals in a up-down orientation, single electrons will reside in each carbon that shares a double bond. the electron pair will be present in the two orbitals above the remaining two carbons. These electron pairs are part of the conjugated system, ie they are part of the ring and can therefore delocalize around the ring. The lone pairs would NOT delocalize if they are not part of the ring structure (ie the lone pairs of pyridine), but in this case they are.

http://www.chemtube3d.com/model/aromatic_heterocycles/pyrrole/pyrrole-stucture.jpg
Click there for the oribital structure of pyrolle.

Also, whether a molecule is planar or not is also another requirement for aromaticity so this question does ask about planar geometry even if it specifically say it since its asking about aromaticity

Hey, I wrote the same 😛

 

[KittySquared]

The first structure IS aromatic because it has 10 pi electrons.

Why? The 3 double bonds contributed 6 pi electrons total.

The lone pairs on the top 2 WILL participate in contributing to the pi electrons count because the carbons they're attached to do not contribute to double bonds. If those carbons don't participate in double bonds, those lone pairs WILL be considered as pi electrons.

Hence, you have 6pi from the 3 double bonds + 4pi from the 2 carbons' lone pairs.

And the pi electron count are aromatic in the following numbers:

2, 6, 10, 14, 18.... etc. So you got 10e- therefore it's aromatic.

EXCEPTION: If you have 2 lone pairs attached to an atom in the ring that does not participate in a double bond, then ONLY 1 of the lone pairs will be in the P orbitals.

e.g. :N:=
So this nitrogen in the ring is not going to contribute either of its e- pairs to the pi electron count, whereas

:N:-

will contribute 1 of its electron pairs as pi electrons, thus 2 pi electrons.

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I hope this makes sense. Feel free to correct me if anyone sees flaws.

 

[orgohacks]

I stand by my answer this is not aromatic, and if you work out all the resonance forms, you will notice that in each resonance system you can only delocalize 1 of the lone pairs, therefore the whole electron system is NOT delocalized and therefore there is no resonance

It's aromatic. Tom Katz at Columbia made the dianion back in 1960 (J Am Chem Soc 1960 p. 3784) and showed the following:
1) it is unusually easy to form this dianion through reduction of cyclooctatetraene with a reducing agent (potassium) and the dianion has unusual stability relative to other dianionic species
2) the NMR spectrum shows a single resonance, inconsistent with what would be expected from an alternating bond structure
3) NMR also shows a downfield shift consistent with a ring current, which also implies that the molecule is flat.

This follows from the prediction that molecules with 4n+2 electrons are aromatic, since 10 is a Huckel number.

Where are the 10 electrons? They're in p orbitals. Just like pyrrole and furan have a p orbital with 2 electrons, in the case of the cyclooctatriene dianion you'll have a resonance form with 2 p orbitals each containing an electron pair. the rest of the electrons can be drawn as being in double bonds.

Maybe one way it would help you to see the resonance is if you draw a monosubstituted cyclooctatetraene. If you start drawing resonance forms you'll clearly see how the negative charge moves around the ring.


hope this helps - James

 

[americanpierg]

EXCEPTION: If you have 2 lone pairs attached to an atom in the ring that does not participate in a double bond, then ONLY 1 of the lone pairs will be in the P orbitals.

e.g. :N:=
So this nitrogen in the ring is not going to contribute either of its e- pairs to the pi electron count, whereas

:N:-

will contribute 1 of its electron pairs as pi electrons, thus 2 pi electrons.

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I hope this makes sense. Feel free to correct me if anyone sees flaws.

:N:= the nitrogen atom here would actually have a negative charge becuase its associated with 6 electrons instead of 5. i dont know how this work but most of the time that one lone pair would indeed delocalize over and create a triple bonded neutral nitrogen (5 total electrons on nitrogen...hence why nitrogen likes making triple bonds)

As for why lone pairs on an atom that is already associated with a double bond doesnt delocalize, its because that p orbital is already being occupied with the double bond electron, and therefore the electron pair is located in an orbital that is NOT associated with the ring. Take pyridine for example. Since the lone nitrogen is double bonded to a carbon, there is already a single electron in its p orbital. the lone electron pair cannot occupy that same orbital, and therefore it is not part of the ring structure and cannot be delocalized (in pictures, that electron pair is shown as occupying the side orbital that sticks "out" of the page).

 

[dhk5]

i found this
http://faculty.swosu.edu/tim.hubin/OrgIILectures/Och15lec2.ppt ..if you go to slide 5..it tells you that it is aromatic and goes into other examples if that helps