113zami

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I thought any cyclo compound with 8 carbons or higher would not be planar, but i just found in my text that cyclononatetraenyl anion ( cyclononane with 4 conjugated double bonds and the last carbon has lone pairs) is aromatic,

how can it be aromatic, its not even planar
how can you find if something is planar or not?
 

doc3232

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I thought any cyclo compound above 8 carbons would not be planar, but i just found in my text that cyclononatetraenyl anion ( cyclononane with 4 conjugated double bonds and the last carbon has lone pairs) is aromatic,

how can it be aromatic, its not even planar
how can you find if something is planar or not?
Ok, so this is the deal,
Nitrogen, Oxygen, and sulfur can all donate electrons (if they have spares). Also Carbons with electron pairs can do the same.
But I do have a a question, where is the lone pair?
But whats important is you understand that it would be theoretically aromatic nonetheless.
Hence, it would have a total of 8+2 electrons which works with 4n+2 rule.
So chemists are a bit lenient, it doesn't have to be completely planar. As long as most of it is somewhat planar (when you go above 6 carbons, all molecules start flopping).

Got it?
 

userah

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so pretty much if it follows the 4n + 2 rule when you go above 6 carbons, it's going to be planar and thus aromatic?
 
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113zami

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this is how it looks like, i know it satisfys 4n+2 but how can you tell if something with 8 carbons or higher is planar or not? that's my question, there are other molecules given in the book which are above 8 carbons, satisfy 4n+2 and are non aromatic, because they're not planar, so how can you tell??
 

userah

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alright from my knowledge, all the carbons must be sp2 hybridized. If there are any carbons that are anything else, then it's non planar and thus non aromatic. In the example you gave, it is planar because all the carbons are sp2 hybridized; even the carbon with the lone pair electrons. those lone pair electrons are what make it sp2 hybridized as opposed to if they weren't there, that carbon would be sp3 hybridized.

think of it this way. for something to be sp2 hybridized, usually it must mean it has to be conjugated (double bond). in this case, that one specific carbon can't be "conjugated" because you can't have two double bonds next to each other so instead, that carbon has those lone pair electrons, which make it conjugated in a sense and thus make it sp2 hybridized. let me know if that clarifies for you.
 

doc3232

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ok, that explains, I misread it, but what I said is still true.
The electron pair on the last carbon allows it to still have a pi orbital in a sense.
The other carbons have the double bonds to satisfy that rule.
 
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113zami

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all the carbons must be sp2 hybridized. If there are any carbons that are anything else, then it's non planar and thus non aromatic. .
if what you're saying is true then why is conjugated cyclooctatetraene nonplanar and thus non aromatic? all the carbons are sp2
 
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joh020340

if what you're saying is true then why is conjugated cyclooctatetraene nonplanar and thus non aromatic? all the carbons are sp2
Because it doesn't follow Huckels rule of pie electrons must = 4n+2 to be aromatic. 8 pie electrons is antiaromatic (4n). Aromatic must have 2,6,10,14,18 pie electrons and so on in addition to the conjugated planar stuff.
 
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113zami

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Because it doesn't follow Huckels rule of pie electrons must = 4n+2 to be aromatic. 8 pie electrons is antiaromatic (4n). Aromatic must have 2,6,10,14,18 pie electrons and so on in addition to the conjugated planar stuff.
I did realize that too, it should be ANTIaromatic because it satisfys 4n NOT NONaromatic, but my textbook says its NONaromatic not antiaromatic, nonaromatic and antiaromatic are not the same thing you know
 

BlackSails

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If there are any carbons that are anything else, then it's non planar and thus non aromatic.
Fullerene and Paracyclophanes (rather, anything above [4] or [5]paracyclophane) are nonplanar and aromatic.

Aromatic compounds just need "good" overlap between all adjacent orbitals.
 

userah

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Fullerene and Paracyclophanes (rather, anything above [4] or [5]paracyclophane) are nonplanar and aromatic.

Aromatic compounds just need "good" overlap between all adjacent orbitals.
haha i'm sure there are exceptions but it's found to generally be the case that if it's sp2 hybridized, then it's planar and thus aromatic.
 

bruin432

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it non aromatic because it takes on the "tub" conformation, making it stable, but still less stable than benzene because it lacks the tight overlap between the p orbital. So while it only satisfies the 4n, because of the tub conformtion, it actually is a stable molecule, so it is no longer considered anti-aromatic, just non aromatic. hope that makes sense:)