About Aromatic compound,,,,!!!

[potbelly]

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Hi all,

I have one questions.

A 8 carbon-ring with 4 conjugated double bonds is considered as anti-aromatic or non-aromatic? Kaplan BB says it is anti aromatic but topscore says it is non aromatic.

Any can explain this clearly?????????????


Thanks.

 

[Vicviper]

There's Huckel's rule, part of which is anything aromatic needs to have 4n+2 pi bonding electrons or something like that. Since it has 4 double bonds, that's 8 electrons, so it doesn't match 4n+2, it's 4n, so it's anti.

Pretty sure that's right!

EDIT: I'm wrong, see my posts below, it's non-aromatic.

 

[notjeff]

I believe its non aromatic. In nature it exists as a tub shape(instead of a ring) so it is not planar.

 

[peyman2002]

Yep BB is right, based on 4n+2 its an anti aromatic compound.

 

[potbelly]

I believe its non aromatic. In nature it exists as a tub shape(instead of a ring) so it is not planar.

I think you are right.. Thanks.. Since this compound is not planar, it can't be anti-aromatic even though it has 4n pi electrons... Thus it is non- aromatic as you said! Thank you!

 

[Vicviper]

Yeah, I think he is right that it's non-planar. Now that I'm thinking about what the structure looks like, I'm pretty sure I remember that as one of the ones that's an exemption from 4n+2 because of it's shape. So, I think I change my answer to non-aromatic!

 

[smurf528]

that structure would be anti-aromatic. In order for a cyclic compound to be aromatic or anti-aromatic it should follow these 4 rules:
1) Conjugation of double bonds.
2) each atom in the ring must have a free p-orbital
3) continuous overlap of the free p-orbitals
4) Huckel's Rule: 4n+2= aromatic, 4n= anti-aromatic.

If the ring doesn't follow any of the first three than it is non-aromatic. A cyclooctane with four conjugated double bonds follows the first three rules and the 4n rule therefore it is antiaromatic.

 

[peyman2002]

why do u guys make it difficult!
as smurf confirmed it too its anti aromatic.
just to assure u, it has 8 sides with 4 conj double bonds. it does not have any sp3 carbons thus it has an uninterrupted pi system. and it follows the 4n formula.

 

[Vicviper]

http://en.wikipedia.org/wiki/File:COT-all-cis.png

It's IUPAC name is 1,3,5,7-Cyclooctatetraene. Take a look at that picture, it shows that it is non-planar, and for it to be either anti-aromatic OR aromatic, it must be planar. Thus, it is non-aromatic. See what I mean?

 

[Sublimation]

I just took my orgo 2 exam on aromaticity and that my friends in non aromatic. when u have a ring that big with all cis bonds, especially if its a 8annalene or above it will end up taking on a more favorable conformation, one that is non aromatic.

 

[smurf528]

okay. This has got me a bit frustrated so I did some reading out of my OChem textbook. cyclooctatetraene can be either anti or non depending if Huckel's rule applies or not. This is what the text says:

"Cyclooctatetraene would be antiaromatic if Huckel's rule applied, so the conjugation of its double bonds is energetically unfavorable. Remember that Huckel's rule applies to a compound only if there is a continuous ring of overlapping p orbitals, usually in a planar system. Cyclooctatetraene is more flexible than cyclobutadiene, and it assumes a nonplanar "tub" conformation that avoids most of the overlap between adjacent pi bonds. Huckel's rule simply does not apply." Organic Chemistry. 6th Edition, 2006. Wade L.G.

Huckel's Rule = antiaromatic
No Huckel's Rule = nonaromatic