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Which of the following is NOT aromatic?

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3dObject
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Joined: 24 May 2006
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PostPosted: Sat Jul 01, 2006 1:39 pm    Post subject: Which of the following is NOT aromatic? Reply with quote

The answer to the following question is given as choice C. But I think that choice D could also be a valid answer because the Carbon with the plus sign is not sp2 hybridized. Is my reasoning correct?



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adrian
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PostPosted: Sat Jul 01, 2006 9:26 pm    Post subject: Aromaticity rules Reply with quote

Please remember the three rules for aromaticity:
(4n+2) pi electrons
Continuos conjugation
plane molecule
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3dObject
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PostPosted: Sat Jul 01, 2006 9:49 pm    Post subject: But... Reply with quote

I thought there was a rule that also said that all the atoms in the ring must be sp2 hybridized.

Is that not correct? If it is, then is it due to the plus charge "moving around" the ring if you will due to resonance, and therefore ensuring all the carbons will be sp2 hybridized?
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adrian
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PostPosted: Sun Jul 02, 2006 10:36 am    Post subject: but to but... Reply with quote

OK, continuos cojugation involve sp2 hybridization.. Let's analyse :
a: 6 pi electrons (=4x1+2)
b: 2 pi electrons (=4x0+2)
c: 4 pi electrons (non aromatic)
d: 6 pi electrons
e: 6 pi electrons

For the d structure: we have 6 pi electrons in the benzene ring. The missing electron is from the sp2 orbital and is not participate to the continuos conjugation. In my opinion, such a cation is not stable and that is just a theoretical problem. More on the problem of aromaticity you will find working on Huckel equations.
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elmyhalily13
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PostPosted: Fri Mar 23, 2007 5:20 pm    Post subject: Reply with quote

I had been this question during my studies. But still couldn't understand how to differentiate whether they are aromatic or not. Is there any principle or law to distinguish them?
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pdavis68
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PostPosted: Mon Mar 26, 2007 8:46 pm    Post subject: Reply with quote

Here's the deal with D, I believe. D has a positive charge, so it's short 2 electrons. Therefore, in D, a hydrogen left and took the electron pair with it.

It still has its other electrons involved in the conjugated pi system, so these are unaffected. So it retains its aromaticity.

Does that make sense?
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kristoph
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PostPosted: Fri Apr 06, 2007 2:51 pm    Post subject: Reply with quote

Quote:
It still has its other electrons involved in the conjugated pi system, so these are unaffected. So it retains its aromaticity.


But thats exactly what would make it non-aromatic. By losing those two electrons it will behave differently, similar to B. I am not sure of the answer, but I think it is one of those abnormal qualities of benzene that allows it to still be aromatic?
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RobJim
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PostPosted: Thu May 17, 2007 6:22 am    Post subject: Reply with quote

Those of you who posted after adrian don't seem to have understood xyr explanation. Aromatic systems always have 4n+2 pi electrons, or 2n+1 pi bonds, with n being zero or a positive integer. Because C has 4 pi electrons (2 pi bonds) it is not aromatic;

4(0)+2 = 2
4(1)+2 = 6

There's no way to get 4 from 4n+2 if n's an integer. All the others have the proper number of pi electrons/bonds.
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