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Topic: Resonance and aromatic compounds  (Read 7300 times)

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Offline Ak

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Resonance and aromatic compounds
« on: July 14, 2008, 01:18:44 PM »
"Consider the following structures.  Some of them have resonance structures that are aromatic.  Identify those compounds and in each case show the resonance structure that is aromatic."

I was given this question by my professor.  I decided to draw out the resonance structures of the first structure before continuing on with the rest of the question.  I have attempted the first structure and my work is attached.  My first question is, are there any more resonance structures or am I wrong on any of them.  My second question is, can anyone understand the third sentance, ill ask my prof but that cant happen until tuesday. Thanks

Offline azmanam

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Re: Resonance and aromatic compounds
« Reply #1 on: July 14, 2008, 03:12:54 PM »
Bottom left is missing a charge.

As for the third sentence.  I think the prof's trying to point out that when you draw some of the charge separated compounds, some of the rings in the resonance structure no longer obey Huckel's rule.  These resonance structures would technically be non-aromatic.  I think that's what the prof means.
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Offline Ak

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Re: Resonance and aromatic compounds
« Reply #2 on: July 15, 2008, 02:08:26 PM »
thanks azmanam...and just to make sure, none of the resonance structures i drew are aromatic as they also have to be sp2 hybridized, right?

Offline azmanam

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Re: Resonance and aromatic compounds
« Reply #3 on: July 15, 2008, 04:48:23 PM »
Assuming I'm reading your prof's question correctly, I count 4 valid, aromatic resonance structures.

Be careful with hybridization of charged carbon atoms.  Assume an sp3 carbon atom with 4 substituents.  We would all say that it has a tetrahedral geometry.  If you remove one of the substituents, but leave behind the lone pair of electrons - that is, if you make it a carbanion with 3 substituents and a full -1 charge (as in a deprotonation)- the carbon atom remains tetrahedral. 

On the other had, if you remove one of the substituents and the lone pair of electrons leaves with the substutent - that is, if you make it a carbocation with 3 substituents and a full +1 charge - the carbon atom rehybridizes from an sp3 carbon atom to an sp2 carbon atom.  This analysis is the same whether you start with an sp3 or an sp2 carbon atom.

That being said, look at your various carbocations and make sure you have identified the hybridization correctly for each.
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Offline Ak

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Re: Resonance and aromatic compounds
« Reply #4 on: July 15, 2008, 10:26:40 PM »
im sorry but i just dont see how there are 4 aromatic compounds as the pentane ring always has a carbon that has 2 hydrogens and 2 other carbons attached to it, and would therefore be sp3.  I know carbocations are sp2, but the positive charge never reaches that 1 carbon.

Offline spirochete

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Re: Resonance and aromatic compounds
« Reply #5 on: July 15, 2008, 10:47:54 PM »
I disagree I only see 3 total aromatic structures there, the first three.  The other three valid structures have cycles containing an ambiguous amount of electrons, because the third pi bond is outside the ring and cannot participate in the aromatic cycle. 

Azaman is correct though that in every structure all six carbons in the six membered ring are SP2 hybridized.

Offline azmanam

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Re: Resonance and aromatic compounds
« Reply #6 on: July 16, 2008, 07:45:38 AM »
I guess I was counting the original structure as one in my answer.  Excluding the original structure, there are 3 other resonance that can be drawn.

The electrons do not have to be delocalized over every atom in a molecule for the molecule to be considered aromatic.  As long as there is at least one ring that obeys Huckels rule, the compound can be classified aromatic.
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Offline sjb

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Re: Resonance and aromatic compounds
« Reply #7 on: July 16, 2008, 07:47:22 AM »
To make it 4 resonance structures take the carbonyl as either the double bond, or the charge-separated hybrid each in one pair, and then in each pair simply push the double bonds in the benzene ring one way or the other.

I think...

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