Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Dev on September 27, 2012, 11:19:25 AM
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Aniline - which is an aromatic ring with NH2 group on one of it's Carbons
If I need to determine how good a base Aniline is, as compared to a simple cyclohexane with an NH2, the only difference i can see is of course the resonance.
But it seems to me that the lone pair on N and also the positive charge that will appear on N after acid/base reaction, are in resonance with the pi bonds of the benzene ring only from one side and not from the other side.
If i take a particular resonating structure,
(P=pi bond S=sigma bond)
If i move in a particular direction - P-S- (NH2)-S-S-P
The first 2 sigmas are the same C-N bond. The 2 sigmas in a row should break the conjugation.
So, if I include the lone pair (or positive charge) of Nitrogen in the resonance, it is broken.
Which means that the 3 pi bonds (only) in the ring are in conjugation and responsible for aromaticity, and any charge on nitrogen can't get delocalised.
What is wrong with this reasoning?
Another reason for why N is not counted in the resonance: if i count pi electrons, along with N's lone pair, there are 8 and without the lone pair there are 6.
And 6 is more favorable because it follows Huckel's rule.
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But it seems to me that the lone pair on N and also the positive charge that will appear on N after acid/base reaction, are in resonance with the pi bonds of the benzene ring only from one side and not from the other side.
No, the +ve charge on the anilinium cation is not resonance stabilised. Once the N is protonated, a C=N bond can no longer be formed - if you try to do it you will see that it would require a pentavalent N.
But it seems to me that the lone pair on N and also the positive charge that will appear on N after acid/base reaction, are in resonance with the pi bonds of the benzene ring only from one side and not from the other side.
I do not understand what you mean here.
If i take a particular resonating structure,
(P=pi bond S=sigma bond)
If i move in a particular direction - P-S- (NH2)-S-S-P
I can't make head nor tail of the rest of your post. Please draw.
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Hi Dan!
You are right, a positive charge on N won't delocalise.
I will reframe my question:
Is N's lone pair is resonance with the double bond?
I have attached a .png file.
The Check means the double bond and lone pair are in conjugation
The Cross means they aren't.
If there are 2 sigma consecutive sigma bonds, that means conjugation is broken which means resonance shouldn't exist.
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It's been a long time since I drew these.
Have a look and give me a comment.
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Alright, so then what I'm infering from this is that conjugation need not go around in a full loop and end where it began.
Another question: Will you count N's lone pair electrons in the no. of pi electrons?
If you do, then the ring become anti-aromatic.
Why don't we count it, (if we don't)?
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These are the resonance structures that contribute. Three of them if you accept for the sake of argument that the left and right sides are different.
The N lone pair is not counted in the number of pi electrons therefore the ring is not anti-aromatic.
These are only resonance structures, which we draw on paper to try and explain certain reactivity or properties, if they actually exist, well no one really knows. They probably do not exist as such, but no doubt someone can produce a picture of electron density being higher at certain points of the molecule.
Have a look here:
http://www.rsc.org/chemistryworld/2012/09/bond-order-ibm-afm-microscopy-buckyball (http://www.rsc.org/chemistryworld/2012/09/bond-order-ibm-afm-microscopy-buckyball)