Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: herrhansen on April 03, 2011, 10:45:31 AM
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Hello.
Does anyone know which of the nitrogens on 6-aminopyridine that will be first protonated if making the HCl salt? If they share electrons, then both could be an option as I see it?
Hope anyone knows. Thanks.
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It should be 2-aminopyridine exactly. Since it exists in two tautomeric forms it is difficult to predict which nitrogen will be protonated first.
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Yeah, it should be 2-aminopyridine, unless there is a functional group at the 1- position already.
Anyway, assuming that it is 2-aminopyridine, I think it is possible that either one of them could be protonated.
Protonation of the amino group would seem favourable because it is an sp3 Nitrogen, as compared to the N in the ring, which is sp2 hybridized.
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi54.tinypic.com%2F2j4dcm9.gif&hash=417f2f1f12917db384d1ccfa4ce59e3df3abdadb)
However , as can be seen in the above pic, the aromaticity of the pyridine ring is retained when the amino group is protonated, but the positive charge on the N remains localized to it. On the other hand, delocalization of the positive charge is possible when the N in the ring is protonated, but at the cost of aromaticity (or that's what I figured)
So, it really is a tough call. But I don't know if I'm right. Can someone please check my answer?
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Thanks so much for Quick answers. It was good answers. The tautomeric forms should be 1000:1 favoring the amine form.
But an elementary analysis suggested protonation of both which I dont understand.
It would be hard to have two cations next to each other.
Schrødinger i Am sorry about the snack. I tried to give you one. I Will dó it again after 24h. Sorry.
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protonation occurs at the nitrogen in the ring. This is analagous to DMAP which is nucleophilic at the nitrogen of the pyridine rather than the amine. Protonation of the lone pair on pyridine does not disturb aromaticity as the lone pair is orthogonal to the pi system and therefore does not contribute to aromaticity. further the positive charge is stabilized by resonance from the amine.
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Thanks G-Bones.
But in DMAP they are placed para to each other. Here they are placed close. I was wondering if they could share electron pairs?
Thanks.
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Comparison of DMAP and 2-aminopyridine is a bit too far-fetched.
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Thanks G-Bones.
But in DMAP they are placed para to each other. Here they are placed close. I was wondering if they could share electron pairs?
Thanks.
If they both have full octets, how could they possibly share electron pairs?
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The amino nitrogen is sp2 hybridized due to conjugation with the ring -- it does NOT have sp3 hybridization!
The key here is to draw resonance structures. If you protonate on the ring nitrogen, you can draw a resonance structure where the charge is transferred to the amino group but every atom still retains its full octet. Now try this with the amine protonated, it doesn't work.
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Comparison of DMAP and 2-aminopyridine is a bit too far-fetched.
It seems like in the context of the question, its reasonable. why is it too far-fetched?
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Sorry g-bones, my fault. My brain was blocked ;D. You´re absolutely right.
The answer to the question is obviously very simple. Just count the number of resonance structures obtained from protonation of pyridine N vs. amine N. The one with the more resonance structures wins...
@herrhansen --> for further understanding:
A good example in this context is DBU. Which one of the N´s will be protonated in DBU?
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Thanks for the replies.
I guess it would be protonated at the double bonded nitrogen then. Since it would stabilized by the double bond?
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Yes, but:
stabilization comes from delocalization of the positive charge. In case of protonation at the double-bonded nitrogen there are two possible mesomeric structures. Protonation of the tertiary amine is less stabilized since the positive charge can not be delocalized. Write it down on a sheet of paper also for the pyridine and clearity will overwhelm you...