ttttang
Registered: 09/29/09
Posts: 4
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Reply with quote | #1 |
I was wondering why the Nitrogens outside of the ring are sp2 rather than sp3. I thought that there were 4 regions of electron density, so shouldn't it be sp3? I don't quite understand it.
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akt01
Registered: 10/10/09
Posts: 10
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Reply with quote | #2 |
The Nitrogens outside the ring contain lone pairs that can form pi bonds with the carbons that are part of the ring. In doing so, it causes the nitrogens outside the ring to be sp2. |
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elham90794
Registered: 09/26/09
Posts: 50
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Reply with quote | #3 |
Also conjugation will cause all atoms attached to an sp2 atom to be sp2 hybridized as well and lie in the same plane. |
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lgarza7
Registered: 10/26/09
Posts: 2
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Reply with quote | #4 |
If the Nitrogens lie outside the ring and are too far to be involved in conjugation, there is no incentive to use sp2 hybridization and will remain in sp3 with its four attachments.
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jeff_usui
Registered: 10/07/09
Posts: 7
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Reply with quote | #5 |
You have to remember the order of priority:
Energy -> geometry -> hybridization
In other words, whatever structure yields the lowest energy for the entire molecule will win out regardless of what it ends up looking like. Since the sp2 carbons that those nitrogen are attached to are conjugated, the nitrogen will become conjugated as well despite the fact that the torsional tension would prefer pyramidal geometry. In order for the pi orbitals to line up in the same plane with the other pi bonds, the nitrogens form pi orbitals that are perpendicular to the plane of the ring. The geometry of the nitrogen change the hybridization of them from sp3, which would be predicted from torsion, to sp2, which is predicted by conjugation. |
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