DAT O.Chem 2016 Question #71

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tony729

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I am a bit confused on the hybridization of the nitrogen, to me it looks like it would be sp3 since there are 3 bonds and a lone pair.
 
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Due to Resonance, there is electrons delocalization in the 5 membered ring, and that is why it its not sp3 hybridized (the one pair does not just stay with Nitrogen). Another way to look at this is Aromaticity. The 5 membered ring is Aromatic with conjugated double bond system, less than 7 members in the ring and odd number of electrons pair ( only 1 pair is counted to determine Aromaticity) and hence, the member in the cyclic ring cannot be sp3 hybridized.
 
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Yes, remember that electrons in s-orbital (smaller orbital that is closer to nucleus) experience greater nuclear charge and hence cannot "per say" delocalize!
 
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tony729 said:
I am a bit confused on the hybridization of the nitrogen, to me it looks like it would be sp3 since there are 3 bonds and a lone pair.
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Not a chance. This N is sp2 because if you consider the RESONANCE structure, it will reveal a molecule that has a N with significant double bond character. For practice, I want you to look up the structure of an AMIDE. You will see a N with 3 bonds and a lone pair. This N is also sp2 since the amide has significant structural contribution possessing double bond character.

Hope this helps.

Dr. Romano
 
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