What is the hybridization if it has resonance?

[deleted647690]

If you had an amide, the hybridization for the nitrogen would be sp^3 because it has two bonded Hydrogens, a bond to the carbonyl carbon, and a lone pair.
However, if you drew the resonance structure, the N would be sp^2

So do you say that the nitrogen is sp^2 or sp^3? Based on what I've seen in the BR book, it assumes the smallest hybridization, so sp^2. But in my EK1001 orgo practice, it looks like it would be sp^3.

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[Mr. otcoD]

If you had an amide, the hybridization for the nitrogen would be sp^3 because it has two bonded Hydrogens, a bond to the carbonyl carbon, and a lone pair.
However, if you drew the resonance structure, the N would be sp^2

So do you say that the nitrogen is sp^2 or sp^3? Based on what I've seen in the BR book, it assumes the smallest hybridization, so sp^2. But in my EK1001 orgo practice, it looks like it would be sp^3.

From what I understood is that you would always assume the smallest hybridization you can. Like the oxygen in the amide: normally (with no charge) it has sp2. But with a negative formal charge, it has sp3. When thinking about the hybrid of a molecule, the electron is literally going from the nitrogen to the oxygen back to N back to O etc. Its so delocalized that the molecule assumes the smallest hybridization in order to account for an incoming electron charge.

 

[aldol16]

You should understand that resonance forms are not distinct forms of compound - the compound exists in all of these resonance forms at once (a superposition, if you will) and so if a lone pair can delocalize, it is delocalized. So it will be sp2. This is why something like aniline is not very basic.