delocalizable pi electrons

[Astra]

Could someone explain what this means? I understand what this means in terms of reactions, but in terms of actual atoms and theory, I do not understand.

If possible, please also explain how it relates Peptide bonds.

Thank you very much!

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[aldol16]

Well, a pi system is one in which you have a series of electrons that are in p orbitals. So basically, sp3 atoms cannot do this while sp2 and sp atoms can. The quintessential example is, of course, benzene. The p orbitals all undergo pi overlap and therefore they are "delocalized" in a ring made of p orbitals, i.e. the pi system.

With regard to peptide bonds, the lone pair on the nitrogen is in a p orbital because the nitrogen is sp2 hybridized (you should be able to tell orbital hybridization just by looking at it now). So, what it can do is donate into the pi antibonding orbital of the carbonyl (made from mixing the carbon and oxygen p orbitals), breaking the C=O pi bond and forming a C=N pi bond. This results in resonance delocalization, which stabilizes the amide. However, it should be noted (since I am a chemist) though perhaps beyond the scope of the MCAT, that breaking an amide bond is relatively simple, meaning that the resonance form is not a huge stabilizing force. Kicking out the amine is the RDS of peptide hydrolysis. So the resonance stabilizes the amide but not by that much.

 

[Astra]

Well, a pi system is one in which you have a series of electrons that are in p orbitals. So basically, sp3 atoms cannot do this while sp2 and sp atoms can. The quintessential example is, of course, benzene. The p orbitals all undergo pi overlap and therefore they are "delocalized" in a ring made of p orbitals, i.e. the pi system.

With regard to peptide bonds, the lone pair on the nitrogen is in a p orbital because the nitrogen is sp2 hybridized (you should be able to tell orbital hybridization just by looking at it now). So, what it can do is donate into the pi antibonding orbital of the carbonyl (made from mixing the carbon and oxygen p orbitals), breaking the C=O pi bond and forming a C=N pi bond. This results in resonance delocalization, which stabilizes the amide. However, it should be noted (since I am a chemist) though perhaps beyond the scope of the MCAT, that breaking an amide bond is relatively simple, meaning that the resonance form is not a huge stabilizing force. Kicking out the amine is the RDS of peptide hydrolysis. So the resonance stabilizes the amide but not by that much.

In this case, the Nitrogen is sp2 hybridized due to the double bond resonance from the carbonyl?

normally, Nitrogen with 3 groups attached would be sp3 correct? ( 1 lone pair, 3 bonds. 4 orbitals needed. s p p p , sp3)

 

[aldol16]

In this case, the Nitrogen is sp2 hybridized due to the double bond resonance from the carbonyl?

normally, Nitrogen with 3 groups attached would be sp3 correct? ( 1 lone pair, 3 bonds. 4 orbitals needed. s p p p , sp3)

Whether the nitrogen lone pair is in a sp3 or p orbital generally depends on whether the specific factors that are inherent in the molecule. For instance, the nitrogen lone pair on indane is in a p orbital because it can delocalize into the pi system. In contrast, the lone pair on a basic amine that cannot delocalize is in an sp3 orbital. It all depends and you have to assess each case molecule-by-molecule.

 

[Astra]

Whether the nitrogen lone pair is in a sp3 or p orbital generally depends on whether the specific factors that are inherent in the molecule. For instance, the nitrogen lone pair on indane is in a p orbital because it can delocalize into the pi system. In contrast, the lone pair on a basic amine that cannot delocalize is in an sp3 orbital. It all depends and you have to assess each case molecule-by-molecule.

As for this case though, the nitrogen atom is sp3 correct?

 

[aldol16]

As for this case though, the nitrogen atom is sp3 correct?

Which case? The case of a peptide bond? In the case of a peptide bond, the nitrogen lone pair is in a p orbital so it can delocalize. Again, that's why amide bonds are pseudo-stable.

 

[Astra]

Which case? The case of a peptide bond? In the case of a peptide bond, the nitrogen lone pair is in a p orbital so it can delocalize. Again, that's why amide bonds are pseudo-stable.

ohhhh. Normally Nitrogen would be Sp3 but since it can form a resonance structure it is sp2. makes sense

 

[aldol16]

ohhhh. Normally Nitrogen would be Sp3 but since it can form a resonance structure it is sp2. makes sense

Basically (no pun intended). Nitrogen is actually very common in the sp2 form because it's often found in heterocycles where it can delocalize into the pi system. But if it can't delocalize, then sp3 would be more stable because it gives s character to all valence orbitals.