Amide in a Ring

[wheatthinners]

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[aldol16]

Are you talking about an amide in which the nitrogen atom is part of a heterocyclic, aromatic ring? Also, are you talking about electron-donation in the ring itself or donation of electron density towards the carbonyl? At that point, the lone pair that was on the nitrogen has been delocalized into the aromatic ring and therefore exhibit reduced donation towards the carbonyl. So if you were to, say, do an IR, you would see that your C-N bond length got longer as compared to a free amide whereas your C=O bond got shorter.

 

[NextStepTutor_2]

If an amide is in a ring, that means it's sp2 hybridized since the electrons go into a pi bond.

Does that mean the amide is no longer an electron donating group but becomes an electron withdrawing group?

Thanks.

If an amide is in a ring, that means it's sp2 hybridized since the electrons go into a pi bond.

Does that mean the amide is no longer an electron donating group but becomes an electron withdrawing group?

Thanks.

AFAIK, being in a ring does not guarantee a N atom is sp2 hybridized at all. An amide by definition howver, is a nitrogen adjacent to a pi bonded carbon.

This is a sp3 hybridized Nitrogen.


Generally, N atoms that are part of aromatic rings (pyridine, pyrrole) will have planer, sp2 hybridization.

In contrast, atropine, coniine, have stereogenic pyramidal nitrogen atoms in their structural formulas. Think of the non-bonding electron pair as a fourth substituent on a sp3 hybridized nitrogen.

I would wager that the MCAT will not ask you to differentiate between very similar structures, especially with the new MCAT having so little organic chem on it, and what little there is, is very simplified.

With regards to amides, specifically, if you draw a typical amide, the nitrogen will look sp3. But amides have a resonance structure where the C - O bond is single, and the C - N bond is double. So the nitrogen is actually sp2.
Anything that is or can be sp2 hybridized is stabilized when next to π bonds (i.e. oxygen, nitrogen, cations, radicals and anions). A good rule of thumb for the MCAT is that any atom that can be sp2 will be sp2 when next to a double bond (like the amide nitrogen next to the carbonyl)

Hope this helps, good luck!

 

[wheatthinners]

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[aldol16]

In regards to the electron donating/withdrawing, I was wondering as to how it relates with carbocation stability. As far as I know, for sp3 amides, the carbonyl side of the amide will be electron withdrawing and stabilize a carbocation, while the Nitrogen side of the amide will be electron donating and destabilize a carbocation.

It all depends on where the carbocation is. In this case, the carbonyl group's electron-withdrawing nature will destabilize any carbocations near it.