[johnj7]

This question is going to get a bit confusing, as its hard to describe with words.

Its simply a conceptual question, and I probably don't see the answer because I don't understand electron orbitals as well as I should.

My ochem professor defined conjugation as requiring 3 parallel, adjacent, overlapping p orbitals.  I can see this easily in something like 1,3-butene and benzene.

But what about between the C=O, H-N-R2, in a segment of a polypeptide?
How come this is not considered a conjugated segment?
1. O in the C=O is sp2 ... giving one p orbital
2. C in the C=O is sp2... giving another p orbital
3. I learned that H-N-R2 can have sp2 hybridization so it can enjoy the extra stabilization in resonance / conjugation
so if there are 3 p orbitals, why can't it be conjugated?  does the p orbital of nitrogen not overlap with the others?  

[AWK]

N has a nonbonded electron pair on unhybridized p orbital. pi-bond in CO if formed from unhybridized p electrons of C and O.

[johnj7]

sorry, its probably real basic,
but could you elaborate? I couldn't see where you were going with that

[AWK]

Just draw a correct Lewis structure for eq CH₃CONHCH₃

[johnj7]

Right, I drew the correct lewis structure.
I see the nonbonded electron pair on unhybridized p orbital of N, I see the pi bond formed in the carbonyl, I see that there's resonance, and I can see the possible resonance structures, but my prof told me it wasn't considered "conjugation."

Why not?

[AWK]

My ochem professor defined conjugation as requiring 3 parallel, adjacent, overlapping p orbitals.

This is the case!

May be you drew a bit wrong  Lewis structure

[Vidya]

in polypetides --CONH2
   there is no conjugation because it needs p orbitals which can overlap to help in delocalization. On N we have sp3 orbitals and lone pairs are in sp3 orbitals and O we have sp2 orbitals and one of the lone pair is in sp2 orbitals.During resonance N becomes sp2 -lone pairs come in p orbitals and oxygen lone pairs in sp3 orbital. So no overlapping of p orbitals and hence no conjugation.
I would be more able to help you with the structures.Hope  it is now clear to you .

[johnj7]

AH finally!
I understand!
vidya kaushik, so you're saying that you can't have both the O being sp2 hybridized and N being sp2 hybridized at the same time, thus no p orbital overlap. Would that be correct?

[johnj7]

actually this is what my professor responded in email

...The definition of conjugation would be the extra
stabilization provided by 3 parallel, overlapping, adjacent p orbitals.  I was
wondering then, would free amino acids count as "conjugated" molecules?  Namely the
carbonyl oxygen p orbital, the carbonyl carbon p orbital, and the alpha amino nitrogen p
orbital?

Response:  No, because the amino nitrogen lone pair is not adjacent to the carbonyl.

could you make sense of this?

[Vidya]

You are right free aminoacids are not conjugated because there is a sp3 orbital between N lone pair and carbonyl p orbitals.

[johnj7]

sorry to be so nitpicky, but do you know what my prof means when he said
"no, because the amino nitrogen lone pair is not adjacent to the carbonyl" ?

[macman104]

sorry to be so nitpicky, but do you know what my prof means when he said
"no, because the amino nitrogen lone pair is not adjacent to the carbonyl" ?

That is what vidya is saying above.  You have a carbon that is sp3 hybridized inbetween the nitrogen lone pair, and the carbonyl carbon (which is sp2 hybridized).

[johnj7]

macman, are you sure?

i wasn't referring to the chiral carbon in a free amino acid, but the carbon involved in the peptide bond.
so there is no sp3 carbon in between the nitrogen lone pair and carbonyl carbon

[nox]

johnj7, a FREE amino acid is one that is NOT part of a peptide, it is a standalone, individual amino acid on its own, there are NO peptide bonds whatsoever and no conjugation is possible due to the presence of the alpha-carbon which is sp3 hybridized

however, in a peptide (with 2 or more amino acids), the peptide bond is conjugated since all 3 atoms (C, N, O) have p orbitals that are parallel and adjacent to each other

hope the following diagram will clear things up

[spirochete]

This is really a question of semantics.  You already understand that there's resonance there. The question is does it qualify as conjugation.  

By your professor's definition the peptide bond is conjugated!  It does contain 3 adjacent P orbitals: The P orbital on oxygen, the P orbital on carbon and the P orbital where the lone pair is.  Note that this lone pair has to be in a P orbital to be delocalized through resonance.

The wikipedia definition agrees with your professor, although he doesn't agree with himself!

I checked my most advanced textbook and it defines conjugations more specifically.  It defines conjugation as only involving standard pi bonds, not lone pairs.  By this definition you actually need four adjacent P orbitals.  The only way to make three adjacent P orbitals using alkene/carbonyl etc pi bonds is to make a cumulated system, which are surprisingly unstable and form a special class of molecules.  Look up cumulation on wikipedia for an explanation if you aren't familiar.

To summarize, different sources say different things.  Your professor is the only one who's completely wrong because he doesn't even understand his own definition.