[Ayudaporfa]

I have a piperidine ring attached to my molecule and a phenyl group as well. Why are all the Cs on the piperidine in different environments? (=> 5 environments on my piperidine), while my phenyl has 4? (4 of the Cs are in pairs of 2 different environments)

Spedctrum: http://imgur.com/Ze0gTU8

Molecule: http://imgur.com/smXNwRx

I know all the assignments.  I know the ones on the piperidine ring are
A: 24.47, B:25.55, D: 26.62, C: 41.44 and E: 42.68;
C and E should be the same environment, and so should B and D, but they ar eon different peaks?

I know HNMR better than CNMR,

Thanks for your help

[kriggy]

Show the molecule and spectra and then we can talk. Without the structure, we can just guess

[Ayudaporfa]

I've updated it.

[Babcock_Hall]

What do you know about the freedom of rotation of the C-N bond within an amide?

[Ayudaporfa]

What do you know about the freedom of rotation of the C-N bond within an amide?

Well I know that the lone pair on the N can delocalise into the carbonyl C, giving it a partial double bond character. This limits rotation about that bond. .. And Im not too sure what implications this has on the Cs in the ring.. Im guessing the fact the the N is Sp2 hybridized also limits the rotation of the Cs on the ring right next to it ?

Or then.. when the lone pair is delocalized, the positive charge on the N causes it to pull the electrons from the ring more towards it as it is quite electronegative, and this affects the spins on the carbons to different extents? (I dont know, total guess...)

[Babcock_Hall]

Suppose for a moment (just for the sake of argument) that there was no rotation at all around the C-N bond.  Then the environments of C and E would be different, would they not?  Nucleus C is much closer to the carbonyl oxygen than E is, for example.  The truth is slightly more complicated, but that might not be important for right now.

[Ayudaporfa]

Suppose for a moment (just for the sake of argument) that there was no rotation at all around the C-N bond.  Then the environments of C and E would be different, would they not?  Nucleus C is much closer to the carbonyl oxygen than E is, for example.  The truth is slightly more complicated, but that might not be important for right now.

Yeah, I should of seen that actually. The O does in fact play a role.. So if i base my self on that, my assignment is a little inversed, right? As C is closer to the O, it should be more deshielded, which means its actually the highest shift of the ones I stated, so it would be: C: 42.68 and E: 41.44 ;  D:25.55, and B: 26.62.

The thing is though, Ive used a software to help me with the assignment of the peaks (Spartan '14 if you know it), and it tells me that E is actually the more deshielded one.. Ive done a bit of reading, and it said that carbon just like Hnmr is affected by deshielding, but also it states "more strongly by the shielding from an abundance of electrons in non-spherical p-orbitals". Based on this, would it be correct to say that, although nucleus C is closer to the carbonyl oxygen, the angle [Carbonyl C]-N-nucE is 123degrees, while the angle [Carbonyl-C]-N-nucC is 116degrees, makes it easier for electrons in E to interact with N (as its more similar to that sp2 angle) while in C the angle is smaller and maybe less convenient?

Or maybe the software is actually wrong lol.

Im sorry if this is wrong and sounds stupid, I've never really done CNMR before. Please either do explain it to me, or refer me to a good source, thanks.

Im writing my final year project and I would like to include it this.

[Ayudaporfa]

I guess thats just wrong then ?

[Irlanur]

the chemical shift difference of E and C ist pretty small and I don't think you can really assign with the data you have. Also you the effect of "(de)shielding" of the nuclei by electrons is usually neglectable for 13C in organic molecules. The paramagnetic contribution (mainly comming from low lying electronic states -> small HOMO-LUMO difference) is usually the most relevant one. A simple way to assign this (if it's needed) would be a NOESY.

[Ayudaporfa]

Alright, thanks!

[pgk]

What you see in the 13C-NMR spectrum, is the mixture of the two conformers of piperidine ring, with axial and equatorial N-alcoyl substitution, respectively. That’s why peaks that correspond to same carbons, have slightly different shifts and different intensities. The explanations for these differences, are given above.

[pgk]

QUESTION: Equilibrium of conformers is very fast at room temperature. Therefore, distinguish of conformers by NMR spectroscopy demands temperatures, as low as - 60oC. Then, what happens here and so, both conformers of piperidine ring are observed at ordinary temperature?

[Irlanur]

it depends on the exact frequency difference and the reaction rate: if the exchange is very fast compared to the frequency difference you just see the average.

[Babcock_Hall]

Dimethylformamide is a well studied example.  Scroll about 2/3 of the way down:  http://131.104.156.23/lectures/CHEM_207/CHM_207_NMR.htm

[pgk]

The equilibrium rate of piperidine conformers, is very fast. But in this case and exceptionally, the equilibrium of N-alcoyl-piperidine is “frozen’ by the intermolecular Hydrogen bond between the carbonyl group and the hydroxyl group at the β-position and thus it can be observed at room temperature.
The shift difference of methyl protons of DMF, is also due to the magnetic anisotropy. Methyl protons of DMF are chemically equivalent but they are not magnetically equivalent due to the neighboring with carbonyl group that is rich in electrons. Above 120oC, the rotation of dimethylamine group is very fast and thus, the shift difference is not observed anymore.