[Dan]

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.

The amide bond is planar. N is sp² hybridised - axial/equatorial disposition applies only to sp³ centres, the acyl substituent occupies a position of intermediate torsional angle (analogous to the O in cyclohexanone). I think this argument is invalid. 

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.

For this molecule:



Despite my point above, let's entertain this possibility for the purposes of argument. I see no reason why an intramolecular H-bond between the amide carbonyl and the β-OH would slow the conformational equilibrium of the piperidine ring. Even if it did, the two slowly interconverting isomers you propose would differ in chemical shift to some extent for all signals, because what you are suggesting is two resolvable conformational diastereoisomers. These would be observed as two different molecules (i.e. we should see doubling of all peaks to some extent - particularly the amide C), which is not what is observed.

Conversely, as has been said, the restricted rotation of amide bonds is well documented, routinely observed at room temperature, and fully accounts for the observed inequivalence of the piperidine CH₂s in the spectrum.

[pgk]

Rotation between carbons atoms 2 and 3 is not free due to the formation of intermolecular hydrogen bond. The carbon atoms of axial piperidine(a) are close to the carbonyl group and thus magnetically protected but the signal of carbon A (13C-NMR ) is less affected. However, the equatorial piperidine conformer (b) is predominant and that’s why the corresponding 13C-NMR signals are more intense.
Piperidine conformer equilibrium is difficult, if the carbonyl and hydroxyl groups are antiparallel, due to the steric hindrance of the phenyl ring.
(For simplicity reasons, the methyl group at carbon atom 2 is not represented, hereby.)
Question: Why is the equatorial piperidine conformer (b) predominant?

[Dan]

the equatorial piperidine conformer (b) is predominant

Can you explain to me why you think this amide bond, contrary to the vast majority of amides, is not planar?

That restricted rotation of the amide C-N results in the effect under discussion is well known, it has nothing to do with piperidine - it would be observed for an acyclic tertitary amide as well (e.g. the dimethyl amide, see DMF as the simplest example).

[pgk]

Good remark and a good chance for clarifications.
Nitrogen atom in molecules, has a pyramidal structure that also might be considered as a tetrahedral one if the electron pair is taken in account. Therefore, substitution of heterocyclic nitrogen can be either axial or equatorial.
Nitrogen atom that belongs to an amide bond, “tends” to have a planar structure rather than a pyramidal one. This is due to the partial delocalization of the nitrogen electron pair to the carbonyl(partial hybridization) that forms an iminium enol conjugate. How planar the amide bond depends on the substitution of both the carbon and the nitrogen that finally reflects on the contribution of the amide form and the iminium enol form, to the conjugation respectively. For example, peptide bond is very planar.
On the flip side, the real structure is captured by the spectra.
By the 1H-MNR spectrum of DMF or the formamide, the amide form is captured and not the iminium enol (otherwise, protons splitting would be observed. Why?), nor the hybridization (otherwise, the formyl proton would be exchanged and disappeared by D2O. Why?). Besides, the IR spectrum would appear: a broad weak peak around 2000cm-1 (corresponding to the iminium cation), a strong peak at 1630-1650cm-1 (corresponding to the imine) and strong broad peak at 1600-1650 cm-1 (corresponding to the enol anion) or overlapped average peaks (corresponding to the hybridization), instead of the carbonyl peak (≈ 1700cm-1). Consequently, the methyl groups are syn- and anti- to the carbonyl rather than cis- and trans- to a double bond. The latter means that the magnetic anisotropy of the syn-methyl, is due to the neighboring with the carbonyl group and not to the cis-, trans- isomerism that has no anisotropic effect.
The 13C-NMR spectra of DMF (methyl carbons) and dimethylacetamide (methyl and methylene carbons) have a similar appearance with respect to the magnetic anisotropy but not the ones of n-formylpiperidine and N-acetylpiperidine where 3- and 5- piperidine carbons are magnetically equivalents, while the corresponding 2- and 6- carbons are magnetically inequivalent (because contrary to ethyl groups, the rotation of ring carbons is synchronized).
Tacking in account all above, the magnetic anisotropy that is induced by the amide bond, fails to explain the different 13C-NMR signals of B and D carbons of the given compound. Thus, the only reasonable explanation remains the “freezing” of the conformers equilibrium. 

[Irlanur]

I feel this discussion is way overcomplicated and I still don't see where this should add to the discussion.

[Dan]

Nitrogen atom in molecules, has a pyramidal structure that also might be considered as a tetrahedral one if the electron pair is taken in account. Therefore, substitution of heterocyclic nitrogen can be either axial or equatorial.

Only if sp³

Nitrogen atom that belongs to an amide bond, “tends” to have a planar structure rather than a pyramidal one.

The vast majority are planar, including, in my opinion, this one. Amides that do not adopt a planar amide bond are unable to for reasons of extreme strain - such as cases where N is at the bridgehead position (where sp² hybridization of the N violates Bredt's rule) or in other cases of extreme steric strain (e.g. amides of 2,2,6,6-tetramethylpiperidine). The steric clashing associated with piperidine methylene groups is very similar to methyl groups. If you are suggesting the steric bulk of the piperidine prevents overlap of the N lone pair with the C=O (i.e. sp² hybridization) in N-acylpiperidines, but this overlap can be achieved for N,N-dimethyl or -diethyl amides, I don't think you can reasonably explain this with steric arguments. I am not going to take this from you on faith - show me evidence that piperidine amide bonds are not planar.

Consequently, the methyl groups are syn- and anti- to the carbonyl rather than cis- and trans- to a double bond. The latter means that the magnetic anisotropy of the syn-methyl, is due to the neighboring with the carbonyl group and not to the cis-, trans- isomerism that has no anisotropic effect.

So you are saying that the N in N-acetyl and N-formylpiperidine is sp³? Can you please cite a reference providing peer-reviewed evidence and/or discussion of this phenomenon?

The 13C-NMR spectra of DMF (methyl carbons) and dimethylacetamide (methyl and methylene carbons) have a similar appearance with respect to the magnetic anisotropy but not the ones of n-formylpiperidine and N-acetylpiperidine where 3- and 5- piperidine carbons are magnetically equivalents, while the corresponding 2- and 6- carbons are magnetically inequivalent (because contrary to ethyl groups, the rotation of ring carbons is synchronized).

If you look up the data, you see this is simply not true. All 5 of the piperidine C (including C3 and C5) in both N-formyl and N-acetylpiperidine are inequivalent (just as all 5 piperidine CH₂s are inequivalent in the OP's spectrum):

13C NMR of N-acetylpiperidine: 21.44 (Me). 21.52 ppm (CH2), 25.54 ppm (CH2), 26.46 (CH2), 42.50 (CH2), 47.48 (CH2), 168.77 (CO) - from J. Phys. Chem. B, 2008, 112, 7885

13C NMR of N-formylpiperidine: (400 MHz, CDCl3) δ 24.59, 24.99, 26.48, 40.53, 46.75, 160.58 - from Synth. Commun., 2011, 41, 476 (see supporting info)

And also, the amide stretch for N-acetylpiperidine is 1645 cm^-1 - from Tet. Lett., 2011, 52, 2722 (see supporting info). This is normal amide range and is evidence of strong N-C=O overlap (i.e. sp² N).

You have still failed to convince me that: 1. Piperidine amides have an sp³ N; 2. That the rotation of the amide N-C bond is not restricted by N-C=O overlap; 3. That the inequivalence of the piperidine methylenes in 13C NMR are not due to restricted C-N rotation. I would be interested to see some evidence.

I don't think there is much more I can add to this discussion.

[pgk]

OK, I agree.