# Can an amide nitrogen be a hydrogen bond acceptor?

Can $\ce{N}$ (i.e. main chain $\ce{NH}$s of recognised residues in a peptide or protein) be a hydrogen bond acceptor? It is a well known fact that the main chain $\ce{NH}$s can be a hydrogen bond donor (as observed in the α-helix), but can they act as a hydrogen bond acceptor? If yes, help me with an example or even better a reference. And, if no, why can't they act as an acceptor, after all it's an electronegative atom?

• see if this reference is help full :pubs.rsc.org/en/content/articlehtml/1989/p2/p29890001355 "The extremes of both scales are charted. Alkyl thiols and amines are negligible as proton donors; correspondingly, π-donor hetero-atoms as e.g. in esters and amides are negligible acceptors" Dec 15 '15 at 14:05
• @DavePhD It’s very hard to actually extract data from that article that rules out acceptors from amide nitrogens — since they measure thermodynamic data, any H bonds will probably be accepted by the amide carbonyl which gladly accepts them and there is no way to distinguish.
– Jan
Dec 15 '15 at 22:50
• All the electrons in an amide need to do is induce any dipole in a H- bond donor. If you want an example look at the macromolecule Kevlar as the sheer volume and number of these interactions give it its properties and not just in one plane. When heated it is a liquid crystal. I would look for colligative effects as sources. You've already indicated amide bonds as donors, look at other polymers. Dec 16 '15 at 1:09
• According to the references below, amide nitrogen can be a weak hydrogen bond acceptor: pubs.acs.org/doi/abs/10.1021/acs.jpca.7b11013 ncbi.nlm.nih.gov/pubmed/27166805 Sep 5 '19 at 0:43

If I understood you correctly, you are talking about the peptide bond nitrogens ($$\ce{R-C(=O)-\color{red}{N}H-R}$$). This is, reduced to its significant chemical functional group an amide, more precisely a carboxylic amide.

The amide nitrogen technically has a lone pair and thus technically could function as a hydrogen bond acceptor when viewed alone. However, this lone pair is actually significantly delocalised towards the carbonyl bond. If we ignore the two $$\ce{R}$$ residues, we could write this the following way:

$$\ce{O=C-N-H <-> ^{-}O-C=N^{+}-H}$$

The lone pair is not really available as a lone pair on the nitrogen, but forms a π-bond to the neighbouring carbonyl carbon to a significant extent. Thus, these nitrogens cannot be hydrogen bond acceptors.

Further, the fact that a Grignard reagent cannot react with an amide gives great support for the fact that nitrogen’s electrons are delocalized onto the carboxylate.

Finally, thank you Orthocresol for supplying me with the paper given at the end of this post. The authors performed some quantum chemical calculations to extract the energy difference between free formamide and water molecules and various conceivable hydrogen bonds between them. The strongest stabilisation can be found in an amide–amide hydrogen bond of the following type:

$$\ce{H2N-CH=O\bond{...}H-NH-CO-H}$$

At around $$-40~\mathrm{kJ \cdot mol^{-1}}$$ for a $$\ce{H\bond{...}O}$$ distance of $$285~\mathrm{pm}$$ and a $$\ce{C=O\bond{...}H}$$ angle of $$60^\circ$$.

$$\ce{H2N-CH=O\bond{...}H-OH}$$

For the interaction of a water molecule donor and the carbonyl oxygen as depicted just above, the stabilisation is given as up to $$-27~\mathrm{kJ \cdot mol^{-1}}$$ for a distance of $$280~\mathrm{pm}$$ and a $$60^\circ$$ $$\ce{C=O\bond{...}H}$$ angle.

$$\ce{OHC-HN-H\bond{...}OH2}$$

For this interaction where the amide nitrogen is donor and water is acceptor, the stabilisation was calculated to $$\Delta E \approx - 30~\mathrm{kJ \cdot mol^{-1}}$$ — achieved twice. Once for the geometry $$d(\ce{H\bond{...}O}) = 280~\mathrm{pm}$$ and $$\angle(\ce{H\bond{...}O-H2}) = 30^\circ$$ (the angle is measured between the $$\ce{H-O-H}$$ plane and the amide plane), and once for $$d = 280~\mathrm{pm}$$ and $$\angle = 15^\circ$$.

$$\ce{OCH-H2N\bond{...}H-OH}$$

This is the one you are looking for. These energy differences are all given as positive. The lowest is $$\Delta E \approx +0.8~\mathrm{kJ \cdot mol^{-1}}$$ for a distance $$d(\ce{H\bond{...}N}) = 360~\mathrm{pm}$$$$1.3$$ times the previous distances. (The angle is not given but from the picture we can assume a $$90^\circ$$ angle between the amide’s plane and the $$\ce{N\bond{...}H}$$ axis.) For the $$280~\mathrm{pm}$$ which brought the most stabilisation in the other cases and which thus seems to be the ideal hydrogen bond distance, they report $$\Delta E \approx + 30~\mathrm{kJ \cdot mol^{-1}}$$ — this is about the dimension of the stabilisation of the previous hydrogen bond only in a destabilising manner!

These studies let us conclude that it is safe to say the amide nitrogen will never be a hydrogen bond acceptor.

Reference:
A. Johansson, P. Kollman, S. Rothenberg, J. McKelvey J. Am. Chem. Soc. 1974, 96, 3794. DOI: 10.1021/ja00819a013.

• Hi, RE: "The lone pair is not really available as a lone pair on the nitrogen, but forms a π-bond to the neighbouring carbonyl carbon to a significant extent. Thus, these nitrogens cannot be hydrogen bond acceptors." A silly question probably, but does this reasoning extend to aniline also? Jan 30 '18 at 13:33
• @GaurangTandon Somewhat, yes, but aniline’s lone pair is not involved in resonance as much.
– Jan
Feb 2 '18 at 16:03
• I understand you mean to say that it cannot be predicted theoretically accurately? Feb 2 '18 at 16:05
• @GaurangTandon It can be predicted. The six-membered benzene ring of aniline is pretty stable in itself so the additional nitrogen lone pair doesn’t add much. The $\ce{C=O}$ double bond benefits much more because it is fully localised by itself.
– Jan
Feb 2 '18 at 16:09
• Sorry for disturbing, but by the same reasoning, can we say that the solubility of benzyl alcohol in water is more than that of phenol? (because the latter has weaker H-bonds, due to a delocalised lone pair, as compared to the former) Feb 2 '18 at 16:26