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It is known that hydrogen bonding results in the broadening of peaks in the infrared spectrum of a molecule. Hydrogen bonding results in a spectrum of different bond lengths with different stretching frequencies. This results in a the functional group giving peaks of a whole continuous range of vibrational frequencies and hence, we see a broad peak. Broad peaks are most often observed from $\ce {2500}$ to $\ce {3000 cm^{-1}}$ for the stretching of carboxylic acid functionalities. Similarly, hydrogen bonding also results in alcohol $\ce {O-H}$ stretching giving broad peaks in the range of $\ce {3200}$ to $\ce {3600 cm^{-1}}$, albeit the degree of broadening is much less for alcohols. Amines are also able to form rather strong hydrogen bonds. However, amine $\ce {N-H}$ stretches often are not seen as broad peaks and are in fact, often sharp. Why is this so?

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The sharpness or broadness of a stretch in IR spectra depends on extent of Hydrogen bonding present in the molecule. Basically, if it undergoes immense intermolecular hydrogen bonding, the peaks tend to be broader and the lesser the hydrogen bonding becomes, the sharper the peaks get in the spectra.

Now, if you recall the criteria needed for stronger hydrogen bonding, the more electronegative the atom attached to hydrogen is the better is the extent of hydrogen bonding as the more electronegative atom will be able to create more $\delta +$ve cahrge on hydrogen which is better for stronger hydrogen bonding. $\ce{O}$ is far more electronegative than $\ce{N}$. $(\chi_o = 3.44, \chi_N = 3.04)$. So, the extent of hydrogen bonding is much much lesser with $\ce{N}$ being the attached atom than $\ce{O}$. (If you want to get an idea how strong the hydrogen bonding is in case of $\ce{O}$ than $\ce{N}$, simply consider $\ce{H2O}$ and $\ce{NH3}$, due to hydrogen bonding the former has its boiling point at $100^o C$, while that for the latter is $-33.4^oC$.) But anyways Hydrogen bonding is there in amines, only the extent is lesser.

However if you can somehow increase the $\delta+$ charge on hydrogen, by attaching some electron withdrawing group to the nitrogen, like in the case of amides, there you will be again able to see broad $\ce{N-H}$ stretches as now it can undergo stronger hydrogen bond formation, as $\ce{N}$ pulls the electron density more towards itself due to resonance., and $\ce{H}$ becomes electron deficient, and peak becomes broad. Here is the IR spectra of $\ce{N-methylacetamide}$ from the book Introduction to Spectroscopy which shows broad $\ce{N-H}$ stretches due to the reason mentioned above.

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In this spectra you can also see the $\ce{C=O}$ stretch is also relatively broader than observed generally. The reason is again extent of intermolecular hydrogen bonding, which is greater due to higher electronic density on the $\ce{O}$. But apart from that, the general reason for broadening the spectra is the extent of intermolecular hydrogen bonding, which is somewhat lesser for normal amines.

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  • $\begingroup$ I think with regards to amides, the reason for the broadening may be the "tautomerism" rather than hydrogen bonding. Due to strong delocalisation of the N lone pair, there may be some non-negligible tautomerism going on. $\endgroup$ – Tan Yong Boon Mar 31 at 2:10

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