N–H Absorptions in Amines and Amides

Introduction

Amines and amides contain N–H bonds, and like O–H bonds they absorb in the high-wavenumber region of an infrared spectrum. But N–H bands carry a special bonus: the number of peaks tells you how many hydrogen atoms are on the nitrogen. A primary amine gives two bands, a secondary amine gives one and a tertiary amine gives none. This page explains why, and how to tell N–H from O–H.

Core explanation

Position and shape. N–H stretches absorb at about 3300–3500 cm⁻¹, overlapping the alcohol O–H region. Nitrogen is less electronegative than oxygen, so N–H bonds are less polar and form weaker hydrogen bonds. As a result, N–H bands are usually less intense and sharper than O–H bands. They are often described as medium-intensity bands with a pointed shape.

Primary amines: two bands. A primary amine, R–NH₂, has two N–H bonds on the same nitrogen. They do not vibrate independently; they couple together. In the symmetric stretch both bonds lengthen at the same time; in the asymmetric stretch one lengthens as the other shortens. These two modes have slightly different energies, giving two bands roughly 70–100 cm⁻¹ apart, for example near 3380 and 3300 cm⁻¹ in a simple alkylamine. The asymmetric stretch is the higher of the two.

Secondary amines: one band. A secondary amine, R₂NH, has only one N–H bond, so there is only one stretching mode and one band, usually weak, near 3300–3350 cm⁻¹.

Tertiary amines: none. R₃N has no N–H bond at all, so there is no band in this region. A tertiary amine must be recognised by other evidence, such as C–N absorptions in the fingerprint region or other techniques.

Amides. Amides show the same counting rule. A primary amide, R–CONH₂, gives two N–H bands (often near 3350 and 3180 cm⁻¹ in the solid, lowered by strong hydrogen bonding); a secondary amide, R–CONHR′, gives one. What sets amides apart is a strong C=O stretch, called the amide I band , at a relatively low 1630–1690 cm⁻¹. Primary and secondary amides also show an N–H bending band, the amide II band, near 1550–1640 cm⁻¹.

N–H bending in amines. Primary amines show an NH₂ scissoring bend near 1580–1650 cm⁻¹, which can be mistaken for a C=C band if you are not careful.

Step-by-step reasoning

To analyse a band between 3300 and 3500 cm⁻¹:

1. Decide whether it is broad and strong (typical of O–H) or sharper and medium (typical of N–H). 2. Count the peaks: two suggests NH₂, one suggests NH. 3. Look for a strong C=O band; near 1630–1690 cm⁻¹ it points to an amide. 4. If there is no C=O band, the compound is likely an amine. 5. Remember that a tertiary amine shows no N–H band at all.

Visual explanation

Picture the region above 3000 cm⁻¹ for three compounds. Propylamine shows a pair of neat, twin dips like two teeth. Diethylamine shows a single small tooth. Triethylamine shows a flat baseline, with only C–H dips just below 3000 cm⁻¹.

Real-world analogy

Two people on a seesaw can move up together or alternately; the two motions feel different and have different rhythms. Two N–H bonds on one nitrogen behave the same way, producing two distinct vibrations, whereas a lone rider on a swing has only one rhythm.

Real-world example

Paracetamol contains a secondary amide group. Its IR spectrum shows a single N–H band near 3320 cm⁻¹, a broad phenol O–H band, and a strong amide I C=O band near 1650 cm⁻¹. Pharmaceutical laboratories use this pattern to confirm the identity of tablets and detect counterfeit products.

Why?

Why are N–H bands weaker than O–H bands? Infrared intensity depends on how much the dipole moment changes during the vibration. N–H is less polar than O–H because nitrogen is less electronegative than oxygen, so stretching it changes the dipole moment less and absorbs less radiation.

Common misconception

"Two peaks near 3400 cm⁻¹ mean the molecule has two separate N–H groups." The two bands of a primary amine come from one NH₂ group vibrating in two coupled ways, not from two different nitrogen atoms.

Worked example

Question: A compound C₃H₇NO shows two medium bands at 3350 and 3180 cm⁻¹ and a strong band at 1660 cm⁻¹. Identify its functional group and suggest a structure.

Reasoning: Two bands in the N–H region indicate an NH₂ group. The strong band at 1660 cm⁻¹ is a low-wavenumber C=O, typical of an amide I band. A primary amide with three carbons is propanamide.

Answer: Primary amide; propanamide, CH₃CH₂CONH₂.

Quick check

1. How many N–H stretching bands would you expect in the spectrum of the secondary amine diethylamine? Answer: One, because it has only one N–H bond.

Exam focus

When asked to distinguish amines, state the number of bands and link it to the number of N–H bonds. When distinguishing N–H from O–H, mention both shape (sharper) and intensity (weaker). For amides, quote the low C=O wavenumber as extra evidence.

Advanced insight

The amide C=O absorbs at lower wavenumber than a ketone because the nitrogen lone pair is delocalised into the carbonyl group. This resonance gives the C=O bond partial single-bond character, lowering its force constant. The same delocalisation makes the C–N bond partly double, which is why amide bonds in proteins are planar and rigid.

Summary

N–H stretches appear at about 3300–3500 cm⁻¹ and are usually sharper and weaker than O–H bands. Primary amines and amides give two bands from coupled symmetric and asymmetric stretches; secondary ones give one; tertiary amines give none. Amides also show a strong, relatively low C=O band at 1630–1690 cm⁻¹.

Practice questions

1. Explain why a primary amine shows two N–H stretching bands. Answer: The two N–H bonds of the NH₂ group couple into a symmetric and an asymmetric stretch, which have different energies and so absorb at different wavenumbers. 2. How could IR distinguish propylamine from trimethylamine? Answer: Propylamine shows two N–H bands near 3300–3400 cm⁻¹; trimethylamine has no N–H bond and shows no band there. 3. State two differences between the IR spectra of ethanamide and ethylamine. Answer: Ethanamide shows a strong C=O band near 1650–1690 cm⁻¹ and generally lower, broader N–H bands due to strong hydrogen bonding; ethylamine has no C=O band. 4. Give two features that help distinguish an N–H band from an O–H band. Answer: N–H bands are usually sharper and less intense than the broad, strong O–H bands, and a primary N–H shows a characteristic pair of peaks.