InfraRed

INFRARED STRETCHING ABSORPTIONS

Functional Group	Bond	IR Range (cm^-1)
Alkane	[C–H]	2850 – 2960
Alkane	[C–C]	1100 – 1300
Alkene	[C=C]	1640 – 1680
Alkyne	[≡C–H]	3300
Alkyne	[C≡C]	2100 – 2260
Alcohol	[O–H]	3400 – 3650
Alcohol / Ether / Ester	[C–O]	1000 – 1300
Amine	[C–N]	1030 – 1230
Amine	[N–H]	3300 – 3500
Carbonyl	[C=O]	1680 – 1750
Carboxylic acid	[RCOO–H]	2500 – 3100
Nitrile	[C≡N]	2210 – 2260
Alkyl Chloride	[C–Cl]	600 – 800

The bond between two atoms is not rigid but expands and contracts like a spring. The reported bond length is simply the average of the fully expanded and fully contracted "spring". The general positions of IR absorptions can be understood by considering the masses of the two bonded atoms and the length of the bond . . . . for a bond between A and X, A–X (1 spring) is longer than A=X (2 springs) which is longer than A≡X (3 springs). As the bond contracts and expands, its motion can be described using a sine wave. When IR energy is passed through the molecule, the energy with the same wavelength as a bond's stretching motion is absorbed. The absorbed energy increases the amplitude of the stretching motion, but the wavelength is unchanged. If a detector is positioned behind the molecule, it will "detect" the absorption of IR energy at the wavelength of a particular bond's stretching motion and create a peak to indicate this phenomenon.

The table to the right gives the IR absorption range for the stretching motions of common single, double and triple bonds. The absorption position is given in reciprocal centimeters, cm^-1. If the wavelength is expressed in the typical units of meters . . . .

Wavenumber (cm^-1) = 1

λ × 1 cm

1E-2 m

To see the various IR motions and absorptions, use the radio buttons to change the top atom (C, N or O), the bond type (single, double or triple) and the bottom atom (H, C, N, O, Cl).

Single Bonds:

C–H vs N–H vs O–H . . . . since Hydrogen has the smallest elemental mass (1 amu), the spring expands very little. This creates a molecular motion with a short wavelength and high energy, E = hc
λ .
C–C vs C–N vs C–O . . . . replacing the Hydrogen with C, N or O increases the mass by a factor of 12, 14 or 16. This causes the spring to expand significantly creating a molecular motion with a long wavelength and low energy.

Double Bonds:

C=C vs C=N vs C=O . . . . adding a second "spring" creates a molecular motion with a shorter wavelength (higher Energy) than the corresponding single bond.

Triple Bonds:

C≡C vs C≡N . . . . adding a third "spring" creates a molecular motion with a shorter wavelength (higher Energy) than the corresponding double bond.

In addition to the stretching motions metioned above, molecules can also have other vibrations that absorb energy in the IR region of the Electromagnetic spectrum. Each of the vibrations listed below display a molecular motion with a specific wavelength . . . .

Symmetrical stretching

Antisymmetrical stretching

In-plane bending

Out-of-plane bending

Monosubstituted:	690–710 cm^–1	1,2,4-Trisubstituted:	780–830 cm^–1
Monosubstituted:	730–770 cm^–1	1,2,4-Trisubstituted:	870–900 cm^–1

o-Disubstituted:	735–770 cm^–1	1,2,3-Trisubstituted:	670–720 cm^–1
		1,2,3-Trisubstituted:	750–790 cm^–1
m-Disubstituted:	690–710 cm^–1
m-Disubstituted:	810–850 cm^–1	1,3,5-Trisubstituted:	660–700 cm^–1
			830–900 cm^–1
p-Disubstituted:	810–840 cm^–1	1,3,5-Trisubstituted:

Aromatic compounds show strong absorptions in the 660 to 900 cm^–1 range due to C–H out-of-plane bending.

These absorptions can be used to distinguish between mono-, di- and tri-substituted benzenes.

In addition, further discrimination between di-substituted (o, m or p) and tri-substituted (1,2,4 or 1,2,3 or 1,3,5) patterns can also be derived from the IR absorptions of C–H out-of-plane bending motions as shown in the table to the right.