Welcome to
Infrared Spectroscopy
By H.T.Wong
IR & NMR to provide information about the
structure of a molecule.
 Modern instrumental technique to
understand the structures of molecules
 IR-vibrations of atoms
 NMR-magnetic properties of atomic nuclei
 Different functional groups have
characteristic absorption wavenumbers.

Notes on IR Spectroscopy
ABSORPTION OF INFRARED
RADIATION
 The atoms in molecules are not static, but
vibrate about their equilibrium positions.
When molecules absorb infrared radiation,
the absorbed energy causes an increase in
the amplitude of the vibration of the bonded
atoms.

The molecule is in an excited vibrational state.
The absorbed energy is then dissipated as heat
when the molecule returns to the ground state.
more
vigorous
vibration
IR
higher
energy
level
n
tio
dia
ra
Energy
absorbed
A particular type of bond between a certain pair of atoms will
have a characteristic frequency of vibration. The natural
frequency of vibration lies in the infrared region (1.20  1013
- 1.20  1014 Hz) of the electromagnetic spectrum.
Absorption of infrared radiation causes the atoms in
molecules to vibrate more vigorously.

IR spectrum is obtained by measuring the absorption of IR
radiation of a sample at different frequencies. The
absorption is usually quoted in wavenumber (cm-1) which is
the reciprocal of the wavelength (1/l). Since frequency ()
and wavelength (l) are related by the equation, c=l (where
c is the speed of light), wavenumber will be a direct measure
of frequency.
Electromagnetic Spectrum
Wavelength / m
X-rays
-rays
Ultraviolet
Visible Infrared
Frequency / Hz
Energy
decreasing
Microwaves
Radiowaves
Relationship between Frequency, Wavelength and
Wavenumber





c = l
c = velocity of light (ms-1)
 = frequency (Hz or s-1)
l = wavelength (m)
In IR spectroscopy, the frequency of radiation is
often expressed as wavenumber (cm-1) :
 wavenumber = 1/l
where
Frequency(Hz)
1.2  1013
4.5  1013
1.2  1014

2.50  10-6
Wavenumber- (cm-1 )
400
1500
4000
Example: The wavenumber of a vibration
frequency with the value 4.5  1013 Hz can
be calculated as follows:
13 Hz  4.5  1013 s-1
 Since 4.5  10
wavenumber= 4.5  1013 / 3  108  102
=1500 cm-1


Wavelength-l (m)
2.50  10-5
6.67  10-6
Photo 1: Double Beam Dispersive Infrared Spectrometer
Photo 2: Fourier Transform Infrared (FTIR)
Spectrophotometer
Photo 4: Apparatus for Handling Samples for Infrared
Spectroscopy
FEATURES OF AN INFRARED
SPECTRUM

An infrared spectrum is a plot of per cent
transmittance (or absorbance) against
wavenumber (frequency or wavelength). A
typical infrared spectrum is shown below.
Infrared Spectrum of Propanal
100
90
Per cent transmittance
80
70
60
50
40
30
20
10
C=O
0
4600
3800
3000
2200
1800
1400
Wavenumber (cm-1)
1000
800
600
400
A 100 per cent transmittance in the spectrum implies no
absorption of IR radiation. When a compound absorbs IR
radiation, the intensity of transmitted radiation decreases. This
results in a decrease of per cent transmittance and hence a dip in
the spectrum. The dip is often called an absorption peak or
absorption band.

Different types of groups of atoms (C-H, O-H, N-H, etc…)
absorb infrared radiation at different characteristic
wavenumbers.
INTERPRETATION OF IR SPECTRA

In general, the IR spectrum can be split into four regions for
interpretation:

4000 - 2500 cm-1: Absorption of single bonds formed by
hydrogen and other elements e.g. C-H, O-H, N-H
2500 - 2000 cm-1: Absorption of triple bonds e.g. C≡C, C≡N
2000 - 1500 cm-1: Absorption of double bonds e.g. C=C, C=O
1500 - 400 cm-1: This region often consists of many different,
complicated bands. This part of the spectrum is unique to each
compound and is often called the fingerprint region. It is rarely
used for identification of particular functional groups.



The region between 4000 and 1500 cm-1 of an
infrared spectrum is often used for the
identification of various functional groups.

Different functional groups have characteristic
absorption wavenumbers.
A correlation table
providing information is shown below. It could help
to interpret an infrared spectrum.
Table 1: Correlation Table of Functional Groups and IR
Absorption Wavenumbers










Bond
C=C
C=O
C≡C
C≡N
O-H
C-H
O-H
N-H
Characteristic Range
Wavenumber (cm-1)
Alkenes
1610 to 1680
Aldehydes, ketones, acids, esters
1680 to 1750
Alkynes
2070 to 2250
Nitriles
2200 to 2280
Acids (hydrogen-bonded)
2500 to 3300
Alkanes, alkenes, arenes
2840 to 3095
Alcohols, phenols(hydrogen-bonded)
3230 to 3670
Primary amine
3350 to 3500
If the infrared spectrum of an unknown compound shows an
absorption peak in the region of 3200 to 3700 cm-1, it is
reasonable to assume that the compound may contain either
O-H or

N-H group.
The following examples show how to make use of the
characteristic absorption wavenumbers to identify the functional
groups present in organic compounds. Fig.1 shows a portion of
the IR spectrum of ethanol.
Fig.1 IR spectrum of ethanol

The peak at 2950 cm-1 corresponds to the
absorption of C-H group and that at 3340
cm-1 corresponds to the absorption of O-H
group. Hence, a O-H group is present in
the ethanol.
Fig. 2 shows a portion of the IR spectrum of
methanoic acid.
The peak at 1710 cm-1 corresponds to the
absorption of C=O group and that at 3100 cm-1
corresponds to the absorption of O-H group.
Hence, both the C=O and O-H groups are
present in the methanoic acid.

It should be noted that the absorption of the
O-H group in alcohols and carboxylic acids
does not usually appear as a sharp peak.
Instead, a broad band is observed because
the vibration of O-H group is complicated
by the hydrogen bonding.
A SUMMARY OF STRATEGY FOR
INDENTIFICATION OF FUNCTIONAL GROUPS
1. Look at the high-wavenumber end of the
spectrum (above 1500 cm-1) and concentrate
initially on the major absorption peaks.
 2.
Classify the major absorption peaks in the
spectrum as follows:

(a) X-H absorption
(4000 - 2500 cm-1)
 (b) triple bond absorption
(2500 - 2000 cm-1)
 (c) double bond absorption (2000 - 1500 cm-1)

3.
For each absorption peak, short-list the
possibilities using a correlation table or chart.
(Note: Do not expect to be able to assign every
absorption peak in the spectrum.)

4. Place as much reliance on negative evidence as on
positive evidence, e.g. if there is no absorption peak in
1600-1800 cm-1 region, it is likely that a carbonyl
group is absent in the test sample.
LIMITATIONS OF THE USE OF IR
SPECTROSCOPY IN THE IDENTIFICATION OF
ORGANIC COMPOUNDS
1. Some infrared absorptions have very close
wavenumbers and the peaks coalesce.
 2.
NOT all vibrations give rise to strong absorption
peaks.
 3.
NOT all absorption peaks in a spectrum can be
associated with a particular bond or part of the
molecule.

4. Intermolecular interactions in molecules
complicate infrared spectra.
 5.
Infrared spectroscopy can only provide limited
structural information of molecule

Characteristic Absorptions
of Carbonyl Groups










Wavenumber / cm-1
RCOCl
Acyl chloride
1815 - 1790
RCOOR’ Aliphatic ester
1750 - 1730
RCHO
Aliphatic aldehyde
1740 - 1720
RCOOH Aliphatic acid
1725 - 1700
RCOR’
Aliphatic ketone
1725 - 1700
ArCHO Aromatic aldehyde
1715 - 1695
ArCOR Aromatic ketone
1700 - 1680
ArCOAr Diaromatic ketone
1670 - 1650
RCONH2 Aliphatic amide
1680 - 1640

(Source: Modern Chemical Techniques, The Royal Society of Chemistry, Cambridge, 1992.)



Wavenumber (cm-1)
2970
1640



Bond in Hex-1-ene
C-H
C=C
Per cent transmittance
100
80
60
40
20
4000
3500
3000
2500
2000
Wavenumber (cm




Wavenumber (cm-1)
3300
2900 to 2950
2200




-1
1800
)
Bond in hex-1-yne
C-H *
C-H
CC
1600




Wavenumber (cm-1)
2980
2830, 2720
1730




Bond in propanal
Aliphatic C-H
C-H of CHO
C=O
Per cent transmittance
100
80
60
40
20
4000
3500
3000
2500
2000
1800
1600
Wavenumber (cm -1 )



Wavenumber (cm-1)
2880 to 2980
1720



Bond in 3-methylbutan-2-one
C-H
C=O
Per cent transmittance
100
80
60
40
20
4000
3500
3000
2500
2000
Wavenumber (cm




Wavenumber (cm-1)
3080
1720
1635




-1
1800
1600
)
Bond in hex-5-en-2-one
C-H
C=O
C=C




Wavenumber (cm-1)
3330
2940, 2820
1030




Bond in methanol
O-H
C-H
C-O




Wavenumber (cm-1)
3380
3050
1220




Bond in phenol
O-H
Aromatic C-H
C-O



Wavenumber (cm-1)
3100
1710



Bond in methnoic acid
O-H
C=O
Per cent transmittance
100
80
60
40
20
4000
3500
3000
2500
2000
Wavenumber (cm



Wavenumber (cm-1)
2900 to 2950
1740



-1
1800
1600
)
Bond in butyl ethanoate
C-H
C=O
Per cent transmittance
100
80
60
40
20
4000
3500
3000
2500
Wavenumber (cm



Wavenumber (cm-1)
3300 to 3400
2850 to 2950



2000
-1
1800
)
Bond in butylamine
N-H
C-H
1600



Wavenumber (cm-1)
3000
2250



propanenitrile
C-H
CN
1.
Explain how you can distinguish between each
isomer in the following pairs of compounds by
infrared absorption peaks.
(a)
(b)
(c)
H2C=CHCH2OH and (CH3)2C=O
HCCCH2CH2CH3 and H2C=CHCH2CH=CH2
CH3CH2CH2NH2 and (CH3)3N
a) C=C 1610-1680

O-H 3230-3670
 b) CC 2070-2250
 c) N-H 3350-3500


a) C=O 1680-1750
b) C=C 1610-1680
 c) no

2. An unknown compound with molecular formula
C3H6O shows no absorption peak in the region of
1700 to 1750 cm-1 of its infrared spectrum.
Which one of the following is likely to be the
unknown compound?
propanone; propanal; prop-2-en-1-ol
(A)
(B)
(C)
 No
C=O
 (A)
& (B) impossible
 (C) is the unknown
P e r c e n t tra n s m itta n c e
3.
An organic compound Z, with relative molecular mass
below 100, has the following composition by mass.
C 66.7%, H 11.1% and O 22.2%.
(a) Determine the molecular formula of Z.
(b) A portion of the infra-red (IR) spectrum of Z is shown below:
100
80
60
40
20
4000
3500
3000
2500
2000
W a v e n u m b e r / c m -1
1500
Using the IR spectrum and the result from (a), deduce
two possible structures of Z, each belonging to a different
homologous series.









a)
no of mol:
C
H
O
66.7/12 11.1/1 22.2/16
5.56
11.1
1.39
simplest ratio: 4
8
1
empirial formula is C4H8O
let the molecular formula be (C4H8O)n
rel. mol. mass: (4x12+8x1+16) x n < 100
n=1
molecular formula is C4H8O
 Wavenumber
2950

1720
 CH3CH2CH2CHO
 CH3CH2COCH3

Bond
C-H
C=O
P e r c e n t tr a n s m itta n c e
4. The infrared spectra in Figure 1 and 2 represent an ester
and an alkyne. Identify the peaks marked with “ * ” on each
spectrum and hence determine which spectrum represents
which compound.
Which of the following is an ester or alkyne?
Figure 1:
100
80
60
40
*
20
*
4000
3500
3000
2500
2000
W a v e n u m b e r (c m -1 )
1800
1600
Figure 2:
Per cent transmittance
100
80
*
60
*
40
20
4000
3500
3000
2500
2000
Wavenumber (cm -1 )
1800
1600
 Figure
1:
 2920cm-1 : C-H
 1720cm-1 : C=O
 an ester
 Figure
2:
 2920cm-1 : C-H
 2180cm-1 : CC
 an alkyne
5. An alkene A (C5H10), on ozonolysis, gave two different
compounds, B and C with molecular formulae C3H6O and
C2H4O respectively. Compound B has an infrared
spectrum as shown in Figure 3 below:
Per cent transmittance
100
80
60
40
20
4000
3500
3000
2500
2000
Wavenumber (cm -1 )
1800
1600
When B and C were separately treated with acidified potassium
dichromate(VI) solution, B did not react but C gave D with
molecular formula C2H4O2. D gave effervescence when mixed
with sodium hydrogencarbonate.
Deduce possible structures for A, B, C and D.
A:
2-methylbut-2-ene
 B : propanone
 C : ethanal
 D : ethanoic acid