Semester Dec – Apr 2010
INFRARED ABSORPTION
SPECTROSCOPY
LEARNING OUTCOMES

By the end of this topic, students should be able to:







Explain the principles and the working mechanism of infrared
(IR) absorption spectroscopy
Identify the molecular species that absorb IR radiation
Interpret IR spectrum
Explain stretching and bending vibrations in relation to IR
absorption
Determine unknown qualitatively using IR absorption
Draw a schematic diagram of a conventional IR instrument
and a fourier transform IR instrument and explain the
function of each component of the instrument
Differentiate between a dispersive IR instrument and a FTIR
2
spectrometer
INFRARED SPECTROSCOPY
Mostly for qualitative analysis
 Absorption spectra is recorded as transmittance
spectra
 Absorption in the infrared region arise from
molecular vibrational transitions
 Absorption at specific wavelengths
 Thus, IR spectra provides more specific
qualitative information
 IR spectra is called “fingerprints”

- because no other chemical species will have
identical IR spectrum

3
COMPARISON BETWEEN TRANSMITTANCE (UPPER) VS
ABSORBANCE (LOWER) PLOT
The transmittance
spectra provide better
contrast btw
intensities of strong
and weak bands
compared to
absorbance spectra
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ELECTROMAGNETIC SPECTRUM
Energy of IR photon insufficient to cause electronic
excitation but can cause vibrational excitation

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INTRODUCTION

Comparison between UV-vis and IR
Energy:
 Frequency:
 Wavelength:

UV > vis > IR
UV > vis > IR
UV < vis < IR
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INFRARED SPECTROSCOPY



Infrared (IR) spectroscopy deals with the interaction of
infrared radiation with matter
IR spectrum provides:
 Important information about its chemical nature
and molecular structure
IR applicability:
 Analysis of organic materials
 Polyatomic inorganic molecules
 Organometallic compounds
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
IR region of EM spectrum:
 λ: 780 nm – 1000 μm
 Wavenumber: 12,800 – 10cm-1
IR region subdivided into 3 subregions:
1. Near IR region (Nearest to the visible)
- 780 nm to 2.5 μm (12,800 to 4000 cm-1)
2. Mid IR region
- 2.5 to 50 μm (4000 – 200 cm-1)
3. Far IR region
- 50 to 1000 μm (200 – 10cm-1)
N
E
A
R
M
I
D
infrared

F
A
R
8
8

When IR absorption occur?
1.IR absorption only occurs when IR radiation interacts
with a molecule undergoing a change in dipole
moment as it vibrates or rotates.
2.Infrared absorption only occurs when the incoming IR
photon has sufficient energy for the transition to the
next allowed vibrational state
Note: If the 2 rules above are not met, no absorption
can occur
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WHAT HAPPEN WHEN A MOLECULE ABSORBS
INFRARED RADIATION?

Absorption of IR radiation corresponds to energy
changes on the order of 8 to 40 kJ/mole.
- Radiation in this energy range corresponds to stretching
and bending vibrational frequencies of the bonds in
most covalent molecules.


In the absorption process, those frequencies of IR
radiation which match the natural vibrational
frequencies of the molecule are absorbed.
The energy absorbed will increase the amplitude
of the vibrational motions of the bonds in the
molecule.
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

NOT ALL bonds in a molecule are capable of absorbing
IR energy. Only those bonds that have change in
dipole moment are capable to absorb IR radiation.
The larger the dipole change, the stronger the
intensity of the band in an IR spectrum.
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
What is a dipole moment?

is a measure of the extent to which a separation exists
between the centers of positive and negative charge within
a molecule.
δ-
O
H
δ+
H
δ+
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 In
heteronuclear diatomic molecule, because
of the difference in electronegativities of the
two atoms, one atom acquires a small positive
charge (q+), the other a negative charge (q-).
 This
molecule is then said to have a dipole
moment whose magnitude, μ =qd
distance of separation of the charge
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MOLECULAR SPECIES THAT ABSORB
INFRARED RADIATION
 Compound
absorb in IR region
Organic compounds, carbon monoxide
 Compounds
DO NOT absorb in IR region
O2, H2, N2, Cl2
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IR VIBRATIONAL MODES
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Molecular vibration
divided
into
back &
forth
movement
stretching
involves
change in
bond angles
bending
wagging
scissoring
symmetrical
asymmetrical
rocking
in-plane
vibration
twisting
out of
plane
vibration
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STRETCHING
17
BENDING
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SAMPLE HANDLING TECHNIQUES
 Gases

evacuated cylindrical cells equipped with suitable
windows
 Liquid
sodium chloride windows
 “neat” liquid

 Solid
Pellet (KBr)
 Mull

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LIQUID


a drop of the pure (neat) liquid is squeezed between two
rock-salt plates to give a layer that has thickness 0.01mm
or less
2 plates held together by capillary mounted in the beam
path
What is meant by “neat” liquid?


Neat liquid is a pure liquid that do not contain any solvent or
water.
This method is applied when the amount of liquid is small
or when a suitable solvent is unavailable
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SOLID SAMPLE PREPARATION

There are three ways to prepare solid sample for IR
spectroscopy.

Solid that is soluble in solvent can be dissolved in a
solvent, most commonly carbon tetrachloride CCl4.

Solid that is insoluble in CCl4 or any other IR solvents
can be prepared either by KBr pellet or mulls.
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PELLETING
(KBr PELLET)
Mixing the finely ground solid sample with
potassium bromide (KBr) and pressing the
mixture under high pressure (10,000 – 15,000 psi)
in special dye.
 KBr pellet can be inserted into a holder in the
spectrometer.

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MULLS
Formed by grinding 2-5 mg finely powdered
sample, presence 1 or 2 drops of a heavy
hydrocarbon oil (Nujol)
 Mull examined as a film between flat salt plates


This method applied when solid not soluble in an
IR transparent solvent, also not convenient
pelleted in KBr
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What is a mull

A thick paste formed by grinding an insoluble solid
with an inert liquid and used for studying spectra of
the solid
What is Nujol

A trade name for a heavy medicinal liquid paraffin.
Extensively used as a mulling agent in spectroscopy
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INSTRUMENTATION
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IR INSTRUMENT
 Dispersive

sequential mode
 Fourier

spectrometers
Transform spectrometers
simultaneous analysis of the full spectra range using
inferometry
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IR INSTRUMENT (DISPERSIVE)

Important components in IR dispersive
spectrometer
1
source
lamp
2
sample
holder
Source:
- Nernst glower
- Globar source
- Incandescent wire
3
λ
selector
4
detector
5
signal processor
& readout
Detector:
- Thermocouple
- Pyroelectric transducer
- Thermal transducer
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RADIATION SOURCES
 generate
a beam with sufficient power in the
λ region of interest to permit ready detection
& measurement
 provide continuous radiation; made up of all
λ’s with the region (continuum source)
 stable output for the period needed to
measure both P0 and P
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SCHEMATIC DIAGRAM OF A DOUBLE BEAM
INFRARED SPECTROPHOTOMETER
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FTIR
FOURIER TRANSFORM INFRARED
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FTIR
Why is it developed?
to overcome limitations encountered with the
dispersive instruments
 especially slow scanning speed; due to individual
measurement of molecules/atom
 utilize an interferometer

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
Interferometer
Special instrument which can read IR frequencies
simultaneously
 faster method than dispersive instrument
 interferograms are transformed into frequency spectrums
by using mathematical technique called Fourier
Transformation

FT
Calculations
interferograms
IR spectrum
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COMPONENTS OF FOURIER TRANSFORM INSTRUMENT
- majority of commercially available Fourier transform infrared
instruments are based upon Michelson interferometer
3
4
1
2
5
6
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Advantages (over dispersive instrument)
high sensitivity
 high resolution
 speed of data acquisition ( data for an entire spectrum can
be obtained in 1 s or less)

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INTERPRETATION
INFRARED SPECTRA
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INFRARED SPECTRA
 IR
spectrum is due to specific structural
features, a specific bond, within the molecule,
since the vibrational states of individual
bonds represent 1 vibrational transition.
 e.g.
IR spectrum can tell the molecule has an
O-H bond or a C=O or an aromatic ring
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INFRARED SPECTRA
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HOW TO INTERPRET INFRARED SPECTRA?
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How to analyze IR spectra
1. Begin by looking in the region from 4000-1300.
Look at the C–H stretching bands around 3000:
Indicates:
Are any or all to the right of 3000?
alkyl groups (present in most organic
molecules)
Are any or all to the left of 3000?
a C=C bond or aromatic group in the
molecule
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2. Look for a carbonyl in the region 1760-1690.
If there is such a band:
Indicates:
Is an O–H band also present?
a carboxylic acid group
Is a C–O band also present?
an ester
Is an aldehyde C–H band also
present?
an aldehyde
Is an N–H band also present?
an amide
Are none of the above present?
a ketone
(also check the exact position of the carbonyl band for clues as to the type
of carbonyl compound it is)
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3. Look for a broad O–H band in the region 3500-3200 cm-1.
If there is such a band:
Indicates:
Is an O–H band present? an alcohol or phenol
4. Look for a single or double sharp N–H band in the region 3400-3250
cm-1.
If there is such a band:
Indicates:
Are there two bands?
a primary amine
Is there only one band?
a secondary amine
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5. Other structural features to check for:
Indicates:
Are there C–O stretches?
an ether (or an ester if there is a
carbonyl band too)
Is there a C=C stretching band?
an alkene
Are there aromatic stretching bands?
an aromatic
Is there a C≡C band?
an alkyne
Are there -NO2 bands?
a nitro compound
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How to analyze IR spectra



If there is an absence of major functional group bands in the
region 4000-1300 cm-1 (other than C–H stretches), the compound
is probably a strict hydrocarbon.
Also check the region from 900-650 cm-1. Aromatics, alkyl halides,
carboxylic acids, amines, and amides show moderate or strong
absorption bands (bending vibrations) in this region.
As a beginning student, you should not try to assign or interpret
every peak in the spectrum. Concentrate on learning the major
bands and recognizing their presence and absence in any given
spectrum.
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44
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ALKANE
H
H
H
H
C
C
C
H
H
H
n
H
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C-H
CH2
at
CH3
CH2
cm-1
Stretch for sp3 C-H around 3000 – 2840 cm-1.
Methylene groups have a characteristic bending absorption
approx 1465 cm-1
Methyl groups have a characteristic bending absorption at
approx 1375 cm-1
The bending (rocking) motion associated with four or more
CH2 groups in an open chain occurs at
about 720 47
ALKENE
H H
C
C
H H
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ALKENE
=C-H
=C-H
C=C
Stretch for sp2 C-H occurs at values greater than 3000 cm-1.
out-of-plane (oop) bending occurs in the range 1000 – 650 cm-1
stretch occurs at 1660 – 1600 cm-1;
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often conjugation moves C=C stretch to lower frequencies
and increases the intensity
ALKYNE
HC CH
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ALKYNE
CH
C C
Stretch for sp C - H occurs near 3300 cm-1.
Stretch occurs near 2150 cm-1; conjugation moves
stretch to lower frequency.
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AROMATIC RINGS
C H
C C
C H
Stretch for sp2 C-H occurs at values greater than 3000 cm-1.
Ring stretch absorptions occur in pairs at 1600 cm-1 and
1475 cm-1.
Bending occurs at 900 - 690cm-1.
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AROMATIC RINGS
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C-H Bending ( for Aromatic Ring)
The out-of-plane (oop) C-H bending is useful in order to assign the
positions of substituents on the aromatic ring.
Monosubstituted rings
•this substitution pattern always gives a strong absorption near 690
cm-1. If this band is absent, no monosubstituted ring is present. A
second strong band usually appears near 750 cm-1.
Ortho-Disubstituted rings
•one strong band near 750 cm-1.
Meta- Disubstituted rings
•gives one absorption band near 690 cm-1 plus one near 780 cm-1. A
third band of medium intensity is often found near 880 cm-1.
Para- Disubstituted rings
- one strong band appears in the region from 800 to 850 cm-1.
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Ortho-Disubstituted rings
C H
Bending observed as one strong band near 750 cm-1.
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Meta- Disubstituted rings
C H
- gives one absorption band near 690 cm-1 plus one near 780 cm-1.
A third band of medium intensity is often found near 880 cm-1.
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Para- Disubstituted rings
C H
- one strong band appears in the region from 800 to 850 cm1.
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ALCOHOL
H H
H C
H OH H
C OH
H C C C H
H H
H H H
Primary alcohol 10
Secondary alcohol 20
CH3
H3C C OH
CH3
Tertiary alcohol 30
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ALCOHOL
O-H
The hydrogen-bonded O-H band is a broad peak at 3400 – 3300 cm-1.
This band is usually the only one present in an alcohol that has not
been dissolved in a solvent (neat liquid).
C-O-H Bending appears as a broad and weak peak at 1440 – 1220 cm-1 often
obscured by the CH3 bendings.
C-O
Stretching vibration usually occurs in the range 1260 – 1000 cm-1.
This band can be used to assign a primary, secondary or tertiary
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structure to an alcohol.
PHENOL
OH
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PHENOL
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ETHER
R O R'
C-O
The most prominent band is that due to C-O stretch, 1300 – 1000 cm-1.
Absence of C=O and O-H is required to ensure that C-O stretch
is not due to an ester or an alcohol.
Phenyl alkyl ethers give two strong bands at about 1250 – 1040 cm-1,
while aliphatic ethers give one strong band at about 1120 cm-1.
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CARBONYL COMPOUNDS
cm-1
1810
1800
Anhydride
Amide
(band 1) Chloride
1760
Acid
1735
Anhydride
(band 2)
1725
Ester
1715
Aldehyde
1710
Ketone
1690
Carboxylic
acid
Normal base values for the C=O stretching vibrations for carbonyl groups
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A. ALDEHYDE
R
C H
O
R
C H
O
Ar
C H
O
C-H
C=O stretch appear in range 1740-1725 cm-1 for normal
aliphatic aldehydes
Conjugation of C=O with phenyl; 1700 – 1660 cm-1 for
C=O
and 1600 – 1450 cm-1 for ring (C=C)
Stretch, aldehyde hydrogen (-CHO), consists of weak
bands, one at 2860 - 2800 cm-1 and
the other at 2760 – 2700 cm-1.
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B. KETONE
R
C R'
O
R
C R'
O
Ar
C R'
O
C=O stretch appear in range 1720-1708 cm-1 for
normal aliphatic ketones
Conjugation of C=O with phenyl; 1700 – 1680 cm-1 for C=O
and 1600 – 1450 cm-1 for ring (C=C)
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C. CARBOXYLIC ACID
R
C OH
O
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D. ESTER
R C O R
O
R C O R
O
Ar
C O R
O
C–O
C=O stretch appear in range 1750-1735 cm-1 for normal
aliphatic esters
Conjugation of C=O with phenyl; 1740 – 1715 cm-1 for C=O
and 1600 – 1450 cm-1 for ring (C=C)
Stretch in two or more bands, one stronger and broader than
the other, occurs in the range 1300 – 1000 cm-1
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E. AMIDE
O
R C N
10
H
O
R C N
H
20
H
R
O
R C N
30
R
R
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AMIDE
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F. ACID CHLORIDE
O
R
C
Cl
C O
Stretch appear in range 1810 -1775 cm-1 in conjugated
chlorides. Conjugation lowers the frequency to 1780 – 1760
cm-1
C
Stretch occurs in the range 730 -550 cm-1
Cl
Acid chloride show a very strong band for the C=O group.
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F. ANHYDRIDE
R
O
O
C
C
R
O
C O
Stretch always has two bands, 1830 -1800 cm-1 and 1775 – 1740 cm-1,
with variable relative intensity.
Conjugation moves the absorption to a lower frequency. Ring strain
(cyclic anhydride) moves absorptions to a higher frequency.
C
Stretch (multiple bands) occurs in the range 1300 -900 cm-1
O
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AMINE
R N
H
H
10
H
R N R
R
N R
R
20
30
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AMINE
N–H
N–H
Stretching occurs in the range 3500 – 3300 cm-1.
Primary amines have two bands.
Secondary amines have one band: a vanishingly weak one for
aliphatic compounds and a stronger one for aromatic secondary
amines.
Tertiary amines have no N – H stretch.
Bending in primary amines results in a broad band in the range
1640 – 1560 cm-1.
Secondary amines absorb near 1500 cm-1
N–H
Out-of-plane bending absorption can sometimes be observed
near 800 cm-1
C–N
Stretch occurs in the range 1350 – 1000 cm-1
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PRIMARY AMINE
Secondary Amine
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TERTIARY AMINE
Aromatic Amine
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