Infrared spectroscopy (IR spectroscopy or vibrational spectroscopy) involves the interaction of infrared radiation with matter. It covers a range of techniques, mostly based on absorption spectroscopy. As with all spectroscopic techniques, it can be used to identify and study chemicals

1. INFRARED ABSORPTION
SPECTROSCOPY

2. 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
2

3. Comparison between transmittance (upper) vs absorbance (lower) plot
3
The transmittance spectra
provide better contrast btw
intensities of strong and
weak bands compared to
absorbance spectra

4. Electromagnetic Spectrum
4
 Energy of IR photon insufficient to cause electronic excitation but
can cause vibrational excitation

5. INTRODUCTION
• Comparison between UV-vis and IR
– Energy: UV > vis > IR
– Frequency: UV > vis > IR
– Wavelength: UV < vis < IR
5

6. 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
6

7. • 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)
7
infrared
N
E
A
R
M
I
D
F
A
R
7

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
8

9. 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.
9

10. • 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.
10

11. • 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.
11
O
H H
δ+
δ-
δ+

12. • 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
12
distance of separation of the charge

13. 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
13

14. IR Vibrational Modes
14

15. Molecular vibration
divided
into
stretching bending
back & forth
movement
involves
change in bond
angles
symmetrical asymmetrical
scissoring
rocking twisting
wagging
in-plane
vibration
out of
plane
vibration
15

16. STRETCHING
16

17. BENDING
17

18. Sample Handling Techniques
• Gases
– evacuated cylindrical cells equipped with suitable windows
• Liquid
– sodium chloride windows
– “neat” liquid
• Solid
– Pellet (KBr)
– Mull
18

19. 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
19

20. 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.
20

21. 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.
21

22. 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
22

23. 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
23

24. Instrumentation
24

25. IR Instrument
• Dispersive spectrometers
– sequential mode
• Fourier Transform spectrometers
– simultaneous analysis of the full spectra range
using inferometry
25

26. IR Instrument (Dispersive)
• Important components in IR dispersive
spectrometer
26
source
lamp
sample
holder
λ
selector
detector
signal processor
& readout
1 2 3 4 5
Source:
- Nernst glower
- Globar source
- Incandescent wire
Detector:
- Thermocouple
- Pyroelectric transducer
- Thermal transducer

27. 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
27

28. Schematic Diagram of a Double Beam Infrared
Spectrophotometer
28

29. FTIR
Fourier Transform Infrared
29

30. 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
30

31. • 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
31
FT
Calculations
interferograms
IR spectrum

32. Components of Fourier Transform Instrument
32
- majority of commercially available Fourier transform infrared instruments
are based upon Michelson interferometer
1
3
2
4
5
6

33. 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
33

34. Infrared Spectroscopy
• Infrared Spectroscopy
– Energy Absorption and Vibration
• IR electromagnetic radiation is just less energetic than visible light
• This energy is sufficient to cause excitation of vibrational energy levels
• Wavelength (l) = 2.5-16.7 x 10-6 m
• n = wavenumbers. Larger n = higher energy
• Excitation depends on atomic mass and how tightly they are
bound
– Hooke’s Law for 2 masses connected by a spring
– C—H Bond: Reduced Mass = (12+1)/(12x1) = 13/12 = 1.08
– C—C Bond: Reduced Mass = (12+12)/(12x12) = 24/144 = 0.167
kcal/mol101cm4000-600
1~ 1
 
l

21
21 )(~
mm
mm
fk


k = constant
f = force constant =
bond strength
m-term = reduced mass
sJ10x6.626ConstantsPlanck'h
hc
hc
hE
1
m/s10x3.0λυc
34-
8





l

l


35. 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)
35

36. Interpretation
Infrared Spectra
36

37. Vibrational modes leading to IR absorptions:
Many possible absorptions per molecule exist: stretching, bending,…

39. B.Using IR in Organic Chemistry
1.Functional Groups have characteristic IR absorptions
2.Fingerprint Region (600-1500 cm-1) is unique for every
molecule and lets us match an unknown with a known
spectrum

40. Regions of the Infrared Spectrum
• 4000-2500 cm-1 N-H,
C-H, O-H (stretching)
– 3300-3600 N-H, O-H
– 3000 C-H
• 2500-2000 cm-1 CC and C  N
(stretching)
• 2000-1500 cm-1 double bonds
(stretching)
– C=O 1680-1750
– C=C 1640-1680 cm-1
• Below 1500 cm-1 “fingerprint”
region

41. 12.8 Infrared Spectra of Some Common Functional Groups
Alkanes, Alkenes, Alkynes
• C-H, C-C, C=C, C  C have characteristic peaks
– absence helps rule out C=C or C  C

43. 4.IR of Alkenes
a. Alkene C—H absorbs at higher energy than alkanes because the
force constant is stronger than alkanes (sp2 hybridization)
b.Substitution pattern of alkenes give characteristic absorptions
» Terminal alkenes give 910, 990 cm-1
» Geminal disubstituted gives 890 cm-1
» trans disubstituted gives 970 cm-1
C C
R
H
H
H
C C
R
H
H
R
C C
H
H
R
R

47. IR: Aromatic Compounds
• Weak C–H stretch at 3030 cm1
• Weak absorptions 1660 - 2000 cm1 range
• Medium-intensity absorptions 1450 to 1600
cm1

49. IR: Alcohols and Amines
• O–H 3400 to 3650 cm1
– Usually broad and intense
• N–H 3300 to 3500 cm1
– Sharper and less intense than an O–H

52. IR: Carbonyl Compounds
• Strong, sharp C=O peak 1670 to 1780 cm1
• Exact absorption characteristic of type of
carbonyl compound
– 1730 cm1 in saturated aldehydes
– 1705 cm1 in aldehydes next to double bond or aromatic ring

53. C=O in Ketones
• 1715 cm1 in six-membered ring and acyclic ketones
• 1750 cm1 in 5-membered ring ketones
• 1690 cm1 in ketones next to a double bond or an aromatic ring
• 1735 cm1 in saturated esters
• 1715 cm1 in esters next to aromatic ring or a double bond
C=O in Esters

56. Infrared Spectra
56

57. How to Interpret Infrared Spectra?
57

58. 58
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
1. Begin by looking in the region from 4000-1300.
Look at the C–H stretching bands around 3000:
How to analyze IR spectra

59. 59
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)

60. 60
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

61. 61
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

62.  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.
62
How to analyze IR spectra

63. 63

64. 64

65. IR Correlation Table
65

66. ALKANE
66
C C
H
H
H
H
H C H
H
H
n

67. C-H Stretch for sp3 C-H around 3000 – 2840 cm-1.
CH2 Methylene groups have a characteristic bending absorption at
approx 1465 cm-1
CH3 Methyl groups have a characteristic bending absorption at
approx 1375 cm-1
CH2 The bending (rocking) motion associated with four or more CH2
groups in an open chain occurs at about 720 cm-1
67

68. ALKENE
68
C C
H H
H H

69. =C-H Stretch for sp2 C-H occurs at values greater than 3000 cm-1.
=C-H out-of-plane (oop) bending occurs in the range 1000 – 650 cm-1
C=C stretch occurs at 1660 – 1600 cm-1;
often conjugation moves C=C stretch to lower frequencies
and increases the intensity
ALKENE
69

70. ALKYNE
HC CH
70

71. 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.
ALKYNE
71

72. AROMATIC RINGS
C H Stretch for sp2 C-H occurs at values greater than 3000 cm-1.
C C Ring stretch absorptions occur in pairs at 1600 cm-1 and 1475 cm-1.
C H Bending occurs at 900 - 690cm-1.
72

73. AROMATIC RINGS
73

74. 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.
74

75. Bending observed as one strong band near 750 cm-1.C H
Ortho-Disubstituted rings
75

76. 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.
C H
76

77. Para- Disubstituted rings
- one strong band appears in the region from 800 to 850 cm-1.C H
77

78. ALCOHOL
C
CH3
CH3
H3C OH
Primary alcohol 10
Secondary alcohol 20
Tertiary alcohol 30
C C OH
H
H
H
H
H
C C C
H
H
H
OH
H
H
H
H
78

79. 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 structure to
an alcohol.
79

80. PHENOL
OH
80

81. PHENOL
81

82. 82

83. 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.
83

84. 84

85. CARBONYL COMPOUNDS
cm-1
1810 1800 1760 1735 1725 1715 1710 1690
Anhydride Acid Anhydride Ester Aldehyde Ketone Carboxylic Amide
(band 1) Chloride (band 2) acid
Normal base values for the C=O stretching vibrations for carbonyl groups
85

86. A. ALDEHYDE
R C
O
H
R C
O
H
Ar C
O
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)
C-H Stretch, aldehyde hydrogen (-CHO), consists of weak
bands, one at 2860 - 2800 cm-1 and
the other at 2760 – 2700 cm-1.
86

87. 87

88. B. KETONE
C R'R
O
C R'R
O
C R'Ar
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)
88

89. 89

90. C. CARBOXYLIC ACID
R C OH
O
90

91. 91

92. D. ESTER
R C
O
O R
R C
O
O R
Ar C
O
O R
C – O Stretch in two or more bands, one stronger and broader than
the other, occurs in the range 1300 – 1000 cm-1
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)
92

93. 93

94. E. AMIDE
R C N
O
H
H
R C N
O
H
R
R C N
O
R
R10
20
30
94

95. AMIDE
95

96. F. ACID CHLORIDE
96
C Cl
O
R
C O
C Cl
Stretch appear in range 1810 -1775 cm-1 in conjugated chlorides.
Conjugation lowers the frequency to 1780 – 1760 cm-1
Stretch occurs in the range 730 -550 cm-1
Acid chloride show a very strong band for the C=O group.

97. F. ANHYDRIDE
97
C O C
O O
RR
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 O Stretch (multiple bands) occurs in the range 1300 -900 cm-1

98. R N
H
H
R
H
N R
R N R
R
AMINE
98
10 20
30

99. AMINE
99
Out-of-plane bending absorption can sometimes be observed
near 800 cm-1
Stretch occurs in the range 1350 – 1000 cm-1
N – H
Bending in primary amines results in a broad band in the range
1640 – 1560 cm-1.
Secondary amines absorb near 1500 cm-1
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.
N – H
N – H
C – N

100. Primary Amine
100
Secondary Amine