The document discusses infrared (IR) spectroscopy, how it works, and how to analyze IR spectra. It explains that IR spectroscopy analyzes the interaction of infrared radiation with molecules and can provide information about chemical structure. The document covers topics like sample preparation, instrumentation, functional group identification, and interpreting spectra to determine structural features like alkyl groups, carbonyls, and aromatic rings.

2. 2

3.  Mostly for qualitative analysis .
 Absorption spectra is recorded as
transmittance .
 Absorption in the infrared region arise from
molecular vibrational transitions
 Absorption for every substance are at specific
wavelengths where IR spectra provides more
specific qualitative information.
 IR spectra is called “fingerprints”
because no other chemical species will have
similar IR spectrum.
3

4. The transmittance
spectra provide
better contrast
between intensities
of strong and weak
bands compared to
absorbance
spectra.
4

5. Energy of IR photon insufficient to cause electronic
excitation but can cause vibrational excitation 5

6. 6

7.  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 for…..
 Analysis of organic materials
 Polyatomic inorganic molecules
 Organometallic compounds
7

8. IR region subdivided into 3 sub-regions
A.Near
IR region (Nearest to the visible)
780 nm to 2.5 μm (12,800 to 4000 cm-1)
le
visi b N
B. Mid IR region E
A
2.5 to 50 μm (4000 – 200 cm-1) R
M
infrared
I
D
C. Far IR region
F
50 to 1000 μm (200 – 10cm-1) A
R
e
w av
cro
mi 8

9. 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.
No absorption can occur if both rules 9

10.  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. 10

11.  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.
11

12. is a measure of the extent to which a
separation exists between the centers
of positive and negative charge within
a molecule.
δ-
O
δ+
H H
δ+
12

13.  In heteronuclear diatomic molecule,
because of the difference in
electronegativities of the two atoms, one
atom acquires a small positive charge (δ+),
the other a negative charge (δ-).
 This molecule is then said to have a dipole
moment whose magnitude, μ = qd
distance of separation of the charge
13

14. A. Compound absorb in IR region
Organic compounds, carbon
monoxide
B. Compounds DO NOT absorb in
IR region
O2, H2, N2, Cl2
14

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

16. STRETCHING
16

17. 17

18. BENDING
18

19. 1. Gases
 Using evacuated cylindrical cells
equipped with suitable windows.
1. Liquid
 sodium chloride windows.
 “neat” liquid
1. Solid
 Pellet (KBr)
 Mull
19

20.  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.
20

21. What is meant by “neat” liquid?
Neat liquid is a pure liquid that do not contain
any solvent or water.
Neat liquid method is applied when the amount
of liquid is small or when a suitable solvent is
unavailable.
21

22. There are 2 ways to prepare solid
sample for IR spectroscopy.
1. Solid that is soluble in solvent . The
most commonly IR solvent is carbon
tetrachloride, CCl4.
2. Solid that is insoluble in CCl 4 or any
other IR solvents can be prepared
either by KBr pellet or Mulls.
22

23. KBr PELLET
 The finely ground solid sample is mixed with
potassium bromide (KBr). The mixture is
pressed under high pressure (10,000 –
15,000 psi) in special die to form a pellet.
 KBr pellet then can be inserted into a holder
in the IR spectrometer.
23

24. MULLS
 2-5 mg finely powdered sample is ground
(grind) together with the presence 1 or 2
drops of a heavy hydrocarbon oil called
Nujol to form a Mull.
 Mull is then examined as a film between flat
salt plates.
 Mulls method is applied when solid not
soluble in an IR transparent solvent
and solid is not convenient to be
pelleted with KBr.
24

25. What is 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.
25

26. 26

27. Dispersive spectrometers
sequential mode
Fourier Transform spectrometers
simultaneous analysis of the full spectra
range using inferometry.
27

28. Important components in IR dispersive
spectrometer
1 2 3 4 5
source sample λ signal processor
detector & readout
lamp holder selector
Detector:
Source:
- Thermocouple
- Nernst glower
- Pyroelectric transducer
- Globar source
- Thermal transducer
- Incandescent wire
- Nichrome wire
28

29.  Generate a beam with sufficient
power in the λ region of interest to
permit ready detection & measurement.
 Provide continuous radiation which
made up of all λ’s with the region
(continuum source).
 Provide stable output for the period
needed to measure both P 0 and P.
29

30. 30

31. 31

32. Why FTIR is developed?
 To overcome limitations
encountered with the
dispersive instruments.
 Dispersive IR
spectrophotometer has slow
scanning speed due to
measurement of individual
molecules/atom.
 It utilize the use of an 32

33. 33

34. 34

35. 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
35

36. Majority of commercially available FTIR instruments
are based upon Michelson interferometer.
3
4
1
5 2
6
36

37. Advantages FTIR
 High sensitivity.
 High resolution.
 Quick data acquisition ( data for an
entire spectrum can be obtained in 1
s or less).
37

38. 38

39.  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.
 From IR spectrum we could predict
the present of atoms or group of
atoms or functional groups such as the
present of an O-H bond or a C=O or an
aromatic ring.
39

40. 40

41. 41

42. 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 alkyl groups (present in
of 3000? most organic molecules)
Are any or all to the left of a C=C bond or aromatic
3000? group in the molecule
42

43. 2. Look for a carbonyl in the region
1760-1690. If there is such a band:
Indicates
a carboxylic acid
Is an O–H band also present?
group
Is a C–O band also present? an ester
Is an aldehyde C–H band also
an aldehyde
present?
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)
43

44. 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
44

45. 5. Other structural features to check for
Indicates
an ether (or an ester if there
Are there C–O stretches?
is a carbonyl band too)
Is there a C=C stretching
an alkene
band?
Are there aromatic
an aromatic
stretching bands?
Is there a C≡C band? an alkyne
Are there -NO2 bands? a nitro compound
45

46. 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
46
spectrum.

47. 47

48. 48

49. H H H
H C C C H
H H H
n
49

50. CH Stretch for sp3 C-H around 3000 – 2840 cm-1.
CH 2 Methylene groups have a characteristic bending absorption
at approximate 1465 cm-1
CH 3 Methyl groups have a characteristic bending absorption at
approximate 1375 cm-1
CH 2 The bending (rocking) motion associated with four or more
CH2 groups in an open chain occurs at about 720 cm -1
50

51. H H
C C
H H
51

52. ALKENE
=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.
52

53. ALKYNE
HC CH
53

54. ALKYNE
CH Stretch for sp C - H occurs near 3300 cm-1.
C C Stretch occurs near 2150 cm-1; conjugation moves stretch to
lower frequency.
54

55. AROMATIC
RINGS
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
C C 1475 cm-1.
C H Bending occurs at 900 - 690cm-1.
55

56. AROMATIC
RINGS
56

57. 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 57
- one strong band appears in the region from 800 to 850 cm -1.

58. Ortho-Disubstituted rings
C H Bending observed as one strong band near 750 cm-1.
58

59. Meta- Disubstituted rings
- gives one absorption band near 690 cm-1 plus one near 780
C H cm-1. A third band of medium intensity is often found near 880
cm-1.
59

60. Para- Disubstituted rings
C H - one strong band appears in the region from 800 to 850
cm-1.
60

61. ALCOHOL
H H
H OH H
H C 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
61

62. 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.
62

63. PHENOL
OH
63

64. PHENOL
64

65. 65

66. 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.
66

67. 67

68. CARBONYL
COMPOUNDS
cm-1
1810 1800 1760 1735 1725 1715 1710 1690
Anhydride Acid Chloride Anhydride Ester Aldehyde Ketone Carboxylic acid
Amide
(band 1) (band 2)
Normal base values for the C=O stretching vibrations for
carbonyl groups.
68

69. ALDEHYDE
R C H
O
R C H C=O stretch appear in range 1740-1725 cm-1 for
O normal aliphatic aldehydes
Ar C H Conjugation of C=O with phenyl; 1700 – 1660 cm-1 for C=O
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.
69

70. 70

71. KETONE
R C R'
O
R C R' C=O stretch appear in range 1720-1708
O cm-1 for normal aliphatic ketones
Ar C R' Conjugation of C=O with phenyl at 1700 –
O 1680 cm-1 for C=O
and 1600 – 1450 cm-1 for ring (C=C)
71

72. 72

73. CARBOXYLIC
ACID
R C OH
O
73

74. 74

75. ESTER
R C O R
O
R C O R C=O stretch appear in range 1750-1735 cm-1 for
O normal aliphatic esters
Ar C O R Conjugation of C=O with phenyl; 1740 – 1715 cm -1
O for C=O
and 1600 – 1450 cm-1 for ring (C=C)
C – O Stretch in two or more bands, one stronger and
one broader than the other,
occurs in the range 1300 – 1000 cm-1
75

76. 76

77. AMIDE
O O O
H H R
R C N R C N R C N
H R R
10
2
0
30
77

78. AMIDE
78

79. O
R C Cl
Stretch appear in range 1810 -1775 cm-1 in
C O conjugated chlorides. Conjugation lowers the
frequency to 1780 – 1760 cm-1
C Cl Stretch occurs in the range 730 -550 cm -1
Acid chloride show a very strong band for the C=O group.
79

80. O O
R C O C R
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
80

81. H
R N R
Secondary amine , 20
H R N R
R N
H R
Primary amine, 10 Tertiary amine, 30
81

82. 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.
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
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
82

83. Secondary Amine
83

84. Aromatic Amine
84