INFRARED SPECTROSCOPY -FUNCTIONAL GROUP DETECTION FROM IR SPECTRUM

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

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

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)
B. Mid IR region
2.5 to 50 μm (4000 – 200 cm-1)
C. Far IR region
50 to 1000 μm (200 – 10cm-1)
8
visible
infrared
microwave
NE
AR
MI
D
F
AR

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
above are not met.
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.
12
O
δ-
δ+H H
δ+

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
13
distance of separation of the charge

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. A molecule can move via vibration, rotation
and translation (3 degree of freedom)
Polyatomic molecules containing N atoms
will have 3N degree of freedom
Molecules containing 3 atoms, two groups of
the triatomic molecules may be
distinguished; linear and non linear
Eg: CO2 (OCO) and H2O (HOH)
15

16. 2349 cm-1
667 cm-1,
degenerate energy
3650 cm-1,
Symmetric
stretching
3756 cm-1,
Asymmetric
stretching
1595 cm-1,
bending

17. 17
Type of degree of freedom Linear Non
linear
Translational 3 3
Rotational 2 3
Vibrational 3N-5 3N-6
Total 3N 3N

18. Molecular vibration
divided into
back & forth
movement
involves change
in bond angles
stretching bending
symmetrical asymmetrical
scissoring
rocking twisting
wagging
in-plane
vibration
out of
plane
vibration
18

19. STRETCHING
19

20. 20

21. BENDING
21

22. 1. Gases
 Using evacuated cylindrical cells
equipped with suitable windows.
1. Liquid
 sodium chloride windows.
 “neat” liquid
1. Solid
 Pellet (KBr)
 Mull
22

23.  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.
23

24. 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.
24

25. 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 CCl4 or any
other IR solvents can be prepared
either by KBr pellet or Mulls.
25

26. 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.
26

27. 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.
27

28. 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.
28

29. 29

30. Dispersive spectrometers
sequential mode
Fourier Transform spectrometers
simultaneous analysis of the full spectra
range using inferometry.
30

31. Important components in IR dispersive
spectrometer
31
1 2 3 4 5
source
lamp
sample
holder
λ
selector
detector
signal processor
& readout
Source:
- Nernst glower
- Globar source
- Incandescent wire
- Nichrome wire
Detector:
- Thermocouple
- Pyroelectric transducer
- Thermal transducer

32.  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 P0 and P.
32

33. 33

34. 34

35. 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
35

36. 36

37. 37

38. 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.
38
FT
Calculations
interferograms IR spectrum

39. Majority of commercially available FTIR instruments
are based upon Michelson interferometer.
39
1
3
2
4
5
6

40.  High sensitivity.
 High resolution.
 Quick data acquisition ( data for an
entire spectrum can be obtained in 1
s or less).
40
Advantages FTIR

41. 41

42.  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.
42

43. 43

44. Overtone and Combination Bands
Overtone bands – are multiplies of the
fundamental absorption frequency
First overtone band will appear at twice the
wavenumber of the fundamental.
Combination bands – arises when two
fundamental bands absorbing at v1 and v2
absorbs energy simultaneously. The resulting
band will appear at v1+v2 wavenumber
44

45. Example:
A molecule has strong fundamental bands as
follows:
C-H bending at 730cm-1
C-C stretching at 1400 cm-1
C-H stretching at 2950 cm-1.
Determine the wavenumbers of possible
combination bands and the first overtones.
45

46. Fermi resonance
-leads to two bands appearing close together
when one is expected
When an overtone or a combination band has
the same frequency as or similar frequency to
a fundamental, two bands appear, split at
either side of the expected value and are of
equal intensity – called as Fermi doublet
46

47.  Coupling
 Give rise to the complexity of the IR spectrum
 Vibration in the skeleton of the molecules become
coupled
 Bands can no longer be assigned to one bond.
 Very common when adjacent bonds have similar
frequencies.
 Commonly occurs between C-C, C-O and C-N
stretching, C-H rocking and C-H wagging
 The vibrational mode is observed at different
frequencies
47

48. 48

49. 49
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

50. 50
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)

51. 51
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

52. 52
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

53. 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. 53

54. 54

55. 55

56. 56
H
H
H C C
C H
H
H
H
H
n

57. CH Stretch for sp3 C-H around 3000 – 2840 cm-1.
CH2 Methylene groups have a characteristic bending absorption
at approximate 1465 cm-1
CH3 Methyl groups have a characteristic bending absorption at
approximate 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 57

58. 58
H H
C C
H H

59. 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.
59

60. ALKYNE
HC CH
60

61. CH
C C
ALKYNE
Stretch for sp C - H occurs near 3300 cm-1.
Stretch occurs near 2150 cm-1; conjugation moves stretch to
lower frequency.
61

62. 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.
62

63. AROMATIC
RINGS
63

64. 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. 64

65. Ortho-Disubstituted rings
Bending observed as one strong C H band near 750 cm-1.
65

66. 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
66

67. Para- Disubstituted rings
- one strong band appears in the region from 800 to 850
cm-1.
C H
67

68. ALCOHOL
CH3
H3C C
OH
CH3
H
Primary alcohol 10
H
Secondary alcohol 20
Tertiary alcohol 30
C C OH
H
H
H
H
C C C
H
H
OH
H
H
H
H
68

69. 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.
69

70. PHENOL
OH
70

71. PHENOL
71

72. 72

73. 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.
73

74. 74

75. 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.
75

76. 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.
76

77. 77

78. 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 at 1700 –
1680 cm-1 for C=O
and 1600 – 1450 cm-1 for ring (C=C)
78

79. 79

80. CARBOXYLIC ACID
R C OH
O
80

81. 81

82. ESTER
R C
O
O R
R C
O
O R
Ar C
O
O R
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)
C – O Stretch in two or more bands, one stronger and
one broader than the other,
occurs in the range 1300 – 1000 cm-1
82

83. 83

84. AMIDE
O
R C N
H
H
O
R C N
H
R
O
R C N
R
10 R 20 30
84

85. AMIDE
85

86. 86
O
C Cl
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.

87. 87
O O
R R
C O C
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

88. R N
H
H
R
HN
R
R N R
R
88
Primary amine, 10
Secondary amine , 20
Tertiary amine, 30

89. 89
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.
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
N – H
N – H
C – N

90. 90
Secondary Amine

91. 91
Aromatic Amine