M Pharma Student's Guide to IR Spectroscopy

PRESENTED BY,
PRATIKSHA C CHANDRAGIRIVAR
M PHARMA 1ST SEM
DEPT, OF PHARMACEUTICS
HSK COP BAGALKOT
FACILITATED BY,
Dr. B.S.KITTUR
HOD AND PROFESSOR
DEPT. OF PHARMACHEMISTRY
HSK COP BAGALKOT
IR SPECTROSCOPY 1

IR SPECTROSCOPY 3
CONTENTS:
1. INTRODCTION
2. PRINCIPLE
3. THEORY
4. MODES OF MOLECULAR VIBRATIONS
5. SAMPLE HANDLING
6. INSTRUMENTATION
7. FACTORS INFLUENCING VIBRATIONAL
FREQUENCIES
8. APPLICATIONS

INTRODUCTION:
DEFINITION:
Infrared spectroscopy is a spectroscopy which
is concerned with the study of absorption of
infrared radiation, its properties, characteristics
etc,. and also its interaction with matter.
◊ It is also known as vibrational spectroscopy.
◊ Infrared (IR) radiation lies in the part of the
electromagnetic spectrum i.e. between the visible
and microwave regions.
IR SPECTROSCOPY 4

◊ Any compound, either organic or inorganic, if it have
covalent bond then it have a ability to absorb some
frequencies of electromagnetic radiation in the IR
region of the electromagnetic spectrum.
◊ This absorption of infrared radiation (quantised)
causes the various bands in a molecule to stretch
and bend with respect to one another.
◊ Because of this reason it is main tool for the
identification of compounds.
IR SPECTROSCOPY 6

In any molecule, there are group of atoms or
a atom is connected by bonds. These bonds are
analogous to springs and not rigid in nature. Because
of the continuous motion of the molecule, they
maintain some vibrations with some frequency,
characteristics to every portion of molecule. This is
called natural frequency of vibration.
When energy in the form of infrared
radiation is applied and if that
Applied infrared frequency = Natural frequency of
vibration
IR SPECTROSCOPY 7

Then absorption of IR radiation takes place and peak is
observed.
Example : Infrared vibrations of ethanol
IR SPECTROSCOPY 8

Every bond or portion of a molecule or functional
group requires different frequency of absorption.
Hence characteristic peak is observed for every
functional group or part of the molecule.
In other words, IR spectra is nothing but a finger
print region of a molecule.
Example: IR spectra of Ibuprofen
IR SPECTROSCOPY 9

The different IR regions are:
1. Near IR region = 0.8µ to 2.5µ
2. Mid IR region = 2.5 to 15µ
3. Far IR region = 15µ to 200µ
IR SPECTROSCOPY 10

Example:
The natural frequency of vibration HCL molecule is
about 8.7 × 1013 sec-1(2890 cm-1).
When IR radiations permitted to pass through a sample
of HCL and transmitted radiation is analysed by the
IR spectrometer.
It is observed that the part of radiation which has a
frequency of 8.7 × 1013 sec-1 has been observed by
HCL molecule where as the remaining frequencies of
the radiation are transmitted.
Thus, the frequency 8.7 × 1013 sec-1 is characteristics
of HCL molecule.
IR SPECTROSCOPY 11

In pharmaceutical analysis, we use mid IR region
of wavelength 2.5 to 15µ ( 2.5 to 25µ) or in terms of
wavenumbers we can say it as 400cm-1 to 4000cm-1.
In IR spectra we use wavenumbers instead of
wavelengths for mentioning the characetristic peak,
Because wave numbers are larger values and easy to
handle than wavelengths which will show only small
differences between functional groups.
o Wave number is nothing but the number of waves
present per cm. which can be calculated from the
wavelength.
1
----------------------- × 104 = wavelength per cm or cm-1
wavelength in µ
IR SPECTROSCOPY 12

For a molecule to absorb IR radiation, it has to fulfill
certain requirements,
Those are,
a) Correct wavelength of radiation
b) Electric dipole
a) Correct wavelength of radiation:
 A molecule absorbs radiation only when the natural
frequency of vibration of some part of a molecule
( i.e. atoms or group of atoms comprising it) is the
same as the frequency of radiation.
IR SPECTROSCOPY 13

b) Electric dipole:
 This is another condition for a molecule to absorb IR
radiation.
 A molecule can only absorb IR radiation when its
absorption causes a change in its electric dipole
(dipole moment).
 A molecule is said to have electric dipole when there
is a slight positive and a slight negative electric
charge on its component atoms.
 When the molecule having electric dipole is kept in
the electric field , it is the same as that when the
molecule is kept in a beam of IR spectroscopy.
IR SPECTROSCOPY 14

 on the other hand , when the rate of vibration of
the charged atoms in a molecule is slow , there will
be weak bands in the IR spectrum.
 As the symmetrical diatomic molecules like O2 and
N2, do not posses electric dipole , they can not be
excited by infrared radiation and for this they do
not give rise to IR spectra.
 Carbon dioxide and water vapour do in air do absorb
in the molecule , but these do not effect IR spectra
taken on a double beam instrument , as they are
fairly weak and cancel out between the sample beam
and reference beam.
IR SPECTROSCOPY 15

This field exerts forces on the electric
charges in the molecules. Opposite charges will
experience forces in opposite directions. This tends
to decrease separation.
 But in the case of electric field of IR radiation
changes its polarity periodically, it means that the
spacing between the charged atoms (electric dipole)
of the molecule also changes periodically.
 When these charged atoms vibrate , they absorb IR
radiation from the radiation source.
 If the rate of vibration at charged atom in a
molecule is fast, the absorption of radiation is
intense and thus, the IR spectrum will have intense
absorption band.
IR SPECTROSCOPY 16

 There is no change in the dipole moment is produced
by the carbon-carbon double bond stretching of the
symmetrical molecule like ethylene. Since there is
no change in dipole moment ,the bond does not
absorb infrared radiation.
 On the other hand, substituition of a bromine for a
hydrogen atom to form bromoethylene which
destroys the symmetry around the double bond.
The stretching of the double bond now generates
a significant change in the dipole moment and strong
absorbance in the infrared is observed.
IR SPECTROSCOPY 17

The closer atoms in a molecule are to each other,
the greater will be the strength of the dipole ,
faster will be the rate of change of the dipole , the
higher will be the frequency of vibration , and the
more intense will be the absorption of radiation.
IR SPECTROSCOPY 18

In order to understand the theory of
absorption spectroscopy, one has to understand the
phenomena of vibration rotational spectra.
 Let us consider a diatomic molecule associated with
a dipole moment.
 The vibratory motion of the nuclei of diatomic
molecule may be similar to the vibration of linear
harmonic oscillator.
 In such oscillator, the force tending to restore an
atom to its original state is proportional to the
displacement of the vibratory atom from the
original position (Hooke’s law).
IR SPECTROSCOPY 19

Suppose the bond between the two nuclei of a diatomic
molecule is distorted from its equilibrium length re
to a new length r.
Then the restoring forces on each atom of the
diatomic molecule will be given by
m1 d2 r1 = -K(r - re) …………… (1)
d t2
m2 d2 r2 = -K(r - re) ……………. (2)
d t2
Where, K = proportionality constant and known as force
constant, it is regarded as a measure of the
stiffness of the bond.
IR SPECTROSCOPY 21

r1 and r2 are the positions of atom 1 and 2 relative to
the centre of gravity of the molecule.
We know,
m2
r1 = ------------ r -------(3)
m1 + m2
m1
r2 = ------------ r --------(4)
m1 + m2
Where, m1 and m2 are the masses of two atoms of a
vibrating dia-atomic molecule.
IR SPECTROSCOPY 22

On substituting equation(3) and equation(1) we get,
m1 m2 d2 r
---------- ----- = - K( r - re) ………….(5)
m1 + m2 dt2
As re is a constant, its differentiation with respect to time
t will be zero, i.e.,
d2 r d2 (r - re)
------ = ------------ …………..(6)
dt2 dt2
Substituition of the above equation in equation(5) yields
m1 m2 d2(r - re)
--------- ------------ = - K( r - re) ………….(7)
m1 + m2 dt2
IR SPECTROSCOPY 23

m2
Put r – re = x and --------- = µ ..........(8)
m1 + m2
on substituting equation
d2 x
µ ------ = - K x ………….(7)
dt2
Or d2 x
µ ------ + K x = 0 ………….(8)
dt2
Or d2 x
µ ------ + w2x = 0 ………….(9)
dt2
IR SPECTROSCOPY 24

Where, w2 = K/µ or w = √ K/µ …………..(10)
Equation(9) is the expression of simple harmonic
motion with frequency of vibration (v) as follows
v = w/2∏ = 1/2∏ √ K/µ
Where,
K is the force constant expressed in dynes per cm
and
µ is the reduced mass of the system.
IR SPECTROSCOPY 25

According to the character of
vibration, normal vibrations can be divided in to two
principle group they are
1. Stretching vibrations
2. Bending vibrations
1.Stretching vibrations:
In this type of vibrations, the atoms
move essentially along the bond axis so that the
bond length increases or decreases periodically, i.e.,
at regular intervals.
IR SPECTROSCOPY 26

 As this type of vibrations corresponds to one
dimensional motion.
 During stretching vibrations, bond angles change
only if it to do so by the centre of gravity resisting
displacement.
Stretching vibrations are of two types:
a) Symmetric vibration
b) Asymmetric vibration
IR SPECTROSCOPY 27

a) Symmetric vibration:
In this type , the movement of the atoms
with respect to a particular atom in a molecule is in
the same direction.
Fig.
IR SPECTROSCOPY 28

b) Asymmetric vibration:
In these vibrations, one atom approaches the
central atom while the other departs from it.
Fig.
IR SPECTROSCOPY 30
:

2.Bending vibrations:
In this type of vibrations, the positions of
the atoms change with respect to the original bond
axis.
 We know that more energy is required to stretch a
spring than that required to bend it.
 We can safely say that stretching absorptions of a
bond appear at high frequencies (high energy) as
compared to the bending of the same bond.
 This kind of vibration is also know as deformations.
IR SPECTROSCOPY 32

There are mainly two types of bending
vibrations are there those are:
A. In-plane deformation vibrations
B. Out-of plane deformation vibrations
A. In-plane deformation vibrations:
 In these vibrations , there is change in bond angle
takes place.
 Bending of bond angle takes place within the same
plane.
IR SPECTROSCOPY 33

It is further can be divided into two types, viz.,
1. Scissoring
2. Rocking
1.Scissoring:
 In this type, two atoms approach each other.
 Here, bond angle decreases.
Fig.
IR SPECTROSCOPY 34

2. Rocking:
 In this type, the movement of the atoms takes place
in the same directions.
 Here bond angle is maintained, but both bonds
moves within the plane.
fig,
IR SPECTROSCOPY 36

2.Out-of plane deformation vibrations:
These vibrations takes place outside the
plane of molecule.
It is further can be divided into two types viz.,
a) Wagging
b) Twisting
IR SPECTROSCOPY 38

a) wagging:
Two atoms move “up and down” the plane with
respect to the central atom(i.e., moves one side of
the plane).
Fig.
IR SPECTROSCOPY 39

b) Twisting:
In this type, one of the atoms moves up
the plane while the other moves down the plane with
respect to the central atoms.
Fig.
IR SPECTROSCOPY 41

As infrared spectroscopy has been used for
the characterizaton of solid, liquid or gases , it is
evident that samples of different phases have to be
handled.
 Different forms of samples treated differently.
 The only common point to the sampling of different
phases is that the material containing the sample
must be transparent to IR radiation.
 So we selected the salts like NaCl or KBr.
IR SPECTROSCOPY 44

Sample handling involves following techniques:
A. Sampling of solids
B. Sampling of liquids
C. Sampling of gases
D. Sampling of solutions
A. Sampling of solids:
There are four techniques are generally
employed for preparing solid samples viz.,
1. Solids run in solution
2. Solid films
3. Mull technique
4. Pressed pellet technique
IR SPECTROSCOPY 45

 Solids are dissolved in a non-aqueous solvent which
do not have chemical interaction with the solvent
and also which do not absorb in the studied ranged.
 A drop of the solution is placed on a alkali metal
disk and the solvent is allowed to evaporate, leaving
a thin film of the solute, or the entire solution is
placed in a liquid sample cell.
 If the solution of solid can be prepared in a suitable
solvent then the solution is run in one of the cells
for liquids.
IR SPECTROSCOPY 46

 But this method cannot be used for all solids
because suitable solvents are limited in number and
there is no single solvent which transparent
throughout the IR region.
 When the investigating solids are in solutions, the
absorption due to solvent has to be compensated by
keeping the solvent in a cell of same thickness as
that containing in the sample , in the path of the
reference beam of double-beam spectrometer.
IR SPECTROSCOPY 47

If a solid is amorphous in nature the
sample is deposited on the surface of a KBr or NaCl
cell by evaporation of a solution of the solid.
This techniques useful for rapid
qualitative analysis but becomes useless for carrying
out quantitative analysis.
IR SPECTROSCOPY 48

In this technique,
 The finely ground solid sample is mixed with Nujol
(mineral oil) to make a thick paste of which is then
made to spread between IR transmitting windows.
 This is then mounted in a path of infrared beam and
the spectrum is run.
 This Nujol can be used with hexachlorobutadiene
while taking in Nujol mull for clear bands.
 This method is good for qualitative analysis.
IR SPECTROSCOPY 49

Disadvantage:
1. It shows absorption maxima only at 2915, 1462, 1376
and 719 cm-1 .
2. Polymorphic changes, degradation and other changes
may occur during grinding.
3. It cannot be used for quantitative analysis.
IR SPECTROSCOPY 52

 In this technique a small amount of finely ground
solid samples is intimately mixed with about 100
times its weight of powdered potassium bromide.
 The finely ground mixture is then passed under very
high pressure in a press (at least 25,000 p sig) to
from a small pellet (about 1-2 mm thick and 1 cm in
diameter).
 The resulting pellet transparent to IR radiation and
is run as such.
IR SPECTROSCOPY 53

Device for processing potassium bromide discs
IR SPECTROSCOPY 54

Working:
 A device for the pressing the mixture of KBr and
solid sample to forma a pellet is shown in the figure.
 The powder (KBr + sample) is introduced as shown,
then the upper screw ‘A’ is tightened until the
powder is compressed into a thin disc.
 After compressing the sample, one removes the
bolts (A and A’) and places the steel cylinder with
the sample disc inside it in the path of the beam of
infrared spectrometer and a ‘blank’ potassium
bromide pellet of the reference beam.
IR SPECTROSCOPY 55

 The best advantage of this over Nujol Mull method
is that the use of KBr eliminates the problem of
bands which appear in the IR spectrum due to the
mulling agent.
 KBr pellets can be stored for long periods of time.
 As the concentration of sample can be suitably
adjusted in the pellets, it can be used for
quantitative analysis.
 The resolution of the spectrum in the KBr is
superior to that obtained with mulls.
IR SPECTROSCOPY 56

 It always has a band at 3450 cm-1 , from the OH
group of moisture present (always) in the sample.
 The high pressure involved during the formation of
pellets may bring about polymorphic changes in
crystallinity in the samples, specially if inorganic
complexes are there, which may cause complications
in IR spectrum.
 In some cases substituition by bromide may occurs.
 This method is not successful for some polymers
which are difficult to grind with KBr.
IR SPECTROSCOPY 57

B. Sampling of liquids:
 Samples that are liquids at room temperature are
usually put frequently with no preparation , into
rectangular cells made up of NaCl , KBr or ThBr and
their IR spectra are obtained directly.
IR spectrum of Hexachlorobuta-diene-1,3
IR SPECTROSCOPY 58

Method :
Step-1 : A drop of liquid sample is placed on the top
of sodium chloride plate.
Step-2: Place another sodium chloride plate on it.
Step-3: The pair of sodium chloride plates enclosing
the liquid film is then placed in the path of sample
beam.
IR SPECTROSCOPY 59

 The sample thickness should be so selected that the
transmittance should lies between 15-20%.
 For most liquids , it will be a layer of 0.01-0.05m in
thickness.
 If a cell possesses good quality windows , flat and
parallel, its thickness can be calculated by the
formula:
2t = N/w1 – w2
Where , t = Thickness
N = the number of fringes between wave
number w1 and w2
IR SPECTROSCOPY 60

For double beam work following conditions are applied:
1. “Matched cells” are employed.
2. One cell will contain the sample while the other cell
will have a solvent used in sample.
3. These matched cells must have same thickness.
4. They should be protected from moisture because
dissolve in water.
5. With similar conditions, organic liquid samples must
be dried before pouring into cells.
6. If sample contain water, then the plates must be
constructed with calcium flouride.
IR SPECTROSCOPY 61

Method:
 The gaseous sample is introduced into a ‘gas cell’
 Usually they are about 10 cm long but they may be
up to 1m also.
 Multiple reflections can be used to make the
effective path length as long as 40m. , so that
constituents of gas can be determined.
 The gas must not be react with cell windows or
reflecting surfaces.
 The walls of the cell are made up of sodium chloride.
IR SPECTROSCOPY 62

 Very few organic compound can be examined as
gases.
IR SPECTROSCOPY 63

Complications:
 The low frequency rational changes in the gaseous
phase often split the high frequency vibrational
bands.
IR SPECTROSCOPY 64

 It is the most convenient to determine the spectrum
in solutions.
 Excellent solvents are those which have poor
absorptions of their own.
 Important solvent that are used in IR technique are:
a) chloroform
b) Carbon tetra chloride
c) Carbon disulphide
 Water can not be used as a solvent as it absorbs in
several regions of the infra-red spectrum.
IR SPECTROSCOPY 65

Method:
1. The sample under analysis is dissolved in a solvent.
2. Its 1-5% solution is placed in a solution cell
consisting of transparent windows which is made of
alkali metal halide.
3. A second cell containing the pure solvent is placed in
the path of the reference beam to cancel out solvent
interferences.
4. For careful analysis ,the path length of reference
cell should be less than that of the solution cell.
IR SPECTROSCOPY 69

There are mainly three types of instrumentation is
done for IR spectroscopy:
1. Dispersive instrumentation
2. Fourier transform instrumentation
3. Non dispersive instrumentation
But commonly used are first two.
IR SPECTROSCOPY 70

It usually contains following parts viz.,
1. Source 10.Slits
2. Mirror
3. Sample cell
4. Reference cell
5. Beam chopper
6. Motor
7. Monochromator
8. Detector
9. Amplifier and recorder
IR SPECTROSCOPY 72

1.Sources :
 The commonly used sources in dispersive infrared
spectrophotometers are
i. Incandescent lamp
ii. Nernst glower
iii. Globar source
iv. Pressurized mercury arc.
2.Mirrors :
 These helps to convert beam of radiations into
parallel radiation of same intensity.
IR SPECTROSCOPY 73

3. Sample cell :
 There is a rugged window material for cuvette which
is transparent and invert over this region.
 The alkali halides are widely used like KBr , NaCl,
and ThBr which are transparent at wavelength as
long as 625 cm-1 .
 Silver chloride is preferred if sample is water
soluble or moist.
4. Reference cell :
 It contains the solvent omitting the sample.
 We can say it as a blank.
IR SPECTROSCOPY 74

5. Beam chopper :
 These are the rotating sector that passes the two
beam produced by the device alternately to the
gratings or prisms(in older model).
6. Motor :
 It controls the whole process technically by
providing energy.
7. Monochromator :
 Usually gratings are used as monochromator in
newer models of dispersive IR spectrophotometer.
 Prisms are used in older one.
IR SPECTROSCOPY 75

8. Detector :
 Here thermocouple detectors are used.
9.Amplifier and recorder :
 The response are amplified by amplifier and
recorded in the form of graphical data.
10. Slits :
 It allows beam to pass through.
IR SPECTROSCOPY 76

1. The instrument produces a beam of infrared
radiation from a hot wire.
2. By means of mirrors , these rays are divided into two
parallel beams of equal intensity.
3. One beam fall on sample cell and other on reference.
4. The these beams pass into the monochromator where
dispersion of each beam takes place and results into
a continuous spectrum of frequencies of infrared
light.
5. Beam chopper present in monochromator are a
rotating sector helps in passing these two beams
alternately to diffraction grating.
IR SPECTROSCOPY 77

5. The slowly rotating diffraction grating varies the
frequency or wavelength of radiation reaching the
thermocouple detector.
6. Then the detector senses the ration between the
intensities of the reference and sample beams,
7. In this way, the detector determines which
frequencies have been absorbed by the sample and
which frequencies are un effected by the light
passing through the sample.
8. After the signal from the detector is amplified, the
recorder draws the resulting spectrum of the
sample on chart.
IR SPECTROSCOPY 78

 Less sensitivity.
 Less accurate.
 Many moving parts results in to mechanical slippage.
 Slow scan makes the instrument too slow for
monitoring systems under going rapid changes.
Example : Gc effluents
IR SPECTROSCOPY 79

The present modern instruments works on different
principle viz.,
The design of the optical pathway produces a pattern
called an interferogram.
The interferogram is a complex signal ,but its wave like
pattern contains all the frequencies that make up
the infrared spectrum.
It is a plot of intensity versus time.
But chemist is interested in a spectrum which is a plot
intensity versus frequency.
IR SPECTROSCOPY 80

An mathematical operation known as Fourier
transform can help to achieve it, i.e., separates the
individual absorption frequencies from
interferogram and produces a spectrum that is
identical to other two instruments.
 It is commonly called as FT-IR.
 There single beam and double beam FT-RI but we
commonly use double beam.
IR SPECTROSCOPY 81

Working of instrument:
1. The FT-IR uses an interferometer to process energy
sent to the sample.
2. In the interferometer, the source energy passes
through a beam splitter, a mirror placed at 45⁰ angle
to the incoming radiation , which allows the incoming
radiation to pass through but separates it into two
perpendicular beams, one un deflected , the other
oriented at 90⁰ angle.
3. The beam which is oriented at 90⁰ goes to a
stationary or fixed mirror and returned to the beam
splitter.
IR SPECTROSCOPY 84

4. The un deflected beam goes to a moving mirror and
is also returned to the beam splitter.
5. The motion of the mirror causes the pathlength that
he the second beam traverses to vary.
6. When the two beams meet at the beam splitter,
they recombine , but the pathlength
differences(differing wavelength content) of the
two beams cause both constructive and destructive
interferences.
7. The combined beam containing these interference
patterns is nothing but interferogram.
8. This interferogram contains all of the radiative
energy coming from the source and has a wide range
of wavelengths.
IR SPECTROSCOPY 85

9. The interferogram generated by combining the two
beams is oriented toward the sample by the beam
splitter.
10. As it passes through sample , the sample
simultaneously absorbs all of the
wavelength/frequencies that are found in its
infrared spectrum.
11. The modified interferogram signal that reaches the
detector contains information about the amount of
energy that was absorbed at every wavelength.
12. Then the computer compares the modified
interferogram to a reference laser beam to have a
standard of comparison.
IR SPECTROSCOPY 86

13. The final interferogram contains all of the
information in one time – domain signal.
14. A mathematical process called a Fourier transform
must be implemented by the computer to extract
the individual frequencies that were absorbed and
to reconstruct and plot what we recognize as a
typical infrared spectrum.
Advantages:
 They have better signal- to – noise ratios than other
IR instruments,
 They provide high resolution(‹0.1 cm-1).
 They are highly accurate and reproducible.
 Theoratical advantage is that they better optics to
provide energy,
IR SPECTROSCOPY 87

Following four factors are responsible for shifting the
vibrational frequencies viz.
A. Coupled vibrations and Fermi resonance
B. Electronic effects
C. Hydrogen bonding
D. Bond angles
IR SPECTROSCOPY 88

Coupled vibrations:
We expect only one stretching for C-H bond
but
i. When point comes to group like methylene it
contains
- CH2 - , we can see two absorption which corresponds
to symmetric and asymmetric vibration.
i.e.,
IR SPECTROSCOPY 89

In this type of case, asymmetric vibrations
always occur at higher wave number compared with
symmetric vibrations.
Such kind of vibrations are called coupled vibrations.
ii. Another case is that,
coupled vibration occurs in methyl group i.e., - CH3 –
IR SPECTROSCOPY 90

Here reason we can give that , as we know that two
different vibrational levels have nearly the same
energy.
Then if such coupled vibration occur then a mutual
perturbation of energy may occur ,that results in
shifting one towards lower frequency and another
towards higher one, which further results in
alteration or substantial in higher intensed bands.
IR SPECTROSCOPY 91

Fermi resonance :
 This was first observed by Enrico Fermi in case of
carbon dioxide.
 This kind of resonance is seen in certain lactones
and cycloketones.
 A type of resonance that occurs in coupled
pendulum such resonance is called Fermi resonance.
Explaination:
Here a molecule transfers its energy from fundamental
to overtone and back again.
Quantum mechanically, resonance pushes the two levels
apart and mixes their character so that each level
becomes partly fundamental and partly overtone in
character. Results in pair of transitions of equal
intensity.
IR SPECTROSCOPY 92

Example :
Carbon dioxide is a linear triatomic molecule and four
fundamental vibrations are expected for it.
Out of these , symmetric stretching vibration is infra-
red inactive since it produces in dipole moment.
IR SPECTROSCOPY 93

For symmetric stretching , Raman spectrum shows a
strong band at 1337 cm-1 .
The two bending vibrations ate equivalent and absorb
at the same frequency of 667.3 cm-1 .
The over tone of this is 1334.6 cm-1 i.e., 2 × 667.3cm-1 .
Thus , Fermi resonance takes place resulting in the
shift of first level towards higher frequency.
The mutual perturbation of 1337 cm-1 (fundamental)
and 1334.6 cm-1 (overtone) gives rise to two bands
at 1285.5 cm-1 and 1388.3 cm-1 having intensity ratio
of
1: 0.9 .
IR SPECTROSCOPY 94

Changes in frequencies of a particular group can be
occurs due to change in the particular substituents
of neighbourhood functional group.
Reason is that influence of electronic effects like
inductive effect, mesomeric effect and field
effects etc.,
These effects can not be isolated from one another
and combination of these one of them can only be
estimated approximately.
Under the influence of these effects , the force
constant of bond strength changes and its
absorption frequency shifts from the normal value.
IR SPECTROSCOPY 95

Introduction of alkyl group cause +I effect which
results in the lengthening and weakening of bonds
takes place and hence force constant is lowered and
wave number decreases.
Let us compare the wavenumbers of c=o, absorptions
by using following compounds:
1. Formaldehyde= HCHO = 1750cm-1
2.Aacetaldehye = CH3CHO = 1745cm-1
3. Acetone = CH3COCH3 = 1715cm-1
NOTE: Aldehydes have higher wave number than that
of ketones.
IR SPECTROSCOPY 96

The introduction of an electronegative atom or group
causes –I effect which results in the bond order to
increase.
Thus , the force constant increases and hence the wave
number of absorption rises.
Now let us consider wave numbers of absorptions in
following compounds:
1. Acetone = CH3COCH3 = 1715cm-1
2. Chloroacetone = CH3COCH2Cl = 1725cm-1
3. Dichloroacetone = CH3COCHCl2 = 1740cm-1
4. Tetrachloroacetone = Cl2CH-CO-CHCl2 = 1750,
1778cm-1
IR SPECTROSCOPY 97

In most of cases , mesomeric effects works along with
inductive effects and we cannot ignore that.
Conjugation lowers the absorption frequency of C=O
stretching whether the conjugation is due to 𝛼, 𝛽-
unsaturation or due to an atomic ring.
In most of cases, inductive effect dominates over
mesomeric effect while reverse holds for other
cases.
Mesomeric effects causes lengthening or the
weakening of a bond leading in the lowering of
absorption frequency.
IR SPECTROSCOPY 98

Consider the following compounds:
In these two cases, -I effect is dominated by
mesomeric effect and thus, the absorption
frequency falls.
IR SPECTROSCOPY 99

Consider in ortho substituted compounds, the lone
pairs of electrons on two atoms influence each
other through space interactions and change the
vibrational frequencies of the both groups .this
effect is called Field effect.
Example: orthohaloacetophenone
IR SPECTROSCOPY 100

Explaination:
The non bonding electrons present on oxygen atom and
halogen atom causes electrostatic repulsions.
This causes a change in the state of hybridisation of
C=O group and also makes it to go out of the plane
of the double bond.
Thus, the conjugation is diminished and absorptions
occurs at a higher wave number.
Thus, for such ortho substituted compound s, cis
absorbs(field effect) at a higher frequency as
compared to the trans isomer.
IR SPECTROSCOPY 101

Hydrogen bonding brings about remarkable downward
frequency shifts.
How means, stronger the hydrogen bonding , greater is
the absorption shift towards lower wave number
than normal value.
Two types of hydrogen bonding can be readily
distinguished in infra-red technique.
- Inter molecular hydrogen bonds give rise to broad
bands.
- Intra molecular hydrogen bands are sharp and well
defined.
IR SPECTROSCOPY 102

covalent bond (strong)
intermolecular
attraction(weak)
IR SPECTROSCOPY 103

Inter molecular hydrogen bonding are Concentration
dependent. Hence absorption frequency of them
are less.
 The hydrogen bonding in O-H androgen d N-H are
special cases viz.,
 As nitrogen atom is less electronegative than that
of an oxygen atom , hydrogen bonding in amines is
weaker than that of alcohols, thus the frequency
shift is less towards amines.
Example:
Amines shows N-H stretching at 3500 cm-1 in dil.
Solutions while in condensed spectra absorption is
at 3300 cm-1 .
IR SPECTROSCOPY 104

 In aliphatic alcohols , a sharp band appears at
3650cm-1 in dilute solutions due to free O-H group
while a broad band is noticed at 3350cm-1 due to
hydrogen O-H group.
 Alcohols are strongly hydrogen bonded in condensed
phases.
 These are usually associated as dimers and polymers
which results in the broadening of bands at lower
absorption frequencies.
 In vapour state or inert solvents, molecules exist in
free state and absorbs strongly at 3650cm-1 .
IR SPECTROSCOPY 105

Alcohols are written in the following resonating
structures:
IR SPECTROSCOPY 106

Explaination: structure III is the hybrid of structures
I and II. This results in the lengthening of the
original O-H group. The electrostatic force of
attraction with which hydrogen atom of one
molecule is attracted by the oxygen atom of another
molecule makes it easier to pull hydrogen away from
oxygen atom. Thus, small energy will be required to
stretch such a bond (O-H) and hence absorption
occurs at a lower wave number.
IR SPECTROSCOPY 107

 It has been found that the highest C=O frequencies
arise in strained cyclobutanones.
 This can be explain terms of bond angles strain.
 The C-CO-C angle is reduced below the normal angle
of 120⁰ and this leads to increased s-character in
the C=O bonds.
 Greater s-character causes shortening of C=O
structure occurs at higher frequencies.
 In case the bond angle is pushed outwards above
120⁰, the opposite effect operates.
 Due to this reason di-tert-butyl ketone absorbs at
1697cm-1 (low) as a result of C=O str.
IR SPECTROSCOPY 108

1. Identification of functional group and structural
elucidation.
2. Identification of drug substances.
3. Identifying impurities in drug sample.
4. Study of hydrogen bonding.
5. Study of polymers.
6. Identify ratio of cis-trans isomers in a mixture of
compounds.
7. Quantitative analysis.
8. To find out difference between hydrogen bonding.
IR SPECTROSCOPY 109

1. Text book of pharmaceutical analysis by Dr. Ravi
Sankar – 4th edition.
2. Instrumental methods of chemical analysis by
Gurdeep R Chatwal, Sham K Anand – 6th edition.
3. Elementary organic spectroscopy by Y.R.Sharma –
multicolour edition.
4. Spectroscopy by Lampman, Pavia, Kriz and Vyvyan –
4th edition and 3rd edition.
5. Principles of instrumental analysis by Skoog, Holler
and Crouch – 6th edition and 8th edition
6. Spectrometric identification of organic compounds
by Robert.M.Silverstein, Francis.X.Webster and
David.J.Kiemle – 7th edition.
IR SPECTROSCOPY 110

M Pharma Student's Guide to IR Spectroscopy

Recomendados

Recomendados

Más contenido relacionado

La actualidad más candente

La actualidad más candente (20)

Similar a M Pharma Student's Guide to IR Spectroscopy

Similar a M Pharma Student's Guide to IR Spectroscopy (20)

Más de Pratiksha Chandragirivar

Más de Pratiksha Chandragirivar (17)

Último

Último (20)

M Pharma Student's Guide to IR Spectroscopy