Infrared spectroscopy is a technique used to identify functional groups in a compound based on the vibrational transitions of bonds observed in the infrared region of the electromagnetic spectrum. The most common technique is absorption spectroscopy, where infrared radiation is passed through a sample and the frequencies at which absorption occurs are measured. For a molecule to absorb infrared radiation, the vibration must cause a change in the dipole moment of the molecule. There are two main types of vibrations observed - stretching vibrations which change bond lengths, and bending vibrations which change bond angles. Stretching vibrations are further divided into symmetrical and asymmetrical types. Bending vibrations include in-plane and out-of-plane types such as scissoring,

1. Infrared Spectroscopy
By: Dr. P. R. Padole

2. Infrared Spectroscopy
Infrared spectroscopy - a technique used to
identify information about the
functional groups present in the compound.
• The most common technique used is absorption
spectroscopy.

3. Spectroscopy is an instrumentally aided study of
the interactions between matter (sample being
analyzed) and energy (any portion of the
electromagnetic spectrum)
EMR ANALYTE SPECTROPHOTOGRAPH
1. UV-Visible radiations--------excitation of electrons---------UV-visible spectrum
2. IR-radiations------------------vibration changes in electrons--------IR spectrum
3. Radio frequency---------------spin rotational changes-------------N.M.R spectrum
Conc. should be
lower

4. Principle of IR spectroscopy:
Q.1) Describe the principle or theory of IR spectra?
Q.2) Explain, how the IR spectrum are presented (Diagram expected). Also name the solvents
used to scan the IR spectra. (W-12, 2 Mark)
Q.3) In which region of electromagnetic spectrum the vibrationals transitions are observed?
(S-16, ½ Mark)
(a) U.V. (b) Microwave (c) IR (d) Visible

5. Principle of IR spectroscopy:
Defination: The spectra produced due to,
(i) Absorption of IR radiation by molecule &
(ii) Vibrational energy level transition,
is called as Vibrational spectra or IR spectra.
It is observed in IR region in spectral range 4000-667 cm-1.
When IR absorption occur?
Rule-1) IR absorption only occurs when IR radiation
interacts with a molecule undergoing a change in dipole
moment as it vibrates or rotates.
Rule-2) Infrared absorption only occurs when the
incoming IR radiation has sufficient energy for the
transition to the next allowed vibrational state
Note: If the Rule-1 & 2, above are not met, no absorption can occur.

6. PRINCIPLE OF IR SPECTROSCOPY
 Molecules are made up of atoms linked by
chemical bonds. The movement of atoms and
the chemical bonds look like spring and balls
(vibration).
 This characteristic vibration are called
Natural frequency of vibration.
 Applied infrared frequency = Natural frequency
of vibration
 Change in dipole moment is required

7. When a wave of infrared light encounters this oscillating EM field
generated by the oscillating dipole of the same frequency, the
two waves couple, and IR light is absorbed
The coupled wave now vibrates with twice the amplitude
IR beam from
spectrometer
EM oscillating wave
from bond vibration
“coupled”
wave

8. Infrared Absorption
• When IR radiation is passed through molecule, the radiation of definite frequency
(frequency equal to vibrational frequency of molecule) is absorbed by the
molecule.
• The energy absorbed too small to cause rotational excitation.
• But the energy absorbed is sufficient to increase amplitude of vibration.
• So, the molecule is excited and undergoes vibrational transition from one
vibrational energy level to higher vibrational energy level within same electronic
energy level.

9. Principle of IR spectroscopy
Covalent bond in molecule behave as tiny
spring
The atom will not remain in fix motion with respect to
each other but the avg. distance remain same
So the vibration motion is occurred
When internal vibrational energy of molecule
matches with energy of externally applied IR,
quantized

10. Molecules
absorb IR
Molecule excited from
lower to the higher vibrational level
within same electronic energy level.
i.e. Increases the amplitude of vibration
P.S.Kalsi; Spectroscopy Of Organic Compounds, Sixth Edition:2004,
Page no. 65-68.

11. Infrared Spectra
• The intensity of remaining transmitted light (It) is measure by
IR spectrophotometer (detector).
• It is compared with intensity of incident light (Io); to get
absorbance (A) or percentage transmittance (%T) of the
substance (molecule).
• Thus, the intensity of light absorbed by the molecule is measured interms of %T
verses frequency or wave number (in cm-1) of IR region.
• The graph of %T verses frequency or wave number (in cm-1) of radiation passed
is recorded on the charge.
It is called as IR Spectra

12. • The frequency of vibration producing IR spectra is given by
cm-1
Where,
• ʋ is frequency of vibration or frequency of spectral line or position of
spectral line in cm-1.
• K is the force constant of bond. It is measured bond strength
• μ is reduced mass of molecule

13. Types of Vibrational modes:
• Only those vibrational modes (vibration) which produces change
in dipole moment of molecule, absorbs IR radiation and show IR
spectra.
• So, the dipole moment of molecule must change during
vibration.
Q.1) What are the essential conditions for a compound to absorb IR radiation?
(S-11 & W-13, 2 Mark)
Q.2) Give reasons: Chlorine molecule does not absorb IR radiation. (S-11, 2 Mark)
Q.3) What are the conditions or absorption in IR region? (W-11, 2 Mark)
Q.4) Give the selection and essential condition (requirement) for a molecule to show IR
spectra?
Q.5) Give the necessary condition for a molecule to absorb IR radiation?
(or to become IR active).
Q.6) What are the conditions (requirement) for a substance to absorb in the IR region?
Which of the following compounds are IR active? (W-19, 4 Mark)
(i) CO2 (ii) H2O & (iii) NH3

14. Dipole moment:
For a molecule to show infrared absorpsions it must possess a
specific feature: an electric dipole moment of the molecule must
change during the vibration.
A dipole moment, µ is defined as, ‘the product of magnitude of
charge on any one of the atoms and the distance between them’.
Mathematically, dipole moment is given by
µ = q . r or µ = q .d (Unit: C.m)
Where,
q = Charge on any one of the atoms
r (or d) = distance between the atoms.
When these charge atoms vibrates, they shows change in
dipole moment, & molecule absorb IR radiation.

15. Essential conditions for a compound to absorb IR radiation:
The vibration frequency of the molecule must match (equal) with frequency of
IR radiation.
The dipole moment of molecule must change during vibration.
What is mean by IR Active Compounds??
If the vibration transition in molecule is capable of
change in dipole moment so the molecule is said to be
IR active.
e.g. I-Cl, C=O,CHCl3, H2O,etc.
What is mean by IR Inactive Compounds??
If the vibration transition in molecule is not producing
change in dipole moment so the molecule is said to be
IR inactive.
e.g. H2, Cl2, N2, O2, C=C, etc.

16. IR region:
0.8 µm (800nm) to 1000 µm
(1mm)
Near IR:
λ=0.8-2.5 µm
ʋ=12500-4000 cm-1
Middle IR:
λ=2.5-15 µm
ʋ= 4000 - 667 cm-1
Far IR:
λ=15-200 µm
ʋ= 667 - 50 cm-1
Most of the analytical applications are confined to the middle
(mid) IR region because absorption of organic molecules is
high in the middle (mid) IR region.
Approximate range of IR region:
Q.1) What is approximate range of mid IR region? (W-09, 1 Mark)
Q.2) Give the approximate range of IR region in cm-1. (S-10, 1 Mark)
Q.3) Give approximate range of IR (mid) region. (W-12, 1 Mark)

17. Name the radiation source in
IR spectroscopy
Radiation
source
Incandescent
lamp
Nernst
Glower
Mercury
Arc
Globar
source
Q.1) Name the common source of IR radiation. (W-11, 1 Mark)
Q.2) Name the radiation source in IR spectroscopy. (S-12 & S-14, 1 Mark)
Q.3) Write at least one source of Infra red radiation. (W-12, 1 Mark))

18. SOURCE of RADIATION:
•COMMONLY USED
•
•GLOBAR FILAMENT & NERNST GLOWER
Globar - resistance rod of silicon carbide
Globar - for longer wavelengths
Nernst glower- a spindle of rare earth oxide
(thorium,zirconium,etc)
Nernst glower- for shorter wavelengths

22. What is a vibration in a molecule?
Any change in shape of the molecule-
stretching of bonds, bending of bonds, or
internal rotation around bonds called vibration.

23. Vibration or Vibrational motion:
Defination:
 The periodic displacement of bonded atom
about it’s equilibrium position in the
molecule along bond axis or
perpendicular to bond axis is called as
vibration or vibrational motion.
=

24. Fundamental or Normal modes of Vibration:
Defination:
 The vibration motion in which all atom of
the molecule vibrate in phase
simultaneously and with same
frequency is called as normal or
fundamental mode of vibration.

26. Types of vibrational modes:
Q.1) Explain the various types of fundamental (normal) modes of vibration.
Q.2) Define and explain: (i) Stretching vibration & (ii) Bending vibration.
Q.3) Name the various type of vibration?
Q.4) What is bending vibration? (W-09, 1 Mark)
Q.5) Identify type of vibrations indicated in each of the following figures. (S-11, 2 Mark)
Q.6) Explain various types of stretching vibrations with suitable example. (W-11, 3 Mark)
Q.7) What are bending vibrations? Indicate in-plane and out-plane bending vibrations by
taking suitable example. (W-13, 2 Mark)
Q.8) Identify type of vibrations indicated in each of the following figures. (S-14, 2 Mark)
Q.9) Illustrate with diagrams the different types of bending vibrations. (S-13 & W-17, 4 Mark)
Q.10) What are stretching vibrations? (W-15, 1 Mark)
Q.11) Describe different types of bending vibrations in IR spectroscopy. (W-15, 4 Mark)
Q.12) Discuss the types of mode of vibrations in IR spectroscopy. (W-16, 4 Mark)
Q.13) Describe types of vibrational modes in IR spectroscopy. (W-18, 4 Mark)
Q.14) Explain the following terms with diagram: (S-19, 4 Mark)
(i) Scissoring & (ii) Twisting

27. MOLECULAR VIBRATIONS
Bending
vibrations
Stretching
vibrations
 Vibration or oscillation
along the line of bond
 Change in bond length
 Occurs at higher energy:
4000-1250 cm-1
a) Symmetrical stretching
b) Asymmetrical stretching
• Vibration not along the
line of bond
• Bond angle is altered
• Occurs at low energy:
1400-667 cm-1
a) In plane bending
b) Out plane bending

28. 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
IR
VIBRATIONAL
MODES:
involves
change in bond
length

29. NUMBER OF VIBRATIONAL MODES IN A MOLECULE:
Symmetric Stretching Asymmetric Stretching
Scissoring
Rocking
Wagging Twisting

30. STRETCHING VIBRATION: (BOND LENGTH CHANGES) (2
BONDS INCREASE OR DECREASE IN LENGTH SIMULTANEOUSLY)
Defination: The fundamental mode of vibration in which,
 Bond length (internuclear distance between two atoms)
changes (increases or decreases) periodically.
 Bond angle remain constant,
is called as stretching vibration.
 The molecule containing N number of total atom has (N-1)
number of stretching vibration.
 There are two types of stretching vibration.

31. STRETCHING VIBRATIONS
Asymmetrical
Stretching
vibrations
Symmetrical
Stretching
vibrations
 All bond stretched and
compressed alternately in
symmetrical manner.
 having permanent
dipole moment is IR
active
 having zero dipole
moment is IR inactive
• One bond stretched while
other bond is compressed
and vice - versa.
• In this, one bond length is
increased and other is
decreased
• Asymmetrical stretching
vibration is IR active

32. Example: Symmetrical stretching vibration in CH2 group
Example: Asymmetrical stretching vibration in CH2 group

33. a) Symmetrical stretching:
2 bonds increase or decrease in bond length
simultaneously.
H
H
C
STRECHING VIBRATIONS

34. b) Asymmetrical stretching
 In this, one bond length is increased and other
is decreased.
H
H
C

35. Bending vibrations:
(bond angle changes)
1Scissoring: In this type, two atoms
approach each other with respect to central atom.
2Rocking: In this type, the movement of atoms
takes place in the same direction with respect to central atom.
3
Wagging: In this type, two atoms move “up and down”
the plane with respect to central atom
4
Twisting: In this type, one of the atom moves up the plane while
the other moves down the plane with respect to central atom
Types of
Bending
Vibration

36. a) In plane bending
i. Scissoring:
 2 atoms approach each other
 Bond angles are decrease
H
H
CC

37. a) In plane bending
ii. Rocking:
 Movement of atoms take place in the same
direction with respect to central atom.
(Bonds swing in one direction in a plane)
 .
H
H
CC

38. b) Out plane bending
i. Wagging:
 2 atoms move to one side of the plane.
 They move up and down the plane.
H
H
CC

39. b) Out plane bending
ii. Twisting:
 One atom moves above the plane and another
atom moves below the plane.
H
H
CC

40. Company
LOGO
Fundamental modes of
vibration
PRP

41. Company name
Fundamental modes of vibration:
Type of
molecule
containing
Total “N”
atom
Total
degrees of
freedom
Translational
degrees of
freedom
Rotational
degrees of
freedom
Vibrational
degrees of
freedom
Linear
molecule
3 N 3 2 3N - 5
Non-linear
molecule 3 N 3 3 3N - 6
IR

42. Company name
Fundamental modes of vibration:
 The number of degrees of freedom is equal to the sum of
the coordinates necessary to locate all the atoms of the
molecule in space.
 Three coordinates (X, Y and Z) are required the position of
the atom P in space.
 Degree of freedom for the single atom P is equal to Three.
Because three coordinates are required to locate the atom
in space.
IR

43. Company name
Fundamental modes of vibration:
 If a single atom restricted to movement in a plane, then translational
degrees of freedom is Two, because only two coordinates (X, Y) would
be required to fix the position of the atom in the plane.
 If there are two atoms in space, total translational degrees of
freedom = 6 (2x3). Each atom would have three translational degrees
of freedom because three coordinates would be required to fix the
position of each atom.
 If there are N atoms in space, total translational degrees of freedom
= 3 N (N x 3). Each atom would have three translational degrees of
freedom because three coordinates would be required to fix the
position of each atom.
IR

44. Company name
Fundamental modes of vibration:
 Atoms have three degrees of freedom, all of which are
translational.
 When atoms combine to form molecules, no degrees of
freedom are lost. That is, the total number of degrees of
freedom of the molecule will be 3N
Where, N is the number of atoms in the molecule.
• The “3N” degrees of freedom of the molecule will be made
up of rotational, vibrational and translational degrees of
freedom.
IR

45. Company name
Fundamental modes of vibration:
Rotational degrees of freedom result from the
rotation of a molecule about an axis through
the center of gravity. These rotations result in
a degree of freedom only if the positions of the
atoms in space change during the rotation.
IR

46. Company name
Fundamental modes of vibration:
 For examples:
(i) For Linear molecule (Rotation about two axes (X and Y) will
result in a change in the position of the atoms);
Rotational degrees of freedom = 2
(ii) For Non-linear molecule (Rotation about all (three) axes
(X, Y and Z) will result in a change in the position of the atoms);
Rotational degrees of freedom = 3
IR

47. Company name
Fundamental modes of vibration:
 All degrees of freedom not accounted for by translational and
rotational are vibrational degrees of freedom.
3 N degrees of freedom =
= translational + rotational + vibrational
Therefore,
Vibrational degrees freedom =
= 3 N - (Translational + Rotational)
IR

48. Company name
Fundamental modes of vibration:
Type of
molecule
containing
Total N
atom
Total
degrees of
freedom
Translation
al degrees
of freedom
Rotational
degrees of
freedom
Vibrational
degrees of
freedom
Linear
molecule
3 N 3 2 3N - 5
Non-linear
molecule 3 N 3 3 3N - 6
IR

49. Company name

50. Company name
Q.1) Calculate fundamental modes of vibrations in each case of following molecules:(W-12, 3 Mark)
(i) Ammonia, (ii) Cyclohexanol & (iii) Benzene.
Q.2) What are the fundamental modes of vibration in water molecule? (W-13, 2 Mark)
Q.3) Calculate the number of vibrational modes in H2O. Explain it’s IR spectrum. (S-13, 4 Mark)
Q.4) Calculate the number of vibrational degrees of freedom in the following compounds:(S-14, 4 )
(i) CO2, (ii) H2O, (iii) H-C≡C-H & (iv) NH3
Q.5) Calculate the vibrational degrees of freedom (fundamental modes of vibration) for following
molecules in IR spectroscopy. (S-15, 4 Mark)
(i) CO2 & (ii) NH3
Q.6) Calculate fundamental modes of vibrations in each of the following molecules: (S-16, 4 Mark)
(i) NO (ii) CO2 (iii) CH4 & (iv) Benzene.
Q.7) Calculate the number of fundamental modes of vibrations for the following molecules:
(S-17, 4 Mark)
(i) Water (H2O) (ii) Ammonia (NH3) (iii) Carbon dioxide (CO2) & (iv) Benzene (C6H6).
Q.8) Calculate the vibrational degrees of freedom for the following molecules in IR spectroscopy:
(i) CO2 (ii) NH3 (iii) Benzene (C6H6) & (iv) CH4 (S-19, 4 Mark)
Q.9) Water molecule shows _____ different modes of vibrations. (W-19, ½ Mark)
(a) two (b) three (c) four (d) five
Q.10) Calculate the number of vibrational degrees of freedom in the following compounds:
(i) NH3 & (ii) H2O (W-19, 4 Mark)

51. Company name
Fundamental modes of vibration:
IR
Sr. No. Molecule
Total no. of
atoms (N)
Geometry of
the molecule
Applicable
formula
Fundamental
modes of
vibration
(FMV)
1. NO 2 Linear 3N – 5 1
2. CO2 3 Linear 3N – 5 4
3. H2O 3 Non-linear 3N – 6 3
4. NH3 4 Non-linear 3N – 6 6
5. CH4 5 Non-linear 3N – 6 9
6. C6H6 12 Non-linear 3N – 6 30
7. H-C≡C-H 4 Linear 3N – 5 7
8. Cyclohexanol 19 Non-linear 3N – 6 51
9. CS2

52. Company name
Fundamental modes of vibration:
IR
Sr.
No.
Molecule
Total no.
of atoms
(N)
Geometry
of the
molecule
Applicabl
e formula
Fundament
al modes
of
vibration
(FMV)
1. NO
2. CO2
3. H2O
4. NH3
5. CH4
6. C6H6
7. H-C≡C-H
8. Cyclohexanol
9. CS2

53. IR region:
0.8 µm (800nm) to 1000 µm
(1mm)
Near IR:
λ=0.8-2.5 µm
ʋ=12500-4000 cm-1
Middle IR:
λ=2.5-15 µm
ʋ= 4000 - 667 cm-1
Far IR:
λ=15-200 µm
ʋ= 667 - 50 cm-1
Most of the analytical applications are confined to the middle
(mid) IR region because absorption of organic molecules is
high in the middle (mid) IR region.
Approximate range of IR region:
Q.1) What is approximate range of mid IR region? (W-09, 1 Mark)
Q.2) Give the approximate range of IR region in cm-1. (S-10, 1 Mark)
Q.3) Give approximate range of IR (mid) region. (W-12, 1 Mark)

54. Company name
Mid IR region or Spectral Range:
1
Functional group region:
( 2.5 to 7.7 µm
i.e.,4000-1300 cm-1)
The common functional
group show IR absorption
band in this region due to
O-H, N-H, C=O and C-H
stretching vibration of
functional group.
So, it is called as functional
group region.
2
Finger print region:
(7.7-11 µm
i.e.,1300 to 909 cm-1)
In this region each
compound shows its
unique (characteristics)
absorption band. Similar
to finger prints. This
region is very useful for
sample comparison.
3
Aromatic region:
(11-15 µm
i.e, 909-667 cm-1)
The aromatic compound show
absorption band in this region
due to aromatic character.
This region also useful in
determination of substitution
patterns on aromatic
compounds such as ortho,
meta, para substitution.
IR
Q.1) What is meant by finger print region in IR spectroscopy? (S-15 & W-16, 1 Mark)
Q.2) Define the term: Finger print region. (S-18, 2 Mark)
Q.3) Give the range of fingerprint region in IR Spectroscopy. (W-18, 1 Mark)

55. Company name
Mid IR region or Spectral Range:
1
Functional group region:
( 2.5 to 7.7 µm
i.e.,4000-1300 cm-1)
The common functional
group show IR absorption
band in this region due to
O-H, N-H, C=O and C-H
stretching vibration of
functional group.
So, it is called as functional
group region.
2
Finger print region:
(7.7-11 µm
i.e.,1300 to 909 cm-1)
In this region each
compound shows its
unique (characteristics)
absorption band.
Similar to finger prints.
This region is very useful
for sample comparison.
3
Aromatic region:
(11-15 µm
i.e, 909-667 cm-1)
The aromatic compound show
absorption band in this region
due to aromatic character.
This region also useful in
determination of substitution
patterns on aromatic
compounds such as ortho,
meta, para substitution.
IR C-C
C-O
C-N
Q.1) What is meant by finger print region in IR spectroscopy? (S-15 & W-16, 1 Mark)
Q.2) Define the term: Finger print region. (S-18, 2 Mark)
Q.3) Explain: Aromatic region in IR spectroscopy. (S-18, 2 Mark)
Q.4) Give the range of fingerprint region in IR Spectroscopy. (W-18, 1 Mark)
Q.5) The range of finger print region is __________.
(S-19, ½ Mark)

56. Company name
Hydrogen bonding:
IR

57. Hydrogen Bonding
These three bonds all have;
• A strong permanent dipole
• A hydrogen atom
• An atom with lone pair electrons
The three types of bonds which give
molecules significant hydrogen bonding
are; (i) N – H (ii) O – H (iii) F – H

58.  Examples

59. Intermolecular Forces
O -
H
H

60.  Intermolecular hydrogen bond:
 Hydrogen bond formed between two molecules
 Intramolecular hydrogen bond:
 Hydrogen bond formed between
two different atoms in the
same molecule
 Intermolecular hydrogen bond is stronger
than van der Waals’ forces
N
O
O
O
H

61. INTRAMOLECULAR
HYDROGEN BONDING
When hydrogen bonding exists within the molecule
it is called intramolecular hydrogen bonding. In such
type of hydrogen bonding two groups of the same
molecule link through hydrogen bond, forming a
stable five or six membered ring structure e.g.,
salicylaldehyde, o-chlorophenol, acetylacetone,
ethylacetoacetate etc.

62. This intramolecular hydrogen bonding was
first called chelation (after the Greek word
"Chela" meaning, claw) because in the same
molecule the formation of a ring hydrogen
bonding is a pincer like action resembling
the closing of a Crab's claw. Some more
examples of intramolecular hydrogen
bonding are :

63. Company name
Mid IR region or Spectral Range:
Functional Group Region
4000 – 1300 cm-1 Finger Print Region
1300 - 909 cm-1
Aromatic Region
909 - 667 cm-1

64. Company name
Eight spectral region to identify the compound:
S.No. Wave number, cm-1 Band causing absorption
1 3750 - 3000 O-H, N-H stretching
2 3300 - 2900
-C≡C-H, >C=C<H, Ar-H (C-H
stretching)
3 3000 - 2700
-CH3, -CH2-, >C-H, -CHO
(C-H stretching)
4 2400 - 2100 C≡C, C≡N stretching
5 1900 - 1650
C=O (acids, aldehydes, ketones,
amides, esters, anhydride, acid
halides) stretching
6 1675 - 1500
>C=C< (aliphatic and aromatic),
>C=N- stretch
7 1475 - 1300 ≡C-H bending
8 1000 - 650 >C=C<H, Ar-H bending (Out of plane)

65. Company name
Identification of characteristic absorption bands
S.No. Wave number, cm-1 Band causing absorption
1 3750 - 3000 O-H, N-H stretching
1.1 3700 - 3500
Free (non hydrogen bonded) O-H
stretching
1.2 3500 Free phenol O-H stretching
1.3 3450 - 3200 Hydrogen bonded O-H stretching
1.4 3500 - 3300
Non-bonded amines (Free) N-H
stretching
1.5 3500 - 3100 Bonded amine N-H stretching
Note that: Primary amines shows two
bands, 2o amines & imines show only
one band, and 3o amines show no bands.
1.6 3500 - 3300 Amides also show N-H stretching
1.7 3000 - 2500 Carboxylic acid (-COOH)

66. Company name
Identification of characteristic absorption bands
S.No. Wave number, cm-1 Band causing absorption
2 3300 - 2900
-C≡C-H, >C=C-H, Ar-H
(C-H stretching)
2.1 3030 Ar-H stretch
2.2 3300 C≡C-H stretch
2.3 3040 - 3010 C=C-H stretch
3 3000 - 2700
-CH3, -CH2-, ΞC-H, -CHO
(C-H stretching)
3.1 2960 & 2870 -CH3 (two bands)
3.2 2930 & 2850 -CH2- (two bands)
3.3 2890 >C-H
3.4 2720 -CHO (C-H stretching)

67. Company name
Identification of characteristic absorption bands
S.No. Wave number,
cm-1
Band causing absorption
4 2400 - 2100 C≡C, C≡N stretching
4.1 2140 - 2100 H-C≡C-H
4.2 2260 - 2190 H-C≡C-R’
4.3 No absorption
R-C≡C-R (Symmetrical
vibration causes no change in the
dipole moment)
4.4 2260 - 2240 R-C≡N

68. Company name
Identification of characteristic absorption bands
S.No. Wave number, cm-1 Band causing absorption
5 1900 - 1650
C=O (acids, aldehydes, ketones,
amides, esters, anhydride,
acid halides) stretching
5.1 1740 - 1720 Aldehydes i.e. R-CHO
5.2 1725 - 1705 Acids i.e., R-COOH
5.3 1725 - 1705 Ketones i.e., R-CO-R
5.4 1740 - 1710 Esters i.e., R-COO-R
5.5 1815 -1720 Acid halides i.e., R-COX
5.6
1850 – 1800 &
1780 - 1740
Anhydrides (Two bands separated by
approximately 60 cm-1)
5.7 1700 - 1640 Amides i.e., R-CONH2

69. Company name
Identification of characteristic absorption bands
S.No. Wave number, cm-1 Band causing absorption
6 1675 - 1500
>C=C<
(aliphatic and aromatic),
>C=N- stretch
6.1 1680 - 1620 >C=C< (aliphatic)
6.2 1500 - 1400 >C=C< (aromatic)
6.3 1690 - 1640 >C=N-
6.4 1630 - 1575 -N=N-
7 1475 - 1300 >C-H bending

70. Company name
Identification of characteristic absorption bands:
S.No. Wave number, cm-1 Band causing
absorption
8 1000 - 650
>C=C<H, Ar-H bending
(Out of plane)
8.1 990 & 910 RCH=CH2 (C=C-H bending)
8.2 690 RCH=CRH (cis)
8.3 970 RCH=CRH (trans)
8.4 890 R2C=CH2
8.5 840 -790 R2C=CHR
Substituted Benzene: Type of Substitution
8.6 750 & 700
Mono-substituted aromatic
(5 adjacent H)
8.7 750 Ortho aromatic (4 adjacent H)
8.8 810 - 780 Meta aromatic (3 adjacent H)
8.9 850 - 800 Para aromatic (2 adjacent H)

71. Company name

72. Company name

73. Company name

74. Company name
Problems on Interpretation of
Infrared Spectra:
Q.1) In which IR spectral region, the following compound be
expected to absorb light? What bond gives rise to each
absorption?
IR
CH3CH2C-H
O
S.No. Wave number,
cm-1
Band causing absorption
1. 3000 - 2700 CH3, CH2 and aldehyde proton
(C-H stretch)
2. 2960 & 2870 -CH3 (two bands)
3. 2930 & 2850 -CH2- (two bands)
4. 2720 (->C-H stretching) aldehyde proton
5. 1740 - 1720 Aldehydes
6. 1475 - 1300 >C-H bending
 
R
O
HR
O
H

75. Company name
Problems on Interpretation of
Infrared Spectra:
Q.2) In which IR spectral region, the following compound be
expected to absorb light? What bond gives rise to each
absorption?
IR
 
R
O
HR
O
H
N-H
O
S.No. Wave number,
cm-1
Band causing absorption
1. 3500 – 3300
Or 3500 – 3100
Or 3500 - 3300
Non-bonded amines (Free) N-H stretching
Bonded amines N-H stretching
Amides also show N-H stretching
2. 2930 & 2850 -CH2- (two bands)
3.
1725 – 1705
Or 1700 - 1640
Ketones i.e., R-CO-R
or Amides i.e., R-CONH2
4. 1475 - 1300 >C-H bending
 
R
O
R

76. Company name
Problems on Interpretation of
Infrared Spectra:
Q.3) An IR spectra shows a strong absorption band at 1715 cm-1,
the functional group likely to be present is >C=O .
(W-16, ½ Mark)
(a) -CH=CH- (b) >C=O (c) –NH2 (d) -NO2
IR
 
R
O
HR
O
H
S.No. Wave number,
cm-1
Band causing absorption
1. 1725 - 1705 Ketones, >C=O
 
R
O
R
1900 - 1650
C=O (acids, aldehydes,
ketones, amides, esters,
anhydride, acid halides)
stretching

77. Company name
Problems on Interpretation of
Infrared Spectra:
Q.4) In which region of IR, absorption bands of stretching
vibration occur for the following functional groups?
(W-16, 4 Mark)
(i) C=O (ii) =C-H (iii) –N-H (iv) C≡C
IR
 
R
O
HR
O
H
S.No. Wave number, cm-1 Band causing
absorption
1. 1900-1650 >C=O
2. 3040 - 3010 C=C-H
3.
3500 – 3300 Non-bonded
3500 – 3100 Hydrogen-bonded
-N-H
4. 2400 - 2100 C≡C
 
R
O
R

78. Company name
Problems on Interpretation of
Infrared Spectra:
Q.5) In which region of IR, absorption bonds of stretching
vibrations occur for the following functional groups?
(S-17, 4 Mark)
(i) -C=O (ii) -N-H (iii) -C-H (iv) –C=C-
IR
 
R
O
HR
O
H
S.No. Wave number,
cm-1
Band causing absorption
1. >C=O
2. -N-H
3. -C-H
4. –C=C-
 
R
O
R

79. Company name
Problems on Interpretation of
Infrared Spectra:
Q.6) Write the appropriate frequency range in cm-1 for the
following functional groups: (W-19, 4 Mark)
(i) –N-H (ii) =CH (iii) -C≡N (iv) >C=O
IR
 
R
O
HR
O
H
S.No. Wave number,
cm-1
Band causing absorption
1. -N-H
2. =C-H
3. -C-≡H
4. >C=O
 
R
O
R

80. Company
LOGO
Structure of H2O molecule:
Q.1) Explain different types of stretching vibrations in tri-atomic molecule.
(S-13, 3 Mark)
Q.2) Calculate the number of vibrational modes in H2O. Explain it’s IR spectrum.
(S-13, 4 Mark)
Q.3) Explain the structure of water molecule on the basis of IR spectroscopy.
(S-15 & W-19, 4 Mark)
Q.4) Describe IR spectrum of H2O molecule. (W-15, 4 Mark)
Q.5) What tupes of vibrational modes are expected in H2O molecule?
Discuss its spectrum. (W-18, 4 Mark)

81. Company name
Structure of Organic Compounds:
Structure of H2O molecule:
 The total fundamental mode of vibration of
non-linear, H2O molecule are (3N-6) =
= [(3 x 3) – 6] = 3
 Out of three fundamental mode of vibration;
The molecule containing N number of total
atom has (N-1) number of stretching vibration.
i.e., (3-1) = 2 vibration are stretching vibration
and one vibration is bending vibration.
 Out of two stretching vibration; one is
symmetrical stretching vibration and another is
asymmetrical stretching vibration.
IR

82. Company name
Structure of H2O molecule:
The modes of vibrations of water molecule are represented as below:
H2O molecule is polar having certain dipole moment.
It’s dipole moment changes in each vibration modes.
So, all vibrations are IR active.

83. Company name
Structure of H2O molecule:
The modes of vibrations of water molecule are represented as below:
So, IR spectrum of water molecule shows three absorption bands.
One for bending vibration at 1600 cm-1,
second band due to symmetrical stretching at 3650 cm-1 and
third band due to asymmetrical stretching vibration at 3760 cm-1.
1600 cm-1
3650 cm-1
3760 cm-1

84. LOGOwww.themegallery.com
Structure of CO2 molecule:
Q.1) Discuss IR spectrum of carbon dioxide molecule. (W-09, 3 Mark)
Q.2) How many modes of vibrations the CO2 molecule will have? Which of them will be IR active?
(S-10 & W-14, 2-4 Mark)
Q.3) Give reasons: Carbon dioxide molecule shows only two IR bands in it’s IR spectrum inspite of it’s four
fundamental modes of vibrations.. (S-11, S-12 & W-13, 2 Mark)
Q.4) Describe IR spectrum of CO2 molecule. (W-14, 4 Mark)
Q.5) Calculate the number of vibrational modes in CO2. Discuss it on the basis of IR spectrum. (S-18, 4 Mark)
Q.6) Explain the structure of CO2 molecule on the basis of IR spectroscopy. (W-19, 2 Mark)

85. Company name
Structure of CO2 molecule:
The modes of vibrations of CO2 molecule are represented as below:
Structure of CO2 molecule: (Linear molecule)
 The Carbon dioxide (CO2) is a linear molecule.
 CO2 molecule has fundamental modes of vibration =(3N-5)
= [(3 x 3) - 5] = 4.
 Out of four fundamental mode of vibration;
The molecule containing N number of total atom has (N-1)
number of stretching vibration.
i.e., (3-1) = 2 vibration are stretching vibration and two
are bending vibration.
 Out of two stretching vibration; one is symmetrical
stretching vibration and another is asymmetrical stretching
vibration.

86. Company name
Structure of CO2 molecule:
The modes of vibrations of CO2 molecule are represented as below:
The remaining three vibrations are IR active because that produce
changes in dipole moment of CO2 molecule.
The two bending vibrations have same energy, because they are
degenerate (doubly).
So, IR spectrum of CO2 molecule shows two absorption bands.

87. Company name
Structure of CO2 molecule:
The modes of vibrations of CO2 molecule are represented as below:
So, IR spectrum of CO2 molecule shows
two absorption bands.
2350 cm-1
Bending
667cm-1
Asymmetrical

88. Company name
Problems:
Q.1) Which of the following vibrational modes
are IR active or inactive? (S-18, 4 Mark)
(i) Symmetric CO2 stretching
(ii) Antisymmetric CO2 stretching
(iii) Symmetric H2O stretching
(iv) H2O bending

89. Company name
Choice of solvent in IR spectra:
The essential condition or requirements (properties)
for a good solvent in IR spectroscopy are as given
below:
 The solvent should be transparent to IR radiation
( i.e., It should not be absorbed in IR region).
 It should not interact with the solute.
 It should be pure and the solute should be in it.
 It should be less polar or non polar.
Examples: CS2 and CCl4 are good solvents in IR
spectroscopy.
IR

90. Company name
Distinguish following compounds
on the basis of IR spectroscopy:
Q.1) How would the infrared spectra of the
following compounds differ:
(W-09, 3 Mark)
IR
(i) CH3
COCH3
& CH2
=CH-C-CH3
O
(ii) CH3
-C-OH & C6
H5
COCH3
O
(iii)
NH2
&
NH-C-CH3
O

91. Company name
Distinguish following compounds
on the basis of IR spectroscopy:
IR
S.NO. Band causing
absorption,
in cm-1
Compound A Compound B
1. 3040 - 3010 C=C-H stretch
2. 2960 & 2870 -CH3 (two bands) -CH3 (two bands)
3. 2930 & 2850 -CH2- (two bands)
4. 1725 - 1705 Ketones i.e., R-CO-R Ketones i.e., R-CO-R
5. 1680 - 1620 >C=C< (aliphatic)
6. 1475 - 1300 >C-H bending >C-H bending
(i) CH3
COCH3
& CH2
=CH-C-CH
O
(ii) CH3
-C-OH & C6
H5
COCH3
O
(iii)
NH2
&
N
(i) CH3
COCH3
& CH2
=CH-C-CH3
O
(ii) CH3
-C-OH & C6
H5
COCH3
O
(iii)
NH2
&
NH-C
O

92. Company name
Distinguish following compounds
on the basis of IR spectroscopy:
IR
S.NO. Band causing
absorption,
in cm-1
Compound A Compound B
1.
2.
3.
4.
5.
6.
(i) CH3
COCH3
& CH2
=CH-C-CH3
O
(ii) CH3
-C-OH & C6
H5
COCH3
O
(iii)
NH2
&
NH
(i) CH3
COCH3
& CH2
=CH-C-CH3
O
(ii) CH3
-C-OH & C6
H5
COCH3
O
(iii)
NH2
&
NH

93. Company name
Distinguish following compounds
on the basis of IR spectroscopy:
IR
S.NO. Band causing
absorption,
in cm-1
Compound A Compound B
1.
2.
3.
4.
5.
6.
(i) CH3
COCH3
& CH2
=CH-C-CH3
O
(ii) CH3
-C-OH & C6
H5
COCH3
O
(iii)
NH2
&
NH-C-CH3
O

94. Company name
Distinguish following compounds
on the basis of IR spectroscopy:
IR
Q.12) Differentiate the following pairs of compounds on the
basis of IR spectroscopy. (S-17 & S-19, 4 Mark)
(i) Acetone & Ethanol
(ii) Acetamide & Acetic acid
S.NO. Band causing
absorption,
in cm-1
Compound A
Acetone
Compound B
Ethanol
CH3CH2OH
1.
2.
3.
4.
5.
6.
(i) CH3
COCH3
& CH2
=CH-C-CH3
O
(ii) CH3
-C-OH & C6
H5
COCH3
O
(iii)
NH2
&
NH-C-C
O

95. Company name
Distinguish following compounds
on the basis of IR spectroscopy:
IR
Q.12) Differentiate the following pairs of compounds on the
basis of IR spectroscopy. (S-17 & S-19, 4 Mark)
(ii) Acetamide & Acetic acid
S.NO. Band causing
absorption,
in cm-1
Compound A
Acetamide
CH3CONH2
Compound B
Acetic acid
CH3COOH
1.
2.
3.
4.
5.
6.

96. Company name
Distinguish following compounds
on the basis of IR spectroscopy:
IR
Q.13) Differentiate the following pairs of compounds on the
basis of IR spectroscopy. (W-17, 4 Mark)
(i) Acetaldehyde & Acetone
S.NO. Band causing
absorption,
in cm-1
Compound A
Acetaldehyde
CH3CHO
Compound B
Acetone
CH3COCH3
1.
2.
3.
4.
5.
6.

97. Company name
Distinguish following compounds
on the basis of IR spectroscopy:
IR
Q.13) Differentiate the following pairs of compounds on the
basis of IR spectroscopy. (W-17, 4 Mark)
(ii) Acetamide & Acetic acid
S.NO. Band causing
absorption,
in cm-1
Compound A
Acetamide
CH3CONH2
Compound B
Acetic Acid
CH3COOH
1.
2.
3.
4.
5.
6.

98. Company name
Distinguish following compounds
on the basis of IR spectroscopy:
IR
Q.14) Differentiate between following pairs on the basis of IR
spectroscopy. (W-18, 4 Mark)
S.NO. Band causing
absorption,
in cm-1
Compound A
Ethyl bromide
CH3CH2Br
Compound B
Ethanol
CH3CH2OH
1.
2.
3.
4.
5.
6.
(i) CH3CH2Br and CH3CH2OH
(ii) CH3COCH3 and CH3CONH2

99. Company name
Distinguish following compounds
on the basis of IR spectroscopy:
IR
Q.14) Differentiate between following pairs on the basis of IR
spectroscopy. (W-18, 4 Mark)
S.NO. Band causing
absorption,
in cm-1
Compound A
Acetone
CH3COCH3
Compound B
Acetamide
CH3CONH2
1.
2.
3.
4.
5.
6.
(i) CH3CH2Br and CH3CH2OH
(ii) CH3COCH3 and CH3CONH2

100. Company name
Distinguish following compounds
on the basis of IR spectroscopy:
IR
Q.10) How will you distinguish following compounds on the
basis of IR spectroscopy? (W-14, 4 Mark)
S.NO. Band causing
absorption,
in cm-1
Compound A
Ethanol
CH3CH2OH
Compound B
Dimethyl ether
CH3OCH3
1.
2.
3.
4.
5.
6.
(i) CH3-CH2-OH & CH3-O-CH3
(ii) CH3-CO-CH3 & CH3-COOH

101. Company name
Distinguish following compounds
on the basis of IR spectroscopy:
IR
Q.10) How will you distinguish following compounds on the
basis of IR spectroscopy? (W-14, 4 Mark)
S.NO. Band causing
absorption,
in cm-1
Compound A
Acetone
CH3COCH3
Compound B
Acetic acid
CH3COOH
1.
2.
3.
4.
5.
6.
(i) CH3-CH2-OH & CH3-O-CH3
(ii) CH3-CO-CH3 & CH3-COOH

102. Company name
Distinguish following compounds
on the basis of IR spectroscopy:
IR
Q.9) Distinguish the following pairs of compounds on the basis
of their IR spectra? (W-14, 4 Mark)
(i) CH3-CH2-CHO & CH3-CO-CH3
(ii) CH3-CH2-OH & CH3-O-CH3
S.NO. Band causing
absorption,
in cm-1
Compound A Compound B
1.
2.
3.
4.
5.
6.

103. Company name
Distinguish following compounds
on the basis of IR spectroscopy:
IR
Q.8) How would you differentiate the following compounds on
the basis of IR spectra? (W-13, 3 Mark)
Q.7) How will you distinguish following compounds on the
basis of IR spectroscopy? (W-13, 3 Mark)
(i) CH3-C-CH3 & CH3-CH2-C-OH
(ii) CH3
-CH2
-OH & CH3
-O-CH3
O O
(i) C6
H5
-COOH & C6
H5
CHO
(ii) CH3
CH2
OH & CH3
-C-CH3
O

104. Company name
Distinguish following compounds
on the basis of IR spectroscopy:
IR
Q.6) Using IR spectrum of individual compound, how will you
distinguish following pair of compounds? (W-12, 3 Mark)
(i) Acetaldehyde and acetone
(ii) Methyl amine and dimethyl amine
Q.5) How the I.R. spectra of the following compounds differ?
(S-12, 3 Mark)
(i) CH3
-CH2
-NH & CH3
-CH2
-NH-CH3
(ii) CH3-CH2-C-CH3 & CH3-CH2-C-OH
O O

105. Company name
Distinguish following compounds
on the basis of IR spectroscopy:
IR
Q.4) On the basis of infrared spectra, how will you differentiate
the following compounds? (W-11 & S-14, 2 Mark)
Q.3) How the following compounds can be differentiated on
the basis of infrared spectroscopy?
(S-10 & S-14, 3 Mark)
(i) C6H5OH & C6H5COOH
(ii) CH3
-C-CH3
& CH3
-O-CH3
O
(i) CH3
-CH2
-C-CH3
& CH3
-CH2
-C-OH
O O
(ii) C6
H5
-NH2
& C6
H5
-C-N(CH3
)2
O

106. Company name
Distinguish following compounds
on the basis of IR spectroscopy:
IR
Q.2) How the infrared spectra of the following compounds
differ? (S-10 & W-14, 3 Mark)
(i) CH3-CH2-OH & CH3-O-CH3
(ii) CH3-CH2-NH2 & CH3-CH2-N-CH3
H
(iii) CH3-CH2-C-CH3 & CH2=CH-C-CH3
O O

109. I
R
R
I

110. A small truth to
make life 100%

112. Hard Work
H+A+R+D+W+O+R+K
8+1+18+4+23+15+18+11 = 98%
Knowledge
K+N+O+W+L+E+D+G+E
11+14+15+23+12+5+4+7+5 = 96%

113. Love
L+O+V+E
12+15+22+5 = 54%
Luck
L+U+C+K
12+21+3+11 = 47%
(why is it that most of us think this is the most important ???)

114. Then what makes 100%?
Is it Money? ... NO ! ! !
M+O+N+E+Y
13+15+14+5+25 = 72%
Is it Leadership? ... NO ! ! !
L+E+A+D+E+R+S+H+I+P
12+5+1+4+5+18+19+9+16 = 89%

115. Every problem has a solution,
but only if we change our
attitude.
To go to the top and achieve
100%
We really need to go further...

116. ATTITUDE
A+T+T+I+T+U+D+E
1+20+20+9+20+21+4+5 = 100%
It is OUR ATTITUDE towards Life and
Work that makes OUR Life = 100%

117. Change Your Attitude …
… and You Change Your Life ! ! !

119. LOGO

120. LOGO

124. I
R
R
I