Spectroscopy

1. Dr. Mausumi Chattopadhyaya
Assistant Professor
Department of Chemistry
mausumi@nitc.ac.in
+919433710271

2. Reference Books
1. Fundamentals of Molecular Spectroscopy, 4th Edition,
by Colin N. Banwell and Elaine M. McCash
2. Introduction to Spectroscopy, 5th Edition, Donald L.
Pavia, Gary M. Lampman, George S. Kriz
3. Organic Spectroscopy By William Kemp
4. Spectroscopy Of Organic Compounds By P S Kalsi
5. Elementary Organic Spectroscopy By Y R Sharma

3. 3
IR Range of EM Spectrum
X-Ray Ultraviolet Visible Infrared Microwave Radio
Near IR Middle IR Far IR
0.8 2.5 25 250
12500 4000 400 40
E = 1-10 kcal/mol
Bond Vibrations (most useful I.R. region)

4. 4

5. 5
D
E
T
E
C
T
O
r

6. ------------------------------------------------------------------------------------------
 IR Spectroscopy is concerned with the study of absorption of
IR radiation, which causes vibrational transitions in the
molecule due to the change in the dipole moment. Hence IR
spectroscopy is also known as Vibrational spectroscopy.
 Used in structural elucidation to determine the functional
groups.

7. 7
H
3
C
C
H
3
O

8. 8
Theory of IR Spectra - Vibrational Motion in Covalent Bond
Normal Vibration Vibration having Absorbed Energy
A bond will absorb radiation of a frequency similar to its
vibration(s)

9. 9
Types of Bond Motions
Symmetric
Asymmetric
Bending
Stretching
In Plane Bending
Out of Plane Bending

10. 10
H H
C
Symmetric Stretching
H H
C
Asymmetric Stretching

11. 11
Scissoring Rocking Twisting Wagging
In Plane Out Of Plane
Bending

12. ------------------------------------------------------------------------------------------
The energy of a diatomic molecule:
The two atoms in a molecule are settle at a mean inter-nuclear
distance such that these forces are just balanced and the total energy
of the whole system is minimum.

13. 13
Electric Field
Induced
Stretching-Out
Stretching-In
Induced Electric Field Interacts with the EM Radiation
Fluctuating Dipole
of Polar Molecule
Cl H

14. 14
Electromagnetic
Radiation Frequency 𝝑
The vibrating bond will absorb EM radiation (in IR Range)
only when its frequency is same as the frequency of the
oscillating dipole.
Frequency
Absorbed
Each Vibrational mode with change in dipole moment of the
system leads to absorption of frequency in IR range
Fluctuating Dipole of
Polar Molecule
Frequency 𝜗 IR spectra

15. 15
Criteria for IR Absorption
Dipole Moment of the bond must change during
Vibrational Motion
e.g. H2, N2, O2
IR-Inactive
e.g. H2O, SO2, HCl
IR-active

16. 16
Though CO2 is non polar it is IR active
Change in Dipole
Change in Dipole
No Change in Dipole
Example CO2

17. 17
Internuclear distance
0
Energy
req
SHO
Simple Harmonic Oscillator (SHO)
Morse Energy Diagram
In classical mechanics, a harmonic
oscillator is a system that, when
displaced from its equilibrium
position, experiences a restoring
force F proportional to the
displacement x:
If F is the only force acting on the
system, the system is called a
simple harmonic oscillator (SHO)

18. 18
For diatomic molecule behaving as SHO
Frequency of a vibrational stretching
Evib = v +
1
2
é
ë
ê
ù
û
úhu0
u0 =
1
2p
k
m
u0 = cu
u =
1
2pc
k
m
m =
m1m2
m1 + m2
Reduced Mass
k= force constant
c=velocity
= wave number,
unit cm-1
u
Two conclusions can be drawn from the above
equation:-
u µ k Means stronger bonds absorb in higher
frequency range
u µ 1/m Means heavier atoms absorb frequency
of low wave number
v is the vibrational quantum numbers, unit Joules, v=0,1,2,3……
Hooke’s law

19. 19
Problems
1. What is the order of decreasing vibrational
frequency for C — H & C — D?
2. What is the correct increasing order of stretching
frequencies for C ≡ C, C = C and C — C?

20. 20
 A molecule can never have
zero vibrational energy.
 The energy of the molecule
corresponding to v = 0 level
is known as Zero Point
Energy.
Energy levels of SHO are equally spaced
v=0
v=1
v=2
v=3
v=4
v=5
v=6
Selection rule for a SHO undergoing
vibrational changes
Energy
Internuclear distance
E =
1
2
hn0
E =
3
2
hn0
E =
5
2
hn0
E =
7
2
hn0
……
E =
1
2
hn0,v = 0 ZERO point energy
Dv =±1
Evib = v+
1
2
é
ë
ê
ù
û
úhu0
in Joules
ev =
E
hc
= v+
1
2
é
ë
ê
ù
û
ún in cm-1

21. 21
------------------------------------------------------------------------------------------
Energy
Internuclear distance
DEvib = v+1
( )+
1
2
é
ë
ê
ù
û
úhu0 - v+
1
2
é
ë
ê
ù
û
úhu0
v
v+1
v+2
v+….

22. 22
The Anharmonic Oscillator (AHO)
Real molecules do not obey laws of SHO for large amplitude of
vibrations.
Dissociation
Energy
v=0
v=1
v=2
v=3
v=4 De
Energ
y
Internuclear Distance
The selection rules for
anharmonic oscillator
v = 1, 2, …. etc
evib (2) = 2+
1
2
é
ë
ê
ù
û
úwe - 2+
1
2
é
ë
ê
ù
û
ú
2
wexe
evib (1) = 1+
1
2
é
ë
ê
ù
û
úwe - 1+
1
2
é
ë
ê
ù
û
ú
2
wexe
evib (0) = 0+
1
2
é
ë
ê
ù
û
úwe - 0+
1
2
é
ë
ê
ù
û
ú
2
wexe
v = 0,1,2,3...........
xe
we
anharmonicity constant
Equilibrium oscillation
frequency in cm-1
evib = v+
1
2
é
ë
ê
ù
û
úwe - v+
1
2
é
ë
ê
ù
û
ú
2
we ce in cm-1

23. 23
Calculate the Zero Point Energy??
evib = v+
1
2
é
ë
ê
ù
û
úwe - v+
1
2
é
ë
ê
ù
û
ú
2
we ce
evib = v+
1
2
é
ë
ê
ù
û
úwe 1- v+
1
2
æ
è
ç
ö
ø
÷ce
ì
í
î
ü
ý
þ
in cm-1
in cm-1

24. 24
Probable Transitions
v = 0 ®v =1,Dv =1
Intense Absorption (Fundamental Absorption)
De =we 1-2ce
( )cm-1
Case 1:
Case 2:
v = 0 ®v = 2,Dv = 2
With small intensity (First Overtone)
De = 2we 1-3ce
( )cm-1
Case 3:
v = 0 ®v =3,Dv =3
Negligible intensity (Second Overtone)
De = 3we 1-4ce
( )cm-1

25. 25
Q1. The fundamental and first overtone transitions of NO are
centred at 1876 cm-1 and 3724 cm-1 respectively.
Evaluate the equilibrium vibrational frequency, the
anharmonicity constant and the zero point energy.
------------------------------------------------------------------------------------------
Q2. The spectrum of HCl, shows a very intense absorption at
2886 cm-1 a weak one at 5668 cm-1 and a very weak one at
8347 cm-1. Calculate the equilibrium frequency (𝝎𝒆) and
anhormonicity constant (𝝌𝒆).
Ans. 𝝌𝒆 = 0.0174, we = 2990 cm-.
De =we 1- 2ce
( )cm-1
De = 2we 1-3ce
( )cm-1
evib = v+
1
2
é
ë
ê
ù
û
úwe 1- v+
1
2
æ
è
ç
ö
ø
÷ce
ì
í
î
ü
ý
þ

26. 26
Frequency (ϑ)
Fundamental
First overtone Second overtone
Frequency (𝜗)
Harmonic oscillator Anharmonic oscillator
v=0
v=2
v=1
v=3
v=4
E
Bond distance Bond distance
First Overtone
Fundamental
v=0
v=1
v=2
E
hv

27. ------------------------------------------------------------------------------------------
27
Degree of freedom: The number of directions in which atom can
move freely is defined as its degree of freedom.
A molecule composed of N atoms. The total number of
coordinates to specify the locations of N atoms is 3N (e.g. x,
y, z per atom). Says that the molecule has 3N degree of
freedom.
3N = degrees of freedom of translational + rotational +
vibrational motions
For linear molecules rotational motions = 2
For non linear molecule rotational motions = 3

28. 28
The degree of freedom of vibrational motion
No. of stretching motions of linear or non linear molecule is N-1
No. of bending motions for
linear molecules = 2N-4,
Non-linear molecules = 2N-5
For linear molecule 3N-5
Non-linear molecule 3N-6
} Fundamental vibrations

29. ------------------------------------------------------------------------------------------
29

30. ------------------------------------------------------------------------------------------
30
 IR band shapes come in various forms. Two of the most common
are narrow and broad. Narrow bands are thin and pointed, like a
dagger. Broad bands are wide and smoother.
 A typical example of a broad band is that displayed by O-H bonds,
such as those found in alcohols and carboxylic acids, as shown
below.

31. 31
------------------------------------------------------------------------------------------
Band Depends on Frequency
Position Reduced mass of
atoms
Lighter atoms – higher
frequency
Bond strength Strong – higher frequency
Strength Change in dipole
moment
Large dipole moment gives
strong absorption
Width Hydrogen bonding Strong H-bond gives wide
peak

32. ------------------------------------------------------------------------------------------
32
 Not all covalent bonds display bands in the IR spectrum. Only
polar bonds do so. These are referred to as IR active.
 Strongly polar bonds such as carbonyl groups (C=O) produce
strong bands.
 Medium polarity bonds and asymmetric bonds produce
medium bands.
 Weakly polar bond and symmetric bonds produce weak or non
observable bands.
Symmetric bond that has identical or nearly identical groups on
each end will not absorb in the infrared

33. • Stronger bonds have a larger force constant K and vibrate at higher
frequencies than weaker bonds.
• bonds between atoms of higher masses (larger reduced mass) vibrate
at lower frequencies than bonds between lighter atoms.

34. • Bending motions occur at lower energy (lower frequency) than the
typical stretching motions because of the lower value for the bending
force constant K.
• Hybridization affects the force constant K, also sp>sp2>sp3

35. 35
Group Frequencies

36. 36
Bond Class of compound Range (cm-1)
C-C Alkane 1200 - 700
C=C Alkene 1680 - 1620
Alkyne 2130 - 2150
C-H Alkane (C-sp3) 2965 - 2850
Aromatic (C-sp2) > 3000
Infrared Spectroscopy Table

37. 37
Bond Class of compound Range (cm-1)
Aldehyde 1740 – 1720
Ketone 1725 - 1705
Carboxylic acid 1725 - 1700
Amide 1700 - 1630

38. 38
Bond Class of compound Range (cm-1)
O-H Alcohol (monomer) 3650 - 3590
O-H…O Alcohol (H-bonded) 3420 – 3200
O-H Carboxylic acid
(H-bonded)
3300 - 3250
C-O Alcohols, Esters,
Acids and Ethers
1300 - 1000

39. 39
Bond Class of compound Range (cm-1)
N-H Amine/Amide 3500 (approx.)
CN Nitrile 2260 - 2240
C-X X = Chloride 800 - 600
X = Bromide 600 - 500
X = Iodide 500 (approx.)

40. 4000 cm-1 2700 cm-1
2000 cm-1 1600 cm-1 400 cm-1
40
IR Absorption Range

41. Fingerprint region: complex and
difficult to interpret reliably.
Focus your analysis on this region
(Functional group region). This is where
most stretching frequencies appear.
Looking at a Spectrum
 Fingerprint region is rich in Bending Vibration of C-C, C-H, C-N, C-O
etc.

42. 42
Hydrocarbons

43. Note the strong bands in the 3000-2850 cm-1 region due to C-H stretch.
The C-H scissoring (1470 cm-1), methyl rock (1383 cm-1), and long-chain
methyl rock (728 cm-1) are noted on this spectrum.

47. 47
Carboxylic Acid

48. 48
Amines
z
z

50. 50
Factors Influencing the Vibrational Frequency of C=O Bond
Inductive Effect
 Addition of +I groups result in weakening of CO bond and the 𝜗𝑐𝑜
value decreases.
Reason: Decrease in Bond Order and hence the Bond strength
H
O
H
1750 cm-1
H
3
C
O
H
1745 cm-1
H
3
C
O
C
H
3
1715 cm-1

51. 51
 Introduction of -I-groups results in increase in bond
order of CO and hence the absorption frequency (𝝑
value) increases.
H
3
C
O
C
H
3
1715 cm-1 1725 cm-1 1740cm-1
1750 cm-1
1750 cm-1
O
C
H
C
l
C
l
C
H
C
l
C
l
O
C
H
C
l
C
l
H
3
C
O
C
H
2
C
l
H
3
C

52. 52
Mesomeric Effect
 Conjugation lowers the 𝝑𝒄𝒐value by weakening the bond.
 Mode of conjugation enhances, the 𝝑𝒄𝒐 value further
decreases.
1706 cm-1
H
3
C
C
O
C
H
3
1715 cm-1
No Conjugation
H
3
C
C
O
C
H
C
H
2
Conjugation
1693 cm-1
C
O
H
3
C
Extension of Conjugation

53. • Resonance also affects the strength and length of a bond and
hence its force constant K.
• Thus, whereas a normal ketone has its C=O stretching vibration at
1715 cm−1; a ketone that is conjugated with a C=C double bond
absorbs at a lower frequency, near 1675 to 1680 cm−1
• That is because resonance lengthens the C=O bond distance and
gives it more single-bond character.
No two molecules will give exactly the
same IR spectrum (except enantiomers).

54. 54
1. Problem: In the IR spectra, Benzamide and Phenyl Acetate
showed 𝝑𝒄𝒐 stretching frequency at 1663 cm-1 and 1730 cm-1
respectively. Explain the reason for difference in their peak
position.
2. Problem: Arrange the following compounds according to
their 𝝑𝒄𝒐 absorption frequency: 1677 cm-1, 1700 cm-1, 1770
cm-1.
C O
H3C
NO2
C O
H3C
O
C O
H3C
CH3

55. 55
Hydrogen Bonding
 H-bonding brings downward frequency shifts.
 Stronger the H-bonding, greater is the absorption shift
towards lower frequency.
 Inter-molecular H-bonding gives rise to broad band and
strong peak.
 Intra-molecular H-bonding gives rise to sharp and defined
peak.

56. 56
Intra and Inter molecular Hydrogen bonding

57. Effect of dilution on Hydrogen bonding - effect on O-H
stretching
Intermolecular hydrogen bonding weaken the O-H bond,
thereby shifting the band to lower frequency.
Neat/concentrated Diluted with CCl4 Very diluted, with CCl4
Intermolecular Hydrogen bonding

58. Intramolecular H bonding
Intramolecular-H bonding shifts the OH band to lower
frequency
“Does not depend on the concentration”

60. 60
1058 cm-1
1109 cm-1

61. 61
1220 cm-1

62. 62
Explain why,
1. In dilute solution, Amines show stretching
frequency at 3650 cm-1, while a broad band is
appeared at 3300 cm-1 in condense phase spectra.
Explain why?

63. 63
Explain why,
2. The spectrum of glycol in CCl4 shows two peak at
3644 and 3612 cm-1.
No Intra but Intermolecular
H-bonding will be there
No Intermolecular H-bonding
H
H
O
O
H H
H
H
A n t i
O
H
H
O
H H
H
H
Syn
I n t r a m o l e c u l a r H - b o n d i n g

64. 𝝑𝑪𝑶 ~ 1700-1630 cm-1 (Sharp & Strong) in aldehydes, ketones,
carboxylic acids, esters, etc.
64
CO

65. Esters show a strong absorption between 1750 - 1730 cm-1 due to
C=O bond stretching.
O
O
R
65

66. O-H STRETCH
C=O STRETCH
O-H STRETCH
C=O STRETCH
AND
ALCOHOL
ALDEHYDE
CARBOXYLIC
ACID
3300-2500 cm-1
1725-1630 cm-1
3300-2500 cm-1
1725-1630 cm-1

69. Look Again !!!!!!!!

70. How to analyze IR spectra ???

71. How to analyze IR spectra ???
(also check the exact position of the carbonyl band for clues as to the
type of carbonyl compound it is)

72. Look for a single or double sharp N–H band in the region 3400-3250 cm-1
If there is such a band

73. C=O 1706 cm-1
C=O 1693 cm-1
C=O 1695 cm-1 C=O 1730 cm-1
Explain why the C=O stretching vibration occurs at the following
values compared to normal keto group ???

74. IR Spectroscopy : Carbonyl compounds
All carbonyl compounds absorb in the region 1760-1665 cm −1 due to the
stretching vibration of the C=O bond

76. 76
Ans:

77. 77
DBE or double bond equivalent or level of unsaturation is the number of unsaturations
present in a organic molecule. The term unsaturation mean a double bond or a ring system.
For instance , in benzene there are 3 double bonds and 1 ring which gives us 4 DBE.
Moreover a triple bond can be regarded as DBE=2. It must be noted that X is the total
number of halogens, namely Cl, Br, F and I present in the structure. Presence of an oxygen
atom does not effect the DBE calculation.
Double bond equivalent

79. Ans: Double bond equivalents/Degree of unsaturation can be
calculated using the following equation.

81. Ans:

83. Ans:

86. Ans:

88. Ans:

89. 89

90. 90

91. 91

92. 92