IR spectroscopy . P.K.Mani, BCKV, West Bengal, India

Basic Principles of Infra Red Spectroscopy and usefulness

1. Introduction and Principle of
IR Spectrophotometry
Dr.P.K.Mani
[email protected]
BCKV, West Bengal,
India, 09477465968

2. Origin of IR spectra
Atoms or atomic groups in a molecules are in continuous motion with
respect to one another. IR spectra originate from the difference modes of
vibration and rotation of a molecule, whereas the UV-visible absorption
bands are primarily due to electronic transition.
In order to absorb IR radiation, a molecule must undergo a net change in
dipole moment as a consequence of its vibrational or rotational motion. The
dipole moment is determined by the magnitude of the charge difference and
the distance between the two centers of charge.
The change in bond length or angle due to vibrational or rotational
motion must cause a net change in the dipole moment of the molecule.
No net change in dipole moment occurs during the vibration or rotation of
homonuclear species such as O
2, N
2, or Cl
2 ; consequently, such compounds
cannot absorb in the IR.
Vibrational modes which do not involve a change in dipole moment are
said to be IR-inactive. With exception of a few compounds of this type, all
molecular species exhibit IR-active.

3. IR ABSORPTION BY MOLECULES
Molecules with covalent bonds may absorb IR radiation
Molecules absorb radiation when a bond in the molecule vibrates
at the same frequency as the incident radiant energy
Absorption is quantized
Molecules move to a higher energy state
Molecules vibrate at higher amplitude after absorption
IR radiation is sufficient enough to cause rotation and vibration
Radiation between 1 and 100 µm will cause excitation to higher
vibrational states
Radiation higher than 100 µm will cause excitation to higher
rotational states

4. Ball and stick figure of an ethanol
molecule. But exactly what is the
ball, and for that matter, what is the
stick? An atom doesn’t really look
like a ball, nor does a chemical bond
look like a stick, right?
In a covalently bonded diatomic molecule, vibratory motion of the
atoms causes compression and extension of the bond like a
mechanical spring obeying Hook’s law

5. Principle of IR spectroscopy
•Molecules are made up of atoms linked by chemical
bonds. The movement of atoms and the chemical bonds
like spring and balls (vibration)
•This characteristic vibration are called Natural frequency
of vibration.

6. IR SPECTROSCOPY
The IR region is divided into
Near-IR (NIR): 750 nm – 2500 nm
Mid-IR: 2500 nm – 20000 nm
Far-IR: 20000 nm – 400000 nm

7. DIPOLE MOMENT (µ)
- The repetitive changes in µ makes it possible for polar
molecules to absorb IR radiation
- Symmetrical molecules do not absorb IR radiation since
they do not have dipole moment (O
2
, F
2
, H
2
, Cl
2
)
- Diatomic molecules with dipole moment are IR-active
(HCl, HF, CO, HI)
- Molecules with more than two atoms may or may not be
IR active depending on whether they have permanent

net dipole moment
µ = Q x r
Q = charge and r = distance between charges
- Asymmetrical distribution of electrons in a bond renders the bond
polar

8. Molecular vibrations
There are 2 types of vibrations:
1.Stretching vibrations
2.Bending vibrations
1.Stretching vibrations:
Vibration or oscillation along the line of bond
•Change in bond length resulting from change in interatomic
distance (r)
Occurs at higher energy: 4000-1250 cm
-1
2 types:
a)Symmetrical stretching
b)Asymmetrical stretching

9. a) Symmetrical stretching:
2 bonds increase or decrease in length simultaneously.
H
H
C
Symmetrical stretching is IR-inactive (no change in µ)

10. b) Asymmetrical stretching
•in this, one bond length is increased and other is
decreased.
H
H
C

11. PRINCIPAL MODES OF VIBRATION
Bending
- Change in bond angle or change in the position of a group of
atoms with respect to the rest of the molecule
Bending Modes
- Scissoring and Rocking
- In-plane bending modes (atoms remain in the same plane)
- Wagging and Twisting
Out-of-plane (oop) bending modes (atoms move out of plane)

12. Bending vibrations
•Vibration or oscillation not along the line of bond
•These are also called as deformations
•In this, bond angle is altered
•Occurs at low energy: 1400-666 cm
-1
2 types:
a)In plane bending: scissoring, rocking
b)Out plane bending: wagging, twisting

13. a) In plane bending
i.Scissoring:
•This is an in plane blending
•2 atoms approach each other
•Bond angles are decrease
H
H
CC

14. ii.Rocking:
•Movement of atoms take place in the same
direction.
H
H
CC

15. b) Out plane bending
i. Wagging:
•2 atoms move to one side of the plane. They move
up and down the plane.
ii.Twisting:
•One atom moves above the plane and another
atom moves below the plane.
H
H
CC
H
H
CC

16. Molecular vibrations
• How many vibrations are possible (=fundamental vibrations)?
A molecule has as many degrees of freedom as the total degree of
freedom of its individual atoms. Each atom has three degrees of
freedom (corresponding to the Cartesian coordinates), thus in an N-
atom molecule there will be 3N degree of freedom.
In molecules, movements of the atoms are constrained by interactions
through chemical bonds.
Translation - the movement of the entire molecule while the positions of the
atoms relative to each other remain fixed: 3 degrees of translational freedom.
Rotational transitions – interatomic distances remain constant but the
entire molecule rotates with respect to three mutually perpendicular axes :
3 rotational freedom (nonlinear), 2 rotational freedom (linear).
Vibrations – relative positions of the atoms change while the
average position and orientation of the molecule remain fixed.

17. PRINCIPAL MODES OF VIBRATION
3N-6 possible normal modes of vibration
N = number of atoms in a molecule; Degrees of freedom = 3N
H
2
O for example
- 3 atoms; - Degrees of freedom = 3 x 3 = 9
- Normal modes of vibration = 9-6 = 3

18. • Vibrations which do not change the dipole moment are Infrared
Inactive (homonuclear diatomics).
Selection Rules
The energy associated with a quantum of light may be transferred to
the molecule if work can be performed on the molecule in the form of
displacement of charge.
Selection rule:
A molecule will absorb infrared radiation if the change in
vibrational states is associated with a change in the dipole
moment (m) of the molecule.
µ = qr
q: electrical charge, r: directed distance of that charge from some defined
origin of coordinates from the molecule.
Dipole moment is greater when electronegativity difference between the
atoms in a bond is greater.
Some electronegativity values are:
H 2.2; C 2.55; N 3.04; O 3.44; F 3.98; P 2.19; S 2.58; Cl 3.16

19. w =
1
2p
(k/m)
1/2 w = oscillation frequency
m = reduced mass
k = force constant of the bond
v =
1
2pc
(k/m)
m =
m
x
m
y
m
x
+ m
y
Oscillation frequency of a vibrating diatomic molecule (XY) following
simple harmonic motion
Mass of atom X and Y respectively.
To convert frequency to wave number we
must divide W by the velocity of light
Hz
cm
-1
1/2
E
ν =
2
(ν + ) hω
1
Vibrational energies (Ev) like all other molecular energies are quantized and the
allowed vibrational energies for any particular system may be calculated from the
Schrodinger eqn
ν= vibrational quantum number

20. w =
1
2p
(k/m)
1/2
Molecule k / aJÅ
-2
F
2
(F−F) 4.45
O
2
(O=O) 11.41
N
2
(N≡N) 22.41
Several Force Constants
Note: IR spectra are typically presented in units called wavenumber, or more
correctly, reciprocal centimetres (cm
-1
). Increasing wavenumber corresponds to
increasing frequency.

21. The IR spectrum is basically a plot
of transmitted (or absorbed)
frequencies vs. intensity of the
transmission (or absorption).
Frequencies appear in the
x-axis in units of inverse
centimeters (wavenumbers), and
intensities are plotted on the
y-axis in percentage units.
IR SPECTRUM IN ABSORPTION MODE
IR SPECTRUM IN TRANSMISSION MODE
The graph shows a spectrum in
transmission mode.
This is the most commonly used
representation and the one
found in most chemistry and
spectroscopy books. Therefore we
will use this representation.

22. IR bands can be classified as strong (s), medium (m), or
weak (w),
depending on their relative intensities in the infrared spectrum.
A strong band covers most of the y-axis.
A medium band falls to about half of they-axis, and
a weak band falls to about one third or less of the y-axis.
CLASSIFICATION OF IR BANDS

23. Group frequencies
With certain functional or structural groups, it has been found that their
vibrational frequencies are nearly independent of the rest of the
molecule – group frequencies.
Carbonyl group 1650 to 1740 cm
-1
various aldehydes and
ketones
For many groups involving only two atoms, the approximate
frequency of the fundamental vibration can be calculated from a simple
harmonic oscillator model.
Calculations show that for most groups of interest, characteristic
frequencies of stretching vibrations should lie in the region 4000 to
1000 cm
-1
. In practical, the region from 4000 to 1300 cm
-1
is often called
the group frequency region.
The presence of various group vibrations in the IR spectrum
is of great assistance in identifying the absorbing molecule.

25. Note that the blue colored sections above the dashed line refer to stretching vibrations, and the
green colored band below the line encompasses bending vibrations.
The complexity of infrared spectra in the 1450 to 600 cm
-1
region
makes it difficult to assign all the absorption bands, and because of the
unique patterns found there, it is often called the fingerprint region.
Absorption bands in the 4000 to 1450 cm
-1
region are usually due to
stretching vibrations of diatomic units, and this is sometimes called the
group frequency region.

26. Fingerprint region
In the region from » 1300 to 400 cm
-1
, vibrational frequencies
are affected by the entire molecule, as the broader ranges for
group absorptions in the figure below – fingerprint region.
Absorption in this fingerprint region is characteristic of the
molecule as a whole. This region finds widespread use for
identification purpose by comparison with library spectra.

28. How to approach the analysis of an IR spectrum
1. Is a carbonyl group present ? C=O 1820~1660 cm
–1
(strong absorption)
2. If C=O is present, check the following types. (If absent, go to 3)
Acids -- is OH also present ? -- OH 3400~2400 cm
–1
(broad absorption)
Amides-- is NH also present ?-- NH 3500 cm
–1
(medium absorption)
Esters -- is C-O also present ? -- C-O 1300~1000 cm
–1
(strong absorption)
Anhydrides -- have two C=O absorptions near 1810 and 1760 cm
–1
.
Aldehydes -- is aldehyde CH present ? -- Two weak absorptions near 2850
and 1760 cm
–1
.
Ketones -- The above 5 choices have been eliminated.
3. If C=O is absent
Alcohols / Phenols-- check for OH-- OH 3600~3300 cm
–1
(broad
absorption) C-O 1300~1000 cm
–1
.
Amines -- check for NH -- NH 3500 cm
–1
. (medium absorption)
Ethers -- Check for C-O (and absence of OH) -- 1300~1000 cm
–1
.

29. OH stretch
CH stretch
CH bend
CO stretch
OH bend CC
stretch
The IR Spectrum of Ethanol
Bands representing common groups don’t always appear in exactly
the same position, as the local environment has an effect on the force
constant. Eg. Conjugation has the effect of lowering the observed
frequency.

30. How to approach the analysis of an IR spectrum
4. Double bonds and / or aromatic rings
C=C 1650 cm
–1
(weak absorption)
aromatic ring 1650~1450 cm
–1

aromatic and vinyl CH 3000 cm
–1

5. Triple bonds CºN 2250cm
–1
(sharp absorption)
C º C 2150cm
–1
(sharp absorption)
acetylenic CH 3300 cm
–1

6. Nitro group --two strong absorptions at
1600~1500 cm
–1
and 1390~1300 cm
–1

7. Hydrocarbons -- none of the above are found, CH 3000 cm
–1

(major absorption)

31. Wavenumber / cm
-1
StrengthVibrational mode
900 w C-C stretch
1080 s C-O stretch
1260 m O-H bend
1400 m C-H bend
2800-3000 s C-H stretch
3650 m O-H stretch
The IR Spectrum of Ethanol;
Tabulating IR data

32. IR Spectroscopy
Identifying Functional Groups
C Hstretch
ALKANES
~2850-2950 cm
-1
C Hbend~1350-1450 cm
-1
CH2 rock ~720 cm
-1
C Hstretch
ALKENES
~3000-3100 cm
-1
C Hbend~800-1000 cm
-1
C stretch~1600-1700 cm
-1
C
C Hstretch
ALKYNES
~3250-3350 cm
-1
C Hbend~630 cm
-1
C stretch~2100 cm
-1
C
O Hstretch
ALCOHOLS
~3200-3650 cm
-1
O Hbend~1330-1420 cm
-1
C stretch~1000-1260 cm
-1
O
C Cstretch
AROMATICS
~1600 & 1400-1500 cm
-1
C Hstrecth~3000 cm
-1

33. Identifying Functional Groups
O Csym. stretch
ETHERS
~1050 cm
-1
O Casym. stretch~1250 cm
-1
MOLECULES WITH CARBONYL GROUPS (C=O)
N Hstretch
AMINES
~3250-3450 cm
-1
N Hbend~1600-1650 cm
-1
C stretch~1000-1250 cm
-1
N
C
C
C
O
R G
Functional group
Ketone
Aldehyde
Carboxylic Acid
Acid Chloride
Acid Fluoride
-G
-H
-OH
-Cl
-F
-R
~1720-1740
~1750-1770
~1775-1815
~1870
~1680-1720
cm
-1
Functional group
Ester/Lactone
Amide/Lactam
-G
-NR
-OR
~1650-1700
~1735-1750
cm
-1
O
OR
O
R
~1750 & 1815
Note: Conjugation in ANY of these systems results in a lowering of the
carbonyl stretching frequency!

34. Understanding & Identifying Molecular Structure
IR Spectroscopy
The origin of broad -OH and -NH bands.
gas
liquid
Hydrogen bonding results in lower
electron density at each oxygen,
thus lowering the force constant, k,
thus lowering (& broadening) the
frequency for the mode.
H
3C
CH
2
O
H
H
O
CH
2
H
3C
H
O
CH
2
H
3C
H
O
CH
2
H
3C
H
O
CH
2
H
3C
H
O
CH
2
H
3C

35. Understanding & Identifying Molecular Structure
Q. The two IR spectra on the right
correspond to two different
molecules sharing the same
molecular formula; C
3
H
6
O.
a)Identify which is an alcohol and
which is a ketone.
b)Propose molecular structures
for these two molecules!
C
O
CH
3
H
3C
H
C
C
H
2
H2C
OH

36. Understanding & Identifying Molecular Structure
Q. The three IR spectra on the right
correspond to three different molecules
all with a C
3
carbon chain but different
degrees of unsaturation.
a)Identify which of these is propane,
propene and propyne.
b) Label each peak with the relevant
vibrational mode.
Satisfy yourself that some features
unambiguously identify some kinds of
functional groups
20003000 1000
cm
-1
4000

37. Absorption Regions
Irtutord.exe

39. IR spectrum of n-butanal (n-butyraldehyde).

40. THE FINGERPRINT REGION
Although the entire IR spectrum can be used as a fingerprint for the purposes
of comparing molecules, the 600 - 1400 cm
-1
range is called the fingerprint
region.
This is normally a complex area showing many bands, frequently overlapping
each other. This complexity limits its use to that of a fingerprint, and should be
ignored by beginners when analyzing the spectrum. As a student, you should
focus your analysis on the rest of the spectrum, that is the region to the left of
1400 cm
-1
.

41. y axis is %T or A
x axis is wavenumber (or wavelength)
I
o
® sample ® I
T = I/I
o
%T = 100 I/I
o
T transmission / transmittance
A = -log T
A absorbance

(no units)
(Note A (but not T) µ concentration)
IR spectrum

43. IR SPECTRUM OF ALKANES
Alkanes have no functional groups. Their IR spectrum displays only C-C
and C-H bond vibrations. Of these the most useful are the C-H bands,
which appear around 3000 cm
-1
.
Since most organic molecules have such bonds, most organic
molecules will display those bands in their spectrum.

44. Besides the presence of C-H bonds, alkenes also show sharp, medium
bands corresponding to the C=C bond stretching vibration at about
1600-1700 cm
-1
.
Some alkenes might also show a band for the =C-H bond stretch, appearing
around 3080 cm
-1
IR SPECTRUM OF ALKENES

45. The most prominent band in alkynes corresponds to the carbon-carbon triple bond. It
shows as a sharp, weak band at about 2100 cm
-1
. The reason it’s weak is because
the triple bond is not very polar. In some cases, such as in highly symmetrical alkynes, it
may not show at all due to the low polarity of the triple bond associated with those
alkynes.
Terminal alkynes, that is to say those where the triple bond is at the end of a carbon
chain, have C-H bonds involving the sp carbon (the carbon that forms part of the triple
bond). Therefore they may also show a sharp, weak band at about 3300 cm
-1
corresponding to the C-H stretch.
Internal alkynes, that is those where the triple bond is in the middle of a carbon chain, do
not have C-H bonds to the sp carbon and therefore lack the aforementioned band.
unsymmetrical
terminal alkyne
(1-octyne)
symmetrical
internal alkyne
(4-octyne).

46. Carbonyl compounds are those that contain the C=O functional group. In aldehydes,
this group is at the end of a carbon chain, whereas in ketones it’s in the middle of the
chain. As a result, the carbon in the C=O bond of aldehydes is also bonded to another
carbon and a hydrogen, whereas the same carbon in a ketone is bonded to two other
carbons.
Aldehydes and ketones show a strong, prominent, stake-shaped band around
1710 - 1720 cm-1 (right in the middle of the spectrum). This band is due to the
highly polar C=O bond. Because of its position, shape, and size, it is hard to
miss.
Because aldehydes also contain a C-H bond to the sp2 carbon of the C=O bond,
they also show a pair of medium strength bands positioned about 2700 and 2800
cm-1. These bands are missing in the spectrum of a ketone because the sp2
carbon of the ketone lacks the C-H bond.

47. IR SPECTRUM OF AN ALCOHOL
The most prominent band in alcohols is due to the O-H bond, and it appears as a
strong, broad band covering the range of about 3000 - 3700 cm-1. The sheer size
and broad shape of the band dominate the IR spectrum and make it hard to miss.

48. IR SPECTRUM OF A CARBOXYLIC ACID
because it has both the O-H bond and the C=O bond.
Therefore carboxylic acids show a very strong and broad band covering a
wide range between 2800 and 3500 cm-1 for the O-H stretch. At the same
time they also show the stake-shaped band in the middle of the spectrum
around 1710 cm-1 corresponding to the C=O stretch.

49. IR SPECTRUM OF
A NITRILE
In a manner very similar to alkynes, nitriles show a prominent band around 2250
cm
-1
caused by the CN triple bond. This band has a sharp, pointed shape just
like the alkyne C-C triple bond, but because the CN triple bond is more polar, this
band is stronger than in alkynes.