INFRARED SPECTROSCOPY

THE ELECTROMAGNETIC SPECTRUM

High Frequency (n) Low

High Energy Low

X-RAY ULTRAVIOLET INFRARED MICRO- RADIO FREQUENCY
WAVE

Nuclear
Vibrational

Ultraviolet Visible magnetic
infrared

resonance

2.5 mm 15 mm 1 m 5 m
200 nm 400 nm 800 nm

BLUE RED

short Wavelength (l) long

Types of Energy Transitions in Each Region
of the Electromagnetic Spectrum

REGION ENERGY TRANSITIONS

UV/Visible Electronic
Infrared Vibrational- rotational
Microwave Rotational

Radio Frequency Nuclear Spin
(NMR)
ESR Electron Spin

INFRA RED REGIONS
Region Wavelength Wavenumber

λ[μm] [cm-1]
Near IR 0.78-2.5 12800-4000
Mid IR 2.5-50 4000-200

(2.5-15)
Functional 2.5-7.7 4000-1300

group
Finger Print 7.7-11.0 1300-900

Aromatic 11.0-15.0 900-670
Far IR 50-1000 200-10

4

Basic principles of IR spectroscopy

•Atoms are in continuous motion.

Quantized vibrational states & quantized rotational states.

Absorption of IR radiation occurs in molecules that have small
energy differences between various vibrational& rotational
states.

Each functional group present in a molecule has its own
natural vibrational frequency

The frequency of vibration exactly matches the natural
vibrational frequency of the molecule-absorption of radiation
takes place.

3/2/2021 5

CRITERIA FOR IR ABSORPTION

•The frequency of IR radiation should exactly
match with the natural vibrational frequency
of the molecule.

•A molecule must undergo a net change in
dipole moment

Eg:HCl
δ+ δ-

H Cl

7

DIPOLE MOMENT

Only bonds which have significant dipole moments will absorb
IR radiation. R R

R C C R
Bonds which do not absorb IR include:
•Symmetrically substituted alkenes and alkynes R R

•Symmetric diatomic molecules C C
+ + d+

O
O

H-H Cl-Cl – – d-

Oscillating dipoles couple and
energy is transferred

STRONG ABSORBERS

d-
The carbonyl group is one O
of the strongest absorbers d+

C

Also O-H and C-O bonds O H C O

infrared beam

C C
+ + d+

O
O

– – d-

oscillating dipoles couple and
energy is transferred

Infrared Spectroscopy
•Atoms in a molecule do not maintain fixed positions -they
vibrateback and forth.

•The bond length is an average value.

•A diatomic molecule, e.g. H-Cl can only undergo a
STRETCHING vibration.

•More complex molecules exhibit both STRETCHING and
BENDING vibrations.

•Vibrational motion is excited within a singleelectronic
state by infrared light.

BOND VIBRATIONAL ENERGY LEVELS
Bonds do not have a fixed distance.
They vibrate continually even at 0oK (absolute).
The frequency for a given bond is a constant.
Vibrations are quantized as levels.
The lowest level is called the zero point energy.

Vibrational

E modes

n •Fundamental
Bond

e Vibration
dissociation

r bands

g energy
Vibrational •Overtone

y bands
energy levels

•Combination
bands

zero point energy

rmin rmax
distance

ravg (Average bond length)

Molecular vibrations
Two major types :

STRETCHING

C C

BENDING C

C
C

both of these types are “infrared active”
( excited by infrared radiation )

Molecular vibrations

Stretching- Symmetric
– Assymmetric

Bending –
In-plane- Scissoring, Rocking

Out of plane- Wagging, Twisting

3/2/2021 14

TYPES OF VIBRATION

Stretching Vibration Bending Vibration

A. Symmetrical Stretching 1. In plane Blending
B. Asymmetrical Stretching Scissoring and Rocking

2. Out plane Blending
Wagging and Twisting

MOLECULAR VIBRATIONS
Stretching vibrations : Symmetric and asymmetric

Symmetric Stretch Asymmetric Stretch Bend

Bending vibrations:
In-plane-Scissoring and Rocking
Out of plane- Wagging and Twisting

THE UNIT USED ON AN IR SPECTRUM IS
WAVENUMBERS ( n )

n = wavenumbers (cm-1)

1
n =

l(cm) l = wavelength (cm)

c = speed of light
n = frequency = nc

c = 3 x 1010 cm/sec

or

1 c cm/sec 1
n (= ) c = =

l l cm sec

Wavenumbers is directly proportional to frequency

Plot of I.R. Spectra: % T vs Wavenumber

%
T

Wave number

ν= c / λ =c x n
ν =frequency
c = speed of light = 3 x 1010 cm/sec
λ= wavelength (cm)
ν =Wavenumber in cm-1 18

Compress : Bond length
decreases

m1 m2

Stretching : Bond length
Increases

Molecule as per Hooke’s Law

The energy of a vibrational level is given by the equation

E =(v+1/2) hν
Where v+ quantum number of the energy level involved in the transition

Mostly single energy level transitions take place in IR spectroscopy
v0-v1, v1-v2 and so on

For a transition from v0-v1

∆ E = E1-E0 Overtone bandsv2

v1

E0= (0+1/2) hv = ½ hv v0
E1 =(1+1/2) hv = 3/2 hv

∆ E = 3/2 hv -1/2 hv = hv Fundamental absorption

This transition results in absorption of energy and generation of
Fundamental absorption bands in IR spectroscopy
Fundamental absorption bands are the major bands observed for each
functional group in the molecule

Higher Order transitions are also possible

v0-v2, v0-v3 and so on

For a transition from v0-v2

∆ E = E2-E0

E0= (0+1/2) hv = ½ hv
E2 =(2+1/2) hv = 5/2 hv

∆ E = 5/2 hv -1/2 hv = 2hv

This transition results in requires higher energy and results in the generation
of Overtone bands ( Overtone of the Fundamental absorption band)in IR
spectroscopy

Overtone absorption bands are rarely observed in IR spectroscopy as they
require higher energy and are observed at a multiple of the frequency of
the fundamental absorption band

THE EQUATION OF A n = frequency
in cm-1

SIMPLE HARMONIC
OSCILLATOR

HOOKE’S LAW c = velocity of light

( 3 x 1010 cm/sec )

1 K
n = K = force constant

2pc m in dynes/cm

where

m C C > C C > C C
1 m2

m = multiple bonds have higher K’s
m1 + m2

m = atomic masses
This equation describes the
vibrations of a bond. m = reduced mass

HOOKE’S LAW Larger K, higher
frequency

1 K
n =

2pc m Larger atom masses,
where lower frequency
K = force constant

in dynes/cm increasing K
m1 m2

m
= m1 + m C==C > C=C > C-C

2

m= reduced mass 2150 1650 1200

m= atomic mass

increasing m

C-H > C-C > C-O > C-Cl > C-Br
3000 1200 1100 750 650

Vibrational modes

Each atom has 3 degrees of freedom

A molecule of n atoms has 3n degrees of freedom

Nonlinear molecules 3n-6 and linear molecules
3n-5 degree of freedom

3n-6 and 3n-5 are called normal modes

3/2/2021 24

Linear molecule:
CO2 exhibits 2 absorption peaks- at 2330/cm and
667/cm.

C=O stretching vibrations of aliphatic ketone is
1700/cm.

If no coupling interaction in CO2, absorption peak of
CO2 and aliphatic ketones will be the same.

3/2/2021 25

CO2 is a linear molecule.

It has 4 normal modes of vibration.

2 fundamental stretching vibrations-an
asymmetrical and a symmetrical .

2 fundamental bending vibrations-
scissoring vibrations.

3/2/2021 26

•Here interaction between the two stretching vibrations
can occur since the bonds involved are associated with a
common atom.

The symmetric vibration is IR inactive since there is no
change in dipole moment.

The asymmetric vibration causes a periodic change in
charge distribution and absorption at 2330/cm occurs

The bending vibrations are the resolved components at 900

to one another of the bending motion in all possible planes
around the bond axis.

The two vibrations are identical in energy& produce peak at
667/cm.

3/2/2021 27

Factors that reduces the expected
theoretical number of bands:

The symmetry of molecule is such that no change in dipole
results from a particular vibration.

The energies of 2or more vibrations are identical or nearly
identical.

Fundamental bands that are too weak to be observed.

The vibrational energy is in a wavelength region beyond the
range of the instrument. ie,those fall outside of 4000-400
cm-1

3/2/2021 28

More number of peaks than expected:

.Due to overtone bands

Combination of bands:

The frequency of combination bands is
approx. the sum or difference of the 2
fundamental frequencies.

 Encountered when a photon excites 2
vibrational modes simultaneously bands.

This occurs when a quantum of energy is
absorbed by 2 bonds rather than one.

3/2/2021 29

Vibrational coupling (Fermi Resonance)

Coupling between two fundamental vibrations

C-H STRETCHING VIBRATION : Methyl and Methylene
H H H H H

C H
C C C

H
Symmetric Stretch ~2872 cm-1 H H

Symmetric Stretch ~2853 cm-1
H

C H H H H H
C

H C C
H H

Asymmetric Stretch ~2962 cm-1
Asymmetric Stretch ~2962 cm-1

Also observed incase of NO2 and NH2 groups

CO2: Coupling of two C=O stretching vibrations, C=O stretch at 2350 cm-1
O O

Carboxylic anhydrides: O O

CH3 C C CH
CH3 C C CH 3

3
O

O

Symmetric Stretch ~1760cm-1 Asymmetric Stretch~1820cm-1

Vibrational coupling (Fermi Resonance)

Amides: O H

R C N R

Amide-I band :C=O stretch
C-N stretch is of similar frequency to N-H bend. Interaction between these two
vibrations gives rise to two bands , one at higher ν and one at lower ν than
uncoupled vibration.
Amide II band: N-H bend at 1550-1510 cm -1
Amide III band : C-N stretch at 1250cm-1(w)

Coupling between a fundamental vibration and an overtone
O

Aldehydes:
CH3 C H

Fermi resonance between the Fundamental aldehydic C-H stretch(2850 cm-1) and
the first overtone of the aldehydic C-H bend(1390 cm-1) : Doublet between 2850-
2720 cm-1)

Typical IR Absorption Regions
• Functional Group region : 3600-1300cm-1

1.Hydrogen stretching : 3600-2700 cm-1

2. Triple Bond : 2500-2000 cm-1 3. Double Bond region: 2000-1500cm-1

• Finger print region : 1500-650 cm-1

O-H C-H C N C=O C=N C-Cl
Very C-O

N-H C C few C=C
C-N

bands C-C
N=O N=O *

4000 2500 2000 1800 1650 1500 650

FREQUENCY (cm-1)

BASE VALUES These are

(+/- 10 cm-1) the minimum
number of
values to
memorize.

O-H 3600
N-H 3400
C-H 3000

C N 2250
C C 2150

C=O 1715
C=C 1650

C O ~1100 large range

The C-H stretching region
BASE VALUE = 3000 cm-1

UNSATURATED

•C-H sp stretch in Alkynes ~ 3300 cm-1

•C-H sp2stretch C=C-H in Alkenes/Aromatics> 3000 cm-1

3000 divides

•C-H sp3 stretch in Alkanes < 3000 cm-1

SATURATED

•C-H aldehyde, two peaks
~ 2850 and 2750 cm-1

STRONGER BONDS HAVE LARGER FORCE CONSTANTS
AND ABSORB AT HIGHER FREQUENCIES

Increasing frequency (cm-1)

3300 3100 3000 2900 2850 2750

==C-H =C-H -C-H -CH=O
(weak)

sp sp2 sp3 aldehyde

Increasing s character in bond

Increasing CH Bond Strength

Increasing force constant K

CH BASE VALUE = 3000 cm-1

C-H STRETCHING VIBRATIONS
METHYL GROUP METHYLENE GROUP

Three C-H bonds share a central carbon Two C-H bonds share a central carbon
(hydrogens attached to the same carbon) H H H H

H C C C
C H H H

H Symmetric Stretch ~2853 cm-1

Symmetric Stretch ~2872 cm-1 H H H H
H C C C

C H H H

H
Asymmetric Stretch ~2926 cm-1

Asymmetric Stretch ~2962 cm-1

METHYLENE GROUP BENDING VIBRATIONS

Scissoring Wagging

H ~1465 cm-1
H H H ~1250 cm-1

C C C
H H

H ~720 cm-1
H H H ~1250 cm-1

C C C
H H

Rocking Twisting

in-plane out-of-plane

Bending
Vibrations

ALKANE C – H stretching: 3000-2850 cm-1

C – H bending(scissoring) : 1450-1375 cm-1

C – H bending(rocking) : 725-720 cm-1

Alkenes
C–H alkene stretching vibrations : 3100-3000cm-1

C=C stretching vibrations are seen at 1680-1600cm-1

Out of plane C–H bending vibrations between 1000-650 cm-1

Alkynes
C≡C stretching vibrations : 2260 -2100 cm-1

C–H stretching vibrations : 3300 -3270 cm-1

C–H bending vibrations : 700-610 cm-1

Aromatics
Aromatic C–H stretching :3100-3000 cm-1

Weak combination and overtone bands :
2000-1665 cm-1

Ring C–H oop bend: 900-675 cm-1

Aromatic C–H stretching :3100-3000 cm-1
AROMATIC

Weak combination and overtone bands : 2000-1665 cm-1

Ring C–H oop bend: 900-675 cm-1

Toluene

str
Ar-H str CH3

C=C
benzene

CH
3

Ar-H oop

The O-H stretching region

• O-H 3600 cm-1 (alcohol, free)

• O-H 3300 cm-1 (alcohols & acids,

H-bonding)
broadens

shifts

FREE H-BONDED

3600 3300

HYDROGEN-BONDED HYDROXYL“FREE” HYDROXYL

Many kinds of OH bonds of different
Distinct bond has a

lengths and strengths
well-defined length

This leads to a broad absorption.
R R and strength.

O H O CCl4
H H CCl4

R O
R O O R O CCl4

H
H R H

O CCl4
R H CCl4

Longer bonds are weaker and lead to
lower frequency.

Solvent molecules
Hydrogen bonding occurs in concentrated surround but do not
solutions (undiluted alcohol ) hydrogen bond.

Occurs in dilute solutions of alcohol /
in an “inert” solvent like CCl4.

Alcohols
O–H stretching vibrations( Dilute samples): ~ 3600cm-1 ALCOHOL
O–H stretching vibrations( Neat samples): 3400-3200 cm-1

C–O stretching vibrations occur at 1260 -1000 cm-1

O–H bending vibrations occur at 1420-1330 cm-1

Carboxylic aids
O–H stretching: 3300-2700( broad band)
C=O absorption bands at 1720 -1670 cm-1

C – O stretching occur at 1320-1210cm-1

O–H bending at 1440-1395 cm-1

The N-H stretching region
N-H 3400 -3250 cm-1

• Primary amines give two peaks

H H
N N

H H
Symmetric, 3330-3250 cm-1 Asymmetric, 3400- 3300cm-1

• Secondary amines give one peak 3350-3310cm-1

• Tertiary amines give no peak
• N–H bending vibrations occur in the 1650-1580 cm-1

• C –N stretching: 1250-1020 cm-1

The Triple bond stretching region: 2500-2000cm-1

• C == N 2260-2220 cm-1 (2250 cm-1)
• C = C 2260-2120 cm-1 (2150 cm-1

= )

•The cyano group : Strong, sharp peak due to its large
dipole moment.
•The carbon-carbon triple bond: Sharp peak, but it is
often weak due to a lack of dipole.
•Symmetric alkynes: IR inactive due to center of a
symmetry.

R C C R

THE CARBONYL STRETCHING REGION

• Amides Carboxylic acids Aldehydes Ketones Esters
• Acid Chlorides Anhydrides

• 1800 to 1650 cm-1 – Right In The Middle Of The
Spectrum

• The base value is 1715 cm-1 (ketone)
• The bands are very strong !!! due to large C=O

dipole moment.
• C=O is often one of the strongest peaks in the

spectrum

C=O IS SENSITIVE TO ITS ENVIRONMENT

Each different kind of C=O absorbs at a different frequency
Acid Carboxylic
chloride Ester Aldehyde Ketone acid Amide

O O O O O O

R C R C R C R C R C R C
Cl OR’ H R OH NH2

1800 1735 1725 1715 1710 1690

Anhydride

O O BASE

R C VALUE
O C R

1810 and 1760 THESE VALUES ARE
( Two peaks ) WORTH LEARNING

all are +/- 10 cm-1

FACTORS THAT INFLUENCE THE C=O ABSORPTION

•INDUCTIVE EFFECT
O

Electron-donating groups
R C

A weaken the carbonyl

Lower its absorption frequency
R = Me, Et, etc.
Aldehydes , Ketones

O
Electron-withdrawing groups

B
C X strengthen the carbonyl

Raises its absorption frequency
X = F, Cl, Br, OR
Acid halides, Esters

O

R C

RESONANCOE EFFECTS and HYDROGEN BONDING
C X

–
O O

+ Resonance
C Y C Y weakens the carbonyl

Y = N, O, or C=C, Lowers its absorption frequency
C Amides (Note the lengthening of

the C=O bond! )

O H O Hydrogen bonding

D lengthens and weakens

R C the C=O bond

Lowers its absorption frequency
R

HOW THESE FACTORS AFFECT C=O FREQUENCY
STRETCHING VIBRATIONS

Acid Carboxylic
chloride Ester Aldehyde Ketone acid Amide

B A C
O O O O O O

R C R C R C R C R C R C
Cl OR’ H R OH NH2

1800 1735 1725 1715 1710 1690
D

Anhydride

O O
BASE

R C O C R VALUE

1810 and 1760
( Two peaks ) A E-donating C Resonance

B E-withdrawing D H-bonding

CONFIRMATION OF FUNCTIONAL GROUP

Every type of carbonyl compound shows other peaks
you can look to confirm your conclusion based on C=O
frequency alone.

=O at 1710 cm-1
O C=O at 1725 cm-1 O C

Also look for OH
Also look for aldehyde CH

R C R C (H-bonded, lower
2850and 2750 cm-1

H ncy) and
2851(unsym doublet) O H Freque

C-O ~1200 cm-1

O C=O at 1690 cm-1 O C=O at 1735 cm-1

R C R C
Also look for Also look for two

N H
NH peaks at O R’ C-O at 1200 and

H 3400-3300 cm-1 1000 cm-1

Ketones have C=O at 1715 cm-1 and no NH, OH, C-O or -CHO
Anhydrides have two C=O peaks near 1800 cm-1 and two C-O

CARBOXYLIC ACID DIMER

Lowers
frequency
of C=O

O H O
C R and also

R C of O-H

O H O

Strong hydrogen-bonding in the dimer weakens the O-H and
C=O bonds and leads to broad peaks at lower frequencies.

Aldehyedes
Aldehyde C=O stretching:1740-1720 cm-1

C–H stretching: Doublet between 2850-2720
cm-1

Esters
Ester C=O stretching:1750-1735 cm-1

C–O stretching: 1300-1000 cm-1

Conjugation of C=O with C=C

• Conjugation of a carbonyl with a C=C bond shifts
values to lower frequencies

• For aldehydes, ketones and esters, subtract about 25-
30 cm-1 for conjugation with C=O

• Conjugated ketone = 1690 to 1680 cm-1

• Conjugated ester = 1710 to 1700 cm-1

CONJUGATION LOWERS THE FREQUENCY OF C=O
AND ALSO OF C=C

Resonance lengthens
(weakens) C=O

–
O

O
+

R C CH CH2 R C CH CH2

O O
C=C is also

R C R R C CH CH2 lengthened

….. and
1715 1690 cm-1 polarized !

lowered

CONJUGATION AND RING SIZE EFFECTS

O O O O O

1815 1780 1745 1715 1705 1690

RING STRAIN CONJUGATION
O O

R C R C
R CH CH2

O
Normal
aliphatic R C
ketones

The C=C stretching region (Alkenes and
Aromatics)

• C=C str: Weak band at 1675-1600 (1650 cm-1 )

• C=C benzene ring: Peak(s) near 1600 and 1400 cm-1

• When C=C is conjugated with C=O, it is stronger and
comes at a lower frequency.

Nitro compounds
The N=O stretching region
Sym N=O stretch- 1350 cm-1

Asymmetric N=O stretch- 1550 cm-1

Finger print region :The C-O stretching
region

• The C-O band appears in the range of 1300
to 1000 cm-1

• Look for one or more strong bands
appearing in this range!

• Ethers, alcohols, esters and carboxylic
acids have C-O bands

HALIDES: The C-X stretching region

• C-Cl : 785 to 540 cm-1, Often hard to find
amongst the fingerprint bands!!

• C-Br and C-I :Below 500 cm-1

• Outside the useful range of infrared
spectroscopy

• C-F bonds : 1000-1400 cm-1, can be found
easily, but are not that common.

=C-H OUT OF PLANE BENDING (OOP)

PLANE
above

H

ALKENES H

H
below

H Also with
BENZENES

BENZENES 10 11 12 13 14 15 m

Monosubstituted 730-770 s s

Disubstituted
ortho 770-735 s

meta 750-810 s 680-725 s

para 790-850 s

RING H’s
Trisubstituted OOPS

1,2,4 m s

1,2,3 s m

1,3,5 s m
YOU DO NOT NEED TO MEMORIZE
THE ALKENE AND AROMATIC OOP

1000 900 800 700 cm-1
ABSORPTION CHARTS !

YES NO MAKING
C=O present ?

DECISIONS

YES YES

Anhydride 2 C=O Peaks OH present ? Alcohol

Acid OH present ? NH present ? Amine

Amide NH present ? C-O present ? Ether

Ester C-O present ? C==N present ? Nitrile

Aldehyde CHO present ? C==C present ? Alkyne

Ketone C=C present ? Alkene
NO

(benzene ?) Aromatic

NO2 present ? Nitro

C-X present ? Halides

How to Use an Infrared Spectrum

•Molecular formula: Calculate index of hydrogen deficiency

•Check for carbonyl: Note any shift from 1715 cm-1

•Check for O-H, N-H

•Check for triple bonds

•Check for C=C, benzene rings

•Look below 1550 cm-1 check for C-O and Nitro group

Go back over spectrum for refinements
•Check the C-H region for aldehydes
and for peaks above 3000 cm-1 (alkenes , aromatics and terminal
alkynes)

Final Summary: The Minimum You Need To Know

BASE VALUES EXPANDED CH
3000

OH 3600
3300 3100 2900 2850

NH 3400 2750
C-H =C-H -C-H -CHO

CH 3000

CH2 and CH3 bend : 1465 and 1365

C N 2250
1800 1735 1725 1715 1710 1690

C C 2150 Aldehyde Acid
Acid Ester Ketone Amide

C=O 1715 chloride

Anhydride : 1810 and 1760
C=C 1650

EXPANDED C=O

C-O 1100 Benzene C=C : between 1400 and 1600

Remember also the effects of H-bonding, conjugation and ring size.

THANK YOU