Infrared Spectroscopy (IR) - Overview & Interpretation

By
Manoj Prajapati & Moumita Halder
M.S.(Pharm), Medicinal Chemistry
National Institute of Pharmaceutical Education And Research (NIPER)
S.A.S. Nagar, Punjab

Basic concepts
 IR Spectroscopy is a qualitative analytical technique that helps to
identify the functional groups in a molecule.
 When the IR radiation irradiated to the molecule the part of
molecule which have functional group absorbs it, as a result of
this absorbance vibrational and rotational changes observed in
the molecule.
 After the excitation of molecular vibration, these molecules
comes back to their original state by releasing that energy of
certain wave number which is recorded as transmittance on the
spectrophotometer.

REGION WAVE
LENGTH
λ (μm)
WAVE NUMBER
υ (cm-1)
FREQUENCY
RANGE
Hz
NEAR 0.78 - 2.5 12800 - 4000 3.8x1014-1.2x1014
MIDDLE 2.5 - 50 4000 - 200 1.2x1014 - 6x112
FAR 50 - 1000 200 -10 6x1012- 30x1011
MOST USED 2.5 - 15 4000 - 670 1.2x1014-2x1013
IR-Region: 12,800 - 10 cm-1
1.Near IR----Carbohydrates and Proteins
2.Middle IR-----Organic molecules—functional groups
3.Far IR—in-organic –co-ordination bonds & quaternary ammonium compounds

Characteristic of molecule
Correct wavelength of radiation
In order to absorb the electromagnetic radiation for a molecule the
frequency of the incident radiation matches the natural frequency of
the vibration, the IR photon is absorbed and the amplitude of the
vibration increases.
Change in dipole moment
A molecule is said to have an electric dipole when there is a slight
positive and a slight negative charge on its component of atom. The
dipole moment of the molecule must change as a result of a molecular
vibration. Symmetric molecules (or bonds) do not absorb IR radiation
since there is no dipole moment.

Characteristic of molecule
• If the dipole moment of a molecule would not change i.e.
as in Symmetric stretching the absorption spectra of radiation cannot be
obtained. Such spectra is called as Forbidden or Inactive IR Spectra.
•If the molecule vibrate asymmetrically, the change in its dipole moment
takes place so absorption spectra of this molecule can be obtained. This
is called Active IR spectra.
H
H
C
H
H
C

IR Spectrum
 There are two type of IR Spectra from which we can obtain the information
about the quality of molecule .
1. The Functional Group region: Identifies the functional group with the consequence
of changing stretching vibrations. Ranges from 4000 to 1600 cm-1.
2. The Fingerprint region: Identifies the exact molecule with the consequence of
changing bending vibrations. Ranges from 1600 to 625cm-1.
Focus your analysis on this region. Functional group region: This
is where most stretching frequencies appear.
Fingerprint region: complex and difficult to
interpret reliably.

How to analyze IR Spectrum
• Pay the most attention to
the strongest
absorptions:
• Pay more attention to the peaks to the left of the fingerprint
region (>1250 cm-1)
• Note the absence of certain peaks
– -C=O
– -OH
– -NH2
– -C≡N
– -NO2

The intensity of the IR “peaks” is
proportional to the change in
dipole moment that a bond
undergoes during a vibration.
C=O bonds absorb strongly.
C=C bonds generally absorb much
less.
Comparison of intensities of absorption peak of C=O and C=C

N-H absorption usually has one or
two sharp absorption bands of
lower intensity, O-H gives a broad
absorption peak
Comparison of shape of absorption peak of O-H and N-H

► 3600—3000cm-1
---OH, --NH2 , >NH, C-H.
► 3200—3000cm-1
C-H, Ar— C-H.
►3000—2500 cm-1
--C—H of methyl/methelene
asymmetric stre. --C—H, --COOH
►2300—2100 cm-1
Alkynes 2210---2100
Cyanides 2260—2200
Isocyanides 2280—2250
►1900—1650 cm-1
strong bands--- >c=o---1725—1760
anhydrides ----- 1850---1740
Imides ------ two broad band at 1700
Functional
group
region
11
General guidelines for IR

► 1650--1000cm-1
confirms ---
esters, alcohol, ethers. Nitro
► 1000—800 cm-1
C— Cl, C-Br
► 800—710cm-1
meta substituted benzene
► 770—730cm-1
strong mono substituted benzene.
► 710—665cm-1
ortho, Para, benzene.
Finger print
region
General guidelines for IR interpretation

Functional group Wavenumber (cm-1)
sp3 C-H str ~3000-2800
sp2 C-H str ~3100-3000
Aromatic C-H str 3150-3050
sp C-H str ~3300
O-H str ~3650-3200
S-H str 2600-2550
C=S str 1200-1050
S=O str 1050
O-H str in COOH ~3400
N-H str ~3500-3300
Aldehyde C-H str ~2700, ~2800
Characteristic IR Wavenumbers

Functional group Wavenumber (cm-1)
C=C isolated ~1640-1680
C=C conjugated ~1620-1640
C=C aromatic ~1600
C≡N 2260-2240
C≡C 2250-2100
C=O ester ~1730-1740
C=O aldehyde ~1740-1720
C=O ketone ~1725-1705
C=O carboxylic acid ~1725-1700
C=O amide ~1640-1680
Characteristic IR Wavenumbers

How to approach analysis of spectra
When analysing the spectra of unknown compound ,
concentrate first on determining the presence or absence
of few major functional group .
The C=O , O-H , C-O , C=C , C≡C , C≡N, and NO2 peaks are
the most conspicuous and give immediate structural
information if they are present.

1. Look if carbonyl group (C = O) is present ?
The C = O group gives strong absorption in the region 1820-1600cm-1
the peak is strong and medium width.
2. If C=O is present , check for the presence of the following groups
Acids Is O-H present?
Broad absorption near 3400cm-1
Amides Is N-H also present?
Medium absorption near 3400cm-1
Esters Is C-O also present?
Strong intensity absorptions near 1810 and 1760cm-1
Anhydride
s
Two C=O absorptions near 1810 and 1760cm-1
Aldehydes Is aldehyde C-H present?
Two weak absorptions near 2850 and 2750cm-1
Ketones The preceding five choices have been eliminated

3. If C=O is absent
Alcohols, Phenols
Check for O-H
Broad absorption near 3400-3300cm-1
Confirm this by finding C-O near 1300-1000cm-1
Amines
Check for N-H
Medium absorption near 3400cm-1
Ethers
Check for C-O near 1300-1000cm-1(absence of O-H near
3400cm-1)

4. Double bond and / or Aromatic ring :
(i) C=C is a weak absorption near 1650 cm-1 .
(ii) Medium to strong absorption in the region of 1600 - 1550cm-1
often imply an aromatic ring.
(iii) Confirm the double bond or aromatic ring by consulting the C-H
region, if C-H stretch occurs to the left of 3000cm-1 ,then it is
aromatic or vinyl . If C-H stretch occurs to right of 3000cm-1 then it
is aliphatic.

5. Triple bond :
(i) C ≡ N is a medium sharp peak near 2250cm-1
(ii) C ≡ C is a weak sharp peak near 2150 cm-1
(iii) Check also for acetylinic C-H near 3300cm-1 , which gives an
idea if the triple bond is placed terminally.

6. Nitro group :
Two strong absorption at 1600 - 1500cm-1 and 1390 - 1300cm-1
7. Hydrocarbon :
None of the preceding is found.
Major absorptions are in C-H region near 3000cm-1

IR spectrum of Alkanes
C–H stretch from 3000–2850 cm-1
C–H bend or scissoring from 1470-1450 cm-1
C–H rock, methyl from 1370-1350 cm-1
C–H rock, methyl, seen only in long chain alkanes, from 725-720 cm-1
IR spectrum of decane IR spectrum of n-hexane

IR spectrum of Alkenes
Alkenes show many more peaks than alkanes
(a) = C - H stretch for sp2 C-H occurs at region slightly greater than 3000cm-1
(b) = C - H out of plane (oop) bending occurs in ranges of 1000-650cm-1.
(b) Medium bands corresponding to the C=C bond stretching vibration at
about 1600-1700 cm-1

C-H out of plane bending (oop) absorbs at 1000 – 650 cm-1
Often very strong absorptions
 Can be used to determine type of substitution:
Monosubstituted gives two peaks near 990 and 910 cm-1
1,2-disubstituted (cis) gives one strong band near 700 cm-1
1,2-disubstitued (trans) gives band near 970 cm-1
A monosubstituted alkene gives two strong
peaks near 990 and 910 cm-1
A cis 1,2-disubstituted alkene gives one strong
band near 700 cm-1
C=C stretch is much less intense than for the
monosubstituted

C=C stretch occurs in region of 1670 – 1640 cm-1
Can be used to determine type of substitution
Symmetrically substituted does not absorb at all
A cis isomer absorbs more strongly than a trans isomer (cis is less symmetrical
than trans)
Simple monosubstituted absorbs at 1640 cm-1
Cis isomer – more intense C=C stretch
Single large peak at 700 cm-1 (indicates cis isomer)
Trans isomer – less intense C=C stretch
Band near 970 cm–1 (indicates trans isomer)

IR spectrum of Alkynes
•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 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.

Comparison between an unsymmetrical terminal alkyne (1-octyne) and
symmetrical internal alkyne (4-octyne)

IR spectrum of Nitriles
In a manner very similar to alkynes, nitriles show a prominent band around 2250 cm-1
caused by the 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.
C N

IR spectrum of Aromatics
C-H stretch occurs between 3050 and 3010 cm-1
C-H out-of-plane bending occurs at 900 – 690 cm-1
(useful for determining type of ring substitution)
C=C stretching often occurs in pairs at 1600 cm-1 and 1475 cm-1
Overtone and combination bands occur between 2000 and 1667 cm-1
Monosubstituted rings give strong absorptions at 690 cm-1 and 750 cm-1

Ortho substituted rings give one strong band at 750 cm-1
Meta substituted rings gives bands at 690 cm-1, 780 cm-1, and sometimes a third
band of medium intensity at 880 cm-1
Para substituted rings give one band from 800 to 850 cm-1

IR spectrum of 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.
Bond Frequency (cm-1) Intensity
O-H (free) 3600-3650 Weak
O-H (H bonded) 3200-3500 Medium, broad
C-O 1000-1250 Medium

Infrared Spectroscopy (IR) - Overview & Interpretation

IR spectrum of Amine
The most characteristic band in amines is due to the N-H bond stretch, and it appears as a
weak to medium, somewhat broad band (but not as broad as the O-H band of alcohols). This
band is positioned at the left end of the spectrum, in the range of about 3200 - 3600 cm-1.
Primary amines have two N-H bonds, therefore they typically show two spikes that make
this band resemble a molar tooth. Secondary amines have only one N-H bond, which makes
them show only one spike, resembling a canine tooth. Finally, tertiary amines have no N-H
bonds, and therefore this band is absent from the IR spectrum altogether.

Dibutyl amine – Secondary amine
N-methyl aniline – Secondary aromatic amine

IR spectrum of Aldehyde and Ketone
•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.

Comparison of spectra of aldehyde and ketone

Effect of conjugation in Aldehyde
1740 – 1725 cm-1 for normal aliphatic aldehyde
1700 – 1680 cm-1 for conjugation with double bond
1700 – 1660 cm-1 for conjugation with phenyl group
Conjugation decreases the C-O bond order and
therefore decreases the stretching frequency

OH O
O OH
H
O
Cyclobutanol 2-butanone Ethyl vinyl ether 2 methyl-2 propen 1-ol 2 methyl propanal
Usefulness of infrared absorption spectroscopy: Five C4H8O isomers

Carbonyl Compounds - Aldehydes and Ketones
Three factors are known to perturb the carbonyl stretching
frequency:
1) Conjugation with a double bond or benzene ring lowers the
stretching frequency.
O O
O
O
H
O
1716 cm-1 1685 cm-1 1716 cm-1 1683 cm-1 1678 cm-1

2) Incorporation of the carbonyl group in a small ring (5,4, or 3)
Raises the stretching frequency.
O
O
O
1748 cm-1 1783 cm-1 1850 cm-1

H3C
H
O
Cl3C
H
O
O O
H3CO
1729 cm-1 1768 cm-1 1683 cm-1 1674 cm-1
Electron withdrawing substituent Electron donating substituent
3) Changing an alkyl substituent of a ketone for an electron releasing
or withdrawing group.

IR spectrum of Carboxylic acid
A carboxylic acid functional group combines the features of alcohols and ketones because it
has both the O-H bond and the C=O bond.
Carboxylic acids show a very strong and broad band covering a wide range between 2800
and 3500 cm-1 for the O-H stretch.
They also show the stake-shaped band in the middle of the spectrum around 1710 cm-1
corresponding to the C=O stretch.

IR spectrum of Amide
The amide functional group combines the features of amines and ketones because it has
both the N-H bond and the C=O bond.
Amides show a very strong, somewhat broad band at the left end of the spectrum, in the
range between 3100 and 3500 cm-1 for the N-H stretch.
They also show the stake-shaped band in the middle of the spectrum around 1710 cm-1 for
the C=O stretch. As with amines, primary amides show two spikes, whereas secondary
amides show only one spike.

IR spectrum of Ester
Esters show a vey strong band for the C=O group that appears in the range of 1750-
1735cm-1 for simple aliphatic esters. The C=O band is shifted to lower frequency
when it is conjugated to C=C or phenyl group.

1750 – 1735 cm-1 for normal aliphatic esters
1740 – 1750 cm-1 if carbonyl carbon conjugated with an alkene
1740 – 1715 cm-1 if carbonyl carbon conjugated with aromatic
Effect of conjugation in Ester
Methyl benzoate – aromatic group adjacent to C=O group
Vinyl acetate – alkene group adjacent to C=O group

IR spectrum of Ether
Ether show prominent C-O stretching band at 1300 to 1000cm-1.
Absence of C=O and O-H is required to ensure that C-O stretch is
not due to an ester and phenol .phenyl alkyl ether gives two strong
bands at 1250cm-1 and 1040cm-1.
Alkyl Ether Aryl alkyl Ether

APPLICATIONS OF IR SPECTROSCOPY
i. Identification of functional group and structure elucidation
ii. Identification of substances
iii. Studying the progress of the reaction
iv. Detection of impurities
v. Quantitative analysis
Quantitative IR absorption methods differ somewhat from ultraviolet/visible molecular
spectroscopic methods because of the greater complexity of the spectra, the narrowness of
the absorption bands, and the instrumental limitations of infrared instruments.
Quantitative data obtained with IR instruments are generally significantly inferior in quality
to data obtained with ultraviolet/visible spectrophotometers.

Tips for interpretation of IR for unknown structure
 Always place relines to negative information evidence i.e., absence of band
at
1900 cm-1---1600 cm-1----absence of >C=O, >CHO
 Always starts from higher frequency end of the spectrum.
 Absence of band at 880 cm-1—650 cm-1 indicates absence of aromatic ring.
 For easy identification go for fingerprint and functional group region.
 Finger print region range is 1400 cm-1--900 cm-1. In this region if
absorbance band is present the groups esters, alcohols, ethers, nitro are
Confirmed.
 Functional region range is 4000 cm-1---1400 cm-1.amines, alcohols, aromatic
rings, carboxylic acids, alkynes, alkanes, alkenes, anhydrides, imides, etc,
may be confirmed.
 Stretching vibrations at 4000 cm-1----600 cm-1.
 Bending vibrations at 1500 cm-1-----500 cm-1.

Example for interpretation of IR for known structure
Acetaminophen
(4-acetamido-Phenol)
A. N-H Amide----3360 cm -1 .
B. Phenolic—OH -- 3000 cm -1 --3500 cm -1
C. C—H Stretching---3000 cm-1 .
D. Aromatic overtone ----1840 cm-1 --1940 cm -1
E. >C=O Amide stretching -----1650 cm -1
F. Aromatic C=C stretching--- 1608 cm -1 .
G. N-H Amide bending ----1568 cm -1
H. Aromatic C=C stretching ----1510 cm -1 .
I. >C—H bending --------810 cm -1
HN
OH
C
O
CH3
A
B
C
D
E
F
G
H
I

Infrared Spectroscopy
Common Functional Groups
Questions

A) benzyl alcohol
B) 2,4,6-cycloheptaheptatrien-1-one
C) acetophenone
D) benzaldehyde
E) phenylacetic acid
C7H6O
Question #1
Identify the compound from the IR above.

Question #2
C10H12O
A) 2,4,5-trimethylbenzaldehyde
B) p-allylanisole
C) 2-allyl-4methylphenol
D) 1-phenyl-2-butanone

Question #3
C3H4O
A) cyclopropanone
B) propynol
C) acrylaldehyde
D) propenoic acid

Question #4
A) methylbenzoate
B) o-hydroxyacetophenone
C) o-toluic acid
D) p-anisaldehyde
C8H8O2

T H A N K
Y O U
In case any query, contact at
Email: prajapatidaksha13@gmail.com

Infrared Spectroscopy (IR) - Overview & Interpretation

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