An IR spectrum is a plot of percent transmittance (or absorbance) against wavenumber (frequency or wavelength). The interpretation of IR Spectra helps in the characterization of the unknown organic compound.

1. Dr. Ashwani Dhingra
Associate Professor
GGSCOP, Yamunanagar

2. Basic concepts
• IR Spectroscopy is a qualitative
analytical technique that helps to
indicate mainly the functional group
of a molecule.
• When the IR radiation irradiated to
the molecule the part of molecule
which is functional absorbs it, as a
result of this absorbance the
molecular vibration increases.
• 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.

3. Regions of IR radiation

4. 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.
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 -
6x1012
FAR 50 - 1000 200 -10 6x1012-
30x1011
MOST
USED
2.5 - 15 4000 - 670 1.2x1014-
2x1013

5. 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.

6. 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

7. 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.

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

9. Comparison of
intensities of
absorption peak
of C=O and C=C
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.

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

11. General guidelines for IR interpretation
► 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

12. General guidelines for IR interpretation
► 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

13. Characteristic
IR Wave
numbers
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

14. 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 Wave
numbers

15. How to approach analysis of IR 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.

16. 1. Look if
carbonyl grou
p (C = O)
is present?
The C = O group gives strong absorption in the
region 1820-1600cm-1 the peak is strong and
medium width.
• 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
Anhydrides 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

17. 2. 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)

18. (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.
3. Double
bond and/or
aromatic
ring :

19. 4. 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.

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

21. 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

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

23. 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

24. 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)

25. 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.

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

27. 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

28. 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

29. 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

30. 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

32. 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.

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

34. 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.

35. Comparison of spectra of aldehyde and ketone

36. 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

37. 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

38. 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

39. 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

40. 3) Changing an alkyl substituent of a ketone for an electron
releasing or withdrawing group.
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

41. 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.

42. 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.

43. 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.

44. 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

45. 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

46. APPLICATIONS
OF IR
SPECTROSCOPY
• Identification of functional group
and structure elucidation
• Identification of substances
• Studying the progress of the
reaction
• Detection of impurities
• 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.

47.  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.
Tips for interpretation of IR for
unknown structure

48. Example for interpretation of IR for known
structure
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
Acetaminophen
(4-acetamido-Phenol)
B
H
G
A
E
C
D
F
I
HN
OH
C
O
CH3

49. Dr. Ashwani Dhingra
Associate Professor
GGSCOP, Yamunanagar