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2
 Mostly for qualitative analysis. 
 Absorption spectra is recorded as 
transmittance. 
 Absorption in the infrared regi...
4 
The transmittance 
spectra provide 
better contrast 
between intensities 
of strong and weak 
bands compared to 
absorb...
5 
Energy of IR photon insufficient to cause electronic 
excitation but can cause vibrational excitation
6
 Infrared (IR) spectroscopy deals with 
the interaction of infrared radiation with 
matter. 
 IR spectrum provides….. 
...
IR region subdivided into 3 sub-regions 
A.Near IR region (Nearest to the visible) 
780 nm to 2.5 μm (12,800 to 4000 cm-1)...
1. IR absorption only occurs when IR radiation 
interacts with a molecule undergoing a 
change in dipole moment as it vibr...
 Absorption of IR radiation corresponds to energy 
changes on the order of 8 to 40 kJ/mole. 
Radiation in this energy ra...
 NOT ALL bonds in a molecule are capable 
of absorbing IR energy. Only those bonds 
that have change in dipole moment are...
is a measure of the extent to which a 
separation exists between the centers 
of positive and negative charge within 
a mo...
 In heteronuclear diatomic molecule, 
because of the difference in 
electronegativities of the two atoms, one 
atom acqui...
A. Compound absorb in IR region 
Organic compounds, carbon 
monoxide 
B. Compounds DO NOT absorb in 
IR region 
O2, H2, N2...
A molecule can move via vibration, rotation 
and translation (3 degree of freedom) 
Polyatomic molecules containing N at...
2349 cm-1 
667 cm-1, 
degenerate energy 
3650 cm-1, 
Symmetric 
stretching 
3756 cm-1, 
Asymmetric 
stretching 
1595 cm-1,...
17 
Type of degree of freedom Linear Non 
linear 
Translational 3 3 
Rotational 2 3 
Vibrational 3N-5 3N-6 
Total 3N 3N
Molecular vibration 
divided into 
back & forth 
movement 
involves change 
in bond angles 
stretching bending 
symmetrica...
STRETCHING 
19
20
BENDING 
21
1. Gases 
 Using evacuated cylindrical cells 
equipped with suitable windows. 
1. Liquid 
 sodium chloride windows. 
 “...
 a drop of the pure (neat) liquid is squeezed 
between two rock-salt plates to give a layer 
that has thickness 0.01mm or...
What is meant by “neat” liquid? 
Neat liquid is a pure liquid that do not contain 
any solvent or water. 
Neat liquid meth...
There are 2 ways to prepare solid 
sample for IR spectroscopy. 
1. Solid that is soluble in solvent . The 
most commonly I...
KBr PELLET 
 The finely ground solid sample is mixed with 
potassium bromide (KBr). The mixture is 
pressed under high pr...
MULLS 
 2-5 mg finely powdered sample is ground 
(grind) together with the presence 1 or 2 
drops of a heavy hydrocarbon ...
What is Mull 
A thick paste formed by grinding an 
insoluble solid with an inert liquid and used 
for studying spectra of ...
29
Dispersive spectrometers 
sequential mode 
Fourier Transform spectrometers 
simultaneous analysis of the full spectra 
ran...
Important components in IR dispersive 
spectrometer 
31 
1 2 3 4 5 
source 
lamp 
sample 
holder 
λ 
selector 
detector 
s...
 Generate a beam with sufficient 
power in the λ region of interest to 
permit ready detection & measurement. 
 Provide ...
33
34
Why FTIR is developed? 
 To overcome limitations 
encountered with the 
dispersive instruments. 
 Dispersive IR 
spectro...
36
37
Interferometer 
 Special instrument which can read IR 
frequencies simultaneously. 
 Faster method than dispersive instr...
Majority of commercially available FTIR instruments 
are based upon Michelson interferometer. 
39 
1 
3 
2 
4 
5 
6
 High sensitivity. 
 High resolution. 
 Quick data acquisition ( data for an 
entire spectrum can be obtained in 1 
s o...
41
 IR spectrum is due to specific structural 
features, a specific bond, within the 
molecule, since the vibrational states...
43
Overtone and Combination Bands 
Overtone bands – are multiplies of the 
fundamental absorption frequency 
First overton...
Example: 
A molecule has strong fundamental bands as 
follows: 
C-H bending at 730cm-1 
C-C stretching at 1400 cm-1 
...
Fermi resonance 
-leads to two bands appearing close together 
when one is expected 
When an overtone or a combination ...
 Coupling 
 Give rise to the complexity of the IR spectrum 
 Vibration in the skeleton of the molecules become 
coupled...
48
49 
How to analyze IR spectra 
1. Begin by looking in the region from 
4000-1300. Look at the C–H stretching 
bands around...
50 
2. Look for a carbonyl in the region 
1760-1690. If there is such a band: 
Indicates 
Is an O–H band also present? 
a ...
51 
3. Look for a broad O–H band in the 
region 3500-3200 cm-1. If there is 
such a band: 
Indicates 
Is an O–H band prese...
52 
5. Other structural features to check for 
Indicates 
Are there C–O stretches? 
an ether (or an ester if there 
is a c...
How to analyze IR 
spectra 
 If there is an absence of major functional 
group bands in the region 4000-1300 cm-1 
(other...
54
55
56 
H 
H 
H C C 
C H 
H 
H 
H 
H 
n
CH Stretch for sp3 C-H around 3000 – 2840 cm-1. 
CH2 Methylene groups have a characteristic bending absorption 
at approxi...
58 
H H 
C C 
H H
ALKENE 
=C-H Stretch for sp2 C-H occurs at values greater than 3000 cm-1. 
=C-H out-of-plane (oop) bending occurs in the r...
ALKYNE 
HC CH 
60
CH 
C C 
ALKYNE 
Stretch for sp C - H occurs near 3300 cm-1. 
Stretch occurs near 2150 cm-1; conjugation moves stretch to ...
AROMATIC 
RINGS 
C H Stretch for sp2 C-H occurs at values greater than 3000 cm-1. 
C C Ring stretch absorptions occur in p...
AROMATIC 
RINGS 
63
C-H Bending ( for Aromatic 
Ring) 
The out-of-plane (oop) C-H bending is useful in order to assign the 
positions of subst...
Ortho-Disubstituted rings 
Bending observed as one strong C H band near 750 cm-1. 
65
Meta- Disubstituted rings 
- gives one absorption band near 690 cm-1 plus one near 780 
cm-1. A third band of medium inten...
Para- Disubstituted rings 
- one strong band appears in the region from 800 to 850 
cm-1. 
C H 
67
ALCOHOL 
CH3 
H3C C 
OH 
CH3 
H 
Primary alcohol 10 
H 
Secondary alcohol 20 
Tertiary alcohol 30 
C C OH 
H 
H 
H 
H 
C C...
ALCOHOL 
O-H The hydrogen-bonded O-H band is a broad peak at 3400 – 3300 cm-1. 
This band is usually the only one present ...
PHENOL 
OH 
70
PHENOL 
71
72
ETHER 
R O R' 
C-O The most prominent band is that due to C-O stretch, 
1300 – 1000 cm-1. 
Absence of C=O and O-H is requi...
74
CARBONYL 
COMPOUNDS 
cm-1 
1810 1800 1760 1735 1725 1715 1710 1690 
Anhydride Acid Chloride Anhydride Ester Aldehyde Keton...
ALDEHYDE 
R C 
O 
H 
R C 
O 
H 
Ar C 
O 
H 
C=O stretch appear in range 1740-1725 cm-1 for 
normal aliphatic aldehydes 
Co...
77
KETONE 
R C R' 
O 
R C R' 
O 
Ar C R' 
O 
C=O stretch appear in range 1720-1708 
cm-1 for normal aliphatic ketones 
Conjug...
79
CARBOXYLIC ACID 
R C OH 
O 
80
81
ESTER 
R C 
O 
O R 
R C 
O 
O R 
Ar C 
O 
O R 
C=O stretch appear in range 1750-1735 cm-1 for 
normal aliphatic esters 
Co...
83
AMIDE 
O 
R C N 
H 
H 
O 
R C N 
H 
R 
O 
R C N 
R 
10 R 20 30 
84
AMIDE 
85
86 
O 
C Cl 
R 
C O 
C Cl 
Stretch appear in range 1810 -1775 cm-1 in 
conjugated chlorides. Conjugation lowers the 
frequ...
87 
O O 
R R 
C O C 
C O Stretch always has two bands, 1830 -1800 cm-1 and 1775 – 
1740 cm-1, with variable relative inten...
R N 
H 
H 
R 
HN 
R 
R N R 
R 
88 
Primary amine, 10 
Secondary amine , 20 
Tertiary amine, 30
89 
Stretching occurs in the range 3500 – 3300 cm-1. 
Primary amines have two bands. 
Secondary amines have one band, a va...
90 
Secondary Amine
91 
Aromatic Amine
INFRARED SPECTROSCOPY
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INFRARED SPECTROSCOPY

INFRARED SPECTROSCOPY
-FUNCTIONAL GROUP DETECTION FROM IR SPECTRUM

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INFRARED SPECTROSCOPY

  1. 1. 2
  2. 2.  Mostly for qualitative analysis.  Absorption spectra is recorded as transmittance.  Absorption in the infrared region arise from molecular vibrational transitions  Absorption for every substance are at specific wavelengths where IR spectra provides more specific qualitative information.  IR spectra is called “fingerprints” because no other chemical species will have similar IR spectrum. 3
  3. 3. 4 The transmittance spectra provide better contrast between intensities of strong and weak bands compared to absorbance spectra.
  4. 4. 5 Energy of IR photon insufficient to cause electronic excitation but can cause vibrational excitation
  5. 5. 6
  6. 6.  Infrared (IR) spectroscopy deals with the interaction of infrared radiation with matter.  IR spectrum provides…..  Important information about its chemical nature and molecular structure  IR applicability for…..  Analysis of organic materials  Polyatomic inorganic molecules  Organometallic compounds 7
  7. 7. IR region subdivided into 3 sub-regions A.Near IR region (Nearest to the visible) 780 nm to 2.5 μm (12,800 to 4000 cm-1) B. Mid IR region 2.5 to 50 μm (4000 – 200 cm-1) C. Far IR region 50 to 1000 μm (200 – 10cm-1) 8 visible infrared microwave NE AR MI D F AR
  8. 8. 1. IR absorption only occurs when IR radiation interacts with a molecule undergoing a change in dipole moment as it vibrates or rotates. 2. Infrared absorption only occurs when the incoming IR photon has sufficient energy for the transition to the next allowed vibrational state. No absorption can occur if both rules above are not met. 9
  9. 9.  Absorption of IR radiation corresponds to energy changes on the order of 8 to 40 kJ/mole. Radiation in this energy range corresponds to stretching and bending vibrational frequencies of the bonds in most covalent molecules.  In the absorption process, those frequencies of IR radiation which match the natural vibrational frequencies of the molecule are absorbed.  The energy absorbed will increase the amplitude of the vibrational motions of the bonds in the molecule. 10
  10. 10.  NOT ALL bonds in a molecule are capable of absorbing IR energy. Only those bonds that have change in dipole moment are capable to absorb IR radiation.  The larger the dipole change, the stronger the intensity of the band in an IR spectrum. 11
  11. 11. is a measure of the extent to which a separation exists between the centers of positive and negative charge within a molecule. 12 O δ- δ+H H δ+
  12. 12.  In heteronuclear diatomic molecule, because of the difference in electronegativities of the two atoms, one atom acquires a small positive charge (δ+), the other a negative charge (δ-).  This molecule is then said to have a dipole moment whose magnitude, μ = qd 13 distance of separation of the charge
  13. 13. A. Compound absorb in IR region Organic compounds, carbon monoxide B. Compounds DO NOT absorb in IR region O2, H2, N2, Cl2 14
  14. 14. A molecule can move via vibration, rotation and translation (3 degree of freedom) Polyatomic molecules containing N atoms will have 3N degree of freedom Molecules containing 3 atoms, two groups of the triatomic molecules may be distinguished; linear and non linear Eg: CO2 (OCO) and H2O (HOH) 15
  15. 15. 2349 cm-1 667 cm-1, degenerate energy 3650 cm-1, Symmetric stretching 3756 cm-1, Asymmetric stretching 1595 cm-1, bending
  16. 16. 17 Type of degree of freedom Linear Non linear Translational 3 3 Rotational 2 3 Vibrational 3N-5 3N-6 Total 3N 3N
  17. 17. Molecular vibration divided into back & forth movement involves change in bond angles stretching bending symmetrical asymmetrical scissoring rocking twisting wagging in-plane vibration out of plane vibration 18
  18. 18. STRETCHING 19
  19. 19. 20
  20. 20. BENDING 21
  21. 21. 1. Gases  Using evacuated cylindrical cells equipped with suitable windows. 1. Liquid  sodium chloride windows.  “neat” liquid 1. Solid  Pellet (KBr)  Mull 22
  22. 22.  a drop of the pure (neat) liquid is squeezed between two rock-salt plates to give a layer that has thickness 0.01mm or less.  2 plates held together by capillary mounted in the beam path. 23
  23. 23. What is meant by “neat” liquid? Neat liquid is a pure liquid that do not contain any solvent or water. Neat liquid method is applied when the amount of liquid is small or when a suitable solvent is unavailable. 24
  24. 24. There are 2 ways to prepare solid sample for IR spectroscopy. 1. Solid that is soluble in solvent . The most commonly IR solvent is carbon tetrachloride, CCl4. 2. Solid that is insoluble in CCl4 or any other IR solvents can be prepared either by KBr pellet or Mulls. 25
  25. 25. KBr PELLET  The finely ground solid sample is mixed with potassium bromide (KBr). The mixture is pressed under high pressure (10,000 – 15,000 psi) in special die to form a pellet.  KBr pellet then can be inserted into a holder in the IR spectrometer. 26
  26. 26. MULLS  2-5 mg finely powdered sample is ground (grind) together with the presence 1 or 2 drops of a heavy hydrocarbon oil called Nujol to form a Mull.  Mull is then examined as a film between flat salt plates.  Mulls method is applied when solid not soluble in an IR transparent solvent and solid is not convenient to be pelleted with KBr. 27
  27. 27. What is Mull A thick paste formed by grinding an insoluble solid with an inert liquid and used for studying spectra of the solid. What is Nujol A trade name for a heavy medicinal liquid paraffin. Extensively used as a mulling agent in spectroscopy. 28
  28. 28. 29
  29. 29. Dispersive spectrometers sequential mode Fourier Transform spectrometers simultaneous analysis of the full spectra range using inferometry. 30
  30. 30. Important components in IR dispersive spectrometer 31 1 2 3 4 5 source lamp sample holder λ selector detector signal processor & readout Source: - Nernst glower - Globar source - Incandescent wire - Nichrome wire Detector: - Thermocouple - Pyroelectric transducer - Thermal transducer
  31. 31.  Generate a beam with sufficient power in the λ region of interest to permit ready detection & measurement.  Provide continuous radiation which made up of all λ’s with the region (continuum source).  Provide stable output for the period needed to measure both P0 and P. 32
  32. 32. 33
  33. 33. 34
  34. 34. Why FTIR is developed?  To overcome limitations encountered with the dispersive instruments.  Dispersive IR spectrophotometer has slow scanning speed due to measurement of individual molecules/atom.  It utilize the use of an 35
  35. 35. 36
  36. 36. 37
  37. 37. Interferometer  Special instrument which can read IR frequencies simultaneously.  Faster method than dispersive instrument.  Interferograms are transformed into frequency spectrums by using mathematical technique called Fourier Transformation. 38 FT Calculations interferograms IR spectrum
  38. 38. Majority of commercially available FTIR instruments are based upon Michelson interferometer. 39 1 3 2 4 5 6
  39. 39.  High sensitivity.  High resolution.  Quick data acquisition ( data for an entire spectrum can be obtained in 1 s or less). 40 Advantages FTIR
  40. 40. 41
  41. 41.  IR spectrum is due to specific structural features, a specific bond, within the molecule, since the vibrational states of individual bonds represent 1 vibrational transition.  From IR spectrum we could predict the present of atoms or group of atoms or functional groups such as the present of an O-H bond or a C=O or an aromatic ring. 42
  42. 42. 43
  43. 43. Overtone and Combination Bands Overtone bands – are multiplies of the fundamental absorption frequency First overtone band will appear at twice the wavenumber of the fundamental. Combination bands – arises when two fundamental bands absorbing at v1 and v2 absorbs energy simultaneously. The resulting band will appear at v1+v2 wavenumber 44
  44. 44. Example: A molecule has strong fundamental bands as follows: C-H bending at 730cm-1 C-C stretching at 1400 cm-1 C-H stretching at 2950 cm-1. Determine the wavenumbers of possible combination bands and the first overtones. 45
  45. 45. Fermi resonance -leads to two bands appearing close together when one is expected When an overtone or a combination band has the same frequency as or similar frequency to a fundamental, two bands appear, split at either side of the expected value and are of equal intensity – called as Fermi doublet 46
  46. 46.  Coupling  Give rise to the complexity of the IR spectrum  Vibration in the skeleton of the molecules become coupled  Bands can no longer be assigned to one bond.  Very common when adjacent bonds have similar frequencies.  Commonly occurs between C-C, C-O and C-N stretching, C-H rocking and C-H wagging  The vibrational mode is observed at different frequencies 47
  47. 47. 48
  48. 48. 49 How to analyze IR spectra 1. Begin by looking in the region from 4000-1300. Look at the C–H stretching bands around 3000. Indicates Are any or all to the right of 3000? alkyl groups (present in most organic molecules) Are any or all to the left of 3000? a C=C bond or aromatic group in the molecule
  49. 49. 50 2. Look for a carbonyl in the region 1760-1690. If there is such a band: Indicates Is an O–H band also present? a carboxylic acid group Is a C–O band also present? an ester Is an aldehyde C–H band also present? an aldehyde Is an N–H band also present? an amide Are none of the above present? a ketone (also check the exact position of the carbonyl band for clues as to the type of carbonyl compound it is)
  50. 50. 51 3. Look for a broad O–H band in the region 3500-3200 cm-1. If there is such a band: Indicates Is an O–H band present? an alcohol or phenol 4. Look for a single or double sharp N–H band in the region 3400-3250 cm-1. If there is such a band: Indicates Are there two bands? a primary amine Is there only one band? a secondary amine
  51. 51. 52 5. Other structural features to check for Indicates Are there C–O stretches? an ether (or an ester if there is a carbonyl band too) Is there a C=C stretching band? an alkene Are there aromatic stretching bands? an aromatic Is there a C≡C band? an alkyne Are there -NO2 bands? a nitro compound
  52. 52. How to analyze IR spectra  If there is an absence of major functional group bands in the region 4000-1300 cm-1 (other than C–H stretches), the compound is probably a strict hydrocarbon.  Also check the region from 900-650 cm-1. Aromatics, alkyl halides, carboxylic acids, amines, and amides show moderate or strong absorption bands (bending vibrations) in this region.  As a beginning student, you should not try to assign or interpret every peak in the spectrum. Concentrate on learning the major bands and recognizing their presence and absence in any given spectrum. 53
  53. 53. 54
  54. 54. 55
  55. 55. 56 H H H C C C H H H H H n
  56. 56. CH Stretch for sp3 C-H around 3000 – 2840 cm-1. CH2 Methylene groups have a characteristic bending absorption at approximate 1465 cm-1 CH3 Methyl groups have a characteristic bending absorption at approximate 1375 cm-1 CH2 The bending (rocking) motion associated with four or more CH2 groups in an open chain occurs at about 720 cm-1 57
  57. 57. 58 H H C C H H
  58. 58. ALKENE =C-H Stretch for sp2 C-H occurs at values greater than 3000 cm-1. =C-H out-of-plane (oop) bending occurs in the range 1000 – 650 cm-1 C=C stretch occurs at 1660 – 1600 cm-1; often conjugation moves C=C stretch to lower frequencies and increases the intensity. 59
  59. 59. ALKYNE HC CH 60
  60. 60. CH C C ALKYNE Stretch for sp C - H occurs near 3300 cm-1. Stretch occurs near 2150 cm-1; conjugation moves stretch to lower frequency. 61
  61. 61. AROMATIC RINGS C H Stretch for sp2 C-H occurs at values greater than 3000 cm-1. C C Ring stretch absorptions occur in pairs at 1600 cm-1 and 1475 cm-1. C H Bending occurs at 900 - 690cm-1. 62
  62. 62. AROMATIC RINGS 63
  63. 63. C-H Bending ( for Aromatic Ring) The out-of-plane (oop) C-H bending is useful in order to assign the positions of substituents on the aromatic ring. Monosubstituted rings •this substitution pattern always gives a strong absorption near 690 cm-1. If this band is absent, no monosubstituted ring is present. A second strong band usually appears near 750 cm-1. Ortho-Disubstituted rings •one strong band near 750 cm-1. Meta- Disubstituted rings •gives one absorption band near 690 cm-1 plus one near 780 cm-1. A third band of medium intensity is often found near 880 cm-1. Para- Disubstituted rings - one strong band appears in the region from 800 to 850 cm-1. 64
  64. 64. Ortho-Disubstituted rings Bending observed as one strong C H band near 750 cm-1. 65
  65. 65. Meta- Disubstituted rings - gives one absorption band near 690 cm-1 plus one near 780 cm-1. A third band of medium intensity is often found near 880 cm-1. C H 66
  66. 66. Para- Disubstituted rings - one strong band appears in the region from 800 to 850 cm-1. C H 67
  67. 67. ALCOHOL CH3 H3C C OH CH3 H Primary alcohol 10 H Secondary alcohol 20 Tertiary alcohol 30 C C OH H H H H C C C H H OH H H H H 68
  68. 68. ALCOHOL O-H The hydrogen-bonded O-H band is a broad peak at 3400 – 3300 cm-1. This band is usually the only one present in an alcohol that has not been dissolved in a solvent (neat liquid). C-O-H Bending appears as a broad and weak peak at 1440 – 1220 cm-1 often obscured by the CH3 bendings. C-O Stretching vibration usually occurs in the range 1260 – 1000 cm-1. This band can be used to assign a primary, secondary or tertiary structure to an alcohol. 69
  69. 69. PHENOL OH 70
  70. 70. PHENOL 71
  71. 71. 72
  72. 72. ETHER R O R' C-O The most prominent band is that due to C-O stretch, 1300 – 1000 cm-1. Absence of C=O and O-H is required to ensure that C-O stretch is not due to an ester or an alcohol. Phenyl alkyl ethers give two strong bands at about 1250 – 1040 cm-1, while aliphatic ethers give one strong band at about 1120 cm-1. 73
  73. 73. 74
  74. 74. CARBONYL COMPOUNDS cm-1 1810 1800 1760 1735 1725 1715 1710 1690 Anhydride Acid Chloride Anhydride Ester Aldehyde Ketone Carboxylic acid Amide (band 1) (band 2) Normal base values for the C=O stretching vibrations for carbonyl groups. 75
  75. 75. ALDEHYDE R C O H R C O H Ar C O H C=O stretch appear in range 1740-1725 cm-1 for normal aliphatic aldehydes Conjugation of C=O with phenyl; 1700 – 1660 cm-1 for C=O and 1600 – 1450 cm-1 for ring (C=C) C-H Stretch, aldehyde hydrogen (---CHO), consists of weak bands, one at 2860 - 2800 cm-1 and the other at 2760 – 2700 cm-1. 76
  76. 76. 77
  77. 77. KETONE R C R' O R C R' O Ar C R' O C=O stretch appear in range 1720-1708 cm-1 for normal aliphatic ketones Conjugation of C=O with phenyl at 1700 – 1680 cm-1 for C=O and 1600 – 1450 cm-1 for ring (C=C) 78
  78. 78. 79
  79. 79. CARBOXYLIC ACID R C OH O 80
  80. 80. 81
  81. 81. ESTER R C O O R R C O O R Ar C O O R C=O stretch appear in range 1750-1735 cm-1 for normal aliphatic esters Conjugation of C=O with phenyl; 1740 – 1715 cm-1 for C=O and 1600 – 1450 cm-1 for ring (C=C) C – O Stretch in two or more bands, one stronger and one broader than the other, occurs in the range 1300 – 1000 cm-1 82
  82. 82. 83
  83. 83. AMIDE O R C N H H O R C N H R O R C N R 10 R 20 30 84
  84. 84. AMIDE 85
  85. 85. 86 O C Cl R C O C Cl Stretch appear in range 1810 -1775 cm-1 in conjugated chlorides. Conjugation lowers the frequency to 1780 – 1760 cm-1 Stretch occurs in the range 730 -550 cm-1 Acid chloride show a very strong band for the C=O group.
  86. 86. 87 O O R R C O C C O Stretch always has two bands, 1830 -1800 cm-1 and 1775 – 1740 cm-1, with variable relative intensity. Conjugation moves the absorption to a lower frequency. Ring strain (cyclic anhydride) moves absorptions to a higher frequency. C O Stretch (multiple bands) occurs in the range 1300 -900 cm-1
  87. 87. R N H H R HN R R N R R 88 Primary amine, 10 Secondary amine , 20 Tertiary amine, 30
  88. 88. 89 Stretching occurs in the range 3500 – 3300 cm-1. Primary amines have two bands. Secondary amines have one band, a vanishingly weak one for aliphatic compounds and a stronger one for aromatic secondary amines. Tertiary amines have no N – H stretch. Out-of-plane bending absorption can sometimes be observed near 800 cm-1 Stretch occurs in the range 1350 – 1000 cm-1 N – H Bending in primary amines results in a broad band in the range 1640 – 1560 cm-1. Secondary amines absorb near 1500 cm-1 N – H N – H C – N
  89. 89. 90 Secondary Amine
  90. 90. 91 Aromatic Amine

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