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

4

The transmittance spectra provide better contrast between intensities of strong and weak bands compared to absorbance spectra.

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….. Important information about its chemical

nature and molecular structure

IR applicability for….. Analysis of organic materials Polyatomic inorganic molecules Organometallic compounds

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

infrared

microwave

visible

NEAR

MID

FAR

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

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

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

is a measure of the extent to which a separation exists between the centers of positive and negative charge within a molecule.

12

O

H Hδ+

δ-

δ+

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

A. Compound absorb in IR regionOrganic compounds, carbon monoxide

B. Compounds DO NOT absorb in IR regionO2, H2, N2, Cl2

14

Molecular vibration

divided into

stretching bending

back & forth movement

involves change in bond angles

symmetrical asymmetrical

scissoring

rocking twisting

wagging

in-plane vibration

out of plane

vibration

15

STRETCHING

16

17

BENDING

18

1. Gases Using evacuated cylindrical cells

equipped with suitable windows.2. Liquid

sodium chloride windows. “neat” liquid

3. Solid Pellet (KBr) Mull

19

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.

20

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.

21

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.

22

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.

23

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.

24

What is MullA thick paste formed by grinding an insoluble solid with an inert liquid and used for studying spectra of the solid.

What is NujolA trade name for a heavy medicinal liquid paraffin. Extensively used as a mulling agent in spectroscopy.

25

26

Dispersive spectrometerssequential mode

Fourier Transform spectrometerssimultaneous analysis of the full spectra range using inferometry.

27

Important components in IR dispersive spectrometer

28

sourcelamp

sampleholder

λselector

detectorsignal processor

& readout

1 2 3 4 5

Source:

- Nernst glower

- Globar source

- Incandescent wire

- Nichrome wire

Detector:

- Thermocouple

- Pyroelectric transducer

- Thermal transducer

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.

29

30

31

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 interferometer.

32

33

34

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.

35

FTCalculations

interferograms IR spectrum

36

Majority of commercially available FTIR instruments are based upon Michelson interferometer.

1

3

2

4

56

High sensitivity. High resolution. Quick data acquisition ( data for an

entire spectrum can be obtained in 1 s or less).

37

Advantages FTIR

38

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.

39

40

41

42

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

1. Begin by looking in the region from 4000-1300. Look at the C–H stretching bands around 3000.

How to analyze IR spectra

43

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)

44

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

45

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

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.

46

How to analyze IR spectra

47

48

49

C C

H

H

H

H

H C H

H

H

n

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

50

51

C C

H H

H H

=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.

ALKENE

52

ALKYNE

HC CH

53

CH

C C

Stretch for sp C - H occurs near 3300 cm-1.

Stretch occurs near 2150 cm-1; conjugation moves stretch to lower frequency.

ALKYNE

54

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.

55

AROMATIC RINGS

56

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.

57

Bending observed as one strong band near 750 cm-1.C H

Ortho-Disubstituted rings

58

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

59

Para- Disubstituted rings

- one strong band appears in the region from 800 to 850 cm-1.

C H

60

ALCOHOL

C

CH3

CH3

H3C OH

Primary alcohol 10

Secondary alcohol 20

Tertiary alcohol 30

C C OH

H

H

H

H

HC C C

H

H

H

OH

H

H

H

H

61

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.

62

PHENOL

OH

63

PHENOL

64

65

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.

66

67

CARBONYL COMPOUNDS

cm-1

1810 1800 1760 1735 1725 1715 1710 1690Anhydride 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.

68

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=Oand 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.

69

70

KETONEC R'R

OC R'R

O

C R'Ar

O

C=O stretch appear in range 1720-1708 cm-1 for normal aliphatic ketonesConjugation of C=O with phenyl at 1700 – 1680 cm-1 for C=O and 1600 – 1450 cm-1 for ring (C=C)

71

72

CARBOXYLIC ACID

R C OH

O

73

74

ESTERR C

O

O R

R C

O

O R

Ar C

O

O R

C – O Stretch in two or more bands, one stronger and one broader than the other, occurs in the range 1300 – 1000 cm-1

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=Oand 1600 – 1450 cm-1 for ring (C=C)

75

76

AMIDE

R C N

OH

HR C N

OH

RR C N

OR

R1020

30

77

AMIDE

78

79

C Cl

O

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.

80

C O C

O O

RR

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

R NH

H

RHN R

R N R

R

81

Primary amine, 10

Secondary amine , 20

Tertiary amine, 30

82

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

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.

N – H

N – H

C – N

83

Secondary Amine

84

Aromatic Amine