CHM260 - IR


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CHM260 - IR

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  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. The transmittancespectra providebetter contrastbetween intensitiesof strong and weakbands compared toabsorbancespectra. 4
  4. 4. Energy of IR photon insufficient to cause electronicexcitation but can cause vibrational excitation 5
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  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-regionsA.Near IR region (Nearest to the visible) 780 nm to 2.5 μm (12,800 to 4000 cm-1) le visi b NB. Mid IR region E A 2.5 to 50 μm (4000 – 200 cm-1) R M infrared I DC. Far IR region F 50 to 1000 μm (200 – 10cm-1) A R e w av cro mi 8
  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 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 aseparation exists between the centersof positive and negative charge withina molecule. δ- O δ+ H H δ+ 12
  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 distance of separation of the charge 13
  13. 13. A. Compound absorb in IR region Organic compounds, carbon monoxideB. Compounds DO NOT absorb in IR region O2, H2, N2, Cl2 14
  14. 14. Molecular vibration divided into back & forth involves change movement in bond angles stretching bending wagging scissoringsymmetrical asymmetrical rocking twisting out of in-plane plane vibration vibration 15
  15. 15. STRETCHING 16
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  17. 17. BENDING 18
  18. 18. 1. Gases  Using evacuated cylindrical cells equipped with suitable windows.1. Liquid  sodium chloride windows.  “neat” liquid1. Solid  Pellet (KBr)  Mull 19
  19. 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
  20. 20. What is meant by “neat” liquid?Neat liquid is a pure liquid that do not containany solvent or water.Neat liquid method is applied when the amountof liquid is small or when a suitable solvent isunavailable. 21
  21. 21. There are 2 ways to prepare solidsample 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 CCl 4 or any other IR solvents can be prepared either by KBr pellet or Mulls. 22
  22. 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
  23. 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
  24. 24. 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. 25
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  26. 26. Dispersive spectrometers sequential modeFourier Transform spectrometers simultaneous analysis of the full spectra range using inferometry. 27
  27. 27. Important components in IR dispersivespectrometer 1 2 3 4 5source sample λ signal processor detector & readout lamp holder selector Detector:Source: - Thermocouple - Nernst glower - Pyroelectric transducer - Globar source - Thermal transducer - Incandescent wire - Nichrome wire 28
  28. 28.  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 P 0 and P. 29
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  31. 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 32
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  34. 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. FT Calculationsinterferograms IR spectrum 35
  35. 35. Majority of commercially available FTIR instrumentsare based upon Michelson interferometer. 3 4 1 5 2 6 36
  36. 36. Advantages FTIR High sensitivity. High resolution. Quick data acquisition ( data for an entire spectrum can be obtained in 1 s or less). 37
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  38. 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
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  41. 41. How to analyze IR spectra1. Begin by looking in the region from 4000-1300. Look at the C–H stretching bands around 3000. IndicatesAre any or all to the right alkyl groups (present inof 3000? most organic molecules)Are any or all to the left of a C=C bond or aromatic3000? group in the molecule 42
  42. 42. 2. Look for a carbonyl in the region 1760-1690. If there is such a band: Indicates a carboxylic acid Is an O–H band also present? group Is a C–O band also present? an ester Is an aldehyde C–H band also an aldehyde present? 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 tothe type of carbonyl compound it is) 43
  43. 43. 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 phenol4. 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 44
  44. 44. 5. Other structural features to check for Indicates an ether (or an ester if thereAre there C–O stretches? is a carbonyl band too)Is there a C=C stretching an alkeneband?Are there aromatic an aromaticstretching bands?Is there a C≡C band? an alkyneAre there -NO2 bands? a nitro compound 45
  45. 45. 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 46 spectrum.
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  48. 48. H H HH C C C H H H H n 49
  49. 49. CH Stretch for sp3 C-H around 3000 – 2840 cm-1.CH 2 Methylene groups have a characteristic bending absorption at approximate 1465 cm-1CH 3 Methyl groups have a characteristic bending absorption atapproximate 1375 cm-1CH 2 The bending (rocking) motion associated with four or more CH2 groups in an open chain occurs at about 720 cm -1 50
  50. 50. H HC CH H 51
  51. 51. 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 -1C=C stretch occurs at 1660 – 1600 cm-1; often conjugation moves C=C stretch to lower frequencies and increases the intensity. 52
  52. 52. ALKYNEHC CH 53
  53. 53. ALKYNE CH Stretch for sp C - H occurs near 3300 cm-1.C C Stretch occurs near 2150 cm-1; conjugation moves stretch to lower frequency. 54
  54. 54. AROMATIC RINGSC H Stretch for sp2 C-H occurs at values greater than 3000 cm-1. Ring stretch absorptions occur in pairs at 1600 cm-1 andC C 1475 cm-1. C H Bending occurs at 900 - 690cm-1. 55
  55. 55. AROMATIC RINGS 56
  56. 56. C-H Bending ( for AromaticRing)The out-of-plane (oop) C-H bending is useful in order to assign thepositions of substituents on the aromatic ring.Monosubstituted rings•this substitution pattern always gives a strong absorption near 690cm-1. If this band is absent, no monosubstituted ring is present. Asecond 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. Athird band of medium intensity is often found near 880 cm -1.Para- Disubstituted rings 57- one strong band appears in the region from 800 to 850 cm -1.
  57. 57. Ortho-Disubstituted rings C H Bending observed as one strong band near 750 cm-1. 58
  58. 58. Meta- Disubstituted rings - gives one absorption band near 690 cm-1 plus one near 780C H cm-1. A third band of medium intensity is often found near 880 cm-1. 59
  59. 59. Para- Disubstituted ringsC H - one strong band appears in the region from 800 to 850 cm-1. 60
  60. 60. ALCOHOL H H H OH HH C C OH H C C C H H H H H HPrimary alcohol 10 Secondary alcohol 20 CH3 H3C C OH CH3 Tertiary alcohol 30 61
  61. 61. ALCOHOLO-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 thathas not been dissolved in a solvent (neat liquid).C-O-H Bending appears as a broad and weak peak at 1440 – 1220 cm-1often 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 tertiarystructure to an alcohol. 62
  62. 62. PHENOL OH 63
  63. 63. PHENOL 64
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  65. 65. ETHER R O RC-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
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  67. 67. CARBONYL COMPOUNDS cm-11810 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
  68. 68. ALDEHYDE R C H OR C H C=O stretch appear in range 1740-1725 cm-1 for O normal aliphatic aldehydesAr C H Conjugation of C=O with phenyl; 1700 – 1660 cm-1 for C=O 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. 69
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  70. 70. KETONE R C R O R C R C=O stretch appear in range 1720-1708 O cm-1 for normal aliphatic ketonesAr C R Conjugation of C=O with phenyl at 1700 – O 1680 cm-1 for C=O and 1600 – 1450 cm-1 for ring (C=C) 71
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  74. 74. ESTER R C O R OR C O R C=O stretch appear in range 1750-1735 cm-1 for O normal aliphatic estersAr C O R Conjugation of C=O with phenyl; 1740 – 1715 cm -1 O 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 75
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  76. 76. AMIDE O O O H H RR C N R C N R C N H R R 10 2 0 30 77
  77. 77. AMIDE 78
  78. 78. O R C Cl Stretch appear in range 1810 -1775 cm-1 in C O conjugated chlorides. Conjugation lowers the frequency to 1780 – 1760 cm-1 C Cl Stretch occurs in the range 730 -550 cm -1Acid chloride show a very strong band for the C=O group. 79
  79. 79. O O R C O C RC 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 80
  80. 80. H R N R Secondary amine , 20 H R N RR N H RPrimary amine, 10 Tertiary amine, 30 81
  81. 81. N–H 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 Bending in primary amines results in a broad band in the range 1640 – 1560 cm-1. Secondary amines absorb near 1500 cm-1N–H Out-of-plane bending absorption can sometimes be observed near 800 cm-1C–N Stretch occurs in the range 1350 – 1000 cm-1 82
  82. 82. Secondary Amine 83
  83. 83. Aromatic Amine 84