Ir spectroscopy
- 2. Infrared spectroscopy • Mostly for qualitative analysis • Absorption spectra is recorded as transmittance spectra • Absorption in the infrared region arise from molecular vibrational transitions • Absorption at specific wavelengths • Thus, IR spectra provides more specific qualitative information • IR spectra is called “fingerprints” • - because no other chemical species will have identical IR spectrum 2
- 3. Comparison between transmittance (upper) vs absorbance (lower) plot 3 The transmittance spectra provide better contrast btw intensities of strong and weak bands compared to absorbance spectra
- 4. Electromagnetic Spectrum 4 Energy of IR photon insufficient to cause electronic excitation but can cause vibrational excitation
- 5. INTRODUCTION • Comparison between UV-vis and IR – Energy: UV > vis > IR – Frequency: UV > vis > IR – Wavelength: UV < vis < IR 5
- 6. INFRARED SPECTROSCOPY • 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: – Analysis of organic materials – Polyatomic inorganic molecules – Organometallic compounds 6
- 7. • IR region of EM spectrum: – λ: 780 nm – 1000 μm – Wavenumber: 12,800 – 10cm-1 • IR region subdivided into 3 subregions: 1. Near IR region (Nearest to the visible) - 780 nm to 2.5 μm (12,800 to 4000 cm-1) 2. Mid IR region - 2.5 to 50 μm (4000 – 200 cm-1) 3. Far IR region - 50 to 1000 μm (200 – 10cm-1) 7 infrared N E A R M I D F A R 7
- 8. • When IR absorption occur? 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 Note: If the 2 rules above are not met, no absorption can occur 8
- 9. What happen when a molecule absorbs infrared radiation? • 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. 9
- 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. 10
- 11. • What is a dipole moment? – is a measure of the extent to which a separation exists between the centers of positive and negative charge within a molecule. 11 O H H δ+ δ- δ+
- 12. • In heteronuclear diatomic molecule, because of the difference in electronegativities of the two atoms, one atom acquires a small positive charge (q+), the other a negative charge (q-). • This molecule is then said to have a dipole moment whose magnitude, μ =qd 12 distance of separation of the charge
- 13. Molecular Species That Absorb Infrared Radiation • Compound absorb in IR region Organic compounds, carbon monoxide • Compounds DO NOT absorb in IR region O2, H2, N2, Cl2 13
- 15. 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
- 16. STRETCHING 16
- 17. BENDING 17
- 18. Sample Handling Techniques • Gases – evacuated cylindrical cells equipped with suitable windows • Liquid – sodium chloride windows – “neat” liquid • Solid – Pellet (KBr) – Mull 18
- 19. LIQUID – 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 What is meant by “neat” liquid? • Neat liquid is a pure liquid that do not contain any solvent or water. – This method is applied when the amount of liquid is small or when a suitable solvent is unavailable 19
- 20. Solid sample preparation • There are three ways to prepare solid sample for IR spectroscopy. – Solid that is soluble in solvent can be dissolved in a solvent, most commonly carbon tetrachloride CCl4. – Solid that is insoluble in CCl4 or any other IR solvents can be prepared either by KBr pellet or mulls. 20
- 21. PELLETING (KBr PELLET) • Mixing the finely ground solid sample with potassium bromide (KBr) and pressing the mixture under high pressure (10,000 – 15,000 psi) in special dye. • KBr pellet can be inserted into a holder in the spectrometer. 21
- 22. MULLS • Formed by grinding 2-5 mg finely powdered sample, presence 1 or 2 drops of a heavy hydrocarbon oil (Nujol) • Mull examined as a film between flat salt plates • This method applied when solid not soluble in an IR transparent solvent, also not convenient pelleted in KBr 22
- 23. What is a 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 23
- 25. IR Instrument • Dispersive spectrometers – sequential mode • Fourier Transform spectrometers – simultaneous analysis of the full spectra range using inferometry 25
- 26. IR Instrument (Dispersive) • Important components in IR dispersive spectrometer 26 source lamp sample holder λ selector detector signal processor & readout 1 2 3 4 5 Source: - Nernst glower - Globar source - Incandescent wire Detector: - Thermocouple - Pyroelectric transducer - Thermal transducer
- 27. Radiation Sources • generate a beam with sufficient power in the λ region of interest to permit ready detection & measurement • provide continuous radiation; made up of all λ’s with the region (continuum source) • stable output for the period needed to measure both P0 and P 27
- 28. Schematic Diagram of a Double Beam Infrared Spectrophotometer 28
- 30. FTIR Why is it developed? – to overcome limitations encountered with the dispersive instruments – especially slow scanning speed; due to individual measurement of molecules/atom – utilize an interferometer 30
- 31. • 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 31 FT Calculations interferograms IR spectrum
- 32. Components of Fourier Transform Instrument 32 - majority of commercially available Fourier transform infrared instruments are based upon Michelson interferometer 1 3 2 4 5 6
- 33. Infrared Spectra • 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. • e.g. IR spectrum can tell the molecule has an O-H bond or a C=O or an aromatic ring 33
- 34. Infrared Spectroscopy • Infrared Spectroscopy – Energy Absorption and Vibration • IR electromagnetic radiation is just less energetic than visible light • This energy is sufficient to cause excitation of vibrational energy levels • Wavelength (l) = 2.5-16.7 x 10-6 m • n = wavenumbers. Larger n = higher energy • Excitation depends on atomic mass and how tightly they are bound – Hooke’s Law for 2 masses connected by a spring – C—H Bond: Reduced Mass = (12+1)/(12x1) = 13/12 = 1.08 – C—C Bond: Reduced Mass = (12+12)/(12x12) = 24/144 = 0.167 kcal/mol101cm4000-600 1~ 1 l 21 21 )(~ mm mm fk k = constant f = force constant = bond strength m-term = reduced mass sJ10x6.626ConstantsPlanck'h hc hc hE 1 m/s10x3.0λυc 34- 8 l l
- 35. Advantages (over dispersive instrument) – high sensitivity – high resolution – speed of data acquisition ( data for an entire spectrum can be obtained in 1 s or less) 35
- 37. Vibrational modes leading to IR absorptions: Many possible absorptions per molecule exist: stretching, bending,…
- 39. B.Using IR in Organic Chemistry 1.Functional Groups have characteristic IR absorptions 2.Fingerprint Region (600-1500 cm-1) is unique for every molecule and lets us match an unknown with a known spectrum
- 40. Regions of the Infrared Spectrum • 4000-2500 cm-1 N-H, C-H, O-H (stretching) – 3300-3600 N-H, O-H – 3000 C-H • 2500-2000 cm-1 CC and C N (stretching) • 2000-1500 cm-1 double bonds (stretching) – C=O 1680-1750 – C=C 1640-1680 cm-1 • Below 1500 cm-1 “fingerprint” region
- 41. 12.8 Infrared Spectra of Some Common Functional Groups Alkanes, Alkenes, Alkynes • C-H, C-C, C=C, C C have characteristic peaks – absence helps rule out C=C or C C
- 43. 4.IR of Alkenes a. Alkene C—H absorbs at higher energy than alkanes because the force constant is stronger than alkanes (sp2 hybridization) b.Substitution pattern of alkenes give characteristic absorptions » Terminal alkenes give 910, 990 cm-1 » Geminal disubstituted gives 890 cm-1 » trans disubstituted gives 970 cm-1 C C R H H H C C R H H R C C H H R R
- 47. IR: Aromatic Compounds • Weak C–H stretch at 3030 cm1 • Weak absorptions 1660 - 2000 cm1 range • Medium-intensity absorptions 1450 to 1600 cm1
- 49. IR: Alcohols and Amines • O–H 3400 to 3650 cm1 – Usually broad and intense • N–H 3300 to 3500 cm1 – Sharper and less intense than an O–H
- 52. IR: Carbonyl Compounds • Strong, sharp C=O peak 1670 to 1780 cm1 • Exact absorption characteristic of type of carbonyl compound – 1730 cm1 in saturated aldehydes – 1705 cm1 in aldehydes next to double bond or aromatic ring
- 53. C=O in Ketones • 1715 cm1 in six-membered ring and acyclic ketones • 1750 cm1 in 5-membered ring ketones • 1690 cm1 in ketones next to a double bond or an aromatic ring • 1735 cm1 in saturated esters • 1715 cm1 in esters next to aromatic ring or a double bond C=O in Esters
- 57. How to Interpret Infrared Spectra? 57
- 58. 58 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
- 59. 59 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)
- 60. 60 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
- 61. 61 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
- 62. 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. 62 How to analyze IR spectra
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- 66. ALKANE 66 C C H H H H H C H H H n
- 67. C-H Stretch for sp3 C-H around 3000 – 2840 cm-1. CH2 Methylene groups have a characteristic bending absorption at approx 1465 cm-1 CH3 Methyl groups have a characteristic bending absorption at approx 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 67
- 68. ALKENE 68 C C H H H H
- 69. =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 69
- 70. ALKYNE HC CH 70
- 71. 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 71
- 72. 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. 72
- 74. 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. 74
- 75. Bending observed as one strong band near 750 cm-1.C H Ortho-Disubstituted rings 75
- 76. 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 76
- 77. Para- Disubstituted rings - one strong band appears in the region from 800 to 850 cm-1.C H 77
- 78. ALCOHOL C CH3 CH3 H3C OH Primary alcohol 10 Secondary alcohol 20 Tertiary alcohol 30 C C OH H H H H H C C C H H H OH H H H H 78
- 79. 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. 79
- 80. PHENOL OH 80
- 81. PHENOL 81
- 82. 82
- 83. 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. 83
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- 85. CARBONYL COMPOUNDS cm-1 1810 1800 1760 1735 1725 1715 1710 1690 Anhydride Acid Anhydride Ester Aldehyde Ketone Carboxylic Amide (band 1) Chloride (band 2) acid Normal base values for the C=O stretching vibrations for carbonyl groups 85
- 86. A. 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. 86
- 87. 87
- 88. B. KETONE C R'R O C R'R O C R'Ar O C=O stretch appear in range 1720-1708 cm-1 for normal aliphatic ketones Conjugation of C=O with phenyl; 1700 – 1680 cm-1 for C=O and 1600 – 1450 cm-1 for ring (C=C) 88
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- 90. C. CARBOXYLIC ACID R C OH O 90
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- 92. D. ESTER R 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 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=O and 1600 – 1450 cm-1 for ring (C=C) 92
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- 94. E. AMIDE R C N O H H R C N O H R R C N O R R10 20 30 94
- 95. AMIDE 95
- 96. F. ACID CHLORIDE 96 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.
- 97. F. ANHYDRIDE 97 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
- 98. R N H H R H N R R N R R AMINE 98 10 20 30
- 99. AMINE 99 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