2 Learning OutcomesBy the end of this topic, students should be able to:Explain the principles and the working mechanism of infrared (IR) absorption spectroscopyIdentify the molecular species that absorb IR radiationInterpret IR spectrumExplain stretching and bending vibrations in relation to IR absorptionDetermine unknown qualitatively using IR absorptionDraw a schematic diagram of a conventional IR instrument and a fourier transform IR instrument and explain the function of each component of the instrumentDifferentiate between a dispersive IR instrument and a FTIR spectrometer
3 Infrared spectroscopy Mostly for qualitative analysisAbsorption spectra is recorded as transmittance spectraAbsorption in the infrared region arise from molecular vibrational transitionsAbsorption at specific wavelengthsThus, IR spectra provides more specific qualitative informationIR spectra is called “fingerprints”- because no other chemical species will have identical IR spectrum
4 Comparison between transmittance (upper) vs absorbance (lower) plot The transmittance spectra provide better contrast btw intensities of strong and weak bands compared to absorbance spectra
5 Electromagnetic Spectrum Energy of IR photon insufficient to cause electronic excitation but can cause vibrational excitation
6 INTRODUCTION Comparison between UV-vis and IR Energy: UV > vis > IRFrequency: UV > vis > IRWavelength: UV < vis < IR
7 INFRARED SPECTROSCOPY Infrared (IR) spectroscopy deals with the interaction of infrared radiation with matterIR spectrum provides:Important information about its chemical nature and molecular structureIR applicability:Analysis of organic materialsPolyatomic inorganic moleculesOrganometallic compounds
8 IR region of EM spectrum: λ: 780 nm – 1000 μm Wavenumber: 12,800 – 10cm-1IR 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)visibleNEARMIDinfraredFAR8microwave
9 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 stateNote: If the 2 rules above are not met, no absorption can occur
10 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.
11 NOT ALL bonds in a molecule are capable of absorbing IR energy 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.
12 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.δ-δ+δ+
13 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 =qddistance of separation of the charge
14 Molecular Species That Absorb Infrared Radiation Compound absorb in IR regionOrganic compounds, carbon monoxideCompounds DO NOT absorb in IR regionO2, H2, N2, Cl2
16 Molecular vibration divided into stretching bending wagging scissoring back & forth movementinvolves change in bond anglesstretchingbendingwaggingscissoringsymmetricalasymmetricalrockingtwistingin-plane vibrationout of plane vibration
20 LIQUIDa drop of the pure (neat) liquid is squeezed between two rock-salt plates to give a layer that has thickness 0.01mm or less2 plates held together by capillary mounted in the beam pathWhat 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
21 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.
22 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.
23 MULLSFormed 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 platesThis method applied when solid not soluble in an IR transparent solvent, also not convenient pelleted in KBr
24 What is a mull What is Nujol A thick paste formed by grinding an insoluble solid with an inert liquid and used for studying spectra of the solidWhat is NujolA trade name for a heavy medicinal liquid paraffin. Extensively used as a mulling agent in spectroscopy
26 IR Instrument Dispersive spectrometers Fourier Transform spectrometers sequential modeFourier Transform spectrometerssimultaneous analysis of the full spectra range using inferometry
27 IR Instrument (Dispersive) Important components in IR dispersive spectrometer51234signal processor& readoutsourcelampsampleholderλselectordetectorDetector:- Thermocouple- Pyroelectric transducer- Thermal transducerSource:- Nernst glower- Globar source- Incandescent wire
28 Radiation Sourcesgenerate a beam with sufficient power in the λ region of interest to permit ready detection & measurementprovide 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
29 Schematic Diagram of a Double Beam Infrared Spectrophotometer
31 FTIR Why is it developed? to overcome limitations encountered with the dispersive instrumentsespecially slow scanning speed; due to individual measurement of molecules/atomutilize an interferometer
32 InterferometerSpecial instrument which can read IR frequencies simultaneouslyfaster method than dispersive instrumentinterferograms are transformed into frequency spectrums by using mathematical technique called Fourier TransformationFTCalculationsinterferogramsIR spectrum
33 Components of Fourier Transform Instrument - majority of commercially available Fourier transform infrared instruments are based upon Michelson interferometer341256
34 Advantages (over dispersive instrument) high sensitivityhigh resolutionspeed of data acquisition ( data for an entire spectrum can be obtained in 1 s or less)
36 Infrared SpectraIR 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
39 How to analyze IR spectra Begin by looking in the region fromLook 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
40 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 groupIs a C–O band also present?an esterIs an aldehyde C–H band also present?an aldehydeIs an N–H band also present?an amideAre 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)
41 Indicates: Is an O–H band present? an alcohol or phenol Indicates: 3. Look for a broad O–H band in the region 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 cm-1.If there is such a band:Indicates:Are there two bands?a primary amineIs there only one band?a secondary amine
42 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 alkeneAre there aromatic stretching bands?an aromaticIs there a C≡C band?an alkyneAre there -NO2 bands?a nitro compound
43 How to analyze IR spectra If there is an absence of major functional group bands in the region cm-1 (other than C–H stretches), the compound is probably a strict hydrocarbon.Also check the region from 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.
47 C-H Stretch for sp3 C-H around 3000 – 2840 cm-1. CH2 Methylene groups have a characteristic bending absorption at approx 1465 cm-1CH3 Methyl groups have a characteristic bending absorption at approx 1375 cm-1CH2 The bending (rocking) motion associated with four or more CH2 groups in an open chain occurs at about 720 cm-1
49 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 frequenciesand increases the intensity
54 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 ringsthis 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 ringsone strong band near 750 cm-1.Meta- Disubstituted ringsgives 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.
55 Ortho-Disubstituted rings Bending observed as one strong band near 750 cm-1.
56 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.
57 Para- Disubstituted rings - one strong band appears in the region from 800 to 850 cm-1.
59 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 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.
63 ETHERC-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 stretchis 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.
65 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) acidNormal base values for the C=O stretching vibrations for carbonyl groups
66 A. ALDEHYDEC=O stretch appear in range cm-1 for normal aliphatic aldehydesConjugation 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 cm-1 andthe other at 2760 – 2700 cm-1.
72 D. ESTERC=O stretch appear in range cm-1 for normal aliphatic estersConjugation of C=O with phenyl; 1740 – 1715 cm-1 for C=Oand 1600 – 1450 cm-1 for ring (C=C)C – O Stretch in two or more bands, one stronger and broader than the other, occurs in the range 1300 – 1000 cm-1
76 Acid chloride show a very strong band for the C=O group. F. ACID CHLORIDEStretch appear in range cm-1 in conjugated chlorides. Conjugation lowers the frequency to 1780 – 1760 cm-1Stretch occurs in the range cm-1Acid chloride show a very strong band for the C=O group.
77 F. ANHYDRIDEStretch always has two bands, 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.Stretch (multiple bands) occurs in the range cm-1
79 AMINE 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 – HBending in primary amines results in a broad band in the range1640 – 1560 cm-1.Secondary amines absorb near 1500 cm-1N – HN – HOut-of-plane bending absorption can sometimes be observednear 800 cm-1Stretch occurs in the range 1350 – 1000 cm-1C – N