1 / 55

Dong

Dong. -. Sun Lee / cat. -. lab. /. SWU. 2010-Fall Version. Chapter 26 B. Molecular Absorption Spectrometry IR spectrometry. Infrared absorption spectrometry. 1) The IR regions of spectrum

thyra
Télécharger la présentation

Dong

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Dong - Sun Lee / cat - lab / SWU 2010-Fall Version Chapter 26 B Molecular Absorption Spectrometry IR spectrometry

  2. Infrared absorption spectrometry 1) The IR regions of spectrum Designation Wavelength Frequency(Hz) Wave number (cm–1) Transition Near IR 780~2500nm 1.2~3.8×1014 12,800 ~4,000 Molecular vibration Overtone region Mid IR 2.5~50m 6×1012 ~1.2×10144,000 ~ 200 Molecular vibration (Fundamental region) Conjugation region 2,500 ~ 2,000 Triple bond 2,000 ~1,540 Double bond Group frequency 4,000~1,300 Functional group Finger print region 1,300 ~ 650 Complete molecule Far IR 50 ~ 1000 m 3×1011 ~6×1012 200 ~ 10 Molecular rotation

  3. 2) Origin of IR spectra Atoms or atomic groups in a molecules are in continuous motion with respect to one another. IR spectra originate from the difference modes of vibration and rotation of a molecule, whereas the UV-visible absorption bands are primarily due to electronic transition. In order to absorb IR radiation, a molecule must undergo a net change in dipole moment as a consequence of its vibrational or rotational motion. The dipole moment is determined by the magnitude of the charge difference and the distance between the two centers of charge. The change in bond length or angle due to vibrational or rotational motion must cause a net change in the dipole moment of the molecule. No net change in dipole moment occurs during the vibration or rotation of homonuclear species such as O2, N2, or Cl2 ; consequently, such compounds cannot absorb in the IR. Vibrational modes which do not involve a change in dipole moment are said to be IR-inactive. With exception of a few compounds of this type, all molecular species exhibit IR-active.

  4. C-C stretching 1165 cm–1 C=O stretching 1730 cm–1 -C-H bending 1460 cm–1 1365 cm–1 C-H stretching of CH3 2960 cm–1 2870 cm–1 C-H stretching of CHO 2720 cm–1 Vibrations and characteristic frequencies of acetaldehyde.

  5. IR spectra of acetaldehyde.

  6. 3) Types of vibration 1) Stretching (or valency ) vibration :  Symmetric Asymmetric 2) Bending ( or deformation ) vibration :  In-plane bending Scissoring Rocking Out of plane Wagging Twisting 3) Breathing of ring compounds Vibrational modes for methylene group(a) and breathing vibration for a ring compound (b).

  7. 4) Mechanical model of stretching vibration Let us consider the vibration of a mass attatched to a spring that is hung from an immovable object. If the mass is displaced a distance y from its equilibrium position by application of a force along the axis of the spring, the restoring force is proportional to the displacement (Hooke’s law). F = –ky Where F is the restoring force and k is the force constant, which depends upon the stiffness of the spring. The potential energy E, is a maximum when the spring is stretched or compress to its maximum amplitude A, and decreases parabolically to zero at the rest or equilibrium position. dE = –Fdy = kydy  dE = k  ydy E = ½k y2 The vibration frequency vm , of the oscillation is dependent upon the force constant and reduced mass  . v m = (1/2)(k / ) = (1/2) {k (m1m2) / (m1+m2)}  v = (1/2c)(k / ) = 5.3×10–12 (k / )

  8. Reduced mass and force constants for various atom pairs.

  9. 5) Vibrational modes Fundamental ( normal ) vibration modes 1) Non-linear molecule : 3n – 6 vibrational modes 3 possible rotational modes 2) Linear molecule : 3n – 5 vibrational modes 2 possible rotational modes where is the number of atoms in the molecule, and 3n cartesian coordinates are called as degree of freedom . Example linear molecule : CO2 : 3n – 5 = 3 ×3 – 5 = 4 non-linear molecule : H2O : 3n – 6 = 3 ×3 – 6 = 3 HCHO : 3n – 6 = 3 ×4 – 6 = 6

  10. Illustration of vibrational modes in H2O and CO2.

  11. IR spectrum of H2O and CO2. Single and double beam spectra of atmospheric water vapor and CO2.

  12. Vibrational modes for formaldehyde.

  13. IR spectra of formaldehyde.

  14. Instrumentation of IR spectrometer Dispersive IR spectrometer Single beam is not very practical because of the absorption of IR radiation by atmospheric H2O and CO2. Double beam Sample cell is usually placed in front of the monochromator to minimize the effects of IR emission and stray radiation from the cell compartment. Detecting method Optical null system Ratio recording system Nondispersive IR spectrometer Filter photometer Dielectric filter spectrometer Special purpose spectrometer Fourier Transform IR spectrometer Interferometer

  15. Components of dispersive IR spectrometer Region of electromagnetic spectrum Near IR Mid IR Far IR Wavenumber (cm–1) 12,500 4,000 200 10 Wavelength (m) 0.8 2.5 50 1,000 Source of radiation Tungsten filament Nernst glower, Globar, High-pressure lamp or coil of nichrome wire mercury-arc lamp Optical system One or two Two to four plane Double beam quartz prisms or diffraction gratings grating for use prism grating with either a foreprism to 700 m ; double monochromator monochromator or interferometer IR filters for use to 1000 m Detector Photoconductive Thermopile, Golay cells thermister, or pyroelectric semiconductor

  16. Optical null double beam IR spectrometer

  17. Fourier transform IR spectroscopy FT techniques are possible because the units of time and frequency are inversely related. A function in the time domain can be transformed into its equivalent function in the frequency domain. The mechanism by which the instrument generates the time domain signal depends on the form of spectroscopy. IR radiation can be analyzed by means of a scanning Michelson interferometer. Fourier analysis is a procedure in which a curve is decomposed into a sum of sine and cosine terms, called a Fourier series. y = a0 sin(0x)+b0 cos(0 x)+a1sin(1x)+ b1cos(1 x) + a2sin(2x)+ b2cos (2 x) + …… =  [ an sin(nx) + bn cos (n  x)] where = 2 /(x2–x1)

  18. A curve to be decomposed into a sum of sine and cosine terms by Fourier analysis. Fourier series reconstruction of the curve in left Fig. Solid line is the original curve and dashed lines are made from a series of n=0 to n=2, 4 or 8 in the Fourier series equation : y =  [ an sin(nx) + bn cos (n  x)]

  19. The Nobel Prize in Physics 1907Albert Abraham Michelson, (December 19, 1852 - May 9, 1931), was born in Strzelno, Poland (then Strelno, Provinz Posen Kingdom of Prussia). He came to the United States with his parents when he was two years old. Michelson was an American physicist known for his work on the measurement of the speed of light. In 1907 he received a Nobel prize for physics. http://nobelprize.org/physics/laureates/1907/michelson-bio.html

  20. Interferometry The heart of a Fourier transform infrared specrtophotometer is the interferometer. Radiation from the source at the left strikes a beamsplitter, which transmits some light and reflects some light. For the sake of this discussion, consider a beam of monochromatic radiation. (In fact, the Fourier transform spectrophotometer uses a continuum source of infrared radiation, not a monochromatic source.) For simplicity, suppose that the beamsplitter reflects half of the light and transmits half. When light strikes the beamsplitter at point O, some is reflected to a stationary mirror at a distance OS and some is transmitted to a movable mirror at a distance OM. The rays is transmitted and half is reflected. One recombined ray travels in the direction of the detector, and another heads back to the source.

  21. Schematic diagram of Michelson interferometer. Detector response as a function of retardation (= 2[OM – OS] ) is shown for the case of monochromatic incident radiation of wavelength .

  22. Michelson Interferometer The Michelson interferometer produces interference fringes by splitting a beam of monochromatic light so that one beam strikes a fixed mirror and the other a movable mirror. When the reflected beams are brought back together, an interference pattern results. Precise distance measurements can be made with the Michelson interferometer by moving the mirror and counting the interference fringes which move by a reference point. The distance d associated with m fringes is d = m/2 http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/michel.html

  23. In general, the paths OM and OS are not equal, so the two waves reaching the detector are not in phase. If the two waves are in phase, they interfere constructively to give a wave with twice the amplitude. If the waves are one-half wavelength (180°) out of phase, they interfere destructively and cancel. For any intermediate-phase difference, there is partial cancellation. The difference in pathlength followed by the two waves in the interferometer is 2(OM-OS). This difference is called the retardation, . Constructive interference occurs whenever  is an integral multiple of the wavelength () of the light. A minimum appears when  is a half-integral multiple of . If mirror M moves away from the beamsplitter at a constance speed, light reaching the detector goes through a sequence of maxima and minima as the interference alternates between constructive and destructive phases.

  24. A graph of output light intensity versus retardation, , is called an interferogram. If the light from the source is monochromatic, the interferogram is a simple cosine wave: I() = B()cos(2π/ ) = B()cos(2π ) where I() is the intensity of light reaching the detector and and  is the wavenumber (=1/ ) of the light. Clearly, I is a function of the retardation, . B() is a constant that accounts for the intensity of the light source, efficiency by beamsplitter (which never gives exactly 50% reflection and 50% transmission), and response of the detector. All these factors depend on . In the case of monochromatic light, there is only one value of .

  25. Interferograms produced by different spectra

  26. Figure a) shows the interferogram produced by monochromatic radiation of wavenumber o=2㎝-1. The wavelength (repeat distance) of the interforogram can be seen in the figure to be  =0.5㎝, which is equal to 1/ o = 1/(2㎝-1). Figure b) shows the interferogram that results from a source with two monochromatic waves (o = 2 and o = 8㎝-1) with relative intensities 1:1. There is a short wave oscillation ( = 1/8㎝) superimposed on a long wave oscillation ( = 1/2㎝). The interferogram is a sum of two terms: I() = B1cos(2π 1) + B2cos(2π 2) where B1 = 1, 1 = 2㎝-1, B2 = 1, and 2 = 8㎝-1. Fourier analysis decomposes a curve into its component wavelengths. Fourier analysis of the interferogram in Figure a) gives the (trivial) result that the interferogram is made from a single wavelength function, with  = 1/2㎝. Fourier analysis of the interferogram in Figure b) gives the slightly more interesting result that the interferogram is composed of two wavelengths ( = 1/2㎝ and  = 1/8㎝) with relative contributions 1:1. We say that the spectrum is the Fourier transform of the interferogram.

  27. The interferogram in Figure c) is a less trivial case in which the spectrum consists of an absorption band centered at o = 4 ㎝-1. The interferogram is the sum of contributions from all source wavelengths. The Fourier transform of the interferogram in Figure c) is indeed the third spectrum in Figure c). That is, decomposition of the interferogram into its component wavelength gives back the band centered around o = 4 ㎝-1. Fourier analysis of the interferogram gives back the intensities of its component wavelengths. The interferogram in Figure d) is obtained from the two absorption bands in the spectrum at the left. The Fourier transform of this interferogram gives back the spectrum to its left.

  28. Michelson interferometer http://www.3dimagery.com/michelsn.html Interference pattern created by Michelson interferometer

  29. A two dimensional representation of the interference of two monochromatic wavefronts of the same frequency. Diagram of a Michelson interferometer.

  30. Formation of interferograms at the output of the Michelson interferometer.

  31. Spectrum of a continuum light source. • Inteferogram of the light source in (a) produced at the output of the Michelson interferometer.

  32. http://www.infrared-analysis.com/info1.htm

  33. He-Ne Layout of Fourier transform infrared spectrometer. Often, benchtop instruments purge the FT-IR spectrometer with an inert gas or dry, CO2-free air to reduce the background absorption from water vapor and CO2.

  34. Most FT-IR spectrometers are of the single beam type. To obtain the spectrum of sample, the background spectrum is first obtained by FT of the interferogram from background (solvent, ambient water, and carbon dioxide). This is normally a measurement with no sample in the beam. Next, the sample spectrum is obtained. Finally, the ratio of the single beam sample spectrum to that of the background spectrum is calculated, and absorbance or transmittance versus wavelength or wavenumber is plotted

  35. (a) Interferogram obtained from a typical FTIR spectrometer for methylene chloride. (b) IR spectrum of methylene chloride produced by the Fourier transformation of the data in (a).

  36. Sample Preparation In general the amount of sample necessary to obtain a good IR spectrum is the order of 1 to 5 mg (sample/KBr = 1~5mg/100mg). Since almost all substances absorb IR radiation at some wavelengths, cell window materials, cell pathlengths, and solvents must be carefully chosen for the wavelength region and sample of interest. Solid substances Solid state forces such as intermolecular hydrogen bonding render such spectra somewhat unreliable for diagnostic purposes. 1) Sample must be finely ground so that the particle size is smaller than the wavelength(1m) of IR radiation. Otherwise pronounced scattering of the incident light occurs. 2) These small particles must now be suspended in a medium of similar refractive index. A) Mulls Mulls are normally prepard by grinding a few mg of the powdered sample with an agate(alumina) mortar and pestle. A few drops of the mineral oil (Nujol; medicinal paraffin: refined mixture of saturated hydrocarbons) are then added. Grinding is continued in the presence of the oil until a smooth paste is obtained. A small amount of the resulting paste is then spread between two polished NaCl plates and placed in the spectrometer. Nujol shows absorption in the region near 2950 cm–1 for  (CH), at 1450 cm–1 for asy (methylene and methyl group CH) and 1380 cm–1 for sym (methyl group CH). If Nujol absorption is severe in a region of interest, chlorinated(hexachlorobutadiene) or fluorinated(Fluorolube) oils can be used.

  37. B) KBr pellet 1 mg of sample is mixed with 100 mg of dry KBr (spectroscopic grade) powder in a mortar, the mixture is then compressed under ~60MPa(60atm: 5000~10,000 Kg at 5 mmHg) in a die to form a transparent pellet(=disc) pellet. And the pellet is mounted in a suitable holder and then can be placed directly into the spectrometer. Properly made pellets are quite clear and the KBr is transparent in the IR region out of ~25 cm–1. Many substances tend to react with KBr under pressure or even while mixing. Thus, with unknown samples it is usually wise to obtain a spectrum of the material in a mull as well for comparison purposes. In addition, KBr is quite hygroscopic and the spectra obtained are difficult to reproduce. While mulls and pellets are satisfactory for qualitative analysis, neither technique is well suited for quantitative analysis.

  38. Infrared transmitting materials

  39. Pure liquid(neat) substances A drop of the pure liquid is placed between two NaCl plates which are then clamped together in a demountable cell. Spectra of pure liquids often show strong intermolecular hydrogen bonding and association effects. Solution samples The first problem when using solution samples for IR spectrometry is to find a suitable solvent. Choice of solvent depends on the region of the spectrum of most interest. By using “window areas”, that is, transparent areas of the solvent, the whole spectrum may be covered. For instance, the most common use of carbon tetrachloride is from 4000 to 1300 cm–1 and for carbon disulfide, 1300 to 660 cm –1. NaCl cells are employed, the most useful thickness being 0.1 mm and 0.5 mm.

  40. M2000 FT-IR spectrometers http://www.midac.com/m_series.htm

  41. How to Operate MIDAC Spectrometer The program that we are using to operate spectrometer is called LAB CALCTo start Lab Calc from Windows • Open File Manager • Find an lc (lc stands for LAB CALC) directory • On the right side of the File Manager window find a file named lc.exe and press Enter • When MIDAC FT-IR screen appears press any key • Alternative way to start Lab Calc from Windows • In Program Manager find a START UP icon • In Start Up window find MS-DOS FT-IR icon and click on it • When MIDAC FT-IR screen appears press any key • Before running any samples you have to set up parameters • When Lab Calc screen appears press F2 key (F2 = Menu) • After pressing this key next screen appears and you will see the following menu at the bottom of the screen • http://patsy.hunter.cuny.edu/GStud/pevsner/midac.htm

  42. Environment Collect Arithmetic I-Peak File Draw Plot Text Quit 3 Environment will be highlighted and you will also see a submenu directory Template+ parms Mode Display Limits Axes FileSave Windows Status Collar - Choose Template and press Enter - Another (pink) submenu appears: choose Master Method press Enter - Then you will see a yellow submenu choose STD-IR and press enter

  43. 4. Press F2 key again -Choose Mode Display/Paged and press Enter 5. Press F2 key -Choose Directory A pink "Enter Default Directory Window" will appear Type a directory in which you want to store you data. For example, if I want to store the data in my file I would type c:\alex . Spectra of my samples will automatically be stored in this directory. There is a directory called U761 where your spectra can be stored. Each group should also create their own subdirectory in U761 and stored their files in there. For example, suppose I was assigned to the first group. I would create a subdirectory called one in the directory U761 . Therefore when it comes to choosing a default directory I would type C:\U761\one. 6. Press F2 key - Choose Filesave/Autosave and press Enter You done with Environment, press à to highlight Collect and press Enter. On the bottom of the screen you will see the following menu

  44. Name Memo Type Gain Resolution Scans Align Begin • Remember before you run any samples you have to take a spectrum of background. Background is also called reference. • Highlight Name and press Enter • - Type the name of your reference 4. Highlight Memo • -Type background or reference • 5. Highlight Type • -Choose Reference. The sample that you will run know will be taken as the reference. You have to take spectrum of the reference only once. Computer will automatically store reference spectrum in its memory. Every time you run your sample, computer will use the last background spectrum that you took as the reference. • 6. Highlight Gain and type 0 7. Highlight Resolution, choose 2 cm • 8. Highlight Scans and type 10 • 9. Now you ready to take run a spectrum. Highlight Begin and press Enter • When Spectrometer finished scanning, a screen with the spectrum will appear. In the lower right corner of the window you will see the question • Return to Collect ? YesNO • If you want to play with the spectrum choose NO, if don’t choose Yes

  45. You have the spectrum of a background. Now you ready to take the spectrum of your analyte. 1. Do through the same steps as you did for reference except one thing 2. When you get to the Type choose Absorbance 3. Gain, Resolution and number of Scans will be the same as before To Quit Lab Calc -Press F2 -Highlight Quit/Yes Processing of spectra is done on another computer, therefore you data files have to be copied on the floppy disk. To do that -Open File Manager and find your directory -On the right side of the screen you will see the files that are stored in your directory. All of them have spc extension. Highlight the files you want to copy. - From File menu choose Copy , type b:\ and press Enter

  46. How to approach the analysis of an IR spectrum 1. Is a carbonyl group present ? C=O 1820~1660 cm–1 (strong absorption) 2. If C=O is present, check the following types. (If absent, go to 3) Acids is OH also present ? OH 3400~2400 cm–1 (broad absorption) Amides is NH also present ? NH 3500 cm–1 (medium absorption) Esters is C-O also present ? C-O 1300~1000 cm–1 (strong absorption) Anhydrides have two C=O absorptions near 1810 and 1760 cm–1. Aldehydes is aldehyde CH present ? Two weak absorptions near 2850 and 1760 cm–1 . Ketones The above 5 choices have been eliminated. 3. If C=O is absent Alcohols / Phenols check for OH OH 3600~3300 cm–1 (broad absorption) C-O 1300~1000 cm–1 . Amines check for NH NH 3500 cm–1 . (medium absorption) Ethers Check for C-O (and absence of OH) 1300~1000 cm–1 . 4. Double bons and / or aromatic rings C=C 1650 cm–1 (weak absorption) aromatic ring 1650~1450 cm–1 aromatic and vinyl CH 3000 cm–1 5. Triple bonds CN 2250cm–1(sharp absorption) C  C 2150cm–1(sharp absorption) acetylenic CH 3300 cm–1 6. Nitro group two strong absorptions at 1600~1500 cm–1and 1390~1300 cm–1 7. Hydrocarbons none of the above are found, CH 3000 cm–1 (major absorption)

More Related