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Title: anal chem II / IR spectrometry Subject: IR spectrometry FT-IR Author: dslee Last modified by: user Created Date: 8/27/1998 4:46:32 AM – PowerPoint PPT presentation

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Title: anal chem II / IR spectrometry


1
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 (cm1) Transition Near IR
7802500nm 1.23.81014 12,800
4,000 Molecular vibration
Overtone region Mid IR 2.550?m
61012 1.21014 4,000 200
Molecular vibration (Fundamental region)

Conjugation region 2,500 2,000
Triple bond

2,000 1,540 Double bond

Group frequency 4,0001,300
Functional group
Finger print region
1,300 650 Complete
molecule Far IR 50 1000 ?m
31011 61012 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
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
(Hookes 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 ½ k 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) / (m1m2) ? v
(1/2?c)?(k / ?) 5.31012 ?(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
(cm1) 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. Fou
rier 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 n0 to
n2, 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 (
2OM 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/ph
yopt/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(2p?/ ?) B(?)cos(2p ??) 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 ?o2?-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 11.
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(2p ?1 ?) B2cos(2p ?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 11. 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
  1. Spectrum of a continuum light source.
  2. Inteferogram of the light source in (a) produced
    at the output of the Michelson interferometer.

32
http//www.infrared-analysis.com/info1.htm
33
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
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36
(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).
37
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38
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
15mg/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 cm1 for ? (CH), at 1450 cm1
for ?asy (methylene and methyl group CH) and
1380 cm1 for ?sym (methyl group CH). If Nujol
absorption is severe in a region of interest,
chlorinated(hexachlorobutadiene) or
fluorinated(Fluorolube) oils can be used.
39
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 500010,000 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 cm1. 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.
40
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41
Infrared transmitting materials
42
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 cm1 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.
43
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44
M2000 FT-IR spectrometers http//www.midac.com/m
_series.htm
45
  • How to Operate MIDAC Spectrometer The program
    that we are using to operate spectrometer is
    called LAB CALC To 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.h
    tm

46
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
47
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
48
  • 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 ? Yes NO
  • If you want to play with the spectrum choose NO,
    if dont choose Yes

49
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  
50
How to approach the analysis of an IR spectrum 1.
Is a carbonyl group present ? CO
18201660 cm1 (strong absorption) 2. If CO is
present, check the following types. (If absent,
go to 3) Acids is OH also
present ? OH 34002400 cm1 (broad
absorption) Amides is NH also
present ? NH 3500 cm1 (medium absorption)
Esters is C-O also present ?
C-O 13001000 cm1 (strong absorption)
Anhydrides have two CO absorptions near
1810 and 1760 cm1. Aldehydes is aldehyde
CH present ? Two weak absorptions near 2850 and
1760 cm1 . Ketones The above 5
choices have been eliminated. 3. If CO is
absent Alcohols / Phenols check for
OH OH 36003300 cm1 (broad
absorption) C-O 13001000 cm1 . Amines
check for NH NH
3500 cm1 . (medium absorption) Ethers
Check for C-O (and absence
of OH) 13001000 cm1 . 4. Double bons and /
or aromatic rings
CC 1650 cm1 (weak absorption)
aromatic ring 16501450 cm1
aromatic and vinyl CH 3000
cm1 5. Triple bonds C?N
2250cm1 (sharp absorption) C ? C
2150cm1 (sharp absorption)
acetylenic CH 3300 cm1 6.
Nitro group two strong absorptions at
16001500 cm1 and 13901300 cm1 7.
Hydrocarbons none of the above are
found, CH 3000 cm1 (major absorption)
51
IR spectrum of n-butanal (n-butyraldehyde).
52
Frequency of various group vibrations in the
group frequency region and in the fingerprint
region.
53
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55
Q n A Thanks
Home page http//mail.swu.ac.kr/dslee Electr
onic mail dslee_at_swu.ac.kr
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