Transcript anal chem II / IR spectrometry

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
Designation
Wavelength
Frequency(Hz)
Near IR
780~2500nm 1.2~3.8×1014
Wave number (cm–1)
Transition
12,800 ~4,000
Molecular vibration
4,000 ~ 200
Molecular vibration
Overtone region
Mid IR
2.5~50m
6×1012 ~1.2×1014
(Fundamental region)
Conjugation region
Far IR
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
200 ~ 10
Molecular rotation
50 ~ 1000 m 3×1011 ~6×1012
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.
-C-H bending
1460
cm–1
C-C stretching
C=O stretching
1165 cm–1
1730 cm–1
1365 cm–1
C-H stretching
C-H stretching
of CH3
of CHO
2960 cm–1
2870
2720 cm–1
cm–1
Vibrations and characteristic frequencies of acetaldehyde.
IR spectra of acetaldehyde.
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).
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 = ½ k y2
The vibration frequency vm , of the oscillation is
dependent upon the force constant and reduced mass  .
v m = (1/2)(k / )
 v = (1/2c)(k / )
= (1/2) {k (m1m2) / (m1+m2)}
= 5.3×10–12 (k / )
Reduced mass and force constants for various atom pairs.
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
Illustration of vibrational modes in H2O and CO2.
IR spectrum of H2O and CO2.
Single and double beam spectra of atmospheric water vapor and CO2.
Vibrational modes for formaldehyde.
IR spectra of formaldehyde.
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
Components of dispersive IR spectrometer
Region of electromagnetic spectrum
Near IR
Wavenumber (cm–1)
Wavelength (m)
Source of radiation
12,500
0.8
Tungsten filament
lamp
Optical system
Mid IR
Far IR
4,000
200
10
2.5
50
1,000
Nernst glower, Globar,
High-pressure
or coil of nichrome wire
mercury-arc lamp
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
IR filters
Detector
interferometer
for use to 1000 m
Photoconductive
Thermopile,
Golay
cells
thermister, or
pyroelectric
semiconductor
Optical null double beam IR spectrometer
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)
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)]
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
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.
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 .
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
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.
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 .
Interferograms
produced by different
spectra
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.
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.
Michelson interferometer
http://www.3dimagery.com/michelsn.html
Interference pattern created by Michelson
interferometer
Diagram of a Michelson interferometer.
A two dimensional representation of
the interference of two
monochromatic wavefronts of the
same frequency.
Formation of interferograms at the output of the Michelson interferometer.
(a) Spectrum of a continuum light source.
(b) Inteferogram of the light source in (a) produced at the output of the
Michelson interferometer.
http://www.infrared-analysis.com/info1.htm
Layout of Fourier transform
infrared spectrometer.
He-Ne
Often, benchtop instruments
purge the FT-IR spectrometer
with an inert gas or dry, CO2free air to reduce the
background absorption from
water vapor and CO2.
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
(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).
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.
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.
Infrared transmitting materials
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.
M2000 FT-IR spectrometers
http://www.midac.com/m_series.htm
How to Operate MIDAC Spectrometer The program that we are using to operate
spectrometer is called LAB CALC
To start Lab Calc from Windows
1.
Open File Manager
2.
Find an lc (lc stands for LAB CALC) directory
3.
On the right side of the File Manager window find a file named lc.exe and press
Enter
4.
When MIDAC FT-IR screen appears press any key
Alternative way to start Lab Calc from Windows
1.
In Program Manager find a START UP icon
2.
In Start Up window find MS-DOS FT-IR icon and click on it
3.
When MIDAC FT-IR screen appears press any key
Before running any samples you have to set up parameters
1.
When Lab Calc screen appears press F2 key (F2 = Menu)
2.
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
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
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
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.
1.
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 don’t choose Yes
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
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
OH
3600~3300 cm–1 (broad absorption) C-O 1300~1000 cm–1 .
Alcohols / Phenols
check for OH
Amines
check for NH
Ethers
Check for C-O (and absence of OH)
NH 3500 cm–1 . (medium absorption)
1300~1000 cm–1 .
4. Double bons and / or aromatic rings
C=C
1650 cm–1 (weak absorption)
aromatic and vinyl CH
5. Triple bonds
CN
aromatic ring 1650~1450 cm–1
3000 cm–1
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–1 and 1390~1300 cm–1
7. Hydrocarbons
none of the above are found, CH 3000 cm–1 (major absorption)
IR spectrum of n-butanal (n-butyraldehyde).
Frequency of various group vibrations in the group frequency region
and in the fingerprint region.
Q n A
Thanks
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