Ch313-Lecture08

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Chapter 16
Infrared Absorption Spectroscopy
An IR spectrum contains information about the functional groups in
a molecule, and this is used to uniquely identify the compound.
UV
Vis
400 nm
Near IR
780 nm
Mid IR
Far IR
2500 nm
50,000 nm
or 4000 cm-1
or 200 cm-1
Wavenumbers (cm-1) are used since they are directly proportional to energy –
E  hν 
hc
λ
e.g. convert 2.5 µm to wavenumbers (cm-1)
Typical Infrared Spectrum
“overtones”
C-H
C=O
Higher energy vibrations
C-C
Mechanical Model of a Stretching Vibration in a Diatomic Molecule
Treats the vibrating bond
like a spring with a given
“stiffness” --
ν 
1
k
2π
μ
= frequency of vibration
k = force constant (spring “stiffness”)
µ = “reduced mass”
m 1m 2
m1  m 2
Quantum Mechanical Treatment of Normal Modes
1. Only certain vibrations are “allowed”
2. The vibrational quantum number v = 0, 1, 2, 3……
3. Only transitions between adjacent energy levels are possible, i.e. v =  1
4. The frequency of the photon has to equal the frequency of the vibration
(see next slide)
5. The molecule must have a change in “dipole moment” as a result of the vibration
Dipole moment = charge difference X separation distance
-
+
Homonuclear diatomics never have an Infrared spectrum. Why?
Frequency of absorbed light  frequency of vibration
+
+
+
+
-
S1
+
v =  1
-
IR Absorption if
the vibration
results in a
change in dipole
moment
So
Common Types of Vibrations (“Normal Modes”)
Higher
energy
modes
Lower
energy
modes
The number of normal modes = 3N-6
For linear molecules it’s 3N-5
Example – predict the normal modes of CO2
Which normal modes are
Infrared active?
Example 16-1, p.436
Calculate the approximate wavenumber and wavelength of the
fundamental absorption peak due to the stretching vibration of a
C=O group (k = 1.0 x 103 N/m)
ν
1
k
2π
μ
Force Constants increase
with Bond Strength
Bond
Type*
Force
Constant,
k (N/m)
Wavenumber
(cm-1)
Bond
Energy
(kJ/mol)
C-C
5 x 102
800-1200
347
C=C
10 x 102
1600
620
CΞC
15 x 102
2100
812
Bond
Type*
Reduced
mass (kg)
Wavenumber
(cm-1)
Bond
Energy
(kJ/mol)
C-H
1.55 x 10-27
3000
414
C-N
1.07 x 10-26
1000-1350
276@
C-O
1.14 x 10-26
1000-1300
351@
(a).
(b)
(a).as  then frequency
(b) When masses approx.
equal, then see peak at same
wavenumbers
* all bonds are stretches
@ consider these as being approx. equal
IR Sources
Output from a Nernst Glower
1. Nichrome – Ni/Cr alloy, resistively heats
up and emits IR, 1100 K
2. Globar – SiC rod, 1500 K
3. Nernst Glower – electrically heated rare
earth-oxide ceramic, 2000 K
cm-1
Optics – IR spectra are measured in the mid-IR, so
have to use halides such as NaCl and KBr
UV
Vis
400 nm
Near IR
780 nm
Mid IR
Far IR
2500 nm
50,000 nm
or 4000 cm-1
or 200 cm-1
Never use aqueous
samples, or water to clean
salt plates
Sample Handling – a few drops of “neat” sample between “salt plates”
Solutions
Detectors for the Mid-IR – 200-4000cm-1
1. Pyroelectric – crystal of DTGS (Deuterated Triglyceine Sulfate)
Nice link:
http://www.doitpoms.ac.uk/tlplib/pyroelectricity/printall.php?question=2&type=1
1. DTGS maintains polarization when heated to
just below the Curie Point
2. Above the Curie Point, the permanent
polarization of the DTGS crystal disappears.
3. The closer to the Curie Point, the more
responsive the detector
4. Much faster response so can take a spectrum
much more quickly (FTIR)
Detectors for the Mid-IR – 200-4000cm-1
2. Photoconduction - Mercury-Cadmium-Tellurium (MCT)
1. Semiconductor-based
Liquid N2 @ 77K
2. Cooled to 77 K using liquid N2
3. Much faster response so can take a
spectrum much more quickly (FTIR)
http://www.newport.com
http://www.boselec.com/pdf/Laser_Focus_World.PDF
Definition of the Fourier Transform ()  ( )    f ( t )  


f (t )e
 i t
i
dt
1
  2f

The Inverse Fourier Transform f (t )  
1
 ( )  
1
2


 ( ) e
i t
d

The functions f(t) and () are called "transform pairs"
f(t)  ()
time (t)  frequency (s-1, Hz)
e.g. FTIR
distance (cm)  wavenumber (cm-1)
background
interferogram
sample
interferogram
background
spectrum = Po
sample
spectrum = P
P/Po =
IR spectrum
e.g. FT-NMR
time (s)  ppm (Hz)
Free Induction Decay
(FID) Signal (time)
NMR
Spectrum
(ppm, Hz)
Fourier Transforms "invert dimensionality" –
signal domain
FT domain
Instrument
time (t)
frequency (s-1, Hz)
oscilloscope
mirror
distance (cm)
wavenumber (cm-1)
FTIR
free induction
decay (t)
ppm (from Hertz)
FT-NMR
Fast Fourier Transform (FFT) on computers -
•Cooley-Tukey algorithm
•very efficient on a PC
fixed mirror
¼
IR source
+
+
+
-
-
movable mirror →
50/50 beamsplitter
D
interferogram
/4
2 /4 3/4 4/4
mirror distance, x (cm) →
If polychromatic radiation enters the interferometer, then each
wavelength produces a separate interferogram. The output from the
interferometer will therefore be a superposition of all wavelengths -
1  1
2  2
3  3
←mirror distance (-x, cm)
mirror distance (+x, cm) →
signal at
detector S(x)  A ( 1 ) cos 4 x  1   A ( 2 ) cos 4 x  2   A ( 3 ) cos 4 x  3   ....
N


A ( i ) cos 4  x  i 
i 1



A ( i ) cos 4  x  i d 

this term contains the IR spectrum
Taking the Fourier Transform of S(x) results in the IR spectrum, A() -
A ( ) 


S ( x ) cos 4  x  i dx

Depends on the distance the mirror moves in the Michelson Interferometer –
the further it moves, the greater the resolution.
 
1
2x

1

"retardation“ = 2x
minimum detectable
difference in
wavenumber
x = distance mirror moves
e.g. What length of mirror drive in a Michelson Interferometer is required to
separate 20.34 and 20.35 m?
Fourier Transform Infrared (FTIR) Spectrometers
The Multiplex (Fellgett's) Advantage - entire spectrum obtained virtually
instantaneously; results in a higher S/N because during the time a scanning
instrument is slowly obtaining the spectrum, a multiplex instrument such as an
FTIR can acquire 100’s pf spectra. Averaging these extra spectra causes the
increased S/N (see next slide) and allows quantitative IR.
S/N increases as
N where N = number of "resolution elements"
e.g. scan range 400 - 4000 cm-1 at a resolution of 2 cm-1
2 cm-1
4000 cm-1
400 cm-1
N 
( 4000  400 cm
2 cm
1
1
)
 1800 "resolution elements"
so the S/N increases by 1800  40 X IF the FTIR can obtain the entire
spectrum for the same amount of time a slower, scanning instrument requires to
acquire only one resolution element.
Signal-to-Noise Ratio (S/N)
Single-Beam – run background first (see next slide) and store in memory (Po)
Higher energy throughput (Jacquinot’s Advantage) and no stray light problems
Typical IR background spectrum (Po)
inexpensive (~ $25K)
range = 350 – 7800 cm-1 (29 to 1.3 m)
max resolution = 4 cm-1
scan time = as fast as 1 sec
detector = DTGS
expensive (~ $100K)
range = 10 – 50,000 cm-1 (1000 m to 200 nm)
max resolution < 0.01 cm-1
scan time = can be minutes at high-res
detector(s) = DTGS, MCT
Advantages over Scanning Instruments
1. fast scan times
2. higher S/N because of signal averaging
3. higher resolution
4. because there are no slits, there’s a higher energy throughput
(Jacquinot’s Advantage); means larger signals for the same
concentration (therefore lower LODs)
5. no stray light problems
Disadvantages
1. higher cost than scanning instruments
2. difficult to align the interferometer (automation helps)
3. IR optics water soluble (beamsplitter made of KBr)
Chapter 17
Applications of IR Spectrometry
1. Qualitative Analysis
"group frequency region"
"fingerprint region"
2. Quantitative Analysis
Much less sensitive (i.e.
higher LODs) in the IR
compared to the UV-Vis
because of 1. lower source powers than in UV-Vis (in FTIR Jacquinot's
Advantage partially offsets)
2. detectors suffer from thermal noise (i.e. larger
backgrounds or blank)
3. salt plates have a very short path length (A = bc)
Signal averaging with FTIR's has improved the S/N enough
to allow more sensitive quantitative work in the IR.
3. Diffuse Reflectance Accessory
Typically used for powdered samples, i.e. forensic drug analyses
Kubelka-Munk Units = converts the reflectance spectrum to the
equivalent of an absorbance spectrum.
R’ = sample refectivity/KBr reflectivity
f(R’) = (1 - R’)2/2 R’ = k/S
k = 2.303  C
 = the molar absorptivity
C is the sample concentration
S = "scattering coefficient"
For a sample to follow a linear relationship between f(R’) and concentration,
the following criteria must be met:
1. the sample must be diluted in a non-absorbing matrix such as KBr or KCl.
2. the scattering coefficient S must remain constant over the entire spectrum.
3. there must be no specular, or regular reflectance off the surface of the
sample.
4. Attenuated Total Internal Reflectance (ATR) Accessory
Samples - conventional (solutions, liquids, etc) as well as powders,
pastes, suspensions, colloids
Total Internal Reflection and the "evanescent wave" -
5. Infrared Microscopy

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