10. Other Spectroscopies
1.IR
2.NMR
VIBRATIONAL CHROMOPHORES
Any bond can act as a spring which can be described as the balance
Between the force due to acceleration and to the restoring force of the spring
F  ma
Set equal
F   ky
d2y
 ky  ma
a
2
dt
 d 2 y
 ky  m 2 
 dt 
d
2
cosat   a
dt
2
2
cosat 
Need a function which reinvents itself
d cosat 
dt
  sinat 
d cosat 
dt
d
2
d at 
dt
 a sinat 
cosat   d  a sinat   a cos at d at 
 
dt 2
dt
dt
 d 2 y
 ky  m 2 
 dt 
d
 d y
ky
 2 
m
 dt 
2
2
y  cosat 
cosat   a
dt
2
2
cosat 
k
a 
m
2
1
m 
2
a is a frequency
a  2 m
Natural mechanical frequency
2 m 
k
m
k
m
The frequency of the oscillation is
Related to the bond constant, k,
And m, which is a mass term
m1m2
m
m1  m2
1
m 
2
Bond
Single
Double
Triple
Example: Calculate double bond band for
R-C=O
k
m
1mole
 12 g  
  kg 
 26
mc  

2
x
10
kg




 mole   6x1023 atoms   103 g 
k(Newton/m)
500
1000
1500
1mole
 16g  
  kg 
 26
mO  

2
.
7
x
10
kg




 mole   6x1023 atoms   103 g 
N
kg  m kg
 2
 2
m s m s
1
m 
2
m
2 x10
2 x10
 26
 26

kg 2.7 x10  26 kg
 
kg  2.7 x10
kg
1000 2
s

 26
11
. x10 kg
 26

kg

 11
. x10  26 kg
9.09 x10 28
13 1
 4.8 x10
6.28s
s
1
m 
2
kg
1000 2
s

 26
11
. x10 kg
9.09 x10 28
1
 4.8 x1013
6.28s
s
For IR usually reported as reciprocal cm
  c
1 

 c
1

cm
1
s  1,600cm 1
cm
s
4.8 x1013
3x1010
The average reported value is 1690-1760
1600 cm-1
What is the predicted frequency of C-O?
1
m 
2
Bond
Single
Double
Triple
k(Newton/m)
500
1000
1500
  c
1 

 c
1

cm
Example: Calculate single bond band for
R-C-O
k
m
1
2
1
s  1131
, cm 1
cm
s
3.393x1013
3x1010
m 
500
kg
s2
11
. x10  26 kg
 3.393x1013
1
s
What would
Be this frequency?
1600 cm-1
1100 cm-1
1
m 
2
Bond
Single
Double
Triple
k
m
k(Newton/m)
500
1000
1500
Example: Calculate single bond band for
R-N-Pb
1mole
 14 g  
  kg 
 26
mN  

  3   2.33x10 kg
23
 mole   6x10 atoms   10 g 
1mole
 207.2 g  
  kg 
 25
mPb  

  3   3.45x10 kg
23
 mole   6x10 atoms   10 g 
m
  c
1 

 c
2.33x10
2.33x10
1
m 
2
 26
 26

kg 3.45x10  25 kg
 
kg  3.45x10
 25

kg

 2.19 x10  26 kg
kg
500 2
13 1
s
 2.404 x10
 26
s
2.19 x10 kg
1

cm
2.404 x1013
3x1010
cm
s
1
s  8016
. cm 1
800 cm-1
Bonds
R-N-Pb
R-C-O
R-C=O
1600 cm-1
1100 cm-1
cm-1
800
1100
1600
R-CO=O ?
Bonds
R-N-Pb
R-C-O
R-C=O
cm-1
800
1100
1600
R-CO=O ?
Our calculations
Lead EDTA spectra 2009 data
100
90
Pb-N
80
60
50
Them
cm-1
350
400
450
595
710
800
845
918
975
1020
1085
1140
1200
1265
1280
1335
1406
1435
1450
1600
2841
2936
Us 2009
strength
breadth cm-1
m,b
m,b
409
w,sh
432
509
555
w,b
640
w,b
721
w,b
806
w,b
vs
914
w,b
957
w
1040
w
1090
m,b
1160
w
1230
m,b
vw,sh
s
1320
vs
1380
m,sh
m,sh
1490
m,b
1550
2360
2850
s
2930
strength
breadth
vw,sh
Their assignments
40
Pb-O
Pb-N stretch
30
m,sh
m,sh
20
m,sh
m,sh
m,sh
10
m,sh
w
m
w
m
m
w
s,sh
0
C-C acetate stretch
4000
3500
3000
2500
2000
cm-1
C-N stretch
CH2 wag in CH2COOH
s,sh
s,b
carbonyl
carbonyl
carbonyl
carbonyl
vs,sh
vs,b
C-H in HCOOC-H in CH2,COO-
1500
1000
500
0
normalized %T
70
480
Cu(0.71)
cm-1 (M-N bond)
470
460
450
440
430
Pb(1.32)
420
We once tried looking for the Pb-N
Changes but this region is too far into the
“fingerprint region” to get good data
0
0.5
1
1.5
2
2.5
3
q/r
Higher charge density = higher frequency
2009 Data, 0, 0.25, 0.5, 0.75, and 1 fraction Pb EDTA
120
100
60
40
20
0
4500
4000
3500
3000
2500
2000
cm-1
1500
1000
500
0
%T
80
Bonds
R-N-Pb
R-C-O
R-C=O
cm-1
800
1100
1600
R-CO=O ?
Pb-N bond
424, cm-1
CO=O, 1750cm-1
Pure EDTA, Na
What would influence the degree of the frequency shift of CO=O?
Hint – who would have a stronger attraction for the electrons on O?
More energy
1750cm-1
Ionic bonding at C-O;
Pulls electrons from
C=O, creating bond
1.5 order
Covalent type bonding (H) at CO
Leaves double bond character at
CO=O
lead
Greater electrostatic attraction
1
m 
2
m1m2
m
m1  m2
k
m
Bond
Single
Double
Triple
k(Newton/m)
500
1000
1500
Lower reduced mass has higher frequency
Higher
energy
Cm-1
Bonds
3600-3200
-O-H
3500-3300
-N-H
3100-3010
R=CH-H
3100-2970
R=C-H
-C-H
2280-2210
-C=N
2260-2100
-C=C
1760-1690
-C=O
1680-1500
-C=C
1360-1180
-C-N
1300-1050
-C-O
What do you observe?
Group
Frequency
Region
1
m 
2
k
m
m1m2
m
m1  m2
Cm-1
Bond
3600-2700
X-H
2700-1900
X=Y
1900-1500
X=Y
1500-500
X-Y
“fingerprint” region
Bond
Single
Double
Triple
k(Newton/m)
500
1000
1500
H3C
CH3
H3C
CH3
H3C
H
H3C
OH
H3C
CH3
H3C
CH3
Which is which?
Can you identify all of them?
http://orgchem.colorado.edu/hndbksupport/specttutor/irchart.html
Where is the expected vibrational absorption band for
R-C=O?
We predicted a band at 1600 cm-1 for R-C=O
Is this the only band we should observe?
IR bands Observed
1.
2.
3.
4.
5.
Degrees of freedom set total possible bands
Requires a change in dipole moment
Requires a high molar absorptivity
Requires good instrumental “window”
Complicated by passing of overtones
Degrees of freedom
Bands observed = total DF – (translation + rotational)
3N
- ( 3 + (2 or 3))
Possible Bands
Linear molecule: 3N-5
Non-linear:
3N-6
EXAMPLE
How many bands should we observe for O=C=O?
Linear molecule: 3N-5 = 9-5=4
Each possible band
must have a change in dipole moment
should not be degenerate
1. O=C=O
symmetric stretch
Which have a change in dipole
Moment?
2. O=C=O
asymmetric stretch
Which are (if any) are degenerate?
3. O=C=O
scissoring
4. O=C=O
scissoring
EXAMPLE
How many bands should we observe for O=C=O?
Linear molecule: 3N-5 = 9-5=4
Each possible band
must have a change in dipole moment
should not be degenerate
No dipole moment change
1. O=C=O
symmetric stretch
2. O=C=O
asymmetric stretch
3. O=C=O
scissoring
4. O=C=O
scissoring
Degenerate – only one is observed
Total of two bands!
Animation associated with spectra:
http://www.succeedingwithscience.com/labmouse/chemistry_a2/2906.php?Labmou
seOnline=191f6382b179f1a28fb4da8c34523817
EXAMPLE
How many bands should we observe for H-O-H?
Non-Linear molecule: 3N-6 = 9-6=3
Each possible band
must have a change in dipole moment
should not be degenerate
O
H
H
Symmetric stretch
O
H
H
O
H
H
Asymmetric stretch
scissoring
Animation of water vibrations
http://www.lsbu.ac.uk/water/vibrat.html
http://www.lsbu.ac.uk/water/vibrat.html
Fused silica Si-O
Typical region of IR interest!
http://www.laseroptik.de/?Substrates:Transmission_Curves:%26%23150%3B_IR-FS
Can not use the quartz materials we were using in UV-Vis
Quartz vibrational bands overlap
C-H bonds
Some times we observe more bands than predicted.
Why?
FT instrument
f det ector
2 mirror  1  1
 
   2 mirror
  n

n
1
Light that passes at 100 cm-1 will also pass at 200, 300, and 400 cm-1
That means that it could excite absorption at 400 cm-1!
Source light
1st order
cm-1
100
200
300
400
800
2nd order
cm-1
200
400
600
800
1600
3RD ORDER
cm-1
300
600
900
1200
2400
4th order
cm-1
400
800
1200
1600
3200
Our calculations
R-N-Pb
800
R-C-O
1100
R-C=O
1600
Our calculated band at 800 cm-1 could be observed at 200 4th order;
400 2nd order and 800 first order!
INSTRUMENTATION
Generic
Source
Cell
Monochromator
Detector
Spatial Arrangment Components
Readout
Solvents
not = UV-Vis
not = UV-Vis
not = UV-Vis
not = UV-Vis
not = UV-Vis
IR SOURCES:
Nernst Glower (Blackbody)
Tungsten-filament bulb (household)
Hg arc
Tunable CO2 laser
IR DETECTORS
What is the main problem in using a UV-Vis detector here?
IR DETECTORS
1. Thermal
What kind of problem can you
Anticipate here?
a. Thermocouple
Bi/Sb/Bi
Voltage difference
6 to 8 mV/uW
b. Bolometer or thermister
Change in resistance on T change
2. Pyroelectric
1. Move charge with temperature
= capacitor
= capacitor current with T

3
NH  CH2  COO

3

NH    CH2    COO
3. Photoconductivity
low bandgap semiconductor
e
LUMO
+
HOMO

DESIGN OF INSTRUMENT
1.
2.
3.
4.
Double Beam vs Single Beam?
To Chop or not to Chop?
Whither the sample?
To FT or not to FT?
An old (not FT instrument): Describe what you see
UV-Vis:
Source - Monochrom – Sample – Detector
Place the sample after the monochromator to decrease the
Total power on the sample. Energy of individual frequencies
Is large enough to remove an electron (break bonds) and
Decompose the sample
IR
Source – Sample – Monochrom – Detector
Sample not decomposed. Placing the monochromator
After the sample can help prevent scattered radiation
From entering the detector.
To FT or not to FT?
f det ector
 2velocity mirrow 
 



c
Mirror velocity in a typical instrument is 0.01 to 10 cm/s
Do we FT the instrument or scan through each individual wavelength?
Why?
1. Optical resolution
2. Controls mirror
3. Eliminates Phase Shift
Why should eliminating phase
Shift be so important?
Describe this instrument in terms of
Various functional parts
Instrument
Single Beam
FT
Range
1.3 to 29
0.4 to 1000
Resolution
4
8 to 0.01
Cost
16k
150k
http://www.wooster.edu/chemistry/analytical/ftir/ftir.swf
Shows how the FT is generated by the movable mirror
Time
1s
>1min
SAMPLE HANDLING
1. Gases
Effusion of volatile liquid through a pin hole
2. Solution: need a cell that minimizes IR activity
of background
Background?
solvents (H2O)
Air (H2O, CO2)
windows (Si-O band in quartz)
One way to minimize solvent effects is to minimize
the total amount of solvent
Use a very thin cell (small value of b)
How could we then determine the value of b?
Interference Fringes
2.5
2.5
2b  N
22
film
1.5
1.5
Amplitude
Amplitude
11
0.5
0.5
00
00
0.5
0.5
11
1.5
1.5
-0.5
-0.5
-1
-1
-1.5
-1.5
-2
-2
-2.5
-2.5
Film
Film Thickness
Thickness
22
2.5
2.5
2b  N
1
film
 film
1
 film1
1
 film2
N1  N 2
N

2b
N1

2b
N2

2b
 1
1
 N  2b
 
 1 2 
Peak at
1900
1400
900
Example
N
1
2
3
Count the peaks between two known frequencies
1900
1700
1500
1300
1100
900
700
500
cm-1
1
1 

 3  1  2b 1900
 900


cm
cm 
1
b
cm
1000
N 1  N 2   N  2b  1   2 
1900
1700
1500
1300
cm-1
1100
Try this one
900
700
500
SOLID SAMPLES
Crush sample into a solid matrix (could also use as the “window”
What kind of solid material would you chose and why?
HINT
Bond
Single
Double
Triple
Ionic
Use an ionic solid to avoid background
In the region of interesting covalent bonds
force constant
500 N/m
1000N/m
1500N/m
????
KBr
One difficulty here is the fact
That the windows are now
Water soluble
Any other considerations important?
Hint consider the scattered light equation
1
m 
2
Bond
Single
Double
Triple
k
m
k(Newton/m)
500
1000
1500
  c
1 

 c
Example: Calculate single bond band for
R-C-F;
1mole
 12 g  
  kg 
 26
mc  

2
x
10
kg




 mole   6x1023 atoms   103 g 
1mole
 19  
  kg 
 26
mF  

31
.
x
10
kg




 mole   6x1023 atoms   103 g 
m
m 
1
2
2 x10
2 x10
 26
 26

kg 31
. x10  26 kg
 
kg  31
. x10
 26

kg

 122
. x10  26 kg
kg
13 1
s2

3
.
22
x
10
s
122
. x10  26 kg
500
1
3.393x10
1
s  1080cm 1

cm
cm
3x1010
s
13
The C-F bond is shifted out of the
Region of the C-H bond.
3100-3010 cm-1
In your lab you used PTFE for your IR “plate”
Polytetrafluorethylene
100
90
2009 student data
80
http://www.internation
alcrystal.net/polycard.
htm
60
"blank PTFE"
50
40
normalized %T
70
30
1100
3100 (C-H)
1030
20
10
1210
3700
3200
2700
2200
1700
1150
1200
700
0
200
cm-1
1080 predicted C-F
120
100
60
%T
80
40
20
2000
1800
1600
1400
1200
cm-1
1000
800
600
0
400
Pretty hard to pull
Understandable changes
Out of here
Why is the “far” IR hard instrumentally?
Three primary reasons
1. Overlapping orders of light create a “forest” of bands
2. Intensity of the light source is low so we are trying to measure changes in small
signals
3. The beamsplitters used in the FT instrument do not transmit both forms of
polarized light equivalently
4. The detectors must be sensitive to very low amounts of light – implies that any
environmental noise will be a problem.
Here is an article detailing problems and solutions to creating Far IR instruments
http://infrared.phy.bnl.gov/pdf/homes/homes_01ins.pdf
B
A
Moving mirror
Fixed mirror
C
Beam
splitter
IR source
detector
Frequency of light
Constructive interference occurs when
1
AC  BC  n
2
From “really basic optics”
 f det ector
2 mirror  light
cn
AC  BC 
n/2
cm-1
0.5
overtonesin cm-1
1
n
2
Light at these frequencies
Will appear at 400 cm-1
1
1.5
2
2.5
3
3.5
4
4.5
5
100
125
150
175
200
225
250
275
300
325
350
375
400
425
450
475
500
525
550
575
600
625
650
675
700
725
750
775
800
825
850
875
900
925
950
66.66667
83.33333
100
116.6667
133.3333
150
166.6667
183.3333
200
216.6667
233.3333
250
266.6667
283.3333
300
316.6667
333.3333
350
366.6667
383.3333
400
416.6667
433.3333
450
466.6667
483.3333
500
516.6667
533.3333
550
566.6667
583.3333
600
616.6667
633.3333
50
62.5
75
87.5
100
112.5
125
137.5
150
162.5
175
187.5
200
212.5
225
237.5
250
262.5
275
287.5
300
312.5
325
337.5
350
362.5
375
387.5
400
412.5
425
437.5
450
462.5
475
40
50
60
70
80
90
100
110
120
130
140
150
160
170
180
190
200
210
220
230
240
250
260
270
280
290
300
310
320
330
340
350
360
370
380
33.33333
41.66667
50
58.33333
66.66667
75
83.33333
91.66667
100
108.3333
116.6667
125
133.3333
141.6667
150
158.3333
166.6667
175
183.3333
191.6667
200
208.3333
216.6667
225
233.3333
241.6667
250
258.3333
266.6667
275
283.3333
291.6667
300
308.3333
316.6667
28.57143
35.71429
42.85714
50
57.14286
64.28571
71.42857
78.57143
85.71429
92.85714
100
107.1429
114.2857
121.4286
128.5714
135.7143
142.8571
150
157.1429
164.2857
171.4286
178.5714
185.7143
192.8571
200
207.1429
214.2857
221.4286
228.5714
235.7143
242.8571
250
257.1429
264.2857
271.4286
25
31.25
37.5
43.75
50
56.25
62.5
68.75
75
81.25
87.5
93.75
100
106.25
112.5
118.75
125
131.25
137.5
143.75
150
156.25
162.5
168.75
175
181.25
187.5
193.75
200
206.25
212.5
218.75
225
231.25
237.5
22.22222
27.77778
33.33333
38.88889
44.44444
50
55.55556
61.11111
66.66667
72.22222
77.77778
83.33333
88.88889
94.44444
100
105.5556
111.1111
116.6667
122.2222
127.7778
133.3333
138.8889
144.4444
150
155.5556
161.1111
166.6667
172.2222
177.7778
183.3333
188.8889
194.4444
200
205.5556
211.1111
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
105
110
115
120
125
130
135
140
145
150
155
160
165
170
175
180
185
190
um
100
125
150
175
200
225
250
275
300
325
350
375
400
425
450
475
500
525
550
575
600
625
650
675
700
725
750
775
800
825
850
875
900
925
950
100
80
66.66667
57.14286
50
44.44444
40
36.36364
33.33333
30.76923
28.57143
26.66667
25
23.52941
22.22222
21.05263
20
19.04762
18.18182
17.3913
16.66667
16
15.38462
14.81481
14.28571
13.7931
13.33333
12.90323
12.5
12.12121
11.76471
11.42857
11.11111
10.81081
10.52632
200
250
300
350
400
450
500
550
600
650
700
750
800
850
900
950
1000
1050
1100
1150
1200
1250
1300
1350
1400
1450
1500
1550
1600
1650
1700
1750
1800
1850
1900
cm-1
200
400
800
1400
1800
2000
Why is the “far” IR hard instrumentally?
Three primary reasons
1. Overlapping orders of light create a “forest” of bands
2. Intensity of the light source is low so we are trying to measure changes in
small signals
3. The beamsplitters used in the FT instrument do not transmit both forms of
polarized light equivalently
4. The detectors must be sensitive to very low amounts of light – implies that any
environmental noise will be a problem.
Planck’s Blackbody Law
Stefan-Boltzmann Law
I  T4

 8h 3 


3
 c

Intensity
exp
h
kT
1
= Energy density of radiation
h= Planck’s constant
C= speed of light
k= Boltzmann constant
T=Temperature in Kelvin
= frequency
0
500
1000
1500
2000
nm
 max
b

T
Wien’s Law
From: “Really Basic Optics”
2500
3000
3500
 1
   3
 
8h
exp
hc
 kT
1
1. As  ↓(until effect of exp takes over)
2. As T,exp↓, 
Why is the “far” IR hard instrumentally?
Three primary reasons
1. Overlapping orders of light create a “forest” of bands
2. Intensity of the light source is low so we are trying to measure changes in small
signals
3. The beamsplitters used in the FT instrument do not transmit both forms of
polarized light equivalently
4. The detectors must be sensitive to very low amounts of light – implies that any
environmental noise will be a problem.
The intensity of light (including it’s component polarization) reflected as compared
to transmitted (refracted) can be described by the Fresnel Equations
p TE
perpendicular
parallel
s
TM
Medium 1
interface
Angle of transmittence
Is controlled by
The density of
Polarizable electrons
In the media as
Described by Snell’s Law
From “really basic optics”
Medium 2
T
Perpendicular will transmitted
to a greater degree than
parallel at the interface since it
oscillates into the second
medium
R  r 
R/ /  r/ / 
2
2
 i cosi  t cost 

 
 i cosi  t cost 
  t cos  i   i cos  t 

 
  i cos  i   t cos  t 
2
T  t  
2
2
T/ /  t / / 
2


2i cosi

 
 i cosi  t cost 


2i cosi

 
 i cost  t cosi 
The amount of light reflected depends upon the Refractive indices and
the angle of incidence.
We can get Rid of the angle of transmittence using Snell’s Law
sin  i  t

sin  t  i
Since the total amount of light needs to remain constant we also know that
R//  T//  1
R  T  1
From “really basic optics”
2
Therefore, given the two refractive
Indices and the angle of incidence can
Calculate everything
2
Consider and air/glass interface
i
0.8
“Take-home message” can select
For different polarization of light by
Controlling the angel of incidence
0.7
Perpendicular
Transmittance
0.6
0.5
0.4
0.3
Here the transmitted parallel light is
Zero! – this is how we can select
For polarized light!
0.2
Parallel
0.1
0
0
10
20
30
40
Angle of incidence
From “really basic optics”
This is referred to as the polarization
angle
50
Notice that as you change
The angle of the beam splitter
You can have very large
Asymmetry in the light
Reflected which leads to
“false absorbance”
http://infrared.phy.bnl.gov/pdf/homes/homes_01ins.pdf
Why is the “far” IR hard instrumentally?
Three primary reasons
1. Overlapping orders of light create a “forest” of bands
2. Intensity of the light source is low so we are trying to measure changes in small
signals
3. The beamsplitters used in the FT instrument do not transmit both forms of
polarized light equivalently
4. The detectors must be sensitive to very low amounts of light – implies
that any environmental noise will be a problem.
Where does the noise come from?
Pyroelectric detector frequency
Response is
Signal  VHz
cm

V  optical velocity  0.316 

s 
 
www.rsc.org/ej/PC/2000/b001200i/
1
radiation
cm
cm 

Signal   0.316  Hz

s 
400cm-1
10 cm-1
Range for Far IR
cm  
`1 

Signal   0.316   400  Hz ~ 125Hz

s 
cm 
cm   `1 

Signal   0.316   10  Hz ~ 3Hz

s   cm 
Room noise from the electric lines is 60 Hz which smack dab in the middle
So we will get a lot of room noise also
8. Other Spectroscopies
1.IR
2.NMR
Excellent sites for information on NMR can be found within the Analytical Sciences
Digital Library: www.ASDLib.org
Below are some great ones found there
http://www.cem.msu.edu/~reusch/VirtualText/Spectrpy/nmr/nmr1.htm
http://www-keeler.ch.cam.ac.uk/lectures/understanding/chapter_5.pdf
Nuclear Magnetic Resonance
External Magnetic Field, Bo, causes the energy associated with the spin to split
spin
+
Causes a
Local magnet
Orange=
magnetic moment
Green equals external
Magnetic field
http://www.cit.gu.edu.au/~s55086/qucomp/qubit.html
NMR active species
NMR active
Odd mass nuclei have fractional spins
1H, 13C, 19F I = ½
17O I = 5/2
Even mass nuclei with odd numbers of protons and neutrons
have integral spins
2H, 14N I = 1
Even mass nuclei with even number protons and neutrons
have zero spin
12C, 16O I = 0
I = spin quantum number ½ is active
1H
19F
31P
13C
magnetic moment in magnetons (5.05078x10-27 J/T)
2.7927
2.6233
1.1305
0.7022
What are the spin numbers of the four isotopes of
Lead shown below?
Mass
204
206
207
208
Average % Relative Abundance
1.36
25.4
22.7
52.1
I?
1H
2H
11B
13C
17O
19F
29Si
31P
207Pb
%
Natl Abund
99.9844
0.0156
81.17
1.108
0.037
100
4.7
100
22.7
What did
We get
From
Preceding
Slide?
spin
I
0.50
1.00
1.50
0.50
2.50
0.50
0.50
0.50
5.05078e-27 J/T
10e7 rad/(Ts)
nuclear Magnetons magnetogyric ratio
2.7927
26.753
0.8574
4,107
2.688
-0.7022
6,728
-1.893
-3,628
2.6273
25,179
-0.5555
-5,319
1.1305
10,840
6


Example 1H
 

 27 J 
5
.
05078
x
10

  2rad 
T

magnetons

magnetons
cycle






6.626x10
 34

J s I
 
 2
hI
rad    

  4.8 x10 7
 

T  s  I 
 107 rad 

7 rad   2.7927magnetons 
8 rad
   4.8x10
 26.7

  2.67 x10





Ts
0.5
Ts
 Ts 
Calculate the nuclear magnetic moment for Pb207

c


m=-1/2β
3x108
E  h
m
s
frequency
1
s
v  c
 h 
E   
B
 2  o
m=-1/2β m=+1/2
 1 
     Bo
 2 
Bo
m=+1/2

c


Larmor
frequency
c2
Bo
All describe the splitting
Of nuclear states equally
Well BUT…. One is more
convenient
Splitting of Orbitals as a function of Applied magnetic field
80
4.50E+08
70
4.00E+08
1H Wavelength m
3.00E+08
 1 
B
 2  o
50
 
40

c


2.50E+08
c2
Bo
2.00E+08
30
1.50E+08
tv
FM
radio
1H Frequency Hz
3.50E+08
60
20
1.00E+08
10
5.00E+07
0
0.00E+00
0
1
2
3
4
5
6
7
8
9
10
Bo, Tesla
Because there is a linear relationship between the applied magnetic
Field and the splitting frequency is the preferred unit
Signal
m=-1/2β
250
1H
200
300
MHz
150
300 MHz
Frequency, MHz
100
13C
50
0
0
1
2
3
4
5
6
7
8
9
10
-50
13C
-100
-150
m=+1/2
-200
1H
-250
Bo, Tesla
Hold the Magnetic field constant and scan the frequency while monitoring Absorption ……
Where will we find the various resonances?
Change table
 1 
B
 2  o
 
Tesla
1H
2H
11B
13C
17O
19F
29Si
31P
207Pb
%
Natl Abund
99.9844
0.0156
81.17
1.108
0.037
100
4.7
100
22.7
spin
I
0.50
1.00
1.50
0.50
2.50
0.50
0.50
0.50
7.05
4.7
2.35
5.05078e-27 J/T
10e7 rad/(Ts)
resonant proton frequency
nuclear Magnetons magnetogyric ratio 300 MHz 200 MHz 100 MHz
2.7927
26.753
300
0.8574
4,107
46.1
2.688
-0.7022
6,728
75.5
-1.893
-3,628
2.6273
25,179
283
-0.5555
-5,319
1.1305
10,840
122
6
The designation of an instrument as 300, 200, or 100 MHz NMR is
Referring to the natural resonance of the proton under the application
Of magnets varying between 7.05 to 2.35 Tesla
As an exercise in preparation for an exam you should fill in the table.
Where will we find 207Pb resonances?
300 MHz
MHz
Why are the sensitivities so low?
1
0.9
0.8
1H
0.001
0.0009
0.0008
0.7
19F
0.0007
Relative Signal
Relative Signal
1H
2H
13C
19F
31P
Frequency Relative Sensitivity
300
1
46.1
1.45E-06
75.5
1.76E-04
283
8.30E-01
122
6.63E-02
Different atoms have different magnetogyric
Ratios and therefore absorb at different frequencies
0.6
0.0006
0.0005
0.0004
13C
0.0003
0.5
0.0002
0.0001
0
0
0.4
50
100
150
200
250
300
350
Frequency (in MHz)
0.3
0.2
31P
0.1
0
0
50
100
150
200
250
100 MHz
Frequency, (in MHz)
300
350
Argument on the size of signals that follows is from Atkins, Phys. Chem. p. 459, 6th Ed
Stimulated Emission
*
w  BN o
o
Photons can stimulate
Emission just as much
As they can stimulate
Absorption
(idea behind LASERs
Stimulated Emission)
w  BN *
The rate of stimulated event is described by :
Where w =rate of stimulated emission or absorption

Is the energy density of radiation already present at the frequency of
the transition
B= empirical constant known as the Einstein coefficient for stimulated
absorption or emission
N* and No
are the populations of upper state and lower states

can be described by the Planck equation for black body radiation at some T

 8h 3 


3
 c 
exp
h
kT
1
In order to measure absorption it is required that the
Rate of stimulated absorption is greater than the
Rate of stimulated emission
wabsorption  BN o  wenussion  BN *
N o  N *
If the populations of * and o are the same the net absorption is zero as a photon is
Absorbed and one is emitted
Scale the difference to the total population to get relative signal one can expect
To get:
No  N * No  N *
Signal 

N total
No  N *
Boltzman distribution can help us describe this
N*
 exp
No
Signal 

E
kt
 exp
N o  N o exp
N o  N o exp
Signal 
1  exp
1  exp

hc
kT

hc
kT


h
kT
h
kT
h

kT

1  exp
1  exp

h
kT

h
kT
v  c
Signal 
1  exp
1  exp


hc
kT
h=6.626x10-34 Js
k= 1.381x10-23 J/K
c=3x108m/s
hc
kT
In a UV-Vis experiment, 400 nm, at room temperature (298 K)

N*
 exp
N0
 6.626 x10
 34

m

J  s  3 x 108 

s


 23 J 
9
 1.381 x 10
  298 K  400 x 10 m

K
1  exp 120.7
Signal 
1
120.7
1  exp

N*
 exp 120.7
No
Signal relative to the concentration
Is 1
For an NMR experiment the exponential term is larger
For an NMR experiment
N*
 exp
No

E
kt
 exp

h
kT
 exp

 

h
Bo 
 2

kT
h=6.626x10-34 Js
k= 1.381x10-23 J/K
c=3x108m/s
For a proton at 4.7 Tesla, magnetogyric ratio of 2.67x108rad/(Ts), at Room Temp, 298 K


1 

 6.626 x 10  34 J  s  2 .67 x 108
  4 .7 

T s 
N*
 exp
No
1.381x10
 23 J
K
 298 K 
 exp  0.000202
1  exp  .0002 1  0.9998002
5
Signal 


6
.
66
x
10
1  exp  .0002 1  0.9998002
NMR signal relative to the concentration is 6.66x10-5
Relatively WEAK!
In order to get a signal that is measureable a relaxation experiment is performed
Consider, for example, a temperature jump
Small
Energy
Gap
Means
High
N*
population
N*
 exp
No
heat

E
kt
 exp

h
kT
Monitor
decay
As T (hc/kT)↓ so exp so N*  so
Signal based on decay 
We can accomplish a “temperature jump” by altering the applied
External field
250
which changes the energy difference between the
Ground and excited states and alters the populations of
the two
200
1H
150
Frequency, MHz
100
13C
50
0
0
1
2
3
4
5
6
7
8
9
10
-50
13C
-100
-150
-200
Magnetic Pulse populates
A range of energy states (frequencies)
1H
-250
Bo, Tesla
Signal will be the sum of all the various protons that respond to those different
frequencies
Consider two frequencies summed but not too far apart – a “beat” frequency
Will result
E  sin( A)  sin B
E  2 sin(
1
1
A

B
cos(


 A  B)
2
2
This is our signal up to
This point
2.5
2.5
2
2
1.5
1.5
1
amplitude
amplitude
1
0.5
0.5
0
0
100
200
300
400
500
600
700
-0.5
0
0
100
200
300
400
-0.5
500
600
700
800
-1
-1.5
-1
-2
-1.5
-2.5
time
-2
-2.5
time
But~remember it will decay!
800
Consider the decay involved in a temp jump experiment
In many types of experiments this is done by performing a temperature jump
Which creates a greater population of the excited state: consider the reaction
At Temp1
k f ,T 1

M*
M k
,bT 1



d M
dt
At equilibrium
d M
dt
Therefore

  k fT 1  M   k bT 1  M *
0


*
k fT 1 M equilibrium  k bT 1 M equilibrium

Increase the temperature instantaneously, the temp 1 equilibrium concentrations
Must change to the new temp 2 equilibrium values with a rate driven by the new
Rate constants



*
k fT 2 M equilibrium,T 2  k bT 2 M equilibrium
,T 2

The old equilibrium values deviate from the new equilibrium by some value
x
M
equilibrium ,T 1
  x M
and
M *
equilibrium ,T 1
equilibrium ,T 2
  x M *

equilibrium ,T 2

The concentration of M therefore changes as:
d M 
dt
d M 
dt
 
  k T 2 x  M equilibrium,T 2

  k  x   M *
bT 2

equilibrium,T 2

  k T 2 x   k T 2 M equilibrium,T 2  k bT 2 x  k bT 2 M *equilibrium,T 2


recall



*
k fT 1 M equilibrium  k bT 1 M equilibrium

so
d M 
dt

  k T 2 x  k bT 2 x   x k fT 2  k bT 2
First order reaction with solution of
x  xo exp

t


Represents the decay
In the signal after
A jump
1

 k f  kb
http://www.cit.gu.edu.au/~s55086/qucomp/qubit.html
Our “decay” represents a flip in the magnetic moment.
How can we easily represent a flip?
Define an xy plane
Describe as the sum of two vectors which cancel
In the xy plane and sum value in the z plane
Flip of the magnetic moment is described as
The motion of two vectors
Longitudinal Relaxation, T1
M z ( t )  M eq  exp

t
T1
Related to motion of molecules
Which usually is not very interesting to us
Motion of molecules
In addition there is the “fanning out
Of the spins” in the xy plane
Transverse Relaxation, T2
T2 < = T1
Note that this image
Suggests out of phase
Signals implying phase
Manipulation of the acquired data
M y ( t )  exp

t
T2
What is this component of the signal due to?
Due to the fact that the local magnetic
Field is not entirely in the z direction due
To contributions from neighboring nuclei
Spin/spin
Coupling
Lifetime of
T2
spin
+
Causes a
Local magnet
Local magnetic
Fields associated with
Electrons alter the magnitude
Of the external field experienced
spin
+
Causes a
Local magnet
Coupling
Described by
Constant, J,
 1  Which is a measure
     Bo Of the energy
 2 
Of the interaction
What we have “learned” so far:
1. Pushed sample into an excited state which will emit a radiofrequency
response on return to equilibrium
signal  Sin( A)
Radiofrequency
2. Spin spin coupling can also alter the energy levels as described by a coupling
constant, J.
3.
Spin spin coupling results in fixed lifetime of the response, leading to an
exponential decay of the signal

exp
t
T2
4. The net signal in the time domain is described by:

Sin( A) exp
t
T2
1.5
Free Induction Decay (FID) due to T2
1
FID= Free Induction Decay
Amplitude
0.5
0
0
50
100
150
200
250
300
350
400
-0.5
-1
The longer the signal lasts
The larger T2
-1.5
Sample frequency t (s)

Sin( A) exp
t
T2
Decay of signal
We learned (a long time ago) that a square wave can be modeled as
The sum of a series of sine waves which can be “uncovered” by an FT
F f  



f  t  exp  2jft dt
Digitally filter the high frequency
FT
subtract
15000
10000
Amplitude
5000
0
0
0.2
0.4
0.6
-5000
-10000
-15000
Time (s)
0.8
1
1.2
Anti-FT
An exponential decay can also be described by the sum of a series of sine
Waves.:
1.2
1
Amplitude
0.8

exp
0.6
t
T2
0.4
120
0.2
0
100
0
200
400
600
800
1000
1200
Time
Amplitude
80
60
40
20
0
1
4
7
10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67
frequencies
1.5
Free Induction Decay (FID) due to T2
60
1
50
40
0
0
50
100
150
200
250
300
350
400
-0.5
Amplitude
Amplitude
0.5
30
20
-1
10
-1.5
0
1
Sample frequency t (s)

Sin( A) exp
t
T2
4
7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73
Frequences
FT
Decay of signal

FT of Sin( A) exp
Why did we “all of a sudden” get some frequencies
On the left side of the peak?
This distribution is fit by a Lorentzian population
t
T2
For those interested: how a Lorentzian population is derived mathematically
NMR peak shapes are NOT described Gaussian, but Lorentzian distribution
Probability of
Length x as a function
Of b and theta

 bdx 
d   2

 b  x2 
b
x
x
tan  
b
P( x )
 x
 b
  tan 1  

 1
d  
x2

1 2

b
The cumulative probability over all angles is

1 
b

  2
2
  b   x  m 
Where m is the peak location and b
Is the half width at half height

 dx
  
 b 


Another form is


a

y  
2
 c   x  b 
In this form a is the half
Peak width and b is the
Location of the peak


a

y  
2
 c   x  b 
Where c is a/H and h is the height of the peak
This can be converted to a quadratic
1 x 2 2bx
b2


c
y
a
a
a
Lorentzian Peak shape
In this form
a is the half Peak width
b is the Location of the peak
The signal width at half height is related to this
lifetime
a  1/ 2 
1
T2
Calc. the transverse relaxation time if a peak has a half height width of 10 Hz
0.7
1
PG 
exp
 2
2
x b


0.5
2 1
0.4
Lorentzian
PL 
 a
 
 2
1

Lorentzian is sharper
0.6
Gaussian
 x  b 2
 a
 
 2
2
0.3
0.2
0.1
0
-5
-4
-3
-2
-1
0
1
2
3
4
5
What have we learned so far that affects instrumentation?
1.
2.
3.
4.
5.
6.
7.
8.
External magnetic field
Signal depends on exact magnetic field so it will need to be uniform
Need to pulse the magnetic field (width of pulse is related to number of frequencies)
Signal will involve several frequencies (beats) due to chemical information
Need to use FT to find those frequencies.
Signal will decay (causes a Lorentzian shape)
Decay will have a phase component so need to worry about phasing of the signal
Want to be able to capture the peak at half height because it gives chemical
Information.
Typical Experiment:
0. Insert sample and lock
1. Shim (tune) the magnetic field to be constant within the sample cell
Starting population ratio set by starting applied magnetic field; temperature
2. Jump the population (Pulse applied magnetic field) by applying a square
wave through the coil.
3. Watch the magnetization decay: cause small electric currents in a coil
surrounding the sample.
4. Currents are time based and are the sum of all the frequencies
Set a) acquisition time and b) sampling frequency
5. Repeat as desired to increase the S/N, allowing a delay time to equilibrate
6. Convert from time to frequency by FTing the measured electric signal
7. Display the signals as a function of frequency.
8. Adjust for the phase of the T2 signal
9. Scale the frequencies relative to some standard compound.
http://images.google.com/imgres?imgurl=http://nmr.chem.ualberta.ca/nmr_news/figures/283_8360.JPG&imgrefurl=http://nmr.chem.
ualberta.ca/AOVNMR_course/chapter_5.htm&h=1338&w=460&sz=132&hl=en&start=42&tbnid=8mLZIhy1rkvZM:&tbnh=150&tbnw=52&prev=/images%3Fq%3Dvarian%2BNMR%2Bscreen%2Bimage%26start%3D40%26gbv%3
D2%26ndsp%3D20%26hl%3Den%26sa%3D
Insert Sample
Monitor 2H signal as a way of checking
for magnetic field drift
Here you attempt to get a nice
large signal for 2H (46.1 MHz)
http://images.google.com/imgres?imgurl=http://nmr.chem.ualberta.ca/nmr_news/figures/283_8360.JPG&imgrefurl=http://nmr.chem.
ualberta.ca/AOVNMR_course/chapter_5.htm&h=1338&w=460&sz=132&hl=en&start=42&tbnid=8mLZIhy1rkvZM:&tbnh=150&tbnw=52&prev=/images%3Fq%3Dvarian%2BNMR%2Bscreen%2Bimage%26start%3D40%26gbv%3
D2%26ndsp%3D20%26hl%3Den%26sa%3D
Locking
Monitor this
Signal as a check
On the stability
Of the homogeneity
Of the magnetic field
Typical Experiment:
0. Insert sample and lock
1. Shim (tune) the magnetic field to be constant within the sample cell
Starting population ratio set by starting applied magnetic field; temperature
2. Jump the population (Pulse applied magnetic field) by applying a square
wave through the coil.
3. Watch the magnetization decay: cause small electric currents in a coil
surrounding the sample.
4. Currents are time based and are the sum of all the frequencies
Set a) acquisition time and b) sampling frequency
5. Repeat as desired to increase the S/N, allowing a delay time to equilibrate
6. Convert from time to frequency by FTing the measured electric signal
7. Display the signals as a function of frequency.
8. Adjust for the phase of the T2 signal
9. Scale the frequencies relative to some standard compound.
http://bouman.chem.georgetown.edu/nmr/nuts/shims.htm
Shimming involves tuning the magnetic field across the sample so that it is
Homogeneous – in effect you are “tuning” the x axis of the experiment
http://images.google.com/imgres?imgurl=http://nmr.chem.ualberta.ca/nmr_news/figures/283_8360.JPG&imgrefurl=http://nmr.chem.
ualberta.ca/AOVNMR_course/chapter_5.htm&h=1338&w=460&sz=132&hl=en&start=42&tbnid=8mLZIhy1rkvZM:&tbnh=150&tbnw=52&prev=/images%3Fq%3Dvarian%2BNMR%2Bscreen%2Bimage%26start%3D40%26gbv%3
D2%26ndsp%3D20%26hl%3Den%26sa%3D
Shimming
Manipulates
The magnetic
Field in
Multiple dimensions
To achieve
consistency
One way to ensure that your sample has as homogeneous a magnetic
Field as you can get is to:
Spin the sample
rmn.iqfr.csic.es/guide/man/bsms/chap5.2.htm
Link worked in 2008, source of material, but no longer works
Typical Experiment:
1. Shim (tune) the magnetic field to be constant within the sample cell
Starting population ratio set by starting applied magnetic field; temperature
2. Jump the population (Pulse applied magnetic field) by applying a square
wave through the coil.
3. Watch the magnetization decay: cause small electric currents in a coil
surrounding the sample.
4. Currents are time based and are the sum of all the frequencies
Set a) acquisition time and b) sampling frequency
5. Repeat as desired to increase the S/N, allowing a delay time to equilibrate
6. Convert from time to frequency by FTing the measured electric signal
7. Display the signals as a function of frequency.
8. Adjust for the phase of the T2 signal
9. Scale the frequencies relative to some standard compound.
Things YOU control:
a. Shimming
b. Pulse width
The pulse width will control the range of frequencies
That can be sampled.
It will have to change with respect to different
1proton resonance frequencies
250
1H
200
150
Frequency, MHz
100
15000
10000
13C
50
0
0
1
2
3
4
5
7
8
9
10
13C
5000
Amplitude
6
-50
-100
0
0
0.2
0.4
0.6
0.8
1
1.2
-150
-200
-5000
1H
-250
-10000
Bo, Tesla
-15000
Time (s)
Example of Pulse Width
Assume 500 MHz NMR. The proton frequency range at this field strength is 5000 to
7000 Hz. Will a pulse width of 8 uz sample that range?
8s pulse
TYPICALLY ASSUME MINIMUM 5 FREQUENCIES
y square (t ) 
4
1
1

 sin2ft   sin6ft   sin10ft .....


3
5
 cycles   1 
f base  
 

 s   2 pulse 
At a minimum we should get the 5th frequency above this so
That the bandwidth would be
15000
10000
Amplitude
5000
0
0
0.2
0.4
0.6
-5000
-10000
-15000
Time (s)
0.8
1
1.2
 5 
f 

 2 pulse 
For our example this is:
5


f 
 312500 Hz
6 
 2 x8 x10 s 
Typical Experiment:
1. Shim (tune) the magnetic field to be constant within the sample cell
Starting population ratio set by starting applied magnetic field; temperature
2. Jump the population (Pulse applied magnetic field) by applying a square
wave through the coil.
3. Watch the magnetization decay: cause small electric currents in a coil
surrounding the sample.
4. Currents are time based and are the sum of all the frequencies
Set a) acquisition time and b) sampling frequency
5. Repeat as desired to increase the S/N, allowing a delay time to equilibrate
6. Convert from time to frequency by FTing the measured electric signal
7. Display the signals as a function of frequency.
8. Adjust for the phase of the T2 signal
9. Scale the frequencies relative to some standard compound.
Number of points is important for 2 reasons
aliasing (Nyquist frequency minimum)
and 2^x for FT processing
1.5
1.5
1
1
Aliasing
0.5
1.5
0.5
0
0
200
400
600
800
1000
1200
0
0
20
40
60
80
100
120
140
160
180
-0.5
-0.5
-1
1
-1
-1.5
-1.5
0.5
0
0
200
400
600
800
1000
1200
-0.5
-1
-1.5
Sampling of a sine wave at too low number of points results in some
Psychedelic types of patterns, need at least three points to describe a sine wave
200

 

2 xsweep width  sw
1
Digital Re solution  
 

acquisition
time
at
number
ofdata
po
int
s
np





 

number of data points (np): 4K
Fourier number (fn): 4K
acquisition time: 0.35 sec
zero-filling: none
number of data points (np): 8K
Fourier number (fn): 8K
acquisition time: 0.70 sec
zero-filling: none
number of data points (np): 16K
Fourier number (fn): 16K
acquisition time: 1.40 sec
zero-filling: none

 

2 xsweep width  sw
1
Digital Re solution  
 

acquisition
time
at
number
ofdata
po
int
s
np





 

number of data points (np): 16K
Fourier number (fn): 16K
acquisition time: 1.40 sec
zero-filling: none
number of data points (np): 24K
Fourier number (fn): 32K
acquisition time: 2.00 sec
zero-filling: 1.33 x
number of data points (np): 24K
Fourier number (fn): 64K
acquisition time: 2.00 sec
zero-filling: 2.66 x
Typical Experiment:
1. Shim (tune) the magnetic field to be constant within the sample cell
Starting population ratio set by starting applied magnetic field; temperature
2. Jump the population (Pulse applied magnetic field) by applying a square
wave through the coil.
3. Watch the magnetization decay: cause small electric currents in a coil
surrounding the sample.
4. Currents are time based and are the sum of all the frequencies
Set a) acquisition time and b) sampling frequency
5. Repeat as desired to increase the S/N, allowing a delay time to equilibrate
6. Convert from time to frequency by FTing the measured electric signal
7. Display the signals as a function of frequency.
8. Adjust for the phase of the T2 signal
9. Scale the frequencies relative to some standard compound.
pulse
acquire
delay
http://www-keeler.ch.cam.ac.uk/lectures/understanding/chapter_5.pdf
This link is still good, 2009
This chapter describes the mathematics of the instrument
In terms of “getting” the signal.
http://www.chemistry.nmsu.edu/Instrumentation/NMSU_NMR300_J.html
Typical Experiment:
1. Shim (tune) the magnetic field to be constant within the sample cell
Starting population ratio set by starting applied magnetic field; temperature
2. Jump the population (Pulse applied magnetic field) by applying a square
wave through the coil.
3. Watch the magnetization decay: cause small electric currents in a coil
surrounding the sample.
4. Currents are time based and are the sum of all the frequencies
Set a) acquisition time and b) sampling frequency
5. Repeat as desired to increase the S/N, allowing a delay time to equilibrate
6. Convert from time to frequency by FTing the measured electric signal
7. Display the signals as a function of frequency.
8. Adjust for the phase of the T2 signal
9. Scale the frequencies relative to some standard compound.
pulse
acquire
delay
http://www-keeler.ch.cam.ac.uk/lectures/understanding/chapter_5.pdf
This chapter describes the mathematics of the instrument
In terms of “getting” the signal.
http://images.google.com/imgres?imgurl=http://nmr.chem.ualberta.ca/nmr_news/figures/283_8360.JPG&imgrefurl=http://nmr.chem.
ualberta.ca/AOVNMR_course/chapter_5.htm&h=1338&w=460&sz=132&hl=en&start=42&tbnid=8mLZIhy1rkvZM:&tbnh=150&tbnw=52&prev=/images%3Fq%3Dvarian%2BNMR%2Bscreen%2Bimage%26start%3D40%26gbv%3
D2%26ndsp%3D20%26hl%3Den%26sa%3D
Free Induction Decay of an NMR signal in a 1D experiment with a 50 msec expansion showing the
digitization of the signal.
http://edison.mbi.ufl.edu/bch6745/lecture3.pdf
A 90 degree pulse followed by turning on the detector
leads to a Free Induction Decay (FID)
(J) and altered
frequency
The signal decays
Different decay rates can be seen
NMR FID of Methyl -D-Arabinofuranoside in CD3CN.
Collected at 11.7 T by Jim Rocca in AMRIS.
13C
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9 sec
http://edison.mbi.ufl.edu/bch6745/lecture3.pdf
Expansion of previous FID
Most of the signal in the fast
relaxing peak is gone by around
100 ms.
The longer T2* the greater the resolution
(narrower the peaks)
0.02
0.04
0.06
0.08
0.10
0.12
Peak to peak spacing is
just under 50 ms,
contains J spin spin
energy coupling
0.14
0.16
0.18
Now we can see complicated fine structure
0.20
sec
http://edison.mbi.ufl.edu/bch6745/lecture3.pdf
Fourier Transform of previous FID
Some peaks are bigger
than others (CD3CN)
The FID had several different
frequencies. These are called
chemical shifts.
0.02
0.04
0.06
0.08
0.10
0.12
120
0.14
0.16
0.18
0.20
110
This peak has other
stuff going on.
sec
100
90
80
70
60
50
40
30
20
10
ppm
http://edison.mbi.ufl.edu/bch6745/lecture3.pdf
Expansion of big peaks
The width of this peak is about 10 Hz. This
is R2* (or 1/T2*). The * indicates that it is
the natural linewidth + experimental sources
of inhomogeneities.
Scalar (J) coupling.
This J is just over
20 Hz.
The width of
these peaks is
around 1 Hz.
They come from
the slowly
relaxing part of
the FID.
The value for the J
coupling is the inverse of
the spacing between beats
on the FID!
120
110
100
90
80
70
60
50
40
30
20
10
ppm
Typical Experiment:
1. Shim (tune) the magnetic field to be constant within the sample cell
Starting population ratio set by starting applied magnetic field; temperatur
2. Jump the population (Pulse applied magnetic field) by applying a square
wave through the coil.
3. Watch the magnetization decay: cause small electric currents in a coil
surrounding the sample.
4. Currents are time based and are the sum of all the frequencies
Set a) acquisition time and b) sampling frequency
5. Repeat as desired to increase the S/N, allowing a delay time to equilibrate
6. Convert from time to frequency by FTing the measured electric signal
7. Display the signals as a function of frequency.
8. Adjust for the phase of the T2 signal
9. Scale the frequencies relative to some standard compound.
What is effect of phase shifting on FT?
Typically measure only one
Of the phases OR
Convert both to a single phase
Starting FT data point
(0 o)
50
1.2
1
40
0.8
0.6
30
0.2
0
0
50
100
150
200
250
300
350
20
400
-0.2
Amplitude
Amplitude
Signal (sin)
0.4
-0.4
-0.6
-0.8
-1
Starting
FT data
Point 90o
Time
10
0
1
5
9 13 17 21 25 29 33 37 41 45 49 53 57 61 65 69 73 77 81 85 89 93
93
-10
Starting FT data point -20(180o)
-20
-30
-30
Frequency
Frequency
Typical Experiment:
1. Shim (tune) the magnetic field to be constant within the sample cell
Starting population ratio set by starting applied magnetic field; temperatur
2. Jump the population (Pulse applied magnetic field) by applying a square
wave through the coil.
3. Watch the magnetization decay: cause small electric currents in a coil
surrounding the sample.
4. Currents are time based and are the sum of all the frequencies
Set a) acquisition time and b) sampling frequency
5. Repeat as desired to increase the S/N, allowing a delay time to equilibrate
6. Convert from time to frequency by FTing the measured electric signal
7. Display the signals as a function of frequency.
8. Adjust for the phase of the T2 signal
9. Scale the frequencies relative to some standard compound.
Nuclear Magnetic Resonance
External Magnetic Field, Bo, causes the energy associated with the spin to split
spin
Local magnetic
Fields associated with
Electrons alter the magnitude
Of the external field experienced
+
Causes a
Local magnet
 1 
B
 2  o
 
Shielding
Effect from
Local electrons
 1 
 B 1   
 2  o
 
 1 
 B 1   
 2  o
 
400   1 

  1  
    
B
1



 o 
 Bo  1   Sample
 
TMS 
  2  
  2  


 0
  1  
    
 Bo   Sample   TMS \
  2  
300


  0.3
Frequency (MHz)
200
100
  0.8
0
0
2
4
6
8
10
12
14
  0.8
16
-100
  0.3
-200
-300
CH3
H3C Si CH3
 0
CH3
-400
Bo, Tesla
The change in the absorption frequency
Gives information about the chemical environment
The change is compared to
A standard (TMS)
http://www.cem.msu.edu/~reusch/VirtualText/Spectrpy/nmr/nmr1.htm
  1  
    
 Bo   Sample   TMS \
  2  


400
The magnitude of the shielding effect
Depends directly on the External Applied
Magnetic Field
300
 0
  0.3
Frequency (MHz)
200
100
0
0
2
4
6
8
10
12
14
  0.8
16
-100
-200
-300
 
  0.8
To be able to compare between
Instruments the data is normalized
By the external magnetic field resonant
Frequency for a non shielded electron
-400 1  
 Bo   Sample   TMS \
 
  2  

  1  
 Bo 
 
  2  



Bo, Tesla
  Sample   TMS
  0.3
 0
These normalized differences
Are typically 10-6 in magnitude
So the numbers are multiplied by
106 to avoid working with small numb
   Sample   TMS or other reference 10 6
C-H
Chemical shift
O-H
Less electron
Density
Less shielding
Aryl protons experience
Larger local magnetic
Field rather than lower
So it is seen as less
shielding
More shielding
http://edison.mbi.ufl.edu/bch6745/lecture3.pdf
Chemical Shifts
Chemical shifts are influenced by the electronic environment. Therefore, they are
diagnostic for particular types of molecular structures. The following figure
indicates average ranges of chemical shifts for different types of molecules.
Table from:
http://www.cem.msu.edu/~reusch/OrgPage/nmr.htm
Here is nice correlation chart from the web
http://www.chem.uic.edu/web1/ocol/spec/HTable.htm
Link still works 2009
Try Problem 1:
http://www.chem.uic.edu/web1/ocol/spec/NMR.htm
# of Peaks Observed in Proton NMR
2 NI  1 # peaks
# of atoms
Proton I= 1/2
N  1  # peaks
2+1
CH3-CH2-OH
3+1
Split these
Protons into
4 peaks
1+1
Split
These
Protons
Into 3 peaks
Split these protons
Into 2
2x4=8
Expected peak heights (if not split) 3: 2:1
For more problems see
http://www.chem.uic.edu/web1/ocol/spec/NMR3.htm
CH3-CH2-OH
3 chemical environments
Proportion of protons (3:2:1) is correct
Expected splitting (by order of peak
Size) was expected to be 3:(4x2):3
Splitting pattern observed:
3: 4: 1
4 instead of 8 tells us what?
Keep thinking – we will come
Back to this
Try Problems 2&3
Correction to 3:C2H4O
http://www.chem.uic.edu/web1/ocol/spec/NMR.htm
How many peaks for EDTA?
O
O
-
O
O
-
N
4x2=8
2x2=4
O
N
O
-
O
O
-
pH
11.3
O
O
-
O
10.3
O
-
9.3
8.3
N
7,3
O
N
O
-
O
-
O
How many peaks for Pure EDTA?
2 with relative intensity of 8:4 or
2:1
6.3
5.3
Starting at pH 11.3 and moving more acidic
,protonation of N can
Be seen in the shift of peak location, pKa
Values for NH3 around 9.2
What happened to the peaks outlined
In red here?
O
O
-
O
O
pH
-
N
11.3
O
10.3
N
O
-
O
-
9.3
O
8.3
7,3
O
O
O
-
O
N
+
O
-
O
H
H
N
-
O
6.3
O
-
+
5.3
O
O
N
O
-
O
N
-
O
O
O
O
-
a  1/ 2 
O
O
-
O
O
+
N
-
H
O
N
O
-
O
O
-
The broader peaks at ~ pH 9-10 mean
That the lifetime is SHORTER.
O
O
-
H
O
O
O
-
+
-
N
O
-
N
O
1
T2
O
-
What might make the lifetime shorter at
That pH?
O
O
-
O
O
pH
-
N
11.3
O
10.3
N
O
-
O
-
9.3
O
O
O
-
O
O
N
+
O
-
O
H
N
H
-
8.3
O
O
O
O
-
O
-
O
+
+
N
N
N
O
-
H
O
O
O
O
O
-
-
O
O
N
-
O
-
-
O
O
H
O
-
O
-
O
O
-
N
6.3
O
N
O
O
O
O
Why do we choose this model instead
Of?
O H
O
O
HO
Would not
Expect much
Effect here
a’
N
a
O
N
O
-
O
O
7,3
-
+
-
-
5.3
a  1/ 2 
1
T2
1. pH at which we observe the effects
2. Both types of protons are affected
3. Would expect splitting of carboxylic
arm protons due to different spatial
orientation of protons
O
O
-
O
O
pH
-
N
11.3
O
N
O
-
O
-
10.3
O
O
O
-
O
O
N
+
O
-
O
H
N
H
-
O
O
O
O
-
O
-
O
+
+
N
-
O
-
-
H
O
O
O
O
N
O
-
O
N
-
O
-
-
9.3
O
O
H
O
N
O
-
O
O
O
-
8.3
+
-
N
O
N
O
O
O
O
-
7,3
6.3
pKa EDTA 1.99; 2.67; 6.16; 10.2
5.3
a  1/ 2 
1
T2
And remember we just did
This one
We predicted 3 peaks for this OH group
We predicted 8 peaks (4x2) for the CH2
group what happened?
As our EDTA example suggested
We are looking at conformational
changes
Lines can be broader if the local chemical environment is shifting
Due to the dynamics of equilibrium
Peter Atkins 6th Ed. Physical Chemistry, Freeman
Fast = single line, mean of both
Intermediate = very broad
Slow = two lines
Predicted Number of Peaks in addition to the main proton peaks
For the Pb-EDTA compound from the 207 Pb
2 N * I   1  # peaks
O
O
-
O
O
-
4x2=8
N
I Pb 207  1 / 2
2x2=4
O
N
O
1
 

2
1
*

1


  Pb 207 2 
  2 peaks


-
O
-
O
Expect to that EDTA bound to 207 should split into two peaks
For each type of proton
300 MHz instrument
 
1
  1  
 Bo   Sample   TMS \
 

 2  
signal scaled to largest PbEDTA peak
0.9


  1  
 Bo 
 
  2  

  Sample   TMS

What is the coupling, J, in Hz?
0.8
(1.72-1.66)x10-6)*300x106)=18 Hz
J – coupling of
207 Pb to H
0.7
What is the lifetime?
0.6
Peak width at half height =
(1.69-1.68)x10-6)*300x106)= 3 Hz
0.5
0.4
Lifetime = 1/((3.14)*3(1/s))=106ms
1.68 1.69
0.3
1.66
0.2
1.72
0.1
0
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
-1 X ppm scaled to standard =0
a  1/ 2 
1
T2
Using the area related to spin coupling
Of Pb207 what is the % 207 in the sample?
0.7
1
PG 
exp
 2
2
x b


0.5
2 1
0.4
Lorentzian
PL 
 a
 
 2
1

Lorentzian is sharper
0.6
Gaussian
 x  b 2
 a
 
 2
2
0.3
0.2
0.1
0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Deconvolution of NMR peaks – assumed Gaussian
0.7
Calculated relative areas
1
0.383
0.289
signal scaled to largest PbEDTA peak
0.6
0.5
0.4
% Pb207  100 *
Pb207
 0.383  0.289 

 100  40%
Pb207  other  1  0.383  0.289 
0.3
Expected value = 21%
0.2
0.1
0
1.4
1.45
1.5
1.55
1.6
1.65
-1 X ppm scaled to standard =0
1.7
1.75
1.8
Lorentzian Decovolution for the areas
0.7
0.6
Relative areas by Lorentzian:
Main peak:
1
Side 1
0.0959
Side 2
0.0966
0.5
Signal
0.4
Pb207
 0.0959  0.0966 
% Pb207  100 *

.
 100  1614%
Pb207  other  1  0.0959  0.0966 
0.3
0.2
0.1
0
1.5
1.55
1.6
1.65
1.7
1.75
1.8
-0.1
ppm (reversed)
I cheated to account for the poor baseline = due to T2* which is an instrumental
300 MHz instrument
1
Suggests that the protons on the carboxylic arms
Are not-equivalent
signal scaled to largest PbEDTA peak
0.9
0.8
Could be due to poor shimming
On z
But! Would expect these features
At both peaks!
0.7
0.6
0.5
What about these features?
0.4
1.66
1.72
0.3
0.2
Spinning side bands?
Could be due to poor
Shimming of x and y
0.1
0
1
1.1
1.2
1.3
1.4
1.5
-1 X ppm scaled to standard =0
1.6
1.7
1.8
Increased splitting pattern
Can be due to differences
Of orientation of the protons
This supposition is increased
By the very nearly observable
Splitting in the central peak
O
O
O
H
+
-
N
+
O
a’
a
O
O
Pb
N
O
a’
a
O
Non-equivalent
Protons due to binding
Peak Width
Lifetime depends upon
a) chemistry (for example a deprotonation reaction

HA 
 A
 H
b) rotation of the solution lattice of molecules (T1)
c) spin/spin exchange (T2)
A*  A
 A  A*
PbEDTA6C. N .  H 


Pb 2   HEDTA 3
PbEDTA6C. N .  2 L 
 PbEDTA4 C . N . L2
These reactions can reduce
The lifetime of Pb207 which
In the complex. Since Pb207 is
Causing the splitting a reduction
In it’s lifetime can eliminate the
signal
O
N
N
Pb
O
Excess Pb2+
O
O
O
Excess EDTA
O
O
N
N
N
Pb
O
O
Pb
N
N
O
O
O
N
O
O
Pb O
O
O
N
N
Pb
O
O
O
Pure EDTA
1:1 Pb-EDTA
1:2 Pb EDTA
Excess EDTA
Adding more EDTA
Main peak splits
Removes the main
Due different orientation Peak split (protons
Of protons around Pb
Around Pb now look
The same).
The splitting
Due to Pb207 also can
(It might also be that
Be lost
Lead is not complexing at this
High pH)
Application of lead isotope analysis in shooting incident investigations.
Zeichner, Arie;
Ehrlich, Sarah; Shoshani, Ezra; Halicz, Ludwik. Division of Identification and Forensic Science,
Israel Police National Headquarters, Jerusalem, Israel. Forensic Science International
(2006), 158(1), 52-64. Publisher: Elsevier Ltd., CODEN: FSINDR ISSN: 0379-0738. Journal
written in English. CAN 145:57143 AN 2006:202948 CAPLUS
Abstract
A study was conducted to examine the potential of the considerable variability of the lead isotope
compns. in bullets (projectiles) and primers in shooting incident investigations. Multiple-collector
inductively coupled plasma mass spectrometry (MC-ICP/MS) was used to analyze lead isotopic
compns. in projectiles, cartridge cases, firearms discharge residues (FDR) in barrels of firearms
and in the gunshot entries. .22 caliber plain lead and plated ammunition and 9 mm Luger full metal
jacket (FMJ) ammunition were employed in shooting expts. using semiautomatic pistols. Cotton
cloth served as the target material and two firing distances were tested; 1 cm (near contact) and 2
m distances. It was obsd. that various mech. or chem. means of cleaning do not completely
remove lead deposits ("lead memory") from barrels of firearms. Nonetheless, it was shown that
anal. of lead isotopic compn. may provide valuable evidence in investigating specific scenarios of
shooting incidents. For instance in a shoot-out where several firearms and ammunition brands are
involved, it may be feasible to point out which ammunition and/or firearm caused a particular
gunshot entry if the ammunition brands involved (bullets and primers) differ considerably in their
lead isotopic compn.
Origin assignment of unidentified corpses by use of stable isotope ratios of
light (bio-) and heavy (geo-) elements - A case report.
Rauch, Elisabeth;
Rummel, Susanne; Lehn, Christine; Buettner, Andreas. Institute of Forensic
Medicine, Ludwig-Maximilians-University Munich, Munich, Germany. Forensic
Science International (2007), 168(2-3), 215-218. Publisher: Elsevier Ltd., CODEN:
FSINDR ISSN: 0379-0738. Journal written in English. CAN 147:316041 AN
2007:474376 CAPLUS
Abstract
An unknown male body was found near an expressway in Germany. As different
criminalistic and forensic methods (e.g. tooth status, fingerprint or DNA-anal.) could not
help to identify the person, multielement stable isotope investigations were applied.
The combined anal. of stable isotope ratios of light (H, C, N) and heavy elements (Pb,
Sr) on the man's body tissues supported to assign him to Romania. The case report
demonstrates an application of multielement-isotope anal. in the forensic fields and its
potential.
END HERE
“Coalesance of the two lines occurs when the lifetime of a conformation, tao,
Gives rise to a linewidth that is comparable to the difference of resonance frequencies,”
Peter Atkins 6th Ed. Physical Chemistry, Freeman

2
  
Example, when the chemical shifts differ by 100Hz what is the maximum
Lifetime of a single conformation?
The rate of interconversion is the inverse of the lifetime.
What is the rate of interconversion?
Notice that the magnetic
Moment can be decomposed
Into two components along
The y and x directions.
http://www.physics.sjsu.edu/becker/physics51/mag_field.htm
http://www.urmc.rochester.edu/smd/Rad/MRIweb/mri_html/Animation_spin01.gif
http://www.papimi.gr/Ilika%20kimata.jpg
http://home.tiscali.nl/physis/deHaasPapers/DiracEPR/DiracEPR.html