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Nuclear Magnetic Resonance Spectroscopy!
Sample
Magnet
Magnetization
Response
Detection
Data
Fourier!
transformation
Spectrum!
Storage
NMR Spectroscopy!
Nuclear Magnetism!
NMR is a manifestation of nuclear spin angular momentum (P).!
P = h[ I (I + 1)]1/2!
I = angular momentum quantum number!
h= Planck’s constant/2!!
!
Possible Values of I:!
1.!
I = 0, mass number (A) and atomic number (Z) are even!
2.!
I = half-integral value, A is odd.!
3.!
I = integral value, A is even and Z is odd.!
!
Nuclear Magnetism!
For NMR:
1.!
I = 0 are NMR-inactive
2.!
I > 1/2, nuclei possess electric quadrupole moment due to
non-spherical nuclear charge distribution. The lifetime
of the magnetic states for quadrupoles in solution are
much shorter than for I = 1/ 2 This results in line
broadening and the can be more difficult to study.
3.!
I = 1/2 include 1H, 13C, 15N and 31P.
!
Nuclear Magnetism!
Due to restrictions of quantum mechanics only one of the
three Cartesian coordinates can be specified.
Pz = hm
m = (-I, -I + 1, …., I – 1, I) – magnetic quantum number.
!
Nuclear Magnetism!
Iz has 2I + 1 values of m. This behavior is called
directional quantization.
In the case I=1/2
In the case I=1!
Nuclear Magnetism!
Nuclei with non-zero spin angular momentum also possess
nuclear magnetic moment:
!="P
-" in part determines the receptivity of a nucleus in NMR
spectroscopy. It is a constant for a given nuclide and it is
referred to as the gyromagnetic ratio.
Therefore:
µz = m"h
- No simple model can predict or explain the actual magnetic
moments of nuclei.
Nuclear Magnetism!
More Quantitatively:
µ = g ehI
4!Mc
M = mass
e = charge uniformly spread over the surface
c = speed of light.
Properties of some nuclei of importance in NMR
Nuclear Magnetism!
•! In the absence of an external magnetic field:
2I + 1 states of m are equivalent
•! In the presence of an external magnetic field, the
spin states of a nucleus have energies given by:
E = µz B0 = - Iz" hB0
B0 = static magnetic field strength.
Nuclear Magnetism!
The projection of angular momentum of the nuclei
onto the z-axis of the laboratory frame results in 2I +
1 equally spaced energy levels, which are referred to
as the Zeeman levels or Zeeman States.
In the case I=1/2
In the case I=1!
Nuclear Magnetism!
For Iz = 1/2 we have m = +1/2 and m = -1/2 in which
µz is either parallel or anti-parallel to the field
direction.
B0!
Nuclear Magnetism!
In quantum mechanics:
m = + 1/2 = spin function #$
m = - 1/2 = spin function %$
**Since we are restricted to single quantum
transitions:
$
$
$&m = ± 1
Nuclear Magnetism!
!
!
!
!
!
!
!
&E = "hB0!
Nuclear Magnetism!
What Effects &E and why is this important?
•!Gyromagnetic ratio
•!Strength of static magnetic field
At equilibrium the energy states are not equally populated
and the relative populations is given by the Boltzmann
distribution.
( " Em
Nm
= exp&&
N
' k BT
I
%
##
$
( " Em %
##
BT $
! exp&&' k
m=" I
In our case, Em = -mh"B0
!
& mh)B0
= exp$$
% k BT
#
!!
"
I
& mh)B0
exp $$
% k BT
m=( I
'
#
!!
"
Nuclear Magnetism!
Since m!"B0 << kbT the exponential can be expanded to
first order using Taylor series.
!
!
!
!
!
!
!
& mh*B0
' $$1 +
k BT
%
#
!!
"
& mh*B0
' $$1 +
k BT
%
#
!!
"
I
& mh*B 0
$$1 +
k BT
m=) I %
(
(2 I + 1)
•! exp (-E/k T) " 1 – E/k T
•! denominator sums to " 2I + 1
#
!!
"
Nuclear Magnetism!
Here is an example for I = 1/2
N#
N"
= e $ %E
k bT
& 1 $ %E k b T = 1 $ !hB0 k b T
k: Boltzmann constant (1.3805 x 10-16 erg K-1)
T: temperature (K)
h: Plank constant (1.0546 x 10-27 erg S)
B0: magnetic field strength
(Tesla, T; 100 MHz ! 2.35 T)
": gyromagnetic ratio (107 Hz T-1)
Nuclear Magnetism!
Example:
At T = 300 K and B0 = 5.875 T (250 MHz),
1H:
N$
N#
= 1"
26.7519 ! 10 7 ! 1.0546 ! 10 "27 ! 5.875
1.3805 ! 10 "16 ! 300
N " $ (1 # 0.00004 )N ! = 0.99996 N !
For B0 = 18.8 T (800MHz),
N " # 0.99987 N !
Nuclear Magnetism!
In classical representation the nuclear dipole process about the
magnetic field direction with Larmor Frequency:
'o = "B0 (Radians/sec)
( = '0/2! = "B0 /2! (Hertz or cycles/sec)
The z-component is given by the magnetic moment:
µz = "Pz = m"h
Nuclear Magnetism!
When m = ± 1/2; the nuclear dipoles precess around a
double cone with a half-angle of the cone is 54.7°.
Due to the differences in populations of the energy states
there is macroscopic magnetization along the magnetic field
direction.
Nuclear Magnetism!
For a macroscopic sample:
The x and y components are random and sum to zero so:
I
M 0 = #h
! mN
m
m=" I
But we know that :
N m = N exp(mh"B0 k b T )
! exp(mh"B
0
k bT )
Doing the math:
M 0 = N! 2 h 2 B0 I (I + 1) (3k b T )
Nuclear Magnetism!
So the macroscopic magnetization (M0) depends on:
•! The External Magnetic Field- B0
•! The angular momentum quantum number- I
•! The gyromagnetic ratio- "2
•! The temperature- 1/T
Note: I and " are specific to the nuclei of interest.
Nuclear Magnetic Resonance (NMR)!
What happens in NMR:
To induce NMR transitions a radiofrequency pulse is
applied to the sample. The pulse is generated by a linear
alternating electromagnetic field (B1) along a transverse
axis. This results in M0 being rotated into the XY plane.
Resonance Condition:
h(1 = &E
Figure 1-6
Nuclear Magnetic Resonance!
Larmor frequency : (1 = (L = |" /2!| B0
At 11.7 T : for 1H" = 2.675 x 108 (0 = 500 MHz
for 13C" = 0.6726 x 108 (0 = 125 MHz
Nuclear Magnetic Resonance!
Pulsed NMR:
- All nuclei are excited simultaneously by a radiofrequency
pulse.
- Pulse angle is proportional to the pulse width (duration) and
pulse power (Magnitude of B1 field).
-The coil is arranged so that this field is perpendicular to the
applied field (in the x-y plane).
Nuclear Magnetic Resonance!
The oscillating magnetic field is equivalent to two counterrotating magnetization vectors:
We can use this pair of counter-rotating vectors as an
equivalent representation of the rf signal.
Nuclear Magnetic Resonance!
We will represent the rf along the x-direction as two
vectors with the same magnitude (B1).
•! one vector rotates
clockwise B1 (r).
•! one vector rotates
counter-clockwise B1 (l)
Nuclear Magnetic Resonance!
Only one of the two components is capable of interacting
with the processing nuclear dipoles (B1).
The sample magnetization is
static along the z-axis.
What happens when the rf
magnetization (which is moving)
interacts with the sample
magnetization?
Nuclear Magnetic Resonance!
The Rotating Frame:
The experimentally detected signal is subtracted from the
carrier frequency. We chose a set of coordinates that rotate
along with the nuclear precession.
Nuclear Magnetic Resonance!
!
!
ROTATING FRAME
Nuclear Magnetic Resonance!
The Rotating Frame:
How can we use this to look at rf pulses (B1)?
- In the rotating frame B1 is at right angles to M.
- The net result is to produce a torque acting around B1 at
a speed depending on the field strength.
Nuclear Magnetic Resonance!
The Rotating Frame:
How does this look in our two frames of reference?
Nuclear Magnetic Resonance!
The Rotating Frame:
What happens to M as we increase the length of the pulse?
Nuclear Magnetic Resonance!
The Rotating Frame:
How des this effect the NMR signal?
Our receiver is oriented with its axis along the y-direction.
Nuclear Magnetic Resonance!
)!= pulse angle or pulse flip-angle.
Four common pulses and their phases (*):
Nuclear Magnetic Resonance!
Nuclear Magnetic Resonance!
Nuclear Magnetic Resonance!
Pictorial representation of phase coherence:
Free Induction Decay (FID) and Fourier
Transformation (FT)!
FID
T2
E
Y
X
AQ
YO
Time
+
FT
+
+
Free Induction Decay (FID) and Fourier
Transformation (FT)!
Fourier Transformation (FT):!
#
g ( w) =
$
f (t )e i!t dt
"#
e "i!t = cos !t + i sin !t
real part
(Absorption)
imaginary part
(Dispersion)
- dispersion component is removed by phase cycling.
Free Induction Decay (FID) and Fourier
Transformation (FT)!
FIGURE 1.15
90MHz 1H NMR spectrum of methyl iodide CH3I(1);
One pulse, spectral width 1200 Hz, 8 K data points,
acquisition time 0.8s. A: time domain spectrum (FID);
the generator frequency is almost exactly equal to the
Resonance frequency of the sample; B: frequency
domain spectrum obtained by Fourier transformation of
A.
FIGURE 1.16
22.63 MHz 13C NMR spectrum of methanol 13CH3OH (2);
Solvent: D2O, 17 pulses, spectral width 1000 Hz, 8 Kne
pulse, spcectral width 1200 Hz, 8 K data points, A: Time
domain spectrum (FID); B: frequency domain spectrum
obtained by Fourier transformation of A. This consists of a
quarter, as the 13C nucleus is coupled to the three protors of
the methyl group.
Signal Detection!
Sensitivity of the NMR Signal$
+!= electromagnetic induction force in a detection coil
(Faradays Law of Induction).
Procession of a bulk magnetic moment about a static field
yields a time varying magnetic field. This produces an
induced electromotive force.
Signal Detection!
# ! dM (t ) dt = "M 0 B
M 0 = N" 2 h 2 B0 I (I + 1) (3k b T )
"M 0 B = N" 3 h 2 B02 I (I + 1) (3k b T )
Higher field = higher signal (lose some to noise)
For two nuclei with same value of I their sensitivity is
related by :
(" a
" b )3 for 1H vs 13C ! (1 4 )3 = 0.0159
a = 13C
b = 1H
Properties of some nuclei of importance in NMR
Relaxation!
T1: spin-lattice relaxation time (longitudinal)!
T2: spin-spin relaxation time (transverse)!
!
-! After we remove the pulse, the magnetization vector
returns to equilibrium.!
-! Bloch assumed that the processes were first-order and
defined them by T1 and T2:!
dMz/dt = - Mz – Mo/T1!
so!
Mz = M0(1-e-t/T1)!
dMx,/dt = -Mx,/T2!
dMy,/dt = -My,/T2!
Relaxation!
!
!
!
!
On Resonance:!
T1 Relaxation!
Experimental determination:
T1 Relaxation!
1H:
T1 Relaxation!
13C:
T1 Relaxation!
13C:
T1 Relaxation!
How do we determine the T1 value?!
!
!M0 - Mz
= Ae--/T1!
After a 180° pulse during the inversion recovery
experiment M0 = -Mz so A = 2M0:!
!M0 - Mz
= 2M0e--/T1!
!ln (M0 - Mz) = ln2M0 - -/T1!
!
Substituting I for M where I is intensity of signal:!
ln (I0-Iz) = ln2I0 - -/T1!
Tzero = T1 ln2!
T1 values can vary greatly from one nuclei to the next in
the same molecule.!
T1 Relaxation!
Need to wait 5T1 periods for system to return to
equilibrium so:!
!
!(5 T1 - 180°x' - - - 90°x' - FID)n !
!
Why measure T1?!
! !/2 pulse elicits maximal NMR signal at thermal
equilibrium.!
- for repetitive signal averaging this is not always the best
pulse angle.!
- need to wait about 5T1 for z-magnetization to return to
equilibrium.!
- more efficient to repeat sooner.!
T1 Relaxation!
Ernst Angle:!
!
!Cos#E = e -Tr/T1!
- Tr = the delay between pulses.!
- best value of # actually depends on offsets of signals in
spectrum but this is an average value.!
If stick with !/2 pulses then:!
- the optimum repetition rate is Tr = 1.27 T1!
- in this case you are about 80% efficient.!
T1 Relaxation!
Influence of protons on 13C T1 values:!
!
- main contribution is via dipole-dipole
interactions, so directly bonded protons can have
large effects.!
- increasing the number of protons shortens the
T1 although the correlation is not always direct.!
Influence of molecular size on T1:!
- T1 values decrease as the size of the molecules. !
T2 Relaxation!
After 90°x' pulse Mz = 0 and net magnetization is along the
y-axis. The population of N# and N% are equal . !
Phase Coherence: a small proportion of nuclear dipoles are
bunched along the y-axis.!
T2 - determines how fast Mx' and My' return to zero.!
T2 Relaxation!
T2 Relaxation!
How do we view this?:!
-Nuclei change from one energy state to the other and we lose
the phase coherence.!
- this is not the main contribution.!
Magnetic Field inhomogeneities: there is a small &B0
throughout the sample which leads to fanning out of signals.!
-fanning out process causes signals that are chemically
equivalent to process with slightly different Larmor
frequencies.!
- the net result is an increase in the line width of the various
signals.!
By definition though
T1 . T2!
T2 Relaxation!
Experimental Determination of:!
- We want to measure a true value which is independent of the
&B0 fanning out process.!
Spin-echo experiment:!
After initial 90°x' pulse you apply a 180°x' pulse at set interval -.!
At
!2- - resultant transverse magnetization is -y' direction.!
!4- - magnetization is y' direction.!
!6- - magnetization is -y' direction.!
Etc……we get an echo with alternating phase at intervals of 2-.!
However, the net magnetization is decreasing with each echo
due to true spin-spin relaxation.!
T2 Relaxation!
T2 Relaxation!
!
-! the 180°x' pulse turns the Ay' component into the (-y')
direction but has no effect on the Ax'. The angle
between vector A and the y'-axis is unchanged if we
disregard sign.!
- the effect of a 180°x' pulse is that vectors A and B
undergo reflections in the x', z' plane.!
!
T2 Relaxation!
Quantitatively:!
!
!
!My' = A e -t/T2!
!
At t=0 , A=M0!
!
!
!
!ln My' = lnMo - t/T2!
!
since I # My'!
!
!
!ln I(t) = ln I0 - t/T2!
!
Plot I(t) vs t where I(t) is intensities at each echo. Slope of
line is -1/T2.!
!
T2 Relaxation!
For the most part we do not concern ourselves with T2
measurement.!
- Of more consequence is T2* which takes into account the
inhomogeneity of the magnetic field.!