B - Molecular Imaging Center

SCUOLA NAZIONALE
DI RISONANZA MAGNETICA NUCLEARE
Concetti di base
Parametri NMR
Esperimento 1D-FT NMR
Torino 23-27 Settembre 2013
Stefano Mammi
Dipartimento di Scienze Chimiche
Università di Padova
[email protected]
RIFERIMENTI BIBLIOGRAFICI
M. Levitt “Spin Dynamics” ”, 2nd ed., John Wiley & Sons, 2008.
J. Keeler, “Understanding NMR Spectroscopy”, 2nd ed., John Wiley &
Sons, 2010.
T. D. W. Claridge, “High-Resolution NMR Techniques in Organic
Chemistry”, Pergamon Press, 1999.
J. Cavanagh, “Protein NMR spectroscopy: principles and practice”, 2nd
ed., Elsevier, 2007.
http://www.cis.rit.edu/htbooks/nmr
http://www.chem.queensu.ca/FACILITIES/NMR/nmr/webcourse
http://www-keeler.ch.cam.ac.uk/lectures/index.html
1
RIFERIMENTI BIBLIOGRAFICI
H. Günther, "NMR Spectroscopy", 2nd ed., John Wiley & Sons, 1995.
J. K. M. Sanders and B. K. Hunter, "Modern NMR Spectroscopy", 2nd
ed., Oxford University Press, 1994.
H. Friebolin, "Basic One- and Two-Dimensional NMR Spectroscopy",
VCH, 1991.
R. M. Silverstein and F. X. Webster, "Identificazione Spettroscopica di
Composti Organici", Casa Editrice Ambrosiana, 1999.
A. E. Derome, "Modern NMR Techniques for Chemistry Research",
Pergamon Press, 1987.
The Nobel Prize in Physics 1943
"for his contribution to the
development of the molecular ray
method and his discovery of the
magnetic moment of the proton"
The Nobel Prize in Physics 1944
"for his resonance method for
recording the magnetic
properties of atomic nuclei"
Otto Stern
Isidor Isaac Rabi
The Nobel Prize in Physics 1952
"for their development of new methods
for nuclear magnetic precision measurements
and discoveries in connection therewith”
Felix Bloch
Edward Mills Purcell
2
The Nobel Prize in Chemistry 1991
"for his contributions to the
development of the methodology of
high resolution nuclear magnetic
resonance (NMR) spectroscopy"
The Nobel Prize in Chemistry 2002
"for his development of nuclear magnetic
resonance spectroscopy for determining
the three-dimensional structure of
biological macromolecules in solution”
Richard R. Ernst
Kurt Wüthrich
1/2 of the prize
The Nobel Prize in Physiology or Medicine 2003
"for their discoveries concerning magnetic resonance imaging”
Paul C. Lauterbur
Sir Peter Mansfield
The Nobel Prize in Physics 2003
”for pioneering contributions to the theory of superconductors”
Alexei A. Abrikosov
1/3 of the prize
Vitaly L. Ginzburg
1/3 of the prize
3
I = h I(I+1)
- I ≤ m ≤ +I
I z = hm
z
B0
dI
dt
µ = γI
µz = γhm
µ
I
ω0
dI = µ × B
0
dt
ω0 = -γB0
E = −µB0
E = −γhmB0
I=1/2
M0
B0
Pβ
Pα
=e
−
∆E
kT
M0 = Σ µ = M z
α: m = +½
µα = +½γh
Eα = −½γhB0
∆E = γhB0 = hω0 = hν0
Mx = M y = 0
β:
m = −½
µβ = −½γh
Eβ = +½γhB0
4
Pβ
Pα
=e
−
∆E
kT
M0 = Σ µ = M z
∆E = γhB0 = hω0 = hν0
the new arrival!
Season greetings - 2003
CERM, University of Florence
5
Which Elements or Molecules
are NMR Active?
• Any atom or element with an odd number of
neutrons and/or an odd number of protons
• Any molecule with NMR active atoms
• 1H - 1 proton, no neutrons, AW = 1
• 13C - 6 protons, 7 neutrons, AW =13
• 15N - 7 protons, 8 neutrons, AW = 15
• 19F - 9 protons, 10 neutrons, AW = 19
• 31P - 15 protons, 16 neutrons, AW = 31
http://www.bruker.de/guide/eNMR/chem/NMRnuclei.html
6
Different Isotopes Absorb at
Different Frequencies
15N
2H
13C
31P
19F
1H
50 MHz
77 MHz
125 MHz
200 MHz
470 MHz
500 MHz
low frequency
high frequency
NMR Parameters for some nuclei
Isotope
1
1/2
2
1
1/2
1/2
H
H
C
15
N
13
19
NMR Resonance Frequencies
Sensitivity
Relative Absolute
(MHz) at a field of (T):
4.6975
9.395 14.0926
99.98
1
1
200
400
600
1.50E-02 9.65E-03 1.45E-06 30.701 61.402 92.102
Spin Nat. Ab. %
1.108 1.59E-02 1.76E-04 50.288 100.577 150.864
0.37 1.04E-03 3.85E-06 20.265 40.531 60.796
100
0.83
0.83 188.154 376.308 564.462
1/2
F
1/2
100 6.63E-02 6.63E-02 80.961 161.923 242.884
P
Relative Sensitivity = at the same field and for the same number of nuclei
Absolute Sensitivity = (relative sensitivity * natural abundance)
31
7
z
B0
B0
M
z
Bx(t) = Bx * cos (ω
ωot)
M
y
x
B1
y
x
x
y
time
BR (t ) = B1 / 2(i cos ωrf t + j sin ωrf t )
B x (t ) = B1 cos ω rf t
BL (t ) = B1 / 2(i cos ωrf t − j sin ωrf t )
8
So how does that help?
z, B0
ω0 = ωL
ωrf ( B1 )
− ωrf ( B1 )
x
z, B0
ω0 = ωL
ωrf ( B1 )
9
The Rotating Frame
z, B0
ω0 = ωL
ω rf ( Bω1 )rf =( Bω1 )L
x
y
time
z’ = z
z
B0
M
M
y
–ω0
+ω0
y’
ω0
x’
x
LAB.
2ω0
(neglected)
ROTATING
10
z’ = z
z’ = z
M
z’ = z
M
y’
y’
B1
x’
x’
M
B1
z’ = z
θ
M
y’
x’
y’
θ = π/2
x’
z
z
B0
M
ω0
x
y
ω0t
y
ω0
x
90° Pulse
ν = γBo/2π
22
11
90°x
FT
relax.
Preparation
Detection
z
90°x
y
t2
ω0
x
+∞
F(ω) = ∫ f ( t )eiωt dt
−∞
M 0 = M z ∝ Pα0 − Pβ0 = δ
Pα0 = (N + δ ) 2
Pβ0 = (N − δ ) 2
Pα + Pβ = N
M z = M 0 cos θ ∝ δ cos θ = Pα − Pβ
z
Mcosθ
θ
M
y
x
Pα = (N + δ cos θ ) 2
Pβ = (N − δ cos θ ) 2
If θ = 90° ⇒ cosθ = 0 ⇒ Pα = Pβ = N/2
12
Relaxation
90° Pulse
Effect on Population:
90° Pulse
M 0 = M z`0 = ∑ µ z0
Mz = 0
Relaxation
90° Pulse
Effect on Phase:
90° Pulse
M xy0 = ∑ µ xy0 = 0
M y = ∑ µ z0 = M 0
M xz = 0
13
Relaxation
Excitation
• Different populations
• No phase coherence
(
• Equal populations
• Partial phase coherence
Relaxation
dM z M 0 − M z
=
dt
T1
M z = M 0 1 − e − t T1
T1
T2
dM xy
dt
)
=−
M xy
T2*
M xy = M 0ze − t T2
*
Bloch Equations
dJ( t )
= M(t ) × B(t )
dt
Multiply by γ
dM( t )
= M(t ) × γB(t )
dt
14
Bloch Equations
dM ( t )
= M(t ) × γB(t ) − R( M(t ) − M0 )
dt
Relaxation Matrix
M (t ) − M 0
dM z (t )
= γ [ M x (t ) B y (t ) − M y B x (t )] − z
dt
T1
Magnetization
along the z-axis
dM x (t )
M (t )
= γ [ M y (t ) Bz (t ) − M z B y (t )] − x
dt
T2
Magnetization
along the x-axis
dM y (t )
Magnetization
along the y-axis
dt
= γ [ M z (t ) Bx (t ) − M x Bz (t )] −
M y (t )
T2
ROTATING FRAME OF REFERENCE
faster
slower
15
z, B0
ω0 = ωL
ω rf ( B1 ) = ω L
z, B0
ω0 − ωrf
ω rf ( B1 ) < ω 0
∆B = (ω0 − ω rf ) / γ = B0 − ωrf / γ
Ω=(ω0-ωrf)
∆B= -Ω/γ
16
Bloch Equations in the Rotating Frame
Ω = −γB0 − ω rf
(where B0=Bz and is not time-dependent)
dM z (t )
M (t ) − M 0
= γ [ M x (t ) B yr (t ) − M y Bxr (t )] − z
dt
T1
dM x (t )
M (t )
= −ΩM y (t ) − γM z B yr (t ) − x
dt
T2
dM y (t )
dt
= γM z (t ) Bxr (t ) + ΩM x −
M y (t )
The “r”
superscript
refers to a
magnetic field
in the rotating
frame
T2
Bloch Equations in the Rotating Frame
Example: 90° pulse along x
dM z (t )
M (t ) − M 0
= γ [ M x (t ) B yr (t ) − M y Bxr (t )] − z
dt
T1
dM z (t )
= −γM y Bxr (t )
dt
dM x (t )
M (t )
= −ΩM y (t ) − γM z B yr (t ) − x
dt
T2
dM x (t )
=0
dt
dM y (t )
M y (t )
dM y (t )
T2
dt
dt
= γM z (t ) Bxr (t ) + ΩM x −
= γM z (t ) Bxr (t )
17
Bloch Equations in the Rotating Frame
Example: Free Precession
dM z (t )
M (t ) − M 0
= γ [ M x (t ) B yr (t ) − M y Bxr (t )] − z
dt
T1
dM x (t )
M (t )
= −ΩM y (t ) − γM z B yr (t ) − x
dt
T2
dt
= γM z (t ) Bxr (t ) + ΩM x −
dM x (t )
M (t )
= −ΩM y (t ) − x
dt
T2
M y (t )
dM y (t )
T2
dt
= +ΩM x −
M y (t )
T2
Typical 1H NMR Spectrum
Absorbance
dM y (t )
dM z (t )
M (t ) − M 0
=− z
dt
T1
18
Chemical Shift
O
N
C6H5−CH2−O−C−CH3
e
Bo
σBo
H
H
H
H
H
H
|
— C —H
|
H
H
|
—C—
|
H
8
7
6
5
δ(ppm) =
TMS
4
3
2
1
0
ω − ωTMS
× 10 6
ω0
“Chemical” Shift
1951
19
ω = −γ (B0 + Bloc) = ω0 + ωloc
∆E·∆t≈h
h∆ν·∆t≈h
∆ν≈1/∆t
For 1H at B0 = 11.744 T
(ω0 = 500 MHz):
if
∆ν ≥ ±2500 Hz
(10 ppm)
then
∆t ≤ 400 µs
Effective Field Strength
If a spin is not on resonance, B0 cannot be completely neglected:
Beff = ∆B2 + B12
ωeff = Ω 2 + ω12
20
We have two fields!
z
B0
z
B(t)
B
B0 – ωrf/γ
y’
y
B1
B1(t)
x’
x
Move into rotating frame (sit on the carousel)
Flip Angle and Field Strength
z’ = z
z’ = z
M
M
y’
x’
z’ = z
y’
B1
x’
M
B1
z’ = z
θ
x’
M
y’
x’
y’
θ = π/2
ω1 = −γB1 = 2πν1 = (π/2)/PW90
ν1 =
γB1
1
=
2π 4PW90
21
Effective Field Strength
γ
1
B1 =
2π
PW (360o )
γB1eff = (γB1 ) 2 + ( ∆ω ) 2
γB1eff
∆ω
θ
γB1
If (γB1)/2π = 25000 Hz
[PW(360°) = 40 µs]
and
∆ν = ±2500 Hz
(10 ppm @ 500 MHz)
then
(γB1eff )/2π = 25125 Hz
Rotation under an Offset
22
Offset Excitation
Effect of a 90° Pulse
Effect of a 180° Pulse
Offset Excitation
Rotation under a 180° Pulse
for Ω = 0.2 − 0.6 ω1
Effect of a 180° Pulse
23
DE
P1 = 1/BW
PL1
AQ = DW·TD
Acquisition Time
D1
RG
Repetition Time
Precession and Relaxation
1
1
0.75
0.5
0.5
0.25
Mx
1
2
3
4
5
t
My
1
2
3
4
5
t
-0.25
-0.5
-0.5
-0.75
1
0.8
0.75
Mz
Mz
0.6
0.5
1
0.25
0.5
0
-1
0.4
0
-0.5
0.2
0
My
Mx
-0.5
0.5
1
t
A. S. Edison
University of Florida
24
Detection
1
1
0.5
0.5
0.2
0.4
0.6
0.8
1
0.2
0.4
0.6
0.8
1
-0.5
-0.5
analog
-1
-1
1
0.5
20
40
60
80
100
-0.5
digital
-1
Nyquist: ∆t = 1/(2·νmax)
A
B
C
ω0 = ωB
Static:
z
z
y
90°−x
x
z
y
t2
x
Rotating (ω0 = ωB):
z
y’
x’
A. S. Edison
University of Florida
y
x
z
90°−x
z
y’
x’
ω ≈ 108 Hz
t2
y’
x’
ω ≈ 103 Hz
25
500 MHz ± 2500 Hz
± 2500 Hz
NMR
Signal
ADC
Det.
Computer
Memory
ω0 (reference)
500 MHz
Nyquist: ∆t = 1/(2·νmax)
100 µs
1/2· νmax = 10-4 s
DW = 1/SW
νmax = 5.0 kHz
ν = 2.0 kHz
ν = 4.5 kHz
ν = 2.0 kHz
ν = 8.0 kHz
26
Effect of Incorrect Sampling Rate
z
x
M0cos(ωt)
y
+∞
M0sin(ωt)
ω0t
ω0
F(ω) = ∫ f ( t )eiωt dt
−∞
eiωt = cos(ωt ) + i ⋅ sin (ωt )
+∞
Re[F(ω)] = ∫ f ( t ) cos(ωt )dt
−∞
+∞
Im[F(ω)] = ∫ f ( t )sin (ωt )dt
−∞
27
z
x
M0cos(ωt)
+∞
M0sin(ωt)
ω0t
+∞
F (ω) = A(ω) + iD(ω)
A(ω) =
−∞
eiωt = cos(ωt ) + i ⋅ sin (ωt )
ω0
y
F(ω) = ∫ f ( t )eiωt dt
Re[F(ω)] = ∫ f ( t ) cos(ωt )dt
−∞
1
(ω − Ω)2 T22 + 1
+∞
Im[F(ω)] = ∫ f ( t )sin (ωt )dt
−∞
D(ω) =
(ω − Ω)T2
(ω − Ω)2 T22 + 1
Phase Correction
F (ω) = A(ω) + iD(ω)
F (ω ) = exp(iφinstr ){A(ω ) + iD (ω )}
Re[F (ω )] = A(ω )
Re[F (ω )] = {cos(φinstr ) A(ω ) − sin (φinstr )D(ω )}
Im[F (ω )] = D (ω )
Im[F (ω )] = {cos(φinstr )D(ω ) + sin (φinstr )A(ω )}
28
Pulses and Phases
My
y
y
t
x
x
Mx
y
y
t
x
x
29
Quadrature Phase Detection
Quadrature Phase Detection
PSD
ADC
Computer
Memory
A
0° reference
NMR
Signal
ω0
90° reference
PSD
ADC
Computer
Memory
B
30
The Redfield Trick
PSD
∆t = DW/2 = 1/2·SW
0° reference
ω0
NMR
Signal
ADC
Computer
Memory
(+ + − −)
90° reference
PSD
Each subsequent point is phase-shifted by 90°
The period of this added frequency is 4·∆t
The added frequency is therefore SW/2
The Redfield Trick
Original SW:
Added frequency:
Final SW:
−νmax ↔ +νmax
SW/2 = νmax
0 ↔ +2νmax
31
How does an NMR receiver work?
20 MHz
620 MHz
600 MHz
* sin
Sample
Cable
V~ µV
1st mixer
preamplifier
I.F. (20MHz)
Probe Coil
10 kHz spectral window =
599,995,000.0 – 600,005,000.0 Hz
V ~ mV
2nd mixer
V ~ Volt
amplifier
* cos
Audio (+-kHz)
1st ADC
0o
2nd ADC
90o
-5,000.0 - + 5,000.0 Hz
19,995,000.0 – 20,005,000.0 Hz
• The amplifiers – mixers amplify both the signal and the noise!!
• We do not gain S/N with these components!!
Problems…
If the two ADCs are not perfectly matched we have “quad images”
32
Problems with
Quadrature Phase Detection
z
z
y
M0sin(ωt)
y
t2
M0cos(ωt)
90°y
ωt
x
x
z
z
y
90
x
t2
-M0sin(ωt)
°
−x
ωt
y
M0cos(ωt)
x
33
z
z
y
ωt
t2
-M0sin(ωt)
90°−y
x
x
z
z
y
90
x
-M0cos(ωt)
y
t2
-M0cos(ωt)
ωt
°
x
y
M0sin(ωt)
x
CYCLOPS
34
The “Direct Digital” receiver…
620 MHz
600 MHz
Sample
Cable
V~ µV
1st mixer
preamplifier
I.F. (20MHz)
V ~ mV
ADC
We digitize directly at
the I.F.
Probe Coil
19,995,000.0 – 20,005,000.0 Hz
10 kHz spectral window = 599,995,000.0 – 600,005,000.0 Hz
•Using fast ADCs we can digitize at the 20 MHz I.F.
•There are no positive and negative frequencies and there is no need to distinguish
between them.
•Quad images and other artifacts are not there at all since the mechanism that is
generating them is absent!
Some spectra…
Spectra acquired with a single transient
Quad
Images
Quadrature Receiver
16-bit ADC
with oversampling
This is the
peak you
want
Varian DDR
Center
Spike
35
FID Clipping
The effects of Relaxation
FT
FT
36
The effects of Relaxation
ν
ν
NMR parameters and relaxation in a
simple 1D experiment
f.i.d.
f.i.d.
rd
aq
Acquisition time
Recycle delay
37
Length of Acquisition
S(t) ∝ exp[-(t/T2)]
∆ν = 1/(πT2)
Acquire until S(t) = 0.01•S(0)
if T2 ≈ 0.15s and S(0) = 1
0.01 = exp(-tmax/0.15)
tmax = 0.7s
Acquisition time vs T2
I
e
−t
T2
2π
(ω I − ω1 )
t
I
t
38
Recycle delay vs T1
I
−t
I
e T2
t
−t
e T2
t
2π
(ω I − ω1 )
2π
(ω I − ω1 )
Zero-Filling
spectral width (Hz)
DR
DR==SW
SW//SI
SI
data size (n. of points)
39
Zero-Filling
spectral width (Hz)
DR
DR==SW
SW//SI
SI
data size (n. of points)
Fourier Pairs
+∞
F(ω) = ∫ f ( t )e iωt dt
−∞
40
Truncation and Apodization
Truncation and Apodization
41
Window Functions:
Sensitivity Enhancement
S = A exp{-t/T2} exp{-at} = A exp{-t (a+ 1/T2)}
a = 1/T2 (matched filter)
RESOLUTION
Resolution is the ability to distinguish between two frequencies
AQ = DW*TD
No resolution
AQ = DW*TD’
Resolution
DIGITAL
RESOLUTION
42
Window Functions:
Resolution Enhancement
S = A*exp{-t/T2} exp{-at}exp{-bt2}
a = -1/T2 ; b>0
Other Window Functions:
Sine Bell and Sine Bell Squared
S = A*exp{-t/T2} sin{(π-φ)t/tmax +φ}
S=A
S = A*exp{-t/T2} sin2{(π-φ)t/tmax +φ}
43
Linear Prediction
Integration
O
C6H5−CH2−O−C−CH3
52
TMS
32
21
8
7
6
5
4
3
2
1
0
44
Spin-Spin Coupling
H
H
H
H
|
|
|
|
C-Y
X
C - CH
Z
X
Z
C - CH2
X
C - CH3
Z
X
Z
J
singlet
AX System,
doublet
triplet
quartet
E/h = −ΣiνimI(i) +ΣΣi<jJijmI(i) mI(j)
J=
≠0
ββ
ββ
A2
A2
X2
αβ
βα
X1
A1
αα
Eαα = − ½νA − ½νX + ¼JAX
Eαβ = − ½νA + ½νX − ¼JAX
Eβα = + ½νA − ½νX − ¼JAX
Eββ = + ½νA + ½νX + ¼JAX
¼JAX
X2
αβ
βα
X1
A1
αα
EA1 = Eβα − Eαα = νA − ½JAX
EA2 = Eββ − Eαβ = νA + ½JAX
EX1 = Eαβ − Eαα = νX − ½JAX
EX2 = Eββ − Eβα = νX + ½JAX
45
Origin of Scalar Coupling
H
H
C
αβ (and βα) levels are
stabilized
C
J>0
H
H
C
C
αα (and ββ) levels are
destabilized
Origin of Scalar Coupling
αβ (and βα) levels are
stabilized
J>0
αα (and ββ) levels are
destabilized
46