Concetti di base Parametri NMR Esperimento 1D-FT NMR · 1 SCUOLA NAZIONALE DI RISONANZA MAGNETICA...
Transcript of Concetti di base Parametri NMR Esperimento 1D-FT NMR · 1 SCUOLA NAZIONALE DI RISONANZA MAGNETICA...
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SCUOLA NAZIONALE
DI RISONANZA MAGNETICA NUCLEARE
Torino 23-27 Settembre 2013
Stefano MammiDipartimento di Scienze Chimiche
Università di [email protected]
Concetti di baseParametri NMR
Esperimento 1D-FT NMR
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/nmrhttp://www.chem.queensu.ca/FACILITIES/NMR/nmr/webcourse
http://www-keeler.ch.cam.ac.uk/lectures/index.html
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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 1952"for their development of new methodsfor nuclear magnetic precision measurements and discoveries in connection therewith”
Felix Bloch Edward Mills Purcell
The Nobel Prize in Physics 1944"for his resonance method for recording the magneticproperties of atomic nuclei"
Isidor Isaac Rabi
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"
Otto Stern
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The Nobel Prize in Chemistry 1991"for his contributions to the development of the methodology of high resolution nuclear magnetic resonance (NMR) spectroscopy"
Richard R. Ernst
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”
Kurt Wüthrich 1/2 of the prize
The Nobel Prize in Physics 2003”for pioneering contributions to the theory of superconductors”
Alexei A. Abrikosov Vitaly L. Ginzburg1/3 of the prize 1/3 of the prize
The Nobel Prize in Physiology or Medicine 2003"for their discoveries concerning magnetic resonance imaging”
Paul C. Lauterbur Sir Peter Mansfield
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B0
I
µ
ω0
dIdt
z
mIz
hγ=γ= µµ
mI
ImI- 1)I(II
zh
h
=
+≤≤+=
000 B- BdtId γ=ω×= µ
B0
M0
I=1/2 E = −µB0
E = −γhmB0
α: m = +½
µα = +½γhEα = −½γhB0
β: m = −½
µβ = −½γhEβ = +½γhB0
kT
E
eP
P ∆−
α
β =
M0 = Σµ = Mz
Mx = My = 0
∆E = γhB0 = hω0 = hν0
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∆E = γhB0 = hω0 = hν0
kT
E
eP
P ∆−
α
β =
M0 = Σµ = Mz
the new arrival!
CERM, University of FlorenceSeason greetings - 2003
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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
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Different Isotopes Absorb at Different Frequencies
low frequency high frequency
15N 2H 13C 19F 1H
50 MHz 77 MHz 125 MHz 200 MHz 470 MHz 500 MHz
31P
Isotope Spin Nat. Ab. %Relative Absolute
4.6975 9.395 14.09261H 1/2 99.98 1 1 200 400 6002H 1 1.50E-029.65E-031.45E-06 30.701 61.402 92.10213C 1/2 1.1081.59E-021.76E-04 50.288 100.577 150.86415N 1/2 0.371.04E-033.85E-06 20.265 40.531 60.79619F 1/2 100 0.83 0.83 188.154 376.308 564.46231P 1/2 1006.63E-026.63E-02 80.961 161.923 242.884
Relative Sensitivity = at the same field and for the same number of nucleiAbsolute Sensitivity = (relative sensitivity * natural abundance)
Sensitivity NMR Resonance Frequencies (MHz) at a field of (T):
NMR Parameters for some nuclei
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Bx(t) = B x * cos ( ωωωωot)
x
y
z
MB0 B0
x
y
z
M
B1
x
y
time
1( ) cosx rfB t B tω=
1( ) / 2( cos sin )R rf rfB t B t tω ω= +i j
1( ) / 2( cos sin )L rf rfB t B t tω ω= −i j
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Lωω =0
z, B0
The Rotating Frame
)( 1Brfω Lrf B ωω =)( 1
x
y
time
+ω0–ω0
M
x
y
zB0
LAB.
2ω0(neglected)x’
y’
M
z’ = z
ω0
ROTATING
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x
y
z
M
B0
ω0
x’
y’
z’ = z
M
x’
y’
z’ = z
M
B1
x’
y’
z’ = z
B1
M θ
x’
y’
z’ = z
M
θ = π/2
x
y
z
ω0
ω0t
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νννν = γγγγBo/2ππππ
90° Pulse
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FTrelax.
°x90
Preparation Detection
x
y
z
°x90 t2
ω0
∫+∞
∞−
ω=ω dte)t(f)(F ti
x
y
z
MθMcosθ
δ=−∝= βα00
z0 PPMM ( ) 2NP0 δ+=α ( ) 2NP0 δ−=β
βα −=θδ∝θ= PPcoscosMM 0z
NPP =+ βα
( ) 2cosNP θδ+=α
( ) 2cosNP θδ−=β
If θ = 90° ⇒ cosθ = 0 ⇒ Pα = Pβ = N/2
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Relaxation 90° Pulse
Effect on Population:
90° Pulse
Mz = 0∑== 000zzMM µ
Relaxation 90° Pulse
Effect on Phase:
90° Pulse
000 ==∑ xyxyM µ 00 MM zy ==∑µ 0=xzM
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Relaxation
• Different populations• No phase coherence
• Equal populations• Partial phase coherence
Excitation
Relaxation
T1
T2
*2Tt0
zxy eMM −=
*2
xyxy
T
M
dt
dM−=
1
z0z
T
MM
dt
dM −=
( )1Tt0z e1MM −−=
Bloch Equations
d t
dtt t
JM B
( )( ) ( )= ×
d t
dtt t
MM B
( )( ) ( )= × γMultiply by γ
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Bloch Equations
d t
dtt t
MM B
( )( ) ( )= × γ
1
0
T
)()]()()([
)( MtMtBMtBtM
dt
tdM zxyyx
z −−−= γ Magnetization
along the z-axis
2T
)()]()()([
)( tMtBMtBtM
dt
tdM xyzzy
x −−= γ Magnetization along the x-axis
2T
)()]()()([
)( tMtBMtBtM
dt
tdM yzxxz
y −−= γ Magnetization along the y-axis
d t
dtt t t M
MM B R M
( )( ) ( ) ( ( ) )= × − −γ 0
Relaxation Matrix
ROTATING FRAME OF REFERENCE
faster slower
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Lωω =0
z, B0
)( 1Brfω Lrf B ωω =)( 1
01)( ωω <Brf
z, B0
γωγωω //)( 00 rfrf BB −=−=∆
rfωω −0
Ω=(ω0-ωrf) ∆B= -Ω/γ
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Bloch Equations in the Rotating Frame
rfB ωγ −−=Ω 0
1
0
T
)()]()()([
)( MtMtBMtBtM
dt
tdM zrxy
ryx
z −−−= γ The “r” superscript refers to a magnetic field in the rotating frame
2T
)()()(
)( tMtBMtM
dt
tdM xryzy
x −−Ω−= γ
2T
)()()(
)( tMMtBtM
dt
tdM yx
rxz
y −Ω+= γ
(where B0=Bz and is not time-dependent)
Bloch Equations in the Rotating Frame
Example: 90° pulse along x
)()(
tBMdt
tdM rxy
z γ−=
0)( =
dt
tdM x
)()()(
tBtMdt
tdM rxz
y γ=
1
0
T
)()]()()([
)( MtMtBMtBtM
dt
tdM zrxy
ryx
z −−−= γ
2T
)()()(
)( tMtBMtM
dt
tdM xryzy
x −−Ω−= γ
2T
)()()(
)( tMMtBtM
dt
tdM yx
rxz
y −Ω+= γ
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Bloch Equations in the Rotating Frame
Example: Free Precession
1
0
T
)()]()()([
)( MtMtBMtBtM
dt
tdM zrxy
ryx
z −−−= γ
2T
)()()(
)( tMtBMtM
dt
tdM xryzy
x −−Ω−= γ
2T
)()()(
)( tMMtBtM
dt
tdM yx
rxz
y −Ω+= γ
1
0
T
)()( MtM
dt
tdM zz −−=
2T
)()(
)( tMtM
dt
tdM xy
x −Ω−=
2T
)()( tMM
dt
tdM yx
y −Ω+=
Typical 1H NMR Spectrum
Abs
orba
nce
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8 7 6 5 4 3 2 1 0
Chemical ShiftO
C6H5−CH2−O−C−CH3
H H
H
H
H
H|
— C —|H
H|
— C —H|H
TMS
6
0
TMS 10)ppm( ×ωω−ω=δ
N
eBo
σBo
“Chemical” Shift
1951
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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:
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2eff B∆BB += 2
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eff ω+Ω=ω
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x’
y’
z
B0
B1(t)
B(t)
x
y
z
B1
B
We have two fields!
Move into rotating frame (sit on the carousel)
B0 – ωωωωrf/γγγγ
x’
y’
z’ = z
M
x’
y’
z’ = z
M
B1
x’
y’
z’ = z
B1
M θ
x’
y’
z’ = z
M
θ = π/2
Flip Angle and Field Strength
ω1 = −γB1 = 2πν1 = (π/2)/PW90
90
11 4PW
1
2π
γBν ==
22
)360(
1
2 1 oPWB =
πγ 22
11 )()( ωγγ ∆+= BBeff
γγγγB1
∆ω∆ω∆ω∆ωγγγγB1
eff
θ
If (γB1)/2π = 25000 Hz [PW(360°) = 40 µs]and ∆ν = ±2500 Hz (10 ppm @ 500 MHz)
then (γB1eff )/2π = 25125 Hz
Effective Field Strength
Rotation under an Offset
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Offset Excitation
Effect of a 90° Pulse Effect of a 180° Pulse
Offset Excitation
Effect of a 180° PulseRotation under a 180° Pulse for Ω = 0.2 − 0.6 ω1
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D1
Repetition Time
DEP1 = 1/BW
PL1
AQ = DW·TDAcquisition Time
RG
Precession and Relaxation
A. S. EdisonUniversity of Florida
1 2 3 4 5
-0.5
0.5
1
1 2 3 4 5
-0.75
-0.5
-0.25
0.25
0.5
0.75
1
Mx My
0.2
0.4
0.6
0.8
Mz
-1-0.5
0
0.5
1
-0.5
0
0.5
1
0
0.25
0.5
0.75
1
-1-0.5
0
0.5
1
Mz
Mx
My
t
t t
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Detection
A. S. EdisonUniversity of Florida
0.2 0.4 0.6 0.8 1
-1
-0.5
0.5
1
0.2 0.4 0.6 0.8 1
-1
-0.5
0.5
1
analog
20 40 60 80 100
-1
-0.5
0.5
1
digital
∆t = 1/(2·νmax)Nyquist:
A B C
°−x90 t2
x
y
z
ω ≈ 108 Hz
Static:
Rotating (ω0 = ωB):
x
y
z
x
y
z
°−x90 t2
x’
y’
z
ω ≈ 103 Hzx’
y’
z
x’
y’
z
ω0 = ωB
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Det. ADCNMRSignal
ω0 (reference)
ComputerMemory
500 MHz ± 2500 Hz
500 MHz
± 2500 Hz
∆t = 1/(2·νmax) DW = 1/SW
1/2· νmax = 10-4 s νmax = 5.0 kHz
ν = 2.0 kHz
ν = 4.5 kHz
ν = 2.0 kHz
ν = 8.0 kHz
100 µs
Nyquist:
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Effect of Incorrect Sampling Rate
∫+∞
∞−
ω=ω dte)t(f)(F ti
[ ] ∫+∞
∞−
ω=ω dt)tcos()t(f)(FRe
[ ] ∫+∞
∞−
ω=ω dt)t(sin)t(f)(FIm
)t(sini)tcos(e ti ω⋅+ω=ω
z
x
yω0
ω0t
M0sin(ωt)
M0cos(ωt)
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∫+∞
∞−
ω=ω dte)t(f)(F ti
[ ] ∫+∞
∞−
ω=ω dt)tcos()t(f)(FRe
[ ] ∫+∞
∞−
ω=ω dt)t(sin)t(f)(FIm
)t(sini)tcos(e ti ω⋅+ω=ω
1
12
22 +Ω−ω
=ωT
A)(
)(
122
22
+Ω−ωΩ−ω=ωT
TD
)()(
)(
)()()( ω+ω=ω iDAF
z
x
yω0
ω0t
M0sin(ωt)
M0cos(ωt)
Phase Correction
)()()( ω+ω=ω iDAF ( ) )()(exp)( ωωφω iDAiF instr +=
[ ] )()(Re ωω AF =
[ ] ( ) ( ) )(sin)(cos)(Im ωφωφω ADF instrinstr +=
[ ] ( ) ( ) )(sin)(cos)(Re ωφωφω DAF instrinstr −=
[ ] )()(Im ωω DF =
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Quadrature Phase Detection
Quadrature Phase Detection
PSD ADC
PSD ADC
NMRSignal ω0
0° reference
90° reference
ComputerMemory
A
ComputerMemory
B
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The Redfield Trick
PSD
PSD
ADCNMRSignal ω0
0° reference
90° reference
ComputerMemory(+ + − −)
∆t = DW/2 = 1/2·SW
Each subsequent point is phase-shifted by 90°The period of this added frequency is 4·∆tThe added frequency is therefore SW/2
The Redfield Trick
Original SW: −νmax ↔ +νmax
Added frequency: SW/2 = νmax
Final SW: 0 ↔ +2νmax
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1st mixer preamplifierI.F. (20MHz)
2nd mixeramplifier
Audio (+-kHz)
1st ADC 0o
Sample
Probe Coil
Cable
2nd ADC 90o
600 MHz
10 kHz spectral window =599,995,000.0 – 600,005,000.0 Hz
19,995,000.0 – 20,005,000.0 Hz
-5,000.0 - + 5,000.0 Hz
620 MHz 20 MHz
* sin
* cosV~ µV V ~ mV V ~ Volt
• The amplifiers – mixers amplify both the signal and the noise!!
• We do not gain S/N with these components!!
How does an NMR receiver work?
Problems…
If the two ADCs are not perfectly matched we have “quad images”
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Problems with Quadrature Phase Detection
x
y°y90
t2
z
x
yωt
M0sin(ωt)
M0cos(ωt)
z
z
x
y°−x90
t2
z
x
yωt
M0cos(ωt)
-M0sin(ωt)
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The “Direct Digital” receiver…
1st mixer preamplifierI.F. (20MHz)
Sample
Probe Coil
Cable
600 MHz
10 kHz spectral window = 599,995,000.0 – 600,005,000.0 Hz
19,995,000.0 – 20,005,000.0 Hz
620 MHz
V~ µV V ~ mV ADC We digitize directly at the I.F.
•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…
Quadrature Receiver16-bit ADC with oversampling
Varian DDR
Quad Images
Spectra acquired with a single transient
This is the
peak you
want
Center
Spike
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ν ν
The effects of Relaxation
NMR parameters and relaxation in a
simple 1D experiment
f.i.d. f.i.d.
aq
rd
Acquisition time
Recycle delay
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Length of AcquisitionS(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
I
tt
I
tt
Acquisition time vs T2
2T
t
e−
)(
2
1ωωπ−I
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Recycle delay vs T1
I
t
2T
t
e−
)(
2
1ωωπ−I
I
t
2T
t
e−
)(
2
1ωωπ−I
Zero-Filling
DR = SW / SIDR = SW / SI
spectral width (Hz)
data size (n. of points)
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Zero-Filling
DR = SW / SIDR = SW / SI
spectral width (Hz)
data size (n. of points)
Fourier Pairs
∫+∞
∞−
ω=ω dte)t(f)(F ti
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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
DIGITALRESOLUTION
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Window Functions:Resolution Enhancement
S = A*exp-t/T2 exp-atexp-bt2a = -1/T2 ; b>0
Other Window Functions:Sine Bell and Sine Bell Squared
S = A*exp-t/T2 sin( π-φ)t/tmax +φ
AS=
S = A*exp-t/T2 sin2( π-φ)t/tmax +φ
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Spin-Spin Coupling
C - Y C - CH C - CH2 C - CH3
H|
H|
H|
H|
singlet doublet triplet quartet
X ZX Z X Z X ZJ
AX System, J = 0 E/h= −ΣiνimI(i)
A2
X1
X2
A1
αα
ββ
βααβ
Eαα = − ½νA − ½νX
Eαβ = − ½νA + ½νX
Eβα = + ½νA − ½νX
Eββ = + ½νA + ½νX
EA1 = Eβα − Eαα = νA
EA2 = Eββ − Eαβ = νA
EX1 = Eαβ − Eαα = νX
EX2 = Eββ − Eβα = νX
A2
X1
X2
A1
αα
ββ
βααβ
+ΣΣi<jJijmI(i) mI(j)
+ ¼JAX
− ¼JAX
− ¼JAX
+ ¼JAX
− ½JAX
+ ½JAX
− ½JAX
+ ½JAX
≠
¼JAX