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Chem 526Chem 526
NMR for Analytical Chemists
Lecture 7Lecture 7
(Chapter 3)
Homework 3 #1 (Due 9/20): Explain what the following operation or function means. What is the purpose for each item?
(1) Shimming
(2) Lock
(3) Spinning
(4) Window Function
(5) Fourier Transform( )
(6) Phasing
(7) Probe Tuning
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Homework 3 #2Q 1. Plot vs A() for = 100 and =10
Q 2. Plot f vs D() for = 100 and =10
)(
S
Q 3. Plot f vs A() for /2 = 200 and =10
Q 4. Plot f vs A() for /2 = 200 and =40
Q 5. Plot f vs D() for = 200 and =40
2222)(
iS
22)(
A
22
)(
D
Calculation of FT (p139-140)• S’(t) = exp(it)exp(-t)exp(i) ( =1/T2)
• [3.14b]
exp)()(0
titsdtS
Calculate FT of s’(t) in [3.14b] for = /2 as S’() and CalculateReal(S’()) and Imag(S’())
S() =
)(exp0
ttidt
0
)(exp1
ttii
=
=
2222
i
i
1
A() D()
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0 04
0.06
0.08
0.1
0.12
A1(w)
-0.06
-0.04
-0.02
0
0.02
0.04
-200 -100 0 100 200 300 400
A1(w)
D2(w)
A5(w)
Q 1. Plot vs A() for = 100 and =10
Q 2. Plot f vs D() for = 100 and =10
Q 5. Plot f vs D() for = 200 and =40
Announcement 1• How is the preliminary analysis of the unknown
(1H & 13C NMR) for Homework 4 (Due 9/22)?
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Chemical Shifts• Electron near nuclei creates extra field –B0
–B0(1 + )
Higher Frequency Lower Frequency
7Lower Shielding Higher Shielding(Higher Field)(Lower Field)
Induction Effects in 1H shifts
1H chemical shifts of CHnCl4-n(ppm) CH 0 23CH4 0.23 CH3Cl 3.05CH2Cl2 5.3CHCl3 7.27
1H shifts of CH3X (ppm)
Cl C-H3
e-
8
3 (pp )
F OH NH2 H Me3Si
4.26 3.38 2.47 0.23 0.0
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Analysis Example: 13C and 1H NMR of
diethyl ether (CH3CH2)2O
Q1. 13C NMR: Which is CH3?
(ppm)(CH3CH2)2O( 3 2)2
(Ppm) = [Shift from TMS (Hz) ]/ [NMR Frequency (MHz)]
Q2. 1H NMR: Which is CH3?
13C Chemical shift additivity principle
• δ = -2.5 +Ajnj,
• Aj denotes a substituent parameter for• Aj denotes a substituent parameter for neighbouring chemical groups
Substituent parameters for C and CH, CH2, CH3,
α β γ
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9.1 9.4 -2.5
CH3-CH2-CH2-CH2-CH3CH3: -2.5 + 9.1 + 9.4 -2.5 = 13.4 (13.9)
Exp
CH2: -2.5 + 2x9.1 + 9.4 -2.5 = 22.6 (22.8)
CH2: -2.5 + 2x9.1+ 2x9.4 = 34.5 (34.7)
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13C aliphatic shifts of ethyl-ether• CH3-CH2-O-CH2-CH3
CH3 : -2.5+9.1+ 8 -2.5 = 12.1 ppmCH : 2 5 + 9 1 + 58 +9 4 2 5 = 71 5 ppmCH2 : -2.5 + 9.1 + 58 +9.4 -2.5 = 71.5 ppm
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13C aliphatic shifts of ethyl-ether• CO2H-CH2-NH2 (Glycine)CH2 : -2.5 + 29 + 21 =47.5 ppm
• CO2H-CH(CH3)-NH3 (Alanine)CH : -2.5 + 29 + 21 +9.1=56.6 ppm
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2D NMRA. 2D Heteronuclear Correlation NMR (HETCOR) The basic idea of 2D HETCOR NMR is to specify a peak with two frequencies I andS. The sequence is given as an extension of polarization The sequence is given as an extension of polarization transfer experiment, but there is the t1 period for indirectly detecting 1H signal.
Sxcos(It1)
+ Iysin(It1)Ixcos(It1)
13
Array of FIDs are taken for different t1 period
Sxcos(St2)+Sysin(St2)cos(It1)
Ch4.1 (See p275)
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2D FT Processing
• So we obtained
s(t1, t2) = cos(It1)exp(iSt2).
• First step is Complex FT on t2
s(t1, ω2) = cos(It1)A(ω2 -S)
• Second, Cosine FT on t1s(t ) A( )A( )
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s(t1, ω2) = A(ω1 -I)A(ω2 -S)
2D COSY for Coupled A-X system
A
1
X A
A
X
X 2
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COSY
TOCSY(total correlation spectroscopy)
Q1. Do you expect a cross peak between A and B if A and B form a dipeptide like A-B?
Q2. Do you expect a cross peak between A and B if A and B are a mixture of amino acids?
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Water Suppression
Rectangular Pre-saturation
Jump & Return
Jump & Return
3.7.1 Presaturation (p223)
• Irradiate weakly (1/2 ~ 50 Hz) for a long ti (1 2 ) Q Wh t ld h ?time (1-2 s) Q. What would happen?
• Advantage: Simple
• Disadvantage:
-Requires a good shimming
- Exchangeable 1H (such as amide 1H) may be also suppressed
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3.2 Data Acquisition
• FID Generated cos(0t)
Mixing to lower frequency (Real/Imag+Filter)
cos(0 - RF ) sin(0 - RF )
Digitization
Store to the computer
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Display
3.2.3 Quadrature Detection (p132)
cos(wRFt)cos(RFt)cos(0t)= cos(RF+ 0)t /2
+ cos (RF – 0)t/2
cos(w0t) Low pass
Low pass
22
Q. Explain how the experimental scheme for data acquisition &detection (So called quadrature detection) works
sin(wRFt)
sin(RFt)cos(0t)= sin(RF+ 0)t/2
+ sin (RF – 0)t/2
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3.2.1 Sampling (p124)
After digitization, a time-domain signal is digitized ass(tn) = s(nt) A/D Conversion
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Nyquist frequency: fn = 1/(2t) defines the highest-frequency sinusoidal that can be reproduced from the digitized data. S(t) = s(kt)sinc[2fn(t-kt)],
where sinc(x)=sin(x)/x D/A Conversion
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What happens if the frequency is higher than fn?
• First note that 2π(2fn∆t) = 2πFrom this, cos[2π(2fnM+ ν0)k∆t] = cos[2π 2fnMk∆t+ 2π ν0k∆t] = cos[2πMk + 2π ν0k∆t] = cos[2π ν0k∆t] (k is an integer)
(2fnM+ ν0) and ν0 can’t be distinguished by the digital sampling
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sampling.
Hence, the signal shows up at νa
νa = (2fnM+ ν0), where -fn ≤ νa ≤ fn
-fn fn
Q. Where does the signal va show up?v0
Aliasing
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νa-fn fn
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3.2.3 Quadrature Detection (p132)
cos(wRFt)
cos(RFt)cos(0t)= cos(RF+ 0)t /2
+ cos (RF – 0)t/2
cos(w0t)
sin(w t)
Low pass RF Fileter+ Band Path Audio Filter
Low pass RF Filter + Band Path Audio Filter
Remove Aliasing of Noise
27Experimental Scheme for Data Acquisition &Detection (Quadrature detection)
sin(wRFt) sin(RFt)cos(0t)= sin(RF+ 0)t/2
+ sin (RF – 0)t/2
Digital resolution & Zero Filling
• Actual digital resolution in the obtained t ispectrum is
1/(N*∆t) = 1/tmax
∆t = sampling interval and N is data points in the FID
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the FID
Hence, the digital resolution can be enhanced by extending tmax.
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Ch 3.4 Pulse Effects
• We will cover – 1H Decoupling (3.4.3)
– Off-Resonance Effects (3.4.1)
– Composite Pulse (3.4.2)
– Selective Pulse (3.4.4)
– Water Suppression (3 4 5)
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Water Suppression (3.4.5)
1H decoupling
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13C spectrum of ethylbenzene
Q. Why are there no 1H-13C J splitting?
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J coupling
The local field that the S spin experiences due to a J coupling between I and S spins is denoted by
2πJIZ = 2πJmI.Z I
The energy due to the J coupling is
EJ = 2πJmISZ = 2πJIZSZ.
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What is 1H decoupling?
• The idea is to exchange and states rapidly for the I spin so that the splitting is averaged. p p g g
If <mI> = 0
<EJ> = 2πJ<IZ>SZ
= 2πJ<mI>SZ = 0
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• If you remember IZ – πIX -IZUnder continuous RF field matched at 1H frequency,
< IZ> = <mI> = 0
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Chemical Exchange (Ch5. p400)
A B
k
kA Bk
10 s-1
100s-1
450s-1
160 Hz
33
450s
1000s-1
5000s-1
1H Spectrum of Leu-enkephalin in DMSO
(Tyr-Gly-Gly-Phe-Leu)Broadeningby chemical
D h i
yexchange
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Amide Phe
From “high-resolution NMR techniques in organic chemistry” by Claridge
Dephasing period
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How strong 1H decoupling must be?
• The idea is to exchange and states rapidly for the I spin so that the splitting israpidly for the I spin so that the splitting is averaged.
From the chemical exchange example,
IZ – πIX -IZ IZ πIX IZ
must be faster than J coupling
However, there is a problem in 1H decoupling Heating by rf
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1H spectrum of L tryptophan in 400 MHz
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Off-resonance Effects & Effective Field
= - BHow is the magnetization evolved by the effective field ?
eff = - Beff the effective field ?
37
eff = (2 + 12)1/2
eff = - Beff Near On-Resonance
eff = - Beff Off-Resonance
eff = (2 + 12)1/2
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3.4.1 Off-Resonance Effects Excitation by /2-pulsefor different offsets
R ( ) R () R ( )R ()Ry(,) ~ Rz() Ry()Rz()
39
Iz after pulse
|IX|2 + |Iy|21/2 after pulse
Excitation by -pulsefor different offsets
Off-resonance Effects in 1H Decoupling
• Scaled coupling is J c,where = Ω/Ω2 + (γB1)21/2where c Ω/Ω + (γB1)
c = 0 for the on-resonance case Ω=0.However, when Ω~γB1, c = 0.707.
On the other hand, if you use a strong B1 field, you can bake your sample because of theyou can bake your sample because of the heat generated by a decoupling RF field. (W B1
2)
What is solution ?
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Resonance Offset Dependence of Effective Coupling
Resonance-Offset Dependence
0 91
0 20.30.40.50.60.70.80.9
lc c
DecouplingNot Effective
00.10.2
0 0.5 1 1.5 2 2.5 3omega/gB1
DecouplingEffective
3.4.3 Composite Pulse (p174)
• Composite pulse compensates non-ideal ff t f ff d RFeffects of off-resonance and RF-
inhomogeneity over sample i.e. B1(1±).
• One idea is to add an extra pulse that does not function in the ideal case butdoes not function in the ideal case, but compensates for the non-ideal effects.
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Off-resonance dependence of composite rotation
β = 90º β’ = 240º
From “Principles of NMR in one and two dimension” by Ernst et al.
3.5.2 Super cycle for Composite Pulse Decoupling (p204)
• R = 90x180-x270x
• R = 90-x180x270-xR 90 x180x270 x
• If 90 = 1, R = 1x2-x3x
RRRR WALTZ4
RRRRRRRR WALTZ8RRRRRRRR WALTZ8
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Offset (Ω) dependence of effective coupling constants
Fig. 3.28
Effects of Super cycle
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Computer optimized sequence Garp-1
3.4.4 Selective Pulse (p179)Ideal Selective /2-pulse
0.2
0.4
0.6
0.8
1
0
-1-0.8
-0.6
-0.4
-0.2 0
0.2
0.4
0.6
0.8
1
1.5
Ideal Selective -pulse
-1.5
-1
-0.5
0
0.5
-1-0.8
-0.6
-0.4
-0.2 0
0.2
0.4
0.6
0.8
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Typical Shaped Pulse for Selective Excitation
Gaussian Sinc
Hermitian
Selective 90° PulseMX MY MZ
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Water Suppression
Rectangular Pre-saturation
Jump & Return
Jump & Return
3.7.1 Presaturation (p223)
• Irradiate weakly (1/2 ~ 50 Hz) for a long ti (1 2 ) Q Wh t ld h ?time (1-2 s) Q. What would happen?
• Advantage: Simple
• Disadvantage:
-Requires a good shimming
- Exchangeable 1H (such as amide 1H) may be also suppressed
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3.7.2 Jump-Return
90y ------ ------ 90-y
In case of on-resonance (H2O signal)
IZ – 90y IX – IZ – 90-y IZ
In case of some-what off-resonance (Other signals)
Water Signal Undetected!
IZ – 90y IX – Ixcos + Iysin
– 90-y IZcos + Iysin
Other Signals Detected!
Spin Lock & Field Gradient for Water Suppression
SLy
Water /2x
Non-selective /2X
SLy
Non-selective /2X Non-selective X
FG FG
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H exchange rate kex (p 224)From “NMR of Proteins and Nucleic Acids” By K. Wuthrich
Lo
g(k
ex )
(m
in-1
)
Indole
pH and Temperature dependence of kex
(min)
From:http://www.hxms.com/
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pH = 4.1 25 º C
H/D Exchange of BPTI
Use of DMSO
2D 1H/15N HSQC spectrum of β2m amyloid fibril in 95 % DMSO/5 % D2O Goto et al. Nature Structural Biology 2002
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Dissolved into DMSOAfter 8 days incubation of amyloid In D2O
D2O DMSO
HH
H
D2O
DD
H
D
DH
T
H/D Exchange Pattern
Amyloid Fibril Native
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Fast 2D NMR and Real-time Monitoring of H/D exchange in protein folding
PNAS 2007PNAS 2007Brutscher et al.