Transcript 投影片 1 - Academia Sinica

Lecture 7 Two-dimensional NMR
(A, A)
(A, X)
Cross-peak
F1
(X, A)
(x, X)
Diagonal
F2
Interpretation of peaks in 2D spectrum
Need mixing time to transfer magnetization to see cross peaks !
1H
Allows interaction
to take place
excitation
General scheme:
To keep track of
magnetization
(Signal not recorded)
1H
Experiment:
t1 = 0
Get a series of FIDs with incremental t1 by
a time . Thus, for an expt with n traces t1
For the traces will be 0, , 2, 3, 4 ----(n-1), respectively. We will obtain a series
of n 1D FID of S1(t1, t2). Fourier transform
w.r.t. t2 will get a series of n 1D spectra
S2(t1, 2). Further transform w.r.t. t1 will
get a 2D spectrum of S3(1, 2).
t1 = 
Spectral width in the t1 (F1) dimension
will be
SW = 1/
t1 = 5
FT
Signal contains info
due to the previous
three steps
F2 = 1
F2 = 2
t1 = 2
F2 = 1
t1 = 3
t1 = 4
t1 = n
F2 = 3
F2 = 4
F2 = N
FT
Review on product operator formalism:
1. At thermal equilibrium: I = Iz
2. Effect of a pulse (Rotation):
exp(-iIa)(old operator)exp(iIa) = cos (Old operator) + sin (new operator)
= - Iy for 1tp = 90o
3. Evolution during t1 :
(free precession) (rotation w.r.t. Z-axis):
Product operator for two spins: Cannot be treated by vector model
Two pure spin states: I1x, I1y, I1z and I2x, I2y, I2z
I1x and I2x are two absorption mode signals and
I1y and I2y are two dispersion mode signals.
These are all observables (Classical vectors)
Coupled two spins: Each spin splits into two spins
Anti-phase magnetization: 2I1xI2z, 2I1yI2z, 2I1zI2x, 2I1zI2y
(Single quantum coherence)
(Not observable)
Double quantum coherence: 2I1xI2x, 2I1xI2y, 2I1yI2x, 2I1yI2y (Not observable)
Zero quantum coherence: I1zI2z (Not directly observable)
Including an unit vector, E, there are a total of 16 product operators in a weaklycoupled two-spin system. Understand the operation of these 16 operators is
essential to understand multi-dimensional NMR expts.
Example 1: Free precession of spin I1x of a coupled two-spin system:
Hamiltonian: Hfree = 1I1z + 2I2z
No effect
= cos1tI1x + sin1tI1y
Example 2: The evolution of 2I1xI2z under a 90o pulse about the y-axis
applied to both spins:
Hamiltonian: Hfree = 1I1y + 1I2y
Evolution under coupling:
Hamiltonian: HJ = 2J12I1zI2z
Causes inter-conversion of in-phase and
anti-phase magnetization according to the
Diagram, i.e. in-phase  anti-phase and
anti-phase  in-phase according to the rules:
Must have only
one component in
the X-Y plane !!!
Coherence order: Only single quantum coherences are observables
Single quantum coherences (p = ± 1): Ix, Iy, 2I1zI2y, I1yI2z, 2I1xI2z …. etc
Zero quantum coherence: Iz, I1z2z
Raising and lowering operators: I+ = ½(Ix + iIy); I- = 1/2 (Ix –i-Iy)
Coherence order of I+ is p = +1 and that of I- is p = -1
 Ix = ½(I+ + I-); Iy = 1/2i (I+ - I-) are both mixed states contain order
p = +1 and p = -1
For the operator: 2I1xI2x we have:
2I1xI2x = 2x ½(I1+ + I1-) x ½(I2+ + I2-) = ½(I1+I2+ + I1-I2-) + ½(I1+I2- + I1-I2+)
P = +2
P = -2
P=0
P=0
The double quantum part, ½(I1+I2+ + I1-I2-) can be rewritten as:
(Pure double quantum state)
Similar the zero quantum part can be rewritten as:
½(I1+I2- + I1-I2+) = ½ (2I1xI2x – 2I1yI2y)
(Pure zero quantum state)
2D-NOESY of two spins w/ no J-coupling:
Consider two non-J-coupled spin system:
1. Before pulse:: Itotal =
Let us focus on spin 1 first:
2. 90o pulse (Rotation):
3. t1 evolution:
4. Second 90o pulse:
5. Mixing time: Only term with Iz can transfer energy thru chemical exchange. Other
terms will be ignored. This term is frequency labelled (Dep. on 1 and t1). Assume a
fraction of f is lost due to exchange. Then after mixing time (No relaxation):
6. Second 90o pulse:
7. Detection during t2:
The y-magnetization =
Let
A1(2) = FT[cos1t2] is the absorption signal at 1 in F2 and A2(2) = FT[os2t2] as the
absorption mode signal at 2 in F2. Thus, the y-magnetization becomes:
Thus, FT w.r.t. t2 give two peaks at 1 and 2 and the amplitudes of these two peaks are
modulated by (1-f)cos1t1 and fcos1t1, respectively.
FT w.r.t. t1 gives:
where A11 = FT[cost] is the absorption mode signal at 1 in F1.
 Starting from spin 1 we observe two peaks at (F1, F2) = (1, 1) and (F1, F2) = (1, 2)
(Diagonal)
(Cross peak)
 Similarly, if we start at spin 2 we will get another two peaks at: (F1, F2) = (2, 2) and
(F1, F2) = (2, 1)
 Thus, the final spectrum will contain four peaks at (F1, F2) = (1, 1), (F1, F2) = (1, 2),
(F1, F2) = (2, 1), and (F1, F2) = (2, 2)
 The diagonal peaks will have intensity (1-f) and the off-diagonal peaks will have
intensities f, where f is the fraction magnetization transferred, which is usually < 5%.
1H
Allows interaction
to take place
excitation
General scheme:
To keep track of
magnetization
(Signal not recorded)
1H
Experiment:
Get a series of FIDs with incremental t1 by
a time . Thus, for an expt with n traces t1
For the traces will be 0, , 2, 3, 4 ----(n-1), respectively. We will obtain a series
of n 1D FID of S1(t1, t2). Fourier transform
w.r.t. t2 will get a series of n 1D spectra
S2(t1, 2). Further transform w.r.t. t1 will
get a 2D spectrum of S3(1, 2).
Spectral width in the t1 (F1) dimension
will be
SW = 1/
t1 = 0
FT(t1)
Signal contains info
due to the previous
three steps
F2 = 1
t1 = , cos1
F2 = 2
t1 = 2, cos12
F2 = 1
t1 = 3, cos13
t1 = 4, cos14
t1 = 5, cos15
F2 = 3
F2 = 4
cos4
FT
t1 = n, cos1n
F2 = N
FT()
7.4. 2D experiments using coherence transfer through J-coupling
7.4.1. COSY:
After 1st 90o pulse:
t1 evolution:
J-coupling:
Effect of the second pulse:
(p=0, unobservable)
(p=0 or ±2)
(unobservable)
(In-phase, dispersive)
(Anti-phase)
(Single quantum coh.)
The third term can be rewritten as:
Thus, it gives rise to two dispersive peaks at 1 ± J12 in F1 dimension
Similar behavior will be observed in the F2 dimension, Thus it give a double
dispersive line shape as shown below.
The 4th term can be rewritten as:
Two absorption peaks of opposite signs (anti-phase) at 1 ± J12 in F1 dimension will be
observed.
Similar anti-phase behavior will be observed in F2 dimension, thus multiplying F1 and F2
dimensions together we will observe the characteristic anti-phase square array.
 Use double-quantum filtered COSY (DQF-COSY)
Double-quantum filtered-COSY (DQF-COSY):
Using phase cycling to select only the double quantum
term (2) can be converted to single quantum for
observation. (Thus, double quantum-filtered)
P = 2
P = -2
P = 0
P = 0
Rewrite the double quantum term as:
The effect of the last 90o pulse:
Anti-phase absorption
diagonal peak
Anti-phase absorption
cross peak
Heteronuclear correlation spectroscopy
1. Heteronuclear Multiple Quantum Correlation (HMQC):
For spin 1, the chemical shift evolution is totally
refocused at the beginning of detection. So we need to
analyze only the 13C part (spin 2)
J-coupling
J-coupling
After 90o 1H pulse:
At the end of :
13C
- I1y =
= 2I1xI2z
for  = 1/2J12
After 2nd 90o pulse:
The above term contains both zero and double quantum coherences.
Multiple quantum coherence is not affected by J coupling. Thus, we
need to consider only the chemical shift evolution of spin 2.
J-coupling during 2nd :
evolution J-coupling
Phase cycling: If the 1st 90o pulse is applied alone –X axis the final term will also change
sign. But those which are not bonded to 13C will not be affected. Those do two expt with
X- and –X-pulses alternating and subtract the signal will remove unwanted signal.
2. Heteronuclear Multiple-Bond Correlation (HMBC):
In HMQC optimal  = 1/2J = 1/(2x140) = 3.6 ms. In order to detect long range coupling
of smaller J one needs to use longer , say 30-60 ms (For detecting quaternary carbon
which has no directly bonded proton).
 Less sensitive due to relaxation.
3. Heteronuclear Single Quantum Correlation (HSQC)
Too complex to analyze in detail for every terms.
 Need intelligent analysis, i.e. focusing only on terms that lead to observables.
W/ or w/o DCPL
7.6 Lineshape and frequency discrimination: