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CHEM3013 NMR Spectroscopy
Gottfried Otting
How to measure an NMR spectrum
- lock 4
- shim 5
- tuning/matching 8
- pulse calibration 8
How to maximise the signal-to-noise ratio 9
2D NMR
- the 5 truly important 2D experiments (small molecules) 11
- example spectra of menthol 13
The art of processing a 1D NMR spectrum
- Quadrature detection 22
- How can an NMR spectrometer collect complex data points
with a single coil 24
- Fourier transformation 27
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- Phase correction 31
- Window multiplication 36
- Zero filling 41
- Baseline correction 42
Understanding 2D NMR
- The NOESY experiment 44
- Phase cycling 49
- The supertrick of 2D NMR 49
- t1-noise 54
- Sensitivity of 2D NMR 55
- How long does it take to record a 2D NMR spectrum? 56
- Quadrature detection in the indirect dimension 60
The 13C-HSQC experiment 62
- Spin-echo 63
- In-phase and antiphase magnetisation 65
- Magnetisation transfer through scalar couplings 66
- Spin-echoes with selective 180o pulses 68
- Reverse INEPT 69
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- Broadband decoupling 70
The 13C-HMBC experiment 71
DQF-COSY 73
- Understanding the multiplet fine-structure of COSY cross-peaks
76
Two footnotes to NOESY
- NOE build-up curve 79
- Chemical exchange in NOESY 80
INADEQUATE the ultimate for tracing carbon chains 83
Pulsed field gradients 85
Selective pulses 87
Appendix on the product operator formalism 89
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How to measure an NMR spectrum
To measure a 1H NMR spectrum on a high-field NMR spectrometer,
we need to
lock, shim, tune and match, and determine the 90 degree
pulse.
Lock:
The magnetic field of the spectrometer has to be extraordinarily
stable, because
the frequency differences between the nuclei are so small.
NMR magnets need to stand clear of magnetic perturbations (any
magnetic tools,
cars parked outside the NMR room, lifts etc.) Even a small
thermal expansion of
the outside of the magnet by a change in temperature (worst
case: the sun shining
on the magnet) causes a noticeable change in the magnetic
field.
The lock is an electronic circuit that continuously measures the
2H NMR
spectrum of solvent to determine the exact value of the current
magnetic field. If it
detects a change in magnetic field, it either changes the
frequency of the
electronics (as in modern NMR spectrometers) or (in conventional
spectrometers)
Problem 1: a) Assume that a spectral resolution of 0.4 Hz is
adequate. How many
ppm are 0.4 Hz on a 400 MHz NMR spectrometer?
b) To put this in perspective: if Mt Everest is about 10,000 m
high,
what height is 0.1 ppm of Mt Everest?
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adjusts the magnetic field strength by increasing or decreasing
the current through
a helper magnetic coil, which is positioned at the very inside
of the magnet in the
room temperature section.
The lock signal usually only reports the height of the
(strongest) peak found in the
2H NMR spectrum.
Shimming:
While the lock can make the magnetic field stable, the field
also needs to be
homogeneous, i.e. the same across the entire sample. This is
achieved by
numerous shim coils in the room temperature section of the
magnet (room
temperature shims).
Problem 2: An external lock is a small vial containing a
deuterated solvent
built into the probe head, obviating the need for using
deuterated solvent for the NMR measurement. What then is the
point of using deuterated solvents?
If the spectrometer knows which solvent it is locking on, it can
use this for
spectrum calibration!
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The axis of the magnet is defined as the z-direction. Changing
the z-shim means
that the magnetic field is made stronger at one end of the
sample than at the other,
with a linear gradient along the z-axis.
For the z2-shim, the gradient is quadratic (like a
parabola).
There are z3, z4, z5, z6, z7, z8 shims
The shims are designed to be orthogonal to each other, but in
practice changing
one will affect others as well.
On-axis shims or spinning shims are the shims z z8.
Off-axis shims or non-spinning shims are all shims that contain
x and y
components (e.g. x, y, xy, x2-y2, xyz, xz, yz2, etc.)
Shimming a magnet involves a multi-dimensional search which can
take very
long (hours) even if the shims are perfectly orthogonal. It
involves maximising the
height of the lock signal. The most critical shims are those
along the z-axis,
because that is where the sample is longest. Spinning (=
rotation of the NMR tube
at about 20 Hz) can average inhomogeneities in the off-axis
directions (but leads
Problem 3: a) In which direction does the magnetic field point
that is generated by the
x-shim?
b) In which direction does the x-shim change the magnetic field
gradient?
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to spinning sidebands in the NMR spectrum). For good off-axis
shims, the lock
signal does not drop very much when spinning is turned off.
Examples of the effects of ill-adjusted on-
axis shims on the line shape.
Fortunately, shimming has been automated with the help of pulsed
field gradients
(PFG, to be discussed later).
Note: taking a sample out of the magnet and re-inserting it into
the magnet
changes its position - good idea to shim the on-axis shims
again!
For a good shim, the 1H NMR line width of CHCl3 is < 8 Hz at
the height of
the 13C satellites (without spinning).
! !
!!
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Tuning and Matching:
On high-field NMR spectrometers, the resonance circuit
delivering the pulses and
detecting the signal must be tuned to be sensitive to the Larmor
frequency. Two
adjustable capacitors achieve the necessary fine-tuning, one of
them for tuning,
the other for matching. The optimum setting minimizes reflected
power, leading
to the shortest 90 degree pulse lengths and, most importantly,
the best sensitivity.
90 degree pulse:
The biggest signal in the NMR spectrum is obtained after a 90o
pulse (if only a
single pulse is applied). The 90o pulse length, delivered with
the same power,
depends on the solvent. In particular, 90o pulses are longer for
salty samples. If an
experiment requires an exact 90o pulse, it needs to be
determined experimentally.
Problem 4: a) Determine the 90o pulse by performing a series
of
experiments with systematically increased pulse lengths.
How do you expect the spectrum to change?
b) Why is it easier to measure the 180o or the 360o pulse
than
the 90o pulse?
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How to maximize the signal-to-noise ratio
- Average over more scans: S/N ~ Sqrt(ns) (ns = number of
scans)
- Choose the optimal recovery (interscan) delay: too long (>
5T1) wastes time, too
short and not enough equilibrium magnetisation has
recovered.
- Ernst angle: combining a short recovery delay (e.g. 0.8 s)
with a short (e.g. 30o)
pulse angle yields a better S/N than waiting longer between
scans and using a
90o pulse. (To be precise: cosopt = exp(-T/T1), where opt is the
optimal flip
angle of the pulse and T is the recovery delay. Only helpful if
we can guess T1.)
- Use the optimal acquisition time (= duration of the FID): once
the FID has
decayed to the level of white noise, there is no point in
recording it for much
longer.
- Use an adequate receiver gain.
- Apply window multiplication before Fourier transform: the
matched window
function multiplies the FID with a function that follows the
envelope of the FID.
If the FID decays exponentially, the matched window function is
an
exponentially decaying function that doubles the line width.
Problem 5: A spectrum contains one signal of 5 Hz line width
(full line width
measured at half height) and one signal of 10 Hz line width.
Following (i) multiplication with a window function that leads
to 7
Hz line broadening and (ii) FT, how wide will the lines be?
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- Use a bigger magnet: S/N ~ 2.5 ( = Larmor frequency)
- Use a cryoprobe: cooling the detection coil and preamplifier
to about 20 K
reduces the white noise picked up by the electronics.
- Minimize unnecessary salt content.
- Changing the temperature: the equilibrium magnetisation is
greater at lower
temperature, but the effect needs very low temperatures (liquid
N2) to become
noticeable. For large molecules (MW > 2000) the line widths
tend to be
narrower at increased temperatures, i.e. raising the temperature
may help.
- Accelerate T1 relaxation by adding paramagnetic compounds
(e.g. O2).
Drawback: paramagnetic compounds also accelerate T2 relaxation
(leading to
broader lines)
- Make a more concentrated sample!
Problem 6: Given a spectrum that displays a moderate
signal-to-noise ratio. If it
was recorded with 1 scan, how many scans will it take to double
the
S/N ratio? If it was recorded with 10000 scans, how many scans
will
it take to double the S/N ratio?
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2D NMR
Tabulated chemical shifts and coupling constants are not
enough:
- Chem. shifts depend on solvents, temperature, pH etc.
- Coupling constants are difficult to resolve under conditions
of spectral overlap
and strong coupling (established data bases often refer to
spectra recorded on
low-field NMR spectrometers)
2D spectra actually dont take long to record (minutes).
The 5 truly important 2D experiments (small molecules)
[13C, 1H]-HSQC (heteronuclear single-quantum coherence)
- Short name: 13C-HSQC
- Correlates: 13C with 1H chemical shifts
- Magnetisation transfer: via 1JHC
[13C,1H]-HMBC (heteronuclear multiple-bond correlation)
- Short name: 13C-HMBC
- Correlates: 13C with 1H chemical shifts
Not much beats the clarity of an assigned [1H,13C]-HSQC
spectrum!
Not to mention the many parameters that are accessible only by
2D NMR
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- Magnetisation transfer: via JHC (1JHC suppressed)
DQF-COSY (double-quantum filtered correlation spectroscopy)
- Short name: COSY
- Correlates: 1H with 1H chemical shifts
- Magnetisation transfer: via JHH
TOCSY (total correlation spectroscopy)
- Correlates: 1H with 1H chemical shifts
- Magnetisation transfer: via one or several JHH couplings
NOESY (nuclear Overhauser effect spectroscopy)
- Correlates: 1H with 1H chemical shifts
- Magnetisation transfer: via NOEs (i.e. through-space
dipole-dipole relaxation)
- detects internuclear distances < 5
! Albert W. Overhauser 1925-2011
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Below are example spectra of menthol:
- The cross-peaks correlate the 13C spins with the directly
bonded 1H spins.
- As in most 2D NMR spectra, when plotting the 1D NMR spectra
along both
axes the cross-peaks appear at the intersection between peaks in
the 1D 13C-
NMR spectrum and the 1D 1H-NMR spectrum.
- Positive cross-peaks (plotted with multiple contour lines) are
from CH and
CH3 groups.
- Negative cross-peaks (plotted with a single contour line) are
from CH2 groups.
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- Cross-peaks are generated by 2JHC and 3JHC couplings
- Cross-peaks due to 1JHC couplings are suppressed, but may be
observed with
weak intensities. The JHC couplings are not decoupled in the 1H
dimension!
Problem 7: a) How many cross-peaks does a CH2 group with
degenerate 1H
chemical shifts show in the 13C-HSQC spectrum?
b) Where do the 13C-HSQC cross-peaks appear for a CH2 group
with non-degenerate 1H chemical shifts?
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- Mixed absorptive and dispersive line shape, hence best
presented in
magnitude mode: I = Sqrt(IRe2 + IIm2), where IRe is the signal
in the real part of
the spectrum and IIm is the signal in the imaginary part of the
spectrum. More
about this later.
Problem 8: Can the 13C chemical shifts of quaternary carbons be
determined
from HMBC spectra? What is required?
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- The cross-peaks correlate the 1H spins via JHH couplings.
- The cross-peaks appear at the intersection between peaks in
the 1D 1H-NMR
spectra.
- The cross-peaks have positive and negative multiplet
components. (Here:
positive peaks plotted with a single contour line, negative
peaks with multiple
contour lines.)
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- JHH usually needs to be > 1 Hz to produce a cross-peak.
Cross-peaks tend to
be more intense for large JHH couplings.
- The spectrum is symmetric about the diagonal.
- Peaks on the diagonal are called diagonal peaks.
- The vertical noise bands around 0.85 and 0.95 ppm are
so-called t1 noise
arising from spectrometer instabilities. (More about this
later.)
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- The cross-peaks correlate the 1H spins via one or multiple JHH
couplings
within a spin-system.
- The cross-peaks appear at the intersection between peaks in
the 1D 1H-NMR
spectra.
- Cross-peaks tend to be positive, but on closer inspection
reveal impure line
shapes.
- The spectrum is symmetric about the diagonal.
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- A cross-peak is weak (or absent) if too many JHH couplings are
required to
link the two 1H spins, or if one of the JHH couplings in the
chain is very small.
A spin system is a set of 1H spins that are all directly or
indirectly linked by
JHH couplings. Example: ethylbenzene. The aromatic ring presents
one spin
system. If no JHH coupling can be observed between the ethyl
group and the
phenyl ring, the ethyl group presents a second spin system.
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- Cross-peaks arise from NOEs and scalar couplings (JHH).
- Cross-peaks from JHH are called zero-quantum peaks or
J-cross-peaks.
They appear at the position of COSY cross-peaks with funny
phases.
- Like in COSY, the spectrum is symmetric about the diagonal.
For small
molecules (MW < 2000), the diagonal peaks are positive and
the cross-peaks
are negative.
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- t1 noise is prominent, because NOESY cross-peaks are weak (a
few percent of
diagonal peaks).
The NMR spectra of menthol are actually trickier to assign than
one might think
for such a small molecule. Note, however, the excellent
resolution in the 13C-
HSQC spectrum: all cross-peaks are nicely separated, providing a
MUCH better
fingerprint of the compound than a 1D 1H-NMR spectrum! So, its
worth
assigning the 13C-HSQC spectrum.
Confusing! Positive NOEs give rise to negative cross-peaks in
NOESY!
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The art of processing a 1D NMR spectrum
A standard 1D 1H-NMR spectrum is recorded as a free induction
decay (FID)
following a pulse to generate transverse (x and y)
magnetisation. Fourier
transformation of the FID yields the NMR spectrum. There is,
however, more to it
(what did you expect?) the spectrum must be phase corrected to
get a purely
absorptive spectrum and it is usually important to multiply the
FID with a
window function prior to Fourier transformation. Then there is
the issue of
zero filling. Finally, baseline corrections help calculate
reliable integrals.
To understand whats going on requires complex numbers,
trigonometric
functions (sine and cosine) and some idea about the Fourier
transform. It is not
difficult.
Quadrature detection
First, each data point in the FID actually comes from two
measurements that are
combined into a complex number. The real part contains one
measurement, the
imaginary part contains the other measurement. The two
measurements are of the
transverse magnetisation, measured as the projection onto the x
and y axis,
respectively, in the rotating frame (more about this below).
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The projection onto the x axis is cos.
The projection onto the y axis is sin.
is the angle between the x axis and the magnetisation
vector.
= t = 2t, where is the Larmor frequency (in radians per second),
t the time
(in seconds) and the Larmor frequency in Hz (= s-1).
By measuring the projections onto both the x and y axes, we can
accurately tell,
which way the magnetisation vector points. Collect a second
complex data point a
short time (dwell time) later and we know the sense of
precession (clockwise or
anti-clockwise).
In summary, by digitising as complex points, we simultaneously
collect a cosine
and a sine-modulated signal. This is called quadrature
detection.
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How can an NMR spectrometer collect complex data points with a
single coil
The NMR spectrometer detects the precessing magnetisation by a
single coil, i.e.
the signal is indistinguishable from a linearly polarized
magnetisation oscillating
at the Larmor frequency (400 MHz, if the spectrometer is a 400
MHz NMR
spectrometer).
Footnote: physicists established that = -B0, i.e. the Larmor
frequency is
negative for nuclear spins with positive gyromagnetic ratio
(most spins,
such as 1H, 13C, 19F, 31P, etc.). The negative sign means that
the sense of
precession, represented by a vector along the axis of
precession, is opposite
to the direction of the external magnetic field B0. The minus
sign appears to
originate from cumbersome definitions, but it's the reason why
we plot the
frequency axes (along with ppm values) from right to left!
Chemists being
chemists, however, the minus sign has long been dropped:
For good measure, 15N has a negative , but chemists always plot
15N-NMR
spectra still with frequencies (and ppm values) increasing from
right to left
NMR spectroscopists simply dont care about the sign of . On a
400 MHz
NMR spectrometer (as in the figure), we just care about the 1H
NMR
spectrum found in a 10 ppm window!
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400 MHz constitutes the spectrometer frequency 0 in the example
above. 1H on a
400 MHz NMR spectrometer has a Larmor frequency of about 400
MHz
(varying by only a few ppm between different 1H spins and from
the spectrometer
frequency 0). When recording the NMR spectrum, we need to know
whether the
actual Larmor frequency is greater or smaller than the
spectrometer frequency 0
(which defines the frequency of the rotating frame). What
matters is the offset
= 0. There will be spins with positive and negative , if 0 is in
the middle
of the NMR spectrum (which is important to distribute the
radio-frequency power
of the pulses across the entire spectrum as uniformly as
possible).
The current induced by the precessing magnetisation corresponds
to a linear
polarised oscillation, e.g. cos(t). 400 MHz is very hard to
digitize. Hence, the
signal is first mixed down to audiofrequency. A mixer is an
electronic device
that multiplies two signals. The NMR spectrometer delivers a
pair of signals,
modulated with a cosine or sine, respectively: cos(0t) and
sin(0t). The product
yields sum and difference frequencies:
cos(t) cos(0t) = 0.5[cos(0t + t) + cos(0t - t)]
and
cos(t) sin(0t) = 0.5[sin(0t + t) + sin(0t - t)]
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Low-pass filters remove the sum frequencies (about 800 MHz) and
let the
audiofrequencies = 0 pass.
Each of the two signals, modulated by cos(t) and sin(t), are
then digitised by
analogue-to-digital converters (ADC), which forward the results
to the
computer.
The phases of a pulse or of detection (x or y) refer to the
rotating frame.
A 90oy pulse is simply delivered with a phase shift of 90o
relative to a 90ox pulse.
The spectrometer frequency 0 defines the frequency of the
rotating frame.
The offset can be positive, negative or zero.
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Fourier transformation
Mr Fourier claimed that he could decompose any function into a
sum of sine and
cosine functions. Fourier transformation simply means a plot of
the amplitudes of
the requisite sine and cosine functions (which have, of course,
different
frequencies) versus the frequency.
Actually, two plots. One is for the cosine functions, the other
for the sine
functions. We call one the real part, the other the imaginary
part. Making both
plots is referred to as complex FT.
Jean Baptiste Joseph Fourier
1768-1830
Fourier transforms are straightforward to calculate: multiply
the FID with a cosine
function of a specific frequency (a test cosine function, if you
like) and calculate
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the integral. If the integral is zero, this particular cosine
function is not contained
in the FID.
Remember, the integral of a cosine function is zero (there are
as many positive as negative areas in the function). The integral
of a squared cosine function, however, is finite. So, if our test
cosine function is in any way present in the FID, the integral will
be non-zero. For test cosine functions with the wrong frequency,
the integral will be zero. Fouriers idea was to test the FID with
cosine functions that systematically increase in frequency, then
plot the integral values found against the frequencies tested. Once
its been done with cosine functions to get the real part, it can be
repeated with sine functions to get the imaginary part. Voil!
The FT transforms the FID from the time domain to the frequency
domain.
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A minor complication:
The FID is complex:
The FT is also complex (real part produced by a cos-FT,
imaginary part produced
by a sin-FT). In effect, this situation produces 4 FTs: a cos-FT
and a sin-FT of the
real part of the FID, and a cos-FT and a sin-FT of the imaginary
part of the FID.
It turns out that adding the result of the cos-FT of the
cosine-modulated signal to the result of the sin-FT of the
sine-modulated signal yields sign discrimination, i.e. we can tell
whether the frequency of the NMR signal relative to the
spectrometer frequency was positive or negative. (The above applies
to the real part of the Fourier transformed spectrum. For the
imaginary part, we add the result of the sin-FT of the
cosine-modulated signal to the result of the cos-FT of the
sine-modulated signal.)
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Description in mathematical terms:
The complex signal (=FID):
FT of the complex signal:
The bubbles highlight the real part, the remainder is the
imaginary part. Note how
the imaginary number i = Sqrt(-1) neatly arranges for the right
combinations of
the results of the 4 FTs. Only the real part is displayed on the
computer
screen. The imaginary part is needed for phase correction.
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Phase correction The FT of an exponentially decaying signal S(t)
= cos(t).exp(-t/T2) is a
Lorentzian. The cos-FT of a cosine-modulated signal delivers an
absorptive
Lorentzian, whereas the sin-FT of the same signal delivers a
dispersive
Lorentzian:
A() describes the absorption lineshape.
D() describes the dispersion lineshape.
is the frequency of the NMR signal.
T2 is the transverse relaxation time.
A conventional 1D NMR spectrum displays only
absorption lineshapes.
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One has to be very lucky to find a purely cosine-modulated
signal in the real part
of the FID, so that the real part of the FT delivers a purely
absorptive spectrum.
All usual cases require a phase correction after the FT to
produce a spectrum with
absorptive lineshapes.
Problem 9: Calculate the full linewidth at half height in Hz for
an absorption
peak from the equation on the previous page.
Solution to problem 9: FWHH = 1/(T2) The shorter T2, the broader
the line.
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From James Keeler Understanding NMR Spectroscopy:
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Phase correction is simply a linear combination of the data
points in the real and
imaginary part of the spectrum:
S() = exp(icorr)S0()
where S0() is the complex spectrum (i.e. both real and imaginary
part) before
phase correction and corr is the phase of the correction. The
equation defines the
zero-order phase correction.
To find the optimal phase, NMR software allows interactive phase
correction.
Sometimes one needs a frequency dependent phase correction or
so-called first-
order phase correction. The illustration below is again from the
book
Understanding NMR Spectroscopy by James Keeler.
Problem 10: Use exp(i) = cos + isin to show that the new real
part of the spectrum
becomes after phase correction SRe() = cos(corr)S0Re() -
sin(corr)S0Im()
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First-order phase corrections become necessary, when the
magnetisation had the
chance to precess for a short while before the first point of
the FID is recorded.
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Window multiplication
The quality of a NMR spectrum can be greatly enhanced by
multiplying the FID
with a so-called window function prior to FT. A window function
is a function
that starts where the FID starts and ends where the FID ends.
Quite trivial.
An exponentially decaying function emphasises the signal at the
beginning of the
FID and increasingly decreases the noise towards the end of the
FID, thus
resulting in improved signal-to-noise ratio. The best S/N is
obtained by the
matched window function, which follows the envelope of the FID
(and therefore
doubles the linewidths). The decay constant L is usually entered
as a line
broadening parameter LB. The single LB value can be optimal only
for signals
that are LB Hz wide after an FT without window
multiplication.
Problem 11: Show that, if the FID is multiplied by an
exponential function exp(-Lt),
the resulting linewidth (FWHH) after FT will be broader by L
Hz.
Hint: check out the solution to problem 9.
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Window multiplications can also be used to make the lines in the
spectrum
narrower. This can be achieved by
using Gaussian multiplication. This
function attenuates the FID at the
beginning, emphasising its later parts.
The function has two parameters. GB
determines the position of the maximum as a fraction of the
acquisition time (0 <
GB
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The point of this function is to bring the end of the FID
smoothly to zero. This is
important if recording of the FID ended before it disappeared in
the noise. The
effect of truncation is well illustrated in Keelers book
Understanding NMR:
Sometimes multiplication by the cosine window still leaves small
wiggles behind.
In this case, try the squared cosine window.
Problem 12: a) What does the squared cosine window look
like?
b) Which window function generates more linebroadening, the
cosine or
the squared cosine window?
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Zero filling (or get something for nothing)
The Fourier transform converts N complex data points in the FID
into N complex
data points in the frequency domain. N/2 of those are in the
real part (the
absorptive part of the spectrum) and half of them in the
imaginary part (the
dispersive part of the spectrum). As we plot only the real part,
this would discard
the information from half of the data in the FID!
The situation is rescued by appending N zeros to the end of the
FID (= two times
zero filling), resulting in N points in the real part after
FT.
Magically, this improves the spectral resolution by doubling the
number of data
points per Hz in the NMR spectrum and it adds no noise (zeros
are noise-free)!
The original points are still present in the spectrum, but zero
filling adds a new
data point between each pair of original points.
We can also play the trick on a grander scale: appending 3N
zeros to the end of
the FID (=four times zero filling) produces 4 times more points
in the real part
of the FID. 3 new data points appear between each pair of
original points.
Spooky? Where do the new data come from? It can be shown that
the information
content no longer increases after two times zero filling,
because more zero filling
simply places new points between existing ones by an
interpolation algorithm. In
fact, after two times zero filling, the imaginary part of the
spectrum can be deleted
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because it can be reconstructed at any time from the real part
by a so-called
Hilbert transform.
From Keelers book:
Baseline correction
Measuring integrals will be problematic, if the baseline has a
non-zero offset,
slope or bend like a washing line. This can be fixed by
subtracting a function that
follows the baseline. This can be done by fitting a polynomial
(y = a + bx + cx2 +
dx3 + ) manually or automatically. Manual corrections solve the
problem that
automatic routines may have with distinguishing very broad peaks
from baseline.
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Understanding 2D NMR
A 2D peak displays a Lorentzian lineshape in every cross-section
taken along the
F1 or F2 frequency axis.
The first 2D NMR experiment (2-pulse COSY) was proposed 1971 by
the Belgian
physicist Jean Jeneer at an AMPERE summer school in former
Jugoslavia.
Richard R. Ernst received the Nobel Prize in Chemistry in 1991
for the
development of FT NMR and 2D NMR.
Jean Jeneer (left) and
Richard Ernst (right)
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The NOESY experiment
2D NMR experiments require more than a single radiofrequency
(rf) pulse. The
NOESY pulse sequence is the simplest experiment for
understanding 2D NMR
spectroscopy. It consists of three 90o pulses separated by
delays:
Any 2D NMR experiment (the NOESY experiment is no exception) is
performed
as a series of FIDs (recorded during t2) which are identical
except for systematic
incrementation of the delay t1. t1 is called the evolution time.
m is the mixing
time; it is a constant delay. t2 is the detection period. The
time between the last
90o pulse and the first 90o pulse of the next scan is the
recovery delay.
(typical delays in NOESY: t1 incremented from 0 to 50 ms in
increments of, e.g.,
100 s. m: anywhere between 50 ms and 1 s. t2: acquisition time
as in a usual
FID.)
As each FID is digitized, the result presents a 2D data matrix,
one axis being t2,
the other representing t1. This data matrix is the time domain
of the spectrum.
The 2D spectrum is obtained by Fourier transform of the
time-domain data matrix
along the t1 and t2 axes. (For simplicity, the t2-dimension is
transformed first. It
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45
actually doesnt matter which dimension is transformed first).
The resulting 2D
data matrix presents the frequency domain of the spectrum (this
is the spectrum
one would plot). The axes are labelled with ppm just like
conventional 1D NMR
spectra. Window functions, zero filling, phase corrections and
baseline corrections
are applied in both dimensions independently.
Diagonal- and cross-peaks:
If the magnetisation precesses with the same frequency during t2
as during t1, it
gives rise to a diagonal peak. If it precesses with different
frequencies, it gives
rise to a cross-peak. Cross-peaks require the transfer of
magnetisation between A-
spins (say, the blue one) and B-spins (say, the red one) during
the mixing time m
by NOE. The NOE results in exchange of longitudinal
magnetisation between A-
spins and B-spins, i.e. a fraction of A-spin magnetisation ends
up as B-spin
magnetisation and vice versa.
The figure below explains the NOESY pulse sequence with
magnetisation vectors
for two spins, A (blue) and B (red). The three 90o pulses are
delivered with the
phases 1, 2 and 3, respectively. By default, 1 = 2 = 3 = x. For
simplicity, we
start with A-spin magnetisation only.
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46
Step by step explanation of the different time points:
1) Equilibrium A-spin magnetisation. Also called longitudinal or
z-magnetisation.
It is parallel to the axis of the magnetic field B0.
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47
2) Transverse magnetisation. After a 90ox pulse, it is
y-magnetisation.
Sense of rotation: by convention, all rotations are
right-handed.
2 ways of picturing this:
a) Point the thumb of your right hand in the direction of the
rotation axis -> the
other fingers indicate the direction of the actual rotation.
b) Draw the right-handed Cartesian coordinate system: z-axis up,
y-axis to the
right, x-axis coming out of the plane towards you.
Picture yourself in the origin of the coordinate system. Looking
outwards
along the rotation axis (here: the x-axis, indicated by the B1
vector), apply a
clockwise rotation (here: Mz becomes My).
OBS: negative rotation frequencies are described by rotation
vectors pointing in
negative directions!
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48
3) Precession during the delay t1 results in transverse
magnetisation somewhere in
the x-y plane.
4) The 2nd 90ox pulse converts the magnetisation in the x-y
plane into
magnetisation in the x-z plane.
5) Only the z-component of the magnetisation after the 2nd 90ox
pulse is retained
(achieved by phase cycling, explained below), i.e. any
transverse
magnetisation components are discarded. During the mixing time
m, part of
the A-spin magnetisation (blue) is transferred to the B-spin
(red) by NOE.
6) The 3rd 90ox pulse turns z-magnetisation into
y-magnetisation, which
generates the FID during the detection period t2.
After each scan, a recovery delay is required to re-establish
equilibrium
magnetisation (by T1 relaxation) before the next scan.
Problem 13: In the figure explaining the NOESY pulse sequence,
does the blue vector
precess with positive or negative frequency during the evolution
time t1?
Remember: = 2, where is the frequency in Hz. Note that the
original
parameters measured are frequencies in Hz. Another good name for
the axes is
F1 and F2. Ultimately, I label the axes 1 and 2 when they
display ppm values.
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49
Phase cycling:
Phase cycling involves the summation/subtraction of scans
recorded with different
phases of the pulses. (All other parameters are kept the same.)
In our example, the
phase of the 3rd 90o pulse is changed from x to x to obtain a
second scan, where
the magnetisation of interest points along the y-axis rather
than the y-axis. By
subtracting the result of this second scan from that of the
first scan (which was
recorded with a 90ox pulse rather than a 90o-x pulse), the
signals add up.
In contrast, this phase cycle subtracts any x-component of the
magnetisation
present at the end of m (because x-magnetisation is unaffected
by the rotation
around the x-axis and, hence, doesnt change its sign).
Any y-component of the magnetisation present at the end of m is
turned into
longitudinal magnetisation by the 90ox pulse. Longitudinal
magnetisation does not
precess and therefore does not induce a current in the coil
(i.e. it is not detected).
The supertrick of 2D NMR:
The amplitude of the FID varies as a function of t1. What is the
frequency of
variation? Easy apply a Fourier transformation to find out!
In the hypothetical experiment where we start from A-spin
magnetisation only,
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50
the FIDs look as shown below (after FT(t2 -> 2), i.e. after
transforming each FID
recorded with different t1 delay into a spectrum, i.e.
converting the time domain t2
into the frequency domain 2.
The result of the FT(t2 -> 2) transformation is illustrated
above for four different
FIDs recorded with different t1-times:
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51
i) t1= 0. The spectrum contains a peak for the A-spins and a
smaller one for the B-
spins (which originates from the NOE during m).
ii) t1 so that At1 = /2, i.e. the A-spin magnetisation precessed
by precisely 90o
during t1. No magnetisation is observable, because
x-magnetisation present at
the end of t1 remains x magnetisation after the 2nd 90ox pulse
and, hence, is
destroyed by phase cycling.
iii) t1 so that At1 = , i.e. the A-spin magnetisation precessed
by precisely 180o
during t1. The A spin magnetisation simply changes sign by the
end of t1.
Therefore, all signals resulting from this magnetisation are
inverted.
iv) t1 so that At1 = 3/2, i.e. the A-spin magnetisation
precessed by precisely
270o during t1. This turns it into magnetisation aligned with
the x-axis at the
end of t1 which is destroyed by phase cycling.
Clearly, the t1-2 data matrix contains signals at the 2
frequencies of the A-spin
and the B-spin which oscillate in sign and magnitude as a
function of t1 with the
frequency A. Looking along the t1 axis, the signal intensities
observed at 2 = A
Problem 14: In the figure above, why does the peak amplitude of
the B-spin oscillate
with the frequency A? Hint: consider the origin of the
B-spin
magnetisation.
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52
are modulated like a damped cosine function. The cosine function
arises from the
fact that of the magnetisation precessing during t1, only the
projection onto the y-
axis is selected. The dampening arises from T2 relaxation during
t1. The signal of
the B-spin oscillates with the same frequency as the signal of
the A-spin (namely
A).
Apply the supertrick of 2D NMR: FT(t1 -> 1)
Fourier transform of the t1-2 data matrix in the t1 dimension
converts the t1 axis
into the 1 axis. This transforms the modulation of the signal
amplitudes into
signals at 1 = A. There is one diagonal peak (at 1 = 2 = A) and
one cross-
peak (at 1 = A / 2 = B).
Starting from B-spin magnetisation:
At the start of each scan, not only A-spin magnetisation is
present, but also B-spin
magnetisation. Of course, the pulse sequence works in the same
way for the B-
spins as for A-spins: B-spin magnetisation precesses during t1,
part of it gets
transferred to A-spin magnetisation during m by NOE, etc. The
same pictures can
be drawn, exchanging blue and red, and A and B. The complete
NOESY spectrum
thus ends up with two diagonal peaks (at 1 = 2 = A and at 1 = 2
= B) and
two cross-peaks. The NOESY spectrum is symmetric about the
diagonal, because
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53
the likelihood of magnetisation transfer from A-spins to B-spins
is the same as for
that from B-spins to A-spins.
Problem 15: Where in the spectrum will the cross-peak be that
originates from B-spin
magnetisation at the start of the NOESY experiment?
Advanced Problem I: A NOESY spectrum is recorded of a sample
containing 90%
H2O. The solvent peak is so huge that it was suppressed by
selective
irradiation of the water resonance prior to the NOESY pulse
sequence (to
prevent the build-up of any equilibrium magnetisation of the
water).
Unfortunately, this also saturates any resonance of the solute
that
happens to be at the water frequency. Will it be possible to
observe
NOESY cross-peaks with those solute signals anyway?
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54
t1-noise:
What happens if the magnetic field or frequency of the
electronics is not perfectly
stable during a 2D NMR experiment? If the spectrum randomly
shifts a little in
frequency from FID to FID, or its amplitude changes slightly,
the traces taken
along the t1 axis of the 2D data matrix will become noisy. The
Fourier transform
of noise is noise. Hence, noise bands appear in the F1
dimension. To minimize t1-
noise, the room temperature should be stable to within 0.5
degrees and no
bicycles, trolleys, cars must come anywhere near the magnet.
Problem 16: a) Why are the F1 and F2 dimensions often referred
to as the indirect and
direct dimensions, respectively?
b) Why is there no such thing as t2-noise?
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55
Sensitivity of 2D NMR:
For a single scan, a good signal-to-noise (S/N) ratio is
obtained by recording the
signal until it starts to disappear in the noise.
In a two-dimensional NMR experiment, every point of the 2D data
matrix
contributes to the S/N ratio of the final two-dimensional peak.
As a 2D NOESY
spectrum is typically recorded with about 512 data points in the
indirect
dimension, resonances become observable that are below the level
of noise in
every single FID of the NOESY experiment.
NOTE: the sensitivity and resolution of a NMR spectrum depends
on
the acquisition time tmax, not on the spectral width. For the
same tmax value, there
is no gain in sensitivity for using a smaller spectral width (=
longer increments)!
This holds in F1 and F2. Far too many people quote the spectral
width or the
number of data points, which says nothing about spectral
resolution and
sensitivity. t1max and t2max are the relevant experimental
parameters!
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56
How long does it take to record a 2D NMR spectrum?
If sensitivity is not an issue, this question boils down to
a) how many FIDs do we need to record (I)
b) how many scans do we need per FID
c) how long do we need to wait between scans
d) how many FIDs do we need to record (II)
The short answer is minutes. Now for the long answer
How many FIDs do we need to record (I):
The more points we record, the better the spectral resolution
will be. Here is an
example of a 1D NMR spectrum, with more data points available in
b than in a:
The maximal acquisition times in the t1 and t2 dimensions are
called t1max and t2max,
where t2max is simply the acquisition time of each FID. As we
need to wait for the
recovery of equilibrium magnetisation between scans anyway, we
can use the
entire recovery delay for acquisition without making the 2D
experiment any
longer.
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57
Doubling t1max, however, doubles the duration of the 2D NMR
experiment! With
the next NMR user breathing down your neck, you may decide that
some lesser
resolution in the indirect dimension is perfectly acceptable
How many scans do we need per FID:
At least one.
If undesired magnetisation components need to be removed by
phase cycling, the
minimal number of scans is determined by the length of the phase
cycle. In most
experiments, phase cycles can be replaced by pulsed field
gradients (to be
discussed later). In this case, a single scan per FID can be
just fine.
How long do we need to wait between scans:
At least t2max.
The real answer depends on the relaxation time T1. We cannot
afford to wait 5*T1.
With a recovery delay T = T1, we already get 70% of the
equilibrium
magnetisation. In fact, usual recovery delays are even shorter
(~ 1.5 s). Under
those circumstances, the experiment starts with steady-state
magnetisation rather
than equilibrium magnetisation. To prevent the very first scan
from starting with
full equilibrium magnetisation, the 2D experiment is initiated
with a number of
dummy scans to establish the steady state. (Dummy scans perform
the pulse
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58
sequence without storing the FID.)
The remaining problem is that different phase cycles feed
different amounts of
magnetisation into the detection period t2, i.e. the steady
state ends up being not so
steady after all, leading to artefact peaks in the 2D NMR
spectrum. This can be
fixed by randomising all magnetisation right after each scan.
This excludes the
FID from the recovery delay, so the cleaner spectrum comes at
the expense of
somewhat lesser S/N.
How many FIDs do we need to record (II):
The spectral resolution (measured in Hz per point) in the
indirect dimension is
determined only by t1max (apart from zero filling). Here are two
tricks to come to
the same t1max value with fewer FIDs.
1) Use a longer t1-increment (dwell time).
According to the Nyquist theorem, however,
this results in a correspondingly smaller
spectral width. If a peak happens to land
Problem 17: Why are NOESY spectra symmetric about the diagonal
in principle, but
the cross-peaks appear elongated in the F1 dimension in
practice?
! Harry Nyquist 1889 - 1976
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59
outside the spectral width, it will still appear in the
spectrum, but at the wrong
frequency. This effect is called folding or aliasing.
The picture shows, how the red signal will falsely appear to
have the
frequency of the blue signal, if sampled at the blue time
points.
The position of the folded signals can, however, be predicted.
There is also a
recipe for folding in such a way that the folded signals acquire
the opposite
sign. The diagonal of a 2D NOESY spectrum appears like this:
2) Use the original t1-increment and sample up to t1max, but
omit some of the FIDs
on the way. Then go on to reconstruct the missing points from
the recorded data
points. Then Fourier transform the repaired data set as
usual.
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60
This method has become the topic of active research in the past
few years, as
computers have become sufficiently fast to perform the
reconstruction in
reasonable time. It works fine so long as the number of
experimentally recorded
FIDs comfortably exceeds the number of peaks in each F1
cross-section.
Quadrature detection in the indirect dimension
The spectrometer frequency (the carrier frequency) is routinely
placed in the
centre of the spectrum, as this optimally distributes the pulse
power across the
entire spectrum. This means, however, that we must distinguish
Larmor
frequencies that are positive or negative relative to the
carrier frequency. Like in
1D NMR, we need to record complex data points in the indirect
dimension, i.e. for
each t1 value, we need a data point that belongs to the
cosine-modulated signal
and one that belongs to the sine-modulated signal. How do we get
this?
Returning to the scheme of the NOESY pulse sequence, the FID
will be maximal
for t1 = 0 and pulse phases 1 = 2 = x. Including all t1 values,
the FIDs resulting
from spin A will be amplitude-modulated with cos(At1). If we
perform the same
experiment with 1 = y and 2 = x, there will be no signal for t1
= 0. With this
phase setting, the FIDs resulting from spin A will be
amplitude-modulated with
sin(At1). Voil, we have achieved quadrature detection!
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61
The upshot is that, omitting some of the FIDs to save time is
fine, but each t1-
data point still needs to be recorded as a complex data point,
i.e. with the two
different phase settings of 1 and 2.
Problem 18: The function sin(At1) is zero for t1 = 0 and maximal
for t1 = /2. Using
the vector description of the NOESY experiment, show that 1 = y
and 2
= x results in amplitude modulation of the FIDs that is
phase-shifted by
90o relative to the situation where 1 = 2 = x.
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62
The 13C-HSQC experiment
The HSQC pulse sequence looks more complicated than it is:
Conventions in writing pulse sequences:
Narrow bar: 90o hard pulse
Wide bar: 180o hard pulse
The decoupling block is a train of 180o pulses of much lower
amplitude than the
hard pulses.
Pulse phases are x unless indicated otherwise.
The experiment is an ingenious way of sensitivity
enhancement by starting from 1H magnetisation, detecting 1H
magnetisation, but in between transferring the magnetisation
to the heteronucleus (13C) and letting it precess during t1
to
measure its frequency indirectly. The pulse sequence was
invented in 1980 by Geoffrey Bodenhausen. It produces
cross-peaks with purely absorptive lineshapes that are singlets
in the indirect
Geoffrey
Bodenhausen *1951
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63
dimension and split only by JHH couplings in the 1H
dimension.
The HSQC experiment encompasses concepts that are important in
many NMR
experiments: spin-echo, in-phase and antiphase magnetisation,
magnetisation
transfer through scalar couplings, refocussing of heteronuclear
couplings,
broadband decoupling.
Spin-echo:
The spin-echo sequence is (t 180o t). It refocusses chemical
shifts, but the J-
couplings continue to evolve during the entire duration 2t.
Consider chemical shift evolution first. The figure below
illustrates the
refocussing effect for the three different spins f, i and s:
At the end of the spin-echo, all magnetisation vectors are again
aligned, as if no
chemical shift evolution had occurred.
13C-HSQC spectra are the most sensitive way of measuring
13C-chemical shifts!
Problem 19: Show that a 180oy instead of a 180ox pulse would
also refocus the spins.
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64
Now consider J-coupling evolution for a simple doublet (ignoring
chem. shift
evolution). We describe the doublet by two vectors (blue and
red) precessing with
slightly different frequencies.
This happens to the doublet in the spin-echo (note: vectors
depict the
magnetisation of only a single doublet, the doublet of the
coupling partner of
course exists but is not shown):
So, in the case of the doublet in a spin-echo, the 180oy pulse
appears to do
nothing! This is the combined result from two effects of the
180o pulse: (a)
flipping the magnetisation vectors by 180o about the y-axis and
(b) inverting
longitudinal magnetisation (equivalent to interconverting the
and states of the
coupling partner), which amounts to repainting the red vector
blue and the blue
vector red.
The upshot is that the spin-echo always refocuses the chemical
shifts, but not the
J-couplings so long as the 180o pulse hits the coupling partner
as well.
Remember: doublets arise, because the coupling partner can be in
the state or .
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65
In-phase and antiphase magnetisation:
A special situation arises, when the two vectors representing
the doublet
components point in opposite directions. This is called
antiphase magnetisation,
in contrast to the situation after excitation by a single 90o
pulse, when both vectors
point in the same direction (this is called in-phase
magnetisation). Antiphase
magnetisation is the key to magnetisation transfer in COSY,
HSQC, HMBC and
TOCSY spectra.
How long does it take to convert in-phase magnetisation into
antiphase
magnetisation? The frequency difference between the two doublet
components is
J. Hence, it takes 1/J seconds for a full circle and 1/(2J)
seconds for the two
vectors to accumulate a 180o phase difference.
Starting from in-phase y-magnetisation, a ( 180oy ) spin-echo
with 2 =
1/(2J) produces antiphase magnetisation aligned with the x-axis
(remember, we
dont need to worry about any chemical shift evolution in a spin
echo).
Problem 20: Starting from in-phase y-magnetisation, show that
the doublet turns into
antiphase x-magnetisation after a ( 180oy ) spin-echo with 2 =
1/(2J).
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66
During free precession, in-phase magnetisation converts to
antiphase
magnetisation which converts back to in-phase magnetisation and
so on.
Fortunately, 1JHC couplings are all very similar (about 130-160
Hz), so that a
single spin-echo delay of 1/(2JHC) converts all C-H spin pairs
more or less
quantitatively into antiphase magnetisation.
NMR spectrum of an antiphase doublet
Magnetisation transfer through scalar couplings:
Here are three ways of depicting the same antiphase
magnetisation:
Left panel: as before (the two doublet components are shown with
different
colours)
Centre: uses a different colour code: the two doublet components
are both painted
red (as they belong to the same type of spins (e.g. spin A),
just located in different
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67
molecules. The spin state of the (blue) coupling partner is in
the molecules for
which the vector points along the positive y-axis, and in the
molecules for which
the vector points along the negative y-axis.
Right panel: same as centre panel, except that its a perspective
drawing and the
spin states of the blue coupling partner are indicated by
vectors along the z-axis.
From the representation of the right panel one would expect that
a 90ox pulse
would convert antiphase magnetisation of the red spin into
antiphase
magnetisation of the blue spin. This is indeed the case!
The purpose of the first part of the HSQC pulse sequence is
to
transfer antiphase 1H magnetisation to antiphase 13C-
magnetisation. This pulse sequence element is commonly
referred to as INEPT (insensitive nuclei enhanced by
polarisation transfer).
Magnetisation transfer via scalar couplings:
A 90o pulse applied to both spins converts magnetisation of spin
A (that is
antiphase with respect to spin B) into magnetisation of B (that
is antiphase with
respect to spin A).
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68
Spin-echoes with selective 180o pulses:
The next pulse sequence element in the HSQC experiment is the
t1-evolution
delay with a 180o pulse applied half way through the delay:
13C-magnetisation (antiphase with respect to 1H) is precessing
freely. Half-way
through, the spin-states of 1H are inverted. Therefore, all JHC
couplings are
refocused by the end of t1 and only the chemical shift evolution
of 13C dephases
the magnetisation.
For completeness, lets also consider this version of a
spin-echo:
Advanced Problem II: Why is it necessary that the phase of the
second 90o(1H) pulse
in the HSQC pulse sequence is y (and not x)? Hint: this pulse
must
convert transverse 1H vectors into longitudinal (z) vectors.
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69
In this sequence, the chemical shift evolution is refocused for
13C (but not for 1H).
All JHC couplings are refocused because the 180o pulse does not
invert the spin
states of the 1H spins; its a simple spin-echo for 13C spins,
irrespective of whether
the different 13C precession frequencies arise from different
chemical shifts or
peak splittings due to JHC. This explains, why the INEPT
experiment needs 180o
pulses on both 1H and 13C spins! (Otherwise the JHC couplings
would be refocused
to in-phase magnetisation, which cannot be transferred to the
heteronucleus.)
Reverse INEPT:
The last part of the HSQC pulse sequence is called reverse
INEPT because it looks like the initial part in reverse.
The 90o(1H,13C) pulses at the start of the reverse INEPT
convert transverse 13C-magnetisation (antiphase with respect
to 1H) back to 1H-magnetisation (antiphase with respect to
13C).
The following spin-echo refocuses the antiphase to in-phase
magnetisation.
Advanced Problem III: Following free precession of
13C-magnetisation during the
evolution time t1, the 13C-magnetisation is anywhere in the
transverse plane.
The 90ox(13C) pulse of the reverse INEPT changes y- but not
x-magnetisation.
How does this produce peak amplitudes modulated by cos(Ct1)?
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70
Broadband decoupling:
During data acquisition (t2), each 1H-resonance would be split
into a doublet by
1JHC unless we decouple. Broadband decoupling is a train of 180o
pulses to change
the 13C-spin states rapidly between and . The decoupled signal
appears in the
centre of the doublet.
Problem 21: Too much pulse power during broadband decoupling
fries the sample,
amplifiers and probehead. Use the condition for fast chemical
exchange k
>> A - B to estimate a rate with which the 180o pulses
must be delivered
for decoupling!
Problem 22: a) How does broadband decoupling affect antiphase 1H
magnetisation?
b) Why do small multiple-bond JHC couplings not give rise to
cross-peaks
in the 13C-HSQC spectrum?
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71
The 13C-HMBC experiment
Here is the basic pulse sequence of the 13C-HMBC experiment:
The delay 1/(2JHC) is tuned to small 2J and 3J couplings. During
this delay, both
chemical shifts and couplings evolve. As a result of the
chemical shift evolution,
the phases of the magnetisation detected during t2 are messy (a
mixture of
absorption and dispersion that depends on the Larmor frequency
of the
resonance). The F2 dimension is often presented in magnitude
mode: I = Sqrt(IRe2
+ IIm2), where IRe is the signal in the real part of the
spectrum and IIm is the signal
in the imaginary part of the spectrum.
The 13C-dimension is relatively clean. JHC couplings are
refocused during t1, but
JHH couplings evolve (and result in correspondingly broader
peaks in the F1
dimension).
Problem 23: a) Why are JCC couplings during t1 not a problem at
natural isotope
abundance?
b) Why is decoupling during acquisition not a good idea in the
HMBC
experiment?
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72
Cross-peaks due to 1JHC couplings are not welcome in HMBC
spectra. They are
usually suppressed by a 90o(13C) pulse combined with suitable
phase cycling to
kill all antiphase magnetisation present after the delay
1/(21JHC):
Note that this pulse sequence is not much longer than the one
before, because
1/(21JHC) is much shorter compared with 1/(2nJHC) (n >
1).
Problem 24: If you suspect the nJHC coupling to be really small,
which delay could you
adjust to have a better chance of observing a cross-peak?
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73
DQF-COSY
Below is the DQF-COSY spectrum of sucrose from the library
generated by
Teodor Parella at
http://triton.iqfr.csic.es/guide/manualw.html
DQF means double-quantum filter
COSY stands for 2D correlated spectroscopy
The spectrum is symmetric about the diagonal. Each cross-peak
has an equal
number of positive and negative multiplet components.
Cross-peaks have antiphase lineshapes for the active couplings
(i.e. the coupling
generating the cross-peak) and in-phase splittings for the
passive couplings (i.e.
any couplings with third spins).
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74
The insert demonstrates that cross-peaks can appear only at
intersections where
there are signals in the 1D NMR spectra. No signal in the 1D, no
cross-peak!
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75
The zoom shows that proton 2 couples with 3, 3 couples with 4,
etc.
The magnitude of the active couplings can be estimated from the
separation
between the antiphase components.
The diagonal peaks are not purely absorptive (hence, use the
cross-peaks for
phase correction!)
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76
Understanding the multiplet fine-structure of COSY
cross-peaks
Cross-peaks have antiphase lineshapes for the active couplings
(i.e. the coupling
generating the cross-peak) and in-phase splittings for the
passive couplings (i.e.
any couplings with third spins).
The pictures underneath are reproduced from Neuhaus et al., Eur.
J. Biochem.
151, 257-273 (1985).
For an AX spin-system, each cross-peak is an antiphase square
(Panel A). The
peak separation shows the active coupling constant JAX.
For an AMX spin-system, each cross-peak consists of an antiphase
square that is
replicated for each passive coupling. The peak separation in the
antiphase square
(for example, in the A-X cross-peak) still shows the active
coupling constant
(JAX). The square pattern is replicated by the passive couplings
JAM and JMX.
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77
(Disregard the unusual directions of the 1 and 2 frequencies the
argument
remains unchanged by changing the directions of the 1 and 2
axes)
For a simple antiphase doublet:
The maxima of the two Lorentzian lines composing the antiphase
doublet are
closer than the maxima of the resulting antiphase doublet
(dashed lines). The
cancellation of peak intensity in the centre of the antiphase
doublet means that it
has overall smaller peak heights and the apparent coupling
constant is too large.
Beware: the anti-phase splitting observed in a COSY cross-peak
often is not an
accurate measurement of the active coupling, because of
cancellation effects.
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78
In fact, when the coupling constant is significantly smaller
than the line width, the
same splitting is observed irrespective of the actual coupling
constant:
Problem 25: Imagine an in-phase doublet with a fixed coupling
constant J. In a thought
experiment, allow the line width LW to increase gradually, going
from
LW
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Two footnotes to NOESY
NOE build-up curve
The longer the mixing time, the bigger the NOE cross-peaks (if
it werent for
relaxation during the mixing time). Long mixing times have the
problem of spin-
diffusion, i.e. the magnetisation can continue to migrate from
spin A to B to C to
D etc.
When the NOE is negative (i.e. the cross-peaks are positive as
in big molecules),
multiple spin-diffusion pathways easily conspire to produce
cross-peaks between
nuclear spins that are more than 5 apart.
When the NOE is positive (negative cross-peaks), spin-diffusion
can lead to
positive cross-peaks even though the molecule is small. Here is
a calculation of
this situation for a linear spin system as a function of the
mixing time t. The
magnetisation starts from spin A and simple T1 relaxation has
been neglected
(from Neuhaus and Williamson, The Nuclear Overhauser Effect, VCH
1989).
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80
The NOE from spin A to spin B is positive, the relayed NOE from
A to C (via B)
is negative and the double-relay (A to D via B and C) is again
positive.
Chemical exchange in NOESY
NOESY spectra detect not only NOEs, but slow chemical exchange
between two
NMR signals also leads to cross-peaks in the NOESY spectrum.
Chemical
exchange always produces cross-peaks of the same sign as the
diagonal, see the
example of N,N-dimethylacetamide below.
Problem 26: (a) How would T1 relaxation during the mixing time
change the NOE
build-up curves?
(b) How can one distinguish direct NOEs from spin-diffusion?
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81
2.02.5
2.0
2.5
D2(1H)/ppm2.02.5
D2(1H)/ppm
D1(1H)/ppm
CH3-C(O)-N(CH3)2
The spectrum on the right is the NOESY spectrum with a mixing
time m = 60 ms,
plotted at fairly high contour levels to highlight the different
peak heights of the
cross-peaks and diagonal peaks. If m is short, the exchange rate
kex of the methyl
groups can easily be calculated:
kex = (IC/ID)/m
where IC is the intensity (volume!) of the cross-peak and ID the
intensity of the
diagonal peak. (Strictly speaking, ID should be measured in a
spectrum with zero
mixing time. The equation is valid only if T1 relaxation and
spin-diffusion during
the mixing time can be neglected, which is OK for short mixing
times.)
The spectrum on the left was recorded with m = 4 s (!). The
chemical exchange
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82
has equilibrated the intensities of the diagonal and
cross-peaks. The NOE cross-
peaks with the acetyl group are negative (and much weaker very
low contour
levels had to be plotted to make the NOEs visible).
Footnote: some people refer to experiments performed to measure
chemical
exchange as EXSY, but its the same pulse sequence as NOESY.
Problem 27: Why are the NOEs between the acetyl and amide
methyl-groups of the
same intensity after a mixing time of 4 s, even if the distances
are
different? Hint: compare with spin-diffusion.
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13C-INADEQUATE the ultimate for tracing carbon chains
incredible natural abundance double quantum transfer
experiment
The INADEQUATE experiment displays double-quantum frequencies in
the F1
(the vertical) dimension, i.e. the cross-peaks appear at the sum
of the chemical
shifts of the two coupling partners. The double-quantum
coherence is between
two neighbouring 13C spins. At natural isotopic abundance, there
is a chance of
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84
only 1 in 10000 that a molecule has two neighboring 13C spins.
Therefore, the
INADEQUATE experiment is VERY insensitive. The attraction of the
experiment
lies in the possibility of unambiguous assignments of the
signals of quaternary
13C-spins (i.e. even those 13C can be assigned that have no 1H
coupling partner,
putting them out of reach in a 13C-HMBC spectrum).
The experiment was invented in 1980 by Ad Bax and Ray
Freeman. It is ideal for tracing the carbon chain of organic
molecules.
Why would one choose the complicated representation of
the 13C-INADEQUATE and not record a [13C,13C]-COSY
instead? Because the sensitivity is 2 times better than that
of
a DQF-COSY.
Ad Bax *1956
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Pulsed field gradients
Narrow lineshapes can be obtained only with very homogeneous B0
fields. Much
can be gained, however, from a linear field gradient can be
turned on and off
quickly. This pulsed field gradient (PFG) is delivered by an
additional coil in the
probehead. A PFG applied along the z-axis renders the Larmor
frequency
dependent on the z-coordinate, so that after a period of free
precession in the
presence of the PFG the magnetisation is completely dephased
(and, hence,
unobservable). By applying a PFG of the same duration but
opposite sign
refocuses the effect of the PFG. More often, PFGs of the same
sign are applied in
a spin-echo:
Remember: a spin-echo refocuses chemical shifts
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86
PFGs are particularly useful for 1H-detected 13C- and 15N-NMR
experiments at
natural isotopic abundance. Obviously, most protons do not
contribute in these
experiments and merely generate t1-noise, if the phase cycle
cannot eliminate
them completely because of limited stability of the
spectrometer.
Due to their 4-fold lower Larmor frequency, 13C-spins rephase 4
times more
slowly than 1H-spins under the influence of a PFG. We can thus
use a PFG to
rephase the magnetisation of 13C and rephase it with a 4 times
smaller PFG after
the magnetisation has been transferred to 1H. 1H magnetisation
that does not talk
to 13C will not rephase and therefore remain unobservable. Great
way for
minimising t1-noise and other artifacts in NMR spectra.
Many of the small-molecule NMR experiments with PFGs have been
pioneered
by James Keeler. They have made 2D NMR much faster.
OBS: each time a PFG is applied, the lock must be switched off
temporarily
(something that is written into the pulse program).
James Keeler
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Selective pulses
The excitation profile of a pulse is closely related to (though
not identical with)
the Fourier transform of the pulse shape. The Fourier transform
of a rectangle is
the sinc function (sin(x)/x). To deliver the maximal power in
the shortest time,
hard pulses have a rectangular shape. The longer a rectangular
pulse, the narrower
the central excitation band becomes, until it no longer covers
the entire NMR
spectrum. For a very long pulse, it may cover only a single NMR
resonance, but
there would be excitation sidebands as shown in the figure below
(produced by
Max Keniry, RSC).
The trick with the Gaussian pulse shape is that it avoids
excitation sidebands,
delivering more selectivity for the same pulse duration.
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88
The selective pulses can be used to perform 2D NMR experiments
using a very
narrow spectral width in the indirect dimension, corresponding
to a very large
increment of the evolution time and, hence, a short overall
experimental time. It is
even quite possible to excite a single resolved resonance,
reducing the 2D NMR
experiment to a 1D experiment. (See Kessler et al., Magn. Reson.
Chem. 29, 527-
557 (1991).)
Selective pulses are also used to suppress intense solvent
signals, without
affecting the rest of the NMR spectrum. (See Piotto et al., J.
Biomol. NMR 2,
661-665 (1992).)
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Appendix
The Product Operator Formalism
In 1984, Richard Ernst published a formalism that made it
possible for the first
time to predict the outcome of NMR pulse sequences in a simple
way (Srensen,
Eich, Levitt, Bodenhausen, Ernst (1984) Product operator
formalism for the
description of NMR pulse experiments. Progress NMR Spectr. 16,
163-192).
The article itself is a bit difficult to read, but it boils down
to a few simple rules
summarized below.
By way of introduction, assume a 2-spin system with spins A and
B. Start from
spin A. At equilibrium, we have longitudinal magnetisation along
the z-axis, Az.
Following a 90ox pulse, we have magnetisation along the y axis,
-Ay.
Note 1: Ay represents all multiplet components of spin A (in
this case its just a
doublet), i.e. during free precession, the different multiplet
components of -Ay
will fan out in the transverse plane.
Note 2: Immediately after the 90o pulse all multiplet components
of A are still
aligned. This is in-phase magnetisation. If we were to apply a
pulse on B, this
would not have any influence on the components of A.
In the product operator formalism, in-phase magnetisation is
represented by Ax
and Ay.
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90
Antiphase magnetisation: best represented by this picture:
and denoted AyBz in the product operator formalism (i.e. spin A
is transverse
along the y-axis and the orientation of the two double
components depend on the
spin state of B in half the molecules itll be , in the other
half ). If we apply,
e.g., a 180o pulse on B, this would invert the A
magnetisation!
Free precession:
As spin A precesses, it converts from in-phase to antiphase
magnetisation and
back again under the influence of the J-coupling:
-Ay -Ay cos(Jt) + AxBz sin(Jt)
AxBz AxBz cos(Jt) + Ay sin(Jt)
For t = 1/(2J), the conversion between in-phase and antiphase
magnetisation is
complete.
On top of that, there is the influence of chemical shift
evolution:
Ax Ax cos(t) + Ay sin(t)
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Ay Ay cos(t) - Ax sin(t)
If the offset from the spectrometer frequency is zero ( = 0),
there is no chemical
shift evolution. The way to remember the x and y and signs is to
look at the
projections of a single vector onto the x- and y-axis:
Let the vector move anti-clockwise and it will change its
direction: from x to y to
x to y to x etc. If the vector started on x, it will now be
described by
x cos + y sin
i.e. the sum of the projections onto the x- and y-axes. If it
started on y, it will be
described by
-y cos + x sin
and so on.
Chemical shift evolution + coupling evolution:
In the general case, precession is due to chemical shifts AND
couplings. This is
easily calculated by doing one first, then the other.
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Chem. shift evolution: -Ay -Ay cos(t) + Ax sin(t)
Coupling evolution: -Ay cos(t)cos(Jt) + AxBz cos(t)sin(Jt)
+ Ax sin(t)cos(Jt) + AyBz sin(t)sin(Jt)
where the colours track the origins of the terms.
A standard formula compilation shows that
cosA.cosB = 0.5[cos(A+B) + cos(A-B)]
sinA.sinB = 0.5[cos(A-B) - cos(A+B)]
sinA.cosB = 0.5[sin(A+B) + sin(A-B)]
cosA.sinB = 0.5[sin(A+B) - sin(A-B)]
Advanced Problem: Using the trigonometric relationships above,
show that Ax creates
a spectrum with two components as expected for an in-phase
doublet. Hint:
only in-phase magnetisation induces a current in the detection
coil.
Advanced Problem: Show that AxBz creates a spectrum with a
positive and a negative
doublet component, as expected for an antiphase doublet. Hint:
describe the
evolution into in-phase magnetisation under the influence of
chemical shift
and coupling evolution.
Hence, we obtain signals at the frequencies J, as expected for a
doublet.
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Names:
Az: longitudinal magnetisation of spin A
Ax: in-phase x-magnetisation of spin A
Ay: in-phase y-magnetisation of spin A
AxBz: antiphase x-magnetisation of spin A, or more specifically,
x-magnetisation of
spin A antiphase with respect to spin B
AyBz: antiphase y-magnetisation of spin A, or more specifically,
y-magnetisation of
spin A antiphase with respect to spin B
AxBx, AyBy, AxBy and AyBx: two-spin coherence of spins A and
B
AzBz: longitudinal two-spin order of spins A and B
In larger spin systems, three-spin product operators may
appear:
AxBzCz: x-magnetisation of spin A, in antiphase with respect to
the spins B and C
AxBxCz: two-spin coherence of spins A and B, in antiphase with
respect to spin C
AxBxCx: three-spin coherence
AzBzCz: longitudinal three-spin order.
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Calculate a COSY spectrum
The COSY pulse sequence (without double-quantum filter) is a
two-pulse
sequence:
90ox t1 90ox t2
We start with equilibrium magnetisation Az in a two-spin
system.
90ox: Az -Ay
t1-evolution (as shown above): -Ay cos(At1)cos(Jt1) + AxBz
cos(At1)sin(Jt1)
+ Ax sin(At1)cos(Jt1) + AyBz sin(At1)sin(Jt1)
90ox: -Ay -Az (unobservable z-magnetisation)
AxBz -AxBy (unobservable two-spin coherence)
Ax Ax (in-phase magnetisation of A: diagonal peak)
AyBz AzBy (antiphase magnetisation of B: cross-peak)
Hence, the diagonal peak and cross-peak are 90o out of phase,
i.e. if the cross-
peak is absorptive, the diagonal peak is dispersive (producing a
bad baseline
because dispersive peaks are so broad). This is true in both
dimensions, because
sin(At1)cos(Jt1) is an odd function (point symmetry with respect
to zero),
whereas sin(At1)sin(Jt1) is an even function (mirror symmetry
with respect to
zero).
In this way, the outcome of most NMR pulse sequences can be
calculated.
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Note: the product operator formalism described above applies to
spin systems
only. In the literature, terms containing 2 or more spins are
multiplied by
normalisation factors (two-spin terms are multiplied by 2,
three-spin terms by 4,
etc.) to maintain the magnitude of the product operators when A
= , B= , etc.
Thus, when in-phase magnetisation interconverts with anti-phase
magnetisation:
Ax 2AyBz, and 2AxBz Ay. These normalisation factors complicate
the
writing without contributing to understanding.