NMR Otting PartA Spectroscopy Course

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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)


- Magnetisation transfer: via one or several JHH couplings

NOESY (nuclear Overhauser effect spectroscopy)


- 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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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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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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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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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.

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,

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:

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.

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

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?

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?

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!

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.

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

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

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.

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!

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.

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

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.

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 .

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).

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

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).

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.

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)?

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?

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?

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?

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).

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!

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!)

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.

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.

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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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?

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

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.

83

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

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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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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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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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.

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