The vector model of NMR

Question 1

Nuclear magnetic moments in the absence of an applied field

Answer 1

Randomly orientated

No net magnetisation

Question 2

Nuclear magnetic moments in an applied field, B0

Answer 2

The individual spin states, ml, lose their energetic degeneracy and populate new energy levels according to a Boltzmann distribution
Consequently there is an excess of nuclear magnetic moments aligned with the applied field, leading to a net magnetisation, M0, which is represented as a vector

Question 3

M0

Answer 3

Net/bulk magnetisation of the sample
Can be represented as a vector
Points along the direction of the applied field B0

Question 4

The rotating frame in NMR experiments

Answer 4

In a magnetic field B0, the net magnetisation vector M precesses about the axis of the applied field at the Larmor frequency and is inclined at a precession cone angle
To simplify the magnetisation in an NMR experiment, we move from the static laboratory frame x, y, z to the rotating frame x’, y’, z’. which rotates about the z axis in the same direction and at the same rate as B1
Frame rotation matches the Larmor frequency
Z’ is inclined to the z-axis at the precession cone angle

Question 5

Advantage of using the rotating frame method

Answer 5

Allows us to analyse what happens to M/to individual spins when the B1 magnetic field is applied at a frequency matching the Larmor frequency

Question 6

The vector model of NMR

Answer 6

Considers what happens to M during the experiment as viewed in the rotating frame - basically we sit on the vector and observe the stationary origin

Question 7

SW

Answer 7

Sweep/spectral width/window about the offset, in ppm
Refers to the chemical shift range over which the data is recorded
e.g. a sweep width of 2400 Hz covers 12 ppm of chemical shift if the 1H Larmor frequency is 200 MHz (2400Hz/200MHz = 12ppm)

Question 8

What determines the centre of a spectrum?

Answer 8

The offset frequency i.e. the precise spectrometer frequency

Question 9

Chemical shift range observed in the final spectrum

Answer 9

Extends from -0.5SW to +0.5SW
e.g. an offset frequency of 5 ppm with a spectral window of 10 ppm will yield a spectrum that starts at 0 ppm and ends at 10 ppm

Question 10

Rf pulse

Answer 10

Radiofrequency pulse
Has a characteristic frequency (= spectrometer frequency) that depends on the nucleus you wish to observe and the magnetic field strength of the spectrometer
Excites the nuclei, which then emit Rf during the acquisition time, giving rise to an NMR signal in the form of an exponentially decaying sine wave (=FID)

Question 11

How are NMR spectrometers generally named?

Answer 11

After the frequency at which hydrogen atoms resonate

e.g. a 500 will cause H atoms to resonate at approx. 500 MHz

Question 12

Why is an Rf pulse that excites spins at only one frequency not desirable?

Answer 12

Because the NMR frequencies are spread out over a range of frequencies, called the range of chemical shifts

Question 13

Why are short Rf pulse lengths used in FT-NMR?

Answer 13

Short pulse widths irradiate a wide range of different frequencies due to the Heisenberg Uncertainty Principle
A pulse length of 10 us causes the Rf power to be distributed over 1/pw = 1/0.00001 = 100,000 Hz of frequencies

Question 14

Heisenberg Uncertainty Principle

Answer 14

The time it takes for a system to change significantly multiplied by the uncertainty in energy is always greater than h/2pi

Question 15

The Heisenberg Uncertainty Principle in relation to NMR

Answer 15

As the pulse length is shortened, the uncertainty in the frequency results in a larger field of excitation
See green equation in notes

Question 16

When would a longer, lower energy pulse be useful in NMR?

Answer 16

A longer, lower energy pulse will have less frequency spread so can be used for frequency-selective excitation or saturation

Question 17

Pulse width

Answer 17

Refers to TIME
i.e. units are us
= the amount of time the pulse of Rf energy is applied to the particular sample in order to flip the bulk magnetisation from the z-axis into the xy-plane

Question 18

Pulse angle

Answer 18

The angle by which the magnetisation (M) is displaced away from the z’-axis towards the y’ axis
Determined by the strength (i.e. time length) of B1

Question 19

B1

Answer 19

Excitation pulse

Question 20

What happens to M after the Rf pulse has been applied?

Answer 20

M precesses about the z’ axis as it relaxes back to its pre-pulse, equilibrium state - i.e. the excited nuclear spins lose energy and drop from the excited state back to the ground state (Boltzmann distribution)

Question 21

What is the observable NMR signal?

Answer 21

The projection of the precession of M onto the x’y’ plane
The receiver coil is orientated along the y’ axis, meaning the intensity of the signal observed is dependent upon the amount of magnetisation in the y’ direction

Question 22

How can maximum signal intensity be achieved?

Answer 22

By using a 90 degree pulse angle

Question 23

When can we expect a 0 intensity signal?

Answer 23

When the pulse angle is 180 degrees

Question 24

How long does relaxation of nuclei take?

Answer 24

Can vary from seconds to hours depending on the relaxation mechanisms available
Nuclei must relax fully before applying the next pulse in order to obtain a true and maximised signal

Question 25

How does relaxation generally occur?

Answer 25

By transfer of magnetisation to other dipoles

Question 26

Dipole-dipole relaxation

Answer 26

Occurs when the sample contains other magnetic dipoles (e.g. magnetic moments of other nuclei/unpaired electrons) that can interact with the nuclear magnetic moment Occurs by two mechanisms: T1 relaxation = spin-lattice relaxation T2 relaxation = spin-spin relaxation

Question 27

Factors affecting the effectiveness of relaxation

Answer 27

Relaxation depends on, and is proportional to, gamma(A)^2 multiplied by gamma(B)^2, where A is the starting nucleus and B is the receiving nucleus i.e. nuclei with larger gyromagnetic ratios will have faster relaxation times than those with smaller gamma

Question 28

Why is relaxation in paramagnetic complexes dominated by the presence of the unpaired electron(s)?

Answer 28

Because the gyromagnetic ratio of an electron is ~1000x larger than that of a proton, so relaxation is very efficient Spectra of paramagnetic complexes often have broad line widths

Question 29

For relaxation to occur...

Answer 29

...dipoles must set up a magnetic field that oscillates at approximately the NMR resonant frequency Molecules can match their frequencies to better or worse extents depending on: Their size/shape The viscosity of the medium The temperature

Question 30

Long relaxation times (i.e. slower relaxation) occur with...

Answer 30

...small molecules, low viscosities and high temperatures

Question 31

Long relaxation times (i.e. inefficient relaxation) give...

Answer 31

...narrower signals

Question 32

Effect of concentration on relaxation

Answer 32

Higher concentration leads to more efficient (faster) relaxation, which gives broader signals This is because dipole-dipole relaxation is a through-space effect and decreases with distance (proportional to r^-6)

Question 33

T1 relaxation

Answer 33

= spin-lattice relaxation = longitudinal relaxation = relaxation in the z-direction Corresponds to the process of (re-)establishing the Boltzmann population distribution of alpha and beta spin states in the magnetic field i.e. corresponds to the 'growing back' of magnetisation in the z' direction

Question 34

T2 relaxation

Answer 34

= spin-spin relaxation = transverse relaxation = relaxation in the xy plane Corresponds to the loss of phase coherence among nuclei/corresponds to the 'fading away' of magnetisation in the x'y' plane Independent of the relaxation of the z' component

Question 35

What determines the line width of an NMR signal?

Answer 35

``` T2 Short T2 = broader lines (T2)^-1 = piW(1/2) OR W(1/2) = 1/piT2 ```

Question 36

W(1/2)

Answer 36

= width at half height

Question 37

What determines the maximum repetition rate during the acquisition of an NMR signal?

Answer 37

T1 Short T1 = magnetisation recovers more rapidly so a spectrum can be acquired in less time i.e. T1 determines how frequently we can pulse the nucleus

Question 38

Sources of line broadening in NMR

Answer 38

Viscous medium Molecule has a large molecular mass (e.g. protein Low temperature (increases viscosity) Nucleus has a quadrupole moment and is lower than cubic symmetry Paramagnetic centres/impurities in the sample Sample inhomogeneity (e.g. poor mixing, presence of solid particles) Instrumental problems e.g. tuning, misshapen NMR tube

Question 39

Why is T2 always shorter than T1?

Answer 39

Because return of magnetisation to the z-direction inherently causes loss of magnetisation in the xy plane

Question 40

How is T1 measured?

Answer 40

By the inversion-recovery experiment

Question 41

Steps in the inversion-recovery experiment

Answer 41

1. Application of a 180degree pulse inverts the bulk magnetisation to the -z axis 2. Immediately after this inversion, the spins begin to recover along the +z axis 3. After a delay (tau), the spin are irradiated with a 90degree pulse - this delay is varied, which produces multiple signals 4. Plotting the intensities of the signals vs the delay (tau) will produce a curve, which can be fit to an exponential function of the form M(tau) = M0 - 2M0e^-tauT1)

Question 42

Value of tau to give a signal with zero intensity

Answer 42

tau = ln2T1

Question 43

Short delay (i.e. short tau)

Answer 43

Gives signals with negative intensities

Question 44

Delay (tau) = ln2T1

Answer 44

Gives signals with zero intensity

Question 45

Long delay (i.e. long tau)

Answer 45

Gives signals with positive intensities

Question 46

Delay time between scans

Answer 46

5 x T1

Question 47

Quadrupolar relaxation

Answer 47

Depends on the size of the nuclear quadrupole and the electric field gradient of the nucleus

Question 48

Electric field gradient

Answer 48

Measures the rate of change of the electric field at an atomic nucleus generated by the electronic charge distribution and the other nuclei Very sensitive to the electronic density in the immediate vicinity of a nucleus Couples with the quadrupole moment of quadrupole nuclei to generate an effect that can be measured by NMR

Question 49

What does the electric field gradient depend on?

Answer 49

The symmetry of the molecule It is smallest when the environment is cubic/highly symmetrical It is largest when the molecule possesses no symmetry

Question 50

When is quadrupole relaxation very efficient?

Answer 50

When the quadrupole moment is large and/or the nuclear environment is far from cubic symmetry This speeds up T2 relaxation and gives the NMR excited states very short lifetimes, leading to Heisenberg broadening and therefore broad lines in the spectrum

Question 51

When are quadrupolar nuclei NMR invisible?

Answer 51

When the lifetimes of the NMR excited states are so short that the resonances are 1000s of Hz wide This gives lines so broad that they are essentially indistinguishable from the baseline noise

Question 52

35Cl NMR of CDCl3 c.f. NaCl

Answer 52

In CDCl3, the environment around Cl is very asymmetrical so gives a broad, ill-defined peak In NaCl, there is only one Cl- (i.e. the environment is as symmetrical as is possible) so the signal is much sharper

Question 53

Spectroscopically useful nuclei

Answer 53

Have quadrupole moments < 0.15 x 10^-28 m^2 | e.g. 2H, 6Li, 10/11B, 17O, 23Na