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NMR background Flashcards

1
Q

NMR spectroscopy records…

A

…the absorption of energy between quantised energy levels

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2
Q

How does nuclear spin arise?

A

From unpaired protons or neutrons in the nucleus

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3
Q

I

A

Spin quantum number

i.e. nuclear spin

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4
Q

What is the spin quantum number (I) for protons and neutrons?

A

I = 1/2

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5
Q

Odd number of protons and odd number of neutrons

A

Spin (I) = integer 1, 2, 3 etc

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6
Q

Even number of protons and odd number of neutrons

A

Spin (I) = 1/2 integer e.g. 13C

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7
Q

Odd number of protons and even number of neutrons

A

Spin (I) = 1/2 integer e.g. 19F

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8
Q

Why do paramagnetic compounds cause problems with NMRs?

A

Because the magnetic moment of an unpaired electron is nearly 1000x larger than that of nuclei
Additional magnetic field leads to large shifts
Effective relaxation mechanism leads to broad NMR signals

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9
Q

I = 0

A

NMR spectroscopy not possible

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10
Q

I = 1/2

A

These nuclei usually give good, easily interpretable results

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11
Q

I > 1/2

A

These nuclei often give broad signals - so generally problematic

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12
Q

When does a nucleus have both a magnetic and a quadrupole moment?

A

When I > 1/2

This is due to the non-spherical nature of the nucleus

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13
Q

Effect of quadrupole nuclei on NMR spectra

A

Broad lines due to rapid relaxation of nuclei

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14
Q

The energy levels involved in NMR spectroscopy arise from the interactions of nuclear spins with:

A
  1. The spectrometer magnetic field, B0
  2. The magnetic fields created by the electrons in the system (this is what leads to chemical shifts)
  3. The magnetic fields created by other NMR-active nuclei in the system
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15
Q

Gyromagnetic ratio

A

Gamma

Governs the strength of the NMR signal

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16
Q

Negative gyromagnetic ratio

A

Usually results in a negative NOE enhancement

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17
Q

Quadrupole moment

A

Q

The smaller the quadrupole moment, the better

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18
Q

What does a spinning nucleus possess?

A

Nuclear spin angular momentum (P)

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19
Q

Nuclear spin angular momentum

A

P
Can be quantised such that P = hbar x root[I(I+1)]
Where hbar = h/2pi and I = spin quantum number

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20
Q

Spinning nucleus

A

A nucleus is a charged body
Therefore a spinning nucleus generates an associated magnetic moment (nuclear magnetic moment, mu) which is a vector quantity

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21
Q

Vector quantity

A

A geometric object with magnitude and direction

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22
Q

Nuclear magnetic moment

A

Mu = gammaP

Where gamma = gyromagnetic ratio (a constant for each nuclide)
P = nuclear spin angular momentum

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23
Q

What does the sensitivity of detection of a nuclide depend on?

A

Gyromagnetic ratio

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24
Q

Large gyromagnetic ratio

A

Nuclei are easy to observe (sensitive)

e.g. 19F

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25
Q

Small gyromagnetic ratio

A

Nuclei are difficult to observe (insensitive)

e.g. 15N

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26
Q

Combining nuclear spin angular momentum and nuclear magnetic moment equations gives…

A

mu = gamma(hbar x root[I(I+1)])

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27
Q

Do different isotopes of the same element have the same values of gamma?

A

No

Different isotopes of the same element have different values of gamma

28
Q

Why do nuclides with I=0 have no NMR signal?

A

I=0 means no nuclear spin angular momentum (P) and therefore no nuclear magnetic moment (mu)
Examples of these nuclei include 12C, 16O, 32S

29
Q

Magnetic quantum number

A

ml
= the component of the nuclear spin (I) along the z-axis
Can take values I, I-1, …, -I (i.e. there are 2I+1 possible values of ml associated with the nuclear spin I)
e.g. I = 3/2 has 4 possible ml values

30
Q

ml states in the absence of a magnetic field…

A

…are all energetically degenerate
Because there is nothing to interact with the nuclear magnetic moment and the energy of the system is not affected by the spatial orientation of the moment

31
Q

Placing a nucleus in a magnetic field

A

If a nucleus with angular momentum P and magnetic moment mu is placed in a magnetic field of strength B0 orientated along the z-axis, the nuclear angular momentum orientates such that:
Pz = ml(hbar)
where ml = magnetic quantum number

32
Q

Different orientations of the nuclear magnetic moment will have…

A

…different magnetic moments and therefore different energies depending on their orientations w.r.t. to the direction of the applied field B0

33
Q

Why are only certain orientations of the magnetic moment (mu) allowed?

A

Due to the quantisation of I
Each orientation corresponds to a different ml value
There are 2I+1 allowed ml values (and an equal number of allowed Pz values)
Therefore there is the same number of discrete energy levels

34
Q

Zeeman splitting

A

The splitting of atomic energy levels in the presence of a magnetic field

35
Q

Energy of a magnetic dipole in a magnetic field of strength/flux density B0

A

E = -muzB0

Where muz is the component of mu along the z-axis (the direction of the field)

E = -gamma(ml)(hbar)(B0)

36
Q

The energy of the nucleus is raised or lowered by an amount proportional to:

A
  1. The gyromagnetic ratio (gamma)
  2. The z-component of the nuclear spin angular momentum (Pz = ml(hbar))
  3. Magnetic field strength (B0)
37
Q

DeltaE

A

Energy difference between ml levels/states

DeltaE = -y(hbar)(B0)

38
Q

What is the selection rule for transitions between nuclear spin states in NMR spectroscopy?

A

Deltaml = +/-1

i.e. allowed transitions occur between adjacent energy levels

39
Q

How many possible transitions are there for a spin I nucleus?

A

2I

Each transition has the exact same energy
Therefore only one resonant frequency for the nucleus is expected, regardless of the value of I for the nuclide

40
Q

How can the energy of a transition be expressed?

A

In terms of a frequency, v

41
Q

Frequency of EM radiation that must be applied to the nucleus in order to detect a transition in NMR spectroscopy

A

v = yB/2pi

DeltaE = E1 - E2 = hv = y(hbar)B

42
Q

Larmor frequency

A

= vL
= resonant frequency
= the frequency of the precession (rotation) of the nucleus around the z-axis of the magnetic field
= depends on the spectrometer’s field strength

43
Q

Angle of cone of precession

A

55degrees44’

44
Q

How many cones of precession are there for I=1/2 nuclei?

A

2, representing ml=+1/2 and ml=-1/2

45
Q

Why do all spins in the nucleus precess at the same frequency?

A

Because vL is independent of ml

46
Q

How do chemical shift and coupling arise?

A

From interactions with the magnetic fields generated from the spin of other nuclides and electrons
These interactions alter the nuclides resonant frequency (often so more than a single resonance is observed)

47
Q

Why do the resonances for a sample recorded in different NMR spectrometers with different magnetic field strengths always exhibit the same chemical shifts?

A

Because chemical shift (ppm) = [DeltavL (Hz) / spectrometer freq. (Hz) x 10^6]
OR
chemical shift (ppm) = DeltavL (Hz) / spectrometer freq. (MHz)

48
Q

Magnetic field (B0) of a typical NMR instrument

A

9.4 Tesla

49
Q

Resonance frequency for a nuclear isotope X

A

vX = (yX/yH) x vH
(using only the magnitude of y, not its sign)

Use this equation for calculating resonant frequencies of other nuclei compared to 1H

50
Q

How are transitions between different energy levels induced experimentally?

A

By irradiating the sample with a superimposed field B1 of the correct quantum energy
i.e. we do not see signals from 1H nuclei in a 31P NMR spectrum - the spectrometer is tuned to the nuclide of choice

51
Q

What happens to the nuclear spins when a sample is irradiated at the Larmor frequency?

A

They are both:
Excited to the next highest energy level (absorption)
Stimulated to relax back to the next lowest energy level (emission)

52
Q

What does the intensity of the observed NMR signal depend on?

A

The difference between the numbers of absorption and emission processes occurring (i.e. the net movement)

53
Q

What does the net movement between spin-states depend on?

A

Because the probabilities of absorption and emission are equal, net movement between spin-states solely depends on the population difference between the upper and lower states
This population difference is given by Boltzmann statistics

54
Q

Population difference between upper and lower states

A

Extremely small (approx. 1 in 30,000 population difference in favour of the lower state)

55
Q

What does a population excess in the lower level mean?

A

Absorption is the dominant process
This means the observed signal is proportional to Na-Nb, as well as to the total number of net spins in the sample (and thus the nuclide concentration in the sample)

56
Q

What happens if the populations in the upper and lower state are exactly equal?

A

No signal is observed (= saturation)
Because there is an equal number of absorptions and emissions
This is used in decoupling experiments

57
Q

Na

A

Lower state

58
Q

Nb

A

Upper state

59
Q

Why do nuclides with low gyromagnetic ratios give weaker signals than those with high gyromagnetic ratios?

A

Because the energy level separation is proportional to the gyromagnetic ratio of the nuclide

60
Q

The sensitivity/receptivity of a nucleus, for a fixed value of B0, is proportional to:

A

y^3[I(I+1)]

61
Q

If the receptivity of 1H is given as 1, the relative receptivity of nucleus X is:

A

(yX / yH)^3 ([Ix(Ix+1) / [Ih(Ih+1)]) (abundance of X / abundance of H)

62
Q

In what ways can the sensitivity of an NMR experiment be increased?

A

Increasing B0 (i.e. using a stronger magnet)
Lowering the sample temperature
(See equations in green box)

More concentrated samples are also advantageous

63
Q

Pulsed NMR

A

Involves applying a sequence of radiofrequency pulses to the sample so that all spectral frequencies are irradiated at once (i.e. all spins excited at once)
These pulses are either pi-pulses (180) or pi/2-pulses (90)
The pulses lead to a FID signal, that is then converted to a ‘real’ spectrum by Fourier Transform (converting time domain to frequency domain)

64
Q

FID

A

Free Induction Decay

65
Q

Advantages of pulsed NMR (c.f. CW-NMR)

A

Very time-efficient

Allows ‘rapid multiple passes’, which allows a good signal to noise ratio to be established

66
Q

Signal to noise ratio in NMR

A

The signal increases by a factor of N for N scans
The noise increases by a factor of rootN for N scans
Therefore the S/N ratio increases/improves by N/rootN (=rootN) for N scans
Therefore, to double the S/N ratio you must do 4x as many scans

67
Q

Why can the S/N ratio not be infinitely increased?

A

Because S/N is only increased by rootN
For a really weak signal, the time required to obtain an acceptable S/N would be impractically long (so would need to increase conc of sample)

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