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Coupling Flashcards

1
Q

Scalar coupling

A

= spin-spin coupling
= J-coupling
Through-bond interaction

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2
Q

How does J-coupling arise?

A

The magnetic moments of other NMR-active nuclei surrounding the observed nucleus produce small fields, in addition to the spectrometer field and the chemical shift fields
i.e. the observed nucleus does not see just one net magnetic field - there is the possibility of several extra fields depending on the number and nature of the surrounding NMR-active nuclei

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3
Q

Why is J-coupling independent of the spectrometer field strength?

A

The magnitude of J-coupling depends only on the interaction between the nuclear magnetic dipoles
(J is constant at different external magnetic field strengths)

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4
Q

Homonuclear coupling

A

Between the same isotope i.e. proton-proton coupling

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5
Q

Heteronuclear coupling

A

Between different isotopes/elements i.e. proton-deuterium coupling

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6
Q

What coupling is observed in an NMR spectrometer?

A

Coupling exists between ALL NMR active nuclei present in a molecule
However, coupling between identical nuclei in identical positions in a molecule is not observed

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7
Q

What are the 2 distinct types of magnetic interaction (i.e. coupling) between nuclei with non-zero spin?

A
  1. Direct interaction (dipole-dipole coupling, D)

2. Indirect/scalar coupling (spin-spin splitting, J)

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8
Q

Direct interaction

A

= dipole-dipole coupling
= through-space coupling
Basically the magnetic effect on nucleus 1 caused by the magnetic field generated by nucleus 2
Approx. 1000x larger than scalar coupling - but its magnitude is affected by the natural abundance of the NMR active nuclei and the distance between the nuclei
Dipolar coupling is completely averaged out by the random motion of molecules in mobile isotropic liquids, meaning no direct effects are seen - i.e. does not lead to multiplets in liquid-state NMR spectra

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9
Q

Properties of J coupling

A

Always reported in Hz
Field-independent
Mutual (i.e. Jax = Jxa)

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10
Q

Why does the magnitude of J coupling fall off rapidly as the number of intervening bonds increases?

A

Because the J coupling effect is usually transmitted through bonding electrons

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11
Q

Through-space mechanism of coupling between two spin 1/2 nuclei, A and X

A

In the absence of X, A resonates at a frequency vA in a particular field Bz
When X is present, the orientation of the spin of X with respect to A will create 4 possible spin states: aa, ab, ba, bb - each of these is a different energy
In the presence of X, A resonates at two different frequencies, separated by 2Bx = J (vA + J/2 and vA - J/2)
And X will resonate at vX + J/2 and vX - J/2

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12
Q

What does the energetic ordering of the ab and ba states depend on?

A

The vA and vX values

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13
Q

NMR selection rule

A

Deltaml = +/- 1
i.e. only aa to ab and ba to bb transitions are allowed (for X)
(for A would be aa to ba and ab to bb)

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14
Q

Why does the ‘through space’ mechanism average spin-spin interactions to zero?

A

In a solution or liquid, the orientations of the spins are totally random which means the field experienced by A due to X averages out to zero without restriction of movement
Coupling is orientation dependent
The dipole orientations of A and X are only fixed relative to each other in a solid

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15
Q

Fermi contact interaction

A

The through-bond magnetic interaction between an electron and an atomic nucleus
Responsible for the appearance of coupling in NMR

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16
Q

How is coupling transmitted?

A

By the polarisation of electrons in bonding MOs

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17
Q

Fermi contact mechanism

A

For A-X

  1. The nuclear spin on A polarises the electron spin to be antiparallel (or parallel if gamma is negative)
  2. This, in turn, polarises the other electron in the bond to be antiparallel as is demanded by Pauli’s exclusion principle
  3. This electron then polarises the nuclear spin on X - which could be aligned with A (parallel spins, unstable, higher energy, J is negative) or be opposed to A (antiparallel spins, lower energy, J is positive)
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18
Q

When is J positive?

A

When the nuclear spins are opposed

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19
Q

Why does the magnitude of the coupling constant decrease as the number of bonds between 2 nuclei increases?

A

Because the ability for efficient polarisation decreases

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20
Q

What does J depend on?

A

Number of bonds

Bond angles

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21
Q

Effect of bond angles on J

A 
Known as the Karplus relationship
Torsional/dihedral angle:
0 degrees = 8 Hz
60 degrees = 2 Hz
90 degrees = 0 Hz
180 degrees = 10 Hz
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22
Q

Effect of gyromagnetic ratio on J

A

For a given pair of nuclei, the magnitude of the scalar coupling is proportional to the product of the gyromagnetic ratios of the nuclei, if all other molecular factors stay the same
J proportional to yAyB

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23
Q

Why can coupling constants only be compared for isotope of the same element? (e.g. H/D)

A

Because only isotopes have the same electron cloud

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24
Q

Comparing coupling constant for isotopes of the same element

A

Where two NMR-active isotopes of a given element exist, e.g. 10B and 11B, the ratio of coupling constants to a third nucleus will be the ratio of the gyromagnetic ratios

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25
Q

Why do (hybrid) molecular orbitals with a high s-character transmit coupling most efficiently?

A

Coupling is transmitted via bonding electrons rather than through space
Therefore, for coupling to occur, the bonding electrons must be able to ‘contact’ both nuclei
Only s-orbitals have non-zero density at the nucleus (i.e. no node) so molecular orbitals with a high s-character will transmit coupling more efficiently

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26
Q

Why are coupling constants generally larger in linear molecules than trigonal/tetrahedral molecules?

A

Linear molecules are sp hybridised, trigonal are sp2 and tetrahedral are sp3
Bonding in linear molecules has greater s-character (50 %) than in trigonal (33 %) and tetrahedral (25 %)

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27
Q

Another factor that contributes to higher coupling constants for linear vs. trigonal/tetrahedral

A

Increase in bond strength from sp3 to sp2 to sp

but higher s-character is the main factor

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28
Q

How is coupling between nuclear spins classified?

A

Into spin systems

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29
Q

When are nuclei termed equivalent nuclei?

A

When they are in identical environments with identical chemical shifts

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30
Q

Nuclei labelled with adjacent letters of the alphabet

A

Non-equivalent nuclei with very similar chemical shifts which are strongly coupled
The difference in their resonance frequencies is similar to their mutual coupling
e.g. (CH3CH2)3Ga = A3B2 spin system (small chemical shift separation so use adjacent letters)

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31
Q

Nuclei labelled with letters far apart in the alphabet

A

Weakly coupled nuclei
The difference in their resonance frequencies greatly exceeds their mutual coupling
e.g CH3CH2F = A3M2X (large chemical shift separation so use letters that are far apart)

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32
Q

Condition for nuclei to be magnetically equivalent

A

When they have the same chemical shift and the same coupling constants with the same partners
i.e. magnetically equivalent nuclei have the same coupling to a nucleus (magnetic inequivalence arises when 2 nuclei have different couplings to the same nucleus)

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33
Q

AMX spin system

A

Consists of 3 nuclei with 3 different chemical shifts and 3 distinct coupling constants (Jam, Jax, Jmx)

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34
Q

Coupling in AXn systems when I = 1/2

A

For n equivalent X nuclei, the resonance of A is split into 2nI+1 (n+1) equally spaced lines, where their relative intensities are given by the (n+1)th row of Pascal’s triangle

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35
Q

Coupling in AMX systems where I = 1/2

A

The resonance of A can be predicted by determining the possible combinations of the spins (Ml values) of M and X
i.e. both up, MupXdown, MdownXup, both down
This leads to 4 lines of equal intensity, because there are 4 non-degenerate arrangements of the M and X spins that are all ~equally likely
The 4 lines are displaced from the chemical shift of A by simple combinations of the coupling to A - gives a doublet of doublets

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36
Q

Coupling in AXn systems, where X has I > 1/2

A 
If A (I = 1/2) is coupled to X (I > 1/2), the resonance of A is split into 2nI+1 equally spaced lines
When n = 1, the intensities of the lines are equal
When n > 1, the intensities are best deduced from the tree-splitting approach
(If n = 2, can use an ml combination table)
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37
Q

Typical multiplet pattern for AMX2 system (Jax > Jam)

A

Triplet of doublets

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38
Q

Typical multiplet pattern for AM2X system (Jax > Jam)

A

Doublet of triplets

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39
Q

Typical multiplet pattern for AM2X2 system (Jax > Jam)

A

Triplet of triplets

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40
Q

Typical multiplet pattern for AX, where X has I = 1

A

3 lines of equal intensity

2nI + 1 = 2(1)(1) + 1 = 3

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41
Q

Typical multiplet pattern for AX, where X has I = 3/2

A

4 lines of equal intensity

2nI + 1 = 2(1)(3/2) + 1 = 4

42
Q

Typical multiplet pattern for AX2, where X has I = 3/2

A

7 lines with relative intensities 1-2-3-4-3-2-1

2nI + 1 = 2(2)(3/2) + 1 = 7

43
Q

How should chemical shift be thought of?

A

In terms of a frequency - the rate of precession of the induced magnetisation

44
Q

Coupling constants

A

Measured in Hz

Correspond to a difference in the rates of precession of the induced magnetisation

45
Q

First-order system

A

Deltav/J&raquo_space; 1
If 2 nuclei are far apart in chemical shift (i.e. there is a large difference in their rates of precession), then the effects of coupling and chemical shift can be easily seen/resolved
The nuclei are weakly coupled and their energy levels are well separated, leading to ‘well-behaved’ multiplets

46
Q

Second-order system

A

DeltaV/J ~ 1 (generally <5)
If the frequency separation (in Hz) between the chemical shifts of different sets of nuclei (A and B) is small, it is similar to the magnitude of the coupling constant (Jab) between the nuclei
This means the nuclear spin energy levels of A and B become close in energy
Thus, significant ‘mixing’ of the wavefunctions of the energy levels occurs, which changes the transition probabilities of each of the 4 NMR lines
Complicated spectral signals are observed (second-order spectrum)

47
Q

Zero-order spectrum

A

No coupling

Deltav/J tends to infinity (because J = 0)

48
Q

When do (groups of) nuclei form first-order multiplets?

A

If the chemical shift differences between the nuclei are large compared to the coupling constants between them

49
Q

AB spin system (I = 1/2)

A

As Deltav approaches Jab, the inner lines become more allowed (stronger) and the outer lines become less allowed (weaker)
The multiplets ‘lean’ towards each other - “roofing” of signals
The roofing becomes more pronounced as the chemical shift difference between the coupled multiplets gets smaller, until the inner lines become coincident at the mutual chemical shift as a ‘pseudo-singlet’ - the nuclei are still coupled with a finite Jab value, but all signs of this are lost when Deltav is negligible

50
Q

Deltav for chemically equivalent nuclei

A

Deltav = 0 for chemically equivalent nuclei

Therefore, irrespective of whether they are magnetically equivalent or not, no coupling is observed between them

51
Q

When are second-order spectra more common?

A

When the spectrometer field is low

Second-order spectra can, more often than not, be converted to first-order spectra by increasing the field strength

52
Q

Why can second-order spectra generally be resolved at higher spectrometer fields?

A

Chemical shift difference in Hz (i.e. Deltav) is implicitly field-dependent (chemical shift difference in ppm i.e. Deltadelta is field-independent)
This means that for e.g. two 1H resonances separated by 1 ppm, they are 100 Hz apart on a 100 MHz spectrometer, but are 500 Hz apart on a 500 MHz spectrometer
Increasing field strength simplifies the spectrum (separates peaks)

53
Q

Isotopomers

A

Compounds that differ only in the identity of the particular isotopes of specific atoms present in the sample

54
Q

NMR spectrum of isotopomers

A

In order to predict the NMR spectrum of a isotopic sample, all the different possible combinations of isotopes that may arise must be taken into account
The spectrum will consist of several overlaid spectra

55
Q

Advantage of decoupling for low sensitivity nuclei

A

e.g. 13C
Can help increase the intensity of the signals (rather than an already weak signal being split into several smaller, even weaker signals)
Simplifies the spectrum

56
Q

How is decoupling of 1H achieved?

A

By powerful broad-band irradiation of the whole 1H frequency range (“broad-band decoupling”)
This saturates the 1H nuclei - equalises the populations in the upper and lower energy levels so there is no longer a population difference

57
Q

“Side-effect” of decoupling

A

Nuclear Overhauser Effect of neighbouring decoupled 1H results in a signal enhancement of the 13C signal by up to 200%

58
Q

Factors affecting magnitude of NOE signal enhancement

A

Distance between C/H (inversely related to distance)
Number of attached protons - NOE in 13C spectra increases with an increasing number of attached protons on the carbon - i.e. CH3 > CH2 > CH, and quaternary C have no NOE
Sign of y - the nuclei must have the same sign for enhancement. Nuclei with y of opposite sign (e.g. y is -ve for 15N) lead to a decrease in signal intensity

59
Q

NOE

A

Nuclear Overhauser Effect
The change in intensity of the signal from one spin, when the spin transition of another nucleus is perturbed from its equilibrium population e.g. by saturation or inversion

60
Q

Use of NOE

A

NOE is particularly useful for increasing the signal intensities (i.e. the sensitivity) of low y and low abundance nuclei when they are (scalar/dipolar) coupled to high y and high abundance nuclei
e.g. 13C coupled to 1H

61
Q

Drawback of NOE / why does standard 13C{1H} not allow the use of signal integration?

A

Intensity of signals depends on the magnitude of NOE - different NOE enhancement leads to variations in intensities of up to 200 %
Therefore intensity is no longer proportional to the number of C atoms
(can no longer integrate)

62
Q

Why can decoupling and NOE be separated?

A

Because they act at different timescales
Decoupling of neighbouring nuclei is instantaneous
NOE builds up and decays slowly (on the T1 timescale, seconds)
This means one can be “swtiched on” in the absence of the other if the timing of the decoupler is chosen carefully

63
Q

Gated decoupling

A

Decoupler is on between acquisitions and off during acquisition
Observe multiplets (i.e. there is no decoupling) with full NOE enhancement
(draw)

64
Q

Inverse gated decoupling

A

Decoupler is off between acquisitions and on during acquisition
Observe the decoupled spectrum without NOE enhancement
(draw)

65
Q

How can quantitative decoupled 13C NMR spectra be obtained?

A

By turning off NOE

i.e. inverse gated decoupling

66
Q

Why are 13C NMR spectra rarely quantitative, in addition to NOEs?

A 
The intensity of standard 13C spectra depends strongly on T1, which is variable - fast-relaxing nuclei (e.g. CH3) are much more intense than slow-relaxing nuclei (e.g. quaternary C)
Relaxation times (T1) can be very long (minutes), meaning scans are typically repeated before the signal completely relaxes (otherwise the delay times would be prohibitively time consuming)
This means some of the maximal signal intensity is lost
However, exponential signal recovery means that most of the magnetisation relaxes after a short time, so the larger number of scans more than compensates for the intensity lost from a single scan
67
Q

How can changing pulse angle improve the intensity of 13C spectra?

A

Intensity can be further improved using 30 degree pulses instead of 90 degrees
30 degree pulses give signals of 50% intensity, but recover ~3x faster
This means 3 scans using 30degree pulses can be taken in the time taken for 1 90degree scan and give a 50% more intense signal

68
Q

Ernst angle

A

Optimal pulse angle for the excitation of a particular spin that gives the maximal signal intensity in the least amount of time
cos(angle) = e^(-r/T1), where r = repetition time

69
Q

Conditions for inverse gated decoupling

A

Need to wait long enough between scans to allow for full relaxation of NOE
Generally the delay time needs to be >5x longest T1

70
Q

Limitation of inverse gated decoupling

A

T1 relaxation time can be minutes (e.g. for quaternary C) which means the experiment can become very long
Very concentrated samples are required in order to get good spectra in few scans - but the experiment still typically takes hours/days rather than minutes

71
Q

How can relaxation times be shortened?

A

By the use of relaxation agents

i.e. paramagnetic compounds

72
Q

Effect of paramagnetic compounds on relaxation

A

Paramagnetic compounds result in much faster relaxation, not only within the same molecule but also in neighbouring molecules (“outer sphere relaxation”)

73
Q

Ideal properties of relaxation agent

A

No specific interactions with the molecule - the paramagnetic relaxation agent needs to be stable (e.g. no free coordination sites)
Isotopic spin distribution i.e. no orientation dependence, equal magnetism in all directions

These properties mean that the only effect on the observed molecule is faster relaxation with no paramagnetic shift

74
Q

Typical paramagnetic relaxation agents

A

Cr(acac)3
Fe(acac)3
Both soluble in CDCl3, used mainly for the faster relaxation of slow relaxing nuclei e.g. used in quantitative 13C

Gadolinium complexes (f7) are mainly used as MRI contrast agents (relaxation of water)

75
Q

Magnetic susceptibility

A

The measure of magnetisation of a material/compound when placed in an applied magnetic field

76
Q

Evans method for magnetic susceptibility

A

= using 1H NMR to measure the magnetic susceptibility of a solution
A paramagnetic compound will shift the solvent signal in 1H NMR
Can measure this change in chemical shift relative to that of the pure solvent
(In most cases, the effect of the diamagnetic susceptibilities can be neglected)
Can use an equation to approximate magnetic moment

77
Q

Different types of fluxional/dynamic processes that can occur on the NMR timescale

A

Intramolecular processes = conformational changes/ligand interconversion
Intermolecular processes = exchange of bound and free ligands

78
Q

31P spectrum of PF5 at low/room/high temp

A

PF5 undergoes Berry pseudo-rotation
Low temp = 1 resonance, 1-3-3-1 quartet for JPFe of 1-2-1 triplets for JPFa
Room temp = 1 broad resonance (all F interchanging on NMR timescale)
High temp = 1 resonance, AX5 sextet showing weighted mean JPF

79
Q

Rate of exchange so that individual resonances (i.e. separate peaks) can be seen

A

Must be less than 2piDeltav, where Deltav is the difference between the resonance frequencies of the 2 sites (in Hz)

80
Q

Why can IR spectroscopy distinguish each site in very fast exchanging systems, whereas NMR is sensitive to relatively slow motions?

A

NMR resonances of exchanging sites differ by up to a few thousand Hz, whereas a band difference of 1cm^-1 in IR spectroscopy corresponds to 3x10^10 Hz
IR spectroscopy can look at timescales of «10^-10 s, which means that, in very fast exchanging systems, IR can distinguish each site
NMR can only look at exchange processes in the ms-us range so is sensitive to quite slow motions

81
Q

NMR spectrum of DMF

A

Gives 2 Me signals at low temperatures
DMF is a planar molecule - each Me is in a different environment, either cis or trans to O
As temperature is increased, rate of exchange becomes faster so the 2 signals coalesce to give one signal

82
Q

Slow exchange regime

A

When the rate of exchange between sites is much slower than the NMR timescale (i.e. much slower than the difference in resonance frequencies of the sites), a separate resonance is observed for each site

83
Q

Broadening of resonances in the slow exchange regime

A

Some broadening will be seen depending on how much time nuclei spend at each site (Heisenberg uncertainty principle)
Faster rate of exchange = broader peaks
No exchange = natural linewidth

84
Q

Equation for rate constant from a slow exchange regime

A

Deltaw = (w-w0) = k/pi = 1/pi(tau)
Therefore k = piDeltaw

Where tau = lifetime of the site (inverse of rate constant)
And Deltaw = increase in w1/2 above the natural linewidth

85
Q

Fast exchange regime

A

When the rate of exchange between sites is much faster than the NMR timescale (i.e. much faster than the difference in resonance frequencies of the sites), a single resonance is observed at a weighted average resonance position

86
Q

Linewidth becomes narrower as…

A

…rate of exchange increases
At very fast exchange, the system behaves as though a single site existed
This is the natural linewidth, wf

87
Q

Equation for rate constant from a fast exchange regime

A

Deltaw = (w-wf) = pi(deltav)^2/2k = pi(tau)(deltav)^2/2
Therefore k = pi(deltav)^2/2Deltaw

Where tau = lifetime of the site (inverse of rate constant), Delta w = increase in w1/2 above natural linewidth and deltav = separation of the 2 peaks in Hz (i.e. difference in resonance frequencies) in the slow exchange regime

88
Q

Coalescence temperature/point

A

The point at which the resonances of the exchanging sites have just merged into a single resonance
Somewhere in between the fast and slow exchange limits

89
Q

When does coalescence occur?

A

For two equally populated sites, coalescence occurs when tau = root2/pi(deltav)
Therefore k = pi(deltav)/root2

90
Q

Heisenberg broadening (/Heisenberg’s uncertainty principle)

A

The shorter the lifetime (tau) of a particle in a particular stationary state, the greater the degree of uncertainty in the energy of that state
And therefore the greater the uncertainty of any resonance frequency corresponding to a transition energy (deltaE) involving such states

deltaEdeltatau >= h/2pi

91
Q

Heisenberg broadening in I=1/2 vs quadrupole nuclei

A

I=1/2 nuclides have relatively long lifetimes so display sharp lines (narrow linewidths)
Quadrupolar nuclides have short lifetimes due to efficient relaxation so display broad lines

92
Q

Heisenberg broadening in fluxional systems

A

Fluxional systems display broad resonances because the lifetimes of species in the individual positions are relatively short
The lines sharpen as lifetime in the ‘freely exchanging’ state increases

93
Q

Heisenberg broadening in viscous liquids

A

NMR of viscous liquids gives broad lines

T1 is long but T2 is short (efficient magnetic coupling to neighbouring spins)

94
Q

Lineshape analysis

A

Analyse the lineshapes of the peaks due to exchanging sites at various temperatures and fit the observed curves to theoretical ones
Allows accurate calculation of linewidths and chemical shifts, which can then be used to calculate rate constants
Then if the exchange rates (rate constants) are known over a range of temperatures, then parameters such as DeltaH, DeltaS and DeltaG can be calculated

95
Q

Equations for calculating thermodynamic/kinetic parameters for an exchange process

A

DeltaG = DeltaH - TDeltaS

k = (kbT/h)e^(DeltaG/RT)

Giving -Rln(kh/kbT) = DeltaH/T - DeltaS

Substituting these values for constants gives

ln(k/T) = 23.76 - DeltaH/RT + DeltaS/(R)

96
Q

Method for determining thermodynamic/kinetic parameters for an exchange process

A
  1. Take spectra at various temperatures
  2. Peak fit spectra with simulated bands in order to obtain the parameters for calculating the rate constant of the exchange process (linewidths and chemical shift)
  3. Plot ln(k/T) vs. 1/T
    (a) Slope = -DeltaH/R
    (b) Intercept = 23.76 + DeltaS/R
97
Q

Errors in NMR method for thermodynamic parameters

A

There are inherent errors in measuring temperature inside the NMR probe (+/- 2K)
Which leads to uncertainties in DeltaG of +/- 1 kJmol-1 (ok) and DeltaS of +/- 20 J mol-1 K-1 :(

98
Q

How to estimate DeltaG for a simple dynamic process in the absence of lots of data

A

Just use the coalescence point

DeltaG = -RTln(kh/kbT), where k = pi(deltav)/root2

99
Q

Fluxional processes can be either…

A

…intermolecular (ligand dissociation and re-association) or intramolecular (ligand reorganisation)

100
Q

How can we determine whether a fluxional process is inter- or intramolecular?

A

From heteroatom coupling constants
i.e. for an intermolecular process, if an X-Y bond is being broken, then J(XY) coupling is not observed at the fast exchange limit

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