Structural Biology - NMR

Question 1

What are the main applications of NMR?

Answer 1

analytical (checking correct product)
chemical structure
3D structure determination (NOEs)
Molecular interactions (chemical shift + NOEs)
Molecular dynamics
NMR imaging
Metabolomics (metabolic profiling of organisms from blood or urine)
Mainly used for Organic Compounds

Question 2

What is nuclear spin (I) ?

Answer 2

Spin is the rotation about an axis
Spin can be parallel or antiparallel to the magnetic field
Spin is quantised and comes in values of multiples of 1/2
number of states of angular momentum = 2I + 1
protons, unpaired electrons and neutrons have spin (I) of 1/2 –> have 2 states of angular momentum –> magnetic field is pointing in parallel and antiparallel directions

Question 3

How is nuclear spin important in NMR?

Answer 3

Sample is placed in a strong magnetic field, so spins align either parallel or antiparallel to the field
NMR detects the nucleus
Nuclei are not elementary particles, their spin is dependent on the number of protons and neutrons
Even neutrons+protons have 0 spin
Odd protons+neutrons have integral spin numbers
Odd protons, even neutrons have 1/2 spin (H1, C13, N15, F19, P31)
So only an odd number of protons or neutrons or both have nonzero nuclear spin which means they are suitable for NMR

Question 4

Explain the energy levels of spins

Answer 4

Sample is placed in a strong magnetic field, so spins align either parallel or antiparallel to the field
Parallel spin has lower energy than antiparallel spin
Radiofrequency pulse is applied which can induce transition between these two energy states
Planks law associates frequency of magnetic field with energy
Strength of magnetic field determines energy gap
Energy=Magnetogyric ratioplanks constant/2pilab magnetic field
Magnetogyric ratio is the strength of the magnetic field and is correlated to the size of the nucleus
Small proton=fast spin=high magnetic field=high magnetogyric ratio=higher energy gap=higher sensitivity

Question 5

Advantages of NMR

Answer 5

Using radiofrequency is safe

Question 6

Explain frequently used nuclei and their spins

Answer 6

H1: spin of 1/2, most sensitive due to small nucleus and high abundance
C13/N15: spin of 1/2, low sensitivity due to big nucleus and low abundance
Deuteron (2H) is NMR active, spin of 1

Question 7

What is the larmor procession of nuclear spins?

Answer 7

Nuclei with nonzero spin act as tiny magnets with a magnetic moment associated with their spin
Antiparallel/parallel is at a slightly offset angle
Residual force is trying to make the magnetic field fully parallel
Instead precesses in an external magnetic field (rotate around the direction of the field) at the larmor frequency
Larmor (precessional) frequency is directly correlated to frequency for absorption
Forms resonance in NMR
When radiofrequency pulse is applied at Larmor frequency, it interacts with nuclear spins, leads to transition between energy states, absorption of radiation and gives NMR signals
energy levels further apart = higher Larmor frequency

Question 8

Explain the macroscopic magnetisation of a sample

Answer 8

Net magnetisation when placed in the magnetic field
Net magnetisation vector (M) is the sum of the individual magnetic moments (parallel/antiparallel)
Parallel spin is of lower energy so populate the sample first so there is more parallel spins

Question 9

Explain the effect of a radiofrequency pulse

Answer 9

All particles are in the laboratory magnetic field (B0), Z
A short pulse of radiofrequency is applied at 90 degrees to the Z direction
Magnetisation vector (M) flips to the Y axis
Common flip angles are 90 degrees or 180 degrees
Resonance occurs when the frequency of the RF pulse matches the Larmor frequency of the nuclear spins
Energy is absorbed by nuclear spins, causing spin flips from +1/2 (parallel, low energy) to -1/2 (antiparallel, high energy)
Magnetisation vector precesses about the Z axis/undergoes Larmor precession
Precession of the nuclear spins generates an oscillating magnetic field which is detected as an NMR signal
Eventually macroscopic magnetisation returns to equilibrium and realigns with B0

Question 10

Explain the use of a short radiofrequency pulse

Answer 10

A short pulse is not a precise frequency, it contains many different frequencies in it
It is a short sine wave with zeros (no frequency) after
The shorter the pulse, the more frequency information in it
A short pulse can excite resonances of all nuclear spins in a sample at the same time
A short pulse allows precision when flipping magnetisation vector from Z to Y plane

Question 11

Explain pulse Fourier transform in NMR

Answer 11

Fourier pairs: rectangular pulse in frequency domain matches sine function in the time domain
Rotating magnetisation vector leads to a change in magnetic field which induces a current in the receiver coil - called free induction decay
Sine wave with decreased magnitude as the magnetisation vector realigns with B0
FID is recorded as a function of time
Fourier transform is used to convert the time domain into the frequency domain (frequency=1/time)

Question 12

What are the different parameters of an NMR spectrum?

Answer 12

Chemical shift - frequency axis along bottom
Integral - area under peak
Scalar coupling constant (J) - separation between lines
Relaxation times (T1 and T2) - time taken to get back to Z/B0
Dipolar coupling (D) - don’t see in spectrum
Nuclear Overhauser Effect (NOE) - don’t see in spectrum

Question 13

Explain chemical shift in terms of the electron magnetic field

Answer 13

Electrons circulate about the direction of an applied magnetic field
As they carry a negative charge it generates a magnetic field
In total: Lab magnetic field B0, Nuclear magnetic moment, Electron magnetic field
Direction of spin depends on electron’s location relative to the nucleus
At nucleus: opposite direction as B0
Out of nucleus: same direction as B0
Electron magnetic field causes shielding so the nucleus feels a smaller magnetic field and decreases frequency
Nearby electronegative groups withdraw electrons away from nucleus which reduces shielding (deshielding) so nucleus experiences a higher magnetic field and increases frequency

Question 14

Explain chemical shift in terms of frequency scale

Answer 14

The chemical environment around a nucleus affects the local magnetic field which affects the resonance frequency
Chemical shift is how much the resonance frequency deviates from the reference frequency of tetramethylsilane (TMS)
Frequency scale is expressed as a fraction of the lab magnetic field, normalised scale to magnetic field strength

Question 15

Explain chemical shift in terms of ring currents

Answer 15

Pi electrons circulate in response to an applied magnetic field (move around all 6 groups)
Ring current generates a local magnetic field which opposes the applied magnetic field
Protons at the edge of the ring are not shielded (deshielded) and feel a larger magnetic field
Aromatic ring protons behave like electronegative groups, so will resonate at high frequency and higher chemical shift

Question 16

Explain the properties of TMS

Answer 16

TMS is at 0 frequency/ppm
Reference for every molecule
Tetrahedral symmetry with 4 methyl groups
3 protons in the 4 methyl group are in the same chemical environment
Gives ONE signal

Question 17

Chemical shift equivalence

Answer 17

Nuclei that are interchangeable by a symmetry operation (Rotation, reflection, inversion) have the same chemical shift
Ex. Rotation in TMS methyl groups
Nuclei that have rapid exchange on the NMR timescale have the same chemical shift
Ex. Protons of hydroxyl/amine groups can swap
Gives an average chemical shift

Question 18

Chemical shift ranges for different groups

Answer 18

Less shielded and higher frequency on left of scale
More shielded and lower frequency on right of scale
Alkanes: 0.5 - 1.5 ppm
Alkenes: 4.5 - 6.5 ppm
Alkynes: 1.5 - 2.5 ppm
Aromatic protons: 6.5 - 8.5 ppm
Alcohols: 0.5 - 5.0 ppm
Aldehydes: 9.0 - 10.0 ppm
Ketones: 2.0 - 3.0 ppm
Amide: 5.5 - 8.5ppm

Question 19

How to interpret basic NMR spectrum

Answer 19

Chemical shift used to identify the type of proton
Number of peaks in the spectrum correspond to the number of chemical environments
Area under the peak (integral) is proportional to the number of spins (protons) in that peak
Ratio between area under peaks gives number of protons in each peak
Splitting pattern is number of protons in neighbouring environment + 1

Question 20

Scalar coupling in two spin systems

Answer 20

Scalar coupling (J) is the spin-spin interaction between two neighbouring non equivalent 1/2 spin nuclei
Parallel/antiparallel state of proton A will affect state of neighbouring proton X
Can only consider 3 bond couplings
Sometimes 4 if there is a double or triple bond
Two states/magnetic environments of A so proton X forms a doublet (and vice versa)
Doublet: one other proton adjacent (parallel or antiparallel)
Triplet: two protons adjacent (2 antiparallel, 2 parallel or 1 parallel 1 antiparallel)
N + 1 = splitting pattern
Separation between 2 peaks is of 10Hz difference (very small)

Question 21

Scalar coupling to exchangeable OH/NHs

Answer 21

Sample dissolved in organic solvent
Ex. for drug testing pharma
Small amount of D2O (deuterium: water with hydrogen isotope H2) will replace any exchangeable group so peaks will disappear
Used to see which groups are exchangeable
But usually groups are dissolved in water
Protons on water will replace any exchangeable group so these will not be on the spectrum
OH protons (unless strongly hydrogen bonded) are never on spectrum (high exchange rate)
Amides/protons bound to nitrogen can be seen (low exchange rate)

Question 22

Spin-lattice and spin-spin relaxation (T1 and T2)

Answer 22

T1 and T2 are Relaxation Times after applying the RF pulse
T1 (spin-lattice) is enthalpic: recovery of Z magnetization
At 0 (right after pulse) there is no magnetization
Any change in magnetization in Z has to be a flip of spins (parallel to antiparallel)
T1 is equal to T2 in small molecules
T1 is much longer than T2 for large molecules

T2 (spin-spin) is entropic: loss of magnetization in X, Y plane
Anything in X, Y can have the same number of antiparallel/parallel spins
Loosing of order/coherence of spins
T2 is directly related to line width
Slow in small molecules, fast in large molecules
Small molecules move fast in solution = long T2 = long FID / magnetization is lost slowly = narrow lines
Big molecules = short T2 = short FID = broad lines

Question 23

Dipolar coupling

Answer 23

Cannot be seen in NMR spectrum
Direct, through space magnetic interaction from nucleus to nucleus
Proton A and B are far apart in the chemical structure
but close in space (5-6 Angstroms)
Interaction depends on distance (up to 6 angstroms) and the angle the internuclear vector suspends with laboratory magnetic field
Usually angle averages to zero by molecular tumbling (so it is not seen in NMR)
Dipolar coupling is dependant on 1/r^3

Question 24

Nuclear Overhauser effect (NOE)

Answer 24

Measure of dipolar coupling
Through space effect
It is dependant on 1/r^6
The change in intensity of an NMR signal when the transition of another is saturated (equalise population of upper and lower spin states)
Not visible in spectrum
Saturate B so A is no longer coupled = affects relaxation time of A = change in intensity

Question 25

2D NMR advantages

Answer 25

1D spectrum: not enough frequency space in spectrum to separate all of the peaks (very congested), used to show small peptides/metabolites
2D NMR: uses 2 frequency axes to increase separation and show signals, analogous to 2D SDS page, prevents overlap

Question 26

Carbon-13 NMR Problems

Answer 26

Low abundance (1.1%)
low magnetic moment (magnetogyric ratio is 1/4 of 1H)
long relaxation times
longer waiting time between pulses due to long relaxation time
Integrals do not correspond to number of carbons (would have to decouple protons by saturating all protons, this would change signal of C13)

Question 27

Carbon-13 NMR Advantages

Answer 27

Spin is 1/2
Has an odd mass number so has spin
2 states (antiparallel and parallel)
No effective 13C-13C coupling
Large chemical shift range (>200ppm)
Easy to interpret

Question 28

Carbon-13 NMR coupling constants

Answer 28

Scalar coupling constants (J) are constant and independent of the magnetic field
Main couplings are 13C - 1H and 1H - C - 13C

Question 29

Carbon-13 NMR spectrum

Answer 29

Off resonance decoupling or fully decoupled spectrum
Off resonance decoupling: C13 nuclei are only split by protons attached directly to them (C13-H1). Removes 2 bond coupling (H1-C-C13). Signal for C is the number of protons attached with N protons + 1 rule. CH3 is a quartet, C=O is a singlet
Fully decoupled proton: irradiate with proton frequencies to saturate H1 signals. Gives singlets for all C. Spin-spin coupling is eliminated. Sharp signals, high resolution.
Deshielding from electronegative groups increases chemical shift
The area under each peak corresponds to number of carbon atoms

Question 30

How is a 2D NMR recorded?

Answer 30

Apply 90 degree pulse
Wait for evolution time (T1)
Plot 1D signal and use Fourier transform to get second time axis
Apply another train of pulses to mix 90 degree magnetization into its neighbouring nuclei
Measure acquisition time (T2)
Based on FID
Detect signal again after mixing
Cross peaks show correlation between frequencies on two frequency axes: Either: J (scalar coupling) through bonds OR D (NOE) dipolar coupling through space

Question 31

2D H1-H1 Correlation Spectroscopy (COSY)

Answer 31

0 - 10 ppm on both axes
Diagonal peak: represents chemical shift/frequency of proton on both axes
Off diagonal peak/correlation peak: correlates to it’s coupled partner
4 dots = 2 peaks (doublet)
Only shows SCALAR couplings (J) / through bond couplings
look at image in notes

Question 32

2D - H1-H1NOESY spectra

Answer 32

Intra-residue cross peaks
NOE coupling/through space couplings
measures which protons are close to each other in space

NOE cross peak patterns give 3D conformational information
Weak peaks (small dots) = protons are further apart
Intense peaks (large dots) = protons are closer together (2-3 angstroms)
Pattern depends on the conformation of the protein
Image in notes

Question 33

2D - HMQC

Answer 33

No diagonal (unlike COSY and NOESY)
Correlates the C13 spectrum with the H1 spectrum
Only directly bonded C13 - H1 coupling will give cross peaks
Can go up to 10s of kDa (works for large molecules/proteins)

Question 34

2D - HMBC

Answer 34

Correlates the C13 spectrum with the H1 spectrum
Shows one bond couplings (direct C13 to H1) AND 2 bond couplings (C13-C-H1)
High resolution
Similar to COSY information