NMR Flashcards

Question

2D NMR

Answer 1

A

All elementary particles possess an intrinsic angular momentum (I) known as spin e.g. an electron has spin; it is spinning on an axis
Spin is quantised and comes in properties of ½
I = the spin quantum number:
The most basic is ½ in 2 possible states (+ve and -ve)
If I = 1 then you have 3 possible states (-1, 0, +1)
Protons, electrons and neutrons possess spin where I = ½ (+ve and -ve states)
Different nuclei have different spin quantum numbers:
Nuclei with even number of protons and neutrons = 0 spin
Nuclei with odd number of protons and neutrons = integral spin numbers (1, 2, etc)
All the rest have ½ integral spin (½, 1 ½, 2 ½)
A moving charge creates a magnetic field
All nuclei (except 0 spin) have a magnetic field at the atomic level –not a macroscopic property

Answer 2

A

There can never be 0 magnetic field as we have the Earth’s magnetic field; there is always a slight energy difference between the two spin states
- Dipole = has a direction associated with it
- Strength of the applied magnetic field (Bo) determines the energy gap of a given nuclei; Bo is in the z-axis
- Protons can align with Bo = α state (½); or against Bo = ß state (- ½); α and ß align oppositely
- α = lower energy (parallel), more stable as more protons occupy that state
- ß = higher energy (antiparallel), less stable due to lower population
- Higher Bo = even fewer ß spin states occupying higher levels due to the bigger difference in energy levels
- There is continuous flipping between the two energy states, but there is always have the population difference
- You absorb net energy as you promote some states into the higher level
- Equal populations in energy levels = no signal, thus need to allow to come back to equilibrium before recording spectroscopy again
- Can use UV light to transition between higher and lower energy levels if we match the frequency of our electromagnetic radiation correctly (~Planks Law) –bigger magnet (Bo) needs higher frequency to induce these transitions
- Higher magnetic fields (Bo) = larger energy gap (∆E) = higher sensitivity; greater separation between the two states and fewer population upper level

Answer 3

A

γ = magnetogyric ratio i.e. how large the magnetic field that’s associated with it
The higher the magnetogyric ration the higher the magnetic moment and field associated with it
Correlates with the property that smaller nuclei spin much faster than larger nuclei as they have higher magnetic moments
∆E = energy difference between α and ß spin states

∆E = hv (h = Planks constant; v = frequency Hz)
V = [γ/qπ x Bo] (Bo = strength of applied magnetic field; always proportional to ∆E)

Rearrange 1 and 2:

∆E = h [γ/qπ x Bo]
(remember Hz to MHz divide by 1 x 10⁶

Answer 4

A

Proton = the most sensitive nucleus
C13/N15 = low abundance, therefore low sensitivity
Deuteron = most NMR active but not be used because: spectra is more complex due to multiple states; much less sensitive; spectra are broader and more challenging to interpret at atomic resolution
Deuteron has spin because it has one proton and one neutron in its nucleus, both have parallel spin thus + ½ values = total spin of 1
C12 = most abundant isotope but not NMR active; has spin 0 because it has 6 protons and 6 neutrons occupying 3 energy levels (2P + 2N each) which all have + ½ or – ½

Answer 5

A

Put sample in a tube into a magnetic field
Irradiate it with the appropriate electromagnetic radiation = radiofrequency range

Continuous wave NMR = previous method where you apply a continuous wave and sweep through the frequency
Very laborious and inefficient process
1st proton NMR spectrum = paraffin wax by Bloch and Purcell in 1945
There are many different types of protons, so they decided to observe just one signal which only tells us the technique works, but not for individual protons –need a more sophisticated way of interacting with our spins but need to understand the Larmor precession beforehand

Answer 6

A

Larmor precession = the magnetic moment (μ) associated with a spinning spherical charge will precess in an external magnetic field (Bo)
- Absorption occurs when we match the energy difference i.e when the radiofrequency matches the Larmor frequency
- All protons have a spin angular momentum thus a magnetic field
- Even though the magnetic field is aligned in a direction, the nuclear magnetic moment vector cannot align perfectly parallel thus it is drawn at a tilt
- This is because of quantum mechanics; there is a small angle suspended between the main magnetic field and nuclear magnetic moment so there is still a bit of force invoked between the two; this force is translated into a precession –the magnetic moment will precess around the direction of the main magnetic field
- Precessional frequency is dependent on the interaction between the main magnetic field with the magnetic moment; the stronger forces the faster it will go around
  -Protons have a larger magnetic moment so will precess faster than C13 nuclei

Answer 7

A

Phase = where the spin is on its Larmer precession (x, y, z…)
In the sample at equilibrium, system will have random phases evenly distributed about the Larmer precession cone; each nucleus has a different phase
Magnetic fields can be added as they are vector properties; opposite directions subtract, same direction add
Bulk/net magnetisation vector = all individual magnetic moments with spin added together; aligned along z
Bulk magnetisation (Mo) is aligned with the magnetic field even though the individual ones are not; can visualize the magnetic properties of the whole sample
Once all nuclei are in the ß spin state, the net magnetisation vector is going to spiral away from Bo field into the x-y plane

Answer 8

A

Use a short pulse because a continuous wave that is infinitely long has a single precise frequency associated with it
- Applying a short square pulse truncates the frequency; it excites all of the spins in the sample at the same time
- Can design pulse to just excite specific part of spectrum e.g. amides
- The longer the pulse gets, the reduced range you get until it is infinite thus a single frequency
- If you truncate a pulse (i.e. 10 ms), it no longer has very precise frequencies, rather a long range of frequencies which you exploit to get excitation of the whole sample
- Radiofrequency pulse applied to the x or y -axis (e.g in x-axis)
- Radiofrequency pulse has an oscillating magnetic field associated with it; it is 90º to net magnetization; it will try and rotate net magnetization about the direction of the pulse (e.g. towards the y-axis)
- Calibrate system to leave the bulk magnetization in the y-axis which is what the pulse length will be; in y-plane because it is where the detector is placed to detect
- As soon as the magnetisation gets into the x-y plane, it induces a current in the coil; get a big initial current/signal when the pulse drops into the y-axis
- When the pulse is switched off, the perturbed system magnetic field exerts its force on the bulk magnetization and tries to bring it back to z where it is back at equilibrium
- The pulse is doing two things
  1. Causing transitions between the two energy levels; this is because magnetization on y means you have equal populations of α and ß so the spins have flipped
  2. Making all of the spins become coherent in the x-y plane –this coherence is what is detected; rather than their phases being randomly distributed about the Larmor precession
- All spins have different frequencies at different times; detector measures all the spins at once
- Understanding EM waves:
- Sinusoidal = detector is on y so when magnetization is on x it is 0, then goes to -y, then 0 again
- Signal intensities drop because the whole magnetization is moving away from the x-y plane; they keep decreasing until it is 0, the point at which it is back to z
- Free induction decay (FID) = all of the frequencies of the sample in one wave; decaying radiofrequency signal generated in the receiver coil

Answer 9

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Can find out what frequency range is in the pulse by Fourier pairs
converting time domain into frequency domain with a Fourier transform
the longer the pulse, the narrower frequency range

Answer 10

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Apply a pulse  get FID  get frequency domain
Takes about 1s
Cannot understand what kind of frequencies there are by looking at a complex waveform but when transferred to frequency domain you get highly resolved signals

Answer 11

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Chemical shift (δ)
Integral
Scalar coupling (J)
Relaxation times (T1 & T2)
Dipolar coupling (D)
Nuclear Overhauser effect (NOE)

Answer 12

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Chemical shift is the frequency axis along the bottom
- Tells you what type of proton it is; methyl group in an aliphatic/aromatic environment? electronegative?
- Electrons circulate about the direction of the laboratory magnetic field within its constraints within the orbital causing a small shielding local magnetic field at the nucleus thus reducing the laboratory MF slightly where the energy gap gets smaller and the frequency changes
- Direction of the magnetic field lines outside the nucleus are in the same direction as the laboratory magnetic field; at the nucleus, they are in the opposite direction
- Frequency of nucleus is directly correlated to the local field induced by electron movement; dependent on the chemical environment ie. known as the chemical shift
- Main laboratory magnetic field is very precise; made by manufacturers
- Electronegative groups have a very big effect; they withdraw electrons away from the system to reduce the shielding effect
- Electropositive groups (e.g. Si) push electrons towards the nucleus thus increasing the shielding effect
- TMS = tetramethylsilane (silicon with 4 methyl groups) used as a reference in NMR; has one of the lowest and most shielded chemical shifts pushing electrons towards the methyl group
- In an aromatic/delocalised system, where you have double bonds that are conjugated, the electrons are delocalised over a larger molecular area
- Electrons circulate in π over the whole ring in response to an applied MF and creates a local MF which will have a much larger effect over a wider area
- Deshielded = not shielded anymore because the electronic field is adding to the laboratory MF; deshielding type local field resonate at very high frequencies
- CH in aliphatic environment is at one end of the spectrum (shielded)
- CH in aromatic ring is at the other end (deshielded)
- Hydrophobic core in proteins is filled with 3 major aromatics (Tyr, Phe, Trp) with delocalised electrons; also have lots of other hydrophobics which are packed against the aromatic rings ie. methyl group packed against the face gets a shielding effect; packed against the edge get a deshielding effect thus a downfield signal (left)
- In protein cores, get ring current shifts which are unique signatures for protein structure as it tells you how methyl groups are arranged around aromatic systems
- Peaks in proton NMR spectrum between 6.4 and 8.8 ppm are usually indicators of aromatic rings
- For example, in NRM spectra, you look at the chemical shift
  ie. 1.2ppm; you look it up and it tells you that it is probably a methyl group or a CH2, with no electronegativity
  ie. 2ppm tells you it has some electronegativity around it
  Ie. 4 ppm has a lot of electronegativity
- If we take two protons one near an electronegative atom and one without, and pulse them with an RF pulse, we will get two signals in our FID, one oscillating at a high Larmor frequency (the one that is deshielded, next to electronegative) and one oscillating at a low Larmor frequency (the one that is shielded)

Answer 13

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Area under the peak gives you the relative numbers of nuclei
Integrate spectrum and can read off what the integral is; tells you how many protons there are
CH2 has twice the intensity of CH; CH3 has 3x the intensity of CH
Usually 1, 2, or 3 protons returned by the signal
After the integral, use the chemical shift table:
ie. methyl group at 1.1ppm- it is a purely aliphatic methyl group with no electronegativity near it
ie. CH2 at 4ppm means there must be a nearby electronegative atom directly bound to the C
Integrals work for H1 but not so true for C13

Answer 14

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Helpful in determining chemical structure in NMR
Peaks are not the same: singlet, doublet, triplet, quartet
Adjacent non-equivalent spin ½ nuclei will experience a spin-spin interaction
Scalar coupling is not directly from spin to spin, it is mediated through electrons in bonds (up to 4 bonds)
Scalar coupling is a weak interaction
Two electrons in a bond: one is antiparallel and one parallel to get pairing
Electrons communicate the different possibilities of pairing
(n+1) rule = number of lines in a peak is always +1 than the number of hydrogens on the neighbouring carbon
Coupling pattern tells you what is next to it
For protons to couple, they must be within three bonds of each other
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