NMR Spectroscopy Flashcards

1
Q

Learning outcomes

A
  • Describe how nuclear spins are affected by magnetic field and be able to explain what happens when radio frequency radiation is absorbed
  • Be able to determine the number of proton or hydrogen NMR signals expected from a compound given its structure
  • Be able to calculate the approximate chemical shift of hydrogens in a compound, and be able to assign peaks in the NMR spectrum to specific H in a compound
  • Be able to determine the splitting pattern of peaks in a H NMR spectrum of a compound given its structure
  • Be able to interpret NMR spectra integrations
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2
Q

What are the three sequential step principles of NMR

A

The principle of NMR usually involves three sequential steps:

  1. The alignment (polarization) of the magnetic nuclear spins in an applied, constant magnetic field B0. (The nuclear magnetic moment is the magnetic moment of an atomic nucleus and arises from the spin of the protons and neutrons. It is mainly a magnetic dipole moment; the quadrupole moment does cause some small shifts in the hyperfine structure as well.)
  2. The perturbation of this alignment of the nuclear spins by a weak oscillating magnetic field, usually referred to as a radio frequency (RF) pulse. The oscillation frequency required for significant perturbation is dependent upon the static magnetic field (B0) and the nuclei of observation.
  3. The detection of the NMR signal during or after the RF pulse, due to the voltage induced in a detection coil by precession of the nuclear spins around B0. After an RF pulse, precession usually occurs with the nuclei’s intrinsic Larmor frequency and, in itself, does not involve transitions between spin states or energy levels.[
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3
Q

What are some NMR applications?

A
  • Covalent structure elucidation- analytical chemistry
  • Dynamic processes- reaction kinetics, equilibria
  • Structural studies- proteins, protein-ligand, nucleic acids, carbohydrates
  • Metabolomics- urine, plasma, CSF, tissue extracts, plant extracts etc
  • Drug design- structure activity relationships
  • Medicine- MRI
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4
Q

What technique would you use to identify between these compounds given the information given?

A

NMR

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5
Q

NMR related nobel prize winners

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6
Q

What type of information does NMR give?

A

Nuclear magnetic resonance (NMR) spectroscopy may be the most powerful method of gaining structural information about organic compounds

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7
Q

What does NMR involve?

A
  • NMR involves an interaction between electromagnetic radiation (radio frequencies) and the nuclei of atoms whilst inside a magnetic field
  • Here we will focus on hydrogen H although other nuclei such as C, N, F and P can also be used, but are less important
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8
Q

What can the structure (connectivity) of a molecule affect?

A
  • The structure (connectivity) of a molecule affects how the radiation interactions with and between each magnetically active nucleus in a molecule
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9
Q

What must there be for NMR to work correctly?

A

A magnetic field

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10
Q

How do the protons and neurons behave within the nucleus?

A

As if they are spinning

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11
Q

If the atomic number is ODD or the total number of neutrons and protons is ODD what will the atoms have? What are some examples?

A
  • If the atomic number is ODD or the total number of neurons and protons is an ODD number, the atoms will have a net nuclear spin
  • Examples: 1H,13C,15N, 19F and 31P
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12
Q

What does the spinning charge in the nucleus create?

A

A magnetic moment

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13
Q

If the atomic number is** EVEN** and the number of neurons is also EVEN then what is seen? Give some examples

A
  • If the atomic number is EVEN and the number of neurons is also EVEN then there is no net nuclear spin, no magnetic moment, and no NMR effect
  • Examples: 12C, 16O and 18O
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14
Q

What about the following nuclei? 2H (deuterium), 17O, 14N

A
  • What about the following nuclei? 2H (deuterium), 17O, 14N have a net nuclear spin and an NMR effect, but they are quadrupolar nuclei with a very fast relaxation
  • Quadrupolar= fast relaxations
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15
Q

When is the magnetic moment produced?

A

During spin

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16
Q

Whats the arrangements of the magnetic moment in individual nuclei and when exposed to an external magnetic field?

A
  • The magnetic moments of individual nuclei are randomly arranged, but when exposed to an external magnetic field, these magnetic moments will align both with and against the external field
  • Magnetic field applied to a B0
  • Alpha and beta arrangements. Beta is aligned against the magnetic field and alpha is aligned with the field.
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17
Q

As the magnetic field increases what does as well?

A
  • No energy difference when field is 0
  • Energy gap increases as magnetic field does
  • More in ground state than excited state when magnetic field is greater. The greater the field the greater the sensitivity
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18
Q

Tell me about nuclear shielding and what causes it

A
  • The magnetic moment of the electrons produces a magnetic field around the nucleus and opposes the effect of the external field
  • The magnetic field “felt” at the nucleus is therefore not the same as the applied field
  • This effect is known as **shielding **
  • Different nuclei will be shielding to different extents
  • The more shielding a nucleus is with electron density, the smaller the alpha to beta energy gap
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19
Q

How/ Why does chemical shift occur. Use this example with ethanol

A
  • O is electron negative and pulls electron density and pulls electron away from C. C then pulls electrons away from H as a result. More than those H away from the C.
  • Chemical shift axis shown above, a frequency axis measure in parts per million
  • Si is an electropositive group, pushing e- in and making more positive
  • OH not split because H are stable but still exchange (labile?)

more electronegative effect shifts left

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20
Q

What are the effects of halides on chemical shift?

A
  • Chlorine is an electronegative atom
  • Chlorine Deshields the proton
  • It takes electron density away from carbon its bonded to
  • Which in turn takes more electron density from hydrogen
  • Proton is de-shielded- higher frequency in the NMR spectrum
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21
Q

Whats the effects of Pi electrons on chemical shift?

A
  • Pi electrons oppose magnetic field
  • H more than B0
  • Around 7 on chemical shift axis when a benzene ring is present
  • The H reinforces the magnetic field on C=C as well as in benzene ring
22
Q

How can you estimate chemical shifts?

A
23
Q

Proton NMR correlation table

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24
Q

When measuring chemical shift what are some references which can be used?

A
  • Ideally, we would like to measure the chemical shift relative to a proton with no electron around it
  • This is impractical
  • We can use a reference compound that come to one end of the spectrum and measure all signals relative to the reference
  • Two common reference compounds are tetramethyl silane (TMS) for chloroform solutions, and** sodium d4- trimethylsilylpropionic acid (TSP) for aqueous solutions
    **
25
Q

How is reference values for chemical shifts used?

A
  • So, data can be reproducible then a reference is used which in this case is TSP
  • Then separation from TSP from reference value is measured to work out difference
  • Divide it by reference frequency which in this case is 127,000,000
  • Change to ppm units
  • Each ppm corresponds to Frequency of spectrometer (from this graph)
  • Now have a scale irrespective of field strength
26
Q

Typical set up of an NMR lab

A
27
Q

Block diagram of an NMR system

A
  • High power, fast (about 20 millionths of a second)
  • Weak signal so needs an amplifier
  • Temperature of sample affects results so NMR are temperature controlled
28
Q

How does one acquire a 1H NMR spectrum and what is produced?

A
29
Q

Example of a real NMR FID and spectrum of valine dissolved in D2O

A
30
Q

When collecting NMR data what is presented and what does it represent?

A
  • NMR spectroscopy as a technique is generally insensitive
  • The signal-to-noise ratio (SNR) measures how easily peaks can be distinguished from the noise
  • To increase the SNR, we repeatedly pulse with RF and collect the FIDs and add them together
31
Q

Chemical structures

A
32
Q

What are homotopic protons?

A
  • We need to consider if any protons are equivalent as we need to count up how many proton environments we have
  • We can substitute a hydrogen for a deuterium and check if a chiral centre has been produced
  • The substitution does not produce a stero centre i.e., it remains achiral
  • These protons are called homotropic
  • Homotropic atoms or groups have identical chemical shifts under all conditions
  • Chiral centre= mirror image will not superimpose (structures look identical but mirror)
  • Achiral= Hydrogens are equivalent. If H substituted with deuterium, then the molecule is not chiral
33
Q

What are enantiomeric groups

A
  • H atoms on C3 are diastereotopic
  • Substitution with D, creates a second chiral centre
  • Because there is already a chiral centre in the molecule at C2, diasterotopic
  • Chiral centre effects chemical shifts in spectrum so make sure to make note if present
  • Compounds are different so produce different chemical shifts
  • Count number of environments there are (difference could indicate a chiral centre effecting shifts)
34
Q

How many environments in ethanol?

A
35
Q

How many environments for ethylbenzene?

A
36
Q

Multiplicity results from magnetic effects that protons have on each other, consider Ha and Hb attached to ethane how will they be aligned? How would this appearkon the NMR

A
  • Multiplicity results from magnetic effects that protons have on each other, consider Ha and Hb
    * Proton spins align with and against an external magnetic field
  • Hb will be aligned with the magnetic field in some of the molecules and against the field in other molecules
  • Some Hb atoms have a slightly shielding effect on Ha and others have a slight de-shielding effect
  • This makes the NMR spectrum appear to be split into 2 lines- a doublet
  • H/proton spins can align with or against the field (Ha and Hb)
  • Hb: Against is higher energy than with the field
    About half of molecules in sample are with the field and half against. As practically 50:50 then the signal appears as a doublet
    Same affect seen for Ha as Hb
37
Q

How would this molecule appear on the NMR spectrum and why

A
  • Consider an example where there are two protons on the adjacent carbon
  • There are three possible affects the Hb protons have on Ha
  • Both spins align against the field, both spins align with the field, or one spin aligns with the field and 1 spin aligns against the field (2orentiations)
  • The produces an NMR peak for Ha that is a triplet in the ratio 1:2:1
  • Ha will affect Hb as well. Ha splits Hb into 2 and Hb splits Ha into 3
38
Q

How does this appear on the NMR spectrum and why

A
  • Taking this idea one stage further by considering a proton Ha now adjacent to a methyl group of 3 equivalent Hb protons
  • This gives rise to a quartet peak with a ratio of 1:3:3:1
39
Q

What can you determine by analysing the splitting pattern of a signal in the 1H NMR spectrum?

A
  • By analysing the splitting pattern of a signal in the 1H NMR spectrum, you can determine the number of equivalent protons on adjacent carbons
  • This is a simplified splitting
40
Q

What are the different types of splitting patterns seen in an 1H NMR spectrum?

The splitting or coupling sizes are also known as what and how are they measured?

A
  • The splitting patterns we see often in the 1H NMR spectrum are shown below
  • The splitting or coupling sizes, also known as J-coupling or spin-spin coupling are measured in Hz
41
Q
  • If we now go back to our spectrum of ethanol, the splitting patterns for -CH2 and -CH3 now become clear
  • However, the -OH has remained as a single peak- why is this?
A
  • The -OH group is liable, in that the -H can easily exchange with any residual water H’s, and this happens at a fast rate which eliminates any coupling, so the signal remains as a singlet. We see the same effect for -NH2 and -NH-R
  • If the solvent used is deuterium oxide (D2O) instead of deuteron-chloroform (CDCl3) then we would not see any signal at all from the exchangeable -H’s as they would be replaced with -D’s
42
Q

How do we measure coupling constants?

A
  • you can see in the figure that we measure coupling constants as the separation between peaks
  • this is the coupling constant or J-value is the distance between peaks of a splitting pattern measured in Hz
  • the splitting on one signal is the same as the splitting on the coupling group
  • The coupling constant will be constant even if an NMR instrument with a stronger or weaker magnetic field is used
43
Q

What are coupling constants measured in and what can they be useful in determining?

A
  • Coupling constants are measured in Hz. Values can be useful in determining a chemical structure. They are a through bond effect
44
Q

What are some coupling constants, their ranges what it represents

A
  • The number before the “J” indicated how many bonds the coupling is observed through
  • For example, 2JHH means that the coupling is from one proton to another and through 2 bonds i.e., they are attached to the same carbon. Previously we have seen 3JHH couplings in the case of ethanol, where the value is about 7 Hz
  • Approximate range of values are as follows
    o **2JHH -9 to -21 Hz
    o 3JHH 0 to 10 Hz
    o 4JHH 0 to 2 Hz
    o 5JHH roughly 0 Hz **
  • Proton coupling constants can vary quite widely as a function of structure
  • Generally the values reduce the more bonds the coupling goes through
  • function of separation of two H
45
Q

What are some characteristics of an 1H NMR spectrum?

A
  • An NMR spectrum contain a lot of structural information
    **1. Look for chiral centre first
    2. Number of signals
    3. Signal location-chemical shifts
    4. Signal shape- coupling or splitting pattern
    5. Signal area- integration
    **
46
Q

What is one of the most useful features of an NMR spectrum?
What does it represent?

A
  • One of the most useful features on an NMR spectrum is the **peak areas are quantitative and relative to the number of protons giving rise to a signal **
  • The NMR software will calculate the area of each peak represented by a step curve
  • The curve heights represent the peak areas or integration curve
47
Q

How are the integrations values of the peaks calculated?

A
  • We set the smallest integration value i.e., 27.0 to 1 by dividing by 27.0
  • Then calculate the ratio of the other peaks i.e., divide all integrations by 27.0
  • This gives 2 integral values at about 1.0. However, 2 of the integration values are about 1.5 and we should have whole number as we have whole numbers of protons
48
Q

How are integer peak integration values calculated?

A
  • We now have 4 signals with approximately whole number integrals 2:3:2:3
  • The peak at 0.92 ppm with an integral of 3 is a methyl group and the splitting in the same as the peat at 1.6ppm which is a CH2.
  • We would expect a quartet splitting from a methyl group and we see more splitting i.e., 6 peaks, and this is due to the splitting with the CH2 group at 2.4 ppm.
  • It has a higher frequency than the CH2 at 1.6 and is attached to O
  • The singlet peak at 2.14 is a methyl group attached to O
49
Q

Peak integration

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50
Q

How many peaks would you expect to see for the following compounds

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51
Q

Fill in the boxes

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52
Q

Fill in the boxes

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