NMR

1. MC-511, Part 3
Nuclear Magnetic Resonance (NMR)
Partial lecture notes
1

2. Suggested reading material…
• Spectrometric identification of organic compounds by
Silverstein, R.M.; et al (the Bible of spectroscopy, with some
good problems at the end).
• Modern NMR spectroscopy, a Guide for Chemists, by Snyder (a
highly advanced text book, but very helpful).
• http://orgchem.colorado.edu/hndbksupport/spect.html - a website
run by University of Colorado, which gives detailed theory and
some good problems in NMR and IR.
• http://arrhenius.rider.edu/nmr/nmr_tutor/selftests/h1/h1_fs_soln.h
tml - more problems.
• http://www.chem.ucla.edu/cgi-bin/webspectra.cgi - some more
problems!
Note: This list will be updated! 2

3. Spin I…
• Electrons – spin about their own axes
– Spin quantum number of + ½ or – ½.
– Effect of electron spin – magnetic moment, also called a magnetic
dipole (direction?)
– Remember – a charged body spinning about its own axis generates a
magnetic dipole (moment) along its axis.
• Similarly, some nuclei (not all!) have spin!
– Examples: 1H, 13C, 19F, 31P, etc.
– Nucleus – positive charge; hence spinning charge generates a magnetic
dipole.
– Hence, each nucleus acts as a tiny magnet.
3

4. Spin II…
• Spin (nuclear or electronic) determined by the spin quantum
number, S.
– The rules for determining net spin:
• If # of neutrons and protons are even – no spin.
• If # of neutrons + protons is odd, the nucleus has a half integer spin
(1/2, 3/2, 5/2…)
• If # of neutrons + protons is even, the nucleus has an integer spin
(1, 2, 3…)
– Number of spin states (or overall spin) given by formula (2S + 1).
– Examples: consider a hydrogen atom, 1H.
• Only one proton in the nucleus
• Hence, sum of protons and neutrons = 1
• Nuclear spin = 1/2.
• Number of spin states = 2(1/2) + 1
• +1/2 and -1/2
4

5. Spin III
– For 2H, S = 1; # of spin states = 3
– Nuclei with S = 0 are not NMR active (examples, 12C, 18F, 18O, etc.)
• Spin states:
– A nucleus of spin ½, can have two possible orientations.
– In absence of a magnetic field, these orientations will be of equal
energy.
– In a magnetic field, the magnetic moment created by the spinning
charge can line up with or against the field.
– Alignment with the magnetic field is a lower energy state ( state) than
against ( state).
– Difference in the energy between these two states depends on the
strength of the applied magnetic field.
– Population of energy levels governed by Boltzmann distribution: there
is always a finite excess of nuclei in the lower energy state than in the
higher energy state.
5

6. NMR phenomenon I…
• Imagine a nucleus of spin ½ in a magnetic field, in the lower
energy level.
• In a magnetic field, the axis of rotation of the nucleus will
precess around the magnetic field.
– Precess – change in orientation of the rotation axis of a rotating body.
6

7. NMR phenomenon II…
• If energy is now absorbed by the nucleus, the magnetic moment
is now „flipped‟ so that it now opposes the applied field (higher
energy state) – resonance!
• This absorbed energy depends on the applied magnetic field –
quantized!
7

8. Transition energy I…
• Magnetic moment of the nucleus is proportional to its spin, S.
– Where, = magnetic moment,
= „magnetogyric‟ or „gyromagnetic‟ ratio, a
fundamental nuclear constant, dependent on nucleus.
h = Planck‟s constant
• Energy of a particular energy level is;
B
– Where, B is strength of magnetic field at the nucleus (not equal to the
applied field).
8

9. Transition energy II…
• The difference in energy levels (transition energy) can be
obtained from :
• Features of the equation:
– If magnetic field, B increases, E increases.
– Greater the , greater E.
– is the ratio of magnetic moment to angular momentum.
– If increases, precession increases.
– Hence, greater energy required for the „flip‟.
9

10. Relaxation I…
• Only a small proportion of nuclei in the state can get excited
and absorb radiation.
• At some point, the population in the and states become equal.
• No further absorption of radiation – saturated spin system.
• Relaxation – return of nuclei to the lower energy state.
– Spin-lattice relaxation:
• The NMR sample – called as lattice.
• Nuclei in the lattice are in vibrational and rotational motions, giving
rise to magnetic field, called lattice field.
• If nuclear precession frequency is equal in phase and frequency to
lattice field, the nucleus in the state can transfer its energy to
lattice and return to state.
• Results in a slight warming of the sample.
10

11. Relaxation II…
– Spin-spin relaxation:
– Interact with neighboring nuclei with identical precession frequencies.
– However, nuclei in both states can interact!
– No net change in populations, but lifetime of a nucleus in the state
will decrease – line broadening in the spectrum – not good!
• Relaxation time T1:
– Average lifetime of nuclei in the higher energy state.
– Depends on of the nucleus and mobility of the lattice.
– As mobility increases, vibrational and rotational frequencies increase,
increasing the probability of interaction with excited nuclei.
11

12. Chemical shifts I…
• Magnetic field at nucleus is not equal to the applied field.
• Electrons around the nucleus shield it from the applied field.
• Nuclear shielding – difference between applied magnetic field
and field at the nucleus.
• Consider s-electrons in a molecule:
– Symmetry?
– Circulate in the applied field – produce a magnetic field, which opposes
the applied field.
– Applied field strength must increase, for the nucleus to resonate.
– Upfield shift, also called as diamagnetic shift.
12

13. Chemical shifts II…
– If the electron density around the nucleus is reduced considerably
(how?), applied field strength must decrease for resonance.
– Nuclear deshielding, also called downfield shift.
– Electrons in p-orbitals have no spherical symmetry.
– They produce comparatively large magnetic fields at the nucleus.
– Deshielding or paramagnetic shift.
• In proton NMR, p-orbitals play no part (why?).
– Small range of chemical shift (10 ppm) observed.
– Effect of s-electrons on chemical shift – look at substituted methanes.
– CH3X – as X becomes more electronegative, what happens to:
• Shielding?
• Chemical shift?
13

14. Chemical shift IV…
• If two scientists want to compare data using two different field
strengths, that correction has to be applied.
• Hence, chemical shifts!
• Definition: nuclear shielding in an applied magnetic field.
• A function of the nucleus and its environment.
• Measured relative to a reference compound.
• For 1H and 13C NMR, usually use TMS (Me4Si) as internal
standard.
• We will see later about use of NMR solvents as internal
standards.
14

15. Chemical shift VI…
• Information on what kinds of protons are present in the molecule.
– Aromatic, aliphatic, primary, secondary, tertiary, vinylic, allylic,
benzylic, acetylenic, adjacent to halogens or hetero atoms, etc.
• Shifts observed due to shielding and deshielding of nuclei.
• Circulation of electrons nearby can generate induced magnetic
field (what kind of compounds?) that can either reinforce or
oppose the applied magnetic field:
– When induced magnetic field opposes – shielding.
– Shielded protons require higher applied field for resonance – upfield
shift.
– When induced magnetic field reinforces – deshielding.
– Deshielded protons require lower applied field for resonance –
downfield shift.
15

16. Measurement of chemical shift…
• Denoted by , units are ppm.
• Reference material: TMS (tetramethylsilane)
• Why TMS?
– Inert, wouldn‟t react with the sample!
– All equivalent protons, so only one signal.
– Volatile (b.p. = 27 0C).
– Soluble in common organic solvents.
– Can be used as an external standard when using D2O as solvent.
– Low electronegativity of silicon, hence highly shielded protons.
– Most compounds absorb downfield to TMS.
– Large number protons means only a drop of standard is required.
16

17. Calculation of chemical shift…
• Denoted by units are ppm.
• If a proton absorbs at 60 Hz in a 60 MHz instrument,
= 1 ppm
• is independent of operating frequency of instrument. The same
signal above will absorb at 100 Hz in a 100 MHz instrument.
Thus,
= 1 ppm
• Remember, depends on the environment of the nucleus!
• All these calculations are done by the computer.
17

18. Magnetic anisotropy I…
• Anisotropy – non-uniform.
• Non-uniform magnetic field.
• Recall:
– Circulation of electrons nearby can generate induced magnetic fields
that can either reinforce or oppose the applied magnetic field
• Nearby protons can experience 3 fields:
– Applied field
– Shielding field of the valence electrons
– Field due to the system
• Depending on the position in the third field, the proton can be:
– Shielded (smaller )
– Deshielded (larger )
18

19. Magnetic anisotropy II…
• Acetylene:
– Shape of molecule?
– Triple bond symmetrical about the axis.
– If axis is aligned with the magnetic field, electrons of the triple bond
circulate perpendicular to applied field.
– Induce their own magnetic field, opposing the applied field.
– Protons lie along the magnetic axis – the induced field shields them.
– Hence, for acetylenic protons are more upfield than expected.
19

20. Magnetic anisotropy III…
• Now, let‟s look at benzene:
– “Ring current effect” in play here.
Field lines aligned with
applied field
– So, aromatic protons are deshielded, more downfield then expected and
hence larger .
– A proton held directly above or below the ring would be heavily
shielded.
20

21. Magnetic anisotropy IV…
• Finally, let‟s look at ethylene:
– Double bond oriented perpendicular to the applied field.
– electrons circulating at right angles.
– Induced magnetic field lines are parallel to the external field at the
location of the alkene protons.
– Hence, downfield shift.
21

22. Number of signals…
• Indicates the kinds of protons in the molecule.
• Equivalent protons: protons in the same environment.
• Non-equivalent protons: protons in different environments.
• Example and some problems:
Cl H
* * Cl
H
O
F O
22

23. Kinds of protons I…
• Homotopic protons:
– Proton, when substituted by a deuterium, leads to the same structure.
– Always equivalent, and will give one signal in the NMR spectrum.
Ha D Ha
F F F
Hb F Hb F D F
Replace Ha Replace Hb
Same compounds, hence Ha
and Hb are homotopic.
23

24. Types of protons II…
• Enantiotopic protons:
– Proton, when substituted by a deuterium, leads to a pair of enantiomeric
structures.
– Appear to be equivalent and usually, give one signal.
– In a chiral environment, can be made non-equivalent and give two
signals.
Ha D Ha
F F F
Hb Cl Hb Cl D Cl
Replace Ha Replace Hb
Enantiomers, hence Ha and Hb
are enantiotopic.
24

25. Kinds of protons III…
• Diastereotopic protons:
– Proton, when substituted by deuterium, leads to a pair of diastereomeric
structures.
– Not equivalent, and usually, give two signals in the spectrum.
Ha D Ha
F F F
Hb R* Hb R* D R*
R* is a chiral center, which Replace Ha Replace Hb
undergoes no change.
Diastereomers, hence Ha and
Hb are diastereotopic.
25

26. Splitting I…
• Just chemical shift information alone wouldn‟t be useful.
• Splitting of peaks is what adds extra value to NMR.
• Splitting due to 1H-1H coupling, also called spin-spin coupling or
J coupling.
• How does it work?
– Imagine a molecule with two different protons, HA and HB.
– How many signals would you expect?
– HA feels the presence of HB and vice-versa.
– Recollect, these protons are tiny magnets, oriented with or against the
applied magnetic field.
– When HB reinforces the magnetic field, HA feels a slightly stronger
field; when it opposes the applied field, HA feels a slightly weaker field.
– So, we see two signals for HA.
– Same can be applied for HB.
26

27. Splitting II…
• How does it work?
– Overall, we see two „doublets‟ for the two kinds of protons.
– So, when there is only one proton adjacent, we see 2 peaks due to that
proton.
For this line, H B is lined up For this line, H B is lined up
w ith the magnetic f ield against the magnetic f ield
(adds to the overall (subtracts from the overall
magnetic field, so the line magnetic field, so the line
comes at higher frequency) comes at low er f requency) HA HB
C C
HA HB
HA is split into tw o lines because HB is split into tw o lines because
it f eels the magnetic field of H B. it f eels the magnetic field of H A.
27

28. Splitting III…
• When there is more than one proton in the neighboring carbon?
– More lines!
– Consider Cl2CHCH2Cl.
– Look at CH – it „feels‟ the two protons from CH2.
– The 2 protons:
• Both are aligned with the field
• Both oppose the field
• One proton is aligned and the other is against the field
• The reverse of the above case
– Because the two protons in CH2 are the same, the last two cases add up.
– Hence, CH has three lines in the ratio of 1:2:1 (triplet).
28

29. Splitting IV…
Applied
field
Signal from
uncoupled proton
Ho
Spin combinations for
adjacent -- CH2
Spin-spin coupling: coupling with two protons give a 1 : 2 : 1 triplet.
29

30. Splitting V…
• Generally, the “n+1” rule is followed.
• If there are “n” (equivalent) protons in the neighboring carbon(s),
the proton of interest will be split into n+1 peaks.
• Intensity of the lines is dictated by “Pascal‟s triangle”.
• Example:
– Doublet – 1:1
– Triplet – 1:2:1
– Quartet – 1:3:3:1
– Pentet – 1:4:6:4:1
• Applies only to simple systems!
• Most „real world‟ systems are much more complex!
30

31. Pascal‟s triangle…
31

32. NMR spectrum of ethanol…
32

33. NMR spectrum of ethyl acetate…
H O
H H H
H O H
H H
33

34. Coupling constants I…
• The line separation within a given multiplet is the coupling
constant.
• Measure of interaction between a pair of protons – structural
information!
• Indicated by J.
• Units of J is Hz, and is magnetic field independent.
• Example:
34

35. Coupling…
• Interaction between „related‟ protons.
• Three types:
• Vicinal coupling
• Geminal coupling
• Long range coupling
35

36. Vicinal coupling…
• Denoted as 3J.
• Coupling transmitted through three bonds.
• Magnitude depends on the dihedral angle:
– Maximum (about 16 Hz) at 180 0
– About 10 Hz at 0 0
– Minimum (close to 0 Hz) at 90 0.
– To get the dihedral angle, draw the Newman structures and compute the
angle.
– Staggered conformation – 60 or 180 0; eclipsed – 0 0; Gauche - ~ 60 0.
36

37. Vicinal coupling II…
CH3 H CH3CH3
H H H
H H
H H H
H
H H H
H CH3 HH
H
H HH
Br H H H
H Br Br Br
37

38. Karplus curve…
• Variation of coupling constant with change in dihedral angle.
• Remember: these values are approximate!
38

39. Vicinal coupling – problem…
• Based on the Karplus curve, predict the approximate coupling
constants of the indicated protons in the following molecules:
– Trans-1,2-dimethylcyclohexane. Assume that the two methyl groups
are axial. Predict 3J between hydrogens on C1 and C2.
– Trans-1,2-dimethylcyclohexane. Assume that the two methyl groups
are equatorial. Predict 3J between hydrogens on C1 and C2
– Cis-2-butene – 3J between hydrogens on C2 and C3.
– Trans-2-butene – 3J between hydrogens on C2 and C3
39

40. Geminal coupling…
• Denoted as 2J.
• Coupling between protons on the same carbon.
• Note: these two protons must be non-equivalent!
• Again, value of J depends on H-C-H coupling.
• Normal values – about 10 – 18 Hz; at about 125 0, 2J = 0;
maximum at about 100 0, 2J = 35 Hz.
• Particularly important in terminal vinyl systems.
a
H
b
H
40

41. Long range coupling…
• Coupling beyond three bonds (> 3J).
• Normally, observed up to 4 – 5 bonds.
• With polyalkynes, this can be observed as far as 9 bonds!
• Typical coupling constants are in the range 0 – 4 Hz.
• Two types:
– Allylic coupling
– W coupling
41

42. Allylic coupling…
c
H
a
b H
H a
H a
H
• Why are Hb and Hc non-equivalent?
– 4Jab = 3 Hz
– 4Jac = 3.5 Hz
42

43. 'W' coupling…
a
H H
a
H a
H
b
b
O H H
H b
H
4J
ab (meta) = 1 – 3 Hz 4J
ab = 0 – 2 Hz Bicyclo[2.2.1]hexane
4J = 7 Hz
ab
43

44. More couplings…
• Consider 1,1,2-trichloropropane:
H Cl Cl
H Cl
H H H
– Look at the proton on C2.
– Expected splitting pattern is pentet (why?).
– Can get a pentet, only if J1-2 and J2-3 are identical.
– Actually, get a quartet of doublet (why?).
– The larger coupling is mentioned first.
– Denoted as: 4.30, 1H, J = 6.6, 3.8 Hz.
44

45. 1,1,2-trichloropropane…
45

46. Coupling tree…
46

47. Another example…
• Consider trimethylsilyl ethylene:
SiMe 3 H
H H
– Ignore the methyls in the silyl group.
– How many kinds of protons in the double bond?
– How are they split?
– Approximate coupling constants?
47

48. Integration…
• The area under each peak is obtained by integration of the signal.
• Proportional to the number of hydrogen nuclei giving rise to the
signal.
• Sometimes, integral shown as a step function at the top of each
peak, with the height proportional to the area.
• Error in integration can be high – up to 10 %; depends upon
instrument optimization.
• Usually, all integration done by the instrument / computer.
• Normalized values are shown.
• Integration gives a measure of the proton count, adjusted for
molecular symmetry.
48

49. Molecular symmetry I…
• Consider the spectrum of 2-butanone:
– Symmetry?
– Can get the actual proton count: 3 + 2 + 3.
49

50. Molecular symmetry II…
• Consider diethyl ether, CH3CH2OCH2CH3, a total of 10 protons.
– Symmetry?
– Two peaks in the ratios 3 : 2.
50

51. Leaning of peaks I…
• Consider ethanol (again!)
51

52. Leaning of peaks II…
• Triplet not a „perfect‟ triplet; quartet not a „perfect‟ quartet.
• Coupled peaks lean towards each other.
• Sometimes, helpful in complex systems.
52

53. Advantage of higher field I…
• Separation of different sets of protons is proportional to field
strength.
• However, coupling constants do not change!
• Consider spectrum of benzyl alcohol, recorded at 90 and 400
MHz.
• 90 MHz:
– A broad strong signal at d = 7.24 ppm; characteristic of aromatic
protons.
– Chemical shifts of the 5 protons are identical; no spin coupling is
observed.
• 400 MHz:
– The aromatic peaks are more dispersed.
– Spin coupling of adjacent protons are now seen.
53

54. Advantage of higher field II…
http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/nmr/nmr2.htm 54

55. Advantage of higher field III…
• In an instrument of field strength X MHz, the distance between
two units on the scale equals X Hz.
• Example:
– In a 90 MHz instrument, this difference is 90 Hz.
– In a 400 MHz instrument, this difference is 400 Hz.
• However, J and remain the same!
• Hence, greater field strength translates to greater dispersion.
– Dispersion – resonances with different chemical shifts are further apart.
55

56. Structural elucidation…
• To determine structure – a suggested approach:
– Usually, molecular formula, IR, NMR and MS information will be
given.
– For now, only molecular formula and NMR!
– Calculate the degree of unsaturation from the molecular formula.
– Look at the NMR spectrum to determine the connectivity.
– Draw some possible structures and see if they “work” with the data
given.
– Approach this as a jigsaw puzzle, where you have all the pieces of
information – just need to put them together in the correct order!
– It is a lot of fun!
56

57. Degree of unsaturation I…
• Also known as index of hydrogen deficiency.
• Can determine the number of rings, double or triple bonds…
• Doesn‟t give the exact number rings / double / triple bonds.
• Sum of number of rings and double bonds + twice the number of
triple bonds.
• Formula:
– where,
• H = # hydrogens
• X = # halogens
• N = # nitrogens
57