1. NMR Nuclear Magnetic Resonance Spectroscopy p. 83
a hydrogen nucleus (a proton) has a charge, spread
over the surface
a spinning charge produces a magnetic moment m
(a vector = direction + magnitude) along rot axis
which produces a magnetic field, just like a bar magnet

2. this moment interacts with an external field, just like a magnet
p. 83
this moment interacts with an external field,
just like a magnet N B0 S
BUT only certain orientations are allowed, which depends upon the nuclear spin quantum number I

3. The magnetic moment, m is proportional to the spin quantum number I
m = g I (h/2p)
where g = Magnetogyric ratio = magnetic moment
angular momentum
is different for each nucleus (a constant)
I is allowed only certain values and these values
can range from mI = –I to +I in integer steps ONLY
Our proton is simple as it has I = ½, so only two values of mI are permitted, +1/2 and -1/2

4. In ‘bar magnet’ terms, these values of mI correspond to
the magnet aligned with and opposed to the external field or B0 =
mI = +1/ /2
However, these two states ONLY have different energy
in a non-zero external field, B0 E B0
DE = hn
B0 = 0
p. 83/84

5. we make a nucleus ‘flip’ its spinB0
DE = hn
so, if we put ‘our proton’ in a strong magnetic field, B0, there are TWO nuclear spin states:
lower energy = aligned with field (mI = +1/2)
higher energy = opposed to field (mI = -1/2)
A transition will occur at a frequency n
this is in the radiofrequency range, ~108Hz (100 MHz)
and this is the basis of the NMR (or MRI) experiment
we make a nucleus ‘flip’ its spin

6. However, we do NOT really have little bar magnets!!!
p. 84
But this frequency, n is the same as what we have
called the ‘flip frequency’
NOTE: n is proportional to B0 – more about that later

7. The spin number, I depends on the isotope
x = number of protons + neutrons = mass x E Y
Y = number of protons = charge = Atomic #
e.g. 11H has mass of 1 and charge of 1
21H = D has mass 2, charge 1
mass 2 = 1 proton + 1 neutron
IF x and y are both even, I = 0, NOT NMR ACTIVE
e.g C O 42He
Common isotopes

8. If charge (y) is odd and mass (x) is even, I = integer
p. 85
If charge (y) is odd and mass (x) is even, I = integer
e.g. 21D and 147N both have I = 1
If mass is odd, I = Integer/2
11H, 136C F P all have I = 1/2 IMPORTANT
3517Cl I = 3/2; O I = 5/2 too difficult for now

9. Number of spin states p. 85
Nuclei have spin states of magnetic quantum number
MI = +I, (I-1), (I-2)...-I
where total number of states is then 2I+1
so if I = ½ , # states = 2, and MI = +1/2 and -1/2
if I = 1, # = 3, MI = +1, 0, -1 case for Deuterium
if I = 3/2, # = 4, MI = +3/2, +1/2, -1/2, -3/2 case for 11B

10. p. 86
I = ½ 1H
I = 3/2
11B

11. This frequency is the Larmor frequency, where:
p. 86 E B0
DE = hn
B0 = 0
This frequency is the Larmor frequency, where:
gB0
n = 2p
so irradiation of nucleus with frequency n causes a transition from lower to upper state which we call
‘flipping the spin’
This frequency depends on B0 (field) and g (nucleus type)

12. Earth’s magnetic field is ca. 0.5 G at surface n = 2p
gB0
Earth’s magnetic field is
ca. 0.5 G at surface
n = 2p
For 1H, g = (radian MHz per Tesla)
Then when B0 = 2.35 Tesla, n = 100MHz
[1 Tesla = 10,000 gauss = 10kGauss]
so for each Tesla of B0, n = MHz
so for each 100 MHz you need 2.35 Tesla of B0
Our largest departmental NMR is referred to as a 500 MHz
instrument because this is the resonance frequency of 1H
at its magnetic field of Tesla (cost ~ 500K$)

13. Magnetogyric ratios nucleus g grel 1H 267.1 1 19F 251.8 0.94
p. 86
Magnetogyric ratios
nucleus g grel
(relative freq to 1H) 1H
19F
31P
11B
13C
29Si 2H
n, MHz

14. For different nuclei: n1 = k g1 B1 n2 = k g2 B2 k = 1/2p n1 g1 B1 = so
So if know five of these we can calculate other but note that B1 = B2 for the same instrument

15. Known by what PROTON frequency they operate at
NMR Instruments
Known by what PROTON frequency they operate at
e.g. we have 300, 360 and 500 MHz instruments
these all can run carbon spectra, and since grel for 13C
= 0.25, these same instruments operate at
75, 90 and 125MHz respectively when running
carbon spectra
[magnetic field strength is essentially fixed for each instrument]
If all protons absorbed at exactly the same frequency n the NMR experiment would NOT be very useful!
gB0
n = 2p

16. Fortunately, this is not the case: electrons have an
p. 87
Fortunately, this is not the case: electrons have an
associated magnetic field too so small differences
in the electronic environment can cause one 1H nucleus
to resonate at a slightly different n than another
gB0
n =
alcohol 2p
Essentially, the applied field is slightly modified by the local electronic environment: n is a very sensitive probe of the local field B and we can measure radiofrequencies very accurately
Bactual = B0 + Blocal

17. e- SHIELDING: local electronic effects on Bnet Bo Binduced
p. 88
SHIELDING: local electronic effects on Bnet e- Bo
Binduced
Electrons in bonds circulate in a magnetic field creating an opposing magnetic field Binduced that cancels part of Bo
More e- means a bigger Binduced and a smaller Bnet
so to reach resonance, you would need to apply a
higher Bo (or, if Bo is fixed, a lower ν is required)

18. p. 88/89
CHEMICAL SHIFT, d
Variations in n caused by local effects are on the order of
1 part in 105 to 107 of the ‘normal’ 1H resonant n
We COULD just quote the absolute resonant frequencies
for a given 1H type BUT:
The differences would be tiny and unwieldy to quote:
e.g vs MHz
b) Absolute frequencies vary linearly with B0:
e.g vs on a 250 vs. 500 MHz
machine

19. p. 88
CHEMICAL SHIFT, d
A better way: measure all resonances RELATIVE to a standard and divide by B0 (or more simply the standard resonance frequency for a given nucleus at a particular B0) to remove the field dependence
The Standard: TMS = tetramethylsilane = Si(CH3)4 = 0
Sample peak n - TMS peak n Hz
d =
= ppm =
Spectrometer n in MHz
MHz
shift in Hz n
TMS
sample

20. The shift from TMS is thus
p. 89
CHCl3 = 7.2 ppm
The shift from TMS is thus
7.2 x spectrometer operating frequency in MHz
e.g. at 100MHz = 7.2 x 100 = 720 Hz
at 300MHz = 7.2 x 300 = 2160 Hz
at 500MHz = 7.2 x 500 = 3600 Hz
Thus the absolute shift depends upon the instrument,
but the chemical shift does not (7.2 ppm)
Higher frequency (larger field) instruments
are more sensitive, and more expensive!

21. SUMMARY
Nuclei have spin I (depends on # protons/neutrons)
Spinning charges = magnetic field
Nuclear spin I quantized: only mI states from +I,..,-I
allowed, each with a different magnetic moment
mI states same energy unless placed in an external
magnetic field:
+ mI (spin aligned with ext. field) lower energy than - mI
NMR experiment probes n required to ‘flip’ spin: cause
nuclear spin state to change (eg. mI = 1/2 to -1/2)
n depends on ext. field strength but also on tiny
variations in electron density (because e- are spinning
charges too) in the locale of the nucleus: we can tell
something about the nuclear environment from n

22. p. 90
Deshielded
Downfield
Low field
TMS
5.7
3.5
7.3
3.7

23. = integration = area under peak is proportional to NUMBER of H’s = 1:3
7.3
3.7
= integration = area under peak
is proportional to NUMBER of H’s = 1:3
it is a RATIO, i.e 1:3 or 2:6 or 3:9 etc

24. = integration = area under peak is proportional to NUMBER of H’s = 2:3
TMS
= integration = area under peak is proportional to NUMBER of H’s = 2:3
it is a RATIO, i.e 2:3 or 4:6 or 6:9 etc

25. p. 91
d 1
d 3.5
size ratio
3:1

26. SPIN-SPIN COUPLING F---PCl2
Both P and F have spin I = ½ (we can ignore Cl here)
high n
low field high field
low n

27. THE COUPLING CONSTANT J
nJXY where n = number of bonds between coupled the nuclei
X = nucleus that is being observed
Y = nucleus that is coupling to X (often the order of XY or YX is arbitrary)
4JFH
3JHH
2JPF

28. d = chemical shift = centre of all lines (ppm)
J = coupling constant = separation of lines (Hz)
s, d, t, q = quartet, pentet, sextet, septet, octet
nonet, decet

29. **cannot see coupling between identical nucleiHA HB
HA=HB
**cannot see coupling between identical nuclei

30. More nuclei – AX2 e.g. PF2Cl
The P can see both fluorines up
(twice as likely) one up, one down
both down
so there are THREE lines,
the middle one has
twice the intensity

31. TREE DIAGRAMS AX2 A A split by X JAX split by X again
Can always construct pattern as a tree, by splitting
one nucleus at a time – when nuclei are same, e.g.
X above, lines fall on top of each other and simplify
pattern

32. Spectrum of A: (2nI + 1) lines 4+1 = 5 lines AX
AX2
AX3
AXn
AX4
Spectrum of A: (2nI + 1) lines
4+1 = 5 lines
I = spin of X; n = number of X atoms
For I = ½ , = n + 1 lines
1:4:6:4:1
intensities

33. Numerically, we can get intensities from Pascal’s triangle:1
1+5 =6 1
5+10
=15
10+10
=20
5+1 =6
10+5
=15 1
so, intensities of a septet are 1:6:15:20:15:6:1

34. Example A p. 95 3 neighbors, so 3+1=4 lines 2 neighbors, so
1:3:3:1
2 neighbors, so
2+1 = 3 lines
Br----CH2-----CH3
1:2:1

35. Example B p. 95 CHCl2—CHCl---CHCl2 2 identical neighbors, so triplet
1 neighbor, so doublet

36. Example C p. 96 If J’s are different, best to draw a tree, e.g. PHFCl
You need to know J’s, assume 1JPF > 1JPH for now
In 31P
Spectrum: P
1JPF
P split by F
1JPH
1JPH
split again by H d
so spectrum is a doublet of doublets

37. Example D p. 96 PH2F P P split by F split again by 2H get a doublet of
triplets

38. For AXYZ draw tree by starting with largest J then next largest J
p. 96
For AXYZ
draw tree by starting with largest J
then next largest J
then smallest J
then lines will cross over each other the least!
Consider CHF2CH3
H is triplet of quartets
H2 is
doublet
13CF3CH2CH3
of quartets
of quartets

39. SIZE OF COUPLING CONSTANTS
Table, manual p. 97
Generally
1J >> 2J >> 3J > 4J > 5J > 6J
Ball park figures!!
200Hz Hz Hz Hz Hz
but there are plenty of exceptions
>CHAHB ~15
=CHAHB~2 *
only P, F,....
1 Hz
0 Hz
* For H-H = 0, except
in pi-bonds
10-18Hz
7Hz

40. Example E p. 98 C5H8F4O C5H8F4O - C5H12O = 0 DBE
Integration = 1:1: H total
Peak at d 6 is 1H t of t
(triplet of triplets)

41. C5H8F4O Integration = 1:1:6 ---8H total so t of t must be caused
p. 98
C5H8F4O
Integration = 1:1: H total
so t of t must be caused
by F’s
arranged 2 + 2
Other H signals
are singlets
There is a large coupling and a
small coupling
Large Coupling ~ 1ppm
=60Hz = 2J = H-C-F2
small coupling ~0.1ppm
=6Hz = 3J = H-C-C-F2
so we have CHF2-CF2- group
not split by anything else

42. Integration = 1:1:6 ---8H total CHF2—CF2--
p. 98
C5H8F4O
Integration = 1:1: H total
CHF2—CF2--
6H’s are identical and are
NOT split by neighbors
we have 3C’s, O, 7H’s
to find
X, Y not H or F otherwise
would see coupling 3J
X—C(CH3)2 Y
One of these must be CHF2CF2

43. Integration = 1:1:6 ---8H total
p. 98
C5H8F4O
CHF2—CF2--
Integration = 1:1: H total
C—C(CH3)2
now groups left to attach
are H and CHF2CF2-- O so
CHF2CF2—C(CH3)2
we will learn later that OH
usually do not couple OH

44. You can now do
ASSIGNMENT 4