2. Unit One
Part 9
nmr spectroscopy
equivalent hydrogens
integral vs. number of hydrogens
how do we know all
this info about
molecules shape,
interactions etc?
3. Part 9
Unit One
nmr spectroscopy
equivalent hydrogens
integral vs. number of hydrogens
well a lot of our knowledge comes from
looking at molecules using various forms
of spectroscopy. Today we’ll look at nmr
and next lecture ir and mass
spectrometry...
4. ENERGY
electromagnetic
radiation describes a
form of energy...
5. ENERGY
it is a continuum of energy
that is arbitrarily divided into a
number of regions (the most
obvious being visible light)
6. ENERGY
low energy EM radiation has a long
wavelength and low frequency (ask
the physicists) whilst high energy is
small wavelength and high
frequency
7. can electromagnetic
radiation interact with
molecules?
a stupid question...you
know what the answer
must be otherwise why
have I introduced this
topic?
13. different
molecules
interact
...water in clouds blocks infrared
energy (which makes us hot) but
with
does not block ultraviolet energy
different
(which burns us)
energy
15. E2
if we take a molecule in its
resting state (low energy or
happily sitting around
chilling) and...
E1
molecule in energy
state E1
16. ...irradiate the molecule
with energy some will be
absorbed causing the
molecule...
E2
absorption of
energy
E1
molecule in energy
state E1
17. ...to become excited or to
rise to a higher energy
level...
E2
absorption of
energy
E1
molecule in energy
state E1
18. molecule in energy
state E2 or ‘excited’
the amount absorbed E2
will depend on what we
have changed within the
molecule...we can
measure this...
E1
19. molecule in energy
state E2 or ‘excited’
eventually the molecule will relax
E2
and emit energy as it returns to
its resting state...we can
measure this energy as well...
emission of
energy
E1
20. ...so, by measuring the energy
absorbed or emitted we can
‘see’ changes in the molecule.
So, what can we see?
what part of the
molecule is effected?
22. ultraviolet-visible (UV)
high energy UV radiation
can excite an electron between orbitals
...this tells us about the conjugation within
the molecule (multiple bonds separated
UV by one single bond)
excites an electron
24. infrared (IR)
IR has slightly less energy and
alters the vibration of bonds within
a molecule...this can tell us about
the functional groups within that
molecule
vibrates bonds
O
H
26. nuclear magnetic resonance
(NMR)
NMR uses low energy radio waves
to alter the spin of the nucleus of certain atoms...
this can give us a lot of useful information about
the position of C and H within the molecule. It is
by far the most useful form of spectroscopy we
routinely use
C-H framework
H H H H
H C C H
C O C
H H H H
30. this beast is an MRI...it is
identical to an NMR machine
except it spins a magnet
around a patient and an NMR
spins the sample...
31. nobody likes the
I guess it’s called MRI and not
word ‘nuclear’
NMR as no patient would allow a
doctor to insert them in a
machine with the word “nuclear”
in the title
32. so what is nmr?
...but seriously...what
does an NMR (or MRI)
do?
33. hydrogen
a hydrogen atom comprises of
one proton and one electron. for
nmr we are primarily interested
with the single proton of the
nucleus...
+ve
–ve
atom
34. hydrogen
the positively charged proton is
spinning and this creates a
magnetic field (is it the left or
right-handed rule?)
nucleus spins
35. sample
in a molecule (or collection of molecules)
we have lots of protons (or hydrogens) all
spinning in various directions...but if we
apply a magnetic field to the sample...
random orientation
36. sample β-spin
state
MAGNETIC FIELD
ENERGY
α-spin
state
aligned
37. sample β-spin
state
MAGNETIC FIELD
ENERGY
...then all the protons
will align with the
field...
α-spin
state
aligned
38. sample β-spin
state
MAGNETIC FIELD
ENERGY
if we give the protons energy
we can excite them so that they
oppose the field (these are the
only two states they can exist
in once in the field)
α-spin
state
aligned
39. sample β-spin
state
MAGNETIC FIELD
ENERGY
α-spin
state
excited
40. sample β-spin
state
MAGNETIC FIELD
the energy required to do this
is in the radio wave frequency
(so not much)
ENERGY
α-spin
state
excited
41. β-spin
sample
state
take away the energy source and
the protons will relax (become
ENERGY
aligned with the field once more)
α-spin
state
relaxes
42. β-spin
sample
state
ENERGY
record energy
emitted
when they do this they emit
energy as a radio wave...which
we can measure
α-spin
state
relaxes
43. β-spin
sample
state
ENERGY
record energy
emitted
now, the cool bit is...
the amount of energy they will
emit depends on the strength of
the magnetic field that they
experience...
α-spin
state
relaxes
44. hydrogen
each proton has an electron
spinning around it...this is a
spinning charge so it too
produces a magnetic field...
+ve
–ve
atom
45. β-spin
sample
state
ENERGY
record energy
emitted
...so, the amount of energy emitted is
influenced by the electrons around the
proton or the chemical environment. As
bonds are electrons we now have
information about bonds / structure of
the molecule
α-spin
state
relaxes
48. O CH3
H3C N
N
H
O N N
CH3
caffeine
CH3
H3C Si CH
3
H3C
TMS
NMR tells us about
the individual H in a
molecule...
49. O CH3
H3C N
N in caffeine there are
H four different kinds of H
N (or proton) depending
O N on where they are...
CH3
caffeine
CH3
H3C Si CH
3
H3C
TMS
50. O CH3
H3C N
N
H
O N N
CH3
caffeine
so the NMR
spectrum has four
CH3
peaks...one for each
kind of H H3C Si CH
3
H3C
TMS
51. O CH3
H3C N
N
H
O N N
CH3
caffeine
CH3
H3C Si CH
3
H3C
TMS
52. deshielded shielded
H H H H
H
C X C C C C C
H
alkenyl X = O, N allylic alkyl
aromatic or halide
8 6 4 2 0
low field predict roughly
we can high field δ scale
where a peak will be found
in the H NMR spectrum as
it is related to the high
low electron electrons
concentration of electron
density H
near each density
53. deshielded shielded
if there are lots of H
H H H H
electrons near to an H
X C
then it C set to be
is C C C C
H
shielded...
alkenyl X = O, N allylic alkyl
aromatic or halide
8 6 4 2 0
low field high field δ scale
high
low electron electron
density density
54. deshielded shielded
if there aren’t many
electrons near the H it is
H
H H deshielded...H H
C X C C C C C
H
alkenyl X = O, N allylic alkyl
aromatic or halide
8 6 4 2 0
low field high field δ scale
high
low electron electron
density density
55. electron withdrawing
groups pull electrons away
causing objects to be
deshielded and once
again...
deshielded shielded
H H H H
H
C X C C C C C
H
alkenyl X = O, N allylic alkyl
aromatic or halide
8 6 4 2 0
low field high field δ scale
high
low electron electron
density density
56. deshielded shielded
H H...proving it is useful to H
H
H
C X know about polarisation C C
C C C of
H bonds etc
alkenyl X = O, N allylic alkyl
aromatic or halide
8 6 4 2 0
low field high field δ scale
high
low electron electron
density density
57. deshielded shielded
H H H H
H
C X C C C C C
H
alkenyl X = O, N allylic alkyl
aromatic or halide
8 6 4 2 0
low field high field δ scale
high
low electron electron
density density
58. correlation table
Type of Type of Type of
δ (ppm) δ (ppm) δ (ppm)
hydrogen hydrogen hydrogen
H2
C CH3 0.70–1.30 C C Ph 2.60 C CH2 4.60–5.00
H2 H2 C C C
C C C 1.20–1.35 C C I 3.10–3.30 H 5.20–5.70
C
H2
C C C 1.40–1.65 C C Br 3.40 CHCl2 5.80–5.90
H
H2
H3C C C 1.60–1.90 C C Cl 3.50 Ph H 6.60–8.00
H3C C O O
2.10–2.60 H2C O 3.50–3.80 9.50–9.70
C C H
O
H3C N 2.10–3.00 H3C O 3.50–3.80 10.00–13.00
C C OH
Ph CH3 2.20–2.50 Ar OH 4.00–8.00 R OH 0.50–5.50
H2
C C H 2.40–2.70 C C F 4.30–4.40
59. correlation table
Type of Type of Type of
δ (ppm) δ (ppm) δ (ppm)
hydrogen hydrogen hydrogen
H2
C CH3 0.70–1.30 C C Ph 2.60 C CH2 4.60–5.00
H2 H2 C C C
C C C 1.20–1.35 C C I 3.10–3.30 H 5.20–5.70
C
H2
C C C 1.40–1.65 C C Br 3.40 CHCl2 5.80–5.90
H
H2
H3C C C 1.60–1.90 C C Cl 3.50 Ph H 6.60–8.00
H3C C O O
2.10–2.60 H2C O 3.50–3.80 9.50–9.70
C C H
shows where we would O
H3C N expect 2.10–3.00
to find peaks in our O
H3C 3.50–3.80 10.00–13.00
C C OH
NMR data...you will always
Ph CH3 be given a table like this OH
2.20–2.50 Ar 4.00–8.00 R OH 0.50–5.50
for your exams...
H2
C C H 2.40–2.70 C C F 4.30–4.40
60. chemically equivalent hydrogen
O
H S
CH3
H H
this is DMSO
(dimethylsulfoxide)...it has
six hydrogen atoms...do
we see six peaks?
62. chemically equivalent hydrogen
O O O
H S H S H S
CH3 CH3 CH3
H H H H H H
all the hydrogens are chemically
equivalent...if we rotate the C–C bond we
see we can put each H in exactly the
same place...so we can not tell them
apart and neither can the nmr. They are
identical
64. chemically equivalent hydrogen
CH3
N
H3C CH3
the peak is slightly
shifted compared to
DMSO as N does not
pull the electrons
towards it as much as
the S=O group
65. two chemical environments
CH3
H H mesityl has two peaks as
it has one set of H
attached to the aromatic
H3C CH3 ring and one set attached
to methyl groups
H
66. three chemical environments
O
δ 3.42 ppm δ 2.15 ppm
O
H3C CH3
H H
δ 4.03 ppm
three chemical
environments (methyl ether,
methyl ketone and the one
in the middle) so three
peaks
67. three chemical environments
O
δ 3.42 ppm δ 2.15 ppm
O
H3C CH3
H H this peak shifted down field
(deshielded) as the ketone is electron
δ 4.03 ppm
withdrawing and the electronegative
O is electron withdrawing so low
electron density near the two H
68. integration (hydrogen counting)
O when the NMR prints a spectrum
in normally has a second bit of
O information attached...this blue
H3C CH3 line (although the colour is
H H unimportant)
69. integration (hydrogen counting)
O
O
H3C CH3
H H
this is called the
integral and its height
is proportional to the
area of each peak
71. integration (hydrogen counting)
O
O so without any knowledge of polarity
H3C CH3 (electron withdrawing groups) we know that
this peak must be due to the central CH2 as
H H it is smaller than the other two peaks (which
are caused by 3 H each)
3x 3x
2x
73. integration (hydrogen counting)
H3C O CH3
we have 6 x H on the
two CH3 groups so they
must be the peak near
1 ppm whilst we have 4
x H near the O so they
are the deshielded H at 6x
3.5 4x
74. integration (hydrogen counting)
H3C O CH3
6x
4x
but note...it is only a
ratio...ethanol
(CH3CH2OH) would give
a very similar spectrum
as the ratio 3:2 is the
same as 6:4
75. integration (hydrogen counting)
H3C O CH3
also note how broad the
peaks are...this is
because nmr is a lot
more complicated than
I’ve shown... 6x
4x
78. integration (hydrogen counting)
H H
H
H
H
H
H
H
H H
Cheeky example...
The H and H turn out to
be similar so overlap or
6x
come at the same
broad as two 4x point...we only see one
different H broad peak
82. integration (hydrogen counting)
HHH
H H
H H
H H
H O
HH
H
H H
H H
H H H 18x
H
HHH
3x
2x
1x
Hopefully, you can see that there
are four chemically equivalent hydrogens. (in
real speak we say they have the ‘same
environment’)
84. integration (hydrogen counting)
H
H H
H N H
the green H is most deshielded
(further down field) as the nitrogen is
electronegative and is pulling the
2x 2x electrons away from the H and
towards itself...
1x
85. integration (hydrogen counting)
H
H H
H N H
so why is the red H more
shielded than the blue H even
though it is closer to the
2x 2x electronegative N?
1x
86. integration (hydrogen counting)
H
H H
the shielding effect tends to
alternate around the ring. The C with
H N H
the green H is deshielded (δ+) so it drags
electrons towards itself meaning the C with the
red H is now slightly shielded (δ–). As the red
CH has had the electrons pulled onto it,
they must have come from somewhere...
so the blue H is δ+...
2x 2x
1x
87. integration (hydrogen counting)
HH
H H H
H H
H H
H H H
H H H
H H
H HH
11x
9x
very broad (>1ppm) as 7
different H (& t-Bu group)
88. integration (hydrogen counting)
4 HH
1
H4 2 H H
5 H H
6 H H
H H H
2
H 3H H
7
H 5 H3
H HH
there are only seven chemically
equivalent hydrogens around the ring because it
is symmetrical. The axial H are different from the
11x
equatorial H. Hopefully the numbering shows
this... 9x
very broad (>1ppm) as 7
different H (& t-Bu group)
90. signal splitting
H3C O CH3
if we look closer the
spectrum actually
shows seven lines (a
quartet and a triplet)
6x
4x
91. signal splitting
H3C O CH3
6x
4x
this is due to peak splitting
and it tells me a lot about
the structure of the
molecule...luckily you don’t
need to know about it...yet...
92. a real 1H NMR spectrum
H
O
looks nasty but
S is very useful!
Tol
H