5.33 Lecture Notes,Magnetic Resonance Spectroscopy
In our discussion of spectroscopy,we have shown that absorption of E.M,
radiation occurs on resonance,When the frequency of applied E.M,field matches the
energy splitting between two quantum states,
Magnetic resonance differs from these other methods in the sense that we need to
immerse the same in a magnetic field in order to see the levels that we probe with an
external (rf or μwave) field,(Two fields,Static magnetic and E.M.)
We will be probing the energy levels associated with the spin angular momentum
of nuclei and electrons,NMR--nuclear magnetic resonance and ESR/EPR--electron spin
resonance,
Angular momentum,
In our treatment of rotational energy levels,we said that the energy levels
depended on the rotational angular momentum,L,which was quantized,
L
2
= null
2
J (J + 1) J = 0,1,2… rot, quant,number
Degeneracy of J was (m
J
= 0,…,± J)→(2 J + 1)
We related L
2
to the energy levels
L
2
E
rot
=
2I
∝ BJ (J + 1)
Actually,all angular momentum is quantized,
If a particle can spin,it has A.M,and quantized E levels,
In particular,we also have to be concerned with the spin of individual nuclei and
electrons,
5.33 Lecture Notes,Magnetic Resonance Spectroscopy Page 1
You already know that electrons have…
ORBITAL angular momentum
2
M
2
= nullnull (null + 1)
null = 0,1,2… orbital angular momentum quantum number
degeneracy of orbitals,2null + 1 from…
m
null
=?null
1
…,+null magnetic quantum number
m
null
represents the quantization of the
components of M, z
M
null = 1
M
Z
= m
null
null
+ 1null
2 M null =
(How we choose
z doesn’t matter until we
apply a magnetic field.)
0null
1null
Now,the angular momentum that we are concerned with is,
Electron Spin Angular Momentum
S
2
= null
2
s (s + 1) s,electron spin quantum number =?
1
2
for each unpaired e
SS
z
= m
s
null m,±
1
(?S,?+ 1,…,+S)
s
2
one unpaired e
5.33 Lecture Notes,Magnetic Resonance Spectroscopy Page 2
Nuclear spin angular momentum
I,nuclear spin quantum numberI
2
= null
2
II + 1)(
I
z
= m
I
null m
I
,? I,? I + 1,…,I
null What is I? Each proton/neutron has a spin quantum number of 1/2,
null Spin of many nucleons add up vectorily to give I,
null Spins pair up so that even number of spins are paired → I = 0,
=1,For even number of protons plus even number of neutrons,I0,
12
C,
16
O
2,For one unpaired nucleon → I =
1
2
m
I
= ±
1
2
degeneracy 2 I + 1 = 2
1
H,
13
C,
15
N
So the proton and electron are similar—both spin
1
2
particles, We’ll talk about
these two particles more specifically…
3,Two unpaired nucleons → I = 1 m
I
= 0,±1
2
H,
14
N
5.33 Lecture Notes,Magnetic Resonance Spectroscopy Page 3
For spin? particles (I or S =?) there are two degenerate energy states,
When you put these in a magnetic field you get a splitting,
null A low energy state aligned with the field,and
null A high energy state aligned against the field
Classical picture,
Think of our electron or nucleus as a charged particle with angular momentum,M,
A circulating charge produces a magnetic field,
This charge possesses a magnetic dipole moment μ that can be affected by an applied
magnetic field,
Mm ( v × r )=
μ= Qv × r ) / 2c
M
r
μ
m,Q
v
(
The dipole lies along M and the strength of μ is
proportional to μ,
μ=
Q
M ≡ γ? M
2mc
γ is the gyromagnetic ratio
Quantum,
e
μ=
e
For electrons,
e
2m
e
c
gS
g factor (2.0023)
=?γ
e
S
+e
N N
For nuclei,
μ=
2m
N
c
gI
g
N
= 5.6 for
1
H
=γ
N
I
g is a relativistic quantum mechanical correction,
5.33 Lecture Notes,Magnetic Resonance Spectroscopy Page 4
If we immerse this system in a magnetic field,B,which is oriented along the z axis
The interaction potential is E
int
=?μ? B
The spins align along the magnetic field,E
int
=?μ
z
B
SSince μ=?γ
ez
and μ=?γ
N
I
z
,
z z
using S
z
= nullm
s
and I
z
= nullm
I
we have
1
For electrons,E
int
=γ
e
nullm
s
B m
s
=±
2
1
For nuclei,E
int
=?γ
N
nullm
I
B m
I
=±
2
So,as we increase the magnetic field strength,the two energy levels? originally
degenerate – split,one increasing in energy and one decreasing in energy,This is known
as the Zeeman effect,
Nuclei Electrons
Eint
m
I
=?1/2
E?
m
I
= +1/2
B
E
m
s
= +1/2
0
=?γ
N
nullB?E =γ
e
nullB
m
s
=?1/2
0
Now we have a system that can absorb E.M,radiation on resonance,?E=hν
ν is the applied frequency (in the radio frequency range)
Selection Rules,?m
s
= ±1 ;?m
I
= ±1
γB
Typically sweep B and hold ν constant,ν=
2π
(Larmor frequency)
5.33 Lecture Notes,Magnetic Resonance Spectroscopy Page 5
Typical numbers,
Nuclear Magnetic Resonance Electron Spin Resonance
|μ
H
| = 2.8 × 10
-23
(±?) erg/gauss |μ
e
| =?1.8 × 10
-20
(±?) erg/gauss
for B = 10 kG,ν = 42.6 MHz ν = 28 GHz
Typical,300 MHz/70 kG 9.5 GHz/3.4 kG
Typical population difference?N = (Ν
N
+
)/(Ν
+ N
+
) ≈ 0.008%
The Chemical Shift
Thus far you would think that all
1
H absorb at same frequency,Yet,in practice different
nuclei absorb at different frequencies,
The resonance frequency depends on the,effective” magnetic field that a proton feels,
This can differ for different types of
1
H due to local electron currents that counteract the
applied field,→ Shielding
B
eff
= B
app
(1?σ)
Eint
σ,screening constant
typically 10
-6
0
E
int
=?γ
N
nullm
I
B
app
(1?σ)
γB
ν=
app
(1?σ)
0 Bapp
E
No shielding
With shielding
(shifted to lower ν)
2π
Shift of frequency due to screening,Chemical Shift
5.33 Lecture Notes,Magnetic Resonance Spectroscopy Page 6
Typically you measure the chemical shift due to screening relative to a standard (TMS),
γB
app
ν?ν
ref
=
2π
(σ
ref
σ
i
)
i
Measure positions in δ,ppm,10
6
δ
i
=
ν
i
ν
ref
≈ (σ
ref
σ
i
)
ν
ref
The magnitude of the shielding depends on how the motion of electrons modifies the
local field,(Interaction between field and electron angular momentum)
Qualitatively,proton NMR spectra can be interpreted by considering electronegativity of
bound functional groups,Greater E.N,draws electrons away from
1
H,lowering the
resonance field and giving a larger δ,
Other nuclei have larger chemical shifts because there are more electrons,
Couplings between Nuclear and Electron Spins
Based on the effects of screening (chemical shift),we would expect one line for each
nucleus in an NMR spectrum,
Actually there is usually additional structure,with each line split into several others,
(Both in NMR and ESR)
null Splittings arise when different magnetic spins on nuclei and/or electrons interact
with each other,
null The interactions change the spin energies to give new lines,
null Understanding the interaction allows you to reveal structural information such as
connectivity,
The couplings between two magnetic dipoles can be written as E
coupling
∝μ
1
μ
2
Couplings between unpaired electrons (a,b,c…) and nuclear spins (i,j,k…) are,
nuclear-nuclear (“J”) coupling μ
N
(i)?μ
N
( j)=γ
N
(i)γ
N
( j) I
i
I
j
electron-electron (“fine”) coupling μ
e
(a )?μ
e
( b)=γ γ
ee
S
a
S
b
electron-nuclear (“hyperfine”) coupling μ
e
(a )?μ
N
(i)=γγ
N
(i) S
a
I
ie
5.33 Lecture Notes,Magnetic Resonance Spectroscopy Page 7
NMR
Interaction between protons,
iE
coupling
= null J
ij
m
I
() m
I
( j) m
I
=±
J,nuclear spin coupling constant; typically 1-10 Hz,
1
2
Total nuclear interaction energy with spins,
E
int
= E
nuclear Zeeman
+ E
J coupling
i i i=?nullB
app ∑
γ
N
() m
I
()(1?σ
i
)+ null
∑
J
ij
m
I
() m
I
( j)
i i≠ j
ESR
Dominant effect,hyperfine coupling
E
coupling
= null a
bj
m
s
() m
I
( j)b
a,hyperfine coupling constant; typically 1-100 MHz
Total interaction energy of electron and nuclear spins,
E
int
= E
electron Zeeman
+ E
hyperfine coupling
+ E
nuclear Zeeman
+ E
J coupling
b b=?nullB
app ∑
γ
e
() m
s
()(1?σ
b
)+ null
∑∑
a
bj
m
s
( b) m
I
(i)+null
b i,j,..,a,b..,
Sum of electron Zeeman effect and hyperfine interaction,Nuclear Zeeman
energy and J coupling is relatively small in comparison,
ESR Example,H atom? 1 proton and 1 electron
E
int
= E
electron Zeeman
+ E
hyperfine coupling
+
=?nullBm
nuclear Zeeman
E
γ
es
+ null a m
s
m
I
With no coupling,two states,with splitting as before,
5.33 Lecture Notes,Magnetic Resonance Spectroscopy Page 8
With coupling,
s
1
m
2
=± ;
I
1
m
2
=± and
s
1
mm
4
= ±
B
Eint
0
/4null
/4?null
B1 B2
ms mI
1/ 2+
1/ 2+
1/ 2?
1/ 2?
1/ 2+
1/ 2+
1/ 2?
1/ 2?
E
ms=1
mI=0
At zero field,two states
1
E
4
=+ null and
1
E
4
=? null
With increasing field,degenerate states split,sweeping field with constant
resonance frequency?E=hν,we see resonances at two fields,B
1
and B
2
,
1
e
Ea 2
B
=
γnull
and
2
e
Ea 2
B
+
=
γnull
so that the field splitting gives the hyperfine coupling,( )
2 e
a B=γ?null
Alternatively,you could stay in a fixed field B and sweep the rf frequency?E,and
you would observe two resonances,
e
E/ B a 2? γ ±null
Again the frequency splitting gives the hyperfine coupling,
I
a a
By
1
B
=
5.33 Lecture Notes,Magnetic Resonance Spectroscopy Page 9
Splitting Patterns
For multiple nuclei coupled to an unpaired electron,we can expect each hyperfine
interaction to split the remaining transition into a pair of peaks split by a,The overall
spectrum can be predicted diagrammatically by a pattern of splittings in which one
electron resonance is sequentially split in frequency by each hyperfine coupling
interaction,
Two inequivalent couplings a
1
and a
2
,Two equivalent couplings (a
1
= a
2
),
5.33 Lecture Notes,Magnetic Resonance Spectroscopy Page 10
In our discussion of spectroscopy,we have shown that absorption of E.M,
radiation occurs on resonance,When the frequency of applied E.M,field matches the
energy splitting between two quantum states,
Magnetic resonance differs from these other methods in the sense that we need to
immerse the same in a magnetic field in order to see the levels that we probe with an
external (rf or μwave) field,(Two fields,Static magnetic and E.M.)
We will be probing the energy levels associated with the spin angular momentum
of nuclei and electrons,NMR--nuclear magnetic resonance and ESR/EPR--electron spin
resonance,
Angular momentum,
In our treatment of rotational energy levels,we said that the energy levels
depended on the rotational angular momentum,L,which was quantized,
L
2
= null
2
J (J + 1) J = 0,1,2… rot, quant,number
Degeneracy of J was (m
J
= 0,…,± J)→(2 J + 1)
We related L
2
to the energy levels
L
2
E
rot
=
2I
∝ BJ (J + 1)
Actually,all angular momentum is quantized,
If a particle can spin,it has A.M,and quantized E levels,
In particular,we also have to be concerned with the spin of individual nuclei and
electrons,
5.33 Lecture Notes,Magnetic Resonance Spectroscopy Page 1
You already know that electrons have…
ORBITAL angular momentum
2
M
2
= nullnull (null + 1)
null = 0,1,2… orbital angular momentum quantum number
degeneracy of orbitals,2null + 1 from…
m
null
=?null
1
…,+null magnetic quantum number
m
null
represents the quantization of the
components of M, z
M
null = 1
M
Z
= m
null
null
+ 1null
2 M null =
(How we choose
z doesn’t matter until we
apply a magnetic field.)
0null
1null
Now,the angular momentum that we are concerned with is,
Electron Spin Angular Momentum
S
2
= null
2
s (s + 1) s,electron spin quantum number =?
1
2
for each unpaired e
SS
z
= m
s
null m,±
1
(?S,?+ 1,…,+S)
s
2
one unpaired e
5.33 Lecture Notes,Magnetic Resonance Spectroscopy Page 2
Nuclear spin angular momentum
I,nuclear spin quantum numberI
2
= null
2
II + 1)(
I
z
= m
I
null m
I
,? I,? I + 1,…,I
null What is I? Each proton/neutron has a spin quantum number of 1/2,
null Spin of many nucleons add up vectorily to give I,
null Spins pair up so that even number of spins are paired → I = 0,
=1,For even number of protons plus even number of neutrons,I0,
12
C,
16
O
2,For one unpaired nucleon → I =
1
2
m
I
= ±
1
2
degeneracy 2 I + 1 = 2
1
H,
13
C,
15
N
So the proton and electron are similar—both spin
1
2
particles, We’ll talk about
these two particles more specifically…
3,Two unpaired nucleons → I = 1 m
I
= 0,±1
2
H,
14
N
5.33 Lecture Notes,Magnetic Resonance Spectroscopy Page 3
For spin? particles (I or S =?) there are two degenerate energy states,
When you put these in a magnetic field you get a splitting,
null A low energy state aligned with the field,and
null A high energy state aligned against the field
Classical picture,
Think of our electron or nucleus as a charged particle with angular momentum,M,
A circulating charge produces a magnetic field,
This charge possesses a magnetic dipole moment μ that can be affected by an applied
magnetic field,
Mm ( v × r )=
μ= Qv × r ) / 2c
M
r
μ
m,Q
v
(
The dipole lies along M and the strength of μ is
proportional to μ,
μ=
Q
M ≡ γ? M
2mc
γ is the gyromagnetic ratio
Quantum,
e
μ=
e
For electrons,
e
2m
e
c
gS
g factor (2.0023)
=?γ
e
S
+e
N N
For nuclei,
μ=
2m
N
c
gI
g
N
= 5.6 for
1
H
=γ
N
I
g is a relativistic quantum mechanical correction,
5.33 Lecture Notes,Magnetic Resonance Spectroscopy Page 4
If we immerse this system in a magnetic field,B,which is oriented along the z axis
The interaction potential is E
int
=?μ? B
The spins align along the magnetic field,E
int
=?μ
z
B
SSince μ=?γ
ez
and μ=?γ
N
I
z
,
z z
using S
z
= nullm
s
and I
z
= nullm
I
we have
1
For electrons,E
int
=γ
e
nullm
s
B m
s
=±
2
1
For nuclei,E
int
=?γ
N
nullm
I
B m
I
=±
2
So,as we increase the magnetic field strength,the two energy levels? originally
degenerate – split,one increasing in energy and one decreasing in energy,This is known
as the Zeeman effect,
Nuclei Electrons
Eint
m
I
=?1/2
E?
m
I
= +1/2
B
E
m
s
= +1/2
0
=?γ
N
nullB?E =γ
e
nullB
m
s
=?1/2
0
Now we have a system that can absorb E.M,radiation on resonance,?E=hν
ν is the applied frequency (in the radio frequency range)
Selection Rules,?m
s
= ±1 ;?m
I
= ±1
γB
Typically sweep B and hold ν constant,ν=
2π
(Larmor frequency)
5.33 Lecture Notes,Magnetic Resonance Spectroscopy Page 5
Typical numbers,
Nuclear Magnetic Resonance Electron Spin Resonance
|μ
H
| = 2.8 × 10
-23
(±?) erg/gauss |μ
e
| =?1.8 × 10
-20
(±?) erg/gauss
for B = 10 kG,ν = 42.6 MHz ν = 28 GHz
Typical,300 MHz/70 kG 9.5 GHz/3.4 kG
Typical population difference?N = (Ν
N
+
)/(Ν
+ N
+
) ≈ 0.008%
The Chemical Shift
Thus far you would think that all
1
H absorb at same frequency,Yet,in practice different
nuclei absorb at different frequencies,
The resonance frequency depends on the,effective” magnetic field that a proton feels,
This can differ for different types of
1
H due to local electron currents that counteract the
applied field,→ Shielding
B
eff
= B
app
(1?σ)
Eint
σ,screening constant
typically 10
-6
0
E
int
=?γ
N
nullm
I
B
app
(1?σ)
γB
ν=
app
(1?σ)
0 Bapp
E
No shielding
With shielding
(shifted to lower ν)
2π
Shift of frequency due to screening,Chemical Shift
5.33 Lecture Notes,Magnetic Resonance Spectroscopy Page 6
Typically you measure the chemical shift due to screening relative to a standard (TMS),
γB
app
ν?ν
ref
=
2π
(σ
ref
σ
i
)
i
Measure positions in δ,ppm,10
6
δ
i
=
ν
i
ν
ref
≈ (σ
ref
σ
i
)
ν
ref
The magnitude of the shielding depends on how the motion of electrons modifies the
local field,(Interaction between field and electron angular momentum)
Qualitatively,proton NMR spectra can be interpreted by considering electronegativity of
bound functional groups,Greater E.N,draws electrons away from
1
H,lowering the
resonance field and giving a larger δ,
Other nuclei have larger chemical shifts because there are more electrons,
Couplings between Nuclear and Electron Spins
Based on the effects of screening (chemical shift),we would expect one line for each
nucleus in an NMR spectrum,
Actually there is usually additional structure,with each line split into several others,
(Both in NMR and ESR)
null Splittings arise when different magnetic spins on nuclei and/or electrons interact
with each other,
null The interactions change the spin energies to give new lines,
null Understanding the interaction allows you to reveal structural information such as
connectivity,
The couplings between two magnetic dipoles can be written as E
coupling
∝μ
1
μ
2
Couplings between unpaired electrons (a,b,c…) and nuclear spins (i,j,k…) are,
nuclear-nuclear (“J”) coupling μ
N
(i)?μ
N
( j)=γ
N
(i)γ
N
( j) I
i
I
j
electron-electron (“fine”) coupling μ
e
(a )?μ
e
( b)=γ γ
ee
S
a
S
b
electron-nuclear (“hyperfine”) coupling μ
e
(a )?μ
N
(i)=γγ
N
(i) S
a
I
ie
5.33 Lecture Notes,Magnetic Resonance Spectroscopy Page 7
NMR
Interaction between protons,
iE
coupling
= null J
ij
m
I
() m
I
( j) m
I
=±
J,nuclear spin coupling constant; typically 1-10 Hz,
1
2
Total nuclear interaction energy with spins,
E
int
= E
nuclear Zeeman
+ E
J coupling
i i i=?nullB
app ∑
γ
N
() m
I
()(1?σ
i
)+ null
∑
J
ij
m
I
() m
I
( j)
i i≠ j
ESR
Dominant effect,hyperfine coupling
E
coupling
= null a
bj
m
s
() m
I
( j)b
a,hyperfine coupling constant; typically 1-100 MHz
Total interaction energy of electron and nuclear spins,
E
int
= E
electron Zeeman
+ E
hyperfine coupling
+ E
nuclear Zeeman
+ E
J coupling
b b=?nullB
app ∑
γ
e
() m
s
()(1?σ
b
)+ null
∑∑
a
bj
m
s
( b) m
I
(i)+null
b i,j,..,a,b..,
Sum of electron Zeeman effect and hyperfine interaction,Nuclear Zeeman
energy and J coupling is relatively small in comparison,
ESR Example,H atom? 1 proton and 1 electron
E
int
= E
electron Zeeman
+ E
hyperfine coupling
+
=?nullBm
nuclear Zeeman
E
γ
es
+ null a m
s
m
I
With no coupling,two states,with splitting as before,
5.33 Lecture Notes,Magnetic Resonance Spectroscopy Page 8
With coupling,
s
1
m
2
=± ;
I
1
m
2
=± and
s
1
mm
4
= ±
B
Eint
0
/4null
/4?null
B1 B2
ms mI
1/ 2+
1/ 2+
1/ 2?
1/ 2?
1/ 2+
1/ 2+
1/ 2?
1/ 2?
E
ms=1
mI=0
At zero field,two states
1
E
4
=+ null and
1
E
4
=? null
With increasing field,degenerate states split,sweeping field with constant
resonance frequency?E=hν,we see resonances at two fields,B
1
and B
2
,
1
e
Ea 2
B
=
γnull
and
2
e
Ea 2
B
+
=
γnull
so that the field splitting gives the hyperfine coupling,( )
2 e
a B=γ?null
Alternatively,you could stay in a fixed field B and sweep the rf frequency?E,and
you would observe two resonances,
e
E/ B a 2? γ ±null
Again the frequency splitting gives the hyperfine coupling,
I
a a
By
1
B
=
5.33 Lecture Notes,Magnetic Resonance Spectroscopy Page 9
Splitting Patterns
For multiple nuclei coupled to an unpaired electron,we can expect each hyperfine
interaction to split the remaining transition into a pair of peaks split by a,The overall
spectrum can be predicted diagrammatically by a pattern of splittings in which one
electron resonance is sequentially split in frequency by each hyperfine coupling
interaction,
Two inequivalent couplings a
1
and a
2
,Two equivalent couplings (a
1
= a
2
),
5.33 Lecture Notes,Magnetic Resonance Spectroscopy Page 10
