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NMR
Nuclear Magnetic Resonance
Proton NMR
Index NMR-basics
Anisotropy of Aromatic compounds: in plane and above
H
H H
H H
CH3
H H dring 7.27-6.95 ppm
dMe -0.51 ppm
H H
CH3
H H
H H H H
H
H H H
dring 8.14-8.64 ppm H H
dMe -4.25 ppm H
H
H
H
H
dOUTSIDE 9.28 ppm H H
dINSIDE -2.99 ppm H H
Anisotropy: Aromatic
Electronic effects Deshielded
+
H2C H2C
-
O O
H3C H3C
O
H H COOEt H COOEt
7.10 ppm O 6.28 ppm 6.83 ppm
H H COOEt EtOOC H
O
O
O 5.93 ppm H
6.10 ppm H
7.07 ppm H H
6.38 ppm
7.71 ppm H H
6.28 ppm
Electronic effects: conjugation with carbonyl
3
O O
7 5 2 1
6 4
8 7 6 5 4 3
Electronic effects: conjugation with carbonyl
O
1
2
deshielded 4 3
6 5
7.75
6.20
8 7 6 5 4 3 2
Electronic effects: conjugation with heteroatom
+
O
O H
H
-
C
H
H shielded
S
6.06 ppm S
H
H H
5.48 ppm H
5.81 ppm
O O
6.22 ppm H
H H
5.78 ppm 4.82 ppm H
Electronic effects: no conjugation with heteroatom
O
5
4 3
2 1
6.5 6.0 5.5 5.0 4.5 4.0 3.5
Electronic effects: conjugation with heteroatom
O
5
4 3
2 1
shielded
2.65 2.60 2.55
6.35 6.30 4.95 4.90
7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5
Electronic effects: conjugation with carbonyl
o m O CH3
p
deshielded
8.0 7.5 deshielded
8 7 6 5 4 3 2
Electronic effects: conjugation with heteroatom
H3C
O
Shielded
shielded
o
m
p
7.3 7.2 7.1 7.0 6.9 6.8
7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5
Electronic effects: conjugation with heteroatom
H3C
NH
m o Shielded
p
shielded
7.0 6.5
7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5
Aromatic: inductive effect and resonance effect
F
7.5 7.4 7.3 7.2 7.1 7.0 6.9
Cl
7.4 7.3 7.2 7.1
Br
7.6 7.5 7.4 7.3 7.2 7.1
Hydrogen bond
Protons on Heteroatoms
• OH, NH, SH
– Exchangeable (with D2O)
– Hydrogen bonding
– On Nitrogen (14N), as the spin state of that
nuclei is 1, there can be partial coupling that
produce broaden lines. There can be also full
coupling that would produce 3 lines of equal
intensity (I=1 has 3 orientations in a magnetic
field)
Protons on Heteroatoms
• OH
– Aliphatic d 0.5-4.0 ppm (depend on Concentration)
– Intramolecular hydrogen bonding deshield OH and render it
less sensitive to concentration
• Usually OH exchange rapidly (no coupling with
neighbors
• In DMSO or Acetone, the exchange rate is slower =>
there is coupling with neighbors
• Phenols : d 7.5-4.0 ppm
Intramolecular bond 12-10 ppm
• Carboxylic Acids: Exist as Dimers 13.2-10 ppm
H2O signal moves with temperature
H2O
OH in DMSO
CH3-CH2-OH
OH CH2
qd
(CH3)2 -CH-OH
OH
CH
Protons on Heteroatoms
• NH : 14N: I=1 => 2I+1 lines
• NH has different rate of exchange
• 14N can relax quickly. Depending on relaxation rate,
heteronuclear coupling will be visible or produce
broadened peaks.
• R-NH : Aliphatic amines => rapid exchange
– Sharp singlets : no coupling to N: d~3-0.5 ppm
• R-NH: Amides, Pyrroles, Indoles, Carbamates
– NH broad
– CHa shows coupling the NH
NH
Amide
Protonated Amines
Formamide
H{14N}-NMR
H-CO-NH2
H-NMR
NH: Amide, Pyrrole Indole
d: 8.5-5.0 ppm
In Amides: Slow rotation can show different isomers
O O
H CH3
H N H N
CH3 H
In Amine Salt:
• Moderate Rate of exchange => broad peaks ~ d 8.5-6.0 ppm
•CHa => show coupling to NH+
Sometimes broad [NHx+] consist of 3 broad hump due to 14N coupling
1J ~ 50 Hz
NH
SH
• Slow exchange
SH couple to CHa
• When shaken with D2O, SH Disapear
d ~ 1.6 – 1.2 ppm Aliphatic SH
d ~ 3.6 – 2.8 ppm Aromatic SH
Chemical Shift and
Coupling
An example: C10H12O2 I = 10 + 1 – 12/2 = 5
O CH3
CH3
O-CH2-CH3
Me-C= J=7 Hz
J=7 Hz
J=8 Hz H3C O
2H 3H 3H
O O CH3
2H 2H
O Me-C=C
H3C CH3
X 4 = 12
H3C CH3
O
Scalar coupling: Coupling through bond
n=0 1
1 1 1 2nI + 1 lines
2 1 2 1
3 1 3 3 1
4 1 4 6 4 1 a
5 1 5 10 10 5 1
6 1 6 15 20 15 6 1 doublet
H3C CH3
b a
b
septet
2.70
2.5 2.0 1.5
Scalar coupling: Coupling through bond
n=0 1
1 1 1 2nI + 1 lines
2 1 2 1
3 1 3 3 1 m o
4 1 4 6 4 1
5 1 5 10 10 5 1
6 1 6 15 20 15 6 1 p b
a
m o Br
p
a 2 x triplet b
7.50 7.40 7.30 7.20 7.10 7.00
3.40 3.30 3.20 3.10 3.00
7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0
Scalar coupling: Coupling through bond
n=0 1
1 1 1 2nI + 1 lines
2 1 2 1
3 1 3 3 1
4 1 4 6 4 1
5 1 5 10 10 5 1
6 1 6 15 20 15 6 1
c 2 x triplet
1 quintet
a
b
a Br
c
b
3.0 2.5 2.0
Scalar coupling: Coupling through bond C7 H14 O2 I = 7 -14/2 + 1 = 1
2nI + 1 lines
d (ppm) Int mult J (Hz) COMMENT
n=0 1 0.9 3H triplet 7 CH3->(CH2)
1 1 1 1.1 3H triplet 7 CH3->(CH2)
2 1 2 1 1.35 2H sixtet 7 CH2 (CH3, CH2)
3 1 3 3 1
1.55 2H quintet 7 CH2 (CH2, CH2)
4 4 6 4 1
1
5 1 5 10 10 5 1 2.3 2H quartet 7 =C- CH2 (CH3)
6 1 6 15 20 15 6 1 4.1 2H triplet 7 CH2 -O (CH2)
CH3 CH2 O CH2 2 x triplet
Roof effect CH2 CH2 2.3 CH3
0.9 ppm 1.1 3H
Triplet: 1.35
O 3H
2H
Quartet:
2H Quintet:
Sixtet:
2H 2H
4.0 3.5 3.0 2.5 2.0 1.5 1.0
Scalar coupling: Coupling through bond
n=0 1
1 1 1 2nI + 1 lines
2 1 2 1
3 1 3 3 1
4 4 6 4 1
1
5 1 5 10 10 5 1
6 1 6 15 20 15 6 1
H3C 3 O 5 2 x triplet
1 2 4 CH3
6
6
Triplet: 1
O
4
Quartet:
5 Quintet:
3 Sixtet:
2
4.0 3.5 3.0 2.5 2.0 1.5 1.0
Common first order spin system 2nI + 1 lines
Ha Hb Hb Ha Hb
C C Hb C C C Hb
Hb Hb
Ha Hb
C C Hb
Hb Ha Hb
C C C
Ha Hb
C C Hb
Hb
Common first order spin system 2nI + 1 lines
Jab = Jab’
Hc Ha Hb
Ha Hb
Jab Jab C C C Hb
C C Hb’
qd Hb
Hb’ Ha Hb
C C C
Jab’
td
Hc’ Ha Hb
Hb’ Ha Hb
C C C Hb’
C C C Hb
Geminal
Coupling
Vicinal
Coupling
3J => Perch
3J => tool 1
3J => Mestrec tool
Ha
Using Vicinal
Coupling to establish
isomer
Jab
Jad
Jac
Long Range
Coupling
Long Range coupling
H
4 H 4J
5 = 1.07 Hz
H1-H3 H C C C H
5J
H1-H4 = 1.21 Hz 4J
5J H-H = 9 Hz
H1-H5 = 0.95 Hz
N 5J
H4-H7 = 0.67 Hz
H C C C C H
H
H 7 5J
1
H-H = 3 Hz
H
H C C C H
4J = 1-2 Hz
H H-H
4J = 3 Hz
H-H
H H C C C C H
4J = 1.1 Hz
H-H
5J = 3 Hz
H-H
H
Spin System in Pople notation
Structural Unit Spin system Partial spectrum
-CH2-CH3 A 3X2
2.5 2.0 1.5
-CH-CH3 A3X
3.0 2.5 2.0 1.5
CH2-CH2-CH3 A3M2X2
2.0 1.5 1.0
Each chemical shift is represented by a letter (far way letter for very
large shift difference – compare with the size of the coupling)
Dn
Second Order spectra: Dn / J
AB instead of AX J J
5.0
As the difference in shift become smaller- 1 2 3 4
compare with the size of the coupling the 4.0
outer peaks become smaller in intensity
3.0
nA and nB : center of gravity of doublet 2.0
Chemical shift
Dn = (1-4) * (2-3) 1.0
SpinWorks => load AB
0.5
AB-Spectra
AMX C6 H4 O5 N2
I = 6 - 4/2 + 2/2 +1
I= 6
Phenyl = 4 I
NO2 = 1 I
A2X and A2B
SpinWorks => load A2B
AMX
AMX
OMe
Substituants : H 7.58 ppm
OMe
2 OMe (~ 3.9 ppm)
CHO (~ 9.8 ppm) H
CHO
6.83 ppm
H
J J J 7.2 ppm
Meta Ortho Ortho
Para Meta Para
CHO
H 7.58 ppm
OMe
H
OMe
6.83 ppm
H
7.2 ppm
meta bromo nitro benzene
Br
H H
B A
Calculated shifts
O
dHA=8.44 dHB=7.82 dHC=7.31 dHD=8.19
+
H N
C
-
H O
D
HA
HB
HD HC
8.5 8.0 7.5
AFMX
C5 H4 N Br
I = 5 – 4/2 – 1/2 +1/2 +1
I = 4 (aromatic ring)
d
J
Assignment of 1H NMR of: cartilagineal
Me
CHO
H Cl
Cl H
H 7 5 3 1
8 6 4 2 H
Me Cl 9
H
O H H
Jcis=10.5 dd H-5 dd
H-3 ddd 7 3J
4,5 = 8.5
H
CHO-9 3J =15.5
3,4
8 R
6 4J = 1.0
J = 2.0 Hz H-1(s) 4J J =17 Hz 3,5
3,5 =1.0 H trans
4J =2.0
4,9
3J
3,4=15.5
3J =8.5
4,5
H-4 dd
Complicated proton spectra : CH3-CH2-S-PF2
Almost quintet
3J
PH
3J 3J
HH HH
4J
FH
t t
Identifying 31P
couplings
{31P}
H R
C C
P P
H H
dd
P
NMR – From Spectra to Structures An Experimental approach
Second edition (2007) Springler-Verlag
Terence N. Mitchellm Burkhard Costisella
Identifying 31P couplings: another example
Ph, 2H
CH3
1H
1H CH2
NMR – From Spectra to Structures An Experimental approach
Second edition (2007) Springler-Verlag
Terence N. Mitchellm Burkhard Costisella
P31 NMR
Identifying 31P couplings: another example
NMR – From Spectra to Structures An Experimental approach
Second edition (2007) Springler-Verlag
Terence N. Mitchellm Burkhard Costisella
H-nmr P31 decoupled
1H CH2
To identify a compound: PF215NHSiH3
Use as many techniques as possible
Proton nmr spectra is difficult to analyze with so many J’s
But with 19F, 15N and 31P spectra it’s easier (get heteronuclear J)
To identify a compound: PF215NHSiH3
Use as many techniques as possible
Using decoupler : easier analysis
Another example H{X}
Changing the solvent
Changing solvent can be used to improve
dispersion of chemical shifts
C6D6
CDCl3
Changing the solvent Me
Me
CH2
H2 C
=CH2
OH Me
CH2
C6D6
CH-OH
=CH2 CH2
ABX
CDCl3 CH-OH
Decoupling Me
Me
CH2
H2 C
=CH2
OH Me
CH2
CDCl3 ABX
CH-OH
CH2
AB
Spin-Spin
Decoupling
dq dq
dd
NMR – From Spectra to Structures An Experimental approach
Second edition (2007) Springler-Verlag
Terence N. Mitchellm Burkhard Costisella
Homo decoupling
JPH
JHH
Decoupling H-1 glucose derivative
H-2
H-1
Several Decoupling
NOE
nOe
NOE: applying gB2 to the A of an AX spin system
bb bb
X2 A2 {A} X2 A2
ba ab ba ab
A1 X1 A1 X1
aa aa
X1 Dp = 2 X1 Dp = 2
X2 Dp = 2 X2 Dp = 2
Immediately after irradiation, there is NO change in the intensity of X
Turning on the Decoupler do not change population of the X transition
NOE: relaxation with double quantum pathway W2
probability (positive NOE)
bb bb
X2 A2 {A} X2 A2
ba ab ba ab
A1 X1 A1 X1
aa X1 Dp = 2 aa
X2 Dp = 2
delay
bb bb
X2 A2 X2 A2
… T1
ba ab ba W2 ab
A1 X1 Dec. continue A1 X1
aa aa
X1 Dp = 3 After W2 relaxation, there is a net increase in
X2 Dp = 3 the intensity of X (50%)
Relaxation takes time to establish a new equilibrium: T1 process
NOE: Relaxation with zero quantum pathway W0
probability (negative NOE)
bb bb
X2 A2 {A} X2 A2
ba ab ba ab
A1 X1 A1 X1
aa X1 Dp = 2 aa
X2 Dp = 2 delay
bb bb
X2 A2 X2 A2
… T1
ba W0 ab ba W0 ab
A1 X1 Dec. continue A1 X1
aa aa
X1 Dp = 1 After W0 relaxation, there is a net decrease in
X2 Dp = 1 the intensity of X (50%) negative NOE
Relaxation takes time to establish a new equilibrium: T1 process
NOE: summary of relaxation pathways
bb W1: probability of single quantum
X2 W2 A2 relaxation do not create nOe
ba W0 ab
A1 X1 A new population ditribution is generated by
aa relaxation through dipole-dipole relaxation :
double quantum and zero quantum pathway
W2 and W0
If W2 is efficient (small molecule – fast motion large frequency )
Level increase level increase also with decoupler continuing
W2 pathway yield positive nOe
If W0 is efficient (large molecule – slow motion small freq. Diff.)
Level increase level increase also with decoupler continuing
W0 pathway yield negative nOe
NOE difference: nOe-d
NOE is a kinetic effect: need delay ~ T1 Cl Me
It take time to develop C C
It takes time to decay Me
H
d1 AQ
irr
Dec on frq
irr control
d1 AQ
Dec off frq nOe
difference
NOE
H3C H
{Me –cis} => +19%
H3C COOH {Me –trans} => -2%
{Ha} => +45%
Choosing a structure by nOe
OH Ph
H 1 {OH}
6 2 CH3
5 3
H 4 H
OMe
{OMe}
OMe Ph
H 1
6 2 CH3
5 3
H 4 H
H6 H3
H5
OH
NMR – From Spectra to Structures An Experimental approach
Second edition (2007) Springler-Verlag
Terence N. Mitchellm Burkhard Costisella
NOEd
Cl
O
CH3
CH P O
CH3
O
Cl
NOEd example
Organometallic compounds Proton - NMR
Increasing the 1 s orbital density increases the shielding
M=C M = Si M = Ge
MH4 0.1 3.2 3.1
MH3I 2.0 3.4 3.5
MH3Br 2.5 4.2 4.5
MH3Cl 2.8 4.6 5.1
(MH3)2O 3.2 4.6 5.3
MH3F 4.1 4.8 5.7
Shift to low field when the metal is heavier (SnH4 - d = 3.9 ppm)
Proton – NMR : Chemical shift
• Further contribution to shielding / deshielding is the anisotropic magnetic
susceptibility from neighboring groups (e.g. Alkenes, Aromatic rings ->
deshielding in the plane of the bound)
• In transition metal complexes there are often low-lying excited electronic
states. When magnetic field is applied, it has the effect of mixing these to
some extent with the ground state.
• Therefore the paramagnetic term is important for those nuclei themselves =>
large high frequency shifts (low field). The protons bound to these will be
shielded (d => 0 to -40 ppm) (these resonances are good diagnostic. )
• For transition metal hydride this range should be extended to 70 ppm!
• If paramagnetic species are to be included, the range can go to 1000 ppm!!
Exchange : DNMR – Dynamic NMR
NMR is a convenient way to study rate
of reactions – provided that the lifetime
of participating species are comparable
to NMR time scale (10-5 s)
H
H
H
GeMe3
H
H
At low temperature, hydrogens form an
A2B2X spin system
At higher temperature germanium hop
from one C to the next
Index NMR-basics NMR-Symmetry
