Download Introduction to NMR spectroscopy Nuclei of isotopes which possess

Introduction to NMR spectroscopy
Nuclei of isotopes which possess an odd number of protons, an odd number of neutrons, or both,
have a nuclear spin quantum number, I, such that, I = 1/2n, where n is an integer 0,1,2,3...etc.
Since atoms have charge, a spinning nucleus generates a small electric current which in turn creates
a finite but small magnetic field. The magnetic dipole, of the nucleus varies with each element.
Thus, any nucleus with a spin can be observed by NMR spectroscopy.
When a nucleus with a spin is placed in a magnetic field the nuclear magnet experiences a torque
which tends to align it with the external field just like a compass needle in the earth’s magnetic
field.
For a nucleus with a spin of 1/2, there are two allowed orientations of the nucleus: parallel to the
field (low energy) and against the field (high energy). Since the parallel orientation is lower in
energy, this state is slightly more populated than the anti-parallel, high energy state.
spin = –1/2
Turn on
mangetic field
Field splits system
into two allow spin states
spin = +1/2
Direction of
Applied Field
Page 1
Energy
NMR spectroscopy
The energy difference between the state is proportional to the magnetogyric ratio of the nucleus and
the magnitude of the applied field as shown in the figure and equation below.
Applied Magnetic Field
Planck's constant
9.532 x10-14 Kcal/mol
ΔE =
h X νH
2π
Magnetogyric ratio
(26,753 radian/gauss-sec for protons)
Happ
Applied magnetic field (variable)
70460 gauss for a routine machine
An applied magnetic field of roughly 7000 gauss corresponds to energy difference of 2.85 x 10-5
Kcal/mol for protons, an extremely small difference
If you have 2,000,000 protons in such a field roughly 999,995 would be spin down aligned against
the field and 1,000,005 would be spin up with the field.
Since E = hν, the frequency of electromagnetic radiation needed to promote the nuclei from the
lower energy to higher energy state, for the condition described in the equation above corresponds
to a frequency of = 300 x 106 sec-1 or 300 MHz which is FM radio frequency
Page 2
In the NMR experiment a radio frequency applied to molecules is applied to the nucleus in the
magnetic field.
apply RF corresponding
to exactly ΔE and a spin is
promoted to the higher
energy state
spin = –1/2
ΔE
spin = +1/2
Peak corrsponding to
absorption of energy
Intensity
Energy of Spin States
When he RF radiation energy exactly matches the energy gap between the energy level for the spin
aligned with the magnetic field and the spin antiparallel to the magnetic field the nuclei can absorb
the energy and be promoted from the lower energy state to the higher energy state.
Chemical Shift
When this condition is met the nuclei are said to be in resonance with the applied field and this
absorption of radiation leads to a peak in a spectrum.
Page 3
NMR and chemical structure.
Four types of information from an NMR spectra make it an extremely valuable technique for
determining chemical structure.
1. The number of sets of peaks: This corresponds to the number of magnetically (chemically)
different types of protons.
2. Chemical Shift: The key to NMR spectroscopy is that the exact energy at which this peak
comes (which can be measured extremely accurately) is dependent upon the local environment
of the proton.
3. Splitting: The number of peaks observed for a given proton is dependent upon the number and
type of protons in the vicinity of the one that is being observed.
4. Integration: The area of the peak is proportional to the number of protons that are magnetically
equivalent.
The NMR spectrum of methyl isopropyl ether (CH3 -OCH(CH3 )2 ) is shown below:
Note that there are three sets of resonances. One has seven peaks, one has one peak and one has
two peaks. They show up in different places in the spectrum and they have different integrated
areas under each of the resonances.
It is the combination of all of this information that enables us to assign the structure to the
molecule.
Page 4
Chemical Shift
Naively you might expect that all nuclei of a given type would undergo the spin-flip transition at
exactly the same applied frequency in a given magnetic field. However, electrons in the molecule
have small magnetic fields associated with them which tend to oppose the applied field.
Electron Cloud
Happ
Induced Current
Induced Magnetic field
at nucleus
These magnetic fields are said to screen the nuclei from the full strength of the applied field.
The greater the electron density, the greater this 'shielding' will be, hence nuclei which are in electron
rich environments will undergo transition at a higher applied field than nuclei in electron poor
environments.
Thus, the relevant field at the nucleus which determine the energy of the transition now becomes
Heff = Happ + Hshielding
Remember that Hshielding is opposite in sign to Happ.
The resulting shift in the NMR signal for a given nuclei is referred to as the chemical shift. For 1 H
the chemical shift is given relative to the absorption of tetramethyl silane ((CH3 )4 Si)(TMS).
The chemical shift in Hz is proportional to the operating frequency and is given by
Chemical shift (Hz) = νH – νTMS.
Page 5
The delta scale for chemical shifts
Chemical shifts are reported in a manner that is independent of frequency by normalizing both the
frequency of the observed hydrogens and those for TMS hydrogens by the operating frequency of
the instrument as defined by the equation shown below:
Chemical Shift (δ) = (νH – νTMS (Hz))/operating frequency (Hz) x 106 ppm
The factor of 106 is introduced into the equation to give a simple whole number scale for
convenience.
In this scale TMS is arbitrarily assigned a value of zero. TMS is used because its protons are more
highly shielded than those observed in most common organic molecules and because it is
chemically inert. For 1 H NMR, the δ scale generally extends from 0-12 ppm and then each
chemically inequivalent nucleus has it own chemical shift that is independent of the operating
frequency of the NMR.
The ppm scale is plotted from δ = 0 on the right and with increasingly deshielded nuclei (increasing
δ) to the left. Peaks to the left are referred to a “downfield” of peak to the right since it would
require a smaller applied field for them to be in resonance with a fixed excitation frequency.
• Electronegative groups are "deshielding" and thus protons on carbons in the vicinity of
electronegative atoms are deshielded and appear at higher chemical shifts. The more
electronegative the group the further down field the proton will be.
•
If the electronegative group is directly attached to the carbon the effect is greatest and if it is
attached to the α carbon it is somewhat damped and more damped when attached to the β
carbon and so on.
•
If multiple electronegative groups are attached then there is an additive (although not strictly
additive) deshielding effect on the proton.
Protons on oxygen or nitrogen have highly variable chemical shifts which are sensitive to
concentration, solvent, temperature, etc.
Page 6
Anisotropic Effects
The π-system of alkenes, aromatic compounds and carbonyls strongly deshield attached protons
and move them "downfield" to higher δ values.
This is due to an effect known as anistropic shielding and deshielding.
Happ
Happ
O
H
H
H
Happ
H
Happ
So now the total magnetic field is given by:
Heff = Happ + Hshielding + Hanisotropic shielding
Thus, for aromatic rings, olefins and carbonyl compounds this additional magnetic field adds to
Happ and for terminal acetylenes the field substracts from Happ.
Page 7
Proton NMR Chemical Shifts for Common Functional Groups
Page 8
Functional Group and NMR shifts
δ
Cyclopropane
C3 H6
0.22
Ethane
CH3 -CH3
0.88
Ethylene
CH2 =CH2
5.84
Acetylene
HC≡ CH
2.88
Benzene
C6 H6
7.27
Propene
CH2 =CH-CH3
1.71
Propyne
CH≡ C-CH3
1.80
Acetone
CH3 -CO-CH3
2.17
Cyclohexane
C6 H1 2
1.44
Methyl chloride
CH3 Cl
3.10
Methylene chloride
CH2 Cl2
5.30
Chloroform
CHCl3
7.27
Ethanol
CH3 CH2 OH
1.22
CH3 CH2 OH
3.70
CH3 CH2 OH
2.58
CH3 -COOH
2.10
CH3 -COOH
8.63
CH3 -CHO
2.20
CH3 -CHO
9.80
(CH3 CH2 )2 O
1.16
(CH3 CH2 )2 O
3.36
CH3 COOCH2 CH3
2.03
CH3 COOCH2 CH3
1.25
CH3 COOCH2 CH3
4.12
Trimethylamine
N(CH3 ) 3
2.12
Triethylamine
N(CH2 CH3 )3
2.42
Toluene
C6 H5 -CH3
2.32
Benzaldehyde
C6 H5 -CHO
9.96
Acetic acid
Acetaldehyde
Diethyl ether
Ethyl acetate
Page 9
Chemical Shift Table II.
Starting Value of Shift Positions: Methyl, δ = 0.90 methylene, δ =1.20 methine, δ =1.55 Alpha
refers to the substituent being on the same carbon as the hydrogen and for β it is one removed.
Functional Group R
Chlorine
Bromine
Iodine
Aryl
-HC=O and -RC=O
-CO2 H and -CO2 R
-C=C-O-H and -O-Alkyl
-O-Aryl
-OCO-Alkyl
Amine (R=nitrogen)
-C≡ N
-NO2
-C≡ C-
Type of Hydrogen
CH3 -CH2 -CHCH3 -CH2 -CHCH3 -CH2 -CHCH3 -CH2 -CHCH3 -CH2 -CHCH3 -CH2 -CHCH3 -CH2 -CHCH3 -CH2 -CHCH3 -CH2 -CHCH3 -CH2 -CHCH3 -CH2 -CHCH3 -CH2 -CHCH3 -CH2 -CHCH3 -CH2 -CH-
Page 10
Alpha Shift
2.30
2.30
2.55
1.80
2.15
2.20
1.30
1.95
2.70
1.45
1.45
1.35
1.25
1.10
0.95
1.20
1.00
0.95
0.90
0.75
1.25
2.45
2.30
2.20
2.95
3.00
3.30
2.90
2.95
3.45
1.25
1.40
1.35
1.10
1.10
1.05
3.50
3.15
3.05
0.90
0.80
0.35
Beta Shift
0.60
0.55
0.15
0.80
0.60
0.25
1.10
0.60
0.35
0.35
0.55
---0.25
0.30
--0.25
0.30
--0.05
0.10
--0.35
0.15
---0.40
0.45
---0.40
0.45
---0.20
0.15
--0.45
0.40
--0.65
0.85
--0.15
0.05
---
Integration
One further feature of the proton NMR is the fact that the intensity of the absorbance of a given
class of nuclei (with a certain chemical shift) is proportional to the number of protons giving rise to
the signal; that is, the area under a given peak (the integration) is directly proportional to the number
of that type of proton in the molecule.
Integrations are typically given as simplest whole-number ratios, hence, acetic acid, CH3 COOH, will
have two peaks in the proton NMR, one at δ ~ 2, area = 3, and a second at δ ~ 12, area = 1. t-butyl,
methylether, (CH3 )3 COCH3 , will also have two peaks in the proton NMR, one at δ ~ 1, area = 3,
and a second at δ ~ 3.5, area = 1 (the relative areas or both peaks are the same, but each one
represents three hydrogens).
Equivalence
Homotopic hydrogens: If you have a methylene group, of the form X- CH2 -X and you replace one
of hydrogens of the CH2 with a dummy atom and then you independently replace the other
hydrogen of the CH2 group with a dummy atom, then the two molecules thus created will be
identical to each other, the hydrogens are said to be homotopic. The protons on a methyl group are
homotopic.
Enantiotopic: If you have a methylene group, of the form X- CH2 -Y and you replace one of the
hydrogens of the CH2 with a dummy atom and then you independently replace the other hydrogen
of the CH2 group with a dummy atom, the two molecules thus created will be enantiomers of each
other, the protons are said to be enantiotopic. In a nonchiral environment enantiotopic protons are
equivalent. However, in a chiral environment such as a chiral solvent, they can, in principle, have
different chemical shifts.
Diastereotopic: If you have a methylene group, for example, in a chiral molecule X- CH2 -Y* (where
the * indicates that Y is a chiral group) and you replace one of the hydrogens of the CH2 with a
dummy atom and then you independently replace the other hydrogen of the CH2 group with a
dummy atom the two molecules thus created will be diastereomers to each other thus, in principle,
the two hydrogens should have different chemical shifts.
Page 11
Spin-Coupling/Splitting
Proton NMR
Diethyl ether, exhibits two sets of protons in the NMR spectrum; a CH3 group around δ = 1, and a
CH2 group around δ 4 (shifted downfield by its proximity to the electronegative oxygen).
The NMR spectrum of diethyl ether does not have simply two peaks, rather, has two sets of peaks,
one with four lines of relative intensity 1:3:3:1 which in total integrate to two protons and one set of
three lines of relative intensity 1:2:1 which in total integrate to three protons.
This multiplicity is due to the phenomena known as spin coupling and arises because of interaction
of among proton magnetic field mediated by bonding electrons.
To understand this better let's first examine a simple case in which one has X2 CHa–CHb Y2 .
Each proton has spin 1/2 and to first approximation there is an equal probability of its spin being
aligned with (or aligned against) the applied magnetic field.
Page 12
Coupling in X2 CHa –CHbY2
So if we observe Ha, for 50% of the molecules in the sample Hb is aligned with the field:
•
thus the total field is slightly larger and
•
thus for these molecules the resonance will occur at slightly higher frequency
For the other 50% of the molecules in the sample Hb is aligned against the field:
•
thus the total field is slightly smaller
•
thus for these molecules the resonance will occur at slightly higher frequency
As a result, we observed two peaks for Ha whose total integrated area adds to 1.
Similarly we observed two peaks for Hb whose total integrated area adds to 1.
The energy difference between the two peaks for Ha in Hz is the coupling constant (noted as J):
• because there are two peaks Ha would be called a doublet
•
when two nuclei are coupled to each other the coupling constants are identical
•
thus Hb would also be a doublet
•
furthermore, the energy difference in Hz between the two peaks for Hb would equal that of Ha in
Hz since Jab = Jba
Thus, spin–spin coupling is very useful to provide information about the connectivity in organic
molecules since:
• protons on adjacent atoms can couple to each other
• and if there are multiple occurances of this happening in the spectra one can match up pairs of
coupled protons by checking to see which set have identical coupling constants
As we will see shortly, depending upon the distance and orientation of the protons and the
hybridization of the carbons to which they are attached the coupling constants between protons have
distinctive values.
Page 13
The ethyl group
Now lets examine a slightly more complicated case of an ethyl group as shown below:
HA HB
HA
C
C
α-spin
β-spin
O
HA HB 2
JAB
Contribution to magnetic
field felt by HA from HB's
H
H
JAB
Contribution to magnetic
field felt by HA from
spectrometer
Resultant net magnetic
field felt by HA
HA HB
HA
C
C
HA HB
O
2
JBA
Contribution to magnetic
field felt by HB from HA's
H
JBA
JBA
H
Contribution to magnetic
field felt by HB from
spectrometer
Resultant net magnetic
field felt by HB
In the spectrum for diethyl ether, the CH3 group is split by the two protons on the adjacent CH2
group into three peaks (a triplet).
This results from the adjacent spin either being +1/2, +1/2 or +1/2, –1/2 or –1/2, +1/2
–1/2, –1/2.
or
Notice that for the middle two spin possibilities, the magetic fields associated with each of the two
spins exactly cancel, and that there are two possible ways for this two occur:
+1/2, –1/2 or –1/2, +1/2.
Likewise, the absorbance for the CH2 is split by the three protons on the methyl group into (n + 1)
= 4 peaks (a quartet). The two ethyl groups in diethyl ether show as a single absorbance since the
Page 14
molecule has a plane of symmetry, that is, both ethyl groups are chemically and magnetically
equivalent:
• typical values for J are typically less than 20 Hz
• for simple splitting the number of peaks observed for a proton next to n equivalent protons is n
+1
• the intensity of the lines for the observed resonance can be shown to be that as the prefactors
from the binomial expansion, (1 + x)n , which follows Pascals triangle, shown below.
n=0
1
2
3
4
5
6
many
1
1
1
1
1
2
3
singlet
doublet
triplet
quartet
pentet
hextet
septet
multiplet
1
3
1
1
4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
mess
Page 15
s
d
t
q
p
h
sp
m
Splitting Trees
The effects of splitting are often described using a splitting tree which depicts the original
absorbance being split by a coupling constant J into (n + 1) peaks.
This is illustrated for the example below:
note peak heights
not to absolute scale
HC
CH2
Jab
Jab Jab
CH2
CH
CH2 peaks need to
be 2X
A proton can be coupled to two inequivalent protons in which case the magnetic field coupling
between each pair would not necessarily be equal. This is illustrated below:
HaC
CHb
CHc
Jab
Jab
Jbc
Jbc
Jbc
Jab
CHa
CHc
Jab
CHb
In this case, you can see that for Hb coupling to Ha would yield a doublet with coupling constant Jab.
When we consider the effect of coupling to Hc each of these lines are further split into two lines
separated by an amount corresponding to Jbc.
Such a pattern is called a double of doublets.
Page 16
More complex situations
What happens to a doublet of doublet as Jab– Jbc decreases. This is illustrated below.
Jab = Jbc
decreasing J ab – Jbc
The simple coupling patterns we have seen so far assume the difference in chemical shift between
the peaks (in Hz) is much larger than the coupling constants between the hydrogens.
When this condition is met the spectrum is said to have first-order coupling. However, when Δδab is
not much larger than Jab then the coupling is said to be non-first-order and the appearance of the
spectrum changes. In some cases quite dramatically.
Jab
Jab
Δδab >> Jab
Jab
Jab
Δδab > Jab
Jab
Jab
Δδab ~ Jab
Δδab = 0
In particular, the inner lines of the pattern gain intensity at the expense of the outer line as shown
below. The origin of this evolution is in quantum mechanics, which is beyond the scope of the
course.
These patterns are quite common and being able to recognize them coupled with a knowledge of
which hydrogen atoms have similar chemical shift is a useful tool to make assignments, since you
now have two pieces of information (i.e. the two hydrogen atoms are likely adjacent and have
similar environments).
Note that unlike chemical shift coupling constants are due to the internal magnetic field of the
protons and thus coupling constants are independent of spectrometer frequency.
Page 17
Things are not always ideal in the real world
A splitting tree such as this is useful in understanding more complex splitting patterns, such as
those that occur in Br-CH2 CH2 CH2 -OD, as shown below:
As shown by the coupling tree, the spectrum is predicted to consist of 9 peaks centered around
δ 1.53. The actual NMR, as shown above, shows only five peaks in this region.
This is because the central seven peaks are only separated by 2 Hz, and the peaks are not infinitely
narrow. As a result some overlap and appear as single peaks.
Page 18
Common 1 H NMR Splitting Patterns
The examples given above represent only the simplest and most common coupling patterns seen in
the 1 H NMR.
A simple ethyl group displays a quartet and a triplet in the ratio 2:3; the chemical shift of the CH2
group is sensitive to the attached substituent and typically varies between δ 4 (for oxygen) to δ 2
(for a carbonyl).
An isopropyl group displays a septet (7 peaks) and a doublet in the ratio 1:6; again, the chemical
shift of the CH group is sensitive to the attached substituent and typically varies between δ 4 (for
oxygen) to δ 2 (for a carbonyl).
A 1,4-disubstituted aromatic compound displays two doublets in the ratio of 1:1, typically in the
region around δ 7.
Although the chemical shift of a monosubstituted aromatic compound should give rise to three sets
of peaks (integration ratio 2:2:1), often only a single peak of integrated intensity 5 is observed
around δ = in the 1 H NMR.
This is because the differences in chemical shift are typically small, as are the coupling constants.
Page 19
Table of coupling constants between Hydrogen (given in Hz)
H
H
C
C C
H
H
12-20
0-3.5
H
H
H
H
C C
C C
2-9
6-14
C H
H
C C
C C
H
H
11-18
4-10
H
C H
C C
O
C C
H
H
10-13
C C
H
1-3
3-7
H
C
C
H
2-3
Ha
H
H
H
C H
C
He
He
H
Ha
H
H
7-10
2-3
0.1-
Page 20
Jaa = 10-13
Jae = 2-5
Jee = 2-5
Use of deuterium in NMR
Although deuterium has a nuclear spin, deuterium NMR and proton NMR require different
operating frequencies.
•
Deuterium NMR absorptions are not detected under the conditions used for proton NMR.
•
The coupling constants for proton-deuterium splitting are very small.
•
Deuterium substitution can be used to simplify NMR spectra and assign resonances.
•
Solvents used in NMR spectroscopy must either be devoid of protons, or their protons must
not have NMR absorptions that obscure the sample absorptions.
NMR spectra of dynamic systems
•
•
•
The spectrum of a compound involved in a rapid equilibrium is a single spectrum that is the
time-average of all species involved in the equilibrium.
The time-averaging effect of NMR is not limited to simple conformational equilibria; the spectra
of molecules undergoing any rapid process, such as a chemical reaction, are also averaged by
NMR spectroscopy.
The assignment of the OH proton can be confirmed by adding a drop of D2 O to the NMR
sample tube shaking and taking the spectra. The OH protons rapidly exchange with the protons
of D2 O to form OD groups on the alcohol and thus become invisible to NMR.
NMR of alcohols and amines
The chemical shift of the OH proton in an alcohol depends on the degree to which the alcohol is
involved in hydrogen bonding under the conditions that the spectrum is determined.
• The presence of water, acid or base causes collapse of the OH resonance to a single line and
obliterates all coupling associated with this proton.
• This effect is caused by a phenomenon called chemical exchange.
• This type of behavior is quite general for alcohols, amines, and other compounds with a proton
bonded to an electronegative atom.
• Rapidly exchanging protons do not show spin-spin splitting with neighboring protons.
• Acid and base catalyze this exchange reaction, accelerating it enough that splitting is no longer
observed. In the absence of acid or base, this exchange is much slower, and splitting of the OH
protons and neighboring protons is observed.
• The assignment of the OH proton can be confirmed by adding a drop of D2 O to the NMR
sample tube shaking and taking the spectra. The OH protons rapidly exchange with the protons
of D2 O to form OD groups on the alcohol and thus become invisible to NMR.
Page 21
Introduction to 1 3C NMR
Any nucleus with a nuclear spin can be studied by NMR spectroscopy; for a given magnetic field
strength, different nuclei absorb energy in different frequency ranges.
The NMR spectroscopy of carbon is called carbon NMR or CMR; the only isotope of carbon that
has a nuclear spin is 1 3C.
• The relative abundance of 1 3C occurs in 1.1% natural abundance.
•
The magnetogyric ratio for the 1 3C nucleus is about 0.25 of the 1 H nucleus and the
sensitivity goes as this ratio to the third power which by itself would lead to a 1 3C
nucleus having 0.159 the sensitivity of the 1 H nucleus.
•
Because of these two factors, 1 3C spectra are about 0.0002 times as intense as 1 H
spectra.
•
For weak signals, signal to noise is increased by averaging out the signal from multiple
scans. The noise averages to zero and the signal increases.
•
Signal to noise increases as (# of scans)0.5. Given then relative sensitivity to 1 H NMR
you can see that many scans could be needed.
•
Special instrumental techniques have been devised for obtaining such weak spectra.
•
Coupling (splitting) between carbons generally is not observed. The reason is the low
natural abundance of carbon (1.1%), making it unlikely that two 1 3C nuclei will reside
next to each other in a given molecule.
•
Attached protons, however, with a spin of 1/2, will couple with the
generate spin-coupling.
•
The signal from the carbon will be split into (n + 1) peaks, where n is the number of
attached protons.
Page 22
13
C nucleus to
13
C Chemical shfts
The range of chemical shifts is large compared to that in proton NMR; trends in carbon chemical
shifts parallel those for proton chemical shifts.
Chemical shifts in CMR are more sensitive to small changes in chemical environment.
Characteristics of chemical shifts for carbons:
• simple methyl carbons in alkanes, or attached to an alkene, of aryl group δ 15-30
•
simple methylene, methine and quarternary carbons, carbons in alkanes attached to an
alkene of aryl group δ 20-60
•
•
characteristic chemical shifts for carbons attached to electronegative atoms
oxygen (in ethers or the single bonded oxygen in an ester) δ 40-80,
•
nitrogen (in amines or the single bonded nitrogen in an amide) δ 25-27
•
chlorine δ 20-60
•
bromine δ 10-50
•
iodine δ –15 -25
•
alkyne δ 65-85,
•
nitrile δ 110-130
•
alkene carbons to δ 100-150
•
aromatic carbons δ 110-170
•
amide (C=O) δ 160-180
•
ester (C=O) δ 160-190
•
ketone and aldehyde (C=O) 190-220
Page 23
13
C coupling and decoupling
Splitting of 1 3C resonances by protons (1 3C – 1 H splitting) is so large relative to differences in
chemical shifts that is can be difficult to distinguish between different peaks and peaks that are split
by coupling.
• Spectra can be obtained in proton noise-decoupling mode wherein which the sample is
irradiated at a second frequency which promotes all of the protons in the molecule to high spin
states, disallowing the spin-coupling process. All of the split 1 3C peaks are thereby reduced to
sharp singlets.
• While this operating mode provides less information than the non-decoupled mode, it is
commonly used because it results in a significant signal enhancement due to a phenomena
known as the "Nuclear Overhauser Effect" (NOE).
• NOE enhancement is not uniform, hence the integration of a 1 3C NMR spectrum is typically
meaningless.
Spectra in which proton coupling has been eliminated are called proton-decoupled carbon NMR
spectra.
•
•
The DEPT (distortionless enhancement by polarization transfer) technique yields separate
spectra for methyl, methylene and methine carbons, and each line in these spectra corresponds
to a line in the complete CMR spectrum; lines in the complete CMR spectrum that do not
appear in the DEPT spectra are assumed to arise from quaternary carbons.
CMR spectra are generally not integrated because the instrumental technique used for taking
spectra, gives relative peak integrals that are governed by factors other than the number of
carbons.
Page 24