Reich
Chem 345
Univ. Wisconsin, Madison
NMR Spectroscopy
Chemical Shifts
Chemical shifts have their origin in the circulation of electrons induced by the magnetic field, which reduces the
actual field at the nucleus. Thus a higher magnetic field has to be applied to achieve resonance. Different types of
protons in a molecule are surrounded by different electron densities, and thus each one sees a slightly different
magnetic field.
•H
Bo
eA
B = Bo - Be
(magnetic field at nucleus)
νo = γB/2π
(Larmor precession frequency of H A)
Be
The Larmour precession frequency νo depends on the magnetic field strength. Thus at a magnet strength of 1.41
Tesla protons resonate at a frequency of 60 MHz, at 2.35 Tesla at 100 MHz, and so on. Although Hz are the
fundamental energy unit of NMR spectroscopy, the use of Hz has the disadvantage that the position of a peak is
dependent on the magnetic field strength. This point is illustrated by the spectra of 2-methyl-2-butanol shown below at
several different field strengths, plotted at a constant Hz scale.
Effect of Spectrometer Magnetic Field Strength
c
CH3 b a
HO C CH2 CH3
CH3
d
c
Me4Si
220 MHz
400
300
200
100
0
c
a
d
b
100 MHz
200
100
0
100
0
60 MHz
For this reason, the distance between the reference signal (Me 4Si) and the position of a specific peak in the
spectrum (the chemical shift) is not usually reported in Hz, but rather in dimensionless units of δ, which is the same on
all spectrometers.
δ =
(Frequency shift from Me4Si in Hz)
(Spectrometer frequency, MHz)
1
1
H Chemical Shifts
H
H
X=O,Cl,Br
X H
H
O
H
H
X=N,S
X H
H
O
Alkanes
H
R
H
10
9
8
7
6
5
4
3
2
1
0
δ ppm
Downfield
Deshielded
High frequency
Bo decreases
Bo increases
νo increases
νo decreases
Upfield
Shielded
Low frequency
The ranges above provide an estimate of the chemical shift for simple molecules, but don't help very much when
there are multiple substituents. A simple scheme can be used to estimate chemical shifts of protons on sp 3 carbons.
Use the base shift for methyl groups. CH2 groups, and CH groups, and add to these the increments for each α
substituent:
Base Shift
CH3
0.9
CH2
1.2
CH
1.5
Increment
OC(=O)R
OR
Br
Cl
Aryl
C(=O)R
C=C
3.0
2.3
2.2
2.4
1.4
1.0
1.0
Base shift CH:
α Ph:
α OH:
1.5
1.4
2.3
Calculated:
Observed:
5.3
4.8
2
OH
Ph
Cl
Base shift CH2:
α Cl:
1.2
2.4
Calculated:
Observed:
3.6
3.65
Representative Proton Chemical Shift Values (δ -4 to 6)
(Me3Si)3Si-Te-H
Li
Ph-Te-H
-8.8
Me-Te-H
(Me3Si)3Si-Se-H
H
-5.5
-2
-2.1
-2.2
-2.3
H-Se-CH2Ph
0
-0.1
-2.4
-2.5
-0.3
-0.4
-2.7
-2.8
-2.9
-0.5
-0.6
-0.7
-3.1
-3
-3.2
-3.3
-3.4
-3.5
-3.6
-3.7
-3.8
-3.9
-4
-1.2
-1.3
-1.4
-1.5
-1.6
-1.7
-1.8
-1.9
-2
(CH3)2Mg
(Me3Si)3Si-S-H
CH3Li
-0.2
-2.6
-0.8
-0.9
-1.1
-1
CH3
N≡C
CH3
CH3
CH3-CH2-CH3
(CH3)3CCl
CH3-C≡N
(CH3)3COH
CH3CH2Br
CH3-CO2Me
CH3CH2I
2
1.9
1.8
HO
H
H
H
1.6
1.5
1.4
1.3
1.2
1.1
0.9
1
O
H-C≡C-S-Ph
O
CH3-CO2CH3
(CH3O)2CH2
O
Ph-C-C≡C-H
CH3O-Ph
Br-CH2CH2-Br
CF3CH2OH
H
O
CH3CH2OH
(CH3)4Sn
(CH3CH2)2CO (CH3)3CH
H
1.7
(CH3)4C
CH3CH2OH
H
(CH3)3CBr
(CH3)4Si
0.8
N
H
N
H
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
(Me2N)3P=O
CH3CCl3
H
(CH3)2N-Ph
Ph-CH2-CH3
(CH3)2O
CH3OH
CH3CH2I CH3Cl
H-C≡C-Ph
3.4
3.2
2.9
CH3Br
(CH3)3N
O
(CH3CH2)2CO
(CH3)2S
CH3
O
H-C≡C-H
(CH3CH2)3N
Ph-CH3
CH3I
CH3
CH3
O
4
3.9
3.8
3.7
3.6
3.5
3.3
3.1
3
H
H
H
(CH3O)2CH2
ClCH2CH2Cl
CHCl2CO2Me
6
5.9
5.8
H
H Ph
5.7
5.6
5.5
2.6
H
HC(OEt)3
2.5
2.4
H
H
H
H
OAc
Ph
H-SnMe3
CH2Cl2
H
OAc
5.4
5.3
H
5.2
H
5.1
5
H
H
PhCH2Br
3
4.9
4.7
HO
H
H
H
CH3F
H
4.8
4.6
2.1
2
(CH3)2CHCl
O
CH2Br2
2.2
H CH3NO2
Me3Si
H2SiPh2
PhCH2Cl
H
2.3
H
N2
H
H
2.7
SiMe3
H
O
2.8
4.5
4.4
4.3
4.2
O
Me-C-OCH2CH3
4.1
4
Representative Proton Chemical Shift Values (δ 6 to 12)
O
HCCl3
N
O
H
Me
H
H
O
AcO
H
H
H
H
H
O
H
OMe
H
H
CH3
O
OMe
8
7.9
H
7.8
HO2C
H
7.7
7.6
7.5
7.4
H
HO
H
H
H
H
H
EtO2C
H
H
H
Ph
7.3
7.2
H
H
7.1
CO2Et
6.9
7
6.8
6.7
6.6
6.5
6.4
H
CCl3CCl2H
6.3
6.2
6.1
6
O
OMe
NO2
O
H
O
H
H
O
Ph
Me
10
9.9
O2N
NO2
H
H
9.8
9.7
9.6
9.5
9.4
NMe2
NMe
9.3
9.2
9.1
8.9
9
O
8.8
8.7
N
H
8.6
8.5
H
H
Ph
8.4
8.3
8.2
8.1
8
10.4
10.3
10.2
10.1
10
N
H
H
H
H
11.1
N
O
H
H
O
H
OMe
OMe
O
S
t
Bu
OH
CH3CO2H
H
H
O
O
12
11.9
11.8
11.7
11.6
11.5
11.4
11.3
11.2
11.1
11
10.9
10.8
10.7
10.6
10.5
O
Se
iPr3Si
H
17.3
H
Me
S
O
H
O
H
14.9
14
13.9
O
13.8
H
HgH
O
16.7
13.7
13.6
13.5
13.4
13.3
13.2
13.1
13
4
12.9
12.8
12.7
12.6
12.5
12.4
12.3
12.2
12.1
12
Reich
Chem 345
Integration of NMR Spectra - Number of Protons
NMR is unique among common spectroscopic methods in that signal intensities are directly proportional to the
number of nuclei causing the signal (provided certain conditions are met). In other words, all absorption coefficients
for a given nucleus are identical. This is why proton NMR spectra are routinely integrated, whereas IR and UV spectra
are not. A typical integrated spectrum is shown below, together with an analysis.
O
30 20 10 0
Hz
26.5 mm
16.2 mm
11.8 mm
8.5
8.0
7.5
7.0
6.5
6.0
5.5
5.0
4.5
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.0
If given the molecular formula (C9H10O), there are 10H in molecule
Total area: 26.5 + 11.8 + 16.2 = 54.5 mm
Thus 5.5 mm per H
26.5 / 5.5 = 4.86 i.e. 5H
11.8 / 5.5 = 2.16 i.e. 2H
16.2 / 5.5 = 2.97 i.e. 3H
The vertical displacement of the integral gives the relative number of protons. It is not possible to determine the
absolute numbers without additional information (such as a molecular formula). Sometimes a numeric value will be
given, or sometimes, as in the example above, you have to measure the distance with a ruler. In this example, if we
add up all of the integrals, we get 54.5; dividing by the number of hydrogens in the molecular formula gives 5.5 mm per
H. We can then directly estimate the number of protons corresponding to each multiplet by rounding to the nearest
integer. It is generally possible to reliably distinguish signals with intensities of 1-8, but it becomes progressively
harder to make a correct assignment as the number of protons in a multiplet increases beyond 8, because of the
inherent inaccuracies in the method.
The two parts of aromatic proton integral at δ 7.5 - 8.0 can be separately measured as a 2:3 ratio of ortho to
meta+para protons.
5
Coupling - Splitting of NMR Signals
H
Hα
C
→
→
If two protons have different chemical shifts and are within 3 bonds of each other (geminal or vicinal) then the
protons will be coupled to each other: the signal will be split into a doublet (two lines separated by the coupling
constant J) due to two magnetic orientations of the other proton. When there are two, three, or more neighbors,
additional splittings can be observed
H
C
Hβ
C
C
J1
J2
E
s
d
t
s
d
dd
dd
s
d
t
Two equal couplings.
Two different couplings.
When all of the couplings to a given proton are the same, then regular multiplets are formed, with the intensities
shown below:
# of Vicinal
H atoms
Intensities
Pascal's triangle)
Called:
Examples:
0
1
singlet
X-CH3 X-CH2-CH2-X C6H6
1
1 1
doublet
X2CH-CH3 X2CH-CHY2
2
1 2 1
triplet
X-CH2-CH3 X2CH-CH2-CHX2
3
1 3 3 1
quartet
X-CH2-CH3 X-CH2-CH-CHX2
4
1 4 6 4 1
pentet
X-CH2-CH-CH2-X CH3-CH2-CHX2
5
1 5 10 10 5 1
sextet
CH3-CH2-CH2-X CH3-CHX-CH2-R
6
1 6 15 20 15 6 1
heptet
X-CH(CH3)2 (X-CH2)3CH
7
1 7 21 35 35 21 7 1
octet
CH-CH(CH3)2
8
1 8 29 56 70 56 29 8 1
nonet
XCH2-CH(CH3)2
triplet n = 2
quartet n = 3
pentet n = 4
sextet n = 5
heptet n = 6
However, when some of the coupling constants are different, then more complicated multiplets are seen. The
simplest type is the doublet of doublets (dd) which arises from one proton coupled to two neighboring protons by
different coupling constants.
6
Coupling Constants
Coupling constants J vary widely in size, but the vicinal couplings in acyclic molecules that we are mostly going to
be interested in are usually 7 Hz. The leading superscript ( 3J) indicates the number of bonds between the coupled
nuclei.
H
C
2
J = 2-15 Hz
Typical: -12 Hz
C
H
H
C
H
vicinal
geminal
3
C
J = 2-20 Hz
H
4
C
Typical: 7 Hz
C
J = 0-3 Hz
H
long-range
One situation where the size of J provides important information is in the vicinal coupling across double bonds,
where trans couplings are always substantially larger than cis couplings.
H
H
H
J = 14 - 18 Hz
H
J = 8 - 12 Hz
There are also a few situations where coupling across 4 bonds are observed in NMR spectra. This is rarely seen
across single bonds, but small couplings (typically 1-3 Hz) are seen when there are intervening double or triple
bonds.
H
H
H
H
H
H
H
H
4
Meta
J = 2 to 3 Hz
4
Allylic
J = 0 to 3 Hz
4
7
Propargylic
J = 2 to 4 Hz
4
Allenic
J = 6 to 7 Hz
Reich
Chem 345
NMR Spectra with no Coupling
C5H8O4
300 MHz 1H NMR Spectrum
Solv: CDCl3
Source: Aldrich Spectra Viewer/Reich
O
O
MeO
OMe
Dimethyl malonate
5.91
2.00
10
9
8
7
6
C4H8O2
300 MHz 1H NMR Spectrum
Solv: CDCl3
Source: Aldrich Spectral Viewer/Reich
5
ppm
4
3
2
1
0
O
OMe
Methoxyacetone
3.16
3.01
2.05
10
9
8
7
6
5
ppm
8
4
3
2
1
0
Reich
Chem 345
Absence of Splitting between Equivalent Protons
Protons that have the same chemical shift do not show spin-spin splitting. Thus the CH 2 groups of both
1,2-dimethoxy- and 1,2-dibromoethane are singlets, whereas those of Br-CH 2CH2-OCH3, where there is significant
chemical shift between the CH2 groups, are two triplets
C2H4Br2
300 MHz 1H NMR Spectrum in CDCl3
Source: Aldrich Spectral Viewer/Reich
Br
Br
1,2-Dibromoethane
10
9
8
7
6
5
ppm
4
3
2
1
0
Problem R-18U C3H7BrO
300 MHz 1H NMR spectrum in CDCl3
Source: Aldrich Spectra Viewer/Reich
Br
O
1-Methoxy-2-bromoethane
3.8
10
9
3.7
8
3.6
7
3.5
6
3.4
5
4
3
2
C4H10O2
300 MHz 1H NMR Spectrum
Solv: CDCl3
Source: Aldrich Spectral Viewer/Reich
1
0
OMe
MeO
1,2-Dimethoxyethane
6.24
4.00
10
9
8
7
6
5
ppm
9
4
3
2
1
0
Reich
Chem 345
Simple Coupling Patterns
Problem R-18N C2H3Cl3
300 MHz 1H NMR spectrum in CDCl3
Source: Aldrich Spectra Viewer/Reich
Cl
Cl
Cl
1,1,2-Trichloroethane
2.15
1.00
5.80 5.75 5.70
10
9
8
7
4.00 3.95 3.90
6
ppm 5
Problem R-18H C2H4Cl2
300 MHz 1H NMR spectrum in CDCl3
Source: Aldrich Spectra Viewer/Reich
4
3
2
1
0
Cl
Cl
1,1-Dichloroethane
2.86
1.00
5.95 5.90 5.85
10
9
8
7
6
2.10 2.05 2.00
ppm 5
4
3
2
1
0
Problem R-18G C2H5
300 MHz 1H NMR spectrum in CDCl3
Source: Aldrich Spectra Viewer/Reich
Br
Bromoethane
1.45
1.00
3.5
10
9
8
7
6
3.4
5
10
1.70 1.65
4
3
2
1
0
Reich
Chem 345
Simple Coupling Patterns
Problem R-18F C3H7Br
300 MHz 1H NMR spectrum in CDCl3
Source: Aldrich Spectra Viewer/Reich
Br
1-Bromopropane
1.9
1.8
1.05 1.00
1.49
1.03
1.00
3.40 3.35
10
9
8
7
6
Problem R-18E C3H7Br
300 MHz 1H NMR spectrum in CDCl3
Source: Aldrich Spectra Viewer/Reich
5
ppm
4
3
2
1
0
Br
2-Bromopropane
5.93
4.4
4.3
1.75 1.70 1.65
4.2
1.00
10
9
8
7
6
5
ppm
11
4
3
2
1
0
Practice Problems
Reich
Chem 345
Problem R-18Q: C5H10O2
300 MHz 1H NMR Spectrum in CDCl3
Source: Aldrich Spectral Viewer/Reich
30
20
10
0
Hz
1.7
1.00
1.6
0.96
0.66
0.64
0.95 0.90
2.35 2.30 2.25
3.70 3.65
10
9
8
7
6
5
4
3
2
1
0
ppm
Problem R-18C C10H12O
300 MHz 1H NMR spectrum in CDCl3
Source: Aldrich Spectra Viewer/Reich
30
20
10
0
Hz
2.50 2.45 2.40
2.51
1.05 1.00 0.95
1.52
1.00
1.00
7.35 7.30 7.25 7.20 7.15
3.70 3.65 3.60
10
9
8
7
6
5
ppm
12
4
3
2
1
0
Reich
Chem 345
Size of Coupling Constants
Vicinal coupling across double bonds shows a strong stereochemical dependence, with cis couplings (typically 10
Hz) always being less than trans couplings (typically 15 Hz).
H
O
Cl
30
20
10
0
Hz
1870.3
H
1883.9
HO
2244.9
2258.5
Problem R-18P C3H3ClO2
300 MHz 1H NMR spectrum in CDCl3
Source: Aldrich Spectra Viewer/Reich
J = 13.6 Hz
7.4
12
11
7.2
10
9
7.0
8
6.8
7
6
6.6
5
4
3
30
HO
20
10
0
6.2
1
0
1
0
Hz
H
1871.4
1879.9
H
2055.6
2063.7
2
Cl
O
Problem R-18Q C3H3ClO2
300 MHz 1H NMR spectrum in CDCl3
Source: Aldrich Spectra Viewer/Reich
6.4
J = 8.1 Hz
7.0
12
11
6.8
10
9
6.6
8
6.4
7
6
13
6.2
5
4
3
2
OH and NH Protons
The chemical shifts of OH and NH protons vary over a wide range depending on details of sample concentration and
substrate structure. The shifts are very strongly affected by hydrogen bonding, with strong downfield shifts of H-bonded
groups compared to free OH or NH groups. Thus OH signals tend to move downfield at higher substrate concentration
because of increased hydrogen bonding (see the spectra of ethanol below).
Pure ethanol
×
×
OH proton
×
10% EtOH in CCl4
5% EtOH in CCl4
×
0.5% EtOH in CCl4
5.0
4.5
4.0
3.5
3.0
ppm
2.5
2.0
1.5
1.0
0.5
There is a general tendency for the more acidic OH and NH protons to be shifted downfield. This effect is in part a
consequence of the stronger H-bonding propensity of acidic protons, and in part an inherent chemical shift effect. Thus
carboxylic amides and sulfonamides NH protons are shifted well downfield of related amines, and OH groups of phenols
and carboxylic acids are downfield of alcohols.
Chemical Shift Ranges of OH, NH and SH Protons:
Except for alcohols, the shifts are for dilute solutions in CDCl3
R-SO2NH2
R=CF3
R-NH3+
R-CO2H
12
R-SH
Ar-NH2
R-NH2
Ar-OH
R-SO3H
13
Ar-SH
O
R-CNH2 R=CH
3
11
10
9
8
7
6
δ ppm
14
concentrated
R-OH
5
3
4
dilute
2
1
0
Reich
Chem 345
Second Order Effects
When two sets of protons that are coupled to each other are relatively close in chemical shift (i.e. when the chemical shift
between them is similar in size to the coupling between them) simple multiplets are no longer formed. A commonly observed
effect is that the intensities of the lines no longer follow the simple integer ratios expected - the multiplets "lean" towards
each other. In other words, the lines of the multiplet away from the chemical shift of other proton (outer lines) become
smaller and lines closer (inner lines) become larger. This can be seen in the marked triplets below (see next page for a
simpler example). The leaning becomes more pronounced as the chemical shift difference between the coupled multiplets
becomes smaller.
C3H7BrO
300 MHz 1H NMR spectrum in CDCl3
Source: Aldrich Spectra Viewer/Reich
Br
O
"leaning"
3.8
10
9
"leaning"
3.7
8
3.6
7
3.5
3.4
6
5
4
3
2
1
0
In addition there may be more lines than that predicted by the multiplet rules. A nice example is provided by the
compound below. For the BrCH 2CH2O group the two methylenes at δ 3.48 and δ 3.81 have a relatively large chemical
shift separation, and they form recognizable triplets (although with a little leaning). For the MeOCH 2CH2O group the
chemical shift between the CH2 groups is small, and the signals are a complicated multiplet with only a vague
resemblance to a triplet.
C5H11BrO2
300 MHz 1H NMR spectrum in CDCl3
Source: Aldrich Spectra Viewer/Reich
3.8
10
9
8
O
3.7
7
Br
O
3.6
3.5
6
3.4
5
ppm
15
4
3
2
1
0
The "leaning" or "roof" effect
Why equivalent protons do not show coupling
νAB = 50 Hz
HA HB
C C
When the two protons are well separated
in chemical shift, each one is a doublet due
to coupling with the neighboring proton
A
B
νAB = 40 Hz
A
B
As the chemical shift becomes smaller, the
the two peaks closest to each other (the
inner peaks) become larger, and the outer
peaks become smaller
νAB = 30 Hz
A
B
νAB = 20 Hz
A
B
νAB = 10 Hz
A
B
Eventually the outer peaks disappear, and
the inner peaks merge in to one - one sees
only a singlet. So it is not that protons with
the same shift don't couple, it is that the
peaks that would show us the coupling (the
outer peaks) have all disappeared.
νAB = 3 Hz
AB
50
40
30
20
10
0
Hz
-10
-20
-30
-40
16
Coupling to Different Protons
So far, we have seen only spectra where all of the couplings to a proton are the same, so that simple multiplets like triplets,
quartets, etc are formed. However, there are many circumstances where a proton may be coupled to two protons by different
coupling constants, leading not to a triplet, but to a doublet of doublets. One common situation of this type occurs in aromatic
compounds, where both ortho and meta couplings are large enough to see, but the ortho coupling (8 Hz) is much larger than the
meta (2 Hz). The para coupling is usually too small to see. This is thus one of the important exceptions to the rule that protons
separated by more than 3 bonds do not show coupling.
J1
J2
J1 = J2
dd
t
Problem R-23D C7H6BrNO2
300 MHz 1H NMR spectrum in CDCl3
H3
H4
Br
H6
H3
CH3
30
Jortho (coupling to H3)
20
10
0
Hz
Jmeta (coupling to H6)
H6
H4
O 2N
8.1
9
8.0
8
7.9
7
6
ppm
7.8
5
7.7
4
3
2
1
0
Other situations where protons separated by more than 3 bonds show coupling also involve intervening π bonds (double or
triple bonds). Such couplings are typically smaller than the 7 Hz often seen for 3-bond couplings. See if you can assign the
signals in the spectrum below, and identify the couplings.
Problem R-27L C5H8O2
250 MHz 1H NMR spectrum in CDCl3
Source: Adam Fiedler/Reich
H
O
CH3
O
CH3
3.17
H
30
20
10
0
Hz
3.04
7.05
7.00ppm
6.95
6.90
1.01
1.00
5.90 ppm
5.85
8
7
6
5
5.80
ppm
3.75ppm
3.70
4
17
3
1.90
ppm 1.85
2
1
0
Diastereotopic Effects
Diastereotopic protons are defined as two protons which have identical connectivity to the rest of the molecule, but have
different chemical shifts because of some stereochemical feature of the molecule. The situation is simple with gem-alkene
protons - it is easy to see how they are different. However, it is more complicated for sp3 carbons.
H
H
H
Br
Br
These two prrotons are diastereotopic
Cl
H
H
H
These two prrotons are diastereotopic
It turns out that CH2 groups in any molecule that has a true asymmetic center (a center of chirality) anywhere in the
molecule will be diastereotopic (see the substitution test in the text book). An typical example is 1,2-dibromopropane (NMR
below). Rotation around the 1,2-C-C bond does not actually interchange the environment of the two hydrogens. To convince
yourself of this, make two models of 1,2-dibromopropane, and put both in the same conformation. In one mark one of the
hydrogens at C1, in the other mark the other one. Then see if you put the two marked hydrogens in exactly the same
environment by rotating the bonds (this is the substitution test done with models).
You can see that 1,2-dibromopropane has four sets of signals, with the two protons of the CH2 group separated by about
0.3 ppm. Not only are the shifts of the two C-1 protons different, but the coupling constant to the C-2 proton is also different.
The C-1 H-C-H 2-bond coupling (to the other proton at C-1) is accidentally nearly the same as the H-C-C-H 3-bond coupling
(to the proton at C-2) for one of the protons at C-1. This gives the triplet at δ 3.55 Some people call these "apparent triplets"
because the two couplings are certainly different, but apparently not by much. For the other proton at C-1 the H-C-C-H
coupling is much smaller, and so a dd is seen at δ 3.86. The proton at C-2 is pretty complicated - it is actually a doublet of
doublets of quartets (ddq) from coupling to the two different protons at C-1 and the methyl group at C-3.
Problem R-22C C3H6Br2
300 MHz 1H NMR spectrum in CDCl3
Source: ASV/Reich
Br
Br
30
20
10
0
Hz
1.8
4.3
4.2
4.1
4.0
3.9
3.8
3.7
3.6
3.5
2.90
1.00
5
0.94
4
0.89
3
2
ppm
18
1
0
Effect of Electron Donating and Withdrawing Substituents on NMR Chemical Shifts
Problem R-19C (C6H7N)
300 MHz 1H NMR spectrum in CDCl3
Source: Aldrich Spectra Viewer/Reich
3.13
NH2
2.00
8.5
8.0
7.5
7.0
6.5
6.0
C7H8O
300 MHz 1H NMR spectrum in CDCl3
Source: Aldrich Spectra Viewer/Reich
3.00
2.00
OMe
8.5
8.0
7.5
7.0
6.5
6.0
8.0
7.5
7.0
6.5
6.0
7.0
6.5
6.0
C6H5Cl
300 MHz 1H NMR spectrum in CDCl3
Source: Aldrich Spectra Viewer/Reich
Cl
8.5
C6H5NO2
300 MHz 1H NMR spectrum in CDCl3
Source: Aldrich Spectra Viewer/Reich
3.12
2.00
NO2
8.5
8.0
7.5
ppm
19
Reich
Chem 345
Isomeric Methoxynitrobenzenes
Problem R-19B (C7H7NO3)
300 MHz 1H NMR spectrum in CDCl3
Source: Aldrich Spectra Viewer/Reich
8.2
8.1
10
8.0
9
7.9
8
7.8
7.7
7
7.6
6
7.5
5
7.4
7.3
4
7.2
3
7.1
7.0
2
6.9
1
0
ppm
8.2
8.1
10
8.0
9
7.9
8
7.8
7.7
7
7.6
6
7.5
5
7.4
7.3
4
7.2
3
7.1
7.0
2
6.9
1
0
ppm
8.2
10
8.1
8.0
9
7.9
8
7.8
7
7.7
7.6
6
7.5
5
ppm
20
7.4
4
7.3
7.2
3
7.1
2
7.0
6.9
1
0
Reich
Chem 345
Three Isomeric Trichlorobenzenes
Problem R-19A Three isomers of C6H3Cl3
300 MHz 1H NMR spectrum in CDCl3
Source: Aldrich Spectra Viewer/Reich
30
20
10
0
7.6
10
9
8
7
6
10
9
9
8
8
7
7
7.5
7.4
5
7.6
10
Hz
7.5
7.3
4
7.4
7.1
3
7.3
6
5
7.5
7.4
7.3
6
5
4
21
7.2
4
7.0
2
7.2
3
7.1
1
7.0
2
7.2
3
6.9
0
6.9
1
0
1
0
7.1
2
Reich
Chem 345
Carbon-13 NMR Spectroscopy
Chemical Shift Ranges:
N
C
150-170
O
C
C
200-218
X
Carboxylic esters
acids, amides
O
C
Me4Si
C=C
C
C
Ketones, aldehydes
C
220
13
200
180
160
140
120
C
R3C-O-
N
100
δ
Alkyl halide
alkyl amine
80
Alkanes
60
40
20
0
-20
C Chemical shifts of some simple compounds:
H
C
CH2=N2
Me3COH
Me2CHOH
CH3CH2OH
CH2=C=CH2
MeOH
HC≡CH
Me3N
Me2O +NMe4
HOCH=CH2 CDCl3
88.0
90
100
77.8
69.6
73.9 73.2
70
80
61.2 55.6
64.6
58.2
N
MeCl
O
Me-Me
Me4Si
CH4
MeSH
MeI
MeBr
MeLi
50.2
47.6
27.8
18.2
25.223.1
39.7
50
60
H
C O
H
30
40
20
10.3 5.9
0.0 -2.9
6.5 2.5
-2.1
10
-13.2
-10
0
-20.0
-20
δ
O
Me + Me
N
O
CH2=C=CH2
H
H
C O
H
206.2
199.6
210
190
180
HC(OEt)3
H2C=N-Me
⊕
177.0
200
NH2
CH2=CH2
- +
C≡N-Me
OMe
194.0
211.7
220
H
O
H
O
H
CO3H-
O=C=NMe
160
170
150
δ
22
NC-Me
HOCH=CH2
169.9 164.9 160 156.7 149.0
167.9
158.2
128.5
123.2 117.7 113.9
127.2 121.7
140
Li
130
120
110
102.6
100
Reich
Chem 345
Carbon-13 NMR Spectroscopy
O
Shown below are the 13C NMR spectra of three isomers of C6H10O:
OH
Determine the expected numbers of carbons for each isomer.
Determine which spectrum corresponds to which compound
Identify the types of carbon signals, do a rough assignment
19.0
25.1
32.0
SiMe4
160
140
120
100
ppm
80
60
200
180
160
140
120
100
ppm
80
60
200
180
160
140
120
100
ppm
80
60
40
40
42.1
211.8
23
20
40
0
-20
20
0
-20
20
0
-20
19.6
180
52.1
200
24.6
130
27.0
25.0
131
CDCl3
130.2
130.0
65.4
O