Nmr2 pl

Nuclear Magnetic Resonance Spectroscopy
1
Presented
By
Uttam Prasad Panigrahy

• Nuclear magnetic resonance spectroscopy is a powerful
analytical technique used to characterize organic
molecules by identifying carbon-hydrogen frameworks
within molecules.
• Two common types of NMR spectroscopy are used to
characterize organic structure: 1H NMR is used to
determine the type and number of H atoms in a
molecule; 13C NMR is used to determine the type of
carbon atoms in the molecule.
• The source of energy in NMR is radio waves which
have long wavelengths, and thus low energy and
frequency.
• When low-energy radio waves interact with a molecule,
they can change the nuclear spins of some elements,
including 1H and 13C.
2
Introduction to NMR Spectroscopy

• When a charged particle such as a proton spins on its axis, it
creates a magnetic field. Thus, the nucleus can be considered
to be a tiny bar magnet.
• Normally, these tiny bar magnets are randomly oriented in
space. However, in the presence of a magnetic field B0, they
are oriented with or against this applied field. More nuclei are
oriented with the applied field because this arrangement is
lower in energy.
• The energy difference between these two states is very small
(<0.1 cal).
3

4
• In a magnetic field, there are now two energy states
for a proton: a lower energy state with the nucleus
aligned in the same direction as B0, and a higher
energy state in which the nucleus aligned against B0.
• When an external energy source (hn) that matches the
energy difference (DE) between these two states is
applied, energy is absorbed, causing the nucleus to
“spin flip” from one orientation to another.
• The energy difference between these two nuclear spin
states corresponds to the low frequency RF region of
the electromagnetic spectrum.

• Thus, two variables characterize NMR: an applied
magnetic field B0, the strength of which is measured
in tesla (T), and the frequency n of radiation used for
resonance, measured in hertz (Hz), or megahertz
(MHz)—(1 MHz = 106 Hz).
5

6
• The frequency needed for resonance and the applied
magnetic field strength are proportionally related:
• NMR spectrometers are referred to as 300 MHz
instruments, 500 MHz instruments, and so forth,
depending on the frequency of the RF radiation used
for resonance.
• These spectrometers use very powerful magnets to
create a small but measurable energy difference
between two possible spin states.

7

8
• Protons in different environments absorb at slightly
different frequencies, so they are distinguishable by
NMR.
• The frequency at which a particular proton absorbs is
determined by its electronic environment.
• The size of the magnetic field generated by the
electrons around a proton determines where it absorbs.
• Modern NMR spectrometers use a constant magnetic
field strength B0, and then a narrow range of
frequencies is applied to achieve the resonance of all
protons.
• Only nuclei that contain odd mass numbers (such as 1H,
13C, 19F and 31P) or odd atomic numbers (such as 2H and
14N) give rise to NMR signals.

9
1H NMR—The Spectrum
• An NMR spectrum is a plot of the intensity of a peak against its chemical shift,
measured in parts per million (ppm) .

10
• NMR absorptions generally appear as sharp peaks.
• Increasing chemical shift is plotted from left to right.
• Most protons absorb between 0-10 ppm.
• The terms “upfield” and “downfield” describe the
relative location of peaks. Upfield means to the right.
Downfield means to the left.
• NMR absorptions are measured relative to the position
of a reference peak at 0 ppm on the d scale due to
tetramethylsilane (TMS). TMS is a volatile inert
compound that gives a single peak upfield from typical
NMR absorptions.

11
• The chemical shift of the x axis gives the position of an NMR
signal, measured in ppm, according to the following equation:
• By reporting the NMR absorption as a fraction of the NMR
operating frequency, we get units, ppm, that are
independent of the spectrometer.
• Four different features of a 1H NMR spectrum provide
information about a compound’s structure:
a. Number of signals
b. Position of signals
c. Intensity of signals.
d. Spin-spin splitting of signals.

12
1H NMR—Number of Signals
• The number of NMR signals equals the number of different
types of protons in a compound.
• Protons in different environments give different NMR signals.
• Equivalent protons give the same NMR signal.
• To determine equivalent protons in cycloalkanes and alkenes,
always draw all bonds to hydrogen.

13

• In comparing two H atoms on a ring or double bond, two
protons are equivalent only if they are cis (or trans) to
the same groups.
14

• Proton equivalency in cycloalkanes can be determined
similarly.
15

16
1H NMR—Position of Signals
• In the vicinity of the nucleus, the magnetic field generated by
the circulating electron decreases the external magnetic field
that the proton “feels”.
• Since the electron experiences a lower magnetic field
strength, it needs a lower frequency to achieve resonance.
Lower frequency is to the right in an NMR spectrum, toward a
lower chemical shift, so shielding shifts the absorption upfield.

17
• The less shielded the nucleus becomes, the more of
the applied magnetic field (B0) it feels.
• This deshielded nucleus experiences a higher magnetic
field strength, to it needs a higher frequency to
achieve resonance.
• Higher frequency is to the left in an NMR spectrum,
toward higher chemical shift—so deshielding shifts an
absorption downfield.
• Protons near electronegative atoms are deshielded, so
they absorb downfield.

18

19

20

• Protons in a given environment absorb in a predictable region in
an NMR spectrum.
21
1H NMR—Chemical Shift Values

• The chemical shift of a C—H bond increases with
increasing alkyl substitution.
22

Methylene Methine
23
Calculating 1H NMR—Chemical Shift Values
• The chemical shift of a C—H can be calculated with a
high degree of precision if a chemical shift additivity table is
used.
• The additivity tables starts with a base chemical shift value
depending on the structural type of hydrogen under consideration:
CH3 CH
2
CH
0.87 Base Chemical Shift ppm 1.20 ppm 1.20 ppm

24
• The presence of nearby atoms or groups will effect the base
chemical shift by a specific amount:
• The carbon atom bonded to the hydrogen(s) under consideration
are described as alpha (a) carbons.
• Atoms or groups bonded to the same carbon as the hydrogen(s)
under consideration are described as alpha (a) substituents.
• Atoms or groups on carbons one bond removed from the a carbon
are called beta (b) carbons.
• Atoms or groups bonded to the b carbon are described as alpha
(a) substituents.
b a
(Hydrogen under C C H consideration)

25
Added Chemical Shifts
Substituent Type of Hydrogen -Shift -Shift
C C CH3 0.78 ---
CH2 0.75 -0.10
CH --- ---
Y
RC C C
[Y = C or O] CH3 1.08 ---
Aryl- CH3 1.40 0.35
CH2 1.45 0.53
CH 1.33 ---
Cl- CH3 2.43 0.63
CH2 2.30 0.53
CH 2.55 0.03
Br- CH3 1.80 0.83
CH2 2.18 0.60
CH 2.68 0.25
I- CH3 1.28 1.23
CH2 1.95 0.58
CH 2.75 0.00
OH- CH3 2.50 0.33
CH2 2.30 0.13
CH 2.20 ---
RO- (R is saturated) CH3 2.43 0.33
CH2 2.35 0.15
CH 2.00 ---
O
R–CO
or ArO CH3 2.88 0.38
CH2 2.98 0.43
CH 3.43 ---
(ester only)
O
R–C
CH3 1.23 0.18
where R is alkyl, aryl, OH, CH2 1.05 0.31
OR', H, CO, or N CH 1.05 ---
H
H
C C H (Hydrogen under consideration)
H
H
Cl
b a
Base Chemical Shift = 0.87 ppm
no a substituents = 0.00
one b -Cl (CH3) = 0.63
TOTAL = 1.50 ppm
H
H
C C H (Hydrogen under consideration)
H
H
Cl
a b
one a -Cl (CH2) = 2.30
no b substituents = 0.00
TOTAL = 3.50 ppm

• In a magnetic field, the six p electrons in benzene circulate
around the ring creating a ring current.
• The magnetic field induced by these moving electrons reinforces
the applied magnetic field in the vicinity of the protons.
• The protons thus feel a stronger magnetic field and a higher
frequency is needed for resonance. Thus they are deshielded
and absorb downfield.
26

• In a magnetic field, the loosely held p electrons of the
double bond create a magnetic field that reinforces the
applied field in the vicinity of the protons.
• The protons now feel a stronger magnetic field, and
require a higher frequency for resonance. Thus the
protons are deshielded and the absorption is downfield.
27

• In a magnetic field, the p electrons of a carbon-carbon triple
bond are induced to circulate, but in this case the induced
magnetic field opposes the applied magnetic field (B0).
• Thus, the proton feels a weaker magnetic field, so a lower
frequency is needed for resonance. The nucleus is shielded and
the absorption is upfield.
28

29

30
1H NMR—Chemical Shift Values)

1H NMR of Methyl Acetate
O
R C
O
H3C C O
one a = 2.88 ppm
TOTAL O = 3.75 ppm
CH3
O
one a C
R
= 1.23 ppm
TOTAL = 2.10 ppm

2,3-Dimethyl-2-Butene
(Hydrogen under consideration)
one a (CH3) = 0.78 ppm
H2C CH
TOTAL = 1.65 ppm

• The area under an NMR signal is proportional to the
number of absorbing protons.
• An NMR spectrometer automatically integrates the area
under the peaks, and prints out a stepped curve
(integral) on the spectrum.
• The height of each step is proportional to the area under
the peak, which in turn is proportional to the number of
absorbing protons.
• Modern NMR spectrometers automatically calculate and
plot the value of each integral in arbitrary units.
• The ratio of integrals to one another gives the ratio of
absorbing protons in a spectrum. Note that this gives a
ratio, and not the absolute number, of absorbing
protons.
33
1H NMR—Intensity of Signals

34
1H NMR—Intensity of Signals

39
1H NMR—Spin-Spin Splitting
• Consider the spectrum below:

1H - 1H Coupling
You’ll notice in the spectra that we’ve seen that the signals don’t appear as
single lines, sometimes they appear as multiple lines. This is due to 1H - 1H
coupling (also called spin-spin splitting or J-coupling). Here’s how it works:
Imagine we have a molecule which contains a proton (let’s call it HA) attached
to a carbon, and that this carbon is attached to another carbon which also
contains a proton (let’s call it HB). It turns out that HA feels the presence of HB.
Recall that these protons are tiny little magnets, that can be oriented either with
or against the magnetic field of the NMR machine. When the field created by HB
reinforces the magnetic field of the NMR machine (B0 ) HA feels a slightly
stronger field, but when the field created by HB opposes B0, HA feels a slightly
weaker field. So, we see two signals for HA depending on the alignment of HB.
The same is true for HB, it can feel either a slightly stronger or weaker field due
to HA’s presence. So, rather than see a single line for each of these protons, we
see two lines for each.
HA HB
C C
HA HB
HA is sp lit into two lines b ecause
it feels the magnetic field of HB.
HB is sp lit into two lines b ecause
it feels the magnetic field of HA.
For this line, HB is lined up
with the magnetic field
(adds to the overall
magnetic field, so the line
comes at higher frequency)
For this line, HB is lined up
against the magnetic field
(sub tracts from the overall
magnetic field, so the line
comes at lower frequency)

More 1H - 1H Coupling
What happens when there is more than one proton splitting a
neighboring proton? We get more lines. Consider the molecule below
where we have two protons on one carbon and one proton on another.
HA HB
C C
Note that the signal p roduced
b y HA + HA ' is twice the size
HA' HA + HA' HB
HA and HA ' app ear at the same
chemical shift b ecause they are
in identical environments
They are also sp lit into two lines
(called a doub let) b ecause they
feel the magnetic field of HB.
HB is sp lit into three lines
because it feels the magnetic
field of HA and HA '
of that p roduced b y HB

Spin-Spin Splitting in 1H NMR Spectra
• Peaks are often split into multiple peaks due to magnetic
interactions between nonequivalent protons on adjacent carbons,
The process is called spin-spin splitting
• The splitting is into one more peak than the number of H’s on the
adjacent carbon(s), This is the “n+1 rule”
• The relative intensities are in proportion of a binomial distribution
given by Pascal’s Triangle
• The set of peaks is a multiplet (2 = doublet, 3 = triplet, 4 =
quartet, 5=pentet, 6=hextet, 7=heptet…..)
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
singlet
doublet
triplet
quartet
pentet
hextet
heptet

Rules for Spin-Spin Splitting
• Equivalent protons do not split each other
• Protons that are farther than two carbon atoms apart do not
split each other

45
If Ha and Hb are not equivalent, splitting is observed when:
Splitting is not generally observed between protons separated by
more than three s bonds.

• Spin-spin splitting occurs only between nonequivalent protons
on the same carbon or adjacent carbons.
46
The Origin of 1H NMR—Spin-Spin Splitting
Let us consider how the doublet due to the CH2 group on
BrCH2CHBr2 occurs:
• When placed in an applied field, (B0), the adjacent proton
(CHBr2) can be aligned with () or against (¯) B0. The likelihood
of either case is about 50% (i.e., 1,000,006 vs 1,000,000¯).
• Thus, the absorbing CH2 protons feel two slightly different
magnetic fields—one slightly larger than B0, and one slightly
smaller than B0.
• Since the absorbing protons feel two different magnetic fields,
they absorb at two different frequencies in the NMR spectrum,
thus splitting a single absorption into a doublet, where the two
peaks of the doublet have equal intensity.

47
The frequency difference, measured in Hz, between two peaks
of the doublet is called the coupling constant, J.
J

48
Let us now consider how a triplet arises:
• When placed in an applied magnetic field (B0), the adjacent
protons Ha and Hb can each be aligned with () or against (¯) B0.
• Thus, the absorbing proton feels three slightly different
magnetic fields—one slightly larger than B0(ab). one slightly
smaller than B0(¯a¯b) and one the same strength as B0 (a¯b).

49
• Because the absorbing proton feels three different magnetic
fields, it absorbs at three different frequencies in the NMR
spectrum, thus splitting a single absorption into a triplet.
• Because there are two different ways to align one proton with B0,
and one proton against B0—that is, a¯b and ¯ab—the middle peak
of the triplet is twice as intense as the two outer peaks, making
the ratio of the areas under the three peaks 1:2:1.
• Two adjacent protons split an NMR signal into a triplet.
• When two protons split each other, they are said to be coupled.
• The spacing between peaks in a split NMR signal, measured by the
J value, is equal for coupled protons.

50

Whenever two (or three) different sets of adjacent protons are
equivalent to each other, use the n + 1 rule to determine the
splitting pattern.
53

equivalent to each other, use the n + 1 rule to determine the
splitting pattern.
54

not equivalent to each other, use the n + 1 rule to determine the
splitting pattern only if the coupling constants (J) are identical:
55
Free rotation around C-C bonds averages
a a
c
b
coupling constant to J = 7Hz
Jab = Jbc

not equivalent to each other, use the n + 1 rule to determine the
splitting pattern only if the coupling constants (J) are identical:
56
a
c
b
c
Jab = Jbc

57
1H NMR—Structure Determination

58

59

60

61
The Nature of NMR Absorptions
• 1H NMR of methyl acetate has 2 equivalent kinds of H’s so shows 2 peaks
• Electrons in neighboring bonds shield or expose nuclei from magnetic field
H’s on C next to electron
withdrawing C=O
H’s on C next to electron
withdrawing O
 Intensity of 1H NMR peak is proportional to # of equivalent H’s

 13C NMR of methyl acetate has 3 kinds of C’s so shows 3 peaks
 Electrons in neighboring bonds shield nuclei from magnetic field
62
C next door to electron
C next to ewleitchtdroranw ing C=O
withdrawing O
C of
C=O
 Intensity of 13C NMR peak is not related to # of equivalent C’s

Example:
• How many signals would you expect each
to have in its 1H and 13C spectra?
1H 13C
63
CH3
CH3
C C
CH3
CH3
O
CH3
CH3

Solution:
• How many signals would you expect each
to have in its 1H and 13C spectra?
64
CH3
CH3
C C
CH3
CH3
O
CH3
CH3
1H 13C
CH3
CH3
C C
CH3
CH3
CH3 CH3
C C
CH3
CH3
1 2
3 5
O
CH3
CH3
H H
H
H
H
H
H H
C
C
C
C
C
C
O
CH3
CH3

Shift = relative energy of resonance
65
13.3 Chemical Shifts
Downfield =
deshielded
(more exposed to
magnet)
Upfield =
shielded
(more protected
from magnet)
CH3
tetramethylsilane [TMS]
Reference point Si
H3C CH3
CH3

13.4 Signal Averaging & FT-NMR
Carbon-13: (only carbon
isotope with a nuclear spin)
Natural abundance =1.1%
of C’s so sample is very
dilute in this isotope
66
Single
run
Average of
200 runs
 Sample measured using
repeated accumulation of data
and averaging of signals,
incorporating pulse and the
operation of Fourier transform
(FT-NMR)
 All signals are obtained
simultaneously using a broad
pulse of energy and resonance
recorded
 Frequent repeated pulses give
many sets of data that are
averaged to eliminate noise
 Fourier-transform of averaged
pulsed data gives spectrum

67
13.5 13C NMR Spectroscopy
• Each signal shows different types of environments of carbon
• 13C resonances are 0 to 220 ppm downfield from TMS
• C’s shift downfield (deshield) if next to electron-withdrawing
– Like O, N, X (halogens)
sp3 C signal ~
d 0 to 90
sp2 C ~
d 110 to
220
C(=O) at low field,
d 160 to 220

68
13C NMR Example: 2-butanone

13C NMR Example: p-bromoacetopheone
69

Learning Check:
Assign resonances in the given 13C spectrum of methyl
propanoate
70
O
1 2 3 4
CH3 O C CH2 CH3

Solution:
Assign resonances in the given 13C spectrum of methyl
propanoate
71
O
1 2 3 4
CH3 O C CH2 CH3

1H and 13C NMR compared:
both give us information about the number of
chemically nonequivalent nuclei
(nonequivalent hydrogens or nonequivalent
carbons)
both give us information about the
environment of the nuclei (hybridization state,
attached atoms, etc.)
it is convenient to use FT-NMR techniques for
1H; it is standard practice for 13C NMR

13C requires FT-NMR because the signal for a
carbon atom is 10-4 times weaker than the
signal for a hydrogen atom
a signal for a 13C nucleus is only about 1% as
intense as that for 1H because of the
magnetic properties of the nuclei, and
at the "natural abundance" level only 1.1% of
all the C atoms in a sample are 13C (most are
12C)

13C signals are spread over a much wider
range than 1H signals making it easier to
identify and count individual nuclei
Figure 13.20 (a) shows the 1H NMR spectrum
of 1-chloropentane; Figure 13.20 (b) shows
the 13C spectrum. It is much easier to identify
the compound as 1-chloropentane by its 13C
spectrum than by its 1H spectrum.

Figure 13.20(a) (page 511)
ClCH2
1H
CH3
ClCH2 2CH2CH2CH2CH3
2 3
10.0 9.0 8.0 7.0 6.0 5.0 4.0 3.0 2.0 1.0 0
Chemical shift (d, ppm)

Figure 13.20(b) (page 511)
ClCH2
13C
2CH2CH2CH2C
H3
a separate, distinct
peak appears for
ceaarbcohn osf the 5 CDCl33
200 180 160 140 120 100 80 60
40 20 0

13.15
13C Chemical Shifts
are measured in ppm (d)
from the carbons of TMS

13C Chemical shifts are most affected by:
• hybridization state of carbon
• electronegativity of groups attached to carbon

Examples (chemical shifts in ppm from TMS)
23
138
sp3 hybridized carbon is more shielded than sp2

OH
61
O
202
sp3 hybridized carbon is more shielded than sp2

OH
23 61
an electronegative atom deshields the carbon
to which it is attached

O
138 202
an electronegative atom deshields the carbon
to which it is attached

Table 13.3 (p 513)
Type of carbon Chemical shift (d),
ppm
RCH33 0-35
R2
R3
R4
2CH2
3CH
4C
15-40
25-50
30-40

Table 13.3 (p 513)
ppm
RCH3 0-35
ppm
3 RC CR 65-90
R22CH2
R3
15-40
25-50
R22C CR22 100-150
3CH
110-175
R4 4C 30-40

Table 13.3 (p 513)
ppm
RCH2
RCH2
RCH2
RCH2
RCH2
2Br
2Cl
2NH2
2OH
2OR
20-40
25-50
35-50
50-65
50-65

Table 13.3 (p 513)
ppm
RCH2 20-40
ppm
2Br O
RCH2
RCH2
2Cl
RCOR
160-185
2NH2 35-50 25-50 O
RCH2
RCH2
2OH
2OR
50-65
50-65
RCR 190-220

13.16
13C NMR and Peak Intensities
Pulse-FT NMR distorts intensities of signals.
Therefore, peak heights and areas can be
deceptive.

Figure 13.21 (page 514)
CH 3 7 carbons give 7
signals, but
intensities are not
equal
OH
200 180 160 140 120 100 80 60 40 20 0

Peaks in a 13C NMR spectrum are typically
singlets
13 C—13C splitting is not seen because the
probability of two 13C nuclei being in the same
molecule is very small.
13C—1H splitting is not seen because spectrum
is measured under conditions that suppress
this splitting (broadband decoupling).

13.18
Using DEPT to Count the Hydrogens
Attached to 13C
Distortionless Enhancement
of Polarization Transfer

Measuring a 13C NMR spectrum involves
1. Equilibration of the nuclei between the lower
and higher spin states under the influence of
a magnetic field
2. Application of a radiofrequency pulse to give
an excess of nuclei in the higher spin state
3. Acquisition of free-induction decay data
during the time interval in which the equilibrium
distribution of nuclear spins is restored
4. Mathematical manipulation (Fourier transform)
of the data to plot a spectrum

Steps 2 and 3 can be repeated hundreds of times
to enhance the signal-noise ratio
2. Application of a radiofrequency pulse to give
an excess of nuclei in the higher spin state
3. Acquisition of free-induction decay data
during the time interval in which the equilibrium
distribution of nuclear spins is restored

In DEPT, a second transmitter irradiates 1H
during the sequence, which affects the appearance
of the 13C spectrum.
some 13C signals stay the same
some 13C signals disappear
some 13C signals are inverted

Figure 13.23 (a) (page 516)
O
CCH2
CHCH
2CH2CH2CH3
CH2 2
O CH CH2
CH2
2
2 CH3
C
C
200 180 160 140 120 100 80 60
40 20 0

Figure 13.23 (a) (page 516)
O
CCH2
CHCH
2CH2CH2CH3
CH3 3
CH
CH2 CH2 2 CH2
2
2
200 180 160 140 120 100 80 60
40 20 0

Figure 13.23 (b) (page 516)
O
CCH2
CHCH
2CH2CH2CH3
CH3 3
CH
CH and CH3
unaffected
C and C=O nulled CH22 CHCH2
2
CH2 inverted 100 80 60
2
40 20 0

99
13.6 DEPT 13C NMR
• DEPT (distortionless enhancement by polarization transfer)
 Normal spectrum shows all C’s then:
 Obtain spectrum of all C’s except quaternary
(broad band decoupled)
 Change pulses to obtain separate information for
CH2, CH
 Subtraction reveals each type

DEPT 13C NMR
Normal spectrum shows all C’s:
(Difficult to Assign so many C’s)
DEPT-90: shows
only CH’s
Quaternary C’s
don’t show
(Can now narrow our assignments)
100
CH3
C
C
C
C
H OH
C
CH3
CH3
H
H H
H H
1
2
3
4
5
6
7
8
5 2
6

DEPT 13C NMR
DEPT-135:
Positive =shows CH’s
and CH3’s
Negative =shows CH2’s
(Can narrow assignments even further)
101
CH3
C
C
C
C
H OH
C
CH3
CH3
H
H H
H H
1
2
3
4
5
6
7
8
5 2
6
1
7,
8

DEPT 13C NMR
DEPT-135:
Positive =shows CH’s
and CH3’s
Negative =shows CH2’s
(Can narrow assignments even further)
102
CH3
C
C
C
C
H OH
C
CH3
CH3
H
H H
H H
1
2
3
4
5
6
7
8
7,
8
5 2 1
6
2
4

103
13.7 Uses of13C NMR: Example
• Evidence for product of elimination of 1-chloro-methyl cyclohexane
Cl
CH3 CH2
CH3
or
KOH
ethanol
Expect 7 different
C’s;
5 sp3 resonances d 20-50
Expect 5 different
C’s;
CH3

13.8 1H NMR & Proton Equivalence
• Proton NMR is much more sensitive than 13C and the active
nucleus (1H) is nearly 100 % of the natural abundance
• Shows how many kinds of nonequivalent hydrogens are in
a compound
• Theoretical equivalence can be predicted by seeing if
replacing each H with “X” gives the same or different
outcome
• Equivalent H’s have the same signal while nonequivalent
are different
104
– There are degrees of nonequivalence

105
Nonequivalent H’s
• If replacement of each H with “X” gives a different constitutional isomer
then the H’s are in constitutionally heterotopic environments and will
have different chemical shifts
• – they are nonequivalent under all circumstances

106
Equivalent H’s
• Two H’s that are in identical environments (homotopic) have the same
NMR signal
• Test by replacing each with X if they give the identical result, they are
equivalent (homotopic)

107
Enantiotopic Distinctions
• If H’s are in environments that are mirror images of each other, they are
enantiotopic
• Replacement of each H with X produces a set of enantiomers
• The H’s have the same NMR signal (in the absence of chiral materials)

108
Diastereotopic Distinctions
• In a chiral molecule, paired hydrogens can have different
environments and different shifts
• Replacement of a pro-R hydrogen with X gives a different diastereomer
than replacement of the pro-S hydrogen
• Diastereotopic hydrogens are distinct chemically and spectrocopically
*

Learning Check:
• Identify sets of H’s as
Unrelated (U), homotopic (H), enantiotopic (E), or diasterotopic (D)
109

Solution:
• Identify sets of H’s as
Unrelated (U), homotopic (H), enantiotopic (E), or diasterotopic (D)
110
D D
D
E
D H

111
13.9 Chemical Shifts in 1H NMR
• Proton signals range from d 0 to d 10
• Electronegative atoms cause downfield shift H’s on
sp3 C
Higher field
H’s on sp2 C
Lower field

Nmr2 pl

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