John E. McMurry
www.cengage.com/chemistry/mcmurry
Chapter 13
Structure Determination:
Nuclear Magnetic
Resonance Spectroscopy
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Learning Objectives
(13.1)
 Nuclear magnetic resonance spectroscopy
(13.2)
 The nature of NMR absorptions
(13.3)
 The chemical shift
(13.4)
 Chemical shifts in 1H NMR spectroscopy
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Learning Objectives
(13.5)
 Integration of 1H NMR absorptions: Proton
counting
(13.6)
 Spin–spin splitting in 1H NMR spectra
(13.7)
 1H NMR spectroscopy and proton equivalence
(13.8)
 More complex spin–spin splitting patterns
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Learning Objectives
(13.9)
 Uses of 1H NMR spectroscopy
(13.10)
 13C NMR spectroscopy: Signal averaging and
FT–NMR
(13.11)
 Characteristics of 13C NMR spectroscopy
(13.12)
 DEPT 13C NMR spectroscopy
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Learning Objectives
(13.13)
 Uses of 13C NMR spectroscopy
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Nuclear Magnetic Resonance
Spectroscopy
 Nuclei are positively charged and interact with
an external magnetic field denoted by B0
 Magnetic rotation of nuclei is random in the
absence of a magnetic field

In the presence of a strong magnet, nuclei adopt
specific orientations
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Nuclear Magnetic Resonance
Spectroscopy
 When exposed to a certain frequency of
electromagnetic radiation, oriented nuclei
absorb energy and causes a spinflip from a
state of lower energy to higher energy

Nuclear magnetic resonance - Nuclei are in
resonance with applied radiation
 Frequency that causes resonance depends on:



Strength of external magnetic field
Identity of the nucleus
Electronic environment of the nucleus
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Nuclear Magnetic Resonance
Spectroscopy
 Larmor equation

Relation between


Resonance frequency of a nucleus
Magnetic field and the magnetogyric ratio of the
nucleus
 γ 
ν =   B0
 2π 
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Worked Example
 Calculate the amount of energy required to spin-
flip a proton in a spectrometer operating at 300
MHz

Analyze if the increase of spectrometer frequency
from 200 MHz to 300 MHz increases or
decreases the amount of energy necessary for
resonance
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Worked Example
 Solution:
8
c 3.0 × 10 m / s
λ= =
; ν = 300 MHz = 3.0 ×108 Hz
ν
ν
8
c 3.0 × 10 m / s
λ= =
= 1.0 m
8
ν
3.0 × 10 Hz
1.20× 10-4 kJ / mol
E=
= 1.20 ×10-4 kJ / mol
1.0

Increasing the spectrometer frequency from 200
MHz to 300 MHz increases the amount of energy
needed for resonance
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The Nature of NMR Absorptions
 Absorption frequencies differ across 1H and
13C
molecules
 Shielding: Opposing magnetic field produced
by electrons surrounding nuclei to counteract
the effects of an external magnetic field



Effect on the nucleus is lesser than the applied
magnetic field
Beffective = Bapplied – Blocal
Individual variances in the electronic environment
of nuclei leads to different shielding intensities
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Figure 13.3 - NMR Spectrum of
1H and 13C
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Working of an NMR
Spectrometer
 Organic sample dissolved in a suitable solvent is
placed in a thin glass tube between the poles of
a magnet
 1H and 13C nuclei respond to the magnetic field
by aligning themselves to one of the two
possible orientations followed by rf irradiation
 Constant and varied strength of the applied field
causes each nucleus to resonate at a slightly
varied field strength
 Absorption of rf energy is monitored by a
sensitive detector that displays signals as a
peak
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Figure 13.4 - Operation of a
Basic NMR Spectrometer
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NMR Spectrometer
 Time taken by IR spectroscopy is about 10–13 s
 Time taken by NMR spectroscopy is about 10–3
s

Provides a blurring effect that is used in the
measurement of rates and activation energies of
vary fast processes
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Worked Example
 Explain why 2-chloropropene shows signals for
three kinds of protons in its 1H NMR spectrum
 Solution:

2-Chloropropene has three kinds of protons

Protons b and c differ because one is cis to the
chlorine and the other is trans
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The Chemical Shift
 The left segment of the chart is the downfield

Nuclei absorbing on the downfield have less
shielding as they require a lower field for
resistance
 The right segment is the upfield

Nuclei absorbing on the upfield have more
shielding as they require a higher field strength
for resistance
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Figure 13.5 - The NMR Chart
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The Chemical Shift
 Chemical shift is the position on the chart at
which a nucleus absorbs
 The delta (δ) scale is used in calibration of the
NMR chart
1 δ = 1 part-per-million of the spectrometer
operating frequency
Observed chemical shift (number of Hz away from TMS)
δ=
Spectrometer frequency in MHz
 The delta scale is used as the units of
measurement can be used to compare values
across other instruments

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Worked Example
 The 1H NMR peak of CHCl3 was recorded on a
spectrometer operating at 200 MHz providing
the value of 1454 Hz

Convert 1454 Hz into δ units
 Solution:
Observed chemical shift (number of Hz away from TMS)
δ=
Spectrometer frequency in MHz
1454Hz
δ=
= 7.27 δ for CHCl3
200MHz
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Chemical Shifts in 1H NMR
Spectroscopy
 Chemical shifts are due to the varied
electromagnetic fields produced by electrons
surrounding nuclei
 Protons bonded to saturated, sp3-hybridized
carbons absorb at higher fields
 Protons bonded to sp2-hybridized carbons
absorb at lower fields
 Protons bonded to electronegative atoms
absorb at lower fields
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Table 13.2 - Regions of the 1H
NMR Spectrum
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Worked Example
 CH2Cl2 has a single 1H NMR peak

Determine the location of absorption
 Solution:
 For CH2Cl2 , δ = 5.30

The location of absorption are the protons
adjacent to the two halogens
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Integration of 1H NMR
Absorptions: Proton Counting
 In the figure, the peak caused by (CH3)3C–
protons is larger than the peak caused by –OCH
protons
 Integration of the area under the peak can be
used to quantify the different kinds of protons in
a molecule
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Worked Example
 Mention the number of peaks in the 1H NMR
spectrum of 1,4-dimethyl-benzene (para-xylene
or p-xylene)

Mention the ratio of peak areas possible on
integration of the spectrum
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Worked Example
 Solution:



There are two absorptions in the 1H NMR
spectrum of p-xylene
The four ring protons absorb at 7.05 δ and the six
methyl-groups absorb at 2.23 δ
The peak ratio of methyl protons:ring protons is
3:2
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Worked Example
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Spin-Spin Splitting in 1H NMR
Spectra
 Multiplet: Absorption of a proton that splits into
multiple peaks


The phenomenon is called spin-spin splitting
Caused by coupling of neighboring spins
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Spin-Spin Splitting in 1H NMR
Spectra
 Alignment of –CH2Br proton spins with the
applied field can result in:

Slightly larger total effective field and slight
reduction in the applied field to achieve resonance
 There is no effect if one of the –CH2Br proton
spins aligns with the applied field and the other
aligns against it
 Alignment of –CH2Br proton spins against the
applied field results in:

Smaller effective field and an increased applied
field to achieve resonance
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Figure 13.8 - The Origin of Spin-Spin
Splitting in Bromoethane
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Spin-Spin Splitting in 1H NMR
Spectra
 n + 1 rule: Protons that exhibit n + 1 peaks in
the NMR spectrum possess

n = number of equivalent neighboring protons
 Coupling constant is the distance between
peaks in a multiplet
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Spin-Spin Splitting in 1H NMR
Spectra
 It is possible to identify multiplets in a complex
NMR that are related


Multiplets that have the same coupling constant
can be related
Multiplet-causing protons are situated adjacent to
each other in the molecule
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Rules of Spin-Spin Splitting
 Chemically equivalent protons do not show spin-
spin splitting
 The signal of a proton with n equivalent
neighboring protons is split into a multiplet of n +
1 peaks with a coupling constant
 Two groups of photons coupled together have
the same coupling constant, J
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Worked Example
 The integrated 1H NMR spectrum of a
compound of formula C4H10O is shown below

Propose a structure
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Worked Example
 Solution:




The molecular formula (C4H10O) indicates that
the compound has no multiple bonds or rings
The 1H NMR spectrum shows two signals,
corresponding to two types of hydrogens in the
ratio 1.50:1.00, or 3:2
Since the unknown contains 10 hydrogens, four
protons are of one type and six are of the other
type
The upfield signal at 1.22 δ is due to saturated
primary protons
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Worked Example





The downfield signal at 3.49 δ is due to protons
on carbon adjacent to an electronegative atom in this case, oxygen
The signal at 1.23 δ is a triplet, indicating two
neighboring protons
The signal at 3.49 δ is a quartet, indicating three
neighboring protons
This splitting pattern is characteristic of an ethyl
group
The compound is diethyl ether,
CH3CH2OCH2CH3
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1H
NMR Spectroscopy and
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
comparing structures formed by replacing each
H with X gives the same or different outcome
 Equivalent H’s have the same signal while
nonequivalent are different

There are degrees of nonequivalence
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1H
NMR Spectroscopy and
Proton Equivalence
 One use of 1H NMR is to ascertain the number
of electronically non-equivalent hydrogens
present in a molecule
 In relatively small molecules, a brief look at the
structure can help determine the kinds of
protons present and the number of possible
NMR absorptions

Equivalence or nonequivalence of two protons
can be determined by comparison of structures
formed if each hydrogen were replaced by an X
group
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1H
NMR Spectroscopy and
Proton Equivalence
 Possibilities

If the protons are chemically unrelated and nonequivalent, the products formed by substitution
would be different constitutional isomers
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1H
NMR Spectroscopy and
Proton Equivalence

If the protons are chemically identical, the same
product would be formed despite the substitution
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1H
NMR Spectroscopy and
Proton Equivalence

If the hydrogens are homotopic but not identical,
substitution will form a new chirality center

Hydrogens that lead to formation of enantiomers
upon substitution with X are called enantiotopic
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1H
NMR Spectroscopy and
Proton Equivalence

If the hydrogens are neither homotopic nor
enantiotopic, substitution of a hydrogen at C3 would
form a second chirality center
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Worked Example
 How many absorptions will (S)-malate, an
intermediate in carbohydrate metabolism have
in its 1H NMR spectrum? Explain
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Worked Example
 Solution:


Because (S)-malate already has a chirality
center(starred), the two protons next to it are
diastereotopic and absorb at different values
The 1H NMR spectrum of (S)-malate has four
absorptions
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More Complex Spin-Spin
Splitting Patterns
 Some hydrogens in a molecule possess
accidentally overlapping signals

In the spectrum of toluene (methylbenzene), the
five aromatic ring protons produce a complex,
overlapping pattern though they are not
equivalent
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More Complex Spin-Spin
Splitting Patterns
 Splitting of a signal by two or more
nonequivalent kinds of protons causes a
complication in 1H NMR spectroscopy
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Figure 13.14 - Tree Diagram for the C2
proton of trans-cinnamaldehyde
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Worked Example
 3-Bromo-1-phenyl-1-propene shows a complex
NMR spectrum in which the vinylic proton at C2
is couples with both the C1 vinylic proton (J = 16
Hz) and the C3 methylene protons (J = 8 Hz)

Draw a tree diageam for the C2 proton signal and
account for the fact that a live-line multiplet is
observed
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Worked Example
 Solution:

C2 proton couples with vinylic proton (J = 16) Hz


C2 proton’s signal is split into a doublet
C2 proton also couples with the two C3 protons
(J = 8 Hz)

Each leg of the C2 proton doublet is split into a triplet
to produce a total of six lines
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Worked Example
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Uses of 1H NMR Spectroscopy
 The technique is used to identify likely products
in the laboratory quickly and easily

NMR can help prove that hydroboration-oxidation
of alkenes occurs with non-Markovnikov
regiochemistry to yield the less highly substituted
alcohol
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Figure 13.15 - 1H NMR Spectra of
Cyclohexylmethanol
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Worked Example
 Mention how 1H NMR is used to determine the
regiochemistry of electrophilic addition to
alkenes

Determine whether addition of HCl to 1methylcyclohexene yields 1-chloro-1nethylcyclohexane or 1-chloro-2methylcyclohexane
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Worked Example
 Solution:

Referring to 1H NMR methyl group absorption


The unslpit methyl group in the left appears as a
doublet in the product on the right
Bonding of a proton to a carbon that is also bonded
to an electronegative atom causes a downfield
absorption in the 2.5–4.0 region
 1H
NMR spectrum of the product would confirm
the product to be 1-chloro-1-methylcyclohexane
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13C
NMR Spectroscopy: Signal
Averaging and FT–NMR
 Carbon-13 is the only naturally occurring carbon
isotope that possesses a nuclear spin, but its
natural abundance is 1.1%
 Signal averaging and Fourier-transform NMR
(FT–NMR) help in detecting carbon 13
 Due to the excess random electronic
background noise present in 13C NMR, an
average is taken from hundreds or thousands of
individual NMR spectra
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Figure 13.16 - Carbon-13 NMR
Spectra of 1-Pentanol
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13C
NMR Spectroscopy: Signal
Averaging and FT–NMR
 Spin-spin splitting is observed only in 1H NMR


The low natural abundance of 13C nucleus is the
reason that coupling with adjacent carbons is
highly unlikely
Due to the broadband decoupling method used
to record 13C spectra, hydrogen coupling is not
seen
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Characteristics of 13C NMR
Spectroscopy

13C
NMR provides a count of the different
carbon atoms in a molecule
 13C resonances are 0 to 220 ppm downfield
from TMS
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Characteristics of 13C NMR
Spectroscopy
 General factors that determine chemical shifts



The electronegativity of nearby atoms
The diamagnetic anisotropy of pi systems
The absorption of sp3-hybridized carbons and sp2
carbons
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Figure 13.18 - Carbon-13 Spectra of 2butanone and para-bromoacetophenone
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Worked Example
 Classify the resonances in the
13C
spectrum of
methyl propanoate, CH3CH2CO2CH3
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Worked Example
 Solution:

Methyl propanoate has four unique carbons that
individually absorb in specific regions of the 13C
spectrum
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DEPT 13C NMR Spectroscopy
 DEPT-NMR (distortionless enhancement by
polarization transfer)
 Stages of a DEPT experiment



Run a broadband-decoupled spectrum
Run a DEPT-90
Run a DEPT-135
 The DEPT experiment manipulates the nuclear
spins of carbon nuclei
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Figure 13.20 – DEPT-NMR Spectra
for 6-methyl-5-hepten-2-ol
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Uses of 13C NMR Spectroscopy
 Helps in determining molecular structures


Provides a count of non-equivalent carbons
Provides information on the electronic
environment of each carbon and the number of
attached protons
 Provides answers on molecule structure that IR
spectrometry or mass spectrometry cannot
provide
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Figure 13.21 - 13C NMR Spectrum of
1-methylcyclohexane
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Worked Example
 Propose a structure for an aromatic
hydrocarbon, C11H16, that has the following 13C
NMR spectral data:



Broadband decoupled: 29.5, 31.8, 50.2, 125.5,
127.5, 130.3, 139.8 δ
DEPT-90: 125.5, 127.5, 130.3 δ
DEPT-135: positive peaks at 29.5, 125.5, 127.5,
130.3 δ; negative peak at 50.2 δ
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Worked Example
 Solution:

Calculate the degree of unsaturation of the
unknown compound


Look for elements of symmetry


C11H16 has 4 degrees of unsaturation
7 peaks appearing in the 13C NMR spectrum indicate
a plane of symmetry
According to the DEPT-90 spectrum, 3 of the
kinds of carbons in the aromatic ring are CH
carbons
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Worked Example

The unknown structure is a monosubstituted
benzene ring with a substituent containing CH2
and CH3 carbons
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Summary
 Nuclear magnetic resonance spectroscopy or
NMR is the most important spectroscopic
technique used in the determination of molecular
structure
 Magnetic nuclei such as 1H and 13C spin-flip
from a lower energy state to a higher energy
state when they absorb radiofrequency waves
 Each 1H or 13C nucleus possesses a unique
electromagnetic field that causes it to resonate
at different values of the applied field causing
peaks whose exact position is termed a chemical
shift
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Summary
 Delta (δ) is the unit of calibration in NMR charts
 Tetramethylsilane (TMS) is a reference point on
the NMR chart
 TMS absorption that occurs at the right-hand
(upfield) side of the chart is assigned a value of
0δ
 Fourier-transform NMR (FT–NMR)
spectrometers are used to obtain 13C spectra
using broadband decoupling of proton spins
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Summary
 Electronic integration of the area under each
absorption peak in 1H NMR spectra is used to
determine the number of hydrogens that cause
each peak
 Neighboring nuclear spins can couple to cause
the spin-spin splitting of NMR peaks into
multiplets
 The NMR signal of a hydrogen neighbored by n
equivalent adjacent hydrogens splits into n + 1
peaks (the n + 1 rule) with coupling constant J
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