Nuclear Magnetic Resonance Spectroscopy

PRESENTED BY
DEVISHA TATINENI
PHARMACEUTICAL CHEMISTRY
M.PHARM
1

CONTENTS
 Introduction
 Fundamental principles of NMR
 Interpretation
Chemical shift
 Number of signals
 Spin-Spin coupling: Splitting of signals
 Coupling constant
 Integrals
2

NMR Spectroscopy
 Nuclear Magnetic Resonance is a branch of
spectroscopy in which radio frequency waves induce
transitions between magnetic energy levels of nuclei
of a molecule.
3

The Electromagnetic Spectrum
High Frequency
Longer Wavelength
4

The frequency of radio waves lies between 107 and 108cps
The energy of radio frequency ( rf ) radiation can be
calculated by using the equation :
E = h v
h = Planck ̕ s constant = 6.6 × 10-27 erg sec
v = frequency = 107 _ 108 cps(cycles per sec).
5

E = 6.6 × 10_27 × 107 ( or 108 ergs)
= 6.6 × 10-20 ( or 6.6 × 10-19 ergs )
Energy of rf radiation is very small to vibrate, rotate , or
excite an atom or molecule. But this energy is
sufficient to affect the nuclear spin of the atoms of a
molecule.
6

The nuclei of some atoms have a property called “SPIN”.
NUCLEAR SPIN
These nuclei behave as if
they were spinning.
This is like the spin property
of an electron, which can have
two spins: +1/2 and -1/2 .
Each spin-active nucleus has a number of spins defined by
its spin quantum number, I.
The number of Spin states = 2I + 1.
7

 If the number of neutrons and the number of protons are
both even, then the nucleus has NO spin. 12C, 16O ,32S etc.
 If the number of neutrons plus the number of protons is
odd, then the nucleus has a half-integer spin (i.e. 1/2, 3/2,
5/2) 1H,19F, 31P
 If the number of neutrons and the number of protons are
both odd, then the nucleus has an integer spin (i.e. 1, 2, 3)
2H, 14N
8

1
H
2
H
12
C
13
C
14
N
16
O
31
P
32
S
15
N
19
FElement
Nuclear spin
quantum
number ( I )
Number of
spin states
1/2 1 0 0 01/2 1
2 3 1 2 3 1
1/2
2 1
1/2
2
1/2
2
9

Principle
 NMR spectroscopy is the interaction of magnetic field with spin
of nuclei and then absorption of radio frequency. For example,
the nucleus of proton 1H+ has two spin rotations : clockwise
rotation with a spin quantum number I = +½ and |
counterclockwise rotation with a spin quantum number I = - ½
 The number of spin sates is 2I+1 which is 2x (1/2) +1 = 2 state
10

NUCLEAR SPIN STATES - HYDROGEN NUCLEUS
+ 1/2 - 1/2
The two states
are equivalent
in energy in the
absence of a
magnetic or an
electric field.
+ +
The spin of the positively
charged nucleus generates
a magnetic moment vector, m.
m
m
TWO SPIN STATES
11

 Without the magnetic field the spin states of nuclei
possess the same energy, and energy level transition is
not possible.
 When a magnetic field is applied, the separate levels
and radio frequency radiation can cause transitions
between these energy levels.
12

no difference in absence of magnetic field
proportional to strength of external magnetic field
Energy Differences Between Nuclear Spin States
+
+
E E '
increasing field strength
13

Some important relationships in NMR
The frequency of absorbed
electromagnetic radiation
is proportional to
the energy difference between
two nuclear spin states
which is proportional to
the applied magnetic field
Units
Hz
kJ/mol
(kcal/mol)
tesla (T)
14

Magnetic properties of nuclei
 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.
15

 But when magnetic field is applied, the proton (H)
posses spin & their own magnetic field align
themselves either or opposite to magnetic field.
 For e.g. 1H has +1/2 & -1/2 spin state, the proton (H)
have +1/2 spin state align themselves with field
(Lower energy) and with -1/2 spin state align
opposite to field (Higher energy).
16

Change in spin state energy separation with increase by applied
magnetic field ,B0
19

External Magnetic Field
When placed in an external field, spinning
protons act like bar magnets.
=>
20

THE “RESONANCE” PHENOMENON
absorption of energy by the
spinning nucleus
21

Nuclear Spin Energy Levels
Bo
+1/2
-1/2
In a strong magnetic
field (Bo) the two
spin states differ in
energy.
aligned
unaligned
N
S
22

Two Energy States
The magnetic fields of
the spinning nuclei
will align either with
the external field, or
against the field.
A photon with the right
amount of energy can
be absorbed and
cause the spinning
proton to flip.

N SN S
Add
Energy
N SS N
Aligned = Low Energy
Excited state = High energy
N SS N
Energy
Released
Back to low energy ground state
24

 According to the quantum theory, a spinning nucleus
can only have values for the spin angular momentum
given by the equation :
Spin angular momentum = [I(I+1)]1/2 h / 2µ
I = Spin quantum number
h= Planck̕s constant
25

 But µ = γ × spin angular momentum
µ = magnetic moment of the nucleus
γ = gyro magnetic ratio
 If a nucleus having a magnetic moment is introduced into a
magnetic field , H0, the two energy levels become separate
corresponding to m I = -1/2 (anti-parallel to the direction of
magnetic field) and m I = +1/2 (parallel to the direction of
magnetic field).
26

 For a nucleus with I = 1/2, the energies E1 and E2 for the
two states with m I = +1/2 and m I = -1/2 , respectively,
are
E1 = -1/2 [ γ h / 2µ ] H0
E2 = + 1/2 [ γ h / 2µ ] H0
27

 When the nucleus absorbs energy, the nucleus will be
promoted from the lower energy state E1, to the
higher energy state E2 by absorption of energy , ∆E,
equal to the energy difference, E2 – E1 .
 It means that the absorption of energy ∆E changes
the magnetic moment from the parallel state m I =
+1/2 to the anti-parallel state (m I = -1/2).
29

 The frequency ν at which energy is absorbed or
emitted is given by Bohr̕ s relationship:
ν = E2 – E1 / h
ν = 1/2 [ γ h / 2µ ] H0 + 1/2 [ γ h / 2µ ] H0 / h ( or )
ν= γ / 2µ H0
 This is the Larmor equation
30

The Larmor Equation!!!
γ
n =
2µ
Ho
g is a constant which is different for
each atomic nucleus (H, C, N, etc)
gyromagnetic
ratio g
strength of the
magnetic field
frequency of
the incoming
radiation that
will cause a
transition

 From the Larmor equation :
that the frequency absorbed or emitted by a nucleus
in moving from one energy level to another is directly
proportional to the applied magnetic field.
32

 When a nucleus is placed in a system where it
absorbs energy, it becomes excited. It then loses
energy to return to the unexcited state .It absorbs
energy and again enters an excited state.
 This nucleus which alternatively becomes excited and
unexcited is said to be in a state of resonance.
33

WHEN A SPIN-ACTIVE HYDROGEN ATOM IS
PLACED IN A STRONG MAGNETIC FIELD
….. IT BEGINS TO PRECESS
A SECOND EFFECT OF A STRONG MAGNETIC FIELD

 Precessional frequency:- May be defined as
revolutions per second made by the magnetic moment
vector of the nucleus around the external magnetic
field Bo & Alternatively precessional frequency of
spinning bar magnet may be defined as equal to the
frequency of EMR in megacycles per second necessary
to induce a transition from one spin state to another
spin state.
35

 The frequency of precession (proton) is directly proportional to
the strength of applied magnetic field.
 If the precessing nucleus is irradiated with electromagnetic
radiation of the same frequency as that frequency of
precession nucleus, then
 Energy is absorbed
 The nuclear spin is flipped from spin state +1/2 (with the
applied field) to -1/2 (against the applied field).
36

N
S
w Nuclei precess at
frequency w when
placed in a strong
magnetic field.
hn
If n = w then
energy will be
absorbed and
the spin will
invert.
NUCLEAR
MAGNETIC
RESONANCE
NMR
RADIOFREQUENCY
radiation

Interpreting Proton NMR
Spectra

TYPES OF INFORMATION
FROM THE NMR SPECTRUM
1. Each different type of hydrogen gives a peak
or group of peaks (multiplet).
3. The integral gives the relative numbers of each
type of hydrogen.
2. The chemical shift (ᵟ, in ppm) gives a clue as
to the type of hydrogen generating the peak
(alkane, alkene, benzene, aldehyde, etc.)
4. Spin-spin splitting gives the number of hydrogens
on adjacent carbons.
5. The coupling constant J also gives information
about the arrangement of the atoms involved.

Chemical Shift (Position of Signals)
 The utility of NMR is that all protons do not show
resonance at same frequency because, it is surrounded by
particular no. of valence electrons which vary from atom
to atom so, they exist in slightly different electronic
environment from one another.
 Position of signals in spectrum help us to know nature of
protons i.e. aromatic, aliphatic, acetylinic, vinylic, adjacent
to electron releasing or withdrawing group. Thus they
absorb at different field strength.
41

42
 The relative energy of resonance of a particular nucleus
resulting from its local environment is called chemical shift.
 NMR spectra show applied field strength increasing from
left to right.
 Left part is down field right part is up field.
 Nuclei that absorb on up field side are strongly shielded.

43
 Let’s consider the just the proton (1H) NMR
Ho
’
> HoHo
Increasing magnetic field strength
Increased shielding of nucleus
Downfield Upfield
Bare proton
Proton in organic molecule

45
Measuring Chemical Shift
 Numeric value of chemical shift: difference between
strength of magnetic field at which the observed
nucleus resonates and field strength for resonance
of a reference.

 For measuring chemical shifts of various protons in a
molecule, the signal of TMS is taken as reference.
 The NMR signal for particular protons will appear at
different field strength compared to signal from TMS.
 This difference in the absorption position of the
proton with respect to TMS signal is called as
chemical shift (δ – value).
46

TMS
shift in Hz
0
Si CH3CH3
CH3
CH3
tetramethylsilane
“TMS”
reference compound
n
Rather than measure the exact resonance position of a
peak, we measure how far downfield it is shifted from TMS.
Highly shielded
protons appear
way upfield.
Chemists originally
thought no other
compound would
come at a higher
field than TMS.
downfield

 TMS (Tetra methyl silane) is most commonly used as IS in NMR
spectroscopy. Due to following reasons;
 It is chemically inert and miscible with a large range of solvents.
 Its twelve protons are all magnetically equivalent.
 Its protons are highly shielded and gives a strong peak even small
quantity.
 It is less electronegative than carbon.
 It is highly volatile and can be easily removed to get back sample.
48

chemical
shift
= d =
shift in Hz
spectrometer frequency in MHz
= ppm
This division gives a number independent
of the instrument used.
parts per
million
the “chemical shift” in the following way:

 The alternative system which is generally used for defining the
position of resonance relative to the reference is assigned tau ()
scale.
 = 10 - d
 A small numerically value of d indicates a small downfield shift while
large value indicates a large downfield shift.
 A small value of  represents a low field absorption and a high value
indicates a high field absorption.
 scale

SHIELDING AND DESHIELDING
 Rotation of electrons (π) to nearby nuclei generate
field that can either oppose or strong the field on
proton.
 If magnetic field is oppose applied magnetic field on
proton, that proton said shielded proton and if field is
strong the applied field then, proton feels high
magnetic field strength and such proton called as
Deshielded proton.
51

 So, shielded proton shifts absorption signal to right
side (upfield) and deshielded proton shifts absorption
signal to left side (down field) of spectrum.
 So, electric environment surrounding proton tells us
where proton shows absorption in spectrum.
52

 The electrons around the proton create a magnetic
field that opposes the applied field. Since this reduces
the field experienced at the nucleus, the electrons
are said to shield the proton.

Shielded Protons
Magnetic field strength must be increased for a
shielded proton to flip at the same frequency.
=>

 The protons are shielded by the electrons that surround
them. In an applied magnetic field, the valance electrons of
the protons are caused to circulate. This circulation, called
a local diamagnetic current, generates a counter magnetic
field that opposes the applied magnetic field. This effect,
which is called diamagnetic shielding or diamagnetic
anisotropy.
55

 The counter field that shields a nucleus diminishes the
net applied magnetic field that the nucleus
experiences. As a result, the nucleus precesses at a
lower frequency.
 This means that it also absorbs radiofrequency
radiation at this lower frequency.
56

 When the secondary fields produced by the circulating
electrons oppose the applied field at a particular
nucleus in the molecule, it means that effective field
experienced by the nucleus is less than the applied
field. This is known as positive shielding and the
resonance position moves up field in NMR spectrum.
57

 If the secondary field produced by the circulating
electrons reinforces the applied field, the position of
resonance moves downfield. This is known as negative
shielding.
58

 shielded/upfield:
higher electron density requires a stronger field for
resonance
 deshielded/downfield:
lower electron density requires a weaker field for
resonance
(using a constant radiofrequency)

Chemical shift depends upon following parameters:
1. Inductive effect
2. Hybridization
3. Anisotropic effect
4. Hydrogen bonding
60

ELECTRONEGATIVITY – CHEMICAL SHIFT
Compound CH3X
Element X
Electronegativity of X
Chemical shift d
CH3F CH3OH CH3Cl CH3Br CH3I CH4 (CH3)4Si
F O Cl Br I H Si
4.0 3.5 3.1 2.8 2.5 2.1 1.8
4.26 3.40 3.05 2.68 2.16 0.23 0
Dependence of the Chemical Shift of CH3X on the Element X
deshielding increases with the
Electronegativity of atom X
TMSmost
deshielded

highly shielded
protons appear
at high field
“deshielded“
protons appear
at low field
deshielding moves proton
resonance to lower field
C HCl
Chlorine “deshields” the proton,
that is, it takes valence electron
density away from carbon, which
in turn takes more density from
hydrogen deshielding the proton.electronegative
element
DESHIELDING BY AN ELECTRONEGATIVE ELEMENT
NMR CHART
d- d+
d- d+

Substitution Effects on Chemical Shift
CHCl3 CH2Cl2 CH3Cl
7.27 5.30 3.05 ppm
-CH2-Br -CH2-CH2Br -CH2-CH2CH2Br
3.30 1.69 1.25 ppm
most
deshielded
most
deshielded
The effect decreases
with increasing distance.
The effect
increases with
greater numbers
of electronegative
atoms.

 Electron withdrawing groups– reduces electron density around
the proton Deshielding.
Examples: F, Cl, Br, I, OH, NH2, NO2
 Electron releasing groups- increases electron density around
the proton Shielding.
Examples: Alkyl groups
66

 Hybridization
sp3 Hydrogens ( no electro negative elements and Π bonded
groups )
30 > 20 > 10 > cyclopropane
67

 sp2 Hydrogens:
In an sp2 C-H bond, the carbon atom has more s character
(33% s), which effectively renders it more electronegative
than an sp3 carbon (25% s).
 If the sp2 carbon atom holds its electrons more tightly, this
results in less shielding for the H nucleus than in an sp3
bond.
 Another effect anisotropy.
68

 sp Hydrogens:
On the basis of hybridization, acetylenic proton to have a
chemical shift greater than that of vinyl proton.
But chemical shift of acetylenic proton is less than that of
vinyl proton.
 Finally sp2 > sp > sp3 .
69

Anisotropic Effect
Due to the presence of pi bonds
70

 Anisotropic effects constitute shielding and deshielding
effects on the proton because of induced magnetic fields in
other parts of the molecule which operate through space.
 For example if a magnetic field is applied to a molecule
having п electrons, these electrons begin to circulate at
right angles to the direction of the applied field thereby
producing induced magnetic field.
71

 The effect of this field on the nearby proton has been
found to depend upon the orientation of the proton
with respect to the Π bond producing the induced
field.
72

Secondary magnetic field
generated by circulating 
electrons deshields aromatic
protons
Circulating  electrons
Ring Current in BenzeneRing Current in Benzene
Bo
Deshielded
H H fields add together

C=C
HH
H H
Bo
ANISOTROPIC FIELD IN AN ALKENE
protons are
deshielded
shifted
downfield
secondary
magnetic
(anisotropic)
field lines
Deshielded
fields add

Bo
secondary
magnetic
(anisotropic)
field
H
H
C
C
ANISOTROPIC FIELD FOR AN ALKYNE
hydrogens
are shielded
Shielded
fields subtract

Protons attached to sp2 hybridized carbon are less shielded than
those attached to sp3 hybridized carbon.
H H
HH
H
H
C C
HH
H H
CH3CH3
d 7.3 ppm d 5.3 ppm d 0.9 ppm

Acidic Hydrogens &
Hydrogen bonding
77

Acidic hydrogens:
Both resonance and the Electronegativity effect of
oxygen withdraw electrons from the acid proton.
78

HYDROGEN BONDING DESHIELDS PROTONS
O H
R
O R
HHO
R
The chemical shift depends
on how much hydrogen bonding
is taking place. ( in concentrated solution)
Hydrogen bonding lengthens the
O-H bond and reduces the valence
electron density around the proton
- it is deshielded and shifted
downfield in the NMR spectrum.

O
C
O
R
H
H
C
O
O
R
Carboxylic acids have strong
hydrogen bonding - they
form dimers.
O
O
O
H
CH3
In methyl salicylate, which has strong
internal hydrogen bonding,
Notice that a 6-membered ring is formed.
SOME MORE EXTREME EXAMPLES

APPROXIMATE CHEMICAL SHIFT RANGES (ppm) FOR SELECTED TYPES OF PROTONS
R-CH3 0.7 - 1.3
R-C=C-C-H 1.6 - 2.6
R-C-C-H 2.1 - 2.4
O
O
RO-C-C-H 2.1 - 2.5
O
HO-C-C-H 2.1 - 2.5
N C-C-H 2.1 - 3.0
R-C C -C-H 2.1 - 3.0
C-H 2.3 - 2.7
R-N-C-H 2.2 - 2.9
R-S-C-H 2.0 - 3.0
I-C-H 2.0 - 4.0
Br-C-H 2.7 - 4.1
Cl-C-H 3.1 - 4.1
RO-C-H 3.2 - 3.8
HO-C-H 3.2 - 3.8
R-C-O-C-H 3.5 - 4.8
O
R-C=C-H
H
6.5 - 8.0
R-C-H
O
9.0 - 10.0
R-C-O-H
O
11.0 - 12.0
O2N-C-H 4.1 - 4.3
F-C-H 4.2 - 4.8
R3CH 1.4 - 1.7
R-CH2-R 1.2 - 1.4 4.5 - 6.5
R-N-H 0.5 - 4.0 Ar-N-H 3.0 - 5.0 R-S-H
R-O-H 0.5 - 5.0 Ar-O-H 4.0 - 7.0
R-C-N-H
O
5.0 - 9.0
1.0 - 4.0R-C C-H 1.7 - 2.7

YOU DO NOT NEED TO MEMORIZE THE
PREVIOUS CHART
IT IS USUALLY SUFFICIENT TO KNOW WHAT TYPES
OF HYDROGENS COME IN SELECTED AREAS OF
THE NMR CHART
aliphatic
C-H
CH on C
next to
pi bonds
C-H where C is
attached to an
electronegative
atom
alkene
=C-H
benzene
CH
aldehyde
CHO
acid
COOH
2346791012 0
X-C-H
X=C-C-H
MOST SPECTRA CAN BE INTERPRETED WITH
A KNOWLEDGE OF WHAT IS SHOWN HERE

Number of signals
 In a given molecule the protons with different environments
absorb at different applied field strengths whereas the protons
with identical environments absorb at the same field strength.
 A set of protons of identical environments are known as
equivalent protons while the protons with different
environments are known as non-equivalent protons.
 The number of signals in a PMR spectrum tells us how many
kinds of protons are present in a given molecule.
83

 Prediction of different kinds of protons:
suppose each hydrogen or proton in the molecule to be
substituted by some other atom Z. If the substitution of
either of two protons by Z is expected to furnish the same
product or enantiomeric products (i.e. mirror images), the
two protons would be chemically and magnetically
equivalent, otherwise not.
84

 Ethyl chloride
CH3-CH2-Cl
 Substitution of a methyl proton
CH2Z-CH2-Cl
 Substitution of a methylene proton
CH3-CHZ-Cl
So methyl protons are not equivalent to methylene
protons.
85

Methyl Acetate
(Two NMR signals)

Ethyl benzene
(Three NMR signals)

Wrong
Methyl cyclopropane
(Three signals)

 2- chloropropene having three methyl and two
vinylic protons.
 The methyl protons are non-equivalent to vinylic
protons but equivalent amongst themselves. In
case of the vinylic protons, replacement of either
of the two would yield one of a pair of
diastereomeric products.
96

The area under a peak is proportional
to the number of hydrogens that
generate the peak.
Integration = determination of the area
under a peak
INTEGRATION OF A PEAK
Not only does each different type of hydrogen give a
distinct peak in the NMR spectrum, but we can also tell
the relative numbers of each type of hydrogen by a
process called integration.

55 : 22 : 33 = 5 : 2 : 3
The integral line rises an amount proportional to the number of H in each peak
METHOD 1
integral line
integral
line
simplest ratio
of the heights
Benzyl Acetate

Benzyl Acetate (FT-NMR)
assume CH3
33.929 / 3 = 11.3
33.929 / 11.3
= 3.00
21.215 / 11.3
= 1.90
58.117 / 11.3
= 5.14
Actually : 5 2 3
METHOD 2
digital
integration
Modern instruments report the integral as a number.
CH2 O C
O
CH3
Integrals are
good to about
10% accuracy.

Each different type of proton comes at a different place .
NMR Spectrum of Phenyl acetone

Often a group of hydrogens will appear as a multiplet
rather than as a single peak.
SPIN-SPIN SPLITTING
Multiplets are named as follows:
Singlet Quintet
Doublet Septet
Triplet Octet
Quartet Nonet
This happens because of interaction with neighboring
Hydrogens.

 The sub peaks are due to spin-spin splitting and are
predicted by the n+1 rule.
 Each type of proton senses the number of equivalent
protons (n) on the carbon atom(s) next to the one to which
it is bonded, and its resonance peak is split into (n+1)
components.
114

C C
H H
H
C C
H H
H
two neighbors
n+1 = 3
triplet
one neighbor
n+1 = 2
doublet
singlet
doublet
triplet
quartet
quintet
sextet
septet
MULTIPLETSthis hydrogen’s peak
is split by its two neighbors
these hydrogens are
split by their single
neighbor

EXCEPTIONS TO THE N+1 RULE
IMPORTANT !
Protons that are equivalent by symmetry
usually do not split one another
CH CHX Y CH2 CH2X Y
no splitting if x=y no splitting if x=y
1)
2) Protons in the same group
usually do not split one another
C
H
H
H or C
H
H
more
detail
later

3) The n+1 rule applies principally to protons in
aliphatic (saturated) chains or on saturated rings.
EXCEPTIONS TO THE N+1 RULE
CH2CH2CH2CH2CH3
CH3
Hor
but does not apply (in the simple way shown here)
to protons on double bonds or on benzene rings.
CH3
H
H
H
CH3
NO
NO
YES YES

SOME COMMON SPLITTING PATTERNS
CH2 CH2X Y
CH CHX Y
CH2 CH
CH3 CH
CH3 CH2
CH3
CH
CH3
( x = y )
( x = y )

SOME EXAMPLE SPECTRA
WITH SPLITTING

NMR Spectrum of Bromoethane
CH2CH3Br

NMR Spectrum of 2-Nitropropane
CCH3 CH3
N
H
O O
+
-
1:6:15:20:16:6:1
in higher multiplets the outer peaks
are often nearly lost in the baseline

NMR Spectrum of Acetaldehyde
offset = 2.0 ppm
CCH3
O
H

INTENSITIES OF
MULTIPLET PEAKS
PASCAL’S TRIANGLE

PASCAL’S TRIANGLE
Intensities of
Multiplet peaks

THE ORIGIN OF
SPIN-SPIN SPLITTING
HOW IT HAPPENS

C C
H H
C C
H HA A
upfielddownfield
Bo
THE CHEMICAL SHIFT OF PROTON HA IS
AFFECTED BY THE SPIN OF ITS NEIGHBORS
50 % of
molecules
50 % of
molecules
At any given time about half of the molecules in solution will
have spin +1/2 and the other half will have spin -1/2.
aligned with Bo opposed to Bo
neighbor aligned neighbor opposed
+1/2 -1/2

C C
H H
C C
H H
one neighbor
n+1 = 2
doublet
one neighbor
n+1 = 2
doublet
SPIN ARRANGEMENTS
The resonance positions (splitting) of a given hydrogen is
affected by the possible spins of its neighbor.

C C
H H
H
C C
H H
H
two neighbors
n+1 = 3
triplet
one neighbor
n+1 = 2
doublet
SPIN ARRANGEMENTS
methylene spins
methine spins

three neighbors
n+1 = 4
quartet
two neighbors
n+1 = 3
triplet
SPIN ARRANGEMENTS
C C
H H
H
H
H
C C
H H
H
H
H
methyl spins
methylene spins

J J
J
J J
THE COUPLING CONSTANT
The coupling constant is the distance J (measured in Hz)
between the peaks in a simple multiplet.
J is a measure of the amount of interaction between the
two sets of hydrogens creating the multiplet.
C
H
H
C H
H
H
J

 The coupling constant is a measure of how strongly a
nucleus is affected by the spin states of its neighbor.
 Coupling constant is expressed in Hertz (Hz).
137

100 MHz
200 MHz
123456
123
100 Hz
200 Hz
200 Hz
400 Hz
J = 7.5 Hz
J = 7.5 Hz
7.5 Hz
7.5 Hz
Coupling constants are
constant - they do not
change at different
field strengths
The shift is
dependant
on the field
ppm
FIELD COMPARISON
Separation
is larger

NOTATION FOR COUPLING CONSTANTS
The most commonly encountered type of coupling is
between hydrogens on adjacent carbon atoms.
C C
HH This is sometimes called vicinal coupling.
It is designated 3J since three bonds
intervene between the two hydrogens.
Another type of coupling that can also occur in special
cases is
C H
H
2J or geminal coupling
Geminal coupling does not occur when
the two hydrogens are equivalent due to
rotations around the other two bonds.
( most often 2J = 0 )
3J
2J

Couplings larger than 2J or 3J also exist, but operate
only in special situations.
Couplings larger than 3J (e.g., 4J, 5J, etc) are usually
called “long-range coupling.”
C
C
C
H H
4J , for instance, occurs mainly
when the hydrogens are forced
to adopt this “W” conformation
(as in bicyclic compounds).
LONG RANGE COUPLINGS

C C
H H
C C
H
H
C C
HH
C
H
H
6 to 8 Hz
11 to 18 Hz
6 to 15 Hz
0 to 5 Hz
three bond 3J
two bond 2J
three bond 3J
three bond 3J
SOME REPRESENTATIVE COUPLING CONSTANTS
Hax
Hax
Heq
Heq
Hax,Hax = 8 to 14
Hax,Heq = 0 to 7
Heq,Heq = 0 to 5
three bond 3J
trans
cis
geminal
vicinal

C
H
C H
4 to 10 Hz
H C C C
H
0 to 3 Hz four bond 4J
three bond 3J
C C
C H
H
0 to 3 Hz four bond 4J
H
H
cis
trans
6 to 12 Hz
4 to 8 Hz
three bond 3J
Couplings that occur at distances greater than three bonds are
called long-range couplings and they are usually small (<3 Hz)
and frequently nonexistent (0 Hz).

Interpretation of some
functional groups
143

Alkanes
 Alkanes have three types of hydrogens – methyl,
methylene and methyne.
144

Alkenes
146
 Alkenes have two types of hydrogens – vinyl (those attached
directly to the double bond) and allylic hydrogens (those
attached to the α- carbon, the carbon atom attached to the
double bond).

Aromatic compounds
 Aromatic compounds have two types of hydrogens –
aromatic ring hydrogens (benzene ring hydrogens) and
benzylic hydrogens (those attached to an adjacent
carbon atom).
148

R R = alkyl (only)
Apparently the ring current
equalizes the electron density
at all the carbons of the ring
and, therefore, at all of the
hydrogen atoms.
In monosubstituted rings with an alkyl substituent
all ring hydrogens come at the same place in the NMR
spectrum.
ALKYL-SUBSTITUTED RINGS

NMR Spectrum of Toluene
CH3
5
3

X = OH, OR,
O R O R O R O R
+ + +
:
..
:
- -
-
: : : :
..
Electron-donating groups
shield the o-, p- positions due to
resonance (see below).
.. ..
.. ..
..
..
..
X
.. unshared pair
ester
Electronegative elements with unshared pairs shield
the o- and p- ring positions, separating the hydrogens
into two groups.
NH2, NR2,
-O(CO)CH3
SUBSTITUENTS WITH UNSHARED PAIRS

O CH3
Anisole (400 MHz)
2 3
shielded
CH3
The ring protons in
toluene come at
about 7.2 ppm at
the red line.
Compare:

C
HH
R
OC
R
O
HH
Only the o- protons are in range for this effect.
The same effect is sometimes seen with C=C bonds.
When a carbonyl group is attached to the ring the
o- protons are deshielded by the anisotropic
field of C=O
THE EFFECT OF CARBONYL SUBSTITUENTS

C
CH3
O
HH
Acetophenone (90 MHz)
2 3
3
deshielded
CH3
The ring protons in
toluene come at
about 7.2 ppm at
the red line.
Compare:

1,4-Disubstituted benzene rings will show
a pair of doublets, when the two groups
on the ring are very different
an example:
1-iodo-4-methoxybenzene
para-Disubstitution

NMR Spectrum of
1-iodo-4-methoxybenzene
OCH3I
CHCl3 impurity
2 2
3

NMR Spectrum of
1-bromo-4-ethoxybenzene
OCH2CH3Br
4
2
3

X
Y
X
X'
X
X
X = Y X ~ X’ X = X
THE p-DISUBSTITUTED PATTERN CHANGES AS THE
TWO GROUPS BECOME MORE AND MORE SIMILAR
all H
equivalent
All peaks move closer.
Outer peaks get smaller …………………..… and finally disappear.
Inner peaks get taller…………………………. and finally merge.
same groups

NMR Spectrum of
1-amino-4-ethoxybenzene
OCH2CH3H2N
4
2 2
3

NMR Spectrum of p-Xylene
(1,4-dimethylbenzene)
CH3CH3
4
6

NMR
3/15/2016 165
m o/p
 The Amino group is Electron Donating
and Activates the ring.
 Increases electron density around
Ortho & Para protons relative to Meta.
 Chemical Shift, d, of ring protons is up field,
decreased ppm
o
m
m
op
2 Amino
Protons
Aniline (C6H7N)

3/15/2016 166
 Nitro group is electron withdrawing
and deactivates the ring.
 Protons in ring are deshielded moving
Chemical Shift downfield.
 Magnetic Anisotropy causes the Ortho
protons to be more deshielded than
the Para & Meta protons.
p m
o
o
m
m
p
o
Nitrobenzene (C6H5NO2)

3/15/2016 167
Ha Ha’
Hb
Hb’
Ha&Ha’Hb&Hb’
2 Amino Protons
 Para Di-Substituted Benzene ring
 Ha & Ha’ have same Chemical Shift
 Hb & Hb’ have same Chemical Shift
 Ha is split into doublet by Hb
 Hb is split into doublet by Ha
 Two sets of peaks produced by relative
Electronegativity of Amino & Cl groups
P-Chloroaniline (C6H6ClN)

Alkynes
 Alkynes have two types of hydrogens – acetylenic
hydrogen and α hydrogens found on carbon atoms
next to the triple bond.
168

Alkyl halides
 Alkyl halides have one type of hydrogen – α hydrogen
(the one attached to the same carbon as the halogen).
170

Alcohols
 Alcohols have two types of hydrogens – hydroxyl
proton and the α hydrogens (those on the same carbon
as the hydroxyl group).
172

Ethers
 Ethers have one type of hydrogen – the α hydrogens
(those attached to the α carbon, which is the carbon
atom attached to the oxygen).
174

Amines
 Amines have two types of hydrogens – those attached
to the nitrogen (the hydrogens of the amino group)
and those attached to the α carbon (the same carbon
to which the amino group is attached).
176

Nitriles
 Nitriles have one type of hydrogen – the α hydrogens
(those attached to the same carbon as the cyano
group)
178

Aldehydes
 Aldehydes have two types of hydrogens – aldehyde
hydrogen and the α hydrogens (those attached to the
same carbon as the aldehyde group).
180

Ketones
 Ketones have one type of hydrogens – hydrogens
attached to the α carbon.
182

Esters
 Esters have two types of hydrogens – those on the
carbon atom attached to the oxygen atom in the
alcohol part of the ester and those on the α carbon in
the acid part of the ester ( that is, those attached to the
carbon next to the C=O group).
184

Carboxylic acids
 Carboxylic acids have two types of hydrogens – acid
proton (the one attached to the –COOH group) and
the α hydrogens (those attached to the same carbon as
the carboxyl group).
186

NMR Spectrum of
2-Chloropropanoic Acid
offset = 4.00 ppm
COOH
C
O
OHCH
Cl
CH31
1
3
~12 ppm

Amides
 Amides have three types of hydrogens – hydrogens
attached to nitrogen, α hydrogens attached to the
carbon atom on the carbonyl side of the amide group,
and hydrogens attached to a carbon atom that is also
attached to the nitrogen atom.
188

Nitroalkanes
 Nitroalkanes have one type of hydrogens – α hydrogens,
those hydrogen atoms that are attached to the same
carbon atom to which the nitro group is attached.
190

EXAMPLE SPECTRA
FOR DISCUSSION
PART ONE

CH3 C
O
CH2CH3
NMR 07
Methyl Ethyl Ketone
1

NMR 08
Ethyl Acetate
2
CH3 C
O
O CH2CH3
Compare the methylene shift to that of Methyl Ethyl Ketone (previous slide).

a-Chloropropionic Acid
NMR 043
CH C
O
OH
Cl
CH3

t-Butyl Methyl Ketone NMR 034
C
O
CCH3 CH3
CH3
CH3
(3,3-dimethyl-2-butanone)

NMR 051-Nitropropane5
CH3CH2CH2 N
O
O
+
-

NMR 171,3-Dichloropropane6
Cl CH2CH2CH2 Cl

EXAMPLE SPECTRA
FOR DISCUSSION
PART TWO

NMR 09a-Bromobutyric Acid7
C
O
CH OH
Br
CH3CH2

NMR 14Phenylethyl Acetate8
CH2CH2 O C
O
CH3

NMR 18Ethyl Succinate9
O C
O
CH2CH2 C
O
OCH3CH2 CH2CH3

NMR 19Diethyl Maleate10
O C
O
CH3CH2 C C CH2CH3O
O
C
H H

NMR 12
Benzyl Alcohol
12
CH2 OH

13 NMR 13n-Propyl Alcohol
CH3CH2CH2 OH

Phenacetin (C10H13NO2)
3/15/2016 211
n+1=0+1=1
(singlet)
n+1=3+1=4
(quartet)
2H
1H
3H
P-Disubstitution
n+1=2+1=3
(triplet)
3H
4H

SPECTRUM OF ETHYL IODIDE (CH3CH2I)--

• Example: para-methoxypropiophenone
Each of two types of equivalent
Ar-H protons is split by its
neighbor (n = 1): a doublet
The -OCH3 protons are
unsplit (n = 0): a singlet
The 2 equivalent O=CCH2 protons
are split by 3 –CH3 protons (n = 3):
a quartet and vice-versa, a triplet

One triplet
1.2d
One quartet
3.5d
CH3CH2OCH2CH3

Benzyl acetate
 Integrals (Signal Area) (Con’t)
 NMR Spectrum – Benzylacetate (
3/15/2016 216
h1
h2
h3
(a)
(b)(c)

Four slides demonstrating a process for
interpreting an NMR Spectra
3/15/2016 217
3H
 Chemical Shift (d) in PPM
Note: Magnetic Field (Ho) increases 

Sextet
Quintet
2 protons see 4 protons
 5 peaks (quintet) produced
3 protons see 2 protons
 3 peaks (triplet) produced
1 proton sees 5 protons
 6 peaks (sextet) produced
3 protons see 1 proton
 2 peaks (doublet) produced
3H
Mono-substituted
Benzene Ring
Doublet Triplet
From chemical shifts, peak integration
values, and splitting patterns, develop
substructures for each signal.

3/15/2016 219
Solve the Puzzle

References
 Instrumental Methods of Chemical Analysis by
Gurdeep R. Chatwal, Sham K. Anand.
 Spectroscopy by Pavia, Lampman, Kriz, Vyvyan.
 www.google.com
221

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