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Education

Education

Education

Education

## Description

CONTENTS

e Introduction
e Fundamental principles of NMR
e Interpretation

¥ Chemical shift
Y Number of signals
¥ Spin-Spin coupling: Splitting of signals
Y Coupling constant
V Integrals

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NMR Spectroscopy

e Nuclear Magnetic Resonance is_ a branch of

spectroscopy in which radio frequency waves induce

transitions between magnetic energy levels of nuclei

of a molecule.

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_ Highest energy Lowest energy
Wavelength (nm) a Longer Wavelength

10°? 10° 10° 104 10° 108 19! 10!
| | | | | | | | | | | | | | |
T } T I
I I | |

Gamma |
ray X-ray violet

|’z Infrared Microwave ! Radio frequency

| } > | |
1 i l l

1 | | | | | | | | | | | |
10° 19!8 1o!° 10!4 10! 10! 108 10° 107

High Frequency —€— Frequency (s“’)

400 500 600 700 750 nm

Visible region

The Electromagnetic Spectrum
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The frequency of radio waves lies between 107 and 10®cps

calculated by using the equation :

E=hy

h = Planck’s constant = 6.6 x 1027 erg sec

v = frequency = 107- 10° cps(cycles per sec).

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E = 6.6 x 10-77 x 107 ( or 10° ergs)
= 6.6 x 10-° ( or 6.6 x 10-9 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.

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~ NUCLEAR SPIN

The nuclei of some atoms have a property called “SPIN”.

(a These nuclei behave as if

they were spinning.

See
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 = 21 +1

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e If the number of neutrons and the number of protons are

both even, then the nucleus has NO spin. »C, «O ,32S etc.

e 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, 3:P

e 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, yN

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Element H 2H 13 14 15 16 197 31p 32g

Nuclear spin
quantum Le ee ee eS eA ee
number ( J )

Number of Z 3 1 2 3 Z 1 2 DN
spin states

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Principle

e NMR spectroscopy is the interaction of magnetic field with spin

of nuclei and then absorption of radio frequency. For example,

the nucleus of proton ,H* has two spin rotations : clockwise

rotation with a spin quantum number I = +% and |

counterclockwise rotation with a spin quantum number I = – 4

e The number of spin sates is 2]+1 which is 2x (1/2) +1 = 2 state

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The two states
iv are equivalent

in energy in the
+1/2 -1/2 absence of a

magnetic or an

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e Without the magnetic field the spin states of nuclei

possess the same energy, and energy level transition is

not possible.

e When a magnetic field is applied, the separate levels

between these energy levels.

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Energy Differences Between Nuclear Spin S

oO

28 |
OY
= increasing field strength ASSEN

no difference in absence of magnetic field
proportional to strength of external magnetic field

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= Some important relationships in NMR

Units

The frequency of absorbed
Hz

is proportional to

the energy difference between kJ/mol

two nuclear spin states (kcal/mol)

which is proportional to

the applied magnetic field tesla (T)

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Magnetic properties of nuclel

° 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.

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® 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).

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A spinning proton re

crea tmagenestic field.

o

The nuclear | are

oriwieth nor tagaeinsdt B ..

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Nuclear Spin
Copyright © The McGraw-Hill Companies, inc. Permission required for reproduction or display

O s

Ho

(a) No external magnetic field (b) Apply external magnetic field #¢

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Higher

energy state Spin —4
(aligned against
the applied field)

Lower

energy state Spin +4

(aligned with
the applied field)

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

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External Magnetic Field
When placed in an external field, spinning

protons act like bar magnets.

GG
lower energy higher energy
more stable less stable

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THE “RESONANCE PHENOMENON

absorption of energy by the
spinning nucleus

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N

> -1/2
unal igned In a strong magnetic

field (B,) the two
spin states differ in

energy.

+1/2

aligned
B

O
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Two Energy States

The magnetic fields of wo

(| B state —

the spinning nuclei ‘ bit p Es

Y
will align either with
the external field, or Bo hv = AE

against the field.
A

A photon with the right
( | ) & state—

amount of energy can : Le

be absorbed and
cause the spinning
proton to flip.

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Excited state = High energy

N S

7) s 7

Energy Released

Aligned = Low Energy Back to low energy ground state

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e 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)]2/2 h / 2u

| = Spin quantum number

h= Plancks constant

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® But U=y~x spin angular momentum

LL = magnetic moment of the nucleus

y = gyro magnetic ratio

e If a nucleus having a magnetic moment is introduced into a

magnetic field , Hy the two energy levels become separate

corresponding to m , = -1/2 (anti-parallel to the direction of

magnetic field) and m, = +1/2 (parallel to the direction of

magnetic field).

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e Fora nucleus with I = 1/2, the energies E, and E, for the
two states with m , = +1/2 and m , = -1/2, respectively,
are
B= r/o |v hy |

E,=+1/2|yh2/p] H,

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——
T=1/2,m=+1/2,-1/2

Energy

No field

m=- 1/2

a

a a

m=+1/2 ‘
“§

BS
m=+1/2

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° When the nucleus absorbs energy, the nucleus will be

promoted from the lower energy state E,, to the

higher energy state E, by absorption of energy , AE,

equal to the energy difference, E, —E, .

e It means that the absorption of energy AE changes

the magnetic moment from the parallel state m , =

+1/2 to the anti-parallel state (m , = -1/2).

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e The frequency v at which energy is absorbed or
emitted is given by Bohr’s relationship:
v=E,-E/h

v=i1/2|yh/2p|H +1/2|vyh/sutH, /h (Cor)

Ve tl
° This is the Larmor equation

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freq uency of 6yromagnetic
the coming —— Y ratio y

tr, ansiti on ou
strength of the
Magnetic field

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® 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.

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e When a nucleus is placed in a system where it

absorbs energy, it becomes excited. It then loses

energy and again enters an excited state.

© This nucleus which alternatively becomes excited and

unexcited is said to be in a state of resonance.

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at es

” ASECOND EFFECT OF A STRONG MAGNETIC FIELD

WHEN A SPIN-ACTIVE HYDROGEN ATOM IS
PLACED IN A STRONG MAGNETIC FIELD

as IT BEGINS TO PRECESS

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°e 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.
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° The frequency of precession (proton) is directly proportional to

the strength of applied magnetic field.

e 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).

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Applied magnetic field
+

r — –
‘ –

Precessional .”~ ‘

orbit a ‘
‘ *

4 aov”

Spinning
nucleus

_——
~

~
~~

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ee ee

Nuclei precess at

frequency @ when

placed in a strong

Of
NUCLEAR

If V = @ then
energy will be MAGNETIC

RESONANCE

absorbed and NMR

invert. :
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Interpreting Proton NMR
Spectra

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TYPES OF INFORMATION
FROM THE NMR SPECTRUM

1. Each different type of hydrogen gives a peak
or group of peaks (multiplet).

2. The chemical shift (°, in ppm) gives a clue as
to the type of hydrogen generating the peak
(alkane, alkene, benzene, aldehyde, etc.)

3. The integral gives the relative numbers of each
type of hydrogen.

4. Spin-spin splitting gives the number of hydrogens

5. The coupling constantJ also gives information
about the arrangement of the atoms involved.

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~ 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.

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e 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.

e Nuclei that absorb on up field side are strongly shielded.

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° Let’s consider the just the proton (1H) NMR

7 Bare proton

Proton in organic molecule |

| |
aon } H,’ >H,

Increasing magnetic field strength ————>>
Increased shielding of nucleus © ———>

<d{_— Downfield Upfield ————>

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10 9 6 5 4 a O ppm
— Low-field Direction of field sweep —— High-field

Intensity of absorption ——>

~-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.

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® 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 (6 — value).

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~ Rather than measure the exact resonance position of a
peak, we measure how far downfield it is shifted from TMS.

reference compound
tetramethylsilane

a TM S ”

Highly shielded
protons appear
way upfield.

TMS Chemists originally

shift in Hz thought no other
compound would

downfield
come at a higher

field than TMS.

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e 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.

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the “chemical shift” in the following way:
parts per
million

chemical _ 5 shift in Hz = ppm
shift _ spectrometer frequency in MHz

This division gives a number independent
of the instrument used.

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e The alternative system which is generally used for defining the
position of resonance relative to the reference is assigned tau (Tt)
scale.

t™=10-5

e Asmall numerically value of 6 indicates a small downfield shift while
large value indicates a large downfield shift.

e Asmall value of t represents a low field absorption and a high value
indicates a high field absorption.

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SHIELDING AND DESHIELDING

® Rotation of electrons (mt) 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.
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° 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.

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e 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.

H H, ’ Hepp = Hp – He

spectrometer
field

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ln

Shielded Protons

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

electrons
shielding

( @ 60 MHz
BY
absorbs

Bo By Bo
14,092.0 gauss 14,092.0 gauss 14,092.3 gauss

naked proton shielded proton stronger applied field =>
absorbs at 14,092.0G!| __ feels less than 14,092.0G compensates for shielding

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e 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.

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e 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.

e This means that it also absorbs radiofrequency

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e 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.

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e 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.

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higher electron density requires a stronger field for

resonance

lower electron density requires a weaker field for

resonance

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Chemical shift depends upon following parameters:

Inductive effect

Hybridization

Anisotropic effect

Hydrogen bonding

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Inductive effect

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ELECTRONEGATIVITY — CHEMICAL SHIFT

Dependence of the Chemical Shift of CH,X on the Element X

Compound CH,X CHF CHLOH CHC] CH{Br CH CH, (CH,),Si

Element X F O Cl Br I H Si

Electronegativity of X AOA 35 3.1 2.8 25 eS

Chemical shift 6 4.26 3.40 2:05 260 dO 023 0

=
TMS

most 7

deshielded deshielding increases with the
Electronegativity of atom X

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CH,CH,CI © The H, protons are deshielded because they are closer to the electronegative
4. hi, Cl atom, so they absorb downfield from H,.

BrCH.CH,F © Because F is more electronegative than Br, the H, protons are more
vi he deshielded than the H, protons and absorb further downfield.

CICH,CHCI, © The larger number of electronegative Cl atoms (two versus one) deshield H,,
A, he more than H,, so it absorbs downfield from H,.

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— Ee

eet

=e

Chlorine “deshields” the proton,
C] —— C H that is, it takes valence electron

density away from carbon, which
in turn takes more density from

electronegative hydrogen deshielding the proton.
element

NMR CHART

“deshielded“ highly shielded
protons appeal protons appear

at low field at high field

<<._

deshielding moves proton
resonance to lower field

DESHIELD ING BY AN ELECTRONEGATIVE ELEMENT
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oi
at

Substitution Effects on Chemical Shift

most

deshielded The effect
CHC, CH,Cl, CHC increases with

7:27 5.30 3-05 ppm greater numbers
of electronegative

atoms.

most
deshielded -CH,-Br -CH,-CH,Br -CH,-CH,CH,Br

3.30 1.69 125 ppm

The effect decreases

with increasing distance.
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e Electron withdrawing groups— reduces electron density around

the proton Deshielding.

Examples: F, Cl, Br, |, OH, NH, NO,

e Electron releasing groups- increases electron density around

the proton Shielding.

Examples: Alkyl groups

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sp? Hydrogens ( no electro negative elements and N bonded

groups )

0 20 Se cyclopropane

H

a CHo CH3 i

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° sp? Hydrogens:

In an sp? C-H bond, the carbon atom has more s character

(33% s), which effectively renders it more electronegative

than an sp3 carbon (25% s).

e If the sp carbon atom holds its electrons more tightly, this

results in less shielding for the H nucleus than in an sp3

bond.

e Another effect anisotropy.

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>» 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 sp? >sp > sp3.

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Anisotropic Effect

Due to the presence of pi bonds

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e 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.

e For example if a magnetic field is applied to a molecule

having m electrons, these electrons begin to circulate at

right angles to the direction of the applied field thereby

producing induced magnetic field.

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° The effect of this field on the nearby proton has been

found to depend upon the orientation of the proton

with respect to the II bond producing the induced

field.

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‘ | f
Circulating 7 electrons

Deshiel
ded

es

Secondary Magnetic field
generated by Circulating 7
electrons deshields aromatic

protons
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protons are

See deshielded

Deshhiieldee d

we = shifted

C=C

“secondary
magnetic
(anisotropic)
field lines

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eS i eee SS

~ ANISOTROPIC FIELD FOR AN AL KYNE ~

aS secondary

Z magnetic
(anisotropic)

Shielded field

hydrogens

are shielded
fields subtract

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jccanaieenamaicaRaliietiaiciaiab

} Protons attached to sp? hybri dized carbon are | less shielded than
those attached to sp? hybridized carbon.

H

H HH = n

‘ — f CH,CH,

_ – a

H

57.3 ppm 55.3 ppm 60.9 ppm

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Acidic Hydrogens &
Hydrogen bonding

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Acidic hydrogens:

° O

= ——
OH OH

Both resonance and the Electronegativity effect of
oxygen withdraw electrons from the acid proton.

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ee
———

“HYDROGEN BONDING DESHIELDS PROTONS
The chemical shift depends

R on how much hydrogen bonding
| is taking place. ( in concentrated solution)
O—-

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.

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D—O—-x

ma

0

— =

~~ SOME MORE EXTREME EXAMP LES
mE a O. Carboxylic acids have strong

R—C C—R hydrogen bonding – they

> O H G form dimers.

In methyl salicylate, which has strong
internal hydrogen bonding,

Noticew wthwa.Dtu loaM,i0x.-commembered ring is formed.

APPROXIMATE CHEMICAL SHIFT — (ppm)

FOR SELECTED TYPES OF PROTONS

PR CH. 0.7 – 1.3 R-N- C a a Sepa ae Tat: C H

R-CH,-R 1.2 – 1.4 R-S-C-H 2.0 – 3.0 4.5 – 6.5

CH 1.4 -1. I

– CCH : ne ¢ – =
a aan ee Br-C-H 2.7-4.1
T ee

R-C-C-H2.1- 2.4 Cl-C-H 31-4.1

TI RO-C-H 3.2 – 3.8 R- INH

RO-C-C-H 2.1 = 2.5 ae 3 Se
0 -C- 3.2 – 3. i

; O R-C-H
HO-C-C-H 2.1- 2.5 | . O – 10.0

: R-C-O-C-H 3.5- 4.8
N=C-CH 2.1 – 3.0 i

; O,N-C-H 4.1- 4.3 ei
R-C=0-Ca 2.1 – 3.0 ‘ 0. =

F-C-H 4.2-4.8 11.0 – 12.0

¢ – CH 23-27
R-N-H 0.5 – 4.0 Ar-N-H 3.0 – 5.0 R-S-H

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

C-H where C is CHC

acid aldehyde |benzene | alkene anaes Wen next to aliphatic
COOH | CHO CH =C-H See ni bonds C-H

X.CH X=C-C-H
12 10 9 7 4 a

MOST SPECTRA CAN BE INTERPRETED WITH

A KNOWLEDGE OF WHAT IS SHOWN HERE

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Number of signals

e 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.

e A set of protons of identical environments are known as

equivalent protons while the protons with different

environments are known as non-equivalent protons.

e The number of signals in a PMR spectrum tells us how many

kinds of protons are present in a given molecule.

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e 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.

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e Ethyl chloride

CH3-CH2-Cl
e Substitution of a methyl proton
CH2Z-CH2-Cl

e Substitution of a methylene proton

CH3-CHZ-Cl
So methyl protons are not equivalent to methylene
protons.

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Prediction of Signal
Number

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(one signal)

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Benzene

(one signal)

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(Two NMR signals)

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Methyl Acetate

(Two NMR signals)

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Ethyl benzene

(Three NMR signals)

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Propane2-ol

(3 NMR signals)

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(Three signals)

wrong |

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(4 signals)

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e 2- chloropropene having three methyl and two
vinylic protons.

e 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.

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One signal

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H2C _. OC H3

Two signals

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O

AH3.C——C

OCH,

Two signals

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A.C ——O

Two signals

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wwwww..DDuulloM iix.ccoomm

CH>
>

HC O
7 Three signals

H3C

www.wD.uDluolMo iMxi.xc.coomm 103

Three signals

www.wD.uDluolMo iMxi.xc.coomm 104

Four signals

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INTEGRATION

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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 bya
process called integration.

Integration = determination of the area
under a peak

The area under a peak is proportional
to the number of hydrogens that

generate the peak.
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Benzyl Acetate
eee —

The integral line rises an amount proportional to the number of H in each peak
p

S00 490 200 200 Ore
i He

. 32.5 DW.

METHOD 1

integral line

+++

4 | A + sa =

; 3 a eee simplest ratio

29 = 33 w=ww.DueloMix.c om ; 2 of the heights

nN

Benzyl cetate (FT -NMR)
Actually : “~ @

58.117 / 11.3 21.215 / 11.3 33-929 / 11.3
= 5-14 = 1.90 = 3.00

METHOD 2

digital

integration
4:

Integrals are

to os 00 10% accuracy.
78 70 65 60 65 50 45 40 35 30 25 20

PPM

Modern instruments report the integral as a number.
wwwww.D.DuluoloMMixix..ccoomm

INTEGRAL

| T

| 4° _| sp oton
s AES el . te sh ielrs

Wp ? | |¢ ? _® r
o=— me. Teen aes go a 10 wares

25 »

15 ‘o 5 4

—T— i
Z

f

AN eas [ b

| N

| Pec
r

4

. at ok!

ee SS i ee a 20 ae;
Each different type of proton comes at a different place .

NMR Spectrum of Phenyl acetone
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SPIN-SPIN SPLITTING

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Often a group of hydrogens will appear as a multiplet
rather than as a single peak.

Multiplets are named as follows:

Singlet Quintet
Doublet Septet
Triplet Octet

Quartet Nonet

This happens because of interaction with neighboring
Hydrogens.

wwww..DDuulloMiix.ccoomm

1, 1, 2-Trichloroethane

www.wD.uDluolMo iMxi.xc.coomm 13

e The sub peaks are due to spin-spin splitting and are

predicted by the n+1 rule.

e 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.

wwwww.D.DuluoloMMixix..ccoomm 114

n+1 RULE

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80 70 60 50 40 30 20 10 OPPM

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this hydrogen’s peak these hydrogens are | MULTIPLETS
is split by its two neighbors split by their single

neighbor

Z singlet
doublet

i triplet

i quartet
H quintet

sextet
two neighbors
N+1 = 3 one neighbor septet

n+1 = 2

wwwww.D.DuluoloMMixix..ccoomm

PACEPHONS 10 Tet eerie) Le

IMPORTANT !

1) Protons that are equivalent by symmetry

usually do not split one another

El
X-CH+CH :

-Y X-CH2*CH2-Y

no eplitting if x=y no aolithina if x=y

2) Protons in the same group

usually do not split one another

d H

| more
= H Zs gee se detail

H H later

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w.Dul i com

rar ret 4 S f ~—T TT’ fw} : r a
& fr ; C = i J a, ’ il | KU Se

3) The n+irule applies principally to protons in
aliphatic (saturated) chains or on saturated rings.

YES YES

but does not apply (in the simple way shown here)
to protons on double bonds or on benzene rings.

tae NO or NO
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wwwww..DDuulloMiix.ccoomm

} Bb UU VY |Be |

ee
X-CH-CH-Y

(x4y)

|
=CHs-CH= CH3-CHo—

I
X-CHs-CHo-Y

(x4y)

N
WwWwWw .-DDullooMMixi’x . coor m

SOME EXAMPLE SPECTRA

WITH SPLITTING

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wwwww..DDuulloMiix.ccoomm

in higher multiplets the outer peaks
1:6:15:20:16:6:1 : ;

www.DuloMi_ x.co omm__are often nearly lost in the baseline
w .DuloMix.c J

~ NMR Spectrum of Acetaldehyde
400 200 100

ay
GHg- Cw

C | ] Core
; } ae | |
; Lid |

– >—_+—+ ; eH : :

er

a

offset = 2.0 ppm
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8-3-#-3-3)

PASCAL’S TRIANGLE

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Singlet

Doublet

Triplet

Quartet

Quintet 1 4 6 4 1

Sextet 1 5 10 10 5 1

Pascal’s triangle. Septet 1 6 15 20 15 6 1

Intensities of

Multiplet peaks

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THE ORIGIN OF

SPIN-SPIN SPLITTING
HOW IT HAPPENS

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H, H, –

R= C,

Cs R R Cy Cs R

R R R R

ise isnsaiasaceimil ees
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Chemical shift of proton A Chemical shift of proton A
in X-type molecules —————»> <—————_ in Y-type molecules

(deshielded—proton B has spin (shielded—proton B has spin
aligned with applied field) opposed to applied field)

— ——_——— §

Chemical shift of proton A if proton
B were absent

wwwww.D.DuluoloMMixix..ccoomm 130

~-AAFFECTED BY THE SPIN OF ITS NEIGHBORS

aligned with B,, opposed to B,,

50 % of 50 % of
molecules | | molecules

vA 1
C= 2 =

downfield upfield
neighbor aligned neighbor opposed

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.

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one neighbor one neighbor

The resonance positions (splitting) of a given hydrogen is

affected by the possible spins of its neighbor.
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one —
Nn+1 =
doublet =>

methine spins
methylene spins

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~ SPI N ARRANG
three neighbors two ae

n+1 = 4 n+1 =
Gz – triplet ‘=>

a 144
Ht thy
ith vst n 1

on

methylene spins

methyl spins wwwww.D.DuluoloMMixix..ccoomm

7

a

>

<~_—

<=

THE COUPLING CONSTANT

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7 . 4
eee Ti

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.

wwwww.D.DuluoloMMixi x..ccoomm

e The coupling constant is a measure of how stronglya
nucleus is affected by the spin states of its neighbor.

e Coupling constant is expressed in Hertz (Hz).

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100 Hz
Coupling constants are
constant – they do not
change at different

field strengths

200 MHz

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NOTATION FOR COUPELING:CONSTANTS
~ The most commonly encountered type of coupling is
between hydrogens on adjacent carbon atoms.

(H) This is sometimes called vicina! coupling.
] It is designated *J since three bonds

C=C. intervene between the two hydrogens.

Another type of coupling that can also occur in special
cases is

Se – Ef ( most often 2J = 0 )

; | Geminal coupling does not occur when
J the two hydrogens are equivalent due to

rotations around the other two bonds.

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Couplings larger than 7J or 7J also exist, but operate
only in special situations.

HX p= CNH)

4] , for instance, occurs mainly
when the hydrogens are forced
(as in bicyclic compounds).

Couplings larger than 3] (e.g., 4J, 5J, etc) are usually

called “long-range coupling.”
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SOME REPRESENTATIVE COUPLING CONS
——_

H H
paral sé 6 to 8 Hz three bond 3]

A.
trans C=C. 1 to 18 Hz three bond 3J

H

cis| See C=C oe 6to15Hz ~ thre3 e bond . 3]

/ H
geminal 2 oto5 Hz two bond “J

Hax Hax,Hax = 8 to 14

Heq Hax,Heq = 0 to 7 threebond 43J

Hed Hax Heq,Heq=o0to5

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cis 6to 12: Hz

J
trans 4to8Hz

H
H

me 4 to 10 Hz three bond 3)

C=

H
‘c=0* o to 3 Hz four bond 4y

GC—H

H—-C=C—C. 0 to3 Hz four bond 4]

H

Couplings that occur at distances greater than three bonds are
called long-range couplings and they are usually small (<3 Hz)
and frequently nonexistent (o Hz).

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Interpretation of some
functional groups

www.wD.uDluolMo iMxi.xc.coomm 143

~ Alkanes

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

CHEMICAL SHIFTS COUPLING BEHAVIOR
R-CH; 0.7-1.3 ppm ~CH-CH- *J=7-8 Hz

R-CH,-R_ ~—1.2-1.4 ppm

R3CH 1.4-1.7 ppm

wwwww.D.DuluoloMMixix..ccoomm 144

_— — oa ae

= — _ – — — = — a

7 —_ = – _ —- — ee. — _ —

aba. ~ octane. — ~ a
_ — — —

CH3(CHo}eCHs

10

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-Alkenes

e Alkenes have two types of hydrogens — vinyl (those attached
directly to the double bond) and allylic hydrogens (those
attached to the a- carbon, the carbon atom attached to the
double bond).

COUPLING BEHAVIOR
CHEMICAL SHIFTS

H—C=C—H >J rans ~ 11-18 Hz
C=C—-H 4.5-6.5 ppm Jeis ~ 6-15 Hz

—C-C-H 2] = 0-1 Hz

C=C—C-—H 1.6~2.6 ppm H
eee ae 4J= 0-3 Hz

H
wwwww.D.DuluoloMMixix..ccoomm 146

Fr -— -— ~ —~ ~~ 2-methyl-l-pentene –

Ane
Be = “ena

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

CHEMICAL SHIFTS
COUPLING BEHAVIOR

(1 6.5—8.0 ppm Mg = J-10 Hz

H “Ime=t 2a-3 Hz
“Ipg= 01H

CH— 2.3-2.7 ppm H

wwwww.D.DuluoloMMixix..ccoomm 148

a aaa = eer, ee ea ae

Pic, esas a
D

www.wD.uDluolMo iMxi.xc.coomm 149

MONOSUBSTITUTED RINGS

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–ALKYL-SUBSTITUTED RINGS

In monosubstituted rings with an alkyl substituent
all ring hydrogens come at the same place in the NMR
spectrum.

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.

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NMR Spectru of Toluene

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SUBSTITUENTS WITH UNSHARED PAIRS =
~Electronegative elements with unshared pairs shield

the o- and p- ring positions, separating the hydrogens
into two groups.

unshared pair
X

Electron-donating groups
shield the o-, p- positions due to
resonance (see below).

X = OH, OR, s + +
Co-R ‘O-R Ls ‘q-A

NH,, NR,, C) : O} §

-O(CO)CH,

—— ester
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Compare:

Orn
The ring protons in
toluene come at

the red line.

3 | 3

| . | : | | : | , | , | ; , | , |

a shicided : z wwwww..DDuullo M;i ix.ccoomm ! 7

THE EFFECT OF CARBONYL SUBSTITUENTS=——
When a carbonyl group is attached to the ring the
o- protons are deshielded by the anisotropic
field of C=O

ee ee

Only the o- protons are in range for this effect.

The same effect is sometimes seen with C=C bonds.

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Acetophenone (90 MHz) eee =

H H
Compare:

a ro
The ring protons in
toluene come at

the red line.

T | T | T | T | T ] 1 1

7 6 5 4 : 2 i 5

deshielded
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tO

Oo —

para -DISUBSTITUTED RINGS

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para-Disubstitution

1,4-Disubstituted benzene rings will show
a pair of doublets, when the two groups
on the ring are very different

lh
an example:

1-iodo-4-methoxybenzene

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NMR Spectrum of
1-iodo-4-methoxybenzene

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NMR Spectruomf

ocrs

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8-8 -8-8-8)

THE p-DISUBSTITUTED PATTERN CHANGES AS THE 92
————

_-FWO GROUPS BECOME MORE AND MORE SIMILAR|

All peaks move closer.
Outer peaks pet smaller: …5.c50.csssccccscecsans and finally disappear.
Inner peaks get taller……………….csscss0es: and finally merge.

okidd
bam bom all H

equivalent

X AY Kex’ X=X
www.wD.uDluolMoiMxi.xc.coomm Same grou ps

NMR Spectrum of
1-amino-4-ethoxybenzene

1 i l i a Ee ae aa eee Be Bn AS ee ee
80 72 60 50 40 30 20

Seber de Ge T ude ned I eee oe ve ee + ce a ee a

wwwww.D.Duluo loMMixix..ccoomm

8-88-88,

‘i

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

rr es he a

ato 400 300 200 100 oces

+190
| 6

50 rmty

CHs Sen
|

a
4 ;

|

it pb t SRSEERaiSens: ; Cesarsnseness
; |

et gee ee 4 aera t r T T vr Tt
00 70 60 50 40 30 20 19

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R – Alkyl Group
pseeee

——-.._._Produces a Singlet because all remaining
ring protons have identical chemical
shifts.

x – Electron Donating Group
Produces more electron density around
the O/P protons than the Meta protons,

m Ee Pp
producing separate signals.

xX – Electron Withdrawing Group
Decreases electron density around
the Of/P protoms more than the Meta
protons, producing separate signals.

xX=Y – Withdrawing groups such as Nitro,
Carbonyp, or other Double Bonds
influenced by anisotropy produce amore
pronounced withdrawing effect on the
Ortho protoms than the M/ FP protons.

xX(CcIBr.p X – Rings with Electron VWithdrawing Groups,
such as electronegative Halides, are more
influenced by the electron donating
resonance effect than the withdrawing
effect. The Ortho protons receive more
electron density than the miép protons.

p-Disubstitution
(a)

(a) -xX = Y (Same substituent)
(b) – Electronegativity x ¥ Y
(c) – Electronegativity X similar to Y

Mote: in patterns (b) & (c), four signals are

Produced because H, splits H, into a

doublet and H, splits H, into a

doublet.
The (b) & (c) patterns are diffeent
because the a & a’ (b & b”) protoms
are not magnetically equivalent.

3/15/2016 wwwww.D.DuluoloMMixix..ccoomm 164

Ch GP > GP

The Amino group is Electron Donating NH
and Activates the ring. ?

Increases electron density around
Ortho & Para protons relative to Meta.

Chemical Shift, 5, of ring protons is up field, i Pp O
decreased ppm mA

Aniline (CgH7N)
m_ jo/p

|

2 Amino
Protons

0
.

Lee 0°

Nitrobenzene (C,H,NO,)

= 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.

whwewe.‘DlDeuuloiM iix..com

‘P-Chloroaniline (CgHgCIN)

2 Hb& Hb’ Ha&Ha = # Para Di-Substituted Benzene ring
mF – = Ha & Ha’ have same Chemical Shift
™ . #” Hb & Hb’ have same Chemical Shift

cl = 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

2 Amino Pro tons

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Alkynes

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

CHEMICAL SHIFTS COUPLING BEHAVIOR

C=C—-H 1.7-2.7 ppm H-C=C-C-H +*J=2-3Hz

C=C-CH- 1.6-2.6 ppm

wwwww.D.DuluoloMMixix..ccoomm 168

– ye Rd – me 8

= 3 CH3CHoCH»—C =CH –
—_— _ —_ _ — _ ~— —_—

u i o

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Alkyl halides

e Alkyl halides have one type of hydrogen – « hydrogen
(the one attached to the same carbon as the halogen).

CHEMICAL SHIFTS

—CH-—I 2.0—4.0 ppm

—CH—Br 2.7—4.1 ppm

—CH—Cl 3.1—4.1 ppm

=—CH—F 4.2—4.8 ppm

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l-chlorobutane is
——— —+

abcd. 1 triplet

_CHCHYCHCHG
ee oo — ——-

sextet

| k at .

S s

www.wD.uDluolMo iMxi.xc.coomm 171

Alcohols

° Alcohols have two types of hydrogens — hydroxyl
proton and the a hydrogens (those on the same carbon
as the hydroxyl group).

CHEMICAL SHIFTS COUPLING BEHAVIOR

C—OH 0.5-5.0ppm CH-OH Nocoupling
(usually), or
37=5 Hz

CH-—O-H — 3.2-3.8 ppm

wwww..DDuulloMiix.ccoomm

~— Sea EenEa oo

doublet
Ln cepsieti, iets yes Game: vee oe CH – ss ssins

oat ee

oo CHsCHCH,OH
nian = sais ene so ans i= bd aman ate ara

4 triplet

Ae

3 1

wwwww.D.Duluolo MMixix..ccoomm 173

Ethers

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

CHEMICAL SHIFTS
R-O-CH- 32-3p8pm

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butyl methyl ether.

singlet _ |

——— CHCHCHyCHOCH; d
abce@ G0

triplet}
of multiplets

00
—————— oot –

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Amines

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

CHEMICAL SHIFTS COUPLING BEHAVIOR

R—N—-H 0.5—4.0 ppm

C—N-H 3J=0 Hz
|

—CH-—N- 2.22.9 ppm H

( )-xn 3.0-5.0 ppm

wwwww.D.DuluoloMMixix..ccoomm 176

— propylam i ne, —

__ CHyCH,CHaN
abdc

10

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Nitriles

e Nitriles have one type of hydrogen – the a hydrogens
(those attached to the same carbon as the cyano
sroup)

CHEMICAL SHIFTS

—-CH-C=N ss. 2.1-3.0 ppm

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valeronitrile. |

fea ee eee

‘tiget | ‘oe 7
cocina pip ipemainegtianemnaah sas

~—CHCHSCH,CH,CN cn

a dD ¢ a

———— —
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“Aldehydes

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

CHEMICAL SHIFTS COUPLING BEHAVIOR

R—CHO 9.0-10.0ppn —CH-CHO = “%J=13Hz

R-—CH-—CH=0 ~— 2.1-2.4 ppm

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Somr ere . er – eee me ee —

S acd | : doublet triplet
CH CHCH-C—H 4/8

|
| Coun wee =k a

en
coc |

—~

peels a a le t il

10 9 8 7 6 9 Q 3 0

www.wD.uDluolMo iMxi.xc.coomm 181

Ketones

e Ketones have one type of hydrogens — hydrogens
attached to the a carbon.

CHEMICAL SHIFTS

ae ets 2.1-2.4 ppm

R

wwww..DDuulloMiix.ccoomm

ace | da
CHCH,CH—C—CH

pence CH at eats AER pae

– *b
ee we re -__—_ oS

— a ‘ ee

10 g 8 7 6 § –

www.wD.uDluolMo iMxi.xc.coomm 183

“Esters

e 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 a carbon in
the acid part of the ester ( that is, those attached to the
carbon next to the C=O group).

CHEMICAL SHIFTS

R-—CH—COOR — 2.1-2.5 ppm

R-COO-CH— = 3.5-4.8 ppm

wwwww.D.DuluoloMMixix..ccoomm 184

_——

Le

| doublet

0 CH mace d _

¢ |. a]
– OCC oe amt

doublet ,
3 CHs— _ qo

i a 3 eos Awww:

_ =” mtt i let

ee an

: a ff
: 8 7 6 4 5 2 | 0

wwwww..DDuullo Miix.ccoomm

Carboxylic acids

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

CHEMICAL SHIFTS
R-COOH 11.0-12.0 ppm

—CH—-COOH 2.1-2.5 ppm

wwwww.D.DuluoloMMixix..ccoomm 186

NMR Spectrum o
2-Chloropropanoic Acid

60 70

offset = 4.00 ppm
wwwww..DDuullo Miix.ccoomm

“Amides

e Amides have three types of hydrogens — hydrogens
attached to nitrogen, a 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.

CHEMICAL SHIFTS

R(CO)—N—-H 5.0-9.0 ppm

—CH—CONH- 2.1—2.5 ppm

R(CO)—N—-CH 2.2—2.9 ppm

wwwww.D.DuluoloMMixix..ccoomm 188

‘H Spectrum of butyramide

0 _
abc | d triplet triplet

CHyCHCH—C—NHy “”

d
a

oe __ .

w . u o i . o
10 9 8 7 6 5 4 3

wwwD.DlulMo Mxixc.comm 189

Nitroalkanes

e Nitroalkanes have one type of hydrogens — a hydrogens,
those hydrogen atoms that are attached to the same
carbon atom to which the nitro group is attached.

—CH-NOQ, 41-43 ppm

wwwww.D.DuluoloMMixix..ccoomm 190

—abed
CHsCH,CH,CH,NO,
— – — _ — — -—_- —- — — — — a

10 g 8 7 6 9 4 3 2 1 0

wwwww..DDuull oMiix.ccoomm

EXAMPLE SPECTRA

FOR DISCUSSION

PART ONE

wwwww.D.DuluoloMMixix..ccoomm

——

Methyl Ethyl Ketone

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Ethyl Acetate

Compare the methylene shift to that of Methyl Ethyl Ketone (previous slide).

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OL-( hloropropionic Acid

wwww..DDuulloMiix.ccoomm

t-Butyl Methy4 Ketone
yr (3,3-dimethyl-2-butanone)

wwwww..DDuulloMiix.ccoomm

I-Nitroprepane___ TT.

O
CH3CH>CH>-I ¥

O

wwww..DDuulloMiix.ccoomm

[,3-Dichioroprepane____

CI-CH2GH2CH2-Cl

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EXAMPLE SPECTRA

FOR DISCUSSION

PART TWO

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wwwww..DDuulloMiix.ccoomm

~ Phenylethyl Acetate———___

wwwww..DDuulloMiix.ccoomm

Ethyl Succinare ——__—

O O
I |

C H3C H»5 =“O=C=C HoCH» -C-O-CH»sCH3

wwwww..DDuullo Miix.ccoomm

/

| -——-+
tT

i anh
ees 4—-+

+
+

= 40 = Ph +4

Tt + 4 eee ene 2
| . ted–

bast “4 o? te’ 4–

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—)
<j

OD
ro a

4)
ay

fp

*??
b>tu4

; ‘ >=
: > >

bbnL: s —_
: 14 -_-

; i
~_*—&— e+=— e +

; 1.4 + bows
nah o>

~ Ethanob————.

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n-Propyl Alcoho+—____—

CH3CHsCH>-OH

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an 0 6c 0 an a0 o 9 OPP

The ‘H NMR spectrum of 2-nitropr(o6p0 aMnHze),
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Integration = [5 2 2 2 a

il teat Jh
fr a

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“wer « So

13.0 12.5 12.0 11.5 PPM

_| s__| ,|}
ae 2 St et oe Tc a ee ee ee Se ee ee eee ee oa ae ee ee Se Se

7 6 5 = 3 2 1 PPM

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Phenacetin (C,,H,3;NO;)

3H n+1=0+1=1

// nl (singlet)

3H
n+1=2+1=3

P-Disubstitution (triplet)

4H

yo oH |
n+1=3+1=4

1H (quartet)

| : | ; | : | : | , | : | ; | : | | ; | : |

ey. 10 9 3 e 6 => 4 3 2 1 0

PRM
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‘The ‘H NMR spectrum of ethy! iodide (60 MHz).

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eee —— am =a

eee as

e Example: para-methoxypropiophenone

The -OCHs protons are
unsplit (n = 0): a singlet

Each of two types of equivalent The 2 equivalent O=CCH, protons
Ar-H protons is split by its are split by 3 -CH, protons (n = 3):
neighbor (n = 1): a doublet a quartet and vice-versa, a triplet

I
|

|

|
|

10 9 8 7 6 5 4

Chemical shift (6)

Intensity

+

CH,CH,OCH,CH,

4 <—

One quartet Oné triplet

; i E |
|

g LAA |

5 We
TMS

10 9 8 a 6 5 4 1 0 ppm

Chemical shift (6)

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| &) S Protons es
a. Protoss

|
h, 32.5 Div

b> | 2? Protons

ppm
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Four slides demonstrating.a,process for
interpreting an NMR Spectra

3H

5H
eH

4H

CTT TTT yee ye TT TUTTTTITTYTTIrryerrerry rr r yet errryrrirryrrrryttrry

isd Baha 5 5 4 3 2 { PPM
< Chemical Shift (5) in PPM

Note: Magnetic Field (H,) increases >

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F rom chemical shifts, peak integration 3H
values, and splitting patterns, develop
substructures for each signal.

Mono-substituted 5H
Benzene Ring

Try “TT

7 6
| H

Ga “f |
H

CH,
2 protons see 4 protons 3 protons see 2 protons

. 5 peaks (quintet) produced . 3 peaks (triplet) produced

1 proton sees 5 protons 3 protons see 1 proton
. 6 peaks (sextet) produced ”. 2 peaks (doublet) produced

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Solve the Puzzl
e

HOH H H

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2-Phenylbutane

S OLC

H 4
C
hH-C

C H

H;

Cpe :

7 6 3 4 oe. 2 { PPM

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~ References

© Instrumental Methods of Chemical Analysis by
Gurdeep R. Chatwal, Sham K. Anand.

© Spectroscopy by Pavia, Lampman, Kriz, Vyvyan.