NUCLEAR MAGNETIC
RESONANCE SPECTROSCPY
A guide for A level students
TMS
5
4
3
2
1
KNOCKHARDY PUBLISHING
0
d
2015
SPECIFICATIONS
KNOCKHARDY PUBLISHING
NMR SPECTROSCOPY
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...
www.knockhardy.org.uk/sci.htm
Navigation is achieved by...
either
clicking on the grey arrows at the foot of each page
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NMR SPECTROSCOPY
CONTENTS
• Prior knowledge
• What is nmr?
• Origin of spectra
• Tetramethylsilane
• Chemical shift
• Resolution
• Multiplicity - splitting patterns
• Integration
• OH signals and the use of D2O
• Working out spectra
• Questions on proton nmr
• Carbon-13 nmr
NMR SPECTROSCOPY
Before you start it would be helpful to…
• know the names and structures of organic functional groups
• find the structures of isomers given the molecular formula
PREVIEW
WHAT IS NMR AND WHAT DOES AN NMR SPECTRUM TELL YOU?
Proton nuclear magnetic resonance spectroscopy provides...
• information about the hydrogen atoms in molecules
How does it work?
• involves the interaction of materials with the LOW ENERGY RADIO WAVES
It provides the information by...
spinning a sample of the compound in a magnetic field
• hydrogen atoms in different environments respond differently to the field
• each different environment of hydrogen produces a signal in a different position
• the area under each peak / signal is proportional to the number of hydrogens
• signal can be split according to how many H’s are on adjacent atoms
NMR SPECTROSCOPY – ORIGIN OF SPECTRA
All nuclei possess charge and mass. Those with either an odd mass number or an
odd atomic number also possess spin. This means they have angular momentum.
POSSESS SPIN
1
H
1
2
H
1
DON’ T POSSESS SPIN
12
C
6
A nucleus without spin cannot be detected by
nuclear magnetic resonance spectroscopy.
b
ENERGY
A spinning nucleus such as 1H
behaves as a spinning charge and
generates a magnetic field. It can
be likened to a bar magnet. When it
is placed in an externally applied
field it can align with, or against,
the field. The energy difference
between the two states (DE)
depends on the applied field.
13
C
6
19
F
9
31
P
15
aligned against the field
DE = h
a
aligned with the field
NMR SPECTROMETERS
The sample is spun round in the field of a large electromagnet and a radio-frequency (RF)
field is applied. The magnetic field is increased and the excitation or “flipping” of nuclei
from one orientation to another is detected as an induced voltage resulting from the
absorption of energy from the RF field.
THE BASIC ELEMENTS
OF AN NMR
SPECTROMETER
RADIOFREQUENCY
OSCILLATOR
An nmr spectrum is the plot of the induced voltage against the sweep of the field. The
area under a peak is proportional to the number of nuclei “flipping”
Not all hydrogen nuclei absorb energy at the same field strength at a given frequency;
the field strength required depends on the environment of the hydrogen.
By observing the field strength at which protons absorb energy, one can deduce
something about the structure of a molecule.
NMR SPECTROSCOPY
INTERPRETATION OF SPECTRA
NMR spectra provide information about the structure of organic molecules from the ...
•
•
•
•
number of different signals in the spectrum
position of the signals (chemical shift)
intensity of the signals
splitting pattern of the signals
NMR SPECTROSCOPY
INTERPRETATION OF SPECTRA
NMR spectra provide information about the structure of organic molecules from the ...
•
•
•
•
number of different signals in the spectrum
position of the signals (chemical shift)
intensity of the signals
splitting pattern of the signals
NMR SPECTROSCOPY
INTERPRETATION OF SPECTRA
NMR spectra provide information about the structure of organic molecules from the ...
•
•
•
•
number of different signals in the spectrum
position of the signals (chemical shift)
intensity of the signals
splitting pattern of the signals
OBTAINING SPECTRA
• a liquid sample is placed in a tube which spins in a magnetic field
• solids are dissolved in deuterated solvents (CDCl3) or solvents without H’s (CCl4 )
[solvents with hydrogen atoms in them will produce peaks in the spectrum]
• TMS, tetramethylsilane, (CH3)4Si, is added to provide a reference signal
• when the spectrum is run, it can be integrated to find the relative peak areas
• spectrometers are now linked to computers to analyse data and store information
TETRAMETHYLSILANE - TMS
PROVIDES THE REFERENCE SIGNAL
• non-toxic liquid - SAFE TO USE
• inert - DOESN’T REACT WITH COMPOUND BEING ANALYSED
• has a low boiling point - CAN BE DISTILLED OFF AND USED AGAIN
• all the hydrogen atoms are chemically equivalent - PRODUCES A SINGLE PEAK
• twelve hydrogens so it produces an intense peak - DON’T NEED TO USE MUCH
• signal is outside the range shown by most protons - WON’T OBSCURE MAIN SIGNALS
• given the chemical shift of d = 0
• the position of all other signals is measured relative to TMS
The molecule contains four
methyl groups attached to a
silicon atom in a tetrahedral
arrangement. All the hydrogen
atoms are chemically equivalent.
CHEMICAL SHIFT
• each proton type is said to be chemically shifted relative to a standard (usually TMS)
• the chemical shift is the difference between the field strength at which it absorbs and
the field strength at which TMS protons absorb
• the delta (d) scale is widely used as a means of reporting chemical shifts
d
•
•
•
•
Observed chemical shift (Hz) x 106
ppm (parts per million)
Spectrometer frequency (Hz)
=
the chemical shift of a proton is constant under the same conditions (solvent, temperature)
the TMS peak is assigned a value of ZERO (d = 0.00)
all peaks of a sample under study are related to it and reported in parts per million
H’s near to an electronegative species are shifted “downfield” to higher d values
H
Approximate
chemical shifts
- C-X
ROH
-CHO
-COOH
13
12
11
-C=CH10
9
The actual values
depend on the
environment
- C-H
8
7
6
5
TMS
4
DOWNFIELD - ‘deshielding’
3
2
1
0
d
LOW RESOLUTION - HIGH RESOLUTION
• low resolution nmr gives 1 peak for each environmentally different group of protons
• high resolution gives more complex signals - doublets, triplets, quartets, multiplets
• the signal produced indicates the number of protons on adjacent carbon atoms
LOW RESOLUTION SPECTRUM OF 1-BROMOPROPANE
LOW RESOLUTION - HIGH RESOLUTION
• low resolution nmr gives 1 peak for each environmentally different group of protons
• high resolution gives more complex signals - doublets, triplets, quartets, multiplets
• the signal produced indicates the number of protons on adjacent carbon atoms
HIGH RESOLUTION SPECTRUM OF 1-BROMOPROPANE
The broad
peaks are split
into sharper
signals
The splitting pattern depends on the number of hydrogen atoms on adjacent atoms
MULTIPLICITY (Spin-spin splitting)
• low resolution nmr gives 1 peak for each environmentally different group of protons
• high resolution gives more complex signals - doublets, triplets, quartets, multiplets
• the signal produced indicates the number of protons on adjacent carbon atoms
Number of peaks = number of chemically different H’s on adjacent atoms + 1
1 neighbouring H
2 peaks “doublet”
1:1
2 neighbouring H’s
3 peaks “triplet”
1:2:1
3 neighbouring H’s
4 peaks “quartet”
1:3:3:1
4 neighbouring H’s
5 peaks “quintet”
1:4:6:4:1
Signals for the H in an O-H bond are unaffected by hydrogens on adjacent atoms - get a singlet
MULTIPLICITY (Spin-spin splitting)
• low resolution nmr gives 1 peak for each environmentally different group of protons
• high resolution gives more complex signals - doublets, triplets, quartets, multiplets
• the signal produced indicates the number of protons on adjacent carbon atoms
Number of peaks = number of chemically different H’s on adjacent atoms + 1
0 neighbouring H’s
1 neighbouring H
2 neighbouring H’s
3 neighbouring H’s
4 neighbouring H’s
signal isn’t split
signal split into
1 peak
2 peaks
3 peaks
4 peaks
5 peaks
“singlet”
“doublet” ratio = 1:1
“triplet”
1:2:1
“quartet”
1:3:3:1
“quintet”
1:4:6:4:1
MULTIPLICITY (Spin-spin splitting)
• low resolution nmr gives 1 peak for each environmentally different group of protons
• high resolution gives more complex signals - doublets, triplets, quartets, multiplets
• the signal produced indicates the number of protons on adjacent carbon atoms
Number of peaks = number of chemically different H’s on adjacent atoms + 1
0 neighbouring H’s
1 neighbouring H
2 neighbouring H’s
3 neighbouring H’s
4 neighbouring H’s
signal isn’t split
signal split into
1 peak
2 peaks
3 peaks
4 peaks
5 peaks
“singlet”
“doublet” ratio = 1:1
“triplet”
1:2:1
“quartet”
1:3:3:1
“quintet”
1:4:6:4:1
PASCAL’S TRIANGLE
1
It is interesting to note the relationship between the
successive peak ratios. It follows the pattern found in
Pascal’s triangle.
1
1
1
Each number in the series is the sum of the two numbers
above it in the triangle
What would be the pattern for 6 neighbouring hydrogens?
PRESS THE SPACE BAR FOR THE ANSWER
1
1
2
3
4
1
3
6
1
4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
MULTIPLICITY (Spin-spin splitting)
Splitting patterns are worked out by considering the effect adjacent, chemically
different protons have on another signal in a given environment.
The spin of the proton producing the signal is affected by each of the two forms of the
adjacent proton.
One orientation augments/enhances its field and the other opposes/reduces it. This
is done by calculating the various possible combinations of alignment of adjacent
protons.
HOWEVER
Signals for the H in an O-H bond are not affected
by hydrogens on adjacent atoms so are not split
MULTIPLICITY (Spin-spin splitting)
ANALOGY
Imagine you had an opinion on something. If
nobody influenced you, your opinion would be the
same. However if another person had a view on the
topic, they would either agree or disagree with you.
Their ideas would either enhance what you thought
or diminish it. There would be two possibilities of
equal chance.
If there were two people offering views they could
either be both for it (1 possibility) , both against (1
possibility) or one could be in favour and the other
against (2 possibilities). There would be three
possibilities of relative chance 1:2:1
CONTENTS
FOR AGAINST
MULTIPLICITY (Spin-spin splitting)
O adjacent H’s
There is no effect
MULTIPLICITY (Spin-spin splitting)
O adjacent H’s
There is no effect
1 adjacent H
can be aligned either with a or against b the field
there are only two equally probable possibilities
the signal is split into 2 peaks of equal intensity
MULTIPLICITY (Spin-spin splitting)
O adjacent H’s
There is no effect
1 adjacent H
can be aligned either with a or against b the field
there are only two equally probable possibilities
the signal is split into 2 peaks of equal intensity
2 adjacent H’s
more possible combinations
get 3 peaks in the ratio 1 : 2 : 1
MULTIPLICITY (Spin-spin splitting)
O adjacent H’s
There is no effect
1 adjacent H
can be aligned either with a or against b the field
there are only two equally probable possibilities
the signal is split into 2 peaks of equal intensity
2 adjacent H’s
more possible combinations
get 3 peaks in the ratio 1 : 2 : 1
3 adjacent H’s
even more possible combinations
get 4 peaks in the ratio 1 : 3 : 3 : 1
EXPLAIN THE THEORY BEHIND THE SPLITTING PATTERN FOR 4 ADJACENT H’s
MULTIPLICITY (Spin-spin splitting)
4 adjacent H’s
gives 5 peaks in the ratio 1 : 4 : 6 : 4 : 1
INTEGRATION
•
•
•
•
the area under a signal is proportional to the number of hydrogen atoms present
an integration device scans the area under the peaks
the ratio of relative areas is displayed on the spectrum
historically, lines on the spectrum showed the relative abundance of each H type
By measuring the distances between the integration lines one can
work out the simple ratio between the various types of hydrogen.
before integration
NOTICE THAT THE O-H SIGNAL IS ONLY A SINGLET
(see later for an explanation of this)
after integration
INTEGRATION – HISTORICAL PRESENTATION
Measure the
distance between
the top and
bottom lines.
Compare the
heights from
each signal and
make them into a
simple ratio.
Computers now
do the integration
automatically
HOW TO WORK OUT THE SIMPLE RATIOS
• Measure how much each integration line rises as it goes of a set of signals
• Compare the relative values and work out the simple ratio between them
• In the above spectrum the rises are in the ratio... 1:2:3
IMPORTANT: It doesn’t provide the actual number of H’s in each environment, just the ratio
O-H bonds and splitting patterns
• The signal due to the hydroxyl (OH) hydrogen is a singlet ... there is no splitting
• H’s on OH groups do not couple with adjacent hydrogen atoms
Arises because the H on the OH, rapidly exchanges with protons on other molecules
(such as water or acids) and is not attached to any particular oxygen long enough to
register a splitting signal.
O-H bonds and splitting patterns
• The signal due to the hydroxyl (OH) hydrogen is a singlet ... there is no splitting
• H’s on OH groups do not couple with adjacent hydrogen atoms
Arises because the H on the OH, rapidly exchanges with protons on other molecules
(such as water or acids) and is not attached to any particular oxygen long enough to
register a splitting signal.
OH hydrogens are always seen as a
singlet ... there is no splitting
This is a quartet despite the fact that
there are 4 H’s on adjacent atoms the H on the OH doesn’t couple
O-H bonds and the D2O shake
As has been pointed out, the signal due to the hydroxyl (OH) hydrogen is a singlet.
It is possible to identify which signal is caused by the H of an O-H group by doing a
‘D2O shake’. A small amount of deuterium oxide D2O, a form of water, is added to the
sample and the spectrum is re-run. Any signal due to O-H proton disappears.
The H in the O-H bond changes places with a deuterium atom, 2H or D, from D2O
Deuterium doesn’t exhibit nuclear magnetic resonance under the conditions used for
proton nmr so the signal is removed to another part of the spectrum.
.
O-H bonds and the D2O shake
As has been pointed out, the signal due to the hydroxyl (OH) hydrogen is a singlet.
It is possible to identify which signal is caused by the H of an O-H group by doing a
‘D2O shake’. A small amount of deuterium oxide D2O, a form of water, is added to the
sample and the spectrum is re-run. Any signal due to O-H proton disappears.
The H in the O-H bond changes places with a deuterium atom, 2H or D, from D2O
Deuterium doesn’t exhibit nuclear magnetic resonance under the conditions used for
proton nmr so the signal is removed to another part of the spectrum.
before shaking with D2O
O-H bonds and the D2O shake
As has been pointed out, the signal due to the hydroxyl (OH) hydrogen is a singlet.
It is possible to identify which signal is caused by the H of an O-H group by doing a
‘D2O shake’. A small amount of deuterium oxide D2O, a form of water, is added to the
sample and the spectrum is re-run. Any signal due to O-H proton disappears.
The H in the O-H bond changes places with a deuterium atom, 2H or D, from D2O
Deuterium doesn’t exhibit nuclear magnetic resonance under the conditions used for
proton nmr so the signal is removed to another part of the spectrum.
.
before shaking with D2O
after shaking with D2O
H atoms attached to the N in amines also interchange with deuterium
NMR SPECTROSCOPY
When is a hydrogen chemically different?
TWO SIGNALS
Quartet and triplet :- ratio of peak areas = 3 : 2
1
2
3
BUTANE
4
Carbons 1 & 4 are the similar and so are carbons 2 & 3
so there are only two different chemical environments.
The signal for H’s on carbon 2 is a quartet - you ignore
the two neighbours on carbon 3 because they are
chemically identical.
NMR SPECTROSCOPY
When is a hydrogen chemically different?
TWO SIGNALS
Quartet and triplet :- ratio of peak areas = 3 : 2
1
2
3
4
BUTANE
Carbons 1 & 4 are the similar and so are carbons 2 & 3
so there are only two different chemical environments.
The signal for H’s on carbon 2 is a quartet - you ignore
the two neighbours on carbon 3 because they are
chemically identical.
TWO SIGNALS
both singlets :- ratio of peak areas = 2 : 1
ETHANE-1,2-DIOL
Hydrogens on OH groups only give singlets. The signal
for H’s on each carbon are not split, because
- H’s on the neighbouring carbon are chemically
identical... and
- H’s on adjacent OH groups do not couple.
NMR SPECTROSCOPY - SUMMARY
An nmr spectrum provides several types of information :number of signal groups tells you
chemical shift
peak area (integration)
multiplicity
the number of different proton environments
the general environment of the protons
the number of protons in each environment
how many protons are on adjacent atoms
In many cases this information is sufficient to deduce the structure of an organic
molecule but other forms of spectroscopy are used in conjunction with nmr.
NMR SPECTROSCOPY
HOW TO WORK OUT AN NMR SPECTRUM
1.
2.
3.
4.
Get the formula of the compound
Draw out the structure
Go to each atom in turn and ask the ‘census’ questions
Work out what the spectrum would look like ... signals due to H’s nearer
electronegative atoms (Cl,Br,O) are shifted downfield to higher d values
THE BASIC “CENSUS”
Ask each hydrogen atom to...
- describe the position of the atom on which it lives
- say how many hydrogen atoms live on that atom
- say how many chemically different hydrogen atoms live
on adjacent atoms
BUT, REMEMBER THAT
H atoms on OH groups
- ONLY PRODUCE ONE PEAK
- DON’T COUNT AS A NEIGHBOUR
NMR SPECTROSCOPY
“CENSUS” QUESTIONS
- describe where each hydrogen lives
- say how many hydrogens live on that atom
- say how many chemically different hydrogen
atoms live on adjacent atoms
1
2
3
1-BROMOPROPANE
ATOM
UNIQUE DESCRIPTION OF THE POSITION
OF THE HYDROGEN ATOMS
H’S ON THE
ATOM
CHEMICALLY DIFFERENT
H’S ON ADJACENT ATOMS
SIGNAL
SPLIT INTO
1
On an end carbon, two away
from the carbon with the
bromine atom on it
3
2
2+1 = 3
2
On a carbon atom second from
the end and one away from the
carbon with the bromine atom
2
3+2 = 5
5+1 = 6
3
On an end carbon atom which
also has the bromine atom on it
2
2
2+1 = 3
Spectrum of 1-bromopropane
CHEMICAL SHIFTS
1
2
3
3 environments = 3 signals
Triplet
Sextet
Triplet
d = 3.4
d = 1.9
d = 1.0
Signal for H’s on carbon
3 is shifted furthest
downfield from TMS due
to proximity of the
electronegative halogen
TMS
5
4
3
2
1
0
d
Spectrum of 1-bromopropane
INTEGRATION
1
2
3
Area ratio from relative
heights of integration
lines = 2 : 2 : 3
Carbon 1
Carbon 2
Carbon 3
2
3
TMS
3
2
2
2
5
4
3
2
1
0
d
Spectrum of 1-bromopropane
SPLITTING
1
1
2
3
SPLITTING PATTERN
Carbon 1
Chemically different
hydrogen atoms on
adjacent atoms = 2
TMS
2+1 =3
The signal will be a
TRIPLET
5
4
3
2
1
0
d
Spectrum of 1-bromopropane
SPLITTING
2
1
2
3
SPLITTING PATTERN
Carbon 2
TMS
Chemically different
hydrogen atoms on
adjacent atoms = 5
5+1 =6
The signal will be a
SEXTET
5
4
3
2
1
0
d
Spectrum of 1-bromopropane
SPLITTING
3
1
2
3
SPLITTING PATTERN
Carbon 3
TMS
Chemically different
hydrogen atoms on
adjacent atoms = 2
2+1 =3
The signal will be a
TRIPLET
5
4
3
2
The signal is shifted furthest away (downfield) from TMS as the
hydrogen atoms are nearest the electronegative bromine atom.
1
0
d
Spectrum of 1-bromopropane
SPLITTING
3
1
2
2
1
3
3 environments = 3 signals
1 Triplet
2 Sextet
3 Triplet
d = 1.0
d = 1.9
d = 3.4
3 H’s
2 H’s
2 H’s
TMS
Signal for H’s on carbon
3 is shifted furthest
downfield from TMS due
to proximity of the
electronegative halogen
5
4
3
2
1
0
d
Spectrum of 1-bromopropane
1
2
3
TMS
4
SUMMARY
Peaks
Shift
Splitting
Integration
3
2
1
0
d
Three different signals as there are three chemically different protons.
Signals are shifted away from TMS signal, are nearer to the halogen.
Signals include a triplet (d = 1.0) sextet (d = 1.8) triplet (d = 3.4)
The integration lines show that the ratio of protons is 2:2:3
The signals due to the protons attached to carbon ...
C1 triplet
C2 sextet
C3 triplet
(d = 1.0)
(d = 1.8)
(d = 3.4)
coupled to the two protons on carbon C2
coupled to five protons on carbons C1 and C3
coupled to the two protons on carbon C2
( 2+1 = 3 )
( 5+1 = 6 )
( 2+1 = 3 )
NMR SPECTROSCOPY
SUPPLEMENTARY QUESTIONS
1. Why is proton nmr more useful for the investigation of organic compounds ?
2. What other nucleus found in organic compounds is investigated using nmr ?
3. What compound is used as the internal reference for proton nmr chemical shifts ?
How many peaks does it produce and at what delta (d) value does it appear ?
4. What uses have been made of nuclear magnetic resonance in other scientific areas ?
Supplementary Questions - Answers
SEE NEXT
FOR hydrogen
ANSWERSatoms.
1. Because organic compounds
tend PAGE
to contain
2. Carbon 13
3. Tetramethylsilane (TMS) gives a strong single peak at d = 0
4. Magnetic resonance imaging in body scanners
NMR SPECTROSCOPY
SUPPLEMENTARY QUESTIONS
1. Why is proton nmr more useful for the investigation of organic compounds ?
2. What other nucleus found in organic compounds is investigated using nmr ?
3. What compound is used as the internal reference for proton nmr chemical shifts ?
How many peaks does it produce and at what delta (d) value does it appear ?
4. What uses have been made of nuclear magnetic resonance in other scientific areas ?
Supplementary Questions - Answers
1. Because organic compounds tend to contain hydrogen atoms.
2. Carbon 13
3. Tetramethylsilane (TMS) gives a strong single peak at d = 0
4. Magnetic resonance imaging in body scanners
WHAT IS IT!
C2H5Br
ANSWER
WHAT IS IT!
C2H3Br3
ANSWER
WHAT IS IT!
C2H4Br2
ANSWER
WHAT IS IT!
C6H12
ANSWER
WHAT IS IT!
C2H4O2
ANSWER
WHAT IS IT!
C4H8O2
ANSWER
WHAT IS IT!
C3H6O
ANSWER
WHAT IS IT!
C3H6O
ANSWER
WHAT IS IT!
C4H8O
ANSWER
WHAT IS IT!
C8H16O2
ANSWER
WHAT IS IT!
C11H16
IN THIS SPECTRUM,
PROTONS ON AN
AROMATIC (BENZENE)
RING HAVE BEEN
SHOWN AS A SINGLET
ANSWER
WHAT IS IT!
C8H10
IN THIS SPECTRUM,
PROTONS ON AN
AROMATIC (BENZENE)
RING HAVE BEEN
SHOWN AS A SINGLET
ANSWER
WHAT IS IT!
C8H10
IN THIS SPECTRUM,
PROTONS ON AN
AROMATIC (BENZENE)
RING HAVE BEEN
SHOWN AS A SINGLET
ANSWER
WHAT IS IT!
C9H12
IN THIS SPECTRUM,
PROTONS ON AN
AROMATIC (BENZENE)
RING HAVE BEEN
SHOWN AS A SINGLET
ANSWER
WHAT IS IT!
C6H10O3
ANSWER
WHAT IS IT!
C4H8Br2
ANSWER
CARBON-13
NMR SPECTROSCOPY
CARBON-13 NMR SPECTROSCOPY
After hydrogen, the most useful atom providing information is carbon-13.
Natural carbon contains about 1% of this isotope so the instruments for its detection
need to be sensitive and spectra will take longer to record.
Only the chemical shift is important as each spectrum gives only single lines for each
chemically equivalent carbon.
CARBON-13 NMR SPECTROSCOPY
After hydrogen, the most useful atom providing information is carbon-13.
Natural carbon contains about 1% of this isotope so the instruments for its detection
need to be sensitive and spectra will take longer to record.
Only the chemical shift is important as each spectrum gives only single lines for each
chemically equivalent carbon.
Environment
C - C (alkanes)
C - C=O
C - Cl or C - Br
C - N (amines)
C - OH
C = C (alkenes)
aromatic C’s (benzene rings)
C=O (esters, acids, amides)
C=O (aldehydes, ketones)
Chemical shift / d
10 - 35
10 - 35
30 - 70
35 - 65
50 - 65
115 - 140
125 - 150
160 - 185
190 – 220
Carbon-13 nmr has wide applications in the study of natural
products, biological molecules and polymers.
CARBON-13 NMR SPECTRA
Isomers
of C3H7Br
H H H
H Br H
HCCCBr
HCCCH
H H H
H H H
3 peaks
all three carbons are different
2 peaks
the two outer carbons are similar
CARBON-13 NMR SPECTRA
Isomers
of C3H7Br
H H H
H Br H
HCCCBr
HCCCH
H H H
H H H
3 peaks
all three carbons are different
2 peaks
the two outer carbons are similar
Ethanol
C2H5OH
H
COH
H
H
H
H  C  C  OH
H
H
H
HC
H
This is where the
proton nmr
spectrum of ethanol
would be on the
same scale.
CARBON-13 NMR SPECTRA
The carbon-13 spectrum of 2-methylbutane (CH3)2CHCH2CH3
chemically
equivalent
carbon atoms
H CH3 H
H
HCCCCH
H
H H
H
There are four
chemically different
carbon atoms in the
molecule so there
are four peaks in the
C-13 nmr spectrum.
NO SPLITTING WITH C-13
ONLY ONE PEAK FOR
EACH CARBON
Other isomers of C5H12
pentane
2,3-dimethylpropane
CH3CH2CH2CH2CH3
(CH3)4C
3 peaks
2 peaks
CARBON-13 NMR SPECTRA - QUESTIONS
How many peaks would you expect there to be in the carbon-13 spectrum of…
• butane
• 2-methylpropane
CH3CH2CH2CH3
CH3CH(CH3)CH3
•
•
•
•
•
CH3CH2CH2CHO
CH3COCH2CH3
CH3COCH2CH2CH3
CH3CH2COCH2CH3
C6H12
butanal
butanone
pentan-2-one
pentan-3-one
cyclohexane
CARBON-13 NMR SPECTRA - QUESTIONS
How many peaks would you expect there to be in the carbon-13 spectrum of…
• butane
• 2-methylpropane
CH3CH2CH2CH3
CH3CH(CH3)CH3
2
2
•
•
•
•
•
CH3CH2CH2CHO
CH3COCH2CH3
CH3COCH2CH2CH3
CH3CH2COCH2CH3
C6H12
4
4
5
3
1
butanal
butanone
pentan-2-one
pentan-3-one
cyclohexane
19
CARBON-13 NMR SPECTRA - QUESTIONS
Identify the isomers of C4H8O
CARBON-13 NMR SPECTRA - QUESTIONS
Identify the isomers of C4H8O
A butanal
B butanone
C 2-methylpropanal
CARBON-13 NMR SPECTRA - QUESTIONS
Identify the isomers of C6H12
CARBON-13 NMR SPECTRA - QUESTIONS
Identify the isomers of C6H12
X hex-1-ene or hex-2-ene
2-methylpent-1-ene
3-methylpent-1-ene
2-methylpent-2-ene
3-methylpent-2-ene
Y cyclohexane
Z 2,3-dimethylbut-2-ene
or
or
or
or
or
REVISION CHECK
What should you be able to do?
Recall the how an nmr spectrum is produced
Explain and understand the origin of chemical shift
Explain and understand the purpose of integration
Explain and understand the purpose of shaking with D2O
Recall the differences between high and low resolution spectra
Explain and understand the origin of splitting patterns
Interpret and explain a simple proton or carbon-13 nmr spectrum
NUCLEAR MAGNETIC
RESONANCE SPECTROSCPY
THE END
© 2008, 2015 JONATHAN HOPTON & KNOCKHARDY PUBLISHING
START
AGAIN
WHAT IS IT!
C2H5Br
NEXT SPECTRUM
WHAT IS IT!
C2H3Br3
NEXT SPECTRUM
WHAT IS IT!
C2H4Br2
NEXT SPECTRUM
WHAT IS IT!
C6H12
NEXT SPECTRUM
WHAT IS IT!
C2H4O2
NEXT SPECTRUM
WHAT IS IT!
C4H8O2
NEXT SPECTRUM
WHAT IS IT!
C3H6O
NEXT SPECTRUM
WHAT IS IT!
C3H6O
NEXT SPECTRUM
WHAT IS IT!
C4H8O
NEXT SPECTRUM
WHAT IS IT!
C8H16O2
NEXT SPECTRUM
WHAT IS IT!
C11H16
NEXT SPECTRUM
WHAT IS IT!
C8H10
NEXT SPECTRUM
WHAT IS IT!
C8H10
NEXT SPECTRUM
WHAT IS IT!
C9H12
NEXT SPECTRUM
WHAT IS IT!
C6H10O3
NEXT SPECTRUM
WHAT IS IT!
C4H8Br2
NEXT SLIDE