IR Spectroscopy
2
IR Absorption
The major use of infrared spectroscopy is in determining the structures of
organic compounds. In an infrared spectrometer, infrared radiation in the range
400–4000cm−1 is passed through a sample.
The printout of the spectrum then shows which frequencies (wavenumbers)
are absorbed.
Infrared spectra are always looked at with the baseline (representing 100%
transmittance, i.e. zero absorbance) at the top.
The troughs (usually called ‘bands’) thus represent wavenumbers at which
radiation is absorbed.
3
IR Spectrum of Propanone
4
IR Spectrum of Propanone
The infrared spectrum can be used to determine the bonds present in a
molecule.
−1
Thus, in the region above 1500 cm in the infrared spectrum of propanone
there are two bands, corresponding to the C–H stretch and the C=O stretch.
The region below 1500 cm−1 is called the ‘fingerprint region’ and is
characteristic of the molecule as a whole. Comparison of the spectrum in the
fingerprint region with spectra in databases of infrared spectra can be used to
identify the molecule.
5
IR Spectrum of Propanone
For example, the infrared spectra of butanone and propanone can be
distinguished using the fingerprint region. They both show very similar bands
in the region above 1500cm−1 because they have the same functional group,
but they have different fingerprint regions.
he
6
IR Spectrum of Propanone
We are interested in identifying the bonds/functional groups in an organic
molecule.To
a good
approximation
the
variousbonds
bonds
in observed
a molecule
can be
The important
correlations
between
different
and
absorptions
are shown to
invibrate
Figure 9.22.
Hydrogen bonding
considered
independently
of each broadens
other. the absorption peaks of
−OH groups in alcohols, and even more so in carboxylic acids.
The wavenumbers at which some bonds vibrate are shown below.
Wavenumber ranges
–1
–1
4000 cm
–1
2500 cm
C
O
N
H
H
H
single bond
stretching
vibrations
–1
1900 cm
C
C
C
N
triple bond
stretching
vibrations
1500 cm
C
C
C
O
double bond
stretching
vibrations
–1
650 cm
fingerprint
region
7
Wavenumbers
So how do
absorbed?
We are usually interested in identifying the bonds/functional groups
we
connect
molecular
features
to
wavenumbers
in an organic molecule. To a good approximation the various bonds in a
molecule can be considered to vibrate independently of each other. The
Take
a
look
at
this
table:
wavenumbers at which some bonds vibrate are shown in Table A2.
Examiner’s tip
A table of infrared st
frequencies is given
IBO Chemistry Dat
The values in that ta
slightly from the on
here – you should u
in the data booklet f
examination.
8
Wavenumbers
Note: These values are a very close approximation to the actual
wavenumbers absorbed by di erent parts of the molecule. The
surrounding environment of each type of bond determines the exact
wavenumber absorbed.
We can use an infrared spectrum to identify the bonds present in a
molecule but cannot always distinguish between functional groups.
For example, we could identify the presence of C=O in a molecule but
ff
would not be able to distinguish between an aldehyde and a ketone.
9
Skill Check
Which of the following molecules will have an infrared band in
the 1700 to 1750cm−1 region?
A. but-2-ene
B. propanal
C. CH3CH2CH(OH)CH3
10 Butanoic Acid
This is the IR spectrum for butanoic acid. The ‘broad band’ absorption around
3000 cm-1 is characteristic of carboxylic acids:
also occurs in the
−1
00 cm , but this
bscured by the
ption.
the O–H band is
onding between
nfrared spectrum
xylic acid in the
H band is much
11 Butanoic Acid
We first of all look at the region above 1500 cm−1.
We can identify the C=O stretch, as this absorption band occurs in the 1700–
−1
1750 cm region.
The very broad absorption band between about 2400 and 3400cm−1 is due to
the O–H stretch in carboxylic acids and is very characteristic of those molecules.
The functional group of a carboxylic acid contains a C–O bond, and therefore we
should now look in the ngerprint region to con rm the presence of an absorption
in the region 1000–1300cm−1 which is, indeed, the case.
12 Butanoic Acid
If there were no band in this region, we would have to review our hypothesis
that the molecule is a carboxylic acid.
peaks in the 1000–1300 cm
13 Propan-1-ol
−1
region but no C–O.
The infrared spectrum of propan-1-ol is shown in Figure A12.
We can identify
O–H (about 3350cm−1) and
C–H (about 2900–
3000cm−1) bonds. A C–O
bond should also be
present, and we can see that
there is a band in the region
−1
1000–1300cm .
Again, by comparison of the bands with the values in Table A2, we
−1
−1
can identify O–H (about 3350 cm ) and C–H (about 2900–3000 cm )
bonds. A C–O bond should also be present, and we can see that there is a
14 Propan-1-ol
peaks in the 10
The infrared sp
Figure 1
CH3COO
0
0
20
20
Absorbance / %
242
Absorbance / %
15
Using the data in Table 18.3, note the broad bands
in Figures 18.8 and 18.9 arising from the O H groups
Ethanol
involved in hydrogen bonding in the alcohol and in the
40
60
80
100
4000
40
60
80
3500
3000
2500
2000
1500
–1
Wavenumber / cm
1000
500
100
400
nds
oups
16
n the
00
Figure 18.9 The infra-red spectrum of ethanoic acid,
Ethyl
ethanoate
CH3COOH.
0
Absorbance / %
20
40
60
80
500
100
4000
3500
3000
2500
2000
1500
–1
Wavenumber / cm
1000
500
harp
a-red
bonding
hols
O bond
peak.
acid and
nds
carboxylic acid. Contrast the width of these peaks with
the
sharp
peak
of
the
carbonyl
group
in
the
ester,
ethyl
Ethanoic Acid
ethanoate in Figure 18.10.
0
Absorbance / %
see that
y we 17
infraout the
r 29).
20
40
60
80
100
4000
3500
3000
2500
2000
1500
–1
Wavenumber / cm
1000
500
Absorbanc
Absorbance
absorbances
havehave
characteristic
widths
(broad
or sharp
absorbances
characteristic
widths
(broad
or sharp
40 40
peaks)
and
intensities
(strong
or
weak)
on
the
infra-red
peaks) and intensities (strong or weak) on the infra-red
spectrum.
For For
example,
the the
presence
of hydrogen
bonding
spectrum.
example,
presence
of hydrogen
bonding
60 60
in alcohols
makes
the absorbance
of the
O OH bonds
H bonds
in alcohols
makes
the absorbance
of the
and and
carboxylic
acids
broad.
By contrast,
the the
C CO bond
O bond
carboxylic
acids
broad.
By contrast,
80 80 Cambridge International AS Level Chemistry
in carbonyl
groups
has has
a strong,
sharp
absorbance
peak.
in carbonyl
groups
a strong,
sharp
absorbance
peak.
LookLook
at the
spectra
of ethanol,
ethanoic
acidacid
andand
at infra-red
the infra-red
spectra
of ethanol,
ethanoic
100100
3500 3000
3000 2500
2500 2000
2000 1500
1500 1000
1000 500500
40004000 3500
ethanoate
shown
in Figures
18.8–18.10.
ethylethyl
ethanoate
shown
in Figures
18.8–18.10.
Wavenumber
/ –1
cm–1
Wavenumber
/ cm
Using
the data
in Table
broad
bands
Using
the data
in Table
18.3,18.3,
notenote
the the
broad
bands
These values will usually be given to you. You can see that
Figure
18.9
infra-red
spectrum
of ethanoic
acid,
Figure
18.9
TheThe
infra-red
spectrum
of ethanoic
acid,
H groups
in Figures
arising
from
in Figures
18.818.8
andand
18.918.9
arising
from
the the
O OH groups
absorption
bands
overlap
considerably.
That
is why we
CH
COOH.
CH
COOH.
3 3
involved
in hydrogen
bonding
in the
alcohol
in the
involved
in hydrogen
bonding
in the
alcohol
andand
in the
need to use a variety of techniques, such as NMR, infra0 0
0 0 red spectroscopy and mass spectrometry, to work out the
structure of a new organic compound (see Chapter 29).
20 20
20 20
As well as their wavenumber bands, particular
absorbances have characteristic widths (broad or sharp
40 40
40 40 peaks) and intensities (strong or weak) on the infra-red
spectrum. For example, the presence of hydrogen bonding
60 60
60
60
makes the absorbance of the O H bonds in alcohols
242
242
and carboxylic acids broad. By contrast, the C O bond
80 80
80 80
in carbonyl groups has a strong, sharp absorbance peak.
100 Look at the infra-red spectra of ethanol, ethanoic acid and
100 100
100
3500 3000
3000
2500
2000 1500
1500 1000
1000 500500
4000 40003500350030003000250025002000200015001500 10001000 500 500
40004000
3500
2500
2000
ethyl
ethanoate
shown
in
Figures
18.8–18.10.
–1
Wavenumber
Wavenumber
/ –1
cm–1
Wavenumber
/ cm/–1cm
Wavenumber
/
cm
Using the data in Table 18.3, note the broad bands
Figure
infra-red
spectrum
of ethanol,
Figure
infra-red
spectrum
offrom
ethyl
ethanoate,
2OH.
Figure
18.818.8
The The
infra-red
spectrum
of ethanol,
CH3CH
CH32CH
OH.
in18.10
Figures
18.8
andspectrum
18.9
arising
the
O H groups
Figure
18.10
TheThe
infra-red
of ethyl
ethanoate,
3COOCH
2CH3.
CH3CH
COOCH
involved
2CH3. in hydrogen bonding in the alcohol and in the
ethanol
0
Absorbance / %
20
40
60
80
20
40
60
80
242
0
20
40
60
80
0
20
40
60
40
60
100
4000
3500
3000
2500
2000
1500
Wavenumber / cm–1
1000
500
Figure 18.9 The infra-red spectrum of ethanoic acid,
CH3COOH.
ethanoic acid
0
20
40
60
80
100
4000
20
80
Absorbance / %
0
Absorbance / %
0
B
B
Absorbance / %
Absorbance / %
A
0
20
7 Look
at two
the two
infra-red
spectra
below:
7 Look
at the
infra-red
spectra
below:
A
ethyl ethanoate
carboxylic acid. Contrast the width of these peaks with
the sharp peak of the carbonyl group in the ester, ethyl
ethanoate in Figure 18.10.
Absorbance / %
Absorbance / %
Absorbance / %
Absorbance / %
QUESTION
QUESTION
Absorbance / %
Absorbance / %
18 Comparing IR Spectrums
40
60
80
3500
3000
2500
2000
1500
Wavenumber / cm–1
1000
500
80 Figure 18.8 The infra-red spectrum of ethanol, CH3CH2OH.
100
4000
3500
3000
2500
2000
1500
Wavenumber / cm–1
1000
Figure 18.10 The infra-red spectrum of ethyl ethanoate,
CH3COOCH2CH3.
500
C “ O in RCHO
1730
C “ O in RCO2H
1720
C “ O in R2CO
1715
C “ C in alkenes
1650
C “ C in arenes
1600 and 1500
C—O
1100–1250
19 Skill Check
C “ C double bonds
transmittance/%
double bonds to oxygen
80
60
40
Compounds T and U are isomers with 20the molecular formula
single bonds
Worked
C3H6example
O2. Suggest
their structures based
on
the
spectra
shown
0
Compounds T and U are isomers with the molecular formula C3H6O2. Suggest their 4000
structures based on the spectra shown in Figures 29.15 and 29.16.
below:
80
60
40
20
0
4000
3000
1500
2500
2000
wavenumber/cm–1
1000
500
3000
2500
2000
1500
wavenumber/cm–1
1000
500
100
29_15 Cam/Chem AS&A2
Barking Dog Art
80
Figure 29.16 IR spectrum of U
transmittance/%
transmittance/%
100
3500
60
40
20
3500
100
29_15 Cam/Chem AS&A2
Barking Dog Art
80
3000
1500
2500
2000
wavenumber/cm–1
IR Spectra for
Compound T
1000
500
0
4000
3500
IR Spectra for
Both
T and
show a C“O absorption in their spectrum at about 1700–1800 cm
Barking
DogUArt
Compound
U
–1
C¬O absorption at about 1250 cm . T shows a broad hydrogen-bonded O¬H
Answer
29_16 Cam/Chem AS&A2
–1
20 Skill Check
Both T and U show a C=O absorption in their spectrum at
1700-1800 cm-1, and a C—O absorption at about 1250cm-1.
T shows ‘broad band’ absorption around 3000 cm-1, indicating the
presence of an O-H bond, while U shows no O-H band.
So, T is CH3CH2CO2H (propanoic acid) and U could either be the
ester CH3CO2CH3 (methyl ethanoate) or the ester HCO2CH2CH3
(ethyl ethanoate).
21
C “ C double bonds
single bonds
single bonds
C “ C in arenes
C
“ C in arenes
C—O
1600 and 1500
1600
and 1500
1100–1250
C—O
1100–1250
Worked example
Worked example
Compounds T and U are isomers with the molecular formula C3H6O2. Suggest their
Compounds
T and
arespectra
isomersshown
with the
molecular
formula
C3H6O2. Suggest their
structures based
onUthe
in Figures
29.15
and 29.16.
structures based on the spectra shown in Figures 29.15 and 29.16.
T
transmittance/%
transmittance/%
100
100
R) spectrum of T
R) spectrum of T
80
80
60
60
40
40
20
20
0
04000
4000
U
3000
3000
1500
2500
2000
wavenumber/cm
2500
2000 –1 1500
wavenumber/cm–1
1000
1000
500
500
3000
3000
2500
2000
1500
wavenumber/cm
2500
2000 –1 1500
wavenumber/cm–1
1000
1000
500
500
100
29_15 Cam/Chem AS&A2
100
Barking
Dog Art AS&A2
29_15 Cam/Chem
80 Dog Art
Barking
80
transmittance/%
transmittance/%
m of U
m of U
3500
3500
60
60
40
40
20
20
0
04000
4000
3500
3500
Answer
29_16 Cam/Chem AS&A2
–1
3
2
CH3COOCH2CH3.
22 Skill Check
QUESTION
7 Look at the two infra-red spectra below:
0
0
20
20
Absorbance / %
Absorbance / %
Which one of the infra-red spectra is that of butanone and which one is of
B
A
butan-2-ol?
40
60
80
100
4000
40
60
80
3500
3000
2500 2000 1500
Wavenumber / cm–1
1000
500
100
4000
3500
3000
2500 2000 1500
Wavenumber / cm–1
a Which one of the infra-red spectra is that of butanone and which one is of butan-2-ol?
b Explain your reasoning in part a.
1000
500
O
H
hydrogen bonded in alcohols, phenols
3230–3550
23 Skill Check
O
H
free
3580–3670
One of the three spectra labelled A to C below is produced when ethanal is analysed in an infra-red
spectrophotometer:
Which one of the infra-red spectra is that of ethanal? Give three reasons.
50
0
A
100
4000
3000
2000
1500
1000
Wavenumber / cm–1
500
Absorbance / %
0
Absorbance / %
Absorbance / %
0
50
C
100
4000
50
B
100
4000
3000
2000
1500
1000
Wavenumber / cm–1
500
3000
2000
1500
1000
Wavenumber / cm–1
500
converted to compound W by reagent X. Use the spectra in Figures 29.17 and 29.18
identify the functional groups present in V and W, and suggest the identity of reagen
24 Skill Check
transmittance/%
100
80
60
Compound V (C3H6O) gives aTechniques
silver mirror
when
warmed
with
Tollens’
of analysis 40
reagent. It can be converted to compound W20by reagent X. Use the spectra
below
to
identify
the
functional
groups
present
in
V
and
W,
and
suggest
the
0
Now try this
1500
1000
500
4000
3500
3000
2500
2000
identity
reagent
Compound
V (C Hof
O) gives
a silver mirrorX.
when warmed with Tollens’ reagent. It can be
wavenumber/cm
–1
3 6
converted to compound W by reagent X. Use the spectra in Figures 29.17 and 29.18 to
Figure 29.17 Infrared (IR) spectrum of V
identify the functional groups present in V and W, and suggest the identity of reagent X.
V
80
transmittance/%
transmittance/%
100
60
40
20
0
4000
W
29_17 Cam/Chem AS&A2
100 Dog Art
Barking
80
60
40
20
3500
3000
2500
2000
wavenumber/cm–1
Figure 29.17 Infrared (IR) spectrum of V
29_17 Cam/Chem AS&A2
1500
1000
500
0
4000
3500
3000
2500
2000
wavenumber/cm–1
Figure 29.18 IR spectrum of W
29_18 Cam/Chem AS&A2
1500
1000
500
(2)
b) 1-bromopropane
(2)
25 Skill Check
c) Propanone
(2)
100
Relative abundance/%
a) Cyclohexane
80
2 Oxidation gives
of butan-1-ol
gives with
a product
the spectrum shown
Oxidation of butan-1-ol
a product
the with
infrared
infrared spectrum shown in Figure 15.14. Use the data
60
below. Use the data
sheet
to
interpret
the
spectrum.
State
the
reagents
and
sheet to interpret the spectrum. State the reagents and
that
used
carry
out the oxidation
of
conditions that conditions
were used
towere
carry
outtothe
oxidation
of the alcohol
and
40 give
the
alcohol
and
give
your
reasons.
(4)
your reasons.
80
0
Transmittance/%
100
20
0
60
M
Figure 15.15 !
40
b) Write an equatio
molecular ions.
20
0
4000
10
3500
3000
2500
2000
1500
Wavenumber/cm–1
1000
500
c) i) Write an equ
ion fragment
26 Skill Check
The infra-red spectrum of compound A is shown below.
–1 give about compound
What
information
does
the
absorption
at
1690
cm
•
A?
•
• The infra-red spectrum of compound A does not show a broad absorption
–1
There is a sharp absorption at 2950 cm . What could this be due to?
at about 3300 cm–1. What information does give about compound A?
27
DERN
ANALYTICAL CHEMISTRY
100
80
60
40
B data booklet to
for:
20
[2]
[2]
hich could be used
mpounds. Explain
be used.
[2]
which has the
elength in cm.
3. The infra-red spectrum of compound A is shown below.
% transmittance
ool for identifying
rs at the molecular
d (ir) radiation and
or ir absorption to
re particularly
anic molecules. [4]
[2]
0
3000
(a) What information does the
give about compound A?
2000
1000
wavenumber / cm–1
–1
absorption at 1690 cm
[1]
(b) There is a sharp absorption at 2950 cm –1. What could
this be due to?
[1]
(c) The infra-red spectrum of compound A does not show
28 Skill Check
Infrared spectroscopy is a powerful tool for identifying organic compounds.
List the absorption regions expected for:
(a)ethanoic acid.
(b)methylmethanoate.
29 Skill Check
Identify the absorption listed in (a) which could be used to distinguish
between these two compounds. Explain why the other absorptions could
not be used.
CH
OH
NH
C C
C N
single bond
stretching
vibrations
triple bond
stretching
vibrations
Note
30 Skill Check
Most organic molecules contain C–H
bonds. As a result, most organic
compounds have a peak in their
infrared spectrum at around 3000 cm–1.
C C
CO
double bond
stretching
vibrations
fingerprint
region
The following are the infrared spectra of ethanol, ethanal and ethanoic acid.
Figure 15.4 !
Main regions of the infrared spectrum and important correlations between bonds and
observed absorptions.
(a)Which spectrum belongs to which compound?
–1?
(b)Why do all three spectra
have
a
peak
at
around
3000
cm
Molecules with several atoms can vibrate in many ways because the vibrations
Data
4000
B
c
Transmittance/%
A
Transmittance/%
b
Transmittance/%
a
of one bond affect others close to it. The complex pattern of vibrations can be
used ashave
a ‘fingerprint’
to peaks
be matched
the recorded infrared
spectrum in
(c)Why do two of the spectra
broad
atagainst
wavenumbers
between
a database.
3000 and 3500cm–1?
C
D
2000
Wavenumber/cm –1
1000
600
4000
2000
Wavenumber/cm –1
E
F
G
1000
600
4000
2000
Wavenumber/cm –1
1000
600
2 Certain molecules absorb infrared (IR) radiation.
31 Skill Check
H CO
a Consider the molecules:
2
HCl N2 O2
Select the molecule(s) from this list that can absorb IR radiation and explain your choice.
[3]
A student recorded the IR spectrum of a compound, X. He knew that X was
[4]
one of the following compounds. Deduce which of the molecules is X and
IR spectroscopy is one of the techniques that is used to determine the structure of organic molecules.
your
by
A explain
student recorded
thechoice
IR spectrum
of areference
compound, X: to the spectrum.
b Explain why there are three absorption bands in the IR spectrum of SO2 but only two in the
IR spectrum of CO2.
c
The student knew that X was one of the following compounds.
Deduce which of the molecules is X and explain your choice by reference to
32 Environmental Concerns
Small molecules in the atmosphere (especially CO2, CH4, H2O and
CFCs) are responsible for the greenhouse e ect: they absorb
infrared radiation that is emitted from the surface of the Earth,
preventing it from being lost to space.
Consequently, the amount of heat lost is less than that gained from
ff
solar radiation, and the Earth warms up.
fi
33 Environmental Concerns
• IR spectroscopy works quickly and accurately to monitor
pollutants, including nitrogen dioxide, sulfur dioxide, carbon
monoxide and carbon dioxide, as well as more than a hundred
VOCs (volatile organic compounds) and low-level ozone.
• Scientists can use the characteristic wavelengths of infrared
radiation absorbed by the molecules of the pollutants to identify
them. They can also analyse the intensity of the absorptions to
nd the concentration of each pollutant present in a sample.
34 Environmental Concerns
• Monitored over a period of time, this data provides useful information on the
effectiveness of pollution control measures introduced locally and on a
global level.
Mass Spectrometry
36 Mass Spectrometry
Is used to determine the structures of organic compounds:
1. finding the molecular formula of a compound by measuring the mass of its
molecular ion to a high degree of accuracy
2. finding the number of carbon atoms in a molecule by measuring the
abundance ratio of its molecular ion (M) peak and the M+1 peak
37 Mass Spectrometry
3. finding whether a compound contains chlorine or bromine atoms, and if so,
how many of each, by measuring the abundance ratios of the M & M+2
peaks
4. working out the structure of a molecule by looking at the fragments
produced when an ion decomposes inside a mass spectrometer
38 Mr Using Mass Spectrometry
If we vaporise an organic molecule and subject it to the ionising conditions
inside a mass spectrometer, the mass/charge ratio (m/e) for the molecular ion
can be measured, and hence the relative molecular mass can be found.
Analysing the molecular ion
If we vaporise an organic molecule and subject it to the ionising co
39 Mr Using
Mass
Spectrometry
a mass spectrometer, the mass/charge ratio (m/e) for the molecular
measured,
and
hence
the
relative
molecular
mass
can
be
found.
For example, one of the non-bonding electrons on the oxygen atom of
For example, one of the non-bonding electrons on the oxygen at
propanone can be removed by electron bombardment, to give an ionised
can be removed by electron bombardment, to give an ionised mole
molecule:
••
••
••
O
O
e
C
CH 3
CH 3
•
2e
C
CH3
The m/e ratio for the resulting molecular ion is 58.
CH3
The m/e ratio for the resulting molecular ion is (3 × 12 + 6 × 1 + 16) :
Using very high resolution mass spectrometry, we can measu
an accuracy of five significant figures (1 part in 100 000). By this me
and 43, as
40 Mr of Propanone
Figure 29.8 Mass spectrum of propanone
100
43
80
60
40
15
20
58
0
10
20
30
40
mass number
50
60
relative abundance
relative abundance
100
Figu
80
60
40
20
0
41 The M+1 Peak
Carbon has two stable isotopes, carbon-12 and carbon-13.
C-12 makes up 98.9% of all the carbon atoms and C-13 makes up the other
1.1%.
This means that out of every 100 methane (CH4) molecules, about 99
molecules will be 12CH4 and just one molecule will be 13CH4.
42 The M+1 Peak
For ethane, C2H6, the chances of a molecule containing one 13C atom will
have increased to about 2 in 100,
because each C atom has a chance of 1 in 100 to be 13C, and there are two of
them.
43 The M+1 Peak
If you look to the right of the molecular ion, M+, you can usually see a much
smaller peak, called an M+1 peak.
This is caused by the molecular ion with a C-13 atom replacing one of the
C-12 atoms.
44 The M+1 Peak
If the compound has just one C atom, this extra peak will be just 1.1% of the
height of the main molecular ion peak, because only 1.1% of the molecules will
have a C-13 atom.
But if the compound has two C atoms, the extra peak will be 2.2% of the
height of the main molecular ion peak, because there are two chances that the
molecule has a C-13 atom in it.
Introduction to organic chemistry
45 The M+1 Peak
If the compound has just one C atom, this extra peak
If there are three C atoms, the extra peak
will
be
3.3%
of
the
height,
and
so
on.
height of the main molecular ion peak, because only
will have a C-13 atom. But if the compound has two C
You can use these M + 1 peaks to work
the number
of C atoms
in themolecular ion
willout
be 2.2%
of the height
of the main
two
chances
that
the
molecule
has
a
C-13
atom
in
it.
parent molecule.
the extra peak will be 3.3% of the height, and so on.
Relative abundance
100
M
You can use these M + 1 peaks to work out the numb
parent
(see
Figure
13.13).
The M+1
peakmolecule
is 6.6% of
the
height
of the M peak.
M + 2 peaks
79
81
The compound
contains
6.6/1.1
=
6
C
atoms
Bromine has two stable isotopes, Br and Br, whi
50
6.6
0
M+1
Relative mass
abundant. This means that a molecule containing on
peaks for the molecular ion, separated by two mass
79
81
contains Br; the M + 2 peak contains Br. The pea
equal heights, so equal-sized M + 2 peaks are a tell-t
containing bromine.
35
37
46 The M+1 Peak Summary
The M peak of a molecule is due to all carbon atoms in the molecule being 12C.
and the M+1 peak is due to one carbon atom being 13C and the rest 12C.
By measuring the ratio of the M to M+1 peaks, we can thus work out the
number of carbon atoms the molecule contains.
The ratio approximately is 100 : 1.1(n)
where n is the number of carbon atoms.
47 Skill Check
The molecular ion peak of a compound has an m/e value of 136, with a relative
abundance of 17%, and an M+1 peak at m/e 137 where the relative
abundance is 1.5%. How many carbon atoms are in the molecule?
48 Skill Check
A compound contains C, H and O atoms. Its mass spectrum has a peak at m/e
132 with a relative abundance of 43.9 and a peak at m/e 133 with a relative
abundance of 2.9.
Calculate the number of carbon atoms in each molecule, and suggest its
molecular formula.
49 M & M+2 Peaks
Bromine has two stable isotopes, 79Br and 81Br, which have almost equal
abundance.
This means that a molecule containing one Br atom shows two peaks for the
molecular ion, separated by two mass units.
The M peak contains 79Br and the M+2 peak contains 81Br.
The peaks are of roughly equal heights, so equal-sized M and M+2 peaks are
a clear sign for compounds containing one bromine atom in their formula.
third the height of the M peak. This is because natural chlorine is made up of
35
37
about 75% Cl and 25% Cl, that is a 3 : 1 ratio.
50 Mass Spectrum for 1-Bromopropane
43
100
Relative abundance
s
80
60
40
20
0
0
40
80
Relative mass
Fig 13.14 The mass spectrum for 1-bromopropane, CH3CH2CH2Br
120
51 M & M+2 Peaks With Chlorine
Chlorine has two stable isotopes, 35Cl and 37Cl, which have abundance in the
3 1 ratio.
This means that a molecule containing one Cl atom shows two peaks for the
molecular ion, separated by two mass units.
The M peak contains 35Cl and the M+2 peak contains 37Cl.
:
In this case the M peak is 3 times larger than the M+2 peak.
52 M & M+2 Peaks
The mass spectrum of a compound containing one of these elements should
therefore show two molecular ions, one with an m/e value two mass units
higher than the other.
:
:
The ratio of the M/(M+2) peak should reflect the natural abundances given in
the table, i.e. 3 1 for chlorine and 1 1 for bromine.
53 M & M+2 Peaks of Chloromethane
the mass spectrum of CH3Cl will have peaks:
for CH335Cl+ at m/e 50 (12 + 3 + 35 = 50) and
for CH337Cl+ at m/e 52 (12 + 3 + 37 = 52)
:
The relative abundances of the two peaks will be in the ratio 3 1 which is the
ratio of the two Cl isotopes.
54 M & M+2 Peaks Of Bromoethane
There are two peaks for the molecular ion of C2H5Br,
one for the molecule containing the isotope 79Br (M): C2H579Br
the other for the one with the
81Br isotope (M + 2): C2H581Br
Because the two isotopes are of similar abundance, the peaks are of similar
height.
81Br
40
60
80
molecular ion contains...79Br
20
Abundance %
100
55 M & M+2 Peaks Of Bromoethane
0
10
20
30
40
50
60
70
80
90
100
110 120
130 140
m/e
Rel
peaks separated by two mass units. The peak with the lower value of m/z is
26
three times higher than the peak with the higher value of m/z.
the most abundant ion.
66
49
Two
Compounds
Containing
Halogen
Atoms
56
51
Bromine consists largely of two isotopes, Br and Br, in roughly equal
79
81
0
proportions. If a molecule contains one bromine atom, the molecular
ion
0
20
shows up as two peaks of roughly equal intensity separated by two mass units.
64
100
40
m/z
60
80
100
15
100
94
Relative intensity/%
Relative intensity/%
28
29
27
50
66
49
26
50
79
51
0
0
20
40
m/z
60
80
100
Figure 19.4 The mass spectra of two compounds containing halogen atoms.
100
15
94
High-resolution mass spectrometry
0
47
28
0
20
40
m/z
60
80
100
An example of the [M + 2] peak is shown on the mass
spectrum of chlorobenzene (Figure 29.42).
57 M & M+2 Peaks Example
112
[C6H535Cl]+
100
c How many peaks would
beyond the molecular io
dibromoethane? What w
mass-to-charge ratios an
relative to the molecula
13
peaks due to C.)
Relative abundance (%)
80
Applications of the mass spe
77
[C6H5]+
60
40
51
+
[C
H
]
4 3
+
20
0
114
[C6H537Cl]+
[C4H2]
35
30
56
50
70
90
Mass-to-charge ratio, m/e
110
To identify the components in a m
gas–liquid chromatography (GLC)
liquid chromatography (HPLC) ap
mass spectrometer.
This combined technique is very
solutes that can be separated with a
on a GLC column can be identifie
the mass spectrometer without the
Identification is by comparing the
58 Skill Check
80
60
29
40
122 124
20
Abundance %
100
43
79 81
0
M/E
10
20
30
40
50
60
70
80
90
100
110
120
130
140
59 Skill Check
17
(c) At one time, bromomethane, CH3Br, was widely used to control insect pests in agricultural
crops and timber. It is now known to break down in the stratosphere and contribute to
the destruction of the ozone layer.
Samples can be screened for traces of bromomethane by subjecting them to mass
spectrometry.
(i)
Which peak(s) would show the presence of bromine in the compound?
..................................................................................................................................
(ii)
How could you tell by studying the M and M+2 peaks that the compound contained
bromine rather than chlorine?
..................................................................................................................................
..................................................................................................................................
F
Exam
U
60 Analysing Molecular Fragments
If the ionising electron beam in a mass spectrometer has enough energy, the
molecular ions formed by the loss of an electron can undergo bond fission,
and molecular fragments are formed.
Some of these will carry the positive charge, and therefore appear as further
peaks in the mass spectrum.
61
molecular fragments are formed (see Figure 29.7). Some of these will carr
positive charge,
and therefore
appear as further
peaks in the mass spectru
Analysing
Molecular
Fragments
of Propanone
d from
+
O
C
H3C
(a) (b)
homolytic cleavage
at (a)
CH3
heterolytic cleavage at (b)
O
O
CH3
+
+
C
CH3
(m/e = 43)
C
H3C
+
+
CH3
(m/e = 15)
62 Analysing Molecular Fragments of Propanone
We therefore exp
and
43,
as
well
a
We therefore expect the mass spectrum of propanone to contain peaks at m/e
= 15 and 43, Figure
as well as29.8
the molecular
ion peak
58
Mass spectrum
of at
propanone
100
43
80
60
40
15
20
58
0
10
20
30
40
mass number
50
60
relative abundance
relative abundance
100
Figure 29.
80
60
40
20
0
63 Analysing Molecular Fragments of Propanal
The fragmentation pattern can readily distinguish between isomers. Compare
the following the mass spectrum of propanal.
Here there is no peak at m/e = 15, nor one at m/e = 43. Instead, there is a peak
at m/e = 57 and several from m/e = 26 to 29.
64 Analysing Molecular Fragments of Propanal
We therefore expect the mass spectrum of propanone to contain pea
Here there
is no
at m/e
15,molecular
nor one at m/e
43. Instead,
thereFigure
is a peak
and
43,peak
as well
as =the
ion =peak
at 58 (see
29.8).
at m/e = 57 and several from m/e = 26 to 29.
panone
50
Figure 29.9 Mass spectrum of propanal
58
60
relative abundance
100
29
80
60
40
58
20
57
0
10
20
30
40
mass number
50
60
65 Fragment Patterns
We can think of fragmentation in two ways: either we can look at the ion
formed when the molecular ion breaks apart or we can look at the group lost
from the molecular ion.
Consider the mass spectrum of propanoic acid shown in in the next slide.The
molecular ion peak occurs at m/e = 74, so the relative molecular mass is 74.
There is a peak in the spectrum at m/z = 57, which corresponds to the loss of
OH (mass 17) from the molecular ion. The fragment responsible for the peak
at m/e = 57 is thus (C2H5CO2H − OH), that is C2H5CO.
66 Fragment Patterns of Propanoic Acid
67 Fragment Patterns of Propanoic Acid
The peak at 45 corresponds to the loss of 29 from the molecular ion. A group
with mass 29 is C2H5, and therefore we can deduce that C2H5 is lost from
C2H5CO2H+ (M+) to form this peak. The peak at 45 is thus due to the CO2H+
ion.
The peak at 29 is due to the C2H5+ ion, which is formed by loss of CO2H from
C2H5CO2H+ (M+).
The peak at 29 is due to the C2H5 ion, which is formed by loss of
+
COOH from C2H5COOH .
Common
Fragment
Ions
68The
formulas of some common fragment ions are shown in Table A3.
69 Common Fragments Lost
Peaks may also be formed when the fragments are lost from the molecule.
Peaks may also be formed when the fragm
In that case the m/e value of the peak in the spectrum
In that will
casebe
the=m/z value of the peak in th
relative molecular mass (Mr) − the mass of t
relative molecular mass (Mr)−the mass of the fragment.
Examiner’s tip
Groups lost from the
molecular ion do not need a
positive charge, but any species
that forms a peak in the mass
spectrum must have a positive
charge.
−
−
−
−
−
35
+
37
+
(C2H5 Cl ) and 66 (C2H5 Cl ).
70 Mass Spectrum of Chloroethane
Fig 13.11 Mass spectrum of a compound X, molecular formula C3H6O. The table identifies the
fragment causing the major peaks. (There are several smaller peaks which have been omitted.)
71 Mass Spectrum of Butanone
43
Relative abundance
72
0
29
27
57
15
10
20
30
40
Relative mass
50
60
70
72 Skill Check
The following shows the mass spectra of two compounds with the molecular
formula C2H4O2. One is methyl methanoate, and the other is ethanoic acid.
Decide which is which?
Figure 29.11 shows the mass spectra of two compounds with the molecular formula
C2H
O
.
One
is
methyl
methanoate,
and
the
other
is
ethanoic
acid.
Decide
which
is
w
4
2
73
by assigning structures to the major fragments whose m/e values are indicated.
relative abundance
a 100
31
80
60
29
40
60
20
15
0
10
20
30
mass number
40
50
60
0
74
20
30
mass number
40
50
43
100
relative abundance
b
10
60
60
45
80
60
40
28
20
0
10
20
30
mass number
40
50
60