INFRARED SPECTROSCOPY
THEORY
A molecule is not a rigid assemblage of atoms. The atoms, even in the solid state, vibrate about
an equilibrium position. Each atom vibrates with a frequency which depends on its mass and the
length and strength of any bonds which it has formed. There are two kinds of fundamental
vibrations for molecules, stretching and bending (or deformation).
Stretching
In stretching vibrations, the distance between the atoms increases or decreases, but the atoms
remain in the same bond axis.
symmetrical stretching
asymmetrical stretching
Bending
In bending vibrations, the position of the atom changes relative to the original bond axis.
+
+
+
-
scissoring
rocking
wagging
twisting
(in-plane bending vibrations)
(out-of-plane bending vibrations)
+ and – signify vibrations above and below the plane of the paper respectively
Each of the various stretching and bending vibrations of a bond occur at a certain particular
frequency, in the range 1.20 x 1013 to 1.20 x 1014 Hz, which corresponds to a frequency in the
infrared region of the electromagnetic spectrum. Therefore, when infrared radiation of that same
frequency falls on the molecule, energy is absorbed and the amplitude of that vibration
increases. Only those vibrations which cause a change in the dipole moment of the molecule
will result in the absorption of infrared radiation.
So that the numbers involved are more convenient, absorption is usually quoted at a particular
wavenumber (the reciprocal of the wavelength in cm of the absorbed radiation), which has the
unit cm-1.
Bending vibrations generally require less energy and therefore occur at lower wavenumbers
than stretching vibrations. Stretching vibrations occur in the order of bond strengths. Thus, a
triple bond absorbs radiation of a higher wavenumber than the equivalent double bond, which in
turn absorbs radiation of a higher wavenumber than the equivalent single bond.
TOPIC 13.15: ANALYTICAL TECHNIQUES 1
THE INFRARED SPECTROMETER
An infrared spectrometer has two beams of radiation from the same source, one passing
through the sample, the other passing through a reference cell.
Sample
Chart recorder
Detector
glowing
source
Reference
Monochromatic
grating
A rotating segmented mirror takes information from the sample and reference beams
alternately. If a particular frequency is absorbed by the sample, less radiation is transmitted, and
the detector compares the intensity of radiation of each wavelength passing through the sample
with the intensity passing through the reference cell.
When the spectrum is determined, a calibration line is usually recorded on the paper. For this
purpose, the absorption peak of polystyrene at 1603cm -1 is commonly used.
Preparing Samples
Glass absorbs infrared radiation. Therefore, cells for use in an infrared spectrometer have to be
made from some other material, which is transparent to infrared. Sodium chloride and
potassium bromide are frequently used. Potassium bromide is preferred, because sodium
chloride absorbs radiation below 625cm-1.
Gases
Gases are placed in a special cell, typically 10cm in length.
Liquids
Liquids are examined in the form of a thin film, formed by sandwiching a small drop of liquid
between sodium chloride discs.
Solutions
Solutions, usually in solvents such as CCl4 and CHCl3, are placed in cells with a path length
between 0.01mm and 0.5mm.
Solids
A disc can be made by finely grinding the solid with KBr in a mortar. The powder is then placed
in a circular die and compressed to give a transparent disc which is used to record the
spectrum.
Alternatively, a paste can be made from the solid and a long-chain liquid hydrocarbon called
Nujol. This mull is then placed between sodium chloride discs.
TOPIC 13.15: ANALYTICAL TECHNIQUES 2
INTERPRETATION OF INFRARED SPECTRA
Infrared spectra are valuable aids in the identification of unknown compounds.
A complex molecule has a large number of vibrational modes. Some of these vibrations (mainly
above 1500cm-1) are associated with individual bonds or functional groups, while others (below
1500cm-1) are considered as vibrations of the whole molecule.
The absorption associated with a particular individual bond or functional group always occurs at
approximately the same wavenumber, irrespective of the molecule involved.
e.g. the carbonyl (C=O) absorption in the following aliphatic ketones:
butanone
1718cm-1
cyclohexanone
1715cm-1
Structural changes close to the vibrating bond can cause small shifts.
e.g. the carbonyl absorption in the following classes of compounds:
aliphatic ketones
1725-1705cm-1
aromatic ketones
1700-1680cm-1
aliphatic aldehydes
1740-1720cm-1
aromatic aldehydes
1715-1695cm-1
esters
1750-1735cm-1
aliphatic carboxylic acids 1725-1700cm-1
acyl chlorides
1795cm-1
The infrared spectrum can conveniently be split into four regions for interpretation.
4000-2500 cm-1
absorption due to stretching vibrations of single bonds to hydrogen
e.g. C-H, O-H, N-H
2500-2000 cm-1
absorption due to stretching vibrations of triple bonds
e.g. C=C, C=N
2000-1500 cm-1
absorption due to stretching vibrations of double bonds
e.g. C=C, C=O
1500-400 cm-1
absorption due to stretching vibrations of some single bonds
e.g. O-H, C-O
absorption due to various bending vibrations
The region 1500-400 cm-1 is not mainly used for recognising individual bond absorptions. This
region contains absorption bands which are characteristic of the compounds under test and no
other compound. This region is therefore called the fingerprint region. The fingerprint region
can be used to confirm the identity of a substance by comparison with the infrared spectrum of
an authentic sample.
TOPIC 13.15: ANALYTICAL TECHNIQUES 3
Correlation Table for Infrared Spectroscopy
4600
4200
3800
aliphatic C-H stretch
3400
3000
2600
2200
2000
1800
N-H stretch
amines (v)
1600
1400
1200
1000
C-N stretch
C-H bend
aromatic nitro cpds. aliphatic (v)
C-H stretch
C C-H (sh)
C N stretch
aliphatic nitriles
C=O stretch (vs)
C-H stretch
alkenes (sh)
C
C-H stretch
aromatic ring (w-m)
C stretch (v)
C-Br stretch
haloalkanes
2600
C=C stretch
Ar-C=C (w-m)
Abbreviations used:
w = weak
m = medium
s = strong
vs = very strong
v = variable
br = broad
sh = sharp
C-N stretch
aromatic nitro cpds.
N-H bend (v)
2200
2000
1800
1600
1400
Wavenumber / cm-1
TOPIC 13.15: ANALYTICAL TECHNIQUES 4
C-Cl stretch
haloalkanes
C=C stretch
aromatic (m)
C-H stretch (2 bands)
aldehydes
3000
C-N bend
amines
C=C stretch
alkenes (v)
C-H stretch
alkanes (v)
3400
C-H bend
C=C-H (m-s)
C-C skeletal vibration (w)
O-H stretch
ROH (sh)
3800
C-H bend
C C-H (s)
N-O stretch
aromatic nitro cpds.
H-bonded O-H stretch (br)
4200
400
C-O stretch
esters, alcohols, carboxylic acids
O-H stretch
COOH (w, br)
4600
600
C-O-C asymmetric stretch (vs)
free O-H stretch (sh)
aromatic C-H stretch (w)
800
Fingerprint region
1200
1000
800
600
400
Characteristic Infrared Absorption Ranges
Wavenumber range/ cm-1
Functional Group
Vibration
3750 – 3200
3500 – 3300
3500 – 3140
3300 – 2500
3095 – 3010
3080 – 3030
2962 – 2853
2900 – 2700 (2 bands)
2260 – 2215
2260 – 2140
1850 – 1800
1790 – 1740
1795
1750 – 1735
1740 – 1720
1730 – 1717
1725 – 1705
1725 – 1700
1700 – 1680
1700 – 1630
1669 – 1645
1650 – 1590
1600, 1500 & 1450
1570 – 1500
1550 – 1330
1485 – 1365
1370 – 1300
1310 – 1100
880 – 700
800 – 600
600 – 500
alcohol & phenol
amine
amide
carboxylic acid
alkene
arene
alkane
aldehyde
nitrile
alkyne
carboxylic anhydrides
carboxylic anhydrides
acyl chloride
ester
aldehyde
aryl ester
ketone
carboxylic acid
aryl carboxylic acid
amide
alkene
amide & amine
arene
aromatic nitro compound
aromatic nitro compound
alkane
aromatic nitro compound
ester, alcohol, carboxylic acid
arene
chloroalkane
bromoalkane
O-H stretch
N-H stretch
N-H stretch
O-H stretch
C-H stretch
C-H stretch
C-H stretch
C-H stretch
C=N stretch
C=C stretch
C=O stretch
C=O stretch
C=O stretch
C=O stretch
C=O stretch
C=O stretch
C=O stretch
C=O stretch
C=O stretch
C=O stretch
C=C stretch
N-H bend
C=C stretch
C-N stretch
N-O stretch
C-H bend
C-N stretch
C-O stretch
C-H bend
C-Cl stretch
C-Br stretch
TOPIC 13.15: ANALYTICAL TECHNIQUES 5
MASS SPECTROMETRY
Mass spectrometry is a useful technique for determining the structure of a compound. It is about
1000 times more sensitive than either infrared or NMR spectroscopy and requires only a
microgram or less of the sample. The basics of this topic were covered in module CHM1.
BASIC FEATURES
1. A substance is vaporised. For gases and volatile liquids this is done simply by injecting them
into the ionisation chamber, which is at a very low pressure. Less volatile liquids and solids
are introduced into the mass spectrometer through a heated inlet.
2. The molecules in the vapour are bombarded in the ionisation chamber with a beam of high
energy electrons. This results in:
a) the formation of positively charged molecular ions
b) the breaking of chemical bonds and the formation of a range of lighter fragment ions
3. The ions are accelerated by an electrostatic field and collimated into a narrow beam. In
practice, the ions generated in 2) have a range of kinetic energies. If the path through the
electrostatic field is curved, the ions are deflected according to their kinetic energies, and
only those of a certain energy can pass through the exit slit. This improves the resolution of
the instrument. These double resolution instruments can detect mass differences of 1 in 10 6.
4. The ions are passed through a magnetic field. Here the ions are separated according to their
mass:charge ratio (m/z). In practice most ions carry a single charge. For a given charge,
those ions with the lightest mass are deflected most.
vaporised sample
(+)
ionisation chamber
electron
gun
(+) accelerating electric field
(-)
heavier ions
lighter ions
recorder
magnetic field
ions of intermediate mass
amplifier
ion detector
to vacuum pump
5. The separated streams of ions are focused in turn on to a detecting device by varying the
magnetic field. This leads finally to a plot of the signal from the recorder against m/z being
drawn out on a pen recorder or recorded photographically.
TOPIC 13.15: ANALYTICAL TECHNIQUES 6
The mass spectrum consists of a series of sharp peaks; the abundance of an ion is proportional
to its peak height. It is usual to convert the spectrum to a stick diagram in which only the major
peaks are drawn, the height of the line representing a percentage of the height of the highest
peak.
e.g. stick diagram of the mass spectrum of ethanol.
Relative abundance %
[CH2OH]+
100
80
60
parent ion
m/e 46
[C2H5O]+
40
[C2H5]+
[C2H3]+
[C2H5OH]+
20
[M+1]+
10
20
30
40
50
m/z
Applications of Mass Spectrometry
1. Mass spectrometry can be used for the determination of the isotopic composition of a
sample of an element and hence for the measurement of relative atomic mass.
2.
3. A mass spectrum can be used to determine relative molecular masses of compounds since
the line with the highest m/z value is usually the molecular ion, M+. There is always a very
small peak at (M+1) for organic compounds which is due to the 13C isotope, present with a
natural abundance of 1.1%.
4. The structure of a molecule can be determined from the fragmentation pattern. Once the
molecular ion has been formed, it will usually decompose into smaller fragments. By looking
at the major peaks in the mass spectrum, it is possible to decide which fragments have been
lost and thence to build up a picture of the original molecule. Occasionally a rearrangement
of an ion may take place.
TOPIC 13.15: ANALYTICAL TECHNIQUES 7
Common Losses from Molecular Ions
Ion
M-15
M-16
M-17
M-18
M-27
M-28
M-29
M-30
M-31
M-32
M-36
M-41
M-42
Group being lost
CH3
O
NH2
OH
H2O
HCN
CO
C2H4
CHO
C2H5
OCH2
NO
OCH3
CH3OH
HCl
C3H5
CH2CO
C3H6
M-44
M-45
M-46
CO2
COOH
C2H5OH
NO2
Possible inference
-NO2
-CONH2
alcohol
alcohol, aldehyde, ketone
nitrile
quinone
aromatic ethyl ether (ArOC2H5)
ethyl ester (RCOOC2H5)
propyl ketone (RCOC3H7)
aldehyde
ethyl ketone (RCOC2H5)
aromatic methyl ether (ArOCH3)
aromatic nitro compound
methyl ester (RCOOCH3)
methyl ester (RCOOCH3)
primary haloalkane
propyl ester (RCOOC3H7)
methyl ketone (RCOCH3)
aromatic ethanoate (CH3COOAr)
butyl ketone (RCOC4H9)
aromatic propyl ether (ArOC3H7)
carboxylic acid, ester, acid anhydride
carboxylic acid
ethyl ester (RCOOC2H5)
aromatic nitro compound
Aromatic compounds often produce fragment ions at m/z 77 (C6H5+) and m/z 91 (C6H5CH2+).
TOPIC 13.15: ANALYTICAL TECHNIQUES 8
Interpretation of the Mass Spectrum
If the likely structure of the compound is known, it can usually be confirmed by considering the
fragmentation pattern. It is not always possible to distinguish between isomers e.g. the different
disubstituted benzenes.
If the structure is unknown, the mass spectrum can be compared with libraries of mass spectra
which are commercially available on computer software. The mass spectra of all compounds
having the same Mr (or molecular formula) as the unknown can be called up and compared.
A compound containing one atom of chlorine will give two molecular ions which are two units
apart. The higher m/z peak will have an intensity one third that of the lower m/z peak. This is
because chlorine contains the isotopes 35Cl and 37Cl in the ratio 3:1.
A compound containing one atom of bromine will give two molecular ions which are two units
apart. Both peaks will be equal in intensity. This is because bromine contains two isotopes 79Br
and 81Br in the ratio 1:1.
High-Resolution Mass Spectrometry
In the Atomic Structure topic, we learned that a mass spectrometer can be used to work out
isotopic abundances and hence relative atomic masses. It was also noted that the m/z value of
the molecular ion peak (the peak with the highest m/z value) can give the Mr of a compound.
A more sensitive high-resolution mass spectrometer can measure the mass of a molecular
ion to several decimal places. Using very precise relative atomic masses, molecules with similar
relative molecular masses can be distinguished from each other.
Element
Hydrogen
Carbon
Nitrogen
Oxygen
Chlorine
Symbol
H
C
N
O
Cl
Relative atomic mass Ar
1 decimal place 4 decimal places
1.0
1.0079
12.0
12.0107
14.0
14.0067
16.0
15.9994
35.5
35.4532
For example, methanal and ethane have the same Mr values when determined via in low
resolution mass spectrometry:
H2C=O
Mr = (12 x 1) + (1 x 2) + (16 x 1) = 30.0
C2H6
Mr = (12 x 2) + (1 x 6)
TOPIC 13.15: ANALYTICAL TECHNIQUES 9
= 30.0
However, using a more sensitive high resolution mass spectrometer a difference is observed:
H2C=O
Mr = (12.0107x1) + (1.0079x2) + (15.9994x1)
= 30.0259
C2H6
Mr = (12.0107x2) + (1.0079x6)
= 30.0688
The difference is very small but is enough to uniquely identify each compound.
The high resolution mass spectrometer enables very precise measurements of masses to be
made. Since the mass defect varies from element to element, the masses of ions with the same
nominal mass show small differences if precise measurements are made. E.g. for a nominal m/z
of 28:
Ion
Precise mass
CO+
N2+
CH2N+
C2H4+
27.994914
28.006148
28.018724
28.031300
Thus it is possible with a high resolution spectrum to determine not only the molecular mass but
also the molecular formula.
High-Resolution Mass Spectrometry – Example Questions
Q1. The molecular ion peak in a high-resolution mass spectrum has an m/z ratio
of 58.0789. The substance is either butane or propanone. Calculate the exact
molecular mass of both substances to 4 d.p. and then deduce which it is.
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TOPIC 13.15: ANALYTICAL TECHNIQUES 10
Q2. An unknown substance (X) was put through a high resolution mass
spectrometer. The resulting molecular ion peak had a value of m/z = 60.0518 to 4
d.p.. An infrared absorption spectrum showed an intense absorption at 1740cm-1
and a broad absorption at 2500-3000cm-1. The substance was then heated with
acidified potassium dichromate and no colour change was noted.
Deduce the identity of X and explain your answer.
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Q3. Unlabelled samples of glucose (C6H12O6) and aspirin (C9H8O4) were mixed
up. Explain and show how high resolution mass spectrometry could distinguish
between the two.
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TOPIC 13.15: ANALYTICAL TECHNIQUES 11