WM4 Instrumental analysis
The 3 key instrumental techniques
How do we know that salicylic acid contains –
OH and –COOH groups?
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Mass spectroscopy (m.s.).
Infrared (i.r.) spectroscopy.
Nuclear magnetic resonance (n.m.r.)
spectroscopy.
Making use of infrared spectroscopy

Any unidentified, new substance has its i.r. spectrum
recorded.
6.4 Infrared spectroscopy
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Energy in molecules is quantised.
In i.r., molecules are exposed to radiation
between 1014Hz – 1013Hz (wavelengths
2.5µm -15 µm). Remember: c = λ v
Bonds vibrate and stretch (pull apart then
push together again) as they absorb energy.
Vibrational changes of CO2
Asymmetric stretch
Symmetric stretch
bending
Infrared spectra: signals = stretches
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An IR spectrum hangs down from a baseline
(100% transmittance = no absorbtion).
The signals (look like ‘icicles’) on an IR
spectra correspond to bonds absorbing a
packet of energy and vibrating more.
The –OH and –CO bonds in salicylic acid
absorb energy at specific wavelengths (λ)/µm
and so wavenumbers (1/ λ)/ cm-1.
Generally, particular bonds can be matched
to a particular absorption region.
Bond
Location
Wavenumber/cm-1
Intensity
C-H
Alkanes
2850 -2950
M-S
C=C
Alkenes
1620 – 1680
M
Arenes
Sev peaks 1450-1650
Variable
Alkynes
2100 – 2260
M
C=O
Aldehydes
1720 – 1740
S
C-O
Alcohols, ethers
phenols
1050 - 1300
S
C-F
fluoroalkanes
1000 – 1400
S
O-H
Alcohols
3600 – 3640
S
N-H
1o amines
3300 - 3500
M-S
C
C
Interpreting spectra: 2-ethylbut-1-ene
?
Interpreting spectra: propanone
Regions in the IR spectrum where
typical absorptions occur
Absorption range/cm-1 Bonds responsible
Examples
4000 – 2500
Single bonds to hydrogen
O-H, C-H, N-H,
2500 – 2000
Triple bonds
C
2000 – 1500
Double bonds
C=C, C=O
Below 1500
(fingerprint region)
Various
C-O,
(not used to ID functional groups) C-X (halogen)
C, N
N
Label these regions on the previous two examples; sketch the structures and
link them to the main signals. Do this for the following examples, too.
Examples of infrared spectra
Butane
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Strong
absorption at
2970 cm-1
characteristic of
C-H stretching
in aliphatic
compounds.
No indication of
any functional
groups.
Examples of infrared spectra
Methylbenzene
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2 types of C-H
absorption ~3000
cm-1 (above = CH on benzene;
below = C-H on
methyl group).
No indication of
any functional
groups.
Absorption pattern
~700 cm-1 is
typical of a
benzene ring with
a substituted
group.
Examples of infrared spectra
Benzoic acid
 A sharp
absorption at
3580 cm-1 is
due to O-H
bond.
 A strong
absorption at
1760 cm-1 is the
C=O group.
 Position of C-H
absorption
suggests an
aromatic
compound.
Summary of IR spectroscopy
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An IR spectrum measures the extent to
which electromagnetic radiation is
transmitted through a sample of substance.
Frequency ranges absorbed give clues about
functional groups which are present.
IR spectrum of salicylic acid gives evidence
of C=O and –OH groups.
Evidence from nuclear magnetic
resonance (n.m.r.) spectroscopy.
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This technique helps to determine structure,
as it investigates the different environments
in which (hydrogen) nuclei are situated. The
n.m.r. spectrum for salicylic acid shows
signals for the different environments of the 6
protons:
One proton in a –COOH environment.
One proton in a phenolic –OH environment.
Four protons attached to a benzene ring.
Absorption
n.m.r. spectrum for salicylic acid
12
11
The evidence so far….
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A combination of i.r. and n.m.r. spectroscopy
shows that salicylic acid has an –OH group
and –COOH group both attached to a
benzene ring; we can now rename it
HYDROXYBENZOIC ACID.
However, it could be one of 3 possible
isomers: 2-hydroxybenzoic acid, 3hydroxybenzoic acid and 4-hydroxybenzoic
acid.
Mass spectroscopy can determine which
isomer we have.
The mass spectrum of salicylic acid
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Signals correspond to positively charged ions formed
from the parent compound, and fragment ions.
The fragmentation pattern
is characteristic of a particular
compound…the fragment at
120 can only come from 2hydroxybenzoic acid…can
you see why?
mass
Parent or
molecular ion
Fragmentation
Positive ions in a mass spectrometer can
break down into ‘building blocks’.
+
Example: 2-ethoxybutane.

CH3-CH2-CH-O-CH2CH3
?
(M=87)
Loss of
CH3
Loss and
rearrangement of
CH3CH=CHCH3
?
CH3-CH2-CH-O-CH2CH3
+
+
CH3-CH2-CH
?
CH3 (M=102)
+
HO-CH-CH3 (M=45)
?
+
CH3-CH2 (M=29)
Loss of
CH3-CH2
(M=87)
?
CH3
Positively charged fragments form.
Mass difference suggests functional groups.
For each fragmentation,
one product has a
positive charge:
M+
A+ + B
A + B+
The most stable ion
usually forms.
Mass
Difference
Group that is
suggested
15
CH3
17
OH
28
C=O or C2H4
29
C2H5
43
COCH3
45
COOH
77
C6H5
Isotope peaks: heights are in the same
ratio of abundance for particular elements.
Pairs of peaks correspond to
isotopes of 35Cl and 37Cl in the
ratio of 75%:25% ie. 3:1.
Highlight these.
mass
Now it’s over to you!
Do activity WM4: use accurate Mr values,
isotope peaks and a database to lead you to
the formula of salicylic acid.
It shows you how chemists use fragmentation
patterns to deduce or confirm a molecular
structure.
 Do assignments 1 and 2 C.S. p110-111
 Do ‘Problems for 6.5’ on mass spectrometry,
C.I., p145-146.
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