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
Instrumental Analysis Techniques
• Chromatography
– TLC
(Thin Layer Chromatography)
– HPLC (High Pressure Liquid Chromatography)
– GC/GLC (Gas Liquid Chromatography)
• Spectroscopy and spectrometry
–
–
–
–
–
UV-Vis
AAS
IR
NMR
MS
(UltraViolet & Visible Spectroscopy)
(Atomic Absorption Spectroscopy)
(InfraRed Spectroscopy)
(Nuclear Magnetic Resonance Spectroscopy)
(Mass Spectrometry)
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
InfraRed Spectroscopy (IR)
• Used to identify types of bonds and functional groups
• Specific energy in infrared region is absorbed by different
types of bonds as they change vibrational states
ie stretching or bending
• Data Book Table 7 lists key absorbances of IR radiation
• X-axes is measured in wavenumbers (cm-1) and the scale
usually runs from 4000 cm-1 to 400 cm-1
• The region below 1400 cm-1 is known as
“fingerprint region” and can uniquely identify a compound
• Y-axes is usually measured in % transmittance
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
IR Spectrum
Fingerprint region
• Each ‘peak’ represents a different bond type
absorbing IR energy
• Full interpretation of every peak will NOT be asked
• IR can only be used to uniquely identify a
compound if an IR library is used for comparison
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
IR Spectrum – Key peaks
C=O (carbonyl) of an
ester or carboxylic acid
O-H of an
alcohol
O-H of a
carboxylic acid
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SDBSWeb : http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, Date of access May ‘08

IR Spectroscopy
Sample Q11
Which one of the following compounds will show an absorption band in
the infrared spectrum at about 3500 cm-1
CH3
H3C
C
CH3
OH
H3C
C
CH3
CH3
A.
B.
CH3
Cl
H3C
CH 2
O
CH3
H3C
C
CH3
CH3
C.
D.
Data Book Table 7, p7
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
Nuclear Magnetic Resonance
Spectroscopy (NMR)
• Used to identify chemical environment of either
hydrogens or carbons - can give clues to structure
• Radio wave energy causes nuclei of 1H or 13C atoms
to ‘flip’ spin states
• Data Book Tables 5 + 6 lists chemical shifts of
1H or 13C atoms caused by radio waves
• X-axes of NMR spectrum measured in ppm
relative to an internal standard, tetramethyl silane (TMS),
which produces a peak at 0 ppm
• Y-axes is usually not given a scale
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1H
NMR Spectrum
13C
NMR Spectrum
1H
NMR Spectrum

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
Interpreting 1H NMR Spectra
• Four key pieces of information on 1H NMR spectrum
Number of
peak regions
Number of different 1H
environments
Splitting pattern of
peak regions
Number of adjacent 1H in
different environments
(n+1 rule)
Location of
peak regions
Nearby atoms influencing 1H
- causing chemical shift
of peak region
Ratio of areas under
peak regions
Number of 1H present in a
particular environment
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
Interpreting 1H NMR Spectra
Two peak regions mean only two
hydrogen environments present
Integration ratio of
2:3 so 2Hs causing
left peak region
and 3Hs causing
right peak region
Molecular formula = C2H5Cl
3H
Cl
H
C
H
H
Chemical shift
from
Data Book
C
H
2H
H
TMS
Peak region split into four
(quartet, n+1) so these
two hydrogens are next to
3 other hydrogens [n=3]
Peak region split into three
(triplet, n+1) so these three
hydrogens are next to 2
other hydrogens [n=2]9

Interpreting 13C NMR Spectra
• Same as for 1H NMR except:
– Peak regions never split
– Chemical shifts are different
– Ratio of 13C atoms not possible
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
13C
NMR spectroscopy
Sample Q12
The structures of the two amino acids, glycine and alanine are shown below.
O
O
H2N
CH 2
C
glycine alanine
OH
H2N
CH
C
OH
CH3
The 13C NMR spectra can be used to uniquely identify each amino acid. Glycine
and alanine will produce 13C NMR spectra with the following number of peaks.
A.
B.
C.
D.
1 and 2
2 and 2
2 and 3
3 and 3
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
Mass Spectrometry (MS)
• Technique does not involve absorption of energy
• Used to identify:
– molecular mass of organic compounds (M+)
– possible structure of compounds
(base peak and other fragments)
– isotopic abundance of elements
• Generates cations (atoms or molecules)
• Sorts cations on basis of different mass to charge ratio
- m/z ratio, using magnetic field
• Y-axes on mass spectrum written as relative intensity
or abundance of cation
• The X-axes measures m/z ratio
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
MS - Instrumental set-up
Image sourced from
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Heinemann 2 Commons et al. 3ed
MS – Ionisation equations
Example – Show the reaction for ionisation of methane
Two valid equations
CH4(g) + e- → CH4+(g) + 2eor
e-
CH4(g) → CH4+(g) + e-
Note :- Electrons are always shown with no states
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
Interpreting Mass Spectra
Base peak
The most
stable cation
formed
M+ peak
The relative
mass of the
original
molecule
• Charged organic molecules fragment into smaller species
• Each peak represents detected fragment with specific m/z
SDBSWeb : http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, date of access May ‘08)
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
Interpreting Mass Spectra
• M+ ion (parent ion) gives relative mass of compound
• Fragmentation produces BOTH
– charged cation fragment
– uncharged fragment
m/z value
detected
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17
29
31
32
45
(detected)
and
(lost / undetected)
Fragment
DETECTED
[CH ]+
Difference in
mass units
Fragment
LOST
15
CH3
[OH]+
[CH2CH3]+, [CHO]+
[CH O]+
17
OH
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CH2CH3, CHO
31
CH3O
[CH3OH]+
[C2H5O]+, [COOH]+
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CH3OH
3
3
45
C2H5O,
COOH
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
Interpreting Mass Spectra
Sample Q13
Determine the fragment that must have been lost from the molecular ion
to account for the peak at m/z = 31
Peak at m/z = 31
Difference in
mass units = 15
fragment lost
is CH3
M+ peak
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• m/z difference between peaks shows size of fragment lost

Overview of instruments
Sort these Technique
correctly
UV-Vis
13C
NMR
Determines / Detects
Qual
Quant
metals


Vis

UV-Vis
coloured solutions
Energy
used
organic compounds

GC
organic compounds
(heat sensitive)


- n/a -
HPLC
organic compounds
(heat stable)


- n/a -
functional groups / bond types
in organic compounds

Infra
red
AAS
hydrogen environment
in organic compounds

Radio
wave
1H
carbon environment
in organic compounds

Radio
wave
molecular ion and mass of
fragments in organic molecules

TLC
Mass spec
IR
NMR
- n/a -
- n/a 1818
-

Unit 3 AoS 2Organic chemical pathways
• Naming organic molecules
• Understanding organic reactions
– addition, substitution, oxidation, condensation
• Fractional distillation
• Biomolecules – reactions and uses
– formation, hydrolysis, identification of functional groups
•
•
•
•
Lipids (triglycerides), biodiesel
Carbohydrates, bioethanol
Proteins, 1, 2, 3 structure, enzymes, protein markers
DNA, gel electrophoresis, applications for forensics
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
Biodiesel
• Biofuel manufactured from triglycerides in plants
i.e. vegetable oils
• Ideally carbon neutral fuel
• Triglycerides hydrolysed with KOH into glycerol and
fatty acids
• Fatty acids then converted into methyl ester biodiesel
by reaction with methanol
O
H2C
HC
O
O
O
C
O
R
C
R
H3C
O
C
O
R
R
H3C
O
C
R
H3C
O
O
C
R
H2C
O
H
HC
O
H
H2C
O
H
O
KOH / CH3OH
O
H2C
C
+
• Issues - land needed to grow plants to produce
vegetable oils which could be used for food crops
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
Bioethanol
• Biofuel manufactured from carbohydrates
(sugars and starch) in plants
• Ideally carbon neutral fuel
• Sugars fermented to produce 10% – 20% (v/v) ethanol
yeast enzymes
C6H12O6(aq)

2CH3CH2OH(aq) + 2CO2(g)
• Product distilled to produce 95% ethanol, dried to produce
final product which is 99.7% pure
• 5% petrol added to ‘poison’ the “pure” alcohol – foul taste
• Issues - land needed to grow plants to produce
sugar cane which could be used for food crops
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
Protein markers
• The body can release particular proteins as a
result of
– Disease
– Heart attack
• Monitoring and assaying for proteins can
therefore allow detection of these conditions
Extract from VCAA June 2008
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
Using protein
markers of disease
to rationally design
new drugs
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
DNA
• DNA – deoxyribonucleic acid
– genetic map of all living things
– contains elements C, H, N, O and P
– polymer made from nucleotide monomers
– each nucleotide made from
• phosphate group
• sugar (deoxyribose in DNA)
• base (adenine, thymine, guanine or cytosine)
(A)
(T)
(G)
(C)
(Structures are found in VCE Data Book Table 10)
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
DNA – component molecules
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
Formation of DNA
Formation of a single molecule of DNA involves
linking nucleotides via condensation reactions
NH2
N
OH
O
P
O
N
O
NH2
N
O
O
OH
N
O
N
P
O
N
N
O
N
O
H
NH2
NH2
OH
O
P
-H2O
N
O
N
O
O
H
P
O
N
N
N
O
N
N
O
N
O
O
OH
H
nucleotide
OH
H
Start of a DNA strand
Hydrolysis of DNA requires one water molecule to
separate each nucleotide from a strand
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
Formation of DNA double helix
• When DNA double helix formed, nitrogeneous bases
on each strand base pair up in specific way
•
Complementary base pairs are A = T and
CH3
NH2
N
N
N
guanine and cytosine
have three hydrogen
bonds between the bases
O
.
HN
N
GC
N
O
N
O
.
adenine and thymine
have two hydrogen
bonds between the bases
.
N
NH2
HN
N
N
H2N
N
O
.
Note - A = T link is weaker than G  C link
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
•
Formation of DNA double helix
.
thymine
CH3
O
adenine
NH2
.
N
O
HN
N
O
N
O
N
N
N
O
N
O
NH2
phosphate
units
HN
N
.
N
H2N
O
N
guanine
O
cytosine
sugar
.
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
Formation of DNA double helix
•
Double strand is often represented in simplified
form as:
.
.
.
G
C
A
T
C
G
G
C
T
A
C
G
.
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
DNA Analysis
• Analysis of DNA is commonly performed by chopping
up DNA using restriction enzymes and using
Gel electrophoresis to identify fragments
• DNA fragments are all negatively charged due to
phosphate group in DNA
• Size of fragments commonly measured in kb
(ie 1000’s of bases)
– E.g. a fragment which is 6.4 kb is made up of
6400 bases in length
• In forensics, a pattern of fragments from a sample
can be compared with those from a suspect
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
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
Gel Electrophoresis
-ve charge applied
• smaller and more
highly charged
fragments move
faster
Direction of fragment movement
• negatively charged
fragments move to
the positive end of
the gel
+ve charge applied
Reference materials used as
basis of size comparison
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
Sample Q
On the diagram shown below, draw in the bonds that form between
adenine and thymine base pair as they would exist in the DNA double
helix, and then identify the type of bonding you have drawn.
CH3
O
H
H
N
NH
NH
N
N
O
N
The type of bonding formed between bases is
NH
hydrogen bonding
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
Sample Q
A piece of double stranded DNA, which is known to have 100 base pairs,
is found to contains 40 cytosine bases.
Determine the number of adenine bases in this piece of DNA.
If the DNA has 100 base pairs the DNA must have a
total of 200 bases present.
If 40 are cytosine bases, there must also be 40 guanine bases.
Together, giving 80 G and C bases out of the 200 total present.
The remaining 120 bases must be the A -T pairs, which means
there would be 60 or each.
Answer – There are 60 adenine bases present in this fragment.
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