instrumental analysis techniques
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11
• Chromatography– TLC (Thin Layer Chromatography)
– HPLC (High Pressure Liquid Chromatography)
– GC/GLC (Gas Liquid Chromatography)
• Spectroscopy and spectrometry– UV-Vis (UltraViolet & Visible Spectroscopy)
– AAS (Atomic Absorption Spectroscopy)
– IR (InfraRed Spectroscopy)
– NMR (Nuclear Magnetic Resonance Spectroscopy)
– MS (Mass Spectrometry)
Instrumental Analysis Techniques
22
• 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 statesie 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
InfraRed Spectroscopy (IR)
33
• 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
IR Spectrum
Fingerprint region
44
O-H of an O-H of an alcoholalcohol
O-H of a O-H of a carboxylic acidcarboxylic acid
C=O (carbonyl) of an C=O (carbonyl) of an ester or carboxylic acidester or carboxylic acid
SDBSWeb : http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, Date of access May ‘08
IR Spectrum – Key peaks
55
IR SpectroscopySample Q11Sample Q11
Data Book Table 7, p7
Which one of the following compounds will show an absorption band in the infrared spectrum at about 3500 cm-1
A. B. C. D.
C
CH3
CH3
CH3
OH C
CH3
CH3
CH3
Cl C
CH3
CH3
CH3
CH3CH2CH3 O CH3
66
• 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 ppmrelative to an internal standard, tetramethyl silane (TMS), which produces a peak at 0 ppm
• Y-axes is usually not given a scale
Nuclear Magnetic Resonance Spectroscopy (NMR)
77
1H NMR Spectrum
1H NMR Spectrum
13C NMR Spectrum
88
• Four key pieces of information on 1H NMR spectrum
Interpreting 1H NMR Spectra
Number of peak regions
Splitting pattern of peak regions
Location of peak regions
Ratio of areas under peak regions
Number of different 1H environments
Number of adjacent 1H in different environments
(n+1 rule)
Nearby atoms influencing 1H - causing chemical shift
of peak region
Number of 1H present in a particular environment
99
Two peak regions mean only two Two peak regions mean only two hydrogen environments presenthydrogen environments present
Peak region split into four Peak region split into four (quartet, n+1) so these (quartet, n+1) so these
two hydrogens are next to two hydrogens are next to
3 other hydrogens [n=3]3 other hydrogens [n=3]
Peak region split into three Peak region split into three (triplet, n+1) so these three (triplet, n+1) so these three
hydrogens are next to 2 hydrogens are next to 2 other hydrogens [n=2]other hydrogens [n=2]
2H
3H
Interpreting 1H NMR Spectra
TMS
Integration ratio of Integration ratio of 2:3 so 2Hs causing 2:3 so 2Hs causing
left peak region left peak region and 3Hs causing and 3Hs causing right peak regionright peak region
Chemical shift Chemical shift from from
Data BookData Book
C C
ClH
HH H
H
Molecular formula = C2H5Cl
1010
• Same as for 1H NMR except:
– Peak regions never split
– Chemical shifts are different
– Ratio of 13C atoms not possible
Interpreting 13C NMR Spectra
1111
13C NMR spectroscopySample Q12Sample Q12
The structures of the two amino acids, glycine and alanine are shown below.
glycine alanine
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. 1 and 2
B. 2 and 2
C. 2 and 3
D. 3 and 3
CH2 CNH2
O
OH CH CNH2
O
OH
CH3
1212
• 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
Mass Spectrometry (MS)
1313
MS - Instrumental set-up
Image sourced from Heinemann 2 Commons et al. 3ed
141414
MS – Ionisation equationsExample – Show the reaction for ionisation of methane
Two valid equations
CH4(g) + e- → CH4+
(g) + 2e-
or
CH4(g) → CH4+
(g) + e-
Note :- Electrons are always shown with no states
e-
151515
Interpreting Mass Spectra
• 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)
`
Base peakBase peak
The most The most stable cation stable cation
formedformed
MM++ peak peak
The relative The relative mass of the mass of the
original original moleculemolecule
1616
• M+ ion (parent ion) gives relative mass of compound
• Fragmentation produces BOTH
– charged cation fragment (detected) and
– uncharged fragment (lost / undetected)
Interpreting Mass Spectra
m/z value detected
Fragment DETECTED
1515 [CH3]+
1717 [OH]+
2929 [CH2CH3]+, [CHO]+
3131 [CH3O]+
3232 [CH3OH]+
4545 [C2H5O]+, [COOH]+
Difference in mass units
Fragment LOST
1515 CH3
1717 OH
2929 CH2CH3, CHO
3131 CH3O
3232 CH3OH
4545 C2H5O, COOH
1717
Interpreting Mass Spectra
• m/z difference between peaks shows size of fragment lostlost
Peak at m/z = 31Peak at m/z = 31
MM++ peak peak
Sample Q13Sample Q13Determine the fragment that must have been lost from the molecular ion to account for the peak at m/z = 31
Difference in mass units = 15 fragment lost
is CH3
1818
Technique Determines / Detects Qual Quant Energy used
metals coloured solutions
organic compounds - n/a -
organic compounds(heat sensitive) - n/a -
organic compounds(heat stable) - n/a -
functional groups / bond typesin organic compounds hydrogen environment in organic compounds carbon environment
in organic compounds molecular ion and mass of
fragments in organic molecules - n/a -
Overview of instruments
VisVis
UV-VisUV-Vis
Infra Infra redred
Radio Radio wavewave
Radio Radio wavewave
HPLCHPLC
AASAAS
GCGC
11H NMRH NMR
IRIR
TLCTLC
1313C NMRC NMR
UV-VisUV-Vis
Sort these Sort these correctlycorrectly
Mass specMass spec
1919
Unit 3 AoS 2- Organic 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
2020
• 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
• Issues - land needed to grow plants to produce vegetable oils which could be used for food crops
H2C
HC
H2C
O
O
O
C R
O
C R
O
C R
O
C R
O
OCH3 H2C
HC
H2C
O
O
O
H
H
H
C R
O
OCH3
C R
O
OCH3
KOH / CH3OH
+
Biodiesel
2121
• Biofuel manufactured from carbohydrates (sugars and starch) in plants
• Ideally carbon neutral fuel
• Sugars fermented to produce 10% – 20% (v/v) ethanol
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
Bioethanol
yeast enzymes
2222
• 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
Protein markers
Extract from VCAA June 2008
2323
Using protein markers of disease to rationally design new drugs
2424
• 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)
DNA
2525
DNA – component molecules
2626
Formation of a single molecule of DNA involves linking nucleotides via condensation reactions
Hydrolysis of DNA requires one water molecule to separate each nucleotide from a strand
nucleotide Start of a DNA strand
OO
HOH
N
N
N
N
NH2
P
OH
O
O
-H2O
OO
HO
N
N
N
N
NH2
P
O
O
O
OO
HOH
N
N
N
N
NH2
P
O
OO
O
HOH
N
N
N
N
NH2
P
OH
O
O
Formation of DNA
2727
• When DNA double helix formed, nitrogeneous bases on each strand base pair up in specific way
• Complementary base pairs are A = T and G C
Note - A = T link is weaker than G C link.
N
N
NH2
O
.N
N
N
N
NH2
NNH
N
N
NH2
O.
NNH
O
O
CH3
.
adenine and thymine have two hydrogen bonds between the
bases
guanine and cytosine have three hydrogen bonds between the
bases
Formation of DNA double helix
2828
•
N
N
NH2
OO
N
N
N
N
NH2
O
.
. NNH
O
O
CH3
O
NNH
N
N
NH2
OO
.
.
phosphate units
sugar
adenine
thymine
cytosine
guanine
Formation of DNA double helix
2929
• Double strand is often represented in simplified form as:
Formation of DNA double helix
.
.
G
A
C
G
T
C
.
.
C
T
G
C
A
G
3030
• Analysis of DNA is commonly performed by chopping up DNA using restriction enzymes and usingGel 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
DNA Analysis
3131
3232
-ve charge applied
+ve charge applied
• negatively charged fragments move to the positive end of the gel
• smaller and more highly charged fragments move faster
Reference materials used as basis of size comparison
Gel Electrophoresis
Directio
n o
f fragm
ent m
ovem
ent
Directio
n o
f fragm
ent m
ovem
ent
3333
Sample QSample 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.
The type of bonding formed between bases is
N
N
NH
N
NH H
NH NH
O
O
CH3
hydrogen bonding
3434
Sample QSample 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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