instrumental analysis techniques

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)


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


• 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


2929

• Double strand is often represented in simplified form as:


.

.

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

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.

instrumental analysis techniques

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