Instrumental analysis techniques
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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)

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Instrumental analysis techniques


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


Infrared spectroscopy ir

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 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


Ir spectrum

Fingerprint region

IR Spectrum

  • 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 key peaks

IR Spectrum – Key peaks

C=O (carbonyl) of an ester or carboxylic acid

O-H of an alcohol

O-H of a carboxylic acid


SDBSWeb : (National Institute of Advanced Industrial Science and Technology, Date of access May ‘08

Ir spectroscopy
IR Spectroscopy

Sample Q11

Which one of the following compounds will show an absorption band in the infrared spectrum at about 3500 cm-1

A. B. C. D.

Data Book Table 7, p7


Nuclear magnetic resonance spectroscopy nmr

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 of1H 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


1 h nmr spectrum

1H NMR Spectrum

13C NMR Spectrum

1H NMR Spectrum


Interpreting 1 h nmr spectra

Interpreting 1H NMR Spectra

Number of peak regions

Number of different 1H environments

  • Four key pieces of information on 1H NMRspectrum

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 shiftof peak region

Ratio of areas under peak regions

Number of 1H present in a particular environment


Interpreting 1 h nmr spectra1

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




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]

Chemical shift from Data Book


Interpreting 13 c nmr spectra

Interpreting 13C NMR Spectra

  • Same as for 1H NMR except:

    • Peak regions never split

    • Chemical shifts are different

    • Ratio of 13C atoms not possible


13 c nmr spectroscopy
13C NMR spectroscopy

Sample 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


Mass spectrometry ms
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

Ms instrumental set up
MS - Instrumental set-up

Image sourced from Heinemann 2 Commons et al. 3ed


Ms ionisation equations
MS – Ionisation equations

Example – Show the reaction for ionisation of methane

Two valid equations

CH4(g) + e- → CH4+(g) + 2e-


CH4(g) → CH4+(g) + e-

Note :- Electrons are always shown with no states




Interpreting mass spectra
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 : (National Institute of Advanced Industrial Science and Technology, date of access May ‘08)


Interpreting mass spectra1
Interpreting Mass Spectra


M+ ion (parent ion) gives relative mass of compound

Fragmentation produces BOTH

  • charged cation fragment (detected) and

  • uncharged fragment (lost / undetected)

Interpreting mass spectra2
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


m/z difference between peaks shows size of fragment lost

Overview of instruments
Overview of instruments

Sort these correctly



Infra red

Radio wave

Radio wave






Mass spec





Unit 3 aos 2 organic chemical pathways

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



  • 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



  • 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

yeast enzymes


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


Using protein markers of disease to rationally design new drugs



  • 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 – component molecules


Formation of DNA

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

Start of a DNA strand



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 G  C

    Note - A = T link is weaker than G  C link

guanine and cytosine have three hydrogen bonds between the bases

adenine and thymine have two hydrogen bonds between the bases


Formation of DNA double helix



phosphate units





Formation of DNA double helix

  • Double strand is often represented in simplified form as:


DNA Analysis

  • 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



Direction of fragment movement

Gel Electrophoresis

-ve charge applied

  • negatively charged fragments move to the positive end of the gel

  • smaller and more highly charged fragments move faster

+ve charge applied

Reference materials used as basis of size comparison


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.

The type of bonding formed between bases is

hydrogen bonding


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.