1. Spectroscopy Workshop
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The Queen’s University of Belfast
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2. Workshop Content Spectroscopy overview Ultra-violet/visible (UV-vis)
Infra-Red (IR)
Nuclear Magnetic Resonance (NMR)
Mass Spectrometry
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3. Spectroscopy
In spectroscopy, transitions between different energy levels within atoms and molecules are recorded and then used to give information on chemical structure.
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4. The range of energies that can be used for spectroscopy is very large and spans a large proportion of the electromagnetic spectrum.
Visible
X-Rays
Gamma Rays UV IR
Radio
Microwave
10- 11
10- 9
10- 7
10- 5
10- 3
10- 1
10 3 10
Wavelength (cm)
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5. In a typical experiment, the molecules or atoms start at lower energy and go to a higher energy level upon absorption of radiation of appropriate wavelength.
After DE
Before
Absorption
Energy
Energy
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6. Absorption can only occur when the energy of the radiation (calculated from the frequency or wavelength) matches the energy gap.
If there are several different upper levels (and there usually are) then several transitions will be observed.
After
After
After
Energy
Before
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7. For current purposes we look only at: UV/visible ( highest energy)
Infra red (intermediate)
Radio frequency (lowest energy).
But in all cases :
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8. To record a spectrum, sweep through the appropriate range of energies and look for absorption at particular values.
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9. Interpretation depends on the energy range investigated.
Absorption gives peaks, when these have been measured this gives the energy gaps within the sample. These can then be related to structure.
Interpretation depends on the energy range investigated.
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10. UV/visible Spectroscopy
Chemical compounds are coloured because they absorb visible light.
In general, even organic compounds that are colourless will absorb UV light.
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11. Absorption of visible light
Where has the energy that was within the photons gone to ?
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12. In UV/visible spectroscopy the energy of the absorbed photon is used is used to drive the molecule into an excited electronic state.
In the excitation the energy of the whole molecule increases.
After DE
Before
Energy
Absorption
Energy
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13. This overall change is typically due to promotion of a single electron from a lower to higher energy orbital. The energy of the transition depends on the gap between the two orbitals.
In organic compounds which have only single bonds between the atoms the excitation energy is very high- lies in deep UV.
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14. Even if have a simple p bond, the excitation from highest occupied to lowest unoccupied orbitals still lies in the UV.
This excitation gives a dramatic decrease in bond order due to excitation from
a bonding to
an anti-bonding orbital.
e.g. ethene
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15. If we have a highly conjugated molecule the energy separation between the orbitals is smaller.
Excitation of the electron thus has a proportionately smaller effect and requires less energy- energy gap may lie in the visible region.
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16. Again note that lowest energy transition may lie in visible.
Bonding
Anti-bonding
Energy
Orbitals of Butadiene
Again note that lowest energy transition may lie in visible.
But we can also excite to higher orbitals with sufficiently energetic (UV) photons.
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17. With increasing conjugation, the decreasing energy gap is reflected by absorption at longer wavelengths.
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18. The structures of many coloured compounds show they are very extensively conjugated.
beta-Carotene
trans-Crocetin
Xanthopterin
16,17-DimethoxyViolanthrone
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19. Substituents added to the compound may alter the energy of the orbitals which e- is excited from or to.
Auxochromes: substituents that alter the wavelength or intensity of the absorption due to the chromophore
ORANGE
PURPLE
BLUE
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20. Changes in chemical composition can give rise to pronounced colour changes since this can dramatically alter the energies of the orbitals involved in the transitions e.g. pH indicators.
Phenolphthalein HO O- HO O-
-2H+
pink
colourless
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21. Methyl orange
red
orange-yellow H+
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22. Summary
Absorption of UV-vis radiation occurs via excitation of electrons from filled to unfilled orbitals i.e. they are electronic transitions.
Molecules have characteristic absorption spectra.
The absorption can lead to coloured materials.
pH Indicators use the change in colour between the acid and alkali forms of the molecules.
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23. IR Spectroscopy Origin of the absorption The spectrometer The spectra
Organic compounds
Example problem
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24. Origin of IR absorptions
CO2
Atoms within a molecule are never still. They vibrate in a variety of ways (modes).
Atoms may be considered as weights connected by springs.
Each vibrational mode has its own resonant frequency.
symmetric stretch
asymmetric stretch
bending
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25. Appropriate energy for this is infra-red
If the vibrational mode involves a change in molecular dipole moment, the vibration can be induced by absorption of a photon - it is ‘IR-active’
Appropriate energy for this is infra-red
symmetric stretch
no dipole
no dipole
asymmetric stretch
change in dipole - IR active
bending
change in dipole - IR active
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26. The IR spectrometer
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27. The symmetric stretch is not IR active
CO2 IR spectra
2800
2400
2000
1600
1200
800
400
100
Transmittance /%
The bigger the change in dipole, the more intense the absorption
Stretching higher energy than bending
Wavenumber /cm-1
The symmetric stretch is not IR active
(no change in dipole)
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28. IR spectra of organic compounds
More complex:
2000
4000
3000
1500
1000
Wavenumber/cm-1
500
Ethyl ethanoate (CH3COOCH2CH3)
C=O bond
C-O stretch
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29. But functional groups have characteristic frequencies
1000
Wavenumber / cm- 1
650
1500
2000
3000
4000 N H C O Cl
(all types)
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30. Four regions in the spectrum: Wavenumber / cm-1
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31. Example problem
Identify two main functional groups present in the compound which gave this spectrum
Explain why infrared radiation is absorbed by molecule HCl but not by molecules H2 and Cl2.
Explain what occurs in the HCl molecule when infrared radiation is absorbed.
The simplified infrared spectrum below is that of an organic compound.
Identify two main functional groups on the spectrum.
This compound has composition by mass C, 67.9%; H, 5.7%; N, 26.4%, and Mr of 53. Suggest a structural formula for the compound.
4000
3600
3200
2800
2400
2000
1900
1800
1700
1600
Wavenumber / cm-1 10 20 30 40 50 60 70 80 90
100
Transmittance / %
C=C
C=O?
C=N?
C-H
CC
CN?
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32. Combine this information with the following data to deduce its structure
Mr 53
Likely structure:
Cyanoethene C
So, formula = C3H3N
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33. Summary
Absorption of IR can occur if a vibrational mode is associated with a change in dipole.
Functional groups have characteristic absorption frequencies.
In combination with other analytical data, the structure of an organic compound can often be deduced.
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34. NMR Spectroscopy The Basis of NMR Spectroscopy The Spectrometer
Chemical Shifts
Signal Intensity and Integration
Coupling Constants
Example Spectra
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35. The Basis of NMR Spectroscopy
Atomic nuclei behave like small bar
magnets as a result of their charge
and spin.
In the presence of an applied magnetic field the spin states have different energy and the magnetic moment can align with or against the applied field.
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36. The difference in energy between the two spin states is
dependent on the external magnetic field strength.
Irradiation of a sample with radio frequency energy
corresponding to the spin state separation (DE) will excite
nuclei in the +½ state to the higher energy –½ state.
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37. The 1H NMR Experiment
For example, consider a water sample in a T
external magnetic field irradiated by 100 MHz radiation.
If the magnetic field is increase to T the water
protons will at some point absorb rf energy (DE) and a
resonance signal will appear,
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38. The Chemical Shift
Not all protons give resonance signals at the same field frequency. Electrons move in response to the applied field and generate a secondary magnetic field which opposes the applied field. The secondary field shields the nucleus from the applied field and nuclei in different environments resonate at different frequencies.
The difference in resonance frequency is measured as a chemical shift, units d
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39. Proton Chemical Shift Ranges
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40. Signal Intensity
The relative area of the absorption signals can provide
valuable structural information.
The area under a peak is proportional to the number of
a given type of nuclei in the molecule.
MEK
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41. The keto-enol equilibrium ratio of 2,4-pentandione can
determined by 1H NMR spectroscopy
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42. Spin-Spin Coupling
The applied magnetic field experienced by a proton Ha will be modified by the local field produced by its neighbouring Hb
Ha modifies the field at Hb by aligning with or against the
applied field and and gives 2 resonant frequencies for Hb (doublet)
Similarly Hb modifies the field at Ha in 3 different ways (triplet)
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43. Splitting pattern can provide valuable structural information
Chemically equivalent protons act as a group and a peak
due to n adjacent protons is split into n+1 lines, with a
coupling constant J
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44. 1H NMR Spectrum of
Ethyl Acetate
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45. 1H NMR Spectrum of
1,3-Dichloropropane
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46. Example Problem
Given the formula, deduce what you can about the structure
Integration corresponds to 2H : 2H : 3H
A triplet must correspond to 2 near neighbour protons
A sextet corresponds to 5 near neighbour protons
Therefore CH2, CH2 and CH3 groups are present
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47. Solution
Connectivity can be deduced to be
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48. Summary
NMR spectroscopy involves irradiating a sample with radio frequency radiation
Protons in different chemical environments have different chemical shifts d
Protons in different environments can couple to each other with a coupling constant J
The combination of chemical shifts and coupling constants provides valuable structural information
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49. Mass Spectrometry The basic principles Applications
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50. What is a mass spectrometer ?
A mass spectrometer is an instrument which produces charged particles (ions) from chemical substances under analysis.
It then uses magnetic and/or electric fields to separate those ions and to measure their mass.
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51. Mass Spectrometer Schematic
Sample
Introduction
Data
Output
Inlet
Data
System
Ion
Source
Mass
Analyzer
Ion
Detector
Vacuum
Pumps
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52. Ion Generation School of Chemistry ~70 Volts Electron Collector (Trap)
Neutral
Molecules
Positive Ions +
Inlet
Repeller _ _ + + +
To Analyzer + + + +
Electrons e- e- e- _
Filament
Extraction
Plate
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53. The magnetic field exerts a force on these fast-moving ions and causes them to move in a circular path, the radius of which is dependent upon their mass to charge ratio (m/z) and speed. 
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54. Magnetic Mass Separation
Correct m/z ratio
ion detected
Ion
Source
Detector S N
Electromagnet
ion not detected
m/z too small
ion not detected
m/z too large
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55. Applications Mass spectrometers are used for all kinds of
chemical analyses:
- Chemical analysis (Chemical Research)
- Environmental analysis - Analysis of petroleum products
- Trace metals
- Biological materials
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56. How is mass spectral information used?
Let us use water (H2O) as an example.
If a beam of electrons is directed through water vapour in the source of a mass spectrometer, some of the electrons will hit water molecules and knock off an electron, producing charged ions from the water:
H2O + 1 (fast) electron  [H2O]+ + 2 electrons
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57. Electron impact on a water molecule
Some of the collisions between water molecules and electrons will be so hard that the water molecules will be broken into fragments. For water, those fragments will be [OH]+, O+, and H+ with the following masses:
1 = H+
16 = O+
17 = [OH]+
18 = [H2O]+
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58. (mass-to-charge ratio)
Mass Spectrum of Water 18
[H2O]+
Relative
Abundance 17
[OH]+ 1 H+ O+ 16
Mass
(mass-to-charge ratio)
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59. Examples
Alcohols
Pentan-3-ol
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60. An alcohol's molecular ion is small or non-existent
An alcohol's molecular ion is small or non-existent.  Cleavage of the C-C bond next to the oxygen usually occurs.  A loss of H2O may occur as in the spectra below. 59
m/z(parent ion) = 88
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61. Alkanes
Hexane
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62. Molecular ion peaks are present, possibly with low intensity.
The fragmentation pattern contains clusters of peaks 14 mass units apart (which represent loss of (CH2)n CH3).
m/z(parent ion) = 86 15 43 71 57 29
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63. Aromatics
Naphthalene
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64. Molecular ion peaks are strong due to the stable structure.
100 80 60 40 20
mass / charge (m/z)
relative abundance
128
120
140
m/z(parent ion) = 128
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65. Esters
Ethylethanoate
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66. Fragments appear due to bond cleavage next to C=O (alkoxy group loss, -OR) and hydrogen rearrangements.
100 80 60 40 20
mass / charge (m/z)
relative abundance
-OCH2CH3
-C2H3 43 45 88 61 H
m/z(parent ion) = 88
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67. Halo-organics
Chloroethene
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68. Isotopes are shown by mass spectrometry
The natural abundance of each isotope gives characteristic fragmentation
e.g. 35Cl:37Cl is in a 3:1 ratio therefore the peaks containing Cl are in a 3:1 ratio and separated by 2 mass units 27
m/z(parent ion) = 62/64 64 62 35 37
100 80 60 40 20 30 50
H2C = CH-Cl 26
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69. Summary
Mass spectrometry involves the ionisation of molecules and atoms.
The mass spectrometer measures the mass to charge ratio.
On ionisation the molecule can break up giving fragments of different m/z ratios .
Each molecule has a characteristic fragmentation pattern which can be used to identify the molecule.
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