1. Structural Organic Chemistry
Spectroscopy and beyond!
Structural Organic Chemistry

2. Structural analysis Elemental microanalysis Mass spectrometry
(combustion analysis)
Mass spectrometry
Infrared spectroscopy
Nuclear magnetic resonance
X-ray crystallography

3. Elemental microanalysis
Used to determine the masses of the elements in a sample of an organic compound
This information is used to calculate the empirical formula

4. Sample of organic compound (known weight)
burned in an oxygen atmosphere
measure weights of carbon dioxide, water, nitrogen and sulphur dioxide produced.
Output from the analysis
% composition for C, H, N and S

5. Divide to give whole number ratio
Analysis of an alcohol shows it to contain 37.5% carbon and 12.5% hydrogen.
What is the formula?
element
carbon
hydrogen
oxygen
% by weight
37.5
12.5 50
Divide by atomic mass
Ratio of atoms
37.5 = 3.125 12
12.5 = 12.5 1
50 = 3.125 16
Divide to give whole number ratio
Whole number ratio
= 1
3.125
12.5 = 4 4
The formula for this compound is CH4O.

6. Mass spectrometry Used to determine accurate molecular masses
Used to indicate structural features of an organic compound.

7. Mass spectrometry

8. A mass spectrometer consists of:
an injector, often at a high temperature to vaporise the sample;
Stage 1: Ionisation
The atom is ionised by knocking one or more electrons off to give a positive ion. This is true even for things which you would normally expect to form negative ions (chlorine, for example) or never form ions at all (argon, for example). Mass spectrometers always work with positive ions
Stage 2: Acceleration
The ions are accelerated so that they all have the same kinetic energy.
Stage 3: Deflection
The ions are then deflected by a magnetic field according to their masses. The lighter they are, the more they are deflected.
The amount of deflection also depends on the number of positive charges on the ion - in other words, on how many electrons were knocked off in the first stage. The more the ion is charged, the more it gets deflected.
Stage 4: Detection
The beam of ions passing through the machine is detected electrically.

9. Investigation of the mass spectrum can give information about:
the molecular mass = peak at highest mass/z (if the parent, M+ ion is present)
the structure of the sample, from the distribution of the fragment ions.
Where very accurate masses are available from the mass spectrometer, these alone can often provide chemical formulae for the parent and fragment ions.

10. What the mass spectrometer output looks like
The output from the chart recorder is usually simplified into a "stick diagram". This shows the relative current produced by ions of varying mass/charge ratio.
Molecular mass

11. Mass spectrum of naphthalene
Structure from fragmentation pattern
Mass spectrum of naphthalene
The stable molecular ion, C10H8+, gives a very large response at m/z 128

12. CH3 CH3 CH CH2 CH2 CH3 2-methylpentane 15 71 43 43 57 29 71 15
This molecular ion fragments easily  this is why the peak is small
The peak with the highest m/z value is often the molecular ion
The base peak in a mass spectrum is the most abundant peak, assigned an arbitrary abundance of 100%
m/z
Relative abundance

13. Mass difference
Suggested group 15
CH3 17 OH 28
C=O or C2H4 29
C2H5 31
CH3O 45
COOH 77
C6H5
 Fragment masses

14. Compound A has molecular formula, C3H6O
Structure from fragmentation pattern
Compound A has molecular formula, C3H6O
The mass difference from 58 to 43 is 15
A methyl group is likely to be responsible for this

15. Some high resolution mass spectrometers provide an accuracy of 1 in 106 a.m.u, so that very accurate masses can be obtained for the peaks in the mass spectrum.
If you were asked for the molecular masses of ethane (C2H6) and methanal (HCHO), the data booklet information would give This is a 1 decimal place figure using an average of the masses of the isotopes normally present in these compounds on the Earth.
A mass spectrometer would separate the individual molecules containing different isotopes; 12C1H216O would appear as a different peak from 13C1H216O, one mass unit higher. The first use of mass spectrometry was to measure the masses of isotopes of elements.
A high resolution mass spectrometer would be able to distinguish between the two compounds above because their accurate masses are not the same.

16. This question concerns analysis of a colourless, pleasant-smelling liquid, containing only carbon, hydrogen and oxygen. (Compound L) Question 1 When g of a liquid is completely burned in excess of oxygen, g of CO2 and g of H2O are produced. What is the empirical formula?
Take me to the Answer

17. C4H4O This formula suggests an aromatic compound.
ANSWER 1
FIRST  need to find mass of each element present in g of L.
This allows moles of each element to be calculated  empirical formula.
g of CO2 contains 0.288g Carbon (know 44g CO2 has 12g C)
0.216g H2O contains 0.024g Hydrogen (know 18g H2 O has 2g H)
The mass of oxygen is obtained by subtracting the
mass of C and H from the total mass given for L in the
question. Mass of Oxygen = 0.096g
Use these mass to find the moles of each element present
Find whole number ratio of moles
Answer to 1
C4H4O This formula suggests an aromatic compound.
Questions 2 and 3

18. Questions 4, 5 and 6 Answers to 2 and 3 A2 136
Question 2 Use the mass spectrum of the compound L to find out the m/z value of the highest m/z ion? Question 3 Does this ion's mass agree with the empirical formula; i.e. is this mass a simple multiple of the empirical formula?
Questions 4, 5 and 6
A2 136
A3 yes this mass is 2x the mass of the empirical formula
Answers to 2 and 3

19. Question 4 What is the molecular formula for the compound
Question 4 What is the molecular formula for the compound? Question 5 What is the m/z value of the base peak? Question 6 From the mass difference from the molecular ion to base peak, what fragment might have been lost?

20. Question 7 What is a likely structure for the ion of m/z 77
Question 7 What is a likely structure for the ion of m/z 77? Question 8 What is a possible fragment composition lost when the m/z 105 ion forms the m/z 77 ion? Question 9 These three fragments (m/z ) fortunately add to the molecular mass (136). So what is a likely structure for the compound? Question 10 Name compound L
Answer to 10
methyl benzoate

21. Infrared spectroscopy
Used to identify functional groups
in organic compounds

22. Infrared spectroscopy
Infra-red radiation causes parts of the molecule to vibrate.
The wavelengths which are absorbed and cause
vibrations depend on the type of chemical bond and
the groups or atoms at the ends of these bonds.

23. The source of infrared radiation has its beam split into two.
One part of the beam passes through the sample; the other through a reference cell, which might contain the solvent.

24. The monochromator grating scans the wavelengths prior to a detector which compares the intensity of the two beams.

25. The amplified signal is plotted as % transmission or absorbance.

26. Infrared spectroscopy
How radiation and matter (molecules) interact.
Simple diatomic molecules ( eg halogens) have only one bond, which can vibrate only by stretching and compressing.
The two halogen atoms can pull apart and then push together.
The frequency of this vibration depends on the mass of the two atoms and the strength of the bond.

27. Infrared spectroscopy
Stronger bond

28. Infrared spectroscopy
Heavier atoms

29. Infrared spectroscopyH H
Asymmetrical stretching C H H
Symmetrical stretching C

30. Infrared spectroscopyH H
Bending or stretching C H H
Rocking or in plane bending C

31. Infrared spectroscopy- + H H
twisting or out-of -plane bending C + + H H
wagging or out-of -plane bending C

32. Infrared spectroscopy
These vibrations absorb different frequencies of infra-red radiation
The absorption is usually given in
wavenumbers (units cm-1)
Remember
check units
e.g. speed of light m s-1
but
wavenumber cm-1

33. Infrared spectroscopy
E = Lhc
This equation is used to calculate the energy (in kJ mol-1) associated with a peak at a specific wavenumber.
This equation was introduced in Unit 1.
L = Avogadro’s Constant (mol-1 )
c = speed of light (m s-1)
 = wavenumber (cm-1)
h = Planck’s Constant J s)

34. Infrared spectroscopy
How this equation is derived
E = Lhc
Remember
C =  x 
(m s-1) (m) (s-1)
 = C 
Remember
 = 1 
Remember
E = Lh
E = LhC 
E = Lhc

35. Nuclear magnetic resonance spectroscopy (NMR)
Used to investigate
the different types of hydrogen (and carbon) atom environments in organic compounds
how many atoms there are in each of these environments
Only hydrogen atom environments will be covered in Ad H Chemistry 1H

36.  NMR spectrometer

37. 1H
Discovered many nuclei behave as though they were spinning about an axis
Any spinning charged particle  magnetic field
(behaves like a little magnet)
In a magnetic field the nuclei will orientate themselves – some in same direction as the magnetic field, others will oppose the magnetic filed

38. Higher energy nuclei

39. nuclei in a strong magnetic field+ + + + +
electromagnetic radiation of a suitable frequency + + + + + + + +
Some nuclei absorb energy and flip to the higher energy quantum state.
The energy needed for this transition is in the radio frequency region of the spectrum
(60 MHz to 1000 MHz)
When the nuclei fall to the lower energy state, the energy is emitted again and can be detected.

40. Electrons also spin  produce their own magnetic field round the nucleus.
This will shield the 1H nucleus from the full effects of a magnetic field.
In an organic molecule the density of the electron cloud will vary from one part of a molecule to another.
Different hydrogen nuclei will experience different magnetic fields and have different ∆E between spin states
Radiation absorbed/emitted by an 1H nucleus will depend on the local environment of that nuclei within the molecule.
This variation is called CHEMICAL SHIFT

41. Very strong magnetic fields are required for NMR, so that superconducting electromagnets cooled in liquid nitrogen are used.
The sample is dissolved in CDCl3 or CD3COCD3 ("deuterated" solvents with hydrogen replaced by deuterium (D or 2H)).
Sample has no 1H nuclei
Absolute values are difficult to obtain, so values are obtained by reference to a standard arbitrarily assigned the δ value 0. the values are measured in the ppm.
The standard is tetramethylsilane (TMS), which has an NMR signal well away from those found in most organic molecules.

42. The two signals are said to have different chemical shifts.
In this case the hydrogen atom in the -OH group is shifted more than those in -CH3
The -OH group hydrogen feels the effect of the external field more.

43. Benzene has 6 identical H atoms
 one only peak on NMR spectrum
Ethanal has 2 H atoms in 2 different environments
 two peaks on NMR spectrum
Ethanol has 6 H atoms in 3 different environments
 three peaks on NMR spectrum

44. Size of peaks on NMR spectrum can be used to find the number of each type of H atom.
Use of integration curve makes this easier

45. Note:
the relative size of the two signals from CH3 and OH
Ratio is 3:1

46. X-ray crystallography
Used to reveal the 3D structure
Involves the scattering of X rays by electrons

47. X-ray crystallography
The wavelength of X-rays is comparable to interatomic distances in molecules (around m).
When X-rays pass through the regular 3D array of atoms in a crystal, a symmetrical arrangement of diffraction spots is produced, which can be used to determine the precise 3D structure.
diffraction
the bending or spreading out of waves, eg sound or light, as they pass round the edge of an obstacle or through a narrow aperture

48. X-ray crystallography
Electron density map

49. X-ray crystallography
Electron density maps are produced from the positions and intensities of the "spots" in a diffraction pattern, produced when a crystal of an organic compound is exposed to X-rays of a single wavelength.
The precise location of each atom in the molecule can be determined from the electron density map.
Since heavier atoms have more electrons than lighter ones each atom in a molecule can be identified. Hydrogen atoms have a low electron density and so are not easily detected by X-rays.

50. Section 5
Medicines

51. Medicines
Section 5

52. Medicines
Drugs are substances that change the biochemical processes in the body.
Medicines are drugs that have a beneficial effect on the body

53. Historical Development
The first medicines were plant extracts
Plant extracts were analysed to identify the pharmacologically active ingredients.
Where practicable these compounds or their derivatives were synthesised.
Aspirin was developed this way.

54. How a medicine functions
Most medicines work by binding to a receptor.
Receptors are usually protein molecules.
Receptors on the surface of a cell interact with small biologically active molecules.
Receptors can also be enzymes which catalyse chemical reactions (catalytic receptors).

55. How a medicine functions
Effective medicines work by binding to the receptor site.
They either mimic the response of the active molecule or block the effect of the active molecule.

56. How a medicine functions
If the medicine mimics the natural active molecule, it will stimulate the same response and activate the biological response. In this case, the medicine is classified as an agonist.
An agonist is a drug that binds to a receptor, triggering or increasing a particular activity in that cell

57. How a medicine functions
Salbutamol is an agonist which mimics adrenaline. It switches on receptors which lead to dilation (widening) of the airways. It can be used to treat asthma attacks.

58. How a medicine functions
If the medicine binds onto the receptor site and does not switch it on, it prevents the action of the body's natural active molecule and is classed as an antagonist.
An antagonist is a drug which binds to a receptor without stimulating cell activity and prevents any other substances from occupying that receptor.

59. How a medicine functions
Propranolol is an antagonist. It blocks the receptors in the heart that are stimulated by adrenaline. Propranolol is a β-blocker and is used to relieve high blood pressure.

60. How a medicine functions
Pharmacophore = the minimum structural feature that gives pharmacological activity
The shape of the pharmacophore complements the shape of the receptor site – this allows it to fit into the receptor.
The functional groups on both are positioned to interact and bind the medicine to the receptor.

61. How a medicine functions
Etorphine (a synthetic opiate) is almost 100 times as potent as morphine and is used in veterinary medicine to immobilize large animals.