Analical chemistry ok1294986152

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Analical chemistry ok1294986152

  1. 1. A1 Analytical techniques
  2. 2. A2 Principles of spectroscopy
  3. 3. The Electromagnetic Spectrum:
  4. 4.
  5. 5. Absorption spectra have parts ‘missing’. In the diagram below the troughs indicate energy absorbed by the molecules
  6. 6. Observing with a hand held spectroscope gives you evidence for line emission spectra similar to the one shown below Here light is being emitted as the excited electrons drop down to lower ‘allowed’ energy levels.
  7. 7.
  8. 8. • a beam of infra red radiation is passed through the sample • a similar beam is passed through the reference cell • the frequency of radiation is varied • bonds vibrating with a similar frequency absorb the radiation • the amount of radiation absorbed by the sample is compared with the reference • the results are collected, stored and plotted The Infra-red Spectrophotometer
  9. 9. SYMMETRIC STRETCHING BENDING AND STRETCHING IN WATER MOLECULES
  10. 10. ASYMMETRIC STRETCHING BENDING AND STRETCHING IN WATER MOLECULES
  11. 11. BENDING AND STRETCHING IN WATER MOLECULES BENDING
  12. 12. Vertical axis Absorbance the stronger the absorbance the larger the peak Horizontal axis Frequency wavenumber (waves per centimetre) / cm -1 Wavelength microns (m); 1 micron = 1000 nanometres INFRA RED SPECTRA - INTERPRETATION
  13. 13. FINGERPRINT REGION • organic molecules have a lot of C-C and C-H bonds within their structure • spectra obtained will have peaks in the 1400 cm -1 to 800 cm -1 range • this is referred to as the “fingerprint” region • the pattern obtained is characteristic of a particular compound the frequency of any absorption is also affected by adjoining atoms or groups.
  14. 14. IR SPECTRUM OF A CARBONYL COMPOUND • carbonyl compounds show a sharp, strong absorption between 1700 and 1760 cm -1 • this is due to the presence of the C=O bond
  15. 15. IR SPECTRUM OF AN ALCOHOL • alcohols show a broad absorption between 3200 and 3600 cm -1 • this is due to the presence of the O-H bond
  16. 16. WHAT IS IT! O-H STRETCH C=O STRETCH ALCOHOL ALDEHYDE CARBOXYLIC ACID One can tell the difference between alcohols, aldehydes and carboxylic acids by comparison of their spectra. O-H STRETCH C=O STRETCH AND
  17. 17. A4 Mass spectrometry
  18. 18.
  19. 19. An outline of what happens in a mass spectrometer
  20. 20. Peaks appear due to characteristic fragments (e.g. 29 due to C 2 H 5 + ) and differences between two peaks also indicates the loss of certain units (e.g. 18 for H 2 O, 28 for CO and 44 for CO 2 ). Many of the fragments do not show up themselves on the spectrum because they are not ions.
  21. 21. FRAGMENTATION PATTERNS OF FUNCTIONAL GROUPS <ul><li>Interpretation of thousands of spectra has shown that many classes of organic compound show characteristic fragmentation patterns due to their functional groups. It is often possible to identify the type of compound from its spectrum by looking at the ... </li></ul><ul><li>• position of peaks </li></ul><ul><li>• differences between major peaks. </li></ul><ul><li>Alkanes The mass spectra of these simple hydrocarbons have peaks at m/z values corresponding to the ions produced by breaking C-C bonds. </li></ul>
  22. 22. 2,6-dimethyloctane
  23. 23. A5 Nuclear magnetic resonance (NMR) spectroscopy
  24. 24. INTERPRETATION OF SPECTRA <ul><li>NMR spectra provide information about the structure of organic molecules from the </li></ul><ul><li>number of different signals in the spectrum </li></ul><ul><li>position of the signals (chemical shift) </li></ul><ul><li>splitting pattern of the signals(Higher level) </li></ul><ul><li>intensity of the signals </li></ul>
  25. 25. LOW RESOLUTION - HIGH RESOLUTION LOW RESOLUTION SPECTRUM OF 1-BROMOPROPANE <ul><li>low resolution nmr gives 1 peak for each environmentally different group of protons </li></ul><ul><li>high resolution gives more complex signals - doublets, triplets, quartets, multiplets </li></ul><ul><li>the signal produced indicates the number of protons on adjacent carbon atoms </li></ul>
  26. 26. How do you identify the different H environments in a molecule? <ul><li>For example CH 3 .CH 2 .OH </li></ul><ul><li>There are 3 environments </li></ul><ul><li>Three H’s associated with CH3 </li></ul><ul><li>Two H’s associated with CH2 </li></ul><ul><li>And 1 H associated with OH </li></ul>
  27. 27. <ul><li>the area under a signal is proportional to the number of hydrogen atoms present </li></ul><ul><li>an integration device scans the area under the peaks </li></ul><ul><li>lines on the spectrum show the relative abundance of each hydrogen type </li></ul><ul><li>By measuring the distances between the integration lines one can </li></ul><ul><li>work out the simple ratio between the various types of hydrogen . </li></ul><ul><li> before integration after integration </li></ul>INTEGRATION NOTICE THAT THE O-H SIGNAL IS ONLY A SINGLET
  28. 28. INTEGRATION <ul><li>HOW TO WORK OUT THE SIMPLE RATIOS </li></ul><ul><li>Measure how much each integration line rises as it goes of a set of signals </li></ul><ul><li>Compare the relative values and work out the simple ratio between them </li></ul><ul><li>In the above spectrum the rises are in the ratio... 1:2:3 </li></ul>IMPORTANT: It doesn’t provide the actual number of H’s in each environment, just the ratio Measure the distance between the top and bottom lines. Compare the heights from each signal and make them into a simple ratio.
  29. 29. MRI <ul><li>In 2003 two scientists, Paul C Lauterbur (1929-2007) and Peter Mansfield (1933- ) won the Nobel Prize in Physiology or Medicine for their work n developing the use of NMR in diagnostic medicine. Because the protons in water, lipids, carbohydrates, etc give different signals, depending upon their environment, an image of the whole body can be built up by placing the patient inside the magnet of a large NMR machine. This is known as magnetic resonance imaging </li></ul>
  30. 30. How NMR is used in body scanners
  31. 31. A6 Atomic absorption (AA) spectroscopy
  32. 32. This is what an Atomic Absorption spectroscopic machine looks like
  33. 33. This is the basic set up for AAS
  34. 34. Schematically
  35. 35. Schematically
  36. 36. UV-vis Spectroscopy and Beer’s Law <ul><li>UV-vis spectroscopy provides useful qualitative and quantitative information about a chemical sample. Since electromagnetic radiation in the ultraviolet and visible range causes electrons to be promoted between energy levels, the shape and wavelength of maximum absorbance of a UV-vis spectrum can help chemists identify a particular chemical. UV-vis spectra can also provide quantitative information about a particular chemical through a relationship called Beer’s Law. Beer’s Law states that the amount of light absorbed by a chemical is directly related to the concentration of the chemical in a solution. See equation 1. </li></ul><ul><li>A = ε bc </li></ul><ul><li>Equation 1 </li></ul>
  37. 37. A = ε bc In this equation, the absorbance (A) of the solution is directly proportional to the concentration (c) in moles/Liter, the pathlength (b) in cm-1 and the molar absorptivity (e ) in Lcm-1mol-1. Under the conditions of this experiment, the molar absorptivity and the path length are constant. Thus the absorbance of a solution is directly proportional to the concentration of solution. This concentration is expressed in molarity or moles/Liter.
  38. 38. This equation and relationship between absorbance and concentration can be used to of several samples of a particular chemical on the x-axis and the corresponding maximum absorbance on the y-axis, a Beer’s Law Plot is obtained. 
  39. 39. The Plot can then be used to determine an unknown concentration of the same chemical. At dilute concentrations, many chemical solutions obey Beer’s Law, resulting in a straight line plot with the general equation for a straight line y = mx + b .
  40. 40. To find the unknown concentration of a solution, use the equation for a straight line and solve for x. y = mx + b Sample calculation: Using the Beer's Law Plot above, determine the unknown concentration of a solution with an absorbance reading of .600. In this example, the slope (m) is 6.414 and the intercept ( b) is -7.99 x 10 -3. Solving for x, .600 = 6.414 x - 0.00799 x = 0.095 M
  41. 41. A7 Chromatography
  42. 42. What is Chromatography ? Chromatography is a technique for separating mixtures into their components in order to analyze, identify, purify, and/or quantify the mixture or components. Separate <ul><li>Analyze </li></ul><ul><li>Identify </li></ul><ul><li>Purify </li></ul><ul><li>Quantify </li></ul>Components Mixture
  43. 43. Uses for Chromatography <ul><li>Chromatography is used by scientists to: </li></ul><ul><li>Analyze – examine a mixture, its components, and their relations to one another </li></ul><ul><li>Identify – determine the identity of a mixture or components based on known components </li></ul><ul><li>Purify – separate components in order to isolate one of interest for further study </li></ul><ul><li>Quantify – determine the amount of the a mixture and/or the components present in the sample </li></ul>
  44. 44. Definition of Chromatography <ul><li> Detailed Definition: </li></ul><ul><li>Chromatography is a laboratory technique that separates components within a mixture by using the differential affinities of the components for a mobile medium and for a stationary adsorbing medium through which they pass. </li></ul><ul><li>Terminology : </li></ul><ul><ul><li>Differential – showing a difference, distinctive </li></ul></ul><ul><ul><li>Affinity – natural attraction or force between things </li></ul></ul><ul><ul><li>Mobile Medium – gas or liquid that carries the components ( mobile phase ) </li></ul></ul><ul><ul><li>Stationary Medium – the part of the apparatus that does not move with the sample ( stationary phase ) </li></ul></ul>
  45. 45. Uses for Chromatography <ul><li>Real-life examples of uses for chromatography: </li></ul><ul><li>Pharmaceutical Company – determine amount of each chemical found in new product </li></ul><ul><li>Hospital – detect blood or alcohol levels in a patient’s blood stream </li></ul><ul><li>Law Enforcement – to compare a sample found at a crime scene to samples from suspects </li></ul><ul><li>Environmental Agency – determine the level of pollutants in the water supply </li></ul><ul><li>Manufacturing Plant – to purify a chemical needed to make a product </li></ul>
  46. 46. Definition of Chromatography <ul><li>Simplified Definition : </li></ul><ul><li>Chromatography separates the components of a mixture by their distinctive attraction to the mobile phase and the stationary phase. </li></ul><ul><li>Explanation : </li></ul><ul><ul><li>Compound is placed on stationary phase </li></ul></ul><ul><ul><li>Mobile phase passes through the stationary phase </li></ul></ul><ul><ul><li>Mobile phase solubilizes the components </li></ul></ul><ul><ul><li>Mobile phase carries the individual components a certain distance through the stationary phase, depending on their attraction to both of the phases </li></ul></ul>
  47. 47. Types of Chromatography <ul><li>Liquid Chromatography – separates liquid samples with a liquid solvent (mobile phase) and a column composed of solid beads (stationary phase) </li></ul><ul><li>Gas Chromatography – separates vaporized samples with a carrier gas (mobile phase) and a column composed of a liquid or of solid beads (stationary phase) </li></ul><ul><li>Paper Chromatography – separates dried liquid samples with a liquid solvent (mobile phase) and a paper strip (stationary phase) </li></ul><ul><li>Thin-Layer Chromatography – separates dried liquid samples with a liquid solvent (mobile phase) and a glass plate covered with a thin layer of alumina or silica gel (stationary phase) </li></ul>Types of Chromatography
  48. 48. Principles of Paper Chromatography <ul><li>Capillary Action – the movement of liquid within the spaces of a porous material due to the forces of adhesion, cohesion, and surface tension. The liquid is able to move up the filter paper because its attraction to itself is stronger than the force of gravity. </li></ul><ul><li>Solubility – the degree to which a material (solute) dissolves into a solvent. Solutes dissolve into solvents that have similar properties. (Like dissolves like) This allows different solutes to be separated by different combinations of solvents. </li></ul><ul><li>Separation of components depends on both their solubility in the mobile phase and their differential affinity to the mobile phase and the stationary phase. </li></ul>
  49. 49. Developing the Chromatograms
  50. 50. Observing the Chromatograms Concentration of Isopropanol 0% 20% 50% 70% 100%
  51. 51. Kinds of Chromatography 1. Paper Chromatography   2. Column Chromatography   3. Thin-layer Chromatography
  52. 52. LIQUID COLUMN CHROMATOGRAPHY A sample mixture is passed through a column packed with solid particles which may or may not be coated with another liquid. With the proper solvents, packing conditions, some components in the sample will travel the column more slowly than others resulting in the desired separation.
  53. 53.
  54. 54. Column Chromatography
  55. 55. <ul><li>The blue compound is obviously more polar than the yellow one - it perhaps even has the ability to hydrogen bond. You can tell this because the blue compound doesn't travel through the column very quickly. That means that it must adsorb more strongly to the silica gel or alumina than the yellow one. The less polar yellow one spends more of its time in the solvent and therefore washes through the column much faster. </li></ul><ul><li>The process of washing a compound through a column using a solvent is known as elution . The solvent is sometimes known as the eluent . </li></ul>
  56. 56. A8HL Visible and ultraviolet (UV-Vis) spectroscopy
  57. 57. COLOURED IONS Theory ions with a d 10 (full) or d 0 (empty) configuration are colourless ions with partially filled d-orbitals tend to be coloured it is caused by the ease of transition of electrons between energy levels energy is absorbed when an electron is promoted to a higher level the frequency of light is proportional to the energy difference ions with d 10 (full) Cu + ,Ag + Zn 2+ or d 0 (empty) Sc 3+ configuration are colourless e.g. titanium(IV) oxide TiO 2 is white colour depends on ... transition element oxidation state ligand coordination number A characteristic of transition metals is their ability to form coloured compounds
  58. 58. SPLITTING OF 3d ORBITALS Placing ligands around a central ion causes the energies of the d orbitals to change Some of the d orbitals gain energy and some lose energy In an octahedral complex, two ( z 2 and x 2 -y 2 ) go higher and three go lower In a tetrahedral complex, three ( xy, xz and yz ) go higher and two go lower Degree of splitting depends on the CENTRAL ION and the LIGAND The energy difference between the levels affects how much energy is absorbed when an electron is promoted. The amount of energy governs the colour of light absorbed. 3d 3d OCTAHEDRAL TETRAHEDRAL
  59. 59. COLOURED IONS Absorbed colour nm Observed colour nm VIOLET 400 GREEN-YELLOW 560 BLUE 450 YELLOW 600 BLUE-GREEN 490 RED 620 YELLOW-GREEN 570 VIOLET 410 YELLOW 580 DARK BLUE 430 ORANGE 600 BLUE 450 RED 650 GREEN 520 The observed colour of a solution depends on the wavelengths absorbed Copper sulphate solution appears blue because the energy absorbed corresponds to red and yellow wavelengths. Wavelengths corresponding to blue light aren’t absorbed. WHITE LIGHT GOES IN SOLUTION APPEARS BLUE ENERGY CORRESPONDING TO THESE COLOURS IS ABSORBED
  60. 60. COLOURED IONS a solution of copper(II)sulphate is blue because red and yellow wavelengths are absorbed white light blue and green not absorbed
  61. 61. COLOURED IONS a solution of copper(II)sulphate is blue because red and yellow wavelengths are absorbed
  62. 62. Ultraviolet (UV) Spectroscopy – Use and Analysis Of all the forms of radiation that go to make up the electromagnetic spectrum UV is probably the most familiar to the general public (after the radiation associated with visible light which is, for the most part, taken for granted). UV radiation is widely known as something to be aware of in hot weather in having a satisfactory effect of tanning the skin but which also has the capacity to damage skin cells to the extent that skin cancer is a direct consequence of overexposure to UV radiation. This damage is associated with the high energy of UV radiation which is directly related to its high frequency and its low wavelength (see the equations below). This is designed to introduce you to UV spectroscopy and give you enough information to ensure that you understand how you can use the technique as a quantitative as appose to a qualitative method.. c =  E = h  E = (hc)/  E  1/  E = energy; c = speed of light;  = wavelength;  = frequency; h = Planck’s constant
  63. 63. Ultraviolet (UV) Spectroscopy – Use and Analysis When continuous wave radiation is passed through a prism a diffraction pattern is produced (called a spectrum) made up of all the wavelengths associated with the incident radiation. When continuous wave radiation passes through a transparent material (solid or liquid) some of the radiation might be absorbed by that material. This slide is part automatically animated – if animation does not occur click left hand mouse button. Radiation source Diffraction prism Spectrum Transparent material that absorbs some radiation Spectrum with ‘gaps’ in it If, having passed through the material, the beam is diffracted by passing through a prism it will produce a light spectrum that has gaps in it (caused by the absorption of radiation by the transparent material through which is passed). The effect of absorption of radiation on the transparent material is to change is from a low energy state (called the ground state) to a higher energy state (called the excited state). The difference between all the spectroscopic techniques is that they use different wavelength radiation that has different associated energy which can cause different modes of excitation in a molecule. For instance, with infra red spectroscopy the low energy radiation simply causes bonds to bend and stretch when a molecule absorbs the radiation. With high energy UV radiation the absorption of energy causes transition of bonding electrons from a low energy orbital to a higher energy orbital. The energy of the ‘missing’ parts of the spectrum corresponds exactly to the energy difference between the orbitals involved in the transition.
  64. 64. Ultraviolet (UV) Spectroscopy – Use and Analysis    *  * n Occupied Energy Levels Unoccupied Energy Levels The bonding orbitals with which you are familiar are the  -bonding orbitals typified by simple alkanes. These are low energy (that is, stable). Next (in terms of increasing energy) are the  -bonding orbitals present in all functional groups that contain double and triple bonds ( e.g. carbonyl groups and alkenes). Higher energy still are the non-bonding orbitals present on atoms that have lone pair(s) of electrons (oxygen, nitrogen, sulfur and halogen containing compounds). All of the above 3 kinds of orbitals may be occupied in the ground state. Two other sort of orbitals, called antibonding orbitals, can only be occupied by an electron in an excited state (having absorbed UV for instance). These are the  * and  * orbitals (the * denotes antibonding). Although you are not too familiar with the concept of an antibonding orbital just remember the following – whilst electron density in a bonding orbital is a stabilising influence it is a destabilising influence (bond weakening) in an antibonding orbital. Antibonding orbitals are unoccupied in the ground state UV A transition of an electron from occupied to an unoccupied energy level can be caused by UV radiation. Not all transitions are allowed but the definition of which are and which are not are beyond the scope of this tutorial. For the time being be aware that commonly seen transitions are  to  * which correctly implies that UV is useful with compounds containing double bonds. A schematic of the transition of an electron from  to  * is shown on the left. Increasing energy
  65. 65. Ultraviolet (UV) Spectroscopy – The Output The output from a UV scanning spectrometer is not the most informative looking piece of data!! It looks like a series of broad humps of varying height. An example is shown below. Decreasing wavelength in nm Increasing absorbance * *Absorbance has no units – it is actually the logarithm of the ratio of light intensity incident on the sample divided by the light intensity leaving the sample. There are two particular strengths of UV (i) it is very sensitive (ii) it is very useful in determining the quantity of a known compound in a solution of unknown concentration. It is not so useful in determining structure although it has been used in this way in the past. The concentration of a sample is related to the absorbance according to the Beer Lambert Law which is described above. A = absorbance ; c = concentration in moles l -1 ; l = pathlength in cm ;  = molar absorptivity (also known as extinction coefficient ) which has units of moles -1 L cm -1 . Beer Lambert Law A =  .c.l
  66. 66. Ultraviolet (UV) Spectroscopy – Analysing the Output Handling samples of known concentration If you know the structure of your compound X and you wish to acquire UV data you would do the following. Prepare a known concentration solution of your sample. Run a UV spectrum (typically from 500 down to 220 nm). From the spectrum read off the wavelength values for each of the maxima of the spectra (see left) Read off the absorbance values of each of the maxima (see left). Then using the known concentration (in moles L -1 ) and the known pathlength (1 cm) calculate the molar absorptivity (  ) for each of the maxima. Finally quote the data as follows (for instance for the largest peak in the spectrum to the left and assuming a concentration of 0.0001 moles L -1 ).  max = 487nm A= 0.75  = 0.75 /(0.001 x 1.0) = 7500 moles -1 L cm -1 Determining concentration of samples with known molar absorptivity (  ). Having used the calculation in the yellow box to work out the molar absorptivity of a compound you can now use UV to determine the concentration of compound X in other samples (provided that these sample only contain pure X). Simply run the UV of the unknown and take the absorbance reading at the maxima for which you have a known value of  . In the case above this is at the peak with the highest wavelength (see above). Having found the absorbance value and knowing  and l you can calculate c. This is the the principle used in many experiments to determine the concentration of a known compound in a particular test sample – for instance monitoring of drug metabolites in the urine of drug takers; monitoring biomolecules produced in the body during particular disease states Beer Lambert Law A =  .c.l
  67. 67. Organic molecules that contain a double bond can absorb UV
  68. 68. phenolphthalein
  69. 69. Retinol
  70. 70. Other conjugated systems that absorb in the visible UV
  71. 71. Typical absorbtions
  72. 72. A9HL Nuclear magnetic resonance (NMR) spectroscopy
  73. 73. <ul><li>non-toxic liquid - SAFE TO USE </li></ul><ul><li>inert - DOESN’T REACT WITH COMPOUND BEING ANALYSED </li></ul><ul><li>has a low boiling point - CAN BE DISTILLED OFF AND USED AGAIN </li></ul><ul><li>all the hydrogen atoms are chemically equivalent - PRODUCES A SINGLE PEAK </li></ul><ul><li>twelve hydrogens so it produces an intense peak - DON’T NEED TO USE MUCH </li></ul><ul><li>signal is outside the range shown by most protons - WON’T OBSCURE MAIN SIGNALS </li></ul><ul><li>given the chemical shift of  = 0 </li></ul><ul><li>the position of all other signals is measured relative to TMS </li></ul>TETRAMETHYLSILANE - TMS The molecule contains four methyl groups attached to a silicon atom in a tetrahedral arrangement. All the hydrogen atoms are chemically equivalent. PROVIDES THE REFERENCE SIGNAL
  74. 74.
  75. 75. Multiplicity & splitting
  76. 76. Number of peaks = number of chemically different H’s on adjacent atoms + 1 1 neighbouring H 2 peaks “doublet” 1:1 2 neighbouring H’s 3 peaks “triplet” 1:2:1 3 neighbouring H’s 4 peaks “quartet” 1:3:3:1 4 neighbouring H’s 5 peaks “quintet” 1:4:6:4:1 Signals for the H in an O-H bond are unaffected by hydrogens on adjacent atoms - get a singlet MULTIPLICITY (Spin-spin splitting) <ul><li>low resolution nmr gives 1 peak for each environmentally different group of protons </li></ul><ul><li>high resolution gives more complex signals - doublets, triplets, quartets, multiplets </li></ul><ul><li>the signal produced indicates the number of protons on adjacent carbon atoms </li></ul>
  77. 77. LOW RESOLUTION - HIGH RESOLUTION HIGH RESOLUTION SPECTRUM OF 1-BROMOPROPANE The broad peaks are split into sharper signals The splitting pattern depends on the number of hydrogen atoms on adjacent atoms <ul><li>low resolution nmr gives 1 peak for each environmentally different group of protons </li></ul><ul><li>high resolution gives more complex signals - doublets, triplets, quartets, multiplets </li></ul><ul><li>the signal produced indicates the number of protons on adjacent carbon atoms </li></ul>
  78. 78. INTEGRATION <ul><li>HOW TO WORK OUT THE SIMPLE RATIOS </li></ul><ul><li>Measure how much each integration line rises as it goes of a set of signals </li></ul><ul><li>Compare the relative values and work out the simple ratio between them </li></ul><ul><li>In the above spectrum the rises are in the ratio... 1:2:3 </li></ul>IMPORTANT: It doesn’t provide the actual number of H’s in each environment, just the ratio Measure the distance between the top and bottom lines. Compare the heights from each signal and make them into a simple ratio.
  79. 79. CONTENTS NMR SPECTROSCOPY “ CENSUS” QUESTIONS - describe where each hydrogen lives - say how many hydrogens live on that atom - say how many chemically different hydrogen atoms live on adjacent atoms 1-BROMOPROPANE ATOM UNIQUE DESCRIPTION OF THE POSITION OF THE HYDROGEN ATOMS H’S ON THE ATOM CHEMICALLY DIFFERENT H’S ON ADJACENT ATOMS SIGNAL SPLIT INTO 1 On an end carbon, two away from the carbon with the bromine atom on it 3 2 2+1 = 3 2 On a carbon atom second from the end and one away from the carbon with the bromine atom 2 3+2 = 5 5+1 = 6 3 On an end carbon atom which also has the bromine atom on it 2 2 2+1 = 3 1 2 3
  80. 80. A10HL Chromatography
  81. 81. This is what a HPLC machine looks like
  82. 82. What does a HLPC chromatogram look like?
  83. 83. Chromatogram of Orange Juice Compounds

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