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Structural Determination of Organic Compounds

Chapter 38. Structural Determination of Organic Compounds. 38.1 Introduction 38.2 Isolation and Purification of Organic Compounds 38.3 Qualitative Analysis of Elements in an Organic Compound 38.4 Determination of Empirical and Molecular Formulae from Analytical Data

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Structural Determination of Organic Compounds

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  1. Chapter 38 Structural Determination of Organic Compounds 38.1Introduction 38.2Isolation and Purification of Organic Compounds 38.3Qualitative Analysis of Elements in an Organic Compound 38.4Determination of Empirical and Molecular Formulae from Analytical Data 38.5Chemical Tests for Functional Groups 38.6Use of Infra-red Spectroscopy in the Identification of Functional Groups

  2. 38.1 Introduction (SB p.137) The determination of the structure of an organic compound involves four steps: • Isolation and purification of the compound • Qualitative analysis of the elements present in the compound • Determination of the molecular formula of the compound • Determination of the functional group present in the compound

  3. 38.2 Isolation and Purification of Organic Compounds (SB p.138) Filtration • The solid/ liquid mixture to be filtered is poured onto the folded filter paper in the filter funnel • Liquids are allowed to pass through the filter paper (known as filtrate) while insoluble solids (known as residue) are retained on the filter paper

  4. 38.2 Isolation and Purification of Organic Compounds (SB p.139) Centrifugation • The tubes containing undissolved solids are spun around very rapidly and are thrown outwards in the centrifuge • The denser solid collects as a lump at the bottom of the tube with the clear liquid above it • The liquid can be removed by decantation

  5. 38.2 Isolation and Purification of Organic Compounds (SB p.139) Crystallization Crystallization by Making and Cooling a Hot Concentrated Solution • The solubility of most solids increase with a rise of temperature • As a hot concentrated solution cools, it cannot hold all of its dissolve solute. The excess solute separates out as crystals

  6. 38.2 Isolation and Purification of Organic Compounds (SB p.140) Crystallization by Evaporating a Solution at Room Temperature • When the solution becomes saturated after evaporation, further evaporation causes crystallization to occur • The crystallization process will be slow if it occurs at room temperature

  7. 38.2 Isolation and Purification of Organic Compounds (SB p.141) Solvent Extraction • To dissolve out a component from a mixture with a suitable solvent • An aqueous solution containing the organic product is usually shaken with diethyl ether in a separating funnel. The ether layer is then run off • Repeated extraction with fresh ether to extract any organic products remaining • The ether portions are combined and dried, and distilled away to obtain the organic products

  8. 38.2 Isolation and Purification of Organic Compounds (SB p.141) Simple Distillation • To separate a liquid from a solution of a liquid and a non-volatile solid or liquid • Only the liquid vapourizes to form vapours which condense to liquid on cold surface which collected as distillate • Anti-bumping granules are added to prevent bumping

  9. 38.2 Isolation and Purification of Organic Compounds (SB p.142) Fractional Distillation • To separate a liquid from a mixture of two or more miscible liquids which have large difference in boiling point • Fractionating column provides a large surface area for condensation and vapourization of the mixture to occur

  10. 38.2 Isolation and Purification of Organic Compounds (SB p.143) Steam Distillation • Many liquid or even solid compounds which are immiscible with water can be purified by distillation in a current of steam

  11. 38.2 Isolation and Purification of Organic Compounds (SB p.143) • Principle of steam distillation: • the total vapour pressure of an immiscible mixture is equal to the sum of the vapour pressures of individual components the water and the compound will distil at a lower temperature than the boiling point of either one of them

  12. 38.2 Isolation and Purification of Organic Compounds (SB p.144) Sublimation • Direct change of a solid to vapour on heating or vapour to solid on cooling • A mixture of two compounds is heated in an evaporating dish, one compound sublimes. The vapour changes back to solid on a cool surface while the other compound is not affected and remains in the evaporating dish

  13. 38.2 Isolation and Purification of Organic Compounds (SB p.144) Chromatography • To separate a complex mixture of substances • As the various components are being absorbed or partitioned at different rates, they move upwards to different extent • The ratio of the distance travelled by the substance to the distance travelled by the solvent is known as Rf value, which is the characteristic of the substance

  14. 38.2 Isolation and Purification of Organic Compounds (SB p.146) A summary of different techniques of isolation and purification

  15. 38.2 Isolation and Purification of Organic Compounds (SB p.146) Tests for Purity Determination of Melting Point • Some of the dry solid is placed in a thin-walled glass melting point tube. The tube is attached to a thermometer • The temperature at which the solid melts is its melting point • Pure solid has a sharp melting point • Impure solid melts gradually over a wide temperature range

  16. 38.2 Isolation and Purification of Organic Compounds (SB p.147) Determination of Boiling Point • The boiling point of a liquid can be determined by using the simple distillation apparatus • The temperature at which the liquid boils steadily is its boiling point • The boiling point of a pure liquid is quite sharp, but varies with changes in external pressure • The presence of non-volatile solutes such as salts raises the boiling point of a liquid

  17. 38.3 Qualitative Analysis of Elements in an Organic Compound (SB p.147) Carbon and Hydrogen • Carbon and hydrogen can be detected by heating a small amount of the substance with Cu2O • Carbon and hydrogen would be oxidized to CO2 and H2O respectively • CO2 turns lime water milky, and H2O turns anhydrous CoCl2 paper pink

  18. 38.3 Qualitative Analysis of Elements in an Organic Compound (SB p.147) Halogens, Nitrogen and Sulphur • Sodium fusion test is performed to test for the presence of halogens, nitrogen and sulphur • The compound under test is fused with a small piece of sodium metal in a small combustion tube and then heated strongly • The product of the test are extracted with water and analyzed • During sodium fusion, halogens, nitrogens and sulphur in the organic compound are converted to sodium halide, sodium cyanide and sodium sulphide respectively

  19. 38.3 Qualitative Analysis of Elements in an Organic Compound (SB p.148)

  20. 38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.148) • Quantitative analysis is to find out the percentage composition by mass of the compound • Methods to determine the mass of the elements: • 1. Carbon and hydrogen • The organic compound is burnt in oxygen. • The carbon dioxide and water vapour formed are respectively absorbed by KOH solution and anhydrous CaCl2. • The increases in mass in KOH solution and CaCl2 represent the masses of carbon dioxide and water vapour formed respectively

  21. 38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.149) • 2. Nitrogen • The organic compound is heated with excess Cu2O • The NO and NO2 formed are passed over hot Cu and the volume of N2 formed is measured

  22. 38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.149) • 3. Halogens • The organic compound is heated with fuming HNO3 and excess AgNO3 solution • The mixture is allowed to cool and then water is added • The dry AgX formed is weighed

  23. 38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.149) • 4. Sulphur • The organic compound is heated with fuming HNO3 • After cooling, barium nitrate solution is added • The dry barium sulphate formed is weighed

  24. 38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.149) • The empirical formula of the compound can be calculated after the determination of percentage composition by mass of a compound • The molecular formula can be calculated after knowing relative molecular mass and the empirical formula

  25. 38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.149) • The empirical formula of a compound is the formula which shows the simplest whole number ratio of the atoms present in the compound • The molecular formula of a compound is the formula which shows the actual numberof each kind of atoms present in a molecule of the compound

  26. 38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.149) Example 38-1 On quantitative analysis, an organic compound was found to contain 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Calculate the empirical formula of the compound.(R.a.m.: H = 1.0, C = 12.0, O = 16.0) Answer

  27. 38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.149) Solution: Let the mass of the compound be 100 g. Then,mass of C in the compound = 40.0 gmass of H in the compound = 6.7 gmass of O in the compound = 53.3 g ∴ The empirical formula of the organic compound is CH2O.

  28. 38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.150) Example 38-2 An organic compound Z has the following composition by mass: (a) Calculate the empirical formula of compound Z. (b) If compound Z is found to have a relative molecular mass of 60, determine the molecular formula of compound Z. (R.a.m.: H = 1.0, C = 12.0, O = 16.0) Answer

  29. 38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.150) Solution: Let the mass of the compound be 100 g. Then,mass of C in the compound = 60.0 gmass of H in the compound = 13.33 gmass of O in the compound = 26.67 g ∴ The empirical formula of compound Z is C3H8O.

  30. 38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.150) Solution: The molecular formula of the compound is (C3H8O)n. Relative molecular mass of (C3H8O)n = 60 n  (12.0 3 + 1.0 8 + 16.0) = 60 n = 1 ∴ The molecular formula of compound Z is C3H8O.

  31. 38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.150) Example 38-3 An organic compound is found to contain carbon, hydrogen and oxygen only. On complete combustion, 0.15 g of this compound gives 0.22 g of carbon dioxide and 0.09 g of water. If the relative molecular mass of this compound is found to be 60, determine the molecular formula of this compound.(R.a.m.: H = 1.0, C = 12.0, O = 16.0) Answer

  32. 38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.151) Solution: Let the empirical formula of the compound be CxHyOz.Mass of C in CxHyOz = Mass of C in CO2Mass of H in CxHyOz = Mass of H in H2OMass of O in CxHyOz = Mass of CxHyOz– mass of C in CxHyOz– mass of H in CxHyOz Relative molecular mass of CO2 = 12.0 + 16.0  2 = 44.0 Mass of C in 0.22 g of CO2 = 0.22 g  = 0.06 g Mass of H in 0.09 g of H2O = 0.09 g  = 0.01 g

  33. 38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.151) Solution: Mass of O in the compound = (0.15 – 0.06 – 0.01) g = 0.08 g The simplest whole number ratio of x, y and z can be determined by following the steps in the table below. ∴ The empirical formula of the organic compound is CH2O.

  34. 38.4 Determination of Empirical and Molecular Formulae from Analytical Data (SB p.151) Solution: The molecular formula of the compound is (CH2O)n. Relative molecular mass of (CH2O)n = 60 n  (12.0 + 1.0  2 +16.0) = 60 n = 2 ∴ The molecular formula of the compound is C2H4O2.

  35. 38.5 Chemical Tests for Functional Groups (SB p.152) • The molecular formula does not give enough information on the structure of the compound ∵ compounds having the same molecular formula may have widely different arrangement of atoms and even different functional groups e.g.

  36. 38.5 Chemical Tests for Functional Groups (SB p.152) • ∴the following step is to find out the functional group(s) present so as to deduce the actual arrangement of atoms in the molecule • The presence of certain functional groups in an organic compound can be detected by simple chemical tests

  37. 38.5 Chemical Tests for Functional Groups (SB p.152)

  38. 38.5 Chemical Tests for Functional Groups (SB p.152)

  39. 38.5 Chemical Tests for Functional Groups (SB p.152)

  40. 38.5 Chemical Tests for Functional Groups (SB p.152)

  41. 38.5 Chemical Tests for Functional Groups (SB p.153)

  42. 38.5 Chemical Tests for Functional Groups (SB p.153)

  43. 38.5 Chemical Tests for Functional Groups (SB p.154)

  44. 38.5 Chemical Tests for Functional Groups (SB p.154)

  45. 38.5 Chemical Tests for Functional Groups (SB p.156) Example 38-4 An organic compound is found to have an empirical formula of CH2O and a relative molecular mass of 60. It reacts with sodium hydrogencarbonate to give a colourless gas which turns lime water milky. (a) Calculate the molecular formula of the compound. Solution: (a) Let the molecular formula of the compound be (CH2O)n. Relative molecular mass of (CH2O)n = 60 n  (12.0 + 1.0  2 + 16.0) = 60 n = 2 ∴ The molecular formula of the compound is C2H4O2. Answer

  46. 38.5 Chemical Tests for Functional Groups (SB p.156) Solution: (b) The compound reacts with sodium hydrogencarbonate to give carbon dioxide gas (which turns lime water milky), indicating that the compound contains a carboxyl group (–COOH). Eliminating the –COOH group from the molecular formula of C2H4O2, the atoms left are one carbon and three hydrogen atoms. This obviously shows that a methyl group (–CH3) is present. The structural formula of the compound is therefore CH3COOH. (c) The IUPAC name for the compound is ethanoic acid. Example 38-4 (cont’d) (b) Deduce the structural formula of the compound. (c) Give the IUPAC name for the compound. (R.a.m.: H = 1.0, C = 12.0, O = 16.0) Answer

  47. 38.5 Chemical Tests for Functional Groups (SB p.156) Example 38-5 15 cm3 of a gaseous hydrocarbon were mixed with 120 cm3 of oxygen which was in excess. The mixture was exploded and after cooling, the residual volume was 105 cm3. On adding concentrated potassium hydroxide solution, the volume decreased to 75 cm3. (a) Calculate the molecular formula of the compound, assuming all volumes were measured under room temperature and pressure. Answer

  48. 38.5 Chemical Tests for Functional Groups (SB p.156) Solution: (a) Let the molecular formula of the compound be CxHy. Volume of hydrocarbon reacted = 15 cm3 Volume of unreacted oxygen = 75 cm3 Volume of oxygen reacted = (120 – 75) cm3 = 45 cm3 Volume of carbon dioxide formed = (105 – 75) cm3 = 30 cm3 Volume of CxHy reacted : volume of CO2 formed = 1 : x = 15 : 30 ∵ ∴ x = 2

  49. 38.5 Chemical Tests for Functional Groups (SB p.156) Solution: Volume of CxHy reacted : volume of O2 reacted = 1 : = 15 : 45 ∵ ∴ y = 4 ∴ The molecular formula of the compound is C2H4

  50. 38.5 Chemical Tests for Functional Groups (SB p.156) Solution: (b) C2H4 belongs to the alkene series. (c) The structural formula of the hydrocarbon is: Example 38-5 (cont’d) (b) To which homologous series does the hydrocarbon belong? (c) Give the structural formula of the hydrocarbon. Answer

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