Organic identification

this ppt describes how to Identify Organic Compound in lab by classical method

1. Practical organic Identification By Lec.Mzgin Mohammed Ayoob [email protected]

2. Contents Qualitative and Quantitative Physical properties . Solubility tests. Elemental analysis. Classification tests. Preparation of derivatives

3. Qualitative analysis of an organic compound is  to determine what elements are present in the compound Qualitative Analysis of an Organic Compound

4. Carbon and Hydrogen Tests for carbon and hydrogen in an organic compound are usually unnecessary  an organic compound must contain carbon and hydrogen

5. Halogens, Nitrogen and Sulphur Halogens , nitrogen and sulphur in organic compounds can be detected  by performing the sodium fusion test

6. Contents Qualitative and Quantitative Physical properties . Solubility tests. Elemental analysis. Classification tests. Preparation of derivatives

7. The physical properties of a compound include its colour , odour , density , solubility , melting point and boiling point The physical properties of a compound depend on its molecular structure Structural Information from Physical Properties

8. Structural Information from Physical Properties e.g.  Hydrocarbons have low densities , often about 0.8 g cm –3  Compounds with functional groups have higher densities

9. Contents Qualitative and Quantitative Physical properties . Solubility tests. Elemental analysis. Classification tests. Preparation of derivatives

10. solubility

11. Organic compound Water Soluble Insoluble Soluble Insoluble Ether Next slide

12. Organic compound Water Insoluble Soluble NaOH Soluble HCl Insoluble NaHCO3 Insoluble Soluble A polar molecule and molecules with larger parts Despite polar Group being present Insoluble

13. Neutral compound

14. Contents Qualitative and Quantitative Physical properties . Solubility tests. Elemental analysis. Classification tests. Preparation of derivatives

15. Halogens, Nitrogen and Sulphur The compound under test is  fused with a small piece of sodium metal in a small combustion tube  heated strongly The products of the test are extracted with water and then analyzed

16. Halogens, Nitrogen and Sulphur During sodium fusion ,  halogens in the organic compound is converted to sodium halides  nitrogen in the organic compound is converted to sodium cyanide  sulphur in the organic compound is converted to sodium sulphide

17. Element Observation Halogens, as chloride ion (Cl - ) A white precipitate is formed. It is soluble in excess NH 3 ( aq ). bromide ion (Br - ) A pale yellow precipitate is formed. It is sparingly soluble in excess NH 3 (aq). iodide ion (I - ) A creamy yellow precipitate is formed. It is insoluble in excess NH 3 ( aq ). Results for halogens, nitrogen and sulphur in the sodium fusion test

18. Results for halogens, nitrogen and sulphur in the sodium fusion test Element Material used Observation Nitrogen,as cyanide ion (CN - ) A mixture of iron(II) sulphate and iron(III) sulphate solutions A blue-green colour is observed. Sulphur, as sulphide ion (S 2- ) Sodium pentacyanonitrosylferrate(II) solution A black precipitate is formed

19. Contents Qualitative and Quantitative Physical properties . Solubility tests. Elemental analysis. Classification tests. Preparation of derivatives

20. Structural Information from Chemical Properties The next stage is  to find out the functional group(s) present  to deduce the actual arrangement of atoms in the molecule

21. Organic compound Test Observation Saturated hydrocarbons • Burn the saturated hydrocarbon in a non-luminous Bunsen flame • A blue or clear yellow flame is observed Chemical tests for different groups of organic compounds

22. Organic compound Test Observation Unsaturated hydrocarbons (C = C, C  C) • Burn the unsaturated hydrocarbon in a non-luminous Bunsen flame • A smoky flame is observed • Add bromine in 1,1,1-trichloroethane at room temperature and in the absence of light • Bromine decolourizes rapidly • Add 1% (dilute) acidified potassium manganate(VII) solution • Potassium manganate (VII) solution decolourizes rapidly Chemical tests for different groups of organic compounds

23. Organic compound Test Observation Haloalkanes (1°, 2° or 3°) • Boil with ethanolic potassium hydroxide solution, then acidify with excess dilute nitric(V) acid and add silver nitrate(V) solution • For chloroalkanes , a white precipitate is formed • For bromoalkanes , a pale yellow precipitate is formed • For iodoalkanes , a creamy yellow precipitate is formed Chemical tests for different groups of organic compounds

24. Organic compound Test Observation Halobenzenes • Boil with ethanolic potassium hydroxide solution, then acidify with excess dilute nitric(V) acid and add silver nitrate(V) solution • No precipitate is formed Chemical tests for different groups of organic compounds

25. Chemical tests for different groups of organic compounds Organic compound Test Observation Alcohols (  OH) • Add a small piece of sodium metal • A colourless gas is evolved • Esterification: Add ethanoyl chloride • The temperature of the reaction mixture rises • A colourless gas is evolved

26. Chemical tests for different groups of organic compounds Organic compound Test Observation Alcohols (  OH) • Add acidified potassium dichromate(VI) solution • For 1° and 2° alcohols, the clear orange solution becomes opaque and turns green almost immediately • For 3° alcohols, there are no observable changes

27. Chemical tests for different groups of organic compounds Organic compound Test Observation Alcohols (  OH) • Iodoform test for: Add iodine in sodium hydroxide solution • A yellow precipitate is formed

28. Chemical tests for different groups of organic compounds Organic compound Test Observation Alcohols (  OH) • Lucas test: add a solution of zinc chloride in concentrated hydrochloric acid • For 1° alcohols, the aqueous phase remains clear • For 2° alcohols, the clear solution becomes cloudy within 5 minutes • For 3° alcohols, the aqueous phase appears cloudy immediately

29. Chemical tests for different groups of organic compounds Organic compound Test Observation Ethers (  O  ) • No specific test for ethers but they are soluble in concentrated sulphuric(VI) acid 

30. Organic compound Test Observation Aldehydes ( ) • Add aqueous sodium hydrogensulphate (IV) • Crystalline salts are formed • Add 2,4-dinitrophenylhydrazine • A yellow, orange or red precipitate is formed • Silver mirror test: add Tollens’ reagent (a solution of aqueous silver nitrate in aqueous ammonia) • A silver mirror is deposited on the inner wall of the test tube Chemical tests for different groups of organic compounds

31. Organic compound Test Observation Ketones ( ) • Add aqueous sodium hydrogensulphate(IV) • Crystalline salts are formed (for unhindered ketones only) • Add 2,4-dinitrophenylhydrazine • A yellow, orange or red precipitate is formed • Iodoform test for: Add iodine in sodium hydroxide solution • A yellow precipitate is formed Chemical tests for different groups of organic compounds

32. Organic compound Test Observation Carboxylic acids ( ) • Esterification: warm the carboxylic acid with an alcohol in the presence of concentrated sulphuric (VI) acid, followed by adding sodium carbonate solution A sweet and fruity smell is detected • Add sodium hydrogencarbonate The colourless gas produced turns lime water milky Chemical tests for different groups of organic compounds

33. Organic compound Test Observation Esters ( ) No specific test for esters but they can be distinguished by its characteristic smell A sweet and fruity smell is detected Chemical tests for different groups of organic compounds

34. Organic compound Test Observation Acyl halides ( ) Boil with ethanolic potassium hydroxide solution, then acidify with excess dilute nitric(V) acid and add silver nitrate(V) solution For acyl chlorides, a white precipitate is formed For acyl bromides, a pale yellow precipitate is formed For acyl iodides, a creamy yellow precipitate is formed Chemical tests for different groups of organic compounds

35. Organic compound Test Observation Amides ( ) Boil with sodium hydroxide solution The colourless gas produced turns moist red litmus paper or pH paper blue Chemical tests for different groups of organic compounds

36. Organic compound Test Observation Amines ( NH 2 ) 1 o aliphatic amines: dissolve the amine in dilute hydrochloric acid at 0 – 5 o C, then add cold sodium nitrate(III) solution slowly Steady evolution of N 2 (g) is observed 1 o aromatic amines: add naphthalen-2-ol in dilute sodium hydroxide solution An orange or red precipitate is formed Chemical tests for different groups of organic compounds

37. Organic compound Test Observation Aromatic compounds ( ) Burn the aromatic compound in a non-luminous Bunsen flame A smoky yellow flame with black soot is produced Add fuming sulphuric(VI) acid The aromatic compound dissolves The temperature of the reaction mixture rises Chemical tests for different groups of organic compounds

38. Contents Qualitative and Quantitative Physical properties . Solubility tests. Elemental analysis. Classification tests. Preparation of derivatives

39. THE PREPARATION OF DERIVATIVES CARBOXYLIC ACIDS, ACID ANHYDRIDES, ACID HALIDES

40. Carboxylic acid S- Benzylthiuronium halide undergoes reaction with the salt of the carboxylic acid to yield the corresponding S- benzylthiuronium salt

41. p- nitrobenzyl or the 4-bromophenacyl esters are preferred. These are prepared by treating the salts of the acids with either 4-nitrobenzyl chloride or 4 bromophenacyl bromide

42. ALCOHOLS The most general derivatives of primary and secondary alcohols are the phenylurethanes and I- naphthylurethanes

43. 4-Nitrobenzoates or 3,5-dinitrobenzoates

44. ALDEHYDES AND KETONES 2,4-dinitrophenylhydrazones, phenylhydrazones , and 4-nitrophenylhydrazones

45. AMIDES Indirect :- The most general method for chemically characterizing primary amides consists in hydrolyzing them with alkali to the salt of the carboxylic acid and ammonia. Direct derivative ( Xanthydrol ) not common

46. ESTERS

47. AMINES Arylsulfonamides are used as derivatives for amines

48. A secondary amine reacts with 4-toluenesulfonyl chloride to give the rylsulfonamide , which is usually insoluble . Picric acid undergoes reaction with tertiary amines to yield the picrates .

49. ETHERS-ALIPHATIC ETHERS-AROMATIC

50. PHENOLS

51. 4-nitrobenzoates , or 3,5dinitrobenzoates

52. NITRILES NITRO COMPOUNDS

53. Question styles Deduce the structure for the unknown compound(C 4 H 6 O 4 ) according to the following observation 1-Burn with blue flame 2- No color with KMnO 4 3- Give a positive test with KI/KIO 3 4- Has weak peaks at 2826cm -1 and 2955cm -1 How you can distinguish each of pair compound (classically and instrumentally) 1- aldehyde and ketone. What are the roles for each of the following reagents or tests? Prepare the derivative for each of the following compounds

54. Contents Qualitative and Quantitative Physical properties . Solubility tests. Elemental analysis. Classification tests. Preparation of derivatives

55. Compound A : insoluble in water; insoluble in sulfuric acid; sodium fusion, followed by silver nitrate treatment, gave a white precipitate. Compound B : insoluble in water; quickly decolorized bromine; did not react with acetyl chloride; did not form a precipitate with 2,4-dinitrophenylhydrazine; did not form a precipitate when treated with excess iodine in aqueous sodium hydroxide solution. Compound C : insoluble in water; soluble in sulfuric acid; sodium fusion, followed by silver nitrate treatment, gave no precipitate. Compound D : soluble in water; gave a rapid reaction with acetyl chloride; did not form a precipitate with 2,4-dinitrophenylhydrazine; formed a yellow precipitate with a noxious odor when treated with excess iodine in aqueous sodium hydroxide solution. Choices: 2-propanol, cyclohexane, chlorocyclohexane , cyclohexene, diethyl ether.

56. Complete the following reactions

57. What are the necessary information to determine the structure of an organic compound? Answer Molecular formula from analytical data, functional group present from physical and chemical properties, structural information from infra-red spectroscopy and NMR spectrometry Back

58. (a) Why is detection of carbon and hydrogen in organic compounds not necessary? (b) What elements can be detected by sodium fusion test? Answer (a) All organic compounds contain carbon and hydrogen. (b) Halogens, nitrogen and sulphur Back

59. Let the mass of the compound be 100 g. Then, mass of carbon in the compound = 40.0 g mass of hydrogen in the compound = 6.7 g mass of oxygen in the compound = 53.3 g ∴ The empirical formula of the organic compound is CH 2 O. Carbon Hydrogen Oxygen Mass (g) 40.0 6.7 53.3 Number of moles (mol) Relative number of moles Simplest mole ratio 1 2 1 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. Answer

60. An organic compound Z has the following composition by mass: (a) Calculate the empirical formula of compound Z. (b) If the relative molecular mass of compound Z is 60.0, determine the molecular formula of compound Z. Answer Element Carbon Hydrogen Oxygen Percentage by mass (%) 60.00 13.33 26.67

61. (a) Let the mass of the compound be 100 g. Then, mass of carbon in the compound = 60.00 g mass of hydrogen in the compound = 13.33 g mass of oxygen in the compound = 26.67 g ∴ The empirical formula of the organic compound is C 3 H 8 O. Carbon Hydrogen Oxygen Mass (g) 60.00 13.33 26.67 Number of moles (mol) Relative number of moles Simplest mole ratio 3 8 1

62. (b) The molecular formula of the compound is (C 3 H 8 O) n . Relative molecular mass of (C 3 H 8 O) n = 60.0 n × (12.0 × 3 + 1.0 × 8 + 16.0) = 60.0 n = 1 ∴ The molecular formula of compound Z is C 3 H 8 O.

63. Relative molecular mass of CO 2 = 12.0 + 16.0 × 2 = 44.0 Mass of carbon in 0.22 g of CO 2 = 0.22 g × = 0.06 g Relative molecular mass of H 2 O = 1.0 × 2 + 16.0 = 18.0 Mass of hydrogen in 0.09 g of H 2 O = 0.09 g × = 0.01 g Mass of oxygen in the compound = (0.15 – 0.06 – 0.01) g = 0.08 g An organic compound was found to contain carbon, hydrogen and oxygen only. On complete combustion, 0.15 g of this compound gave 0.22 g of carbon dioxide and 0.09 g of water. If the relative molecular mass of this compound is 60.0, determine the molecular formula of this compound. Answer

64. ∴ The empirical formula of the organic compound is CH 2 O. Carbon Hydrogen Oxygen Mass (g) 0.06 0.01 0.08 Number of moles ( mol ) Relative number of moles Simplest mole ratio 1 2 1

65. Let the molecular formula of the compound be (CH 2 O) n . Relative molecular mass of (CH 2 O) n = 60.0 n × (12.0 + 1.0 × 2 + 16.0) = 60.0 n = 2 ∴ The molecular formula of the compound is C 2 H 4 O 2 . Back

66. Why does the solubility of amines in water decrease in the order: 1 o amines > 2 o amines > 3 o amines? Answer The solubility of primary and secondary amines is higher than that of tertiary amines because tertiary amines cannot form hydrogen bonds between water molecules. On the other hand, the solubility of primary amines is higher than that of secondary amines because primary amines form a greater number of hydrogen bonds with water molecules than secondary amines.

67. The empirical formula of an organic compound is CH 2 O and its relative molecular mass is 60.0. It reacts with sodium hydrogencarbonate to give a colourless gas which turns lime water milky. Calculate the molecular formula of the compound . (b) Deduce the structural formula of the compound . (c) Give the IUPAC name for the compound . Answer (a) Let the molecular formula of the compound be (CH 2 O) n . Relative molecular mass of (CH 2 O) n = 60.0 n  (12.0 + 1.0  2 + 16.0) = 60.0 n = 2 ∴ The molecular formula of the compound is C 2 H 4 O 2 .

68. The compound reacts with sodium hydrogencarbonate to give a colourless gas which turns lime water milky. This indicates that the compound contains a carboxyl group ( COOH). Eliminating the  COOH group from the molecular formula of C 2 H 4 O 2 , the atoms left are one carbon and three hydrogen atoms. This obviously shows that a methyl group ( CH 3 ) is present. Therefore, the structural formula of the compound is: (c) The IUPAC name for the compound is ethanoic acid.

69. Thank you Any Questions