34 34.1 34.2 34.3 34.4 34.5 34.6 34.7 34.8 34.9 1 Structural Determination of Organic Compounds Introduction Isolation and Purification of Organic Compounds Tests for Purity Qualitative Analysis of Elements in an Organic Compound Determination of Empirical Formula and Molecular Formula from Analytical Data Structural Information from Physical Properties Structural Information from Chemical Properties Use of Infra-red Spectrocopy in the Identification of Functional Groups Use of Mass Spectra to Obtain Structural Information The general steps to determine the structure of an organic compound 2 Isolation and Purification of Organic Compounds 3 Technique 1. Filtration To separate an insoluble solid from a liquid (slow) 2. Centrifugation To separate an insoluble solid from a liquid (fast) To separate a solid from other solids based on their different solubilities in suitable solvent(s) To separate a component from a mixture with a suitable solvent 3. Recrystallization 4. Solvent extraction 5. Distillation 4 Aim To separate a liquid from a solution containing non-volatile solutes Technique Aim 6. Fractional distillation To separate miscible liquids with widely different boiling points 7. Steam distillation To separate liquids which are immiscible with water and The mixture boils decompose easily below their below 100C b.p. 8. Vacuum distillation ditto 9. Sublimation To separate a mixture of solids in which only one can sublime 10. Chromatography 5 To separate a complex mixture of substances (large/small scale) Tests for Purity • If the substance is a solid, its purity can be checked by determining its melting point • If it is a liquid, its purity can be checked by determining its boiling point 6 34.2 Isolation and Purification of Organic Compounds (SB p.78) Isolation and Purification of Organic Compounds • The selection of a proper technique depends on the particular differences in physical properties of the substances present in the mixture 7 34.2 Isolation and Purification of Organic Compounds (SB p.78) Filtration • To separate an insoluble solid from a liquid particularly when the solid is suspended throughout the liquid • The solid/liquid mixture is called a suspension 8 34.2 Isolation and Purification of Organic Compounds (SB p.78) Filtration The laboratory set-up of filtration 9 34.2 Isolation and Purification of Organic Compounds (SB p.78) Filtration • There are many small holes in the filter paper allow very small particles of solvent and dissolved solutes to pass through as filtrate • Larger insoluble particles are retained on the filter paper as residue 10 34.2 Isolation and Purification of Organic Compounds (SB p.79) Centrifugation • When there is only a small amount of suspension, or when much faster separation is required Centrifugation is often used instead of filtration 11 34.2 Isolation and Purification of Organic Compounds (SB p.79) Centrifugation • The liquid containing undissolved solids is put in a centrifuge tube • The tubes are then put into the tube holders in a centrifuge A centrifuge 12 34.2 Isolation and Purification of Organic Compounds (SB p.79) Centrifugation • The holders and tubes are spun around at a very high rate and are thrown outwards • The denser solid is collected as a lump at the bottom of the tube with the clear liquid above 13 34.2 Isolation and Purification of Organic Compounds (SB p.79) Crystallization • Crystals are solids that have a definite regular shape smooth flat faces and straight edges • Crystallization is the process of forming crystals 14 34.2 Isolation and Purification of Organic Compounds (SB p.79) 1. Crystallization by Cooling a Hot Concentrated Solution • To obtain crystals from an unsaturated aqueous solution the solution is gently heated to make it more concentrated • After, the solution is allowed to cool at room conditions 15 34.2 Isolation and Purification of Organic Compounds (SB p.79) 1. Crystallization by Cooling a Hot Concentrated Solution • The solubilities of most solids increase with temperature • When a hot concentrated solution is cooled the solution cannot hold all of the dissolved solutes • The “excess” solute separates out as crystals 16 34.2 Isolation and Purification of Organic Compounds (SB p.79) 1. Crystallization by Cooling a Hot Concentrated Solution Crystallization by cooling a hot concentrated solution 17 34.2 Isolation and Purification of Organic Compounds (SB p.80) 2. Crystallization by Evaporating a Cold Solution at Room Temperature • As the solvent in a solution evaporates, the remaining solution becomes more and more concentrated eventually the solution becomes saturated further evaporation causes crystallization to occur 18 34.2 Isolation and Purification of Organic Compounds (SB p.80) 2. Crystallization by Evaporating a Cold Solution at Room Temperature • If a solution is allowed to stand at room temperature, evaporation will be slow • It may take days or even weeks for crystals to form 19 34.2 Isolation and Purification of Organic Compounds (SB p.80) 2. Crystallization by Evaporating a Cold Solution at Room Temperature Crystallization by slow evaporation of a solution (preferably saturated) at room temperature 20 34.2 Isolation and Purification of Organic Compounds (SB p.80) Solvent Extraction • Involves extracting a component from a mixture with a suitable solvent • Water is the solvent used to extract salts from a mixture containing salts and sand • Non-aqueous solvents (e.g. 1,1,1trichloroethane and diethyl ether) can be used to extract organic products 21 34.2 Isolation and Purification of Organic Compounds (SB p.80) Solvent Extraction • Often involves the use of a separating funnel • When an aqueous solution containing the organic product is shaken with diethyl ether in a separating funnel, the organic product dissolves into the ether layer 22 34.2 Isolation and Purification of Organic Compounds (SB p.80) Solvent Extraction The organic product in an aqueous solution can be extracted by solvent extraction using diethyl ether 23 34.2 Isolation and Purification of Organic Compounds (SB p.80) Solvent Extraction • The ether layer can be run off from the separating funnel and saved • Another fresh portion of ether is shaken with the aqueous solution to extract any organic products remaining • Repeated extraction will extract most of the organic product into the several portions of ether 24 34.2 Isolation and Purification of Organic Compounds (SB p.80) Solvent Extraction • Conducting the extraction with several small portions of ether is more efficient than extracting in a single batch with the whole volume of ether • These several ether portions are combined and dried the ether is distilled off leaving behind the organic product 25 34.2 Isolation and Purification of Organic Compounds (SB p.81) Distillation • A method used to separate a solvent from a solution containing non-volatile solutes • When a solution is boiled, only the solvent vaporizes the hot vapour formed condenses to liquid again on a cold surface • The liquid collected is the distillate 26 34.2 Isolation and Purification of Organic Compounds (SB p.81) Distillation The laboratory set-up of distillation 27 34.2 Isolation and Purification of Organic Compounds (SB p.81) Distillation • Before the solution is heated, several pieces of anti-bumping granules are added into the flask prevent vigorous movement of the liquid called bumping to occur during heating make boiling smooth 28 34.2 Isolation and Purification of Organic Compounds (SB p.81) Distillation • If bumping occurs during distillation, some solution (not yet vaporized) may spurt out into the collecting vessel 29 34.2 Isolation and Purification of Organic Compounds (SB p.81) Fractional Distillation • A method used to separate a mixture of two or more miscible liquids 30 34.2 Isolation and Purification of Organic Compounds (SB p.82) Fractional Distillation The laboratory set-up of fractional distillation 31 34.2 Isolation and Purification of Organic Compounds (SB p.82) Fractional Distillation • A fractionating column is attached vertically between the flask and the condenser a column packed with glass beads provide a large surface area for the repeated condensation and vaporization of the mixture to occur 32 34.2 Isolation and Purification of Organic Compounds (SB p.82) Fractional Distillation • The temperature of the escaping vapour is measured using a thermometer • When the temperature reading becomes steady, the vapour with the lowest boiling point firstly comes out from the top of the column 33 34.2 Isolation and Purification of Organic Compounds (SB p.82) Fractional Distillation • When all of that liquid has distilled off, the temperature reading rises and becomes steady later on another liquid with a higher boiling point distils out • 34 Fractions with different boiling points can be collected separately 34.2 Isolation and Purification of Organic Compounds (SB p.82) Sublimation • Sublimation is the direct change of a solid to vapour on heating, or a vapour to solid on cooling without going through the liquid state 35 34.2 Isolation and Purification of Organic Compounds (SB p.82) Sublimation • A mixture of two compounds is heated in an evaporating dish • One compound changes from solid to vapour directly The vapour changes back to solid on a cold surface • The other compound is not affected by heating and remains in the evaporating dish 36 34.2 Isolation and Purification of Organic Compounds (SB p.83) Sublimation A mixture of two compounds can be separated by sublimation 37 34.2 Isolation and Purification of Organic Compounds (SB p.83) Chromatography 38 • An effective method of separating a complex mixture of substances • Paper chromatography is a common type of chromatography 34.2 Isolation and Purification of Organic Compounds (SB p.83) Chromatography The laboratory set-up of paper chromatography 39 34.2 Isolation and Purification of Organic Compounds (SB p.83) Chromatography • A solution of the mixture is dropped at one end of the filter paper 40 34.2 Isolation and Purification of Organic Compounds (SB p.83) Chromatography • The thin film of water adhered onto the surface of the filter paper forms the stationary phase • The solvent is called the mobile phase or eluent 41 34.2 Isolation and Purification of Organic Compounds (SB p.83) Chromatography • When the solvent moves across the sample spot of the mixture, partition of the components between the stationary phase and the mobile phase occurs 42 34.2 Isolation and Purification of Organic Compounds (SB p.83) Chromatography • As the various components are being adsorbed or partitioned at different rates, they move upwards at different rates • The ratio of the distance travelled by the substance to the distance travelled by the solvent known as the Rf value a characteristic of the substance 43 34.2 Isolation and Purification of Organic Compounds (SB p.84) A summary of different techniques of isolation and purification Technique (a) Filtration Aim To separate an insoluble solid from a liquid (slow) (b) Centrifugation To separate an insoluble solid from a liquid (fast) (c) Crystallization To separate a dissolved solute from its solution (d) Solvent To separate a component from a extraction mixture with a suitable solvent (e) Distillation To separate a liquid from a solution containing non-volatile solutes 44 34.2 Isolation and Purification of Organic Compounds (SB p.84) A summary of different techniques of isolation and purification Technique (f) Fractional distillation (g) Sublimation Aim To separate miscible liquids with widely different boiling points To separate a mixture of solids in which only one can sublime (h) To separate a complex mixture of Chromatography substances Check Point 34-2 45 34.4 Qualitative Analysis of Elements in an Organic Compound 46 34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86) Qualitative Analysis of an Organic Compound • Qualitative analysis of an organic compound is to determine what elements are present in the compound 47 34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86) Carbon and Hydrogen • Tests for carbon and hydrogen in an organic compound are usually unnecessary an organic compound must contain carbon and hydrogen 48 34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86) Carbon and Hydrogen • Carbon and hydrogen can be detected by heating a small amount of the substance with copper(II) oxide • Carbon and hydrogen would be oxidized to carbon dioxide and water respectively • Carbon dioxide turns lime water milky • Water turns anhydrous cobalt(II) chloride paper pink 49 34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86) Halogens, Nitrogen and Sulphur • Halogens, nitrogen and sulphur in organic compounds can be detected by performing the sodium fusion test 50 34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86) 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 51 34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86) 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 52 34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86) Results for halogens, nitrogen and sulphur in the sodium fusion test Element Material used Observation Halogens, as Acidified silver nitrate solution chloride ion (Cl-) A white precipitate is formed. It is soluble in excess NH3(aq). bromide ion (Br-) A pale yellow precipitate is formed. It is sparingly soluble in excess NH3(aq). A creamy yellow precipitate is formed. It is insoluble in excess NH3(aq). iodide ion (I-) 53 34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86) Results for halogens, nitrogen and sulphur in the sodium fusion test Element Material used Observation Nitrogen,as cyanide ion (CN-) A mixture of A blue-green colour is iron(II) sulphate observed. and iron(III) sulphate solutions Sulphur, as sulphide ion (S2-) Sodium pentacyanonitr osylferrate(II) solution A black precipitate is formed Check Point 34-4 54 34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data 55 34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87) Quantitative Analysis of an Organic Compound • After determining the constituent elements of a particular organic compound perform quantitative analysis to find the percentage composition by mass of the compound the masses of different elements in an organic compound are determined 56 34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87) 1. Carbon and Hydrogen • The organic compound is burnt in excess oxygen • The carbon dioxide and water vapour formed are respectively absorbed by potassium hydroxide solution and anhydrous calcium chloride 57 34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87) 1. Carbon and Hydrogen • The increases in mass in potassium hydroxide solution and calcium chloride represent the masses of carbon dioxide and water vapour formed respectively 58 34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87) 2. Nitrogen • The organic compound is heated with excess copper(II) oxide • The nitrogen monoxide and nitrogen dioxide formed are passed over hot copper the volume of nitrogen formed is measured 59 34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87) 3. Halogens • The organic compound is heated with fuming nitric(V) acid and excess silver nitrate solution • The mixture is allowed to cool then water is added the dry silver halide formed is weighed 60 34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87) 4. Sulphur • The organic compound is heated with fuming nitric(V) acid • After cooling, barium nitrate solution is added the dry barium sulphate formed is weighed 61 34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87) Quantitative Analysis of an Organic Compound • After determining the percentage composition by mass of a compound, the empirical formula of the compound can be calculated 62 34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87) Quantitative Analysis of an Organic Compound The empirical formula of a compound is the formula which shows the simplest whole number ratio of the atoms present in the compound 63 34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87) Quantitative Analysis of an Organic Compound • When the relative molecular mass and the empirical formula of the compound are known, the molecular formula of the compound can be calculated 64 34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.88) Quantitative Analysis of an Organic Compound The molecular formula of a compound is the formula which shows the actual number of each kind of atoms present in a molecule of the compound 65 34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.88) Example 34-5A Example 34-5B Check Point 34-5 66 34.6 Structural Information from Physical Properties 67 34.6 Structural Information from Physical Properties (SB p.89) Structural Information from Physical Properties • 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 68 34.6 Structural Information from Physical Properties (SB p.89) Structural Information from Physical Properties • From the physical properties of a compound, obtain preliminary information about the structure of the compound 69 34.6 Structural Information from Physical Properties (SB p.89) 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 70 34.6 Structural Information from Physical Properties (SB p.89) Structural Information from Physical Properties • The densities of most organic compounds are < 1.2 g cm–3 • Compounds having densities > 1.2 g cm–3 must contain multiple halogen atoms 71 34.6 Structural Information from Physical Properties (SB p.90) Physical properties of some common organic compounds Organic compound Hydrocarbo ns (saturated and unsaturated) 72 Density at 20oC Melting point and boiling point Solubility In water In nonor highly polar polar organic solvents solvents All • Generally low but Insoluble increases with have number of carbon densities atoms in the molecule < 0.8 g cm–3 • Branched-chain hydrocarbons have lower boiling points but higher melting points than the corresponding straight-chain isomers Soluble 34.6 Structural Information from Physical Properties (SB p.90) Physical properties of some common organic compounds Organic compound Density at 20oC Aromatic Between hydrocarbons 0.8 and 1.0 g cm–3 73 Melting point and boiling point Generally low Solubility In water In nonor highly polar polar organic solvents solvents Insoluble Soluble 34.6 Structural Information from Physical Properties (SB p.90) Physical properties of some common organic compounds Organic Density at compound 20oC Haloalkanes 74 • 0.9 - 1.1 g cm–3 for chloroalkanes • >1.0 g cm–3 for bromoalkanes and iodoalkanes Melting point and boiling point • Higher than alkanes of similar relative molecular masses ( haloalkane molecules are polar) • All haloalkanes are liquids except halomethanes • Both the m.p. and b.p. increase in the order: RCH2F < RCH2Cl < RCH2Br < RCH2I Solubility In water In nonor highly polar polar organic solvents solvents Insoluble Soluble 34.6 Structural Information from Physical Properties (SB p.90) Physical properties of some common organic compounds Organic compound Density at 20oC Melting point and boiling point Alcohols • Simple • Much higher than alcohols are hydrocarbons of liquids and similar relative alcohols with molecular masses ( formation of > 12 carbons are hydrogen bonds waxy solids between alcohol molecules) 75 Solubility In water or highly polar solvents In nonpolar organic solvents • Lower members: Completely miscible with water ( formation of hydrogen bonds between alcohol molecules and water molecules) Soluble 34.6 Structural Information from Physical Properties (SB p.90) Physical properties of some common organic compounds Organic compound Density at 20oC Melting point and boiling point Alcohols • All simple • Straight-chain alcohols alcohols have have higher b.p. than the densities corresponding < 1.0 g cm–3 branched-chain alcohols 76 Solubility In water or highly polar solvents In nonpolar organic solvents • Solubility Soluble decreases gradually as the hydrocarbon chain lengthens 34.6 Structural Information from Physical Properties (SB p.91) Physical properties of some common organic compounds Organic compound Density at 20oC Carbonyl • <1.0 g cm–3 for aliphatic compcarbonyl ounds compounds (aldehydes and ketones) 77 Melting point and boiling point Higher than alkanes but lower than alcohols of similar relative molecular masses (Molecules of aldehydes or ketones are held together by strong dipole-dipole interactions but not hydrogen bonds) Solubility In water or highly polar solvents In nonpolar organic solvents • Lower Soluble members: Soluble in water ( the formation of hydrogen bonds between molecules of aldehydes or ketones and water molecules) 34.6 Structural Information from Physical Properties (SB p.91) Physical properties of some common organic compounds Organic compound Density at 20oC Carbonyl • > 1.0 g cm–3 for aromatic compcarbonyl ounds compounds (aldehydes and ketones) 78 Melting point and boiling point Solubility In water or highly polar solvents In nonpolar organic solvents • Solubility decreases gradually as the hydrocarbon chain lengthens Soluble 34.6 Structural Information from Physical Properties (SB p.91) Physical properties of some common organic compounds Organic compound Carboxylic acids 79 Density at 20oC Melting point and boiling point • Lower members have densities similar to water • Methanoic acid has a density of 1.22 g cm–3 Higher than alcohols of similar relative molecular masses ( the formation of more extensive intermolecular hydrogen bonds) Solubility In water or highly polar solvents • First four members are miscible with water in all proportions • Solubility decreases gradually as the hydrocarbon chain lengthens In nonpolar organic solvents Soluble 34.6 Structural Information from Physical Properties (SB p.91) Physical properties of some common organic compounds Organic compound Esters 80 Density at 20oC Lower members have densities less than water Melting point and boiling point Slightly higher than hydrocarbons but lower than carbonyl compounds and alcohols of similar relative molecular masses Solubility In water or highly polar solvents Insoluble In nonpolar organic solvents Soluble 34.6 Structural Information from Physical Properties (SB p.91) Physical properties of some common organic compounds Organic compound Amines 81 Density at 20oC Melting point and boiling point Most amines • Higher than have alkanes but lower densities than alcohols of less than similar relative water molecular masses Solubility In water or highly polar solvents • Generally soluble • Solubility decreases in the order: 1o amines > 2o amines > 3o amines In nonpolar organic solvents Soluble 34.6 Structural Information from Physical Properties (SB p.91) Physical properties of some common organic compounds Organic compound Amines 82 Density at 20oC Melting point and boiling point • 1o and 2o amines are able to form hydrogen bonds with each other but the strength is less than that between alcohol molecules (NH bond is less polar than O H bond) Solubility In water or highly polar solvents In nonpolar organic solvents 34.6 Structural Information from Physical Properties (SB p.91) Physical properties of some common organic compounds Organic compound Amines 83 Density at 20oC Melting point and boiling point • 3o amines have lower m.p. and b.p. than the isomers of 1o and 2o amines ( molecules of 3o amines cannot form intermolecular hydrogen bonds) Solubility In water or highly polar solvents In nonpolar organic solvents 34.6 Structural Information from Physical Properties (SB p.92) Example 34-6 84 Check Point 34-6 34.7 Structural Information from Chemical Properties 85 34.7 Structural Information from Chemical Properties (SB p.93) Structural Information from Chemical Properties • The molecular formula of a compound does not give enough clue to the structure of the compound • Compounds having the same molecular formula may have different arrangements of atoms and even different functional groups 86 34.7 Structural Information from Chemical Properties (SB p.93) Structural Information from Chemical Properties • e.g. The molecular formula of C2H4O2 may represent a carboxylic acid or an ester: 87 34.7 Structural Information from Chemical Properties (SB p.93) 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 88 34.7 Structural Information from Chemical Properties (SB p.93) Chemical tests for different groups of organic compounds Organic compound Saturated hydrocarbons 89 Test Observation • Burn the • A blue or clear saturated yellow flame is hydrocarbon observed in a nonluminous Bunsen flame 34.7 Structural Information from Chemical Properties (SB p.93) Chemical tests for different groups of organic compounds Organic compound Test Observation Unsaturated • Burn the unsaturated • A smoky flame is hydrocarbons hydrocarbon in a observed (C = C, non-luminous Bunsen C C) flame • Add bromine in 1,1,1- • Bromine trichloroethane at decolourizes rapidly room temperature and in the absence of light 90 • Add 1% (dilute) acidified potassium manganate(VII) solution • Potassium manganate(VII) solution decolourizes rapidly 34.7 Structural Information from Chemical Properties (SB p.93) Chemical tests for different groups of organic compounds Organic compound Haloalkanes (1°, 2° or 3°) 91 Test Observation • Boil with ethanolic • For chloroalkanes, a potassium hydroxide white precipitate is solution, then acidify formed with excess dilute • For bromoalkanes, a nitric(V) acid and add pale yellow silver nitrate(V) precipitate is formed solution • For iodoalkanes, a creamy yellow precipitate is formed 34.7 Structural Information from Chemical Properties (SB p.93) Chemical tests for different groups of organic compounds Organic Test Observation compound Halobenzenes • Boil with ethanolic • No precipitate is potassium hydroxide formed solution, then acidify with excess dilute nitric(V) acid and add silver nitrate(V) solution 92 34.7 Structural Information from Chemical Properties (SB p.94) Chemical tests for different groups of organic compounds Organic compound Alcohols ( OH) 93 Test • Add a small piece of sodium metal • Esterification: Add ethanoyl chloride Observation • A colourless gas is evolved • The temperature of the reaction mixture rises • A colourless gas is evolved 34.7 Structural Information from Chemical Properties (SB p.94) Chemical tests for different groups of organic compounds Organic compound Alcohols ( OH) 94 Test • Add acidified potassium dichromate(VI) solution Observation • For 1° and 2° alcohols, the clear orange solution becomes opaque and turns green almost immediately • For 3° alcohols, there are no observable changes 34.7 Structural Information from Chemical Properties (SB p.94) Chemical tests for different groups of organic compounds Organic compound Alcohols ( OH) Test • Iodoform test for: Add iodine in sodium hydroxide solution 95 Observation • A yellow precipitate is formed 34.7 Structural Information from Chemical Properties (SB p.94) Chemical tests for different groups of organic compounds Organic compound Alcohols ( OH) 96 Test • Lucas test: add a solution of zinc chloride in concentrated hydrochloric acid Observation • 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 34.7 Structural Information from Chemical Properties (SB p.94) Chemical tests for different groups of organic compounds Organic compound Ethers ( O ) 97 Test • No specific test for ethers but they are soluble in concentrated sulphuric(VI) acid Observation 34.7 Structural Information from Chemical Properties (SB p.94) Chemical tests for different groups of organic compounds Organic Test Observation compound Aldehydes • Add aqueous sodium • Crystalline salts are hydrogensulphate(IV) formed ( ) • Add 2,4• A yellow, orange or dinitrophenylhydrazine red precipitate is formed • Silver mirror test: add Tollens’ reagent (a solution of aqueous silver nitrate in aqueous ammonia) 98 • A silver mirror is deposited on the inner wall of the test tube 34.7 Structural Information from Chemical Properties (SB p.94) Chemical tests for different groups of organic compounds Organic Test compound Ketones • Add aqueous sodium hydrogensulphate(IV) Observation • Crystalline salts are formed (for unhindered ketones ) only) • Add 2,4• A yellow, orange or dinitrophenylhydrazine red precipitate is formed ( • Iodoform test for: 99 Add iodine in sodium hydroxide solution • A yellow precipitate is formed 34.7 Structural Information from Chemical Properties (SB p.95) Chemical tests for different groups of organic compounds Organic Test compound Carboxylic • Esterification: warm acids the carboxylic acid with an alcohol in the ( ) presence of concentrated sulphuric(VI) acid, followed by adding sodium carbonate solution • Add sodium hydrogencarbonate 100 Observation • A sweet and fruity smell is detected • The colourless gas produced turns lime water milky 34.7 Structural Information from Chemical Properties (SB p.95) Chemical tests for different groups of organic compounds Organic Test compound Esters • No specific test for esters but they can be ( ) distinguished by its characteristic smell 101 Observation • A sweet and fruity smell is detected 34.7 Structural Information from Chemical Properties (SB p.95) Chemical tests for different groups of organic compounds Organic Test compound Acyl halides • Boil with ethanolic potassium hydroxide ( ) solution, then acidify with excess dilute nitric(V) acid and add silver nitrate(V) solution 102 Observation • 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 34.7 Structural Information from Chemical Properties (SB p.95) Chemical tests for different groups of organic compounds Organic Test compound Amides • Boil with sodium hydroxide solution ( ) 103 Observation • The colourless gas produced turns moist red litmus paper or pH paper blue 34.7 Structural Information from Chemical Properties (SB p.95) Chemical tests for different groups of organic compounds Organic Test Observation compound Amines • 1o aliphatic amines: • Steady evolution of dissolve the amine in N2(g) is observed (NH2) dilute hydrochloric acid at 0 – 5 oC, then add cold sodium nitrate(III) solution slowly • 1o aromatic amines: • An orange or red add naphthalen-2-ol in precipitate is formed dilute sodium hydroxide solution 104 34.7 Structural Information from Chemical Properties (SB p.95) Chemical tests for different groups of organic compounds Organic Test compound Aromatic • Burn the aromatic compounds compound in a nonluminous Bunsen ( ) flame • Add fuming sulphuric(VI) acid 105 Observation • A smoky yellow flame with black soot is produced • The aromatic compound dissolves • The temperature of the reaction mixture rises 34.7 Structural Information from Chemical Properties (SB p.96) Example 34-7A Example 34-7C 106 Example 34-7B Check Point 34-7 The END 107 34.1 Introduction (SB p.77) 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 mass spectrometry Back 108 34.2 Isolation and Purification of Organic Compounds (SB p.84) For each of the following, suggest a separation technique. (a) To obtain blood cells from blood (b) To separate different pigments in black ink (c) To obtain ethanol from beer (d) To separate a mixture of two solids, but only one sublimes (e) To separate an insoluble solid from a liquid 109 (a) (b) (c) (d) (e) Centrifugation Chromatography Fractional distillation Sublimation Filtration Back Answer 34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.87) (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 110 34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.88) 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 111 34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.88) Back 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 Mass (g) Carbon Hydrogen Oxygen 40.0 6.7 6.7 6 .7 1.0 53.3 53.3 3.33 16.0 6.7 2 3.33 3.33 1 3.33 Number of 40.0 3.33 moles (mol) 12.0 Relative 3.33 1 number of 3.33 moles Simplest 1 2 1 mole ratio 112 ∴ The empirical formula of the organic compound is CH2O. 34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.88) An organic compound Z has the following composition by mass: Element Percentage by mass (%) Carbon Hydrogen Oxygen 60.00 13.33 26.67 (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 113 34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.89) (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 Carbon Hydrogen 60.00 13.33 Number of moles (mol) 26.67 26.67 60.00 13.33 1.67 5 13.33 16.0 12.0 1.0 Relative number of moles 5 3 1.67 13.33 8 1.67 Mass (g) Oxygen 1.67 1 1.67 Simplest 3 8 1 mole ratio 114 ∴ The empirical formula of the organic compound is C3H8O. 34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.89) (b) The molecular formula of the compound is (C3H8O)n. Relative molecular mass of (C3H8O)n = 60.0 n × (12.0 × 3 + 1.0 × 8 + 16.0) = 60.0 n =1 ∴ The molecular formula of compound Z is C3H8O. Back 115 34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.89) 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 116 34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.89) Relative molecular mass of CO2 = 12.0 + 16.0 × 2 = 44.0 12.0 Mass of carbon in 0.22 g of CO2 = 0.22 g × 44.0 = 0.06 g Relative molecular mass of H2O = 1.0 × 2 + 16.0 = 18.0 Mass of hydrogen in 0.09 g of H2O = 0.09 g × 2.0 18.0 = 0.01 g Mass of oxygen in the compound = (0.15 – 0.06 – 0.01) g = 0.08 g 117 34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.89) Mass (g) Number of moles (mol) Relative number of moles Simplest mole ratio Carbon Hydrogen Oxygen 0.06 0.06 0.005 12.0 0.01 0.01 0.01 1.0 0.08 0.08 0.005 16.0 0.005 1 0.005 0.01 2 0.005 1 0.005 1 0.005 2 ∴ The empirical formula of the organic compound is CH2O. 118 1 34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.89) Let the molecular formula of the compound be (CH2O)n. Relative molecular mass of (CH2O)n = 60.0 n × (12.0 + 1.0 × 2 + 16.0) = 60.0 n =2 ∴ The molecular formula of the compound is C2H4O2. Back 119 34.6 Structural Information from Physical Properties (SB p.92) Back Why do branched-chain hydrocarbons have lower boiling points but higher melting points than the corresponding straight-chain isomers? Answer Branched-chain hydrocarbons have lower boiling points than the corresponding straight-chain isomers because the straight-chain isomers are being flattened in shape. They have greater surface area in contact with each other. Hence, molecules of the straightchain isomer are held together by greater attractive forces. On the other hand, branched-chain hydrocarbons have higher melting points than the corresponding straight-chain isomers because branched-chain isomers are more spherical in shape and are packed more efficiently in solid state. Extra energy is required to 120break down the efficient packing in the process of melting. 34.6 Structural Information from Physical Properties (SB p.92) Back Why does the solubility of amines in water decrease in the order: 1o amines > 2o amines > 3o 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. 121 34.6 Structural Information from Physical Properties (SB p.92) Match the boiling points 65oC, –6oC and –88oC with the compounds CH3CH3, CH3NH2 and CH3OH. Explain your answer briefly. Answer 122 34.6 Structural Information from Physical Properties (SB p.92) Back Compounds Boiling point (°C) CH3CH3 –88 CH3NH2 –6 CH3OH 65 Ethane (CH3CH3) is a non-polar compound. In pure liquid form, ethane molecules are held together by weak van der Waals’ forces. However, both methylamine (CH3NH2) and methanol (CH3OH) are polar substances. In pure liquid form, their molecules are held together by intermolecular hydrogen bonds. As van der Waals’ forces are much weaker than hydrogen bonds, ethane has the lowest boiling point among the three. Besides, as the O H bond in alcohols is more polar than the N H bond in amines, the hydrogen bonds formed between methylamine molecules are weaker than those formed between methanol molecules. Thus, methylamine has a lower boiling 123 point than methanol. 34.6 Structural Information from Physical Properties (SB p.92) (a) Butan-1-ol boils at 118°C and butanal boils at 76°C. (i) What are the relative molecular masses of butan-1ol and butanal? (ii) Account for the higher boiling point of butan-1-ol. (a) (i) (ii) 124 The relative molecular masses of butan-1-ol and butanal are 74.0 and 72.0 respectively. Butan-1-ol has a higher boiling point because it is able to form extensive hydrogen bonds with each other, but the forces holding the butanal molecules together are dipole-dipole interactions only. Answer 34.6 Structural Information from Physical Properties (SB p.92) (b) Arrange the following compounds in order of increasing solubility in water. Explain your answer. Ethanol, chloroethane, hexan-1-ol Answer (b) The solubility increases in the order: chloroethane < hexan-1-ol < ethanol. Both hexan-1-ol and ethanol are more soluble in water than chloroethane because molecules of the alcohols are able to form extensive hydrogen bonds with water molecules. Molecules of chloroethane are not able to form hydrogen bonds with water molecules and that is why it is insoluble in water. Hexan-1-ol has a longer carbon chain than ethanol and this explains why it is less soluble in water than ethanol. 125 34.6 Structural Information from Physical Properties (SB p.92) (c) Explain why (CH3)3N (b.p.: 2.9°C) boils so much lower than CH3CH2CH2NH2 (b.p.: 48.7°C) despite they have the same molecular mass. Answer (c) They are isomers. The primary amine is able to form hydrogen bonds with the oxygen atom of water molecules, but there is no hydrogen atoms directly attached to the nitrogen atom in the tertiary amine. 126 34.6 Structural Information from Physical Properties (SB p.92) Back (d) Match the boiling points with the isomeric carbonyl compounds. Compounds: Heptanal, heptan-4-one, 2,4-dimethylpentan-3-one Boiling points: 124°C, 144°C, 155°C (d) 127 Compound Boiling point (oC) Heptanal 155 Heptan-4-one 144 2,4-Dimethylpentan-3-one 125 Answer 34.7 Structural Information from Chemical Properties (SB p.96) The empirical formula of an organic compound is CH2O and its relative molecular mass is 60.0. It reacts with sodium hydrogencarbonate to give a colourless gas which turns lime water milky. (a) Calculate the molecular formula of the compound. Answer (a) Let the molecular formula of the compound be (CH2O)n. Relative molecular mass of (CH2O)n= 60.0 n (12.0 + 1.0 2 + 16.0) = 60.0 n =2 ∴ The molecular formula of the compound is C2H4O2. 128 34.7 Structural Information from Chemical Properties (SB p.96) The empirical formula of an organic compound is CH2O and its relative molecular mass is 60.0. It reacts with sodium hydrogencarbonate to give a colourless gas which turns lime water milky. (b) Deduce the structural formula of the compound. Answer 129 34.7 Structural Information from Chemical Properties (SB p.96) (b) 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 C2H4O2, the atoms left are one carbon and three hydrogen atoms. This obviously shows that a methyl group ( CH3) is present. Therefore, the structural formula of the compound is: 130 34.7 Structural Information from Chemical Properties (SB p.96) The empirical formula of an organic compound is CH2O and its relative molecular mass is 60.0. It reacts with sodium hydrogencarbonate to give a colourless gas which turns lime water milky. (c) Give the IUPAC name for the compound. (c) The IUPAC name for the compound is ethanoic acid. Back 131 Answer 34.7 Structural Information from Chemical Properties (SB p.96) 15 cm3 of a gaseous hydrocarbon were mixed with 120 cm3 of oxygen which was in excess. The mixture was exploded. 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 the volumes were measured under room temperature and pressure. (b) To which homologous series does the hydrocarbon belong? (c) Give the structural formula of the hydrocarbon. 132 Answer 34.7 Structural Information from Chemical Properties (SB p.97) (a) Let the molecular formula of the compound be CxHy. Volume of CxHy 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 CxHy + (x + y )O2 xCO2 + y H2O 2 4 Volume of CxHy reacted : Volume of CO2 formed = 1 : x = 15 : 30 1 15 x 30 x=2 133 34.7 Structural Information from Chemical Properties (SB p.97) (a) Volume of CxHy reacted : Volume of O2 reacted = 1 : ( x y ) 4 = 15 : 45 1 15 45 y (2 ) 4 y 2 3 4 y=4 The molecular formula of the compound is C2H4. (b) C2H4 belongs to alkenes. (c) The structural formula of the hydrocarbon is: Back 134 34.7 Structural Information from Chemical Properties (SB p.97) Answer 20 cm3 of a gaseous organic compound containing only carbon, hydrogen and oxygen were mixed with 110 cm3 of oxygen which was in excess. The mixture was exploded at 105oC and the volume of the gaseous mixture was 150 cm3. After cooling to room temperature, the residual volume was reduced to 90 cm3. On adding concentrated potassium hydroxide solution, the volume further decreased to 50 cm3. (a) Calculate the molecular formula of the compound, assuming that all the volumes were measured under room temperature and pressure. (b) The compound is found to contain a hydroxyl group ( OH) in its structure. Deduce its structural formula. (c) Is the compound optically active? Explain your answer. 135 34.7 Structural Information from Chemical Properties (SB p.97) (a) Let the molecular formula of the compound be CxHyOz. Volume of CxHyOz reacted = 20 cm3 Volume of unreacted oxygen = 50 cm3 Volume of oxygen reacted = (110 - 50) cm3 = 60 cm3 Volume of carbon dioxide formed = (90 - 50) cm3 = 40 cm3 Volume of water (in the form of steam) formed = (90 - 50) cm3 = 40 cm3 y CxHyOz + (x + - z )O2 xCO2 + y H2O 4 2 2 Volume of CxHyOz reacted : Volume of CO2 formed = 1 : x = 20 : 40 1 20 x 40 x=2 136 34.7 Structural Information from Chemical Properties (SB p.98) y (a) Volume of CxHyOz reacted : Volume of H2O formed = 1 : = 20 : 60 2 2 20 y 60 y=6 y z Volume of CxHyOz reacted : Volume of O2 reacted = 1 : ( x ) 4 2 = 20 : 60 1 20 y z 60 (x - ) 4 2 6 z 2 3 4 2 z=1 The molecular formula of the compound is C2H6O. 137 34.7 Structural Information from Chemical Properties (SB p.98) Back (b) As the compound contains a OH group, the hydrocarbon skeleton of the compound becomes C2H5 after eliminating the OH group from the molecular formula of C2H6O. The structural formula of the compound is: (c) The compound is optically inactive as both carbon atoms in the compound are not asymmetric, i.e. both of them do not attach to four different atoms or groups of atoms. 138 34.7 Structural Information from Chemical Properties (SB p.99) (a) A substance contains 42.8% carbon, 2.38% hydrogen, 16.67% nitrogen by mass and the remainder consists of oxygen. (i) Given that the relative molecular mass of the substance is 168.0, deduce the molecular formula of the substance. (ii) The substance is proved to be an aromatic compound with only one type of functional group. Give the names and structural formulae for all isomers of the substance. Answer 139 34.7 Structural Information from Chemical Properties (SB p.99) (a) (i) Let the mass of the compound be 100 g. Mass (g) Carbon Hydrogen Nitrogen Oxygen 42.8 2.38 16.67 38.15 Number of 38.15 42.8 16.67 2.38 2.38 3.57 1.19 2.38 moles (mol) 12.0 16 . 0 14 . 0 1 .0 Relative 2.38 3.57 2.38 1.19 2 3 number of 2 1 1.19 1.19 1.19 1.19 moles Simplest mole ratio 3 2 1 ∴ The empirical formula of the compound is C3H2NO2. 140 2 34.7 Structural Information from Chemical Properties (SB p.99) (a) (i) (ii) 141 Let the molecular formula of the compound be (C3H2NO2)n. Molecular mass of (C3H2NO2)n = 168.0 n × (12.0 × 3 + 1.0 × 2 + 14.0 + 16.0 × 2) = 168.0 ∴ n =2 ∴ The molecular formula of the compound is C6H4N2O4. 34.7 Structural Information from Chemical Properties (SB p.99) (b) 30 cm3 of a gaseous hydrocarbon were mixed with 140 cm3 of oxygen which was in excess, and the mixture was then exploded. After cooling to room temperature, the residual gases occupied 95 cm3 by volume. By adding potassium hydroxide solution, the volume was reduced by 60 cm3. The remaining gas was proved to be oxygen. (i) Determine the molecular formula of the hydrocarbon. (ii) Is the hydrocarbon a saturated, an unsaturated or an aromatic hydrocarbon? Answer 142 34.7 Structural Information from Chemical Properties (SB p.99) (b) (i) Volume of hydrocarbon reacted = 30 cm3 Volume of unreacted oxygen = (95 – 60) cm3 = 35 cm3 Volume of oxygen reacted = (140 - 35) cm3 = 105 cm3 Volume of carbon dioxide formed = 60 cm3 CxHy + (x + y )O2 xCO2 + y H2O 2 4 Volume of CxHy reacted : Volume of CO2 formed = 1 : x = 30 : 60 1 30 x 60 x=2 143 34.7 Structural Information from Chemical Properties (SB p.99) (b) (i) (ii) 144 Volume of CxHy reacted : Volume of O2 reacted = 1 : ( 2 y ) = 30 : 105 4 1 30 y 105 (x ) 4 y 30 (2 ) 105 4 y=6 The molecular formula of the compound is C2H6. From the molecular formula of the hydrocarbon, it can be deduced that the hydrocarbon is saturated because it fulfils the general formula of alkanes CnH2n+2. 34.7 Structural Information from Chemical Properties (SB p.99) (c) A hydrocarbon having a relative molecular mass of 56.0 contains 85.5% carbon and 14.5% hydrogen by mass. Detailed analysis shows that it has two geometrical isomers. (i) Deduce the molecular formula of the hydrocarbon. (ii) Name the two geometrical isomers of the hydrocarbon. (iii) Explain the existence of geometrical isomerism in the hydrocarbon. Answer 145 34.7 Structural Information from Chemical Properties (SB p.99) (c) (i) Let the mass of the compound be 100 g. Mass (g) Number of moles (mol) Relative number of moles Simplest mole ratio Carbon Hydrogen 85.5 14.5 85.5 7.125 12.0 7.125 1 7.125 1 14.5 14.5 1 .0 14.5 2 7.125 2 ∴ The empirical formula of the compound is CH2. 146 34.7 Structural Information from Chemical Properties (SB p.99) Back (c) (i) Let the molecular formula of the hydrocarbon be (CH2)n. Molecular mass of (CH2)n = 56.0 n × (12.0 + 1.0 × 2) = 56.0 n =4 ∴ The molecular formula of the hydrocarbon is C4H8. (ii) (iii) Since but-2-ene is unsymmetrical and free rotation of but-2-ene is restricted by the presence of the carbon-carbon double bond, geometrical isomerism exists. 147 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.102) What is the relationship between frequency and wavenumber? Answer The higher the frequency, the higher the wavenumber. Back 148