34 34.1 34.2 34.3 34.4 34.5 34.6 34.7 34.8 1 34.9 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 New Way Chemistryto forObtain Hong KongStructural A-Level 3B Information 34.1 Introduction 2 New Way Chemistry for Hong Kong A-Level 3B 34.1 Introduction (SB p.77) Introduction • The determination of the structure of an organic compound involves: 1. Isolation and purification of the compound 2. Qualitative analysis of the elements present in the compound 3. Determination of the molecular formula of the compound 4. Determination of the functional group present in the compound 3 New Way Chemistry for Hong Kong A-Level 3B 34.1 Introduction (SB p.77) Introduction The general steps to determine the structure of an organic compound Check Point 34-1 4 New Way Chemistry for Hong Kong A-Level 3B 34.2 Isolation and Purification of Organic Compounds 5 New Way Chemistry for Hong Kong A-Level 3B 34.2 Isolation and Purification of Organic Compounds (SB p.78) Isolation and Purification of Organic Compounds • These techniques include: 1. Filtration 2. Centrifugation 3. Crystallization 4. Solvent extraction 5. Distillation 6 New Way Chemistry for Hong Kong A-Level 3B 34.2 Isolation and Purification of Organic Compounds (SB p.78) Isolation and Purification of Organic Compounds • These techniques include: 5. Fractional distillation 6. Sublimation 7. Chromatography 7 New Way Chemistry for Hong Kong A-Level 3B 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 8 New Way Chemistry for Hong Kong A-Level 3B 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 9 New Way Chemistry for Hong Kong A-Level 3B 34.2 Isolation and Purification of Organic Compounds (SB p.78) Filtration The laboratory set-up of filtration 10 New Way Chemistry for Hong Kong A-Level 3B 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 11 New Way Chemistry for Hong Kong A-Level 3B 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 12 New Way Chemistry for Hong Kong A-Level 3B 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 13 New Way Chemistry for Hong Kong A-Level 3B 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 14 New Way Chemistry for Hong Kong A-Level 3B 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 15 New Way Chemistry for Hong Kong A-Level 3B 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 16 New Way Chemistry for Hong Kong A-Level 3B 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 17 New Way Chemistry for Hong Kong A-Level 3B 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 18 New Way Chemistry for Hong Kong A-Level 3B 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 19 New Way Chemistry for Hong Kong A-Level 3B 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 20 New Way Chemistry for Hong Kong A-Level 3B 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 21 New Way Chemistry for Hong Kong A-Level 3B 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 22 New Way Chemistry for Hong Kong A-Level 3B 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 23 New Way Chemistry for Hong Kong A-Level 3B 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 24 New Way Chemistry for Hong Kong A-Level 3B 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 25 New Way Chemistry for Hong Kong A-Level 3B 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 26 New Way Chemistry for Hong Kong A-Level 3B 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 27 New Way Chemistry for Hong Kong A-Level 3B 34.2 Isolation and Purification of Organic Compounds (SB p.81) Distillation The laboratory set-up of distillation 28 New Way Chemistry for Hong Kong A-Level 3B 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 29 New Way Chemistry for Hong Kong A-Level 3B 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 30 New Way Chemistry for Hong Kong A-Level 3B 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 31 New Way Chemistry for Hong Kong A-Level 3B 34.2 Isolation and Purification of Organic Compounds (SB p.82) Fractional Distillation The laboratory set-up of fractional distillation 32 New Way Chemistry for Hong Kong A-Level 3B 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 33 New Way Chemistry for Hong Kong A-Level 3B 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 34 New Way Chemistry for Hong Kong A-Level 3B 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 • 35 Fractions with different boiling points can be collected separately New Way Chemistry for Hong Kong A-Level 3B 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 36 New Way Chemistry for Hong Kong A-Level 3B 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 37 New Way Chemistry for Hong Kong A-Level 3B 34.2 Isolation and Purification of Organic Compounds (SB p.83) Sublimation A mixture of two compounds can be separated by sublimation 38 New Way Chemistry for Hong Kong A-Level 3B 34.2 Isolation and Purification of Organic Compounds (SB p.83) Chromatography 39 • An effective method of separating a complex mixture of substances • Paper chromatography is a common type of chromatography New Way Chemistry for Hong Kong A-Level 3B 34.2 Isolation and Purification of Organic Compounds (SB p.83) Chromatography The laboratory set-up of paper chromatography 40 New Way Chemistry for Hong Kong A-Level 3B 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 41 New Way Chemistry for Hong Kong A-Level 3B 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 42 New Way Chemistry for Hong Kong A-Level 3B 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 43 New Way Chemistry for Hong Kong A-Level 3B 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 44 New Way Chemistry for Hong Kong A-Level 3B 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 45 New Way Chemistry for Hong Kong A-Level 3B 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 46 New Way Chemistry for Hong Kong A-Level 3B 34.3 Tests for Purity 47 New Way Chemistry for Hong Kong A-Level 3B 34.3 Tests for Purity (SB p.84) 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 48 New Way Chemistry for Hong Kong A-Level 3B 34.3 Tests for Purity (SB p.85) Determination of Melting Point • To determine the melting point of a solid, some of the dry solid is placed in a thin-walled glass melting point tube 49 • The tube is attached to a thermometer • The temperature at which the solid melts is its melting point New Way Chemistry for Hong Kong A-Level 3B 34.3 Tests for Purity (SB p.85) Determination of Melting Point Determination of the melting point of a solid using an oil bath 50 New Way Chemistry for Hong Kong A-Level 3B 34.3 Tests for Purity (SB p.85) Determination of Melting Point • A pure solid has a sharp melting point melting occurs within a narrow temperature range (usually less than 0.5°C) • An impure solid does not have a sharp melting point melts gradually over a wide temperature range 51 New Way Chemistry for Hong Kong A-Level 3B 34.3 Tests for Purity (SB p.85) Determination of Melting Point • The presence of impurities lowers the melting point of a solid • Melting point is a useful indication of the purity of a substance 52 New Way Chemistry for Hong Kong A-Level 3B 34.3 Tests for Purity (SB p.85) Determination of Boiling Point • The boiling point of a liquid can be determined by using the distillation apparatus • The temperature at which the liquid boils steadily is its boiling point • A flammable liquid should be heated in a water bath, instead of heated with a naked flame 53 New Way Chemistry for Hong Kong A-Level 3B 34.3 Tests for Purity (SB p.85) Determination of Boiling Point • The boiling point of a pure liquid is quite sharp • The presence of non-volatile solutes such as salts raises the boiling point of a liquid 54 New Way Chemistry for Hong Kong A-Level 3B 34.4 Qualitative Analysis of Elements in an Organic Compound 55 New Way Chemistry for Hong Kong A-Level 3B 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 56 New Way Chemistry for Hong Kong A-Level 3B 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 57 New Way Chemistry for Hong Kong A-Level 3B 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 58 New Way Chemistry for Hong Kong A-Level 3B 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 59 New Way Chemistry for Hong Kong A-Level 3B 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 60 New Way Chemistry for Hong Kong A-Level 3B 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 61 New Way Chemistry for Hong Kong A-Level 3B 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). iodide ion (I-) 62 A creamy yellow precipitate is formed. It is insoluble in excess New Way Chemistry for Hong Kong A-Level 3B NH3(aq). 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 63 New Way Chemistry for Hong Kong A-Level 3B 34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data 64 New Way Chemistry for Hong Kong A-Level 3B 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 65 New Way Chemistry for Hong Kong A-Level 3B 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 66 New Way Chemistry for Hong Kong A-Level 3B 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 67 New Way Chemistry for Hong Kong A-Level 3B 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 68 New Way Chemistry for Hong Kong A-Level 3B 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 69 New Way Chemistry for Hong Kong A-Level 3B 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 70 New Way Chemistry for Hong Kong A-Level 3B 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 71 New Way Chemistry for Hong Kong A-Level 3B 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 72 New Way Chemistry for Hong Kong A-Level 3B 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 73 New Way Chemistry for Hong Kong A-Level 3B 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 74 New Way Chemistry for Hong Kong A-Level 3B 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 75 New Way Chemistry for Hong Kong A-Level 3B 34.6 Structural Information from Physical Properties 76 New Way Chemistry for Hong Kong A-Level 3B 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 77 New Way Chemistry for Hong Kong A-Level 3B 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 78 New Way Chemistry for Hong Kong A-Level 3B 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 79 New Way Chemistry for Hong Kong A-Level 3B 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 80 New Way Chemistry for Hong Kong A-Level 3B 34.6 Structural Information from Physical Properties (SB p.90) Physical properties of some common organic compounds Organic compound Hydrocarbo ns (saturated and unsaturated) 81 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 New Way Chemistry for Hong Kong A-Level 3B 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 82 Melting point and boiling point Generally low Solubility In water In nonor highly polar polar organic solvents solvents Insoluble New Way Chemistry for Hong Kong A-Level 3B Soluble 34.6 Structural Information from Physical Properties (SB p.90) Physical properties of some common organic compounds Organic Density at compound 20oC Haloalkanes 83 Melting point and boiling point Solubility In water In nonor highly polar polar organic solvents solvents • 0.9 - 1.1 • Higher than alkanes of Insoluble g cm–3 similar relative for molecular masses ( haloalkane chloroalkanes molecules are polar) • >1.0 g • All haloalkanes are cm–3 for liquids except bromohalomethanes alkanes • Both the m.p. and b.p. and increase in the order: iodoRCH2F < RCH2Cl < New Way Chemistry for Hong Kong A-Level 3B alkanes RCH2Br < RCH2I 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) 84 Solubility In water or highly polar solvents • Lower members: Completely miscible with water ( formation of hydrogen bonds between alcohol molecules and New Way Chemistry for Hong Kong A-Level 3B water molecules) In nonpolar organic solvents 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 85 Solubility In water or highly polar solvents In nonpolar organic solvents • Solubility Soluble decreases gradually as the hydrocarbon chain lengthens New Way Chemistry for Hong Kong A-Level 3B 34.6 Structural Information from Physical Properties (SB p.91) Physical properties of some common organic compounds Organic compound Density at 20oC Melting point and boiling point Carbonyl • <1.0 g cm–3 for aliphatic compcarbonyl ounds compounds (aldehydes and ketones) 86 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) New Way Chemistry for Hong Kong A-Level 3B 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) 34.6 Structural Information from Physical Properties (SB p.91) Physical properties of some common organic compounds Organic compound Density at 20oC Melting point and boiling point Carbonyl • > 1.0 g cm–3 for aromatic compcarbonyl ounds compounds (aldehydes and ketones) 87 Solubility In water or highly polar solvents In nonpolar organic solvents • Solubility decreases gradually as the hydrocarbon chain lengthens Soluble New Way Chemistry for Hong Kong A-Level 3B 34.6 Structural Information from Physical Properties (SB p.91) Physical properties of some common organic compounds Organic compound Carboxylic acids 88 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 Solubility In water or highly polar solvents • First four members are miscible with water in all proportions • Solubility decreases gradually as the hydrocarbon chain New Way Chemistry for Hong Kong A-Level 3B lengthens Higher than alcohols of similar relative molecular masses ( the formation of more extensive intermolecular hydrogen bonds) 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 89 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 New Way Chemistry for Hong Kong A-Level 3B 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 90 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 New Way Chemistry for Hong Kong A-Level 3B 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 91 Density at 20oC Melting point and boiling point Solubility In water or highly polar solvents • 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) New Way Chemistry for Hong Kong A-Level 3B In nonpolar organic solvents 34.6 Structural Information from Physical Properties (SB p.91) Physical properties of some common organic compounds Organic compound Amines 92 Density at 20oC Melting point and boiling point Solubility In water or highly polar solvents • 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) New Way Chemistry for Hong Kong A-Level 3B In nonpolar organic solvents 34.6 Structural Information from Physical Properties (SB p.92) Example 34-6 93 Check Point 34-6 New Way Chemistry for Hong Kong A-Level 3B 34.7 Structural Information from Chemical Properties 94 New Way Chemistry for Hong Kong A-Level 3B 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 95 New Way Chemistry for Hong Kong A-Level 3B 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: 96 New Way Chemistry for Hong Kong A-Level 3B 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 97 New Way Chemistry for Hong Kong A-Level 3B 34.7 Structural Information from Chemical Properties (SB p.93) Chemical tests for different groups of organic compounds Organic compound Saturated hydrocarbons 98 Test Observation • Burn the • A blue or clear saturated yellow flame is hydrocarbon observed in a nonluminous Bunsen flame New Way Chemistry for Hong Kong A-Level 3B 34.7 Structural Information from Chemical Properties (SB p.93) Chemical tests for different groups of organic compounds Organic Test Observation compound 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 99 • Add 1% (dilute) • Potassium acidified potassium manganate(VII) manganate(VII) solution decolourizes New Way Chemistry for Hong Kong A-Level 3B solution 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°) 100 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 New Way Chemistry for Hong Kong A-Level 3B 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 101 New Way Chemistry for Hong Kong A-Level 3B 34.7 Structural Information from Chemical Properties (SB p.94) Chemical tests for different groups of organic compounds Organic compound Alcohols ( OH) 102 Test Observation • 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 New Way Chemistry for Hong Kong A-Level 3B 34.7 Structural Information from Chemical Properties (SB p.94) Chemical tests for different groups of organic compounds Organic compound Alcohols ( OH) 103 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 New Way Chemistry for Hong Kong A-Level 3B 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: Observation • A yellow precipitate is formed Add iodine in sodium hydroxide solution 104 New Way Chemistry for Hong Kong A-Level 3B 34.7 Structural Information from Chemical Properties (SB p.94) Chemical tests for different groups of organic compounds Organic compound Alcohols ( OH) 105 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 New Way Chemistry for Hong Kong A-Level 3B 34.7 Structural Information from Chemical Properties (SB p.94) Chemical tests for different groups of organic compounds Organic compound Ethers ( O ) 106 Test • No specific test for ethers but they are soluble in concentrated sulphuric(VI) acid New Way Chemistry for Hong Kong A-Level 3B 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) 107 • A silver mirror is deposited on the inner wall of the test tube New Way Chemistry for Hong Kong A-Level 3B 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: 108 • A yellow precipitate is formed Add iodine in sodium New Way Chemistry for Hong Kong A-Level 3B hydroxide solution 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 109 Observation • A sweet and fruity smell is detected • The colourless gas produced turns lime water milky New Way Chemistry for Hong Kong A-Level 3B 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 110 Observation • A sweet and fruity smell is detected New Way Chemistry for Hong Kong A-Level 3B 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 111 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 New Way Chemistry for Hong Kong A-Level 3B 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 ( ) 112 Observation • The colourless gas produced turns moist red litmus paper or pH paper blue New Way Chemistry for Hong Kong A-Level 3B 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 113 New Way Chemistry for Hong Kong A-Level 3B 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 114 Observation • A smoky yellow flame with black soot is produced • The aromatic compound dissolves • The temperature of the reaction mixture rises New Way Chemistry for Hong Kong A-Level 3B 34.7 Structural Information from Chemical Properties (SB p.96) Example 34-7A Example 34-7C 115 Example 34-7B Check Point 34-7 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectrocopy in the Identification of Functional Groups 116 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.99) The Electromagnetic Spectrum • Electromagnetic radiation has dual property i.e. the properties of both wave and particle • Can be described as a wave occurring simultaneously in electrical and magnetic fields • Can also be described as consisting of particles called quanta or photons 117 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.99) The Electromagnetic Spectrum • All electromagnetic radiation travels through vacuum at the same velocity, 3 108 m s-1 • The relationship between the frequency () of an electromagnetic radiation, its wavelength () and velocity (c) is: ν c λ 118 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.100) The Electromagnetic Spectrum • The energy of a quantum of electromagnetic radiation is directly related to its frequency: E = h where h is the Planck constant (i.e. 6.626 10-34 J s). 119 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.100) The Electromagnetic Spectrum • ν c λ the energy of a quantum of electromagnetic radiation is inversely proportional to its wavelength: E hc λ 120 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.100) The Electromagnetic Spectrum • Electromagnetic radiation of long wavelength has low energy • Electromagnetic radiation of short wavelength has high energy 121 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.100) The Electromagnetic Spectrum • Visible light has wavelength between 400 nm and 800 nm • Infra-red radiation has wavelength between 800 nm and 300 m 122 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.100) The Electromagnetic Spectrum Regions of the electromagnetic spectrum 123 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.100) The Electromagnetic Spectrum • When electromagnetic radiation falls onto a hydrogen atom, the electron in the hydrogen atom will absorb a definite amount of energy • The electron is excited from the ground state to a higher energy level 124 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.100) The Electromagnetic Spectrum • The electron is unstable at a higher energy level it will fall back to a lower energy level • Excess energy is given out in the form of electromagnetic radiation 125 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.100) The Electromagnetic Spectrum • The radiation emitted has the frequency as shown by the following relationship: E = E2 – E1 = h = hc λ 126 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.100) The Electromagnetic Spectrum • The atomic spectrum of hydrogen is originated from electron transitions between energy levels in a hydrogen atom 127 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.101) The Electromagnetic Spectrum • In the case of molecules, the absorption of energy can cause the excitation of electrons increase the extent of vibration of the bonds and the speed of rotation of the molecule • This is the basis of infra-red spectroscopy 128 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.101) Infra-red Spectroscopy • Organic compounds absorb electromagnetic radiation in the IR region of the spectrum IR radiation does not have sufficient energy to cause the excitation of electrons 129 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.101) Infra-red Spectroscopy • IR radiation causes atoms and groups of atoms of organic compounds to vibrate with increased amplitude about the covalent bonds that connect them 130 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.101) Infra-red Spectroscopy • These vibrations are quantized the compounds absorb IR radiation of a particular amount of energy only 131 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.101) Infra-red Spectroscopy Effect of absorption of IR radiation on vibration of atoms in a molecule 132 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.101) Infra-red Spectroscopy • Infra-red spectrometer is used to measure the amount of energy absorbed at each wavelength of the IR region An infra-red spectrometer 133 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.101) Infra-red Spectroscopy • A beam of IR radiation is passed through the sample the intensity of the emergent radiation is carefully measured • The spectrometer plots the results as a graph called infra-red spectrum shows the absorption of IR radiation by a sample at different frequencies 134 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.101) Infra-red Spectroscopy • The IR radiation is usually specified by its wavenumber (unit: cm-1) the reciprocal of wavelength • Frequency and wavelength are related by the equation c = Wavenumber is a direct measure of frequency 135 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.101) Infra-red Spectroscopy • Covalently bonded atoms have only particular vibrational energy levels the levels are quantized 136 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.101) Infra-red Spectroscopy • When the compound absorbs IR radiation of the exact energy required (or a particular wavelength or a particular frequency) the excitation of a molecule from one vibrational energy level to another occurs only 137 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.101) Infra-red Spectroscopy • Molecules can vibrate in a variety of ways • Two atoms joined by a covalent bond can undergo a stretching vibration where the atoms move back and forth as if they were joined by a spring A stretching vibration of two atoms 138 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.102) Infra-red Spectroscopy A variety of stretching and bending vibrations 139 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.102) Infra-red Spectroscopy • The frequency of a given stretching vibration of a covalent bond depends on the masses of the bonded atoms and the strength of the bond • Lighter atoms vibrate at higher frequencies than heavier ones 140 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.102) Infra-red Spectroscopy • 141 The stretching vibrations of single bonds involving hydrogen (C H, O H and N H) occur at relatively high frequencies Bond Range of wavenumber (cm-1) CH 2840 – 3095 OH 3230 – 3670 NH 3350 – 3500 Characteristic absorption wavenumbers of some single bonds in infra-red spectra New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.102) Infra-red Spectroscopy • Triple bonds are stronger and vibrate at higher frequencies than double bonds Bond Range of wavenumber (cm-1) CC 2070 – 2250 CN 2200 – 2280 C=C 1610 – 1680 C=O 1680 – 1750 Characteristic absorption wavenumbers of some double bonds and triple bonds in infra-red spectra 142 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.102) Infra-red Spectroscopy • The IR spectra of even relatively simple compounds contain many absorption peaks • The possibility of two different compounds having the same IR spectrum is very small • An IR spectrum has been called the “fingerprint” of a compound 143 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.103) Use of IR Spectrum in the Identification of Functional Groups • An IR spectrum is a plot of percentage of transmittance against wavenumber of IR radiation 144 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.103) Use of IR Spectrum in the Identification of Functional Groups The IR spectrum of hex-1-yne 145 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.103) Use of IR Spectrum in the Identification of Functional Groups • 100% transmittance in the spectrum implies no absorption of IR radiation 146 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.103) Use of IR Spectrum in the Identification of Functional Groups • When a compound absorbs IR radiation, the intensity of transmitted radiation decreases results in a decrease in percentage of transmittance a dip in the spectrum often called an absorption peak or absorption band 147 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.103) Use of IR Spectrum in the Identification of Functional Groups • In general, an IR spectrum can be split into four regions for interpretation purpose 148 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.103) The four regions of an IR spectrum Range of wavenumber (cm-1) Interpretation 400 – 1500 • Often consists of many complicated bands • Unique to each compound • Often called the fingerprint region • Not used for identification of particular functional groups 1500 – 2000 Absorption of double bonds, e.g. C = C, C = O 2000 – 2500 Absorption of triple bonds, e.g. C C, C N 2500 – 4000 149 Absorption of single bonds involving New Way Chemistry for Hong Kong A-Level 3B hydrogen, e.g. C H, O H, N H 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.104) Use of IR Spectrum in the Identification of Functional Groups • The region between 4 000 cm-1 and 1 500 cm-1 is often used for identification of functional groups from their characteristic absorption wavenumbers 150 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.104) Characteristic range of wavenumbers of covalent bonds in IR spectra Compound Bond Alkenes Aldehydes, ketones, acids, esters Alkynes Nitriles Acids (hydrogen-bonded) C=C C=O Characteristic range of wavenumber (cm-1) 1610 – 1680 1680 – 1750 CC CN OH 2070 – 2250 2200 – 2280 2500 – 3300 Alkanes, alkenes, arenes Alcohols, phenols (hydrogen-bonded) CH OH 2840 – 3095 3230 – 3670 Primary amines NH 3350 – 3500 151 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.104) Interpretation of IR Spectra 1. Butane The IR spectrum of butane 152 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.104) 1. Butane Wavenumber (cm-1) Intensity 2968 2890 Very strong Medium 1468 Strong Indication CH stretching C H bending Interpretation of the IR spectrum of butane 153 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.105) 2. cis-But-2-ene The IR spectrum of cis-but-2-ene 154 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.105) 2. cis-But-2-ene Wavenumber (cm-1) Intensity Indication 3044 Very strong 3028 Very strong C H stretching (sp2 C H) 2952 Very strong 1677 Medium 1657 Medium 1411 Strong C H stretching (sp3 C H) C = C stretchinh C H bending Interpretation of the IR spectrum of cis-but-2-ene 155 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.105) 3. Hex-1-yne The IR spectrum of hex-1-yne 156 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.105) 3. Hex-1-yne Wavenumber (cm-1) Intensity 3313 Very strong 2963 Very strong 2938 Very strong 2874 Strong 2119 Strong C C stretching 1468 Strong 1445 Medium C H bending (sp C H) C H bending (sp3 C H) 157 Indication C H stretching (sp C H) C H stretching (sp3 C H) New Way Chemistry for Hong Kong A-Level 3B Interpretation of the IR spectrum of hex-1-yne 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.106) 4. Butanone The IR spectrum of butanone 158 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.106) 4. Butanone Wavenumber (cm-1) Intensity 2983 Strong 2925 Strong 1720 1416 Indication C H stretching Very strong C = O stretching Medium C H bending (shifted as adjacent to C = O) Interpretation of the IR spectrum of butanone 159 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.107) 5. Butan-1-ol The IR spectrum of butan-1-ol 160 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.107) 5. Butan-1-ol Wavenumber (cm-1) 3330 Intensity Broad band 2960 Medium 2935 Medium 2875 Medium Indication OH stretching CH stretching Interpretation of the IR spectrum of butan-1-ol 161 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.107) 6. Butanoic Acid The IR spectrum of butanoic acid 162 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.107) 6. Butanoic Acid Wavenumber (cm-1) 3100 1708 Intensity Indication Broad band O H stretching Strong C=O stretching Interpretation of the IR spectrum of butanoic acid 163 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.108) 6. Butanoic Acid • The absorption of the O H group in alcohols and carboxylic acids does not usually appear as a sharp peak a broad band is observed the vibration of the O H group is complicated by the hydrogen bonding formed between the molecules 164 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.108) 7. Butylamine The IR spectrum of butylamine 165 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.108) 7. Butylamine Wavenumber (cm-1) Intensity Indication 3371 Strong N H stretching 3280 Strong 2960 – 2875 Weak 1610 Medium N H bending 1475 Medium C H bending C H stretching Interpretation of the IR spectrum of butylamine 166 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.108) 8. Butanenitrile The IR spectrum of butanenitrile 167 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.109) 8. Butanenitrile Wavenumber (cm-1) 2990 – 2895 Intensity 2246 Very strong 1420 1480 Strong Strong Strong Indication CH stretching CN stretching CH bending Interpretation of the IR spectrum of butanenitrile 168 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.109) Strategies for the Use of IR Spectra in the Identification of Functional Groups 1. Focus at the IR absorption peak at or above 1500 cm–1 Concentrate initially on the major absorption peaks 169 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.109) Strategies for the Use of IR Spectra in the Identification of Functional Groups 2. For each absorption peak, try to list out all the possibilities using a table or chart Not all absorption peaks in the spectrum can be assigned 170 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.109) Strategies for the Use of IR Spectra in the Identification of Functional Groups 3. The absence and presence of absorption peaks at some characteristic ranges of wavenumbers are equally important the absence of particular absorption peaks can be used to eliminate the presence of certain functional groups or bonds in the molecule 171 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.109) Limitation of the Use of IR Spectroscopy in the Identification of Organic Compounds 1. Some IR absorption peaks have very close wavenumbers and the peaks always coalesce 2. Not all vibrations give rise to strong absorption peaks 172 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.109) Limitation of the Use of IR Spectroscopy in the Identification of Organic Compounds 3. Not all absorption peaks in a spectrum can be associated with a particular bond or part of the molecule 4. Intermolecular interactions in molecules can result in complicated infra-red spectra 173 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.110) Example 34-8 174 Check Point 34-8 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information 175 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.113) Mass Spectrometry • One of the most sensitive and versatile analytical tools • More sensitive than other spectroscopic methods (e.g. IR spectroscopy) • Only a microgram or less of materials is required for the analysis 176 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.113) Mass Spectrometry In a mass spectrometric analysis, it involves: 1. the conversion of molecules to ions 2. separation of the ions formed according to their mass-to-charge (m/e) ratio m is the mass of the ion in atomic mass units and e is its charge 177 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.113) Mass Spectrometry • Finally, the number of ions of each type (i.e. the relative abundance of ions of each type) is determined • The analysis is carried out using a mass spectrometer 178 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114) Mass Spectrometry Components of a mass spectrometer 179 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114) Mass Spectrometry In the vaporization chamber, • the sample is heated until it vaporizes changes to the gaseous state 180 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114) Mass Spectrometry • The molecules in the gaseous state are bombarded with a beam of fast-moving electrons Positively-charged ions called the molecular ions are formed One of the electrons of the molecule is knocked off 181 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114) Mass Spectrometry • Molecular ions are sometimes referred to as the parent ion 182 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114) Mass Spectrometry • one of the electrons is removed from the molecules during the ionization process the molecular ion contains a single unpaired electron the molecular ion is not only a cation, it is also a free radical 183 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114) Mass Spectrometry • e.g. if a molecule of methanol (CH3OH) is bombarded with a beam of fast-moving electrons the following reaction will take place: 184 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114) Mass Spectrometry • The molecular ions formed in the ionization chamber are energetically unstable undergo fragmentation • Fragmentation can take place in a variety of ways depend on the nature of the particular molecular ion 185 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114) Mass Spectrometry • The way that a molecular ion fragments give us highly useful information about the structure of a complex molecule 186 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114) Mass Spectrometry • The positively charged ions formed are then accelerated by electric field and deflected by magnetic field causes the ions to arrive the ion detector • 187 The lighter the ions, the greater the deflection New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114) Mass Spectrometry 188 • Positively charged ions of higher charge have greater deflection • Ions with a high m/e ratio are deflected to smaller extent than ions with a low m/e ratio New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115) Mass Spectrometry • In the ion detector, the number of ions collected is measured electronically • The intensity of the signal is a measure of the relative abundance of the ions with a particular m/e ratio 189 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115) Mass Spectrometry • The spectrometer shows the results by plotting a series of peaks of varying intensity each peak corresponds to ions of a particular m/e ratio • The graph obtained is known as a mass spectrum 190 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115) Mass Spectrum • 191 Generally published as bar graphs New Way Chemistry for Hong Kong A-Level 3B . Mass spectrum of methanol 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115) Mass Spectrum Corresponding ion H3C+ m/e ratio 15 H CO+ 29 H2C = OH+ 31 CH3OH 32 Interpretation of the mass spectrum of methanol 192 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115) Formation of Fragments • The molecular ions formed in the ionization chamber are energetically unstable Some of them may break up into smaller fragments Called the daughter ions 193 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115) Formation of Fragments • These ionized fragments are accelerated and deflected by the electric field and magnetic field • Finally, they are detected by the ion detector and their m/e ratios are measured explains why there are so many peaks appeared in mass spectra 194 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Fragments • The peak at m/e 31 the most intense peak • Arbitrarily assigned an intensity of 100% Called the base peak Mass spectrum of methanol 195 Corresponds to the most common ion New Way Chemistry for Hong Kong A-Level 3B formed 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Fragments • The peak at m/e 31 corresponds to the ion H2C = OH+ formed by losing one hydrogen atom from the molecular ion 196 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Fragments • The ion H2C = OH+ is a relatively stable ion the positive charge is not localized on a particular atom it spreads around the carbon and the oxygen atoms to form a delocalized system 197 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Fragments • The peak at m/e 29 corresponds to the ion HC O+ formed by losing two hydrogen atoms from the ion H2C = OH+ 198 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Fragments • The ion HC O+ has two resonance structures: 199 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Fragments • The peak at m/e 15 corresponds to the ion H3C+ formed by the breaking of the C O bond in the molecular ion 200 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Fragments Mass spectrum of pentan-3-one 201 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Fragments Corresponding ion m/e ratio CH3CH2+ 29 CH3CH2CO+ 57 CH3CH2COCH2CH3 86 Interpretation of the mass spectrum of pentan-3-one 202 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116) Formation of Fragments • The fragmentation pattern of pentan-3-one is summarized below: 203 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118) Example 34-9A Example 34-9B Example 34-9C 204 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.121) Fragmentation Pattern 1. Straight-chain Alkanes • Simple alkanes tend to undergo fragmentation by the initial loss of a • CH3 to give a peak at M – 15 This carbocation can then undergo stepwise cleavage down the alkyl chain 205 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.121) 1. Straight-chain Alkanes • 206 Take hexane as an example: New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.121) 2. Branched-chain Alkanes • Tend to cleave at the “branch point” more stable carbocations are formed 207 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.121) 2. Branched-chain Alkanes • 208 e.g. New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122) 3. Alkyl-substituted Aromatic Hydrocarbons • Undergo loss of a hydrogen atom or alkyl group yield the relatively stable tropylium ion • Gives a prominent peak at m/e 91 209 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122) 3. Alkyl-substituted Aromatic Hydrocarbons • 210 e.g. New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122) 4. Aldehydes and Ketones • Frequently undergo fragmentation by losing one of the side chains generate the substituted oxonium ion often represents the base peak in the mass spectra 211 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122) 4. Aldehydes and Ketones 212 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122) 5. Esters, Carboxylic Acids and Amides • Often undergo cleavage that involves the breaking of the C X bond form substituted oxonium ions as shown below: (where X = OH, OR, NH2, NHR, NR2) 213 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122) 5. Esters, Carboxylic Acids and Amides • For carboxylic acids and unsubstituted amides, characteristic peaks at m/e 45 and 44 are observed respectively 214 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123) 6. Alcohols • In addition to the loss of a proton and the hydroxyl radical, alcohols tend to lose one of the alkyl groups (or hydrogen atoms) form oxonium ions 215 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123) 6. Alcohols • For primary alcohols, the peak at m/e 31, 45, 59 or 73 often appears depends on what the R1 group is 216 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123) 7. Haloalkanes • Haloalkanes simply break at the C X bond 217 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123) 7. Haloalkanes • In the mass spectra of chloroalkanes, two peaks, separated by two mass units, in the ratio 3 : 1 will be appeared 218 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123) 7. Haloalkanes • In the mass spectra of bromoalkanes, two peaks, separated by two mass units, having approximately equal intensities will be appeared 219 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123) Check Point 34-9 220 New Way Chemistry for Hong Kong A-Level 3B The END 221 New Way Chemistry for Hong Kong A-Level 3B 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 222 New Way Chemistry for Hong Kong A-Level 3B 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 223 (a) (b) (c) (d) (e) Centrifugation Chromatography Fractional distillation Back Sublimation New Way Chemistry for Hong Kong A-Level 3B Filtration 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 224 New Way Chemistry for Hong Kong A-Level 3B 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 225 New Way Chemistry for Hong Kong A-Level 3B 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 226 ∴ The empirical formula ofWay the organic compound is3B CH2O. New Chemistry for Hong Kong A-Level 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 227 New Way Chemistry for Hong Kong A-Level 3B 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 228 ∴ The empirical formula ofWay the organic compound is3B C3H8O. New Chemistry for Hong Kong A-Level 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 229 New Way Chemistry for Hong Kong A-Level 3B 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 230 New Way Chemistry for Hong Kong A-Level 3B 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 231 New Way Chemistry for Hong Kong A-Level 3B 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. 232 New Way Chemistry for Hong Kong A-Level 3B 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 233 New Way Chemistry for Hong Kong A-Level 3B 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 234break down the efficient New Way Chemistryinforthe Hongprocess Kong A-Level packing of 3B 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. 235 New Way Chemistry for Hong Kong A-Level 3B 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 236 New Way Chemistry for Hong Kong A-Level 3B 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 237 point than methanol.New Way Chemistry for Hong Kong A-Level 3B 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) 238 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. New Way Chemistry for Hong Kong A-Level 3B 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. 239 New Way Chemistry for Hong Kong A-Level 3B 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. 240 New Way Chemistry for Hong Kong A-Level 3B 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) 241 Compound Boiling point (oC) Heptanal 155 Heptan-4-one 144 2,4-Dimethylpentan-3-one 125 New Way Chemistry for Hong Kong A-Level 3B 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. 242 New Way Chemistry for Hong Kong A-Level 3B 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 243 New Way Chemistry for Hong Kong A-Level 3B 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: 244 New Way Chemistry for Hong Kong A-Level 3B 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 245 New Way Chemistry for Hong Kong A-Level 3B 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. 246 New Way Chemistry for Hong Kong A-Level 3B 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=2 247 New Way Chemistry for Hong Kong A-Level 3B 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 248 New Way Chemistry for Hong Kong A-Level 3B 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. 249 New Way Chemistry for Hong Kong A-Level 3B 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=2 250 New Way Chemistry for Hong Kong A-Level 3B 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. 251 New Way Chemistry for Hong Kong A-Level 3B 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. 252 New Way Chemistry for Hong Kong A-Level 3B 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 253 New Way Chemistry for Hong Kong A-Level 3B 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. 254 New Way Chemistry for Hong Kong A-Level 3B 2 34.7 Structural Information from Chemical Properties (SB p.99) (a) (i) 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. (ii) 255 New Way Chemistry for Hong Kong A-Level 3B 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 256 New Way Chemistry for Hong Kong A-Level 3B 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=2 257 New Way Chemistry for Hong Kong A-Level 3B 34.7 Structural Information from Chemical Properties (SB p.99) (b) (i) (ii) 258 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. New Way Chemistry for Hong Kong A-Level 3B 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 259 New Way Chemistry for Hong Kong A-Level 3B 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. 260 New Way Chemistry for Hong Kong A-Level 3B 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. 261 New Way Chemistry for Hong Kong A-Level 3B 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 262 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.110) An organic compound with a relative molecular mass of 72.0 was found to contain 66.66% carbon, 22.23% oxygen and 11.11% hydrogen by mass. A portion of its infra-red spectrum is shown below. 263 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.110) (a) Determine the molecular formula of the compound. (b) Deduce two possible structures of the compound, each of which belongs to a different homologous series. Answer 264 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.110) (a) Let the mass of the compound be 100 g. Then, mass of carbon in the compound = 66.66 g mass of hydrogen in the compound = 11.11 g mass of oxygen in the compound = 22.23 g Mass (g) Number of moles (mol) Relative number of moles Simplest mole ratio 265 Carbon Hydrogen Oxygen 66.66 11.11 22.23 11.11 66.66 22.23 11.11 5.56 1.39 1 .0 12.0 16.0 1.39 5.56 11.11 1 4 8 1.39 1.39 1.39 4 8 1 ∴ The empirical formula of the compound is C H O. New Way Chemistry for Hong Kong A-Level4 3B8 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.110) Let the molecular formula of the compound be (C4H8O)n. Relative molecular mass of (C4H8O)n = 72.0 n × (12.0 × 4 + 1.0 × 8 + 16.0) = 72.0 ∴ n =1 ∴ The molecular formula of the compound is C4H8O. 266 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.110) (b) From the IR spectrum, it can be observed that there are absorption peaks at 2 950 cm–1 and 1 700 cm–1. The absorption peak at 2 950 cm–1 corresponds to the stretching vibration of the C H bond, and the absorption peak at 1 700 cm–1 corresponds to the stretching vibration of the C = O bond. Since there is only one oxygen atom in the molecule of the compound, we can deduce that the compound is either an aldehyde or a ketone. If it is an aldehyde, its possible structure will be: 267 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.110) (b) If it is a ketone, its possible structure will be: Back 268 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.111) (a) An organic compound X forms a silver mirror with ammoniacal silver nitrate solution. Another organic compound Y reacts with ethanoic acid to give a product with a fruity smell. The portions of infra-red spectra of X and Y are shown below. 269 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.111) Sketch the infra-red spectrum of a carboxylic acid based on the IR spectra of X and Y. Answer 270 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.111) (a) From the information given, X would be an aldehyde and Y would be an alcohol. Comparing the structures of an aldehyde and an alcohol with that of a carboxylic acid, some common features are found between the two. In the IR spectrum of a carboxylic acid, it is expected that it contains the characteristic O — H (similar to the alcohol) and C = O (similar to the aldehyde) absorption peaks. Thus, peak values at around 3300 cm–1 and 1720 cm–1 are expected. A broad band at around 3300 cm–1 is observed due to the complication of the stretching vibration of the O — H group by hydrogen bonding and it overlaps with the absorption of the C — H bond in the 2950 – 2875 cm–1 region. 271 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.111) (a) The infrared spectrum of a carboxylic acid is as follows: 272 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.112) Answer (b) The infra-red spectra of two organic compounds A and B are shown below. Decide which compound could be an alcohol. Explain your 273 answer briefly. New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.112) (b) Compound B could be an alcohol. From the two spectra given, compound B shows a broad band at 3300 cm–1 and several peaks at 2960 – 2875 cm–1. This broad band corresponds to the complication of the stretching vibration of the O — H bond by hydrogen bonding occurring among alcohol molecules. 274 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.112) (c) The table below shows the characteristic absorption wavenumbers of some covalent bonds in infra-red spectra. Bond C=O OH Range of wavenumber (cm-1) 1680 – 1750 2500 – 3300 CH NH 2840 – 3095 3350 – 3500 Answer 275 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.112) Sketch the expected infra-red spectrum for an amino acid with the following structure: 276 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.112) (c) The infra-red spectrum of the amino acid is shown as follows: 277 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.112) (d) A portion of the infra-red spectrum of an organic compound X is shown below. To which homologous series does it belong? Answer 278 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.112) (d) In the IR spectrum of compound X, the wide absorption band at 3500 – 3000 cm–1 corresponds to the stretching vibration of the O — H bond. Besides, the absorption peak at 1760 – 1720 cm–1 corresponds to the stretching vibration of the C = O bond. Therefore, compound X is a carboxylic acid. 279 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.112) (e) A portion of the infra-red spectrum of an organic compound Y is shown on the right. Identify the functional groups that it contains. Answer 280 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.112) (e) From the IR spectrum of compound Y, the two peaks in the 3300 – 3180 cm–1 region show that the compound contains the –NH2 group. Besides, the sharp peak at 1680 cm–1 implies that the compound also contains the C = O bond. 281 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.113) (f) An organic compound Z with a relative molecular mass of 88.0 was found to contain 54.54% carbon, 36.36% oxygen and 9.10% hydrogen by mass. A portion of its infra-red spectrum is shown below: 282 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.113) (i) Determine the molecular formula of compound Z. (ii) Based on the result from (i), draw two possible structures of the compound, each of which belongs to a different homologous series. (iii) Using the information from the IR spectrum, name the homologous series that compound Z belongs to. Explain your answer. Answer 283 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.113) (f) (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 Oxygen 54.54 9.10 36.36 9.10 54.54 36.36 9 . 10 4.55 2.27 1 . 0 12.0 16.0 9.10 4.55 2.27 4 2 1 2.27 2.27 2.27 2 4 1 ∴ The empirical formula of the compound is C2H4O. 284 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.113) (f) (i) Let the molecular formula of the compound be (C2H4O)n. Relative molecular mass of (C2H4O)n = 88.0 n × (12.0 × 2 + 1.0 × 4 + 16.0) = 88.0 n =2 ∴ The molecular formula of the compound is C4H8O2. (ii) 285 New Way Chemistry for Hong Kong A-Level 3B 34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.113) (f) (iii) From the IR spectrum of compound Z, the absorption peak at 3200 – 2800 cm–1 corresponds to the stretching vibration of the C — H bond. Besides, the absorption peak at 1800 – 1600 cm–1 corresponds to the stretching vibration of the C = O bond. The absence of the characteristic peak of the O — H bond in the 3230 – 3670 cm–1 region indicates that compound Z is an ester. Back 286 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118) Answer The mass spectrum of pentan-2-one (CH3COCH2CH2CH3) is shown below: What ions do the peaks at m/e 86, 71 and 43 represent? 287 New Way Chemistry for Hong Kong A-Level 3B Explain your answer. 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118) The relative molecular mass of pentan-2-one is 86. Therefore, the peak at m/e 86 corresponds to the molecular ion of pentan-2-one. When the C1 C2 bond is broken, the ion CH3CH2CH2CO+ (m/e = 71) is formed. When the C2 C3 bond is broken, the ion CH3CO+ (m/e = 43) is formed. Back 288 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118) The mass spectrum of hydrocarbon X is shown below: 289 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118) (a) What is the relative molecular mass of hydrocarbon X? (b) Which peak is the base peak? (c) How many mass units is the base peak less than the peak for the molecular ion? (d) Deduce the structures of hydrocarbon X. (e) Explain the peak at m/e 43. (f) Propose the fragmentation pattern of the molecular ion which gives rise to the peaks at m/e 58, 43, 29 Answer and 15. 290 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118) (a) (b) (c) (d) 291 The relative molecular mass of hydrocarbon X is 58.0. The base peak is at m/e 43. 15 mass units Since the compound is a hydrocarbon, the molecular formula of the compound must be CxHy. From the relative molecular mass of the compound (i.e. 58.0), we can deduce that the compound contains 4 carbon atoms only. (If the compound contains 5 carbon atoms, the relative molecular mass would be more than 12.0 × 5 = 60.0). The number of hydrogen atoms in the compound is (58.0 - 12.0 × 4 = 10) 10. Therefore, the hydrocarbon is butane. New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118) Back (e) The peak at m/e 43 is 15 mass units less than the molecular ion. This suggests that a methyl group is lost during the fragmentation of the molecular ion. The peak at m/e 43 corresponds to CH3CH2CH2+. (f) 292 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.120) An organic compound is investigated. The structural formula of this compound is shown below: 293 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.120) The mass spectrum of the compound is shown below: Interpret the peaks at m/e 134, 119, 91 and 43. 294 New Way Chemistry for Hong Kong A-Level 3B Answer 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.120) Back The peak at m/e 134 corresponds to the molecular ion. The peak at m/e 119 corresponds to the ion that is 15 mass units less than the molecular ion. This suggests that a methyl group is lost from the molecular ion. The peak at m/e 91 is the base peak, which corresponds to the ion C6H5CH2+ . The peak at m/e 43 corresponds to the ion CH3CO+. 295 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123) Why would the molecular ion compound have two peaks, separated by two mass units, in the ratio 3 : 1? Answer Chlorine has two isotopes, chlorine-35 and chlorine-37. Their relative abundances are in the ratio of 3 : 1. Back 296 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123) Why would the molecular ion of a bromine-containing compound have two peaks, separated by two mass units, having approximately equal intensities? Answer Bromine has two isotopes, bromine-79 and bromine-81. Their relative abundances are in the ratio of 1 : 1. Back 297 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.124) (a) What is base peak in a mass spectrum? Why is the m/e of the base peak not the molecular mass of the compound? Answer (a) The base peak is the most intense peak in a mass spectrum. It represents the most stable ion formed during fragmentation or the ion that can be formed in various ways during fragmentation of the molecular ion. As molecular ions are usually unstable and will undergo fragmentation, they do not normally show up as base peaks in mass spectra. 298 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.124) (b) The following is the mass spectrum of bromomethylbenzene (benzyl bromide). Interpret the peaks at m/e = 172, 170 and 91. Answer 299 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.124) (b) The relative molecular mass of bromomethylbenzene (benzyl bromide) is 171.0. However, as bromine contains equal abundances of the 79Br and 81Br isotopes, the spectrum shows two small peaks of equal intensity at m/e = 172 and 170. The base peak at m/e = 91 is due to the formation of the ion C6H5CH2+. 300 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.124) Answer (c) Study the following spectrum carefully and deduce what group of organic compound it is. The compound has a relative molecular mass of 114. 301 New Way Chemistry for Hong Kong A-Level 3B 34.9 Use of Mass Spectra to Obtain Structural Information (SB p.124) (c) The base peak is at m/e = 57 which may be an oxonium ion or a carbocation. This is a mass spectrum of a ketone, an aldehyde or a hydrocarbon. Back 302 New Way Chemistry for Hong Kong A-Level 3B