34

Structural Determination of Organic Compounds

1 34.1

34.2

34.3

34.4

34.5

34.6

34.7

34.8

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 34.9

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 the compound of

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

4 The general steps to determine the structure of an organic compound

Check Point 34-1

New Way Chemistry for Hong Kong A-Level 3B

5

34.2

Isolation and Purification of Organic Compounds

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

10 The laboratory set-up of filtration 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 13 A centrifuge 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 aqueous solution from an unsaturated  the solution is gently heated to make it more concentrated • After, the solution is allowed to cool room conditions at 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 dissolved solutes all of the • 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

18 Crystallization by cooling a hot concentrated solution 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 to form for crystals 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

21 Crystallization by slow evaporation of a solution (preferably saturated) at room temperature 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. 1,1,1 trichloroethane 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 the ether layer into 23 New Way Chemistry for Hong Kong A-Level 3B

34.2

Isolation and Purification of Organic Compounds (SB p.80)

Solvent Extraction

24 The organic product in an aqueous solution can be extracted by solvent extraction using diethyl ether 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

28 The laboratory set-up of distillation 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 heating bumping to occur during  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

32 The laboratory set-up of fractional distillation 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 Fractions with different boiling points can be collected separately 35 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 cold surface on a • 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

38 A mixture of two compounds can be separated by sublimation New Way Chemistry for Hong Kong A-Level 3B

34.2

Isolation and Purification of Organic Compounds (SB p.83)

Chromatography

• • An effective method of separating a complex mixture of substances Paper chromatography type of chromatography is a common 39 New Way Chemistry for Hong Kong A-Level 3B

34.2

Isolation and Purification of Organic Compounds (SB p.83)

Chromatography

40 The laboratory set-up of paper chromatography 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 of the filter paper adhered onto the surface 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 R f 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 Aim

(a) Filtration 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 extraction (e) Distillation 45 To separate a component from a mixture with a suitable solvent To separate a liquid from a solution containing non-volatile solutes 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 Aim

(f) Fractional distillation (g) Sublimation To separate miscible liquids with widely different boiling points To separate a mixture of solids in which only one can sublime (h) Chromatography To separate a complex mixture of substances 46

Check Point 34-2

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 The tube is attached to a thermometer The temperature at which the solid melts is its melting point 49 New Way Chemistry for Hong Kong A-Level 3B

34.3

Tests for Purity (SB p.85)

Determination of Melting Point

50 Determination of the melting point of a solid using an oil bath 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 0.5

° C ) (usually less than • An impure solid does not melting point have a sharp  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 quite sharp liquid is • 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

55

34.4

Qualitative Analysis of Elements in an Organic Compound

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 compounds can be detected in organic  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 water of the test are extracted with 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 chloride ion (Cl bromide ion (Br ) iodide ion (I 62 ) ) Acidified silver nitrate solution A white precipitate is formed. It is soluble in excess NH 3 (aq) .

A pale yellow precipitate is formed. It is sparingly soluble in excess NH 3 (aq) .

A creamy yellow precipitate is formed. It is insoluble in excess New Way Chemistry for Hong Kong A-Level 3B NH 3 (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 iron(II) sulphate and iron(III) sulphate solutions A blue-green observed.

colour is Sulphur, as sulphide ion (S 2 ) Sodium pentacyanonitr osylferrate(II) solution A black precipitate formed is 63

Check Point 34-4

New Way Chemistry for Hong Kong A-Level 3B

64

34.5

Determination of Empirical Formula and Molecular Formula from Analytical Data

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 compound of the  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 anhydrous calcium chloride and 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 nitrate solution and excess silver • 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 can be calculated of the compound 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

76

34.6

Structural Information from Physical Properties

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 Density at 20 o C Melting point and boiling point Solubility In water or highly polar solvents In non polar organic solvents

Insoluble Soluble Hydrocarbo ns (saturated and unsaturated) All have densities < 0.8 g cm –3 • Generally low increases with but number of carbon atoms in the molecule • Branched-chain hydrocarbons have lower boiling points but higher melting points than the corresponding 81

34.6

Structural Information from Physical Properties (SB p.90) Physical properties of some common organic compounds

Organic compound

Aromatic hydrocarbons

Density at 20 o C

Between 0.8 and 1.0 g cm –3

Melting point and boiling point

Generally low

Solubility In water or highly polar solvents In non polar organic solvents

Insoluble Soluble 82 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

Halo alkanes 83

Density at 20 o C Melting point and boiling point Solubility In water or highly polar solvents

• 0.9 - 1.1 g cm –3 for chloro alkanes • Higher than alkanes similar relative molecular masses (  haloalkane molecules are polar ) of • >1.0 g cm –3 for bromo • All haloalkanes are liquids except halomethanes alkanes • Both the m.p.

and b.p.

and increase in the order: iodo alkanes RCH 2 F < RCH 2 Cl < New Way Chemistry for Hong Kong A-Level 3B RCH 2 Br < RCH 2 I Insoluble

In non polar organic solvents

Soluble

34.6

Structural Information from Physical Properties (SB p.90) Physical properties of some common organic compounds

Organic comp ound

Alcohols

Density at 20 o C Melting point and boiling point

• Simple alcohols are liquids and alcohols with > 12 carbons are waxy solids • Much higher than hydrocarbons of similar relative molecular masses (  formation of hydrogen bonds between alcohol molecules)

Solubility In water or highly polar solvents In non polar organic solvents

• Lower members : Completely miscible with water (  formation of hydrogen bonds between alcohol molecules Soluble 84 molecules)

34.6

Structural Information from Physical Properties (SB p.90) Physical properties of some common organic compounds

Organic comp ound Density at 20 o C

Alcohols • All simple alcohols have densities < 1.0 g cm –3

Melting point and boiling point

• Straight-chain alcohols have higher b.p.

than the corresponding branched-chain alcohols

Solubility In water or highly polar solvents In non polar organic solvents

• Solubility decreases gradually as the hydrocarbon chain lengthens Soluble 85 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 comp ound Density at 20 o C Melting point and boiling point

Carbonyl comp ounds (alde hydes and ketones) • <1.0 g cm –3 for aliphatic carbonyl compounds 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) 86

Solubility In water or highly polar solvents In non polar organic solvents

• Lower members: Soluble in water (  the formation of hydrogen bonds between molecules of aldehydes or ketones and water Soluble

34.6

Structural Information from Physical Properties (SB p.91) Physical properties of some common organic compounds

Organic comp ound Density at 20 o C

Carbonyl comp ounds (alde hydes and ketones) • > 1.0 g cm –3 for aromatic carbonyl compounds

Melting point and boiling point Solubility In water or highly polar solvents In non polar organic solvents

• Solubility decreases gradually as the hydrocarbon chain lengthens Soluble 87 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 comp ound

Carbo xylic acids

Density at 20 o C

• Lower members have densities similar to water • Methanoic acid has a density of 1.22 g cm –3

Melting point and boiling point

Higher than alcohols of similar relative molecular masses (  the formation of more extensive intermolecular hydrogen bonds )

Solubility In water or highly polar solvents In non polar organic solvents

• First four members are miscible with water in all proportions • Solubility decreases gradually as the hydrocarbon chain Soluble 88

34.6

Structural Information from Physical Properties (SB p.91) Physical properties of some common organic compounds

Organic comp ound

Esters

Density at 20 o C Melting point and boiling point Solubility In water or highly polar solvents In non polar organic solvents

Insoluble Soluble Lower members have densities less than water Slightly higher than hydrocarbons but lower than carbonyl compounds and alcohols of similar relative molecular masses 89 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 comp ound

Amines

Density at 20 o C Melting point and boiling point

Most amines have densities less than water • Higher than alkanes but lower than alcohols of similar relative molecular masses

Solubility In water or highly polar solvents In non polar organic solvents

• Generally soluble • Solubility decreases the order: in 1 o 2 o 3 o amines > amines > amines Soluble 90 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 comp ound Density at 20 o C Melting point and boiling point Solubility In water or highly polar solvents In non polar organic solvents

Amines • 1 o and 2 o amines are able to form hydrogen bonds with each other but the strength is less than that between alcohol molecules ( N  H bond is less polar than O  H bond) 91 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 comp ound Density at 20 o C Melting point and boiling point Solubility In water or highly polar solvents In non polar organic solvents

Amines • 3 o amines have lower m.p. and b.p

. than the isomers of 1 o and 2 o amines (  molecules of 3 o amines cannot form intermolecular hydrogen bonds ) 92 New Way Chemistry for Hong Kong A-Level 3B

34.6

Structural Information from Physical Properties (SB p.92) 93

Example 34-6

Check Point 34-6

New Way Chemistry for Hong Kong A-Level 3B

94

34.7

Structural Information from Chemical Properties

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 structure give enough clue to the of the compound • Compounds having the same molecular formula  may have atoms different arrangements of 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 C 2 H 4 O 2 represent a carboxylic acid or an may 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 Test • Burn the saturated hydrocarbon in a non luminous Bunsen flame Observation • A blue or clear yellow flame is observed 98 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 Test Observation

Unsaturated hydrocarbons (C = C, C  C) 99 • Burn the unsaturated hydrocarbon in a non-luminous Bunsen flame • A smoky flame is observed • Add bromine in 1,1,1 trichloroethane at room temperature and in the absence of light • Bromine decolourizes rapidly • Add 1% (dilute) acidified potassium • Potassium manganate(VII) manganate(VII) solution New Way Chemistry for Hong Kong A-Level 3B solution rapidly decolourizes

34.7

Structural Information from Chemical Properties (SB p.93) Chemical tests for different groups of organic compounds

Organic compound Test Observation

Haloalkanes (1 ° , 2 ° or 3 ° ) • Boil with ethanolic potassium hydroxide solution, then acidify with excess dilute nitric(V) acid and add silver nitrate(V) solution • For chloroalkanes , a white precipitate is formed • For bromoalkanes , a pale yellow precipitate is formed • For iodoalkanes , a creamy yellow precipitate is formed 100 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 Test Observation

Halobenzenes • Boil with ethanolic potassium hydroxide solution, then acidify with excess dilute nitric(V) acid and add silver nitrate(V) solution • No precipitate formed is 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)

Test

• Add a small piece of sodium metal • Esterification : Add ethanoyl chloride • A

Observation

colourless gas evolved • The temperature of the reaction mixture rises is • A colourless gas evolved is 102 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) • Add

Test

acidified potassium dichromate(VI) solution • For

Observation

1 ° and 2 alcohols ° , the clear orange solution becomes opaque and turns green almost immediately • For 3 ° alcohols , there are no observable changes 103 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: • A

Observation

yellow is formed precipitate 104 Add iodine in sodium hydroxide solution 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

compound

Alcohols (  OH) 105

Organic

•

Test

Lucas test : add a solution of zinc chloride in concentrated hydrochloric acid New Way Chemistry for Hong Kong A-Level 3B

Observation

• For 1 ° alcohols , the aqueous phase remains clear • For 2 ° alcohols , the clear solution becomes cloudy within 5 minutes • For 3 ° alcohols , the aqueous phase appears cloudy immediately

34.7

Structural Information from Chemical Properties (SB p.94) Chemical tests for different groups of organic compounds

Organic compound

Ethers (  O  ) •

Test

No specific test soluble in concentrated for ethers but they are sulphuric(VI) acid

Observation

 106 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 Test

Aldehydes • Add aqueous sodium hydrogensulphate(IV) ( ) • Add 2,4 dinitrophenylhydrazine • Crystalline salts formed • A

Observation

are yellow, orange or red precipitate formed is • Silver mirror test : add Tollens’ reagent (a solution of aqueous silver nitrate in aqueous ammonia) • A silver mirror is deposited on the inner wall of the test tube 107 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 Test Observation

Ketones ( ) • Add aqueous sodium hydrogensulphate(IV) • Add 2,4 dinitrophenylhydrazine • Crystalline salts are formed (for unhindered ketones only) • A yellow, orange or red precipitate is formed • Iodoform test for: • A yellow precipitate is formed Add iodine in sodium 108

34.7

Structural Information from Chemical Properties (SB p.95) Chemical tests for different groups of organic compounds

Organic compound Test Observation

Carboxylic acids ( ) 109 • Esterification : warm the carboxylic acid with an alcohol in the presence of concentrated sulphuric(VI) acid , followed by adding sodium carbonate solution • A sweet and fruity smell is detected • Add sodium hydrogencarbonate • The colourless gas produced turns lime water milky 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 compound Test

Esters ( ) • No specific test for esters but they can be distinguished by its characteristic smell • A

Observation

sweet and fruity smell is detected 110 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 compound Test

Acyl halides ( ) • Boil with ethanolic potassium hydroxide solution, then acidify with excess dilute nitric(V) acid and add silver nitrate(V) solution • For

Observation

acyl chlorides white precipitate formed , a is • For acyl bromides , a pale yellow precipitate is formed • For acyl iodides , a creamy yellow precipitate is formed 111 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 compound Test Observation

Amides ( ) • Boil with sodium hydroxide solution • The colourless gas produced turns moist red litmus paper or pH paper blue 112 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 compound Test Observation

Amines  NH 2 ) • 1 o aliphatic amines: dissolve the amine in dilute hydrochloric acid at 0 – 5 o C , then add cold sodium nitrate(III) solution slowly • 1 o aromatic amines: add naphthalen-2-ol in dilute sodium hydroxide solution • Steady evolution of N 2 (

g

) is observed • An orange or red precipitate is formed 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 compound Test Observation

Aromatic compounds ) • Burn the aromatic compound in a non luminous Bunsen flame • A smoky yellow flame with black soot is produced • Add fuming sulphuric(VI) acid • The aromatic compound dissolves • The temperature of the reaction mixture rises 114 New Way Chemistry for Hong Kong A-Level 3B

34.7

Structural Information from Chemical Properties (SB p.96)

Example 34-7A

Example 34-7B

115

Example 34-7C

Check Point 34-7

New Way Chemistry for Hong Kong A-Level 3B

116

34.8

Use of Infra-red Spectrocopy in the Identification of Functional Groups

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 particle and • 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  10 8 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

123 Regions of the electromagnetic spectrum 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 onto a hydrogen atom , falls  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 = E 2 – E 1 =

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 bonds the extent of vibration of the 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

132 Effect of absorption of IR radiation on vibration of atoms in a molecule 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 133 An infra-red spectrometer 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 carefully measured is • 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 by the equation

c

=  are related  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 undergo a stretching vibration can where the atoms move back and forth joined by a spring as if they were 138 A stretching vibration of two atoms 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

139 New Way Chemistry for Hong Kong A-Level 3B A variety of stretching and bending vibrations

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

• The stretching vibrations of single bonds involving hydrogen (C  H, O  H and N  H) occur at relatively high frequencies 141

Bond

C  H O  H N  H

Range of wavenumber (cm

2840 3230 – 3095 – 3670 3350 – 3500 Characteristic absorption wavenumbers of some single bonds in infra-red spectra New Way Chemistry for Hong Kong A-Level 3B

-1 )

34.8

Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.102)

Infra-red Spectroscopy

142 • Triple bonds are stronger higher frequencies and vibrate at than double bonds

Bond

C  C C  N

Range of wavenumber (cm -1 )

2070 2200 – 2250 – 2280 C = C 1610 – 1680 C = O 1680 – 1750 Characteristic absorption wavenumbers of some double bonds and triple 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

• 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

145 The IR spectrum of hex-1-yne 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 , 147  the intensity of transmitted radiation decreases  results in a decrease in percentage of transmittance  a dip in the spectrum  often called an absorption peak absorption band New Way Chemistry for Hong Kong A-Level 3B or

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 )

400 – 1500

Interpretation

1500 – 2000 2000 – 2500 2500 – 4000 149 • Often consists of many complicated bands • Unique to each compound • Often called the fingerprint region • Not used for identification of particular functional groups Absorption of double bonds , e.g. C = C, C = O Absorption of triple bonds , e.g. C  C, C  N Absorption of single bonds involving 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 1 500 cm -1 is often used for and  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) Alkanes, alkenes, arenes Alcohols, phenols (hydrogen-bonded) C = C C = O C C O C O      C N H H H

Characteristic range of wavenumber (cm -1 )

1610 – 1680 1680 – 1750 2070 2200 2500 2840 3230 – 2250 – 2280 – 3300 – 3095 – 3670 N  H 3350 – 3500

34.8

Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.104)

Interpretation of IR Spectra 1. Butane

152 The IR spectrum of butane 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

153 Wavenumber (cm -1 ) 2968 2890 Intensity Very strong Medium Indication C  H stretching 1468 Strong C  H bending Interpretation of the IR spectrum of butane 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

154 The IR spectrum of cis-but-2-ene 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

155 3044 3028 2952 1677

-1 ) Intensity

Very strong Very strong Very strong Medium

Indication

( C 

sp

2 H stretching C  H) ( C 

sp

3 H stretching C  H) C = C stretchinh 1657 Medium 1411 Strong C  H bending Interpretation of the IR spectrum of cis-but-2-ene 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

156 The IR spectrum of hex-1-yne 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 2963 2938 Very strong Very strong Very strong

Indication

( C 

sp

H stretching C  H) ( C 

sp

3 H stretching C  H) 157 2874 Strong 2119 Strong C  C stretching 1468 Strong ( C 

sp

H bending C  H) 1445 Medium ( C 

sp

3 H bending 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

158 The IR spectrum of butanone 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 )

2983

Intensity

Strong

Indication

C  H stretching 159 2925 Strong 1720 Very strong C = O stretching 1416 Medium C  H bending (shifted as adjacent to C = O) Interpretation of the IR spectrum of butanone 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

160 The IR spectrum of butan-1-ol 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 2960

Intensity

Broad band Medium

Indication

O  H stretching C  H stretching 2935 Medium 161 2875 Medium Interpretation of the IR spectrum of butan-1-ol 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

162 The IR spectrum of butanoic acid 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 ) Intensity Indication 3100 1708 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

165 The IR spectrum of butylamine 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 )

3371

Intensity

Strong

Indication

N  H stretching Strong 3280 2960 – 2875 1610 Weak Medium C  H stretching N  H bending 166 1475 Medium C  H bending Interpretation of the IR spectrum of butylamine 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

167 The IR spectrum of butanenitrile 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

168

Wavenumber (cm -1 )

2990 – 2895 2246 1420

Intensity

Strong Very strong Strong

Indication

C  H stretching C  N stretching C  H bending 1480 Strong Interpretation of the IR spectrum of butanenitrile 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 absorption peaks major 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 peaks and presence of absorption 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 absorption peaks give rise to strong 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

Check Point 34-8

174 New Way Chemistry for Hong Kong A-Level 3B

175

34.9

Use of Mass Spectra to Obtain Structural Information

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 analytical tools and versatile • More sensitive than other spectroscopic methods (e.g. IR spectroscopy ) • Only a microgram or less required for the analysis of materials is 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

179 Components of a mass spectrometer 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 molecular ions called the are formed  One of the electrons is knocked off of the molecule 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 (CH 3 OH) 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 The lighter deflection the ions, the greater the 187 New Way Chemistry for Hong Kong A-Level 3B

34.9

Use of Mass Spectra to Obtain Structural Information (SB p.114)

Mass Spectrometry

• • Positively charged ions of higher charge have greater deflection Ions with a high

m

/

e

to smaller extent ratio are deflected than ions with a low

m

/

e

ratio 188 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 electronically collected is measured • The intensity of the signal is  a measure of the relative abundance the ions with a particular

m

/

e

ratio of 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 intensity of varying  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

• Generally published as bar graphs .

191 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

H 3 C + H  CO +

m/e ratio

15 29 H 2 C = OH + CH 3 OH 31 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 deflected by the electric field and 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 H 2 C = 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 H 2 C = OH + is a relatively stable ion   the positive charge a particular atom is not localized on  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

HC  O + 29 corresponds to the ion  formed by losing two hydrogen atoms from the ion H 2 C = 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  structures: O + has two resonance 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

ion H 3 C + 15 corresponds to the  formed by the breaking of the C  bond in the molecular ion O 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

201 Mass spectrum of pentan-3-one 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

CH 3 CH 2 + CH 3 CH 2 CO + 29 57 CH 3 CH 2 COCH 2 CH 3 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 is summarized below: of pentan-3-one 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 • CH 3 at M – 15 to give a peak  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

• Take hexane as an example: 206 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

• e.g.

208 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 group of a hydrogen atom or alkyl  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

• e.g.

210 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: 213 (where X = OH, OR, NH 2 , NHR, NR 2 ) 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 amides , and unsubstituted  characteristic peaks at

m

/

e

are observed respectively 45 and 44 214 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 hydroxyl radical , and the  alcohols tend to lose one of the  groups (or  hydrogen atoms ) alkyl  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 appears often  depends on what the R 1 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  bond X 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) 220

Check Point 34-9

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) 222

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

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

(a) (b) (c)

Back

Answer

(d) 223 (e) Centrifugation Chromatography Fractional distillation Sublimation Filtration New Way Chemistry for Hong Kong A-Level 3B

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 Carbon Hydrogen Oxygen Mass (g) Number of moles (mol) 40.0

40.0

 3 .

33 12.0

6.7

 6 .

7 1.0

6.7

53.3

53.3

 3 .

33 16.0

Relative number of moles 3.33

3.33

 1 6.7

3.33

 2 3.33

3.33

 1 Simplest mole ratio 1 2 1 The empirical formula of the organic compound is CH 2 O.

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 60.00

Hydrogen 13.33

Oxygen 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 Oxygen Mass (g) Number of moles (mol) 60.00

13.33

26.67

60.00

 5 12.0

13.33

 13 .

33 1.0

26.67

 1 .

67 16.0

Relative number of moles 5 1.67

 3 13.33

1.67

 8 1.67

1.67

 1 Simplest mole ratio 3 8 1 The empirical formula of the organic compound is C 3 H 8 O.

34.5

Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.89) (b) The molecular formula of the compound is (C 3 H 8 O)

n

.

Relative molecular mass of (C 3 H 8 O)

n n

× (12.0 × 3 + 1.0 × = 60.0

8 + 16.0) = 60.0

n

= 1 ∴ The molecular formula of compound Z is C 3 H 8 O.

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 CO 2 = 12.0 + 16.0 × 2 = 44.0

Mass of carbon in 0.22 g of CO 2 = 0.22 g × = 0.06 g 12.0

44.0

Relative molecular mass of H 2 O = 1.0 × 2 + 16.0

= 18.0

Mass of hydrogen in 0.09 g of H 2 O = 0.09 g × = 0.01 g 2.0

18.0

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 0.06

0.06

 0 .

005 12.0

0.005

0.005

1  1 Hydrogen 0.01

0.01

 1.0

0 .

01 0.01

0.005

2  2 ∴ The empirical formula of the organic compound is CH 2 O.

Oxygen 0.08

0.08

 0 .

005 16.0

0.005

0.005

 1 1 232 New Way Chemistry for Hong Kong A-Level 3B

34.5

Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.89) Let the molecular formula of the compound be (CH 2 O)

n

.

Relative molecular mass of (CH 2 O)

n n

× (12.0 + 1.0 × = 60.0

2 + 16.0) = 60.0

n

= 2 ∴ The molecular formula of the compound is C 2 H 4 O 2 .

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?

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 straight chain 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 234 packed more efficiently in solid state. Extra energy is required to

34.6

Structural Information from Physical Properties (SB p.92)

Back

235 Why does the solubility of amines in water decrease in the order: 1 o amines > 2 o amines > 3 o amines?

Answer

The solubility of primary and secondary amines is higher than that of tertiary amines because tertiary amines cannot form hydrogen bonds between water molecules. On the other hand, the solubility of primary amines is higher than that of secondary amines because primary amines form a greater number of hydrogen bonds with water molecules than secondary amines.

New Way Chemistry for Hong Kong A-Level 3B

34.6

Structural Information from Physical Properties (SB p.92)

Match the boiling points 65 o C, – 6 o C and – 88 o C with the compounds CH 3 CH 3 , CH 3 NH 2 and CH 3 OH. 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 CH 3 CH 3 CH 3 NH 2 CH 3 OH Boiling point ( ° C) –88 –6 65 Ethane (CH 3 CH 3 ) is a non-polar compound. In pure liquid form, ethane molecules are held together by weak van der Waals’ forces. However, both methylamine (CH 3 NH 2 ) and methanol (CH 3 OH) 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  polar than the N  H bond in alcohols is more 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-1 ol and butanal?

(ii) Account for the higher boiling point of butan-1-ol.

Answer

(a) (i) The relative molecular masses of butan-1-ol (ii) 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.

238 New Way Chemistry for Hong Kong A-Level 3B

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

(b) The solubility increases in the order: chloroethane <

Answer

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 (CH 3 ) 3 N (b.p.: 2.9

°

C) boils so much lower than CH 3 CH 2 CH 2 NH 2 (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

Answer

(d) Compound Heptanal Heptan-4-one 2,4-Dimethylpentan-3-one Boiling point ( o C) 155 144 125 241 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 CH 2 O and its relative molecular mass is 60.0. It reacts with sodium hydrogencarbonate to give a colourless gas which turns lime water milky.

(a) Calculate the molecular formula of the compound.

Answer

(a) Let the molecular formula of the compound be (CH 2 O)

n

.

Relative molecular mass of (CH 2 O)

n =

60.0

n

 (12.0 + 1.0  2 + 16.0) = 60.0

n

= 2 ∴ The molecular formula of the compound is C 2 H 4 O 2 .

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 CH 2 O and its relative molecular mass is 60.0. It reacts with sodium hydrogencarbonate to give a colourless gas which turns lime water milky.

(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 C 2 H 4 O 2 , the atoms left are one carbon and three hydrogen atoms. This obviously shows that a methyl group (  CH 3 ) is present. Therefore, the structural formula of the compound is: 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 CH 2 O and its relative molecular mass is 60.0. It reacts with sodium hydrogencarbonate to give a colourless gas which turns lime water milky.

(c) Give the IUPAC name for the compound.

Answer

(c) The IUPAC name for the compound is ethanoic acid.

Back

245 New Way Chemistry for Hong Kong A-Level 3B

34.7

Structural Information from Chemical Properties (SB p.96)

15 cm 3 of a gaseous hydrocarbon were mixed with 120 cm of oxygen which was in excess. The mixture was exploded. After cooling, the residual volume was 105 cm adding concentrated potassium hydroxide solution, the volume decreased to 75 cm 3 .

3 . On 3 (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.

Answer

246 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 C x H y .

Volume of C x H y reacted = 15 cm 3 Volume of unreacted oxygen = 75 cm 3 Volume of oxygen reacted = (120 - 75) cm 3 = 45 cm 3 Volume of carbon dioxide formed = (105 - 75) cm 3 C x H y + (x + )O 4 2  xCO 2 y 2 2 O = 30 cm 3 Volume of C x H y  1 x  15 30  x = 2 reacted : Volume of CO 2 formed = 1 : x = 15 : 30 247 New Way Chemistry for Hong Kong A-Level 3B

34.7

Structural Information from Chemical Properties (SB p.97) (a) Volume of C x H y reacted : Volume of O 2 = 15 : 45  1 ( 2  y 4 )  15 45   2  y 4  3 y = 4 The molecular formula of the compound is C 2 H 4 .

(b) C 2 H 4 belongs to alkenes.

(c) The structural formula of the hydrocarbon is:

Back

248 New Way Chemistry for Hong Kong A-Level 3B 4

34.7

Structural Information from Chemical Properties (SB p.97)

Answer

20 cm 105 o 3 of a gaseous organic compound containing only carbon, hydrogen and oxygen were mixed with 110 cm oxygen which was in excess. The mixture was exploded at C and the volume of the gaseous mixture was 150 cm reduced to 90 cm 3 . On adding concentrated potassium 3 of 3 After cooling to room temperature, the residual volume was hydroxide solution, the volume further decreased to 50 cm . 3 .

(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 C x H y O z .

Volume of C x H y O z reacted = 20 cm 3 Volume of unreacted oxygen = 50 cm 3 Volume of oxygen reacted = (110 - 50) cm 3 = 60 cm 3 Volume of carbon dioxide formed = (90 - 50) cm 3 = 40 cm 3 Volume of water (in the form of steam) formed = (90 - 50) cm 3 C x H y O z = 40 cm 3 y + (x + 4 z 2 )O 2  xCO 2 y 2 2 O Volume of C x H y O z  1  20 x 40  x = 2 reacted : Volume of CO 2 formed = 1 : x = 20 : 40 250 New Way Chemistry for Hong Kong A-Level 3B

34.7

Structural Information from Chemical Properties (SB p.98) (a) Volume of C x H y O z  2  20 y 60  y = 6 y reacted : Volume of H 2 O formed = 1 : = 20 : 60 2 Volume of C x H y O z reacted : Volume of O 2 ( x  y 4  z 2 )    = 20 : 60 1 (x  y 4 z 2 )  20 60 2  6 4  z  3 2 z = 1 The molecular formula of the compound is C 2 H 6 O.

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 C 2 H 5   OH group from the molecular formula of C after eliminating the 2 H 6 O. 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.

Carbon Hydrogen Nitrogen Oxygen Mass (g) Number of moles (mol) Relative number of moles Simplest mole ratio 42.8

42 .

8  3 .

57 12 .

0 3 1 .

.

57 19  3 2.38

16.67

38.15

2 .

38  2 .

38 1 .

0 16 .

67  1 .

19 14 .

0 38 .

15  2 .

38 16 .

0 2 1 .

.

38 19  2 1 1 .

.

19 19  1 2 .

38  2 1 .

19 3 2 1 2 ∴ The empirical formula of the compound is C 3 H 2 NO 2 .

254 New Way Chemistry for Hong Kong A-Level 3B

34.7

Structural Information from Chemical Properties (SB p.99) (a) (i) Let the molecular formula of the compound be (C 3 H 2 NO 2 )

n

.

n

× Molecular mass of (C 3 H 2 NO 2 )

n

(12.0 × 3 + 1.0 × 2 + 14.0 + 16.0 × = 168.0

2) = 168.0

∴ ∴

n

= 2 The molecular formula of the compound is C 6 H 4 N 2 O 4 .

(ii) 255 New Way Chemistry for Hong Kong A-Level 3B

34.7

Structural Information from Chemical Properties (SB p.99)

(b) 30 cm cm 3 3 of oxygen which was in excess, and the mixture was then exploded. After cooling to room temperature, the residual gases occupied 95 cm reduced by 60 cm 3 3 by volume. By adding potassium hydroxide solution, the volume was . The remaining gas was proved to be oxygen.

of a gaseous hydrocarbon were mixed with 140 (i) Determine the molecular formula of the hydrocarbon.

256

(ii) Is the hydrocarbon a saturated, an unsaturated or an aromatic hydrocarbon?

Answer

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 cm 3 Volume of unreacted oxygen = (95 – 60) cm 3 = 35 cm 3 Volume of oxygen reacted = (140 - 35) cm 3 = 105 cm 3 Volume of carbon dioxide formed = 60 cm 3 C x H y + (x + )O 4 2  xCO 2 + H 2 2 O Volume of C x H y reacted : Volume of CO 2 formed = 1 : x = 30 : 60  1 x  30 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) Volume of C x H y 2  reacted : Volume of O 2 4 reacted  ( x 1  y 4 )  30 105 30  ( 2    y 4 )  105 y = 6 The molecular formula of the compound is C 2 H 6 .

From the molecular formula of the hydrocarbon, it can be deduced that the hydrocarbon is saturated because it fulfils the general formula of alkanes C

n

H 2

n

+2 .

258 New Way Chemistry for Hong Kong A-Level 3B

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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.

Carbon Hydrogen Mass (g) Number of moles (mol) Relative number of moles Simplest mole ratio 85.5

85 .

5  7 .

125 12 .

0 7 .

125 7 .

125  1 1 14 7 .

14.5

5 1 .

0 14 .

5 .

125  14 .

5 2  ∴ The empirical formula of the compound is CH 2 .

2 260 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 (CH

2 )

n

.

Molecular mass of (CH 2 )

n n

× (12.0 + 1.0 × = 56.0

2) = 56.0

n

= 4 ∴ The molecular formula of the hydrocarbon is C 4 H 8 .

(ii) 261 (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.

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) 262 What is the relationship between frequency and wavenumber?

Answer

The higher the frequency, the higher the wavenumber.

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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 265 Carbon Hydrogen Oxygen Mass (g) Number of moles (mol) Relative number of moles Simplest mole ratio 66.66

66 .

66 12 .

0 5 .

56  5 .

56  4 1 .

39 4 11.11

∴ The empirical formula of the compound is C 4 H New Way Chemistry for Hong Kong A-Level 3B 8 O.

22.23

11 .

11  11 .

11 1 .

0 11 .

11  1 .

39 8 8 22 .

23 16 .

0 1 .

39  1 .

39  1 1 .

39 1

34.8

Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.110) Let the molecular formula of the compound be (C 4 H 8 O) n .

Relative molecular mass of (C 4 H 8 O) n n × (12.0 × 4 + 1.0 × = 72.0

8 + 16.0) = 72.0

∴ ∴ n = 1 The molecular formula of the compound is C 4 H 8 O.

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 cm –1 and 1 700 cm –1 . The absorption peak at 2 950 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: 268

Back

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

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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.

273

Decide which compound could be an alcohol. Explain your 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

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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 O



H C



N



H H Range of wavenumber (cm -1 ) 1680 – 1750 2500 – 3300 2840 – 3095 3350 – 3500

Answer

275 New Way Chemistry for Hong Kong A-Level 3B

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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

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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?

278 New Way Chemistry for Hong Kong A-Level 3B

Answer

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.

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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.

280 New Way Chemistry for Hong Kong A-Level 3B

Answer

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 –NH 2 Besides, the sharp peak at 1680 cm –1 group. implies that the compound also contains the C = O bond.

281 New Way Chemistry for Hong Kong A-Level 3B

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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

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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

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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.

Carbon Hydrogen Oxygen Mass (g) Number of moles (mol) Relative number of moles Simplest mole ratio 54.54

54 .

54 12 .

0 4 .

55  4 .

55  2 2 .

27 2 9.10

9 .

10  9 .

10 1 .

0 9 .

10 2 .

27  4 4 ∴ The empirical formula of the compound is C 2 H 4 O.

36.36

36 .

36  2 .

27 16 .

0 2 .

27 2 .

27  1 1 284 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 (C 2 H 4 O)

n

.

Relative molecular mass of (C 2 H 4 O)

n n

× (12.0 × 2 + 1.0 × = 88.0

4 + 16.0) = 88.0

n

= 2 ∴ The molecular formula of the compound is C 4 H 8 O 2 .

(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 (CH 3 COCH 2 CH 2 CH 3 ) is shown below: What ions do the peaks at

m/e

86, 71 and 43 represent?

287

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 C 1  C 2 bond is broken, the ion CH 3 CH 2 CH 2 CO + When the C 2  C 3 bond is broken, the ion CH 3 CO ( +

m/e

( = 71) is formed.

m/e

= 43) is formed.

288

Back

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 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) The relative molecular mass of hydrocarbon X is 58.0.

(b) The base peak is at

m/e

43.

(c) 15 mass units (d) Since the compound is a hydrocarbon, the molecular formula of the compound must be C

x

H

y

. 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. 291 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 CH 3 CH 2 CH 2 + .

(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

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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.

Answer

294 New Way Chemistry for Hong Kong A-Level 3B

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 C 6 H 5 CH 2 + . The peak at

m/e

43 corresponds to the ion CH 3 CO + .

295 New Way Chemistry for Hong Kong A-Level 3B

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Use of Mass Spectra to Obtain Structural Information (SB p.123) 296 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.

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New Way Chemistry for Hong Kong A-Level 3B

34.9

Use of Mass Spectra to Obtain Structural Information (SB p.123) 297 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.

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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

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Use of Mass Spectra to Obtain Structural Information (SB p.124)

(b) The following is the mass spectrum of bromomethylbenzene (benzyl bromide).

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 79 Br and 81 Br 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 C 6 H 5 CH 2 + .

300 New Way Chemistry for Hong Kong A-Level 3B

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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) 302 (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.

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New Way Chemistry for Hong Kong A-Level 3B