Structural Determination of Organic Compounds

1

Structural Determination Structural Determination of Organic Compoundsof Organic Compounds

34.134.1 IntroductionIntroduction

34.234.2 Isolation and Purification of Organic CompoundsIsolation and Purification of Organic Compounds

34.334.3 Tests for PurityTests for Purity

34.434.4 Qualitative Analysis of Elements in an Organic CompoundQualitative Analysis of Elements in an Organic Compound

34.534.5 Determination of Empirical Formula and Molecular Determination of Empirical Formula and Molecular

Formula from Analytical DataFormula from Analytical Data

34.634.6 Structural Information from Physical PropertiesStructural Information from Physical Properties

34.734.7 Structural Information from Chemical PropertiesStructural Information from Chemical Properties

34.834.8 Use of Infra-red Spectrocopy in the Identification of Use of Infra-red Spectrocopy in the Identification of

Functional GroupsFunctional Groups

34.934.9 Use of Mass Spectra to Obtain Structural InformationUse of Mass Spectra to Obtain Structural Information

3344

2

The general steps to determine the structure of an organic compound

3

Isolation and Isolation and Purification Purification of Organic of Organic

CompoundsCompounds

4

Technique Aim

1. Filtration To separate an insoluble solid from a liquid (slow)

2. Centrifugation To separate an insoluble solid from a liquid (fast)

3. Recrystallization To separate a solid from other solids based on their different solubilities in suitable solvent(s)

4. Solvent extraction

To separate a component from a mixture with a suitable solvent

5. Distillation To separate a liquid from a solution containing non-volatile solutes

5

Technique Aim

6. Fractional distillation

To separate miscible liquids with widely different boiling points

7. Steam distillation To separate liquids which are immiscible with water and decompose easily below their b.p.

8. Vacuum distillation ditto

9. Sublimation To separate a mixture of solids in which only one can sublime

10. Chromatography To separate a complex mixture of substances (large/small scale)

The mixture boils below 100C

6

• 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

Tests for PurityTests for Purity

7

34.2 Isolation and Purification of Organic Compounds (SB p.78)

Isolation and Purification Isolation and Purification of Organic Compoundsof Organic Compounds

• The selection of a proper technique

depends on the particular differences in physical properties of the substances present in the mixture

8


FiltrationFiltration

• 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



The laboratory set-up of filtration

10



• 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


CentrifugatioCentrifugationn• When there is only a small amount of

suspension, or when much faster separation is required

Centrifugation is often used instead of filtration

12


CentrifugatioCentrifugationn

• The liquid containing undissolved solids is put in a centrifuge tube

• The tubes are then put into the tube holders in a centrifuge

A centrifuge

13

34.2 Isolation and Purification of Organic Compounds (SB p.79)CentrifugatioCentrifugationn• 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


CrystallizatioCrystallizationn

• Crystals are solids that have

a definite regular shape

smooth flat faces and straight edges

• Crystallization is the process of forming crystals

15


1. Crystallization by Cooling a Hot 1. Crystallization by Cooling a Hot Concentrated SolutionConcentrated Solution

• To obtain crystals from an unsaturated aqueous solution

the solution is gently heated to make it more concentrated

• After, the solution is allowed to cool at room conditions

16



• The solubilities of most solids increase with temperature

• When a hot concentrated solution is cooled

the solution cannot hold all of the dissolved solutes

• The “excess” solute separates out as crystals

17



Crystallization by cooling a hot concentrated solution

18


2. Crystallization by Evaporating a 2. Crystallization by Evaporating a Cold Solution at Room TemperatureCold 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



• If a solution is allowed to stand at room temperature,

evaporation will be slow

• It may take days or even weeks for crystals to form

20



Crystallization by slow evaporation of a solution (preferably saturated) at room

temperature

21


Solvent Solvent ExtractionExtraction

• 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



• Often involves the use of a separating funnel

• When an aqueous solution containing the organic product is shaken with diethyl ether in a separating funnel,

the organic product dissolves into the ether layer

23



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

ether

24



• 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



• 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


DistillatioDistillationn

• 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



The laboratory set-up of distillation

28



• Before the solution is heated,

several pieces of anti-bumping granules are added into the flask

prevent vigorous movement of the liquid called bumping to occur

during heating

make boiling smooth

29



• If bumping occurs during distillation,

some solution (not yet vaporized) may spurt out into the collecting vessel

30


Fractional DistillationFractional Distillation

• A method used to separate a mixture of two or more miscible liquids

31



The laboratory set-up of fractional distillation

32



• 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


Fractional Fractional DistillationDistillation• 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


Fractional Fractional DistillationDistillation• 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


SublimatioSublimationn

• 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


SublimatioSublimationn• A mixture of two compounds is heated in an

evaporating dish

• One compound changes from solid to vapour directly

The vapour changes back to solid on a cold surface

• The other compound is not affected by heating and remains in the evaporating dish

37


SublimatioSublimationn

A mixture of two compounds can be separated by sublimation

38


ChromatographChromatographyy

• An effective method of separating a complex mixture of substances

• Paper chromatography is a common type of chromatography

39



The laboratory set-up of paper chromatography

40



• A solution of the mixture is dropped at one end of the filter paper

41



• The thin film of water adhered onto the surface of the filter paper forms the stationary phase

• The solvent is called the mobile phase or eluent

42



• 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


ChromatographChromatographyy• As the various components are being

adsorbed or partitioned at different rates,

they move upwards at different rates

• The ratio of the distance travelled by the substance to the distance travelled by the solvent

known as the Rf value

a characteristic of the substance

44


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

To separate a component from a mixture with a suitable solvent

(e) Distillation To separate a liquid from a solution containing non-volatile solutes

A summary of different techniques of isolation and purification

45


Technique Aim

(f) Fractional distillation

To separate miscible liquids with widely different boiling points

(g) Sublimation To separate a mixture of solids in which only one can sublime

(h) Chromatography

To separate a complex mixture of substances

A summary of different techniques of isolation and purification

Check Point 34-2Check Point 34-2

46

34.34.44 Qualitative Qualitative

Analysis of Analysis of Elements in an Elements in an

Organic Organic CompoundCompound

47

34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86)

• Qualitative analysis of an organic compound is

to determine what elements are present in the compound

Qualitative Analysis of Qualitative Analysis of an Organic Compoundan Organic Compound

48


Carbon and Carbon and HydrogenHydrogen• Tests for carbon and hydrogen in an

organic compound are usually unnecessary

an organic compound must contain carbon and hydrogen

49


Carbon and Carbon and HydrogenHydrogen• 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

50


Halogens, Nitrogen and Halogens, Nitrogen and SulphurSulphur• Halogens, nitrogen and sulphur in organic

compounds can be detected

by performing the sodium fusion test

51


Halogens, Nitrogen and Halogens, Nitrogen and SulphurSulphur• The compound under test is

fused with a small piece of sodium metal in a small combustion tube

heated strongly

• The products of the test are extracted with water and then analyzed

52


Halogens, Nitrogen and Halogens, Nitrogen and SulphurSulphur

• 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

53


Element Material used Observation

Halogens, as Acidified silver nitrate solution

chloride ion (Cl-) A white precipitate is formed. It is soluble in excess NH3(aq).

bromide ion (Br-) A pale yellow precipitate is formed. It is sparingly soluble in excess NH3(aq).

iodide ion (I-) A creamy yellow precipitate is formed. It is insoluble in excess NH3(aq).

Results for halogens, nitrogen and sulphur in the sodium fusion test

54


Results for halogens, nitrogen and sulphur in the sodium fusion test

Element Material used Observation

Nitrogen,as

cyanide ion (CN-)

A mixture of iron(II) sulphate and iron(III) sulphate solutions

A blue-green colour is observed.

Sulphur, assulphide ion (S2-)

Sodium pentacyanonitrosylferrate(II) solution

A black precipitate is formed


55

34.34.55Determination of Determination of

Empirical Empirical Formula and Formula and

Molecular Molecular Formula from Formula from

Analytical DataAnalytical Data

56

34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87)

• After determining the constituent elements of a particular organic compound

perform quantitative analysis to find the percentage composition by mass of the compound

the masses of different elements in an organic compound are determined

Quantitative Analysis of Quantitative Analysis of an Organic Compoundan Organic Compound

57


1. Carbon and 1. Carbon and

HydrogenHydrogen• The organic compound is burnt in excess oxygen

• The carbon dioxide and water vapour formed are respectively absorbed by

potassium hydroxide solution and anhydrous calcium chloride

58


1. Carbon and 1. Carbon and

HydrogenHydrogen• The increases in mass in potassium hydroxide solution and calcium chloride represent

the masses of carbon dioxide and water vapour formed respectively

59


2. 2.

NitrogenNitrogen• 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

60


3. 3.

HalogensHalogens• The organic compound is heated with

fuming nitric(V) acid and excess silver nitrate solution

• The mixture is allowed to cool

then water is added

the dry silver halide formed is weighed

61


4. 4.

SulphurSulphur• The organic compound is heated with

fuming nitric(V) acid

• After cooling,

barium nitrate solution is added

the dry barium sulphate formed is weighed

62



• After determining the percentage composition by mass of a compound,

the empirical formula of the compound can be calculated

63



The empirical formula of a compound is the formula which shows the simplest whole number ratio of the atoms present in the compound

64



• When the relative molecular mass and the empirical formula of the compound are known,

the molecular formula of the compound can be calculated

65



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

66


Example 34-5AExample 34-5A Example 34-5BExample 34-5B


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34.34.66 Structural Structural

Information Information from Physical from Physical

PropertiesProperties

68

34.6 Structural Information from Physical Properties (SB p.89)

• The physical properties of a compound include its colour, odour, density, solubility, melting point and boiling point

• The physical properties of a compound depend on its molecular structure

Structural Information from Structural Information from Physical PropertiesPhysical Properties

69



• From the physical properties of a compound,

obtain preliminary information about the structure of the compound

70


Structural Information from Structural Information from Physical PropertiesPhysical Properties• e.g.

Hydrocarbons have low densities, often about 0.8 g cm–3

Compounds with functional groups have higher densities

71



• 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

72


Organic compound

Density at 20oC

Melting point and boiling point

Solubility

In water or highly

polar solvents

In non-polar

organic solvents

Hydrocarbons (saturated and unsaturated)

All

have densities < 0.8 g cm–3

• Generally low but increases with number of carbon atoms in the molecule

• Branched-chain hydrocarbons have lower boiling points but higher melting points than the corresponding straight-chain isomers

Insoluble Soluble

Physical properties of some common organic compounds

73


Organic compound

Density at 20oC


Solubility

In water or highly

polar solvents

In non-polar

organic solvents

Aromatic hydrocarbons

Between 0.8 and 1.0 g cm–3

Generally low Insoluble Soluble


74


Organic compound

Density at 20oC


Solubility

In water or highly

polar solvents

In non-polar

organic solvents

Halo-alkanes

• 0.9 - 1.1 g cm–3 for chloro-alkanes

• >1.0 g cm–3 for bromo-alkanes and iodo-alkanes

• Higher than alkanes of similar relative molecular masses ( haloalkane molecules are polar)

• All haloalkanes are liquids except halomethanes

• Both the m.p. and b.p. increase in the order: RCH2F < RCH2Cl < RCH2Br < RCH2I

Insoluble Soluble


75


Organic comp-ound

Density at 20oC


Solubility

In water or highly polar

solvents

In non-polar

organic solvents

Alcohols • 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)

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

Soluble


76


Physical properties of some common organic compoundsOrganic

comp-ound

Density at 20oC


Solubility


solvents

In non-polar

organic solvents

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

• Straight-chain alcohols have higher b.p. than the corresponding branched-chain alcohols

• Solubility decreases gradually as the hydrocarbon chain lengthens

Soluble

77



comp-ound

Density at 20oC


Solubility


solvents

In non-polar

organic solvents

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)

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

Soluble

78



comp-ound

Density at 20oC


Solubility


solvents

In non-polar

organic solvents

Carbonyl

comp-ounds

(alde-hydes

and ketones)

• > 1.0 g cm–3 for aromatic carbonyl compounds


Soluble

79



comp-ound

Density at 20oC


Solubility


solvents

In non-polar

organic solvents

Carbo-xylic

acids

• Lower members have densities similar to water

• Methanoic acid has a density of 1.22 g cm–3

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

• First four members are

miscible with water in all proportions


Soluble

80



comp-ound

Density at 20oC


Solubility


solvents

In non-polar

organic solvents

Esters Lower members

have densities less than water

Slightly higher than hydrocarbons but lower than carbonyl compounds and

alcohols of similar relative molecular masses

Insoluble Soluble

81



comp-ound

Density at 20oC


Solubility


solvents

In non-polar

organic solvents

Amines Most amines have densities less than water

• Higher than alkanes but lower than alcohols of similar relative molecular masses

• Generally soluble

• Solubility decreases in the order:

1o amines > 2o amines > 3o amines

Soluble

82



comp-ound

Density at 20oC


Solubility


solvents

In non-polar

organic solvents

Amines • 1o and 2o amines are able to form hydrogen bonds with each other but the strength is less than that between alcohol molecules (NH bond is less polar than O H bond)

83



comp-ound

Density at 20oC


Solubility


solvents

In non-polar

organic solvents

Amines • 3o amines have lower m.p. and b.p. than the isomers of 1o and 2o amines ( molecules of 3o amines cannot form intermolecular hydrogen bonds)

84


Example 34-6Example 34-6 Check Point 34-6Check Point 34-6

85

34.34.77 Structural Structural

Information Information from Chemical from Chemical

PropertiesProperties

86

34.7 Structural Information from Chemical Properties (SB p.93)

Structural Information from Structural Information from Chemical PropertiesChemical Properties

• The molecular formula of a compound

does not give enough clue to the structure of the compound

• Compounds having the same molecular formula

may have different arrangements of atoms and even different functional groups

87



• e.g.The molecular formula of C2H4O2 may represent a carboxylic acid or an ester:

88



• The next stage is

to find out the functional group(s) present

to deduce the actual arrangement of atoms in the molecule

89


Organic compound

Test Observation

Saturated

hydrocarbons

• Burn the saturated hydrocarbon in a non-luminous Bunsen flame

• A blue or clear yellow flame is observed

Chemical tests for different groups of organic compounds

90


Organic compound

Test Observation

Unsaturated hydrocarbons (C = C, C C)

• Burn the unsaturated hydrocarbon in a non-luminous Bunsen flame

• A smoky flame is observed

• Add bromine in 1,1,1-trichloroethane at room temperature and in the absence of light

• Bromine decolourizes rapidly

• Add 1% (dilute) acidified potassium manganate(VII) solution

• Potassium manganate(VII) solution decolourizes rapidly


91


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


92


Organic compound

Test Observation

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

• No precipitate is formed


93


Chemical tests for different groups of organic compoundsOrganic

compoundTest Observation

Alcohols ( OH)

• Add a small piece of sodium metal

• A colourless gas is evolved

• Esterification: Add ethanoyl chloride

• The temperature of the reaction mixture rises

• A colourless gas is evolved

94




Alcohols ( OH)

• Add acidified potassium dichromate(VI) solution

• For 1° and 2° alcohols, the clear orange solution becomes opaque and turns green almost immediately

• For 3° alcohols, there are no observable changes

95




Alcohols ( OH)

• Iodoform test for:

Add iodine in sodium hydroxide solution

• A yellow precipitate is formed

96



Alcohols ( OH)

• Lucas test: add a solution of zinc chloride in concentrated hydrochloric acid

• For 1° alcohols, the aqueous phase remains clear

• For 2° alcohols, the clear solution becomes cloudy within 5 minutes

• For 3° alcohols, the aqueous phase appears cloudy immediately


97



Ethers ( O )

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


98


Organic compound

Test Observation

Aldehydes

( )

• Add aqueous sodium hydrogensulphate(IV)

• Crystalline salts are formed

• Add 2,4-dinitrophenylhydrazine

• A yellow, orange or red precipitate is formed

• Silver mirror test: add Tollens’ reagent (a solution of aqueous silver nitrate in aqueous ammonia)

• A silver mirror is deposited on the inner wall of the test tube


99


Organic compound

Test Observation

Ketones

( )

• Add aqueous sodium hydrogensulphate(IV)

• Crystalline salts are formed (for unhindered ketones only)

• Add 2,4-dinitrophenylhydrazine

• A yellow, orange or red precipitate is formed

• Iodoform test for:

Add iodine in sodium hydroxide solution

• A yellow precipitate is formed


100


Organic compound

Test Observation

Carboxylic acids

( )

• Esterification: warm the carboxylic acid with an alcohol in the presence of concentrated sulphuric(VI) acid, followed by adding sodium carbonate solution

• A sweet and fruity smell is detected

• Add sodium hydrogencarbonate

• The colourless gas produced turns lime water milky


101


Organic compound

Test Observation

Esters

( )

• No specific test for esters but they can be distinguished by its characteristic smell

• A sweet and fruity smell is detected


102


Organic compound

Test Observation

Acyl halides

( )

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

• For acyl chlorides, a white precipitate is formed

• For acyl bromides, a pale yellow precipitate is formed

• For acyl iodides, a creamy yellow precipitate is formed


103


Organic compound

Test Observation

Amides

( )

• Boil with sodium hydroxide solution

• The colourless gas produced turns moist red litmus paper or pH paper blue


104


Organic compound

Test Observation

Amines

(NH2)

• 1o aliphatic amines: dissolve the amine in dilute hydrochloric acid at 0 – 5 oC, then add cold sodium nitrate(III) solution slowly

• Steady evolution of N2(g) is observed

• 1o aromatic amines: add naphthalen-2-ol in dilute sodium hydroxide solution

• An orange or red precipitate is formed


105


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


106


Example 34-7AExample 34-7A Example 34-7BExample 34-7B

Example 34-7CExample 34-7C Check Point 34-7Check Point 34-7

107

The END

108

34.1 Introduction (SB p.77)

What are the necessary information to determine the structure of an organic compound? AnswerMolecular formula from analytical data,

functional group present from physical and

chemical properties, structural information from

infra-red spectroscopy and mass spectrometry

Back

109


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 liquidAnswer(a) Centrifugation

(b) Chromatography

(c) Fractional distillation

(d) Sublimation

(e) Filtration

Back

110


(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

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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

112


Let the mass of the compound be 100 g. Then,

mass of carbon in the compound = 40.0 g

mass of hydrogen in the compound = 6.7 g

mass of oxygen in the compound = 53.3 g

∴ The empirical formula of the organic compound is CH2O.

Carbon Hydrogen Oxygen

Mass (g) 40.0 6.7 53.3

Number of moles (mol)

Relative number of moles

Simplest mole ratio

1 2 1

33.312.040.0

7.61.06.7

33.316.053.3

13.333.33

23.336.7

13.333.33

Back

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An organic compound Z has the following composition by mass:

(a) Calculate the empirical formula of compound Z.

(b) If the relative molecular mass of compound Z is 60.0, determine the molecular formula of compound Z. Answer

Element Carbon Hydrogen

Oxygen

Percentage by mass (%)

60.00 13.33 26.67

114


(a) Let the mass of the compound be 100 g. Then,

mass of carbon in the compound = 60.00 g

mass of hydrogen in the compound = 13.33 g

mass of oxygen in the compound = 26.67 g

∴ The empirical formula of the organic compound is C3H8O.


Mass (g) 60.00 13.33 26.67



Simplest mole ratio

3 8 1

512.060.00

33.131.0

13.33 67.1

16.026.67

81.67

13.33 1

1.671.67

31.67

5

115


(b) The molecular formula of the compound is (C3H8O)n.

Relative molecular mass of (C3H8O)n = 60.0

n × (12.0 × 3 + 1.0 × 8 + 16.0) = 60.0

n = 1

∴ The molecular formula of compound Z is C3H8O.

Back

116


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

117


Relative molecular mass of CO2 = 12.0 + 16.0 × 2 = 44.0

Mass of carbon in 0.22 g of CO2 = 0.22 g ×

= 0.06 g

Relative molecular mass of H2O = 1.0 × 2 + 16.0

= 18.0

Mass of hydrogen in 0.09 g of H2O = 0.09 g ×

= 0.01 g

Mass of oxygen in the compound = (0.15 – 0.06 – 0.01) g

= 0.08 g

44.012.0

18.02.0

118


∴ The empirical formula of the organic compound is CH2O.


Mass (g) 0.06 0.01 0.08



Simplest mole ratio

1 2 1

005.012.00.06

01.01.00.01

005.016.00.08

20.0050.01

10.0050.005

10.0050.005

119


Let the molecular formula of the compound be (CH2O)n.

Relative molecular mass of (CH2O)n= 60.0

n × (12.0 + 1.0 × 2 + 16.0) = 60.0

n = 2

∴ The molecular formula of the compound is C2H4O2.

Back

120


Why do branched-chain hydrocarbons have lower boiling points but higher melting points than the

corresponding straight-chain isomers?

AnswerBranched-chain hydrocarbons have lower boiling points than the

corresponding straight-chain isomers because the straight-chain

isomers are being flattened in shape. They have greater surface

area in contact with each other. Hence, molecules of the 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

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

break down the efficient packing in the process of melting.

Back

121


Why does the solubility of amines in water decrease in the order:

1o amines > 2o amines > 3o amines?

AnswerThe 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.

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122


Match the boiling points 65oC, –6oC and –88oC with the compounds CH3CH3, CH3NH2 and CH3OH. Explain your answer briefly.

Answer

123


Compounds Boiling point (°C)

CH3CH3 –88

CH3NH2 –6

CH3OH 65

Ethane (CH3CH3) is a non-polar compound. In pure liquid form, ethane

molecules are held together by weak van der Waals’ forces. However,

both methylamine (CH3NH2) and methanol (CH3OH) are polar

substances. In pure liquid form, their molecules are held together by

intermolecular hydrogen bonds. As van der Waals’ forces are much

weaker than hydrogen bonds, ethane has the lowest boiling point

among the three. Besides, as the O H bond in alcohols is more

polar than the N H bond in amines, the hydrogen bonds formed

between methylamine molecules are weaker than those formed

between methanol molecules. Thus, methylamine has a lower boiling

point than methanol.

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124


(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 and butanal are 74.0 and 72.0

respectively.

(ii) 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.

125


(b) Arrange the following compounds in order of increasing solubility in water. Explain your answer.

Ethanol, chloroethane, hexan-1-olAnswer(b) The solubility increases in the order: chloroethane <

hexan-1-ol < ethanol. Both hexan-1-ol and ethanol are

more soluble in water than chloroethane because

molecules of the alcohols are able to form extensive

hydrogen bonds with water molecules. Molecules of

chloroethane are not able to form hydrogen bonds with

water molecules and that is why it is insoluble in water.

Hexan-1-ol has a longer carbon chain than ethanol and

this explains why it is less soluble in water than ethanol.

126


(c) Explain why (CH3)3N (b.p.: 2.9°C) boils so much lower than CH3CH2CH2NH2 (b.p.: 48.7°C) despite they have the same molecular mass. Answer

(c) They are isomers. The primary amine is able to form

hydrogen bonds with the oxygen atom of water

molecules, but there is no hydrogen atoms directly

attached to the nitrogen atom in the tertiary amine.

127


(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°CAnswer(d)

1252,4-Dimethylpentan-3-one

144Heptan-4-one

155Heptanal

Boiling point (oC)Compound

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128


The empirical formula of an organic compound is CH2O and its relative molecular mass is 60.0. It reacts with sodium hydrogencarbonate to give a colourless gas which turns lime water milky.

(a) Calculate the molecular formula of the compound.

Answer(a) Let the molecular formula of the compound be (CH2O)n.

Relative molecular mass of (CH2O)n= 60.0

n (12.0 + 1.0 2 + 16.0) = 60.0

n = 2

∴ The molecular formula of the compound is C2H4O2.

129



(b) Deduce the structural formula of the compound.

Answer

130


(b) The compound reacts with sodium hydrogencarbonate to give a

colourless gas which turns lime water milky. This indicates that

the compound contains a carboxyl group ( COOH). Eliminating

the COOH group from the molecular formula of C2H4O2, the

atoms left are one carbon and three hydrogen atoms. This

obviously shows that a methyl group ( CH3) is present.

Therefore, the structural formula of the compound is:

131



(c) Give the IUPAC name for the compound.Answer(c) The IUPAC name for the compound

is ethanoic acid.

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132


15 cm3 of a gaseous hydrocarbon were mixed with 120 cm3 of oxygen which was in excess. The mixture was exploded. After cooling, the residual volume was 105 cm3. On adding concentrated potassium hydroxide solution, the volume decreased to 75 cm3.

(a) Calculate the molecular formula of the compound, assuming all the volumes were measured under room temperature and pressure.

(b) To which homologous series does the hydrocarbon belong?

(c) Give the structural formula of the hydrocarbon.

Answer

133


(a) Let the molecular formula of the compound be CxHy.

Volume of CxHy reacted = 15 cm3

Volume of unreacted oxygen = 75 cm3

Volume of oxygen reacted = (120 - 75) cm3 = 45 cm3

Volume of carbon dioxide formed = (105 - 75) cm3 = 30 cm3

CxHy + (x + )O2 xCO2 + H2O

Volume of CxHy reacted : Volume of CO2 formed = 1 : x = 15 : 30

x = 2

2y

4y

3015

x1

134


(a) Volume of CxHy reacted : Volume of O2 reacted = 1 : ( )

= 15 : 45

y = 4

The molecular formula of the compound is C2H4.

(b) C2H4 belongs to alkenes.

(c) The structural formula of the hydrocarbon is:

4y

x

4515

)4y

2(

1

34y

2

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135


20 cm3 of a gaseous organic compound containing only carbon, hydrogen and oxygen were mixed with 110 cm3 of oxygen which was in excess. The mixture was exploded at 105oC and the volume of the gaseous mixture was 150 cm3. After cooling to room temperature, the residual volume was reduced to 90 cm3. On adding concentrated potassium hydroxide solution, the volume further decreased to 50 cm3.

(a) Calculate the molecular formula of the compound, assuming that all the volumes were measured under room temperature and pressure.

(b) The compound is found to contain a hydroxyl group ( OH) in its structure. Deduce its structural formula.

(c) Is the compound optically active? Explain your answer.

Answer

136


(a) Let the molecular formula of the compound be CxHyOz.

Volume of CxHyOz reacted = 20 cm3

Volume of unreacted oxygen = 50 cm3


Volume of carbon dioxide formed = (90 - 50) cm3 = 40 cm3

Volume of water (in the form of steam) formed

= (90 - 50) cm3 = 40 cm3

CxHyOz + (x + - )O2 xCO2 + H2O

Volume of CxHyOz reacted : Volume of CO2 formed = 1 : x = 20 : 40

x = 2

2y

4y

2z

4020

x1

137


(a) Volume of CxHyOz reacted : Volume of H2O formed = 1 : = 20 : 60

y = 6

Volume of CxHyOz reacted : Volume of O2 reacted = 1 :

= 20 : 60

z = 1

The molecular formula of the compound is C2H6O.

2y

6020

y2

)2z

4y

x(

6020

)2z

-4y

(x

1

32z

46

2

138


(b) As the compound contains a OH group, the hydrocarbon

skeleton of the compound becomes C2H5 after eliminating the

OH group from the molecular formula of C2H6O. The structural

formula of the compound is:

(c) The compound is optically inactive as both carbon atoms in the

compound are not asymmetric, i.e. both of them do not attach to

four different atoms or groups of atoms.

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139


(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

140


(a) (i) Let the mass of the compound be 100 g.

∴ The empirical formula of the compound is C3H2NO2.

Carbon Hydrogen Nitrogen Oxygen

Mass (g) 42.8 2.38 16.67 38.15



Simplest mole ratio

3 2 1 2

57.30.128.42

38.20.138.2

19.10.14

67.16 38.2

0.1615.38

319.157.3

219.138.2

119.119.1

219.138.2

141


(a) (i) Let the molecular formula of the compound be (C3H2NO2)n.

Molecular mass of (C3H2NO2)n = 168.0

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

∴ n = 2

∴ The molecular formula of the compound is C6H4N2O4.

(ii)

142


(b) 30 cm3 of a gaseous hydrocarbon were mixed with 140 cm3 of oxygen which was in excess, and the mixture was then exploded. After cooling to room temperature, the residual gases occupied 95 cm3 by volume. By adding potassium hydroxide solution, the volume was reduced by 60 cm3. The remaining gas was proved to be oxygen.

(i) Determine the molecular formula of the hydrocarbon.

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

143


(b) (i) Volume of hydrocarbon reacted = 30 cm3

Volume of unreacted oxygen = (95 – 60) cm3 = 35 cm3


Volume of carbon dioxide formed = 60 cm3

CxHy + (x + )O2 xCO2 + H2O

Volume of CxHy reacted : Volume of CO2 formed

= 1 : x = 30 : 60

x = 2

2y

4y

6030

x1

144


(b) (i) Volume of CxHy reacted : Volume of O2 reacted

= 1 : ( ) = 30 : 105

y = 6

The molecular formula of the compound is C2H6.

(ii) From the molecular formula of the hydrocarbon, it can

be deduced that the hydrocarbon is saturated because it

fulfils the general formula of alkanes CnH2n+2.

4y

2

10530

)4y

x(

1

105)4y

2(30

145

(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


146


(c) (i) Let the mass of the compound be 100 g.

∴ The empirical formula of the compound is CH2.

Carbon Hydrogen

Mass (g) 85.5 14.5



Simplest mole ratio

1 2

125.70.125.85

5.140.15.14

1125.7125.7

2125.7

5.14

147


(c) (i) Let the molecular formula of the hydrocarbon be (CH2)n.

Molecular mass of (CH2)n = 56.0

n × (12.0 + 1.0 × 2) = 56.0

n = 4

∴ The molecular formula of the hydrocarbon is C4H8.

(ii)

(iii) Since but-2-ene is unsymmetrical and free rotation of but-

2-ene is restricted by the presence of the carbon-carbon double

bond, geometrical isomerism exists.

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148

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

What is the relationship between frequency andwavenumber?

AnswerThe higher the frequency, the higher the

wavenumber.

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Structural Determination of Organic Compounds