1 structural determination of organic compounds 34.1introduction 34.2isolation and purification of...

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

Use of Infra-red Use of Infra-red Spectrocopy in Spectrocopy in

the the Identification of Identification of

Functional Functional GroupsGroups

8

Infra-red Infra-red SpectroscopySpectroscopyArises from absorption of IR radiation by organic compounds

Causes atoms and groups of atoms of organic compounds to vibrate with increased amplitude about the covalent bonds that connect them

9

Two basic modes : stretching and bending

Two atoms joined by a covalent bond can undergo a stretching vibration where the atoms move back and forth as if they were joined by a spring

A stretching vibration of two atoms

Modes of vibrationsModes of vibrations

10

Symmetrical stretching : -

Asymmetrical stretching : -


11

Bending (scissoring) : - two modes

Behind paperOut of paper


12


Symmetric stretching

Asymmetric stretching

An in-plane bending

(scissoring)

An out-of-plane bending

(twisting)

Rocking

Wagging

13

The frequency() of a given stretching vibration of a covalent bond depends on

μk

νBond strength

Reduced mass of the system

yx

yx

mm

mmμ

For a diatomic molecule X – Y

14

Bond strength

k

C – C < C = C < C C

C – O < C = O

Increasing frequency of vibration

μk

ν

15

Masses of bonding atoms

C – H O - H N - H

μk

ν

Chemical bonds containing H atoms have high frequency of vibration due to the small mass of H

16

• Molecular vibrations are quantized

the molecules absorb IR radiation of a particular amount of energy only.

Infra-red Infra-red SpectroscopySpectroscopy

• Only IR radiation with the same frequency as the vibrational frequency can be absorbed by the molecules.

E = h

17


E = h

After the absorption of a quantum of energy (h), the amplitude of vibration but the frequency of vibration () remains unchanged.

18

Frequency (Hz / s1)

Wavelength (m)

The electromagnetic spectrumThe electromagnetic spectrum

UV / visible / Near IR electronic transition

E = hL = 400 – 1000 kJ mol1

19

Frequency (Hz / s1)

Wavelength (m)


Mid IR vibrational transition

E = hL = 5 – 40 kJ mol1

20

Frequency (Hz / s1)

Wavelength (m)


Far IR / microwave rotational transition

E = hL 0.02 kJ mol1

21

Frequency (Hz / s1)

Wavelength (m)


translational transitionE = hL 0 kJ mol1

22

• Different functional groups (e.g. C=O, O-H) have different vibration frequencies

They have characteristic absorption frequencies


• Functional groups can be identified from their characteristic absorption frequencies

23

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

3350 – 3500N H

3230 – 3670O H

2840 – 3095C H

Range of wavenumber (cm-1)

Bond

Characteristic absorption wavenumbers of some single bonds in infra-red spectra


24

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

1680 – 1750C = O

1610 – 1680C = C

2200 – 2280C N

2070 – 2250C C

Range of wavenumber (cm-1)

Bond

Characteristic absorption wavenumbers of some double bonds and triple bonds in infra-

red spectra


25

• An IR spectrum is obtained by scanning the sample with IR radiations from

1.21013 Hz to 1.21014 Hz


Or,

• % transmittance rather than absorbance is displayed

Wavenumber : 400 cm1 to 4000 cm-1

c

1)(cmwavenumber 1

26

Dips(peaks) show absorptions by functional groups

C=C

C

HC

H

27

Fingerprintregion

Band region

28

• Infra-red spectrometer is used to

measure the amount of energy absorbed at each wavelength of the IR region

An infra-red spectrometer


30

• A beam of IR radiation is passed through the sample

the intensity of the emergent radiation is carefully measured

• The spectrometer plots the results as a graph called infra-red spectrum

shows the absorption of IR radiation by a sample at different frequencies


31

• When a compound absorbs IR radiation,

the intensity of transmitted radiation decreases

results in a decrease in percentage of transmittance

a dip in the spectrum

often called an absorption peak or absorption band

Use of IR Spectrum in the Use of IR Spectrum in the Identification of Functional GroupsIdentification of Functional Groups

32

• In general, an IR spectrum can be split into four regions for interpretation purpose


33

Range of wavenumber

(cm-1)

Interpretation

400 – 1500 • Often consists of many complicated bands (stretching and bending)

• Unique to each compound• Often called the fingerprint region• Not used for identification of

particular functional groups

1500 – 2000 Absorption of double bonds,e.g. C = C, C = O

2000 – 2500 Absorption of triple bonds, e.g. C C, C N

2500 – 4000 Absorption of single bonds involving hydrogen, e.g. C H, O H, N H

34

• The region between 4 000 cm-1 and 1 500 cm-1 is often used for

identification of functional groups from their characteristic

absorption wavenumbers


35

Compound Bond Characteristic range of wavenumber (cm-1)

Alkenes C = C 1610 – 1680

Aldehydes, ketones, acids, esters

C = O 1680 – 1750

Alkynes C C 2070 – 2250

Nitriles C N 2200 – 2280

Acids (hydrogen-bonded) O H 2500 – 3300

Alkanes, alkenes, arenes C H 2840 – 3095

Alcohols, phenols (hydrogen-bonded)

O H 3230 – 3670

Primary amines N H 3350 – 3500

What is the characteristic range of wavenumber of C=N bond?Bond strength/wavenumber : C=C < C=N < C=O

36

Strategies for the Use of IR Spectra in Strategies for the Use of IR Spectra in the Identification of Functional the Identification of Functional GroupsGroups

1. Focus at the IR absorption peak at or above 1500 cm–1

Concentrate initially on the major absorption peaks

37

2. The absence and presence of absorption peaks at some characteristic ranges of wavenumbers are equally important

the absence of particular absorption peaks can be used to eliminate the presence of certain functional groups or bonds in the molecule

Strategies for the Use of IR Spectra in Strategies for the Use of IR Spectra in the Identification of Functional the Identification of Functional GroupsGroups

38

Limitation of the Use of IR Spectroscopy Limitation of the Use of IR Spectroscopy in the Identification of Organic in the Identification of Organic CompoundsCompounds

1. Some IR absorption peaks have very close wavenumbers and the peaks always coalesce

2. Not all vibrations give rise to strong absorption peaks

39

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

Limitation of the Use of IR Spectroscopy Limitation of the Use of IR Spectroscopy in the Identification of Organic in the Identification of Organic CompoundsCompounds

40

1610-1680 cm1

1680-1750 cm1

Absorption Slightly above

3000 cm1

AlkeneArene

Y

N

Y

C

H

C

H

C=C bonds in benzene are weaker ~1610 cm1

41

1680-1750 cm1

Broad2500-3300 cm1

YY

2070-2280 cm1

N

Aldehyde, Ketone, Ester

N

Y

Carboxylic acid

Containoxygen

C

O

OH

C

O

H

C

O

C

O

O

Aldehyde : lower C-H absorption at 2720-2820 cm-1

42

2070-2280 cm1 Containnitrogen

Y

Nitrile2200-2280 cm1

Y

Alkyne2070-2250 cm1

N

N

3200-3700 cm1

Additional sharp peak ~3250 cm1 for termnal alkyne, -CC-H

43

Containnitrogen

Y

Amine3350-3500 cm1

Y

NAlkane

2840-3000 cm1

N

3200-3700 cm1

Containoxygen

Y

Alcohol, phenol3230-3700 cm1

Single peak for 2 amine; double peaks for 1 amine

44

3300

3100

Several peaks (2840-3000 cm1) due todifferent modes of Symmetrical & asymmetrical stretching

Different modes of C-H bending (~1400 cm1)

weak

medium

45

C

H

C

C

46

Q.82

47

Q.82

48

Q.82

49

Q.82

C

C

C

H

50

Q.82

51

Q.83

52

Q.83

53

Q.83

2720- 28202840-

3000

2840-3000

54

Q.83

2720- 28202840-

3000

2840-3000

55

Q.84 C3H5N IOU = 3 + 1 -5/2 + ½ = 2

CC, CN, diene

CC, CN

56

C

H

H

H

C

H

H

C NC CH C N H

HH

H

H C

H

H

C C N

H

H

CC, CN

C-H1 amine

57

Q.85 C6H10O IOU = 6 + 1 -10/2 = 2

Ozonolysis gives CO2 terminal C=CH2

58

Q.85 C6H10O IOU = 6 + 1 -10/2 = 2

+ve iodoform test C O

H3C

H3C C

OH

H

59

Q.85 C6H10O IOU = 6 + 1 -10/2 = 2

No broadband within 3200-3600 H3C C

O

=C-H

C=C

C=O

-C-H

60

C

O

H3C *

C

O

H3C

C

O

H3C

61

H

C

C H

O

C O

62

C

C

C

H

C

H

63

H

C

C C

C C H

64

H

C

1 amine

No C-H stretching above 3000 cm1

No C=C-H

N-H bendin

g

65

O H

C

H

C O

C

CBroad band with peak at about 2900 cm1

C O

66

H

C

C O

C O

67

H

C

C

C

1 amine

C

H

68

H

C

C N

69

O H

C

C

C OBroad band with peak at about 2900 cm1

C O

70

H

C

O H

C C

C C H

C O

Broad band with peak at about 3300 cm1

71

H

C

C C

N C H

C N

C O

72

H

C

C

H

C

C

C O

C O

73

H

CC

H

C

C

74

C

H

C

C

No aliphatic C-H stretching

C H

O

C O

75

C

HC

C

C N


76

H

C

O H

C C

C C H

C O


77

H

C

O HC O


78

H

C

O HC O


C O

79

H

C

1 amine

No C-H stretching above 3000 cm1

No C=C-H

N-H bendin

g

80

H

C

O HC O


81

H

CC O

82

H

C

C N

83

H

C

O H

C O


84

H

C

C

H

C

C

C O

C O

85

H

CO H

C C

C C H

C O


86

H

C

C C

C C H

87

H

C

O HC O


C O

88

H

C

C O

C O

89

H

C

O H C O


C O

90

H

C

C O

91

H

C

C H

O

C O

92

O H

C

H

C

C

C O



93

H

C

O H

C O


94

H

CC O

95

H

C

C

C

C

H

96

C O

H

C

C

C

2 amine

C

H

97

4. 4. ButanoneButanone

The IR spectrum of butanone

98

4. 4. ButanoneButanoneWavenumber (cm-1) Intensity Indication

2983 Strong C H stretching

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

99

5. Butan-1-5. Butan-1-olol

a broad band is observed

the vibration of the O H group is complicated by the hydrogen

bonding formed between the molecules

100

5. Butan-1-5. Butan-1-ololWavenumber

(cm-1)Intensity Indication

3330 Broad band O H stretching

2960 Medium C H stretching

2935 Medium

2875 Medium

Interpretation of the IR spectrum of butan-1-ol

101

6. Butanoic 6. Butanoic AcidAcid

a broad band is observed

the vibration of the O H group is complicated by the hydrogen

bonding formed between the molecules

102

6. Butanoic 6. Butanoic AcidAcidWavenumber (cm-

1)Intensity Indication

3100 Broad band O H stretching

1708 Strong C = O stretching

Interpretation of the IR spectrum of butanoic acid

103

7. 7. ButylaminButylaminee

The IR spectrum of butylamine

104

7. 7. ButylaminButylamineeWavenumber (cm-

1)Intensity Indication

3371 Strong N H stretching

3280 Strong

2960 – 2875 Weak C H stretching

1610 Medium N H bending

1475 Medium C H bending

Interpretation of the IR spectrum of butylamine

105

8. 8. ButanenitrileButanenitrile

The IR spectrum of butanenitrile

106

8. 8. ButanenitrileButanenitrile

Wavenumber (cm-1)

Intensity Indication

2990 – 2895 Strong C H stretching

2246 Very strong C N stretching

1420 Strong C H bending

1480 Strong

Interpretation of the IR spectrum of butanenitrile

107

Example 34-8Example 34-8 Check Point 34-8Check Point 34-8

108

34.34.99 Use of Mass Use of Mass

Spectra to Spectra to Obtain Obtain

Structural Structural InformationInformation

109

34.9 Use of Mass Spectra to Obtain Structural Information (SB p.113)

Mass Mass SpectrometrySpectrometry

• One of the most sensitive and versatile analytical tools

• More sensitive than other spectroscopic methods (e.g. IR spectroscopy)

• Only a microgram or less of materials is required for the analysis

110


Mass Mass SpectrometrySpectrometryIn 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

111



• 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

112



Components of a mass spectrometer

113



In the vaporization chamber,

• the sample is heated until it vaporizes

changes to the gaseous state

114

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

SpectrometrySpectrometry• The molecules in the gaseous state are

bombarded with a beam of fast-moving electrons

Positively-charged ions called the molecular ions are formed

One of the electrons of the molecule is knocked off

115


SpectrometrySpectrometry

• Molecular ions are sometimes referred to as the parent ion

116


SpectrometrySpectrometry 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

117


SpectrometrySpectrometry• e.g.

if a molecule of methanol (CH3OH) is bombarded with a beam of fast-moving electrons

the following reaction will take place:

118



• 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

119



• The way that a molecular ion fragments

give us highly useful information about the structure of a complex molecule

120


Mass Mass SpectrometrySpectrometry• 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 the ions, the greater the deflection

121



• Positively charged ions of higher charge have greater deflection

• Ions with a high m/e ratio are deflected to smaller extent than ions with a low m/e ratio

122


Mass Mass SpectrometrySpectrometry• In the ion detector,

the number of ions collected is measured electronically

• The intensity of the signal is

a measure of the relative abundance of the ions with a particular m/e ratio

123



• The spectrometer shows the results by

plotting a series of peaks of varying intensity

each peak corresponds to ions of a particular m/e ratio

• The graph obtained is known as a mass spectrum

124


SpectrumSpectrum• Generally published as bar graphs.

Mass spectrum of methanol

125


SpectrumSpectrum

Corresponding ion m/e ratio

H3C+ 15

H CO+ 29

H2C = OH+ 31

CH3OH 32

Interpretation of the mass spectrum of methanol

126

34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115)Formation of Formation of

FragmentsFragments

• The molecular ions formed in the ionization chamber are energetically unstable

Some of them may break up into smaller fragments

Called the daughter ions

127


FragmentsFragments• These ionized fragments are accelerated

and deflected by the electric field and magnetic field

• Finally, they are detected by the ion detector and

their m/e ratios are measured

explains why there are so many peaks appeared in mass spectra

128


FragmentsFragments

Mass spectrum of methanol

• The peak at m/e 31

the most intense peak

• Arbitrarily assigned an intensity of 100%

Called the base peak

Corresponds to the most

common ion formed

129


FragmentsFragments• The peak at m/e 31

corresponds to the ion H2C = OH+

formed by losing one hydrogen atom from the molecular ion

130


FragmentsFragments• The ion H2C = OH+ is a relatively stable ion

the positive charge is not localized on a particular atom

it spreads around the carbon and the oxygen atoms to form a delocalized system

131


FragmentsFragments• The peak at m/e 29 corresponds to the ion

HC O+

formed by losing two hydrogen atoms from the ion H2C = OH+

132


FragmentsFragments• The ion HC O+ has two resonance

structures:

133


FragmentsFragments• The peak at m/e 15 corresponds to the

ion H3C+

formed by the breaking of the C O bond in the molecular ion

134


FragmentsFragments

Mass spectrum of pentan-3-one

135


FragmentsFragments

Corresponding ion m/e ratio

CH3CH2+ 29

CH3CH2CO+ 57

CH3CH2COCH2CH3 86

Interpretation of the mass spectrum of pentan-3-one

136


FragmentsFragments• The fragmentation pattern of pentan-3-one

is summarized below:

137

Example 34-9AExample 34-9A Example 34-9BExample 34-9B

Example 34-9CExample 34-9C


138

34.9 Use of Mass Spectra to Obtain Structural Information (SB p.121)Fragmentation Fragmentation

PatternPattern1. Straight-chain 1. Straight-chain AlkanesAlkanes• Simple alkanes tend to undergo

fragmentation by

This carbocation can then undergo stepwise cleavage down the alkyl chain

the initial loss of a • CH3 to give a peak at M – 15

139

34.9 Use of Mass Spectra to Obtain Structural Information (SB p.121)1. Straight-chain 1. Straight-chain

AlkanesAlkanes• Take hexane as an example:

140


2. Branched-chain 2. Branched-chain AlkanesAlkanes

• Tend to cleave at the “branch point”

more stable carbocations are formed

141


2. Branched-chain 2. Branched-chain AlkanesAlkanes• e.g.

142


3. Alkyl-substituted Aromatic 3. Alkyl-substituted Aromatic HydrocarbonsHydrocarbons• Undergo loss of a hydrogen atom or alkyl

group

yield the relatively stable tropylium ion

• Gives a prominent peak at m/e 91

143


3. Alkyl-substituted Aromatic 3. Alkyl-substituted Aromatic HydrocarbonsHydrocarbons

• e.g.

144


4. Aldehydes and 4. Aldehydes and KetonesKetones• Frequently undergo fragmentation by losing

one of the side chains

generate the substituted oxonium ion

often represents the base peak in the mass spectra

145


4. Aldehydes and 4. Aldehydes and KetonesKetones

146

34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122)5. Esters, Carboxylic Acids and 5. Esters, Carboxylic Acids and

AmidesAmides• Often undergo cleavage that involves the

breaking of the C X bond

form substituted oxonium ions as shown below:

(where X = OH, OR, NH2, NHR, NR2)

147

34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122)5. Esters, Carboxylic Acids and 5. Esters, Carboxylic Acids and

AmidesAmides• For carboxylic acids and unsubstituted

amides,

characteristic peaks at m/e 45 and 44 are observed respectively

148

34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123)6. 6.

AlcoholsAlcohols• In addition to the loss of a proton and the

hydroxyl radical,

alcohols tend to lose one of the alkyl groups (or hydrogen atoms)

form oxonium ions

149


AlcoholsAlcohols• For primary alcohols,

the peak at m/e 31, 45, 59 or 73 often appears

depends on what the R1 group is

150


HaloalkaneHaloalkaness • Haloalkanes simply break at the C X

bond

151


HaloalkaneHaloalkaness

• In the mass spectra of chloroalkanes,

two peaks, separated by two mass units, in the ratio 3 : 1 will be appeared

152


7. 7. HaloalkaneHaloalkaness

• In the mass spectra of bromoalkanes,

two peaks, separated by two mass units, having approximately equal intensities will be appeared

153


Check Point 34-9Check Point 34-9

154

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

155


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

156



The laboratory set-up of filtration

157



• 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

158


CentrifugatioCentrifugationn• When there is only a small amount of

suspension, or when much faster separation is required

Centrifugation is often used instead of filtration

159


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

160

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

161


CrystallizatioCrystallizationn

• Crystals are solids that have

a definite regular shape

smooth flat faces and straight edges

• Crystallization is the process of forming crystals

162


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

163



• 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

164



Crystallization by cooling a hot concentrated solution

165


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

166



• 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

167



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

temperature

168


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

169



• 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

170



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

ether

171



• 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

172



• 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

173


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

174



The laboratory set-up of distillation

175



• 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

176



• If bumping occurs during distillation,

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

177


Fractional DistillationFractional Distillation

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

178



The laboratory set-up of fractional distillation

179



• 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

180


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

181


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

182


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

183


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

184


SublimatioSublimationn

A mixture of two compounds can be separated by sublimation

185


ChromatographChromatographyy

• An effective method of separating a complex mixture of substances

• Paper chromatography is a common type of chromatography

186



The laboratory set-up of paper chromatography

187



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

188



• 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

189



• When the solvent moves across the sample spot of the mixture,

partition of the components between the stationary phase and the mobile phase

occurs

190


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

191


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

192


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


193

34.34.44 Qualitative Qualitative

Analysis of Analysis of Elements in an Elements in an

Organic Organic CompoundCompound

194

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

195


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

196


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

197


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

compounds can be detected

by performing the sodium fusion test

198


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

199


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

200


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

201


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


202

34.34.55Determination of Determination of

Empirical Empirical Formula and Formula and

Molecular Molecular Formula from Formula from

Analytical DataAnalytical Data

203

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

204


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

205


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

206


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

207


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

208


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

209



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

the empirical formula of the compound can be calculated

210



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

211



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

the molecular formula of the compound can be calculated

212



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

213




214

34.34.66 Structural Structural

Information Information from Physical from Physical

PropertiesProperties

215

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

216



• From the physical properties of a compound,

obtain preliminary information about the structure of the compound

217


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

218



• 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

219


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

220


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


221


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


222


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


223


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

224



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

225



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

226



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

227



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

228



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

229



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)

230



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)

231


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

232

34.34.77 Structural Structural

Information Information from Chemical from Chemical

PropertiesProperties

233

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

234



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

235



• The next stage is

to find out the functional group(s) present

to deduce the actual arrangement of atoms in the molecule

236


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

237


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


238


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


239


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


240


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

241




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

242




Alcohols ( OH)

• Iodoform test for:

Add iodine in sodium hydroxide solution

• A yellow precipitate is formed

243



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


244



Ethers ( O )

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


245


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


246


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


247


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


248


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


249


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


250


Organic compound

Test Observation

Amides

( )

• Boil with sodium hydroxide solution

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


251


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


252


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


253



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

254

The END

255

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

256


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

257


(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

258


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

259


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 7.6

1.06.7 33.3

16.053.3

13.333.33 2

3.336.7 1

3.333.33

Back

260


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

261


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




∴ 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 33.13

1.013.33 67.1

16.026.67

81.67

13.33 11.671.67 3

1.675

262


(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

263


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

264


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

265


∴ 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 01.0

1.00.01 005.0

16.00.08

20.0050.01 1

0.0050.005 1

0.0050.005

266


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

267


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

268


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.

Back

269


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

Answer

270


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.

Back

271


(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.

272


(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.

273


(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.

274


(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

Back

275


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


276



(b) Deduce the structural formula of the compound.

Answer

277


(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:

278



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

is ethanoic acid.

Back

279


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

280


(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

281


(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

Back

282


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

283


(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

284


(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

285


(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.

Back

286


(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

287


(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 38.2

0.138.2 19.1

0.1467.16 38.2

0.1615.38

319.157.3 2

19.138.2 1

19.119.1 2

19.138.2

288


(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)

289


(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

290


(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

291


(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

292

(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


293


(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 5.14

0.15.14

1125.7125.7 2

125.75.14

294


(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.

Back

295

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.

Back

296


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.

297


(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

298


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




∴ The empirical formula of the compound is C4H8O.


Mass (g) 66.66 11.11 22.23



Simplest mole ratio

4 8 1

56.50.1266.66 11.11

0.111.11 39.1

0.1623.22

439.156.5 8

39.111.11 1

39.139.1

299


Let the molecular formula of the compound be (C4H8O)n.


n × (12.0 × 4 + 1.0 × 8 + 16.0) = 72.0

∴ n = 1

∴ The molecular formula of the compound is C4H8O.

300


(b) From the IR spectrum, it can be observed that there are absorption

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

cm–1 corresponds to the stretching vibration of the C H bond, and

the absorption peak at 1 700 cm–1 corresponds to the stretching

vibration of the C = O bond. Since there is only one oxygen atom in

the molecule of the compound, we can deduce that the compound is

either an aldehyde or a ketone.

If it is an aldehyde, its possible structure will be:

301


(b) If it is a ketone, its possible structure will be:

Back

302


(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.

303


Sketch the infra-red spectrum of a carboxylic acid based on the IR spectra of X and Y.

Answer

304


(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.

305


(a) The infrared spectrum of a carboxylic acid is as follows:

306


(b) The infra-red spectra of two organic compounds A and B are shown below.

Decide which compound could be an alcohol. Explain your answer briefly.

Answer

307


(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.

308


(c) The table below shows the characteristic absorption wavenumbers of some covalent bonds in infra-red spectra.

Bond Range of wavenumber (cm-1)

C = O 1680 – 1750

O H 2500 – 3300

C H 2840 – 3095

N H 3350 – 3500Answer

309


Sketch the expected infra-red spectrum for an amino acid with the following structure:

310


(c) The infra-red spectrum of the amino acid is shown as follows:

311


(d) A portion of the infra-red spectrum of an organic compound X is shown below. To which homologous series does it belong?

Answer

312


(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.

313


(e) A portion of the infra-red spectrum of an organic compound Y is shown on the right. Identify the functional groups that it contains.

Answer

314


(e) From the IR spectrum of compound Y, the two peaks in the 3300 –

3180 cm–1 region show that the compound contains the –NH2 group.

Besides, the sharp peak at 1680 cm–1 implies that the compound

also contains the C = O bond.

315


(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:

316


(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

317


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

∴ The empirical formula of the compound is C2H4O.


Mass (g) 54.54 9.10 36.36



Simplest mole ratio

2 4 1

55.40.1254.54 10.9

0.110.9 27.2

0.1636.36

227.255.4 4

27.210.9 1

27.227.2

318


(f) (i) Let the molecular formula of the compound be (C2H4O)n.


n × (12.0 × 2 + 1.0 × 4 + 16.0) = 88.0

n = 2


(ii)

319


(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

320


The mass spectrum of pentan-2-one (CH3COCH2CH2CH3) is shown below:

What ions do the peaks at m/e 86, 71 and 43 represent? Explain your answer.

Answer

321


The relative molecular mass of pentan-2-one is 86. Therefore, the peak

at m/e 86 corresponds to the molecular ion of pentan-2-one. When the

C1 C2 bond is broken, the ion CH3CH2CH2CO+ (m/e = 71) is formed.

When the C2 C3 bond is broken, the ion CH3CO+ (m/e = 43) is formed.

Back

322


The mass spectrum of hydrocarbon X is shown below:

323


(a) What is the relative molecular mass of hydrocarbon X?

(b) Which peak is the base peak?

(c) How many mass units is the base peak less than the peak for the molecular ion?

(d) Deduce the structures of hydrocarbon X.

(e) Explain the peak at m/e 43.

(f) Propose the fragmentation pattern of the molecular ion which gives rise to the peaks at m/e 58, 43, 29 and 15. Answer

324


(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 CxHy. From the relative molecular mass of the

compound (i.e. 58.0), we can deduce that the compound contains 4

carbon atoms only. (If the compound contains 5 carbon atoms, the

relative molecular mass would be more than 12.0 × 5 = 60.0). The

number of hydrogen atoms in the compound is (58.0 - 12.0 × 4 =

10) 10. Therefore, the hydrocarbon is butane.

325


(e) The peak at m/e 43 is 15 mass units less than the molecular ion.

This suggests that a methyl group is lost during the fragmentation of

the molecular ion. The peak at m/e 43 corresponds to CH3CH2CH2+.

(f)

Back

326


An organic compound is investigated. The structural formula of this compound is shown below:

327


The mass spectrum of the compound is shown below:

Interpret the peaks at m/e 134, 119, 91 and 43. Answer

328


The peak at m/e 134 corresponds to the molecular ion. The peak at m/e

119 corresponds to the ion that is 15 mass units less than the molecular

ion. This suggests that a methyl group is lost from the molecular ion. The

peak at m/e 91 is the base peak, which corresponds to the ion C6H5CH2+

. The peak at m/e 43 corresponds to the ion CH3CO+.

Back

329


Why would the molecular ion compound have two peaks, separated by two mass units, in the ratio 3 :

1?

AnswerChlorine has two isotopes, chlorine-35

and chlorine-37. Their relative

abundances are in the ratio of 3 : 1.

Back

330


Why would the molecular ion of a bromine-containing

compound have two peaks, separated by two mass units, having approximately equal intensities?

AnswerBromine has two isotopes, bromine-79

and bromine-81. Their relative

abundances are in the ratio of 1 : 1.

Back

331


(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.

332


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

Interpret the peaks at m/e = 172, 170 and 91. Answer

333


(b) The relative molecular mass of bromomethylbenzene (benzyl

bromide) is 171.0. However, as bromine contains equal

abundances of the 79Br and 81Br isotopes, the spectrum shows

two small peaks of equal intensity at m/e = 172 and 170. The

base peak at m/e = 91 is due to the formation of the ion

C6H5CH2+.

334


(c) Study the following spectrum carefully and deduce what group of organic compound it is. The compound has a relative molecular mass of 114.

Answer

335


(c) The base peak is at m/e = 57 which may be an oxonium ion or a

carbocation. This is a mass spectrum of a ketone, an aldehyde

or a hydrocarbon.

Back

1 structural determination of organic compounds 34.1introduction 34.2isolation and purification of...

Documents

c slide

purity slide

c c c c o c

organic compound slide

small mass of h slide

twisting rockingwagging

h atoms

atoms modes of vibrations