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Suggested Teaching Scheme


The following suggested teaching schemes are for teachers’ reference only. Teachers may revise them based on the time-tabling arrangement of their own schools.

Scheme 1: Chemistry to be studied in Secondary 3, 4, 5 and 6In many schools, the Chemistry curriculum is studied in Secondary 3, 4, 5 and 6. Although the distribution of periods varies from school to school, the total number of periods for the curriculum is generally around 4�6. A possible distribution of periods is as follows:

A possible distribution of periods for S3, S4, S5 and S6

S3 S4 S5 S6

Number of teaching weeks per year 28 28 28 �6

Number of periods per week 2 5 5 5

Total number of periods per year 56 �40 �40 80

Total number of periods for the curriculum 4�6

Suggested teaching scheme for the curriculum

Level ContentSuggested number

of period(s)

S3(56 periods)

Topic � Planet Earth �2

Topic 2 Microscopic World I 44

S4(�40 periods)

Revision on laboratory safety �

Topic 3 Metals 39

Topic 4 Acids and Bases 45

Topic 5 Redox Reactions, Chemical Cells and Electrolysis 4�

Topic 6 Microscopic World II �4

S5(�40 periods)


Topic 7 Fossil Fuels and Carbon Compounds 32

Topic 8 Chemistry of Carbon Compounds 45

Topic 9 Chemical Reactions and Energy �3

Topic �0 Rate of Reaction �6

Topic �� Chemical Equilibrium �8

Topic �2 Patterns in the Chemical World �5

S6(80 periods)


Topic �3 Industrial Chemistry 39

Topic �4 Materials Chemistry 39

Topic �5 Analytical Chemistry 40

Schools taking investigative study need to allocate an extra of 30 periods for the curriculum.

Only 2 out of 3 Only 2 out of 3

2

Topic 8 Chemistry of Carbon Compounds

Scheme 2: Chemistry to be studied in Secondary 4, 5 and 6In some schools, the Chemistry curriculum is studied in Secondary 4, 5 and 6. The total number of periods for the curriculum is generally around 360. A possible distribution of periods is as follows:

A possible distribution of periods for S4, S5 and S6

S4 S5 S6

Number of teaching weeks per year 28 28 �6

Number of periods per week 5 5 5

Total number of periods per year �40 �40 80

Total number of periods for the curriculum 360

Suggested teaching scheme for the curriculum

Level ContentSuggested number

of period(s)

S4(�40 periods)

Topic � Planet Earth 8

Topic 2 Microscopic World I 3�

Topic 3 Metals 32

Topic 4 Acids and Bases 36

Topic 5 Redox Reactions, Chemical Cells and Electrolysis 33

S5(�40 periods)


Topic 6 Microscopic World II �3

Topic 7 Fossil Fuels and Carbon Compounds 29

Topic 8 Chemistry of Carbon Compounds 4�

Topic 9 Chemical Reactions and Energy �2

Topic �0 Rate of Reaction �5

Topic �� Chemical Equilibrium �6

Topic �2 Patterns in the Chemical World �3

S6(80 periods)


Topic �3 Industrial Chemistry 39

Topic �4 Materials Chemistry 39

Topic �5 Analytical Chemistry 40

Schools taking investigative study need to allocate an extra of 30 periods for the curriculum.

Only 2 out of 3 Only 2 out of 3

3


Suggested number of periods for Topic 8

Chemistry forTotal number

of periodsSuggested number of periods for each unit

S3–S6(Scheme �)

45

Unit 29 An introduction to the chemistry of carbon compounds

Unit 30 IsomerismUnit 3� Typical reactions of selected functional groupsUnit 32 Synthesis of carbon compoundsUnit 33 Important organic substances

67

�57

�0

S4–S6(Scheme 2)

4�

Unit 29 An introduction to the chemistry of carbon compounds

Unit 30 IsomerismUnit 3� Typical reactions of selected functional groupsUnit 32 Synthesis of carbon compoundsUnit 33 Important organic substances

56

�37

�0

4


Organic chemistry is a very important branch of chemistry. Based on the previous topic ‘Fossil Fuels and Carbon compounds’, this topic aims to provide a deeper understanding of carbon compounds, including their systematic names, structural characteristics, physical properties and reactions, and the interconversions of carbon compounds.

Unit 29 introduces selected homologous series and the functional group which members of each series contains. Students are expected to give systematic names of alkanes, alkenes, haloalkanes, alcohols, aldehydes and ketones, carboxylic acids, esters, unsubstituted amides and primary amines, with not more than eight carbon atoms in their carbon chains.

Students should also understand the effects of functional groups and length of carbon chains on the boiling point and water solubility of carbon compounds.

Unit 30 covers the basic concepts of isomerism, including structural isomerism, geometrical isomerism and enantiomerism.

In Unit 3�, students learn the typical reactions of various functional groups, including the reagents, reaction conditions, products and observations. They should be able to make use of some chemical methods to distinguish different functional groups and to identify unknown carbon compounds. They should also be able to predict major products of reactions between alkenes and hydrogen halides using Markovnikov’s rule.

Unit 32 focuses on the synthesis of carbon compounds. Students should be aware of the most important application of organic chemistry, i.e. the synthesis of useful carbon compounds through interconversions between different functional groups. They should be able to plan simple synthetic routes. Common separation and purification methods for organic products would also be discussed. To further their understanding, laboratory preparation of �-bromobutane would be used for illustration.

Unit 33 looks at the chemistry and chemical structures of some important organic substances, including aspirin, detergents, nylons, polyesters, carbohydrates, lipids, proteins and polypeptides.

Teaching Plan

5

Teaching Plan

Unit 33Important organic

substances

Unit 32Synthesis of carbon

compounds

Unit 31Typical reactions

of selectedfunctional groups

Chemistry of Carbon Compounds

Unit 29An introduction tothe chemistry of

carbon compounds

Unit 30Isomerism

Organization of the topic

6


Section Key point(s)Suggested task(s) for

studentsRemark

Total number of period = 1

29.� Functional groups: centre of reactivity

• What a functional group is

• Functional groups which members of some homologous series contain

• Activity 29.� — Building molecular models of compounds in different homologous series

• Practice 29.�

29.2 Naming alkanes and alkenes

• Revising the naming of alkanes and alkenes learnt in Topic 7

• Practice 29.2

29.3 IUPAC rules of naming carbon compounds

• The IUPAC rules of naming

• Names of parent chains• Numerical prefixes• Names of substituents

29.4 Naming haloalkanes • Naming haloalkanes containing the halogeno functional group:

– F – Cl – Br – I

• Practice 29.3


29.5 Naming alcohols • Naming alcohols with the general formula CnH2n+�OH

• Naming polyhydric alcohols

• Practice 29.4

29.6 Naming aldehydes and ketones

• Naming aldehydes (general formula RCHO where R is an alkyl or aryl group or hydrogen)

• Naming ketones (general formula RCOR�, where R and R� are alkyl or aryl groups)

• Practice 29.5

Unit 29 Naming and physical properties of carbon compounds

Continued on next page

7

Teaching Plan


studentsRemark

29.7 Naming carboxylic acids

• Naming carboxylic acids (general formula RCOOH where R is an alkyl or aryl group or hydrogen)

• Practice 29.6


29.8 Naming esters • Naming an ester based on the alcohol and carboxylic acid from which the ester is derived

• Practice 29.7

29.9 Naming amides • Naming amides with an unsubstituted –NH2 group

29.�0 Naming amines • Naming primary amines • Practice 29.8


29.�� Intermolecular forces and physical properties of carbon compounds

• Strength of intermolecular forces in a carbon compound depends on

– the functional group it contains

– the length of its carbon chain

29.�2 Physical properties of haloalkanes

• Boiling point• Solubility in water

29.�3 Physical properties of alcohols


• Refer to the following

animation for the hydrogen bonding between molecules of ethanol and water:

http://chemmovies.unl.edu/ChemAnime/ETOHWD/ETOHWD.html (accessed July 20�4)

29.�4 Physical properties of aldehydes and ketones



8



studentsRemark

Total number of periods = 2 (Scheme 1), total number of period = 1 (Scheme 2)

29.�5 Physical properties of carboxylic acids


29.�6 Physical properties of esters


29.�7 Physical properties of amides


29.�8 Physical properties of amines


• Discussion

29.�9 Common names of carbon compounds

• Common names or trivial names of some carbon compounds

• Library Search & Presentation — Common names of some carbon compounds

9

Teaching Plan


studentsRemark

Total number of periods = 2

30.� Isomerism • Structural isomerism• Stereoisomerism

30.2 Structural isomerism • Chain isomerism• Position isomerism• Functional group

isomerism

• Activity 30.� — Building molecular models of structural isomers

• Practice 30.�

• Isomerism


website for the different isomers of pentane:

http://resources.hkedcity.net/resource_detail.php?rid=435840695 (accessed July 20�4)


30.3 Cis-trans isomerism • Cis-trans isomerism exhibited by compounds containing a C=C bond

• Explaining the difference in melting point / boiling point / water solubility of two cis-trans isomers

• Activity 30.2 — Building molecular models of cis-trans isomers

• Practice 30.2

Total number of periods = 3 (Scheme 1), total number of periods = 2 (Scheme 2)

30.4 Chirality • What a chiral object is

Unit 30 Isomerism


�0



studentsRemark

30.5 Enantiomers • What enantiomers are • Characteristic of a

simple chiral molecule• Looking at the chiral

molecule of CHFClBr and lactic acid

• Identifying chiral carbons in chiral compounds

• Chirality• Refer to the following

website for a chirality game:

http://nobelprize.org/nobel_prizes/chemistry/laureates/200�/illpres/game.html (accessed July 20�4)

• Identifying isomers

30.6 Test for chirality — plane of symmetry

• Determining whether a molecule has a plane of symmetry

• Activity 30.3 — Building models of some molecules and determining whether the molecules are chiral

• Practice 30.3

30.7 Distinguishing the enantiomers of a chiral compound

• How enantiomers of a chiral compound perturb plane-polarized light

• Polarimeter for measurement of rotation of plane-polarised light

• Practice 30.4• Chemistry Magazine

— The thalidomide tragedy


video clip on how enantiomers of a chiral compound perturb plane-polarized light:

http://www.youtube.com/watch?v=uD9j3nbaHsE&feature=related (accessed July 20�4)

��

Teaching Plan


studentsRemark


3�.� Introduction • Importance of planning synthetic routes for the synthesis of new molecules from readily available molecules

3�.2 Important reactions of alkanes

• Combustion• Reaction with halogens

— substitution reactions


website for an animation showing the substitution reaction between methane and bromine:

http://resources.hkedcity.net/resource_detail.php?rid=�4327�527 (accessed July 20�4)


3�.3 Addition reactions of alkenes

• Addition of hydrogen to alkenes in the presence of catalysts

• Addition of halogens to alkenes and test for unsaturation with aqueous bromine

• Addition of hydrogen halides to alkenes and using Markovnikov’s rule to predict the major product

• Practice 3�.�


3�.4 Substitution reactions of haloalkanes

• Hydrolysis of haloalkanes to form alcohols

Unit 31 Typical reactions of selected functional groups


�2



studentsRemark


3�.5 Reactions of alcohols

• Primary, secondary and tertiary alcohols

• Substitution reactions of alcohols with halides

• Elimination reactions — dehydration of alcohols to form alkenes

• Oxidation of alcohols

• Practice 3�.2• Practice 3�.3• Practice 3�.4• Activity 3�.� —

Studying the properties of alcohols

• Activity 3�.2 — Oxidizing ethanol to ethanoic acid and testing the ethanoic acid produced

• Practice 3�.5• Problem solving

• Preparing ethene

and investigating its properties

• Oxidizing ethanol• Refer to the following

website for principles and applications of an alcohol breathalyzer:

http://www.howstuffworks.com/gadgets/automotive/breathalyzer3.htm (accessed July 20�4)

Total number of periods = 2 (Scheme 1), total number of period = 1 (Scheme 2)

3�.6 Reactions of aldehydes and ketones

• Oxidation of aldehydes and ketones

• Reduction of aldehydes and ketones

• Activity 3�.3 — Studying the properties of propanal and propanone

• Practice 3�.6

Total number of periods = 3 (Scheme 1), total number of periods = 2 (Scheme 2)

3�.7 Reactions of carboxylic acids

• Reaction with alkalis and hydrogencarbonates

• Esterification / Condensation reaction

• Reduction• Amides from carboxylic

acids

• Activity 3�.4 — Studying the reaction between ethanol and ethanoic acid

• Activity 3�.5 — Identifying unknown carbon compounds

• Practice 3�.7

• Reacting ethanol with

ethanoic acid• Refer to the following

website for an animation showing an esterification reaction:

http://resources.hkedcity.net/resource_detail.php?rid=�47872�266 (accessed July 20�4)


�3

Teaching Plan


studentsRemark


3�.8 Hydrolysis of esters • Hydrolysis of esters in aqueous acid / alkali

• Obtaining the products after the hydrolysis of an ester in alkaline solution

3�.9 Hydrolysis of amides • Hydrolysis of amides in aqueous acid / alkali

• Practice 3�.8 • Typical reactions

�4



studentsRemark


32.� Planning a synthesis • Deploying suitable reactions and functional group interconversions to alter the groups attached to a basic carbon skeleton

• Percentage yield of a product

• Choosing the reagents

• Discussion • Practice 32.�

32.2 Two-step synthetic routes

• Working backwards from the target molecule until a suitable starting material can be found

• Simple two-step synthetic routes

• Practice 32.2 • Important reactions in

carbon chemistry


32.3 More complicated synthetic routes

• Synthetic routes with three or more steps

• Practice 32.3


32.4 Laboratory preparation of simple carbon compounds

• Planning• Carrying out the

reaction to obtain crude product

• Separating the crude product from the reaction mixture

• Purifying and drying the product

• Measuring the percentage yield of the product

Unit 32 Synthesis of carbon compounds


�5

Teaching Plan


studentsRemark

32.5 Common separation and purification methods in carbon compound preparation

• Common separation and purification methods for liquid products

– distillation – fractional distillation – liquid-liquid extraction• Common separation

and purification method for solid products

– re-crystallization


32.6 Preparing �-bromobutane in the laboratory

• Reaction of butan-�-ol with a mixture of sodium bromide and concentrated sulphuric acid

• Separating the crude product from the reaction mixture

• Purifying and drying the product

• Calculating the percentage yield of the product

• Activity 32.� — Preparing and purifying 2-chloro-2-methylpropane

• Practice 32.4• Find & Share

— Synthetic routes of important organic substances

• Preparing crude 2-

chloro-2-methylpropane

�6



studentsRemark


33.� Introduction • Raising the awareness of the importance of organic substances in daily life

33.2 Aspirin — a common painkiller

• Functional groups acetylsalicyclic acid contains

• Uses and problems of aspirin tablets


website for an additional information about aspirin:

http://health.howstuffworks.com/medicine/medication/aspirin.htm (accessed July 20�4)


33.3 Detergents • Detergents can decrease surface tension of water

• Soapy detergents• Soapless detergents

• Reference website with

information about inventors:

http://inventors.about.com/library/inventors/blsoap.htm (accessed July 20�4)

33.4 How do detergents help water to clean?

• Structure of a typical anionic detergent

• Activity 33.� — Investigating the properties of detergents

• Surface tension of

water and using soap or detergent as a wetting agent

33.5 The wetting and emulsifying properties of detergents in relation to their structures

• The wetting property of detergents

• The emulsifying property of detergents

Unit 33 Important organic substances


�7

Teaching Plan


studentsRemark

33.6 The cleaning action of detergents

• How does a detergent help to clean

• Practice 33.� • The emulsifying process

of a detergent


website for an animation illustrating the cleaning action of a detergent:

http://resources.hkedcity.net/resource_detail.php?rid=��7�602452 (accessed July 20�4)


33.7 Making soaps and soapless detergents

• Making soaps from fats or oils in school laboratory

• Manufacture of soapless detergents from hydrocarbons obtained from petroleum

• Activity 33.2 — Preparing a soap and testing its properties

33.8 Fats and oils • Structure of triglyceride, glycerol, fatty acid and its salt and equation for saporification

• Practice 33.2• Chemistry Magazine

— Environmental issues related to the use of detergents

• A greener alternative

to biodegradable detergents


33.9 Polyesters • Monomers for producing polyesters

• Condensation polymerization

• Uses of poly(ethylene terephthalate)

• Practice 33.3


�8



studentsRemark


33.�0 Nylons • Monomers for producing nylons

• Condensation polymerization

• Uses of nylons

• Activity 33.3 — Preparing nylon

• Practice 33.4

• Preparing nylon


website for an animation showing the preparation of nylon from its monomers:

http://www.lstlcw.edu.hk/t9544/animation/con_polymerization/tc_con_poly.html (accessed July 20�4)

�9

Teaching Notes


page 3N1

Identifying functional groups in compounds

Examination questions often ask students to identify the functional groups in unfamiliar compounds.

e.g.

Compound Functional groups present

Tamiflu

CH2CH3

CH3

CH3CH2

CH2CH3

CH

NH2

C

NH

O

O

C

OO

carbon-carbon double bondester functional groupamine functional groupamide functional group(ether linkage, not taught in the curriculum)

Main constituent of a vegetable oil

H2C O C(CH2)7CH

O

HC O C(CH2)7CH

O

H2C O C(CH2)7CH

O

CHCH2CH CH(CH2)4CH3

CHCH2CH CH(CH2)4CH3

CHCH2CH CH(CH2)4CH3

carbon-carbon double bondester functional group

Teaching Notes

20


Unit 30 Isomerism

page 48N1

Chain isomers of C5H12

Chain isomers of C5H�2 are shown below:

CH3

CH3

CH3

H C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

H

pentane

H C

H

H

C

H

H

C

H

C

H

H

H

2-methylbutane

H C

H

H

C C

H

H

H

2,2-dimethylpropane

A common mistake is to draw another structure as shown below:

CH3

H C

H

H

C

H

C

H

H

C

H

H

H

This is actually 2-methylbutane and all that has happened is that the molecule has been turned around.


To draw the chain isomers of C6H�4, start with a six-carbon chain. Then systematically make the parent chain one carbon atom shorter. Arrange the remaining carbon atoms in each case on the chain in as many different ways as possible without extending the length of the parent chain.

CH3CH2CH2CH2CH2CH3

hexane

• Isomers of C6H�4 with five carbon atoms in the parent chain

CH3CHCH2CH2CH3

CH3

2-methylpentane

CH3CH2CHCH2CH3

CH3

3-methylpentane

• Isomers of C6H�4 with four carbon atoms in the parent chain

CH3CCH2CH3

CH3

CH3

2,2-dimethylbutane

CH3 CH3

CHCH3CH3CH

2,3-dimethylbutane

2�

Teaching Notes


To draw the chain isomers of C8H�8, start with an eight-carbon chain. Then systematically make the parent chain one carbon atom shorter. Arrange the remaining carbon atoms in each case on the chain in as many different ways as possible without extending the length of the parent chain.

CH3CH2CH2CH2CH2CH2CH2CH3

octane

• Isomers of C8H�8 with seven carbon atoms in the parent chain

CH3CHCH2CH2CH2CH2CH3

CH3

2-methylheptane

CH3CH2CHCH2CH2CH2CH3

CH3

3-methylheptane

CH3CH2CH2CHCH2CH2CH3

CH3

4-methylheptane

• Isomers of C8H�8 with six carbon atoms in the parent chain

CH3CH2CHCH2CH2CH3

CH2CH3

3-ethylhexane

CH3CCH2CH2CH2CH3

CH3

CH3

2,2-dimethylhexane

CH3CH2CCH2CH2CH3

CH3

CH3

3,3-dimethylhexane

CH3 CH3

CHCH2CH2CH3CH3CH

2,3-dimethylhexane

CH3CH2CH

CH3 CH3

CHCH2CH3

3,4-dimethylhexane

CH3CHCH2CHCH2CH3

CH3 CH3

2,4-dimethylhexane

CH3CHCH2CH2CHCH3

CH3 CH3

2,5-dimethylhexane

• Isomers of C8H�8 with five carbon atoms in the parent chain

CHCH3CH3CH CH

CH3CH3 CH3CH3CH CH2CH3

CH3

CH3

CH3

CH3CCH2CHCH3

CH3

CH3 CH3

CH3C CHCH2CH3

CH3CH3

CH3

CH3CH2 CH2CH3

CH3

CH2

CH3

CH3CH

CH3 CH2

CHCH2CH3

CH3

2,2,4-trimethylpentane 2,3,3-trimethylpentane

C

2,3,4-trimethylpentane

3-ethyl-2-methylpentane 3-ethyl-3-methylpentane

C

2,2,3-trimethylpentane

22


• Isomer of C8H�8 with four carbon atoms in the parent chain

CH3

CH3

CH3

CH3

CH3

CH3

2,2,3,3-tetramethylbutane

C C

page 53N2

Isomers having a certain molecular formula

Examination questions may ask students to give all the isomers having a certain molecular formula.

e.g. isomers having the molecular formula C4H8

Structural isomers Cis-trans isomers

CH3CH=CHCH3

CH3CH3

H HC C

CH3

CH3

H

HC C

CH2=CHC2H5

CH2=C(CH3)2

C C

C C

H

H

HH

H H

H

H

CH3

C

CCH

H

H

H

H

23

Teaching Notes

e.g. isomers having the molecular formula C3H4F2

Structural isomers Cis-trans isomers

CHF=CFCH3

CH3

F

H

FC C

CH3F

H FC C

CF2=CHCH3

CHF=CHCH2F

CH2F

F

H

HC C

CH2FF

H HC C

CH2=CHCHF2

CH2=CFCH2F

F

C

CCH

H

H

F

H

F

C

CCH

H

F

H

H

24


page 54N3

Melting points of alkanes

The melting points of alkanes follow a similar trend to boiling points. That is, the larger the molecule of an alkane is, the higher is its melting point.

There is one significant difference between boiling points and melting points. Solids have a more rigid and fixed structure than liquids. This rigid structure requires energy to break down. Solid structures that are better packed together will require more energy to break.

For alkanes, this can be seen from the following information.

Compound Formula Boiling point (°C) Melting point (°C)

Pentane CH3(CH2)3CH3 36 –�30

Hexane CH3(CH2)4CH3 69 –95

Heptane CH3(CH2)5CH3 98 –9�

Octane CH3(CH2)6CH3 �26 –57

Nonane CH3(CH2)7CH3 �5� –54

Decane CH3(CH2)8CH3 �74 –30

2,2,3,3-tetramethylbutane (CH3)3C–C(CH3)3 �06 �00

Notice that the boiling points of the straight-chain alkanes (pentane through decane) increase rather smoothly with molecular mass, but alkanes with odd-numbered carbon atoms have a lower melting point trend than the alkanes with even-numbered carbon atoms. This is because alkanes with even-numbered carbon atoms pack well in the solid phase, forming a well-organized structure, which requires more energy to break. Alkanes with odd-numbered carbon atoms pack less well and so the ‘more loosely’ organized solid structure requires less energy to break.

Note that the last compound 2,2,3,3-tetramethylbutane, an isomer of octane, is nearly spherical and has an exceptionally high melting point (only 6 °C below the boiling point).

25

Teaching Notes

The melting points of branched-chain alkanes can be either higher or lower than those of the corresponding straight-chain alkanes, again depending on the ability of the alkane in question to pack well in the solid phase. This is particularly true for 2-methyl isomers, which often have melting points higher than those of the linear analogs.

page 55N4

Comparing the melting points of isomeric compounds

Examination questions often ask students to compare the melting points of isomeric compounds.

e.g.

NO2

OH OH

NO2

4-nitrophenol114 °C

2-nitrophenol45 °Cmelting point:

Only 2-nitrophenol can form intramolecular hydrogen bonds.

N

O

O

O

H

It forms less intermolecular hydrogen bonds than 4-nitrophenol. Hence the melting point of 2-nitrophenol is lower than that of 4-nitrophenol.

page 59N6

Types of isomerism exhibited by a compound

Students need to distinguish clearly the different types of isomerism exhibited by a given compound.

For example, butan-2-ol (CH3CH(OH)C2H5) can exhibit enantiomerism as it contains a chiral carbon. It can also exhibit structural isomerism.

Three structural isomers of butan-2-ol are shown below:

CH3

OH

CH3

OHOHH C

H

H

C

H

H

C

H

H

C

H

H

H C

H

H

C

H

C

H

H

H C

H

H

C C

H

H

H

26


page 59N7

Chirality of ring compounds

Examination questions may also ask about the chirality of complex ring compounds.

Look at methylcyclohexane and 2-methylcyclohexanone.

Methylcyclohexane is achiral because no carbon atom in the molecule is bonded to four different groups. The C� carbon is bonded to a –CH3 group, to a –H atom and to C2 and C6 of the ring. The C6–C5–C4 ‘substituent’ is equivalent to the C2–C3–C4 ‘substituent’. Thus methylcyclohexane is achiral.

Another way of reaching the same conclusion is to realize that methylcyclohexane has a plane of symmetry passing through the –CH3 group and through C� and C4 of the ring.

OCH3H

1

6

5

4

32

CH3H

2

3

4

5

61

methylcyclohexane(achiral)

2-methylcyclohexanone(chiral)

2-methylcyclohexanone has no plane of symmetry and is chiral because C2 is bonded to four different groups: a –CH3 group, a –H atom, a –COCH2– ring bond (C�) and a –CH2CH2– ring bond (C3).

page 60N8

Marking chiral carbons on the structures of unfamiliar compounds

Examination questions may ask students to mark chiral carbons with asterisks on the structures of unfamiliar compounds.

e.g. Tamiflu Aspartame

CH2CH3

CH3

CH3CH2

CH2CH3

CHNH2

C

NH

O

OC

OO

* **

NH

CH2

CH

CH2

CH

CO2H

H2N OCH3

O

C

O

C* *

27

Teaching Notes

page 61N9

Biological properties of enantiomers

Many natural products which contain a stereocentre exist in nature as a single enantiomer. This is particularly important for amino acids and carbohydrates. Hence all living things (including humans) are composed of chemicals which are present as a single enantiomer.

Since enantiomers react differently with other chiral compounds, whenever a chiral compound is eaten, drunk, smelt, ingested, etc. then the two enantiomers may have different biological effects as they will interact differently with the chiral proteins, polysaccharides and nucleic acids they encounter. One example of this is the naturally occurring terpene called carvone. One of the enantiomers of this compound has smells of spearmint, whilst the other has smells of caraway. They smell different because the receptors in the nose are composed of proteins which are built from single enantiomers of naturally occurring amino acids, so the two enantiomers of carvone interact differently with them.

The different biological properties of enantiomers are commonly explained by a three point binding model. In this model, the biological receptor for the chiral compound is assumed to have a fixed geometry and to contain groups capable of interacting with the chiral compound.

If the receptor contains only one or two groups that can interact with the chiral molecule, then both enantiomers of the chiral compound will be able to bind to the receptor as shown (figure a). In this case, the two enantiomers will exhibit the same biological properties.

If, however, the receptor contains three or more groups that can interact with the compound, then only one enantiomer of the compound will be able to bind to all three groups and elicit a biological response (figure b). In this case, the two enantiomers will have different biological properties.

28


page 63N11

Classifying the relationship between two molecules

Examination questions often ask students to classify the relationship between two molecules.

e.g.

CH3

and

CH3

CH3

are structural isomers.

CH3

CH3

HOOC

HOOC and

COOH

CH3

HOOC

H3C are structural isomers.

H3C

H H

CC C

H3C CH3

and

H3C

H

H

CC C

H3C

CH3 are cis-trans isomers.

CH2 CH

Br

H

C CH3 and

CH2 CH

H

Br

C CH3 are enantiomers.

H3C

H3C Cl

BrC C and

H3C

H3C

Cl

BrC C are identical molecules.

CH2CH3

Cl

Cl

and

CH2CH3

Cl

Cl

are identical molecules.

29

Teaching Notes


page 84N3

Deducing the structure of a compound based on some given information

Examination questions may ask students to deduce the structure of a compound based on the type of isomerism it can exhibit and its reactions.

Given information Deductions

• Molecular formula C6H�2

• It has a pair of enantiomers.• It loses its chiral centre after hydrogenation.

• The compound should be an alkene as it can undergo hydrogenation.

• The compound should have a chiral carbon.• Structures of the compound:

H2C=CH

C2H5

CH3

C HHC=CH2

C2H5

H3C

CH

• Molecular formula C6H�2

• It reacts with Br2 to give a single compound.• It reacts with HBr to give a single achiral compound.

• The compound should be an alkene as it can undergo addition reactions.

• Each carbon atom in the double bond should have the same substituents.

Br2

(CH3)2C=C(CH3)2 (CH3)2CBrC(CH3)2Br single compound

HBr

(CH3)2C=C(CH3)2 (CH3)2CHC(CH3)2Br single achiral compound

30


page 86N4

Preparation of alcohols

Different ways of preparing alcohols are summarized in the chart below.

Industrial production of ethanol

Industrial ethanol is generally produced by the catalytic hydration of ethene according to the following equation:

H C

H

H

C

H

H

O H(g)CC

H

H

H

H

(g) + H2O(g)catalyst

300 °C, 65 atm

The reaction is carried out at 300 °C and a pressure of 65 atmospheres. The catalyst used is either concentrated sulphuric acid or concentrated phosphoric acid. The process produces a mixture of ethanol and water. Ethanol can be separated by fractional distillation.

Making ethanol by fermentation

The making of alcohol was one of the first chemical reactions carried out by people. The alcohol in alcoholic drinks is ethanol.

Ethanol can be produced by the fermentation of sugars and starches. In beer making, starch is extracted from barley by soaking in hot water. The starch is then heated to 55 °C with malt. The malt contains enzymes that break down the starch to maltose (C�2H22O��).

3�

Teaching Notes

Finally, yeast is added to the maltose. Yeast contains enzymes which break down the maltose first to glucose and then to ethanol and carbon dioxide.

enzymes in yeast

C�2H22O��(aq) + H2O(l) 2C6H�2O6(aq) maltose glucose

enzymes in yeast

C6H�2O6(aq) 2C2H5OH(aq) + 2CO2(g) glucose ethanol carbon dioxide

Wine is made by fermenting a sweet sugary liquid extracted from grapes.

Fermentation is the slow breakdown of large organic molecules into small molecules by microorganisms in the absence of oxygen.

Under normal conditions, the fermentation can proceed until the ethanol concentration reaches about �5% by volume. At this concentration the yeast dies and the fermentation stops. We can increase the concentration of ethanol by fractional distillation.

Fermentation will be discussed in Topic �3 Industrial Chemistry.

page 86N5

Reaction of sodium with alcohols

Sodium reacts violently with water to form hydrogen and sodium hydroxide solution. During the reaction, an O–H bond is broken:

2H2O(l) + 2Na(s) 2NaOH(aq) + H2(g)

Ethanol reacts in the same way with sodium, to produce hydrogen and sodium ethoxide. During the reaction, the sodium fizzes and produces a colourless solution. The reaction is slower than that of sodium with water, because the O–H bond in ethanol is harder to break than the O–H bond in water.

2C2H5OH(l) + 2Na(s) 2C2H5O–Na+(alc)◀ + H2(g)

Sodium reacts with other alcohols, to produce hydrogen and an alkoxide. The name of the alkoxide is derived from the alkyl chain of the alcohol.

2ROH(l) + 2Na(s) 2RO–Na+(alc) + H2(g)

The alcohol in this example acts as a very weak acid, since it loses a proton. RO–Na+ is a sodium alkoxide.

This also explains why sodium cannot be stored in alcohols.

◀ The state symbol (alc) represents an alcoholic solution.

32


page 104N10

Chemical tests for distinguishing between compounds

Examination questions often ask students to give chemical tests for distinguishing between different compounds.

Example 1

Distinguishing between the following liquids: �-bromopropane, but-�-amine, cyclohexene and propanone

Add water to the liquids. Both but-�-amine and propanone can mix with water in all proportions.

Add a piece of pH paper to the aqueous solutions. Solution of but-�-amine is alkaline while that of propanone is not.

Add aqueous bromine to the two compounds which are immiscible with water. Only cyclohexene can decolorize aqueous bromine.

Example 2

Distinguishing between

CH2I

and

CH2Cl

Heat each compound with NaOH(aq). Then acidify each mixture with HNO3(aq) and followed by the addition of AgNO3(aq).

CH2I

will give a yellow precipitate (AgI).

CH2Cl

will give a white precipitate (AgCl).

33

Teaching Notes

Example 3

Distinguishing between OH

CH3 and CH2OH

Warm with K2Cr2O7 / H3O+.

OH

CH3 has no observable change (tertiary alcohol cannot be oxidized).

CH2OH is oxidized. The orange dichromate solution turns green.

Example 4

Distinguishing between CHO

and O

Warm with K2Cr2O7 / H3O+.

O

has no observable change (ketone cannot be oxidized).

CHO

is oxidized. The orange dichromate solution turns green.

Example 5

Distinguishing between the following liquids:

CH3CH2OH (CH3)3COH CH3(CH2)3NH2 CH3(CH2)7OH

CH3(CH2)3NH2 is the only alkaline compound with a fishy smell.

CH3(CH2)7OH is the only compound which is immiscible with water.

CH3CH2OH is readily oxidized by K2Cr2O7 / H3O+ while (CH3)3COH is not. When CH3CH2OH is warmed with

acidified potassium dichromate solution, the dichromate solution changes from orange to green.

More chemical tests for distinguishing between carbon compounds will be discussed in Topic �5 Analytical Chemistry.

34


page 106N11

Deducing the structure of a compound based on some given information

Examination questions may ask students to deduce the structure of a compound based on the type of isomerism it can exhibit and its reactions.


A(C4H6O) gives B(C4H6O2) upon oxidation.

Oxidation of A adds one more oxygen atom and thus the reaction is an oxidation of aldehyde to carboxylic acid.

Possible structures of A Possible structures of B

OHH C

H

C

H

C

H

H

C

O

H H C

H

C

H

C

H

H

C

O

OHH C

H

H

C

H

C

H

C

O

H H C

H

H

C

H

C

H

C

O

C(C5H�2O) can exist as a pair of enantiomers.

It gives steamy fumes when reacted with PCl5.

C gives HCl when reacted with PCl5. This shows that C is an alcohol.

Possible structures of C:

OH

H C

H

H

C

H

H

C

H

H

C

H

C

H

H

H

CH3 OH

H C

H

H

C

H

C

H

C

H

H

H

CH3

OHH C

H

H

C

H

H

C

H

C

H

H


35

Teaching Notes


An acyclic compound R has a linear structure. R can be converted to S and then to T: LiAlH4 H2 / PdC7H�2O C7H�4O C7H�6O R S T

• R exists as a mixture of cis-trans isomers;

• S has a chiral carbon;• T does not have any chiral carbon.

T should be an alcohol as it is formed by the reduction of R.

R should have a C=O group.

S should have a C=C bond as it can undergo hydrogenation to form T.

As T does not have any chiral carbon, the structure of T should be

OH

H C

H

H

C

H

H

C

H

H

C

H

C

H

H

C

H

H

C

H

H

H

The structure of S may be

OH

H C

H

H

C

H

C

H

C

H

or

C

H

H

C

H

H

C

H

H

H

OH

H C

H

H

C

H

C

H

C

H

C

H

H

C

H

H

C

H

H

H

The structure of R may be

or

H C

H

H

C

H

C

H

C

O

C

H

H

C

H

H

C

H

H

H

H C

H

H

C

H

C

H

C

O

C

H

H

C

H

H

C

H

H

H

36


page 108N12

Reactions of carboxylic acids with sodium carbonate and sodium hydrogencarbonate

Sodium carbonate and sodium hydrogencarbonate will react with aqueous solutions of carboxylic acids to form carbon dioxide.

e.g.

Na2CO3(s) + 2CH3COOH(aq) 2CH3COO–Na+(aq) + H2O(l) + CO2(g)

NaHCO3(s) + CH3COOH(aq) CH3COO–Na+(aq) + H2O(l) + CO2(g)

page 108N13

Preparation of esters

Examination questions often ask about the preparation of an ester.

• Students should be able to state clearly the conditions required for esterification to take place, heating a carboxylic acid with an alcohol in the presence of concentrated sulphuric acid.

• Concentrated sulphuric acid acts as a catalyst, NOT as an oxidizing agent.

• Given the names of a carboxylic acid and an alcohol, students should be able to deduce the systematic name / structural formula of the ester formed, or vice versa.

• Students should be able to draw the experimental set-up for heating carboxylic acid and alcohol under reflux.

• Remind students to distinguish between the preparation of an ester and the oxidation of an alcohol. The preparation of an ester requires a catalyst (concentrated sulphuric acid) while the oxidation of an alcohol requires an oxidizing agent (e.g. acidified potassium dichromate solution).

page 108N14

Esters can also be produced by the reaction of acid chlorides (also called acyl chlorides) with alcohols. The reaction occurs rapidly and does not require an acid catalyst.

page 108N16

Comparison between esterification and neutralization

Very often students are confused by the two processes: ‘esterification’ and ‘neutralization’. Actually they are very different processes. A comparison of these two processes is shown in the table below.

Comparing esterification and neutralization

Esterification Neutralization

� Reaction between a carboxylic acid and an alcohol2 Reversible3 Slow process, catalyst needed4 A covalent compound, ‘ester’, is formed

� Reaction between any acid and base2 Irreversible3 Quick process, no catalyst needed4 An ionic compound, ‘salt’, is formed

37

Teaching Notes

page 109N17

Examination questions often give carbon compounds with similar structures (such as CH3H

O

C O and

OHCH3

O

C ) and ask students to compare them.

• Whether they have the same molecular formula / relative molecular mass.

• Whether they are soluble in water.

• Whether they have the same odour.

• Whether they have the same boiling point.

• Whether they have the same chemical properties.

To tackle these questions, first identify the homologous series to which each compound belongs (such as alcohol, carboxylic acid or ester). Then answer according to the general properties of the series concerned.


page 153N2

Preparation of benzoic acid

Preparation of benzoic acid (C6H5COOH) involves heating methyl benzoate (C6H5COOCH3) with excess sodium hydroxide solution under reflux for some time. The resultant mixture contains sodium benzoate and methanol.

+ NaOH

COOCH3

+ CH3OH

COO–Na+

Obtain crude benzoic acid from the resultant mixture by adding excess mineral acid.

(aq) + H+(aq)

COO–Na+

(s)

COOH

The crude sample of benzoic acid can be purified by re-crystallization from water.

• Use the minimum amount of hot water to dissolve the crude solid product. Filter to remove the insoluble impurities.

• Allow the filtrate to cool slowly for crystals to re-form. Collect the crystals by filtration and dry them.

38



page 177N3

Cationic detergents

Cationic detergents are a much smaller group of detergents. They have a positively charged end. In many cationic detergents, the positively charged end is a quaternary ammonium salt.

CH2 CH2 CH2 CH2 CH2 CH2CH3

CH3

N+ CH3Cl–

CH3 CH2 CH2 CH2 CH2 CH2

+ Cl–

Cationic detergents are not suitable for cleaning glass and china because their positive ends are attracted to the surface of the glass, leaving their non-polar ends sticking out forming a greasy layer. However, this strong attraction to negatively charged surfaces can be an advantage in certain situations. Many fabrics acquire a negative charge when they become wet. Cationic detergents in the form of fabric softeners adsorb onto the surface of the fibres forming a waxy coating. The coating reduces static and tangling and makes fabrics feel softer.

Hair also becomes negatively charged when it is washed. Consequently scales on the hair stick out, making the hair dull and difficult to manage. Hair conditioners are cationic detergents. The positive ends are attached to the hair with the non-polar ends sticking out to create a thin oily layer. This makes the hair shiny and more manageable. Cationic detergents are also mildly antiseptic and are frequently used in nappy washes and domestic disinfectants.

39

Teaching Notes

Non-ionic detergents

Non-ionic detergents have uncharged polar groups within the molecule. Usually these polar groups have an oxygen atom linking two non-polar groups. This oxygen atom forms hydrogen bonds with water molecules, while the non-polar ends adsorb into grease and oils as shown below.

Non-ionic detergents do not form as much foam as other detergents and so are used in dishwashers and front loading washing machines. They can also be added to cationic and anionic detergents to reduce the amount of lathering.

CH2CH2

OCH2

O CH2

CH2

CH2

CH3CH2CH2

CH2

CH2

CH3

O

O

O

O

O

OO

H

H

O

H

HO

H

HO

HH

O

H

H

OH

HO

R

R

RR

RR

R

R

R

R

page 182N6

Examination questions often ask about triesters formed by the condensation of glycerol and carboxylic acids.

To draw the structure of such triesters, just follow the rule for working out the structure of a simple ester producing from an alcohol and a carboxylic acid.

R1H O

O

CR O H + H2O+R1R O

O

C

40


page 184N7

Reactions of the alkenyl chain of unsaturated fatty acids

The double bonds of the carbon chains of fatty acids undergo characteristic alkene addition reactions.

Br

Br Br

Br

CH3(CH2)nCHCH(CH2)mCOOHCH3(CH2)nCH=CH(CH2)mCOOH

+

Br2

in organic solvent

CH3(CH2)nCHCH(CH2)mCOOH

H

CH3(CH2)nCHCH(CH2)mCOOHH2

Ni catalyst

H H

CH3(CH2)nCHCH(CH2)mCOOHHBr

H

page 184N8

Products from the hydrolysis of triglycerides

Examination questions often ask students to give the products formed from the complete hydrolysis of triglycerides existing in natural fats and oils.

e.g.

OH

OH

OH

CH2

CH

CH2 (CH2)16CH3

(CH2)7CH

(CH2)16CH3

CH(CH2)7CH3

O C

O

O C

O

O C

OCH2

CH

CH2

+ CH3(CH2)16COOH

+

completehydrolysis

CH3(CH2)7

H

H

(CH2)7COOHC C

CH3(CH2)7

H

(CH2)7COOH

HC C

4�

Teaching Notes

page 185N9

Advantages and limitations of soaps

Advantages

� Soaps are good cleansers in soft water.

2 Soaps are non-toxic. When sewage containing soaps is drained into the river or the sea, water lives are not harmed.

3 Soaps are biodegradable.

4 Soaps are only mildly alkaline. They seldom cause skin allergy.

Limitations

� Soaps cannot be used in strongly acidic solutions. At pH values much lower than 7, soaps are converted to carboxylic acids. Carboxylic acids of high relative molecular masses are insoluble in water. The cleaning property of soaps is thus lost.

O–(aq) + H+(aq)R C

soap

O

OH(s)R C

carboxylic acid

O

2 Soaps react with calcium ions and magnesium ions in hard water to give scum. This problem can be solved by removing the hardness of water.

Advantages of soapless detergents

� Soapless detergents do not form scum with hard water.

2 Soapless detergents do not form precipitate in strongly acidic solutions. Thus their cleaning action is not affected.

3 The production of soaps requires oils or fats. Vegetable oils and animal fats are more expensive than hydrocarbons obtained from petroleum for making soapless detergents. Natural fats and oils can be saved as food.

4 Soapless detergents can be made to serve a specific purpose. To make a particular kind of detergent, a certain kind of hydrophobic tail may be chosen or a certain hydrophilic head may be used. Different grades of washing powders and liquid detergents are made in this way.

42


page 185N10

Comparing soaps and soapless detergents

Examination questions often require students to know clearly the similarities and differences between soaps and soapless detergents.

• raw materials for their production

• their structural difference

• their emulsifying properties

• their biodegradability

page 193N17

Polymers to monomers

Examination questions often give the structure of a polymer and ask students to deduce the structure(s) of the monomer(s) used to produce the polymer.

Look at the polymer structure. Does the polymer backbone contain only carbon atoms?

YesThe polymer is an addition polymer.

NoThe polymer is a condensation polymer.

• Identify the repeating unit. (This is usually after every second carbon in the backbone.)

• Draw the structure of the repeating unit.• Put a double bond between the carbon atoms to

find the monomer.

• Look for repeats of functional groups in the polymer structure.

• Identify the repeating unit, starting in the centre of identical functional groups.

• Draw the structure of the repeating unit; split the two fragments that come from different monomers at the functional group.

• Restore the structures of the monomers.

Refer to Example �. Refer to Example 2.

Example 1

What is the monomer used to prepare the polymer with the structure shown?

Cl

C

H

C

H

C

H

C

H

C

HH

C

H

H H H

Cl

C

H

C

HCl Cl

43

Teaching Notes

• The backbone has only carbon atoms, so it is an addition polymer.

• Identify the repeating unit, and mark the polymer.

Cl

C

H

C

H

C

H

C

H

C

HH

C

H

H H H

Cl

C

H

C

HCl Cl

• Draw the repeating unit. Put a double bond between the carbon atoms to obtain the structure of the monomer.

Cl

monomerrepeating unit

C

H

C

H

C

H

C

H

H H

Cl

Example 2

What are the monomers used to prepare the polymer with the structure shown?

C N

H

O

N NC

H

O

(CH2)4 C C

O

H

(CH2)4 N N

H

O

(CH2)4 C

O

H

(CH2)4

• The backbone contains nitrogen atoms as well as carbon atoms, so it is a condensation polymer.

• Look for the functional groups. Obtain the repeating units as shown (notice that some groups are ‘reversed’).

C N

H

O

N NC

H

O

(CH2)4 C C

O

H

(CH2)4 N N

H

O

(CH2)4 C

O

H

(CH2)4

functionalgroup

reversed functionalgroup

reversed functionalgroup

44


The repeating unit has a full functional group as shown, which can be split to give two fragments.

N N

H

(CH2)4 C C

O

H

(CH2)4

O

functional group

repeating unit

N N

H

(CH2)4 C C

O

H

(CH2)4

O

• Add ‘H’ or ‘OH’ to the free bonds, as shown, to give the structures of the monomers.

add 2 OH groupsadd 2 H atoms

another monomerone monomer

N N

H

(CH2)4 C C

O

H

(CH2)4

O

NH N H

H

(CH2)4 CHO C OH

O

H

(CH2)4

O

45

Suggested Answers

page 1

� Hydroxyl group

2 C6H5OH; refer to a similar formula in Fig. 25.9 of Topic 7 Fossil Fuels and Carbon Compounds

3 Reduction


Practice

P29.1 page 4

Carbonyl group

C=C bond

–COOH group

P29.2 page 4

a) 2,3-dimethylbutane

CH3

H C

H

H

C

CH3

H

C

CH34

321

H

b) 3-ethylpentane

c) but-2-ene

d) 3-methylhex-3-ene

P29.3 page 9

� a) trichloromethane

b) 2-chlorohexane

c) 2,2-dibromo-3-chloropentane

Suggested Answers

46


d) 2-chloro-2-methylbutane

H3C CH3

Cl

C2H5

C21

3,4

e) 3-bromobut-�-ene

H Br

CH3

H C

H

C C H1 2 3

4

2 a)

C

H

H

Cl

Cl C

Cl

H

b)

C

H

H

CH3

CH3

Br C

H

H

C

H

H

C H

c)

C

H

H

Cl

H C

H

H

C C

H

H

P29.4 page 12

� a) methylpropan-�-ol

b) 2-methylbutan-2-ol

H3C CH3

C2H5

OH

C2 1

3,4

c) 4-phenylpentan-�-ol

d) 4-ethylcyclohexanol

e) but-3-en-2-ol

47

Suggested Answers

2 a)

C

H

H

OH

H C

H

H

C

H

H

C

H

C

H

H

H

b)

C

H

H

C

H

H

OH

c)

C

OH

H

OH

H C

H

H

C

H

C

H

H

H

d)

C

HOH

H C

H

H

C

H

H

C

H

CH

H

P29.5 page 15

� a) 3-methylbutanal

b) 4-methylpentan-2-one

c) hexan-3-one

d) 3-methylbenzaldehyde

2 a)

HCH3CH

CH3

C

O

b)

CH2CH3CH3CH

CH3

C

O

c) O

48


P29.6 page 16

� a) 3-methylpentanoic acid

b) hexanoic acid

c) butanedioic acid

d) 3-phenylpropanoic acid

2 a)

OHCH3CCH2

CH3

CH3

C

O

b)

CH2CH2CH2CH2HO C

O

OHC

O

c) COOH

CH3

P29.7 page 19

� a) methyl butanoate

b) ethyl methanoate

c) 2-propyl benzoate

2 a)

CH2CH2CH3H C

O

O

b)

CH2CH3CH2 C

O

O

49

Suggested Answers

P29.8 page 20

� a) butanamide

b) methylpropanamide

c) butan-�-amine

d) 3-phenylpropan-�-amine

2 a)

CH3CH2CHCH2C

CH3

NH2

O

b)

CH3CHCH3

NH2

Discussion page 35

The order of increasing boiling point of the five compounds is:

CH3CH2CHO < CH3CH2CH2CHO < CH3CH2CH2CH2NH2 < CH3CH2CH2CH2OH < CH3CH2CH2COOH propanal butanal butan-�-amine butan-�-ol butanoic acid

The boiling point of a compound depends on the strength of its molecular attractions.

The intermolecular attractions in but-�-amine, butan-�-ol and butanoic acid are mainly hydrogen bonds while those in propanal and butanal are van der Waals’ forces.

Hydrogen bonds are stronger than van der Waals’ forces.

Thus, the boiling points of butan-�-amine, butan-�-ol and butanoic acid are higher.

As nitrogen is less electronegative than oxygen, the N–H bond in butan-�-amine is less polar than the O–H bond in butanol. As a result, the hydrogen bonds in butan-�-amine are not so strong as those in butan-�-ol. Thus, the boiling point of butan-�-amine is lower than that of butan-�-ol.

Butanoic acid can for more extensive hydrogen bonds than butan-�-ol. Thus, the boiling point of butanoic acid is higher than that of butan-�-ol.

Compared with a propanal molecule, a butanal molecule is larger and contains more electrons. Thus, the van der Waals’ forces in butanal are stronger and the boiling point of butanal is higher than that of propanal.

50


page 36Find & Share

Common names of some carbon compounds

Structural formula of compound Common name Origin of the common name

HCOOH formic acidfrom the Latin word for ant, formica.(Ants can spray attackers with methanoic acid.)

CH3COOH acetic acid from the Latin word for vinegar, acetum

COOH

benzoic acidderived from gum benzoin (the resinous product from styrax trees), which was for a long time the only source for benzoic acid.

HCHO formaldehyde the common name of an aldehyde is often derived from the name of the acid it forms.• formaldehyde derived from formic acid;• acetaldehyde derived from acetic acid.

CH3CHO acetaldehyde

CH3COCH3 acetone

common names for ketones can be derived by naming the two alkyl or aryl groups bonded to the carbonyl group as separate words followed by the word ketone.• acetophenone (i.e. phenylethanone)• diethyl ketone (i.e. pentan-3-one)

CH3CH(OH)CH3 isopropyl alcohol

common names for alcohols usually take the name of the corresponding alkyl group and add the word ‘alcohol’, e.g. methyl alcohol and ethyl alcohol. Propyl alcohol may be n-propyl alcohol or isopropyl alcohol, depending on whether the hydroxyl group is bonded to the 1st or 2nd carbon of the propane chain.

CHCl3 chloroform

chloroform was named and chemically characterized in 1834 by the French chemist Jean-Baptiste Dumas [Fr. chloroforme, from chlor- ‘chlorine’ + formique ‘formic (acid)’].

5�

Suggested Answers

pages 41–46Unit Exercise

� Homologous series Formula of

functional groupExample (structural formula

and IUPAC name)Name General formula

Alkenes CnH2n CC CH2H2C

ethene

Haloalkanes RX –XCH3CH2Cl

chloroethane

Alcohols ROH –OHCH3CH2OH

ethanol

Aldehydes RCHOC H

O

CH3 C H

O

ethanal

Ketones RCOR1

C

O

CH3 C CH3

O

propanone

Carboxylic acids (H or R)COOHC OH

O

CH3 C OH

O

ethanoic acid

Esters (H or R)COOR1

C O

O

CH3 C O CH3

O

methyl ethanoate

Amides RCONH2C NH2

O

CH3 C NH2

O

ethanamide

Amines RNH2 –NH2CH3CH2NH2

ethanamine

52


2 a) No

The first one is a carboxylic acid while the second one is an ester.

b) No

The first one is a ketone while the second one is an aldehyde.

c) Yes

They are both alcohols.

d) No

The first one is an amide while the second one is an amine.

3 a) Any one of the following:

CH3 CH3

OH

OH

OH

OHH C

H

H

C

H

H

C

H

H

C

H

H

H C

H

H

C

H

H

C

H

C

H

H

H

H C

H

H

C

H

C

H

H

H C

H

H

C C

H

H

H

b) Some possible structures are shown below:

C2H5

H C

H

H

C

H

H

C

H

H

C

H

H

C

H

C

H

H

H C

H

H

C

H

H

C

H

C

H

C

H

H

C

H

H

H

CH3

H C

H

H

C

H

H

C

H

H

C

H

C

H

C

H

H

H

H C

H

H

C

H

H

C

H

H

C C

H

H

H C

H

H

C

H

H

C C

H

H

53

Suggested Answers

c) Any one of the following:

H C

H

H

C

H

H

C

O

C

H

H

C

H

H

H H C

H

H

C

O

C

H

H

C

H

H

C

H

H

H

d) Any one of the following:

H C

H

H

C

H

H

C

H

H

C

H

H

C

O

O C

H

H

H H C

H

H

C

H

H

C

H

H

C

O

O C

H

H

C

H

H

H

H C

H

H

C

H

H

C

O

O C

H

H

C

H

H

C

H

H

H H C

H

H

C

O

O C

H

H

C

H

H

C

H

H

C

H

H

H

H C

O

O C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

H

4 D Both CH3COOC2H5 and HCOOC3H7 are esters.

5 B HCOOCH3 is methyl methanoate. It is a methyl ester.

6 B

7 C

8 B

9 D C6H5OH has a very high carbon to hydrogen ratio. It is most likely to burn with a smoky flame.

�0 C CH3COOCH2CH3 is an ester. It is most likely to have a fruity smell.

54


�� B Options B, C and D — Compared with two other compounds (C2H5CH(CH3)2 and C(CH3)4), the molecular shape of CH3CH2CH2CH2CH3 allows the greatest surface contact between molecules. The van der Waals’ forces in CH3CH2CH2CH2CH3 is the strongest.

�2 B Propanal molecules can form hydrogen bonds with water molecules.

�3 C Options C and D — Molecules of CH3CH2CH2 — O — H and CH3CH2CH2CH2 — O — H can form hydrogen bonds with water molecules.

CH3CH2CH2 — O — H is more soluble in water than CH3CH2CH2CH2 — O — H.

�4 A OHCH3 CONH

�5 B

CH2 OH

OHH C

COH

C HHO

C OHH

C OHH

�6 D

�7 A (2) X is an aldehyde while Y is a ketone. They belong to different homologous series.

(3) Both X and Y are soluble in water.

�8 D (�) X is an ester while Y is a carboxylic acid. Thus, they belong to different homologous series.

(3) Hydrogen bonds exist in Y but not in X. Thus, the boiling point of X is lower than that of Y.

55

Suggested Answers

�9 a) 3-bromo-�-chlorobutane

Cl

Br

C C C C

H H

H

H H H

H

H

1234

b) �-chloro-4-methylpentane

CH3

CH3

ClC C C C

H H

H

H H

H

H

1234

5

c) �,�-dibromobut-2-ene

Br

Br C C C C

H H

H

H

H

H

1 2 3 4

d) 3-methylpentan-3-ol

C2H5

OH

H3C C2H5C3 1,2

4,5

e) but-2-en-�-ol

C C C C

H H H

HO

H

H

H

H

1 2 3 4

f) phenylethanal

C C

H O

H

H

12

56


g) 2-methylpentan-3-one

H3C CH3

CH3

C C C

H O H

H

12345

h) propanedioic acid

HO CH2 OHC C

O O

123

i) ethyl benzoate

C2H5 O C

O

j) 2-methylbutan-�-amine

CH3

C C C NH2

H H

CH3

H H H

1234

20 a) D: pentane-2,4-diol

E: phenylmethanol

b) A

c) D

2� a)

I C

H

H

C

H

H

C C

H H

C

H

H

H

57

Suggested Answers

b) OH

HC C

H

C

H

C

H

H

H

Br

c) CHO

d) H

H

H C

H

H

C

O

C

H

C

H

C

H

H

H

e)

OHCH2 C

O

f)

H C

H

H

C

O

O CH(CH3)2

g)

(CH3)2CH C

O

O CH2CH3

h)

C

H

H

H C

H

H

C

H

H

C

H

H

C

O

NH2

22 Alcohols that are low in the homologous series (methanol, ethanol and propan-�-ol) are miscible with water in all proportions.

An alcohol molecule is made up of two parts — the non-polar hydrocarbon part and the polar –OH group.

The water solubility of an alcohol is due to the hydrogen bonding formed between the –OH group of the alcohol molecule and water molecules.

As the number of carbon atoms in an alcohol molecule increases, the influence of the –OH group becomes less significant compared with that of the increasingly large hydrocarbon part. Properties of the higher alcohols tend more and more towards those of the corresponding alkanes.

Hence the water solubility of alcohols decreases as the number of carbon atoms in their molecules increases.

58


23 The boiling point of a compound depends on the strength of its intermolecular attractions.

a) The boiling point of a compound depends on the strength of its intermolecular attractions.

The intermolecular attraction in ethanol are mainly hydrogen bonds while those in ethane are van der Waals’ forces.


Thus, ethanol has a higher boiling point than ethane.

b) In water, there are two hydrogen atoms attached to the oxygen atom. On the other hand, there is only one hydrogen atom attached to the oxygen atom in ethanol. Water can form two hydrogen bonds per molecule while ethanol can only form one hydrogen bond per molecule. Hence the intermolecular forces in water are stronger than those in ethanol, and the boiling point of water is higher than that of ethanol.

Thus, water has a higher boiling point than ethanol.

c) Both ethanol and butanol have one –OH group for hydrogen bond formation. Then boiling point difference is due to the difference in the strength of van der Waals’ forces between molecules.

Compared with an ethanol molecule, a butan-�-ol molecule is larger and contains more electrons. Thus, the van der Waals’ forces in butan-�-ol are stronger and butan-�-ol has a higher boiling point than ethanol.

24 CH3(CH2)3Cl (79 °C) < CH3(CH2)3Br (�0� °C) < CH3(CH2)3OH (��8 °C) < CH3(CH2)2COOH (�64 °C)

The boiling point of a compound depends on the strength of its molecular attractions.

The intermolecular attractions in CH3(CH2)3OH and CH3(CH2)2COOH are mainly hydrogen bonds while those in CH3(CH2)3Cl and CH3(CH2)3Br are van der Waals’ forces.


Thus, the boiling points of CH3(CH2)3OH and CH3(CH2)2COOH are higher.

Both CH3(CH2)3OH and CH3(CH2)2COOH have one –OH group for hydrogen bond formation. Then boiling point difference is due to the difference in polarity of molecules.

CH3(CH2)2COOH is more polar than CH3(CH2)3OH. Thus, the boiling point of CH3(CH2)2COOH is higher than that of CH3(CH2)3OH.

Compared with a CH3(CH2)3Cl molecule, a CH3(CH2)3Br molecule is larger and contains more electrons. Thus, the van der Waals’ forces in CH3(CH2)3Br are stronger and the boiling point of CH3(CH2)3Br is higher than that of CH3(CH2)3Cl.

25 a) Mass of carbon in �.44 g of W = 3.52 g x �2.044.0

= 0.960 g

Mass of hydrogen in �.44 g of W = �.44 g x 2.0

�8.0 = 0.�60 g

Mass of oxygen in �.44 g of W = (�.44 – 0.960 – 0.�60) g = 0.320 g

59

Suggested Answers

Carbon Hydrogen Oxygen

Mass of element in the compound

0.960 g 0.�60 g 0.320 g

Number of moles of atoms that combine

0.960 g�2.0 g mol–� = 0.0800 mol

0.�60 g�.00 g mol–� = 0.�60 mol

0.320 g�6.0 g mol–� = 0.0200 mol

Simplest whole number ratio of atoms

0.0800 mol0.0200 mol

= 4.000.�60 mol0.0200 mol

= 8.000.0200 mol0.0200 mol

= �.00

∴ the empirical formula of W is C4H8O.

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

Relative molecular mass = n(4 x �2.0 + 8 x �.0 + �6.0) = 72n

∴ 72n = 72.0 n = �

∴ The molecular formula of W is C4H8O.

b) Any TWO of the following:

C

H

H

H C

H

H

C

H

butanal

H

C

O

H C

H

CH3

H C

H

methylpropanal

H

C

O

H C

H

H

H C

H

H

C

O

C

H

butanone

H

H

60


Unit 30 Isomerism

Practice

P30.1 page 52

�

CH3CH2CH2CH2CH3

pentane

H C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

H H C

H

H

C

H

H

C

CH3

H

C

H

H

H

CH3CH(CH3)CH2CH3

2-methylbutane

CH3C(CH3)2CH3

2,2-dimethylpropane

H C

H

H

H C

H

H

H C

H

H

C C

H

H

H

2 a) position isomers

b) functional group isomers

c) chain isomers

P30.2 page 55

� a) Structural formula of �,2-dibromoethane:

C

Br Br

H

C

H

The compound can exhibit cis-trans isomerism. Two different atoms are attached to each carbon of the C=C bond.

cis-1,2-dibromoethene

C CH

Br

H

Br

trans-1,2-dibromoethene

C CH

Br

Br

H

6�

Suggested Answers

b) Structural formula of hex-�-ene:

C

H

H

H C

H

H

C

H

H

C

H

H

C

H

C

H

H

The compound does not exhibit cis-trans isomerism. One of the carbon of the C=C bond has identical hydrogen atoms attached to it.

2 The melting point of a substance depends on intermolecular attraction and molecular symmetry.

The intermolecular forces of both isomers is van der Waals’ forces and they are of comparable strength.

The second isomer is more symmetrical so it better fit into a solid lattice.

∴ It has a higher melting point.

P30.3 page 60

� CH3CH2CH2CH(OH)CH3 represent two enantiomers. It has a chiral carbon atom.

CH3CH2CHClCH2CH3 and CH3CH2CH2C(OH)2CH2CH2Cl have no enantiomers. They have no chiral carbon.

2 CHH2C

CH2C

C CH3

H2C

H3C

C CH

O

*

P30.4 page 64

a) i) Z

ii)C

H

CHO

HC C

H

H

CHOC

b) i) W

ii)

CH2CH3CH3CH2 OHH

CH3

C

HOH

CH3

C

62


iii) No. They have the same boiling point.

Chemistry Magazine page 64

The thalidomide tragedy

�

CHCH

CH

C

C

CH

COO N

H

CH

C

CH2 C

O

N

C

CH2

O

structure of thalidomide

chiral carbon

2 • It reduces the possible side effects of the product. If thalidomide had been used as the ‘correct’ single enantiomer, morning sickness would have been prevented without the deformities caused by the other enantiomers.

• It reduces the risk of companies being taken to court for negligence.

• It improves pharmacological activity and therefore reduces the quantity needed for each dose. When a drug containing a mixture of two enantiomers prescribed, half of the drug is often wasted in the body because only one of the isomers has the desired therapeutical effect. Making a drug containing one enantiomers reduces the required dose by half.

• If the synthesis can be planned so that only the desired enantiomer is produced at each stage, the overall process is much more economical and there is no need to discard half the product.

3 Any ONE of the following:

• Safety test on drugs for use during pregnancy

• Test to see if the drug can pass through placenta

• Longer period of testing / longer clinical tests

• Testing on pregnant animals

63

Suggested Answers


� a) structural isomers

b) stereoisomers

c) chain isomers

d) position isomers

e) CH3CH2CH2CH2OH and CH3CHCH2CH3

OH

f) CH3CH2C OH and

O

CH3C O CH3

O

g) cis-trans isomers

h) enantiomers

i) CH3CH3 Br

Cl

H

C and

BrCl

H

C

2 a) B, E

b) Any ONE of the following:

• A, B, E (Any TWO)

• C, F

3 a) carbon-carbon double bond

carboxyl group

b)

CH3

CH3C C

H

CH2CH

C

CH2COOH

H3C C

H

H

In this carbon-carbon double bond, identical methyl groups are attached to one of the carbon atoms.

In this carbon-carbon double bond, identical hydrogen atoms are attached to one of the carbon atoms.

64


4 a) The molecule contains one carbon atom bonded to four different groups of atoms.

b)

NH2H2N HCOOH

CH3

C

HHOOC

CH3

C

5 B Structural isomers of C5H�2 are shown below:

pentane

H C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

H H C

H

H

C

H

H

C

CH3

H

C

H

H

H

2-methylbutane

H C

H

H

C

CH3

CH3

C

H

H

H

2,2-dimethylpropane

6 A The structural isomers that are carboxylic acids with the formula C4H8O2 are shown below.

CH3CH2CH2 OHC

O

CH3CH

CH3

OHC

O

7 C

8 C Option A — Both CH3COOCH2CH3 and CH3CH2COOCH3 contain an ester functional group

( C O

O

).

Option B — Both (CH3)2CHCH(CH3)2 and (CH3)3CCH2CH3 are alkanes.

Option C — H2NCH2CH2CH2COOH contains –NH2 and –COOH functional groups.

H2NCH2COOCH2CH3 contains –NH2 and –COO functional groups.

Option D — Both ClCH2CH2CH=CH2 and CH3CH=CHCH2Cl contain carbon-carbon double bond.

9 D

65

Suggested Answers

�0 B

CHCOOH

CH2COOH

CH(OH)COOH

chiral carbons

�� A

CC

OHHO

OOH

CH CH CH2OHO C

chiral carbons

�2 D CHCl=CHCHClCH3 contains one chiral carbon. It has enantiomers.

CHCl CHCHClCH3

chiral carbon

CHCl=CHCHClCH3 has two different atoms or groups on each carbon atom of double bond. It has cis-trans isomers.

�3 C

�4 C Option C — The two isomers are enantiomers. They have identical boiling point.

�5 A (�) Both compounds have the molecular formula C3H6O2. Thus, they have the same relative molecular mass.

(2) They are structural isomers.

(3) CH3CH2COOH is miscible with water in all proportions but CH3COOCH3 is not.

�6 C (�) X and Y are cis-trans isomers.

(2) Both are optically inactive.

(3) The melting point of X (the cis isomer) is lower than that of Y (the trans isomers).

�7 D The two isomers are enantiomers.

66


�8 D

andBr

H

CCI

CI

(1)

Br

H

CCI

CI

and

(2)

CC

HCl

ClH

CC

ClCl

HH

and

are identical molecules.

are cis-trans isomers.

are structural isomers.C C C OH

H(3) H H

H

H H H

C C C H

H H H

H

H OH H

�9 a) Structural isomer — compounds with the same molecular formula but differs in the order in which atoms are linked

Name of isomer — 2,4-dimethylhexane

Empirical formula — C4H9

b) The molecule of the branched chain isomer is more compact. The molecular surface area available for coming into contact with neighbouring molecules is smaller.

20

isomer 5

NO2

CH3

O2N

NO2

isomer 6

NO2

CH3

2� a) Different boiling point

b) i) compounds with the same molecular formula but differs in the order in which atoms are linked

ii) 3-methylbut-�-ene

iii) Any ONE of the following:

CH3

CH3 CH3

67

Suggested Answers

22 a) cis-trans isomerism occurs in some alkenes because

• rotation about the C=C bond is restricted;

• there are different groups on each carbon atoms of the C=C bond.

b) C, E

23 a)

H2C

CH

CHO

CH2

HC

CHO

b) i) Stereoisomers have their atoms linked in the same way, but they differ in the spatial arrangement of their atoms.

ii) cis-trans isomerism

iii) H

CC

CHO

H

24 Answers for the HKASLE question are not provided.

25 a) i)

HOOC

CH

CH2

H2C

CHC

C

O

O

N

H

H2N

CH3

O

ii) Carbonyl group

Ester functional group

Amide functional group

Amine functional group

b) Adverse side effect / toxicity / irritation

68


26 a) Enantiomerism

b) CH(OH)CH3*

*

c)

CH3CH3 OHHC

HOH C

d) They rotate the plane of polarization of plane-polarized light in opposite directions.

e) No. They have the same boiling point.

27 a) identical

b) enantiomeric

c) structurally isomeric

d) identical

e) cis-trans isomeric

28 a) CH3CH2CH=CH2

b) CH3CH2CH=CHCH2CH3

c) Any ONE of the following:

CH3CH2 C

O

CH2CH3C

O

d)

OH

H C

H

H

C

H

H

C

H

C

H

H

H

69

Suggested Answers


Practice

P31.1 page 84

� a) CH3

CH3

H

C2H5

HBrC C CH3 C

CH3

Br

C

H

H

C2H5

b)

CH3CH2CH2CHCH3

Cl

2 a) Structural isomerism

b)CH3

CH3

Br2 (in organic solvent)H2C C H C

H

Br

C

CH3

Br

CH3

H3C

H

CH3

H

Br2 (in organic solvent)C C H3C C

H

Br

C

H

Br

CH3

c) i) cis-trans isomerism

ii) H3C

H

H

CH3

C C

P31.2 page 87

a)

tertiary alcohol

H C

H

H

C

H

H

C

CH3

OH

C

H

H

C

H

H

H

70


b)

primary alcohol

H C

H

H

C

H

H

C

H

H

C

H

H

OH

c) OH

secondary alcohol

P31.3 page 91

a) 2-iodobutane

b) bromocyclohexane

c) �-chloropentane

P31.4 page 96

� a)H

H

CH2CH3

CH3

dehydrationC CH C

H

H

C

OH

CH3

C

H

H

CH3

alkene X

CH3

CH3

CH3

H

dehydrationC CH C

H

H

C

OH

CH3

C

H

H

CH3

alkene Y

b) Alkene X has two alkyl groups attached to the carbon atoms of the C=C bond while alkene Y has three. Alkene Y is more highly substituted alkene and hence the major product.

2 a) Dehydration / elimination

b)

CH2CH CH2CH H

CH3

C CH CH3

HC

7�

Suggested Answers

c) CH2CH CH2 is the minor product Z.

This alkene has only one alkyl group attached to the carbon atoms of the C=C bonds while the other two are the more substituted alkenes and thus the major dehydration products.

d) cis-trans isomers

P31.5 page 104

a) Isomer B

b) i) H+(aq), Cr2O7

2–(aq)Isomer D CH3CH2 C

O

CH3

ii) The reaction mixture turns form orange to green.

c) Any ONE of the following:

• Isomer A

CH3CH2CH2 C

O

H

• Isomer C

CH3 C

H

CH3

C

O

H

P31.6 page 107

� B: CH3CH2CH2OH

C: CH3CH2COOH

2 a) Functional group isomerism

b) Warm each isomer with acidified potassium dichromate solution.

Isomer X turns the orange dichromate solution green.

There is no observable change for isomer Y.

72


P31.7 page 113

� a) Methanol in the reaction mixture is flammable.

b) An insoluble layer observed.

A fruity smell is detected.

c) i) CH3CH2CH2 C

O

O CH3 methyl butanoate

ii) ester

2 a) Mass of oxygen in �.53 g of Z = (�.53 – 0.900 – 0.�50) g = 0.480 g



0.900 g 0.�50 g 0.480 g


0.900 g�2.0 g mol–� = 0.0750 mol

0.�50 g�.00 g mol–� = 0.�50 mol

0.480 g�6.0 g mol–� = 0.0300 mol


0.0750 mol0.0�50 mol

= 5.000.�50 mol0.0�50 mol

= �0.000.0300 mol0.0�50 mol

= 2.00

∴ the empirical formula of Z is C5H�0O2.

Let (C5H�0O2)n be the molecular formula of Z.

Relative molecular mass = n(5 x �2.0 + �0 x �.0 + 2 x �6.0) = �02n

∴ �02n = �02.0 n = �

∴ The molecular formula of Z is C5H�0O2.

73

Suggested Answers

b) Test 2 — Z is not a carboxylic acid

Test 3 — Z does not contain C=C bond.

CH3CH2CH2 C

O

O CH3

methyl butanote

(CH3)2CH C

O

O CH3

methyl methylpropanoate

CH3CH2 C

O

O CH2CH3

ethyl propanoate

CH3 C

O

O CH2CH2CH3

1-propyl ethanoate

CH3

or

or

or

or C

O

O CH(CH3)2

2-propyl ethanoate

Other possible structure is also acceptable

P31.8 page 117

a) O– Na+(aq) + CH3CH2OH(aq)CH3CH2 C

O

O CH2CH3(l) + NaOH(aq) CH3CH2 C

O

74


b)

c) Separate the products by fractional distillation.

Ethanol is obtained as distillate. Sodium propanoate remains in the solution.

Add excess dilute sulphuric acid / hydrochloric acid to the sodium salt to obtain propanoic acid.

Problem Solving page 104

Add some acidified aqueous potassium dichromate to each liquid and shake.

Only CH3CH2OH and CH3(CH2)7OH change the colour of the solution from orange to green.

CH3CH2OH is miscible with water so a homogenous layer is formed, while CH3(CH2)7OH forms is immiscible with water so two immiscible layer is formed.

To the other two liquid, add some aqueous silver nitrate and shake.

Only CH3(CH2)2Br forms a creamy precipitate, while the (CH3)3COH shows no observable changes.


� a) haloalkane

b) addition

c) hydrolysis

d) substitution

75

Suggested Answers

e) heating with excess conc. H2SO4

f) dehydration / elimination

g) oxidation

h) carboxylic acid

i) � LiAlH4 / ethoxyethane

2 H3O+

j) reduction

k) ketone

l) oxidation

m) � LiAlH4 / ethoxyethane

2 H3O+

n) reduction

o) alkane

p) halogen / UV light or heat

q) substitution

r) conc. H2SO4

s) esterification

t) H3O+

u) hydrolysis

v) strong acid

w) hydrolysis

2 a) H2C=CHCH2CH3, CH3CH=CHCH3, (CH3)2C=CH2

b) CH3CH(OH)CH2CH3

c) CH3COOCH2CH3

d) CH3CH2CH2CHO

e) CH3CH2CH2COONH2, (CH3)2CHCOONH2

76


3

H2 / Ni Br2

mixture of isomers

CC

CH3

CH3H

H

HBr

H C

H

H

C

H

CH3

CH3 H C

H

Br

C

Br

CH3

CH3

H C

H

Br

C

H

CH3

CH3 H C

H

H

C

Br

CH3

CH3

4

K2Cr2O7(aq) /H2SO4(aq)

reflux

acid / heat

K2Cr2O7(aq) /H2SO4(aq)

distil

CH3COOH / H2SO4

reflux

CH3CH2CH2CH2OH

CH3CH2CH2COOH CH3CH2CH=CH2

CH3CH2CH2CHO CH3COOCH2CH2CH2CH3

77

Suggested Answers

5 Test for C=C

Any ONE of the following:

• Add aqueous bromine

The aqueous bromine changes from yellow-brown to colourless quickly.

• Add cold acidified permanganate solution

The permanganate solution changes from purple to colourless quickly.

Test for –COOH


• Add sodium hydrogencarbonate solution.

Effervescence occurs.

• Test with moist blue litmus paper.

The blue litmus paper turns red.

6 C

7 D 2-methylpentan-2-ol has three alkyl groups attached to the carbon atom bearing the hydroxyl group. Thus, it is a tertiary alcohol.

C

OH

CH3

C

H

H

2-methylpentan-2-ol

H C

H

H

C

H

H

C

H

H

H

8 A

9 D

�0 C

�� D

78


�2 D Option A — HCOOC2H5(l) + NaOH(aq) HCOO–Na+(aq) + C2H5OH(aq)

Option B — HCOOC2H5(l) + H2O(l) H3O

+

HCOOH(aq) + C2H5OH(aq)

Option C — CH3COOCH3(l) + NaOH(aq) CH3COO–Na+(aq) + CH3OH(aq)

Option D — CH3COOCH3(l) + H2O(l) H3O

+

CH3COOH(aq) + CH3OH(aq)

�3 A

�4 C (2) X reacts with HBr to give the products shown below.

CH3CH=CHCH2OH HBr CH3CH2CHBrCH2OH + CH3CHBrCH2CH2OH

(3) CH3CH=CHCH2OH has two different atoms or groups on each carbon atom of the double bond. It can exhibit cis-trans isomerism.

�5 C

�6 D

�7 B

CH3CHCH3CCCH3

H

H

C

OH

H

H

H

CHCH2CH3

dehydration

CH3CH2CHCH3CCCH3

H

H

C

OH

H

H

H

CHCH3

dehydration

�8 A

�9 a) CH2CH2OH

Na2Cr2O7 / H3O+

heat

CH2COOH

b) COOHH3C CH2OHH3C1 LiAIH4 / ethoxyethane

2 H3O+

c)

CH3CH2 CH2CH3OC

O

CH3CH2 CH3CH2OH+O–Na+C

O

NaOH(aq)

79

Suggested Answers

d)

CH3 NH2C

O

CH3 NH3+O–Na+C

O

NaOH(aq)

e)

CH CHCH

O

NaBH4 / H2O

CH CHCH2OH

f) CH3 CH2Cl

Cl2

UV light

g)

OCCH3

O

COOH

OH

COO–Na+

OH–(aq)

heatCH3COO

–Na++

h) CH3CH2CH2OH CH3CH CH2

excess conc. H2SO4

180 °C

20 a) i) Reagent: hydrogen

Catalyst: Nickel / Palladium / Platinum

ii) �,2-dibromoethane

H C

H

Br

C

H

Br

H

iii) From purple to colourless

b) The bromine atom can add to either carbon atom of the C=C bond, so forming either �-bromopropane or 2-bromopropane.

2� a) 3-bromo-3-methylpentane

b)

CH3Br

H C

H

H

C

H

C

H

C C

H

H

H

H

H

position isomerism

80


22 a) i) Aqueous sodium hydroxide, heat under reflux

ii) Substitution

iii)

H C

H

H

C

H

C

H

H H

C Br

H

H

H C

H

H

C

H

C

H

H H

C OH

H

H

OH– (aq)

b) Concentrated sulphuric acid is a strong oxidizing agent. It oxidizes potassium iodide to iodine.

23 a) 2-methylpentan-2-ol

b)

C CCH2CH3

H

H3C

H3C

c) position isomerism

24 Answers for the HKCEE question are not provided.

25 a) Sodium / potassium dichromate solution

dilute sulphuric acid

b)

26 a) organic product: CH3CH3 C

O

equation: (CH3)2CHOH + [O] (CH3)2CO + H2O

8�

Suggested Answers

b) i) Methylpropan-2-ol

ii) Ester

iii) C(CH3)3H3C C

O

O

27 a) i)

δ–O hydrogen bond

H CH2CH2CH3

δ+

δ–O

H CH2CH2CH3

δ+

lone pair

ii) The boiling point of a compound depends on the strength of its molecular attractions.

The intermolecular attractions in propan-�-ol are mainly hydrogen bonds while those in propene are van der Waals’ forces.


Thus, propan-�-ol has a higher boiling point than propene.

b) i) elimination

ii) Any ONE of the following:

• Concentrated sulphuric acid / phosphoric acid

Heat / reflux

• Aluminium oxide / silica / pumice / porous pot

Heat, propan-�-ol vapour passes over the catalyst

c) i)

HH C

H

H

C

H

H

C

Br Br

H

HH C

H

H

C

H

C

H

H

or

ii)

HH C

H

H

C

H

H

C

H

H

d) Platinum

82


28 a) methylpropan-2-ol

b)

OH

c) i) Compounds with the same molecular formula but differ in the order in which atoms are linked.

ii)

OHC

H

H

H C

H

H

C

H

H

C

H

H

OH or H C

H

H

C

CH3

H

C

H

H

d) Ethane-�,2-diol has hydroxyl groups that can form hydrogen bonds with water molecules.

e)

CH2CH2CH3 C

O

O O C

O

CH3

f) i)C C

H

CH3

H

H3CC C

H

CH3

H3C

H


iii) CH3CH2CH=CH2

29 a) Aluminium oxide / silica / pumice / porous pot

b)

83

Suggested Answers

c) (CH3)2C=CHCH3 CH2=C(CH3)(CH2CH3)

d) These compounds do not exhibit cis-trans isomerism. One of the carbon of the C=C bond has identical hydrogen atoms attached to it.

e) i) Chain isomerism

ii) Warm with acidified potassium dichromate solution.

Compound Y will change the dichromate solution from orange to green. There is no observable change for compound X.

30 CH2OH

CH2Br

COOH

CONH2

W:

X:

Y:

Z:

3� Reagent for reduction — lithium tetrahydridoaluminate in anhydrous ethoxyethane, followed by hydrolysis using dilute acid

CH3COCHO + 4[H] CH3CH(OH)CH2OH

Reagent for oxidation — acidified potassium dichromate solution, the dichromate solution changes from orange to green in the process

CH3COCHO + [O] CH3COCOOH

32 a) CH3CH2CH2CH2OH

b) Reaction A — react with lithium tetrahydridoaluminate in anhydrous ethoxyethane, followed by hydrolysis using dilute acid

Reaction B — heat with acidified potassium dichromate solution under reflux

Reaction C — warm with ethanol in the presence of concentrated sulphuric acid

Reaction D — Any ONE of the following:

• Heat the ester under reflux with a dilute mineral acid catalyst.

• Heat the ester with sodium hydroxide solution. Add excess mineral acid to the butanoate to obtain the butanoic acid.

84


33 a) Compound XH3CCH=CHCH(OH)CH3

pent-3-ene-2-ol

Compound Y

CH3CH2CH2CH2 HC

Opentanal

Compound Z

CH3CH2CH2 CH3C

Opentan-2-one

b) Functional group isomerism

c) i)C C

CH(OH)CH3

H

H

H3CC C

H

CH(OH)CH3

H

H3C


iii) Restricted rotation about the C=C bond

d) Warm each compound with acidified potassium dichromate solution.

Compound Y turns the dichromate solution from orange to green.

There is no observable change for compound Z.

34 a) i) Steamy / misty / white fumes is observed.

Hydrogen chloride is formed.

ii) The purple permanganate solution turns colourless.


• Addition of –OH groups to the C=C bond to form a diol.

• Oxidation of the –OH group to form an aldehyde or a carboxylic acid.

iii) The yellow-brown aqueous bromine turns colourless.

Additon of Br atoms / OH groups to the C=C bond to form a dibromo compound / bromoalcohol.

b) Any ONE of the following:

C

H

H

H C

H

Br

C

H

Br

C

H

H

O H C

H

H

H C

H

Br

C

H

O

H

C

H

H

O H C

H

H

H C

H

O

C

H

Br

H

C

H

H

O H

85

Suggested Answers

35 From (3), Y contains –COOH group. This suggests that X also contains –COOH group.

From (�) and (2), X contains C=C group, and two different groups on each carbon of C=C.

∴ Possible structures of X: C

H

COOH

HC C

H

H

COOHC

Possible structure of Y: CH2CH2COOH

36 a) i) speed up the reaction / reduce time taken

ii) to condense vapours and prevent loss of volatile substances

b) i) C3H7OH, C4H9COOH

ii) sulfuric acid

37 a) i)

CH3CH2 CH3OH(aq)+O–Na+(aq)C

O

CH3CH2 NaOH(aq)+CH3(l)C

O

ii) Hydrolysis

b) i) Air condenser

ii) Serving as a fractionating column for separating methanol from the products formed.

c) Methanol

d) To ensure even boiling.

e) B

f) Any one of the following:

H OCH2CH2CH3

propyl methanoate

C

O

CH3 OCH2CH3

ethyl ethanoate

C

O

H OCH(CH3)2

methylethyl methanoate

C

O

86


38 a) Warm with acidified potassium dichromate solution.

CH2CH2CHO

will change the dichromate solution from orange to green.

There is no observable change for

CCH2CH3

O

.

b) Mix with aqueous bromine.

CH2=CHCH2OH will change the solution from yellow-brown to colourless.

There is no observable change for CH3CCH3

O

.

c) Mix with sodium carbonate solution.

Effervescence occurs for CH3CH2COOH.

There is no observable change for CH3COCH2OH.

39 a) Mass of carbon in �.45 g of X = 3.30 g x �2.044.0

= 0.900 g

Mass of hydrogen in �.45 g of X = �.35 g x 2.0

�8.0 = 0.�50 g

Mass of oxygen in �.45 g of X = (�.45 – 0.900 – 0.�50) g = 0.400 g



0.900 g 0.�50 g 0.400 g


0.900 g�2.0 g mol–� = 0.0750 mol

0.�50 g�.00 g mol–� = 0.�50 mol

0.400 g�6.0 g mol–� = 0.025 mol


0.0750 mol0.0250 mol

= 3.000.�50 mol0.0250 mol

= 6.000.0250 mol0.0250 mol

= �.00

∴ the empirical formula of X is C3H6O.

87

Suggested Answers

Let (C3H6O)n be the molecular formula of X.

Relative molecular mass = n(3 x �2.0 + 6 x �.0 + �6.0) = 58n

∴ 58n = 58.0 n = �

∴ The molecular formula of X is C3H6O.

b) X gives a negative result when warmed with acidified K2Cr2O7(aq), so X is not aldehyde and primary and secondary alcohol.

X is CH3 CH3

C

O

.

c) CH3CH2CHO

40 a) Aqueous sodium hydroxide

b) i)

HH C

H

H

C

OH

H

C

H

H

ii) propan-2-ol

c) i) ethanol

ii) Heat under reflux.

Use excess oxidizing agent / K2Cr2O7.

d) concentrated sulphuric acid

e)

CH(CH3)2CH3 C

O

O

f) Reaction II

88



Practice

P32.1 page 137

� Refer to the chart on page �58 of the textbook.

2 CH2BrCH2Br

H2C=CH2

CH3CH2OH

CH3CH3

CH3CH2Br

CH3CHO

CH3COOH CH3CONH2CH3COOC2H5

Br2 (in organic solvent)addition reaction

H2SO4 or H3PO4 or Al2O3, heatdehydration

Br2

UV light or heatsubstitution

reaction

H2 / Pt catalyst

addition reaction

HCl or HBr o

r HI

addition reactio

n

HBrsubstitution reaction

NaOH(aq), reflux substitution reaction

NH3(aq), heatamide synthesis

acid or alkalihydrolysis

C2H5OHconc. H2SO4 catalyst

esterification

acid or alkalihydrolysis

K2Cr2O7 /

H3O+

oxidatio

n

1 LiAlH4 /

ethoxyethane

2 H3O+

reduction

K2Cr2O7 / H3O+

oxidation1 LiAlH4 / ethoxyethane2 H3O

+

reduction

K2Cr2O

7 / H3O +

oxidation

89

Suggested Answers

P32.2 page 140

a) CH3CH2COOH CH3CH2CH2OH CH3CH2CH2Cl1 LiAlH4 / ethoxyethane

2 H3O+

reflux with conc. HCl + ZnCl2 catalyst; or

mix with PCl5; orreflux with SOCl2

b) CH2CH2OH

K2Cr2O7 / H3O+

reflux

CH2COOH CH2COOCH3

CH3OH

conc. H2SO4, heat

c) OH

conc. H2SO4 or H3PO4

heat

H2

Pd catalyst

P32.3 page 143

a)

NaOH(aq)

reflux

K2Cr2O7 / H3O+

heat

HCl(g)

CH=CH2 CHClCH3 CH(OH)CH3 CCH3

O

b) CH3CH2CH2Br CH3CH2CH2OHNaOH(aq)

refluxCH3CH2COOH

K2Cr2O72–(aq), H+(aq)

heat

CH3CH2CONH21 NH3(aq)

2 heat

c) NaOH(aq)

refluxCH2COOCH3

H+(aq)CH2COO– Na+ CH2COOH

CH2CH2OH1 LiAlH4 / ethoxyethane

2 H3O+

90


P32.4 page 155

a) i) CH3CH2OH(l) + CH3COOH(l) CH3COOCH2CH3(l) + H2O(l)

ii) Concentrated sulphuric acid acts as a catalyst for the esterification reaction.

b) The distillate contains traces of ethanoic acid and perhaps some sulphuric acid. Sodium carbonate solution is added to remove these.

c) The density of the organic layer (ethyl ethanoate) is lower than that of the aqueous layer. Therefore we can separate the organic layer from the aqueous layer by running off the lower aqueous layer.

d) Anhydrous calcium chloride is a drying agent and it can remove any remaining traces of water from the product.

e) i) Number of moles of CH3CH2OH used = 60.0 g

46.0 g mol–� = �.30 mol

Number of moles of CH3COOH used = 50.0 g

60.0 g mol–� = 0.833 mol

ii) According to the equation for the esterification reaction, � mole of CH3CH2OH reacts with � mole of CH3COOH to give � mole of CH3COOCH2CH3.

During the reaction, 0.833 moles of CH3COOH would react with 0.833 moles of CH3CH2OH. Therefore CH3CH2OH was in excess. The amount of CH3COOH limited the amount of ester produced.

Theoretical yield of ester = 0.833 mol x 88.0 g mol–� = 73.3 g

Percentage yield of ester = 45.0 g73.3 g

x �00% = 6�.4%

∴ the percentage yield of ethyl ethanoate was 6�.4%.

Discussion page 136

This discussion aims at raising students’ awareness of the principles of green chemistry. See the following list of green chemistry principles the discussion questions help to convey.

� Safer auxiliary substances — Whenever possible, avoid the use of auxiliary substances such as solvents and separating agents. When these substances are needed, they should be non-toxic.

2 Energy efficiency — Energy efficiency should be considered when designing and producing a product.

3 Safer auxiliary substances — Whenever possible, avoid the use of auxiliary substances such as solvents and separating agents. When these substances are needed, they should be non-toxic.

4 Safer auxiliary substances — Whenever possible, avoid the use of auxiliary substances such as solvents and separating agents. When these substances are needed, they should be non-toxic.

5 Use of renewable resources — When possible, the feedstocks should come from a renewable resource.

9�

Suggested Answers

page 156Find & Share

Synthetic routes of important organic substances

Aspartame

Aspartame is 200 times as sweet as sucrose. Aspartame is utilized as a low-calorie sweetener in soft drinks, salad dressings, ready-made meals, table-top sweeteners, and pharmaceuticals.

NH

O

OCH3

NH2

O

O

HO

structure of aspartame

The three main reactants used in the synthesis of aspartame are

• phenylalanine,

• methanol and

• aspartic acid.

Ester synthesis

Treatment of phenylalanine with methanol in the presence hydrochloric acid causes an esterification reaction with the carboxyl group on the phenylalanine. The product is the methyl ester of phenylalanine.

NH2

O

OH + CH3OH

NH2

O

OCH3

Amide synthesis

The final goal is to react the methyl ester of phenylalanine with aspartic acid to give the dipeptide structure. A series of reactions is required so that only the carboxyl group near to the amino group of the aspartic acid is reacted. The carboxyl group on the side chain is protected so that it does not react.

NH2

O

OCH3

NH2

O

OH

O

aspartamemethyl ester of phenylalanineaspartic acid

protect

+

NH

O

OCH3

NH2

O

O

HOHO

92


References:

http://www.chemcases.com/nutra/nutra2.htm

http://www.elmhurst.edu/~chm/vchembook/549aspartame.html

(accessed July 20�4)

Aspirin

Aspirin is one of the most widely used medications to reduce fever and is also used as a painkiller. It is a derivative

of salicylic acid (OH

COOH

).

structure of aspirin

OCOCH3

COOH

The synthesis of aspirin from salicylic acid results in the formation of an ester functional group, and therefore, is an esterification reaction. Salicylic acid reacts rapidly with ethanoic anhydride in the presence of an acid catalyst to produce aspirin and ethanoic acid. Sulphuric acid or phosphoric acid is often used to catalyze the reaction.

Ethanoic anhydride is used instead of ethanoic acid because it reacts faster with salicylic acid.

OHOH

COOH

+50 °C

acidcatalyst

O

COOH

CH3H3C C

O

O C

O

CH3C

O

+H3C C

O

Reference:

http://homepages.ius.edu/DSPURLOC/C�22/asp.htm


Ibuprofen

Ibuprofen is one of the commonly used anti-inflammatory agents. It is used to relieve the symptoms of a wide range of illnesses such as headaches, backache, period pain, dental pain, neuralgia, rheumatic pain, muscular pain, migraine, cold and flu symptoms and arthritis.

CH3COOH

structure of ibuprofen

H3C

CH3

93

Suggested Answers

Boots process

The Boots process is an older commercial process of obtaining ibuprofen developed by the Boot Pure Drug Company. The Boots Company patented the process in �960s. This synthesis consists of six steps and resulted in unwanted by-products.

CH3N

H3C

CH3

OH

CH3COOH

ibuprofenH3C

CH3

CH3H

H3C

CH3CH3

H3C

OO

CH3

CO2C2H5H3C

H3C

AlCl3 NaOC2H5

CH3O

H3C

CH3

CH3C N

H3C

CH3

H3C

H3C

O

O

O

H3O+

NH2OH

COOC2H5Cl

H

H

O

H

H

O

H

H

Source: http://www.scheikundeinbedrijf.nl/content/Modules/Modulenaam/Files/case.pdf


94


Celanese process

In the mid eighties when the patent on ibuprofen was about to run out, other companies began developing new synthetic methods to produce ibuprofen. During this time, the Hoechst Celanese Corporation and the Boots Company agreed to a joint venture resulting in the BHC Company.

The Celanese process (hereafter known as the green synthesis) has only three steps and gives fewer harmful by-products.

CH3COOH

ibuprofenH3C

CH3

CH3H

H3C

CH3CH3

H3C

O

HF

OH

CH3CH3

H3C

H3C

H3C

O

O

O

H2

Raney nickel

Pd

CO

Source: http://www.scheikundeinbedrijf.nl/content/Modules/Modulenaam/Files/case.pdf


References:

http://www.rsc.org/learn-chemistry/resources/chemistry-in-your-cupboard/nurofen/4-6

http://www.chm.bris.ac.uk/motm/ibuprofen/synthesis.htm

http://www.scheikundeinbedrijf.nl/content/Modules/Modulenaam/Files/case.pdf



� a) Purifying

b) drying

c) the percentage yield

d) distillation

95

Suggested Answers

e) fractional distillation

f) liquid-liquid extraction

g) re-crystallization

2 A: H2 / Pt catalyst

B: Br2 (in organic solvent)

C: aqueous bromine

D: cold dilute KMnO4 / OH–

E: HI

3 A: HBr(g)

B: reflux with NaOH(aq)

C: reflux with NaBr and H2SO4, or reflux with red phosphorus and Br2

D: excess conc. H2SO4, �80 °C; or Al2O3, 300 °C

E: reflux with CH3COOH and conc. H2SO4

F: reflux with NaOH(aq)

G: K2Cr2O7 / H3O+

H: � LiAlH4 / ethoxyethane

2 H3O+

4 A: Cl2(g), UV

B: reflux with SOCl2 or PCl5, or reflux with HCl and ZnCl2 catalyst

C: K2Cr2O7 / H3O+

D: � LiAlH4 / ethoxyethane

2 H3O+

E: NaOH(aq)

F: reflux with CH3OH and conc. H2SO4

G: reflux with NaOH(aq), separate the products by fractional distillation, obtain the acid by adding excess mineral acid to the salt of carboxylic acid

H: NH3(aq)

I: heat

J: reflux with NaOH(aq), obtain the acid by adding excess mineral acid to the sodium propanoate

96


5 B The order of the boiling points of the three alcohols is X < Y < Z.

Thus Y is more volatile than Z.

6 C

CH3CO OCH2CHCH2CH3

this part comes from CH3COOH

this part comes from CH3CH2CHCH2OH

CH3

CH3

7 B CH3CH=CHCH3 HCl(g) CH3CH2CHClCH3

8 D

9 D

�0 D (2) and (3) When heated with excess concentrated H2SO4,

CH2CH2OH

undergoes dehydration to

produce

CH CH2

.

�� A

�2 B (�) Heating ethanol, sodium dichromate solution and dilute sulphuric acid under reflux produces ethanoic acid, NOT ethanal.

(3) Ethene and bromine react quickly at room temperature. There is NO need to heat under reflux.

�3 a) Reagent(s) and reaction condition(s): HBr(g)

Type of reaction: addition

b) Reagent(s) and reaction condition(s): expose a mixture of CH3CH2CH3 and Br2 to ultraviolet light or heat

Type of reaction: substitution

c) Reagent(s) and reaction condition(s): reflux with NaOH(aq); substitution

Type of reaction: hydrolysis

d) Reagent(s) and reaction condition(s): 20% H2SO4, heat

Type of reaction: dehydration / elimination

97

Suggested Answers

e) Reagent(s) and reaction condition(s): reflux with conc. HCl + ZnCl2 catalyst; or mix with PCl5; or reflux with SOCl2

Type of reaction: substitution

f) Reagent(s) and reaction condition(s): � LiAlH4 / ethoxyethane

2 H3O+;

Type of reaction: reduction

g) Reagent(s) and reaction condition(s): NaOH(aq), heat


h) Reagent(s) and reaction condition(s): � reflux with NaOH(aq)

2 separate the products by fractional distillation

3 obtain the acid by adding excess mineral acid to the salt of carboxylic acid


�4 a)

Br2

UV light

CH2CH3

NaOH

reflux

CHBrCH3

conc. H3PO4

heat

CH(OH)CH3 CH2 CH2

b) OH

CH3CCH2CH2CH3

1 LiAlH4 / ethoxyethane

2 H3O+

excess conc. H2SO4

180 °C

O

CH3CHCH2CH2CH3 CH3CH=CHCH2CH3

c)

CH3CH2CH2COH1 LiAlH4 / ethoxyethane

2 H3O+

K2Cr2O7 / H3O+

gentle heat, distil

O

CH3CH2CH2CH

O

CH3CH2CH2CH2OH

d)1 LiAlH4 / ethoxyethane

2 H3O+

CH2COOH reflux with conc. HCl + ZnCl2 catalyst;

or mix with PCl5;or reflux with SOCl2

CH2CH2OH CH2CH2Cl

e) CH2=CHClK2Cr2O7 / H3O

+

gentle heat, distil

reflux with

NaOH(aq)

H2(g) /

Pt catalystCH3CH2OH CH3CHOCH3CH2Cl

f)

OHOH

CH3CHCH2CH3CHOHCH3

excess conc. H2SO4, 180 °C;

or Al2O3, 300 °C

cold dilute KMnO4 /

OH–CH3CH=CH2

98


g)

1 LiAlH4 / ethoxyethane

2 H3O+

reflux with CH3COOH

and conc. H2SO4

CH2COOH CH2CH2OH CH2CH2OCCH3

O

h) CH3COO– NH4+C2H5OH

K2Cr2O7 / H3O+

reflux

NH3(aq) heatCH3CONH2CH3COOH

�5 Answers for the HKDSE question are not provided.

�6 a) i) According to the equation, � mole of butan-�-ol gives � mole of �-bromobutane.

For a 80% yield, 0.�25 mole of butan-�-ol gives 0.�25 mole x 80% (i.e. 0.�00 mole) of �-bromobutane.

ii) Mass of 0.�25 mol of butan-�-ol = 0.�25 mol x 74.0 g mol–�

= 9.25 g

Volume of butan-�-ol = 9.25 g

0.8� g cm–3

= ��.4 cm3

iii) Minimum mass of potassium bromide = 0.�25 mol x ��9.0 g mol–�

= �4.49 g

b)

c) i) The lower layer contains �-bromobutane because �-bromobutane is denser.

ii) To neutralize / remove the remaining hydrochloric acid.

The reaction between hydrochloric acid and sodium hydrogencarbonate solution produces carbon dioxide gas.

99

Suggested Answers

d) i) Distillation / fractional distillation

ii) Any ONE of the following:

• Measure the boiling point of the product and compare with data book value.

• The product boils over a very small temperature range.

�7 Answers for the HKASLE question are not provided.

�8 a) Sodium / potassium dichromate solution

b) From orange to green

c)

d) Any TWO of the following:

• the yield would be reduced. / Reactants and products would be lost.

• complete oxidation could not occur.

• The vapour is flammable / harmful / irritant / toxic.

e) The vapour is cooled and condensed.

f) i) Anhydrous calcium chloride / anhydrous magnesium sulphate / anhydrous sodium sulphate

ii) The filter paper would absorb some of the product.

g) i) Number of moles of propan-�-ol = �0.0 g

60.� g mol–�

= 0.�66 mol

According to the equation, � mole of propan-�-ol gives � mole of propanoic acid,

i.e. number of moles of propanoic acid = 0.�66 mol

Maximum mass of propanoic acid formed = 0.�66 mol x 74.� g mol–�

= �2.3 g

�00


ii) Mass of propanoic acid obtained = 6.0 cm3 x 0.99 g cm–3

= 5.94 g

Percentage yield = 5.94 g�2.3 g

x �00%

= 48.3%

h) CH3CH2CHO

Only partial oxidation occurs / incomplete oxidation / product removed as formed.

�9 a)

b)

(l) + NaOH(aq)

COOCH2CH3

+ CH3CH2OH(aq)

COO– Na+

c) ethanol

d) Add dilute sulphuric acid / hydrochloric acid to the solution remaining in the flask.

Then filter.

e) Dissolve the crude sample in minimum amount of hot solvent.

Filter while hot.

Allow the filtrate to cool for crystals to re-form. Filter off the crystals.

�0�

Suggested Answers

f) Number of moles of ethyl benzoate = 3.30 g

�50.0 g mol–�

= 0.0220 mol

According to the equation, � mole of C6H5COOCH2CH3 gives � mole of C6H5COOH upon hydrolysis,

i.e. number of moles of C6H5COOH = 0.0220 mol

Theoretical yield of C6H5COOH = 0.0220 mol x �22.0 g mol–�

= 2.68 g

Percentage yield of C6H5COOH = �.90 g2.68 g

x �00%

= 70.9%

20 a) Shaking and the heat of the hands may cause the liquid inside the separating funnel to vaporize. Pressure may build up inside the funnel.

An explosion may occur if the tap is not opened regularly for pressure release.

b) Without releasing the pressure, the liquid in the separating funnel will not drain out of the funnel.

c) The product may not separate out as crystals if too much solvent is used.


Practice

P33.1 page 180

a) Detergent I

b)

CH3(CH2)15CH2 O–K+

O

C

hydrophilicpart

hydrophobicpart

c) Before shaking an oil-water mixture, the hydrocarbon ‘tails’ of detergent particles are soluble in the oil and the anionic ‘heads’ are soluble in the water.

Upon shaking, oil droplets form. Each droplet is surrounded by detergent particles with the anionic ‘heads’ in the water.

Repulsion between the anionic heads prevents the oil droplets from coming together again. Thus, the oil droplets remain suspended in the water. An emulsion is formed.

d) i) CH3(CH2)�5COOH

ii) 2H+(aq) + CO32–(aq) CO2(g) + H2O(l)

�02


P33.2 page 184

a) i)

C17H35

C17H33

C17H29

O C

O

O C

O

O C

O

H2C

HC

chiral carbonH2C

ii) H2 / Pt catalyst

iii) X has one chiral carbon and hence optically active.

Y does not have any chiral carbons and hence optically inactive.

Thus, the conversion involve a change in optical activity.

b) i)

C17H35

C17H35 + 3NaOH + 3C17H35COO– Na+

C17H35

O C

O

O C

O

O C

O

H2C

HC

H2C

OH

OH

OH

HC

H2C

H2C

ii) A detergent helps water to remove dirt by

• its ability to act as a wetting agent; and

• its emulsifying action.

P33.3 page 191

�

C

O

C

O

CH2 CH2 CH2 CH2 OO

2 a) Step 1 — Cr2O72–(aq) / H+(aq)

Step 2 — cold alkaline KMnO4

Step 3 — conc. H2SO4, heat

b) PET is a polyester. It undergoes hydrolysis in the presence of acid / alkali to give its monomers.

�03

Suggested Answers

P33.4 page 195

� Strong and lightweight

2 a) HOOC COOH

b) H2NCH2CH2NH2

Chemistry Magazine page 185

Environmental issues related to the use of detergents

� This detergent is biodegradable.

Producing this detergent can save food.

2 Algae cut off the light supply. Plants underneath the surface cannot carry out photosynthesis and die. They used up dissolved oxygen upon decomposition and causes smells and plagues.

Deprived of shelter and food, the fish larvae starve.


� a) i) B

ii) A

b)

2 When we add a detergent solution to a table cloth stained with grease, the hydrocarbon ‘tails’ of the detergent particles dissolve in the grease. The anionic ‘heads’ of the detergent particles are insoluble in the grease and remain outside.

The surrounding water molecules attract the anionic ‘heads’ and lift the grease off the surface.

By stirring, the grease (which carries the dirt particles) breaks up into tiny droplets suspended in the water. These tiny droplets cannot come together again due to the repulsion between the anionic ‘heads’ of detergent particles on their surfaces. An emulsion forms.

Rinsing washes away the greasy suspension, and leaves the surface clean.

�04


3 a) A hexanedioic acid

B hexane-�,6-diamine

b)

H

N

O

C (CH2)4 C

O

H

N (CH2)6 + 2n H2O

n HOOC(CH2)4COOH + n H2N(CH2)6NH2

n

c) Condensation polymerization

4 a) O

C

O

C O CH2 OCH2

b) i) O

CHO

O

C OH CH2HO OHCH2

ii) Condensation polymerization

c) Fibres used to make clothes

Bottles for carbonated drinks

d) By hydrolysis in acidic environment.

5 B

6 D

7 B

2n

OHHOn HO+ OHC

O

C

O

CH2 CH2

C

O

C

O

n

CH2 CH2 OO

n

+ H2O

8 C

9 B

�05

Suggested Answers

�0 A

�� B (�) The detergent is a soapless one. It is manufactured from chemicals derived from petroleum.

(3) The Na+ part is NOT involved in the cleansing process.

�2 C

�3 A

�4 B (�) X is a condensation polymer.

(2) X is nylon-6,6. It is NOT used in making packaging materials.

H +

H2O

HO

N (CH2)6C C

O

N(CH2)4

H nO

H

HOH N (CH2)6 N H

H

H

+

H2O

+

H2O

HO...... ......C (CH2)4

O

C

O

repeated condensation

H2O released

�5 a) i) Oxidation

ii) Acidified potassium dichromate solution

b) Heat ethanoic acid and compound Y in the presence of concentrated sulphuric acid.

c) i)

OCOCH3

COO–Na+

ii) The ion-dipole interactions between –COO– and H2O are stronger than the hydrogen bonds between –COOH and H2O.

�6 Answers for the HKCEE question are not provided.


�06


�8 detergent molecule has a hydrophobic region and a hydrophilic region / detergent molecule has a polar head and a non-polar tail

tail can form intermolecular forces with fat / non-polar region can form intermolecular forces with fat / hydrophobic region can attract fat molecules

head can form intermolecular forces with water / polar region can form intermolecular forces with water / hydrophilic region can attract water molecules


20 a) butane-�,4-diol

b)

OHHOHO +OHC

O

C

O

CH2 CH2 CH2 CH2

C

O

C

O

n

CH2 CH2 CH2 CH2 OO

c)

C O

O

d) They contain two functional groups.

2� a)

O

C

O

C OHHO

HO CH2 CH2 OH

b) ethane-�,2-diol

c) condensation polymerisation

22 a)

HO OHC

O

(CH2)4 C

O

N

H

(CH2)6 NH

H

H

�07

Suggested Answers

b)

HO OHC

O

(CH2)4 C

O

N

H

(CH2)6 NH+

H

H

N

H

(CH2)6C

O

N(CH2)4 C

HO n

c) amide

d) Any TWO of the following:

• more corrosion resistant

• needs no lubricant

• strong but light in weight

• can be moulded to different shapes easily

23 a)

n

+ H2O2n

HO+ OHC

O

C

O

N

H

NH

H

H

C

O

C

O

N

H

N

H

n n

b)

C

O

N

H

c) During the production of Kevlar, the monomer molecules X and Y react to form polymer molecules, releasing water molecules. A condensation reaction occurs.

A condensation reaction is a chemical reaction in which two or more molecules react together to form a larger molecule with the elimination of a small molecule.

�08


24 a) i)

C

C6H5

H

O C

H

C6H5

C

O

O C ............

O

ii)

C

COOC2H5COOC2H5

CH3

C

H

H

C

CH3

C

H

H

............

b) i)

CH3 C

H

OH

CH3

ii)

C

O

CH3 OHC

H

CH3

iii) Mandelic acid has an extra –COOH group.

It forms more hydrogen bonds with water molecules.

δ+δ+

δ+

δ+

δ–

δ–

δ+

δ+

δ–

δ+ δ–

δ–

iv) To test for adverse side effects / toxicity / irritation.

�09

Suggested Answers

pages 207–215Topic Exercise

� B

2 B

3 B Butan-2-ol has two alkyl groups attached to the carbon atom bearing the hydroxyl group. Thus, it is a secondary alcohol.

OH

C H

H

H

H C

H

H

C

H

H

C

butan-2-ol

H

4 D CH3CHBrCH=CHCl contains one chiral carbon. It has enantiomers.

CH3CHBrCH CHCl

chiral carbon

CH3CHBrCH=CHCl has two different atoms or groups on each carbon atom of double bond. It has cis-trans isomers.

5 A CH3CH CH2(g) + [O] + H2O(l) CH3CH(OH)CH2OH(aq)

from acidified potassium permanganate solution

6 B CH3CH2OH(l) + PCl5(s) CH3CH2Cl(l) + POCl3(l) + HCl(g)

7 C

8 D

9 A

�0 B (�) X and Y are a pair of enantiomers.

(3) Both X and Y are optically active.

��0


�� B

�2 D

�3 A

�4 D

�5 C

�6 A


�8 a) Structural isomer — compounds with the same molecular formula but differs in the order in which atoms are linked

Stereoisomers — compounds with their atoms linked in the same way but differs in the spatial arrangement of their atoms.

b) cis-trans isomerism occurs in some alkenes because

• rotation about the C=C bond is restricted;


c) Let (CH2)n be the molecular formula of compounds F, G and H.

Relative molecular mass = n(�2.0 + 2 x �.0) = �4n

∴ �4n = 70.0 n = 5

∴ The molecular formula of compounds F, G and H is C5H�0.

Structures of F and G:

C C

C2H5

H H

CH3

C C

C2H5

H CH3

H

Structure of H:

C C

CH3CH2CH2

H H

H



��

Suggested Answers

2� a) i)

C C

CH2H3C

H CH2 OH

H

ii) • rotation about the C=C bond is restricted;


b)

CH2H3C CH2 C

Cl

H

C

H

H

ClCH2H3C CH2 C

H

H

C

H

H

Cl

c) i) Acidified potassium dichromate solution

ii) Carbonyl group

d) i) Boil a liquid with a vertical condenser.

ii)

OH

carboxyl group

C

O

22 a) Isomer B


• There are two alkyl groups attached to the carbon bearing the –OH group.

• There are one hydrogen atom attached to the carbon bearing the –OH group.

b) Isomer C

because it is a tertiary alcohol / there is no hydrogen atom to be removed from the carbon bearing the –OH group.

c) Isomer B

CH3CH2CCH3

O

d) Isomers A and D

e) Steamy / misty / white fumes

f) C4H9Cl + POCl3 + HCl

��2


23 a) HOCH2CHO + 3[O] HOOCCOOH + H2O

b) HOCH2CH2OH

24 Answers for the HKDSE question are not provided.

25 Answers for the HKALE question are not provided.


27 a) C=C bond

Hydroxyl group

b) i)

Any three

HO

R

ii) Enantiomers are mirror images that are not superposable.

c) i)

OCH3

R

C

O

ii) Ethanoic acid, concentrated sulphuric acid

d) C24H3�NO + 2Br2 C24H3�NOBr4

e) Use the minimum amount of hot solvent to dissolve the solid.

Filter to remove the insoluble impurities.

Allow the filtrate to cool slowly for crystals to form.

Filter off the crystals.

Wash and dry.

f) Is the drug safe?

Does the drug do the job it is designed to do?

Is it better than the standard treatment being used?

��3

Suggested Answers

28 a) Any ONE of the following:

• Add sodium hydrogencarbonate solution.

Only compound F gives effervescence.

• Test with moist blue litmus paper.

Only compound F turns the blue litmus paper red.

b) Warm with acidified potassium dichromate solution.

Only compound J turns the orange dichromate solution green.



3� Answers for the HKDSE question are not provided.




35 a) Condensation polymerization

b)

N

H

(CH2)6 NH

H

H H HC

O

(CH2)4O C O

O

c)

C

O

(CH2)4 C

O

N

H

(CH2)6 N

H

d) There are hydrogen bonds between the polymer chains of nylon.

Only weak instantaneous dipole-induced dipole attractions exist between the polymer chains of polythene.

Hence nylon is stronger than polythene.

e) A hole would appear because the acid hydrolyzes the amide linkages in nylon.

��4


f) Method I:

The waste contains dichromate ions which is toxic.

Method II:

Concentrated nitric acid is a strong acid. Discharge of the waste into waterways leads to environmental pollution.

suggested teaching scheme - stmgssintranet.stmgss.edu.hk/~ted/ccy/new_edition/book3banswer.pdf ·...

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