CONTENTS
STPM Scheme of Assessment ii
TERM 1
CHAPTER Atoms, Molecules and Stoichiometry 1
1.1 Fundamental Particles of an Atom 1
1 1.2 Relative Atomic, Isotopic, Molecular and Formula Masses 2
1.3 The Mole and the Avogadro Constant 11
CHAPTER Electronic Structure of Atoms 21
2 2.1 Electronic Energy Level of Atomic Hydrogen 21
2.2 Atomic Orbitals: s, p and d 29
2.3 Electronic Configuration 31
2.4 Classification of Elements in the Periodic Table 39
CHAPTER Chemical Bonding 42
3.1 Ionic Bonding 42
3 3.2 Covalent Bonding 45
3.3 Metallic Bonding 58
3.4 Intermolecular Forces 62
CHAPTER States of Matter 70
4.1 Gases 70
4 4.2 Liquids 88
4.3 Solids 94
4.4 Phase Diagrams 99
CHAPTER Reaction Kinetics 105
5.1 Rate of Reaction 105
5 5.2 Rate Law 109
5.3 The Effect of Temperature on Reaction Kinetics 112
5.4 The Role of Catalysts in Reactions 118
5.5 Order of Reactions and Rate Constants 121
CHAPTER Equilibria 134
6.1 Chemical Equilibria 134
6 6.2 Ionic Equilibria 143
6.3 Solubility Equilibria 156
6.4 Phase Equilibria 165
STPM Model Paper 962/1 178
v
TERM 2
CHAPTER Chemical Energetics 183
7 7.1 Enthalpy Changes of Reaction, ΔH 183
7.2 Hess’ s Law 188
7.3 Born-Haber Cycle 195
7.4 The Solubility of Solids in Liquids 202
CHAPTER Electrochemistry 206
8.1 Half-cell and Redox Equations 206
8 8.2 Standard Electrode Potential 208
8.3 Non-standard Cell Potentials 218
8.4 Fuel Cells 221
8.5 Electrolysis 222
8.6 Applications of Electrochemistry 233
CHAPTER Periodic Table: Periodicity 236
9.1 Physical Properties of Elements of Period 2 and Period 3 236
9 9.2 Reactions of Period 3 Elements with Oxygen and Water 240
9.3 Acidic and Basic Properties of Oxides and Hydrolysis of Oxides 243
CHAPTER Group 2 252
10.1 Selected Group 2 Elements and Their Compounds 252
10 10.2 Anomalous Behaviour of Beryllium 262
10.3 Uses of Group 2 Compounds 269
CHAPTER Group 14 271
11 11.1 Physical Properties of Group 14 Elements 271
11.2 Tetrachlorides and Oxides of Group 14 Elements 274
11.3 Relative Stability of +2 and +4 Oxidation States of Group 14
Elements 282
11.4 Silicon, Silicone and Silicate 287
11.5 Tin Alloys 293
CHAPTER Group 17 295
12.1 Physical Properties of Selected Group 17 Elements 295
12 12.2 Reactions of Selected Group 17 Elements 298
12.3 Reactions of Selected Halide Ions 305
12.4 Industrial Applications of Halogens and Their Compounds 309
vi
CHAPTER Transition Elements 313
13.1 Physical Properties of First Row Transition Elements 313
13 13.2 Chemical Properties of First Row Transition Elements 315
13.3 Nomenclature and Bonding of Complexes 329
13.4 Uses of First Row Transition Elements and Their Compounds 332
STPM Model Paper 962/2 335
TERM 3
CHAPTER Introduction to Organic Chemistry 342
14 14.1 Bonding of the carbon atoms: The Shapes of Ethane, Ethene,
Ethyne, and Benzene Molecules 342
14.2 General, Empirical, Molecular and Structural Formulae of Organic
Compounds 343
14.3 Functional Groups: Classification and Nomenclature 346
14.4 Isomerism: Structural and Stereoisomerism 347
14.5 Free Radicals, Nucleophiles and Electrophiles 350
14.6 Molecular Structure and its Effect on Physical Properties 353
14.7 Inductive and Resonance Effect 355
CHAPTER Hydrocarbons 360
15.1 Alkanes 360
15 15.2 Alkenes 365
15.3 Arenes 380
CHAPTER Haloalkanes 391
16 16.1 Physical Properties of Haloalkanes 391
16.2 Nucleophilic Subtitution of Haloalkanes 391
16.3 Elimination Reaction 406
16.4 Mechanism of Nucleophilic Substitution 410
16.5 Reactivity of Primary, Secondary and Tertiary Haloalkanes 413
16.6 Reactivity of Chlorobenzene and Chloroalkanes in Hydrolysis
Reactions 413
16.7 Organometallic Compounds 417
16.8 Uses of Haloalkanes 419
16.9 Effects of CFC on the Environment 419
CHAPTER Hydroxy Compounds 421
17.1 Introduction to Hydroxy Compounds 421
17 17.2 Alcohols 422
17.3 Phenols 442
vii
CHAPTER Carbonyl Compounds 451
18.1 Introduction to Carbonyl Compounds 451
18 18.2 Physical Properties of Carbonyl Compounds 451
18.3 Reactions of Carbonyl Compounds 452
18.4 Carbohydrates 471
CHAPTER Carboxylic Acids and their Derivatives 472
19.1 Carboxylic Acid 472
19 19.2 Acyl Chlorides 481
19.3 Esters 488
19.4 Amides 493
CHAPTER Amines, Amino Acids and Proteins 499
20.1 Amines 499
20 20.2 Amino Acids 510
20.3 Protein 522
CHAPTER Polymers 523
21.1 Synthetic Polymers 523
21 21.2 Condensation Polymerisation 524
21.3 Addition Polymerisation 527
21.4 Classification of Polymers 532
STPM Model Paper 962/3 533
Answers
539
viii
Chapter Haloalkanes
16
16.1 Physical Properties of Haloalkanes
Section A Multiple-choice Questions
Question 1
Haloalkanes are more reactive than alkanes. Why?
A Haloalkane molecules are larger.
B The molar mass of haloalkanes are higher.
C The difference in electronegativity between halogan atoms and hydrogen
is large.
D The carbon-halogen bonds are stronger.
Answer: C
Alkanes are non-polar because the difference in electronegativity between C
and H is small, whereas that between C and X (halogen) is larger. As a result,
the C!X is polar and hence more reactive (especially towards polar reagents
such as OH–, CN– and NH3).
δ+ δ–
C!X + Nu– !: C!Nu + X–
16.2 Nucleophilic Subtitution of Haloalkanes
Section A Multiple-choice Questions
Question 1
Cl F
A B D H
H C C
C
Br H
Which of the bonds A, B, C or D in the above structure is the most difficult to Term
break?
Answer: D 3
391
Chemistry Term 3 STPM Chapter 16 Haloalkanes
Due to the small size of the fluorine atom, the C!F bond has the highest
bond dissociation energy.
Bond C!H C!F C!Cl C!Br
Bond energy / kJ mol–1 413 467 346 290
Question 2
With which of the following reagents, chlorocyclohexane does not undergo
substitution?
A Aqueous NaOH
B Ethanolic KOH
C Concentrated NH3
D Ethanolic KCN
Answer: B
With ethanolic KOH, chlorocyclohexane undergoes elimination to produce
cyclohexene.
Cl
H + KCl + H2O
+ KOH !Eth!an!ol/H!ea:t
Question 3
One of the possible products when bromocyclohexane is treated with an
ethanolic solution of sodium hydroxide is an ether with the structure below:
OC2H5
What type of reaction is this? C Nucleophilic substitution
A Neutralisation D Elimination
B Electrophilic addition
Answer: C
Term Ethanol reacts with sodium hydroxide to form the ethoxide ion:
C2H5OH + NaOH !: C2H5O– + Na+ + H2O
3 The C2H5O– is a nucleophile and substitutes the Br in bromocyclohexane.
Br + C2H5O– OC2H5 + Br –
392
Chemistry Term 3 STPM Chapter 16 Haloalkanes
Question 4
Consider the reactions below:
C3H7I + 2NH3 !: C3H7NH2 + NH4I I
C3H7I + H2O !: C3H7OH + KI II
The rate of reaction for reaction I is larger than that for reaction II at the same
temperature. This is because
A NH3 is a stronger nucleophile than H2O.
B NH3 is a base but H2O is amphoteric.
C the NH4+ needs less energy to form.
D NH3 is a stronger electrophile than H2O.
Answer: A
Both reactions involve substitution by nucleophiles, NH3 and H2O. NH3 is a
stronger base, and hence a stronger nucleophile than H2O.
[Note: A base (:NH3) is also a nucleophile.]
Question 5
1,1-dibromoethane is boiled with aqueous sodium hydroxide. The final organic
product is
A 1,1-ethanediol C Ethanal
B 1,2-ethanediol D Ethanoic acid
Answer: C
Br O
& '
CH3!&C!H + 2NaOH !: CH3!C!H + 2NaBr + H2O
Br
1,1-ethanediol is not formed because it is unstable and will decompose
simultaneously to ethanal.
Br OH
& &
CH3!&C! H + 2NaOH !: CH3!&C!H + 2NaBr
Br OH
Term
O! H O
& '
CH3!&C!H !: CH3!C!H + H2O 3
OH
393
Chemistry Term 3 STPM Chapter 16 Haloalkanes
Question 6
Compound X has the structure below:
Br
CHCH3
&
Br
What product is formed when X is treated with aqueous sodium hydroxide?
A OH C OH
CHCH3 CHCH3
& &
Br OH
B Br D ONa
CHCH3 CHCH3
& &
OH ONa
Answer: B
The Br that is bonded directly to the ring carbon is unreactive towards
electrophiles (OH–). This is due to the delocalisation of the non-bonding pair
of electrons on the bromine atom into the benzene ring. Thus, only the Br in
the side chain is hydrolysed to !OH.
Question 7
A reaction scheme involving benzyl chloride is shown below:
CH2Cl
Ethanolic KCN. Heat X Dilute H2SO4. Heat Y
What is Y?
A COOH C CONH2
Term
B CH2COOH D CH2CH2NH2
3
394
Answer: B Chemistry Term 3 STPM Chapter 16 Haloalkanes
CH2Cl
+ KCN CH2CN
+ KCl
CH2CN CH2COOH
+ 2H2O + H+ + NH4+
Question 8 CH2CH(I)CH3 CH2CH(NH2)CH3
CH"CHCH3
HI NH3
What type of reactions are involved in the above scheme?
A Free radical addition and nucleophilic addition.
B Electrophilic substitution and nucleophilic substitution.
C Electrophilic addition and nucleophilic substitution.
D Both are nucleophilic substitution.
Answer: C
The first reaction is an addition reaction to an alkene involving electrophiles
(electron-loving species). The second reaction is nucleophilic substitution of a
haloalkane by the nucleophile, NH3.
Question 9
The reaction between chloroethane and aqueous sodium hydroxide is slower
when compared to iodoethane with aqueous sodium hydroxide. Why is it so?
A The C!I bond is more polarised than the C!Cl bond.
B The Cl– ion is smaller than the I– ion.
C The size of Cl– ion is smaller than I– ion.
D Chlorine is smaller than bromine.
Answer: D Term
EExxaamm Tips 3
Hydrolysis involves the breaking of the carbon-halogen bond.
C!X + OH– !: C!OH + X–
395
Chemistry Term 3 STPM Chapter 16 Haloalkanes
Due to the smaller size of Cl, the C!Cl bond is stronger than the C!I bond
and requires more energy to break.
Section B Structured Questions
Question 10
Given the following haloalkanes:
1-Chlorobutane, 1-Bromobutane, 1-Iodobutane
(a) Arrange the three compounds in the order of increasing reactivity towards
nucleophilic substitution. Explain your answer.
(b) Suggest an experiment to verify your answer in (a).
Answer:
(a) 1-Chlorobutane < 1-Bromobutane < 1-Iodobutane.
The nucleophilic substitution involves the breaking of the carbon-halogen,
C!X bond:
R!X + Nu– !: R!Nu + X–
The C!X bond in 1-halobutane gets weaker from C—Cl to C—Br to
C—I as the size of the halogen atom increases.
(b) EExxaamm Tips
Precipitate the halogens as silver halides.
Warm the 1-haloalkanes separately with ethanolic silver nitrate and note
the time for the appearance of a precipitate.
R!X + H2O + Ag+ !: R!OH + H+ + AgX(s)
The time taken for the appearance of the silver halide precipitate is an
indication of the rate of the nucleophilic substitution.
The time taken for the appearance of the precipitate increases in the
order:
1-Iodobutane < 1-Bromobutane < 1-Chlorobutane
Term Question 11
(a) Using appropriate equations describe the mechanism of the hydrolysis of
1-iodopropane with aqueous sodium hydroxide.
(b) Suggest how 1-iodopropane can be converted to
(i) CH3CH2CH2NH2 (ii) CH3CH2CH2CH2NH2 (iii) CH3CH2CH2COOH
3
396
Answer: Chemistry Term 3 STPM Chapter 16 Haloalkanes
(a)
H H
C H3CH2CH2I + OH– !: HO– C I (Slow)
&
C2H5
H H
HO– C I !: CH3CH2CH2OH + I– (Fast)
&
C2H5
(b) (i) Heat excess 1-iodopropane with concentrated ammonia in a closed
vessel.
CH3CH2CH2I + 2NH3 !: CH2NH2 + NH4I
(ii) EExxaamm Tips
The final product has one carbon atom more than the starting material.
As a result, it must involve a nitrile (RCN) intermediate.
Heat iodopropane with an ethanolic solution of potassium cyanide.
CH3CH2CH2I + KCN !: CH3CH2CH2CN + KI
Warm the nitrile obtained with lithium aluminium hydride in ether.
CH3CH2CH2CN + 4[H] !: CH3CH2CH2CH2NH2
(iii) Heat iodopropane with an ethanolic solution of potassium cyanide.
CH3CH2CH2I + KCN !: CH3CH2CH2CN + KI
Boil the nitrile obtained with dilute sulphuric acid.
CH3CH2CH2CN + 2H2O + H+ !: CH3CH2CH2COOH + NH4+
Question 12
2-Iodopropane reacts with sodium hydroxide under two different sets of
conditions to give two different products X , C3H8O and Y, C3H6.
(a) Draw the structure for X and Y.
(b) Describe the conditions necessary for the formation of X and Y.
(c) Suggest two simple chemical tests to differentiate between X and Y.
Answer: Term
(a) X: CH3—&CH—CH3
OH 3
Y: CH3!CH"CH2
397
Chemistry Term 3 STPM Chapter 16 Haloalkanes
(b) X: Aqueous NaOH. Heat.
Y: Ethanolic NaOH. Heat.
(c) EExxaamm Tips
X is an alcohol and Y is an alkene.
Test 1: Add PCl5 at room temperature.
X: Steamy white fumes released.
CH3!&CH !CH3 + PCl5 !: CH3!CH!CH3 + POCl3 + HCl(g)
&
OH Cl
Y: No visible change observed.
Test 2: Add Br2 dissolved in CCl4.
X: No visible change observed.
Y: Reddish brown colour decolourises.
CH3!CH"CH2 + Br2 !: CH3!CHBr!CH2Br
Question 13
Iodoethane is boiled with aqueous sodium hydroxide. Excess nitric acid is then
added to the resulting solution followed by ethanolic silver nitrate. A yellow
precipitate is formed.
(a) Name the yellow precipitate and give its formula.
(b) Write an equation for the reaction taking place when iodoethane is boiled
with sodium hydroxide.
(c) Why does the resulting solution have to be acidified before adding ethanolic
silver nitrate?
Answer:
(a) Silver iodide, AgI.
(b) C2H5I + NaOH !: C2H5OH + NaI
(c) To neutralise the excess sodium hydroxide. Otherwise the sodium
hydroxide will react with silver nitrate to give a grayish precipitate of
silver oxide and interfere with the observations.
Term 2OH– + 2Ag+ !: Ag2O(s) + H2O
3
398
Chemistry Term 3 STPM Chapter 16 Haloalkanes
Question 14
1-chlorobutane reacts with concentrated ammonia to produce butanamine.
(a) Write an equation for the reaction.
(b) Name the type of mechanism involved. Write the rate equation for the
reaction.
(c) What is the effect to the rate of reaction if
(i) the concentration of ammonia is increased?
(ii) 1-chlorobutane is replaced with 1-iodobutane?
Answer:
(a) CH3CH2CH2CH2Cl + 2NH3 !: CH3CH2CH2CH2NH2 + NH4Cl
(b) It follows SN2 mechanism.
Rate = k[Ammonia][1-chlorobutane]
(c) (i) EExxaamm Tips
The rate is directly proportional to the concentration of ammonia.
Rate increases.
(ii) Rate increases. (Because the C!I bond is weaker than the C!Cl
bond.)
Section C Essay Questions
Question 15
(a) What do you understand by the term nucleophiles?
(b) Describe the mechanism when 1-iodoethane is boiled under reflux with
aqueous sodium hydroxide.
(c) What would happen to the rate of reaction if the concentration of sodium
hydroxide is increased? Explain your answer.
Answer:
(a) A nucleophile is a species that donates a lone pair of electrons to another
species to form a coordinate bond. A nucleophile is an electron-rich
species that attacks regions of low electron density.
(b) The mechanism involved is SN2, which stands for bimolecular nucleophilic Term
substitution. The nucleophile in this case is the OH– ion from the
dissociation of NaOH.
NaOH !: Na+ + OH–
3
Due to the higher electronegativity of I, the C!I bond is polarised as
shown below:
399
Chemistry Term 3 STPM Chapter 16 Haloalkanes
H
&
C H3!δ+&C!Iδ–
H
The OH– is attracted to the partially positive C and form a five-coordinated
C atom intermediate.
CH3 &CH3
C!I !Slo:w HO– C I
OH– H H H
H
In the intermediate, the C!O bond is in the process of forming while
the C!I bond is in the process of breaking. Eventually, the C!O bond
is totally formed and the C!I bond is totally broken to give the final
product.
&CH 3 &CH3
HO– C I !Fas:t C + I–
H H H OH
H
(c) The rate will increase.
The rate determining step involves both OH– and the iodobutane molecule.
The rate equation is given by:
Rate = k[OH–][1-iodoethane]
Thus, increasing the concentration of NaOH will increase the rate of
collision between OH– and 1-iodoethane. This will lead to an increase in
the rate of reaction.
Term Question 16
Suggest a chemical test to differentiate between the following pairs of
compounds.
(a) 1-Chlorobutane and 1-iodobutane
(b) 2-Chlorobutane and 2-chlorobutene
(c) Chlorobenzene and chlorocyclohexane
(d) 2-Iodo-2-methylpropane and 1-iodobutane
(e) Chlorobenzene and benzylchloride
3
400
Jawapan
Kertas Model STPM 962/1
Section A 3. A 4. D 5. B
1. B 2. D 8. B 9. D 10. B
6. B 7. A 13. D 14. C 15. C
11. C 12. A
Section B Giant Giant ionic Giant Simple
16. (a) Solid metallic Ionic covalent covalent
Covalent van der
type Sodium Silicon Waals and/
chloride dioxide or Hydrogen
Bonding Metallic bonds
Example Copper Water
(b) (i) Cl N Cl Cl B Cl
Cl Cl
(ii) NBCCll33:: Trigon pyramidal
Trigonal planar
1 7 . ((ab)) N43N9.1HH2g44++gNiisowaNnailHldBror2eønraneactastetcewstdsait-whLpiort4owh3.t1r9o52yn0.a×t5coig5dN0Na.H5nH2dg–4.CNNl.HH42C– ils=a Brønsted-Lowry base.
5.33 g NH4Cl
VThoelulmimeitoifnNg Hre3agpernotduiscNeda N=H423..192 × 2 × 22.4 dm3
= 4.73 dm3
Section C
1 8. (a) The assumptions are:
The size of the gas molecules is negligible compared to the volume of
the container where the gas is placed. There is no intermolecular forces
(attractive or repulsive) between the gas molecules.
These assumptions are valid for a real gas only if the gas is under very low
pressure and/or very high temperature.
539
Chemistry STPM Answers
When the pressure on the gas is high, the molecules are pushed very close
to one another and the volume occupied by the gas is small. In such case
the total volume of the gas molecules cannot be ignored when compared
to the volume of the container where the gas is placed. However, at very
low pressure, the volume occupied by the gas is so large that the total
volume of the gas molecules is negligible.
At low temperatures, the kinetic energy of the gas molecules is low
and the intermolecular forces between the molecules become significant.
However, at high temperatures, the kinetic energy of the gas molecules is
very high and the intermolecular forces between them can be ignored.
(b) (i) Composition by mass of nitrogen = 100 – 12.5
= 87.5%
87.5 12.5
Mol ratio of N : H = 14 : 1
= 6.25 : 12.5
=1:2
ELEemqtuptahitrieiocmnalooflfoerdcmeucluaolrmafopofromsXituiilsoanNb:He 2(.NH2)n.
(NH2)n ➝ n N2 + nH2
2
From the equation:
( )12.5 cm3 of X would produce 12.5
( )12.5 n + n cm3 product
2
n2 n ==
n+ 37.5
(ii) 2
MThoeleocxuidlaartifoonrmnuulma boefrXoifsHNi2nH4X.
= +1.
Let the oxidation number of N = y.
2y + (1 × 4) = 0
y = –2
(iii) Lewis diagram of X is:
HH
H NN H
(c) Toluene is a hydrocarbon. The intermolecular forces are the van der
Waal’s forces.
Ethanol on the other hand contains the highly polar O—H bond.
CH3 — CH2 — O — H
The intermolecular forces are the van der Waal’s forces and hydrogen
bonds.
540