Q&A STPM CHEMISTRY ( PENGGAL 1.2.3 )

CONTENTS
CONTENTS

STPM Scheme of Assessment ii

TERM 1

CHAPTER Atoms, Molecules and Stoichiometry 1
1 1.1 Fundamental Particles of an Atom 1 2
1.2 Relative Atomic, Isotopic, Molecular and Formula Masses
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
29
2.2 Atomic Orbitals: s, p and d
2.3 Electronic Configuration 31
2.4 Classification of Elements in the Periodic Table 39


CHAPTER Chemical Bonding 42
3 3.1 Ionic Bonding 42
45
3.2 Covalent Bonding
3.3 Metallic Bonding 58
3.4 Intermolecular Forces 62


CHAPTER States of Matter 70
4 4.1 Gases 70
4.2 Liquids
88
4.3 Solids 94
4.4 Phase Diagrams 99


CHAPTER Reaction Kinetics 105
5 5.1 Rate of Reaction 105
109
5.2 Rate Law
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 6.1 Chemical Equilibria 134
6.2 Ionic Equilibria
143
6.3 Solubility Equilibria 156
6.4 Phase Equilibria 165

STPM Model Paper 962/1 178


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TERM 2

CHAPTER Chemical Energetics 183
7 7.1 Enthalpy Changes of Reaction, ΔH 183
188
7.2 Hess’ s Law
7.3 Born-Haber Cycle 195
7.4 The Solubility of Solids in Liquids 202


CHAPTER Electrochemistry 206
8 8.1 Half-cell and Redox Equations 206
208
8.2 Standard Electrode Potential
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 9.1 Physical Properties of Elements of Period 2 and Period 3 236
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 10.1 Selected Group 2 Elements and Their Compounds 252
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
274
11.2 Tetrachlorides and Oxides of Group 14 Elements
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 12.1 Physical Properties of Selected Group 17 Elements 295
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




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CHAPTER Transition Elements 313
13 13.1 Physical Properties of First Row Transition Elements 313
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, 342
Ethyne, and Benzene Molecules
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 15.1 Alkanes 360
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 17.1 Introduction to Hydroxy Compounds 421
422
17.2 Alcohols
17.3 Phenols 442



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CHAPTER Carbonyl Compounds 451
18 18.1 Introduction to Carbonyl Compounds 451
451
18.2 Physical Properties of Carbonyl Compounds
18.3 Reactions of Carbonyl Compounds 452
18.4 Carbohydrates 471



CHAPTER Carboxylic Acids and their Derivatives 472
19 19.1 Carboxylic Acid 472
19.2 Acyl Chlorides
481
19.3 Esters 488
19.4 Amides 493



CHAPTER Amines, Amino Acids and Proteins 499
20 20.1 Amines 499
20.2 Amino Acids
510
20.3 Protein 522


CHAPTER Polymers 523
21 21.1 Synthetic Polymers 523
524
21.2 Condensation Polymerisation
21.3 Addition Polymerisation 527
21.4 Classification of Polymers 532


STPM Model Paper 962/3 533


Answers 539






















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Chapter

16 Haloalkanes








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 NH 3 ).
–
–
δ+ δ–
C!X + Nu !: C!Nu + X –
–
16.2 Nucleophilic Subtitution of Haloalkanes

Section A Multiple-choice Questions

Question 1
Cl F
B D
A
H  C  C  H
C

Br H
Which of the bonds A, B, C or D in the above structure is the most difficult to Term
break?
3
Answer: D




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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 413 467 346 290
–1


Question 2
With which of the following reagents, chlorocyclohexane does not undergo
substitution?
A Aqueous NaOH
B Ethanolic KOH
C Concentrated NH 3
D Ethanolic KCN

Answer: B
With ethanolic KOH, chlorocyclohexane undergoes elimination to produce
cyclohexene.
Cl
H
Ethanol/Heat
+ KOH !!!!: + KCl + H 2 O


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:

OC 2 H 5
What type of reaction is this?
A Neutralisation C Nucleophilic substitution
B Electrophilic addition D Elimination


Answer: C
Ethanol reacts with sodium hydroxide to form the ethoxide ion:
C 2 H 5 OH + NaOH !: C 2 H 5 O + Na + H 2 O
–
+
Term
The C 2 H 5 O is a nucleophile and substitutes the Br in bromocyclohexane.
–
3
Br + C 2 H 5 O – OC 2 H 5 + Br –

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Chemistry Term 3 STPM Chapter 16 Haloalkanes
Question 4

Consider the reactions below:
C 3 H 7 I + 2NH 3 !: C 3 H 7 NH 2 + NH 4 I I
C 3 H 7 I + H 2 O !: C 3 H 7 OH + KI II
The rate of reaction for reaction I is larger than that for reaction II at the same
temperature. This is because
A NH 3 is a stronger nucleophile than H 2 O.
B NH 3 is a base but H 2 O is amphoteric.
C the NH 4 needs less energy to form.
+
D NH 3 is a stronger electrophile than H 2 O.

Answer: A
Both reactions involve substitution by nucleophiles, NH 3 and H 2 O. NH 3 is a
stronger base, and hence a stronger nucleophile than H 2 O.
[Note: A base (:NH 3 ) 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
& '
CH 3 !C!H + 2NaOH !: CH 3 !C!H + 2NaBr + H 2 O
&
Br
1,1-ethanediol is not formed because it is unstable and will decompose
simultaneously to ethanal.
Br OH
& &
CH 3 !C!H + 2NaOH !: CH 3 !C!H + 2NaBr
&  &
Br OH Term

O! H O
& '
CH 3 !C!H !: CH 3 !C!H + H 2 O 3
&
OH


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Chemistry Term 3 STPM Chapter 16 Haloalkanes
Question 6

Compound X has the structure below:
Br

CHCH 3
&
Br
What product is formed when X is treated with aqueous sodium hydroxide?
A OH C OH


CHCH 3 CHCH 3
& &
Br OH
B Br D ONa

CHCH 3 CHCH 3
& &
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:
CH 2 Cl

Ethanolic KCN. Heat Dilute H 2 SO 4 . Heat
X Y
What is Y?
A COOH C CONH 2





B CH 2 COOH D CH 2 CH 2 NH 2
Term
3





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Chemistry Term 3 STPM Chapter 16 Haloalkanes
Answer: B
CH 2 Cl CH 2 CN

+ KCN + KCl


CH 2 CN CH 2 COOH

+ 2H 2 O + H + + NH 4 +



Question 8

CH"CHCH 3 CH 2 CH(I)CH 3 CH 2 CH(NH 2 )CH 3
HI NH 3


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, NH 3 .

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
Exam Tips
Exam Tips
Hydrolysis involves the breaking of the carbon-halogen bond. 3
C!X + OH !: C!OH + X –
–

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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) Exam Tips
Exam 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 + H 2 O + 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


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) CH 3 CH 2 CH 2 NH 2 (ii) CH 3 CH 2 CH 2 CH 2 NH 2 (iii) CH 3 CH 2 CH 2 COOH
Term
3





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Chemistry Term 3 STPM Chapter 16 Haloalkanes
Answer:
(a) H H

CH 3 CH 2 CH 2 I + OH !: HO –  C  I (Slow)
–
&
C 2 H 5
H H
HO –  C  I !:  CH 3 CH 2 CH 2 OH + I (Fast)
–
&
C 2 H 5
(b) (i) Heat excess 1-iodopropane with concentrated ammonia in a closed
vessel.
CH 3 CH 2 CH 2 I + 2NH 3 !: CH 2 NH 2 + NH 4 I
(ii) Exam Tips
Exam 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.
CH 3 CH 2 CH 2 I + KCN !: CH 3 CH 2 CH 2 CN + KI
Warm the nitrile obtained with lithium aluminium hydride in ether.
CH 3 CH 2 CH 2 CN + 4[H] !: CH 3 CH 2 CH 2 CH 2 NH 2
(iii) Heat iodopropane with an ethanolic solution of potassium cyanide.
CH 3 CH 2 CH 2 I + KCN !: CH 3 CH 2 CH 2 CN + KI
Boil the nitrile obtained with dilute sulphuric acid.
+ +
CH 3 CH 2 CH 2 CN + 2H 2 O + H !: CH 3 CH 2 CH 2 COOH + NH 4

Question 12
2-Iodopropane reacts with sodium hydroxide under two different sets of
conditions to give two different products X , C 3 H 8 O and Y, C 3 H 6 .
(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: CH 3 —CH—CH 3
&
OH 3
Y: CH 3 !CH"CH 2



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Chemistry Term 3 STPM Chapter 16 Haloalkanes
(b) X: Aqueous NaOH. Heat.
Y: Ethanolic NaOH. Heat.

(c) Exam Tips
Exam Tips
X is an alcohol and Y is an alkene.

Test 1: Add PCl 5 at room temperature.
X: Steamy white fumes released.
CH 3 !CH!CH 3 + PCl 5 !: CH 3 !CH!CH 3 + POCl 3 + HCl(g)
& &
OH Cl
Y: No visible change observed.
Test 2: Add Br 2 dissolved in CCl 4 .
X: No visible change observed.
Y: Reddish brown colour decolourises.
CH 3 !CH"CH 2 + Br 2 !: CH 3 !CHBr!CH 2 Br


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) C 2 H 5 I + NaOH !: C 2 H 5 OH + 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.
2OH + 2Ag !: Ag 2 O(s) + H 2 O
–
+
Term
3





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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) CH 3 CH 2 CH 2 CH 2 Cl + 2NH 3 !: CH 3 CH 2 CH 2 CH 2 NH 2 + NH 4 Cl
(b) It follows S N 2 mechanism.
Rate = k[Ammonia][1-chlorobutane]
(c) (i) Exam Tips
Exam 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 S N 2, which stands for bimolecular nucleophilic
substitution. The nucleophile in this case is the OH ion from the Term
–
dissociation of NaOH.
+
NaOH !: Na + OH –
Due to the higher electronegativity of I, the C!I bond is polarised as 3
shown below:



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Chemistry Term 3 STPM Chapter 16 Haloalkanes
H
&
CH 3 ! C!I δ–
δ+
&
H
The OH is attracted to the partially positive C and form a five-coordinated
–
C atom intermediate.
CH 3 CH 3
&
Slow
C!I !: 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 CH 3
&  &
Fast
HO – C I !: 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.


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
Term
3





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Jawapan




Kertas Model STPM 962/1
Section A
1. B 2. D 3. A 4. D 5. B
6. B 7. A 8. B 9. D 10. B
11. C 12. A 13. D 14. C 15. C
Section B
16. (a) Solid Giant Giant Simple
type metallic Giant ionic covalent covalent
van der
Waals and/
Bonding Metallic Ionic Covalent
or Hydrogen
bonds
Sodium Silicon
Example Copper Water
chloride dioxide
(b) (i) Cl N Cl Cl B Cl

Cl Cl
(ii) NCl : Trigon pyramidal
3
BCl : Trigonal planar
3
17. (a) NH ion donates a proton to NH .
+
–
4
2
NH is a Brønsted-Lowry acid and NH is a Brønsted-Lowry base.
–
+
2
4
(b) 39 g NaNH reacts with 50.5 g NH Cl.
2
4
4.12 g will react with 4.12 × 50.5 g NH Cl = 5.33 g NH Cl
39 4 4
The limiting reagent is NaNH .
2
4.12
Volume of NH produced = × 2 × 22.4 dm 3
3 39
= 4.73 dm 3
Section C
18. (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.
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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
Empirical formula of X is NH .
2
Let the molecular formula be (NH )n.
2
Equation of decomposition:
n
(NH )n ➝ N + nH 2
2
2
2
From the equation:
(
12.5 cm of X would produce 12.5 n + n ) cm product
3
3
2
(
12.5 n + n ) = 37.5
2
n = 2
Molecular formula of X is N H .
4
2
(ii) The oxidation number of H in X = +1.
Let the oxidation number of N = y.
2y + (1 × 4) = 0
y = –2
(iii) Lewis diagram of X is:
H H
H N N 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.
CH — CH — O — H
3 2
The intermolecular forces are the van der Waal’s forces and hydrogen
bonds.


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