Chemistry_Anyflip Sem 2

PENERBIT
ILMU
BAKTI SDN. BHD.

Contents

PENERBITChapter 1 Reaction Kinetics 1
ILMU 8
BAKTI SDN. BHD.1.1 Reaction Rate 10
1.2 Collision Theory 16
1.3 Factors Affecting Reaction Rate
Summative Practice 1

Chapter 2 Thermochemistry

2.1 Concept of Enthalpy 19

2.2 Calorimetry 24

2.3 Hess’s Law 28

2.4 Born-Haber Cycle 31

Summative Practice 2 36

Chapter 3 Electrochemistry 38
51
3.1 Galvanic Cell 59
3.2 Electrolytic Cell
Summative Practice 3

Summative Assessment Test (UPS 1) 62

Chapter 4 Introduction to Organic Chemistry

4.1 Molecular and Structural Formulae 65

4.2 Functional Groups and Homologous Series 69

4.3 Isomerism 72

4.4 Reactions of Organic Compounds 76

Summative Practice 4 81

Chapter 5 Hydrocarbons

5.1 Alkanes 84

5.2 Alkenes 95

Summative Practice 5 107

Chapter 6 Benzene and Its Derivatives 109
112
6.1 Introduction 116
6.2 Nomenclature of Benzene Derivatives 123
6.3 Chemical Properties of Benzene
Summative Practice 6

Summative Assessment Test (UPS 2) 125

Chapter 7 Haloalkanes (Alkyl Halides)

7.1 Introduction 128

7.2 Chemical Properties of Haloalkanes 131

Summative Practice 7 139
PENERBIT
ILMUChapter 8 Hydroxy Compounds
BAKTI SDN. BHD.
8.1 Introduction 141

8.2 Physical Properties of Alcohols 144

8.3 Preparation of Alcohols 146

8.4 Chemical Properties of Alcohols 149

8.5 Phenol 153

Summative Practice 8 156

Chapter 9 Carbonyl Compounds

9.1 Introduction 158

9.2 Preparation of Carbonyl Compounds 161

9.3 Chemical Properties of Carbonyl Compounds 164

Summative Practice 9 180

Chapter 10 Carboxylic Acids and Its Derivatives

10.1 Introduction 183

10.2 Physical Properties of Carboxylic Acids 186

10.3 Preparation of Carboxylic Acids 190

10.4 Chemical Properties of Carboxylic Acids 193

Summative Practice 10 203

Summative Assessment Test (UPS 3) 206

Chapter 11 Amines 208
211
11.1 Introduction 215
11.2 Physical Properties of Amines 218
11.3 Preparation of Amines 228
11.4 Chemical Properties of Amines
Summative Practice 11

Chapter 12 Amino Acids

12.1 Introduction 231

12.2 Chemical Properties of Amino Acids 235

Summative Practice 12 240

Semester Examination 243
Answers 248

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• Complete answers

1CHAPTER BAKTI SDN. BHD.

Reaction Kinetics

1.1 Reaction Rate

Learning Outcomes

• Define reaction rate.
• Write differential rate equation.
• Determine reaction rate based on differential rate equation of a reaction.
• Define rate law, order of reaction and half-life.
• Write rate law with respect to the order of reaction.
• Write the integrated rate equation for zero, first and second order reactions.
• Determine the order of reaction using initial rate method, half-life based on

graph of concentration against time and linear graph method based on the
integrated rate equation and rate law.
• Perform calculation using integrated rate equations involving single reactant.
ILMU ConcentrationSmart Tips!

Products

PENERBIT 1 The definition of reaction rate is the change of concentration of reactant or Reactants
product with time.
Time
2 The following is the mathematical formula for reaction rate.
Figure 1.1
Rate = (Final concentration) – (Initial concentration)
Time The rate of reaction
∆(Concentration) expressed in terms of
= ∆t reactant concentration
is negative because the
3 The graph of concentration of reactant and product versus time in Figure 1.1 reactant decreases with
shows that the concentration of the reactant decreases, while the concentration time.
of the product increases with time. The rate of reaction
expressed in terms of
4 The following general equation represents a chemical reaction. product concentration
is positive because the
aA + bB cC + dD product increases with
time.
Rate of reaction = – 1 d[A] = – 1 d[B] = + 1 d[C] = + 1 d[D]
a dt b dt c dt d dt

1

 Reaction Kinetics Example 1

2 The balanced equation for the conversion of a compound T into compound R is

as below. T → 2R

The concentration of T is 4.14 mol dm–3 at t = 30 s and 2.15 mol dm–3 at t = 120 s.
BAKTI SDN. BHD.
What is the average rate of the reaction during this time interval?

Solution
Average rate of reaction = (4.14 – 2.15) mol dm–3 = 0.022 mol dm–3 s–1

(120 – 30) s

Chapter 1 Example 2

The balanced equation for the breakdown of ozone into oxygen is as follows.
2O3(g) → 3O2(g)

(a) Write the differential rate equation for the reaction.
(b) If the rate of appearance of oxygen, O2 is 6.0 × 10–5 M s–1, what is the rate

of disappearance of ozone, O3?

Solution

(a) The differential rate equation is:

Rate = – 1 d[O3] =+ 1 d[O2]
2 dt 3 dt

(b) Given that + d[dOt2] = 6.0 × 10–5 M s–1.
ILMU d[O3] = + d[O2]
– 1 dt 1 dt
2 3

– d[O3] = 2 d[O2]
dt 3 dt

– d[O3] =2(6.0 × 10–5) = 4.0 × 10–5 M s–1
dt 3
PENERBIT
So, the rate of disappearance of O3 is 4.0 × 10–5 M s–1

Rate Law

1 Rate law is an equation which relates the rate of a reaction to the
concentration of the reactants with each term raised to some power.

2 Consider the balanced general equation below.

aA + bB cC + dD
The rate law is written as:

Rate = k[A]m[B]n

where k = rate constant
m = order of reaction with respect to reactant A
n = order of reaction with respect to reactant B

3 The order of reaction with respect to a reactant can only be determined
experimentally.

Reaction Kinetics 

4 The order of a reaction with respect to a particular reactant is defined as the
power to which the concentration of the reactant is raised in the rate law.
A reaction can be zero order, first order or second order with respect to a
reactant.

5 The overall order of a reaction is the sum of the individual orders. For
example, if the reaction rate is given by rate = k[A]m[B]n, the overall order of
the reaction is m + n.

6 Half life, t 1 is the amount of time needed for the concentration of a reactant to
2
decrease by half compared to its initial concentration.
PENERBIT
ILMUExample 3 Chapter 1
BAKTI SDN. BHD.
What is the half-life of Np-240 if the mass of a 50 g sample was found to be
3.125 g after 4 hours?

Solution

t1 t1 t1 t1
50 g 2→ 25 g 2→ 12.5 g 2→ 6.25 g 2→ 3.125 g

4t 1 = 4 hours

2

t 1 = 1 hour

2

7 Table 1.1 summarises the characteristics of zero order, first order and second order reactions.

Table 1.1 Zero order, first order and second order reactions; consider the reaction as: A → product

Order Zero order First order Second order
Definition A chemical reaction in
A chemical reaction in A chemical reaction in which the rate of reaction
Rate law is proportional to the
(single-reactant which the rate of reaction which the rate of reaction square of concentration of
reactions) the reactant
Plot of rate is constant and independent is proportional to the
versus [A] Rate = k[A]2
of the concentration of the concentration of the
Rate
reactant reactant

Rate = k Rate = k[A]

Rate Rate

0 [A] 0 [A] 0 [A]

Unit of rate M time–1 Time–1 M–1 time–1
constant, k mol dm–3 time–1 dm3 mol–1 time–1
mol L–1 time–1 L mol–1 time–1
Integrated rate
equation [A]0 – [A] = kt ln[A]0 – ln[A] = kt 1 1
[A] [A]0
ln [A]0 = kt – = kt
[A]

3

 Reaction Kinetics

Half-life t1 = [A]0 t1 = ln 2 t1 = 1
Plot of [A] 2k k k[A]0
2 2 2

[A] [A] [A]

versus t [A]0 [A]0 [A]0

[A]0 t2 = 1 t1 [A]0BAKTI SDN. BHD.t1 = t2 = t1 [A]0 t2 = 2t1
[A2]0 2 [A2]0 [A2]0
40 t 2
40 t1 t t 40 t1 t
1 t

2
t2 
















t2

Half-life method • As [A]0 is halved, t1 is • As [A]0 is halved, t 1 • As [A]0 is halved, t1 is
halved. remains constant 2 doubled.
2 2

Chapter 1 • t 1 is directly • t1 is independent of [A]0. • t 1 is inversely

2 2 2

proportional to [A]0. proportional to [A]0.

Linear graph [A] In [A] 1
method [A]0 In [A]0 [A]
based on the
integrated rate

equation and

rate law 1
[A]0
t t t

[A] = –kt + [A]0 ln[A] = –kt + ln[A]0 1 = kt + 1
[A] [A]0
y = mx + c
y = mx + c y = mx + c

Note: t1 = first half-life, t2 = second half-life
ILMU
Example 4

Consider the following reaction given by the equation:
S + Y → Product

PENERBIT Experiment [S]0 (mol dm–3) [Y]0 (mol dm–3) Initial rate (mol dm–3 s–1)
0.2 0.2
1 0.4
2 0.2 0.4 1.6
3 0.8
0.4 0.2

Determine
(a) the order of reaction with respect to each reactant.
(b) the rate law for this reaction.
(c) the overall order of reaction.

Solution

(a) The rate equation is: Rate = k[S]m [Y]n


To find n, choose the experiments where [S] remains constant.

Substitute the values from experiments 1 and 2 into the rate equation.

0.4 = k(0.2)m (0.2)n ➀

1.6 = k(0.2)m (0.4)n ➁

4

Divide (2) by (1). Reaction Kinetics 

1 2 01..64 =0.4 n Alternative
0.2 Method

4 = 2n Example 4 shows how
to determine the order
n = 2 of reaction using the
initial rate method, while
[P]P(molEdm-3)NERBIT Therefore, the reaction is second order with respect to Y. Example 5 uses the
ILMU graphical method.
BAKTI SDN. BHD. To find m, choose the experiments where [Y] remains constant.
5
Compare experiments 1 and 3.

0.4 = k(0.2)m (0.2)n ➀

0.8 = k(0.4)m (0.2)n ➂

Divide (3) by (1). Chapter 1

1 2 00..84 =0.4 m
0.2

2 = 2m

m = 1

Therefore, the reaction is first order with respect to S.

(b) The rate law is:

Rate = k[S][Y]2

(c) The overall order of reaction = 1 + 2 = 3

Example 5

The concentration of reactant P in a decomposition reaction is given in the
following table. Determine the order of reaction.

Time (min) [P] (mol dm–3)
0 24.0
30 14.1
50 12.2
70 9.4
120 5.1
160 4.5

Solution

Plot the graph of [P] versus time. From the graph, the second half-life is twice
the length of the first half-life.

30
25
20
15
10

5
0 20 40 60 80 100 120 140 160 180 Time (min)

 Reaction Kinetics Example 6

6 The decomposition of reactant A into product B is a zero order reaction. The initial
concentration of A is 0.065 M and the rate constant at 25 °C is 1.5 × 10–3 M s–1.
Calculate
(a) the half-life for the reaction.
(b) the concentration of A after 7 s.
(c) the time required for the concentration of A to decrease by 80% from its

initial concentration.
BAKTI SDN. BHD.
Solution

Chapter 1 (a) The equation for the half-life of a zero order reaction is t1 = [2Ak]0.

2

t1 = [A]0 = 2 × 0.065 M s–1 = 21.7 s
2k 1.5 × 10–3 M
2

(b) Use the integrated rate equation for zero order reactions, [A]0 – [A] = kt.

[A] = –kt + [A]0 = – (1.5 × 10–3 M s–1)(7 s) + 0.065 M = 0.0545 M

(c) When the concentration of A has decreased by 80%, the remainder is 20%.

[A] = 0.065 M × 20 = 0.013 M
100

[A] = –kt + [A]0
0.013 M = – (1.5 × 10–3 M s–1)(t) + 0.065 M

t = 34.67 s

Example 7ILMU

The rate constant for the first-order decomposition of compound A at 450 °C is
9.1 × 10−3 s−1.
(a) Determine the half-life of the reaction.
(b) How long will it take for 80.0% of a sample of A to decompose?

PENERBIT Solution

(a) t1 = ln 2
k
2

t1 = 9.1 ln 2 s–1 = 76.17 s
× 10–3
2

(b) Although the initial concentration, [A]0 is not provided, but the
information that stated 80.0% of the sample has decomposed is enough to

solve this problem. 20% of A remains if 80% of A has decomposed.

Assume [A]0 = x. So, the remaining concentration, [A] = 0.2x.

ln [[AA]]0 = kt

ln x = (9.1 × 10−3 s−1)(t)
0.2x

t = 176.9 s

Example 8 Reaction Kinetics 

The rate constant for the reaction, M → N + O is 5.2 × 10–5 dm3 mol–1 s–1. 7
The initial concentration of M is 0.5 mol dm–3.
Calculate
(a) the half-life of the reaction.
(b) the time required for 75% of M to have reacted.
PENERBIT
ILMUSolution
BAKTI SDN. BHD.
The reaction is a second order reaction since the unit of the rate constant is

dm3 mol–1 s–1.

(a) t1 = 1 = 5.2 × 10–5 dm3 1 s–1 (0.5 mol dm–3) Chapter 1
k[M]0 mol–1
2

= 3.846 × 104 s

(b) 25% of M remains if 75% of M has reacted.

Concentration of M if 75% has reacted
=12050 × 0.5 mol dm–3
= 0.125 mol dm–3

[A1] – [A1]0 = kt

0.1125 – 1 = (5.2 × 10–5)(t)
0.5

t = 1.154 × 105 s

Quick Check 1.1

1 Consider the reaction below.
F2(g) + 2I–(aq) → I2(s) + 2F–(aq)

The concentration of I– at 0 min = 0.35 mol L–1.
After 15 minutes, the concentration of I– is 0.22 mol L–1.
Calculate the rate of formation of I2.

2 Given the rate of change in the concentration of hydrogen is 4.5 × 10–5 M s–1
for the reaction below.
N2(g) + 3H2(g) → 2NH3(g)

What is the rate of change in the concentration of ammonia?

3 The following reaction,

A → 2B

is a zero order reaction. The rate constant of this reaction is 1.59 × 10–3 M s–1.
The half-life for the reaction is 10.5 s.
(a) Sketch the plot of the concentration of A versus time for this reaction.
(b) What is the initial concentration of A?
(c) What is the concentration of A after 6 s?

Summative Practice 1

Objective Questions BAKTI SDN. BHD. 5 Calculate the rate constant of a first order
reaction which has a half-life of 1 min 30 s.
1 What is the order of reaction if the graph of 1 A 7.7 × 10–2 s
[A] B 7.7 × 10–3 s
versus time is a straight line? C 14.14 × 10–2 s
D 14.14 × 10–3 s
A Second

Chapter 1 B Third

C Zero 6 Given the value of k for the reaction, A → B is
0.035 M–2 s–1. What is the order of the reaction?
D First A First
B Zero
2 The rate of a first order reaction is C Second
1.45 × 10–2 mol L–1 min–1 at 0.5 M concentration D The order of reaction cannot be identified
of reactant. The half-life of the reaction is
A 0.525 min 7 What is the unit of the rate constant of a second
B 10.27min
C 23.9 min order reaction?
D 239 min
A s–1 C L2 mol s–1

3 Which of the plots show a zero-order reaction? B L2 mol–2 s–1 D L mol–1 s–1

A C 8 Given the equation of a reaction below.

ILMU 2NO(g) + O2(g) → 2NO2(g)
Which of the statement is true about the
ln [A] [A] equation?
A The rate of formation of NO2 is same as the
t t
rate of consumption of O2
B D B The rate of formation of NO2 is same as the

rate of consumption of NO
[A] [A] C The rate of oxygen consumption is higher
PENERBIT
tt than the rate of formation of NO2
D The rate of consumption of NO is the same
4 The rate of a first order reaction is
1.45 × 10–2 min–1 when the concentration of the as the rate of O2 consumption
reactant is 0.3 mol L–1. What is the half-life of
this reaction? 9 Which of the following statements is true based
A 14.3 min on the collision theory?
B 28.6 min I The reaction rate is directly proportional to
C 47.8 min the collision frequency
D 95.6 min II Temperature affects the rate of reaction
III Particle size does not influence the rate of
reaction
A I only
B I and II
C II and III
D I, II and III

16

10 The activation energy of an exothermic reaction, Reaction Kinetics 
A → B, is 90 kJ mol–1 and the heat of reaction is
–210 kJ mol–1. What is the activation energy of 13 Suppose that a plate of rice turns stale in about
the reaction, B → A, in kJ mol–1? 3 hours at a temperature of 28 °C but it takes
A 90 about 42 hours at 18 °C. Calculate the activation
B 120 energy for this process.
C 210 A 192 kJ mol–1
D 300 B 83.5 kJ mol–1
PENERBIT C 21.0 kJ mol–1
ILMU 11 Which of the following is a form of the D 20.5 kJ mol–1 Chapter 1
BAKTI SDN. BHD.Arrhenius equation that shows the relationship
between the rate constant, k and temperature T ? 14 Choose the statement which is false about
catalysts.
1 2A ln k2 = Ea 1 – 1 A Catalysts reduce the activation energy for a
k1 R T1 T2 reaction
B A small amount of catalyst can accelerate a
–ln1 2B k2 = Ea 1 – 1 reaction
k1 R T1 T2 C Catalysts do not appear in the balanced
equation
ln1 2C k2 = Ea 1 + 1 D Catalysts do not alter the reaction pathway
k1 R T1 T2
15 What is/are the factor(s) that reaction rates
1 2D ln k2 = Ea 1 – 1 depend on?
k1 R T2 T1 I Temperature
II Concentration
12 The following are graphs of rate constant, k III Catalyst
against temperature, T. Which graph shows the IV Particle size
Arrhenius equation? A I only
A C B I and II
C I, II and III
k k D I, II, III and IV

B T T

D

k k
T T

17

 Reaction Kinetics

Structured Questions
1 Consider the following chemical reaction.

2NO(g) + Cl2(g) → 2NOCl(g)
The initial rates of reaction at different concentrations of NO and Cl2 are recorded in Table 1.
BAKTI SDN. BHD.
Experiment [NO] (M) [Cl2] (M) Initial reaction rate (M s–1)
1 0.034 0.012 3.4 × 10–4

2 0.017 0.012 8.5 × 10–5

Chapter 1 3 0.017 0.048 3.4 × 10–4

Table 1
Determine the order of reaction with respect to Cl2 and to NO, the overall order of reaction and
the rate law for the reaction.

2 If the rate constant of a first order reaction is 3.47 × 10–3 s–1, what is the half-life, t 1 of the
2
reaction?

3 The half-life, t 1 of a first order reaction is 0.15 s. What is the rate constant?
2

4 Determine the rate constant when the half-life, t 1 of a first order reaction is 0.98 s.
2

5 The radioactive isotope iodine-131 decays to xenon-131 with a rate constant of 0.138 d−1.
Radioactive decay is a first order reaction.
(a) How many days will it take for 85% of a sample of iodine-131 to decay to Xe-131 if the
initial concentration is 0.6 M?
(b) Determine the half-life for the reaction.
ILMU
6 Table 2 shows the data concerns the dimerisation of C4H6. Determine the rate constant of this
reaction.

PENERBIT Experiment Time (s) Concentration of C4H6 (mol L–1)
1 0 0.01011
2 0.00503
3 1 600 0.00337
4 3 200 0.00253
4 800

Table 2

18

Summative Assessment Test (UPS 1)

Time: 1 hour
[17 marks]
Answers all questions.
PENERBIT
ILMU 1 Consider the reaction: I The molecules of H2(g), I2(g) and HI(g) must
BAKTI SDN. BHD. collide with one another
A → Product
II The molecules of H2(g) and I2(g) must
Which of the following is correct, if the reaction collide with one another with the correct
is zero order with respect to A? orientation
A When the concentration of A doubles, the
III The molecules of H2(g) and I2(g) must
reaction rate doubles collide with kinetic energies equal to or
B When the concentration of A doubles, the greater than the activation energy

reaction rate quadruples A I only
C The reaction rate is directly proportional to B I and II
C II and III
the concentration of A D I, II and III
D When the concentration of A doubles, the
4 Which of the following statements is true
reaction rate remains unchanged regarding the effect on reaction rate when the
temperature of a reaction increases?
2 A tertiary haloalkane with the molecular formula A Larger volumes are occupied by gases at
C5H11Br is converted to an alcohol as follows: higher temperatures causing the reaction rate
to increase
C5H11Br + NaOH → C5H11OH + NaBr B The activation energy decreases with
increase in temperature which causes the
The reaction rate triples when the concentration reaction rate to be higher
of C5H11Br is tripled while keeping the C More collisions occur with the correct
concentration of NaOH constant. However, orientation at higher temperatures causing
the reaction rate remains constant when the the reaction rate to increase
concentration of NaOH is doubled while keeping D More reactant molecules collide with kinetic
the concentration of C5H11Br constant. energy equal to or greater than the activation
What is the rate law for the reaction? energy which causes the reaction rate to
A Rate = k[NaOH] increase
B Rate = k[C5H11Br]
C Rate = k[C5H11OH] 5 The Arrhenius equation is k = Ae– Ea
D Rate = k[C5H11Br] [NaOH]
RT.
3 Hydrogen reacts with iodine according to the
following equation: Based on the Arrhenius equation, what is/are

H2(g) + I2(g) → 2HI(g) the factor(s) affecting the rate constant of a

Not all collisions between H2 and I2 molecules reaction?
would result in the formation of hydrogen iodide,
HI(g). I Temperature
What is/are the condition(s) required for a
collision to result in the formation of hydrogen II Catalyst
iodide?
III Gas constant

A I only

B I and II

C II and III

D I, II and III

62

Summative Assessment Test (UPS 1) 

6 Which of the following is/are the standard 9 Which of the following is the equation for the
conditions for a reaction? second electron affinity of Z?
I The concentration of all solutions is 1 M A Z(g) + e– → Z–(g)
II The temperature is 298.15 K B Z(g) → Z+(g) + e–
III The pressure of all gases is 1 atm C Z–(g) + e– → Z2–(g)
A I only D Z+(g) → Z2+(g) + e–
B I and II
C II and III
D I, II and III
10 The lattice energies of MgO and MgF2 are given
below.
PENERBIT
ILMU 7 Which of the following equations represents theIonic compoundMgO MgF2
BAKTI SDN. BHD.
standard enthalpy change of the formation of Lattice energy (kJ mol–1) –3 950 –2 920

CuSO4•5H2O? 9 Table 1
2
A Cu(s) + S(s) + O2(g) + 5H2(g) → The lattice energy of MgO is more exothermic
than MgF2 because
CuSO4•5H2O(s) ∆H°f = –2 278 kJ mol–1 A the charge of the oxide ion is higher than the

B Cu(s) + S(s) + 2O2(g) + 5H2O(l) → fluoride ion. Therefore, stronger ionic bonds
CuSO4•5H2O(s) ∆H°f = –2 278 kJ mol–1 are formed between magnesium ions and
oxide ions.
C Cu2+(aq) + SO42–(aq) + 5H2O(l) → B the charge of the fluoride ion is higher than
the oxide ion. Therefore, stronger ionic
CuSO4•5H2O(s) ∆H°f = –2 278 kJ mol–1 bonds are formed between magnesium ions
Cu2+(aq) + SO42–(aq) + 5H2(g) + 5 and fluoride ions.
D 2 O2(g) → C the oxide ion has a smaller ionic radius than
the fluoride ion. Therefore, stronger ionic
CuSO4•5H2O(s) ∆H°f = –2 278 kJ mol–1 bonds are formed between magnesium ions UPS 1
and oxide ions.
8 An experiment was performed to determine the D the fluoride ion has a smaller ionic radius
specific heat capacity of metal A. than the oxide ion. Therefore, stronger ionic
bonds are formed between magnesium ions
Metal A and fluoride ions.

Figure 1 11 Which of the following is not true about the
standard hydrogen electrode?
The data collected is as shown below: A It is a reference electrode which is arbitrarily
assigned the standard electrode potential
Mass of metal A = V g value of 0.0 V
B It is the half-cell used to measure the ease
Temperature rise = W °C with which another half-cell gains electrons
under standard conditions
Heat absorbed by metal A = X J C It is the half-cell used to measure the
standard electrode potential of other half
Based on the definition of specific heat capacity, cells by measuring the potential difference
between it and the half-cell of interest
determine which expression can be used to D It consists of a platinum electrode immersed
in an aqueous solution containing 1.0 M H+
calculate the specific heat capacity of metal A. ions with a stream of hydrogen gas at 1 atm
bubbling over the surface of the platinum
A X J g–1 °C–1 C XV J g–1 °C–1 electrode at 298.15 K
VW W

B VWX J g–1 °C–1 D XW J g–1 °C–1
V

63

 Summative Assessment Test (UPS 1)

12 Table 2 shows the standard electrode potentials The following can increase the cell potential of
of three half-cells. the galvanic cell except …
A replacing the iron half-cell with a copper
Half-cell E° (V)
Ag+(aq) + e– → Ag(s) +0.80 half-cell.
Ba2+(aq) + 2e– → Ba(s) –2.90 B replacing the magnesium half-cell with a
Zn2+(aq) + 2e– → Zn(s) –0.76
calcium half-cell.
C increasing the concentration of the iron(II)

sulphate solution to 5.0 M.
D increasing the concentration of the

magnesium sulphate solution to 5.0 M.
PENERBIT UPS 1 Table 2
ILMU
BAKTI SDN. BHD.Based on the standard electrode potential values, 15 Which of the following is not true regarding the
hydrogen-oxygen fuel cell?
arrange the metals in order of increasing strength A The fuel cell is very polluting
B Theoretically, the fuel cell can continuously
as reducing agents. supply electricity
A Ag , Ba , Zn C At the anode, hydrogen gas is oxidised to
B Zn , Ba , Ag form hydrogen ions
C Ag , Zn , Ba D At the cathode, oxygen gas is reduced under
D Ba , Zn , Ag acidic conditions to form water molecules

13 Consider the following standard electrode
potentials:

Half-cell E° (V) 16 What are the products formed at the anode
Fe3+(aq)/Fe2+(aq) +0.74 and cathode during the electrolysis of aqueous
Cu2+(aq)/Cu(s) +0.34 sodium sulphate?
Sn4+(aq)/Sn2+(aq) +0.15
Ce3+(aq)/Ce(s) –2.33 Cathode Anode

Table 3 A Sodium metal Sulphur

Based on the standard electrode potentials given, B Hydrogen gas Oxygen gas
determine which of the following reactions is
spontaneous? C Sodium metal Oxygen gas
A Cu(s) + Sn4+(aq) → Cu2+(aq) + Sn2+(aq)
B Ce3+(aq) + 3Fe2+(aq) → Ce(s) + 3Fe3+(aq) D Hydrogen gas Sulphur
C 3Cu2+(aq) + 2Ce(s) → 2Ce3+(aq) + 3Cu(s)
D Cu2+(aq) + 2Fe2+(aq) → Cu(s) + 2Fe3+(aq)
17 Figure 2 shows a simple apparatus set-up for

carrying out the electrolysis of water.

Hydrogen gas

Oxygen gas

14 A galvanic cell is prepared using an iron nail Cathode
immersed in 1.0 M iron(II) sulphate solution
and a magnesium strip immersed in 1.0 M Anode
magnesium sulphate solution.
Water
Half-cell E° (V) Battery Figure 2
Mg2+(aq)/Mg(s) –2.37
Fe2+(aq)/Fe(s) –0.44 Based on Faraday’s first law of electrolysis, the
Cu2+(aq)/ Cu(s) +0.34 products at the anode and cathode increase when …
Ca2+(aq)/Ca(s) –2.87 A the volume of water is increased.
B the time of electrolysis is increased.
Table 4 C the concentration of water is increased.
D water is replaced with sodium sulphate.

64

6CHAPTERPENERBIT
ILMU
Benzene and Its BAKTI SDN. BHD.
Derivatives

6.1 Introduction

Learning Outcomes

• Describe:
(a) aromaticity
(b) Kekulé structure
(c) resonance structure of benzene.

Benzene Smart Tips!

1 Benzene, C6H6 is a colourless liquid at room temperature. Its boiling point is Addition or reduction
80 °C and its melting point is 6 °C. reaction will disrupt the
ring structure of benzene
2 Benzene is a hydrocarbon because it is composed of carbon and hydrogen and thus cause the loss
only. of aromaticity, whereas
substitution reactions
3 Benzene is an aromatic hydrocarbon or arene. Aromatic compounds are cyclic allow benzene to maintain
compounds where delocalisation of electrons enhances the stability of the its aromaticity.
molecule.

4 Benzene’s molecular formula, C6H6 suggests a high degree of unsaturation
due to a low hydrogen index compared to the corresponding alkane (C6H14).
Generally, unsaturated hydrocarbons are more reactive than saturated
hydrocarbons. However, benzene is particularly stable and does not undergo
addition reactions.

5 Instead benzene usually undergoes substitution reactions where one of its
hydrogen atoms is replaced by another atom or group.

109

 Benzene and Its Derivatives

Kekulè Structure

1 The structure of benzene was first introduced by August Kekulè in 1866. This
structure consisted of six carbon atoms which form a ring and each carbon
atom is also attached to one hydrogen atom.

2 Kekulè proposed that the ring contains three double bonds alternating with
BAKTI SDN. BHD.
single bonds. H

HH

Chapter 6 HH
H
ILMU
Figure 6.1 Kekulè structure of benzene
PENERBIT
3 However, there were problems with the Kekulè structure of benzene. Firstly, if
benzene contains double bonds, it would be expected to undergo the reactions
of an alkene but it does not. For example, it does not undergo the addition
reactions like an alkene.

4 Secondly, if the benzene ring consists of double bonds and single bonds,
the ring would be irregular. A single bond is longer than a double bond. The
Kekulè structure implies a ring with alternating shorter and longer sides
whereas benzene is a perfectly regular hexagon in which all the C–C bonds
are of the same length. The C–C bond length of benzene is intermediate
between that of a single bond and a double bond.

5 Thirdly, benzene is more stable than predicted by the Kekulè structure. The
enthalpy for the hydrogenation of cyclohexene (which has one double bond)
to form cyclohexane is –120 kJ mol–1. If benzene had three double bonds, an
enthalpy change of –360 kJ mol–1 would be expected for benzene. However,
the enthalpy change is only about –208 kJ mol–1, much less than predicted and
even less than the enthalpy change of hydrogenation for 1,3-cyclohexadiene
which has only two double bonds.

Resonance Structure of Benzene

1 The structure of benzene is said to be a resonance hybrid of the two Kekulè
structures shown in Figure 6.2. Notice that the positions of the single and
double bonds are different in the two structures. The actual structure of
benzene is neither structure A nor B, nor does benzene rapidly shift back and
forth between the two structures. The actual structure is a hybrid of the two.

110 AB
Figure 6.2 Resonance hybrid of two Kekulè structures

2 All the carbon atoms in the benzene ring are sp2-hybridised and all the C–C
bonds are equal, with a bond length between that of a single bond and a
double bond. Each carbon atom has an unpaired electron in a p orbital. The p
orbitals overlap with each other and the unpaired electrons can move around
the entire ring, that is, the electrons are delocalised (Figure 6.3). So, the
benzene ring is often drawn as shown in Figure 6.4.

Benzene and Its Derivatives 

HH

H—C C————C C—H Delocalised pi electron
C————C cloud above and below the
plane of the carbon ring

PENERBIT HH
ILMUFigure 6.3 Delocalisation of pi electrons in benzene
BAKTI SDN. BHD.
Figure 6.4 Representation of the benzene ring Chapter 6

3 In order to be classified as aromatic compound, the molecules must have the
following criteria:
(a) The molecule must be cyclic.
Example:



Cyclic Non-cyclic

(b) The molecule must have planar structure.
Example:

H
H

Planar Non-planar

(c) All atoms in the cycle are sp2-hybridised.
Example:

:– H
H

All sp2-hybridised atoms Has sp3-hybridised atom

(d) The molecule has 4n+2 electrons (Hückelʼs Rule). Where n = 0, 1, 2, 3, …
Example:



6 electrons 12 electrons

Quick Check 6.1

1 What makes benzene an aromatic hydrocarbon?
2 Draw the Kekulè structure of benzene.
3 Draw the structure of benzene as a resonance hybrid.

111

 Benzene and Its Derivatives

6.2 Nomenclature of Benzene Derivatives

Learning Outcomes

• Give the names of benzene derivatives according to IUPAC nomenclature for:
(a) monosubstituted benzenes
(b) disubstituted benzenes
(c) tri- and tetrasubstituted benzenes

• Give the structures of benzene derivatives.
• Apply IUPAC rules to name compounds with C6H5– (phenyl) or C6H5CH2–

(benzyl) as substituents.
BAKTI SDN. BHD.
Chapter 6 Smart Tips! Monosubstituted Benzenes

Examples of 1 Monosubstituted benzenes have one substituent attached to the benzene ring.
monosubstituted benzene: The substituent can be a functional group. The names of monosubstituted
benzenes may be used as the base name or parent name of more highly
F substituted benzene derivatives.

Fluorobenzene 2 Table 6.1 gives the list of important benzene derivatives arranged in decreasing
Cl order of priority. This means that if there are two functional groups, the
functional group higher up in the list will determine the parent name.
Chlorobenzene
Br Table 6.1 Benzene derivatives

No Parent name Structure Substituent Formula Priority
1 Benzoic acid O
Carboxyl C7H6O2 Highest
HO (–COOH)

ILMU O Carbonyl C₇H₆O
Bromobenzene 2 Benzaldehyde
H group, C=O
Smart Tips! 3 Phenol bonded to H

PENERBIT To name benzene 4 Aniline OH Hydroxyl C6H5OH
derivatives with more (–OH)
than one substituent, 5 Methylbenzene
it would be helpful to (Toluene) NH2 Amino C6H7N
memorise the parent (–NH2) C7H8
names, structures and
priority order in Table 6.1. CH3 Methyl
(–CH3)

6 Nitrobenzene NO2 Nitro (–NO2) C6H5NO2
Lowest

112

Benzene and Its Derivatives 

Disubstituted Benzenes Smart Tips!

1 Disubstituted benzenes have two substituents attached to the benzene ring. The older style prefixes
2 Two chlorine atoms are attached to the benzene ring in the following ortho-, meta- and para
may also be used instead
compound. The name of the compound is 1,4- dichlorobenzene, of numbers to show the
para-dichlorobenzene or p-dichlorobenzene. relative positions of two
substituents.
Cl • Ortho: 1,2
• Meta: 1,3
C1 • Para: 1,4

C4

Cl
3 Naming disubstituted benzenes begins with identifying the parent name. The

parent name is determined based on the functional group priority given in
Table 6.1.
4 Carbon atom containing the functional group of the parent name is designated
as C1 if the parent name has a special name.
5 The parent name of the following compound is phenol. This is because,
the –OH functional group has higher priority than the –CH3 group. The
carbon bearing the –OH group is numbered as C1. So the compound is called
2-methylphenol or also known as o-methylphenol.

OH
C1 CH3

C2
PENERBIT Chapter 6
ILMU
BAKTI SDN. BHD.
6 The parent name of the following compound is nitrobenzene. This compound
is 2-bromonitrobenzene, also called o-bromonitrobenzene.

Br
2 NO2

1

Tri- and Tetrasubstituted Benzenes

1 Tri- and tetrasubstituted benzenes have three and four substituents attached
to the benzene ring, respectively.

The carbon atoms in the ring must be numbered in a way that the locations
of the substituents are the lowest possible.

CH2CH4 CH2CH3
6 1 2 CH3 1
2 6 CH3

5 3 3 5
4 4

 

Substituents at C1, C2 and C5 Substituents at C1, C3 and C6

113

 Benzene and Its Derivatives

CH2CH3 CH2CH3
6 3 2 1 CH3
5 1 CH3

4 2 4 6
3 5

BAKTI SDN. BHD. 

Substituents at C1, C4 and C6 Substituents at C1, C2 and C4

Thus, the name of the compound is 2-methyl-4-isopropyl-1-methylbenzene

2 Like the disubstituted benzenes, the parent name is based on functional group

Chapter 6 priority.

Example:

COOH OH NO2

NH2 NO2 CH3

O2N NO2 O2N NO2

3,5-dinitrobenzoic acid 2-amino-5-nitrophenol 2,4,6-trinitrotoluene

Example 1

Name the following compounds according to IUPAC rules.

(a) NO2 (b) CH3 (c) NH2

Cl I Cl
ILMU
CH3 Cl

PENERBIT Solution

(a) 2-chloro-3-nitro-methylbenzene
(b) 1-chloro-3-iodo-2-methylbenzene
(c) 3-chloroaniline or m-chloroaniline

Phenyl Group and Benzyl Group

1 The terms phenyl and benzyl are used for the C6H5– and C6H5CH2– groups
respectively.

Phenyl CH2
Benzyl

114

Benzene and Its Derivatives 

2 For example, chlorobenzene is also known as phenyl chloride and
(bromomethyl)benzene is also called benzyl bromide.

Cl Br

PENERBITPhenyl chloride Benzyl bromide
ILMU(chlorobenzene)((bromomethyl)benzene)
BAKTI SDN. BHD.
3 When the benzene ring is attached to a carbon chain with six or more carbon Chapter 6
atoms or if the chain contains the principal functional group, the benzene
ring is treated as the substituent and phenyl is used as part of the name of the
compound.

CH2CH2CH2OH C | H3 CH = CH2
C—OH
|
CH3

3-phenyl-1-propanol 2-phenyl-2-propanol Phenylethene

Quick Check 6.2

1 Which of these two benzene derivatives has a higher functional group
priority, aniline or toluene?

2 Name the following benzene derivatives according to IUPAC nomenclature.

(a) Cl (c) CH3
HO CH2CH3

NH2

(b) Br Br (d) CH2CH2OH

115

 Benzene and Its Derivatives

6.3 Chemical Properties of Benzene

Learning Outcomes

• Explain the electrophilic aromatic substitution reaction of benzene:
(a) Nitration
(b) Halogenation
(c) Friedel-Crafts alkylation
(d) Friedel-Crafts acylation

• Illustrate the mechanism for electrophilic aromatic substitution of benzene:
(a) Nitration
(b) Halogenation
(c) Friedel-Crafts alkylation

• Explain the influence of ortho-para and meta directing substituents towards
electrophilic aromatic substitution reaction.

• Predict the product of electrophilic aromatic substitution of monosubstituted
benzene.

• Explain the following reactions of alkylbenzene:
(a) Oxidation with hot acidified KMnO4 or K2Cr2O7
(b) Halogenation (free radical substitution)
Chapter 6 BAKTI SDN. BHD.

Electrophilic Aromatic Substitution Reactions of Benzene

1 The delocalised electrons of the benzene ring make benzene highly attractive
to electrophiles – species that seek electrons.

2 Benzene is resistant to addition reactions but undergoes substitution reactions
which allow it to maintain its delocalised electron system by only replacing a
hydrogen atom with another atom or group of atoms.

3 The general mechanism of the electrophilic substitution reactions of benzene
consists of three steps.

Step 1: Formation of an electrophile
ILMU

PENERBIT Step 2: Electrophilic addition

• The electrophile (E+) is attracted to the delocalised pi electron system of
the benzene ring and forms a bond with a carbon atom by accepting two pi
electrons.

• Note that the hydrogen atom shown attached to the benzene ring is not from
some other compound but was already present and is purposely drawn to
explain the next stage of the reaction.

• An arenium ion is formed.

H

+ E+ E

+

Arenium ion

116

Benzene and Its Derivatives 

Step 3: Deprotonation of the arenium ion

• Once the arenium ion is formed, it quickly reacts with the base by losing a
proton to restore the stable delocalised electron system of the benzene ring.

H E

E + :B– + H:B

Base
PENERBIT
ILMUNitration of benzene Chapter 6
BAKTI SDN. BHD.
1 Nitration happens when one (or more) of the hydrogen atoms of the benzene
ring is replaced by a nitro group, –NO2.

2 Nitration of benzene is carried out by reacting benzene with a mixture of nitric
acid, HNO3 and sulphuric acid, H2SO4.

Step 1: The electrophile, NO2+ (nitronium ion) is formed by the reaction
between HNO3 and H2SO4.
Formation of electrophile

HO—NO2 + H—OSO3H

H+O| —NO2 + – OSO3H
H

H2O + +NO2 + – OSO3H
Electrophile

Step 2: The NO2+ ion acts as the electrophile which attacks the delocalised pi
electrons of the benzene ring, forming an arenium ion.
Formation of arenium ion

H +H H H
+ +NO2 NO2 + NO2 + NO2

Arenium ion

Step 3: Then, the HSO4– ion removes a hydrogen from the benzene ring to
form sulphuric acid again (regeneration of catalyst).
Deprotonation of arenium ion

+H NO2 + H—OSO3H
NO2 + – OSO3H

Nitrobenzene

117

Summative Practice 6

PENERBITObjective Questions 7 What is the IUPAC name of the compound below?
ILMU NH2
BAKTI SDN. BHD. 1 What is the molecular formula of benzene?
Cl l
A C6H5 C C6H7
B C6H6 D C6H8

2 Benzene is an … C arene Br Chapter 6
A alkene D alkane Figure 1
B alkyl A 4-bromo-2-chloro-6-iodoaniline
B 4-bromo-6-chloro-2-iodoaniline
3 Which statement about benzene is true? C 2-chloro-4-bromo-6-iodoaniline
A Substitution reactions would cause the loss D 2-iodo-4-bromo-6-chloroaniline
of its aromaticity
B Benzene is a reactive hydrocarbon 8 What is the IUPAC name of the compound below?
C Benzene has three double bonds O–
D None of the above
|

l N+
O

4 Which of the following is one of the problems Br
with the Kekulè structure of benzene? Figure 2
A Alternating double and single bonds would
make the hexagon structure irregular A 3-bromo-5-iodonitrobenzene
whereas benzene is a regular hexagon B 5-bromo-3-iodonitrobenzene
B The carbons are arranged in a hexagon with C 1-bromo-3-iodo-5-nitrobenzene
each carbon forming a bond to only one D 1-bromo-5-iodo-3-nitrobenzene
hydrogen
C Benzene does not have a hexagonal structure 9 What is the electrophile in the nitration of
D None of the above
benzene?

5 When toluene is heated under reflux with A HNO3 C NO2
acidified KMnO4, a white precipitate is formed. B NO3– D NO2+
What is the reaction which has occurred?
A Reduction 10 What is the function of deprotonation in
B Addition electrophilic addition reactions of benzene?
C Oxidation A To make the benzene ring stable again
D Halogenation B To accept the electrophile
C To attract the base
D None of the above

6 What is the main product when benzene 11 What is a suitable reagent for this reaction?
is reacted with bromine in the presence of O
aluminium tribromide? OH
A o-cresol
B Bromobenzene A KMnO4,H+ C HNO3
C Tetrabromoaluminium B SOCl2 D AlCl3
D meta-bromobenzyl bromide

123

 Benzene and Its Derivatives 14 What makes benzene an aromatic compound?
12 What is a suitable catalyst for this reaction? A It contains double bonds
B It is an unsaturated hydrocarbon
CH3 C It has electrons that are delocalised
throughout the benzene ring
+ CH3—Br + HBr D All of the above

A Potassium permanganate BAKTI SDN. BHD. 15 Which of the following are the major products of
B Aluminium tribromide
C Thionyl chloride the nitration of methylbenzene?
D Bromine
I CH3 II CH3 III CH3
NO2
13 How many electrons form the delocalised
Chapter 6
electron system in benzene? NO2

A 1 C 4 NO2 C II and III
A I only D I, II and III
B 3 D 6 B I and II

Structured Questions

1 (a) Why does benzene cannot undergo addition reactions?

(b) Give the IUPAC name of the following compounds.

(i) Br (ii) CH3 NO2 (iii) CH2CH2OH
NO2

Br

ILMU
NO2
PENERBIT
(c) Draw the structure of the following compounds.
(i) Phenylethene
(ii) 1-phenyl-1,2-ethanediol
(iii) Benzyl chloride

2 Besides using an alkyl halide in the presence of a Lewis acid as catalyst, benzene can also
be alkylated with an alkene in the presence of a suitable catalyst. Isopropylbenzene can be
produced from benzene and propene in the presence of phosphoric acid.
Based on steps (a), (b) and (c) below, illustrate the mechanism of the reaction.
(a) A carbocation is formed from the reaction between propene and phosphoric acid.
(b) The carbocation acts as the electrophile which attacks the benzene ring.
(c) H2PO4– removes a hydrogen from the benzene ring.

3 (a) Identify the major products when phenol undergoes nitration.
(b) How is the resonance-stabilised arenium ion intermediate formed during the nitration of
benzene?
(c) Explain why nitration tends to occur in the ortho and para positions with phenol.

4 (a) Dimethylbenzene is also known as xylene. Name the products formed when m-xylene and
p-xylene are oxidised with H2CrO4.

(b) Name the organic product from the reaction of ethylbenzene with KMnO4.
(c) Predict the major products when phenol reacts with bromine in the presence of aluminium

tribromide. Give a reason for your answer.

124

PENERBIT 11CHAPTER
ILMU
BAKTI SDN. BHD.Amines

11.1 Introduction

Learning Outcomes

• Give the names of aliphatic and aromatic amines according to IUPAC
nomenclature (parent chain < C10).

• State the common names of amines with parent chain < C5.
• Give the structural formulae of amines.
• Classify primary, secondary, and tertiary amines.

1 Amines are organic derivatives of ammonia, NH3 with one, two, or all three
hydrogen atoms replaced by alkyl or aryl groups. Amines that do not have
a benzene ring bonded to the nitrogen atom are called aliphatic amines.
Amines that have at least one benzene ring bonded directly to the nitrogen
atom are called aromatic amines.

2 Amines are classified as primary (1°), secondary (2°), or tertiary (3°)
depending on the number of alkyl or aryl groups bonded to the nitrogen atom.

Table 11.1 Classes of amines

Class of amine Primary Secondary Tertiary
1 2 3
Number of alkyl or
aryl groups bonded to R—N—H R—N—Rʹ R—N—Rʹ
nitrogen, N atom | | |
H H Rʺ
Structure (R may be
alkyl or aryl group)

208

Nomenclature of Amines Amines 

1 (a) In IUPAC nomenclature, aliphatic primary amines are named as 209
alkanamines, that is, the suffix -e of the corresponding alkane is replaced
by the suffix -amine. For example, methane becomes methanamine and
ethane becomes ethanamine.

(b) In the case of common names, primary amines are named as alkylamines.

Table 11.2 Examples of aliphatic primary amines
PENERBIT
ILMUStructureCH3—NH2CH3CH2—NH2H2N—
BAKTI SDN. BHD.
IUPAC name Methanamine Ethanamine Cyclopentanamine Chapter 11
Common name Methylamine Ethylamine Cyclopentylamine

2 The IUPAC rules for naming primary amines is the same as for other organic
compounds.
(a) The longest continuous carbon chain containing the amino group
(–NH2) is selected as the parent chain.
(b) The carbon atoms of the parent chain are numbered beginning from the
end nearest to the –NH2 group.
(c) The substituents are listed in alphabetical order.

12

C 7 H36CH25CH24CCHH233CC| HHCNHH22CH2CH3
3-propyl-2-heptanamine

3 Note that if a substituent with higher priority than the amino group is present
(such the hydroxyl group), the compound is no longer named as an amine.
HOCH2CH2CH2NH2 is 3-amino-1-propanol, not 3-hydroxy-1-propanamine.

4 For cyclic amines, the carbon bearing the –NH2 group is designated as C1 and
the other carbon atoms in the ring are numbered in the direction that would
give the other substituents the lowest possible numbers.

NH2

CH3
3-methylcyclohexanamine

5 Aliphatic secondary and tertiary amines have more than one alkyl group
bonded to the nitrogen atom.
(a) In IUPAC nomenclature, the longest alkyl chain determines the parent
name, while the other alkyl group or groups are referred to as substituents,
with an italicised letter N placed before each substituent.
(b) For the common names, the alkyl groups are listed in alphabetical order. If
there are identical groups, the prefixes di- and tri- are used.

 Amines Table 11.3 Examples of aliphatic secondary amines

210 Structure CH3—NH—CH2CH3 CH3—NH—CH3
IUPAC name N-methylethanamine N-methylmethanamine
Common name Ethylmethylamine Dimethylamine

Table 11.4 Examples of aliphatic tertiary aminesBAKTI SDN. BHD.

Structure C H3CH2CH2—N| —CH3 CH3CH2—N| —CH3
CH2CH3 CH3

Chapter 11 IUPAC name N-ethyl-N-methylpropanamine N,N-dimethylethanamine

Common name Ethylmethylpropylamine Ethyldimethylamine

6 (a) The simplest aromatic amine is phenylamine which is a primary amine.

—NH2

IUPAC name: Phenylamine
Common name: Aniline

The common name aniline has been retained by IUPAC and may also be
used in IUPAC nomenclature.

(b) Hence ‘aniline’ can be used as the base name for secondary and tertiary
aromatic amines in IUPAC nomenclature.
ILMU
Table 11.5 Examples of secondary and tertiary aromatic amines

Class of Structure and IUPAC name
amine

PENERBIT Secondary —NHCH3 —NHCH3
amine N-methylaniline
CH3
N-methyl-3-methylaniline

Tertiary C| H3 C| H2CH3
amine —N—CH2CH2CH3 —N—CH2CH3

N-methyl-N-propylaniline CH3
N,N-diethyl-3-methylaniline

Quick Check 11.1 Amines 

1 Identify the classes of the following amines and name them according to 211

IUPAC nomenclature.

(a) (d)
H2N— —CH2CH3 C H3CH2N| —
PENERBIT
ILMU CH3 CH2CH3
BAKTI SDN. BHD.
(b) C| H3 (e) (CH3)3N
CH3CH2NHCH2CH2CHCH3
Chapter 11
(c) (f)
CH3NH— —CH3 NH2

2 Draw the structural formulae of the following amines and amino compound.
(a) Ethylisopropylmethylamine
(b) tert-butylamine
(c) N-methyl-2-butanamine
(d) 3-aminobenzoic acid

11.2 Physical Properties of Amines

Learning Outcomes

• Compare the physical properties of amines:
(a) Boiling points of
• primary, secondary and tertiary amines
• amines and alkanes, haloalkanes, alcohols, carbonyl compounds and
carboxylic acids of comparable molar mass
(b) Solubility of primary, secondary and tertiary amines

Boiling Points

1 The boiling points of primary, secondary and tertiary amine isomers are
different even though they have the same molecular mass. The boiling points
of 1-butanamine (primary amine), N-ethylethanamine (secondary amine) and
N,N-dimethylethanamine (tertiary amine) are shown in Table 11.6.

11.3 Preparation of Amines Amines 

Learning Outcomes 215

• Explain the preparation of:
(a) aromatic amines by reduction of nitro compounds using Zn/HCl, Sn/HCl
or Fe/HCl
(b) primary aliphatic amines by reduction of nitriles using LiAlH4 followed by
H2O or H2 with catalyst
(c) primary, secondary, and tertiary amines by reduction of amides using
LiAlH4 followed by H2O
(d) primary alkyl and aryl amines by Hofmann degradation of primary
amides
PENERBIT Chapter 11
ILMU 1 Aromatic amines can be prepared by the reduction of nitro compounds using
BAKTI SDN. BHD.
the following reducing agents:

(a) Zn/HCl (b) Fe/HCl (c) Sn/HCl

Examples: —NH2
—NO2 Zn/H+

NO2 NH2
—NH2
—NO2 Sn/H+
NH2
NO2

HO HO

—COOH Fe/H+ —COOH

NO2 NH2

2 Primary aliphatic amines can be prepared from the reduction of nitrile
compounds using the following reducing reagents:
(a) Hydrogen gas in the presence of a catalyst (such as Ni or Pt or Pd)
(b) LiAlH4 followed by hydrolysis (H3O+)

R—CN 1. LiAlH4 RCH2NH2
2. H3O+

 Amines Examples:

216 CH3CH2CH2CN 1. LiAlH4 CH3CH2CH2CH2NH2
2. H3O+

Cl Cl
C H3C| HCH2 C| HCH3 H2, Pt CH3C| HCH2C| HCH3
CN CH2NH2BAKTI SDN. BHD.

3 Reduction of amides using LiAlH4, followed by hydrolysis (H3O+) can be
used to form all the three classes of amine depending on the amide used.

Chapter 11 (a) O H 1. LiAlH4 H H
|| | 2. H3O+ | |
R—C—N—H R—C—N—H
|
H

1° amide 1° amine

(b) O H H H
|| | 1. LiAlH4 | |
R—C—N—Rʹ 2. H3O+ R—C—N—Rʹ

|
H

2° amide 2° amine

(c) O Rʺ H Rʺ
|| | 1. LiAlH4 | |
ILMU R—C—N—Rʹ 2. H3O+ R—C—N—Rʹ

|
H

3° amide 3° amine

Examples:

PENERBIT O 1. LiAlH4 CH3CH2CH2—N| —H
|| 2. H3O+ H
C H3CH 2C—N| —H
H

O 1. LiAlH4 CH3CH2CH2—N| —CH3
|| 2. H3O+ H
C H3CH 2C—N| —CH3
H

O
C H3CH 2C|| —CN| H—3 CH3 12.. HLi3AOl+H4 CH3CH2CH2—NC| —H3CH3

4 Hofmann degradation of primary amides can be used to form primary Amines 

amines with one carbon atom less than the amide. The primary amide is 217

reacted with halogen (such as Cl2 or Br2) in alkaline condition and loses the
carbonyl group. The general equation is:

O X2/OH–
||
R—C—NH2
PENERBIT RNH2
ILMU
BAKTI SDN. BHD.Primary amidePrimary amine

Examples: Cl2/OH– CH3NH2
O
|| Chapter 11
CH3—C—NH2

O Br2/OH– CH3 CH2 NH2
||
CH3 CH2—C—NH2

O Br2/OH– —NH2
||

—C—NH2

Quick Check 11.3

1 Suggest suitable reagents for the preparation of the following amines and

state the method of amine preparation.

(a) Aniline from nitrobenzene

(b) O
||
C H3CH2 —C—N| —CH2CH3 CH3CH2CH2N| CH2CH3
CH3 CH3

(c) O CH3CH2C| HNH2
|| CH3

C H3CH2C| HC —NH2
CH3

2 Complete the following reaction schemes with the structural formulae of

compounds P and Q.

(a) O
P Cl2/OH– CH3C| H—C|| —NH2 1. LiAlH4
2. H3O+ Q

CH3

(b) Q
Zn/H+

P Cl2/OH– —NH2

Summative Practice 11

Objective Questions BAKTI SDN. BHD. 6 Which of the following has a lower boiling point
than N-methyl-2-propanamine?
1 Name the following compound according to A Butanol
IUPAC nomenclature. B Butanoic acid
C 2-butanamine
—N| —CH2CH3 D N,N-dimethylethanamine
CH2CH2CH3
Chapter 11 7 Which of the following is used to prepare aniline
A N-phenyl-N-ethylpropanamine from nitrobenzene?
B N-phenyl-N-propylethanamine A Sn, H+
C N-ethyl-N-propylbenzene B Cl2, OH–
D N-ethyl-N-propylaniline C NaBH4, CH3OH
D NaNO2, HCl at 0 to 5 °C

2 What is the gas released when primary aliphatic 8 Which of the following reagents will produce a

amines react with nitrous acid? green precipitate when reacted with C6H5N(CH3)2?

A Oxygen C Ammonia A Hinsberg reagent C Bromine water

B Hydrogen D Nitrogen B Nitrous acid D Phenol

3 Which of the following is used to prepare aILMU 9 Choose the correct observations for the reaction
benzenediazonium salt during dye preparation? of nitrous acid with 1°, 2° and 3° aliphatic amines.
A Fe, H+
B Phenol 1° amine 2° amine 3° amine
C NaNO2, HCl at 0 to 5 °C
D conc. HNO3, conc. H2SO4 at 50 to 55 °C A Gas bubbles Yellow oil Clear
solution
4 Which of the following shows the presence of
aniline? B Gas bubbles Clear White
A Formation of a green precipitate with nitrous solution precipitate
acid
B Formation of a white precipitate with C Clear Yellow oil Green
bromine water solution precipitate
C Formation of a white precipitate with
Hinsberg reagent
D Formation of an orange precipitate with
benzenediazonium chloride
PENERBIT D White White Clear

precipitate precipitate solution

10 Which of the following observations will be
obtained with 2° aliphatic amines in the Hinsberg
test?

Benzenesulphonyl Acidification
chloride in KOH
5 Which of the following compounds has the
highest solubility in water? A Clear solution White precipitate
A CH3CH2NHCH3
B CH3CH2NH2 B White precipitate Clear solution
C CH3NHCH3
D CH3N(CH3)2 C White precipitate White precipitate
remains
228
D No observable change Clear solution

Amines 

11 What is the product of the following reaction? 14 An amine is prepared from a nitrile as follows:

—N| —CH2CH3 NaNO2, HCl C| H3 C| H3
0 – 5 °C H3C—C| —CN H3C—C| —CH2NH2
CH2CH2CH3

CH3 CH3
PENERBITA O =N— —N| —CH2CH3
ILMU CH2CH2CH3 Which of the following can be used in this
BAKTI SDN. BHD.
B N=O reaction?

I NaBH4 in methanol
II H2, Pt
III Cl2, OH–
O =N— —N| —CH2CH3 IV LiAlH4 followed by hydrolysis
O=N CH2CH2CH3 A I and II C II and III Chapter 11

B III and IV D II and IV

C +H Cl– 15 The 3° aliphatic amine shown below is reacted
—N| —CH2CH3 with nitrous acid.
CH2CH2CH3
C| H3 NaNO2, HCl
D C l– N=N+— —N| —CH2CH3 CH3CH2—N| 0 – 5 °C
CH2CH2CH3 CH3

12 Which of the following is arranged in increasing Which of the following are the products from the
order of boiling point?
I N,N-dimethylethanamine , reaction?
N-ethylethanamine , 1-butanamine
II 2-butanamine , butanol , butanoic acid I C| H3
III 2-chlorobutane , butanone , 2-butanamine C H3CH2—N| C+—l– N=OH
A I only CH3
B I and II
C I and III II C| H3
D I, II and III C H3CH2—N| C+—l– H
CH3

13 Which of the following is arranged in decreasing III C| H3
C H3CH2—N| C+—l– N=O
order of solubility in water? CH3

I 1-butanamine . N-ethylethanamine .

N,N-dimethylethanamine IV C| H3
C H3CH2—N| —N=O
II Ethanamine . 2-propanamine . CH3

3-butanamine . 2-pentanamine

III N-methylethanamine .

N,N-dimethylmethanamine . 1-propanamine

A III only C II and III A I and II C II and III
B I and IV D III and IV
B I and II D I, II and III

229

 Amines BAKTI SDN. BHD.

Structured Questions
1 The following equation shows a reaction.

—NH2 Br2, H2O

Chapter 11 (a) Draw the structure of the product formed.
(b) State the observation for the reaction.
(c) Identify the purpose of this reaction.

2 Compound R has the molecular formula C3H9N.
(a) Draw three isomers of compound R which belong to different classes of amines and name
them according to IUPAC nomenclature.
(b) Write the equations and state the observations for the test to distinguish the isomers in
2(a).
(c) Compare the boiling points and solubility in water of the isomers in 2(a).

3 The following is a reaction scheme.

O
P Cl2, OH– CH3CH2CH2C|| NH2 X CH3CH2CH2CH2NH2

Y

Q H3O+ CH3CH2CH2CN
Δ

ILMU (a) Identify reagents X and Y and draw the structures of compounds P and Q. Name the
process for each reaction.

(b) Suggest a reagent to identify the class of compound P and write the equation for the
reaction.

(c) Compare the boiling points of compounds P and Q.

4 (a) Draw the structures of ethanoic acid, 2-propanol and 1-propanamine.
(b) Arrange ethanoic acid, 2-propanol, and 1-propanamine in order of increasing boiling
point. Explain your answer.

5 Write the equations for the following.
(a) Synthesis of aniline from benzene
(b) Formation of diazonium salt from nitrobenzene
(c) Formation of a dye from benzenediazonium chloride
PENERBIT

230

Semester Examination

Time: 2 hours
[80 marks]

Answers all questions.

1 The rate constant of the decomposition reaction of dinitrogen pentoxide is 4.82 × 10–4 s–1
at 45 °C.
PENERBIT
ILMU2N2O5(g) 4NO2(g) + O2(g)
BAKTI SDN. BHD.
(a) Sketch the graph of ln [N2O5] against time.
(b) If the initial concentration of N2O5 is 1.63 × 10–1 mol dm–3, calculate
(i) the concentration of N2O5 after 20 minutes.
(ii) the half-life of the reaction in minutes.
(c) When the temperature of the reaction increases, predict what will happen to the reaction

rate.
(d) Calculate the activation energy of the reaction if the rate constant of the reaction increases

to 5.35 × 10–3 s–1 at 65 °C.
[10 marks]

2 (a) In an experiment, a student neutralised 12.50 mL of 2.0 M sulphuric acid solution with
25.10 mL of 2.0 M potassium hydroxide solution. The initial average temperature of
the solutions was 29.5 °C. During the experiment, the highest temperature recorded was
38.5 °C. If the heat capacity of the calorimeter is 152.3 J °C–1, determine

(i) the amount of heat released.
(ii) the heat of neutralisation of the reaction.

[5 marks]
(b) Figure 1 shows an energy cycle diagram which includes the enthalpy change of formation

of potassium sulphide.

M + S(s) ∆H1 = –408.8 kJ mol–1 K2S(s)
∆H4 = +279.0 kJ mol–1
∆H2 = +178.0 kJ mol–1

2K(g) S(g)

∆H5 = –200.0 kJ mol–1

∆H3 S–(g) ∆H7 = –1 959.4 kJ mol–1

∆H6 = +456.0 kJ mol–1

2K+(g) + S2–(g)

Figure 1
(i) Name the enthalpy change represented by ∆H6.
(ii) Identify species M.

243

 Semester Examination

(iii) Calculate the first ionisation energy of potassium.
(iv) Determine whether the lattice energy of Na2S will have a larger or smaller magnitude

than the lattice energy of K2S. Explain your answer.
[5 marks]

3 (a) The standard reduction potentials, E° of three half-cells are given below.BAKTI SDN. BHD.

MnO4–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(l) E° = +1.51 V
Cr2O72–(aq) + 14H+(aq) + 6e– → 2Cr3+(aq) + 7H2O(l) E° = +1.33 V
Cl2(g) + 2e– → 2Cl–(aq) E° = +1.36 V

(i) Arrange the species in order of increasing strength of oxidising agents. Explain your
answer.

(ii) Explain whether potassium permanganate or potassium dichromate solution could
oxidise aqueous chloride solution to chlorine gas.
[4 marks]

(b) Gold plating is often used in the electronics industry to protect the copper conductors from
corrosion. Calculate the mass of gold deposited, if a current of 2.0 A is passed through
Au3+ ions for 2 hours.
[3 marks]

Semester Examination 4 (a) Hydrocarbons A and B have the same molecular formula of C4H8 and are functional group
isomers. Both A and B can react with bromine under different conditions to give different
products.

(i) The following is the structure of compound C.

Br
ILMU
Compound C

If compound C is one of the bromination products of A, draw the most stable intermediate
formed during bromination of A.

(i) If the oxidation of B with hot concentrated acidified KMnO4 produces carbon dioxide
and carboxylic acid, draw the structure of the products formed when B reacts with
HBr and HBr in hydrogen peroxide respectively.

(ii) State the conditions for bromination of A and B respectively.
(iii) Suggest a chemical test to differentiate A and B.

(b) Give the monomer for polymer X.
PENERBIT
|
———CH—CH2——
244
n

Polymer X

[6 marks]

Semester Examination 

5 Figure 1 shows a reaction scheme of the conversion of benzene to various other compounds.

C H3C| HCH2CH3 COOH
|

F conc. HNO3, D E
Br2,
FeBr3 conc. H2SO4
50 – 55 °C
PENERBIT
ILMU Br2,
BAKTI SDN. BHD.uv light

GH
(major product)

Figure 1

(a) Identify the reagents and reaction conditions represented by D and E.
(b) Draw the structural formulae of F, G and H.
(c) Figure 2 shows the electrophile formed during the formation of F. Show the mechanism

for the formation of F.

+ Semester Examination

NO2
Electrophile

Figure 2

[7 marks]

6 Haloalkane J with molecular formula, C4H9Cl can be converted into other compounds as
shown in the reaction scheme below.

L KOH, ethanol C4H9Cl Mg, M
ref lux Haloalkane J dry ether
1. O
K ||
C
(CH3)3COCH3 H H
Figure 3
2. H3O+
N

(a) Haloalkane J can be converted into 2-methoxy-2-methylpropane using reagent K.
Suggest reagent K.

(b) Write the structural formulae of L, M and N.
(c) Write the mechanism for the formation of K.

[7 marks]

7 Compounds P and Q are branched alcohols with the molecular formula of C5H12O. P
decolourises acidified potassium permanganate, but Q does not. The reaction of P with
acidified potassium permanganate gives compound R. When P is treated with alkaline iodine,
a yellow precipitate is formed.
(a) Deduce the structures of compounds Q and R.
(b) Suggest a chemical test to differentiate P and Q. State the observations involved.
[5 marks]

245

 Semester Examination

8 Compounds S and T are monosubstituted isomers with the molecular formula of C8H8O.
Both S and T give yellowish orange precipitates when treated with 2,4-dinitrophenylhydrazine,
but only S gives brick-red precipitate and silver mirror when reacted with Fehling’s reagent
and Tollen’s reagent respectively.
(a) Deduce the structural formulae of S and T.
(b) Write the structural formulae of the products formed when S reacts with

(i) lithium aluminium hydride, followed by acid hydrolysis.
(ii) sodium bisulphite, NaHSO3.

(c) Ozonolysis of alkene U produces compound S and propanone. Draw the structure of
alkene U.
[7 marks]
BAKTI SDN. BHD.
9 (a) Arrange the following compounds in descending order of boiling point. Explain your
answer.

Pentane Butanoic acid Butanol

Semester Examination (b) Compound V with molecular formula of C5H10O2 has fruity smell. Compound V is
formed when compound U is heated with ethanol in the presence of a trace amount of
concentrated sulphuric acid.

(i) Write the structural formulae of compounds U and V.
(ii) Compound U can be synthesised from ethanol. Suggest another synthetic pathway

other than via Grignard reagent for the formation of compound U from ethanol.
(c) Kevlar is a strong and heat-resistant synthetic polymer with many applications such as

making bicycle tyres and bulletproof vests.
ILMU
Kevlar has the following structure:

O

PENERBIT H
|
N

O n
N
|
H

Draw the structure of one of the monomers of Kevlar.

[10 marks]

246

Semester Examination 

10 Aniline and propanamine have the following structural formulae.

NH2

CH3CH2CH2NH2
PENERBIT
ILMUAnilinePropanamine
BAKTI SDN. BHD.
Figure 4

(a) Compare the basicity of aniline and propanamine.
(b) Show how propanamine is synthesised from butanamide.
(c) Show how the azo compound, 4-(phenyldiazenyl)phenol is formed from aniline.

—N —OH
N—

4-(phenyldiazenyl)phenol Semester Examination
(d) Suggest a chemical test to differentiate aniline and propanamine.

[7 marks]

11 Lysine (2,6-diaminohexanoic acid) and alanine (2-aminopropanoic acid) are examples of
α-amino acids with isoelectric point at pH 9.7 and 6.0 respectively. An alkene and compound
W and X are formed when alanine reacts with nitrous acid. Lysine and alanine can also react
with one another in the mole ratio of 1:1 to form dipeptide, Ala-Lys.
(a) Draw the structure of alanine at pH 4.0.
(b) Draw the structural formulae of compound W, compound X and the dipeptide Ala-Lys.
[4 marks]

247

Answers

PENERBITSummative Assessment Test (UPS 1) ln 4.82 × 10–4
ILMU 5.35 × 10–3
BAKTI SDN. BHD. 1 D 2 B 3 C 4 D 5 B
6 D 7 A 8 A 9 C 10 A Ea 1 1
11 B 12 C 13 C 14 D 15 A J mol–1 273.15)K 273.15)K
16 B 17 B 1 2 = –
8.314 K–1 (65 + (45 +

Ea = 107.6 kJ mol–1 [2]
2 (a) (i) qreaction = qsolution + qcalorimeter
Summative Assessment Test (UPS 2)
qreaction = mc∆T + C∆T
1 C 2 D 3 A 4 D 5 A = (37.60 g)(4.18 J g–1 °C–1)(38.5 – 29.5) °C +
6 A 7 C 8 C 9 C 10 B
11 A 12 A 13 D 14 D 15 B (152.3 J °C–1)(38.5 – 29.5) °C
16 A 17 B 18 D 19 C
= 2 785.2 J

Amount of heat released = 2.785 kJ [2]

Summative Assessment Test (UPS 3) (ii) 21 H2SO4(aq) + KOH(aq) →

1 A 2 B 3 B 4 B 5 B 1 K2SO4(aq) + H2O(l)
6 C 7 B 8 A 9 B 10 A 2
11 D
Number of moles of H2SO4
2.0 × 12.50
Semester Examination = 1 000 = 0.025 mol

1 (a) Number of moles of KOH

In [N2O5] = 2.0 × 25.10 = 0.0502 mol
1 000


In [N2O5]0

t (s) 12 mol H2SO4 ≡ 1 mol KOH

[2] ∴ 0.025 mol H2SO4 requires 0.050 mol KOH.
Number of moles of KOH available
(b) (i) ln [N2O5]t = –kt + ln [N2O5]0 = 0.0502 mol . number of moles of KOH

ln [N2O5]20 = –(4.82 × 10–4)(20 × 60) + needed (0.050 mol)

ln (1.63 × 10–1) So, KOH is the excess reactant and H2SO4 is
the limiting reactant. [1]
[N2O5]20 = 0.09141 mol dm–3
[2] 21 mol H2SO4 ≡ 1 mol H2O
(ii) t 21 = lnk2 ln 2 0.025 mol H2SO4 ≡ 0.050 mol H2O [1]
= 4.82 × 10–4 s–1

= 1 438 s = 23.97 min [2] 0.050 mol H2O ≡ 2.785 kJ

(c) When temperature increases, the kinetic energy of 1 mol H2O ≡ 2.785 = 55.7 kJ
0.050
the molecules increases. Hence, the frequency of

collision increases and more molecules have kinetic Heat of neutralisation of the reaction

energy equal to or greater than the activation = –55.7 kJ mol–1 [1]

energy. Therefore, the reaction rate increases. [2] (b) (i) Second electron affinity of sulphur [1]

1 2(d) lnk1 Ea
k2 = R 1–1 (ii) K(s) [1]
T2 T1
(iii) ∆H1 = ∆H2 + ∆H3 + ∆H4 + ∆H5 + ∆H6 + ∆H7

–408.8 = +178.0 + ∆H3 + 279.0 – 200.0 +
456.0 – 1 959.4

∆H3 = +837.6

248

Answers 

∆H3 = 2 × IE1 of potassium (iii) The bromination of A is free radical

IE1 of potassium = +837.6 kJ mol–1 substitution reaction which requires the
2
presence of sunlight (uv light) or heat, while

= +418.8 kJ mol–1 [1] the bromination of B is an electrophilic

(iv) The lattice energy of Na2S will have a larger addition reaction which does not require the
magnitude than K2S because the Na+ ion has
a smaller ionic radius (size) than the K+ ion. presence of sunlight (uv light) or heat. [1]
PENERBIT
ILMU (iv) Baeyer’s test (using cold dilute alkaline
BAKTI SDN. BHD.
Hence, the ionic bonds between Na+ ions and KMnO4) or test with bromine in CH2Cl2 or

S2– ions are stronger. [2] bromine water. [1]

3 (a) (i) Cr2O72– (aq) , Cl2(g) , MnO4–(aq) (b)

The more positive the standard reduction

potential value, the greater the tendency for a |
CH=CH2
species to undergo reduction and, hence the [1]

stronger it is as an oxidising agent. [2] 5 (a) D: CH3CHClCH2CH3, AlCl3 [1]

(ii) For potassium permanganate and chloride E: KMnO4/H+, ∆ [1]

ions: (b) F G H

E°cell = E°cathode – E°anode N | O2 N| O2 C H3C| BrCH2CH3

E°cell = +1.51 V – (+1.36 V)

E°cell = +0.15 V

For potassium dichromate(VI) and chloride Br [3]
(c) Step 2: Formation of arenium ion
ions: +H
NO2
E°cell = E°cathode – E°anode HH

E°cell = +1.33 V – (+1.36 V) + + NO2 + NO2

E°cell = –0.03 V [1] + NO2→

Arenium ion

Based on the E°cell value, potassium Step 3: Removal of proton
permanganate solution can oxidise chloride
ions to chlorine gas as the E°cell value is H NO2
positive (reaction is spontaneous), while + NO2 + –OSO3H + H2SO4
potassium dichromate solution cannot because [2]
the E°cell value is negative (non-spontaneous 6 (a) Reagent K: CH3OH [1] Answers
reaction). [1] (b)
LM N
(b) Q = It = (2.0 A)(2 × 60 × 60) s = 14 400 C [1] (CH3)2C=CH2 (CH3)3CMgCl (CH3)3CCH2OH
Cathode: Au3+ + 3e– → Au
3 mol e– ≡ 3 × 96 500 C ≡ 1 mol Au (c) SN1 [3]
14 400 C ≡ x

x = 14 400 C × 1 mol Au = 0.04974 mol Au [1]
3 × 96 500 C

Mass of Au = 0.04974 mol × 197.0 g mol–1 = 9.799 g (CH3)3C—Cl (CH3)3C+ + Cl–
[1]
+
4 (a) (i)
(CH3)3C+ + HO–CH3 (CH3)3C—O| —CH3
H

[1] +
(ii) aCPpP brrrHeoossedd3eCnuuncHcccette2ffCoooofHrrfhmmBhyeeyrdddCdrorHwwog3ghhe eeennnnppBBeerrroreeoxaaxiccdidttesse::wwiitthh HHBBrr iinn tt[hh1ee] (CH3 )3C—OH| —CH3(C+HH3)3OC–—COH—3 CH3
CH3CH2CH2CH2Br [1] +

+ HO| —CH3
H
[3]

249

PENERBIT
ILMU
BAKTI SDN. BHD.