CADANGAN JAWAPAN PSPM 2 20172018

CHEMISTRY SK026 PSPM II 2017/2018

SECTION A [60 marks]
Answer all questions in this section.

1. a) The standard enthalpies of formation of NO and NO2 are 90.37 and 33.80 kJ
mol-1, respectively.
i. Give the definition of standard enthalpy of formation

Standard enthalpy of formation is heat change when 1 mol of
compound/substance is formed from its elements under standard
conditions (25oC, 1 atm)

ii. By referring to 1(a)i), explain why the standard enthalpy of formation
for elements is zero.
Elements are formed from their respective atoms/elements.

iii. Write the thermochemical equations that show the formation of NO
and NO2.

½ N2 (g) + ½ O2 (g) → NO (g) ΔH= +90.37 kJ

½ N2 (g) + O2 (g) → NO2 (g) ΔH= +33.80 kJ

iv. Determine the enthalpy change of the following reaction using an
energy cycle diagram.

2NO + O2 2 NO2

2NO(g) + O2 (g) ΔH 2 NO2 (g)

O2 (g) 2 O2 (g)
ΔH = 2 x (+90.37) kJ ΔH = 2 x (+33.80) kJ

N2 (g)

ΔH = - (2 x 90.37) + (2 x 33.80) = -113.14 kJ

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CHEMISTRY SK026 PSPM II 2017/2018

b) An aqueous solution of lead (II) nitrate was electrolyzed using a current of 2.0
A for 30 minutes. Determine the volume of gas evolved at the anode at room
temperature.

Q = It
= 2.0 x (30 x 60)
= 3600 C

Half equation of anode: 2 H2O (l) O2 (g) + 4H+ (g) + 4 e-
4 mol e- ≡ 4F ≡ 4 (96500) C ≡ 1 mol O2

386000 C ≡ 1 mol O2
If 3600 C ≡ 9.326 X 10-3 mol O2
Volume of oxygen gas = 9.326 X 10-3 mol X 24 dm3 mol-1

= 0.2238 dm3

2. Reaction of a reactive bromine radical with compound A forms a monobrominated
product B, while compound C reacts with bromine gas in dichloromethane to
produce compound D.

Compound A 3-tert-butylhexane Compound C 3-tert-butyl-3-hexene

a) Draw the structural formulae for compounds A and C.

CH3CH2CH2CHCH2CH3 CH3CH2CH CCH2CH3
H3C C CH3 H3C C CH3
CH3
CH3
Compound A Compound C

b) Identify which of the compounds in 2(a) exhibits chirality. Draw a pair of
enantiomer using 3-dimensional formulae for the identified compound.
Compound A

CH3 CH3
CH3C CH3 CH3C CH3

C CH 2CH 3 CH3 CH2 C CH2CH2CH3
CH3CH2CH2H H

mirror

A pair of enantiomer

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CHEMISTRY SK026 PSPM II 2017/2018

c) Name the type of reaction involved in the formation of B and D.
Compound B: Free radical substitution reaction

Compound D : Electrophilic addition reaction

d) Draw the structures of B and D.

Br Br
CH3CH2CH2C CHC2 H3 CH3CH2CH CH CH2CH3

H3C C CH3 H3C C CH3
CH3
CH3
Compound B Compound D

e) By using curved arrows show the initiation and propagation steps for the
formation of B.

Initiation: Br Br uv 2Br.

Propagation: Br
H

CH3CH2CH2C CH2CH3 CH3 CH2 CH2C CH2CH3
H3C C CH3
H3C C CH3 + H Br

CH3 CH3

CH3CH2CH2C CH2CH3 Br Br Br + Br
H3C C CH3 CH3CH2CH2C CH2 CH3

CH3 H3C C CH3

CH3

[4 marks]
f) Suggest a chemical test to distinguish between A and C. State the

observations and write the chemical equation involved.
[3 marks]

Chemical test: Bromine test

Reagent and condition: Br2, CH2Cl2, in the dark

Observation:-

Compound C : Reddish brown colour of bromine decolourised

Compound A: Reddish brown color of bromine remain unchanged

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CHEMISTRY SK026 PSPM II 2017/2018

Equation:

H Br2, CH2Cl2 Br
CH3CH2CH2C CH2CH3 in the dark CH3CH2CH2C CH2CH3

H3C C CH3 H3C C CH3

CH3 CH3

@ Baeyer’s test @ Bromine water

3. a) Treatment of 2-bromo-2-methylbutane with water yields compounds
E. The dehydration of E gives compounds F and G.
i. State the reagent and reaction condition used in the dehydration of E.

conc. H2SO4, heated @ Δ

ii. Draw the structural formulae of compounds E, F and G.

E: OH F/G: C CH3 F/G:
CH3 CH2C CH3 CH3 CH CH3
CH3 CH3CH2C CH2
CH3

iii. Indicate the major product formed in the dehydration reaction.
Explain your answer.

Major product:

CH3CH CCH3
CH3

The product follows Saytzeff’s rule

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CHEMISTRY SK026 PSPM II 2017/2018

iv. Draw the mechanism for the formation of E.

Step 1: Formation of carbocation +CH3CH2C+CH3 Br-
Br
CH3
CH3CH2C CH3
CH3 H
O+ H
Step 2: Nucleophilic attack
CH3CH2C CH3
+CH3CH2C+CH3 H2O
CH3 CH3

Step 3: Loss of H+ OH
H
O+ H

+CH3 CH2C CH3 H2O +CH3CH2C CH3 H3O+
CH3
CH3

b) 3-phenylpentene reacts with acidified water to yield two compounds, H and J. The
formation of compound H follows Markovnikov’s rule whereas carbocation
rearrangement forms compound J.

i. Draw the structural formulae for compounds H and J.

H: OH J:

HO

ii. Suggest the product formed when compound H is oxidised.

O

iii. What will happen when compound J is oxidised with acidified KMnO4
solution? Explain your answer.

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CHEMISTRY SK026 PSPM II 2017/2018

No changes@ No reaction @ Purple colour of KMnO4 remain unchanged.
This is because J is 3o alcohol @ no hydrogen attached to carbinol carbon @

no benzylic hydrogen.

4. a) 3-methyl-2-phenyl-2-butanol can be prepared from the reaction between
carbonyl compounds and Grignard reagents.

i. Draw the structure and state the class of this alcohol.

CH3 CH3
CH3 C C
, tertiary alcohol
H OH

ii. Draw all the possible pairs of carbonyl compounds and Grignard
reagents that can be used to form 3-methyl-2-phenyl-2-butanol.

CH3 +CH3 MgCl
CH3 C C

HO

CH3 + CH3 MgCl
CH3C C

HO

+O C CH3 CH3

CH3CH
MgCl

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CHEMISTRY SK026 PSPM II 2017/2018

b) The reaction of 1-bromopentane is shown in the scheme below.
1-bromopentane

K KOH

CN N
PCC KMnO4, H+, Δ
H3O+
L PQ
NaOH
OH

R
M

i. Draw the structures of L, M, N, P, Q and R.

HO O OH
O
O N
L M ONa
H
OH O
O O R
P Q

ii. State reagent K.

KCN

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CHEMISTRY SK026 PSPM II 2017/2018

Section B

5. N2O5 decomposes to NO2 and O2 according to the following chemical equation:

2N2O5 (g) 4NO2 (g) + O2 (g)

TABLE 1 shows the changes in N2O5 concentration against time when the reaction is
performed at 50oC.

TABLE 1

[N2O5]/M 1.00 0.77 0.60 0.45 0.35 0.27

Time/min 0 1 2 3 4 5

Determine the order and the rate constant for the reaction using a linear graph
method. From the graph, what is the half-life of the reaction?

Time/min [A] ln[A]
0 1.00 0.000
1 0.77 -0.261
2 0.60 -0.511
3 0.45 -0.799
4 0.35 -1.050
5 0.27 -1.309

0.000 Time/min

0123456

-0.200

-0.400

ln[A] -0.600

-0.800

-1.000

-1.200

-1.400

Graph ln[A] against time give a straight line. Therefore, this reaction is first order
reaction.
k = - gradient = 0.271 min-1 (show tangent line on the graph)

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CHEMISTRY SK026 PSPM II 2017/2018

Half life (show dotted lines ------ )

t1/2 = 2.56 min

@ t1/2 = = 2.56 min


Given the Ea of this reaction is 58 kJ mol-1 , calculate the rate constant at 65oC.
Compare the rate constants at the two temperatures and explain it in terms of
collision theory.

ln ( ) = ( − )



ln ( . ) = . ( − )

.

= 0.707 min-1
At higher temperature, rate constant is larger.

At higher temperature, the molecules are more energetic @ move faster @average
kinetic energy increase. Frequency of effective collisions increases. Fraction of
molecules with energy greater than Ea increases. Number of effective collisions
increases. The rate of reaction increases. Hence, the rate constant increases.

Describe how the addition of catalyst affects the rate of the reaction.

Catalyst increases the rate of reaction @ speeds up the reaction by lowering the Ea

6. a) Hexane, C6H14 , can exist in many isomeric forms. Draw all structural isomers
of hexane. Is there any chiral isomer? Explain your answer.

Name the type of bond cleavage when isomers of hexane undergo
halogenation reaction in the presence of light. Write an equation to show this
bond cleavage.

No chiral isomers. No chiral centre exist @ no carbon that attached to four
different groups.
Bond cleavage: Homolytic cleavage

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CHEMISTRY SK026 PSPM II 2017/2018

Equation to show the cleavage: + HBr

H + Br

@ any isomers from 6(a)

b) Hydrocarbon S, C8H8, reacts with hydrogen in the presence of platinum catalyst to
yield T, C8H10. Compound U is formed when T is heated with alkaline potassium
permanganate solution, followed by hydrolysis. The reaction of S with hydrogen
bromide gives V, whereas ozonolysis of S produces W and methanal, H2C=O. Draw
the structures of S to W. Give the name of the reaction for the conversion of S to T.
Write all chemical equations involved.

HC CH2 H2C CH3 COOH CH3 HC O
HC CH3

ST U VW

Conversion S to T: Hydrogenation

HC CH2 H2C CH3

H2
Pt

H2C CH3 COOH

i) KMnO4, OH-, D

ii) H3O+

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CHEMISTRY SK026 PSPM II 2017/2018

HC CH2 Br
HC CH3

HBr

HC CH2 HC O

i) O3 + O CH2
ii) Zn, H2O

7. a) Aromatic ring is less reactive towards chlorination than alkene. By using
suitable equations, explain this statement.

FeCl3 Cl

+ Cl2

+H2C CH2 CH2Cl2 HC CH Cl
Cl
Cl2

@ any alkene

Benzene reacts with chlorine gas in the presence of Lewis acid to yield
substituted product @ benzene undergoes electrophilic aromatic
substitution reaction.

Alkene reacts with chlorine gas to yield addition product @ alkene
undergoes electrophilic addition reaction.

Aromatic ring has delocalized electron which stabilizes the ring thus less
susceptible to electrophilic attack.

Outline a synthetic pathway from benzene to benzylamine.

CH2NH2

benzylamine

CH3 CH2Br CH2NH2

CH3Cl Br2 excess NH3
AlCl3 uv


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CHEMISTRY SK026 PSPM II 2017/2018

b) Grignard reagent is a versatile tool in synthetic organic chemistry. Using
bromocyclopentane as a starting material, show how Grignard reagent, X, is
synthesised.

Reaction of X with water produces compound Y while treatment in carbon
dioxide followed by hydrolysis forms compound Z. 3-Methyl-2-butanone
reacts with X and hydrolyses to yield compound AA. Draw the structural
formulae of compounds Y, Z and AA and write the chemical equations
respectively.

Br Mg MgBr
X
dry ether

H3C CH3
CH OH
COOH
C
CH3

YZ AA

MgBr H2O, H+

MgBr i) CO2 COOH
ii) H3O+

MgBr O

i) CH3C CH(CH3)2

ii) H3O+

8. a) A group of students runs one simple experiment to distinguish three different
classes of aliphatic amines and aniline. BB, CC and DD are primary, secondary
and tertiary aliphatic amines respectively. By using specific examples of BB,
CC and DD, suggest a chemical test to differentiate between:

i. BB, CC and DD.

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CHEMISTRY SK026 PSPM II 2017/2018

CH3CH2 NH2 CH3CH2 NH CH3 CH2 N CH3
BB CH3
CH3
CC DD

Nitrous acid test
Reagent and conditions: NaNO2, HCl, 0-5oC

@ Hinsberg test

ii. aliphatic amine and aniline.

Test: Bromine water
Reagent : Br2, H2O

State the observations for 8(a)i) and 8(a)ii).

8(a)i)
Observation:
BB: bubble of gas formed.
CC: yellow oil solution formed.
DD: clear solution formed.

8(a)ii).

Observation:

Aliphatic amine : no white precipitate formed

Aniline : white precipitate formed.

Arrange BB, CC and aniline in order of increasing basicity. Explain your
answer.

Increasing basicity: aniline < BB < CC

Alkyl group in BB and CC is electron donating group which stabilize
the positive charge @ increase the electron density on nitrogen.

Aniline is weaker base due to delocalization of electron in the
benzene ring destabilize the positive charge on nitrogen.

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CHEMISTRY SK026 PSPM II 2017/2018

b) The pI value for valine, (CH3)2CHCH(NH2)COOH is 5.96. Draw the structures of
valine at pI value, pH 2.39 and pH 9.32 respectively. Valine reacts with 2,2-
dimethylpropanol in the presence of acid catalyst to yield compound EE.
Draw the structure of EE.

CH3 O CH3 O CH3 O

CH CHC O- CH CHC OH CH CHC O-
CH3 NH+3 CH3 NH3+ CH3 NH2

pI pH 2.39 pH 9.32

CH3 O CH2 CH3
C
CH CH C O CH3 CH3
CH3 NH3+

EE

c) Polymer can be prepared either by condensation polymerisation or addition
polymerisation. Give one (1) example of synthetic polymer for each type of
polymerisation. By using your choice of condensation polymer, state the
respective monomers.

Condensation polymerization:

Example: Kevlar @ nylon 6 @ nylon 6,6 @ Dacron

Kevlar
Monomer: HOOCC6H4COOH and H2NC6H4NH2

Nylon 6
Monomer: HOOC(CH2)5NH2

Nylon 6,6
Monomer: HOOC(CH2)4COOH and H2N(CH2)6NH2

Addition polymerisation:
Polyethylene @ polyvinyl chloride @ polystyrene

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