PSPM Tutorial

ANSWER PSPM 2 20142015

Section A i. Lattice energy defined as energy required to completely separate 1 mol of
1. a) ii. ionic compound into gaseous ions.

Na(s) + 1/2Br2(g) -360 kJ NaBr(s)

107 kJ 97 kJ

Na(g) Br(g) Lattice Energy
496 kJ -324 kJ

b) i. Na+(g) + Br_(g)

By using Hess’s law
∆Hf = ∆Hsubl + ∆Ha + IE + EA +∆ Hlatiice
−360 = 107 + 97 + 496 + (−324) + ∆Hlatiice
∆Hlatiice = −736 kJ mol−1

Anode: Zn(s)  Zn2(aq)  2e
Cathode: Cu2(aq)  2e  Cu(s)
Overall equation: Zn(s)  Cu2(aq)  Zn2(aq)  Cu(s)

ii. Eo  Eo  Eo
cell cathode anode

 E  Eo o
Cu2 Cu Zn2 Zn (s)

= (0.34 V) – (- 0.76 V)

= 1.10 V

iii. Given:
Ecell = 1.17 V
[Cu2+] = 4.95 M

Ecell  Eo  0.0592 log Zn2 
cell n Cu2 

1.17 V   1.10 V  0.0592 log Zn2 
2
 4.95 M 

[Zn2+] = 0.0214 M

1

IJ-JI
“The struggle you are in today is developing the strength you need tomorrow.”

ANSWER PSPM 2 20142015

2. a) 2,2,4,4-tetramethylpentane < 2,2-dimethylheptane< 4-ethylheptane < nonane
Explanation:
 All molecules has same molecular weight but different molecular surface area.
 Nonane is unbranched molecule, but 4-ethylheptane, 2,2-dimethylheptane and
2,2,4,4-tetramethylpentane are branched molecules.
 Number of branched in 4-ethylheptane < 2,2-dimethylheptane < 2,2,4,4-
tetramethylpentane.
 Molecular surface area of of nonane > 4-ethylheptane > 2,2-dimethylheptane >
2,2,4,4-tetramethylpentane.
 Strength of Van der Waals forces between molecules nonane > 4-ethylheptane >
2,2-dimethylheptane > 2,2,4,4-tetramethylpentane.
 Energy needed to overcome the forces between molecules nonane > 4-
ethylheptane > 2,2-dimethylheptane > 2,2,4,4-tetramethylpentane.
 Therefore, boiling points of nonane > 4-ethylheptane > 2,2-dimethylheptane >
2,2,4,4-tetramethylpentane.

b) i.

Br2, CH2Cl2 CH3CH2CH2 + CH3 CH CH3
Br Br
CH3CH2 CH3 h

ii.
CH3 CHCH3
Br

Explanation:
Involves a more stable free radical, 2o free radical.

iii. Free radical substitution mechanism
Initiation step:

Br Br uv 2Br

Propagation step: H
H

+H3C C H Br +H3C C HBr

CH3 CH3

H H

+H3C C Br Br +H3C C Br Br

CH3 CH3

2

IJ-JI
“The struggle you are in today is developing the strength you need tomorrow.”

ANSWER PSPM 2 20142015

Termination step: Br Br

Br + Br

Br + CH3 H
C CH3
H H3C C Br
CH3

H H3C H3C H CH3
H3C C C C CH3
+ C CH3 CH3 H

CH3 H

c) Excess O2:
CH3CH(CH3)CH2CH2CH2CH3 19 2O2  6CO2  7H2O

Limited O2:
CH3CH(CH3)CH2CH2CH2CH3 132O2  6CO  7H2O

3

IJ-JI
“The struggle you are in today is developing the strength you need tomorrow.”

ANSWER PSPM 2 20142015

3. a) i. Based on table given, the relative rate of SN2 reactions towards class of alkyl
halides are as follow:

Methyl > 1o > 2o > 3o
Explanation:
 Class of alkyl halides inversely proportional to relative rate of reaction.
 Methyl alkyl halides undergo SN2 easily compared to 3o alkyl halides due

to high steric effect in 3o alkyl halides.
 Thus, relative rate in methyl alkyl halides is higher.

ii. Compound A:
Br

CH3 CH2CHCH3

iii. Value of B will be < 0.03.
Explanation:
 2-bromo-2-methylpentane has bulky alkyl group attached to carbon-α. It
has high steric effect (30 alkyl halide)
 It does not follow SN2 mechanism.
 Thus, relative rate will be the lowest among those alkyl halides.

iv. Nucleophilic substitution reaction

C: D: E:
CH3 CH2 OCH3
CH3 CH2 CN CH3 CH2 NH2

b) i. 3-methyl-3-pentanol

ii. Alkene:

CH3 CH2

C CHCH3

H3C

Reagents: H2O, H+

iii. Chemical test: Lucas Test

Observation:
1-pentanol (1o alcohol): no cloudy solution forms
2-pentanol (2o alcohol): cloudy solution forms within 5-10 minutes
F (3o alcohol): cloudy solution appears immediately

4

IJ-JI
“The struggle you are in today is developing the strength you need tomorrow.”

ANSWER PSPM 2 20142015

4. a) i. Conversion of butanal to 1-butanol

CH3 CH2CH2C H i) LiAlH4 CH3 CH2CH2CH2 OH
O ii) H3O+

ii. Conversion of butanal to butanoic acid

CH3 CH2CH2C H i) KMnO4, OH-, heat O
O ii) H3O+ CH3 CH2CH2C OH

b) Conversion of propanone to isopropylcyclopentane via Grignard reagent.

MgBr OH CH3 C CH2 CH3 CHCH3
O CH3 C CH3 H2, Ni

i. , dry ether conc.H2SO4
CH3C CH3 ii. H 3O+ 

c) Reason:
 Methanoic acid has carbonyl functional group.
 It shows reducing properties when react with Tollen’s reagent.

Observation:
 formed silver mirror precipitate at the wall of test tube.

Equation: Ag(NH3)2+ CO2 + H2O + Ag
silver mirror formed
O
H C OH

d) i. react with COOH ii. react with COOH iii. react COOH
(reduction)
(neutralization) (Nucleophilic subt.)
CH2OH
OO

C O-K+ C
Cl

CH3 CH3 CH3

iv. react with COOH (esterification)

O

C
OCH3

CH3

5

IJ-JI
“The struggle you are in today is developing the strength you need tomorrow.”

ANSWER PSPM 2 20142015

Section B 1st order with respect to KI
5. a) zero order with respect to HCl
Overall order of reaction = 2
b)
rate = k [H2O2][I−]
1.0×10−4 M s−1= k(0.1 M)(0.1 M)

k = 0.01 M−1s−1

G = [H2O2]
rate = k [H2O2][I−]
1.0×10−4 M s−1= (0.01 M−1s−1)(G)(0.1 M)
G = 0.2 M

H = [I−]
rate = k [H2O2][I−]
2.0×10−4 M s−1=(0.01 M−1s−1)(0.1 M)(H)
H = 0.1 M

J = rate of reaction
J = k [H2O2][ [I−]
J = (0.01 M−1s−1)(0.2 M)(0.2 M)

= 4.0×10−4 M s−1

Based on Arrhenius equation,

 Ea

k  Ae RT

 Rate constant, k is directly proportional to absolute temperature, T.

 Ea

 When T is high, term e RT will be high, hence, k will high.

Exothermic reaction

Energy profile diagram,

energy, kJ mol -1

activated complex

Ea N2(g) + NO 2(g)
N2O(g) + NO(g) reaction progress

H = -110 kJ

6

IJ-JI
“The struggle you are in today is developing the strength you need tomorrow.”

ANSWER PSPM 2 20142015

ln k1  Ea  1  1 
k2 R  T2 T1 
 

0.0234  Ea  1  1  1
0.750 8.314 J mol1 K1  
 ln  773.15  673.15 K

Ea = 150.03 kJ mol−1

Ea for reversed reaction(endothermic reaction)
Ea = 150.03 kJ mol−1 + 110 kJ mol−1

= 260.03 kJ mol−1

7

IJ-JI
“The struggle you are in today is developing the strength you need tomorrow.”

ANSWER PSPM 2 20142015

6. 3 isomers of 2o alkyl halide

CH3 CH2CH2CHCH3 CH3 CH2 CHCH2 CH3 Br
Br Br CH3 CHCHCH3

CH3

Each molecule undergo dehydrohalogenation to give 4 alkenes

K: L: M / N:
CH3CH2CH2 CH
CH3C CHCH3 CH2 C CH2 CH3

CH3 CH3 CH2

M / N:

CH3CH2 CH CHCH3

Product that can exist as geometrical isomers (cis-trans)
N:
CH3CH2 CH CHCH3

Reason:
 Each C at C=C attached to two different group of atoms.

Equation for ozonolysis of K

i) O3 OO
ii) Zn,H2O
CH3 C CHCH3 +CH3 C CH3 H C CH3
CH3

L react with HBr gives P
P:

CH3

CH3 C CH2CH3

Br

Electrophilic addition mechanism of L to P

CH3 slow H CH3

+CH3 CH2 C CH2 H Br CH3 C C CH3 + Br-
H
+

CH3 CH3
CH3 C CH2 CH3
CH3 C CH2CH3 Br -
+ Br

L react with HBr, H2O2 (Anti-Markovnikov)
Br
CH2 CHCH2CH3

CH3

8

IJ-JI
“The struggle you are in today is developing the strength you need tomorrow.”

ANSWER PSPM 2 20142015

7. a) Molecular structure of S, T, U, V and W

S: alkylbenzene T: 3o haloalkane U: 1o haloalkane

CH3 CH3

CHCH3 C CH3 CHCH2 Cl

Cl CH3

V: benzoic acid W: 3o alcohol
COOH CH3
CCH3

OH

To form S,
Benzene will undergo Friedel Crafts Alkylation through electrophilic substitution
mechanism.

Step 1: AlBr3 H H AlBr4-
H3C C Br+ AlBr 3
H +H3C C+
CH3
+H3C C Br CH3

CH3

Step 2:

H CH3 CH3 CH3 CH3
H H H
+ C+
+ CHCH3 CHCH3 CHCH3
H3C
+ +

Step 3:

CH3 CH3
H CHCH3

+ +CHCH3 AlBr4- + AlBr3 + HBr

b) 2-bromo-2-methylpropane > 2-bromopropane > 1-bromopropane
Reason:
 The reactivity of alkyl halide towards SN1 mechanism is depends on type of
carbocation form in the mechanism.
 2-bromo-2-methylpropane is 3o alkyl halide which will forms a stable 3o
carbocation. Hence, easily undergo SN1.
 1-bromopropane and 2-bromopropane forms 1o and 2o carbocation respectively.
 Compare to 3o carbocation, 1o and 2o carbocation are less stable.
 Thus, 1-bromopropane and 2-bromopropane less reactive towards SN1.

9

IJ-JI
“The struggle you are in today is developing the strength you need tomorrow.”

ANSWER PSPM 2 20142015

Methanolysis of 2-bromo-2-methylpropane

CH3 CH3
CH3CCH3
+ +CH3OH  CH3C CH3 HBr
Br
OCH3

SN1 mechanism

Step 1:

CH3 slow CH3
H3C C Br
+H3C C+ Br -

CH3 CH3

Step 2:

CH3 CH3 CH3
C+
H3C + CH3 OH H3C C CH3 +H3C C CH3 HBr
CH3 O+
CH3 H O CH3

Br-

10

IJ-JI
“The struggle you are in today is developing the strength you need tomorrow.”

ANSWER PSPM 2 20142015

8. a) Preparation of aniline from benzene NO2 NH2
SnCl2/H+
conc. HNO3, conc.H2SO4
50-55oC

Formation of azo compound from aniline

NH2 N+ N
NaNO2, HCl

0 - 5 °C

N+ N OH

+ NN OH
0 - 5 °C

azo compound

Cyclohexanamine is more basic than aniline
Reason:
 Cyclohexanamine has lower the pKb value.
 Cyclohexyl is electron donating group. So it increases electron density on

nitrogen atom.
 In aniline, phenyl group is electron withdrawing group which delocalized the

electron density of nitrogen atom.

b) Structure of alanine at pH12 (basic medium)
HO

H3C C C O-

NH2

Isoelectronic point is pH at which amino acid exists primarily in its neutral
form.

Structure of X and Y. Y:
X: H

H H3C C COOCH2CH3

H3C C COOH +

+- NH3

NH3 Cl

c) Monomers of Kevlar
OO

Cl C C Cl H2N NH2

Condensation polymerisation

11

IJ-JI
“The struggle you are in today is developing the strength you need tomorrow.”

ANSWER PSPM2 20152016

SECTION A

1 (a)(i) Heat released by reaction = Heat absorbed by water + Heat absorbed by calorimeter

@ - qrxn = (qwater + qbomb)
@ qrxn = – (qwater + qbomb)

qrxn = - (mcΔT + CΔT)
= - (780 g x 4.18 J/g °C x 11.1 °C) + (870 J/°C x 11.1 °C)

= - 45847 J
= - 45.85 kJ

(a)(ii) Mass of 1 mole of C6H7O2 = [7(12) + 6(1) +2(16)] g
= 122 g

2.34 g C6H7O2 released 45.85 kJ of heat,

122 g C6H7O2 released = (45.85 kJ x 122 g mol-1 )
2.34 g

= 2390 kJ mol-1

∴ ΔHc = –2390 kJ mol-1

(b)(i)
Anode : Zn(s) →Zn2+(aq) + 2e

Cathode : Fe2+(aq) + 2e → Fe(s)

(b)(ii) E°cell = E°cathode – E°anode
= [ – 0.44 – (– 0.76)]

= + 0.32 V Ecell = E°cell – 0.0592 log [Zn]2+
Using Nernst Equation; n [Fe2+]

= 0.32 – 0.0592 log (0.1)
2 (0.5)

= +0.34 V

1

ANSWER PSPM2 20152016

2. (a)(i) (CH3)2CBrCH2CH3

(a)(ii) Free radical substitution

(a)(iii) Initiation step: hv 2Br.
Br Br

Propagation step: H3C
H3C
. +CH3CH2C H Br
CH3CH2C H + . Br

H3C H3C
H3C
H3C CH3CH2C

. +CH3CH2C Br Br Br + .Br

H3C H3C

Termination step:

CH3 CH3 CH3 CH3
CH3CH2C C CH2CH3
. + .CH3CH2C CCH2CH3
CH3 CH3
CH3 CH3

CH3 CH3
CH3CH2C Br
.CH3CH2C + .Br

CH3 CH3

Br. + .Br Br Br

2

ANSWER PSPM2 20152016

(b)(i) B : CH2CH3
CH3 CH CH2 CH2CH3

C : CH2CH3
CH3CBrCH2CH2CH3

D : CH2CH3
Br CH2C(OH)CH2CH2CH3

E : CH2CH3
Br CH2CBr CH2CH2CH3

(b)(ii) Alkyl halide : (Cl @ Br)

CH2CH3
Br CH2CHCH2CH2CH3

CH2CH3
CH3CBr CH2CH2CH3

CH2CH3
CH3CHCHBr CH2CH3

CHBr CH3
CH3 CH CH2 CH2 CH3

Reagent :
EtOH / Ethanolic KOH @ Ethanolic NaOH @ NaH
(b)(iii) Baeyer’s test

Product :

CH2CH3
HO CH2C(OH) CH2CH2CH3

The decolorised of purple color @ the appearance of brown precipitate.

3

ANSWER PSPM2 20152016

3. (a)(i) CH3CH2CH2-NH2

(a)(ii) CH3CH2CH2-OCH3

(a)(iii) CH3CH2CH2-CN

(a)(iv) CH3CH2CH2-OH

(b) CH2 CH3 CH2 CH3
CH2 CH3 NC C Br

- + C Br slow fast NC C + -
:Br
CN

H HH H
H
H

transition state

(c)(i) G : CH3CH2CH2CH2Cl

H : CH3CH2CH2CH2MgCl

(c)(ii) X : CO2
Y : H3O+ @ H2O/H+ @ H2O,H+

(c)(iii) i. HBr, H2O2

ii. NaOH (aq)

(c)(iv) G CN- CH3CH2CH2CH2CN H3O+ CH3CH2CH2CH2COOH

@ KCN @ H2O, H+

@ NaCN

4

ANSWER PSPM2 20152016

4. (a)(i) K : CH3CH2CHClCOOH ; L : CH3CHClCH2COOH

(a)(ii) CH3CH2CHClCOOH is more acidic than CH3CHClCH2COOH
The position of Cl increases the acidity due to inductive effect.
The acidity decreases as the distance between Cl and carboxyl group increases.

(a)(iii) CH3 CH CHCOOK

(b)(i)

N: HO O CH2CH2CH3

CH3

O
P:

-
O

(b)(ii) Q : CHI3
R : CH3COCl, AlCl3 @ other Lewis acid catalyst

S : LiAlH4, T : H3O+ @ H2O / H+ @ S : NaBH4 T : CH3OH

(b)(iii) Brady’s test
Observation
M : Gives yellow / orange / yellowish orange precipitate
U : No yellow / orange / yellowish orange precipitate form @
no observable change

O2N NO2 O2N NO2
RH RH

O + H2N N NN

R R

OR

Lucas test
Observation M : No observable change

U : Cloudy solution formed

CH(OH)CH3 ZnCl2 / HCl CH(Cl)CH3

5

ANSWER PSPM2 20152016

SECTION B
5. (a)

Time/s 0 200 400 600 1200 1800 3000
[H2O2] 2.32 2.01 1.72 1.49 0.98 0.62 0.25
ln [H2O2] 0.84 0.70 0.54 0.40 -0.02 -0.48 -1.39

FIRST METHOD: Plot ln [H2O2] versus time
Graph obtained is a straight line @ y-mx + c
The decomposition of H2O2 is a first order reaction

 k   slope@ y1  y2
 x1  x2

slope   7.46 x 10 4

half life, t1  ln2
2
k

 ln2 s 1
7.46 x 10 4

 928.6 s

SECOND METHOD: Plot [H2O2] versus time
From the graph, t1/2 (1) is almost equal to t1/2 (2)
The decomposition of H2O2 is a first order reaction

t 1 (1)  t1 (2)
t1  2 2
2
2

 970 s

ln  H 2O2    kt  ln  H 2O2 
t 0

 ln    1600
2.32  ln H 2O2 t  7.46 x104

 H2O2  e0.352

t

 0.72 M

% decomposition  2.32  0.72 x100
2.32

 69.8%

6

(b) Lower temperature, T ANSWER PSPM2 20152016
No. of molecules 1 T1<T2

Higher temperature, T
2

Molecules with enough kinetic
energy to react

E
a Kinetic energy

As temperature increase,
 More molecules move at higher speeds @ K.E of molecules
 More molecules have energy ≥ Ea
 The number of effective collision increase, thus the reaction rate increases

7

ANSWER PSPM2 20152016

6. CH3 CHCH2 CH2 CH3 V : CH3CH=CHCH2CH3

OH W : CH3CH2CH2CH=CH3
2-pentanol Major product : V, follow Saytzeff’s rule

Geometrical isomers :

HH H CH2 CH3
CC
CC

H3C CH2 CH3 H3C H

cis isomer trans isomer

.. H+ +
: OH
CH3 CH2 CH2 CHCH3 :OH2

CH3 CH2 CH2 CHCH3
protonated alcohol

+O..H2 +
CH3 CH2 CH2 CHCH3
CH3CH2CH2CHCH3 + H2O
2o carbocation

H + + HSO4- CH3 CH2 C C CH3
CH3 CH2 C HH
C CH2
H major product
HH
CH3 CH2 CH2 C CH2
H

minor product

ZZ : CH3CH2CHO @ CH3CHO
AA : CH3CHO @ CH3CH2CHO

Reaction equation : i. O3 O O H
CH3 CH CH CH2 CH3 ii. Zn, H2O @ (CH3)2S CH3 C
+H CH3 CH2 C

CH3 CH2 CH2 X Mg CH3 CH2 CH2 MgX i. CH3CHO CH3 CH2 CH2 CHCH3
dry ether OH
ii. H3O+

8

7. (a) Electrophilic aromatic substitution ANSWER PSPM2 20152016

CH2CH3 COOH O
BB : CC : C OCH2CH3
DD :

Chemical equation for the formation of BB :

CH3CH2Cl CH2CH3
AlCl3

Chemical equation for the formation of DD :

O

ethanol

COOH H+ C OCH2CH3 + H2O

Mechanisme :
Step 1 :

+CH3 CH2 Cl AlCl3 +

CH3 CH2

Step 2 : H CH2 CH3 H CH2 CH3 H CH2 CH3

+ ++

+ CH3 CH2 +

Step 3 : CH2 CH3 CH2CH3
H
+
AlCl- +

9

ANSWER PSPM2 20152016

(b) CH3 O

alkyl bromide, EE + nucleophile CH3CH2C O CCH3

H

CH3
EE = 2alkyl halide = CH3CH2C Br

H

O

nucleophile = CH3C -
O

CH3 CH2CH3 FF is an ether yield from the reaction
FF = CH3CH2C O between alkyl halide and alcohol

H

H H
C
GG = C Saytzeff's rule : more highly
CH3 substituted product

H3C

Reaction equation :

CH3 O CH3
CH3CH2C O
+CH3CH2C Br CH3C - CH2CH3
O H

H

CH3 CH3 CH2CH3
CH3CH2C O
+CH3CH2C Br CH3CH2OH
H
H

CH3 KOH H H
CH3CH2C Br ethanol C C

H H3C CH3

10

8. (a) ANSWER PSPM2 20152016

R2NH more basic than RNH2
The presence of electron donating substituent R, increase the inductive effect /
increase e density on N, increase basicity.
Aniline, aromatic amine, less basic compare to aliphatic amine.
Electrons delocalized into the benzene ring of aniline thus reducing basicity.

NH2 NaNO2 / HCl + OH
N2
0 - 5oC NN OH
NaOH, 0oC

p-(phenylazo)phenol

(b) B H2N CHCOOH
A H2N CH2COOH
2-aminoethanoic acid CH3
2-aminopropanoic acid

OO

AA H2N CH2C N CH2COOH BB H2N CH C N CHCOOH
H CH3 H CH3

O O
AB H2N CH2C N CHCOOH BA H2N CHC N CH2 COOH

H CH3 CH3 H

peptide bond
(c) Homopolymer – a polymer formed from a single type of monomer.

HH
CC
H Cl n

Copolymer – a polymer formed from the joint of two different types of monomer.

OO

N (CH2)4 N C (CH2)4 C

HH n

nylon 6,6

11

CHEMISTRY UNIT
KOLEJ MATRIKULASI MELAKA
ANSWERS - PSPM SK026 2016/2017 SESSION

SECTION A i. Heat release when 1 mole of ethanol is burned completely in excess oxygen at
1. (a)

standard condition 25oC and 1 atm.

ii. nethanol = 3.68 = 0.08 mol
46.0 /

Heat releasing by reaction = Heat absorbed by calorimeter

qsys = qrxn + qcal + qwater
-qrxn = qcal

= Ccal∆T
= 1.98 kJ/oC (66 – 25oC)
= 81.18 kJ

0.08 mol C2H5OH ≡ 81.18 kJ
1 mol C2H5OH ≡ 1014.8 kJ
∴ ∆Hoc = - 1014.8 kJ/mol

(b) Anode : ( Al(s) Al3+(aq) + 3e ) 2

Cathode : ( Ni2+(aq) + 2e Ni(s) ) 3

Overall : 2 Al(s) + 3 Ni2+(aq) 2 Al3+(aq) + 3 Ni(s)

Eocell = Eocathode - Eoanode
= -0.25 V – (- 1.66 V)
= + 1.41 V

Ecell = Eocell – 0.0592 log [ 3+]2
[ 2+]3
[ ]2
Ecell = +1.14 V – 0.0592 log [1.50]3
6

∴ x= 5.90 M

(c) Anode : Sn(s) Sn2+(aq) + 2e

Cathode : Sn2+(aq) + 2e Sn(s)

At anode Sn atom is oxidized to formed Sn2+ ion and becomes thinner.
At cathode Sn2+ ion is reduced to formed Sn atom and deposited on iron nail.

ANSWERS – SK026 PSPM 2016/2017 Page 1 of 12

2. (i) B: CH3 OH
CH3 C CHCH3

CH3

Protonation of alcohol:

CH3 OH + H CH3 H + OH
CH3 C CH CH3 O+ H CH3 C O+ H H

CH3 H CH3 CH CH3

Formation of carbocation:

CH3 H CH3 + H
CH3 C O+ H CH3 C +CHCH3 OH

CH3 CH CH3 CH3

Formation of alkene:

CH3 H OH CH3 H
CH3 C +C C H + H CH3 C C C H + O+ H

CH3 H H CH3 H H H

(ii) C: CH3 CH2
D: CH3 C CH

CH3

CH3
CH3 C CH2 CH3

CH3

(iii) E: CH3 Br
CH3 C CH CH3

CH3

Secondary @ 2 haloalkane @ alkyl halide

ANSWERS – SK026 PSPM 2016/2017 Page 2 of 12

(iv)

Test Bromination test D
Reagent Br2, CH2Cl2 in the dark The reddish brown colour of bromine
Compound remains.
Observation C

The reddish brown colour of bromine
disappears.

Equation CH3 CH3
CH3 C CH CH3 C CHBr CH3
CH2 + Br2 CH2Cl2
CH3 reddish CH3

brown

@

Test Baeyer’s test
Reagent
Compound dilute KMnO4, OH, cold
Observation
C D
Equation The purple colour of KMnO4 remains.
The purple colour of KMnO4 disappears
and brown precipitate forms.

CH3 OH-, cold CH3 OH + MnO2(s)
CH3 C CH CH2 + KMnO4 CH3 C CHCH2 brown precipitate

CH3 purple CH3 OH

3. (a) i. G: CH3CH2CH2CH2Cl @ CH3CH2CH2CH2Br

H: CH3CH=CHCH3

J: O
CH3 CH2 CH2 C OH

ii. i)CH3CH2CH2MgCl, dry ether @ CH3CH2CH2MgBr, dry ether
ii)H2O,H+ @ H3O+

iii. C-C double bond

iv.

Test Lucas test 2-methyl-2-butanol
Reagent conc. HCl, conc. ZnCl2
Compound The solution turns cloudy
Observation F immediately.
The solution does not turn cloudy

Equation CH3 conc.HCl, conc. ZnCl2 CH3
CH3CH2CH2C CH3 CH3CH2CH2C CH3
+ H2O
OH Cl
insoluble in water

ANSWERS – SK026 PSPM 2016/2017 Page 3 of 12

v. Pyridinium chlorochromate in CH2Cl2 @ PCC, CH2Cl2

(b) Bimolecular nucleophilic substitution @ SN2

-

H - H H -
C Br + O + Br
slow H O C Br fast HO C
H d+ d- H
CH3 CH2 CH2 H
H CH2 CH2 CH3 CH2 CH2 CH3

transition state

4. (a) M: CH3CH2CHO
N: NNH2
CH3 CH2 CH
Q: KMnO4, H+,  @ K2Cr2O7, H+,  @ Na2Cr2O7, H+,  @ H2CrO4, 
R: H+, KCN @ H+, NaCN

(b) i. CH3CH2COCl

ii. CH3CH2CH2OH

iii. CH3CH2COONa

iv. CH3CH2CONH2

(c) The boiling point of M < K < L

The boiling point of a compound depends on the strength of intermolecular forces. Stronger
intermolecular forces require more energy to separate the molecules in the liquid state in
order for it to boil.

The forces acting between M molecules are London dispersion forces and dipole-dipole
forces. If K exists as trans-isomer, it is a non-polar molecule and has only London dispersion
forces between molecules. But if K exists as cis-isomer it is a polar molecule and has both
London dispersion forces and dipole-dipole forces between molecules. However, the boiling
point of K (Mr = 84) is higher than M (Mr = 58). This is because much higher molecular
weight makes polarizability more significant in K and causes it to have much stronger
London dispersion forces.

L (Mr = 74) have higher boiling point than K because the difference in molecular weight is
smaller but L is able to form hydrogen bond between molecules. The hydrogen bonds
between L molecules are also very strong because the presence of two hydrogen bonds
between two molecules enables them to form a dimer.

ANSWERS – SK026 PSPM 2016/2017 Page 4 of 12

(d) Observation:
The formation of silver mirror.

Chemical equation:

O Ag(NH3)2+(aq) O
CH3 CH2 C O- + Ag(s)
CH3 CH2 CH OH-

@

O AgNO3(aq), NH3(aq) O
CH3 CH2 C O- + Ag(s)
CH3 CH2 CH OH-

The chemical test to confirm the presence of carbonyl group:
Brady’s test @ reaction with 2,4-dinitrophenylhydrazine @ reaction with 2,4-dNPH

SECTION B

5. (a) 0 10 20 30 40 50 60
4.50 3.10 2.38 1.92 1.60 1.40 1.22
Time (min)
[ S ], M 0.22 0.32 0.42 0.52 0.63 0.71 0.82
1/ [ S ], M-1

− =

[ ] [ ]

0.32 – 0.22 = k (10 min)

∴ k = 0.01 M-1min-1

=
[ ]


=1

0.01(4.50)

∴ t1/2 = 22 min

∴ Rate = k [S]2

Based on graph ,
1/ [S] = 0.470 M-1

[S] = 2.13 M
∴ Concentration reactant S at 1500s is 2.13M

ANSWERS – SK026 PSPM 2016/2017 Page 5 of 12

1/[S] versus time

1.100

1.000

0.900

0.800

0.700

0.600

1/[S] (M-1) 0.500
0.470
d[S] = 0.820  0.222
0.400 dt = 60  0

0.300

0.200

0.100

0.000 10 20 25 30 40 50 60
0
Page 6 of 12
-0.100 Time (min)

ANSWERS – SK026 PSPM 2016/2017

(b) Rate 1 = k [W]

Rate2 k [W]

2.25 x 10−3 = k (0.33)
Rate 2 k (0.85)

∴ Rate 2 = 5.80 x 10-3 mol dm-3s-1

ln ( 1) = ( 1 − 1 )

2 2 1

ln ( 1.74 1100−−55) = (1 − 1)
8.314 /
6.61 308.15 298.15

∴ Ea = 101.95 kJ/mol

The two factors that affect reaction rate are:

i) Concentration
 When the concentration of reactants increase,
 The number of molecules per unit volume increase
 The number of effective collision increase
 Thus the reaction rate increases.
Particle size
ii) Size of reacting particle decreases
 Total surface area exposed for reaction increases
 The reaction becomes faster
 Thus the reaction rate increases.


6. (a) 2 types of bond cleavage ; i) Homolytic cleavage
ii) Heterolytic cleavage

Heterolytic cleavage Homolytic cleavage
Homolytic cleavage occurs in a non-
Heterolytic cleavage occurs in a polar polar covalent bond involving two
covalent bond between atoms of atoms of the same or similar
different electronegativity. A single electronegativity. A single bond breaks
bond breaks unsymmetrically in which symmetrically in which each atom
the more electronegative atom attracts attracts the shared electrons equally
the shared electrons stronger thus gain thus each atom gain an electron of the
both of the shared electrons and form shared electrons.
an anion
• Free radicals are formed.
• The products are cation and anion.

ANSWERS – SK026 PSPM 2016/2017 Page 7 of 12

H Cl + NH3 - + H NH+3
Cl

The bond cleavage occurs in HCl is heterolytic cleavage .Chlorine is a more electronegative
atom than hydrogen thus attracts the shared electrons stronger and gain both of the shared
electrons. So, the bond breaks unsymmetrically and form chloride ion and hydrogen ion. The
lone pair at nitrogen atom of the ammonia molecule occupies the empty orbital of hydrogen
ion thus form ammonium ion, NH4+.

(b) Y is reactive because it contains C-C double bond which is an electron rich site. The  bond is
weaker than the  bond and also more expose to electrophilic attack.

Therefore Y undergoes electrophilic addition reaction.

Y: CH3 CH2 C CH3
CH3 CH

Z: Br
CH3 CH2 C CH3
CH3 CH2

AA: CH3 CH2 CHCH3
CH3 CH2

BB: KMnO4, H+, 

Stereoisomerism of Y: CH3 CH2 CH3
CC
CH3 CH2 H
CC H3C H
trans-isomer
H3C CH3
cis-isomer

7. (a) CC: CH3
EE: COOH
FF: CH2Cl

ANSWERS – SK026 PSPM 2016/2017 Page 8 of 12

GG:
CH2OCH3

DD: KMnO4, H+, 

Initiation step:
Cl Cl uv 2Cl

Termination steps:

Cl + Cl Cl2

H Cl H
C+ C Cl
H H

HH HH
C+C CC
HH HH

Name of the reaction that convert benzene to CC: Friedel-Craft alkylation

Step 1:

d+ d- d+ d- +-
H3C Cl AlCl3
H3C Cl + AlCl3

Step 2: +H + H H
+- CH3 CH3 + CH3

H + H3C Cl AlCl3

-
+ Cl AlCl3

ANSWERS – SK026 PSPM 2016/2017 Page 9 of 12

Step 3: CH3 + HCl + AlCl3
CH3 + HCl + AlCl3
+H -
+ Cl AlCl3

CH3
@

+H -
+ Cl AlCl3

CH3

(b) CH3CH2CH2Cl Mg, dry ether CH3CH2CH2MgCl

HH: CH3CH2CH2CN
JJ: CH3CH2CH2CH2NH2
KK: CH3CH2CH2COOH
LL: CH3 CH2 CH2 CHCH3

OH

CH3 CH2 CH2 C N i. LiAlH4 CH3 CH2 CH2 CH2 NH2
ii. H2O

@

CH3 CH2 CH2 C N H2, Pt @ Pd @ Ni CH3 CH2 CH2 CH2 NH2

@

CH3 CH2 CH2 C N NaBH4, methanol CH3 CH2 CH2 CH2 NH2

CH3 CH2 CH2 MgCl i. CH3CHO, dry ether CH3 CH2 CH2 CH CH3
ii. H2O, H+ @ H3O+ OH

ANSWERS – SK026 PSPM 2016/2017 Page 10 of 12

8. (a) conc. HNO3, conc. H2SO4
D
NO2
50-55oC

H2, Pt @ Pd @ Ni
@
i. Sn, HCl
ii. OH-

@ @ i. Fe, HCl
ii. OH-
i. Zn, HCl
ii. OH-

NH2

NH2 NaNO2, HCl + + NaCl + H2O
0 C-5 C
NN
Cl-
MM

+ OH NN OH + HCl
NN
N N+
Cl-

NN has intense colour and commonly used as dyes

(b) Isoleucine: CH3 O

CH3 CH2 CH CH C OH

NH2

At pH 2.0 CH3 O

CH3 CH2 CH CH C OH
NH+3

At pH 12.0 CH3 O
CH3 CH2 CHCH C
O-
NH2

ANSWERS – SK026 PSPM 2016/2017 Page 11 of 12

At pI CH3 O

CH3 CH2 CHCH C O-
NH+3

CH3 O

CH3 CH2 CH CH C O CH2 CH3
NH+3

Name of the reaction: Esterification

(c) Repeating unit is the portion of molecule the repeats itself in a polymer.
Monomer is small organic compounds that can be covalently bonded to each
other in a repeating pattern.

Repeating unit:
Cl

CH2 C CH2 CH
Cl Cl

Monomers:
CH2=CCl2 and CH2=CHCl

ANSWERS – SK026 PSPM 2016/2017 Page 12 of 12

PSPM 2 2017/2018

Section A

1. (a) i) 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) Elements are formed from their respective atoms/elements.

iii) ½ N2 (g) + ½ O2 (g) NO (g) ΔH= +90.37 kJ mol-1
½ N2 (g) + O2 (g) NO2 (g) ΔH= +33.80 kJ mol-1

iv) 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

(b) 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. (a)

H3C CH2 CH2CH CH2CH3 H3C CH2CH C CH2CH3
H3C C CH3 H3C C CH3

H3C H3C
Compound C
Compound A

(b) Compound A

CH3 CH3

H3C CH3 H3C CH3

C CH2CH3 H3CH2C C CH2CH2CH3
H
H3CH2CH2C
H

mirror

A pair of enantiomer

(c) Compound B: Free radical substitution reaction
Compound D : Electrophilic addition reaction

(d) Br
H3C CH2CH2 C CH2CH3
Br
H3C CH2 CH2C CH2CH3 H3C C CH3

H3C C CH3 H3C

H3C Compound D
Compound B

(e)

Initiation: Br Br uv 2Br .

Propagation: Br
H

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

H3C H3C

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

H3C

(f)_ 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 colour of bromine remain unchanged

Equation:

H3C H Br2, CH2Cl2 H3C Br
CH2 CH2C CH2CH3 in the dark CH2 CH2C CH2CH3
H3C C CH3 H3C C CH3

H3C H3C

@ Baeyer’s test @ Bromine water

3. (a) i) conc. H2SO4, heated @ Δ F/G: F/G:
ii) H3C CH C CH3 H3C CH2 C CH2
CH3 CH3
E: OH
H3C CH2 C CH3

CH3

iii) Major product:
H3C CH C CH3

CH3
The product follow Saytzeff’s rule.

iv) +H3C CH2 C+CH3 Br-
Step 1: Formation of carbocation
Br CH3
H3C CH2 C CH3

CH3
Step 2: Nucleophilic attack

H3C CH2 C+CH3 H2O H
O+ H
+CH3
H3C CH2 C CH3

CH3

Loss of H+ OH
H
O+ H

+H3C CH2 C CH3 H2O +H3C CH2 C CH3 H3O+
CH3
CH3

(b) i) OH J:
H:

HO

ii)

O

iii) 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) i)

CH3
H3C C C CH3

H OH , tertiary alcohol

ii)

CH3 + MgCl
H3C C C CH3

HO

CH3 + H3C MgCl
H3C C C

HO

+O C CH3 CH3

H3C CH
MgCl

4.b) i)

HO O OH

O O N
L M ONa

H OH O
O O R
P Q

ii)KCN

Section B

5.

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)

Half life (show dotted lines )

t1/2 = 2.56 min


@ t1/2 = = 2.56 min

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.

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

6. (a)

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

H + Br + HBr

@ any isomers from 6(a)

(b) H2C CH3 COOH
HC CH2

S T U
HC O
Br
HC CH3

VW
Conversion S to T: Hydrogenation

HC CH2 H2C CH3

H2
Pt

H2C CH3 COOH

i) KMnO4, OH-, 

ii) H3O+

HC CH2 Br
HC CH3

HBr

HC CH2 HC O

i) O3 + O CH2
ii) Zn, H2O

7. (a)

FeCl3 Cl

+ Cl2

CH2Cl2

+H2C CH2 Cl2 HC CH Cl
Cl
@ 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.

CH3 CH2Br CH2NH2

CH3Cl Br2 excess NH3
AlCl3 uv 

(b)

Br MgBr
X
Mg
dry ether

COOH OH

YZ AA

MgBr H2O, H+

MgBr i) CO2 COOH
ii) H3O+

MgBr O

i)H3C C CH(CH3)2 OH

ii) H3O+

8. a) H3C CH2 NH H3C CH2 N CH3
CH3 CH3
H3C CH2 NH2
BB CC DD

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

Observation:
BB: bubble of gas formed.
CC: yellow oil solution formed.
DD: clear solution formed.
@ Hinsberg test

ii) Test: Bromine water

Reagent : Br2, H2O

Observation:

Aliphatic amine: no white precipitate formed

Aniline : white precipitate formed.

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.

b)

CH3 O CH3 O CH3 O

CH CH C O- CH CH OH CH CH C O-
NH+3 NH2
H3C H3C NH+3 H3C

pI pH 2.39 pH 9.32

CH3 O CH2 CH3

CH CH C O C
NH+3 H3C CH3
H3C

EE

(c) 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

PSPM 2 2018/19 ANSWER

1. (a) i) Given : − [ 2 5] = 1.25 x 10-2 M min-1



+ [ 2] = − 1 [ 2 5]

2

= 1 (1.25 10−2)

2

= 6.25 x 10-3 M min-1

ii) ln [N2O5]o - ln [N2O5] = kt
ln (100) – ln (75) = k (10)
k = 0.0287 min-1

iii) 1/2 = ln 2


ln 2

= 0.0287

= 24.15 min

(b) i) ln 1 = ( 1 − 1 )

2 2 1

1.74 10−5 = ( 1 - 1 )
308.15 298.15
In 6.61 10−5 8.314

Ea = 101950 J mol-1

= 101.95 kJ mol-1

ii) In k

m= -Ea/R

1 / T (K-1)

From Arrhenius equation : k = A −
In k = ln A - 1



y=c+mx

Thus, by plotting graph ln k against 1/T, a straight line graph with negative
slope form. Activation energy can be determined from gradient of the graph.

Gradient, m = −



Ea = - m x R

2. (a) NaOH (aq) + HCl (aq) NaCl (aq) + H2O (l)

-qreleased = +qabsorbed

-qrxn = qsoln
= mscs∆T

= (50) (4.18) (7.0)

= 1463 J

No. of mole HCl and NaOH = MV/1000

= (1.0)(25) /1000

= 0.025 mol

Thus, mole of H2O = 0.025 mol

0.025 mol H2O = - 1463 J

1 mol H2O = - 58.52 k J
∆Hn = - 58.52 k J mol-1

∆Hf = -414 kJ

(b) i) 2Na (s) + ½ O2 (g) Na2O (s)

∆Ha= ? x 2 ∆Ha= 249 kJ

2Na (g) O (g)

IE = 496kJ x 2 EA1 = -141 kJ ∆Hlattice = -2478 kJ
O- (g)

EA2 = 750 kJ

2Na+ (g) + O2- (g)

ii) ∆Hf Na2O = (∆Ha Na x 2) + (IE Na x 2) + ∆Ha O + EA1 O + EA2 O + ∆Hlattice
-414 = (∆Ha Na x 2) + (496 x 2) + 249 -141 + 750 -2478
∆Ha Na = 107 kJmol-1

3. (a) Anode : Zn (s) Zn2+ (aq) + 2e-

Cathode : 2Ag+ (aq) + 2e- Ag (s)

Overall : Zn (s) + 2 Ag+ (aq) Zn2+ (aq) + Ag (s)

n=2

Ecell = Eocell - 0.0592 log [ 2+]
[ +]2

0.0592
1.58 = 1.56 - 2 log (0.15)2

x = 4.75 x 10-3 M

(b) i) 1 F ≡ 96500 C

0.025 F = 2412.5 C

Q = It

2412.5 = 2.00 t

t = 1206.25 s

= 20.1 min

ii) 2H2O (l) O2 + 4H+ (aq) + 4e-
1 mol O2 ≡ 4 F ≡ 4 (96500) C
Thus, 1 mol O2 ≡ 386 000 C
6.25 x 10-3 mol O2 = 2412.5 C

Volume of O2 = 6.25 x 10-3 (22.4)
= 0.14 L

4. (a) CH3 CH2 Cl CH3 CH3
(b) Cl Cl Cl

Cl Cl uv +Cl Cl
+CH3 HCl
CH3 Cl

H+

5. (a) B: 1-methylcyclohexene @ methylcyclohexene
C: toluene @ methylbenzene

(b)

CH3 O

i. KMnO4 , OH-,  OH
O
ii. H3O+ H3C COOH

B

CH3
i. KMnO4 , OH-, 

ii. H3O+

C

B undergoes oxidation that occur cleavage of double bond and produce carboxylic
acid and ketone.

C undergoes oxidation that occur on side chain of benzene since contains of
benzylic hydrogen and produce benzoic acid.

(c) Hydrogenation of alkene @ Reduction

6. (a) D: (CH3)3CBr
E: (CH3)3COH
F: CH3CH2CH2CH2Br
G: CH3CH2CH2CH2OH

(b) Reagent for formation of E: H2O
Reagent for formation of G: NaOH (aq)

(c) Step 1: Formation of Carbocation

H3C CH3 H3C CH3 +-
C Br C+ Br
CH3
CH3

Step 2: Nucleophilic attack

CH3 .. CH3
+H3C C+ :O H C O+ H
CH3 H
H H3C

CH3

Loss of H+

CH3 CH3

+H3C C O+ H .. +H3C C O H3O+
CH3 H :O H

H CH3 H

7. (a) Hydroxyl
(b)

+H3C CH2 CH2CH2 Br Mg dry ether H3C CH2 CH2CH2 MgBr

H3C CH2 CH2CH2 MgBr i. CH3CH2CHO , dry ether H3C CH2CH2CH2 CH CH2 CH3
ii. H3O+ OH

(c) J: CH3CH2COOH
Function of sulphuric acid : acts as catalyst.

8. (a) Boiling point of L < N < M
(b)
L, M and N has similar molecular weight. L has lower boiling point that N and M
since L only can form weak Van der Waals forces between its molecules. N and M
can form hydrogen bonds between its molecules. Van der Waals forces weakest
than hydrogen bonds. Thus, less energy needed to break Van der Waals forces
between L molecules.

N and M can form hydrogen bonds between its molecules but M can form stable
hydrogen bonded dimer between its molecules. Thus, more energy needed to break
stable hydrogen bonded dimer of M than hydrogen bond of N.

P: CH3CH2CH2CH=CHCH2CH2CH3

T: CH3CH2CH2CH=NOH

Q: NaBH4 in CH3OH @ i) LiAlH4 ii) H3O+

R: KMnO4, H3O+ , heat (or any oxidising agent)

S: i. CH3MgBr, dry ether

ii. H3O+