SKEMA PSPM 2 2012_2013

SK026 : ANSWER 2012/2013

1 a) Li+(g) + Cl-(g) LiCl(s) ∆H= - 846 kJmol-1

LiCl(s) Li+(aq) + Cl-(aq) ∆H= - 37 kJmol-1

Li+(g) Li+(aq) ∆H= - 510 kJmol-1

∆H1 = - 37 kJmol-1

LiCl(s) Li+(aq) + Cl-(aq)

∆H2 = 846 kJmol-1 ∆H3 = - 510 kJmol- ∆H4 4

1 1
1
Li+(g) + Cl-(g) 1
1
∆H1 = ∆H2 + ∆H3 + ∆H4 1
-37 = 846 + (-510) + ∆H4 1
∆H4 = -373 kJmol-1 1
1
b) i. Cathode : Ag+(aq) + e Ag(s)
1
Anode : Mg(s) Mg2+(aq) + 2e
1
ii. Mg(s) | Mg2+(aq, 1.0M) ‖ Ag+(aq, 1.0 M) | Ag(s) 1

iii. Ag+ 1

c) Q = It
= 2.00A x (30x60)s
= 3600 C

O2(g) + 4H+(aq) + 4e 2H2O(l)

4e = 4F = 386000 C produce 1 mol of O2
3600 C produces 3600 C 1mol = 9.33 x 10-3 mol of O2
386000 C

PV = nRT
nRT 9.33 x 10−3 mol x 0.08206 atm.L.mol−1.K −1 x 298.15 K

V= =
P (1 atm)

= 0.23 L

SK026 : ANSWER 2012/2013

2 a) hexane > 2,2-dimethylbutane > butane 1
1
• Hexane and 2,2-dimethylbutane have same number of C. Hexane has
highest boiling point than 2,2-dimethylbutane because hexane is the straight 1
chain, greater surface area and stronger Van der Waals forces.

• Butane has the lowest boiling point only 4 C , smaller surface area and
weakest Van der Waals forces.

b) i. Structural isomer of A

CH3 CH3
CH3

CH3

ii. Isomer can exist as geometrical isomer

Both isomer can exist as geometrical isomer because
- Restricted rotation due to cyclic structures
- Carbon atom bonded to two different atom or group of atoms

CH3 H HH

H CH3 CH3 CH3
trans isomer cis isomer

CH3 H CH3 CH3

H CH3 HH
trans isomer cis isomer

c) i. Structures of the products

Cl

CH3CCH3 CH3CHCH2Cl
CH3 CH3

2

SK026 : ANSWER 2012/2013 uv Cl
CH2Cl2 CH3CCH3
ii. Free radical substitution
iii. Most stable free radical formed CH3

●
CH3CCH3

CH3

iv. Reaction equations

CH3CHCH3 + Cl2
CH3

3 a) i. 1-butanol is less acidic than phenol because the electrons on oxygen
atom are localized.

ii. CH3CH(OH)COOH has higher boiling point than CH3COCOOH
because can form more hydrogen bond

iii. 2-methyl-2-propanol cannot undergo oxidation because tertiary alcohol
and no hydrogen bonded to C that attached to OH group.

iv. 2-hexanol gives yellow precipitate with iodine in NaOH solution
because it contain methyl alcohol group –CH(OH)CH3.

v. Both phenol and 1-propanol react with sodium but only phenol
dissolves in aqueous NaOH because

b) B MgBr
C
D CH3
CCH3
OH

CN

E CH2NH2

F COOH

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SK026 : ANSWER 2012/2013

c) i. Oxidation test or Lucas test

ii. Observations for oxidation test using K2Cr2O7

Alcohol G
Orange solutions of K2Cr2O7 turn to green.

Acohol H
Orange solutions of K2Cr2O7 remain.

iii. Chemical equation for alcohol G

CH3CHCH2OH i. K2Cr2O7 CH3CHCOOH
CH3 ii. H3O+, ∆ CH3
ZnCl2

O H3O+ CH3CHCH2OH
iv. CH3

CH3CH-MgBr + H-C-H

CH3

4 a) i. Structure for J and K

J CH2 CHCH3
OH

SO3- Na+
K CH2CCH3

OH

ii. Nucleophile involved in the formation of J

iii. Chemical test to differentiate between I and J
Oxidation test, Lucas test or Brady’s Test,

Compound I

O i. K2Cr2O7 no reaction
CH2 CCH3 ii. H3O+, ∆

ZnCl2

Observation : Orange solutions of K2Cr2O7 remain.

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SK026 : ANSWER 2012/2013

Compound J i. K2Cr2O7 O
ii. H3O+, ∆ CH2 CCH3
CH2 CHCH3 ZnCl2
OH

Observation : Orange solutions of K2Cr2O7 turn to green.

iv. I can be synthesized from alkyl halide

CH2 CHCH3 OH- CH2 CHCH3
Br OH

i. K2Cr2O7
ii. H3O+, ∆
ZnCl2
O
CH2 CCH3

b) 1-butanol < butanoic acid < 4-chlorobutanoic acid < 2-chlorobutanoic acid <
2,3-dichlorobutanoic acid

1-butanol has the lowest acidity because the electrons on oxygen atom are
localized.

Butanoic acid more acidic than 1-butanol because the electrons can delocalized
between 2 oxygen atoms.

4-chlorobutanoic acid and 2-chlorobutanoic acid more acidic than butanoic
acid because present of electron withdrawing group (EWG). 2-chlorobutanoic
more acidic than 4-chlorobutanoic the distance of EWG near to carboxyl
group.

2,3-dichlorobutanoic acid has the highest acidity because present more EWG
group.

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SK026 : ANSWER 2012/2013

5 a) Let rate = k [L]m[M]n

rate 2 = 0.008 = k (0.002)m (0.004)n
rate 1 0.002 k (0.001)m (0.004)n

4 = 2m
m =2
The order of reaction respect to L is 2.

rate 4 = 0.016 = k (0.004)m (0.002)n
rate 3 0.008 k (0.004)m (0.001)n

2 = 2n
n =1
The order of reaction respect to M is 1.

So, rate = k [X]2 [Y]

From exp1: 0.002
k=

(0.001)2 (0.004 )

k = 2.20 x 10-4 M-2 min-1 or
5 x 105 M-2 s-1

b) Effect of temperature and activation energy, Ea on the rate constant,k of a
reaction using Arrhenius equation.

k = Ae-Ea/RT or ln k = ln A− Ea 1
R T 

Increasing the temperature causes the term e-Ea/RT to be larger thus increasing
the value of k.

Decreasing the value of activation energy, Ea causes the term e-Ea/RT to be
larger thus increasing the value of k.

Effect of temperature on reaction rate using Maxwell-Boltzmann distribution
curve.

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SK026 : ANSWER 2012/2013

Fraction of collisions Ea
Low T High T

Collision energy
When temperature increase, the kinetic energy of molecules increases.
The number of molecules having minimum energy equal or greater than
activation energy, Ea increases.
The number of effective collisions also increases.
The rate of reaction increases.

6 P, C4H8 – alkene or cycloalkane
Q, R, S react with bromine in CH2Cl2 to produces alkyl halide – alkene
T, U do not react with bromine in CH2Cl2 – cycloalkane
Boiling point increase I order : S < Q < R
S CH2=C(CH3)CH3
Q CH2= CHCH2CH3
R CH3CH=CHCH3
T CH3

U
Isomers that give optically active product when react with HCl

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SK026 : ANSWER 2012/2013

Q CH2= CHCH2CH3

R CH3CH=CHCH3

Chemical equation and mechanism involved

Cl

+H2C CHCH2 CH3 HCl CH3 CHCH2 CH3

Cl

+CH3 CH CHCH3 HCl CH3 CHCH2 CH3

+CH3 CH CHCH3 H Cl CH3 C+ CH2 CH3
H
Cl-

Cl

CH3 CCH2 CH3
H

S has the lowest boiling point because
✓ S is a branched alkene and has lower surface area so weaker van der Waals
forces.

Chemical test: KMnO 4, OH - OH
H2C CHCH2 CH3 cold
+CH2 CHCH2 CH3 MnO 2

OH

The purple colour of KMnO4 decolourised and brown precipitate formed.

CH3 KMnO 4, OH - no reaction
cold

No observable changes of KMnO4.

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SK026 : ANSWER 2012/2013

7 a) The aromaticity of cyclopenta−1,3−diene anion.
✓ Cyclopenta−1,3−diene anion has 6  electrons and follow Huckel’s

rule with n=1.
✓ All its carbon atoms are sp2 hybridised and the  electrons are

delocalised in the unhybridised o prbitals.

Resonance structures:
-

-

-

Benzene will undergo electrophilic aromatic substitution with nitric acid,
HNO3, in presence of H2SO4.

Mechanism:

Step 1 : Formation of electrophile

+HNO 3 H2SO4 + +NO + -
2 HSO 4 H2O

Step 2: Formation of arenium ion NO2 NO2
+ H H

+ +
+ NO2+
NO2
H
+

Step 3: Loss of H+

+ NO2 NO2
H + H3O+

H2O

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SK026 : ANSWER 2012/2013

b) SN : nucleophilic substitution

RX + Nu- RNu + X-

Example : CH3Cl + OH- CH3OH + Cl-

Halogen will substituted by Nu

Alcohol W becomes cloudy immediately when reacted with a Lucas reagent.
W is 3o alcohol.

Br OH

CH3 CH2CCH3 CH3 CH2CCH3

CH3 CH3
V W

Mechanism for the conversion of V to W : SN1 mechanism

Br slow +CH3 C+ CH2 CH3 Br -
CH3 CCH2 CH3
CH3
CH3

CH2 CH3 f ast CH2 CH3
CH3 C O+ H
CH3 C+ + HO H
CH3 H
CH3

CH2 CH3 CH2 CH3
CH3 C OH
+CH3 C O+ H HO H
CH3
CH3 H

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SK026 : ANSWER 2012/2013

8 a) Aniline < N,N-dimethylethanamine < propylamine < N-methylethanamine

Aniline is least basic due to the lone pair on N delocalise into the benzene ring.
Therefore, less available to accept proton.

N,N-dimethylethanamine is 3o amine, less basic than propylamine and
N-methylethanamine due steric effect and difficult for N to accept proton.

N-methylethanamine is most basic due to presence of methyl group, increase
availability of electrons on N to accept proton.

Amine that can be prepared through Hoffmann’s degradation.

Aniline and propylamine.

Equation: NH2
CONH 2

Br 2
NaOH(aq)

CH3 CHCONH 2 Br 2 CH3 CHNH2
CH3 NaOH(aq) CH3

Chemical test:

Hinsberg’s test or nitrous acid

CH3 CHNH2 HNO 2 H2C + + +CH CH3CH OH CH3CH Cl N2
CH3 0 - 5 oC
CH3 CH3 CH3

Bubble gas released.

CH3 CH2 NH HNO 2 CH3 CH2 N N O
CH3 0 - 5 oC CH3

Yellow oil formed.

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SK026 : ANSWER 2012/2013

b) Peptide bond. CH2 CH2COOH
Structure:

NH2 CH C NH CHCOOH
O CH2 CH2COOH

Zwitterion: O
H3N+CH C O-

CH2 CH2COOH

It is acidic due to one free carboxyl group.

c) O HO CH3 OH
Cl C Cl C

X CH3
Y

Condensation polymerisation.
Usage: bullet-proof windows, crash helmets
CH3COOCOOCOCH3

OO O
CH3C O C O C CH3

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