SKEMA PSPM 2 2015_2016

SK026 : ANSWER 2015/2016

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

Heat released by reaction, qrxn = − (qwater + qbomb)

qrxn = − (qwater + qbomb)
= (mc∆T + C∆T)
= (780 g x 4.18 J g-1 oC-1 x 11.1 oC) + (870 J oC-1 x 11.1 oC)
= 45847 J
= 45.9 kJ

(a)(ii) Mass of 1 mol C7H6O2 = 7(12) + 6(1) + 2(16) = 122 g

2.34 g released 45.9 kJ = 2393 kJ mol-1
122 g released 45.9 kJ x 122 g

2.34 g

Enthalpy of combustion of C7H6O2 = −2393 kJ mol-1

(b)(i) Anode : Zn(s) Zn2+(aq) + 2e-
(b)(ii) Cathode : Fe2+(aq) + 2e Fe(s)

Eocell = Eocathode − Eoanode
= −0.44 − (−0.76)

= + 0.32 V

Nernst equation:

Ecell = Eocell - RT ln [Zn2+]
nF [Fe2+]

(8.314)(293.15) 0.1
= 0.32 − 2 (96500) ln 0.5
= + 0.34 V

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SK026 : ANSWER 2015/2016

2 (a)(i) (CH3)2CBrCH2CH3
(a)(ii) Free radical substitution

(a)(iii) Initiation step:

Br Br uv 2Br

Propagation step:

H + Br + H Br

+ Br Br +Br Br

Termination step: Br2

+Br Br

+

+ Br Br

(b)(i) CH2 CH3 CH2 CH3
CH3 CHCH2 CH2 CH3 CH3 C CH2 CH2 CH3

(B) Br
(C)

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SK026 : ANSWER 2015/2016

CH2 CH3 CH2 CH3
CH2 C CH2 CH2 CH3 CH2 C CH2 CH2 CH3
Br Br
Br OH
(D) (E)

(b)(ii) CH2 CH3 CH2 CH3
CH2 CHCH2 CH2 CH3 or CH3 C CH2 CH2 CH3
Br
Br

CH2 CH3 Br CHCH3
CH3 CHCHCH2 CH3 or CH3 CHCH2 CH2 CH3

Br

Reagent:
EtOH / Ethanolic acid KOH @ Ethanolic NaOH @ NaH

(b)(iii) Baeyer’s test

CH2 CH3
CH2 CCH2 CH2 CH3
OH OH

The decolourised of purple colour @ the apperance of a brown precipitate.

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SK026 : ANSWER 2015/2016

3 (a)(i) CH3CH2CH2NH2
(a)(ii) CH3 CH2 CH2 O CH3

(a)(iii) CH3CH2CH2CN
(a)(iv) CH3CH2CH2OH

(b) SN2

CH 2CH 3 slow CH 2CH 3

+ H C Br CN.... C.... Br

CN - H HH

transition state

CH 2CH 3

NC C H
H

(c)(i) G : CH3CH2CH2CH2Cl
H : CH3CH2CH2CH2MgCl

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

(c)(iii) i. HBr, H2O2
ii. NaOH

(c)(iv)

CN- CN
Cl H3O+

COOH

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SK026 : ANSWER 2015/2016

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

(a)(ii) Cl Cl
CH3 CHCH2 COOH
is more acidic than

CH3 CH2 CHCOOH

Reason:
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 CHCOO -K+
(b)(i)
O CH2 CH2 CH3
HO O CHI 3
C
C
CH3 O- K+

(N) (P) (Q)

(b)(ii) R : CH3COCl, AlCl3

S : LiAlH4 T : H3O+ @ H2O / H+

or S : NaBH4 T : CH3OH

(b)(iii) Brady’s test
Observation : M : Yellowish orange precipitate formed
N : No yellowish orange precipitate formed

NO 2 NO 2 NO 2 NO 2
R R

H H
NN
+O H2N N
R
R

or Lucas test
Observation : M : No observable change

N : Cloudy solution formed

CHCH3 ZnCl 2 / HCl CHCH3
OH Cl

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SK026 : ANSWER 2015/2016

5 (a) Plot a graph ln[H2O2]t versus time

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

ln [H2O2]

time / s

*straight line *4 point on the line *slope

Straight line graph, y = mx + c
First order reaction for the decomposition of H2O2.

= − = 2 − 1
2 − 1
= −7.30 10−4

, 1⁄2 = ln 2
=
=
ln 2

7.30 10−4 −1
949.5 s

ln [H2O2]t = − k t + ln [H2O2]o
ln (2.32) − ln [H2O2]t = (7.30 x 10-4) (1600)
= e -0.326
[H2O2]t =
0.72 M

% decomposition = 2.32 – 0.72 x 100 = 68.9%
2.32

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SK026 : ANSWER 2015/2016

(b) Number of collisions

T1 (K) T1 < T2 Shape
T2 (K) Axis
Label (start origin)
Ea

Ea Kinetic energy
As temperature increases:

More molecules move at higher speeds / kinetic energy of molecules increases.

More molecules have energy  Ea

The number of effective collsion increases, thus reaction rate increases.

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SK026 : ANSWER 2015/2016

6 W : H2C CHCH2CH2CH3
2−pentanol V : CH3CH CHCH2CH3

Major product, V follow Saytzeff’s rule

Geometrical isomer : H H H3C H
C C
CC

H3C CH2 CH3 H CH2 CH3
trans-isomer
cis-isomer

OH +OH 2

+

H

protonated alcohol

+H

Z : CH3CH2CHO H base-
carbocation H
AA : CH3CHO

i. O 3 O
ii. (CH 3)2S
+O

H

Cl Mg MgCl
dry ether

i. HCH3O3C+HO
ii.

OH

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SK026 : ANSWER 2015/2016

7 (a) Electrophilic aromatic substitution

O O
C O CH2 CH3
CH2 CH3 C OH

(BB) (CC) (DD)
Formation of BB: CH2 CH3
CH 3CH 2Cl
Formation of DD: AlCl 3 (BB)
COOH
O

CH 3CH 2Cl +C OCH2 CH3 H2O
H+

(DD)

Mechanism:
Step 1

CH3 CH2 +Cl AlCl 3 +CH3 CH+2 AlCl -
4

+

+ CH3 CH+2 + +

H
+

AlCl -
4

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SK026 : ANSWER 2015/2016

7 (b) CH3 O
CH3 CH2CH O CCH3
+alkyl bromide Nu

(EE)

2o alkyl halide O O-
CH3 CH3 C

CH3CH2CH Br

CH3
CH3 CH2C O CH2 CH3

H
(FF)

is an ether yield from the reaction between alkyl halide and alcohol

HH

CC

H3C CH3

(GG)
Saytzeff’s rule: more highly substituted product

CH3 O CH3 O
CH3 CH2CH Br CH3 CH2C O CCH3
+ CH3 C O-
H
@
CH3 COO - Na+

CH3 CH3
CH3 CH2C O CH2 CH3
+CH3 CH2CH Br CH3 CH2 OH
H

CH3 HH
CH3 CH2CH Br
KOH CC
ethanol

H3C CH3

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SK026 : ANSWER 2015/2016

8 (a) R2NH more basic than RNH2.

Presence of electron donating substituent, R,
increase the inductive effect,
increasing basicity.

Aniline, aromatic amine less basic compare to aliphatic amine.

Electron delocalised into the benzene ring of aniline. Thus, reducing the
basicity.

Structure @ name : aliphatic amine / aromatic amine 1+1
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NH2 NaNO 2 / HCl NH+2 Cl _
0 - 5 oC OH

NaOH,
0 oC

NN OH
p-(phenylazo)phenol

(b) NH2CH2COOH
2−aminoethanoic acid

CH3

H2N COOH H COOH
2−aminopropanoic acid N COOH

H2N O
H
H2N N

O

SK026 : ANSWER 2015/2016

H
N COOH
H2N

O
peptide bond

H COOH
N
H2N

O

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

HH
CC
H Cl

n

Copolymer
a polymer from the joined of two different types of monomer

OO n
HH
N (CH2)6 N C (CH2)4 C

Nylon 6,6

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