SKEMA PSPM 2 2014_2015

SK026 : ANSWER 2014/2015

1 (a) (i) Lattice energy
(a) (ii) Lattice energy is the energy released when 1 mole of ionic compound is
formed from its constituent gaseous ions.
Enthalpy

Na+ (g) + Br (g) EA = −324 kJ
∆Ha = + 97 kJ Na+ (g) + Br- (g)

Na+ (g) + ½ Br2 (g)
IE = +496 kJ

Na (g) + ½ Br2 (g) ∆Hlattice = ?
∆Hsub = + 107 kJ

0 Na (s) + ½ Br2 (g)
∆Hf = −360 kJ
NaBr (s)

∆Hf = 107 + 496 + 97 + (−324) + ∆Hlattice
∆Hlattice = −736 kJ mol-1

(b) (i) Zn(s)  Zn2+(aq)  Cu2+(aq)  Cu(s)

anode : Zn(s) Zn2+(aq) + 2e
cathode : Cu2+(aq) + 2e Cu(s)

overall : Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
= 1.10 V
(b) (ii) Eocell = Eocathode − Eoanode
(b) (iii) = 0.34 − (−0.76)

Ecell = Eocell − 0.0592 log [Zn2+]
n [Cu2+]

1.17 = 1.10 − 0.0592 log [Zn2+]

2 4.95

0.07 = −0.0296 log [Zn2+]

4.95

[Zn2+]
log 4.95 = −2.365

[Zn2+] = antilog (−2.365) = 4.315 x 10-3

4.95

[Zn2+] = 0.02136 M

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SK026 : ANSWER 2014/2015

2 (a) 2,2,4,4−tetramethylpentane < 2,2−dimethylheptane < 4−ethylheptane <
nonane

Nonane – straight chain / unbranched molecules
− larger surface area, the strength of Van der Walls forces increases
− highest boiling point

2,2,4,4−tetramethylpentane – branches increase / many substituent
− smaller surface area / molecule more compact, the strength

of Van der Waals decreases
− lower boiling point

(b) (i) Br

(b) +CH3CH2 CH3 Br2 +uv CH3CH2 CH2 Br
(ii) @ CH3 CH CH3
Br
(b) (iii) 2−bromopropane
CH3 CHCH3
Initiation step:

Br Br uv 2Br

Propagation step: H
H CH3 C

+CH3C H Br + H Br

CH3 CH3
H
H + Br Br +Br Br
CH3 C CH3 C

CH3 CH3

Termination step:

+Br Br Br2

H + H HH
CH3 C C CH3 CH3 C C CH3
CH3
CH3 CH3 CH3

H + Br H
CH3 C CH3 C Br

CH3 CH3

2

SK026 : ANSWER 2014/2015

3 (a) (i) Relative rate of SN2 reaction is higher for primary alkyl halide compared to
tertiary alkyl halide because 1o is less hindered @ 3o is more hindered or vice

versa.

(a) (ii) 2−bromobutane @ Br

CH3 CHCH2CH3
(a) Zero @ less than 0.03

(iii)

Tertiary alkyl halide does not favor SN2 due to bulkiness (steric hindrance)

(a) (iv) C : CH3CH2CN
D : CH3CH2NH2

E : CH3CH2OCH3
(b) (i) F : 3−methyl−3−pentanol

(b) Alkene :

(ii) CH3 CH2 C CH CH3 CH3 CH2 CH CH CH2 CH3 CH2 C CH2CH3

or or

CH3 CH3 CH2

3-methyl-2-pentene 3-methyl-1-pentene 2-ethyl-1-butene

Reagent : H2O, H+ @ H2O, H3O+ @ H2O, H2SO4
(b) (iii) Lucas test

1−pentanol : no observable change
2−pentanol : cloudy solution formed within 5−10 minutes

F : cloudy solution formed immediately

3

SK026 : ANSWER 2014/2015

4 (a) (i) i. LiAlH4 @ NaBH4, CH3OH @ H2 / Pd @ Pt @ Ni @

ii. H3O+
Pd−C

(a) (ii) @ Na2Cr2O7 , H+ , @ H2CrO4 , H+ ,
KMnO4 , H+ ,

(b) OH

MgBr CH3 C CH3

O i. dry ether
ii. H 2O
+

conc. H2SO4
heat

CH3 C CH2 CH3 C CH3 CH3 CH CH3
@ @

H2 / Pt
CH3 CH CH3

…or…

4

SK026 : ANSWER 2014/2015

O i. LiAlH 4 OH
ii. H 3O+ H3C C H
H3C CH3
CH3

PCl 3 @ PCl 5 @ SOCl 2 @ PBr 3

MgCl Mg Cl
H3C C H dry ether H3C C H

CH3 CH3

O HO CH(CH 3)2

+ i. dry ether
ii. H 2O

conc. H2SO4
heat

CH3 CH CH3 CH3 C CH3 CH3 CH CH3
@
H2 / Pt

(c) Methanoic acid has carboxyl and carbonyl functional group thus
it shows reducing properties in reaction with Tollen’s reagent.

Observation: silver mirror formed / grey precipitate

O [Ag(NH 3)2]+ OH - Ag
HC OH

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SK026 : ANSWER 2014/2015

5 (a) KI : First order HCl : Zero order Overall : Second order reaction

G = 0.2 mol dm-3
H = 0.1 mol dm-3
J = 4.0 x 10-4 mol dm-3 s-1

1.0 x 10-4 = k (0.1) (0.1)
k = 0.01 dm3 mol-1 s-1

Yes, the value of k will increase.

(b) ln  k1  = Ea  1 − 1 
k2 R T2 T1

ln  0.0234 = Ea  1 − 1 
 0.750  8.314  773 673

Ea = 150 kJ mol-1

energy

Activated complex

Ea = +150 kJ mol-1

N2O (g) + NO (g) 1
1
∆H = −110 kJ mol-1 1
N2 (g) + NO2 (g) 1
1

progress of reaction

∆H = Ea forward – Ea reverse

Ea for the reverse reaction = Ea forward – ∆H
= 150 – (−110)

= 260 kJ mol-1

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SK026 : ANSWER 2014/2015

6 Isomers of 2o alkyl halide : 2-bromopentane
CH3 CH2 CH2CH CH3 3-bromopentane
Br 2-bromo-3-methylbutane
CH3 CH2 CH CH2CH3
Br

CH3
CH3 CHCH CH3

Br

Products of dehydrohalogenation reaction :
CH3

K : CH3 C CH CH3

CH3
L : CH3 CHCH CH2

M : CH3 CH2 CH CH CH3 can be interchangeable
N : CH3 CH2 CH2CH CH2

Product exists as geometrical isomer : M

Reason : M has 2 different groups of atoms at each C=C

Ozonolysis of K : i. O 3 CH3
CH3 ii. Zn / H 2O
+CH3C O O CH CH3
CH3 C CH CH3 i. O 3
(K) ii. (CH 3)2S

@

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SK026 : ANSWER 2014/2015

CH3 HBr
CH3 CHCH CH2

(L)

CH3 CH3 CH3

CH3 C CH2CH3 + +CH3 CHCH CH3 CH3 CHCH2CH2

Br Br Br

(P) (Q) (R)

2-bromo-2-methylbutane 2-bromo-3-methylbutane 1-bromo-3-methylbutane

P / Q / R : are interchangeable

Mechanism : H Br CH3

CH3 CH3 C CH CH3
CH3 CHCH CH2 H +

(L) rearrangement

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

3o carbocation

Product of reaction with HBr / H2O2 :
CH3

CH3 CHCH2CH2 Br

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SK026 : ANSWER 2014/2015

7 (a) V: COOH
S: OH
Cl
T: W:

U:
Cl

Name of reaction : Friedel-Crafts Alkylation

Mechanism : AlBr 3 + + AlBr -
Br 4

+

+ +

+ + +

H
+

AlBr -
4

(b) 2-bromo-2-methylpropane > 2-bromopropane > 1-bromopropane

2-bromo-2-methylpropane is a tertiary alkyl halide which will form the most
stable tertiary carbocation.

CH3

+CH3C Br CH 3OH +(CH3)3COCH 3 HBr

CH3

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SK026 : ANSWER 2014/2015

Mechanism : CH 3OH

CH3 slow + + Br-
CH3 C Br (CH 3)3C
3o carbocation
CH3

f ast

+ +(CH3)3COCH 3 HBr
(CH3)3COCH 3

H Br-

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SK026 : ANSWER 2014/2015

8 (a) NO 2 NH 2

HNO 3 Sn
H2SO4 H+

NH 2 +

HNO 2 NN
-5 oC OH

NN OH

Cyclohexanamine is more basic.

Cyclohexyl group is EDG, so it increases electron density which is localised at
the
N atom hence increasing basicity.

Phenyl group is EWG with the resonance effect which delocalised electron
density at the N atom hence decreasing basicity.

(b) H
H2N C COO -

CH3
Isoelectric point, pI :
pH at which the amino acid has no net charge.

H
X : H3N+ C COOH

CH3
H

Y : H2N C COOCH 2CH 3

CH3

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SK026 : ANSWER 2014/2015

(c) Monomers :

O

H3C C C OH

O
1,4-benzenedicarboxylic acid

O

Cl C C Cl

O
1,4-benzenedicarbonyl chloride

H2N NH2

1,4-diaminobenzene

Condensation polymerisation.

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