SKEMA PSPM 2 2011_2012

SK026 : ANSWER 2011/2012

1. a) i. Energy released when 1 mole of substance is completely burnt in
oxygen at standard conditions ( 298 K @ 25⁰C and 1 atm pressure).

ii. Mol of C6H5COOH = 0.0262 mol

1 mole of C6H5COOH released 3226.7 kJ

0.0262 mole of C6H5COOH released 84.631 kJ

qrxn = mwcw∆T + Ccal∆T
= (2.0 kg x 4.18 kJkg-1 ⁰C-1 x 3.8⁰C) + (Ccal x 3.8⁰C )

Ccal = 13.9 kJ ⁰C-1

b) i. Oxidation (anode) : Mg (s) Mg2+ (aq) + 2e
Reduction (cathode) : Cd2+(aq) + 2e Cd(s)

ii. E°cell = E°cathode – E°anode
= -0.40 V – (-2.37 V)

= +1.97 V

iii. Ecell = E°cell - 0.0592 log [Mg 2+ ]
n [Cd 2+ ]

1.54 V = +1.97 V - 0.0592 (1.00M )
2 log [Cd 2+ ]

[Cd2+] = 2.97 x 10-15 M

iv. Voltage will be reduced

2. a) i. A = Br2 , hv @ Cl2, hv

B = CH3CH(Br)CH2CH3 @ CH3CH(Cl)CH2CH3

C = NaOH (aq) @ KOH(aq)

ii. X2 =Br2 @ Cl2

Initiation hv 2X●
XX

Propagation

H

CH3CH2CHCH3 + ●X ●
CH3CH2CHCH3 + HX

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SK026 : ANSWER 2011/2012

● X + ●X
CH3CH2CHCH3 + X X CH3CH2CHCH3

Termination ●X X
● CH3CH2CHCH3

CH3CH2CHCH3 +

X● + ●X X2 CH3
CH3CH2CHCHCH2CH3
●●
CH3CH2CHCH3 + CH3CH2CHCH3

CH3

b) i. CH3CH2CH2CH=CH2 + CH3 CH2CH=CHCH3

ii. Saytzeff Rule Cl
iii. CH3CH2CH2CH=CH2 + Cl2 CH3CH2CH2CHCH2Cl

CH3CH2CH2CH=CH2 + H2O H+ OH
CH3CH2CH2CHCH3
+
CH3CH2CH2CH=CH2 + Br2 H2O OH
CH3CH2CH2Cl HCH2Br

3. a) E = CH3CH2CH2Br +

F = CH3CH2COOH

G = CH3CH2CH2O-Na+

H = CH3CH2CH2OCH3

b) Na2Cr2O7 / H+ or KMnO4 / H+

c) Lucas test

Compounds D conc. HCl no observable change
CH3CH2CH2OH ZnCl2

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SK026 : ANSWER 2011/2012

2-methyl-2-butanol

OH conc. HCl Cl
CH3CH2CCH3 ZnCl2 CH3CH2CCH3

CH3 CH3

Observation : Cloudy solution formed immediately

*Other tests
Oxidation test using acidified KMnO4 or Na2Cr2O7

d) Bimolecular nucleophilic substitution @ SN2

e) HH

H H3CO CH Br
H
CH 2CH 3
C Br Transition state

CH 3O- H3CH 2C

H

H3CO C
H

CH 2CH 3

f) Ether
g) (CH3CH2COO)2Ca
4. a) i. I = CH3CH2CH2CH2CN

J = CH3CH2CH2CH2COOH
ii. pentanoic acid
b) i. CH3CH2CH2CH2CONHCH3
ii. CH3CH2CH2CH2CH2OH
iii. CH3CH2CH2CH2COCl

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SK026 : ANSWER 2011/2012

iv. CH3CH2CH2CH2COOCOCH3 @
OO

CH3CH2CH2CH2C O CCH3
c) i. NaBH4, H3O+ or

i) LiAlH4 ii) H3O+ or
H2, Ni/Pt/Pd

ii. KCN / H+ or NaCN / H+ or CN- / H+

iii. CH3COCl, AlCl3

d) Benzoic acid, 4-nitrophenol, phenol, cyclohexanol
decending order acidity

C6H5 COO- is the most stable since the negative charge is delocalized between
two electronegative oxygen atoms.

O2N-C6H5 CO- is less stable than C6H5 COO- since negative charge is
dispersed on one oxygen atom and less electronegative carbon atoms, and
dispersed farther by electron withdrawing group of -NO2.

C6H5 CO- is less stable than O2N-C6H5 CO- since it has no electron
withdrawing group, -NO2.

C6H11O- is the least stable since the negative charge is localized on one oxygen
atom only.

5. a) Order of reaction is first order.

The unit of rate constant is year-1 which is characteristic of first order reaction.
@ Decomposition of radioactive is first order reaction.

ln [K ] = kt
[K ]t

ln (3.5x10−6 ) = (3.17year−1)(0.5year)
[ K ]t
[K ]t = 7.174 x10 −7 molL−1

Concentration of element K after 6 months = 7.17 x 10-7 molL-1

ln [K ] = kt
[K ]t

3.5x10 −6 = (3.17 year −1 )t
ln 1.75 x10 −6

t = 0.22year

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SK026 : ANSWER 2011/2012

Time taken for concentration reduce to 1.75 x 10-6 mol L-1 = 0.22 year
b) Activation energy is minimum energy required to initiate a chemical reaction.

Energy profile diagram
Energy

Ea = 165 kJmol-1

H2(g) + I2(g)
∆H = -9.48 kJmol-1

2HI (g)

Reaction progress

Ea for reverse reaction = 165 kJmol-1 + 9.48 kJmol-1
= 174.48 kJmol-1

ln k = ln A − Ea
RT

ln 138 = ln A − 165 x10 3
8.314 x973

A = 9.956x1010

ln k = ln 9.956 x1010 − 174 .48 x10 3
8.314 x973

k = 42.75Lmol−1s−1

6. Combustion of 2-methyl-pentane

CH3CHCH2CH2CH3 + 19/2O2 6CO2 + 7H2O
CH3

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SK026 : ANSWER 2011/2012

Dehydrohalogenation of L

Br KOH CH3C=CHCH2CH3 + CH2=CCH2CH2CH3
CH3CH2OH CH3 CH3
CH3CCH2CH2CH3 M N
CH3
L

Oxidative cleavage of M

CH3C=CHCH2CH3 KMnO4/H+ CH3C=O + CH3CH2COOH
CH3 ∆ Q
CH3
Structure S & R P
OH
O=CCH2CH2CH3
CH3CCH2CH2CH3
CH3 CH3
S R

Mechanism formation of S

CH3C=CHCH2CH3 + + +
CH3 H ӦH CH3 CCH2CH2CH3

H CH3

H3O+ OH :ӦH2 :ӦH2
+ CH3 CCH2CH2CH3
+
CH3 HӦH
CH3 CCH2CH2CH3

CH3

Synthesis of Q from ethane NaCN CH3CH2CN
CH3CH3 Br2/CH2Cl CH3CH2Br H3O+
2 light

CH3CH2COOH

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SK026 : ANSWER 2011/2012

7. CH2CH3 U: Cl
T: CHCH3

V: CH2CH2Cl W: CH2COOH

X : CH(OH)CH3

AA : CrO3/ H+ @ K2Cr2O7/ H+ @ KMnO4/H+
BB : i) excess I2 , NaOH ii) H3O+
Name of the reaction for the formation of T – Friedel-Crafts alkylation

Step 1 : Formation of electrophile

δ+ δ-- +
CH3CH2 + AlCl4-
CH3CH2Cl + AlCl3 CH3CH2 Cl AlCl3

Step 2: Formation of arenium ion + CH2CH3 CH2CH3
+ H
H
+ CH2CH3 + CH2CH3
H
+

Step 3: Loss of H+ CH2CH3
+ CH2CH3 + HCl + AlCl3
H AlCl4-

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SK026 : ANSWER 2011/2012

Hydrolysis of U – SN1 mechanism
Nucleophile Z – CH3CH2O-

Step 1 :

Cl slow +
CHCH3 CHCH3 + Cl-

Step 2 +
CH ӦH
+CHCH3 + :Ӧ H fast CH3 H
H
:Ӧ H
-H+ H

CH OH
CH3

Enantiomer of Y

C6H5 H5C6

C H3CH 2CO C
H OCH 2CH 3 H

CH3 CH3

8. a) CC : 2-amino-3-phenylpropanoic acid

NH3+

Zwitterion of CC CH2CHCOO-

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SK026 : ANSWER 2011/2012

DD : NH3+Cl-
CH2CHCOOH

Cl

CH2CHCOOH

EE : @
OH

CH2CHCOOH

CH=CHCOOH
@

Peptide bond
OH

H2NCHC NCHCOOH
FF : CH2 CH2

b) O O
HOOC C Cl
COOH Cl C

@
OO

H3CO C C OCH3

HO CH2CH2 OH

Polymerisation process - condensation polymerisation

Terelyne is used as fibres in textiles or film / indoor slippers / sleeping bags /
drinking bottles / fibre in pillow.

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SK026 : ANSWER 2011/2012

c) CH3 CH3
NH2
NH2 H3C NH2
1º amine

NHCH3
2º amine
Use HNO2 to differentiate primary and secondary amines.

CH3 NaNO2, HCl CH3
NH2 0-5ºC N2+Cl-

Observation : clear solution and no bubble gas released

NHCH3 NaNO2, HCl NO
0-5ºC N CH3

Observation : yellow oily liquid

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