Topic
8
Chemistry of Carbon Compounds
Part A Unit-based exercise
Unit 29 An introduction to the chemistry of carbon compounds
Fill in the blanks
1
carbon-carbon double bond 2
hydroxyl
3
carbonyl
4
carboxyl
5
O
C
6
O
C
O
N
7
permanent dipole-permanent dipole attractions
8
hydrogen bonds
True or false
9
F
10 T
11 F
12 F 13 T
14 F
15 T
16 T
17 T
18 T
Multiple choice questions
19 B
20 A
21 A
22 D
23 B
24 C
25 C
26 D
27 D
28 D
29 A
30 C
31 A
32 D
33 C
34 C
35 B
36 D
37 B
38 C
39 A
40 B
41 D
42 A
43 B
44 D
45 A
46 C
47 D
48 C
49 C
50 B
51 D
52 A
53 B
54 B
55 D
56 C
57 D
58 C
59 C
60 C
61 D
62 A
63 C
64 B
65 B
66 C
67 A
68 B
69 B
70 D
71 B
72 D
73 C
74 B
75 A
76 D
Unit 30 Isomerism
Fill in the blanks
1
structural
2
stereoisomers
3
Enantiomers
4
chiral
5
opposite; optically active
True or false
6
T
7
F
8
F
9
T
10 T
11 F
12 T
13 F
14 T
15 F
Multiple choice questions
16 B
17 D
18 B
19 B
20 C
21 D
22 A
23 D
28 C
25 B
26 C
27 B
28 C
29 C
30 D
31 B
32 A
33 B
34 B
35 B
36 B
37 B
38 A
39 D
40 B
41 C
42 C
43 B
44 A
45 A
46 D
47 B
48 B
49 C
50 D
51 B
52 A
53 B
54 A
55 B
56 B
57 C
58 C
59 B
60 B
61 A
62 A
63 C
64 B
65 C
66 D
67 C
68 B
69 D
70 A
71 D
72 D
73 C
74 C
Unit 31 Typical reactions of selected functional groups
Fill in the blanks
1
addition / hydrogenation 2
substitution / hydrolysis 3
substitution 4
dehydration / elimination 5
oxidation 6
reduction 7
esterification / condensation 8
hydrolysis
True or false
9
T
10 F
11 F
12 F
13 T
14 T
15 T
Multiple choice questions
16 C
17 C
18 B
19 D
20 D
21 C
22 C
23 A
24 C
25 B
26 D The reaction mixture is heated in a flask connected to a reflux condenser. In the reflux condenser, all the
vapour is condensed back into the flask. This prevents any loss of the reaction mixture.
27 B
28 C
29 C
30 D
31 A
32 A
33 D
34 C Option A — The acidified potassium dichromate solution acts as an oxidizing agent.
Option B — Propanoic acid is the final product.
Option C — Propan-1-ol is first oxidized to propanal and then to propanoic acid.
Option D — The reaction mixture changes from orange to green as the dichromate ions are reduced to
chromium(III) ions.
35 C
36 B
37 C
38 B
39 C
40 A
41 B
42 C
43 A
44 D
45 A
46 D
47 C
48 A
49 C
50 C
51 D The reaction mixture is heated in a flask connected to a reflux condenser. In the reflux condenser, all the
vapour is condensed back into the flask. This prevents any loss of the reaction mixture and favours the
oxidation of ethanol to propanoic acid rather than to propanal.
52 D
53 B
54 D Ethanol in the wine is oxidized to ethanoic acid. The acid makes the wine tastes sour.
55 B
56 C
57 C
58 D
59 B
60 C
61 A
62 C
63 D
64 A
65 B
66 A An ester is formed when a carboxylic acid reacts with an alkanol. The reaction is known as
esterification.
67 D The rate of esterification is slow. Hence concentrated sulphuric acid is used as a catalyst to speed up the
reaction.
O
68 B CH3
C
O
O
carboxylic acid
H+H
O
+
alcohol
CH3
CH3
C
ester
O
CH3 + H2O
+ water
69 D The flavours and fragrances of many fruits are due to a mixture of natural esters. Esters are also used in
artificial flavourings.
70 B
71 C
72 D
73 C
10
74 A
75 B
76 C
77 B
78 A
79 A
80 C
81 C
82 B
83 B
84 D
85 B
86 A
87 A
88 D
89 C Compound X is an alcohol and it undergoes oxidation in the reaction. The reaction mixture changes from
orange to green as the dichromate ions are reduced to chromium(III) ions.
90 A
91 D
92 A The compound is ethyl ethanoate.
(3) It is produced from ethanol and ethanoic acid.
93 B
94 D
95 A
96 D
11
97 A
98 B
99 D
100A
101B
102C
103B
104C
105C
106C The reaction between ethanoic acid and ethanol is esterification.
107D The rate of esterification is slow. A catalyst such as concentrated sulphuric acid is used to speed up the
reaction.
108B
109A
110D
Unit 32 Synthesis of carbon compounds
Fill in the blanks
12
1
a) i) Br2 (in organic solvent)
(1)
ii) Br2 (aq)
(1)
ii) H2 / Ni catalyst
(1)
iii) HBr(g)
(1)
–
b) i) conc. H2SO4 or conc. H3PO4 or Al2O3, heat
(1)
ii) reflux with conc. HCl, ZnCl2 catalyst; or mix with PCl5; or reflux with SOCl2 (1)
iii) reflux with NaBr + conc. H2SO4; or reflux with red P + Br2 (1)
iv) reflux with Nal + conc. H3PO4; or reflux with red P + I2
(1)
–
iv) MnO4 / OH (1)
+
v) heat with K2Cr2O7 / H3O , distil off the product
(1)
vi) CH3COOH, conc. H2SO4, heat
(1)
c) i) reflux with K2Cr2O7 / H3O+ (1)
+
ii) 1 LiAlH4 / ethoxyethane; 2 H3O (1)
iii) concentrated H2SO4, heat
(1)
iv) reflux with NaOH(aq)
(1)
v) 1 NH3(aq); 2 heat (1)
vi) HCl(aq), heat (1)
vii)reflux with NaOH(aq) (1)
True or false
2
F
3
T
4
T
5
F
6
F
Multiple choice questions
7
A
8
A
9
D
10 A
11 C
12 D
13 C
14 D
15 C
16 D
17 C
18 C
19 C
13
20 C
21 D
22 B
23 D
24 B
25 B
26 D
27 D
28 B
29 A
30 D
31 C
32 D
33 C
34 C
35 A
36 A
37 C
38 C
39 B
40 B
41 A
42 D
43 C
44 A
14
45 A
46 D
47 B
48 B
49 A
50 B
51 C
52 A
53 C
54 C
Unit 33 Important organic substances
Fill in the blanks
1
fats; oils; saponification
2
calcium; magnesium
3
scum 4
non-biodegradable 5
hydrogen 6
condensation 7
monosaccharides
8
glucose; frustose
9
triglyceride
10 amino acid
True or false
11 T
12 F
15
13 F
14 F
15 T
16 T
17 F
18 T
19 T
20 F
Multiple choice questions
21 B
22 C
23 D The hydrocarbon ‘tail’ is soluble in grease and oil, i.e. hydrophobic. The ionic ‘head’ is soluble in water,
i.e. hydrophilic.
24 C When fats or oils are heated with an alkali, they are hydrolyzed first to form glycerol and carboxylic
acids. The acids then react with the alkali (sodium hydroxide or potassium hydroxide) to form sodium or
potassium salts which are soaps. Such reactions are called saponification.
25 A Concentrated sodium chloride solution lowers the solubility of soap in water. The soap separates from
the solution and floats on the surface.
26 D
27 B
28 A Detergents I and IV are soaps. They would form scum with hard water.
29 C Detergent III is non-biodegradable.
30 B Detergents II and III are soapless detergents.
31 B
32 C
O
33 D The ionic ‘head’ of a soap particle is usually a
16
C
O– group.
34 A Option A — Sea water is hard water and contains a relatively high concentration of magnesium salts.
Major elements found in sea water
Element
Content (parts per million, ppm)
Chlorine
19 500
Sodium
10 770
Magnesium
1 290
Sulphur
905
A soap solution forms scum with sea water. A proper lather cannot form until the soap has
reacted with all the magnesium salts in the sea water.
Option B — The tap water in Hong Kong is soft water.
35 A
36 D The calcium carbonate deposit can be removed by an acid. Vinegar contains ethanoic acid.
CaCO3(s) + 2H+(aq)
Ca2+(aq) + CO2(g) + H2O(l)
37 B
38 D
39 B
40 D
41 B
42 A
43 A
44 A
45 D
46 A
47 D
48 D
49 B
50 A
17
51 B (1) Hard water contains a great amount of dissolved calcium and magnesium salts.
(2) Sea water contains a relative high concentration of magnesium salts. Therefore sea water is hard
water.
(3) The tap water in Hong Kong is soft water.
52 D (1) The ionic ‘head’ of a soapless detergent particle is usually a sulphonate –SO2–O– or sulphate
–O–SO2–O– group.
53 B
54 A
55 A (1) When the phosphates go into rivers and seas, they become the nutrients of the algae. As a result,
algae will grow suddenly. Sometimes algae may grow so that they form a thick layer on the water
surface. Light cannot enter the deeper level of the water. As a result, the plants underneath cannot
carry out photosynthesis and they die.
56 B
57 A
58 C (2) Nylon is a condensation polymer.
59 B
60 B (1) Poly(ethylene terephthalate) is common polyester. It is formed by the condensation of a diol and a
dioic acid.
(2) Poly(ethylene terephthalate) is a thermoplastic. There is no covalent bond between polymer chains.
61 D
62 D
63 A
64 C
65 B
66 D
67 B
68 B The correct explanation is that soap acts as an emulsifying agent.
69 D The hydrocarbon tail of soaps is hydrophobic. Soaps are made from natural fats and oils.
18
70 C Bacteria use up oxygen in the water during the decomposition of detergents. This makes the water of
some streams, rivers and seas smells badly due to oxygen depletion.
71 D Detergents with branched-chain hydrocarbons are non-biodegradable. They cause more harm to the
environment when compared to those with straight-chain hydrocarbons.
72 A
73 B Soapless detergents do not form scum with either soft water or hard water.
74 D
75 C Nylon softens upon heating.
76 D
77 C
Part B Topic-based exercise
Multiple choice questions
1
C
2
D
3
C
4
B
5
A
6
C
7
A
8
A
9
A
10 A
11 B
12 A
19
13 B
14 D
15 D
16 C
17 A
18 B
19 B
20 C
21 C
22 B
23 C
24 B
25 C
26 C
27 D
28 C
29 B
30 D
31 C
32 C
33 D
34 D
35 C
36 C
20
37 D
38 B
39 D
40 D
41 C
42 C
43 C
44 B
45 A
46 C
47 B
48 A
49 B
50 D
51 B
52 A
53 C
54 C
55 D
56 A
57 C
58 B
59 C
60 B
21
61 D
62 A
63 A
64 D
65 D
66 A
67 A
68 C
69 D
70 B
71 C
72 D
73 A
74 C
75 A
76 C
77 D
78 B
79 D
80 A
81 B
82 B
83 D
84 A
22
85 A
86 B
87 C
88 A
Short questions
89 a) 4-methylpent-2-ene
(1)
b) 3-bromobut-1-ene
(1)
c) 2-chloro-2-methylhexane
(1)
d) pentan-2-ol
(1)
e) 2-ethylbutan-1-ol
(1)
f) hex-3-enal
(1)
g) 3-oxobutanoic acid
(1)
h) 4-methylpentan-2-one (1)
i) 2-methylpropyl methanoate
(1)
j) 2-aminopropanoic acid
(1)
90 a) CH 3
CH3
CH3CHCH2CHCH3
b)
c)
CH3
(1)
(1)
CH2CH3
O
C
CH3
(1)
d)
O
CH3CH2CH
C
CH3
e) CH2=CHCONH2
O
CH2CH3
(1)
(1)
23
91 a) functional group isomers
(1)
b) chain isomers
(1)
c) position isomers
(1)
d) position isomers
(1)
e) functional group isomers
(1)
f) functional group isomers
(1)
92 a) no
(1)
b) yes
(1)
c) no
(1)
d) no
(1)
e) yes
(1)
f) yes
(1)
93 a)
24
H
b)
H
c)
H
e)
H
f)
H
H
H
H
Cl
H
C
C
C
C*
C
H
H
H
H
H
H
H
OH H
C
C
C*
C
H
H
H
H
H
H
CH3 H
H
C
C
C*
C
H
H
H
H
H
CH3
C
C
C*
H
H
H
H
Br
C
C*
H
H
CHO
C
H
H
(1)
(1)
H (1)
COOH
(1)
(1)
2-chloropentane
(1)
butan-2-ol
(1)
3-methylpent-1-ene
(1)
2-methylbutanoic acid
(1)
2-bromopropanal
(1)
94 a) propane (1)
(1)
b) ethanoic acid
c) ethyl ethanoate
(1)
d) butanamine
(1)
e) cyclohexene
(1)
f) butanal
(1)
95 a) and b)
Type of reaction
involved
Test
Observation(s)
Shake a few drops of X with
aqueous bromine.
The aqueous bromine changes from yellowbrown to colourless quickly. (1)
addition
(1)
Add phosphorus pentachloride
to X. Test any gas evolved with
moist blue litmus paper.
Steamy fumes are observed. The moist blue litmus paper turns red. (1)
(1)
substitution (1)
Add acidified potassium
dichromate solution to X
and heat the mixture.
The acidified potassium dichromate solution
changes from orange to green. (1)
oxidation (1)
96 a) Any one of the following:
• Add solid sodium hydrogencarbonate (or solution). (1)
(1)
• Use moist blue litmus paper / pH paper for testing. (1)
(1)
b) Any one of the following:
• Add aqueous bromine.
(1)
(1)
• Add cold acidified dilute potassium permanganate solution.
(1)
(1)
• Heat with acidified potassium dichromate solution.
(1)
(1)
X gives a gas that turns limewater milky.
X turns moist blue litmus paper / turns pH paper orange (or red). Y turns aqueous bromine from yellow-brown to colourless quickly.
Y turns the permanganate solution from purple to colourless quickly.
Y turns the dichromate solution from orange to green.
25
97 a) Cl2 / ultraviolet; light
(1)
b)
c) red P + I2; reflux
(1)
d) conc. H2SO4 heat
(1)
e)
(1)
f) CH3CH(OH)CH2CH3
(1)
g) 1 NH3; 2 heat
(1)
h) 1 LiAlH4 / ethoxyethane; 2 H3O+
(1)
i) NaOH(aq); heat
(1)
j)
CH3
(1)
COOH
COOH
OH
(1)
98 a) CH3CH(OH)CH2OH
H3CH2C
b)
c)
C
C
Br
H
C
C
D
Br
d)
e)
f) CH3CH2CH2I
g)
(1)
CH2CH3
H
H
H
26
CH2CH3
H3C
(1)
CH3
(1)
O
(1)
Cl
(1)
(1)
CH3
CH3
C
CHCH2CH3
(1)
OH
h)
CH3
i)
(1)
CH3CH2O
NH2
+ CH3COO–Na+
H
99 a) i)
(1)
(1)
O
C
H
C
OH
HO
C
H
H
C
OH
H
C
OH
ii) Carbonyl group
(1)
Hydroxyl group
(1)
b) i) Carboxyl group
(1)
H
O
HO
C
C
CH3
ii)
H2N
(1)
CH2OH
H
O
C
C
H
O
N
C
C
H
CH3
O
OH
(1)
H
O
C
C
OH
iii) Any one of the following:
• Peptide link / amide functional group (1)
• Ester functional group (1)
CH3
(1)
CH3
Structured questions
100a)
H
CH3CH2
C
CH2
H
(1)
CH3
C
C
H
CH3
H
(1)
CH3
C
C
CH3
CH3
H
(1)
CH3
C
CH2
(1)
27
b) i)
H
C
C
CH3
H
CH3
H
and
C
C
CH3
CH3
H
are cis-trans isomers.
(1)
ii) In order for the intermolecular forces to work well, the molecules must be able to pack together
efficiently in the solid.
(1)
Molecule of the trans isomer has a more regular and symmetrical structure than molecule of the cis
isomer. (1) Molecules of the trans isomer can pack more compactly in the solid.
More heat is required to overcome the intermolecular forces between the molecules.
Hence the trans isomer has a higher melting point than the cis isomer.
(1)
101 a) 2-hydroxybenzaldehyde
(1)
b) Position isomerism
(1)
c) X can form intramolecular hydrogen bonds.
(1)
Y forms more intermolecular hydrogen bonds than X does.
(1)
Thus the melting point of Y is much higher than that of X.
102a) Any one of the following:
28
• Warm each compound with acidified potassium dichromate solution. (1)
A turns the dichromate solution from orange to green.
(1)
There is no observable change for B.
(1)
• Warm each compound with acidified potassium permanganate solution.
(1)
A turns the purple permanganate solution colourless.
(1)
There is no observable change for B.
(1)
b) Put about 2 cm3 of ethanol and 1 cm3 of silver nitrate solution in each of two test tubes.
(1)
Place the test tube in a water bath at 60 °C.
(1)
Add several drops of C and D separately to each test tube.
A yellow precipitate forms rapidly in the test tube containing D.
(1)
A white precipitate forms slowly in the test tube containing C.
(1)
c) Warm each compound with acidified potassium dichromate solution. (1)
F turns the dichromate solution from orange to green. (1)
There is no observable change for E. (1)
d) Add aqueous bromine to each compound separately.
(1)
G decolorizes aqueous bromine quickly.
(1)
There is no observable change for H.
(1)
103a) i) hydroxyl group
(1)
ii) carbonyl group
(1)
iii) carboxyl group
(1)
b) 1 LiAlH4 / ethoxyethane
c) Excess concentrated sulphuric acid
(1)
(1)
d) V e) Any one of the following:
• Warm each compound with acidified potassium dichromate solution.
(1)
U turns the dichromate solution from orange to green. (1)
There is no observable change for Y.
(1)
• Mix each compound with phosphorus pentachloride.
(1)
U gives steamy fumes of hydrogen chloride.
(1)
There is no observable change for Y.
(1)
f) Heat V and Z under reflux in the presence of concentrated sulphuric acid.
2 H3O+
(1)
180 °C
(1)
K2Cr2O7 / H3O+
(1)
O
g) V (1)
CH3CH2CH2C
(1)
CH3
O
*
CH2CH(OH)CH2CH3
CHCH2CH3
(1)
(1)
(1)
29
104a) i) Condenser
(1)
ii) Compound X is flammable.
(1)
iii) To ensure even boiling.
(1)
iv) Propanoic acid
(1)
v) CH3CH2CH2OH
vi) As an oxidizing agent
vii)
[O]
CH3CH2COOH
(1)
(1)
UIFSNPNFUFS
XBUFSPVU
DPOEFOTFS
SFBDUJPONJYUVSF
BOUJCVNQJOH
HSBOVMFT XBUFSJO
HFOUMFIFBU
BRVFPVTTPMVUJPO
PGQSPQBOPJDBDJE
30
(1 mark for a correct set-up; 1 mark for correct labels; 1 mark for the correct direction of water flow
in condenser; award 0 mark if the set-up is not workable)
(3)
b) i) Ethyl propanoate
(1)
ii) As a catalyst
(1)
iii) Heating under reflux can reduce the loss of volatile reactants / products by evaporation.
(1)
105a) i) substitution
(1)
ii) dehydration / elimination
(1)
iii) addition
(1)
b) i) Cl2 (1)
(1)
ii) sodium hydroxide solution
(1)
(1)
iii) excess concentrated sulphuric acid (1)
(1)
c) i) Structural isomers are compounds with the same molecular formula,
(1)
(1)
ultraviolet light / sunlight
reflux / heat
180 °C
but differ in the order in which atoms are linked.
ii)
Alcohol
Classification
CH3CH2CH2CH2OH
OH
primary
(0.5)
secondary
(0.5)
CH3CH2CHCH3
CH3
CH3CHCH2OH
primary
(1)
(0.5)
CH3
tertiary
CH3COH
CH3
(1)
(0.5)
d) Any one of the following:
• HCl(g) (1) in the presence of ZnCl2 catalyst (1)
• reflux with concentrated hydrochloric acid (1) using ZnCl2 as catalyst (1)
• reflux (1) with sulphur dichloride oxide (1)
• mix (1) with phosphorus pentachloride (1)
e) i) Dehydration
(1)
ii) CH3CH=CHCH3 (1)
106a) Carbon-carbon double bond
(1)
(1)
b)
H C
C
H
H
H C
C
Br
Br
Hydroxyl group
H
H
CH3
CH2CH2OH
3-methylbutan-1-ol
(1)
2,3-dibromo-3-methylbutan-1-ol
(1)
(1)
CH3
CH2CH2OH
(1)
c) i) 3-methylbut-3-enoic acid
(1)
ii) Heat under reflux
(1)
(1)
with acidified potassium dichromate solution
31
iii) H
d)
H
C
C
H
CH2CHO
C
C
H
107a)
H
32
CH3
CH3
or 3-methylbut-3-enal
+
2HBr (1)
(1)
H
H
CH3
C
C
H
Br
CH2CH2OH
H
H
C
C
H
H
CH2CH2Br
+
H2O (1)
Br
b) i) Heat under reflux
(1)
(1)
ii) React with aqueous ammonia to form an ammonium salt.
(1)
(1)
c) i) Substitution
(1)
ii) Oxidation
(1)
iii) Reduction
(1)
d) CH3COOH > CH3CH2OH > CH3CH2NH2
(1)
The boiling point of a compound depends on its molecular attractions.
(1)
All the compounds can form hydrogen bonds.
Molecules of CH3COOH can form more extensive hydrogen bonds than those of CH3CH2OH.
Thus the boiling point of CH3COOH is higher than that of CH3CH2OH.
Since oxygen is more electronegative than nitrogen, the O–H bond in CH3CH2OH is more polar than the
N–H bond in CH3CH2NH2.
As a result, stronger hydrogen bonds exist between molecules of CH3CH2OH.
Thus the boiling point of CH3CH2OH is higher than that of CH3CH2NH2.
with acidified potassium dichromate solution
Evaporate the water and subsequently heat the dry salt.
(1)
(1)
(1)
108a) Carbon-carbon double bond
(1)
(1)
b) i) Stereoisomers have their atoms linked in the same way,
(1)
(1)
ii) Geometrical isomerism
Carbonyl group
but they differ in the spatial arrangement of their atoms.
(1)
iii) Due to the restricted rotation about a carbon-carbon double bond.
(1)
iv) Identical methyl groups attach to one of the carbon atoms of the C=C bond.
(1)
c) i)
(H3C)2C
ii) Primary alcohol
(1)
(1)
iii) 1 LiAlH4 / ethoxyethane
iv) (1)
(H3C)2C
(2)Esterification
CHCH2CH2C
CHCH2OH
(1)
CH3
There is one alkyl group attached to the carbon bearing the –OH group. 2 H3O+
(1)
O
CHCH2CH2C
CHCH2
O
C
CH3
CH3
(1)
109a)
CH3
CH3CHCH
2Br
(1)
b)
CH3
C
CH2
CH3
methylpropene
(1)
(1)
c)
i)
CH3
CH3CBrCH3
2-bromo-2-methylpropane
(1)
(1)
ii) Structural isomers are compounds with the same molecular formula,
(1)
(1)
iii) CH3CH2CH2CH2Br
(1)
(1)
but differ in the order in which atoms are linked.
CH3CH2CHBrCH3
110a)
H
H
H
C
C
C
Br
Br
H
H
(1)
33
b) X, Y and Z are formed from the dehydration of alcohol W.
Thus they are alkenes. (1)
X and Y react with bromine to give 1,2-dibromo-1-phenylpropane, i.e.
C
C
H
H
C
C
CH3
C
C
C
Br
Br
H
H .
CH3
(1)
Z reacts with bromine to give 2,3-dibromo-1-phenylpropane, i.e.
H
H
H
C
C
C
H
Br
Br
H .
It can be deduced that the structure of Z is as follows:
H
H
H
C
C
C
H
H
W undergoes dehydration to give X, Y and Z. It can be deduced that the structure of W is as follows:
(1)
H
H
H
C
C
C
H
OH
H
H
c) X and Y exhibit geometrical isomerism.
(1)
(1)
d) i) The purple permanganate solution becomes colourless quickly.
34
H
H
(1)
H
It can be deduced that the structures of X and Y are as follows:
H
H
X, Y and Z exhibit structural / position isomerism.
ii)
H
H
H
C
C
C
OH
OH
H
(1)
(1)
H
(1)
111a) i) Suppose we have 100 g of compound X, so there are 40.0 g of carbon, 6.65 g of hydrogen and
53.3 g of oxygen.
Carbon
Hydrogen
Oxygen
Mass of
element
in the
compound
40.0 g
6.65 g
53.3 g
Number of
moles of
atoms that
combine
40.0 g
= 3.33 mol
12.0 g mol–1
6.65 g
= 6.65 mol
1.0 g mol–1
53.3 g
= 3.33 mol
16.0 g mol–1
(1)
3.33 mol
= 1 3.33 mol
6.65 mol
= 2
3.33 mol
3.33 mol
= 1 3.33 mol
(1)
Simplest
ratio of
atoms
ii) Let (CH2O)n be the molecular formula of X.
Relative molecular mass of X =
=
i.e. 30n =
n =
∴ the molecular formula of X is C3H6O3.
b) Effervescence occurs when X is mixed with sodium hydrogencarbonate solution. Thus it should be a
carboxylic acid and contain a –COOH group.
(1)
X undergoes esterification with ethanoic acid. Thus it should contain a –OH group.
(1)
X is chiral. Thus it should contain at least one chiral carbon.
(1)
Structure of X:
c)
H
∴ the empirical formula of X is CH2O.
H
n(12.0 + 2 x 1.0 + 16.0)
30n
90.0
3
H
OH
O
C
C
C
H
H
OH
chiral carbon
2-hydroxypropanoic acid
H
O
O
C
C
C
H
(1)
(1)
(1)
OH
(1)
35
O
d)
CH3
H
C
O
C
C
O
C
H
CH3
O
Ester functional group
(1)
(1)
112Since W can be hydrolyzed to give an acid X and a neutral compound Y, so W should be an ester.
Relative molecular mass of W = 10 x 12.0 + 12 x 1.0 + 2 x 16.0 = 164.0
Relative molecular mass of Y = 164.0 + 18.0 – 122.0 = 60.0 Let the molecular formula of Y be CnH2n+1OH.
(1)
(1)
i.e. 12n + 2n+1 + 16.0 + 1.0 = 60.0
n = 3
∴ the molecular formula of Y is C3H7OH (1) and that of X is C7H6O2 or C6H5COOH. (1)
COOH
.
(1)
The structure of X is
Y can be oxidized to Z, so Z should be an aldehyde or a ketone.
As Z does not react with acidified potassium dichromate solution, thus Z should be a ketone.
(1)
Z is propanone, CH3COCH3.
(1)
OH
∴ Y is a secondary alcohol, CH3CHCH3 .
(1)
COOCH(CH3)2
W is
.
113a) CH3CH2CH2CH2Br
K2Cr2O7 / H3O
reflux
36
(1)
NaOH(aq)
reflux
CH3CH2CH2CH2OH
(1)
+
(1)
CH3CH2CH2COOH
(1)
CH2CH2OH conc. H2SO4
b)
heat
CH
(1)
(1)
CHBrCH3
HBr(g)
(1)
CH2
CH(OH)CH3
NaOH(aq)
reflux
(1)
(1)
(1)
O
CCH3
K2Cr2O7 / H3O+
reflux
c)
(1)
CH3CHCH3
conc. H2SO4
heat
cold alkaline
d)
(1)
CH3CHCH2OH
(1)
OH
CH2CH2Br
e)
f)
(1)
(1)
(1)
H3O+
CH2CONH2
Cl
reflux
(1)
(1)
CH2COOH
K2Cr2O7 / H3O+
CH2COOCH3
CH3OH / conc. H2SO4
heat
CH2CH2OH
NaOH(aq)
reflux
CH2
(1)
OH
dilute KMnO4
CH3CH
heat
CH2COOH
(1)
(1)
OH
NaOH(aq)
reflux
(1)
Br2 (in organic solvent)
(1)
(1)
1 LiAlH4 / ethoxyethane
2 H3O+
CH2CH2OH
(1)
conc. H2SO4 or conc. H3PO4
heat
(1)
(1)
Br
Br
37
114a) Pressure builds up in the set-up when the mixture is heated. It was dangerous to conduct an experiment
using a closed system. An explosion may occur.
(1)
Modification – add a receiver adaptor between the condenser and the round-bottomed flask.
(1)
b) i) Use a teat pipette to remove the bottom layer. / Use a separating funnel and run off the bottom
layer.
(1)
ii) To dry the organic layer.
iii) Any one of the following:
• Add aqueous bromine to the distillate.
(1)
(1)
• Add cold acidified dilute potassium permanganate solution to the distillate.
(1)
(1)
iv) • Add phosphorus pentachloride to the distillate. (1)
(1)
(1)
The yellow-brown aqueous bromine changes to colourless quickly.
The purple permanganate solution changes to colourless quickly.
• Steamy fumes would be observed. 115a) The process is exothermic.
(1)
b) i) To remove the excess dilute sulphuric acid.
(1)
ii)
BRVFPVTMBZFS
CSPNPQSPQBOFMBZFS
38
(1 mark for correct drawing of separating funnel with tap; 1 mark for showing 1-bromopropane
layer on bottom; award 0 mark if the funnel is not workable)
(2)
iii) To prevent pressure building up due to the formation of carbon dioxide gas.
c) To dry the product.
(1)
d) Use an electric heating mantle / a water bath / an oil bath for heating
(1)
(1)
because propan-1-ol / 1-bromopropane is flammable.
(1)
e) i)
UIFSNPNFUFS
XBUFSPVU
TUJMMIFBE
DPOEFOTFS
GSBDUJPOBUJOH
DPMVNO
SFDFJWFS
BEBQUPS
HMBTTCFBET
XBUFSJO
PSHBOJD
MBZFS
BOUJCVNQJOH
HSBOVMFT
EJTUJMMBUF
IFBU
(1 mark for correct set-up; 1 mark for correct labels; 1 mark for correct drawing of fractionating
column; 1 mark for correct direction of water flow in condenser; award 0 mark if the set-up is not
workable)
(4)
ii) 70 °C – 72 °C f) i) Number of moles of propan-1-ol =
According to the equation, 1 mole of propan-1-ol gives 1 mole of 1-bromopropane.
Theoretical yield of 1-bromopropane = 0.152 mol x 122.9 g mol–1
= 18.7 g
Percentage yield of 1-bromopropane =
ii) Any one of the following:
• The reaction is incomplete.
(1)
• Some of the propan-1-ol dehydrate to form an alkene.
(1)
• Some of the product stay in the aqueous layer.
(1)
(1)
9.12 g
–1
60.0 g mol
= 0.152 mol
10.8 g
x 100%
18.7 g
= 57.8% (1)
(1)
116a) Methanol (1) and propanoic acid (1)
b) As a catalyst
c) CH3OH(l) + CH3CH2COOH(l)
d)
H
(1)
H
H
O
C
C
C
H
H
CH3CH2COOCH3(l) + H2O(l) (1)
H
O
C
H
H
(1)
39
e) Methanol is flammable and it catches fire easily.
f) Any two of the following:
g) B (1)
h) Any one of the following:
(1)
A pleasant smell could be detected. (1) / An insoluble layer formed on the sodium carbonate solution.
(1) / Effervescence occurred. (1)
O
H
O
OCH2CH2CH3 (1) propyl methanoate (1) CH3
C
C
OCH2CH3 (1)
ethyl ethanoate (1)
O
H
C
i)
H
H
OCH(CH3)2 (1)
methylethyl methanoate (1)
H
H
H
O
C
C
C
C
H
H
H
H
CH3 O
C
C
H
H
C
OH
(1)
butanoic acid (1)
(1)
methylpropanoic acid (1)
OH
117a)
XBUFSPVU
SFGMVYDPOEFOTFS
XBUFSJO
NJYUVSFPGFUIZM
CFO[PBUFBOE
FYDFTT/B0)BR
BOUJCVNQJOH
HSBOVMFT
IFBU
40
(1 mark for correct set-up; 1 mark for correct direction of water flow in condenser; award 0 mark if the
set-up is not workable)
(2)
b) To prevent any loss of the volatile substance / ester by evaporation.
(1)
c) To convert sodium benzoate into benzoic acid.
(1)
d) Dissolve the crude sample in minimum amount of hot water.
(1)
Filter the mixture while hot.
(1)
Allow the filtrate to cool and collect the crystals by filtration.
(1)
e) i) Number of moles of ethyl benzoate =
According to the equation, 1 mole of ethyl benzoate gives 1 mole of benzoic acid on hydrolysis.
Theoretical yield of benzoic acid = 0.0300 mol x 122.0 g mol–1
= 3.66 g Percentage yield of benzoic acid =
ii) Hydrolysis is not complete. / Benzoic acid is slightly soluble in water.
4.50 g
–1
150.0 g mol
= 0.0300 mol
2.29 g
x 100%
3.66 g
= 62.6%
(1)
(1)
(1)
118a) 3-methylbutyl ethanoate
(1)
b) (CH3)2CH(CH2)2OH
3-methylbutan-1-ol
(1)
(1)
c) Reagent Y is concentrated sulphuric acid.
(1)
(1)
d) Esterification
e) Any two of the following:
• Isoamyl acetate
(1)
• Acetic acid
(1)
• Water
(1)
d) • Shake the mixture with sodium hydrogencarbonate solution until no more gas is evolved.
(1)
• Extract the isoamyl acetate from the organic layer using ethoxyethane.
(1)
• Remove the ethoxyethane by simple distillation.
(1)
Heat the reaction mixture under reflux.
119a) i) Cause bleeding in the stomach lining
(1)
(1)
b) i) 2-hydroxybenzoic acid
(1)
ii) Heat with methanol
(1)
(1)
ii) Paracetamol
and a little concentrated sulphuric acid.
41
iii)
COO–Na+
(1)
OH
8.28 g
–1
138.0 g mol
= 0.0600 mol
c) Number of moles of salicylic acid used =
Number of moles of aspirin obtained =
Percentage conversion =
d) To show the presence of the –COOH group, add sodium hydrogencarbonate solution to the
compound.
(1)
Evolution of colourless gas bubbles indicates the presence of the –COOH group.
(1)
To show the presence of the ester group, heat the compound with dilute acid.
(1)
Smell of vinegar is noted.
(1)
e) i) Ca(OH)2
ii) Aspirin is a covalent compound with intramolecular hydrogen bonds. This reduces the extent of
hydrogen bonding with water, thus making aspirin insoluble in water. (1)
The large hydrophobic benzene ring also leads to low water solubility.
(1)
The calcium salt is ionic. Strong interactions exist between water and the ions.
(1)
(1)
7.20 g
–1
138.0 g mol
= 0.0400 mol
(1)
0.0400 mol
x 100%
0.0600 mol
= 66.7%
(1)
(1)
120a) Saponification
42
(1)
O
b)
H2C
O
C
O
C17H33
HC
O
C
O
C17H33
H2C
O
C
C17H33
c) Label C
+
3NaOH
H2C
OH
HC
OH
H2C
OH
+
3C17H33COO–Na+
(2)
(1)
d) Any one of the following:
• Wear safety glasses.
(1)
• Wear protective gloves.
(1)
e) To lower the solubility of soap in water.
(1)
f) To prevent the soap from dissolving in the water.
(1)
g) Perfume (1) and colouring (1)
h) Not suitable
(1)
Sea water contains a lot of magnesium ions and calcium ions. The soap would react with the metal ions
to form scum and thus reduce the cleaning effectiveness of soap.
(1)
121a) Any two of the following:
• Hydrocarbons obtained from petroleum
(1)
• Concentrated sulphuric acid
(1)
• Sodium hydroxide
(1)
b) i)
O
O
S
O–
ii) CH3(CH2)10CH2 —
c) When mixed with a mixture of water and paraffin oil, the hydrophobic parts of the detergent particles
dissolve in the oil while the hydrophilic parts dissolve in water.
(1)
Upon shaking, oil droplets form. Each drop is surrounded by detergent particles with the hydrophilic parts
in the water.
(1)
The hydrophilic parts carry negative charges and prevent the oil droplets from coming together again.
Hence an emulsion is formed.
(1)
d) Detergent X can emulsify the crude oil into droplets and
(1)
(1)
e) i) Yes
f) i) Calcium ion
(1)
(1)
ii) Detergent X can function well in hard water
(1)
(1)
O
the droplets can be carried out away by water currents.
(1)
(1)
(1)
ii) During the decomposition of the detergent, bacteria use up oxygen in the water. This makes the water
smell badly due to oxygen depletion.
(1)
Magnesium ion
because it does not form scum with hard water.
43
122a) i) Carboxyl group
(1)
(1)
b) Sodium hydroxide solution
c)
ii) Hydroxyl group
O
O
C
C
(1)
O
CH2
CH2
O
(1)
d) Condensation polymerization
(1)
e) Permanent dipole-permanent dipole attractions between PET polymer chains are stronger
(1)
(1)
f) i) Hydrolysis
ii) HO
than the instantaneous dipole-induced dipole attractions between polythene polymer chains.
(1)
CH2
Na+O–
CH2
OH
O
O
C
C
(1)
O–Na+
(1)
123a i) The molecule of valine contains one carbon atom bonded to four different groups of atoms.
(1)
ii)
H
(CH3)2HC
C
H
COOH
NH2
HOOC
H2N
(1)
C
CH(CH3)2
(1)
NJSSPS
iii) (1)Rotation of plane of plane-polarized light (1)
(1)
b)
(2)Rotate the plane of plane-polarized light in opposite directions.
H2N
44
O
CH
C
CH(CH3)2
N
CH(CH3)2 O
CH
COOH
H
H2N
CH3
CH
C
(1)
CH3
N
H
CH
COOH
(1)
CH3
c)
H3N+CHCOOH
(1)
CH3
CHCOO–
H2N
(1)
124a) Amine functional group (1), carboxyl group (1), amide functional group (1), ester functional group (1)
b)
O
*
CH
H2N
O
C
NH
CH2
*
CH
C
OCH3
CH2
CO2H
c)
(2)
H
O
H2N
C
C
CH2COOH
OH
(1)
d) Silicon dioxide increases the bulk mass of the sweetener.
e) Phenylketonurics are unable to metabolize the phenylalanine produced from the hydrolysis of
aspartame.
(1)
COOH
H2N
CHCH2
(1)
A high level of phenylalanine in the brain is extremely harmful and sometimes fatal.
125a) Condensation polymerization
b)
N
(CH2)6
(1)
N
H
H
(1)
O
H
(1)
H
H
(1)
O
C
(CH2)4
C
O
O
H
(1)
45
c)
H
N
O
(CH2)6
N
C
(CH2)4
H
C
O
d) Any one of the following:
Lighter in weight
(1)
Does not corrode easily
(1)
e) There are hydrogen bonds between the polymer chains of nylon.
Only weak instantaneous dipole-induced dipole attractions exist between the polymer chains of
polythene.
(1)
Hence nylon is stronger than polythene.
f) A hole would appear
(1)
(1)
g) Method I:
The waste contains dichromate ions which is toxic.
Method II:
Concentrated nitric acid is a strong acid. Discharge of the waste into waterways leads to environmental
pollution.
(1)
because the acid hydrolyzes the amide linkages in nylon.
(1)
(1)
(1)
126a) Any two of the following:
46
Burning fossil fuels produces a lot of air pollutants.
(1)
The resources of fossil fuels are limited.
(1)
Supply and price of petroleum may become unstable in times of political unstability in petroleum-producing
countries. (1)
Petroleum spills during drilling or transportation cause major damage to ecosystems both on land and at
sea.
(1)
b) Burning ethanol returns carbon dioxide
(1)
(1)
c) More energy can be obtained.
d) Any two of the following:
which has been recently removed from the atmosphere during the photosynthesis of plants.
Farmers use fossil-fuel-powered equipment to plant, maintain and harvest the corn.
(1)
(1)
Fossil-fuel-powered machines are required to process the corn into ethanol.
(1)
Ethanol is transported to collection points via fossil-fuel-powered transport.
(1)
e) Any one of the following:
This may lead to diversion of investment from food production, resulting in increased food prices.
Ethanal is also produced during the combustion of ethanol. It can harm vegetation, irritate the skin and
eyes, and damage the lungs at high concentrations.
(1)
f) As a solvent (1)
(1)
127Any three of the following properties:
Physical properties
• Hydrogen bonds exist between ethanol molecules.
• Due to the hydrogen bonding between ethanol molecules and water molecules,
(1)
(1)
Chemical properties
• Ethanol reacts with hydrogen halides to form haloethanes.
OR
• Ethanol can be dehydrated to form ethene.
• Ethanol can be oxidized to ethanal, and then to ethanoic acid.
• Ethanol reacts with carboxylic acid in the presence of concentrated sulphuric acid to form an ester. (1)
For example, heating ethanol with methanoic acid in the presence of concentrated sulphuric acid gives
the ester ethylmethanoate.
(1)
(1)
Thus the boiling point of ethanol is higher than those of alkanes / haloalkanes of similar relative molecular
masses. (1)
ethanol is miscible with water in all proportions.
(1)
Bromoethane can be prepared by heating a mixture of ethanol, sodium bromide and concentrated sulphuric
acid under reflux. Concentrated sulphuric acid reacts with sodium bromide to produce hydrogen bromide.
Hydrogen bromide then reacts with ethanol to give bromoethane.
(1)
Bromoethane can be prepared by heating a mixture of ethanol, red phosphorus and bromine under
reflux. Phosphorus first reacts with bromine to give phosphorus tribromide. The tribromide then reacts
with ethanol to give bromoethane.
(1)
(1)
This is done by heating ethanol at 180 °C with excess concentrated sulphuric acid / passing ethanol
vapour over aluminium oxide at 300 °C.
(1)
(1)
This is done by heating ethanol with acidified potassium dichromate solution under reflux. The reflux
condenser prevents any loss of the reaction mixture and favours the oxidation of ethanol to ethanoic acid
rather than to ethanal.
(1)
(3 marks for organization and presentation)
47
128Add water to the liquids.
(1)
Both butanoic acid and propanone can mix with water in all properties.
(1)
Add a piece of pH paper to the aqueous solutions.
(1)
Solution of butanoic acid is acidic while that of propanone is not.
(1)
Add aqueous bromine to the remaining two compounds.
(1)
Only cyclohexene can decolorize the aqueous bromine quickly.
(1)
OR
Add silver nitrate solution to the remaining two compounds.
(1)
Only 1-bromobutane gives a pale yellow precipitate slowly.
(1)
(3 marks for organization and presentation)
129Soaps are mainly made from the reaction between fats (or vegetable oil) and sodium hydroxide
solution. (1)
48
Soapless detergents are made from hydrocarbons obtained from petroleum, concentrated sulphuric acid and
sodium hydroxide.
(1)
Both of them have good cleaning ability in soft water.
(1)
However, soaps form scum and lose cleaning ability in hard water while soapless detergents do not.
(1)
Branch-chained soapless detergents are non-biodegradable. They form thick foam which prevents oxygen
from dissolving into water and leads to the death of aquatic life.
(1)
Soaps and straight-chained soapless detergents are biodegradable. When they are decomposed by bacteria in
rivers or sea, oxygen in water is used up. This leads to oxygen depletion and thus causes death of aquatic
lives. (1)
In general, soaps and straight-chained soapless detergents cause fewer pollution problems than branch-chained
soapless detergents.
(3 marks for organization and presentation)