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CM TB solutions C22

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Chemistry Matters for GCE ‘O’ Level (2 Edition): Full Solutions to Textbook Questions
Chapter 22
Chapter 22
Alkanes and Alkenes
Test Yourself 22.1 (page 429)
1. 2x + 2 = 26. Therefore, x = 12.
2. 1 and 2
Test Yourself 22.2 (page 432)
1. 1 mol of the alkane → 2CO2 + 3H2O
Therefore, 1 mol of the alkane contains 2 carbon atoms and 6 hydrogen atoms. The molecular
formula of the alkane is C2H6.
2. 3 mol of chlorine is required to react with 1 mol of methane.
CH4(g) + 3Cl2(g)
UV light
CHCl3(g) + 3HCl(g)
3. X is ethane.
Y is methylpropane (or 2-methylpropane).
Z is butane.
© 2013 Marshall Cavendish International (Singapore) Private Limited
22.1
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Chemistry Matters for GCE ‘O’ Level (2 Edition): Full Solutions to Textbook Questions
Chapter 22
Test Yourself 22.3 (page 435)
1. (a) C3H6
(b)
(c) ‘Dot and cross’ diagram of propene:
2. (a) They contain only carbon and hydrogen atoms.
(b) P belongs to the alkane homologous series. Q and R belong to the alkene homologous series.
(c) Q and R have the same number of carbon atoms. They each contain one carbon–carbon double
bond. (Any one answer)
(d) Q and R are isomers. They have the same molecular formula, C4H8. However, their double
bonds are in different places, giving them different structural formulae.
Test Yourself 22.4 and 22.5 (pages 441–442)
1. (a) P, methylpropene
Q, methylpropane
(b) Isomers of P:
But-1-ene
Isomer of Q:
But-2-ene
Butane
© 2013 Marshall Cavendish International (Singapore) Private Limited
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Chemistry Matters for GCE ‘O’ Level (2 Edition): Full Solutions to Textbook Questions
Chapter 22
2. It can be prepared by the addition of HCl to ethene.
→ 3. (a) Bromination. P is propene.
(b) Substitution reaction. Q is ethane.
(c) Hydration. R is ethene.
4. (a) (i)
(ii)
(iii)
(b) (i)
(ii)
X is C8H18.
Y is C3H6.
Z is H2.
X
Y
Test Yourself 22.6 (page 443)
(a) P is ethane.
Q is ethene.
(b) P contains only carbon–carbon and carbon–hydrogen single bonds, but Q contains a carbon–carbon
double bond.
Test Yourself 22.7 (page 444)
1. Add aqueous bromine to each of the compounds. Aqueous bromine remains unchanged when
added to saturated fats, but is decolourised when added to polyunsaturated fats.
2. Margarine is produced by the hydrogenation of vegetable oil at a temperature of 200°C in the
presence of a nickel catalyst.
IT Learning Room (page 444)
(a)
Source of fat
Butter
Margarine
Olive oil
Peanut oil
Coconut oil
Saturated fat/%
63
35
14
17
87
Monounsaturated
fat/%
26
35
73
46
6
Polyunsaturated
fat/%
4
5
11
32
2
(b) (i) Olive oil
(ii) Coconut oil
© 2013 Marshall Cavendish International (Singapore) Private Limited
22.3
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Chemistry Matters for GCE ‘O’ Level (2 Edition): Full Solutions to Textbook Questions
Chapter 22
Get it Right (page 445)
(a) False. Alkanes are called saturated hydrocarbons because they contain only carbon–carbon single
covalent bonds.
(b) False. Butane and butene are not isomers. They have different chemical formulae.
(c) True
(d) False. The reaction of alkanes with bromine is a substitution reaction.
(e) False. The first member of the alkene homologous series is ethene.
(f) True
(g) False. The catalytic cracking of long-chain alkanes produces short-chain alkenes and short-chain
alkanes or hydrogen.
(h) False. Alkenes (but not alkanes) react rapidly with bromine water at room temperature.
(i) False. Polyunsaturated fats contain two or more carbon–carbon double bonds.
(j) True
Let’s Review (pages 446–448)
Section A: Multiple-Choice Questions
1. A
2. C
3. C
4. C
5. C
6. D
7. B
8. D
Section B: Structured Questions
1. (a) An alkane is a saturated hydrocarbon with the general formula CnH2n+2.
(b) Relative molecular mass of J = 72
Therefore, 12n + 2n + 2 = 72
n=5
The molecular formula of J is C5H12.
(c) (i) Substitution reaction
(ii) The reddish-brown colour of bromine disappears slowly. White fumes of HBr are formed
slowly.
(iii) C5H12 + Br2 → C5H11Br + HBr
2. (a)
(b)
(c)
(d)
They are compounds containing hydrogen and carbon atoms only.
P, Q and R
The melting and boiling points increase as the molecular sizes increase.
Two isomers of Q:
3. (a) (i)
(ii)
(b) (i)
(ii)
X is ethene.
Y is steam.
I: Hydrogenation; II: Hydration
Reaction I: 200°C; nickel catalyst
Reaction II: 300°C and 60 atm; phosphoric(V) acid
4. (a) Catalytic cracking
(b) A contains only single bonds. It is a saturated hydrocarbon. B contains carbon–carbon double
bonds. It is an unsaturated hydrocarbon.
© 2013 Marshall Cavendish International (Singapore) Private Limited
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Chemistry Matters for GCE ‘O’ Level (2 Edition): Full Solutions to Textbook Questions
Chapter 22
5. (a) C3H8
(b) Propane
(c) 2C3H8(g) + 10O2(g) → 6CO2(g) + 8H2O(g)
6. (a) Two
(b)
H H H H
H
| |
|
|
|
H – C = C – C = C – H + 2Br2 → Br – C –
|
H
Br Br H
|
|
|
C – C – C – Br
|
|
|
H H H
X
Y
(c) The molecular formula of Y is C4H6Br4.
The Mr of Y = (4 × 12) + (6 × 1) + (4 × 80) = 374
4 ! 30
Percentage of bromine in Y =
× 100 = 85.6%
374
Section C: Free-Response Questions
1. (a) (i) Difference in b.p. of pentane and butane = 36.1 – (–0.5) = 36.6°C
(ii) Difference in b.p. of undecane and dodecane = 216 – 196 = 20°C
(b) This hydrocarbon is not an alkane. Based on answer (a), the average increase in boiling point for
the increase of one carbon atom per molecule
36.6 + 20
=
= 28.3°C
2
If Z is an alkane with 6 carbon atoms, its boiling point is expected to be about 28.3°C above the
boiling point of pentane, that is, 64.4°C and not 80.1°C.
(c) (i) The general formula of alkanes is CnH2n+2.
Hence, the molecular formula of dodecane = C12H(2×12)+2 = C12H26.
(ii) Mr of dodecane = (12 × 12) + (26 × 1) = 170
12 ! 12
Percentage of carbon in dodecane =
× 100 = 84.7%
170
2. (a) (i)
(ii)
(b) (i)
(ii)
1.0 mol of hexene contains one C=C bond, so 1.0 mol of hexene reacts with 1.0 mol of
hydrogen. Thus, the heat of hydrogenation should be the same as that of butene, that is,
–127 kJ.
1.0 mol of hexadiene contains two C=C bonds, so 1.0 mol of hexadiene reacts with 2.0 mol
of hydrogen. Thus, the heat of hydrogenation should be 2(–127) kJ = –254 kJ.
Ethene reacts with steam under the conditions 300°C, 60 atm, and in the presence of the
catalyst phosphoric(V) acid (H3PO4). The product is ethanol.
C2H4(g) + H2O(g) → C2H5OH(l)
volume of gas
0.72
Number of moles of ethene used =
= 0.03 mol
=
molar volume
24
Number of moles of ethanol formed = 0.03 mol
Mr of ethanol (C2H5OH) = (2 × 12) + (5 × 1) + 16 + 1 = 46
Mass of ethanol = 0.03 × 46 = 1.38 g
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Chemistry Matters for GCE ‘O’ Level (2 Edition): Full Solutions to Textbook Questions
Chapter 22
26
= 0.026 mol
1000
(ii) 1.0 mol of the alkene reacts with 1.0 mol of bromine.
∴ Number of mol of alkene reacted = 0.026 mol
mass
1.46
0.026 =
=
molar mass
molar mass
∴ Molar mass of X = 56.2 g/mol
(b) The general formula of an alkene is CnH2n.
12n + 2n = 56.2
n=4
The molecular formula is C4H8.
(c) Straight-chain alkene:
3. (a) (i)
Number of mol of Br2 = 1.0 !
Branched-chain alkene:
(d) (i)
+ H2 →
Methylpropane
(ii) Molar mass of methylpropane (C4H10) = (12 × 4) + (10 × 1) = 58
Mass of 0.1 mol of methylpropane = 0.1 × 58 = 5.8 g
4. (a)
Name
Molecular
formula
ethyne
C 2H 2
–84
propyne
C 3H 4
–23
butyne
C 4H 6
8
Displayed formula
Boiling
point/°C
© 2013 Marshall Cavendish International (Singapore) Private Limited
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Chemistry Matters for GCE ‘O’ Level (2 Edition): Full Solutions to Textbook Questions
Chapter 22
(b) (i) 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)
(ii) An acetylene–oxygen flame burns at a higher temperature than a propane–oxygen flame.
(iii) Air contains about 21% oxygen. Pure oxygen is used to increase the flame temperature.
(c) The boiling points of the alkynes increase as their molecular sizes increase. This is because as
the molecular size increases, the attractive forces between the molecules become stronger.
(d) (i) CnH2n–2
(ii) The molecular formula of P is C6H(2×6)–2, i.e. C6H10.
(e) The reddish-brown colour of bromine will be decolourised. This is because butyne is an
unsaturated hydrocarbon.
© 2013 Marshall Cavendish International (Singapore) Private Limited
22.7
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