Carbon Compounds MCQ Ans
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Multiple Choice Answers 1
C (1) is incorrect. For compounds in the same homologous series, they show a steady variation in physical properties as the number of carbon atoms increases. For example, the boiling point always increases with the increase in the number of carbon atoms.
2
D
3
A Since the members of a homologous series have similar structures, they show similar chemical properties.
4
D The molecules of alcohols and carboxylic acids are held together by hydrogen bonds as well as van der Waals’ forces. More energy is required to overcome hydrogen bonds. Therefore, alcohols and carboxylic acids have higher boiling points than alkanes with a similar relative molecular mass.
5
A They belong to the same homologous series because they have the same general formula, C 2 H 2n .
6
C 1-chloropropane and 2-chlorobutane have the same general formula, C n H 2n+1 Cl and they differ
Condensed formulae are the simple forms of structural formulae.
from each other by a CH 2 group. 7
A Alkanes contain single bonds only.
8
D
9
A 2,2-dimethylpropane is a branched-chain alkane. It is more rounded and has a smaller surface area than pentane which is a straight-chain alkane. Therefore, the dispersion forces among 2,2-dimethylpropane molecules are weaker than those among pentane molecules.
10
C
11
A Pentanal is an aldehyde. Its structure does not contain any hydrogen atoms attached to the highly electronegative atoms such as N, O and F atoms. Therefore, pentanal cannot form hydrogen bonds among their molecules.
12
C Both molecules of propanoic acid and propan-1-ol are held together by hydrogen bonds as well as van der Waals’ forces. However, propanoic acid has more extensive intermolecular hydrogen bonds
The molecular formula of ethyl ethanoate is C 4 H 8 O 2 .
than propan-1-ol because both C=O group and OH group can participate in hydrogen bond formation. The molecules of methyl ethanoate are held together by dipole-dipole forces and dispersion forces only. 13
D When more than one substituent is present, the parent chain is numbered from the end closest to the first substituent. Then the substituents are arranged in alphabetical order.
14
A Both molecules of propanoic acid and ethanamide are held together by hydrogen bonds as well as van der Waals’ forces. However, ethanamide has more extensive intermolecular hydrogen bonds than propanoic acid. Since each ethanamide molecule contains two slightly positive hydrogen atoms and two lone pairs of electrons on the oxygen atom, it can form more intermolecular hydrogen bonds. The molecules of methyl ethanoate are held together by dipole-dipole forces and dispersion forces only.
15
D
16
D Each amide molecule contains two slightly positive hydrogen atoms and two lone pairs of electrons on the oxygen atom, so it can form more intermolecular hydrogen bonds. Therefore, more energy is needed to overcome the intensive hydrogen bonds during the boiling process.
17
A Only unsubstitued amides are acid derivatives. It can be formed from the reaction between a
carboxylic acid and ammonia. Unsubstituted amides can form more intermolecular hydrogen bonds than carboxylic acids with a similar relative molecular mass. Carboxylic acids have stronger intermolecular hydrogen bonds than primary amine molecules. Therefore, the correct order of decreasing boiling point is unsubstituted amides > carboxylic acids > primary amines. 18
C
For C, all compounds are haloalkanes and they have the same general formula, C n H 2n+1 Cl.
19
B Both propan-1-ol and propanoic acid can form hydrogen bonds with water molecules, so they are soluble in water. However, chloromethane cannot form hydrogen bonds with water molecules, so it is insoluble in water.
20
B CH 3 CCl 3 cannot form hydrogen bonds with water molecules, so it is insoluble in water. However, the others can form hydrogen bonds with water molecules, so they are soluble in water.
21
D
22
A Referring to the general formula of ketones, RCOR’, ketone molecules do not contain any slightly positive hydrogen atoms attached to some highly electronegative atoms e.g. N, O and F. Therefore, ketone molecules do not form hydrogen bonds between each other.
23
D
24
D
25
A Propan-2-ol and butanoic acid are soluble in water because their molecules can form hydrogen bonds with water molecules.
26
B In each ester molecule, there is no hydrogen atom attached to highly electronegative atoms e.g. N, O and F atoms. Therefore, there are no hydrogen bonds among the ester molecules. The ester molecules are held together by dispersion forces as well as dipole-dipole forces.
27
C 2-methylpropanamide is an unsubstituted amide because the two hydrogen atoms attached to the nitrogen atom are not replaced by other atoms or groups of atoms.
28
A Alkanes and alkenes have lower boiling and melting points than other organic compounds with similar relative molecular masses.
29
B In a condensed formula, single bonds are usually omitted. However, carbon-carbon multiple bonds e.g. C=C and C≡C must be written.
30
B Both molecules of butanoic acid and butanol are held together by hydrogen bonds as well as van der Waals’ forces while the molecules of butanone are held together by dipole-dipole forces as well as dispersion forces only. Since the strength of hydrogen bonds > dipole-dipole forces > dispersion forces, the boiling points of butanoic acid and butanol are higher than that of butanone. Butanoic acid has more extensive intermolecular hydrogen bonds than butanol, so butanoic acid has a higher boiling point than butanol.
31
D Each amide molecule contains two slightly positive hydrogen atoms and two lone pairs of electrons on the oxygen atom, so it can form a maximum of four hydrogen bonds.
32
B If only one hydrogen atom attached to the nitrogen atom of ammonia is substituted by an alkyl group, it is a primary amine.
33
D
34
A
Propanoic acid has more extensive intermolecular hydrogen bonds than propan-1-ol.
Since all molecules can form hydrogen bonds with water molecules, they can dissolve in water.
35
B
36
C
37
C For 2-methylpropanamide and 2,3-dichloropropan-1-ol, there are hydrogen bonds between the lone pair on the highly electronegative oxygen atom and the slightly positive hydrogen atom in another molecule.
38
D
39
A
40
B
41
B
42
D
43
A
44
C
45
D
46
D
47
D
48
D The carbonyl oxygen atom of propanone can form two hydrogen bonds with water molecules. The amide group of propanamide can form four hydrogen bonds with water molecules. The amine group of propan-1-amine can form three hydrogen bonds with water molecules.
49
A
50
D
Carbon-carbon double bonds only take the suffix form (-en) and are used with other suffixes.
51
B
The compound has functional groups of COOH and CONH 2 .
52
C When carbon compounds contain more than one functional group, the order of priority determines which group is named as prefix or suffix. Since aldehydes have higher priority than ketones in naming,
The general formula for unsubstituted amides is RCONH 2 , where R is an alkyl group.
CHO group takes the suffix form while the others take the prefix forms. In addition, prefixed substituents are arranged in alphabetical order. 53
C There are seven carbon atoms (hepta-) and two double bonds (dien) in the longest chain. One double bond occurs between C-2 and C-3 while another between C-5 and C-6 (2,5-dien). In addition, a C=O group, a Cl group and a OH group are attached to C-4 (4-oxo-), C-3 (3-chloro-) and C-6 (6-hydroxy) respectively. Finally, there is a COOH group at C-1 (-oic acid).
54
C
55
C
56
C It has two functional groups namely amino group and carboxyl group. ‘2-’ is not needed since the amino group is always at C-2 when it is the prefix.
57
B The functional group of carboxylic acids has the highest priority in naming. Therefore, it takes the suffix form while the others take the prefix forms. Prefixed substituents are arranged in alphabetical order.
58
D Halogen atoms must take the prefix form.
59
B In this case, the functional group of amide has the higher priority than the others in naming. Therefore, it takes the suffix form while the others take the prefix forms. Prefixed substituents are
Carbon-carbon double bonds only take the suffix form (-en).
arranged in alphabetical order. 60
A
61
B The functional group of carboxylic acid is the highest priority group in naming. Therefore, it takes the suffix form while the others take the prefix forms. The prefixed substituents are arranged in alphabetical order.
62
A (2) is incorrect. When organic compounds contain more than one functional group, the order of priority determines their homologous series. (3) is incorrect. Two organic compounds with the molecular mass differing by CH 2 must belong to the same homologous series.
63
B
The structural formula of diol is
. The IUPAC name for the compound
is 4-hydroxypentanal, not 5-formylpent-2-ol.
The IUPAC name for the compound
is
2-chloro-3-hydroxybutanoic acid, not 3-carboxy-3-chlorobutan-2-ol. 64
C Carboxylic acid is the highest priority group in naming, so it takes the suffix form, with all others taking the prefix forms.
65
D
66
B
67
B The functional group of carboxylic acids has the highest priority in naming, so it takes the suffix form while the others take the prefix forms.
68
D The functional group of carboxylic acids has the highest priority in naming, so it takes the suffix form and determines the homologous series.
69
A
70
D
71
C
72
C
73
A The alcohol molecules are held together by hydrogen bonds as well as van der Waals’ forces while the alkane molecules are held together by van der Waals’ forces only. More energy is required to overcome hydrogen bonds. Therefore, alcohols have higher boiling and melting points than alkanes with a similar relative molecular mass.
74
C Haloalkanes are polar but most of them are insoluble in water. Because the amount of energy released from the formation of intermolecular forces between most haloalkane molecules and water molecules cannot compensate for the hydrogen bonds among water molecules.
75
D The boiling points of carboxylic acids are higher than those of alcohols with a similar relative molecular mass because carboxylic acids have more extensive intermolecular hydrogen bonds than
alcohols. 76
A The haloalkane molecules are held together by dipole-dipole forces as well as dispersion forces while the alkane molecules are held together by dispersion forces only. Since the strength of dipole-dipole forces > dispersion forces, more energy is needed to separate the haloalkane molecules during boiling.
77
A
78
A
79
D The acyclic structural isomers of C 5 H 10 are shown as follows:
,
,
,
, 80
D A pair of structural isomers has the same relative molecular mass because they have the same molecular formula.
81
D Ethanoic acid is a carboxylic acid while methyl methanoate is an ester. Since they have different functional groups, they have different chemical properties.
82
D All of them have the same molecular formula C 4 H 8 O.
83
D Both ethanoic acid and methyl methanoate have the same molecular formula, so they have the same relative molecular mass. They have different functional groups, so they have different chemical properties.
84
A Both methyl propanoate and 4-hydroxybutanal have the same molecular formula C 4 H 8 O 2 . However, the molecular formula of butane-1,4-diol is C 4 H 10 O 2 , not C 4 H 8 O 2 .
85
D Butanoic acid is reduced to butan-1-ol. Butan-2-ol is the position isomer of butan-1-ol because it has the same molecular formula as butan-1-ol (i.e. C 4 H 10 O). They only differ in the position of the functional group.
86
A Isomers are compounds with the same molecular formula but different arrangements of atoms in space. Therefore, they have the same empirical formula but different structural formula. Besides, isomers may belong to different homologous series, so they may have different general formulae.
87
D They are position isomers because they differ only in the position of the functional groups. Therefore, they have the same functional groups and similar chemical properties.
88
D For (1), cyclohexane cannot decolorize bromine in 1,1,1-trichloromethane. For (2), isomers with different structures have different melting points or boiling points. For (3), 3-methylpent-1-ene is a chiral compound.
89
C The structural formulae of the structural isomers of C 4 H 8 are shown below:
,
,
,
90
D
91
C The structural isomers of C 3 H 6 Cl 2 are shown as follows:
and
All of them have the molecular formula C 5 H 10 .
,
,
,
92
B
Propyl ethanoate has the molecular formula C 5 H 10 O 2 , not C 5 H 10 O.
93
A A pair of geometrical isomers ‘may’ be optically active.
94
D (1) is incorrect because if either one of the doubly-bonded carbon atoms is attached to two identical atoms or groups of atoms, no cis-trans isomers are possible. For (3), a pair of cis-trans isomers can also exhibit enantiomerism, e.g. 4-chloropent-2-ene can exhibit both cis-trans isomerism and enantiomerism.
95
C The structural formula of 3,6-dibromo-4-methylhept-4-en-2-ol is shown as follows:
96
B In 2-methylpent-2-ene, one of the doubly-bonded carbon atoms is attached to two identical groups of atoms (CH 3 ), so it could not exhibit geometrical isomerism.
97
A Both 2-aminopropanal and 2,3-dihydroxybutanedioic acid have chiral carbon atoms in their molecules.
2-aminopropanal
2,3-dihydroxybutanedioic acid 98
B
99
B
The carbon atom with a hydroxyl group attached to it is a chiral carbon.
100
B Chiral carbon atom is a carbon atom which has four different atoms or groups of atoms attached to it. (1) is correct because CH 3 CH=CHCH 2 CHO does not contain any chiral carbon atoms. (2) is incorrect because this compound has a functional group of CHO instead of a hydroxyl group. (3) is correct because the 2 hydrogen atoms are on the same side of the the C=C double bond.
101
D (1) is incorrect because it has two identical atoms (i.e. H) attached to one of the doubly-bonded carbon atoms.
102
C
Vitamin E has three chiral carbon atoms.
Vitamin E is soluble in fats or oils only because it has a long hydrocarbon chain. 103
B (1) is a pair of geometrical isomers. (3) is a pair of enantiomers. The two compounds in (2) are identical.
104
A Isomers are compounds with the same molecular formula, so they must have the same relative molecular mass. If a compound has two identical atoms or groups of atoms attached to the same doubly-bonded carbon atom, it does not show geometrical isomerism even it contains a carbon-carbon double bond. A mixture of a pair of enantiomers can rotate the plane of polarized light if they are not present in the same amount. Compounds with the same functional group at different positions are called position isomers.
105
C Enantiomers are chiral molecules with at least one carbon atom attached to four different atoms or group of atoms. There is no plane of symmetry in the chiral molecules.
106
B In (CH 3 ) 2 C=CH 2 , each doubly-bonded carbon atom is attached to two identical atoms (H) or groups of atoms (CH 3 ), so it could not exhibit geometrical isomerism.
107
B In CH 2 =CHCHBrCH 3 , the C-3 carbon atom is attached to four different atoms or groups of atoms, so it is a chiral molecule and exhibits enantiomersim.
108
C 2-methylpropanal does not have any carbon atoms attached to four different atoms or groups of atoms, so it is not a chiral molecule.
109
D The functional group isomers are: and
The geometrical isomers are:
and The enantiomers are:
and 110
B In CH 3 CH 2 CH 2 CH(CH 3 )COCH 3 , the C-3 carbon atom is attached to four different atoms or groups of atoms, so it is a chiral molecule and exhibits enantiomerism.
111
B The trans isomer has a higher melting point than the cis isomer because the trans isomer has a higher packing efficiency in the solid state. If one of the doubly-bonded carbon atoms of an organic compound is attached to two identical atoms or groups of atoms, no geometrical isomers are possible.
112
D One of the carbon atoms is attached to four different groups, so compound X has a chiral carbon atom. Compound X can form hydrogen bonds with the water molecules, so it is soluble in water.
113
C
114
B The molecules of cis-2,3-dibromobut-2-ene are held together by dipole-dipole forces as well as dispersion forces while the molecules of trans-2,3-dibromobut-2-ene are held together by dispersion forces only. Therefore, they have different intermolecular forces which lead to different boiling points.
115
C One of the planes of symmetry cuts from the hydrogen atom and divides the molecule into two equal halves while the other plane of symmetry cuts from the chlorine atom.
116
D
117
A A pair of enantiomers can rotate the plane of polarized light to different directions. A pair of enantiomers has the same boiling point because they have identical intermolecular forces.
118
B Only enantiomers are mirror images of one another. A pair of eanantiomers which are present in different amounts can rotate the plane of polarized light.
119
C
120
B An achiral molecule does not contain any chiral carbon atoms. Butan-1-ol is an achiral molecule because it does not contain any chiral carbon atoms. A chiral molecule cannot be superimposed on its own mirror image.
121
D
All of them have a carbon atom attached to four different atoms or groups of atoms.
122
B
Only 1-chloro-2-methylbutane has a chiral carbon atom, so it can exhibit enantiomerism.
123
D Both 2-chloro-2-fluorobutane and 5-chloro-3,4-dimethylpentan-1-ol have chiral carbon atoms in their molecules.
2-bromobut-2-ene has two geometrical isomers, cis-2-bromobut-2-ene and trans-2-bromobut-2-ene.
All of them have chiral carbon atoms in their molecules, so they could exhibit enantiomerism.
2-chloro-2-fluorobutane
5-chloro-3,4-dimethylpentanoic acid 124
B
They differ in the types of functional group present.
125
C Both trans-5-chloro-5-fluoropent-2-ene and 2-chloro-2-fluorobutane have chiral carbon atoms in their molecules.
126
C
127
D Geometrical isomers have the same molecular formula, so they have the same relative molecular mass.
128
A The C-3 carbon atom of 3-bromobut-1-ene is attached to four different atoms or groups of atoms. There is a chiral carbon atom in the molecule of 3-bromobut-1-ene. Therefore, 3-bromobut-1-ene could exhibit enantiomerism.
129
A
130
C Chain isomers have different physical properties. For example, pentane and 2-methylbutane are chain isomers. Their strengths of intermolecular forces are different, so they have different boiling points.
131
B 2-methylbutane is a branched-chain alkane which is more rounded and has a smaller surface area than pentane, so the dispersion forces between their molecules are weaker. Therefore, less energy is needed to separate the molecules during boiling. Pentane and 2-methylbutane are chain isomers because they differ in the length of the main carbon chain.
132
C Position isomers differ only in the position of the functional group(s) but not the types of functional group present.
133
C Some isomers have different functional groups, e.g. functional group isomers, so they have different chemical properties.
134
A
135
C The mixture of the pair of enantiomers does not rotate plane-polarized light only if the two enantiomers are present in the same amount.
136
A
137
A When methane is in excess, chloromethane is the major product.
138
D The product formed from the above reaction is a mixture of chloroalkanes, including chloromethane, dichloromethane, trichloromethane and tetrachloromethane. Chloroform is the trivial name of trichloromethane.
139
A
140
D
Geometrical isomerism is a type of stereoisomerism.
The reaction only takes place in the presence of sunlight which supplies energy to break the ClCl
bonds in chlorine molecules. 141
D
142
B According to Markovnikov’s rule, the hydrogen atom in hydrogen chloride is added to the carbon atom of the carbon-carbon double bond that already carries the larger number of hydrogen atoms. Then the chlorine atom is added to the carbon atom carrying fewer hydrogen atoms.
143
C (1) is incorrect. A finely divided metal catalyst such as Pt, Pd or Ni is often used to speed up the hydrogenation.
144
C According to Markovnikov’s rule, the hydrogen atom in hydrogen bromide is added to the carbon atom of the carbon-carbon double bond that already carries the larger number of hydrogen atoms. Then the bromine atom is added to the carbon atom carrying fewer hydrogen atoms.
145
B According to Markovnikov’s rule, the hydrogen atom in HBr is added to the carbon atom of the carbon-carbon double bond that already carries the larger number of hydrogen atoms. The major products of the reactions for A, C and D should be CH 3 CHBrCH 3 , Br 2 CHCH 3 and (CH 3 ) 3 CCHBrCH 3 respectively.
146
C According to Markovnikov’s rule, the hydrogen atom in hydrogen chloride is added to the carbon atom of the carbon-carbon double bond that already carries the larger number of hydrogen atoms. Then the chlorine atom is added to the carbon atom carrying fewer hydrogen atoms. Therefore, the major product is 2-chlorobutane.
147
D According to Markovnikov’s rule, the hydrogen atom in hydrogen bromide is added to the carbon atom of the carbon-carbon double bond that already carries the larger number of hydrogen atoms. Then the bromine atom is added to the carbon atom carrying fewer hydrogen atoms. Therefore, the major product is 2,5-dibromo-2,5-dimethylhexane.
148
B
149
A (3) is a substitution reaction.
150
B According to Markovnikov’s rule, the major product of the reaction between propene and hydrogen chloride is 2-chloropropane.
151
C A is incorrect. Propene can decolorize acidified potassium permanganate solution but not acidified potassium dichromate solution. B is incorrect. Propene can be prepared by treating propan-1-ol with
Both of them are addition reactions.
concentrated sulphuric acid at 180C. D is incorrect. According to Markovnikov’s rule, propene reacts with hydrogen chloride to give 2-chloropropane as the major product. 152
C
According to Markovnikov’s rule, the major product of the reaction is 2-chloro-2-methylpropane.
153
C
Hydrogenation of alkenes is exothermic. This is because the energy released during the formation
of new CH bonds is more than the energy needed for partially breaking the original C=C bonds. 154
B
155
D
According to Markovnikov’s rule, the major product of the reaction is 2,3-diiodobutane.
156
B
(2) is a substitution reaction between butane and chlorine.
157
B For (2), OH ions are in aqueous solution e.g. sodium hydroxide solution.
158
C CH 3 CH 2 CH 2 CH 3 undergoes substitution reaction with bromine in sunlight to give CH 3 CH 2 CH 2 CH 2 Br. CH 3 CH 2 CH 2 CH 2 Br undergoes substitution reaction with hydroxide ions to give
CH 3 CH 2 CH 2 CH 2 OH. 159
A (3) is an addition reaction between but-2-ene and hydrogen.
160
B Since the strength of CCl bond is stronger than that of CI bond, the reaction between 2-chlorobutane and NaOH(aq) should be slower.
161
C The chlorine atoms in chloromethane are substituted by hydroxide ions, so the reaction mixture contains chloride ions which react with silver ions to form silver chloride, a white precipitate.
162
B
163
D Propan-1-ol is a primary alcohol which is first oxidized to propanal (aldehyde) and is then further oxidized to propanoic acid (carboxylic acid).
164
D Primary alcohols are first oxidized to aldehydes and are then further oxidized to carboxylic acids. Secondary alcohols are oxidized to ketones which are resistant to be further oxidized. Tertiary alcohols are resistant to oxidation.
165
B Butan-1-ol is dehydrated to but-1-ene by treating with concentrated sulphuric acid at 180C. Therefore, compound X is but-1-ene. Then, but-1-ene undergoes hydrohalogenation to give two products, 1-chlorobutane and 2-chlorobutane. According to Markovnikov’s rule, 2-chlorobutane is the major product.
166
B
167
B Only primary alcohols can be oxidized to carboxylic acids. Secondary alcohols are oxidized to ketones and tertiary alcohols do not undergo oxidation under the same conditions.
168
D
169
A Butan-2-ol is a secondary alcohol, so it can be oxidized to butanone (ketone) only. Butanone is resistant to further oxidation, so it is left as the end product.
170
A Pentan-1-ol is a primary alcohol, so it is first oxidized to pentanal and is then further oxidized to pentanoic acid. The intermediate product – pentanal could be collected by special experimental set-up.
171
B
172
C Since methanol is a primary alcohol, it is first oxidized to methanal and is then further oxidized to methanoic acid.
173
D
174
A Firstly, KBr reacts with concentrated H 2 SO 4 to give HBr and KHSO 4 . Then, HBr formed reacts with CH 3 CH 2 OH to give CH 3 CH 2 Br.
175
A According to Markovnikov’s rule, the hydrogen atom of hydrogen chloride should be added to the carbon atom of the carbon-carbon double bond that already carries the larger number of hydrogen
(2) is the reagent used to convert alkene to dihaloalkane.
(2) is an addition reaction.
atoms. Besides, the OH group is substituted by Cl. Therefore, the major product should be 1,2,2-trichlorobutane. The compound reacts with acidified potassium dichromate solution to give a carboxylic acid. 176
B
(1) is a primary alcohol, (2) is a secondary alcohol and (3) is a tertiary alcohol.
177
A Hydrogen bromide should be made by heating solid potassium bromide with concentrated sulphuric acid.
178
A (2) is incorrect because concentrated sulphuric acid should be used instead of the dilute one for the
dehydration of an alcohol. (3) is incorrect. Since compound X is a secondary alcohol, it is only oxidized to a ketone which is resistant to further oxidation. 179
C
180
A (1) and (2) are common oxidizing agents.
181
A (3) is incorrect because ketones are resistant to oxidation.
182
B Lithium aluminium hydride in dry ether has no reaction with alkenes. It is used to reduce aldehydes to alcohols.
183
B Pentan-2-one is a ketone which is resistant to oxidation, therefore it does not react with acidified potassium dichromate solution.
184
A LiAlH 4 is insoluble in common organic solvents, it is used as a solid reducing reagent.
185
B
186
C For (2), ketones are reduced to secondary alcohols. For (3), the reactions between haloalkanes and sodium hydroxide solution could give primary alcohols, secondary alcohols and tertiary alcohols.
187
D
188
D The ketone is reduced to a diol.
189
B
190
C There are two types of reactions. One is a substitution of the bromine atom by a hydroxide ion and
(2) is incorrect. The ketone group should also be reduced to a secondary alcohol,
The ketone is reduced to an alcohol.
the other is a neutralization between COOH group and sodium hydroxide. 191
C Lithium aluminium hydride in dry ether is used to reduce propanoic acid to propan-1-ol. Therefore, compound A is propan-1-ol. Then propan-1-ol is heated with a mixture of phosphorus and bromine to give 1-bromopropane.
192
B Propanal does not react with concentrated sulphuric acid and thionyl chloride. Lithium aluminium hydride in dry ether is used to reduce propanal to propan-1-ol.
193
A Lithium aluminum hydride in dry ether is used to reduce carboxylic acids, aldehydes and ketones to alcohols.
194
B CH 3 COCH 2 CH 3 is a ketone and (CH 3 ) 3 COH is a tertiary alcohol. Both of them are resistant to oxidation. Only primary and secondary alcohols can be oxidized by acidified potassium dichromate solution.
195
D CH 3 CH 2 CH 2 Cl undergoes substitution with hydroxide ions to give CH 3 CH 2 CH 2 OH. Then, CH 3 CH 2 CH 2 OH is oxidized by acidified K 2 Cr 2 O 7 to give CH 3 CH 2 COOH.
196
A (3) only produces a secondary alcohol.
197
A
198
B
199
B
200
D
Butanoic acid is reduced to butan-1-ol.
201
D Since butan-1-ol is a primary alcohol, it can be oxidized to butanoic acid. The products formed from the acid hydrolysis of butyl methanoate are methanoic acid and butan-1-ol.
202
D
The acid hydrolysis does not go to completion because it is a reversible reaction.
203
D
There are acid hydrolysis and alkaline hydrolysis of an ester respectively.
204
C An amide is hydrolysed to give a carboxylate ion and ammonia by heating under reflux with sodium hydroxide solution.
205
A There are two types of reactions. One is a substitution of the chlorine atom by a hydroxide ion and the other is an alkaline hydrolysis of an amide to give a carboxylate ion and ammonia.
206
D
207
B CH 3 CH 2 CONH 2 undergoes alkaline hydrolysis to give carboxylate ion, CH 3 CH 2 COO.
208
C Propan-2-ol reacts with an oxidizing agent to give a ketone. However, ketones are resistant to further oxidation.
209
C
210
B
211
A
212
A The reaction may give 2 products, 2-chloro-2-methylbutane and 2-chloro-3-methylbutane. However, according to Markovnikov’s rule, 2-chloro-2-methylbutane is the major product.
213
C
214
C Ethanol can be oxidized by acidified potassium dichromate solution. Ethanol is a primary alcohol, so it can be oxidized to ethanoic acid.
215
D Carboxylic acids can be reduced to alcohols by mixing with LiAlH 4 in dry ether first, the reaction mixture is then treated with a dilute acid. LiAlH 4 must be used in dry ether solvent because it reacts violently with water. Besides, LiAlH 4 is a strong reducing agent.
216
C The synthetic route is:
Heating ethyl butanoate with dilute acid under reflux will give butanoic acid and ethanol.
An alcohol is converted to an alkene by dehydration, not oxidation.
Alkane Haloalkane Alcohol Carboxylic acid Amide. 217
218
B
The synthetic route is:
A The synthetic route is:
219
A The synthetic route is: 2-bromobutane butan-2-ol butanone.
220
D 1,2-dichloropropane reacts with hydroxide ions to give propane-1,2-diol. Then propane-1,2-diol is oxidized by acidified potassium dichromate solution and the primary alcohol is converted to a carboxylic acid while the secondary alcohol is converted to a ketone. The synthetic route is
221
C
The synthetic route is amide carboxylic acid primary alcohol alkene alkane.
222
B
The synthetic route is
223
C
The synthetic route is
224
C
The synthetic route is
225
B The reagent A is concentrated sulphuric acid which dehydrates an alcohol to form an alkene. According to Markovnikov’s rule, the major product is 2-bromopropane.
226
D
The bromine and chlorine atoms are substituted by the hydroxide ions.
227
B
The synthetic route is alkane haloalkane alcohol carboxylic acid amide.
228
B
The synthetic route is
229
D
The synthetic route is
230
D
The synthetic route is
231
D (1) involves a reduction of a carboxylic acid to an alcohol which requires a dilute acid. (2) involves an oxidation of an alcohol to a carboxylic acid which requires an acid to acidify potassium dichromate solution. (3) involves an acid hydrolysis of an ester which requires dilute HCl.
232
B
233
A Haloalkane alcohol carboxylic acid
234
C
235
D
(1) and (3) are regarded as alkaline hydrolysis. (2) is an amide formation. The synthetic route is CH 3 CH 2 CH 2 OH CH 3 CH=CH 2 CH 3 CH 2 CH 3 .
The synthetic route is: CH 3 CH 2 CONH 2 CH 3 CH 2 COOH CH 3 CH 2 CH 2 OH CH 3 CH=CH 2 236
A Butan-2-ol undergoes dehydration at 180C. Both but-1-ene and but-2-ene are formed.
237
B The chlorine atom of 2-chloropropane is substituted by hydroxide ions to give propan-2-ol. The secondary alcohol is oxidized by acidified K 2 Cr 2 O 7 to give ketone.
238
D
239
C In reaction A, the alcohol is dehydrated to give an alkene. In reaction B, a hydrogen halide is added across a carbon-carbon double bond, converting an alkene to a haloalkane.
240
A Ethanol is oxidized by acidified potassium dichromate solution to give ethanal and then ethanoic acid. However, ethanol does not react with ethanoic acid to give ethyl ethanoate.
241
D
The synthetic route is
242
B
The synthetic route is
243
C
The synthetic route for the conversion of propene to propanone is:
The synthetic route is
244
C
The synthetic route for the conversion of ethene to ethanamide is:
245
B
The synthetic route for the conversion of 2-bromo-2-methylbutane to 2-methylbutane is:
246
A No catalyst is used in the reaction. To prepare ethanal from ethanol, the reaction mixture should be warmed to a temperature that is above the boiling point of ethanal but below that of ethanol. Then, ethanal is distilled out as soon as it is formed. Ethanol is flammable, so the reaction mixture should be heated in a water bath.
247
C
The synthetic route is
248
B
The synthetic route is
249
B
The synthetic route is
250
D
The synthetic route is
Since excess Cl 2 in organic solvent is used during the conversion of Q to R, all hydrogen atoms of ethane are substituted by chlorine atoms to give CCl 3 CCl 3 .
251
D
The synthetic route is
Since all hydrogen atoms of ethane are substituted by chlorine atoms to give CCl 3 CCl 3 , excess Cl 2 in organic solvent should be used during the conversion of CH 3 CH 3 to CCl 3 CCl 3 . 252
D
The synthetic route is
Since all hydrogen atoms of ethane are substituted by chlorine atoms to give CCl 3 CCl 3 , excess Cl 2 in organic solvent should be used during the conversion of CH 3 CH 3 to CCl 3 CCl 3 . 253
C
The synthetic route is
254
D
The synthetic route is
255
C
The synthetic route is
256
A The synthetic route is
257
D
The synthetic route is
258
D
The synthetic route is
259
C
The synthetic route is
260
C
The synthetic route is
261
B
The synthetic route is
262
C
The synthetic route is
263
B (1) involves reducing CH 3 CHO to CH 3 CH 2 OH by using LiAlH 4 which is a reducing agent. (3) involves reducing CH 3 COOH to CH 3 CH 2 OH by using LiAlH 4 which is a reducing agent.
264
B
265
C
266
C Propanone and propanal can be reduced to a secondary alcohol and a primary alcohol respectively in a single step. Propene reacts with acidified potassium permanganate solution to give propane-1,2-diol
The synthetic route is alkane haloalkane alcohol aldehyde.
in a single step. However, the synthetic route for converting propane to an alcohol is alkane haloalkane alcohol. 267
D
The synthetic route is
268
D
The synthetic route is
269
C The overall reaction is CH 3 CH 2 OH + HBr CH 3 CH 2 Br + H 2 O No. of moles of CH 3 CH 2 OH used = No. of moles of HBr used =
23.0 g (12.0 2 1.0 6 16.0) g mol 1
= 0.500 mol
48.5 g = 0.600 mol (1.0 79.9) g mol 1
From the equation, the mole ratio of CH 3 CH 2 OH to HBr = 1 : 1, ∴ CH 3 CH 2 OH is a limiting reagent. From the equation, the mole ratio of CH 3 CH 2 OH to CH 3 CH 2 Br = 1 : 1, ∴ no. of moles of CH 3 CH 2 Br obtained = 0.500 mol Theoretical mass of CH 3 CH 2 Br obtained = 0.500 mol × (12.0 × 2 + 1.0 × 5 + 79.9) g mol1 = 54.5 g 32.7 g 100% = 60.0% Percentage yield of CH 3 CH 2 Br = 54.5 g 270
D The mole ratio of ethanol to ethanoic acid is 1 : 1, ∴ the theoretical no. of moles of ethanoic acid obtained = 1 mol The theoretical mass of ethanoic acid obtained = 1 mol × (12.0 × 2 + 1.0 × 4 + 16.0 × 2) g mol1 = 60.0 g The actual mass of ethanoic acid obtained = 0.8 mol × (12.0 × 2 + 1.0 × 4 + 16.0 × 2) g mol1 = 48.0 g 48.0 g 100% = 80.0% Percentage yield of ethanoic acid = 60.0 g
271
Ethanol is a primary alcohol which is first oxidized to ethanal and is then further oxidized to ethanoic acid. Acidified KMnO 4 is also an oxidizing agent. D The overall percentage yield = 80% × 75% × x% ∴ 36% = 80% × 75% × x%.
272
A The overall yield = 80% × 60% × 50% = 24%
273
C
The overall equation is
No. of moles of ethanol used =
4.6 g = 0.1 mol (12.0 2 1.0 6 16.0) g mol 1
From the equation, the mole ratio of ethanol to ethene is 1 : 1, ∴ no. of moles of ethene obtained = 0.1 mol Theoretical mass of ethene obtained = 0.1 mol × (12.0 × 2 + 1.0 × 4) g mol1 = 2.8 g 1.4 g Percentage yield of ethene = 100% = 50% 2.8 g 274
B The overall yield = 90% × 80% × 50% × 40% = 14.4%
275
C The overall equation is CH 3 CH 2 Cl + OH CH 3 CH 2 OH + Cl 70.0 g No. of moles of chloroethane used = = 1.09 mol (12.0 2 1.0 5 35.5) g mol 1 From the equation, the mole ratio of chloroethane to ethanol is 1 : 1, ∴ no. of moles of ethanol obtained = 1.09 mol Theoretical mass of ethanol obtained = 1.09 mol × (12.0 × 2 + 1.0 × 6 +16.0) = 50.1 g 27.0 g 100% = 53.9% The percentage yield of ethanol = 50.1 g
276
D
277
B The overall equation is CH 2 = CH 2 + HCl CH 3 CH 2 Cl. 28.0 g No. of moles of ethene used = = 1 mol (12.0 2 1.0 4) g mol 1 From the equation, the mole ratio of ethene to chloroethane is 1 : 1, ∴ no. of moles of chloroethane obtained = 1 mol Theoretical mass of chloroethane obtained = 1 mol × (12.0 × 2 + 1.0 × 5 + 35.5) g mol1 = 64.5 g 38.0 g Percentage yield of chloroethane = 100% = 58.9% 64.5 g
278
C The overall yield = 70% × x% × 80% = 28%.
279
C The overall equation is
No. of moles of butan-1-ol =
12.0 g = 0.162 mol (12.0 4 1.0 10 16.0) g mol 1
No. of moles of ethanoic acid =
10.2 g = 0.170 mol (12.0 2 1.0 4 16.0 2) g mol 1
From the equation, the mole ratio of butan-1-ol to ethanoic acid = 1 : 1, ∴ butan-1-ol is the limiting reagent. From the equation, the mole ratio of butan-1-ol to butyl ethanoate = 1 : 1, ∴ no. of moles of butyl ethanoate obtained = 0.162 mol Theoretical mass of butyl ethanoate obtained = 0.162 mol × (12.0 × 6 + 1.0 × 12 + 16.0 × 2) g mol1 = 18.8 g 5.8 g × 100% = 30.9% Percentage yield of butyl ethanoate = 18.8 g 280
C The overall equation is
No. of moles of methyl ethanoate used =
10.5 g = 0.142 mol (12.0 3 1.0 6 16.0 2) g mol 1
From the equation, the mole ratio of methyl ethanoate to ethanoic acid is 1 : 1, ∴ theoretical no. of moles of ethanoic acid obtained = 0.142 mol Theoretical mass of ethanoic acid obtained = 0.142 mol × (12.0 × 2 + 1.0 × 4 + 16.0 × 2) g mol1 = 8.52 g The equation for the reaction between ethanoic acid and sodium carbonate solution is 2CH 3 COOH + Na 2 CO 3 2CH 3 COONa + H 2 O + CO 2 No. of moles of Na 2 CO 3 used = 0.5 mol dm3 × 0.1 dm3 = 0.05 mol From the equation, the mole ratio of CH 3 COOH to Na 2 CO 3 is 2 : 1, ∴ no. of moles of CH 3 COOH obtained = 0.05 mol × 2 = 0.10 mol Mass of CH 3 COOH obtained = 0.10 mol × 60.0 g mol1 = 6.0 g 6.0 g Percentage yield of ethanoic acid = 100% = 70.4% 8.52 g 281
C The overall equation is
The mass of methanoic acid used = 10 cm3 × 1.2 g cm3 = 12.0 g 12.0 g No. of moles of methanoic acid used = = 0.261 mol (12.0 1.0 2 16.0 2) g mol 1 The mass of ethanol used = 10 cm3 × 0.8 g cm3 = 8.0 g
No. of moles of ethanol used =
8.0 g = 0.174 mol (12.0 2 1.0 6 16.0) g mol 1
From the equation, the mole ratio of methanoic acid to ethanol is 1 : 1, ∴ ethanol is the limiting reagent. From the equation, the mole ratio of ethanol to ethyl methanoate is 1 : 1, ∴ theoretical no. of moles of ethyl methanoate obtained = 0.174 mol Theoretical mass of ethyl methanoate obtained = 0.174 mol × (12.0 × 3 + 1.0 × 6 + 16.0 × 2) g mol1 = 12.9 g 8.5 g 100% = 65.9% Percentage yield of ethyl methanoate = 12.9 g 282
D The condenser must be kept open to the atmosphere, otherwise, pressure will be built up inside the condenser.
283
D
284
D Sodium carbonate solution reacts and removes any acidic substances in the distillate. Calcium chloride reacts and removes any unreacted ethanol in the distillate. Anhydrous calcium chloride is a drying agent that removes remaining traces of water from the distillate.
285
B Concentrated sulphuric acid is used as a catalyst in the laboratory preparation of ester. It also removes water, driving the equilibrium of esterification to the product side.
286
B
287
B The reflux condenser should be kept open to the atmosphere during heating, otherwise, a high pressure will be built up inside.
288
D Carbohydrates are composed of carbon, hydrogen and oxygen with H and O in the ratio of 2 : 1, as in water. The ratio of H to O in C 7 H 15 O 7 is not 2 : 1.
289
B (2) is incorrect. Glucose contains an aldehyde group while fructose contains a ketone group, so they are functional group isomers.
290
A Simple sugars such as glucose and fructose are soluble in water.
291
B (1) is correct. Glucose has 4 chiral carbon atoms while fructose has 3 chiral carbon atoms.
glucose
fructose
(2) is incorrect. They are functional group isomers because glucose contains an aldehyde group while fructose contains a ketone group. (3) is correct. Both of them have many OH groups which can form hydrogen bonds with water molecules, so they are soluble in water. 292
A
293
B
294
A
295
B They contain a higher proportion of triglycerides derived from long-chain saturated fatty acids.
296
A
297
A Palmitic acid is a saturated fatty acid.
298
D Animal fats contain a higher percentage of saturated triglycerides derived from long-chain saturated fatty acids. Since cholesterol dissolves well in saturated triglycerides, eating too much animal fats will raise blood cholesterol level.
299
B (2) is incorrect. If the hydrocarbon chain of a fatty acid contains carbon-carbon double bonds, the fatty acid is unsaturated. For (3), since fatty acids contain COOH, they can be regarded as carboxylic acids.
300
B The carbon-carbon double bonds in olive oil adopt the cis configuration. This causes the triglyceride molecules to pack less efficiently. Therefore, olive oil has a lower melting point and is usually a liquid at room temperature.
301
D D is incorrect. If R 1 is not equal to R 3 , the vegetable oil below is optically active.
302
A Glucose in open chain has an aldehyde group which can be oxidized by acidified potassium dichromate solution. (3) is incorrect. Only starch reacts with iodine to give a dark blue solution.
303
C (1) is correct. The molecule has a chiral carbon atom.
(2) is incorrect. It is not a polymer, but a product formed by condensation between glycerol and fatty acids. (3) is correct. After hydrogenation R 1 , R 2 and R 3 become the same hydrocarbon chains. 304
C
305
D Proteins are long polypeptide chains formed from a large number of amino acid molecules joined by amide linkages. When two amino acid molecules are joined together by amide linkage, water is eliminated.
306
A
307
B (2) is incorrect. Glycine is an amino acid with no chiral carbon atom. The structure of glycine is shown below:
308
A (2) is incorrect. Protein is a condensation polymer. (3) is incorrect. Glycine is an amino acid without any chiral carbon atoms. The structure of glycine is
309
A Each amino acid molecule has a central carbon atom, which forms covlaent bonds with an amino (NH 2 ) group, a carboxyl (COOH) group and a side-chain specific to that amino acid.
310
A The functional groups of aspirin include benzene ring, carboxyl group and ester group.
311
C
312
B
313
D For (3), low dose of aspirin prevents stroke as it has a blood-thinning effect.
314
B The functional groups of aspirin include benzene ring, carboxyl group and ester group.
315
B
316
C The functional groups of aspirin include benzene ring, carboxyl group and ester group.
317
B
318
D
319
B Soapy detergents are made from fats or oils.
320
A Soapless detergents form lather in hard water because the ionic heads of soapless detergent particles do not form participate with magnesium ions.
321
D Detergents work by reducing the surface tension of water, enabling it to wet things more effectively, and by emulsifying grease.
322
C Soapy detergents are always alkaline and made from fats and oils.
323
A Glycerol is an alcohol while aspirin is an ester.
324
B Soapy detergents are made by the saponification of animal fats and vegetable oils.
325
B (1) is incorrect. A detergent anion contains a hydrophilic head and a hydrophobic tail. (3) is incorrect. The ionic head of soapy detergent anion is always a carboxylate group while the ionic head of soapless detergent anion is usually a sulphonate group or a sulphate group.
326
B Saponification is used to prepare soapy detergents. It hydrolysed animal fats and vegetable oils in the presence of sodium hydroxide solution to produce glycerol and soapy detergent.
327
A For (2), the ionic head (SO 3 ) of soapless detergent particles does not form precipitate with either calcium or magnesium ions. For (3), the stearate ions in soaps combine with calcium or magnesium ions to form scum only, so no lather is formed.
328
A When a vegetable oil is hydrolysed with sodium hydroxide solution, glycerol (propane-1,2,3-triol) and salt of fatty acids (soapy detergents) are formed.
329
D A is incorrect. Ester would be formed if an alchohol and a carboxylic acid are heated under reflux with concentrated H 2 SO 4 . B is incorrect. Soapy detergents react with calcium ions and magnesium ions in hard water to form an insoluble scum. C is incorrect. They do not decompose in acids. D is correct. CH 3 (CH 2 ) 16 COO + H+ CH 3 (CH 2 ) 16 COOH.
330
B A detergent particle contains an ionic head (either anion or cation) and a long hydrophobic tail.
331
A
332
A
333
A The detergents can emulsify grease because of its emulsifying property, not its wetting property.
334
D The ionic heads of the detergent anions are hydrophilic, so they are soluble in water. The hydrocarbon tails are hydrophobic, so they are soluble in oil. Oil droplets are negatively charged because detergent anions spread over the surface of the droplet.
335
A Soapy detergents form scum instead of lather in hard water because soap anions form insoluble substance with calcium and/or magnesium ions. Therefore, soaps do not work well in hard water.
336
C
337
A Chemicals obtained from petroleum are the starting materials for making soapless detergents.
338
B Soaps do not work well in hard water because the soap anions form insoluble substance with calcium and/or magnesium ions. Only soaps are made from fats or oils. Soapless detergents are derived from petroleum.
339
A Only soapy detergents are made from animal fats while soapless detergents are derived from petroleum.
340
B (1) and (3) are incorrect. Soapy detergents react with calcium ions and/or magnesium ions in hard water to form scum, no lather is formed.
341
C For a detergent anion, the ionic head is soluble in water, but not in oil, so it is said to be hydrophilic (water-loving). The hydrocarbon tail is soluble in oil, but not in water, so it is said to be hydrophobic (water-hating).
342
C The ionic head of soapy detergents is always a carboxylate group (COO).
343
C The hydrophilic ionic heads dissolve in water while the hydrophobic hydrocarbon tails dissolve in oil. The negatively charged oil droplets repel each other and cannot join together.
344
C The ionic heads of the detergent anions dissolve in water while the hydrocarbon tails dissolve in oil. The negatively charged oil droplets repel each other and cannot join together, so they disperse throughout the water.
345
D
346
A (3) is an organic acid which is not a detergent. Soapy detergents are sodium or potassium salts of long-chain carboxylic acids. Soapless detergents are sodium salts of long-chain alkylbenzenesulphonate or alkylsulphate.
347
B They do not contain any carboxylate groups (COO) and do not behave as soaps because they have different ionic heads.
348
D A detergent particle should have one ionic head and one hydrocarbon tail.
349
B A is a soapless detergent particle. C and D are not detergent particles.
350
A B and C are not detergent particles. D is a soapy detergent particle.
351
A Detergent is a wetting agent which helps water spread over the surface. B is impossible. C is the shape of water droplet on the piece of cloth without detergent. D is not possible as the bottom of the water droplet should be flat.
352
C Detergent is an emulsifying agent which helps to stabilize the oil-water emulsion. A and B are incorrect because water and oil will not separate in the presence of detergent. D is incorrect because an emulsion should be formed instead of a homogeneous solution.
353
A The ionic heads of detergent anions dissolve in the water and the hydrocarbon tails dissolve in the oil droplets.
354
A Detergents are able to lower the surface tension of a liquid, so they are known as surfactants. Only soapy detergents form scum in hard water while soapless detergents do not.
355
A Soaps are salts but not organic acids, covalent molecules or polymers.
356
B Soapy detergents are sodium or potassium salts of long-chain carboxylic acids usually with 12 to 20 carbon atoms. A is incorrect because the hydrocarbon tail is too short, so it dissolves well in water but not in oil. C is an organic acid without any detergent properties because the molecule does not have an ionic head. D is a soapless detergent.
357
D Sea water contains lots of ions including calcium ions and magnesium ions. Most of the soapy detergent particles form insoluble scum in sea water, so only little lather is formed.
358
C The creamy yellow solid is the sodium salts of long-chain carboxylic acids the soaps.
359
A Butter is an animal fat which can be hydrolysed by sodium hydroxide solution to give a soapy detergent.
360
B (2) is incorrect. Only very small amount of heat is evolved in the dissolution of sodium hydroxide. The grease (triester) is hydrolysed by sodium hydroxide solution and gives soluble sodium salts. The process is saponification. The soluble sodium salts is the soap which also helps to remove the remaining grease.
361
C Bath soaps are made from animal fats or vegetable oils and potassium hydroxide solution. Bath soaps are softer and milder but more expensive when compared to laundry soaps which are made from animal fats or vegetable oils and sodium hydroxide solution.
362
D Soapy detergents change to insoluble long-chain carboxylic acids in acidic medium. Soapy detergents form scum in sea water which contains a considerable concentrations of calcium ions and/or magnesium ions. Soapy detergents form scum in hard water. Therefore, soapy detergents do not work well in acidic medium, sea water and hard water.
363
B Alkali is used to make detergents, so it is not added to the washing powders.
364
D The soapy detergent anions react with hydrogen ions in acidic medium to form insoluble carboxylic acid (CH 3 (CH 2 ) 16 COOH). Since sea water contains a considerable concentrations of calcium ions and/or magnesium ions, the soapy detergent anions react with those ions to form insoluble scum such as (CH 3 (CH 2 ) 16 COO) 2 Ca and (CH 3 (CH 2 ) 16 COO) 2 Mg.
365
A Phosphates are water softeners which can remove magnesium ions and calcium ions. Carbonate ions in washing soda also react with magnesium ions and calcium ions to form insoluble salts. Ca2+(aq) + CO 3 2(aq) CaCO 3 (s) and Mg2+(aq) + CO 3 2(aq) MgCO 3 (s).
366
A Sea water contains a considerable concentrations of calcium ions and/or magnesium ions which form insoluble substances with soaps.
367
C Soap reacts with acid to form a carboxylic acid which is insoluble in water and comes out as white precipitate.
368
C
369
is
neither
a
soap
nor
a
soapless
detergent.
is a soapy detergent. B The detergent particles can attach to the grease with dirt and the grease is lifted up from the surface to form an emulsion. The grease is then suspended in water. Detergent particles actually decrease the cohesive forces between water molecules to increase the wetting ability of water.
370
B Brine is saturated sodium chloride solution. The addition of brine to the product mixture is to decrease the solubility of soap in water, so that the soap is separated from the solution and floats on the surface.
371
B Soapy detergents react with calcium and/or magnesium ions to form insoluble scum.
372
A 240 < 12.0n + 2.0n + 1.0 + 12.0 + 16.0 × 2 + 23.0 < 245 240 < 14.0n + 68.0 < 245 172 < 14.0n < 177 12.3 < n < 12.6 Therefore, n = 12 as it is an integer.
373
C Lime water contains calcium ions which react with soap anions to form white precipitate i.e. the scum.
374
C
375
B The sodium hydroxide solution in pipe cleaners reacts with the grease in the pipes to form soap which is soluble in water and further helps to remove the remaining grease. Sodium hydroxide is highly corrosive.
376
C It is the by-product formed in the process of saponification.
377
C Paraffin oil is not an animal fat or a vegetable oil. It is a mixture of hydrocarbons. The other are either animal fats or vegetable oils which can be used in saponification.
378
B Saponification is an alkaline hydrolysis of animal fats or vegetable oils (triesters) by using sodium hydroxide solution to give sodium salts of long-chain carboxylic acids.
379
A Only Soapy detergents form scum in hard water. Bath soaps are soapy detergents.
380
D It is a soapless detergent, so it is made from chemicals obtained from petroleum and it forms lather instead of scum in hard water.
381
A Polyesters contain ester linkages.
382
D
383
C (1) is incorrect. Nylon is a polyamide. (3) is incorrect. The correct general repeating unit of nylon is
.
384
C Nylon is a polyamide containing many amide linkages.
385
A PET is linked by ester linkages.
386
D Condensation is a type of reaction in which two or more molecules join together to form a large molecule, with the elimination of small molecules e.g. H 2 O or HCl. During the synthesis of nylon or polyesters, their monomers are joined together by eliminating water.
387
D Proteins are condensation polymers of amino acids. Nylon 6,6 is a condensation polymer of hexanedioyl chloride and hexane-1,6-diamine. Terylene is a condensation polymer of a diol and a dioic acid.
388
B A polyester is usually formed between a diol and a dioic acid. (2) is not a diol. (3) contains OH group and COOH group which undergo condensation to give a polyester.
389
C
390
C The first statement is incorrect. Each fructose molecule contains two terminal OH groups which can be oxidized to give carboxylic acids.
391
C The first statement is incorrect. Fructose has an ketone group.
392
C Glycerol contains three hydroxyl groups, so it is very soluble in water.
393
A The carbon-carbon double bonds in vegetable oils adopt cis configuration. This causes the triglyceride molecules to pack less efficiently. As a result, vegetable oils have lower melting points and are usually liquids at room temperature.
394
C Soapless detergents can be acidic, neutral and alkaline.
395
A
396
C Soaps are always alkaline. However, soapless detergents can be acidic, neutral, or alkaline.
397
A Since the surface tension of water is reduced by the detergent, water spreads over the surface and wets the cloth more easily.
398
A
399
D Soapless detergents are sodium salts of long-chain alkylbenzenesulphonate or alkylsulphate. Soapless detergents are made from chemicals obtained from petroleum.
400
B Detergent is an emulsifying agent which can stabilize an oil-water emulsion, so oil and water can mix together. Detergent is a wetting agent which helps water spread over the surface and wet it more easily.
401
A
402
C When a detergent is mixed with water, lather is formed. The lather contains air rather than any gases produced.
403
A
404
A
405
A
406
B Soapless detergents work well in hard water because they do not form insoluble scum with calcium and/or magnesium ions.
407
A
408
C Soapy detergents do not work well in acidic water because long-chain carboxylic acids are formed
which are insoluble in water. 409
D Both statements are incorrect. Since sea water contains considerable concentrations of magnesium ions and calcium ions, soap anions react with these ions to form insoluble scum. Therefore, the soapy detergents lose their cleaning powers in sea water.
410
A
411
B Both nylon and polyesters are insoluble in water because they have long hydrocarbon chains.
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