Carbon Compounds SQ Ans

SQ Answers 1

(a) hydrogen bond

[2] (b)

hydrogen bonds

[2]

(c) Propanoic acid has a higher boiling point than propan-1-ol. [1] Propanoic acid contains the carboxyl group, . [1] Both and OH groups can participate in hydrogen bond formation. [1] As a result, it has more extensive intermolecular hydrogen bonds than propan-1-ol. [1] More energy is needed to separate its molecules during the boiling process. [1] (d)

[1] Propyl propanoate [1] 2

(a) It belongs to alcohols. [1] (b) Propane-1,2,3-triol [1] (c) Glycerol is denser than water [1] because it contains more than one OH group. [1] (d) Glycerol has a higher boiling point than propan-1-ol. [1] Since each glycerol molecule contains three OH groups, glycerol has more extensive intermolecular hydrogen bonds than propan-1-ol. [1] More energy is needed to overcome the extensive hydrogen bonds between the glycerol molecules during boiling. [1]

3

(a)

butane [1]

butan-1-ol [1]

butanoic acid [1]

2-methylpropane [1] (b) Butane and 2-methylpropane [1] (c) 2-methylpropane < Butane < Butan-1-ol < Butanoic acid [1] Molecules of butanoic acid and butan-1-ol are held together by hydrogen bonds as well as van der Waals’ forces. [1] However, butanoic acid has more extensive intermolecular hydrogen bonds than butan-1-ol because both C=O group and OH group can participate in hydrogen bond formation. [1] Therefore, more energy is needed to overcome the extensive intermolecular hydrogen bonds among butanoic acid molecules than that among butan-1-ol molecules. [1] The molecules of butane and 2-methylpropane are held together by weak dispersion forces only. [1] Since the strength of hydrogen bonds > dispersion forces, the boiling point of butan-1-ol is higher than that of butane and 2-methylpropane. [1] 2-methylpropane is a branched-chain alkane which has a smaller surface area than butane. [1] The dispersion forces between 2-methylpropane molecules are weaker. [1] Therefore, 2-methylpropane has a lower boiling point than butane. 4

(a) NH 2 (CH 2 ) 4 NH 2 : Butane-1,4-diamine [1] NH 2 (CH 2 ) 5 NH 2 : Pentane-1,5-diamine [1] (b)

Dimethylamine [1]

Trimethylamine [1]

(c)

hydrogen bonds

[2] (d)

hydrogen bonds

5

[2]

(a) It belongs to esters because it has a functional group of COO. [2] (b)

[1] (c) Yes, it is. Because it is made by the reaction of butanoic acid, which is a carboxylic acid. [2] (d) water out

water in pear-shaped flask a mixture of ethanol, butanoic acid and conc. H 2 SO 4

anti-bumping granule

heat Correct set-up [1]

Correct labelling [2] (e) The molecules of hexanoic acid are held together by hydrogen bonds as well as van der Waals’ forces while the molecules of ethyl butanoate are held together by weak van der Waals’ forces only. [2] Since the strength of hydrogen bonds is greater than that of van der Waals’ forces, more energy is needed to overcome strong hydrogen bonds between hexanoic acid molecules during boiling process. [2] 6

(a) (i) (ii) (iii) (iv) (b)

Compounds X and Z [2] Carbon-carbon double bond [1] They belong to alkenes. [1] The general formula is C n H 2n . [1]

[1] or

7

[1]

(a) Hydrogen gas [1] (b) The general formula is C n H 2n O 2 . [1] (c) They belong to carboxylic acids. [1] . [1] (d) The functional group is Test the compounds with sodium carbonate solution. [1] They will give colourless gas. [1] (e) The third member is propanoic acid. [1] The fourth member is butanoic acid. [1]

8

(a)

Hex-2-ene [1]

Hexan-3-ol [1]

Hexanal [1]

Hexan-2-amine [1] (b) Hexan-3-ol, hexanal and hexan-2-amine are soluble in water because they can form hydrogen bonds with water molecules. [2] (c) The increasing order of boiling point: Hex-2-ene < Hexanal < Hexan-2-amine < Hexan-3-ol [1] The molecules of hexan-2-amine and hexan-3-ol are held together by hydrogen bonds as well as van der Waals’ forces while the molecules of hex-2-ene and hexanal are held together by van der Waals’ forces only. [2] Therefore, the boiling points of hexan-2-amine and hexan-3-ol are higher than those of hex-2-ene and hexanal. The molecules of hexanal are held together by dipole-dipole forces as well as dispersion forces while the molecules of hex-2-ene are held together by dispersion forces only. [2] Therefore, the boiling point of hexanal is higher than that of hex-2-ene. Both hexan-2-amine and hexan-3-ol have hydrogen bonds among their molecules. However, the nitrogen atom on hexan-2-amine is not as electronegative as the oxygen atom on hexan-3-ol, so hexan-2-amine has a lower boiling point than hexan-3-ol. [1] 9

(a) C n H 2n+1 COOH [1] (b) C n H 2n+1 COOH = 88.0 [1] 12.0n + 12.0 + 16.0 × 2 + 2.0n + 2.0 = 88.0 14.0n + 46.0 = 88.0 14.0n = 42.0 n = 3 [1] ∴ the molecular formula of the carboxylic acid is C 3 H 7 COOH. [1] (c)

[1] Butanoic acid [1]

[1] 2-methylpropanoic acid [1] 10

(a) Chloroethane and bromoethane belong to haloalkanes. [1] Ethanol belongs to alcohols. [1] Ethene belongs to alkenes. [1]

(b) Ethanol is soluble in water. [1] The OH group of ethanol molecules can form hydrogen bonds with water molecules while the other compounds cannot. [1] (c) The order of boiling point is: ethene < chloroethane < bromoethane < ethanol. [1] The ethanol molecules are held together by hydrogen bonds as well as van der Waals’ forces. [1] The chloroethane and bromoethane molecules are held together by dipole-dipole forces as well as dispersion forces. [1] As the molecular size of Br atom > Cl atom and the strength of dispersion forces between haloalkane molecules depends on the molecular size, [1] the boiling point of bromoethane is higher than that of chloroethane. At last, the ethene molecules are held together by dispersion forces only. [1] Since the strength of hydrogen bonds > dipole-dipole forces > dispersion forces, the boiling point of ethanol > bromoethane > chloroethane > ethene. [1] 11

(a) It contains OH group and CHO group. [2] (Accept ‘ROR group’ as answer.) (b) It belongs to aldehydes. [1] (c) It is soluble in water [1] because it contains OH group and CHO group which can form hydrogen bonds with water molecules. [1] (d) It is used as an ingredient in perfume. [1]

12

(a) CH 3 CH 2 CH 2 CHO [1] (b) CH 3 CH=CH 2 + H 2 + CO  CH 3 CH 2 CH 2 CHO [1] (c) Flammable [1] (d) His statement is correct. [1] CH 3 CH 2 CH 2 COH would be misinterpreted as an alcohol if butanal is expressed in this condensed form. [1] Thereofore, it should be written as CH 3 CH 2 CH 2 CHO. [1]

13

(a) The organic compounds that belong to the same homologous series have the same general formula. Each member of the same homologous series differs from the next member of the series by a CH 2  group. [2] (b) (i) They show a steady variation in physical properties as the number of carbon atoms increases. [1] (ii) They have similar chemical properties. [1] (c) (i) Functional group:

[1] It belongs to aldehydes. [1] (ii) Functional group: [1] It belongs to ketones. [1] (iii) Functional group:

[1] It belongs to unsubstituted amides. [1] 14

(a)

Fluoromethane

Chloromethane

Bromomethane

Iodomethane [4]

(b) Iodomethane [1] (c) Fluoromethane, chloromethane and bromomethane [3] (d) The boiling points of these compounds are higher than that of methane. [1] These compounds are haloalkanes, as halogens are more electronegative than carbon, all carbon-halo-gen bonds are polar. [1] Since both dipole-dipole forces and dispersion forces exist between these molecules, they have higher boiling points than methane. [1] 15

(a) (i) It is a primary amine because one hydrogen atom in NH 3 is replaced by one alkyl group. [2] (ii) It is a secondary amine because two hydrogen atoms in NH 3 are replaced by two alkyl groups. [2] (iii) It is a tertiary amine because all hydrogen atoms in NH 3 are replaced by three alkyl groups. [2] (b) (i) The nitrogen atom is not as electronegative as the oxygen atom. [1] (ii)

hydrogen bond

[2] (c) It is because hydrogen bonds are formed between two ethanoic acid molecules to give a dimer. [1]

hydrogen bond [2]

16

(a) Compound A belongs to aldehydes. [1] Compound B belongs to ketones. [1] Compound C belongs to carboxylic acids. [1] Compound D belongs to alcohols. [1] (b) Compounds C and D. [2] (c) Compound C > Compound D > Compound B > Compound A [1] Both molecules of compounds C and D are held together by hydrogen bonds as well as van der Waals’ forces while the molecules of compounds A and B are held together by weak van der Waals’ forces only. [2] Therefore, compounds C and D have higher boiling points than compounds A and B. Compound C has more extensive intermolecular hydrogen bonds than compound D, so compound C has a higher boiling point than compound D. [1] Since the molecules of compound B have larger molecular sizes than that of compound A, van der Waals’ forces among the molecules of compound B are stronger. [1] Therefore, compound B has a higher boiling point than compound A. (d) (i) Compounds C and D. [2] (ii) Methyl ethanoate [1] (iii) The boiling point of the ester is similar to those of compounds A and B, as there are no hydrogen bonds among the ester molecules. [2]

17

(a)

[1] (b) 2-methylpropanoic acid [1] (c)

hydrogen bonds

18

[3]

(a) CH 3 (CH 2 ) 2 NH 2 belongs to primary amines. [1] CH 3 COOH belongs to carboxylic acids. [1] (b) For CH 3 (CH 2 ) 2 NH 2 , there are hydrogen bonds between the lone pair on the electronegative nitrogen atom and the slightly positive hydrogen atom in another molecule. [2] For CH 3 COOH, there are hydrogen bonds between the lone pair on the electronegative oxygen atom and the slightly positive hydrogen atom in another molecule. [2] (c) CH 3 COOH has a higher boiling point than CH 3 (CH 2 ) 2 NH 2 . [1]

Although both compounds have hydrogen bonds between their molecules, the nitrogen atom is not as electronegative as the oxygen atom. [1] As a result, the hydrogen bonds between CH 3 COOH molecules are stronger than that between CH 3 (CH 2 ) 2 NH 2 , more energy is needed to overcome the stronger hydrogen bonds among CH 3 COOH molecules during boiling. [1] 19

(a) (i)

[1] (ii) CH 3 C(CH 3 ) 2 CH 3 [1] (b) (i)

[1] (ii) CH 2 =CHC(CH 3 )=CH 2 [1] (c) (i)

[1] (ii) CH 2 ClCBr=CH 2 [1] (d) (i)

1] (ii) CH 3 (CH 2 ) 2 CO 2 H [1] (e) (i)

[1] (ii) CH 3 COCH 3 [1] 20

(a)

[1]

[1]

3-methylbutanone

2-methylbutanoic acid

[1]

[1]

2-methylbutan-1-ol

2-methylbutanal

(b) 2-methylbutanoic acid and 2-methylbutan-1-ol [2] (c) 2-methylbutanoic acid has the highest boiling point. [1] It contains the carboxyl group, . [1] Both and OH groups can participate in hydrogen bond formation. [1] As a result, it has more extensive intermolecular hydrogen bonds than 2-methylbutan-1-ol. [1] More energy is needed to separate its molecules during the boiling process. [1] 21

(a) (b) (c) (d) (e) (f)

22

(a)

1,3-dichloropropanone [1] 2,2-dichloro-3-formylpropanoic acid [1] 2-fluorobutanedial [1] Butyl 4-chloropent-2-enoate [1] 1,2-diaminopropane-1,3-diol [1] Ethanedioic acid [1]

[1] (b)

[1]

(c)

[1] (d)

[1] (e)

[1] (f)

[1] 23

(a) (i) (ii) (iii) (iv) (b) (i)

2-aminopropanoic acid [1] 4-bromo-3-methylpentanal [1] 2-amino-3-hydroxypropanoic acid [1] 2,3-dioxopentanoic acid [1]

[1] (ii)

[1] (iii)

[1]

(iv)

[1] 24

(a) (i)

[1] (ii) CH 3 CCH 3 =CHCOCH 3 [1] (b) (i)

[1] (ii) CH 2 OHCH 2 CH=CHCH 3 [1] (c) (i)

[1] (ii) CH 3 COCOCH 3 [1] (d) (i)

[1] (ii) NH 2 CHOHCH 2 COCHO [1] (e) (i)

[1] (ii) CH 3 CH 2 CH 2 NH 2 [1] 25

(a) The trivial name for CH 3 COOH is acetic acid. [1] The trivial name for CH 3 COCH 3 is acetone. [1] The trivial name for CH 3 CH(OH)CH 3 is isopropyl alcohol. [1]

The trivial name for CHCl 3 is chloroform. [1] The trivial name for HCHO is formaldehyde. [1] (b)

CH 3 COOH [1]

CH 3 COCH 3 [1]

CH 3 CH(OH)CH 3 [1]

CHCl 3 [1]

HCHO [1] (c) The molecules of CH 3 COOH are held together by hydrogen bonds as well as van der Waals’ forces while the molecules of HCHO are held together by weak van der Waals’ forces only. [2] Since the strength of hydrogen bonds > van der Waals’ forces, more energy is needed to break the strong hydrogen bonds among CH 3 COOH molecules during boiling. [2] 26

(a) (i)

[1]

Ethyl ethanoate [1]

[1]

Methyl propanoate [1]

[1] Propyl methanoate [1] (ii) They are position isomers. [1] (b) (i)

[1]

2-methylpropanoic acid [1]

[1]

Butanoic acid [1]

(ii) They are chain isomers. [1] (c) A and B are functional group isomers. [1] 27

(a) But-2-ene exhibits geometrical isomerism but no enantiomerism. [1] Each doubly-bonded carbon atom is attached to two different atoms or groups of atoms (H and CH 3 ), so geometrical isomers are possible. [1] Besides, it does not have any chiral carbon atom, so no enantiomers are possible. [1]

[1] and

[1]

(b) 2,4-dimethylpentenedioic acid exhibits both geometrical isomerism and enantiomerism. [1] Each doubly-bonded carbon atom is attached to two different atoms or groups of atoms, so geometrical isomers are possible. [1] Besides, it has a chiral carbon atom, so enantiomers are possible. [1]

[1] and

[1]

[1] and

[1]

(c) 2-chlorobutane exhibits enantiomerism but no geometrical isomerism. [1] It does not contain any carbon-carbon double bond, so no geometrical isomers are possible. [1] Besides, it has a chiral carbon atom, so enantiomers are possible. [1]

[1] and 28

[1]

(a) CH 3 CHClCH=CH 2 exhibits enantiomerism but no geometrical isomerism. [1] One of the doubly-bonded carbon atoms is attached to two identical hydrogen atoms, so no

geometrical isomers are possible. [1] Besides, it has a chiral carbon atom, so enantiomers are possible. [1]

[1] and

[1]

(b) CH 3 CH=CHCHCH 3 COOH exhibits both geometrical isomerism and enantiomerism. [1] Each doubly-bonded carbon atom is attached to two different atoms or groups of atoms, so geometrical isomers are possible. [1] Besides, it has a chiral carbon atom, so enantiomers are possible. [1]

[1] and

[1] and 29

[1]

[1]

(a)

[1] (b) His statement is incorrect. [1] Since the pair of enantiomers of mecoprop has the same intermolecular forces, they have very similar physical properties, e.g. boiling point and melting point. [1] Therefore, it is impossible to distinguish them by simply comparing their boiling points and melting points. (c) His statement is partly correct. [1] The plane of polarized light is not rotated by the mixture of the pair of enantiomers of mecoprop only if they are present in the same amount. [1] 30

(a) 2,3-dihydroxybutanedioc acid [1] (b)

[2]

(c) The enantiomers of tartaric acid can rotate the plane of polarized light to different directions. [1] Therefore, they can be distinguished by measuring their optical activities using a polarimeter. [1] 31

(a) Pent-2-enedioic acid [1] (b) Stereoisomerism occurs when there are isomers in which atoms are joined in the same order but have different spatial arrangements. [1] (c) Glutaconic acid exhibits geometrical isomerism. [1] (d)

[1] (e) Trans-glutaconic acid has a higher melting point. [1] It has a more regular and symmetrical structure than cis-glutaconic acid. [1] Therefore, it has a higher packing efficiency in the solid state. [1] The intermolecular forces holding the molecules together are thus stronger. [1] More energy is needed to overcome these intermolecular forces during the melting process. (f) Trans-glutaconic acid is more soluble in water. [1] For trans-glutaconic acid, the two carboxyl groups are pointing to the opposite directions. Intermolecular hydrogen bondings are formed between trans-glutaconic acid molecules and water molecules. [1]

Intermolecular hydrogen bonds [2] On the contrary, for cis-glutaconic acid, due to the close proximity of the two carboxyl groups, intramolecular hydrogen bonds can be formed readily. [1] Less intermolecular hydrogen bonds can be formed between cis-glutaconic acid molecules and water molecules.

Intramolecular hydrogen bond

[2]

32

(a) Measure the optical activity of compound A by using a polarimeter. [1] (b) It exhibits enantiomerism. [1] (c) Since it reacts with NaHCO 3 solution, it contains a COOH group. [1] Since it is optically active, it has a chiral carbon atom. [1] The structural formula of compound A is:

[1] (d) The enantiomers of compound A were present in the same amount, so the optical activity of each enantiomer was cancelled out. [2] 33

(a) Compound A contains a carbon-carbon double bond. [1] (b)

(i)

[1]

and

[1] Trans-hex-3-ene [1]

Cis-hex-3-ene [1] (ii)

[1] and

[1]

(c)

[1] 34

(a) The reaction should be carried out in the presence of sunlight. [1] (b) (i)

[1] 1-bromobutane [1]

and

[1] 2-bromobutane [1]

(ii) C 4 H 9 Br exhibits position isomerism. [1] 1-bromobutane and 2-bromobutane are position isomers. [1] Besides, it exhibits enantiomerism. [1] The C-2 carbon atom in 2-bromobutane is attached to four different atoms or groups of atoms, so it has a chiral carbon atom in the molecule and exhibits enantiomerism. [1] (c)

[1] or 35

[1]

(a) 2-hydroxypropanoic acid [1] (b) (i) Lactic acid exhibits enantiomerism. [1] The central carbon atom of lactic acid is attached to four different groups. [1] There is a chiral carbon atom in the lactic acid molecule. [1] (ii)

[1] and

[1]

(c) His statement is incorrect. [1] Since the pair of enantiomers of lactic acid has identical intermolecular forces, they have identical physical properties. [1] Therefore, it is impossible to distinguish them by simply comparing their boiling points and melting points. Instead, the pair of enantiomers can rotate the plane of polarized light to different directions. [1] Therefore, they can be distinguished by measuring their optical activities using a polarimeter. [1] 36

(a) 2-aminopropanoic acid [1] (b) (i) Alanine exhibits enantiomerism. [1] The central carbon atom of alanine is attached to four different groups. [1] There is a chiral carbon atom in the alanine molecule. [1] (ii)

[1] and

[1]

(c) Since the pair of enantiomers can rotate the plane of polarized light to different directions, [1] it is possible to distinguish them by measuring their optical activities using a polarimeter. [1]

37

(a) Isomers are compounds with the same molecular formula but different arrangements of atoms in space. [1] (b) (i)

[1]

[1]

[1]

(ii)

[1] and

[1]

38 [1]

[1]

[1]

[1]

[1] 39

(a) The amino acid molecule is chiral. [1] The central carbon atom is attached to four different atoms or groups of atoms. Therefore, the amino acid molecule has a chiral carbon atom. [1]

[1] (b) The amino acid molecule exhibits enantiomerism [1] because it has a chiral carbon atom. [1] (c)

[1] and

[1]

(d) Since the enantiomers of the amino acid can rotate the plane of polarized light to different directions, [1] they can be distinguished by measuring their optical activities using a polarimeter. [1] 40

(a) (i)

[2] (ii) 3-methylpent-1-yne could exhibit enantiomerism [1] because it has a chiral carbon atom. [1] (b) (i)

[2] (ii) 2-methylbutanal could exhibit enantiomerism [1] because it has a chiral carbon atom. [1] (c) (i)

[2] (ii) 3-methylpentan-2-one could exhibit enantiomerism [1] because it has a chiral carbon atom. [1] 41

(a) Enantiomerism occurs when there are two isomers that are mirror images of one another but are non-superimposable. [2] (b) The 2,4-dichloro-1,5-difluoropentan-3-one molecule has two chiral carbon atoms. [1] (c)

[3] (d)

[1] and 42

[1]

(a)

[1] Cis-1,2-difluoroethene [1]

[1] Trans-1,2-difluoroethene [1] (b) Cis-1,2-difluoroethene has a higher boiling point. [1] It has two polar CF bonds pointing to the same side. [1] As a result, the molecule has a net dipole moment [1]. There are dipole-dipole forces as well as dispersion forces among the cis-1,2-difluoroethene molecules. [1] Thus, more energy is needed to separate the molecules during the boiling process. [1] (c) Trans-1,2-difluoroethene has a higher melting point. [1] It has a more regular and symmetrical structure than cis-1,2-difluoroethene. [1] As a result, the molecules of trans-1,2-difluoroethene can pack more closely in the solid state. [1] The intermolecular forces holding the molecules together are thus stronger. [1] Therefore, more energy is needed to break these intermolecular forces during the melting process. [1] 43

His statement is not completely correct. [1] The organic compound with the molecular formula C 3 H 7 OCl could exhibit structural isomerism as well as enantiomerism. [2] The structural formulae of all possible structural isomers of C 3 H 7 OCl are shown as follows:

1-chloropropan-1-ol [1]

2-chloropropan-1-ol [1]

3-chloropropan-1-ol [1]

1-chloropropan-2-ol [1]

2-chloropropan-2-ol [1] Referring to the above structural formulae, 1-chloropropan-1-ol, 2-chloropropan-l-ol and 1-chloropropan-2-ol could exhibit enantiomersim because they have chiral carbon atoms in their molecules. [1] 44

(a) 3-bromobut-1-ene and but-2-enoic acid could undergo addition reactions [2] because both of them have carbon-carbon double bonds. [1] (b) (i) But-2-enoic acid and 1-chloropropan-1-ol [2] (ii) CH 3 CH=CHCOOH + CH 3 CH 2 CHClOH CH 3 CH=CHCOOCHClCH 2 CH 3 + H 2 O [1] (c) But-2-enoic acid could exhibit geometrical isomerism. [1] Each doubly-bonded carbon atom is attached to two different atoms or groups of atoms, so geometrical isomers are possible. [1] The structural formulae of the cis and trans isomers of but-2-enoic acid are shown as follows:

cis-but-2-enoic acid [1]

trans-but-2-enoic acid [1]

(d) 3-bromobut-1-ene and 1-chloropropan-1-ol could exhibit enantiomerism. [2] Both of them have chiral carbon atoms in their molecules, so they could exhibit enantiomerism. [1] The three-dimensional representations of enantiomers of 3-bromobut-1-ene are shown as follows:

[1] and

[1]

The three-dimensional representations of enantiomers of 1-chloropropan-1-ol are shown as follows:

[1] and 45

(a) (b) (c) (d)

46

(a) (i)

[1]

The above reaction is an addition reaction. [1] Compound A is 2-bromobutane and compound B is 1-bromobutane. [2] Compound A exhibits enantiomerism because it has a chiral carbon atom. [2] They are position isomers. [1]

Markovnikov’s rule states that when a molecule of HX is added to an alkene, the hydrogen atom is added to the carbon atom of the carbon-carbon double bond that already carries the larger number of hydrogen atoms. [2] (ii) The major product is 2-iodopropane [1] and the minor product is 1-iodopropane. [1] (b) (i) Compound X is propan-2-ol. [1] (ii) CH 3 CHICH 3 + NaOH  CH 3 CH(OH)CH 3 + NaI [1] (iii) Reagent Y is concentrated sulphuric acid. [1]

47

(a) (i)

Ethanol is mixed with acidified potassium dichromate solution and the reaction mixture is heated under reflux gently. [2] (ii) The reaction mixture changes from orange to green. [1] The orange dichromate ions are reduced to green chromium(III) ions. [1] (b) Ethanol is mixed with acidified potassium dichromate solution. [1] The reaction mixture is warmed to a temperature that is above the boiling point of ethanal but below that of ethanol. [1] Ethanal is distilled out as soon as it is formed, preventing it from further contact with the oxidizing agent. [1]

48

(a) CH 3 CH 2 CH 2 OH + HCl  CH 3 CH 2 CH 2 Cl + H 2 O [1] (b) The upper layer contained 1-chloropropane [1] and the lower layer contained concentrated hydrochloric aicd. [1] 1-chloropropane was insolube in water. [1] Besides, it had a lower density than water, so it was formed over the aqueous layer. [1] (c) His suggestion was incorrect. [1] It is because concentrated sulphuric acid readily oxidizes iodide to iodine. [1] The hydrogen iodide cannot be generated to react with the alcohol during the reaction. [1]

49

(a) Stage 1: Cl 2 and sunlight [2] Stage 2: Dilute sodium hydroxide solution [1]

Stage 3: Concentrated sulphuric acid at 180C [2] (b) Stages 1 and 2 are substitutions. [2] Stage 3 is dehydration. [1] (c) Use excess cyclohexane for the reaction in Stage 1. [1] (d) By passing the hydroxycyclohexane vapour over a heated catalyst of aluminium oxide or pumice stones at a temperature of 350C. [1] 50

(a) It was because propan-1-ol was first oxidized to propanal and it was then further oxidized to propanoic acid. [1] (b) The reaction mixture of propan-1-ol and acidified potassium dichromate solution was warmed to a temperature that is above the boiling point of propanal but below that of propan-1-ol. [2] Propanal was distilled out as soon as it was formed, avoiding any further contact with the oxidizing agent. [1] (c) Acidified potassium permanganate solution [1]

51

(a) Mix the two compounds with bromine solution (dissolved in an organic solvent) in darkness respectively. [1] Only propene could decolorize bromine solution while propane could not. [1] (b) Mix the two compounds with acidified potassium dichromate solution respectively. [1] Only propanal could change acidified potassium dichromate solution from orange to green while propanone could not. [1] (c) Mix the two compounds with acidified potassium dichromate solution respectively. [1] Then, heat the mixture gently under reflux. [1] Only butan-2-ol could change acidified potassium dichromate solution from orange to green while 2-methylpropan-2-ol could not. [1]

52

(a) (i)

Reagent A is lithium aluminium hydride in dry ether. [1] Reagent B is dilute sulphuric acid/dilute hydrochloric acid. [1] (ii) Reaction 1 is a reduction. [1] (iii) Propan-2-ol could be separated from the product mixture by fractional distillation. [1] (b) (i) Reagent C is concentrated sulphuric acid. [1] (ii) Reaction 2 is a dehydration. [1]

53

(a) Compound A is 2-bromopropane. [1] Compound B is propan-2-ol. [1] Compound C is propanone. [1] (b) (i) The reaction mixture changed from orange to green during the reaction. [1] It is because orange dichromate ions were reduced to green chromium(III) ions. [1] (ii) The reflux condenser should be kept open to the atmosphere during the reaction. [1] Water should enter the condenser from the lower outlet and leave from the upper outlet. [1]

(iii) His statement is incorrect. [1] Propanone is resistant to further oxidation, so it is the end product of the oxidation. [1] 54

(a) Compound A is propanoic acid. [1] Reagent B is acidified potassium dichromate solution. [1] (b) (i) Compound C is ethanol and compound D is sodium bromide. [2] (ii) CH 3 CH 2 Br + NaOH  CH 3 CH 2 OH + NaBr [1] (c) (i) Ethyl propanoate [1] (ii) Concentrated sulphuric acid [1] and heating under reflux. [1] (d) His suggestion is incorrect. [1] Propanone is resistant to oxidation, so it is not oxidized to propanoic acid. [1] Therefore, no ester can be formed. [1]

55

(a)

[1] and

[1]

(b) They are position isomers. [1] (c) Concentrated sulphuric acid [1] (d) CH 3 CH 2 CH 2 OH + 2[O]  CH 3 CH 2 COOH + H 2 O [1] CH 3 CH(OH)CH 3 + [O]  CH 3 COCH 3 + H 2 O [1] 56

(a) The two compounds are heated with acidified potassium dichromate solution respectively. [1] Butan-2-ol reacts with acidified potassium dichromate solution and changes the reaction mixture from orange to green. [1] However, 2-methylpropan-2-ol does not react with acidified potassium dichromate solution. It does not cause any colour change. [1] (b) The two compounds are heated with acidified potassium dichromate solution respectively. [1] Propanal reacts with acidified potassium dichromate solution and changes the reaction mixture from orange to green. [1] However, propanone does not react with acidified potassium dichromate solution. It does not cause any colour change. [1] (c) The two compounds are mixed with sodium hydroxide solution and then followed by acidified silver nitrate solution respectively. [2] Chloroethane gives a white precipitate [1] while bromoethane gives a creamy precipitate. [1]

57

(a) Butan-1-ol is oxidized to butanal. [1] Butan-2-ol is oxidized to butanone. [1] (b) The two compounds are heated with acidified potassium dichromate solution respectively. [1] Propanal reacts with acidified potassium dichromate solution and changes the reaction mixture from orange to green. [1] However, propanone does not react with acidified potassium dichromate solution. It does not cause any colour change. [1]

(c) The propan-2-ol molecules are held together by hydrogen bonds while the propanone molecules are held together by dipole-dipole forces and dispersion forces. [2] Since the strength of hydrogen bonds > dipole-dipole forces > dispersion forces, more energy is needed to separate propan-2-ol molecules during the boiling process. [2] (d) Propanone can be separated from the mixture by distillation. [1] The mixture is warmed to a temperature that is above the boiling point of propanone but below that of propan-2-ol. [2] 58

(a) Reagent A is lithium aluminium hydride. [1] Reagent B is dilute sulphuric acid/dilute hydrochloric acid. [1] (b) Compound C is 2-methylpropan-1-ol. [1] The condensed formula is (CH 3 ) 2 CHCH 2 OH. [1] (c) Reagent A reacts violently with water, so dry ether can prevent it from contacting water or atmospheric moisture to give an explosive reaction. [2]

59

(a) (i) (ii) (iii) (b) (i) (ii)

60

(a) (i)

Reagent A is lithium aluminium hydride in dry ether. [1] Reagent B is dilute sulphuric acid/hydrochloric acid. [1] (ii) Compound C is butan-1-ol. [1] (iii) His suggestion is incorrect. [1] Butanoic acid should be first mixed with lithium aluminium hydride in dry ether. [1] Then the reaction mixture is treated with dilute sulphuric acid/dilute hydrochloric acid. [1] Otherwise, an explosion caused by mixing reagents A and B directly may occur. [1] (b) Reagent D is concentrated sulphuric acid. [1] (c) (i) The two products were 1-chlorobutane and 2-chlorobutane. [2] (ii) The major product was 2-chlorobutane. [1] 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. [1] The chlorine atom is added to the carbon atom carrying fewer hydrogen atoms. [1] Therefore, the major product is 2-chlorobutane.

61

(a) Compound A can react with sodium carbonate solution, so it has a COOH group. It is a carboxylic acid. [1]

Butanoic acid [1] and 2,2-dimethylpropanol [1] Concentrated sulphuric acid [1] and heating [1] 2,2-dimethylpropyl butanoate [1] Propanone [1] and butanoic acid [1] Lithium aluminium hydride in dry ether [1] and dilute sulphuric acid/dilute hydrochloric acid [1] (iii) Propanone is reduced to propan-2-ol. [1] Butanoic acid is reduced to butan-1-ol. [1]

The general formula of carboxylic acid is C n H 2n+1 COOH. Then, 12.0(n+1) + 2n + 2.0 + 32.0 = 60.0 [1] n=1 The molecular formula of compound A is C 2 H 4 O 2 . [1] ∴ compound A is ethanoic acid. [1] (b) 2CH 3 COOH(aq) + Na 2 CO 3 (aq)  2CH 3 COONa(aq) + CO 2 (g) + H 2 O(l) [2] (c) (i) Compound B is propyl ethanoate. [1] (ii) Concentrated sulphuric acid [1] and heating under reflux. [1] 62

(a) Compound A was propene. [1] It reacted with bromine to form 1,2-dibromopropane which was colourless. [1] Compound B was propanoic acid. [1] It reacted with sodium carbonate solution to give colourless carbon dioxide gas. [1] Compound C was propan-1-ol. [1] It reacted with neither bromine nor sodium carbonate solution. [1] (b) (i) Propanoic acid [1] and propan-1-ol [1] (ii) Concentrated sulphuric acid [1] and heating under reflux [1] (iii) Propyl propanoate [1]

63

(a) Hydrolysis is a chemical reaction in which a compound is broken down by reaction with water. [1] (b) (i) Sodium hydroxide solution [1] (ii) It acts as a catalyst to speed up the hydrolysis. [1] (c) The products are propan-1-ol and sodium methanoate. [2] (d) It is used to make soaps. [1]

64

(a) (i) Compound A is propanoic acid and compound B is ammonium ion. [2] (ii) Reaction 1 is an acid hydrolysis. [1] (b) (i) Reagent C is lithium aluminium hydride in dry ether. [1] Reagent D is dilute sulphuric acid/dilute hydrochloric acid. [1] (ii) Reaction 2 is a reduction. [1]

65

(a) Compound A:

[1] Compound B:

[1] Compound C:

[1] (b) Reaction 1 is an oxidation. [1] Reaction 2 is an acid hydrolysis. [1] Reaction 3 is an esterification. [1] 66

## (a) (i) Concentrated sulphuric acid at 180C [2] (ii) Acidified potassium dichromate solution and heating [2] (iii) Thionyl chloride [1] (b) (i)

[1] and

[1]

(ii)

[1] 67

(a) Compound A: Methylpropanal [1] Compound B: Methylethyl ethanoate [1] Compound C: Ethanamide [1] (b)

[2]

[2]

[2] (c) Hydrolysis is a chemical reaction in which a compound is broken down by reaction with water. [1]

The reactions involving compounds B and C respectively are hydrolysis. [2] (d)

[3] 68

(a) Compound A is propanamide. [1] Its structural formula is

[1] (b) The two products are sodium propanoate [1] and ammonia. [1] 69

(a)

[5] (b)

[5] (c)

[7]

70

(a)

[7] (b)

[7]

71

(a) [3] (b) [3] (c)

[3] (d)

[3] (e)

[3] 72

(a) (i) 1-chloropropane [1] (ii)

[6] (b) (i) 2-chloropropane [1] (ii)

[4]

73

(a)

[3] (b) (i) SOCl 2 and NH 3 [2] (ii) 2-oxopropanamide [1] 74

(a) 3-methylhex-2-enoic acid [1] (b) (i) It exhibits geometrical isomerism. [1]

(ii)

[1]

and

[1]

cis-3-methylhex-2-enoic acid [1] trans-3-methylhex-2-enoic acid [1] (c) Heating a mixture of CH 3 CH 2 CH 2 C(CH 3 )=CHCOOH and CH 3 OH in the presence of concentrated sulphuric acid. [3] 75

(a) Compound A is propanoic acid. [1] Compound B is methanol. [1] Compound C is methyl propanoate. [1] (b) Concentrated sulphuric acid [1] (c) Propanal [1]

76

(a) It is an optically active compound, so one of the carbon atoms is attached to four different atoms or groups of atoms. [1] The structure of compound A is

[1] (b) (i) (ii)

Concentrated H 2 SO 4 and at 180C [2]

[1] [1] [1] (c) (i) Oxidation [1] (ii) Acidified potassium dichromate solution and heating under reflux [2] (d) Compound B < Compound C < Compound A [1] The molecules of compound A are held together by hydrogen bonds as well as van der Waals’ forces. [1] The molecules of compound B are held together by dispersion forces only. [1] The molecules of compound C are held together by dipole-dipole forces as well as dispersion forces. [1] Since the strength of hydrogen bonds > dipole-dipole forces > dispersion forces, the boiling point of compound A > compound C > compound B. [1] 77

(a)

[3]

(b)

[3]

(c)

[3] 78

(a) (i) (ii) (iii) (b) (i) (ii) (iii)

79

(a)

Addition reaction [1] Substitution reaction [1] Oxidation [1] HCl [1] Dilute NaOH [1] Acidified K 2 Cr 2 O 7 [1]

[3] (b)

[3]

80

(a) (i)

[1] (ii) 2CH 3 CH 2 COOH(aq) + Na 2 CO 3 (aq)  2CH 3 CH 2 COONa(aq) + H 2 O(l) + CO 2 (g) [2]

(b) (i)

[1] (ii) Ethanoic acid and methanol [2] (c) (i)

[2] (ii) 2-oxopropanoic acid [1] 81

(a)

[3] (b)

[3]

82

(a)

[5] (b)

[3] (c)

[3]

83

(a) (i)

Compound A [1]

Compound B [1] (ii) They are position isomers. [1] (b) (i)

Compound C is propanoic acid. [1] It contains COOH group which can react with Na 2 CO 3 (aq) to give carbon dioxide. [1] Compound D is propanone. [1] It does not contain COOH group, so it does not react with Na 2 CO 3 (aq). [1]

(ii) 2CH 3 CH 2 COOH(aq) + Na 2 CO 3 (aq)  2CH 3 CH 2 COONa(aq) + CO 2 (g) + H 2 O(l) [2] (c) (i) Propyl propanoate [1] (ii) [1] 84

(a) Propanone [1] (b)

[6] 85

(a)

[3] (b)

[3] 86

(a)

[3] (b) Percentage yield of a product =

68.0% =

actual mass of the product obtained  100% theoretical mass of the product obtained

117.2 g 100% [1] theoretical mass of the product obtained

Theoretical mass of 1,2-dibromo-3-methylbutane obtained = 172.4 g [1] Theoretical no. of moles of 1,2-dibromo-3-methylbutane =

172.4 g (12.0  5  1.0 10  79.9  2) g mol1

= 0.750 mol [1]

From the equation, the mole ratio of 3-methylbutan-2-ol to 1,2-dibromo-3-methylbutane is 1 : 1, ∴ no. of moles of 3-methylbutan-2-ol required = 0.750 mol [1] ∴ mass of 3-methylbutan-2-ol required = 0.75 mol × (12.0 × 5 + 1.0 × 12 + 16.0) g mol1 = 66.0 g [1] 87

(a) Step 1: Dilute HCl and heating [2] Step 2: LiAlH 4 in dry ether and dilute HCl [2] Step 3: Concentrated H 2 SO 4 at 180C [2] Step 4: HBr [1] (b) Any THREE of the following: Availability of starting materials and reagents/Rate of reaction/Percentage yield/Number of steps/By-products formed [3] (c) The overall equation for step 3 is CH 3 CH 2 CH 2 OH  CH 3 CH=CH 2 + H 2 O [1]

No. of moles of propan-1-ol used =

60.0 g (12.0  3  1.0  8  16.0) g mol 1

= 1 mol [1] From the equation, the mole ratio of propan-1-ol to propene is 1 : 1, ∴ no. of moles of propene obtained = 1 mol Theoretical mass of propene obtained = 1 mol × (12.0 × 3 + 1.0 × 6) g mol1 = 42.0 g [1] 21.0 g 100% = 50% [1] The percentage yield of propene = 42.0 g 88

(a)

compound X [1]

aspirin [1] (b) (i)

[2] (ii) Esterification [1] (iii) Compound Y is a colourless volatile liquid and immiscible with water. [2] (c) The overall equation is

No. of moles of compound X used =

8.0 g (12.0  7  1.0  6  16.0  3) g mol 1

= 0.0580 mol [1] From the equation, the mole ratio of compound X to aspirin is 1 : 1, ∴ no. of moles of aspirin obtained = 0.0580 mol Theoretical mass of aspirin obtained

= 0.0580 mol × (12.0 × 9 + 1.0 × 8 + 16.0 × 4) g mol1 = 10.4 g [1] 6.7 g Percentage yield of aspirin = × 100% = 64.4% [1] 10.4 g 89

(a) [3] (b) (i) Concentrated sulphuric acid and heating under reflux [2] (ii) The ester is ethyl propanoate. [1] (c) The overall percentage yield = 70% ∴ 67% × 42% = 19.7% [1]

90

(a) (i) Methanol and propanoic acid [2] (ii) Methanol belongs to alcohols. [1] Propanoic acid belongs to carboxylic acids. [1] (b)

[2] (c) water out

water in pear-shaped flask a mixture of methanol, propanoic acid and conc.

anti-bumping granule

heat Correct drawing [2] Correct labelling [1] (d) Increase the amount of methanol. [1] Since the reaction is reversible, when there is an increase in the concentration of methanol, the equilibrium position will shift to the product side, more compound A can be obtained. [1] (e) (i) Sodium carbonate reacts and removes any acidic substances, e.g. unreacted propanoic acid in

(f)

91

the distillate. [1] (ii) Calcium chloride reacts and removes any unreacted methanol in the distillate. [1] (iii) Anhydrous calcium chloride removes remaining traces of water from the distillate. [1] Sodium carbonate solution and calcium chloride solution will react to form calcium carbonate. Excess methanol and propanoic acid cannot be removed. The newly formed calcium carbonate will make the distillate become turbid. [2]

(a) (i) But-2-ene [1] (ii)

[1] cis-but-2-ene [1]

[1] trans-but-2-ene [1] (b) Addition reaction [1] (c) (i) The reaction mixture changed from orange to green. [1] (ii) water out

water in pear-shaped flask butan-2-ol + acidified potassium dichromate solution

anti-bumping granule

heat Correct drawing [2] Correct labelling [1] 92

(a) The direction of water in and water out was reversed. [1] There was no anti-bumping granule. [1] The condenser should not be stoppered. [1] (b) (i)

[2] (ii) Propyl methanoate [1] (c) The overall equation is

No. of moles of methanoic acid =

2.0 g (12.0 1  1.0  2  16.0  2) g mol 1

= 0.0435 mol 2.5 g = 0.0417 mol No. of moles of propan-1-ol = (12.0  3  1.0  8  16.0) g mol 1 From the equation, the mole ratio of methanoic acid to propan-1-ol = 1 : 1, ∴ propan-1-ol is the limiting reagent. From the equation, the mole ratio of propan-1-ol to propyl methanoate = 1 : 1, ∴ no. of moles of propyl methanoate obtained = 0.0417 mol [1] Theoretical mass of propyl methanoate obtained = 0.0417 mol × (12.0 × 4 + 1.0 × 8 + 16.0 × 2) g mol1 = 3.67 g [1] 2.0 g  100% = 54.5% [1] Percentage yield of propyl methanoate = 3.67 g 93

(a) A mixture of propan-1-ol and excess acidified potassium dichromate solution is heated under reflux for 20 to 30 minutes. [2] (b) CH 3 CH 2 CH 2 OH + 2[O]  CH 3 CH 2 COOH + H 2 O [1] (c) Propanoic acid is separated from the product mixture by distillation. [1] Then, the distillate which contains an aqueous solution of propanoic acid is redistilled and the fraction that boils between 140C and 142C is collected. [2] (d) The overall equation is 3CH 3 CH 2 CH 2 OH + 2Cr 2 O 7 2 + 16H+  3CH 3 CH 2 COOH + 4Cr3+ + 11H 2 O The mass of propan-1-ol used = 20.0 cm3 × 0.8 g cm3 = 16.0 g 16.0 g No. of moles of propan-1-ol used = (12.0  3  1.0  8  16.0) g mol 1 = 0.267 mol [1] From the equation, the mole ratio of propan-1-ol to propanoic acid is 1 : 1, ∴ no. of moles of propanoic acid obtained = 0.267 mol Theoretical mass of propanoic acid obtained = 0.267 mol × (12.0 × 3 + 1.0 × 6 + 16.0 × 2) g mol1

= 19.8 g [1] Percentage yield of propanoic acid =

94

14.5 g 100% = 73.2% [1] 19.8 g

(a) Assume there were 100 g of maltose, then there would be 42.1 g of carbon, 6.4 g of hydrogen and 51.5 g of oxygen. C

H

O

42.1

6.4

51.5

No. of moles of atoms (mol)

42.1  3.5 12.0

6.4  6.4 1.0

51.5  3.2 16.0

Relative no. of moles

3.5  1.1 [1] 3.2

6.4  2 [1] 3.2

3.2  1 [1] 3.2

Masses (in g)

∴ the empirical formula of maltose is CH 2 O. [1]

(b) Condensation reaction [1] (c) (i) Aldehyde group and hydroxyl group [2] (ii)

[1] (iii) There are 4 chiral carbon atoms in a glucose molecule. [1]

[1]

95

(a)

glucose [1] (b) (c) (d) (e)

fructose [1]

Glucose belongs to aldehydes. [1] Fructose belongs to ketones. [1] Glucose and fructose are functional group isomers. [1] Glucose and fructose are added to acidified potassium dichromate solution respectively. [1]

The aldehyde group of glucose is oxidized by acidified potassium dichromate solution and the solution changes from orange to green. [1] However, there is no reaction between fructose and acidified potassium dichromate solution and thus no observable changes are made. [1] 96

(a) (i)

If the hydrocarbon chain of the fatty acid contains no carbon-carbon double bond, it is a saturated fatty acid. [1] If the hydrocarbon chain of the fatty acid contains one or more carbon-carbon double bonds, it is an unsaturated fatty acid. [1] (ii) Palmitic acid is a saturated fatty acid. [1] (iii)

[1] (b) The carbon-carbon double bonds in animal fats adopt trans configuration. [1] This causes the triglyceride molecules to pack more efficiently. [1] As a result, animal fats have higher melting points and are usually solids at room temperature. [1] 97

(a) Linoleic acid is an unsaturated fatty acid because its hydrocarbon chain contains carbon-carbon double bonds. [2] (b) The mole ratio of glycerol to linoleic acid is 1 : 3. [1] (c)

[1] (d) Add bromine water to the triglyceride mentioned in (b). [1] If the bromine water is decolorized, the triglyceride molecules should be unsaturated. [1] (e) The carbon-carbon double bonds in corn oil adopt the cis configuration. [1] This causes the triglyceride molecules to pack less efficiently. [1] As a result, corn oil has a lower melting point and is a liquid at room conditions. [1]

98

(a) Glycerol [1] (b) Salts of fatty acid [1] (c) The detergents are wetting agents which reduce the surface tension of water and enable water to wet the surface of clothes thoroughly. [1] The hydrocarbon tails of detergent anions dissolve in grease while the ionic heads of detergent anions dissolve in water. [1] Water molecules attract the ionic heads of detergent anions, lifting up the grease from the surface of clothes. [1] By stirring, the grease forms tiny negatively charged oil droplets which repel each other. [1] (d) The biodegradable detergents can be broken down by bacteria in water. During the process, bacteria would use up dissolved oxygen in water. As a result, some marine organisms may suffocate to death. [2]

99

(a) Condensation reaction [1] (b) (i) 2-aminoethanoic acid [1] (ii) 2-aminopropanoic acid [1] (c)

[1]

[1]

[1]

[1] (d) (i)

[1] (ii)

[1]

100

(a) Amide linkages/peptide linkages [1] (b) Dilute sodium hydroxide solution/hydrochloric acid and heating [2] (c)

[1]

[1]

[1] (d)

[1] There is no chiral carbon atom in the amino acid molecule, so it does not exhibit enantiomerism. [1] 101

(a)

[3] (b) The functional groups of aspirin include benzene ring, carboxyl group and ester group. [3] (c) Aspirin has an anti-inflammatory effect and an anti-platelet effect. [2] (d) Stomach upset and ulcer [1] or increased bleeding [1] 102

(a)

[1] (b) Dilute hydrochloric acid/Dilute sulphuric acid [1] (c) The overall yield of the conversion = 70% × 95% = 66.5% [1]

(d) Stomach upset and ulcer/increased bleeding [1] (e) Panadol/Paracetamol [1] 103

(a)

[1] (b)

[2] (c) It has an anti-inflammatory effect and an anti-platelet effect. [2] (d) Since aspirin is acidic, large dosage will cause stomach upset and ulcer. Magnesium hydroxide is used to neutralize the acidity. [2] 104

(a) Aspirin [1] (b) Salicylic acid and ethanoic acid [2] (c) It causes stomach upset and ulcer. [1] Use alternative medicine like Panadol. [1]/Aspirin is used with magnesium hydroxide. [1]

105

(a) Acetylsalicylic acid [1] (b)

[1] (c) Any TWO of the following: Carboxyl group [1] Ester group [1] Benzene ring [1] (d) Any TWO of the following: Relieving pain [1] Reducing inflammation and fever [1] Reducing the risk of heart attack [1] (e) Any TWO of the following:

Stomach ulcer sufferers [1] Liver disease sufferers [1] Children under the age of 12 [1]

106

(a)

[2] (b) Ester linkage [1] (c) (i) If the hydrocarbon chain contains no carbon-carbon double bond, it is a saturated fatty acid. [1] If the hydrocarbon chain contains one or more carbon-carbon double bonds, it is an unsaturated fatty acid. [1] (ii) Stearic acid is a saturated fatty acid. [1] (d) (i) CH 3 (CH 2 ) 16 COONa+ [1] (ii) It is a soapy detergent. [1] 107

(a) (i)

Stearic acid is a saturated fatty acid because its hydrocarbon chain contains no carbon-carbon double bond. [2]

(ii)

[1] (b) (i) Sodium hydroxide solution [1] and boiling [1] (ii) Add the detergent into a solution containing calcium ions or magnesium ions. [1] If lather is observed, the detergent is a soapless detergent. [1] If scum is observed, the detergent is a soapy detergent. [1] 108

(a) (i) Three [1] (ii)

[1] (iii) Water [1] (b) (i) Coconut oil is boiled with sodium hydroxide solution. [1] (ii)

[2]

109

A detergent enables water to wet the object thoroughly. [1] The hydrophobic tails of detergent anions dissolve in grease. [1] Water molecules attract the hydrophilic heads of detergent anions, lifting up the grease from the surface. [1] By stirring, the grease is broken down into tiny droplets, forming an emulsion. [1]

110

(a) Sulphate group/OSO 3  [1] (b) Petroleum [1] (c)

foam tiny oil droplets emulsion Correct drawing [1] Correct labelling [1] (d) Yes, it can. It is because soapless detergent will not combine with Ca2+ ions or Mg2+ ions to form scum in sea water. [2]

111

(a)

[2] (b) (i)

[1] (ii)

[2] (c) (i)

Soapy detergents are made from fats or oil while soapless detergents are derived from petroleum. [2] Soapy detergents are sodium/potassium salts of long-chain carboxylic acids while soapless detergents are sodium salts of long-chain alkylbenzenesulphonate or alkylsulphate. [2] (ii) The detergent formed in (b)(ii) is a soapy detergent. [1] (iii) The detergent enables water to wet the clothes thoroughly. [1] The hydrophobic tails of detergent anions dissolve in grease. [1] The hydrophilic heads of detergents anions are attracted by water molecules, then the grease is lifted off from the surface of clothes. [1] By stirring, the grease forms tiny droplets which disperse throughout water to form an emulsion. [1] The lather produced when mixing detergent and water also helps to suspend grease which can then be easily washed away. [1] (iv) No, it does not. [1] It is because the detergent anions form insoluble substance with calcium and/or magnesium ions. [1]

112

(a) Soaps: Sodim stearate/CH 3 (CH 2 ) 16 COONa+ [1] Soapless detergents: Sodium alkylbenzenesulphonate/ CH 3 (CH 2 ) 11 (C 6 H 4 )SO 3 Na+/Sodium alkylsulphate/CH 3 (CH 2 ) 11 OSO 3 Na+ [1] (b) Soaps: animal fats or oils and sodium hydroxide solution [2] Soapless detergents: chemicals obtained from petroleum, concentrated sulphuric acid and sodium hydroxide solution [3] (c) The soaps and soapless detergents are added to a beaker containing magnesium sulphate solution respectively. [1] Scum is formed in a beaker containing soaps. [1] It is because soap anions form insoluble substance with magnesium ions. [1] The chemical equation for the reaction between soap anions and magnesium ions is: Mg2+(aq) + 2CH 3 (CH 2 ) 16 COO(aq)  (CH 3 (CH 2 ) 16 COO) 2 Mg(s) [1] Lather instead of scum is formed in a beaker containing soapless detergents. [1] It is because the ionic heads of soapless detergent particles do not form precipitate with magnesium ions. [1]

113

(a) A detergent is a substance which helps water to clean things better. [1] (b) Animal fats and vegetable oils. [2] (c) A soapy detergent particle contains an ionic group (the head)/a carboxylate group [1] and a hydrocarbon chain (the tail). [1] The ionic group is hydrophilic/water-loving/soluble in water. [1] The hydrocarbon chain is hydrophobic/water-hating/insoluble in water/soluble in oil. [1] (d) The ionic heads (usually negatively charged) will dissolve in water and the hydrocarbon tails will dissolve in oil. [1] Therefore, the surfaces of all oil droplets are negatively charged. [1] The oil droplets repel each other and cannot join together. [1]

114

(a) It is a soapless detergent. [1] (b) Petroleum [1] (c) A detergent particle should have an ionic group (head) [1] which is soluble in water [1] and a hydrocarbon chain (tail) [1] which is soluble in grease or oil. [1] (d) Wetting property and emulsifying property [2] (e) water oil droplet Correct drawing [1]

Correct labeling [1] 115

(a) (i)

Compounds A and C are soapy detergent particles. [2] Because both of their ionic heads are

carboxylate group (COO). [1] (ii) Animal fats and vegetable oils [2] (iii) Any ONE of the following: Bath soaps/laundry soaps/liquid soaps [1] (b) (i) Compound B is a soapless detergent particle. [1] Because its ionic head is a sulphonate group (SO 3 ). [1] (ii) Petroleum [1] (iii) Any ONE of the following: Washing powder/washing-up liquids/shampoos [1] 116

(a) (i)

Hard water is water that contains considerable concentrations of calcium and/or magnesium ions. [1] (ii) It is because soapy detergent anions form insoluble substances with calcium and/or magnesium ions. [1] Therefore, they will lose their cleaning action. Ca2+(aq) + 2CH 3 (CH 2 ) 16 COO(aq)  (CH 3 (CH 2 ) 16 COO) 2 Ca(s) [1] (iii) Washing soda contains carbonate ions which react with calcium ions to form insoluble calcium carbonate. [1] Therefore, calcium ions are removed and the hardness of water is reduced. [1]

Ca2+(aq) + CO 3 2(aq)  CaCO 3 (s) [1] (b) (i) In acidic solution, soapy detergents anions react with hydrogen ions to form long-chain carboxylic acids which are insoluble in water. [1] Therefore, they become precipitates. [1] (ii) Washing soda is sodium carbonate which removes hydrogen ions in the acidic solution, thus acidity of water is reduced. [1] (c) Stirring helps to lift up the grease from the surface of dishes and clothes. [1] It also helps to break down the grease into small oil droplets for easy dispersion in water. [1] (d) 12.0(n+1) + 23.0 + 16.0 × 2 + 2.0n + 1 = 306 [1] 14.0n + 68 = 306 n = 17 [1] 117

(a) CH 3 (CH 2 ) 14 COONa [1] (b) It is used to salt out the soap from the solution. [1] The addition of saturated sodium chloride solution can reduce the solubility of the soap. [1] (c) White creamy solid is formed and it floats on the surface. [1] (d) It is to wash away the water soluble impurities such as unreacted alkali, sodium chloride and glycerol formed. [1]

118

(a) It is a soapy detergent. [1] Because its ionic head is carboxylate group. [1]

(b) The ionic head of the detergent particle is hydrophilic [1] while the hydrocarbon tail is hydrophobic. [1] (c) His suggestion is incorrect. [1] Since sea water contain considerable concentrations of calcium ions and/or magnesium ions, the soapy detergent anions can form insoluble scum with these ions. [1] Therefore, the soapy detergent will lose its cleaning power. [1] 119

(a) Z should be a soapy detergent. [1] It is the only one which forms scum with magnesium ions. [1] Only soapy detergents form scum with magnesium ions. [1] (b) X could cause skin allergies. [1] It has a low pH value of 3. It is too acidic for our skin. [1] (c) Y. [1] It has a pH value of 7 which is as neutral as water. It could not cause skin allergies. [1] It does not form scum in hard water, so it can work well in hard water. [1] It is biodegradable, so it causes less environmental problems. [1]

120

(a) Amide linkage [1] (b)

[1]

[1] OR

[1]

(c) Condensation reaction [1] (d) Water [1] or HCl [1] 121

## (a) The general repeating unit of nylon:

R and R’ are hydrocarbon chains [1] The general repeating unit of polyester:

R and R’ are hydrocarbon chains [1] (b) There are amide linkages in nylon. [1] There are ester linkages in polyesters. [1] (c) Nylon: water-proof/tough/high tensile strength [1] Polyesters: resistant to water and chemicals/tough/strong/smooth/lightweight [1] (d) Nylon: making fishing ropes/fishing lines/fishing nets/tennis racket strings/ stockings [1] Polyesters: making fibres/clothes/lightweight sails/bottles for soft drinks [1]