Pra-U STPM Chemistry Term 3 2018 CB039348c

Chemistry Term 3 STPM Chapter 21 Polymers 292 21 Quick Check 21.1 1 Terylene can also be obtained through the polymerisation of ethane-1,2-diol and the dimethyl ester of hexanedioic acid, CH3COO(CH2)4COOCH3. (a) During the polymerisation process, what other molecules are eliminated? (b) Write an equation to represent the polymerisation. 2 Draw a repeating unit of the polyamide formed by the polymerisation of (a) 6-aminohexanoic acid (b) a mixture of 1,6-hexadiamine and 1,6-hexadioyl chloride 3 Nylon-6,10 is an example of polyamide. (a) Draw the structures for the two monomers of nylon-6,10. (b) Draw a repeating unit of nylon-6,10. 21.4 Addition Polymerisation 1 Addition polymerisation is the process where monomers are joined with one another to form polymeric chains without the elimination of other smaller molecules. 2 The monomers for addition polymers are usually alkenes or their derivatives which contain carbon-carbon double bonds. Or: & & & & n !CRC! !: !!(C!C)n!! & & 3 Note that the empirical formulae of addition polymers are the same as those of their respective monomers. 4 Some examples of addition polymers are given below. Poly(ethene), PE (or Polyethylene) 1 Poly(ethene) or polythene (PE) is produced by the polymerisation of ethene. 2 When ethene gas, under a pressure of 1200 atm is heated to about 200 °C in the presence of traces of oxygen, a soft clear solid called poly(ethene) or polythene (PE) is formed. nCH2RCH2 !: !!(CH2!CH2)n!! Ethene Poly(ethene) !! !! CRC !! !! CRC !! !! CRC !: !! !! !! !! !! !! !C!C!C!C!C!C! Empirical formula = molecular formula 2008/P1/Q40 2011/P1/Q40 2009/P2/Q10(b) 2010/P2/Q4(c),9(b) 2010/P1/Q50 2012/P1/Q40 2013/P3/Q17(b)

Chemistry Term 3 STPM Chapter 21 Polymers 293 21 3 The poly(ethene) produced by this method (discovered by ICI in 1933) is highly branched. As a result, the polymer chains cannot be closely packed. 4 As a result, the poly(ethene) produced has a low melting/ softening point (about 105 °C) and a low density (0.92 g cm–3), and is easily deformed and softens in boiling water. 5 This form of poly(ethene) is called low density poly(ethene), LDPE, and is used generally for making plastic bag or sheets for wrapping, squeeze bottles and electrical insulation materials. 6 Another method for the manufacture of poly(ethene) was developed by Karl Ziegler and Giulio Natta in 1953 by using a catalyst called the Ziegler-Natta catalyst. This allows the polymerisation to take place at 1 atm and 60 °C. 7 The poly(ethane) produced by this method is made up mostly of linear polymeric chains. This allows the chains to be more closely packed to form orderly structures. 8 As a result, the poly(ethene) produced is more crystalline, has higher density (0.96 g cm–3) and higher melting/softening point (130 °C to 140 °C). This is known as high density poly(ethene), HDPE. 9 HDPE is stronger and harder than LDPE and is used mainly for molding more rigid articles such as bottles and milk crates. Poly(propene)/Polypropylene 1 Propene, like ethene, can be polymerised to give poly(propene) or polypropylene. nCH3!CHRCH2 !: !!(CH!CH2)n!! & CH3 Low density polyethene (LDPE) Info Chem An example of Ziegler-Natta catalyst is a mixture of trialkylaluminium, R3Al and titanium tetrachloride, TiCl4. Exam Tips Exam Tips As the chains are of different lengths and with different degree of branching, poly (ethene) does not melt at a fixed temperature. Instead, it melts over a range of temperature as the interchain attractive forces are of different strength.

Chemistry Term 3 STPM Chapter 21 Polymers 294 21 2 Due to the presence of the methyl side groups, the polymeric chains produced can have one of the following structures: (a) !CH!CH2!CH!CH2!CH!CH2!CH!CH2! & & & & CH3 CH3 CH3 CH3 In this form, all the CH3 groups are on the same side of the chain. This is called an isotactic polymer. (b) CH3 CH3 & & !CH!CH2!CH!CH2!CH!CH2!CH!CH2! & & CH3 CH3 In this form, the position of the CH3 groups alternates above and below the chain. This is called syndiotactic polymer. (c) CH3 & !CH!CH2!CH!CH2!CH!CH2!CH!CH2! & & & CH3 CH3 CH3 The third form of polypropene, the arrangement of the CH3 groups is random. This is called an atactic polymer. Atactic poly(propene) is rubbery, amorphous and relatively weak. 3 Free radical polymerisation of propene always produces the atactic form. However, by using Ziegler-Natta catalyst, stereospecific poly(propene) can be formed. 4 In the presence of Ziegler-Natta catalyst, lower temperature and pressure can be used for the polymerisation process. 5 The presence of CH3 groups increases the intermolecular van der Waals forces between the polymer chains, but makes the chain more difficult to pack. Hence, it has low density (0.90 g cm–3) but higher melting/softening point (140 °C). 6 Polypropene is used to make ropes, moulds, bottles, kitchenware, carpets, battery containers and for apparatus that needs to be steam sterilised. Poly(chloroethene) 1 Poly(chloroethene) or polyvinyl chloride (PVC) is produced from chloroethene or vinyl chloride. nCH2RCH !: !![CH2!CH]n!! & & Cl Cl Isotactic form Syndiotactic form Atactic form Stereospecific polymer Uses of polypropene The C!Cl bond is polar.

Chemistry Term 3 STPM Chapter 21 Polymers 295 21 2 The presence of polar C!Cl bonds increases the intermolecular forces between chains making PVC considerably stronger and harder. 3 PVC is used to make curtain, artificial leather goods, pipes, cables and gramophone records. Other Examples of Addition Polymers The table below lists some other addition polymers and their major uses. Polymer Monomer Repeating unit Main uses Polyethene CH2RCH2 (Ethene) !![CH2!CH2 ]!! Plastic bags, bowls, kitchenware, buckets, bottles Polytetrafluoroethene CF2RCF2 (Tetrafluoroethene) !![CF2!CF2 ]!! Non-stick pans, electric insulators Polypropene CH3CHRCH2 (propene) !![CH!CH2 ]!! & CH3 Packaging films, ropes, bottles, moulds, battery case, carpets Polyvinyl chloride CH2RCHCl (Chloroethene) !![CH!CH2 ]!! & Cl Artificial leather goods, gutters, vinyl records, pipes Polystyrene [Styrofoam] !CHRCH2 (Phenylethene/ Styrene) !![CH!CH2 ]!! Packaging materials, egg trays, heat insulators, toys Perspex CH2RC!CH3 & COOCH3 (Methyl 2-methylpropenoate) CH3 & !![CH2!C!!]! & COOCH3 Substitute for glass, car headlights, aeroplane windows, optical instruments H & 9CH29Cδ+9 & Clδ– Uses of PVC 2009/P1/Q39 Quick Check 21.2 1 The monomer of Orlon is acrylonitrile. CH2RCHC#N (a) Give the IUPAC name for the monomer. (b) Draw the structure of Orlon showing three repeating units. (c) Would you expect Orlon to be stronger or weaker than poly(ethene)? Explain your answer. 2 An addition polymer is obtained by copolymerisation between the following two monomers: CH2RCHCl and CHRCH2 By drawing a repeating unit each, show two different structures possible for the polymer.

Chemistry Term 3 STPM Chapter 21 Polymers 296 21 3 The repeating unit of two polymers, X and Y are shown below: !!CH2!CH!!!! NH2 C ! O R X !!!!CH!CH2!!! n n Y (a) Draw the structures of the monomers used to make polymers X and Y. (b) Which of the two is suitable for the storage of concentrated aqueous solution of sulphuric acid? Give your reason. 4 Part of the structure of a copolymer is shown below: !!CH2!CH!CH2!CH2!CH2!CH!CH2!CH2!! & & C#N C#N Draw the structures of its two monomers. Polymerisation of Dienes 1 Dienes are organic compounds with two carbon-carbon double bonds in their structure. 2 A simple example is 1,3-butadiene: CH2RCH!CHRCH2 3 Like alkenes, dienes undergo addition polymerisation as shown below: Or, !![!CH2!CHRCH!CH2!!!]n The polymer is named poly(1,3-butadiene) or simply as polybutadiene. 4 Polybutadiene is used in making tyre treads and coating resins. 5 Unlike polyalkenes, there are still carbon-carbon double bonds in the polymeric chain of polydiene. As a result, polydienes can undergo cross-linking such as vulcanisation (to be discussed later.) CH2RCH!CHRCH2 !!CH2!CHRCH!CH2!CH2!CHRCH!CH2!CH2!CHRCH!CH2! CH2RCH!CHRCH2 CH2RCH!CHRCH2 There are still CRC bonds in the polymer. Uses of polybutadiene The C"C bond can be attacked by ozone or other oxidising agents. 2016/P3/Q15 2017/P3/Q15

Chemistry Term 3 STPM Chapter 21 Polymers 297 21 Quick Check 21.3 The monomer of neoprene is 2-chloro-1,3-butadiene (also called chloroprene). (a) Draw the structure of chloroprene. (b) What is the IUPAC name for neoprene? (c) Draw a repeating unit of neoprene. 21.5 Classification of Polymers Generally, polymers can be classified under three broad categories: thermoplastics, thermosetting plastics and elastomers. Thermoplastics 1 A thermoplastic, also known as a thermosoftening plastic is a polymer that becomes soft (and thus moldable) when heated and hardens when cooled. 2 The process of softening and hardening can be repeated over many times. 3 Thermoplastics are linear polymers with no cross link between the polymer chains. The chains are held together by weak van der Waals forces only. Heating Cooling 4 On heating, the chains drift further apart and soften. On cooling, the reverse occurs. 5 Examples of thermoplastics are poly(ethene), polyvinyl chloride, (PVC), poly(propene), polystyrene and polybutadiene. Thermosetting Plastics 1 A thermosetting plastic, also known as a thermoset is a polymer that once set cannot be soften through heating. It is like cement, the process cannot be reversed once it is harden. 2 There are extensive cross links (via strong covalent bonds) between the polymer chains during the polymerisation process. These cross links prevent the polymer chains from moving apart when heated. 3 When heated strongly, the polymer decomposes. 4 An example is bakelite, a copolymer between methanal and phenol. 2014/P3/Q15

Chemistry Term 3 STPM Chapter 21 Polymers 298 21 CH2 CH2 OH CH2 OH CH2 CH2 OH CH2 OH OH CH2 OH CH2 OH CH2 OH OH 5 Other examples of thermosets are vulcanised rubber, melamine resins and epoxy resins. Elastomers 1 An elastomer is a polymer that can be deformed or stretched when a force is applied to it. However, it reverts back to its original shape or length when the force is removed. In other words, an elastomer has elasticity properties. 2 In the raw state, the polymer chains are coiled and twisted with a few cross links between the chains. Stretched Force released 3 A certain amount of cross links has to be present, otherwise the chains will slide over one another and will not go back to its original state. 4 However, too many cross links will end up with a thermosetting plastic. Natural Rubber 1 Natural rubber is an elastomer. That is, it lengthens when stretched. When the force is released, it reverts back to its original length. 2 The monomer of natural rubber is 2-methylbuta-1,3-diene or isoprene.

Chemistry Term 3 STPM Chapter 21 Polymers 299 21 C C H !! ! !! RR !! RR CH3 CH2 CH3 CH2 or CH2RC!CHRCH2 3 The isoprene molecules undergo addition polymerisation to give poly(isoprene). nCH2RC!CHRCH2 !: !![!CH2!CRCH!CH2!!!]n & & CH3 CH3 4 However, the poly(isoprene) polymeric chains can exist in two forms: The poly-cis-isoprene and poly-trans-isoprene depending on the relative spatial arrangement of the two !CH2 groups. 5 The poly-cis-isoprene has the following structure where the two !CH2 groups bonded to the unsaturated carbon atoms are on the same side. CRRC !! !! !! !! CH2 CH3 CH2 H n 6 The trans-form has the following structure: CRRC !! !! !! !! CH2 CH3 CH2 H n 7 The cis-form is derived from the Hevea brasiliensis trees, while the trans-form is derived from the gutta-percha trees. 8 The trans-form is non-elastic and is of little use. It is used mainly for the manufacture of golf balls and the outer covering of underwater cables. The cis-form on the other hand is elastic and is much more useful. 9 The cis-form can be extended to about 10 times its length and returned to its original size. This is because in its raw form, rubber is a tangle of polymeric chains which are coiled and twisted. The chains straighten when stretched, and spring back to the original form when the force is released. Isoprene: Polyisoprene: n cis(polyisoprene): n trans(polyisoprene): n Rubber is an elastomer.

Chemistry Term 3 STPM Chapter 21 Polymers 300 21 Random coils Stretched Force released Stretched chains 10 However, in the raw state, there is no cross-link between the polymer chains. It becomes soft and sticky when hot, hard and brittle when cold. At the same time, it is sensitive to oxidising agent such as ozone in the atmosphere. As a result, its use is limited. 11 In 1838, an American, Charles Goodyear discovered that these defects could be overcome by heating raw rubber with sulphur which forms disulphide cross-links between polymeric chains. H H S CH!! CH!! ! S! ! ! ! ! ! CH3 !!CH2 CRC ! ! ! ! CH3 !!CH2 CRC Disulphide link 12 The disulphide links prevent the polymeric chains from slipping past one another when the rubber is stretched, making it more elastic. 13 Vulcanised rubber is harder, stronger and more resistant to atmospheric oxidation. 14 However, too much vulcanisation would make the rubber hard and brittle. 15 Vulcanised rubber is used mainly to make tyres, footwear, mattresses, gloves, elastic bands, tubings, toys and belts. Example 21.2 Explain why rubber is an excellent material to manufacture surgical gloves. Solution Natural rubber is impermeable to water. Hence, body fluids and blood cannot penetrate the glove and infect the medical personnel. Vulcanisation of rubber The disulphide bridge Uses of vulcanised rubber

Chemistry Term 3 STPM Chapter 21 Polymers 301 21 Production of Rubber from Latex 1 Latex is an aqueous emulsion of negatively charged rubber molecules. Negative charged protein membrane Rubber molecule 2 The negative charge on the rubber molecules causes them to repel one another. 3 When acids such as methanoic (formic) acid or ethanoic (acetic) acid is added to latex, the H+ from the acids neutralises the negative charge on the rubber molecules. The molecules then approach one another and coagulate to form a solid mass. 4 The solid mass is removed from the coagulating tank. They are pressed into sheets by passing them through rollers to remove water. 5 The rubber sheets are then smoked and dried. Example 21.3 Raw rubber is sometimes exported in the form of latex. Ammonia is added to the latex to prevent it from coagulating. Explain the chemistry behind this method. Solution Latex is an aqueous suspension of negative charged rubber particles. Addition of ammonia, which supplies the negative charged OH– , further prevents the rubber particles from coming together. NH3(aq) + H2O(l) L NH4 +(aq) + OH– (aq) Synthetic Rubber 1 The first synthetic rubber was produced by the polymerisation of 2-chlorobuta-1,3-diene. CH2RCH!CRCH2 !: !![!CH2!CHRC!CH2!!]n ! & & Cl Cl The polymer is named neoprene. 2 Neoprene is resistant to most chemicals. It is used to make hose for petrol and containers for corrosive liquids. 3 Since then, many types of synthetic rubber are made from 1,3-butadiene and its derivatives. Latex is a colloid. Formic acid is used to coagulate latex. Uses of neoprene Exam Tips Exam Tips Coagulation can also occur through the action of bacteria.

Chemistry Term 3 STPM Chapter 21 Polymers 302 21 4 The most important in terms of quantity produced is the styrene-butadiene rubber (SBR). 5 SBR is a copolymer between styrene (phenylethene) and 1,3-butadiene in the ratio of 1 : 3. CHRCH2 + CH2RCH!CHRCH2 !![!CH!CH2!CH2!CHRCH!CH2!]!! 6 SBR can be vulcanised in the same way as natural rubber. It is used mainly to make car tyres, footwears and carpet backing. Uses of Polymers 1 Polymers have become part and parcel of our lives that it is difficult to visualise the world without polymers. 2 We use polymers in almost every aspect of our lives. Plastics for containers, food wrappers, pipes, plastic bottles, toys, carpets, footwear, artificial leather (PVC), electrical insulators, non-stick pans, artificial silk, synthetic fabrics, ropes, bullet-proof vests, vehicle tyres, packaging materials and gloves. The list can go on and on. Effects of Polymer Disposal on the Environment 1 Most synthetic polymers (commonly called plastics) are nonbiodegradable. They cannot be decomposed by bacteria or other forms of decomposers. When discarded, they stay in the earth for decades if not centuries. 2 These polymers release chemicals that are hazardous both to humans and the environment. 3 Burning of polymers in dump sites also releases toxic chemicals (such as carbon monoxide, hydrogen cyanide and dioxin) into the atmosphere causing acid rain and air pollution. 4 Indiscriminate dumping of polymers into rivers and seas causes water pollution and blockage of the water-ways causing floods. 5 Some of these polymers are ingested by sea-birds, fish and other aquatic animals causing death due to chocking. 6 One way to overcome these problems is to invent biodegradable polymers that can be decomposed by bacteria or water. Uses of SBR 2015/P3/Q15 INFO Environmental Impact of Plastics

Chemistry Term 3 STPM Chapter 21 Polymers 303 21 SUMMARY SUMMARY 1 Polymers are long-chain molecules produced by linking smaller molecules called monomers. 2 Addition polymers are polymers that are formed by the linking of monomers without the elimination of other smaller molecules. 3 Examples of addition polymers are poly (ethene), poly(styrene), PVC, poly(propene). 4 Monomers for addition polymers have CRC bonds in their molecules. 5 Condensation polymers are polymers formed by linking monomers with the elimination of other smaller molecules. 6 Examples of condensation polymers are polyamides (nylons) and polyesters (terylene). 7 The monomer for natural rubber is isoprene or 2-methylbuta-1,3-diene. 8 Vulcanisation is the process where the rubber chains are linked by disulphide bonds. 9 Thermoplastics, sometimes called thermosoftening plastics are polymers that become soft on heating, and harden when cool. This process of softening and hardening can be repeated over and over again. 10 Thermosetting plastics or thermosets are threedimensional polymers consisting of extensive cross links which are formed during the polymerisation process. They harden on cooling and cannot be softened by heating. 11 Elastomers are polymers which stretch when pulled, and return to their original length when the force is released. STPM PRACTICE 21 Objective Questions 1 A part of a polymer chain is given below: O H N N H O n Which is not true about the polymer? A It has a low softening point. B It is a copolymer. C HCl is eliminated during its formation. D It is a condensation polymer. 2 A thermosetting plastic is a polymer having the following property. A Its melting point is below 0 °C. B It can be softened on heat and hardened on cooling, and the process can be repeated. C It is chemically inert and non-biodegradable. D Once hardened, it cannot be soften by heat. 3 The repeating unit of a polymer is shown below. ! CH2 ! CH !! n CH2 CH2 !! n CH2 ! CH !n C R H H C This is an example of A homopolymer B copolymer C a condensation polymer D a thermosetting polymer 4 Which of the following statements is not correct regarding natural rubber? A It softens when heated and hardens when cooled. B It can be formed by the polymerisation of isoprene. C It is an elastomer. D Vulcanised rubber is more elastic than natural rubber. 5 What happen when a thermosetting polymer is heated? A It becomes harder B The cross-links between the polymer chains break C It softens D It turns into a liquid

Chemistry Term 3 STPM Chapter 21 Polymers 304 21 6 Which of the following is not a naturally occurring polymer? A Carbohydrate C Terylene B Protein D Rubber 7 A repeating unit of polymer X is shown below. ! CH ! CH2 ! CH2 ! CH"CH ! CH2 ! CH ! CH2 !! C6 H5 C # N Which statement is not true about X? A It is a co-polymer. B It is a thermosetting plastic. C It can exist in the cis and trans forms. D It reacts with ozone. 8 Which of the following is true regarding natural rubber? A It is a condensation polymer. B It is also called isoprene. C It is also called poly(cis-methylbuta-1, 3-diene). D It is resistant to oxidation. 9 Which of the following pairs of monomers could undergo condensation polymerisation? A CH3CHRCH2 and HOCH2CH2OH B ClCO(CH2)4COCl and H2N(CH2)4NH2 C CH3COOH and C2H5OH D C H 2 R CH(CH 3 )CH R C H 2 a n d CH2RCHCl 10 Which of the following polymers are thermoplastic? I Polystyrene II Nylon III Bakelite A I and II C II and III B I and III D I, II and III 11 The repeating unit of a synthetic polymer is shown below: 99[9CH29C"CH9CH2999]n & Cl What of the following statements are true about the polymer? I Its monomer is CH2"C9CH"CH2. & Cl II It is susceptible to oxidation. III It is a thermosetting plastic. A I and II B I and III C II and III D I, II and III 12 Tetrafluoroethene (Teflon) is a polymer used in non-sticking cooking utensils. Which of the following statements are true about Teflon? I It is a condensation polymer. II It is chemically inert. III It is thermally stable. A I and II B I and III C II and III D I, II and III 13 Part of the structure of a polymer is shown below: 9HN(CH2)5CONH(CH2)5CONH(CH2)5CO9 Which of the following statements are true of the polymer? I It is an example of a homopolymer. II It will decompose in the presence of a mineral acid. III It is a polyester. A I and II C II and III B I and III D I, II and III 14 Which of the following is not true of Bakelite? A It is a thermosetting plastic. B It has a cross-linked structure. C It softens on heating. D It is a copolymer. 15 Which of the following polymers can be attacked by atmospheric ozone? A Poly(isoprene) C Nylon-6,6 B Poly(ethene) D Poly(styrene) 16 Which of the following statements about the polymerisation of ethene using Ziegler-Natta catalyst is not true? A It produces high density of poly(ethene). B The poly(ethene) consists of mainly of linear chains. C It requires a low temperature. D It requires a high pressure.

Chemistry Term 3 STPM Chapter 21 Polymers 305 21 17 Poly(tetrafluoroethene), also known as 'Teflon' is used to coat cooking utensils. Which of the following statements about poly(tetrafluoroethene) are true? I It is a condensation polymer. II It is chemically inert. III It is brittle. IV It is heat resistant. A I and III B II and III C II and IV D II, III and IV 18 Part of the structural formula of a polymer is shown below: 99[9CH9CH29(CH29CH"CH9CH2)3999]n Which statement is not true of the polymer? A It is a thermosetting polymer. B It can be vulcanised. C It is formed via addition polymerisation. D It is a copolymer. Structured and Essay Questions 1 (a) What do you understand by ‘addition polymerisation’? (b) (i) Ethene can be polymerised to form polyethene. Draw a repeating unit for polyethene. (ii) In what way is polyethene different from ethene physically? (c) In an experiment, ethene is forced into a container of volume 1.5 dm3 until the pressure reaches 10 atm at 298 K. A solid catalyst is then introduced into the container. After 25 minutes, the temperature in the container increases to 348 K and the pressure decreases to 4 atm. (i) Is the polymerisation an exothermic or endothermic process. (ii) Explain why the pressure of the system decreases. (iii) Calculate the percentage of conversion of ethene to polyethene after 25 minutes. State any assumptions you make in your calculations. 2 (a) Explain the term ‘condensation polymerisation’. (b) Explain why the empirical formula of the polymer is not the same as those of the monomers. (c) 1,6-diaminohexane, H2N(CH2)6NH2 and hexanedioic acid, HOOC(CH2)4COOH can be used to prepare nylon. (i) Draw a repeating unit for the nylon. (ii) Name the nylon. (iii) Explain why the nylon produced is stronger than polyethene. 3 (a) Give an example each for the following substances. (i) A naturally occurring polymer (ii) A synthetic polymer (b) Nylon-6,6 is an example of a condensation polymer. (i) Draw the structure of the monomers of nylon-6,6. (ii) Show schematically how the two monomers can be prepared from 1,4-dibromobutane. (iii) State an important use for nylon-6,6. 4 Consider the following compounds: (I) CH2RCHCOOCH3 (II) CH3CH2CHCH3 & OH (III) Cl!C! !C!Cl ∫ ∫ O O

Chemistry Term 3 STPM Chapter 21 Polymers 306 21 (IV) HO!(CH2)4!OH (V) CH3CH2COOH Which of the compounds (I) to (V) are used in the manufacture of the following polymers? Draw a repeating unit for each of the polymer formed. (a) An addition polymer (b) A condensation polymer 5 (a) Polymerisation of propene can produce three different types of poly(propene). (i) Draw the structure of each type of poly(propene). Then, name the type of polymers that are produced. (ii) What are the main uses of poly(propene)? (b) (i) Explain the differences between thermosetting and thermoplastic materials. Then, give an example for each of them. (ii) What is the major structural difference between thermosetting plastics and thermoplastics? 6 Terylene is prepared from benzene-1,4-dicarboxylic acid and ethane-1,2-diol. (a) Give the structural formulae of these two monomers. (b) Draw a repeating unit of Terylene. (c) What linkage is present in Terylene. (d) Suggest how 1,4-dimethylbenzene can be converted into benzene-1,4-dicarboxylic acid. Write a balanced equation for the reaction. (e) What mass of benzene-1,4-dicarboxylic acid can be obtained from 12 kg of 1,4-dimethylbenzene if the conversion is only 75% efficient? 7 The first synthetic rubber produced is neoprene. It is obtained through the polymerisation of chloroprene. CH2RCH — CRCH2 & Cl (a) Give the IUPAC name for chloroprene. (b) Draw a repeating unit of neoprene. (c) Like natural rubber, neoprene can be vulcanised. (i) Name a substance used in vulcanising rubber. (ii) Draw a diagram to show the structure of vulcanised neoprene. (d) Neoprene is mechanically stronger than poly(1,3-butadiene). (i) Draw a repeating unit for poly(1,3-butadiene). (ii) Explain why neoprene is stronger than poly(1,3-butadiene). 8 The monomer of natural rubber is isoprene. CH2RC — CHRCH2 & CH3 (a) Give the IUPAC name for isoprene. (b) Polymerisation of isoprene produces two types of poly(isoprene): the cis form and the trans form. (i) Draw diagram to show a repeating unit of cis-poly(isoprene) and trans-poly(isoprene). (ii) What is the major difference between the cis and the trans form? (c) Natural rubber in its raw state does not have much use. Explain why it is so. (d) The problem stated in (c) can be overcome by a process called vulcanisation. (i) How is vulcanisation carried out? (ii) What are the differences between vulcanised rubber and unvulcanised rubber? (iii) State what would happen if too much vulcanising agent is used.

Chemistry Term 3 STPM Chapter 21 Polymers 307 21 9 Ethene can be polymerised to produce either low-density polyethene (LDPE) or high density polyethene (HDPE) depending on the conditions used. (a) State how LDPE is produced. (b) State how HDPE is produced. (c) State the uses of LDPE and HDPE. 10 Kevlar is a polymer formed between 1,4-benzenedicarboxylic acid (terephthalic acid) and 1,4-benzenediamine (p-phenylenediamine). (a) Draw the structure of the two monomers of Kevlar. (b) Draw a repeating unit of Kevlar. (c) Kevlar is strong and yet light. Give one major use of Kevlar. 11 Terylene is formed by the copolymerisation of 1,4-benzenedicarboxylic acid and ethane-1,2-diol. (a) Draw the structure of the two monomers of Terylene. (b) Draw a repeating unit of Terylene. (c) To what class of polymer does Terylene belongs to? (d) Terylene is used in making fabrics. Explain why Terylene dissolves slowly when place in dilute mineral acids. 12 Write balanced equations, including reaction conditions, how you would synthesise nylon-6,6 from 1,4-diiodobutane. 13 Ethene gas is forced into a 1500 cm3 container until its pressure is 8.5 × 105 Pa at 298 K. A catalyst is then introduced into the vessel. After some time, the temperature in the vessel increases to 323 K and the pressure drops to 1.0 × 105 Pa. (a) Explain what happens in the experiment. (b) Calculate the percentage conversion of ethene into poly(ethene) at the particular time. State any assumption(s) you made in the calculation. 14 Polymers can be classified as thermoset or thermoplastic. (a) What do you understand by thermoset and thermoplastic? (b) What is the major structural difference between the two types of polymers? (c) One example of a thermoplastic is obtained via the polymerisation of the following molecule: C#N CH2"C C9OC2H5 ' O (i) Draw a repeating unit for the polymer formed. (ii) Name the type of polymerisation involved.

308 Section A [15 marks] Bahagian A [15 markah] Answer all questions in this section. Jawab semua soalan dalam bahagian ini. 1 Ethyne has the molecular formula of H!C;C9H. Which orbital of the carbon atom overlaps with the 1s orbital of hydrogen to form the C9H σ bond? Etuna mempunyai formula molekul H!C;C!H. Orbital yang manakah pada atom karbon yang bertindih dengan orbital 1s pada atom hidrogen untuk membentuk ikatan σ C!H? A sp3 C 2p B sp D 2s 2 The resonance structures of benzene, C6 H6 are shown below: Struktur resonans untuk benzena, C6 H6 ditunjukkan di bawah. H H H H H H H H H H H H Which of the following statements is not true regarding the resonance structures? Pernyataan yang manakah antara berikut tidak benar tentang struktur-struktur resonans itu? A They have the same arrangement of atoms. Mereka mempunyai susunan elektron yang sama. B They have the same energy. Mereka mempunyai tenaga yang sama. C They are in rapid equilibrium with one another. Mereka berada dalam keseimbangan pesat antara satu sama lain. D They have the same number of electrons. Mereka mempunyai bilangan elektron yang sama. 3 Metoprolol is a cardioselective β1-adrenergic blocking agent used for the treatment of mild hypertension. It has the following structure. Metoprolol adalah agen pencegah penyakit kardioselektif β1 yang digunakan untuk rawatan darah tinggi yang sederhana. Ia mempunyai struktur berikut. CH3 H3 CO OH O H N CH3 Metoprolol is classified as Metoprolol dikelaskan sebagai suatu A a ketone keton B a carboxylic acid asid karboksilik C a phenol fenol D an amine amina 4 Which of the following compounds has the lowest pKa value? Sebatian yang manakah antara berikut mempunyai nilai pKa yang paling rendah? A ClCH2 COOH B FCH2 OH C CH3 CH2 OH D CO2 H STPM Model Paper (962/3)

Chemistry Term 3 STPM STPM Model Paper (962/3) 309 5 What is the best choice of reagent(s) to perform the following transformation? Apakah reagen yang paling sesuai digunakan untuk melaksanakan transformasi berikut? O O A Hot, concentrated alkaline KMnO4 KMnO4 beralkali pekat dan panas B LiAlH4 in ether followed by boiling with dilute H2 SO4 LiAlH4 dalam eter diikuti oleh pendidihan dengan H2 SO4 cair C O3 followed by hydrolysis in the presence of zinc O3 diikuti oleh hidrolisis dalam kehadiran zink D Ethanolic KCN with heat KCN beretanol yang panas 6 Ethanol reacts with sodium metal to produce hydrogen gas. Etanol bertindak balas dengan logam natrium untuk membebaskan gas hidrogen. 2C2 H5 OH + 2Na 9: 2C2 H5 ONa + H2 Which of the following best describes the reaction? Yang manakah antara berikut menjelaskan tindak balas ini? A Redox Redoks B Neutralisation Peneutralan C Electrophilic addition Penambahan elektrofilik D Free radical substitution Penukargantian radikal bebas 7 Ethanal and propanone can be distinguished using Etanal dan propanon boleh dibezakan dengan menggunakan A iodine in sodium hydroxide iodin dalam natrium hidroksida B 2,4-dinitrophenylhydrazine 2,4-dinitrofenilhidrazin C Tollen’s reagent Reagen Tollen D bromine in concentrated potassium hydroxide bromin dalam kalium hidroksida pekat 8 What is the major product obtained from the following sequence of reactions? Apakah hasil utama yang diperoleh daripada turutan tindak balas berikut? Toluene 9: K2 Cr2 O7 /H+ 9: Br2 /FeBr3 Toluena A CH3 Br B COOH Br C COOH Br D CH2 Br Br 9 Fructose is an isomer of glucose. Its structural formula is shown below. Fruktosa adalah suatu isomer glukosa. Ia mempunyai struktur yang berikut. CH2 OH & C " O & HO ! C ! H & H ! C ! OH & H ! C ! OH & CH2 OH

Chemistry Term 3 STPM STPM Model Paper (962/3) 310 Which statement about fructose is not correct? Antara pernyataan berikut, yang manakah tidak benar tentang fruktosa? A It has three chiral centres. Ia mempunyai tiga pusat kiral. B It is classified as a polyhydroxyl carbonyl compound. Ia dikelaskan sebagai sebatian polihidroksi karbonil. C It gives a yellow precipitate when warmed with an alkaline solution of iodine. Ia memberikan suatu mendakan kuning apabila dipanaskan dengan larutan iodin beralkali. D It gives an orange precipitate with 2,4-dinitrophenylhydrazine. Ia memberikan suatu mendakan jingga dengan 2,4-dinitrofenilhidrazin. 10 Compared to chlorocyclohexane, chlorobenzene is rather inert towards nucleophilic substitution. Which of the following statements best explains the inertness of chlorobenzene towards nucleophilic substitution? Berbanding dengan klorosikloheksana, klorobenzena adalah lengai terhadap penukargantian nukleofilik. Ini adalah disebabkan A The C9Cl bond in chlorobenzene is stronger than that in chlorocyclohexane. ikatan C9Cl dalam klorobenzena lebih kuat daripada yang berada dalam klorosikloheksana. B Chlorobenzene is unsaturated. korobenzena adalah sebatian tak tepu. C The benzene ring is an electronwithdrawing group. gelang benzena merupakan kumpulan penarik elektron. D Chlorobenzene itself is a nucleophile. klorobenzena sendirinya adalah suatu nukleofil. 11 Which reagent would react with ammonia to produce phenylacetamide? Reagen yang manakah antara berikut bertindak balas dengan ammonia untuk membentuk fenilasetamida? A Benzoic acid Asid benzoik B Phenylethanoic acid Asid feniletanoik C Phenylethanoyl chloride Feniletanoil klorida D Phenylethanoate Feniletanoat 12 The structures of poly(styrene) and poly(ethane) are shown below. Struktur poli(stirena) dan poli(etena) adalah ditunjukkan di bawah. ! CH2 ! CH !n ! C ! C ! H & H & & H & H n Which statement is correct? Pernyataan yang manakah adalah benar? A Both polymers are planar. Kedua-dua polimer adalah bersatah. B Poly(styrene) is more resistant to biodegradation than poly(ethane). Poli(stirena) lebih lengai terhadap biodegradasi berbanding dengan poli(etena). C Poly(styrene) is a condensation polymer while poly(ethane) is an addition polymer. Poli(stirena) merupakan polimer kondensasi manakala poli(etena) adalah polimer penambahan. D Poly(styrene) is a thermosetting plastic while poly(ethane) is a thermoplastic. Poli(stirena) merupakan plastik termoset manakala poli(etena) adalah termoplastik.

Chemistry Term 3 STPM STPM Model Paper (962/3) 311 13 Compound X has the following properties: Sebatian X mempunyai ciri-ciri berikut: • Reacts with bromine water to produce a white precipitate. Membentuk mendakan putih dengan air bromin. • No effervescence with nitrous acid at 2 °C. Tiada pembuakan berlaku dengan asid nitrus pada 2 °C. • Produces a white fume with thionyl chloride. Membentuk wasap putih dengan tionil klorida. Compound X could be Sebatian X mungkin A NH2 COOH C HO CH2 NH2 B HO NH2 D CH2 NH2 COOH 14 Which of the following would be produced when propanamide is heated with phosphorous(V) oxide? Yang manakah antara berikut akan diperolehi apabila propanamida dipanaskan dengan fosforus(V) oksida? A Propylamine Propilamina B Ethylamine Etilamina C Propylpropanoate Propilpropanoat D Ammonium propanoate Ammonium propanoat 15 Which statement about fats is not correct? Pernyataan yang manakah antara berikut adalah tidak benar tentang lemak? A They are triesters of glycerol. Lemak adalah triester bagi gliserol. B They are known as triglycerides. Lemak dikenali sebagai trigliserida. C All fats are unsaturated. Semua lemak adalah tepu. D They are insoluble in water. Lemak tidak larut dalam air. Section B [15 marks] Bahagian B [15 markah] Answer all questions in this section. Jawab semua soalan dalam bahagian ini. 16 An alcohol has the molecular formula of C4 H10O. Suatu alkohol mempunyai formula molekul C4 H10O. (a) Draw all the structural formulae of the alcohol. [3 marks] Lukis formula struktur semua isomer alkohol itu. [3 markah] (b) Which of the above isomers is resistant to oxidation? [1 mark] Yang manakah antara alkohol-alkohol di atas adalah rintang terhadap pengoksidaan? [1 markah] (c) One of the isomers in (a) reacts with an alkaline solution of iodine to produce a yellow precipitate. Identify the isomer. [1 mark] Salah satu alkohol di (a) bertindak balas dengan larutan beralkali iodin untuk membentuk suatu mendakan kuning. Kenalpasti alkohol itu. [1 markah] (d) Which isomer on dehydration produces two isomeric alkenes? Draw the structures of the two isomeric alkenes. [2 marks] Isomer yang manakah apabila dihidratkan menghasilkan dua alkena berisomerik? Lukis struktur kedua-dua alkena berisomerik itu. [2 markah]

Chemistry Term 3 STPM STPM Model Paper (962/3) 312 (e) Which isomer can be prepared by the reaction between a suitable ketone and a suitable Grignard’s reagent? Briefly explain your answer. [2 marks] Isomer yang manakah boleh disediakan melalui tindak balas antara suatu keton yang sesuai dengan reagen Grignard yang sesuai. Jelaskan jawapan anda. [2 markah] 17 Benzocaine is often used as a local anesthetic and has the structure below. Benzocaine sering digunakan sebagai anestatik tempatan dan mempunyai struktur beikut. H2 N!C6 H4 !COOC2 H5 (a) Give the IUPAC name of benzocaine. [1 mark] Berikan nama IUPAC benzocaine. [1 markah] (b) Benzocaine can be synthesized from toluene using the schematic pathway shown below. Benzocaine boleh disintesiskan daripada toluena melalui skema tindak balas berikut. I II III IV C6 H5 CH3: O2 N!C6 H4 !CH3: O2 NC6 H4 COOH : O2 NC6 H4 COOC2 H5: Benzocaine Complete the table below for the reagents and conditions necessary for each of the four steps above. [4 marks] Lengkapkan jadual berikut untuk reagen dan keadaan tindak balas yang diperlukan untuk setiap langkah di atas. [4 markah] Step Langkah Reagent(s) Reagen Condition(s) Keadaan I II III IV (c) Why is benzocaine insoluble in water but dissolves in the stomach when ingested?[1 mark] Mengapakah benzocaine tidak larut dalam air tetapi larut dalam perut apabila dimakan? [1 markah] Section C [30 marks] Bahagian C [30 markah] Answer any two questions from this section. Jawab mana-mana dua soalan daripada bahagian ini. 18 This question is about the aldehyde, ethanal, CH3 CHO. Soalan ini adalah tentang aldehid etanal, CH3 CHO. (a) Show how you would convert ethanal into Tunjukkan bagaimana anda dapat menukarkan etanal kepada (i) 1-phenyl-2-chloropropane, CH3 CH(Cl)CH2 C6 H5 [3 marks] 1-fenil-2-kloropropana, CH3 CH(Cl)CH2 C6 H5 [3 markah] (ii) propylamine, CH3 CH2 CH2 NH2 [4 marks] propilamina, CH3 CH2 CH2 NH2 [4 markah]

Chemistry Term 3 STPM STPM Model Paper (962/3) 313 (b) Ethanal reacts with hydrogen cyanide in the presence of a little potassium cyanide as a catalyst to produce 2-hydroxypropanenitrile. Etanal bertindak balas dengan hidrogen sianida dalam kehadiran sedikit kalium sianida sebagai mangkin untuk menghasilkan 2-hidroksipropananitril. (i) State the type of reaction mechanism involved. [1 mark] Nyatakan jenis mekanisme tindak balas yang berlaku. [1 markah] (ii) Describe the reaction mechanism and explain why potassium cyanide is considered as a catalyst. [4 marks] Terangkan mekanisme tindak balas itu dan jelaskan mengapa kalium sianida dianggap sebagai mangkin. [4 markah] (c) 2-Hydroxypropanenitrile is optically active. 2-Hidroksipropananitril adalah aktif secara optik. (i) What do you understand by ‘optical activity’? [1 mark] Apakah yang anda faham dengan ‘aktiviti optik’? (ii) Draw the structure of the two optical isomers of 2-hydroxypropanenitrile. [2 marks] Lukis kedua-dua isomer optik untuk 2-hidroksipropananitril. 19 A neutral compound, A has the following composition by mass: Suatu sebatian A yang neutral mempunyai komposisi mengikut jisim seperti berikut: C: 81.6%; H: 4.80%; N: 13.6% 10.3 g of a sample of A occupies a volume of 1.76 dm3 at 200 kPa and 150 °C. 10.3 g suatu sampel A menempati isi padu 1.76 dm3 pada 200 kPa dan 150 °C. A can be reduced to B, C7 H9 N which reacts with dilute hydrochloric acid to produce a colourless solution from which a white crystalline solid, C, C7 H10NCl can be isolated. Treatment of B with a warm solution of freshly prepared nitrous acid gives a colourless gas and compound D, C7 H8 O. D undergoes oxidation with acidified potassium dichromate producing E, C7 H6 O2 . Treatment of E with thionyl chloride produces F, C7 H5 OCl which reacts with D to give G, C14H12O2 . F reacts with excess ammonia to produce H, C7 H7 NO. A boleh diturunkan kepada B, C7 H9 N yang bertindak balas dengan asid hidroklorik untuk membentuk suatu larutan tak berwarna dari mana suatu hablur putih, C, C7 H10NCl boleh diasingkan. B ditindakbalaskan dengan suatu larutan asid nitrus untuk membebaskan suatu gas tak berwarna dan sebatian D, C7 H8 O. D mengalami pengoksidaan dengan larutan berasid kalium dikromat untuk membentuk E, C7 H6 O2 . E bertindak balas dengan tionil klorida untuk menghasilkan F, C7 H5 OCl yang bertindak balas dengan D untuk membentuk G, C14H12O2 . F bertindak balas dengan ammonia berlebihan untuk membentuk H, C7 H7 NO. (a) Determine the empirical formula of A. [1 mark] Tentukan formula empirik A. [1 markah] (b) Determine the molecular formula of A. [2 marks] Tentukan formula molekul A. [2 markah] (c) Other than using litmus, how can you show that A is neutral? State the reagent used and the expected observations. [2 marks] Selain menggunakan kertas litmus, bagaimanakah anda boleh menunjukkan bahawa A adalah neutral? Nyatakan reagen yang digunakan dan pemerhatian yang dijangkakan. [2 markah] (d) Identify compounds A to G giving your reasoning. [7 marks] Kenal pasti sebatian-sebatian A hingga G dengan memberikan panaakulan anda. [7 markah]

Chemistry Term 3 STPM STPM Model Paper (962/3) 314 (e) A, B and H are nitrogen-containing compounds. Suggest a simple scheme to show how you would distinguish them. Write equations where appropriate. [3 marks] A, B dan H adalah sebatian-sebatian yang mengandungi nitrogen. Cadangkan suatu skema ringkas untuk membezakan antara sebatian-sebatian itu. Tuliskan persamaan-persamaan tindak balas pada mana yang sesuai. [3 markah] 20 (a) What do you understand by ‘addition polymerisation’? [2 marks] Apakah yang anda faham dengan istilah ‘pempolimeran penambahan’? [2 markah] (b) Under suitable conditions, ethene polymerises to poly(ethene) which is used mainly as ‘plastic bags’. State one similarity and one difference between ethene and poly(ethene). [2 marks] Di bawah keadaan yang sesuai, etena boleh dipolimerkan menjadi poli(etena) yang digunakan terutamanya sebagai ‘beg plastik’. Nyatakan satu persamaan dan satu perbezaan antara etena dengan poli(etena). [2 markah] (c) In an experiment, ethene gas was forced into an evacuated container of capacity 1.0 dm3 , until the pressure was 25.0 kPa at 25 °C. A suitable catalyst was then introduced and the mixture was left standing. After 30 minutes, the temperature went up to 32.0 °C and the pressure dropped to 12.2 kPa. Dalam suatu eksperimen, gas etena dilalukan ke dalam suatu bekas yang berisipadu 1.0 dm3 , sehingga tekanan mencapai 25.0 kPa pada 25 °C. Satu mangkin yang sesuai kemudian ditambahkan dan campuran itu dibiarkan. Selepas 30 minit, suhu menaik kepada 32.0 °C dan tekanan jatuh kepada 12.2 kPa. (i) Write an equation for the polymerisation of ethene. [1 mark] Tuliskan persamaan tindak balas bagi pempolimeran etena. [1 markah] (ii) Explain why the temperature increases during the polymerisation process. [2 marks] Jelaskan mengapa suhu meningkat semasa pempolimeran itu berkalu. [2 markah] (iii) Calculate the average rate of polymerisation for the first 30 minutes. State any assumption(s) you make in your calculations. [3 marks] Hitung kadar purata pempolimeran pada 30 minit yang pertama. Nyatakan sebarang anggapan yang anda buat dalam pengiraan anda. [3 markah] (d) Another addition polymer is poly(chloroethene), or PVC. Suatu polimer penambahan yang lain ialah poli(kloroetena), atau PVC. (i) Draw the repeating unit of PVC. [1 mark] Lukis unit ulangan bagi PVC. [1 markah] (ii) Between poly(ethene) and PVC, which is expected to have higher tensile strength? Explain your answer. [3 marks] Antara poli(etena) dengan PVC, yang manakah dijangkakan mempunyai kekuatan tegangan yang lebih tinggi. Jelaskan jawapan anda. [3 markah] (iii) State one use of PVC. [2 marks] Nyatakan satu kegunaan PVC. [2 markah]

315 Periodic Table of Elements 1HHydrogen 1 1HHydrogen 1 Proton number 1234567 Period 3Li Lithium 7 4 Be Beryllium 9 11 Na Sodium 23 12 Mg Magnesium 24 19KPotassium 39 20 Ca Calcium 40 37 Rb Rubidium 85.5 38 Sr Strontium 88 55 Cs Caesium 133 56 Ba Barium 137 57 – 71 Lanthanoids 89 – 103 Actinoids 87 Fr Francium 88 Ra Radium 105 Db Dubnium 106 Sg Seaborgium 107 Bh Bohrium 108 Hs Hassium 109 Mt Meitnerium 110 Ds Darmstadtium 111 Rg Roentgenium 112 Cn Copernicium 113 Nh Nihonium 114 Fl Flerovium 115 Mc Moscovium 116 Lv Livermorium 21 Sc Scandium 4539YYttrium 89 22 Ti Titanium 4840 Zr Zirconium 9172 Hf Hafnium 178.5 23VVanadium 5141 Nb Niobium 9373 Ta Tantalum 181 24 Cr Chromium 5242 Mo Molybdenum 9674 WTungsten 184 25 Mn Manganese 5543 Tc Technetium 9875 Re Rhenium 186 26 FeIron 5644 Ru Ruthenium 101 76 Os Osmium 190 27 Co Cobalt 5945 Rh Rhodium 103 77 Ir Iridium 192 28 Ni Nickel 5946 Pd Palladium 106 78 Pt Platinum 195 29 Cu Copper 6447 Ag Silver 108 79 AuGold 197 30 ZnZinc 6548 Cd Cadmium 112 80 Hg Mercury 201 31 Ga Gallium 7049 InIndium 115 81 Tl Thallium 204 32 Ge Germanium 7350 SnTin 119 82 PbLead 207 33 As Arsenic 7551 Sb Antimony 122 83 Bi Bismuth 209 34 Se Selenium 7952 Te Tellurium 128 84 Po Polonium 209 35 Br Bromine 8053IIodine 127 85 At Astatine 210 117 Ts Tennessine 118 Og Organesson 36 Kr Krypton 8454 Xe Xenon 131 86 Rn Radon 222 5BBoron 1113 Al Aluminium 27 6CCarbon 1214 Si Silicon 28 7NNitrogen 1415P Phosphoros 31 8OOxygen 1616SSulphur 32 9FFluorine 1917 Cl Chlorine 35.5 2 He Helium 410 Ne Neon 2018 Ar Argon 40 58 La Lanthanum 139 90 Ac Actinium 59 Ce Cerium 140 91 Th Thorium 232 60 Pr Praseodymium 141 92 Pa Protactinum 231 61 Nd Neodymium 144 93UUranium 238 62 Pm Promethium 94 Np Nepturium 63 SmSamarium 150 95 Pu Plutonium 64 Eu Europium 152 96 Am Americium 65 Gd Gadolinium 157 97 CmCurium 66 Tb Terbium 159 98 Bk Berkelium 67 Dy Dysprosium 162 99 Cf Californium 68 Ho Holmium 165 100 Es Elinsteinium 69 Er Erbium 167 101 FmFermium 70 TmThulium 169 102 Md Mendelevium 71 Yb Ytterbium 173 103 No Nobelium 71 Lu Lutetium 175 103 Lr Lawrencium 104 Rf Rutherfordium Symbol of element Name of element Relative atomic mass Transition elements Lantanide series Actinide series I1 II2 3 4 5 6 7 8 9 10 Group 11 12 III 13 IV14 V15 VI 16 VII 17 VIII 18 Metals Key: Semi-metals Non-metals

Chemistry Term 3 STPM Appendix 316 Chemical Constants Speed of light in vacuum c = 3.0 × 108 m s–1 Planck constant h = 6.63 × 10–34 J s = 3.99 × 10–13 kJ mol–1 s Molar gas constant R = 8.31 J K–1 mol–1 Faraday constant F = 9.65 × 104 C mol–1 Rydberg constant RH = 1.097 × 107 m–1 Avogadro constant L, NA = 6.02 × 1023 mol–1 Mass of proton mp = 1.67 × 10–27 kg Mass of neutron mn = 1.67 × 10–27 kg Mass of electron me = 9.11 × 10–31 kg Electronic charge e = –1.60 × 10–19 C Ionic product of water Kw = 1.0 × 10–14 mol2 dm–6 (at 298 K) Molar volume of gas Vm = 22.4 dm3 (at s.t.p.) = 24.0 dm3 (at room conditions) Specific heat capacity of water c = 4.18 J g–1 K–1 p = 3.142 ln x = 2.303 log10 x Electronegativity (Pauling’s Scale) 19 K 0.8 37 Rb 0.8 55 Cs 0.7 87 Fr 0.7 20 Ca 1.0 38 Sr 1.0 56 Ba 0.9 1 H 2.1 3 Li 1.0 11 Na 0.9 4 Be 1.5 12 Mg 1.2 21 Sc 1.3 39 Y 1.2 57 La 1.1 22 Ti 1.5 40 Zr 1.4 72 Hf 1.3 23 V 1.6 41 Nb 1.6 73 Ta 1.5 24 Cr 1.6 42 Mo 1.8 74 W 1.7 25 Mn 1.5 43 Tc 1.9 75 Re 1.9 26 Fe 1.8 44 Ru 2.2 76 Os 2.2 27 Co 1.9 45 Rh 2.2 77 Ir 2.2 28 Ni 1.9 46 Pd 2.2 78 Pt 2.2 29 Cu 1.9 47 Ag 1.9 79 Au 2.4 30 Zn 1.6 48 Cd 1.7 80 Hg 1.9 31 Ga 1.6 49 In 1.7 81 Tl 1.8 32 Ge 1.8 50 Sn 1.8 82 Pb 1.9 83 Bi 1.9 34 Se 2.4 52 Te 2.1 84 Po 2.0 35 Br 2.8 33 As 2.0 51 Sb 1.9 53 I 2.5 85 At 2.2 36 Kr 54 Xe 86 Rn 5 B 2.0 13 Al 1.5 6 C 2.5 14 Si 1.8 15 P 2.1 8 O 3.5 16 S 2.5 7 N 3.0 9 F 4.0 17 Cl 3.0 2 He 10 Ne 18 Ar 88 Ra 0.9

Chemistry Term 3 STPM Appendix 317 Ionisation Energies (kJ mol–1) Z Name 1st 2nd 3rd 4th 5th 6th 7th 1 H 1310 2 He 2370 5250 3 Li 519 7300 11 800 4 Be 900 1760 14 800 21 000 5 B 799 2420 3660 25 000 32 828 6 C 1090 2350 4610 6220 37 830 42 270 7 N 1400 2860 4590 7480 9440 53 270 64 360 8 O 1310 3390 5320 7450 10 990 13 320 71 330 9 F 1680 3370 6040 8410 11 020 15 160 17 870 10 Ne 2080 3950 6150 9290 12 180 15 240 20 000 11 Na 494 4560 6940 9540 13 350 16 610 21 110 12 Mg 736 1450 7740 10 500 13 630 17 800 21 700 13 Al 577 1820 2740 11 600 14 830 18 380 23 290 14 Si 786 1580 3230 4360 16 090 15 P 1060 1900 2920 4960 3270 21 270 16 S 1000 2260 3390 4540 7010 8500 27 110 17 Cl 1260 2300 3850 5150 6540 9360 11 020 18 Ar 1520 2660 3950 5770 7240 8780 12 000 19 K 418 3070 4600 5860 20 Ca 590 1150 4940 6480 21 Sc 632 1240 2390 7110 8840 10 720 22 Ti 661 1310 2720 4170 9570 11 520 23 V 648 1370 2870 4600 6290 12 360 24 Cr 653 1590 2990 4770 6690 8740 25 Mn 716 1510 3250 5190 6990 9200 26 Fe 762 1560 2960 5400 7240 9600 27 Co 757 1640 3230 5100 7670 9840 28 Ni 736 1750 3390 5400 7280 10 400 29 Cu 745 1960 3550 5690 7710 9940 30 Zn 908 1730 3828 5980 7970 10 400 31 Ga 577 1980 2960 6190 32 Ge 762 1540 3300 4390 9020 35 Br 140 2080 3460 4850 5760 8550 9940 38 Sr 548 1060 4120 5440 50 Sn 707 1410 2940 3930 6980 53 I 1010 18 470 2040 4030 56 Ba 502 966 3390 82 Pb 716 1450 3080 4080 6640

Chemistry Term 3 STPM Appendix 318 Confirmatory Test for Functional Groups Alkenes 1 Decolourisation of bromine in non-polar solvent (e.g. CCl4, cyclohexane). For example, In cyclohexane CH2"CH2 + Br2 !!!!!: CH2!CH2 & & Br Br 2 Decolourisation of acidified potassium manganate(VII). For example, KMnO4 CH2"CH2 + [O] + H2O !!!: CH2!CH2 & & OH OH Benzene 1 Warm the benzene with a mixture of concentrated nitric acid and concentrated sulphuric acid. Pour the resulting mixture into cold water. A layer of yellowish oil is formed. H NO2 H2SO4/55 °C + HNO3 !!!!!: + H2O 2 Mix benzene and bromine in the dark. Add in a little anhydrous aluminium chloride. The reddish-brown colour of bromine fades slowly and a white fume is formed. H Br AlCl3/Room temperature + Br2 !!!!!!!!!: + HBr Alkyl Benzene 1 Warm the alkyl benzene with an acidified solution of potassium manganate(VII). The purple colour of KMnO4 is decolourised. For example, CH3 COOH + 3[O] !!: + H2O

Chemistry Term 3 STPM Appendix 319 Haloalkanes 1 Boil the haloalkane with aqueous sodium hydroxide. Cool. Add a slight excess of dilute nitric acid, followed by aqueous silver nitrate. Observe the colour of the precipitate formed. R!Cl + H2O + Ag+ !: R!OH + H+ + AgCl (White precipitate) R!Br + H2O + Ag+ !: R!OH + H+ + AgBr (Cream precipitate) R!I + H2O + Ag+ !: R!OH + H+ + AgI (Yellow precipitate) Alcohols 1 Add phosphorus(V) chloride, PCl5, or thionyl chloride, SOCl2 into the alcohol. Effervescence occurs with the release of white fumes. R!OH + PCl5 !: R!Cl + POCl3 + HCl R!OH + SOCl2 !: R!Cl + SO2 + HCl 2 Warm the alcohol with acidified potassium manganate(VII). Class of alcohol Primary Secondary Tertiary Observation Purple to colourless Purple to colourless No reaction 3 Warm the alcohol with concentrated hydrochloric acid in the presence of zinc. Note the duration for the solution to turn cloudy. R!OH + HCl !: R!Cl + H2O Class of alcohol Primary Secondary Tertiary Duration Longer A few minutes A few seconds Phenols 1 Add aqueous bromine into the phenol. A white precipitate is formed. OH OH + 3Br2 !: + 3HBr Br Br Br 2 Add aqueous iron(III) chloride into the phenol. A blue/purple solution is formed.

Chemistry Term 3 STPM Appendix 320 Carbonyl Compounds (Aldehydes and Ketones) 1 Add 2,4-dinitrophenylhydrazine. A coloured precipitate is formed. For example, NO2 NO2 H H C"O + H2N!NH! !NO2 !: C"N!NH! !NO2 + H2O CH3 CH3 Aldehydes 1 Warm the aldehyde gently with Tollen's reagent. A silver mirror is formed on the inner wall of the test tube. For example, CH3CHO + 2[Ag(NH3)2]+ + 2OH– !: CH3COO– + 2Ag + H2O + 3NH3 + NH4 + ∆ 2 Boil the aldehyde with Fehling's solution. A reddish-brown precipitate is formed. For example, CH3CHO + 2Cu2+ + 5OH– !: CH3COO– + Cu2O + 3H2O Methyl-ketone and Ethanal 1 Warm with a solution of iodine in sodium hydroxide. A yellow precipitate is formed. CH3!C!R + 3I2 + 4NaOH !: CHI3 + RCOONa + 3NaI + 3H2O ' ∆ O (where R = H, alkyl or aryl) Carboxylic Acids 1 Add aqueous sodium bicarbonate or sodium carbonate into the carboxylic acid. Effervescence occurs. RCOOH + NaHCO3 !: RCOONa + H2O + CO2 2RCOOH + Na2CO3 !: 2RCOONa + H2O + CO2 Methanoic Acid (Formic Acid) 1 Add acidified potassium manganate(VII). The purple colour turns colourless. HCOOH + [O] !: H2O + CO2 Ethanedioic Acid (Oxalic Acid) 1 Add acidified potassium manganate(VII). The purple colour fades slowly. H2C2O4 + [O] !: 2CO2 + H2O Acyl Chlorides 1 Add water. White fumes are given off. RCOCl + H2O !: RCOOH + HCl

Chemistry Term 3 STPM Appendix 321 2 Add aqueous silver nitrate. A white precipitate is formed instantly. RCOCl + H2O + Ag+ !: RCOOH + AgCl + H+ Amides 1 Warm with aqueous sodium hydroxide. A gas that turns moist red litmus paper blue is given off. RCONH2 + NaOH !: RCOONa + NH3 Primary Amines and Phenylamines 1 Warm with nitrous acid (A mixture of sodium nitrite and dilute HCl). Effervescence occurs. RNH2 + HNO2 !: R!OH + H2O + N2 Phenylamines 1 Add bromine water. A white precipitate is formed. NH2 NH2 + 3Br2 !: + 3HBr Br Br Br 2 Add phenylamine to nitrous acid at 2 °C (cooled in ice), followed by a solution of phenol in sodium hydroxide. A bright yellow solid is formed. !NH2 + HNO2 + H+ !: !N+#N + 2H2O !N+#N + H! !OH !: !N"N! !OH + H+

322 Addition polymerisation Pempolimeran penambahan A process where small molecules (called monomers) are joined with one another to form long-chain molecules without the elimination of other smaller molecules Addition reaction Tindak balas penambahan A reaction where two or more molecules are combined to form a larger molecule Amino acid Asid amino A molecule that contains the basic 9NH2 group and the acidic 9COOH group in its structure Carbanion Karbanion An anion where the negative charge is on a carbon atom Carbohydrates Karbohidrat Naturally occurring polyhydroxylated carbonyl compounds Carbonium ion Ion karbonium A cation where the positive charge is on a carbon atom Catenation Katenasi The ability of the atoms of an element to form covalent bonds with one another Condensation polymerisation Pempolimeran kondensasi A process where small molecules (called monomers) are joined with one another to form long-chain molecules with the elimination of other smaller molecules Copolymer Kopolimer A polymer that is made up of two or more different monomers Cracking Peretakan A process where large hydrocarbon molecules are broken into smaller molecules Elastomer Elastomer A polymer that stretches when pulled and returns to its original length when the force is released Electrophile Elektrofil A reagent that accepts an electron pair from another substance to form a coordinate (dative) bond Elimination reaction Tindak balas penyingkiran A reaction where two or more atom or group of atoms are removed from a molecule Empirical formula Formula empirik It shows the simplest whole number ratio of atoms of each element present in the molecule of a compound Enantiomerism Enantiomerisme See optical isomerism Enantiomers Enantiomer Compounds with the same molecular formula but are mirror images and are not superimposable Essential amino acids Asid amino perlu Amino acids that cannot be synthesised by the body and have to be obtained through food supplements Fatty acids Asid lemak Long-chain carboxylic acids that normally exist as the esters of glycerol(1,2,3-propanetriol)

Chemistry Term 3 STPM Glossary 323 Fractional distillation Penyulingan berperingkat A process for separating liquid components in a liquid mixture based on their different boiling points Free radical Radikal bebas A neutral atom or molecule containing an unpaired electron. Free radicals are formed by the heterolytic fission of a covalent bond. Functional group Kumpulan berfungsi An atom or a group of atoms that gives the chemical characteristic of an organic molecule Geometrical isomerism Keisomeran geometri Molecules with the same molecular formula, the same structural formula but with different spatial arrangement of atoms or group of atoms in the molecule due to the presence of C = C bonds or cyclic structures that restrict free rotation Grignard reagent Bahan uji Grignard An organic compound that contains the polar carbon-magnesium bond Homopolymer Homopolimer A polymer that is formed from one type of monomer only Hybridisation Penghibridan The mixing of two or more non-equivalent atomic orbitals to form a set of equivalent hybrid orbitals Inductive effect Kesan induktif An experimentally observable effect of the transmission of charge through a carbon chain in a molecule by electrostatic induction Isoelectric point Titik isoelektrik The pH at which an amino acid molecule has no net charge Isomerism Keisomeran The occurrence of two or more compounds with the same molecular formula Markovnikov’s rule Peraturan Markovnikov In the addition of H9X to an unsymmetrical alkene, the hydrogen from HX adds to the unsaturated carbon atom that has the greatest number of hydrogen atoms bonded to it. Molecular formula Formula molekul It shows the actual number of the different types of atoms (indicated by their symbols) in one discrete molecule of a substance. Monosaccharides Monosakarida Carbohydrates that cannot be converted into smaller sugar molecules by hydrolysis Nucleophile Nukleofil A reagent which donates lone pair electrons to another substance to form a coordinate bond Optical isomerism Keisomeran optik Molecules with the same molecular formula but are mirror images to one another. The isomers are non-superimposable and they rotate plane-polarised light in opposite directions Phenol Fenol An organic compound where the 9OH group is bonded to one of the carbon atoms of the benzene ring

Chemistry Term 3 STPM Glossary 324 Pi (π) bond Ikatan pi (p) A covalent bond formed by the sideways overlapping of atomic orbitals (usually the p orbitals). The electron density of the bond is concentrated above and below the plane containing the nuclei of the two bonded atoms. Polymerisation Pempolimeran A process where small molecules (called monomers) are joined to form long-chain molecules Ring-activating group Kumpulan pengaktif gelang An electron-donating group that makes the benzene ring more reactive towards electrophiles Ring-deactivating group Kumpulan pendeaktif gelang An electron-withdrawing group that makes the benzene ring less reactive towards electrophiles Sigma (σ) bond Ikatan sigma A covalent bond formed by the end-to-end overlapping of atomic orbitals. The electron density of the bond is concentrated in the region between the nuclei of the bonding atoms sp hybridisation Penghibridan sp The mixing of an s and a p orbital to form two equivalent hybrid orbitals sp2 hybridisation Penghibridan sp2 The mixing of an s and two p orbitals to form three equivalent hybrid orbitals sp3 hybridisation Penghibridan sp3 The mixing of an s and three p orbitals to form four equivalent hybrid orbitals Stereoisomerism Kestereoisomeran The occurrence of different compounds with the same molecular formula, same structural formula, but with different spatial arrangement of atoms Structural isomerism Keisomeran struktur The occurrence of two or more different compounds with the same molecular formula but different structural formula Structural isomers Isomer struktur Compounds with the same molecular formula but different structural formula Substitution reaction Tindak balas penukargantian A reaction where an atom or a group of atoms in a molecule is replaced by another atom or group of atoms Thermoplastics Termoplastik Polymers that soften when heated and harden when cooled sufficiently. The process can be repeated indefinitely. Thermosetting plastic Plastik termoset A polymer that once hardened cannot be softened by heat Vulcanisation Pemvulkanan The process where rubber polymer chains are joined by disulphide links Zaitsev’s rule Peraturan Zaitsev In the elimination of HX from a haloalkane, the more highly substituted alkene product predominates. Zwitterion Zwiterion An internal salt of an amino acid

325 Chapter 14 Introduction to Organic Chemistry Quick Check 14.1 1 (a) C:H:O = 3.3 : 6.6 : 3.3 = 1 : 2 : 1 Empirical formula = CH2O (b) (CH2O)n = 90.1 n = 3 Molecular formula = C3H6O2 2 (a) No. of moles of CO2 = 1.76 ——44 = 0.04 No. of moles of H2O = 0.72 ——18 = 0.04 C : H = 0.04 : (2 × 0.04) = 1 : 2 Empirical fomula = CH2 (b) (CH2)n = 70 n = 5 Molecular formula = C5H10 Quick Check 14.2 1 No 2 No 3 Yes 4 No 5 No 6 No 7 No Quick Check 14.3 1 (a) No (b) No (c) Yes, CH3*CH(OH)NH2 (d) No (e) Yes, H I * (f) No Quick Check 14.4 1 (a) Electrophile (b) Nucleophile (c) Free radical (d) Electrophile (e) Electrophile (f) Nucleophile (g) Nucleophile 2 (c), (d) and (e) Revision Exercise 14 Objective Questions 1 B 2 C 3 B 4 B 5 D 6 D 7 C 8 D 9 B 10 A 11 C 12 A Structured and Essay Questions 1 (a) The occurrence of two more compounds having the same structural formula but with different spatial arrangement of the atoms or groups. (b) Geometrical isomerism and optical isomerism (c) Optical isomers: CH CHRCH!CH2Br BrCH2!CHRCH 3 CH3 C C H H OH OH Geometrical isomers: H H OH CH CH2Br 3!CH CRC H CH2Br OH CH3!CH H CRC cis-isomer trans-isomer (d) The two optical isomers. 2 (a) CH3!CH2!CHRCH2 1-butene CH3!CHRCH!CH3 2-butene CH3!CHRCH2 2-methylpropene & CH3 (b) (i) Geometrical isomerism CH3 CH3 H H CRC cis-2-butene CH3 H H CH3 CRC trans-2-butene (ii) Boiling point. The cis-isomer will have a higher boiling point than the trans-isomer. 3 (a) (i) A free radical is a species that has an unpaired electron. Example is the chlorine free radical, Cl•. (ii) A nucleophile is a species that can donate a lone pair electrons to another species to form a coordinate bond. Examples are NH3, CN– and OH– . (iii) An electrophile is a species that accepts a lone pair electrons from another species to form a coordinate bond. Examples are AlCl3, Cl+. (b) Tertiary free radicals have three alkyl groups that are electron-releasing which helps to stabilise the free radical. Whereas, secondary free radicals have two electron-releasing alkyl groups, and primary free radicals have one alkyl group. Chapter 15 Hydrocarbons Quick Check 15.1 1 CH3CH2CH2CH2CH2CH3 CH3CH2CH2!CH!CH3 & CH3 CH3 & CH3CH2!CH!CH2!CH3 CH3 CH3 & & CH3!CH!CH!CH3 CH3 & CH3!CH2!C!CH3 & CH3 2 CH3  CH3!C!CH2!CH2!CH3  CH3 ANSWERS

Chemistry Term 3 STPM Answers 326 Primary carbon Secondary carbon Quaternary carbon Primary hydrogen Secondary hydrogen 4 2 1 12 4 Quick Check 15.2 1 (a) Butane < hexane < nonane (b) 3,3-dimethylhexane < 2-methylheptane < octane Quick Check 15.3 1 (a) Cl & CH3!C!Cl Cl!CH2!CH2Cl & H (b) 1,2-dichloroethane. Linear chain compound. 2 (a) CH3 & CH3!C!CH3 & Cl (b) Cl2 !!: 2Cl• CH3 CH3 & & CH3!C!CH3 + • Cl !!: CH3!• C!CH3 + HCl & H CH3 CH3 & & CH3! • C!CH3 + C l2 !!: CH3!C!CH3 + • C l & Cl 2 • Cl !!: Cl2 Quick Check 15.4 1 CxHy + (x + y —4 ) O2 !!: xCO2 + y —2 H2O Volume of CO2 = 10x = 30 x = 3 10(x + y —4 ) = 50 y = 8 The hydrocarbon is C3H8. Quick Check 15.5 1 (a) C22H46 !!: C10H20 + C6H12 + C6H14 (b) C22H46 !!: C4H8 + C6H12 + + C5H12 + C7H14 Quick Check 15.6 1 (b) H H H CH3CH2 CH2CH3 CH3CH2 CRC CH2CH3 H CRC (d) H H H CH3CH2CH2 CH2CH3 CH3CH2CH2 CRC CH2CH3 H CRC (e) CH3 H CH3 CH3CH2CH2 CH2CH3 CH3CH2CH2 CRC CH2CH3 H CRC Quick Check 15.7 1 (a) CH3—CH2—CH—CH3 | CH3 (b) CH3 | CH3—CH—C—CH3 | | Br Br (c) CH3 | CH3—CH2—C—CH3 | OH (d) CH3 | CH3—CH2—C—CH3 | I (e) CH3 | CH3—CH—C—CH3 | | Br OH (f) CH3 | CH3—CH2—C—CH3 | OH Quick Check 15.8 1 (a) C OH HOCH2 CH2CH3 (b) OH CH2OH (c) HOOC—CH2CH2CH2CH2—COOH 2 (a) CH2==CH—CH2CH==CH2 (b) CH3CH2—C==CHCH3 | CH2CH3 (c) CH2==CH2 (d) CH3CH==CHCH3 (e) CH3—CH—CH==CHCH3 | CH3 Quick Check 15.9 1 CH2—CH—CH2—CH2—CH—CH2OH | | | OH OH OH 2 CH(OH)CH2OH OH OH

Chemistry Term 3 STPM Answers 327 Quick Check 15.10 1 (a) O O || || CH3—C—H and H—C—H (b) O O || || 2H—C—H + H—C—CH2CH2—C—H || O (c) O O || || H—C—C—CH2CH2CH2CH2—C—H + H—C—H || || O O 2 (a) C H C H (b) CH2 Quick Check 15.11 1 CH3CH2!CH!CH2 2 R C O 3 OSO3H Quick Check 15.12 1 2H2SO4+ HNO3 L NO2 + + 2HSO4 – + H3O+ + CH CH3 3 + NO2 + !!: H NO2 !!: CH3 NO2 + H+ Quick Check 15.13 CH3 + Cl !!: CH2 + HCl • • CH2 + Cl2 !!: CH2Cl Cl2 !!: 2Cl 2Cl !!: Cl2 + Cl • • • • Quick Check 15.14 1 (a) CH3CH2CH2CH2CH2CH2! + 18 [O] COOH !!: + 5CO2 + 6H2O (b) C2H5 CH3 + 9[O] HOOC COOH + CO2 + 3H2O !!: 2 (a) Acidified KMnO4. Heat. Benzene: No Reaction. Methylbenzene: Purple colour decolourised. (b) Bromine in tetrachloromethane. Benzene: No Reaction. Cyclohexene: Reddish-brown colour decolourised. (c) Acidified KMnO4, heat. Pass vapour/gas into lime water. Methylbenzene: No reaction with lime water. Ethylbenzene: Lime water turns cloudy. (d) Bromine in tetrachloromethane. Ethyl benzene: No reaction. Phenylethene: Reddish-brown colour decolourised. 3 (a) CH3 + CH3Cl !!: + HCl + H2O AlCl3 CH3 NO2 CH3 + HNO3 !!: H2 SO4 COOH NO2 NO2 CH3 + 3[O] !!!!: KMnO4 /H+ Δ (b) CH3 + CH3Cl !!: + HCl + H2O AlCl3 COOH NO2 COOH + HNO3 !!: + H2O H2 SO4 CH3 COOH + 3[O] !!!!: KMnO4 /H+ Δ (c) + CH3!CH!CH3 + HCl !!: AlCl3 Cl O CH3!CH! CH3 !C!CH3 + HCl !!: AlCl3 CH3!CH! CH3 CH + CH3!C!Cl 3!CH! CH3 R O R

Chemistry Term 3 STPM Answers 328 Revision Exercise 15 Objective Questions 1 A 2 C 3 C 4 D 5 A 6 B 7 A 8 C 9 C 10 D 11 D 12 D 13 B 14 C 15 C 16 D 17 C 18 D 19 B 20 A 21 B 22 D 23 B 24 B 25 D 26 B 27 B 28 D 29 C 30 B 31 C 32 C 33 D 34 B 35 A 36 D 37 A 38 C 39 D Structured and Essay Questions 1 (a) Fractional distillation (b) (i) Zeolite (ii) Lower cost, as it requires lower pressure and temperature. (iii) C7H16 (c) C7H16 !: C7H8 + 4H2 or CH3 CH2 CH2 CH2 CH2 CH2 CH3 !: CH3 + 4H2 (d) (i) + Cl2 + HCl CH2 CH Cl 3 UV light + KCN + KCl CH2 CH2 Cl CN Ethanol Δ + 4[H+] CH2 CH2 NH2 CH2 CN LiAlH4 /ether H+ / Δ (ii) + Cl2 + HCl CH2 CH Cl 3 UV light + + HCl + NaOH + NaCl CH2 Cl CH2 OH Aqueous Δ CH3 ! C ! OCH2 ' O CH3 ! C ! Cl ' O CH2 OH Δ 2 (a) Presence of ultraviolet light or bright sunlight. • (b) Cl2 !!: 2Cl • • CH3!CH2!CH3 + Cl !: CH3!CH—CH3 + HCl • • CH3!CH!CH3 + Cl2 !: CH3!CH!CH3 + Cl | Cl • 2Cl !!: Cl2 (c) Production of 2-chloropropane involves the formation of a secondary free radical that is more stable. • (d) 2 CH3!CH!CH3 !!: CH3!CH!CH!CH3 & & CH3 CH3 3 (a) Zeolite (b) P is an alkene. The structure of P: CH3 CH3 CH3 CH3 C== O O ==C →    C===C CH3 H CH3 H P Ozonolysis of P: CH3 CH3 CH3 CH3 C==C + O3 + H2O Zn + H+ → C==O + O==C + H2O2 CH3 H CH3 H 4 (a) Carbon dioxide gas being absorbed by the sodium hydroxide. CO2(g) + 2NaOH(aq) !!: Na2CO3(aq) + H2O(1) (b) To ensure complete combustion of the hydrocarbon to carbon dioxide and water. y y (c) CxHy + (x + ––)O2 !!: xCO2 + –– H2O 4 2 Volume of CO2 produced = 10x = 50.0 ∴ x = 5 Volume of oxygen used = 10(x + —y 4 ) = 100 – (75 – 50) = 75 ∴ y = 10 Empirical formula of A = C5H10 (d) Molecular formula = C5H10 (e) CH3!CH2!CH2!CHRCH2 (I) CH3!CH2!CHRCH!CH3 (II) CH3!CH2!CRCH2 (III) & CH3 CH3!CH!CHRCH2 (IV) & CH3 CH3!CRCH!CH3 (V) & CH3 (f) CH3!CH2!CRCH2 + [O] !: CH3!CH2!C!CH3 + H2O & ∫ CH3 O 5 (a) No, this is because there are two hydrogen atoms bonded to one of the unsaturated carbon atoms. (b) C12H26 !!: C3H6 + C9H20 (c) CH3!CHRCH2 + [O] + H2O !!: CH3!CH!CH2OH & OH [B] CH3!CHRCH2 + H2SO4 !!: CH3!CH!CH3 & OSO3H [D] CH3!CH!CH3 + H2O !!: CH3!CH!CH3 + H2SO4 & & OSO3H OH [E]

Chemistry Term 3 STPM Answers 329 CH3!CHRCH2 + 5[O] !!: CH3COOH + CO2 + H2O [C] (d) CH3 H & & !![C!!C]n!! & & H H 6 (a) The occurrence of two or more compounds having the same structural formula but with different spatial arrangement of the atoms/groups due to the presence of carbon-carbon double bond that restrict free rotation. (b) (i) No. of moles of X = ——— 360 24 000 = 0.015 mol 0.84 Relative molecular mass of X = ––––– = 57.2 0.015 338 No. of moles of H2 = –––––– = 0.0141 mol 24 000 ∴ X has one CRC bond in its structure. The general formula of X is CnH2n. CnH2n = 57.2 ∴ n = 4 ∴ Molecular formula is C4H8. (ii) CH3!CHRCH!CH3 (X) CH3!*CH!*CH!CH3 (Y) & & Br Br CH3!*CH!*CH!CH3 (Z) & & OH OH (c) CH3 CH3 H H CRC CH3 H H CH3 CRC cis-isomer trans-isomer 7 (a) Cold, dilute acidified KMnO4 CH3!CHRCH2 + [O] + H2O !!: CH3!CH!CH2OH & OH (b) Boiling with concentrated acidified KMnO4. CH3!CHRCH!CH3 + 4[O] !!: 2CH3COOH 8 (a) (i) CH3!CHRCH2 + HBr !!: CH3!CH2!CH2Br (I) CH3!CHRCH2 + HBr !!: CH3!CH!CH3 (II) & Br (ii) Isomer (II) is the major product. It involves the production of a more stable secondary carbonium ion intermediate. (b) (i) Optical isomers are mirror images that are not superimposable. (ii) Br H H Br C C CH3 CH3 CH2Br CH2Br (iii) They differ in their direction of rotation of plane-polarised light. 9 (a) Mole ratio of C : H = ——85.7 12 : 14.3 = 1 : 2 Empirical formula = CH2 No. of moles of the hydrocarbon = —––— 100 22 400 = 4.46 × 10–3 mol 0.25 Relative molecular mass of hydrocarbon = –––––––––– 4.46 × 10–3 = 56.0 (CH2)n = 56 ∴ n = 4 Molecular formula = C4H8 (b) CH3!CH2!CHRCH2 (I) CH3 CH3 H H CRC (II) CH3 H H CH3 CRC (III) CH3 (IV) & CH3!CRCH2 (c) Cyclobutane CH2!!CH2 & & & & CH2!!CH2 10 The first step in the reaction is the formation of a carbonium intermediate: CH2RCH2 + HI !!: CH3!+CH2 + I– The carbonium ion then reacts with the I– to form iodoethane as the only product. CH3!+CH2 + I– !!: CH3CH2I But in aqueous sodium chloride, three anions are present; Cl– , I– and OH– . Hence, the carbonium ion can combine with any of the three anions to produce three products. CH3!+CH2 + I– !!: CH3CH2I CH3!+CH2 + Cl– !!: CH3CH2Cl CH3!+CH2 + OH– !!: CH3CH2OH On boiling, the iodo-compound and chloro-compound are hydrolysed to ethanol. CH3!CH2I + H2O !!: CH3CH2OH + HI CH3!CH2Cl + H2O !!: CH3CH2OH + HCl 11 (a) (i) CH3 CH3 (ii) Warm both hydrocarbons with bromine dissolved in tetrachloromethane. Pentene: The reddish brown colour of bromine is decolourised. 1,2-Dimethylcyclopentane: No visible change

Chemistry Term 3 STPM Answers 330 (b) (i) Substitution reaction Cl2 + Cl + HCl 2Cl UV Cl2 2Cl CH3 ! CH ! CH3 CH3 ! C ! CH3 + Cl2 + Cl CH3 ! C ! CH3 CH3 ! C ! CH3 Cl (ii) CH3 ! CH ! CH2 Cl Because it involves the formation of a primary free radical intermediate which is less stable. 12 (a) An electrophile is an electron-deficient species which accepts a lone pair electrons from another species to form a coordinate bond. Electrophiles are Lewis acids. (b) CH2==CH2 + Br2 ⎯→ BrCH2—CH2Br (c) Mechanism: CH2==CH2 δ+Br——Brδ– → CH2——+CH2 + Br– | Br CH2——+CH2 + Br– → CH2——CH2 | | | Br Br Br 13 (a) A is cyclohexane. (b) B is 1,4-dimethylbenzene (or any other dimethylbenzene isomers). CH3 COOH Oxidation !!!: CH3 COOH (c) C is 1-butene. H & CH3!CH2!CHRCH2 + HBr !: CH3CH2!*C!CH3 & Br (d) D is 2-methyl-2-butene CH3 O & Oxidation ∫ CH3!CRCH!CH3 !!: CH3!C!CH3 + CH3COOH 14 (a) React benzene with chloromethane in the presence of aluminium chloride. CH3 AlCl3 + CH3Cl !!: + HCl Boiling methylbenzene with acidified KMnO4. CH3 COOH + 3[O] !!:   + H2O (b) Prepare benzoic acid as in (a). Bubble chlorine gas through benzoic acid in the presence of aluminium chloride. COOH COOH + Cl2 !: + HCl Cl [Note: The !COOH is a 3-directing group.] (c) Prepare methylbenzene as in (a). React bromine with methylbenzene in the presence of aluminium chloride. CH3 CH3 AlCl3 + Br2 !!: + HBr Br [Note: The !CH3 group is a 2- and 4-directing group.] Boiling 4-bromomethylbenzene with acidified KMnO4. CH3 COOH + 3[O] !!: + H2O Br Br 93.3 6.7 15 (a) C : H = –––– : ––– 12 1 = 1.16 : 1 = 7 : 6 Empirical formula of A is C7H6. 2.4 (b) No. of moles of A = ––– = 0.1 mol 24 18 Mr of A = ––– = 180 0.1 (C7H6)n = 180 ∴ n = 2 Molecular formula of A = C14H12 (c) (i) B is: H H & & CRC C is: CRCH2 (ii) CRC + 4[O] !: 2 H H B C COOH C O CRCH2 + 4[O]: R + CO2 + H2O

Chemistry Term 3 STPM Answers 331 CRC + 4[O] !: 2 H H B C COOH C O CRCH2 + 4[O]: R + CO2 + H2O (d) Isomer B C*!C H H Br H 16 (a) Nitrogen oxides: Produced by the reaction between oxygen and nitrogen (in the air intake) under high pressure and high temperature in the engine compartment. Volatile hydrocarbons: From unburnt petrol. [Note: Petrol is a mixture of volatile hydrocarbons, both aliphatic and aromatic] (b) (i) Platinum, rhodium and palladium. (ii) 2NxOy ! : xN2 + yO2 CxHy + (x + y 4 )O2 !: xCO2 + y 2 H2O (c) Manufacture more efficient internal combustion engines. Increase the proportion of branched-chain hydrocarbon in the petrol. [The more branched-chain hydrocarbon in a petrol, the higher its octane number]. 17 (a) Cl!Cl UV light !!!: 2Cl• Cl• + H!CH3 !!: H!Cl + •CH3 Cl!Cl + •Cl3 !!: CH3!Cl + Cl• Cl• + Cl• !!: Cl2 (b) CH2RCH2 + HI !!: CH3!CH2 + + I– CH3!CH2 + + I– !!: CH3CH2I (c) HNO3 + 2H2SO4 L NO2 + + 2HSO4 – + H3O+ H H NO2 + NO2 + !!: + H NO NO2 2 !!: + H+ + Chapter 16 Haloalkanes Quick Check 16.1 1 (a) 3° (b) 3° (c) 2° (Active) (d) 2° (e) 1° (f) 3° (Active) (g) 1° (h) 1° (i) 3° Quick Check 16.2 1 1-chloropropane. The molecule is polar. 2 It cannot form hydrogen bonds with water. Quick Check 16.3 1 CH3!CH!CH3 2 C!H & ' OH O 3 CH3 OH 4 HO OH Quick Check 16.4 1 (a) CH3 CH3 & & A: CH3!CH!CN B: CH3!CH!COOH (b) NC CN HOOC COOH C: D: (c) E: !CH2!CH2!CH! CN F: !CH2!CH2!CH! CH2NH2 2 (a) CH3 + CH3Cl AlCl !!:3 + HCl CH3 CH2Cl + Cl2 UV light !!!: + HCl CH2Cl CH2CN + KCN Ethanol !!!: ∆ + KCl CH2CN CH2COOH + 2H2O + H+ Dilute H !!!:2SO4 ∆ + NH4 + (b) CH3!CH!CH3 AlCl3 + CH3!CH!CH3 !: + HCl & Cl + HCl CH3!CH!CH3 CH3!C!CH3 Cl + Cl2 !!!: UV light + KCl CH3!C!CH3 CN CH3!C!CH3 Cl + KCN !!!: Ethanol Δ + NH4 + CH3!C!COOH CH3 CH3!C!CN CH3 + 2H2O + H+ !!!: Dilute H2 SO4 Δ

Chemistry Term 3 STPM Answers 332 CCl4 (c) CH2RCH2 + Br2 !!: Br!CH2!CH2!Br Ethanol Br!CH2CH2!Br + 2KCN !!∆ !: NC!CH2CH2!CN + 2KBr Dilute H2SO4 NC!CH2CH2!CN + 4H2O + 2H+ !!!!!: HOOC!CH2!CH2!COOH + 2NH4 + (d) Cl CN + KCN Ethanol !!!: ∆   + KCl CN CH2NH2 + 4[H] !!!: LiAlH4 Quick Check 16.5 1 (a) CH2RCH2 (b) CH3!CHRCH2 (c) CH3!CH2!CHRCH2 (d) (e) CH3 (f) No reaction. (g) CHRCH2 2 (a) UV light CH3CH3 + Cl2 !!!: CH3!CH2Cl + HCl Ethanol CH3!CH2Cl + KOH !!∆ !: CH2RCH2 + KCl + H2O (b) CH3!CH2!CHRCH2 + HCl !!: CH3!CH2!CH!CH2 & Cl Ethanol CH3!CH2!CH!CH3 + KOH !!∆ !: & Cl CH3!CHRCH!CH3 + KCl + H2O (c) Ethanol CH3!CH!CH2Cl + KOH !!∆ !: & CH3 CH3!CRCH2 + KCl + H2O & CH3 CH3 Conc. H2SO4 & CH3!CRCH2 + H2O !!!!: CH3!C!CH3 & H2O, ∆ & CH3 OH (d) + KOH !!!: + KBr + H2O Ethanol Δ Br Br Br + Br2 !!!: CCl4 Quick Check 16.6 1 (a) SN2 (b) SN1 (c) SN1 Quick Check 16.7 1 Cl 2 Cl C!CH3 ∫ O CH2CH2CH3 3 Cl Cl Quick Check 16.8 1 (a) CH3 & CH3!CH!CH2OH (b) CH2CH2OH (c) CH3 & C!CH2!CH!CH2!CH3 & & CH3 OH (d) CH3 CH3 & & C!!CH2!!C & & CH3 OH 2 O C!CH3 + MgBr R Revision Exercise 16 Objective Questions 1 D 2 D 3 B 4 C 5 A 6 B 7 C 8 B 9 A 10 D 11 D 12 B 13 C 14 A 15 A 16 C 17 D Structured and Essay Questions 1 (a) Cl Cl (b) Warm both the liquids separately with ethanolic silver nitrate. Chlorocyclohexane will give a white precipitate after 1 to 2 minutes, while there is no reaction with chlorobenzene. 2 (a) CH2Cl CH3 Cl (X) (Y)

Chemistry Term 3 STPM Answers 333 (b) + HCl CH3 + CH3Cl !!!: AlCl3 + HCl CH CH2Cl 3 + Cl2 !!!: UV light 3 (a) (i) Aqueous NaOH: Nucleophilic substitution (ii) Ethanolic NaOH: Elimination (b) (i) CH3 OH CH3 CH (ii) 3 OH CH3 (c) CH3 CH3 + + Cl– Cl Slow CH3 CH3 + + OH– OH Fast 4 (a) (i) C2H5OH + KBr + H2SO4 !∆ : C2H5Br + KHSO4 + H2O (ii) C2H5OH + PCl5 !!: C2H5Cl + POCl3 + HCl Ethanol, heat C2H5Cl + KCN !!!!: C2H5CN + KCl C2H5CN + 2H2O + H+ !∆ : C2H5COOH + NH4 + 29.7 64.1 (b) (i) C : H : Pb = –––– : 6.2 : –––– 12 207 = 8 : 20 : 1 Molecular formula is C8H20Pb. (ii) C2H5 & C2H5!Pb!C2H5 & C2H5 (iii) As a petrol additive to prevent ‘pre-ignition’ which leads to ‘knocking’. 5 (a) + Cl2 CH3 ! CH! CH3 + Cl2 CH3 ! CH! CH3 UV + HCl (W) CH3 ! C* ! CH2 Cl H UV + HCl (X) CH3 ! C ! CH3 Cl + NaOH + NaCl (Y) CH3 ! C ! CH2 OH H CH3 ! C ! CH2 Cl H OH Cl + Cl2 CH3 ! CH! CH3 UV + HCl (X) CH3 ! C ! CH3 Cl + NaOH + NaCl (Y) CH3 ! C ! CH2 OH H CH3 ! C ! CH2 Cl H + NaOH + NaCl (Z) CH3 ! C ! CH3 OH CH3 ! C ! CH3 Cl (b) Mechanism: Free radical substitution Production of W: + Cl CH3 ! CH ! CH3 + Cl2 CH3 ! CH ! CH2 + HCl CH3 ! CH ! CH2 + Cl CH3 ! CH ! CH2 Cl Cl2 2Cl 2Cl Cl2 Production of X: + Cl CH3 ! CH! CH3 + Cl2 CH3 ! C! CH3 + HCl CH3 ! C ! CH3 + Cl Cl CH3 ! C ! CH3 Cl2 2Cl 2Cl Cl2 The major product is X as it involves a more stable tertiary free radical intermediate compared to the less stable primary free radical in the production of W. (c) + KOH H CH3 ! C ! CH2 Cl CH3 ! C " CH2 Ethanol Δ + KCl + H2 O + HCl CH3 ! C " CH2 Cl CH3 ! C ! CH3 Δ [NOTE: The addition follows the Markovnikov’s rule] 6 (a) CH2Cl CH3 CH3 CH3 Cl Cl Cl (I) (II) (III) (IV)

Chemistry Term 3 STPM Answers 334 CH2OH (b) (I) !: (c) (i) Isomer (I) (ii) CH2Cl CH2CN + KCN Ethanolic, heat !!!!: + KCl CH2CN CH2COOH + 2H2O + H+ Heat !: + NH4 + 7 (I) CH3!CH2!CH2!CH2Cl (II) CH3!CH2!CH!CH3 & Cl (III) CH3!CH!CH2Cl & CH3 (IV) CH3 & CH3!C!CH3 & Cl (a) I, III, IV (b) II (c) IV (d) IV (e) I, III 8 (a) Mol ratio C : H : Cl = 71.2/12 : 7.72 : 21.08/35.5 = 5.93 : 7.72 : 0.593 = 10 : 13 : 1 The empirical formula is C10H13Cl (C10H13Cl)n = 168.5 n = 1 The molecular formula is C10H13Cl (b) A is a tertiary haloalkane (undergoing SN1) with a chiral centre (optically active). Structure of A: CH3 ! C ! CH2 CH3 (A) * (C) (B) ether + NaOH Cl CH3 ! C ! CH2 CH3 Cl + NaCl CH3 ! C ! CH2 CH3 OH + Mg CH3 ! C ! Cl CH2 CH3 CH3 ! C ! MgCl CH2 CH3 CH3 ! C ! MgCl + CH3 CH2 ! C ! H + H2 O + Mg(OH)Cl CH2 CH3 CH3 ! C !!! C ! CH2 CH3 CH2 CH3 H OH ' O (c) CH3 CH3 H3 C H (trans-isomer) ! C " CHCH (D) 3 C RC H H3 C CH3 (cis-isomer) C RC 9 (a) CH3CH2CH2CH2I + KOH Ethanolic, heat !!!!!: CH3CH2!CHRCH2 + KI + H2O CH3CH2!CHRCH2 + HI !: CH3CH2!CHCH3 & I (b) CH3CH2CH2OH Excess conc. H2SO4, heat !!!!!!!!: CH3!CHRCH2 + H2O H2SO4, H2O CH3!CHRCH2 + H2O !!!: CH3!CH!CH3 & OH (c) Cl + Cl2 AlCl3 !!: + HCl (d) CH3 + CH3Cl AlCl3 !!: + HCl CH3 CH2Cl + Cl2 UV light !!: + HCl Ethanolic, heat (e) CH3CH2CH2I + KCN !!!: CH3CH2CH2CN + KI CH3CH2CH2CN + 2H2O + H+ !∆ : CH3CH2CH2COOH + NH4 + 10 (a) Grignard’s reagent is an organomagnesium compound. (b) CH3CH2CH2I + Mg Ether !!: CH3CH2CH2MgI (c) (i) CH3CH2CH2MgI + HCHO + H2O !!∆ !: CH3CH2CH2CH2OH + Mg(OH)I (ii) CH3CH2CH2MgI + CH3COCH3 + H2O !∆ : CH3 & CH3CH2CH2!C!CH3 + Mg(OH)I & OH 11 (a) (i) CI C F F CI Tetrahedral Hybridisation: sp3

Chemistry Term 3 STPM Answers 335 (ii) 3CCl4 + SbF3 9: 3CCl2F2 + 2SbCl3 (iii) Freon-12 is non-toxic, non-flammable, chemically inert, colourless, odourless, thermally stable and easily liquefied by moderate pressure. (b) (i) Ozone is formed in the stratosphere via the following processes: Molecular oxygen absorbs ultra violet radiation from the Sun and splits into oxygen atoms: O2(g) 9: 2[O](g) The oxygen atom combines with another oxygen molecule to produce ozone: O2(g) + [O](g) 9: O3(g) The ozone formed is in turn destroyed by absorbing ultra-violet radiation from the Sun and decomposes into oxygen molecule and oxygen atom: O3(g) 9: O2(g) + [O](g) The oxygen atom formed then combines with another oxygen molecule to regenerate ozone: O2(g) + [O](g) 9: O3(g) The whole cycle is repeated. There is thus a dynamic equilibrium between the production and destruction of ozone: 3O2 N 2O3 Both the forward and reverse reactions involve absorption of ultra-violet radiation from the Sun. (ii) Freon, which is chemically inert, when released into the atmosphere stays unaffected and drifts into the stratosphere where ultra-violet radiation cleavage the C9Cl bond to produce chlorine free radicals: CCl2F2 9: Cl9CCF2 + Cl. The chlorine free radical then undergoes the following reactions: O3 + C. l 9: O2 + C. lO C . lO + O 9: C . l + O2 The overall reaction is the destruction of ozone O3 + [O] 9: 2O2 Each chlorine free radical is able to ‘destroy’ 100,000 ozone molecules before it is removed by other reactions. (iii) Ozone is a greenhouse gas. It keeps the Earth warm by absorbing the heat reflected from the Earth surface and prevent it from re-radiating into space. [Other greenhouse gases are water, carbon dioxide, methane and nitrogen oxides] 12 (a) I: CH3CH2CH2CH2Cl II: CH3CH2CHCH3 & Cl III: CH3—CH—CH2Cl & CH3 CH3 | IV: CH3—C—CH3 | Cl (b) Isomer II is optically active. There is a chiral carbon in the molecule. (c) (i) Isomer II (ii) CH3CH2CH=CH2 CH3 CH3 C == C H H CH3 H C == C H CH3 13 (a) Nucleophilic substitution unimolecular, SN1. CH3 CH3 | | CH3—C—CH3 Slow !!: CH3—C+—CH3 + I– | I CH3 CH3 | | CH3—C+—CH3 + OH– Fast !!: CH3—C—CH3 | OH (b) Reaction axis Energy (c) The rate will be slower. The rate-determining step involves the breaking of the carbon-halogen bond. Due to the smaller size of the chlorine atom, the C—Cl bond is stronger and needs more energy to break. Chapter 17 Hydroxy Compounds Quick Check 17.1 1 CH3CH2CH2CH2CH2CH3 < CH3CH2CH2CH2CH2OH < HOCH2CH2CH2CH2OH CH3CH2CH2CH2CH2CH3: van der Waals force CH3CH2CH2CH2CH2OH: Hydrogen bonding HO—CH2CH2CH2CH2—OH: More extensive hydrogen bond 2 Methanol: Intermolecular hydrogen bond Ethene: van der Waals force 3 Phenol: Intermolecular hydrogen bond Methylbenzene: van der Waals force Quick Check 17.2 1 Ethanol can form hydrogen bond with water. Hexanol: Presence of a large hydrophobic hydrocarbon group. 2 1-propanol < 1,2-propanediol < 1,2,3-propanetriol The more !OH group in the molecule, the more hydrogen bonds it can form with water. 3 1,2-propanediol. It has 2 !OH groups.

Chemistry Term 3 STPM Answers 336 Quick Check 17.3 1 (a) 1° (b) 2° (c) 3° (d) 3° (e) 2° Quick Check 17.4 1 OH CH2OH < CH3!CH!CH3 < & OH 2 (a) 2-fluoroethanol (b) Flourine is an electron-withdrawing group and helps to weaken the O!H bond. Quick Check 17.5 1 HO!CH2!CH2!OH + 2Na !!: Na+–O!CH2!CH2!O– Na+ + H2 62 g = 22.4 dm3 ∴ 23 g = —– 23 62 × 22.4 = 8.31 dm3 2 (a) No. of moles of H2 = —–6 24 = —1 4 mol No. of moles of alcohol = 2 × —1 4 = —1 2 mol ∴ Relative molecular mass = 30 × 2 = 60 ∴ Molecular formula = C3H7OH (b) CH3!CH2!CH2OH CH3!CH!CH3 & OH 3 (a) CH3OH + Cl2 !!: CH2Cl!OH + HCl (b) HO!CH2CH2!OH + 2Na !!: Na+–O!CH2CH2!O– Na+ + H2 (c) 2HO!CH2!CH!CH2OH + 6Na !!: & OH 2Na+–O!CH2!CH!CH2O– Na+ + 3H2 & O– Na+ CH3 CH3 & & (d) 2CH3!C!CH3 + 2Na !: 2CH3!C!CH3 + H2 & & OH O– Na+ (e) Br OH OH Br + Br2 !!!: 4 Bromine in tetrachloromethane 1-cyclohexenol: Reddish-brown colour decolourised 1-cyclohexanol: No reaction Quick Check 17.6 1 (a) C!O O R O O ∫ ∫ (b) CH3!O!C!CH2CH2!C!O!CH3 O O ∫ ∫ (c) CH2!O!C!C!O!CH2 O CH3 ∫ & (d) C!O!CH!CH3 2 (a) HCOOH and CH3OH (b) COOH and CH3OH (c) HOOC!CH2CH2!COOH and CH3OH (d) COOH and CH2RCH!CH2OH O ∫ 3 CH2!C!18O!CH2CH3 Quick Check 17.7 1 (a) HCOOH (b) No reaction (c) COOH (d) OR (e) No reaction Quick Check 17.8 1 (a) No reaction (e) CHRCH2 (b) CH3!CHRCH2 (f) No reaction (c) CH3 (g) CH3 & CH3!CRCH2 (d) No reaction (h) CH2RCH!CHRCH2 Quick Check 17.9 1 (a) Negative (e) Negative (b) Positive (f) Negative (c) Negative (g) Negative (d) Positive 2 CH3CH2CH2CH2OH : Negative CH3!CH!CH2!CH3 : Positive & OH CH3 & CH3!C!CH3 : Negative & OH CH3 & CH3!C!CH2OH : Negative & H Quick Check 17.10 (a) No (b) Yes (c) Yes (d) No (e) No

Chemistry Term 3 STPM Answers 337 Quick Check 17.11 1 OH Br CH3 Br Br 2 C!O O R 3 O– Na+ NO2 O2N NO2 4 + HCl OH CH CH2Cl 3 OH + Cl2 !!!: Revision Exercise 17 Objective Questions 1 D 2 C 3 D 4 D 5 C 6 C 7 D 8 C 9 B 10 C 11 B 12 D 13 B 14 B 15 D 16 C 17 C Structured and Essay Questions 1 C : H : O = —–– 64.9 12 : 13.5 : —–– 21.6 16 = 4 : 10 : 1 Empirical formula of A = C4H10O Using pV = nRT (200 × 103 )(52.4 × 10–6) = —–– 0.25 M × 8.31 × 373 M = 74.4 (C4H10O)n = 74.4 n = 1 Molecular formula of A = C4H10O D is CH3 & CH3!C!CH3 & Br E is CH3 & CH3!C!CH3 & OH A is CH3!CH!CH2OH & CH3 B is CH3!CH!C!H & ∫ CH3 O C is CH3!CRCH2 & CH3 2 (a) A = CH!CH3 & OH (b) B = CH2RCH!CH!CH3 & CH3 C = CH3!CH!CH!CH3 & & Br CH3 D = CH3!CH!CH!CH3 & & OH CH3 E = CH3!C!CH!CH3 ∫ & O CH3 3 (a) KMnO4/H+ CH3CH2CH2OH + 2[O] !!!!: ∆ CH3CH2COOH + H2O (b) Excess conc. H2SO4, heat CH3CH2CH2OH !!!!!!!!: CH3!CHRCH2 + H2O CH3!CHRCH2 + HCl !!: CH3!CH!CH3 & Cl CH3!CH!CH3 + KCN Ethanolic, heat !!!!: CH3!CH!CH3 + KCl & & Cl CN Dilute H2SO4, heat CH3!CH!CH3 + 2H2O + H+ !!!!!!: & CN CH3!CH!CH3 + NH4 + & COOH 4 (a) Phenol is a weak acid. It is not strong enough to react with sodium carbonate. (b) Cyclohexanol is a secondary alcohol whereas phenol is a tertiary alcohol. There is no oxidisable hydrogen atom in phenol. (c) The benzene ring is an electron-withdrawing group that weakens the O!H bond in phenol. (d) Phenol reacts with NaOH to form the water soluble sodium phenoxide, but cyclohexanol (a weaker acid than water) does not. 5 (a) * (A): CH3 !CH2 !CH!CH3 s OH (C): CH3 !CHRCH!CH3 (E): CH3 !C!H ' O (G): H!C!H ' O CH3 s (I): CH3 !CH2 !C!COOH s OH (B): CH3 !CH2 !C!CH3 ' O (D): CH3 !CH2 !CHRCH2 (F): CH3 !CH2 !C!H ' O CH3 s (H): CH3 !CH2 !C!CN s OH * (A): CH3 !CH2 !CH!CH3 s OH (C): CH3 !CHRCH!CH3 (E): CH3 !C!H ' O (G): H!C!H ' O CH3 s (I): CH3 !CH2 !C!COOH s OH (B): CH3 !CH2 !C!CH3 ' O (D): CH3 !CH2 !CHRCH2 (F): CH3 !CH2 !C!H ' O CH3 s (H): CH3 !CH2 !C!CN s OH * (A): CH3 !CH2 !CH!CH3 s OH (C): CH3 !CHRCH!CH3 (E): CH3 !C!H ' O (G): H!C!H ' O CH3 s (I): CH3 !CH2 !C!COOH s OH (B): CH3 !CH2 !C!CH3 ' O (D): CH3 !CH2 !CHRCH2 (F): CH3 !CH2 !C!H ' O CH3 s (H): CH3 !CH2 !C!CN s OH * (A): CH3 !CH2 !CH!CH3 s OH (C): CH3 !CHRCH!CH3 (E): CH3 !C!H ' O (G): H!C!H ' O CH3 s (I): CH3 !CH2 !C!COOH s OH (B): CH3 !CH2 !C!CH3 ' O (D): CH3 !CH2 !CHRCH2 (F): CH3 !CH2 !C!H ' O CH3 s (H): CH3 !CH2 !C!CN s OH (b) CH3 !CH2 !CH!CH3 !!!: CH3 !CH2 !C!CH3 + H2 s ' OH O (A) (B) CH3 !CH2 !CH!CH3 !!!: CH3 !CHRCH!CH3 + H2 O s OH (A) (C) CHCHCHCH:CHCHCHCHHOCu Δ H2 SO4 Δ H2SO4

Chemistry Term 3 STPM Answers 338 CH3 !CH2 !CH!CH3 !!!: CH3 !CH2 !C!CH3 + H2 s ' OH O (A) (B) CH3 !CH2 !CH!CH3 !!!: CH3 !CHRCH!CH3 + H2 O s OH (A) (C) CH3 !CH2 !CH!CH3 !!!: CH3 !CH2 !CHRCH2 + H2 O s OH (A) (D) Cu Δ H2 SO4 Δ H2 SO4 Δ CH3 !CH2 !CH!CH3 !!!: CH3 !CH2 !C!CH3 + H2 s ' OH O (A) (B) CH3 !CH2 !CH!CH3 !!!: CH3 !CHRCH!CH3 + H2 O s OH (A) (C) CH3 !CH2 !CH!CH3 !!!: CH3 !CH2 !CHRCH2 + H2 O s OH (A) (D) Cu Δ H2 SO4 Δ H2 SO4 Δ CH3 !CH"CH!CH3 + O3 + H2 O !!: (D) 2CH3 !C!H + H2 O2 < O (E) CH3 !CH2 !CH"CH2 + O3 + H2 O !!: (D) O O < < CH3 !CH2 !C!H + H!C!H + H2 O2 (F) (G) CN e CH3 !CH2 !C!CH3 + HCN !!!: CH3 !CH2 !C!CH3 < e O OH (B) (H) CH3 e CH3 !CH2 !C!CN + 2H2 O + H+ !!: e OH (H) CH3 e CH3 !CH2 !C!COOH + NH e OH (I) Zn Δ Zn Δ NaCN Δ Δ + 4 (c) NaCN!: Na+ + CN– CN s CH3 !CH2 !C!CH3 + CN– !!: CH3 !CH2 !C!CH3 ' s O O– CN CN s s CH3 !CH2 !C!CH3 + H!CN !!: CH3 !CH2 !C!CH3 s s O– OH Slow NaCN!: Na+ + CN– CN s CH3 !CH2 !C!CH3 + CN– !!: CH3 !CH2 !C!CH3 ' s O O– CN CN s s CH3 !CH2 !C!CH3 + H!CN !!: CH3 !CH2 !C!CH3 s s O– OH Slow NaCN!: Na+ + CN– CN s CH3 !CH2 !C!CH3 + CN– !!: CH3 !CH2 !C!CH3 ' s O O– CN CN s s CH3 !CH2 !C!CH3 + H!CN !!: CH3 !CH2 !C!CH3 s s O– OH Slow 6 (a) Structural formula of A: H s !C!CH3 s OH (b) I: Concentrated H2SO4 followed by boiling with water. II: Benzoyl chloride. Heat. (c) Structure of B: !CH2 CH2 OH To differentiate between A and B: Heat both compounds in separate test tubes with an alkaline solution of iodine (iodine dissolved in NaOH): A: A yellow precipitate (iodoform) is formed. B: No yellow precipitate is formed. H s !C!CH3 + 4I2 + 6NaOH s OH !: CHI3 + !COO– Na+ + 5NaI + 5H2 O H s !C!CH3 + 4I2 + 6NaOH s OH !: CHI3 + !COO– Na+ + 5NaI + 5H2 O 7 (a) C : H : O = —– 60 12 : 13.4 : —–– 26.6 16 =3 : 8 : 1 Formula of A = C3H8O Compound Structure A CH3CH2CH2OH B CH3!CHRCH2 C CH3!CH!CH3 & OH (b) CH3CH2CH2OH + 2[O] !!: CH3COOH + H2O CH3!CH!CH3 + [O] !: CH3!C!CH3 + H2O & ∫ OH O (c) Warm with alkaline iodine. A: No reaction. C: Yellow precipitate. 8 (a) Compound Structure A C2H5—O—C2H5 B CH3—CH—CH2CH3 | OH C CH3CH2CH2CH2OH D CH3 | CH3—C—CH3 | OH (b) Isomer B (c) E are: CH3 CH3 H H CRC CH3 H H CH3 CRC 9 (a) Ethanol can form hydrogen bonds with water molecule. Although phenol too can form hydrogen bonds with water molecule, the presence of a large hydrophobic C6H5 group makes it insoluble.

Chemistry Term 3 STPM Answers 339 (b) Ethanol has the CH3CH(OH)! group, but phenol does not. CH3CH(OH)H + 4I2 + 6NaOH → CHI3 + HCOONa + 5NaI + 5H2O (c) (i) OH NO2 (ii) Electrophilic substitution (d) Ethanol: White fumes of hydrogen chloride evolved. CH3CH2OH + PCl5 → C2H5Cl + POCl3 + HCl Phenol: No reaction 10 (a) I: PCl5, room conditions II: Excess concentrated H2SO4, 180 °C III: Acidified KMnO4, heat (II) CH3 | CH2==C—CH3 (b) (I) CH3 | CH3—C—CH3 | Cl (III) No reaction 11 (a) OH C* OH H CH3 HO C* HO CH3 H (b) Insoluble. Due to the presence of the hydrophobic benzene ring. (c) (i) Na+–O CHCH3 OH (ii) Na+–O CHCH3 O– Na+ (iii) HO C CH3 O (iv) HO C ONa + CHI3 O (v) HO Br Br CHCH3 OH 12 (a) Mol ratio: C : H : O = 64.9 12 : 13.5 1 : 21.6 16 = 5.4 : 13.5 : 1.35 = 4 : 10 : 1 Empirical formula = C4H10O Let the molecular formula = (C4H10O)n \ 74n = 74 n = 1 Molecular formula of Y is C4H10O. (b) Y liberates white fumes with PCl5. This shows that Y is an alcohol. I: CH3CH2CH2CH2OH 1-butanol II: CH3CH2—CH—CH3 2-butanol | OH III: CH3 | CH3—CH—CH2OH 2-methyl-1-propanol IV: CH3 | CH3—C—CH3 2-methyl-2-propanol | OH (c) The isomer concerned is a secondary alcohol: 2-butanol CH3CH2—CH—CH3 + [O] → CH3CH2—C—CH3 + H2O | || OH O (d) The isomer is a tertiary alcohol: 2-methyl-2-propanol CH3 | CH3—C—CH3 | OH There is no hydrogen atom attached to the —OH carrying carbon atom. 13 (a) Reaction with PCl5 shows that X is an alcohol. X is a tertiary alcohol because it is resistance to oxidation. Therefore X is: CH3 | CH3—C—CH3 | OH (b) React iodomethane with magnesium in dry ether to form the Grignard reagent. CH3I + Mg → CH3MgI Reacting the Grignard reagent with propanone followed by hydrolysis with dilute sulphuric acid. CH3 | CH3MgI + CH3—C—CH3 + H2O → CH3—C—CH3 + Mg(OH)I || | O OH Chapter 18 Carbonyl Compounds Quick Check 18.1 1 (a) 4-methyl-2-pentanone (b) 2-propylcyclohexanone (c) 2-methyl-3-bromo-2-butenal (d) 2-methylcyclopentanone (e) 1-phenyl-5-hydroxy-3-hexanone 2 (a) CH2I!C!CH3 ∫ O (b) CHO CH3 CH3

Chemistry Term 3 STPM Answers 340 (c) O ∫ CH3!CH!CH2!C!H & OH (d) CH3O & ∫ CH3!CH2!CH2!C!C!H & CH3 (e) CH3 & CH2RC!C!CH3 ∫ O Quick Check 18.2 1 (a) H!C!H (b) CH3!C!H ∫ O ∫ O (c) No reaction (d) C!CH3 O R 2 (a) H!C!H ∫ O O O ∫ ∫ (b) CH3!CH2!C!H + H!C!H (c) C!H+ H!C!H O R O R (d) H!C!CH2CH2CH2CH2!C!H ∫ ∫ O O (e) CH3!C!CH2CH2CH2!C!H ∫ ∫ O O Quick Check 18.3 1 (a) CH3OH (b) CH!CH3 OH (c) HOCH2!CH2CH2!CH!CH3 & OH (d) HOCH2CH2CH2OH 2 (a) RO (b) CH2!C!H O R (c) CH3!C!CH2!CH2!C!CH2CH3 ∫ ∫ O O (d) C!H O Cl R Quick Check 18.4 CRN!NH CH3 NO2 NO2 Quick Check 18.5 1 (a) OH & H!C!CN & H (b) C!CN H OH (c) OH & CH3!CH!CH2!CH2!C!CN & & Cl H (d) C CN OH (e) HO CN CH3 & 2 (a) CH3!C!CH3 + HCN !!: CH3!C!CN ∫ & O OH (b) NaCN !!: Na+ + CN– CH3 & CH3!C!CH3 + CN– !!: CH3!C!CN ∫ & O O– CH3 CH3 & & CH3!C!CN + HCN !!: CH3!C!CN + CN– & & O– OH (c) As a catalyst Quick Check 18.6 1 (a) Negative (b) Positive (c) Negative (d) Negative (e) Positive (f) Positive (g) Negative (h) Negative (i) Positive (j) Positive Quick Check 18.7 1 OH & HOCH2!(CHOH)4!CH!CN 2 HOCH2!(CHOH)4!CH2OH 3 O ∫ Cl!CH2!(CH)4!C!H & Cl

Chemistry Term 3 STPM Answers 341 4 HOOC!C!C!C!C!COOH ∫ ∫ ∫ ∫ O O O O Revision Exercise 18 Objective Questions 1 A 2 B 3 A 4 C 5 A 6 D 7 A 8 C 9 B 10 B 11 B 12 C 13 C 14 A 15 D 16 B 17 B Structured and Essay Questions 1 (a) A: CH3 CHO (b) B: CH!CH3 OH C: C!CH3 O R (c) D: CH3!C!CH3 ∫ O (d) CH3CH2!C!CH3 CH3CH2!CH!CH3 ∫ & O OH (E) (F) 2 (a) Warm with Tollen’s reagent. Propanal: Silver mirror is formed Propanone: No reaction (b) Warm with alkaline iodine. Methanal: No reaction Ethanal: Yellow precipitate (c) Warm with alkaline iodine. 3-pentanone: No reaction 2-pentanone: Yellow precipitate (d) Warm with alkaline iodine. Propanone: Yellow precipitate Cyclohexanone: No reaction 3 (a) C : H : O = 8 : 8 : 1 Empirical formula = C8H8O (b) Using pV = nRT (101 × 103)(0.1 × 10–3) = 0.305 M × 8.31 × 480 M = 120 Molecular formula = C8H8O (c) Triiodomethane (iodoform), CHI3 (d) C!CH3 O (X) (Y) (Z) R CH!CH3 CHRCH2 OH (e) C C H H H 4 (a) LiAlH4 CH3!CH2!C!CH3 + 2[H] !!: CH3!CH2!CH!CH3 ∫ & O OH CH3!CH2!CH!CH3 + PCl5 !!: & OH CH3!CH2!CH!CH3 + POCl3 + HCl & Cl CH3 & (b) CH3!C!CH3 + HCN !!: CH3!C!CN ∫ & O OH CH3 CH3 & Dilute H2SO4 & CH3!C!CN + 2H2O + H+ !: CH3!C!COOH + NH4 + & & OH OH 5 (a) C : H = 1 : 2 (CH2)n = 56 ∴ n = 4 Molecular formula of A = C4H8 A is CH3!CH2!CHRCH2 B is CH3!CH2!CH!CH3 & Br C is CH3!CH2!CH!CH3 & OH D is CH3!CH2!C!CH3 ∫ O (b) CH3CH2CHCH3 + 4I2 + 6NaOH !!: & OH CHI3 + CH3CH2COONa + 5NaI + 5H2O CH3CH2CCH3 + 3I2 + 4NaOH !!: ∫ O CHI3 + CH3CH2COONa + 3NaI + 3H2O (c) An orange precipitate. CRN!NH! !NO2 CH3 CH3CH2 NO2 6 (a) CN & CH3!CH2!C!CH3 + HCN !: CH3!CH2!C!CH3 ∫ & O OH (b) NaCN !!: Na+ + CN– CN & CH3!CH2!C!CH3 + CN– !!: CH3!CH2!C!CH3 ∫ & O O–