Chemistry of Carbon Compounds

Topic 8 hemistry of arbon ompounds Unit 29 An introduction to the chemistry of carbon compounds Unit 30 Isomerism Unit 31 Typical reactions of selected functional groups Unit 32 Synthesis of carbon compounds Unit 33 Important organic substances

Key o ncepts An introduction to the chemistry of carbon compounds omologous series Systematic naming Effects of functional groups and chain length on physical properties Synthesis of carbon compounds Synthetic routes for carbon compounds Preparation of simple carbon compounds hemistry of arbon ompounds Typical reactions of selected functional groups Reactions of alkanes, alkenes, haloalkanes, alcohols, aldehydes, ketones, carboxylic acids, esters and amides Isomerism Structural isomerism chain isomerism, position isomerism and functional group isomerism Stereoisomerism geometrical isomerism and enantiomerism Important organic substances Aspirin Soaps and soapless detergents ylons and polyesters arbohydrates Lipids Proteins

Topic 8 hemistry of arbon ompounds Unit 29 An introduction to the chemistry of carbon compounds Unit 29 An introduction to the chemistry of carbon compounds 29.1 29.12 & 29.21 29.1 The value of medicines: longer and healthier lives 29.2 Functional groups: centre of reactivity 29.3 aming alkanes and alkenes 29.4 aming carbon compounds with one type of functional group 29.5 aming haloalkanes 29.6 aming alcohols 29.7 aming aldehydes and ketones 29.8 aming carboxylic acids omologous series 1 The following table summarizes the nomenclature of compounds in various homologous series. General formula Functional group it contains Alkanes n 2n+2 Alkenes n 2n omenclature add appropriate prefix to ane replace ane of the corresponding alkane with ene Structural formula 3 3 2 3 IUPA name 2-methylbutane 3 2 propene 29.9 aming esters 29.10 aming amides 29.11 aming amines 29.12 aming compounds with more than one type of functional group 29.13 Intermolecular forces and physical properties of carbon compounds aloalkanes RX X (X = F, l, Br or I) Alcohols R add the name of halogeno functional group as prefix to the corresponding alkane replace the last letter e of the corresponding alkane with ol l 2-chloropropane 3 3 3 2 ethanol 29.14 Physical properties of haloalkanes 29.15 Physical properties of alcohols 29.16 Physical properties of aldehydes and ketones Aldehydes R replace the last letter e of the corresponding alkane with al 3 2 propanal 29.17 Physical properties of carboxylic acids 29.18 Physical properties of esters 29.19 Physical properties of amides Ketones RR 1 replace the last letter e of the corresponding alkane with one 3 3 propanone 29.20 Physical properties of amines 29.21 ommon names of carbon compounds arboxylic acids R replace the last letter e of the corresponding alkane with oic acid 3 2 2 butanoic acid Esters RR 1 the name consists of two separate words, the first word comes from the alcohol, the second word comes from the acid 3 3 methyl ethanoate Amides R 2 (unsubstituted) replace the oic acid ending of the corresponding acid by amide 3 2 ethanamide Amines R 2 (primary) replace the last letter e of the corresponding alkane with amine 3 2 methanamine

Topic 8 hemistry of arbon ompounds Unit 29 An introduction to the chemistry of carbon compounds 2 Sometimes there is more than one functional group in one compound. This kind of compound should be named according to the following order of precedence of functional groups: > > 2 > > > > 2 > = Give the IUPA names of the following compounds. a) 2 The longest continuous chain containing the carbon bearing the group may T appear in a straight line. Questions often ask about the names of such structures. b) 3 2 ( 3 ) 2 (1 mark) (1 mark) 3 3,4 2 5 2 1 3 2-methylbutan-2-ol D T spell amine as ammine. D T confuse amine and amide. The general formula of primary amine is R 2, while that of amide is ( or R) 2. Students may need to identify the functional groups present in unfamiliar compounds. 3 2 2 3 3 oseltamivir 2 2 3 Besides the ether linkage, functional groups present in oseltamivir include: = bond; amide functional group; amine functional group; and ester functional group. c) 3 = 2 (1 mark) d) 3 ( 2 ) 2 (1 mark) a) phenylethanoic acid (1) b) 2-methylpropyl ethanoate (1) c) but-2-en-1-ol (1) d) 3-aminobutanoic acid (1) 29.13 29.20 for (a), the acid. (b) group is a phenyl group. It is attached to the ethanoic 3 2 ( 3 ) 2 this part comes from ethanoic acid this part comes from 2-methylpropan-1-ol (c) The double bond takes the form -en-. (d) is the principle functional group. The 2 group is named as a prefix. 1 The physical properties ( the boiling point and water solubility) of a carbon compound are affected by a) the functional group it contains; b) the length of the carbon chain in molecules.

Topic 8 hemistry of arbon ompounds Unit 29 An introduction to the chemistry of carbon compounds 2 The following table summarizes the physical properties of members of some homologous series. omologous series Intermolecular forces Physical properties omologous series aloalkanes Intermolecular forces permanent dipole-permanent dipole attractions between molecules δ+ δ 3 l δ+ δ 3 l δ+ δ 3 l permanent dipole-permanent dipole attraction polar molecules can interact with water molecules h y d r o g e n b o n d i n g b e t w e e n molecules 3 2 2 3 Physical properties boiling points higher than those of alkanes of similar relative molecular masses slightly soluble in water boiling points much higher than those of alkanes of similar relative molecular masses Aldehydes and ketones permanent dipole-permanent dipole attractions between molecules 3 3 δ+ δ 3 3 δ+ δ 3 3 δ+ δ permanent dipole-permanent dipole attraction hydrogen bonding between aldehyde / ketone molecules and water molecules 3 3 hydrogen bond boiling points higher than those of alkanes of similar relative molecular masses aldehydes and ketones with less carbon atoms show appreciable water solubility Alcohols hydrogen bond hydrogen bonding between alcohol molecules and water molecules 3 2 alcohols with less carbon atoms are miscible with water in all proportions alcohols with a long carbon chain in their molecules are much less soluble in water h y d r o g e n b o n d i n g b e t w e e n molecules; more extensive than that in alcohols 3 hydrogen bond 3 boiling points higher than those of alcohols of similar relative molecular masses 3 2 hydrogen bond arboxylic acids hydrogen bonding between acid molecules and water molecules the first four acids are miscible with water in all proportions 3 hydrogen bond

10 Topic 8 hemistry of arbon ompounds Unit 29 An introduction to the chemistry of carbon compounds 11 omologous series Intermolecular forces Physical properties omologous series Intermolecular forces Physical properties Esters permanent dipole-permanent dipole attractions between molecules hydrogen bonding between ester molecules and water molecules 3 2 3 boiling points are about the same as those of aldehydes and ketones of similar molecular masses simple esters are very soluble in water hydrogen bonding between molecules of primary amines; hydrogen bonding less strong than that in alcohols 3 2 2 3 2 3 boiling points of primary amines higher than those of alkanes but generally lower than those of alcohols of similar relative molecular masses hydrogen bond Amines hydrogen bond h y d r o g e n b o n d i n g b e t w e e n molecules boiling points are high h y d r o g e n b o n d i n g b e t w e e n primary amine molecules and water molecules primary amines with less carbon atoms are very soluble in water 3 2 3 hydrogen bond 3 Amides hydrogen bond hydrogen bonding between amide molecules and water molecules simple amides are very soluble in water Questions may ask students to arrange carbon compounds in order of increasing boiling point. 3 ( 2 ) 2 3 < 3 ( 2 ) 3 3 < 3 ( 2 ) 3 l < 3 ( 2 ) 3 Attractions weak instantaneous dipole- permanent dipole- hydrogen between induced dipole attractions permanent dipole bonds molecules attractions The following carbon compounds are miscible with water: 3 methanol, ethanol and propan-1-ol; ethanal and propanone; methanoic acid, ethanoic acid, propanoic acid and butanoic acid. hydrogen bond

12 Topic 8 hemistry of arbon ompounds Unit 30 Isomerism 13 ompounds W, X, Y and Z are all colourless liquids. Suggest how you would distinguish the four compounds from each other. 3 2 ( 3 ) 3 3 ( 2 ) 2 Br 3 ( 2 ) 7 W X Y Z (5 marks) Distinguishing the liquids by water solubility Add water to the liquids. Both 3 2 and ( 3 ) 3 are miscible with water. (1) Distinguishing the two liquids which are miscible with water Warm each of these two liquids with acidified potassium dichromate solution. (1) nly 3 2 turns the dichromate solution from orange to green. (1) Distinguishing the two liquids which are not miscible with water Warm each of the two liquids which are not miscible with water with Ag 3 (aq) in ethanol. (1) nly 3 ( 2 ) 2 Br gives a creamy precipitate slowly. (1) Unit 30 Isomerism 30.1 Isomerism 30.2 Structural isomerism 30.3 Geometrical isomerism 30.4 Physical properties of geometrical isomers 30.5 hirality 30.6 Enantiomers 30.7 Test for chirality plane of symmetry 30.8 Properties of enantiomers In questions of this type, it is common to distinguish the compounds by their water solubility before other chemical tests. ydrocarbons such as cyclohexane and cyclohexene often appear in this type of questions. They are insoluble in water.

14 Topic 8 hemistry of arbon ompounds Unit 30 Isomerism 15 30.1 30.2 The following charts show the classification of isomers. isomers different compounds that have the same molecular formula When ask about the type of isomerism, give the precise type, instead of just stating structural isomerism. 3 and 3 The type of isomerism involved is functional group isomerism. Questions often ask about methods for distinguishing between isomeric compounds. structural isomers atoms are linked in different orders structural isomers stereoisomers atoms are linked in the same way but with different spatial arrangements 3 and 3 can be distinguished by a physical method comparing their boiling points / melting points; 3 has a higher boiling point / melting point. a spectroscopic method comparing their IR spectra; 3 has a broad and chain isomers isomers with the same functional group but different carbon skeletons 3 2 2 2 3 and 3 3 2 3 position isomers isomers with the same carbon skeleton and functional group, but the position of the functional group is different 3 2 2 2 and 3 2 3 functional group isomers isomers with the same molecular formula but different functional groups 3 2 and 3 3 strong absorption at 3 230 3 670 cm 1. onsider the isomeric compounds X and Y shown below: compound X compound Y a) ame the type of isomerism involved. (1 mark) b) Which of the above compounds has a higher melting point? Explain. (3 marks) a) Position isomerism (1) b) The melting point of compound Y is higher than that of compound X. (1) nly compound X can form intramolecular hydrogen bonds. (1) ompound Y forms more intermolecular hydrogen bonds than compound X does. (1)

16 Topic 8 hemistry of arbon ompounds Unit 30 Isomerism 17 Questions often ask students to compare the melting point / volatility of isomeric compounds. ompound X forms intramolecular hydrogen bonds due to the close proximity of the group and group. onsider the melting points and boiling points of cis-1,2-dichloroethene and trans- 1,2-dichloroethene. ompound Melting point ( ) Boiling point ( ) cis-1,2-dichloroethene 80 60 hydrogen bond trans-1,2-dichloroethene 50 48 Explain why a) cis-1,2-dichloroethene has a higher boiling point; (3 marks) b) trans-1,2-dichloroethene has a higher melting point. (2 marks) 30.3 30.4 1 Stereoisomers that have a different arrangement of their atoms in space due to the restricted rotation about a carbon-carbon double bond are geometrical isomers. 2 ompounds with two different groups attached to each carbon of the double bond have two alternative structures, which are geometrical isomers. 3 3 cis-isomer and 3 3 trans-isomer 3 Geometrical isomers normally have similar chemical properties. owever, their physical properties are often quite different. a) The boiling point of a compound depends on its intermolecular attractions. (1) In a molecule of the cis isomer, the dipole moments of the two polar l bonds reinforce each other. Thus, the molecule has a net dipole moment. Thus, these molecules are held together by permanent dipole-permanent dipole attractions. (1) In a molecule of the trans isomer, the dipole moments of the two polar l bonds cancel each other. Thus, the molecule has no dipole moment. Thus, these molecules are held together by weaker instantaneous dipole-induced dipole attractions. (1) b) In addition to intermolecular attractions, the melting point of a compound depends also on the degree of compactness of molecules in the solid state. (1) The cis isomer has a lower degree of symmetry. It fits into a crystalline lattice relatively poor and thus has a lower melting point. (1) Questions often ask students to compare the melting points and boiling points of geometrical isomers. 3 3 3 3 m.p. 102 m.p. 19 3 Br 3 l and are T 3 l 3 Br geometrical isomers. They are identical molecules. ne of the carbon atoms of the double bond has two methyl groups attached to it. Their intermolecular attractions are van der Waals forces of comparable strength. The trans isomer has a higher melting point because it is more symmetrical.

18 Topic 8 hemistry of arbon ompounds Unit 30 Isomerism 19 30.5 30.8 1 A chiral molecule is defined as one that is not superposable on its mirror image. A chiral molecule and its mirror image are called a pair of enantiomers. 2 Most simple chiral molecules contain one carbon atom bonded to four different atoms or groups of atoms. Such a carbon atom is called a chiral carbon. 3 A pair of enantiomers have the same physical property and chemical property, except a) their behaviour towards plane-polarized light; and b) their reactions with chiral reagents. Questions often ask about enantiomers. Questions may ask students to mark chiral carbons on the structures of unfamiliar compounds. 2 3 Tamiflu 2 3 Aspartame * * 2 3 a) 3 (1) (1) b) They rotate the plane of polarization of a beam of plane-polarized light to opposite directions. (1) 3 Students need to give good drawings of three-dimensional structures. Use the conventions commonly used in the representation of three-dimensional structures. Questions often ask students to suggest a different physical property between a pair of enantiomers. ne of them turns the plane of polarization of a beam of plane-polarized light clockwise, while the other anticlockwise. 3 2 * * * 2 2 2 2 3 D T confuse the terms chiral and achiral. ompound X has the following structural formula: ( 3 ) The above structural formula can represent two stereoisomers. a) Draw three-dimensional structures of the two stereoisomers. (2 marks) b) State a physical property which is different for the two stereoisomers. (1 mark)

20 Topic 8 hemistry of arbon ompounds Unit 31 Typical reactions of selected functional groups 21 Unit 31 Typical reactions of selected functional groups 31.1 Introduction 31.2 Important reactions of alkanes 31.3 Addition reactions of alkenes 31.4 Addition of hydrogen to alkenes 31.5 Addition of halogens to alkenes 31.6 Addition of hydrogen halides to alkenes 31.7 Substitution reactions of haloalkanes 31.8 Reactions of alcohols 31.9 Reactions of aldehydes and ketones 31.10 Reactions of carboxylic acids 31.11 ydrolysis of esters 31.12 ydrolysis of amides 31.1 31.7 1 The figure below summarizes the addition reactions of alkenes. R 2 diol cold alkaline dilute potassium permanganate solution R 2 3 Br(g) alkane R 2 alkene 2(g) Pt catalyst Br 2 (in organic solvent) Br 2(aq) RBr 2 Br dibromoalkane R 3 R 2 Br R 2 2 Br + + Br major product minor product bromoalcohol bromoalkanes RBr 2 Br dibromoalkane 2 Markovnikov s rule for addition reaction of an asymmetric alkene: When a molecule A adds to an asymmetric alkene, the major product is the one in which the hydrogen atom attaches itself to the carbon atom already carrying the larger number of hydrogen atoms. 3 Substitution reactions of haloalkanes alkaline hydrolysis of haloalkanes R X a(aq) reflux R

22 Topic 8 hemistry of arbon ompounds Unit 31 Typical reactions of selected functional groups 23 Questions often ask students to predict the major product of an addition reaction involving an asymmetrical alkene. 3 3 3 Br Br 3 3 3 Put about 2 cm 3 of ethanol and 1 cm 3 of silver nitrate solution in each of two test tubes. (1) Place the test tube in a water bath at 60. (1) Add several drops of compounds X and Y separately to each test tube. A yellow precipitate forms rapidly in the test tube containing compound Y. (1) A white precipitate forms slowly in the test tube containing compound X. (1) 2 5 3 2 5 Br Br 2 5 2 5 3 Questions often ask about chemical tests for distinguishing haloalkanes. I I 31.8 31.9 Questions may ask students to deduce the structure of a compound based on the type of isomerism it can exhibit and its reactions. Given information: Molecular formula 6 12 It has a pair of enantiomers. It loses its chiral centre after hydrogenation. Deductions: The compound should be an alkene as it can undergo hydrogenation. The compound should have a chiral carbon. Structures of the compound: 2 = 2 5 3 3 2 5 = 2 3 7 l 1-chloropropane 3 7 Br 1-bromopropane 3 7 I 1-iodopropane reflux with conc. l + Znl 2 catalyst; or mix with Pl 5; or reflux with Sl 2 reflux with abr + conc. 2S 4; or reflux with red P + Br 2 reflux with ai + conc. 3P 4; or reflux with red P + I 2 3 2 propene 3 7 propan-1-ol (1 alcohol) 3 + conc. 2S 4 excess conc. 2S 4, 180 ; or Al 2 3, 300 3 3 7 propyl ethanoate K 2r 2 7 / 3 + distil off the propanal 1 LiAl 4 / ethoxyethane 2 3 + K 2r 2 7 / 3 +, reflux; or KMn 4 / 3 +, reflux 1 LiAl 4 / ethoxyethane 2 3 + 3 2 propanal K 2r 2 7 / 3 +, heat 3 2 propanoic acid Describe, by giving reagent(s) and stating observations, how you could distinguish between the following two compounds using a simple test tube reaction. 3 2 3 and 3 2 3 3 () 3 propan-2-ol (2 alcohol) K 2r 2 7 / 3 + reflux 1 LiAl 4 / ethoxyethane 2 3 + 3 3 propanone l compound X I compound Y (4 marks) ( 3 ) 3 methylpropan-2-ol (3 alcohol) K 2r 2 7 / 3 + reflux no reaction

24 Topic 8 hemistry of arbon ompounds Unit 31 Typical reactions of selected functional groups 25 The formation of hydrogen chloride fumes upon reaction with phosphorus pentachloride is a test for the presence of a hydroxyl group in a compound. The relative ease with which alcohols undergo dehydration shows the following order: Ease of dehydration: 3 > 2 > 1 alcohol Questions often ask about the oxidation of alcohols: the oxidizing agent required; the name(s) of the product(s). Remember that LiAl 4 will T act upon = bonds. The transformation of to 2 involves 2 steps: reduction by LiAl 4, followed by treatment with 3 +. Describe, by giving reagent(s) and stating observations, how you could distinguish between the following two compounds using a simple test tube reaction. 3 compound X and 2 compound Y (2 marks) 31.10 31.12 3 2 propanoic acid + 3 2 ethanol 3 2 2 3 ethyl propanoate 2 / 3 + a solution 3 2 a + sodium propanoate + 3 2 ethanol 3 2 + conc. 2S 4 1 LiAl 4 / ethoxyethane 2 3 + 3 2 2 propan-1-ol 3 2 propanoic acid 3 2 propanoic acid 2 / 3 + aqueous 3 3 2 + 4 ammonium propanoate heat 3 2 2 propanamide a solution 3 2 a + sodium propanoate Warm each compound with acidified potassium dichromate solution. (1) nly compound Y turns the dichromate solution from orange to green. (1) The oxidation reaction is commonly used to distinguish a primary / secondary alcohol from a tertiary alcohol; an aldehyde from a ketone. Test ompound Warm with K 2 r 2 7 / 3 + Alcohols Aldehydes Ketones 1 and 2 alcohols clear orange solution turns green almost immediately 3 alcohols no observable change clear orange solution turns green no observable change When propan-1-ol and propanoic acid are heated under reflux to produce an ester, the ester can be separated from the reaction mixture by fractional distillation or using a separating funnel. Given the structure of an ester, students should be able to analyze the ester and deduce the alcohol and carboxylic acid forming the ester. The active ingredient of a superglue has the following structure: It is an ester formed from 3 and 3.

26 Topic 8 hemistry of arbon ompounds Unit 32 Synthesis of carbon compounds 27 Questions often give carbon compounds with similar structures (such Unit 32 Synthesis of carbon compounds as 3 and 3 ) and compare them. Whether they have the same molecular formula / relative molecular mass. Whether they are soluble in water. Whether they have the same odour. Whether they have the same boiling point. Whether they have the same chemical properties. To tackle these questions, first identify the homologous series to which each compound belongs. Then answer according to the general properties of the series concerned. 32.1 Introduction 32.2 Planning a synthesis 32.3 Problems in devising a synthesis 32.4 Laboratory preparation of simple carbon compounds 32.5 Preparing 1-bromobutane in the laboratory il of wintergreen is a common ester. Salicylic acid can be obtained from it in two steps. 3 oil of wintergreen Step 1 a sodium salt of salicylic acid + 3 Step 2 salicylic acid a) Give the reagent and condition used in Step 1. (2 marks) b) Suggest a structure for the sodium salt of salicylic acid formed in Step 1. (1 mark) c) Suggest a reagent that can be used in Step 2. (1 mark) a) Sodium hydroxide solution (1) eat under reflux (1) b) a + (1) c) Dilute sulphuric acid / dilute hydrochloric acid (1) To draw the structure of sodium salt of salicyclic acid in (b), just use the general equation for the hydrolysis of an ester for deduction. R + a R a + + R 1 R 1 Students may also work backwards from the structure of salicyclic acid to deduce the structure of its sodium salt.

28 Topic 8 hemistry of arbon ompounds Unit 32 Synthesis of carbon compounds 29 32.1 32.3 c) l (aq) Steps for devising a synthesis: heat (1) (0.5) a) identify an immediate precursor to the target molecule; b) continue until the starting molecule is reached. target molecule 1st precursor 2nd precurso starting molecule conc. 2S 4 heat or conc. 3P 4 heat (1) (0.5) Br 2 (in organic solvent) (1) Br Br The syntheses discussed would T involve a change in the length of the carbon chain. Students should be familiar with the reactions of organic functional groups in order to suggest workable synthetic routes for the transformations. In part (a), the immediate precursor of the target molecule (an alcohol) may be a carboxylic acid or a haloalkane. As a carboxylic acid can be obtained from the starting molecule (an amide) via hydrolysis, so the synthesis can be done in the two steps shown. Many conversions involve utline a syntheticc route, in T MRE TA TREE STEPS, to accomplish each of the following conversions. For each step, give the reagent(s), the conditions and the structure of the product. a) 3 2 2 2 3 2 2 2 (3 marks) b) 2 3 3 (4 marks) 32.4 32.5 the hydrolysis of haloalkanes to alcohols; the oxidation of alcohols to ketones and carboxylic acids; the reduction of aldehydes, ketones and carboxylic acids to alcohols. The laboratory preparation of a carbon compound involves five main stages: c) l Br Stage 1 planning the preparation; Br (4 marks) Stage 2 carrying out the reaction to produce the desired product; Stage 3 separating the crude product from the reaction mixture; a) 3 2 2 2 2 / 3 + heat (1) 3 2 2 (1) Stage 4 purifying and drying the product; Stage 5 calculating the percentage yield of the product. 1 LiAl 4 / ethoxyethane 2 3 + (1) 3 2 2 2 ommon separation and purification methods for products b) 2 3 Br 3 Type of product Separation and purification method to employ simple distillation Br 2 Liquid product fractional distillation UV light or heat liquid-liquid extraction (1) (0.5) Solid product re-crystallization () 3 3 (aq) heat (1) (0.5) K 2r 2 7 / 3 + reflux (1)

30 Topic 8 hemistry of arbon ompounds Unit 32 Synthesis of carbon compounds 31 D T confuse an experimental set-up for fractional distillation with that for simple distillation. Questions often ask students to describe the re-crystallization procedure for the purification of a crude solid product. a) 2-chloro-2-methylpropane can be prepared by reacting methylpropan-2-ol with concentrated hydrochloric acid. ( 3 ) 3 (l) + l(aq) ( 3 ) 3 l(l) + 2 (l) This preparation follows the steps outlined below: Step 1 Step 2 Step 3 Step 4 Step 5 Shake excess concentrated hydrochloric acid with methylpropan-2-ol in a separating funnel. Separate the organic layer from the aqueous layer. Wash the crude product with 10% sodium hydrogencarbonate solution. Dry the product with a suitable reagent. Purify the dried product to remove the remaining methylpropan-2-ol. The table below lists some information of methylpropan-2-ol and 2-chloro-2- methylpropane: ompound methylpropan-2-ol 2-chloro-2-methylpropane Density (g cm 3 ) 0.78 0.84 Boiling point ( ) 82 51 Water solubility miscible very slightly soluble a) Draw a diagram of the separating funnel after the reaction has taken place in Step 1, labelling clearly the aqueous layer and organic layer. (2 marks) b) Suggest E advantage of using a separating funnel to carry out Step 1. (1 mark) c) utline the experimental procedure for washing the crude product in Step 3. Include the necessary safety precautions. (4 marks) d) Suggest a suitable reagent for drying in Step 4. (1 mark) e) ame a method for purifying the dried product in Step 5. (1 mark) organic layer aqueous layer b) Any one of the following: Allow better mixing of the reactants by vigorous shaking of the separating funnel. (1) Allow easy isolation of the product by draining out the lower aqueous layer from the separating funnel. (1) c) Add slowly 10% sodium hydrogencarbonate solution. Wait until no effervescence occurs. (1) Stopper and shake the separating funnel vigorously. (1) During shaking, open the tap of the separating funnel regularly to release the pressure built inside the funnel. (1) Remove the aqueous layer. Repeat the washing procedure until no effervescence occurs upon the addition of sodium hydrogencarbonate solution. (1) d) Anhydrous calcium chloride / sodium sulphate (1) e) Simple distillation / fractional distillation (1) Questions often ask about the purposes of different steps in the preparations of carbon compounds. When giving a reagent for drying, remember to include the word anhydrous. (2)

32 Topic 8 hemistry of arbon ompounds Unit 33 Important organic substances 33 Unit 33 Important organic substances 33.1 Introduction 33.2 Aspirin from herbal remedy to modern drug 33.3 Analyzing aspirin tablets by back titration 33.4 Detergents 33.5 ow do detergents help water to clean? 33.6 The wetting and emulsifying properties of detergents in relation to their structures 33.7 The cleaning action of detergents 33.8 Manufacture of soaps and soapless detergents 33.9 The cleaning abilities of soaps and soapless detergents in hard water 33.10 ylons 33.11 Polyesters 33.12 arbohydrates 33.13 Lipids 33.14 Proteins and polypeptides 33.1 33.9 & 33.13 1 a) The structure of the active ingredient of aspirin tablets, acetylsalicylic acid, is shown below. It contains two functional groups. 3 carboxyl group ester group b) The percentage by mass of acetylsalicylic acid in aspirin tablets can be determined by back titration. 2 a) There are two types of detergents: i) soap made from natural fats and oils; and ii) soapless detergents made from chemicals derived from petroleum. b) The structure of a typical anionic detergent is shown below: hydrophobic hydrocarbon tail hydrophilic anionic head c) A detergent helps water to remove dirt by i) the ability to act as a wetting agent; and ii) the emulsifying action. 3 Fats and oils are mixed triglycerides. 4 Soap is made by boiling natural oils or fats with a strong alkali, usually sodium hydroxide or potassium hydroxide solution.

34 Topic 8 hemistry of arbon ompounds Unit 33 Important organic substances 35 Students should be able to draw the correct structure of the ester linkage in a triglyceride (see the structure shown below). The three carboxylic acids bearing long chains are attached to a glycerol backbone. When fats and oils are heated with sodium hydroxide solution, they are hydrolyzed first to form glycerol and carboxylic acids. The acids then react with the alkali to form sodium carboxylates, which are soaps. Such reactions are called saponification. R a) Animal fat is one of the raw materials in the production of detergent X. i) ame another raw material that is needed in the production of detergent X from an animal fat. (1 mark) ii) ame the type of reaction that takes place in the production of detergent X. (1 mark) iii) Write a chemical equation for the reaction involved in the preparation. (1 mark) b) An oil tanker was wrecked and spilt a lot of crude oil in the sea. State whether detergent X is suitable for treating the oil spill. Explain your answer. (3 marks) triglyceride in fat or oil Questions often give the structure of a soap / soapless detergent and ask about its properties. Whether it is made from natural fats and oil or from hydrocarbons obtained from petroleum. Whether it forms scum in hard water / sea water. Whether it is biodegradable. ydrogenation of vegetable oils produces margarines. The structure of the main chemical constituent of an animal fat is shown below: R R + 3a alkali ( 2 ) n 3 ( 2 ) n 3 glycerol + 3R a + sodium carboxylate (soap) a) i) Sodium hydroxide (1) ii) Saponification / alkaline hydrolysis (1) iii) ( 2 ) n 3 ( 2 ) n 3 ( 2 ) n 3 + 3a + 3 3 ( 2 ) n a + b) ot suitable (1) Sea water contains a lot of metal ions, such as calcium ions and magnesium ions. (1) Detergent X will react with these metal ions to form scum. (1) (1) ( 2 ) n 3

36 Topic 8 hemistry of arbon ompounds Unit 33 Important organic substances 37 Questions often ask about the purpose of adding concentrated sodium chloride solution in soap preparation. oncentrated sodium chloride solution is added to salt out the soap produced. Questions often ask students to give the products formed from the complete hydrolysis of triglycerides existing in natural fats and oils. 2 ( 2 ) 16 3 2 ( 2 ) 7 ( 2 ) 16 3 ( 2 ) 7 3 complete hydrolysis 2 2 c) The repeating unit of poly(ethylene terephthalate) is from dicarboxylic acid 2 2 from diol 3 a) Glucose can exist in open-chain and cyclic forms. b) The open-chain form of glucose contains an aldehyde group and five hydroxyl groups. 4 a) Fructose can exist in open-chain and cyclic forms. b) The open-chain form of fructose contains a keto group and five hydroxyl groups. 5 a) When two amino acid molecules undergo condensation reaction, a water molecule is eliminated and a peptide link forms. The product is a dipeptide. R R 1 + 3 ( 2 ) 16 + 3 ( 2 ) 7 ( 2 ) 7 + amino acid 1 amino acid 2 + 3 ( 2 ) 7 ( 2 ) 7 peptide link (or amide group) R R 1 33.10 33.12 & 33.14 + 2 a dipeptide 1 a) ylons are polyamides which contain the amide linkage b) ylons are formed by condensation polymerization.. b) The peptide links in peptides and proteins can be hydrolyzed to release the individual amino acids. c) The repeating unit of nylon-6,6 is ( 2 ) 6 ( 2 ) 4 from diamine from dicarboxylic acid 2 a) Polyesters contain the ester linkage. b) Polyesters are formed by condensation polymerization. A condensation reaction is a reaction in which two or more molecules react together to form a larger molecule with the elimination of a small molecule such as water. ylon and poly(ethylene terephthalate) are thermoplastics as well as condensation polymers. T all thermoplastics are addition polymers. Questions often give the repeating unit of a polymer and ask about information concerning the polymer. whether it is an addition polymer or a condensation polymer; whether it is formed from one monomer or two different monomers.

38 Topic 8 hemistry of arbon ompounds Unit 33 Important organic substances 39 Twaron is a heat resistant, high strength fibre used in protective clothing. A short section of the structure of Twaron is shown below. a) Draw the repeating unit of Twaron. (1 mark) b) Draw the structures of the two monomers that could be used to prepare Twaron. (2 marks) c) Explain why Twaron has a great strength. (1 mark) Given the structure of a polymer, students should be able to deduce the structure(s) of monomer(s) used to produce the polymer. Look at the polymer structure. If the polymer backbone contains only carbon atoms, then the polymer is an addition polymer produced from one monomer. 6 5 6 5 from the following monomer: 6 5 is an addition polymer formed 6 5 a) b) 2 2 (1) (1) c) There are strong hydrogen bonds between chains of Twaron. (1) (1) If the polymer backbone contains other atoms, then the polymer is a condensation polymer, probably produced from two monomers (see Twaron shown above). Some condensation polymers are produced from one monomer. the repeating unit of polylactide is shown below: 3 It is a condensation polymer formed from one monomer, lactic acid. 3 lactic acid