Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 8 The following ions are Bronsted-Lowry bases. C2H5O– I CH3CCO– II C6H5O– III C6H5CCO– IV Which is the correct sequence of the species according to the ascending order of their pKb values? A I, II, III, IV C I, IV, III, II B IV, II, III, I D I, III, II, IV 9 The structural formula of cyanobenzene is shown below: C # N What is the type of hybridization formed by the nitrogen atom in the molecule? A None C sp2 B sp D sp3 10 What type of stereoisomerism (if any) is shown by the following molecule? CH3 (CH2 )5 (CH2 )7 COOH H H C"C A Geometrical B Optical C Geometrical and optical D None 11 2-Fluroethanoic acid is a stronger acid than 2-chloroethanoic acid because A the C—F bond is stronger B the C—F bond is more polar C F is more electronegative than Cl D the 2-fluroethanoate ion can be stabilised through resonance but the 2-chloroethanoate ion cannot 12 The structure of molecule X is shown below: O O H3C H3C H2C H3C CH3 CH3 H H H O How many chiral carbon atoms are there is this molecule? A 3 C 5 B 4 D 6 Structured and Essay Questions 1 (a) Explain the term stereoisomerism. (b) What are the two main types of stereoisomerisms? (c) Draw all the possible stereoisomers for the following compounds. CH3!CH!CHRCH!CH2Br & OH (d) Which of the two isomers will have the same boiling point? 2 (a) Give the formulae of the three structural isomers of an alkene with the molecular formula of C4H8. (b) One of the above isomers shows a type of stereoisomerism. (i) Draw the structure of the stereoisomers and name them. (ii) Suggest how these stereoisomers can be distinguished. 3 (a) Using suitable examples, explain what is meant by (i) free radical (ii) nucleophile (iii) electrophile (b) The relative stability of free radicals increases in the order: primary < secondary < tertiary. Explain the trend. 42
CHAPTER HYDROCARBONS 15 Concept Map Alkanes Alkenes Cycloalkanes Physical Properties of Alkanes Nomenclature • Straight-chain • Branched-chain Nomenclature • Straight-chain • Branched-chain Cycloalkenes • Cycloalkenes • Dienes, trienes and polyenes Reactions of Alkanes • Halogenation • Combustion Reactions of Alkenes • Electrophilic addition • Oxidation Air Pollution • Reducing pollution Structural and cis-trans Isomerism Physcial Properties of Alkenes Importance of Ethene in Industry Polymerisation of Alkenes Natural Source of Hydrocarbon – Crude Oil • Fractional distillation • Cracking • Reforming Arenes Reactions of Benzene • Nitration • Acylation • Substituent effect • Sulphonation • Halogenation • Hydrogenation • Alkylation • Oxidation Reactions of Alkylbenzene • Involving the benzene ring • Involving the side chain Physical Properties of Benzene Alkylbenzene Nomenclature Structural Isomerism in Arenes Uses of Arenes Benzene
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 Learning earning Outcomes Students should be able to: Alkanes • write the general formula for alkanes; • explain the construction of the alkane series (straight and branched), and IUPAC nomenclature of alkanes for C1 to C10; • describe the structural isomerism in aliphatic alkanes and cis-trans isomerism in cycloalkanes; • state the physical properties of alkanes; • define alkanes as saturated aliphatic hydrocarbons; • name alkyl groups derived from alkanes and identify primary, secondary, tertiary and quatenary carbons; • explain the inertness of alkanes towards polar reagents; • describe the mechanism of free radical substitution as exemplified by the chlorination of methane (with particular reference to the initiation, propagation and termination reactions); • describe the oxidation of alkanes with limited and excess oxygen, and the use of alkanes as fuels; • explain the use of crude oil as a source of aliphatic hydrocarbons; • explain how cracking reactions can be used to obtain alkanes and alkenes of lower Mr from larger hydrocarbon molecules; • discuss the role of catalytic converters in minimising air pollution by oxidising CO to CO2 and reducing NOx to N2; • explain how chemical pollutants from the combustion of hydrocarbon affect air quality and rainwater as exemplified by acid rain, photochemical smog and greenhouse effect. Alkenes • write the general formula for alkenes; • name alkenes according to the IUPAC nomenclature and their common names for C1 to C5; • describe structural and cis-trans isomerism in alkenes; • state the physical properties of alkenes; • define alkenes as unsaturated aliphatic hydrocarbons with one or more double bonds; • describe the chemical reactions of alkenes as exemplified by the following reactions of ethene: (i) addition of hydrogen, steam, hydrogen halides, halogens, bromine water and concentrated sulphuric acid, (ii) oxidation using KMnO4, O2/Ag, (iii) ozonolysis, (iv) polymerisation; • describe the mechanism of electrophilic addition in alkenes with reference to Markovnikov’s rule; • explain the use of bromination reaction and decolourisation of MnO4 − ion as simple tests for alkenes and unsaturated compounds; • explain briefly the importance of ethene as a source for the preparation of chloroethane, epoxyethane, ethane-1,2-diol and poly(ethene). Arenes • name aromatic compounds derived from benzene according to the IUPAC nomenclature, including the use of ortho, meta and para or the numbering of substituted groups to the benzene ring; • describe the structural isomerism in arenes; • describe the chemical reactions of arenes as exemplified by substitution reactions of haloalkanes and acyl chloride (Friedel-Crafts reaction), halogen, conc. HNO3/conc. H2SO4 and SO3 with benzene and methylbenzene (toluene); • describe the mechanism of electrophilic substitution in arenes as exemplified by the nitration of benzene; • explain why benzene is more stable than aliphatic alkenes towards oxidation; • describe the reaction between alkylbenzene and hot acidified KMnO4; • determine the products of halogenation of methylbenzene (toluene) in the presence of (i) Lewis acid catalysts, (ii) light; • explain the inductive effect and resonance effect of substituted groups (−OH, −Cl, −CH3, −NO2, −COCH3, −NH2) attached to the benzene ring towards further substitutions; • predict the products in an electrophilic substitution reaction when the substituted group in benzene is electron accepting or electron donating; • explain the uses of arenes as solvents; • recognise arenes as carcinogen. 44
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 15.1 Alkanes 1 Alkanes are saturated hydrocarbons. 2 A hydrocarbon is a compound that contains only carbon and hydrogen atoms in its structure. 3 A saturated hydrocarbon contains only single bonds in its molecule. 4 Alkanes are also called saturated aliphatic hydrocarbons because higher members of the series have physical properties similar to those compounds found in animal fats and plant oils (from the Greek word, aleiphar, meaning oil or fat). 5 The general formula of alkanes is CnH2n+2 (where n ≥ 1). 6 All the carbon atoms in alkane molecules undergo sp3 hybridisation. 7 All the bond angles in alkane molecules are ≈109.5°. 8 Examples of alkanes are petrol, kerosene, paraffin wax and domestic cooking gas. 9 The first member of the alkane series is methane, CH4. It has the following structure: H H H H C Nomenclature of Straight-chain Alkanes 1 Alkanes can be divided into linear-chain alkanes and branchedchain alkanes. 2 Examples of linear chain and branched chain alkanes are CH3!CH2!CH2!CH2!CH3 (Linear-chain alkane) CH3!CH!CH2!CH3 & CH3 (Branched-chain alkane) 3 The prefixes that correspond to the number of carbon atoms in a straight chain molecule are given below: Number of carbon atom 1 2 3 4 5 6 7 8 9 10 Prefix Meth Eth Prop But Pen Hex Hep Oct Non Dec 45
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 4 The names for the first twenty linear chain alkanes are given in the table below. Molecular formula Structural formula Name CH4 CH4 Methane C2H6 CH3CH3 Ethane C3H8 CH3CH2CH3 Propane C4H10 CH3CH2CH2CH3 Butane C5H12 CH3CH2CH2CH2CH3 Pentane C6H14 CH3CH2CH2CH2CH2CH3 Hexane C7H16 CH3CH2CH2CH2CH2CH2CH3 Heptane C8H18 CH3CH2CH2CH2CH2CH2CH2CH3 Octane C9H20 CH3CH2CH2CH2CH2CH2CH2CH2CH3 Nonane C10H22 CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3 Decane C11H24 CH3(CH2)9CH3 Undecane C12H26 CH3(CH2)10CH3 Dodecane C13H28 CH3(CH2)11CH3 Tridecane C14H30 CH3(CH2)12CH3 Tetradecane C15H32 CH3(CH2)13CH3 Pentadecane C16H34 CH3(CH2)14CH3 Hexadecane C17H36 CH3(CH2)15CH3 Heptadecane C18H38 CH3(CH2)16CH3 Octadecane C19H40 CH3(CH2)17CH3 Nonadecane C20H42 CH3(CH2)18CH3 Eicosane (Notice that all their names end with -ane.) 5 Alkyl radicals are formed when one hydrogen atom from an alkane molecule is removed. For example, the alkyl radical for methane, CH4 has the formula of CH3. 6 The table below lists the IUPAC names of the first ten straight chain alkyl groups. Alkyl group IUPAC name CH3! Methyl CH3CH2! Ethyl CH3CH2CH2! Propyl CH3CH2CH2CH2! Butyl CH3CH2CH2CH2CH2! Pentyl CH3CH2CH2CH2CH2CH2! Hexyl CH3CH2CH2CH2CH2CH2CH2! Heptyl CH3CH2CH2CH2CH2CH2CH2CH2! Octyl CH3CH2CH2CH2CH2CH2CH2CH2CH2! Nonyl CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2! Decyl 46
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 7 The table below lists the IUPAC and common names of some branched chain alkyl groups. Alkyl group IUPAC name Common name CH3CH! & CH3 1-methylethyl Isopropyl CH3CH2CH2CH2! Butyl n-butyl CH3CHCH2CH3 & 1-methylpropyl Sec-butyl CH3 & CH3!C! & CH3 1,1-dimethylethyl Tert-butyl CH3!CH!CH2! & CH3 2-methylpropyl Isobutyl Nomenclature of Branched-chain Alkanes 1 The first three members of the alkanes can form only a linear chain. CH4 Methane CH3CH3 Ethane CH3CH2CH3 Propane 2 Branching begins with alkanes with four or more carbon atoms. For example, there are two structures possible for the hydrocarbon with the molecular formula of C4H10. One, a linear molecule and the other, a branched molecule. CH3CH2CH2CH3 and CH3CHCH3 & CH3 They are structural isomers. 3 The number of isomers increases with the number of carbon atoms in the molecule. For example, C6H14 has 5 isomers, C10H22 has 75 isomers and C30H62 has 4 111 846 763 isomers (that’s more than 4 billion isomers). 4 The IUPAC name for branched chain alkanes consists of a parent (root) name which indicates the longest carbon chain in the molecule. 5 Each substituent (the alkyl groups) is given a name and a number. The number indicates the carbon atom in the parent chain to which the substituent is bonded. 6 If more than one number can be assigned to the substituent, the lowest number would be adopted. 7 For example, the name of the following alkane: 1 2 3 4 5 6 (Wrong) 6 CH3!5 CH2!4 CH2!3 CH!2 CH2!1 CH3 (Correct) & CH3 is 3-methylhexane and not 4-methylhexane. 47
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 8 Sometimes, the longest chain is not written in a linear form. Take the following example: 1 CH3!2 CH!CH3 & 3 CH2!4 CH2!5 CH2!6 CH!CH2!CH3 & 7 CH2!8 CH2!9 CH2!10CH3 The longest chain (indicated by numbers) contains 10 carbon atoms, though they are not arranged in a linear form. The root/parent name is decane. The substituents are methyl (at carbon-2) and ethyl (at carbon-6). Hence, the IUPAC name is 2-methyl-6-ethyldecane. 9 The above structure can be redrawn as 1 2 3 4 5 6 7 8 9 10 CH3!CH!CH2!CH2!CH2!CH!CH2!CH2!CH2!CH3 & & CH3 CH2CH3 Cycloalkanes 1 There is another class of saturated hydrocarbons with the general formula of CnH2n (where n ≥ 3). 2 They are cyclic compounds. They are called cycloalkanes. 3 The IUPAC name for cycloalkane is obtained by adding prefix cyclo- to the name of the alkane with the same number of carbon atoms. 4 Examples of cycloalkanes are: Formula Structural formula Skeletal formula Name C3H6 H2C CH2 CH2 H2C CH2 CH2 Cyclopropane C4H8 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 Cyclobutane C5H10 CH2 CH2 CH2 H2C CH2 CH2 CH2 CH2 H2C CH2 Cyclopentane C6H12 CH2 CH2 H2C CH2 CH2 CH2 CH2 CH2 H2C CH2 CH2 CH2 Cyclohexane The skeletal structure is Info Chem Cycloalkanes have the same general formula as alkenes. 48
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 Quick Check 15.1 1 Draw all possible isomers for an alkane with the molecular formula of C6H14. 2 How many primary, secondary, tertiary and quaternary carbon and hydrogen atoms are there in 2,2-dimethylpropane? Structural Isomerism in Alkanes and cis-trans Isomerism in Cycloalkanes 1 Alkanes with four or more carbon atoms exhibit structural isomerism as shown below: (a) C4H10 CH3CH2CH2CH3 Butane CH3—CH—CH3 2-methylpropane & CH3 (b) C5H12 CH3CH2CH2CH2CH3 Pentane CH3CH2—CH—CH3 2-methylbutane & CH3 CH3 & CH3—C—CH3 2,2-dimethylpropane & CH3 2 The number of structural isomers increases as the number of carbon atoms increases. Alkane C6H14 C7H16 C15H32 Number of isomers 5 9 4347 3 Certain cycloalkanes exhibit geometrical (cis-trans) isomerism. An example is 1,2-dimethylcyclohexane. CH3 H CH3 H CH3 H H cis-1,2-dimethylcyclohexane trans-1,2-dimethylcyclohexane CH3 2015/P3/Q18(a)(i) 49
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 Physical Properties of Alkanes Boiling and Melting Points 1 The melting and boiling points of some linear chain alkanes are given in the table below. Alkane Melting point/oC Boiling point/oC Physical state CH4 –182 –164 Gas C2H6 –183 –88 Gas C3H8 –190 –42 Gas C4H10 –138 0 Gas C5H12 –130 36 Liquid C6H14 –95 69 Liquid C7H16 –90 98 Liquid C8H18 –57 126 Liquid C9H20 –51 151 Liquid C10H22 –30 174 Liquid 2 The boiling and melting points are measures of the strength of the attractive forces that hold the molecules together. 3 Alkanes are non-polar compounds. The intermolecular force is the weak van der Waals force. Little energy is required to break or weaken the van der Waals forces. As a result, they have relatively low melting and boiling points. 4 The first four alkanes are gases. Alkanes from C5H12 to C17H36 are liquids. Alkanes from C18H38 onwards are solids. 5 Generally, the melting and boiling points increase with the number of carbon atoms in the molecule. This is because the strength of the van der Waals force increases with the size and total number of electrons in the molecule. 6 The boiling points of the three structural isomers for the alkane with a molecular formula of C5H12 are given in the table below. Alkane Structural formula Boiling point/oC Pentane CH3CH2CH2CH2CH3 36 2-methylbutane CH3CHCH2CH3 & CH3 28 2,2-dimethylpropane CH3 & CH3!C!CH3 & CH3 10 7 Branched chain alkanes are more spherical and are smaller in size compared to their linear counterparts. Thus, their boiling points are lower. 8 The size of the molecule decreases further with increasing branching, causing a further decrease in the boiling/melting point. Pentane: 2-methylbutane: 2,2-dimethylpropane: 50
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 Quick Check 15.2 1 Arrange the alkanes in each of the following sets in order of increasing boiling points. (a) Butane, nonane and hexane (b) Octane, 2-methylheptane and 3,3-dimethylhexane Density and Solubility in Water 1 All alkanes are less dense than water. The density of alkanes increases as the molecular mass increases. Alkane Pentane Hexane Octane Nonane Decane Molecular formula C5H12 C6H14 C8H18 C9H20 C10H22 Density/g cm–3 0.626 0.659 0.703 0.718 0.730 2 Generally, branched chain isomers have lower densities than linear chain isomers. This is because branched chain isomers are less compactly packed compared to the linear chain isomers. For example: Alkane Density/g cm–3 Hexane 0.659 2,2-dimethylbutane 0.649 3 Being non-polar, alkanes are insoluble in water but are soluble in non-polar solvents such as benzene and tetrachloromethane. Reactions of Alkanes Inertness of Alkanes 1 Generally, alkanes and cycloalkanes are unreactive towards most reagents (especially polar reagents). 2 This is to be expected because they are non-polar (the difference in the electronegativity between carbon and hydrogen is very small), and their molecules contain only strong sigma bonds. 3 Alkanes and cycloalkanes are not affected by acids, alkalis, oxidising agents and reducing agents. 4 However, under suitable conditions, alkanes react with oxygen and halogens (chlorine and bromine). Halogenation 1 Alkanes and cycloalkanes do not react with halogen in the dark. 2 However, if a mixture of alkane and halogen is exposed to ultraviolet light, the hydrogen atoms in the alkane molecule will be substituted by halogen atoms to form haloalkanes. The general equation is UV light R!H + X2 !!!!: R!X + HX (where R = alkyl group; CnH2n+1) All alkanes are less dense than water. The density of alkanes increases with molecular mass. Branched-chain isomers are less dense than linear-chain isomers. Hexane: 2,2-dimethylbutane: The C!H bond is strong and not polarised. Substitution reaction 2008/P1/Q29 2012/P1/Q30 2016/P3/Q18(a)(b)(c) 2017/P3/Q18(a) 2015/P3/Q5 51
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 3 Such a reaction is called a photochemical reaction, since light is needed before the reaction can take place. It proceeds via a free radical mechanism. 4 When a mixture of chlorine gas (yellowish-green in colour) and methane is exposed to UV light, the colour of chlorine slowly fades and white fumes of hydrogen chloride are liberated. Chloromethane is produced. CH4(g) + Cl2(g) !!: CH3Cl(g) + HCl(g) 5 In the presence of excess chlorine, further substitution occurs with the formation of dichloromethane, trichloromethane and tetrachloromethane. CH3Cl + Cl2 !!: CH2Cl2 + HCl Dichloromethane CH2Cl2 + HCl !!: CHCl3 + HCl Trichloromethane (chloroform) CHCl3 + Cl2 !!: CCl4 + HCl Tetrachloromethane The extent of polysubstitution depends on the relative amount of methane and chlorine present. 6 Reaction with bromine proceeds at a slower rate. The reddishbrown colour of bromine is decolourised and white fumes of hydrogen bromide are liberated. For example, the reaction between cyclohexane produces bromocyclohexane. H(l) + Br2(l) !: Br(l) + HBr(g) Cyclohexane Bromocyclohexane Mechanism of Free Radical Substitution 1 The substitution reaction between methane and chlorine proceeds via free radicals. 2 The mechanism is divided into three stages. (a) Initiation In the presence of ultraviolet light, the Cl!Cl bond undergoes homolytic fission to form chlorine free radicals. uv light Cl!!Cl !!!!: 2Cl• ∆H° = +242 kJ (b) Propagation The chlorine-free radical is a very reactive species, it abstracts an electron from the C!H bond in the methane molecule producing hydrogen chloride and a methyl free radical. Cl• + H!!CH3 !!: H!Cl + • CH3 The methyl free radical reacts with another molecule of chlorine to regenerate chlorine free radical and chloromethane is formed. Info Chem Light is not a catalyst in the reaction because it is not regenerated at the end of the reaction. Polysubstitution Tetrachloromethane is also known as carbon tetrachloride. Info Chem To get a good yield of monosubstituted haloalkane, the alkane must be in excess. Reaction with bromine is slower. Info Chem Every steps produce a new free radical. 2017/P3/Q18(a)(ii) 52
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 Cl!!Cl + • CH3 !!: CH3!Cl + Cl• The chlorine free radical produced then attacks another CH4 molecule and the whole process repeats itself. These two steps constitute a chain reaction. Each chlorine free radical can produce about 10 000 molecules of chloromethane before it is consumed in a chain termination step. (c) Termination The chain reaction terminates when two free radicals combined to form a molecule and no free radical is generated. The possible termination steps are Cl• + Cl• !!: Cl2 Cl• + • CH3 !!: CH3Cl H3C• + • CH3 !!: CH3!CH3 [Note: Some ethane is produced by the chain termination step.] 3 In the absence of UV light, free radical substitution can also occur if the mixture is heated to a high temperature. The high temperature will cause cleavage of the Cl!Cl bond to produce free radicals. Cl2 !!Δ !: 2Cl• 4 Sometimes, a free radical initiator such as benzoyl peroxide is used. When heated, benzoyl peroxide dissociates to form phenyl free radicals. C6H5!C!O!O!C!C6H5 !!Δ !: 2 • C6H5 + 2CO2 ∫ ∫ O O The benzoyl-free radical (represented as R) then attacks a chlorine molecule to produce the chlorine free radical. • R + Cl!Cl !!: R!Cl + Cl• 5 In the monochlorination of an unsymmetrical alkane such as propane, two possible products are formed. They are 1-chloropropane and 2-chloropropane. CH3CH2CH3 + Cl2 !!: CH3CH2CH2Cl + HCl 1-chloropropane (Minor product) CH3CH2CH3 + Cl2 !!: CH3CHCH3 + HCl & Cl 2-chloropropane (Main product) Each propagation step is made up of 2 distinct steps No free radical is generated. Thermal cleavage of Cl!Cl bond Note: Another free radical initiator is tetraethyllead (C2H5)4Pb → 4 • C2H5 + Pb 53
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 6 However, 2-chloropropane is the major product, while 1-chloropropane is the minor product. This is explained by the mechanism below. (a) The first step is the homolytic fission of the chlorine molecule to produce chlorine free radicals. UV light Cl!Cl !!!!: 2Cl• (b) The chlorine free radical then attacks the propane molecule in two possible ways. CH3CH2CH3 + Cl• !!: CH3CH2CH2 • + HCl (I) CH3CHCH3 + Cl• !!: CH3! • CH!CH3 + HCl (II) & H (c) Reaction (II) is preferred because it produces a secondary free radical which is more stable than the primary free radical from reaction (I). (d) The secondary free radical then reacts with chlorine to produce 2-chloropropane. CH3 CH3 & & CH3!C!CH3 + Cl2 !!: CH3!C!CH3 + Cl• • & Cl Example 15.1 Predict the major monosubstituted products formed when the following alkanes react with bromine in the presence of ultraviolet light. (a) Butane (b) 2-methylpropane Solution (a) CH3!CH!CH2!CH3 (b) & Cl CH3 & CH3!C!CH3 & Cl Secondary free radical is more stable than primary free radical. Quick Check 15.3 1 With excess chlorine, ethane reacts with chlorine in the presence of ultraviolet light to produce dichloroethane. (a) Draw the structures of all the possible isomers of dichloroethane. (b) Which isomer is expected to have the highest boiling point? Explain your answer. 2 2-methylpropane and chlorine react in the presence of UV light to produce 2-chloro-2-methylpropane. (a) Draw the structural formula of 2-chloro-2-methylpropane. (b) With the aid of balanced equations, describe the mechanism of the reaction. 54
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 Oxidation (or Combustion) 1 Alkanes do not react with oxygen at room temperature. However, they burn in excess oxygen when ignited with a naked flame or a spark to produce carbon dioxide and water. The general equation for the combustion is CxHy + (x + y —4 )O2 !!: xCO2 + y —2 H2O 2 This reaction is highly exothermic. As a result, alkanes such as natural gas (methane, CH4), liquefied petroleum gas (LPG), diesel and gasoline are used extensively as fuels for energy and power. 3 The heat of combustion of some alkanes is given below. CH4 + 2O2 !: CO2 + 2H2O ΔH° = –890 kJ mol–1 CH3CH2CH3 + 5O2 !: 3CO2 + 4H2O ΔH° = –2220 kJ mol–1 C8H18 + —25 2 O2 !: 8CO2 + 9H2O ΔH° = –5471 kJ mol–1 [Heat of combustion is the heat released when one mole of a substance is burned in excess oxygen under standard conditions.] 4 However, if the supply of air is limited, incomplete combustion will occur producing some carbon monoxide. For example: 2C3H8 + 7O2 !!: 6CO + 8H2O 5 Combustion data can be used to determine the empirical or molecular formula of a hydrocarbon. Example 15.2 The heat of combustion of octane and 2-methylheptane are –5470 kJ mol–1 and –5466 kJ mol–1. Explain the observation. Solution The equations for the combustion process are CH3(CH2)6CH3 + —25 2 O2 : 8CO2 + 9H2O –5470 kJ mol–1 CH3CH(CH2)4CH3 + —25 2 O2 : 8CO2 + 9H2O –5466 kJ mol–1 & CH3 In both the reactions, the type and number of bonds broken and formed are almost the same. As a result, they have almost the same value of heat of combustion. Heat is liberated during combustion. The combustions are highly exothermic. Definition of heat of combustion Info Chem Carbon monoxide is poisonous. It can poison the haemoglobin in our blood. 55
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 Example 15.3 The combustion of octane is highly exothermic (–5470 kJ mol–1). However, the reaction does not occur at room temperature. (a) Draw an energy profile for the combustion of octane. (b) Explain why the reaction does not occur at room temperature. Solution (a) Energy CH3(CH2)6CH3 + —25 2 O2 –5470 kJ 8CO2 + 9H2O (b) The reaction does not occur at room temperature because the activation energy of the reaction is very high. At room conditions, the reactants do not have enough energy to overcome the activation energy. As a result, heat has to be supplied first before the reaction can take place. Example 15.4 10 cm3 of an hydrocarbon, W is mixed with 100 cm3 of oxygen (in excess) and the mixture sparked. After cooling, 85 cm3 of gas remains. The volume is reduced to 65 cm3 when the remaining gas is bubbled through concentrated sodium hydroxide solution. Determine the molecular formula of W. [All volumes are measured under the same conditions.] Solution Let the hydrocarbon be CxHy. The equation of combustion is CxHy + (x + y —4 )O2 !!: xCO2 + y —2 H2O Volume of gas absorbed by NaOH is carbon dioxide. ∴ Volume of CO2 produced = 85 – 65 = 20 cm3 From the equation, 10 cm3 of W will produce 10x cm3 of CO2 10x = 20 ∴ x = 2 The gas remaining after passing through NaOH is oxygen. ∴ Volume of oxygen reacted = 100 – 65 = 35 cm3 From the equation, 10 cm3 of W requires 10(x + y —4 ) cm3 O2 10(x + y —4 ) = 35 ∴ y = 6 The molecular formula of W is C2H6. 56
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 Quick Check 15.4 1 10 cm3 of a gaseous hydrocarbon requires 50 cm3 of oxygen for a complete combustion to produce 30 cm3 of carbon dioxide. Calculate the molecular formula of the hydrocarbon. [All volumes are measured under the same conditions.] Natural Source of Hydrocarbon – Crude Oil 1 The major source of alkanes is crude oil or petroleum. 2 Petroleum is a liquid mixture of thousands of compounds, most of them hydrocarbons, which are formed from the decomposition of ancient plankton under heat and pressure in the Earth's crust over millions of years. 3 The petroleum industry is concerned with the separation of the thousands of hydrocarbon in crude oil to produce useable products. 4 The different components in crude oil are separated by fractional distillation. Fractional Distillation of Crude Oil 1 Fractional distillation is a separation process based on the different boiling points of the compounds. 2 The crude oil is separated into fractions. Each fraction consists of a mixture of hydrocarbon which boils over a limited range of temperature. 3 The table below lists the different fractions obtained from crude oil. Fraction Molecular size Boiling point range/°C Uses Petroleum gas C1 to C4 < 20 Fuel for domestic gas cookers and for heating Light petroleum C5 to C6 20 to 60 Organic solvent Light naphtha C6 to C7 60 to 100 Organic solvent Petrol C5 to C12 40 to 205 Fuel for motor vehicles Kerosene C12 to C18 175 to 325 Fuel for jet engines Gas oil/Diesel C18 to C25 275 to 400 Fuel for diesel engines Lubricating oil C20 to C34 > 400 Lubricants Bitumen > C34 Solid residue For road surfacing and roofing 4 Beside being used as fuels, some of these fractions are used to manufacture thousands of other compounds such as plastics, detergents, paints, polymers, synthetic rubber and medicines. 5 The petroleum fractions with small molecular size (C1 to C12) are in greater demand than the heavier fractions. 6 The lighter fractions are easier to vaporise and are therefore more useful fuels. 7 In the petrochemical industry, the heavier fractions are broken down into lighter fractions by a process called cracking and reforming. What is fractional distillation? VIDEO Fractional Distillation 57
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 Cracking 1 Cracking is a process where large hydrocarbon molecules (from crude oil) are broken down into smaller and more volatile molecules. 2 There are two types of cracking: thermal cracking and catalytic cracking. 3 Thermal cracking uses high temperature and high pressure to bring about the cracking process. 4 Catalytic cracking uses suitable catalysts for the process, which can be carried out at lower temperature and pressure. The catalysts used are usually alumina (aluminium oxide), silica (silicon dioxide) or zeolites. 5 Catalytic cracking produces more branched chain alkanes than thermal cracking. 6 The cracking process can be summarised as Cracking Large molecule alkane !!!: Small molecule alkanes + alkene 7 For example: C20H42 !!: C10H22 + 2C5H10 Decane Pentene Or C20H42 !!: C8H18 + C7H14 + C5H10 Octane Heptene Pentene Reforming 1 Reforming is a process where alkanes are converted into arenes such as benzene. 2 Reforming is carried out under heat, pressure and the presence of catalysts such as platinum, rhodium and iridium. 3 Examples of reforming are: CH3CH2CH2CH2CH2CH3 !!: + 4H2 Hexane Benzene CH3CHCH2CH2CH2CH3 !!: !CH3 + 4H2 & CH3 2-methylhexane Methylbenzene 4 Arenes are used extensively in the manufacture of other important compounds. What is ‘cracking’? The process of breaking longchain hydrocarbons into shorter ones. Catalytic cracking uses lower temperature and pressure. Exam Tips Exam Tips Catalytic cracking produces more branched-chain hydrocarbons. Info Chem Alkenes are used as feed-stock in the polymer industry. Quick Check 15.5 1 Write balanced equations to show the cracking of C22H46 into (a) three fractions (b) four fractions 2009/P2/Q9(b) 2016/P3/Q16(a) 58
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 Air Pollution 1 Most internal combustion engines run on petrol which is a mixture of alkanes with molecular formulae ranging from C5H12 to C12H26 2 However, burning of petrol contributes to the environmental pollution as the exhaust fumes contain unburned petrol, carbon monoxide, carbon dioxide as well as oxides of nitrogen. 3 No engine works with 100% efficiency. Hence, some of the unburned hydrocarbons (including benzene which is highly toxic) are released into the atmosphere. These hydrocarbon molecules exist in our atmosphere as suspended particles and give rise to photochemical smog which causes eye irritation and respiratory illness. 4 Carbon monoxide is the by-product of incomplete combustion of hydrocarbons. If inhaled, it combines with haemoglobin in the blood making haemoglobin unable to transport oxygen from the lungs to the cells. 5 Carbon dioxide is one of the components of the atmosphere. However, too high concentration of carbon dioxide (which is a greenhouse gas) will cause global warming and dramatic climate change which will affect all living organisms on Earth. 6 At the high temperature in the engine compartment, nitrogen and oxygen (from the air) will react to form oxides of nitrogen. N2(g) + O2(g) !!: 2NO(g) Nitrogen monoxide When released into the atmosphere, nitrogen monoxide reacts rapidly with oxygen to form nitrogen dioxide. 2NO(g) + O2(g) !!: 2NO2(g) The oxides of nitrogen cause irritation to the eye and respiratory system. 7 Oxides of nitrogen dissolve in the water vapour in the atmosphere to form acid rain which causes damage to buildings, changes the pH of soil and waterways which in turn lead to destruction of plants and animals. 2NO2(g) + H2O(l) !!: HNO2(aq) + HNO3(aq) 2NO2(g) + H2O(l) + 1 – 2 O2(g) !!: 2HNO3(aq) 8 Nitrogen dioxide also catalyses the oxidation of atmospheric sulphur dioxide to sulphur trioxide, which is another contributor to acid rain. SO2(g) + NO2(g) !!: SO3(g) + NO(g) NO(g) + 1 – 2 O2(g) !!: NO2(g) SO3(g) + H2O(l) !!: H2SO4(aq) SO2(g) + 1 – 2 O2(g) + H2O(l) !!: H2SO4(aq) 2014/P3/Q16 The major source of atmospheric pollution Photochemical smog Global warming Nitrogen monoxide is colourless. NO2 is a brown gas. Acid rain Catalytic oxidation of atmospheric sulphur dioxide Exam Tips Exam Tips Oxides of nitrogen also contribute to photochemical smog. 59
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 9 The main source of atmospheric sulphur dioxide are volcanic eruption and burning of fossil fuels. Catalytic Converter 1 One important way to reduce pollution by exhaust fumes is by fitting a catalytic converter to the exhaust system before the gases are expelled into the atmosphere. 2 A typical catalytic converter consists of a ceramic honeycomb structure coated with platinum, palladium and rhodium which act as catalysts. 3 In the converter, carbon monoxide and unburned hydrocarbons are catalytically oxidised. 2CO(g) + O2(g) !!: 2CO2(g) CxHy + (x + y —4 )O2 !!: xCO2 + y —4 H2O 4 Oxides of nitrogen are catalytically reduced to nitrogen. 2NO(g) + 2CO(g) !!: N2(g) + 2CO2(g) CxHy, CO, NO Catalytic converter CO2, H2O, N2 15.2 Alkenes 1 Alkenes are unsaturated aliphatic hydrocarbons with the general formula of CnH2n (where n ≥ 2). 2 They are unsaturated due to the presence of one or more carboncarbon double bonds, CRC in the molecule. 3 The carbon atoms that are joined by the double bond undergo sp2 hybridisation. The bond angle around the unsaturated carbon atoms is about 120°. 4 Compared to alkanes, alkenes have fewer hydrogen atoms. 5 The simplest alkene is ethene, C2H4 or CH2RCH2. 6 The molecular orbital model of ethene is discussed in Chapter 14. 7 The carbon-carbon double bond in ethene (or other alkenes) is made up of a s-bond and a p-bond: C C C C H H H H H H H H σ σ Exam Tips Exam Tips • The honeycomb structure is to increase the surface area of the catalyst. • Use of electric cars to cut down atmospheric pollution 2014/P3/Q16 60
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 Nomenclature of Straight-chain Alkenes 1 Alkenes can be divided into straight-chain alkenes and branchedchain alkenes. 2 Examples of straight-chain and branched-chain alkanes are: CH3!CH2!CH2!CHRCH2 CH2RC!CH2!CH3 (Straight-chain alkene) & CH3 (Branched-chain alkene) 3 In the IUPAC system, alkenes are named by substituting the suffix ‘ane’ of corresponding alkanes with ‘ene’. 4 Choose the longest carbon chain that contains the double bond. This forms the root/parent name of the alkene. 5 Number the longest chain that contains the double bond in the direction that gives the unsaturated carbon atoms the lowest possible numbers. 6 Indicate the location of the double bond using the number of its first carbon. 7 Examples are: CH2RCH2 Ethene (Ethylene) CH3CHRCH2 Propene (Propylene) CH3CH2CHRCH2 1-butene (1-butylene) CH3CHRCHCH3 2-butene CH3CH2CH2CHRCH2 1-pentene (1-pentylene) CH3CHRCHCH2CH3 2-pentene CH3CH2CH2CH2CHRCH2 1-hexene CH3CH2CH2CHRCHCH3 2-hexene CH3CH2CHRCHCH2CH3 3-hexene (The names in brackets are common names.) Nomenclature of Branched-chain Alkenes 1 The first three members of alkenes form only straight chain. 2 Branching begins with alkenes with four or more carbon atoms. For example, there are three structures possible for alkenes with the molecular formula of C4H8. CH3CH2CHRCH2 1-butene CH3CHRCHCH3 2-butene CH3CRCH2 2-methylpropene (Isobutylene) & CH3 3 The IUPAC name for branched-chain alkenes consists of a parent (root) name which indicates the longest carbon chain containing the double bond. 61
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 4 Each substituent (the alkyl groups) is given a name and a number. The number (beginning from the end closer to the double bond) indicates the carbon atom in the parent chain to which the substituent is bonded. 5 For example, the name of the following alkene: 6 5 4 3 2 1 [Wrong] 1 2 3 4 5 6 [Correct] CH2RCH!CH2!CH!CH2!CH3 & CH3 is 4-methyl-1-hexene and not 3-methyl-5-hexene. Example 15.5 Draw the structures for the following alkenes. (a) 2-methyl-1-hexene (c) 2-methyl-2-butene (b) 2-ethyl-4-methyl-1-pentene Solution (a) CH2RCCH2CH2CH2CH3 (c) CH3!CRCH!CH3 & & CH3 CH3 CH3 & (b) CH2RC!CH2!CH!CH3 & CH2CH3 Cycloalkenes 1 Cycloalkenes are unsaturated hydrocarbons with the general formula of CnH2n–2 (where n ≥ 3). 2 In naming a cycloalkene, the carbon in the ring that carries the double bond is designated as carbon 1. 3 The ring is then numbered starting from carbon 1, going across the double bond, in the direction that would give the substituent encounter first the lowest number possible. 4 Examples are: Structure Skeletal structure Name C3H4 Cyclopropene C4H6 Cyclobutene C5H8 Cyclopentene C6H10 Cyclohexene Numbering begins from the end closer to the double bond. Info Chem Cycloalkenes have the same general formula as alkynes. The side chains are named in alphabetical order. For example, ethyl followed by methyl. 62
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 5 The following compound: CH3 3(2) (3)4 2(1) 1 is named as 4-methyl-1-cyclohexene and not 3-methyl-1- cyclohexene. Example 15.6 Write the IUPAC names for the following cycloalkenes. (a) CH3 CH3CH2 (b) CH3 CH3 Solution (a) 4-ethyl-1-methylcyclohexene (b) 1,6-dimethylcyclohexene (not 1,2-dimethylcyclohexene) Dienes, Trienes and Polyenes 1 Alkenes with two carbon-carbon double bonds in their structure are called dienes (general formula of CnH2n-2), and those with three carbon-carbon double bonds are called trienes and so on. 2 The names of some alkenes with more than one carbon-carbon bonds are given below. Structure Name CH2RCHCHRCH2 1,3-butadiene CH2RCCHRCH2 & CH3 2-methyl-1,3-butadiene (isoprene) CH2RCHCHRCHCHRCH2 1,3,5-hexatriene Structural and cis-trans Isomerism in Alkenes 1 Alkenes exhibit two types of isomerism: structural and cis-trans isomerism. 2 The first two alkenes (C2H4, C3H6) do not exhibit isomerism, either structural or cis-trans. 3 Alkenes with four or more carbon atoms in their molecules exhibit both structural and cis-trans isomerism. 4 An example is butene, C4H8, which has three structural isomers. CH3CH2—CH=CH2 1-butene CH3—CH=CH—CH3 2-butene CH3—C=CH2 & 2-methylpropene CH3 1,3-butadiene 2-methyl-1,3-butadiene 1,3,5-hexatriene Info Chem The general formula of trienes is CnH2n – 4. 63
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 5 However, 2-butene can exist in two isomeric forms: cis-2-butene and trans-2-butene. C H H CH3 CH3 RRC C H CH3 CH3 H RRC cis-2-butene trans-2-butene 6 Alkenes are also isomeric with cycloalkanes since both have the same molecular formula of CnH2n. For example, butene (C4H8) is isomeric with cyclobutane (C4H8). CH2!CH2 & & CH2!CH2 Example 15.7 Name the following alkenes. (a) C H H H3C CH2CH3 RRC (b) C CH3CH2 CH3 CH2CH3 H RRC Solution (a) cis-2-pentene (Both H on the same side.) (b) trans-3-methyl-3-hexene (Both C2H5 on the opposite side.) Example 15.8 The boiling point of cis-2-butene is 3.7 °C, while that of trans-2- butene is 0.9 °C. Explain the difference in their boiling points. Solution Boiling point is a measure of the strength of the intermolecular forces that hold the molecules together in the liquid state. The intermolecular forces in both isomers are the van der Waals forces. CH3 C H H CH3 CH3 RRC C H CH3 H RRC cis-2-butene trans-2-butene However, the dipole moments in trans-2-butene cancel out one another but not in cis-2-butene. As a result, the van der Waals force in cis-2-butene is stronger. Exam Tips Exam Tips cis-2-butene is slightly polar but trans-2-butene is nonpolar. 64
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 Quick Check 15.6 1 Draw all possible geometrical isomers, if any, for the following alkenes. (a) 2-methyl-2-butene (d) 3-heptene (b) 3-hexene (e) 4-methyl-3-heptene (c) 4-methyl-1-hexene Physical Properties of Alkenes 1 Like alkanes, alkenes are non-polar compounds with weak van der Waals forces between the molecules. 2 The lower members are mostly gases or liquids. 3 Due to the smaller number of hydrogen atoms in the molecule, alkenes have lower melting/boiling points than their corresponding alkanes. 4 Alkenes are non-miscible with water, but dissolve readily in nonpolar organic solvents. Reactions of Alkenes 1 In contrast to alkanes, alkenes are more reactive. This is due to the presence of the carbon-carbon double bond. 2 The C!C bond strength is about 347 kJ mol–1 and that of the CRC bond is 612 kJ mol–1. Hence at first sight, we would expect alkenes to be less reactive than alkanes. However, this is not so for three reasons: (a) A carbon-carbon double bond is made up of a sigma (s) bond and a pi (π) bond. The strength of the s bond is about 347 kJ mol–1. Hence, the π bond will have an energy of (612 – 347) = 265 kJ mol–1. (b) From the above, it is clear that a π bond is weaker than a s-bond. Hence, it is relatively easy to break the π bond while leaving the s bond intact. This is what happens in most reactions involving alkenes. It is very rare for the CRC bond to break completely. (c) Secondly, due to the high electron density around the unsaturated carbon atoms, they are readily attacked by electrophiles. (d) Finally, rupture of the weaker π bond is replaced by 2 stronger σ-bonds. This makes the product more stable. 3 The characteristic reaction of alkenes is electrophilic addition where a reagent adds across the double bond in such a way that the π bond is broken, and in its place two new s bonds are formed. & & C C R + X!Y !: !C!C! & & X Y 4 Other reactions are oxidation and polymerisation. Info Chem A C=C bond is stronger than a C–C bond. C C 347 kJ mol–1 fi265 kJ mol–1 !! Info Chem The weak p bond is replaced by two stronger σ bonds. Thus, addition reactions are energetically feasible. 2017/P3/Q2 65
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 Electrophilic Addition Hydrogenation 1 When an alkene, together with hydrogen gas is passed over heated nickel (or platinum), an alkane is produced. & & C C R + H!H !: !C!C! & & H H 2 Examples are: CH2RCH2 + H2 !!: CH3CH3 Ethene Ethane CH3CHRCH2 + H2 !!: CH3CH2CH3 Propene Propane CH3CRCHCH2CH2CH3 + H2 !: CH3CHCH2CH2CH2CH3 & & CH3 CH3 2-methyl-2-hexene 2-methylhexane + H2 !: H H Cyclohexene Cyclohexane 3 Hydrogenation of alkenes is sometimes called catalytic reduction. 4 In industry, catalytic hydrogenation is used to convert edible oils (which are polyunsaturated) into solid fats such as margarine. By controlling the degree of hydrogenation, the margarine can be made soft or hard as required. 5 Hydrogenation can be used quantitatively to determine the number of carbon-carbon double bonds in one molecule of an alkene. One mole of a monounsaturated alkene would react with one mole of hydrogen gas, while one mole of an alkene consisting of two double bonds per molecule would require two moles of hydrogen gas. Example 15.9 Write the structure of the product formed when 1,3-butadiene is hydrogenated using nickel catalyst. Solution The structure of 1,3-butadiene is CH2RCH!CHRCH2 The equation of the reaction is CH2RCH!CHRCH2 + 2H2 !: CH3CH2CH2CH3 2008/P1/Q29 2010/P1/Q29 2014/P3/Q6 2011/P1/Q28 2015/P3/Q19(a) 2012/P1/Q4(b) 2016/P3/Q4 2017/P3/Q6 66
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 Example 15.10 2.72 g of an alkene (Mr = 68) reacts with 1.79 dm3 of hydrogen measured at s.t.p. (a) Determine the number of carbon-carbon double bond in the alkene. (b) Draw the structures of the possible isomers of the alkene. Solution 2.72 (a) Number of moles of alkene = —— 68 = 0.04 mol 1.79 Number of moles of hydrogen = ——22.4 =0.08 mol 1 mole of the alkene requires 2 moles of H2. Hence, the alkene is a diene. The general formula of a diene is CnH2n–2. 12n + (2n – 2)(1) = 68 ∴ n = 5 The molecular formula of the alkene is C5H8. (b) The possible isomers of the alkene are CH3!CHRCH!CHRCH2 1,3-pentadiene [Note: 1,3-pentadiene can exist in the cis and trans form.] CH3 & CH2RC!CHRCH2 2-methyl-1,3-butadiene Halogenation 1 Alkenes react with bromine and chlorine (in an inert solvent such as tetrachloromethane) in the dark to form dihaloalkanes. & & C C R + X ! X !: !C!C! & & X X CH3CHRCHCH3 + Cl2 !: CH3CH!CHCH3 & & Cl Cl 2,3-dichlorobutane 2 When the halogen used is bromine, the reddish-brown colour of bromine is decolourised. This serves as a simple test for the presence of carbon-carbon double bond in a molecule. Exam Tips Exam Tips Decolourisation of bromine serves as a test for the CRC bond. 2009/P1/Q29, Q32 2009/P2/Q9 2015/P3/Q18(a)(iii) 2013/P3/Q4 67
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 CH3!CHRCH2 + Br2 !: CH3!CH!CH2Br & Br 1,2-dibromopropane + Br2 : Br Br Cyclohexene 1,2-dibromocyclohexane CH2RCHCHRCH2 + 2Br2!: CH2Br!CHBr!CHBr!CH2Br 1,3-butadiene 1,2,3,4-tetrabromobutane Example 15.11 Suggest a simple chemical test to differentiate between cyclohexane and cyclohexene. Solution Add bromine (dissolved in tetrachloromethane) to both compounds separately in the dark. The one that decolourises bromine is cyclohexene, while the other is cyclohexane. Addition of Hydrogen Halides 1 Hydrogen halides (either in the gaseous state or in concentrated aqueous solution) react with alkenes to produce haloalkanes. & & C C R + H!X !: !C!C! & & H X 2 The reactivity increases in the order HCl < HBr < HI. HCl requires the presence of aluminium chloride as a catalyst for the reaction to take place at an appreciable rate. HBr and HI react rapidly with alkenes at room temperature. 3 The reaction involves the breaking of the hydrogen-halogen bond in the hydrogen halide. Going from HCl to HI, the bond strength decreases as the size of the halogen atom increases leading to an increase in the reactivity. Bond H!Cl H!Br H!I Bond energy/kJ mol–1 431 366 299 4 Examples are: CH2RCH2 + HI !: CH3CH2I Ethene Iodoethane CH3CHRCHCH3 + HBr !: CH3CH!CHCH3 2-butene & & H Br 2-bromobutane Reactivity: HI > HBr > HCl Info Chem Addition with symmetrical alkene produces a single product. 68
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 5 Ethene and 2-butene are symmetrical alkenes. The molecule can be divided into two exact halves through the carbon-carbon double bond. Symmetrical alkenes produce only one addition product with hydrogen halides. 6 However, for unsymmetrical alkenes, two addition products are possible. For example, the reaction of propene with hydrogen bromide would produce 1-bromopropane and 2-bromopropane as shown below. CH3!CHRCH2 + H!Br !: CH3!CH2CH2Br 1-bromopropane CH3!CHRCH2 + H!Br !: CH3!CH!CH3 & Br 2-bromopropane 7 In such cases, one of the products is the major product. It is produced in larger quantity than the other. 8 Markovnikov’s rule is used to predict which of the two alkenes is the major product. Markovnikov rule states that, when an unsymmetrical reagent is added to an unsymmetrical alkene, the more electropositive atom or group of the reagent will be attached to the unsaturated carbon atom that has the most number of hydrogen atoms bonded to it. 9 In hydrogen bromide molecule, the hydrogen atom is more electropositive than the bromine atom. Hence, when HBr is added to propene, the H from HBr would attach itself to carbon-1 which has two H atoms instead of carbon-2 which has only one H atom. The Br– ion then attaches to carbon-2 to produce 2-bromopropane as the major product. CH3!CHRCH2 + Hδ+!Brδ– !: CH3!CH!CH3 2 1 & Br 2-bromopropane Addition of water 1 Alkenes do not react with water under normal conditions. 2 However, alkenes react with steam in the presence of concentrated phosphoric acid at elevated temperature and pressure to produce alcohols. & & C C R + H!OH !: !C!C! & & H OH CH2"CH2 CH39CH"CH9CH3 Info Chem Addition to an unsymmetrical alkene produces two different products. Another statement of Markovnikov’s rule: In the addition reaction between a protic acid and an alkene, the acidic hydrogen attaches to the carbon with fewer alkyl substituents and the halide group attaches to the carbon with more alkyl substituents. ‘Like attacks like’ 2017/P3/Q19 69
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 3 The addition follows Markovnikov rule. Hence, hydration of propene produces 2-propanol as the major product. H3PO4, 300 °C, 60 atm CH3CHRCH2 + H2O !!!!!!!: CH3!CH!CH3 & OH 4 Industrially, direct hydration of ethene is an important method of producing ethanol. H3PO4, 300 °C, 60 atm CH2RCH2 + H2O !!!!!!!: CH3!CH2OH Addition of Concentrated Sulphuric Acid 1 This is an indirect method for the hydration of alkenes to form alcohols. 2 The alkene is bubbled slowly into cold concentrated sulphuric acid. Markovnikov addition takes place to produce an alkyl hydrogensulphate. & & C C R + H2SO4 !: !C!C! & & H OSO3H 3 When the hydrogen sulphate is diluted with water and warmed, an alcohol is produced and sulphuric acid is regenerated. & & & & !C!C! + H2O !: !C!C! + H2SO4 & & & & H OSO3H H OH 4 The structural formula of the alkyl hydrogensulphate is C H O O !C O S OH ! ! ! ! ! ! ! !R R 5 For example: CH3CHRCH2 + H2SO4 !: CH3!CH!CH3 & OSO3H CH3!CH!CH3 + H2O !: CH3!CH!CH3 + H2SO4 & & OSO3H OH CH3 H CH3 OH H H + H2O Example: Markovnikov addition Indirect hydration of alkenes 70
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 Example 15.12 Draw the structure of the alcohol produced when 2-methyl-propene is subjected to hydration by concentrated sulphuric acid followed by warming with water. Solution The structure of 2-methylpropene is CH3!CRCH2 & CH3 Addition of H2SO4 follows the Markovnikov rule. CH3 & CH3!CRCH2 + H2SO4 !: CH3!C!CH3 & & CH3 OSO3H Hydrolysis produces 2-methyl-2-propanol. CH3 CH3 & & CH3!C!CH3 + H2O !: CH3!C!CH3 + H2SO4 & & OSO3H OH Addition of Bromine Water 1 Bromine dissolves sparingly in water to produce what is known as ‘bromine water’. Br2 + H2O !: HBr + HOBr 2 An alkene reacts with bromine water, at room temperature, to produce a bromo-alcohol. & & C C R + HOBr !: !C!C! & & Br OH The addition follows Markovnikov rule. 3 Oxygen is more electronegative than bromine, the HOBr molecule is polarised as shown below. HOd–!!Brd+ 4 As a result, the reaction between propene and bromine water will produce 1-bromo-2-propanol as the major product as predicted by Markovnikov rule. CH3CHRCH2 + HOBr !: CH3!CH!CH2Br & OH 2008/P1/Q35 Info Chem HOBr is named as bromic(I) acid. Orange colour of the bromine water is decolourised. Info Chem Oxygen is more electronegative than bromine. 71
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 5 Similarly with chlorine water, CH3CHRCH2 + HOCl !: CH3!CH!CH2Cl & OH 1-chloro-2-propanol Quick Check 15.7 1 Give the structural formulae of the products formed when 2-methyl-2-butene reacts with (a) hydrogen in the presence of platinum, (b) bromine dissolved in tetrachloromethane, (c) steam in the presence of phosphoric acid, (d) hydrogen iodide, (e) bromine water, (f) concentrated sulphuric acid followed by hot water. Oxidation of Alkenes 1 Unlike alkanes, alkenes can be oxidised by oxidising agents such as potassium manganate(VII) and potassium dichromate(VI). 2 The products of oxidation depend on the conditions of the reaction and the type of alkenes involved. Mild Oxidation 1 Alkenes decolourises cold, dilute acidified potassium manganate(VII) to produce diols (compounds with two !OH groups). & & C C R + [O] + H2O !: !C!C! & & OH OH 2 In the reaction, the purple colour of KMnO4 is decolourised. This is another test (called the Baeyer test) for the presence of carboncarbon double bond in a molecule. 3 If acidified potassium dichromate(VI) is used instead of potassium manganate(VII), the orange colour of the dichromate changes to green. 4 An example is the oxidation of ethene to ethane-1,2-diol. KMnO4 CH2RCH2 + [O] + H2O !!!: CH2!CH2 & & OH OH 2011/P2/Q4(a) Alkenes can be oxidised, but alkanes cannot. Decolourisation of KMnO4 is another test for the CRC bond. 2013/P3/Q6 2016/P3/Q4, Q16(b) 72
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 Vigorous Oxidation 1 When an alkene is treated with hot, concentrated acidified KMnO4, the carbon-carbon double bond breaks to produce ketones, carboxylic acids or carbon dioxide, depending on the groups that are attached to the double bond. 2 Terminal alkene, that is alkene with the RCH2 group is oxidised to CO2 and H2O. Hot, concentrated KMnO4 R CH2 !!!!!!!!!: CO2 + H2O 3 Alkene with the R C!H group is oxidised to carboxylic acid. & R Hot, concentrated KMnO4 R C!H & R !!!!!!!!!: R!C!OH ∫ O 4 Alkene with the RC!R group is oxidised to ketone. & R Hot, concentrated KMnO4 R C!R & R !!!!!!!!!: R!C!R ∫ O 5 Summary: The RCH2 group is oxidised to carbon dioxide and water. The RC!R group is oxidised to R!C!OH & ∫ H O (carboxylic acid) The RC!R is oxidised to a ketone, R!C!R & ∫ R O 6 Examples are: KMnO4, Heat CH2RCH2 + 6[O] !!!!!: 2CO2 + 2H2O KMnO4, Heat CH3CHRCH2 + 5[O] !!!!!: CH3C!OH + H2O + CO2 ∫ O Ethanoic acid (Acetic acid) CH3H O & & KMnO4, Heat ∫ CH3!CRC!CH3 + 3[O] !!!: CH3!C!CH3 + CH3COOH CH3CH3 O & & KMnO4, Heat ∫ CH3!CRC!CH3 + 2[O] !!!: 2 CH3!C!CH3 The CRC bond breaks to produce two fragments. 2015/P3/Q6 2017/P3/Q5 73
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 7 Oxidation by hot, concentrated KMnO4 can be used to determine the structure of an alkene. To get the structure of the alkene, all the oxygen atoms from the products are removed and the two carbonyl carbon atoms are joined by a double bond. 8 For example: Alkene Y !!!!: CH3CHCOOH + KMnO4, Heat !C!CH3 & ∫ CH3 O 2-methylpropanoic acid 1-cyclohexylethanone The structure of Y is obtained as follows: & & & & & & CH3 O!H CH3 CH3 H CH3 CH3CH!CR O O RC! :CH3CH!CRC! 2-cyclohexyl-4-methyl-2-pentene Quick Check 15.8 1 Write structural formulae for the products formed when the following alkenes are oxidised under the following conditions. Cold dilute KMnO4 (a) CH2RCCH2CH3 !!!!!!: & (b) Cold dilute KMnO4 RCH2 !!!!!!: (c) Hot concentrated KMnO4 !!!!!!!!: 2 Deduce the structure of the following alkenes. Cold dilute KMnO4 (a) A !!!!!!: CH2OHCH(OH)CH2CH(OH)CH2OH Hot concentrated KMnO4 (b) B !!!!!!!: (CH3CH2)2CRO + CH3COOH Hot concentrated KMnO4 (c) C !!!!!!!!: CO2 Hot concentrated KMnO4 (d) D !!!!!!!!: CH3COOH Hot concentrated KMnO4 (e) E !!!!!!!!: CH3COOH + CH3!CH!COOH & CH3 74
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 With Oxygen 1 In the absence of a catalyst, alkenes burn in excess oxygen to produce carbon dioxide and water. For example: CH2RCH2 + 3O2 !: 2CO2 + 2H2O CH3CHRCHCH3 + 6O2 !: 4CO2 + 4H2O 2 However, if a mixture of alkene and oxygen is passed over heated silver as catalyst, epoxyalkane (also known as alkene oxide) is produced. C & & RC + ––O2 !!!: !C!!C! O 1 Ag, 200 °C 2 3 When the epoxyalkane is diluted with water and warmed, a diol is formed. & & !C!!C! + H2O !!!: & & & & !C!!C! O OH OH 4 For example, ethene reacts with oxygen in the presence of silver to form epoxyethane or ethene oxide. + ––O2!!!!: CH2!!CH2 O CH2RCH2 1 2 Ag, 200 °C 15 atm When diluted with water and warmed, ethane-1,2-diol is produced. CH2!!CH2 !!: CH2!!CH2 O + H2O & OH & OH 5 Ethane-1,2-diol is used in the automobile industry as a coolant and anti-freeze and also in the manufacture of polyesters. Quick Check 15.9 Give the formula of the final products formed when each of the following alkenes is reacted with oxygen in the presence of silver catalyst followed by warming with water. 1 CH2RCH!CH2CH2!CHRCH2 2 CHRCH2 Combustion of alkenes Catalytic oxidation of alkenes 75
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 Ozonolysis 1 Ozone, O3 is an allotrope of oxygen. 2 In the upper atmosphere (stratosphere), it acts as a filter to prevent harmful ultraviolet radiation from reaching the Earth's surface. 3 In the lower atmosphere, it is a pollutant. Ozone is a powerful oxidising agent and causes damage to mucus and respiratory tissues in animals and tissues in plants. 4 When ozone gas is bubbled through a solution of alkene in an inert solvent (such as cyclohexane) at 20 °C, an ozonide is formed. 5 When the ozonide is warmed with dilute sulphuric acid in the presence of zinc, it decomposes to produce two carbonyl compounds. 6 The overall reaction is the cleavage of carbon-carbon double bond: O3, Zn/H+ C == C !!!!:C = O + O = C 7 An advantage of using ozone, instead of hot, concentrated potassium permanganate or potassium dichromate to cleave the carbon-carbon bond is that the carbonyl compounds obtained will not be further oxidised to carboxylic acids. 8 For example, CH3 & Zn/H+ CH3!C = CH!CH3 + O3 + H2O !!: 2-methyl-2-butene O O ∫ ∫ CH3!C!CH3 + CH3!C!H + H2O2 Propanone (Acetone) Ethanal (Acetaldehyde) Zn/H+ + O3 + H2O !: H—C—CH2CH2CH2CH2—C—H + H2O2 ∫ ∫ O O Cyclohexene Hexane-1,6-dial 9 Ozonolysis is a useful reaction to deduce the structure of an alkene. For example, an alkene, X on ozonolysis produces propanal and ethanal. O3, Zn/H+ X !!!: CH3CH2—C—H + CH3—C—H ∫ ∫ O O C C + H2O2 O O C C C O O C+ O Ozonide O3 H2O/Zn Info Chem Ozone denatured rubber by breaking the C=C bonds in the polymer. 76
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 The structure of X is found by removing the oxygen atoms and joining the oxygen-carrying carbon atoms by a double bond. CH3CH2—C = O O = C—CH3 !: CH3CH2—C = C—CH3 & & & & H H H H X is pentene. Example 15.13 Draw the structural formulae for the products of the following alkenes when undergo ozonolysis. (a) Ethene (b) 2-butene (c) 3-methyl-1-butene (d) 1-methylcyclohexene Solution (a) H—C—H ∫ O (b) CH3—C—H ∫ O (c) H—C—H + CH3—CH—C—H ∫ & ∫ O CH3 O (d) CH3—C—CH2—CH2—CH2—CH2—C—H ∫ ∫ O O Example 15.14 Ozonolysis of an alkene, Y produces propanone as the only product. Deduce the structure of Y. Solution The structure of propanone is CH3—C—CH3. ∫ O CH3—C==O O==C—CH3 !: CH3—C==C—CH3 & & & & CH3 CH3 CH3 CH3 Y is 2,2-dimethyl-1-butene 77
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 Quick Check 15.10 1 Draw the structure of the organic products when the following alkenes undergo ozonolysis. (a) CH3CH=CH2 (b) CH2=CH—CH2CH2—CH=CH2 (c) CH CH2 2 Deduce the structure of the alkene which would produce the following organic products when subjected to ozonolysis. (a) C OH (b) O HC O H + only Polymerisation of Alkenes 1 Being unsaturated, alkenes undergo addition polymerisation to produce addition polymers. 2 For example, ethene can be polymerised to poly(ethene), a plastic. nCH2RCH2 !: ![!CH2!CH2!]n ! The polyethene produced has a relative molecular mass of between 104 to 106 . Mechanism of Electrophilic Addition and Markovnikov Rule 1 The first step in the addition reaction between propene and HBr is the attack of the partial positive charged hydrogen atom with one of the two unsaturated carbon atom to form a carbonium ion intermediate. 2 The attack can result in the formation of either a primary or a secondary carbonium ion. CH3CHRCH2 + fi+H!!Brfi– !: CH3CH2CH2 + Br– + (Primary carbonium ion) CH3CHRCH2 + fi+H!!Brfi– !: CH3!C!CH3 + Br– + & H (Secondary carbonium ion) 2009/P2/Q9(a) The attacking species is the electrophile, H+. INFO Markovnikov’s Rule and Electrophilic Addition 78
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 3 However, the secondary carbonium ion is more stable than the primary carbonium ion because there are two ‘electron-releasing methyl groups which help to stabilise the ion. Hence, the secondary carbonium ion is preferred. This results in the formation of 2-bromopropane as the major product. Br + & CH3!C!CH3 + Br– !: CH3!C!CH3 & & H H Importance of Ethene in Industry 1 Ethene can be polymerised to produce poly(ethene) which is used to make plastic bags, kitchen wares, bowls, bottles, toys and buckets. 2 Hydration of ethene produces ethanol which is used extensively as a solvent and as starting material for the manufacture of other organic compounds. 3 Ethane-1,2-diol (which is produced from ethene) is used as antifreeze and coolant, and in the manufacture of polyesters as well as detergents. 4 Ethene is used to manufacture chloroethane. CH2 = CH2 + HCl !: C2H5Cl One major use of chloroethane is to produce tetraethyllead (TEL), an anti-knock additive for petrol. 4Na + Pb + 4C2H5Cl !: (C2H5)4Pb + 4NaCl 15.3 Arenes 1 Arenes are a class of compounds called aromatic hydrocarbons with a general formula of CnH2n–6 (where n ≥ 6). 2 The first member of arenes is benzene. 3 Benzene, with a molecular formula of C6H6 was first isolated by Michael Faraday in 1825. 4 The molecular formula of benzene suggests that it is highly unsaturated and would have chemical properties similar to that of the alkenes, which is addition reaction. 5 Surprisingly, it is not so. Benzene is rather inert and does not react with the common reagents that attack alkenes. 6 For example, benzene does not decolourise acidic KMnO4 (i.e. it is not oxidised), nor does it react with bromine in the dark. KMnO4 Br2, in CCl4, Dark C6H6 !!: No reaction C6H6 !!!!!!: No reaction This shows that there are no ‘true’ carbon-carbon multiple bonds in the benzene molecule. Info Chem Secondary carbonium ion is more stable than a primary carbonium ion. 79
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 7 However, under suitable conditions, benzene undergoes substitution, acting as if it is a saturated hydrocarbon, like an alkane. 8 For example, in the presence of anhydrous aluminium chloride, benzene reacts with chlorine to give chlorobenzene, where one of the hydrogen atoms in the C6H6 molecule is substituted by chlorine. AlCl3 C6H6 + Cl2 !!: C6H5Cl + HCl Nomenclature of Arenes 1 Monosubstituted alkylbenzenes are named as derivatives of benzene. For example: CH3 CH2CH3 CH3!CH!CH3 Methylbenzene Ethylbenzene Isopropylbenzene (Toluene) (Cumene) 2 For polysubstituted alkylbenzenes, the positions of the substituents are located by numbering the benzene ring as follows: Y 1 (o) 6 2 (o) (m) 5 3 (m) 4 (p) 3 Positions 2 and 6 are equivalent and are sometimes named ortho (o–). Positions 1 and 4 are equivalent and are sometimes named para (p–). Positions 3 and 5 are equivalent and are sometimes named meta (m–). 4 Examples are: CH3 CH3 C2H5 CH3 CH3 1,2-dimethylbenzene 3-methylethylbenzene (or o-methyltoluene or o-xylene) (or m-methylethylbenzene) The common name of methylbenzene is toluene. Cumene is also named as 2-phenylpropane or 1-methylethylbenzene. The common name for dimethylbenzene is xylene. 80
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 CH3 C2H5 CH3 2-ethylmethylbenzene (or o-ethyltoluene) 5 Sometimes, an arene is more conveniently named as phenylsubstituted alkane. For example: CH3 & CH3CH2CCH2CH2CH3 CH2CHRCHCH2 1-phenyl-2-butene 3-methyl-3-phenyl-hexane CH2CH2CH2CH2 1,4-diphenylbutane Structural Isomerism in Arenes 1 Polysubstituted benzene exhibits structural isomerism, especially in the positions of the substituents. 2 For example, there are three structural isomers for dimethylbenzene (C8H10): CH3 CH3 CH3 CH3 CH3 CH3 1,2-dimethylbenzene 1,3-dimethylbenzene 1,4-dimethylbenzene 3 Note that dimethylbenzene (C8H10) is also isomeric with ethylbenzene. CH2CH3 81
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 Physical Properties of Benzene 1 Benzene is a colourless liquid at room conditions with a characteristic smell. 2 The melting point and boiling point of benzene are 6 °C and 80 °C respectively. 3 Benzene is immiscible with water but dissolves readily in organic solvents. 4 Due to the high carbon content, benzene burns with a smoky and luminous flame. 5 Benzene is toxic and is a known carcinogen. Prolong exposure to benzene can cause anaemia and leukaemia. Reactions of Benzene 1 Due to the stability of the benzene ring (with the ring system of electrons), the main reaction of benzene is substitution. 2 Benzene is inert towards oxidising agents such as potassium permanganate (KMnO4) or potassium dichromate (K2Cr2O7). 3 Due to the high electron density of the ring, benzene is readily attacked by electrophiles. It undergoes electrophilic substitution. It can be represented by the general equation below. H E + E+ !!: + H+ Nitration of Benzene 1 Nitration is the reaction in which a nitro group !NO2 is introduced into the molecule. 2 Benzene reacts with a mixture of concentrated nitric acid and concentrated sulphuric acid (known as the nitrating mixture) to form nitrobenzene (a yellowish oil) if the temperature is kept below 55 °C. H NO2 Concentrated H2SO4, 55 °C + HNO3 !!!!!!!!!: + H2O 3 The electrophile is the nitronium (or nitryl) ion, NO2 +. It is generated by the reaction between concentrated nitric acid and concentrated sulphuric acid. 2H2SO4 + HNO3 L 2HSO4 – + H3O+ + NO2 + Info Chem % of carbon in benzene 12 × 6 = × 100% 78 = 92.3% A carbon atom requires 4 times as much O2 for combustion compared to a H atom. Benzene does not decolourise KMnO4. One H atom is substituted by the electrophile. 2008/P1/Q29 2010/P1/Q29 2017/P3/Q6 2011/P1/Q31 2015/P3/Q20(c) 2009/P1/Q49 2011/P1/Q29 82
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 4 The NO2 + ion attacks the π-electron of the benzene ring to form an intermediate where the positive charge is delocalised over the remaining five carbon atoms. H H NO2 Concentrated H2SO4, 55 °C + NO2 + !!!!!!!!!: + 5 Elimination of H+ from the intermediate returns the ring of delocalised π-electrons and nitrobenzene is produced. H NO2 NO2 !!: + H+ + 6 The reaction between concentrated nitric acid and concentrated sulphuric acid is an acid-based reaction. The HNO3 molecule acts as a base and accepts a proton from H2SO4 to form protonated nitric acid. HNO3 + H2SO4 !!: H2NO3 + + HSO4 – .. + ORN!O!H + H!HSO4 !: ORN!O!H + HSO4 – p p & O O H Dissociation of the protonated species produces nitronium ion: H2NO3 + !!: NO2 + + H2O 7 In excess of nitric acid and elevated temperature, further nitration of nitrobenzene occurs to produce dinitrobenzene, a yellow solid. C6H5NO2 + HNO3 !!: C6H4(NO2)2 + H2O 8 However, there are three possible position isomers of dinitrobenzene: NO2 NO2 NO2 NO2 CH3CH3 NO2 1,2-dinitrobenzene 1,3-dinitrobenzene NO2 1,4-dinitrobenzene 9 The major product is determined by the substituent effect. [Refer pg. 89] Exam Tips The positive charge in the benzene ring is delocalised through resonance. O H + N—O O H p + O=N : O + H2O 3 possible isomers of dinitrobenzene 83
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 Halogenation 1 Benzene reacts with halogen (chorine and bromine), in the dark, in the presence of a ‘halogen carrier’ (or Lewis acid) such as anhydrous aluminium chloride or iron powder to produce halobenzene. 2 An example is the reaction of benzene with chlorine in the presence of aluminium chloride at room temperature to produce chlorobenzene with the release of white fumes of hydrogen chloride. H Cl AlCl3, Room temperature + Cl2 !!!!!!!!: + HCl 3 The electrophile in the reaction is the Cl+ ion, which is generated by the reaction between chlorine and aluminium chloride. AlCl3 + Cl2 !!: Cl+ + AlCl4 – 4 In this reaction, aluminium chloride acts as a Lewis acid by accepting a lone pair of electrons from the chlorine molecule. Cl Cl & & Cl!Al + • •Cl!Cl !!: Cl!Al;Cl– + Cl+ & & Cl Cl 5 If iron powder is used as the catalyst, the electrophile is generated via the following process. 2Fe + 3Cl2 !!: 2FeCl3 FeCl3 + Cl2 !!: Cl+ + FeCl4 – 6 The Cl+ ion attacks the π-electrons of the benzene ring to form an intermediate, where the positive charge is delocalised over the other five carbon atoms as shown below. H H Cl + Cl+ !!: + A catalyst is required. x xx x x xx xx xx xx xx xx xxxx xx xxxx Cl Cl xx – Cl Cl Al 2008/P1/Q48 2011/P1/Q30 2015/P3/Q7 2017/P3/Q4 84
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 7 Elimination of H+ from the intermediate returns the ring of delocalised π-electrons and chlorobenzene is produced. H Cl Cl !!: + H+ + H+ + AlCl4 – !!: HCl + AlCl3 8 Bromine reacts the same way as chlorine. H Br AlCl3, room temperature + Br2 !!!!!!!!!: + HBr 9 However, under the same conditions, iodine does not react with benzene. Alkylation 1 In alkylation (also called Friedel-Crafts alkylation), one of the hydrogen atoms in the benzene ring is substituted by an alkyl group. 2 This is a convenient way to prepare alkyl benzene from benzene. 3 In the presence of a Lewis acid such as aluminium chloride, in the dark, benzene reacts with an alkyl halide (or haloalkane) to produce alkyl benzene. H R AlCl3, room temperature + R!X !!!!!!!!!: + HX 4 For example: H CH3 AlCl3, room temperature + CH3Cl !!!!!!!!!: + HCl Info Chem The Friedel-Crafts reaction is a set of reactions developed by Charles Friedel and James Craft in 1887 to attach substituents to a benzene ring. White fumes of HCl are released. 2012/P1/Q31 2015/P3/Q18(b) 2006/P1/Q31 2008/P1/Q39 85
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 5 The electrophile in this case is the carbonium ion, CH3 +, which is generated by the following reaction. AlCl3 + CH3Cl !!: CH3 + + [AlCl4]– The CH3 + ion attacks the π-electrons of the benzene ring to form an intermediate. H H CH3 + CH3 + !!: + 6 Elimination of H+ from the intermediate returns the ring of delocalised π-electrons and methylbenzene is produced. H CH3 CH3 !!: + H+ + H+ + AlCl4 – !!: HCl + AlCl3 7 With chloroethane, ethylbenzene is produced. CH2CH3 AlCl3 + CH3CH2Cl !!!!: + HCl When heated with zinc powder to about 650 °C, dehydrogenation occurs to produce phenylethene (common name: styrene). CH2CH3 CHRCH2 !!Δ !: + H2 Phenylethene is used to produce poly(phenylethene) or polystyrene. Info Chem Polystyrene: !! ( CH!CH2!!) n ! 86
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 Example 15.15 What is the structure of the product formed when benzene is treated with 2-iodopropane in the presence of anhydrous aluminium chloride? Give the formula of the electrophile. Solution The product is 2-phenylpropane. CH3!CH!CH3 The electrophile is: H & CH3!C!CH3 + Acylation 1 Acylation is the substitution of an acyl group into a benzene ring. R!C! ∫ O 2 Benzene is mixed with acyl chloride (also known as acid chloride) in the presence of anhydrous aluminium chloride and warmed to about 80 °C to produce aromatic ketones. H O O ∫ ∫ + R!C!Cl !!: R!C! + HCl Acyl chloride Aromatic ketone 3 For example: H O O ∫ ∫ + CH3!C!Cl !!: CH3!C! + HCl Ethanoyl chloride Phenyl ethanone (Methylphenyl ketone) 4 The electrophile is the ion which is produced by the following reactions. R!C+ ∫ O R!C!Cl + AlCl3 L R!CRO + [AlCl4]– ∫ + O Exam Tips Exam Tips A convenient way to prepare aromatic ketones. 2017/P3/Q18(a) Info Chem Benzene reacts with propene in the presence of phosphoric acid and aluminum chloride to produce 2-phenylpropane. + CH3–CH=CH2 p CH3–CH–CH3 & 87
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 Example 15.16 Using balanced equations, describe the mechanism of the reaction between benzene and ethanoyl chloride in the presence of anhydrous aluminium chloride. Solution CH3COCl + AlCl3 !: CH3!+ CRO + AlCl4 – H H COCH3 CH3!+ CRO + !!: + H COCH3 + !!: !C!CH3 + H+ ∫ O H+ + AlCl4 – !!: HCl + AlCl3 Sulphonation 1 Benzene reacts with oleum (or fuming sulphuric acid) when warmed to produce benzenesulphonic acid. SO3H + SO3 !!: 2 The attacking species is the sulphur trioxide molecule, SO3, which is present in oleum. Due to the polarisation of the sulphuroxygen bond (oxygen being more electronegative), the sulphur atom in the SO3 molecule acquires a partial positive charge and acts as the electrophile that attacks the benzene ring. H H SO3– SO3H + !!: !!: O O O R RSδ R+ + H SO3– + Info Chem Oleum is concentrated sulphuric acid saturated with sulphur trioxide. Info Chem Structural formula of benzenesulphonic acid: !S!OH O O R R 88
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 3 If concentrated sulphuric acid is used instead of oleum, then a higher temperature is required for the reaction to occur. 4 When diluted with water and warmed, benzenesulphonic acid is hydrolysed to phenol. SO3H OH + H2O !!: + H2SO4 This is the easiest way of preparing phenol in the laboratory. 5 Benzenesulphonic acid is used to manufacture detergents. As the detergents contain !SO3H group, they lather (form bubbles) even in hard water as the calcium and magnesium salts of sulphonic acid are soluble in water. An example is: CH3(CH2)11 SO3 – Na+ Hydrogenation 1 When the vapour of benzene and hydrogen is passed over heated nickel, hydrogenation takes place and cyclohexane is produced. + 3H2 !!: 2 In this aspect, benzene acts as if it is an alkene. Quick Check 15.11 Give the structures of the monosubstituted products formed when benzene is treated with the following under suitable conditions. 1 CH3CH2CHCH3 2 !C!Cl 3 Concentrated sulphuric acid & ∫ Cl O Substituent Effect 1 When a substituted benzene, C6H6Y, undergoes further substitution, the substituted group (Y) originally in the substituted compound will influence the rate and location of attack on the benzene ring of the incoming electrophile. Info Chem Hard water that does not lather with soap is due to the presence of dissolved Ca2+ and Mg2+ salts. Manufacture of detergents 2012/P1/Q29 2013/P3/Q5 89
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 2 Y can be a ring-activating group or ring-deactivating group. 3 Ring-activating groups are electron donors (positive inductive effect). These groups increase the electron density of the benzene ring and make it more reactive towards electrophilic substitution compared to benzene itself. 4 Ring-deactivating groups are electron acceptors (negative inductive effect). These groups decrease the electron density of the benzene ring and make it less reactive towards electrophilic substitution compared to benzene itself. 5 Ring-activating groups are usually single atom or saturated groups (with single bonds in their structure). They are ortho and para directing. They direct incoming electrophiles to the ortho (carbon-2 and carbon-6) and para (carbon-4) position. This is because they increase the electron density at the 2-, 4- and 6- positions. 6 Examples of ortho and para directors are: R (alkyl), !OH, !NH2, halogens and !OR (alkoxide). 7 Ring-deactivating groups are usually unsaturated groups containing either double bonds or triple bonds in their structures. They are meta (carbon-3 and carbon-5) directing. This is because they decrease the electron density at the 2-, 4- and 6- positions, thereby making the 3- and 5- positions relatively richer in electrons. 8 Examples of meta directors are: !N!O, !C!OH, !C!, !C#N, !C!OR, ∫ ∫ ∫ ∫ O O O O !C!Cl, !C!NH2 ∫ ∫ O O 9 Coming back to our previous question regarding further nitration of nitrobenzene. As the nitro group, !NO2, is a ring deactivator, further nitration of nitrobenzene requires more st r ingent conditions, and the produc t for med is 1,3-dinitrobenzene. NO2 NO2 Concentrated H2SO4, >55 °C + HNO3 !!!!!!!!!: + H2O NOCH23 Y p Electron-donating group Y q Electron-accepting group Info Chem –NO2 is electron-withdrawing. It exhibits negative inductive effect. It is more difficult to introduce a second –NO2 group into nitrobenzene because the –NO2 group is a ring-deactivating group. 90
Chemistry Term 3 STPM Chapter 15 Hydrocarbons 15 10 Further nitration would produce 1,3,5-trinitrobenzene. NO2 O2N NO2 Example 15.17 Explain why ring-activating group, such as !OH, is 2-, 4- and 6-directing. Solution The structure of phenol is as shown below. The Kekulé structure is used for this explanation. :O H The lone pair electrons on the oxygen atom can shift to form a carbon-oxygen double bond, causing the migration of the π electrons in one of the carbon-carbon double bonds giving rise to the resonance structures as shown below. fi :O H fi fi L O+ H – fi L RR L O+ H – RR fi O+ H – fi RR The delocalisation causes an increase in the electron density on C-2, C-4 and C-6. As a result, they are more susceptible to attack by electrophiles than C-3 and C-5. +O H – – – RR Info Chem 1,3,5-trinitrobenzene is an explosive. It is a light yellow solid. 91