Pra-U STPM Chemistry Term 3 2018 CB039348c

3 PELANGI BESTSELLER Biology Term 3 STPM Chapter 14 Taxonomy and Biodiversity PELANGI MCHEMISTRY STPM Text CHEMISTRY CHEMISTRY STPM Text INFO VIDEO Peter Yip 3 TERM

1 3 TERM First Published 2018 ISBN: 978-983-00-9000-9 Printed in Malaysia by UG Press Sdn. Bhd. 40, Jalan Pengasah 15/13, 40000 Shah Alam, Selangor Darul Ehsan. Tel: 03-8922 3993 Fax: 03-8926 1223 / 8920 2366 E-mail: [email protected] Enquiry: [email protected] Please log on www.pelangibooks.com/errata for up-to-date adjustments to the contents of the book (where applicable) Peter Yip CHEMISTRY CHEMISTRY STPM Text

PREFACE Pre-U STPM Text Chemistry Term 3 is written based on the latest syllabus prepared by the Malaysian Examinations Council (MEC). The book is designed and well organised with the following features to help students understand the concepts taught. Analysis of STPM Papers (2015 – 2017) v Chapter 2015 2016 2017 A B C A B C A B C 14 Introduction to Organic Chemistry 3 3 3 15 Hydrocarbons 2 1 1 1 3 16 Haloalkanes 1 1 3 2 1 17 Hydroxy Compounds 2 1 2 1 18 Carbonyl Compounds 1 1 —4 2 1 1 19 Carboxylic Acids and their Derivatives 2 1 1 —4 2 1 1 —2 1 1 —2 20 Amines, Amino Acids and Proteins 2 1 1 —2 2 1 —2 4 1 —2 1 21 Polymers 2 1 1 Total 15 2 3 15 2 3 15 2 3 STPM Scheme of Assessment Term of Study Paper Code and Name Type of Test Mark (Weighting) Duration Administration First Term 962/1 Chemistry Paper 1 Written Test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 1 to 6. 60 (26.67%) 15 15 30 1—1 2 hours Central assessment Second Term 962/2 Chemistry Paper 2 Written Test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 7 to 13. 60 (26.67%) 15 15 30 1—1 2 hours Central assessment Third Term 962/3 Chemistry Paper 3 Written Test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 14 to 21. 60 (26.67%) 15 15 30 1— 1 2 hours Central assessment 962/5 Chemistry Paper 5 Written Practical Test 3 compulsory structured questions to be answered. 45 (20%) 1—1 2 hours Central assessment First, Second and Third Terms 962/4 Chemistry Paper 4 School-based Practical Assessment 13 compulsory experiments and one project to be carried out. 225 to be scaled to 45 (20%) Throughout the three terms School-based assessment vi Chemistry Term 3 STPM Chapter 21 Polymers 21 CHAPTER POLYMERS 21 Concept Map Learning earning Outcomes Students should be able to: • state the examples of natural and synthetic polymers; • define monomer, polymer, repeating unit, homopolymer and copolymer; • identify the monomers in a polymer; • describe condensation polymerisation as exemplified by terylene and nylon-6,6; • describe addition polymerisation as exemplified by poly(ethene)/polyethylene/ polythene, poly(phenylethene)/polystyrene and poly(chloroethene)/polyvinylchloride; • state the role of Ziegler-Natta catalyst in the addition polymerisation process; • explain the classification of polymers as thermosetting, thermoplastic and elastomer; • identify isoprene (2-methylbuta-1,3-diene) as the monomer of natural rubber; • describe the two isomers in poly(2-methylbuta-1,3- diene) in terms of the elastic cis form (from the Hevea brasiliensis trees) and the inelastic trans form (from the gutta-percha trees); • state the uses of polymers; • explain the difficulty in the disposal of polymers; • outline the advantages and disadvantages of dumping polymer-based materials in rivers and seas. Polymers Uses of Polymers Effects of Polymer Disposal on the Environment Addition Polymerisation • Examples of addition polymers: – Poly(ethene), PE – Polypropylene – Poly(chloroethene) • Polymerisation of dienes • Natural rubber • Production of rubber from latex • Synthetic rubber Condensation Polymerisation • Polyamides • Polyesters STPM Scheme of Assessment Latest STPM scheme of Assessment starting 2012. Concept Map Provides an overall view of the concepts learnt in the chapter. Learning Outcomes A list of subtopics that students will learn in each chapter. ii Analysis of STPM Papers Shows past yearsʼ STPM questions according to the chapters.

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 14 (b) On the other hand, the phenyl group is an electronwithdrawing group (negative inductive effect). N: H H This reduces the electron density of the lone pair electrons and makes them less reactive towards H+. Example 14.8 Arrange the following compounds in the order of decreasing acid strength. Explain your answer. Chloroethanoic acid, bromoethanoic acid, iodoethanoic acid Solution Acid strength: Chloroethanoic acid > bromoethanoic acid > iodoethanoic acid The structures of the three acids are: Cl←CH2COOH Br←CH2COOH I←CH2COOH The electronegativity of the halogen decreases in the order: Cl > Br > I due to the increase in the size of the halogen atom with increasing proton number. As a result, the inductive effect decreases in the order: Cl > Br > I leading to the decrease in the acid strength in the same order. SUMMARY SUMMARY 1 Organic chemistry is the study of carbon compounds. 2 A homologous series is a group or family of compounds with the following characteristics: • They can be represented by a general formula. • The molecular formula of each member differs from their preceding member by a CH2 group. • They show a gradual change in their physical properties as the relative molecular mass increases. • They have the same functional group and hence have similar chemical properties. • They can be prepared using similar methods. 3 A functional group is an atom or group of atoms that gives the chemical properties of the compound. 4 The empirical formula shows the simplest ratio between the different types of atoms in the molecule of a compound. 5 The molecular formula gives the actual number of the different types of atoms in the molecule of a compound. 6 The structural formula gives the molecular formula of a compound and the way the atoms are bonded to one another. 7 Isomerism is the occurrence of two or more compounds with the same molecular formula. INFO Acidity and Inductive/ Resonance Effect 40 Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 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: H3 O O C 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. 42 138 Chemistry Term 3 STPM Chapter 16 Haloalkanes 16 16 STPM PRACTICE Objective Questions 1 2-chloro-2-methylpropane and 1-chlorobutane are separately boiled with aqueous sodium hydroxide. Which of the following can be used to differentiate the products of the reactions? A Phosphorus pentachloride B Tollens reagent C Alkaline iodine D Acidified potassium manganate(VII) 2 Which of the following reagents does not react with both benzyl chloride and chlorobenzene under suitable conditions? A Magnesium in ether B Sodium metal C Bromine in the presence of iron(III) bromide D Potassium hydroxide 3 A reaction scheme is shown below. KCN, ethanol Dilute H2SO4 CH3CH2CH2I X Y What are the structures of X and X Y ? Y A CH3CH2CH2I – K+ CH3CH2CH2SO3H B CH3CH2CH2CN CH3CH2CH2COOH C CH3CH2CH2CN CH3CH2CH2COO– K+ D CH3CHCH & 3 CN CH3CHCH & 3 4 Which of the following is COOH not a property of ‘Freons’? A They are chemically inert. B They are easily liquefied by pressure. C They have a pungent smell. D They are volatile. 5 Freons (chlorofluorocarbon) are used as cleaning agents. Which of the bonds in the structure below has the lowest bond energy? H H & & A & B & C Cl !!C !!C !!F & & & & H H 6 When 1,2-diiodoethane is boiled under reflux with aqueous sodium hydroxide, ethane-1,2- diol is produced. What is the product formed when 1,1-diiodoethane is boiled under reflux with aqueous sodium hydroxide? A Ethane-1,1-diol B Ethanal C Ethanoic acid D No reaction 7 What is the pro duc t for med w hen 1,1,1-trichloroethane is boiled under reflux with aqueous sodium hydroxide? A CH3COOH B CH3CH C 2ONa CH3COONa D CH3COCH3 8 Which reaction is a nucleophilic substitution reaction? A Benzene and a mixture of concentrated nitric acid and concentrated sulphuric acid at 55 °C B Chloromethylbenzene and ethanolic potassium cyanide C Propanone and aqueous hydrogen cyanide in the presence of sodium cyanide D Chlorocyclohexene with acidified potassium manganate(VII) 9 Which compound reacts with methylmagnesium iodide to produce 3-methyl-2-butanol? A CH3CH2COCH3 B CH3COCH3 C CH3CH2CH D (CH 2CHO 3)3CH 10 2OH Which statement about the compound below is not correct? CH3CH2CH2CH2MgI A It contains the carbon-magnesium covalent bond. B It can be used to prepare alcohols. C It reacts vigorously with water. D It is a convinient source of carbonium ion. D Example Gives step-by-step solutions to explain the example. QR Code Info/Video Scan QR code using mobile smartphone devices to access related information or videos Online Quick Quiz Scan QR code for self-assessment at the end of each chapter 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 such as benzene. is a process where alkanes are converted into arenes 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. Tips Exam Tips Exam 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 Info Chem Provides extra information that relates to the subtopics learnt. Past-year Questions tagging Enables students to know the topics frequently tested in the exam. Exam Tips Provides helpful tips for students in answering exam questions. Notes Provides explanation in a simple way to ease students’ understanding. Summary Summarises key concepts learnt in the chapter. STPM Practice A variety of examinationtype questions to check students’ understanding of the chapters learnt. Side Column Notes Provides additional information or pointers to improve students’ understanding. iii Quick Check Provides short questions for students to test their understanding of the concepts learnt in the subtopics.

308 Section A [15 marks] Bahagian A [15 markah] Answer all questions in this section. Jawab semua soalan dalam bahagian ini. 1 Ethyne has the molecular formula of H!C;C9H. Which orbital of the carbon atom overlaps with the 1s orbital of hydrogen to form the C9H σ bond? Etuna mempunyai formula molekul H!C;C!H. Orbital yang manakah pada atom karbon yang bertindih dengan orbital 1s pada atom hidrogen untuk membentuk ikatan σ C!H? A sp3 C 2p B sp D 2s 2 The resonance structures of benzene, C6 H6 are shown below: Struktur resonans untuk benzena, C6 H6 ditunjukkan di bawah. H H H H H H H H H H H H Which of the following statements is not true regarding the resonance structures? Pernyataan yang manakah antara berikut tidak benar tentang struktur-struktur resonans itu? A They have the same arrangement of atoms. Mereka mempunyai susunan elektron yang sama. B They have the same energy. Mereka mempunyai tenaga yang sama. C They are in rapid equilibrium with one another. Mereka berada dalam keseimbangan pesat antara satu sama lain. D They have the same number of electrons. Mereka mempunyai bilangan elektron yang sama. 3 Metoprolol is a cardioselective β1-adrenergic blocking agent used for the treatment of mild hypertension. It has the following structure. Metoprolol adalah agen pencegah penyakit kardioselektif β1 yang digunakan untuk rawatan darah tinggi yang sederhana. Ia mempunyai struktur berikut. CH3 H3 CO OH O H N CH3 Metoprolol is classified as Metoprolol dikelaskan sebagai suatu A a ketone keton B a carboxylic acid asid karboksilik C a phenol fenol D an amine amina 4 Which of the following compounds has the lowest pKa value? Sebatian yang manakah antara berikut mempunyai nilai pKa yang paling rendah? A ClCH2 COOH B FCH2 OH C CH3 CH2 OH D CO2 H STPM Model Paper (962/3) 322 Addition polymerisation Pempolimeran penambahan A process where small molecules (called monomers) are joined with one another to form long-chain molecules without the elimination of other smaller molecules Addition reaction Tindak balas penambahan A reaction where two or more molecules are combined to form a larger molecule Amino acid Asid amino A molecule that contains the basic 9NH2 group and the acidic structure 9COOH group in its Carbanion Karbanion An anion where the negative charge is on a carbon atom Carbohydrates Karbohidrat Naturally occurring polyhydroxylated carbonyl compounds Carbonium ion Ion karbonium A cation where the positive charge is on a carbon atom Catenation Katenasi The ability of the atoms of an element to form covalent bonds with one another Condensation polymerisation Pempolimeran kondensasi A process where small molecules (called monomers) are joined with one another to form long-chain molecules with the elimination of other smaller molecules Copolymer Kopolimer A polymer that is made up of two or more different monomers Cracking Peretakan A process where large hydrocarbon molecules are broken into smaller molecules Elastomer Elastomer A polymer that stretches when pulled and returns to its original length when the force is released Electrophile Elektrofil A reagent that accepts an electron pair from another substance to form a coordinate (dative) bond Elimination reaction Tindak balas penyingkiran A reaction where two or more atom or group of atoms are removed from a molecule Empirical formula Formula empirik It shows the simplest whole number ratio of atoms of each element present in the molecule of a compound Enantiomerism Enantiomerisme See optical isomerism Enantiomers Enantiomer Compounds with the same molecular formula but are mirror images and are not superimposable Essential amino acids Asid amino perlu Amino acids that cannot be synthesised by the body and have to be obtained through food supplements Fatty acids Asid lemak Long-chain carboxylic acids that normally exist as the esters of glycerol(1,2,3-propanetriol) 325 Chapter 14 Introduction to Organic Chemistry Quick Check 14.1 1 (a) C:H:O = 3.3 : 6.6 : 3.3 = 1 : 2 : 1 Empirical formula = CH2O (b) (CH2O)n = 90.1 n = 3 Molecular formula = C3H6O2 2 (a) No. of moles of CO2 = 1.76 ——44 = 0.04 No. of moles of H2O = 0.72 ——18 = 0.04 C : H = 0.04 : (2 × 0.04) = 1 : 2 Empirical fomula = CH2 (b) (CH2)n = 70 n = 5 Molecular formula = C5H10 Quick Check 14.2 1 No 2 No 3 Yes 4 No 5 No 6 No 7 No Quick Check 14.3 1 (a) No (b) No (c) Yes, CH3*CH(OH)NH2 (d) No (e) Yes, H I * (f) No Quick Check 14.4 1 (a) Electrophile (b) Nucleophile (c) Free radical (d) Electrophile (e) Electrophile (f) Nucleophile (g) Nucleophile 2 (c), (d) and (e) Revision Exercise 14 Objective Questions 1 B 2 C 3 B 4 B 5 D 6 D 7 C 8 D 9 B 10 A 11 C 12 A Structured and Essay Questions 1 (a) The occurrence of two more compounds having the same structural formula but with different spatial arrangement of the atoms or groups. (b) Geometrical isomerism and optical isomerism (c) Optical isomers: CHRCH!CH2Br BrCH2!CHRCH CH3 CH3 C H C H OH OH Geometrical isomers: H H OH CH2Br CH3!CH CRC CH2Br H OH H CH3!CH CRC cis-isomer trans-isomer (d) The two optical isomers. 2 (a) CH3!CH2!CHRCH2 1-butene CH3!CHRCH!CH3 2-butene CH3!CHRCH2 2-methylpropene & CH3 (b) (i) Geometrical isomerism CH3 CH3 H H CRC cis-2-butene H CH3 CH3 H CRC trans-2-butene (ii) Boiling point. The cis-isomer will have a higher boiling point than the trans-isomer. 3 (a) (i) A free radical is a species that has an unpaired electron. Example is the chlorine free radical, Cl•. (ii) A nucleophile is a species that can donate a lone pair electrons to another species to form a coordinate bond. Examples are NH3, CN– and OH– . (iii) An electrophile is a species that accepts a lone pair electrons from another species to form a coordinate bond. Examples are AlCl3, Cl+. (b) Tertiary free radicals have three alkyl groups that are electron-releasing which helps to stabilise the free radical. Whereas, secondary free radicals have two electron-releasing alkyl groups, and primary free radicals have one alkyl group. Chapter 15 Hydrocarbons Quick Check 15.1 1 CH3CH2CH2CH2CH2CH3 CH3CH2CH2!CH!CH3 & CH3 CH3 & CH3CH2!CH!CH2!CH3 CH3 CH3 & & CH3!CH!CH!CH3 CH3 & CH3!CH2!C!CH3 & CH3 ANSWERS 358 Chemistry Term 3 STPM Index Index A Addition polymerisation 292 Aldose 197 Alkane 45 Arenes 79 Atactic form 294 Azo-dye 263 B Baeyer test 72 Benzene, structure 11 Benzenediazonium ion, reaction of 263 Benzylic carbon 97 C Carbanions 29 Carbohydrate 197 Carbon atom, classification 20 Carbonium ion 29 Catalytic convertor 60 Catenation 3 cis-trans isomerism 23 Condensation polymerisation 289 Coupling reaction 263 Cracking 58 Crude oil, fractional distillation 57 Cumene, process 173 Cyanohydrin 192 Cycloalkane 49 D DDT 133 Dettol 174 Dextrorotatory 26 D-isomer 26 2,4-dinitrophenylhydrazine 190 E Elastomer 298 Electrophile 31 Electrophilic addition, alkenes 66 Electrophilic substitution, arenes 82 Empirical formula 12 Enantiomers 24 Ethanedioic acid 220 F Fatty acid 209 Fehling’s solution 196 Formic acid 220 Free radical 27 Free radical substitution, mechanism 52 Fermentation 166 Friedel-Craft reaction 85 Fructose 197 G Geometrical isomerism 23 Glucose 197 Grignard’s reagent 128 H Haloalkanes 108 Halogen carrier 84 HDPE 293 Homologous series 18 Hybridisation 5 I Inductive effect, negative 34 Inductive effect, positive 34 Iodoform test 164, 196 Isomerism 21 Isotactic form 294 STPM Model Paper A model paper that follows the latest STPM exam format is provided for practice. Glossary Gives a list of important terms to ease the students’ understanding of their meanings. Answers Complete answers are provided. Index Provides a list of terms to enable easy and direct access to the subtopics. iv

Analysis of STPM Papers (2015 – 2017) v Chapter 2015 2016 2017 A B C A B C A B C 14 Introduction to Organic Chemistry 3 3 3 15 Hydrocarbons 2 1 1 1 3 16 Haloalkanes 1 1 3 2 1 17 Hydroxy Compounds 2 1 2 1 18 Carbonyl Compounds 1 1 —4 2 1 1 19 Carboxylic Acids and their Derivatives 2 1 1 —4 2 1 1 —2 1 1 —2 20 Amines, Amino Acids and Proteins 2 1 1 —2 2 1 —2 4 1 —2 1 21 Polymers 2 1 1 Total 15 2 3 15 2 3 15 2 3

STPM Scheme of Assessment Term of Study Paper Code and Name Type of Test Mark (Weighting) Duration Administration First Term 962/1 Chemistry Paper 1 Written Test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 1 to 6. 60 (26.67%) 15 15 30 1—1 2 hours Central assessment Second Term 962/2 Chemistry Paper 2 Written Test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 7 to 13. 60 (26.67%) 15 15 30 1—1 2 hours Central assessment Third Term 962/3 Chemistry Paper 3 Written Test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 14 to 21. 60 (26.67%) 15 15 30 1— 1 2 hours Central assessment 962/5 Chemistry Paper 5 Written Practical Test 3 compulsory structured questions to be answered. 45 (20%) 1—1 2 hours Central assessment First, Second and Third Terms 962/4 Chemistry Paper 4 School-based Practical Assessment 13 compulsory experiments and one project to be carried out. 225 to be scaled to 45 (20%) Throughout the three terms School-based assessment vi

CONTENTS Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 14 INTRODUCTION TO ORGANIC CHEMISTRY 1 14.1 Bonding of the Carbon Atoms 3 14.2 Empirical, Molecular and Structural Formulae of Organic Compounds 12 14.3 Functional Groups: Classification and Nomenclature 18 14.4 Isomerism: Structural and Stereoisomerism 21 14.5 Free Radicals, Nucleophiles and Electrophiles 27 14.6 Molecular Structure and its Effect on Physical Properties 32 14.7 Inductive and Resonance Effect 34 Summary 40 STPM Practice 14 41 QQ 42 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 15 HYDROCARBONS 43 15.1 Alkanes 45 15.2 Alkenes 60 15.3 Arenes 79 Summary 99 STPM Practice 15 100 QQ 107 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 16 HALOALKANES 108 16.1 Classification 109 16.2 Nomenclature 110 16.3 Structural and Optical Isomerism in Haloalkanes 112 16.4 Physical Properties of Haloalkanes 112 16.5 Nucleophilic Substitution of Haloalkanes 114 16.6 Elimination Reaction 121 16.7 Mechanism of Nucleophilic Substitution 122 16.8 Reactivity of Primary, Secondary and Tertiary Haloalkanes 124 16.9 Reactivity of Chlorobenzene and Chloroalkanes in Hydrolysis Reactions 127 16.10 Organometallic Compounds 128 16.11 Uses of Haloalkanes 132 16.12 DDT 133 16.13 Effects of CFC on the Environment 134 Summary 137 STPM Practice 16 138 QQ 142 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 17 HYDROXY COMPOUNDS 143 17.1 Introduction to Hydroxy Compounds 144 17.2 Alcohols 150 17.3 Phenols 166 vii

Summary 175 STPM Practice 17 176 QQ 180 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 18 CARBONYL COMPOUNDS 181 18.1 Introduction 182 18.2 Nomenclature 183 18.3 Structural and Optical Isomerism in Carbonyl Compounds 185 18.4 Physical Properties 186 18.5 Preparation of Carbonyl Compounds 187 18.6 Reactions of Carbonyl Compounds 189 18.7 Carbohydrates 197 Summary 199 STPM Practice 18 200 QQ 204 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 19 CARBOXYLIC ACIDS AND THEIR DERIVATIVES 205 19.1 Carboxylic Acids 207 19.2 Acyl Chloride 221 19.3 Esters 228 19.4 Amides 234 Summary 240 STPM Practice 19 242 QQ 247 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 20 AMINES, AMINO ACIDS AND PROTEINS 248 20.1 Amines 250 20.2 Amino Acids 265 20.3 Proteins 277 Summary 278 STPM Practice 20 280 QQ 286 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 21 POLYMERS 287 21.1 Introduction 288 21.2 Synthetic Polymers 288 21.3 Condensation Polymerisation 289 21.4 Addition Polymerisation 292 21.5 Classification of Polymers 297 Summary 303 STPM Practice 21 303 QQ 307 • STPM Model Paper (962/3) 308 • Appendix 315 • Glossary 322 • Answers 325 • Index 358 viii

CHAPTER Functional Groups: Classification and Nomenclature Chemical Formulae • Empirical formula • Molecular formula • Structural formula Hybridisation Involving Carbon Atoms • sp, sp2, sp3 hybridisation Free Radicals, Nucleophiles and Electrophiles Molecular Structure and its Effect on Physical Properties Inductive and Resonance Effect • Negative • Positive Introduction to Organic Chemistry Bonding of the Carbon Atoms Isomerism • Structural isomerism • Stereoisomerism INTRODUCTION TO 14 ORGANIC CHEMISTRY Concept Map

Learning earning Outcomes Students should be able to: Bond ing of the carbon atoms: shapes of ethane, ethene, ethyne and benzene molecules • use the concept of sp3, sp2 and sp hybridisations in carbon atoms to describe the bonding and shapes of molecules as exemplified by CH4, C2H4, C2H2 and C6H6; • explain the concept of delocalisation of π electrons in benzene ring. General, empirical, molecular and structural formulae of organic compounds • state general, empirical, molecular and structural formulae of organic compounds; • determine the empirical and molecular formulae of organic compounds. Functional groups: classification and nomenclature • describe classification of organic compounds by functional groups and nomenclature according to the IUPAC rules for the following classes of compounds: (i) alkanes, alkenes, alkynes and arenes (ii) haloalkanes (iii) alcohols (including primary, secondary and tertiary) and phenols (iv) aldehydes and ketones (v) carboxylic acids and their derivatives (acyl chlorides, amides and esters) (vi) primary amines, amino acids and protein Isomerism: structural and stereoisomerism • define structural and stereoisomerism (geometrical and optical); • explain the meaning of chiral centre in the optical isomerism; • classify isomers as structural, cis-trans and optical isomers; • identify chiral centres and/or cis-trans isomerism in the molecule of given structural formula; • deduce possible isomers for an organic compound of known molecular formula. Free radicals, nucleophiles and electrophiles • describe homolytic and heterolytic fissions; • define the terms free radical, nucleophile and electrophile; • explain that nucleophiles such as OH−, NH3, H2O, Br−, I− and carbanion are Lewis bases; • explain that electrophiles such as H+, NO2 +, Br2, AlCl3, ZnCl2, FeBr3, BF3 and carbonium ion are Lewis acids. Molecular structure and its effect on physical properties • describe the relationship between the size of molecules in the homologous series and the melting and boiling points; • explain the forces of attraction between molecules (van der Waals forces and hydrogen bonding). Inductive and resonance effect • explain inductive effect which can determine the properties and reactions of functional groups; • use inductive effect to explain why functional groups such as −NO2, −CN, −COOH, −COOR, C=O, −SO3H, −X (halogen), −OH, −OR, −NH2, −C6H5 are electron acceptors whereas R(alkyl) is an electron donor; • explain how the concept of induction can account for the differences in acidity between CH3COOH, ClCH2COOH, Cl2CHCOOH and Cl3CCOOH; between ClCH2CH2CH2COOH and CH3CH2CHClCOOH; • use the concept of resonance to explain the differences in acidity between CH3CH2OH and C6H5OH, as well as the differences in basicity between CH3NH2 and C6H5NH2. 2

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 14.1 Bonding of the Carbon Atoms 1 Organic chemistry is a branch of chemistry dealing with the study of the compounds of carbon, excluding substances such as carbon dioxide, carbon monoxide, metallic carbides, cyanides, cyanates, carbonates and hydrogen carbonates. 2 Today, there are more than 10 million known organic compounds and new ones are discovered or synthesised almost daily. 3 What is so unique about the carbon atom that it is possible for it to form such a vast variety of compounds? 4 Carbon is the first member of the Group 14 elements in the Periodic Table. 5 Due to its small size, the carbon atom can form very strong covalent bonds with other elements and also with atoms of its own kind. The latter is known as catenation. 6 Other than forming single bonds with other atoms, carbon also form strong carbon-carbon double bonds and carbon-carbon triple bonds, resulting in long chains or ring structures of different sizes and complexities. Type of bond Bond energy/kJ mol–1 C!C 350 CRC 610 C#C 840 !C!C!C!C!C! Straight chain !C!CRC!C!C#C! Branched chain & C & Or Ring/cyclic structure C C C C C C 7 Due to its ability to catenate, it is able to form an immense diversity of compounds, from the simplest methane (with one carbon atom) to nucleic acids (such as DNA) which have hundreds of thousands of carbon atoms. 8 However, not all organic compounds are derived from living organisms. Many synthetic organic compounds such as medicines, dyes, polymers, and pesticides are prepared in the laboratory. Organic chemistry is the study of carbon compounds. ‘Organic’ means ‘derived from living things’. Carbon can form very strong carbon-carbon bonds. Info Chem Carbon also forms double bond and triple bond with other elements. Example: O ' 9C9, 9C#N Synthetic organic compounds are prepared in the laboratory. 3

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 9 In the valence bond theory, a covalent bond is formed when two atoms approach each other until their atomic orbitals (each containing one unpaired electron) overlap with one another. 10 Take the example of the hydrogen molecule, H2. H x H H H H H x x Atomic orbitals Overlap H2 molecule 11 The electronic configuration of carbon is 2.4 or 1s 2 2s 2 2p2 . 1s 2s 2p u u n n 12 Even though the configuration shows that the carbon atom has only two unpaired electrons, carbon exhibits tetravalency in almost all of its compounds. 13 This is made possible when one of the paired 2s electrons is promoted to one of the empty 2p orbitals. 1s 2s 2p C* [C*: Excited carbon atom] u n n n n 14 As a result, carbon now has four unpaired electrons which it can use to form single bonds or multiple bonds with other atoms. Examples are: H & H!C!H & H H OH & & H!CRC!H H!C#C!H 15 Let’s look at the methane molecule more closely. Due to carbon having two types of orbitals (2s and 2p) to form bonds, we might expect the methane molecule to have two types of carbonhydrogen bonds. One s–s bond and three s–p bonds. 1s 2s 2p C* u n n n n • H • H • H • H 16 However, experiments show that all the carbon-hydrogen bonds in the methane molecule are identical and are directed towards the corners of a tetrahedron. 17 To explain these observations, Linus Pauling put forward the hybridisation theory in 1931. H & s–s bond H!C!H & s–p bond H one s–s bond and three s–p bonds Info Chem The 2 unpaired electrons must have opposite spin. 4

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 Hybridisation Involving Carbon Atoms 1 Hybridisation is the mixing of two or more non-equivalent atomic orbitals to give a set of equivalent hybrid orbitals. 2 The number of hybrid orbitals formed is equal to the total number of atomic orbitals that took part in the process. 3 An analogy to hybridisation is the mixing of 50 cm3 of redcoloured water and 50 cm3 of blue-coloured water. The result is 100 cm3 of pink-coloured water. Red Blue Pink 4 Hybridisation of two atomic orbitals of 50 kJ mol–1 and 70 kJ mol–1 respectively produces two hybrid orbitals of 60 kJ mol–1 each. 50 70 Energy/kJ mol–1 60 60 Atomic orbitals Hybrid orbitals 5 The main type of hybridisation in carbon atom involves the 2s and 2p orbitals. The 2s orbital has a lower energy than the 2p orbitals. 2p Energy 2s 6 When an s-orbital mixes with a p-orbital, the result is two sp hybrid orbitals of the same shape and energy. + 2 s-orbital p-orbital sp-orbital 2p Energy sp sp 2s sp hybrid orbitals What is hybridisation? 2014/P3/Q1 2015/P3/Q1 2017/P3/Q1 VIDEO Hybridisation 5

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 7 There are three types of hybridisation involving the s and p orbitals: sp3 hybridisation, sp2 hybridisation and sp hybridisation. sp3 Hybridisation 1 sp3 hybridisation involves the mixing of one s-orbital and three p-orbitals to form four equivalent sp3 hybrid orbitals. 1s 2s 2p C* u n n n n 1s sp3 sp3 sp3 sp3 C* u n n n n 2 The four sp3 hybrid orbitals (each containing one electron) direct themselves towards the 4 corners of a tetrahedron with bond angles of 109.5°. Info Chem sp3 is pronounced as ‘s-p-three’. z + + z y x s z y x p x z y x p y + + _ _ z y x p z + _ y x – z y x + – z y x – + z y x – + 6

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 3 Overlap of the four sp3 orbitals with the 1s orbitals of four hydrogen atoms produces the methane molecule, CH4. C H H H H 109.5° sp3 sp3 sp3 1S sp3 1S 1S 1S σ σ σ σ This produces four equivalent carbon-hydrogen sigma (σ) bonds. 4 Another example of sp3 hybridisation is found in ethane, C2H6. C H H C H H H H sp2 Hybridisation 1 sp2 hybridisation involves the mixing of an s-orbital and two p-orbitals to form three sp2 hybrid orbitals. 1s 2s 2p C* u n n n n 1s sp2 sp2 sp2 2p C* u n n n n Info Chem sp2 is pronounced as ‘s-p-two’. Unhybridised 2p orbital 7

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 2 The three sp2 hybrid orbitals arrange themselves in a trigonplanar structure with bond angles of 120°. z y x s z y x px z y x py + _ _ z y x + – z y x + – z y + x + + 120° z y x – + + + 3 However, there is a 2p orbital not involved in the hybridisation process. This unhybridised 2p orbital is placed perpendicular to the plane containing the three sp2 hybrid orbitals [Figure (a)]. Unhybridised p orbital Planar sp2 sp2 sp2 p 120° C C H H H H Sigma (σ) bonds (a) (b) σ σ σ σ σ 4 Let us consider the overlap that leads to the formation of the ethene molecule, C2H4. When two sp2 hybridised carbon atoms approach one another, they form a carbon-carbon σ bond with one another. The remaining sp2 hybrid orbitals overlap with the 1s orbital of four hydrogen atoms to form four carbon-hydrogen σ bonds [Figure (b)]. 120° sp2 sp2 sp2 Unhybridised 2p-orbital 8

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 5 The two unhybridised 2p orbitals then overlap sideways to form a carbon-carbon π bond. C C C C H H H H H H H H σ σ 6 The ethene molecule is represented as H H H H C Cσ "π 7 As might be expected, the carbon-carbon bond in ethene is stronger and shorter than the carbon-carbon bond in ethane because it involves the sharing of four electrons instead of two. Molecule Carbon-carbon bond length/nm Carbon-carbon bond energy/kJ mol–1 Ethane, C2H6 0.154 376 Ethene, C2H4 0.133 611 8 However, the carbon-carbon double bond in ethene is considerably less than twice as strong as the carbon-carbon single bond in ethane. This is because the overlap which results in the formation of the π bond is not as extensive compared to the overlap in the σ bond. sp Hybridisation 1 Carbon atoms can also share three electrons with one another to form a triple bond as in the ethyne (acetylene) molecule, C2H2. 2 This is brought about by sp hybridisation where a 2s orbital mixes with a 2p orbital to form two sp hybrid orbitals. 1s 2s 2p C* u n n n n 1s sp sp 2p C* u n n n n In sp hybridisation, there are two unhybridised 2p orbitals. H x x C C x H x x x H H Exam Tips Exam Tips A π bond is weaker than a σ bond. x x x x x H C C H Info Chem A double bond consists of one σ-bond and one π-bond. 9

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 3 The two sp hybrid orbitals are linear, while the two unhybridised 2p orbitals are perpendicular to one another and also to the plane containing the sp hybrid orbitals. + + z y x s z y x px z y – + – + x z y x + z y + – x sp sp p C p 180° 180° 4 When two sp-hybridised carbon atoms approach one another, the sp hybrid orbitals overlap head-on to form a carbon-carbon σ bond. Overlap of the remaining two sp orbitals with the 1s orbital of two hydrogen atoms forms carbon-hydrogen σ bonds. sp sp sp sp C C H 1s H 1s 5 The four unhybridised 2p orbitals then overlap sideways to give two carbon-carbon π bonds. H C C H 6 There are now altogether six electrons that are shared between the two carbon atoms giving rise to a carbon-carbon triple bond. 7 The carbon-carbon bond length in ethyne is 0.120 nm, with a bond energy of 835 kJ mol–1. Info Chem A triple bond consists of one σ-bond and two p-bonds. H9C#C9H 10

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 Structure of Benzene 1 Benzene, first isolated by Michael Faraday in 1825, has a molecular formula of C6H6, which suggests a high degree of unsaturation. 2 Each carbon atom in the benzene molecule undergoes sp2 hybridisation. Each carbon atom then forms three σ bonds with two adjacent carbon atoms and a hydrogen atom to give a hexagonal ring structure. H C C C C C C H H H H H 1s sp2 sp2 sp2 3 There is one unhybridised 2p orbital on each carbon atom. These six unhybridised 2p orbitals overlap sideways to give two ‘rings’ of delocalised π electron clouds, one above and the other below the plane containing the hexagonal ring. H H H H H H 4 All the carbon-carbon bonds in the benzene molecule are identical, with bond length of 0.139 nm (which is intermediate between that of the carbon-carbon single bond, 0.154 nm, and the carbon-carbon double bond, 0.133 nm). 5 The delocalised π electron clouds in the benzene molecule is usually represented by a circle within the hexagonal ring structure. 6 Sometimes, the hydrogen atoms are omitted and the benzene molecule is represented as Ring of delocalised π–electrons 7 The ring of delocalised π electrons stabilises the benzene molecule. 2p sp2 sp2 sp2 C H H H H H H 2013/P3/Q1 11

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 14.2 Empirical, Molecular and Structural Formulae of Organic Compounds Empirical Formula 1 The empirical formula of an organic compound shows the simplest integer ratio between the different types of atoms (or elements) in a molecule of the compound. 2 Take the example of ethanoic acid. The ratio between the three different types of atoms are C : H : O = 1 : 2 : 1 The empirical formula of ethanoic acid is CH2O. 3 The empirical formula of a compound can be determined if we know the composition by mass of the different types of atoms (or elements) present in one molecule of the compound or from combustion data. This is illustrated by the following examples. Example 14.1 An organic compound has the following composition by mass: Carbon, 40.0%; Hydrogen, 6.7%; Oxygen, 53.3% Calculate the empirical formula of the compound. Solution 40 6.7 53.3 Mole ratio of C : H : O = —– : —— : —— 12 1 16 = 3.33 : 6.7 : 3.33 = 1 : 2 : 1 The empirical formula is CH2O. Example 14.2 When a hydrocarbon is burned completely in oxygen, 2.64 g of carbon dioxide and 0.54 g of water is produced. Calculate the empirical formula of the hydrocarbon. Solution Number of moles of CO2 produced = —2.64— = 0.060 mol 44 ∴ Number of moles of carbon atom = 0.060 mol Number of moles of H2O produced = —0.54— = 0.030 mol 18 ∴ Number of moles of hydrogen atom = 2(0.030) = 0.060 mol Mole ratio of C : H = 0.060 : 0.060 = 1 : 1 The empirical formula is CH. 12

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 Molecular Formula 1 The molecular formula of an organic compound shows the actual number of the different types of atoms (or elements) in one molecule of the compound. 2 For example, the molecular formulae of ethanoic acid, ethane and glucose are C2H4O2, C2H6 and C6H12O6, while their empirical formulae are CH2O, CH3 and CH2O respectively. 3 The molecular formula is a simple multiple of the empirical formula. 4 The molecular formula can be determined from the empirical formula if we know the molar mass (or relative molecular mass) of the compound. Example 14.3 Give the empirical formula of the following compounds. C2H6; C6H12O6; C3H6; C2H6O; C3H6O2; C2H4 Solution Compound Empirical formula C2H6 CH3 C6H12O6 CH2O C3H6 CH2 C2H6O C2H6O C3H6O2 C3H6O2 C2H4 CH2 Example 14.4 A gaseous hydrocarbon, X has the following composition by mass: C, 88.9%; H, 11.1% 1.08 g of X occupies a volume of 490.4 cm3 at 101 kPa and 298 K. (a) Calculate the empirical formula of X. (b) Determine the molecular formula of X. Solution (a) Mole ratio of C : H = —88.9— : 11.1 12 = 7.4 : 11.1 = 1 : 1.5 Or = 2 : 3 Empirical formula of X is C2H3. (b) Using pV = nRT (101 × 103 )(490.4 × 10–6) = n × 8.31 × 298 n = 0.020 mol 1.08 Relative molecular mass of X = —–— = 54.0 0.020 Exam Tips Molecular formula is a simple multiple of the empirical formula. 2014/P3/Q2 2016/P3/Q1 13

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 Let the molecular formula be (C2H3)n (12 × 2 + 3)n = 54 27n = 54 ∴ n = 2 Molecular formula of X = C4H6 Example 14.5 An organic compound, Y, has the following composition by mass: C, 68.9%; H, 4.9%; O, 26.2% When 2.5 g of Y is vaporised at 200 °C and 101 kPa, the vapour occupies a volume of 797.5 cm3 . (a) Calculate the empirical formula of Y. (b) Determine the molecular formula of Y. Solution (a) Mole ratio of C : H : O = —68.9— : 4.9 : —26.2— 12 16 = 5.74 : 4.9 : 1.64 = 3.5 : 3 : 1 Or = 7 : 6 : 2 Empirical formula of Y is C7H6O2. (b) Using pV = nRT (101 × 103 )(797.5 × 10–6) = n × 8.31 × (200 + 273) n = 0.0205 mol 2.5 Relative molecular mass of Y = ——— = 120.0 0.0205 Let the molecular formula be (C7H6O2)n (12 × 7 + 6 + 16 × 2)n = 120 n = 0.98 < 1 Molecular formula of Y is C7H6O2. Quick Check 14.1 1 An organic compound, Q (relative molecular mass = 90.1) has the following composition by mass: C, 40.0%; H, 6.6%; O, 53.4% (a) Calculate the empirical formula of Q. (b) Determine the molecular formula of Q. 2 Complete combustion of 1.2 g of an organic compound W produces 1.76 g of carbon dioxide and 0.72 g of water. Given that the relative molecular mass of W is 70, determine (a) the empirical formula of W, (b) the molecular formula of W. 14

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 Structural Formula 1 The structural formula of a compound shows the actual number of the different type of atoms (or elements) in a molecule of the compound and how the atoms are connected (or bonded) to one another. 2 There are essentially three types of structural formulae: Expanded structural formula, condensed structural formula and skeletal structural formula. Expanded Structural Formula 1 The expanded structural formula (or displayed formula) shows all the covalent bonds between the atoms. 2 The bonds can be single bonds or multiple bonds (double and triple bonds). 3 When writing the expanded structural formula, it is worth remembering that carbon is tetravalent. 4 The expanded structural formulae of ethane, 2-propanol, 2-methylpropene and ethanoic acid are shown below. H H H O & & & ∫ H!C!C!H H!C!C!O!H & & & H H H Ethane Ethanoic acid H H H H H & & & & & H!C!C!C!H H!C!CRC!H & & & & & H O H H & & H!C!H H & 2-propanol H 2-methylpropene Condensed Structural Formula 1 In the condensed structural formula, carbon-carbon single bonds are usually not shown, but double bonds and triples bonds are shown. 2 For example: CH3CH3 CH3CHCH3 & OH 2-propanol CH3COH ∫ O Ethanoic acid Ethane 15

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 3 Sometimes, certain single bonds are shown for clarity purpose. For example: CH3!CH!CH3 CH3!CRCH2 & CH3 2-methylpropene CH3!C!OH & ∫ OH O 2-propanol Ethanoic acid 4 If there are several similar groups in the molecule of a compound, parentheses and subscripts are used to represent them. For example: (a) Octane, C8H18 CH3CH2CH2CH2CH2CH2CH2CH3 or CH3(CH2)6CH3 (b) 2-methyl-2-propanol CH3 & CH3!C!CH3 or (CH3)3COH & OH Skeletal Structural Formula 1 Skeletal structural formula (also known as line-angle or stick structural formula) does not show carbon and hydrogen atoms. 2 However, other functional groups that contain hydrogen atoms (such as !OH, !COOH or !NH2) must be shown. 3 The bonds are shown by straight lines. A single line for single bond, double lines for double bonds and triple lines for triple bonds. 4 Carbon atoms occupy the beginning and end of a line or when two or more lines meet unless it is stated otherwise. 5 When a line ends with nothing, the group at the end of it is CH3. For example: !! !!CH3 (a) (b) 6 When two lines intercept, the group at the intersection is CH2. For example: CHCH2 2 (a) (b) 7 When three lines intercept, the group at the intersection is CH. For example: CHCH (a) (b) 16

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 8 When four lines intercept, the group at the intersection is a carbon atom. C (a) (b) 9 For example, the skeletal structural formula for 2-methylpropene is CH3!CRCH2 & CH3 (a) (b) 10 The skeletal structure for cyclohexene, C6H12 is H C C H H H H C RR C C H H H H H C (a) (b) 11 Skeletal formulae (especially for cyclic compounds) have a distinct advantage that they can be drawn quickly, and the emphasis is on the functional groups rather than the carbon backbone. 12 Other examples of skeletal structures are shown below. Hexane CH3CH2CH2CH2CH2CH3 (a) (b) 2-pentanol CH3CH2CH2CHCH3 & OH OH (a) (b) Propanoic acid CH3CH2C!OH ∫ O O OH (a) (b) 17

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 Ethylcyclopentane CH2CHCH3 2CH3 (a) (b) 1-methylcyclohexene CH3 CH3 (a) (b) 1,2-dichlorohexane CH3CH2CH2CH2CHCH2Cl & Cl Cl Cl (a) (b) 14.3 Functional Groups: Classification and Nomenclature 1 Due to the immense quantity of organic compounds (whether natural or synthetic), it is impossible to study the properties of each and every one of them. 2 To facilitate the study, chemists group organic compounds with similar properties together to form a family called homologous series, much like when the elements are classified in Groups in the Periodic Table. 3 The members of a particular homologous series are called homologs and have the following properties: (a) They can be represented by a general formula. For example, all alkanes can be represented by CnH2n+2, where n is an integer. (b) Each member differs from the preceding one by a !CH2! group. For example, the first member in the arene series has the molecular formula of C6H6. Hence, the molecular formula of the second member is (C6H6 + CH2) = C7H8. (c) They have similar chemical properties. This is because homologs of the same series have the same functional group. (i) A functional group is an atom or a group of atoms in a molecule that gives the compound its chemical properties. 2009/P1/Q28 2015/P3/Q2 Characteristics of a homologous series Definition of functional group 18

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 (ii) The properties of a functional group usually do not change irrespective of where the functional group is found. (iii) Hence, to a large extent, the chemistry of organic compounds is determined by the type of functional groups they contain. (d) There is a gradual change in their physical properties (e.g. melting point, boiling point, density and solubility) as the size of the molecules increases. (e) They can be prepared by similar methods. 4 The table below lists some of the common homologous series. Name General formula Functional group Name of functional group IUPAC nomenclature Example Alkane CnH2n + 2 C H Hydrogen atom Alkane C2H6 Ethane Alkene CnH2n C C Carbon-carbon double bond Alkene C2H4 Ethene Arene CnH2n – 6 R Benzene ring – CH3 Methylbenzene Alkyne CnH2n – 2 C C Carbon-carbon triple bond Alkyne C2H2 Ethyne Haloalkane CnH2n + 1X X Halogen atoms Haloalkane or alkyl halide C2H5Br Bromoethane Alcohol CnH2n + 1OH or CnH2n + 2O OH Hydroxy Alkanol C2H5OH Ethanol Aldehyde CnH2nO C O H Carbonyl Alkanal CH3CHO Ethanal Ketone CnH2nO C O Carbonyl Alkanone CH3COCH3 Propanone Carboxylic acid CnH2nO2 or CnH2n + 1COOH COOH Carboxyl Alkanoic acid CH3COOH Ethanoic acid Ester CnH2nO2 COOR Ester Alkyl alkanoate CH3COOCH3 Methylethanoate Acyl chloride CnH2n – 1OCl C Cl O Acyl chloride Alkanoyl chloride CH3COCl Ethanoyl chloride Amide CnH2n + 1NO C NH2 O Amide Alkanamide CH3CONH2 (Ethanamide) Amine CnH2n + 3N NH2 Primary amine Alkanamine C2H5NH2 Ethanamine Nitrile CnH2n – 1N C N Cyanide or nitrile Alkanenitrile CH3CN Ethanenitrile 19

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 Classification of Carbon Atoms 1 The carbon atom in an organic molecule can be classified as primary, secondary, tertiary or quaternary depending on the number of other carbon atoms that are bonded to it. 2 If the carbon atom concerned is bonded to one other carbon atom, it is known as a primary carbon atom. 3 If it is bonded to two other carbon atoms, it is known as a secondary carbon atom and so on. 4 The hydrogen atom that is attached to the primary carbon atom is called the primary hydrogen atom. The hydrogen atom that is attached to the secondary carbon atom is called the secondary hydrogen atom and so on. Note that there are no quaternary hydrogen atoms. CH3 CH3 H & & & CH3!C !C !C!C H3 & & & CH3 H H Quaternary Tertiary Secondary Primary 5 Haloalkanes and alcohols are classified as primary, secondary or tertiary depending on the class of carbon atom to which the functional groups are attached. For example: Primary haloalkane Primary alcohol CH3!CH2Cl CH3!CH2OH Secondary haloalkane Secondary alcohol CH3!CH!CH3 CH3!CH!CH3 & & Cl OH Tertiary haloalkane Tertiary alcohol CH3 CH3 & & CH3!C!CH3 CH3!C!CH3 & & Cl OH Note that there are no quaternary haloalkanes or alcohols. 2011/P1/Q31 2016/P3/Q3 Info Chem There are no quaternary hydrogen atoms. 20

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 Example 14.6 How many primary, secondary, tertiary and quaternary carbon and hydrogen atoms are there in the following molecule? CH3 CH3 CH3 C CH2 CH CH3 CH3 Solution Atom Primary Secondary Tertiary Quaternary Carbon 5 1 1 1 Hydrogen 15 2 1 0 14.4 Isomerism: Structural and Stereoisomerism 1 Isomerism refers to the occurrence of two or more different compounds having the same molecular formula. 2 The different compounds with the same molecular formula are called isomers. (Not to be confused with isotopes, which are atoms with the same proton number but different nucleon number.) 3 There are two main types of isomerism in organic compounds: Structural isomerism and stereoisomerism. 4 Stereoisomerism is further classified as geometrical (or cis-trans) and optical isomerism. Structural Isomerism 1 In structural isomerism, the compounds have the same molecular formula but different structural formula. 2 Structural isomers have the same molecular formula but their atoms are joined differently. 3 Structural isomerism can be subcategorised into chain isomerism, position isomerism and functional group isomerism. Chain Isomerism 1 Chain isomers have the same functional group and are members of the same homologous series, but the carbon skeletal backbone is different. 2 They have similar chemical properties but different physical properties. Different compounds with the same molecular formula. Isomerism Structural Stereo Optical Geometrical Same functional group 2014/P3/Q4 2015/P3/Q18(a) 2016/P3/Q2 21

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 3 For example, there are two alkanes with the molecular formula of C4H10: CH3CH2CH2CH3 Butane (b.p. = 0 °C) CH3CHCH3 2-methylpropane (b.p. = –12 °C) & CH3 4 In butane, the four carbon atoms are joined to form a linear chain whereas in 2-methylpropane, one of the carbon atoms is found in the side chain. Positional Isomerism 1 Positional isomers have the same functional group and the same carbon skeletal backbone. They differ from one another in the positioning of the functional group. 2 Positional isomers are members of the same homologous series. They generally have similar chemical properties but different physical properties. 3 For example, there are two positional isomers for the compound with the molecular formula of C3H7Cl. CH3CH2CH2Cl Cl Cl 1-chloropropane CH3CHCH3 2-chloropropane & Cl Cl Cl 4 Another example is CH3 NH2 CH3 NH2 4-amino-methylcyclohexane CH3 NH2 CH3 NH2 2-amino-methylcyclohexane Functional Group Isomerism 1 Functional group isomers have the same molecular formula but different functional groups. 2 Functional group isomers have different physical and chemical properties as they are from different homologous series. Skeletal structures of butane and 2-methylpropane: and Same functional group Different functional groups INFO Structural Isomerism 22

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 3 Examples are: CH3CH2!C!OH and CH3!C!OCH3 ∫ ∫ O O Propanoic acid (An acid) Methyl ethanoate (An ester) CH3CH2OH and CH3!O!CH3 Ethanol (An alcohol) Dimethyl ether (An ether) CH3!C!Cl and CH2Cl!C!H ∫ ∫ O O Ethanoyl chloride 2-chloroethanal (An acid chloride) (An aldehyde) Stereoisomerism 1 Stereoisomers have the same molecular formula and the same structural formula, but are different from one another due to the spatial arrangement of the atoms or groups. 2 There are two types of stereoisomerism, geometrical isomerism and optical isomerism. Geometrical Isomerism 1 Geometrical isomerism (also called cis-trans isomerism) occurs in compounds in which free rotation about a bond is restricted. 2 This happens in compounds with carbon-carbon double bonds or compounds with cyclic structures. 3 Geometrical isomerism are common in alkenes and substituted alkenes. For example, 2-butene has two geometrical isomers: cis-2-butene and trans-2-butene. C H H CH3 CH3 RRC C H H CH3 CH3 RRC C H H CH3 CH3 RRC C H H CH3 CH3 RRC cis-2-butene trans-2-butene (b.p. = 4 °C) (b.p. = 1 °C) 4 In the cis-isomer, the two methyl groups are on the same side of the carbon-carbon double bond, whereas in the trans-isomer, the two methyl groups are on the opposite sides of the double bond. 5 Another example is 1,2-diiodocyclohexane: I H H H I I H I I H H H I I H I cis-1,2-diiodocyclohexane trans-1,2-diiodocyclohexane Acids release CO2 from carbonates but esters do not. Alcohols react with PCl5 but ethers do not. Acid chlorides fume in moist air but aldehydes do not. Spatial arrangement of the atoms/groups is different. INFO Geometrical Isomerism 23

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 6 The cis-isomer and the trans-isomer are not interconvertable through twisting or rotating. 7 For a compound to exhibit geometrical isomerism, it cannot have identical atoms or groups bonded to the same unsaturated carbon atom. 8 For example, CH3CH=CCl2 does not show geometrical isomerism. C H CH3 Cl RRC Cl Cl Cl C H CH3 RRC C H CH3 Cl RRC Cl Cl Cl C H CH3 RRC The two structures above are interconvertable simply by flipping them. 9 Geometrical isomers usually have similar chemical properties (as they are from the same homologous series) but different physical properties. Exam Tips Exam Tips Cannot have 2 identical atoms/groups attached to the same unsaturated carbon atom. Quick Check 14.2 Determine if the following compounds exhibit geometrical isomerism. 1 CH3CH2CH2CH2CH3 6 I H CH3 7 I H CH3 2 CH2RCH2 3 CH3CHRCHCH3 4 CH3CH2CHRCH2 5 CH2RCH!CHRCH2 Optical Isomerism 1 Optical isomerism (sometimes called enantiomerism) occurs in compounds with the same molecular formula and same structural formula, but are mirror images of one another, and are non-superimposable. 2 Optical isomers or enantiomers occur in pairs. 3 An analogy to optical isomerism is the hand glove model. A lefthand glove looks excatly like the right-hand glove. They are mirror images. However, the left-hand glove can only be worn on the left hand and not the right hand. 4 Optical isomerism occurs in compounds with one or more chiral (or asymmetric) carbon atoms in their molecules. There are no planes of symmetry in their molecules. 5 A chiral carbon is a carbon atom that is bonded to four different atoms/groups. The chiral carbon is usually identified by an asterisk (*). 2010/P1/Q28 The two isomers are mirror images. What is a chiral carbon? Info Chem Optical activity: The ability of a substance to rotate plane-polarised light 24

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 For example, 2-bromobutane: or CH3CH2 H CH3 Br CH3CH2 C* CH3 H Br C* CH3CH2 H CH3 Br CH3CH2 C* CH3 H Br C* The four different atoms/groups are: H, CH3, CH3CH2 and Br. 6 The two optical isomers of 2-bromobutane are shown below. CH3CH2 H CH3 Br C* CH3 H CH2CH3 Br C* Mirror 7 The structure on the left can be turned, rotated or flipped in any direction, but only two of the groups can superimpose on one another, not all four. Hence, they are two different compounds. 8 Another way to repesent the structures of the two optical isomers is by using the Fischer projection (which is in two dimension) as shown below. CH3 CH3 CH2CH3 CH2CH3 H Br Br H Mirror 9 One important rule when using the Fischer projection is that, the molecules can be slided onto one another and rotated, but they cannot be lifted out of the plane of the paper. 10 Enantiomers are called optical isomers because they can rotate plane-polarised light. The angle through which the planepolarised light is rotated is called the specific rotation, α. 11 One isomer will rotate the light clockwise (or to the right) and the other isomer (its mirror image) will rotate it anti-clockwise (or to the left). The angles through which the light is rotated are the same but in the opposite direction. Specific rotation The chiral carbon is located at the intersection of the two lines. INFO Optical Isomerism 25

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 12 The direction in which the light is rotated can only be determined through experiment using a polarimeter. 13 The isomer that rotates the light in the clockwise direction is said to be dextrorotatory, and is known as the D-isomer. The angle of rotation is given a positive (+) value. 14 The isomer that rotates the light in the anti-clockwise direction is said to be laevorotatory, and is known as the L-isomer. The angle of rotation is given a negative (–) value. 15 A mixture containing equimolar quantity of the D and L isomers is optically inactive because they cancel out each other’s rotation. Such a mixture is called a racemic mixture. 16 If a compound has more than one chiral carbon in its molecule, the total number of optical isomers is 2n . Where n is the number of chiral carbon. 17 For example, 2-bromo-3-iodopentane has four optical isomers because its molecule contains two chiral carbons. H H H & & & CH3!C*!C*!C!CH3 & & & Br I H 18 Optical isomers have identical physical (boiling point, melting point and density) and chemical properties except in the reactions with optically active reagents. 19 One isomer may react readily with the optically active reagent, while its mirror image may not, or reacts only very slowly with the same reagent. This is especially true for biological reactions involving enzymes (which are optically active). 20 Optical isomers can be separated by a technique called resolution. Example 14.7 What is the molecular formula of the first member of the alkanes to exhibit optical isomerism? Solution The structural formula of the alkane is 3-methylhexane: H & CH3CH2CH2!C*!CH2CH3 & CH3 The molecular formula is C7H16. Dextrorotatory: Plane polarised light Laevorotatory: Plane polarised light Racemic mixture Exam Tips Exam Tips No. of optical isomers = 2n where n = number of chiral carbon Info Chem Optical isomers react differently with optically active reagents. 26

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 Quick Check 14.3 1 Which of the following compounds have chiral carbon in their structure? For those compounds, mark the chiral carbon with an asterisk (*). (a) CH3CH2CHCH2CH3 (e) I H CH3 & Cl (b) CH2RCHCH2OH (c) CH3CH(OH)NH2 (d) (f) I H CH3 14.5 Free Radicals, Nucleophiles and Electrophiles 1 Organic reactions involve the breaking of covalent bonds in the reactant molecules, and the formation of new bonds in the products. 2 Covalent bonds are formed when electrons are shared between atoms. 3 There are two ways in which a covalent bond can break: Homolytic fission and heterolytic fission. Homolytic Fission: Free Radicals 1 Take the case of the chlorine molecule, Cl2. The two chlorine atoms are bonded by a pair of electrons (donated by each of the two chlorine atoms) that are shared between them. In homolytic fission, each of the chlorine atoms regains the electron that was shared. xx xx x xx xx xx x x x x + or Cl Cl Cl Cl Cl Cl 2Cl• 2 Each new species formed (in this case, two chlorine atoms) has one unpaired electron and is called a free radical. 3 Other examples of free radicals are: CH3!CH3 !!: 2H3C• (methyl free radical) HO!OH !!: 2HO• (hydroxyl free radical) C6H5!C!O!O!C!C6H5:2 • C6H5 + 2CO2 (Phenyl free radical) && && O O 27

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 4 A free radical is very reactive. It will snatch an electron from another species so that its lone electron is paired. However, in doing so, it generates another free radical in the process. For example, the chlorine free radical can react with a methane molecule to produce a methyl free radical. H3C!H + Cl• !!: H3C• + HCl × • Stability of Free Radicals 1 Free radicals are very reactive species. They usually exist only as intermediates in chemical reactions. 2 The stability of a free radical depends on the electron density on the radical. The higher the electron density, the more stable the free radical. 3 Hence, free radicals are stabilised by electron releasing groups such as alkyl groups. 4 The more alkyl groups that are bonded to a carbon atom, the more stable is the radical. This is because the ‘electron releasing’ effect of the alkyl group helps to reduce the electron deficiency of the free radical and makes them less reactive. [Note: The more reactive a species, the less stable it is.] 5 Free radicals can be classified as primary, secondary and tertiary depending on the number of alkyl groups bonded to the carbon carrying the lone electron. Primary free radical H H & & R : C• Example: CH3 : C• & & H H Secondary free radical R CH3 p p R : C• Example: CH3 : C• & & H H Tertiary free radical R CH3 p p R : C• Example: CH3 : C• q q R CH3 6 The stability of the free radicals increases in the order: • CH3 < primary < secondary < tertiary Free radicals are very reactive. Info Chem Electron-releasing group stabilises the free radicals. Info Chem 1 Reactivity ∝ ———— stability Info Chem There are no quaternary free radicals. 28

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 Heterolytic Fission: Carbonium Ions and Carbanions 1 In heterolytic fission of a covalent bond, both the bonding electrons go to one of the atoms only. This results in the formation of an ion pair. 2 The atom or group that loses both the bonding electrons becomes a cation, and the atom or group that gains the two bonding electrons becomes an anion. For example: xx xx xx xx xx xx x + x Cl Cl Cl+ Cl– 3 An example of heterolytic fission is the reaction between 2-methyl-2-chloropropane and aqueous sodium hydroxide. CH3 CH3 & NaOH & CH3!C!!Cl !!!: CH3!C+ + :Cl– & & CH3 CH3 4 The cation in the above example, where the positive charge is on the carbon atom is called a carbonium ion or carbocation. 5 Heterolytic cleavage of a covalent bond can sometimes result in a species where the negative charge is localised on the carbon atom. Such species is called a carbanion. Examples are CH3CH2 – and HC#C– . 6 The ethyne carbanion is produced when ethyne dissolves in liquid ammonia in the presence of sodium amide, Na+NH2 – . HC#C!H + NH2 – !!: HC#C– + NH3 7 Another example of carbanion is from the decomposition of Grignard's reagents. R—MgX !!: R– + Mg2+ + X– For example, C2H5MgI !!: C2H5 – + Mg2+ + I– Stability of Carbonium Ions 1 Like the free radicals, carbonium ions can be classified as primary, secondary and tertiary depending on the number of alkyl groups that are bonded to the positively-charged carbon atom. 2 The more alkyl groups that are bonded to the carbonium ion, the more stable it is. This is because the electron-releasing effect of the groups help to reduce the electron deficiency of the cation. Cl9Cl 9: Cl+ + Cl  – • Carbonium ion and carbanion 29

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 Primary carbonium ion H H & & R : C+ Example: CH3 : C+ & & H H Secondary carbonium ion R CH3 p p R : C+ Example: CH3 : C+ & & H H Tertiary carbonium ion R CH3 p p R : C+ Example: CH3 : C+ q q R CH3 3 The stability of the carbonium ions increases in the order: CH3 + < primary < secondary < tertiary Stability of Carbanions 1 Carbanions are destabilised by the presence of electron-releasing groups. 2 Electron-releasing groups increase the electron density of the negative-charged carbon atom making it more reactive and hence less stable. Primary carbanion H H & & R : C– Example: CH3 : C– & & H H Secondary carbanion R CH3 p p R : C– Example: CH3 : C– & & H H Tertiary carbanion R CH3 p p R : C– Example: CH3 : C– q q R CH3 3 The stability of the carbanions increases in the order: Tertiary < secondary < primary < CH3 – Exam Tips Electron-donating groups reduce the electron deficiency of the carbonium ion and thereby stabilise it. Info Chem Electron-releasing group stabilises the carbonium ion. Exam Tips Exam Tips Electron-donating groups increase the electron density of the carbanion and thereby destabilise it. 30

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 Nucleophiles and Electrophiles Other than free radicals, the other two common reagents that are encountered in organic reactions are nucleophiles and electrophiles. Nucleophiles 1 Nucleophiles are species with a lone pair electrons that can be donated to other atoms which are electron deficient to form coordinate bonds. 2 Nucleophiles are electron-rich species and act as a Lewis base (electron pair donor). 3 Nucleophiles are usually represented by (Nu:) to emphasise the lone pair electrons. 4 Examples of nucleophiles are Cl– , OH– , NH3, H2O, CN– and carbanions. xx xx xxxx xx x xx xx xx x x x x x x x x xx xx xx O H H H H H H N N Cl– – C O– 5 In organic reactions, nucleophiles attack carbon atoms which are electron-deficient or carrying a partial positive charge. 6 An example is the reaction between iodopropane and aqueous sodium hydroxide. The nucleophile in this reaction is the OH– ion. H H & & CH3CH2!Cδ+!I + :OH– !: CH3CH2!C!OH + I – & & H H (The partial positive charge on the carbon atom is due to the polarisation of the carbon-iodine bond resulting from the difference in the electronegativity of carbon and iodine.) Electrophiles 1 Electrophiles are species that have electron deficient sites. These include positive ions such as carbonium ions, NO2 +, CH3C+O, Cl+, H+ and electron-deficient compounds such as AlCl3, FeBr3, BF3 and SO3. Nucleophile: Nucleus loving Exam Tips Exam Tips Nucleophiles are Lewis bases. |δ+ δ– —C+⎯→ I | Polarisation of C––I bond Electrophile: Electron loving 31

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 Oδ– Oδ– Oδ– RR RR Sδ RR+ Electron-deficient site x x x F F F B Can still accept 2 electrons to achieve octet configuration (Due to the high electronegativity of oxygen, the sulphur-oxygen bond is polarised with the sulphur atom acquiring a partial positive charge.) 2 Electrophiles are Lewis acids. They react by accepting lone pair electrons to form coordinate bonds. For example: Cl H Cl H & & & & Cl!Al + :N!H !!: Cl!Al ; :N!H & & & & Cl H Cl H Electrophile Nucleophile (Lewis acid) (Lewis base) 14.6 Molecular Structure and its Effect on Physical Properties 1 Most organic compounds are gases, liquids with low boiling points, or solids with low melting points. 2 This is because the intermolecular forces between the molecules of organic compounds are the weak van der Waals forces. 3 The strength of the van der Waals forces increases with the size of the molecule and the number of electrons in the molecule. For example: Alkanes Compound CH4 C2H6 C3H8 C4H10 C5H12 C6H14 Boiling point/ °C –164 –88 –42 0 36 69 Haloalkanes Compound CH3Cl C2H5Cl C3H7Cl Boiling point/ °C –24 12 51.5 4 Thus, in any particular homologous series, there is a gradual increase in the boiling point and melting point as the size of the molecule increases. Exam Tips Electrophiles are Lewis acids. Info Chem Lewis acids are electron-pair acceptors. Lewis bases are electron-pair donors. 32

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 5 For a group of isomeric compounds, the melting point and boiling point decrease with the extension of branching. 6 Take the example of the two isomers below: Molecule Structural formula Boiling point/ °C Pentane CH3CH2CH2CH2CH3 36 2,2-dimethylpropane CH3 | CH3—C—CH3 | CH3 10 7 The skeletal formulae of the two molecules are shown below: Pentane 2,2-dimethylpropane 2,2-dimethylpropane is more spherical while pentane is more 'linear'. 8 The molecules of pentane can approach one another closer than the molecules of 2,2-dimethylpropane. van der Waals forces van der Waals forces The larger surface area of contact between pentane molecules results in the van der Waals forces in pentane to be stronger than that of 2,2-dimethylpropane. 9 However, for organic molecules (such as alcohols, amines and carboxylic acids) that can form intermolecular hydrogen bonding, their melting and boiling points are higher than expected because more energy is required to break the stronger hydrogen bonding. For example, Molecule Structure Mr Nature of intermolecular force Boiling point/°C Butane CH3CH2CH2CH3 58 Van der Waals 0 Propanamine CH3CH2CH2NH2 59 Hydrogen bond 49 1-propanol CH3CH2CH2OH 60 Hydrogen bond 97 Ethanoic acid CH3COOH 60 Hydrogen bond 118 33

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 14.7 Inductive and Resonance Effect Negative Inductive Effect 1 When a carbon atom is bonded to another atom/element other than carbon itself, the covalent bond will be polarised due to the difference in the electronegativity between carbon and the other atom/group. 2 The polarity of the carbon atom in the polarised bond depends on the electronegativity of the other atom/group. 3 For example, due to the higher electronegativity of chlorine, the carbon-chlorine bond is polarised as shown below: & !δ+C!!Clδ– & The carbon atom acquires a partial positive charge. 4 The polarisiation can also be shown as & !δ+C!: Clδ– & 5 Since, chlorine atom attracts the bonding electrons towards itself (and acquires a partial negative charge) it is said to exhibit negative inductive effect. 6 They are also known as ‘electron-pulling’ or ‘electronwithdrawing’ groups. 7 Examples of atoms/groups with negative inductive effect are !NO2, !C #N, !COOH, !COOR (ester), C RO (carbonyl), !SO3H (sulphonic acid), !NH2 (amine), !OH (alcohol), !OR (ether) and C6H5!(phenyl group). Positive Inductive Effect 1 If a carbon atom is bonded to an atom/group that is less electronegative than carbon, then the covalent bond will be polarised in such a way that the carbon atom acquires a partial negative charge. & !δ–C!!Y δ+ & 2 The polarisation can also be represented as & !δ–C ;! Y δ+ & Info Chem Electronegativity of C and Cl is 2.5 and 3.0 respectively. Info Chem • Cl acquires a partial negative charge. It is said to exhibit negative inductive effect. • All halogens are more electronegative than carbon. Info Chem Y acquires a partial positive charge. It is said to have positive inductive effect. 2013/P3/Q2 34

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 3 Examples are: & !δ–C ;! Mgδ+ (as in the Grignard’s reagent) & & !δ–C ;! Liδ+ (as in lithium organometallic compounds) & 4 Since these atoms/groups ‘push’ electrons to the carbon atom and acquire a partial positive charge, they are said to exhibit positive inductive effect. 5 They are also known as ‘electron-pushing’ or ‘electron donating’ groups. 6 All alkyl groups are electron-donating groups. They possess positive inductive effect. & R : Cδ–! & 7 This arises because hydrogen (electronegativity = 2.1) is less electronegative than carbon (electronegativity = 2.5). As a result, the carbon-hydrogen bond is sightly polarised as shown below. H p H : Cδ– : C q H This makes the carbon atom slightly negatively-charged, and hence can donate electron to another carbon atom to which it is bonded as shown below. & CH3 : C! & 1 Classify the following spesies as a free radical, electrophile or nucleophile. (a) CH3 + (e) H3O+ (b) C2H5 – (f) NaNH2 (c) NO2 (g) F– (d) BF3 2 Which of the following species exhibit negative inductive effect? (a) C2H5! (c) !NO2 (b) (d) !NH2 (e) !COOH Quick Check 14.4 35

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 Effect of Inductive Effect on the Properties of Compounds Induction and the Strength of Organic acids (A) Ethanoic acid and chloroethanoic acids 1 The general formula of organic acids (also known as carboxylic acids) is: O ∫ R!C!O!H 2 In aqueous solutions, the molecule dissociates according to the equation: O O ∫ ∫ R!C!O!H + H2O R!C!O– + H3O+ Or, simply as: O O ∫ ∫ R!C!O!H R!C!O– + H+ 3 The strength of the carboxylic acids is measured by the acid dissociation constant, Ka. RCOOH RCOO– + H+ [RCOO– ][H+] Ka = —————— [RCOOH] The higher the Ka value, the stronger is the acid. 4 The Ka values for ethanoic acid and chloroethanoic acids are given in the table below: Compound Name Ka/mol dm–3 CH3COOH Ethanoic acid 1.8 × 10–5 ClCH2COOH Chloroethanoic acid 1.4 × 10–3 Cl2CHCOOH Dichloroethanoic acid 3.3 × 10–2 Cl3CCOOH Trichloroethanoic acid 2.0 × 10–1 The strength of the acids is: Cl3CCOOH > Cl2CHCOOH > ClCH2COOH > CH3COOH 5 Chlorine is more electronegative than carbon. It exhibits negative inductive effect as shown below: O ∫ CH3!C!O!H O ∫ ClCH2CO!H O ∫ ClCHCO!H Cl Cl O ∫ ClCCO!H Cl 2010/P1/Q48 2010/P2/Q4(a) 2015/P3/Q4 36

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 6 The electron-withdrawing effect of chlorine weakens the O—H bond in the chloroethanoic acid molecules making it easier to break compared to ethanoic acid. As a result, the chloroethanoic acids are stronger acid than ethanoic acid. 7 The more chlorine atoms in the molecule, the stronger is the electron-withdrawing effect, and the stronger is the acid. (B) Chlorobutanoic acids 1 Acid dissociation constants for the three isomers of chlorobutanoic acid are given below: Compound Name Ka/mol dm–3 2-chlorobutanoic acid CH3―CH2―CH—COOH Cl 1.5 × 10–3 3-chlorobutanoic acid CH3―CH—CH2—COOH Cl 1.1 × 10–4 4-chlorobutanoic acid Cl—CH2—CH2—CH2—COOH 3.0 × 10–5 2 The further the chlorine atom from the —OH group of the acid molecule, the weaker is the acid. 3 This is because the inductive effect decreases with the distance between the chlorine atom and the —OH group, which is expected. CH3!CH2!CHCOOH Cl CH3!CHCH2COOH Cl ClCH2CH2CH2COOH Effect of Resonance on the Strength of Acids and Bases Difference in Acidity between Ethanol and Phenol 1 Acid dissociation constants for ethanol and phenol are given in the table below: Alcohol Structure Ka/mol dm–3 Ethanol CH3CH2OH 1.00 × 10–16 Phenol OH 1.33 × 10–10 Phenol is a stronger acid than ethanol. Info Chem Bromine is less electronegative than chlorine. The negative inductive effect of bromine is lower. As a result, bromoethanoic acid is weaker than chloroethanoic acid. BrCH2COOH < Cl–CH2COOH 2014/P3/Q5 2017/P3/Q3 37

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 2 Ethanol and phenol dissociate according to the equations below to produce the ethoxide ion and phenoxide ion respectively. C2H5O—H C2H5O– + H+ Ethoxide ion OH O– + H+ Phenol Phenoxide ion 3 The relative strength of the two compounds can be explained by the stability of the anions formed (Resonance effect). 4 The more stable the anion, the more likely that it will form and this will increase the dissociation of the hydroxyl compounds. 5 The negative charge on the oxygen atom in the phenoxide ion can be delocalised into the benzene ring. O– O– 6 The delocalisation decreases the charge density of the oxygen atom in the phenoxide ion making it less readily recombine with H+. 7 On the other hand, the negative charge of the ethoxide ion cannot undergo delocalisation. This makes it more readily recombine with H+. As a result, ethanol is a weaker acid than phenol. 8 Alternative explanation (Inductive effect): Phenyl group is an electron-withdrawing group that reduces the electron density of the O—H bond. O H C2H5:O!H On the other hand, ethyl group is an electron-donating group that increases the electron density of the O—H bond. As a result, the O—H bond in phenol is weaker than that in ethanol. This makes phenol a stronger acid than ethanol. 38

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 Difference in the Basicity between Methylamine and Phenylamine 1 The structures of methylamine and phenylamine (a.k.a. aniline) are shown below: NH2 Phenylamine CH3NH2 Methylamine 2 Base dissociation constants of methylamine and phenylamine are given in the table below: Compound Formula Kb/mol dm–3 Methanamine CH3—NH2 4.4 × 10–4 Phenylamine NH2 4.2 × 10–10 3 Both compounds are Lewis bases as they use the lone pair electrons on the nitrogen atom to combine with H+ ions to form substituted ammonium ions. N: + H+ ⎯→ N:→H+ For example, CH3NH2 + H+ ⎯→ CH3NH3 + NH2 + H+ ⎯→ NH3 + 4 The strength of the organic bases depends on the availability of the lone pair electrons to react with H+. 5 The lone pair electrons in phenylamine can delocalise into the benzene ring as shown below: H H N H H As a result, it is less available to react with H+. 6 On the other hand, the lone pair electrons in methylamine cannot undergo delocalisation. Hence, it is more available to react with H+ making it a stronger base than phenylamine. 7 Alternative explanation (Inductive effect): (a) The methyl group in methylamine is an electron-donating group (positive inductive effect): .. CH3 →NH2 This increases the electron density of the lone pair electrons and makes them more susceptible to attack by H+. Exam Tips Exam Tips The electron-releasing CH3 stabilises the CH3 +NH3 ion, while the electron-withdrawing C6H5 destabilises the phenylammonium ion. 39

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 (b) On the other hand, the phenyl group is an electronwithdrawing group (negative inductive effect). N: H H This reduces the electron density of the lone pair electrons and makes them less reactive towards H+. Example 14.8 Arrange the following compounds in the order of decreasing acid strength. Explain your answer. Chloroethanoic acid, bromoethanoic acid, iodoethanoic acid Solution Acid strength: Chloroethanoic acid > bromoethanoic acid > iodoethanoic acid The structures of the three acids are: Cl←CH2COOH Br←CH2COOH I←CH2COOH The electronegativity of the halogen decreases in the order: Cl > Br > I due to the increase in the size of the halogen atom with increasing proton number. As a result, the inductive effect decreases in the order: Cl > Br > I leading to the decrease in the acid strength in the same order. SUMMARY SUMMARY 1 Organic chemistry is the study of carbon compounds. 2 A homologous series is a group or family of compounds with the following characteristics: • They can be represented by a general formula. • The molecular formula of each member differs from their preceding member by a CH2 group. • They show a gradual change in their physical properties as the relative molecular mass increases. • They have the same functional group and hence have similar chemical properties. • They can be prepared using similar methods. 3 A functional group is an atom or group of atoms that gives the chemical properties of the compound. 4 The empirical formula shows the simplest ratio between the different types of atoms in the molecule of a compound. 5 The molecular formula gives the actual number of the different types of atoms in the molecule of a compound. 6 The structural formula gives the molecular formula of a compound and the way the atoms are bonded to one another. 7 Isomerism is the occurrence of two or more compounds with the same molecular formula. INFO Acidity and Inductive/ Resonance Effect 40

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry 14 STPM PRACTICE 14 Objective Questions 8 Structural isomerism occurs in the compounds with the same molecular formula but with different structural formula. 9 Geometrical isomerism occurs in the compounds with the same molecular formula and same structural formula, but are different in the spatial arrangement of the atoms or groups due to the presence of carbon-carbon bonds or cyclic structure that restrict free rotation. 10 Optical isomerism occurs in the compounds with the same molecular formula and same structural formula, but are different in the spatial arrangement of the atoms or groups due to the presence of a chiral carbon atom. 11 A free radical is a species that has an unpaired electron in its structure. It is usually formed from the homolytic cleavage of a covalent bond. 12 Nucleophiles are species with a lone pair electrons that can be donated to other atoms which are electron deficient to form coordinate bonds. Nucleophiles are Lewis bases. 13 Electrophiles are Lewis acids. They react by accepting lone pair electrons to form coordinate bonds. 14 Stability of free radicals: 3° > 2° > 1° 15 A carbonium ion is an ion where the carbon atom acquires a positive charge. 16 Stability of carbonium ions: 3° > 2° > 1° 17 A carbanion is an ion where the carbon atom acquires a negative charge. 18 Stability of a carbanion: 1° > 2° > 3° 1 How many π bonds are in the molecule of propanenitrile, C2H5CN? A 1 C 3 B 2 D 8 2 Which of the following is not a nucleophile? A OH– C NO2 + B NH3 D NaCN 3 Consider the following structure of an organic compound, X? O O O CH3 HO X can be classified as any of the following except as A a ketone B an aldehyde C an alcohol D an ester 4 Which of the following exhibits geometrical isomerism? A C3H8O2 C H2C2O4 B C2H2F2 D C4H9OH 5 Which compound does not contain tertiary carbon atom? A Phenol B 1-Methylcyclohexanol C Cyclohexenol D Cyclohexanol 6 The organic compound that shows optical isomerism is A 2-methyl-2-propanol B propanenitrile C 2-methylbenzoic acid D 2-butanol 7 Which molecule has carbon atoms with sp and sp2 hybridisation? A C6H6 B CH3CH2CH2CHCH3 & CH3 C CH2=CH–CH–CH3 D C6H5NH2 41