Chemistry Term 3 STPM Chapter 19 Carboxylic Acids and their Derivatives 242 19 STPM PRACTICE 19 Objective Questions 1 Esterification between an acid and an alcohol is an example of A nucleophilic substitution B electrophilic substitution C nucleophilic addition D electrophilic addition 2 Which of the following is not true about ethanoyl chloride, CH3COCl? A It can be prepared by the action of thionyl chloride on ethanol. B It reacts with excess ammonia to produce a primary amide. C It reacts with alkaline phenol to produce an ester. D It does not react with tertiary amines. 3 Which of the following processes will not produce a carboxylic acid or its salt? A Boiling an alkyl nitrile with aqueous sodium hydroxide B Reacting a Grignard's reagent with a ketone C Boiling an alkyl cyanide with dilute sulphuric acid D Oxidation of alkylbenzene with hot, acidified potassium permanganate 4 An ester, X has the structural formula of C6H5COOCH(CH3)CH3. What are the products formed when X reacts with lithium aluminium hydride in dry ether followed by dilute sulphuric acid with heat? A C6H5OH + CH3CH(OH)CH3 B C6H5CH2OH + CH3CH2CH2OH C C5H5CH2OH + CH3CH(OH)CH3 D C6H5OH + CH3CH2CH2OH 5 Which of the following compounds will decolourise an acidic solution of potassium manganate(VII)? I HOOCCH2COOH II HCOOH III CH3 COOH A I C II and III B II D I, II and III 6 Which reagent will react with benzoic acid to produce benzoyl chloride? A Concentrated hydrochloric acid B Chlorine in the presence of ultraviolet light C Thionyl chloride D Chlorine in the presence of anhydrous aluminium chloride 7 Compound X has the following structure. O O N CH2 OH X can be prepared by the reaction between A C O O OH + H2 N — CH2 — C — Cl B C O Cl + CH3 — NH — CH2 — COOH C C O NH — CH3 + CH3 — COOH D C O OCH3 + CH3 — NH — COOH 8 Which statement is not correct? A Benzoic acid is insoluble in cold water but soluble in hot water. B Fatty acids are long-chain carboxylic acids. C Methanoic acid gives a yellow precipitate with 2,4-dinitrophenylhydrazine. D Oxalic acid reacts with excess concentrated sulphuric acid without leaving any solid residue.
Chemistry Term 3 STPM Chapter 19 Carboxylic Acids and their Derivatives 243 19 9 Acid X has the following structure. OH COOH Which of the following is not true of X? A It gives a white precipitate with bromine water. B It decolourises bromine in cyclohexane. C It forms an ester with 2-propanol. D It turns the colour of acidified potassium dichromate from orange to green. 10 Which of the following compounds will produce benzoic acid when heated with an alkaline solution of potassium manganate(VII) followed by an excess of dilute sulphuric acid? I OH II CH2OH III CH2CH3 IV C(CH3)3 A II B I and II C II and III D II, III and IV 11 Which of the following statements are true? I Benzoic acid is stronger than cyclohexanecarboxylic acid. II 2-chloropropanoic acid is stronger than ethanoic acid. III 4-nitrobenzoic acid reacts with sodium carbonate but 4-chlorobenzoic acid does not. A I and II C II and III B I and III D I, II and III 12 The skeletal structural formula of compound X is shown below. O O CH3 CH3 Which statement is true about X? A It decolourises bromine in tetrachloromethane. B It reacts with lithium aluminium hydride followed by acidic hydrolysis to produce two primary alcohols. C It is optically active. D It does not react with warm acidified KMnO4 solution. 13 Which reaction does not produce a carboxylic acid? A Ethylmagnesium iodide and propanone B Oxidation of toluene C Boiling propanenitrile with dilute sulphuric acid D Boiling ethylethanoate with dilute hydrochloric acid 14 Esterification between an acid and an alcohol is an example of A nucleophilic substitution B electrophilic substitution C nucleophilic addition D electrophilic addition 15 Which of the following is not true during saponification? A The product is the sodium salt of longchain carboxylic acids B Hot, dilute sulphuric acid is used C Sodium chloride is added in the final stage D One of the raw material is fat or oil 16 Aspirin can be produced from salicylic acid via the following reaction. OH COOH OCOCH3 COOH !!:X What is X? A CH3OH C CH3COCl B CH3COOH D CH3COCH3
Chemistry Term 3 STPM Chapter 19 Carboxylic Acids and their Derivatives 244 19 17 Which reagent does not react with ethanamide? A NaOH B HNO2 C CH3COCl D 2,4-dinitrophenylhydrazine 18 Each of the following chlorine compounds reacts separately with an ethanolic solution of silver nitrate. I II III Chlorobenzene Benzoyl chloride Chlorocyclohexane The correct sequence of reactivity of the three compounds in ascending order is A I, III, II B II, III, I C III, I, II D I, II, III 19 Which of the following compounds reacts most vigorously with cold water? A C6H5COOH + C6H5OH B C6H5COOH + C6H5ONa C C6H5COCl + C6H5OH D C6H5COCl + C6H5ONa 20 Which reagents will react with the following compound? CH=CHCH2COOCH3 I Hot, acidified potassium manganate(VII) II Lithium aluminium hydride III Aqueous sodium hydroxide A I C II and III B I and II D I, II and III 21 Which reagent does not react with benzamide, C6H5CONH2? A Nitrous acid B Phosphorus(V) chloride C Sodium hydroxide D Chlorine in the presence of aluminium chloride 22 Which of the following compounds will react with aqueous sodium carbonate to release carbon dioxide gas? I C6H5COOH II C6H5COCl III C6H5COOC2H5 A I B II C I and II D I, II and III 23 Which of the following would be produced when the compound, C6H5COOCH2CH2CH3 is treated with lithium aluminium hydride, LiAlH4, in ether? A C6H5CH2OH and CH3CH2CH2OH B C6H5OH and CH3CH2CH2CH2OH C C6H5COOH and CH3CH2CH2OH D C6H5CH2OH and CH3CH(OH)CH3 Structured and Essay Questions 1 Malonic acid is used in the manufacture of polyester. The structure of malonic acid is shown below. HOOC—CH2—COOH (b) Give the IUPAC name for malonic acid. (b) State the physical state of malonic acid at room conditions. (c) Write a reaction scheme how you would convert ethanoic acid to malonic acid. (d) Malonic acid dissociates in two steps with pKa1 = 2.83 and pKa2 = 5.69 (i) Write equations for the dissociation processes. (ii) What is the pKa for the following process? HOOCCH2COOH 4 – OOCCH2COO– + 2H+ 2 When chlorine is bubbled into ethanoic acid, boiling under reflux in the presence of ultraviolet light, chloroethanoic acid is produced. (a) (i) Write a balanced equation for the reaction. (ii) Using balanced equations, describe the mechanism of the reaction.
Chemistry Term 3 STPM Chapter 19 Carboxylic Acids and their Derivatives 245 19 (b) Is chloroethanoic acid a stronger or weaker acid compared to ethanoic acid? Explain your answer. (c) Write equations to show the reactions between chloroethanoic acid and (i) phosphorus(V) chloride (ii) 1-propanol in the presence of concentrated sulphuric acid (iii) lithium aluminium hydride in ether 3 (a) Draw the full structural formulae for 1-chlorobutane, chlorobenzene and ethanoyl chloride. (b) Describe an experiment to compare the rate of hydrolysis of the three chlorine-containing compounds. State the expected results. (c) Describe the mechanism for the hydrolysis of 1-chlorobutane. 4 A neutral compound X, C9H10O2 undergoes reduction to form Y, C2H6O and Z, C7H8O. Both Y and Z react with PCl5 to produce acidic white fumes. Y produces a yellow precipitate when warmed with iodine in aqueous sodium hydroxide. Z decolourises hot acidified KMnO4. When the final solution is cooled, a colourless crystalline solid is obtained. (a) Draw the structural formulae of X, Y and Z. (b) Name the reagent(s) used in the reduction of X and write a balanced equation for the reduction. (c) What is the colourless crystalline solid obtained at the end of the experiment? Write an equation for its formation. (d) Explain the observation for the reaction between Y and alkaline iodine. (e) Draw the structures of the products formed when Y and Z react with PCl5. 5 An organic compound A (relative molecular mass = 120.5) is optically active and fumes in moist air. A has the following composition by mass: C, 49.8%; H, 7.5%; O, 13.2%; Cl, 29.5% When A is warmed with water, compound B is formed. (a) Determine the molecular formula of A. (b) Draw the structure of A and indicate the chiral carbon with an asterisk ‘*’. (c) Determine the structural formula of B. (d) Suggest how you would convert B back to A. (e) A reacts with ethanol to produce a sweet smelling liquid C. (i) What homologous series does C belongs to? (ii) Write a balanced equation for the reaction between A and ethanol. 6 An ester W with molecular formula of C6H12O2 is optically active. When W is boiled with aqueous sodium hydroxide followed by acidification, compound X, C2H4O2 and an alcohol Y are formed. Alcohol Y is optically active and gives a yellow precipitate with alkaline iodine. Oxidation of Y produces Z which is optically inactive. (a) Identify compounds W, X, Y and Z. (b) Other than using alkaline iodine, suggest how you would differentiate between X and Y. (c) State how would Z reacts (if any) with (i) alkaline iodine (ii) 2,4-dinitrophenylhydrazine (iii) Tollen’s reagent 7 A sweet-smelling organic liquid X has the molecular formula of C10H12O, hydrolysis of X by aqueous sodium hydroxide followed by acidification produces an acid and an alcohol Y. Y reacts with bromine water to produce a white precipitate. When X is reduced using lithium aluminium hydride, two alcohols Y and Z are formed. Z reacts with excess of hot, concentrated sulphuric acid to produce 2-methylpropene. (a) Identify X, Y and Z. (b) Is X optically active? Explain your answer. (c) 2-methyl-2-propene can undergo polymerisation. Draw a repeating unit of the polymer formed.
Chemistry Term 3 STPM Chapter 19 Carboxylic Acids and their Derivatives 246 19 8 An ester has the molecular formula of C5H10O2. (a) Draw all possible isomers of the ester, which on hydrolysis produced ethanoic acid as one of its products. (b) One of the isomer in (a), X on hydrolysis produces ethanoic acid and compound A. Oxidation of A gives B which reacts with 2,4-dinitrophenylhydrazine but has no reaction with Tollen’s reagent. (i) Name the oxidising agent used in the oxidation of A. (ii) Name the functional group present in B. (iii) Draw the structure of B. (iv) Determine the structure of X. 9 The structure of acetylsalicylic acid is shown below. C O OH & O C ' CH3 O (a) Give the structure of two compounds that could be used to prepare acetylsalicylic acid. (b) Comment on the solubility of acetylsalicylic acid in water at room temperature. (c) Draw the structural formula of the major organic product formed when acetylsalicylic acid is treated with the following reagents. (i) Thionyl chloride (ii) Cold, dilute sodium hydroxide (iii) Hot, dilute sodium hydroxide (iv) Lithium aluminium hydride followed by acidic hydrolysis 10 The structure of an ester, X is shown below: !C!OCHCH3 & C ' CH3 (a) Name the ester according to the IUPAC name. (b) By using a reaction scheme, show the three-step synthesis to produce X from benzene. (c) Draw the structural formulae of the organic compounds obtained when X is treated with LiAlH4 in ether, followed by acidic hydrolysis. (d) Is X soluble in water? Briefly explain your answer. 11 The molecular formula of compound X is C3H4OCl2. When ethanol is added to X, compound Y, C5H9O2Cl is formed. Boiling X with aqueous sodium hydroxide followed by acidification produces Z, C3H6O3, which is optically active. (a) Identify compounds X, Y and Z. (b) When Z is warmed with alkaline iodine, a yellow precipitate is obtained. (i) Give the name and formula of the yellow precipitate. (ii) Write a balanced equation for the reaction between Z and alkaline iodine. (c) Draw the structures for the two optical isomers of Z. 12 Compound A (relative molecular mass = 80) has the empirical formula of C2H2N. A on reduction produces B with the empirical formula of C2H6N. A on boiling with dilute sulphuric acid produces an acid C, which reacts with ethanol to produce an ester D, with the relative molecular mass of 174. Identify compounds A, B, C and D.
Chemistry Term 3 STPM Chapter 19 Carboxylic Acids and their Derivatives 247 19 13 Aspirin has antiplatelet property that prevent the binding together of platelets and is used long term, in low doses to prevent heart attacks and blood clot in people with high risk. Aspirin is prepared via the reaction between salicylic acid and ethanoyl chloride. C"O OH OH Salicylic acid (a) Give the IUPAC name of salicylic acid. (b) Starting from iodomethane, show how you would convert iodomethane to ethanoyl chloride. (c) Write a balanced equation for tha manufacture of aspirin from salicylic acid. (d) Which compound would be more soluble in water, salicylic acid or aspirin? Explain your answer. (e) Describe a chemical test to distinguish salicylic acid from aspirin. (f) Suggest how you would modify the aspirin molecule to increase its solubility in water. Explain your reasoning. 14 (a) Starting with 1-iodopropane, suggest how you would convert it to butanenitrile. Give the reagents and state the conditions for the reaction. (b) When butanenitrile is boiled with dilute sulphuric acid, compound X is produced. X reacts with phosphorus(V) chloride to form Y with the released of a dense white fume. When phenol in sodium hydroxide is added to Y, a sweet smelling liquid Z is produced. (i) What functional group is present in X? (ii) Give the name of the dense white fume. (iii) Identify X, Y and Z. (c) Draw the structures of the organic products formed when Z reacts with lithium aluminium hydride. 15 (a) By giving the reagents and conditions of the reactions, show how you would convert benzoic acid to benzoyl chloride, C6H5COCl. (b) Benzoyl chloride fumes in moist air. (i) Name the white fume. (ii) Write an equation to show the reaction taking place. (c) Write equations for the reactions between benzoyl chloride and the following compounds. (i) Phenol, C6H5OH (ii) Phenylamine, C6H5NH2 (iii) Dimethylamine, CH3NHCH3 16 A sweet-smelling liquid, A (relative molecular mass = 116) has the following composition by mass: C: 62.0%; H: 10.3%; O: 27.7% A is neutral to litmus. When A is boiled with dilute sulphuric acid, compound B, C2H4O2, and an alcohol C is produced. B reacts with aqueous sodium carbonate to release carbon dioxide. Alcohol C is optically active and gives a yellow precipitate with alkaline iodine. When C is warmed with acidified potassium dichromate, D is produced, which also gives a yellow precipitate with alkaline iodine. Draw the structures of compounds A, B, C and D. Write equations for all the reactions involved.
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins CHAPTER AMINES, AMINO ACIDS 20 AND PROTEINS Concept Map Amines Uses of Amines Nomenclature Structural and Optical Isomerism in Amines Classification • Primary • Secondary • Tertiary Physical Properties Preparation of Amines • Reduction of nitriles and amides • Reduction of nitro-compounds Chemical Reactions of Amines • Basicity of amines • Reactions – Mineral acids – Benzenediazonium salts – Nitrous acid – Aniline with bromine water – Acyl chlorides Amino Acids and Proteins Nomenclature Structural and Optical Isomerism in α-Amino Acids Acid-base Properties of Amino Acids • Reactions with bases, acids, nitrous acid and esterification Proteins • Primary structure of protein and the peptide linkage • Hydrolysis of protein • Biological importance of proteins Peptide Linkage • Hydrolysis of polypeptides Physical Properties of Amino Acids Formation of Zwitterions
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins Learning earning Outcomes Students should be able to: Amines • write the general formula of amines; • name amines according to the IUPAC nomenclature and their common names; • describe structural and optical isomerism in amines; • state the physical properties of amines; • classify amines into primary, secondary and tertiary amines; • explain the relative basicity of ammonia, ethanamine and phenylamine (aniline) in terms of their structures; • describe the preparation of ethanamine by the reduction of nitriles, and phenylamine by the reduction of nitrobenzene; • explain the formation of salts when amines react with mineral acids; • differentiate primary aliphatic amines from primary aryl (aromatic) amines by their respective reactions with nitric(III) acid (nitrous acid) and bromine water; • explain the formation of dyes by the coupling reaction of diazonium salt as exemplified by the reaction of benzenediazonium chloride with phenol. Amino acids • write the structure and general formula of α-amino acids; • name α-amino acids according to the IUPAC nomenclature and their common names; • describe structural and optical isomerism in amino acids; • state the physical properties of α-amino acids; • describe the acid and base properties of α-amino acids; • describe the formation of zwitterions; • explain the peptide linkage as amide linkage formed by the condensation between two or more α-amino acids as exemplified by glycylalanine and alanilglycine. Protein • identify the peptide linkage in the primary structure of protein; • describe the hydrolysis of proteins; • state the biological importance of proteins. 249
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 250 20.1 Amines 1 Nitrogen is the fourth most common element (after carbon, hydrogen and oxygen) in organic compounds. 2 It is found in amines, amino acids, proteins and nucleic acids. 3 Amines (with the general formula of CnH2n + 3N) are compounds where the nitrogen atom is bonded to a carbon atom. Nomenclature 1 The IUPAC system for naming amines is to replace the suffix ‘-e’ of the corresponding alkanes with ‘amine’. 2 The chain is numbered from the end closest to the amine group. 3 For example: CH3NH2 Methanamine (Methylamine) CH3CH2NH2 Ethanamine (Ethylamine) ClCH2CH2 NH2 2-chloroethanamine (2-chloroethylamine) 1-phenylethanamine (1-phenylethylamine) !C!CH3 H NH2 ! ! H2NCH2CH2CH2CH2CH2CH2NH2 1,6-hexanediamine 4 The simplest aromatic amine, where the amino group is attached directly to the benzene ring is named phenylamine (or Aniline). NH2 5 Other aromatic amines are named as derivatives of aniline. For example: !!NH2 CH3 CH3 HO!! !!NH2 2-methylaniline 2-methyl-4-hydroxyaniline 6 Sometimes, the amine is named as the amino derivative of an alkane/arene. For example: CH3!!CH!!CH3 & & NH2 NH2 COOH 2-aminopropane 2-aminobenzoic acid 2016/P3/Q13
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 251 7 Secondary and tertiary amines are commonly named as N-substituted primary amines. For unsymmetrical amines, the larger group is taken to be the parent amine. For example: H CH3 & & CH3!!N!!CH3 CH3!!N!!CH3 Dimethylamine Trimethylamine H CH3 & & CH3!!N!!CH2CH3 N!!CH3 N-methylethanamine N,N-dimethylaniline H & N!!CH3 N-methylaniline Quick Check 20.1 1 Name the following compounds and classify them into primary, secondary or tertiary amine. (a) CH3(CH2)5NH2 (e) (b) H2N(CH2)4NH2 (c) CH2!!CH2NH2 N CH3 CH3 (d) NO2 NH2 2 Write the structural formulae for the following amines and classify them as primary, secondary or tertiary amine. (a) Triphenylamine (e) 2-chloro-2-aminobutane (b) Benzylamine (f) 2-aminohexanoic acid (c) 2-amino-1-hexanol (d) 1-phenyl-2-aminopropane (common name: amphetamine) Structural and Optical Isomerism in Amines 1 For primary amines, structural isomerism is exhibited by amines with three or more carbon atoms in their molecules. An example is C3H7NH2: CH3CH2CH2NH2 Propanamine CH3—CH—CH3 2-aminopropane & NH2
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 252 2 For aromatic amines, structural isomerism arises from the different positions of amino group with respect to the other substituents. For example, NH2 CH3 NH2 CH3 NH2 CH3 2-methylaniline 3-methylaniline 4-methylaniline 3 Primary amines are also isomeric with secondary and tertiary amines. For example, CH3CH2CH2NH2 Propanamine (Primary) CH3CH2—N—CH3 N-methylethanamine (Secondary) & H CH3—N—CH3 N,N,N-trimethylamine (Tertiary) & CH3 4 Optical isomerism occurs in primary amines with four or more carbon atoms in their molecules. For example, CH3 C2H5 C* NH2 H C* H2N C2H5 CH3 H Physical Properties 1 Due to the difference in the electronegativity of nitrogen and hydrogen, the N!H bond in amines is polarised. 2 The N!H bond is more polarised than the C!X bond, but less polar than the O!H bond. R!!N!!H !! H : δ– δ+ δ+ 3 Amines are able to form intermolecular hydrogen bonds with its own molecules. Info Chem Methylaniline is isomeric with phenylmethanamine: 9CH2NH2
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 253 H!!N: R H H!!N: R H Hydrogen bond As a result, their boiling points are higher than the corresponding alkanes but lower than that of the alcohols. Compound Relative molecular mass Boiling point/°C C2H6 30 –88 CH3NH2 31 –6 CH3OH 32 65 4 Methanamine and ethanamine are gases at room conditions. Higher amines are either liquids or solids. The boiling points of some amines are given in the table below. Compound Formula Boiling point/°C Methanamine (Methylamine) CH3NH2 –6 Ethanamine (Ethylamine) CH3CH2NH2 17 Propanamine (Propylamine) CH3CH2CH2NH2 49 Butanamine (Butylamine) CH3CH2CH2CH2NH2 78 Phenylamine (Aniline) NH2 184 5 However, the boiling points of isomeric secondary and tertiary amines are lower than expected. The table below compares the boiling points of three isomeric amines. Amine Formula Class Molar mass/g mol–1 Boiling point/°C Propanamine (Propylamine) CH3CH2CH2NH2 Primary 59 49 N-methylethanamine CH3CH2 NCH3 & H Secondary 59 37 N,N-dimethylmethanamine CH3!N!CH3 & CH3 Tertiary 59 4 Tertiary amines have no hydrogen bonded to the nitrogen atom and thus cannot form intermolecular hydrogen bonds. R R & & R!N: R!N: No hydrogen bonding possible & & R R van der Waals force 6 Primary amines have two hydrogen atoms bonded to nitrogen compared to only one hydrogen for the case of the secondary amines. Exam Tips Exam Tips Oxygen is more electronegative than nitrogen. As a result, the intermolecular hydrogen bonds in alcohols are stronger than those between amines. 2° amines form less hydrogen bonds. 3° amines cannot form hydrogen bond.
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 254 As a result, primary amines can form more extensive intermolecular hydrogen bonds than secondary amines. Therefore, the boiling point for isomeric amines increases generally in the order: Tertiary < secondary < primary 7 Amines with low molecular mass are soluble in water because they can form hydrogen bonds with water. O H H H H H!!N: R H O Hydrogen bond 8 Phenylamine (aniline) and other amines (especially tertiary amines) with large hydrophobic hydrocarbon groups are only slightly soluble in water. 9 Amines have a ‘rotten fish’ smell. Quick Check 20.2 1 Arrange the following isomeric amines in the order of increasing boiling points. CH3CH2CH2NH2, CH3CH2NHCH3, (CH3)3N Explain your answer. 2 Arrange the following amines in the order of increasing solubility in water. CH3CH2CH2NH2, CH3CH2NHCH3, NH2 Classification 1 Amines can be classified as primary, secondary or tertiary depending on the number of alkyl groups that are bonded to the nitrogen atom. 2 The structures of the three classes of amines are shown below: R—N—H R—N—R R—N—R & & & H H R Primary amine Secondary amine Tertiary amine 3 The following is not an amine. It is a quaternary ammonium salt: R & R—N+—R & R It is a tetraalkylammonium ion.
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 255 Chemical Reactions of Amines 1 Like ammonia, amines are weak bases. An aqueous solution of amine has pH >7. 2 An amine behaves as a base by donating its lone-pair electrons on the nitrogen atom to a H+ ion to form a coordinate bond or by accepting a proton from an acid to form an alkylammonium ion. R!N: + H!OH L + OH– R!N:H !! H ! H H ! H + R!N: + H+ !!: R!N:H !! H ! H H ! H + 3 Compare the similarity with that of ammonia. H!N:!! + H!OH L H!N:H + OH– H ! H H ! H + H!N: + H+ !!: H!N:H !! H ! H H ! H + Basicity of Amines 1 The strength of a weak base is measured by its base dissociation constant, Kb. RNH2 + H2O L RNH3 + + OH– [RNH3 +][OH– ] Kb = –––––––––––––– [RNH2] The higher the value of Kb, the stronger the base. 2 The table below lists the Kb values for ammonia and some amines. Compound Formula Kb/mol dm–3 Ammonia NH3 1.8 × 10–5 Methanamine CH3NH2 4.4 × 10–4 Ethanamine CH3CH2NH2 5.4 × 10–4 N-methylmethanamine CH3!N!CH3 & H 5.0 × 10–4 Amines can be considered as alkyl-substituted ammonia. 2009/P2/Q10(a) 2012/P1/Q38 2017/P3/Q20
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 256 Compound Formula Kb/mol dm–3 N,N-dimethylmethanamine CH3!N!CH3 & CH3 6.5 × 10–5 Phenylmethanamine CH2NH2 2.2 × 10–5 Phenylamine (Aniline) NH2 4.2 × 10–10 3 Aliphatic amines are stronger bases than ammonia (Kb > 1.8 × 10–5). This is because the alkyl group that is bonded to the nitrogen atom is an electron-donating group. H & R:N: & H This increases the electron density on the nitrogen atom and makes the lone-pair electrons more readily react with H+ compared to ammonia. 4 Aromatic amines (e.g. aniline) are a weaker base than ammonia. This is because the benzene ring is an electron-withdrawing group. H & ;N: & H As a result, the lone-pair electrons are more tightly held by the nitrogen atom and are less readily donated to H+. 5 Furthermore, the lone-pair electrons can also delocalise into the benzene ring making it less available for protonation. 6 Electron-donating substituents increases the basicity of aromatic amines, while electron-withdrawing substituents decreases the basicity of aromatic amines. H & D : 9N: & H D = Electron-donating group H & W; 9N: & H W = Electron-withdrawing group Info Chem The electron-releasing alkyl group also stabilises the alkylammonium cation: H & R:N+9H & H Info Chem The electron-withdrawing benzene ring destabilises the benzeneammonium cation: H & ;N+9H & H Exam Tips Exam Tips • Electron-donating group increases the basicity of phenylamine. • Electron-withdrawing group decreases the basicity of phenylamine. Phenylamines are weaker base than aliphatic amines.
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 257 7 The table below lists the Kb values for phenylamine (aniline) and substituted phenylamine. Compound Structure Kb /mol dm–3 Phenylamine (aniline) NH¨ 2 4.2 × 10–10 4-hydroxyphenylamine (4-hydroxyaniline) HO: NH2 3.0 × 10–9 4-nitrophenylamine (4-nitroaniline) O2N; NH¨ 2 9.5 × 10–14 Preparation of Amines Reduction of Nitriles 1 Nitriles are reduced to primary amines. RCN + 4[H] !!: RCH2NH2 2 The reducing agents used are (a) hydrogen gas in the presence of nickel or platinum, (b) lithium aluminium hydride in ether, followed by acidic hydrolysis. 3 For example: CH3CN + 2H2 !!: CH3CH2NH2 Ethanitrile Ethanamine (Ethylamine) CN + 2H2 !!: CH2NH2 Cyanobenzene Phenylmethanamine Reduction of Amides 1 Lithium aluminium hydride or hydrogen in the presence of a suitable catalyst reduces amides to amines. The class of amines produced depends on the class of amides used. & Reduction & !!C!!N!! !!9!: !!CH2!!N!! ∫ O 2 Reduction of a primary amide produces a primary amine. R!C!NH2 + 4[H] !!: RCH2NH2 + H2O ∫ O For example: CH3CONH2 + 4[H] !!: CH3CH2NH2 + H2O Ethanamide Ethanamine INFO Basicity of Amines
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 258 3 Reduction of a secondary amide produces a secondary amine. R!C!N!R + 4[H] !!: R!CH2!N!R + H2O ∫ & & O H H For example: CH3!C!N!CH3 + 4[H] !!: CH3CH2!N!CH3 + H2O ∫ & & O H H N-methylethanamide N-methylpropanamine 4 Reduction of a tertiary amide produces a tertiary amine. R!C!N!R + 4[H] !!: R!CH2!N!R + H2O ∫ & & O R R For example: CH3!C!N(CH3)2 + 4[H] !!: CH3CH2!N!CH3 ∫ & O CH3 N,N-dimethylethanamide N,N-dimethylethanamine Reduction of Nitro-compounds 1 Nitro-compounds are organic compounds with the !NO2 group. 2 Phenylamine is usually prepared by the reduction of nitrobenzene using a mixture of tin and concentrated hydrochloric acid. !NO2 + 6[H] !!: NH2 + 2H2O Nitrobenzene Phenylamine Example 20.1 Suggest how you would convert benzene into phenylamine. Solution React benzene with a mixture of concentrated nitric acid and concentrated sulphuric acid at temperature not exceeding 55 °C to produce nitrobenzene. H NO2 + HNO3 !!: + H2O Preparation of arylamines
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 259 Reduce nitrobenzene with a mixture of tin and concentrated hydrochloric acid. NO2 NH2 + 6[H] !!: + 2H2O Quick Check 20.3 1 Suggest how you would carry out the following preparations. (a) Ethylamine from iodoethane (b) Phenylmethylamine from methylbenzene (c) Propylamine from iodoethane Reactions of Amines with Mineral Acids 1 Amines react with dilute mineral acids to form ionic salts which can be crystallised out as colourless crystals. RNH2 + HX !!: RNH3 +X– For example: CH3CH2NH2 + HCl !!: CH3CH2NH3 +Cl– Ethanamine Ethylammonium chloride NH2 + HCl !!: NH3 +Cl– Phenylamine Phenylammonium chloride 2 The salts are decomposed by strong bases to regenerate the amines. CH3CH2NH3 +Cl– + OH– !!: CH3CH2NH2 + H2O + Cl– NH3 +Cl– + OH– !: NH2 + H2O + Cl– Example 20.2 Phenylamine is insoluble in water. However, it is completely miscible with dilute hydrochloric acid. When dilute sodium hydroxide is added to the solution of phenylamine in dilute acid, two layers separate out. Explain the observations. Exam Tips Exam Tips Amines are weaker base than NaOH. 2017/P3/Q12
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 260 Solution Phenylamine is insoluble in water due to the presence of the large hydrophobic phenyl group. On the addition of dilute hydrochloric acid, phenylamine reacts to form a water soluble salt. NH2(l) + HCl(aq) !: NH3 +Cl– (aq) Phenylammonium chloride On the addition of sodium hydroxide, phenylamine is regenerated and two layers separate out. NH3 +Cl– (aq) + NaOH(aq) !!: NH2(l) + H2O(l) + NaCl(aq) Quick Check 20.4 1 Draw the structures of the salts formed in the following reactions. (a) (d) (b) (c) 2 The pKb values for aniline and nitroaniline are given in the table below. Compound Structure pKb Aniline NH2 9.37 2-nitroaniline NH2 NO2 11.5 4-nitroaniline NO2 NH2 13.0 Explain the difference in the pKb values of the three compounds. 3 Suggest how you would separate a mixture containing aniline and nitrobenzene. !!NH2 + HCl CH3!!NH!!CH3 + H2SO4 !!N!! + HCl !! NH + HCl
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 261 Reaction with Nitrous Acid 1 Nitrous acid, HNO2 is unstable and is readily oxidised by air to nitric acid, HNO3. O N OH RR : 2 Nitrous acid is usually generated in the solution by the reaction of dilute hydrochloric acid on sodium nitrite. NaNO2(aq) + HCl(aq) !!: NaCl(aq) + HNO2(aq) 3 Nitrous acid reacts with amines in different ways depending on the class of amines involved. Nitrous Acid with Primary Amines 1 Nitrous acid reacts with primary amines upon warming to produce alcohols with the release of nitrogen gas (effervescence). R!N H2 + ORN!OH !!: R!OH + N2 + H2O Or RNH2 + HNO2 !!: ROH + N2 + H2O 2 Examples are: CH3CH2CH2NH2 + HNO2 !!: CH3CH2CH2OH + N2 + H2O Propanamine 1-propanol NH2 + HNO2 !: OH + N2 + H2O Phenylamine Phenol 3 Primary aliphatic amines also produce a little alkene as side products in its reaction with nitrous acid. This is because primary aliphatic amines react with nitrous acid to give a highly unstable aliphatic diazonium ion as intermediate. RNH2 + HNO2 + H+ !!: RN+#N + 2H2O The diazonium ion decomposes spontaneously even at a low temperature to a carbonium ion and nitrogen gas. R!N+#N !!: R+ + N2 The carbonium ion then reacts with water to form alcohol. R+ + H2O !!: R!OH + H+ Or the carbonium ion can eliminate a H+ ion to form an alkene. For example: CH3!CH2 + !!: CH2RCH2 + H+ HNO2 + —1 2 O2 !!: HNO3 x O x x O x H N Production of alkenes as side products 2008/P2/Q4(a) 2017/P3/Q9
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 262 4 Primary aromatic amines such as phenylamine react with nitrous acid to produce benzenediazonium ion which is stable at temperature below 5 °C (no effervescence). <5 °C NH2 + HNO2 + H+ !: N+#N + 2H2O Phenylamine Benzenediazonium ion 5 The benzenediazonium is stablised through delocalisation of the positive charge into the benzene ring. No such delocalisation is possible for non-aromatic amines. + !N#N 6 The benzenediazonium salts are very useful intermediates in the synthesis of other aromatic compounds. Nitrous Acid with Secondary Amines 1 Secondary amines, whether aliphatic or aromatic react with nitrous acid to produce oily yellow liquids called N-nitrosoamines. 2 There is no liberation of nitrogen gas (no effervescence) during the reaction. This process is called nitrosation. Rfi R R!N ! !H + HO!NRO !: N!NRO + H2O Rfi 3 For example: CH3 CH3 CH3!N!CH3 + HNO2 ! !: N!NRO + H2O H N-methylmethanamine N-methyl-N-nitrosomethanamine (Dimethylamine) (N-nitrosodimethylamine) N!CH3 + HNO2 !: N!NRO + H2O & & H CH3 N-methylphenylamine N-methyl-N-nitrosophenylamine Nitrous Acid with Tertiary Amines 1 Tertiary aliphatic amines react with nitrous acid to form colourless water soluble salt. Hence, no visible change is observed during the reaction. R3N + HNO2 !!: R3NH+NO2 – Production of benzenediazonium salt 2010/P2/Q4(b) Info Chem Nitrosoamines are carcinogens. 2014/P3/Q13 Exam Tips Exam Tips 9CH2N2 cannot be stabilised through resorance due to the presence of the 9CH2 group +
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 263 For example: (CH3)3N + HNO2 !!: (CH3)3NH+NO2 – Trimethylamine Trimethylammonium nitrite 2 Nitrous acid provides a convenient laboratory method to distinguish between the different classes of amines. This is summarised in the table below. Class of amine Observation with nitrous acid Primary Effervescence at room temperature Secondary Yellowish oil Tertiary No visible change Reactions Involving Benzenediazonium Salts 1 Benzenediazonium ion is unstable at temperature above 5 °C. For example, when a solution containing benzenediazonium chloride is warmed gently, nitrogen gas is evolved and phenol is formed. N+#N + H2O !Δ : OH + N2 + H+ 2 Benzenediazonium ion is seldom isolated. In the solid state, it will explode when warmed or when subjected to slight shock. It is usually left in solution and used immediately by adding appropriate reagents before it decomposes. 3 When warmed, benzenediazonium chloride decomposes to produce a positive charge benzene ring which is then attacked by the nucleophiles present in the reagents. N+#N !Δ : + + N2 + + :Nu– !: Nu 4 The above mechanism is that of nucleophilic substitution. Coupling Reactions 1 Benzenediazonium ion undergoes coupling reaction with aromatic amines and phenols to yield a class of compounds called azo compounds. 2 The azo compounds are brightly-coloured solids at room conditions and are used extensively as dyes (Azo-dyes). 3 Coupling usually takes place at the para position of the aromatic amines or phenols. However, if the para position is not available, then coupling will occur at the ortho position. No observable change. To differentiate between 1°, 2° and 3° amines Coupling occurs at the para position. Nucleophilic substitution 2010/P1/Q38 2015/P3/Q8 2014/P3/Q13 2016/P3/Q20(b)
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 264 4 Coupling between benzenediazonium chloride and phenol produces a bright orange solid. 5 Coupling with N,N-dimethylphenylamine produces a bright yellow solid. N+#N + H N(CH3)2 !!: NRN N(CH3)2 + H+ N,N-dimethylazobenzene Reaction of Aniline (Phenylamine) with Bromine Water 1 When bromine water is added to aniline (phenylamine) at room temperature, the reddish-brown colour of bromine water is decolourised and a white precipitate of 2,4,6-tribromophenylamine is formed. + 3Br2 + 3HBr NH2 Br Br Br !: NH2 2,4,6-tribromophenylamine Phenylamine 2 Similarly with 2-methylphenylamine. + 2Br2 + 2HBr NH2 CH3 CH3 Br Br !: NH2 2-methyl-4,6-dibromophenylamine 2-methylphenylamine 3 This serves as a simple method to differentiate phenylamine from non-phenylamine. N+#N + H OH !: NRN OH + H+ 4-hydroxyazobenzene Exam Tips Exam Tips Test for phenylamine Exam Tips Exam Tips Phenol also gives the same observation. OH Br Br Br 2010/P2/Q10(b) 2013/P3/Q16(b) 2017/P3/Q13
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 265 4 Phenylamine also reacts with chlorine water to produce a white precipitate of 2,4,6-trichlorophenylamine. 5 This is an example of electrophilic substitution. The !NH2 group is a powerful ring-activator. It activates the 2, 4 and 6 position of the benzene ring to be easily attacked by electrophiles. (Note: Phenol also gives a white precipitate with bromine water.) Quick Check 20.5 1 Suggest a simple chemical test to differentiate between (a) aniline and phenylmethylamine, (b) aniline and phenol, (c) ethylamine and ethanamide. 2 9.3 g of a sample of aniline is added to excess bromine water. 22.6 g of a white precipitate is obtained. Calculate the percentage purity of the sample of aniline. Uses of Amines 1 Amines are used to manufacture azo-dye. 2 Amines are widely used as drugs such as antihistamine. 3 Amines and their derivatives are used as tranquillisers, analgesics, bactericides and pesticides. 4 Amines are used to manufacture polymers such as nylon and polyurethane. !![!NH!(CH2)6!NH!C!(CH2)4!C!!!]n ∫ ∫ O O Nylon-6,6 20.2 Amino Acids 1 Amino acids are organic compounds with both amino group, !NH2 and carboxyl group, !COOH in their structure. 2 Amino acids have the general formula of HOOC!R !NH2 3 Amino acids are the basic building blocks for proteins. 4 The amino acids in proteins are α-amino acids. These are amino acids in which both the carboxyl group and the amino group are bonded to the same carbon atom. R!!C!!COOH NH2 !! !! α-carbon H NH2 & & Cl Cl Cl 2016/P3/Q14 2017/P3/Q16(b)
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 266 Nomenclature 1 Amino acids are named as the amino derivatives of carboxylic acids. However, their original non-IUPAC names are commonly used. 2 The simplest α-amino acid is aminoethanoic acid or glycine, H2NCH2COOH. H!C!COOH NH2 H ! ! 3 Other examples of α-amino acids are: CH3!C!COOH NH2 H ! ! 2-aminopropanoic acid (Alanine) !CH2!C!COOH NH2 H! ! 2-amino-3-phenylpropanoic acid (Phenylalanine) H2NCH2CH2CH2CH2!C!COOH NH2 H! ! 2,6-diaminohexanoic acid (Lysine) HOOC!CH2!C!COOH NH2 H ! ! 2-aminobutanedioic acid (Aspartic acid) 4 Amino acids are classified as neutral, acidic or basic depending on the relative number of amino group and carboxyl group they have. The following is not an α-amino acid. & & !C!C!COOH & & NH2 INFO Amino Acids and Protiens
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 267 5 If the number of amino groups and carboxyl groups are the same, the aqueous solutions of amino acids are near neutral (pH 7). Examples are glycine and alanine. H!C!COOH NH2 H! ! CH3!C!COOH NH2 H! ! Glycine Alanine 6 If the number of carboxyl groups exceeds the number of amino groups, the aqueous solution of these amino acids will be acidic. An example is: HOOC!CH2!C!COOH NH2 H ! ! 2-aminobutanedioic acid 7 If the number of amino groups exceeds the number of carboxyl groups, the aqueous solution of these amino acids will be basic. An example is: H2NCH2CH2CH2CH2!C!COOH NH2 H! ! 2,6-diaminohexanoic acid 8 The names, abbreviations and structures of the 20 α-amino acids in protein are given below. Name Structure Alanine Ala & & H3C!C!COOH H NH2 Arginine Arg H H & & H2N!C!N!CH2!CH2!CH2!C!COOH ∫ & NH NH2 Asparagine Asn O H ∫ & H2N!C!CH2!C!COOH & NH2
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 268 Name Structure Aspartic acid Asp H & HOOC!CH2!C!COOH & NH2 Cysteine Cys H & HS!CH2!C!COOH & NH2 Glutamic acid Glu H & HOOC!CH2!CH2!C!COOH & NH2 Glutamine Gln O H ' & H2N!C!CH2!CH2!C!COOH & NH2 Glycine Gly H & H!C!COOH & NH2 Histidine His & & & HCRC!CH2!C!COOH H N NH2 C H R ! NH Isoleucine Ile CH3H & & H3C!CH2!C!C!COOH & & H NH2 Leucine Leu & & CH!CH2!C!COOH H NH2 H3C H3C Lysine Lys H & H2N!CH2!CH2!CH2!CH2!C!COOH & NH2 Methionion Met H & H3N!S!CH2!CH2!C!COOH & NH2 Phenylalanine Phe H & CH2!C!COOH & NH2
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 269 Name Structure Proline Pro H & HN!C!COOH & & H2C CH2 CH2 Serine Ser H & HO!CH2!C!COOH & NH2 Threonine Thr OH H & & H3C!C!C!COOH & & H NH2 Tryptophan Trp & & & ! !!C!CH2!C!COOH H N H NH2 ∫ CH Tyrosine Tyr & & HO! !CH2!C!COOH H NH2 Valine Val && & H3C!CH!C!COOH 3C HH NH2 Quick Check 20.6 1 Predict whether the aqueous solution of the following amino acid is acidic, basic or neutral. (a) (c) !CH2!C!COOH NH2 H! ! HOOC!CH2CH2!C!COOH NH2 H! ! (b) H2N!C!CH2!C!COOH NH2 H! ! O R Info Chem Proline is the only a-amino acid where the amino group is a secondary amine.
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 270 Structural and Optical Isomerism in α-amino Acids 1 All the 20 α-amino acids found in proteins are optically active except aminoethanoic acid (glycine). This is because aminoethanoic acid does not have a chiral carbon in its structure. H & H2N9C9COOH & H 2 The other 19 α-amino acids have a chiral centre in their respective molecules. For example, alanine (2-aminopropanoic acid) and asparagine (2-aminobutanedioic acid). H & H2N9C*9COOH Alanine & CH3 H & HOOC—CH2—C*—COOH Asparagine & NH2 3 α-amino acids also exhibit structural isomerism with other amino acids. For example, asparagine is isomeric with 2-amino-2-methylpropanedioic acid which is not optically active. CH3 & HOOC—C—COOH & NH2 Physical Properties of Amino Acids 1 Amino acids are all crystalline solids at room conditions. They are soluble in water but insoluble in non-polar solvents. 2 The presence of —NH2 and —COOH groups enable the amino acids to form hydrogen bonds with water molecules. As a result, they are soluble in water but not in non-polar solvents. Example 20.3 Predict whether the aqueous solutions of the following amino acids will be neutral, acidic or basic. (a) H2NCH2COOH (b) H2N!(CH2)4!CH!COOH & NH2 2017/P3/Q14
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 271 (c) HOOC!CH2!CH!COOH & NH2 Solution (a) Neutral. It contains the same number of !NH2 and !OH groups. (b) Basic. It has more !NH2 group than !COOH group. (c) Acidic. It has more !COOH group than !NH2 group. Asid-base Properties of Amino Acids With the presence of both amino group and carboxyl group in their structures, amino acids have both acidic and basic characteristics. They are amphoteric. Reaction with Bases 1 Amino acids react with bases to form salts. H2N!R!COOH + OH– !!: H2N!R!COO– + H2O Examples are: H2NCH2COOH + NaOH !!: H2NCH2COO– Na+ + H2O Sodium aminoethanoate H2N!CH!COOH + NaOH !: H2N!CH!COO– Na+ + H2O & & CH2 CH2 2 Amino acids react with sodium carbonate to liberate carbon dioxide gas. 2H2N!R !COOH + Na2CO3 !: 2H2N!R!COO– Na+ + CO2 + H2O For example: 2H2NCH2COOH + Na2CO3 !: 2H2NCH2COO– Na+ + CO2 + H2O Esterification 1 Amino acids react with alcohols in the presence of concentrated sulphuric acid as a catalyst to produce esters. H2N!R!COOH + R9OH !!: H2N!R!COOR9 + H2O Acid-based reaction Reactions of acid and amine 2012/P1/Q39, Q50 2013/P3/Q14 2015/P3/Q20(b)
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 272 2 For example: H2NCH2COOH + C2H5OH !!: H2NCH2COOC2H5 + H2O Ethyl aminoethanoate With Phosphorus Pentachloride or Thionyl Chloride 1 Amino acids react with PCl5 or SOCl2 with the liberation of white fumes of hydrogen chloride. H2N!R!COOH + PCl5 !: H2N!R!COCl + POCl3 + HCl H2N!R!COOH + SOCl2 !: H2N!R!COCl + SO2 + HCl 2 For example: H2NCH2COOH + PCl5 !: H2NCH2COCl + POCl3 + HCl H2N!CH!COOH + SOCl2 !: H2N!CH!COCl + SO2 + HCl & & CH2 CH2 Reaction with Mineral Acids The basic amino group reacts with acids to form salts. HOOC!R!NH2 + H+ !!: HOOC!R!NH3 + For example: HOOC!CH2!NH2 + HCl !!: HOOC!CH2!NH3 +Cl– Reaction with Nitrous Acid Amino acids react with nitrous acid to liberate nitrogen gas, and the !NH2 group is replaced by !OH. HOOC!R!NH2 + HNO2 !: HOOC!R!OH + N2 + H2O For example: HOOCCH2NH2 + HNO2 !: HOOC!CH2!OH + N2 + H2O Hydroxyethanoic acid H2N!CH!COOH + HNO2 !: HO!CH!COOH + N2 + H2O & & CH2 CH2 2-hydroxy-3-phenylpropanoic acid Acid-based reaction
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 273 With Acyl Chlorides Amino acids react with acyl chlorides or acid anhydrides to form substituted amides. R9COCl + H2N!R!COOH !: R9!C!N!R!COOH + HCl ∫ & O H For example: Quick Check 20.7 Draw the structure of the product formed for the reaction between H2N!CH!CH2CH2COOH and & COOH (a) sodium hydroxide (e) zinc (b) dilute hydrochloric acid (f) ethanoyl chloride (c) phosphorus(V) chloride (g) nitrous acid (d) sodium carbonate (h) ethanol in the presence of concentrated sulphuric acid Formation of Zwitterions 1 A zwitterion is a neutral molecule with a positive and a negative charge at different locations within the molecule. It is sometimes called an internal salt. 2 Amino acids are classical examples of zwitterions. Amino acid molecules contain basic —NH2 group and acidic —COOH group. 3 These two groups can undergo intramolecular acid-base reaction where the proton of the —COOH group is transferred to the nitrogen atom of the —NH2 group. H H & & NH29C9COOH H3N+9C9COO– & & R R An example is glycine (2-aminoethanoic acid): H2NCH2COOH H3N+CH2COO– 4 Other examples are: HOOC9CH29CH9COO– Aspartic acid & NH3 + H3N+9CH9COO– Alanine & CH3 CH3COCl + H2N!CH2!COOH !: CH3!C!N!CH2!COOH + HCl ∫ & O H Exam Tips Exam Tips The overall charge of any zwitterion is zero. 2012/P2/Q9(a) 2015/P3/Q17(b)
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 274 5 In the solid state, all α-amino acids exist in the zwitterion form with strong ionic bonds holding the 'ions' together. As a result, all α-amino acids are crystalline solids with high melting points. 6 The 'ionic' character is responsible for their solubility in water. Example 20.4 The amino acid, lysine has the following structure: H2N(CH2)4CHCOOH & NH2 (a) Draw the zwitterion formed by lysine. (b) Draw the predominant structure when solid lysine is dissolved in 1.0 M hydrochloric acid. (c) Draw the predominant structure when solid lysine is dissolved in 1.0 M sodium hydroxide. Solution (a) H2N9(CH2)49CH9COO– & NH3 + (b) In acidic solution, both the NH2 and COO– groups get protonated. H3N+—(CH2)4—CH—COOH & +NH3 (c) In alkaline solution, the +NH3 group gets deprotonated. H2N—(CH2)4—CH—COO– & NH2 The Peptide Linkage 1 Under suitable conditions, an acid reacts with an amine to form an amide through the elimination of a water molecule. This is called condensation. For example: CH3COOH + C2H5NH2 !!: CH3!C!N!C2H5 + H2O ∫ & O H N-ethylethanamide 2 The !C!N! group is known as a peptide bond or a peptide link. ∫ & O H Alternative explanation: In acidic solution, the NH2 groups get protonated: H2N(CH2)4CHCOOH + 2H+ : & NH2 H3 +N(CH2)4CHCOOH &+NH3 In alkaline solution, the COOH group gets deprotonated: H2N(CH2)4CHCOOH + OH– : & NH2 H2N(CH2)4CHCOO– + H2O & NH2 2008/P2/Q4(b) 2015/P3/Q17(c) 2011/P1/Q39 2014/P3/Q20 2015/P3/Q17(a)
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 275 3 Two amino acid molecules can undergo condensation to form a dipeptide. For example: The dipeptide glycine-alanine is abbreviated as Gly-Ala. 4 However, the dipeptide Gly-Ala still has a carboxyl and amino groups in its structure. As a result, it can still undergo further condensation to form a polypeptide with the repeating unit of !! [ N!CH2!C!N!CH!C !! ] & ∫ & & ∫ n H O H CH3 O 5 By convention, the name of polypeptides are written beginning from the amino acid with the free !NH2 group and proceeding towards the amino acid with the free !COOH group. 6 The amino acid with the free !NH2 group is called N-terminal amino acid and the amino acid with the free !COOH group is called C-terminal amino acid. 7 For example, the following dipeptide: H2N!CH!C!!N!CH2!COOH H ! CH3 ! O R N-terminal amino acid (Alanine) C-terminal amino acid (Glycine) is named as alanil-glycine or Ala-Gly. 8 But the following dipeptide: H2N!CH2!C!!N!CH!COOH H ! CH3 ! O R N-terminal amino acid (Glycine) C-terminal amino acid (Alanine) is named as glycyl-alanine or Gly-Ala. H2N!CH2!COOH + H2N!CH!COOH !: H2N!CH2!C!N!CH!COOH + H2O & ∫ & & CH3 O H CH3 Glycine Alanine Glycine-alanine Polypeptide Exam Tips Exam Tips Name of peptides: The amino group preceeds the acid group. Exam Tips Exam Tips Number of amino acid residues in a polypeptide = number of peptide link + 1
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 276 Quick Check 20.8 1 Draw the structures of dipeptide seryl-alanine, Ser-Ala and alanil-serine, Ala-Ser. 2 Name the following tripeptide. H2N!CH2!C!N!CH!C!N!CH!COOH H ! H ! CH2 !! CH3 ! O R O R Hydrolysis of Polypeptides 1 Polypeptide chains can be hydrolysed into its component amino acids by boiling it with dilute mineral acids such as sulphuric acid, or dilute alkali such as sodium hydroxide. 2 During hydrolysis, the peptide linkage between the amino acids breaks. O H ∫ & H+ ∫ & + H2O !!: !COOH + !NH2 !!C!!N!! 3 The amino acid residues can then be separated and analysed using electrophoresis or other separating techniques. Quick Check 20.9 1 Draw the structural formulae of the amino acid residues obtained when the following polypeptide is boiled with excess dilute sulphuric acid. H2N!CH2!C!N!CH!C!N!CH!COOH H ! H ! CH2 !! CH3 ! O R O R 2011/P1/Q50 2012/P2/Q9(c) 2013/P3/Q17(b) 2015/P3/Q12
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 277 20.3 Proteins Primary Structure of Protein and the Peptide Linkage 1 Proteins are linear polymers of α-amino acid residues. The amino acids are joined by peptide bonds to form long-linear chains of polypeptides. Amino acid Amino acid Amino acid Amino acid Amino acid Peptide linkage 2 The general formula of a protein chain is: H R O H R' O & & ' & & ' 99[9N9CH9C9N9CH9C99]n 9 Peptide link 3 Protein molecules can contain from about fifty to tens of thousands of amino acid residues in their structure. For example, the protein ribonuclease has 124 amino acid residues in the polypeptide chain. 4 The type of amino acids and the sequence they are joined via the peptide linkage give the primary structure of the protein. 5 The primary structure of ribonuclease is shown below: H2N Lys 1 7 12 10 20 72 70 65 60 58 21 84 26 30 90 119 95 40 100 124 120 80 110 HOOC 50 41 Glu Thr Ala Ala Ala Lys Phe Glu ArgGlnHisMetAsp Ser Ser Thr Ser Ala Ala Ser Ser Ser Asn Tyr Cys Asn Gln Met Met Lys Ser Arg Asn Leu Thr Lys Asp Arg Cys Lys Pro Val Asn Phe Thr Ser Glu His Val Leu Ala Asp Val Gln Ala Val Cys Ser Gln Lys Asn Val Ala Cys Lys Asn Gly Gln Thr Asn Cys Tyr Gln Ser Tyr Ser Thr Met Ser Ile Thr Asp Cys Arg Glu Thr Gly Ser Ser Lys Tyr Pro Asn Cys Ala Tyr Lys Thr Lys GlnAlaAsn Thr His Ile He Val Ala Cys Glu Gly Asn Pro Tyr Val Pro Val His Phe Asp Val Ser Ala 6 Each protein has its own primary structure.
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 278 SUMMARY SUMMARY 1 Amines are compounds with !NH2 group. 2 Phenylamines are compounds where the !NH2 group is attached directly to the benzene ring. 3 Amines can be classified as primary, secondary and tertiary. 4 Amines are weak bases. 5 Aromatic amines (such as aniline) are weaker base than aliphatic amines. 6 The presence of electron-releasing groups increases the basic strength of amines. 7 The presence of electron-withdrawing groups decreases the basic strength of amines. 8 All primary amines give effervescence with nitrous acid at room temperature. 9 Phenylamines can react with nitrous acid at 5 °C to produce benzenediazonium ion. 10 Phenylamines react with aqueous bromine or aqueous chlorine to give a white precipitate. Hydrolysis of Protein 1 The peptide linkages in a protein chain can undergo hydrolysis to regenerate respective amino acids. For example, Hydrolysis H2N9CH9C9N9CH9C9OH 999: & ' & & ' R O H R' O H2N9CH9COOH + H2N9CH9COOH & & R R' 2 The hydrolysis of protein can be carried out by three methods: (a) Boiling with dilute mineral acid such as sulphuric acid. (b) Boiling with aqueous sodium hydroxide or potassium hydroxide. (c) Action of enzyme at about 37 °C. Biological Importance of Proteins 1 Proteins are found in abundance in the human body. 2 Proteins are found in hormones, haemoglobin, enzymes, antibodies, skin, hair, connecting tissues and many others. 3 Proteins are essential for life. Haemoglobin carries oxygen from the lungs to the cells. Enzymes catalyse biochemical reactions which otherwise would be too slow to sustain life at body temperatures. Antibodies destroy foreign microorganisms in our system.
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 279 Reactions of Primary Amines Reagent Product Remarks Dilute mineral acids, room temperature RNH3 + Acid-base reaction Aliphatic primary amines are stronger base than NH3. HNO2, warm ROH + N2 Effervescence R9COCl, room temperature R9!C!NH!R ∫ O Acylation Formation of substituted amide Reactions of Phenylamine Reagent Product Remarks Dilute mineral acids, Room temperature NH3 + Acid-base reaction Phenylamine is a weaker base than NH3. HNO2(aq), warm OH + N2 Effervescence N2 gas is liberated. HNO2(aq), < 5 °C N2 + Formation of benzenediazonium ion No effervescence Br2(aq), room temperature NH2 Br Br Br White precipitate of 2,4,6-tribromophenylamine Reactions of Benzenediazonium Ion Reagent Product Remarks H2O, heat OH – Phenol in NaOH NRN OH Coupling reaction Coloured solid is formed. 11 Amino acids are compounds with two functional groups: The amino group, !NH2; and the carboxyl (or carboxylic acid) group, !COOH. 12 Amino acids are crystalline solids with high melting points and are soluble in water. 13 In solid state, they exist as zwitterions. 14 α-amino acids have the amino group attached to the α-carbon in the molecule. 15 All amino acids are optically active except glycine (aminoethanoic acid). 16 Amino acids show the typical reactions of carboxylic acids and primary amines. 17 Proteins are polypeptides consisting of many amino acid molecules joined by peptide bonds. 18 Hydrolysis of protein yields their amino acid residues. 19 The primary structure of a protein refers to the sequence how the amino acids are joined in the polypeptide chains.
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 280 STPM PRACTICE 20 Objective Questions 1 A peptide has the structural formula of H2NCH2CONHCH(CH3)CONHCH(CH2OH) COOH. The following compounds are formed from the hydrolysis of the peptide except A H2NCH2COOH B H2NCOCH(CH2OH)COOH C H2NCH(CH3)COOH D H2NCH(COOH)CH2OH 2 The structural formulae of three organic bases are shown below: CH3 !N!CH3 & CH3 NH2 CH2 NH2 I II III Which is the correct sequence of the bases according to the decreasing order based on their pKb values? A I, II, III C III, II, I B II, III, I D I, III, II 3 Which of the following is true regarding aniline and cyclohexylamine? A Cyclohexylamine is a primary amine whereas aniline is a tertiary amine. B Aniline is a st ronger base than cyclohexylamine. C They react differently with nitrous acid at room temperature. D Cyclohexylamine has a higher boiling point than aniline. 4 Which of the following aqueous solution decomposes readily when warmed? A CH3CH2CHRCHCl B NH2 C N2 +HSO4 – D CN 5 Which of the following is true regarding ethylamine? A It is a weaker base than ammonia. B It is a weaker base than methylamine. C It can be produced by reducing ethanenitrile using lithium aluminium hydride. D It reacts with concentrated nitric acid to release nitrogen gas when heated. 6 Wh i c h s t a t e m e n t i s t r u e a b o u t HOOCCH2COOH and H2NCH2COOH? A H2NCH2COOH is optically active but HOOCCH2COOH is not. B Both can act as buffer solutions. C HOOCCH2COOH is soluble in dilute NaOH but H2NCH2COOH is not. D Both are solid at room conditions. 7 Which of the following statements regarding aniline (phenylamine) is not true? A It liberates ammonia when warmed with aqueous sodium hydroxide. B It liberates nitrogen gas when warmed with nitrous acid. C It gives a white precipitate when reacted with chlorine water. D It is soluble in aqueous sodium hydroxide but not in dilute sulphuric acid. 8 Benzenediazonium chloride reacts with an alkaline solution of phenol to produce W. W is used as a A dye C flavour enhancer B food preservative D herbicide 9 Some reactions of compound X is listed in the table below: Reagent Observation Phosphorus(V) chloride White fumes Nitrous acid Effervescence Sodium carbonate Effervescence X could be A H2NCH(CH3)COOH B CH3NHCH2COOH
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 281 C H2N COOH D CONH2 CH2OH 10 Wh a t w o u l d b e p r o d u c e d w h e n (CH3)2C"CHC#N reacts with hydrogen gas in the presence of nickel catalyst? A (CH3)2CHCH2C#N B (CH3)2C"CHCH2NH2 C (CH3)2CHCH2CH2NH2 D CH3CH2CH3 and CH3CN 11 Which of the following reagents can be used to differentiate between aniline and 2-aminobutane? I Bromine water II Chlorine water III Aqueous sodium hydroxide IV Dilute sulphuric acid A I and II C I, II and III B III and IV D II, III and IV 12 Compound X has the following properties: • It releases white steamy fumes with thionyl chloride • It reacts with nitrous acid at room temperature with the release of a colourless gas. • It does not form a white precipitate with bromine water What is the structural formula of X? A H2N9C6H49COCH3 B H2NCH29C6H49COOCH3 C H2NCH2CH29C6H49CH2OH D H2N9C6H49OH 13 Which of the following statements regarding amino acids is not correct? A All amino acids found in protein are α-amino acids. B Amino acids are crystalline solid at room conditions. C All amino acids dissolved in water to form an acidic solution. D Amino acids are amphoteric. 14 Which of the following compounds is formed when 4-cyanoniline is boiled with dilute sodium hydroxide? A NH3 + CN C NH2 COO– B NH3 + COO– D NH2 COOH 15 Niacin together with nicotinamide makes up a group of vitamin called the B3 complex. O N OH Niacin O N NH2 Nicotinamide Which of the following is true about both niacin and nicotinamide? A Both decolourises alkaline KMnO4 on heating. B Both react with hot, aqueous NaOH. C They are both amino acids. D Both are optically active. 16 An amino acid, Y has the following structure. H2NCH2CHCH2CHCOOH & OH In what respect is Y different from an amino acid obtained from the hydrolysis of protein? A Y contains a secondary alcohol group. B Y is not an α-amino acid. C Y is insoluble in water. D Y does not form a zwitterion. 17 Which compound has the highest melting point? A CH3CH2CH2OH B CH3CH2COOH C H2NCH2CH2COOH D CH3CH2COONH2
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 282 18 An amino acid, W, has the structure below: HOOC9CH9CH29CH9NH2 & & CH3 COOH Which reagent will not react with W? A PCl5 B CH3COCl C HNO2 D KMnO4 19 An organic compound X reacts with sodium nitrite in dilute hydrochloric acid at 5°C to form an intermediate Y. When Y is added to phenol, a bright yellow solid is formed. What could be X and Y? X Y A NH2 NH3 +Cl– B CH3 NH2 CH3 N2 + C H2N-CH2 CH3 CH2N2 +Cl CH – 3 D NH2 N2 +Cl– 20 Aminoethanoic acid is similar to ethanoic acid in its reaction with A nitrous acid B dilute hydrochloric acid C aqueous sodium carbonate D aqueous bromine 21 Which of the following statements about amino acids are true? I All amino acids are α-amino acids. II In the solid state, they exist as zwitterions. III Their aqueous solutions are acidic. IV They act as buffers. A I and II C I, II and III B II and IV D II, III and IV 22 The structural formula of an organic compound is shown below. CH3CH29CH9CH29CH9CH3 & & COOH NH2 This compound A is a component of proteins B decolourises aqueous bromine C is an amino acid D optically inactive 23 The structural formula of an organic compound is shown below. CH3—CH—CH—NH2 & & CH3 COOH Which of the following statements are true of the compound? I It is a solid at room conditions. II It is optically active. III It is amphoteric. A I and II C II and III B I and III D I, II and III Structured and Essay Questions 1 Two amino acids, phenylalanine and glutamic acid have the structures below. & CH2 & H2 N!C!COOH & H NH2 & HO2 C!CH2 !CH2 !C!CO2 H & H Phenylalanine Glutamic acid (a) Name phenylalanine and glutamic acid according to the IUPAC name. (b) Which of the two amino acids is more acidic? (c) Draw the structure of phenylalanine at pH 3. (d) Draw the structure of the zwitterion of glutamic acid.
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 283 (e) Under suitable conditions, phenylalanine can undergo condensation with glutamic acid to form a dipeptide. (i) Define dipeptide linkage. (ii) Draw the structure of the dipeptide formed between phenylalanine and glutamic acid. 2 By using a reaction scheme, state the reagents and reaction conditions necessary to convert benzene to 4-hydroxybenzoic acid. 3 With reference to ammonia, methylamine and phenylamine (aniline) discuss the factors that influence the relative basic strength of the three compounds. 4 Explain the following observations. Write equations where necessary. (a) Ethylamine and phenylamine (aniline) both reacts with nitrous acid at 5 °C. However, effervescence is observed only in ethylamine but not in phenylamine (aniline). (b) When ethylamine is warmed with a mixture of hydrochloric acid and sodium nitrate(III), a mixture consisting of nitrogen gas, ethene and ethanol is produced. 5 Compound A, C6H7N is basic. When A is treated with aqueous bromine followed by ethanoyl chloride, compound B, C8H6NOBr3 is formed. However, if A is first treated with ethanoyl chloride then followed by aqueous bromine, compound C, C8H8NOBr is produced. (a) Identify A, B and C. (b) Write equation to show the reactions of A with the following reagents. (i) Warm nitrous acid (ii) Nitrous acid at 5 °C 6 The structural formula of compound X is shown below. CH3CH(CH3)9C6H49NH2 (a) Give the IUPAC name of X. (b) X reacts with a freshly prepared solution of nitrous acid, cooled to 2 °C to form Y. When Y is added to a solution of phenol, a bright coloured product, Z is formed. Give the structures of Y and Z. Write a balanced equation for the production of Z. (c) Write a reaction scheme to show how X could be synthesized from benzene. 7 Consider the reaction scheme below. CH3CHCH3 CH3COCH3 + A NO2 !: !: I !!: O2 /H+ Heat IV NH2 !:II N2 + Cl– !:III (a) State the reagents and conditions necessary to bring about conversions I, II, III and IV. (b) Draw the structure for compound A. (c) A reacts with the product of conversion III in a coupling reaction. (i) Draw the structure of the product formed. (ii) What is the physical characteristic of the product in (c)(i)? (d) Suggest the mechanism for conversion I.
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 284 8 A neutral compound, A, C7H5N, reacts with lithium aluminium hydride in dry ether to give compound B, C7H9N, which reacts with dilute hydrochloric acid to form a salt. B reacts with cold aqueous nitrous acid to give a gas and C, C7H8O. The resulting solution when added to alkaline phenol, no coloured precipitates are formed. When C is warmed with acidified potassium dichromate(VI), compound D, C7H6O2 is produced. Compound D is also produced when A is boiled under reflux with dilute sulphuric acid. D reacts with phosphorus(V) chloride gives E, C7H5ClO, which reacts with C to give F, C14H12O2. E also reacts with benzene in the presence of anhydrous aluminium chloride to give G, C13H10O. G gives a coloured precipitate with 2,4-dinitrophenylhydrazine. Identify compound A to G and write equations for the reactions taking place. 9 Two isomers X and Y have the following structural formulae. CONH2 NH2 OH COOH X Y (a) X and Y are soluble in both dilute sulphuric acid and dilute aqueous sodium hydroxide. Give the formulae of the species formed when X and Y dissolves in hydrochloric acid and sodium hydroxide. (b) When compound X is boiled with aqueous sodium hydroxide, a gas that turns moist red litmus blue is released. (i) Name the gas. (ii) Write a balanced equation for the reaction. (c) Show how you would synthesise Y from toluene. 10 Compound A, C7H9N dissolves in dilute hydrochloric acid, from which a colourless crystal B, C7H10NCl can be isolated. A dissolves in fresh nitrous acid at 2 °C to produce C, C7H7N2Cl. When C is mixed with water and the mixture is heated, effervescence occurs and compound D, C7H10O is formed. When C is added to alkaline phenol, a bright coloured solid E, C13H12N2O, is formed. (a) Draw the structures of A, B, C, D and E. (b) Write the chemical equations for all the reactions involved. (c) State what happens when aqueous bromine is added to A. (d) State a use of solid E. (e) Name the mechanism involved in the formation of E. 11 An aromatic compound with molecular formula C7H9N has two isomers X and Y which are weakly basic. X reacts with aqueous bromine to give a white precipitate while Y does not. Both X and Y are soluble in dilute hydrochloric acid. (a) Identify X and Y, then write equations for the reactions involved. (b) Which is a stronger base, X or Y? Explain your answer. (c) When X is dissolved in nitrous acid at 2 °C and an alkaline solution of phenol is added, a bright coloured solid Z is produced. Draw the structure of Z. Name one use for Z. 12 Compound X has the following structure: NH2 HO COOC2H5
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 285 (a) How would you confirm the presence of the —NH2 and —OH group in X? (b) Would you expect X to be soluble or insoluble in water? Explain your answer. (c) Write the structural formulae of the product(s) formed when X is reacted with the following reagents: (i) Cold, dilute sodium hydroxide (ii) Hot, dilute hydrochloric acid (iii) Ethanoyl chloride 13 N-benzylmethanamine has the following structure: CH299NHCH3 (a) Write equations to show how you would synthesise N-benzylmethanamine from benzene. (b) Give the formula of the product formed when N-benylmethanamine reacts with dilute hydrochloric acid. (c) What would you observe when N-benzylmethanamine is warmed with dilute nitrous acid? 14 pKb for some bases are given below: pKb values: 4.74, 9.34, 3.36 Compound CH3NH2 C6H5NH2 NH3 pKb Complete the table and explain your answer. 15 Consider the following reaction scheme: Sn/HCl HNO2/5 °C NaOH I III II A B C NO2 OH (a) State the reagents and reaction conditions for step I. (b) Name the reaction involved in step I. (c) Draw the structural formulae of A and B. (d) State the reagents and reaction conditions for step II. What observable change(s) occur in step II? (e) Draw the structural formula of compound C and state one use of compound C. (f) Name the reaction involved in step III. 16 The structures of three amino acid are shown below: H & CH3 !C!COOH & NH2 (Alanine, Ala) H & HOOC!CH2 !C!COOH & NH2 (Aspartic acid, Asp) H & CH2 !C!COOH & NH2 (Phenylalanine, Phe)
20 Chemistry Term 3 STPM Chapter 20 Amines, Amino Acids and Proteins 286 (a) Give the IUPAC name of the three amino acids. (b) Discuss the solubility of the amino acids in water and cyclohexane. You may use alanine in your answer. (c) Condensation between the three amino acids above would produce tripeptides. (i) How many tripeptides are possible from the condensation of the three amino acids? (ii) Draw the structures of the following tripeptides: Ala-Asp-Phe and Asp-Phe-Ala (d) Suggest how you would convert the tripeptides back to their amino acid components. (e) Amino acids in one of the buffer system in human blood. Using one of the above amino acid explain the buffering action of the amino acid. 17 The formula of phenylalanine is as follows: CH2CHCOOH NH2 (a) Give the IUPAC name of phenylalanine. (b) In the solid state, phenylalanine exists in the form of a zwitterion. Draw the structure of the zwitterion. (c) Phenylalanine is amphoteric. Write two balanced equations to illustrate the amphoteric nature of phenylalanine. (d) Draw a repeating unit for a dipeptide of phenylalanine. 18 Enzymes are polypeptides. (a) What is meant by a polypeptide? (b) Draw the repeating unit in a polypeptide. (c) Enzymes are readily denatured in strong alkaline or acidic conditions. Explain why it is so. 19 (a) Aminoethanoic acid is a component of protein. Between aminoethanoic acid and ethanoic acid, which would you expect to have a higher melting point? Explain your reasoning. (b) Write a reaction scheme to show how aminoethanoic acid can be prepared from methanamine, CH3NH2. (c) The structure of a polypeptide is shown below. H2N C C O CH2C C C H H H N O H H OH O C OH C N CH2 CH3 H C C O N H H O Identify the amino acids formed when the polypeptide is boiled with dilute sulphuric acid.
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
Chemistry Term 3 STPM Chapter 21 Polymers 288 21 21.1 Introduction 1 Polymers are long-chain macromolecules consisting of a large number of smaller molecules (called monomers) linked together through chemical reactions. 2 The molecular mass of polymers are by comparison very much higher compared to common organic compounds, and typically range from about 10 000 g mol–1 to 1 000 000 g mol–1. 3 Generally, polymers can be divided into two main groups: natural polymers and synthetic polymers. 4 Natural polymers (sometimes called biopolymers) are synthesised by living organisms, while synthetic polymers are synthesised in the laboratory. 5 Examples of natural polymers are carbohydrates, proteins, rubber, cotton, silk, nuclei acids and many more. 6 Examples of synthetic polymers are plastics and nylons. 21.2 Synthetic Polymers 1 Synthetic polymers are produced by the process called polymerisation. 2 Polymerisation is the process in which small molecules (called monomers) link with one another to form long-chain molecules called polymers. nA !: !A!A!A!A!A!A!A! A typical polymer chain can contain as much as 105 monomer units. 3 All polymers can be represented by a repeating unit. This is the atom or group that when repeated would give the structure of the polymer. For example, the polymer above can be represented by the repeating unit: !![!A!!!]n 4 The above polymer is called a homopolymer. It is made up of one type of monomer only. 5 Polymerisation can also take place between different monomers. These are called copolymers. For example, nA + nB !: !![!A!B!!!]n Definition of polymerisation Repeating unit Homopolymer and co-polymer 2015/P3/Q14
Chemistry Term 3 STPM Chapter 21 Polymers 289 21 6 Polymers can be further classified as linear, branched or network polymers. Linear polymer Branched polymer 7 Network polymers are formed when linear chains or branched chains are joined together by covalent bonds through processes such as vulcanisation. Network polymers are sometimes called cross-linked polymers. For example, !A!A!A!A!A!A!A! & & X X & & !A!A!A!A!A!A!A! 21.3 Condensation Polymerisation 1 In condensation polymerisation, monomers joined to form long linear-chain polymers through the elimination of other smaller molecules such as water or hydrogen chloride. 2 As a result, the empirical formulae of condensation polymers are different from those of their monomers. 3 The monomers involved in condensation polymerisation must have difunctional groups. 4 Examples of naturally occurring condensation polymers are protein, starch and cellulose. Polyamides 1 Examples of polyamides are nylons. They are formed by the condensation reaction between a dicarboxylic acid and a diamine. O H O H ∫ & ∫ & !!C!OH + H!N!! !: !!C!N!! + H2O 2009/P1/Q40 2010/P2/Q4(c) 2013/P3/Q15, Q17(a) VIDEO Polymers
Chemistry Term 3 STPM Chapter 21 Polymers 290 21 2 The first nylon synthesised is nylon-6,6. This was produced by the reaction between 1,6-hexanedioic acid and 1,6-hexanediamine. H!O!C!(CH2)4!C!OH + H!N!(CH2)6!N!H !!!!!: ∫ ∫ & & O O H H !![C!(CH2)4!C!N!(CH2)6!N]n!! ∫ ∫ & & O O H H –(2n – 1)H2O 3 This polymer is called nylon-6,6 because each monomer of the polymer consists of six carbon atom each. (Note: The first digit refers to the number of carbon atoms in the diamine, and the second digit refers to the number of carbon atoms in the dicarboxylic acid.) 4 Another example is nylon-6,10. It is formed from 1,6-hexanediamine and 1,10-decanedioic acid. HO!C!(CH2)8!C!OH + H!N!(CH2)6!N!H !!!!!: ∫ ∫ & & O O H H !![C!(CH2)8!C!N!(CH2)6!N]n!! ∫ ∫ & & O O H H –(2n – 1)H2O 5 Nylons are used mainly as a substitute for silk and fabric. Nylons are also used to make ropes, carpets and gears. Example 21.1 Explain why nylon-6,6 has higher melting/softening point than poly(ethene). Solution Poly(ethene) is a polyalkane. The intermolecular forces between polymeric chains are the weak van der Waals forces. On the other hand, nylon-6,6 is a polyamide. Due to the polar nature of the !NH and !CO bonds, the intermolecular forces between polymeric chains are the stronger hydrogen bonds. CRRO H!!N H!!N CRRO δ– δ+ As a result, nylon-6,6 has higher tensile strength, and a higher melting/ softening point than poly(ethene). Uses of nylon !CH2!CH2!CH2!CH2! !CH2!CH2!CH2!CH2! van der Waals forces
Chemistry Term 3 STPM Chapter 21 Polymers 291 21 Polyesters 1 Esters are formed by the reaction between a carboxylic acid and an alcohol. For example, CH3COOH + CH3CH2OH L CH3COOCH2CH3 + H2O 2 A polyester is formed through the reaction between a dicarboxylic acid and a diol. HO!C!R!C!OH + H!O!R!O!H !: ∫ ∫ O O !![!C!R!C!O!R9!O!!!]n ∫ ∫ O O 3 An example of polyester is Terylene. The monomers for Terylene are ethane-1,2-diol and 1,4-benzenedicarboxylic acid. HO!C C!OH + H!O!CH2CH2!O!H ∫ ∫ O O !!!!!: !![!C C!O!CH2CH2!O!!!]n ∫ ∫ O O –(2n – 1)H2O 4 Terylene is marketed under the brand name ‘Dacron’. It is used as synthetic fabric in replace of wool and cotton. 5 Sometimes, an acyl chloride is used in place of the dicarboxylic acid in the production of polyesters. For example, O ∫ Cl!C C!Cl + H!O!CH2CH2!O!H !!!: ∫ O O ∫ !![!C C!O!CH2CH2!O!!!]n ∫ O –(2n – 1)HCl 2011/P2/Q4(a)(ii) Uses of terylene