Chemistry Term 3 STPM Answers 342 CN CN & & CH3!CH2!C!CH3 + HCN !: CH3!CH2!C!CH3 + CN– & & O– OH Sodium cyanide acts as a catalyst. (c) Optical isomerism. CH OH HO 3CH2 CH2CH3 C* C* CH3 CH3 CN NC (d) OH & CH3CH2!C!COOH & CH3 7 (a) 3-phenyl-2-propenal (b) Geometrical isomerism H H CHO CRC H CHO H CRC Cis-isomer Trans-isomer (c) (i) Aqueous HCN in the presence of NaCN !CHRCH!C!CN H OH (ii) Warm with Tollen’s reagent !CHRCH!COO– Na+ (iii) Bromine in tetrachloromethane !CH!CH!CHO Br Br (iv) Mixture of concentrated nitric acid and sulphuric acid !CHRCH!CHO NO2 (v) Lithium aluminium hydride in ether !CHRCH!CH2OH (d) Add PCl5 A: No reaction Product: White fumes liberated 8 (a) CH3 & CH3!C!CN (Nucleophilic addition) & OH (b) CRN!NH! !NO2 CH3 H NO2 (Condensation) (c) CH3CH2CH2COO– Na+ + CHI3 (Oxidation) (d) CH2RCH!CH2OH (Reduction) 9 (a) The methyl-ketone group (b) CH3CH2CH2!C!CH3 CH3!CH!C!CH3 ∫ & ∫ O CH3 O (I) (II) CH3CH2—C—CH2CH3 ∫ O (III) (c) Isomer (I) CH3CH2CH2!C!CH3 + 4[O] !!: ∫ O CH3CH2CH2COOH + CO2 + H2O 10 (a) CH2OH!(CHOH)4!C!H ∫ O (b) As a source of energy (c) Heat with Fehling’s solution. The aldehyde group will reduce Fehling’s solution to give a reddish-brown precipitate of copper(I) oxide. (d) (i) Reduction occurs. CH2OH!(CHOH)4!CH2OH (ii) Orange precipitate formed. CRN!NH! !NO2 CH2OH(CHOH)4 H NO2 11 (a) CH2CH2OH H OH (c) No reaction (b) CH2COO– O R (d) CH2COOH O R 12 (a) 3-Oxo-4-hexenoic acid. (b) (i) CN & CH3 !CH"CH!C*!CH2 COOH & OH (ii) Yes. There is a chiral carbon (C*) in the molecule. (iii) Nucleophilic addition (c) (i) C2 H5 & CH3 !CH"CH!C!CH2 COOH & OH (ii) CH3 !CH"CH!CH!CH2 CH2 OH & OH (iii) CH3 !CH2 !CH2 !CH!CH2 COOH & OH 13 (a) CH3—CH"CH2 + H2O(g) → CH3—CH—CH3 | OH
Chemistry Term 3 STPM Answers 343 CH3—CH—CH3 + [O] KMnO4 / H+ / Heat → CH3—C—CH3 + H2O | || OH O (b) CH3CH2CH2CH2I + NaOH(aq) Heat → CH3CH2CH2CH2OH + NaI CH3CH2CH2CH2OH + [O] K2Cr2O7/H+/Warm → CH3CH2CH2—C—H + H2O ' O (c) + CH3COCl Warm —C—CH3 + HCl ' O (d) CH3CH==CHCH3 + O3 + H2O Zn/H+/Warm → 2CH3—C—H + H2O2 ' O Chapter 19 Carboxylic Acids and their Derivatives Quick Check 19.1 O ∫ 1 CH3CH2CH2CH2CH2!C!H 2 CH3CH2CH2CH2CH2CH2OH 3 HO!CH2CH2CH2CH2CH2CH2!OH Quick Check 19.2 1 + POCl3 + HCl OH + PCl5 !!: Cl + KCl Cl + KCN !!: Ethanol CN Δ + NH4 + CN + 2H2O + H+ !!!!: Dilute H COOH 2 SO4 Δ CH3 CH3 & & 2 CH3!C!CH3 + PCl5 !: CH3!C!Cl + POCl3 + HCl & & OH CH3 CH3 CH3 & Ethanol & CH3!C!Cl + KCN !!∆ !: CH3!C!CN + KCl & & CH3 CH3 CH3 CH3 & & CH3!C!CN + 2H2O + H+ Dilute H2SO4 →∆ CH3!C!COOH + NH4 + & & CH3 CH3 Quick Check 19.3 1 Cyclohexanoic acid < 4-aminobenzoic acid < 2-methylbenzoic acid < Benzoic acid < 2-nitrobenzoic acid 2 Acid strength: Cl3CCOOH > Cl2CHCOOH > ClCH2COOH > CH3COOH Presence of the electronegative chlorine atoms weakens the O!H bond. Quick Check 19.4 1 (a) COO– Na+ Cl (b) COOH (c) COO– Na+ Na+–O (d) COO– Na+ Cl 2 (a) H2C2O4 + Mg !!: MgC2O4 + H2 (b) HOOC!COOH + 2K !: K+–OOC!COO– K+ + H2 (c) HOOC!COOH + 2NH3 !!: NH4 +OOC!COO– NH4 + (d) H2C2O4 + Ca(OH)2 !!: CaC2O4 + 2H2O Quick Check 19.5 O ∫ (a) CH3!CH2!CH2!C!Cl (b) Cl!C!C!Cl ∫ ∫ O O O ∫ (c) CH3!CH!C!Cl & Cl (d) Cl!C C!Cl O R O R Quick Check 19.6 1 (a) C!OCH3 O R (b) C2H5!O!C!C!OC2H5 ∫ ∫ O O (c) CH3!C!O O R (d) HO C!CH!CH3 CH3 ! O R Quick Check 19.7 (a) CH3OH (b) CH2CH2OH (c) CH2RCHCH2OH (d) HOCH2CH2CH2CH2CH2CH2OH (e) CHRCH!CH2OH
Chemistry Term 3 STPM Answers 344 (f) CH3!CH!CH2OH & OH (g) CH2OH Quick Check 19.8 (a) Acidified KMnO4, heat Ethanoic acid: No reaction Ethanedioic acid: Purple colour decolourises (b) Aqueous mercury(II) chloride Ethanedioic acid: No reaction Methanoic acid: White precipitate Quick Check 19.9 1 (a) COO– Na+ (b) COOH (c) CONH2 (d) C!N CH3 C2H5 O R (e) C!NH O R (f) C!O O R Quick Check 19.10 1 (a) H!C!O!CH2 O R (b) C2H5!O!C!C!O!C2H5 O R O R (c) C!O!CH2! O R O R CH2!O!C! (d) C!O!CH2! O R CH3 2 (a) HCOOH + CH3OH (b) CH3CH2CH2COOH + CH3CH2OH (c) COOH + CH3! OH 3 (a) CH3CH2COOCHCH3 + NH3 !!: & CH3 CH3CH2CONH2 + CH3!CH!CH3 & OH (b) C!O! CH2OH + !!: OH + 4[H] O R 4 O ∫ CH3!C!O!CH2!CH!CH3 & CH3 O ∫ CH3!CH!C!OCH2CH3 & CH3 Quick Check 19.11 1 (a) CH3COO– Na+ + CH3!C!NH! NH2 + NaOH !!: O R (b) CH3COOH+ CH3!C!NH! NH3 + + H2O + H+ !!: O R (c) CH3CH2!NH! CH3!C!NH! + H2O + 4[H] !!: O R (d) CH2Cl!C!NH! CH3!C!NH! + HCl + Cl2 !!: O R O R 2 (a) O O ∫ ∫ CH3!C!Cl + 2NH3 !!: CH3!C!NH2 + NH4Cl (b) C!Cl + 2CH3NH2 C!NHCH3 + CH3NH3 + Cl– !!: O R O R 3 Boiling with dilute H2SO4. CH3CH2CH2CONH2 + H2O + H+ !: CH3CH2CH2COOH + NH4 +
Chemistry Term 3 STPM Answers 345 Revision Exercise 19 Objective Questions 1 A 2 A 3 B 4 C 5 C 6 C 7 B 8 C 9 A 10 C 11 A 12 B 13 A 14 A 15 B 16 C 17 D 18 A 19 D 20 D 21 B 22 C 23 A Structured and Essay Questions 1 (a) Propanedioic acid (b) Solid (c) CH3COOH + CI2 !!: UV light CH2ClCOOH + HCl CH2CICOOH + KCN !!!!: Ethanol Heat NCCH2COOH + KCl NCCH2COOH + 2H2O + H+ !!!!!!!!: Dilute sulphuric acid.Heat HOOCCH2COOH + NH4 + (d) (i) HOOCCH2COOH(aq) + H2O(l) 4 HOOCCH2COO– (aq) + H2O+(aq) pKa = 2.83 HOOCCH2COO– (aq) + H2O(l) 4 – OOCCH2COO– (aq) + H3O+(aq) pKa = 5.69 (ii) pKa = 2.83 + 5.69 = 8.52 2 (a) (i) CH3COOH + Cl2 → CH2Cl—COOH + HCl (ii) Cl2 → 2Cl• Cl• + CH3COOH → •CH2COOH + HCl •CH2COOH + Cl2 → CH2Cl—COOH + Cl• 2Cl• → Cl2 (b) Stronger. This is due to the electron-withdrawing effect (negative induction) of the more electronegative chlorine atom: Cl←CH2—C—O—H ' O This weakens the O—H bond and makes it easier to break compared to ethanoic acid. (c) (i) ClCH2COOH + PCl5 → ClCH2COCl + POCl3 + HCl (ii) ClCH2COOH + CH3CH2CH2OH → ClCH2COOCH2CH2CH3 + H2O (iii) ClCH2COOH + 4[H] → ClCH2CH2OH + H2O 3 (a) CH3CH2CH2CH2Cl 1-chlorobutane Cl Chlorobenzene CH3!C!Cl Ethanoyl chloride ∫ O (b) Add ethanolic silver nitrate at room temperature. Ethanoyl chloride: White precipitate formed immediately 1-chlorobutane: White precipitate formed after a few minutes Chlorobenzene: No reaction (c) CH3CH2CH2CH2Cl + OH– !!: CH3CH2CH2CH2OH + Cl– HO– C slow fast Cl CH2CH2CH3 H H HO– C Cl CH2CH2CH3 H H !!: 4 (a) X: !C!OCH2 CH3 ' O Y: CH3 CH2 OH Z: !CH2 OH (b) Lithium aluminium hydride in ether, followed by warming with dilute sulphuric acid. (c) Benzoic acid !CH2 OH + 2[O] !!!!: !COOH+ H2 O KMnO4 /H+ Δ (d) Y has the following functional group in its structure that reacts with alkaline iodine to produce triiodomethane, a yellow solid. H s CH3 !C! s OH CH3CH2OH + 4I2 + 6NaOH !: CHI3(s) + HCOONa + 5NaI + 5H2O (e) !CH2 Cl CH3CH2CI and 5 (a) C : H : O : Cl = 5 : 9 : 1 : 1 Empirical formula = C5H9OCl (C5H9OCl)n = 120.5 n = 1 Molecular formula of A = C5H9OCl (b) H O & ∫ CH3CH2!C*!C!Cl & CH3 (c) H & CH3CH2!C*!COOH & CH3 (d) Add PCl5. (e) (i) Ester (ii) H O & ∫ CH3CH2!C!C!Cl + C2H5OH !!: & CH3 CH3CH2!CH!COOC2H5 + HCl & CH3
Chemistry Term 3 STPM Answers 346 6 (a) W: CH3!C!O!CH!CH2CH3 ∫ & O CH3 X: CH3COOH Y: CH3!*CH!CH2CH3 & OH Z: CH3!C!CH2CH3 ∫ O (b) Add aqueous sodium carbonate. X: Effervescence Y: No reaction (c) (i) Yellow precipitate CH3CCH2CH3 + 3I2 + 4NaOH !!: ∫ O CHI3 + CH3CH2COONa + 3NaI + 3H2O (ii) Orange precipitate CRN!NH! NO2 CH3 CH3CH2 NO2 (iii) No reaction 7 (a) CH3!CH!C!O! CH3 O R (X) CH3!CH!CH2OH CH3 (Z) (Y) OH (b) No. There are no chiral carbon atoms. (c) CH3H & & !!C! C!! & & CH3H 8 (a) CH3!C!O!CH2CH2CH3 ∫ O CH3!C!O!CH!CH3 ∫ & O CH3 (b) (i) Acidified KMnO4 (ii) Ketone (iii) CH3!C!CH3 ∫ O (iv) CH3!C!O!CH!CH3 ∫ & O CH3 9 (a) O ' !C!OH CH3 !C Cl O Benzoic acid Ethanoyl choride (b) Sparingly soluble [It has an !COOH group that can form hydrogen bonds with water molecules. However, the presence of a large hydrophobic benzene ring reduces its solubility] (c) (i) O CH3 C ' O O CH3 C ' O O O– Na+ C COO– Na+ O– Na+ + CH3 COO– Na+ COOH OH + CH3 CH2 OH (ii) O CH3 C ' O O CH3 C ' O O O– Na+ C COO– Na+ O– Na+ + CH3 COO– Na+ COOH OH + CH3 CH2 OH (iii) O CH3 C ' O O CH3 C ' O O O– Na+ C COO– Na+ O– Na+ + CH3 COO– Na+ COOH OH + CH3 CH2 OH (iv) O CH3 C ' O O CH3 C ' O O O– Na+ C COO– Na+ O– Na+ + CH3 COO– Na+ COOH OH + CH3 CH2 OH 10 (a) 1-(Methylethyl)benzoate (b) + CH3 Cl !!!: + HCl CH3 AlCl3 Δ + 3[O] !!!!: + H2 O CH3 KMnO4 /H+ Δ COOH !COOH + CH3 !CH!CH3 !!!!: !C!O!Cs s OH CConc.H2 SO4 Δ O ' !CH2 OH + CH3 !CH!CH3 s OH + CH3 Cl !!!: + HCl CH3 AlCl3 Δ + 3[O] !!!!: + H2 O CH3 KMnO4 /H+ Δ COOH !COOH + CH3 !CH!CH3 !!!!: !C!O!CHCH3 + H2 s s OH CH3 Conc.H2 SO4 Δ O ' !CH2 OH + CH3 !CH!CH3 s OH + CH3 Cl !!!: + HCl CH3 AlCl3 Δ + 3[O] !!!!: + H2 O CH3 KMnO4 /H+ Δ COOH !COOH + CH3 !CH!CH3 !!!!: !C!O!CHCH3 + H2 O s s OH CH3 Conc.H2 SO4 Δ O ' !CH2 OH + CH3 !CH!CH3 s OH (c) + CH3 Cl !!!: + HCl CH3 AlCl3 Δ + 3[O] !!!!: + H2 O CH3 KMnO4 /H+ Δ COOH !COOH + CH3 !CH!CH3 !!!!: !C!O!Cs s OH CConc.H2 SO4 Δ O ' !CH2 OH + CH3 !CH!CH3 s OH (d) X is insoluble in water. X cannot form hydrogen bonds with water molecules. 11 (a) CH3!CH!C!Cl [X] & ∫ Cl O
Chemistry Term 3 STPM Answers 347 CH3!CH!C!O!C2H5 [Y] & ∫ Cl O CH3!*CH!COOH [Z] & OH (b) (i) Triiodomethane, CHI3 (ii) CH3!CH!COOH + 4I2 + 7NaOH !!: & OH CHI3 + Na+–OOC!COO– Na+ + 5NaI + 6H2O (c) H COOH H OH C C CH3 CH3 OH HOOC 12 Molecular formula of A: (C2H2N)n = 80 ∴ n = 2 Molecular formula of A is C4H4N2. A: NC!CH2CH2!CN B: H2NCH2!CH2CH2!CH2NH2 C: HOOC!CH2CH2!COOH D: C2H5!O!C!CH2CH2!C!O!C2H5 ∫ ∫ O O 13 (a) 2-Hydroxy-benzoic acid. (b) React iodomethane with ethanolic potassium cyanide to form ethanenitrile. Heat CH3I + KCN 9: CH3CN + KI Boiling ethanenitrile with dilute sulphuric acid to produce ethanoic acid. CH3CN + 2H2O + H+ 9: CH3COOH + NH4 + Add thionyl chloride to anhydrous ethanoic acid. CH3COOH + SOCl2 9: CH3COCl + SO2 + HCl (c) O! H !: COOH CH3 !C! CI + ' O O!C!CH3 + HCI ' O COOH (d) Salicylic acid will be more soluble. Salicylic acid has two functional groups, 9COOH and 9OH, where they can form hydrogen bonds with water molecules. On the other hand, Aspirin has only one functional group, 9COOH, to form hydrogen bond with water molecules. (e) Add bromine water to salicylic acid and aspirin separately. Salicylic acid, which has a phenolic group, gives a white precipitate whereas, aspirin does not. COOH OH Br COOH OH + 2Br2 !: + 2HBr Br (f) The solubility of aspirin can be increased by converting the molecule into an ionic salt. This can be brought about by reacting aspirin with calcium carbonate to convert it into the calcium salt of aspirin. COOH O ' O!C!CH3 COO— OH + CaCO3 !:2 + Ca2+ + CO2 2 + H2 O 14 (a) KCN in ethanol, heat CH3CH2CH2I + KCN !!: CH3CH2CH2CN + KI (b) (i) The !OH group (ii) Hydrogen chloride, HCl (iii) X is CH3CH2CH2COOH Y is CH3CH2CH2COCl Z is CH3CH2CH2COOC6H5 (c) CH3CH2CH2CH2OH + C6H5OH 15 (a) PCl5, room temperature C6H5COOH + PCl5 !: C6H5COCl + POCl3 + HCl (b) (i) Hydrogen chloride (ii) C6H5COCl + H2O !: C6H5COOH + HCl (c) (i) C6H5COCl + C6H5OH !: C6H5COOC6H5 + HCl (ii) C6H5COCl + C6H5NH2 !: C6H5CONHC6H5 + HCl (iii) C6H5COCl + CH3NHCH3 !!: C6H5CON(CH3)2 + HCl 16 The mole ratio of C : H : O = 62 12 : 10.3 1 : 27.7 16 = 5.2 : 10.3 : 1.73 = 3 : 6 : 1 Empirical formula of A is C3H6O. Let the molecular formula be (C3H6O)n. 58n = 116 \n = 2 Molecular formula of A is C6H12O2. A is an ester. A is: CH3—C—O—CH—CH2CH3 ' & O CH3 Hydrolysis of A: CH3—C—O—CH—CH2CH3 + H2O → CH3COOH + CH3CH2—CH—CH3 ' & & O CH3 OH [B] [C] Reaction of B with sodium carbonate: 2CH3COOH + Na2CO3 → 2CH3COO– Na+ + H2O + CO2 C has a chiral centre. As a result, it is optically active. H & CH3CH2—*C—CH3 & OH
Chemistry Term 3 STPM Answers 348 Reaction of C with acidified potassium dichromate: CH3CH2—CH—CH3 + [O] → CH3CH2—C—CH3 + H2O & ' OH O [D] Reaction of D with alkaline iodine: CH3CH2COCH3 + 3I2 + 4NaOH → CHI3 + CH3CH2COONa + 3NaI + 3H2O Chapter 20 Amines, Amino Acids and Proteins Quick Check 20.1 1 (a) Hexylamine (1°) (b) 1,4-butanediamine (1°) (c) 2-phenyl-ethanamine (1°) (d) 4-nitroaniline (3°) (e) N,N-dimethylcyclopentylamine (3°) 2 (a) N (3°) (b) CH2NH2 (1°) (c) CH3!CH2!CH2!CH2!CH!CH2OH (1°) & NH2 (d) CH2!CH!CH3 (1°) NH2 (e) Cl & CH3!C!CH2CH3 (1°) & NH2 (f) CH3!CH2!CH2!CH2!CH!COOH (1°) & NH2 Quick Check 20.2 1 (CH3)3N < CH3CH2NHCH3 < CH3CH2CH2NH2 The number of intermolecular hydrogen bonds formed increases in the order: 3° < 2° < 1°. 2 NH2 < CH3CH2NHCH3 < CH3CH2CH2NH2 Quick Check 20.3 1 (a) Heat iodoethane with excess of concentrated ammonia in a sealed tube. C2H5I + 2NH3 !!: C2H5NH2 + NH4I (b) CH3 + Cl2 CH2Cl + HCl UV light CH2Cl + 2NH3 CH2NH2 !!: + NH4Cl !!: Δ Ethanol (c) C2H5I + KCN !!∆ !: C2H5CN + KI LiAlH4 C2H5CN + 4[H] !!∆ !: C2H5CH2NH2 Quick Check 20.4 1 (a) NH3 +Cl– (c) NH2 +Cl– (b) (d) N+ !HCl – 2 Base strength: Aniline > 2-nitroaniline > 4-nitroaniline The !NO2 is an electron-withdrawing group and makes the lone pair electrons on nitrogen atom less readily to combine with H+. The withdrawing effect at the 4-position is stronger than at the 2-position. 3 Add dilute sulphuric acid. Aniline will dissolve in the acid to form a salt. Nitrobenzene is insoluble in sulphuric acid. NH2 + H+ NH3 !!: + NH3 + OH– !!: NH2+ H2O + Cool the solution in ice to precipitate the phenylammonium salt. Treatment with NaOH regenerates aniline. NH2 + H+ NH3 !!: + NH3 + OH– !!: NH2+ H2O + Quick Check 20.5 1 (a) Add bromine water. Aniline: White precipitate Phenylmethylamine: No precipitate (b) Add nitrous acid, warm. Aniline: Effervescence occurs Phenol: No reaction (c) Warm with aqueous NaOH. Ethylamine: No reaction Ethanamide: Gas given off turns lime water chalky. 2 C6H5NH2 + 3Br2 !!: C6H2Br3NH2 + 3HBr 93 g 330 g ∴ Mass of precipitate = 9.3 93 × 330 g = 33.0 g 22.6 ∴ % Purity = –––– × 100% 33.0 = 68.5% Quick Check 20.6 1 (a) Neutral (b) Basic (c) Acidic Quick Check 20.7 (a) H2N!CH!CH2CH2COO– Na+ & COO– Na+ (b) H3N + !CH!CH2CH2COOH & COOH H & CH3!N+!H + HSO4 – & CH3
Chemistry Term 3 STPM Answers 349 (c) H2N!CH!CH2CH2!COCl & COCl (d) H2N!CH!CH2CH2!COO– Na+ & COO– Na+ (e) H2N!CH!CH2CH2COO– & COO– (f) O H ∫ & CH3!C!N!CH!CH2CH2!COOH & COOH (g) HO!CH!CH2CH2!COOH & COOH (h) H & H2N!C!CH2CH2!COOC2H5 & COOC2H5 Quick Check 20.8 1 Seryl-alanine H O H & ∫ & H2N!C!C!N!CH!COOH & & CH2 CH3 & OH Alanil-serine O H H ∫ & & H2N!CH!C!N!C!COOH & & CH3 CH2OH 2 Gly-Ala-Phe Quick Check 20.9 1 H2N!CH2!COOH + H2N!CH!COOH CH3 + H2N!CH!COOH CH2 Revision Exercise 20 Objective Questions 1 B 2 B 3 D 4 C 5 C 6 D 7 A 8 A 9 A 10 C 11 A 12 C 13 C 14 C 15 B 16 B 17 C 18 D 19 B 20 C 21 B 22 C 23 D Structured and Essay Questions 1 (a) Phenylalanine: 2-Amino-3-phenylpropanoic acid Glutamic acid: 2-Aminopentanedioic acid (b) Glutamic acid. It has more !COOH groups that are acidic. (c) !CH2 !CH!COOH s +NH3 +NHs HOOC!CH2 CH2 !C! s H NH2 O COOH s ' s HOOC!CH2 !CH2 !C!!C!N!C!CH2 ! s s s H H H (d) !CH2 !CH!COOH s +NH3 +NH3 s HOOC!CH2 CH2 !C!COO– s H NH2 O COOH s ' s HOOC!CH2 !CH2 !C!!C!N!C!CH2 ! s s s H H H (e) (i) A dipeptide linkage is a covalent bond linking two amino acids with the structure of !CONH!. (ii) !CH2 !CH!COOH s +NH3 +NH3 s HOOC!CH2 CH2 !C!COO s H NH2 O COOH s ' s HOOC!CH2 !CH2 !C!!C!N!C!CH2 ! s s s H H H 2 + HNO3 !!!!!: + H2 O Conc. H2 SO4 55°C H NO2 + 6[H] !!!!: + 2H2 O Sn/HCl Δ NO2 NH2 + HNO2 !!!!!!: + N2 + H2 O NaNO2 + HCl Δ NH2 OH + CH3 Cl !!!!: + HCl AlCl3 Δ OH OH CH3 + 3[O] !!!!: + H2 O KMnO4 /H+ Δ OH OH CH3 COOH + HNO3 !!!!!: + H2 O Conc. H2 SO4 55°C H NO2 + 6[H] !!!!: + 2H2 O Sn/HCl Δ NO2 NH2 + HNO2 !!!!!!: + N2 + H2 O NaNO2 + HCl Δ NH2 OH + CH3 Cl !!!!: + HCl AlCl3 Δ OH OH CH3 + 3[O] !!!!: + H2 O KMnO4 /H+ Δ OH OH CH3 COOH + HNO3 !!!!!: + H2 O Conc. H2 SO4 55°C H NO2 + 6[H] !!!!: + 2H2 O Sn/HCl Δ NO2 NH2 + HNO2 !!!!!!: + N2 + H2 O NaNO2 + HCl Δ NH2 OH + CH3 Cl !!!!: + HCl AlCl3 Δ OH OH CH3 + 3[O] !!!!: + H2 O KMnO4 /H+ Δ OH OH CH3 COOH + HNO3 !!!!!: + H2 O Conc. H2 SO4 55°C H NO2 + 6[H] !!!!: + 2H2 O Sn/HCl Δ NO2 NH2 + HNO2 !!!!!!: + N2 + H2 O NaNO2 + HCl Δ NH2 OH + CH3 Cl !!!!: + HCl AlCl3 Δ OH OH CH3 + 3[O] !!!!: + H2 O KMnO4 /H+ Δ OH OH CH3 COOH + HNO3 !!!!!: + H2 O Conc. H2 SO4 55°C H NO2 + 6[H] !!!!: + 2H2 O Sn/HCl Δ NO2 NH2 + HNO2 !!!!!!: + N2 + H2 O NaNO2 + HCl Δ NH2 OH + CH3 Cl !!!!: + HCl AlCl3 Δ OH OH CH3 + 3[O] !!!!: + H2 O KMnO4 /H+ Δ OH OH CH3 COOH 3 Base strength: Methylamine > ammonia > aniline. The phenyl group is an electron-withdrawing group. It makes the lone pair electrons less available for reaction with H+. On the other hand, the methyl group is an electron-donating group which increases the electron density of the lone pair electrons making them more reactive towards H+. ;!! NH2 CH3 !!: NH2 • • • • 4 (a) At 5 °C, phenylamine forms the benzenediazonium ion which is stable because the positive charge of the ion can be delocalised into the benzene ring. However, no such delocalisation can occur in the ethyldiazonium ion. N2 + C2H5!N2 :!! +
Chemistry Term 3 STPM Answers 350 The ethyldiazonium ion decomposes spontaneously to form ethanol with the release of nitrogen gas. C2H5!N2 + !!: C2H5 + + N2 C2H5 + + H2O !!: C2H5OH + H+ (b) C2H5!N2 + !!: C2H5 + + N2 C2H5 + + H2O !!: C2H5OH + H+ C2H5 + !!: CH2 R CH2 + H+ 5 (a) (b) (i) C6H5NH2 + HNO2 !!: C6H5OH + N2 + H2O (ii) C6H5NH2 + HNO2 + H+ !: C6H5N2 + + 2H2O 6 (a) 4-(1-methylethyl)phenylamine or 4-(1-methylethyl) aniline (b) Y: CH3 !CH! !N s CH3 Z: CH3 !CH! !NRN! !OH s CH3 + 2 Y: CH3 !CH! !N s CH3 Z: CH3 !CH! !NRN! !OH s CH3 + 2 Equation: CH3 !CH! !N2 + !: CH3 !CH! !NRN! !OH + H+ s s CH3 CH3 + OH !N2 + !: CH3 !CH! !NRN! !OH + H+ s CH3 + OH (c) H s CH3 !C!CH3 s H + CH3 !CH!CH3 !!: s Cl + HCl AlCl3 H s CH3 !C!CH3 s + HNO3 !!!!!!!!: + H2 O Conc. H2 SO4 + HNO3 55°C H s CH3 !C!CH3 s NO2 H s CH3 !C!CH3 s H s CH3 !C!CH3 s + 6[H] !!!!: + 2H2 O Sn + HCl Δ NO2 NH2 NH2 NHCOCH3 NHCOCH3 Br A B C Br Br Br 7 (a) I: Concentrated HNO3 and H2SO4 at 55 °C II: Tin and concentrated HCl III: Nitrous acid, 5 °C IV: CH3!CH!CH3 and AlCl3 & Cl (b) OH (c) (i) NRN OH (ii) Bright-coloured solid (d) HNO3 + 2H2SO4 L NO2 + + 2HSO4 – + H3O+ H H + NO2 + !!: NO2 + H NO2 + !!: NO2 + H+ 8 CN A CH2NH2 B COOH D COOCH2 C F G COCl E CH2OH C O R C6H5CN + 4[H] !!: C6H5CH2NH2 A B C6H5CH2NH2 + HCl !!: C6H5CH2NH3 +Cl– B C6H5CH2NH2 + HNO2 !: C6H5CH2OH + N2 B + H2O C C6H5CH2OH + 2[O] !!: C6H5COOH + H2O C D C6H5CN + 2H2O + H+ !!: C6H5COOH + NH4 + A D C6H5COOH + PCl5 !!: C6H5COCl + POCl3 + HCl D E C6H5COCl + C6H5CH2OH !!: C6H5COOCH2C6H5 + HCl E C F C6H5COCl + C6H6 !!: C6H5COC6H5 + HCl E G 9 (a) In acidic solution: X: HO COOH + HOOC NH3
Chemistry Term 3 STPM Answers 351 Y: HO COOH + HOOC NH3 In basic solution: X: COO– NaO COO H – 2N Y: COO– NaO COO H – 2N (b) (i) Ammonia, NH3 (ii) CONH2 HO + NaOH !!: HO COONa + NH3 (c) CH3 CH3 NO2 + HNO3 !!: + H2O H2 SO4 CH3 NO2 CH3 NH2 + 6[H] !!: + 2H2O Sn + HCl CH3 NH2 COOH NH2 + 3[O] !!: + H2O KMnO4 10 (a) (A): CH3 ! !NH2 (C): CH3 ! !N2 Cl– (E): CH3 ! !NRN! !OH + (B): CH3 ! !NH3 Cl– (D): CH3 ! !OH + N2 + HCl + (A): CH3 ! !NH2 (C): CH3 ! !N2 Cl– (E): CH3 ! !NRN! !OH + (B): CH3 ! !NH3 Cl– (D): CH3 ! !OH + N2 + HCl + (A): CH3 ! !NH2 (C): CH3 ! !N2 Cl– (E): CH3 ! !NRN! !OH + (B): CH3 ! !NH3 Cl– (D): CH3 ! !OH + N2 + HCl + (b) CH3 ! !NH2 + HCl !: CH3 ! !NH3 Cl– (A) (B) CH3 ! !NH2 + HNO2 + HCl !: CH3 ! !N2 Cl– + 2H2 O (A) (C) CH3 ! !N2 Cl– + H2 O !: CH3 ! !OH + N2 + HCl (C) (D) CH3 ! !N2 Cl– + H! !OH !: CH3 ! !NR N! !OH +HCl (C) (E) + + + + CH3 ! !NH2 + HCl !: CH3 ! !NH3 Cl– (A) (B) CH3 ! !NH2 + HNO2 + HCl !: CH3 ! !N2 Cl– + 2H2 O (A) (C) CH3 ! !N2 Cl– + H2 O !: CH3 ! !OH + N2 + HCl (C) (D) CH3 ! !N2 Cl– + H! !OH !: CH3 ! !NR N! !OH +HCl (C) (E) + + + + CH3 ! !NH2 + HCl !: CH3 ! !NH3 Cl– (A) (B) CH3 ! !NH2 + HNO2 + HCl !: CH3 ! !N2 Cl– + 2H2 O (A) (C) CH3 ! !N2 Cl– + H2 O !: CH3 ! !OH + N2 + HCl (C) (D) CH3 ! !N2 Cl– + H! !OH !: CH3 ! !NR N! !OH +HCl (C) (E) + + + + CH3 ! !NH2 + HCl !: CH3 ! !NH3 Cl– (A) (B) CH3 ! !NH2 + HNO2 + HCl !: CH3 ! !N2 Cl– + 2H2 O (A) (C) CH3 ! !N2 Cl– + H2 O !: CH3 ! !OH + N2 + HCl (C) (D) CH3 ! !N2 Cl– + H! !OH !: CH3 ! !NR N! !OH +HCl (C) (E) + + + + CH3 ! !NH2 + HCl !: CH3 ! !NH3 Cl– (A) (B) CH3 ! !NH2 + HNO2 + HCl !: CH3 ! !N2 Cl– + 2H2 O (A) (C) CH3 ! !N2 Cl– + H2 O !: CH3 ! !OH + N2 + HCl (C) (D) CH3 ! !N2 Cl– + H! !OH !: CH3 ! !NR N! !OH +HCl (C) (E) + + + + CH3 ! !N2 Cl– + H2 O !: CH3 ! !OH + N2 + HCl (C) (D) CH3 ! !N2 Cl– + H! !OH !: CH3 ! !NR N! (C) (E) + + CH3 ! !NH2 + HNO2 + HCl !: CH3 ! !N2 Cl– + 2H2 O (A) (C) CH3 ! !N2 Cl– + H2 O !: CH3 ! !OH + N2 + HCl (C) (D) CH3 ! !N2 Cl– + H! !OH !: CH3 ! !NR N! !OH +HCl (C) (E) + + + (c) A white precipitate is formed. + 2Br2 !: Br + 2HBr Br NH2 NH2 CH3 CH3 (d) It is used as azo-dye in the dyeing industry. (e) Coupling reaction or electrophilic substitution 11 (a) X is a phenylamine and Y is a phenyl-substituted alkylamine. CH3— —NH2 —CH2NH2 X Y Reaction of X with aqueous bromine: CH3— —NH2 + 2Br2 → CH3 — Br Br —NH2 + 2HBr Reaction of X and Y with dilute HCl: CH3— —NH2 + HCl → CH3— —NH3 +Cl– —CH2NH2 + HCl → —CH2NH3 +Cl– (b) Y is a stronger base than X. The lone pair electrons on the nitrogen atom in X can delocalised into the benzene ring and making them less available for reaction with an acid. (c) CH3— —N==N— —OH Z is used as a dye (azo-dye). 12 (a) —NH2: Add nitrous acid and warmed. Effervescence occurs. —OH: Add aqueous iron(III) chloride. A blue/purple solution is formed. (b) Insoluble. Due to the presence of the large hydrophobic benzene ring. (c) (i) NH2 +Na– O COOC2H5 (ii) NH3 +Cl– HO COOH + C2H5OH
Chemistry Term 3 STPM Answers 352 (iii) NHCOCH3 CH3COO COOC2H5 13 (a) + CH3Cl AlCl3 → —CH3 + HCl —CH3 + Cl2 UV light → Heat —CH2Cl + HCl —CH2Cl + CH3NH2 Heat → —CH2NHCH3 + HCl (b) —CH2—+ NH2—CH3 + Cl– (c) A yellowish oil is obtained. 14 CH3NH2: 3.36 C6H5NH2: 9.34 NH3: 4.74 The lower the value of pKb, the stronger is the base. The basic strength of the three compounds is: CH3NH2 > NH3 > C6H5NH2. CH3 is an electron-donating group that increases the electron density of the lone pair electrons on the nitrogen atom, while C6H5 is an electron-withdrawing group that decreases the electron density of the lone pair electrons on the nitrogen atom. 15 (a) Mixture of concentrated H2SO4 and concentrated HNO3. Temperature not exceeding 55 °C. (b) Electrophilic substitution (c) NH2 N2 + A B (d) HNO2 with heat. Observation: Effervescence. N2 liberated. (e) —N==N— —OH It is used as a dye. (f) Coupling reaction 16 (a) Alanine: 2-Aminopropanoic acid Aspartic acid: 2-Amino-1,4-butanedioic acid Phenylalanine: 2-Amino-3 phenylpropanoic acid (b) The amino acids are soluble in water but insoluble in cyclohexane. Alanine exists as a zwitterion at room conditions: H3 N+ !CH!COO— & CH3 Being ionic, alanine forms strong ion-dipole attraction with water molecules causing it to dissolves in water. (c) (i) The number of polypeptide possible = 3! = 3 × 2 × 1 = 6 (ii) Ala-Asp-Phe: H O H H O H H & ' & & ' & & H2 N!C!C!N!C!!C!N!C!COOH & & & CH3 CH2 COOH CH2 Asp-Phe-Ala: H O H H O H H & ' & & ' & & H2 N!C!!C!N!C!C!N!C!COOH & & & CH2 COOH CH2 CH3 (d) Boiling the tripeptide with dilute sulphuric acid. (e) A buffer solution is a solution whose pH does not change significantly when a little acid or a little base is added to it. Using Alanine as example: On the addition of a little acid, H+: HOOCCH(CH3)NH2 + H+ 9: HOOCCH(CH3)NH3 + On the addition of a little base: H2NCH(CH3)COOH + OH– 9: H2NCH(CH3)COO– + H2O 17 (a) 2-amino-3-phenylpropanoic acid (b) —CH2CHCOO– | NH3 + (c) —CH2CHCOOH —CH2CHCOOH | + H+ → | NH2 NH3 + —CH2CHCOOH —CH2CHCOO– + H2O | + OH– → | NH2 NH2 (d) —CH2CH—NH—C—CHCH2— & ' COOH O 18 (a) A polypeptide is a linear-chain polymer of amino acids. (b) ——[—N—C—CH—]—— & ' & H O R (c) Alkalis or acids will rapture the peptide links in the polypeptide chain. 19 (a) Aminoethanoic acid has a higher melting point than ethanoic acid. In the solid state, aminoethanoic acid exists in the form of zwitterions: H3N+—CH2—COO– The forces between the zwitterion ions are the strong ionic bonds. On the other hand, in the solid state, ethanoic acid exists as discrete covalent molecules with weaker hydrogen bonds between the molecules.
Chemistry Term 3 STPM Answers 353 (b) React methanamine with chlorine in the presence of ultraviolet light. CH3NH2 + Cl2 UV light → Cl—CH2NH2 + HCl Warm the product with ethanolic potassium cyanide. H2N—CH2—Cl + KCN → H2N—CH2—CN + KCl Boil the product with dilute sulphuric acid. H2N—CH2—CN + 2H2O + H+ → H2N—CH2—COOH + NH4 + (c) The amino acids are: H2N—CH2COOH H2N—CH—COOH & CH2COOH H2N—CH—COOH & CH2 & H2N—CH—COOH & CH3 Chapter 21 Polymers Quick Check 21.1 1 (a) CH3OH (b) CH3!O!C!(CH2)4!C!O!CH3 + HO!CH2CH2!OH ∫ ∫ O O O O ∫ ∫ !!C!(CH2)4!C!O!CH2CH2!O !! + CH3OH 2 (a) H O & ∫ !!N!(CH2)5!C !! (b) H H O O & & ∫ ∫ !!N!(CH2)6!N!C!(CH2)4!C!! 3 (a) H2N!(CH2)6!NH2 and HOOC!(CH2)8!COOH (b) H H O O & & ∫ ∫ !!N!(CH2)6!N!C!(CH2)8!C!! Quick Check 21.2 1 (a) Propenenitrile (b) !!CH2!CH!!CH2!CH!CH2!CH!! & & & CN CN CN (c) Stronger. The !C#N group is polar. 2 H Cl !!CH2!C!!CH!CH2!! H Cl !!CH2!C!!CH2!CH!! 3 (a) X: Y: H C!NH2 CH2RC O R CHRCH2 (b) Y. It contains only hydrogen and carbon atoms which are inert towards sulphuric acid. 4 CH2RCH and CH2RCH2 & C#N Quick Check 21.3 (a) CH2RC!CH RCH2 & Cl (b) Poly(2-chlorobuta-1,3-diene) (c) !!CH2!CRCH!CH2!! & Cl Revision Exercise 21 Objective Questions 1 A 2 D 3 B 4 B 5 A 6 C 7 B 8 C 9 B 10 A 11 A 12 C 13 A 14 C 15 A 16 D 17 B 18 A Structured and Essay Questions 1 (a) Addition polymerisation is the process where monomers link together to form long-chain polymers without the elimination of other small molecules. (b) (i) !!(CH2!CH2)n!! (ii) Polyethene is a solid. Ethene is a gas. (c) (i) Exothermic (The temperature increases.) (ii) Using pV = nRT Initially: (10 × 103 )(1.5 × 10–3) = ni × 8.31 × 298 Final:(4 × 103 )(1.5 × 10–3) = nf × 8.31 × 348 nf 4 298 –– = ––– × –––– × 100% ni 10 348 = 34.3% Assumptions: The gas shows ideal behaviour. The vapour pressure of polyethene is negligible.
Chemistry Term 3 STPM Answers 354 2 (a) Condensation polymerisation is the process where monomers link together to form long-chain polymers with the elimination of other smaller molecules. (b) This is because molecules are eliminated when the monomers join to form polymers. (c) (i) !![N!(CH2)6!N!C!(CH2)4!C]n!! & & ∫ ∫ H H O O (ii) Nylon-6,6 (iii) The intermolecular forces between nylon-6,6 chains are the strong hydrogen bonds. On the other hand, the intermolecular forces between polyethene chains are the weak van der Waals forces. 3 (a) (i) Starch (ii) Nylon (b) (i) H2N!(CH2)6!NH2 and HOOC!(CH2)4!COOH (ii) Br!(CH2)4!Br + 2KCN Ethanolic, heat !!!!!: NC!(CH2)4!CN + 2KBr NC!(CH2)4!CN + 8[H] LiAlH4 !!: H2NCH2!(CH2)4!CH2NH2 NC!(CH2)4!CN + 4H2O + 2H+ !!: HOOC!(CH2)4!COOH + 2NH4 + (iii) Substitute for silk 4 (a) (I): !![!CH2!CH!]!! & COOCH3 (b) (III) and (IV) !!C! !C!O!CH2CH2!O!! O R O R 5 (a) (i) (I) (Isotactic) !CH!CH2!CH!CH2!CH!CH2!CH!CH2! & & & & CH3 CH3 CH3 CH3 (II) (Syndiotic) CH3 CH3 & & !CH!CH2!CH!CH2!CH!CH2!CH!CH2! & & CH3 CH3 (III) (Atactic) CH3 & !CH!CH2!CH!CH2!CH!CH2!CH!CH2! & & & CH3 CH3 CH3 (ii) Polypropene is used to make ropes, moulds, bottles, kitchenware, carpets, battery containers and for apparatus that needs to be steam sterilised. (b) (i) Thermosetting materials are polymer which when harden cannot be soften by heating. An example is Bakelite. Thermoplastic are polymers that harden on cooling and become soft on heating. The process can be repeated. An example is polyethene. (ii) There are extensive cross-linking between the chains in thermosetting polymers, while there are no or very little cross-linking in thermoplastic. 6 (a) HOOC! !COOH and HOCH2CH2OH (b) !!C! !C!O!CH2CH2!O!! O R O R (c) The ester link (d) Boiling with alkaline KMnO4, followed by acidification with dilute H2SO4. CH3! !CH3 + 6[O] !!: HOOC! !COOH + 2H2O (e) From the equation: 106 kg !!: 121 kg ∴ 12 kg !!: 12 106 × 121 × 75 100 = 10.27 kg 7 (a) 2-chlorobuta-1,3-diene (b) !![!CH2!CHRC!CH2!!!] & Cl (c) (i) Sulphur (ii) Cl & !!CH2!CRC!CH2!! & S & S & !!CH2!CRC!CH2!! & Cl (d) (i) !![!CH2!CHRCH!CH2!]!! (ii) The carbon-chlorine bond in neoprene is polar. As a result, the intermolecular forces is stronger. 8 (a) 2-methylbuta-1, 3-diene (b) (i) CH2 CH2 CH3 H CRC n CH2 H CH3 CH2 CRC n fi ff fi ff (ii) The cis-form is elastic whereas the trans-form is not. (c) It softens when hot and hardens when cold. (d) (i) Boiling the rubber with sulphur.
Chemistry Term 3 STPM Answers 355 (ii) Vulcanised rubber is more elastic, is harder and stronger and more resistant to atmospheric oxidation. (iii) Too much vulcanisation would make the rubber hard and brittle. 9 (a) Ethene gas with traces of oxygen is compressed under a pressure of 1500 atm and a temperature of 200 °C in the presence of an organic peroxide initiator. (b) Ethene gas is passed through an inert solvent containing a suspension of Ziegler-Natta catalyst at 2 – 6 atm and 60 °C. (c) LDPE: Plastic bags, bowls and squeeze bottles HDPE: Milk bottle crates, ice trays 10 (a) HOOC! !COOH H2N! !NH2 (b) !C! !C!N! O R O R H ! !N! H ! (c) Bullet-proof vest 11 (a) HOOC! !COOH HO!CH2CH2!OH (b) !C! !C!O!CH2CH2!O! O R O R (c) Polyester (Condensation polymer) (d) Hydrolysis of the ester linkage 12 ICH2CH2CH2CH2I + 2KCN Ethanol, heat !!!!!: NC—CH2CH2CH2CH2—CN + 2KI NC—CH2CH2CH2CH2—CN + 4H2O + 2H+ Heat !!!: HOOC(CH2)4COOH + 2NH4 + NC—CH2CH2CH2CH2—CN + 8[H] LiAlH4 in ether, heat !!!!!!!: H2N—(CH2)6—NH2 HOOC(CH2)4COOH + H2N(CH2)6NH2 !: —[—C—(CH2)4—C—N—(CH2)6—NH—]— + H2O ' ' & O O H 13 (a) Ethene polymerises to poly(ethene). The polymerisation process is exothermic. (b) Using: pV = nRT At the beginning of the experiment: (8.5 × 105 )(1500 × 10–6) = n × 8.31 × 298 n = 0.515 mol At time t: (1.0 × 105 )(1500 × 10–6) = n × 8.31 × 323 n = 0.0559 mol Number of moles of ethene polymerised = 0.515 – 0.0559 = 0.459 mol % conversion = 0.459 0.515 × 100% = 89.1% Assumptions: The vapour pressure of poly(ethene) and the catalyst can be ignored. The gas shows ideal behaviours under the conditions of the experiment. 14 (a) Thermoset: Once the polymer hardens, it cannot be soften again by heat. Thermoplastic: The polymer softens on heating and hardens on cooling. The process can be repeated indefinitely. (b) In thermosets, the polymer chains are extensively cross-linked. There are no or very little cross links in thermoplastics. (c) (i) C#N & ——[—CH2——C—]—— & COOC2H5 (ii) Addition polymerisation Specimen Paper Term 3 Section A 1 B 2 C 3 D 4 A 5 C 6 A 7 C 8 B 9 C 10 A 11 C 12 B 13 A 14 A 15 C Section B 16 (a) CH3CH2CH2CH2OH (I) CH3—CH—CH2CH3 (II) | OH * CH3—CH—CH2OH (III) | CH3 CH3 | CH3—C—CH3 (IV) | OH (b) Isomer IV (Tertiary alcohol) (c) Isomer II (Presence of a chiral carbon) (d) Isomer II H H H H H H H H | | | | | | | | H—C"C—C—C—H and H—C—C"C—C—H | | | | H H H H (e) Isomer IV Reaction of Grignard reagent with a ketone produces a tertiary alcohol. 17 (a) Ethyl 4-aminobenzoate (b) Step Reagent(s) Condition(s) I Conc. H2SO4 + Conc. HNO3 55 °C II Acidified K2Cr2O7 Heat III Ethanol + Conc. H2SO4 Reflux IV Tin + Conc. HCl Heat
Chemistry Term 3 STPM Answers 356 (c) In the stomach, benzocaine reacts with the hydrochloric acid in the gastric juice to form a soluble ionic salt. H2N—C6H4—COOC2H5 + H2O !: H3N+—C6H4—COO– + C2H5OH Section C 18 (a) (i) React ethanal with benzylmagnesium chloride in dry ether. Add water and boil. CH3 ! C ! H + ! CH2 MgCl + H2 O ' O !: CH3 ! C ! CH2 ! + Mg(OH)Cl & H & OH Isolate the organic product and add phosphorous(V) chloride. CH3 ! C ! CH2 ! + PCl5 !: CH3 ! C ! CH2 ! + POCl3 + HCl & H & H & Cl & OH (ii) Reduce ethanal to ethanol using lithium aluminium hydride in dry ether. CH3CHO + 4[H] !: CH3CH2OH + H2O Add phosphorous(V) chloride to convert ethanol to chloroethane. CH3CH2OH + PCl5 !: CH3CH2Cl + POCl3 + HCl Heat chloroethane with ethanolic potassium cyanide. CH3CH2Cl + KCN !: CH3CH2CN + KCl Heat the nitrile with lithium aluminium hydride in dry ether followed by acidic hydrolysis. CH3CH2CN + 4[H] !: CH3CH2CH2NH2 (b) (i) Nucleophilic addition (ii) Reaction mechanism: NaCN !: Na+ + :CN– CH3 ! C ! H + :CN– !: CH3 ! C ! CN ' Oδ– δ+ & H & O– CH3 ! C ! CN + HCN !: CH3 ! C ! CN + CN– & & H & H & OH Slow Fast O– In the final step, CN– is regenerated. Hence, it is a catalyst. (c) (i) Optical activity refers to the ability of a molecule to rotate the plane of polarisation of planepolarised light. (ii) H3 C OH mirror H CN C CH3 HO NC H C 19 (a) Mol ratio C : H : N = 81.5/12 : 4.85 : 13.6/14 = 6.79 : 4.85 : 0.9 = 7.5 : 5.3 : 1 The empirical formula is C7H5N. (b) Using the ideal gas equation: pV = nRT (200 × 103 )(1.76 × 10–3) = n × 8.31 × (150 + 273) n = 0.10 mol Molar mass of A = 10.3 0.10 g mol–1 = 103 g mol–1 (C7H5N)n = 103 n = 1 The molecular formula of A is C7H5N. (c) Add aqueous sodium carbonate. No effervescence occurs showing it is non-acidic. Add aqueous ammonium chloride. No gas given off that forms white fume with HCl showing that it is non-basic. (d) A is neutral and contains nitrogen. Hence, it is a nitrile. B is an amine which reacts with acid to give an alkylammonium salt, C. Treatment of B (an amine) with nitrous acid gives a hydroxyl compound, D and nitrogen gas. Oxidation of D, a hydroxyl compound produces a carboxylic acid, E (with two oxygen atoms in its molecule). Treatment of carboxylic acid with thionyl chloride produces an acyl chloride, F. Reaction between D (an alcohol) and F (an acyl chloride) produces an ester, G. Reaction of F (an acyl chloride) with excess ammonia produces an amide, H. The structures of A to H are shown below: A : ! CN E : ! COOH ! F : ! C ! Cl CH2 NH2 B : ! CH2 NH3 Cl C – : D : ! CH2 OH + ' O G : ! C ! O ! CH2 ! ' O ! C ! NH2 H : ' O (e) To A, B and H in separate test tubes, add nitrous acid (NaNO2 + HCl) and heat. A: No visible change B and H: Effervescence occurs (N2 is evolved)
Chemistry Term 3 STPM Answers 357 ! CH2 NH2 + HNO2 !: ! CH2 OH + N2 + H2 O ! CONH2 + HNO2 !: ! COOH + N2 + H2 O To B and H in separate test tubes, add aqueous sodium hydroxide. Heat. Test the gas evolved with moist red litmus paper. B: No change in the colour of the litmus H: Red litmus paper turns blue (NH3 is evolved) ! CONH2 + NaOH !: ! COONa + NH3 20 (a) Addition polymerisation is the process where small molecules (called monomers) are joined to form long chain molecules (called polymers) without the elimination of other smaller molecules. (b) Similarity: Same empirical formula Difference: Ethene is a gas. Poly(ethene) is a solid. (c) (i) nCH2=CH2 !: —[CH2—CH2]n (ii) During polymerisation, a weaker π-bond breaks and is replaced by two stronger σ-bonds. H H H H σ σ σ π C RC !: ! C ! C ! & H H & & H H & More energy is released (when the σ-bonds are formed) than absorbed (to break the π-bond). As a result, the process is exothermic (heat releasing) resulting in the rise in temperature. (iii) Before polymerisation: pV = nRT (25.0 × 103 )(1.0 × 10–3) = n × 8.31 × (25 + 273) n = 1.01 × 10–2 mol After 30 minutes: pV = nRT (12.2 × 103 )(1.0 × 10–3) = n × 8.31 × (32.0 + 273) n = 4.81 × 10–3 mol No. of moles polymerised = (1.01 × 10–2) – (4.81 × 10–3) = 5.29 × 10–3 mol Rate of polymerisation = (5.29 × 10–3) (30 × 60) mol dm–3 s–1 = 2.94 × 10–6 mol dm–3 s–1 (d) (i) H H | | —C—C— | | H Cl (ii) The C—Cl bonds in PVC are polar while the bonds in poly(ethene) are non-polar. The forces of attraction between PVC chains are stronger than that between poly(ethene) chains. As a result, PVC would have a higher tensile strength than poly(ethene). (iii) Making PVC water and sewage piping.
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
359 Chemistry Term 3 STPM Index L Laevorotatory 26 LDPE 293 Lewis acid 32 Lewis base 31 L-isomer 26 Lucas test 160 M Markovnikov’s rule 69 Methanoic acid 220 Methyl-carbonyl compounds 196 Molecular formula 13 Monomer 288 Monosaccharide 197 N Natural rubber 298 Nitrous acid 261 Nucleophile 31 Nucleophilic addition, carbonyl compounds 191 Nucleophilic substitution 114, 159, 223 Nucleophilic substitution, mechanism 122 Nylon 290 O Optical isomerism 24 Oxalic acid 220 Ozone, destruction of 134 Ozonolysis 189 P PAHs 98 Peptide link 274 Perhaloalkanes 110 Photochemical smog 59 Poly(ethene) 292 Protein, primary structure of 277 R Reforming 59 Ring-activating group 90 Ring-deactivating group 90 Saponification 232 SBR 302 Silver mirror test 195 SN1 123 SN2 122 sp hybridisation 9 sp2 hybridisation 7 sp3 hybridisation 6 Specific rotation 25 Stereoisomerism 23 Structural formula 15 Structural isomerism 21 Substituent effect 89 Syndiotic form 294 T Terylene 291 Thermoplastic 297 Thermoset 297 Tollen’s reagent 195 Triiodomethane test 164, 196 V Vulcanisation, of rubber 300 Y Yeast 166 Z Zaitsev rule 163 Ziegler-Natta catalyst 293 Zwitterion 273 Zymase 166
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2 Biology Term 3 STPM Chapter 14 Taxonomy and Biodiversity 14 W.M: RM35.95 / E.M: RM36.95 CB039348c ISBN: 978-983-00-9000-9 Penerbitan Pelangi Sdn. Bhd. (89120-H) TPRE-U STPM Text Chemistry Term 3 is specially designed for students who are sitting for the STPM examination. The comprehensive notes and practices are based on the latest syllabus and exam format set by Majlis Peperiksaan Malaysia. This book will provide you with the necessary skills and strategies to excel in the subject. Mastering MUET Pengajian Am Penggal 1, 2, & 3 Bahasa Melayu Penggal 1, 2, & 3 Biology Term 1, 2, & 3 Physics Term 1, 2, & 3 Chemistry Term 1, 2, & 3 Mathematics (T) Term 1, 2, & 3 Sejarah Penggal 1, 2, & 3 Geografi Penggal 1, 2, & 3 Ekonomi Penggal 1, 2, & 3 Pengajian Perniagaan Penggal 1, 2, & 3 Our Pre-U & STPM Titles: Comprehensive Notes and Practices Useful Features like Concept Maps, Learning Outcomes, Exam Tips, Info Chem and STPM Taggings Summary STPM Practices STPM Model Paper Term 3 Complete Answers FEATURES CHEMISTRY STPM Text CHEMISTRY CHEMISTRY STPM Text 3 TERM