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2007 Nov Paper 1 1. 5. [07N P1 Q01 Relative Masses] 92.2 % 28Si, 4.7 % 29Si and 3.1 % 30Si B C 1st (92.2×28)+(4.7×29)+(3.1×30) 92.2+4.7+3.1 ∆ © Step-by-Step [07N P1 Q02 Mole] C 2nd 3rd 5th 4h 6th moles of Fe3+(aq) 5 850 900 2100 1200 6300 There is a 'big' jump in I.E. between the 5th and 6th I.E. values, showing that the 6th electron is removed from the next inner shell. Hence, element X has 5 valence electrons (Group V element) and so, forms two chlorides, XCl3 and XCl5. (ans) [Ar calculated from options A, C and D are 28.099, 28.068 and 28.071 respectively.] moles of Fe 1 XCl3 I.E. 950 , 1800, 2700, 4800, 6000, 12300 = 28.109 (ans) 2. 07N-1 [07N P1 Q05 Ionisation Energies] From the relative abundance given in option B, Ar of Si = 2007 Nov Paper 1 © Step-by-Step 6. [07N P1 Q06 Bonding] The reaction can be represented as follows: CH3 Fe + 2 Fe3+ → 3 Fe2+ For the resultant solution to contain equal numbers of moles of Fe3+(aq) and Fe2+(aq), an excess of Fe3+(aq) is used; i.e. 3 mol in excess. Hence, the reaction involves 1 mol of Fe and 5 mol of Fe3+(aq), and the resultant mixture contains 3 mol of Fe3+(aq) and 3 mol of Fe2+(aq). (ans) B O C CH3 B has the largest dipole (due to presence of a very electronegative O atom). Cl C has a smaller dipole than B O C because C=O and C–Cl bonds Cl have similar dipoles (since O and Cl are of similar electronegativity). (ans) © Step-by-Step © Step-by-Step 3. [07N P1 Q03 Periodicity] C 7. PCl5 Given: 0.3 mol of XCln gives 0.6 mol of HCl or 1 mol of XCln gives 2 mol of HCl. With excess cold water at room temperature, PCl5 undergoes hydrolysis to give POCl3 and 2HCl. PCl5 + H2O → POCl3 + 2HCl Hence the chloride is PCl5. (ans) Four electrons from each oxygen are involved in forming hydrogen bonds. In ice, each water molecule is hydrogen bonded to two other water molecules. .. H H [07N P1 Q04 Atomic Structure] D B δ+ © Step-by-Step 4. [07N P1 Q07 Solids] Both ions have 20 neutrons in their nuclei. no. of p no. of n no. of e– 36 216 S 16 36 – 16 = 20 16 + 2 = 18 37 17 Cl 17 37 – 17 = 20 17 + 1 = 18 ∴ Both 36S2– and 37Cl – have 20 neutrons in their nuclei. (ans) © Step-by-Step δ– δ+ O : ′′′′′′′′′′′′′ H .. O δ– H .. ′′′′ ′′′′ δ+ ′′′′ H hydrogen δ– bonding O : H .. Hence, 2 lone pairs (or 4 electrons) from each oxygen atom are involved in forming hydrogen bonds. The bond angle around each oxygen atom is 109.5° (four bond pairs, no lone pair). The hydrogen bonds are longer (and weaker) than the O–H covalent bonds. The open structure of ice causes ice to be less dense than water (i.e. ice floats on water). (ans) © Step-by-Step A-Level Solutions – Chemistry 07N-2 8. 2007 Nov Paper 1 11. [07N P1 Q11 Ionic Equilibria] [07N P1 Q08 Electrolysis] C The number of moles of Y deposited is half the number of moles of X deposited. [H+ ] = With YCl2, the reactions involved are: Y 2+(l) + 2e– → Y(s) 2 Cl –(l) → Cl2(g) + 2e– Hence for the same current and the same time, mol of X deposited : mol of Y deposited =2:1 or mol of Y deposited = 12 × mol of X deposited. (ans) © Step-by-Step [07N P1 Q09 Redox] B 3 mol of Tl + = 0.20 × 25 1000 mol of MnO4– = 0.10 × = 5.0 × 10–3 25 1000 pH < 7 H2O(l) H+(aq) + OH–(aq) Kw = [H+ ][OH– ] With XCl, the reactions involved are: 2 X +(l) + 2e– → 2 X(s) 2 Cl –(l) → Cl2(g) + 2e– 9. C = 2.5 × 10–3 = 1.44 × 10−14 = 1.20 × 10–7 mol dm–3 pH = –log10 [H+ ] = –log10 (1.20 × 10–7) = 6.92 Hence, pH < 7. (ans) © Step-by-Step 12. [07N P1 Q12 Titration] 14 12 10 pH 8 6 A 4 2 0 2.5 × 10 = 2 mol of Tl + Since 2 Tl + → 2 Tl 3+ + 4 e–, ∴ 1 mol of MnO4– gains 4 mol of e–, oxidation number of Mn in MnO4– = +7 ∴O.N. of Mn in the reduced form = +7 – 4 = +3 (ans) © Step-by-Step D B C 0 2 4 6 8 10 12 14 16 18 volume of 0.100 mol dm–3 NaOH added / cm3 −3 ∴1 mol of MnO4– reacts with 5.0 × 10 −3 since [H+ ] = [OH– ] Kw For a weak acid HA, HA(aq) H +(aq) + A –(aq) [H + ][ A − ] [HA] Hence, pH = pKa when Ka = [H+ ]; i.e. when [A– ] = [HA]. This occurs at the point of half-neutralisation; i.e. where vol of NaOH added = 5 cm3. (ans) Ka = © Step-by-Step 10. [07N P1 Q10 Redox] 13. [07N P1 Q13 Buffers] B The solution turns brown without effervescence. B H2O2 + 2H+ + 2e– → 2H2O 2 I– → I2 + 2e– +1.77 V –0.54 V H2O2 + 2H+ + 2 I– → I2 + 2H2O +1.23 V o Ecell = +1.77 – 0.54 = +1.23 V Since Ecell o > 0, reaction is feasible. Hence, KI(aq) is oxidised to I2 (shown by solution turning brown) while H2O2 is reduced to water. No effervescence is seen as O2(g) is not formed. (ans) HCO3– The main buffer in blood plasma is the hydrogen carbonate and carbonic acid buffer system, which is linked to the gaseous exchange of CO2 in the lungs. Hence, contaminating H+(aq) ions are removed by the HCO3– ions present in the human body. HCO3– + H+ → H2CO3 (ans) © Step-by-Step © Step-by-Step A-Level Solutions – Chemistry 2007 Nov Paper 1 14. [07N P1 Q14 Kinetics] D 17. [07N P1 Q17 Periodicity] It produces its own catalyst. pH of resulting solution 6.5 solubility in water The graph suggests an autocatalysed reaction, which is slow at first and then becomes more rapid as the catalyst (Mn2+) is produced in the reaction. very slow: no catalyst [MnO4–] fast: catalyst present slow: reactants dissolves B MgCl2(s) + aq → [Mg(H2O)6]2+ + 2Cl – [Mg(H2O)5(OH)]+ + H+ [Mg(H2O)6]2+ MgCl2 (an ionic chloride) dissolves in water with slight hydrolysis (due to the larger polarising power of Mg2+ ions) to give a slightly acidic solution (pH 6.5). (ans) © Step-by-Step used up 0 (ans) time 0 07N-3 18. [07N P1 Q18 Bonding] © Step-by-Step D 15. [07N P1 Q15 Periodicity] C the copper ion in CuO In CuO, Cu2+ has configuration [Ar] 3d9 and so, has a single unpaired electron. (ans) The melting points of the elements decrease steadily. © Step-by-Step Across the period, the melting points of the elements do not decrease steadily but vary as shown. 19. [07N P1 Q19 Isomerism] B CH3CH(OH)CO2H melting point Si Compound X is optically active and so, has a chiral carbon (marked by *). * CH CH(OH)CO H Mg Al Na S 3 P 1 mol of X liberates 1 mol of H2 when reacted with Na suggests that X contains two –OH groups. Ar Cl 11 12 13 14 15 16 17 18 proton no. 2 (ans) CH3CHCO2H + 2Na → CH3CHCO2– Na+ + H2 © Step-by-Step O– Na+ OH (ans) © Step-by-Step 16. [07N P1 Q16 Group VII] inside the cold tube black deposit A • with aqueous silver nitrate white precipitate Iodides react with c. H2SO4 to give HI, which is oxidised by conc. H2SO4 to I2 (seen as black deposit in the cold tube). – – I + H2SO4 → HI + HSO4 8HI + H2SO4 → 4I2 + H2S + 4H2O • C It is optically active. Bupropion has a chiral carbon (marked by *) and so, is optically active. CH3 * CH3C NH CH(CH3) CO CH3 2° amine Chlorides react with c. H2SO4 to give HCl, which forms a white precipitate of AgCl when bubbled into aqueous silver nitrate. Cl – + H2SO4 → HCl + HSO4– Ag+(aq) + Cl –(aq) → AgCl(s) (ans) © Step-by-Step A-Level Solutions – Chemistry 20. [07N P1 Q20 Isomerism] ketone Cl Bupropion has a secondary amine (not amide) and a ketone group. It has no reaction with aq. NaOH (since aryl chloride does not undergo hydrolysis). (ans) © Step-by-Step 07N-4 2007 Nov Paper 1 21. [07N P1 Q21 Alkenes] [93N P4 Q22] B 25. [07N P1 Q25 Halogen Derivatives] CH3COCH2CH3 CH3CH=CHCH3 + ½O2 A catalyst CH3CH2C–CH3 but-2-ene O (ans) © Step-by-Step 22. [07N P1 Q22 Alkenes] C bromoethane Bromoethane, CH3CH2Br, reacts with ammonia to give ethylamine, CH3CH2NH2. δ– δ+ .. NH3 + CH3CH2–Br heat CH3CH2NH2 + HBr – a nucleophilic substitution reaction. (ans) CH3CHCH2I © Step-by-Step Cl 26. [07N P1 Q26 Alcohols] Propene undergoes electrophilic addition. δ+ δ– CH3CH=CH2 + I–Cl C slow + CH3CHCH2I + Cl fast CH3CHCH2I Cl (ans) © Step-by-Step Alcohols with molecular formula C5H12O are: CH3CH2CH2CH2CH2OH pentan-1-ol Butane, CH3CH2CH2CH3, which does not contain any halogen atoms will not destroy ozone and so, can be used safely as a replacement for CFCs. (ans) © Step-by-Step OH CH3 3-methylbutan-2-ol CH3 3-methylbutan-1-ol CH3CH2CHCH2CH3 OH pentan-3-ol CH3 CH3C–CH2OH CH3 CH3 2,2-dimethylpropan-1-ol 24. [07N P1 Q24 Phenols] C C6H5O– Na+ Solid Z dissolves readily in water to give a weakly alkaline solution suggests that it is a salt of a strong base and a weak acid. Hence, Z is C6H5O– Na+, which is formed from NaOH (strong base) and phenol (weak acid). (ans) © Step-by-Step OH pentan-2-ol CH3CH CHCH3 CH3CHCH2CH2OH CH3CH2CH2CH3 CH3CH CH2CH2CH3 CH3CH2CHCH2OH CH3 2-methylbutan-1-ol 23. [07N P1 Q23 Halogen Derivatives] D 2 – CH3CH2C–CH3 OH 2-methylbutan-2-ol Only the two alcohols with CH3CH(OH)– group give yellow precipitate of CHI3 with alkaline aqueous iodine. (ans) © Step-by-Step 27. [07N P1 Q27 Carbonyl Compounds] C CH3COCI3 CH3COCI3 NaOD NaOD D2 O D2 O O CH3C + CDI3 O– Na+ yellow ppt – a triiodomethane (iodoform) reaction. (ans) © Step-by-Step A-Level Solutions – Chemistry 2007 Nov Paper 1 28. [07N P1 Q28 Amino Acids] C 31. [07N P1 Q31 Solids] [89J P3 Q31] It reacts with ethanoyl chloride to form an ester. CO2H N-methyl-D-aspartic acid has CH2 a chiral carbon (marked by *) and so, is optically active. CH3NHCHCO2H * The –CO2H group reacts with ethanol (to form an ester) and it reacts with PCl5 to form an acyl chloride. CO2H 07N-5 CO2C2H5 + 2C2H5OH CH2 CH2 c. H2SO4 heat CH3NHCHCO2H CH3NHCHCO2C2H5 graphite sodium 2 3 In graphite, each C atom is bonded covalently to three other carbon atoms, using three of the valence electrons. The fourth valence electron is available for π bonding between adjacent carbon atoms, resulting in delocalisation throughout the layer. Sodium has a giant metallic structure, which consists of a lattice of sodium ions surrounded by a sea of delocalised electrons moving randomly throughout the lattice. (ans) © Step-by-Step CO2H CH2 COCl + PCl5 CH3NHCHCO2H 32. [07N P1 Q32 Energetics] CH2 CH3NHCHCOCl N-methyl-D-aspartic acid has no alcohol group and so, cannot react with ethanoyl chloride to form an ester. (ans) © Step-by-Step 29. [07N P1 Q29 Amines] B o ∆Hr o ∆Hr = ∑∆Hf (products) – ∑∆Hf (reactants) = ∆Hf o for Ca(OH)2(s) – 2∆Hf o for H2O(l) Further information needed is ∆Hf o for H2O(l). (ans) Order of base strength: C2H5NH2 > NH3 > C6H5NH2 © Step-by-Step ∴ strength of conjugate acid: C2H5NH3+ Cl – < NH4+ Cl – < C6H5NH3+ Cl – Hence C6H5NH3+ Cl – is most acidic in aqueous solution. (ans) © Step-by-Step 30. [07N P1 Q30 Amino Acids] D H2NCHCO2– CH2 CO2– At pH 10 (alkaline medium), the acidic –CO2H groups react with the OH– ions present to give the corresponding carboxylate anion. H2NCHCO2– + 2H2O CH2 CH2 CO2H CO2– (ans) © Step-by-Step A-Level Solutions – Chemistry Ca(s) + 2H2O(l) → Ca(OH)2(s) + H2(g) Since ∆Hf o is zero for both Ca(s) and H2(g), ∴∆Hf o for Ca(OH)2(s) = ∆Hr + 2∆Hf o for H2O(l) C6H5NH3+ Cl – H2NCHCO2H + 2OH– ∆Hf o for H2O(l) 1 33. [07N P1 Q33 Kinetics] 2 3 x y z 1 2 –1 –2 +3 +6 Given: rate = k [P] [Q] x ⇒ k = rate x [P][Q] units of rate units of k = (units of concentration)1+ x -3 -1 = mol dm -3 s1+ x (mol dm ) = mol–x dm3x s–1 Given: units of k = moly dmz s–1 By inspection, y = –x and z = 3x Hence, options 2 and 3. (ans) © Step-by-Step 07N-6 2007 Nov Paper 1 34. [07N P1 Q34 Group II] [80N P3 Q35] 1 37. [07N P1 Q37 Amides] The ionic radius of the M 2+ ion increases from calcium to barium. 1 Ionic radius increases down the group (from Ca to Ba) as each succeeding element has one more shell of electrons. Hence, the outer electrons are progressively further away from the nucleus. Hydration energy of M 2+ ion decreases as size of M 2+ increases. M(g) → M 2+(g) + 2e– 2 • chlorine bromine Br Br Br • – O2 + 4H + 4e → 2H2O [Fe(CN)6]3– + e– → [Fe(CN)6]4– Cr3+ + e– → Cr2+ Fe(OH)3 + e– → Fe(OH)2 + OH– On heating with NaOH(aq), both phenol and amide groups react – phenol undergoes neutralisation while amide undergoes hydrolysis. Hence, two moles of NaOH are used up per mole of capsaicin. Product: CH3O A solution of potassium hexacyanoferrate(III). Na+ –O E o value +1.23 V +0.36 V –0.41 V +1.07 V When left to stand in the atmosphere, only solutions with E o value less positive than +1.23 V will be oxidised. Hence, Cr2+(aq) will be oxidised to Cr3+(aq), while Fe(OH)2 which is formed when FeSO4 reacts with NaOH, will be oxidised to Fe(OH)3. However, [Fe(CN)6]3– remains unchanged (chemically stable) because it is already in its highest oxidation state. (ans) CH2NHCO(CH2)4CH–CHCH(CH3)2 HO 36. [07N P1 Q36 Transition Elements] + With Br2 in organic solvent, both phenol and alkene groups react – phenol undergoes electrophilic substitution while alkene undergoes electrophilic addition. Hence, three atoms of bromine is incorporated into the molecule. Product: CH3O © Step-by-Step 1 alkene phenol 35. [07N P1 Q35 Group VII] Fe3+ + e– → Fe2+ E o = +0.77 V – – E o = +1.36 V Cl2 + 2e → 2Cl – – Br2 + 2e → 2Br E o = +1.07 V – – E o = +0.54 V I2 + 2e → 2I To oxidise Fe2+(aq) to Fe3+(aq), the oxidising agent must have a E o value more positive than +0.77V. Hence, only chlorine and bromine will oxidise Fe2+(aq) to Fe3+(aq). (ans) amide HO © Step-by-Step 1 2 CH2 NHCO (CH2)4 CH=CH CH(CH3)2 CH3O ∆H = 1st I.E. + 2nd I.E. ∆H decreases from Ca to Ba since ionisation energy decreases down a group. (ans) Addition of bromine in an organic solvent causes three atoms of bromine to be incorporated into the molecule. On heating with NaOH(aq), two moles of NaOH are used up per mole of capsaicin. • CH2NH2 + Na+ –O2C(CH2)4CH=CHCH(CH3)2 On heating under reflux with conc. KMnO4/H+, one of the possible reaction is oxidation with cleavage of C=C bond. (CH3)2CHCHO formed initially is further oxidised to give (CH3)2CHCO2H. CH3O CH2NHCO(CH2)4CH=CHCH(CH3)2 HO CH3O CH2NHCO(CH2)4CO2H + (CH3)2CHCO2H HO (ans) © Step-by-Step © Step-by-Step A-Level Solutions – Chemistry 2007 Nov Paper 1 38. [07N P1 Q38 Phenols] 1 2 • • • 40. [07N P1 Q40 Carboxylic acids] Its aqueous solution is acidic. It can exist in optically active forms. Its aqueous solution phenol is acidic due to the presence of an acidic O N 2 phenol group. OH CH3 CHCH2CH3 * • An ester is formed when the phenol group reacts with ethanoyl chloride (and not with ethanol). (ans) 2 3 CO2CH3 OCOCH3 CN OCOCH3 In 2, salicylic acid is obtained by acid hydrolysis of both ester groups. Dinoseb has a chiral NO2 carbon (marked by *) and so, can exist in optically active forms. CO2CH3 OCOCH3 H O/H+ 2 heat • CN OCOCH3 H O/H+ 2 39. [07N P1 Q39 Carbonyl Compounds] heat • 2,4-dinitrophenylhydrazine reagent alkaline aqueous iodine • H NO2 C O + H2 N COCl NO2 N Cl CH3 propanone CO2H + NH4+ OH + CH3CO2H In 1, the aryl chloride group is not hydrolysed because the aryl–Cl bond is strengthened by the overlapping of the p-orbital of Cl with the π orbitals of the benzene ring. Hence, the hydrolysis reaction is With 2,4-DNPH, propanone (a carbonyl compound) reacts to give an orange precipitate. CH3 CO2H + CH3OH OH + CH3CO2H In 3, salicylic acid is obtained by acid hydrolysis of both the nitrile and ester groups. © Step-by-Step 1 2 07N-7 H2O/H + CO2H Cl + HCl heat 2,4-DNPH and salicylic acid is not formed. (ans) N C CH3 • © Step-by-Step H NO2 CH3 NO2 + H2O N orange precipitate With alkaline aqueous I2, propanone (which has CH3C=O group) reacts to give a yellow precipitate of CHI3. O CH3C + 3I2 + NaOH CH3 propanone warm O CHI3 + CH3 yellow ppt C + 3HI O– Na+ [07N P1 MCQ Key] Q. Key Q. Key Q. Key Q. Key 1 2 3 4 5 B C C D C 11 12 13 14 15 C B B D C 21 22 23 24 25 B C D C A 31 32 33 34 35 C D C D B 6 7 8 9 10 B B C B B 16 17 18 19 20 A B D B C 26 27 28 29 30 C C C B D 36 37 38 39 40 D B B B C © Step-by-Step • Only aldehydes react with Fehling's reagent. Hence, there is no reaction with both propanone (a ketone) and pentyl ethanoate (an ester). (ans) © Step-by-Step A-Level Solutions – Chemistry 07N-8 1. 2007 Nov Paper 2 2007 Nov Paper 2 2. [07N P2 Q01 Mole / Gases] (a) alkanes. (ans) [07N P2 Q02 Group VII] (a) (i) Cl2 + 2NaOH → NaCl + NaClO + H2O (ans) (b) 2 C14H30 + 43 O2 → 28 CO2 + 30 H2O (ans) [or C14H30 + 43 2 O2 → 14 CO2 + 15 H2O] (c) (i) mass of C14H30 burnt = (10.8 × 10700) kg = 10.8 × 10700 tonnes 1000 = 115.56 tonnes = 116 tonnes (ans) (ii) Mr of C14H30 = (14 × 12.0) + (30 × 1.0) = 198.0 Mr of CO2 = 12.0 + (2 × 16.0) = 44.0 From the equation in (b), (2 × 198.0) tonnes of C14H30 gives (28 × 44.0) tonnes of CO2. (28 × 44.0) ∴mass of CO2 produced = × 116 (2 ×198.0) = 361 tonnes = 361000 kg (ans) (d) n = (ii) 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O (ans) (b) (i) reactant products species oxidation number Cl2 NaCl 0 –1 species NaClO oxidation number +1 (ans) (ii) In this reaction, Cl2 undergoes disproportionation; i.e. Cl2 is simultaneously oxidised to NaClO and reduced to NaCl. (ans) (c) (i) H2O2 → O2 + 2H+ + 2e– (ans) (ii) H2O2 + 2ClO3– + 2H+ → O2 + 2ClO2 + 2H2O (ans) (iii) H2O2 functions as a reducing agent. (ans) © Step-by-Step (6 ×100 ×103 ) × (670 ×10−6 ) pV = 8.31× (20 + 273) RT = 0.165 mol (ans) [NB. Pressure must be converted into Pa (or N m–2) and volume converted into m3.] 0.165 × 8.31× (5 + 273) (e) p = nRT = V 670 ×10−6 = 569000 Pa = 569 kPa = 5.69 bar (ans) (f) No, the tyres should not be deflated before flight. maximum safe pressure difference between the air inside the tyre and the atmospheric pressure acting on it = 8 – 1.01 = 6.99 bar pressure difference across the wall of the tyre in flight = 5.69 – 0.28 = 5.41 bar Hence, it is not necessary to reduce the air pressure inside the tyre since difference between the internal and external pressure (5.41 bar) is well within the pressure range (6.99 bar) that the tyre can withstand before it will burst. (ans) © Step-by-Step A-Level Solutions – Chemistry 2007 Nov Paper 2 3. (d) (i) stage I: hydrolysis (nucleophilic substitution) (ans) stage II: elimination reaction (ans) [07N P2 Q03 Transition Elements] (a) (i) Co-ordination number of Co = 6 (ans) (ii) No significant colour change is observed because Co2+(aq) is not oxidised by ClO2. E o /V + – – ClO2 + 4H + 5e → Cl + 2H2O +1.50 Co2+(aq) → Co3+(aq) + e– –1.82 2+ + – 07N-9 (ii) Mechanism: nucleophilic substitution with OH– as the nucleophile. slow C Cl 3+ Cl OH C .. OH– Cl ClO2 + 4H + 5Co → Cl + 5Co + 2H2O Θ O O δ+ δ– Cl o Ecell = +1.50 – 1.82 = –0.32 V Since Ecell o < 0, reaction is not feasible. (ans) fast O (iii) Cation P is [Co(NH3)6]2+. (ans) C Cl (iv) [Co(NH3)6]3+ is formed. (ans) OH + Cl – (ans) o ClO2 + 4H+ + 5e– → Cl – + 2H2O [Co(NH3)6]2+ → [Co(NH3)6]3+ + e– 2+ + – E /V +1.50 –0.11 3+ ClO2 + 4H + 5[Co(NH3)6] → Cl + 5[Co(NH3)6] + 2H2O o Ecell = +1.50 – 0.11 = +1.39 V Since Ecell o > 0, reaction is feasible. Hence, [Co(NH3)6]2+ is oxidised by ClO2 to [Co(NH3)6]3+. (ans) © Step-by-Step 4. [07N P2 Q04 Halogen Derivatives] (a) 2CHCl3(l) + O2(g) → 2COCl2(g) + 2HCl(g) ∆H = ∑∆Hf (products) – ∑∆Hf (reactants) –356 = 2 ∆Hf (COCl2) + 2(–92) – 2(–134) – 0 –356 = 2 ∆Hf (COCl2) + 84 ∴ ∆Hf (COCl2) = −356 − 84 2 = –220 kJ mol–1 (ans) (b) This oxidation is initiated by ultraviolet light and involves free-radicals. (ans) (c) (i) • •• • Cl • •× • •• • × O C • • • × ×• • • • •Cl•• (ans) (ii) Cl–C–Cl bond angle is 120°. (ans) A-Level Solutions – Chemistry (e) The organic compound formed is O Cl C OCH2CH3 (ans) [The other Cl atom in phosgene can also be replaced by –OCH2CH3 group.] (f) (i) Reagent: ammonia, NH3 (ans) (ii) Urea can act as a base because the lone pair of electrons on each of the N atoms can accept H+ (from the strong acid) by forming dative bond with it. (ans) (g) (i) 4 CO(NH2)2 + 6 NO2 → 4 CO2 + 8 H2O + 7 N2 (ans) (ii) NO2 in the atmosphere can result in acid rain (by dissolving in water vapour in the air to form nitric acid). (ans) © Step-by-Step 07N-10 5. 2007 Nov Paper 2 (b) (i) [07N P2 Q05 Phenols] (a) reagent OCH3 structural formula of the organic product CH2CHCH2Br CH3C O CH2CH=CH2 (ii) CO2H O + OCH3 Product: (ans) OCH3 [Oxidation with cleavage of C=C bond.] OH OCH3 O2N (c) (i) CH=CHCH2OH CH2CH=CH2 (ans) compound C – (ii) stage I reagent(s) conditions + O Na OCH3 stage II reagent(s) conditions CH2CH=CH2 compound A Product: Tollen's reagent CH3C OH compound C Na OH (ans) [Reaction of phenol and alkene with aq. Br2.] O Product: CH2CHCH2Br Br Product: dilute HNO3 OCH3 Br and compound C CH3COCl OH OH Br compound A, B, C or D / O LiAl H4 (ans) in dry ether (ans) CH3CO2H (ans) heat under reflux with a small amount of conc. H2SO4 (ans) [or CH3COCl at room temperature.] CH=CHC © Step-by-Step OH (ans) A-Level Solutions – Chemistry 2007 Nov Paper 3 1. [07N P3 Q01 Equilibria / Entropy / RX] (a) Kp = pCH CH CH CHO atm–2 (ans) ( pCH CH=CH )( pCO )( pH ) 3 2 3 2 2 2 (i) CH3CH=CH2 + CO + H2 t=0/mol t=0/atm 1 CH3CH2CH2CHO 1 1 0 120 3 120 3 120 3 0 = 40 = 40 = 40 eqm/atm 40 – 39.6 40 – 39.6 40 – 39.6 = 0.4 = 0.4 0 39.6 = 0.4 ∴ partial pressure of CH3CH=CH2 = 0.4 atm (ans) partial pressure of CO = 0.4 atm (ans) partial pressure of H2 = 0.4 atm (ans) pCH CH CH CHO ( pCH CH=CH )( pCO )( pH ) (39.6) = = 619 atm–2 (ans) (0.4)(0.4)(0.4) (ii) Kp = 3 3 2 (ii) • To convert butan-1-ol into 1-bromobutane, react with HBr, which is obtained by heating NaBr with conc. H2SO4. (ans) [or react with PBr3, PBr5 or SOBr2 at room temperature.] • 2 (iii) CH3CH=CH2 + CO + H2 CH3CH2CH2CHO A higher pressure would favour the production of butanal because the forward reaction is accompanied by a reduction in pressure (3 mol of gas converted to 1 mol of gas). By Le Chatelier's principle, an increases in pressure would shift the position of equilibrium to the right (the side with fewer gas molecules) so as to remove some of the excess pressure. (ans) reaction III – LiAl H4 in dry ether. (ans) [reaction II is oxidation, III is reduction] 2 2 07N-11 (c) (i) • reaction II – heat under reflux with acidified K2Cr2O7(aq). (ans) • (b) 2007 Nov Paper 3 To convert butan-1-ol into 2-bromobutane, heat with conc. H2SO4 at 180 °C (to give but-1-ene), followed by reaction with HBr. (ans) (d) (i) With 2-bromobutane, the rate of reaction is slower than that with 2-iodobutane. This is because the C–Br bond is shorter and hence, stronger than C–I bond (since Br atom is smaller than I atom). (ans) (ii) • CH3CH2CHCH3 NaCN in alcohol CH3CH2CHCH3 Br CN LiAl H4 CH3CH2CHCH3 (iv) ∆S for reaction I is negative because there is a decrease in the number of gaseous particles and hence, the system becomes more orderly. (ans) (v) ∆G = ∆H – T ∆S Since ∆S is negative, the term (– T ∆S ) is positive. Hence for ∆G to be negative, ∆H is negative (at least as negative as T ∆S). (ans) (vi) CH3CH=CH2 + CO + H2 → CH3CH2CH2CHO Bonds broken (∆H ) 1 C–C 1 C=C 6 C–H 1 C≡O 1 H–H +350 +610 6(+410) +1077 +436 +4933 Bonds formed 3 C–C 8 C–H 1 C=O ∴ ∆H = +4933 + (–5070) = –137 kJ mol–1 (ans) (∆H ) 3(–350) 8(–410) –740 –5070 CH2NH2 • CH3CH2CHCH3 Br NaOH in alcohol CH3CH=CHCH3 + CH3CH2CH=CH2 CH3CH=CHCH3 + CH3CH2CH=CH2 cold KMnO4(aq) CH3CH–CHCH3 + CH3CH2CH–CH2 OH OH OH OH (ans) (e) Free-radical substitution occurs and any one of the H atoms in butane H H H H may be replaced by Br 1 2 4 3 H C C C C H atom. Replacing any H H H H one of the six H atoms on C-1 and C-4 with Br gives 1-bromobutane, whereas replacing any one of the four H atoms on C-2 and C-3 gives 2-bromobutane. Hence ratio of 1-bromobutane : 2-bromobutane = 6 : 4 or 3 : 2. (ans) © Step-by-Step A-Level Solutions – Chemistry 07N-12 (a) If a white ppt is seen which dissolves in excess NaOH(aq) to give a colourless solution, it suggests that the presence of Al 3+ and so, the sample is Al2O3. [07N P3 Q02 Periodicity / Acid Derivatives] first ionisation energy 2. 2007 Nov Paper 3 Ar P Cl Mg Si Na Al 3+(aq) + 3OH–(aq) → Al (OH)3(s) Al (OH)3(s) + OH–(aq) → Al (OH)4–(aq) (ans) S Al 11 12 13 14 15 16 17 18 proton number First ionisation energy (I.E.) increases from Na to Ar due to the increase in nuclear charge and decrease in atomic size (while shielding effect remains almost the same since the electrons all go into the same shell). (c) NaCl is an ionic chloride and so, dissolves in water without further reaction to give a neutral solution (pH 7). NaCl(s) + aq → Na+(aq) + Cl –(aq) • The discontinuities in the increasing trend are: • • The first I.E. of Al is lower than that of Mg. Mg [Ne] 3s2 Al [Ne] 3s2 3p1 This is because less energy is required to remove a 3p electron in Al than a 3s electron in Mg since the 3p electron is further away from the nucleus and it also experiences slightly better shielding (from the 3s electrons). The first I.E. of S is lower than that of P. P [Ne] 3s2 3px1 3py1 3pz1 S [Ne] 3s2 3px2 3py1 3pz1 This is because less energy is required to remove an electron from the paired 3p electrons in S due to inter-electron repulsion arising from two electrons occupying the same orbital. (ans) (b) Add dilute aqueous acid, e.g. dilute HCl, to the sample of white powder. If there is no reaction, the sample is SiO2. If the white powder dissolves to give a colourless solution, it is either MgO or Al2O3. MgO(s) + 2HCl(aq) → MgCl2(aq) + H2O(l) Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2O(l) • To the resultant solution, add dilute NaOH(aq) dropwise until in excess. If a white precipitate (ppt) is seen which does not dissolves in excess NaOH(aq), it suggests the presence Mg2+ and so, the sample is MgO. Mg2+(aq) + 2OH–(aq) → Mg(OH)2(s) PCl5 is an acidic chloride (simple covalent) and so, undergoes complete hydrolysis in water to give a strongly acidic solution (pH = 1). PCl5 + 4H2O → H3PO4 + 5HCl (ans) (d) (i) One precaution is to perform the experiment in a fume cupboard since pungent gases (SO2 and HCl ) are produced. (ans) [SO2 is poisonous/choking; HCl is acidic.] (ii) SOCl2 + H2O → SO2 + 2HCl (ans) (iii) With AgNO3(aq), a white precipitate of AgCl(s) is seen. Ag+(aq) + Cl –(aq) → AgCl(s) (from HCl ) • With K2Cr2O7(aq), SO2 (a reducing agent) would reduce the orange K2Cr2O7(aq) to green Cr3+(aq) and itself being oxidised to SO42–(aq). Cr2O72– + 2H+ + 3SO2 → 2Cr3+ + 3SO42– + H2O (ans) (e) (i) O O C A C O OH OH C B C Cl Cl (ans) O (ii) I: acid-base reaction (ans) II: nucleophilic substitution (ans) III: hydrolysis reaction (ans) (iii) III: heat under reflux with dilute aqueous H2SO4. (ans) [or heating under reflux with aqueous NaOH.] © Step-by-Step A-Level Solutions – Chemistry 2007 Nov Paper 3 3. [07N P3 Q03 Energetics / Grp II / Carbonyl] (a) (i) Ca2+(g) + 2F(g) + 2e– 07N-13 (iii) Since ∆Hsol is positive (endothermic), the dissolution of CaF2 is, therefore, favoured by high temperature. Hence, CaF2 is more soluble in hot water than in cold. (ans) E (c) (i) The very low melting point (60 °C) suggests that Be(NO3)2 is a covalent compound with a simple molecular structure (due to the high charge density of Be2+ ion). 2nd I.E.(Ca) Ca+(g) + 2F(g) + e– 2(–328) 1st I.E.(Ca) Ca(g) + 2F(g) Ca(g) + F2(g) B.E.(F–F) 0 Ca2+(g) + 2F –(g) • Ca(s) + F2(g) ∆Hat(Ca) = +178 ∆Hf = –1220 L.E. CaF2(s) (ans) By Hess' Law, L.E. = ∆Hf – [∆Hat (Ca) + B.E.(F–F) + 1st I.E. + 2nd I.E. + 2(–328)] = –1220 – [178 + 158 + 590 + 1150 + 2(–328)] = –2640 kJ mol–1 (ans) (ii) L.E. ∝ • • q+ q− (r+ + r− ) (ii) Mg(NO3)2(s) → MgO(s) + 2NO2(g) + • (ans) Ca2+(aq) + 2F–(aq) –543 2(–333) Ca(s) + F2(g) By Hess' Law, ∆Hsol = –543 + 2(–333) – (–1220) = +11 kJ mol–1 (ans) A-Level Solutions – Chemistry Ca ion, being larger in size than Mg ion, has a smaller charge density and hence, weaker polarising power. The NO3– ion is, thus, less polarised by the larger Ca2+ ion. Ca(NO3)2 is, therefore, more stable than Mg(NO3)2 and so, decomposes at a higher temperature. (ans) O O , and CH3C CH3C H Mr 30 CH3 H Mr 44 Mr 58 which are produced in the ratio 1:2:1. (ans) [ 2 (CH3CO2)2Ca + 2 (HCO2)2Ca → CH3COCH3 + HCHO + 2 CH3CHO + 4 CaCO3 ] (ii) The two chemical tests are: 1. Warm each compound separately with alkaline aqueous iodine. Only CH3COCH3 and CH3CHO give yellow precipitate of CHI3. Hence the compound that does not give any such precipitate is HCHO. 2×2.3×10–4 mol dm–3 Ksp = [Ca2+] [F – ]2 = (2.3 × 10–4)(2 × 2.3 × 10–4)2 = 4.87 × 10–11 mol3 dm–9 (ans) –1220 2+ O Ca2+(aq) + 2F –(aq) 2.3×10–4 ∆Hsol 2+ H–C The lattice energy of CaF2 is numerically smaller than that of CaO because the electrostatic forces of attraction between Ca2+ and the singly charged F – ion is weaker than that between the doubly charged Ca2+ and O2– ion. (ans) (ii) CaF2(s) + aq O2(g) (ans) The lattice energy of CaF2 is numerically greater than that of CaCl2 due to the smaller ionic size of F – ion (compared to Cl – ion) while the charge remains the same. (ans) CaF2(s) 1 2 (d) (i) The three carbonyl compounds are: (b) (i) Ksp = [Ca2+] [F – ]2 mol3 dm–9 • The very high melting point (561 °C) suggests that Ca(NO3)2 is an ionic compound with a giant lattice. (ans) 2. Having identified HCHO, heat the remaining two compounds separately with acidified KMnO4(aq). CH3CHO is oxidised to CH3CO2H, and the purple KMnO4(aq) solution is decolourised. CH3COCH3 is not oxidised and so, no such colour change is seen. (ans) [With KMnO4/H+, HCHO is oxidised to give CO2(g), which forms a white ppt with limewater.] © Step-by-Step 07N-14 2007 Nov Paper 3 4. [07N P3 Q04 Ionic Eqm/Kinetics/Aldehyde] (a) (i) HCN(aq) H+(aq) + CN–(aq) Ka = [H + ][CN - ] [HCN] • Let the rate equation be rate = k [(CH3)2CO] [NaCN] [HCN]n Using values from Expt 2 and 4, 1.25 = k (0.050)(0.008)(0.040)n (I) 0.94 = k (0.050)(0.006)(0.050)n (II) k (0.050)(0.008)(0.040)n (I) ÷ (II), 1.25 = 0.94 k (0.050)(0.006)(0.050) n 1.33 = 1.33 × (0.8) n (0.8) n = 1 ∴ n=0 ∴ reaction is zero order w.r.t. HCN. (ans) • rate equation is rate = k [(CH3)2CO] [NaCN] (ans) (ans) (ii) Since the degree of dissociation, α , is very small, [HCN]eqm ≈ [HCN]0 ∴Ka = [H + ]2 since [CN–] = [H+] [HCN] [H+] = K a × [HCN] 4.9 ×10−10 × 0.100 = = 7.00 × 10–6 mol dm–3 (ans) −6 ∴% HCN molecules ionised = 7.00 ×10 × 100 0.100 = 0.00700 % (ans) (iii) HCN H+ + CN– NaCN → Na+ + CN– (I) common ion The addition of NaCN increases the [CN– ]. The presence of this common ion shifts the position of equilibrium (I) to the left and so, less HCN molecules dissociate. Hence the percentage of HCN molecules ionised is decreased. (ans) (iii) Mechanism: nucleophilic addition with CN– as nucleophile. The rate equation shows that only one molecule of (CH3)2CO and one molecule of NaCN are involved in the slow step. HCN being a weak acid is a poor source of CN– ion. Hence, a small amount of NaCN is added to accelerate the reaction. NaCN → Na+ + CN– O– CH3 .. CN– δ- C O slow CH3 δ+ C CH3 CH3 CN + HCN (b) (i) The order of reaction with respect to a given reactant is the power to which the concentration of that reactant is raised to in the experimentally determined rate equation, e.g. In the reaction, aA + bB → products rate equation is rate = k [A]m [B]n where m = order of reaction with respect to A n = order of reaction with respect to B m + n = overall order of reaction. (ans) (ii) Comparing Expt 1 and 2, [HCN] and [NaCN] constant, when [(CH3)2CO] increases from 0.4 to 0.5 mol dm–3 (or factor of 00..45 or 54 ), initial rate also increases by a factor of 1.25 1.00 or 5 4 . st ∴ reaction is 1 order w.r.t. (CH3)2CO. (ans) • Comparing Expt 1 and 3, [(CH3)2CO] and [HCN] constant, when [NaCN] decreases from 0.008 to 0.006 mol dm–3 (or factor of 00..006 008 or 3 4 OH CH3 C CH3 + CN– (ans) CN (iv) Rate of reaction with HCN: CH3CH2 C H O > CH3 C O CH3 Propanal reacts at a faster rate because the carbonyl C atom has a larger δ+ charge (it has one less electron-donating alkyl group) and there is less steric hindrance than in propanone (which has two bulky CH3 groups). (ans) (v) HCN is the reagent for the nucleophilic addition reaction; it acts as a Bronsted acid. NaCN acts as a catalyst in the reaction; it provides the CN– nucleophile initially and is regenerated in the reaction. (ans) ), initial rate also decreases by a factor of 0.75 1.00 or 3 4 . ∴ reaction is 1st order w.r.t. NaCN. A-Level Solutions – Chemistry 2007 Nov Paper 3 OH (c) CH3CH2 C CH3CH2 *C H O + HCN H CN 5. [07N P3 Q05 Proteins] (a) (i) Requirements needed for two molecules to form a hydrogen bond between them are: Optical isomerism occurs since the product obtained has a chiral carbon (marked by *). Isomers D and E are optical isomers. CN CN C C H OH 1. 2. H HO δ+ OH • CH3CH2 hot C H Al2O3 CH3CH=CHCN H CH3 CN cis–isomer • CN H C=C CH3 H trans–isomer Each of the two isomers D and E produce the same mixture of cyanoalkenes because the same reaction (dehydration) takes place in both cases. Both isomers can each lose an H and the OH from adjacent carbon atoms in two different ways, thereby giving the cis and trans compound. (ans) [or since the reagents used are non-chiral, the reactions involved are identical.] © Step-by-Step A-Level Solutions – Chemistry δδ+ δ- H ′′′′′′′′′′′′′ : Hδ + Hδ + N H Dehydration occurs, and the product obtained shows cis-trans isomerism. Isomers F and G are cis-trans isomers. C=C N H H hydrogen bonding CN H an electron-deficient hydrogen atom bonded to a very electronegative atom (such as N, O or F) in one molecule, and a lone pair of electrons on a very electronegative atom (such as N, O or F) in the other molecule. e.g. hydrogen bonds between NH3 molecules .. CH2CH3 CH2CH3 07N-15 δ+ (ans) (ii) The intermolecular forces in carboxylic acids (RCO2H) comprises both van der Waals' forces (between the hydrocarbon, R, chains) and hydrogen bonding (between –CO2H groups), while that between water molecules are mainly hydrogen bonding. • The first four members of δ -O the series of carboxylic H δ δ - ′′′ ′′ + acids are fully Hδ + ′′ δ + O′ soluble in water R C δ- δ+ because they O H ′′′′′′′′′′ form hydrogen Oδ - δ + H bonds with water molecules. • As the carbon chain length increases, the acids become increasingly insoluble because the molecule becomes less polar (or more hydrocarbon-like) and intermolecular van der Waals' forces become predominant. Hence, energy released when weak van der Waals' forces are formed between the acid and water molecules is insufficient to compensate for the energy required to break the strong intermolecular hydrogen bonding between water molecules. Hence, the dissolution of long chain carboxylic acids is endothermic and so, does not occur readily. (ans) Hδ + 07N-16 2007 Nov Paper 3 (b) (i) The primary structure of a protein shows the order (or sequence) of amino acid residues in a protein (or polypeptide chain). (ans) (ii) A section of the protein chain of HSA using serine, cysteine and phenylalanine: O (v) Two other types of side-chain interaction: 1. hydrogen bonding between polar groups, e.g. between serine and lysine (marked by 2. H C H H H O C C N C C H C H H H O N C C N H H O S S S H C CH3 CH3 2 CH2 CH2 CH2OH (ans) H CH2 (iii) A polypeptide chain is held in the shape of an α - helix by hydrogen bonds between the N–H group of each amino acid unit and the fourth C=O group following it along the chain. There are 3.6 amino acids per turn and the pattern repeats itself every five turns. The hydrogen bonds in α - helix are linear and so, are α - helix maximally stable. (ans) (iv) The thiol group in cysteine is oxidised to disulphide in the presence of oxidising agents such as oxygen or hydrogen peroxide. –CH2–SH + HS–CH2– + 1 2 ). van der Waals' attraction between non-polar groups, e.g. between phenylalanine and valine (marked by 2 ). H H C H 1 O2 → –CH2–S–S–CH2– + H2O (ans) [or –CH2–SH + HS–CH2– + [O] → –CH2–S–S–CH2– + H2O or –CH2–SH + HS–CH2– → –CH2–S–S–CH2– + 2[H] 1 (CH2)4NH2 S S (ans) [or ionic bonding between charged groups; e.g. between glutamic acid (RCH2CH2CO2–) and lysine (RCH2CH2CH2CH2NH3+).] (vi) The long carbon chain on stearic acid is hydrophobic. Hence two amino acids which would interact with the long chain on stearic acid are leucine and valine. (ans) [or phenylalanine, all of which interact with stearic acid by hydrophobic (van der Waals' ) interactions.] (vii) Three amino acids likely to interact strongly with water molecules are glutamic acid, lysine and threonine. (ans) [or serine, all of which interact strongly with water molecules by forming hydrogen bonds with water molecules.] © Step-by-Step or –CH2–SH + HS–CH2– → –CH2–S–S–CH2– + H2 ] [NB. the –CH2– chain was required.] • Since there are 36 cysteine residues per molecule of HSA, ∴Within each HSA molecule, there could be 36 2 or 18 disulphide bridges. (ans) A-Level Solutions – Chemistry