NJC_H2_CHEM_P3_Solutions Prelim

NJC SH2 H2 Chemistry Preliminary Exam Paper 3 Mark Scheme 1 (a) (i) 4MnO4 – + 5CH3CH2OH + 12H+ 4Mn2+ + 5CH3COOH + 11H2O (ii) No. of moles of KMnO4 = 0.20 x 17.80/1000 = 0.00356 mol No. of moles of ethanol = 0.00356 x 5/4 = 0.00445 mol Mass of ethanol = 0.00445 x 46 = 0.205 g (iii) Volume of ethanol in beer sample = 0.205/0.789 = 0.259 cm3 Alcohol content of beer = 0.259/10 x 100% = 2.59% (iv) Some ethanol might have evaporated as it is volatile. Or some ethanol might be oxidised by air. (v) End-point is reached when the solution first turn pink. (vi) As the carbon chain length increases, the temporary dipole-induced dipole interactions in alcohol become stronger and the energy released from the formation of hydrogen bonding between alcohol and water is insufficient to overcome the stronger intermolecular forces of attraction between the alcohol molecules. (b) (i) C6H12O6  2CO2 + 2CH3CH2OH (ii) Add I2(aq) and NaOH(aq) to glucose and ethanol in separate test -tubes, warm the mixture. Glucose: No yellow precipitate formed. Ethanol: Yellow precipitate is formed. Equation: CH3CH2OH + 4I– + 6OH–  HCO2 – + CHI3 + 5I– + 5H2O (c) (i) Nucleophilic substitution C Cl H H R' RCH2O - : + - C Cl H H R' RCH2O - CRCH2O H H R' + Cl -

(ii) C C H H H H C C H H H HH Cl C C H H H H C C H H H HH OH H2O(g), conc H3PO4, 65 atm, 300 °C HCl(g) Na(s) C C H H H HH O - C C H H H HH Cl + C C H H H HH O - CH3CH2OCH2CH3 Reagents and conditions of electrophilic addition of water to alkene can be replaced with conc.H2SO4, followed by H2O; warm 2 (a) (i) MgO + SO2  MgSO3 Neutralisation or acid-base reaction (ii) Yes, since Al2O3 is amphoteric in nature, so can react with an acidic oxide such as SO2 to remove it. (b) (i) MgO in water: pH 8 – 9 - Weakly alkaline solution formed as Mg(OH)2 is only slightly soluble in water (due to strong ionic bonds between Mg2+ and OH− ions) Al2O3 in water: pH 7 - Solution is n eutral as Al2O3 is insoluble in water (due to strong ionic bonds between Al3+ and O2− ions). SO2 in water: pH 2 - 3 - Solution is acidic due to the formation of weak acid H2SO3(aq) SO2(g) + H2O(l)  H2SO3(aq) H2SO3 H+ + HSO3  (ii) Al2O3 + 2NaOH + 3H2O  2 Na[Al(OH)4] (c) (i) Step 1: Br2 and anhydrous AlBr3 catalyst Step 2: Al2O3; heat

(ii) Electrophilic substitution Step 1: AlBr3 + Br2 → [AlBr4]− + Br+ Step 2: CH2CH(CH3)CH2OH Br + slow H Br CH2CH(CH3)CH2OH+ Step 3: [AlBr4] - H Br CH2CH(CH3)CH2OH+ .. Br CH2CH(CH3)CH2OH + AlBr3 + HBr (iii) H2 Br O C CH3 (d) (i) [Al(H2O)6]3+(aq) + H2O(l) [Al(H2O)5(OH)]2+(aq) + H3O+(aq) Ka = ]O)[Al(H ]O][H(OH)O)[Al(H +3 62 + 3 +2 52 or ]O)[Al(H ]][H(OH)O)[Al(H +3 62 ++2 52 10−5.01 = (10−2.65)2 [𝐴𝑙(𝐻2𝑂)6 3+]−10−2.65  [Al(H2O)6 3+] = 0.5151 mol dm−3 [Also accept if 10−5.01 = (10−2.65)2 [𝐴𝑙(𝐻2𝑂)6 3+]  [Al(H2O)6 3+] = 0.5129 mol dm−3 AlCl3(aq)  Al3+(aq) + 3Cl(aq) Since Al3+(aq