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)  [Al(H2O)6]3+ ∴ Mass of AlCl3 = (0.5151) ( 20 1000) × 133.5 [or (0.5129) ( 20 1000) × 133.5 = 1.37 g] = 1.38 g (ii) Mg2+ has a lower charge/size ratio tha n Al3+; so will polarise the O -H bond in water ligand to a lesser extent; thus is a weaker acid and a higher pKa value than aq Al3+

(iii) Mg3(PO4)2 (s) 3Mg2+ (aq) + 2PO4 3−(aq) [Mg2+]3 [PO4 3−]2 = (0.10)3[PO4 3−]2 = 1.3 × 10−16 ∴ [PO4 3−]min required for Mg3(PO4)2 to be precipitated = 3.61 × 10−7 mol dm−3 (3 s.f.) AlPO4(s) Al3+(aq) + PO4 3−(aq) [Al3+][PO4 3−] = (0.50) [PO4 3−] = 6.3 × 10−19 ∴ [PO4 3−]min required for AlPO4 to be precipitated = 1.26 × 10−19 mol dm−3 (4 s.f.) To separate the two ions, add PO4 3− till [PO4 3−]  3.61 x 10–7 mol dm–3. Filter the mixture. Mg2+ will remain in the filtrate, while the residue is AlPO4. 3 (a) Mg reacts very slowly with cold water but readily with steam. Ba reacts vigorously with cold water. As Eo(Ba2+/Ba) is more negative, Ba has a higher tendency to be oxidi sed than Mg. Thus the reaction takes place more readily. (b) (i) 2Sr(NO3)2(s) → 2SrO(s) + 4NO2(g) + O2(g) (ii) From Mg to Ba, the ionic radius increases from 0.065 nm to 0.135 nm. As the cations have the same charge, the charge size decreases, leading to a decreasing polari sing power. The large electron cloud of nitrate ion is distorted to a smaller extent down the group and N-O bond is weakened less and less easily broken. Thus thermal stability of nitrates increases from Mg(NO3)2 to Ba(NO3)2. (c) As Ca2+ has a smaller ionic radius than Sr 2+, it has a higher charge size and is more extensively hydrated by water molecules. This produces more drag / resistance and hence its ionic speed is lower than expected. (d) (i) Elimination. (ii) Base (iii) Comparing experiments 1 and 2, when [NaOCH 3] increases 1.5 times and [2-bromopropane] is constant, rate increases 1.5 times, showing that it is a 1st order reaction with respect to NaOCH3. Comparing experiments 1 and 3, when [NaOCH 3] doubles and [2 -bromopropane] increases 4/3 times, the rate increases 8/3 times, showing that it is a 1st order reaction with respect to 2-bromopropane. (can also be shown via mathematical method) Let the n be the order of reaction with respect to (wrt) 2-bromopropane

C CH3 Br CH H H H CH3O+ C C + Br + CH3OH H H H CH3 0.06 (0.100)(0.150) 0.16 (0.200)(0.200) 3 1 3 ()8 2 4 33 ()44 1 n n n n n     rate = k [NaOCH3] [2-bromopropane] rate constant, k = 4 mol–1 dm3 min–1 (iv) marking points:  lone pair of electrons on O  direction of arrows (arrows must start from bond/lone pair of electrons and end on atoms or between two atoms where a pi bond is formed)  only one step (v) Yes, it is consistent as the mechanism shows 1 molecule of 2 -bromopropane reacting with 1 CH3O– ion in the rate-determining step. (vi) I: C–X bond is broken in the rate-determining step. C–Cl bond is stronger than C–Br bond as the C–Cl bond length is greater. When 2 -chloropropane is used, the C –Cl bond is less easily broken and the rate is lower. II: The number of electron -donating alkyl group remains the same in both CH3CH2O– and CH 3O–, ther efore the electron density of the lone pair remains the same and both bases are equally strong. Hence there is no effect on the rate. 4 (a) (i) BaSO4 4Br2(aq) + S2O3 2−(aq) + 5H2O(l)  8Br −(aq) + 2SO4 2−(aq) + 10H+(aq) (ii) Br2 + e− ⇌ 2Br– Eo = +1.07 V I2 + e− ⇌ 2I− Eo = +0.54 V From Eo values, Br 2 is a stronger oxidising agent than I 2 and can oxidise S 2O3 2− to SO4 2−. The oxidation state of sulfur increases from +2 in S2O3 2− to +6 in SO4 2−. Iodine, a weaker O.A, can only oxidise S 2O3 2− to S4O6 2−. The oxidation state of sulfur

increases from +2 in S2O3 2− to +2.5 in S4O6 2−. I2(aq) + 2S2O3 2−(aq)  2I−(aq) + S4O6 2−(aq) Hence there is no white precipitate observed. (b) (i) (ii) Reactions at anode: 2H2O(l)  O2(g) + 4H+(aq) + 4e 2Al(s) + 3/2O2(g)  Al2O3(s) (iii) Volume of Al2O3 = 25.20 x 0.025 = 0.630 cm3 Mass of Al2O3 = 3.95 x 0.630 = 2.49 g No of moles of O2 = 3/2 x No of moles of Al2O3 = 3/2 x 2.489 2 ×27.0 +3 × 16.0 = 0.0366 mol No of moles of electrons passed = 0.0366 x 4 = 0.146 mol Q = 0.146 x 96500 = 14126 C Time needed = 14126  1.8 = 7850 s (c) Information Deduction B is optically active.  B has a chiral carbon B does not react with Na2CO3  B does not have a carboxylic acid. B dissolves slowly in aq. NaOH  B has phenol.  B undergoes neutralisation B reacts with potassium dichromate  B undergoes oxidation.  B has primary or secondary alcohol Pt electrode dilute H2SO4 Al metal

B decolourises aq. Bromine to form white ppt.  B is a phenol.  B undergoes Electrophilic substitution.  From molecular formula of C, one of C-2 or C-4 positions is occupied. B reacts with concentrat