TJC Prelim P3 ANS

OH OH CH2CH=CH2 OH Br Br Br Answers to H2 Chemistry Preliminary Examinations 2009 Paper 3 1 (a) • AlF3 exists as giant ionic lattice with strong ionic bonds between the ions. Large amount of energy of energy is required to overcome the strong electrostatic attraction between oppositely charged ions and hence, a high melting point. • AlCl3 exists as a simple molecular structure . Melting involves overcoming the weak Van der Waals forces of attraction. Little energy is required to overcome the intermolecular forces of attraction and thus, A lCl3 has a lower melting point as compared to NaCl. (b) (i) • High C:H ratio infers that P is an aromatic compound. • P • With anhydrous aluminium chloride P undergoes electrophilic substitution reaction to form Q. As the –OH is an activating group, the substituent will be directed to the ortho or para position. Q OR • P undergoes electrophilic substituition as it decolourises aqueous bromine to form white ppt R. R is 2,4,6-tribromophenol as -OH is a highly activating group. • Q react with aqueous bromine to form white ppt S, with the same no of bromine atoms per molecule of R. As Q has a C-C double bond, 15 OH CH2CH=CH2

CH2CH(OH)CH2Br OH Br Br OH CH2CH=CH2 electrophilic addition takes place as well. This implies that 1 Br will be attached to one of the carbons of the C-C double bonds. Thus, only 2 Br will be attached directly to the benzene. • S S is optically active because of its chiral carbon. The mirror image of S is non-superimposable. (ii) • Electrophilic Substitution • • Mechanism Step 1: Generation of electrophile CH2=CHCH2Cl + AlCl3 CH2=CHCH2 + + [AlCl4]– Step 2: Electrophilic Attack Step 3: Loss of proton + [AlCl4]– + HCl + AlCl3 Any 2 correct steps – Max 1 mark given. 3 correct steps – 2 marks awarded. 16 OH H CH2CH=CH2 + OH CH2CHCH2+ OH H CH2CH=CH2 + slow fast

(c) NCl3 (l) + 3H2O (l) → NH3 (aq) + 3HOCl (aq) PCl3 (l) + 3H2O (l) → H3PO3 (aq) + 3HCl (aq) • Nitrogen is in period 2 and it cannot expand its octet while phosphorus is in period 3 and can expand its octet structure. • P is able to use its energetically accessible 3d subshell for dative bonding with H2O in the reaction/accommodate of the lone pair of electrons from H 2O during the nucleophilic attack of H2O on PCl3. (d) (i) Ksp = [Pb2+][ CrO4 2-] Let the solubility of PbCrO4 be x mol dm-3 PbCrO4 (s) Pb2+ (aq) + CrO42- (aq) Ksp = [Pb2+][ CrO4 2-] 1.69 × 10-14 = x2 x = spK = 1.30 x 10-7 mol dm-3 The solubility of PbCrO4 is 1.30 x 10-7 mol dm-3. (ii) [Pb2+] = 2 4[ ] spK CrO − = -141.69×10 0.010 = 1.69 x 10-12 mol dm-3 2 (a) Order with respect to E = 1 Comparing experiment 1 and 3 x x 1 1 x x 3 3 x [E] [NaOH] (0.20)(0.30) 0.0150 [E] [NaOH] (0.20)(0.20) 0.0150 0.20 0.30 0.0150 0.20 0.20 0.0150 = =     × =       OR When [E] is constant but [OH-] is decreased by 1/3 the rate remains the same 17

• x = 0 Order with respect to F = 1 Comparing experiment 2 and 3 y y 2 2 y y 3 3 y [F] [NaOH] (0.10)(0.40) 0.0080 [F] [NaOH] (0.20)(0.60) 0.0240 0.10 0.40 0.0800 0.20 0.60 0.0240 = =     =       • y = 1 (b) (i) • When the temperature of the reaction increases (pressure remaining the same), the reactant particles gain more kinetic energy and hence move about more rapidly. There is a greater frequency of collision between reactant particles and more reactant particles have energy greater than or equal to the activation energy. • There are more effective collisions between reactant particles giving rise to an increase in the rate of reaction. • 18 No. of molecule s Ea activation energy Energy No. of reactant molecules with energy ≥ Ea at 120oC No. of reactant molecules with energy ≥ Ea at 90oC

(ii) • When the pressure of the reaction increases (temperature remaining the same), the rate of reaction remains the same as the reactants are in the aqueous state and are not affected by pressure. Only the rate of reaction for gaseous reactants are affected by changes in pressure. (c) (i) • Compound F is optically active  Compound F has at least one chiral centre, no plane of symmetry and non-superimposable mirror images. • Compounds E & F do not react with Na metal  Both Compounds E & F do not have –OH groups  Compounds E & F are not alcohols • Compounds E & F give a positive Tollens’ Test  Compounds E & F are both aldehydes • E undergoes nucleophilic substitution with NaOH to form G. Since the reaction is first order with respect to E but zero order with respect to NaOH  The reaction is an SN1 reaction •  E is likely to be a tertiary halogenoalkane and G is a tertiary alcohol • F undergoes electrophilic substitution with NaOH to form H. Since the reaction is first order with respect to F and with respect to NaOH  The reaction is an SN2 reaction •  F is like to be a primary halogenoalkane and H a primary alcohol • E is H3C C C O H CH3 Cl • F is C H O *CCH2Cl CH3 H 19

• G is H3C C C O H CH3 OH • H is C H O *CCH2HO CH3 H [ (ii) • H3C C C O H CH3 Cl + 2Ag+ + 3OH- H3C C C O O- CH3 Cl + 2Ag + 2H2O (d) Mechanism: SN2 nucleophilic substitution •• C Cl H OHC(H3C)HC HHO slow C H CH(CH3)CHOH ClHO fast HO C H CH(CH3)CHO H + Cl 3 (a) (i) At the surface of the sea, pressure of CO2 is 1 atm (normal atmospheric pressure) At 300 m below sea level, the pressure of CO2 = 1 10 300 + = 31 atm 20 δ+ δ– •

• Solubility of CO2 300 m below sea level = 2-103.29 1 31 ×× = 1.02 mol dm-3 (ii) (I) • [H+] = CKa × = 1.02104.5 -7 ×× = 6.77 x 10-4 mol dm-3 • pH = -lg(6.77 x 10-4) = 3.17 (II) • C O x x XX O XOXH (III) • Trigonal planar (iii) (I) • Ka2 = ]CO[H ]][H[HCO 32 - 3 + (II) Ka1 = ][CO ]][H[HCO 2 3 +− = 4.5 x 10-7 mol dm-3 Ka2 = ]CO[H ]][H[HCO 32 - 3 + • = ]CO[H ][CO ][CO ]][H[HCO 32 2 2 - 3 × + = 4.5 x 10-7 x 400 • = 1.80 x 10-4 mol dm-3 21

(III) C O O HO + H+C O OH HO C O OH H3CH2C C O O H3CH2C + H+ • –OH group on the HCO 3 - ion is electron withdrawing and decreases the negative charge density on the COO- group making the ion more stable. • However CH3CH2– group on the CH 3CH2CO2 - ion is electron releasing and increases the negative charge density on the COO - group making the ion less stable. H2CO3 thus dissociates more readily to form the more stable HCO 3 - ion, causing it to be a stronger acid. (b) (i) •• dilute HCl or dilute H2SO4 , reflux NaOH(aq) , reflux (ii) •• In acid hydrolysis CH H3C H3C OH HO P F O CH3 OR In alkaline hydrolysis CH H3C H3C OH and (iii) • Less resistant • P-O bond length is longer than C-O, lower bond energy and easier to break OR P is more susceptible to nucleophilic attack due to the highly electronegative F. 22 and +Na-O P F O CH3

(iv) • Carbon, being a period 2 element does not have energetically accessible vacant d-orbitals to allow it to expand its octet structure and accommodate the electrons from fluorine. (v) •• Any 2 of the following structures (CH3O)PCl2 OR (CH3O)2PCl OR (CH3O)P(Cl)(OH) OR (CH3O)P(OH)2 OR (CH3O)3P OR P(Cl2)(OH) OR P(OH)3 4 (a) (i) MgO (s) + 2HNO3 (aq) → Mg(NO3)2 (aq) + H2O (l) Amount of MgO = 0124.00.163.24 500.0 =+ mol • Amount of HNO3 = 00500.0100.01000 50 =× mol (limiting reagent) • Amount of Mg(NO3)2 produced = 00250.02 005.0 = mol (ii) Mg(OH)2 + 2NH4NO3 → Mg(NO3)2 + 2NH3 + 2H2O • Amount of Mg(NO3)2 = 0101.00.960.283.24 5.1 =++ mol • Volume of NH3 produced = 2420101.0 ×× = 0.485 dm3 (iii) 2Mg(NO3)2 → 2MgO + 4NO2 + O2 Brown gas of NO2 will be observed. (iv) • Ba(NO3)2 is more ther