RVHS Prelim P3 Soln

1 River Valley High School 9729/03 [Turn over 2018 Preliminary Examinations Section A 1 (a) (i) Element E⦵ / V ½F2 + e− ⇌ F− +2.87 ½Cl2 + e− ⇌ Cl− +1.36 ½Br2 + e− ⇌ Br− +1.07 ½ I2 + e− ⇌ I− +0.54 The order of oxidising strength of the halogens, as observed from the E⦵ values above, decreases down the Group. Due to the strong oxidising power of F 2, the oxidation number of Phosphorus +5 in PF5, +5 and +3 in PC l5 and PCl3, +3 in PBr3 and PI3 respectively. [3] (ii) The smaller the p Ka, the stronger the acid, indicating H I is the strongest acid, followed by HBr, HC l and HF. Down the group from F to C l to Br to I, atomic radius increases, effectiveness of orbital overlap between H and X decr eases. The H–X bond becomes increasingly weaker, making it easier to lose the H+. [2] (b) (i) (Acidic) hydrolysis [1] (ii) O and CH3MgBr [2]

2 River Valley High School 9729/03 [Turn over 2018 Preliminary Examinations (iii) [3] (c) (i) Ksp = [Ca2+]3[PO43–]2 [1] (ii) [Ca2+] = 1.14 × 10–7 × 3 = 3.42 × 10–7 mol dm–3 [PO43–] = 1.14 × 10–7 × 2 = 2.28 × 10–7 mol dm–3 Ksp = (3.42 × 10–7)3(2.28 × 10–7)2 = 2.08 × 10–33 mol5 dm–15 [2] (iii) Let the solubility of calciu m phosphate in the presence of potassium phosphate be y. 2.08 × 10–33 = (3y)3(0.15 + y)2 Assume that y is small such that 0.15 + y ≈ 0.15 2.08 × 10–33 = (3y)3(0.15)2 y = 1.51 × 10–11 mol dm–3 [2] (iv) An increase in [SO 42–] causes the equilibrium position of CaSO4 ⇌ Ca 2+ + SO 42– to shift left, decreasing the solubility of CaSO4. Since the solubility of CaSO 4 is lowered, there will be less Ca 2+ from dissolution of CaSO 4. Equilibrium position of Ca3(PO4)2 ⇌ 3Ca2+ + 2PO43– shifts right, increasing the solubility of Ca3(PO4)2. [2] [Total: 18] δ+ δ–

3 River Valley High School 9729/03 [Turn over 2018 Preliminary Examinations 2 (a) (i) NaOH(aq), heat under reflux [1] ( i i ) The C–Cl bond has a partial double bond character as the lone pair of electrons in the p orbital of the C l atom can delocalise into the π electron cloud of the benzene ring. This strengthens the C–C l bond, making it more difficult to break. [1] ( i i i ) [1] ( i v ) Add PCl5 / SOCl2 2,4-dichlorobenzyl alcohol: white fumes (of HCl) observed amylmetacresol: no white fumes observed or Add Br2(aq) 2,4-dichlorobenzyl alcohol: Br2(aq) remains orange amylmetacresol: orange Br2(aq) decolourises or Add neutral FeCl3 2,4-dichlorobenzyl alcohol: no violet complex formed amylmetacresol: violet complex formed [2]

4 River Valley High School 9729/03 [Turn over 2018 Preliminary Examinations (b) Menthol < amylmetacresol < tartaric acid Menthol is the weakest acid. is the least stable anion as the electron-donating alkyl gr oup intensifies the negative charge on the oxygen atom in the anion. Amylmetacresol is more acidic than menthol as the anion is more stable. The negative charge on the oxy gen atom can be delocalised into the benzene ring. Tartaric acid is the most acid ic as the anion is the most stable. The negative charge on the oxygen atom is delocalised over the COO– group, and the delocalisation is mo re effective than that in the anion. [3] (c) (i) Step 1: electrophilic substitution / electrophilic addition Step 2: reduction (Note: for step 1, the catalyst aids in the formation of CH 3CHCH3+ electrophile) [2] ( i i ) 8 enantiomers Proportion = 0.125 / 12.5%/ 1/8 [2]

5 River Valley High School 9729/03 [Turn over 2018 Preliminary Examinations ( i i i ) (Accept: benzoate salt) [1] (d) (i) When 0.00 cm3 of NaOH is added, solution contains only tartaric acid. [H+] = ඥሺ0.0100ሻሺ10ିଶ.଼ଽሻ = 3.59 × 10–3 mol dm–3 pH = –lg(3.59 × 10–3) = 2.45 When 12.50 cm 3 of NaOH is added, solution contains equal concentration of unreacted tartaric acid and tartrate mono-anion, giving rise to a buffer at its maximum buffering capacity. Hence, pH = pKa1 = 2.89. When 50.00 cm 3 of NaOH is added, solution only contains tartrate di-anion which undergoes hydrolysis to form OH–. [Tartrate di-anion] = ( ଶହ.଴଴ ଵ଴଴଴ × 0.0100) ÷ ଻ହ.଴଴ ଵ଴଴଴ = 0.00333 mol dm–3 [OH–] = ටሺ0.00333ሻሺ ଵ.଴଴×ଵ଴షభర ଵ଴షర.రబ ሻ = 9.15 × 10–7 mol dm–3 pOH = –lg(9.15 × 10–7) = 6.04 pH = 14 – 6.04 = 7.96 [4] ( i i ) The solid has a giant ionic lattice st ructure while tartaric acid has a simple covalent structure. More energy is needed to overcome the stronger electrostatic attraction be tween the cations and anions than the weak hydrogen bonds between tartaric acid molecules. [2] (e) (i) Cold dilute acidified / alkaline KMnO4(aq) [1] ( i i ) [1]

6 River Valley High School 9729/03 [Turn over 2018 Preliminary Examinations ( i i i ) The gas is CO2. pV = nRT p(500 × 10–6) = ( ଴.ଷ଺ ସସ.଴)(8.31)(30 + 273) p = 4.12 × 104 Pa [2] [Total: 23]

7 River Valley High School 9729/03 [Turn over 2018 Preliminary Examinations 3 (a) It is a substance that is at a different phase as the reactants, and it speeds up the rate of the reaction by providin g an alternative reaction pathway with a lower activation energy. [2] (b) Zn/Zn2+ half-cell E⦵ = −0.76 V VO3−/ VO2+ half-cell E⦵ = +1.00 V VO2+/ V3+ half-cell E⦵ = +0.34 V V3+/V2+ half-cell E⦵ = −0.26 (VO3−/ VO2+// Zn/Zn2+) E⦵ cell = +1.76 V Zn is able to reduce VO 3− to VO 2+/ reduction of VO 3− to VO 2+by Zn is spontaneous. (VO2+/ V3+// Zn/Zn2+) E⦵ cell = +1.10 V Zn is able to reduce VO 2+ to V 3+/ reduction of VO 2+ to V 3+ b y Z n i s spontaneous. (V3+/ V2+// Zn/Zn2+) E⦵ cell = +0.50 V Zn is able to reduce V3+ to V2+/ reduction of V3+ to V2+ by Zn is spontaneous. Hence, the solution changes from yellow to blue to green to violet. [3] (c) (i) V3+(aq) [1] ( i i ) 5VO2+ + MnO4− + 6H2O → 5VO3− + Mn2+ + 12H+ OR 5VO2+ + MnO4− + H2O → 5VO2+ + Mn2+ + 2H+ [1] ( i i i ) ݊௄ெ௡ைర = 20.63 1000 ×0 .02 =0 .000413݈݋݉ [1]

8 River Valley High School 9729/03 [Turn over 2018 Preliminary Examinations ( i v ) ݊௏యశ +݊௏ைమశ = ଶହ ଵ଴଴଴ ×0 .05 =0 .00125 mol ݊௏యశ =0 .00125ݔ ݊௄ெ௡ைర reacted with VO2+ = ଵ ହݔ ݊௄ெ௡ைర reacted with V3+ = ଶ ହ ሺ0.00125ሻ ଶ ହ ሺ0.00125ሻ + ଵ ହݔ0 .000413 0.0025ݔݔ+0 .002063 ݔ0 .000437 mol In 100 cm3 of solution: ݊௏యశ =4 ሺ0.00125 −0 .000437ሻ =0 .003252 mol ݊௏ைమశݔ0 .001748 mol ݉௓௡ =0 .5[ሺ65.4ሻሺ0.001748ሻ + ሺ65.4ሻሺ2ሻሺ0.003252ሻ] =0 .270 g [4] (d) C:H ratio of A, B, C and D ≈ 1:1 - Contains a benzene ring Compound A undergo oxidative esterification: - A has an aldehyde group / phenol / alcohol group. - B has an ester group (also accept B being ester due to neutral nature) Compound B undergo (alkaline) hydrolysis to form the sodium salt of C - B is a cyclic ester - C has a carboxylic acid and alcohol / phenol group. A, B and C undergo reduction to form D - D contains 2 –OH group, of which one is a primary alcohol. A, B and D can undergo electrophilic substitution with Br2(aq) - A, B and D contains a phenol [7]

9 River Valley High School 9729/03 [Turn over 2018 Preliminary Examinations O O A B C D [Total: 20]

10 River Valley High School 9729/03 [Turn over 2018 Preliminary Examinations Section B Answer one question from this section. 4 (a) (i) Base [1] (ii) SN2 nucleophilic substitution [3] (iii) Phenylamine is a weaker base. The l one pairs of electrons on the –NH 2 group is delocalised into the benzene ri ng, this makes the lone pair less available for protonation. [1] (iv) Step 2: ethanal Step 4: Acidified K2Cr2O7, heat (under reflux) [2] (v) The C=C is electron rich hence C=C does not attract CN– nucleophiles. The carbonyl C in C=O is electron-def icient (or electrophilic) as it is bonded to highly electronegative O atom, attracting CN– nucleophiles. [3] (b) (i) Lattice energy is the enthalpy change when one mole of the solid ionic compound is formed from its consti tuent gaseous ions under standard conditions. [1] δ−δ+