PJC H2 Chemistry 2008 JC2 H2 Chemistry Prelim Paper 2 Suggested Answers

1 Pioneer Junior College JC2 Preliminary Examination 2008 H2 Chemistry Paper 2 Suggested Answers 1 (a) (i) Both N2 and P 4 have simple molecular structure . They are non polar molecules, held together by weak van der waals forces due to induced dipole – induced dipole attractions. A s P 4 has a greater number of electrons / larger electron cloud to be polarised leading to stronger van der waals forces between P 4 molecules. This results in higher melting point in P4, hence P4 exists as solid. (ii) Phosphorus is a relatively big at om with diffused orbitals, side-on overlap of its p orbitals to form π bonds is much less effective than head-on overlap to form sigma bond. (b) (i) N O O O - 120o 3- 109.5o P O O O O Shape of NO3 - is trigonal planar. Shape of PO 4 3- is tetrahedral. (ii) To form NO 4 3-, N must be able to accommodate 10 electrons in its valence shell. Since N is in Period 2, it has no energetically accessible/low lying d orbitals to expand its octet. 2 (a) SrF2(s) + aq ⇌ Sr2+(aq) + 2F-(aq) Solubility of strontium fluoride = 0.073 / 125.6 = 5.81 x 10-4 mol dm-3

2 K sp = [Sr2+]eqm[F-]eqm 2 = (5.81 x 10-4) (2 x 5.81 x 10-4)2 = 7.85 x 10-10 mol3 dm-9 (b) [Sr2+] = (0.0100 x 25.0/1000) ÷ (45.0/1000) = 0.00556 mol dm-3 [F -] = (0.0150 x 20.0/1000) ÷ (45.0/1000) = 0.00667 mol dm-3 Ionic product of SrF 2 = (0.00556) (0.00667)2 = 2.47 x 10-7 mol3 dm-9 Since ionic product is greater than solubility product, precipitate of SrF 2 will form. 3 (a) CH 2C O O - CH 2C O O - NC H 2 CH 2 N CH 2 CH 2 C O O - C O O - edta (b) (i) On dropwise addition of NH 3(aq), blue precipitate of Cu(OH) 2(s) is observed. Equilibrium (1): Cu2+(aq) + 2OH-(aq) ⇌ Cu(OH)2(s) On adding excess NH 3(aq), blue precipitate dissolves to give a deep blue solution due to formation of [Cu(NH3)4(H2O)2]2+ complex. [ Equilibrium (2): Cu(OH) 2(s) + 4NH3(aq) + 2H2O ⇌ [Cu(NH3)4(H2O)2]2+ + 2OH-(aq) By forming the complex, [Cu2+(aq)] decreases, leading to decrease in ionic product of Cu(OH) 2 to the extent of ionic product < K sp. ∴ Precipitate dissolves.

3 O r Equilibrium (2): [ C u ( H 2O)6]2+(aq) + 4NH3(aq) ⇌ [Cu(NH3)4(H2O)2]2+ + 4H2O By forming the complex, [Cu2+(aq)] decreases, position of equilibrium (1) shifts to the left accordi ng to Le Chatelier’s Principle, ∴ precipitate dissolves. (ii) Type of reaction: Ligand exchange [ C u ( N H 3)4(H2O)2]2+ + edta4- → [Cu(edta)]2- + 4NH3 + 2H2O (c) (i) Due to high K c (position of equilibrium lie s more to the right), the poisonous Cd2+ can be removed by solution of edta through complex formation. (ii) Due to relatively similar K c values, Zn 2+ will be removed in addition to Cd2+. This is not desirable as zinc is essential for health. 4 (a) (i) rate = k[H2O2] (ii) 0 rate [H2O2] 1/4a 1/2a 1/8a t1/2 t1/2 t1/2 a t 0 [H2O2]

4 (b) (i) Co(II) acts as homogeneous catalyst due to its ability to exhibit variable oxidation states. (ii) Tartaric acid functions as a ligand (or a complexing agent) to complex with Co(III), thus stabilising it. 5 (a) (i) Reagents & Conditions: NaOH(aq), heat followed by Br 2 (aq) after cooling CH3 CH3 CONHCH2CH3 CH3 CH3 NHCOCH2CH3 and A B To each compound in separate test tubes, add NaOH(aq) and heat. Add Br 2(aq) after cooling the mixture. The test-tube containing A will decolourise reddish brown aqueous bromine with the formation of a white precipit ate. Reddish br own aqueous bromine remains for test tube containing B. CH3 C H3 Br NH2 CH3 C H3 NHCOCH2CH3 CH3 C H3 NH2 CH3 C H3 NH2 + NaOH(aq) + CH 3CH2COO-Na+ + Br2(aq) + HBr

5 (b) NH O NH O and C D To each compound in separate test tubes, add 2,4-dinitrophenylhydrazine. An orange precipitate will form in test-tube containing D and no precipitate observed for test-tube containing C. O NH H H H N N NO2 NO2 H NN NO2 NO2 N H + + H 2O 6 (a) (i) CH2CH(OH)CN NO2 A (ii) Step 1: concentrated HNO 3, concentrated H2SO4, 30°C Step 2: K2Cr2O7, H2SO4(aq), distil Step 3: HCN, trace amount of NaOH or NaCN, 10 – 20 °C Step 4: HC l(aq),heat (b) Step 3: nucleophilic addition Step 4: acid hydrolysis

6 (c) Electrophilic substitution HOH2CH2C NO2 + O2NH HOH2CH2C + NO2 HOH2CH2C + O2NH HOH2CH2C HNO3 + 2H2SO4 NO2 + + 2HSO4 - + H3O+ slow HSO4 - + H2SO4 fast [½] for equation for generation of catalyst 7 (a) Alkene, 3 ° alcohol, phenol and ketone (b) (i) Reddish brown aqueous bromine decolour ises with the formation of a white precipitate. O C H3 C H3 C H3 OH OH O H OH Br O H Br Br Br Br OH (ii) Purple KMnO4 turns green (MnO4 2-) and a brown precipitate of MnO2 is obtained eventually. O C H3 C H3 C H3 OH O - O - O - OH O H OH O H 8 (a) Molar mass of C 15H15NO2 = (12.0 x 15) + (1.0 x 15) + 14.0 + (16.0 x 2) = 241 g mol-1

7 Amount of C 15H15NO2 in 2 capsules = (50 x 10 -3 x 2) / 241 = 0.000415 mol (b) Half-life ≈ 2 hours (c) (i) Solubility of mefenamic acid = (20 x 10 -3) / 241 = 8.30 x 10-5 mol dm-3 (ii) Although mefenamic acid contains carboxylic acid group and amine group that are capable of form ing hydrogen bonds with water molecules, the presence of the tw o large hydrophobic aromatic rings results in its low solubility in water. (d) NaOH(aq) / Na2CO3(aq) at room temperature (e) Overdose = 740 mg kg -1 Mass of mefenamic acid that resu lt in an overdose in a 65 kg patient = 740 x 65 = 48100 mg = 48.1 g Time (hours) 2 hours Amount of mefenamic acid (mole)

8 Since the bioavailability of mefenamic acid is given to be 90%, maximum mass of mefenamic acid that can be ingested before an “overdose” occurs = 48.1 / 0.9 = 53.4 g