AJC H2 Chem 2012 Prelim P2 Soln

1 (a) (i) AgNO3(aq) 2012 H2 Chemistry Prelim Examinations Paper 2 Solutions (ii) NH3(aq) 2 (b) Step Expected Location of each Observations anion 1. Add excess AgNO3(aq). Yellow ppt Cl– and I– exists as Filter AgX in the pptthe mixture. 2. Add excess NH3(aq) to Yellow residue I– exists as AgI in the residue. Colourless filtrate residue Filter the mixture. Cl– remains in filtrate 3. Add HNO3(aq) to the White ppt Cl– exists as AgCl in filtrate. the ppt Filter the mixture after adding excess AgNO3(aq) and after adding excess NH3 to the residue correct identification of ions Yellow ppt with AgNO3 in step 1. White ppt negates the mark. (Note: white ppt is obscured by the yellow ppt.) Yellow residue and colourless filtrate in step 2. White ppt with acid in step 3. FYI The mixture has to be filtered after adding AgNO3 to separate AgCl and AgI from the cations so that insoluble metal hydroxides would not be formed when NH3(aq) is added. (c) To remove other anions (e.g. CO3 2- and SO3 2-) that form insoluble compound with Ag+ (aq). (All nitrates are soluble). (d) Reagent: NH3(aq) and cation: Al3+ Al(OH)3 is insoluble while Zn(OH)2 is soluble in excess NH3(aq) due to complex formation. Al3+ + 3OH– Al(OH)3(s) Zn(OH)2 + 4NH3 [Zn(NH3)4]2+(aq) + 2OH– [allow 2 separate equations showing dissolving of Zn(OH)2] Zn(OH)2 Zn2+ + 2OH– [Zn(H2O)6]2+ + 4NH3 [Zn(NH3)4]2+ + 6H2O

2 (e) Step 1: To 2 cm depth of each of the unknown in a test tube, add a few drops of AgNO3(aq). (allow distilled water – gives white fumes) C6H5Br C6H5CH2Br CH3COBr No ppt No ppt Cream ppt Step 2: To fresh samples of 2 cm depth each of the other 2 compounds in a test tube, add NaOH(aq) and heat gently for 5 minutes. Step 3: To the cooled samples of each of the remaining 2 compounds in a test tube, add excess HNO3(aq), followed by AgNO3(aq). C6H5Br C6H5CH2Br No ppt Cream ppt Appropriate reagents: AgNO 3(aq), NaOH(aq), HNO3(aq) Appropriate conditions : heat (not reflux, warm), (cool), excess Quantities mentioned in all 3 steps: 2 cm depth / 1 – 5 cm3, (a few drops) Correct observations in steps 1 and 3 Alternatives: Add distilled water or dil HCl. Only ethanoyl bromide gives white fumes. Add aqueous silver nitrate and heat. Only ethanoyl bromide and (bromomethyl)benzene give cream ppt.

3 2 (a) (i) Graph B is CO2 at 473 K whilst graph C is SO2 at 298 K. SO2 deviates more than CO 2 at 298 K. SO2 molecules are held by stronger permanent dipole–permanent dipole attractions (or van der Waals forces of attraction) as compared to weaker instantaneous dipole– induced dipole attractions (or van der Waals forces of attraction) between CO2 molecules. CO2 at 473 K deviates less than CO 2 at 298 K. At higher temperature, CO 2 molecules possess higher average kinetic energy and are more able to overcome forces of attraction between the molecules. (ii) p = 2 2 ) ( V a n nbV nRT -- = 2 3 2 53 )105 . 0 ( )687. 0 ( ) 1 ( )10 68. 5105 . 0 ( )298)(31. 8)(1 ( --- - - xxx = 5.588 x 106 – 2.748 x 106 = 2.84 x 106 Pa (iii) Using the ideal gas equation, p = nRT/V = )10x 5 . 0 ( )298)(31. 8)(1 ( 3- = 4.95 x 106 Pa (iv) The actual pressure exerted is much lower than the pressure calculated from the ideal gas equation as forces of attraction between SO2 molecules are not negligible. (b) (i) x: +5; y: 8 (ii) 2- D S S O O O O 2- E S O O O S (iii) Trigonal pyramidal w.r.t. to each S atom. (iv) S O O S O O O O O O 2-

4 (c) (i) Thionyl halides exist as simple covalent molecules held together by van der Waals’ forces of attractions. Across the series from SOF 2 to SOBr 2, as the no. of electrons / size of the electron cloud increases, the strength of van der Waals’ forces of attractions (id–id) between the molecules also increases. As the boiling point increases, the vapour pressure decreases. (ii) 2 points must be above SOCl 2 and SOBr2 2 points must be above HF (iii) HF molecules are held together by stronger hydrogen bonds. Thus, the vapour pressure of HF is lower than that of SOF2. vapour pressure F Cl Br X X X SOF2 SOCl2 SOBr2 HBr HF HCl

5 3 (a) (i) ] [ ] [ ] [ 2 2 2 2 3 OSO SOKc = ) ( ) ( ) ( 22 3 2 2 OSO SO p p p pK = Using pV = nRT p = (n/V)RT = cRT (shown) )RT 1( K )RT 1( ][O][SO ][SO ]RT[O][SO ][SO ]RT) ([O ]RT)([SO ]RT)([SOK c 2 2 2 2 3 2 2 2 2 3 2 2 2 2 3 p = = = = (ii) (iii) The equilibrium [SO3] does not change but equilibrium is reached faster. Hence the effect is due to the addition of a catalyst at t1. A higher temperature will favour the reverse endothermic reaction. Hence the position of equilibrium will shift to the left. The equilibrium concentration of SO 3 will thus be lower. t1 t2

6 (iv) The reaction should be conducted at a high pressure. The forward reaction results in a decrease in the number of moles of gaseous molecules. Hence the position of equilibrium will shift to the right to decrease pressure by producing fewer gas molecules. Thus, increasing the yield of SO3 produced. (b) SO2 has 2 S=O bonds and SO3 has 3 S=O bonds. From reaction II: 4S=O + O=O 6S=O Let x be the bond energy of S=O (4x + (+496)) – 6x = –192 –2x = –192 – 496 x = 344 kJ mol–1 4 (a) (i) CO2 - HO2C NH3 + a–NH3+ group is electron–withdrawing. a–CO2– group being closer/nearer to the a –NH3+ group is more stablised as the negative charge is dispersed to a greater extent. (ii) - correct shape - correctly labeled 3 pH values (that correspond to the 3 pK a values) - correctly labeled equivalence volumes & volumes at MBC (iii) Isoelectric point is the pH at which the amino acid carries no net charge. Accept pH at which the amino acid exists as zwitterionic form. a 1.88 3.65 9.60 Volume of NaOH added / cm3 pH 5 10 15 20 25 30 X

7 CO2 - HO2C NH3 Isoelectric point correctly indicated (with an “X”) on curve. (iv) CO2 - -O2C NH3 + H+ CO2 - HO2C NH3 The (conjugate) base reacts with the added H+ thus maintaining the pH of the solution. (v) n(aspartic acid) present = 0.100 x 100/1000 = 0.01 mol At pH = 3.65 (= pKa of R–group) [HA] = [A–] nHA = nA– = 0.01/2 = 0.005 mol n H+ added = 0.0200 x 50/1000 = 0.001 mol A – + H+ HA nHA present after HCl is added = 0.005 + 0.001 = 0.006 mol nA– present after HCl is added = 0.005 – 0.001 = 0.004 mol Ka = [HA] ]][A [H- + 10–3.65 = )0.150 0.006( )0.150 0.004]( [H + [H+] = 3.358 x 10–4 mol dm–3 pH = 3.474 Change in pH = 3.474 – 3.65 = –0.176 (b) (i) aq NaOH or dilute HCl / H2SO4 heat under reflux for a prolonged period / several hours

8 (ii) ser–asp–tyr–val–gly–ser - correct sequence with 6 a.a. residues Justification of answer: ser–asp–tyr–val–gly–ser ser–asp–tyr–val–gly–ser - Any 2 points N–terminal enzyme Þ 2 tripeptides obtained: ser–asp–tyr & val–gly–ser special reagent Þ 2 peptides obtained; one of these is a dipeptide gly–ser For students: 1. s pecial reagent digests at carboxylic end of val giving gly–ser Þ val–gly–ser 2. enzyme hydrolyses at carboxylic end of tyr Þ tyr–val–gly–ser 3. ser at N–terminal Þ ser–asp–tyr–val–gly–ser

9 5 (a) (i) (101 000) (37.5 x 10–6) = (0.12/M) (8.31) (250+273) Molar mass = 137.7 g mol–1 Mr = 138 (ii) K Elemental Analysis (%) C H O 60.8 4.4 34.8 5.06 4.4 2.18 2.32 2.0 1 n(12 x 2.32 + 2 + 16) = 137.7 => n = 3 C7H6O3 (iii) Phenol (iv) Carboxylic acid (v) 1 mol