MJC H2 CHEM P3 ANS Prelim

No part of this paper should be repro duced for commercial use without the prior written permission of Meridian Junior College. Class Reg Number Candidate Name _____________________________ Meridian Junior College 2009 JC 2 Preliminary Examination Answers H2 Chemistry 9746 14 September 2009 2 hours Paper 3 Free Response Additional Materials Data Booklet Writing paper INSTRUCTION TO CANDIDATES Write your name, class and register number in the spaces provided at the top of this page. Answer any 4 out of 5 questions. Begin each question on a fresh page of writing paper. Fasten the writing pa pers behind the given Cover Page for Questions 1 & 2 and Cover Page for Questions 3, 4 & 5 respectively. Hand in Questions 1 & 2 and 3, 4 & 5 separately. You are advised to spend about 30 min per question only. INFORMATION FOR CANDIDATES The number of marks is given in brackets [ ] at the end of each question or part question. You are reminded of the need for good English and clear presentation in your answers. This question paper consists of 15 printed pages ©chemistry department@meridian jc 1

1 (a) 22 22 () ( ) 1.958 x 0.625 = ( ) 4.042 SO Cl p SO Cl PPK P = 0.303 atm (b) The equilibrium position will shift ri ght towards the endothermic reaction to absorb heat. 0 SO2 SO2 SO2Cl2 SO2Cl2 Time 10 mins 20 mins (c) SOCl2 + H2O SO2 + 2HCl SO2Cl2 + 2H2O H2SO4 + 2HCl  SO 2Cl2 in water yields 3 mol of strong acid in comparison to 2 mole strong acid for SOCl2hence pH is lower. (di) sp 2 hydrisation [1] three sp2 orbitals (ii) Homolytic fission is the breaking of a co valent bond such that each atom retain s only one of the shared electrons. HO + SO2Cl2  HOCl + SO2 + Cl ©chemistry department@meridian jc 2

©chemistry department@meridian jc 3 CH2=CHCH2Cl CH2(Cl) (OH)CH CH 2Cl CH )CH( OH)CH2(CN CN2 CH )CH(OH) CH step 1 step 3 step 2 (iii) 2(COOH 2COOH Step 1: aqueous Br2 or Cl2 Step 2: alcoholic KCN, heat Step 3: aq HCl or H 2SO4, heat (iv) 1 Outline the mechanism that occu rs for the formation of either B or C. CH2Cl CCH H H Br + 2 Concept : A racemic mixture is formed due to equal probability of attack from either plane [1] [1] Br Br + - + :Br- slow CC H H H CH2Cl CH2Cl CCH H H BrBr CH2Cl CCH H H Br + fast + :Br- CH2Br C * CH2Br C* H Br CH2Cl H Br ClH2C

2a(i) Gas liberated - Chlorine gas C lO- (aq) + Cl- (aq) + H2O (l) Cl2 (g) + 2OH- (aq) (ii) Chlorine is only slightly/ moderately solubl e in water as the VDW of attraction is not compatible to the hydrogen bonding between the water molecules (iii) Concentrated H2SO4 is only able to oxidize HI to I2 or Br- to Br2 NaCl + H2SO4  HCl + NaHSO4 NaBr + H2SO4  HBr + NaHSO4 2 HBr + H 2SO4  Br2 + SO2 + 2 H2O or NaI + H2SO4  HI + NaHSO4 8 HI + H 2SO4  4 I2 + H2S + 4 H2O (iv) Thermal stability decreases from HBr to HAt as the covale nt bond strength decreases down the group resulting in the decrease of the bond dissociation energy BE of HI (+299 kJ mol–1) < BE of HBr (+366 kJ mol–1) (bi) E θcell for Rxn 1 = 0.36 V Eθcell for Rxn 2 = 0.30 V Eθcell for Reaction I > Eθcell for Reaction II, hence reaction I is more feasible. (ii) Reagent -HClO3 or KClO3 or NaClO3 eg (ci) Iodine has a large enough atomic size to accommodate the number of O atoms surrounding it in IO5 3- and IO6 5- . Or use steric hinderance factor to explain (ii) Dot and cross diagram for IO6 5- x xx xx xx xxxx xx xxxx x x xx xx xx xxx O O O I O O O xx xx x x x x x x x x x 5- ©chemistry department@meridian jc 4

(iii) 5Ba(IO3)2 Ba 5(IO6)2 + 4I2 + 9O2 Ionic size hence char ge density hence polarising power of of Ca2+ > Ba2+ Hence, Ca2+ has greater ability to distort the anion IO3 - charge/electron cloud Hence, thermal stability of Ca(IO3)2 > Ba(IO3)2. (iv) Overall equation : IO 6 5- + 7I- + 12H+ 4I 2 + 6H2O No of mol IO6 5- = 1/8 (2.20 x 10-3) = 2.75 x 10-4 Mass of sodium iodate (VII) = 2.75 x 10-4 x x 338 = 0.09295g Percentage by mass = 0.09295/0.200 = 46.5% 3(a) alkene, ester,amine ,amide, ether-any 4 out of 5 (b) The phosphate salt is more soluble in water hence more easily absorbed. (c) O -O N H2 O NH2 CH3COO- CH3CH2OH (d) O O NH O O NH2D Br D + ©chemistry department@meridian jc 5

(e) O O OH OH COOH (fi) Nucleophilic substitution (ii) O O O COOCH2CH3 S O O CH3 or O O O COOCH2CH3 S O O CH3 (gi) O O O O (ii) O O O - O O S O O CH3 .. O - S O CH3 O E + ©chemistry department@meridian jc 6

(h) Unusual : The alkene group would have been expected to undergo reduction. (i) CH 3Br , heat under high pressure Compound G would be more basic due to the electron-donating group which increases the electron densit y on the lone pair of the N atom hence making the lone pair more available to accept a proton. (j) Quantity of heat absorbed by water = 70 x 4.18 x 8.4(temp change) = 2457.84 J H neutralisation = - 2457.84 / 0.045 = −54.6 kJ mol−1 Since the Hneutralisation is less exothermic than t hat between strong acid and strong base, can deduce that Tamiflu is a weak base. Some of the energy evolved from the neutralisation process is used to further dissociate the weak base completely. 4(ai) Ka 1 and Ka2 are the acid dissociation constant of the carboxylic acid group and phenolic group respectively. Ka1 is smaller Ka2 because carboxylic acid group is more acidic than the phenolic group. (ii) H+ from the 2nd dissociation is negligible and can be ignored.  + a -5 [H ] = cK 0.05 6.31 x 10 Hence, pH = 2.75 (iii) For acidic buffer pH = -lg 6.31 x 10 -5 +   -4 -4 [6.25 10 /0.03125]lg[6.25 10 /0.03125] = 4.20 ©chemistry department@meridian jc 7

Salt hydrolysis [ COO-Na+HO ] = 1.25 x10-3 / 0.0375 = 3.33 x 10-2 mol dm-3 w b1 a1 KK= K =      14 10 5 11 0 1.58 106.31 10 mol dm-3 - b -2 -10 -6 -3 [OH ]= cK = 3.33×10 ×1.58×10 =2.29×10 mol dm pOH = -lg 2.29 x 10-5 Hence, pH = 14 – 5.63 = 8.36 (iv) pH 12.50 6.25 8.36 4.20 2.75 Volume of NaOH / cm3 ©chemistry department@meridian jc 8

(bi) . Rate = k[ (CD3)3CBr] (ii) Nucleophilic substitution mechanism C CD3 CD3 Br CD3 C CD3 CD3 CD3 +B - slow C CD3 r CD3CD3 C6H5O- C CD3 CD3 CD3 C6H5O fast Step 1 Step 2 (iii) [(CD3)3CBr]/mol dm-3 0.64 0.32 0.08 10 20 30 time/min ©chemistry department@meridian jc 9

(ci) X can be : (accept other logical answers) CC C C C H C CH3CH3 HH H H H Cl H H Cl W : CC C C C H C CH3CH3 HH H H H H H H H Y : CC C C C H C CH3CH3 HH H H H (cii) CC C C C H C CH3CH3 HH H H H H Cl H Cl or CC C C C H C CH3CH3 HH H H H H Cl Cl H (ciii) There is 1 alkene double bond that is capabl e of exhibiting cis-trans isomerism as there are 2 different groups attached