MJC H2 Chem 2013 Prelim P3 Soln

Mark Scheme CONFIDENTIAL ©chemistry@meridian jc 1 1(a) (i) No of moles of NO2 gas produced = 561.01 10 57.6 10 8.31 353    Mr of fexofenadine = -3 1 1.983 10 = 504 (ii) There are significant intermolecular forces of attraction (iii) The R group should be C6H5 since R =77 (b) (i) The hydrochloride salt is more so luble in aqueous solution forming ion- dipole interaction with the water molecules (ii) No of boxes he needs for 5 days = 425 30   = 2 (must be whole no)   (c) (i) CN O Cl         CN N RR HO OH   Compound A Compound B (ii) Electrophilic substitution (iii) NaBH4 in ethanol,r.t.p (iv) COOD O Cl (v) Concentrated H2SO4, heat Suggested Answers for 2013 MJC Prelim P3 939

©chemistry@meridian jc 2 (d) (i) Initiation Br2 2B r Propagation Br CN UV light HBr CN CN Br2 CN Br Br Termination Br CN CN Br     (ii) The formation of the iodo derivative in the propagation step is highly endothermic .   (e) The amine group. The electron donating R gr oup in the amine increase the electron density on the lone pair of electrons on N, making it more available to accept a proton. 2(a) (i) CH3CH(OH)CH3  CH3COCH3 + 2 H+ + 2e CH3CH(OH)CH3 + O2  CH3COCH3 + 2H2O (ii) Eθ = - 0.03 V (iii) The OH- neutralises H + , causing the concentration [H +] to decrease. The equilibrium position will shift left to increase [H +]. Eθ oxid will become less positive. Hence, more positive overall Eθ cell . (iv) Availability of propan-2-ol as source. *Other answers possible (b) (i) [Pt(NH3)4]2+ (aq) + 2e  Pt (s)+ 4NH3 (g). Grey solid formed or pungent ammonia gas evolved ( or state test for ammonia) (ii) No of moles of platinum = ூ் ௡ி = = 2.04 x10-3 Mass of platinum = 2.04 x 10 -3 x 195 = 0.398 g 940

©chemistry@meridian jc 3 (c) (i) P undergoes acid-carbonate reaction / neutralization with NaHCO 3  P contains a carboxylic acid group.  P undergoes electrophilic addition with aq chlorine  P contains an alkene  Q undergoes electrophilic addition and electrophilic substitution with aq chlorine.  Q contains both an alkene and a phenol  Q undergoes oxidation with Tolle n’s reagent but not Fehling solution  P contains an aromatic aldehyde  P and Q undergoes strong oxidation with hot acidified KMnO 4 to give T and U respectively  P has only one R group attached to the benzene ring and Q contains three R groups attached to the benzene . 941

©chemistry@meridian jc 4 (ii) Equation (iii) Compound X is the stronger acid. The distance of Br atom on X is nearer to the COOH/COO - group .Electron-withdrawing effect on –COO - / carboxylate ion hence stabilisation of conjugate base wrt acid for: X > Y. 3(a) (i) HN N N OR HN N N (ii) C O N H (iii) CH3CH2CHO CH3CH2N H C O CH3 CH3CH2COOH CH3CH2NH2 KMnO4,d i lH2SO4 heat HN3,H + cat CH3COCl