NYJC_H2_CHEM_P3_ANS Prelim

1 2011 NYJC Prelim H2 Chemistry 9647/03 Answers 1 (a) (i) I. ½ (CN) 2 + H+ + e HCN E = +0.37 V Cl2 + 2e 2Cl E  = +1.36V E  cell = +1.36 – (+0.37) = +0.99V Overall Equation: Cl 2 + 2HCN (CN)2 + 2H+ + 2Cl-   II. ½ (CN)2 + H+ + e HCN E  = +0.37 V S 4O6 2- + 2e 2S2O3 2- E = +0.09 V E  cell = +0.37 – (+0.09) = +0.28 V Overall Equation: (CN) 2 + 2H+ + S2O3 2-  2 HCN + S4O6 2- III. ½ (CN)2 + H+ + e HCN E  = +0.37 V Cr2O7 2- + 14H+ + 6e 2Cr3+ + 7H2O E = +1.33 V Both are oxidising agents. Hence reaction cannot proceed. (ii) C C N N xx x x x x x xx linear (b) (i) This is because the reaction proceeds with an inversion of configuration. C H (CH2)4CH3 CH3 Cl C H (CH2)4CH3 CH3 NC + Cl -- NC +  To illustrate the inversion in structural formula (3D wedge diagram) (ii) Isomer A: CH3 C (CH2)3CH3 Cl CH2CH3 2

2 Type of reaction: SN1 CH3 C (CH2)3CH3 Cl CH2CH3 + CN - CH3 C + (CH2)3CH3 CH2CH3 + Cl -slow CH3 C + (CH2)3CH3 CH2CH3 CH3 C (CH2)3CH3 CH2CH3 CN   (iii)  Reaction b(i) OR S N2 mechanism is faster.  The rate determining step in b(i) involves NaCN. OR rate = k[NaCN][RX] (c) OH OH O O OH B C K2Cr2O7 , H2SO4 (aq) Heat under reflux HCN NaCN Na in ethanol D heat with concentrated H2SO4 (does not give CO2(g) with Na2CO3) O OH OH CN O OH OH CH2NH2CH3 CH2NH3 + CH3 O O

3 2(a)(i) (There is a plane of symmetry in the molecule or both chiral centres contain the same groups attached to it), the two chiral centres rotate the plane of polarised light to the same extent but in the opposite direction hence cancelling out the optical activity. (ii) Compound A: CH2 CH CH 2 CH2 CH2CH3 CH3 CH3 Compound C: CH3 CC H C H 2 CH3 CH3 CH3 CH3 Compound D: CH3 CC C H 3 CH3 CH3 CH3 CH3 (iii) Isomers A to D are simple molecular compounds with weak van der Waals forces between molecules. However, as we move down the table from A to D, the molecules become increasingly branched, hence they have less surface area of contact with their neighbouring molecules resulting in weaker vdw forces between molecules. Hence less energy is required to break the vdw bonds, resulting in lower b.p. down the table. (iv) 4 molecules + ratio Structural formula mole ratio CH2 CH CH CH 2 CH3CH2 CH3 CH3 Cl 6 3 CH CH CH CH 2 CH3CH3 CH3 CH3Cl 4 2 CH2 CC H C H 2 CH3CH3 CH3 CH3 Cl 2 1

4 CH2 CH CH CH 2 CH3CH3 CH2 CH3 Cl 6 3 (v) CH2 CH C CH 2 CH3CH3 CH3 CH3 CH2CCHCH2CH3 CH3 CH3CH3 OR CHCHCHCH2CH3 CH3 CH3CH3 CH2 CH CH CH CH 3CH3 CH3 CH3 [1st one preferred, because more branched.] (but not CH2 CH CH CH 2 CH2CH3 CH3 CH3 CH2CHCHCH2CH3 CH2 CH3CH3 or any other combination that produces less than 6 branches.) (b) pH = 7.4; pOH = 14-7.4 = 6.6; [OH -] = 10-6.6 = 2.512 × 10-7 moldm-3 K sp of Pb(OH)2 = [Pb2+(aq)][OH-(aq)]2 = 1.43 × 10-20; [Pb 2+(aq)