MJC H2 Chem 2012 Prelim P3 Soln

2012 Meridian JC H2 Chemistry Prelim P3 Suggested Answers 1 (a) Reaction I : NaBH4 in ethanol , rtp Reaction II : PCl5 , rtp Reaction III : LiAlH4 in dry ether , rtp H OO HOOC-C-COOH Cl ClC-CH-C OH A BCl (b) (i)1o and 2o alcohols (ii) Hydrolysis (c) (i)C6H12O6 + 6O2 6CO2 + 6H2O (ii) ∆Hrxn = = [6(-394) + 6(-286)] – [-1275] = - 2805 kJ mol-1 Energy released = 2801 x 0.0500 = 140 kJ (iii) Energy released by combustion = 650 x 4.18 x 44 = 120 kJ (iv) Some heat was lost to the surrounding. (d) (i)M(NO3)2 MO + 2NO2 + ½O2 (ii) Each mole of Group II nitrates release the same amount of gaseous molecules. (iii) Ionic size of cation: Mg2+ < Ba2+ charge density hence polarizing effect of cation: Mg2+ > Ba2+ Therefore Ba(NO3)2 has a higher decomposition temperature. (e) ∆Hsoln (MgCl2) MgCl2 (s) Mg2+ (aq) + 2Cl- (aq) -∆Hf (MgCl2) Mg(s) + Cl2(g) ∆Hat (Mg) BE(Cl – Cl) ∆Hhyd (Mg2+) 2 x ∆Hhyd (Cl-) Mg(g) + 2Cl(g) 1st IE + 2nd IE (Mg) Mg2+(g) + 2Cl(g) + 2e- 2 x 1st EA (Cl) Mg2+(g) + 2Cl-(g) ∆Hhyd (Mg2+) = - 148 - 244 - 736 - 1450 - 2 x (-364) - 2 x (-362) + (- 642) + (-153) hence ∆Hhyd (Mg2+) = -1921 kJ mol-1 copyright chem dept @ meridian jc

copyright chem dept @ meridian jc H3C C C CH3 CH3 CH3 H H W : H3C C C CH2Br CH3 CH3 H H H3C C C CH3 CH3 CH3 H Br Z:Y: 2(a) (i) S hine uv light on the mixture of hydrogen and chlorine gas. (ii) Decolorisation of reddish brown bromine ; white fumes of HBr seen. (iii) Reactivity of halogens with hydrogen decreases down the group. The total bond energy released in forming H -X decreased down the group more significantly than the total bond energy absorbed in breaking X-X and H-H bonds. DHrxn becomes more endothermic down the group, reactivity decrease down the group (b) (i) Limited Br2 gas and uv light (ii) DH = 410 + 151 – 240 - 299 = + 22.0 kJ mol-1 (iii) Endothermic due to the weak C-I bond formed. (iv) (c) (i) Comparing experiments 1 & 3, when [C 6H13Br] was tripled, the initial rate of reaction increased by 3 times, order of reaction wrt C6H13Br is 1. Comparing experiments 1 & 2 Þ [ ][ ] [ ][ ] = 1.30 2.6013.0 26.0 2.60 1.30 b b Þ æö= ç÷èø 2.601 1.30 b Þ b = 0 Rate = k[C6H13Br] (ii) It should be isomer Y since it is a tertiary bromoalkane, it will form a stable carbocation in the slow step of the mechanism due to the presence of three electron donating alkyl group

copyright chem dept @ meridian jc (iii) Mechanism : SN1 (iv) Marking points: · Axes are correctly labeled 3(a) (i) (ii) Cathode: Fe 3+ (l) + 3e à Fe (s); Anode: 2O2- (l) à O2 (g) + 4e (iii) 5 1000 55.8 ´ = I 24 60 60 3 96500 ´´´ ´ ; I = 300 A. (CH3)2CHC(CH3)2Br + OH- (CH3)2CHC(CH3)2+ + Br- + OH- (CH3)2CHC(CH3)2OH + Br- O2 (g) formed Fe (s) formed – + Fe3+ (l) O2- (l) (Pt anode) (Pt cathode) C Br CH3 CH3 (C