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 (CH3)2CH C + CH3 CH3 (CH3)2CHd+ d+ + Br- slow C + CH3 CH3 (CH3)2CH d+ OH- C OH CH3 CH3 (CH3)2CH fast : d-

copyright chem dept @ meridian jc R O H R O H OH OO R: (iv) C + O2 à CO2 ; CO2 + C à 2CO ; Fe2O3 + 3CO à 2Fe + 3CO2 (v) Less / no pollutants of CO or CO2 are formed. (b)(i) Fe(II) is stabilised with respect to Fe( III) in acidic medium compared to basic medium. Fe 2+ in acidic medium is a weaker reducing agent compared to Fe(OH)2 in basic medium. (ii) Fe2+ acts as a homogeneous catalyst Step 1: 2Fe2+ + S2O82- ® 2SO42- + 2Fe3+ Step 2: 2Fe3+ + 2I- ® 2 Fe2+ + I2 (c) (i) Ethoxide ion acts as a base (ii) (Nucleophilic) Addition (iii) (iv) Optical isomerism (v) 4(a) Element X : Sodium / Na Na 2O + H2O → 2NaOH O OH O O N N H NO2 NO2

copyright chem dept @ meridian jc NH CH CH O CH2 NHCOCHCl2 C O CH3 C O H3C O C O CH3 C O O-Na+C O +Na-O Formula of the oxide of Element Y: SO3 Upon addition of Ca(OH)2, white ppt formed could be CaSO4 No of mol of CaSO4 = = 0.09 mol SO3 + Ca(OH)2 → CaSO4 + H2O SO3 +H2O → H2SO4 H2SO4 + Na2O → Na2SO4 + H2O (b) (i) 101 x 1000 × M = 1.60 x 103 x 8.31 × (250+273) Molar mass of gas = 68.8 g mol-1 ≈ 67.5 g mol-1 Hence m = 1, n = 2 (ii) There is significant van der Waals forces of attraction between the ClO2 molecules. (c) (i) A l2+ (g) à Al3+ (g) + e (ii) C is aluminium, as it has the lowest 3 rd IE. The removal of the third electron will lead to the formation of the stable noble gas configuration hence a small amount of energy is required to remove the valence electron. (d) (i) A: B, C: D: (ii) H 2O (g) or steam , conc. H3PO4, 65 atm , 300°C Electrophilic Addition white ppt NH2 C O- O NH2 CH C CH2 NHCOCHCl2 OH

copyright chem dept @ meridian jc (iii) Add 2,4- DNPH to each compound separately and heat. Chloramphenicol-X will not produce orange ppt of hydrazine; it’s isomer will produce orange ppt of hydrazone. 5(a) (i) (ii) Due to the presence of a large hydrophobic benzene rings. (iii) F ormation of ion- dipole interaction with water molecules causes solvation hence drug is better is better absorbed. (b) (i) Free radical substitution (ii) excess (CH 3)3CNH2 , heat (iii) (c) A: Cl O OH C OH H3C COOH N CH3 CH3 Or A can also be 1, 2 di-substituted isomer C ON H O H H F F F H O H O H H Hydrogen bond δ+ δ- δ+ δ+ δ+ δ+ δ- δ- δ+ δ- δ+

copyright chem dept @ meridian jc B: C: (d)(i) Effective nuclear charge increases. The electrostatic force of attraction between nucleus and valence electrons increases hence e nergy required to remove the valence electrons increases. (ii) Li + has the smallest ionic radius among the group I metal ion Less H 2O ligands can be datively bonded to Li + cation, thus the coordination number is 4 and Li(H2O)4+ ions is formed. C OH H3C COOH N CH3 CH3 BrBr C OH H3C COOH +N CH3 CH3 CH3 Br-