CADANGAN JAWAPAN PRA PSPM EC025 KMKK NO SCHEME MARKS 1 (a) i) Rate = k[H2S] [Cl2] 2 ii) Overall order = 3 iii) rate = k[H2S] [Cl2] 2 rate = (2.8x10-2 ) [(2.7x10-2 ) (7.0x10-2 ) 2 = 3.70x10-6 Ms-1 1 1 1 1 1 (b) i) ii) iii) 1 1 1 1 1 1 TOTAL 10
NO SCHEME MARKS 2 (a) 1 1 1 1 2 (b) 1 1 1 1 1 1 @ 1 1 TOTAL 10 3 (a) Ecell = E°cell – 0.0592 log PH2 [Pb2+] 2 [H+ ] 2 0.08 = +0.13 – 0.0592 log (1)(0.60) 2 [H+ ] 2 –log [H+ ] = 0.96 pH = 0.96 1 1 1
NO SCHEME MARKS 3 (b) Q = It = (10.0) (212s) = 2120 C Half-cell equation: 2H2O (l) → O2(g) + 4H+ (aq) + 4e 4(96500)C ≡ 1 mol O2 2120 C ≡ 2120C x 1 mol O2 4(96500)C = 5.5 x 10-3 mol O2 Volume O2 produced = nRT/P = 5.5 x 10-3 mol O2 (8.314)(303) 100 x 103 = 1.4 x 10-4 m3 1 1 1 1 TOTAL 7 4 (a) H3C CH CH2 CH3 CH3 A = H3C CH CH CH3 CH3 Cl B = H2C CCH2CH3 CH3 C = H3C C CH CH3 CH3 D = 1 1 1 1
NO SCHEME MARKS 4(b) H3C CH CH2 CH3 CH3 H3C CH CH CH3 CH3 Cl H3C CH CH CH3 CH3 Cl H2C CCH2CH3 CH3 + H3C C CH CH3 CH3 Cl2 , CH2Cl2 UV KOH , ethanol reflux 1 1 TOTAL 6 5 (a) P = R = S = COOH CH(Br)CH3 @ CH2CH2Br 1 1 1 5 (b) CH3CH2Cl , AlCl3 @ CH3CH2Br , FeBr3 1
NO SCHEME MARKS 5 (c) Equation : + CH3CH2Cl AlCl3 CH2CH3 + HCl STEP 1 CH3CH2Cl + AlCl 3 CH3CH2 + + AlCl 4 - STEP 2 + + CH2CH3 CH + CH2CH3 HC + CH2CH3 CH + CH2CH3 arenium ion STEP 3 C + CH2CH3 H H + AlCl 4 - CH2CH3 + HCl 1 1 1 1 TOTAL MARKS 8 MAX 7 6 (a) Nucleophile: CNStrength of nucleophile: Strong nucleophile 1 1
NO SCHEME MARKS 6 (b) 1 6 (c) Bimolecular nucleophilic substitution reaction (SN2) 1 3 TOTAL MARKS 7 7 (a) Secondary alcohol. 1 7 (b) 4 TOTAL MARKS 5 CH3CH2CCH3 + KCN H Br CH3CH2CCH3 H CN H C H Br H3CH2C CN - slow rate determining step H C H Br CH2CH3 H C H CH2CH3 fast + Br- + NC transition state NC + CH3CH2Br Mg dry ether CH3CH2MgBr i) CH3CH2CH(CH3 )CHO ii) H3O + CH3CH2CHCHCH2CH3 CH3 OH
NO SCHEME MARKS 8 (a) i) O H IUPAC: Cyclohexylmethanal 1 1 ii) O H [Ag(NH3) 2]+ ,OH- O O - + 2Ag(s) Observation: Silver mirror O H O O - + Cu2O (s) Cu 2+ complex, OH- Observation: brick red precipitate 1 1 1 1 (b) Brady’s test Yellow/orange precipitates 1 1 TOTAL MARKS 8 9 (a) H3C CH3 Br2 ,uv H3C CH2Br Mg,dry ether H3C CH2MgBr i..CO2 ii. H3O+ H3C CH2 COOH 1 m 1 m 1 m 1 m 1 m 5
NO SCHEME MARKS 9 (b) 1-butanol 2-bromobutanoic acid 2-chlorobutanoic acid 2,3 – dibromobutanoic acid Explanation: - Butanol is the least acidic because alkyl group strengthen the of -OH bond in butanol. H+ difficult to released - 2-chlorobutanoic acid is more acidic compare to 2- bromobutanoic acid Because bromine is less electronegative compare to chlorine. More alectronegative atom, therefore stronger inductive effect. - 2,3 – dibromobutanoic acid is most acidic because it has more halogen, therefore stronger inductive effect - Stronger inductive effect, more acidic 1 1 1 1 1 TOTAL MARKS 10 10 (a) N,N–dimethylmethanamine< N–methylethanamine < 1 – propanamine N,N–dimethylmethanamine has lowest boiling point because it cannot for hydrogen bonds between molecules. Both N–methylethanamine and 1 – propanamine can form hydrogen bond between molecule. However, 1-propanamine can form more hydrogen bond between molecules than N–methylethanamine. 1 1 1 1
NO SCHEME MARKS 10 (b) Br2 water test @ Nitrous acid test Name test - 1 Both equation s correct – 1 Both observati ons correct -1 TOTAL MARKS 10 NH2 Br2 , H2 O NH2 Br Br Br Observation: White precipitate is formed. CH3 CH2 NH2 Br2 , H2 O Observation: White precipitate does not form. No observable reaction NH2 N + NCl - Observation: No release of bubble gas CH3 CH2 NH2 Observation: Bubble gas released NaNO2 , HCl 0 - 5 o C NaNO2 , HCl 0 - 5 o C CH3 CH2 N NCl + - CH3 CH2 OH + N2 + CH3 CH2 Cl +CH2 =CH2
NO SCHEME MARKS 11 (a) 2 – amino – 3 – hydroxybutanoic acid. 1 (b) 1 (c) 1 (d) 1 C C O - H O H3 N + HO C H CH3 C C O - K + H O H2 N HO C H CH3 C C OH H O ClH3 N+ HO C H CH3