IJC 2008 H2 Chemistry Prelim II Paper2 answers

1 © INNOVA 2008 Answer Scheme 9746/02 Innova JC Preliminary Examination H2 Chemistry Paper 2 Mark Scheme 1 (a) (i) Paracetamol is more toxic as less Paracetamol taken would result in the same toxicity. (ii) Mass of Aspirin taken that would be toxic = 1100 x 50 = 55000 mg Minimum number of Aspirin tablets = 55000/ 325 = 169.23 = 170 (b) (i) It will decompose in moist conditions (undergo hydrolysis). (ii) Acidic hydrolysis Dilute HCl and heat. C O O H O H + C O O H O C O C H H H + C O O C H H H H H2O (iii) Add aq Br2 to both compounds separately. salicylic acid: reddish brown aq Br2 will decolourise. aspirin: reddish brown aq Br2 will not decolourise. (iv) Aspirin is a stronger acid than acetic acid. The negative charge on the anion C O O- O C O C H H H is delocalised into the benzene ring, stabilises it. (c) (i) Standard enthalpy change of combustion of salicylic acid is when one mole of salicylic acid is completely burnt in oxygen at standard conditions of 298 K and 1 atm. (ii) Heat absorbed by bomb calorimeter = 13.62 x 2.5 = 34.05 kJ Amt of salicylic acid = 34.05 /( 3.00 x 10 3) = 0.01135 mol Mass of salicylic acid = 0.01135 x 138 = 1.5663 g Mass of boric acid = 3.56 – 1.5663 = 1.9937 g Mass percentage of boric acid = 1.9937 /3.56 x 100% = 56.0 % 2 (a) N H H N H H 107o Trigonal pyramidal (b) (i) PV = nRT 1.5 x 105 x V = 1 x 8.31 x 300 V = 0.0166 m3 ( or 16.6 dm3) (ii) This is due to the presence of H bonding between the hydrazine molecules , therefore intermolecular forces of attraction becomes significant. (iii) High temperature, low pressure. (iv)

2 © INNOVA 2008 Answer Scheme 9746/02 Pressure 1 Volume 1 Volume (c) (i) Hydrazine has an available lone pair of electrons on the N-atom, hence able to accept protons. (ii) 25 b 24 [OH ][N H ]K [N H ] −+ = (iii) -14 b -9 10K= 3.33 x 10 = 3.00 x 10-6 -6 25 b 24 [OH ][N H ]K 3.00 x 10 = [N H ] − + = [OH-] = 7.775 x 10-4 pOH = 3.11 pH = 10.9 (d) [base] [salt]log pK pOH10b + = 14 – 10 = -lg (3.00 x 10-6) + 10 [salt]log (0.2) [salt] = 6.01 x 10-3 moldm-3 3 (a) (i) Number of protons : ……31……… Number of neutrons: ……37…….. (ii) –––––––– + + + + + + + + Source α Neutrons The deflection remains the same at 3o (b) (i) Gallium oxide has a giant ionic structure with strong electrostatic forces of attraction between the oppositely charged ions. Whereas, gallium chloride has a simple molecular structure with weak van der waals forces of attraction between the molecules. Hence, less energy is required to break the weak van der waals forces of attraction between gallium chloride molecules, the melting point of gallium chloride is lower. (ii) Ga 2O3 (s) + 2NaOH(aq) + H2O Æ 2NaGa[OH]4 (aq) GaCl 3 (s) + 6H2O Æ [Ga(H2O)6]3+ (aq) + 3Cl– (aq) [Ga(H 2O)6]3+ (aq) Æ [Ga(H2O)5(OH)]2+ (aq) + H+ (aq)

3 © INNOVA 2008 Answer Scheme 9746/02 4 (a) C C HN HHC C C C NH2 C HH HH HH HH H H2N C N COOH HHO (b) C C C OH N C HH H H O O + H HH H (c) (i) Position Structure at pH = 7.00 A B H C C H2N COO HH HN C H C C H2N COO OHH CH3 [3] (d) (i) hydrogen bonding (ii) δ +δ −

4 © INNOVA 2008 Answer Scheme 9746/02 (e) Lysine is more basic than tryptophan. (f) (i) C C CH3 ClH H2N C H O Cl (ii) C C N H H NHOOC H H C O CH2CH3 C CH2CH3 O 5 (a) (i) OH COOH+H3N COOH (ii) O-Na+ CH2OHH2N CH3 (iii) OH H2N CH2Cl CH3 Cl Cl H OH Cl H (b) Reagent(s) and Condition(s): ………… Na metal …………………... Observations for : Compound Y: …… no effervescence observed. Compound X: …… effervescence of hydrogen gas observed and it will extinguished burning splint with a ‘pop’ sound.………… (c) Step … I… Explanation: NH 3 is not an electrophile/ it is not electron deficient/ is a nucleophile (d) HCN H + +CN-

5 © INNOVA 2008 Answer Scheme 9746/02 CO H3C C H3C CN O- R R CH3C CN O- + CH3C CN OH + CN-HC N R R CN + - (e) (i) O OH O Cl Intermediate A Intermediate B (ii) Step I Reagents and Conditions: …… aqueous I 2, NaOH (aq), reflux …… Step II Reagents and Conditions: …… HCl(aq)………… Step III Reagents and Conditions: ……… NH 3 heat in a sealed tube End of Paper