IJC 2008 H1 Chemistry Prelim II Paper2 answers

1 © INNOVA 2008 8872/02 Preliminary 2 H1 Chemistry Paper 2 Mark Scheme 1. (a) Trigonal pyramidal (b) ………3N 2H4 Æ 4NH3 + N2 (c) ……… Formation of more stable molecules (d) No of moles of hydrazine = 25/1000 x 1.40 = 0.035 No. of moles of electrons = 4 x 0.035 = 0.140 No. of moles of KMnO4 = 35/1000 x 1 = 0.035 Mole ratio of KMnO4 : electrons = 1:4 Thus, final oxidation number = + 7 – 4 = + 3 Section A (e) (i) H-bonding H2NN H NN H 2 H H δ +δ − H (ii) There are stronger hydrogen bonding exis t between the hydrazine molecules compared to weak van der waalsforces of attraction between the molecules of ethane. Since more energy is required to break stronger H-bonding, the boiling point of hydrazine is higher. 2 (a) (i) Number of protons : ……31…… Number of neutrons: ……37…… Marker’s Comments: • Easy questions but many students made mistakes. • Many students use the Ar of Ga as 69.7 (average value of the different isotopes) given in Data Booklet. • Some students even mistaken Ga for Ge. (ii) –––––––– + + + + + + + + Source α Neutrons II. Deflection is proportional to argch e mass . Since deuterium ion is half the mass of alpha particle and half of it’s charge, the deflection remains the same at 3o.

2 © INNOVA 2008 8872/02 (b) (i) Al2O3 (s) + 2NaOH(aq) + 3H2O Æ 2NaAl[OH]4 (aq) AlCl3 (s) + 6H2O Æ [Al(H2O)6]3+ (aq) + 3Cl– (aq) [Al(H2O)6]3+ (aq) Æ [Al(H2O)5(OH)]2+ (aq) + H+ (aq) (ii) (c) (i) Isoelectronic refers to ions having the same number of electrons. (ii) 1s22s22p6 (iii) s-orbital: Spherical p-orbital: Dumb-bell (iv) Al: 1s22s22p63s23p1 Mg: 1s22s22p63s2 Since less energy is required to remove the valence electron from the higher energy p-orbital of the Al atom compared to the s-orbital of Mg atom. 3 (a) [ ] [] [] 2 3 c 3 22 NHK NH = Units: mol-2dm6 (b) (i) The system will reduce the pressure by shifting equilibrium to the left. (ii) The finely-divided iron acts as catalyst. The catalyst lowers the activation energy of the reaction. Equilibrium is attained faster as it speeds up the rates of both the forward and reverse reactions equally. Hence it has no effect on the position of the equilibrium. (c) (i) step I Reagent(s) and Conditions: … Cl 2 (g),…… FeCl 3 or AlCl3 catalyst ………… step II Reagent(s) and Conditions: … Br 2 in CCl4 ……… uv light……………… (ii) CHOHCH2COO- COO- Cl (iii) Free radical substitution This reaction is not specific

3 © INNOVA 2008 8872/02 4 (a) (i) NH3, heat in sealed tube OH CH2NH2 CH3 (ii) HBr, room temperature OH CH2Cl CH3 H Br H H (b) Reagent(s) and Conditions: Add 2,4-dinitrophenylhydrazine Observations: With Compound X: there is no orange precipitate observed. With Compound Z: there is orange precipitate observed. (c) (i) C C O OH H H C C OH HH H H Intermediate A Intermediate B (ii) step I Reagent(s) and Conditions: …… I 2 (aq), NaOH(aq), reflux………… step II Reagent(s) and Conditions: …… LiAlH 4, dry ether, followed by water…………

4 © INNOVA 2008 8872/02 Section B 1 (a) Kc = [] 3 2 − − ⎡⎤⎣⎦ ⎡⎤⎣⎦ I I I mol-1 dm3 (b) (i) (ii) I - (aq) + I2 (aq) I3 - (aq) Initial amt 0.058 0.080 0 Change –0.0375 –0.0375 +0.0375 (i) Eqm amt 0.0205 0.0425 +0.0375 Eqm conc 0.0205/0.5 0.0425 / 0.5 0.0375/0.5 Kc = [] 3 2 − − ⎡ ⎤⎣ ⎦ ⎡⎤⎣⎦ I I I = 0.0375 0.5 0.0205 0.0425 0.5 0.5 ⎡⎤ ⎢⎥⎣⎦ ⎡⎤ ⎡⎤ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦ = 21.5 mol-1 dm3 (c) (i) Amount of iodine in toluene layer = 0.084 x 500/1000 = 0.042 mol Amount of iodine in aqueous layer = 0.0425 – 0.042 = 5 x 10-4 mol (ii) Partition coefficient = [] 2 2 [I in toluene] I i n w a t e r = ( ) ⎛⎞ ⎜⎟ ⎜⎟⎝⎠ -4 0.084 5x10 5001000 = 84.0 (d) (i) Reagents/conditions: PBr3, heat (ii) Compound K: (CH3)2CHCN Compound L: (CH3)2CHCOOCH2CH3 (e) (i) [O] HCN CH 3CH2CH2CH2OH C 4H8O T S step II Compound S: CH 3CH2CH2CHO Compound T: CH 3CH2CH2CHOH(CN) (ii) nucleophilic addition (f) (i) (ii) The nucleophile for compound G is OH–. Compound H: 9nucleophile is CH 3CH2O–. (iii) Type of reaction : elimination of HCl (iv) Test: add NaOH(aq) to each sample & heat Then add excess HNO 3(aq), followed by AgNO3(aq) Obs:

5 © INNOVA 2008 8872/02 Compound F: white ppt of AgCl Bromobenzene: no ppt. 2 (a) (i) pH is defined as pH = -log [H+] Where [H+] is the concentration of hydrogen ions in aqueous solution. (ii) [H+] of acetylsalicylic acid = (5.62/100) x 0.1 = 5.62 x 10-3 mol dm-3 [1] (iii) Ka = COOH] H O [ ]C ][ H [ 7 2 8 7 2 8 C COO H O−+ [1] = 0.1] [ ] 10 5.62 [2 -3× = 3.16 x 10-4 mol dm-3 [1] (iv) When small amounts of OH - are added, C 8O2H7COOH will react with the OH - and pH changes are resisted. C8O2H7COOH + OH– Æ C8O2H7COO– + H2O When small amounts of H + are added, C 8O2H7COO- will react with the H + and pH changes are resisted. C 8O2H7COO– + H+ Æ C8O2H7COOH (v) Phenolphthalein pH working range of indicator (8-10) coincides with the pH of end point. (b) (i) Amt of salicylic acid = 10 138 = 0.072463 mol Amt of acetic anhydride = 20 102 = 0.19608 mol Limiting reagent = salicylic acid Amt of Aspirin = 0.072463 mol Mass of Aspirin formed = 0.072463 x 180 = 13.0 g (ii) Aspirin decompose to form ethanoic acid which smells like vinegar. (iii) C O O-Na+ O C O C H H H The type of reaction is acid-carbonate. (iv) Dilute HCl and heat.

6 © INNOVA 2008 8872/02 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 (c) (i) MgCO3 + 2HCl Æ MgCl2 + CO2 + H2O (ii) Amt of HCl = 1 100 2 . 0× = 2 x 10-3 mol Amt of MgCO3 needed = 2 x 10-3 / 2 = 1 x 10-3 mol Amt of MgCO3 in one tablet = ( 310 325100 20 −× × )/ 84.3 = 7.7106 x 10-4 mol Number of tablets = 1 x 10-3 / (7.7106 x 10-4) = 1.29 = 2 3(a) 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0 1 02 03 04 05 06 07 08 09 0 1 0 0 1 1 0 Time/min [RCOOR'] mol dm-3 [H+] = 0.20 mol dm-3 [H+] = 0.40 mol dm-3 (b) (i) Since the half lives for the hydrolysis of ester are relatively constant, 31 min and 32 min respectively, the reaction is first order with respect to RCOOR’ (ii) Initial rate when [HCl] = 0.20 mol dm -3 = 1.758 x 10-3 Initial rate when [HCl] = 0.40 mol dm-3 =3.556 x 10-3 When concentration of HCl doubles, in