NJC Prelim P2 ANS

Solutions to 2009 prelim H2 Chemistry P 2 1 1 a i Pe(CH4) = 1.25 atm; Pe(H2S) = 2.5 atm; Pe(CS2) = 1.25 atm; Pe(H2) = 5.0 atm ii Kp = 2 S H CH 4 H CS 24 22 P P P P ) )( ( ) )( ( atm2 iii K p = 100 atm2 iv (I) Equilibrium shifts left. (II) Equilibrium shifts right. Q1b)(i) pH = 5.13 (2 dec pl.) (ii) amt of NH 4 + to be neutralized to form buffer of max buffer capacity = vol of NaOH required = 10 cm 3 pH = pK a = 9.26 Q2a) (i) It is a polypeptide chain joined by amino acid units through peptide bonds. (ii) IIe-Gly-Asp-Glu-Asn-Tyr (b) (i) -CH 2SH + -CH2SH + [O]  -CH2S-SCH2- + H2O -CH2SH + -CH2SH  -CH2S-SCH2- + H2 -CH2SH + -CH2SH  -CH2S-SCH2- + 2[H] (ii) tdid : Valine – phenylalanine: ionic interaction : lysine- glutamic, lysine- aspartic acid H-bonding: Serine – Aspartic acid: (Serine- glutamic acid), (Aspartic – glutamic acid) (c) (i) optimum working pH for this enzyme is 2.5. At pH lower or above 2.5, the existing ionic or H-bonding interactions are destroyed . changing the tertiary structure of protein rendering it inactive. (ii) addition of heavy metal ions which will destroy the disulfide bonds by forming ppt with sulfur. Protein denatured therefore lost its activity. Supply of heat/ high temp also acceptable. (i) CH3CONHC H COOH (CH2)4NHCOCH3

(ii) Both Lysine and compound A can react with alkali to form soluble salt. However, when compound A is formed, it loses its basic property as it forms the amide (not peptide) linkage. No reaction therefore insoluble in acid. However it is still soluble in alkaline medium as it still carries the COOH group that can react to form the soluble salt in alkaline solution. 3(a) (i) Al 2O3 + 6H+  2Al3+ + 3H2O Al2O3 + 2OH + 3H2O  2[Al(OH)4] 2  (b) (i) Ionic Radius Na Mg Al Si P S Cl pH Na2O MgO Al2O3 SiO2 P4O10 7 (ii) Ionic Radius: SO There are two isoelectr onic series in period 3. Within each series, an increasing number of protons results in increasingly stronger nuclear attraction experienced by the valence electrons. Valence electrons are pulled closer to nucleus, leading to a decreasing trend in ionic radius. Anionic series has a larger size than cationic series as the valence electrons are in a higher principal quantum shell, experience weaker nuclear attraction and are less tightly held. pH: Na 2O and MgO are ionic oxides that dissolve completely in water to form basic solution. Al2O3 and SiO2 are insoluble in water pH of solution is that of water. P4O10 and SO3 are covalent oxides that hydrolyse in water to give acidic solution. (c) (i) [Mg(H2O)6]2+ + H2O [Mg(H2O)5(OH)]+ + H3O+

3 S. (ii) H = +4336 kJ mol1 (iii) Formation of both MgCl 2 and MgCl3 give rise to negative S as there is a decrease in the amount of gaseous particles in the system after the reaction. G = H – TS Comparing H of both reactions, formation of MgCl3 will result in a positive G whereas formation of MgCl2 will result in a negative G as G is more dependent on magnitude and sign of H of the reactions since both reactions have a small difference in So formation of MgCl3 will be less feasible than that of MgCl2. 4. (a) (i) K 2Cr2O7, dil. H2SO4, heat with distillation (ii) 2,4-dinitrophenylhydrazine. Positive observation: orange ppt (iii) Cyanide ions can attack the trigonal planar carbonyl carbon from top and bottom with equal probabilities. Hence a racemic mixture is formed. The optical activity of one enantiomer cancels out the optical activity of the other hence the product mixture does not exhibit any optical activity. (b) (i) (CH 3)2CHCH2CH2OCOCH3 + NaOH  (CH3)2CHCH2CH2OH + CH3CO2Na (ii) I. The change in concentration of NaOH per unit time can be tracked by the change in pH of the solution using a pH meter. II. To determine the order wrt to NaOH, measure the change in [NaOH] vs time with large excess of ester. [Ester] is much higher than [NaOH]. Plot [NaOH] vs time graph. From the half-life, order wrt [NaOH] can be determined. To determine the order wrt ester, conduct a second experiment with double [ester], but same [NaOH]. Plot [NaOH] vs time. Compare the ratio is of initial rate of first and second experiment and the ratio of the [ester] in both experiment to determine the order wrt ester. (iii) The intermolecular forces in acid is h ydrogen bonding while in ester is permanent dipole- dipole interaction. Since hydrogen bonding is stronger than permanent dipole-dipole interaction, more energy is required to break the intermolecular forces in acid to change it from liquid to gas. (iv) Stage I: LiAlH 4, dry ether Stage II: excess con. H2SO4, 170°C Stage III: Cold KMnO4, NaOH Compound C: (CH3)2CHCH=CH2