HCI H2 CHEM P3 ANSWERS Prelim

1    Hwa Chong Institution 2011 Prelim Paper 3 Answers 1 (a) (i) At the water surface, the surfactants are orientated such that the hydrophilic heads are exposed to water for favourable ion-dipole interactions, whereas the hydrophobic tails are exposed above the water surface, maintained by dispersion forces. (ii) Mechanical agitation during washing causes oil or grease to be surrounded by surfactant molecules and broken into small droplets so that relatively small micelles are formed. The charged surfaces of the micelles prevent the oil or grease droplets from coalescing again and the micelles are easily washed away. (b) (i) PCl 5 (s), room temperature (ii) acid-base reaction (iii) H2NCH2CH2OH is bifunctional and the acid chloride would be able to react with both the amine and the alcohol to give side products (C 11H23CO2CH2CH2NH2 or C11H23CONHCH2CH2OCOC11H23). (iv) CH3CH2Cl CH2=CH2 CH2BrCH 2OH H2NCH 2CH2OHethanolic NaOH heat Br2 (aq) conc NH 3 heat in a sealed tube (c) Electrophilic Addition NHBr Br  NHBr Br N Br slow fast - HBr N Br H Br Alternatively, Electrophilic Addition (followed by nucleophilic substitution)

2    NHBr Br  NHBr Br NH Br Br   N Br slow fast - H+ (d) (i) Kp  PSO3 2 PSO2 2  PO2 PSO2 = 0.15 atm PO2 = 0.075 atm PSO3 = 8 – 0.15 – 0.075 = 7.775 atm 075.015.0 775.7 2 2  pK = 3.58 x 104 atm–1 (ii) V2O5 acts as a heterogeneous catalyst. It is in a different phase from the reacting mixture. The reactant molecules readily adsorb onto the surface of the catalyst. The adsorption process brings the reactant molecules closer thus increasing their concentrations at the catalyst surface ; adsorption also weakens the strong S=O and O=O bonds ; and allows these molecules to be orientated in the right positions for reaction . The catalysed pathway thus involves lower activation energy. When the reaction is complete, the molecules will desorb and diffuse away from catalyst surface so that the active sites are exposed for further reaction.

3    2 (a) Liquid fuels must be vaporized before combustion can take place. As diesel oil has more C atoms than gasoline and hence a larger electron cloud , the dispersion forces between its molecules are stronger. Its boiling point is higher and hence vaporizes much less easily and therefore is not easily ignited. (b) (i) C2H5OH (l) + H2O (l) + O2 (g)  4H2 (g) + 2CO2 (g) OR 2C2H5OH (l) + 4H2O (l) + O2 (g)  10H2 (g) + 4CO2 (g) (ii) The sign of entropy change should be positive. There is a greater number of moles of gas on the product side of the equation hence there is greater disorder and therefore a greater number of ways to distribute the energy. (iii) Ethanol has more hydrogen atoms and hence can produce more hydrogen fuel per mole of ethanol Methanol is toxic while ethanol is not. (c) (i) H 2 (g) + ½O2 (g)  H2O (g) C2H5OH (l or g) + 3O2 (g)  2CO2 (g) + 3H2O (g) (ii) ∆Hc of H2 = (436 + ½ x 496) – (2 x 460) = - 236 kJ mol–1 ∆Hc of C2H5OH = (410 x 5 + 350 x 1 + 360 x 1 + 460 x 1 + 496 x 3) – (2 x 2 x 740 + 3 x 2 x 460) = - 1010 kJ mol –1 (iii) Since ethanol produces more heat and a greater number of moles of gas per mole of ethanol compared to hydrogen on combustion, ethanol is the more efficient fuel.

4    (d) Summary of the reaction scheme:    Observations Deductions C can exist as a pair of enantiomers but D and E does not show any optical activity. C contains a chiral carbon whereas D and E does not contain chiral carbon (or are symmetrical). 1 mol of C and D reacts with 2 mol of liquid bromine. 1 mol of E reacts with 3 mol of liquid bromine. Electrophilic addition. C and D have 2 C=C, R has 3 C=C. C, D and E react with hydrogen in the presence of nickel to give the same compound. (Catalytic) reduction/hydrogenation occurs. C undergoes oxidative cleavage with acidified KMnO4 to give ethanoic acid and a compound containing 10C. C contains a terminal alkene as there is loss of CO2. D does not contain terminal alkene. C and D contains the CH 3CH=C structure. D undergoes oxidative cleavage with acidified KMnO4 to give ethanoic acid and a compound containing 11C. F reacts with 1 mole of Na2CO3 to produce CO2. Acid-base reaction occurs and F is dibasic (dicarboxylic acid). F reacts with 2 moles of HCN under cold condition with base as the catalyst. Nucleophilic addition occurs and F contains 2 carbonyl groups. Structures of unknown: OO O OH O O O O OH P R C D E F

5    3 (a) Enzymes are highly specific in the type of reactions which they catalyze, without any side-reactions. Enzymes can be denatured by high temperature / have an optimum temperature at which they function most efficiently. Enzymes have an optimum set of pH at which they function most efficiently. Any two properties (bi) Step 1: SODCu2+ + O2   O2 + SODCu+ E cell = +0.42 – (0.33) = +0.75V (> 0, hence feasible) Step 2: SODCu+ + O2  + 2H+  H2O2 + SODCu2+ E cell = +0.89 – (+0.42) = +0.47V (> 0, hence feasible) (bii) α-helix Correctly drawn diagram (helical structure and illustration of hydrogen bond)  Backbone of the polypeptide chain coiled to form a helical arrangement with 3.6 amino acids per turn of helix  Stabilized by hydrogen bonds formed between the peptide C=O and the peptide N-H group four amino acid residues away  Side-chain of each amino acid residue points outwards from the helix (described or shown in diagram)

6    -pleated sheet Correctly drawn diagram (parallel or anti-parallel arrangement and illustration of hydrogen bond)  Segments of polypeptide backbone fold back on itself with adjacent strands running in opposite directions (anti-parallel) or in same direction (parallel)  Stabilized by hydrogen bonds formed between the peptide N-H and C=O groups from adjacent rows (described or shown in diagram)  R-groups perpendicular to plane of sheet / point up and down from plane of sheet (iii)  The guanidinium cation can interfere with the ionic linkages formed between the side chains of the amino acid residues by forming ionic bonds with negatively charged side chains, hence disrupting the tertiary structure.   CO2 NH3+ Ionic linkage between ionic side chains Addition of guanidium chloride CO2 NH3+ Loss of ionic interaction due to formation of ionic bond between guanidinium and carboxylate side chain, leading to loss of tertiary structure. H 2N NH2 NH2  It can also form hydrogen bonds/ ion-dipole interactions with the peptide C=O or N-H groups (secondary structure), and/or polar side chains (tertiary structure) of the amino acid residues, hence disrupting the hydrogen bonding interactions in the secondary and/or tertiary structures of the protein.  +   + parallel   + +  anti-parallel Hydrogen bonding OR

7    C Hydrogen bond formed between peptide C=O and N-H groups Addition of guanidium chloride O N H OH N H C O OH + +   Hydrogen bond formed betw