DHS Prelim_H2_CHEM_P3_ans

© DHS 2016 9647/03 2016 Y6 H2 Chemistry Preliminary Examination Paper 3 (Answer Scheme) 1 (a) (i) Electrophilic substitution (ii) Choose scheme 2 (b) (i) To ensure that [bleach] remains almost constant throughout the reaction so that the rate can be measured with respect to Blue #1 in a pseudo -order reaction. CH2CH3 conc. H2SO4 heat NaOH (aq) 3000C Cl2, uv heat A CH2CH3 CH2CH2Cl CH2CH3

© DHS 2016 9647/03 (ii) From graph hand-plotted, when absorbance decreases from 0.40 to 0.20, t½  78s when absorbance decreases from 0.50 to 0.25, t½  76s Since almost constant t½  77s is observed, the order of reaction wrt to [Blue#1] is 1. OR Reaction is first order with respect to [Blue#1]. (iii) Using t½= ln 2 𝑘⁄ , find value of k for experiment 1. 𝑘 = 𝑙𝑛2 77⁄ = 0.00900 s-1 Compare experiment 1 and 3, when the volume of bleach doubles, rate constant almost doubles (i.e. 0.01498 0.00900⁄ = 1.67 ≈ 2). The reaction is first order with respect to [bleach]. Hence, overall order of reaction is 2. (c) (i) Cl2 + H2O → HCl + HOCl (ii) MgCl2 is an ionic compound which ionises in water to form hydrated Mg 2+ and Cl− ions. As Mg2+ has a high charge and small ionic radius / has a high charge density, it hydrolyses slightly in water to form a weakly acidic solution of pH 6.5 [Mg(H2O)6]2+ + H2O [Mg(H2O)5OH]+ + H3O+ SiCl4 is a covalent compound which hydrolyses in water to form an acidic solution of pH 2. SiCl4 + H2O → Si(OH)4 + 4HCl

© DHS 2016 9647/03 2(a) (i) Na B HH H H x x x+ - (ii) Sodium borohydride is a milder reducing agent and hence is less reactive and will not react as violently with water present. OR Lithium aluminium hydride is a very strong reducing agent and hence is highly reactive and will react violently with water present. (iii)  At 360 K, vanillin is a liquid while vanillyl alcohol is a solid. There is a decrease in the degree of disorder liness, hence the entropy change of reaction at 360 K is negative. (iv)  Ho and So will change as temperature increases.  Go may become more/less negative as temperature increases.  (b) (i) The reaction is exothermic / large amounts of heat will be produced. If it was added all at once and this may cause NaBH4 to decompose. (ii) NaBH4 + 3H2O + HCl → NaCl + H3BO3 + 4H2 (iii) Use excess NaBH4 as it may decompose at high temperatures. Allow reactants to stir for a longer period. (iv) To the aliquot drawn out, add a small amount of 2,4–DNPH. An orange precipitate will be seen if vanillin is present. C OH OCH3 H O NH NO2 NO2 NH2 + C OH OCH3 H NH NO2 NO2N + H2O

© DHS 2016 9647/03 (c) (i) W X Y OH O OH O Br OH O Br Br OH OH O OH Br Br or Step 2: alcoholic KOH, heat under reflux Step 4: phenol, room temperature (ii)  Vanillic acid has a higher boiling point.  Due to proximity of