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 –OH and –COOH groups, 2–hydroxy–3– methoxybenzoic acid is capable of forming intramolecular hydrogen bonding, thus reduces the extent of intermolecular hydrogen bonding formed.  More energy is needed to overcome the more extensive intermolecular hydrogen bonds in vanillic acid. (d) OH O BrMgO (e) (i) 3LiNH2  Li3N + 2NH3 (ii)  Charge density: Li+ ( 06.0 1 ) << Mg2+ ( 065.0 2 )  Polarising power: Li+ << Mg2+ OR Distortion of electron cloud of NH2– by both cations: Li+ << Mg2+  Therefore LiNH 2 will have a higher decomposition temperature than Mg(NH2)2.

© DHS 2016 9647/03 3 (a) (i) Nucleophilic addition (ii) NaOH provides the –OH ions which is a stronger nucleophile for the nucleophilic attack on the carbonyl carbon. (iii) OH + H2O* H2O + *OH The OH – will deprotonate the 18O–enriched water to form the *OH –. *OH– then attacks the methanal via nucleophilic addition reaction. (iv) The carbonyl carbon in aldehyde is less sterically hindered than that in ketones, hence more easily attacked by the nucleophile. OR The carbonyl carbon in aldehyde is more electron deficient ( +) than that of ketones as ketones have an additional electron donating alkyl group. (b) (i) 2–hydroxyethanoic acid has a larger Ka as it is a stronger acid. HCH(OH)COOH HCH(OH)COO– + H+ -–– (1) CH3COOH CH3COO– + H+ -–– (2) The electron –withdrawing –OH group disperses the negative charge on the conjugate base of 2 –hydroxyethanoic acid / 2 –hydroxyethanoate ion hence stabilising the conjugate base/ 2 –hydroxyethanoate ion. The equilibrium position of reaction (1) lies more to the right, producing more H + ions, resulting in 2–hydroxyethanoic acid being a stronger acid. OR Intramolecular hydrogen bond can be formed in the 2–hydroxyethanoate ion between the carboxylate ion and the H of the alcohol group hence stabilising the conjugate base/ 2 –hydroxyethanoate ion. The equilibrium position of reaction (1) lies more to the right, producing more H + ions resulting in 2 – hydroxyethanoic acid being a stronger acid. (ii) If both indicators are used in larger amount, a larger volume of sodium hydroxide will be used to reach the end –point to deprotonate the indicators which are also weak acids. (iii) X = 3.86 X corresponds to the pH of the buffer at maximum buffering capacity. At maximum buffer capacity, pH = pKa (iv)

© DHS 2016 9647/03 (v) First end–point corresponds to the neutralisation of 2–hydroxyethanoic acid. Concentration of 2–hydroxyethanoic acid = ( 17.20 1000 x 1) ÷ 20 1000 = 0.860 mol dm3 Second end–point corresponds to the neutralisation of phenol. Volume of NaOH used for reaction with phenol = 26.80 – 17.20 = 9.60 cm3 Concentration of phenol = ( 9.60 1000 x 1) ÷ 20 1000 = 0.480 mol dm3 (c) (i) H C CH2CH2COOH O H C CH2CH2COOH NH (ii) Strecker’s synthesis produces a racemic mixture and hence, do not display any optical activity while naturally occurring glutamic acid is present as one of the enantiomers and will rotate plane polarised light. (d) (i) (ii) The alpha helix is held in place due to hydrogen bonding. Hydrogen bonding occurs between the peptide –C=O group of the nth amino acid and the peptide –NH group of the (n+4)th amino acid which is in the adjacent turn. (iii) Heat increases thermal vibrations of the protein molecule, disrupting van der Waals’ interactions formed between uncharged R –groups in the tertiary/quaternary structure. The hydrogen bonds between polar R groups in tertiary structure are also broken.

© DHS 2016 9647/03 4 (a) CO2 CF3CO2  CF3 + CO2 + e (b) +3 (c) CO2CH3 CO2CH3 F3C F3C CO2CH3 F3C CO2CH3 CF3 CO2CH3 F3C CO2CH3 CF3 Y X (d) (i) time = 23/20 × 60 = 69 s (ii) n(Y) = 0.401/310 = 0.0012935 mol Since 2e  2CF3  2X  Y, n(e) = 0.0025870 mol Q = (0.0025870)(96500) = 249.65 C I = 249.65/69 = 3.62 A (e) (i) 3 (ii) H CH2CF3 H CO2CH3CO2CH3 CH2CF3 (f) CF3 F3C CONH2 (g) The C–F bonds in the –CF3 group are relatively strong and hence inert to chemical reactions. (h) (i) CO2CH3 CH3 CHF2 (ii) The presence of the methyl group at the C atom with the unpaired electron makes it more sterically hindered. It is more difficult for two bulkier radicals to collide effectively to form the dimer. (i) I CuL+ > CdL+ > NiL+ II log K (Cu2+-CH3COO) > log K (Cu2+-HCOO)

© DHS 2016 9647/03 From the table, the stability of the complexes is in the same order as the availability of the lone pair of electrons on the negatively charged oxygen of the carboxylate anions for dative bonding with the metal ion. Since the lone pair of electrons on the negativ