SAJC H2 CHEM P3 ANS Prelim

1 Solutions for 2015 SAJC H2 Chemistry Prelim Paper 3 1 (a) Organic compounds have various applications. For instance, Ketoprofen is a kind of nonsteroidal anti-inflammatory drug, which can be used as a substitute for aspirin. (i) Friedel Craft Acylation was first discovered by Charles Friedel and James Crafts in 1877. This reaction allowed for the acylation of an aromatic ring with acyl chloride. One such example is given below. Benzophenone, which can be used to synthesise Keto profen, has the structure below. Suggest the reagent needed to convert 2 moles of benzene into 1 mole of benzophenone. [1] COCl2

2 1 (a) (ii) Ketoprofen may be synthesised from benzophenone in the following route. State the reagents and conditions for Steps 1 to 3. Draw the structural formula of A. [4] Step 1: AlCl3 or FeCl3 OR AlBr3 or FeBr3 OR AlBr3 or FeBr3 Step 2: NaOH(aq), heat Step 3: H2SO4 (aq), K2Cr2O7, heat Structure of A: (iii) Ketoprofen is commonly sold as its sodium salt form for better absorption by the body, which is made up of 70% water. Explain, with the aid of a diagram, how this better absorption comes about. [2] The ions in the sodium salt form can form ion-dipole interactions with water , thus enhancing its solubility and absorption by the body.

3 (iv) Nitrogen-containing derivatives of Ketoprofen may be synthesised and some examples are as shown. Describe a simple chemical test to distinguish between compound D and compound E. [2] Add NaOH(aq) to each of the unknown, followed by heating. Add Br2 (aq) at room temperature to each of the resultant products. Orange Br2 (aq) will decolourise, forming a white ppt in the presence of compound E while the orange colour remains in compound D. OR Add NaOH(aq) to each of the unknown, followed by heating. Compound D will produce a colourless, pungent gas which turns moist red litmus paper blue / forms white fumes with a glass rod dipped in concentrated HC l. Compound E does not produce such gas. (v) Ketoprofen can react with cold alkaline HCN to form a product which exists as a mixture of 4 stereoisomers. Give the structure of the product formed. Hence, state the type of isomerism exhibited by this product and draw the stereoisomers. [4] Optical isomerism

4 1 (b) Another use of organic compounds is in the area of food preservation. To prevent food spoilage, the pH of food should not fluctuate too much. Hence, a mixture of lactic acid, CH3CH(OH)COOH, and sodium lactate is commonly added in yoghurt to act as a pH regulator. (i) Explain how the mixture of lactic acid and sodium lactate can function as a pH regulator when an acid is added. Include any relevant equation. [2] Addition of H+: CH3CH(OH)COO- + H+  CH3CH(OH)COOH When a small amount of H + is added, the presence of large reservoir of CH3CH(OH)COO- neutralise the excess H +. Hence, [H +] does not change significantly and pH is kept almost constant.

5 (ii) The pH of yoghurt is usually maintained at about 4.2. Calculate the ratio of the concentrations of lactate ion to lactic acid in yoghurt, given that the pKa of lactic acid is 3.86. [1] Ka = [CH3CH(OH)COO-][H+] / [CH3CH(OH)COOH] = 10-3.86 = 1.380 x 10-4 mol dm-3 [H+] = 10-4.2 = 6.309 x 10-5 mol dm-3 Ratio of [lactate ion] to [lactic acid] = 10-3.86 : 10-4.2 = 2.19 : 1 or 2.19 OR pH = pKa + lg ([lactate ion] / [lactic acid]) 4.2 = 3.86 + lg ([lactate ion] / [lactic acid]) ([lactate ion] / [lactic acid]) = 100.34 = 2.19 (iii) Given that the concentration of lactic acid in a sample of yoghurt is 0.05 mol dm -3 and using your ratio in (b)(ii), calculate the concentration of lactate ion. [1] [lactate ion] = 0.05 x 2.19 = 0.110 or 0.109 mol dm-3 (iv) Perfluorooctanoic acid is a common food contaminant . In yoghurt, this monobasic acid will dissociate completely. Calculate the new pH of the yoghurt when 0.05 mol of perfluorooctanoic acid is added to 5 dm 3 of the yoghurt in b(iii). Hence, calculate the change in pH. [3] No. of moles of lactate ion in the yoghurt = 5 x 0.109 = 0.545 mol No. of moles of lactate ion left after reacting with added H+ = 0.545 – 0.05 = 0.495 mol No. of moles of lactic acid after adding H+ = (5 x 0.05) + 0.05 = 0.3 mol OR CH3CH(OH)COO- + H+  CH3CH(OH)COOH Initial amount / mol 0.545 0.05 5 x 0.05 = 0.25 Change in amount / mol -0.05 -0.05 +0.05 Final amount / mol 0.495 0 0.3 Final concentration / mol dm-3 0.495 / 5 = 0.099 0 0.3 / 5 = 0.06

6 OR CH3CH(OH)COO- + H+  CH3CH(OH)COOH Initial conc./ moldm-3 0.545/5 = 0.109 0.05/5 = 0.01 0.05/5 = 0.01 Change conc./ moldm-3 -0.01 -0.01 +0.01 Final conc. / mol dm-3 0.099 0 0.06 Step  Ka = [CH3CH(OH)COO-][H+] / [CH3CH(OH)COOH] = 10-3.86 = 0.5[H+] / 0.3 pH = - lg(8.282 x 10-5) ≈ 4.08 [1] OR pH = pKa + lg [salt]/[acid] = 3.86 + lg (0.5/0.3) = 4.08 [1] Step  Change in pH = 4.2 – 4.08 = 0.12 [1] [Total: 20 marks] 2 (a) Transition metals are known to have many uses, ranging from materials, medicine to

7 catalysts. Explain what is meant by the term transition metal. [1] A transition element is a d-block element which forms at least one stable ion with an incomplete d sub -shell of electrons / partially filled or incomplete d orbitals of electrons. (b) Aqueous copper( II) sulfate is blue and this is due to the presence of hexaaquacopper(II) ion, [Cu(H2O)6]2+. When 1,2-diaminoethane, H2NCH2CH2NH2, and ammonia are added separately to a sample of copper( II) sulfate solution, the following reactions occur. (Note: “en” represents H2NCH2CH2NH2) Reaction 1: [Cu(H2O)6]2+ (aq) + en (aq) [Cu(en)(H2O)4]2+ (aq) + 2H2O (l) ΔHƟ = - 52 kJ mol-1 Reaction 2: [Cu(H2O)6]2+ (aq) + 2NH3 (aq) [Cu(NH3)2(H2O)4]2+ (aq) + 2H2O (l) ΔHƟ = - 48 kJ mol-1 (i) State the full electronic configuration of Cu2+. [1] 1s2 2s2 2p6 3s2 3p6 3d9 (ii) With reference to relevant bonds, s uggest why the two enthalpy changes are similar in value. [1] Both reactions involve the breaking of 2 C u-O bonds (bond between Cu and H2O) and forming of 2 Cu-N bonds, hence they have similar ΔH values. (iii) By considering the entropy and enthalpy changes of Reaction 1 and Reaction

8 2, suggest and explain the difference in magnitude of the standard Gibbs free energy change of the two reactions. [2] The number of species increased in reaction 1