DHS H2 Chem 2012 Prelim P3 Soln

2 © DHS 2012 9647/03 [Turn over Answer any four questions. 1 The study of organic compounds includes the collection of kinetic data and thermodynamic data of the compounds. (a) The kinetics of the hydrolysis of the ester, CH 3CH2CO2CH3, may be investigated by the following method. CH 3CH2CO2CH3 + H2O ® CH3CH2CO2H + CH3OH In a 1 dm 3 mixture, 0.350 mol of the ester was hydrolysed by heating with water and using hydrochloric acid as catalyst. The following results were obtained. Time/s Concentration of CH3CH2CO2H/mol dm–3 0 0 340 0.105 680 0.185 1080 0.243 1440 0.278 (i) Suggest how the progress of this reaction may be followed in order to obtain the results as stated. Taking samples of reaction at the stated intervals (i.e. at 340 s, 680 s, 1080 s and 1440 s) and quenching it with large amount of cold water. Titrate sample with dilute NaOH and suitable indicator. (ii) By drawing a suitable graph using the data given above, show that the reaction is first order with respect to the ester. It has been found that the hydrolysis reaction is first order with respect to the hydrochloric acid. y = -1E-07x2 + 0.0003x + 0.001 0 0.05 0.1 0.15 0.2 0.25 0.3 0 200 400 600 800 1000 1200 1400 1600 Time/s [RCOOH]/mo ldm The half life of the ester is approximately 640 seconds based on the data. Hence, the order of reaction with respect to the ester is 1 since a constant half life is established.

3 © DHS 2012 9647/03 [Turn over (iii) Deduce the units of the rate constant. Rate = k[ester][HCl] Units for k is mol–1 dm3 s–1 (iv) State and explain the effect of a catalyst on the rate constant. The catalyst increases the rate constant by decreasing the activation energy of the reaction and offers an alternative pathway for reaction. (v) Using suitable bond energy values from the Data Booklet, calculate the ∆H for the hydrolysis of the ester. ∆H = [ 8 ´ 410 + 2 ´ 350 + 2 ´ 360 + 740 + 2 ´ 460] – [8 ´ 410 + 2 ´ 350 + 2 ´ 360 + 740 + 2 ´ 460] = 0 kJ mol–1 (vi) Given that the standard enthalpy change of reaction for the hydrolysis is +7.6 kJ mol –1, suggest a reason for the difference between this given value and the value that you have calculated in (a)(v). The ester is not in gaseous phase which would make the calculation by bond energy inaccurate. [11] (b) Hydrolysis of ester can be achieved in the biologica l system by enzymes known as esterase. Enzymes are proteins that catalysed a specific chemical transformation in the biological system. Such enzymes are generally quaternary proteins. (i) Sketch and explain the graph showing how the rate of hydrolysis c hanges with increasing concentration of the ester. [ester] / mol dm-3 Rate At low concentration of ester, the rate of hydrolysis is roughly proportional to the concentration of ester and therefore the order of reaction with respect to ester is about 1. At moderate concentration of ester, the rate of hydrolysis is no longer proportional to the concentration of ester and therefore the order of reaction with respect to ester is mixed order.

4 © DHS 2012 9647/03 [Turn over At high concentration of ester, the rate of hydrolysis is independent to the concentration of ester and therefore the order of reaction with respect to ester is about 0. (ii) Explain the meaning of quaternary structure of proteins. The quaternary structure of a protein refers to the spatial arrangement of two or more polypeptide chains , held together into a specific geometry by bonding interactions like van der Waals’ forces, hydrogen bonding, ionic bonds and disulfide bridges. (iii) List 2 other major functions of proteins in the body. w As enzymatic c atalysts (e.g. amylase breaks down starch in the digestive system) w As transport molecules (e.g. haemoglobin transports oxygen) w As storage molecules (e.g. iron is stored in the liver as a complex with protein ferritin) w In movement (e.g. proteins are the major component of muscles) w For mechanical support (e.g. skin and bone contain collagen – a fibrous protein) w Mediate cell responses (e.g. rhodopsin is a protein in the eye used for vision) w For immunity / protection against diseases (antibody proteins) w For control of growth and cell differentiation (hormones) (iv) Suggest and explain a chemical method for distinguishing the following pair of esters. You should state the expected observations. HCO2CH2CH3 and CH3CH2CO2CH3 Method 1: Add aqueous NaOH separately to the 2 compounds and heat. Then, add aqueous iodine to both compounds. Observations: Yellow ppt. observed for HCO2CH2CH3. No yellow ppt. observed for CH3CH2CO2CH3. OR Method 2: Add acidified KMnO4 separately to the 2 compounds and heat. Observations: For HCO2CH2CH3, purple colour is decolourised. Effervescence observed. Gas evolved gives white precipitate with limewater. (Gas is CO2.) For CH3CH2CO2CH3, purple colour is decolourised. No effervescence observed. [9] [Total: 20]

5 © DHS 2012 9647/03 [Turn over 2 This question is about the chemistry of the transition metal, nickel and its compound. (a) Explain why the colour of [Ni(NH3)6]2+(aq) is blue. The d orbitals of Ni2+ are split into two groups of different energy levels by NH3 ligands. When white light shines on the complex, a d electron undergoes d–d transition and is promoted to a higher energy d orbital. During the transition, the d electron absorbs in the orange region of the visible spectrum . The blue colour observed is the colour of transmitted light, which is a mixture of remaining wavelengths that are not absorbed. [3] (b) Ni(CO)4 is a compound formed by the reaction between nickel and carbon monoxide. The Mond process was developed by Ludwig Mond to extract and purify nickel from its ores. One of the stages of this process involves the decomposition of Ni(CO) 4 at 227 °C to give nickel as shown in this equation below: Ni(CO)4(g) Ni(s) + 4CO(g) The equilibrium constant, K p, for the equilibrium at 227 °C is 1.01 atm 3. A sample of gaseous Ni(CO)4 was placed in a 2 dm3 evacuated container at 227 °C. At equilibrium, the partial pressure of CO was 2.00 atm. [1 atm = 1.01 x 105 Pa] (i) Sketch the shapes of the hybrid orbitals around the C atom in carbon monoxide. Show 2 sp hybrid orbitals (ii) Write an expression for Kp. Kp = Error! Objects cannot be created from editing field codes. (iii) Calculate the total pressure of the system at equilibrium. (2.00)4 / pNi(CO)4 = 1.01 pNi(CO)4 = 15.842 = 15.8 atm total pressure of system = 15.842 + 2.00 = 17.8 atm (iv) Calculate the mass of Ni(CO)4 placed in the container initially. Ni(CO) 4(g) ⇌ Ni(s) + 4CO(g) Initial p/atm 15.8 + (2.00/4) – 0 = 16.3 Eqm p/atm 15.8 – 2.00

6 © DHS 2012 9647/03 [Turn over Let mass of Ni(CO)4 be m. pV = nRT = (m/M)RT 16.3 x 1.01 x 105 x 2 x 10–3 = {m / [58.7 + 4 x (12.0 + 16.0)]} x 8.31 x (227 + 273) m = 135.27 = 135 g [8] (c) Ni is commonly used in catalytic hydro genation reactions. One such example is given below. Compound A, C 10H12NOCl, has a chiral centre and dissolves in dilute sulfuric acid. It reacts with 2,4– dinitrophenylhydrazine to form an orange precipitate, but does not react with Tollens’ reagent. A reacts with H 2 in the presence of Ni catalyst followed by addition of aqueous bromine to form B, C 10H11NOClBr3. When 1 mol of compound B is heated under reflux with aqueous iodine and excess alkali, followed by careful acidification, compound C, C8H4NO4Br3 is formed together with 2 mol of yellow solid CHI3. Deduce the structures of compounds A, B and C, giving reason