ACJC H2 CHEM P3 Prelim

2 Answer any four questions. 1 (a) Samples of 2-bromo-2-methylpropane were dissolved in dilute aqueous ethanol (80% ethanol and 20% water by volume) and reacted with sodium hydroxide solution. Several experiments were carried out at constant tem perature. The initial rate of reaction was determined in each case. Expt [(CH 3)3CBr] / mol dm-3 [OH-] / mol dm-3 Rate / mol dm-3 s-1 1 0.020 0.010 20.2 2 0.020 0.020 20.2 3 0.040 0.030 40.4 Calculate a value for the rate constant. Hence write the rate equation for the reaction. [3] (b) The hydrolysis of 2-bromopropane is now investigated. Samples were dissolved in dilute aqueous ethanol (80% ethanol and 20% water by volume) and reacted with sodium hydroxide solution. The rate equation is found to be as follows: Rate = 0.24 x 10-5 [CH(CH3)2Br] + 4.7 x 10-5 [CH(CH3)2Br] [OH¯] (i) The rate equation obtained indicates that th e hydrolysis of 2-bromopropane exhibits a mixture of both first order and second order kinetics. Suggest why this may be so. [2] (ii) By deriving an expressi on in terms of [OH -], show that the % rate due to S N2 is - - 4.7[OH ] X 100%4.7[OH ] + 0.24 . [1] (iii) Using the expression derived from (b)(ii), calculate the % rate due to SN2 for various [OH-] of 0.01M, 0.1M and 1.0M. [1] (iv) In light of your answer to (b)(iii) , state how the % rate due to S N2 depends on the concentration of [OH-] and explain why it varies in that manner. [2] (v) By comparing the magnitude of the two rate constants in the rate equation given in (b), comment on the ease of hydrolysis by the two different pathways. The Arrhenius Equation is given as follows: k = A e –Ea / RT [2] (vi) Draw a labelled Maxwell-Boltzmann Curve for this reaction to illustrate the distribution of energy. Based on your answer to (b)(v), indicate the activation energies clearly on the diagram for the two different reaction pathways. [3] ACJC 2011 9647/03/Aug/11 [Turn over

3 (vii) On the same axes, draw the energy profile illustrating both the S N1 kinetics and the SN2 kinetics components observed for this reaction. Indicate the activation energies clearly on the diagram, for the two reaction pathways. [2] (viii) Explain how the rate constant will change if CH(CH 3)2Cl is used instead of CH(CH3)2Br. [2] (ix) Draw the structures of two possible organic by-products that may be formed in this reaction, other than CH(CH3)2(OH). Briefly account for their formations. [2] [Total: 20 marks] ACJC 2011 9647/03/Aug/2011 [Turn over

4 2 Oxalic acid is an organic compound with the formula H 2C2O4. This colourless solid is a dicarboxylic acid. In terms of acid strength, it is about 3,000 times stronger than acetic acid. Its conjugate base, known as oxalate (C 2O4 2−), is a reducing agent as well as a chelating agent for metal cations. Oxalic acid dissociates in water according to the following equations I: HOOC-COOH + H2O HOOC-COO- + H3O+ Ka1 = 5.6 x 10-2 mol dm-3 II: HOOC-COO- + H2O -OOC-COO- + H3O+ Ka2 = 5.4 x 10-5 mol dm-3 (a) (i) Explain why the value of Ka1 is larger than Ka2. [1] (ii) Write expressions for the acid dissociation constants for equation I and II above. [2] (b) (i) The neutralisation between oxalic acid and sodium hydroxide corresponds to the two equations given: HOOC-COOH + NaOH → HOOC-COONa + H 2O step 1 HOOC-COONa + NaOH→ NaOOC-COONa + H2O step 2 A 25 cm 3 sample of 0.100 mol dm -3 of oxalic acid was titrated with 0.100 mol dm -3 of sodium hydroxide. The titration curve for the above titration was given below. 25 50 Vol of NaOH/cm3 pH X Y ACJC 2011 9647/03/Aug/11 [Turn over

5 At the first equivalent point, X, the species formed is HOOC-COO-(aq), which is both an acid and a base where the relevant equilibriums are: HOOC-COO-(aq) + H2O(l) H3O+(aq) + (COO)2 2-(aq) HOOC-COO-(aq) + H2O(l) OH-(aq) + (COOH)2 (aq) It can be shown that in such an instance the [H3O+] at the first equivalent point, X, can be given by expression, [H3O+] = 2 a 1 aK K Using the expression, determine the pH value at point X. [1] (ii) Calculate the pH value at the second equivalent point, Y, given that the [OH -] can be assumed to be entirely due to the hydrolysis: -OOC-COO- + H2O HOOC-COO- + OH- [3] (c) The concentration of oxalic acid in a solution can be determined by an acid-base titration. State another way by which the concentration can be determined volumetrically. [1] (d) Oxalate, the salt from oxalic acid, is able to combine chemically with certain metals commonly found in the human body, it is also able to bond chemically, behaving as bidentate chelating agents, to transition elements, such as iron to form complex ion . A chelating agent is a ligand that is attached to a central metal ion by bonds from two or more donor atoms. [Fe(H 2O)6]3+ + 3 C2O4 2- [Fe(C2O4)3]3- + 6H2O Kstab = 5.00 x 104 (i) What do you understand by the term “bidentate” ligand. [1] (ii) Suggest a reason why the stability constant of the above is greater than 1. [1] (iii) Draw the structure of the complex ion, [Fe(C2O4)3] 3-. State the type of bonds formed between the ligands and the metal ion. Hence suggest the shape for this complex. [3] (e) Oxalic acid was one of the products formed when an aromatic organic compound, A, with molecular formula C 10H10O2 undergoes oxidation with acidified manganate(VII) to form another organic product, B, with the molecular formula C8H8O2. No other organic compound was formed in the oxidation. Compound B reacts readily with 2 moles of Br 2(aq) to form compound E, C8H6O2Br2. Compounds A and B are both soluble in NaOH and both A and B reacts with 2,4 - DNPH. Compound A reacts with acidified dichromate to give an acid, C , C10H10O3. Compound C reacts with SOCl2 to form a sweet-smelling compound D, C10H8O2. Deduce the structures of A, B, C, D and E. Explain your deductions [7] [Total: 20 marks] ACJC 2011 9647/03/Aug/2011 [Turn over

6 3 The lac repressor is a DNA-binding protein which inhibits formation of proteins involved in the metabolism of lactose in bacteria. It is active in the absence of lactose, ensuring that the bacterium only invests energy metabolism of lactose when lactose is present. When lactose becomes available, it is converted into allolactose, which inhibits the lac repressor's DNA binding ability. Structurally, the lac represso