SAJC H2 CHEM P3 Ans Prelim

1 Name: Class: ST ANDREW’S JUNIOR COLLEGE JC2 Preliminary Examinations Chemistry Higher 2 9647/03 16 September 2016 Paper 3 2 hours Candidates answer on separate paper. Additional Materials: Writing paper, Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and civics group on all the work you hand in. Write in dark blue or black pen. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer any four questions. You are reminded of the need for good English and clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 14 printed pages including this page. [Turn over

2 1 Propenoic acid, also commonly known as acrylic acid, is used in the formation of many polymers. (a) Acrylic acid can be synthesised from 2-chloroethanol in the laboratory by the following route. (i) Suggest the structure for intermediate A. [1] (ii) Suggest reagents and conditions for step 2 and for step 3. [2] Step 2: excess conc. H2SO4, 170OC Step 3: H2SO4 (aq), heat (iii) What type of reaction is occurring in step 2? [1] Elimination (b) Acrylic acid is manufactured industrially from the reaction of propene and oxygen gas. (i) Write a balanced equation for this reaction. [1] CH3CH=CH2 + 3/2O2  CH2=CHCO2H + H2O (ii) Propenal was isolated as an intermediate product in the reaction. Draw the structure of propenal. [1] (c) Polyacrylate is used in environmentally -friendly detergents to remove calcium ions and magnesium ions in water. In the manufacture of polyacrylate, acrylic acid was first converted into acrylic chloride, followed by reaction with an alcohol to form a monomer. Polyacrylate is formed when the monomer undergoes polymerisation.

3 (i) Write a balanced equation for the conversion of acrylic acid into acrylic chloride. [1] (ii) Suggest the type of reaction occurring in the for mation of the monomer from acrylic chloride. [1] Nucleophilic substitution/Condensation (iii) Suggest how polyacrylate can remove magnesium ions from water through complex formation. [1] The lone pair of electrons on O of the ester functional group / ligand can form dative bond/coordinate bonds with empty orbitals of the cation. (d) In the kinetic studies of polymerisation, transition element-containing catalysts were used. A copper(I)-containing catalyst is shown below. Cu(I)-nBuPCA catalyst (i) What is the co-ordination number of copper in the Cu(I)-nBuPCA catalyst? [1] 4 (ii) What is the electronic configuration of copper in Cu(I)-nBuPCA? [1] 1s22s22p63s23p63d10

4 (iii) Suggest whether copper in Cu( I)-nBuPCA catalyst can be classified as a transition element. [1] No. Cu+ is not a transition element as it does not have incompletely filled 3d orbitals / does not have partially filled 3d orbitals / has completely filled 3d orbitals. (e) Cu(I) undergoes disproportionation to form Cu and Cu(II). (i) Use relevant data from the Data Booklet to predict the spontaneity of this disproportionation reaction. [2] Cu+ + e– Cu Eθ = +0.52V Cu2+ + e– Cu+ Eθ = +0.15V Eθ cell = (+0.52) − (+0.15) = + 0.37V Eθ cell>0, spontaneous (ii) Describe and explain what you would see when aqueous ammonia is added slowly to an aqueous solution of copper(II) chloride, until the aqueous ammonia is in excess. Write equations for any reactions that occur. [4] When NH3 is added slowly, blue ppt is formed. NH3 + H2O NH4+ + OH‒ [Cu(H2O)6]2+ + 2OH‒ Cu(OH)2 + 6H2O --- Eqm 1 Blue solution blue ppt When NH3 is added in excess, ligand exchange occurs. [Cu(H2O)6]2+ + 4NH3 [Cu(NH3)4(H2O)2]2+ + 4H2O deep blue solution The formation of [Cu(NH 3)4(H2O)2]2+ decreases the concentration of [Cu(H2O)6]2+ and shifts the position of Eqm 1 to left . Hence, the blue ppt dissolves to form a deep blue solution. (iii) Suggest an explanation for the difference in colour between [Cu(H 2O)6]2+ and [CuCl4]2–. [2] The different ligands, H2O and Cl–, result in a different energy gap. Different amount of energy/wavelength of light is absorbed to promote the electron from the lower energy group to the higher energy group. Light reflected/not absorbed is different.

5 [Total: 20] 2 In a Mixed Claisen Condensation Reaction, an ester and a ketone react to form a β-diketone. (a) (i) Draw the β-diketone formed when 2,2 -dimethylcyclohexanone reacts with ethyl ethanoate in the Mixed Claisen Condensation Reaction. 2,2-dimethylcyclohexanone ethyl ethanoate [1] (ii) In not more than 3 steps, propose the synthesis of 2,2 -dimethylcyclohexanone from the following starting material. [5] (b) Due to the presence of two ketone functional groups, β-diketones are a highly valuable substrate in several organic chemistry syntheses. One of the many examples is selective reduction, in which only one of the ketone reduced.

6 Draw the mechanism for the reaction between 4-hydroxypentan-2-one and HCN. [3] Nucleophilic addition HCN H+ + CN‒ NaCN  Na+ + CN‒ OR NaOH + HCN  NaCN + H2O (c) β-diketones exist in equilibrium with a tautomeric form, known as an enol. This is known as keto-enol tautomerisation. The keto-enol tautomerisation of pentan-2,4-dione is shown below. (i) Write an expression for Kc for the keto-enol tautomerisation of pentan-2,4-dione. [1] Kc = [enol]/[keto] (ii) In the case of pentan-2,4-dione, 76% of the mixture exists as an enol at room temperature and pressure. Calculate a value for Kc. [1] ∆H<0

7 Kc = 0.76/0.24 = 3.17 (3sf) (iii) Sketch a graph showing how the rates of the forward and reverse reactions change from the time pure keto is added to the time the reaction reaches equilibrium. Label your two lines clearly. [2] (d) The keto-enol tautomerisation is catalysed by the presence of an inorganic base, such as NaOH. time rate forward reverse teqm

8 (i) State how the presence of NaOH would affect the Kc, as calculated in (c)(ii). [1] No effect. (ii) On the same diagram in (c)(iii), show how the rates of the forward and reverse reactions change for the base catalysed keto-enol tautomerisation. Label your lines clearly. [1] time rate forward reverse forward, catalysed reverse, catalysed teqm teqm (catalysed)

9 (iii) Hence, explain the effect of NaOH on the reaction. [2] The catalyst NaOH provides an alternative pathway of lower activation energy, which increases the number of particles having en ergy more than or equals to the activation energy . Hence, the frequency of effective collision increases and both the rate of the forward and reverse reaction increases / equilibrium reached faster. (e) The enol form is more stable than the keto form due to two reasons Reason I: Intramolecular hydrogen bonding Reason II: Position of p orbitals (i) Draw an appropriate diagram to show how reason I contributes to the stability of the enol form. [1] (ii) State the hybridisation of the carbon atoms at positions 1, 2 and 3. Hence, explain how reason II contributes to the stability of the enol form. [2] sp2 The carbons 1, 2 and 3 have an unhybridised