CJC H2 Chem 2012 Prelim P3 Soln

2 9647/03/CJC JC2 Preliminary Exam 2012 Answer any four questions. 1 This question relates to the chemistry of Be, Mg, Al and their compounds. (a) Beryllium compounds are toxic air pollutants. Inhalation of high levels of beryllium can cause inflammation of the lungs in humans and long-term exposure may cause chronic beryllium disease (berylliosis), in which granulomatous lesions develop in the lung. (i) Given that charge density ∝ ionic charge ionic radius , calculate the relative charge densities of Be2+, Mg2+ and Al3+, using relevant data from the Data Booklet. Be2+: 64.5, Mg2+: 30.8, Al3+: 60.0 (ii) Hence, predict what is observed when aqueous sodium hydroxide is gradually added to aqueous beryllium sulfate until the sodium hydroxide is in an excess . Write equations for all reactions that have taken place. - white ppt, which dissolves in excess NaOH to give a colourless solution Be2+(aq) + 2OH-(aq) ® Be(OH)2(s) Be(OH)2(s) + 2OH-(aq) ® Be(OH)42-(aq) (iii) Suggest the pH of the solution formed when beryllium chloride is dissolved in water. Give your reasoning. pH 3 Be2+ ions have high charge density , which polarises neighbouring H 2O molecules; hence, weakening O— H and H+ lost (iv) Magnesium ions are essential for the action of some enzymes (e.g. alkaline phosphatase found in the liver) by receiving electron pairs from oxygen and nitrogen atoms in the protein. It is thought that beryllium compounds are poisonous because they displace magnesium ions from these enzymes. Suggest a reason why beryllium ions should behave in this way. Be2+ ions have higher charge density (or greater polarising power) than Mg2+; hence has greater tendency to receive electron pairs to form dative covalent bonds. (v) Beryllium chloride may be used as a catalyst in the chlorination of benzene. Suggest a reason why this is possible. O utline the mechanism to show how beryllium chloride is involved in this reaction. [10] In BeCl2, Be atom has only 4 outer electrons and so, is able to act as lone pair acceptor (to generate Cl+ electrophile) BeCl2 + Cl2 à BeCl3- + Cl+

3 9647/03/CJC JC2 Preliminary Exam 2012 Cl+ Cl H Cl H BeCl3 -+ Cl BeCl2 HCl+ + (b) A student carried out a kinetics experiment using a roll of magnesium ribbon that had been exposed to air for some time. He placed a piece of magnesium ribbon of mass 0.12 g into a flask containing 15.0 cm 3 of 1.0 mol dm–3 hydrochloric acid. The progress of the reaction was followed by measuring the pressure of the system at different times. The graph below shows the results of the experiment. (i) Determine, by calculation, the limiting reagent for the experiment. Mg + 2HC l ® MgCl2 + H2 amt of Mg = 3 . 24 12 . 0= 0.00494 mol amt of HCl = 1.0 x 1000 0 . 15= 0.015 mol since Mg º 2 HCl, hence, amt of HCl required for reaction = 2 x 0.00494 = 0.00988 mol < 0.015 mol (initial amount of HCl used ) Hence, Mg is the limiting reagent. (ii) Account for the change in pressure of the system as shown in the graph at points A, B, and from C onwards. C D A 0 B time pressure of the system

4 9647/03/CJC JC2 Preliminary Exam 2012 At A – initially rate is slow; due to layer of oxide/MgO formed on the surface of Mg ribbon due to oxidation in air At B – rapid increase in rate; reaction is exothermic, heat evolved increases rate of reaction C onwards – decrease in rate; as Mg (limiting reagent) is used up [4] (c) An alloy of aluminium and magnesium is used in boat-building. A 1.75 g sample of the alloy was dissolved in the minimum volume of 4 mol dm –3 hydrochloric acid and the solution was then made alkaline by the addition of aqueous sodium hydroxide until no further reaction occurred. The resultant mixture was filtered and the residue, X, rinsed with distilled water, all washings being added to the filtrate, Y. After air drying, 0.18 g of X was obtained. Carbon dioxide was passed into Y and a white solid, Z , which contained aluminium, was collected. Heating Z to constant mass gave a residue of mass 3.16 g. Suggest the identities of X, Y and Z, and determine the percentage composition by mass of the alloy. X – Mg(OH) 2 Y – NaAl(OH)4 Z – Al(OH)3 mass of Mg in Mg(OH)2 = ) 0 . 1 0 . 16 ( 2 3 . 24 3 . 24 + + x 0.18 = 3 . 58 3 . 24x 0.18 = 0.0750 g mass of Al in Al2O3 = ) 0 . 16 ( 3 ) 0 . 27 ( 2 ) 0 . 27 ( 2 + x 3.16 = 0 . 102 0 . 54x 3.16 = 1.67 g % of Mg in alloy = 75 . 1 0750 . 0x 100 = 4.29 % On dissolution in HCl (aq): Mg(s) + 2HCl (aq) à MgCl2 (aq) + H2(g) (and) Al + 3HCl à AlCl3 (aq) + 3/2 H2 (g) On addition of excess NaOH (aq) till no further reaction occurs: Mg2+ + 2OH- à Mg(OH)2(s) Residue X: Mg(OH)2 Al3+ + 3OH- à Al(OH)3 (s) Al(OH)3 (s) + OH- (aq) à Al(OH)4- (aq) Filtrate Y: NaAl(OH)4 [not Al(OH)4-] On addition of CO2 into Y: 2NaAl(OH)4 (aq) + CO2 à 2Al(OH)3(s) + Na2CO3 (aq) White solid Z: Al(OH)3 Heating Z to constant mass: 2Al(OH)3(s) à Al2O3 (s) + 3H2O (aq) residue of mass 3.16g = Al2O3 (s)

5 9647/03/CJC JC2 Preliminary Exam 2012 % of Al in alloy = 75 . 1 67 . 1x 100 = 94.4 % [6] [Total: 20] 2 2-chlorobutane undergoes hydrolysis with NaOH(aq) via two different reaction pathways in the same reaction to form a mixture of two enantiomeric products. CH3CHClCH2CH3 + NaOH → CH3CH(OH)CH2CH3 + NaCl In one of the hydrolysis reaction pathways, only one product is formed and inversion of configuration occurs in the product. For the other reaction pathway, a racemic mixture is formed. (a) In an experiment, one optical isomer of 2-chlorobutane undergoes hydrolysis and two enantiomeric products in a ratio of 95%:5% are formed. (i) Draw the structures of the two enantiomeric products. CH2CH3 OHH3C H CH2CH3 HO CH3 H * * Few students scored full marks for this part. Many students could not represent the enantiomers appropriately. Common errors: 1. No mirror line drawn (or) mirror line drawn as solid line. 2. Enantiomers are not represented as mirror images of each other. 3. Enantiomers are not represented in terms of tetrahedral geometry / 3D configuration. 4. Wedge and dotted line of 3D configuration not drawn in correct direction. 5. -CH 2CH3 often wrongly represented as H3CHC- in enantiomer structures. (ii) One enantiomer is formed in a much higher percentage compared to the other. Explain clearly how this disparity arises by examining the mechanisms of both reaction pathways. You should name both mechanisms involved but an outline of the mechanism is not required.

6 9647/03/CJC JC2 Preliminary Exam 2012 SN2 mechanism CH3CH2 Cl H3C H OH- HO CH2CH3 CH3 H + Cl- H3C CH2CH3 H ClHO S N1 mechanism CH3CH2 H H3C Cld+ d- H CH2CH3 CH3OH- OH- racemic product formed HO CH3 H CH2CH3 OH CH3CH2 H H3C carbocation intermediate Hydrolysis of 2-chlorobutane occurs via both S N2 and S N1 mechanisms. A racemic product is formed via the S N1 mechanism whereas only 1 chiral product is formed during the S N2 mechanism. As such, one of the enantiomers is formed in a greater proportion compared to the other. One of the enantiomers is formed in a much greater percentage as the reaction proceeds largely via the S N2 mechanism that results in the formation of 1 chiral product. (iii) Write a rate equation for the reaction pathway that results in the inversion of the configuration and draw its energy profile diagram, given that the enthalpy change of the hydrolysis is exothermic. rate = k[CH3CHClCH2CH3][OH-] Only 1 product formed; inversion of configuration compared to reactant δ+ δ- ΔH < 0 Ea Energy transition state Reactants (or) CH3CHClCH2CH3 Products (or) CH3CH(OH)CH2CH3 Reaction Pathway

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