AJC H2 Chemistry 9647 N2012 P3 Suggested Solutions

©2013AndersonJC/CHEM 1 H2 Chemistry 9647 N2012 P3 Suggested Solutions 1 (a) Both Al2O3 and MgO behave as basic oxides in the presence of acids. MgO + 2H+ Mg2+ + H2O Al2O3 + 6H+ 2Al3+ + 3H2O Both Al2O3 and P4O10 behave as acidic oxides in the presence of alkalis. P4O10 + 12OH– 4PO43– + 6H2O Al2O3 + 2OH– + 3H2O 2[Al(OH)4]– Al2O3 is amphoteric because it is an ionic oxide with covalent character (acidic) due to the highly charged A l3+ ions distorting the electron cloud of the O 2– ions. Hence Al2O3 has both acidic and basic properties. [1] [1] [1] [1] [1] Comments: Students should emphasise on the interaction between A l3+ and O 2– instead of with H 2O molecules because the question specified Al2O3 instead of Al3+(aq). (b) (i) Al2Cl6 Al Cl Cl Cl Cl Al x x x x x x Cl Cl [1]: correct dot and cross diagram with two dative bonds (circled) between Cl and Al (each with 2 dots or 2 crosses) clearly indicated [1] (ii) Trigonal planar [1] (iii) Increasing the temperature favours the forward endothermic reaction. Position of equilibrium shifts right to absorb the increase of heat energy. This results in an increase in amount of AlCl3 and decrease in amount of Al2Cl6. [1] [1] (iv) mpV RTM= M = pV mRT ))(250x10(1.16x10 1)(500)(1.50)(8.3 65 −= = 215 g mol–1 Average Mr = 215 [1]

©2013AndersonJC/CHEM 2 (v) x = 1 – y 214.9 = 267(1–y) + 133.5y 214.9 = 267 – 267y + 133.5y y = 0.39 x = 0.61 [1] [1] (vi) p(Al2Cl6) = 0.61 x 1.16 x 105 = 70760 Pa p(AlCl3) = 0.39 x 1.16 x 105 = 45240 Pa [1] [1] (vii) 62 3 ClAl 2 AlCl p p pK = 70760 (45240)2 = = 28924 Pa [1]: correct Kp expression [1]: correct answer and units [2] (c) (i) A and B reacts with 2,4–DNPH  aldehyde / ketone functional group is present O O A B [1] each [2] Comments: The reaction is similar to the mechanism of electrophilic aromatic substitution. For B, intramolecular electrophilic substitution occurs as the carbocation is generated at the end of the alkyl side chain. The ketone ring should only join to positions 1 and 2 of the benzene ring. (ii) Warm separate samples of A and B with I2(aq) and NaOH(aq). A will give a yellow precipitate whereas B will not. The products are CHI3 (iodoform) and sodium benzoate. [1] [1] Comments: Hot, acidified KMnO 4 should not be used as both A and B will react with it. A gives benzoic acid and B gives benzene–1,2–dicarboxylic acid.

©2013AndersonJC/CHEM 3 2 (a) (i) [2]: correct components of each half–cell [1]: half–cells connected via a salt bridge and voltmeter [1]: direction of electron flow Note: “a fully labelled diagram”  one electrochemical cell [4] (ii) Measure the potential difference of the Cl2(g)/Cl–(aq) and I3– (aq)/I–(aq) half–cells using a voltmeter. Since chlorine has stronger oxidising power than iodine, cathode will be Pt electrode in th e C l2(g)/Cl–(aq) half -cell. When connected correctly to the voltmeter, the e.m.f. will show as a positive value (of + 0.82 V). [1] [1] (iii) Cl2(g) + 3I–(aq) I3–(aq) + 2Cl–(aq) [or Cl2(g) + 2I–(aq) I2(s) + 2Cl–(aq)] [1] (b) (i) 2Fe3+ + 2I– 2Fe2+ + I2 [1] (ii) Fe3+ + e Fe2+ E = +0.77V [Fe(CN)6]3– + e [Fe(CN)6]4– E = +0.36V • CN– ligands form complexes with both hydrated Fe2+ and Fe3+ ions • Since Eo([Fe(CN)6]3–/[Fe(CN)6]4–) is less positive than Eo(Fe3+/Fe2+) ➢ Fe(CN)6]3– is less easily reduced / a weaker oxidising agent than Fe3+ ➢ C CN– ligands stabilise Fe3+ more than Fe2+ ➢ More Fe 3+ will form a complex with CN – compared to [1] [1] salt bridge Pt(s) [I3–](aq), 1 mol dm–3 [I–](aq), 1 mol dm–3 high-resistance voltmeter V [Cl–](aq), 1 mol dm–3 Cl2(g) at 1 atm, 25 ºC platinised platinum electrode

©2013AndersonJC/CHEM 4 Fe2+ ➢ [Fe3+] in (b)(i) decreases more than [Fe2+] ➢ P.O.E. in (b)(i) shifts to the left. (c) The relative ease of hydrolysis of monohalogenoethanes, C2H5X can be determined by the bond energies of the C–X bond. Since C–I bond is weaker, C 2H5I tends to hydrolyse faster than C2H5Cl. Note: You may refer to the Data Booklet for bond energies of C–I (240 kJ mol –1) and C–Cl (340 kJ mol–1). [1] (d) (i) CH3 OH Cl CH3 OH NH2 C D [1] each [2] (ii) C CCl H H N O CH2CH2OH H NHO O E F [1] each Note: There are 2 considerations to be made for the formation of intermediate E • acid chloride is more reactive than halogenoalkane • amine (with higher nucleophilicity) is more reactive than alcohol The mechanism for formation of F from E C CCl H H N O CH2CH2OH H Na C CCl H H N O CH2CH2O H F The absence of bond strain (6–membered ring) allows F to be formed. [2] (e) (i) Dilute HCl, heat under reflux (or NaOH(aq), heat under reflux) [1] (ii) From hydrolysis of phenylbutazone (using acidic medium) [2]  

©2013AndersonJC/CHEM 5 N N H H H H 2+ HO2C CO2H (using alkaline medium) N N H H O2C CO2 From hydrolysis of phenobarbital (using acidic medium) CO2H CO2H NH4+ + CO2 + H2O (using alkaline medium) CO2 CO2 NH3 + CO32– For your information Step 1: For each amide bond HO C O H N hydrolysed, H and OH are added across the amide bond as shown. [2]

©2013AndersonJC/CHEM 6 NH NH O O O O O OH OH NHH H NHH H C O HO HO Step 2: Consider the medium used to get the final product If alkaline medium is used, the acidic –CO2H will be deprotonated to form –CO2– while an acidic medium will cause the basic –NR2 group present to be protonated to form –NR2H+ (where R can be alkyl, aryl or H).

©2013AndersonJC/CHEM 7 3 (a) (i) Mg(NO3)2 MgO + 2NO2 + 2 1 O2 [1] (ii) Down Group II, cation size increases, thus charge density of cation decreases and polarising power of cation decreases . Hence the electron cloud of the NO 3− ion is distorted to a lesser extent and the N−O covalent bond in NO 3− is weakened to a lesser extent. Therefore ther mal stability of Group II nitrates increases down the group. [1]: stating the trend [1]: explaining the trend [2] (b) (i) M(NO3)2 MO + N2O5 Mechanism: N O- O O M2+ N O O O N OO O N O O O2-M2+ N O O ++ N O O [1] each correct step Note: From the hint in the question that showed the decomposition of CO 32−, you are expected to apply to the NO 3−, which decomposes to form O 2− and NO2+. [2] (ii) N2O5 2NO2 + 2 1 O2 (accept 2N2O5 4NO2 + O2) [1]

©2013AndersonJC/CHEM 8 (iii) N OO O N O O N OO O +slow N OO O N OO O N O O O N O O N OO O N O O O N O O N O O O O+ + N O O O homolytic breaking of N-O bond to give NO2 and NO3 radical dimerisation of NO3 dissociation of dimer via N-O bond cleavage [1] each correct step (need to show curly arrows in order to show clearly which bonds are broken and which are formed) Note: In the first step where NO 2• and NO3• are formed, you need to indicate the unpaired electron on each species. [3] (iv) Slowest step: the first step Given that the reaction is first order wrt N 2O5, only 1 molecule of N 2O5 is involved in the rate−determining step , which is the first step in the mechanism in (iii). [1] (c) (i) HNO3 [1] (ii) N2O5 NO2 + + NO3 - NO2 + NO2 + HNO3 slow fast H NO2 NO3 [1]: correct electron pairs movement, slow/fast step [1]: correct electrophile and intermediate generated [2] Comments: The question stated that N2O5 exists as NO2+ and NO3− in solution, and if you follow this line of reasoning, you would predict that the other product, HNO 3, could