HCI Prelim P3 ANSWERS

1 2015 HCI Prelim H2 Chemistry Paper 3 Answers 1 (a) (i) Heterogeneous catalysis [1] (ii) Optical isomerism [1] [1] (iii) The addition of H 2 to one side (bottom face) of the planar alkene molecule creates a chiral alkane molecule. [1] When the opposite side (top face) of the alkene is now adsorbed to the surface of the Ni catalyst, the addition of H 2 produces a chiral alkane molecule which is the non - superimposable mirror image of the product formed at the bottom face. [1] As there is 50% chance for each side (face) of the alkene to bind to the surface of the catalyst, the two enantiomers are produced in a 1:1 ratio resulting in a racemic mixture. Hence the product mixture does not rotate the plane of plane-polarised light. [1] ================================================================= If a diagram is used, it will substitute the 1st two marks: Show opposite faces adsorbed to the catalyst surface [1] Indicate which carbon forms the chiral alkane and the corresponding enantiomer. [1] (b) (i) H reaction = (B.E. of Bonds Broken) – (B.E. of Bonds Formed) = [B.E (C = C) + B.E. (H – H)] – [B.E (C – C) + B.E. (C – H) x 2] = (610 + 436) – (350 + 410 x 2) = – 124 kJ mol-1 Correct bond energies used [1] Correct equation and answer [1]

2 (ii) H reaction is exothermic. S reaction is negative as there is a decrease in the number of moles of gas in the reaction. [1] Given that G reaction = H reaction – TS reaction G reaction is expected to be negative only at low temperatures and so the reaction would be spontaneous only at low temperatures. [1] (iii) Although the reaction is energetically feasible, the activation energy is very high. Or as the low temperatures are necessary for the reaction to be feasible, the rate of reaction may be very slow. [1] Hence the catalyst provides an alternative pathway with a lower activation energy and the reaction speeds up. [1] (d) (i) A homogeneous catalyst acts in the same phase as the reactants. [1] (ii) Oxidation number of Rh in compound 2: +3 [1] x + (-1) + [(-1) x 2] = 0 x = +3 (working is not required) (iii) The coordination number indicates the number of dative bonds about the central atom or ion. [1] Wilkinson’s catalyst: coordination number = 4 Compound 3: coordination number = 6 [½] for each correct coordination number (iv) Rh is able to exhibit variable oxidation states. [1] The proposed mechanism requires the addition of H 2 to Rh in the form of two hydride ligands, which causes the oxidation state of Rh to increase from +1 to +3. [1] Rh is able to form complexes of variable coordination numbers. [1] The proposed mechanism requires the alkene and H 2 to bind to Rh as ligands before the addition reaction occurs. Or coordination number changes from 4 to 6 from Wilkinson’s catalyst to Compound 3. [1] 2 (a) Chlorine is a greenish yellow gas. Bromine is a reddish-brown liquid. Iodine is a black solid. [1] Down the group, the volatility decreases as the number of electrons in the halogen molecule increases. The size of the electron cloud increases, making the electron cloud more polarizable and hence giving rise to stronger dispersion forces. Thus more energy is required to overcome the stronger dispersion forces between the molecules for them to vaporize. [2] (b) Add a few drops of aqueous AgNO3 to th e sample solution, followed by dilute aqueous ammonia.

3 If a white ppt forms which dissolves in dilute aqueous ammonia, the sample contains C l– ion. Ag+(aq) + Cl– (aq)  AgCl (s) --- white ppt AgCl (s) + 2NH3 (aq)  Ag(NH3)2 + (aq) + Cl– aq) If a cream ppt forms which is insoluble in dilute aqueous ammonia, the sample contains Br – ion. Ag+(aq) + Br – (aq)  AgBr (s) --- cream ppt {AgBr (s) + 2NH 3 (aq)  Ag(NH3)2 + (aq) + Br – (aq)} – this reaction occurs as well but does not decrease the [Ag +] sufficiently to make AgBr dissolve completely since Ksp(AgBr) is lower than Ksp(AgCl). [1] for correct reagents [1] for equation for AgX precipitation and colours of ppt [1] for equation with ammonia and correct observation of solubility OR Add a few drops of Cl2(aq) into the sample solution. If the solution turns from colorless to yellow-orange, the sample contains Br – ion. Cl2(aq) + 2Br – (aq)  Br2(aq) + 2Cl– (aq) If the solution does not turn yellow -orange (OR remains colourless), the sample contains Cl– ion. [1] for correct reagents [1] for correct observations for both ions [1] for redox equation (c) (i) Pyridinium tribromide is a solid and is less volatile, hence it is safer to use compared to bromine which is a fuming liquid and gives off toxic bromine fumes. [1] OR Pyridinium tribromide is a solid, hence it is easier to weigh and use compared to bromine which is a fuming liquid. Pyridinium tribromide is given as an ionic salt, hence students can deduce it is a solid and will be less volatile compared to bromine which is fuming and toxic, and relate its advantage to safety and ease of use. (ii) [1] (iii) Step 1: condensation [1] OR nucleophilic (acyl) substitution or addition-elimination Step 2: electrophilic substitution [1] (iv) In Route 2, the Br 2 reacts with phenylamine. The lone pair on NH 2 delocalises into the benzene ring and increases electron density in the ring. The ring is highly activated towards electrophilic attack , hence it is able to undergo further brominations to give multi-brominated products. [1] In Route 1, the anilide is not as activated as phenylamine as the –NHCOR group is less electron-donating due to the delocalization of the nitrogen lone pair into the – C=O group as well. [1]

4 (v) The bulky –NHCOR group sterically hinders the approach of the electrophile towards the 2nd carbon. Hence substitution occurs mainly on the 4th position. [1] (vi) Phenylamine is the most basic of the three, followed by bromophenylamine and then bromoanilide. [1] Bromoanilide is the least basic. In the amide group, the lone pair on N is delocalized into the C=O group (as well as into the benzene ring) and least (or not) available for protonation. In bromophenylamine and phenylamine, the lone pair on N delocalizes into the benzene ring. However, b romophenylamine is less basic because its Br substituent is electron -withdrawing and decreases the electron density on the benzene ring. Hence the lone pair on N delocalizes into the benzene ring to a greater extent and is less available for protonation, compared to phenylamine. [1] (d) (i) amount of Br2 = 16.0 x 10–3 / (79.9 x 2) = 1.00 x 10–4 mol [Br2] = 1.00 x 10–4 / (150/1000) = 6.67 x 10–4 mol dm–3 [1] The value of KC >1 shows that the equilibrium is product favoured. The concentration of Br– is 0.5/6.67 x 10 –4 = 750 times more than that of Br 2. The large excess of Br – will push the equilibrium position very much to the right , such that the reaction is almost complete. [1] (ii) Assuming the reaction is complete, [Br3 –] at eqm = 6.67 x 10–4 mol dm–3 [Br –] at eqm = 0.5 – 6.67 x 10–4 = 0.499 mol dm–3 [1] (iii) KC = [Br3 –] / [Br2][Br –] 100 = 6.67 x 10–4 / [Br2] (0.499) [Br2] = 1.34 x 10–5 mol dm–3 [1] 3 (a) (i) No. of moles of sulfoxide recovered = 91.3% of theoretical no. of moles ∴ no of moles of thioanisole used = theoretical no of moles of sulfoxide = (9.82/(124.1 + 16.0))/(91.3/100) = (9.82/140.1)/(91.3/100) = 0.07677 mol [1] correct substitution of Mr and % yield ∴ mass of thioanisole used = 0.07677 x 124.1 = 9.53 g (must be 3 s.f.) [1] (ii) Since IO4 – and thioanisole react in a 1:1 ratio, ⇒ no of moles of IO4 – required = 0.076