AJC 2013 A Level Chem P3 solns (2020 version)

2020 ASRJC/CHEM 1 H2 Chemistry 9647 2013 ‘A’ Level P3 Suggested Solutions 1 (a) Halogen (X2) Eo / V F2 + 2e– 2F– +2.87 Cl2 + 2e– 2Cl– +1.36 Br2 + 2e– 2Br– +1.07 I2 + 2e– 2I– +0.54 Down the Group, Eo −/XX2 becomes less positive. Hence, decreasing tendency for X2 to be reduced to X–. The oxidising power of halogens decreases from F2 to I2. [1] [1] (b) (i) 2HX H2 + X2 where X = F, Cl, Br, I [1] (ii) Thermal stability of HX decreases down the Group. (Order of thermal Stability: HF > HCl > HBr > HI) HF and HCl are stable to heat and hence do not decompose. HBr decomposes slightly on heating, brown fumes of Br2 observed. HI decomposes easily, gives out dense purple fumes of I2 on gentle heating. This is because d own the Group, the atomic radius of the halogen (X) increases. Consequently, the bonding electrons are less strongly attracted to the nuclei of H and X and the H−X bond becomes weaker and so, is more easily broken. [1] [1] Comments Most students knew the trend in stabilities of the hydrogen halides. Some wrote a significant amount of irrelevant, factually correct material concerning bond lengths, electronegativities and ionisation energies. A number of students wrote about van der Waals ’ interactions, confusing thermal stability with volatility. (c) (i) Nucleophilic substitution (SN2) C Cl CH3CH2 H H I + − C ClI CH2CH3 H H CI CH2CH3 H H + Cl [1m] for correct curly arrows, lone pair of electrons on I and partial charges [1m] for transition state & inverted final product. [2] Comments A significant number of students drew the attacking species as Na–I (covalent). Several omitted the important lone pair on the iodide ion. Most depicted the partial charges of the C+−Cl– bond, but their curly arrows were not always clearly drawn from the I– lone pair to the + carbon, or from the C–Cl bond to the chlorine. (ii) As NaCl is almost insoluble in propanone (organic solvent), NaCl formed will be precipitated and the concentration of Cl– decreases, causing the position of equilibrium 1 to shift right. Hence, the reaction goes almost to completion. [1]

2020 ASRJC/CHEM 2 (iii) Evidence Deduction (type of reaction & functional group present) A (bromoalkane) when warmed with NaOH(aq) produces alcohol B nucleophilic substitution Alcohol B when heated with excess Na2Cr2O7/H+ gives neutral compound C. B undergoes oxidation to give C. C is a ketone since it is neutral and it i s a product of the oxidation of an alcohol. B is a 2o alcohol.  A is a 2o bromoalkane. A when heated with NaI in propanone, iodoalkane D is formed. nucleophilic substitution Mr of D is 38.2% larger than Mr of A. General formula of A: CxH2x+1Br Mr of A = 12x + 2x + 1 + 79.9 = 14x + 80.9 General formula of D: CxH2x+1I Mr of D = 12x + 2x + 1 + 127 = 14x + 128 0.38280.9) +(14x 80.9) +(14x - 128) +(14x = x = 3  A is a 2o bromoalkane with 3 carbon atoms. OR Let Mr of A be y 0.382y 79.9 - 127 = y = 123.3  Mr of alkyl group in A = 123.3 – 79.9 = 43.4  43  A is a 2o bromoalkane with 3 carbon atoms. [1] [1] A is CH3CHBrCH3. [1]

2020 ASRJC/CHEM 3 (d) (i) Kc = ]][[ ][ 2 3 − − II I units = mol–1 dm3 [1] [1] (ii) No. of moles of I2 = 2.54 / (127 x 2) = 0.0100 mol Concentration of aq. I2 = 0.0100 / (100 / 1000) = 0.100 mol dm–3 Since [I3–(aq)] at equilibrium = 9.98 x 10–2 mol dm–3, x = 9.98 x 10–2 Eqm [I2(aq)] = 0.100 – 9.98 x 10–2 = 2.00 x 10–4 mol dm–3 Eqm [I–(aq)] = 1 – 9.98 x 10–2 = 0.900 mol dm–3 Kc = ) )(0.900 10 x (2.00 )10(9.98 4 2 − − = 554 I2(aq) + I–(aq) I3– Initial conc. 0.100 1.00 0 Change –x –x +x Final conc. 0.100 – x 1 – x x [1] [1] (iii) Kc = ]][[ ][ 2 3 − − II I 554 = ) ](1.00 [ (1.00) 2I [I2(aq)] = 1.81 x 10–3 mol dm–3 [1] (iv) (2 immiscible layers will be observed) Since I2 is much more soluble in hexane than it is in water , [I2(aq)] decreases as it dissolves in hexane (and is removed from the aqueous layer) . Hence, by Le Chatelier’ s P rinciple, the position of equilibrium 2 shifts left. [I3–(aq)] decreases and [I–(aq)] increases. I2(aq) + I–(aq) I3–(aq) –––––– (2) [1] [1] Comments Most students correctly predicted that the effect of hexane on [ I2(aq)]. Some erroneously thought that I2 reacts with hexane, rather than dissolving in it. Some also thought that I– or I3– dissolved in the hexane and came up with incorrect predictions. A significant number did not apply Le Chatelier’s principle correctly, stating that the [I2(aq)] would increase to make up for the iodine that had dissolved in the hexane.

2020 ASRJC/CHEM 4 (v) Eocell = Eored(I3–/I–) – Eoox(Cu2+/Cu) = 0.536 – 0.34 = +0.196 V On addition of hexane to the I3–/I– half cell, the [I3–(aq)] decreases and [I–(aq)] increases as deduced from part (iv). I3–(aq) + 2e– 3I–(aq) By Le Chatelier’ s P rinciple, the position of the above equilibrium shifts left. Hence, the reduction potential for the I3–/I– half cell becomes less positive (than +0.536 V). Therefore, the cell potential will become less positive. [1] [1] Comments Students had to consider two equilibria. It was not always clear from some answers which of the two equilibria students were referring to. Some students suggested there would be no change, as neither I3– nor I– are soluble in hexane. 2 (a) (i) A Bronsted–Lowry base is a proton acceptor. A conjugate acid –base pair contains a pair of Bronsted–Lowry acid and base which differ by the presence or absence of a proton. For example , NH4+ and NH3 is a conjugate acid –base pair, while NH 3 is a Bronsted–Lowry base. [1] [1] (b) Ammonia reacting as: Explanation (i) a base NH3 accepts a proton from H2O forming NH4+ and OH–. [1] (ii) a reducing agent The oxidation state of N increases from –3 in NH3 to 0 in N 2, i.e. NH3 acts as a reducing agent and is itself oxidised in the process. Note: Reaction (ii) is an example of a comproportionation reaction. [1] (iii) an acid NH3 donates a proton to H – forming NH 2– and H 2 in the process. [1] (iv) a reducing agent The oxidation state of N increases from –3 in NH 3 to –2 in N2H4, i.e. NH 3 acts as a reducing agent and is itself oxidised in the process. [1] (v) a nucleophile NH3 possesses a lone pair of electrons which can be donated to the electron deficient (+) (acyl) carbon, thereby undergoing nucl eophilic (acyl) substitution with the acyl chloride functional group. [1]

2020 ASRJC/CHEM 5 (c) Order of basicity: ethylamine > NH3 > phenylamine [1] Ethyl group is electron–donating, thus increasing the electron density on N atom, making the lone pair of electrons on N atom more readily available to accept a proton. Hence, ethylamine is a stronger base than NH3. [1] The lone pair of electrons on N atom is delocalised into the –electron cloud of the benzene ring, making the lone pair less readily availab le to accept a proton, thus phenylamine is a weaker base than NH3. [1] Note: Basicity of amines depends on the availability of the lone pair of electrons on N to accept a proton, H+. (d) (i) 4–methylphenylamine is a stronger base than phenylamine. NH2H3C > NH2 The methyl group is an electron–donating group, thus there is a lesser extent of delocalisation of the lone pair of electrons on N atom into the –electron cloud of the benzene ring. The lone pair is more readily available to acc