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