2016 C1 Promotional Exam Answers

9 2016 HCI C1 H2 Chemistry Promotional Exam / Paper 1 Answers 1 2 3 4 5 6 7 8 C C B B A B B C 9 10 11 12 13 14 15 16 A B B A A C D C 17 18 19 20 D D D C

13 2016 HCI C1 H2 Chemistry Promotional Exam / Paper 2 Answers: 1 (a) (i) Water: Hydrogen Bonding CCl4: Dispersion Forces [½ each] (ii) Water Tetrachloromethane Hydrocarbon Chain Permanent dipole- induced dipole interactions / dispersion forces Dispersion forces Carboxylic acid functional group Hydrogen bonds Permanent dipole- induced dipole interactions / dispersion forces [½ each] (ii) The energy released in forming new interactions between water and caprylic acid is insufficient to compensate for the energy required to break the existing interactions. However, the energy released is sufficient to compensate the energy required when in CCl4. [1] (b) (i) When Br2 approaches the alkene, the electron cloud on Br2 is distorted to form a temporary dipole moment that causes it to be electron deficient. [1] ICl is a polar molecule, the more electronegative Cl causes I to be electron deficient. [1] (ii) The reaction is kept in the dark to prevent the possibility of free radical substitution. [1] (iii) Mechanism for electrophilic addition. Award 1 mark per step. Deduct ½ per mistake (e.g. no charge separation, missing/ wrongly drawn arrows, missing slow step, wrong intermediate) (c) The electrons in the benzene ring are delocalized, hence there is resonance stabilization. If benzene undergoes addition, the stabilization due to resonance is lost and it destabilizes the compound. [1] (d) Step 1: CH3Cl, AlCl3 or FeCl3, heat Step 2: KMnO4. Dilute H2SO4, heat [1 mark per step]

14 2016 HCI C1 H2 Chemistry Promotional Exam / Paper 2 2 (a) (i) [1] (ii) [1] (iii) CnH2nOn [1] (iv) C12H22O11 [1] (b) (i) Circles laevorotatory. The final glucose/fructose mixture is laevorotatory (Afinal is negative). As glucose is dextrorotatory, fructose must be laevorotatory for this to be possible. [1] (half mark for answers which only point out that the observed rotation is becoming less and less positive) (ii) H+ is a catalyst and overall remains unchanged and not used up (or is regenerated), hence its concentration is constant. [1] (iii) Finding half life = 82 min [1 – subtract 0.5 if no construction lines] As the half-life of sucrose is constant at 82 min, the order with respect to sucrose is one (x=1). [1] (iv) 1 2 1ln2 ln2 0.00845min82effk t  [1] for working+value, [1] for units (v) Initial rate = 0.00845 × 0.580 ÷ 2 = 0.00245 mol dm−3 min−1 [1] (vi) The temperature of the solutions could be kept lower than 30 °C. [1] The concentration of HCl used could be lower than 1 mol dm−3. [1] * *

15 2016 HCI C1 H2 Chemistry Promotional Exam / Paper 2 3 (a) (i) There are no/neglig