RI 2024 Alkanes Tutorial Self check answers

-1- Raffles Institution Year 5 H2 Chemistry 2024 Tutorial 10: Hydrocarbons – Alkanes (Suggested Answers) Self-Check Questions – Suggested Answers 1 (a) (i) 3,3-diethyl-4-methylheptane (ii) 4-ethyl-2,2,3,6-tetramethyloctane (iii) 3-ethyl-2,7-dimethyloctane (iv) 2,2,5-trimethylheptane (v) 3,3,4-triethylhexane (b) (i) 2,4,5-trimethylheptane CH3 CH CH2 CH CH CH2CH3 CH3 CH3 CH3 or (ii) 3-ethyl-2,4-dimethylhexane CH3 CH CH CH CH2CH3 CH3 CH3 CH2CH3 or 2 The given compounds have simple molecular structures and can be arranged in the following order of increasing boiling point: 2-methylpentane < hexane < 3,3-dimethylpentane < 2-methylhexane < heptane i.e. (e) < (d) < (c) < (a) < (b) Comparing (c), (a) and (b) vs (e) and (d)  (c), (a) and (b) are seven -carbon alkanes while (e) and (d) are six -carbon alkanes. (c), (a) and (b) have more electrons per molecule and hence, larger and more polarisable electron clouds than (e) and (d). Hence , the instantaneous dipole -induced dipole interactions in (c), (a) and (b) are stronger and more heat is required for boiling these alkanes, resulting in them having a higher boiling point than (e) and (d). Comparing (c), (a) and (b)  the degree of branching decreases from (c) to (a) to (b)  there is increasing surface area of contact between molecules from (c) to (a) to (b)  strength of instantaneous dipole -induced dipole interactions increases from (c) to (a) to (b)  higher boiling point from (c) to (a) to (b) Comparing (e) and (d)  the degree of branching decreases from (e) to (d),  there is greater surface area of contact between molecules of (d),  hence stronger instantaneous dipole-induced dipole interactions in (d)  higher boiling point for (d)

-2- 3 Reaction (1): Combustion + 9 O2 6 CO2 + 6 H2O burn Reaction (2): Free-radical substitution + 2 Cl2 uv light Cl Cl + 2 HCl 4 CnH2n+2 + 2 1n3  O2(g)  n CO2(g) + (n+1) H2O(l) 0 0.5 number of C atoms amount of O2/mol 0 gradient = 1.5 5(a) Volume of unreacted O2 and CO2 = 250 cm3 Volume of unreacted O2 = volume of residual gas (after shaking with NaOH) = 100 cm3 Volume of CO2 produced = 250 – 100 = 150 cm3 From the table, x = 5 And (x + y/4) = 8  y = 12 Hence, the molecular formula of the hydrocarbon is C5H12. 5(b) heat

-3- 6 Answer : C Possible answers: 7 Answer : B 8 The mechanism is free-radical substitution which consists of the following steps: Step 1: Initiation Step 2: Propagation Then (a), (b), (a), (b), …... Step 3: Termination 1 2 3 4 5 1 3 4 2 1 Cl 2 Cl 3 4 Cl Cl 5 Cl Structure Structure 1 Cl 2 Cl 3 4 Cl