RI 2025 Alkanes Tutorial Answers

-1- Raffles Institution Year 5 H2 Chemistry 2025 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- c arbon 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). Co mparing (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), • t here 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 O 2 6 CO2 + 6 H2O burn Reaction (2): Free-radical substitution + 2 Cl2 uv light Cl Cl + 2 HCl 4 CnH2n+2 + 2 1 n 3+ 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 No. of C atoms No. of moles of O2 2 1

-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 CH3CH2 C CH3 CH2 CH3 CH3CH C CH3 CH3 CH3 CH2CH2 C CH3 CH3 CH3 1 2 3 4 5 1 3 4 2 1 Cl 2 Cl 3 4 Cl Cl 5 Cl Structure P Structure Q 1 Cl 2 Cl 3 4 Cl Cl H H + Br Br Br Br + HBr + + Br (a) (b) • • •

-4- (trace amount)

-5- Practice Questions – Suggested Answers 9(a)(i) 9(a)(ii) or C CH3 CH2CH3 CH2CH3 CH3CH2 9(b)(i) CH3(CH2)6CH3 + Br 2 CH 3CH2CHBr(CH2)4CH3 + HBrultraviolet light 9(b)(ii) Orange-red bromine is decolourised and white fumes of HBr which turn damp blue litmus paper red are formed. 9(b)(iii) The mechanism is free-radical substitution Step 1: Initiation Step 2: Propagation Step 3: Termination Note: Skeletal formulae may be used. Step 1: Initiation Step 2: Propagation Then (a), (b), … Step 3: Termination + Br Br ultra-violet light Br Br + Br Br ultra-violet light Br Br

-6- 9(b)(iv) A flash of ultraviolet light provides enough energy for some Br 2 molecules to undergo homolytic bond fission to form Br • radicals. Once the Br • radicals are generated in the initiation step, the chain reaction is started and the propagation steps can proceed successively for many cycles un-aided. 9(b)(v) Consider the 2 propagation steps: The reaction is an example of homogeneous catalysis since the catalyst, Br• • exists in the same phase (dissolved in inert solvent) as the two reactants, octane and Br2. • is consumed in one step and regenerated in another. 9(b)(vi) Use an excess of octane (or use a limiting amount of Br2). 9(b)(vii) ∆H reaction for the 1st propagation step = BE(C–H) – BE(H–I)= +410 – 299 = +111 kJ mol–1 The first propagation step is endothermic due to the breaking of the relatively strong C −H bond and the formation of the weaker H −I bond. This makes it difficult for the chain propagation steps to occur. Thus, the formation of 3-iodooctane is energetically unfavourable.

-7- 10(a) Molar ratio of C to H in alkane F = 83.72 12.0 : (100−83.72) 1.0 = 6.977 : 16.28 = 1 : 2.33 = 3 : 7 Hence, t he empirical formula of F is C3H7. Let the molecular formula of F be (C3H7)m, i.e. C3mH7m. The general formula of an alkane is CnH2n+2. In this case, 3m = n and 7m = 2n + 2 Upon solving the simultaneous equations, m = 2 Hence, t he molecular formula of F is C6H14. F undergoes free-radical substitution reaction with Br2 to form only two mono-brominated products. ⇒ There are 2 types of H atoms in a molecule of F. ⇒ F has to be H is a mono-brominated product and is chiral. ⇒ H contains a chiral carbon atom. ⇒ H is Note: ∗ denotes a chiral carbon atom G is the other mono-brominated product. ⇒ G is 10(b) To form G from F, the Br atom can replace any one of the 2 tertiary hydrogen atoms (denoted by the circle). To form H from F, the Br atom can replace any one of the 12 primary hydrogen atoms (denoted by the box). Hence, based on probability factor (and assuming all the hydrogen atoms have the same reactivity), the approximate molar ratio in which G and H are formed = 2 : 12 = 1 : 6 Note #1: • Can proceed by trial and error to find m. • In this case, m ≠ 1 and m ≠3 • When m = 2, the molecular formula is C 6H14 which fits that of a six-carbon alkane. Note #2: • Let F have n carbon atoms and have the general alkane formula, C nH2n+2. • Then % of carbon (by mass) in F = ( 12n 14n+2 x 100) = 83.72 Hence n = 6.06 = 6 Note #3: • F is not a cycloalkane since percentage of carbon by mass in a cycloalkane is not 83.72%.

-8- 11(a) These reagents do not attack an alkane because 1. A lkanes are non polar. • They do not contain any region of high electron density and thus do not attract electrophilic reagents. • They also do not contain any electron -def icient sites to attract nucleophilic reagents. 2. Alkanes have relatively strong C –C and C–H bonds which do not break under normal conditions. 11(b) To form A, the Cl atom can replace the only 1 tertiary H atom (denoted by the circle) in 2-methylpropane. To form B, the Cl atom can replace any one of the 9 primary H atoms (denoted by the box) in 2-methylpropane. Note: In this example, substituting any primary H in 2- methylpropane results in the same product. This is only true because in this case, the primary H atoms are all equivalent. Combining both the probability factor stated above and the given reactivity ratio factor, the relative ratio of A to B = (1)(21) : (9)(1) = 21:9 = 7:3 11(c) Total number of isomers: 6 12(a)(i) Secondary free radical is more stable than primary free radical. [1] 12(a)(ii) The free radical has a C atom with an unpaired electron/is electron deficient . Since a secondary free radical has 2 electron donating alkyl groups directly attached to this C atom compared to only one in a primar