The problems are on the halogenoalkanes worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Alkanes react with halogens in UV light by free radical substitution, a chain reaction of initiation, propagation and termination. These worked answers correct the 20 problems of the Chemistry 11-12 worksheet on halogenoalkanes, from nucleophiles to CFCs and HFCs: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the equation or method and the result in bold. In mechanisms, check where the curly arrows start and finish, and in calculations check the mole ratios and the units.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB R3.4
Problem 1 Nucleophiles
1. A species that donates a lone pair of electrons to form a covalent bond.2. H₂O, OH⁻ and NH₃ (BF₃ and H⁺ accept electrons; Cl• is a radical).
Problem 2 Primary, secondary or tertiary
Primary; secondary; tertiary; primary (count the carbons attached to the carbon bearing the halogen).Problem 3 Products of bromoethane
1. ethanol, nucleophilic substitution.2. propanenitrile, nucleophilic substitution.
3. ethylamine, nucleophilic substitution.
4. ethene, elimination.
Problem 4 Radicals
1. Breaking of a covalent bond so that each atom takes one of the bonding electrons.2. Br₂ → 2Br•.
3. A species with an unpaired electron.
Problem 5 Chlorinating methane
1. CH₄ + Cl₂ → CH₃Cl + HCl.2. Dichloromethane, trichloromethane and tetrachloromethane, because chlorine keeps replacing hydrogen.
3. CH₄ + 4Cl₂ → CCl₄ + 4HCl.
Problem 6 Bond enthalpies and rate
1. iodo > bromo > chloro > fluoro.2. AgI yellow, AgBr cream, AgCl white; fluoro gives none (AgF is soluble).
3. Bond strength, not polarity, controls the rate: the C–F bond is so strong that breaking it is the slow step.
Problem 7 Ethane and bromine
1. CH₃CH₃ + Br• → CH₃CH₂• + HBr; CH₃CH₂• + Br₂ → CH₃CH₂Br + Br•.2. Br• + Br• → Br₂; CH₃CH₂• + Br• → CH₃CH₂Br; 2CH₃CH₂• → CH₃CH₂CH₂CH₃.
3. Two ethyl radicals sometimes meet and combine, forming butane by termination.
Problem 8 Percentage yield
1. CH₃CHBrCH₃ + NaOH → CH₃CH(OH)CH₃ + NaBr.2. n = 8.00 ÷ 122.9 = 0.06509 mol; mass = 0.06509 × 60.0 = 3.906 g.
3. Yield = 2.81 ÷ 3.906 × 100 = 71.9 %.
Problem 9 Rate data for 1-bromopropane
1. Doubling [RBr] doubles the rate: first order. Tripling [OH⁻] triples the rate: first order.2. rate = k[RBr][OH⁻]; k = 2.0 × 10⁻⁴ ÷ (0.10 × 0.10) = 2.0 × 10⁻² dm³ mol⁻¹ s⁻¹.
3. Both reactants are in the rate equation, so it is SN2.
Problem 10 Ozone destruction
1. Cl• + O₃ → ClO• + O₂ and ClO• + O → Cl• + O₂.2. O₃ + O → 2O₂; Cl• is a catalyst.
3. E = 346 000 ÷ 6.02 × 10²³ = 5.75 × 10⁻¹⁹ J; λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ 5.75 × 10⁻¹⁹ = 346 nm, in the ultraviolet region.
Problem 11 Making an amine
1. 10.89 ÷ 108.9 = 0.1000 mol.2. n(NH₃) = 0.200 mol; mass = 0.200 × 17.0 = 3.40 g.
3. The product amine is also a nucleophile and could react further to give (CH₃CH₂)₂NH; excess NH₃ makes this unlikely.
Problem 12 Elimination
1. CH₃CHBrCH₂CH₃ + KOH → CH₃CH=CHCH₃ + KBr + H₂O.2. but-1-ene, Z-but-2-ene and E-but-2-ene.
3. Elimination uses ethanolic KOH and heat; substitution uses aqueous NaOH, warm.
Problem 13 Silver nitrate test times
1. 40 ÷ 8 = 5.0 times faster.2. The C–I bond is weaker than C–Br and C–Cl, so it breaks most easily and gives halide ions first.
Problem 14 Spot the errors
1. Substitution is stepwise: CH₃Cl, CH₂Cl₂ and CHCl₃ form first.2. OH⁻ is a nucleophile (a lone pair donor).
3. SN2 causes inversion of configuration; SN1 gives the racemic mixture.
Problem 15 Atom economy and mass
1. Mr CH₃Cl = 50.5; reactants = 16.0 + 71.0 = 87.0; atom economy = 50.5 ÷ 87.0 × 100 = 58.0 %.2. n(CCl₄) = 100 ÷ 154.0 = 0.6494 mol; n(Cl₂) = 2.597 mol; mass = 2.597 × 71.0 = 184 g.
Problem 16 Tertiary halogenoalkane
1. Rate is first order in RBr and zero order in OH⁻: rate = k[RBr]; k = 3.0 × 10⁻⁵ ÷ 0.10 = 3.0 × 10⁻⁴ s⁻¹.2. t½ = ln 2 ÷ k = 0.693 ÷ 3.0 × 10⁻⁴ = 2.31 × 10³ s (38.5 min).
3. Three alkyl groups push electrons toward the positive carbon (inductive effect), stabilising the tertiary carbocation.
Problem 17 Stereochemistry and profiles
1. B is SN1: two peaks with a valley, the carbocation intermediate; A has a single peak (one step, SN2).2. SN2 inverts the configuration, giving a single enantiomer, so the product is still optically active.
3. The carbocation is planar, so water attacks equally from both faces, forming a racemic mixture with no net rotation.
Problem 18 Enthalpy of chlorination
1. Broken 413 + 242 = 655; made 346 + 432 = 778; ΔH = 655 − 778 = −123 kJ mol⁻¹.2. Cl• + CH₄ → HCl + •CH₃: 413 − 432 = −19; •CH₃ + Cl₂ → CH₃Cl + Cl•: 242 − 346 = −104; sum −123.
3. A chlorine radical is regenerated at every cycle, so the chain repeats until radicals terminate.
Problem 19 Two-step synthesis
1. Step 1: KCN in ethanol, reflux; step 2: dilute HCl, reflux.2. With water present OH⁻ could compete and form ethanol; ethanol avoids this.
3. CH₃CH₂CN + 2H₂O + HCl → CH₃CH₂COOH + NH₄Cl.
Problem 20 CFCs and HFCs
1. Mr = 12.0 + 19.0 + 106.5 = 137.5; Cl = 106.5 ÷ 137.5 × 100 = 77.5 %.2. E = 467 000 ÷ 6.02 × 10²³ = 7.76 × 10⁻¹⁹ J; λ = 1.989 × 10⁻²⁵ ÷ 7.76 × 10⁻¹⁹ = 256 nm, shorter than for C–Cl.
3. The C–F bond is stronger, and HFCs contain no chlorine to give Cl• radicals that catalyse ozone destruction.
