The problems are on the electrophilic addition worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Alkenes are hydrocarbons with a C=C double bond whose exposed π electrons attract electrophiles. These worked answers correct the 20 problems of the Chemistry 11-12 worksheet on electrophilic addition, from alkene formulae to atom economy of ethanol: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the method first, then the result in bold. Atomic masses used: H 1.0, C 12.0, O 16.0, Cl 35.5, Br 79.9. Check that every equation is balanced and that carbocations are drawn with the charge on the correct carbon.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: A Level 3.3.4
Problem 1 Alkene formulae
1. General formula: CₙH₂ₙ.2. n = 5 gives C₅H₁₀; n = 8 gives C₅H₁₀ and C₈H₁₆.
3. Four carbons, double bond starting on carbon 2: but-2-ene.
Problem 2 Shape of ethene
1. Trigonal planar with three regions of electron density; the angle is about 120°.2. Four C–H σ bonds, one C–C σ bond and one π bond: 5 σ bonds and 1 π bond.
3. The π bond needs the p orbitals on both carbons to be parallel, so rotation would break it. This locks the two CH₂ groups in the same plane.
Problem 3 Spotting unsaturation
1. Add bromine water to each and shake.2. Hex-1-ene turns it from orange-brown to colourless; hexane leaves the orange colour.
3. The product is 1,2-dibromohexane.
Problem 4 Completing equations
1. C₃H₆ + H₂ → C₃H₈ (propane).2. CH₂=CHCH₂CH₃ + Br₂ → CH₂BrCHBrCH₂CH₃, written C₄H₈Br₂.
3. Propane; 1,2-dibromobutane. Each equation has equal numbers of atoms on both sides, so no coefficients are needed.
Problem 5 Electrophile or nucleophile
1. An electrophile accepts an electron pair.2. Electrophiles: H⁺ and Brδ+. Nucleophiles: OH⁻ and CN⁻ (they donate lone pairs).
3. The π bond is a region of high electron density, so it attracts positive species. Negative ions are repelled by it.
Problem 6 Ethene to dibromoethane
1. C₂H₄ + Br₂ → C₂H₄Br₂.2. n = 1.40 / 28.0 = 0.0500 mol.
3. Mole ratio 1 : 1, so m = 0.0500 × 187.8 = 9.39 g.
Problem 7 Hydration of ethene
1. C₂H₄ + H₂O ⇌ C₂H₅OH.2. n = 120 / 24.0 = 5.00 mol; m = 5.00 × 46.0 = 230 g.
3. Only a small fraction reacts in each pass because the reaction is reversible; recycling raises the overall yield and saves raw material.
Problem 8 Propene and hydrogen bromide
1. Major: 2-bromopropane; minor: 1-bromopropane.2. Major route: secondary carbocation CH₃CH⁺CH₃. Minor route: primary CH₃CH₂CH₂⁺.
3. Two alkyl groups push electron density towards the secondary carbon, stabilising it more than the primary ion with one alkyl group. The more stable intermediate forms faster.
Problem 9 Symmetrical or not
1. But-2-ene gives 2-bromobutane only.2. Pent-1-ene gives 2-bromopentane as the major product.
3. In but-2-ene both double-bond carbons are equivalent (each has one H and one CH₃), so either carbon gives the same secondary ion.
Problem 10 Tertiary carbocation
1. H adds to the CH carbon: (CH₃)₂C⁺CH₂CH₃, a tertiary ion.2. Water attacks, then H⁺ is lost: 2-methylbutan-2-ol.
3. The OH is on a carbon bonded to three other carbons: tertiary.
Problem 11 Percentage yield
1. n = 8.4 / 42.0 = 0.200 mol.2. Ratio 1 : 1, so theoretical mass = 0.200 × 122.9 = 24.6 g.
3. Yield = 18.5 / 24.6 × 100 = 75.3 % (using 24.58 g).
Problem 12 Polymer repeat units
1. Propene CH₂=CHCH₃ gives –CH₂–CH(CH₃)–.2. Tetrafluoroethene, CF₂=CF₂.
3. Replace the chain bond by a double bond: CH₂=CH(CN), propenenitrile.
Problem 13 Length of a PVC chain
1. –CH₂–CHCl–, C₂H₃Cl: 24.0 + 3.0 + 35.5 = 62.5 g/mol.2. 1 800 × 62.5 = 112 500 g/mol (1.125 × 10⁵).
3. n = 25 000 / 62.5 = 400 mol, since the monomer and repeat unit have the same formula.
Problem 14 Why addition occurs
1. The π bond holds a region of high electron density that attracts electrophiles.2. Ethane has only σ bonds, with no exposed electrons; the C–C and C–H bonds are strong and non-polar.
3. The weaker π bond is replaced by two strong σ bonds, so the process releases energy and the saturated product is lower in energy.
Problem 15 Disposal of addition polymers
1. Its chains contain only strong non-polar C–C and C–H bonds, which microbes cannot hydrolyse.2. Advantage: energy recovered as heat. Drawback: burning PVC releases HCl, which must be removed by scrubbing.
3. Different polymers melt at different temperatures; mixing them gives a poor product, so sorting keeps the recycled material uniform.
Problem 16 Relative rates of addition
1. Fastest: 2-methylpropene (800), then propene (25), then ethene (1).2. Ethene: CH₃CH₂⁺ (primary); propene: CH₃CH⁺CH₃ (secondary); 2-methylpropene: (CH₃)₃C⁺ (tertiary).
3. More alkyl groups stabilise the ion, lowering the activation energy of the slow step. Tertiary ion gives the fastest reaction.
Problem 17 Finding the errors
1. Error 1: a primary carbocation is shown, but the secondary ion is more stable. Error 2: 1-bromopropane is not the major product.2. Intermediate CH₃CH⁺CH₃; major product 2-bromopropane.
3. The curly arrow must start at the C=C bond and point to the hydrogen, because the π electrons are the nucleophilic site.
Problem 18 Unknown alkene by titration
1. n(Br₂) = 8.00 / 159.8 = 0.0501 mol.2. Alkene : Br₂ = 1 : 1, so M = 2.10 / 0.0501 = 41.9 g/mol.
3. CₙH₂ₙ with 14n = 42 gives n = 3: C₃H₆, propene.
Problem 19 Combustion analysis
1. n(C) = 4.40 / 44.0 = 0.100 mol; n(H) = 2 × 1.80 / 18.0 = 0.200 mol. Ratio 1 : 2, empirical formula CH₂.2. 56 / 14.0 = 4, so C₄H₈.
3. But-1-ene, Z-but-2-ene, E-but-2-ene and 2-methylpropene. But-2-ene (E or Z) gives only 2-bromobutane.
Problem 20 Atom economy of ethanol
1. A: 46.0 / (28.0 + 18.0) × 100 = 100 %.2. B: 46.0 / (108.9 + 40.0) × 100 = 46.0 / 148.9 = 30.9 %.
3. Route A has a far better atom economy and no waste salt; route B needs the halogenoalkane, which itself comes from an alkene, so it adds a step.
