Alkanes, alkenes and reaction mechanisms

What you'll learn

Alkanes and alkenes are the simplest hydrocarbons, but they show two very different kinds of reactivity. Alkanes are saturated and relatively unreactive, reacting mainly by free-radical substitution and combustion. Alkenes contain a reactive C=C double bond and undergo electrophilic addition. This guide covers the structure and properties of alkanes, combustion and its products, free-radical substitution with halogens, the structure of alkenes, electrophilic addition of hydrogen, halogens, hydrogen halides and steam, Markovnikov's rule and carbocation stability, tests for unsaturation, oxidation of alkenes, halogenoalkanes and nucleophilic substitution (Sₙ1 and Sₙ2) and elimination, and the meaning of the key mechanism terms. It links to Trinidad's natural gas and petrochemical industries, where ethane and propane are processed.

CAPE questions require you to draw mechanisms with curly arrows, explain why reactions occur, and predict major products. Practise drawing mechanisms until every arrow, charge and lone pair is in the right place.

Key terms and definitions

Saturated — Containing only single C–C bonds.

Unsaturated — Containing one or more C=C (or C≡C) bonds.

Homolytic fission — Breaking a covalent bond so that each atom takes one electron, forming free radicals.

Heterolytic fission — Breaking a covalent bond so that one atom takes both electrons, forming ions.

Free radical — A species with an unpaired electron.

Electrophile — An electron-pair acceptor, attracted to regions of high electron density.

Nucleophile — An electron-pair donor, attracted to electron-deficient regions.

Carbocation — A positively charged carbon ion with only three bonds.

Curly arrow — Shows the movement of a pair of electrons, from a bond or lone pair to an atom or new bond.

Addition reaction — Two molecules combine to form one product.

Substitution reaction — An atom or group is replaced by another.

Elimination reaction — A small molecule is removed from a larger one, often forming a double bond.

Core concepts

Alkanes: properties and combustion

Alkanes (CₙH₂ₙ₊₂) are non-polar and relatively unreactive because C–C and C–H bonds are strong and non-polar. Boiling points rise with chain length (more London forces); branching lowers boiling point.

Complete combustion gives carbon dioxide and water: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. Incomplete combustion (limited oxygen) gives carbon monoxide (toxic — binds to haemoglobin) and soot (carbon particulates). Burning fuels also produces nitrogen oxides (from the high temperatures in engines) and sulfur dioxide (from sulfur impurities), linking alkanes to air pollution and acid rain.

Free-radical substitution

Alkanes react with chlorine or bromine in the presence of ultraviolet light:

CH₄ + Cl₂ → CH₃Cl + HCl

Mechanism:

1. Initiation: UV light causes homolytic fission: Cl₂ → 2Cl•
2. Propagation:
   • Cl• + CH₄ → CH₃• + HCl
   • CH₃• + Cl₂ → CH₃Cl + Cl• (The chlorine radical is regenerated, creating a chain reaction.)
3. Termination: two radicals combine:
   • Cl• + Cl• → Cl₂
   • CH₃• + Cl• → CH₃Cl
   • CH₃• + CH₃• → C₂H₆

Further substitution produces CH₂Cl₂, CHCl₃ and CCl₄, so a mixture of products forms — a disadvantage for synthesis.

Alkenes: structure

The C=C double bond consists of a σ bond and a π bond. The π bond is a region of high electron density above and below the plane of the molecule, weaker than the σ bond, and easily attacked by electrophiles. Rotation about the double bond is restricted, causing geometric isomerism.

Electrophilic addition

With bromine (test for unsaturation): bromine water changes from orange-brown to colourless when shaken with an alkene: CH₂=CH₂ + Br₂ → CH₂BrCH₂Br.

Mechanism with bromine:

1. As Br₂ approaches the π bond, it becomes polarised (Brᵟ⁺–Brᵟ⁻).
2. The π electrons attack Brᵟ⁺ (curly arrow from the double bond to Br); the Br–Br bond breaks heterolytically (curly arrow to Brᵟ⁻), forming a carbocation and Br⁻.
3. Br⁻ donates a lone pair to the carbocation, forming 1,2-dibromoethane.

With hydrogen halides: CH₂=CH₂ + HBr → CH₃CH₂Br. The H–Br bond is already polar; the π electrons attack Hᵟ⁺, forming a carbocation, which is then attacked by Br⁻.

With steam (hydration): CH₂=CH₂ + H₂O → CH₃CH₂OH, using a phosphoric acid catalyst at around 300 °C and 60–70 atm — an industrial route to ethanol.

With hydrogen (hydrogenation): CH₂=CH₂ + H₂ → CH₃CH₃, using a nickel catalyst at about 150 °C — used to harden vegetable oils into margarine.

Markovnikov's rule and carbocation stability

When an unsymmetrical alkene such as propene reacts with HBr, two products are possible. The major product comes from the more stable carbocation.

Carbocation stability: tertiary > secondary > primary, because alkyl groups push electrons towards the positive carbon (positive inductive effect), spreading out the charge.

Propene + HBr → 2-bromopropane (major, via the secondary carbocation) and 1-bromopropane (minor, via the primary carbocation).

Markovnikov's rule: hydrogen adds to the carbon of the double bond that already has more hydrogen atoms.

Oxidation of alkenes

• With cold, dilute, acidified potassium manganate(VII), alkenes form diols; the purple colour disappears (another test for unsaturation).
• With hot, concentrated, acidified KMnO₄, the double bond is broken completely, forming ketones, carboxylic acids or carbon dioxide depending on the structure — useful for locating the double bond.

Halogenoalkanes: nucleophilic substitution and elimination

The C–X bond is polar (Cᵟ⁺–Xᵟ⁻), so the carbon is attacked by nucleophiles such as OH⁻, CN⁻ and NH₃.

• With aqueous NaOH (heat under reflux): R–X + OH⁻ → R–OH + X⁻ (forms an alcohol).
• With KCN in ethanol: R–X + CN⁻ → R–CN + X⁻ (forms a nitrile, extending the carbon chain by one).
• With excess ethanolic ammonia (heated in a sealed tube): R–X + 2NH₃ → R–NH₂ + NH₄X (forms an amine).

Mechanisms:

• Sₙ2 (primary halogenoalkanes): one step; the nucleophile attacks as the halide leaves, through a transition state; rate = k[RX][OH⁻].
• Sₙ1 (tertiary halogenoalkanes): two steps; the C–X bond breaks first to form a carbocation (slow, rate-determining), then the nucleophile attacks; rate = k[RX].

Reactivity depends on C–X bond strength: C–I is weakest, so iodoalkanes react fastest; C–F is strongest and least reactive.

Elimination: with ethanolic NaOH and heat, OH⁻ acts as a base, removing H⁺ and X⁻ to form an alkene: CH₃CH₂Br + OH⁻ → CH₂=CH₂ + H₂O + Br⁻. Conditions (aqueous vs ethanolic) decide whether substitution or elimination predominates.

CFCs and the ozone layer

Chlorofluorocarbons were once widely used as refrigerants and aerosol propellants. In the upper atmosphere, UV light breaks C–Cl bonds, releasing Cl• radicals, which catalyse the breakdown of ozone in a chain reaction. This is why CFCs were phased out under the Montreal Protocol.

Worked examples

Example 1: Propagation steps

Write the propagation steps for the reaction of ethane with bromine in UV light.

Br• + C₂H₆ → C₂H₅• + HBr C₂H₅• + Br₂ → C₂H₅Br + Br•

Example 2: Major product

Predict the major product when 2-methylpropene reacts with HCl, and explain.

The major product is 2-chloro-2-methylpropane. Addition of H⁺ to the CH₂ carbon forms a tertiary carbocation, which is more stable than the primary alternative because three alkyl groups donate electron density.

Example 3: Substitution or elimination?

Bromoethane is heated with NaOH. What product forms (a) in water, (b) in ethanol?

(a) Ethanol by nucleophilic substitution. (b) Ethene by elimination.

Example 4: Test for unsaturation

Describe a test to distinguish hexane from hex-1-ene.

Shake each with bromine water. Hex-1-ene decolourises the orange-brown bromine water rapidly; hexane does not (in the dark).

Example 5: Describing a mechanism in words

Describe the mechanism for the reaction of propene with hydrogen bromide to form 2-bromopropane.

1. The H–Br bond is polar (Hᵟ⁺–Brᵟ⁻).
2. A curly arrow is drawn from the C=C double bond to the Hᵟ⁺ atom, and another from the H–Br bond to the Br atom, which leaves as Br⁻.
3. The hydrogen adds to the CH₂ carbon (the end carbon), forming a secondary carbocation, CH₃C⁺HCH₃, which is more stable than the primary alternative.
4. A curly arrow is drawn from a lone pair on Br⁻ to the positive carbon, forming 2-bromopropane.

The mechanism is electrophilic addition, with H⁺ (from HBr) acting as the electrophile.

Example 6: Sₙ1 or Sₙ2?

The hydrolysis of 2-bromo-2-methylpropane has the rate equation rate = k[C₄H₉Br]. The hydrolysis of 1-bromobutane has rate = k[C₄H₉Br][OH⁻]. Explain the difference.

2-bromo-2-methylpropane is a tertiary halogenoalkane. Its slow step is the breaking of the C–Br bond to form a stable tertiary carbocation, which does not involve OH⁻, so the rate depends only on the halogenoalkane — an Sₙ1 mechanism. 1-bromobutane is primary; a primary carbocation is too unstable to form, so OH⁻ attacks as Br⁻ leaves in a single step — an Sₙ2 mechanism, in which both species appear in the rate equation. Bulky methyl groups around the tertiary carbon also hinder the backside attack needed for Sₙ2.

Example 7: A synthesis route

Outline how ethene can be converted into ethylamine in two steps, giving reagents and conditions.

1. Ethene + HBr (room temperature) → bromoethane (electrophilic addition).
2. Bromoethane + excess concentrated ammonia in ethanol, heated in a sealed tube → ethylamine (nucleophilic substitution). Excess ammonia reduces further substitution to secondary and tertiary amines.

Reagents and conditions summary

This table is a useful starting point for planning multi-step syntheses.

Common mistakes and how to avoid them

• Curly arrows starting from atoms instead of bonds or lone pairs.
• Forgetting the δ+/δ− dipole on bromine in electrophilic addition.
• Missing the charge on carbocations and the lone pair on the nucleophile.
• Confusing Sₙ1 and Sₙ2 — primary favours Sₙ2, tertiary Sₙ1.
• Mixing up conditions — aqueous NaOH gives substitution; ethanolic NaOH gives elimination.
• Saying alkanes react with bromine water — they need UV light and react by substitution, not addition.

Self-test questions

1. Write the termination step that forms ethane during the chlorination of methane. Answer: CH₃• + CH₃• → C₂H₆.
2. Name the major product when but-1-ene reacts with HBr. Answer: 2-bromobutane, formed via the more stable secondary carbocation.
3. Which reacts fastest with aqueous NaOH: 1-chlorobutane, 1-bromobutane or 1-iodobutane? Explain. Answer: 1-iodobutane, because the C–I bond is the weakest and breaks most easily.
4. State the reagent and conditions needed to convert 2-bromopropane into propene. Answer: Sodium (or potassium) hydroxide dissolved in ethanol, heated — an elimination reaction.

How this links to your Internal Assessment

Organic SBAs may involve testing for unsaturation in local vegetable oils (comparing how much bromine water or iodine solution different oils decolourise), preparing a halogenoalkane from an alcohol, or investigating the rate of hydrolysis of different halogenoalkanes using silver nitrate in ethanol (the time taken for a precipitate to form shows that C–I bonds break faster than C–Cl). Strong reports control temperature and concentrations, use fair timing methods, and explain results in terms of bond enthalpy.

Exam technique for Alkanes, alkenes and reaction mechanisms

• Draw mechanisms with curly arrows, partial charges, lone pairs and intermediate charges clearly.
• Name the mechanism type (free-radical substitution, electrophilic addition, nucleophilic substitution, elimination).
• State conditions (UV light, catalyst, temperature, solvent).
• Use carbocation stability to justify major products.

Quick revision summary

• Alkanes: unreactive; combustion (complete and incomplete); free-radical substitution with UV light (initiation, propagation, termination).
• Alkenes: σ + π bond; electrophilic addition of Br₂, HX, H₂O (H₃PO₄ catalyst) and H₂ (Ni catalyst).
• Bromine water: orange-brown → colourless with alkenes.
• Markovnikov: more stable carbocation (tertiary > secondary > primary) gives the major product.
• KMnO₄: cold dilute → diol; hot concentrated → bond cleavage.
• Halogenoalkanes: nucleophilic substitution (OH⁻, CN⁻, NH₃; Sₙ1/Sₙ2) and elimination (ethanolic OH⁻).
• CFCs release Cl• radicals that destroy ozone.