Predict and explain the products of substitution reactions of alkanes with halogens and addition reactions of alkenes with halogens, hydrogen halides, hydrogen and water

What this dot point is asking

QCAA wants you to predict the products of two key Unit 4 reaction types: free-radical substitution of alkanes by halogens (UV catalysis) and electrophilic addition of alkenes by halogens, hydrogen halides, hydrogen and water. For unsymmetrical alkenes you must apply Markovnikov's rule. The dot point underpins the IA3 organic synthesis pathway question and is examined every year in EA Paper 1 and Paper 2.

The answer

Alkanes and alkenes differ sharply in reactivity. Alkanes are saturated (only single bonds) and only react under harsh conditions; the canonical reaction is halogenation under UV light, proceeding by a free-radical mechanism. Alkenes are unsaturated (one C=C double bond) and react readily by addition across the double bond, breaking the pi bond and forming two new single bonds.

Reactions of alkanes: free-radical substitution

Alkanes react with halogens (Cl2, Br2) in the presence of ultraviolet (UV) light or at high temperature. A hydrogen on the alkane is replaced by a halogen atom.

Generic equation:

Example. Methane and chlorine:

The QCAA syllabus does not require the full radical mechanism (initiation / propagation / termination) but does expect you to recognise:

• UV is the initiation condition. Without UV (or heat), the reaction does not proceed at room temperature.
• Substitution, not addition. A C-H bond is replaced by C-X. Molecular mass changes by (X - H).
• Polysubstitution. With excess halogen, further substitution gives CH2Cl2, CHCl3 and CCl4. QCAA usually specifies "with limited halogen" so monosubstitution dominates.
• HX is the by-product. This is the diagnostic difference from addition.

For longer alkanes, substitution is non-selective (every C-H is in principle replaceable), but secondary C-H bonds react slightly faster than primary. For Unit 4 questions, accept any monosubstituted product as the correct answer unless specified.

Reactions of alkenes: addition

The C=C double bond consists of a sigma bond plus a pi bond. The pi bond is weaker and electron-rich, attracting electrophiles. Addition reactions break the pi bond and form two new single bonds (one to each carbon of the former C=C).

Addition of halogens (Br2, Cl2)

Example. Ethene + bromine:

Product: 1,2-dibromoethane. Reaction is rapid at room temperature without UV (this is the diagnostic difference from alkane halogenation). The reaction is the basis of the bromine-water test for unsaturation: orange-brown bromine water decolourises in the presence of an alkene.

Addition of hydrogen halides (HCl, HBr, HI)

For symmetrical alkenes (ethene, but-2-ene), only one product is possible. For unsymmetrical alkenes (propene, but-1-ene), two products are possible: H can add to either carbon of the C=C. Markovnikov's rule predicts which is major.

Markovnikov's rule. When HX adds to an unsymmetrical alkene, the H attaches to the carbon already carrying more hydrogens; X attaches to the carbon with fewer hydrogens (the more substituted carbon).

Memory aid: "hydrogen-rich gets richer."

Example. Propene + HBr:

H joins C1 (already 2 H), Br joins C2 (had 1 H). The minor product (1-bromopropane) is also formed in trace amount but QCAA expects the major product only.

The underlying reason is that the reaction proceeds via a carbocation intermediate; the more substituted (secondary > primary) carbocation is more stable, so its formation is the kinetically and thermodynamically favoured pathway. QCAA does not require the mechanism but does expect you to invoke "more stable carbocation" if asked to justify.

Addition of hydrogen (hydrogenation)

A nickel or platinum catalyst is required. The double bond is reduced to a single bond, converting the alkene to the corresponding alkane.

Example. Ethene + hydrogen:

Industrial application: catalytic hydrogenation of vegetable oils to produce margarine (partial saturation of triglyceride C=C bonds raises melting point).

Addition of water (acid-catalysed hydration)

A dilute sulfuric acid catalyst (or phosphoric acid in industry) is required. Water adds across C=C; OH goes to the more substituted carbon by Markovnikov's rule (same carbocation logic as HX addition).

Example. Propene + water (H+ catalyst):

This reaction is the industrial route to ethanol from petroleum-derived ethene (the alternative is fermentation from sugars).

Summary table

For alkanes, the corresponding "Br2" reaction differs sharply: requires UV light, proceeds by substitution (not addition), produces HX as a by-product, and is non-selective. This is the cleanest discrimination test between alkanes and alkenes in QCAA stimulus.

Common traps

Writing alkane bromination as addition

Methane + Br2 does NOT give CH2Br2 by addition. Methane has no double bond. Always check saturation before choosing a mechanism.

Forgetting UV for alkane halogenation

No UV, no reaction at room temperature. QCAA marks for the condition explicitly.

Misapplying Markovnikov

The rule applies to unsymmetrical alkenes only and to addition of HX or H2O (not Br2 or Cl2, since both ends of the reagent are identical).

Treating "addition" as identical to "substitution"

Addition preserves all atoms (no by-product); substitution releases HX (or similar). Atom counts diagnose which reaction occurred.

Confusing catalytic hydrogenation with reduction by NaBH4 or LiAlH4

H2 with Ni/Pt reduces C=C bonds but not C=O. Hydride reagents reduce C=O but not isolated C=C. QCAA Unit 4 uses catalytic hydrogenation only.

Examples in context

Example 1. Catalytic cracking at Caltex Lytton refinery. Lytton refinery's fluid catalytic cracker breaks long-chain alkanes (e.g. $\text{C}_{15} \text{H}_{32}$) into shorter, more valuable molecules: $\text{C}_{15} \text{H}_{32} \rightarrow \text{C}_8 \text{H}_{18} + \text{C}_7 \text{H}_{14}$ (octane + heptene). Zeolite catalysts at $500^{\circ}\text{C}$ provide acidic sites that generate carbocations. The shorter alkane goes to petrol blending; the alkene becomes feedstock for polymerisation (LDPE, polystyrene). Cracking is the central refinery operation, transforming low-value heavy ends into high-value light products. It is endothermic but releases a stream of useful alkenes carrying $\text{C} = \text{C}$ reactivity.

Example 2. Bromine-water test at QCAA EA practical. Students differentiate cyclohexane (alkane) from cyclohexene (alkene) using bromine water. Cyclohexane is unreactive: no decolourisation, the substrate is saturated and cannot undergo electrophilic addition. Cyclohexene reacts: $\text{C}_6 \text{H}_{10} + \text{Br}_2 \rightarrow 1,2$-dibromocyclohexane, decolourising the orange water to colourless. The mechanism is electrophilic addition: $\pi$-bond electrons attack $\text{Br}_2$, polarising it, forming a bromonium ion intermediate, then $\text{Br}^-$ attacks the other carbon. Markovnikov's rule is not relevant for symmetric alkenes but matters for $\text{HBr}$ addition.

Try this

Q1. Write balanced equations for (a) complete combustion of propane, (b) free-radical bromination of methane. [3 marks]

• Cue. (a) $\text{C}_3 \text{H}_8 + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2 \text{O}$. (b) $\text{CH}_4 + \text{Br}_2 \rightarrow \text{CH}_3 \text{Br} + \text{HBr}$ (UV/$h\nu$).

Q2. Predict the major product when HBr adds to propene. Justify with Markovnikov's rule. [3 marks]

• Cue. 2-bromopropane major. Markovnikov: H to C with more H, leaving Br on more substituted C (stable secondary carbocation intermediate).

Q3. A student tests four colourless liquids with bromine water in light. (a) Pentane, (b) pent-1-ene, (c) cyclopentane, (d) cyclohexene. Predict observations and identify mechanism for each. [4+4 marks]

• Cue. (a) Slow decolourisation in UV (substitution, free-radical). (b) Fast decolourisation (electrophilic addition). (c) Slow in UV (substitution). (d) Fast (addition).