← Unit 4: Structure, synthesis and design

QLDChemistrySyllabus dot point

Topic 1: Properties and structure of organic materials

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

A focused answer to the QCE Chemistry Unit 4 dot point on alkane and alkene reactivity. Sets out free-radical substitution of alkanes by halogens (UV initiation) and electrophilic addition of alkenes (halogens, hydrogen halides, hydrogen, water) with Markovnikov's rule for unsymmetrical alkenes. Includes the bromine-water test and the IA3 / EA expected products for each.

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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:

R−H+X2→UVR−X+HXR-H + X_2 \xrightarrow{UV} R-X + HX

Example. Methane and chlorine:

CH4+Cl2→UVCH3Cl+HClCH_4 + Cl_2 \xrightarrow{UV} CH_3Cl + HCl

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)

R−CH=CH−R′+X2→R−CHX−CHX−R′R-CH=CH-R' + X_2 \rightarrow R-CHX-CHX-R'

Example. Ethene + bromine:

CH2=CH2+Br2→CH2Br−CH2BrCH_2=CH_2 + Br_2 \rightarrow CH_2Br-CH_2Br

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)

R−CH=CH−R′+HX→R−CH2−CHX−R′ (or isomer)R-CH=CH-R' + HX \rightarrow R-CH_2-CHX-R' \text{ (or isomer)}

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:

CH3−CH=CH2+HBr→CH3−CHBr−CH3 (major: 2-bromopropane)CH_3-CH=CH_2 + HBr \rightarrow CH_3-CHBr-CH_3 \text{ (major: 2-bromopropane)}

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)

R−CH=CH−R′+H2→Ni or PtR−CH2−CH2−R′R-CH=CH-R' + H_2 \xrightarrow{Ni \text{ or } Pt} R-CH_2-CH_2-R'

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:

CH2=CH2+H2→NiCH3−CH3CH_2=CH_2 + H_2 \xrightarrow{Ni} CH_3-CH_3

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)

R−CH=CH−R′+H2O→H+R−CH(OH)−CH2−R′ (or isomer, Markovnikov)R-CH=CH-R' + H_2O \xrightarrow{H^+} R-CH(OH)-CH_2-R' \text{ (or isomer, Markovnikov)}

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):

CH3−CH=CH2+H2O→H+CH3−CH(OH)−CH3 (propan-2-ol)CH_3-CH=CH_2 + H_2O \xrightarrow{H^+} CH_3-CH(OH)-CH_3 \text{ (propan-2-ol)}

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

Summary table

Reactant on alkene Conditions Mechanism Products (general)
X2 (Br2, Cl2) room T, no UV needed electrophilic addition vicinal dihaloalkane
HX (HBr, HCl) room T electrophilic addition haloalkane (Markovnikov)
H2 Ni or Pt catalyst catalytic hydrogenation alkane
H2O dilute H2SO4 catalyst, heat acid-catalysed hydration alcohol (Markovnikov)

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.

In one sentence

Alkanes undergo free-radical substitution with halogens under UV (alkane to haloalkane plus HX), while alkenes undergo addition of halogens, hydrogen halides, hydrogen and water across the C=C double bond, with Markovnikov's rule fixing the regiochemistry of unsymmetrical additions involving HX and H2O.

Past exam questions, worked

Real questions from past QCAA papers on this dot point, with our answer explainer.

2023 QCAA-style4 marksPropene is reacted separately with (a) bromine in CCl4, (b) hydrogen bromide gas, (c) water in the presence of dilute sulfuric acid as a catalyst. For each, write a balanced equation and name the major organic product. Justify the product of (b) using Markovnikov's rule.
Show worked answer →

A 4-mark answer needs the three equations with named products and the Markovnikov justification.

(a) Propene + bromine.

CH3−CH=CH2+Br2→CH3−CHBr−CH2BrCH_3-CH=CH_2 + Br_2 \rightarrow CH_3-CHBr-CH_2Br

Product: 1,2-dibromopropane. Bromine adds across the double bond.

(b) Propene + hydrogen bromide.

CH3−CH=CH2+HBr→CH3−CHBr−CH3CH_3-CH=CH_2 + HBr \rightarrow CH_3-CHBr-CH_3

Major product: 2-bromopropane. Markovnikov's rule: when HX adds to an unsymmetrical alkene, the H attaches to the carbon already bearing more hydrogens, so X attaches to the more substituted carbon. Propene's C2 has one H and C1 has two H; H goes to C1, Br goes to C2.

(c) Propene + water (acid-catalysed).

CH3−CH=CH2+H2O→CH3−CH(OH)−CH3CH_3-CH=CH_2 + H_2O \rightarrow CH_3-CH(OH)-CH_3

Major product: propan-2-ol. Same Markovnikov logic: OH goes to the more substituted carbon (C2).

Markers reward correct balanced equations (single arrow for these addition reactions), correct major product names, and a clear Markovnikov statement that mentions hydrogen distribution.

2022 QCAA-style3 marks(a) Describe the bromine-water test and explain what it detects. (b) Predict the result when bromine water is shaken with hexane, and separately with hex-1-ene.
Show worked answer →

A 3-mark answer needs the test method, what it detects, and the observed result for each compound.

(a) Test method and detection. Bromine water is orange-brown (dilute Br2 in water). Shake a few drops with the unknown organic liquid. If the orange-brown colour disappears (decolourised), an alkene (or any unsaturated C=C or C-triple-bond-C) is present, because bromine adds across the double bond to form a colourless dibromo product. If the colour persists, the compound is saturated.

(b) Hexane. Saturated alkane. No reaction at room temperature in the absence of UV. Bromine water stays orange-brown.

Hex-1-ene. Unsaturated alkene. CH2=CH-(CH2)3-CH3 + Br2 -> CH2Br-CHBr-(CH2)3-CH3 (1,2-dibromohexane). Bromine water decolourises (orange-brown to colourless).

Markers reward the colour description, the explicit "addition across the C=C" mechanism for the alkene, and identification of hexane as unreactive at room temperature without UV. Answers that confuse this with a free-radical substitution under UV miss the point of the test.

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