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Reactions of alkenes and haloalkanes (SACE Stage 2 Chemistry)

Syllabus dot point

“Describe the addition reactions of alkenes and the substitution reactions of haloalkanes, writing equations and products.”

9 min read

Quick answer

Alkenes are unsaturated, so they undergo addition reactions: the C=C\text{C}=\text{C} double bond opens and a small molecule adds across it (hydrogen, a halogen, a hydrogen halide, or water). Haloalkanes are saturated, so they undergo substitution: a nucleophile such as OH−\text{OH}^- replaces the halogen. You must write equations, name products, and use Markovnikov's rule to predict the major product when an unsymmetrical reagent adds to an unsymmetrical alkene.

Jump to a section
  1. What this dot point is asking
  2. Lead worked calculation
  3. Addition reactions of alkenes
  4. Markovnikov's rule
  5. Substitution reactions of haloalkanes
  6. Comparing the two
  7. Why it matters
  8. Exam-style questions

What this dot point is asking

SACE expects you to distinguish addition from substitution, write balanced equations with structural products, and apply Markovnikov reasoning.

Lead worked calculation

Predicting the major hydration product

Predict the major product when but-1-ene (CH2=CHCH2CH3\text{CH}_2=\text{CHCH}_2\text{CH}_3) reacts with water in the presence of dilute acid.

Identify the reaction type

A C=C\text{C}=\text{C} alkene plus water is an addition (hydration) reaction; H\text{H} and OH\text{OH} add across the double bond.

Apply Markovnikov's rule

The carbons of the double bond are C-1\text{C-1} (two hydrogens) and C-2\text{C-2} (one hydrogen). Markovnikov's rule: the -OH\text{-OH} adds to the carbon already bearing fewer hydrogens, here C-2\text{C-2}, and the H\text{H} adds to C-1\text{C-1}.

Write the major product

CH2=CHCH2CH3+H2O→CH3CH(OH)CH2CH3\text{CH}_2=\text{CHCH}_2\text{CH}_3 + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH(OH)CH}_2\text{CH}_3

The major product is butan-2-ol (a secondary alcohol). The minor product would be butan-1-ol.

Addition reactions of alkenes

The double bond is a region of high electron density that opens to let atoms add across it. The four key additions:

The four alkene addition reactions
  • Hydrogenation: H2\text{H}_2 adds with a nickel (or platinum) catalyst to give an alkane. CH2=CH2+H2→CH3CH3\text{CH}_2=\text{CH}_2 + \text{H}_2 \rightarrow \text{CH}_3\text{CH}_3.
  • Halogenation: a halogen adds to give a dihaloalkane. CH2=CH2+Br2→CH2BrCH2Br\text{CH}_2=\text{CH}_2 + \text{Br}_2 \rightarrow \text{CH}_2\text{BrCH}_2\text{Br}. The decolourising of orange bromine water is the test for a C=C\text{C}=\text{C} bond.
  • Hydrohalogenation: a hydrogen halide adds to give a haloalkane. CH2=CH2+HBr→CH3CH2Br\text{CH}_2=\text{CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{CH}_2\text{Br}.
  • Hydration: water adds (acid catalyst) to give an alcohol. CH2=CH2+H2O→CH3CH2OH\text{CH}_2=\text{CH}_2 + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{OH}.

Markovnikov's rule

When an unsymmetrical reagent (such as HBr\text{HBr} or H2O\text{H}_2\text{O}) adds to an unsymmetrical alkene, two products are possible. Markovnikov's rule predicts which dominates.

Markovnikov's rule

The hydrogen of the adding molecule attaches to the double-bond carbon that already has more hydrogen atoms, and the other part (-Br\text{-Br} or -OH\text{-OH}) attaches to the carbon with fewer hydrogens. A handy phrase: "the rich get richer", the carbon with more hydrogens gains the hydrogen. This gives the more stable product as the major one.

Substitution reactions of haloalkanes

Haloalkanes are saturated, so they cannot add; instead the halogen, a good leaving group, is replaced by a nucleophile.

  • Hydrolysis with aqueous hydroxide gives an alcohol: R-X+OH−→R-OH+X−\text{R-X} + \text{OH}^- \rightarrow \text{R-OH} + \text{X}^-.
  • The halogen leaves as a halide ion, and the incoming group takes its place on the same carbon.

This is the reverse logic of addition: addition increases the number of groups on the carbons; substitution swaps one group for another.

Comparing the two

Addition is characteristic of unsaturated compounds (alkenes), where there is a multiple bond to open. Substitution is characteristic of saturated compounds (haloalkanes and alkanes), where a group must be replaced because there is no multiple bond to add across. Identifying whether the starting material is saturated or unsaturated tells you which reaction type to expect.

Applying Markovnikov backwards or to symmetric alkenes

A frequent SACE error is adding the -OH\text{-OH} or -Br\text{-Br} to the wrong carbon. Remember the hydrogen goes to the carbon that already has more hydrogens. Markovnikov only matters for unsymmetrical alkenes and reagents; for ethene or for symmetric reagents like Br2\text{Br}_2 there is only one product. Another error is treating an alkene reaction as substitution (it is addition) or forgetting the catalyst for hydrogenation and hydration.

Why it matters

These reactions convert simple hydrocarbons into alcohols, haloalkanes and polymers, the gateway compounds for synthesising almost all other organic materials. Markovnikov reasoning and the addition-versus-substitution distinction are recurring tools in the synthesis pathways assessed later in this topic.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2022-style5 marks
Propene (CH3CH=CH2\text{CH}_3\text{CH}=\text{CH}_2) reacts in three separate experiments. Write the structural formula and name of the organic product for: (a) reaction with hydrogen over a nickel catalyst; (b) reaction with bromine; (c) reaction with water (acid catalyst), giving the **major** product and justifying it with Markovnikov's rule.
Show worked answer →

(a) Hydrogenation adds H2\text{H}_2 across the double bond: CH3CH2CH3\text{CH}_3\text{CH}_2\text{CH}_3, propane. (1 mark)

(b) Bromine adds across the double bond: CH3CHBrCH2Br\text{CH}_3\text{CHBrCH}_2\text{Br}, 1,2-dibromopropane. (2 marks)

(c) Hydration adds H-OH\text{H-OH}. By Markovnikov's rule the -OH\text{-OH} adds to the more substituted carbon (the one with fewer hydrogens), giving the major product CH3CH(OH)CH3\text{CH}_3\text{CH(OH)CH}_3, propan-2-ol. (2 marks)

SACE 2020-style4 marks
1-bromobutane reacts with aqueous sodium hydroxide. (a) Write the equation and name the organic product. (b) State the type of reaction. (c) 0.685 g0.685\ \text{g} of 1-bromobutane (M=137.0 g mol−1M = 137.0\ \text{g mol}^{-1}) reacts completely; calculate the mass of organic product formed (M=74.0 g mol−1M = 74.0\ \text{g mol}^{-1}).
Show worked answer →

(a) CH3CH2CH2CH2Br+NaOH→CH3CH2CH2CH2OH+NaBr\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br} + \text{NaOH} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} + \text{NaBr}; the product is butan-1-ol. (1 mark)

(b) It is a substitution (nucleophilic substitution / hydrolysis) reaction: the -OH\text{-OH} replaces the -Br\text{-Br}. (1 mark)

(c) n(1-bromobutane)=0.685137.0=5.00×10−3 moln(\text{1-bromobutane}) = \dfrac{0.685}{137.0} = 5.00 \times 10^{-3}\ \text{mol}. The 1:11:1 ratio gives n(butan-1-ol)=5.00×10−3 moln(\text{butan-1-ol}) = 5.00 \times 10^{-3}\ \text{mol}, so m=nM=5.00×10−3×74.0=0.370 gm = nM = 5.00 \times 10^{-3} \times 74.0 = 0.370\ \text{g}. (2 marks)

Practise this

Sources & how we know this

Based on the official SACE Board subject outline. Not yet individually reviewed by a person; always check SACE Board and your school's materials for current rules.

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