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How do alkenes undergo addition and haloalkanes undergo substitution?

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

Addition reactions of alkenes (hydrogenation, halogenation, hydration, hydrohalogenation) and substitution of haloalkanes, with Markovnikov reasoning, equations, and worked SACE-style product-prediction and stoichiometry examples.

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  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

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

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:

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.

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+OHR-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.

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 practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20225 marksPropene (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.
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(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 20204 marks1-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 mol1M = 137.0\ \text{g mol}^{-1}) reacts completely; calculate the mass of organic product formed (M=74.0 g mol1M = 74.0\ \text{g mol}^{-1}).
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(a) CH3CH2CH2CH2Br+NaOHCH3CH2CH2CH2OH+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=5.00×103 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)=5.00×103 moln(\text{butan-1-ol}) = 5.00 \times 10^{-3}\ \text{mol}, so m=nM=5.00×103×74.0=0.370 gm = nM = 5.00 \times 10^{-3} \times 74.0 = 0.370\ \text{g}. (2 marks)

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