Addition reactions of alkenes: WACE Year 12 Chemistry
“Describe the addition reactions of alkenes including hydrogenation, halogenation, hydration and hydrogen halide addition”
The C=C double bond in an alkene is a reactive site. In an addition reaction, a small molecule adds across the double bond, breaking the pi bond and forming two new single bonds, with no atoms left over. Key examples are hydrogenation (adding H2), halogenation (adding Br2 or Cl2), hydration (adding water to form an alcohol) and hydrogen halide addition (adding HBr or HCl).
What this dot point is asking
Alkenes are unsaturated: the carbon-carbon double bond contains a reactive region of electron density. In an addition reaction, that double bond is converted to a single bond as a molecule adds across it.
A reaction in which a molecule adds across a carbon-carbon double (or triple) bond, breaking the multiple bond and forming new single bonds. The whole adding molecule is incorporated, so there is no by-product.
The four key additions
Hydrogenation (adding H2). With a nickel or platinum catalyst, hydrogen adds across the double bond to give the saturated alkane:
This is how unsaturated vegetable oils are hardened into margarine.
Halogenation (adding Br2 or Cl2). A halogen adds across the double bond, with no catalyst or light needed. The rapid decolourising of orange bromine water is the standard test for unsaturation:
Hydration (adding H2O). With steam and an acid catalyst (such as phosphoric acid), water adds across the double bond to form an alcohol. This is an industrial route to ethanol:
Hydrogen halide addition (adding HBr or HCl). A hydrogen halide adds to give a haloalkane:
In every addition reaction the entire adding molecule is incorporated into the product, so there is 100 percent atom economy in principle. This contrasts with substitution, which always releases a small by-product molecule.
Addition to unsymmetrical alkenes
When an unsymmetrical reagent (such as HBr or water) adds to an unsymmetrical alkene (such as propene), two products are possible depending on which carbon gains the hydrogen. The major product is usually the one where the hydrogen adds to the carbon that already has more hydrogens (Markovnikov's rule), giving the more substituted product. At WACE level you should recognise that two products are possible and identify the major one.
Question
Ethene is reacted with steam in the presence of a phosphoric acid catalyst. Name the product and the type of reaction, and explain its industrial importance.
Solution
Water adds across the C=C double bond of ethene. This is a hydration (addition) reaction:
The product is ethanol. The reaction is industrially important because it converts ethene, a cheap product of cracking crude oil, into ethanol used as a solvent, fuel and chemical feedstock. The acid catalyst speeds the reaction without being consumed.
Why the double bond reacts: the pi bond
The reactivity of alkenes comes from the structure of the double bond. A C=C double bond is made of one sigma bond (a strong, direct overlap along the bond axis) and one pi bond (a weaker, sideways overlap of p orbitals above and below the axis). The pi bond is a region of exposed, loosely held electron density, so it readily attacks electron-poor reagents. In an addition reaction it is the pi bond that breaks, leaving the sigma bond intact, while two new sigma bonds form to the adding atoms. This explains why addition is fast and requires milder conditions than the substitution of alkanes, which must break a strong, unreactive sigma bond by a high-energy radical mechanism.
A note on Markovnikov and stability
When an unsymmetrical reagent adds, the orientation is set by the stability of the intermediate. Adding first to the alkene creates a carbocation, and the more stable carbocation (tertiary more stable than secondary, secondary more stable than primary) forms preferentially. The remaining part of the reagent then bonds to that carbon, which is why the major product places the larger group on the more substituted carbon. At WACE level you are expected to identify the major product and link it to this idea of forming the more stable intermediate, rather than to give a full mechanism.
Why this matters
Addition reactions are the way unsaturated hydrocarbons are converted into the more useful functional families: alcohols, haloalkanes and saturated compounds. The same chemistry, repeated, gives addition polymers. These reactions are central to the synthesis pathways and atom economy themes of Unit 4.
The main error is treating an alkene reaction as a substitution; alkenes react by addition (the double bond opens and the whole reagent adds, no by-product), whereas alkanes react by substitution (releasing a small molecule). Another slip is forgetting the catalyst or conditions, for example writing hydration without the acid catalyst and steam. Always match the reaction type to whether the hydrocarbon is saturated or unsaturated.
Exam-style questions
Questions in the style of SCSA 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.
WACE 2022-style6 marksPropene reacts with hydrogen bromide. (a) Draw the structures of the two possible products and name them. (b) State which is the major product and justify your choice using Markovnikov's rule. (c) Write an equation for the reaction of propene with bromine and explain how this reaction is used as a chemical test.Show worked answer →
A 6 mark question rewards both products, the Markovnikov justification, and the bromine test.
(a) Adding across the double bond gives either 2-bromopropane, (Br on the middle carbon), or 1-bromopropane, (Br on the end carbon).
(b) The major product is 2-bromopropane. By Markovnikov's rule the hydrogen adds to the carbon already bearing more hydrogen atoms (the terminal ), so the bromine ends up on the more substituted central carbon, which forms via the more stable secondary carbocation intermediate.
(c) (1,2-dibromopropane). When orange bromine water is added to an alkene it is rapidly decolourised as the bromine adds across the double bond; an alkane (no double bond) does not decolourise it, so this distinguishes unsaturated from saturated hydrocarbons.
Markers reward both named products, 2-bromopropane as major with Markovnikov reasoning, and the decolourising bromine-water test.
WACE 2020-style5 marksCompare the reaction of ethene with the reaction of ethane towards bromine, in terms of the type of reaction, the conditions required, and any by-product. Explain why the alkene is more reactive.Show worked answer →
A 5 mark compare answer needs the reaction types, conditions, by-products, and the reactivity reason.
- Ethene (alkene)
- Reacts by addition: . The reaction is fast at room temperature with no catalyst or light, and there is no by-product because the whole bromine molecule adds across the double bond.
- Ethane (alkane)
- Reacts only by substitution: . It requires ultraviolet light to proceed and releases hydrogen bromide as a by-product.
- Reactivity
- Ethene is more reactive because its carbon-carbon double bond contains an exposed, electron-rich pi bond that is readily attacked, whereas ethane has only strong, non-polar single bonds that must be broken by a high-energy radical mechanism.
Markers reward addition versus substitution, the no-catalyst versus UV-light conditions, the absence versus presence of , and the pi-bond reactivity explanation.