4.3.3Core Organic Chemistry
Addition Reactions of Alkenes
5 min read 5 parts Lesson 13 of 29 Virtual lab
Section 1 of 5Hydrogenation and Halogenation
In every addition reaction the π bond breaks and two new σ bonds form, so a single product results.
Hydrogenation
- An alkene reacts with hydrogen over a nickel catalyst at about 150 °C to give the alkane.
Equation
- C₂H₄ + H₂ → C₂H₆ turns a double bond into a single bond.
Margarine
- Hydrogenation of unsaturated vegetable oils makes them more saturated and so more solid at room temperature.
Bromination
- Alkenes react with bromine at room temperature, for example C₂H₄ + Br₂ → CH₂BrCH₂Br.
Test for unsaturation
- The orange bromine water is rapidly decolourised when shaken with an alkene to form a dibromoalkane.
Section 2 of 5Hydrogen Halides and Steam
Hydrogen Halides and Steam
Hydrogen bromide
- Alkenes react with HBr at room temperature to form bromoalkanes, for example C₂H₄ + HBr → CH₃CH₂Br.
Unsymmetrical alkenes
- Propene with HBr gives two possible products, 2-bromopropane as the major product and 1-bromopropane as the minor product.
Hydration
- Ethene with steam over a phosphoric acid catalyst at about 300 °C and 60–70 atm gives ethanol.
Equation
- C₂H₄(g) + H₂O(g) ⇌ CH₃CH₂OH(g) is reversible and exothermic.
Section 3 of 5Choosing the Product
Symmetrical reagents
- With H₂ and Br₂ the same atom adds to both carbons, so only one product is possible.
Unsymmetrical alkene
- Where both the alkene and the reagent are unsymmetrical, a major and a minor product are possible.
Industrial ethanol
- Hydration of ethene is a source of ethanol with high atom economy, as all reactant atoms end up in the product.
Yield
- A reversible reaction at equilibrium gives a limited conversion per pass, so unreacted ethene is recycled.
| Reagent | Conditions | Product from ethene |
|---|---|---|
| H₂ | Ni, 150 °C | Ethane |
| Br₂ | Room temperature | 1,2-dibromoethane |
| HBr | Room temperature | Bromoethane |
| H₂O (steam) | H₃PO₄, 300 °C, 60–70 atm | Ethanol |
Section 4 of 5Worked Example
Question
- 2.80 g of ethene reacts with excess hydrogen bromide to give 8.00 g of bromoethane.
- Calculate the percentage yield.
- (Ar: H = 1.0, C = 12.0, Br = 79.9)
- Moles of ethene = 2.80 / 28.0 = 0.100 mol.
- The equation C₂H₄ + HBr → C₂H₅Br shows a 1 : 1 ratio, so the theoretical amount of product is 0.100 mol.
- Mr of C₂H₅Br = 24.0 + 5.0 + 79.9 = 108.9, so the theoretical mass is 0.100 × 108.9 = 10.89 g.
- Percentage yield = (8.00 / 10.89) × 100 = 73.5%.
Answer Percentage yield = 73.5% (3 s.f.)
Section 5 of 5Exam Checklist & Tips
4.3.3 – Key Points for the Exam
- Addition reaction A reaction in which two molecules combine to form a single product.
- Hydrogenation The addition of hydrogen across a double bond.
- Hydration The addition of water across a double bond.
- Unsaturation test Decolourising orange bromine water shows a C=C bond is present.
Common Mistakes
- Writing the product of bromine addition as a monobromoalkane instead of a dibromoalkane.
- Giving the condition for hydration as acid and water at room temperature rather than steam with a catalyst.
- Saying bromine water turns colourless to orange; it goes from orange to colourless.
- Forgetting that hydration of ethene is reversible and uses steam, not liquid water.