Alkenes (A-Level) — A-Level Chemistry Revision
Revise Alkenes (A-Level) for A-Level Chemistry. Step-by-step explanation, worked examples, common mistakes and exam-style practice aligned to AQA, Edexcel and OCR.
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Go to AlcoholsWhat is Alkenes (A-Level)?
Alkenes are unsaturated hydrocarbons containing at least one carbon-carbon double bond (C=C), which consists of a sigma (σ) bond and a pi (π) bond. The π-bond is a region of high electron density, making alkenes much more reactive than alkanes. Their key reaction is electrophilic addition, where the double bond breaks and the electrophile adds across the two carbon atoms. This mechanism is used in reactions with hydrogen halides, halogens, and sulfuric acid.
Board notes: All boards require a detailed understanding of the electrophilic addition mechanism. AQA often includes questions on E/Z isomerism and the Cahn-Ingold-Prelog priority rules. Edexcel may focus on the addition polymerisation of alkenes and the properties and uses of the resulting polymers. OCR frequently tests the mechanism including the formation of major and minor products according to Markownikoff's rule.
Step-by-step explanationWorked example
Show the mechanism for the reaction of propene with hydrogen bromide. Step 1: The H-Br bond is polar (Hδ+-Brδ-). The electron-rich C=C double bond attacks the Hδ+, and the H-Br bond breaks. This forms a bromide ion and a carbocation. Step 2: Two possible carbocations can be formed: a primary carbocation or a more stable secondary carbocation. The secondary carbocation (with the positive charge on the central carbon) is the major intermediate. Step 3: The bromide ion (Br-) then attacks the positive carbocation, forming the major product, 2-bromopropane.
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Common mistakes
- 1Confusing the mechanism of electrophilic addition with free-radical substitution. Alkenes undergo addition reactions, while alkanes undergo substitution.
- 2Incorrectly predicting the major product in the addition of an unsymmetrical reagent (like HBr) to an unsymmetrical alkene. According to Markownikoff's rule, the hydrogen atom adds to the carbon atom that is already bonded to the most hydrogen atoms, because this forms the more stable carbocation intermediate.
- 3Forgetting the test for unsaturation. Shaking an alkene with bromine water (orange/brown) will decolourise it as the bromine adds across the double bond.
Alkenes (A-Level) exam questions
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Step-by-step method
Step-by-step explanation
4 steps · Worked method for Alkenes (A-Level)
Core concept
Alkenes are unsaturated hydrocarbons containing at least one carbon-carbon double bond (C=C), which consists of a sigma (σ) bond and a pi (π) bond. The π-bond is a region of high electron density, mak…
Frequently asked questions
What is a pi bond?
A pi (π) bond is formed by the sideways overlap of two p-orbitals, one from each carbon atom in the double bond. The electron density is concentrated above and below the plane of the sigma bond, and this exposed electron density is what makes alkenes susceptible to attack by electrophiles.
What is E/Z isomerism?
E/Z isomerism is a type of stereoisomerism that occurs in alkenes when there are two different groups attached to each carbon atom of the C=C double bond. The 'Z' isomer has the highest priority groups on the same side of the double bond, while the 'E' isomer has them on opposite sides. Priority is determined by atomic number using the Cahn-Ingold-Prelog rules.