Alkenes

AS Level Chemistry · Hydrocarbons. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.

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The idea

An alkene has a C=C double bond made of one σ bond and one π bond. The π bond is a region of high electron density, so it attracts electrophiles (electron-pair acceptors). The π bond breaks and one new atom or group joins each carbon. This is electrophilic addition.

When HX adds to an unsymmetrical alkene, two products are possible. The major product forms through the more stable carbocation. Alkyl groups push electron density towards the positive carbon (the inductive effect), so a tertiary carbocation is more stable than a secondary, and a secondary more than a primary. This explains Markovnikov addition.

Alkenes are also oxidised by acidified KMnO4, and the product depends on the conditions.

Rules to remember

Common mistake

Students explain the major product by saying that hydrogen goes to the carbon with more hydrogen atoms. That is only a way to remember it: the reason the examiner wants is the greater stability of the carbocation intermediate.

Worked example

Hydrogen bromide gas reacts with but-1-ene, CH3CH2CH=CH2. Name the major product.

  1. If H adds to carbon 1, the carbocation is CH3CH2C+HCH3, which is secondary (two alkyl groups on C+).
  2. If H adds to carbon 2, the carbocation is CH3CH2CH2C+H2, which is primary (one alkyl group).
  3. The secondary carbocation is more stable, so most molecules react through it.
  4. Br− then bonds to carbon 2.

Answer: 2-bromobutane, CH3CH2CHBrCH3 (1-bromobutane is the minor product).

Practice questions

Try each one, then open the answer.

1. In the reaction of bromine with ethene, why does the Br–Br bond become polarised as the molecules approach?

  1. A
    the electron-rich π bond of ethene repels the electrons in the Br–Br bond, inducing a dipole
  2. B
    bromine is more electronegative than carbon
  3. C
    the C=C bond is permanently polar
  4. D
    bromine forms free radicals in the presence of ethene
Show answer

Answer: A. Bromine is non-polar until the high electron density of the π bond pushes its bonding electrons away, making the nearer bromine δ+. This δ+ bromine is the electrophile that attacks the double bond, forming a carbocation and Br−.

2. Ethene is shaken with cold, dilute, acidified potassium manganate(VII). What is the organic product?

  1. A
    ethanol
  2. B
    ethanal
  3. C
    ethane-1,2-diol
  4. D
    ethanoic acid
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Answer: C. Cold dilute KMnO4 adds an OH group to each carbon of the double bond, giving the diol, and the purple colour is decolourised. Hot concentrated KMnO4 would break the C=C bond completely.

3. Hydrogen bromide adds to propene. What is the major product, and why?

  1. A
    1-bromopropane, because the primary carbocation is more stable
  2. B
    2-bromopropane, because bromine is attracted to the carbon with more hydrogen atoms
  3. C
    1-bromopropane, because bromine is a large atom
  4. D
    2-bromopropane, because the secondary carbocation intermediate is stabilised by the electron-donating (inductive) effect of two alkyl groups
Show answer

Answer: D. H+ adds first, and it adds to the CH2 end so that the positive charge sits on the middle carbon, where two alkyl groups push electron density towards it. Br− then attacks that more stable secondary carbocation.

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