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Hydrocarbons

Electrophilic addition, Markovnikov's rule and testing for alkenes: testing for a C=C bond

  • 3 things to know
  • AS Unit 2

Overview

Eleven recall questions on electrophilic addition, Markovnikov's rule and testing for alkenes: testing for a C=C bond.

What you need to know

Answer each one before you reveal it.

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Key term

Why is 2-bromopropane called the major rather than the only product of the reaction between propene and HBr?

Key term

How would you show experimentally that a colourless liquid is an alkene rather than an alkane?

Key term

Chlorine also adds to ethene. Write the equation and name the product.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    A sample of 0.560 g of an alkene containing one double bond reacts exactly with 8.00 cm³ of a 1.25 mol dm⁻³ solution of bromine. Determine the molecular formula of the alkene.

    Model answer

    The amount of bromine used is 0.00800 dm³ × 1.25 mol dm⁻³ = 0.0100 mol, and since one C=C bond adds one Br₂ molecule the alkene is also 0.0100 mol. Its molar mass is therefore 0.560 ÷ 0.0100 = 56.0 g mol⁻¹, and as an alkene has the general formula CnH2n with 14.0 g mol⁻¹ per unit, n = 56.0 ÷ 14.0 = 4, so the alkene is C₄H₈.

  2. 2

    Burning 0.100 g of an alkene produces 0.314 g of carbon dioxide and 0.129 g of water. Determine its empirical formula, and its molecular formula given that its relative molecular mass is 56.0.

    Model answer

    The amount of carbon dioxide is 0.314 ÷ 44.0 = 7.14 × 10⁻³ mol, giving 7.14 × 10⁻³ mol of carbon atoms. The amount of water is 0.129 ÷ 18.0 = 7.17 × 10⁻³ mol, giving twice as many hydrogen atoms, 1.43 × 10⁻² mol. The ratio of carbon to hydrogen is 7.14 × 10⁻³ : 1.43 × 10⁻², which is 1 : 2, so the empirical formula is CH₂ with a mass of 14.0. Since 56.0 ÷ 14.0 = 4, the molecular formula is C₄H₈.

  3. 3

    An unknown hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass and has a relative molecular mass of 56.0. Determine its molecular formula and identify the class of compound.

    Model answer

    In 100 g there are 85.7 ÷ 12.0 = 7.14 mol of carbon and 14.3 ÷ 1.0 = 14.3 mol of hydrogen, a ratio of 1 : 2, so the empirical formula is CH₂. As the empirical formula mass is 14.0 and 56.0 ÷ 14.0 = 4, the molecular formula is C₄H₈, which fits the general formula CnH2n and so is an alkene.

  4. 4

    Why is the first step of an electrophilic addition the slow, rate-determining step?

    Model answer

    Forming the carbocation requires the π-bond and the bond in the attacking molecule both to break, so it has a high activation energy, whereas the second step simply brings together an anion and a positively charged carbon, which is fast.

  5. 5

    Chlorine also adds to ethene. Write the equation and name the product.

    Model answer

    The equation is C₂H₄ + Cl₂ → CH₂ClCH₂Cl, and the product is 1,2-dichloroethane.

  6. 6

    Explain why the bromide ion attacks the carbocation rather than the carbocation attacking the bromide ion.

    Model answer

    The bromide ion is the electron-rich species, with lone pairs and a full negative charge, so it acts as the nucleophile and donates a pair of electrons to the electron-deficient positive carbon, which is the electrophile in that step.

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