Halogenoalkanes

AS Level Chemistry · Halogen compounds. 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

In a halogenoalkane the C–X bond is polar, with a δ+ carbon. A nucleophile (a species that donates a lone pair, such as OH−, CN− or NH3) attacks this carbon and the halogen leaves as a halide ion. This is nucleophilic substitution.

There are two mechanisms. In SN2 the nucleophile attacks from the side opposite the halogen while the C–X bond breaks, all in one step. Primary halogenoalkanes react this way. In SN1 the C–X bond breaks first to give a carbocation, which the nucleophile then attacks. Tertiary halogenoalkanes react this way, because three alkyl groups stabilise the carbocation.

The weaker the C–X bond, the faster the reaction.

Rules to remember

Common mistake

Students say chloroalkanes react fastest because C–Cl is the most polar bond. The rate depends on bond strength, not polarity: C–I is the weakest bond, so iodoalkanes react fastest.

Worked example

A halogenoalkane C4H9X is warmed with aqueous silver nitrate in ethanol and gives a cream precipitate. It is known to react by the SN1 mechanism. Identify it.

  1. A cream precipitate is silver bromide, so X is bromine: the formula is C4H9Br.
  2. SN1 is the mechanism of tertiary halogenoalkanes, so the carbon bonded to Br carries three alkyl groups.
  3. With four carbon atoms, the only tertiary structure is (CH3)3CBr.

Answer: 2-bromo-2-methylpropane, (CH3)3CBr.

Practice questions

Try each one, then open the answer.

1. Which statement about the SN2 reaction of bromoethane with hydroxide ions is correct?

  1. A
    a carbocation intermediate forms first
  2. B
    the hydroxide ion attacks the carbon from the side opposite the bromine as the C–Br bond breaks, in a single step
  3. C
    the bromide ion leaves first, then the hydroxide ion attacks
  4. D
    the mechanism is fastest for tertiary halogenoalkanes
Show answer

Answer: B. SN2 is one concerted step through a five-membered transition state; no carbocation forms. Primary halogenoalkanes favour it because the carbon is unhindered and a primary carbocation would be unstable.

2. How is 2-bromo-2-methylpropane, (CH3)3CBr, classified?

  1. A
    primary
  2. B
    secondary
  3. C
    tertiary
  4. D
    it is not a halogenoalkane
Show answer

Answer: C. Classification depends on how many carbon atoms are attached to the carbon bearing the halogen. Here that carbon is bonded to three other carbons, so it is tertiary.

3. Which reaction gives 2-bromopropane as the major product?

  1. A
    propane with HBr(g)
  2. B
    propan-1-ol with HBr(g)
  3. C
    propene with HBr(g) at room temperature
  4. D
    propene with Br2 at room temperature
Show answer

Answer: C. Electrophilic addition of HBr to propene follows Markovnikov: Br goes to the middle carbon. Propane does not react with HBr, propan-1-ol gives 1-bromopropane, and Br2 gives 1,2-dibromopropane.

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