Organic reaction mechanisms

A reaction mechanism shows, step by step, how bonds break and form. Curly arrows show pairs of electrons moving: from a lone pair or a bond, to where a new bond forms or to the atom that takes them. Haloalkanes are attacked by nucleophiles, alkenes by electrophiles, and alkanes by radicals.

Key facts

  • A curly arrow shows a pair of electrons moving. It starts at a lone pair or a bond and points to where the new bond forms, or to the atom that keeps the electrons.
  • A nucleophile is an electron-pair donor (OH⁻, CN⁻, NH₃). An electrophile is an electron-pair acceptor (H⁺, Br₂ once polarised, NO₂⁺). A radical has an unpaired electron.
  • Heterolytic fission: one atom takes both bonding electrons, giving ions. Homolytic fission: each atom takes one electron, giving radicals.
  • Nucleophilic substitution: the C–X bond in a haloalkane is polar (Cδ+–Xδ−). A nucleophile attacks the δ+ carbon and the halide ion leaves. Warm aqueous NaOH gives an alcohol; KCN in ethanol gives a nitrile; excess ammonia gives an amine.
  • Elimination: with hot KOH in ethanol, OH⁻ acts as a base and removes H⁺ from the carbon next to the C–X carbon, forming an alkene.
  • Electrophilic addition: the electron-rich C=C attacks an electrophile such as H–Br. A carbocation forms, then Br⁻ adds. With an unsymmetrical alkene the major product comes from the more stable carbocation: tertiary > secondary > primary.
  • Free-radical substitution: UV light splits Cl₂ into radicals (initiation); chain reactions make the product (propagation); two radicals join (termination).

Worked examples

Example 1

Describe the mechanism of the reaction between ⟦bromoethane⟧ and hydroxide ions.

  1. The C–Br bond is polar: carbon is δ+ because bromine is more electronegative.
  2. Arrow A: a lone pair on the hydroxide ion attacks the δ+ carbon, forming a C–O bond.
  3. Arrow B: the C–Br bond breaks heterolytically; both electrons go to bromine. CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻

Answer: Nucleophilic substitution: OH⁻ attacks the δ+ carbon and Br⁻ leaves; the product is ⟦ethanol⟧.

Example 2

⟦Propene⟧ reacts with hydrogen bromide. Which product is the major one, and why?

  1. The C=C bond attacks the δ+ hydrogen of H–Br (arrow A), and the H–Br bond breaks (arrow B).
  2. If H adds to the end carbon, the positive charge is on the middle carbon: a secondary carbocation, CH₃C⁺HCH₃.
  3. That is more stable than the primary carbocation, because two alkyl groups push electrons towards the positive carbon.
  4. Br⁻ adds to that carbon. CH₃CH=CH₂ + HBr → CH₃CHBrCH₃

Answer: ⟦2-bromopropane⟧, formed from the more stable secondary carbocation

Example 3

Write the steps for the reaction of methane with chlorine in UV light.

  1. Initiation: UV light splits chlorine homolytically. Cl₂ → 2Cl•
  2. Propagation: each step uses one radical and makes another. CH₄ + Cl• → CH₃• + HCl; CH₃• + Cl₂ → CH₃Cl + Cl•
  3. Termination: two radicals join, ending the chain. CH₃• + Cl• → CH₃Cl

Answer: Initiation, two propagation steps and termination