Key Concept
At HL you must be able to draw the full mechanism (with curly arrows) for electrophilic addition reactions of alkenes. These are two-step mechanisms proceeding via a carbocation intermediate.
Mechanism: Ethene + HBr
Step-by-Step Explanation
Step 1 (slow):
- The electron-rich π bond of the alkene acts as a nucleophile
- A curly arrow goes from the π bond to the δ+ H in HBr (new C-H bond forms)
- Simultaneously, the H-Br bond breaks heterolytically (curly arrow from bond to Br)
- Products: a carbocation intermediate + Br⁻
Step 2 (fast):
- Br⁻ acts as a nucleophile and donates a lone pair to the positive carbon
- A new C-Br bond forms → final product
Mechanism: Ethene + Br₂
Induced Dipole Mechanism
Step 1: As Br₂ approaches the C=C, the π electrons repel the Br-Br bonding electrons. This creates an induced dipole (the nearer Br becomes δ+, the further Br becomes δ-).
Step 2: The π bond attacks the δ+ Br. The Br-Br bond breaks heterolytically → carbocation + Br⁻.
Step 3: Br⁻ attacks the carbocation → 1,2-dibromoethane.
Curly Arrow Checklist
- Arrow 1: from the C=C π bond to the δ+ atom of the electrophile
- Arrow 2: from the breaking bond in the electrophile to the more electronegative atom
- Arrow 3 (step 2): from the lone pair on the anion to the C⁺ of the carbocation
- All arrows are double-headed (showing electron pairs, not single electrons)
Think About It
Why is the π bond the part of the alkene that reacts, not the σ bond?
The π bond electrons are held in a region above and below the plane of the molecule, making them more exposed and accessible. The σ bond electrons are localised directly between the two carbon nuclei and are harder to reach. The π bond is also weaker than the σ bond, so it breaks more easily.
Common Exam Mistakes
- Forgetting to show the carbocation intermediate. This is a two-step mechanism, not concerted.
- Drawing the first curly arrow starting from HBr. It must start from the π bond (the alkene is the nucleophile in step 1).
- Not showing the induced dipole on Br₂. Without this, Br₂ has no electrophilic centre.
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