Alkenes and Electrophilic Addition

Alkenes are unsaturated hydrocarbons containing a C=C double bond. The double bond consists of a sigma (σ) bond and a pi (π) bond. The pi bond creates a region of high electron density above and below the plane of the molecule, making alkenes reactive towards electrophiles. This lesson covers the structure, reactions, and mechanisms of alkenes in detail.

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Structure of Alkenes

• General formula: CₙH₂ₙ (for non-cyclic alkenes)
• The C=C consists of a σ bond (head-on overlap of sp² orbitals) and a π bond (sideways overlap of p-orbitals).
• Carbon atoms in the C=C are sp² hybridised, giving a trigonal planar arrangement with bond angles of approximately 120°.
• The π bond prevents free rotation around the C=C (which is why E/Z isomerism occurs).
• The π bond is weaker than the σ bond and is the site of reactivity.

Key Definition: An electrophile is an electron-pair acceptor — a species attracted to regions of high electron density. Examples: H⁺, Br⁺ (or δ+ end of HBr/Br₂), NO₂⁺, carbocations.

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Test for Unsaturation (C=C)

Bromine water test:

• Add bromine water (orange/brown) to the compound.
• If a C=C is present: bromine water is decolourised (orange → colourless).
• If no C=C (e.g. an alkane): bromine water remains orange.

This occurs because Br₂ undergoes electrophilic addition across the C=C.

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Electrophilic Addition Reactions

The general pattern: the π bond of the C=C acts as a nucleophile, attacking an electrophile. This breaks the π bond and forms two new σ bonds (one to each carbon of the former double bond).

Reaction 1: Addition of Hydrogen Halides (HBr)

Overall equation: CH₂=CH₂ + HBr → CH₃CH₂Br (bromoethane)

Mechanism — Electrophilic Addition:

1. The H–Br bond is polar (Br is more electronegative than H), so H is δ+ and Br is δ−.
2. The electron-rich π bond of the C=C is attracted to the δ+ hydrogen.
3. A curly arrow goes from the C=C π bond to the H of HBr (the π bond electrons form a new C–H bond).
4. Simultaneously, a curly arrow goes from the H–Br bond to the Br atom (heterolytic fission of H–Br, giving Br⁻).
5. This forms a carbocation intermediate (a carbon bearing a positive charge) and a bromide ion (Br⁻).
6. A curly arrow goes from a lone pair on Br⁻ to the positively charged carbon, forming the C–Br bond.

Curly arrows (detailed):

• Step 1: Arrow from C=C to H (of HBr); arrow from H–Br bond to Br. This gives a carbocation + Br⁻.
• Step 2: Arrow from lone pair on Br⁻ to C⁺. Product: bromoalkane.

Exam Tip: Always draw the curly arrows starting from the electron-rich species (the C=C or lone pair) pointing towards the electron-poor species. Show the carbocation intermediate clearly with a + charge on the correct carbon.

Reaction 2: Addition of Sulfuric Acid (H₂SO₄)

CH₂=CH₂ + H₂SO₄ → CH₃CH₂OSO₃H (ethyl hydrogensulfate)

Then: CH₃CH₂OSO₃H + H₂O → CH₃CH₂OH + H₂SO₄ (hydrolysis)

This is the industrial route for producing ethanol from ethene. The H₂SO₄ acts as a catalyst — it is regenerated in the second step.

The mechanism for step 1 is electrophilic addition: the C=C attacks the δ+ H of H₂SO₄.

Reaction 3: Addition of Bromine (Br₂)

CH₂=CH₂ + Br₂ → CH₂BrCH₂Br (1,2-dibromoethane)

Mechanism: