IB Chemistry R3.4 R3.4.3
R3.4.3 HL

Electrophilic Addition

Why Are Alkenes Susceptible to Electrophilic Attack?

Alkenes contain a C=C double bond with a pi (π) bond. A region of high electron density above and below the carbon skeleton. This electron-rich region attracts electrophiles (electron-pair acceptors).

Unlike alkanes, alkenes are unsaturated and highly reactive. The π bond is weaker than the σ bond and readily breaks to form new bonds.

The Electrophilic Addition Mechanism

General mechanism (e.g. HBr + alkene)

Step 1. Electrophilic attack:

  1. The electrophile (e.g. HBr) is polarised: Hδ+−Brδ−
  2. The π electrons of C=C attack the Hδ+ end (curly arrow from C=C to Hδ+)
  3. The H−Br bond undergoes heterolytic fissioncarbocation intermediate formed

Step 2. Nucleophilic attack:

  1. The Br⁻ ion (nucleophile) attacks the positively charged carbon (curly arrow from Br⁻ to C⁺)
  2. Final addition product formed. Both atoms added across the double bond

Markovnikov's Rule

The Rule

When HX adds to an unsymmetrical alkene, the H adds to the carbon with more H atoms already attached. This forms the more stable (more substituted) carbocation.

Why? Alkyl groups are electron-donating (inductive effect). They stabilise the positive charge on the carbocation. Stability order: 3° > 2° > 1°.

Example: Propene + HBr

Major Product (Markovnikov)

CH₃CHBrCH₃

2-bromopropane

Via secondary carbocation (more stable)

Minor Product (Anti-Markovnikov)

CH₃CH₂CH₂Br

1-bromopropane

Via primary carbocation (less stable)

Halogenation of Alkenes (Br₂ Test)

Bromine Water Test for Unsaturation

Observation: Bromine water changes from orange → colourless

Mechanism: The π electrons of C=C polarise the approaching Br₂ molecule (induced dipole). The Brδ+ is attacked by the π electrons.

For the IB syllabus, students should be able to show the electrophilic addition mechanism for Br₂ addition to an alkene producing a dihalogenoalkane.

Symmetrical vs Asymmetrical Electrophiles

Symmetrical (e.g. Br₂, H₂)

Only one product possible. Markovnikov's rule does not apply

Br₂ + ethene → 1,2-dibromoethane

Asymmetrical (e.g. HBr, HCl, H₂O)

Markovnikov's rule applies. two possible products (major and minor)

HBr + propene → 2-bromopropane (major)

Summary of Electrophilic Addition Reactions

Reagent Product Type
H₂ (+ Ni, 150 °C)AlkaneHydrogenation
Br₂DihalogenoalkaneHalogenation
HBrHalogenoalkaneHydrohalogenation
H₂O (+ H₃PO₄, 300 °C)AlcoholHydration

Think About It

Why is bromine water used to test for unsaturation, and what is the mechanism?

Br₂ is polarised by the electron-rich C=C bond. The electrophilic Brδ+ attacks the π bond, forming a carbocation intermediate. Br⁻ then attacks → 1,2-dibromoalkane (colourless). The colour change from orange to colourless = positive test for alkenes.

⚠️ Common Exam Mistakes

  • Not drawing the curly arrow starting from the C=C π bond. The electrons come from the double bond, not from the electrophile
  • Applying Markovnikov's rule to symmetrical electrophiles (e.g. Br₂). It only applies to asymmetrical electrophiles like HBr
  • Forgetting that the mechanism has two steps (not one) with a carbocation intermediate
  • Not explaining why the major product forms. Link to carbocation stability (inductive effect of alkyl groups)
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