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In the previous lesson, you learned that alkenes undergo addition reactions. Now we explore how these reactions happen at the molecular level — the mechanism of electrophilic addition. Understanding this mechanism is one of the most important skills for Edexcel A-Level organic chemistry.
An electrophile is an electron-pair acceptor — a species that is attracted to regions of high electron density. The word comes from the Greek for "electron-loving".
Examples of electrophiles include:
The C=C double bond in an alkene is a region of high electron density due to the pi bond. This makes alkenes susceptible to attack by electrophiles.
Before examining specific mechanisms, it is essential to understand what curly arrows mean:
In electrophilic addition, every curly arrow is double-headed (not fish-hook). Fish-hook arrows are only for radical mechanisms.
This is the simplest example and the one you must be able to draw.
The H-Br bond is polar — hydrogen carries a partial positive charge (δ+) and bromine a partial negative charge (δ−). The δ+ hydrogen is the electrophile.
A curly arrow is drawn from the C=C pi bond to the H of HBr. This shows a pair of electrons from the pi bond forming a new C-H sigma bond. Think of it as: the electron-rich pi bond donates its electrons to the electron-poor hydrogen.
Simultaneously, a second curly arrow is drawn from the H-Br bond to the Br atom. This shows the bonding pair moving to bromine as the H-Br bond breaks heterolytically. Bromine takes both electrons and departs as Br⁻.
The result is:
The Br⁻ ion (a nucleophile — an electron-pair donor) attacks the carbocation. A curly arrow is drawn from a lone pair on Br⁻ to the positive carbon.
A new C-Br bond forms, giving the product: bromoethane (CH₃CH₂Br).
Common mistake in mechanism drawing: Students draw the curly arrow from H to the C=C (wrong direction). Remember: electrons flow FROM the electron-rich site (the pi bond) TO the electron-poor site (the δ+ hydrogen). The arrow shows electron movement, not atom movement.
When HBr adds to propene (CH₃CH=CH₂), the electrophile (H⁺) can bond to either carbon of the C=C:
Route A: H bonds to C1 (the terminal CH₂) → carbocation on C2 → 2-bromopropane (major product)
Route B: H bonds to C2 → carbocation on C1 → 1-bromopropane (minor product)
The carbocation formed in Route A is a secondary carbocation (the positive carbon is bonded to two other carbon groups). In Route B, the carbocation is primary (bonded to only one carbon group).
Secondary carbocations are more stable than primary carbocations due to the inductive effect: alkyl groups are electron-releasing (they push electron density towards the positive carbon through sigma bonds), partially stabilising the positive charge.
The stability order is: tertiary > secondary > primary > methyl (CH₃⁺)
Since the more stable carbocation forms preferentially, Route A dominates, and 2-bromopropane is the major product. This is the underlying explanation for Markovnikov's rule: the hydrogen adds to the carbon with more hydrogens because this creates the more stable carbocation on the other carbon.
2-methylpropene is (CH₃)₂C=CH₂.
Route A: H bonds to CH₂ (terminal carbon) → carbocation on the central carbon. The central carbon is bonded to three alkyl groups → tertiary carbocation. This is very stable.
Route B: H bonds to the central carbon → carbocation on CH₂ → primary carbocation. This is very unstable.
Route A strongly dominates. Product: 2-bromo-2-methylpropane (CH₃)₃CBr.
The tertiary carbocation is so much more stable that essentially 100% of the product comes from Route A.
Sulfuric acid (H₂SO₄) can also act as an electrophile. The mechanism is similar:
A curly arrow from the C=C to the H of H₂SO₄. The O-H bond in H₂SO₄ breaks heterolytically, releasing HSO₄⁻.
One carbon bonds to H, and the other becomes a carbocation.
The HSO₄⁻ ion attacks the carbocation. A curly arrow from a lone pair on the oxygen of HSO₄⁻ to the positive carbon.
Product: ethyl hydrogen sulfate (CH₃CH₂OSO₃H)
This intermediate can be hydrolysed with water to produce ethanol.
The addition of bromine to ethene is slightly different because Br₂ is a non-polar molecule. How can a non-polar molecule be an electrophile?
As the Br₂ molecule approaches the electron-rich C=C, the pi electrons repel the electrons in the Br-Br bond. The nearer bromine becomes δ+ and the further one becomes δ−. This induced dipole makes the nearer Br an electrophile.
A curly arrow from the C=C to the nearer Br (δ+). Simultaneously, a curly arrow from the Br-Br bond to the further Br. The Br-Br bond breaks heterolytically.
One carbon bonds to Br. The other carbon becomes a carbocation. Br⁻ is released.
Br⁻ attacks the carbocation. A curly arrow from a lone pair on Br⁻ to the positive carbon.
Product: 1,2-dibromoethane (CH₂BrCH₂Br)
This is the reaction responsible for the decolourisation of bromine water, which serves as the test for alkenes.
When drawing electrophilic addition mechanisms in an exam:
| Reaction | Electrophile | Intermediate | Product |
|---|---|---|---|
| Ethene + HBr | Hδ+ of HBr | Carbocation + Br⁻ | Bromoethane |
| Propene + HBr | Hδ+ of HBr | Secondary carbocation (major) | 2-Bromopropane (major) |
| Ethene + H₂SO₄ | Hδ+ of H₂SO₄ | Carbocation + HSO₄⁻ | Ethyl hydrogen sulfate |
| Ethene + Br₂ | Brδ+ (induced) | Carbocation + Br⁻ | 1,2-Dibromoethane |
| 2-Methylpropene + HBr | Hδ+ of HBr | Tertiary carbocation (very stable) | 2-Bromo-2-methylpropane |
Edexcel 9CH0 specification Topic 6 — Organic Chemistry I, sub-topic 6.4 covers the mechanism of electrophilic addition in detail: drawing curly arrows from the C=C π bond to the electrophile and from the H-X bond to X (heterolytic fission), the carbocation intermediate, attack by the nucleophile (X⁻ or H2O), and the application of carbocation stability to predict regiochemistry (Markovnikov's rule); the special case of bromine and the cyclic bromonium ion intermediate; and the radical addition of HBr in the presence of peroxides (anti-Markovnikov product) — refer to the official specification document for exact wording. Sub-topic 6.4 is examined directly on Paper 2 (mechanism with curly arrows) and synoptically on Paper 3 (kinetics, where the rate-determining step in electrophilic addition is the formation of the carbocation, and the rate is first-order in alkene and first-order in HX). The Edexcel specification expects mechanism diagrams with proper curly-arrow notation; mark schemes routinely deduct for incorrect arrow direction or missing carbocation intermediate.
Question (8 marks):
(a) Draw the full mechanism for the electrophilic addition of HBr to 2-methylpropene ((CH3)2C=CH2). Use curly arrows and identify the carbocation intermediate. (5)
(b) Predict the major product, name it, and explain the regiochemistry by reference to carbocation stability. (3)
Solution with mark scheme:
(a) Step 1 — initial protonation. The C=C π bond attacks the partially-positive H of H-Br; H-Br undergoes heterolytic fission to give H⁺ and Br⁻.
M1 — curly arrow from C=C to H of HBr.
M1 — curly arrow from H-Br σ bond to Br (heterolytic fission).
M1 — tertiary carbocation (CH3)3C+ correctly drawn (the H adds to the less substituted C, placing the + on the more substituted C).
Step 2 — bromide attack. Br⁻ attacks the carbocation:
:Br−→(CH3)3C+→(CH3)3C−Br
M1 — curly arrow from Br⁻ lone pair to C+.
A1 — major product is 2-bromo-2-methylpropane ((CH3)3C−Br).
(b) Step 1 — minor product analysis. If H+ added to the other end of the C=C (to the C with one methyl substituent rather than two), the resulting carbocation would be primary: (CH3)2CH−CH2+, leading to 1-bromo-2-methylpropane.
M1 — minor product identified: 1-bromo-2-methylpropane (anti-Markovnikov).
Step 2 — carbocation stability ranking.
M1 — tertiary > secondary > primary. The tertiary cation is stabilised by the inductive effect of three alkyl groups donating electron density into the empty p-orbital on C+. The primary cation has only one alkyl group and is therefore less stable by ~80 kJ mol−1.
A1 — therefore the major product (Markovnikov) is 2-bromo-2-methylpropane, formed via the more stable tertiary cation; the minor product (~1%) is 1-bromo-2-methylpropane.
Total: 8 marks (M5 A3).
Question (6 marks): Bromine adds to cyclohexene to give trans-1,2-dibromocyclohexane exclusively; no cis product is formed.
(a) Draw the mechanism showing the cyclic bromonium ion intermediate. (3)
(b) Explain the stereochemical outcome (trans-only) by reference to the geometry of the bromonium ion. (3)
Mark scheme decomposition by AO:
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