Alkenes — Electrophilic Addition and Markovnikov's Rule

Spec Mapping — OCR H432 Module 4.1.3 — Alkenes, covering the electrophilic-addition reactions of alkenes with hydrogen (catalytic hydrogenation), with halogens (Br₂, Cl₂), with hydrogen halides (HBr, HCl, HI), and with steam (acid-catalysed hydration), the curly-arrow mechanism for electrophilic addition including the role of the carbocation intermediate, the carbocation stability order (3° > 2° > 1° > methyl), Markovnikov's rule for the regiochemistry of HX addition to unsymmetrical alkenes and its rationalisation via carbocation stability, and the bromine-water test as the qualitative test for the C=C π bond (refer to the official OCR H432 specification document for exact wording).

Electrophilic addition is the central reaction class of alkenes. The C=C π bond — sitting above and below the molecular plane, electron-rich, loosely held — acts as a nucleophile towards approaching electrophiles. The two fragments of the electrophile end up on the two former-alkene carbons; the π bond is sacrificed; the σ bond is preserved; two new σ bonds form. This pattern, repeated millions of times per second in industrial reactors, is the basis of essentially every commodity organic chemical: ethanol (from ethene + steam), polythene (from ethene + initiator), bromoethane and other haloalkanes (from ethene + HBr), the propan-2-ol that becomes the acetone solvent (from propene + steam), and the ethylene oxide that underlies all surfactants. This lesson develops the mechanism step-by-step, introduces Markovnikov's rule — which OCR explicitly names — and grounds the rule in carbocation stability so you can predict products with confidence. The arrow-pushing here is also the mechanistic template for every other electrophilic-addition reaction at A-Level (aldehydes, ketones, carboxylic acids in Year 13).

Key Mechanism — Electrophilic addition of HX to an alkene: (1) Step 1: the C=C π bond attacks the H atom of H–X (which is δ+ because X is more electronegative than H), forming a new C–H σ bond. Simultaneously the H–X bond breaks heterolytically — both electrons go to X, which leaves as X⁻. The result is a carbocation intermediate (the other carbon of the former C=C, now bearing a +1 formal charge) and an X⁻ counterion. (2) Step 2: X⁻ attacks the carbocation using one of its lone pairs, forming a new C–X σ bond. The product is a saturated alkyl halide. Three curly arrows total: π → H ; H–X → X ; X⁻ → C⁺.

1. The general pattern of electrophilic addition

Key Definition — Electrophilic addition: a reaction in which an electrophile adds across a C=C double bond, forming a saturated product. The π bond is broken; one new σ bond is made to each of the two former-alkene carbons; the carbons re-hybridise from sp² (planar) to sp³ (tetrahedral). Net atom economy is 100 % — every atom of the reagent ends up in the product. Compare with substitution (Lesson 6), in which one atom of the substrate leaves as a separate by-product.

Overall pattern:

$\text{C=C} + \text{X-Y} \rightarrow \text{X-C-C-Y}$

The reaction is conceptually simple but has rich consequences. Different X–Y reagents give different functional-group products (haloalkanes from HX or X₂, alkanes from H₂, alcohols from H₂O), and the regiochemistry of unsymmetrical addition (which carbon gets X, which gets Y) is governed by carbocation stability — the heart of Markovnikov's rule.

2. The four key addition reactions you must know

Reagent (X–Y)	Product (from ethene)	Conditions	Industrial use
H₂	Ethane, CH₃CH₃	Nickel catalyst, 150 °C, 5 atm	Hydrogenation of vegetable oils → margarine
Br₂	1,2-dibromoethane, CH₂BrCH₂Br	Room temp, in CCl₄ or H₂O	Bromine-water test for C=C
HBr (HCl, HI)	Bromoethane, CH₃CH₂Br	Room temp, dry gas or conc. solution	Manufacturing haloalkanes
H₂O (steam)	Ethanol, CH₃CH₂OH	Conc. H₃PO₄ catalyst, 300 °C, 60 atm	Industrial production of ethanol

All four are electrophilic-addition reactions. The C=C π bond is the nucleophile; the reagent (or an induced dipole on the reagent) is the electrophile. The mechanism is essentially the same three-arrow template in each case, with the carbocation intermediate playing the central role.

3. Mechanism — Electrophilic addition of HBr to ethene

Let us walk through the mechanism step by step using curly arrows.

Overall reaction: CH₂=CH₂ + HBr → CH₃CH₂Br

3.1 Why HBr is an electrophile

The H–Br bond is polar because Br (Pauling χ = 2.96) is significantly more electronegative than H (χ = 2.20):

$\stackrel{\delta^+}{\text{H}} \textemdash \stackrel{\delta^-}{\text{Br}}$

The hydrogen is the δ+ end — the electrophilic end. (For HCl and HI the picture is similar; for HF the bond is so strong it does not react with alkenes under normal conditions.)

3.2 Step 1 — π bond attacks H

The electron-rich π bond of ethene attacks the δ+ hydrogen of HBr. Simultaneously, the H–Br bond breaks heterolytically — both electrons go to Br, which leaves as Br⁻.

Curly arrows:

Arrow 1: from the C=C π bond (drawn from the middle of the double bond) to the H of HBr (forming a new C–H σ bond).
Arrow 2: from the H–Br bond to the Br (the two bonding electrons end up as a lone pair on Br⁻).

After Step 1:

One of the two carbons (say, the left one) has gained the H and become sp³ tetrahedral (CH₃ centre).
The other carbon (right) has lost its share of the π electrons and is now positively charged, with only six valence electrons — it is a carbocation (C⁺).
Br⁻ has been released into solution as a separate species.

   CH2 = CH2        +        H — Br
       ↓↓                     ↓
   [π to H]                [H-Br bond to Br]

  CH3 — CH2⁺     +    :Br⁻
   (sp³)  (sp², carbocation)

3.3 Step 2 — Br⁻ attacks the carbocation

The Br⁻ anion (a nucleophile with three lone pairs and a full negative charge) attacks the electrophilic carbocation. A new C–Br σ bond forms, giving the final product.

Curly arrow:

Arrow 3: from a lone pair on Br⁻ to the positively charged carbon.

Final product: CH₃CH₂Br (bromoethane).

3.4 Summary

Three curly arrows total:

C=C π → H of HBr (forming new C–H σ).
H–Br bond → Br (heterolytic cleavage; Br⁻ generated).
Br⁻ lone pair → C⁺ (forming new C–Br σ).

graph LR
  A[Ethene + HBr] --> B["Step 1: π attacks δ+ H<br/>HBr breaks heterolytically<br/>gives C⁺ + Br⁻"]
  B --> C["Step 2: Br⁻ attacks C⁺<br/>forms new C-Br σ bond"]
  C --> D[Product: bromoethane]

4. Markovnikov's Rule

When an unsymmetrical alkene (e.g., propene) reacts with an unsymmetrical electrophile (e.g., HBr), two products are in principle possible:

$\text{CH}_3\text{-CH=CH}_2 + \text{HBr} \rightarrow \textbf{CH}_3\text{-CHBr-CH}_3 \text{ (2-bromopropane)} \text{ or } \text{CH}_3\text{-CH}_2\text{-CH}_2\text{Br} \text{ (1-bromopropane)}$

Experimentally, 2-bromopropane is the major product. This regiochemistry is summarised by Markovnikov's rule (Russian chemist Vladimir Markovnikov, 1869):

Markovnikov's Rule: when HX adds to an unsymmetrical alkene, the H attaches to the carbon of the C=C that already bears more hydrogens, and the X attaches to the carbon that bears fewer hydrogens.

Or, the witty memory aid: "the rich get richer" — the carbon already H-rich gains one more.

OCR explicitly names Markovnikov's rule in the specification — you must be able to quote it and explain it.

4.1 Why — carbocation stability

The mechanistic reason is that one of the two possible carbocation intermediates is more stable than the other. The reaction goes through whichever cation is more stable (lower in energy), so that pathway dominates.

Carbocation stability order (A-Level):

$\text{tertiary (3°)} > \text{secondary (2°)} > \text{primary (1°)} > \text{methyl}$

More alkyl groups on the positive carbon → more stable carbocation. The reason:

Alkyl groups are slightly electron-donating via the positive inductive effect (+I) — they push electron density along the σ framework towards the positive carbon, partially neutralising the charge.
Hyperconjugation: each β-C–H σ bond can donate electron density into the empty p orbital of the adjacent carbocation, lowering its energy. A tertiary carbocation has 9 β-C–H bonds (three CH₃ groups); a primary carbocation has only 2 or 3.

4.2 Applying carbocation stability to propene + HBr

Start from CH₃–CH=CH₂ and consider the two possible protonations:

Option A: H attaches to the terminal CH₂ carbon. The positive charge appears on the central carbon, which is attached to two alkyl groups (the –CH₃ and the freshly-formed –CH₂H). Result: CH₃–CH⁺–CH₃ — a secondary carbocation.
Option B: H attaches to the central CH carbon. The positive charge appears on the terminal CH₂ carbon, which has only one alkyl neighbour. Result: CH₃–CH₂–CH₂⁺ — a primary carbocation.

The 2° cation is much more stable than the 1° cation. Option A dominates. Br⁻ attacks the 2° carbocation at the central carbon to give CH₃–CHBr–CH₃ (2-bromopropane) — the Markovnikov product.

4.3 Worked Example — But-1-ene + HBr

But-1-ene: CH₃CH₂CH=CH₂.

Protonating the terminal CH₂ gives a 2° cation on C2: CH₃CH₂CH⁺CH₃.
Protonating C2 gives a 1° cation on C1: CH₃CH₂CH₂CH₂⁺.

The 2° cation is more stable → major product is 2-bromobutane, CH₃CH₂CHBrCH₃ (Markovnikov).

4.4 Worked Example — 2-methylpropene + HBr

2-methylpropene: (CH₃)₂C=CH₂.

Protonating the terminal CH₂ gives a 3° cation on the central C, attached to three methyl groups: (CH₃)₃C⁺.
Protonating the central C gives a 1° cation on the CH₂ carbon: (CH₃)₂CHCH₂⁺.

The 3° cation is dramatically more stable than the 1° (difference ~150 kJ mol⁻¹). The major product (essentially the only product) is 2-bromo-2-methylpropane, (CH₃)₃CBr — pure Markovnikov.

graph TD
  A[Unsymmetrical alkene + HX] --> B[Two possible protonations]
  B --> C[Protonate terminal CH₂:<br/>form more substituted carbocation<br/>= Markovnikov pathway]
  B --> D[Protonate interior C:<br/>form less substituted carbocation<br/>= anti-Markovnikov pathway]
  C --> E[Major product:<br/>X on the more substituted C]
  D --> F[Minor product:<br/>X on the less substituted C]

5. Addition of Bromine (Br₂)

Even though Br₂ is a non-polar molecule (no permanent dipole), it still reacts rapidly with alkenes. How can a non-polar reagent be an electrophile?

5.1 Induced dipole

As a Br₂ molecule approaches the electron-rich π bond of an alkene, the π electrons repel the nearer Br atom's electron cloud, polarising the Br–Br bond into an induced (temporary) dipole:

$\stackrel{\delta^+}{\text{Br}} \textemdash \stackrel{\delta^-}{\text{Br}} \quad \text{(polarised by the approaching C=C π cloud)}$

The δ+ end of Br₂ is now electrophilic and can be attacked by the π bond — exactly as for HBr.

5.2 Mechanism

The π bond attacks the δ+ Br. The Br–Br bond breaks heterolytically — both electrons go to the far Br, which leaves as Br⁻.
The result is a carbocation (formally; at undergraduate level you will learn this is more accurately a 3-membered cyclic bromonium ion, but for OCR A-Level the simpler C⁺ picture is acceptable and is what mark schemes accept).
Br⁻ then attacks the carbocation from the opposite face, giving a 1,2-dibromoalkane with both Br atoms on adjacent carbons.

Product: a 1,2-dibromoalkane. Three curly arrows: π → Br ; Br–Br → Br ; Br⁻ → C⁺.

5.3 The bromine-water test

Orange-brown bromine water + alkene → colourless solution (1,2-dibromoethane is colourless). Alkanes do not decolourise bromine water under these conditions. This is the definitive qualitative test for a C=C π bond. The colour change is fast (a few seconds for ethene) and unambiguous — the most reliable A-Level functional-group test.

6. Hydration — making ethanol industrially

The addition of water (as steam) across C=C gives an alcohol:

$\text{CH}_2{=}\text{CH}_2 + \text{H}_2\text{O} \xrightarrow{\text{conc. H}_3\text{PO}_4, 300\,°\text{C}, 60\,\text{atm}} \text{CH}_3\text{CH}_2\text{OH}$

The H of H₂O attaches to one alkene carbon, OH attaches to the other. The phosphoric-acid catalyst (or sulfuric acid in older plants) protonates the alkene to give a carbocation, which is then attacked by water; loss of H⁺ regenerates the catalyst.

This direct hydration is the industrial route to ethanol from oil-derived ethene. The alternative — fermentation of glucose by yeast — is slower (24-48 h vs minutes), gives lower concentration (max ~14 % before yeast dies), but uses renewable feedstock. The economic balance has shifted over time; the UK currently uses ~60 % direct hydration, ~40 % fermentation. For unsymmetrical alkenes, the product follows Markovnikov's rule — the OH ends up on the more substituted carbon. Propene + H₂O → propan-2-ol (CH₃CHOHCH₃, a 2° alcohol) as the major product.

7. Hydrogenation

Alkenes react with H₂ over a nickel catalyst (150 °C, 5 atm) to give the saturated alkane:

$\text{CH}_2{=}\text{CH}_2 + \text{H}_2 \xrightarrow{\text{Ni, 150\,°C}} \text{CH}_3\text{CH}_3$

Lesson 9: Alkenes — Electrophilic Addition and Markovnikov's Rule