Alkenes — Electrophilic Addition and Markovnikov's Rule

The most important reactions of alkenes are electrophilic additions: the C=C π bond acts as a nucleophile, attacking an electrophile, and the two fragments of the electrophile end up on either side of the former double bond. In this lesson we cover the mechanism in detail and introduce Markovnikov's rule — a key A-Level topic that OCR explicitly names in the specification. This lesson covers OCR A-Level Chemistry A (H432) specification 4.1.3 (d)–(f).

1. The General Pattern of Electrophilic Addition

Key Definition — Electrophilic addition: A reaction in which an electrophile is added across a C=C double bond, forming a saturated product. One σ bond of the original C=C is retained and a new σ bond is formed to each of the two carbons.

Overall:

C=C + X–Y → X–C–C–Y

Key features:

The π bond is broken.
A new C–X and a new C–Y σ bond are formed.
The carbons change from sp² (planar) to sp³ (tetrahedral).
The reaction is an addition (not substitution) — every atom of the reagent ends up in the product.

2. Four Key Addition Reactions You Must Know

Reagent	Product (from ethene)	Conditions	Use
H₂	Ethane, C₂H₆	Nickel catalyst, 150 °C, 5 atm	Hydrogenation (hardening of vegetable oils)
Br₂	1,2-dibromoethane, CH₂BrCH₂Br	Room temp, in CCl₄ or H₂O	Test for C=C (bromine water decolourises)
HBr (HCl, HI)	Bromoethane, CH₃CH₂Br	Room temp, gas or conc. solution	Making haloalkanes
H₂O (steam)	Ethanol, CH₃CH₂OH	Conc. H₃PO₄ catalyst, 300 °C, 60 atm	Industrial production of ethanol

These are all examples of electrophilic addition. The C=C attacks the electrophile.

3. Mechanism — Electrophilic Addition of HBr to Ethene

Let us walk through the mechanism carefully, 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 (H is 2.20, Br is 2.96 on the Pauling scale):

  δ+    δ−
   H — Br

The hydrogen is δ+ and is therefore the electrophilic end.

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:

From the C=C π bond to the H of HBr (forming a new C–H σ bond).
From the H–Br bond to the Br (forming Br⁻).

After Step 1:

One of the two carbons (say, the left one) has gained an H and become sp³ tetrahedral.
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.
Br⁻ has been released into solution.

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

  H3C — CH2⁺     +    :Br⁻
   (sp³)  (sp²⁺, carbocation)

3.3 Step 2 — Br⁻ Attacks the Carbocation

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

Curly arrow: 3. From a lone pair on Br⁻ to the positively charged carbon.

Final product: CH₃CH₂Br (bromoethane).

3.4 Summary with All Curly Arrows

Three arrows total:

C=C π → H of HBr.
H–Br bond → Br.
Br⁻ lone pair → C⁺.

graph LR
  A[Ethene + HBr] --> B[Step 1: π attacks H+<br/>HBr breaks heterolytically<br/>forms C+ and Br-]
  B --> C[Step 2: Br- attacks C+<br/>forms new C-Br 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 possible products can form:

CH₃–CH=CH₂ + HBr → CH₃–CHBr–CH₃ (2-bromopropane) or → CH₃–CH₂–CH₂Br (1-bromopropane)

Experimentally, the 2-bromopropane is the major product. This observation is summarised by Markovnikov's rule:

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.

Or, more memorably: "The rich get richer" — the carbon with more H gets one more H.

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

4.1 Why? Carbocation Stability

The real reason is that one of the two possible carbocation intermediates is more stable than the other. The reaction proceeds via whichever carbocation is more stable (lower in energy).

Carbocation stability order (A-Level):

tertiary (3°) > secondary (2°) > primary (1°) > methyl

The more alkyl groups attached to the positively charged carbon, the more stable the carbocation. This is because:

Alkyl groups are slightly electron-donating (positive inductive effect, +I): they push electron density towards the positive charge, partially stabilising it.
More alkyl groups → more electron donation → more stable cation.

4.2 Applying Carbocation Stability to Propene + HBr

Start from CH₃–CH=CH₂ and consider protonating either carbon:

Option A: H adds to the terminal CH₂ carbon. The positive charge ends up on the central carbon, which is attached to two alkyl groups (the –CH₃ and the –CH₂H).

Result: CH₃–CH⁺–CH₃ — a secondary carbocation.

Option B: H adds to the central CH carbon. The positive charge ends up on the terminal CH₂ carbon, which has only one alkyl neighbour.

Result: CH₃–CH₂–CH₂⁺ — a primary carbocation.

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