Alkenes react primarily via electrophilic addition. The pi bond represents a region of high electron density above and below the plane of the molecule, which is highly attractive to electron-deficient species known as electrophiles.
🔑 Key Principle
During an electrophilic addition reaction, the pi bond breaks and two new single sigma bonds are formed. The overall result is that the unsaturated reactant is converted into a saturated organic product.
An electron-pair acceptor. Electrophiles are electron-deficient species (either positively charged ions or polar molecules with a delta-positive atom) that can react by accepting a pair of electrons.
An organic intermediate ion containing a carbon atom with a positive charge and only three covalent bonds.
Mechanism 1: Reaction with Hydrogen Halides (HBr, HCl, HI)
At room temperature, alkenes undergo electrophilic addition with hydrogen halides to form halogenoalkanes. The reaction has two distinct steps:
- Electrophilic attack: The high electron density of the pi bond attacks the partially positive hydrogen atom (\( \text{H}^{\delta+} \)) in the polar \( \text{H}-\text{X} \) molecule. At the same time, the \( \text{H}-\text{X} \) bond breaks heterolytically, yielding a halide ion (\( \text{X}^- \)) and leaving a carbocation intermediate.
- Halide attack: The halide ion acts as a nucleophile, using a lone pair of electrons to form a coordinate bond with the positively charged carbon atom of the carbocation.
In all mechanisms, curly arrows must start exactly from a region of electron density: either the C=C double bond itself, a covalent bond, or a lone pair. Never draw a curly arrow starting from an atom (like H or C) unless it represents a lone pair on that atom. A common mistake is starting the first arrow from the hydrogen atom.
Asymmetric Alkenes and Markovnikov's Rule
When a hydrogen halide reacts with an unsymmetrical alkene (such as propene), two different products can theoretically form: a major product and a minor product. This is determined by Markovnikov's rule, which is explained by the stability of the intermediate carbocations.
When an unsymmetrical reagent adds to an unsymmetrical alkene, the hydrogen atom attaches to the double-bonded carbon atom that is already bonded to the greater number of hydrogen atoms.
Carbocations are classified based on the number of alkyl groups attached to the positively charged carbon atom:
- Primary (\( 1^\circ \)) carbocation: One alkyl group attached. Least stable.
- Secondary (\( 2^\circ \)) carbocation: Two alkyl groups attached. Moderately stable.
- Tertiary (\( 3^\circ \)) carbocation: Three alkyl groups attached. Most stable.
Stability: Tertiary > Secondary > Primary
This stability order is due to the inductive effect. Alkyl groups are electron-releasing, which means they push electron density towards the positively charged carbon atom. This helps to disperse the positive charge, lowering the overall potential energy and stabilising the carbocation intermediate.
When propene (\( \text{CH}_3-\text{CH}=\text{CH}_2 \)) reacts with \( \text{HBr} \), the first step involves the addition of \( \text{H}^+ \) to one of the double-bonded carbons. This can occur via two pathways:
Pathway A (Major): Hydrogen adds to Carbon 1. This forms a secondary carbocation intermediate (\( \text{CH}_3-\text{CH}^+-\text{CH}_3 \)). The positive charge is on the central carbon, which is stabilised by the inductive electron-donating effect of two methyl groups. Subsequent attack by the bromide ion yields 2-bromopropane.
Pathway B (Minor): Hydrogen adds to Carbon 2. This forms a primary carbocation intermediate (\( \text{CH}_3-\text{CH}_2-\text{CH}_2^+ \)). The positive charge is on the terminal carbon, which has only one alkyl group donating electron density. Subsequent attack by the bromide ion yields 1-bromopropane.
Conclusion: The secondary carbocation is more stable than the primary carbocation because it has more electron-releasing alkyl groups. Therefore, the secondary carbocation intermediate forms faster and is preferred, making 2-bromopropane the major product.
Mechanism 2: Reaction with Halogens (Br2, Cl2, I2)
Alkenes undergo electrophilic addition with halogens at room temperature. Although a halogen molecule like \( \text{Br}_2 \) is non-polar, it behaves as an electrophile in this reaction:
- As the \( \text{Br}_2 \) molecule approaches the high electron density of the alkene pi bond, the electrons in the \( \text{Br}-\text{Br} \) bond are repelled away from the nearer bromine atom.
- This induces a dipole in the bromine molecule, making it polar: \( \text{Br}^\delta+-\text{Br}^\delta- \).
- The pi bond then attacks the \( \text{Br}^\delta+ \) atom, heterolytically breaking the \( \text{Br}-\text{Br} \) bond to form a bromide ion (\( \text{Br}^- \)) and a carbocation intermediate (or a cyclic bromonium ion intermediate, though AQA accepts the acyclic carbocation pathway).
- The bromide ion then attacks the carbocation to form a 1,2-dihalogoalkane (e.g. 1,2-dibromoethane).
Mechanism 3: Reaction with Concentrated Sulfuric Acid (H2SO4)
Alkenes react with concentrated sulfuric acid at room temperature in a highly exothermic reaction. The sulfuric acid acts as the electrophile, reacting as \( \text{H}-\text{OSO}_2\text{OH} \):
- The C=C double bond attacks the partially positive hydrogen of the sulfuric acid molecule. The \( \text{H}-\text{O} \) bond breaks heterolytically, producing a hydrogensulfate ion (\( \text{HSO}_4^- \)) and a carbocation intermediate.
- The hydrogensulfate ion attacks the carbocation to yield an alkyl hydrogensulfate (e.g. ethyl hydrogensulfate).
Hydrolysis to form alcohols: If water is added to the alkyl hydrogensulfate and the mixture is warmed, hydrolysis occurs. This converts the alkyl hydrogensulfate into an alcohol and regenerates sulfuric acid:
\( \text{CH}_3\text{CH}_2\text{OSO}_2\text{OH} + \text{H}_2\text{O} \to \text{CH}_3\text{CH}_2\text{OH} + \text{H}_2\text{SO}_4 \)
In this two-step process, sulfuric acid acts as a catalyst for the overall hydration of ethene to ethanol.
Direct Hydration of Alkenes (Industrial Manufacture of Alcohols)
Industrially, ethanol is manufactured in a single step by the direct hydration of ethene with steam. The reaction is reversible and carried out under specific conditions:
- Reagents: Ethene and steam (water vapour).
- Catalyst: Phosphoric acid (\( \text{H}_3\text{PO}_4 \)) coated on a solid silica support.
- Conditions: Temperature of approximately 300°C and pressure of 60 atm.
\( \text{CH}_2=\text{CH}_2(\text{g}) + \text{H}_2\text{O}(\text{g}) \rightleftharpoons \text{CH}_3\text{CH}_2\text{OH}(\text{g}) \)
Introduction to Addition Polymerisation
Addition polymerisation is a special case of addition reactions. Many alkene molecules (monomers) join together by opening their double bonds to form long-chain saturated molecules (polymers). We will cover this in detail in the next lesson.
Make sure you do not confuse the laboratory method of hydration (reacting with concentrated \( \text{H}_2\text{SO}_4 \) and then adding water) with the industrial method (direct hydration using steam and a phosphoric acid catalyst). You must know the reagent, catalyst, and conditions for both methods.