Reactions of Alkenes: Electrophilic Addition and Mechanism
- Alkenes are hydrocarbons that contain at least one carbon-carbon double bond (C=C).
- This double bond consists of two parts: a strong sigma (σ) bond and a weaker pi (π) bond.
- The pi bond lies above and below the plane of the molecule, making it a region of high electron density.
- This electron-rich area is highly attractive to electrophiles—species that are electron-deficient and seek out electrons to form bonds.
- Electrophiles are molecules or ions that accept a pair of electrons to form a covalent bond.
- When alkenes encounter electrophiles, they undergo electrophilic addition reactions, where the double bond is broken, and new covalent bonds are formed.
Examples include halogen molecules (e.g., $Br_2$, hydrogen halides (e.g., HBr), and protons $H^+$ in acidic solutions.
The pi bond is weaker than the sigma bond, making it the reactive part of the double bond in alkenes.
Electrophilic Addition Reactions of Alkenes
Reaction with Halogens
When alkenes react with halogens like bromine $Br_2$ or chlorine $Cl_2$, the halogen atoms add across the double bond to form a dihalogenoalkane.
$$\text{C}_2\text{H}_4 + \text{Br}_2 \rightarrow \text{C}_2\text{H}_4\text{Br}_2$$
- This reaction is often used as a test for unsaturation because bromine water, which is orange, becomes colorless in the presence of an alkene.
- The disappearance of color indicates that the double bond has reacted with bromine.
Reaction with Hydrogen Halides
- Alkenes also react with hydrogen halides (HX, where X = Cl, Br, I) to form halogenoalkanes.
- If the alkene is unsymmetrical, two possible products can form.
- The major product is determined by Markovnikov's Rule, which states that the hydrogen atom from HX will add to the carbon with the greater number of hydrogen atoms already attached.
- This results in the formation of the more stable carbocation intermediate.
For propene $C_3H_6$, reacting with $HBr$ gives two products:
- 2-bromopropane (major product) via a secondary carbocation intermediate.
- 1-bromopropane (minor product) via a primary carbocation intermediate.
Reaction with Water (Hydration)
In the presence of an acid catalyst (e.g., $H_2SO_4$, alkenes react with water to form alcohols.
$$\text{C}_3\text{H}_6 + \text{H}_2\text{O} \xrightarrow{\text{H}_2\text{SO}_4} \text{C}_3\text{H}_8\text{O}$$
This reaction involves two steps:
- Protonation of the double bond to form a carbocation intermediate.
- Addition of water, followed by deprotonation to produce the alcohol.
The mechanism of the reaction will be covered by HL students in the later sections.
