Electrophilic Addition Reactions (AQA A-Level Chemistry): Revision Notes

3.4.2 Electrophilic Addition Reactions

Alkenes are known for their carbon-carbon double bonds ($C=C$), which create regions of high electron density. This makes alkenes particularly reactive towards electrophiles, leading to electrophilic addition reactions. These reactions occur when an electrophile, a species that accepts electron pairs, reacts with the electron-rich $C=C$ bond of the alkene.

What is an Electrophile?

An electrophile is an atom or molecule that can accept a pair of electrons to form a new covalent bond.

General Mechanism of Electrophilic Addition

The typical electrophilic addition mechanism for alkenes involves two main steps:

Step 1: Polarisation of the Electrophile

• The electron-rich $C=C$ double bond repels electrons in the approaching electrophile, causing polarisation of the electrophile's bond (X-Y).
• For example, in the case of $HBr$, the bond is already polar, with hydrogen having a $δ+$ charge and bromine having a $δ-$ charge. The general reaction using ethene as the alkene and X-Y as the electrophile:

$$CH_2=CH_2 + X-Y \rightarrow CH_2X-CH_2Y$$

Step 2: Formation of a Carbocation Intermediate

• A pair of electrons from the $C=C$ double bond attacks the $δ+$ atom (X) of the electrophile, forming a new covalent bond between one of the carbons and the X atom.
• The X-Y bond breaks heterolytically, where Y takes both bonding electrons, forming Y⁻.
• A carbocation intermediate is formed, where the other carbon of the original double bond is left with a positive charge ($C⁺$).

Step 3: Attack by the Nucleophile

• The $Y⁻$ion (now a nucleophile) donates its lone pair of electrons to the carbocation, forming a new covalent bond.
• This results in the formation of a saturated molecule.

Electrophilic Addition Reactions in Alkenes

Addition of Hydrogen Bromide ($HBr$)

1. Polarisation: The $H-Br$ bond is already polar, with $H$ as $δ+$ and $Br$ as $δ-$.
2. Step 1: The π-electrons from the alkene attack the $δ+$ hydrogen, forming a bond between one carbon of the $C=C$ and $H$. The $H-Br$ bond breaks, releasing $Br⁻$
3. Step 2: A carbocation intermediate is formed at the second carbon of the double bond.
4. Step 3: The $Br⁻$ion acts as a nucleophile and bonds with the carbocation, forming the final addition product bromoalkane.

Addition of Sulfuric Acid ($H₂SO₄$)

1. Polarisation: The $O-H$ bonds in sulfuric acid are polar, with the hydrogen being δ+.
2. Step 1: The alkene attacks the proton ($H⁺$) from $H₂SO₄$, forming a carbocation on the other carbon of the original double bond.
3. Step 2: The negative sulphate ion ($HSO₄⁻$) acts as a nucleophile and bonds with the carbocation.

Addition of Bromine ($Br₂$)

1. Polarisation: As bromine approaches the $C=C$ bond, the π-electrons in the double bond cause induced polarisation of $Br₂$, with one bromine becoming $δ+$ and the other $δ-$.
2. Step 1: The $δ+$ bromine is attacked by the π-electrons from the double bond, and a cyclic bromonium ion is formed.
3. Step 2: The $Br⁻$ion attacks the more positively charged carbon, breaking the cyclic structure and forming a dibromoalkane.