Final answer:
In the electrophilic addition reaction of ethene with HBr, a proton from HBr attacks the double bond of ethene, forming a carbocation intermediate, which is then attacked by a bromide ion to form bromoethane.
Step-by-step explanation:
The electrophilic addition reaction of ethene (CH₂=CH₂) with hydrogen bromide (HBr) involves the formation of a carbocation intermediate and follows these steps:
- Electrophilic attack: The proton (H⁺) from HBr is attracted to the electron-rich π-bond of ethene. This forms a carbocation (positively charged carbon) and a bromide ion (Br⁻).
- Nucleophilic attack: The bromide ion (Br⁻) acts as a nucleophile and attacks the carbocation, resulting in the formation of bromoethane (CH₃-CH₂Br).
Therefore, the overall equation for this reaction is:
CH₂=CH₂ + HBr → CH₃-CH₂Br
The equation for the electrophilic addition reaction of ethene with HBr is:
C2H4 + HBr → C2H5Br
During this reaction, the double bond in ethene (C2H4) is broken and a hydrogen (H) from HBr is added to one of the carbon atoms, resulting in the formation of ethyl bromide (C2H5Br).