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Question:

What happens when bromine attacks CH₂=CH-CH₂-C≡CH?

Solution:

Bromine (Br₂) is an electrophile and will react with the multiple bonds in the molecule CH₂=CH-CH₂-C≡CH. The reaction will occur in two stages:

Stage 1: Reaction with the alkene

The alkene (C=C) is more reactive than the alkyne (C≡C) towards electrophilic addition. Bromine will add across the double bond via electrophilic addition. This results in the formation of a vicinal dibromide.

CH₂=CH-CH₂-C≡CH + Br₂ → CH₂Br-CHBr-CH₂-C≡CH

Stage 2: Reaction with the alkyne (if sufficient bromine is present)

If enough bromine is present, it will also react with the alkyne. However, the reaction with the alkyne is more complex. It involves the formation of a vinyl bromide intermediate, followed by the addition of another molecule of bromine. The final product will be a tetrabromide.

CH₂Br-CHBr-CH₂-C≡CH + Br₂ → CH₂Br-CHBr-CH₂-CBr=CHBr

CH₂Br-CHBr-CH₂-CBr=CHBr + Br₂ → CH₂Br-CHBr-CH₂-CBr₂-CHBr₂

In summary: The reaction of bromine with CH₂=CH-CH₂-C≡CH will primarily involve the addition of bromine to the alkene. If excess bromine is present, it will further react with the alkyne to give a tetrabromide. The exact product distribution will depend on the stoichiometry of the reactants and reaction conditions.