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What is the mechanism for the electrophilic addition reaction between hydrogen bromide and propene? Provide a stepwise explanation of each intermediate and product formed during the reaction.

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The electrophilic addition reaction between hydrogen bromide  HBr  and propene  C3H6  involves a two-step mechanism. Here's a stepwise explanation of the reaction:Step 1: Protonation of the alkeneIn the first step, the hydrogen bromide molecule acts as an electrophile, with the hydrogen atom being partially positive  +  and the bromine atom being partially negative  - . The double bond in propene is a region of high electron density, which attracts the partially positive hydrogen atom of HBr. The electrons in the double bond move towards the hydrogen atom, forming a bond between one of the carbon atoms of the double bond and the hydrogen atom. This results in the formation of a carbocation intermediate and a bromide ion  Br- .There are two possible carbocations that can be formed in this reaction: a primary carbocation  when the hydrogen atom adds to the terminal carbon atom  and a secondary carbocation  when the hydrogen atom adds to the central carbon atom . However, the secondary carbocation is more stable due to hyperconjugation, so it is the major intermediate formed in this reaction.Step 2: Nucleophilic attack by the bromide ionIn the second step, the bromide ion  Br-  acts as a nucleophile and attacks the positively charged carbon atom in the carbocation intermediate. The electrons from the bromide ion form a bond with the carbocation, resulting in the formation of the final product, 2-bromopropane  C3H7Br .In summary, the electrophilic addition reaction between hydrogen bromide and propene proceeds through a two-step mechanism involving the formation of a carbocation intermediate and a nucleophilic attack by the bromide ion. The final product of the reaction is 2-bromopropane.
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