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Using electrophilic addition reactions, explain the mechanism of the addition of HBr to propene and predict the major product formed.

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In the electrophilic addition reaction of HBr to propene, the alkene  propene  acts as a nucleophile, while HBr acts as an electrophile. The mechanism involves a two-step process:Step 1: Protonation of the alkeneIn the first step, the alkene  propene  donates a pair of electrons from its double bond to the hydrogen atom of HBr, forming a bond between the carbon atom and the hydrogen atom. This results in the formation of a carbocation intermediate and a bromide ion  Br- . Propene is an unsymmetrical alkene, so the proton  H+  can add to either of the two carbon atoms involved in the double bond. According to Markovnikov's rule, the proton will preferentially add to the carbon atom that already has more hydrogen atoms, resulting in the more stable carbocation. In the case of propene, the proton will add to the terminal carbon  CH2  with two hydrogens, forming a secondary carbocation:CH3-CH=CH2 + HBr  CH3-CH + -CH3 + Br-Step 2: Nucleophilic attack by the bromide ionIn the second step, the bromide ion  Br-  acts as a nucleophile and attacks the carbocation, donating a pair of electrons to form a bond with the positively charged carbon atom. This results in the formation of the major product, 2-bromopropane:CH3-CH + -CH3 + Br-  CH3-CH Br -CH3So, the major product formed in the electrophilic addition of HBr to propene is 2-bromopropane.
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