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Question 53

The major product in the following conversion is:

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  1. Ether Cleavage ($$\text{S}_{\text{N}}2$$):

    The oxygen atom of the aryl-alkyl ether ($$-\text{O--CH}_3$$) is protonated by $$\text{HBr}$$. The bromide ion ($$\text{Br}^\ominus$$) attacks the smaller alkyl group ($$-\text{CH}_3$$) via an $$\text{S}_{\text{N}}2$$ pathway, breaking the $$\text{O--CH}_3$$ bond to yield a phenol ($$-\text{OH}$$) and bromomethane ($$\text{CH}_3\text{Br}$$). The sturdier sp2 aryl $$\text{C--O}$$ bond remains completely intact.


  2. Electrophilic Addition (Markovnikov's Rule):

    The alkene double bond ($$-\text{CH}=\text{CH}--\text{CH}_3$$) undergoes protonation to generate the more stable secondary carbocation adjacent to the benzene ring, which is heavily stabilized by benzylic resonance. Subsequent attack by the bromide ion ($$\text{Br}^\ominus$$) completes the Markovnikov addition.


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