H3C – H2C – CH2 – O – CH3 Products are

Engineering

Chemistry

Nucleophilic Substitution

Question

H₃C – H₂C – CH₂ – O – CH₃ $\overset{conc . HBr (1 eq .)}{\to}$
Products are

H₃C – CH₂ – CH₂ – OH + CH₃ – Br

H₃C – CH₂ – CH₂ – CH₂ – Br + CH₃–OH

H₃C – CH₂ – CH₂ – Br + CH₃ – Br

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This is an ether cleavage reaction with concentrated HBr. Ethers undergo S_N2 cleavage at the less substituted alkyl group. The given ether is CH₃CH₂CH₂-O-CH₃. The methyl group (CH₃-) is less substituted than the propyl group (CH₃CH₂CH₂-). Therefore, the nucleophilic bromide ion (Br^-) attacks the methyl carbon, leading to the formation of methyl bromide and the corresponding alcohol.

Final Answer: H₃C – CH₂ – CH₂ – OH + CH₃ – Br