Question 34 of 39intermediate🔧 ApplyShort Answer3 marks

Write the equation of the reaction of hydrogen iodide with: (i) 1-propoxypropane (ii) methoxybenzene and (iii) benzyl ethyl ether.

(i)
Answer

CH₃CH₂CH₂-O-CH₂CH₂CH₃ + HI → CH₃CH₂CH₂I + CH₃CH₂CH₂OH (1-propoxypropane reacts with HI to form propyl iodide and propan-1-ol)

(ii)
Answer

C₆H₅-O-CH₃ + HI → C₆H₅OH + CH₃I (Methoxybenzene reacts with HI to form phenol and methyl iodide)

(iii)
Answer

C₆H₅CH₂-O-CH₂CH₃ + HI → C₆H₅CH₂I + CH₃CH₂OH (Benzyl ethyl ether reacts with HI to form benzyl iodide and ethanol)

Explanation

The reaction of ethers with concentrated HI involves protonation of ether followed by nucleophilic attack by iodide. For mixed ethers with two different alkyl groups, when primary or secondary alkyl groups are present, the lower alkyl group forms alkyl iodide via SN2 mechanism. In methoxybenzene (aryl alkyl ether), the aryl-oxygen bond is stronger due to resonance, so cleavage occurs at the methyl side producing phenol and methyl iodide. For benzyl ethyl ether, the benzyl group can form a stable benzylic carbocation, so benzyl iodide is formed preferentially.

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