What is the product formed when benzyl ethyl ether reacts with HI?

The correct answer is Option (4) → Benzyl iodide and ethyl alcohol.

Ethers are cleaved by concentrated HI at high temperature. The acid first protonates the ether oxygen; then the C–O bond on one side breaks, and the iodide ion ends up on that carbon while the other group leaves as an alcohol. Which side breaks depends on the groups:

• two primary or secondary alkyl groups: \( \ce{I^{-}} \) attacks the less hindered carbon (S_N2);

• a tertiary or benzylic group: that group leaves as a stable carbocation (S_N1) and becomes the iodide.

Benzyl ethyl ether is \( \ce{C6H5CH2-O-C2H5} \). After protonation, the bond to the benzyl group breaks, because the benzyl carbocation is stabilised by resonance with the ring; iodide then combines with it:

    \( \displaystyle \ce{C6H5CH2-O-C2H5 + HI -> [C6H5CH2-OH-C2H5]+ + I-} \)

    \( \displaystyle \ce{[C6H5CH2-OH-C2H5]+ -> C6H5CH2+ + C2H5OH} \)

    \( \displaystyle \ce{C6H5CH2+ + I- -> C6H5CH2I} \)

The net reaction:

    \( \displaystyle \ce{C6H5CH2-O-C2H5 + HI -> C6H5CH2I + C2H5OH} \)

So the products are benzyl iodide and ethyl alcohol.

Why the other options are incorrect

• Option (1) Benzyl iodide and methyl alcohol. — Incorrect. The ether has an ethyl group, not a methyl group, on the other side of the oxygen; cleavage does not change the carbon count of a group.

• Option (2) Benzyl iodide and propyl alcohol. — Incorrect. A propyl alcohol would need a three-carbon group on the oxygen.

• Option (3) Benzyl iodide and butyl alcohol. — Incorrect. A butyl alcohol would need a four-carbon group on the oxygen.