A and B , respectively from the following options are:

The correct answer is Option (3) → (1-iodoethyl)benzene and phenol

Ethers are cleaved by concentrated HI. The ether oxygen is first protonated, and then \( \ce{I^{-}} \) attacks a carbon attached to the oxygen. In an alkyl aryl ether the O–aryl bond is stronger (partial double bond character from resonance) and the sp² ring carbon is not attacked, so the bond broken is always O–alkyl. The products are a phenol and an alkyl iodide.

Here the alkyl group is the 1-phenylethyl group. Its carbocation is secondary and benzylic, so it is stabilised by the ring, and the cleavage takes place on the alkyl side:

    \( \ce{C6H5CH(CH3)-O-C6H5 + HI -> C6H5CH(I)CH3 + C6H5OH} \)

So A and B are (1-iodoethyl)benzene and phenol. Even with excess HI, phenol is not converted into iodobenzene, because its C–O bond has partial double bond character and the ring carbon resists substitution.

Why the other options are incorrect

• Option (1) 1-phenylethanol and iodobenzene — Incorrect. This would need the O–aryl bond to break and iodide to substitute on the benzene ring, which does not happen.

• Option (2) (1-iodoethyl)benzene and iodobenzene — Incorrect. The alkyl iodide is right, but phenol does not give iodobenzene with HI.

• Option (4) (1-iodoethyl)benzene and 4-iodophenol — Incorrect. HI does not iodinate the ring of phenol; the phenol is formed unchanged.

NCERT (2026-27 edition):
Cleavage of C–O Bond in Ethers