The major products for the following reaction will be

The correct answer is Option (1) → benzyl iodide and phenol

Ethers are cleaved by hydrogen iodide. The ether oxygen is first protonated, and then the iodide ion breaks one C–O bond. Which C–O bond breaks depends on the groups attached to oxygen:

• O–aryl bond: in an alkyl aryl ether, the bond between oxygen and the benzene ring has partial double bond character (resonance of the O lone pair with the ring) and the ring carbon is sp². This bond is not broken, so the aryl part always ends up as a phenol.

• O–alkyl bond: this bond breaks, and the alkyl group becomes an alkyl iodide. A benzyl group breaks off very easily, because the benzyl carbocation is stabilised by resonance.

The given ether is benzyl phenyl ether, \( \ce{C6H5CH2-O-C6H5} \):

    \( \displaystyle \ce{C6H5CH2-O-C6H5 + HI -> C6H5CH2I + C6H5OH} \)

So the major products are benzyl iodide (\( \ce{C6H5CH2I} \)) and phenol (\( \ce{C6H5OH} \)).

Why the other options are incorrect

• Option (2) benzyl alcohol and benzyl iodide — Incorrect. Both products contain a benzyl group, but the ether has only one benzyl group; its other group is a phenyl group, which gives phenol.

• Option (3) phenol and iodobenzene — Incorrect. Iodobenzene would need the strong O–aryl bond to break, and both products lack the \( \ce{CH2} \) group of the benzyl part.

• Option (4) benzyl alcohol and benzyl iodide — Incorrect. The same pair as Option (2), so it is wrong for the same reason.

Note: It seems that there is a mistake in the original NTA paper. Option (2) and Option (4) are printed as the same pair of structures, benzyl alcohol and benzyl iodide.

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