Q.In the reaction C₆H₅OCH₃ + HI ––(373 K)––> A + B, A and B are:

Cleavage of an ether by HI occurs at the bond that gives the more stable carbocation; for a benzylic ether that is the benzylic C-O bond, giving $C_6H_5CH_2I$.

When an ether such as benzyl methyl ether, $C_6H_5CH_2-O-CH_3$, reacts with excess concentrated HI, the mechanism proceeds by:

1. Protonation of the ether oxygen by $H^+$ to give the oxonium ion $C_6H_5CH_2-\overset{+}{O}H-CH_3$.
2. Cleavage of the C–O bond: since the benzylic carbon can form a resonance-stabilised (benzylic) carbocation $C_6H_5CH_2^+$, whereas the methyl carbon cannot form a stable carbocation, bond breaking occurs preferentially at the benzylic C–O bond (an SN1-like pathway). …

Anisole (methoxybenzene) is cleaved by HI at the alkyl-oxygen bond, not the aryl-oxygen bond, giving phenol + methyl iodide. Correct option (a).

Aryl alkyl ethers such as anisole (C6H5-O-CH3) react with HI by cleaving the C(sp3)-O (O-CH3) bond, because the aryl C-O bond is strong (partial double-bond character from resonance of the lone pair into the ring) and a phenyl cation is very unstable. So iodide attacks the methyl carbon (SN2), taking CH3 as CH3I and leaving the oxygen on the ring as phenol.

Reaction: C6H5-O-CH3 + HI -> C6H5-OH + CH3I

…

Cleavage of the mixed ether anisole ($\text{C}_6\text{H}_5\text{OCH}_3$) with excess HI at 373 K breaks the alkyl–oxygen bond, not the aryl–oxygen bond, giving phenol and methyl iodide.

In unsymmetrical aryl-alkyl ethers, the aryl–oxygen bond has partial double-bond character (due to resonance/conjugation of the O lone pair with the ring) and the aryl carbon is $sp^2$-hybridised, both of which strongly resist nucleophilic ($S_N2$) attack by $\text{I}^-$. The methyl carbon, being $sp^3$ and unhindered, is readily attacked by iodide in an $S_N2$ f …

Cleavage of benzyl phenyl ether with HI breaks the alkyl (benzylic) C–O bond, not the aryl C–O bond, giving benzyl iodide and phenol.

Benzyl phenyl ether, C₆H₅–CH₂–O–C₆H₅, has two different C–O bonds: an aryl–O bond (phenyl ring to oxygen) and a benzylic (alkyl-type) C–O bond (CH₂ to oxygen). When ethers are cleaved by excess HX (here HI), the halide ion attacks the carbon that can best support the mechanism:

• The aryl–O bond has partial double-bond character from resonance delocalisation of the oxygen lone pair into the ring, making it very strong and resistant to nucleophilic attack — this bond essentially never breaks under these conditions. …

Cleavage of anisole (an aryl alkyl ether) by HI attacks the methyl carbon (SN2), giving phenol + methyl iodide, because the aryl–oxygen bond has partial double-bond character and resists cleavage.

Anisole, $C_6H_5-O-CH_3$, on heating with excess HI at 373 K undergoes ether cleavage. In an aryl alkyl ether, the C(aryl)–O bond has partial double-bond character (due to conjugation of the oxygen lone pair with the benzene ring) and is much stronger/harder to break than the C(alkyl)–O bond. Also, an $S_N2$ attack by $I^-$ cannot occur at the $sp^2$ aromatic carbon.

…