Q.Diethyl ether is heated with excess concentrated HI. Give the final organic product(s) and explain why the reaction does not stop at the first cleavage step.

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

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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.

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