Anisole xrightarrow[text{reflux}]{text{excess | vedclass

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Question

(C) When anisole $(C_6H_5OCH_3)$ reacts with excess concentrated $HI$ under reflux conditions,the ether linkage is cleaved. The reaction proceeds via

Vedclass · Accepted Answer

$C_6H_5OH + CH_3I$

Vedclass · Answer

(C) When anisole $(C_6H_5OCH_3)$ reacts with excess concentrated $HI$ under reflux conditions,the ether linkage is cleaved. The reaction proceeds via the protonation of the oxygen atom of the ether,followed by a nucleophilic attack by the iodide ion $(I^-)$ on the less sterically hindered carbon atom (the methyl group) via an $S_N2$ mechanism. This results in the formation of phenol $(C_6H_5OH)$ and methyl iodide $(CH_3I)$. Since $HI$ is in excess,the reaction stops at phenol and methyl iodide,as the $C-O$ bond between the phenyl ring and oxygen cannot be cleaved by $S_N2$ or $S_N1$ mechanisms due to the partial double bond character of the $C-O$ bond in the aromatic ring.

Anisole $\xrightarrow[\text{reflux}]{\text{excess } HI \text{ (conc.)}} \text{ Product}$

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