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(B) The reaction of an alkyl aryl ether with $HI$ involves the protonation of the ether oxygen followed by nucleophilic attack by the iodide ion $(I^-

Vedclass · Accepted Answer

$II$

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(B) The reaction of an alkyl aryl ether with $HI$ involves the protonation of the ether oxygen followed by nucleophilic attack by the iodide ion $(I^-)$.In the case of benzyl phenyl ether $(C_6H_5-O-CH_2-C_6H_5)$,the protonated ether undergoes cleavage.The $C-O$ bond between the oxygen and the benzyl group $(C_6H_5CH_2-)$ is cleaved because the benzyl carbocation is resonance-stabilized,making the benzyl carbon more susceptible to $S_N1$ or $S_N2$ attack by the iodide ion.Thus,the products formed are phenol $(C_6H_5OH)$ and benzyl iodide $(C_6H_5CH_2I)$.This corresponds to option $II$.

What are the major products from the following reaction?
$C_6H_5-O-CH_2-C_6H_5 + HI \xrightarrow{\text{heat}} ?$
$(I) C_6H_5I + C_6H_5CH_2OH$
$(II) C_6H_5OH + C_6H_5CH_2I$
$(III) C_6H_5OCH_2OH + C_6H_5I$
$(IV) C_6H_5OCH_2I + C_6H_5OH$

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What are the major products from the following reaction?$C_6H_5-O-CH_2-C_6H_5 + HI \xrightarrow{\text{heat}} ?$$(I) C_6H_5I + C_6H_5CH_2OH$$(II) C_6H_5OH + C_6H_5CH_2I$$(III) C_6H_5OCH_2OH + C_6H_5I$$(IV) C_6H_5OCH_2I + C_6H_5OH$