Ph-O-CH_2-Ph xrightarrow{HI} A + B will be | vedclass

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(A) In the reaction of benzyl phenyl ether $(Ph-O-CH_2-Ph)$ with $HI$,the cleavage occurs at the $O-CH_2$ bond rather than the $O-Ph$ bond.This is bec

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$Ph-OH + Ph-CH_2-I$

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(A) In the reaction of benzyl phenyl ether $(Ph-O-CH_2-Ph)$ with $HI$,the cleavage occurs at the $O-CH_2$ bond rather than the $O-Ph$ bond.This is because the $C-O$ bond in $Ph-O$ has partial double bond character due to resonance with the benzene ring,making it stronger.The $I^-$ nucleophile attacks the benzyl carbon to form benzyl iodide $(Ph-CH_2-I)$,and the oxygen remains with the phenyl ring to form phenol $(Ph-OH)$.

$Ph-O-CH_2-Ph \xrightarrow{HI} A + B$ will be

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