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(A) The reaction of an ether with excess $HI$ involves the cleavage of the $C-O$ bonds.$1$. The ether is $3$-methoxyphenethyl-$1$-phenylethyl ether.$2

Vedclass · Accepted Answer

$3$-($2$-iodoethyl)phenol + $1$-phenylethyl iodide

Vedclass · Answer

(A) The reaction of an ether with excess $HI$ involves the cleavage of the $C-O$ bonds.$1$. The ether is $3$-methoxyphenethyl-$1$-phenylethyl ether.$2$. The $C-O$ bond attached to the $1$-phenylethyl group is cleaved via an $S_N1$ mechanism because the resulting carbocation is stabilized by the phenyl ring (benzylic carbocation).$3$. The $C-O$ bond of the methoxy group attached to the benzene ring is cleaved to form phenol and methyl iodide $(CH_3I)$.$4$. The $C-O$ bond of the primary alkyl group (phenethyl group) is cleaved to form the corresponding alkyl iodide.$5$. Given the excess $HI$ and heat,the final products are $3$-($2$-iodoethyl)phenol and $1$-phenylethyl iodide.

Predict the major products of the following reaction:
$m$-methoxyphenethyl$-1-$phenylethyl ether + $HI$ (excess) $\xrightarrow{\text{heat}}$ ?

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Predict the major products of the following reaction:$m$-methoxyphenethyl$-1-$phenylethyl ether + $HI$ (excess) $\xrightarrow{\text{heat}}$ ?