Ph - mathop Olimits^{18} - CH_3 xrightarrow{c | vedclass

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(A) The cleavage of ethers with $conc. HI$ follows an $S_N2$ mechanism for primary alkyl groups. In the case of anisole $(Ph-O-CH_3)$,the $C-O$ bond b

Vedclass · Accepted Answer

$Ph - \mathop O\limits^{18} H + CH_3 - I$

Vedclass · Answer

(A) The cleavage of ethers with $conc. HI$ follows an $S_N2$ mechanism for primary alkyl groups. In the case of anisole $(Ph-O-CH_3)$,the $C-O$ bond between the methyl group and oxygen is weaker than the $C-O$ bond between the phenyl group and oxygen due to the partial double bond character of the $Ph-O$ bond. Therefore,the $HI$ attacks the methyl group,leading to the formation of $CH_3I$ and phenol $(Ph-OH)$. Since the oxygen atom in the ether is labeled with $^{18}O$,the label will remain with the phenol product $(Ph - \mathop O\limits^{18} H)$. Thus,the products are $Ph - \mathop O\limits^{18} H$ and $CH_3 - I$.

Column-$I$ (Product)	Column-$II$ (Reagent)
$A$. Mainly $p$-bromoanisole	$i$. Anhydrous $AlCl_3$,$CH_3COCl$
$B$. $4$-nitroanisole	$ii$. $CH_3Cl$,anhydrous $AlCl_3$ in $CS_2$
$C$. $4$-methoxytoluene	$iii$. $Br_2$ in ethanoic acid
$D$. $4$-methoxyacetophenone	$iv$. Mixture of conc. $H_2SO_4$ and $HNO_3$

$Ph - \mathop O\limits^{18} - CH_3 \xrightarrow{conc. HI} \text{products}$

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