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

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Question

(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 $CH_3 - I$.

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

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