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(A) The reaction of an alkyl aryl ether with excess $HI$ involves the cleavage of the $C-O$ bond. In the ether $Ph-O-CH_2-CH_2-(C_6H_4)NO_2$,the oxyge

Vedclass · Accepted Answer

$PhOH + ICH_2CH_2(C_6H_4)NO_2$

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(A) The reaction of an alkyl aryl ether with excess $HI$ involves the cleavage of the $C-O$ bond. In the ether $Ph-O-CH_2-CH_2-(C_6H_4)NO_2$,the oxygen is attached to a phenyl group $(Ph)$ and an alkyl group $(-CH_2CH_2(C_6H_4)NO_2)$. When treated with $HI$,the ether is protonated to form an oxonium ion. The iodide ion $(I^-)$ acts as a nucleophile and attacks the less sterically hindered alkyl carbon. The $C-O$ bond between the oxygen and the alkyl group breaks,resulting in the formation of phenol $(PhOH)$ and an alkyl iodide $(ICH_2CH_2(C_6H_4)NO_2)$. Since $HI$ is in excess,the reaction proceeds to completion,yielding $PhOH$ and $ICH_2CH_2(C_6H_4)NO_2$.

Predict the major product for the following reaction: $Ph-O-CH_2-CH_2-(C_6H_4)-NO_2 + HI \text{ (excess)} \rightarrow \text{Product}$

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