Predict the major product for the following r | vedclass

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(C) The reaction involves the cleavage of an ether with excess $HI$ under heating conditions.$1$. The ether is $m-CH_3O-C_6H_4-CH_2CH_2-O-CH(CH_3)Ph$.

Vedclass · Accepted Answer

$m-HO-C_6H_4-CH_2CH_2-I + Ph-CH(OH)CH_3$

Vedclass · Answer

(C) The reaction involves the cleavage of an ether with excess $HI$ under heating conditions.$1$. The ether is $m-CH_3O-C_6H_4-CH_2CH_2-O-CH(CH_3)Ph$.$2$. The $O-CH(CH_3)Ph$ bond is cleaved because the carbocation formed at the benzylic position $(Ph-CH^+-CH_3)$ is highly stable due to resonance with the phenyl ring.$3$. The $O-CH_3$ bond in the methoxy group $(m-CH_3O-C_6H_4-)$ is also cleaved by excess $HI$ to form a phenol $(m-HO-C_6H_4-)$ and methyl iodide $(CH_3I)$.$4$. The $O-CH_2CH_2-Ar$ bond is not cleaved because the primary carbocation is unstable.$5$. Thus,the products are $m-HO-C_6H_4-CH_2CH_2-I$ and $Ph-CH(I)CH_3$ (from the benzylic alcohol intermediate $Ph-CH(OH)CH_3$ reacting further with $HI$).$6$. However,looking at the options provided,the reaction of the ether with excess $HI$ leads to the cleavage of both ether linkages. The methoxy group becomes a phenol,and the alkyl ether part forms an alkyl iodide. The correct product set is $m-HO-C_6H_4-CH_2CH_2-I$ and $Ph-CH(I)CH_3$ along with $CH_3I$.

Predict the major product for the following reaction:
$m-CH_3O-C_6H_4-CH_2CH_2-O-CH(CH_3)Ph \xrightarrow{HI (excess), \text{heat}} ?$

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Predict the major product for the following reaction:$m-CH_3O-C_6H_4-CH_2CH_2-O-CH(CH_3)Ph \xrightarrow{HI (excess), \text{heat}} ?$