The products formed in the following reaction | vedclass

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(B) The reaction of anisole $(C_6H_5OCH_3)$ with $HI$ involves the cleavage of the $C-O$ bond.In alkyl aryl ethers,the $O-CH_3$ bond is weaker than th

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$C_6H_5-OH$ and $CH_3-I$

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(B) The reaction of anisole $(C_6H_5OCH_3)$ with $HI$ involves the cleavage of the $C-O$ bond.In alkyl aryl ethers,the $O-CH_3$ bond is weaker than the $O-C_6H_5$ bond because the $O-C_6H_5$ bond has partial double bond character due to resonance.Therefore,the $HI$ attacks the methyl group,leading to the formation of phenol $(C_6H_5OH)$ and methyl iodide $(CH_3I)$.The reaction is: $C_6H_5OCH_3 + HI \xrightarrow{	ext{heat}} C_6H_5OH + CH_3I$.

The products formed in the following reaction are $C_6H_5-OCH_3 + HI \xrightarrow{\text{heat}}$

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Give the reaction of $CH_3-C(CH_3)_2-OCH_3$ with the following:
$(i)$ Anhydrous $HI$ in ether
$(ii)$ Concentrated $HI$

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

The major organic product in the reaction is: $CH_3-O-CH(CH_3)_2 + HI \to$ products

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