1-Phenoxypropane is treated with excess of co | vedclass

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(D) Step $1$: Reaction of $1-$phenoxypropane $(C_6H_5-O-CH_2CH_2CH_3)$ with excess conc. $HI$ at $0\,^{\circ}C$. The cleavage of the ether bond occurs

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$n-$propyl chloride $+$ Phenol

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(D) Step $1$: Reaction of $1-$phenoxypropane $(C_6H_5-O-CH_2CH_2CH_3)$ with excess conc. $HI$ at $0\,^{\circ}C$. The cleavage of the ether bond occurs such that the alkyl group forms an alkyl iodide and the phenolic oxygen remains with the phenyl group. Thus,$C_6H_5-O-CH_2CH_2CH_3 + HI \rightarrow C_6H_5OH + CH_3CH_2CH_2I$ (Phenol and $n-$propyl iodide).Step $2$: The mixture of products (Phenol and $n-$propyl iodide) is treated with thionyl chloride $(SOCl_2)$. Thionyl chloride reacts with alcohols to form alkyl chlorides,but it does not react with phenols under standard conditions to form chlorobenzene. Therefore,$n-$propyl iodide reacts with $SOCl_2$ to form $n-$propyl chloride $(CH_3CH_2CH_2Cl)$,while phenol remains unchanged.Final products: $n-$propyl chloride and Phenol.

$1-$Phenoxypropane is treated with excess of conc. $HI$ at $0\,^{\circ}C$ and the mixture of products is treated with thionyl chloride. The products formed are

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