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(B) The reaction of an ether with $HI$ follows the mechanism where the more stable carbocation is formed. For $(CH_3)_3COC_2H_5$,the cleavage occurs t

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$X$ undergoes substitution with water in two steps

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(B) The reaction of an ether with $HI$ follows the mechanism where the more stable carbocation is formed. For $(CH_3)_3COC_2H_5$,the cleavage occurs to form $(CH_3)_3C^+$ and $C_2H_5OH$.Thus,$X = (CH_3)_3CI$ (tert-butyl iodide) and $Y = C_2H_5OH$ (ethanol).$X$ is a tertiary alkyl halide,which undergoes substitution via the $S_{N}1$ mechanism,not $S_{N}2$.$Y$ is a primary alcohol. Primary alcohols do not react with conc. $HCl$ at room temperature; they require $ZnCl_2$ (Lucas reagent) and heating.Reaction of $Y$ $(C_2H_5OH)$ with $Cu / 573 \ K$ is a dehydrogenation reaction that gives an aldehyde $(CH_3CHO)$,not a ketone.$X$ is a tertiary halide,which undergoes substitution with water via the $S_{N}1$ mechanism,which involves two steps (formation of carbocation followed by nucleophilic attack).

The correct statement regarding $X$ and $Y$ formed in the following reaction is
$(CH_3)_3COC_2H_5 \xrightarrow[\Delta]{HI} \text{halide } (X) + \text{alcohol } (Y)$

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The correct statement regarding $X$ and $Y$ formed in the following reaction is$(CH_3)_3COC_2H_5 \xrightarrow[\Delta]{HI} \text{halide } (X) + \text{alcohol } (Y)$