What is Y in the following reaction?C_2H_5I + | vedclass

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(B) Step $1$: The reaction of ethyl iodide $(C_2H_5I)$ with sodium ethoxide $(NaOC_2H_5)$ is a Williamson ether synthesis,which produces diethyl ether

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$C_2H_5I$

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(B) Step $1$: The reaction of ethyl iodide $(C_2H_5I)$ with sodium ethoxide $(NaOC_2H_5)$ is a Williamson ether synthesis,which produces diethyl ether $(X = C_2H_5OC_2H_5)$ and sodium iodide $(NaI)$.$C_2H_5I + NaOC_2H_5 \longrightarrow C_2H_5OC_2H_5 + NaI$Step $2$: The reaction of diethyl ether $(X)$ with excess hydroiodic acid $(HI)$ under heating results in the cleavage of the ether bond to form two molecules of ethyl iodide $(Y = C_2H_5I)$ and water.$C_2H_5OC_2H_5 + 2HI \xrightarrow{\Delta} 2C_2H_5I + H_2O$Therefore,$Y$ is $C_2H_5I$.

What is $Y$ in the following reaction?
$C_2H_5I + NaOC_2H_5 \longrightarrow X + NaI$
$X + 2HI \xrightarrow{\Delta} 2Y + H_2O$

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What is $Y$ in the following reaction?$C_2H_5I + NaOC_2H_5 \longrightarrow X + NaI$$X + 2HI \xrightarrow{\Delta} 2Y + H_2O$