An alcohol X (C_4H_{10}O) reacts with concent | vedclass

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(B) The alcohol $X$ $(C_4H_{10}O)$ reacts with concentrated $HCl$ at room temperature to form $Y$ $(C_4H_9Cl)$. This indicates that $X$ is a tertiary

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$2-$Methylpropene

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(B) The alcohol $X$ $(C_4H_{10}O)$ reacts with concentrated $HCl$ at room temperature to form $Y$ $(C_4H_9Cl)$. This indicates that $X$ is a tertiary alcohol,specifically $2$-methylpropan-$2$-ol (tert-butyl alcohol),because tertiary alcohols react rapidly with $HCl$ at room temperature (Lucas test).The reaction is:$(CH_3)_3C-OH + HCl \rightarrow (CH_3)_3C-Cl + H_2O$When $X$ ($2$-methylpropan-$2$-ol) is passed over heated copper at $573 \ K$,it undergoes dehydration to form an alkene,$2$-methylpropene $(Z)$:$(CH_3)_3C-OH \xrightarrow{Cu, 573 \ K} CH_3-C(CH_3)=CH_2 + H_2O$Therefore,$Z$ is $2$-methylpropene.

An alcohol $X$ $(C_4H_{10}O)$ reacts with concentrated $HCl$ at room temperature to give $Y$ $(C_4H_9Cl)$. Reaction of $X$ with copper at $573 \ K$ gives $Z$. What is $Z$?

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