An alcohol X (C_5H_{12}O) on dehydration give | vedclass

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(B) $1$. Alcohol $X$ $(C_5H_{12}O)$ undergoes dehydration to form alkene $Y$ (major product).$2$. $Y$ reacts with $HBr$ to form $Z$ ($C_5H_{11}Br$,maj

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$2-$methylbutan$-2-$ol and $2-$methylbut$-2-$ene

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(B) $1$. Alcohol $X$ $(C_5H_{12}O)$ undergoes dehydration to form alkene $Y$ (major product).$2$. $Y$ reacts with $HBr$ to form $Z$ ($C_5H_{11}Br$,major product) via Markovnikov addition.$3$. $Z$ undergoes nucleophilic substitution in two steps,which is characteristic of an $S_N1$ mechanism. This implies $Z$ is a tertiary alkyl halide.$4$. Among the options,$2$-methylbutan-$2$-ol is a tertiary alcohol. Its dehydration gives $2$-methylbut-$2$-ene as the major product (Saytzeff product).$5$. Reaction of $2$-methylbut-$2$-ene with $HBr$ gives $2$-bromo-$2$-methylbutane $(Z)$ as the major product,which is a tertiary alkyl halide and follows the $S_N1$ mechanism (two steps).$6$. Thus,$X$ is $2$-methylbutan-$2$-ol and $Y$ is $2$-methylbut-$2$-ene.

An alcohol $X$ $(C_5H_{12}O)$ on dehydration gives $Y$ (major product). Reaction of $Y$ with $HBr$ gave $Z$ ($C_5H_{11}Br$,major product). $Z$ undergoes nucleophilic substitution in two steps. What are $X$ and $Y$?

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