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Consider the following reaction,
the product $Z$, is
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the product $Z$, is
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The correct answer is:
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}$
Key Idea (i) $\mathrm{PBr}_3$ is a halogenating agent, ie, converts $-\mathrm{OH}$ group into $-\mathrm{Br}$.
(ii) Alc. $\mathrm{KOH}$ is a dehydrohalogenating agent.
(iii) $\mathrm{H}_2 \mathrm{SO}_4$ and $\mathrm{H}_2 \mathrm{O}$ converts an olefin into alcohol.
(ii) Alc. $\mathrm{KOH}$ is a dehydrohalogenating agent.
(iii) $\mathrm{H}_2 \mathrm{SO}_4$ and $\mathrm{H}_2 \mathrm{O}$ converts an olefin into alcohol.
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