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Pent-1-ene reacts with diborane to form $\mathrm{X}$. X on oxidation using $\mathrm{H}_2 \mathrm{O}_2$ in the presence of aqueous $\mathrm{NaOH}$ gives $\mathrm{Y}$. Compound $\mathrm{Y}$ is
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The correct answer is:
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}$
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