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$\mathrm{C}_6 \mathrm{H}_5-\mathrm{CH}=\mathrm{CHCHO} \xrightarrow{x} \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}=\mathrm{CHCH}_2 \mathrm{OH}$. In the above sequence $X$ can be
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$\mathrm{NaBH}_4$
$\mathrm{NaBH}_4$ and $\mathrm{LiAlH}_4$ attacks only carbonyl group and reduce it into alcohol group. They do not attack on double bond.
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