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$\mathrm{BF}_3$ reacts with $\mathrm{NaH}$ at $450 \mathrm{~K}$ to form $\mathrm{NaF}$ and $X$. When $X$ reacts with $\mathrm{LiH}$ in diethyl ether, $Y$ is formed. What is $Y$ ?
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The correct answer is:
$\mathrm{LiBH}_4$
$2 \mathrm{BF}_3+6 \mathrm{NaH} \stackrel{450 \mathrm{~K}}{\longrightarrow} 6 \mathrm{NaF}+\mathrm{B}_2 \mathrm{H}_6(X)$
$\mathrm{B}_2 \mathrm{H}_6+2 \mathrm{LiH} \stackrel{\text { Diethyl ether }}{\longrightarrow} 2 \mathrm{LiBH}_4(Y)$
Diborane
Lithium boron hydride
$\mathrm{B}_2 \mathrm{H}_6+2 \mathrm{LiH} \stackrel{\text { Diethyl ether }}{\longrightarrow} 2 \mathrm{LiBH}_4(Y)$
Diborane
Lithium boron hydride
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