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The reaction of $\mathrm{H}_{3} \mathrm{N}_{3} \mathrm{B}_{3} \mathrm{Cl}_{3}$ $(A)$ with $\mathrm{LiBH}_{4}$ in tetrahydrofurane gives inorganic benzene $(B)$. Further, the reaction of $(A)$ with $(C)$ leads to $\mathrm{H}_{3} \mathrm{N}_{3} \mathrm{B}_{3}(\mathrm{Me})_{3}$. than Compounds $(\mathrm{B})$ and $(\mathrm{C})$ respectively, are
Boron nitride and $MeBr$
Borazine and $MeMgBr$
Borazine and $MeBr$
Diborane and $MeMgBr$
Solution
$\mathrm{H}_{3} \mathrm{N}_{3} \mathrm{B}_{3} \mathrm{Cl}_{3} (A)+3 \mathrm{LiBH}_{4} \xrightarrow [(T.H.F.)]{\text { In terahydrofurane }} $$\mathrm{H}_{3} \mathrm{N}_{3} \mathrm{B}_{3} \mathrm{H}_{3}(B)+3LiCl+3BH_3THF$
$\mathrm{H}_{3} \mathrm{N}_{3} \mathrm{B}_{3} \mathrm{Cl}_{3}(A)+3 MeMgBr(C)\rightarrow$$ \mathrm{H}_{3} \mathrm{N}_{3} \mathrm{B}_{3} \mathrm{(CH_3)}_{3}+3MgBrCl$
