The reaction of

Question:

The reaction of $\mathrm{H}_{3} \mathrm{~N}_{3} \mathrm{~B}_{3} \mathrm{Cl}_{3}(\mathrm{~A})$ with $\mathrm{LiBH}_{4}$ in tetrahydrofuran gives inorganic benzene (B). Further, the reaction of $(\mathrm{A})$ with $(\mathrm{C})$ leads to $\mathrm{H}_{3} \mathrm{~N}_{3} \mathrm{~B}_{3}(\mathrm{Me})_{3}$. Compounds (B) and (C) respectively, are:

  1. Borazine and $\mathrm{MeBr}$

  2. Diborane and $\mathrm{MeMgBr}$

  3. Boron nitride and MeBr

  4. Borazine and $\mathrm{MeMgBr}$


Correct Option: , 4

Solution:

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