The reaction of H_{3}N_{3}B_{3}Cl_{3} (A) wit | vedclass

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(B) The reaction of $B$-trichloroborazine $(A)$ with $LiBH_{4}$ in tetrahydrofuran $(THF)$ is a reduction reaction that yields borazine $(B)$,which is

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Borazine and $MeMgBr$

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(B) The reaction of $B$-trichloroborazine $(A)$ with $LiBH_{4}$ in tetrahydrofuran $(THF)$ is a reduction reaction that yields borazine $(B)$,which is known as inorganic benzene: $H_{3}N_{3}B_{3}Cl_{3} + 3LiBH_{4} \rightarrow H_{3}N_{3}B_{3}H_{3} (B) + 3LiCl + 3BH_{3}THF$. The reaction of $(A)$ with a Grignard reagent like $MeMgBr$ $(C)$ results in the methylation of the boron atoms: $H_{3}N_{3}B_{3}Cl_{3} + 3MeMgBr (C) \rightarrow H_{3}N_{3}B_{3}(Me)_{3} + 3MgBrCl$. Thus,$(B)$ is borazine and $(C)$ is $MeMgBr$.

The reaction of $H_{3}N_{3}B_{3}Cl_{3}$ $(A)$ with $LiBH_{4}$ in tetrahydrofuran gives inorganic benzene $(B)$. Further,the reaction of $(A)$ with $(C)$ leads to $H_{3}N_{3}B_{3}(Me)_{3}$. Then,compounds $(B)$ and $(C)$ respectively,are

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