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

  • A
    Boron nitride and $MeBr$
  • B
    Borazine and $MeMgBr$
  • C
    Borazine and $MeBr$
  • D
    Diborane and $MeMgBr$

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