question_answer 57)

A compound X of boron reacts with ammonia on heating to give another compound Y which is called inorganic benzene. The compound X can be prepared by treating \[B{{F}_{3}}\]with lithium aluminium hydride. The compound X and Y are represented by the formulae : (a) \[{{B}_{2}}{{H}_{6}},{{B}_{3}}{{N}_{3}}{{H}_{6}}\]       (b) \[{{B}_{2}}{{O}_{3}},{{B}_{3}}{{N}_{3}}{{H}_{6}}\]                (c)\[B{{F}_{3}},{{B}_{3}}{{N}_{3}}{{H}_{6}}\]                        (d) \[{{B}_{3}}{{N}_{3}}{{H}_{6}},{{B}_{2}}{{H}_{6}}\]

Answer:

  (a) \[{{B}_{2}}{{H}_{6}}+2N{{H}_{3}}\xrightarrow{Low\,temp.}{{B}_{2}}{{H}_{6}}\cdot 2N{{H}_{3}}\] \[3{{B}_{2}}{{H}_{6}}\cdot 2N{{H}_{3}}\xrightarrow{200\,{}^\circ C}\underset{\begin{smallmatrix}  Borazole(y) \\  (lnorganic\,benzene) \end{smallmatrix}}{\mathop{2{{B}_{3}}{{N}_{3}}{{H}_{6}}}}\,+12{{H}_{2}}\] \[4B{{F}_{3}}+3LiAl{{H}_{4}}\xrightarrow{Ether}\underset{(X)}{\mathop{2{{B}_{2}}{{H}_{6}}}}\,+3LiCl+3AlC{{l}_{3}}\]