A) \[{{B}_{2}}{{H}_{6}}+HCl\xrightarrow{{}}{{B}_{2}}{{H}_{5}}Cl+{{H}_{2}}\]
B) \[2{{B}_{2}}{{H}_{6}}+6N{{H}_{3}}\xrightarrow{\Delta }{{B}_{3}}{{N}_{3}}{{H}_{6}}\](borazine)
C) \[{{B}_{2}}{{H}_{6}}+2N{{(C{{H}_{3}})}_{3}}\xrightarrow{{}}2{{(C{{H}_{3}})}_{3}}NB{{H}_{3}}\]
D) \[{{B}_{2}}{{H}_{6}}+6{{C}_{2}}{{H}_{4}}\xrightarrow{{{H}_{3}}{{O}^{+}}}3{{C}_{2}}{{H}_{5}}OH+2B{{(OH)}_{3}}\]
Correct Answer: D
Solution :
Reaction between diborane and alkene are carried out in dry ether under an atmosphere of \[{{N}_{2}}\] because \[{{B}_{2}}{{H}_{6}}\] and the products are very reactive. The products further treated with alkaline \[{{H}_{2}}{{O}_{2}}\]to convert into alcohols. \[{{B}_{2}}{{H}_{6}}+6{{C}_{2}}{{H}_{4}}\xrightarrow[{}]{{}}\underset{reactive}{\mathop{B{{({{C}_{2}}{{H}_{4}})}_{3}}}}\,\xrightarrow[{{H}_{2}}{{O}_{2}}]{alkaline}\] \[3C{{H}_{3}}C{{H}_{2}}OH+{{H}_{3}}B{{O}_{3}}\]
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