Answer:
(A) is diborane, \[{{B}_{2}}{{H}_{6}}\].
It reacts with \[NM{{e}_{3}}\] to form an adduct (B).
\[\underset{(A)}{\mathop{{{B}_{2}}{{H}_{6}}}}\,+2NM{{e}_{3}}\xrightarrow{{}}\underset{(B)}{\mathop{2B{{H}_{3}}\cdot
NM{{e}_{3}}}}\,\]
The (B) on hydrolysis gives boric acid,\[{{H}_{3}}B{{O}_{3}}\].
\[B{{H}_{3}}\cdot
\underset{(B)}{\mathop{NMe}}\,+3{{H}_{2}}O\xrightarrow{{}}\underset{(C)}{\mathop{{{H}_{3}}B{{O}_{3}}}}\,+NM{{e}_{3}}+3{{H}_{2}}\]
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