A compound (A) of boron reacts with NMe3 to give an adduct (B) which on hydrolysis gives a compound (C) and hydrogen gas. Compound (C) is an acid. Identify the compounds A,B and C. Give the reactions involved.
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Solution
As per the question,
A+NMe3→B
B is an adduct.
On hydrolysis,
B+H2O→C+H2(g)
Also stated that C is acidic in nature.
Formation of adduct
Diborane undergoes cleavage reactions with Lewis bases such as Trimethyl nitrogen N(Me)3 to give borane adducts, BH3⋅NMe3
It is an addition compound of borane and Lewis base.
B2H6+2NMe3→2BH3⋅NMe3
Therefore, A is diborane (B2H6).
B is adduct or addition compound called borane adduct i.e. BH2⋅NMe3
Adduct to acid conversion
Compound B on hydrolysis gives boric acid, H3BO3
The reaction is:
BH3⋅NMe3+3H2O→H3BO3+NMe3+3H2
Therefore, the compound C is orthoboric acid (H3BO3) which is not a protonic acid but acts as a Lewis acid.