Ncert Exercises & Solutions

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.

Since, compound (A) of boron reacts with NMe₃ to form an adduct (B) thus, compound (A) is a Lewis acid. Since, adduct (B) on hydrolysis gives an acid (C) and hydrogen gas, therefore, (A) must be B₂H₆ and (C) must be boric acid.

\underset{Diborane (A)}{B_{2} H_{6}} + \underset{Adduct (B)}{2 {NMe}_{3}} \to \underset{Adduct (B)}{2 {BH}_{3} {NMe}_{3}}

\underset{(B)}{{BH}_{3} {NMe}_{3}} + 3 H_{2} O \to \underset{Boric acid (C)}{H_{3} {BO}_{3}} + {NMe}_{3} + 6 H_{2}

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