Solveeit Logo
Login

Question

Chemistry Question on Acids and Bases

B2H6 {B2H6} reacts with (CH3)3N {(CH3)3N} to produce

A

\ceBH3+N−(CH3)3\ce{BH^{+}_3N^{-} (CH_3)_3 }

B

\ceB2H6+N−(CH3)2CH3.BH3\ce{ B_2H^{+}_6 N^{-} (CH_3)_2 CH_3 . BH_3}

C

\ce(CH3)3N+BH3\ce{ (CH_3)_3 N^{+} BH_3}

D

\ceBH3N+(CH3)2CH3BH3\ce{ BH_3 N^{+} (CH_3)_2 CH_3BH_3}

Answer

\ce(CH3)3N+BH3\ce{ (CH_3)_3 N^{+} BH_3}

Explanation

Solution

When B2H6B_2H_6 reacts with (CH3)3N(CH_3)_3N;
2(CH3)3NTrimethyl amine+\underset{\text{Trimethyl amine}}{ {2(CH3)3N}}+ B2H6Diborane−>\underset{\text{Diborane}}{ {B2H6}} {->} 2(CH3)3N+ BH3Complex\underset{\text{Complex}}{ {2(CH3)3 \overset{+}{N} \,BH3}}
in this reaction trimethyl amine, reacts readily with diborane and form a 2(CH3)3 N+ BH32(CH_3)_3\, \overset{+}{N} \,BH_3 complex.