Question

In the reaction, 2X+B_H6 >[BH, (X2)]* (BH,] the amine X is (A) NH3 (B) CH3NH2 (C) (CH3)2NH (D) All of these​

Answer 1

Answer:

In the given reaction,  $2\ X\ +\ B_2H_6\ --->\ [BH_2(X)_2]^+\ \ [BH_4]^-$ the amine X can be option D, All of these·

Explanation:

$B_2H_6$, diborane is the simplest boron hydride, which is the starting material for various other boron hydrides· The boron in the diborane is $sp^{3}$ hybridized and has four hybrid orbitals· It has four terminal $2c-2e^-$ boron-hydrogen bonds and two $3c-2e^-$ hydrogen bridging $B-H-B$ bonds, which are otherwise known as banana bonds· The four-terminal $B-H$ bonds are in the same plane and the two banana bonds are above and below the horizontal plane·

Because of the way that diborane are arranged they undergo a variety of reactions· This is one of the reactions that diborane is reacting with a base·

In the given reaction,

$2\ X\ +\ B_2H_6\ --->\ [BH_2(X)_2]^+\ \ [BH_4]^-$  

diborane undergo an unsymmetrical cleavage around the two $B-H$ bonds on the same side of the $3c-2e^-$, $B-H-B$ bridge bonds to get a product of  $[BH_2(X)_2]^+\ \ [BH_4]^-$· This type of cleavage happened only in the presence of small-sized and strong bases such as $NH_3$ ,  $CH_3NH_2$ , $(CH_3)_2NH$, etc· where the base act as a nucleophile·

In the presence of large size and weak bases such as $N(CH_3)_3$, $OEt_3$, CO, etc· diborane undergo symmetrical cleavage and form an adduct·

Therefore, in the given reaction the amine X can be option D, All of these·