Question

When sodium borohydride (NaBH4) is treated with Iodine (I2), two gaseous products were obtained. One is hydrogen and the other is a highly toxic gas X, which catches fire upon exposure to air. When the gas X is heated with ammonia for a long time, a compound Y of ring structure is obtained. Identify and name X and Y.

Answer 1

2NaBH4 + I2 ➡️ B2H6 + 2NaI + H2

So,

X = B2H6

Y = B3N3H6

Structure of Y is in the attachment:

Answer 2

Answer:

X is diborane and Y is borazine.

Explanation:

The reaction of iodine and sodium borohydride in diglyme gives diborane, which is a toxic gas and hydrogen gas. The reaction is as shown below.

2NaBH₄ + I₂ → B₂H₆ + 2NaI + H₂

Diborane reacts with ammonia depending on the reaction condition. When exposed to ammonia for a longer time it forms Borazine or inorganic benzene.

3B₂H₆ + 6NH₃  → 2B₃N₃H₆ + 12H₂

Thus, X is diborane and Y is borazine.

#SPJ3