Saved Bookmarks
| 1. |
Sodium borohydride upon treatment with idoine produces a Lewis acid (X), which on heating with ammonia produces a cyclic compound (Y) and a colorless gas (Z). X, \( Y \) and \( Z \) are : (A) \( X = BH _{3} ; Y = BH _{3} \). \( NH _{3} ; Z = N _{2} \) (B) \( X = B _{2} H _{6} ; Y = B _{3} N _{3} H _{6} \). \( Z = H _{2} \) (C) \( X = B _{2} H _{6} ; Y = B _{6} H _{6} \). \( Z = H _{2} \) (D) \( X = B _{2} H _{6} ; Y = B _{3} N _{3} H _{6} ; Z = N _{2} \) |
|
Answer» Correct option is (B) X = B2H6 ; Y = B3N3H6 X=B2H6;Y=B3N3H6. Z = H2 2NaBH4 + I2 → \(\underset{X}{B_2H_6}+2NaI+H_2\) 3B2H6 + 6 NH3 → \(\underset{(Y)}{2B_3N_3H_6}+ \underset{(Ƶ)}{12H_2}\) X is B2H6 (Lewis acid) Y = B3N3H6(Cyclic product) (inorganic benzene) Ƶ =H2 Colourless gas. |
|