For the following reaction sequence,
X and Y, respectively, are
To solve this question, we need to understand the reaction sequence given and identify the compounds X and Y.
The reaction sequence is as follows:
- The starting materials are \(3 \text{ NH}_4\text{Cl}\) and \(3 \text{ BCl}_3\).
- These react under heat and in the presence of \(\text{C}_6\text{H}_5\text{Cl}\) to form intermediate 'X'.
- Upon further treatment with \(\text{LiBH}_4\), a reaction forms a cyclic borazine compound typically represented as \(\{HB(NH)\}_3\).
- Finally, compound 'X' reacts with \(3 \text{ H}_2\text{O}\) at room temperature to give compound 'Y'.
Considering the reaction of borazine derivatives with water, it can undergo hydrolysis to form \(\{H(OH)B(NH_2)\}_3\).
Therefore, the correct answers for X and Y, respectively, are \(\{HB(NH)\}_3\) and \(\{H(OH)B(NH_2)\}_3\).
Correct Option: \(\{HB(NH)\}_3\text{ and }\{H(OH)B(NH_2)\}_3\)
Explanation: The multi-step reaction first converts ammonium chloride and boron trichloride into a cyclic compound 'X' under specific conditions. This compound then undergoes hydrolysis to form the 'Y' product, matching the given chemical formulas.
Important Questions from Main Group Elements and Their Compounds
- __________ nitrides are termed by transition + inner transition metals.
- A non-metal ‘X’ occurs in nature in free state. It combines with hydrogen to form a gas, which is colourless and highly toxic. Apart from this, ‘X’ is oxidized to ‘$\text{XO}_2$’ which is a colourless, pungent and irritating gas. ‘X’ represents :
The reaction of $PCl_3$ with $PhLi$ in 1:3 molar ratio yields X as one of the products, which on further treatment with $CH_3I$ gives Y. The reaction of Y with $n-BuLi$ gives product Z. The products X, Y and Z, respectively, are
- Conversion of boron trifluoride to tetrafluoroborate accompanies
The correct statements regarding B among the following
(i) Nuclear spin of 11B is greater than that of 10B
(ii) The polarities of B - H bond and C-H bonds are opposite
(iii) Cross - section of neutron absorption for 10B is much more than that of 11B
(iv) B reacts with boiling aq. NaOH solution to form NaB(OH)4
