
By which of the following reactions is borazine prepared?
(A) ${B_2}{H_6} + N{H_3}\left( {excess} \right) \overset{low temperature}{\rightarrow} $
(B) ${B_2}{H_6} + N{H_3}\left( {excess} \right) \overset{high temperature}{\rightarrow} $
(C) ${{B_2}{H_6} + N{H_3}\left( {excess} \right) \overset{high temperature; ratio; 2NH3:1B2H6}{\rightarrow} }$
(D) None of these
Answer
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Hint:We know that borazine has the same structure like benzene so it is also known as inorganic benzene. This inorganic benzene known as borazine has a molecular formula of ${B_3}{N_3}{H_6}$.
Complete answer:We all know that the diboranes is the simplest and a very important inorganic compound. It is an important reagent in the synthesis of many organic compounds. It is normally prepared in situ. It is a very versatile reagent for the preparation of organoboranes, which is an important intermediate in organic synthesis. Diborane is also a powerful electrophilic reagent used for the preparation of certain functional groups. Diborane was first prepared by Alfred stock by heating magnesium and boron to give magnesium boride and then treated with orthophosphoric acid.
All the boron’s act as a Lewis acid as they can easily accept electron pairs (as the central element boron has electrons less than eight i.e. it contains only six electrons). Due to which they can easily react with ammonia as nitrogen contains lone pairs which can share with boron elements to complete its octet electronic configuration.
The boranes will form different compounds with the ammonia, the final product is depending on the conditions used for the reaction.
The borazine is formed when ammonia and diborane will react in a simple ratio of \[2:1\] at a high temperature.
${{B_2}{H_6} + N{H_3}\left( {excess} \right) \overset{high temperature; ratio; 2NH3:1B2H6}{\rightarrow}} {B_3}{N_3}{H_6}$
Ammonia contains a lone pair of electrons which gets attracted towards the vacant p orbitals of boron and leads to the formation of the final product borazine.
Thus, the correct option is (C) .
Note:Diborane is also a colorless gas and it must be handled with care as it is highly reactive. It catches fire spontaneously and explodes with dioxygen. The heat of combustion is very high. In the laboratory it is handled in a vacuum frame.
Complete answer:We all know that the diboranes is the simplest and a very important inorganic compound. It is an important reagent in the synthesis of many organic compounds. It is normally prepared in situ. It is a very versatile reagent for the preparation of organoboranes, which is an important intermediate in organic synthesis. Diborane is also a powerful electrophilic reagent used for the preparation of certain functional groups. Diborane was first prepared by Alfred stock by heating magnesium and boron to give magnesium boride and then treated with orthophosphoric acid.
All the boron’s act as a Lewis acid as they can easily accept electron pairs (as the central element boron has electrons less than eight i.e. it contains only six electrons). Due to which they can easily react with ammonia as nitrogen contains lone pairs which can share with boron elements to complete its octet electronic configuration.
The boranes will form different compounds with the ammonia, the final product is depending on the conditions used for the reaction.
The borazine is formed when ammonia and diborane will react in a simple ratio of \[2:1\] at a high temperature.
${{B_2}{H_6} + N{H_3}\left( {excess} \right) \overset{high temperature; ratio; 2NH3:1B2H6}{\rightarrow}} {B_3}{N_3}{H_6}$
Ammonia contains a lone pair of electrons which gets attracted towards the vacant p orbitals of boron and leads to the formation of the final product borazine.
Thus, the correct option is (C) .
Note:Diborane is also a colorless gas and it must be handled with care as it is highly reactive. It catches fire spontaneously and explodes with dioxygen. The heat of combustion is very high. In the laboratory it is handled in a vacuum frame.
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