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Standard XII

Chemistry

Physical Properties of Alkenes

The reaction ...

Question

The reaction of $H_{3} N_{3} B_{3} C l_{3} (A)$ with $L i B H_{4}$ in tetrahydrofuran gives inorganic benzene (B). Further, the reaction of (A) with (C) leads to $H_{3} N_{3} B_{3} (M e)_{3}$ . Compounds (B) and (C) respectively, are:

JEE MAIN 2020

Borazine and

M e M g B r

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Boron nitride and

M e B r

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Diborane and

M e M g B r

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Borazine and

M e B r

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Solution

The correct option is A Borazine and $M e M g B r$
$B_{3} N_{3} H_{3} C l_{3} (A) + L i B H_{4} \to B_{3} N_{3} H_{6} B + L i C l + B C l_{3}$

$B_{3} N_{3} H_{3} C l_{3} A + 3 M e M g B r C \to B_{3} N_{3} H_{3} (C H_{3})_{3} + 3 M g B r C l$

So, B is borazine and C is $C H_{3} M g B r$

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JEE MAINS 2020

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The reaction of H3N3B3Cl3(A) with LiBH4 in tetrahydrofuran gives inorganic benzene (B). Further, the reaction of (A) with (C) leads to H3N3B3(Me)3. Compounds (B) and (C) respectively, are: JEE MAIN 2020

The correct option is A Borazine and MeMgBrB3N3H3Cl3(A)+LiBH4→B3N3H6B+LiCl+BCl3 B3N3H3Cl3A+3MeMgBrC→B3N3H3 (CH3)3+3MgBrCl So, B is borazine and C is CH3MgBr

The reaction of $H_{3} N_{3} B_{3} C l_{3} (A)$ with $L i B H_{4}$ in tetrahydrofuran gives inorganic benzene (B). Further, the reaction of (A) with (C) leads to $H_{3} N_{3} B_{3} (M e)_{3}$ . Compounds (B) and (C) respectively, are:

JEE MAIN 2020

The correct option is A Borazine and $M e M g B r$
$B_{3} N_{3} H_{3} C l_{3} (A) + L i B H_{4} \to B_{3} N_{3} H_{6} B + L i C l + B C l_{3}$

$B_{3} N_{3} H_{3} C l_{3} A + 3 M e M g B r C \to B_{3} N_{3} H_{3} (C H_{3})_{3} + 3 M g B r C l$

So, B is borazine and C is $C H_{3} M g B r$