Tag: some important compounds of boron

Questions Related to some important compounds of boron

A compound $A$ of boron reacts with $NMe _3$ to give an adduct $B$ which on hydrolysis gives a compound $C$ and a gas $D$. Compound $C$ is an acid.


Compound $A$ is :

  1. diborane

  2. boric acid

  3. borate salt

  4. none of these


Correct Option: A
Explanation:

     $B _2H _6+2NMe _3\longrightarrow2BH _3.NMe _3$

Diborane (A)                        Adduct (B)

$BH _3.NMe _3+3H _2O\longrightarrow H _3BO _3+NMe _3+3H _2$
                                               Boric acid                      $\downarrow$
                                                    (C)                            (D) gas
Compound $A$ is diborane.

Which of the following are used as catalyst in Friedal-craft reaction?

  1. $AlCl _3$

  2. $SiCl _4$

  3. $BF _3$

  4. $SnCl _4$


Correct Option: A,C,D
Explanation:

$Al{ Cl } _{ 3 },B{ F } _{ 3 },Sn{ Cl } _{ 4 }$ can be used as catalyst in Friedal Craft reaction.

An alkali metal hydride (NaH) react with diborane in 'A' to give a tetrahedral compound 'B' which is extensively used as reducing agent in organic synthesis. The compounds 'A' and 'B' respectively are :

  1. $CH _3COCH _3\,\,\, and \,\,\,B _3N _3H _6$

  2. $(C _2H _5) _2O \,\,\,and\,\,\, NaBH _4$

  3. $C _2H _6\,\,\,and\,\,C _2H _5Na$

  4. $C _6H _6\,\,\,and\,\,\,NaBH _4$


Correct Option: B
Explanation:
Answer:-
When an alkali metal hydride $(NaH)$ react with diborane $({B} _{2}{H} _{6})$ in the presence of ether $({({C} _{2}{H} _{5})} _{2}O)$, a tetrahedral compound (Metal borohydride) is formed which act as a reducing agent in organic synthesis.
$2NaH + {B} _{2}{H} _{6} \; \xrightarrow{{({C} _{2}{H} _{5})} _{2}O} \; \underset{\text{sodium borohydride}}{2NaB{H} _{4}}$
Thus, A is  ${({C} _{2}{H} _{5})} _{2}O$ and B is $NaB{H} _{4}$.

A compound $X$, of boron reacts with $NH _3$ on heating to give another compound $Y$ which is called inorganic benzene. The compound $X$ can be prepared by treating $BF _3$ with lithium aluminum hydride. The compounds $X$ and $Y$ are represented by formula:

  1. $B _2H _6, B _3N _3H _6$

  2. $B _2O _3, B _3N _3H _6$

  3. $BF _3, B _3N _3H _6$

  4. $B _3N _3H _6, B _2H _6$


Correct Option: A
Explanation:
A compound $X$, of boron, reacts with $N{H} _{3}$ on heating to give another compound $Y$ which is called inorganic benzene.
$\underset{X \; (Diborane)}{3{B} _{2}{H} _{6}} + 6N{H} _{3} \longrightarrow 3{[B{H} _{2}{(N{H} _{3})} _{2}]}^{+}{[B{H} _{4}]}^{-} \xrightarrow{heat} \underset{\text{Y (Borazole/Inorganic Benzene)}}{2{B} _{3}{N} _{3}{H} _{6}} + 12{H} _{2}$
$4B{F} _{3} + 3LiAl{H} _{4} \longrightarrow \underset{X}{2{B} _{2}{H} _{6}} +3LiF + 3Al{F} _{3}$

The styx number for $B _2H _6$ is:

  1. $2012$

  2. $2222$

  3. $2002$

  4. $2102$


Correct Option: C

Diborane undergoes cleavage and gives aduct with

  1. CO

  2. $ H _2O $

  3. $ N(Me) _3 $

  4. $ NH _3 $


Correct Option: D