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 :

C H_{3} C O C H_{3} a n d B_{3} N_{3} H_{6}

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(C_{2} H_{5})_{2} O a n d N a B H_{4}

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C_{2} H_{6} a n d C_{2} H_{5} N a

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C_{6} H_{6} a n d N a B H_{4}

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Solution

The correct option is B $(C_{2} H_{5})_{2} O a n d N a B H_{4}$
Answer:-
When an alkali metal hydride $(N a H)$ 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.
$2 N a H + B_{2} H_{6} {(C_{2} H_{5})}_{2} O - ----- \to 2 N a B H_{4} sodium borohydride$
Thus, A is ${(C_{2} H_{5})}_{2} O$ and B is $N a B H_{4}$ .

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