An alkali metal hydride $N a H$ reacts with diborane in $Y$ to give a tetrahedral compound $Z$ which is extensively used as reducing agent in organic synthesis. The $Y$ and $Z$ in the above reaction are:

C_{2} H_{6}, C_{2} H_{5} N H_{2}

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C_{2} H_{5} O C_{2} H_{5}, N a B H_{4}

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N H_{3}, B_{3} N_{3} H_{6}

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C_{3} H_{8}, B_{3} N_{3} H_{6}

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Solution

The correct option is B $C_{2} H_{5} O C_{2} H_{5}, N a B H_{4}$
With the hint of the extensively used reducing agent also given with diborane, we get $Z$ as $N a B H_{4}$ (preparation of Sodium Borohydride) and the solvent used in this reaction is diethyl-ether.

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An alkali metal hydride NaH reacts with diborane in Y to give a tetrahedral compound Z which is extensively used as reducing agent in organic synthesis. The Y and Z in the above reaction are:

The correct option is B C2H5OC2H5, NaBH4With the hint of the extensively used reducing agent also given with diborane, we get Z as NaBH4 (preparation of Sodium Borohydride) and the solvent used in this reaction is diethyl-ether.

An alkali metal hydride $N a H$ reacts with diborane in $Y$ to give a tetrahedral compound $Z$ which is extensively used as reducing agent in organic synthesis. The $Y$ and $Z$ in the above reaction are:

The correct option is B $C_{2} H_{5} O C_{2} H_{5}, N a B H_{4}$
With the hint of the extensively used reducing agent also given with diborane, we get $Z$ as $N a B H_{4}$ (preparation of Sodium Borohydride) and the solvent used in this reaction is diethyl-ether.