Question

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

Answer 1

To proceed with this question, we need to focus on the given hints in the question, first of all keep in mind that the diborane is reacting with a metal hydride and the second thing that the compound Z is the extensive reducing agent.

Complete answer:
So, to answer this question, it is better to write first the skeleton of the chemical reaction:
Hence, the skeleton of the chemical reactions is:
$NaH + {B_2}{H_6}\xrightarrow{Y}Z$
Now, we can easily estimate the products and the solvent.
So, the reaction between the sodium hydride and the diborane will give the product Z as Sodium Borohydride $NaB{H_4}$ , and this is the extensively used reducing agent.
Also, our solvent for the given reaction will be i.e. Y will be diethyl ether i.e. ${C_2}{H_5}O{C_2}{H_5}$ .
Hence, now we have our complete equation:
The equation will be:
$NaH + {B_2}{H_6}\xrightarrow{{{C_2}{H_5}O{C_2}{H_5}}}NaB{H_4}$
Therefore, sodium hydride (metal hydride) reacts with diborane to produce sodium borohydride $NaB{H_4}$ in the presence of diethyl ether ${C_2}{H_5}O{C_2}{H_5}$ .
Hence, on the basis of our discussion, we can easily point out the correct answers.
Therefore, the correct answer is option B i.e. ${C_2}{H_5}O{C_2}{H_5}$ And $NaB{H_4}$ .

Note:
Diethyl ether is a volatile ether compound. Generally it is abbreviated as $E{t_2}O$ . It is a colourless compound with a very sweet smell. Also, it is a flammable liquid and diethyl ether is generally used as a starting material for some engine fuels. It is a non-polar solvent. Diethyl ether has many roles to play like it is used as an inhalation anaesthetic, which is one of the important compounds. Diethyl ether is also used in the production of plastics and a variety of dyes.