$C{H_3}CN\xrightarrow{{Na + {C_2}{H_5}OH}}X$
Then compound X is
(A) $C{H_3}CON{H_2}$
(B) $C{H_3}C{H_2}N{H_2}$
(C) $C{H_2}{H_6}$
(D) $C{H_3}NHC{H_3}$

Hint: Here in this question we have to complete the reaction in which the reactant is methyl cyanide and it has to be reacted as in the presence of a reducing agent named Mendius reagent and from both of these we have to find the product of this reaction. As this type of same reaction will take place with the mixture of sodium with ethanol.

Complete Step by Step Solution:
As we know that, $Na + {C_2}{H_5}OH$ is a reducing agent and the work of reducing agent is to add hydrogen atom with any reactant
By writing full equation we get the full data of reducing agent and get the product which we need in this question,
$C{H_3}CN\xrightarrow{{Na + {C_2}{H_5}OH}}$
As we mentioned above hydrogen is added when reacted with a reducing agent as adding hydrogen in above reaction and balancing the equation from the both side we get the following reaction,
$C{H_3}CN + 4H\xrightarrow{{Na + {C_2}{H_5}OH}}C{H_3}C{H_2}N{H_2}$
As from getting the details from the above equation we get that as both side are equally balanced and we get the product as $C{H_3}C{H_2}N{H_2}$ .
Hence, the correct option is (B).

Note: As per in this question we get that when methyl cyanide is reacted with a reducing agent just like when methyl cyanide is reacted with hydrogen then we get a one degree amine as well as we can also say that ethyl amine. Yes because there are many ways to form amine by using amides as a reactant which is also a very important topic in organic chemistry.