Question

In the above reaction ‘X’ stands for:

A.$N{H_2}$
B.$SnC{l_2}$
C.$Cl$
D.$N{H_4}^ + C{l^ - }$

Answer 1

When nitrobenzene is subjected to reaction with tin ( $Sn$ ) and hydrochloric acid ( $HCl$ ), it undergoes reduction. And the final product we get is aniline, which is the reduced product of nitrobenzene.

Complete answer:
Here we have been given a reaction in which nitrobenzene is subjected to a reaction with tin ( $Sn$ ) and hydrochloric acid ( $HCl$ ). The product formed on the right hand side will be aniline.
Aniline is shown below
Thus, the X in the above reaction stands for $N{H_2}$ . Thus, the correct option is A, $N{H_2}$ .
Now, when nitrobenzene is subjected towards reaction with tin ( $Sn$ ) and hydrochloric acid ( $HCl$ ), it undergoes reduction and forms aniline. This reaction is called reduction of nitrobenzene as nitrobenzene gets reduced to aniline. The oxygen in the nitrobenzene is replaced with hydrogen forming ammonia group
In this reaction tin ( $Sn$ ) and hydrochloric acid ( $HCl$ ) acts as reducing agent, due to which $N{H_2}$ replaces $N{O_2}$ group from the benzene ring which makes it aniline. This reaction can also be performed with iron ( $Fe$ ) and hydrochloric acid ( $HCl$ ), they also act as reducing agents in the reaction forming aniline from nitrobenzene.
When nitrobenzene is subjected to reduction with the help of tin ( $Sn$ ) and hydrochloric acid ( $HCl$ ), water will be formed as a by-product. Also, the hydrochloric acid in this reaction is concentrated.
And hence the correct option is A.

Note:
The reduction of nitrobenzene to aniline can also be done with the help of iron ( $Fe$ ) and hydrochloric acid ( $HCl$ ), and not just tin ( $Sn$ ) and hydrochloric acid ( $HCl$ ). Thus, any of the methods can be used for the reduction of nitrobenzene as per the requirements of the maker.