Question

In acid medium, Nitrobenzene is reduced to Aniline as shown in the reaction $ {C_6}{H_5} - N{O_2} + 6[H] \to {C_6}{H_5} - N{H_2} + 2{H_2}O. $ The reducing agent used in this reaction is.

Answer 1

Hint: A reducing agent is an agent which gets oxidized in redox reactions and donates its electrons to another species. The reducing agent gets oxidized in this process. Some examples of commonly used reducing agents are: $ LiAl{H_4},\,\,Sn/HCl,\,\,Na/alcohol,\,\,{H_2}/Ni $ et cetera.

Complete answer:
Nitrobenzene $ \left( {{C_6}{H_5} - N{O_2}} \right) $ is an aromatic nitro compound in which a nitro group is attached to a phenyl group. It is prepared from Benzene; on electrophilic aromatic substitution reaction, specifically Nitration of Benzene, by using concentrated nitric acid. The formed Nitrobenzene is a precursor to produce Aniline $ \left( {{C_6}{H_5} - N{H_2}} \right). $ Aniline is the simplest aromatic amine. In Aniline, an amino group is attached to a phenyl group.
The reaction we are dealing with involves the reduction of Nitrobenzene to Aniline. The most suitable reducing agent we can use is $ Sn/HCl. $ The reaction is as follows:
 $ 2{C_6}{H_5} - N{O_2}\, + \,\,3Sn\, + \,12HCl \to 2{C_6}{H_5} - N{H_2}\, + \,3SnC{l_4}\, + \,4{H_2}O $
Here, in the reducing agent $ Sn/HCl, $ the $ Sn $ gets oxidized to $ S{n^{4 + }} $ , giving up four electrons to four Chlorine atoms to form $ SnC{l_4}. $ On the formation of this product, $ {H^ + } $ is released. The $ {H^ + } $ produced substitutes Oxygen atoms in Nitrobenzene to form an amino group. This gives us Aniline as the product, and the oxygen bonds with Hydrogen atoms to form water molecules.

Note:
The reduction reaction of nitrobenzene to aniline is an aromatic nitration reaction, which falls under the category of electrophilic aromatic substitution reaction. Other electrophilic aromatic substitution reactions are Sulphonation, Halogenation, Friedel Crafts Alkylation and Friedel Crafts Acylation.