Question

What is the correct sequence of reagent used for converting nitrobenzene into m-dinitrobenzene?

A.

B.

C.

D.

Answer 1

Hint: The aromatic chemical known as nitrobenzene contains a nitro group inside of a benzene ring. The Nitro group is meta-directing. Hence, any forward reaction will be taking place at the meta position of the ring.

Complete Step by Step Solution:
Firstly, we have to do bromination of the nitrobenzene. The reagent used for the reaction is \[B{r_2}/Fe\] reagent. Here, electrophilic substitution takes place. The reaction is as follows:

Image: Bromination of nitrobenzene

After that, we have to convert \[N{O_2}\] group to \[N{H_2}\] group. So, we can use \[Sn/HCl\] reagent. Tin (IV) chloride is created when tin interacts with HCl. Consequently, we may state that \[{H^ + }\] ions are released as a result of the aromatic compound's nitro functional group reaction. The reaction is as follows:

Image: Reduction by tin and HCl reagent

After that, we have to convert it into diazonium salt. The compound reacts with sodium nitrite and HCl in cold conditions to form diazonium salt. The reaction is as follows:

Image: Formation of diazonium salt

Finally, we have to use the Sandmeyer reaction to get the desired product dibromo benzene. A common substitution reaction is the Sandmeyer reaction, which is used to produce aryl halides from aryl diazonium ions. Catalysts for this process include copper salts like chloride, bromide, or iodide ions. The reaction is as follows:

Image: Sandmeyer reaction

The complete reaction is written as follows:

Image: Conversion of m-dinitrobenzene to dibromobenzene
As a result, the correct order of reagents used is option D.

Note: The two-step mechanism reaction is the Sandmeyer reaction. The free radical mechanism is used in the Sandmeyer reaction. The halogen linked to the copper enters the benzene ring during this process.