Question

Which of the following sequences would yield $$m-nitro$$ chlorobenzene (Z) from benzene?
(A) $$Benzene\stackrel{C{l}_2/FeC{l}_3}{\to }(X)\underset{H_{2}S{O}_4}{\overset{HN{O}_3}{\to }}(Z)$$ $$
(B) $$Benzene\stackrel{H_{2}S{O}_4/HN{O}_3}{\to }(Z)$$
(C) $$Benzene\stackrel{H_{2}S{O}_4/HN{O}_3}{\to }(X)\stackrel{C{l}_2/FeC{l}_3}{\to }(Z)$$
(D) All of these

Answer 1

Hint:m-nitro chlorobenzene is an organic compound having formula $C_6{H}_{4}ClN{O}_2$. It is yellow in colour and acts as a precursor in many reactions. Benzene can be converted into m-nitro chlorobenzene by undergoing nitration and addition of chlorine.

Complete step by step answer:
As we know, benzene is $C_6{H}_5$, having structure as shown below

For converting it into m-nitro chlorobenzene, we need to add $$-nitro$$ and $$-chloro$$ groups in benzene. For the same, the reaction is carried out in two steps. In first step, we add nitrating mixture i.e. $HN{O}_3$ and $H_{2}S{O}_4$ to benzene, this step is known as nitration. It leads to the addition of $N{O}_2$ on Benzene. It can be shown as:

In the next step, $FeC{l}_3$ and $C{l}_2$ are added to nitrobenzene. $N{O}_2$ is an electron withdrawing, meta directing group. Therefore, it will add the chlorine at meta position, forming the desired product i.e. $$m-nitro$$ chlorobenzene. The overall reaction is

The meta isomer of nitro chlorobenzene is not active towards nucleophilic substitution at chlorine center. m-nitro chlorobenzene can be reduced to $$3-chloroaniline$$ with $$Fe/HCl$$ mixture, this reaction is called Bechamp reduction. 
Hence, the correct option is C.
Additional information:
m-nitro chlorobenzene is also known as $$3-nitro$$ chlorobenzene. It’s IUPAC name is $$1-chloro3-nitrobenzene$$ . This compound generally does not have any direct application, but it acts as a precursor of other compounds such as $$3-chloroaniline$$ .

Note:
If we do the two above mentioned steps in reverse order i.e. firstly we add $FeC{l}_3$ and $C{l}_2$ into Benzene, it will form chlorobenzene. In the second step, if nitration is to be done we should first note that chlorine is the $$ortho/para$$ directing group. Therefore, it will add the $N{O}_2$ at $$ortho/para$$ position, forming $$o-nitro$$ chlorobenzene or $$p-nitro$$ p-nitro chlorobenzene but not m-nitro chlorobenzene.