Identify the product 'Y' in the given sequence of reactions. (Chlorobenzene reacts with Conc. HNO$_3$ and Conc. H$_2$SO$_4$ to give X (Major). X then reacts with (i) NaOH, 443 K and (ii) H$^+$ to give Y.)
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Nucleophilic aromatic substitution (like replacing -Cl with -OH) on halobenzenes is generally difficult. However, it is greatly facilitated by the presence of strong electron-withdrawing groups (like -NO$_2$) at the ortho and para positions relative to the halogen. The more such groups, the easier the reaction.
Step 1: The first reaction is the nitration of chlorobenzene.
Chlorobenzene reacts with a mixture of concentrated nitric acid (HNO$_3$) and concentrated sulfuric acid (H$_2$SO$_4$). This is an electrophilic aromatic substitution reaction.
The chloro group (-Cl) is an ortho, para-directing group, but it is also deactivating due to its strong -I effect. Nitration will primarily occur at the ortho and para positions.
Due to steric hindrance at the ortho position, the para product is usually the major product. However, under forcing conditions, dinitration can occur. The presence of one deactivating group (-Cl) and one activating/directing group (-NO$_2$) makes the second nitration occur at the other ortho/para position relative to -Cl.
Chlorobenzene $\xrightarrow{\text{Conc. HNO}_3, \text{Conc. H}_2\text{SO}_4}$ 1-chloro-2-nitrobenzene (minor) + 1-chloro-4-nitrobenzene (major).
If the reaction is forced further, dinitration occurs. The -Cl is o,p directing and the -NO$_2$ group is meta-directing.
Starting from 1-chloro-4-nitrobenzene, the next nitration will be directed to the ortho position relative to the -Cl group (which is also meta to the -NO$_2$ group).
This gives 1-chloro-2,4-dinitrobenzene. This is often the major product under strong nitrating conditions. Let's assume X is 1-chloro-2,4-dinitrobenzene.
So, X = 1-chloro-2,4-dinitrobenzene.
Step 2: The second reaction is nucleophilic aromatic substitution.
Compound X (1-chloro-2,4-dinitrobenzene) is treated with aqueous NaOH at 443 K, followed by acidification (H$^+$).
The presence of two strong electron-withdrawing nitro groups (-NO$_2$) at the ortho and para positions strongly activates the benzene ring towards nucleophilic substitution. The C-Cl bond becomes susceptible to attack by nucleophiles like OH$^-$.
The OH$^-$ ion from NaOH will replace the Cl$^-$ ion. This is a nucleophilic aromatic substitution (S$_N$Ar) reaction.
1-chloro-2,4-dinitrobenzene + NaOH $\rightarrow$ Sodium 2,4-dinitrophenoxide + NaCl.
The second step is acidification (ii) H$^+$. The phenoxide ion is protonated to form the phenol.
Sodium 2,4-dinitrophenoxide + H$^+$ $\rightarrow$ 2,4-Dinitrophenol.
So, the final product Y is 2,4-Dinitrophenol. This corresponds to option (A).
