Identify A, B, and C in the following reaction:

The correct answer is Option (2) → A: 2,4-dinitrochlorobenzene; B: 2,4-dinitrophenol; C: 2,4-diaminophenol

The starting compound is $\textbf{p-nitrochlorobenzene}$ (chlorine and nitro group in para position).

The reaction sequence involves:

$\textbf{Nitration}$ using conc. $HNO_3$ / conc. $H_2SO_4$

$\textbf{Nucleophilic substitution}$ using $NaOH$ followed by acidification

$\textbf{Reduction}$ using $Sn/HCl$

Step 1: Nitration

The molecule already contains:

$Cl$ (ortho/para directing)

$NO_2$ (meta directing)

Since the $\textbf{para position relative to Cl is already occupied}$, nitration occurs mainly at the $\textbf{ortho position to chlorine}$, which is also $\textbf{meta to the existing nitro group}$.

Thus the product formed is:

$\textbf{2,4-dinitrochlorobenzene}$

Therefore,

$\textbf{A = 2,4-dinitrochlorobenzene}$

Step 2: Reaction with NaOH

Aryl halides containing strong electron withdrawing groups (like $NO_2$) at ortho/para positions undergo nucleophilic aromatic substitution ($S_NAr$).

Here, $Cl$ is replaced by OH.

Thus:

$\text{2,4-dinitrochlorobenzene} \rightarrow \text{2,4-dinitrophenol}$

Therefore,

$\mathbf{B = \text{2,4-dinitrophenol}}$

Step 3: Reduction (Sn/HCl)

$Sn/HCl$ reduces nitro groups ($NO_2$) to amino groups ($NH_2$).

However, under controlled conditions, one nitro group is reduced preferentially, forming:

2-amino-4-nitrophenol

Therefore,

$\mathbf{C = \text{2-amino-4-nitrophenol}}$

Option 1

This suggests 3,4-dinitro substitution, which is not favored because directing effects place nitration at the 2-position.

Option 2

This assumes both nitro groups are reduced to amino groups, which does not occur in the given reaction conditions.

Option 3

Correctly identifies:

A = 2,4-dinitrochlorobenzene

B = 2,4-dinitrophenol

C = 2-amino-4-nitrophenol

Option 4

This option gives $\textbf{2-nitro-4-aminophenol}$, which would require reduction of the nitro group at a different position.

Incorrect.