Step 1: Identify the first transformation (chlorobenzene to X).
Chlorobenzene under drastic conditions (NaOH, 623 K, 300 atm) undergoes nucleophilic substitution via the benzyne mechanism. This converts chlorobenzene into phenol after acidification with H\(_3\)O\(^+\). Hence, product X is phenol (C\(_6\)H\(_5\)OH). Step 2: Understand the nature of phenol.
Phenol is a strongly activating, ortho/para directing group due to resonance donation of lone pair from oxygen into the benzene ring. This increases electron density at ortho and para positions, making electrophilic substitution easier. Step 3: First nitration step (conc. HNO\(_3\)/H\(_2\)SO\(_4\)).
In nitration mixture, electrophile NO\(_2^+\) is generated. Phenol undergoes rapid electrophilic substitution at ortho and para positions, forming mono and further substituted nitro products due to strong activation. Step 4: Further nitration due to strong activation.
Because phenol strongly activates the ring, repeated nitration occurs at all ortho and para positions relative to –OH group. This leads to substitution at 2, 4, and 6 positions on the ring. Step 5: Formation of final product.
After complete nitration, the product formed is 2,4,6-trinitrophenol, commonly known as picric acid. It contains one –OH group and three –NO\(_2\) groups at activated positions. Step 6: Final verification.
The structure matches option (4), which is 2,4,6-trinitrophenol. This is the major product due to strong activating effect of phenol and multiple nitration under vigorous conditions. Final Answer:
\[
\boxed{\text{2,4,6-trinitrophenol (picric acid)}}
\]
