





The problem asks to identify the final product 'A' of a four-step reaction sequence starting from aniline.
This is a multi-step synthesis that involves several key named reactions in organic chemistry:
Let's trace the transformation through each step of the reaction sequence.
Step 1: Reaction of Aniline with NaNO₂ and HCl
Aniline, a primary aromatic amine, reacts with NaNO₂ and HCl at low temperature to form benzenediazonium chloride. This is the diazotization step.
\[ \text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{NaNO}_2, \text{HCl}, \, 0-5^\circ\text{C}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \]The product of this step is benzenediazonium chloride.
Step 2: Reaction with Cu₂Cl₂ (Sandmeyer Reaction)
The benzenediazonium chloride formed in the first step is then treated with cuprous chloride (Cu₂Cl₂). This is a Sandmeyer reaction, which replaces the diazonium group with a chlorine atom.
\[ \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{Cu}_2\text{Cl}_2} \text{C}_6\text{H}_5\text{Cl} + \text{N}_2 \]The product after the second step is chlorobenzene.
Step 3: Reaction of Chlorobenzene with NaOH at 623 K and 300 atm
The chlorobenzene is subjected to the Dow process. It is treated with aqueous NaOH under high temperature (623 K) and high pressure (300 atm). Under these forcing conditions, a nucleophilic aromatic substitution occurs, and the chlorine atom is replaced by a hydroxyl group. Since the reaction medium is strongly basic, the initially formed phenol is deprotonated to form sodium phenoxide.
\[ \text{C}_6\text{H}_5\text{Cl} + 2\text{NaOH} \xrightarrow{623\,\text{K}, 300\,\text{atm}} \text{C}_6\text{H}_5\text{O}^-\text{Na}^+ + \text{NaCl} + \text{H}_2\text{O} \]The product of this step is sodium phenoxide.
Step 4: Acidification with H⁺
The final step is the acidification of the reaction mixture. The proton source (H⁺) protonates the sodium phenoxide ion to yield the final product, phenol.
\[ \text{C}_6\text{H}_5\text{O}^-\text{Na}^+ \xrightarrow{\text{H}^+} \text{C}_6\text{H}_5\text{OH} \]The final product 'A' is phenol.
Conclusion:
The final product of the entire reaction sequence is phenol. This corresponds to the structure shown in option (2).
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: H2Te is more acidic than H2S.
Reason R: Bond dissociation enthalpy of H2Te is lower than H2S.
In light of the above statements, choose the most appropriate from the options given below:


What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)