The final product in the following sequence of ractions is `C_(6)H_(5)NH_(2) underset(273-278K)overset(NaNO_(2)//HCl)toA overset(C_(6)H_(6)//NaOH)toB`

Options

A. `C_(6)H_(5)N_(2)Cl`
B. `C_(6)H_(5)OH`
C. `C_(6)H_(5)-C_(6)H_(5)`
D. `C_(6)H_(5)N=NOH`

Explanation

To solve the question regarding the sequence of reactions starting from aniline to the final product, we will break down the steps involved in the reactions:

### Step 1: Identify the starting compound
The starting compound is **aniline (C₆H₅NH₂)**.

### Step 2: Reaction with NaNO₂ and HCl
When aniline reacts with sodium nitrite (NaNO₂) in the presence of hydrochloric acid (HCl), it undergoes diazotization to form **Benzene Diazonium Chloride (C₆H₅N₂Cl)**. The reaction can be represented as follows:

\[ \text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + \text{HCl} \rightarrow \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{NaCl} + \text{H}_2\text{O} \]

### Step 3: Reaction with Benzene and NaOH
The next step involves the reaction of the benzene diazonium chloride with benzene (C₆H₆) in the presence of sodium hydroxide (NaOH). This reaction leads to the formation of **biphenyl (C₁₂H₁₄)** and nitrogen gas (N₂) is released. The reaction can be represented as:

\[ \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{C}_6\text{H}_6 \xrightarrow{\text{NaOH}} \text{C}_6\text{H}_5\text{C}_6\text{H}_5 + \text{N}_2 + \text{HCl} \]

### Final Product
Thus, the final product B is **biphenyl (C₁₂H₁₄)**.

### Summary of Steps:
1. Start with aniline (C₆H₅NH₂).
2. React with NaNO₂ and HCl to form benzene diazonium chloride (C₆H₅N₂Cl).
3. React benzene diazonium chloride with benzene and NaOH to form biphenyl (C₁₂H₁₄).