The product B of the following sequence of reactions is
`C_(6)H_(5)CONH_(2) overset(Br_(2))underset(NaOH)toA overset(NaNO_(2)//HCl)underset(ul(" "0-5^(@)C" "))toB`

Options

A.
B.
C.
D.

Answer

To solve the question, we need to follow the sequence of reactions step by step.

### Step 1: Identify the starting compound
The starting compound is given as **C6H5CONH2**, which is **benzamide**.

### Step 2: First Reaction - Hoffmann Degradation
The first reaction involves **Br2** and **NaOH**. This is known as the **Hoffmann degradation reaction**. In this reaction, the amide group (C6H5CONH2) undergoes a transformation where the carbonyl (C=O) and the carbon attached to it are removed, leading to the formation of an amine.

- **Reaction**:
\[ \text{C6H5CONH2} \xrightarrow{\text{Br2, NaOH}} \text{C6H5NH2} + \text{CO2} + \text{HBr} \]
- **Product A**: The product after this reaction is **C6H5NH2**, which is **aniline**.

### Step 3: Second Reaction - Diazotization
The second reaction involves **NaNO2** and **HCl** at a temperature of **0-5°C**. This reaction is used to convert aniline (C6H5NH2) into a diazonium salt.

- **Reaction**:
\[ \text{C6H5NH2} \xrightarrow{\text{NaNO2, HCl}} \text{C6H5N2^+Cl^-} \]
- **Product B**: The product after this reaction is **C6H5N2^+Cl^-**, which is **benzenediazonium chloride**.

### Final Answer
Thus, the final product B of the given sequence of reactions is **benzenediazonium chloride (C6H5N2^+Cl^-)**.

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