Which reagent is used to get iodo benzene from benzene diazonium acid sulphate `[C_(6)H_(5)N_(2)HSO_(4)]`?

Options

A. `CuBr,Delta`
B. Cu power + HI
C. `KI,Delta`
D. None of these

Answer

To convert benzene diazonium acid sulfate \([C_{6}H_{5}N_{2}HSO_{4}]\) into iodobenzene, we can follow these steps:

### Step 1: Understand the Structure
Benzene diazonium salt has the general formula \(C_{6}H_{5}N_{2}^{+}X^{-}\), where \(X\) can be various anions. In our case, \(X\) is \(HSO_{4}^{-}\).

### Step 2: Identify the Desired Reaction
We want to introduce iodine into the benzene ring. The direct introduction of iodine is not straightforward due to the low equilibrium constant of the reaction.

### Step 3: Use the Sandmeyer Reaction
The Sandmeyer reaction is a well-known method for introducing halogens into an aromatic ring. For iodine, we can use potassium iodide (KI) as the reagent.

### Step 4: Reaction Mechanism
When benzene diazonium salt reacts with potassium iodide, the following reaction occurs:
\[ C_{6}H_{5}N_{2}^{+}HSO_{4}^{-} + KI \xrightarrow{\text{heat}} C_{6}H_{5}I + N_{2} + KHSO_{4} \]
In this reaction:
- The diazonium group \((N_{2}^{+})\) is replaced by iodine \((I)\).
- Nitrogen gas \((N_{2})\) is released.
- Potassium hydrogen sulfate \((KHSO_{4})\) is formed.

### Step 5: Conclusion
Thus, the reagent used to convert benzene diazonium acid sulfate into iodobenzene is **potassium iodide (KI)**.

### Final Answer
The correct reagent is **potassium iodide (KI)**.

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