In Balz-Schiemann reaction, benzene diazonium chloride reacts with

Options

A. KI
B. CuCN/KCN
C. `HBF_(4)`
D. `HBF_(4)` and `NaNO_(2),Cu`

Explanation

**Step-by-Step Solution:**

1. **Understanding the Balz-Schiemann Reaction:**
The Balz-Schiemann reaction involves the conversion of benzene diazonium chloride into fluorobenzene. This reaction is significant in organic chemistry for introducing fluorine into aromatic compounds.

2. **Identifying the Reactants:**
In this reaction, benzene diazonium chloride (C6H5N2Cl) is the starting material. The key reagent that reacts with it in the Balz-Schiemann reaction is hydrogen tetrafluoroborate (HBF4).

3. **Reaction Mechanism:**
When benzene diazonium chloride reacts with HBF4, the diazonium group (N2) is replaced by a fluorine atom. The reaction can be summarized as follows:
\[ C6H5N2Cl + HBF4 \rightarrow C6H5F + N2 + BF3 \]
Here, fluorobenzene (C6H5F) is formed, along with nitrogen gas (N2) and boron trifluoride (BF3).

4. **Conclusion:**
Based on the reaction, the correct answer to the question is that benzene diazonium chloride reacts with HBF4 in the Balz-Schiemann reaction to produce fluorobenzene.

**Final Answer:**
Benzene diazonium chloride reacts with HBF4 in the Balz-Schiemann reaction.

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