Fluorobenzene is prepared conveniently

Options

A. by heating benzene diazonium tetrafluoroborate
B. by treating benzene with fluorine
C. by the action of phenol with `SF_4`
D. by treating benzene with `CH_3F` in presence of anhydrous `AlCl_3`

Answer

To prepare fluorobenzene conveniently, we can follow these steps:

### Step-by-Step Solution:

1. **Start with Benzene Diazonium Chloride**:
- We begin with benzene diazonium chloride, which has the formula \( C_6H_5N_2Cl \). This compound is generated from an aromatic amine (like aniline) through a diazotization reaction with nitrous acid (\( HNO_2 \)).

2. **React with Tetrafluoroboric Acid**:
- Next, we treat the benzene diazonium chloride with tetrafluoroboric acid (\( HBF_4 \)). This reaction produces benzene diazonium tetrafluoroborate, represented as \( C_6H_5N_2BF_4 \), along with hydrochloric acid (\( HCl \)).

3. **Heat the Benzene Diazonium Tetrafluoroborate**:
- The obtained benzene diazonium tetrafluoroborate is then subjected to heating. Upon heating, this compound decomposes to yield fluorobenzene (\( C_6H_5F \)), nitrogen gas (\( N_2 \)), and boron trifluoride (\( BF_3 \)).

4. **Final Product**:
- The final product of this reaction is fluorobenzene, which is the desired compound.

### Summary of Reaction:
- The overall reaction can be summarized as follows:
\[ C_6H_5N_2Cl + HBF_4 \xrightarrow{\text{heat}} C_6H_5F + N_2 + BF_3 + HCl \]

### Conclusion:
- Therefore, the most convenient method to prepare fluorobenzene is by treating benzene diazonium chloride with tetrafluoroboric acid and then heating the resulting benzene diazonium tetrafluoroborate.

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