`"CFCl"_(3)` and HF on reaction in presence of `" SbF"_(5)` leads to the formation of:

Options

A. Freon -112
B. Freon - 114
C. freon -11
D. Freon -12

Answer

To solve the question regarding the reaction of CFCl₃ (trichlorofluoromethane) and HF in the presence of SbF₅ (antimony pentafluoride), we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- CFCl₃ (trichlorofluoromethane)
- HF (hydrofluoric acid)
- SbF₅ (antimony pentafluoride, which acts as a Lewis acid)

### Step 2: Understand the Role of SbF₅
SbF₅ is a strong Lewis acid. In this reaction, it will facilitate the process by accepting a pair of electrons, which helps in the generation of a more reactive species. This will assist in the substitution reaction.

### Step 3: Mechanism of Reaction
1. **Formation of a Carbocation**: The Lewis acid (SbF₅) interacts with CFCl₃, which can lead to the formation of a carbocation intermediate. This occurs as one of the chlorine atoms is protonated, making it a better leaving group.

2. **Nucleophilic Attack by HF**: The fluoride ion (F⁻) from HF acts as a nucleophile and attacks the carbocation formed in the previous step.

3. **Elimination of Chlorine**: As the fluoride ion attacks, one of the chlorine atoms (Cl) leaves, resulting in the formation of a new product.

### Step 4: Identify the Product
The product formed from this reaction is CF₂Cl₂ (dichlorodifluoromethane). The structure can be represented as:
- Carbon (C) with two fluorine (F) atoms and two chlorine (Cl) atoms.

### Conclusion
The final product of the reaction between CFCl₃ and HF in the presence of SbF₅ is CF₂Cl₂.

### Final Answer
The product formed is CF₂Cl₂ (dichlorodifluoromethane).

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