The `S_(N)^(2)` reactivity order for halides :-

Options

A. `R–F gt R–Cl gt R–Br gt R–I`
B. `R–I gt R–Br gt R–Cl gt R–F`
C. ` R–Br gt R–l gt R–Cl gt R–F`
D. `R–Cl gt R–Br gt R–F gt R–I`

Answer

To determine the order of SN2 reactivity for the halides Rf, RCl, RBr, and Ri, we need to analyze the characteristics of the leaving groups involved in the SN2 mechanism.

### Step-by-Step Solution:

1. **Understanding the SN2 Mechanism**:
- The SN2 (bimolecular nucleophilic substitution) mechanism involves a single step where the nucleophile attacks the carbon atom from the opposite side of the leaving group, resulting in the inversion of configuration at that carbon center.

2. **Identifying the Halides**:
- The halides in question are: Rf (fluoride), RCl (chloride), RBr (bromide), and Ri (iodide).

3. **Evaluating Leaving Group Ability**:
- The reactivity in SN2 reactions is heavily influenced by the ability of the leaving group to depart. A better leaving group will facilitate a faster reaction.
- The order of leaving group ability for halides generally follows the trend: Iodide (I⁻) > Bromide (Br⁻) > Chloride (Cl⁻) > Fluoride (F⁻).
- This trend is due to the bond strength between the carbon atom and the halide. As we move down the group in the periodic table, the size of the halogen increases, leading to longer and weaker bonds.

4. **Applying the Trend to the Given Halides**:
- **Ri (Iodide)**: Best leaving group, therefore most reactive in SN2.
- **RBr (Bromide)**: Good leaving group, less reactive than Ri but more than RCl and Rf.
- **RCl (Chloride)**: Moderate leaving group, less reactive than RBr.
- **Rf (Fluoride)**: Poor leaving group due to strong C-F bond, therefore least reactive in SN2.

5. **Conclusion**:
- Based on the analysis, the order of SN2 reactivity for the halides is:
- **Ri > RBr > RCl > Rf**

### Final Answer:
The order of SN2 reactivity for the halides is:
**Ri > RBr > RCl > Rf**