Which of the following substrate is most reactive towards methoxide ion `(Me- overset(Theta)(O))`?

Options

A. `CH_(3)- I`
B.
C. `CH_(3) -O - SO_(2) -CF_(3)`
D. `CH_(3) -F`

Answer

To determine which substrate is most reactive towards the methoxide ion (CH₃O⁻), we need to analyze the four given substrates based on their ability to undergo nucleophilic substitution reactions. Here’s a step-by-step solution:

### Step 1: Identify the Nature of the Nucleophile
The methoxide ion (CH₃O⁻) is a strong base and a good nucleophile. It can participate in nucleophilic substitution reactions (SN2) where it attacks an electrophilic carbon atom, leading to the displacement of a leaving group.

**Hint:** Remember that strong nucleophiles favor substitution reactions, especially with good leaving groups.

### Step 2: Analyze the Substrates
We have four substrates to consider:
1. **Methyl iodide (CH₃I)**
2. **Methyl tosylate (CH₃OSO₂)**
3. **Methyl triflate (CH₃OSO₂CF₃)**
4. **Methyl fluoride (CH₃F)**

### Step 3: Evaluate the Leaving Groups
The effectiveness of a nucleophilic substitution reaction often depends on the stability of the leaving group. The better the leaving group, the more reactive the substrate will be. The leaving groups in our substrates are:
- Iodide (I⁻) from methyl iodide
- Tosylate (OSO₂) from methyl tosylate
- Triflate (OSO₂CF₃) from methyl triflate
- Fluoride (F⁻) from methyl fluoride

**Hint:** Compare the stability of the leaving groups. Iodide is a good leaving group, while fluoride is a poor leaving group due to its strong bond with carbon.

### Step 4: Compare the Substrates
1. **Methyl iodide (CH₃I)**: Iodide is a very good leaving group, making this substrate quite reactive.
2. **Methyl tosylate (CH₃OSO₂)**: Tosylate is also a good leaving group, but not as good as iodide.
3. **Methyl triflate (CH₃OSO₂CF₃)**: Triflate is an excellent leaving group due to resonance stabilization and the electron-withdrawing effect of the CF₃ group, making this substrate very reactive.
4. **Methyl fluoride (CH₃F)**: Fluoride is a poor leaving group, making this substrate the least reactive.

**Hint:** Consider the resonance and inductive effects that stabilize the leaving groups.

### Step 5: Conclusion
Among the four substrates, methyl triflate (CH₃OSO₂CF₃) has the best leaving group due to the combined effects of resonance stabilization and the strong electron-withdrawing nature of the trifluoromethyl group. Therefore, it is the most reactive towards the methoxide ion.

**Final Answer:** The substrate that is most reactive towards methoxide ion (CH₃O⁻) is **C. Methyl triflate (CH₃OSO₂CF₃)**.