Which of the following orders regarding the base strength of a leaving group in a reaction of an acid derivative with a nucleophile is correct :

Options

A. `Cl^(o+)gtRCOO^(o+)gtRO^(o+)`
B. `Cl^(o+)gtRO^(o+)gtRCOO^(o+)`
C. `RO^(o+)gtRCOO^(o+)gtCl^(o+)`
D. `RO^(o+)gtCl^(o+)gtRCOO^(o+)`

Answer

To determine the correct order regarding the base strength of a leaving group in a reaction of an acid derivative with a nucleophile, we need to analyze the acidic strength of the conjugate acids corresponding to the leaving groups. Here’s a step-by-step solution:

### Step 1: Identify the Leaving Groups
The leaving groups mentioned in the question are:
- Alkoxide (RO⁻)
- Acetate (RCOO⁻)
- Chloride (Cl⁻)

### Step 2: Determine the Conjugate Acids
For each leaving group, we need to identify its conjugate acid:
- The conjugate acid of alkoxide (RO⁻) is alcohol (R-OH).
- The conjugate acid of acetate (RCOO⁻) is acetic acid (RCOOH).
- The conjugate acid of chloride (Cl⁻) is hydrochloric acid (HCl).

### Step 3: Compare the Acid Strengths
Now we compare the strengths of these acids:
- HCl is a strong acid.
- Acetic acid (RCOOH) is a weak acid.
- Alcohol (R-OH) is weaker than acetic acid.

Thus, the order of acidic strength is:
\[ \text{HCl} > \text{RCOOH} > \text{R-OH} \]

### Step 4: Determine the Base Strengths
The strength of a base is inversely related to the strength of its conjugate acid. Therefore, the weaker the acid, the stronger the conjugate base:
- Cl⁻ is the conjugate base of a strong acid (HCl), so it is the weakest base.
- RCOO⁻ is the conjugate base of a weak acid (RCOOH), so it is a stronger base than Cl⁻.
- RO⁻ is the conjugate base of an even weaker acid (R-OH), making it the strongest base.

### Step 5: Establish the Order of Base Strength
From the above analysis, we can establish the order of base strength:
\[ \text{RO}^- > \text{RCOO}^- > \text{Cl}^- \]

### Conclusion
The correct order regarding the base strength of the leaving groups is:
\[ \text{RO}^- > \text{RCOO}^- > \text{Cl}^- \]

### Final Answer
The correct option is **C: RO⁻ > RCOO⁻ > Cl⁻**.

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