Which one of the following compounds is most acidic

Options

A. `Cl-CH_(2)-CH_(2)-OH`
B.
C.
D.

Explanation

To determine which compound is the most acidic among the given options, we can follow these steps:

### Step 1: Understand Acidity
Acidity is determined by how easily a compound can donate a proton (H⁺). A more stable conjugate base (the anion formed after losing H⁺) indicates a stronger acid.

### Step 2: Analyze the Compounds
We need to analyze the substituents on each compound and how they affect the stability of the conjugate base.

### Step 3: Evaluate Compound A
- **Substituent:** Chlorine (Cl)
- **Effect:** Chlorine has a -I (inductive) effect, which can stabilize the negative charge somewhat but is not as effective as resonance stabilization.

### Step 4: Evaluate Compound B
- **Substituent:** An electron-withdrawing group (e.g., NO₂)
- **Effect:** The electron-withdrawing group will stabilize the negative charge on the conjugate base through resonance, making it more stable and thus more acidic.

### Step 5: Evaluate Compound C
- **Substituent:** A group that can stabilize the conjugate base through resonance.
- **Effect:** The resonance stabilization of the anion will make this compound more acidic than Compound A but less than Compound B.

### Step 6: Evaluate Compound D
- **Substituent:** An electron-donating group.
- **Effect:** This group will destabilize the conjugate base by increasing electron density, making it less acidic.

### Step 7: Compare Acidity
- **Ranking of Acidity:**
- Compound B (most acidic due to strong resonance stabilization)
- Compound C (next most acidic due to resonance)
- Compound A (less acidic due to -I effect)
- Compound D (least acidic due to electron-donating effect)

### Conclusion
Based on the analysis, **Compound B** is the most acidic because it has the strongest electron-withdrawing group that stabilizes the conjugate base effectively.

### Final Answer
**B part is the most acidic.**

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