Arrange in the order of decreasing `K_(a).`
`F-CH_(2)CH_(2)COOH`
`Cl-underset(Cl)underset(|)CH-CH_(2)-COOH`
`F-CH_(2)-COOH`
`Br-CH_(2)-CH_(2)-COOH`
Correct answer is :

Options

A. `QgtSgtPgtR`
B. `PgtRgtSgtQ`
C. `RgtQgtPgtS`
D. `SgtQgtPgtR`

Answer

To arrange the given compounds in the order of decreasing \( K_a \) (which indicates acidity), we need to analyze the effects of the substituents on the acidity of the carboxylic acids. The acidity of a carboxylic acid is influenced by the presence of electron-withdrawing groups (EWGs) and their electronegativity. The more electronegative the substituent, the stronger its electron-withdrawing effect, which increases the acidity.

### Step-by-Step Solution:

1. **Identify the Compounds:**
- \( \text{F-CH}_2\text{CH}_2\text{COOH} \) (Compound P)
- \( \text{Cl-CH}(\text{Cl})\text{CH}_2\text{COOH} \) (Compound Q)
- \( \text{F-CH}_2\text{COOH} \) (Compound R)
- \( \text{Br-CH}_2\text{CH}_2\text{COOH} \) (Compound S)

2. **Analyze the Electron-Withdrawing Effects:**
- **Fluorine (F)** is the most electronegative element, followed by **Chlorine (Cl)**, and then **Bromine (Br)**.
- The presence of these halogens will increase the acidity of the carboxylic acids due to their electron-withdrawing inductive effect.

3. **Consider the Position of Substituents:**
- In compound P, the fluorine is two carbons away from the carboxylic acid group.
- In compound Q, there are two chlorines attached to the same carbon as the carboxylic acid, which will have a stronger effect than a single fluorine.
- In compound R, the fluorine is directly attached to the carbon adjacent to the carboxylic acid, which will enhance its acidity.
- In compound S, bromine is present, which is less effective than fluorine and chlorine.

4. **Rank the Compounds Based on Acidity:**
- Compound R (F-CH₂-COOH) will be the most acidic due to the strong electron-withdrawing effect of fluorine being close to the carboxylic group.
- Compound Q (Cl-CH(Cl)-CH₂-COOH) will be next because of the presence of two chlorines, which are effective EWGs.
- Compound P (F-CH₂CH₂-COOH) will follow as it has one fluorine but is further away from the carboxylic group.
- Compound S (Br-CH₂CH₂-COOH) will be the least acidic due to the presence of bromine, which is a weaker EWG compared to fluorine and chlorine.

5. **Final Order of Decreasing \( K_a \):**
- R > Q > P > S

### Conclusion:
The correct order of decreasing \( K_a \) is:
**R (F-CH₂-COOH) > Q (Cl-CH(Cl)-CH₂-COOH) > P (F-CH₂CH₂-COOH) > S (Br-CH₂CH₂-COOH)**.