The treatment of `CH_(3)CH_(2)COOH` with chlorine in the
presence of phosphorus gives:

Options

A. `CH_(3)CH_(2)COCH_(3)`
B. `CH_(3)CH_(2)CH_(2)Cl`
C. `CH_(3)CH(Cl)COOH`
D. `ClCH_(2)CH_(2)COOH`

Answer

To solve the question regarding the treatment of `CH3CH2COOH` (propanoic acid) with chlorine in the presence of phosphorus, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactant is `CH3CH2COOH`, which is propanoic acid (a carboxylic acid).
- The other reactants are chlorine (`Cl2`) and phosphorus (often used in the form of red phosphorus).

2. **Understand the Reaction Type**:
- The reaction is a type of halogenation known as the Hell-Volhard-Zelinsky (HVZ) reaction. This reaction involves the halogenation of the alpha carbon (the carbon adjacent to the carboxylic acid group) in carboxylic acids.

3. **Mechanism of the HVZ Reaction**:
- In the presence of phosphorus, chlorine will react with the alpha hydrogen of the carboxylic acid.
- The phosphorus helps facilitate the reaction by forming a phosphorus halide, which can then react with the carboxylic acid.

4. **Identify the Alpha Hydrogen**:
- In propanoic acid, the alpha carbon is the carbon next to the carboxylic acid group (the `CH2` group in `CH3CH2COOH`).
- The alpha hydrogen is the hydrogen attached to this alpha carbon.

5. **Halogenation Process**:
- One of the alpha hydrogens is replaced by a chlorine atom from `Cl2`.
- This results in the formation of `CH3CHClCOOH`, where one hydrogen atom from the alpha carbon has been substituted with chlorine.

6. **Final Product**:
- The final product of the reaction is `CH3CHClCOOH`, which is alpha-chloropropanoic acid.

### Conclusion:
The treatment of `CH3CH2COOH` with chlorine in the presence of phosphorus gives `CH3CHClCOOH` (alpha-chloropropanoic acid).