The compound B is :
`CH_(3)CH_(2)COOH overset(Cl_(2))underset(red P)rarrAoverset(Alc. KOH)rarrB`

Options

A. `CH_(3)CH_(2)COCl`
B. `CH_(3)CH_(2)CHO`
C. `ClCH_(2)CH_(2)COOH`
D. `CH_(2)=CHCOOH`

Answer

To determine the compound B formed in the given reaction sequence, we will analyze each step carefully.

### Step 1: Identify the starting compound
The starting compound is propanoic acid, represented as:
\[ \text{CH}_3\text{CH}_2\text{COOH} \]

### Step 2: Reaction with Cl2 and red phosphorus
When propanoic acid reacts with chlorine (Cl2) in the presence of red phosphorus, this is known as the Hell-Volhard-Zelinsky (HVZ) reaction. In this reaction, chlorine replaces one of the hydrogen atoms on the alpha carbon (the carbon adjacent to the carboxylic acid group).

The structure of the compound after this reaction (let's call it compound A) will be:
\[ \text{CH}_3\text{CH}(\text{Cl})\text{COOH} \]
This indicates that one hydrogen atom from the alpha carbon (the second carbon in the chain) has been replaced by a chlorine atom.

### Step 3: Reaction with alcoholic KOH
Next, compound A is treated with alcoholic KOH. This leads to a beta elimination reaction, where a hydrogen atom from the beta carbon (the carbon next to the alpha carbon) and the chlorine atom from the alpha carbon are eliminated.

In this case:
- The alpha carbon is the one with the chlorine (C2).
- The beta carbon is the one next to it (C3).

During the elimination, the hydrogen from the beta carbon (C3) and the chlorine from the alpha carbon (C2) are removed, resulting in the formation of a double bond between these two carbons.

The resulting compound (compound B) will be:
\[ \text{CH}_2=\text{CH}\text{COOH} \]
This represents an alkene with a carboxylic acid group, specifically, it is acrylic acid.

### Conclusion
Thus, the compound B formed in the reaction sequence is:
\[ \text{CH}_2=\text{CH}\text{COOH} \]

### Final Answer
The correct answer is the fourth option, which represents the compound B.

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