`CH_(3) CH_(2) COOH overset (Cl_(2)) underset (Red P) to (A) overset (Alc. KOH) to (B)`
What is (B)?

Options

A. `CH_(3)underset(OH)CHCOOH`
B. `underset(OH)underset(|)CH_(2)CH_(2)COOH`
C. `CH_(2)=CHCOOH`
D. `underset(Cl)underset(|)CH_(2)underset(OH)underset(|)CH-COOH`

Answer

To solve the question step by step, we will analyze the reactions involving the compound CH₃CH₂COOH (propanoic acid) with chlorine in the presence of red phosphorus, followed by treatment with alcoholic KOH.

### Step 1: Identify the Reaction
The first step involves the reaction of propanoic acid (CH₃CH₂COOH) with chlorine (Cl₂) in the presence of red phosphorus (Red P). This is known as the Hell-Volhard-Zelinsky (HVZ) reaction, which is used to halogenate the alpha carbon of carboxylic acids.

### Step 2: Perform the HVZ Reaction
In the HVZ reaction, the alpha hydrogen (the hydrogen atom attached to the carbon adjacent to the carboxylic acid group) is replaced by a chlorine atom. The reaction can be summarized as follows:
- **Starting Material:** CH₃CH₂COOH
- **Reagents:** Cl₂, Red P
- **Product (A):** CH₃CHClCOOH (alpha-chloro propanoic acid)

### Step 3: Elimination Reaction with Alcoholic KOH
The next step involves treating the product (A) with alcoholic KOH. This step typically leads to an elimination reaction where the alpha-chloro group is removed along with the carboxylic acid group, resulting in the formation of an alkene.

The elimination reaction can be summarized as follows:
- **Starting Material (A):** CH₃CHClCOOH
- **Reagents:** Alcoholic KOH
- **Product (B):** CH₂=CHCOOH (acrylic acid)

### Final Answer
Thus, the final product (B) is **CH₂=CHCOOH**, which is acrylic acid.

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