`CH_(3)CH_(2)COOH overset("Red P/HI")rarr` is `overset("alc. KOH")rarr "Product"`. Product

Options

A. `CH_(2)=CHCOOH`
B. `CH_(2)CH_(2)OH`
C. `CH_(3)CH_(2)CN`
D. `CH_(2)CHClCOOH`

Answer

To solve the question regarding the reaction of propanoic acid (CH₃CH₂COOH) with red phosphorus and HI, followed by treatment with alcoholic KOH, we can follow these steps:

### Step 1: Identify the Reaction
The first part of the reaction involves the treatment of propanoic acid (CH₃CH₂COOH) with red phosphorus (P) and hydroiodic acid (HI). This is known as the Hell-Volhard-Zelinsky (HVZ) reaction, which involves the halogenation of the alpha carbon of carboxylic acids.

### Step 2: Halogenation of the Alpha Carbon
In the HVZ reaction, the iodine (I) from HI will be added to the alpha carbon (the carbon adjacent to the carboxylic acid group). Therefore, the structure of the intermediate product after this step will be:

\[ \text{CH}_3\text{CH}(\text{I})\text{COOH} \]

### Step 3: Reaction with Alcoholic KOH
Next, we treat the intermediate product with alcoholic KOH. In this step, the strong base (OH⁻) will facilitate an elimination reaction (E2 mechanism). The base will remove a proton from the beta carbon (the carbon next to the carbon with the iodine), leading to the elimination of the iodine atom.

### Step 4: Final Product Formation
After the elimination reaction, the final product will be:

\[ \text{CH}_3\text{CH}=\text{CHCOOH} \]

This product is 2-propenoic acid (also known as acrylic acid).

### Final Answer
The product formed after the complete reaction is:

\[ \text{CH}_2=\text{CHCOOH} \]