`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 problem, we need to follow the steps of the reactions involving propanoic acid (CH₃CH₂COOH) with red phosphorus and hydrogen iodide (Red P/HI), followed by treatment with alcoholic potassium hydroxide (alc. KOH).

### Step 1: Reaction with Red Phosphorus and Hydrogen Iodide
- **Starting Material**: Propanoic acid (CH₃CH₂COOH)
- **Reagent**: Red phosphorus and hydrogen iodide (Red P/HI)

In this step, propanoic acid undergoes a reaction known as the Hell-Volhard-Zelinsky (HVZ) reaction. This reaction introduces an iodine atom at the alpha position (the carbon adjacent to the carboxylic acid group).

**Reaction**:
\[ \text{CH}_3\text{CH}_2\text{COOH} + \text{Red P/HI} \rightarrow \text{CH}_3\text{CH}(\text{I})\text{COOH} \]

This results in the formation of alpha-iodopropanoic acid (CH₃CH(I)COOH).

### Step 2: Reaction with Alcoholic Potassium Hydroxide
- **Intermediate Product**: Alpha-iodopropanoic acid (CH₃CH(I)COOH)
- **Reagent**: Alcoholic potassium hydroxide (alc. KOH)

In this step, the alcoholic KOH acts as a base and will deprotonate the alpha carbon, leading to the elimination of the iodine atom (leaving group) and forming a double bond between the alpha and beta carbons.

**Reaction**:
1. Deprotonation occurs at the alpha carbon.
2. The bond shifts to form a double bond, and iodine is eliminated.

**Final Product**:
\[ \text{CH}_3\text{CH}=\text{CHCOOH} \]

This product is known as acrylic acid (or prop-2-enoic acid).

### Conclusion
The final product obtained after the two-step reaction is:
\[ \text{CH}_2=\text{CHCOOH} \]

### Final Answer
The product is **CH₂=CHCOOH** (acrylic acid).

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