`CH_3CH_2 COOH overset(Cl_2//Fe)rarr (X) overset(Alc.) underset(KOH)rarr (Y)`
Compound `(Y)` is :

Options

A. `CH_3CH_2 OH`
B. `CH_3 CH_2 CN`
C. `CH_2=CHCOOH`
D. `CH_3 CHClCOOH`

Explanation

To solve the question, we will follow the reactions step by step.

### Step 1: Identify the starting compound
The starting compound is butanoic acid, represented as \( CH_3CH_2COOH \).

### Step 2: Reaction with \( Cl_2 \) in the presence of \( Fe \)
When butanoic acid reacts with chlorine (\( Cl_2 \)) in the presence of iron (\( Fe \)), it undergoes the Hell-Volhard-Zelinsky (HVZ) reaction. In this reaction, the hydrogen atom at the alpha carbon (the carbon adjacent to the carboxylic acid group) is replaced by a chlorine atom.

- The alpha carbon in butanoic acid is the second carbon (the one attached to the carboxylic acid).
- Therefore, the reaction will yield \( CH_3CHClCOOH \).

This compound \( (X) \) is 2-chlorobutanoic acid.

### Step 3: Reaction with alcoholic \( KOH \)
Next, compound \( (X) \) reacts with alcoholic potassium hydroxide (\( KOH \)). In this reaction, the \( HCl \) that was formed during the HVZ reaction is eliminated, leading to the formation of a double bond.

- The elimination of \( HCl \) results in the formation of a double bond between the alpha carbon and the beta carbon.
- Thus, the product \( (Y) \) will be \( CH_2=CHCOOH \), which is acrylic acid.

### Conclusion
The compound \( (Y) \) is acrylic acid, represented as \( CH_2=CHCOOH \).

### Final Answer
Compound \( (Y) \) is acrylic acid.

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