in the reaction
` CH_3 CH_2 COOH overset(P,CI_2) to X overset(KCN ) to Y ` Y is

Options

A. `CH_3 CH_2 CONH_2`
B. `CH_3CH (CN ) COOH `
C. `CH_2 (CN ) CH_2 COOH `
D. `CH_2(CN ) COOH `.

Explanation

To solve the problem step by step, let's break down the reactions involved:

### Step 1: Identify the starting compound
The starting compound is propanoic acid, which has the molecular formula:
\[ \text{CH}_3\text{CH}_2\text{COOH} \]

### Step 2: Reaction with phosphorus and chlorine
When propanoic acid reacts with phosphorus (P) and chlorine (Cl2), it undergoes halogenation at the alpha-carbon (the carbon adjacent to the carboxylic acid group). This reaction introduces a chlorine atom at the alpha position, resulting in the formation of 2-chloropropanoic acid.

The reaction can be represented as:
\[ \text{CH}_3\text{CH}_2\text{COOH} \overset{P, Cl_2}{\rightarrow} \text{CH}_3\text{CHClCOOH} \]

### Step 3: Identify the intermediate product (X)
The product formed after this reaction is:
\[ \text{X} = \text{CH}_3\text{CHClCOOH} \] (2-chloropropanoic acid)

### Step 4: Reaction with potassium cyanide (KCN)
Next, the product X (2-chloropropanoic acid) reacts with potassium cyanide (KCN). In this reaction, the cyanide ion (CN⁻) acts as a nucleophile and attacks the alpha-carbon where the chlorine is attached. Since chlorine is a good leaving group, it will leave, and the cyanide will take its place.

The reaction can be represented as:
\[ \text{CH}_3\text{CHClCOOH} + \text{KCN} \rightarrow \text{CH}_3\text{CH(CN)COOH} + \text{KCl} \]

### Step 5: Identify the final product (Y)
The final product formed after this reaction is:
\[ \text{Y} = \text{CH}_3\text{CH(CN)COOH} \] (2-cyano-propanoic acid)

### Conclusion
Thus, the final answer for Y is:
\[ \text{Y} = \text{2-cyano-propanoic acid} \]

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