In the given reaction `CH_(3)-COOH underset(underset("(iii) "H_(2)O //H^(o+))("(ii) "NaCN))overset("(i) Red "-P+Br_(2))rarr(X)`
'X' will be

Options

A.
B. `COOH-CH_(2)-CH_(2)-COOH`
C.
D.

Answer

To solve the given reaction step by step, we will analyze each part of the reaction involving acetic acid (CH₃COOH) and the reagents provided.

### Step 1: Understanding the Reaction
We start with acetic acid (CH₃COOH). The first step involves treating it with red phosphorus and bromine (Br₂). This is known as the Hell-Volhard-Zelinsky (HVZ) reaction, which is a halogenation reaction at the alpha position of the carboxylic acid.

### Step 2: Halogenation of Acetic Acid
In the presence of red phosphorus and Br₂, the hydrogen atom at the alpha position (the carbon adjacent to the carboxylic acid) is replaced by a bromine atom. Thus, the structure changes from CH₃COOH to:

\[ \text{CH}_2\text{BrCOOH} \]

This compound is known as alpha-bromoacetic acid.

### Step 3: Nucleophilic Substitution with NaCN
Next, we treat the alpha-bromoacetic acid with sodium cyanide (NaCN). In this step, the cyanide ion (CN⁻) acts as a nucleophile and attacks the carbon atom that is bonded to the bromine atom (the electrophilic center). The bromine atom is a good leaving group, allowing the reaction to proceed smoothly.

The product of this reaction will be:

\[ \text{CH}_2\text{CNCOOH} \]

This compound is known as alpha-cyanoacetic acid.

### Step 4: Hydrolysis of the Cyanide Group
The final step involves hydrolysis of the cyanide group in the presence of water (H₂O) and an acid (H⁺). During hydrolysis, the cyanide group (CN) is converted into a carboxylic acid group (COOH). The reaction can be represented as follows:

\[ \text{CH}_2\text{CNCOOH} + H_2O \rightarrow \text{CH}_2\text{COOH} + NH_3 \]

Thus, the final product 'X' after hydrolysis is:

\[ \text{CH}_2\text{COOH} \]

This compound is known as 2-hydroxypropanoic acid or lactic acid.

### Conclusion
The final product 'X' is 2-hydroxypropanoic acid (lactic acid).

### Summary of Steps
1. **Halogenation**: Acetic acid (CH₃COOH) reacts with Br₂ in the presence of red phosphorus to form alpha-bromoacetic acid (CH₂BrCOOH).
2. **Nucleophilic Substitution**: Alpha-bromoacetic acid reacts with NaCN to form alpha-cyanoacetic acid (CH₂CNCOOH).
3. **Hydrolysis**: Alpha-cyanoacetic acid undergoes hydrolysis to yield 2-hydroxypropanoic acid (lactic acid, CH₂COOH).