Which of the following will be obtained when acetic acid is subjected
to HVZ reaction ?

Options

A. `BrCH_(2)COOH`
B. `Br-overset(Br)overset(|)CH-COOH`
C. `Br-overset(Br)overset(|)underset(Br)underset(|)C-COOH`
D. All of these

Explanation

To solve the question regarding the products obtained when acetic acid is subjected to the HVZ (Hell-Volhard-Zelinsky) reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the HVZ Reaction**: The HVZ reaction involves the halogenation of carboxylic acids at the alpha position (the carbon adjacent to the carboxyl group). In this reaction, the alpha hydrogen atoms are replaced by halogen atoms (either chlorine or bromine).

2. **Identify the Structure of Acetic Acid**: Acetic acid has the formula CH₃COOH. The structure can be represented as:
\[ \text{CH}_3-\text{C(=O)}-\text{OH} \]
Here, the alpha carbon is the one bonded to the methyl group (CH₃).

3. **Determine the Number of Alpha Hydrogens**: In acetic acid, there are three alpha hydrogens (the three hydrogens in the CH₃ group).

4. **Apply the HVZ Reaction**: When acetic acid is treated with bromine (Br₂) in the presence of red phosphorus, the alpha hydrogens will be replaced by bromine atoms. The reaction can be represented as:
\[ \text{CH}_3-\text{C(=O)}-\text{OH} \xrightarrow{\text{Br}_2, \text{red P}} \text{BrCH}_2-\text{C(=O)}-\text{OH} + \text{HBr} \]
This shows the formation of alpha-bromoacetic acid.

5. **Consider Multiple Halogenation**: Since there are three alpha hydrogens in acetic acid, if excess bromine is used, all three alpha hydrogens can be replaced:
\[ \text{CH}_3-\text{C(=O)}-\text{OH} \xrightarrow{\text{Br}_2, \text{red P}} \text{Br}_3\text{C}-\text{C(=O)}-\text{OH} + 3\text{HBr} \]
This results in tribromoacetic acid.

6. **Conclusion**: Therefore, when acetic acid is subjected to the HVZ reaction with bromine, we can obtain:
- Monobromoacetic acid (BrCH₂COOH)
- Dibromoacetic acid (Br₂CHCOOH)
- Tribromoacetic acid (Br₃CCOOH)

### Final Answer:
All of the above products (monobromoacetic acid, dibromoacetic acid, and tribromoacetic acid) can be formed, so the answer is "all of these."

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