`CH_(3)-overset(CH_(3))overset("| ")underset(CH_(3))underset("| ")"C "-COOH+Br_(2)overset("Red P")rarr"[Product]"`
The product of the above reaction is -

Options

A. `beta-` Dibromo acid
B. `beta, beta'-` Dibromo acid
C. `beta, beta', beta''- ` Tribormo acid
D. No reaction takes place

Answer

To solve the given question, we need to analyze the reaction of the carboxylic acid with bromine (Br₂) in the presence of red phosphorus (Red P).

**Step 1: Identify the structure of the carboxylic acid.**
The carboxylic acid provided is:
\[ \text{CH}_3-\overset{\text{CH}_3}{|}\underset{\text{CH}_3}{|}\text{C}-\text{COOH} \]
This is 2-methylbutanoic acid, which has the following structure:
```
CH3
|
CH3-C-COOH
|
CH2
```

**Step 2: Understand the reaction conditions.**
The reaction involves Br₂ and red phosphorus. This combination is known to facilitate the formation of alpha-bromo compounds through the Heil-Wollard-Zelensky (HWZ) reaction.

**Step 3: Analyze the presence of alpha hydrogen.**
For the HWZ reaction to occur, there must be alpha hydrogens present on the carbon adjacent to the carboxylic acid group (the carbon that is directly attached to the carboxylic acid). In this case, the alpha carbon is the one attached to the carboxylic acid (COOH), and we need to check if it has any hydrogen atoms.

**Step 4: Determine the presence of alpha hydrogen.**
The structure of the carboxylic acid shows that the alpha carbon (the carbon next to COOH) is bonded to two methyl groups (CH₃) and a hydrogen atom. Therefore, it has alpha hydrogens.

**Step 5: Predict the product.**
Since there are alpha hydrogens present, we can expect that one of these hydrogens will be replaced by a bromine atom (Br) due to the HWZ reaction. The product will be:
\[ \text{CH}_3-\overset{\text{CH}_3}{|}\underset{\text{CH}_3}{|}\text{C}-\text{COOH} \xrightarrow{\text{Br}_2/\text{Red P}} \text{CH}_3-\overset{\text{CH}_3}{|}\underset{\text{CH}_3}{|}\text{CBr}-\text{COOH} \]
This indicates that the product will be 2-bromo-2-methylbutanoic acid.

**Final Answer:**
The product of the reaction is 2-bromo-2-methylbutanoic acid.

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