`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 provided carboxylic acid with bromine in the presence of red phosphorus. This reaction is known as the Hell-Volhard-Zelinsky (HVZ) reaction, which involves the alpha-bromination of carboxylic acids.

### Step-by-Step Solution:

1. **Identify the Structure of the Starting Material**:
The starting material is a carboxylic acid with the structure CH3-CH(CH3)-COOH. This means we have a carboxylic acid with two methyl groups attached to the alpha carbon.

2. **Check for Alpha Hydrogen**:
In the HVZ reaction, the presence of alpha hydrogen is crucial as it allows for the formation of an enol intermediate. An alpha hydrogen is a hydrogen atom attached to the carbon adjacent to the carboxylic acid group. In our case, the alpha carbon (the carbon next to the COOH group) has three methyl groups attached to it (CH3-CH(CH3)-), which means there are no hydrogen atoms available on the alpha carbon.

3. **Conclusion on the Reaction**:
Since there are no alpha hydrogens present in the starting material, the HVZ reaction cannot proceed. Therefore, no bromination will occur, and the reaction will not yield any product.

4. **Final Answer**:
The product of the reaction is "no reaction" or "no product formed" due to the absence of alpha hydrogen.