The addition of `Hbr` to `CH_2 = CHCOOH` results in the formation of.

Options

A. `alpha-`Bromopropanoic acid
B. `beta-`Bromopropanoic acid as the major product
C. None of the above
D. Both (a) and (b) in equal proportion.

Explanation

To solve the problem of what compound is formed when HBr is added to CH₂=CHCOOH (acrylic acid), we can follow these steps:

### Step 1: Identify the Reactants
The reactant is acrylic acid, which has the structure CH₂=CHCOOH. It contains a double bond (C=C) and a carboxylic acid group (COOH).

### Step 2: Understand the Reaction Mechanism
When HBr is added to an alkene, the reaction typically follows Markovnikov's rule. This means that the hydrogen (H) from HBr will attach to the carbon with the most hydrogen atoms already attached, while the bromine (Br) will attach to the carbon with fewer hydrogen atoms.

### Step 3: Formation of Carbocation
In this case, the double bond between the two carbon atoms will break. The addition of HBr leads to the formation of a carbocation. The H+ ion will add to the terminal carbon (the one with two hydrogens), resulting in a more stable secondary carbocation at the other carbon.

### Step 4: Nucleophilic Attack by Bromide Ion
Once the carbocation is formed, the Br- ion (which is negatively charged) will attack the positively charged carbon atom. This will lead to the formation of a new bond between the carbon and bromine.

### Step 5: Final Product Formation
The final product will be a bromo acid. The structure of the product can be represented as CH₂Br-CH₂-COOH. This compound is known as beta-bromo propanoic acid, as the bromine is attached to the beta carbon relative to the carboxylic acid group.

### Conclusion
The addition of HBr to CH₂=CHCOOH results in the formation of beta-bromo propanoic acid.

### Final Answer
The compound formed is **beta-bromo propanoic acid (CH₂Br-CH₂-COOH)**.

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