`CH_2=C(CH_2)_2COOH` and HBr reacts on presence of peroxide to give

Options

A. `BrCH_2CH_2(CH_2)_2COOH`
B. `CH_3CH(CH_2)_2COOH`
C. `CH_3CHCH_2(CH_2)_2COOH`
D. `CH_3CHBr(CH_2)_2COOH`

Answer

To solve the problem of the reaction of the compound \( CH_2=C(CH_2)_2COOH \) with HBr in the presence of peroxide, we can follow these steps:

### Step 1: Identify the Structure of the Compound
The given compound is \( CH_2=C(CH_2)_2COOH \). This can be represented as:

- A double bond between the first carbon \( CH_2 \) and the second carbon \( C \) which is attached to two \( CH_2 \) groups and a carboxylic acid \( COOH \).

### Step 2: Understand the Role of Peroxide
In the presence of peroxide, the reaction follows a free radical mechanism. Peroxide causes homolytic cleavage of HBr, leading to the formation of free radicals.

### Step 3: Homolytic Cleavage of HBr
When HBr undergoes homolytic cleavage, it produces:
- A bromine radical \( Br^\cdot \)
- A hydrogen radical \( H^\cdot \)

### Step 4: Free Radical Addition to the Alkene
The alkene \( CH_2=C(CH_2)_2COOH \) will react with the bromine radical. According to the Anti-Markovnikov rule, the bromine radical will add to the carbon that has more hydrogens.

### Step 5: Determine the Product
The bromine radical will add to the terminal carbon \( CH_2 \) of the double bond, resulting in the formation of the product:

- The double bond will break, and we will have \( CH_2Br \) at one end and the rest of the molecule will remain intact.

The resulting structure will be:
\[ CH_2Br-CH(CH_2)_2-COOH \]

### Step 6: Final Product
The final product after the reaction will be:
\[ CH_2Br-CH(CH_2)_2-COOH \]

### Summary of the Reaction
The reaction of \( CH_2=C(CH_2)_2COOH \) with HBr in the presence of peroxide leads to the formation of \( CH_2Br-CH(CH_2)_2-COOH \).

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