The final product obtained in the reaction :
`CH_(2)=CHCH_(2)CH_(2)CO_(2)Hoverset(Br_(2)//OH^(o+))rarr`

Options

A. `BrCH_(2)-underset(OH)underset(|)CH-CH_(2)CH_(2)CO_(2)H`
B. `HOCH_(2)-CH_(2)CH_(2)CH_(2)CO_(2)H`
C.
D.

Answer

To solve the question regarding the final product obtained from the reaction of 1-pentenoic acid (CH₂=CHCH₂CH₂CO₂H) with bromine (Br₂) in a basic medium, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactant is 1-pentenoic acid, which has a double bond between the first two carbon atoms (C1 and C2) and a carboxylic acid group (CO₂H) at the end of the chain.

2. **Reaction Conditions**:
The reaction is carried out in the presence of bromine (Br₂) and a basic medium (OH⁻). The basic medium will facilitate the removal of the acidic hydrogen from the carboxylic acid group.

3. **Formation of the Carbanion**:
The base (OH⁻) will deprotonate the carboxylic acid, generating a carbanion at the carbon adjacent to the double bond. This results in a negatively charged oxygen and a positively charged carbocation at the carbon where the double bond was.

4. **Ring Formation**:
The positively charged carbocation can react with the negatively charged oxygen, leading to the formation of a cyclic structure. The oxygen atom will form a bond with the carbocation, resulting in a six-membered ring.

5. **Final Product Structure**:
The final product will be a six-membered ring containing one oxygen atom and five carbon atoms. The structure will also include a carbonyl group (C=O) adjacent to the oxygen atom in the ring.

6. **Identify the Correct Answer**:
After analyzing the options provided, the correct answer corresponds to the structure of the six-membered ring with a carbonyl group, which is found in option D.

### Final Product:
The final product of the reaction is a six-membered cyclic compound with one oxygen atom and a carbonyl group.