The IUPAC name of the compound
` CH_3 -CH_2 - overset(O ) overset(||)C-CH_2 -COOH ` is

Options

A. 2-Oxo-but-3-enal
B. 3-Oxo-1-formylpent-4-oic acid
C. 1-Formylpent-4-en-3-one
D. 3-Oxopentanoic acid

Answer

To determine the IUPAC name of the compound `CH₃ - CH₂ - C(=O) - CH₂ - COOH`, we will follow these steps:

### Step 1: Identify the Functional Groups
The compound contains two functional groups:
- A carboxylic acid group (–COOH)
- A ketone group (C=O)

### Step 2: Determine the Main Functional Group
In IUPAC nomenclature, the carboxylic acid group has a higher priority than the ketone group. Therefore, we will consider the carboxylic acid as the main functional group.

### Step 3: Number the Carbon Chain
Since the carboxylic acid is a terminal group, we will start numbering the carbon chain from the carboxylic acid end. The numbering will be as follows:
1. COOH (1)
2. CH₂ (2)
3. C(=O) (3)
4. CH₂ (4)
5. CH₃ (5)

This gives us a total of 5 carbon atoms in the chain.

### Step 4: Name the Carbon Chain
With 5 carbon atoms, the base name of the compound is "pentane".

### Step 5: Identify the Position of the Ketone Group
The ketone group (C=O) is located at the 3rd carbon in the chain. In IUPAC nomenclature, we denote the presence of a ketone group using the prefix "oxo". Since the ketone is at the 3rd position, we will write "3-oxo".

### Step 6: Combine the Names
Now, we combine the names according to IUPAC rules. The suffix for the carboxylic acid is "oic acid". However, since we have an "o" from "oxo" and an "e" from "pentane", we drop the "e" from "pentane" to avoid having two vowels together.

Thus, we get:
- Base name: "pentane"
- Suffix: "oic acid"
- Prefix: "3-oxo"

Putting it all together, we have:
**3-oxo-pentanoic acid**

### Final Answer
The IUPAC name of the compound `CH₃ - CH₂ - C(=O) - CH₂ - COOH` is **3-oxo-pentanoic acid**.

---