The IUPAC name of `CH_(3)-underset(overset(|)COOC_(2)H_(5))(C=CH)-CH_(2)-underset(OH)underset(|)overset(O)overset(||)C` is :

Options

A. 4-ethoxycarbonylpent-3-enoic acid
B. 4-ethanoyloxpent-3-enoic acid
C. 3-ethoxycarbonylbut-2-enecarboxylic acid
D. 3-ethoxycarbonylpent-3-enoic acid

Explanation

To determine the IUPAC name of the compound `CH₃-CH=CH-CH₂-COOH-COOC₂H₅`, we will follow these steps:

### Step 1: Identify the longest carbon chain
The longest carbon chain in the compound includes the carboxylic acid and the ester functional group. Here, we have a total of 6 carbon atoms in the longest chain.

### Step 2: Identify the functional groups
The compound contains:
- A carboxylic acid group (-COOH)
- An ester group (-COOC₂H₅)
- An alkene (C=C) between the second and third carbon atoms.

### Step 3: Number the carbon chain
Number the carbon chain from the end closest to the carboxylic acid to give it the lowest possible number:
1. COOH (carboxylic acid) is on carbon 1.
2. The double bond (alkene) is between carbon 2 and carbon 3.
3. The ester group is attached to carbon 4.

### Step 4: Identify substituents
The substituents include:
- An ethoxy group (-O-C₂H₅) attached to carbon 4.
- The double bond indicates that the compound is an alkene.

### Step 5: Combine the names
- The base name for a 6-carbon carboxylic acid is "hexanoic acid."
- The presence of the double bond (alkene) at carbon 2 modifies the name to "hex-2-enoic acid."
- The ethoxy group at carbon 4 is named as "4-ethoxy."

### Final IUPAC Name
Putting it all together, the IUPAC name of the compound is **4-ethoxy-hex-2-enoic acid**.

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