The IUPAC name of `HOOC-CH=CH-COOH` is:

Options

A. But-2-ene-1,4-dicarboxylic acid
B. But-2-ene-1,4-dioic acid
C. Ethene dicarboxylic acid
D. Ethene dioic acid

Answer

To determine the IUPAC name of the compound `HOOC-CH=CH-COOH`, we can follow these steps:

### Step 1: Identify the structure
The compound consists of a straight chain of four carbon atoms with two carboxylic acid groups (–COOH) and a double bond between the second and third carbon atoms. The structure can be represented as:
```
HOOC-CH=CH-COOH
```

### Step 2: Number the carbon chain
Since the compound is symmetric, we can start numbering from either end. However, we will number from the end closest to the first substituent (the carboxylic acid group). Thus, we can number the carbon atoms as follows:
1. C1: COOH
2. C2: CH (part of the double bond)
3. C3: CH (part of the double bond)
4. C4: COOH

### Step 3: Identify functional groups and their positions
- The carboxylic acid groups (–COOH) are located at positions 1 and 4.
- The double bond is located between carbon atoms 2 and 3.

### Step 4: Name the compound
According to IUPAC nomenclature:
- The base name for a four-carbon alkene is "butene".
- Since there are two carboxylic acid groups, we use the suffix "dioic acid".
- The position of the double bond is indicated by the number of the first carbon involved in the double bond, which is 2.

Putting this all together, the IUPAC name of the compound is:
**But-2-enedioic acid** (also known as fumaric acid).

### Final Answer:
The IUPAC name of `HOOC-CH=CH-COOH` is **But-2-enedioic acid**.

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