If H-atom of `3rd` carbon is replaced by - COOH group in pentane - 1,5 -dioic acid. What will be the IUPAC name of the new compound ?

Options

A. 3-Carboxy pentane -1,5 -dioic acid
B. Pentane -1,3,5 -trioicadic
C. Propane -1,2,3 -tricarboxylic acid
D. All are correct IUPAC name

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Understand the structure of pentane-1,5-dioic acid
Pentane-1,5-dioic acid has a five-carbon chain with carboxylic acid groups (-COOH) at the first and fifth positions. The structure can be represented as:

```
COOH-CH2-CH2-CH2-COOH
```

### Step 2: Identify the carbon to be modified
According to the question, we need to replace the hydrogen atom on the third carbon with another carboxylic acid group (-COOH). The third carbon in the chain is the one that is between the second and fourth carbons.

### Step 3: Draw the new structure
After replacing the hydrogen atom on the third carbon with a -COOH group, the new structure will look like this:

```
COOH-CH2-COOH-CH-COOH
```

This can also be written as:

```
COOH-CH-COOH-COOH
```

### Step 4: Determine the parent chain and number the carbons
In this new compound, we have a total of three carboxylic acid groups. The longest carbon chain that includes all the carboxylic acid groups will be a three-carbon chain (propane). The numbering will be done from one end of the chain to the other, giving us:

1. COOH at position 1
2. COOH at position 2
3. COOH at position 3

### Step 5: Write the IUPAC name
The IUPAC name of the new compound, which is a propane with three carboxylic acid groups, is:

**Propane-1,2,3-tricarboxylic acid**

### Final Answer:
The IUPAC name of the new compound is **Propane-1,2,3-tricarboxylic acid**.

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