The IUPAC name for the compound
`HC -= C - CH_(2) - CH_(2) - COOH` is

Options

A. 4-Pentyn-1-oic acid
B. 1-Pentyn-4-oic acid
C. Pentyn-1-oic acid
D. None of the above

Answer

To determine the IUPAC name for the compound `HC ≡ C - CH₂ - CH₂ - COOH`, we will follow these steps:

### Step 1: Identify the Longest Carbon Chain
The compound has a total of 5 carbon atoms in the main chain. The longest carbon chain is:
- HC (1)
- C (2)
- CH₂ (3)
- CH₂ (4)
- COOH (5)

### Step 2: Identify the Functional Groups
The compound contains a carboxylic acid group (-COOH) and a triple bond (alkyne). The presence of the carboxylic acid takes priority over the alkyne when numbering the carbon chain.

### Step 3: Number the Carbon Chain
We need to number the carbon chain starting from the end nearest to the carboxylic acid group to give it the lowest possible number:
1. COOH (1)
2. CH₂ (2)
3. C (3)
4. C (4)
5. HC (5)

So, the numbering is as follows:
- C1: COOH
- C2: CH₂
- C3: C (triple bond)
- C4: CH₂
- C5: HC

### Step 4: Determine the Parent Name and Suffixes
The parent chain has 5 carbon atoms, which corresponds to "pent". Since there is a triple bond, we will use the suffix "-yne".

### Step 5: Assign Locants
- The carboxylic acid group is on carbon 1 (1-oic acid).
- The triple bond is between carbon 3 and carbon 4 (4-yne).

### Step 6: Combine the Name
We combine the elements of the name:
- Start with the locant of the triple bond: 4
- Then the parent name: pent
- Then the suffix for the triple bond: -yne
- Finally, the locant and suffix for the carboxylic acid: 1-oic acid

Putting it all together, we get:
**4-pent-3-yne-1-oic acid**

### Final IUPAC Name
The IUPAC name for the compound `HC ≡ C - CH₂ - CH₂ - COOH` is **4-pent-3-yne-1-oic acid**.

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