The correct `IUPAC` name of the compound `CH_(2)=CH-CH_(2)-oversetoverset(C_(2)H_(5))(|)(CH)-CH_(3)`

Options

A. `4`-Ethylpent-`1`-ene
B. `2`-Ethylpent-`4`-ene
C. `4`-Methylhex-`1`-ene
D. `3`-Methylhex-`1`-ene

Answer

To determine the correct IUPAC name for the compound \( CH_2=CH-CH_2-CH(C_2H_5)(CH)-CH_3 \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Longest Carbon Chain**:
- The first step is to identify the longest continuous carbon chain in the compound. In this case, we can see that there are 6 carbon atoms in the longest chain.

2. **Number the Carbon Chain**:
- We will number the carbon chain starting from the end closest to the double bond (alkene). This gives us the following numbering:
- Carbon 1: \( CH_2 \) (double bond)
- Carbon 2: \( CH \)
- Carbon 3: \( CH_2 \)
- Carbon 4: \( CH \) (with an ethyl group attached)
- Carbon 5: \( CH \)
- Carbon 6: \( CH_3 \)

3. **Identify Substituents**:
- From the structure, we can identify that there is a methyl group (–CH₃) attached to Carbon 4 and an ethyl group (–C₂H₅) attached to Carbon 4 as well.

4. **Determine the Position of the Double Bond**:
- The double bond is between Carbon 1 and Carbon 2, so we will denote this as "hex-1-ene" because the longest chain has 6 carbons and the double bond starts at Carbon 1.

5. **Combine the Information**:
- Now we can combine all the information:
- The longest chain is hexane (6 carbons).
- The double bond is at position 1.
- There is a methyl group at position 4.
- The correct IUPAC name will be "4-methylhex-1-ene".

### Final Answer:
The correct IUPAC name of the compound is **4-methylhex-1-ene**.

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