Step 1: Understanding the Concept:
The given image is a bond-line representation of an organic molecule containing a double bond and a halogen substituent.
To find the correct IUPAC name, we must identify the longest continuous carbon chain containing the double bond, number it correctly, and list the substituents alphabetically. Step 2: Key Formula or Approach:
1. Identify the principal functional group (alkene double bond) and ensure it gets the lowest possible locant.
2. Find the longest continuous carbon chain that includes the double bond.
3. Number the chain from the end that gives the double bond the lower number.
4. Identify and locate the substituents (bromo, methyl).
5. Assemble the name alphabetically. Step 3: Detailed Explanation:
Let's analyze the bond-line structure based on the correct option (B) and verify it.
The longest chain containing the double bond has 5 carbon atoms. Therefore, the parent name is 'pentene'.
Numbering from right to left (as per the typical bond-line drawing) gives the double bond at carbon 2, making it a 'pent-2-ene'.
If we numbered from left to right, the double bond would be at carbon 3, which violates the lowest locant rule for the principal functional group.
With the chain numbered from right to left:
- Carbon 1 is the \( \text{CH}_3 \) group on the far right.
- Carbon 2 and 3 share the double bond.
- Carbon 4 is bonded to a Bromine atom (Bromo group) and a \( \text{CH}_3 \) group (Methyl group).
- Carbon 5 is the terminal \( \text{CH}_3 \) group on the left.
The substituents are at position 4. We have a '4-bromo' and a '4-methyl'.
Alphabetically, 'bromo' comes before 'methyl'.
Combining these parts gives: 4-Bromo-4-methylpent-2-ene.
This matches Option (B) perfectly. Step 4: Final Answer:
The IUPAC name is 4-Bromo-4-methylpent-2-ene.
