Always remember the priority order: Principal Functional Group $>$ Double/Triple Bonds $>$ Halogens/Alkyl groups. Even though numbering from the right gives the chlorine substituent a lower number (2 instead of 4), the double bond dictates the numbering sequence because it holds a higher priority!
Step 1: Understanding the Question:
The question requires us to determine the correct systematic IUPAC name for a branched unsaturated haloalkane from its structural formula.
Step 2: Key Formula or Approach:
According to IUPAC rules for naming alkenes containing substituents:
1. Identify the longest carbon chain containing the carbon-carbon double bond (C=C).
2. Number the principal chain from the end that gives the double bond the lowest possible locant number (alkene priority over halogen and alkyl branches).
3. List the substituents alphabetically with their respective locant numbers.
Step 3: Detailed Explanation:
Let us analyze the longest chain containing the double bond:
$$\text{CH}_3\text{-CH=C(CH}_3)\text{-CH(Cl)-CH}_3$$
The continuous carbon path stretching from left to right contains 5 carbon atoms, making the parent alkene
pentene.
Numbering from left to right gives the double bond a locant of 2 (-2-ene), and the substituents are at C3 and C4.
Numbering from right to left would give the double bond a locant of 3 (-3-ene). Since 2 is lower than 3, we proceed with left-to-right numbering:
C1: $\text{CH}_3$
C2: CH (start of double bond)
C3: C bearing a methyl group ($-\text{CH}_3$)
C4: CH bearing a chlorine atom ($-\text{Cl}$)
C5: $\text{CH}_3$
Arranging the substituents alphabetically: "4-chloro" comes before "3-methyl".
Combining these parts gives the systematic name:
4-Chloro-3-methylpent-2-ene. This matches option (B).
Step 4: Final Answer:
The IUPAC name of the compound is 4-Chloro-3-methylpent-2-ene, which corresponds to option (B).
