Give the systematic (IUPAC) names of the following alkenes

Give the systematic (IUPAC) names of the following alkenes.
(a) CH2“CH¬CH2¬CH(CH3)2

The Correct Answer and Explanation is :

The given molecular structure is CH₂=CH–CH₂–CH(CH₃)₂. To determine its systematic IUPAC name, we follow these steps:

1. Identify the Longest Carbon Chain Containing the Double Bond:
   The longest continuous chain that includes the double bond has four carbon atoms. This corresponds to “butene,” indicating a four-carbon alkene.
2. Number the Carbon Chain:
   Number the chain from the end nearest to the double bond to assign the lowest possible number to the double bond. Starting from the left (where the double bond is located), the carbon atoms are numbered as follows:

• C1: CH₂= (double bond starts here)
• C2: =CH–
• C3: –CH₂–
• C4: –CH(CH₃)₂

1. Locate and Name Substituents:
   A methyl group (–CH₃) is attached to the fourth carbon atom (C4).
2. Assign Numbers to Substituents:
   The double bond starts at C1, making it a 1-butene. The methyl group is on C4.
3. Combine the Elements into the IUPAC Name:
   The substituent is named first, followed by the base name of the alkene. Thus, the compound is named 4-methylbut-1-ene.

Explanation:

In IUPAC nomenclature, the parent hydrocarbon chain is the longest continuous chain of carbon atoms containing the highest-order functional group, which in this case is the double bond. The chain is numbered to give the double bond the lowest possible locant. Substituents are then identified, and their positions are indicated by the corresponding carbon numbers. The final name is constructed by listing the substituents in alphabetical order (if there are multiple), each preceded by its locant, followed by the base name of the hydrocarbon chain with the appropriate suffix indicating the presence and position of the double bond.

Therefore, CH₂=CH–CH₂–CH(CH₃)₂ is systematically named 4-methylbut-1-ene.