Nomenclature

Naming Alkenes: The IUPAC System

When you name an alkene, the first rule is that the double bond determines everything — the chain you choose, the numbering you use, and the suffix you attach. Unlike alkanes, where the longest carbon chain is simply the parent name, alkenes force you to select the longest chain that includes the double bond. That chain might not be the absolute longest chain in the molecule, but it must contain the $C=C$ unit.

Selecting the Parent Chain

Identify the longest continuous chain of carbon atoms that contains the double bond. This chain becomes the parent alkene. The suffix for alkenes is -ene, which replaces the -ane ending of the corresponding alkane. So an alkane with two carbons (ethane) becomes ethene; with three carbons (propane) becomes propene, and so on.

Numbering the Chain

Number the parent chain from the end that gives the double bond the lowest possible locant (position number). The double-bond locant is the smaller of the two carbon numbers that share the double bond. For example, in a four-carbon chain with the double bond between carbons 2 and 3, the locant is 2, not 3.

If the double bond is equidistant from both ends, then number from the end that gives the first substituent the lower number — the same rule as for alkanes.

The First Member: A Special Case

The formula $C_nH_{2n}$ suggests that the first member of the alkene series should be $CH_2$ (methene, with $n=1$). However, this species is extremely unstable and has a very short lifetime — it does not exist as a stable compound under normal conditions. The first stable member of the alkene series is $C_2H_4$, which has the common name ethylene and the IUPAC name ethene.

IUPAC Names of Simple Alkenes

The table below shows how the rules apply to a few simple alkenes:

Notice the pattern: the locant of the double bond is placed immediately before the -ene suffix, separated by a hyphen. For dienes (two double bonds), the suffix becomes -diene, and both locants are given.

Naming More Complex Alkenes: Worked Examples

The textbook provides several examples that show how to handle branching and multiple double bonds. Let us work through each one carefully.

Problem 9.7 (i): $(CH_3)_2CH-CH=CH-CH_2-CH \equiv CH-CH_3$ with a $C_2H_5$ branch on the third carbon from the right.

The structure is:

CH3-CH-CH=CH-CH2-CH-CH=CH-CH3
   |                    |
  CH3                  C2H5

The longest chain containing both double bonds has 10 carbons. Number from the end nearer the first double bond — the left end gives the first double bond at position 3, the right end gives it at position 7, so left is correct. The double bonds are at positions 3 and 6. There are methyl substituents at positions 2 and 8. The name is 2,8-Dimethyl-3,6-decadiene.

Problem 9.7 (ii): $CH_3-CH=CH-CH=CH-CH=CH-CH_3$

This is a straight chain of 8 carbons with four alternating double bonds. The parent is octane, and with four double bonds it becomes octatetraene. The double bonds are at positions 1, 3, 5, and 7. The name is 1,3,5,7-Octatetraene.

Problem 9.7 (iii): $CH_2=C(CH_2CH_2CH_3)_2$

The central carbon of the double bond has two propyl groups attached. The longest chain containing the double bond has 5 carbons (the double-bond carbon plus one propyl group). The double bond is at position 1. The two propyl groups are identical — they are $n$-propyl groups. The name is 2-n-Propylpent-1-ene.

Problem 9.7 (iv):

CH3-CH2-CH2-CH2-CH2-CH3
   |                 |
CH3-CH-CH=C-CH2-CH-CH3
            |
           CH3

The longest chain containing the double bond has 10 carbons. Number from the end nearer the double bond — the left end gives the double bond at position 4, the right end gives it at position 6, so left is correct. There is an ethyl group at position 4, and methyl groups at positions 2 and 6. The name is 4-Ethyl-2,6-dimethyl-dec-4-ene.

Counting Sigma and Pi Bonds in Alkenes …