How many cis/trans isomers are possible for 2,4,6-octatriene?
The Correct Answer and Explanation is:
The correct answer is 6.
Explanation:
The molecule in question is 2,4,6-octatriene. To determine the number of possible cis/trans isomers (also known as geometric isomers), a step-by-step analysis of its structure is required.
- Identify Stereogenic Centers: The name “2,4,6-octatriene” indicates an eight-carbon chain with three double bonds starting at carbons 2, 4, and 6. The structure is CH₃–CH=CH–CH=CH–CH=CH–CH₃. Cis/trans isomerism is possible around a double bond if each of the two carbons involved in the bond is attached to two different groups.
- The double bond at C2-C3 can be cis or trans.
- The double bond at C4-C5 can be cis or trans.
- The double bond at C6-C7 can be cis or trans.
This gives a total of three locations (n=3) where geometric isomerism can occur.
- Assess Molecular Symmetry: If the molecule were unsymmetrical, the total number of isomers would be 2ⁿ, which would be 2³ = 8. However, 2,4,6-octatriene is a symmetrical molecule. The groups on one side of the central double bond are the same as the groups on the other side. This symmetry reduces the number of unique isomers because some configurations are identical to others or form meso compounds (achiral molecules with stereocenters).
- Calculate the Number of Isomers: For symmetrical molecules with an odd number of stereogenic double bonds (n), the number of possible geometric isomers can be calculated using the formula:Number of isomers = 2ⁿ⁻¹ + 2^((n-1)/2)In this case, n = 3. Applying the formula:
- Number of isomers = 2³⁻¹ + 2^((3-1)/2)
- Number of isomers = 2² + 2^(2/2)
- Number of isomers = 2² + 2¹
- Number of isomers = 4 + 2
- Number of isomers = 6
Therefore, due to the molecule’s symmetry, there are a total of 6 unique cis/trans isomers possible for 2,4,6-octatriene.
