Question

Total geometrical isomers for 2,4- heptadiene

Answer 1

4

Answer 2

The structure of 2,4-heptadiene is $CH_3 - CH = CH - CH = CH - CH_2 - CH_3$.
The double bonds are located at positions 2 (between C2 and C3) and 4 (between C4 and C5).

For a double bond to exhibit geometrical isomerism, each carbon atom of the double bond must be attached to two different groups.

Let's examine the double bond at position 2 (C2=C3): Carbon C2 is attached to a $CH_3$ group and a Hydrogen atom ($H$). These are different. Carbon C3 is attached to a Hydrogen atom ($H$) and a $-CH=CH-CH_2-CH_3$ group. These are different. Since both carbon atoms of the C2=C3 double bond are attached to two different groups, this double bond can exist in either cis (Z) or trans (E) configuration.

Let's examine the double bond at position 4 (C4=C5): Carbon C4 is attached to a $-CH_3-CH=CH-$ group and a Hydrogen atom ($H$). These are different. Carbon C5 is attached to a Hydrogen atom ($H$) and a $-CH_2-CH_3$ group. These are different. Since both carbon atoms of the C4=C5 double bond are attached to two different groups, this double bond can exist in either cis (Z) or trans (E) configuration.

There are two double bonds that can exhibit geometrical isomerism. Let's denote the configuration of the C2=C3 double bond as D1 and the configuration of the C4=C5 double bond as D2. Each double bond can be in E or Z configuration. The possible combinations of configurations are (E,E), (E,Z), (Z,E), and (Z,Z).

We need to determine if any of these isomers are identical due to symmetry. The molecule is $CH_3 - CH = CH - CH = CH - CH_2 - CH_3$. Let's look at the groups at the ends of the conjugated system. At one end, we have a $CH_3$ group attached to C1, and at the other end, we have a $CH_2-CH_3$ group attached to C7. Since the groups at the ends are different ($CH_3$ vs $CH_2CH_3$), the molecule is unsymmetrical.

For an unsymmetrical molecule with $n$ double bonds that can show geometrical isomerism, the total number of geometrical isomers is $2^n$. In this case, $n=2$ (the double bonds at positions 2 and 4). Both double bonds are capable of showing geometrical isomerism. The molecule is unsymmetrical. Therefore, the total number of geometrical isomers is $2^2 = 4$.

The four distinct isomers are:

1. (2E, 4E)-2,4-heptadiene
2. (2E, 4Z)-2,4-heptadiene
3. (2Z, 4E)-2,4-heptadiene
4. (2Z, 4Z)-2,4-heptadiene

These four isomers are all distinct because the molecule is unsymmetrical.

Explanation of the solution:

1. Identify the double bonds in the molecule 2,4-heptadiene ($CH_3-CH=CH-CH=CH-CH_2-CH_3$). The double bonds are at positions 2 and 4.
2. Check if each double bond can exhibit geometrical isomerism. This requires that each carbon of the double bond is attached to two different groups. Both the C2=C3 and C4=C5 double bonds satisfy this condition.
3. Determine if the molecule is symmetrical or unsymmetrical. The molecule has different groups at its ends ($CH_3$ and $CH_2CH_3$), making it unsymmetrical.
4. For an unsymmetrical molecule with $n$ double bonds capable of geometrical isomerism, the total number of isomers is $2^n$. Here, $n=2$.
5. Calculate the total number of isomers: $2^2 = 4$.