Why is (E)-1,2-diisopropylethene more stable than (Z)-1,2-diisopropylethene? Why are they more stable than 1,1-diisopropylethe?

And doesn't the bond angle between the C=C-C carbons increase with (Z)-1,2-diisopropylethene > (E)-1,2-diisopropylethene > 1,1-diisopropylethe? So why isn't the same as (E) being greater than (Z)?

The E (trans-) isomer is usually more stable because it is a less crowded structure (less steric hindrance), and because attractive intermolecular forces involving the H's are more easily formed. Building molecular models of the two kinds of isomers would help you visualize this.

(E)-1,2-diisopropylethene is more stable than (Z)-1,2-diisopropylethene due to steric effects. In (E)-1,2-diisopropylethene, the bulky isopropyl groups are oriented apart from each other on opposite sides of the double bond, leading to minimal steric hindrance. This allows for a more stable molecular conformation.

On the other hand, in (Z)-1,2-diisopropylethene, the isopropyl groups are oriented towards each other on the same side of the double bond, resulting in significant steric hindrance. This steric hindrance destabilizes the molecule, making it less stable than the (E) isomer.

Regarding the stability comparison with 1,1-diisopropylethene, both (E)-1,2-diisopropylethene and (Z)-1,2-diisopropylethene are more stable due to the presence of conjugation between the double bond and the electron-donating isopropyl groups. This conjugation helps to distribute electron density along the double bond and stabilize the molecule.

Now, coming to your second question about the bond angles between the C=C-C carbons, it's important to note that the bond angles in alkene compounds are typically close to the ideal 120 degrees. However, the specific bond angles can vary depending on the presence of different substituents and steric effects.

While it's true that in (Z)-1,2-diisopropylethene, the bond angle between the C=C-C carbons appears larger than in (E)-1,2-diisopropylethene (due to the isopropyl groups being bulkier and causing more molecular strain), this difference in bond angle alone doesn't determine the stability of the isomers.

Overall, the stability of the isomers is primarily influenced by the steric hindrance caused by the specific arrangement of the substituents around the double bond, as well as the presence of conjugation effects.

To understand why (E)-1,2-diisopropylethene is more stable than (Z)-1,2-diisopropylethene, we need to consider the concept of steric hindrance. Steric hindrance refers to the repulsion between atoms or groups of atoms in a molecule that affects their spatial arrangement.

In (E)-1,2-diisopropylethene, the two isopropyl groups (CH3-CH(CH3)2) are located on opposite sides of the double bond. This arrangement minimizes steric hindrance because the bulky isopropyl groups are positioned away from each other, allowing for better dispersion of electron density along the molecule.

In contrast, in (Z)-1,2-diisopropylethene, the two isopropyl groups are located on the same side of the double bond. This arrangement results in significant steric hindrance, as the bulky isopropyl groups are positioned close to each other. The repulsion between these groups destabilizes the molecule and decreases its overall stability.

Regarding 1,1-diisopropylethene, it is less stable than both (E)- and (Z)-1,2-diisopropylethene due to the absence of conjugation. In 1,1-diisopropylethene, the double bond is located between the two isopropyl groups, resulting in significant steric hindrance between them. Additionally, the lack of a conjugated system decreases the stability further.

Now, let's address your question about the bond angle. The bond angle between the C=C-C carbons in (E)-1,2-diisopropylethene, (Z)-1,2-diisopropylethene, and 1,1-diisopropylethene is approximately 120 degrees. The bond angle is determined by the nature of the bond, rather than the stereochemistry (E or Z) of the molecule. Thus, the bond angles remain the same regardless of whether the molecule is (E)-1,2-diisopropylethene, (Z)-1,2-diisopropylethene, or 1,1-diisopropylethene.

In summary, (E)-1,2-diisopropylethene is more stable than (Z)-1,2-diisopropylethene due to lower steric hindrance between the bulky isopropyl groups. Both (E)-1,2-diisopropylethene and (Z)-1,2-diisopropylethene are more stable than 1,1-diisopropylethene because the latter lacks conjugation and experiences significant steric hindrance between the isopropyl groups. The bond angle remains approximately 120 degrees for all three molecules, regardless of their stereochemistry.