What are the C=C-C bond angles for (Z)-1,2-diisopropylethene, (E)-1,2-
diisopropylethene and 1,1-diisopropylethene?
To determine the C=C-C bond angles for the given compounds, we need to consider the arrangement of atoms around the double bond. The geometry of a double bond is typically trigonal planar, meaning the two carbon atoms and the two atoms attached to them all lie in a flat plane.
Let's analyze each compound individually:
1. (Z)-1,2-diisopropylethene:
In this compound, the "Z" notation indicates that the two highest priority substituents on each carbon atom are on the same side of the double bond. The diisopropyl groups make up the highest priority substituents on each carbon atom. Since both diisopropyl groups are on the same side, the compound has a cis configuration. Hence, the bond angles are approximately 120 degrees.
2. (E)-1,2-diisopropylethene:
The "E" notation denotes that the two highest priority substituents on each carbon atom are on opposite sides of the double bond. In this case, the diisopropyl groups are on opposite sides, leading to a trans configuration. Again, we have a trigonal planar arrangement, resulting in bond angles of approximately 120 degrees.
3. 1,1-diisopropylethene:
This compound doesn't possess a double bond. It consists of two isopropyl groups attached to a single carbon atom. Since there is no double bond, there are no C=C-C bond angles to consider.
In summary, for (Z)-1,2-diisopropylethene and (E)-1,2-diisopropylethene, the C=C-C bond angles are approximately 120 degrees due to the trigonal planar geometry of the double bond. However, in 1,1-diisopropylethene, there are no double bonds and therefore no C=C-C bond angles.