If a chlorine atom is lost from 1,2-dichloroethene [C2H2Cl2], would it have a higher or lower dipole?
Isn't 1,2-dichloroethane rather symmetrical? Wouldn't losing a single Cl atom make it more unsymmetrical?
Oh yes so more dipole movement thank you!
To determine if losing a chlorine atom from 1,2-dichloroethene (C2H2Cl2) would result in a higher or lower dipole, we need to consider the molecular structure and polarity.
1,2-dichloroethene has a double bond between the carbon atoms, with two chlorine atoms attached to each carbon atom. Due to the electronegativity difference between carbon and chlorine, the C-Cl bonds are polar, with the chlorine atom being slightly negative and the carbon atom slightly positive. This polarity creates a dipole moment for each C-Cl bond.
When a chlorine atom is lost from 1,2-dichloroethene, it creates a new molecule called vinyl chloride (C2H3Cl). In vinyl chloride, there is only one chlorine atom, which is attached to one of the carbon atoms. The carbon-carbon double bond remains intact.
The dipole moment of a molecule depends on the polarity of its bonds and the molecular structure. In the case of 1,2-dichloroethene, the dipole moments of the C-Cl bonds cancel each other out due to the symmetric structure, resulting in a net dipole moment of zero.
By removing one of the chlorine atoms and forming vinyl chloride, the symmetry is broken. Assuming that the remaining chlorine atom is attached to the same carbon atom that originally had both chlorines, the bond polarity between carbon and chlorine will remain the same. However, since there is no longer a chlorine atom attached to the other carbon atom, the dipole moments will not cancel each other out.
Therefore, by losing a chlorine atom, the molecule will have a higher dipole moment.