Question
Question: Which of the following will show geometrical isomerism. A., the structure is given as follows-
As it is clearly observed, that both the atoms connected via double bond are bonded to different groups, therefore it has a tendency to show geometrical isomerism.
In compound (B), the structure is given as follows-
As it is clearly observed, that both the carbon atoms connected via double bond are bonded to different groups, therefore it has a tendency to show geometrical isomerism.
In compound (C), the structure is given as follows-
As it is clearly observed, both the carbon atoms (1) and (2) are connected to different groups and due to the presence of a ring, the rotation is restricted as well. Therefore, it has a tendency to show geometrical isomerism.
In compound (D), the structure is given as follows-
In the cyclic structures, double bonded carbon atoms are unable to show geometrical isomerism because the terminal groups do not lie in the same plane which is an important condition for a compound to show geometrical isomerism. Moreover, there must be more than one sp3 hybridized carbon atom, which is absent for the given compound. Therefore, it does not have a tendency to show geometrical isomerism.
Hence, options (A), (B) and (C) are the correct answers.
Note:
It is important to note that the necessary conditions for a compound to show geometrical isomerism are as follows:
The rotation must be restricted i.e., the compound must contain or double bond or a ring.
The terminal groups must be present in the same group as of the carbon atoms.