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Question:

How many geometrical isomers are possible in the following two alkenes?

3and4

4and3

3and3

4and4

Solution:

When the ends of alkene containing n double bonds are same , then the number of geometrical isomers=2n + 2p where p=n/2 for even n and (n+1)/2 for odd n
Number of geometrical isomers=22 + 22/2 =3
When the ends of alkene containing n double bonds are different the number of geometrical isomers is 2n thus for the given
Number of geometrical isomers=22=4
Hence, the correct option is D