How many geometrical isomers are possible for the given compound?
`CH_(3)-CH=CH-CH=CH-C_(2)H_(5)`

Options

A. 4 and 4
B. 4 and 3
C. 3 and 3
D. 3 and 4

Answer

D

Explanation

When the ends of alkene containing n double bond are different, the number of geometrical isomers is 2''. Thus for
`CH_(3)-CH=CH-CH=CH-Cl`.
Number of geometrical isomers `=2^(2)=4`
When the ends of alkene containing n double bonds are same, then the number of geometrical isomers `=2^(n-1)+2^(p-1)`
where `p=(n)/(2)` for even n and `(n+1)/(2)` for odd n, thus for
`CH_(3)-CH=CH-CH=CH-CH_(3)`
Number of geometrical isomers
`=2^(2-1)+2^((2)/(2)-1)=2^(1)+2^(0)=2+1=3.`