Number of the geometrical isomers for the molecule

Options

A. 2
B. 4
C. 6
D. 8

Answer

C

Explanation

An organic compound having odd number of double bonds with same terminal groups are present . Number of geometrical isomers = `2^(n+1) + 2^( (n+1)/(2) - 1)`
`n = 3 rArr 2^(3-1) + 2^((3+1)/(2) -1) rArr 2^2 + 2^(2-1) rArr 4 + 2 = 6`