Calculate the number of geometrical isomers in the following polyenes.
(i) `H_(3)C-CH=CH-CH=CH-CH=CH-CH=CH-Br`
(ii) `H_(3)C-CH=CH-CH=CH-CH=CH-CH=CH-CH_(3)`
(iii) `C_(6)H_(5)-CH=CH-CH=CH-CH=CH-C_(6)H_(5)`
(iv) `C_(6)H_(5)-CH=CH-CH=CH-Cl`

Explanation

(i) The molecule has four double bonds and cannot be divided into two equal halves unsymmetrical the number of geometrical isomers `=2^((n-1))+2^((n//2)-1)`
`=2^(3)+2^(1)=8+2=10`
(iii) The molecule has three double bonds (odd number).
The number of geometrical isomers `=2^((n-1))+2^((n+1)/(2)-1)`
`=2^(2)+2^(2-1)=4+2=6`
(iv) The molecule has two double bonds and is unsymeetrical The number of geometrical isomers `=2^(n)=2^(3)=4`