The number of geometrical isomers of $C H_{3} - C H = C H - C H = C H - C H = C H - C H_{3}$ is

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Solution

The correct option is C 6
If the terminal group are same in polyene and the number of double bons is odd then, the number of geometrical isomer
$= 2^{n - 1} + 2^{\frac{n + 1}{2}} - 1$
$= 2^{3 - 1} + 2^{\frac{3 + 1}{2}} - 1$
$= 2^{2} + 2^{1}$
$= 4 + 2$
$= 6$
They are : (1) cis, cis, cis
(2) trans, trans, trans
(3) cis, trans, trans
(4) cis, trans, cis
(5) trans, cis, trans
(6) cis, cis, trans

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