How many geometrical isomers (excluding enantiomers) are possible for the compound given below?

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Solution

The correct option is C 8

Eight geometrical isomers (excluding enantiomers) are possible for the given compound. It contains three $C = C$ double bonds. Hence, total number of geometrical isomers $2^{3} = 8$

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C_{3} H_{5} C l

How many geometrical isomers are possible for the following compound?

C H_{3} - C H = C H - C H = C H - C_{2} H_{5}

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