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  4. How many geometrical isomers are possible for the given compound? CH3 — CH = CH — CH = CH — C2H5

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Q. How many geometrical isomers are possible for the given compound?
$CH_3 — CH = CH — CH = CH — C_2H_5$

Hydrocarbons

Solution:

Since the number of double bonds in the given compound is 2 and the ends are different (i.e. $CH _{3}$ and $C _{2} H _{5}$ ), therefore the total number of geometrical isomers are $2^{2}=4$
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