Number of geometrical isomers of the given co | vedclass

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(B) The given compound is $CH_2=CH-CH=CH-CH=CH-CH_2-CH=CH_2$. First,identify the double bonds that can show geometrical isomerism. $A$ double bond sho

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$4$

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(B) The given compound is $CH_2=CH-CH=CH-CH=CH-CH_2-CH=CH_2$. First,identify the double bonds that can show geometrical isomerism. $A$ double bond shows geometrical isomerism if both carbons of the double bond are attached to two different groups. In the given structure,the terminal double bonds $(CH_2=CH-)$ cannot show geometrical isomerism because one carbon is attached to two identical hydrogen atoms. The internal double bonds are at positions $3$ and $5$. Specifically,the structure is $CH_2=CH-CH=CH-CH=CH-CH_2-CH=CH_2$. Counting the double bonds: $C_1=C_2-C_3=C_4-C_5=C_6-C_7-C_8=C_9$. Double bonds at $C_3=C_4$ and $C_5=C_6$ are capable of showing geometrical isomerism. Thus,there are $n = 2$ geometrical centers. Total number of geometrical isomers $= 2^n = 2^2 = 4$.

Number of geometrical isomers of the given compound: $CH_2=CH-CH=CH-CH=CH-CH_2-CH=CH_2$ is:

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