How many geometrical isomers of the compound | vedclass

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(B) The compound $CH_3-CH=C(Br)-C(Br)=CH-CH_3$ is a symmetrical molecule with $n=2$ double bonds capable of showing geometrical isomerism.For a symmet

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$3$

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(B) The compound $CH_3-CH=C(Br)-C(Br)=CH-CH_3$ is a symmetrical molecule with $n=2$ double bonds capable of showing geometrical isomerism.For a symmetrical molecule with an even number of double bonds $(n=2)$,the number of geometrical isomers is calculated using the formula: $2^{n-1} + 2^{(n/2)-1}$.Substituting $n=2$ into the formula:Number of isomers $= 2^{2-1} + 2^{(2/2)-1} = 2^1 + 2^0 = 2 + 1 = 3$.The three possible geometrical isomers are $(cis, cis)$,$(trans, trans)$,and $(cis, trans)$.

How many geometrical isomers of the compound $CH_3-CH=C(Br)-C(Br)=CH-CH_3$ are possible?

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