CH_3 - CH = C(Br) - C(Cl) = CH - CH_3 How man | vedclass

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(C) The given compound is $CH_3 - CH = C(Br) - C(Cl) = CH - CH_3$. In this molecule,there are two double bonds capable of showing geometrical isomeris

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(C) The given compound is $CH_3 - CH = C(Br) - C(Cl) = CH - CH_3$. In this molecule,there are two double bonds capable of showing geometrical isomerism. Since the molecule is unsymmetrical (the groups attached to the double bonds are different),the number of geometrical isomers is calculated using the formula $2^n$,where $n$ is the number of double bonds. Number of isomers = $2^2 = 4$. The possible isomers are $(E,E)$,$(E,Z)$,$(Z,E)$,and $(Z,Z)$.

$CH_3 - CH = C(Br) - C(Cl) = CH - CH_3$
How many geometrical isomers are possible for this compound?

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$CH_3 - CH = C(Br) - C(Cl) = CH - CH_3$ How many geometrical isomers are possible for this compound?