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Question

(A) The given compound is $CH_3-CH=CH-CH=CH-C_2H_5$.Since the molecule is unsymmetrical,the number of geometrical isomers is given by the formula $2^n

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$4$

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(A) The given compound is $CH_3-CH=CH-CH=CH-C_2H_5$.Since the molecule is unsymmetrical,the number of geometrical isomers is given by the formula $2^n$,where $n$ is the number of double bonds capable of showing geometrical isomerism.Here,both double bonds are capable of showing geometrical isomerism,so $n = 2$.Therefore,the number of geometrical isomers = $2^2 = 4$.The four isomers are:$1$. $cis, cis$$2$. $trans, trans$$3$. $cis, trans$$4$. $trans, cis$

The number of geometrical isomers in case of a compound with the structure $CH_3-CH=CH-CH=CH-C_2H_5$ is

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