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(B) Geometrical isomerism is shown by alkenes where each carbon atom of the double bond is attached to two different groups. $2-$pentene $(CH_3-CH=CH-

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$2-$pentene

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(B) Geometrical isomerism is shown by alkenes where each carbon atom of the double bond is attached to two different groups. $2-$pentene $(CH_3-CH=CH-CH_2-CH_3)$ satisfies this condition as the double-bonded carbons are attached to different groups ($-H$ and $-CH_3$ on one side; $-H$ and $-CH_2CH_3$ on the other). $2-$pentyne $(CH_3-C\equiv C-CH_2-CH_3)$ contains a triple bond,which is linear and does not allow for geometrical isomerism. $2-$methyl propene $(CH_3-C(CH_3)=CH_2)$ has two identical $-CH_3$ groups on one of the double-bonded carbons. $2-$methyl$-2-$butene $(CH_3-C(CH_3)=CH-CH_3)$ also has two identical $-CH_3$ groups on one of the double-bonded carbons. Therefore,only $2-$pentene shows geometrical isomerism.

Which of the following compounds will show geometrical isomerism?

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Which of the following compounds will show geometrical isomerism?

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In the following structure,the double bonds are marked as $I$,$II$,$III$ and $IV$. Geometrical isomerism is not possible at site$(s)$:

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$CH_2 = CH - CH = CH - CH = CH_2$
How many geometrical isomers are possible for this compound?