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(B) For a hydrocarbon to exhibit cis-trans (geometrical) isomerism,there must be a restricted rotation around a double bond $(C=C)$.Each carbon atom o

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$4$

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(B) For a hydrocarbon to exhibit cis-trans (geometrical) isomerism,there must be a restricted rotation around a double bond $(C=C)$.Each carbon atom of the double bond must be attached to two different groups.The smallest acyclic alkene is but$-2-$ene $(CH_3-CH=CH-CH_3)$.In but$-2-$ene,each carbon of the double bond is attached to a hydrogen atom and a methyl group,which are different.Counting the carbon atoms in but$-2-$ene $(CH_3-CH=CH-CH_3)$,we find there are $4$ carbon atoms.Therefore,the smallest acyclic hydrocarbon exhibiting cis-trans isomerism has $4$ carbon atoms.

What is the number of carbon atoms present in the smallest acyclic hydrocarbon capable of exhibiting cis-trans isomerism?

Similar Questions

For which of the following compounds are geometrical isomers possible?

Which of the following cannot show geometrical isomerism?

Which of the following shows geometrical isomers?
$(a)$ $CH_2=CHCl$
$(b)$ $CH_3CH=CH_2$
$(c)$ $CHCl=CHCl$
$(d)$ $(CH_3)_2C=CHC_2H_5$

The incorrect statement regarding the geometrical isomers of $2$-butene is:

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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