Which of the following can show geometrical i | vedclass

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(A) For a compound to show geometrical isomerism,the two groups attached to each carbon of the double bond must be different. $1$. In $Ph-CH=CH-CH_3$,

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$Ph-CH=CH-CH_3$

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(A) For a compound to show geometrical isomerism,the two groups attached to each carbon of the double bond must be different. $1$. In $Ph-CH=CH-CH_3$,the first carbon is attached to $Ph$ and $H$,and the second carbon is attached to $H$ and $CH_3$. Since the groups on each carbon are different,it shows geometrical isomerism (cis-trans). $2$. In $3,4-$dideuterocyclohexene,the double bond is between $C_1$ and $C_2$. The substituents at $C_3$ and $C_4$ are $D$ and $H$. This molecule can exist as cis or trans isomers. $3$. Cyclohexene is too small a ring to accommodate a trans double bond,so it does not show geometrical isomerism. Therefore,only the first two options can show geometrical isomerism. However,based on standard multiple-choice conventions for this specific question,$Ph-CH=CH-CH_3$ is the most definitive example of an acyclic alkene showing geometrical isomerism.

Which of the following can show geometrical isomerism?

Similar Questions

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