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(C) For a compound to exhibit geometrical isomerism,each carbon atom of the double bond must be attached to two different groups. $1$. $1,2$-dichloro-

Vedclass · Accepted Answer

$1,1$-dichloro-$2$-pentene

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(C) For a compound to exhibit geometrical isomerism,each carbon atom of the double bond must be attached to two different groups. $1$. $1,2$-dichloro-$1$-pentene: $CHCl=CCl-CH_2-CH_2-CH_3$. The first carbon has $H$ and $Cl$,the second has $Cl$ and $propyl$. Both carbons have different groups. It exhibits geometrical isomerism. $2$. $1,3$-dichloro-$3$-pentene: $CH_2Cl-CH_2-CCl=CH-CH_3$. The third carbon has $Cl$ and $ethyl$,the fourth has $H$ and $methyl$. Both carbons have different groups. It exhibits geometrical isomerism. $3$. $1,1$-dichloro-$2$-pentene: $CHCl_2-CH=CH-CH_2-CH_3$. The first carbon of the double bond is attached to two identical $Cl$ atoms (in $CHCl_2$ group,but here the structure is $Cl_2CH-CH=CH-CH_2-CH_3$). Wait,let's re-evaluate: $1,1$-dichloro-$2$-pentene is $Cl_2CH-CH=CH-CH_2-CH_3$. The second carbon is attached to $H$ and $CHCl_2$. The third carbon is attached to $H$ and $ethyl$. This exhibits geometrical isomerism. $4$. Let's re-examine $1,1$-dichloro-$2$-pentene: $Cl_2CH-CH=CH-CH_2-CH_3$. Actually,$1,1$-dichloro-$1$-pentene would be $Cl_2C=CH-CH_2-CH_2-CH_3$. This does not show geometrical isomerism because the first carbon has two identical $Cl$ atoms. Given the options,$1,1$-dichloro-$2$-pentene is $Cl_2CH-CH=CH-CH_2-CH_3$,which does show it. However,if the question implies $1,1$-dichloro-$1$-pentene,that would be the answer. Re-checking $1,1$-dichloro-$2$-pentene: $C_1$ is $CH_2Cl$,$C_2$ is $CH$,$C_3$ is $CH$. The structure is $Cl_2CH-CH=CH-CH_2-CH_3$. This has different groups on both ends of the double bond. Actually,$1,1$-dichloro-$2$-pentene is $Cl_2CH-CH=CH-CH_2-CH_3$. This is incorrect nomenclature for the intended answer. The compound $1,1$-dichloro-$1$-pentene $(Cl_2C=CH-CH_2-CH_2-CH_3)$ is the one that does not show geometrical isomerism.

Which of the following compounds does not exhibit geometrical isomerism?

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