Which of the following compounds does not hav | vedclass

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(D) For a cyclic compound to exhibit geometrical isomerism,there must be at least two $sp^3$ hybridized carbon atoms in the ring,each attached to two

Vedclass · Accepted Answer

$1,1,4-$trichlorocyclohexane

Vedclass · Answer

(D) For a cyclic compound to exhibit geometrical isomerism,there must be at least two $sp^3$ hybridized carbon atoms in the ring,each attached to two different groups. In $1,1,4$-trichlorocyclohexane,the carbon at position $1$ is attached to two identical chlorine atoms. Since one of the carbons in the ring is attached to two identical groups,it cannot contribute to geometrical isomerism. Therefore,$1,1,4$-trichlorocyclohexane does not exhibit geometrical isomerism.

Which of the following compounds does not have any geometrical isomer?

Similar Questions

According to the $E$ and $Z$ nomenclature,which isomer is $E$?

Which of the following compounds shows geometrical isomerism?

Which of the following compounds will not show geometrical isomerism across the $\pi$-bond?

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

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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