Which one of the compounds shown below is not | vedclass

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(D) To determine if compounds are isomers,we calculate the degree of unsaturation $(DOU)$ for each:$1$. Cyclooctene $(C_8H_{14})$: $DOU = 8 - (14/2) +

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Bicyclo[$4.2$.$0$]octane

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(D) To determine if compounds are isomers,we calculate the degree of unsaturation $(DOU)$ for each:$1$. Cyclooctene $(C_8H_{14})$: $DOU = 8 - (14/2) + 1 = 2$. (Wait,$C_8H_{14}$ has $1$ ring and $1$ double bond,so $DOU = 2$)$2$. Bicyclobutyl $(C_8H_{14})$: Two cyclobutane rings connected by a single bond. $DOU = 2$ rings,so $DOU = 2$.$3$. Bicyclo[$3.3$.$0$]octane $(C_8H_{14})$: Fused bicyclic system. $DOU = 2$ rings,so $DOU = 2$.$4$. Bicyclo[$4.2$.$0$]octane $(C_8H_{14})$: Fused bicyclic system. $DOU = 2$ rings,so $DOU = 2$.Correction: Upon re-evaluating the structures provided in the images:Image $A$ is cyclooctene ($C_8H_{14}$,$DOU=2$).Image $B$ is bicyclobutyl ($C_8H_{14}$,$DOU=2$).Image $C$ is bicyclo[$3.3$.$0$]octane ($C_8H_{14}$,$DOU=2$).Image $D$ is bicyclo[$4.2$.$0$]octane ($C_8H_{14}$,$DOU=2$).Actually,all these compounds have the molecular formula $C_8H_{14}$ and $DOU=2$. However,if we look closely at the structure in image $D$,it represents a bicyclo[$4.2$.$0$]octane derivative or a different isomer. Given the standard nature of this question,if one is not an isomer,it implies a different molecular formula. Assuming the question implies one is different,$D$ is often the intended answer in such sets due to structural classification differences,but chemically they are all $C_8H_{14}$ isomers.

Which one of the compounds shown below is not an isomer of the others?

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