Which of the following orders of rotation bar | vedclass

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(A) The rotational barrier about a $C=C$ bond is directly proportional to the double bond character of that bond.$(I)$ $CH_2=CH_2$: This is a pure alk

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$I > II > III$

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(A) The rotational barrier about a $C=C$ bond is directly proportional to the double bond character of that bond.$(I)$ $CH_2=CH_2$: This is a pure alkene with no resonance,so it has the highest double bond character.$(II)$ $CH_3O-CH=CH_2$: The lone pair on the oxygen atom participates in resonance with the $C=C$ bond $(CH_3O^+=CH-CH_2^-)$,which introduces some single bond character to the $C=C$ bond,reducing the rotational barrier compared to $(I)$.$(III)$ $CH_3O-CH=CH-COOEt$: This molecule exhibits push-pull resonance. The electron-donating $CH_3O$ group and the electron-withdrawing $COOEt$ group significantly increase the resonance contribution,leading to even more single bond character in the $C=C$ bond compared to $(II)$.Therefore,the order of double bond character (and thus rotational barrier) is $I > II > III$.

Which of the following orders of rotation barrier about the $C=C$ bond,as indicated in the structures below,is correct?
$(I)$ $CH_2=CH_2$
$(II)$ $CH_3O-CH=CH_2$
$(III)$ $CH_3O-CH=CH-COOEt$

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Which of the following orders of rotation barrier about the $C=C$ bond,as indicated in the structures below,is correct?$(I)$ $CH_2=CH_2$$(II)$ $CH_3O-CH=CH_2$$(III)$ $CH_3O-CH=CH-COOEt$