Compare the carbon-carbon double bond rotatio | vedclass

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(B) The rotation around the carbon-carbon double bond requires the breaking of the $\pi$-bond. The energy required for this rotation depends on the st

Vedclass · Accepted Answer

$A > C > B$

Vedclass · Answer

(B) The rotation around the carbon-carbon double bond requires the breaking of the $\pi$-bond. The energy required for this rotation depends on the stability of the resulting zwitterionic species formed upon breaking the $\pi$-bond.In these molecules,the $\pi$-bond breaks to form two rings,one carrying a positive charge and the other a negative charge. The stability of these rings is governed by $H$ückel's rule (aromaticity).For $(A)$: Breaking the bond forms a $5$-membered cation (aromatic,$6\pi$ electrons) and a $3$-membered anion (antiaromatic,$4\pi$ electrons).For $(B)$: Breaking the bond forms a $5$-membered cation (aromatic,$6\pi$ electrons) and a $7$-membered anion (aromatic,$8\pi$ electrons,but actually $7$-membered ring with $8\pi$ electrons is antiaromatic). Wait,let's re-evaluate: $5$-membered ring with $6\pi$ electrons is aromatic. $7$-membered ring with $6\pi$ electrons is aromatic. In $(B)$,we get a $5$-membered aromatic cation and a $7$-membered aromatic anion.For $(C)$: Breaking the bond forms a $3$-membered cation (antiaromatic) and a $7$-membered anion (aromatic). Since $(B)$ forms two aromatic rings,the $\pi$-bond breaks most easily,requiring the least energy. $(C)$ forms an antiaromatic ring,making it the most difficult to break. Thus,the order of rotation energy is $C > A > B$.

Compare the carbon-carbon double bond rotation energy for the following compounds $A$,$B$,and $C$:

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