The following compound

<MultiModal>![Image is a molecular formula, SMILES code is FC(C1=C(C(F)(F)F)C2C(C=C3)C3C1C=C2)(F)F](/media/uploads/fceaeec5-177e-4fed-bb58-c59a7ca0d65a.jpeg)</MultiModal>

After olefin metathesis, polymer $\mathbf{A}$ is obtained (only one double bond in each monomer participates in olefin metathesis). $\mathbf{A}$ spontaneously decomposes into a stable small molecule $\mathbf{B}$ and a well-known polymer $\mathbf{C}$ at room temperature (the polymer chain does not break). Select the correct option:
如下化合物

<MultiModal>![图片为一个分子式，SMILES编码为FC(C1=C(C(F)(F)F)C2C(C=C3)C3C1C=C2)(F)F](/media/uploads/fceaeec5-177e-4fed-bb58-c59a7ca0d65a.jpeg)</MultiModal>

在烯烃复分解后得到高分子$\mathbf{A}$（每个单体只有一根双键参与烯烃复分解），$\mathbf{A}$在室温下会自发分解为稳定的小分子$\mathbf{B}$和一种著名的高分子$\mathbf{C}$（高分子链没有发生断裂），选出正确选项：
All other options are incorrect
$\mathbf{C}$ does not exhibit cis-trans isomerism
$\mathbf{A}$ Spontaneous decomposition is entropically unfavorable, but because it releases stable products, it has a large enthalpic driving force, and therefore can proceed forward.
The molecular weight of small molecule $\mathbf{B}$ is less than 120.
Treating $\mathbf{C}$ with iodine monochloride can enhance conductivity.
$\mathbf{C}$'s monomer has three rotational modes.
其他选项均不正确
$\mathbf{C}$没有顺反异构
$\mathbf{A}$自发分解的过程在熵上不利，但因为放出了稳定的产物，在焓上有巨大驱动力，因此可以正向进行
小分子$\mathbf{B}$的分子量小于120
将$\mathbf{C}$用碘单质处理，可以增强导电性
$\mathbf{C}$的单体有三种转动模式
Observing the molecular structure, it can be seen that the double bond within the four-membered ring in the molecule is very unstable and should be the reaction site for olefin metathesis. Therefore, the monomer structure of polymer $\bf A$ is [R]=CC1C2C(C(F)(F)F)=C(C(F)(F)F)C(C1C=[R])C=C2, where R is the connection site of the polymer monomer.

<Checkpoint>[The monomer structure of polymer $\bf A$ is [R]=CC1C2C(C(F)(F)F)=C(C(F)(F)F)C(C1C=[R])C=C2, where R is the connection site of the polymer monomer](1)</Checkpoint>

<Checkpoint>[The double bond participating in the reaction in olefin metathesis is the double bond within the four-membered ring](1)</Checkpoint>

Subsequently, it was observed that a reverse 4+2 reaction could occur, removing a molecule with a benzene ring structure and obtaining polyacetylene. Therefore, $\bf C$ is polyacetylene, and the structure of $\bf B$ is FC(C1=CC=CC=C1C(F)(F)F)(F)F.

This process generates stable products, reducing tension while producing more molecules, so the reaction entropy is favorable. According to the molecular formula and reaction process, $C$ and $D$ are excluded.

<Checkpoint>[Reverse 4+2 reaction occurs, resulting in $\bf C$ being polyacetylene](1)</Checkpoint>

<Checkpoint>[The structure of $\bf B$ is FC(C1=CC=CC=C1C(F)(F)F)(F)F](1)</Checkpoint>

<Checkpoint>[The number of reaction molecules increases, resulting in an increase in entropy](1)</Checkpoint>

The monomer of polyacetylene is acetylene, which is a linear molecule and has only two rotational modes, excluding $F$.

<Checkpoint>[Acetylene is a linear molecule and does not have three rotational modes](1)</Checkpoint>

Polyacetylene has many double bonds and should have cis-trans isomers, excluding $B$.

<Checkpoint>[Polyacetylene has cis-trans isomers](1)</Checkpoint>

Treating polyacetylene with iodine can result in electron-deficient holes, enhancing conductivity. Therefore, option $E$ is correct, and option $A$ is incorrect.

<Checkpoint>[Treating polyacetylene with iodine can result in electron-deficient holes, enhancing conductivity](1)</Checkpoint>

Therefore, the answer is option E.

<MultiModal>![The monomer structure of polymer $\bf A$ is [R]=CC1C2C(C(F)(F)F)=C(C(F)(F)F)C(C1C=[R])C=C2, where R is the connection site of the polymer monomer; the structure of $\bf B$ is FC(C1=CC=CC=C1C(F)(F)F)(F)F.](/media/uploads/3f357ae9-b202-474d-a5bd-62e34b4066d2.png)</MultiModal>
观察分子结构，可以看到分子中的四元环环内双键部分很不稳定，应为烯烃复分解的反应位点.因此高分子$\bf A$单体结构为[R]=CC1C2C(C(F)(F)F)=C(C(F)(F)F)C(C1C=[R])C=C2，其中R为高分子单体的连接位点。

<Checkpoint>[高分子$\bf A$单体结构为[R]=CC1C2C(C(F)(F)F)=C(C(F)(F)F)C(C1C=[R])C=C2，其中R为高分子单体的连接位点](1)</Checkpoint>

<Checkpoint>[烯烃复分解中参与反应的双键位置为四元环环内双键](1)</Checkpoint>

随后观察到可以发生逆4+2反应，脱去有苯环结构的分子并得到聚乙炔，因此$\bf C$为聚乙炔，$\bf B$结构为FC(C1=CC=CC=C1C(F)(F)F)(F)F。

这个过程生成稳定产物减少张力的同时产生了更多分子，所以反应熵有利，根据分子式和反应过程排除$C$和$D$

<Checkpoint>[发生逆4+2反应，得到$\bf C$为聚乙炔](1)</Checkpoint>

<Checkpoint>[$\bf B$结构为FC(C1=CC=CC=C1C(F)(F)F)(F)F](1)</Checkpoint>

<Checkpoint>[反应分子数增加，为熵增](1)</Checkpoint>

聚乙炔的单体为乙炔，是线性分子，只有两种转动模式，排除$F$

<Checkpoint>[乙炔为线性分子，没有三种转动模式](1)</Checkpoint>

聚乙炔存在很多双键，应当有顺反异构，排除$B$

<Checkpoint>[聚乙炔存在顺反异构](1)</Checkpoint>

将聚乙炔用碘单质处理可以得到缺电子空穴，增强导电性，故选项$E$正确，选项$A$错误

<Checkpoint>[聚乙炔用碘单质处理可以得到缺电子空穴，增强导电性](1)</Checkpoint>

因此答案为选项E

<MultiModal>![高分子$\bf A$单体结构为[R]=CC1C2C(C(F)(F)F)=C(C(F)(F)F)C(C1C=[R])C=C2，其中R为高分子单体的连接位点；$\bf B$结构为FC(C1=CC=CC=C1C(F)(F)F)(F)F。](/media/uploads/3f357ae9-b202-474d-a5bd-62e34b4066d2.png)</MultiModal>