1915 Nobel Prize in Chemistry winner Willstätter first synthesized cyclooctatetraene (hereinafter referred to as COT) through artificial synthesis in the early 20th century, and its synthetic route is shown in the following figure:

<MultiModal>![The starting reactant O=C1CC2N(C)C(C1)CCC2 generates **A** under Na and ethanol conditions, then **A** generates **B** under acetic acid and sulfuric acid conditions, **B** generates **C** under the action of iodomethane and silver oxide with heating, **C** generates **D** under the action of bromine and dimethylamine, and **D** generates COT under the action of iodomethane and silver oxide with heating](/media/uploads/16d98f54-1bfe-487b-b254-f89aa911130b.png)</MultiModal>

With the gradual deepening of research on COT, scientists have discovered that COT has important applications in the synthesis of conductive polymer materials. The following figure shows the synthetic route, where **H** is a common conductive polymer material, obtained from COT through a series of reactions.

<MultiModal>![n molecules of COT react with O=C(OC)C#CC(OC)=O under heating conditions to generate n molecules of **E**, n molecules of **E** generate **F** under the action of Grubbs second-generation catalyst, and **F** generates **H** and n molecules of **G** upon heating](/media/uploads/5280bc66-91f0-45b2-b7cd-83268eff89cc.png)</MultiModal>

Select the correct option from the following
1915 年诺贝尔化学奖获得者 Willstätter 在 20 世纪前期首次通过人工合成的方法合成了环辛四烯（后面简称为 COT），其合成路线如下图所示：

<MultiModal>![起始反应物O=C1CC2N(C)C(C1)CCC2在Na和乙醇条件下生成**A**，随后**A**在乙酸、硫酸条件下生成**B**，**B**在碘甲烷、氧化银作用下加热产生**C**，**C**在溴、二甲胺作用下生成**D**，**D**在碘甲烷、氧化银作用下加热产生COT](/media/uploads/16d98f54-1bfe-487b-b254-f89aa911130b.png)</MultiModal>

随着对于 COT 研究的逐渐深入，科学家们发现 COT 在合成导电高分子聚合材料中有着重要的应用。如下图所示合成路线，**H** 为一种常见导电高分子聚合材料，由 COT 经过一系列反应得到。

<MultiModal>![n分子COT在加热条件下与O=C(OC)C#CC(OC)=O反应生成n分子**E**，n分子**E**在Grubbs二代催化剂作用下生成**F**，**F**经加热生成**H**和n分子**G**](/media/uploads/5280bc66-91f0-45b2-b7cd-83268eff89cc.png)</MultiModal>

选出下列选项中正确的一项
COT can react with U to form a complex with the chemical formula $\mathrm{U(COT)_2}$. This complex oxidizes in oxygen to form $\mathrm{U_3O_8}$. In the chemical equation for this combustion reaction, the formation of one molecule of $\mathrm{U_3O_8}$ requires the consumption of 21.33 molecules of oxygen.
The degree of unsaturation of **C** is 3.
**D** does not contain conjugated double bonds.
In the reaction where **B** generates **C**, a small molecule with an atomicity of 13 is produced, and this small molecule will further react with iodomethane.
**F** has electrical conductivity.
**E** has two rings.
A conjugated system exists in **G**, with a total of 6 atoms participating in the conjugation.
$\mathrm{COT}$ 可以与 $\mathrm{U}$ 生成化学式为$\mathrm{U(COT)_2}$ 的配合物，该配合物在氧气中氧化生成 $\mathrm{U_3O_8}$，在该燃烧反应的化学方程式中，生成一分子 $\mathrm{U_3O_8}$ 需要消耗 21.33 分子氧气
**C** 的不饱和度为3
**D** 中不存在共轭双键
**B** 生成 **C** 的反应中，产生了一原子数为13的小分子，该小分子会进一步与碘甲烷反应
**F** 具有导电性
**E** 有两个环
**G** 中存在共轭体系，共有6个原子参与共轭
The chemical formula of $\mathrm{U}$'s $\mathrm{COT}$ complex is $\mathrm{U(C_8H_8)_2}$. The oxidation of one molecule of $\mathrm{U_3O_8}$ requires the consumption of three molecules of $\mathrm{U(C_8H_8)_2}$. At the same time, the $\mathrm{COT}$ as a ligand is oxidized into 48 molecules of $\mathrm{CO_2}$ and 24 molecules of $\mathrm{H_2O}$. This process consumes a total of 8+48*2+24=128 oxygen atoms, i.e., 64 molecules of oxygen.

<Checkpoint>[The generation of one molecule of $\mathrm{U_3O_8}$ requires the consumption of 64 molecules of oxygen, option A is incorrect](1)</Checkpoint>

In the synthetic route of $\mathrm{COT}$, the carbonyl group is first reduced in sodium ethanol to become a hydroxyl group, yielding OC1CC2N(C)C(C1)CCC2 (**A**); then **A** loses water under acidic conditions to yield CN1C2CCCC1C=CC2 (**B**); in the process of **B** generating **C**, iodomethane reacts with the amino group to generate a quaternary ammonium salt, and silver oxide acts as a base to undergo Hoffmann elimination under heating conditions. Since the amino group in **B** forms two bonds with the main ring, it can be eliminated twice, yielding C1=C\C=C/C=C\CC/1 (**C**), with an unsaturation degree of 4.

<Checkpoint>[The chemical formula of **C** is $\mathrm{C_8H_{10}}$, with an unsaturation degree of 4, option B is incorrect](1)</Checkpoint>

In the Hoffmann elimination reaction using iodomethane, iodomethane will continue to react with the amino group to generate a quaternary ammonium salt, eventually yielding a quaternary ammonium salt with three alkyl groups being methyl groups, followed by elimination, releasing one molecule of trimethylamine, with 13 atoms, and it can further react with iodomethane to produce tetramethylammonium iodide.

<Checkpoint>[The reaction of **B** to generate **C** eliminates one molecule of trimethylamine with 13 atoms, which can further react with potassium iodide, option D is correct](1)</Checkpoint>

**C** reacts with bromine and dimethylamine to obtain **D**, which should be that bromine undergoes addition to the double bond and is then substituted by dimethylamine, and finally undergoes another Hoffmann elimination to obtain $\mathrm{COT}$. The addition reaction of bromine and the substitution of dimethylamine may produce a variety of products, but the product retaining the conjugated double bond is more stable. Therefore, **D** may be CN(C)C1CCC(N(C)C)/C=C\C=C/1 (**D1**) or CN(C)C1C(N(C)C)CC/C=C\C=C/1 (**D2**), where **D2** will obtain a conjugated system after one Hoffmann elimination. At this time, another amino group is on the $sp^2$ carbon, which can no longer be eliminated. The actual structure of **D** is **D1**. But both **D1** and **D2** have a conjugated system.

<Checkpoint>[Addition will tend to obtain more stable products with conjugated systems, option C is incorrect](1)</Checkpoint>

In common organic conductive polymers, all need to have a conjugated main electron system. $\mathrm{COT}$ reacts with dimethyl butynedioate under heating conditions, undergoing a Diels-Alder reaction, possibly yielding O=C(OC)C1=C(C(OC)=O)C2/C=C\C=C/C1C=C2 (**E1**) or an isomer where the conjugated double bond is cyclized into a four-membered ring O=C(OC)C1=C(C(OC)=O)C2C3C=CC3C1C=C2 (**E2**). Among them, **E1** cannot form a polymer under the action of Grubbs catalyst, while **E2** can obtain a polymer through ROMP reaction. Therefore, the structure of **E** is **E1**, containing 3 rings.

<Checkpoint>[The structure of **E** is O=C(OC)C1=C(C(OC)=O)C2C3C=CC3C1C=C2, containing 3 rings, option F is incorrect](1)</Checkpoint>

**E** can be obtained by ROMP reaction to obtain **F**, the structure is shown in the figure below, which does not contain a conjugated main electron system and is not conductive.

<MultiModal>![The polymer structure can be represented as O=C(C1=C([C@H]2[C@@H]([C@@H]([C@@H]1C=C2)C=C)C=C)C(OC)=O)OC. The two monosubstituted alkene groups interact through olefin metathesis, losing one molecule of ethylene, and the polymers are connected by double bonds](/media/uploads/16673350-27fa-4316-ac53-89c6cfa779b2.png)</MultiModal>

<Checkpoint>[**F** does not contain a conjugated main electron system and is not conductive, option E is incorrect](1)</Checkpoint>

**F** can undergo a reverse Diels-Alder reaction under heating conditions, producing dimethyl phthalate (**G**) and the common conductive polymer polyacetylene (**H**). The two ester groups in **G** are conjugated with the benzene ring, with 10 atoms participating in the conjugation.

<Checkpoint>[The two ester groups in **G** are conjugated with the benzene ring, with 10 atoms participating in the conjugation, option G is incorrect](1)</Checkpoint>
$\mathrm{U}$ 的 $\mathrm{COT}$ 配合物化学式为 $\mathrm{U(C_8H_8)_2}$，氧化为一分子 $\mathrm{U_3O_8}$ 需要消耗三分子 $\mathrm{U(C_8H_8)_2}$，同时作为配体的 $\mathrm{COT}$ 被氧化为 48 分子 $\mathrm{CO_2}$ 和 24 分子 $\mathrm{H_2O}$，这一过程共消耗8+48*2+24=128个氧原子，即64分子氧气。

<Checkpoint>[生成一分子 $\mathrm{U_3O_8}$ 需要消耗 64 分子氧气，选项A错误](1)</Checkpoint>

在 $\mathrm{COT}$ 的合成路线中，首先在钠乙醇中发生羰基的还原，变成羟基，得到 OC1CC2N(C)C(C1)CCC2（**A**）；随后 **A** 在酸性条件下失水消除得到CN1C2CCCC1C=CC2（**B**）；在 **B** 生成 **C** 的过程中，碘甲烷与氨基反应生成季铵盐，氧化银作为碱在加热条件下发生 Hoffmann 消除，由于 **B** 中氨基与主环成两根键，因此可以消除两次，得到 C1=C\C=C/C=C\CC/1（**C**），不饱和度为 4。

<Checkpoint>[**C** 的化学式为 $\mathrm{C_8H_{10}}$，不饱和度为 4，选项B错误](1)</Checkpoint>

在使用碘甲烷的 Hoffmann 消除反应中，碘甲烷会保持与氨基反应生成季铵盐，最终得到三个烷基为甲基的季铵盐，随后消除，离去一分子三甲胺，原子数为 13，并可以进一步与碘甲烷反应产生四甲基碘化铵。

<Checkpoint>[**B** 生成 **C** 的反应中消除一分子原子数为 13 的三甲胺，其可进一步与碘化钾反应，选项D正确](1)</Checkpoint>

**C** 在溴和二甲胺反应下得到 **D**，应当是溴与双键发生加成后被二甲胺取代，最后再进行一次 Hoffmann 消除得到 $\mathrm{COT}$。溴的加成反应和二甲胺的取代可能产生多种产物，但是产物中保留共轭双键更加稳定，因此 **D** 可能为 CN(C)C1CCC(N(C)C)/C=C\C=C/1（**D1**）或 CN(C)C1C(N(C)C)CC/C=C\C=C/1（**D2**），其中 **D2** 在发生一次 Hoffmann 消除后会得到共轭体系，此时另一氨基在 $sp^2$ 碳上，无法再进行消除，**D** 的实际结构为 **D1**。但无论是 **D1** 还是 **D2** 都具有共轭体系。

<Checkpoint>[加成会倾向于得到更稳定的带共轭体系的产物，选项C错误](1)</Checkpoint>

在常见有机导电高分子中，都需要具有共轭主电子体系。$\mathrm{COT}$ 在加热条件下与丁炔二酸二甲酯反应，发生 Diels-Alder 反应，可能得到 O=C(OC)C1=C(C(OC)=O)C2/C=C\C=C/C1C=C2（**E1**）或共轭双键环化为四元环的异构体 O=C(OC)C1=C(C(OC)=O)C2C3C=CC3C1C=C2（**E2**）。其中 **E1** 无法在 Grubbs 催化剂的作用下形成聚合物，而 **E2** 可以通过 ROMP 反应得到聚合物，因此 **E** 的结构为 **E1** ，含有 3 个环。

<Checkpoint>[**E** 的结构为 O=C(OC)C1=C(C(OC)=O)C2C3C=CC3C1C=C2，含有 3 个环，选项F错误](1)</Checkpoint>

**E** 可以通过 ROMP 反应得到 **F**，结构如下图所示，其中不含共轭主电子体系，不具有导电性。

<MultiModal>![该聚合物结构可以表示为 O=C(C1=C([C@H]2[C@@H]([C@@H]([C@@H]1C=C2)C=C)C=C)C(OC)=O)OC 相互之间具有的两个单取代烯基发生烯烃复分解失去一分子乙烯，单体间以双键相连构成的聚合物](/media/uploads/16673350-27fa-4316-ac53-89c6cfa779b2.png)</MultiModal>

<Checkpoint>[**F** 不含共轭主电子体系，不具有导电性，选项E错误](1)</Checkpoint>

**F** 在加热条件下可以发生逆 Diels-Alder 反应，产生邻苯二甲酸二甲酯（**G**）和常见导电高分子聚乙炔（**H**），**G** 中的两个酯基与苯环共轭，有 10 个原子参与共轭。

<Checkpoint>[**G** 中的两个酯基与苯环共轭，有 10 个原子参与共轭，选项 G 错误](1)</Checkpoint>