Analyze the mechanism of the reaction in Figure 1, the structure of the intermediate $\mathbf{C}$, and the mechanism of the transformation from $\mathbf{C}$ to the final product:

<MultiModal>![The image shows a multi-step reaction: CC1=C(C(S(=O)(C2=CC=C(C(F)(F)F)C=C2)=O)CC(C3(O)C(O)C(CO3)C)C)C=C(OC)C(O)=C1>$\mathrm{PhI(OCOCF_3)_2, HFIP/DCM}$>[C],[C]>$\mathrm{t-BuOK, THF}$>CC1=CC(C(OC)=CC2=C1CC(C3(O2)C(O)C(C)CO3)C)=O](/media/uploads/230a4f1d-80ee-4c10-8202-687ac58fa56e.png)</MultiModal>

There are the following statements:

1.  Intermediate $\mathbf{C}$ contains three non-aromatic rings.
2.  A large strained structure is formed during the process from intermediate $\mathbf{C}$ to the product.
3.  During the process from intermediate $\mathbf{C}$ to the product, the formation of the seven-membered ring structure occurs simultaneously with the construction of aromaticity.

The option in which all statements are correct and the number of correct statements is the largest is:
分析以下图1反应的机理，中间产物$\mathbf{C}$的结构以及由 $\mathbf{C}$ 转化为最终产物的机理：

<MultiModal>![图片为多步反应：CC1=C(C(S(=O)(C2=CC=C(C(F)(F)F)C=C2)=O)CC(C3(O)C(O)C(CO3)C)C)C=C(OC)C(O)=C1>$\mathrm{PhI(OCOCF_3)_2, HFIP/DCM}$>[C],[C]>$\mathrm{t-BuOK, THF}$>CC1=CC(C(OC)=CC2=C1CC(C3(O2)C(O)C(C)CO3)C)=O](/media/uploads/230a4f1d-80ee-4c10-8202-687ac58fa56e.png)</MultiModal>

有以下几个说法：

1. 中间体$\mathbf{C}$ 含有三个无芳香性的环
2. 中间体$\mathbf{C}$到产物过程中形成了大张力结构
3. 中间体$\mathbf{C}$到产物过程中，七元环环结构形成与芳香性建构同时发生

选项中说法均正确且正确数量最多的是：
All other options are incorrect
1
2
3
1,2
1,3
2,3
1,2,3
其他选项均不正确
1
2
3
1,2
1,3
2,3
1,2,3
Iodine reagent oxidation causes the carbon atom with the lowest intramolecular ionization potential to lose an electron, generating the most stable carbocation. This step occurs at the para position of the phenolic hydroxyl group. The adjacent hydroxyl group undergoes nucleophilic attack with the carbocation, forming a stable six-membered ring structure. The structure of intermediate $\mathbf{C}$ is shown in Figure 2:

<MultiModal>![Fig. 2，图中分子结构以SMILES描述为：O=C1C=C(C)C2(C=C1OC)C(S(=O)(C3=CC=C(C(F)(F)F)C=C3)=O)CC(C)C4(O2)C(O)C(CO4)C](/media/uploads/81db96dd-fe10-46fc-8a33-be3a7de7eb11.png)</MultiModal>

<Checkpoint>[Oxidation occurs at the para position of the phenolic hydroxyl group, and the generated carbocation couples with the adjacent hydroxyl group to form a six-membered ring. The structure of the intermediate is described by SMILES as: O=C1C=C(C)C2(C=C1OC)C(S(=O)(C3=CC=C(C(F)(F)F)C=C3)=O)CC(C)C4(O2)C(O)C(CO4)C](1)</Checkpoint>

Potassium tert-butoxide abstracts the most acidic hydrogen atom within the molecule. This hydrogen atom is located on the carbon bearing the sulfonyl group, and the resulting carbanion can be stabilized by conjugation with the sulfonyl group. The carbanion then undergoes Michael addition with the adjacent $\alpha,\beta$-unsaturated ketone, forming an unstable three-membered ring, as shown in Figure 3:

<MultiModal>![Fig. 3，图中分子以SMILES描述为：[O-]C1=CC2(C)C3(C=C1OC)C(S(=O)(C4=CC=C(C(F)(F)F)C=C4)=O)2CC(C)C5(O3)C(O)C(CO5)C](/media/uploads/b35ca355-0996-4ecf-b52f-be6798f43dc5.png)</MultiModal>

<Checkpoint>[Potassium tert-butoxide causes the carbon bearing the sulfonyl group to form a carbanion, which then undergoes Michael addition to form a three-membered ring. The structure of the intermediate is described by SMILES as: [O-]C1=CC2(C)C3(C=C1OC)C(S(=O)(C4=CC=C(C(F)(F)F)C=C4)=O)2CC(C)C5(O3)C(O)C(CO5)C](1)</Checkpoint>

Observing the product structure, it can be found that a seven-membered ring is ultimately formed in the benzene ring portion. Therefore, the next step is an electrocyclic ring-opening reaction of the unstable three-membered ring, forming a seven-membered ring containing three conjugated double bonds, as shown in Figure 4:

<MultiModal>![Fig. 4，图中分子以SMILES描述为：[O-]C1=C(C=C2C(C(C)=C1)(S(=O)(C3=CC=C(C(F)(F)F)C=C3)=O)CC(C)C4(O2)C(O)C(CO4)C)OC](/media/uploads/c94c614e-1760-4ef6-94c7-f6a0fcd8a47a.png)</MultiModal>

<Checkpoint>[Electrocyclic ring-opening forms a seven-membered ring. The structure of the intermediate is represented by SMILES as: [O-]C1=C(C=C2C(C(C)=C1)(S(=O)(C3=CC=C(C(F)(F)F)C=C3)=O)CC(C)C4(O2)C(O)C(CO4)C)OC](1)</Checkpoint>

In the final step, the carbanion eliminates the sulfonyl anion, regenerating the aromatic ring system.

<Checkpoint>[Elimination of the sulfonyl anion, generating the aromatic ring system](1)</Checkpoint>

In summary, $\bf{C}$ has 3 non-aromatic rings, statement 1 is correct; the process from $\bf{C}$ to the product involves an intermediate containing a highly strained three-membered ring, statement 2 is correct; the process from $\bf{C}$ to the product first undergoes electrocyclic ring-opening to form a seven-membered ring, and then eliminates the sulfonyl anion to make the seven-membered ring acquire aromaticity. The formation of the seven-membered ring structure and the establishment of aromaticity do not occur simultaneously, statement 3 is incorrect. The answer is option E.
碘试剂氧化下分子内电离势最低的碳原子失去电子，生成最稳定的一个碳正离子，该步发生在酚羟基的对位。邻近羟基与该碳正离子发生亲核，形成稳定的六元环结构。中间体$\mathbf{C}$的结构如图2：

<MultiModal>![Fig. 2，图中分子结构以SMILES描述为：O=C1C=C(C)C2(C=C1OC)C(S(=O)(C3=CC=C(C(F)(F)F)C=C3)=O)CC(C)C4(O2)C(O)C(CO4)C](/media/uploads/81db96dd-fe10-46fc-8a33-be3a7de7eb11.png)</MultiModal>

<Checkpoint>[氧化反应发生在酚羟基对位，生成的碳正离子与邻近羟基偶联形成六元环，中间体结构以SMILES描述为：O=C1C=C(C)C2(C=C1OC)C(S(=O)(C3=CC=C(C(F)(F)F)C=C3)=O)CC(C)C4(O2)C(O)C(CO4)C](1)</Checkpoint>

叔丁醇钾攫取分子内酸性最强的氢原子，该氢原子位于磺酰基所在碳上，形成碳负离子可以与磺酰基共轭稳定。碳负离子随后与邻近的$\alpha,\beta$-不饱和酮发生Micheal加成，生成一个不稳定的三元环，如图3：

<MultiModal>![Fig. 3，图中分子以SMILES描述为：[O-]C1=CC2(C)C3(C=C1OC)C(S(=O)(C4=CC=C(C(F)(F)F)C=C4)=O)2CC(C)C5(O3)C(O)C(CO5)C](/media/uploads/b35ca355-0996-4ecf-b52f-be6798f43dc5.png)</MultiModal>

<Checkpoint>[叔丁醇钾使得磺酰基所在碳形成碳负离子，进而发生Micheal加成形成三元环，中间体结构以SMILES描述为：[O-]C1=CC2(C)C3(C=C1OC)C(S(=O)(C4=CC=C(C(F)(F)F)C=C4)=O)2CC(C)C5(O3)C(O)C(CO5)C](1)</Checkpoint>

观察产物结构，可以发现苯环部分最终形成了七元环。因此下一步不稳定三元环发生电开环反应，形成含三根共轭双键的七元环，如图4：

<MultiModal>![Fig. 4，图中分子以SMILES描述为：[O-]C1=C(C=C2C(C(C)=C1)(S(=O)(C3=CC=C(C(F)(F)F)C=C3)=O)CC(C)C4(O2)C(O)C(CO4)C)OC](/media/uploads/c94c614e-1760-4ef6-94c7-f6a0fcd8a47a.png)</MultiModal>

<Checkpoint>[电开环形成七元环，中间体结构以SMILES表示为：[O-]C1=C(C=C2C(C(C)=C1)(S(=O)(C3=CC=C(C(F)(F)F)C=C3)=O)CC(C)C4(O2)C(O)C(CO4)C)OC](1)</Checkpoint>

最后一步碳负离子消除磺酰负离子，重新生成芳香环体系。

<Checkpoint>[消除磺酰负离子，生成芳香环体系](1)</Checkpoint>

综上所述，$\bf{C}$有3个非芳香环，说法1正确；$\bf{C}$到产物过程经历了含高张力三元环的中间体，说法2正确；$\bf{C}$到产物过程先经历电开环形成七元环，再消除磺酰负离子使七元环获得芳香性，七元环环结构形成与芳香性建构没有同时发生，说法3错误。答案选E选项。