As shown in Figure 1, the reactant undergoes the following reaction mechanism to obtain intermediate $\mathbf{A^+}$, intermediate $\mathbf{B}$, and final product $\mathbf{C}$, respectively:

<MultiModal>![Fig. 1, the figure shows a series of consecutive reaction mechanisms. The reactant is described in SMILES as: CS(/C(SC)=C/C(F)(F)F)=O. The reactant reacts with $\mathrm{Tf_2O}$, losing one molecule of $\mathrm{TfO^-}$, and then reacts with phenol, losing one molecule of $\mathrm{TfOH}$, to obtain intermediate $\mathbf{A^+}$. Intermediate $\mathbf{A^+}$ undergoes rearrangement and deprotonation to obtain intermediate $\mathbf{B}$. Intermediate $\mathbf{B}$ undergoes elimination to obtain final product $\mathbf{C}$.](/media/uploads/4b1238ea-dd4a-4ee0-8aae-bddccd10addd.png)</MultiModal>

Speculate on the reaction mechanism and the structures of intermediate $\mathbf{A^+}$, $\mathbf{B}$, and final product $\mathbf{C}$.

The following statements are made:

1.  $\mathbf{A^+}$ contains two carbon-oxygen bonds.
2.  $\mathbf{B}$ contains one sulfur-oxygen bond.
3.  The product of the number of atoms within each six-membered ring or smaller in $\mathbf{C}$, multiplied by the number of carbon-non-carbon non-hydrogen chemical bonds (multiple bonds counted as one bond), is 210.
4.  The oxidation state of the sulfur atom decreases during the entire reaction process.
如图1所示，反应物经历了以下反应机理，分别得到中间体$\mathbf{A^+}$，中间体$\mathbf{B}$和最终产物$\mathbf{C}$：

<MultiModal>![Fig. 1，图中为连续的几步反应机理。反应物以SMILES描述为：CS(/C(SC)=C/C(F)(F)F)=O。反应物与$\mathrm{Tf_2O}$反应，失去一分子$\mathrm{TfO^-}$，然后与苯酚反应，失去一分子$\mathrm{TfOH}$，得到中间体$\mathbf{A^+}$。中间体$\mathbf{A^+}$发生重排，去质子化，得到中间体$\mathbf{B}$。中间体$\mathbf{B}$发生消除，得到最终产物$\mathbf{C}$。](/media/uploads/4b1238ea-dd4a-4ee0-8aae-bddccd10addd.png)</MultiModal>

推测反应机理和中间体$\mathbf{A^+}$、$\mathbf{B}$和最终产物$\mathbf{C}$的结构。

有以下说法：

1. $\mathbf{A^+}$含有两根碳氧键
2. $\mathbf{B}$含有一根硫氧键
3. $\mathbf{C}$中每个六元及以下环内的原子数量乘积再与碳-非碳非氢化学键数量（多重键按一根键计）乘积为210
4. 整个反应过程中硫原子的氧化态降低了
All other options are incorrect
1
2
3
4
1,2
1,3
1,4
2,3
2,4
3,4
1,2,3
1,2,4
1,3,4
2,3,4
1,2,3,4
其他选项均不正确
1
2
3
4
1,2
1,3
1,4
2,3
2,4
3,4
1,2,3
1,2,4
1,3,4
2,3,4
1,2,3,4
In the first step, triflic anhydride can activate the oxygen atom of the reactant, yielding the activated sulfinyl group. It can undergo a nucleophilic substitution reaction with phenol, where phenol attacks the sulfur atom, substituting triflate, to obtain intermediate $\mathbf{A^+}$, the structure of which is shown in Figure 2:

<MultiModal>![Fig. 2, the molecule in the figure is described by SMILES as: C[S+](OC1=CC=CC=C1)/C(SC)=C/C(F)(F)F](/media/uploads/41872858-141b-468a-9f1f-d353574df0d0.png)</MultiModal>

<Checkpoint>[Oxygen atom is activated by triflic anhydride, phenol substitutes $\mathrm{TfOH}$, yielding intermediate $\mathbf{A^+}$ described by SMILES as: C[S+](OC1=CC=CC=C1)/C(SC)=C/C(F)(F)F](1)</Checkpoint>

$\mathbf{A^+}$ contains only one carbon-oxygen bond, statement 1 is incorrect.

According to the question, intermediate $\mathbf{A^+}$ undergoes a rearrangement reaction in the next step. Observing the molecular structure, the sulfur atom is electrophilic, but there is no good nucleophilic site within the molecule. A more likely rearrangement is that the sulfur-oxygen bond undergoes a concerted [3, 3]-sigma migration rearrangement reaction through the double bond and the benzene ring, forming a carbon-carbon bond, yielding the intermediate in Figure 3:

<MultiModal>![Fig. 3, the molecule in the figure is described by SMILES as: C/[S+]=C(C(C1C(C=CC=C1)=O)C(F)(F)F)\SC](/media/uploads/c84d1ed8-9674-4ff9-ba63-214be5211a9d.png)</MultiModal>

<Checkpoint>[Intermediate $\mathbf{A^+}$ undergoes a concerted [3, 3]-sigma migration rearrangement reaction, yielding the intermediate in Figure 3 described by SMILES as: C/[S+]=C(C(C1C(C=CC=C1)=O)C(F)(F)F)\SC](1)</Checkpoint>

This intermediate quickly regains aromaticity to form a hydroxyl group, the hydroxyl group undergoes nucleophilic addition to the carbon-sulfur double bond and deprotonates to obtain intermediate $\mathbf{B}$, the structure of which is shown in Figure 4:

<MultiModal>![Fig. 4, the molecule in the figure is described by SMILES as: CSC1(OC(C=CC=C2)=C2C1C(F)(F)F)SC](/media/uploads/c81954a5-4cba-4c33-9f77-0611a81dade3.png)</MultiModal>

<Checkpoint>[The intermediate in Figure 3 quickly regains aromaticity to form a hydroxyl group, the hydroxyl group undergoes nucleophilic addition to the carbon-sulfur double bond and deprotonates to obtain intermediate $\mathbf{B}$, described by SMILES as: CSC1(OC(C=CC=C2)=C2C1C(F)(F)F)SC](1)</Checkpoint>

$\mathbf{B}$ does not contain a sulfur-oxygen bond, statement 2 is incorrect.

Intermediate $\mathbf{B}$ eliminates a molecule of methanethiol, yielding the aromatic benzofuran structure $\mathbf{C}$, as shown in Figure 5:

<MultiModal>![Fig. 5, the molecule in the figure is described by SMILES as: FC(F)(F)C1=C(OC2=C1C=CC=C2)SC](/media/uploads/0d9016c3-8855-4256-9338-1e0527ac8d74.png)</MultiModal>

<Checkpoint>[Intermediate $\mathbf{B}$ eliminates a molecule of methanethiol, yielding $\mathbf{C}$ described by SMILES as: FC(F)(F)C1=C(OC2=C1C=CC=C2)SC](1)</Checkpoint>

In $\mathbf{C}$, the number of atoms within the six-membered and five-membered rings are 6 and 5, respectively, and the number of carbon-non-carbon non-hydrogen chemical bonds (multiple bonds counted as one bond) is 7. The product of these three is 210, statement 3 is correct.

The sulfur atom of sulfoxide in the reactant forms two carbon-sulfur bonds and one oxygen-sulfur bond, and the final product breaks the oxygen-sulfur bond, leaving only two carbon-sulfur bonds. The entire reaction process is equivalent to the sulfur atom being reduced, and the oxidation state is reduced from +4 to +2, statement 4 is correct.
第一步三氟甲磺酸酐可以活化反应物的氧原子，得到被活化的亚磺酰基。与苯酚可以发生亲核取代反应，苯酚进攻硫原子取代三氟甲磺酸根，得到中间体$\mathbf{A^+}$，结构如图2：

<MultiModal>![Fig. 2，图中分子以SMILES描述为：C[S+](OC1=CC=CC=C1)/C(SC)=C/C(F)(F)F](/media/uploads/41872858-141b-468a-9f1f-d353574df0d0.png)</MultiModal>

<Checkpoint>[氧原子被活化，苯酚取代$\mathrm{TfOH}$，得到中间体$\mathbf{A^+}$以SMILES描述为：C[S+](OC1=CC=CC=C1)/C(SC)=C/C(F)(F)F](1)</Checkpoint>

$\mathbf{A^+}$只含有一根碳氧键，说法1错误。

根据题目，中间体$\mathbf{A^+}$下一步发生重排反应。观察分子结构，硫原子具有亲电性，但是分子内没有较好的亲核位点。更有可能的一种重排是硫氧键通过双键和苯环发生一步[3, 3]-sigma迁移重排反应，形成一根碳碳键，得到图3中间体：

<MultiModal>![Fig. 3，图中分子以SMILES描述为：C/[S+]=C(C(C1C(C=CC=C1)=O)C(F)(F)F)\SC](/media/uploads/c84d1ed8-9674-4ff9-ba63-214be5211a9d.png)</MultiModal>

<Checkpoint>[中间体$\mathbf{A^+}$发生一步[3, 3]-sigma迁移重排反应，得到图3中间体以SMILES描述为：C/[S+]=C(C(C1C(C=CC=C1)=O)C(F)(F)F)\SC](1)</Checkpoint>

该中间体快速重建芳香性形成羟基，羟基亲核加成碳硫双键并去质子化得到中间体$\mathbf{B}$，结构如图4：

<MultiModal>![Fig. 4，图中分子以SMILES描述为：CSC1(OC(C=CC=C2)=C2C1C(F)(F)F)SC](/media/uploads/c81954a5-4cba-4c33-9f77-0611a81dade3.png)</MultiModal>

<Checkpoint>[图3中间体快速重建芳香性形成羟基，羟基亲核加成碳硫双键并去质子化得到中间体$\mathbf{B}$，以SMILES描述为：CSC1(OC(C=CC=C2)=C2C1C(F)(F)F)SC](1)</Checkpoint>

$\mathbf{B}$不含硫氧键，说法2错误

中间体$\mathbf{B}$消除一分子甲硫醇，得到具有芳香性的苯并呋喃结构$\mathbf{C}$，如图 5：

<MultiModal>![Fig. 5，图中分子以SMILES描述为：FC(F)(F)C1=C(OC2=C1C=CC=C2)SC](/media/uploads/0d9016c3-8855-4256-9338-1e0527ac8d74.png)</MultiModal>

<Checkpoint>[中间体$\mathbf{B}$消除一分子甲硫醇，得到$\mathbf{C}$以SMILES描述为：FC(F)(F)C1=C(OC2=C1C=CC=C2)SC](1)</Checkpoint>

$\mathbf{C}$中六元及以下环内原子数分别为6和5，碳-非碳非氢化学键数量（多重键按一根键计）为7，三者乘积为210，说法3正确。

反应物中亚砜基团的硫原子形成两根碳硫键与一根氧硫键，最终产物断开氧硫键只剩两根碳硫键，整个反应过程相当于硫原子被还原，氧化态从+4降低至+2，说法4正确。