Give the products $\mathbf{A}$ of the reaction in Figure 1 and the product $\mathbf{B}$ of the reaction in Figure 2.

<MultiModal>![Fig. 1, the reaction in the figure is described by SMILES as: OC(C=CC=C1)=C1/C(C2=CC=CC=C2)=C3CCCC\3>>[[A]], where the reaction conditions are $\mathrm{HNTf_2 (1mol\%), mCPBA(1.3equiv), EtOAc, rt}$](/media/uploads/3f7809d3-aadd-40cd-9372-5c4f0be854c7.png)</MultiModal>

<MultiModal>![Fig. 2, the reaction in the figure is described by SMILES as: OC(C=CC=C1)=C1/C(C2=C(O)C=CC=C2)=C3CCCC\3>>[[B]], where the reaction conditions are $\mathrm{HNTf_2 (1mol\%), mCPBA(1.3equiv), EtOAc, rt}$](/media/uploads/694409e7-6327-4f99-931c-4b5d1df214d8.png)</MultiModal>

The following statements are made:

1. Only one carbon-carbon single bond cleavage occurs during the formation of both $\mathbf{A}$ and $\mathbf{B}$.
2. $\mathbf{A}$ has three rings of six members or less.
3. $\mathbf{B}$ has five rings of six members or less.
4. The most important reason why the products of Figure 1 and Figure 2 are different is the electron richness of the benzene ring.
给出图1反应的产物$\mathbf{A}$和图2反应产物$\mathbf{B}$。

<MultiModal>![Fig. 1，图中反应以SMILES描述为：OC(C=CC=C1)=C1/C(C2=CC=CC=C2)=C3CCCC\3>>[[A]]，其中反应条件为$\mathrm{HNTf_2 (1mol\%), mCPBA(1.3equiv), EtOAc, rt}$](/media/uploads/3f7809d3-aadd-40cd-9372-5c4f0be854c7.png)</MultiModal>

<MultiModal>![Fig. 2，图中反应以SMILES描述为：OC(C=CC=C1)=C1/C(C2=C(O)C=CC=C2)=C3CCCC\3>>[[B]]，其中反应条件为$\mathrm{HNTf_2 (1mol\%), mCPBA(1.3equiv), EtOAc, rt}$](/media/uploads/694409e7-6327-4f99-931c-4b5d1df214d8.png)</MultiModal>

有以下说法：

1. 生成$\mathbf{A}$和$\mathbf{B}$的过程中都只发生了一次碳碳单键的断裂
2. $\mathbf{A}$中有三个六元及以下的环
3. $\mathbf{B}$中有五个六元及以下的环
4. 图1和图2产物不一样的最重要原因是苯环的富电子程度
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
Figures 1 and 2 share the same principle in the first half of the reaction: $\mathrm{mCPBA}$ introduces an epoxy group onto the reactant double bond.

<Checkpoint>[$\mathrm{mCPBA}$ introduces an epoxy group onto the reactant double bond](1)</Checkpoint>

$\mathrm{HNTf_2}$ is a strong acid, protonating the epoxy group and dissociating into a hydroxyl group, generating the most stable carbocation on the carbon with simultaneous conjugation stabilization of the two benzene rings.

<Checkpoint>[$\mathrm{HNTf_2}$ catalyzes the opening of the epoxide, forming a carbocation at the benzylic position](1)</Checkpoint>

This intermediate structure is similar to the structure after the departure of a hydroxyl group in the Pinacol rearrangement, so a concerted carbon-carbon bond migration occurs in the adjacent position, yielding the intermediate in Figure 3 or the intermediate in Figure 4.

<MultiModal>![Fig. 3, the molecule in the figure is described by SMILES as: OC1=C(C(CCCC2)(C3=CC=CC=C3)C2=O)C=CC=C1](/media/uploads/6c048ec3-40a7-487d-a7fe-20c0b09ef5ad.png)</MultiModal>

<Checkpoint>[The carbocation intermediate undergoes a Pinacol-like rearrangement, and the intermediate generated in the reaction of Figure 1 is represented by SMILES as: OC1=C(C(CCCC2)(C3=CC=CC=C3)C2=O)C=CC=C1](1)</Checkpoint>

<MultiModal>![Fig. 4, the molecule in the figure is described by SMILES as: OC1=C(C(CCCC2)(C3=CC=CC=C3O)C2=O)C=CC=C1](/media/uploads/2cbabf80-97cd-448a-a26b-b442b360d076.png)</MultiModal>

<Checkpoint>[The carbocation intermediate undergoes a Pinacol-like rearrangement, and the intermediate generated in the reaction of Figure 2 is represented by SMILES as: OC1=C(C(CCCC2)(C3=CC=CC=C3O)C2=O)C=CC=C1](1)</Checkpoint>

The carbonyl group of the intermediate can readily form an acetal or hemiacetal structure with the intramolecular hydroxyl group, yielding the final product $\mathbf{A}$ in Figure 5 or the final product $\mathbf{B}$ in Figure 6.

<MultiModal>![Fig. 5, the molecule in the figure is described by SMILES as: OC12CCCCC1(C3=CC=CC=C3)C4=C(O2)C=CC=C4](/media/uploads/ce90456a-2cb1-46fd-809d-8dc547406d12.png)</MultiModal>

<Checkpoint>[The reaction in Figure 1 forms an intramolecular hemiacetal structure, and the final product $\mathbf{A}$ is represented by SMILES as: OC12CCCCC1(C3=CC=CC=C3)C4=C(O2)C=CC=C4](1)</Checkpoint>

<MultiModal>![Fig. 6, the molecule in the figure is described by SMILES as: C1([C@]2(C3=C(O4)C=CC=C3)CCCC[C@]24O5)=C5C=CC=C1](/media/uploads/d640f714-44c9-4e4f-a468-edd809f81e9e.png)</MultiModal>

<Checkpoint>[The reaction in Figure 2 forms an intramolecular hemiacetal structure, and the final product $\mathbf{B}$ is represented by SMILES as: C1([C@]2(C3=C(O4)C=CC=C3)CCCC[C@]24O5)=C5C=CC=C1](1)</Checkpoint>

According to the above mechanism, only one carbon-carbon single bond cleavage occurs in the process of generating $\mathbf{A}$ and $\mathbf{B}$, so statement 1 is correct.
$\mathbf{A}$ has four rings with six or fewer members, so statement 2 is incorrect.
$\mathbf{B}$ has five rings with six or fewer members, so statement 3 is correct.
The most important reason why the products of Figure 1 and Figure 2 are different is that the last step forms an acetal or hemiacetal, which is not related to the electron-richness of the benzene ring.
图1图2反应的前半段原理相同：$\mathrm{mCPBA}$在反应物双键上引入一个环氧基团。

<Checkpoint>[$\mathrm{mCPBA}$在反应物双键上引入一个环氧基团](1)</Checkpoint>

$\mathrm{HNTf_2}$为强酸，质子化环氧基团并解离为羟基，两个苯环同时共轭稳定的碳上生成最稳定的碳正离子。

<Checkpoint>[$\mathrm{HNTf_2}$催化下打开环氧，在苄位形成碳正离子](1)</Checkpoint>

该中间结构与Pinacol重排中一个羟基离去后的结构类似，因此邻位发生一步碳碳键迁移，得到图3中间体或图4中间体。

<MultiModal>![Fig. 3，图中分子以SMILES描述为：OC1=C(C(CCCC2)(C3=CC=CC=C3)C2=O)C=CC=C1](/media/uploads/6c048ec3-40a7-487d-a7fe-20c0b09ef5ad.png)</MultiModal>

<Checkpoint>[碳正离子中间体发生类Pinacol重排，图1反应生成中间体以SMILES表示为：OC1=C(C(CCCC2)(C3=CC=CC=C3)C2=O)C=CC=C1](1)</Checkpoint>

<MultiModal>![Fig. 4，图中分子以SMILES描述为：OC1=C(C(CCCC2)(C3=CC=CC=C3O)C2=O)C=CC=C1](/media/uploads/2cbabf80-97cd-448a-a26b-b442b360d076.png)</MultiModal>

<Checkpoint>[碳正离子中间体发生类Pinacol重排，图2反应生成中间体以SMILES表示为：OC1=C(C(CCCC2)(C3=CC=CC=C3O)C2=O)C=CC=C1](1)</Checkpoint>

中间体羰基恰好可以与分子内的羟基形成缩醛或者半缩醛结构，得到图5最终产物$\mathbf{A}$或图6最终产物$\mathbf{B}$。

<MultiModal>![Fig. 5，图中分子以SMILES描述为：OC12CCCCC1(C3=CC=CC=C3)C4=C(O2)C=CC=C4](/media/uploads/ce90456a-2cb1-46fd-809d-8dc547406d12.png)</MultiModal>

<Checkpoint>[图1反应形成分子内半缩醛结构，最终产物$\mathbf{A}$以SMILES表示为：OC12CCCCC1(C3=CC=CC=C3)C4=C(O2)C=CC=C4](1)</Checkpoint>

<MultiModal>![Fig. 6，图中分子以SMILES描述为：C1([C@]2(C3=C(O4)C=CC=C3)CCCC[C@]24O5)=C5C=CC=C1](/media/uploads/d640f714-44c9-4e4f-a468-edd809f81e9e.png)</MultiModal>

<Checkpoint>[图2反应形成分子内半缩醛结构，最终产物$\mathbf{B}$以SMILES表示为：C1([C@]2(C3=C(O4)C=CC=C3)CCCC[C@]24O5)=C5C=CC=C1](1)</Checkpoint>

根据以上机理，生成$\mathbf{A}$和$\mathbf{B}$的过程中都只发生了一次碳碳单键的断裂，说法1正确。
$\mathbf{A}$中有四个六元及以下的环，说法2错误。
$\mathbf{B}$中有五个六元及以下的环，说法3正确。
图1和图2产物不一样的最重要原因是最后一步形成缩醛或半缩醛，与苯环富电子程度无关。