Figure 1 shows that the reaction proceeds via a free radical mechanism and undergoes an electrocyclic ring-closure process in the reaction.
Deduce the structural formulas of all key intermediates in the reaction.

<MultiModal>![Fig. 1, the reaction in the figure is represented by SMILES as: O=S(N1CC2C=CC=C1)(C2)=O>[CC(C)=O.CC#N]>[H][C@]34NC[C@H](CS5(=O)=O)[C@]35C=C4, where the reaction conditions also include hv, indicating ultraviolet irradiation](/media/uploads/92ddeeca-29fb-4b13-955a-01d449d97810.png)</MultiModal>

The following statements are made:

1.  An intermediate containing a five-membered ring is generated during the reaction (intermediates do not include the reactants and products in Figure 1).
2.  An intermediate containing a four-membered ring is generated during the reaction (intermediates do not include the reactants and products in Figure 1).
3.  The reaction involves the breaking of carbon-carbon single bonds.
4.  The total number of carbon-carbon single bonds formed in the reaction is 2.

The option containing all correct statements and the most correct statements is:
图1反应通过自由基机理进行,并且反应过程中经历了一步电环化过程。
推断该反应所有关键中间体的结构简式。

<MultiModal>![Fig. 1，图中反应以SMILES表示为：O=S(N1CC2C=CC=C1)(C2)=O>[CC(C)=O.CC#N]>[H][C@]34NC[C@H](CS5(=O)=O)[C@]35C=C4，其中反应条件还有hv，表示紫外光照](/media/uploads/92ddeeca-29fb-4b13-955a-01d449d97810.png)</MultiModal>

有以下说法：

1. 反应过程中产生了包含五元环的中间体(中间体不包括图1中的反应物与产物)
2. 反应过程中产生了包含四元环的中间体(中间体不包括图1中的反应物与产物)
3. 反应过程涉及了碳碳单键的断裂
4. 反应中总共形成的碳碳单键数量为2

选项中说法全部正确且有最多正确说法的选项为：
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
Observing the product, it can be found in the reactants. The reaction follows a free radical mechanism, and the $\mathrm{S-N}$ bond is most likely to break first under light irradiation, forming the intermediate shown in Figure 2 (where R represents a free radical).

<MultiModal>![Fig. 2, the molecule is represented by SMILES as: O=S(CC1CN([R])C=CC=C1)([R])=O, where R represents a free radical](/media/uploads/9e56b797-7fc1-427f-8aec-f7edb1862f2c.png)</MultiModal>

<Checkpoint>[$\mathrm{S-N}$ bond cleavage occurs under light irradiation, forming an intermediate represented by SMILES as: O=S(CC1CN([R])C=CC=C1)([R])=O, where R represents a free radical](1)</Checkpoint>

Since the free radical on the seven-membered ring is conjugated with two double bonds, the free radical can re-couple with the sulfur free radical at the other end to form a four-membered ring. The structure of this intermediate is shown in Figure 3. Statement 2 is correct.

<MultiModal>![Fig. 3, the molecule is represented by SMILES as: C1=CC2C(CN=C1)CS2(=O)=O](/media/uploads/b232c4a9-a751-49d7-8193-c78a20e7a73a.png)</MultiModal>

<Checkpoint>[The free radical is coupled to the sulfur atom at the other end through conjugation, forming an intermediate represented by SMILES as: C1=CC2C(CN=C1)CS2(=O)=O](1)</Checkpoint>

This intermediate undergoes a one-step hydrogen migration to obtain the intermediate in Figure 4.

<MultiModal>![Fig. 4, the molecule is represented by SMILES as: O=S1(CC2CNC=CC=C12)=O](/media/uploads/8eec9d97-5468-4fee-baa2-14553198c2e9.png)</MultiModal>

<Checkpoint>[This intermediate undergoes a one-step hydrogen migration to obtain an intermediate, represented by SMILES as: O=S1(CC2CNC=CC=C12)=O](1)</Checkpoint>

The two conjugated double bonds of this intermediate are excited under ultraviolet irradiation, and a symmetry-allowed four-membered ring electrocyclic reaction occurs, yielding the product.

<Checkpoint>[Electrocyclization forms a four-membered ring](1)</Checkpoint>

Based on the above reaction process, it can be judged that statements 1 and 3 are incorrect. Only a carbon-carbon single bond is formed in the electrocyclic reaction, so statement 4 is incorrect.
观察产物，可以发现反应物中的。反应为自由基机理，光照下最可能先发生$\mathrm{S-N}$键断裂，形成图2中间体（其中R表示自由基）。

<MultiModal>![Fig. 2，该分子以SMILES表示为：O=S(CC1CN([R])C=CC=C1)([R])=O，其中R表示自由基](/media/uploads/9e56b797-7fc1-427f-8aec-f7edb1862f2c.png)</MultiModal>

<Checkpoint>[光照下发生$\mathrm{S-N}$键断裂，形成中间体 以SMILES表示为：O=S(CC1CN([R])C=CC=C1)([R])=O，其中R表示自由基](1)</Checkpoint>

由于七元环上的自由基与两个双键共轭，因此该自由基可以与硫自由基在另一端重新偶联，形成一个四元环，该中间体结构如图3。说法2正确。

<MultiModal>![Fig. 3，该分子以SMILES表示为：C1=CC2C(CN=C1)CS2(=O)=O](/media/uploads/b232c4a9-a751-49d7-8193-c78a20e7a73a.png)</MultiModal>

<Checkpoint>[自由基通过共轭在另一端与硫原子偶联，形成中间体以SMILES表示为：C1=CC2C(CN=C1)CS2(=O)=O](1)</Checkpoint>

该中间体经一步氢迁移得到图4中间体。

<MultiModal>![Fig. 4，该分子以SMILES表示为：O=S1(CC2CNC=CC=C12)=O](/media/uploads/8eec9d97-5468-4fee-baa2-14553198c2e9.png)</MultiModal>

<Checkpoint>[该中间体经一步氢迁移得到中间体，以SMILES表示为：O=S1(CC2CNC=CC=C12)=O](1)</Checkpoint>

该中间体两个共轭双键在紫外照射下激发，发生符合对称性的四元环电环化，得到产物。

<Checkpoint>[电环化形成四元环](1)</Checkpoint>

根据以上反应过程可以判断，说法1,3错误。反应中仅在电环化反应形成碳碳单键，说法4错误。