The construction of a nitrogen-containing four-membered ring can be achieved through the reaction route in Figure 1.

<MultiModal>![Fig. 1, the figure shows a three-step continuous reaction. The first step reaction is described by SMILES as: CC1=CC=C(S(/N=C/C2=CC=CC=C2)(=O)=O)C=C1 and C=C[Mg]Cl react to generate [[A]], where the reaction conditions are $\mathrm{THF/Et_2O, r.t.}$ The second step is described by SMILES as: from [[A]] to generate [[B]], where the reaction conditions are $\mathrm{BH_3\cdot SMe_2, then  NaOH, H_2O_2}$. The third step is described by SMILES as: [B] reacts with CC(OC(/N=N/C(OC(C)C)=O)=O)C to generate CC1=CC=C(S(N2CCC2C3=CC=CC=C3)(=O)=O)C=C1, where the reaction conditions are $\mathrm{PPh_3, THF, r.t.}$](/media/uploads/3f3ceb9c-4cb2-4279-8200-f229c8d68299.png)</MultiModal>

Propose the reaction mechanism and the structural formulas of intermediate products $\mathbf{A}$ and $\mathbf{B}$.

The nitrogen-containing four-membered ring can also undergo the subsequent transformation shown in Figure 2:

<MultiModal>![Fig. 2, the reaction in the figure is described by SMILES as: CC1=CC=C(S(N2CCC2C3=CC=CC=C3)(=O)=O)C=C1 and N#CC4=CC=CC=C4 generate [[C]], where the reaction conditions are $\mathrm{BF_3 \cdot OEt_2, DCM, r.t.}$](/media/uploads/a16ae0c3-d60f-4898-8a33-b7832fe6f3c8.png)</MultiModal>

Given the ${ }^{1} \mathrm{H}$ NMR $\left(\mathrm{CDCl}_{3}, 400 \mathrm{MHz}\right) \delta 7.52(\mathrm{~m}, 4 \mathrm{H}), 7.43(\mathrm{~m}, 1 \mathrm{H}), 7.34-7.24(\mathrm{~m}, 7 \mathrm{H}), 7.13(\mathrm{~m}, 2 \mathrm{H}), 4.59(\mathrm{dd}, J=8.56,5.4 \mathrm{~Hz}, 1 \mathrm{H}), 3.95(\mathrm{~m}, 1 \mathrm{H}), 3.87(\mathrm{~m}, 1 \mathrm{H}), 2.43(\mathrm{~s}, 3 \mathrm{H}), 2.19(\mathrm{~m}, 1 \mathrm{H}), 1.78$
$(\mathrm{~m}, 1 \mathrm{H})$ of C. Propose the reaction mechanism and the structure of product $\mathbf{C}$.

There are the following statements:

1. Intermediate product $\mathbf{B}$ contains a secondary hydroxyl group.
2. The mechanism from intermediate product $\mathbf{B}$ to the nitrogen-containing four-membered ring structure is a free radical mechanism.
3. Product $\mathbf{C}$ contains a total of three rings.
4. A total of two new carbon-nitrogen bonds are formed in the process of the nitrogen-containing four-membered ring reacting with benzonitrile to obtain product $\mathbf{C}$.
通过图1反应路线, 可以实现氮杂四元环的构筑。

<MultiModal>![Fig. 1，图中为三步连续反应。第一步反应以SMILES描述为：CC1=CC=C(S(/N=C/C2=CC=CC=C2)(=O)=O)C=C1和C=C[Mg]Cl反应生成[[A]]，其中反应条件为$\mathrm{THF/Et_2O, r.t.}$。第二步以SMILES描述为：从[[A]]生成[[B]]，其中反应条件为$\mathrm{BH_3\cdot SMe_2, then  NaOH, H_2O_2}$。第三步以SMILES描述为：[B]与CC(OC(/N=N/C(OC(C)C)=O)=O)C反应生成CC1=CC=C(S(N2CCC2C3=CC=CC=C3)(=O)=O)C=C1，其中反应条件为$\mathrm{PPh_3, THF, r.t.}$。](/media/uploads/3f3ceb9c-4cb2-4279-8200-f229c8d68299.png)</MultiModal>

推测反应机理和中间产物 $\mathbf{A}$ 和 $\mathbf{B}$ 的结构简式。

该氮杂四元环还能继续发生图2后续转化：

<MultiModal>![Fig. 2，图中反应以SMILES描述为：CC1=CC=C(S(N2CCC2C3=CC=CC=C3)(=O)=O)C=C1和N#CC4=CC=CC=C4生成[[C]]，其中反应条件为$\mathrm{BF_3 \cdot OEt_2, DCM, r.t.}$](/media/uploads/a16ae0c3-d60f-4898-8a33-b7832fe6f3c8.png)</MultiModal>

已知 C 的 ${ }^{1} \mathrm{H}$ NMR $\left(\mathrm{CDCl}_{3}, 400 \mathrm{MHz}\right) \delta 7.52(\mathrm{~m}, 4 \mathrm{H}), 7.43(\mathrm{~m}, 1 \mathrm{H}), 7.34-7.24(\mathrm{~m}, 7 \mathrm{H}), 7.13(\mathrm{~m}, 2 \mathrm{H}), 4.59(\mathrm{dd}, J=8.56,5.4 \mathrm{~Hz}, 1 \mathrm{H}), 3.95(\mathrm{~m}, 1 \mathrm{H}), 3.87(\mathrm{~m}, 1 \mathrm{H}), 2.43(\mathrm{~s}, 3 \mathrm{H}), 2.19(\mathrm{~m}, 1 \mathrm{H}), 1.78$
$(\mathrm{~m}, 1 \mathrm{H})$ 。推测反应机理和产物$\mathbf{C}$的结构。

有以下几种说法：

1. 中间产物 $\mathbf{B}$中含有一个二级羟基
2. 中间产物 $\mathbf{B}$到氮杂四元环结构的机理为自由基机理
3. 产物$\mathbf{C}$总共含有三个环
4. 氮杂四元环与苯腈反应得到产物$\mathbf{C}$的过程中总共新形成了两根碳氮键
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
The first step is the addition of Grignard reagent to electrophilic imine, followed by work-up to obtain intermediate $\mathbf{A}$, whose structure is shown in Figure 3.

<MultiModal>![Fig. 3，图中分子以SMILES描述为：C=CC(C1=CC=CC=C1)NS(C2=CC=C(C)C=C2)(=O)=O](/media/uploads/6873ea1c-1706-4f9f-8679-2d155e375f82.png)</MultiModal>

<Checkpoint>[Grignard reagent undergoes nucleophilic addition to imine, followed by work-up to obtain intermediate $\mathbf{A}$, whose structure is described by SMILES as: C=CC(C1=CC=CC=C1)NS(C2=CC=C(C)C=C2)(=O)=O](1)</Checkpoint>

In the second step, the intramolecular double bond undergoes hydroboration reaction with $\mathrm{BH_3\cdot SMe_2}$ to obtain a terminal borane compound. $\mathrm{H_2O_2}$ oxidizes the borane compound to obtain the terminal hydroxyl intermediate $\mathbf{B}$, whose structure is shown in Figure 4.

<MultiModal>![Fig. 4,图中分子以SMILES描述为：OCCC(C1=CC=CC=C1)NS(C2=CC=C(C)C=C2)(=O)=O](/media/uploads/96a82598-1488-4ab1-b2ae-95bbd5370539.png)</MultiModal>

The hydroxyl group is a primary hydroxyl group, so statement 1 is incorrect.

<Checkpoint>[Hydroboration reaction occurs, and hydrogen peroxide oxidation yields a terminal hydroxyl structure. Intermediate $\mathbf{B}$ is described by SMILES as: OCCC(C1=CC=CC=C1)NS(C2=CC=C(C)C=C2)(=O)=O](1)</Checkpoint>

In the third step, observing the conditions, it can be found that the reaction is an ionic mechanism. In fact, it is a one-step Mitsunobu Reaction. Triphenylphosphine is activated by $\mathrm{DIAD}$ and combines with the hydroxyl group, promoting the departure of the hydroxyl group to form an azetidine ring.

<Checkpoint>[Mitsunobu reaction activates the hydroxyl group to undergo intramolecular substitution reaction to obtain an azetidine ring](1)</Checkpoint>

The mechanism is an ionic mechanism, so statement 2 is incorrect.

The azetidine product opens the large strained ring system under the action of the strong Lewis acid $\mathrm{BF_3 \cdot OEt_2}$, generating a carbocation at the benzylic position.

<Checkpoint>[Under the action of Lewis acid, the four-membered ring opens, forming a carbocation at the benzylic position](1)</Checkpoint>

Benzonitrile may undergo nucleophilic attack through the nitrogen atom or on the benzene ring. According to the NMR results, the number of aromatic hydrogen atoms is calculated to be 14, which can determine that the reaction is the former.

<Checkpoint>[According to the NMR results, it can be determined that the reaction occurs on the nitrogen atom](1)</Checkpoint>

After the nucleophilic attack of the benzonitrile nitrogen atom, the carbon atom connected to it becomes an electrophilic site, which happens to form a six-membered ring structure with another nitrogen atom in the molecule, undergoing a one-step addition reaction to obtain the product $\mathbf{C}$ structure shown in Figure 5.

<MultiModal>![Fig. 5，图中分子以SMILES描述为：CC1=CC=C(S(N2CCC(C3=CC=CC=C3)N=C2C4=CC=CC=C4)(=O)=O)C=C1](/media/uploads/81335f05-21e1-49f1-9e5a-88a664c78307.png)</MultiModal>

<Checkpoint>[Intramolecular nucleophilic addition of nitrogen atom to nitrilium ion to obtain a stable six-membered ring product $\mathbf{C}$, whose structure is described by SMILES as: CC1=CC=C(S(N2CCC(C3=CC=CC=C3)N=C2C4=CC=CC=C4)(=O)=O)C=C1](1)</Checkpoint>

Product $\mathbf{C}$ contains four rings, so statement 3 is incorrect.

A total of two new carbon-nitrogen bonds were formed in the reaction process to obtain product $\mathbf{C}$, so statement 4 is correct.
第一步格氏试剂与亲电亚胺加成，后处理得到中间体 $\mathbf{A}$结构如图3。

<MultiModal>![Fig. 3，图中分子以SMILES描述为：C=CC(C1=CC=CC=C1)NS(C2=CC=C(C)C=C2)(=O)=O](/media/uploads/6873ea1c-1706-4f9f-8679-2d155e375f82.png)</MultiModal>

<Checkpoint>[格氏试剂与亚胺亲核加成，后处理得到中间体 $\mathbf{A}$，结构以SMILES描述为：C=CC(C1=CC=CC=C1)NS(C2=CC=C(C)C=C2)(=O)=O](1)</Checkpoint>

第二步分子内双键与$\mathrm{BH_3\cdot SMe_2}$发生硼氢化反应，得到端基硼化合物。$\mathrm{H_2O_2}$氧化硼化合物得到端基羟基中间体 $\mathbf{B}$，结构如图4。

<MultiModal>![Fig. 4,图中分子以SMILES描述为：OCCC(C1=CC=CC=C1)NS(C2=CC=C(C)C=C2)(=O)=O](/media/uploads/96a82598-1488-4ab1-b2ae-95bbd5370539.png)</MultiModal>

羟基为一级羟基，说法1错误。

<Checkpoint>[发生硼氢化反应，过氧化氢氧化得到端基羟基结构，中间体$\mathbf{B}$以SMILES描述为：OCCC(C1=CC=CC=C1)NS(C2=CC=C(C)C=C2)(=O)=O](1)</Checkpoint>

第三步观察条件可以发现反应为离子机理，实际上是一步光延反应（Mitunobu Reaction），三苯基膦被$\mathrm{DIAD}$活化后与羟基结合，促进羟基离去形成氮杂四元环。

<Checkpoint>[光延反应活化羟基发生分子内取代反应得到氮杂四元环](1)</Checkpoint>

机理为离子机理，说法2错误。

氮杂四元环产物在强路易斯酸$\mathrm{BF_3 \cdot OEt_2}$作用下打开大张力环系，在苄位生成碳正离子。

<Checkpoint>[路易斯酸作用下四元环打开，在苄位形成碳正离子](1)</Checkpoint>

苯腈可能通过氮原子发生亲核，或者在苯环上发生亲核，根据核磁结果计算芳环氢原子数量为14，即可判断反应为前者。

<Checkpoint>[根据核磁结果可以判断反应发生在氮原子上](1)</Checkpoint>

苯腈氮原子亲核进攻后，与其连接的碳原子成为亲电位点，与分子内的另一个氮原子恰好形成六元环结构，发生一步加成反应，得到图5的产物$\mathbf{C}$结构。

<MultiModal>![Fig. 5，图中分子以SMILES描述为：CC1=CC=C(S(N2CCC(C3=CC=CC=C3)N=C2C4=CC=CC=C4)(=O)=O)C=C1](/media/uploads/81335f05-21e1-49f1-9e5a-88a664c78307.png)</MultiModal>

<Checkpoint>[分子内氮原子亲核加成腈鎓离子，得到稳定六元环产物$\mathbf{C}$，结构以SMILES描述为：CC1=CC=C(S(N2CCC(C3=CC=CC=C3)N=C2C4=CC=CC=C4)(=O)=O)C=C1](1)</Checkpoint>

产物$\mathbf{C}$含有四个环，说法3错误。

得到产物$\mathbf{C}$的反应过程中总共新形成了两根碳氮键，说法4正确。