Compounds $\mathbf{C}$ and $\mathbf{D}$ are unstable and react to give the same product $\mathbf{E}$ (Fig. 1, Fig. 2). It is known that $\mathbf{E}$ is a spirocyclic compound.

Predict the structural formulas of the intermediates after the loss of nitrogen from $\mathbf{C}$ and $\mathbf{D}$ and the structure of $\mathbf{E}$.

<MultiModal>![Fig. 1, the reaction diagram for the formation of $\mathbf{E}$ from $\mathbf{C}$, described in SMILES as: [N-]=[N+]1CC2=CC=CC=C2C1>>[[E]]](/media/uploads/72bf8559-bbc2-4f7b-948e-e99aebcae21b.png)</MultiModal>

<MultiModal>![Fig. 2, the reaction diagram for the formation of $\mathbf{E}$ from $\mathbf{D}$, described in SMILES as: C12=CC=CC=C1CN=NC2>> [[E]]](/media/uploads/010de598-a7b3-48bd-86be-b707f424a510.png)</MultiModal>

The following statements are made:

1.  The reaction processes of compound $\mathbf{C}$ and compound $\mathbf{D}$ both pass through the same intermediate belonging to the $C_{2v}$ point group.
2.  $\mathbf{E}$ contains two benzene rings.
3.  The process of generating $\mathbf{E}$ is entirely a free radical mechanism.
4.  $\mathbf{E}$ contains three six-membered rings and contains 4 "$\mathrm{CH_2}$" groups.

The following option with all correct statements and the largest number of correct statements is:
化合物 $\mathbf{C}$ 和化合物 $\mathbf{D}$不稳定，反应得到相同的产物 $\mathbf{E}$ (图1、图2)。已知$\mathbf{E}$为螺环化合物。
推测 $\mathbf{C}$  和 $\mathbf{D}$ 失氮后的中间体的结构简式和$\mathbf{E}$的结构。

<MultiModal>![Fig. 1，图反应为 $\mathbf{C}$ 形成 $\mathbf{E}$的反应，以SMILES描述为：[N-]=[N+]1CC2=CC=CC=C2C1>>[[E]]](/media/uploads/72bf8559-bbc2-4f7b-948e-e99aebcae21b.png)</MultiModal>

<MultiModal>![Fig. 2，图反应为 $\mathbf{D}$ 形成 $\mathbf{E}$的反应，以SMILES描述为：C12=CC=CC=C1CN=NC2>> [[E]]](/media/uploads/010de598-a7b3-48bd-86be-b707f424a510.png)</MultiModal>

有以下说法：

1. 化合物 $\mathbf{C}$ 和化合物 $\mathbf{D}$反应过程中都经过了同一个属于$C_{2v}$点群的中间体
2. $\mathbf{E}$中含有两个苯环
3. 生成$\mathbf{E}$的过程完全为自由基机理
4. $\mathbf{E}$中含有三个六元环，且含有4个“$\mathrm{CH_2}$”基团

下列选项说法全部正确且正确说法数量最多的为：
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 structures of $\mathbf{C}$ and $\mathbf{D}$, it can be found that both have a symmetrical diazo group within the molecule, so they are unstable and easily eliminate nitrogen.

<Checkpoint>[$\mathbf{C}$ and $\mathbf{D}$ easily eliminate nitrogen](1)</Checkpoint>

Since the two carbon-carbon bonds connected to the diazo group are symmetrical, it is difficult to eliminate nitrogen through an ionic mechanism. It is most likely to eliminate nitrogen through a pericyclic reaction-like mechanism. After eliminating one molecule of nitrogen, a radical is formed on each of the two carbons originally connected to the nitrogen atom, and radical coupling may occur. However, a four-membered ring fused with the benzene ring is obtained, which has extremely high tension and is unstable. More likely, the benzene ring directly stabilizes the radical structure through conjugation, resulting in a highly reactive intermediate (as shown in Figure 3). Here, $\mathbf{C}$ and $\mathbf{D}$ generate the same intermediate structure.

<MultiModal>![Fig. 3，The molecule in the figure is described in SMILES as：C=C1C=CC=CC1=C](/media/uploads/12e0a317-0d99-4a76-b210-7804936e57c8.png)</MultiModal>

<Checkpoint>[$\mathbf{C}$ and $\mathbf{D}$ eliminate nitrogen to generate the same intermediate, with the structure: C=C1C=CC=CC1=C](1)</Checkpoint>

The intermediate has a $C_2$ axis and two mirror planes passing through the $C_2$ axis, belonging to the $C_{2v}$ point group. Statement 1 is correct.

<Checkpoint>[The intermediate has a $C_2$ axis and two mirror planes passing through the $C_2$ axis, belonging to the $C_{2v}$ point group](1)</Checkpoint>

According to the question, $\mathbf{E}$ is a spirocyclic compound, indicating that the intermediate is not the final product, which is consistent with the high reactivity of the intermediate. The two intermediates can quickly dimerize and undergo a [4+2] cycloaddition reaction to obtain the next product.

<Checkpoint>[The intermediate is highly reactive and easily dimerizes via a [4+2] addition reaction](1)</Checkpoint>

One molecule of the intermediate acts as a diene, and aromaticity can be reconstructed after the reaction. The other molecule acts as a dienophile, and the reaction may occur on the bridgehead double bond or the six-membered ring. According to the information that product $\mathbf{E}$ is a spirocyclic compound, it can be determined that the reaction occurs on the bridgehead double bond, and the structure of $\mathbf{E}$ is shown in Figure 4:

<MultiModal>![Fig. 4，The molecule in the figure is described in SMILES as：C=C1C=CC=CC21CCC3=CC=CC=C3C2](/media/uploads/e7eceee4-98ff-4086-ac1a-1f55b0036bb2.png)</MultiModal>

<Checkpoint>[The intermediate undergoes [4+2] cycloaddition dimerization to obtain the spirocyclic compound $\mathbf{E}$ with the structure: C=C1C=CC=CC21CCC3=CC=CC=C3C2](1)</Checkpoint>

$\mathbf{E}$ has only one benzene ring, so statement 2 is incorrect. The formation of $\mathbf{E}$ involved a pericyclic reaction mechanism, so statement 3 is incorrect. $\mathbf{E}$ contains three six-membered rings, and there are a total of 4 "$\mathrm{CH_2}$" groups, three on the benzo six-membered ring and one outside the ring, so statement 4 is correct.
观察$\mathbf{C}$和$\mathbf{D}$的结构可以发现，二者分子内均存在一个对称的重氮基团，所以不稳定易脱除氮气。

<Checkpoint>[$\mathbf{C}$和$\mathbf{D}$容易脱除氮气](1)</Checkpoint>

由于重氮基团连接的两根碳碳键对称，因此很难通过离子机理脱除氮气，最有可能通过类似周环反应脱除氮气。脱除一分钟氮气后，在原来与氮原子相连的两个碳上各形成一个自由基，可能发生自由基的偶联，但是得到与苯环并环的四元环，张力极大不稳定。更可能的是苯环直接通过共轭稳定自由基结构，得到高反应性的中间体（如图3）。这里$\mathbf{C}$和$\mathbf{D}$生成的中间体结构相同。

<MultiModal>![Fig. 3，图中分子以SMILES描述为：C=C1C=CC=CC1=C](/media/uploads/12e0a317-0d99-4a76-b210-7804936e57c8.png)</MultiModal>

<Checkpoint>[$\mathbf{C}$和$\mathbf{D}$脱除氮气，生成相同的中间体，结构为：C=C1C=CC=CC1=C](1)</Checkpoint>

该中间体有一个$C_2$轴和穿过$C_2$轴的两个镜面，属于$C_{2v}$点群，说法1正确。

<Checkpoint>[该中间体有一个$C_2$轴和穿过$C_2$轴的两个镜面，属于$C_{2v}$点群](1)</Checkpoint>

根据题干$\mathbf{E}$为螺环化合物，说明该中间体还不是最终产物，这与该中间体具有高反应性符合。两个中间体可以快速二聚，发生[4+2]环加成反应，得到下一步产物。

<Checkpoint>[中间体反应性高，易发生[4+2]加成反应二聚](1)</Checkpoint>

其中一分子中间体作为二烯体，反应后可以重构芳香性。另一分子作为单烯体，反应可能发生在桥头双键或六元环上。根据产物$\mathbf{E}$为螺环化合物的信息，可以确定反应在桥头双键发生，得到$\mathbf{E}$的结构如图4：

<MultiModal>![Fig. 4，图中分子以SMILES描述为：C=C1C=CC=CC21CCC3=CC=CC=C3C2](/media/uploads/e7eceee4-98ff-4086-ac1a-1f55b0036bb2.png)</MultiModal>

<Checkpoint>[中间体发生[4+2]环加成二聚，得到螺环化合物$\mathbf{E}$结构为：C=C1C=CC=CC21CCC3=CC=CC=C3C2](1)</Checkpoint>

$\mathbf{E}$只有一个苯环，说法2错误。生成$\mathbf{E}$的过程中经历了周环反应机理，说法3错误。$\mathbf{E}$含有三个六元环，且包苯并六元环上三个和环外一个共有4个“$\mathrm{CH_2}$”基团，说法4正确。