There is the following reaction:

<MultiModal>![This image describes several steps of an organic cascade reaction. The starting substrate is C/C=C(N=[N+]=[N-])/C1=CC=CC=C1, which reacts with $\rm{NO^+BF_4^-}$ to obtain $\bf{C}$. $\bf{C}$ has two transformation processes: $\bf{C}$ can directly yield $\bf{P_1}$, and $\bf{C}$ can also first transform into $\bf{D}$, and $\bf{D}$ then transforms into $\bf{P_2}$.](/media/uploads/ec0f84ba-c5c7-405d-ae0d-c8dcae591170.png)</MultiModal>

Known:

1. $\bf{C, D}$ are both intermediates with a single positive charge and each contains only one ring.
2. $\bf{P_1, P_2}$ both contain a five-membered ring.

Which of the following statements is correct:
有如下反应：

<MultiModal>![本图描述了几步有机串联反应。最开始的底物为C/C=C(N=[N+]=[N-])/C1=CC=CC=C1，与$\rm{NO^+BF_4^-}$反应得到$\bf{C}$，$\bf{C}$有两个转化过程：$\bf{C}$可以直接得到$\bf{P_1}$，$\bf{C}$也可以首先转化为$\bf{D}$，$\bf{D}$再转化为$\bf{P_2}$。](/media/uploads/ec0f84ba-c5c7-405d-ae0d-c8dcae591170.png)</MultiModal>

已知：

1. $\bf{C,D}$均为带一个正电荷的中间体且均只含一个环。
2. $\bf{P_1,P_2}$均含有五元环。

下列说法正确的是：
All other options are incorrect
$\bf{P_1}$ possesses the linkage $\mathrm{N-C-N}$
$\bf{P_1}$ exhibits a bonded relationship $\mathrm{N-C-O-N}$
$\bf{P_1}$ possesses a linkage relationship $\mathrm{O-C-CH_3}$
$\bf{P_2}$ possesses a linkage relationship $\mathrm{O-C-N-CH_3}$
$\bf{P_2}$ possesses a connectivity relationship $\mathrm{N-O-N-C}$
$\bf{P_2}$ has a linked relationship $\mathrm{N-O-C-N}$
其他选项均不正确
$\bf{P_1}$存在键联关系$\mathrm{N-C-N}$
$\bf{P_1}$存在键联关系$\mathrm{N-C-O-N}$
$\bf{P_1}$存在键联关系$\mathrm{O-C-CH_3}$
$\bf{P_2}$存在键联关系$\mathrm{O-C-N-CH_3}$
$\bf{P_2}$存在键联关系$\mathrm{N-O-N-C}$
$\bf{P_2}$存在键联关系$\mathrm{N-O-N}$
The $\rm{NO^+}$ in $\rm{NO^+BF_4^-}$ is a strong electrophile. The substrate contains an enamine structure, so the carbon atom of the enamine nucleophilically attacks the positively charged nitrogen atom of $\rm{NO^+}$, resulting in a nitroso substituent and forming intermediate $\bf{C}$, with the structure CC(N=O)/C(C1=CC=CC=C1)=N/[N+]#N.

<Checkpoint>[The carbon atom of the enamine nucleophilically attacks the positively charged nitrogen atom of $\rm{NO^+}$](1)</Checkpoint>

<Checkpoint>[The structure of $\bf{C}$ is CC(N=O)/C(C1=CC=CC=C1)=N/[N+]#N](1)</Checkpoint>

$\bf{C}$ readily loses a nitrogen molecule to become a six-electron nitrogen cation; therefore, the nitrogen atom of the enamine now has electrophilic properties. Observing the structure of $\bf{C}$, the departure of nitrogen can be assisted by neighboring group participation;

<Checkpoint>[$\bf{C}$ readily loses a nitrogen molecule, and the nitrogen atom of the enamine now has electrophilic properties](1)</Checkpoint>

<Checkpoint>[The departure of nitrogen can be assisted by neighboring group participation](1)</Checkpoint>

The oxygen atom of the nitroso group in the substrate has nucleophilic properties and can attack the nitrogen atom of the enamine to assist in the departure of nitrogen. At this time, a five-membered ring product will be directly generated without going through other intermediates. Therefore, the product generated through this route is $\bf{P_1}$, with the structure CC1=NON=C1C2=CC=CC=C2.

<Checkpoint>[The oxygen atom of the nitroso group attacks the nitrogen atom of the enamine to assist in the departure of nitrogen, without going through other intermediates](1)</Checkpoint>

<Checkpoint>[The structure of $\bf{P_1}$ is CC1=NON=C1C2=CC=CC=C2](2)</Checkpoint>

In the substrate, the two groups connected to the carbon atom of the enamine can undergo a 1,2-migration reaction to assist in the departure of nitrogen;

<Checkpoint>[The two groups connected to the carbon atom of the enamine can undergo a 1,2-migration reaction to assist in the departure of nitrogen](1)</Checkpoint>

If the phenyl group migrates, a five-membered ring product cannot be obtained, so this possibility is not considered;

<Checkpoint>[If the phenyl group migrates, a five-membered ring product cannot be obtained](1)</Checkpoint>

If the secondary carbon atom connected to the nitroso group migrates, then nitrogen departs, resulting in an $\rm{sp^2}$ carbocation intermediate $\bf{D}$, with the structure CC(N=O)/N=[C+]/C1=CC=CC=C1.

<Checkpoint>[The secondary carbon atom connected to the nitroso group migrates, resulting in an $\rm{sp^2}$ carbocation intermediate](1)</Checkpoint>

<Checkpoint>[The structure of $\bf{D}$ is CC(N=O)/N=[C+]/C1=CC=CC=C1](1)</Checkpoint>

The carbocation of this intermediate is attacked by the oxygen atom of the nitroso group, forming a five-membered ring product $\bf{P_2}$, with the structure CC1=NOC(C2=CC=CC=C2)=N1.

<Checkpoint>[The structure of $\bf{P_2}$ is CC1=NOC(C2=CC=CC=C2)=N1](2)</Checkpoint>

Based on the structures of $\bf{P_2}$ and $\bf{P_1}$, options B-G are all incorrect, and option A is correct.

<MultiModal>![This figure shows the structures of the unknown species involved in this problem, intermediate $\bf{C}$ has the structure CC(N=O)/C(C1=CC=CC=C1)=N/[N+]#N; intermediate $\bf{D}$ has the structure CC(N=O)/N=[C+]/C1=CC=CC=C1; product $\bf{P_1}$ has the structure CC1=NON=C1C2=CC=CC=C2; $\bf{P_2}$ has the structure CC1=NOC(C2=CC=CC=C2)=N1](/media/uploads/9ac7808d-af6d-46bf-af13-64fbefa76fcc.png)</MultiModal>
$\rm{NO^+BF_4^-}$中的$\rm{NO^+}$是强亲电试剂，底物中存在烯胺结构，从而烯胺的碳原子亲核$\rm{NO^+}$的正电性氮原子，得到亚硝基取代基，生成中间体$\bf{C}$，结构为CC(N=O)/C(C1=CC=CC=C1)=N/[N+]#N。

<Checkpoint>[烯胺的碳原子亲核$\rm{NO^+}$的正电性氮原子](1)</Checkpoint>

<Checkpoint>[$\bf{C}$结构为CC(N=O)/C(C1=CC=CC=C1)=N/[N+]#N](1)</Checkpoint>

$\bf{C}$极易离去一个氮气变成六电子氮正离子；故烯胺的氮原子此时具有亲电性。观察$\bf{C}$的结构，氮气的离去可以由邻近基团参与帮助离去；

<Checkpoint>[$\bf{C}$极易离去一个氮气，烯胺的氮原子此时具有亲电性](1)</Checkpoint>

<Checkpoint>[氮气的离去可以由邻近基团参与帮助离去](1)</Checkpoint>

底物中的亚硝基的氧原子具有亲核性，可以进攻烯胺的氮原子帮助氮气离去，此时会直接生成五元环产物，不经过其他中间体，故通过该途径生成的产物为$\bf{P_1}$，结构为CC1=NON=C1C2=CC=CC=C2。

<Checkpoint>[亚硝基的氧原子进攻烯胺的氮原子帮助氮气离去，不经过其他中间体](1)</Checkpoint>

<Checkpoint>[$\bf{P_1}$结构为CC1=NON=C1C2=CC=CC=C2](2)</Checkpoint>

底物中，烯胺的碳原子连接的两个基团可以发生1，2迁移反应协助氮气离去；

<Checkpoint>[烯胺的碳原子连接的两个基团可以发生1，2迁移反应协助氮气离去](1)</Checkpoint>

如果苯基进行迁移，得不到五元环产物，不考虑该可能性；

<Checkpoint>[苯基进行迁移，得不到五元环产物](1)</Checkpoint>

如果连有亚硝基的二级碳原子发生迁移，此时氮气离去，得到$\rm{sp^2}$的碳正离子中间体$\bf{D}$，结构为CC(N=O)/N=[C+]/C1=CC=CC=C1。

<Checkpoint>[连有亚硝基的二级碳原子发生迁移，得到$\rm{sp^2}$的碳正离子中间体](1)</Checkpoint>

<Checkpoint>[$\bf{D}$结构为CC(N=O)/N=[C+]/C1=CC=CC=C1](1)</Checkpoint>

该中间体的碳正离子被亚硝基的氧原子进攻，形成五元环产物$\bf{P_2}$，结构为CC1=NOC(C2=CC=CC=C2)=N1。

<Checkpoint>[$\bf{P_2}$结构为CC1=NOC(C2=CC=CC=C2)=N1](2)</Checkpoint>

根据$\bf{P_2}$和$\bf{P_1}$的结构判断，选项B-G均不正确，选项A正确。

<MultiModal>![该图给出了本题涉及到的未知物种结构，中间体$\bf{C}$结构为CC(N=O)/C(C1=CC=CC=C1)=N/[N+]#N；中间体$\bf{D}$结构为CC(N=O)/N=[C+]/C1=CC=CC=C1；产物$\bf{P_1}$结构为CC1=NON=C1C2=CC=CC=C2；$\bf{P_2}$结构为CC1=NOC(C2=CC=CC=C2)=N1](/media/uploads/9ac7808d-af6d-46bf-af13-64fbefa76fcc.png)</MultiModal>