Pyridine is more prone to aromatic nucleophilic substitution reactions than benzene. The following is the synthetic route of a certain important compound $\bf{X}$:

<MultiModal>![The image describes an organic tandem reaction. The starting substrate $\bf{B}$ is C=CCN, which reacts with O=C1OC(C2=CC=CC=C21)=O and $DMF$ to generate $\bf{C}$. The reaction conditions for $\bf{C}$ to generate $\bf{D}$ are in 2 steps: 1. $9-BBN$; 2. $Pd(OAc)_2,dppr,K_2CO_3,DMF$, and BrC1=NC=C(Br)C=C1. $\bf{D}$ reacts with $NH_2NH_2$ to generate $\bf{E}$, and $\bf{E}$ reacts with $NaNH_2$ to generate $\bf{F}$. The molecular formula of $\bf{F}$ is given in the figure as $C_{11}H_{17}N_3$.](/media/uploads/ea8a8e60-141e-4d68-b385-fa7f43beda16.png)</MultiModal>

The following statements are correct:
吡啶相对于苯环更加容易发生芳香亲核取代反应。下面是某种重要化合物$\bf{X}$的合成路线：

<MultiModal>![该图片描述了一条有机串联反应。最开始底物$\bf{B}$为C=CCN，与O=C1OC(C2=CC=CC=C21)=O和$DMF$反应生成$\bf{C}$。$\bf{C}$生成$\bf{D}$的反应条件分2步：1. $9-BBN$; 2. $Pd(OAc)_2,dppr,K_2CO_3,DMF$,和BrC1=NC=C(Br)C=C1。$\bf{D}$与$NH_2NH_2$反应生成$\bf{E}$，$\bf{E}$与$NaNH_2$反应生成$\bf{F}$，图中给出$\bf{F}$的分子式为$C_{11}H_{17}N_3$。](/media/uploads/ea8a8e60-141e-4d68-b385-fa7f43beda16.png)</MultiModal>

下列说法正确的是：
All other options are incorrect.
The reaction generating $\bf{E}$ is an aromatic nucleophilic substitution reaction.
$\bf{F}$ has three cycles
$\bf{F}$ exists a linkage relationship $N-C-C-N$
$\bf{F}$ has a six-membered ring and a five-membered ring.
$\bf{F}$ does not possess an independent amino group
其他选项均不正确
生成$\bf{E}$的反应为芳香亲核取代反应
$\bf{F}$具有三个环
$\bf{F}$存在键联关系$N-C-C-N$
$\bf{F}$具有一个六元环和一个五元环
$\bf{F}$不具有独立的氨基
The first step of the reaction is relatively simple, which is the Gabriel amidation reaction of the substrate, generating N-substituted phthalimide, the structure of $\bf{C}$ is O=C1N(CC=C)C(C2=CC=CC=C21)=O.

<Checkpoint>[The first step of the reaction is the Gabriel amidation of the substrate](1)</Checkpoint>

<Checkpoint>[The structure of $\bf{C}$ is O=C1N(CC=C)C(C2=CC=CC=C21)=O](1)</Checkpoint>

The second step is a typical Suzuki coupling reaction. First, $9-BBN$ is added to perform hydroboration on the allyl double bond, and then a $Pd(II)$ catalyst is added to couple with the bromine of the substrate 2,5-dibromopyridine. Since the substrate has two bromines, a product in which both bromines are substituted may be obtained (this can also be seen from the chemical formula), and it is speculated that the structural formula of $\bf{D}$ is O=C1C2=C(C=CC=C2)C(N1CCCC3=NC=C(CCCN4C(C(C=CC=C5)=C5C4=O)=O)C=C3)=O.

<Checkpoint>[The second step is a typical Suzuki coupling reaction](1)</Checkpoint>

<Checkpoint>[The substrate has two bromines, so a product in which both bromines are substituted may be obtained](1)</Checkpoint>

<Checkpoint>[It is speculated that the structural formula of $\bf{D}$ is O=C1C2=C(C=CC=C2)C(N1CCCC3=NC=C(CCCN4C(C(C=CC=C5)=C5C4=O)=O)C=C3)=O](1)</Checkpoint>

The reaction of $\bf{D}$ with $N_2H_4$ is the standard method for removing the phthalimide protecting group in Gabriel synthesis, producing an amine. Thus, the structure of $\bf{E}$ is NCCCC1=NC=C(CCCN)C=C1.

<Checkpoint>[The structure of $\bf{E}$ is NCCCC1=NC=C(CCCN)C=C1](1)</Checkpoint>

When $\bf{E}$ is added to sodium amide, sodium amide as a strong base can abstract the proton on the amine group of the substrate; Since the 2/6 positions of pyridine are electron-deficient, the amino anion on the allyl substituent at the 5 position of pyridine can undergo aromatic nucleophilic substitution reaction, and the amino anion attacks the 6 position of pyridine to form a six-membered ring;

<Checkpoint>[The 2/6 positions of pyridine are electron-deficient, and the amino anion can undergo aromatic nucleophilic substitution reaction](1)</Checkpoint>

However, after the formation of the six-membered ring, due to the addition of the substituent at the 6 position, the electron deficiency of pyridine is alleviated, and also because the 3 position is not the most electron-deficient position of pyridine; Therefore, the amino group of the substituent at the 2 position can no longer attack the 3 position to undergo a nucleophilic substitution reaction, so the product $\bf{F}$ only has two six-membered rings; the structure is NCCCC1=NC2=C(CCCN2)C=C1, which just matches the chemical formula.

<Checkpoint>[Due to the addition of the substituent at the 6 position, the electron deficiency of pyridine is alleviated, and also because the 3 position is not the most electron-deficient position of pyridine; Therefore, the amino group of the substituent at the 2 position can no longer attack the 3 position to undergo a nucleophilic substitution reaction](1)</Checkpoint>

<Checkpoint>[$\bf{F}$ only has two six-membered rings; the structure is NCCCC1=NC2=C(CCCN2)C=C1](1)</Checkpoint>

<Checkpoint>[$\bf{F}$ matches the chemical formula, the structure of $\bf{D}$ is speculated to be correct](1)</Checkpoint>

Therefore, options B, C, E, and F are all incorrect; the structure of $\bf{F}$ does not have a N-C-C-N linkage, and option D is incorrect.

In summary, options B-F are all incorrect, and option A is correct.

<MultiModal>![本图片描述了本题涉及未知化合物的结构式，$\bf{C}$结构为O=C1N(CC=C)C(C2=CC=CC=C21)=O，$\bf{D}$的结构式为O=C1C2=C(C=CC=C2)C(N1CCCC3=NC=C(CCCN4C(C(C=CC=C5)=C5C4=O)=O)C=C3)=O，$\bf{E}$结构为NCCCC1=NC=C(CCCN)C=C1，$\bf{F}$结构为NCCCC1=NC2=C(CCCN2)C=C1](/media/uploads/22f18654-1a18-4481-aca7-f88953748321.png)</MultiModal>
反应第一步比较简单，是底物的 Gabriel 酰胺化反应，生成N-取代的邻苯二甲酰亚胺，$\bf{C}$结构为O=C1N(CC=C)C(C2=CC=CC=C21)=O。

<Checkpoint>[反应第一步是底物的 Gabriel 酰胺化](1)</Checkpoint>

<Checkpoint>[$\bf{C}$结构为O=C1N(CC=C)C(C2=CC=CC=C21)=O](1)</Checkpoint>

第二步反应是典型的Suzuki偶联反应。首先加入$9-BBN$，对烯丙基双键进行硼氢化反应，之后加入$Pd(II)$催化剂，与底物2,5-二溴吡啶的溴进行偶联。由于底物有两个溴，从而可能得到两个溴均被取代的产物（这一点也可以从化学式中看出），推测$\bf{D}$的结构式为O=C1C2=C(C=CC=C2)C(N1CCCC3=NC=C(CCCN4C(C(C=CC=C5)=C5C4=O)=O)C=C3)=O。

<Checkpoint>[第二步反应是典型的Suzuki偶联反应](1)</Checkpoint>

<Checkpoint>[底物有两个溴，从而可能得到两个溴均被取代的产物](1)</Checkpoint>

<Checkpoint>[推测$\bf{D}$的结构式为O=C1C2=C(C=CC=C2)C(N1CCCC3=NC=C(CCCN4C(C(C=CC=C5)=C5C4=O)=O)C=C3)=O](1)</Checkpoint>

$\bf{D}$与$N_2H_4$反应，是 Gabriel 合成中脱去邻苯二甲酰亚胺保护基的标准方法，产生氨基。从而$\bf{E}$结构为NCCCC1=NC=C(CCCN)C=C1。

<Checkpoint>[$\bf{E}$结构为NCCCC1=NC=C(CCCN)C=C1](1)</Checkpoint>

$\bf{E}$加入氨基钠，氨基钠作为强碱可以拔去底物氨基上的质子；由于吡啶的2/6号位缺电子，此时位于吡啶5号位的烯丙基取代基上的氨基负离子就可以发生芳香亲核取代反应，氨基负离子进攻吡啶的6号位形成六元环；

<Checkpoint>[吡啶的2/6号位缺电子，氨基负离子可以发生芳香亲核取代反应](1)</Checkpoint>

但是形成六元环后，由于6号位取代基的加入，吡啶缺电子性缓解，也同时因为3号位并非吡啶最缺电子的位置；因此2号位取代基的氨基不再能够进攻3号位发生亲核取代反应，因此产物$\bf{F}$只有两个六元环；结构为NCCCC1=NC2=C(CCCN2)C=C1，刚好符合化学式。

<Checkpoint>[由于6号位取代基的加入，吡啶缺电子性缓解，也同时因为3号位并非吡啶最缺电子的位置；因此2号位取代基的氨基不再能够进攻3号位发生亲核取代反应](1)</Checkpoint>

<Checkpoint>[$\bf{F}$只有两个六元环；结构为NCCCC1=NC2=C(CCCN2)C=C1](1)</Checkpoint>

<Checkpoint>[$\bf{F}$符合化学式，$\bf{D}$结构推测正确](1)</Checkpoint>

由此，选项B,C,E,F均错误；$\bf{F}$的结构没有N-C-C-N的键联关系，选项D错误。

综上，选项B-F均错误，选项A正确。

<MultiModal>![本图片描述了本题涉及未知化合物的结构式，$\bf{C}$结构为O=C1N(CC=C)C(C2=CC=CC=C21)=O，$\bf{D}$的结构式为O=C1C2=C(C=CC=C2)C(N1CCCC3=NC=C(CCCN4C(C(C=CC=C5)=C5C4=O)=O)C=C3)=O，$\bf{E}$结构为NCCCC1=NC=C(CCCN)C=C1，$\bf{F}$结构为NCCCC1=NC2=C(CCCN2)C=C1](/media/uploads/22f18654-1a18-4481-aca7-f88953748321.png)</MultiModal>