The figure describes an organic synthesis route, with the molecular formula of the final product being $\mathrm{C_{12}H_{13}ClN_{4}}$. Regarding the following statements about this organic synthesis route, identify the correct one.

<MultiModal>![The figure shows a multi-step reaction: C1=CC(Cl)=CC=C2CC#N>CCC(=O)OCC.[Na].[O]CC>[A], [A]>C=N=N>[B], [B]>[C]>[D], where [A], [B], [C] are codes, the molecular formula of [C] is $\mathrm{CH_5N_3}$, and no additional information about [D] is provided in the figure](/media/uploads/250ef7a9-a721-4cd8-bc76-5bb4d288c655.png)</MultiModal>
下图描述了一个有机合成路线，最终产物的分子式为$\mathrm{C_{12}H_{13}ClN_{4}}$。关于如下有机合成路线的说法，指出正确的一项。

<MultiModal>![图中为多步反应：C1=CC(Cl)=CC=C2CC#N>CCC(=O)OCC.[Na].[O]CC>[A]，[A]>C=N=N>[B]，[B]>[C]>[D]，其中[A]，[B]，[C]是代号，[C]的分子式为$\mathrm{CH_5N_3}$，图上未给出[D]的其他任何信息](/media/uploads/250ef7a9-a721-4cd8-bc76-5bb4d288c655.png)</MultiModal>
The reaction from the first reactant to $\mathbf{A}$ involves an $S_N1$ mechanism.
$\mathbf{A}$ has 5 more carbon atoms than the first reactant.
The $\mathbf{B}$ molecule contains two rings.
During the process from $\mathbf{A}$ to $\mathbf{B}$, no gas is generated.
In this synthetic route, the basicity of $\mathbf{C}$ is utilized.
The stabilizing mechanisms of reagent $\mathbf{C}$ and carbonate ions differ in origin.
After the reaction of $\mathbf{B}$ with $\mathbf{C}$, the number of rings increases.
In the reaction between $\mathbf{B}$ and $\mathbf{C}$, the electrons in the $\mathbf{B}$ molecule attack $\mathbf{C}$.
From $\mathbf{B}$ to the final product, one molecule of formaldehyde is eliminated.
Initially, the cyano nitrogen atom in the reactants becomes incorporated into the heterocycle in the final product.
The final product contains two sets of π-conjugated systems, and there is a strong interaction between them.
In the structural formula of $\mathbf{B}$, there are no exocyclic double bonds.
The structure after protonation of $\mathbf{C}$ belongs to the $C_\mathrm{3}$ point group.
In the final product molecule, up to 22 atoms are coplanar in specific conformations.
从第一个反应物到$\mathbf{A}$的反应涉及$S_N1$机理。
$\mathbf{A}$比第一个反应物多5个碳原子。
$\mathbf{B}$分子中包含两个环。
从$\mathbf{A}$到$\mathbf{B}$过程中，没有气体产生。
在该合成路线中，使用到了$\mathbf{C}$的碱性。
试剂$\mathbf{C}$与碳酸根的稳定性来源机制不同。
$\mathbf{B}$与$\mathbf{C}$反应后，环的数量增加。
在$\mathbf{B}$与$\mathbf{C}$的反应中，$\mathbf{B}$分子中的电子进攻$\mathbf{C}$。
从$\mathbf{B}$到最终产物，离去一分子甲醛。
一开始反应物中的氰基氮原子在最终产物中进入杂环。
最终产物含有两套π共轭体系，且二者之间有较强的相互作用。
$\mathbf{B}$的结构式中，不存在环外双键。
$\mathbf{C}$质子化后的结构，属于$C_\mathrm{3}$点群。
最终产物的分子中，在特定构象下，最多有22个原子共平面。
The initial reactant is  
<MultiModal>![c1cc(Cl)ccc1CC#N](/media/uploads/3f4d8441-a4d6-41c3-9332-1beae7df0a98.png)</MultiModal>  
Under sodium ethoxide conditions, the proton at the α-position of the cyano group (also the benzylic position) is deprotonated, forming a carbanion that attacks the carbonyl carbon of ethyl propionate. Through an addition-elimination mechanism, the ethoxide ion is displaced, yielding product $\mathbf{A}$.  

<Checkpoint>[Formation of carbanion at the α-position of the cyano group](1)</Checkpoint>  

<Checkpoint>[Addition-elimination mechanism occurs](1)</Checkpoint>  

Option $\mathrm{A}$ is incorrect.  

The structure of $\mathbf{A}$ is  
<MultiModal>![CCC(=O)C(C#N)c1ccc(cc1)Cl](/media/uploads/9c50c15e-4434-413e-9ebe-7cd86a9f5337.png)</MultiModal>  

Comparison reveals that it has three more carbons than the initial substrate, so option $\mathrm{B}$ is incorrect.  

<Checkpoint>[Structure of A is CCC(=O)C(C#N)c1ccc(cc1)Cl](1)</Checkpoint>  

<Checkpoint>[A has three more carbons than the substrate](1)</Checkpoint>  

In the structure of $\mathbf{A}$, one hydrogen atom is simultaneously located at the α-position of the cyano group, the α-position of the ketone carbonyl, and the benzylic position. This hydrogen can be further deprotonated by ethoxide to yield an enolate stabilized by the conjugated system:  
<MultiModal>![CCC(=C(C#N)c1ccc(cc1)Cl)[O-]](/media/uploads/f0ea996f-7f44-4020-9066-6f32a7c25c2b.png)</MultiModal>  

This intermediate is methylated at the oxygen terminus by diazomethane to give $\mathbf{B}$.  

<Checkpoint>[Methylation of the enolate at the oxygen terminus](1)</Checkpoint>  

The structure of $\mathbf{B}$ is  
<MultiModal>![CC/C(=C(\C#N)/c1ccc(cc1)Cl)/OC](/media/uploads/93d8d75d-33bc-425f-af8e-b5cbdc0fbc38.png)</MultiModal>  

<Checkpoint>[$\mathbf{B}$ is CC/C(=C(\C#N)/c1ccc(cc1)Cl)/OC](1)</Checkpoint>  

$\mathbf{B}$ contains only one ring, so option $\mathrm{C}$ is incorrect. The conversion from $\mathbf{A}$ to $\mathbf{B}$ releases nitrogen gas, so option $\mathrm{D}$ is incorrect.  

Examining the structure of $\mathbf{B}$, the exocyclic benzylic carbon and the carbon directly bonded to oxygen are part of a carbon-carbon double bond in the structural formula, so option $\mathrm{L}$ is incorrect.  

<Checkpoint>[The benzylic carbon and the carbon directly bonded to oxygen in $\mathbf{B}$ form a carbon-carbon double bond](1)</Checkpoint>  

$\mathbf{C}$ is guanidine, with the structure:  
<MultiModal>![C(=N)(N)N](/media/uploads/e52bae25-f38c-4ddf-a93a-8461cab76f02.png)</MultiModal>  

<Checkpoint>[$\mathbf{C}$ is C(=N)(N)N](1)</Checkpoint>  

$\mathbf{C}$ is basic and exhibits $\mathrm{Y}$-aromaticity. Its stability source is the same as that of the carbonate ion, so option $\mathrm{F}$ is incorrect.  

<Checkpoint>[C exhibits Y-aromaticity](1)</Checkpoint>  

The most stable conformation of the protonated cation of $\mathbf{C}$ is planar, with the carbon and all nitrogen atoms adopting $\mathrm{sp^2}$ hybridization, and the entire ion belongs to the $D_{\mathrm{3h}}$ point group. Option $\mathrm{M}$ is incorrect.  

<Checkpoint>[The cation of $\mathbf{C}$ has a planar conformation with $D_\mathrm{3h}$ symmetry](1)</Checkpoint>  

Intermediate $\mathbf{B}$ reacts with added $\mathbf{C}$, where a nitrogen atom of $\mathbf{C}$ attacks the carbon atom bonded to oxygen in $\mathbf{B}$ (the β-position of the cyano group). Through an addition-elimination mechanism, the methoxide ion is displaced. Option $\mathrm{H}$ is incorrect.  

<MultiModal>![C1CC(=C(C#N)C2=C1C=C(C=C2)Cl)NC(=N)N](/media/uploads/5fa4ba44-44ca-4533-831d-0029215e9cf8.png)</MultiModal>  

<Checkpoint>[Displacement of the methoxide ion](1)</Checkpoint>  

Subsequently, another nitrogen atom originally from $\mathbf{C}$ attacks the cyano carbon, forming a new six-membered ring. Proton transfer converts the original cyano nitrogen into an amino group. Option $\mathrm{G}$ is correct.  

<Checkpoint>[Reaction of B with C forms a six-membered ring](1)</Checkpoint>  

<MultiModal>![CCc1c(c2ccc(cc2)Cl)c(nc(n1)N)N](/media/uploads/35c61ecf-76af-44d8-b7e9-62d9064b29db.png)</MultiModal>  

<Checkpoint>[Final product is CCc1c(c2ccc(cc2)Cl)c(nc(n1)N)N](1)</Checkpoint>  

The synthetic route does not involve the basicity of guanidine, so option $\mathrm{E}$ is incorrect.  

<Checkpoint>[The synthetic route does not involve the basicity of guanidine](1)</Checkpoint>  

In the formation of the final product, the methoxide ion is displaced, not formaldehyde. Option $\mathrm{I}$ is incorrect.  

From the structure of the final product, the nitrogen atom of the initial substrate's cyano group ends up as an amino group outside the ring. Option $\mathrm{J}$ is incorrect.  

<Checkpoint>[The nitrogen atom of the initial substrate's cyano group forms an amino group in the final product](1)</Checkpoint>  

The final product contains directly linked phenyl and pyrimidine rings, with both *o*-positions of the pyrimidine ring substituted. This creates significant steric hindrance for coplanar alignment between the phenyl and pyrimidine rings, resulting in weak π-conjugation. Thus, option $\mathrm{K}$ is incorrect.  

<Checkpoint>[Steric hindrance prevents coplanarity between phenyl and pyrimidine rings, leading to weak π-conjugation](1)</Checkpoint>  

In the final product, the phenyl ring has 11 coplanar atoms, while the pyrimidine ring and its attached amino groups have 12 coplanar atoms. The ethyl group linked to the pyrimidine ring can contribute 3 coplanar atoms in certain conformations. When the phenyl and pyrimidine rings achieve a coplanar conformation (though the question does not require this conformation to be stable), the molecule can have up to 26 coplanar atoms, exceeding 22. Thus, option $\mathrm{N}$ is incorrect.  

<Checkpoint>[Up to 26 atoms can be coplanar](1)</Checkpoint>
初始反应物为
<MultiModal>![c1cc(Cl)ccc1CC#N](/media/uploads/3f4d8441-a4d6-41c3-9332-1beae7df0a98.png)</MultiModal>
在乙醇钠条件下，拔除氰基α位（同时也是苄位）的质子，形成的碳负离子进攻丙酸乙酯的酯羰基碳，经过加成-消除机理，离去乙醇负离子，得到产物$\mathbf{A}$。

<Checkpoint>[氰基α位形成碳负离子](1)</Checkpoint>

<Checkpoint>[发生加成-消除机理](1)</Checkpoint>

$\mathrm{A}$选项错误。

$\mathbf{A}$的结构为
<MultiModal>![CCC(=O)C(C#N)c1ccc(cc1)Cl](/media/uploads/9c50c15e-4434-413e-9ebe-7cd86a9f5337.png)</MultiModal>

对比看出，其与初始底物相比，多了3个碳，$\mathrm{B}$选项错误。

<Checkpoint>[A结构为CCC(=O)C(C#N)c1ccc(cc1)Cl](1)</Checkpoint>

<Checkpoint>[A比底物多3个碳](1)</Checkpoint>

$\mathbf{A}$的结构中，有一个氢原子同时处于氰基α位、酮羰基α位和苄位，其可以被乙醇负离子进一步去质子化，得到被共轭体系稳定的烯醇负离子
<MultiModal>![CCC(=C(C#N)c1ccc(cc1)Cl)[O-]](/media/uploads/f0ea996f-7f44-4020-9066-6f32a7c25c2b.png)</MultiModal>

该中间体氧端被重氮甲烷甲基化，得到$\mathbf{B}$。

<Checkpoint>[烯醇负离子在氧端甲基化](1)</Checkpoint>

$\mathbf{B}$的结构为
<MultiModal>![CC/C(=C(\C#N)/c1ccc(cc1)Cl)/OC](/media/uploads/93d8d75d-33bc-425f-af8e-b5cbdc0fbc38.png)</MultiModal>

<Checkpoint>[$\mathbf{B}$为CC/C(=C(\C#N)/c1ccc(cc1)Cl)/OC](1)</Checkpoint>

$\mathbf{B}$中只有1个环，$\mathrm{C}$错误。$\mathbf{A}$到$\mathbf{B}$生成氮气，$\mathrm{D}$错误。

观察$\mathbf{B}$的结构，其环外的苄位碳和与氧原子直接相连的碳，在结构式上为碳碳双键，故$\mathrm{L}$选项错误。

<Checkpoint>[$\mathbf{B}$的苄位碳和与氧原子直接相连的一个碳上有碳碳双键](1)</Checkpoint>

$\mathbf{C}$是胍，其结构为
<MultiModal>![C(=N)(N)N](/media/uploads/e52bae25-f38c-4ddf-a93a-8461cab76f02.png)</MultiModal>

<Checkpoint>[$\mathbf{C}为$C(=N)(N)N](1)</Checkpoint>

$\mathbf{C}$具有碱性，具有$\mathrm{Y}$芳香性，其稳定性来源与碳酸根一致，$\mathrm{F}$错误。

<Checkpoint>[C具有Y芳香性](1)</Checkpoint>

$\mathbf{C}$质子化后得到的阳离子的最稳定构象为平面构象，其中的碳原子和所有氮原子均为$\mathrm{sp^2}$杂化，整个离子具有$D_{\mathrm{3h}}$点群。选项$\mathrm{M}$错误。

<Checkpoint>[$\mathbf{C}$的阳离子为平面构象，$D_\mathrm{3h}$点群](1)</Checkpoint>

中间体$\mathbf{B}$与加入的$\mathbf{C}$发生反应，$\mathbf{C}$中的氮原子进攻$\mathbf{B}$中与氧相连的碳原子（氰基的β位），经加成-消除机理，离去甲氧基负离子。$\mathrm{H}$错误。

<MultiModal>![C1CC(=C(C#N)C2=C1C=C(C=C2)Cl)NC(=N)N](/media/uploads/5fa4ba44-44ca-4533-831d-0029215e9cf8.png)</MultiModal>

<Checkpoint>[离去甲氧基负离子](1)</Checkpoint>

随后，原属$\mathbf{C}$中的另一个氮原子进攻氰基碳，生成新的六元环，并通过质子转移，将原本的氰基氮转化为氨基。选项$\mathrm{G}$正确。

<Checkpoint>[B与C反应生成六元环](1)</Checkpoint>

<MultiModal>![CCc1c(c2ccc(cc2)Cl)c(nc(n1)N)N](/media/uploads/35c61ecf-76af-44d8-b7e9-62d9064b29db.png)</MultiModal>

<Checkpoint>[最终产物为CCc1c(c2ccc(cc2)Cl)c(nc(n1)N)N](1)</Checkpoint>

整个合成路线中，不涉及胍的碱性，选项$\mathrm{E}$错误。

<Checkpoint>[合成路线不涉及胍的碱性](1)</Checkpoint>

在最终产物的生成中，离去甲氧基负离子，而非甲醛。选项$\mathrm{I}$错误。

从最终产物的结构可以看到，初始底物的氰基氮原子，在最终产物中形成氨基，不在环内。$\mathrm{J}$选项错误。

<Checkpoint>[初始底物的氰基氮原子最终形成氨基](1)</Checkpoint>

最终产物结构中含有直接相连的苯环和嘧啶环，且嘧啶环的两个*o*位均有取代基，这使得苯环与嘧啶环之间形成共面构象的位阻较大，因此两者之间的π共轭相互作用较弱。故$\mathrm{K}$选项错误。

<Checkpoint>[苯环和嘧啶环共面位阻大，π共轭作用弱](1)</Checkpoint>

在最终产物中，苯环的部分有11个共面原子，嘧啶环以及其连带的两个氨基有12个共面原子，与嘧啶环相连的乙基在特定的构象下，可以有3个原子与嘧啶环共面，当苯环与嘧啶环达到共平面构象（选项中不要求该构象稳定）时，整个分子中至多有26个原子共平面，大于22个，选项$\mathrm{N}$错误。

<Checkpoint>[至多26个原子共面](1)</Checkpoint>