The structure of the Vilsmeier salt cation is shown in the figure, <MultiModal>![CN(C)/C(Cl)=[N+](C)/C](/media/uploads/fa1f6fa6-466b-4196-ac32-e34d54cd85be.png)</MultiModal>

A laboratory attempted to synthesize compound **A** by reacting excess Vilsmeier salt with an equivalent amount of base and o-aminobenzamide, but found that no **A** was formed. Instead, a yellow, low-melting-point, non-crystalline compound **B** was obtained. Liquid chromatography-mass spectrometry (LC-MS) analysis revealed that the molecular weight of **B** was 216.28.

The laboratory then heated compound **B** with a large excess of hydroxylamine hydrochloride and an equivalent amount of sodium acetate in a methanol-water solution, intending to prepare the seven-membered cyclic compound **C** with a molecular weight of 204.23. However, thin-layer chromatography (TLC) development using dichloromethane and petroleum ether containing triethylamine revealed a non-fluorescent compound **D** and a fluorescent compound. LC-MS and <sup>1</sup>H NMR analysis confirmed that the fluorescent compound was **A**. LC-MS analysis showed that the molecular weight of **D** was also 204.23.

X-ray diffraction (XRD) confirmed that compound **D** possesses a hexa-hexacyclic system, which was not the intended structure of **C**. To investigate the reaction mechanism, diethylhydroxylamine was used instead of hydroxylamine hydrochloride and sodium acetate, but no analogs of **D** were formed.

The following options each contain three structural images: the first is the possible structure of **A**, the second is the possible structure of **C**, and the third is the possible structure of **D**. Select the option where all three structures are correct.
Vilsmeier盐的正离子结构如图所示，<MultiModal>![CN(C)/C(Cl)=[N+](C)/C](/media/uploads/fa1f6fa6-466b-4196-ac32-e34d54cd85be.png)</MultiModal>

有实验室想用过量的 Vilsmeier盐和等当量的碱与邻氨基苯甲酰胺反应合成化合物**A**，但是发现没有**A**生成，却得到了黄色低熔点不易结晶的化合物**B**。经过液相质谱仪检测，**B**的分子量为216.28。

该实验室又将化合物**B**与大过量的盐酸羟胺以及与盐酸羟胺等当量的乙酸钠在甲醇的水溶液中加热，拟制备分子量为 204.23 的七元环状化合物**C**，但是经过加有三乙胺的二氯甲烷和石油醚TLC展开发现一无荧光化合物**D**，以及一个具有荧光的化合物。经过液相质谱仪和核磁共振氢谱检测，该具有荧光的化合物就是化合物**A**。经过液相质谱仪检测发现化合物**D**的分子量也是204.23。

经过XRD证实，化合物**D**具有六并六环系，并非我们拟合成的化合物**C**。为了推测该反应的机理，我们用了二乙基羟胺代替盐酸羟胺乙酸钠，发现并没有化合物**D**的类似物生成。 

以下选项均包含三张结构式图片，第一张为可能的**A**的结构，第二张为可能的**C**的结构、第三张为可能的**D**的结构，选出三个结构均正确的选项。
A：<MultiModal>![O=C(N(C)C)NC1=CC=CC=C1C#N](/media/uploads/a0aab231-5b5c-4312-bdfb-3cdaf7a654cd.png)</MultiModal>  

C：<MultiModal>![NC1=NOC(N(C)C)=NC2=CC=CC=C21](/media/uploads/56c8a655-6798-4862-babd-bbb40069e2cd.png)</MultiModal>  

D：<MultiModal>![O/N=C1NC(N(C)C)=NC2=CC=CC=C2\1](/media/uploads/5230512c-826b-4006-8c37-bda94c50d707.png)</MultiModal>
A：<MultiModal>![O=C(N(C)C)NC1=CC=CC=C1C#N](/media/uploads/a0aab231-5b5c-4312-bdfb-3cdaf7a654cd.png)</MultiModal>  

C：<MultiModal>![N=C1C2=CC=CC=C2N=C(N(C)C)NO1](/media/uploads/29559a18-2d98-4478-a230-f2e89d912707.png)</MultiModal>  

D：<MultiModal>![O/N=C1NC(N(C)C)=NC2=CC=CC=C2\1](/media/uploads/5230512c-826b-4006-8c37-bda94c50d707.png)</MultiModal>
A：<MultiModal>![O=C(N(C)C)NC1=CC=CC=C1C#N](/media/uploads/a0aab231-5b5c-4312-bdfb-3cdaf7a654cd.png)</MultiModal>

C：<MultiModal>![NC1=NOC(N(C)C)=NC2=CC=CC=C21](/media/uploads/56c8a655-6798-4862-babd-bbb40069e2cd.png)</MultiModal>

D：<MultiModal>![CN(C)C1=NC2=CC=CC=C2C(ON)=N1](/media/uploads/0fbea86d-da11-4c20-8223-8caa5b733dbc.png)</MultiModal>
A：<MultiModal>![O=C(N(C)C)NC1=CC=CC=C1C#N](/media/uploads/a0aab231-5b5c-4312-bdfb-3cdaf7a654cd.png)</MultiModal>  

C：<MultiModal>![N=C1C2=CC=CC=C2N=C(N(C)C)NO1](/media/uploads/29559a18-2d98-4478-a230-f2e89d912707.png)</MultiModal>  

D：<MultiModal>![CN(C)C1=NC2=CC=CC=C2C(ON)=N1](/media/uploads/0fbea86d-da11-4c20-8223-8caa5b733dbc.png)</MultiModal>
A：<MultiModal>![O=C1C2=CC=CC=C2N=C(N(C)C)N1](/media/uploads/4335b3f2-46ca-41b6-9d10-2ad69c247a05.png)</MultiModal>  

C：<MultiModal>![NC1=NOC(N(C)C)=NC2=CC=CC=C21](/media/uploads/56c8a655-6798-4862-babd-bbb40069e2cd.png)</MultiModal>  

D：<MultiModal>![O/N=C1NC(N(C)C)=NC2=CC=CC=C2\1](/media/uploads/5230512c-826b-4006-8c37-bda94c50d707.png)</MultiModal>
A：<MultiModal>![O=C1C2=CC=CC=C2N=C(N(C)C)N1](/media/uploads/4335b3f2-46ca-41b6-9d10-2ad69c247a05.png)</MultiModal>  

C：<MultiModal>![N=C1C2=CC=CC=C2N=C(N(C)C)NO1](/media/uploads/29559a18-2d98-4478-a230-f2e89d912707.png)</MultiModal>  

D：<MultiModal>![O/N=C1NC(N(C)C)=NC2=CC=CC=C2\1](/media/uploads/5230512c-826b-4006-8c37-bda94c50d707.png)</MultiModal>
A：<MultiModal>![O=C1C2=CC=CC=C2N=C(N(C)C)N1](/media/uploads/4335b3f2-46ca-41b6-9d10-2ad69c247a05.png)</MultiModal>  

C：<MultiModal>![NC1=NOC(N(C)C)=NC2=CC=CC=C21](/media/uploads/56c8a655-6798-4862-babd-bbb40069e2cd.png)</MultiModal>  

D：<MultiModal>![CN(C)C1=NC2=CC=CC=C2C(ON)=N1](/media/uploads/0fbea86d-da11-4c20-8223-8caa5b733dbc.png)</MultiModal>
A：<MultiModal>![O=C1C2=CC=CC=C2N=C(N(C)C)N1](/media/uploads/4335b3f2-46ca-41b6-9d10-2ad69c247a05.png)</MultiModal>  

C：<MultiModal>![N=C1C2=CC=CC=C2N=C(N(C)C)NO1](/media/uploads/29559a18-2d98-4478-a230-f2e89d912707.png)</MultiModal>  

D：<MultiModal>![CN(C)C1=NC2=CC=CC=C2C(ON)=N1](/media/uploads/0fbea86d-da11-4c20-8223-8caa5b733dbc.png)</MultiModal>
None of the above options are correct
A：<MultiModal>![O=C(N(C)C)NC1=CC=CC=C1C#N](/media/uploads/a0aab231-5b5c-4312-bdfb-3cdaf7a654cd.png)</MultiModal>

C：<MultiModal>![NC1=NOC(N(C)C)=NC2=CC=CC=C21](/media/uploads/56c8a655-6798-4862-babd-bbb40069e2cd.png)</MultiModal>

D：<MultiModal>![O/
N=C1NC(N(C)C)=NC2=CC=CC=C2\1](/media/uploads/5230512c-826b-4006-8c37-bda94c50d707.png)</MultiModal>
A：<MultiModal>![O=C(N(C)C)NC1=CC=CC=C1C#N](/media/uploads/a0aab231-5b5c-4312-bdfb-3cdaf7a654cd.png)</MultiModal>

C：<MultiModal>![N=C1C2=CC=CC=C2N=C(N(C)C)NO1](/media/uploads/29559a18-2d98-4478-a230-f2e89d912707.png)</MultiModal>

D：<MultiModal>![O/N=C1NC(N(C)C)=NC2=CC=CC=C2\1](/media/uploads/5230512c-826b-4006-8c37-bda94c50d707.png)</MultiModal>
A：<MultiModal>![O=C(N(C)C)NC1=CC=CC=C1C#N](/media/uploads/a0aab231-5b5c-4312-bdfb-3cdaf7a654cd.png)</MultiModal>

C：<MultiModal>![NC1=NOC(N(C)C)=NC2=CC=CC=C21](/media/uploads/56c8a655-6798-4862-babd-bbb40069e2cd.png)</MultiModal>

D：<MultiModal>![CN(C)C1=NC2=CC=CC=C2C(ON)=N1](/media/uploads/0fbea86d-da11-4c20-8223-8caa5b733dbc.png)</MultiModal>
A：<MultiModal>![O=C(N(C)C)NC1=CC=CC=C1C#N](/media/uploads/a0aab231-5b5c-4312-bdfb-3cdaf7a654cd.png)</MultiModal>

C：<MultiModal>![N=C1C2=CC=CC=C2N=C(N(C)C)NO1](/media/uploads/29559a18-2d98-4478-a230-f2e89d912707.png)</MultiModal>

D：<MultiModal>![CN(C)C1=NC2=CC=CC=C2C(ON)=N1](/media/uploads/0fbea86d-da11-4c20-8223-8caa5b733dbc.png)</MultiModal>
A：<MultiModal>![O=C1C2=CC=CC=C2N=C(N(C)C)N1](/media/uploads/4335b3f2-46ca-41b6-9d10-2ad69c247a05.png)</MultiModal>

C：<MultiModal>![NC1=NOC(N(C)C)=NC2=CC=CC=C21](/media/uploads/56c8a655-6798-4862-babd-bbb40069e2cd.png)</MultiModal>

D：<MultiModal>![O/N=C1NC(N(C)C)=NC2=CC=CC=C2\1](/media/uploads/5230512c-826b-4006-8c37-bda94c50d707.png)</MultiModal>
A：<MultiModal>![O=C1C2=CC=CC=C2N=C(N(C)C)N1](/media/uploads/4335b3f2-46ca-41b6-9d10-2ad69c247a05.png)</MultiModal>

C：<MultiModal>![N=C1C2=CC=CC=C2N=C(N(C)C)NO1](/media/uploads/29559a18-2d98-4478-a230-f2e89d912707.png)</MultiModal>

D：<MultiModal>![O/N=C1NC(N(C)C)=NC2=CC=CC=C2\1](/media/uploads/5230512c-826b-4006-8c37-bda94c50d707.png)</MultiModal>
A：<MultiModal>![O=C1C2=CC=CC=C2N=C(N(C)C)N1](/media/uploads/4335b3f2-46ca-41b6-9d10-2ad69c247a05.png)</MultiModal>

C：<MultiModal>![NC1=NOC(N(C)C)=NC2=CC=CC=C21](/media/uploads/56c8a655-6798-4862-babd-bbb40069e2cd.png)</MultiModal>

D：<MultiModal>![CN(C)C1=NC2=CC=CC=C2C(ON)=N1](/media/uploads/0fbea86d-da11-4c20-8223-8caa5b733dbc.png)</MultiModal>
A：<MultiModal>![O=C1C2=CC=CC=C2N=C(N(C)C)N1](/media/uploads/4335b3f2-46ca-41b6-9d10-2ad69c247a05.png)</MultiModal>

C：<MultiModal>![N=C1C2=CC=CC=C2N=C(N(C)C)NO1](/media/uploads/29559a18-2d98-4478-a230-f2e89d912707.png)</MultiModal>

D：<MultiModal>![CN(C)C1=NC2=CC=CC=C2C(ON)=N1](/media/uploads/0fbea86d-da11-4c20-8223-8caa5b733dbc.png)</MultiModal>
以上选项均不正确
(1)  
Upon observing the substrate, both the amide oxygen and the amino nitrogen of anthranilamide are nucleophilic sites and can react with the Vilsmeier salt.  

<Checkpoint>[Both the amide oxygen and the amino nitrogen of the substrate are nucleophilic sites](1)</Checkpoint>  

The question states that **A** exhibits fluorescence, indicating that **A** should possess a large conjugated system and a relatively rigid skeleton, and cannot consist of only one benzene ring.  

It can be inferred that the formation of compound **A** involves the reaction of the amino group with the Vilsmeier salt. During the formation of **A**, the amino group nucleophilically attacks the Vilsmeier salt, followed by an intramolecular attack by the amide to form a six-membered ring. The elimination of dimethylamine yields **A**, which contains a conjugated system.  

<Checkpoint>[**A** exhibits fluorescence and cannot consist of only one benzene ring](1)</Checkpoint>  

<Checkpoint>[The formation of **A** involves the reaction of the amino group with the Vilsmeier salt](1)</Checkpoint>  

The structure of compound **A** is shown below:  

<MultiModal>![O=C1C2=CC=CC=C2N=C(N(C)C)N1](/media/uploads/4335b3f2-46ca-41b6-9d10-2ad69c247a05.png)</MultiModal>  

<Checkpoint>[The structure of **A** is O=C1C2=CC=CC=C2N=C(N(C)C)N1](2)</Checkpoint>  

If the amide reacts with the Vilsmeier salt first, the amide oxygen nucleophilically attacks the Vilsmeier salt. The Vilsmeier salt acts as a dehydrating agent, formally eliminating water to yield anthranilonitrile and dimethylaminourea. Based on the molecular formula, dimethylaminourea further reacts with the amino group to eliminate water, resulting in **B**.  

<Checkpoint>[The formation of **B** involves the reaction of the amide with the Vilsmeier salt first](1)</Checkpoint>  

The structure of compound **B** is shown below:  

<MultiModal>![CN(C)/C(N(C)C)=N/C1=CC=CC=C1C#N](/media/uploads/8de7e053-1624-409d-b2bf-6d85e283859c.png)</MultiModal>  

<Checkpoint>[The structure of **B** is CN(C)/C(N(C)C)=N/C1=CC=CC=C1C#N](2)</Checkpoint>  

(2)  

Compound **C** is formed by the reaction of compound **B** with hydroxylamine. Based on the molecular weight, **C** contains one more hydroxylamine and one less dimethylamine than **B**, indicating a substitution reaction. Since **C** contains a seven-membered ring, an intramolecular cyclization also occurs. Considering the reactivity of compound **B**, the most electrophilic site is the cyano group, and hydroxylamine should react with the cyano group first.  

<Checkpoint>[Hydroxylamine should react with the cyano group first](1)</Checkpoint>  

Both the nitrogen and oxygen of hydroxylamine are nucleophilic sites, and further analysis requires information about compound **D**.  
The question states: Using diethylhydroxylamine as a reactant fails to yield an analog of **D**, thereby excluding the possibility of nucleophilic attack by the oxygen end of hydroxylamine. Thus, the nitrogen end of hydroxylamine is connected to the carbon atom of the six-membered ring.  

<Checkpoint>[Diethylhydroxylamine as a reactant fails to yield an analog of D, indicating the nitrogen end of hydroxylamine acts as the nucleophilic site](1)</Checkpoint>  

Therefore, the carbon corresponding to the cyano group in **B** is connected to two nitrogen atoms in **C** and **D**, namely the nitrogen of the cyano group and the nitrogen of hydroxylamine.  

<Checkpoint>[The carbon corresponding to the cyano group in **B** is connected to two nitrogen atoms in **C** and **D**](1)</Checkpoint>  

The reaction of the cyano group with hydroxylamine can yield an amidoxime intermediate with the structure N/C(C1=CC=CC=C1/N=C(N(C)C)\N(C)C)=N\O. The guanidine carbon atom in the molecule is an electrophilic site, and either the oxygen atom or the amino nitrogen atom of the amidoxime can act as a nucleophile, forming a seven-membered ring or a six-membered ring, respectively. Eliminating one molecule of dimethylamine yields **C** and **D**, respectively.  

<Checkpoint>[The reaction of the cyano group with hydroxylamine yields the amidoxime intermediate N/C(C1=CC=CC=C1/N=C(N(C)C)\N(C)C)=N\O](1)</Checkpoint>  

<Checkpoint>[The oxygen atom or amino nitrogen atom of the amidoxime can nucleophilically attack the guanidine carbon atom to form a seven-membered/six-membered ring](1)</Checkpoint>  

The structure of compound **C** is shown below:  

<MultiModal>![NC1=NOC(N(C)C)=NC2=CC=CC=C21](/media/uploads/56c8a655-6798-4862-babd-bbb40069e2cd.png)</MultiModal>  

<Checkpoint>[The structure of **C** is NC1=NOC(N(C)C)=NC2=CC=CC=C21](2)</Checkpoint>  

The structure of compound **D** is shown below:  

<MultiModal>![O/N=C1NC(N(C)C)=NC2=CC=CC=C2\1](/media/uploads/5230512c-826b-4006-8c37-bda94c50d707.png)</MultiModal>  

<Checkpoint>[The structure of **D** is O/N=C1NC(N(C)C)=NC2=CC=CC=C2\1](2)</Checkpoint>  

In conclusion, option E is correct.
（一）
观察底物，邻氨基苯甲酰胺的酰胺氧和胺基氮均为亲核位点，均可以与Vilsmeier盐反应。

<Checkpoint>[底物的酰胺氧和胺基氮均为亲核位点](1)</Checkpoint>

题干提到**A**具有荧光，可判断**A**应当具备大共轭体系与较为刚性的骨架，不会仅含一个苯环。

可以想到，化合物**A**的生成涉及胺基与Vilsmeier盐的反应，生成**A**时，胺基将亲核进攻Vilsmeier盐，之后酰胺分子内进攻形成六元环，脱去二甲胺，得到含共轭体系的**A**。

<Checkpoint>[**A**具有荧光，不会仅含一个苯环](1)</Checkpoint>

<Checkpoint>[**A**的生成为胺基与Vilsmeier盐的反应](1)</Checkpoint>

化合物**A**的结构如图所示

<MultiModal>![O=C1C2=CC=CC=C2N=C(N(C)C)N1](/media/uploads/4335b3f2-46ca-41b6-9d10-2ad69c247a05.png)</MultiModal>  

<Checkpoint>[**A**的结构为O=C1C2=CC=CC=C2N=C(N(C)C)N1](2)</Checkpoint>

若酰胺先与Vilsmeier盐反应，则酰胺氧亲核进攻Vilsmeier盐，Vilsmeier盐作为脱水剂形式上脱水得到邻氨基苯甲腈以及二甲氨基脲，根据分子式可得二甲氨基脲还会继续与胺基反应脱水，即得到**B**。

<Checkpoint>[**B**的生成为酰胺先与Vilsmeier盐反应](1)</Checkpoint>

化合物**B**的结构如图所示

<MultiModal>![CN(C)/C(N(C)C)=N/C1=CC=CC=C1C#N](/media/uploads/8de7e053-1624-409d-b2bf-6d85e283859c.png)</MultiModal>

<Checkpoint>[**B**的结构为CN(C)/C(N(C)C)=N/C1=CC=CC=C1C#N](2)</Checkpoint>

（二）

化合物**C**由化合物**B**与羟胺反应生成，根据分子量可以得出**C**比**B**多一个羟胺，少一个二甲胺，故反应为取代反应，由于**C**中含有七元环，还发生了分子内的成环反应。考虑化合物**B**的反应特点，其中最亲电的位点为氰基，羟胺应当先与氰基反应。

<Checkpoint>[羟胺应当先与氰基反应](1)</Checkpoint>

羟胺的氮与氧均是亲核位点，需要结合化合物**D**相关信息进行分析。
题干提到：使用二乙基羟胺无法得到**D**的类似物，从而排除了羟胺使用氧端亲核进攻的选项，故羟胺的氮端与六元环上碳原子相连。

<Checkpoint>[二乙基羟胺作为反应物无法得到D的类似物，从而羟胺的氮端作为亲核位点](1)</Checkpoint>

因此，**B**中氰基对应的碳在**C**、**D**中连接了两个氮，分别是氰基的氮与羟胺的氮.

<Checkpoint>[**B**中氰基对应的碳在**C**、**D**中连接了两个氮](1)</Checkpoint>

氰基与羟胺反应可生成偕胺肟中间体，其结构式为N/C(C1=CC=CC=C1/N=C(N(C)C)\N(C)C)=N\O，分子内胍基碳原子为亲电位点，偕胺肟的氧原子或氨基氮原子均可亲核，分别生成七元环和六元环，脱去一分子二甲胺即分别得到**C**、**D**。

<Checkpoint>[氰基与羟胺反应可生成偕胺肟中间体N/C(C1=CC=CC=C1/N=C(N(C)C)\N(C)C)=N\O](1)</Checkpoint>

<Checkpoint>[偕胺肟的氧原子或氨基氮原子均可亲核胍基碳原子生成七元环/六元环](1)</Checkpoint>

化合物**C**的结构如图所示

<MultiModal>![NC1=NOC(N(C)C)=NC2=CC=CC=C21](/media/uploads/56c8a655-6798-4862-babd-bbb40069e2cd.png)</MultiModal>  

<Checkpoint>[**C**的结构为NC1=NOC(N(C)C)=NC2=CC=CC=C21](2)</Checkpoint>

化合物**D**的结构如图所示

<MultiModal>![O/N=C1NC(N(C)C)=NC2=CC=CC=C2\1](/media/uploads/5230512c-826b-4006-8c37-bda94c50d707.png)</MultiModal>

<Checkpoint>[**D**的结构为O/N=C1NC(N(C)C)=NC2=CC=CC=C2\1](2)</Checkpoint>

综上，E选项符合题意。