Pericyclic reactions are a class of reactions with extremely high application value. Two reactions are given. The first reaction is <MultiModal>![CC#CCCCC1(Br)C2=CC=CC=C2NC1=O reacts in the presence of $\mathrm {Cs_2CO_3}$ to yield CC1=NC2=CC=CC=C2C3=C1CCC3](/media/uploads/4c1fd5bf-e8ba-4eac-8d26-c319184daf54.png)</MultiModal>, in which two intermediates $\mathbf A$ and $\mathbf B$ exist during this transformation.

The second reaction is <MultiModal>![n-Butylamine, n-heptanal, and [C-]#[N+]C(CC1=CC=CC=C1)C(N2CCOCC2)=O react at room temperature in acidic toluene solvent to obtain intermediate $\mathbf X$. Intermediate $\mathbf X$ and O=C(C#CC1=CC=CC=C1)Cl react in the presence of triethylamine at 0 degrees Celsius to obtain intermediate $\mathbf Y$. Intermediate $\mathbf Y$ and O=C(N1C2=CC=CC=C2)C=CC1=O react in toluene solvent at 110 degrees Celsius to obtain the product CCCCCCC1C2=C3C(C(N(C4=CC=CC=C4)C3=O)=O)=C(N5CCOCC5)C(C6=CC=CC=C6)=C2C(N1CCCC)=O](/media/uploads/fa0e90c0-c1fb-410c-ad1f-1c9a78115eb4.png)</MultiModal>

Which of the following statements is correct?
周环反应是一类具有极高应用价值的反应。给出了两个反应，第一个反应为<MultiModal>![CC#CCCCC1(Br)C2=CC=CC=C2NC1=O在$\mathrm {Cs_2CO_3}$存在的条件下反应得到CC1=NC2=CC=CC=C2C3=C1CCC3](/media/uploads/4c1fd5bf-e8ba-4eac-8d26-c319184daf54.png)</MultiModal>，该转化过程中存在两个中间体$\mathbf A$和$\mathbf B$。

第二个反应为<MultiModal>![正丁基胺、正庚醛和[C-]#[N+]C(CC1=CC=CC=C1)C(N2CCOCC2)=O在酸性甲苯溶剂中常温反应得到中间体$\mathbf X$，中间体$\mathbf X$和O=C(C#CC1=CC=CC=C1)Cl在三乙胺、0摄氏度的条件下反应得到中间体$\mathbf Y$，中间体$\mathbf Y$和O=C(N1C2=CC=CC=C2)C=CC1=O在甲苯溶剂中、110摄氏度条件下反应得到产物CCCCCCC1C2=C3C(C(N(C4=CC=CC=C4)C3=O)=O)=C(N5CCOCC5)C(C6=CC=CC=C6)=C2C(N1CCCC)=O](/media/uploads/fa0e90c0-c1fb-410c-ad1f-1c9a78115eb4.png)</MultiModal>


以下说法正确的是
Intermediate $\mathbf A$ contains 3 rings
The intermediate $\mathbf A$ has a chiral carbon.
The intermediate $\mathbf B$ has 13 $\mathrm C$ atoms
There is a carbon-carbon triple bond in the intermediate $\mathbf B$.
There is 1 carbonyl group in the intermediate $\mathbf X$
Intermediate $\mathbf X$ contains two $\mathrm N$ atoms with different nucleophilicities.
The intermediate $\mathbf Y$ contains a γ-lactam structure.
During the process of obtaining the intermediate $\mathbf Y$, $\mathrm {CO}$ is produced.
None of the above options are correct
中间体$\mathbf A$包含3个环
中间体$\mathbf A$有手性碳
中间体$\mathbf B$有13个$\mathrm C$
中间体$\mathbf B$中存在碳碳三键
中间体$\mathbf X$中有1个羰基
中间体$\mathbf X$中有两个不同亲核性的$\mathrm N$原子
中间体$\mathbf Y$中存在γ-内酰胺结构
得到中间体$\mathbf Y$的过程中产生了$\mathrm {CO}$
以上选项均不正确
In the first reaction, the electrons from product N are donated to the benzene ring, $\mathrm {Br}$ departs, and the carbonate neutralizes the $\mathrm H$ on $\mathrm N^+$.


<Checkpoint>[The diene intermediate A is obtained, with the structure <MultiModal>![CC#CCCCC1=C2C=CC=CC2=NC1=O](/media/uploads/83fa9be2-4fee-4d5d-8122-4246e409e69f.png)</MultiModal>](1)</Checkpoint>


Thus, intermediate $\mathbf A$ has only two rings and no chiral carbon.


<Checkpoint>[Intermediate $\mathbf A$ has only two rings and no chiral carbon](1)</Checkpoint>


The alkyne undergoes a DA reaction with the conjugated diene on the five-membered ring of intermediate $\mathbf A$,


<Checkpoint>[Intermediate $\mathbf B$ is obtained, with the structure <MultiModal>![CC1=C2CCCC23C4=CC=CC=C4N1C3=O](/media/uploads/810c712a-238c-4943-af72-785544be5bdd.png)</MultiModal>](1)</Checkpoint>


<Checkpoint>[Intermediate $\mathbf B$ has 14 $\mathrm C$ atoms and no carbon-carbon triple bond](1)</Checkpoint>


In the second reaction,


<Checkpoint>[n-Butylamine reacts with n-heptanal to form an imine by dehydration](1)</Checkpoint>

The imine is nucleophilically attacked by the $\mathrm C$ atom of isocyanide,
<Checkpoint>[The imine is nucleophilically attacked by the $\mathrm C$ atom of isocyanide](1)</Checkpoint>


Subsequently, the $\mathrm O$ atom on the carbonyl group of the isocyanide molecule attacks the carbon of the nitrilium ion, and the keto form transforms into the enol form, yielding an intramolecular cyclization product.

<Checkpoint>[The $\mathrm O$ atom attacks the carbon of the nitrilium ion, undergoing an enolization transformation](1)</Checkpoint>


<Checkpoint>[Intermediate $\mathbf X$ is obtained, with the structural formula <MultiModal>![CCCCCCC(C1=NC(=C(N2CCOCC2)O1)CC3=CC=CC=C3)NCCCC](/media/uploads/64fe5f9a-8a22-4130-b3fd-5f25d2342ea3.png)</MultiModal>](1)</Checkpoint>


<Checkpoint>[Intermediate $\mathbf X$ lacks a carbonyl group and has 3 $\mathrm N$ atoms with different nucleophilicities](1)</Checkpoint>

The $\mathrm N$ atom of the butylamino group in intermediate $\mathbf X$ nucleophilically attacks the acyl chloride group of phenylpropynoyl chloride, resulting in the departure of $\mathrm {Cl}$,


<Checkpoint>[The $\mathrm N$ atom of the butylamino group in intermediate $\mathbf X$ nucleophilically attacks the acyl chloride group of phenylpropynoyl chloride](1)</Checkpoint>


Afterward, the $\mathrm {C=N-O-C=C}$ moiety of the oxazole ring undergoes a DA pericyclic reaction with the alkyne, followed by a retro-DA reaction, eliminating aniline.

<Checkpoint>[The oxazole ring undergoes a DA pericyclic reaction with the alkyne](1)</Checkpoint>


<Checkpoint>[A retro-DA reaction occurs, eliminating aniline and yielding intermediate $\mathbf Y$, with the structural formula <MultiModal>![CCCCCCC1C2=C(C(N1CCCC)=O)C(C3=CC=CC=C3)=C(O2)N4CCOCC4](/media/uploads/b9ca9c5b-2343-42df-90e2-3e43adf2474a.png)</MultiModal>](1)</Checkpoint>




<Checkpoint>[Intermediate $\mathbf Y$ contains an amide structure, eliminating aniline rather than carbon monoxide](1)</Checkpoint>


The final product is G.
在第一个反应中，产物N的电子给到苯环，$\mathrm {Br}$离去，碳酸根中和$\mathrm N^+$上的$\mathrm H$，


<Checkpoint>[得到二烯中间体A，结构为<MultiModal>![CC#CCCCC1=C2C=CC=CC2=NC1=O](/media/uploads/83fa9be2-4fee-4d5d-8122-4246e409e69f.png)</MultiModal>](1)</Checkpoint>


因此中间体$\mathbf A$只有两个环，没有手性碳


<Checkpoint>[中间体$\mathbf A$只有两个环，没有手性碳](1)</Checkpoint>


炔基和中间体$\mathbf A$五元环上的共轭双烯发生DA反应，


<Checkpoint>[得到中间体$\mathbf B$，结构为<MultiModal>![CC1=C2CCCC23C4=CC=CC=C4N1C3=O](/media/uploads/810c712a-238c-4943-af72-785544be5bdd.png)</MultiModal>](1)</Checkpoint>


<Checkpoint>[中间体$\mathbf B$有14个$\mathrm C$原子，没有碳碳三键](1)</Checkpoint>


在第二个反应中，


<Checkpoint>[正丁基胺和正庚醛反应脱去水得到亚胺](1)</Checkpoint>

亚胺被异氰化物的$\mathrm C$原子亲核攻击，
<Checkpoint>[亚胺被异氰化物的$\mathrm C$原子亲核攻击](1)</Checkpoint>


之后异氰化物分子内羰基上的$\mathrm O$原子进攻腈鎓离子的碳，酮式转变为烯醇式，得到分子内环化产物

<Checkpoint>[$\mathrm O$原子进攻腈鎓离子的碳，发生烯醇式转变](1)</Checkpoint>


<Checkpoint>[得到中间体$\mathbf X$，结构式为<MultiModal>![CCCCCCC(C1=NC(=C(N2CCOCC2)O1)CC3=CC=CC=C3)NCCCC](/media/uploads/64fe5f9a-8a22-4130-b3fd-5f25d2342ea3.png)</MultiModal>](1)</Checkpoint>


<Checkpoint>[中间体$\mathbf X$没有羰基，有3个不同亲核性的$\mathrm N$原子](1)</Checkpoint>

中间体$\mathbf X$中的丁基氨基的$\mathrm N$原子亲核进攻苯丙炔酰氯中的酰氯基团，脱去$\mathrm {Cl}$，


<Checkpoint>[中间体$\mathbf X$中的丁基氨基的$\mathrm N$原子亲核进攻苯丙炔酰氯中的酰氯基团](1)</Checkpoint>


之后，噁唑环的$\mathrm {C=N-O-C=C}$部分与炔基发生DA周环反应，再发生逆DA反应，脱去苄基腈

<Checkpoint>[噁唑环与炔基发生DA周环反应](1)</Checkpoint>


<Checkpoint>[再发生逆DA反应，脱去苄基腈，得到中间体$\mathbf Y$，中间体$\mathbf Y$的结构式为<MultiModal>![CCCCCCC1C2=C(C(N1CCCC)=O)C(C3=CC=CC=C3)=C(O2)N4CCOCC4](/media/uploads/b9ca9c5b-2343-42df-90e2-3e43adf2474a.png)</MultiModal>](1)</Checkpoint>




<Checkpoint>[中间体$\mathbf Y$中存在酰胺结构，脱去苄基腈而不是一氧化碳](1)</Checkpoint>


最终结果为G