<MultiModal>![This figure depicts a one-step organic reaction. The substrate is CC1=C(CC(C#CC2=CC=CC=C2)=O)C3=CC=CC=C3N1, which is converted to $\bf A$ under $\rm CF_3COOH, CHCl_3, r.t.$ conditions. $\bf A$ is converted to $\bf B$ under $\rm LDA, THF, -78\degree C to r.t.$ conditions; the substrate can be directly converted to $\bf B$ under $\rm CF_3COH, CHCl_3, MW, 100\degree C$ conditions.](/media/uploads/a5d88a4c-15d8-4f88-9033-bcad40c9fe05.png)</MultiModal>

Regarding the organic reaction shown in the figure above, it is known that:

1. The $\rm ^1H-NMR$ information of $\bf A$ is: $\rm \delta ~7.66(d, 1H), 7.43\sim 7.38(m, 1H), 7.31(t, 1H), 7.23 \sim 7.17(m, 4H), 7.00 \sim 6.96(m, 2H), 6.88(s, 1H), 2.83(d, 1H), 2.73(d, 1H), 2.21(s, 3H)$.

2. $\bf B$ contains a quinoline structure.

3. The molecular weights of the substrate and $\bf A, B$ are all the same.

Which of the following statements is correct:
<MultiModal>![该图描述了一步有机反应。底物为CC1=C(CC(C#CC2=CC=CC=C2)=O)C3=CC=CC=C3N1，其在$\rm CF_3COOH, CHCl_3, r.t.$条件下转化为$\bf A$，$\bf A$在$\rm LDA, THF, -78\degree C to r.t.$条件下转化为$\bf B$；底物在$\rm CF_3COH, CHCl_3, MW, 100\degree C$条件下可直接转化为$\bf B$。](/media/uploads/a5d88a4c-15d8-4f88-9033-bcad40c9fe05.png)</MultiModal>

对上图所示的有机反应，已知：

1. $\bf A$ 的 $\rm ^1H-NMR$ 信息为: $\rm \delta ~7.66(d, 1H), 7.43\sim 7.38(m, 1H), 7.31(t, 1H), 7.23 \sim 7.17(m, 4H), 7.00 \sim 6.96(m, 2H), 6.88(s, 1H), 2.83(d, 1H), 2.73(d, 1H), 2.21(s, 3H)$。

2. $\bf B$ 含有喹啉结构。

3. 底物和 $\bf A,B$ 分子量均一致。

下列说法正确的是：
All other options are incorrect
$\bf A$ contains only one five-membered ring
$\bf A$ does not contain chiral carbon atoms.
The transformation of $\bf A$ to $\bf B$ involves a five-membered-ring intermediate.
$\bf B$ exhibits a bonding relationship: $\rm Ph-C-C-C-CH_3$
$\bf B$ exhibits a bonding relationship: $\rm Ph-C-C-C-N$
其他选项均不正确
$\bf A$ 仅含有一个五元环
$\bf A$ 不存在手性碳原子
$\bf A$ 转化为 $\bf B$ 的过程经历了一个含五个环的中间体
$\bf B$ 存在键联关系：$\rm Ph-C-C-C-CH_3$
$\bf B$存在键联关系：$\rm Ph-C-C-C-N$
Observation of the $^1$H NMR spectrum of $\bf A$: Located in the low field, with a chemical shift greater than 6.5, containing 10 hydrogen atoms, it should be aromatic hydrogen; five of them belong to the phenyl group, and four belong to the benzene ring hydrogen of benzopyrrole. Therefore, it is speculated that $\bf A$ contains a phenyl benzylic $\rm sp^2 C-H$.

<Checkpoint>[$\bf A$ contains a phenyl benzylic $\rm sp^2 C-H$](1)</Checkpoint>

The chemical shift of 2.21 corresponds to methyl hydrogen atoms, and the two d-peaks at 2.83 and 2.73 are clearly methylene peaks. Therefore, the methylene and methyl groups of the substrate did not change.

<Checkpoint>[The methylene and methyl groups of the substrate did not change](1)</Checkpoint>

Considering the reactivity of the substrate, the nucleophilic site is the 3-position of pyrrole, and the electrophilic sites are the carbonyl group and the alkyne bond; the nucleophilic carbonyl group forms an unstable three-membered ring, so the 3-position of pyrrole undergoes nucleophilic attack on the alkyne bond, resulting in a 5-endo-dig reaction to form a five-membered spirocycle. The product structure is CC1=NC2=CC=CC=C2C13C(C4=CC=CC=C4)=CC(C3)=O, which is consistent with the analysis of the $^1$H NMR spectrum, and therefore is structure $\bf A$.

<Checkpoint>[The 3-position of pyrrole undergoes nucleophilic attack on the alkyne bond, resulting in a 5-endo-dig reaction to form a five-membered spirocycle](1)</Checkpoint>

<Checkpoint>[Structure $\bf A$ is CC1=NC2=CC=CC=C2C13C(C4=CC=CC=C4)=CC(C3)=O](1)</Checkpoint>

The transformation of $\bf A$ to $\bf B$ occurs in a basic environment. First, the methylene hydrogen at the α-position of the carbonyl group is abstracted to form an enolate anion, with the structure CC1=NC2=CC=CC=C2C13C(C4=CC=CC=C4)=CC([O-])=C3.

<Checkpoint>[Abstraction of the methylene hydrogen at the α-position of the carbonyl group to form an enolate anion](1)</Checkpoint>

Then, the enolate anion nucleophilically attacks the electrophilic 2-position of pyrrole in the system, forming a five-fused-three-fused-five-membered ring intermediate, with the structure CC12[N-]C3=CC=CC=C3C14C2C(C=C4C5=CC=CC=C5)=O. Therefore, option D is correct.

<Checkpoint>[The enolate anion nucleophilically attacks the electrophilic 2-position of pyrrole in the system, forming a five-fused-three-fused-five-membered ring intermediate](1)</Checkpoint>

<Checkpoint>[Five-fused-three-fused-five-membered ring intermediate, with the structure CC12[N-]C3=CC=CC=C3C14C2C(C=C4C5=CC=CC=C5)=O](1)</Checkpoint>

Since $\bf B$ contains a quinoline structure, the three-membered ring opens to form a benzo six-membered ring, which isomerizes into a quinoline structure; therefore, the structure of $\bf B$ is CC1=NC2=CC=CC=C2C3=C1C(CC3C4=CC=CC=C4)=O.

<Checkpoint>[The three-membered ring opens to form a benzo six-membered ring, which isomerizes into a quinoline structure](1)</Checkpoint>

<Checkpoint>[Structure $\bf B$ is CC1=NC2=CC=CC=C2C3=C1C(CC3C4=CC=CC=C4)=O](1)</Checkpoint>

According to the structure, options B, C, E, and F are all incorrect.

<MultiModal>![$\bf A$ structure is CC1=NC2=CC=CC=C2C13C(C4=CC=CC=C4)=CC(C3)=O；enolate anion, structure is CC1=NC2=CC=CC=C2C13C(C4=CC=CC=C4)=CC([O-])=C3；five-fused-three-fused-five-membered ring intermediate, structure is CC12[N-]C3=CC=CC=C3C14C2C(C=C4C5=CC=CC=C5)=O；$\bf B$ structure is CC1=NC2=CC=CC=C2C3=C1C(CC3C4=CC=CC=C4)=O](/media/uploads/bb271f20-4362-4a08-955f-81ec6fcf466d.png)</MultiModal>
观察 $\bf A$ 的核磁氢谱：位于低场，化学位移大于6.5的含有10个氢原子，应当为芳香氢；其中五个属于苯基，四个属于苯丙吡咯的苯环氢，因此猜测 $\bf A$ 含有苯基苄位的 $\rm sp^2 C-H$。

<Checkpoint>[$\bf A$ 含有苯基苄位的 $\rm sp^2 C-H$](1)</Checkpoint>

化学位移为2.21的为甲基氢原子，2.83和2.73的两个d峰明显为亚甲基峰。故底物的亚甲基和甲基未发生变化。

<Checkpoint>[底物的亚甲基和甲基未发生变化](1)</Checkpoint>

考虑底物反应性，亲核位点为吡咯的3号位，亲电位点为羰基和炔键；亲核羰基形成三元环不稳定，因此吡咯3号位亲核炔键，发生5-endo-dig反应形成五元螺环，产物结构为
CC1=NC2=CC=CC=C2C13C(C4=CC=CC=C4)=CC(C3)=O，该结构与核磁氢谱的分析吻合，因此即为结构 $\bf A$。

<Checkpoint>[吡咯3号位亲核炔键，发生5-endo-dig反应形成五元螺环](1)</Checkpoint>

<Checkpoint>[$\bf A$结构为CC1=NC2=CC=CC=C2C13C(C4=CC=CC=C4)=CC(C3)=O](1)</Checkpoint>

$\bf A$转化为$\bf B$的过程为碱性环境，首先拔除羰基α位的亚甲基氢形成烯醇负离子，结构为CC1=NC2=CC=CC=C2C13C(C4=CC=CC=C4)=CC([O-])=C3。

<Checkpoint>[拔除羰基α位的亚甲基氢形成烯醇负离子](1)</Checkpoint>

之后烯醇负离子亲核体系中亲电的吡咯2号位，形成五并三并五元环的中间体，结构为CC12[N-]C3=CC=CC=C3C14C2C(C=C4C5=CC=CC=C5)=O。故选项D正确。

<Checkpoint>[烯醇负离子亲核体系中亲电的吡咯2号位，形成五并三并五元环的中间体](1)</Checkpoint>

<Checkpoint>[五并三并五元环的中间体，结构为CC12[N-]C3=CC=CC=C3C14C2C(C=C4C5=CC=CC=C5)=O](1)</Checkpoint>

由于$\bf B$含有喹啉结构，因此三元环开环形成苯并六元环，异构化为喹啉结构；故$\bf B$结构为CC1=NC2=CC=CC=C2C3=C1C(CC3C4=CC=CC=C4)=O。

<Checkpoint>[三元环开环形成苯并六元环，异构化为喹啉结构](1)</Checkpoint>

<Checkpoint>[$\bf B$结构为CC1=NC2=CC=CC=C2C3=C1C(CC3C4=CC=CC=C4)=O](1)</Checkpoint>

根据结构可知，选项B,C,E,F均错误。

<MultiModal>![$\bf A$结构为CC1=NC2=CC=CC=C2C13C(C4=CC=CC=C4)=CC(C3)=O；烯醇负离子，结构为CC1=NC2=CC=CC=C2C13C(C4=CC=CC=C4)=CC([O-])=C3；五并三并五元环的中间体，结构为CC12[N-]C3=CC=CC=C3C14C2C(C=C4C5=CC=CC=C5)=O；$\bf B$结构为CC1=NC2=CC=CC=C2C3=C1C(CC3C4=CC=CC=C4)=O](/media/uploads/bb271f20-4362-4a08-955f-81ec6fcf466d.png)</MultiModal>