Pummerer rearrangement typically refers to the chemical reaction of activating the oxygen atom of a sulfoxide with an anhydride, going through a sulfonium ion intermediate, and ultimately generating an α-functionalized sulfide. In the total synthesis of a certain natural product (as shown in the figure below), Pummerer rearrangement was used to construct a new quaternary carbon center.

<MultiModal>![The figure shows a multi-step reaction, which can be described as: [H][C@@]1(CCN(CCS(C2=CC=CC=C2)=O)[C@]13[H])CCC4=C3C5=CC=CC=C5N4C(OC)=O>$(CF_3CO)_2O，CF_3CO_2H$>[A]，[A]>$Raney-Ni，EtOH$>[B]，[B]>$hv，MeOH$>[H][C@@]1(CC(C(OC)=O)=C2N([H])C3=CC=CC=C3[C@]24CCN5CC1)[C@]45[H]。Where [1]>'2'>[3] means compound 1 reacts under the conditions of 2 to obtain compound 3.](/media/uploads/0f296641-db01-444f-8c43-46c18092a322.png)</MultiModal>

Which of the following statements about $\bf{A}$ and $\bf{B}$ are correct:

1.  $\bf{A}$ has three stereochemical centers

2.  $\bf{A}$ and $\bf{B}$ have the same number of rings

3.  A carbon-carbon double bond shift occurred during the generation of $\bf{B}$ from $\bf{A}$

4.  [1,5]-$\sigma$ migration occurred in $\bf{B}$ under light irradiation

5.  Compared with $\bf{B}$, there is no generation or disappearance of stereochemical centers in the final product

6.  An imine structure exists in the intermediate of the formation of the final product from $\bf{B}$
Pummerer重排通常是指用酸酐将亚砜的氧原子活化，经历硫鎓离子中间体，最终生成α-官能团化的硫醚的化学反应。在某天然产物的全合成中（如下图所示），利用Pummerer重排构建了新的季碳中心。

<MultiModal>![图中为多步反应，可以描述为：[H][C@@]1(CCN(CCS(C2=CC=CC=C2)=O)[C@]13[H])CCC4=C3C5=CC=CC=C5N4C(OC)=O>$(CF_3CO)_2O，CF_3CO_2H$>[A]，[A]>$Raney-Ni，EtOH$>[B]，[B]>$hv，MeOH$>[H][C@@]1(CC(C(OC)=O)=C2N([H])C3=CC=CC=C3[C@]24CCN5CC1)[C@]45[H]。其中[1]>'2'>[3]表示化合物1在2的条件下反应得到化合物3。](/media/uploads/0f296641-db01-444f-8c43-46c18092a322.png)</MultiModal>

下列关于 $\bf{A}$ 和 $\bf{B}$ 说法正确的有：

1.$\bf{A}$ 有三个立体化学中心

2.$\bf{A}$ 和 $\bf{B}$ 环的数量相同

3.$\bf{A}$ 生成 $\bf{B}$ 的过程中发生了碳碳双键移位

4.$\bf{B}$ 在光照下发生了[1,5]-$\sigma$迁移

5.最终产物与 $\bf{B}$ 相比不存在立体化学中心的产生或消失

6.$\bf{B}$ 生成最终产物的中间体中存在亚胺结构
1,2,4,5,6
1,2,3,4,5,6
5,6
2,6
2,3,6
1,2,4,6
1,3,4,5
2,4,5
3,4,6
All of the above options are incorrect or the answer is incomplete.
1,2,4,5,6
1,2,3,4,5,6
5,6
2,6
2,3,6
1,2,4,6
1,3,4,5
2,4,5
3,4,6
以上选项均不正确或答案不完全
In the process of the reactant generating $\bf{A}$, firstly, the classical Pummerer rearrangement reaction occurs, that is, the oxygen atom of the sulfoxide nucleophilically attacks trifluoroacetic anhydride, resulting in a sulfonium ion, followed by 1,2-elimination to obtain a carbocation (or an onium ion containing a carbon-sulfur double bond) at the ortho position of the sulfide. Since the indole ring is relatively electron-rich, it can undergo an aromatic electrophilic substitution reaction. The number 3 position of the indole ring nucleophilically attacks the carbocation, constructing a five-membered ring system similar to the product, and generating an imine cation structure, and finally eliminating $\mathrm{H^+}$ to obtain $\bf{A}$, the structure is as follows:

<Checkpoint>[The number 3 position of the indole ring nucleophilically attacks the carbocation, constructing a five-membered ring system similar to the product](1)</Checkpoint>

<MultiModal>![[H][C@@]1(CC=C([C@@](C2=CC=CC=C32)4C(CN5CC1)SC6=CC=CC=C6)N3C(OC)=O)[C@]45[H]](/media/uploads/d20654b1-6677-4610-ae25-ffea6f0e31b5.png)</MultiModal>

<Checkpoint>[The structure of $\bf{A}$ is [H][C@@]1(CC=C([C@@](C2=CC=CC=C32)4C(CN5CC1)SC6=CC=CC=C6)N3C(OC)=O)[C@]45[H]](1)</Checkpoint>

So $\bf{A}$ has four stereochemical centers. Statement 1 is incorrect.

$\bf{A}$ is then reduced by $\mathrm{Raney-Ni}$ to cleave the carbon-sulfur bond, finally yielding one molecule of $\mathrm{PhSH}$ and $\bf{B}$. The structure of $\bf{B}$ is as follows:

<MultiModal>![[H][C@@]1(CC=C([C@@](C2=CC=CC=C32)4CCN5CC1)N3C(OC)=O)[C@]45[H]](/media/uploads/c5c1fbe6-a666-4a1b-8533-07901134fa21.png)</MultiModal>

<Checkpoint>[$\mathrm{Raney-Ni}$ reduces the carbon-sulfur bond](1)</Checkpoint>

<Checkpoint>[The structure of $\bf{B}$ is [H][C@@]1(CC=C([C@@](C2=CC=CC=C32)4CCN5CC1)N3C(OC)=O)[C@]45[H]](1)</Checkpoint>

Because it is reduced, $\bf{B}$ has one less benzene ring than $\bf{A}$, and no carbon-carbon double bond shift occurs in the process of $\bf{A}$ generating $\bf{B}$. Statements 2 and 3 are incorrect.

<Checkpoint>[$\bf{A}$ has one more ring than $\bf{B}$, and no carbon-carbon double bond shift occurs in the process of $\bf{A}$ generating $\bf{B}$](1)</Checkpoint>

Under illumination, the $\mathrm{-COOMe}$ in $\bf{B}$ undergoes a [1,3]-$\sigma$ migration, yielding an imine intermediate, followed by imine-enamine tautomerization to give the final product. In this step, there is no creation or disappearance of stereochemical centers. Therefore, statements 4 is incorrect, and 5 and 6 are correct.

<Checkpoint>[$\bf{B}$ undergoes a [1,3]-$\sigma$ migration under illumination](1)</Checkpoint>

<Checkpoint>[Compared with $\bf{B}$, there is no creation or disappearance of stereochemical centers in the final product](1)</Checkpoint>

<Checkpoint>[There is an imine structure in the intermediate of $\bf{B}$ generating the final product](1)</Checkpoint>

Therefore, statements 5 and 6 are correct, and the answer is C.
在反应物生成 $\bf{A}$ 的过程中，首先发生了经典的Pummerer重排反应，即亚砜的氧原子亲核进攻三氟乙酸酐，得到硫鎓离子，随后发生1，2消除，得到硫醚邻位的碳正离子（或含碳硫双键的鎓离子）。由于吲哚环较富电子，可以发生芳香亲电取代反应，吲哚环三号位亲核进攻碳正离子，构建了类似产物的五元环系，并生成了亚胺正离子结构，最终消除 $\mathrm{H^+}$ 得到 $\bf{A}$，结构如下：

<Checkpoint>[吲哚环三号位亲核进攻碳正离子，构建了类似产物的五元环系](1)</Checkpoint>

<MultiModal>![[H][C@@]1(CC=C([C@@](C2=CC=CC=C32)4C(CN5CC1)SC6=CC=CC=C6)N3C(OC)=O)[C@]45[H]](/media/uploads/d20654b1-6677-4610-ae25-ffea6f0e31b5.png)</MultiModal>

<Checkpoint>[$\bf{A}$结构为[H][C@@]1(CC=C([C@@](C2=CC=CC=C32)4C(CN5CC1)SC6=CC=CC=C6)N3C(OC)=O)[C@]45[H]](1)</Checkpoint>

所以 $\bf{A}$ 有四个立体化学中心。说法1错误。

$\bf{A}$ 随后被 $\mathrm{Raney-Ni}$ 还原碳硫键，最终得到一分子 $\mathrm{PhSH}$ 和 $\bf{B}$，$\bf{B}$ 的结构如下：

<MultiModal>![[H][C@@]1(CC=C([C@@](C2=CC=CC=C32)4CCN5CC1)N3C(OC)=O)[C@]45[H]](/media/uploads/c5c1fbe6-a666-4a1b-8533-07901134fa21.png)</MultiModal>

<Checkpoint>[$\mathrm{Raney-Ni}$ 还原碳硫键](1)</Checkpoint>

<Checkpoint>[$\bf{B}$结构为[H][C@@]1(CC=C([C@@](C2=CC=CC=C32)4CCN5CC1)N3C(OC)=O)[C@]45[H]](1)</Checkpoint>

由于被还原，$\bf{B}$ 比 $\bf{A}$ 少一个苯环，并且 $\bf{A}$ 生成 $\bf{B}$ 的过程中没有发生碳碳双键移位。说法2，3错误

<Checkpoint>[$\bf{A}$ 比 $\bf{B}$ 多一个环，$\bf{A}$ 生成 $\bf{B}$ 的过程中没有发生碳碳双键移位](1)</Checkpoint>

在光照下 $\bf{B}$ 中的 $\mathrm{-COOMe}$ 发生[1,3]-$\sigma$迁移，得到亚胺中间体，随后发生亚胺与烯胺的互变异构得到最终产物。在该步反应中不存在立体化学中心的产生或消失。因此说法4错误，5，6正确

<Checkpoint>[$\bf{B}$ 在光照下发生了[1,3]-$\sigma$迁移](1)</Checkpoint>

<Checkpoint>[最终产物与 $\bf{B}$ 相比不存在立体化学中心的产生或消失](1)</Checkpoint>

<Checkpoint>[$\bf{B}$ 生成最终产物的中间体中存在亚胺结构](1)</Checkpoint>

所以说法5，6正确，答案为C