In the Pummerer rearrangement, when the substrate sulfoxide is replaced with an alkenyl sulfoxide $\bf{A}$, the original Pummerer rearrangement process will not occur. Alkenyl sulfoxide $\bf{A}$ exhibits different reaction properties when reacting with alkenes substituted with different $\rm R$ groups (as shown in the figure below).

<MultiModal>![O=S(CCCS/1)C1=C\C(F)(F)F.C[Si](C)(C/C=C/[R])C>$Tf_2O$, $2,6-t-Bu_2pyridine$>[X]. [X] represents the reaction product.](/media/uploads/9f47fecb-c19d-45e5-b0a3-aa2ac34cf8dd.png)</MultiModal>

When the substrates are different, the reaction product $\bf X$ may be $\bf B, C, D$

When $\mathrm{R = H}$, the molecular formula of the main reaction product $\bf{B}$ is $\mathrm{C_9H_{11}F_3S_2}$; when $\mathrm{R = C_7H_{15}}$, only one product $\bf{C}$ was obtained, and it is known that the formation of $\bf{C}$ involves a pericyclic reaction.

Phenol reacts with alkenyl sulfoxide $\bf{A}$ under the action of $\mathrm{Tf_2O}$ to obtain the spirocyclic compound $\bf{D}$.

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

1. There are two conjugated carbon-carbon double bonds in $\bf{B}$

2. When the substrate is isotopically labeled, i.e., when $\mathrm{R = D}$, the deuterium atom in the main reaction product $\bf{B'}$ is separated from the nearest sulfur atom by three carbon atoms

3. A [3,3]-$\sigma$ migration involving an oxygen atom occurred in the process of generating $\bf{C}$

4. The R group (i.e., $\mathrm{C_7H_{15}}$) in $\bf{C}$ is connected to a primary carbon

5. There are two conjugated carbon-carbon double bonds in $\bf{C}$

6. $\bf{D}$ is chiral
在Pummerer重排反应中，当反应的底物亚砜换为烯基亚砜$\bf{A}$时，原有的Pummerer重排过程将不能发生。烯基亚砜$\bf{A}$在与不同$\rm R$基团取代的烯烃反应时，表现出不同的反应性质（如下图所示）。

<MultiModal>![O=S(CCCS/1)C1=C\C(F)(F)F.C[Si](C)(C/C=C/[R])C>$Tf_2O$, $2,6-t-Bu_2pyridine$>[X]。[X]表示反应产物。](/media/uploads/9f47fecb-c19d-45e5-b0a3-aa2ac34cf8dd.png)</MultiModal>

当底物不同时，反应产物$\bf X$可能为$\bf B, C, D$

$\mathrm{R = H}$ 时，反应主产物 $\bf{B}$ 的分子式为 $\mathrm{C_9H_{11}F_3S_2}$；当 $\mathrm{R = C_7H_{15}}$ 时，反应仅得到了一种产物 $\bf{C}$，已知生成 $\bf{C}$ 的过程涉及周环反应。  
  
苯酚与烯基亚砜 $\bf{A}$ 在 $\mathrm{Tf_2O}$ 的作用下反应，得到了螺环化合物 $\bf{D}$。

下列关于 $\bf{B}$、$\bf{C}$、$\bf{D}$ 说法正确的是：

1.$\bf{B}$ 中存在共轭的两个碳碳双键

2.当用同位素标记底物，即当 $\mathrm{R = D}$ 时，反应主产物 $\bf{B'}$ 中的氘原子与最近的硫原子相隔三个碳原子

3.生成 $\bf{C}$ 的过程中发生了氧原子参与的[3,3]-$\sigma$迁移

4.$\bf{C}$ 中的R基团（即 $\mathrm{C_7H_{15}}$）与一个一级碳相连

5.$\bf{C}$ 中存在共轭的两个碳碳双键

6.$\bf{D}$ 有手性
1,2,4,6
1,2,3,4,6
1,2,3,4,5,6
1,2,6
1,3,5,6
2,4,6
2,4
1,2,3,5,6
1,5,6
3,6
1,2,3,6
2,3,4,5,6
All of the above options are incorrect or the answer is incomplete.
1,2,4,6
1,2,3,4,6
1,2,3,4,5,6
1,2,6
1,3,5,6
2,4,6
2,4
1,2,3,5,6
1,5,6
3,6
1,2,3,6
2,3,4,5,6
以上选项均不正确或答案不完全
The silyl group at the $\beta$ position of an alkene can enhance the nucleophilicity of the carbon-carbon double bond through the hyperconjugation effect of the carbon-silicon single bond on the carbon-carbon double bond. Alkenyl sulfoxide $\bf{A}$ first generates a sulfonium ion under the action of trifluoroacetic anhydride, and then is attacked by the substrate alkene at the terminal carbon-carbon double bond conjugated to the sulfonium ion via an $\mathrm{S_N2'}$ mechanism, leaving the silyl group to obtain a carbon-carbon double bond at the end of the chain, as shown in the following structure:

<MultiModal>![C=CCC(C1=[S+]CCCS1)C(F)(F)F](/media/uploads/9805d989-5b3f-4e93-8f8a-7b0c6c7b59f0.png)</MultiModal>

Finally, a proton is eliminated to form a carbon-carbon double bond conjugated with two sulfur atoms, and the final structure of $\bf{B}$ is as follows:

<Checkpoint>[The sulfonium ion generated from alkenyl sulfoxide $\bf{A}$ is attacked by the substrate alkene via an $\mathrm{S_N2'}$ mechanism](1)</Checkpoint>

<Checkpoint>[Elimination of a proton forms a carbon-carbon double bond conjugated with two sulfur atoms](1)</Checkpoint>

<MultiModal>![C=CC/C(C(F)(F)F)=C1SCCCS\1](/media/uploads/980669f1-2931-47af-891d-74676d19150d.png)</MultiModal>

<Checkpoint>[The structure of $\bf{B}$ is C=CC/C(C(F)(F)F)=C1SCCCS\1](1)</Checkpoint>

The two carbon-carbon double bonds in $\bf{B}$ are not conjugated, statement 1 is incorrect.

When $\mathrm{R = D}$, according to the reaction mechanism, the deuterium atom in the main product $\bf{B'}$ should be separated from the nearest sulfur atom by three carbon atoms, so statement 2 is correct.

When $\mathrm{R = C_7H_{15}}$, the steric hindrance is large, and the generated sulfonium ion is directly nucleophilically attacked by the substrate alkene to obtain a sulfonium ion connected to three carbons. At the same time, the silyl group leaves to obtain a carbon-carbon double bond at the end of the chain, as shown in the following structure:

<MultiModal>![C=CC([S+](CCCS/1)C1=C\C(F)(F)F)CCCCCCC](/media/uploads/8d856d8a-7e53-44f8-bacb-b53cf6bc6faf.png)</MultiModal>

[3,3]-$\sigma$ migration occurs, and a proton is eliminated to form a carbon-carbon double bond conjugated with two sulfur atoms to obtain $\bf{C}$, as shown in the following structure:

<Checkpoint>[When $\mathrm{R = C_7H_{15}}$, the generated sulfonium ion is directly nucleophilically attacked by the substrate alkene](1)</Checkpoint>

<Checkpoint>[A [3,3]-$\sigma$ migration occurs and a proton is eliminated to form a carbon-carbon double bond conjugated with two sulfur atoms](1)</Checkpoint>

<MultiModal>![FC(/C(C/C=C/CCCCCCC)=C1SCCCS/1)(F)F](/media/uploads/a4cec1b0-035c-40b1-aea0-b97480fc34a5.png)</MultiModal>

<Checkpoint>[The structure of $\bf{C}$ is FC(/C(C/C=C/CCCCCCC)=C1SCCCS/1](1)</Checkpoint>

Therefore, the oxygen atom is not involved in the [3,3]-$\sigma$ migration, $\mathrm{C_7H_{15}}$ is connected to a primary carbon, and the two carbon-carbon double bonds in $\bf{C}$ are also not conjugated, so statement 4 is correct, and statements 3 and 5 are incorrect.

When the substrate is replaced with phenol, similarly to the process of generating $\bf{B}$, the ortho position of the electron-rich hydroxyl group of phenol attacks the terminal carbon-carbon double bond conjugated with the sulfonium ion via an $\mathrm{S_N2'}$ mechanism, and then a proton is eliminated to restore aromaticity; the trifluoroacetate anion leaves to obtain a sulfonium ion with a carbon-sulfur double bond-like structure, as shown in the following structure:

<MultiModal>![OC1=C(C(C2=[S+]CCCS2)C(F)(F)F)C=CC=C1](/media/uploads/6bb1a36a-7168-4cb1-9d24-0e9627e07902.png)</MultiModal>

However, since there is no hydrogen in the ortho position at this time to eliminate, the phenolic hydroxyl group attacks the carbon-sulfur double bond to finally obtain the five-membered spirocyclic structure $\bf{D}$, as shown in the following structure:

<Checkpoint>[The ortho position of the electron-rich hydroxyl group of phenol attacks the terminal carbon-carbon double bond conjugated with the sulfonium ion via an $\mathrm{S_N2'}$ mechanism](1)</Checkpoint>

<Checkpoint>[The phenolic hydroxyl group attacks the carbocation to finally obtain the five-membered spirocyclic structure](1)</Checkpoint>

<MultiModal>![FC(C1C2(OC3=C1C=CC=C3)SCCCS2)(F)F](/media/uploads/464c8b0e-e11a-417c-9c3f-82f9d10a62b8.png)</MultiModal>

<Checkpoint>[The structure of $\bf{D}$ is FC(C1C2(OC3=C1C=CC=C3)SCCCS2](1)</Checkpoint>

The benzylic position of the benzene ring in $\bf{D}$ is connected to four different groups. At the same time, there is no mirror plane, center of symmetry, or four-fold improper rotation axis in $\bf{D}$, so $\bf{D}$ is chiral, and statement 6 is correct.

The correct answer is F
烯烃 $\beta$ 位的硅基可以通过碳硅单键对碳碳双键超共轭效应增强碳碳双键的亲核性。烯基亚砜 $\bf{A}$ 首先在三氟乙酸酐的作用下生成硫鎓离子，随后被底物烯烃以 $\mathrm{S_N2'}$ 的机理进攻与硫鎓离子共轭的碳碳双键末端，离去硅基得到链末端的碳碳双键，结构如下：

<MultiModal>![C=CCC(C1=[S+]CCCS1)C(F)(F)F](/media/uploads/9805d989-5b3f-4e93-8f8a-7b0c6c7b59f0.png)</MultiModal>

最后消除氢离子形成与两个硫原子共轭的碳碳双键，最终 $\bf{B}$ 的结构如下：

<Checkpoint>[烯基亚砜 $\bf{A}$ 生成的硫鎓离子被底物烯烃以 $\mathrm{S_N2'}$ 的机理进攻](1)</Checkpoint>

<Checkpoint>[消除氢离子形成与两个硫原子共轭的碳碳双键](1)</Checkpoint>

<MultiModal>![C=CC/C(C(F)(F)F)=C1SCCCS\1](/media/uploads/980669f1-2931-47af-891d-74676d19150d.png)</MultiModal>

<Checkpoint>[$\bf{B}$ 的结构为C=CC/C(C(F)(F)F)=C1SCCCS\1](1)</Checkpoint>

$\bf{B}$ 中两个碳碳双键不共轭，说法1错误

当 $\mathrm{R = D}$ 时，根据反应机理，反应主产物 $\bf{B'}$ 中的氘原子与最近的硫原子应相隔三个碳原子，说法2正确

而当 $\mathrm{R = C_7H_{15}}$ 时位阻较大，生成的硫鎓离子被底物烯烃直接亲核进攻，得到与三个碳相连的硫鎓离子，同时离去硅基得到链末端的碳碳双键，结构如下：

<MultiModal>![C=CC([S+](CCCS/1)C1=C\C(F)(F)F)CCCCCCC](/media/uploads/8d856d8a-7e53-44f8-bacb-b53cf6bc6faf.png)</MultiModal>

发生 [3,3]-$\sigma$迁移并消除氢离子形成与两个硫原子共轭的碳碳双键得到 $\bf{C}$ ，结构如下：

<Checkpoint>[当 $\mathrm{R = C_7H_{15}}$ 时，生成的硫鎓离子被底物烯烃直接亲核进攻](1)</Checkpoint>

<Checkpoint>[发生 [3,3]-$\sigma$迁移并消除氢离子形成与两个硫原子共轭的碳碳双键](1)</Checkpoint>

<MultiModal>![FC(/C(C/C=C/CCCCCCC)=C1SCCCS/1)(F)F](/media/uploads/a4cec1b0-035c-40b1-aea0-b97480fc34a5.png)</MultiModal>

<Checkpoint>[$\bf{C}$ 的结构为FC(/C(C/C=C/CCCCCCC)=C1SCCCS/1](1)</Checkpoint>

所以氧原子没有参与[3,3]-$\sigma$迁移，$\mathrm{C_7H_{15}}$与一个一级碳相连，$\bf{C}$ 中两个碳碳双键也不共轭，说法4正确，3、5错误

当底物换为苯酚时，类似的于生成 $\bf{B}$ 的过程，苯酚富电子的羟基的邻位以 $\mathrm{S_N2'}$ 的机理进攻与硫鎓离子共轭的碳碳双键末端，再消除氢离子恢复芳香性；离去三氟乙酸根负离子，得到带有类似碳硫双键的硫鎓离子，结构如下：

<MultiModal>![OC1=C(C(C2=[S+]CCCS2)C(F)(F)F)C=CC=C1](/media/uploads/6bb1a36a-7168-4cb1-9d24-0e9627e07902.png)</MultiModal>

但由于此时邻位没有氢无法消除，酚羟基进攻碳硫双键最终得到五元螺环结构的 $\bf{D}$，结构如下：

<Checkpoint>[苯酚富电子的羟基的邻位以 $\mathrm{S_N2'}$ 的机理进攻与硫鎓离子共轭的碳碳双键末端](1)</Checkpoint>

<Checkpoint>[酚羟基进攻碳正离子最终得到五元螺环结构](1)</Checkpoint>

<MultiModal>![FC(C1C2(OC3=C1C=CC=C3)SCCCS2)(F)F](/media/uploads/464c8b0e-e11a-417c-9c3f-82f9d10a62b8.png)</MultiModal>

<Checkpoint>[$\bf{D}$ 的结构为FC(C1C2(OC3=C1C=CC=C3)SCCCS2](1)</Checkpoint>

$\bf{D}$ 中苯环苄位连接了四个不同的基团，同时 $\bf{D}$ 中不存在镜面、对称中心、四次映轴，因此 $\bf{D}$ 有手性，说法6正确

正确答案为F