The following substrates, under the action of dichloroethenone, underwent charge-separated intermediates $\bf{G}$ and $\bf{H}$ to generate the five-membered ring lactone $\bf{I}$:

<MultiModal>![O=S(C1=C(C)C(C=C2OCC3=CC=CC=C3)=C(C=C2)N1C(OC(C)(C)C)=O)C(C)C>ClC(Cl)=C=O>[I]. Where [1]>'2'>[3] indicates that compound 1 reacts under the conditions of 2 to obtain compound 3. The formation of product $\bf{I}$ also successively involves intermediates $\bf{G}$ and $\bf{H}$ (i.e., intermediate $\bf{G}$ is first formed, then $\bf{G}$ is converted to $\bf{H}$, and finally $\bf{H}$ is converted to product $\bf{I}$).](/media/uploads/c6d85148-a5bb-4300-815e-6e0317e7470d.png)</MultiModal>

The following are statements about $\bf{G}$, $\bf{H}$, and $\bf{I}$:

1. A pericyclic reaction occurred during the formation of $\bf{G}$

2. $\mathrm{S-O}$ bond cleavage occurred during the process from $\bf{G}$ to $\bf{H}$

3. $\bf{H}$ has only one stereochemical center

4. The process from $\bf{H}$ to $\bf{I}$ has selectivity for generating a thermodynamically stable trans-fused ring

5. $\bf{I}$ has 3 six-membered rings

6. $\bf{I}$ has an $\mathrm{S-C-O}$ linkage

Which of the following options contains all the correct statements?
以下底物在二氯代乙烯酮作用下,经历了电荷分离的中间体 $\bf{G}$ 和 $\bf{H}$ 生成了五元环内酯 $\bf{I}$:

<MultiModal>![O=S(C1=C(C)C(C=C2OCC3=CC=CC=C3)=C(C=C2)N1C(OC(C)(C)C)=O)C(C)C>ClC(Cl)=C=O>[I]。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。在生成产物 $\bf{I}$ 的过程中还依次经历了中间体 $\bf{G}$ 和 $\bf{H}$(即首先反应生成中间体 $\bf{G}$，随后由 $\bf{G}$ 得到 $\bf{H}$，最后由 $\bf{H}$ 得到产物 $\bf{I}$。 )](/media/uploads/c6d85148-a5bb-4300-815e-6e0317e7470d.png)</MultiModal>

以下有关于 $\bf{G}$、$\bf{H}$、$\bf{I}$ 的说法：

1.生成 $\bf{G}$ 的过程中发生了周环反应

2.$\bf{G}$ 到 $\bf{H}$ 的过程中发生了 $\mathrm{S-O}$ 键的断裂

3.$\bf{H}$ 只有一个立体化学中心

4.$\bf{H}$ 到 $\bf{I}$ 的过程中具有生成热力学稳定的反式并环的选择性

5.$\bf{I}$ 中有3个六元环

6.$\bf{I}$ 中有 $\mathrm{S-C-O}$ 的键连关系

下列选项中，包含所有正确说法的选项是：
1,4,5
1,2,4,6
1,2,3,4,5,6
2,3,6
1,2,4,5
1,4
2,3,4,5
1,2,3,4,6
4
1,2,3,4,5
1,2,4
All of the above options are incorrect or the answer is incomplete.
1,4,5
1,2,4,6
1,2,3,4,5,6
2,3,6
1,2,4,5
1,4
2,3,4,5
1,2,3,4,6
4
1,2,3,4,5
1,2,4
以上选项均不正确或答案不完全
Dichloroketene is a strong electrophile and is also prone to cycloaddition reactions due to its structure with cumulative double bonds. Since the reaction first undergoes a charge-separated intermediate $\bf{G}$, the first step is not a pericyclic reaction, but an electrophilic reaction. The site with the strongest nucleophilicity in the substrate is the oxygen atom of the sulfoxide. Therefore, the first step is the nucleophilic addition of the oxygen in the sulfoxide group of the substrate to the carbonyl carbon of dichloroketene, yielding $\bf{G}$, with the following structure:

<Checkpoint>[The first step is not a pericyclic reaction because it first undergoes a charge-separated intermediate $\bf{G}$](1)</Checkpoint>

<Checkpoint>[The oxygen in the sulfoxide group of the substrate undergoes nucleophilic addition to the carbonyl carbon of dichloroketene](1)</Checkpoint>

<MultiModal>![CC(C1=C2C=CC(OCC3=CC=CC=C3)=C1)=C([S+](C(C)C)O/C([O-])=C(Cl)/Cl)N2C(OC(C)(C)C)=O](/media/uploads/08665525-554d-4a55-8c94-a88f2a8129ae.png)</MultiModal>

<Checkpoint>[$\bf{G}$ is CC(C1=C2C=CC(OCC3=CC=CC=C3)=C1)=C([S+](C(C)C)O/C([O-])=C(Cl)/Cl)N2C(OC(C)(C)C)=O](1)</Checkpoint>

No pericyclic reaction occurred during the formation of $\bf{G}$, statement 1 is incorrect.

Subsequently, $\bf{G}$ undergoes a [3,3]-$\sigma$ rearrangement, breaking the $\mathrm{S-O}$ bond and forming a $\mathrm{C-C}$ bond, to give intermediate $\bf{H}$, with the following structure:

<Checkpoint>[Undergoes a [3,3]-$\sigma$ rearrangement](1)</Checkpoint>

<MultiModal>![[O-]C(C(C1(C)C(C=C(C=C2)OCC3=CC=CC=C3)=C2N(/C1=[S+]\C(C)C)C(OC(C)(C)C)=O)(Cl)Cl)=O](/media/uploads/1e775630-cf78-4e1f-8a21-bbb3017b973e.png)</MultiModal>

<Checkpoint>[$\bf{H}$ is [O-]C(C(C1(C)C(C=C(C=C2)OCC3=CC=CC=C3)=C2N(/C1=[S+]\C(C)C)C(OC(C)(C)C)=O)(Cl)Cl)=O](1)</Checkpoint>

The $\mathrm{S-O}$ bond is broken during the process from $\bf{G}$ to $\bf{H}$; $\bf{H}$ has only one stereochemical center, statements 2 and 3 are correct.

Finally, due to conformational constraints, the carboxylate anion in intermediate $\bf{H}$ can only attack the positively charged carbon atom in the carbon-sulfur double bond from the same face, yielding the cis-fused five-membered ring $\bf{I}$, with the following structure:

<Checkpoint>[Due to conformational constraints, the carboxylate anion can only attack the positively charged carbon atom in the carbon-sulfur double bond from the same face](1)</Checkpoint>

<MultiModal>![C[C@]1(C(Cl)2Cl)[C@@](SC(C)C)(OC2=O)N(C3=C1C=C(OCC4=CC=CC=C4)C=C3)C(OC(C)(C)C)=O](/media/uploads/821eda07-3872-499d-bcab-ca3b004ee7df.png)</MultiModal>

<Checkpoint>[$\bf{I}$ is C[C@]1(C(Cl)2Cl)[C@@](SC(C)C)(OC2=O)N(C3=C1C=C(OCC4=CC=CC=C4)C=C3)C(OC(C)(C)C)=O](1)</Checkpoint>

The process from $\bf{H}$ to $\bf{I}$ has selectivity for the formation of a cis-fused ring; $\bf{I}$ has 2 six-membered rings; there is an $\mathrm{S-C-O}$ linkage in $\bf{I}$, statements 4 and 5 are incorrect, and 6 is correct.

Therefore, the correct answer is D.
二氯代乙烯酮是一个强的亲电试剂，由于其累积双键的结构也易发生环加成反应。由于反应首先经历了电荷分离的中间体 $\bf{G}$，因此第一步不是发生周环反应，而是发生亲电反应。而底物中亲核能力最强的位点为亚砜的氧原子。因此，第一步为底物亚砜基团中的氧对二氯代乙烯酮羰基碳进行亲核加成，得到 $\bf{G}$，结构如下：

<Checkpoint>[首先经历了电荷分离的中间体 $\bf{G}$，因此第一步不是发生周环反应](1)</Checkpoint>

<Checkpoint>[底物亚砜基团中的氧对二氯代乙烯酮羰基碳进行亲核加成](1)</Checkpoint>

<MultiModal>![CC(C1=C2C=CC(OCC3=CC=CC=C3)=C1)=C([S+](C(C)C)O/C([O-])=C(Cl)/Cl)N2C(OC(C)(C)C)=O](/media/uploads/08665525-554d-4a55-8c94-a88f2a8129ae.png)</MultiModal>

<Checkpoint>[$\bf{G}$ 是CC(C1=C2C=CC(OCC3=CC=CC=C3)=C1)=C([S+](C(C)C)O/C([O-])=C(Cl)/Cl)N2C(OC(C)(C)C)=O](1)</Checkpoint>

生成 $\bf{G}$ 的过程中没有发生了周环反应，说法1错误

随后 $\bf{G}$ 发生 [3,3]-$\sigma$迁移，断裂 $\mathrm{S-O}$ 键并生成 $\mathrm{C-C}$ 键，得到中间体 $\bf{H}$，结构如下：

<Checkpoint>[发生 [3,3]-$\sigma$迁移](1)</Checkpoint>

<MultiModal>![[O-]C(C(C1(C)C(C=C(C=C2)OCC3=CC=CC=C3)=C2N(/C1=[S+]\C(C)C)C(OC(C)(C)C)=O)(Cl)Cl)=O](/media/uploads/1e775630-cf78-4e1f-8a21-bbb3017b973e.png)</MultiModal>

<Checkpoint>[$\bf{H}$ 是[O-]C(C(C1(C)C(C=C(C=C2)OCC3=CC=CC=C3)=C2N(/C1=[S+]\C(C)C)C(OC(C)(C)C)=O)(Cl)Cl)=O](1)</Checkpoint>

$\bf{G}$ 到 $\bf{H}$ 的过程中发生了 $\mathrm{S-O}$ 键的断裂；$\bf{H}$ 只有一个立体化学中心，说法2、3正确

最后，中间体 $\bf{H}$ 中的羧基负离子由于构象限制只能在同面进攻碳硫双键中的正电性的碳原子，得到顺式的五元并环 $\bf{I}$，结构如下：

<Checkpoint>[羧基负离子由于构象限制只能在同面进攻碳硫双键中的正电性的碳原子](1)</Checkpoint>

<MultiModal>![C[C@]1(C(Cl)2Cl)[C@@](SC(C)C)(OC2=O)N(C3=C1C=C(OCC4=CC=CC=C4)C=C3)C(OC(C)(C)C)=O](/media/uploads/821eda07-3872-499d-bcab-ca3b004ee7df.png)</MultiModal>

<Checkpoint>[$\bf{I}$ 是C[C@]1(C(Cl)2Cl)[C@@](SC(C)C)(OC2=O)N(C3=C1C=C(OCC4=CC=CC=C4)C=C3)C(OC(C)(C)C)=O](1)</Checkpoint>

$\bf{H}$ 到 $\bf{I}$ 的过程中具有生成顺式并环的选择性；$\bf{I}$ 中有2个六元环；$\bf{I}$ 中存在 $\mathrm{S-C-O}$ 的键连关系，说法4、5错误，6正确

因此正确答案为D