<MultiModal>![This figure illustrates 5 organic structural formulas. $\bf{A}$ is N#CC1=CC(CCCC)CC=CC1; $\bf{B}$ is C=C(Br)CC12C=CCC1CC(C(OC)=O)=C(O[Si](C)(C)C)C2; $\bf{C}$ is CC1(C=C2)CC=CC3C1C(OC3)(OC)C2=O; $\bf{D}$ is O=C(CC(C(C)=C)C1)CC2CC1=CCC2; $\bf{E}$ is CC12CCC=C(CCC3)C1C2C3=O.](/media/uploads/925fd954-43cf-4b59-bcf3-db5c76a10b80.png)</MultiModal>

The structures in the above figure are all generated through Cope rearrangement or O-Cope rearrangement; regarding the precursor structural formulas before they undergo rearrangement, the following statements are made:

1.  The rearrangement precursor corresponding to $\bf{A}$ contains a six-membered ring.

2.  The rearrangement precursor corresponding to $\bf{B}$ does not contain a five-membered ring.

3.  The rearrangement precursor corresponding to $\bf{C}$ contains only two six-membered rings and one five-membered ring.

4.  The rearrangement precursors corresponding to $\bf{D，E}$ both contain a hydroxyl group.

5.  The rearrangement precursor corresponding to $\bf{E}$ contains a three-membered ring.

6.  The rearrangement precursor corresponding to $\bf{D}$ reacts with $\rm{O_3, Zn/MeOH}$ to obtain a new structure containing two carbonyl groups.

Among the following options, the one that contains the largest number of correct statement numbers is:
<MultiModal>![本图体现了5个有机结构式。$\bf{A}$为N#CC1=CC(CCCC)CC=CC1；$\bf{B}$为C=C(Br)CC12C=CCC1CC(C(OC)=O)=C(O[Si](C)(C)C)C2；$\bf{C}$为CC1(C=C2)CC=CC3C1C(OC3)(OC)C2=O；$\bf{D}$为O=C(CC(C(C)=C)C1)CC2CC1=CCC2；$\bf{E}$为CC12CCC=C(CCC3)C1C2C3=O。](/media/uploads/925fd954-43cf-4b59-bcf3-db5c76a10b80.png)</MultiModal>

上图的结构都是通过Cope重排或O-Cope重排生成的；关于它们发生重排之前的前体结构式，有下列说法：

1. $\bf{A}$对应的重排前体含有六元环。

2. $\bf{B}$对应的重排前体不含有五元环。

3. $\bf{C}$对应的重排前体只含有两个六元环和一个五元环。

4. $\bf{D，E}$对应的重排前体均含有羟基。

5. $\bf{E}$对应的重排前体含有三元环。

6. $\bf{D}$对应的重排前体与$\rm{O_3, Zn/MeOH}$反应，得到的新结构含有两个羰基。

下列选项中，包含的说法序号全都正确且最多的一项是：
All other options are incorrect
1，3
1，4
2，5
3，5
4，5
5，6
2，6
1，3，4
1，4，5
1，2，6
2，4，5
2，4，6
3，5，6
4，5，6
1，3，4，5
1，4，5，6
1，2，3，4
1，3，5，6
3，4，5，6
1，2，4，5，6
2, 3, 4, 5, 6
1，2，3，4，5，6
1，3，4，6
1，2，4，5
其他选项均不正确
1，3
1，4
2，5
3，5
4，5
5，6
2，6
1，3，4
1，4，5
1，2，6
2，4，5
2，4，6
3，5，6
4，5，6
1，3，4，5
1，4，5，6
1，2，3，4
1，3，5，6
3，4，5，6
1，2，4，5，6
2，3，4，5，6
1，2，3，4，5，6
1，3，4，6
1，2，4，5
The general formula for the Cope rearrangement is:

1,5-dienes undergo a [3,3]$\sigma$ sigmatropic rearrangement under thermal conditions to yield a new 1,5-diene.

<Checkpoint>[Cope rearrangement is a [3,3]$\sigma$ sigmatropic rearrangement that 1,5-dienes undergo under thermal conditions to yield a new 1,5-diene.](1)</Checkpoint>

This reaction forms the $\rm{C1-C6}$ bond, breaks the $\rm{C3-C4}$ bond, and simultaneously migrates the double bond from the 1,5-position to the 3,4-position.

<Checkpoint>[Formation of $\rm{C1-C6}$ bond, breaking of $\rm{C3-C4}$ bond](1)</Checkpoint>

The characteristic of the O-Cope rearrangement compared to the Cope rearrangement is that the substrate contains a hydroxyl group at the 1 or 4 position, and after undergoing a [3,3]$\sigma$ sigmatropic rearrangement, it will produce an enol structure at the 1 or 4 position, thereby converting to a carbonyl group at the 1 or 4 position.

<Checkpoint>[O-Cope rearrangement produces an enol structure at the 1 or 4 position, thereby converting to a carbonyl group at the 1 or 4 position](1)</Checkpoint>

Therefore, the steps for retro-analyzing the Cope rearrangement and O-Cope rearrangement are:

1. Convert the carbonyl group to an enol structure;

2. Find the 1,5-diene structure and number it;

3. Break the $\rm{C1-C6}$ bond, form the $\rm{C3-C4}$ bond, and migrate the double bond.

Therefore, according to the above steps, the precursors corresponding to $\bf{A-E}$ can be deduced as shown in the following figure:

<MultiModal>![$\bf{A1}$ is C=CC1C(C1)(C#N)/C=C/CCCC；$\bf{B1}$ is C[C@]1(C(O[Si](C)(C)C)=C)[C@H](C2)C=C(CC(Br)=C)[C@H]2C1；$\bf{C1}$ is C=CC1C2C(C(C)=CC1C3=O)C3(OC)OC2；$\bf{D1}$ is C=C1C[C@@H]2C[C@@](/C=C\C(C)=C)(O)[C@H]1CC2；$\bf{E1}$ is CC1(C2C1C3(C=C)O)CCC3C2=C。](/media/uploads/447f6ef2-e9f7-4fc7-be77-0051479df824.png)</MultiModal>

The precursor $\bf{A1}$ corresponding to $\bf{A}$ is C=CC1C(C1)(C#N)/C=C/CCCC.

<Checkpoint>[The precursor $\bf{A1}$ corresponding to $\bf{A}$ is C=CC1C(C1)(C#N)/C=C/CCCC.](2)</Checkpoint>

The precursor $\bf{B1}$ corresponding to $\bf{B}$ is C[C@]1(C(O[Si](C)(C)C)=C)[C@H](C2)C=C(CC(Br)=C)[C@H]2C1.

<Checkpoint>[The precursor $\bf{B1}$ corresponding to $\bf{B}$ is C[C@]1(C(O[Si](C)(C)C)=C)[C@H](C2)C=C(CC(Br)=C)[C@H]2C1.](2)</Checkpoint>

The precursor $\bf{C1}$ corresponding to $\bf{C}$ is C=CC1C2C(C(C)=CC1C3=O)C3(OC)OC2.

<Checkpoint>[The precursor $\bf{C1}$ corresponding to $\bf{C}$ is C=CC1C2C(C(C)=CC1C3=O)C3(OC)OC2.](2)</Checkpoint>

The precursor $\bf{D1}$ corresponding to $\bf{D}$ is C=C1C[C@@H]2C[C@@](/C=C\C(C)=C)(O)[C@H]1CC2.

<Checkpoint>[The precursor $\bf{D1}$ corresponding to $\bf{D}$ is C=C1C[C@@H]2C[C@@](/C=C\C(C)=C)(O)[C@H]1CC2.](2)</Checkpoint>

The precursor $\bf{E1}$ corresponding to $\bf{E}$ is CC1(C2C1C3(C=C)O)CCC3C2=C.

<Checkpoint>[The precursor $\bf{E1}$ corresponding to $\bf{E}$ is CC1(C2C1C3(C=C)O)CCC3C2=C.](2)</Checkpoint>

$\bf{D1}$ is cleaved by ozonolysis of the double bond, and because it contains a terminal olefin, it will only produce two carbonyl groups.

According to the structural formula, only statements 4, 5, and 6 are correct.

<Checkpoint>[Statements 4, 5, and 6 are correct](1)</Checkpoint>
Cope重排的通式为：

1,5-二烯类化合物在受热条件下发生[3，3]$\sigma$迁移得到一个新的1,5-二烯。

<Checkpoint>[Cope重排为1,5-二烯类化合物在受热条件下发生[3，3]$\sigma$迁移得到一个新的1,5-二烯。](1)</Checkpoint>

该反应会生成$\rm{C1-C6}$键，断裂$\rm{C3-C4}$键，同时双键从1,5位迁移到3，4位。

<Checkpoint>[生成$\rm{C1-C6}$键，断裂$\rm{C3-C4}$键](1)</Checkpoint>

O-Cope重排相比于Cope重排的特点是底物在1或4号位含有醇羟基，在发生[3，3]$\sigma$迁移后会产生1或4号位的烯醇结构，从而转化为1或4号位的羰基。

<Checkpoint>[O-Cope重排会产生1或4号位的烯醇结构，从而转化为1或4号位的羰基](1)</Checkpoint>

从而逆向分析Cope重排和O-Cope重排的步骤为：

1.将羰基变为烯醇结构；

2.寻找1,5-二烯结构并进行编号；

3.断裂$\rm{C1-C6}$键，生成$\rm{C3-C4}$键，迁移双键。

从而根据上述步骤可以推得$\bf{A-E}$对应的前体如下图所示：

<MultiModal>![$\bf{A1}$为C=CC1C(C1)(C#N)/C=C/CCCC；$\bf{B1}$为C[C@]1(C(O[Si](C)(C)C)=C)[C@H](C2)C=C(CC(Br)=C)[C@H]2C1；$\bf{C1}$为C=CC1C2C(C(C)=CC1C3=O)C3(OC)OC2；$\bf{D1}$为C=C1C[C@@H]2C[C@@](/C=C\C(C)=C)(O)[C@H]1CC2；$\bf{E1}$为CC1(C2C1C3(C=C)O)CCC3C2=C。](/media/uploads/447f6ef2-e9f7-4fc7-be77-0051479df824.png)</MultiModal>

$\bf{A}$对应的前体$\bf{A1}$为C=CC1C(C1)(C#N)/C=C/CCCC。

<Checkpoint>[$\bf{A}$对应的前体$\bf{A1}$为C=CC1C(C1)(C#N)/C=C/CCCC。](2)</Checkpoint>

$\bf{B}$对应的前体$\bf{B1}$为C[C@]1(C(O[Si](C)(C)C)=C)[C@H](C2)C=C(CC(Br)=C)[C@H]2C1。

<Checkpoint>[$\bf{B}$对应的前体$\bf{B1}$为C[C@]1(C(O[Si](C)(C)C)=C)[C@H](C2)C=C(CC(Br)=C)[C@H]2C1。](2)</Checkpoint>

$\bf{C}$对应的前体$\bf{C1}$为C=CC1C2C(C(C)=CC1C3=O)C3(OC)OC2。

<Checkpoint>[$\bf{C}$对应的前体$\bf{C1}$为C=CC1C2C(C(C)=CC1C3=O)C3(OC)OC2。](2)</Checkpoint>

$\bf{D}$对应的前体$\bf{D1}$为C=C1C[C@@H]2C[C@@](/C=C\C(C)=C)(O)[C@H]1CC2。

<Checkpoint>[$\bf{D}$对应的前体$\bf{D1}$为C=C1C[C@@H]2C[C@@](/C=C\C(C)=C)(O)[C@H]1CC2。](2)</Checkpoint>

$\bf{E}$对应的前体$\bf{E1}$为CC1(C2C1C3(C=C)O)CCC3C2=C。

<Checkpoint>[$\bf{E}$对应的前体$\bf{E1}$为CC1(C2C1C3(C=C)O)CCC3C2=C。](2)</Checkpoint>

$\bf{D1}$被臭氧化断裂双键，由于含有末端烯烃，只会产生两个羰基。

根据结构式可知，只有说法4，5，6正确。

<Checkpoint>[说法4，5，6正确](1)</Checkpoint>