For the following polymerization reactions, the following information is available:

<MultiModal>![There are three chemical reactions in the figure. 1 is O=C(C(C)O1)OC(C)C1=O>IC2=CC=C(/C(OB(F)F)=C/C(C3=CC=C(OCCO)C=C3)=O)C=C2>, condition is $\mathrm{Sn(oct)_2,\ 130\degree C}$; 2 is OC1=CC=C(OC2=CC=C(O)C=C2)C=C1.NC3=CC=C(OC4=CC=C(N)C=C4)C=C3>>, condition is $\mathrm{(HCHO)_n},\ CHCl_3,\ NEt_3}$; 3 is OC1=CC=CC=C1C(C(C2=CC=CC=C2O)N)N.O=C[Ar]C=O>>[*], subsequently product $\bf{A}$ in one step generates $\bf{B}$ then in multiple steps generates $\bf{C}$](/media/uploads/057075b4-5a59-4bf3-a938-1858f04fbd26.png)</MultiModal>

1. In reaction 2, the two substrates and formaldehyde react in a 1:1:4 ratio to obtain a chain polymer. The $\mathrm{^1H\ NMR}$ of the product shows that the structure does not contain active hydrogen. Note: This ratio is calculated based on formaldehyde monomers.

2. Reaction 3 obtains a polymer with a classic salen ligand as a side group through condensation and pericyclic reactions.

Which of the following statements is incorrect (select A if all are correct)?
对于以下聚合反应，有如下信息：

<MultiModal>![图中有三个化学反应。1为O=C(C(C)O1)OC(C)C1=O>IC2=CC=C(/C(OB(F)F)=C/C(C3=CC=C(OCCO)C=C3)=O)C=C2>，条件为$\mathrm{Sn(oct)_2,\ 130\degree C}$；2为OC1=CC=C(OC2=CC=C(O)C=C2)C=C1.NC3=CC=C(OC4=CC=C(N)C=C4)C=C3>>，条件为$\mathrm{(HCHO)_n},\ CHCl_3,\ NEt_3}$；3为OC1=CC=CC=C1C(C(C2=CC=CC=C2O)N)N.O=C[Ar]C=O>>[*]，随后产物$\bf{A}$一步生成$\bf{B}$后多步生成$\bf{C}$](/media/uploads/057075b4-5a59-4bf3-a938-1858f04fbd26.png)</MultiModal>

1.反应2中两种底物和甲醛以1:1:4反应，得到链状聚合物。产物的$\mathrm{^1H\ NMR}$显示结构中不含活泼氢。注：该比例按照甲醛单体计算。

2.反应3通过缩合和周环反应得到了一个侧基为经典salen配体的聚合物。

下列说法中错误的是（若均正确，选择A)
All other options are incorrect
In reaction 1, one repeating unit of the polymer product contains 9 atoms.
Reaction 1 is an addition polymerization reaction.
In reaction 2, one repeating unit of the polymeric product contains 6 rings.
In reaction 3, the pericyclic reaction is specifically a [3,3]σ sigmatropic rearrangement.
Reaction 3 is a condensation reaction.
其他选项均不正确
反应1中，高分子产物1个重复单元含有9个原子
反应1为加聚反应
反应2中，高分子产物1个重复单元含有6个环
反应3中，周环反应具体为[3,3]σ迁移反应
反应3为缩聚反应
In the following analysis, when the polymer repeating unit is represented using SMILES, R1 and R2 represent the end groups of the polymer repeating unit.

For reaction 1, IC2=CC=C(/C(OB(F)F)=C/C(C3=CC=C(OCCO)C=C3)=O)C=C2 is the initiator, acting as an end group, initiating the addition polymerization of O=C(C(C)O1)OC(C)C1=O, the repeating unit in the product is [R1]OC(C([R2])C)=O, and the end groups R1 and R2 are hydroxyl and IC1=CC=C(/C(OB(F)F)=C/C(C2=CC=C(OCCO[R])C=C2)=O)C=C1, respectively. Therefore, the repeating unit contains 9 atoms, B is correct; addition polymerization, C is correct.

<Checkpoint>[The repeating unit in the product of reaction 1 is [R1]OC(C([R2])C)=O](1)</Checkpoint>

<Checkpoint>[The end groups R1 and R2 are hydroxyl and IC1=CC=C(/C(OB(F)F)=C/C(C2=CC=C(OCCO[R])C=C2)=O)C=C1, respectively](1)</Checkpoint>

For reaction 2, since the reaction ratio is 1:1:4, it can be determined that 1 molecule of substrate will react with 4 molecules of formaldehyde, undergoing 4 steps of Mannich reaction, forming a polymer in which the repeating unit is CC1=CC=C(OC2=CC=C(N3CC4=CC(OC5=CC6=C(OCN(C6)C)C=C5)=CC=C4OC3)C=C2)C=C1, containing 6 rings, D is correct.

<Checkpoint>[The repeating unit of the product of reaction 2 is CC1=CC=C(OC2=CC=C(N3CC4=CC(OC5=CC6=C(OCN(C6)C)C=C5)=CC=C4OC3)C=C2)C=C1](1)</Checkpoint>

For reaction 3, condensation and pericyclic reactions occurred according to the prompt. $\bf{A}$ is the condensation product OC1=C(C(/N=C/[Ar]C=O)C(/N=C/[Ar]C=O)C2=C(O)C=CC=C2)C=CC=C1.

<Checkpoint>[$\bf{A}$ is OC1=C(C(/N=C/[Ar]C=O)C(/N=C/[Ar]C=O)C2=C(O)C=CC=C2)C=CC=C1](1)</Checkpoint>

$\mathbf{A}\to\mathbf{B}$ is a pericyclic reaction, at this time cope rearrangement occurs in the structure of $\bf{A}$, which can form a salen ligand analog, E is correct. $\bf{B}$ is OC1=C(/C=N/C([Ar]C=O)C(/N=C/C2=CC=CC=C2O)[Ar]C=O)C=CC=C1.

<Checkpoint>[Cope rearrangement occurs in the structure of $\bf{A}$](1)</Checkpoint>

<Checkpoint>[$\bf{B}$ is OC1=C(/C=N/C([Ar]C=O)C(/N=C/C2=CC=CC=C2O)[Ar]C=O)C=CC=C1](1)</Checkpoint>

Subsequently, $\bf{B}\to\bf{C}$ undergoes aldol condensation polymerization, which is a condensation reaction, F is correct. The repeating unit of the product is [R2][Ar]C(/N=C/C1=C(O)C=CC=C1)C(/N=C/C2=C(O)C=CC=C2)[R1].

<Checkpoint>[The repeating unit of the product of reaction 3 is [R2][Ar]C(/N=C/C1=C(O)C=CC=C1)C(/N=C/C2=C(O)C=CC=C2)[R1]](1)</Checkpoint>

Therefore, all statements are correct, choose A.

<MultiModal>![The three reaction processes and products are shown in this figure. The repeating unit in the product of reaction 1 is [R1]OC(C([R2])C)=O, and the end groups R1 and R2 are hydroxyl and IC1=CC=C(/C(OB(F)F)=C/C(C2=CC=C(OCCO[R])C=C2)=O)C=C1, respectively; the repeating unit of the product of reaction 2 is CC1=CC=C(OC2=CC=C(N3CC4=CC(OC5=CC6=C(OCN(C6)C)C=C5)=CC=C4OC3)C=C2)C=C1; in reaction 3, $\bf{A}$ is OC1=C(C(/N=C/[Ar]C=O)C(/N=C/[Ar]C=O)C2=C(O)C=CC=C2)C=CC=C1; $\bf{B}$ is OC1=C(/C=N/C([Ar]C=O)C(/N=C/C2=CC=CC=C2O)[Ar]C=O)C=CC=C1; the repeating unit of the product is [R2][Ar]C(/N=C/C1=C(O)C=CC=C1)C(/N=C/C2=C(O)C=CC=C2)[R1]](/media/uploads/17f56351-6b8f-4cac-9fca-56ac6c2e5116.png)</MultiModal>
以下解析中的高分子重复单元利用SMILES表示时，R1和R2代表高分子重复单元的端基。

对于反应1，IC2=CC=C(/C(OB(F)F)=C/C(C3=CC=C(OCCO)C=C3)=O)C=C2为引发剂，作为端基，引发了O=C(C(C)O1)OC(C)C1=O的加聚，产物中重复单元为[R1]OC(C([R2])C)=O，端基R1和R2分别为羟基和IC1=CC=C(/C(OB(F)F)=C/C(C2=CC=C(OCCO[R])C=C2)=O)C=C1。因此，重复单元含有9个原子，B正确；加聚反应，C正确。

<Checkpoint>[反应1产物中重复单元为[R1]OC(C([R2])C)=O](1)</Checkpoint>

<Checkpoint>[端基R1和R2分别为羟基和IC1=CC=C(/C(OB(F)F)=C/C(C2=CC=C(OCCO[R])C=C2)=O)C=C1](1)</Checkpoint>

对于反应2，由于反应比例为1:1:4，可以判断1分子底物会和4分子甲醛反应，发生4步Mannich反应，形成聚合物中，重复单元为CC1=CC=C(OC2=CC=C(N3CC4=CC(OC5=CC6=C(OCN(C6)C)C=C5)=CC=C4OC3)C=C2)C=C1，含有6个环，D正确。

<Checkpoint>[反应2产物重复单元为CC1=CC=C(OC2=CC=C(N3CC4=CC(OC5=CC6=C(OCN(C6)C)C=C5)=CC=C4OC3)C=C2)C=C1](1)</Checkpoint>

对于反应3，根据提示发生了缩合和周环反应。$\bf{A}$为缩合产物OC1=C(C(/N=C/[Ar]C=O)C(/N=C/[Ar]C=O)C2=C(O)C=CC=C2)C=CC=C1。

<Checkpoint>[$\bf{A}$为OC1=C(C(/N=C/[Ar]C=O)C(/N=C/[Ar]C=O)C2=C(O)C=CC=C2)C=CC=C1](1)</Checkpoint>

$\mathbf{A}\to\mathbf{B}$为周环反应，此时$\bf{A}$的结构中发生cope重排，可以形成salen配体类似物，E正确。$\bf{B}$为OC1=C(/C=N/C([Ar]C=O)C(/N=C/C2=CC=CC=C2O)[Ar]C=O)C=CC=C1。

<Checkpoint>[$\bf{A}$的结构中发生cope重排](1)</Checkpoint>

<Checkpoint>[$\bf{B}$为OC1=C(/C=N/C([Ar]C=O)C(/N=C/C2=CC=CC=C2O)[Ar]C=O)C=CC=C1](1)</Checkpoint>

随后$\bf{B}\to\bf{C}$通过羟醛缩合聚合，为缩合反应，F正确。产物的重复单元为 [R2][Ar]C(/N=C/C1=C(O)C=CC=C1)C(/N=C/C2=C(O)C=CC=C2)[R1]。

<Checkpoint>[反应3产物的重复单元为 [R2][Ar]C(/N=C/C1=C(O)C=CC=C1)C(/N=C/C2=C(O)C=CC=C2)[R1]](1)</Checkpoint>

因此，说法均正确，选择A。

<MultiModal>![三个反应过程及产物如本图所示。反应1产物中重复单元为[R1]OC(C([R2])C)=O，端基R1和R2分别为羟基和IC1=CC=C(/C(OB(F)F)=C/C(C2=CC=C(OCCO[R])C=C2)=O)C=C1；反应2产物重复单元为CC1=CC=C(OC2=CC=C(N3CC4=CC(OC5=CC6=C(OCN(C6)C)C=C5)=CC=C4OC3)C=C2)C=C1；反应3中，$\bf{A}$为OC1=C(C(/N=C/[Ar]C=O)C(/N=C/[Ar]C=O)C2=C(O)C=CC=C2)C=CC=C1；$\bf{B}$为OC1=C(/C=N/C([Ar]C=O)C(/N=C/C2=CC=CC=C2O)[Ar]C=O)C=CC=C1；产物的重复单元为 [R2][Ar]C(/N=C/C1=C(O)C=CC=C1)C(/N=C/C2=C(O)C=CC=C2)[R1]](/media/uploads/17f56351-6b8f-4cac-9fca-56ac6c2e5116.png)</MultiModal>