Polybenzimidazole PBI is a heat-resistant polymer that was studied relatively early. It has a melting point above $400 \degree \mathrm{C}$, and its films and fibers can maintain good mechanical properties up to $300 \degree \mathrm{C}$, but degrade rapidly beyond this temperature. The monomers are aromatic tetraamines and diacids or esters:

<MultiModal>![The reaction of NC1=C(N)C=C(C2=CC(N)=C(N)C=C2)C=C1 and O=C(OC1=CC=CC=C1)C2=CC(C(OC3=CC=CC=C3)=O)=CC=C2 can produce PBI](/media/uploads/dce04acb-b930-49c8-b8f3-7e65fba2749c.png)</MultiModal>

PBI has single bonds remaining in the main chain, which are easily broken when heated at high temperatures. Therefore, all-ladder polymers (molecular structure similar to a ladder) $\mathbf{A}$ are often used in aerospace equipment. It is known that the preparation of $\mathbf{A}$ requires two monomers, and the polycondensation proceeds in two steps. The first step can occur at room temperature without dehydration, and the second step occurs with slight heating, releasing two molecules of water.

<MultiModal>![The repeating unit structure of the polymer monomer is [X1]C(C1=C([X2])C=C(C(N2C3=NC4=C2C=C(N=C([X4])N5[X3])C5=C4)=O)C3=C1)=O, $\mathbf{X_1}$ is connected to $\mathbf{X_3}$; $\mathbf{X_2}$ is connected to $\mathbf{X_4}$.](/media/uploads/df01ba29-c410-4454-a7ba-4b069a48acda.png)</MultiModal>

The following information is available:
1.  The PBI repeating unit has 5 rings.
2.  The entire reaction process for the formation of PBI involves two reaction mechanisms.
3.  Phenyl ester is used here instead of diacid because the PBI synthesis reaction cannot be carried out in an acidic environment.
4.  The two monomers for synthesizing A have a total of 14 degrees of unsaturation.
5.  The first-step intermediate product is insoluble in water.

The option that contains all the correct statements is:
聚苯并咪唑 $\mathrm{PBI}$ 是一种研究得较早的耐热高分子，熔点在 $400 \degree \mathrm{C}$ 以上，薄膜和纤维达 $300 \degree \mathrm{C}$ 仍能保持良好的力学性能，超过这一温度会迅速降解。单体是芳族四元胺和二元酸或酯：

<MultiModal>![NC1=C(N)C=C(C2=CC(N)=C(N)C=C2)C=C1和O=C(OC1=CC=CC=C1)C2=CC(C(OC3=CC=CC=C3)=O)=CC=C2反应可以生成 $\mathrm{PBI}$ ](/media/uploads/dce04acb-b930-49c8-b8f3-7e65fba2749c.png)</MultiModal>

 $\mathrm{PBI}$ 中主链中留有单键，在高温下受热易断裂，所以在宇航设备中常使用全梯形聚合物（分子结构类似梯子） $\mathbf{A}$ 。已知制备 $\mathbf{A}$ 需两个单体，缩聚分两步进行，第一步在室温下即可发生，不脱水，稍加热即可发生第二步反应，脱去两分子水。

<MultiModal>![聚合物单体重复单元结构为[X1]C(C1=C([X2])C=C(C(N2C3=NC4=C2C=C(N=C([X4])N5[X3])C5=C4)=O)C3=C1)=O，$\mathbf{X_1}$ 与 $\mathbf{X_3}$ 相连；$\mathbf{X_2}$ 与 $\mathbf{X_4}$ 相连。](/media/uploads/df01ba29-c410-4454-a7ba-4b069a48acda.png)</MultiModal>

现有以下信息：
1.  $\mathrm{PBI}$ 重复单元中有 $\mathrm{5}$ 个环。
2. 整个 $\mathrm{PBI}$ 生成的反应过程涉及两种反应机理。
3. 这里使用苯酯而不用二元酸是因为 $\mathrm{PBI}$ 合成反应不能在酸性环境下进行。
4. 合成 $\mathrm{A}$ 的两个单体元一共有 $\mathrm{14}$ 个不饱和度。
5. 第一步中间产物不溶于水。

下面包含所有正确说法的选项是：
1
1, 3
1, 4
1, 5
2, 3
1, 3, 4
3, 5
4, 5
3, 4, 5
1, 3, 4, 5
所有说法都正确
1
1, 3
1, 4
1, 5
2, 3
1, 3, 4
3, 5
4, 5
3, 4, 5
1, 3, 4, 5
所有说法都正确
The generation of $\mathrm{PBI}$ involves two reactions:
1. A nucleophilic substitution reaction occurs between an amine group and an ester group, forming an amide bond.
2. Another amine group on the same side nucleophilically attacks the carbonyl group in the amide bond, forming a five-membered ring, with addition forming a hydroxyl group; then a hydroxyl elimination reaction occurs, forming a carbon-nitrogen double bond.

<Checkpoint>[Nucleophilic substitution reaction occurs, amine group reacts with ester group to form amide bond](1)</Checkpoint>

<Checkpoint>[Addition-elimination reaction occurs, amine group adds to carbonyl group, forming hydroxyl group; hydroxyl group undergoes elimination reaction, forming carbon-nitrogen double bond](1)</Checkpoint>

The repeating unit structure of $\mathrm{PBI}$ is:

<MultiModal>![The repeating unit structure of the polymer monomer is [X1]C(N1)=NC2=C1C=C(C3=CC(NC(C4=CC=CC([X2])=C4)=N5)=C5C=C3)C=C2, $\mathbf{X_1}$ is connected to $\mathbf{X_2}$.](/media/uploads/fcecf8d7-ea3f-4b35-af3d-30e5e8a73cb7.png)</MultiModal>

<Checkpoint>[The repeating unit structure of the polymer monomer is [X1]C(N1)=NC2=C1C=C(C3=CC(NC(C4=CC=CC([X2])=C4)=N5)=C5C=C3)C=C2, $\mathbf{X_1}$ is connected to $\mathbf{X_2}$.](2)</Checkpoint>

The phenyl ester is used here instead of a diacid because the polycondensation reaction temperature is high, and decarboxylation easily occurs. Using a phenyl ester can overcome this disadvantage.

<Checkpoint>[Phenyl ester is used instead of diacid because the polycondensation reaction temperature is high, and decarboxylation easily occurs](1)</Checkpoint>

The monomers for the preparation of $\mathbf{A}$ are:

<MultiModal>![The compound on the left is O=C(OC1=O)C2=C1C=C(C(OC3=O)=O)C3=C2; the compound on the right is NC1=CC(N)=C(N)C=C1N](/media/uploads/a8aa87d2-bb0f-49c7-be59-6cf19cf31125.png)</MultiModal>

<Checkpoint>[The monomers for the preparation of $\mathbf{A}$ are: O=C(OC1=O)C2=C1C=C(C(OC3=O)=O)C3=C2 and NC1=CC(N)=C(N)C=C1N](2)</Checkpoint>

The preparation of $\mathbf{A}$ involves two reactions:
1. A nucleophilic substitution reaction occurs between an amine group and a lactone group, forming an amide bond; the five-membered ring opens, forming a carboxylic acid group. At this time, the carboxylic acid group is deprotonated, and the unreacted amine group is protonated, forming a water-soluble zwitterionic intermediate.
2. Another amine group on the same side undergoes dehydration with another carboxylic acid group to form an amide bond; one of the amine groups then nucleophilically attacks the carbonyl group on the other side, forming a five-membered ring, with addition forming a hydroxyl group; then a hydroxyl elimination reaction occurs, forming a carbon-nitrogen double bond.

<Checkpoint>[Nucleophilic substitution reaction occurs, one amine group reacts with one lactone group to form an amide bond; the five-membered ring opens, forming a carboxylic acid group. Carboxylic acid group and amine group undergo proton transfer, forming a zwitterion](1)</Checkpoint>

<Checkpoint>[Another amine group undergoes nucleophilic substitution reaction with carboxylic acid group, forming an amide bond; one amine group undergoes nucleophilic addition reaction with the carbonyl group on the other side, forming a hydroxyl group; then a hydroxyl elimination reaction occurs](1)</Checkpoint>

Therefore, statements 1 and 4 are correct. Select option C.
生成 $\mathrm{PBI}$ 过程经历两个反应：
1. 一个氨基与酯基发生亲核取代反应，生成酰胺键。
2. 同侧另一个氨基亲核进攻酰胺键中的羰基，形成五元环，加成生成羟基；然后发生羟基消去反应，生成碳氮双键。

<Checkpoint>[发生亲核取代反应，氨基与酯基反应生成酰胺键](1)</Checkpoint>

<Checkpoint>[发生加成-消去反应，氨基与羰基加成，生成羟基；羟基发生消去反应，生成碳氮双键](1)</Checkpoint>

 $\mathrm{PBI}$ 的重复单元结构为：

<MultiModal>![聚合物单体重复单元结构为[X1]C(N1)=NC2=C1C=C(C3=CC(NC(C4=CC=CC([X2])=C4)=N5)=C5C=C3)C=C2，$\mathbf{X_1}$ 与 $\mathbf{X_2}$ 相连。](/media/uploads/fcecf8d7-ea3f-4b35-af3d-30e5e8a73cb7.png)</MultiModal>

<Checkpoint>[聚合物单体重复单元结构为[X1]C(N1)=NC2=C1C=C(C3=CC(NC(C4=CC=CC([X2])=C4)=N5)=C5C=C3)C=C2，$\mathbf{X_1}$ 与 $\mathbf{X_2}$ 相连。](2)</Checkpoint>

这里使用苯酯而不用二元酸是因为缩聚反应温度高，容易脱羧，改用苯酯可以克服这个缺点。

<Checkpoint>[使用苯酯而不用二元酸是因为缩聚反应温度高，容易脱羧](1)</Checkpoint>

制备 $\mathbf{A}$ 的单体为：

<MultiModal>![左边化合物为O=C(OC1=O)C2=C1C=C(C(OC3=O)=O)C3=C2；右边化合物为NC1=CC(N)=C(N)C=C1N](/media/uploads/a8aa87d2-bb0f-49c7-be59-6cf19cf31125.png)</MultiModal>

<Checkpoint>[制备 $\mathbf{A}$ 的单体为：O=C(OC1=O)C2=C1C=C(C(OC3=O)=O)C3=C2和NC1=CC(N)=C(N)C=C1N](2)</Checkpoint>

制备 $\mathbf{A}$ 的过程经历两个反应：
1. 一个氨基与一个内酯基发生亲核取代反应，生成酰胺键；五元环打开，形成羧基。这时羧基脱质子化，未反应的氨基质子化，形成可溶于水的内盐中间产物。
2. 同侧另一个氨基与另一个羧基脱水生成酰胺键；其中一个氨基再亲核进攻另一侧羰基，形成五元环，加成生成羟基；然后发生羟基消去反应，生成碳氮双键。

<Checkpoint>[发生亲核取代反应，其中一个氨基与一个内酯基发生反应生成酰胺键；五元环打开，形成羧基。羧基、氨基质子转移，生成内盐](1)</Checkpoint>

<Checkpoint>[另一个氨基与羧基发生亲核取代反应，生成酰胺键；其中一个氨基与另一侧羰基发生亲核加成反应，生成羟基；然后再发生羟基消去反应](1)</Checkpoint>

因此，说法 1, 4 正确。选择选项  C 。