Proteins possess a rich hierarchical structure, and it is this complex and subtle structure that endows proteins with many functions, such as catalysis, transport, and information transfer, making them widely and importantly applied in fields such as biochemistry and medicine. For chemists, the ability to artificially synthesize protein analogs would expand the structural range of biological macromolecules and allow for more precise control in protein modification and other aspects. In 2021, Marcus Weck's group synthesized polyisocyanides that can mimic the secondary and tertiary structures of proteins, and the related results were published on MRC. Note: PMDETA is pentamethyldiethylenetriamine, and the terminal group structure of $\bf{C}$ does not need to be explicitly defined.

<MultiModal>![Reaction of styrene with `CC(Br)C1=CC=CC=C1` in the presence of cuprous bromide and PMDETA at 100°C to obtain $\bf{A}$, $\bf{A}$ first reacts with sodium azide in DMF at room temperature, and then reacts with lithium aluminum hydride in THF to obtain $\bf{B}$, $\bf{B}$ reacts with `C[C@@H](C(=O)NCC(=O)OCC1=C(C(=C(C(=C1F)F)F)F)F)[N+]#[C-]` in DCM in the presence of $[Ni(CN^tBu)_4][ClO_4]_2$ to obtain $\bf{C}$](/media/uploads/3bb60f2e-4eb9-4caf-9c8f-76ad4cfa1433.png)</MultiModal>

The following statements are made:

1. Each molecule of $\bf{B}$ contains 1 $sp^3$ nitrogen atom.

2. $\bf{B}$ contains independent methyl groups and the carbon atom connected to the methyl group is bonded to a heteroatom.

3. A certain segment of $\bf{C}$ can coil into a β-helix-like structure, and the helical structure formed by this segment has a high inversion energy barrier and will not unwind even in strongly polar protic solvents. This segment is a polyisocyanide block.

4. Two segments of $\bf{C}$ have further folded. However, if the monomer added in the step from $\bf{B}$ to $\bf{C}$ is replaced with the following compound, folding will not occur. The main intermolecular force causing this folding is π-π stacking.

<MultiModal>![[C-]#[N+][C@@H](C)C(NCCCCCC(OCC1=CC=CC=C1)=O)=O](/media/uploads/d160877a-b1dc-4263-84c1-13c4710e02bb.png)</MultiModal>

Then the sum of the coefficients of all correct statements is:
蛋白质具有丰富的层次结构，正是这复杂而微妙的结构赋予了蛋白质许多功能，诸如催化、运输、信息传递等等，使得其在生化医药等领域有着广泛且重要的应用。对于化学家来说，如能人工合成出蛋白质的类似物，将扩大生物大分子的结构范围，并在蛋白质修饰等方面可以进行更加精确的控制。2021年，MarcusWeck课题组合成了能够模拟蛋白质二级与三级结构的聚异腈高分子，相关成果发表于MRC上。注：PMDETA为五甲基二亚乙基三胺，无需明确$\bf{C}$的端基结构。

<MultiModal>![苯乙烯与`CC(Br)C1=CC=CC=C1`在溴化亚铜和PMDETA在100℃下反应得到$\bf{A}$，$\bf{A}$先在室温下DMF中与叠氮化钠反应，再在THF中与氢化铝锂反应得到$\bf{B}$，$\bf{B}$在DCM中、$[Ni(CN^tBu)_4][ClO_4]_2$存在下与`C[C@@H](C(=O)NCC(=O)OCC1=C(C(=C(C(=C1F)F)F)F)F)[N+]#[C-]`反应得到$\bf{C}$](/media/uploads/3bb60f2e-4eb9-4caf-9c8f-76ad4cfa1433.png)</MultiModal>

有以下说法：

1.每分子$\bf{B}$中含有1个$sp^3$氮原子

2.$\bf{B}$中存在独立的甲基且甲基所连碳原子与杂原子成键

3.$\bf{C}$的某一链段可以盘曲成β螺旋的类似结构，该链段形成的螺旋结构具有很高的翻转能垒，即使在强极性质子溶剂中也不会解旋。这一链段是聚异腈嵌段

4.$\bf{C}$的两个链段还进一步发生了折叠。然而，如果把$\bf{B}$到$\bf{C}$一步加入的单体换为以下化合物，则折叠无法发生。发生这一折叠的主要分子间作用力是π-π堆积

<MultiModal>![[C-]#[N+][C@@H](C)C(NCCCCCC(OCC1=CC=CC=C1)=O)=O](/media/uploads/d160877a-b1dc-4263-84c1-13c4710e02bb.png)</MultiModal>

则所有正确说法系数和为：
All statements are incorrect.
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2
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9
10
所有说法均不正确
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2
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4
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8
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Generating $\bf{A}$ is a typical atom transfer radical polymerization (ATRP) reaction. Styrene is the monomer, `CC(Br)C1=CC=CC=C1` is the initiator, and cuprous bromide/PMDETA is the catalytic system. First, cuprous bromide reacts with `CC(Br)C1=CC=CC=C1` to generate `C[C]([H])C1=CC=CC=C1` radical (the radical is the carbon atom that was originally bonded to bromine) and cupric bromide, initiating the polymerization. At termination, the polymer radical abstracts the bromine atom from cupric bromide to obtain $\bf{A}$, with the basic structure: `CC(C1=CC=CC=C1)CC(Br)C2=CC=CC=C2` (degree of polymerization is 1), where `[*]CC([*])C1=CC=CC=C1` is the repeating unit.

<MultiModal>![$\bf{A}$，基本结构为：`CC(C1=CC=CC=C1)CC(Br)C2=CC=CC=C2`（聚合度为1），其中`[*]CC([*])C1=CC=CC=C1`为重复单元](/media/uploads/c98e188d-5be0-4133-bc42-a260776bb8b3.png)</MultiModal>

In the process of generating $\bf{B}$, the bromine atom is first replaced by azide, and then reduced by lithium aluminum hydride to generate an amino group, to obtain $\bf{B}$: the basic structure is: `CC(C1=CC=CC=C1)CC(N)C2=CC=CC=C2` (degree of polymerization is 1), where `[*]CC([*])C1=CC=CC=C1` is the repeating unit.

<MultiModal>![$\bf{B}$：基本结构为：`CC(C1=CC=CC=C1)CC(N)C2=CC=CC=C2`（聚合度为1），其中`[*]CC([*])C1=CC=CC=C1`为重复单元](/media/uploads/2f2e7317-1d67-41e4-954b-dc2f70996442.png)</MultiModal>

<Checkpoint>[The basic structure of $\bf{B}$ is: `CC(C1=CC=CC=C1)CC(N)C2=CC=CC=C2` (degree of polymerization is 1), containing 1 amino group, statement 1 is correct](1)</Checkpoint>

The carbon in $\bf{B}$ bonded to the methyl group is bonded to the phenyl group and the methylene group, and is not bonded to heteroatoms.

<Checkpoint>[The carbon in $\bf{B}$ bonded to the methyl group is bonded to the phenyl group and the methylene group, and is not bonded to heteroatoms. Statement 2 is incorrect](1)</Checkpoint>

The amino group in $\bf{B}$ attacks the isocyanide carbon atom, generating a carbanion, which then further attacks the next molecule of isocyanide, resulting in the polymer $\bf{C}$: the basic structure is `CC(CC(N/C([R])=N/[C@H](C(NCCCCCC(OCC1=C(F)C(F)=C(F)C(F)=C1F)=O)=O)C)C2=CC=CC=C2)C3=CC=CC=C3` (the degree of polymerization of both polymer segments is 1), where `[*]CC([*])C1=CC=CC=C1` and `[*]/C([*])=N/[C@@H](C)C(NCCCCCC(OCC1=C(C(F)=C(C(F)=C1F)F)F)=O)=O` are the repeating units, and R represents an unknown end-capping group.

<MultiModal>![$\bf{C}$：基本结构为`CC(CC(N/C([R])=N/[C@H](C(NCCCCCC(OCC1=C(F)C(F)=C(F)C(F)=C1F)=O)=O)C)C2=CC=CC=C2)C3=CC=CC=C3`（两段聚合物聚合度均为1），其中`[*]CC([*])C1=CC=CC=C1`和`[*]/C([*])=N/[C@@H](C)C(NCCCCCC(OCC1=C(C(F)=C(C(F)=C1F)F)F)=O)=O`为重复单元，R表示未知的封端基团](/media/uploads/7f9e462b-180d-4c7b-9871-bc6298d5e218.png)</MultiModal>

The amide groups on the side chains of the polyisocyanide block in $\bf{C}$ can form interchain hydrogen bonds with each other, maintaining a helical structure, and the carbon atoms on the main chain are all $sp^2$ hybridized, which has a certain rigidity, and therefore has a very high inversion energy barrier.

<Checkpoint>[The amide groups on the side chains of the polyisocyanide block in $\bf{C}$ can form interchain hydrogen bonds with each other, maintaining a helical structure, and the carbon atoms on the main chain are all $sp^2$ hybridized, and therefore has a very high inversion energy barrier. Statement 3 is correct](1)</Checkpoint>

Compared with the original monomer, the substituted monomer replaces the pentafluorophenyl group with a phenyl group. The pentafluorophenyl group is an electron-deficient aromatic ring, while the side group of the polystyrene segment is an electron-rich phenyl group. A strong π-π stacking interaction can occur between electron-deficient and electron-rich aromatic rings. After the replacement, this interaction disappears, preventing folding from occurring.

<Checkpoint>[The pentafluorophenyl group is an electron-deficient aromatic ring, and the phenyl group is electron-rich, and a strong π-π stacking interaction can occur between them. Statement 4 is correct](1)</Checkpoint>

Statements 1, 3, and 4 are correct, and the sum is 8, so choose I.
生成$\bf{A}$是一个典型的原子转移自由基聚合（ATRP）反应。苯乙烯是单体，`CC(Br)C1=CC=CC=C1`是引发剂，溴化亚铜/PMDETA是催化体系。首先溴化亚铜与`CC(Br)C1=CC=CC=C1`反应产生`C[C]([H])C1=CC=CC=C1`自由基（自由基为原先与溴成键的碳原子）和溴化铜,引发聚合。终止时，聚合物自由基攫取溴化铜的溴原子，得到$\bf{A}$，基本结构为：`CC(C1=CC=CC=C1)CC(Br)C2=CC=CC=C2`（聚合度为1），其中`[*]CC([*])C1=CC=CC=C1`为重复单元。

<MultiModal>![$\bf{A}$，基本结构为：`CC(C1=CC=CC=C1)CC(Br)C2=CC=CC=C2`（聚合度为1），其中`[*]CC([*])C1=CC=CC=C1`为重复单元](/media/uploads/c98e188d-5be0-4133-bc42-a260776bb8b3.png)</MultiModal>

生成$\bf{B}$的过程中，首先叠氮取代溴原子，随后被氢化铝锂还原产生氨基，得到$\bf{B}$：基本结构为：`CC(C1=CC=CC=C1)CC(N)C2=CC=CC=C2`（聚合度为1），其中`[*]CC([*])C1=CC=CC=C1`为重复单元。

<MultiModal>![$\bf{B}$：基本结构为：`CC(C1=CC=CC=C1)CC(N)C2=CC=CC=C2`（聚合度为1），其中`[*]CC([*])C1=CC=CC=C1`为重复单元](/media/uploads/2f2e7317-1d67-41e4-954b-dc2f70996442.png)</MultiModal>

<Checkpoint>[$\bf{B}$的基本结构为：`CC(C1=CC=CC=C1)CC(N)C2=CC=CC=C2`（聚合度为1），含有1个氨基，说法1正确](1)</Checkpoint>

$\bf{B}$中与甲基成键的碳与苯基和亚甲基成键，不与杂原子成键。

<Checkpoint>[$\bf{B}$中与甲基成键的碳与苯基和亚甲基成键，不与杂原子成键。说法2错误](1)</Checkpoint>

$\bf{B}$中的氨基进攻异腈碳原子，产生碳负离子，随后进一步进攻下一分子异腈，得到聚合物$\bf{C}$：基本结构为`CC(CC(N/C([R])=N/[C@H](C(NCCCCCC(OCC1=C(F)C(F)=C(F)C(F)=C1F)=O)=O)C)C2=CC=CC=C2)C3=CC=CC=C3`（两段聚合物聚合度均为1），其中`[*]CC([*])C1=CC=CC=C1`和`[*]/C([*])=N/[C@@H](C)C(NCCCCCC(OCC1=C(C(F)=C(C(F)=C1F)F)F)=O)=O`为重复单元，R表示未知的封端基团。

<MultiModal>![$\bf{C}$：基本结构为`CC(CC(N/C([R])=N/[C@H](C(NCCCCCC(OCC1=C(F)C(F)=C(F)C(F)=C1F)=O)=O)C)C2=CC=CC=C2)C3=CC=CC=C3`（两段聚合物聚合度均为1），其中`[*]CC([*])C1=CC=CC=C1`和`[*]/C([*])=N/[C@@H](C)C(NCCCCCC(OCC1=C(C(F)=C(C(F)=C1F)F)F)=O)=O`为重复单元，R表示未知的封端基团](/media/uploads/7f9e462b-180d-4c7b-9871-bc6298d5e218.png)</MultiModal>

$\bf{C}$中聚异腈嵌段侧链上的酰胺基团可以彼此形成链间氢键，维持螺旋结构且主链上的碳原子均为 $sp^2$ 杂化，具有一定刚性，因此具有很高的翻转能垒。

<Checkpoint>[$\bf{C}$中聚异腈嵌段侧链上的酰胺基团可以彼此形成链间氢键，维持螺旋结构且主链上的碳原子均为 $sp^2$ 杂化，因此具有很高的翻转能垒。说法3正确](1)</Checkpoint>

替换的单体相比原先的单体，将五氟苯基替换为苯基。五氟苯基是一个缺电子的芳环，而聚苯乙烯链段的侧基是富电子的苯基，缺电子的芳环和富电子的芳环之间可以发生强烈的π-π堆积作用，替换后这种作用消失导致折叠无法发生。

<Checkpoint>[五氟苯基是缺电子的芳环，苯基富电子，相互之间可以发生强烈的π-π堆积作用，说法4正确](1)</Checkpoint>

说法1，3，4正确和为8，选I