The following is the reaction process of modifying polyvinyl chloride through a specific chemical reaction:

<MultiModal>![The figure shows the two-step reaction process. First step: $\mathbf{A}$ reacts with $\mathrm{NaN_3}$ to generate $\mathbf{B}$; Second step: $\mathbf{B}$ reacts with $\mathbf{C}$ to generate $\mathbf{D}$. Among them, the structure of $\mathbf{C}$ is given as a polymer, with the repeating unit `*Si(C)(C)O*`, the end group on the left is `CO*`, and the end group on the right is `*CC#C`, where `*` indicates connection to other parts of the polymer](/media/uploads/709e25a1-48b4-40f2-baab-bbbf3d059df6.png)</MultiModal>

Gel permeation chromatography was used to detect the reactant $\mathbf{B}$, the reactant $\mathbf{C}$, the mixture at the end of the reaction, and the purified product $\mathbf{D}$ involved in the above second step reaction, respectively, and the chromatogram shown in the figure below was obtained.

<MultiModal>![This figure is a gel permeation chromatogram, the abscissa range is 10~25, no ordinate, there are 4 curves (represented by solid line/solid square dots/hollow circle dots/star dots respectively, and the corresponding curves are recorded as A/B/C/D respectively), involving 3 peaks, whose abscissas are approximately 17 (peak 1), 20 (peak 2), 23 (peak 3). Curve A has only one high-intensity peak 3, and the intensity is recorded as 1.0; curve B has only one medium-intensity peak 1, and the intensity is 0.5; curve C has a peak 1 with an intensity of 0.3 and a peak 3 with an intensity of 0.45; curve D has only one peak 2 with an intensity of 0.9. The figure also shows that the corresponding $M_n(\mathrm{g/mol})$ of the four curves A/B/C/D are 640, 192400, 3950, 70800, respectively](/media/uploads/4d9e91aa-af87-451a-8f2e-0ce35e851d7a.png)</MultiModal>

Please correspond the curves in the chromatogram to the four different systems, and combine all the information and calculations in the question to judge the following propositions:

1. The mass of $\mathrm{NaN_3}$ added in the first step reaction is at least $17.8\%$ of the mass of the substrate polyvinyl chloride
2. The molar mass of the by-product of the first step reaction is $58.4\mathrm{~g\cdot mol^{-1}}$
3. The second step reaction can be carried out at room temperature
4. The substituents on the newly formed ring in the second step reaction are in adjacent positions
5. Curve D in the gel permeation chromatogram corresponds to the reactant $\mathbf{B}$ of the second step reaction
6. Curve C in the gel permeation chromatogram corresponds to the product $\mathbf{D}$ of the second step reaction
7. The average number of polydimethylsiloxane side chains connected to each $\mathbf{B}$ molecule in product $\mathbf{D}$ is $190$
8. In the second step reaction, the added reactant $\mathbf{C}$ is $30\%$ more than the reactant $\mathbf{B}$

Calculate the value of $z$:

$$
z = \dfrac{\text{Sum of the squares of the correct proposition numbers}}{\text{Square of the sum of the incorrect proposition numbers}}
$$

Select the correct option, requiring the error between the calculated result and the option to be within $1\%$, otherwise select option A: all other options are incorrect.
以下是通过特定化学反应修饰聚氯乙烯的反应过程：

<MultiModal>![图中展现了两步反应的过程。第一步反应：$\mathbf{A}$ 与 $\mathrm{NaN_3}$ 反应生成$\mathbf{B}$；第二步反应：$\mathbf{B}$ 与 $\mathbf{C}$ 反应生成 $\mathbf{D}$。其中，给出了 $\mathbf{C}$ 的结构为一聚合物，重复单元为 `*Si(C)(C)O*，左侧的端基为 `CO*`，右侧的端基为 `*CC#C`，其中 `*` 表示与聚合物其他部分相连](/media/uploads/709e25a1-48b4-40f2-baab-bbbf3d059df6.png)</MultiModal>

使用凝胶渗透色谱法，对于以上第二步反应所涉及的反应物 $\mathbf{B}$、反应物 $\mathbf{C}$、反应结束时的混合物、纯化后的产物 $\mathbf{D}$ 分别进行了检测，得到如下图所示的色谱图。

<MultiModal>![该图是凝胶渗透色谱图，横坐标的范围在 10~25，无纵坐标，其中有 4 条曲线（分别以实线/实心方框点/空心圆点/星形点表示，其对应的曲线分别记为A/B/C/D），涉及 3 个峰，其横坐标分别大约在17（1号峰）、20（2号峰）、23（3号峰）。曲线 A 在只有一个高强度 3 号峰，强度记为1.0；曲线 B 只有一个中等强度 1 号峰，强度为 0.5；曲线 C 在有强度为 0.3 的 1 号峰与强度为 0.45 的 3 号峰；曲线 D 只有一强度为 0.9 的 2 号峰。图中还给出了A/B/C/D 四条曲线所对应的 $M_n(\mathrm{g/mol})$ 分别为 640、192400、3950、70800](/media/uploads/4d9e91aa-af87-451a-8f2e-0ce35e851d7a.png)</MultiModal>

请将色谱图中的曲线与四种不同体系相对应，并结合题目中所有信息与计算，判断以下命题：

1. 第一步反应加入的 $\mathrm{NaN_3}$ 的质量至少是底物聚氯乙烯质量的 $17.8\%$
2. 第一步反应的副产物的摩尔质量为 $58.4\mathrm{~g\cdot mol^{-1}}$
3. 第二步反应在室温下即可进行
4. 第二步反应新生成环上的取代基处于相邻的位置
5. 凝胶渗透色谱图中曲线 D 对应了第二步反应的反应物 $\mathbf{B}$
6. 凝胶渗透色谱图中曲线 C 对应了第二步反应的产物 $\mathbf{D}$
7. 产物 $\mathbf{D}$ 中每个 $\mathbf{B}$ 分子平均连接的聚二甲基硅氧烷侧链数量为 $190$
8. 第二步反应中，加入的反应物 $\mathbf{C}$ 比反应物 $\mathbf{B}$ 过量 $30\%$

计算以下 $z$ 的值：

$$
z = \dfrac{\text{正确命题序号}^2\text{的和}}{\text{错误命题序号的和}^2}
$$

选择正确的选项，要求计算结果与选项的误差在 $1\%$ 以内，否则选择选项 A：其他选项均不正确。
All other options are incorrect
0.123
0.532
0.791
0.877
1.12
1.52
2.51
3.17
4.32
5.70
其他选项均不正确
0.123
0.532
0.791
0.877
1.12
1.52
2.51
3.17
4.32
5.70
First, it is necessary to observe the two-step reaction in Figure 1. The substrate $\mathbf{A}$ of the first step reaction is polyvinyl chloride. The addition of $\mathrm{NaN_3}$ results in an Sn2 substitution reaction, replacing part or all of the $\mathrm{Cl}$ in the polyvinyl chloride with $\mathrm{N_3}$ to obtain polymer $\mathbf{B}$.

<Checkpoint>[$\mathbf{B}$ is the product of the partial or complete substitution of $\mathrm{Cl}$ in polyvinyl chloride $\mathbf{A}$ with $\mathrm{N_3}$](0.5)</Checkpoint>

The byproduct of this step is $\mathrm{NaCl}$, which has a molecular weight of $58.4\mathrm{~g\cdot mol^{-1}}$. Therefore, proposition 2 is correct.

<Checkpoint>[The byproduct of the first step reaction is $\mathrm{NaCl}$ with a molecular weight of $58.4\mathrm{~g\cdot mol^{-1}}$](1)</Checkpoint>

In the second step reaction, propargyl alcohol-terminated polydimethylsiloxane $\mathbf{C}$ is added. This is a typical condition for the CuACC Click reaction. The conditions for the Click reaction are very mild and generally carried out at room temperature. Therefore, proposition 3 is correct.

<Checkpoint>[The second step reaction is a Click reaction, carried out at room temperature](1)</Checkpoint>

The Click reaction involves the 3+2 reaction of an azide-alkyne, and its product is a 1,4-disubstituted triazole, i.e., two substituents (parts from the polyvinyl chloride backbone and the polysiloxane side chain) are located at the 1 and 4 positions of the triazole ring, respectively, separated by two nitrogen atoms or one carbon atom, and are not adjacent. Adjacent positions refer to 1,2 or 2,3, etc.

<Checkpoint>[The second step reaction product is a 1,4-disubstituted triazole, and the substituents are not adjacent](1)</Checkpoint>

Next, it is necessary to correlate the four curves A/B/C/D in gel permeation chromatography with the reactant $\mathbf{B}$, reactant $\mathbf{C}$, the mixture at the end of the reaction, and the purified product $\mathbf{D}$ involved in the second step reaction, respectively. The figure involves 3 peaks, whose abscissas are approximately 17 (peak 1), 20 (peak 2), and 23 (peak 3), respectively. Curve A has only one high-intensity peak 3, with an intensity of 1.0; curve B has only one medium-intensity peak 1, with an intensity of 0.5; curve C has a peak 1 with an intensity of 0.3 and a peak 3 with an intensity of 0.45; curve D has only a peak 2 with an intensity of 0.9.

The table also shows the number-average molecular weights of the species corresponding to each curve:
| Curve | Number-average molecular weight ($M_n/\mathrm{g\cdot mol^{-1}}$) |
| :--: | :----------------------------: |
|  A   |              640               |
|  B   |             192400             |
|  C   |              3950              |
|  D   |             70800              |

Obviously, the final purified product $\mathbf{D}$ of the reaction should have only one peak, so it can only be curve A or B. Observing the number-average molecular weight (A: $640\mathrm{~g\cdot mol^{-1}}$; B: $192400 \mathrm{~g\cdot mol^{-1}}$), the product $\mathbf{D}$ should be B, which has a larger molecular weight.

<Checkpoint>[Curve B is the purified product $\mathbf{D}$](1)</Checkpoint>

Therefore, peak 1 is the product peak, and curve C, which also has peak 1, is the mixture at the end of the second step reaction, rather than the product $\mathbf{D}$. Therefore, proposition 6 is incorrect.

<Checkpoint>[Curve C is the mixture at the end of the second step reaction](1)</Checkpoint>

Therefore, another peak (peak 3) in curve A and curve C is the same excess reactant $\mathbf{B}$ or $\mathbf{C}$, while curve D containing peak 2 corresponds to the non-excess reactant. Comparing their number-average molecular weights (A: $640\mathrm{~g\cdot mol^{-1}}$; B: $192400 \mathrm{~g\cdot mol^{-1}}$; D: $70800\mathrm{~g\cdot mol^{-1}}$), it is not difficult to find that if peak 3 corresponds to $\mathbf{B}$, then peak 2 corresponds to $\mathbf{C}$, then the average number of $\mathbf{C}$ contained in the product is: $ \dfrac{192400-640}{70800}=2.71 $. This is obviously not a reasonable result, the larger molecular weight should be $\mathbf{B}$ instead of the smaller molecular weight.

Therefore, peak 2 can only correspond to $\mathbf{B}$, then peak 3 corresponds to $\mathbf{C}$, curve A is reactant $\mathbf{C}$, and curve D is reactant $\mathbf{B}$. Therefore, proposition 5 is correct.

<Checkpoint>[Curve A is reactant $\mathbf{C}$, and curve D is reactant $\mathbf{B}$](2)</Checkpoint>

Further, to calculate the average number ($n$) of polydimethylsiloxane side chains attached to each $\mathbf{B}$ molecule, it can be calculated by dividing the difference between the number-average molecular weights of the purified product $\mathbf{D}$ (curve B) and the reactant $\mathbf{B}$ (curve D) by the number-average molecular weight of one polysiloxane $\mathbf{C}$ (curve A):

$$
n = \frac{M_{n, \mathbf{D}} - M_{n, \mathbf{B}}}{M_{n, \mathbf{C}}} = \frac{192400 - 70800}{640} = 190
$$

Therefore, proposition 7 is correct.

<Checkpoint>[The average number of polydimethylsiloxane side chains attached to each $\mathbf{B}$ molecule in product $\mathbf{D}$ is $190$](1)</Checkpoint>

Next, it is necessary to calculate the excess value of reactant $\mathbf{C}$ added in the second step reaction compared to reactant $\mathbf{B}$. In the mixture at the end of the reaction (curve C), its number-average molecular weight is contributed by the generated P2 product and the unreacted excess side chains. Assume that the mixture contains 1 P2 molecule and x unreacted side chain molecules. Then the average molar mass of the mixture is:

$$
M_{n, \text{mixture}} = \frac{(1 \cdot M_{n, \mathbf{D}}) + (x \cdot M_{n,\mathbf{C}})}{1+x}
$$

Substitute the values for calculation:
$$
3950 = \frac{(1 \cdot 192400) + (x \cdot 640)}{1+x}
$$

Solving the equation gives $x \approx 56.9$. This means that for every 190 successfully reacted side chains (attached to $\mathbf{B}$), there are 56.9 unreacted side chains. Therefore, the percentage excess of propargyl-polydimethylsiloxane-monomethyl ether added is:

$$
\text{Percentage excess} = \frac{\text{Moles of unreacted side chains}}{\text{Moles of reacted side chains}} \times 100\% = \frac{56.9}{190} \times 100\% \approx 30\%
$$

Therefore, proposition 8 is correct.

<Checkpoint>[In the second step reaction, the added reactant $\mathbf{C}$ is $30\%$ in excess of reactant $\mathbf{B}$](1)</Checkpoint>

Finally, the percentage of $\mathrm{NaN_3}$ added in the first step reaction relative to the mass of the substrate polyvinyl chloride $\mathbf{A}$ can be calculated. Combining the previous calculation results, it is known that one $\mathbf{B}$ molecule contains at least 190 azides, therefore, the number-average molecular weight of $\mathbf{A}$ is:

$$
M_{n,\mathbf{A}}=M_{n,\mathbf{B}}-190\times(M_{\mathrm{N_3}}-M_{\mathrm{Cl}})=69549.8\mathrm{~g\cdot mol^{-1}}
$$

Therefore, the percentage of $\mathrm{NaN_3}$ relative to the mass of the substrate polyvinyl chloride $\mathbf{A}$ is:

$$
\dfrac{190M_{\mathrm{NaN_3}}}{M_{n,\mathbf{A}}}=17.8\%
$$

Therefore, proposition 1 is correct.

<Checkpoint>[The mass of $\mathrm{NaN_3}$ added in the first step reaction is at least $17.8\%$ of the mass of the substrate polyvinyl chloride](1)</Checkpoint>

Finally, the correct propositions are 1, 2, 3, 5, 7, and 8, and the value of $z$ is:

$$
z= \dfrac{(1^2+2^2+3^2+5^2+7^2+8^2)}{(4+6)^2}=1.52
$$

<Checkpoint>[$z$ has a value of 1.52](1)</Checkpoint>
首先，需要观察图 1 中的两步反应，第一步反应的底物 $\mathbf{A}$ 是聚氯乙烯，加入 $\mathrm{NaN_3} $ 发生 Sn2 取代反应，将聚氯乙烯中的 $\mathrm{Cl}$ 部分或全部取代为 $\mathrm{N_3}$，得到聚合物 $\mathbf{B}$。

<Checkpoint>[$\mathbf{B}$ 为聚氯乙烯 $\mathbf{A}$ 中的 $\mathrm{Cl}$ 部分或全部取代为 $\mathrm{N_3}$ 的产物](0.5)</Checkpoint>

这一步反应的副产物是 $\mathrm{NaCl}$，其分子量为 $58.4\mathrm{~g\cdot mol^{-1}}$。因此，命题 2 正确。

<Checkpoint>[第一步反应的副产物是 $\mathrm{NaCl}$，分子量为 $58.4\mathrm{~g\cdot mol^{-1}}$](1)</Checkpoint>

第二步反应加入了炔丙基醇封端的聚二甲基硅氧烷 $\mathbf{C}$，这是典型的 CuACC Click 反应的条件，Click 反应的条件非常温和，一般均在室温下进行。因此，命题 3 正确。

<Checkpoint>[第二步反应是 Click 反应，在室温下进行](1)</Checkpoint>

Click 反应发生了叠氮炔烃的 3+2 反应，其产物是 1,4-二取代三唑，即两个取代基（来自聚氯乙烯主链和聚硅氧烷侧链的部分）分别位于三唑环的1号位和4号位，它们之间隔着两个氮原子或一个碳原子，不相邻。相邻位置是指1,2或2,3等。

<Checkpoint>[第二步反应产物为 1,4-二取代三唑，取代基并不相邻](1)</Checkpoint>

接下来，需要将凝胶渗透色谱中的 A/B/C/D 四条曲线与第二步反应所涉及的反应物 $\mathbf{B}$、反应物 $\mathbf{C}$、反应结束时的混合物、纯化后的产物 $\mathbf{D}$ 分别进行对应。图中涉及 3 个峰，其横坐标分别大约在17（1号峰）、20（2号峰）、23（3号峰）。曲线 A 在只有一个高强度 3 号峰，强度记为1.0；曲线 B 只有一个中等强度 1 号峰，强度为 0.5；曲线 C 在有强度为 0.3 的 1 号峰与强度为 0.45 的 3 号峰；曲线 D 只有一强度为 0.9 的 2 号峰。

图表中也给出来各个曲线所对应物种的数均分子量：
| 曲线 | 数均分子量 ($M_n/\mathrm{g\cdot mol^{-1}}$) |
| :--: | :----------------------------: |
|  A   |              640               |
|  B   |             192400             |
|  C   |              3950              |
|  D   |             70800              |

显然，该反应最终纯化后的产物 $\mathbf{D}$ 应当只有一个峰，因此只有可能是曲线 A 或 B，观察数均分子量可知（A：$640\mathrm{~g\cdot mol^{-1}}$；B：$192400 \mathrm{~g\cdot mol^{-1}}$），故产物 $\mathbf{D}$ 应该为分子量较大的 B。

<Checkpoint>[曲线 B 为纯化后的产物 $\mathbf{D}$](1)</Checkpoint>

故 1 号峰为产物峰，同样有 1 号峰的曲线 C 为第二步反应结束时的混合物，而非产物 $\mathbf{D}$。因此，命题 6 错误。

<Checkpoint>[曲线 C 为第二步反应结束时的混合物](1)</Checkpoint>

因此，曲线 A 与曲线 C 中另一个峰（3 号峰）为同一个过量的反应物 $\mathbf{B}$ 或 $\mathbf{C}$，而含有 2 号峰的曲线 D 对应不过量的反应物，比较其数均分子量（A：$640\mathrm{~g\cdot mol^{-1}}$；B：$192400 \mathrm{~g\cdot mol^{-1}}$；D：$70800\mathrm{~g\cdot mol^{-1}}$）不难发现，假设 3 号峰对应 $\mathbf{B}$，则 2 号峰对应 $\mathbf{C}$，则产物中含有的 $\mathbf{C}$ 的平均数量为：$ \dfrac{192400-640}{70800}=2.71 $ 这显然不是合理的结果，分子量更大的应该是 $\mathbf{B}$ 而非分子量小的。

因此，只能是 2 号峰对应 $\mathbf{B}$，则 3 号峰对应 $\mathbf{C}$，曲线 A 为反应物 $\mathbf{C}$，曲线 D 为反应物 $\mathbf{B}$。因此，命题 5 正确。

<Checkpoint>[曲线 A 为反应物 $\mathbf{C}$，曲线 D 为反应物 $\mathbf{B}$](2)</Checkpoint>

进一步的，要计算每个 $\mathbf{B}$ 分子上连接的聚二甲基硅氧烷侧链的平均数量 ($n$) ，可以通过纯化的产物 $\mathbf{D}$（曲线B）和反应物 $\mathbf{B}$（曲线D）的数均分子量差，除以一个聚硅氧烷 $\mathbf{C}$（曲线A）的数均分子量来计算：

$$
n = \frac{M_{n, \mathbf{D}} - M_{n, \mathbf{B}}}{M_{n, \mathbf{C}}} = \frac{192400 - 70800}{640} = 190
$$

因此，命题 7 正确。

<Checkpoint>[产物 $\mathbf{D}$ 中每个 $\mathbf{B}$ 分子平均连接的聚二甲基硅氧烷侧链数量为 $190$](1)</Checkpoint>

接下来，要计算第二步反应中，加入的反应物 $\mathbf{C}$ 比反应物 $\mathbf{B}$ 过量值。在反应结束时的混合物（曲线C）中，其数均分子量由生成的P2产物和未反应的过量侧链贡献。假设混合物中含有1个P2分子和 x 个未反应的侧链分子。那么混合物的平均摩尔质量为：

$$
M_{n, \text{混合物}} = \frac{(1 \cdot M_{n, \mathbf{D}}) + (x \cdot M_{n,\mathbf{C}})}{1+x}
$$

代入数值进行计算：
$$
3950 = \frac{(1 \cdot 192400) + (x \cdot 640)}{1+x}
$$

解方程得到 $x \approx 56.9$。这意味着每190个成功反应的侧链（连接在 $\mathbf{B}$ 上），就有 56.9 个侧链未反应。因此，所添加的炔丙基-聚二甲基硅氧烷-单甲醚的过量百分比为：

$$
\text{过量百分比} = \frac{\text{未反应的侧链摩尔数}}{\text{已反应的侧链摩尔数}} \times 100\% = \frac{56.9}{190} \times 100\% \approx 30\%
$$

因此，命题 8 正确。

<Checkpoint>[第二步反应中，加入的反应物 $\mathbf{C}$ 比反应物 $\mathbf{B}$ 过量 $30\%$](1)</Checkpoint>

最终，可以计算第一步反应加入的 $\mathrm{NaN_3}$ 的占底物聚氯乙烯 $\mathbf{A}$ 质量的百分比。结合之前的计算结果，可知一个 $\mathbf{B}$ 分子含有至少 190 个叠氮，因此， $\mathbf{A}$ 的数均分子量为：

$$
M_{n,\mathbf{A}}=M_{n,\mathbf{B}}-190\times(M_{\mathrm{N_3}}-M_{\mathrm{Cl}})=69549.8\mathrm{~g\cdot mol^{-1}}
$$

因此，$\mathrm{NaN_3}$ 的占底物聚氯乙烯 $\mathbf{A}$ 质量的百分比为：

$$
\dfrac{190M_{\mathrm{NaN_3}}}{M_{n,\mathbf{A}}}=17.8\%
$$

因此，命题 1 正确。

<Checkpoint>[第一步反应加入的 $\mathrm{NaN_3}$ 的质量至少是底物聚氯乙烯质量的 $17.8\%$](1)</Checkpoint>

最终，正确的命题有1、2、3、5、7、8，$z$ 的值为：

$$
z= \dfrac{(1^2+2^2+3^2+5^2+7^2+8^2)}{(4+6)^2}=1.52
$$

<Checkpoint>[$z$ 的值为 1.52](1)</Checkpoint>