Using triphenylmethyl tetrakis(pentafluorophenyl)borate ($\mathrm{Ph_3CB(C_6F_5)_4}$) as an initiator, polyvinyl chloride ($-[\mathrm{CH_2CHCl}]_n-$) can react with triethylsilane in chlorobenzene, where most of the chlorine on the polyvinyl chloride is reduced to hydrogen by triethylsilane, and the remaining chlorine is replaced by chlorobenzene, which acts as a solvent, to form $-\mathrm{C_6H_4Cl}$, eventually generating a polymer product similar to ethylene-chlorostyrene copolymer.

The product was dissolved in deuterated tetrachloroethane for high-temperature NMR hydrogen spectroscopy. The spectrum only shows signals in the following chemical shift ($\delta$) ranges:
- $\delta=0$, single peak, peak area is 0.24
- $\delta=0.5\sim3.5$, multiple peaks, peak area is 10.88
- $\delta=6.0$, single peak, peak area is 0.96
- $\delta=6.8\sim7.5$, multiple peaks, peak area is 1.00

Which of the following statements is correct:
以三苯甲基四(五氟苯基)硼酸盐（$\mathrm{Ph_3CB(C_6F_5)_4}$）为引发剂，聚氯乙烯（$-[\mathrm{CH_2CHCl}]_n-$）在氯苯中可以与三乙基硅烷发生反应，其中聚氯乙烯上大部分的氯被三乙基硅烷还原为氢，剩余的氯被作为溶剂的氯苯取代为$-\mathrm{C_6H_4Cl}$，最终生成一种类似于乙烯-氯苯乙烯共聚物的高分子产物。

将产物溶解于氘代四氯乙烷进行高温核磁共振氢谱测试，谱图只在下面几个化学位移（$\delta$）的区间有信号：
- $\delta=0$，单个峰，峰面积为0.24
- $\delta=0.5\sim3.5$，多个峰，峰面积为10.88
- $\delta=6.0$，单个峰，峰面积为0.96
- $\delta=6.8\sim7.5$，多个峰，峰面积为1.00

下列说法正确的是：
All other options are incorrect
The reaction follows a radical mechanism
The proportion of $-[\mathrm{CH_2CH(C_6H_4Cl)}]-$ units in the product is 8.98%
The proportion of $-[\mathrm{CH_2CH(C_6H_4Cl)}]-$ units in the product is 8.42%
If anisole is used as the solvent, a similar reaction can also occur.
The proportion of $\mathrm{-[CH_2CH_2]-}$ units in the product is 82.8%.
其他选项均不正确
该反应为自由基机理
产物中$-[\mathrm{CH_2CH(C_6H_4Cl)}]-$单元的比例为8.98%
产物中$-[\mathrm{CH_2CH(C_6H_4Cl)}]-$单元的比例为8.42%
如果使用苯甲醚作为溶剂，也能发生类似的反应
产物中$\mathrm{-[CH_2CH_2]-}$单元的比例为82.8%
The $\mathrm{Ph_3C^+}$ in $\mathrm{Ph_3CB(C_6F_5)_4}$ can abstract ${\rm H}$ from triethylsilane to generate $\mathrm{Et_3Si^+}$, which as a Lewis acid combines with the chlorine of $-[\mathrm{CH_2CHCl}]_n-$ on polyvinyl chloride to generate a carbocation. If anisole is used as a solvent, the ${\rm O}$ in it will interfere with the formation of silyl cation and carbocation intermediates, preventing the reaction from proceeding normally. Therefore, options B and E are incorrect.

<Checkpoint>[The reaction is a carbocation process](1)</Checkpoint>

<Checkpoint>[When anisole is used as a solvent, the ${\rm O}$ in it will interfere with the formation of silyl cation and carbocation intermediates](1)</Checkpoint>

The carbocation can be reduced to $\mathrm{-H}$ by $\mathrm{HSiEt_3}$, or it can undergo a Friedel-Crafts reaction with chlorobenzene as a solvent, eventually forming a polymer product similar to an ethylene-chlorostyrene copolymer. The product can be represented as:

$$
\mathrm{-[CH_2CH_2]_x-[CH_2CH(C_6H_4Cl)]_y-}
$$

where $x$ is the number of $\mathrm{-[CH_2CH_2]-}$ units, and $y$ is the number of $\mathrm{-[CH_2CH(C_6H_4Cl)]-}$ units.

<Checkpoint>[The chemical formula of the product can be represented as $ \mathrm{-[CH_2CH_2]_x-[CH_2CH(C_6H_4Cl)]_y-} $](1)</Checkpoint>

The high-temperature nuclear magnetic resonance hydrogen spectrum of the product was tested using deuterated tetrachloroethane as a solvent. The spectrum has signals in four chemical shift ($\delta$) ranges, where:
- The peak around $0.0$ comes from the internal standard substance.
- The peak around $6.0$ is the solvent peak of deuterated tetrachloroethane.
- The peak at $0.5\sim3.5$ comes from ${\rm H}$ connected to the alkyl group. There are $4$ such ${\rm H}$ in each ethylene unit and $3$ in each chlorostyrene unit.
- The peak at $6.8\sim7.5$ comes from ${\rm H}$ connected to the benzene ring. There are $4$ such ${\rm H}$ in each chlorostyrene unit.

<Checkpoint>[The peak around $0.0$ comes from the internal standard substance](1)</Checkpoint>

<Checkpoint>[The peak around $6.0$ is the solvent peak of deuterated tetrachloroethane](1)</Checkpoint>

<Checkpoint>[Only the signals at $0.5\sim3.5$ and $6.8\sim7.5$ in the nuclear magnetic resonance hydrogen spectrum come from the product](1)</Checkpoint>

From the peak area of the nuclear magnetic resonance hydrogen spectrum, we know:
$$
4x+3y=kS_{0.5\sim3.5}
$$
$$
4y=kS_{6.8\sim7.5}
$$
Solving the equations:
$$
\frac{y}{x+y}=\frac{4S_{6.8\sim7.5}}{4S_{0.5\sim3.5}+S_{6.8\sim7.5}}=\frac{4\times1.00}{4\times10.88+1.00}=0.0898
$$
Therefore, the proportion of $-[\mathrm{CH_2CH(C_6H_4Cl)}]-$ units in the product is 8.98%, and the proportion of $-[\mathrm{CH_2CH_2}]-$ units is 91.0%.

<Checkpoint>[The proportion of $-[\mathrm{CH_2CH(C_6H_4Cl)}]-$ units in the product is approximately 8.98%](1)</Checkpoint>

<Checkpoint>[The proportion of $-[\mathrm{CH_2CH_2}]-$ units in the product is 91.0%](1)</Checkpoint>
$\mathrm{Ph_3CB(C_6F_5)_4}$中的$\mathrm{Ph_3C^+}$可以攫取三乙基硅烷中的${\rm H}$，生成$\mathrm{Et_3Si^+}$，其作为Lewis酸与聚氯乙烯上$-[\mathrm{CH_2CHCl}]_n-$的氯结合，生成碳正离子。如果使用苯甲醚作为溶剂，其中的${\rm O}$会干扰硅正离子和碳正离子中间体的形成，使反应无法正常发生。从而选项B,E均错误。

<Checkpoint>[反应为碳正离子过程](1)</Checkpoint>

<Checkpoint>[苯甲醚作为溶剂，其中的${\rm O}$会干扰硅正离子和碳正离子中间体的形成](1)</Checkpoint>

该碳正离子既可以被$\mathrm{HSiEt_3}$还原为$\mathrm{-H}$，也可以与作为溶剂的氯苯发生Friedal-Crafts反应，最终生成一种类似于乙烯-氯苯乙烯共聚物的高分子产物，该产物可以表示为：

$$
\mathrm{-[CH_2CH_2]_x-[CH_2CH(C_6H_4Cl)]_y-}
$$

其中$x$为$\mathrm{-[CH_2CH_2]-}$单元的数量，$y$为$\mathrm{-[CH_2CH(C_6H_4Cl)]-}$单元的数量。

<Checkpoint>[产物的化学式可以表示为$ \mathrm{-[CH_2CH_2]_x-[CH_2CH(C_6H_4Cl)]_y-} $](1)</Checkpoint>

以氘代四氯乙烷为溶剂测试产物的高温核磁共振氢谱，谱图在四个化学位移（$\delta$）的区间有信号，其中：
- $0.0$附近的峰来自于内标物质。
- $6.0$附近的峰是氘代四氯乙烷的溶剂峰。
- $0.5\sim3.5$的峰来自于与烷基相连的${\rm H}$，这样的${\rm H}$在每个乙烯单元中有$4$个，在每个氯苯乙烯单元中有$3$个。
- $6.8\sim7.5$的峰来自于与苯环相连的${\rm H}$，这样的${\rm H}$在每个氯苯乙烯单元中有$4$个。

<Checkpoint>[$0.0$附近的峰来自于内标物质](1)</Checkpoint>

<Checkpoint>[$6.0$附近的峰是氘代四氯乙烷的溶剂峰](1)</Checkpoint>

<Checkpoint>[核磁共振氢谱中只有$0.5\sim3.5$和$6.8\sim7.5$的信号来自产物](1)</Checkpoint>

由核磁共振氢谱的峰面积可知：
$$
4x+3y=kS_{0.5\sim3.5}
$$
$$
4y=kS_{6.8\sim7.5}
$$
解得：
$$
\frac{y}{x+y}=\frac{4S_{6.8\sim7.5}}{4S_{0.5\sim3.5}+S_{6.8\sim7.5}}=\frac{4\times1.00}{4\times10.88+1.00}=0.0898
$$
故产物中$-[\mathrm{CH_2CH(C_6H_4Cl)}]-$单元的比例为8.98%，$-[\mathrm{CH_2CH_2}]-$单元的比例为91.0%。

<Checkpoint>[产物中$-[\mathrm{CH_2CH(C_6H_4Cl)}]-$单元的比例约为8.98%](1)</Checkpoint>

<Checkpoint>[产物中$-[\mathrm{CH_2CH_2}]-$单元的比例为91.0%](1)</Checkpoint>