Alkyl iodides ($RCF_2I$) can undergo polymerization with 1,1-difluoroethylene, as shown in the figure below:

In this process, alkyl iodides act as free radical initiating species, undergoing homolytic cleavage of the C-I bond under light conditions to generate the carbon radical $\bf{X}$. $\bf{X}$ initiates the polymerization of 1,1-difluoroethylene, resulting in the polymeric compound $\bf{Y}$.

<MultiModal>![FC(F)(I)[R] undergoes polymerization with C=C(F)F under light conditions, first producing the radical $\bf{X}$: F[C](F)[R], and then obtaining the polymer $\bf{Y}$: The basic structure is `FC(F)(CC(F)(F)I)[R]` (when there is only one repeating unit), where `[*]CC(F)(F)[*]` is the repeating unit](/media/uploads/054ca48f-5188-42ef-8a73-6375bba8945b.jpg)</MultiModal>

PPFR (structure `FC(C(F)(F)F)(C(F)(F)F)[C](C(F)(C(F)(F)F)F)C(F)(C(F)(F)F)C(F)(F)F`, with the central carbon atom being a radical) is a reagent that can generate trifluoromethyl radicals upon heating. The free radical polymerization shown in the figure below yields two polymers, $\bf{Z1}$ and $\bf{Z2}$, in the presence of a catalytic amount of PPFR. It is known that $\bf{Z1}$ and $\bf{Z2}$ have different end groups and the number average molecular weight of $\bf{Z1}$ is approximately twice that of $\bf{Z2}$.

<MultiModal>![1,1-Difluoroethylene and PPFR and `IC(F)(F)C(F)(F)C(F)(F)C(F)(F)I` undergo polymerization at temperatures above 60 degrees Celsius to produce two polymers $\bf{Z1}$ and $\bf{Z2}$](/media/uploads/40b64c5c-22b6-442c-be76-5cd6e38fc65b.png)</MultiModal>

The following statements are made:

1.  In the chain transfer step of the polymer $\bf{Y}$ reaction, the polymer chain can abstract an iodine atom from the alkyl iodide to generate a new $\bf{X}$.

2.  $\bf{Z1}$ has the same end groups on both ends.

3.  $\bf{Z1}$ may have a center of symmetry.

4.  $\bf{Z2}$ does not contain iodine.

Select the sum of the coefficients of all correct statements:
烷基碘化物（$RCF_2I$）可以与1,1-二氟乙烯发生聚合反应，如下图所示：

在该过程中，烷基碘化物作为自由基引发物种，在光照条件下发生C-I键的均裂得到碳自由基$\bf{X}$，$\bf{X}$引发1,1-二氟乙烯的聚合，得到高分子化合物$\bf{Y}$

<MultiModal>![FC(F)(I)[R]在光照条件下与C=C(F)F发生聚合反应，首先在光照下产生自由基$\bf{X}$：F[C](F)[R]，随后得到聚合物$\bf{Y}$：基本结构为`FC(F)(CC(F)(F)I)[R]`（当只有一个重复单元时），其中`[*]CC(F)(F)[*]`为重复单元](/media/uploads/054ca48f-5188-42ef-8a73-6375bba8945b.jpg)</MultiModal>

PPFR（结构为`FC(C(F)(F)F)(C(F)(F)F)[C](C(F)(C(F)(F)F)F)C(F)(C(F)(F)F)C(F)(F)F`，中心碳原子为自由基）是一种加热能产生三氟甲基自由基的试剂。下图所示的自由基聚合在催化量PPFR作用下得到了两种聚合物$\bf{Z1}$和$\bf{Z2}$。已知$\bf{Z1}$和$\bf{Z2}$端基有所不同且$\bf{Z1}$数均分子量约为$\bf{Z2}$的2倍。

<MultiModal>![1,1-二氟乙烯和PPFR和`IC(F)(F)C(F)(F)C(F)(F)C(F)(F)I`在温度高于60摄氏度时产生聚合，得到两种聚合物$\bf{Z1}$和$\bf{Z2}$](/media/uploads/40b64c5c-22b6-442c-be76-5cd6e38fc65b.png)</MultiModal>

有以下说法：

1.聚合物$\bf{Y}$反应的链转移步骤中，聚合物链可以攫取烷基碘化物的碘原子产生新的$\bf{X}$

2.$\bf{Z1}$两端封端基团相同

3.$\bf{Z1}$可能具有对称中心

4.$\bf{Z2}$中不含碘

选出所有正确说法系数的和：
No correct statement
1
2
3
4
5
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9
10
无正确说法
1
2
3
4
5
6
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9
10
Under illumination, the C-I bond of alkyl iodides undergoes homolytic cleavage, generating initiating radicals $RCF_2·$ (i.e., $\bf{X}$) and an iodine radical. $\bf{X}$ attacks 1,1-difluoroethylene to initiate polymerization. Subsequently, chain transfer can occur, where the polymer radical attacks $RCF_2I$ to abstract an iodine atom, generating new $\bf{X}$; or the polymer radical reacts with a molecule of iodine radical, terminating the reaction.

<Checkpoint>[The polymer radical can abstract the iodine atom of alkyl iodide to generate new $\bf{X}$, statement 1 is correct](1)</Checkpoint>

In the reactions producing $\bf{Z1}$ and $\bf{Z2}$, because the free radical polymerization has a narrow molecular weight distribution, and the number-average molecular weight of $\bf{Z1}$ is approximately twice that of $\bf{Z2}$, then $\bf{Z1}$ likely has two polymer segments. First, PPFR is heated to generate trifluoromethyl radicals, which then attack 1,1-difluoroethylene, generating a new radical `F[C](F)CC(F)(F)F`, initiating polymerization until the polymer radical abstracts an iodine atom from `IC(F)(F)C(F)(F)C(F)(F)C(F)(F)I`, generating a new radical `IC(F)(C(F)(C(F)([C](F)F)F)F)F`. At this point, the polymerization of the original polymer terminates, resulting in a basic polymer structure of `FC(F)(I)CC(F)(F)F` (when there is only one repeating unit), where the repeating unit is: `FC(F)([*])C[*]`, which should be $\bf{Z2}$, containing an iodine atom.

<MultiModal>![The figure shows the structure of $\bf{Z2}$, the basic structure is `FC(F)(I)CC(F)(F)F` (when there is only one repeating unit), the repeating unit is: `FC(F)([*])C[*]`](/media/uploads/535ede81-6021-4240-a839-386abadd0c16.png)</MultiModal>

<Checkpoint>[$\bf{Z2}$ needs to abstract an iodine atom during chain termination, therefore it contains iodine, statement 4 is incorrect](1)</Checkpoint>

The newly generated radical then reacts with 1,1-difluoroethylene to generate the radical `IC(F)(C(F)(C(F)(C(F)(C[C](F)F)F)F)F)F`, initiating polymerization. Polymerization terminates when another iodine atom is abstracted from `IC(F)(F)C(F)(F)C(F)(F)C(F)(F)I`. Since `IC(F)(F)C(F)(F)C(F)(F)C(F)(F)I` has two iodine atoms, it can be abstracted twice to initiate two segments of polymerization. Then the final basic structure of the polymer is `FC(F)(C(F)(C(F)(C(F)(CC(F)(I)F)F)F)F)CC(F)(F)I` (when there is only one repeating unit), with two repeating units `FC(F)([*])C[*]`. Because it has two polymerization segments, the number-average molecular weight is approximately twice that of $\bf{Z2}$, which is $\bf{Z1}$, and the structure is shown in the figure below:

<MultiModal>![The basic structure of $\bf{Z1}$ is `FC(F)(C(F)(C(F)(C(F)(CC(F)(I)F)F)F)F)CC(F)(F)I` (when there is only one repeating unit), with two repeating units `FC(F)([*])C[*]`](/media/uploads/6427544e-0c19-4471-a83c-ba5e102301db.png)</MultiModal>

Since the end groups of $\bf{Z1}$ are all iodine atoms, the end groups are the same.

<Checkpoint>[The end groups of $\bf{Z1}$ are all iodine atoms, the end groups are the same, statement 2 is correct](1)</Checkpoint>

The center of $\bf{Z1}$ is `FC(C(F)(C(F)(C(F)([*])F)F)F)([*])F` (the asterisk part connects to other parts of the polymer), which has a center of symmetry. If the degrees of polymerization of the two polymer segments are the same, then $\bf{Z1}$ can have a center of symmetry.

<Checkpoint>[If the degrees of polymerization of the two polymer segments are the same, then $\bf{Z1}$ can have a center of symmetry, statement 3 is correct](1)</Checkpoint>

The correct statements are 1, 2, and 3, the sum is 6, choose G.

In this question, why is the species that initiates polymerization `F[C](F)CC([R])(F)F` (the radical is the carbon atom of the original difluoroethylene connected to two fluorine atoms, R represents other groups) instead of `FC(F)(C(F)(F)[R])[C]([H])[H]` where the radical is the carbon atom of the original difluoroethylene connected to two hydrogen atoms, R represents other groups: These two radicals do not have particularly large differences in stability, but when a carbon atom is connected to two fluorine atoms, the C-I bond is weakened. Therefore, if it is the `FC(F)(C(F)(F)[R])[C]([H])[H]` radical, it can abstract an iodine atom to generate a relatively stable C-I bond instead of further generating new radicals, which will terminate the polymerization. Only the C-I bond generated by the `F[C](F)CC([R])(F)F` radical is relatively weak, which can allow the polymerization to continue by further generating new radicals.

<Checkpoint>[In this question, the species that initiates polymerization is the `F[C](F)CC([R])(F)F` radical](1)</Checkpoint>
在光照下，烷基碘化物的C-I 键均裂，产生引发自由基$RCF_2·$（即$\bf{X}$） 和一个碘自由基，$\bf{X}$进攻1,1-二氟乙烯引发聚合反应。随后可发生链转移，聚合物自由基进攻$RCF_2I$攫取碘原子，产生新的$\bf{X}$；或聚合物自由基与一分子碘自由基反应，终止反应。

<Checkpoint>[聚合物自由基可以攫取烷基碘化物的碘原子产生新的$\bf{X}$，说法1正确](1)</Checkpoint>

在产生$\bf{Z1}$和$\bf{Z2}$的反应中，由于自由基聚合分子量分布较窄，而$\bf{Z1}$数均分子量约为$\bf{Z2}$的2倍，则$\bf{Z1}$中很可能具有两段聚合物结构。首先PPFR加热产生三氟甲基自由基，随后自由基进攻1,1-二氟乙烯，产生新的自由基`F[C](F)CC(F)(F)F`，引发聚合直到聚合物自由基攫取`IC(F)(F)C(F)(F)C(F)(F)C(F)(F)I`的一个碘原子，产生新的自由基`IC(F)(C(F)(C(F)([C](F)F)F)F)F`，此时原先聚合物的聚合终止，得到聚合物的基本结构为`FC(F)(I)CC(F)(F)F`（当只有一个重复单元时），其中重复单元为：`FC(F)([*])C[*]`，应为$\bf{Z2}$，含有碘原子。

<MultiModal>![图中展示出$\bf{Z2}$的结构，基本结构为`FC(F)(I)CC(F)(F)F`（当只有一个重复单元时），重复单元为：`FC(F)([*])C[*]`](/media/uploads/535ede81-6021-4240-a839-386abadd0c16.png)</MultiModal>

<Checkpoint>[$\bf{Z2}$链终止时需要攫取碘原子，因此其中含有碘，说法4错误](1)</Checkpoint>

产生的新自由基随后与1,1-二氟乙烯反应产生自由基`IC(F)(C(F)(C(F)(C(F)(C[C](F)F)F)F)F)F`引发聚合。聚合终止时需要攫取另一`IC(F)(F)C(F)(F)C(F)(F)C(F)(F)I`的一个碘原子。而`IC(F)(F)C(F)(F)C(F)(F)C(F)(F)I`中具有两个碘原子，可被攫取两次引发两段聚合，则最终聚合物的基本结构为`FC(F)(C(F)(C(F)(C(F)(CC(F)(I)F)F)F)F)CC(F)(F)I`（当只有一个重复单元时），具有两段重复单元`FC(F)([*])C[*]`，由于具有两段聚合结构，数均分子量约为$\bf{Z2}$的2倍，是为$\bf{Z1}$，结构如下图所示：

<MultiModal>![$\bf{Z1}$基本结构为`FC(F)(C(F)(C(F)(C(F)(CC(F)(I)F)F)F)F)CC(F)(F)I`（当只有一个重复单元时），具有两段重复单元`FC(F)([*])C[*]`](/media/uploads/6427544e-0c19-4471-a83c-ba5e102301db.png)</MultiModal>

由于$\bf{Z1}$封端均为碘原子，封端基团相同。

<Checkpoint>[$\bf{Z1}$封端均为碘原子，封端基团相同，说法2正确](1)</Checkpoint>

$\bf{Z1}$的中心为`FC(C(F)(C(F)(C(F)([*])F)F)F)([*])F`（星号部分连接聚合物其他部分），具有对称中心，若两段聚合物部分聚合度相同，则$\bf{Z1}$可以具有对称中心。

<Checkpoint>[若两段聚合物部分聚合度相同，则$\bf{Z1}$可以具有对称中心，说法3正确](1)</Checkpoint>

正确的说法为1,2,3，和为6，选G

本题中为何引发聚合的物种为`F[C](F)CC([R])(F)F`（自由基为原二氟乙烯连接两个氟原子的碳，R表示其他基团）而非`FC(F)(C(F)(F)[R])[C]([H])[H]`自由基为原二氟乙烯连接两个氢原子的碳，R表示其他基团）：这两个自由基在稳定性上不具有特别大的区别，但是当碳原子上连接有两个氟原子时C-I键会被削弱。因此若是`FC(F)(C(F)(F)[R])[C]([H])[H]`自由基，可以攫取碘原子后产生相对稳定的C-I键而不是进一步产生新的自由基，会终止聚合。只有`F[C](F)CC([R])(F)F`自由基产生的C-I键相对较弱，可以通过进一步产生新的自由基使聚合进行下去。

<Checkpoint>[本题中引发聚合的物种为`F[C](F)CC([R])(F)F`自由基](1)</Checkpoint>