Under a high-purity nitrogen atmosphere, different monomers were added to a benzene solution of alkyl lithium initiator for reaction. After the reaction, a portion of the solution was transferred to an empty bottle, and cold methanol was added to obtain the high-molecular-weight polymer **1** as shown in the figure below. To the remaining solution, a dichloromethane solution of **2** was added, and the reaction was carried out in a water bath at $25°C$ for $1\ h$, followed by quenching with cold methanol to yield an eight-arm star-shaped block copolymer. The simplest formula of **2** is $C_4H_6Si_2O_3$, where all silicon atoms share the same chemical environment.

The structure of **1** is shown below:
<MultiModal>![The image depicts a block copolymer, where one end of the polymer chain terminates with a sec-butyl group and the other end with a hydrogen atom. The middle section consists of three distinct polymer segments. From left to right, the first segment has a degree of polymerization (m), and its repeating unit is [*:x1]CC(c1ccccc1)[*:x2]. The second segment has a degree of polymerization (x), with a repeating unit of [*:x2]CC=C(C)C[*:x3]. The third segment has a degree of polymerization (y), with a repeating unit of [*:x3]CC(C(C)=C)[*:x4]. Here, [*:x1] represents the connection point between the left-end sec-butyl group and the right end of the first segment. [*:x2] indicates that the right end of the first segment connects to the left end of the second segment. [*:x3] denotes that the right end of the second segment connects to the left end of the third segment. [*:x4] signifies that the right end of the third segment connects to the terminal hydrogen atom.](/media/uploads/dbd6bdc7-de79-43c3-baf5-5e677295244d.png)</MultiModal>

The correct options are:
在高纯氮气氛围下，向烷基锂引发剂的苯溶液加入不同的单体反应。反应结束后，将一部分溶液转移至空瓶中，加入冷的甲醇，得到下图所示高分子聚合物**1**。向剩余溶液中加入**2**的二氯甲烷溶液，于$25°C$水浴中反应$1\ h$，然后加入冷的甲醇淬灭，得到八臂星形嵌段共聚物。**2**的最简式为$C_4H_6Si_2O_3$，其中所有硅原子的化学环境相同。

以下为**1**的结构图：
<MultiModal>![如图是一个高分子嵌段共聚物，该高分子的其中一端的末尾是一个仲丁基，另一端的末尾是氢原子。在高分子中间是3个不同的聚合物片段。从左往右的第一个片段的聚合度是m，链节为[*:x1]CC(c1ccccc1)[*:x2]。第二个片段中，聚合度为x，链节为[*:x2]CC=C(C)C[*:x3]。第三个片段中，聚合度为y，链节为[*:x3]CC(C(C)=C)[*:x4]。在这里[*:x1]代表连接左端仲丁基与右端的第一个片段的连接处。[*:x2]表明第一段的右端会与第二段的左端相连。[*:x3]表明第二段的右端会与第三段的左端相连。[*:x4]表明第三段的右端会与末端的氢原子相连。](/media/uploads/dbd6bdc7-de79-43c3-baf5-5e677295244d.png)</MultiModal>

以下选项正确的是：
The initiator used in the synthesis process is n-butyllithium
There are three types of monomers for synthesizing this block copolymer.
The coordination number of the $Si$ atom in compound **2** is $5$.
In compound **2**, $\frac{1}{3}$ of the $O$ atoms are connected to the $C$ atoms.
If only the spatial distribution of the **Si** atoms in compound **2** is considered, its structure can be approximated as a tetrahedron.
If we only consider the spatial distribution of the **Si** atoms in compound **2**, its structure can be approximately regarded as a cube.
Options A through F are all incorrect
Both options A and B are correct, and all other options are wrong.
Options C and F are both correct, while all other options are incorrect.
合成过程所用引发剂是正丁基锂
合成该嵌段共聚物的单体有3种
化合物**2**中$Si$原子的配位数是$5$
化合物**2**中$\frac{1}{3}$的$O$原子与$C$原子相连
如果只看化合物**2**中的$Si$原子的空间分布，则其结构可以近似看作四面体。
如果只看化合物**2**中的$Si$原子的空间分布，则其结构可以近似看作立方体。
选项A到F全部错误
选项A，B均正确，其余选项全错
选项C，F均正确，其余选项全错
The residue in **1** is a sec-butyl group, so its initiator is sec-butyllithium, making option A incorrect.

<Checkpoint>[The residue in **1** is a sec-butyl group, so its initiator is sec-butyllithium](0.5)</Checkpoint>

The chain structure of polymer **1** shows that it consists of a polystyrene segment and a polyisoprene segment. The polyisoprene segment contains both 1,4-addition and 1,2-addition repeating units, but both units are derived from the same monomer (isoprene). Therefore, the synthesis of this copolymer used two monomers: styrene and isoprene. Option B is incorrect.

<Checkpoint>[By cleaving the intermediate connecting part, styrene monomer and isoprene monomer can be obtained](0.5)</Checkpoint>

In common organosilicon compounds and silicates, silicon atoms are typically tetracoordinated, forming a tetrahedral structure. In the stable structure of **2** in this question, tetracoordination is clearly more reasonable.

Since **2** is the precursor of an eight-arm star-shaped polymer, it can be inferred that the central polyhedron of this polymer has $8$ vertices, and thus the arrangement of the central atoms should resemble a cube. The $SiO$ tetrahedral structure is highly stable and can serve as a stable central framework, providing a stable base point for the attached carbon chains. The $C$ and $H$ atoms act as organic side chains crosslinking with the macromolecular chains. Therefore, the vertices of the central cube can only be occupied by $Si$ or $O$. However, the number of $O$ atoms in the simplest formula is $3$, which is not a factor of $8$ (the number of cube vertices). Thus, the vertices can only be $Si$.

<Checkpoint>[The number of $O$ atoms in the simplest formula is $3$, which is not a factor of $8$. The vertices can only be $Si$](1)</Checkpoint>

Therefore, the molecular formula of compound **2** must be an integer multiple of the simplest formula $C_4H_6Si_2O_3$. To form an eight-arm polymer, a central core with $8$ reactive sites ($8$ vertices) is required, composed of $8$ $Si$ atoms. To include $8$ $Si$ atoms in the molecular formula, the simplest formula must be multiplied by $4$, resulting in $8$ $Si$ atoms and a chemical formula of $C_{16}H_{24}Si_8O_{12}$.

<Checkpoint>[There are $8$ $Si$ atoms, and the chemical formula is $C_{16}H_{24}Si_8O_{12}$](1)</Checkpoint>

With $12$ $O$ atoms, comparing the cubic structure, it can be inferred that the oxygen atoms are approximately distributed along the edges. Alternatively, comparing with the structure of quartz, it can also be concluded that the $O$ atoms are positioned between $Si$ atoms, forming $Si-O-Si$ structures.

<Checkpoint>[The oxygen atoms are approximately distributed along the edges, with $O$ positioned between $Si$ atoms, forming $Si-O-Si$ structures.](2)</Checkpoint>

$C_{16}H_{24}$ divided into 8 parts gives $C_{2}H_{3}$ per part, which is a vinyl group. Verification shows that each $Si$ atom is tetracoordinated.

<Checkpoint>[Each $Si$ atom is tetracoordinated.](1)</Checkpoint>

Thus, options C, D, E, G, H, and I are incorrect, and option F is correct.
**1**中的残基为仲丁基，则其引发剂为仲丁基锂，A错误。

<Checkpoint>[**1**中的残基为仲丁基，则其引发剂为仲丁基锂](0.5)</Checkpoint>

聚合物**1**的链结构显示，它由一个聚苯乙烯链段和一个聚异戊二烯链段构成。聚异戊二烯链段中同时包含了1,4-加成和1,2-加成两种重复单元，但这两种单元均来自同一单体（异戊二烯）。因此，合成该共聚物共使用了苯乙烯和异戊二烯两种单体。B选项错误。

<Checkpoint>[中间连接部分切断，可以得到苯乙烯单体和异戊二烯单体](0.5)</Checkpoint>

在常见的有机硅化合物和硅酸盐中，硅原子通常是四配位的，形成四面体结构。在本题的**2**的稳定结构中，四配位显然更合理。

由于**2**为八臂星形结构聚合物的前体，故可以推得该聚合物中心多面体有$8$个顶点，进而推出中心原子排布应该与立方体类似。$SiO$四面体结构十分稳定，其可以用作稳定的中心框架，为上面的碳链提供稳定的基点。$C,H$原子则作为有机侧链与大分子链交联。由此，中心的立方体顶点只能是$Si$或者$O$。而最简式中$O$的数量为$3$，不是$8$(立方体顶点数)的因数。故顶点处只能为$Si$。

<Checkpoint>[最简式中$O$的数量为$3$，不是$8$的因数。顶点处只能为$Si$](1)</Checkpoint>

因此，化合物**2**的分子式必然是最简式 $C_4H_6Si_2O_3$ 的整数倍。由于形成了八臂聚合物，需要一个具有$8$个反应位点($8$个顶点)的中心核，由$8$个$Si$原子构成。要使分子式中含有$8$个$Si$原子，需要将最简式乘以$4$，所以有$8$个$Si$原子，化学式为$C_{16}H_{24}Si_8O_{12}$

<Checkpoint>[有$8$个$Si$原子，化学式为$C_{16}H_{24}Si_8O_{12}$](1)</Checkpoint>

$12$个$O$，对比立方体结构，可知氧原子近似分布在棱上。或者对比石英的结构，也可以得出$O$处于$Si$原子之间，形成$Si-O-Si$的结构。

<Checkpoint>[氧原子近似分布在棱上，$O$处于$Si$原子之间，形成$Si-O-Si$的结构。](2)</Checkpoint>

$C_{16}H_{24}$分成8份，每一份即是$C_{2}H_{3}$，为乙烯基。检查可得每个$Si$原子均为$4$配位。

<Checkpoint>[每个$Si$原子均为$4$配位。](1)</Checkpoint>

所以C,D,E,G,H,I错误，F正确。