The block copolymer $\mathbf{X}$ is synthesized through the polymerization of two monomers, $\mathbf{A}$ and $\mathbf{B}$, via the following reaction:  

$\mathbf{A}+\mathbf{B}\xrightarrow[toluene, 90\degree \mathrm{C}]{Pd(PPh_3)_2Cl_2}\mathbf{X}$  

$\mathbf{X}$ exhibits strong intermolecular π-π stacking interactions due to non-covalent interactions between the blocks, which lock the molecular conformation and promote coplanarity of the entire conjugated system. Below are several monomer options, each containing two monomers. Please select the monomer pair most likely to achieve conformational locking:  

Monomer $\mathbf{1}$  
<MultiModal>![BrC1=CC(F)=C2C(N([R])C(/C2=C3C4=C(OC\3=O)C=C5C(OC(/C5=C6C7=C(N([R])C\6=O)C(F)=C(Br)C=C7F)=O)=C4)=O)=C1F](/media/uploads/17d1d317-8549-45d1-9394-af403e8bae94.png)</MultiModal>  

Monomer $\mathbf{2}$  
<MultiModal>![FC1=C(Br)C(F)=C2C(N([R])C(/C2=C3C4=C(OC\3=O)C=C5C(OC(/C5=C6C7=C(N([R])C\6=O)C=C(F)C(Br)=C7F)=O)=C4)=O)=C1](/media/uploads/5dd1f61b-d137-4ecb-8b1e-9a21fa068e21.png)</MultiModal>  

Monomer $\mathbf{3}$  
<MultiModal>![BrC1=C(F)C=C2C(N([R])C(/C2=C3C4=C(OC\3=O)C=C5C(OC(/C5=C6C7=C(N([R])C\6=O)C(F)=C(Br)C(F)=C7)=O)=C4)=O)=C1F  
](/media/uploads/d1adeade-c824-4302-86a3-350111d4dd6b.png)</MultiModal>  

Monomer $\mathbf{4}$  
<MultiModal>![FC1=C(Br)C=C2C(N([R])C(/C2=C3C4=C(OC\3=O)C=C5C(OC(/C5=C6C7=C(N([R])C\6=O)C(F)=C(F)C(Br)=C7)=O)=C4)=O)=C1F](/media/uploads/671a98e5-3ea3-4d8d-9b2e-cd0dac6e1ff4.png)</MultiModal>  

Monomer $\mathbf{5}$  
<MultiModal>![C[Sn](C)(C)C1=CSC(C2=CC([Sn](C)(C)C)=CS2)=C1](/media/uploads/40eabc4f-2046-49cb-8f1d-7c12d6dcad69.png)</MultiModal>  

Monomer $\mathbf{6}$  
<MultiModal>![C[Sn](C)(C1=CC=C(S1)C2=CC=C([Sn](C)(C)C)S2)C](/media/uploads/be67041f-2454-469f-aec6-17473eecaf29.png)</MultiModal>  

Monomer $\mathbf{7}$  
<MultiModal>![C[Sn](C)(C1=COC(C2=CC([Sn](C)(C)C)=CO2)=C1)C](/media/uploads/09762964-44c6-498e-ad90-deecfa50d7bf.png)</MultiModal>  

Monomer $\mathbf{8}$  
<MultiModal>![C[Sn](C)(C1=CC=C(O1)C2=CC=C([Sn](C)(C)C)O2)C](/media/uploads/2fea5918-784b-4ad0-bb4d-c8b648e9b1ae.png)</MultiModal>
嵌段共聚物$\mathbf{X}$由两种单体$\mathbf{A}$和$\mathbf{B}$通过下列反应聚合而成：

$\mathbf{A}+\mathbf{B}\xrightarrow[toluene, 90\degree \mathrm{C}]{Pd(PPh_3)_2Cl_2}\mathbf{X}$

$\mathbf{X}$的分子间具有强的π-π堆积作用，这是由于嵌段之间的非共价相互作用使得分子的构象锁定，整个共轭体系趋向共平面。下列给出了若干种单体，每个选项包含其中两个单体，请选择最可能实现构象锁定的单体组合：

单体$\mathbf{1}$
<MultiModal>![BrC1=CC(F)=C2C(N([R])C(/C2=C3C4=C(OC\3=O)C=C5C(OC(/C5=C6C7=C(N([R])C\6=O)C(F)=C(Br)C=C7F)=O)=C4)=O)=C1F](/media/uploads/17d1d317-8549-45d1-9394-af403e8bae94.png)</MultiModal>

单体$\mathbf{2}$
<MultiModal>![FC1=C(Br)C(F)=C2C(N([R])C(/C2=C3C4=C(OC\3=O)C=C5C(OC(/C5=C6C7=C(N([R])C\6=O)C=C(F)C(Br)=C7F)=O)=C4)=O)=C1](/media/uploads/5dd1f61b-d137-4ecb-8b1e-9a21fa068e21.png)</MultiModal>

单体$\mathbf{3}$
<MultiModal>![BrC1=C(F)C=C2C(N([R])C(/C2=C3C4=C(OC\3=O)C=C5C(OC(/C5=C6C7=C(N([R])C\6=O)C(F)=C(Br)C(F)=C7)=O)=C4)=O)=C1F
](/media/uploads/d1adeade-c824-4302-86a3-350111d4dd6b.png)</MultiModal>

单体$\mathbf{4}$
<MultiModal>![FC1=C(Br)C=C2C(N([R])C(/C2=C3C4=C(OC\3=O)C=C5C(OC(/C5=C6C7=C(N([R])C\6=O)C(F)=C(F)C(Br)=C7)=O)=C4)=O)=C1F](/media/uploads/671a98e5-3ea3-4d8d-9b2e-cd0dac6e1ff4.png)</MultiModal>

单体$\mathbf{5}$
<MultiModal>![C[Sn](C)(C)C1=CSC(C2=CC([Sn](C)(C)C)=CS2)=C1](/media/uploads/40eabc4f-2046-49cb-8f1d-7c12d6dcad69.png)</MultiModal>

单体$\mathbf{6}$
<MultiModal>![C[Sn](C)(C1=CC=C(S1)C2=CC=C([Sn](C)(C)C)S2)C](/media/uploads/be67041f-2454-469f-aec6-17473eecaf29.png)</MultiModal>

单体$\mathbf{7}$
<MultiModal>![C[Sn](C)(C1=COC(C2=CC([Sn](C)(C)C)=CO2)=C1)C](/media/uploads/09762964-44c6-498e-ad90-deecfa50d7bf.png)</MultiModal>

单体$\mathbf{8}$
<MultiModal>![C[Sn](C)(C1=CC=C(O1)C2=CC=C([Sn](C)(C)C)O2)C](/media/uploads/2fea5918-784b-4ad0-bb4d-c8b648e9b1ae.png)</MultiModal>
Monomer $\mathbf{1}$ + Monomer $\mathbf{5}$
Monomer $\mathbf{1}$ + Monomer $\mathbf{6}$
Monomer $\mathbf{1}$ + Monomer $\mathbf{7}$
Monomer $\mathbf{1}$ + Monomer $\mathbf{8}$
Monomer $\mathbf{2}$ + Monomer $\mathbf{5}$
Monomer $\mathbf{2}$ + monomer $\mathbf{6}$
Monomer $\mathbf{2}$ + Monomer $\mathbf{7}$
Monomer $\mathbf{2}$ + Monomer $\mathbf{8}$
Monomer $\mathbf{3}$ + Monomer $\mathbf{5}$
Monomer $\mathbf{3}$ + Monomer $\mathbf{6}$
Monomer $\mathbf{3}$ + Monomer $\mathbf{7}$
Monomer $\mathbf{3}$ + Monomer $\mathbf{8}$
Monomer $\mathbf{4}$ + Monomer $\mathbf{5}$
Monomer $\mathbf{4}$ + Monomer $\mathbf{6}$
Monomer $\mathbf{4}$ + Monomer $\mathbf{7}$
Monomer $\mathbf{4}$ + Monomer $\mathbf{8}$
单体$\mathbf{1}$+单体$\mathbf{5}$
单体$\mathbf{1}$+单体$\mathbf{6}$
单体$\mathbf{1}$+单体$\mathbf{7}$
单体$\mathbf{1}$+单体$\mathbf{8}$
单体$\mathbf{2}$+单体$\mathbf{5}$
单体$\mathbf{2}$+单体$\mathbf{6}$
单体$\mathbf{2}$+单体$\mathbf{7}$
单体$\mathbf{2}$+单体$\mathbf{8}$
单体$\mathbf{3}$+单体$\mathbf{5}$
单体$\mathbf{3}$+单体$\mathbf{6}$
单体$\mathbf{3}$+单体$\mathbf{7}$
单体$\mathbf{3}$+单体$\mathbf{8}$
单体$\mathbf{4}$+单体$\mathbf{5}$
单体$\mathbf{4}$+单体$\mathbf{6}$
单体$\mathbf{4}$+单体$\mathbf{7}$
单体$\mathbf{4}$+单体$\mathbf{8}$
<MultiModal>![FC1=CC(/C(C(N2[R])=O)=C3C4=CC5=C(/C(C(O5)=O)=C6C7=C(N([R])C/6=O)C(F)=C(C(F)=C7)C)C=C4OC/3=O)=C2C(F)=C1C8=CC=C(C9=CC(C)=C[S,Se]9)[S,Se]8 The image shows the block copolymer formed by monomer $\mathbf{3}$ and monomer $\mathbf{6}$, with arrows indicating the hydrogen bonds formed between the F atoms in monomer $\mathbf{3}$ and the H atoms in monomer $\mathbf{6}$, as well as the interactions between the F atoms in monomer $\mathbf{3}$ and the S atoms in monomer $\mathbf{6}$.](/media/uploads/0a23da97-e489-45ea-9aec-e081b773747c.png)</MultiModal>  
The two monomers in the question form a block polymer through the Stille coupling reaction.  

<Checkpoint>[The two monomers form a block polymer through the Stille coupling reaction.](1)</Checkpoint>  

The intramolecular interactions that achieve molecular conformational locking in this question include H-F hydrogen bonds and S-F non-covalent interactions.  

<Checkpoint>[The intramolecular interactions that achieve molecular conformational locking include H-F hydrogen bonds and S-F non-covalent interactions.](1)</Checkpoint>  

Since the electron cloud of sulfur is more diffuse and its electronegativity is lower than that of oxygen, the S-F interaction in the molecule is stronger than the O-F interaction. Therefore, using thiophene-containing monomers results in stronger intramolecular interactions.  

<Checkpoint>[The S-F interaction is stronger than the O-F interaction, so thiophene-containing monomers are used.](1)</Checkpoint>  

By analyzing the coupled molecular structure, it can be observed that only when both F atoms are in the ortho position to Br and the $SnMe_3-$ group is at the 2-position of the thiophene can both H-F hydrogen bonds and S-F non-covalent interactions be simultaneously generated.  

<Checkpoint>[Both F atoms must be in the ortho position to Br, and the $SnMe_3-$ group must be at the 2-position of the thiophene.](2)</Checkpoint>  

At this point, the remaining candidate combinations are monomer $\mathbf{2}$ + monomer $\mathbf{6}$ and monomer $\mathbf{3}$ + monomer $\mathbf{6}$.  
Finally, considering the repulsion between the F atom and the carbonyl group in monomer $\mathbf{2}$, the optimal combination for maintaining the planar molecular conformation is monomer $\mathbf{3}$ + monomer $\mathbf{6}$.  

<Checkpoint>[The optimal combination for maintaining the planar molecular conformation is monomer $\mathbf{3}$ + monomer $\mathbf{6}$.](1)</Checkpoint>
<MultiModal>![FC1=CC(/C(C(N2[R])=O)=C3C4=CC5=C(/C(C(O5)=O)=C6C7=C(N([R])C/6=O)C(F)=C(C(F)=C7)C)C=C4OC/3=O)=C2C(F)=C1C8=CC=C(C9=CC(C)=C[S,Se]9)[S,Se]8，图中展示了单体$\mathbf{3}$和单体$\mathbf{6}$形成的嵌段共聚物，用箭头标识了单体$\mathbf{3}$中的F原子与单体$\mathbf{6}$中的H原子形成的氢键、单体$\mathbf{3}$中的F原子与单体$\mathbf{6}$中的S原子的相互作用。](/media/uploads/0a23da97-e489-45ea-9aec-e081b773747c.png)</MultiModal>
题目中的两种单体通过Stille偶联反应形成嵌段聚合物。

<Checkpoint>[两种单体通过Stille偶联反应形成嵌段聚合物。](1)</Checkpoint>

本题目中实现分子构象锁定的分子内相互作用包括H-F氢键和S-F非共价相互作用。

<Checkpoint>[本题目中实现分子构象锁定的分子内相互作用包括H-F氢键和S-F非共价相互作用。](1)</Checkpoint>

由于硫的电子云比氧更弥散且电负性更小，因此分子中的S-F作用比O-F作用更强，故使用包含噻吩的单体能得到更强的分子内相互作用。

<Checkpoint>[S-F作用比O-F作用更强，故使用包含噻吩的单体](1)</Checkpoint>

通过分析偶联后的分子结构，可以看出只有两个F原子都处于Br的邻位，且$SnMe_3-$处于噻吩的2号位时才能同时产生H-F氢键和S-F非共价相互作用。

<Checkpoint>[两个F原子必须都处于Br的邻位，且$SnMe_3-$处于噻吩的2号位](2)</Checkpoint>

这时，备选项只剩单体$\mathbf{2}$+单体$\mathbf{6}$、单体$\mathbf{3}$+单体$\mathbf{6}$。
最后，考虑单体$\mathbf{2}$中F原子与羰基的相互排斥，得到保持分子平面构象的最佳组合是单体$\mathbf{3}$+单体$\mathbf{6}$。

<Checkpoint>[保持分子平面构象的最佳组合是单体$\mathbf{3}$+单体$\mathbf{6}$。](1)</Checkpoint>