In carbon tetrachloride solution,

<MultiModal>![C1(C2(C3=CC=CC=C3)C=CC(C4=CC=CC=C4)(C5=CC=CC=C5)C=C2)=CC=CC=C1](/media/uploads/df5b7c0f-dcb4-49c4-a174-1e06fe78d0b4.png)</MultiModal>

reacts with $\mathrm{Br_2}$ at room temperature for 26h, yielding compound $\mathbf { A } ( \mathrm {C_{30}H_{22}} )$ with a yield of $98 \%$.
If the reactant is changed to

<MultiModal>![COC1=CC=C(C2(C3=CC=C(OC)C=C3)C=CC(C4=CC=C(OC)C=C4)(C5=CC=C(OC)C=C5)C=C2)C=C1](/media/uploads/37ec76a9-7fa2-4811-a6bf-645c160c0eaa.png)</MultiModal>

, the resulting product is $\mathbf{B}(\mathrm {C_{34}H_{26}Br_4O_4})$.

There are the following statements about $\mathbf{A,B}$ and the reaction process:

1. The point groups to which the $\mathbf{A,B}$ molecules belong are the same when in the conformation with the highest symmetry.

2. If $\mathrm{Br_2}$ is replaced with $\mathrm{I_2}$, the point group to which the resulting product $\mathbf{B'}$ belongs in the conformation with the highest symmetry may be different from $\mathbf{B}$.

3. If the reaction time is further extended, the yields of $\mathbf{A,B}$ will remain unchanged.

4. The maximum number of rings in each intermediate that generates $\mathbf{A}$ is 5.

Then, the following options give all the correct options:
在四氯化碳溶液中，

<MultiModal>![C1(C2(C3=CC=CC=C3)C=CC(C4=CC=CC=C4)(C5=CC=CC=C5)C=C2)=CC=CC=C1](/media/uploads/df5b7c0f-dcb4-49c4-a174-1e06fe78d0b4.png)</MultiModal>

与 $\mathrm{Br_2}$ 于室温反应26h，以 $98 \%$ 的产率得到化合物 $\mathbf { A } ( \mathrm {C_{30}H_{22}} )$ 。
若将反应物换为

<MultiModal>![COC1=CC=C(C2(C3=CC=C(OC)C=C3)C=CC(C4=CC=C(OC)C=C4)(C5=CC=C(OC)C=C5)C=C2)C=C1](/media/uploads/37ec76a9-7fa2-4811-a6bf-645c160c0eaa.png)</MultiModal>

，所得产物为 $\mathbf{B}(\mathrm {C_{34}H_{26}Br_4O_4})$ 。  

有下列关于 $\mathbf{A,B}$以及反应过程的说法：

1，$\mathbf{A,B}$分子在对称性最高的构象时所属的点群相同。

2，若将$\mathrm{Br_2}$替换为$\mathrm{I_2}$，所得到的产物$\mathbf{B'}$在对称性最高的构象时所属的点群可能与$\mathbf{B}$不同。

3，若继续延长反应时间，$\mathbf{A,B}$的产率均会保持不变。

4，生成$\mathbf{A}$的各个中间体中环数最多为5。

则下列选项中给出了全部正确选项的是：
All other options are incorrect
1
2
3
4
1，2
1，3
1，4
2，3
2，4
3，4
1，2，3
1，2，4
1，3，4
2，3，4
1，2，3，4
其他选项均不正确
1
2
3
4
1，2
1，3
1，4
2，3
2，4
3，4
1，2，3
1，2，4
1，3，4
2，3，4
1，2，3，4
<MultiModal>![C1(C2(C3=CC=CC=C3)C=CC(C4=CC=CC=C4)(C5=CC=CC=C5)C=C2)=CC=CC=C1](/media/uploads/df5b7c0f-dcb4-49c4-a174-1e06fe78d0b4.png)</MultiModal>

Reacts with $\mathrm{Br_2}$ to initially generate a bromonium ion intermediate:

<MultiModal>![C12C(C3=CC=CC=C3)(C4=CC=CC=C4)C=CC(C5=CC=CC=C5)(C6=CC=CC=C6)C1[Br+]2](/media/uploads/03430afc-a1df-40f2-b562-e4fcc7150b3a.png)</MultiModal>

Subsequently, phenyl migration occurs, opening the three-membered ring to form a carbocation intermediate:

<MultiModal>![BrC1C(C2=CC=CC=C2)(C3=CC=CC=C3)C=C[C+](C4=CC=CC=C4)C1C5=CC=CC=C5](/media/uploads/cd19bb14-f5ec-4cf1-b10e-18cbf44fc509.png)</MultiModal>

Then, another phenyl migration occurs, forming another carbocation intermediate:

<MultiModal>![BrC1[C+](C2=CC=CC=C2)C(C3=CC=CC=C3)C=C(C4=CC=CC=C4)C1C5=CC=CC=C5](/media/uploads/441bbb1b-1aed-4e51-81ce-b511d756a957.png)</MultiModal>

At this point, there are no groups available for migration, so the carbocation loses a proton to form a double bond:

<MultiModal>![BrC1C(C2=CC=CC=C2)=C(C3=CC=CC=C3)C=C(C4=CC=CC=C4)C1C5=CC=CC=C5](/media/uploads/a3fb5b3c-de32-4cc5-aa9a-245785974e11.png)</MultiModal>

Finally, an elimination reaction occurs, forming a stable aromatic ring product:

<MultiModal>![C1(C2=CC=CC=C2)=C(C3=CC=CC=C3)C=C(C4=CC=CC=C4)C(C5=CC=CC=C5)=C1](/media/uploads/6457d8c0-7122-4818-a900-4ce83e592b9b.png)</MultiModal>

<Checkpoint>[Important intermediate 1 for the reaction to generate $\mathbf { A }$: C12C(C3=CC=CC=C3)(C4=CC=CC=C4)C=CC(C5=CC=CC=C5)(C6=CC=CC=C6)C1[Br+]2](1)</Checkpoint>

<Checkpoint>[Important intermediate 2 for the reaction to generate $\mathbf { A }$: BrC1C(C2=CC=CC=C2)(C3=CC=CC=C3)C=C[C+](C4=CC=CC=C4)C1C5=CC=CC=C5](1)</Checkpoint>

<Checkpoint>[Important intermediate 3 for the reaction to generate $\mathbf { A }$: BrC1[C+](C2=CC=CC=C2)C(C3=CC=CC=C3)C=C(C4=CC=CC=C4)C1C5=CC=CC=C5](1)</Checkpoint>

<Checkpoint>[Important intermediate 4 for the reaction to generate $\mathbf { A }$: BrC1C(C2=CC=CC=C2)=C(C3=CC=CC=C3)C=C(C4=CC=CC=C4)C1C5=CC=CC=C5](1)</Checkpoint>

<Checkpoint>[$\mathbf { A }$ is C1(C2=CC=CC=C2)=C(C3=CC=CC=C3)C=C(C4=CC=CC=C4)C(C5=CC=CC=C5)=C1](1)</Checkpoint>

The generation of $\mathbf{B}$ also involves the above process, first producing a product similar to $\mathbf{A}$. Noting that $\mathbf{B}(\mathrm {C_{34}H_{26}Br_4O_4})$ has four bromine atoms, it is speculated that electrophilic substitution has occurred on the benzene ring. Because of the presence of methoxy as an electron-donating group, $\mathbf{B}$ is:

<MultiModal>![BrC1=CC(C2=C(C=C(C(C3=CC(Br)=C(OC)C=C3)=C2)C4=CC=C(OC)C(Br)=C4)C5=CC(Br)=C(OC)C=C5)=CC=C1OC](/media/uploads/7ff4fe16-7deb-414a-bf10-b47353dedc52.png)</MultiModal>

<Checkpoint>[$\mathbf{B}$ is BrC1=CC(C2=C(C=C(C(C3=CC(Br)=C(OC)C=C3)=C2)C4=CC=C(OC)C(Br)=C4)C5=CC(Br)=C(OC)C=C5)=CC=C1OC](1)</Checkpoint>

The point group of molecules $\mathbf{A, B}$ in the conformation with the highest symmetry are both $D_{2h}$, statement 1 is correct.

<Checkpoint>[The point group of molecules $\mathbf{A, B}$ in the conformation with the highest symmetry are both $D_{2h}$](1)</Checkpoint>

If $\mathrm{Br_2}$ is replaced with $\mathrm{I_2}$, the resulting product $\mathbf{B'}$ is structurally similar to $\mathbf{B}$, but due to the larger radius of the iodine atom, it may prevent the two benzene rings from being in the same plane, thereby disrupting the symmetry.

<Checkpoint>[The larger radius of the iodine atom may cause the product to lose its planar structure](1)</Checkpoint>

If the reaction time is further extended, $\mathbf{A}$ is relatively stable and will not continue to react. $\mathbf{B}$ may continue to undergo electrophilic substitution reactions, thus changing the yield.

<Checkpoint>[$\mathbf{B}$ will continue to undergo electrophilic substitution reactions](1)</Checkpoint>

It can be seen from the intermediates that the maximum number of rings in each intermediate that generates $\mathbf{A}$ is 6.

<Checkpoint>[C12C(C3=CC=CC=C3)(C4=CC=CC=C4)C=CC(C5=CC=CC=C5)(C6=CC=CC=C6)C1[Br+]2 has 6 rings](1)</Checkpoint>

Therefore, statements 1 and 2 are correct.
<MultiModal>![C1(C2(C3=CC=CC=C3)C=CC(C4=CC=CC=C4)(C5=CC=CC=C5)C=C2)=CC=CC=C1](/media/uploads/df5b7c0f-dcb4-49c4-a174-1e06fe78d0b4.png)</MultiModal>

与 $\mathrm{Br_2}$反应，首先生成溴鎓离子中间体：

<MultiModal>![C12C(C3=CC=CC=C3)(C4=CC=CC=C4)C=CC(C5=CC=CC=C5)(C6=CC=CC=C6)C1[Br+]2](/media/uploads/03430afc-a1df-40f2-b562-e4fcc7150b3a.png)</MultiModal>

随后发生苯基的迁移，打开三元环形成碳正离子中间体：

<MultiModal>![BrC1C(C2=CC=CC=C2)(C3=CC=CC=C3)C=C[C+](C4=CC=CC=C4)C1C5=CC=CC=C5](/media/uploads/cd19bb14-f5ec-4cf1-b10e-18cbf44fc509.png)</MultiModal>

然后再发生一步苯基的迁移，形成另一个碳正离子中间体：

<MultiModal>![BrC1[C+](C2=CC=CC=C2)C(C3=CC=CC=C3)C=C(C4=CC=CC=C4)C1C5=CC=CC=C5](/media/uploads/441bbb1b-1aed-4e51-81ce-b511d756a957.png)</MultiModal>

此时没有能够发生迁移的基团了，于是失去碳正离子生成双键：

<MultiModal>![BrC1C(C2=CC=CC=C2)=C(C3=CC=CC=C3)C=C(C4=CC=CC=C4)C1C5=CC=CC=C5](/media/uploads/a3fb5b3c-de32-4cc5-aa9a-245785974e11.png)</MultiModal>

最后发生消除反应，形成稳定的芳环产物：

<MultiModal>![C1(C2=CC=CC=C2)=C(C3=CC=CC=C3)C=C(C4=CC=CC=C4)C(C5=CC=CC=C5)=C1](/media/uploads/6457d8c0-7122-4818-a900-4ce83e592b9b.png)</MultiModal>

<Checkpoint>[生成$\mathbf { A }$的反应的重要中间体1：C12C(C3=CC=CC=C3)(C4=CC=CC=C4)C=CC(C5=CC=CC=C5)(C6=CC=CC=C6)C1[Br+]2](1)</Checkpoint>

<Checkpoint>[生成$\mathbf { A }$的反应的重要中间体2：BrC1C(C2=CC=CC=C2)(C3=CC=CC=C3)C=C[C+](C4=CC=CC=C4)C1C5=CC=CC=C5](1)</Checkpoint>

<Checkpoint>[生成$\mathbf { A }$的反应的重要中间体3：BrC1[C+](C2=CC=CC=C2)C(C3=CC=CC=C3)C=C(C4=CC=CC=C4)C1C5=CC=CC=C5](1)</Checkpoint>

<Checkpoint>[生成$\mathbf { A }$的反应的重要中间体4：BrC1C(C2=CC=CC=C2)=C(C3=CC=CC=C3)C=C(C4=CC=CC=C4)C1C5=CC=CC=C5](1)</Checkpoint>

<Checkpoint>[$\mathbf { A }$为C1(C2=CC=CC=C2)=C(C3=CC=CC=C3)C=C(C4=CC=CC=C4)C(C5=CC=CC=C5)=C1](1)</Checkpoint>

$\mathbf{B}$的产生也要经过上述过程，先产生与$\mathbf{A}$类似的产物，注意到$\mathbf{B}(\mathrm {C_{34}H_{26}Br_4O_4})$ 中有四个溴原子，推测为发生了苯环上的亲电取代反应，因为甲氧基给电子基的存在，因此$\mathbf{B}$为：

<MultiModal>![BrC1=CC(C2=C(C=C(C(C3=CC(Br)=C(OC)C=C3)=C2)C4=CC=C(OC)C(Br)=C4)C5=CC(Br)=C(OC)C=C5)=CC=C1OC](/media/uploads/7ff4fe16-7deb-414a-bf10-b47353dedc52.png)</MultiModal>

<Checkpoint>[$\mathbf{B}$为BrC1=CC(C2=C(C=C(C(C3=CC(Br)=C(OC)C=C3)=C2)C4=CC=C(OC)C(Br)=C4)C5=CC(Br)=C(OC)C=C5)=CC=C1OC](1)</Checkpoint>

$\mathbf{A,B}$分子在对称性最高的构象时所属的点群均为$D_{2h}$，说法1正确。

<Checkpoint>[$\mathbf{A,B}$分子对称性最高的构象时所属的点群均为$D_{2h}$](1)</Checkpoint>

若将$\mathrm{Br_2}$替换为$\mathrm{I_2}$，所得到的产物$\mathbf{B'}$与$\mathbf{B}$结构类似，但是由于碘原子半径较大，可能会使两个苯环无法处于同一平面从而破坏了对称性。

<Checkpoint>[碘原子半径较大可能会使产物失去平面结构](1)</Checkpoint>

若继续延长反应时间，$\mathbf{A}$比较稳定，不会继续发生反应，$\mathbf{B}$可能会继续发生亲电取代反应，从而使产率改变。

<Checkpoint>[$\mathbf{B}$会继续发生亲电取代反应](1)</Checkpoint>

从中间体可以看出生成$\mathbf{A}$的各个中间体中环数最多为6。

<Checkpoint>[C12C(C3=CC=CC=C3)(C4=CC=CC=C4)C=CC(C5=CC=CC=C5)(C6=CC=CC=C6)C1[Br+]2中有6个环](1)</Checkpoint>

因此说法1，2正确