Compound $\mathbf{A}$ can be converted into $\mathbf{B}(\mathrm{C_{20}H_{21}IN_2O_2S}),\mathbf{C}(\mathrm{C_{20}H_{21}IN_2O_2S})$ under the action of $\mathrm{I_2}$:

<MultiModal>![Compound $\mathbf{A}$ is C=CCN(S(C1=CC=C(C)C=C1)(=O)=O)CC#CC2=CC=CC=C2N(C)C, the reaction of $\mathbf{A}$ with $\mathrm{I_2}$ can be converted into $\mathbf{B}$ and $\mathbf{C}$, and the reaction of $\mathbf{B}$ with $\mathrm{I_2}$ can also be converted into $\mathbf{C}$](/media/uploads/16c6e1da-84d0-40cd-8c79-88a2ce1c6af1.png)</MultiModal>

The following experiments have been carried out:

|Entry|Solvent|$\mathbf{A}$:$\mathrm{I_2}$|temp($^\circ \mathrm{C}$)|Time(h)|Yield($\mathbf{B}$)|Yield($\mathbf{C}$)|
|---|---|---|---|---|---|---|
|1|DCE|1:1|70|4|92|trace|
|2|DCE|1:2|70|4|0|90|


DCE: dichloroethane.

It is known that $\mathbf{B}$ can be converted into $\mathbf{C}$ in DCE under the same conditions with $\mathrm { 1 eq } ~\mathrm { I } _ { 2 }$.

There are the following statements:

1.  Both $\mathbf{B}$ and $\mathbf{C}$ have three rings.

2.  The number of $"\mathrm{CH_2}"$ groups in $\mathbf{B}$ and $\mathbf{C}$ are the same, and the number of $"\mathrm{CH}"$ groups are different.

3.  The number of rings in the key intermediate for the formation of $\mathbf{B}$ from $\mathbf{A}$ and the key intermediate for the formation of $\mathbf{C}$ from $\mathbf{B}$ are all 3.

4.  If 1 eq of $\mathrm{I_2}$ labeled with a radioisotope is added first to completely convert $\mathbf{A}$ to $\mathbf{B}$, and then 1 eq of unlabeled $\mathrm{I_2}$ is added to convert $\mathbf{B}$ to $\mathbf{C}$, then no radioactivity can be detected in $\mathbf{C}$.

The option that contains all the correct statements is:
化合物$\mathbf{A}$在$\mathrm{I_2}$作用下可以转化为$\mathbf{B}(\mathrm{C_{20}H_{21}IN_2O_2S}),\mathbf{C}(\mathrm{C_{20}H_{21}IN_2O_2S})$：

<MultiModal>![化合物$\mathbf{A}$为 C=CCN(S(C1=CC=C(C)C=C1)(=O)=O)CC#CC2=CC=CC=C2N(C)C，$\mathbf{A}$与$\mathrm{I_2}$作用可以转化为$\mathbf{B}$和$\mathbf{C}$，$\mathbf{B}$与$\mathrm{I_2}$作用也可以转化为$\mathbf{C}$](/media/uploads/16c6e1da-84d0-40cd-8c79-88a2ce1c6af1.png)</MultiModal>

现已经进行了下列实验：

|Entry|溶剂|$\mathbf{A}$:$\mathrm{I_2}$|temp($^\circ \mathrm{C}$)|Time(h)|Yield($\mathbf{B}$)|Yield($\mathbf{C}$)|
|---|---|---|---|---|---|---|
|1|DCE|1:1|70|4|92|trace|
|2|DCE|1:2|70|4|0|90|


DCE：二氯乙烷。

已知$\mathbf{B}$在DCE中以相同条件与 $\mathrm { 1 e q } ~\mathrm { I } _ { 2 }$ 反应可以转化为$\mathbf{C}$。

有下列几个说法：

1，$\mathbf{B，C}$中均有三个环。

2，$\mathbf{B，C}$中$"\mathrm{CH_2}"$基团数目相同，$"\mathrm{CH}"$基团数目不同。

3，$\mathbf{A}$生成$\mathbf{B}$的关键中间体和$\mathbf{B}$生成$\mathbf{C}$的关键中间体中环的数目全部为3。

4，若先加入1 eq被放射性同位素标记的$\mathrm{I_2}$使$\mathbf{A}$完全转化为$\mathbf{B}$后再加入1 eq无放射性同位素标记的$\mathrm{I_2}$使$\mathbf{B}$转化为$\mathbf{C}$，则$\mathbf{C}$中无法检测到放射性。

下列选项中有所有正确说法的选项是：
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
From the reaction ratios given in the table, it can be seen that product $\mathbf{B}$ is formed upon addition of 1 eq of iodine, and product $\mathbf{C}$ is formed upon further addition of 1 eq of iodine. Therefore, the reaction mainly consists of two steps, and the formation of product $\mathbf{C}$ requires more than 1 eq of iodine.

<Checkpoint>[Product $\mathbf{C}$ is not formed when only 1 eq of iodine is added](1)</Checkpoint>

This reaction is a typical sequential reaction starting from an iodonium ion. The triple bond in the substrate first forms an iodonium ion intermediate:

<MultiModal>![C=CCN(S(C1=CC=C(C)C=C1)(=O)=O)CC2=C([I+]2)C3=CC=CC=C3N(C)C](/media/uploads/11014600-e505-4783-849e-5dec8a4c6e5b.png)</MultiModal>

<Checkpoint>[The first intermediate is: C=CCN(S(C1=CC=C(C)C=C1)(=O)=O)CC2=C([I+]2)C3=CC=CC=C3N(C)C](1)</Checkpoint>

Subsequently, amine attack occurs, opening the three-membered ring to give the second intermediate:

<MultiModal>![C=CCN(S(C1=CC=C(C)C=C1)(=O)=O)CC2=C(I)C3=CC=CC=C3[N+]2(C)C](/media/uploads/5f555410-6326-43be-95d9-d58f754c790d.png)</MultiModal>

<Checkpoint>[The second intermediate is C=CCN(S(C1=CC=C(C)C=C1)(=O)=O)CC2=C(I)C3=CC=CC=C3[N+]2(C)C](1)</Checkpoint>

Finally, a reaction occurs in which iodide attacks the methyl group, yielding product $\mathbf{B}$:

<MultiModal>![C=CCN(S(C1=CC=C(C)C=C1)(=O)=O)CC2=C(I)C3=CC=CC=C3N2C](/media/uploads/e15c5b58-bf60-4446-9361-d1cc97eff78c.png)</MultiModal>

<Checkpoint>[$\mathbf{B}$ is C=CCN(S(C1=CC=C(C)C=C1)(=O)=O)CC2=C(I)C3=CC=CC=C3N2C](1)</Checkpoint>

$\mathbf{B}$ can undergo a similar reaction as described above once more. First, an iodonium ion is formed at the double bond:

<MultiModal>![CN1C2=CC=CC=C2C(I)=C1CN(S(C3=CC=C(C)C=C3)(=O)=O)CC4C[I+]4](/media/uploads/322174e5-562c-4fda-90f8-b78022c81c6d.png)</MultiModal>

<Checkpoint>[The third intermediate is CN1C2=CC=CC=C2C(I)=C1CN(S(C3=CC=C(C)C=C3)(=O)=O)CC4C[I+]4](1)</Checkpoint>

Subsequently, the indole ring's carbon at the 3-position attacks the three-membered ring, generating the next intermediate:

<MultiModal>![C[N+]1=C2CN(S(C3=CC=C(C)C=C3)(=O)=O)CC(C2(I)C4=CC=CC=C41)CI](/media/uploads/3c66ad10-7827-4132-bb96-89f3abd68888.png)</MultiModal>

<Checkpoint>[The fourth intermediate is C[N+]1=C2CN(S(C3=CC=C(C)C=C3)(=O)=O)CC(C2(I)C4=CC=CC=C41)CI](1)</Checkpoint>

Finally, an iodide attacks the iodine, eliminating iodine to yield $\mathbf{C}$:

<MultiModal>![CN1C2=CC=CC=C2C3=C1CN(S(C4=CC=C(C)C=C4)(=O)=O)CC3CI](/media/uploads/27414c7c-493e-470d-b961-e82e2428bad8.png)</MultiModal>

<Checkpoint>[$\mathbf{C}$ is CN1C2=CC=CC=C2C3=C1CN(S(C4=CC=C(C)C=C4)(=O)=O)CC3CI](1)</Checkpoint>

1. $\mathbf{B}$ has three rings, and $\mathbf{C}$ has four rings. Incorrect.

2. $\mathbf{B}$ has three $"\mathrm{CH_2}"$ groups and five $"\mathrm{CH}"$ groups, and $\mathbf{C}$ has three $"\mathrm{CH_2}"$ groups and five $"\mathrm{CH}"$ groups. The numbers are the same. Incorrect.

3. The key intermediates in the formation of $\mathbf{B}$ from $\mathbf{A}$ all have 3 rings, and the key intermediates in the formation of $\mathbf{C}$ from $\mathbf{B}$ all have 4 rings. Incorrect.

4. In the last step of the formation of $\mathbf{C}$ from $\mathbf{B}$, the initially present radioactive iodine atom will be transferred to iodine,

<Checkpoint>[Radioactive iodine atoms are transferred to iodine when $\mathbf{B}$ is converted to $\mathbf{C}$](1)</Checkpoint>

The radioactive iodine can then undergo this step again, so radioactivity can also be detected in $\mathbf{C}$. Incorrect.
从表格所给的反应比例可以知道加入1 eq碘单质时生成产物 $\mathbf{B}$，再加入1 eq碘单质才生成产物 $\mathbf{C}$，因此该反应主要分成两步，并且生成产物$\mathbf{C}$是需要大于1 eq的碘单质的。

<Checkpoint>[只加入1 eq碘单质不会生成产物 $\mathbf{C}$](1)</Checkpoint>

这个反应是一个典型的从碘鎓离子开始的连续反应，底物中的三键首先形成碘鎓离子中间体：

<MultiModal>![C=CCN(S(C1=CC=C(C)C=C1)(=O)=O)CC2=C([I+]2)C3=CC=CC=C3N(C)C](/media/uploads/11014600-e505-4783-849e-5dec8a4c6e5b.png)</MultiModal>

<Checkpoint>[第一个中间体为：C=CCN(S(C1=CC=C(C)C=C1)(=O)=O)CC2=C([I+]2)C3=CC=CC=C3N(C)C](1)</Checkpoint>

随后发生胺基进攻，打开三元环得到第二个中间体：

<MultiModal>![C=CCN(S(C1=CC=C(C)C=C1)(=O)=O)CC2=C(I)C3=CC=CC=C3[N+]2(C)C](/media/uploads/5f555410-6326-43be-95d9-d58f754c790d.png)</MultiModal>

<Checkpoint>[第二个中间体为C=CCN(S(C1=CC=C(C)C=C1)(=O)=O)CC2=C(I)C3=CC=CC=C3[N+]2(C)C](1)</Checkpoint>

最后发生碘离子进攻甲基的反应，得到产物 $\mathbf{B}$：

<MultiModal>![C=CCN(S(C1=CC=C(C)C=C1)(=O)=O)CC2=C(I)C3=CC=CC=C3N2C](/media/uploads/e15c5b58-bf60-4446-9361-d1cc97eff78c.png)</MultiModal>

<Checkpoint>[$\mathbf{B}$为C=CCN(S(C1=CC=C(C)C=C1)(=O)=O)CC2=C(I)C3=CC=CC=C3N2C](1)</Checkpoint>

$\mathbf{B}$还可以再发生一次上述类似的反应，先是双键处生成碘鎓离子：

<MultiModal>![CN1C2=CC=CC=C2C(I)=C1CN(S(C3=CC=C(C)C=C3)(=O)=O)CC4C[I+]4](/media/uploads/322174e5-562c-4fda-90f8-b78022c81c6d.png)</MultiModal>

<Checkpoint>[第三个中间体为CN1C2=CC=CC=C2C(I)=C1CN(S(C3=CC=C(C)C=C3)(=O)=O)CC4C[I+]4](1)</Checkpoint>

随后发生吲哚环3号碳对三元环的进攻，生成下一个中间体：

<MultiModal>![C[N+]1=C2CN(S(C3=CC=C(C)C=C3)(=O)=O)CC(C2(I)C4=CC=CC=C41)CI](/media/uploads/3c66ad10-7827-4132-bb96-89f3abd68888.png)</MultiModal>

<Checkpoint>[第四个中间体为C[N+]1=C2CN(S(C3=CC=C(C)C=C3)(=O)=O)CC(C2(I)C4=CC=CC=C41)CI](1)</Checkpoint>

最后发生碘离子对碘的进攻，脱去碘单质后得到$\mathbf{C}$：

<MultiModal>![CN1C2=CC=CC=C2C3=C1CN(S(C4=CC=C(C)C=C4)(=O)=O)CC3CI](/media/uploads/27414c7c-493e-470d-b961-e82e2428bad8.png)</MultiModal>

<Checkpoint>[$\mathbf{C}$为CN1C2=CC=CC=C2C3=C1CN(S(C4=CC=C(C)C=C4)(=O)=O)CC3CI](1)</Checkpoint>

1，$\mathbf{B}$中有三个环，$\mathbf{C}$中有四个环。错误。

2，$\mathbf{B}$中有3个$"\mathrm{CH_2}"$基团，5个$"\mathrm{CH}"$基团，$\mathbf{C}$中有3个$"\mathrm{CH_2}"$基团，5个$"\mathrm{CH}"$基团，数目均相同。错误。

3，$\mathbf{A}$生成$\mathbf{B}$的关键中间体中环的数目全部为3，$\mathbf{B}$生成$\mathbf{C}$的关键中间体中环的数目全部为4。错误。

4，在$\mathbf{B}$生成$\mathbf{C}$最后一步中会将最开始存在的放射性碘原子转移到碘单质中，

<Checkpoint>[$\mathbf{B}$生成$\mathbf{C}$时放射性碘原子转移到碘单质中](1)</Checkpoint>

放射性的碘单质又可以发生该步反应，因此$\mathbf{C}$中也可以检测到放射性。错误