Using $\beta$-aminocrotonate as a substrate, different alkaloid skeletons can be constructed under the catalysis of different Lewis acids:

<MultiModal>![Substrate is O=C(OC(C)(C)C)/C=C/N(CC1=CC=CC=C1)CCC2=CN(C3=CC=CC=C32)[R], reaction conditions are: 1.2 equivalents of VII (structure is FI1OC(C)(C)C2=C1C=CC=C2), LA, solvent, -10 degrees Celsius gradually increased to room temperature, substrate concentration is 0.04 M. When LA=$\rm Cu(OTf)_2$, R=H, solvent=DCM, the product is $\bf C$; when LA=$\rm In(OTf)_3$, R=H, solvent=EtOAc, the product is $\bf D$; when LA=$\rm Sc(OTf)_3$, R=Boc, solvent=DCM, the product is $\bf E$](/media/uploads/f12048ff-1c6b-407b-b02f-82a5759b060f.png)</MultiModal>

It is known that $\bf C$ has a six-fused-five-spiro-six structure, $\bf D$ has a six-fused-five-fused-seven structure, and the molecular formula of $\bf E$ is $\rm C_{25}H_{25}N_2O_4$. $\bf C$, $\bf D$, and $\bf E$ do not contain fluorine, and their formation all involves a key intermediate $\bf X \rm ^{2+}$ that does not contain iodine. The $\rm C$–$\rm I$ bond has never been formed during the reaction. Which of the following statements are correct:

1. $\bf C$ has 10 $\rm sp^2C-H$ bonds
2. If benzene rings are not considered, $\bf D$ has 2 carbon-carbon double bonds
3. $\bf E$ has 4 rings
4. If R=H, the key intermediate $\bf X \rm ^{2+}$ has 16 $\rm sp^3C-H$ bonds
以$\beta$-氨基丙烯酸酯为底物，在不同的Lewis酸催化下可以构筑不同的生物碱骨架：

<MultiModal>![底物为O=C(OC(C)(C)C)/C=C/N(CC1=CC=CC=C1)CCC2=CN(C3=CC=CC=C32)[R]，反应条件为：1.2当量的VII(结构为FI1OC(C)(C)C2=C1C=CC=C2)、LA、solvent、-10摄氏度逐渐升至室温，底物浓度为0.04 M。当LA=$\rm Cu(OTf)_2$，R=H，solvent=DCM时产物为$\bf C$；当LA=$\rm In(OTf)_3$，R=H，solvent=EtOAc时产物为$\bf D$；当LA=$\rm Sc(OTf)_3$，R=Boc，solvent=DCM时产物为$\bf E$](/media/uploads/f12048ff-1c6b-407b-b02f-82a5759b060f.png)</MultiModal>

已知$\bf C$具有六并五螺六结构，$\bf D$具有六并五并七结构，$\bf E$的分子式为$\rm C_{25}H_{25}N_2O_4$。$\bf C$、$\bf D$、$\bf E$均不含氟，它们的形成均经历了不含碘的关键中间体$\bf X \rm ^{2+}$。反应过程中$\rm C$–$\rm I$键没有形成过，下列说法正确的有：

1. $\bf C$中有10根$\rm sp^2C-H$键
2. 如果不考虑苯环，$\bf D$中有2根碳碳双键
3. $\bf E$中有4个环
4. R=H时，关键中间体$\bf X \rm ^{2+}$中有16根$\rm sp^3C-H$键
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
VII is an oxidant. According to the information in the question, oxidation can only be completed by forming $\rm I-O$ bonds. Therefore, the following intermediate is formed first (where "I" represents the iodine atom and its connected groups):

<MultiModal>![IO/C(OC(C)(C)C)=C\C=[N+](CC1=CC=CC=C1)\CCC2=CN(C3=CC=CC=C32)[R]](/media/uploads/76b3a702-2967-4b7b-bec8-7e1ac429da47.png)</MultiModal>

$\bf C$ has a hexacyclopenta[6,5]spiro[6] structure. Only the C3 carbon of indole can undergo nucleophilic attack, then the key intermediate $\bf X \rm ^{2+}$ is obtained (it is not yet certain that it is $\bf X \rm ^{2+}$ at this time; this is written this way for convenience):

<MultiModal>![O=C(OC(C)(C)C)C(C=[N+](CC1=CC=CC=C1)CC2)C32C=[N+]([R])C4=CC=CC=C43](/media/uploads/343ff2f4-5916-46df-9179-27421eae7fb4.png)</MultiModal>

After losing 2 protons, $\bf C$ can be obtained:

<MultiModal>![O=C(OC(C)(C)C)C1=CN(CC2=CC=CC=C2)CCC31C=NC4=CC=CC=C43](/media/uploads/a0a080a2-f91a-4992-97d6-cf409ac60ab0.png)</MultiModal>

<Checkpoint>[$\bf C$ is O=C(OC(C)(C)C)C1=CN(CC2=CC=CC=C2)CCC31C=NC4=CC=CC=C43](1)</Checkpoint>

There are 5+4+1+1=11 $\rm sp^2C-H$ bonds in $\bf C$, statement 1 is incorrect.

To obtain the hexacyclopentahepta[6,5,7] $\bf D$, alkyl migration is required, and after losing 2 protons, $\bf D$ can be obtained:

<MultiModal>![O=C(OC(C)(C)C)C1=CN(CC2=CC=CC=C2)CCC3=C1NC4=CC=CC=C43](/media/uploads/9c9cb768-701f-4284-aa5a-db13d27cfbce.png)</MultiModal>

<Checkpoint>[$\bf D$ is O=C(OC(C)(C)C)C1=CN(CC2=CC=CC=C2)CCC3=C1NC4=CC=CC=C43](1)</Checkpoint>

Ignoring the benzene ring, $\bf D$ has 2 carbon-carbon double bonds, statement 2 is correct.

It can be basically determined that $\bf X \rm ^{2+}$ is O=C(OC(C)(C)C)C(C=[N+](CC1=CC=CC=C1)CC2)C32C=[N+]([R])C4=CC=CC=C43. Considering the benzyl and tert-butyl groups, there are a total of 2+2+2+1+9=16 $\rm sp^3C-H$ bonds, statement 4 is correct.

<Checkpoint>[$\bf X \rm ^{2+}$ is O=C(OC(C)(C)C)C(C=[N+](CC1=CC=CC=C1)CC2)C32C=[N+]([R])C4=CC=CC=C43](1)</Checkpoint>

From the molecular formula of $\bf E$, it can be seen that the tert-butyl group connected to the ester group is removed. Then, the carbonyl oxygen will undergo nucleophilic attack, that is, the carboxyl oxygen in $\bf X \rm ^{2+}$ attacks the imine cation to form a five-membered ring, and after losing a proton, $\bf E$ is obtained:

<MultiModal>![O=C1OC2N(C3=CC=CC=C3C24C1=CN(CC5=CC=CC=C5)CC4)C(OC(C)(C)C)=O](/media/uploads/6c619695-c490-4290-8e52-cb819a0c19f1.png)</MultiModal>

<Checkpoint>[$\bf E$ is O=C1OC2N(C3=CC=CC=C3C24C1=CN(CC5=CC=CC=C5)CC4)C(OC(C)(C)C)=O](1)</Checkpoint>

$\bf E$ contains 5 rings, statement 3 is incorrect.

Statements 2 and 4 are correct, choose J.
VII是一种氧化剂，根据题目信息，只能通过形成$\rm I-O$键来完成氧化，故先形成下面的中间体（此处的“I”表示碘原子及其连接的基团）：

<MultiModal>![IO/C(OC(C)(C)C)=C\C=[N+](CC1=CC=CC=C1)\CCC2=CN(C3=CC=CC=C32)[R]](/media/uploads/76b3a702-2967-4b7b-bec8-7e1ac429da47.png)</MultiModal>

$\bf C$有六并五螺六的结构，只能是吲哚3号碳进行亲核进攻，得到关键中间体$\bf X \rm ^{2+}$（此时还不能确定是$\bf X \rm ^{2+}$，这里为了书写方便才这样写）：

<MultiModal>![O=C(OC(C)(C)C)C(C=[N+](CC1=CC=CC=C1)CC2)C32C=[N+]([R])C4=CC=CC=C43](/media/uploads/343ff2f4-5916-46df-9179-27421eae7fb4.png)</MultiModal>

失去2个质子后可以得到$\bf C$：

<MultiModal>![O=C(OC(C)(C)C)C1=CN(CC2=CC=CC=C2)CCC31C=NC4=CC=CC=C43](/media/uploads/a0a080a2-f91a-4992-97d6-cf409ac60ab0.png)</MultiModal>

<Checkpoint>[$\bf C$为O=C(OC(C)(C)C)C1=CN(CC2=CC=CC=C2)CCC31C=NC4=CC=CC=C43](1)</Checkpoint>

$\bf C$中有5+4+1+1=11根$\rm sp^2C-H$键，说法1错误

要得到六并五并七的$\bf D$，需要经历烷基迁移，之后失去2个质子后可以得到$\bf D$：

<MultiModal>![O=C(OC(C)(C)C)C1=CN(CC2=CC=CC=C2)CCC3=C1NC4=CC=CC=C43](/media/uploads/9c9cb768-701f-4284-aa5a-db13d27cfbce.png)</MultiModal>

<Checkpoint>[$\bf D$为O=C(OC(C)(C)C)C1=CN(CC2=CC=CC=C2)CCC3=C1NC4=CC=CC=C43](1)</Checkpoint>

不考虑苯环$\bf D$有2根碳碳双键，说法2正确

由此基本可以确定$\bf X \rm ^{2+}$为O=C(OC(C)(C)C)C(C=[N+](CC1=CC=CC=C1)CC2)C32C=[N+]([R])C4=CC=CC=C43，考虑苄基和叔丁基，共2+2+2+1+9=16根$\rm sp^3C-H$键，说法4正确

<Checkpoint>[$\bf X \rm ^{2+}$为O=C(OC(C)(C)C)C(C=[N+](CC1=CC=CC=C1)CC2)C32C=[N+]([R])C4=CC=CC=C43](1)</Checkpoint>

由$\bf E$的分子式可知酯基连接的叔丁基脱去了，那么羰基氧会进行亲核进攻，即$\bf X \rm ^{2+}$中羧基氧进攻亚胺正离子形成五元环，失去质子后得到$\bf E$：

<MultiModal>![O=C1OC2N(C3=CC=CC=C3C24C1=CN(CC5=CC=CC=C5)CC4)C(OC(C)(C)C)=O](/media/uploads/6c619695-c490-4290-8e52-cb819a0c19f1.png)</MultiModal>

<Checkpoint>[$\bf E$为O=C1OC2N(C3=CC=CC=C3C24C1=CN(CC5=CC=CC=C5)CC4)C(OC(C)(C)C)=O](1)</Checkpoint>

$\bf E$含5个环，说法3错误

说法2和4正确，选J