<MultiModal>![This diagram depicts a one-pot organic cascade reaction. The substrate Cl/C=C1C(OC(C2=CC=CC=C2)=N/1)=O under hv conditions transforms into O=C1OC(C2=CC=CC=C2)=N/C1=C/Cl; then reacts with $\bf A, \rm EtAlCl_2, DCM$ to produce $\bf X$; $\bf X$ under $\rm p-TSA, MeOH$ conditions produces $\bf Y$; $\bf Y$ under $\rm EtOH, HCl$ conditions produces the final product CC1=C(C)C[C@@H](O)[C@](N)(C(O)=O)C1.](/media/uploads/9aeeea63-278a-4e0e-b5fc-1cb8284f98c9.png)</MultiModal>

Regarding the reaction in the above diagram, it is known that $\bf Y$ contains 17 carbon atoms.

Which of the following statements is correct:
<MultiModal>![本图描述了一步有机串联反应。底物Cl/C=C1C(OC(C2=CC=CC=C2)=N/1)=O在hv条件下转变为O=C1OC(C2=CC=CC=C2)=N/C1=C/Cl；之后与$\bf A, \rm EtAlCl_2, DCM$反应生成$\bf X$；$\bf X$在$\rm p-TSA, MeOH$条件下生成$\bf Y$；$\bf Y$在$\rm EtOH, HCl$条件下生成最终产物CC1=C(C)C[C@@H](O)[C@](N)(C(O)=O)C1。](/media/uploads/9aeeea63-278a-4e0e-b5fc-1cb8284f98c9.png)</MultiModal>

关于上图的反应，已知$\bf Y$含有17个碳原子。本题要求立体化学。

下列说法正确的是：
All other options are incorrect
$\bf A$ The product after treatment with $\rm O_3, MeOH/Zn$ contains one degree of unsaturation.
$\bf X,Y$ both contain the structure of a six-membered ring and a five-membered ring.
$\bf Y$ contains a carboxyl group
All chiral carbons in $\bf X$ are of R configuration.
All chiral carbons in $\bf X$ are of S configuration.
The process from $\bf X$ to $\bf Y$ did not result in a change in the configuration of the chiral center.
The reaction can generate products even without performing the first step under $hv$ conditions.
$\bf Y$ possesses a linkage relationship $\rm O-CH-CH-C=O$
其他选项均不正确
$\bf A$采用$\rm O_3, MeOH/Zn$处理后的产物含有一个不饱和度
$\bf X,Y$均含有六元环并五元环的结构
$\bf Y$含有羧基
$\bf X$中手性碳均为R构型
$\bf X$中手性碳均为S构型
$\bf X$到$\bf Y$的过程没有发生手性中心构型的变化
不进行第一步$hv$条件下的反应同样能生成产物
$\bf Y$存在键联关系$\rm O-CH-C-C=O$
<MultiModal>![$\bf A$ is C=C(C(C)=C)C; the structure of $\bf X$ is Cl[C@H]1CC(C)=C(C)C[C@]12N=C(C3=CC=CC=C3)OC2=O; the structure of $\bf Y$ is O=C(OC)[C@@]12[C@H](OC(C3=CC=CC=C3)=N2)CC(C)=C(C)C1。](/media/uploads/00936895-76ac-4d93-8fc4-ec237921c65c.png)</MultiModal>

Obviously, the first step of light irradiation is the isomerization reaction of the double bond. After isomerization, the $\rm Cl$ atom and the carbonyl group are on the same side of the double bond.

<Checkpoint>[The first step of light irradiation is the isomerization reaction of the double bond](1)</Checkpoint>

Observing the product, it can be seen that the nitrogen and oxygen atoms in the product should originate from the five-membered ring of the substrate according to the bond relationship, and the extra six-membered ring olefin structure of the substrate, according to retrosynthetic analysis, should be a part after the D-A cycloaddition reaction of 2,3-dimethyl-1,3-butadiene. Therefore, $\bf A$ is C=C(C(C)=C)C. It generates O=C(C(C)=O)C under ozonolysis reaction conditions, containing two unsaturated bonds, so option B is incorrect.

<Checkpoint>[The extra six-membered ring olefin structure of the substrate, according to retrosynthetic analysis, should be a part after the D-A cycloaddition reaction of 2,3-dimethyl-1,3-butadiene](1)</Checkpoint>

<Checkpoint>[$\bf A$ is C=C(C(C)=C)C](1)</Checkpoint>

The substrate undergoes a D-A reaction with 2,3-dimethyl-1,3-butadiene, and the dienophile should be the $\rm C=C$ double bond. Since the $\rm Cl$ atom and the carbonyl group are on the same side of the double bond, according to the stereoselectivity of the D-A reaction, the $\rm Cl$ atom and the carbonyl group are on the same side in the product; therefore, the structure of $\bf X$ is Cl[C@H]1CC(C)=C(C)C[C@]12N=C(C3=CC=CC=C3)OC2=O. The two chiral centers of $\bf X$ are 1R,1S, so option E is incorrect.

<Checkpoint>[Since the $\rm Cl$ atom and the carbonyl group are on the same side of the double bond, the $\rm Cl$ atom and the carbonyl group are on the same side in the D-A reaction product](1)</Checkpoint>

<Checkpoint>[The structure of $\bf X$ is Cl[C@H]1CC(C)=C(C)C[C@]12N=C(C3=CC=CC=C3)OC2=O](1)</Checkpoint>

Then, p-toluenesulfonic acid $\rm p-TSA$ and methanol are added, and a transesterification reaction occurs under acidic conditions to generate a methyl ester group and a hydroxyl anion; the hydroxyl anion is now in the trans position to the chloride ion, so an $\rm S_N2$ reaction can occur to form an intramolecular five-membered ring; therefore, the structure of $\bf Y$ is O=C(OC)[C@@]12[C@H](OC(C3=CC=CC=C3)=N2)CC(C)=C(C)C1. (Since the question gives 17 carbons, only a methyl ester group can be generated here.) The two chiral centers of $\bf Y$ are 1R,1S and the configuration of the chiral center has changed, so options F and G are incorrect. $\bf Y$ has a bond relationship of $\rm O-CH-C-C=O$, so option I is correct.

<Checkpoint>[A transesterification reaction occurs under acidic conditions to generate a methyl ester group and a hydroxyl anion](1)</Checkpoint>

<Checkpoint>[The hydroxyl anion is now in the trans position to the chloride ion, so an $\rm S_N2$ reaction can occur to form an intramolecular five-membered ring](1)</Checkpoint>

<Checkpoint>[The structure of $\bf Y$ is O=C(OC)[C@@]12[C@H](OC(C3=CC=CC=C3)=N2)CC(C)=C(C)C1](1)</Checkpoint>

<Checkpoint>[The configuration of the chiral center has changed from $\bf X$ to $\bf Y$](1)</Checkpoint>

According to the structure, $\bf Y$ is a hexafused five-ring system and does not contain a carboxyl group, and $\bf X$ is a hexaspiro five-ring system, so options C and D are incorrect.

Finally, $\bf Y$ hydrolyzes the imine structure under strongly acidic conditions to generate the product, and its stereochemistry meets the requirements.

If the light irradiation is not started to isomerize the double bond, the chlorine atom will be on the same side of the carbonyl group after the D-A reaction, so the $\rm S_N2$ intramolecular cyclization cannot occur and the product cannot be obtained, so option G is incorrect.

<Checkpoint>[Without isomerizing the double bond, the chlorine atom will be on the same side of the carbonyl group after the D-A reaction, so the $\rm S_N2$ intramolecular cyclization cannot occur](1)</Checkpoint>
<MultiModal>![$\bf A$为C=C(C(C)=C)C；$\bf X$的结构为Cl[C@H]1CC(C)=C(C)C[C@]12N=C(C3=CC=CC=C3)OC2=O；$\bf Y$的结构为O=C(OC)[C@@]12[C@H](OC(C3=CC=CC=C3)=N2)CC(C)=C(C)C1。](/media/uploads/00936895-76ac-4d93-8fc4-ec237921c65c.png)</MultiModal>

很明显，第一步光照条件为双键的异构化反应。异构化后，$\rm Cl$原子与羰基处于双键同侧。

<Checkpoint>[第一步光照条件为双键的异构化反应](1)</Checkpoint>

观察产物可知，产物中的氮原子与氧原子根据键联关系应当来源于底物五元环，而底物多出的六元环烯烃结构，根据逆合成分析可知应当为2,3-二甲基-1,3-丁二烯的D-A环加成反应后的部分。因此$\bf A$为C=C(C(C)=C)C。其在臭氧化反应条件下生成O=C(C(C)=O)C，含两个不饱和键，选项B错误。

<Checkpoint>[底物多出的六元环烯烃结构，根据逆合成分析可知应当为2,3-二甲基-1,3-丁二烯的D-A环加成反应后的部分](1)</Checkpoint>

<Checkpoint>[$\bf A$为C=C(C(C)=C)C](1)</Checkpoint>

底物与2,3-二甲基-1,3-丁二烯发生D-A反应，亲双烯体应当为$\rm C=C$双键。由于$\rm Cl$原子与羰基处于双键同侧，因此根据D-A反应的立体选择性可知产物中$\rm Cl$原子与羰基同侧；因此$\bf X$的结构为Cl[C@H]1CC(C)=C(C)C[C@]12N=C(C3=CC=CC=C3)OC2=O。$\bf X$的两个手性中心为1R,1S，选项E错误。

<Checkpoint>[$\rm Cl$原子与羰基处于双键同侧，因此D-A反应产物中$\rm Cl$原子与羰基同侧](1)</Checkpoint>

<Checkpoint>[$\bf X$的结构为Cl[C@H]1CC(C)=C(C)C[C@]12N=C(C3=CC=CC=C3)OC2=O](1)</Checkpoint>

之后加入对甲基苯磺酸$\rm p-TSA$和甲醇，酸性条件下发生酯交换反应生成甲酯基和羟基负离子；羟基负离子此时处于氯离子的反式位置，从而可发生$\rm S_N2$反应形成分子内五元环；因此$\bf Y$的结构为O=C(OC)[C@@]12[C@H](OC(C3=CC=CC=C3)=N2)CC(C)=C(C)C1。（由于题目给出17个碳，此处只能生成甲酯基。）$\bf Y$的两个手性中心为1R,1S且发生了手性中心构型变化，选项F,G错误。$\bf Y$存在键联关系$\rm O-CH-C-C=O$，选项I正确。

<Checkpoint>[酸性条件下发生酯交换反应生成甲酯基和羟基负离子](1)</Checkpoint>

<Checkpoint>[羟基负离子此时处于氯离子的反式位置，从而可发生$\rm S_N2$反应形成分子内五元环](1)</Checkpoint>

<Checkpoint>[$\bf Y$的结构为O=C(OC)[C@@]12[C@H](OC(C3=CC=CC=C3)=N2)CC(C)=C(C)C1](1)</Checkpoint>

<Checkpoint>[$\bf X$到$\bf Y$发生了手性中心构型变化](1)</Checkpoint>

根据结构，$\bf Y$为六并五环系且不含羧基，$\bf X$为六螺五环系，选项C，D错误

最后$\bf Y$在强酸性条件下水解亚胺结构生成产物，其立体化学符合要求。

如果不开始进行光照使得双键异构化，D-A反应后氯原子会处于羰基的同侧，从而无法发生$\rm S_N2$分子内成环因此得不到产物，选项G错误。

<Checkpoint>[不异构化双键，D-A反应后氯原子会处于羰基的同侧，从而无法发生$\rm S_N2$分子内成环](1)</Checkpoint>