<MultiModal>![The image describes a one-pot organic cascade reaction. The substrates are C/C([N+]([O-])=O)=C\C and C=CCC=C, reacting under the action of $\rm SnCl_4$ to obtain $\bf A$; $\bf A$ yields $\bf B$ at $383 ~K$; $\bf B$ yields the final product OC1CC(C)C(N)(C)C(CO)C1 under $\rm H_2, Raney Ni$ conditions.](/media/uploads/9ef3d6f1-05e9-48a8-b254-be917d4a8b1f.png)</MultiModal>

In the reaction above, it is known that the process of generating the final product from $\bf B$ does not involve changes in carbon-carbon bond formation. $\bf A$ is the thermodynamically more stable product.

Which of the following options reflects the correct stereochemistry of the final product?
<MultiModal>![该图片描述了一步有机串联反应。底物为C/C([N+]([O-])=O)=C\C和C=CCC=C，在$\rm SnCl_4$作用下反应得到$\bf A$；$\bf A$在$383 ~K$下得到$\bf B$；$\bf B$在$\rm H_2, Raney Ni$条件下得到最终产物OC1CC(C)C(N)(C)C(CO)C1。](/media/uploads/9ef3d6f1-05e9-48a8-b254-be917d4a8b1f.png)</MultiModal>

上图的反应中，已知由$\bf B$生成最终产物的过程不涉及碳原子成键的变化。$\bf A$为热力学上更稳定的产物。

下列选项中，能反映最终产物正确的立体化学的选项为：
All other options are incorrect
<MultiModal>![O[C@H]1C[C@@H](C)[C@](N)(C)[C@@H](CO)C1](/media/uploads/c25c57de-8f70-4997-aeb3-3a2d608ba034.png)</MultiModal>
<MultiModal>![O[C@H]1C[C@@H](C)[C@@](N)(C)[C@@H](CO)C1](/media/uploads/43393797-88f9-4dd9-993e-17cb16ddb1f9.png)</MultiModal>
<MultiModal>![O[C@H]1C[C@H](C)[C@](N)(C)[C@@H](CO)C1](/media/uploads/a200a6ef-634a-47ea-b447-7bbba859bdea.png)</MultiModal>
<MultiModal>![O[C@@H]1C[C@@H](C)[C@](N)(C)[C@@H](CO)C1](/media/uploads/903c3f07-9877-4926-a3f2-8f7c6ff6c455.png)</MultiModal>
<MultiModal>![O[C@@H]1C[C@H](C)[C@@](N)(C)[C@@H](CO)C1](/media/uploads/7a9e030c-9d7e-4080-9677-a1ab8f489972.png)</MultiModal>
<MultiModal>![O[C@@H]1C[C@H](C)[C@](N)(C)[C@@H](CO)C1](/media/uploads/d0df8630-1071-4d8c-a9af-95d466a6df0f.png)</MultiModal>
其他选项均不正确
<MultiModal>![O[C@H]1C[C@@H](C)[C@](N)(C)[C@@H](CO)C1](/media/uploads/c25c57de-8f70-4997-aeb3-3a2d608ba034.png)</MultiModal>
<MultiModal>![O[C@H]1C[C@@H](C)[C@@](N)(C)[C@@H](CO)C1](/media/uploads/43393797-88f9-4dd9-993e-17cb16ddb1f9.png)</MultiModal>
<MultiModal>![O[C@H]1C[C@H](C)[C@](N)(C)[C@@H](CO)C1](/media/uploads/a200a6ef-634a-47ea-b447-7bbba859bdea.png)</MultiModal>
<MultiModal>![O[C@@H]1C[C@@H](C)[C@](N)(C)[C@@H](CO)C1](/media/uploads/903c3f07-9877-4926-a3f2-8f7c6ff6c455.png)</MultiModal>
<MultiModal>![O[C@@H]1C[C@H](C)[C@@](N)(C)[C@@H](CO)C1](/media/uploads/7a9e030c-9d7e-4080-9677-a1ab8f489972.png)</MultiModal>
<MultiModal>![O[C@@H]1C[C@H](C)[C@](N)(C)[C@@H](CO)C1](/media/uploads/d0df8630-1071-4d8c-a9af-95d466a6df0f.png)</MultiModal>
<MultiModal>![$\bf A$ structure is [O-][N+]1=C(C)C(C)CC(CC=C)O1; the structure of $\bf B$ is CC1(C(CO2)C3)N2OC3CC1([H])C; the product structure is O[C@H]1C[C@@H](C)[C@](N)(C)[C@@H](CO)C1.](/media/uploads/59f33657-263c-46eb-9d74-7d5489426083.png)</MultiModal>

The substrate is obviously a dienophile and a diene, so it undergoes a D-A cycloaddition reaction under the catalysis of Lewis acid $\rm SnCl_4$, so the structure of $\bf A$ is [O-][N+]1=C(C)C(C)CC(CC=C)O1.

<Checkpoint>[The substrate undergoes a D-A cycloaddition reaction under the catalysis of Lewis acid $\rm SnCl_4$](1)</Checkpoint>

<Checkpoint>[The structure of $\bf A$ is [O-][N+]1=C(C)C(C)CC(CC=C)O1](1)</Checkpoint>

$\bf A$ is heated to $\rm 383~K$ and converted to $\bf B$. The process of generating the final product from $\bf B$ does not involve changes in carbon-atom bonding, so it can only be the breaking of the $\rm N-O$ bond; therefore, without considering stereochemistry, the structure of $\bf B$ is such that the nitrogen atom of the final product is connected to two hydroxyl groups, forming a bridged ring structure with the structural formula CC1(C(CO2)C3)N2OC3CC1([H])C.

<Checkpoint>[The process of generating the final product from $\bf B$ is the breaking of the $\rm N-O$ bond](1)</Checkpoint>

<Checkpoint>[The structure of $\bf B$ is such that the nitrogen atom of the final product is connected to two hydroxyl groups, forming a bridged ring structure](1)</Checkpoint>

<Checkpoint>[The structural formula of $\bf B$ is CC1(C(CO2)C3)N2OC3CC1([H])C](1)</Checkpoint>

From this, it can be judged that the reaction of $\bf A$ heating to $\rm 383~K$ to convert to $\bf B$ is an intramolecular [3+2] cycloaddition reaction; the alkene in $\bf A$ undergoes [3+2] cycloaddition with the $C=N^+-O^-$ structure.

<Checkpoint>[The reaction of $\bf A$ converting to $\bf B$ is an intramolecular [3+2] cycloaddition reaction](1)</Checkpoint>

Considering stereochemistry, after the [3+2] cycloaddition, when the bridged ring structure breaks the $\rm N-O$ bond, the generated hydroxymethyl and hydroxyl group were originally on the same side of the bridged ring, so they are in the cis configuration in the final product;

<Checkpoint>[Hydroxymethyl and hydroxyl are in the cis configuration](2)</Checkpoint>

$\bf A$ is the thermodynamic product. Due to steric hindrance effects, the allyl and methyl groups in $\bf A$ should be in the trans conformation. After the [3+2] cycloaddition, the two methyl groups are therefore in the trans conformation; in the final product, they are also in the trans conformation;

<Checkpoint>[In the thermodynamic product $\bf A$, the allyl and methyl groups should be in the trans conformation](1)</Checkpoint>

<Checkpoint>[The two methyl groups are in the trans conformation](2)</Checkpoint>

The amino group and the hydroxymethyl group jointly form a five-membered ring of the bridged ring, located on the same side of the cyclohexane ring, so they should be in cis form in the final product.

<Checkpoint>[The amino group and hydroxymethyl group are in the cis configuration](2)</Checkpoint>

The product structure that satisfies these three stereochemical requirements is O[C@H]1C[C@@H](C)[C@](N)(C)[C@@H](CO)C1, so option B is correct.

<Checkpoint>[The product structure is O[C@H]1C[C@@H](C)[C@](N)(C)[C@@H](CO)C1](1)</Checkpoint>
<MultiModal>![$\bf A$结构为[O-][N+]1=C(C)C(C)CC(CC=C)O1；$\bf B$的结构为CC1(C(CO2)C3)N2OC3CC1([H])C；产物结构为O[C@H]1C[C@@H](C)[C@](N)(C)[C@@H](CO)C1。](/media/uploads/59f33657-263c-46eb-9d74-7d5489426083.png)</MultiModal>

底物很明显为亲双烯体和双烯体，因而在lewis酸$\rm SnCl_4$催化下发生D-A环加成反应，故$\bf A$结构为[O-][N+]1=C(C)C(C)CC(CC=C)O1。

<Checkpoint>[底物在lewis酸$\rm SnCl_4$催化下发生D-A环加成反应](1)</Checkpoint>

<Checkpoint>[$\bf A$结构为[O-][N+]1=C(C)C(C)CC(CC=C)O1](1)</Checkpoint>

$\bf A$加热至$\rm 383~K$转化为$\bf B$，由$\bf B$生成最终产物的过程不涉及碳原子成键的变化，因此只能为$\rm N-O$键的断裂；从而不考虑立体化学，$\bf B$的结构为最终产物的氮原子与两个羟基相连，为桥环结构，结构式为CC1(C(CO2)C3)N2OC3CC1([H])C。

<Checkpoint>[$\bf B$生成最终产物的过程为$\rm N-O$键的断裂](1)</Checkpoint>

<Checkpoint>[$\bf B$的结构为最终产物的氮原子与两个羟基相连，为桥环结构](1)</Checkpoint>

<Checkpoint>[$\bf B$结构式为CC1(C(CO2)C3)N2OC3CC1([H])C](1)</Checkpoint>

由此可判断，$\bf A$加热至$\rm 383~K$转化为$\bf B$的反应为分子内的[3+2]环加成反应；$\bf A$中的烯烃与$C=N^+-O^-$结构进行[3+2]环加成。

<Checkpoint>[$\bf A$转化为$\bf B$的反应为分子内的[3+2]环加成反应](1)</Checkpoint>

考虑立体化学，[3+2]环加成后的桥环结构断裂$\rm N-O$键后，生成的羟甲基与羟基原本位于桥环的同一侧，因此最终产物中处于顺式；

<Checkpoint>[羟甲基与羟基处于顺式](2)</Checkpoint>

$\bf A$为热力学产物，由于空间位阻效应，$\bf A$中烯丙基和甲基应当处于反式构象。在[3+2]环加成后，两个甲基因此处于反式构象；最终产物中，同样处于反式构象；

<Checkpoint>[热力学产物$\bf A$中烯丙基和甲基应当处于反式构象](1)</Checkpoint>

<Checkpoint>[两个甲基处于反式构象](2)</Checkpoint>

氨基与羟甲基共同构成桥环的五元环，位于环己环的同一侧，因此最终产物中应当处于顺式。

<Checkpoint>[氨基与羟甲基处于顺式](2)</Checkpoint>

满足这三个立体化学要求的产物结构为O[C@H]1C[C@@H](C)[C@](N)(C)[C@@H](CO)C1，因此选项B正确。

<Checkpoint>[产物结构为O[C@H]1C[C@@H](C)[C@](N)(C)[C@@H](CO)C1](1)</Checkpoint>