<MultiModal>![This figure describes two organic reactions. The first reaction: the substrate $\bf M$ is O=C(O)[C@H](CCCN=[N+]=[N-])CC1=CC=CC=C1, which reacts under the conditions of $\rm 1. Tf_2O, DCM; 2. KOH, H_2O$, and then reacts with $\rm Ac_2O, DCM$ to obtain two products $\bf A,B$. The second reaction: the substrate $\bf N$ is O=C(Cl)CCC(CN=[N+]=[N-])CC1=CC=CC=C1, yielding product $\bf C$ under the conditions of $\rm TfOH, DCM$.](/media/uploads/15f2d0e1-133d-4129-a55f-5b08919b51b9.png)</MultiModal>

For the reaction in the above figure, it is known that:

1. Products $\bf A, B, C$ are all rearrangement products.
2. $\bf A, B, C$ contain 13, 12, and 12 carbon atoms, respectively, and the number of oxygen atoms is the same.
3. In the reaction mechanism for the formation of $\bf C$, the migrating group is different from the migrating group for the formation of $\bf A$.
4. The process of generating $\bf A, B$ from $\bf M$ goes through two key charged intermediates $\bf 1, 2$, and the process of generating $\bf C$ from $\bf N$ goes through two key charged intermediates $\bf C1, C2$.

This question requires stereochemistry.

The following statement that is correct is:
<MultiModal>![本图描述了两个有机反应。第一个反应：底物$\bf M$为O=C(O)[C@H](CCCN=[N+]=[N-])CC1=CC=CC=C1，其在$\rm 1. Tf_2O, DCM; 2. KOH, H_2O$条件下反应，再与$\rm Ac_2O, DCM$反应得到两种产物$\bf A,B$。第二个反应：底物$\bf N$为O=C(Cl)CCC(CN=[N+]=[N-])CC1=CC=CC=C1，在$\rm TfOH, DCM$条件下生成产物$\bf C$。](/media/uploads/15f2d0e1-133d-4129-a55f-5b08919b51b9.png)</MultiModal>

对于上图的反应，已知：

1. 产物$\bf A,B,C$均为重排产物。
2. $\bf A,B,C$分别含有13，12，12个碳原子，且氧原子数目一致。
3. 生成$\bf C$的反应机理中，迁移的基团与生成$\bf A$的迁移基团不同。
4. 由$\bf M$生成$\bf A，B$的过程经历了两个关键带电中间体$\bf 1,2$，由$\bf N$生成$\bf C$的过程经历了两个关键带电中间体$\bf C1,C2$。

本题要求立体化学。

下列说法正确的是：
All other options are incorrect
$\bf A, B$ both contain two rings.
The stereochemical configuration of the chiral center of $\bf B$ is the S configuration.
$\bf 1$ and $\bf 2$ contain the same types of rings
$\bf C$ and $\bf B$ have inconsistent chemical formulas.
Explaining the transfer selective unavailable steric hindrance effect of generating $\bf C$
其他选项均不正确
$\bf A，B$均含有两个环
$\bf B$的手性中心构型为S构型
$\bf 1$和$\bf 2$含有的环种类相同
$\bf C$与$\bf B$化学式不一致
产生$\bf C$的迁移选择性不可利用空间位阻效应解释
<MultiModal>![clipboard_image_1755075420805](/media/uploads/93c53718-f57b-4da8-acc2-9e76b2da929b.png)</MultiModal>

First, trifluoromethanesulfonic anhydride $\rm Tf_2O$ is added to react with the carboxylate to generate a strongly leaving $\rm OTf$ group, thus activating the carboxyl group.

<Checkpoint>[Carboxyl group is activated by $\rm Tf_2O$ to generate $\rm OTf$ group](1)</Checkpoint>

Observing the structure of the substrate $\rm M$, it can be seen that the azide group in the substrate has strong nucleophilicity and can undergo intramolecular nucleophilic carboxyl group, generating a six-membered ring amide structure; this intramolecular cyclization does not change the configuration of the chiral center, so the structure of the charged intermediate $\bf 1$ is O=C1N([N+]#N)CCC[C@@H]1CC2=CC=CC=C2.

<Checkpoint>[Intramolecular nucleophilic carboxyl group of azide generates amide](1)</Checkpoint>

<Checkpoint>[Structure of $\bf 1$ is O=C1N([N+]#N)CCC[C@@H]1CC2=CC=CC=C2](1)</Checkpoint>

Then observe that this structure has a nitrogen group attached to the nitrogen atom of the six-membered ring amide, which can undergo a Curtius rearrangement reaction to generate an isocyanate; therefore, the structure of the charged intermediate $\bf 2$ is O=C=[N+]1CCC[C@@H]1CC2=CC=CC=C2. $\bf 1$ and $\bf 2$ contain different types of rings, option D is incorrect.

<Checkpoint>[Curtius rearrangement reaction occurs to generate isocyanate](1)</Checkpoint>

<Checkpoint>[Structure of $\bf 2$ is O=C=[N+]1CCC[C@@H]1CC2=CC=CC=C2](1)</Checkpoint>

The isocyanate is attacked by a hydroxyl group under basic conditions to generate an N-carboxylic acid, which easily decarboxylates to form a secondary amine; then acetic anhydride $\rm Ac_2O$ is added to acylate the amine, so the final product is O=C(C)N1CCC[C@@H]1CC2=CC=CC=C2. Since acetic anhydride participates in the reaction, there is one more carbon atom, so this structure is product $\bf A$.

<Checkpoint>[Acetic anhydride $\rm Ac_2O$ acylates the amine](1)</Checkpoint>

<Checkpoint>[$\bf A$ structure is O=C(C)N1CCC[C@@H]1CC2=CC=CC=C2](1)</Checkpoint>

$\bf B$ has the same number of carbon atoms as intermediate $\bf 2$, so it can only be an intramolecular reaction; observing the substrate, it can be seen that the substrate can undergo an aromatic electrophilic reaction, and the benzene ring nucleophilic isocyanate generates a six-membered ring; therefore, the structure of $\bf B$ is O=C1C2=CC=CC=C2C[C@]3([H])N1CCC3. This step also does not change the chiral center, the configuration of the chiral site is R, option C is incorrect. $\bf B$ has three rings, option B is incorrect.

<Checkpoint>[Generation of $\bf B$ is an intramolecular reaction](1)</Checkpoint>

<Checkpoint>[Benzene ring nucleophilic isocyanate generates a six-membered ring](1)</Checkpoint>

<Checkpoint>[$\bf B$ structure is O=C1C2=CC=CC=C2C[C@]3([H])N1CCC3](1)</Checkpoint>

The structure of $\bf N$ is similar. At the beginning of the reaction, $\rm TfOH$ replaces the acyl chloride, and then the azide group nucleophilically attacks the ring, and the resulting intermediate $\bf C1$ is O=C1CCC(CN1[N+]#N)CC2=CC=CC=C2.

<Checkpoint>[Intermediate $\bf C1$ is O=C1CCC(CN1[N+]#N)CC2=CC=CC=C2](1)</Checkpoint>

Since the migrating group that produces $\bf C$ is different, the $\rm C-C$ bond on the carbonyl side migrates in the Curtius rearrangement, so the carbonyl side does not migrate here, but the alkyl group on the other side migrates to obtain the imine cation structure; this intermediate $\bf C2$ is C=[N+](C(CC1)CC2=CC=CC=C2)C1=O.

<Checkpoint>[Alkyl group on the other side migrates to obtain the imine cation structure](1)</Checkpoint>

<Checkpoint>[$\bf C2$ is C=[N+](C(CC1)CC2=CC=CC=C2)C1=O](1)</Checkpoint>

Observing the number of carbon atoms, it can be seen that the reaction to generate $\bf C$ is still an intramolecular reaction, so it can only be that the aromatic ring nucleophilic imine cation obtains a six-membered ring, and the final product $\bf C$ structure is O=C1N2CC3=CC=CC=C3CC2CC1. $\bf C$ and $\bf B$ have the same chemical formula, option E is incorrect.

<Checkpoint>[Reaction to generate $\bf C$ is an intramolecular reaction](1)</Checkpoint>

<Checkpoint>[$\bf C$ structure is O=C1N2CC3=CC=CC=C3CC2CC1](1)</Checkpoint>

After the substrate rearranges, it may generate two kinds of intermediates, isocyanate cation or imine cation, but the isocyanate cation here cannot form a stable reaction conformation with the benzene ring, so the corresponding product cannot be obtained, while the imine cation is very easy to react with the benzene ring, so the final equilibrium shifts to the direction of generating the imine cation. Option F is correct.

<Checkpoint>[Isocyanate cation cannot form a stable reaction conformation with the benzene ring](1)</Checkpoint>

<Checkpoint>[Imine cation is very easy to react with the benzene ring, and the equilibrium shifts to the direction of generating the imine cation](1)</Checkpoint>

Therefore, options B-F are all incorrect, and option A is correct.
<MultiModal>![clipboard_image_1755075420805](/media/uploads/93c53718-f57b-4da8-acc2-9e76b2da929b.png)</MultiModal>

首先加入三氟甲磺酸酐$\rm Tf_2O$与羧酸根反应生成强离去性的$\rm OTf$基团，因此羧基被活化。

<Checkpoint>[羧基被$\rm Tf_2O$活化生成$\rm OTf$基团](1)</Checkpoint>

观察底物$\rm M$的结构可知，底物中叠氮基团具有强亲核性，可以发生分子内亲核羧基，生成六元环酰胺结构；该分子内成环不改变手性中心构型，因此带电中间体$\bf 1$的结构为O=C1N([N+]#N)CCC[C@@H]1CC2=CC=CC=C2。

<Checkpoint>[叠氮基分子内亲核羧基生成酰胺](1)</Checkpoint>

<Checkpoint>[$\bf 1$的结构为O=C1N([N+]#N)CCC[C@@H]1CC2=CC=CC=C2](1)</Checkpoint>

之后观察该结构为六元环酰胺的氮原子上连有氮气基团，可以发生Curtius重排反应生成异氰酸酯；故带电中间体$\bf 2$的结构为O=C=[N+]1CCC[C@@H]1CC2=CC=CC=C2。$\bf 1$与$\bf 2$含有环种类不同，选项D错误。

<Checkpoint>[发生Curtius重排反应生成异氰酸酯](1)</Checkpoint>

<Checkpoint>[$\bf 2$的结构为O=C=[N+]1CCC[C@@H]1CC2=CC=CC=C2](1)</Checkpoint>

异氰酸酯在碱性条件下被羟基进攻生成N-羧酸，该结构容易脱羧生成二级胺；之后加入乙酸酐$\rm Ac_2O$将胺进行酰基化，因此最终产物为O=C(C)N1CCC[C@@H]1CC2=CC=CC=C2，由于乙酸酐参与反应多出一个碳原子，因此该结构为产物$\bf A$。

<Checkpoint>[乙酸酐$\rm Ac_2O$将胺进行酰基化](1)</Checkpoint>

<Checkpoint>[$\bf A$结构为O=C(C)N1CCC[C@@H]1CC2=CC=CC=C2](1)</Checkpoint>

$\bf B$的碳原子数与中间体$\bf 2$相同，因此只能为分子内反应；观察底物可知底物可以发生芳香亲电反应，苯环亲核异氰酸酯生成六元环；因此$\bf B$结构为O=C1C2=CC=CC=C2C[C@]3([H])N1CCC3。该步骤同样对手性中心无改变，手性位点构型为R，选项C错误。$\bf B$有三个环，选项B错误。

<Checkpoint>[生成$\bf B$为分子内反应](1)</Checkpoint>

<Checkpoint>[苯环亲核异氰酸酯生成六元环](1)</Checkpoint>

<Checkpoint>[$\bf B$结构为O=C1C2=CC=CC=C2C[C@]3([H])N1CCC3](1)</Checkpoint>

$\bf N$的结构类似，反应刚开始为$\rm TfOH$对酰氯进行取代后被叠氮基亲核进攻成环，产生的中间体$\bf C1$为O=C1CCC(CN1[N+]#N)CC2=CC=CC=C2。

<Checkpoint>[中间体$\bf C1$为O=C1CCC(CN1[N+]#N)CC2=CC=CC=C2](1)</Checkpoint>

由于产生$\bf C$的迁移基团不同，Curtius重排中迁移羰基一侧的$\rm C-C$键，因此此处不迁移羰基一侧，而是迁移另一侧的烷基，得到亚胺正离子结构；该中间体$\bf C2$为C=[N+](C(CC1)CC2=CC=CC=C2)C1=O。

<Checkpoint>[迁移另一侧的烷基，得到亚胺正离子结构](1)</Checkpoint>

<Checkpoint>[$\bf C2$为C=[N+](C(CC1)CC2=CC=CC=C2)C1=O](1)</Checkpoint>

观察碳原子数可知生成$\bf C$的反应仍旧为分子内反应，因此只能是芳环亲核亚胺正离子得到六元环，最终产物$\bf C$结构为O=C1N2CC3=CC=CC=C3CC2CC1。$\bf C$与$\bf B$化学式一致，选项E错误。

<Checkpoint>[生成$\bf C$的反应为分子内反应](1)</Checkpoint>

<Checkpoint>[$\bf C$结构为O=C1N2CC3=CC=CC=C3CC2CC1](1)</Checkpoint>

底物重排后可能生成异氰酸酯正离子或亚胺正离子两种中间体，但此处的异氰酸酯正离子与苯环间并不能形成稳定的反应构象从而得不到相应的产物，而亚胺正离子则很容易与苯环发生反应，因此最后平衡向生成亚胺正离子的方向移动。选项F正确。

<Checkpoint>[异氰酸酯正离子与苯环间并不能形成稳定的反应构象](1)</Checkpoint>

<Checkpoint>[亚胺正离子则很容易与苯环发生反应，平衡向生成亚胺正离子的方向移动](1)</Checkpoint>

因此选项B-F均错误，选项A正确。