In situ activation of ω-azidocarboxylic acids can induce some interesting reactions. 5-Azido-2-benzylpentanoic acid reacts with oxalyl chloride and tin(IV) chloride to form intermediate $\mathbf C$, which subsequently undergoes two-step reactions to yield intermediate $\mathbf D$ and product $\mathbf E$.

<MultiModal>![The image shows a multistep reaction: O=C(O)C(CCCN=[N+]=[N-])CC1=CC=CC=C1>ClC(C(Cl)=O)=O.Cl[Sn](Cl)(Cl)Cl>[**C**],[**C**]>[**D**],[**D**]>[**E**]
5-Azido-2-benzylpentanoic acid reacts with oxalyl chloride and tin(IV) chloride to form intermediate C, which subsequently undergoes two-step reactions to yield intermediate D and product E](/media/uploads/327affe4-d6d8-4a18-8b11-19ed6ccf8282.png)</MultiModal>

A slight structural modification of the substrate yields 5-azido-4-benzylpentanoic acid. This compound reacts with oxalyl chloride and trifluoromethanesulfonic acid to form the final intermediate $\mathbf F$, which then undergoes one more reaction to produce product $\mathbf G$, distinct from $\mathbf E$.

<MultiModal>![The image shows a multistep reaction: O=C(O)CCC(CN=[N+]=[N-])CC1=CC=CC=C1>ClC(C(Cl)=O)=O.O=S(C(F)(F)F)(O)=O>[**F**],[**F**]>[**G**]
5-Azido-4-benzylpentanoic acid reacts with oxalyl chloride and trifluoromethanesulfonic acid to form the final intermediate F, which then undergoes one more reaction to produce product G](/media/uploads/3ceb6f2b-e777-4aba-ad6c-74923bf7b3c8.png)</MultiModal>

None of the intermediates contain $\mathrm {Cl}$. The following statements are correct:
对$\omega$-叠氮羧酸进行原位活化能够引起一些有趣的反应。5-叠氮基-2-苄基戊酸在草酰氯和四氯化锡的条件下反应得到中间体$\mathbf C$，之后又经过经过两步反应得到中间体$\mathbf D$和产物$\mathbf E$

<MultiModal>![图片中为多步反应：O=C(O)C(CCCN=[N+]=[N-])CC1=CC=CC=C1>ClC(C(Cl)=O)=O.Cl[Sn](Cl)(Cl)Cl>[**C**],[**C**]>[**D**],[**D**]>[**E**]
5-叠氮基-2-苄基戊酸在草酰氯和四氯化锡的条件下反应得到中间体C，之后又经过经过两步反应得到中间体D和产物E](/media/uploads/327affe4-d6d8-4a18-8b11-19ed6ccf8282.png)</MultiModal>


对底物的结构稍作调整可以得到5-叠氮基-4-苄基戊酸，该反应物在草酰氯和三氟甲磺酸的条件下反应得到最后一个中间体$\mathbf F$，再经过一步反应得到产物$\mathbf G$，产物$\mathbf G$与$\mathbf E$不同


<MultiModal>![图片中为多步反应：O=C(O)CCC(CN=[N+]=[N-])CC1=CC=CC=C1>ClC(C(Cl)=O)=O.O=S(C(F)(F)F)(O)=O>[**F**],[**F**]>[**G**]
5-叠氮基-4-苄基戊酸在草酰氯和三氟甲磺酸的条件下反应得到最后一个中间体F，在经过一步反应得到产物G](/media/uploads/3ceb6f2b-e777-4aba-ad6c-74923bf7b3c8.png)</MultiModal>


所有的中间体都不含$\mathrm {Cl}$,以下说法正确的是
The intermediate $\mathbf C$ contains a carboxyl group
The intermediate $\mathbf D$ is a rearrangement product containing two six-membered rings.
The product $\mathbf E$ is a secondary amine.
The intermediate $\mathbf F$ has an isocyanate structure.
The product $\mathbf G$ contains three rings, with the carbonyl group located on the five-membered ring.
The second reaction can form two different rearrangement intermediates, among which the isocyanate cation intermediate reacts with the benzene ring to yield a more stable product.
None of the above options are correct
中间体$\mathbf C$含有羧基
中间体$\mathbf D$是重排产物，含有两个六元环
产物$\mathbf E$为仲胺
中间体$\mathbf F$具有异氰酸酯结构
产物$\mathbf G$含有三个环，其中羰基在五元环上
第二个反应可以形成两种不同的重排中间体，其中异氰酸酯正离子中间体和苯环反应可以得到更稳定的产物
以上选项均不正确
In the first reaction, the azide and carboxylic acid undergo intramolecular cyclization to yield intermediate $\mathbf C$,

<Checkpoint>[The azido group and carboxylic acid undergo intramolecular cyclization to yield intermediate $\mathbf C$, A is incorrect](1)</Checkpoint>

The structural formula of intermediate $\mathbf C$ is O=C1N([N+]#N)CCCC1CC2=CC=CC=C2, <MultiModal>![O=C1N([N+]#N)CCCC1CC2=CC=CC=C2](/media/uploads/b9a9ddac-7730-43dd-8d3a-a9d38c954c3a.png)</MultiModal>

Intermediate $\mathbf C$ undergoes rearrangement to form intermediate $\mathbf D$, which has an isocyanate structure containing a five-membered ring,

<Checkpoint>[Intermediate $\mathbf D$ has an isocyanate structure containing a five-membered ring, B is incorrect](1)</Checkpoint>

The structural formula of intermediate $\mathbf D$ is O=C=[N+]1CCCC1CC2=CC=CC=C2, <MultiModal>![O=C=[N+]1CCCC1CC2=CC=CC=C2](/media/uploads/66b22e7e-bf5b-4252-af96-49f4d6ce2195.png)</MultiModal>

The benzene ring reacts with the isocyanate, undergoing intramolecular cyclization to yield product $\mathbf E$. Since water is not mentioned in the reaction conditions, hydrolysis to produce carbon dioxide and a secondary amine does not occur,

<Checkpoint>[The benzene ring reacts with the isocyanate to yield product $\mathbf E$. Since water is not mentioned in the reaction conditions, a secondary amine is not obtained, C is incorrect](1)</Checkpoint>

The structure of product $\mathbf E$ is C1=CC2=C(C=C1)CN3C(CCC3=O)C2C1=CC2=C(C=C1)C(=O)N3CCCC3C2, <MultiModal>![C1=CC2=C(C=C1)C(=O)N3CCCC3C2](/media/uploads/b254953b-3334-4943-a07f-3cf54adfdd9a.png)</MultiModal>

In the second reaction, the azide and carboxylic acid react and undergo rearrangement to form the imine intermediate $\mathbf F$,

<Checkpoint>[Intermediate $\mathbf F$ has an imine structure, D is incorrect](1)</Checkpoint>

The difference between product $\mathbf E$ and product $\mathbf G$ lies in the preferential migration of the more substituted carbon atom during the rearrangement,

<Checkpoint>[The difference between product $\mathbf E$ and product $\mathbf G$ lies in the preferential migration of the more substituted carbon atom during the rearrangement](1)</Checkpoint>

Since nitrogen gas is released in this reaction, the reaction is irreversible, and G is incorrect,

<Checkpoint>[Since nitrogen gas is released in this reaction, the reaction is irreversible](1)</Checkpoint>

Option F is incorrect.

The structure of intermediate $\mathbf F$ is C=C1CCC([N+]1=C)CC2=CC=CC=C2, <MultiModal>![O=C1CCC([N+]1=C)CC2=CC=CC=C2](/media/uploads/b2b78167-1870-4b6d-b564-048921e2ed52.png)</MultiModal>

The benzene ring reacts with the imine, undergoing intramolecular cyclization to ultimately yield product $\mathbf G$. Product $\mathbf G$ has three rings, and the carbonyl group is located on the five-membered ring,

<Checkpoint>[Product $\mathbf G$ has three rings, and the carbonyl group is located on the five-membered ring, so E is correct](1)</Checkpoint>

The structure of product $\mathbf G$ is O=C1CCC([N+]1=C)CC2=CC=CC=C2, <MultiModal>![C1=CC2=C(C=C1)CN3C(CCC3=O)C2](/media/uploads/01b56cd0-7c51-441e-9cc6-497bbfa3a157.png)</MultiModal>
在第一个反应中，叠氮和羧酸反应发生分子内环化，得到中间体$\mathbf C$，

<Checkpoint>[叠氮基团和羧酸反应发生分子内环化，得到中间体$\mathbf C$，A错误](1)</Checkpoint>

中间体$\mathbf C$的结构式为O=C1N([N+]#N)CCCC1CC2=CC=CC=C2，<MultiModal>![O=C1N([N+]#N)CCCC1CC2=CC=CC=C2](/media/uploads/b9a9ddac-7730-43dd-8d3a-a9d38c954c3a.png)</MultiModal>


中间体$\mathbf C$发生重排得到中间体$\mathbf D$，该中间体具有异氰酸酯结构，含有五元环，


<Checkpoint>[中间体$\mathbf D$具有异氰酸酯结构，含有五元环，B错误](1)</Checkpoint>


中间体$\mathbf D$的结构式为O=C=[N+]1CCCC1CC2=CC=CC=C2，<MultiModal>![O=C=[N+]1CCCC1CC2=CC=CC=C2](/media/uploads/66b22e7e-bf5b-4252-af96-49f4d6ce2195.png)</MultiModal>

苯环和异氰酸酯发生反应，分子内环化，得到产物$\mathbf E$，在该反应条件中没有提到水，不会水解产生二氧化碳和仲氨


<Checkpoint>[苯环和异氰酸酯发生反应，得到产物$\mathbf E$，在该反应条件中没有提到水，不会得到仲氨，C错误](1)</Checkpoint>

产物$\mathbf E$的结构为C1=CC2=C(C=C1)CN3C(CCC3=O)C2C1=CC2=C(C=C1)C(=O)N3CCCC3C2，<MultiModal>![C1=CC2=C(C=C1)C(=O)N3CCCC3C2](/media/uploads/b254953b-3334-4943-a07f-3cf54adfdd9a.png)</MultiModal>

在第二个反应中，叠氮和羧酸发生反应后进行重排，得到亚胺中间体$\mathbf F$，

<Checkpoint>[中间体$\mathbf F$具有亚胺结构，D错误](1)</Checkpoint>

产物$\mathbf E$和产物$\mathbf G$不同的原因在于发生重排时，优先迁移多取代的碳原子

<Checkpoint>[产物$\mathbf E$和产物$\mathbf G$不同的原因在于发生重排时，优先迁移多取代的碳原子](1)</Checkpoint>

由于该反应离去氮气，因此该反应不可逆，G错误

<Checkpoint>[由于该反应离去氮气，因此该反应不可逆](1)</Checkpoint>

F选项错误

中间体$\mathbf F$的结构为C=C1CCC([N+]1=C)CC2=CC=CC=C2，<MultiModal>![O=C1CCC([N+]1=C)CC2=CC=CC=C2](/media/uploads/b2b78167-1870-4b6d-b564-048921e2ed52.png)</MultiModal>

苯环和亚胺发生反应，分子内成环，最终得到产物$\mathbf G$，产物$\mathbf G$具有三个环，并且羰基在五元环上

<Checkpoint>[产物$\mathbf G$具有三个环，并且羰基在五元环上，所以E正确](1)</Checkpoint>
产物$\mathbf G$的结构为O=C1CCC([N+]1=C)CC2=CC=CC=C2,<MultiModal>![C1=CC2=C(C=C1)CN3C(CCC3=O)C2](/media/uploads/01b56cd0-7c51-441e-9cc6-497bbfa3a157.png)</MultiModal>