<MultiModal>![An organic reaction where the substrate is C=CCC(C(OC)=O)(C(OC)=O)C(C(C)C)C[N+]([O-])=O, denoted as A. A reacts under the conditions of DBU, $Ag_2O$, $I_2$, and THF to generate B. B then reacts under the conditions of DBU and THF to produce C and D, both with the molecular formula $C_{13}H_{19}O_6N$.](/media/uploads/fcd9643e-530a-469c-936a-ee62e3aefe9a.png)</MultiModal>  

It is known that B contains a five-membered ring, and C and D share the same molecular formula, $C_{13}H_{19}O_6N$, with no carbon-carbon double bonds. However, D can react in a methanol-water solution of HCl to yield a nitrogen-free compound E. Which of the following statements about the structures of B, C, D, and E (excluding stereochemistry) is correct?
<MultiModal>![一个有机反应，底物是C=CCC(C(OC)=O)(C(OC)=O)C(C(C)C)C[N+]([O-])=O，记为A。A在DBU,$Ag_2O$,$I_2$和THF的条件下生成B。B在DBU和THF的条件下生成C和D，而C和D的分子式都是$C_{13}H_{19}O_6N$。](/media/uploads/fcd9643e-530a-469c-936a-ee62e3aefe9a.png)</MultiModal>

已知B含有五元环，C和D的分子式相同，均为 $C _ { 1 3 } H _ { 1 9 } O _ { 6 } N$ 且均不含碳碳双键，但D可在HCI的甲醇水溶液中反应得到不含氮的化合物E。关于B、C、D、E的结构(不要求立体化学)说法正确的选项是什么？
The degree of unsaturation of B and A is different.
The degree of unsaturation of C is the same as that of B.
D has a structure of existence and ring
D contains a keto-oxime structure
The molecular formula of E is $C_{13}H_{18}O_6$
None of the above options are correct
B和A的不饱和度不同
C的不饱和度和B相同
D存在并环结构
D存在酮肟结构
E的分子式是$C_{13}H_{18}O_6$
以上选项均不对
B=<MultiModal>![ICC1CC(C(OC)=O)(C(OC)=O)C(C1[N+]([O-])=O)C(C)C](/media/uploads/fb28a093-1df1-449d-9d0c-97a29b3cc430.png)</MultiModal>

C=<MultiModal>![CC(C)C1C2(CC2CC1(C(=O)OC)C(=O)OC)[N+](=O)[O-]](/media/uploads/94a639e9-24b0-4c53-84fb-ccd3dba5ac73.png)</MultiModal>

D=<MultiModal>![CC(C)C1C2=[N+]([O-])OCC2CC1(C(=O)OC)C(=O)OC](/media/uploads/2354d80e-e6b4-48d2-83b6-cc30966c6eaa.png)</MultiModal>

E=<MultiModal>![CC(C)C1C(=O)C(CC1(C(=O)OC)C(=O)OC)CO](/media/uploads/54225a6b-6a79-4b74-9adb-6f38a66ed784.png)</MultiModal>

Under alkaline conditions, the α-position of the nitro group in A generates a carbanion, which attacks the carbon-carbon double bond activated by $I_2$, leading to intramolecular cyclization to form B.

<Checkpoint>[Intramolecular cyclization to form B](0.5)</Checkpoint>

Under alkaline conditions, the α-position of the nitro group in B generates a carbanion, which attacks the intramolecular carbon-iodine bond, resulting in intramolecular formation of a three-membered ring to yield C or D. Alternatively, the negatively charged oxygen in the nitro group attacks the carbon-iodine bond, leading to intramolecular formation of a five-membered ring to yield C or D. The product of the former reaction cannot yield a nitrogen-free compound in a methanol-water solution of HCl, so C is obtained. The product of the latter reaction, due to the presence of an electrophilic carbon-nitrogen double bond, can yield a nitrogen-free compound in a methanol-water solution of HCl, resulting in D.

<Checkpoint>[Intramolecular formation of a three-membered ring to yield C](1)</Checkpoint>

<Checkpoint>[Intramolecular formation of a five-membered ring to yield D](1)</Checkpoint>

The hydrolysis process of D converts the carbon-nitrogen double bond into a carbon-oxygen double bond, followed by the elimination of nitrous acid to generate E, with the molecular formula $C_{13}H_{20}O_6$, making option E incorrect.

<Checkpoint>[The molecular formula of E is $C_{13}H_{20}O_6$](1)</Checkpoint>

The process from A to B involves the loss of one π bond but the formation of a ring, leaving the degree of unsaturation unchanged. The process from B to C involves the formation of a ring, increasing the degree of unsaturation by 1. Options A and B are incorrect.

<Checkpoint>[A and B have the same degree of unsaturation](0.5)</Checkpoint>

<Checkpoint>[C has one more degree of unsaturation than B](0.5)</Checkpoint>

D features a fused five-five structure without an oxime moiety, making option C correct and option D incorrect.
B=<MultiModal>![ICC1CC(C(OC)=O)(C(OC)=O)C(C1[N+]([O-])=O)C(C)C](/media/uploads/fb28a093-1df1-449d-9d0c-97a29b3cc430.png)</MultiModal>

C=<MultiModal>![CC(C)C1C2(CC2CC1(C(=O)OC)C(=O)OC)[N+](=O)[O-]](/media/uploads/94a639e9-24b0-4c53-84fb-ccd3dba5ac73.png)</MultiModal>

D=<MultiModal>![CC(C)C1C2=[N+]([O-])OCC2CC1(C(=O)OC)C(=O)OC](/media/uploads/2354d80e-e6b4-48d2-83b6-cc30966c6eaa.png)</MultiModal>

E=<MultiModal>![CC(C)C1C(=O)C(CC1(C(=O)OC)C(=O)OC)CO](/media/uploads/54225a6b-6a79-4b74-9adb-6f38a66ed784.png)</MultiModal>

A在碱性条件下硝基的$\alpha$位生成碳负离子，碳负离子进攻被$I_2$活化的碳碳双键，分子内成环形成B。

<Checkpoint>[分子内成环形成B](0.5)</Checkpoint>

B在碱性条件下硝基的$\alpha$位生成碳负离子，进攻分子内的碳碘键，分子内成三元环形成C或D；另一种反应方式是硝基中带负电的氧进攻碳碘键，分子内成五元环形成C或D。前一种反应产物无法在HCI的甲醇水溶液中反应得到不含氮的化合物，所以得到的是C；后一种反应产物因为有亲电的碳氮双键，可以在HCI的甲醇水溶液中反应得到不含氮的化合物，得到的是D。


<Checkpoint>[分子内成三元环形成C](1)</Checkpoint>


<Checkpoint>[分子内成五元环形成D](1)</Checkpoint>

D的水解过程将碳氮双键变为碳氧双键，再脱去亚硝酸分子，生成E，其分子式为$C_{13}H_{20}O_6$，E选项错误。

<Checkpoint>[E的分子式为$C_{13}H_{20}O_6$](1)</Checkpoint>

A到B的过程减少了一个$\pi$键，但成了一个环，不饱和度不变，B到C的过程成了一个环，不饱和度提升1。A,B选项错误。

<Checkpoint>[A和B的不饱和度相同](0.5)</Checkpoint>

<Checkpoint>[C比B的不饱和度多1](0.5)</Checkpoint>

D存在五并五结构，没有酮肟结构，C选项正确，D选项错误。