<MultiModal>![The image describes a one-pot organic cascade reaction. Substrate $\bf A$ first reacts with CC(C)=O, $NaOEt$; then reacts with $NaBH_4, H_2O$; and finally reacts with $MeNH_2, HCHO, H_2O, AcOH$ to obtain the product CN1C[C@@]([C@@H]2[C@]3([N+]([O-])=O)C1)([N+]([O-])=O)[C@H](OC(C)C2)CC3=O](/media/uploads/0112cfac-ec06-4c17-b58e-4baed88c8ad8.png)</MultiModal>

Which of the following options is correct regarding the structure of the unknown structure $\bf A$ in the above reaction:
<MultiModal>![图片描述了一步有机串联反应。底物$\bf A$先与CC(C)=O，$NaOEt$反应；再与$NaBH_4, H_2O$反应；最后与$MeNH_2, HCHO, H_2O, AcOH$反应，得到产物CN1C[C@@]([C@@H]2[C@]3([N+]([O-])=O)C1)([N+]([O-])=O)[C@H](OC(C)C2)CC3=O](/media/uploads/0112cfac-ec06-4c17-b58e-4baed88c8ad8.png)</MultiModal>

下列选项中，关于上图反应的未知结构$\bf A$的结构，正确的是：
All other options are incorrect
<MultiModal>![O=C1C([N+]([O-])=O)CC([N+]([O-])=O)C=C1](/media/uploads/c6bee7a3-b731-4c5c-946d-dabcd7a6d3fc.png)</MultiModal>
<MultiModal>![O=C1C([N+]([O-])=O)=CC([N+]([O-])=O)C=C1](/media/uploads/494d5a5f-f6c3-47d6-aa92-60ab1ceedcb9.png)</MultiModal>
<MultiModal>![OC1C([N+]([O-])=O)=CC([N+]([O-])=O)C=C1](/media/uploads/5a3b2958-2cc9-4f61-80ca-a2a69e0173f9.png)</MultiModal>
<MultiModal>![OC1=CC=C([N+]([O-])=O)C=C1[N+]([O-])=O](/media/uploads/248c140a-196f-4ee5-b804-3346cbeba07c.png)</MultiModal>
<MultiModal>![OC1=C([N+]([O-])=O)C=CC=C1[N+]([O-])=O](/media/uploads/0ba42a3f-8fd9-411f-97f6-49f7d44f5e8c.png)</MultiModal>
其他选项均不正确
<MultiModal>![O=C1C([N+]([O-])=O)CC([N+]([O-])=O)C=C1](/media/uploads/c6bee7a3-b731-4c5c-946d-dabcd7a6d3fc.png)</MultiModal>
<MultiModal>![O=C1C([N+]([O-])=O)=CC([N+]([O-])=O)C=C1](/media/uploads/494d5a5f-f6c3-47d6-aa92-60ab1ceedcb9.png)</MultiModal>
<MultiModal>![OC1C([N+]([O-])=O)=CC([N+]([O-])=O)C=C1](/media/uploads/5a3b2958-2cc9-4f61-80ca-a2a69e0173f9.png)</MultiModal>
<MultiModal>![OC1=CC=C([N+]([O-])=O)C=C1[N+]([O-])=O](/media/uploads/248c140a-196f-4ee5-b804-3346cbeba07c.png)</MultiModal>
<MultiModal>![OC1=C([N+]([O-])=O)C=CC=C1[N+]([O-])=O](/media/uploads/0ba42a3f-8fd9-411f-97f6-49f7d44f5e8c.png)</MultiModal>
The last step in the formation of the product is obviously a Mannich reaction; the tertiary amine unit $\rm{-CH_2-NCH_3-CH_2-}$ in the product is the structure obtained from the Mannich reaction, so the first step of retrosynthesis requires cleaving this tertiary amine fragment to generate a dicarbanion, O=C1[C-]([N+]([O-])=O)[C@H]2[C-]([N+]([O-])=O)[C@H](OC(C)C2)C1.

<Checkpoint>[The last step in the formation of the product is a Mannich reaction](1)</Checkpoint>

<Checkpoint>[The tertiary amine unit $\rm{-CH_2-NCH_3-CH_2-}$ in the product is the structure obtained from the Mannich reaction](1)</Checkpoint>

<Checkpoint>[Cleave this tertiary amine fragment to generate a dicarbanion with the structure O=C1[C-]([N+]([O-])=O)[C@H]2[C-]([N+]([O-])=O)[C@H](OC(C)C2)C1](1)</Checkpoint>

In fact, the negative charge of the carbanion can be transferred to the nitro group, i.e., the $\rm C=NO_2^-$ double bond structure. Therefore, the structure cleaved in the first step is O=C(C[C@@H]1/C([C@H]/2CC(C)O1)=[N+]([O-])\[O-])C2=[N+]([O-])\[O-].

<Checkpoint>[The structure cleaved in the first step is O=C(C[C@@H]1/C([C@H]/2CC(C)O1)=[N+]([O-])\[O-])C2=[N+]([O-])\[O-]](1)</Checkpoint>

In the next step, sodium borohydride was added. Since it is obvious that the $\rm -O-CHCH_3-CH_2-$ six-membered ring fragment of the product comes from the initially added acetone, consider that sodium borohydride is used to reduce the carbonyl group; however, because the product contains a carbonyl group, the carbonyl group of the product can only be reduced when it is an α,β-unsaturated ketone structure to prevent it from being reduced. Therefore, after the carbonyl group of the acetone structure is reduced to an alcohol hydroxyl group, an intramolecular Michael addition occurs, and the nucleophilic attack on the α,β-unsaturated ketone forms a six-membered ring, which is logical. Therefore, the structure cleaved further is O=C(C=C/C(C/1CC(C)=O)=[N+]([O-])\[O-])C1=[N+]([O-])/[O-].

<Checkpoint>[The $\rm -O-CHCH_3-CH_2-$ six-membered ring fragment of the product comes from acetone](1)</Checkpoint>

<Checkpoint>[Sodium borohydride is used to reduce the carbonyl group](1)</Checkpoint>

<Checkpoint>[After the carbonyl group of acetone is reduced to an alcohol hydroxyl group, an intramolecular Michael addition occurs, and a nucleophilic attack on the α,β-unsaturated ketone forms a six-membered ring](1)</Checkpoint>

<Checkpoint>[The structure cleaved further is O=C(C=C/C(C/1CC(C)=O)=[N+]([O-])\[O-])C1=[N+]([O-])/[O-]](1)</Checkpoint>

The combination of the acetone structure is obviously achieved by converting acetone into an enolate under sodium alkoxide conditions for nucleophilic attack, so cleaving the acetone structure yields a six-membered ring carbocation, with the structure O=C(C=C/C([CH+]/1)=[N+]([O-])\[O-])C1=[N+]([O-])/[O-].

<Checkpoint>[Cleaving the acetone structure yields a six-membered ring carbocation, with the structure O=C(C=C/C([CH+]/1)=[N+]([O-])\[O-])C1=[N+]([O-])/[O-]](1)</Checkpoint>

This structure now carries a negative charge and needs to obtain a hydrogen ion; it is finally found that $\bf A$ contains four hydrogens and can tautomerize to a benzene ring structure, while the nucleophilic attack of the anion generated by acetone is an aromatic nucleophilic substitution reaction.

<Checkpoint>[$\bf A$ contains four hydrogens and can tautomerize to a benzene ring structure](1)</Checkpoint>

<Checkpoint>[The nucleophilic attack of the anion generated by acetone is an aromatic nucleophilic substitution reaction](1)</Checkpoint>

Therefore, the structure of $\bf A$ is OC1=CC=C([N+]([O-])=O)C=C1[N+]([O-])=O, and option E is correct.

<Checkpoint>[The structure of $\bf A$ is OC1=CC=C([N+]([O-])=O)C=C1[N+]([O-])=O](2)</Checkpoint>

<MultiModal>![This figure shows the intermediates of the product being cleaved step by step from left to right. Cleavage of the tertiary amine fragment generates a dicarbanion, with the structure O=C1[C-]([N+]([O-])=O)[C@H]2[C-]([N+]([O-])=O)[C@H](OC(C)C2)C1; tautomerization, to O=C(C[C@@H]1/C([C@H]/2CC(C)O1)=[N+]([O-])\[O-])C2=[N+]([O-])\[O-]; then cleavage of the six-membered ring, with the structure O=C(C=C/C(C/1CC(C)=O)=[N+]([O-])\[O-])C1=[N+]([O-])/[O-]; cleavage of the acetone structure yields a six-membered ring carbocation, with the structure O=C(C=C/C([CH+]/1)=[N+]([O-])\[O-])C1=[N+]([O-])/[O-]; finally, tautomerization yields the structure of $\bf A$ as OC1=CC=C([N+]([O-])=O)C=C1[N+]([O-])=O.](/media/uploads/c9f2543e-1148-4a92-9612-0a98d9e9f317.png)</MultiModal>
生成产物的最后一步很显然为Mannich反应；产物中的三级胺单元$\rm{-CH_2-NCH_3-CH_2-}$为Mannich反应得到的结构，因此逆合成第一步需要切断该三级胺片段，产生双碳负离子，O=C1[C-]([N+]([O-])=O)[C@H]2[C-]([N+]([O-])=O)[C@H](OC(C)C2)C1。

<Checkpoint>[生成产物的最后一步为Mannich反应](1)</Checkpoint>

<Checkpoint>[产物中的三级胺单元$\rm{-CH_2-NCH_3-CH_2-}$为Mannich反应得到的结构](1)</Checkpoint>

<Checkpoint>[切断该三级胺片段，产生双碳负离子，结构为O=C1[C-]([N+]([O-])=O)[C@H]2[C-]([N+]([O-])=O)[C@H](OC(C)C2)C1](1)</Checkpoint>

实际上碳负离子的负电荷可以转移到硝基上，即$\rm C=NO_2^-$双键结构。因此第一步切断的结构为O=C(C[C@@H]1/C([C@H]/2CC(C)O1)=[N+]([O-])\[O-])C2=[N+]([O-])\[O-]。

<Checkpoint>[第一步切断三级胺的结构为O=C(C[C@@H]1/C([C@H]/2CC(C)O1)=[N+]([O-])\[O-])C2=[N+]([O-])\[O-]](1)</Checkpoint>

再向前一步加入了硼氢化钠，由于很明显可以看出产物的$\rm -O-CHCH_3-CH_2-$六元环片段来源于最开始加入的丙酮，考虑硼氢化钠用于还原羰基；但因为产物含有羰基，因此产物的羰基被还原时只能为α，β-不饱和酮结构防止被还原，因此丙酮结构的羰基被还原为醇羟基后发生分子内Micheal加成，亲核进攻α，β-不饱和酮形成六元环，符合逻辑。因此进一步切断的结构为O=C(C=C/C(C/1CC(C)=O)=[N+]([O-])\[O-])C1=[N+]([O-])/[O-]。

<Checkpoint>[产物的$\rm -O-CHCH_3-CH_2-$六元环片段来源于丙酮](1)</Checkpoint>

<Checkpoint>[硼氢化钠用于还原羰基](1)</Checkpoint>

<Checkpoint>[丙酮的羰基被还原为醇羟基后发生分子内Micheal加成，亲核进攻α，β-不饱和酮形成六元环](1)</Checkpoint>

<Checkpoint>[进一步切断的结构为O=C(C=C/C(C/1CC(C)=O)=[N+]([O-])\[O-])C1=[N+]([O-])/[O-]](1)</Checkpoint>

丙酮结构的结合很显然通过将丙酮在醇钠条件下转化为烯醇负离子进行亲核进攻进行，因此切断丙酮结构得到六元环碳正离子，结构为O=C(C=C/C([CH+]/1)=[N+]([O-])\[O-])C1=[N+]([O-])/[O-]。

<Checkpoint>[切断丙酮结构得到六元环碳正离子，结构为O=C(C=C/C([CH+]/1)=[N+]([O-])\[O-])C1=[N+]([O-])/[O-]](1)</Checkpoint>

该结构此时带一个负电荷，需要得到一个氢离子；最终发现$\bf A$含有四个氢，可以互变异构为苯环结构，而丙酮产生的负离子亲核进攻为芳香亲核取代反应。

<Checkpoint>[$\bf A$含有四个氢，可以互变异构为苯环结构](1)</Checkpoint>

<Checkpoint>[丙酮产生的负离子亲核进攻为芳香亲核取代反应](1)</Checkpoint>

因此$\bf A$结构为OC1=CC=C([N+]([O-])=O)C=C1[N+]([O-])=O，选项E正确。

<Checkpoint>[$\bf A$结构为OC1=CC=C([N+]([O-])=O)C=C1[N+]([O-])=O](2)</Checkpoint>

<MultiModal>![本图从左到右为产物逐步切断的中间体。切断三级胺片段，产生双碳负离子，结构为O=C1[C-]([N+]([O-])=O)[C@H]2[C-]([N+]([O-])=O)[C@H](OC(C)C2)C1；互变异构，为O=C(C[C@@H]1/C([C@H]/2CC(C)O1)=[N+]([O-])\[O-])C2=[N+]([O-])\[O-]；之后切断六元环，结构为O=C(C=C/C(C/1CC(C)=O)=[N+]([O-])\[O-])C1=[N+]([O-])/[O-]；切断丙酮结构得到六元环碳正离子，结构为O=C(C=C/C([CH+]/1)=[N+]([O-])\[O-])C1=[N+]([O-])/[O-]；最终通过互变异构得到$\bf A$结构为OC1=CC=C([N+]([O-])=O)C=C1[N+]([O-])=O。](/media/uploads/c9f2543e-1148-4a92-9612-0a98d9e9f317.png)</MultiModal>