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Given that the molecular formula of product $\mathbf{A}$ is $\mathrm{C_{10}H_{10}N_2O_4}$, and it contains 3 rings. Without considering enantiomers, provide the structural formula of $\mathbf{A}$.
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已知产物$\mathbf{A}$的分子式为$\mathrm{C_{10}H_{10}N_2O_4}$，且含有3个环。不考虑对映异构的条件下，试给出$\mathbf{A}$的结构式。
All other options are incorrect
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其他选项均不正确
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First, based on the molecular formula $\mathrm{C_{10}H_{10}N_2O_4}$ of product $\mathbf{A}$, it can be deduced that the reaction consumes 1 equivalent of $\mathrm{PIDA}$

<Checkpoint>[First, based on the molecular formula $\mathrm{C_{10}H_{10}N_2O_4}$ of product $\mathbf{A}$, it can be deduced that the reaction consumes 1 equivalent of $\mathrm{PIDA}$](1)</Checkpoint>

First, $\mathrm{PIDA}$ oxidizes the substrate to obtain an intermediate.

<MultiModal>![O=C1C(C=CC([N+]([O-])=O)=C1)=N](/media/uploads/6057138c-9052-45a3-82ac-a22a3546c4c0.png)</MultiModal>

<Checkpoint>[$\mathrm{PIDA}$ Oxidation intermediate:O=C1C(C=CC([N+]([O-])=O)=C1)=N](1)</Checkpoint>

Then, the intermediate undergoes an inverse electron-demand D-A reaction with an electron-rich diene to obtain the product.

<Checkpoint>[Then, the intermediate undergoes an inverse electron-demand D-A reaction with an electron-rich diene to obtain the *cis*-product.](1)</Checkpoint>

For the diene, the electronegativity of O is greater than that of N, so the N-terminus is relatively more electron-deficient.

<Checkpoint>[For the diene, the electronegativity of O is greater than that of N, so the N-terminus is relatively more electron-deficient.](1)</Checkpoint>

For the dienophile, obviously the C at the *meta*-position of O is more electron-rich.

<Checkpoint>[For the dienophile, obviously the C at the *meta*-position of O is more electron-rich.](1)</Checkpoint>

Therefore, according to the electronegativity, options H and I can be excluded.

<MultiModal>![[H][C@]12OC3=CC([N+]([O-])=O)=CC=C3N[C@@]1([H])CCO2](/media/uploads/c69d0a3b-b91e-411b-a68b-68418dbd9e43.png)</MultiModal>

<Checkpoint>[Product $\mathbf{A}$：[H][C@]12OC3=CC([N+]([O-])=O)=CC=C3N[C@@]1([H])CCO2](1)</Checkpoint>
首先根据产物$\mathbf{A}$的分子式$\mathrm{C_{10}H_{10}N_2O_4}$可以推出该反应消耗了1当量的$\mathrm{PIDA}$

<Checkpoint>[首先根据产物$\mathbf{A}$的分子式$\mathrm{C_{10}H_{10}N_2O_4}$可以推出该反应消耗了1当量的$\mathrm{PIDA}$](1)</Checkpoint>

首先$\mathrm{PIDA}$对底物进行氧化得到中间体

<MultiModal>![O=C1C(C=CC([N+]([O-])=O)=C1)=N](/media/uploads/6057138c-9052-45a3-82ac-a22a3546c4c0.png)</MultiModal>

<Checkpoint>[$\mathrm{PIDA}$ Oxidation intermediate:O=C1C(C=CC([N+]([O-])=O)=C1)=N](1)</Checkpoint>

接着该中间体与富电子双烯体发生逆电子需求的D-A反应得到产物。

<Checkpoint>[接着该中间体与富电子双烯体发生逆电子需求的D-A反应得到顺式产物。](1)</Checkpoint>

对于双烯体而言，O的电负性大于N，因此在N端相对更加缺电子。

<Checkpoint>[对于双烯体而言，O的电负性大于N，因此在N端相对更加缺电子。](1)</Checkpoint>

而对于亲双烯体而言，显然O间位的C上更加富电子。

<Checkpoint>[而对于亲双烯体而言，显然O间位的C上更加富电子。](1)</Checkpoint>

因此根据电性可以排除选项H、I

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<Checkpoint>[产物$\mathbf{A}$：[H][C@]12OC3=CC([N+]([O-])=O)=CC=C3N[C@@]1([H])CCO2](1)</Checkpoint>