Benzaldehyde reacts with a mixture of 98% nitric acid/98% sulfuric acid at $0^\circ \mathrm{C}$ to obtain compound $\mathbf{A}$. Compound $\mathbf{A}$ reacts with <MultiModal>![CC(CC(OCCOC)=O)=O](/media/uploads/b7ff5af8-8fd2-40c1-850e-2b0192f2116c.png)</MultiModal>, <MultiModal>![C1CCCCN1](/media/uploads/5c2f94c4-6d54-4112-98f7-fe4cbda73c9e.png)</MultiModal> in acetic acid in a one-pot reaction to generate compound $\mathbf{B}$. Compound $\mathbf{B}$ reacts with <MultiModal>![C/C(N)=C/C(OC(C)C)=O](/media/uploads/2ed5dde4-02e3-4bd1-822c-001cd574f419.png)</MultiModal> under reflux in isopropyl alcohol solution to obtain compound $\mathbf{C}$, and this transformation proceeds through an electrocyclic reaction. Compound $\mathbf{C}$ reacts with ceric ammonium nitrate to obtain compound $\mathbf{D}$.

Based on the above information, deduce the structures of all compounds and select the correct statement from the following options.
苯甲醛通过与98%硝酸/98%硫酸混合物在$0^\circ \mathrm{C}$下反应得到化合物$\mathbf{A}$。化合物$\mathbf{A}$与<MultiModal>![CC(CC(OCCOC)=O)=O](/media/uploads/b7ff5af8-8fd2-40c1-850e-2b0192f2116c.png)</MultiModal>、<MultiModal>![C1CCCCN1](/media/uploads/5c2f94c4-6d54-4112-98f7-fe4cbda73c9e.png)</MultiModal>在乙酸处理下一锅反应生成化合物$\mathbf{B}$。化合物$\mathbf{B}$与<MultiModal>![C/C(N)=C/C(OC(C)C)=O](/media/uploads/2ed5dde4-02e3-4bd1-822c-001cd574f419.png)</MultiModal>在异丙醇溶液中回流得到化合物$\mathbf{C}$，该转化中经过了电环化反应。化合物$\mathbf{C}$与硫酸铈铵反应可以得到化合物$\mathbf{D}$。

根据以上信息，推断所有化合物的结构，并选择以下选项中说法正确的一个。
Compound $\mathbf{A}$ is <MultiModal>![O=CC1=CC([N+]([O-])=O)=CC([N+]([O-])=O)=C1](/media/uploads/ccd3863a-65ef-4ab5-8f5d-dfaa385593c4.png)</MultiModal>
Compound $\mathbf{A}$ is <MultiModal>![O=CC1=CC=C([N+]([O-])=O)C=C1](/media/uploads/e5cf8164-7a42-4976-a7fb-f0c99d1c3d73.png)</MultiModal>
Compound $\mathbf{B}$ is <MultiModal>![O=C(C)/C(C(OCCOC)=O)=C/C1=CC=CC([N+]([O-])=O)=C1](/media/uploads/bba0724c-19df-40ba-892c-f5e06deb2995.png)</MultiModal>
Compound $\mathbf{B}$ is <MultiModal>![O=C(C)C(C(OCCOC)=O)C(O)C1=CC=CC([N+]([O-])=O)=C1](/media/uploads/4c873013-2bd2-469c-bc31-308f77898611.png)</MultiModal>
Compound $\mathbf{B}$ is <MultiModal>![O=C(C)C(C(OCCOC)=O)C(N1CCCCC1)C2=CC=CC([N+]([O-])=O)=C2](/media/uploads/d48d8cf1-1939-46b4-ae14-be48a5b4ddea.png)</MultiModal>
Compound $\mathbf{C}$ is <MultiModal>![CC1=NC(C)C(C(OC(C)C)=O)=C(OCCOC)/C1=C\C2=CC=CC([N+]([O-])=O)=C2](/media/uploads/02427429-e49d-407a-b088-7fdbc39d7cf8.png)</MultiModal>
Compound $\mathbf{C}$ is <MultiModal>![CC(OC(C1=C(NC(OCCOC)=C(C(C)=O)C1C2=CC([N+]([O-])=O)=CC=C2)C)=O)C](/media/uploads/87e5d828-01d9-4cb3-8878-b4bdd3d97013.png)</MultiModal>
Compound $\mathbf{D}$ is <MultiModal>![CC(OC(C1=C(C2=CC([N+]([O-])=O)=CC=C2)C(C(OCCOC)=O)=C(C)[N+]([O-])=C1C)=O)C](/media/uploads/98c755f4-7676-46df-92da-7cc0e05a66dd.png)</MultiModal>
Compound $\mathbf{D}$ is <MultiModal>![C=C1N=C(C(C(OC(C)C)=O)=C(/C1=C\C2=CC=CC([N+]([O-])=O)=C2)OCCOC)C](/media/uploads/0b4d9e02-a2f9-4750-8c7b-5d514187a8f6.png)</MultiModal>
化合物$\mathbf{A}$为<MultiModal>![O=CC1=CC([N+]([O-])=O)=CC([N+]([O-])=O)=C1](/media/uploads/ccd3863a-65ef-4ab5-8f5d-dfaa385593c4.png)</MultiModal>
化合物$\mathbf{A}$为<MultiModal>![O=CC1=CC=C([N+]([O-])=O)C=C1](/media/uploads/e5cf8164-7a42-4976-a7fb-f0c99d1c3d73.png)</MultiModal>
化合物$\mathbf{B}$为<MultiModal>![O=C(C)/C(C(OCCOC)=O)=C/C1=CC=CC([N+]([O-])=O)=C1](/media/uploads/bba0724c-19df-40ba-892c-f5e06deb2995.png)</MultiModal>
化合物$\mathbf{B}$为<MultiModal>![O=C(C)C(C(OCCOC)=O)C(O)C1=CC=CC([N+]([O-])=O)=C1](/media/uploads/4c873013-2bd2-469c-bc31-308f77898611.png)</MultiModal>
化合物$\mathbf{B}$为<MultiModal>![O=C(C)C(C(OCCOC)=O)C(N1CCCCC1)C2=CC=CC([N+]([O-])=O)=C2](/media/uploads/d48d8cf1-1939-46b4-ae14-be48a5b4ddea.png)</MultiModal>
化合物$\mathbf{C}$为<MultiModal>![CC1=NC(C)C(C(OC(C)C)=O)=C(OCCOC)/C1=C\C2=CC=CC([N+]([O-])=O)=C2](/media/uploads/02427429-e49d-407a-b088-7fdbc39d7cf8.png)</MultiModal>
化合物$\mathbf{C}$为<MultiModal>![CC(OC(C1=C(NC(OCCOC)=C(C(C)=O)C1C2=CC([N+]([O-])=O)=CC=C2)C)=O)C](/media/uploads/87e5d828-01d9-4cb3-8878-b4bdd3d97013.png)</MultiModal>
化合物$\mathbf{D}$为<MultiModal>![CC(OC(C1=C(C2=CC([N+]([O-])=O)=CC=C2)C(C(OCCOC)=O)=C(C)[N+]([O-])=C1C)=O)C](/media/uploads/98c755f4-7676-46df-92da-7cc0e05a66dd.png)</MultiModal>

化合物$\mathbf{D}$为<MultiModal>![C=C1N=C(C(C(OC(C)C)=O)=C(/C1=C\C2=CC=CC([N+]([O-])=O)=C2)OCCOC)C](/media/uploads/0b4d9e02-a2f9-4750-8c7b-5d514187a8f6.png)</MultiModal>
First, the nitration reaction under low-temperature conditions can only occur once, and the aldehyde group is an electron-withdrawing group, occurring at the meta position, so compound $\mathbf{A}$ is <MultiModal>![O=CC1=CC([N+]([O-])=O)=CC=C1](/media/uploads/c2d4aedf-4736-4b91-9506-3f61808165bf.png)</MultiModal>

<Checkpoint>[Compound $\mathbf{A}$ is O=CC1=CC([N+]([O-])=O)=CC=C1](0.5)</Checkpoint>

Next, secondary amine piperidine is added in the process of generating $\mathbf{B}$, which is an acid-catalyzed aldol condensation-like reaction, so compound $\mathbf{B}$ is <MultiModal>![O=C(C)/C(C(OCCOC)=O)=C/C1=CC=CC([N+]([O-])=O)=C1](/media/uploads/bba0724c-19df-40ba-892c-f5e06deb2995.png)</MultiModal>

<Checkpoint>[Compound $\mathbf{B}$ is O=C(C)/C(C(OCCOC)=O)=C/C1=CC=CC([N+]([O-])=O)=C1](1)</Checkpoint>

The ketone carbonyl group of compound $\mathbf{B}$ can undergo condensation with an amino group (the ketone carbonyl group is more electrophilic than the ester carbonyl group and is preferentially attacked), and then combined with the information in the question, an electrocyclic reaction can occur, so compound $\mathbf{C}$ is <MultiModal>![CC(OC(C1=C(NC(C)=C(C(OCCOC)=O)C1C2=CC([N+]([O-])=O)=CC=C2)C)=O)C](/media/uploads/0f71cf45-e273-4318-90ed-65ff6f9dcdfe.png)</MultiModal>

<Checkpoint>[Compound $\mathbf{C}$ is CC(OC(C1=C(NC(C)=C(C(OCCOC)=O)C1C2=CC([N+]([O-])=O)=CC=C2)C)=O)C](1.5)</Checkpoint>

Next, ceric ammonium sulfate is a weak oxidant, which can oxidize the previous heterocycle to an aromatic ring, so compound $\mathbf{D}$ is <MultiModal>![CC(OC(C1=C(C2=CC([N+]([O-])=O)=CC=C2)C(C(OCCOC)=O)=C(C)N=C1C)=O)C](/media/uploads/744b866d-8f47-490d-b9db-f92e05b6f587.png)</MultiModal>

<Checkpoint>[Compound $\mathbf{D}$ is CC(OC(C1=C(C2=CC([N+]([O-])=O)=CC=C2)C(C(OCCOC)=O)=C(C)N=C1C)=O)C](0.5)</Checkpoint>
首先，低温条件下的硝化反应仅能发生一次，且醛基为吸电子基团，发生在间位，故而化合物$\mathbf{A}$为<MultiModal>![O=CC1=CC([N+]([O-])=O)=CC=C1](/media/uploads/c2d4aedf-4736-4b91-9506-3f61808165bf.png)</MultiModal>

<Checkpoint>[化合物$\mathbf{A}$为O=CC1=CC([N+]([O-])=O)=CC=C1](0.5)</Checkpoint>

接着生成$\mathbf{B}$过程中加入仲胺哌啶，为酸催化下的羟醛缩合类反应，故而化合物$\mathbf{B}$为<MultiModal>![O=C(C)/C(C(OCCOC)=O)=C/C1=CC=CC([N+]([O-])=O)=C1](/media/uploads/bba0724c-19df-40ba-892c-f5e06deb2995.png)</MultiModal>

<Checkpoint>[化合物$\mathbf{B}$为O=C(C)/C(C(OCCOC)=O)=C/C1=CC=CC([N+]([O-])=O)=C1](1)</Checkpoint>

而化合物$\mathbf{B}$的酮羰基可以和氨基缩合（酮羰基相较于酯羰基亲电性更强，优先被进攻），并且随后可以结合题目信息，发生电环化反应，故而化合物$\mathbf{C}$为<MultiModal>![CC(OC(C1=C(NC(C)=C(C(OCCOC)=O)C1C2=CC([N+]([O-])=O)=CC=C2)C)=O)C](/media/uploads/0f71cf45-e273-4318-90ed-65ff6f9dcdfe.png)</MultiModal>

<Checkpoint>[化合物$\mathbf{C}$为CC(OC(C1=C(NC(C)=C(C(OCCOC)=O)C1C2=CC([N+]([O-])=O)=CC=C2)C)=O)C](1.5)</Checkpoint>

接下来，硫酸铈铵为弱氧化剂，可以将前面的杂环氧化为芳环，故而化合物$\mathbf{D}$为<MultiModal>![CC(OC(C1=C(C2=CC([N+]([O-])=O)=CC=C2)C(C(OCCOC)=O)=C(C)N=C1C)=O)C](/media/uploads/744b866d-8f47-490d-b9db-f92e05b6f587.png)</MultiModal>

<Checkpoint>[化合物$\mathbf{D}$为CC(OC(C1=C(C2=CC([N+]([O-])=O)=CC=C2)C(C(OCCOC)=O)=C(C)N=C1C)=O)C](0.5)</Checkpoint>