<MultiModal>![The image describes an organic tandem synthesis route. The substrate is CC(CC(OCC)=O)=O, which reacts with O=C1C(N=[N+]=[N-])CCCC1 in the presence of 2.4 equivalents of $LDA$ to generate $\bf{A}$. $\bf{A}$ reacts with $PPh_3, CH_2Cl_2$ to generate $\bf{B}$. $\bf{B}$ reacts with $CF_3COOH$ to generate $\bf{C}$.
](/media/uploads/e82298c9-ec11-468e-bef3-c83e586747f0.png)</MultiModal>

Regarding the structural formula of the final product $\bf{C}$ of the organic synthesis route in the above figure, the correct option is:
<MultiModal>![该图片描述了一个有机串联合成路线。底物为CC(CC(OCC)=O)=O，与O=C1C(N=[N+]=[N-])CCCC1在2.4当量$LDA$存在下反应生成$\bf{A}$，$\bf{A}$与$PPh_3,CH_2Cl_2$反应生成$\bf{B}$，$\bf{B}$与$CF_3COOH$反应生成$\bf{C}$。
](/media/uploads/e82298c9-ec11-468e-bef3-c83e586747f0.png)</MultiModal>

对于上图的有机合成路线的最终产物$\bf{C}$的结构式，正确的选项是：
All other options are incorrect
<MultiModal>![O=C(CC(C1)NC2=C1C=CC=C2)OCC](/media/uploads/bf60a3c0-497e-43d0-bb12-9f339ceb318e.png)</MultiModal>
<MultiModal>![O=C(/C=C(C1)\NC2=C1CCCC2)OCC](/media/uploads/56921b0b-a57d-496a-b5a4-a21472b30359.png)</MultiModal>
<MultiModal>![O=C(CC1=CNC2=C1CCCC2)OCC](/media/uploads/1506bd62-4780-4aeb-81fd-8a203bb9f3ef.png)</MultiModal>
<MultiModal>![O=C(CC1=CC2=C(N1)CCCC2)OCC](/media/uploads/fe8a46e4-81cd-4e7e-b0cb-3c6f82a17874.png)</MultiModal>
<MultiModal>![O=C(C1)C=CNC2=C1CCCC2](/media/uploads/569516cd-2723-41f4-a3ac-046d1396aa65.png)</MultiModal>
<MultiModal>![O=C(C1)CC(NC2=C1CCCC2)=O](/media/uploads/fe055432-0ffe-43ef-8129-0a86d8a92732.png)</MultiModal>
<MultiModal>![CC(C1C(OCC)=O)NC2=C1CCCC2](/media/uploads/1b481f8e-3808-4d00-9b94-7788d584de25.png)</MultiModal>
<MultiModal>![O=C(C1C(C)=O)NC2=C1CCCC2](/media/uploads/3a94f8f8-5c73-4c4f-9202-f70882d8a365.png)</MultiModal>
其他选项均不正确
<MultiModal>![O=C(CC(C1)NC2=C1C=CC=C2)OCC](/media/uploads/bf60a3c0-497e-43d0-bb12-9f339ceb318e.png)</MultiModal>
<MultiModal>![O=C(/C=C(C1)\NC2=C1CCCC2)OCC](/media/uploads/56921b0b-a57d-496a-b5a4-a21472b30359.png)</MultiModal>
<MultiModal>![O=C(CC1=CNC2=C1CCCC2)OCC](/media/uploads/1506bd62-4780-4aeb-81fd-8a203bb9f3ef.png)</MultiModal>
<MultiModal>![O=C(CC1=CC2=C(N1)CCCC2)OCC](/media/uploads/fe8a46e4-81cd-4e7e-b0cb-3c6f82a17874.png)</MultiModal>
<MultiModal>![O=C(C1)C=CNC2=C1CCCC2](/media/uploads/569516cd-2723-41f4-a3ac-046d1396aa65.png)</MultiModal>
<MultiModal>![O=C(C1)CC(NC2=C1CCCC2)=O](/media/uploads/fe055432-0ffe-43ef-8129-0a86d8a92732.png)</MultiModal>
<MultiModal>![CC(C1C(OCC)=O)NC2=C1CCCC2](/media/uploads/1b481f8e-3808-4d00-9b94-7788d584de25.png)</MultiModal>
<MultiModal>![O=C(C1C(C)=O)NC2=C1CCCC2](/media/uploads/3a94f8f8-5c73-4c4f-9202-f70882d8a365.png)</MultiModal>
The first step of the reaction involves a β-dicarbonyl compound as the substrate, which can form an enolate dianion intermediate under the action of two equivalents of strong base $\rm{LDA}$, with the structure [CH2-]C([CH-]C(OCC)=O)=O.

<Checkpoint>[Formation of an enolate dianion intermediate under the action of two equivalents of $\rm{LDA}$, with the structure [CH2-]C([CH-]C(OCC)=O)=O](1)</Checkpoint>

This anion can obviously undergo nucleophilic addition with another substrate containing a ketone carbonyl group. At this point, there is chemoselectivity: the methyl carbanion has a smaller degree of charge delocalization and is more nucleophilic. Under the kinetic product environment of $\rm{LDA}$, it should preferentially nucleophilically attack over the methylene carbanion, resulting in $\bf{A}$ as O=C(CC(OCC)=O)CC1(O)C(N=[N+]=[N-])CCCC1.

<Checkpoint>[The methyl carbanion has stronger nucleophilicity and preferentially nucleophilically attacks over the methylene carbanion](1)</Checkpoint>

<Checkpoint>[$\bf{A}$ is O=C(CC(OCC)=O)CC1(O)C(N=[N+]=[N-])CCCC1](1)</Checkpoint>

The next step is a typical Staudinger reaction. The substrate contains an azide group, which can react with triphenylphosphine to generate a phosphazene ylide. The phosphazene ylide can react with a carbonyl group. Here, chemoselectivity is also involved. The ketone carbonyl group is more electrophilic than the ester carbonyl group, and the intramolecular formation of a seven-membered ring is more difficult than the formation of a five-membered ring. Therefore, the phosphazene ylide attacks the intramolecular ketone carbonyl group, eliminating one molecule of water to form an exocyclic double bond. The structure of $\bf{B}$ should be OC12C(N/C(C2)=C\C(OCC)=O)CCCC1.

<Checkpoint>[The azide group can react with triphenylphosphine to generate a phosphazene ylide](1)</Checkpoint>

<Checkpoint>[The ketone carbonyl group is more electrophilic than the ester carbonyl group](1)</Checkpoint>

<Checkpoint>[Intramolecular formation of a seven-membered ring is more difficult than the formation of a five-membered ring](1)</Checkpoint>

<Checkpoint>[The structure of $\bf{B}$ should be OC12C(N/C(C2)=C\C(OCC)=O)CCCC1](1)</Checkpoint>

In the final step, trifluoroacetic acid is added, creating a strongly acidic environment. The intramolecular tertiary alcohol undergoes elimination, preferentially forming a fused poly-substituted double bond based on selectivity. At this point, an aza-five-membered ring olefin structure can be observed within the molecule. The acidic environment can aromatize it into pyrrole. Therefore, the final product $\bf{C}$ has the structure O=C(CC1=CC2=C(N1)CCCC2)OCC, and only option E is correct.

<Checkpoint>[Alcohol elimination preferentially forms a fused poly-substituted double bond](1)</Checkpoint>

<Checkpoint>[The acidic environment can aromatize it into pyrrole](1)</Checkpoint>

<Checkpoint>[The structure of $\bf{C}$ is O=C(CC1=CC2=C(N1)CCCC2)OCC](2)</Checkpoint>

<MultiModal>![This figure shows the structures of the intermediates and unknown species involved in this problem. At the top of the figure is the enolate dianion intermediate, with the structure [CH2-]C([CH-]C(OCC)=O)=O; at the bottom of the figure are the structures of the unknown species $\bf{A,B,C}$, with SMILES O=C(CC(OCC)=O)CC1(O)C(N=[N+]=[N-])CCCC1; OC12C(N/C(C2)=C\C(OCC)=O)CCCC1; O=C(CC1=CC2=C(N1)CCCC2)OCC.](/media/uploads/afe3c451-1c85-4b26-a7dc-deb725782df4.png)</MultiModal>
反应第一步，底物为β-二羰基化合物，在两当量强碱$\rm{LDA}$作用下可以形成烯醇双负离子中间体，结构为[CH2-]C([CH-]C(OCC)=O)=O。

<Checkpoint>[两当量$\rm{LDA}$作用下形成烯醇双负离子中间体，结构为[CH2-]C([CH-]C(OCC)=O)=O](1)</Checkpoint>

该负离子很显然可以与另一个酮羰基的底物发生亲核加成，此时存在化学选择性：甲基碳负离子的电荷离域程度小，亲核性更强，在$\rm{LDA}$这种动力学产物环境下应当优先于亚甲基碳负离子亲核，从而产生的$\bf{A}$为O=C(CC(OCC)=O)CC1(O)C(N=[N+]=[N-])CCCC1。

<Checkpoint>[甲基碳负离子亲核能力更强，优先于亚甲基碳负离子亲核](1)</Checkpoint>

<Checkpoint>[$\bf{A}$为O=C(CC(OCC)=O)CC1(O)C(N=[N+]=[N-])CCCC1](1)</Checkpoint>

下一步是典型的Staudinger反应，底物存在叠氮基团，可以与三苯基膦反应生成氮叶立德，氮叶立德可以与羰基反应。这里同样涉及化学选择性，酮羰基比酯羰基更加亲电，且分子内成七元环难于成五元环，因此氮叶立德进攻分子内酮羰基，脱去一分子水形成环外双键，$\bf{B}$结构应当为OC12C(N/C(C2)=C\C(OCC)=O)CCCC1。

<Checkpoint>[叠氮基可以与三苯基膦反应生成氮叶立德](1)</Checkpoint>

<Checkpoint>[酮羰基比酯羰基更加亲电](1)</Checkpoint>

<Checkpoint>[分子内成七元环难于成五元环](1)</Checkpoint>

<Checkpoint>[$\bf{B}$结构应当为OC12C(N/C(C2)=C\C(OCC)=O)CCCC1](1)</Checkpoint>

最后一步加入三氟乙酸，为强酸性环境，分子内的三级醇消除，根据选择性优先生成并环多取代双键。此时可以观察到分子内存在氮杂五元环烯烃结构，酸性环境可以芳构化为吡咯。因此最终产物$\bf{C}$结构为O=C(CC1=CC2=C(N1)CCCC2)OCC，只有选项E正确。

<Checkpoint>[醇消除优先生成并环多取代双键](1)</Checkpoint>

<Checkpoint>[酸性环境可以芳构化为吡咯](1)</Checkpoint>

<Checkpoint>[$\bf{C}$结构为O=C(CC1=CC2=C(N1)CCCC2)OCC](2)</Checkpoint>

<MultiModal>![本图给出了本题涉及到的中间体和未知物种结构。图最上方为烯醇双负离子中间体，结构为[CH2-]C([CH-]C(OCC)=O)=O；图下方分别为未知物种$\bf{A,B,C}$的结构，SMILES分别为O=C(CC(OCC)=O)CC1(O)C(N=[N+]=[N-])CCCC1；OC12C(N/C(C2)=C\C(OCC)=O)CCCC1；O=C(CC1=CC2=C(N1)CCCC2)OCC。](/media/uploads/afe3c451-1c85-4b26-a7dc-deb725782df4.png)</MultiModal>