The synthesis of multi-functional group substituted aromatic rings has always been a research hotspot for scientists. The following is the synthetic route of a multi-substituted aromatic compound **G**:

<MultiModal>![The image describes a stepwise organic synthesis process. The substrate structure is O=C(CC1(CC(OCC)=O)OCCO1)OCC. Two equivalents of C=C(/C=C([O-])/C)[O-] are added to react, going through two intermediates **A** and **B** to obtain product **C***. Product **C** is added to sodium hydroxide and refluxed to obtain **D**. **D** is added to hydrochloric acid to obtain **E**. **E** is added to hydrogen iodide to obtain **F**. **F** reacts with selenium dioxide to obtain product **G**.](/media/uploads/45f8cefc-919d-4e07-8d79-48c747febd60.png)</MultiModal>

A protected carbonyl diester undergoes two anionic intermediates, **A** and **B**, to obtain the electrically neutral intermediate product **C**. **C** undergoes four steps of reaction to finally obtain **G**.

Known:
1. **B** has three rings, while **C** has only two rings.
2. **G** is a tricyclic multi-substituted aromatic compound.
3. All unknown structures except **A** have an aromatic ring.
4. **E, F** both have an anthrone skeleton.

The following options are all possible structures of **G**. The correct one is:
多官能团取代芳环的合成一直是科学家们研究的热点问题。下面为一多取代芳香化合物**G**的合成过程：

<MultiModal>![该图片描述了一个有机分步合成过程，底物结构为O=C(CC1(CC(OCC)=O)OCCO1)OCC，加入两当量C=C(/C=C([O-])/C)[O-]发生反应，经历两个中间体**A**，**B**得到产物**C***。产物**C**加入氢氧化钠回流得到**D**，**D**加入盐酸得到**E**，**E**加入碘化氢得到**F**，**F**与二氧化硒反应得到产物**G**。](/media/uploads/45f8cefc-919d-4e07-8d79-48c747febd60.png)</MultiModal>

一个受保护的羰基二酯经历**A**，**B**两个负离子中间体得到电中性中间产物**C**，**C**经历四步反应最终得到**G**。

已知 ：
1. **B**中有三个环，而**C**中只有两个环。
2. **G**为三环多取代芳香化合物。
3. 除**A**外的未知结构均具有芳环。
4. **E，F**均具有蒽酮骨架

下列选项均为可能的**G**的结构，正确的是：
<MultiModal>![OC1=CC(O)=C2C(CC(C=C(C)C=C3O)=C3C2=O)=C1](/media/uploads/5670abfc-418f-4efe-a638-1326c641e19b.png)</MultiModal>
<MultiModal>![OC1=CC(O)=C2C(C(C(C=C(C)C=C3O)=C3C2C)=O)=C1](/media/uploads/8841b2eb-d9d3-4a61-a517-d6c36e52aa44.png)</MultiModal>
<MultiModal>![OC1=C2C(C(C(C=C(O)C=C3C)=C3C2=O)=O)=CC(OCC=O)=C1](/media/uploads/a76c1673-b8b1-44cf-a133-a245f033da3b.png)</MultiModal>
<MultiModal>![OC1=CC(O)=C2C(C(O)C(C=C(O)C=C3C)=C3C2=O)=C1](/media/uploads/73945b12-1aef-46af-8cfd-7e809fde7283.png)</MultiModal>
<MultiModal>![OC1=CC(O)=C2C(C(C(C=C(O)C=C3C)=C3C2=O)=O)=C1](/media/uploads/79c0a58d-4221-4b9b-94fa-96df558acb53.png)</MultiModal>
<MultiModal>![OC1=CC(O)=C2C(C(C(C=C(C)C=C3O)=C3C2=O)=O)=C1](/media/uploads/99c614f0-73bb-4158-b3a5-e565ec13ca38.png)</MultiModal>
<MultiModal>![OC1=C2C(C(C(C=C(C)C=C3O)=C3C2=O)=O)=CC(OCC=O)=C1](/media/uploads/30babcbe-1c7a-4c84-9183-37b77df066cd.png)</MultiModal>
<MultiModal>![OC1=CC(O)=C2C(C(O)=C(CC(C)=CC3=O)C3=C2O)=C1](/media/uploads/6427a409-c1a0-494c-afe8-c544ddacdb68.png)</MultiModal>
<MultiModal>![OC1=CC(O)=C2C(C=C(CC(C)=CC3=O)C3=C2O)=C1](/media/uploads/5b4de69f-4970-4221-aa5a-8db4525b0edd.png)</MultiModal>
<MultiModal>![OC(C(C1=CC2=CC(C)=CC(O)=C2C(O)=C13)=O)=CC3=O](/media/uploads/041b32e4-1422-47f2-90fb-57ea7676250b.png)</MultiModal>
<MultiModal>![OC1=CC(O)=C2C(CC(C=C(C)C=C3O)=C3C2=O)=C1](/media/uploads/5670abfc-418f-4efe-a638-1326c641e19b.png)</MultiModal>
<MultiModal>![OC1=CC(O)=C2C(C(C(C=C(C)C=C3O)=C3C2C)=O)=C1](/media/uploads/8841b2eb-d9d3-4a61-a517-d6c36e52aa44.png)</MultiModal>
<MultiModal>![OC1=C2C(C(C(C=C(O)C=C3C)=C3C2=O)=O)=CC(OCC=O)=C1](/media/uploads/a76c1673-b8b1-44cf-a133-a245f033da3b.png)</MultiModal>
<MultiModal>![OC1=CC(O)=C2C(C(O)C(C=C(O)C=C3C)=C3C2=O)=C1](/media/uploads/73945b12-1aef-46af-8cfd-7e809fde7283.png)</MultiModal>
<MultiModal>![OC1=CC(O)=C2C(C(C(C=C(O)C=C3C)=C3C2=O)=O)=C1](/media/uploads/79c0a58d-4221-4b9b-94fa-96df558acb53.png)</MultiModal>
<MultiModal>![OC1=CC(O)=C2C(C(C(C=C(C)C=C3O)=C3C2=O)=O)=C1](/media/uploads/99c614f0-73bb-4158-b3a5-e565ec13ca38.png)</MultiModal>
<MultiModal>![OC1=C2C(C(C(C=C(C)C=C3O)=C3C2=O)=O)=CC(OCC=O)=C1](/media/uploads/30babcbe-1c7a-4c84-9183-37b77df066cd.png)</MultiModal>
<MultiModal>![OC1=CC(O)=C2C(C(O)=C(CC(C)=CC3=O)C3=C2O)=C1](/media/uploads/6427a409-c1a0-494c-afe8-c544ddacdb68.png)</MultiModal>
<MultiModal>![OC1=CC(O)=C2C(C=C(CC(C)=CC3=O)C3=C2O)=C1](/media/uploads/5b4de69f-4970-4221-aa5a-8db4525b0edd.png)</MultiModal>
<MultiModal>![OC(C(C1=CC2=CC(C)=CC(O)=C2C(O)=C13)=O)=CC3=O](/media/uploads/041b32e4-1422-47f2-90fb-57ea7676250b.png)</MultiModal>
The substrate contains ester groups. The added acetylacetone dianion has two nucleophilic centers: the carbanions at the 1-position and 3-position; however, the nucleophilicity of the 1-position is stronger, leading to nucleophilic substitution with the ester group. Since two equivalents of acetylacetone dianion are added, both ester groups are substituted, forming a symmetrical hexaketone intermediate **A**, with the SMILES CC(CC(CC(CC1(CC(CC(CC(C)=O)=O)=O)OCCO1)=O)=O)=O.

<Checkpoint>[The 1-position of the acetylacetone dianion has stronger nucleophilicity and undergoes nucleophilic substitution with the ester group](1)</Checkpoint>

<Checkpoint>[The structure of **A** is CC(CC(CC(CC1(CC(CC(CC(C)=O)=O)=O)OCCO1)=O)=O)=O](1)</Checkpoint>

**A** has one ring while **B** has three rings, resulting in the formation of two additional rings. The substrate contains numerous $\beta$-dicarbonyl structures. Due to symmetry, one of the triketone structures can form a stable dienolate intermediate CC(CC(CC(CC1(CC([CH-]C([CH-]C(C)=O)=O)=O)OCCO1)=O)=O)=O, which undergoes aldol condensation with another triketone structure to form two six-membered rings. Initially, the first spiro ring O=C(C1)C(C([CH-]C(C)=O)=O)=C(CC(CC(C)=O)=O)CC21OCCO2 is formed, followed by the formation of the second six-membered ring O=C(C1)C(C(C(C(C)=O)=C(CC(C)=O)C2)=O)=C2CC31OCCO3. This intermediate adopts a quinoid structure and readily aromatizes to a phenol, resulting in **B** as O=C(C1)C2=C(O)C(C(C)=O)=C(CC(C)=O)C=C2CC31OCCO3, featuring three rings and one aromatic ring.

<Checkpoint>[The dienolate intermediate CC(CC(CC(CC1(CC([CH-]C([CH-]C(C)=O)=O)=O)OCCO1)=O)=O)=O exists](1)</Checkpoint>

<Checkpoint>[Aldol condensation forms the first six-membered ring O=C(C1)C(C([CH-]C(C)=O)=O)=C(CC(CC(C)=O)=O)CC21OCCO2](1)</Checkpoint>

<Checkpoint>[Aldol condensation forms the second six-membered ring O=C(C1)C(C(C(C(C)=O)=C(CC(C)=O)C2)=O)=C2CC31OCCO3](1)</Checkpoint>

<Checkpoint>[**B** is O=C(C1)C2=C(O)C(C(C)=O)=C(CC(C)=O)C=C2CC31OCCO3](1)</Checkpoint>

From **B** to **C**, one ring is lost, clearly indicating that the five-membered ketal ring is the most prone to ring-opening. Upon ring-opening, the connected six-membered ring also adopts a quinoid structure, leading to aromatization and the formation of a second phenolic hydroxyl group. The structure of **C** is OC1=CC(OCCO)=CC2=C1C(O)=C(C(C)=O)C(CC(C)=O)=C2.

<Checkpoint>[The five-membered ketal ring in **B** undergoes ring-opening to form **C**](1)</Checkpoint>

<Checkpoint>[The structure of **C** is OC1=CC(OCCO)=CC2=C1C(O)=C(C(C)=O)C(CC(C)=O)=C2](1)</Checkpoint>

From **C** to **D**, potassium hydroxide is added under reflux, generating enolate ions in the system. The enolate derived from the ketone substituent adjacent to the phenolic hydroxyl group can conjugate with the aromatic system, while the meta-positioned substituent cannot. Therefore, the enolate adjacent to the phenolic hydroxyl is preferentially formed, undergoing intramolecular aldol condensation with another ketone carbonyl group to yield the six-membered ring product **D**, with the structure OC1=CC(OCCO)=CC2=C1C(O)=C(C(C=C(C)C3)=O)C3=C2. (The selectivity of this aldol condensation is high, as documented in reference 10.1021/ja00435a063.)

<Checkpoint>[The enolate derived from the ketone substituent adjacent to the phenolic hydroxyl group preferentially forms due to conjugation with the aromatic system](1)</Checkpoint>

<Checkpoint>[The structure of **D** is OC1=CC(OCCO)=CC2=C1C(O)=C(C(C=C(C)C3)=O)C3=C2](1)</Checkpoint>

From **D** to **E**, hydrochloric acid is added. **D** features a naphthalene ring fused with a six-membered ring, containing only one complete benzene ring. Under acidic conditions, it readily rearranges into an anthrone structure, which is more stable due to the presence of two complete benzene rings. Thus, the structure of **E** is OC1=CC(OCCO)=CC(CC2=C3C(O)=CC(C)=C2)=C1C3=O.

<Checkpoint>[The naphthalene-fused six-membered ring can rearrange into an anthraquinone under acidic conditions](1)</Checkpoint>

<Checkpoint>[The structure of **E** is OC1=CC(OCCO)=CC(CC2=C3C(O)=CC(C)=C2)=C1C3=O](1)</Checkpoint>

From **E** to **F**, hydroiodic acid is added, clearly serving to remove the protecting group, with ethylene glycol departing to yield a phenolic hydroxyl group. The structure of **F** is OC1=CC(O)=CC(CC2=C3C(O)=CC(C)=C2)=C1C3=O.

<Checkpoint>[Hydroiodic acid is used to remove the ethylene glycol protecting group](1)</Checkpoint>

<Checkpoint>[The structure of **F** is OC1=CC(O)=CC(CC2=C3C(O)=CC(C)=C2)=C1C3=O](1)</Checkpoint>

**G** is the product of the oxidation of **F** by selenium dioxide, where the methylene group in the anthracene intermediate is oxidized to a quinone. If the phenolic hydroxyl group were oxidized, aromaticity would be lost. Thus, the structure of **G** is OC1=CC(O)=CC(C(C2=C3C(O)=CC(C)=C2)=O)=C1C3=O.

<Checkpoint>[Selenium dioxide oxidizes the methylene group in the central six-membered ring of the anthracene intermediate to a quinone](1)</Checkpoint>

<Checkpoint>[The structure of **G** is OC1=CC(O)=CC(C(C2=C3C(O)=CC(C)=C2)=O)=C1C3=O](1)</Checkpoint>

Based on the deduced structure of **G**, option F is correct.

The following two figures visualize the structural formulas involved in the analysis of this problem:

<MultiModal>![This figure depicts the intermediates involved in the formation of **B** from the substrate](/media/uploads/7f56af55-19da-410d-aa8e-651726e12d7e.png)</MultiModal>

<MultiModal>![This figure shows the unknown species involved in the transformation from **C** to **G**, all labeled](/media/uploads/ec4200c5-9a3c-4225-b0e5-4648595eb84b.png)</MultiModal>
底物存在酯基，加入的乙酰丙酮双负离子有两个亲核中心：1号位/3号位碳负离子；但1号位的亲核性更强，与酯基发生亲核取代反应；由于加入两当量乙酰丙酮双负离子，两个酯基都会被取代，生成对称的六酮中间体**A**，SMILES为CC(CC(CC(CC1(CC(CC(CC(C)=O)=O)=O)OCCO1)=O)=O)=O。

<Checkpoint>[乙酰丙酮双负离子1号位的亲核性更强，与酯基发生亲核取代反应](1)</Checkpoint>

<Checkpoint>[**A**结构为CC(CC(CC(CC1(CC(CC(CC(C)=O)=O)=O)OCCO1)=O)=O)=O](1)</Checkpoint>

**A**只有一个环而**B**有三个环，从而生成了两个环；底物含有大量$\beta$-二羰基结构，由对称性，其中一个三酮结构可以形成稳定的双烯醇负离子中间体CC(CC(CC(CC1(CC([CH-]C([CH-]C(C)=O)=O)=O)OCCO1)=O)=O)=O，对另一个三酮结构进行羟醛缩合形成两个六元环；首先生成第一个螺环O=C(C1)C(C([CH-]C(C)=O)=O)=C(CC(CC(C)=O)=O)CC21OCCO2，之后形成第二个六元环O=C(C1)C(C(C(C(C)=O)=C(CC(C)=O)C2)=O)=C2CC31OCCO3。该中间体为醌式结构，易芳构化为酚，从而**B**为O=C(C1)C2=C(O)C(C(C)=O)=C(CC(C)=O)C=C2CC31OCCO3，具有三个环和一个芳环。

<Checkpoint>[存在双烯醇负离子中间体CC(CC(CC(CC1(CC([CH-]C([CH-]C(C)=O)=O)=O)OCCO1)=O)=O)=O](1)</Checkpoint>

<Checkpoint>[羟醛缩合形成第一个六元环O=C(C1)C(C([CH-]C(C)=O)=O)=C(CC(CC(C)=O)=O)CC21OCCO2](1)</Checkpoint>

<Checkpoint>[羟醛缩合形成第二个六元环O=C(C1)C(C(C(C(C)=O)=C(CC(C)=O)C2)=O)=C2CC31OCCO3](1)</Checkpoint>

<Checkpoint>[**B**为O=C(C1)C2=C(O)C(C(C)=O)=C(CC(C)=O)C=C2CC31OCCO3](1)</Checkpoint>

**B**到**C**失去了一个环，很明显五元环缩酮结构最容易开环；开环同时与其连接的六元环也变为醌式结构，从而芳构化生成第二个酚羟基，**C**结构为OC1=CC(OCCO)=CC2=C1C(O)=C(C(C)=O)C(CC(C)=O)=C2。

<Checkpoint>[**B**到**C**五元环缩酮结构开环](1)</Checkpoint>

<Checkpoint>[**C**结构为OC1=CC(OCCO)=CC2=C1C(O)=C(C(C)=O)C(CC(C)=O)=C2](1)</Checkpoint>

**C**到**D**加入氢氧化钾回流，体系中会生成烯醇负离子；酚羟基邻位的酮取代基生成的烯醇负离子能够与芳香体系共轭，而间位的取代基不能，所以优先生成酚羟基邻位的酮取代基的烯醇负离子，与分子内另一个酮羰基进行羟醛缩合得到六元环产物**D**，结构为OC1=CC(OCCO)=CC2=C1C(O)=C(C(C=C(C)C3)=O)C3=C2。（此处羟醛缩合选择性很高，见文献10.1021/ja00435a063）

<Checkpoint>[酚羟基邻位的酮取代基生成的烯醇负离子能够与芳香体系共轭，优先生成](1)</Checkpoint>

<Checkpoint>[**D**结构为OC1=CC(OCCO)=CC2=C1C(O)=C(C(C=C(C)C3)=O)C3=C2](1)</Checkpoint>

**D**到**E**加入盐酸，**D**为萘环并六元环结构，该结构只有一个完整的苯环，酸性条件下容易重排为蒽酮结构，存在两个完整的苯环更稳定；因此**E**结构式为OC1=CC(OCCO)=CC(CC2=C3C(O)=CC(C)=C2)=C1C3=O

<Checkpoint>[萘环并六元环酸性条件可重排为蒽醌](1)</Checkpoint>

<Checkpoint>[**E**结构式为OC1=CC(OCCO)=CC(CC2=C3C(O)=CC(C)=C2)=C1C3=O](1)</Checkpoint>

**E**到**F**加入碘化氢，很明显为脱去保护的反应，乙二醇离去生成酚羟基。**F**结构为OC1=CC(O)=CC(CC2=C3C(O)=CC(C)=C2)=C1C3=O。

<Checkpoint>[碘化氢用于脱去乙二醇](1)</Checkpoint>

<Checkpoint>[**F**结构为OC1=CC(O)=CC(CC2=C3C(O)=CC(C)=C2)=C1C3=O](1)</Checkpoint>

**G**是**F**被二氧化硒氧化的产物，明显为蒽中间结构的亚甲基被氧化为醌，如果酚羟基被氧化则失去芳香性。从而**G**的结构为OC1=CC(O)=CC(C(C2=C3C(O)=CC(C)=C2)=O)=C1C3=O。

<Checkpoint>[二氧化硒可氧化蒽中间六元环的亚甲基为醌](1)</Checkpoint>

<Checkpoint>[**G**的结构为OC1=CC(O)=CC(C(C2=C3C(O)=CC(C)=C2)=O)=C1C3=O](1)</Checkpoint>

根据推出的**G**的结构式，选项F正确。

下面两张图为该题解析的结构式可视化：

<MultiModal>![该图为从底物生成**B**的过程涉及的中间体](/media/uploads/7f56af55-19da-410d-aa8e-651726e12d7e.png)</MultiModal>

<MultiModal>![该图为从**C**生成**G**的过程涉及的未知物种，均有标记](/media/uploads/ec4200c5-9a3c-4225-b0e5-4648595eb84b.png)</MultiModal>