Spiro compounds are often used in the synthesis of complex polycyclic compounds. The following figure describes a reaction of a spiro compound:

<MultiModal>![The image describes an organic multi-step reaction. The substrate is CC1CCC(C(OC)=O)C12CCC(C)=CC2, which reacts with C[Mg]Br and then water is added to generate **A**. **A** reacts with O=CO to generate **B**](/media/uploads/dbf539ce-0d4b-4781-81bc-d1431526cea0.png)</MultiModal>

Known: **B** contains two five-membered rings and one six-membered ring.

Which of the following statements is correct:
螺环化合物常用于合成复杂的多环化合物。下图描述了一个螺环化合物的反应：

<MultiModal>![该图片描述了一个有机多步反应。底物为CC1CCC(C(OC)=O)C12CCC(C)=CC2，其与C[Mg]Br反应后再加入水，生成**A**，**A**与O=CO反应生成**B**](/media/uploads/dbf539ce-0d4b-4781-81bc-d1431526cea0.png)</MultiModal>

已知：**B**含有两个五元环和一个六元环。

下列说法正确的是：
**B** contains 5 chiral carbon atoms.
**B** contains two unsaturation bonds.
<MultiModal>![CC(C)[C+]1C2(CCC(C)=CC2)C(C)CC1](/media/uploads/aa16b05c-e482-4760-b0e0-1ed97277544b.png)</MultiModal>

The above image is the key intermediate for the generation of **B**
<MultiModal>![CC1CCC(C12CCC([CH+]C2)3C)C3(C)C](/media/uploads/c682dc90-b9c0-460a-98fd-57ce99cac818.png)</MultiModal>  

The above image shows the key intermediate for generating **B**.
**B** is completely oxidized to **C** by acidic potassium permanganate solution, then the chemical formula of **C** is $\mathrm{C_{15}H_{24}O_3}$
All other options are incorrect.
**B**中存在5个手性碳原子
**B**中含有两个不饱和键
<MultiModal>![CC(C)[C+]1C2(CCC(C)=CC2)C(C)CC1](/media/uploads/aa16b05c-e482-4760-b0e0-1ed97277544b.png)</MultiModal>

上图是生成**B**的关键中间体
<MultiModal>![CC1CCC(C12CCC([CH+]C2)3C)C3(C)C](/media/uploads/c682dc90-b9c0-460a-98fd-57ce99cac818.png)</MultiModal>

上图是生成**B**的关键中间体
**B**被酸性高锰酸钾溶液完全氧化为**C**，则**C**的化学式为$\mathrm{C_{15}H_{24}O_3}$


其他选项均不正确
The reaction involving the addition of methyl Grignard reagent is relatively straightforward, as the Grignard reagent reacts with an ester to yield a tertiary alcohol.  

<Checkpoint>[Grignard reagent reacts with ester to form tertiary alcohol](1)</Checkpoint>  

Thus, the structure of **A** is  
CC1CCC(C(C)(O)C)C12CCC(C)=CC2.  

<Checkpoint>[Structure of **A** is CC1CCC(C(C)(O)C)C12CCC(C)=CC2](1)</Checkpoint>  

Tertiary alcohols readily undergo dehydration in a strongly acidic environment such as formic acid, forming a stable tertiary carbocation. Therefore, the intermediate leading to **B** includes CC1CCC([C+](C)C)C12CCC(C)=CC2.  

<Checkpoint>[Tertiary alcohol dehydrates in formic acid to form tertiary carbocation](1)</Checkpoint>  

<Checkpoint>[Intermediate of **B** includes CC1CCC([C+](C)C)C12CCC(C)=CC2](1)</Checkpoint>  

**B** contains three rings, indicating an intramolecular cyclization reaction. The carbocation, being a strong electrophile, can only react intramolecularly with the remaining alkene as the nucleophile, leading to ring formation.  

<Checkpoint>[Alkene nucleophilically attacks carbocation, forming a ring intramolecularly.](1)</Checkpoint>  

Given that **B** contains two five-membered rings and one six-membered ring, while **A** already has a spiro ring system (five + six), the newly formed ring must be a five-membered ring. Consequently, the carbon linking the alkene and the carbocation must be a secondary carbon, resulting in the tricyclic intermediate CC1CCC(C(C)2C)C13CC[C+](C)C2C3.  

<Checkpoint>[Newly formed ring is a five-membered ring, so the alkene-linked carbon must be secondary](1)</Checkpoint>  

<Checkpoint>[Tricyclic intermediate formed: CC1CCC(C(C)2C)C13CC[C+](C)C2C3](1)</Checkpoint>  

This intermediate eliminates a hydride ion, following the rule of forming a stable alkene, resulting in an endocyclic double bond. Thus, **B** is CC1CCC(C12CC=C(C3C2)C)C3(C)C.  

<Checkpoint>[**B** is CC1CCC(C12CC=C(C3C2)C)C3(C)C](2)</Checkpoint>  

**B** contains four chiral carbons and one unsaturated bond, so options A and B are incorrect.  

<Checkpoint>[**B** has four chiral carbons and one unsaturated bond, options A and B are incorrect.](1)</Checkpoint>  

The endocyclic double bond in **B** is oxidized by potassium permanganate, cleaving the double bond to form a ketone carbonyl and an aldehyde carbonyl, with the latter further oxidized to a carboxylic acid. The oxidation product has the structure CC1CCC2C(C)(C(CC21CC(O)=O)C(C)=O)C.  

<Checkpoint>[Oxidation product structure: CC1CCC2C(C)(C(CC21CC(O)=O)C(C)=O)C](1)</Checkpoint>  

Its molecular formula is $\mathrm{C_{15}H_{24}O_3}$, so option E is correct.  

<Checkpoint>[Oxidation product molecular formula: $\mathrm{C_{15}H_{24}O_3}$, option E is correct.](0.5)</Checkpoint>  

The intermediate in option C is derived from CC1CCC([C+](C)C)C12CCC(C)=CC2 after a single migration, forming a carbocation. This intermediate is more prone to rearrangement, converting the spiro ring to a fused ring. If the alkene were to nucleophilically attack, the resulting product would have excessive ring strain and would not match the description of **B**, so the tricyclic product cannot form, making option C incorrect.  

<Checkpoint>[Intermediate in C tends to rearrange, converting spiro to fused ring, unable to form tricyclic product, so C is incorrect](1)</Checkpoint>  

Option D represents the product of a tertiary carbocation CC1CCC([C+](C)C)C12CCC(C)=CC2 reacting with the alkene to form a secondary carbocation. However, if the nucleophilic attack occurs at the other carbon, a more stable tertiary carbocation CC1CCC(C(C)2C)C13CC[C+](C)C2C3 forms instead. Thus, it is not the key intermediate, making option D incorrect.  

<Checkpoint>[Carbocation nucleophilic attack favors forming a more stable carbocation: CC1CCC(C(C)2C)C13CC[C+](C)C2C3](1)</Checkpoint>  

<Checkpoint>[Option D is not the key intermediate, so D is incorrect](0.5)</Checkpoint>  

Therefore, option E is correct.  

<Checkpoint>[Therefore, option E is correct.](0.5)</Checkpoint>
加入甲基格氏试剂的反应较为简单，格氏试剂与酯反应生成三级醇。

<Checkpoint>[格氏试剂与酯反应生成三级醇](1)</Checkpoint>

故**A**的结构为
CC1CCC(C(C)(O)C)C12CCC(C)=CC2。

<Checkpoint>[**A**的结构为 CC1CCC(C(C)(O)C)C12CCC(C)=CC2](1)</Checkpoint>

三级醇在甲酸这种强酸性环境下极易脱水生成稳定的三级碳正离子，故生成**B**的中间体包含CC1CCC([C+](C)C)C12CCC(C)=CC2。

<Checkpoint>[三级醇在甲酸环境下脱水生成三级碳正离子](1)</Checkpoint>

<Checkpoint>[生成**B**的中间体包含CC1CCC([C+](C)C)C12CCC(C)=CC2](1)</Checkpoint>

**B**包含三个环，意味着发生了分子内的环化反应。碳正离子作为强亲电物种，分子内只剩下烯烃作为亲核物种了，从而烯烃亲核碳正离子，分子内成环。

<Checkpoint>[烯烃亲核碳正离子，分子内成环。](1)</Checkpoint>

考虑到**B**存在两个五元环和一个六元环，而**A**已经存在五+六的螺环，因此新生成的环为五元环，因此烯烃与碳正离子相连的碳只能为二级碳，形成了三环中间体CC1CCC(C(C)2C)C13CC[C+](C)C2C3。

<Checkpoint>[新生成的环为五元环，因此烯烃与碳正离子相连的碳只能为二级碳](1)</Checkpoint>

<Checkpoint>[形成了三环中间体CC1CCC(C(C)2C)C13CC[C+](C)C2C3](1)</Checkpoint>

该中间体消除一分子氢离子，按照生成稳定烯烃的规则，得到环内双键，从而**B**为CC1CCC(C12CC=C(C3C2)C)C3(C)C

<Checkpoint>[**B**为CC1CCC(C12CC=C(C3C2)C)C3(C)C](2)</Checkpoint>

**B**含有四个手性碳和一个不饱和键，选项A，B错误。

<Checkpoint>[**B**含有四个手性碳和一个不饱和键，选项A，B错误。](1)</Checkpoint>

**B**的环内双键被高锰酸钾氧化，首先双键断裂生成一个酮羰基和一个醛羰基，醛羰基被进一步氧化为羧酸。氧化产物结构式为CC1CCC2C(C)(C(CC21CC(O)=O)C(C)=O)C

<Checkpoint>[氧化产物结构式为CC1CCC2C(C)(C(CC21CC(O)=O)C(C)=O)C](1)</Checkpoint>

其分子式为$\mathrm{C_{15}H_{24}O_3}$，选项E正确。

<Checkpoint>[氧化产物分子式为$\mathrm{C_{15}H_{24}O_3}$，选项E正确。](0.5)</Checkpoint>

选项C的中间体是CC1CCC([C+](C)C)C12CCC(C)=CC2经过一步迁移得到的碳正离子，该中间体更易发生重排，螺环变为并环；若烯烃强行亲核，产物的环张力太大且并不符合**B**的描述，故生成不了三环产物，故C错误。

<Checkpoint>[C的中间体更易发生重排，螺环变为并环，生成不了三环产物，故C错误](1)</Checkpoint>

选项D是三级碳正离子CC1CCC([C+](C)C)C12CCC(C)=CC2与双键亲核生成二级碳正离子的产物，但若亲核位点在另一碳原子上，可生成更稳定的三级碳正离子CC1CCC(C(C)2C)C13CC[C+](C)C2C3，因此不是主要关键中间体，选项D错误。

<Checkpoint>[碳正离子亲核倾向于生成更稳定的碳正离子，生成CC1CCC(C(C)2C)C13CC[C+](C)C2C3](1)</Checkpoint>

<Checkpoint>[选项D不是关键中间体，选项D错误](0.5)</Checkpoint>

因此，E选项正确。

<Checkpoint>[因此，E选项正确。](0.5)</Checkpoint>