Compound $\mathbf{A}$ (structure shown in Figure 1) generates $\mathbf{B}$ with the same molecular formula as $\mathbf{A}$ under acidic conditions. $\mathbf{A}$ reacts with methanol to form two isomeric products, both of which yield compound $\mathbf{C}$ upon reduction with $\mathrm{LiAlH}_{4}$.  

<MultiModal>![Fig. 1. The molecular structure is represented by SMILES: C1=CC=C(C=C1)C2CCC(=O)OC2=O](/media/uploads/dcecdd2b-a863-4326-9fb6-0db0c80a871a.png)</MultiModal>  

Several hypotheses exist regarding the structures of $\mathbf{B}$ and $\mathbf{C}$:  
1. $\mathbf{B}$ contains a highly strained structure.  
2. $\mathbf{B}$ can form a stable isomer under strong alkaline conditions.  
3. All carbon-oxygen bonds in $\mathbf{C}$ are double bonds.  
4. $\mathbf{C}$ contains only one carbon-oxygen double bond.  
5. $\mathbf{C}$ contains no carbon-oxygen double bonds.  

Among the following options, which one includes all correct statements and the maximum number of correct statements?
化合物$\mathbf{A}$(结构如图1所示)在酸性条件下会生成分子式与$\mathbf{A}$相同的$\mathbf{B}$。$\mathbf{A}$与甲醇反应形成互为同分异构体的两个产物，这两个产物分别用 $\mathrm {LiAlH} _ {4}$ 还原均得到化合物$\mathbf{C}$。

<MultiModal>![Fig. 1，图中分子以SMILES表示为：C1=CC=C(C=C1)C2CCC(=O)OC2=O](/media/uploads/dcecdd2b-a863-4326-9fb6-0db0c80a871a.png)</MultiModal>

推测$\mathbf{B}$和$\mathbf{C}$的结构，有以下几种说法：
1. $\mathbf{B}$包含一个大张力结构
2. $\mathbf{B}$在强碱下可以形成稳定的异构体
3. $\mathbf{C}$中所有的碳-氧键均为双键
4. $\mathbf{C}$仅含一根碳氧双键
5. $\mathbf{C}$不含任何碳氧双键

以下选项中全部正确且正确说法最多的是？
All other options are incorrect
1
2
3
4
5
1,2
1,3
1,4
1,5
2,3
2,4
2,5
1,2,3
1,2,4
1,2,5
其他选项均不正确
1
2
3
4
5
1,2
1,3
1,4
1,5
2,3
2,4
2,5
1,2,3
1,2,4
1,2,5
Compound $\mathbf{A}$ is an intramolecular anhydride and may undergo Friedel-Crafts acylation with an aryl group under acidic conditions. Considering that the resulting compound $\mathbf{B}$ has the same molecular formula as compound $\mathbf{A}$, the acylation reaction occurs intramolecularly rather than intermolecularly. Between the two acyl groups as potential reaction sites, one would form a four-membered ring, which is highly strained and unlikely, while the other forms a six-membered ring with a lower energy barrier. Therefore, the structure of compound $\mathbf{B}$ is shown in Figure 2. The molecule does not form a highly strained four-membered ring but instead a less strained six-membered ring, so statement 1 is incorrect. Under strong base conditions, the β-carbon of the carbonyl group can be deprotonated to form a stable enolate anion, so statement 2 is correct.  

<MultiModal>![Fig. 2. The molecule in the figure is described by SMILES as: C1=CC2=C(C=C1)C(=O)CCC2C(=O)O](/media/uploads/2c8bb182-d54d-4fb6-9b85-ff998f5487aa.png)</MultiModal>  

<Checkpoint>[Intramolecular Friedel-Crafts acylation forms a six-membered ring rather than a four-membered ring.](1)</Checkpoint>  

<Checkpoint>[The product is represented by SMILES as C1=CC2=C(C=C1)C(=O)CCC2C(=O)O.](1)</Checkpoint>  

<Checkpoint>[Under strong base conditions, the β-carbon of the carbonyl group can be deprotonated to form a stable enolate anion.](1)</Checkpoint>  

Compound $\mathbf{A}$ is an anhydride and undergoes alcoholysis when reacted with methanol. Since compound $\mathbf{A}$ contains two asymmetric acyl groups, alcoholysis yields two products, each containing a methyl ester group and a carboxyl group. Both groups are directly reduced to alcohols in a single step by the strong reducing agent $\mathrm{LiAlH}_4$, resulting in the same product $\mathbf{C}$, whose structure is shown in Figure 3. The molecule contains no carbon-oxygen double bonds, only carbon-oxygen single bonds, so statements 3 and 4 are incorrect, while statement 5 is correct.  

<MultiModal>![Fig. 3. The molecule in the figure is described by SMILES as: C1=CC=C(C=C1)C(CCCO)CO](/media/uploads/c43000b9-cf46-4760-8041-ff7e0446679d.png)</MultiModal>  

<Checkpoint>[Methanolysis yields two products, each containing a carboxyl group and an ester group, which are directly reduced to alcohols in a single step by the strong reducing agent $\mathrm{LiAlH}_4$, resulting in the same product.](1)</Checkpoint>  

<Checkpoint>[$\mathbf{C}$ is a diol structure, described by SMILES as: C1=CC=C(C=C1)C(CCCO)CO.](1)</Checkpoint>
化合物$\mathbf{A}$为分子内酸酐，在酸性下可能与芳基发生傅-克酰化反应。考虑到反应后得到的化合物$\mathbf{B}$与化合物$\mathbf{A}$分子式相同，说明酰基化反应发生在分子内而非分子间。考虑两个酰基作为反应位点，一个反应形成四元环，张力太大不太可能，另一个形成六元环，反应能垒低，所以化合物$\mathbf{B}$结构如图2。分子并未形成大张力的四元环，而是张力不大的六元环，说法1错误。强碱下羰基β碳可以去质子化，形成稳定的烯醇负离子，说法2正确。

<MultiModal>![Fig. 2，图中分子以SMILES描述为：C1=CC2=C(C=C1)C(=O)CCC2C(=O)O](/media/uploads/2c8bb182-d54d-4fb6-9b85-ff998f5487aa.png)</MultiModal>

<Checkpoint>[分子内傅-克酰基化形成六元环而非四元环，](1)</Checkpoint>

<Checkpoint>[产物以SMILES表示为C1=CC2=C(C=C1)C(=O)CCC2C(=O)O](1)</Checkpoint>

<Checkpoint>[强碱下羰基β碳可以去质子化，形成稳定的烯醇负离子](1)</Checkpoint>

化合物$\mathbf{A}$为酸酐，与甲醇反应发生醇解。化合物$\mathbf{A}$中有两个不对称的酰基，因此醇解生成两个产物，每个产物均有一个甲醇酯基团和一个羧基基团。两个基团在$\mathrm {LiAlH} _ {4}$强还原剂作用下直接一步还原为醇，因此得到同样的产物$\mathbf{C}$，结构如图3。分子中不含碳氧双键，只有碳氧单键，说法3、4错误，5正确。

<MultiModal>![Fig. 3，图中分子以SMILES描述为：C1=CC=C(C=C1)C(CCCO)CO](/media/uploads/c43000b9-cf46-4760-8041-ff7e0446679d.png)</MultiModal>

<Checkpoint>[甲醇醇解得到两个产物，均包含一个羧基一个酯基，在$\mathrm {LiAlH} _ {4}$强还原剂作用下直接一步还原为醇，因此得到同样的产物](1)</Checkpoint>

<Checkpoint>[$\mathbf{C}$为二醇结构，以SMILES描述为：C1=CC=C(C=C1)C(CCCO)CO](1)</Checkpoint>