The product $\mathbf{A}$ $\mathrm{(C_7H_8O_2)}$ was obtained through the following reaction. $\mathbf{A}$ was heated to $70$ $\degree \mathrm{C}$ in $\mathrm{CCl}_4$ solution, and $\mathbf{A}$ underwent intermediate $\mathbf{B}$ and partially converted into its isomer $\mathbf{C}$. The ratio of $\mathbf{A}$, $\mathbf{B}$, and $\mathbf{C}$ at equilibrium was $29: 42: 29$.

<MultiModal>![CC1=C[C@@H](Br)[C@H](OC(C)=O)[C@H](OC(C)=O)[C@H]1Br generates $\mathbf{A}$ under the conditions of $\mathrm{NaOMe}$ and $\mathrm{Et_2O}$](/media/uploads/252b045a-5458-477c-9cf0-0eb42c7bf85b.png)</MultiModal>

The following statements are made:

1. $\mathbf{A}$ contains a carbonyl group.
2. $\mathbf{B}$ has two rings.
3. $\mathbf{A}$ and $\mathbf{C}$ have the same number of rings.
4. $\mathbf{A}$ and $\mathbf{C}$ are enantiomers.
5. The sum of double bonds contained in $\mathbf{A}$, $\mathbf{B}$, and $\mathbf{C}$ is $5$.

The following statement(s) is/are correct:
通过以下反应得到了产物 $\mathbf{A}$ $\mathrm{(C_7H_8O_2)}$ 。将$\mathbf{A}$ 于 $\mathrm{CCl}_4$ 溶液中加热至 $70$ $\degree \mathrm{C}$ ，$\mathbf{A}$ 经历中间体 $\mathbf{B}$ 而部分转化为其同分异构体 $\mathbf{C}$ 。 $\mathbf{A}$ 、$\mathbf{B}$ 、$\mathbf{C}$ 三者在平衡时的比例为 $29: 42: 29$ 。

<MultiModal>![CC1=C[C@@H](Br)[C@H](OC(C)=O)[C@H](OC(C)=O)[C@H]1Br在 $\mathrm{NaOMe}$ 和 $\mathrm{Et_2O}$ 的条件下生成 $\mathbf{A}$ ](/media/uploads/252b045a-5458-477c-9cf0-0eb42c7bf85b.png)</MultiModal>

有以下说法：

1. $\mathbf{A}$ 中有羰基。
2. $\mathbf{B}$ 中有两个环。
3. $\mathbf{A}$ 和 $\mathbf{C}$ 的环个数相等。
4. $\mathbf{A}$ 和 $\mathbf{C}$ 是对映异构体。
5. $\mathbf{A}$ 、$\mathbf{B}$ 、$\mathbf{C}$ 含有双键之和为 $5$ 。

以下说法正确的是：
1, 2
1, 3
2, 4
3, 5
3, 4
4, 5
3, 4, 5
1, 2
1, 3
2, 4
3, 5
3, 4
4, 5
3, 4, 5
The system underwent the following reactions under strong alkaline $\mathrm{NaOMe}$ conditions:
1. The two acetoxy groups in the substrate react with $\mathrm{NaOMe}$ to generate sodium alkoxide and $\mathrm{MeOAc}$.

<Checkpoint>[Acetoxy groups react with $\mathrm{NaOMe}$ to generate sodium alkoxide and $\mathrm{MeOAc}$.](1)</Checkpoint>

2. The oxygen anion in the sodium alkoxide attacks the bromine group on the adjacent carbon atom, and an $\mathrm{S_N2}$ substitution reaction occurs to generate two epoxy groups, yielding compound $\mathbf{A}$.
$\mathbf{A}$ is CC1=C[C@H]2[C@H](O2)[C@H]3[C@@H]1O3.

<Checkpoint>[$\mathrm{S_N2}$ substitution reaction occurs, oxygen anion attacks adjacent bromine group, generating epoxy groups](1)</Checkpoint>

<Checkpoint>[$\mathbf{A}$ is CC1=C[C@H]2[C@H](O2)[C@H]3[C@@H]1O3.](2)</Checkpoint>

Under heating conditions, the epoxy groups in $\mathbf{A}$ are unstable and open to form a stable eight-membered ring compound $\mathbf{B}$ with three double bonds, which is an electrocyclic ring-opening process. This is consistent with $\mathbf{B}$ having the highest proportion among $\mathbf{A}$, $\mathbf{B}$, and $\mathbf{C}$ at equilibrium.
$\mathbf{B}$ is CC1=C\O/C=C\O\C=C/1.

<Checkpoint>[Electrocyclic ring-opening reaction occurs, the epoxy groups in $\mathbf{A}$ open to form the eight-membered ring compound $\mathbf{B}$](1)</Checkpoint>

<Checkpoint>[$\mathbf{B}$ is CC1=C\O/C=C\O\C=C/1.](2)</Checkpoint>

$\mathbf{A}$ and $\mathbf{C}$ are present in equal proportions at equilibrium, suggesting that they have equal energies and similar structures. Given that the electrocyclic ring-opening process from $\mathbf{A}$ to $\mathbf{B}$ is reversible, it can be inferred that $\mathbf{B}$ to $\mathbf{C}$ is an electrocyclic process. Compound $\mathbf{C}$ is CC1=C[C@@H]2[C@@H](O2)[C@@H]3[C@H]1O3. $\mathbf{A}$ and $\mathbf{C}$ are enantiomers.

<Checkpoint>[$\mathbf{A}$ and $\mathbf{C}$ have equal energies and similar structures.](1)</Checkpoint>

<Checkpoint>[$\mathbf{B}$ to $\mathbf{C}$ is an electrocyclic process.](1)</Checkpoint>

<Checkpoint>[$\mathbf{C}$ is CC1=C[C@@H]2[C@@H](O2)[C@@H]3[C@H]1O3](2)</Checkpoint>

Therefore, statements 3, 4, and 5 are correct. Choose option D.
该体系在强碱 $\mathrm{NaOMe}$ 条件下发生了以下反应：
1. 底物中两个乙酰氧基与 $\mathrm{NaOMe}$ 反应生成醇钠和 $\mathrm{MeOAc}$ 。

<Checkpoint>[ 乙酰氧基与 $\mathrm{NaOMe}$ 反应生成醇钠和 $\mathrm{MeOAc}$ 。](1)</Checkpoint>

2. 醇钠中氧负离子进攻临近碳原子上的溴基，发生 $\mathrm{S_N2}$ 取代反应生成两个环氧基，得到化合物 $\mathbf{A}$ 。
$\mathbf{A}$ 为CC1=C[C@H]2[C@H](O2)[C@H]3[C@@H]1O3。

<Checkpoint>[发生 $\mathrm{S_N2}$ 取代反应，氧负离子进攻临近溴基，生成环氧基 ](1)</Checkpoint>

<Checkpoint>[$\mathbf{A}$ 为CC1=C[C@H]2[C@H](O2)[C@H]3[C@@H]1O3。](2)</Checkpoint>

在加热条件下，$\mathbf{A}$ 中的环氧基不稳定，解开形成稳定的具有三个双键的八元环化合物 $\mathbf{B}$ ，为电环化开环过程。这与 $\mathbf{B}$ 在平衡时 $\mathbf{A}$ 、$\mathbf{B}$ 、$\mathbf{C}$ 中占比最高相符。
$\mathbf{B}$ 为CC1=C\O/C=C\O\C=C/1。

<Checkpoint>[发生电环化开环反应，$\mathbf{A}$ 中的环氧基解开形成八元环化合物 $\mathbf{B}$](1)</Checkpoint>

<Checkpoint>[$\mathbf{B}$ 为CC1=C\O/C=C\O\C=C/1。](2)</Checkpoint>

$\mathbf{A}$ 和 $\mathbf{C}$ 在平衡的时候比例相等，暗示它们的能量相等，结构类似。已知 $\mathbf{A}$ 到 $\mathbf{B}$ 的电环化开环过程是可逆的，可以推测 $\mathbf{B}$ 到 $\mathbf{C}$ 为电环化过程。化合物 $\mathbf{C}$ 为CC1=C[C@@H]2[C@@H](O2)[C@@H]3[C@H]1O3。$\mathbf{A}$ 和 $\mathbf{C}$ 为对映异构体。

<Checkpoint>[$\mathbf{A}$ 和 $\mathbf{C}$ 能量相等，结构类似。](1)</Checkpoint>

<Checkpoint>[$\mathbf{B}$ 到 $\mathbf{C}$ 为电环化过程。](1)</Checkpoint>

<Checkpoint>[ $\mathbf{C}$ 为CC1=C[C@@H]2[C@@H](O2)[C@@H]3[C@H]1O3](2)</Checkpoint>

因此说法3, 4, 5正确。选择D选项。