Observe the following reaction:
<MultiModal>![O=C([C@]1(C)CC2)CCCC1=CC2=O and SmI2 react in THF/MeOH solvent at 0°C to obtain A and B, both of which can be converted to C under KOH, MeOH/H2O conditions. C can be converted to D under NaBH4, NaOH, EtOH/H2O conditions.](/media/uploads/927ca69c-f052-4165-b16b-214fb08ef13d.png)</MultiModal>
It is known that both $\bf{A}$ and $\bf{B}$ can be converted to compound $\bf{C}$, and that both $\bf{A}$ and $\bf{B}$ as reaction substrates can yield $\bf{C}$ with similar yields. Both products $\bf{A}$ and $\bf{B}$ contain a three-membered ring. In the $^1$H NMR spectrum of $\bf{C}$, there are two peaks whose integral area corresponds to $3H$.
Which of the following statements is correct:
1. $\bf{A}$ and $\bf{B}$ cannot be converted to each other in the presence of $\mathrm{KOH}$
2. The structure of $\bf{C}$ is shown in the figure
<MultiModal>![O=C1CCC[C@]2(C)[C@@]1(C)CCC(C2)=O](/media/uploads/b8f67f75-060d-4578-9f24-6d81a7e64029.png)</MultiModal>
3. Given that $\bf{D}$ has one ketone carbonyl group, then the structure of $\bf{D}$ should be as shown in the figure
<MultiModal>![C[C@@]12[C@](CCC2=O)(C)[C@@H](O)CCC1](/media/uploads/0bd9ea63-606d-4fdd-b20c-2c7f3c250301.png)</MultiModal>
4. The reason why only one carbonyl group can be selectively reduced from $\bf{C}$ to $\bf{D}$ is that there is a difference in steric hindrance around the two carbonyl groups.
观察下面的反应：
<MultiModal>![O=C([C@]1(C)CC2)CCCC1=CC2=O和SmI2在THF/MeOH溶剂中于0°C下反应得到A和B，它们都可以在KOH，MeOH/H2O的条件下转化为C. C可以在NaBH4, NaOH, EtOH/H2O的条件下转化为D。](/media/uploads/927ca69c-f052-4165-b16b-214fb08ef13d.png)</MultiModal>
已知$\bf{A}$和$\bf{B}$都可以转化为化合物$\bf{C}$，且无论是$\bf{A}$还是$\bf{B}$作为反应底物时都可以以相近的产率得到$\bf{C}$。$\bf{A}$和$\bf{B}$两种产物都含有三元环。$\bf{C}$的核磁共振氢谱中存在两个峰的积分面积对应为$3H$。
下面的说法中正确的是：
1. 在$\mathrm{KOH}$的存在下$\bf{A}$和$\bf{B}$无法互相转化
2. $\bf{C}$的结构如图所示
<MultiModal>![O=C1CCC[C@]2(C)[C@@]1(C)CCC(C2)=O](/media/uploads/b8f67f75-060d-4578-9f24-6d81a7e64029.png)</MultiModal>
3. 已知$\bf{D}$中有一个酮羰基，那么$\bf{D}$的结构应该如图所示
<MultiModal>![C[C@@]12[C@](CCC2=O)(C)[C@@H](O)CCC1](/media/uploads/0bd9ea63-606d-4fdd-b20c-2c7f3c250301.png)</MultiModal>
4. $\bf{C}$到$\bf{D}$中能选择性只还原一个羰基的原因是两个羰基周围的空间位阻存在区别。
All other options are incorrect
2.4.
1.3.
1.2.4.
3.4.
2.3.4.
3.
其他选项均不正确
2.4.
1.3.
1.2.4.
3.4.
2.3.4.
3.
It is known that both $\bf{A}$ and $\bf{B}$ can be converted into compound $\bf{C}$, and whether $\bf{A}$ or $\bf{B}$ is used as the reaction substrate, $\bf{C}$ can be obtained with similar yields, indicating that $\bf{A}$ and $\bf{B}$ can be interconverted in the presence of KOH. 1 Incorrect

<Checkpoint>[$\bf{A}$ and $\bf{B}$ can be interconverted in the presence of KOH](1)</Checkpoint>

Therefore, its structure can be deduced:

<MultiModal>![O=C1CCCC23[C@@]1(C)CCC(O)2C3
OC12CCCC31[C@@]2(C)CCC(C3)=O
](/media/uploads/4f77346a-8af8-4ab1-bedb-3dd7aa2a9e72.png)</MultiModal>

<Checkpoint>[The structure of $\bf{A}$ is O=C1CCCC23[C@@]1(C)CCC(O)2C3, and the structure of $\bf{B}$ is OC12CCCC31[C@@]2(C)CCC(C3)=O](2)</Checkpoint>

The former can undergo ring-opening to obtain the hexafused-pentacyclic structure $\bf{C}$. Contains two methyl groups. 2 Incorrect

<MultiModal>![C[C@@]12[C@](CCC2=O)(C)C(CCC1)=O](/media/uploads/3e6451d2-12f8-45d0-a92c-3eb8a4b8f16b.png)</MultiModal>

<Checkpoint>[The structure of $\bf{C}$ is C[C@@]12[C@](CCC2=O)(C)C(CCC1)=O](1)</Checkpoint>

The hexacyclic carbonyl group is reduced more quickly because it experiences less torsional strain during the reduction process, yielding $\bf{D}$

<MultiModal>![C[C@@]12[C@](CCC2=O)(C)[C@@H](O)CCC1](/media/uploads/0bd9ea63-606d-4fdd-b20c-2c7f3c250301.png)</MultiModal>

<Checkpoint>[The hexacyclic carbonyl group is reduced more quickly because it experiences less torsional strain during the reduction process](1)</Checkpoint>

<Checkpoint>[The structure of $\bf{D}$ is C[C@@]12[C@](CCC2=O)(C)[C@@H](O)CCC1](1)</Checkpoint>

3 Correct, 4 Incorrect.

G is correct.
已知$\bf{A}$和$\bf{B}$都可以转化为化合物$\bf{C}$，且无论是$\bf{A}$还是$\bf{B}$作为反应底物时都可以以相近的产率得到$\bf{C}$ 说明在KOH的存在下$\bf{A}$和$\bf{B}$可以互相转化。1错误

<Checkpoint>[在KOH的存在下$\bf{A}$和$\bf{B}$可以互相转化](1)</Checkpoint>

因此，可以推出其结构：

<MultiModal>![O=C1CCCC23[C@@]1(C)CCC(O)2C3
OC12CCCC31[C@@]2(C)CCC(C3)=O
](/media/uploads/4f77346a-8af8-4ab1-bedb-3dd7aa2a9e72.png)</MultiModal>

<Checkpoint>[$\bf{A}$的结构为O=C1CCCC23[C@@]1(C)CCC(O)2C3 ，$\bf{B}$的结构为OC12CCCC31[C@@]2(C)CCC(C3)=O](2)</Checkpoint>

前者可以开环得到六并五结构的$\bf{C}$。含有两个甲基。2错误

<MultiModal>![C[C@@]12[C@](CCC2=O)(C)C(CCC1)=O](/media/uploads/3e6451d2-12f8-45d0-a92c-3eb8a4b8f16b.png)</MultiModal>

<Checkpoint>[$\bf{C}$的结构为C[C@@]12[C@](CCC2=O)(C)C(CCC1)=O](1)</Checkpoint>

六元环羰基因为在还原过程中受到的扭转张力更小，能更快地被还原，得到$\bf{D}$

<MultiModal>![C[C@@]12[C@](CCC2=O)(C)[C@@H](O)CCC1](/media/uploads/0bd9ea63-606d-4fdd-b20c-2c7f3c250301.png)</MultiModal>

<Checkpoint>[六元环羰基因为在还原过程中受到的扭转张力更小，能更快地被还原](1)</Checkpoint>

<Checkpoint>[$\bf{D}$的结构为C[C@@]12[C@](CCC2=O)(C)[C@@H](O)CCC1](1)</Checkpoint>

3正确，4错误。

G是正确的。