Recently, OL published an article on methodology for the synthesis of a 6-5-5 tricyclic system. The reaction scheme is shown in Figure 1, where the orientation of the hydrogen atom marked with $\mathbf{*}$ is unknown.

<MultiModal>![Fig. 1, the figure shows a two-step sequential reaction. The first step reaction is represented by SMILES as: O=C1CCCC(C2(O[Si](C)(C)C)CCC2)=C1>>[A], the reaction conditions are TMSOTf, DCM, 0℃. The second step reaction is represented by SMILES as: [A]>[R]/C=C/[N+]([O-])=O>O[C@]1(C(C(C23[H*])[R])[N+]([O-])=O)CCC[C@@]31CCCC2=O. The reaction conditions are TiCl4, -78℃](/media/uploads/8418469f-c86c-45b2-aa39-32aab2786ba1.png)</MultiModal>

Deduce the structure of $\mathbf{A}$, considering the orientation of the hydrogen atom marked with $\mathbf{*}$.

There are the following statements:

1.  $\mathbf{A}$ contains three rings of six members or less.
2.  $\mathbf{A}$ contains a carbonyl group located on a six-membered ring.
3.  With the five-membered ring containing $\mathbf{R}$ as the plane, the reaction product with $\mathbf{H^*}$ pointing out of the plane of the paper is more abundant.
4.  With the five-membered ring containing $\mathbf{R}$ as the plane, the reaction product with $\mathbf{H^*}$ pointing into the plane of the paper is more abundant.

Among the following options, the one with all correct statements and the largest number of correct statements is:
近期, OL刊登了一篇构建 6-5-5 三环的方法学文章。其反应模式如图1所示，图中$\mathbf{*}$标记的氢原子朝向未知。

<MultiModal>![Fig. 1，图中为两步连续反应，第一步反应以SMILES记做：O=C1CCCC(C2(O[Si](C)(C)C)CCC2)=C1>>[A]，反应条件为TMSOTf、DCM、0℃。第二步反应以SMILES记做：[A]>[R]/C=C/[N+]([O-])=O>O[C@]1(C(C(C23[H*])[R])[N+]([O-])=O)CCC[C@@]31CCCC2=O。反应条件为TiCl4、-78℃](/media/uploads/8418469f-c86c-45b2-aa39-32aab2786ba1.png)</MultiModal>

推断$\mathbf{A}$的结构，考虑被$\mathbf{*}$标记的氢原子的朝向。

有以下说法

1. $\mathbf{A}$含有三个六元及以下的环
2. $\mathbf{A}$中含有位于六元环上的羰基
3. 以$\mathbf{R}$所在五元环为平面，$\mathbf{H^*}$指向纸面外的反应产物更多
4. 以$\mathbf{R}$所在五元环为平面，$\mathbf{H^*}$指向纸面内的反应产物更多

下列选项中说法全部正确且正确说法数量最多的为：
All other options are incorrect
1
2
3
4
1,2
1,3
1,4
2,3
2,4
1,2,3
1,2,4
其他选项均不正确
1
2
3
4
1,2
1,3
1,4
2,3
2,4
1,2,3
1,2,4
The first step, under the action of a strong Lewis acid, is a Pinacol-like rearrangement reaction, which opens the unstable four-membered ring, forming a more stable five-membered spirocycle, forming intermediate $\mathbf{A}$ as shown in Figure 2.

<MultiModal>![Fig. 2, the molecule is represented by SMILES as: O=C(CCC1)C21C=C(O[Si](C)(C)C)CCC2](/media/uploads/652389fe-dd6e-48fd-87f8-f28ac33e79c2.png)</MultiModal>

<Checkpoint>[Pinacol-like rearrangement occurs to form the intermediate O=C(CCC1)C21C=C(O[Si](C)(C)C)CCC2](1)</Checkpoint>

$\mathbf{A}$ contains only two rings, statement 1 is incorrect. The carbonyl group is located on the five-membered ring, statement 2 is incorrect.

After adding the unsaturated nitroalkene, a Michael addition reaction occurs, yielding a carbanion, which further undergoes an addition reaction with the nearby carbonyl group, and workup yields the final product.

<Checkpoint>[The silyl enol ether further undergoes a Michael addition reaction, and the resulting carbanion is coupled with the nearby carbonyl group to obtain the product given in the question](1)</Checkpoint>

When $\mathbf{A}$ is added to the nitroalkene, the silyl enol ether can initiate an attack from above the six-membered ring, i.e., near the carbonyl group of the five-membered spirocycle, or from below the six-membered ring, i.e., near the methylene group of the five-membered spirocycle. Both the intramolecular carbonyl group and the enol hydroxyl group are bonded to the Lewis acid $\mathrm{TiCl_4}$. The carbonyl group has a larger steric hindrance, while the methylene group contains only two hydrogen atoms and has a smaller steric hindrance. Therefore, the reaction preferentially occurs on the side away from the carbonyl group, resulting in a reaction product with the five-membered ring where $\mathbf{R}$ is located as the plane and $\mathbf{H^*}$ pointing out of the paper. Statement 3 is correct, and statement 4 is incorrect.

<Checkpoint>[Since the carbonyl group is bonded to the Lewis acid $\mathrm{TiCl_4}$ and has a larger steric hindrance than the methylene group, the Michael addition reaction preferentially occurs on the side away from the carbonyl group](1)</Checkpoint>
第一步在强路易斯酸作用下，发生类Pinacol重排反应，打开不稳定的四元环，形成更稳定的五元螺环，形成中间体$\mathbf{A}$如图2。

<MultiModal>![Fig. 2，分子以SMILES表示为：O=C(CCC1)C21C=C(O[Si](C)(C)C)CCC2](/media/uploads/652389fe-dd6e-48fd-87f8-f28ac33e79c2.png)</MultiModal>

<Checkpoint>[发生类Pinacol重排形成中间体O=C(CCC1)C21C=C(O[Si](C)(C)C)CCC2](1)</Checkpoint>

$\mathbf{A}$中仅含有两个环，说法1错误。羰基位于五元环上，说法2错误。

加入不饱和硝基烯烃后，发生Michael加成反应，得到碳负离子，该碳负离子进一步与附近的羰基发生加成反应，后处理得到最终产物。

<Checkpoint>[烯醇硅醚进一步发生Michael加成反应，得到的碳负离子与附近的羰基偶联，得到题中给出的产物](1)</Checkpoint>

在$\mathbf{A}$与硝基烯烃加成时，烯醇硅醚可以从六元环上方即靠近五元螺环羰基处发起进攻，或从六元环下方即靠近五元螺环亚甲基处发起进攻。分子内羰基和烯醇羟基均与路易斯酸$\mathrm{TiCl_4}$结合，羰基空间位阻较大，亚甲基只含有两个氢原子，空间位阻较小，因此反应优先在远离羰基一侧发生，得到以$\mathbf{R}$所在五元环为平面，$\mathbf{H^*}$指向纸面外的反应产物，说法3正确，4错误。

<Checkpoint>[由于羰基与路易斯酸$\mathrm{TiCl_4}$结合，位阻大于亚甲基，因此Michael加成反应优先在远离羰基侧发生](1)</Checkpoint>