Propose a mechanism for the reaction shown in Figure 1, considering whether specific stereochemistry will form at the hydrogen atom marked with *, and the principle of its formation.

<MultiModal>![Fig. 1, The reaction in the figure is represented by SMILES as: CC1=CC=C(N)C=C1.CCCC(O)C(C2=CC=CC=C2)=O>>CCCC(C([H*])(C3=CC=CC=C3)NC4=CC=C(C=C4)C)=O, where the reaction conditions are: SPA (5 mol%), toluene, 4Å MS, 60°C.](/media/uploads/0323b423-d625-404d-b6c9-d2db46143ab8.png)</MultiModal>

The structure of $\mathrm{SPA}$ is shown in Figure 2:

<MultiModal>![Fig. 2, The molecular structure in the figure is described by SMILES as: O=P1(O)OC2=CC=CC3=C2[C@@](CC3)(CC4)C5=C4C=CC=C5O1](/media/uploads/9d23c066-1d8a-4ee0-840a-9938a72e2a97.png)</MultiModal>

There are the following statements:

1.  The reaction of the primary amine with the ketone is later than the elimination reaction
2.  The key step of the reaction involves two isomerization reactions
3.  In order to further increase the proportion of obtaining a product with specific stereochemistry, all of the oxygen atoms in the phosphate catalyst should be esterified with ligands as much as possible, leaving only one $\mathrm{P-OH}$ single bond
4.  Stereoselectivity originates from one step in the elimination reaction
推测图1所示反应的机理，考虑*所标记氢原子处是否会形成特异性立体化学及形成原理。

<MultiModal>![Fig. 1，图中反应以SMILES表示为：CC1=CC=C(N)C=C1.CCCC(O)C(C2=CC=CC=C2)=O>>CCCC(C([H*])(C3=CC=CC=C3)NC4=CC=C(C=C4)C)=O，其中反应条件为：SPA(5 mol%), toluene,  4Å MS, 60°C。](/media/uploads/0323b423-d625-404d-b6c9-d2db46143ab8.png)</MultiModal>

$\mathrm{SPA}$的结构如图2：

<MultiModal>![Fig. 2，图中分子结构以SMILES描述为：O=P1(O)OC2=CC=CC3=C2[C@@](CC3)(CC4)C5=C4C=CC=C5O1](/media/uploads/9d23c066-1d8a-4ee0-840a-9938a72e2a97.png)</MultiModal>

有以下说法：

1. 一级胺与酮的反应晚于消除反应
2. 反应关键步骤涉及了两步异构化反应
3. 为了进一步提升得到特定立体化学产物的比例，应该尽可能让磷酸酯催化剂中的所有氧原子与配体酯化，仅保留一个$\mathrm{P-OH}$单键
4. 立体选择性来源于发生消除反应的某一步
All other options are incorrect
1
2
3
4
1,2
1,3
1,4
2,3
2,4
3,4
1,2,3
1,2,4
1,3,4
2,3,4
1,2,3,4
其他选项均不正确
1
2
3
4
1,2
1,3
1,4
2,3
2,4
3,4
1,2,3
1,2,4
1,3,4
2,3,4
1,2,3,4
Primary amines react with α-hydroxy ketones to form imine compounds, and the imine can undergo tautomerization to form the structure in Figure 3:

<MultiModal>![Fig. 3, the molecule in the figure is represented by SMILES as: CC(C=C1)=CC=C1N/C(C2=CC=CC=C2)=C(CCC)/O](/media/uploads/70f54fa4-d512-47c2-9b3b-c62682ad98de.png)</MultiModal>

<Checkpoint>[Formation of imine, imine tautomerization to form the following structure described by SMILES: CC(C=C1)=CC=C1N/C(C2=CC=CC=C2)=C(CCC)/O](1)</Checkpoint>

This structure is also an enol structure, which can undergo proton transfer isomerization to generate ketone compounds. This process involves the formation of a chiral center, with the proton provided by stereospecific $\mathrm{SPA}$. The $\mathrm{P=O}$ of $\mathrm{SPA}$ can form a hydrogen bond complex with the hydroxyl group, while $\mathrm{P-OH}$ is close to the enol carbon atom, resulting in a one-step concerted proton transfer to generate ketones. In this process, the bulky groups on the phosphate play a stereoselective role, causing protonation to occur only on one side of the double bond plane.

Involving imine and ketone tautomerization, statement 2 is correct.
If the phosphate catalyst does not have $\mathrm{P=O}$, it cannot form a hydrogen-bonded catalytic concerted transition state with the carbonyl group, making it difficult to exert stereoselectivity, statement 3 is incorrect.

<Checkpoint>[$\mathrm{SPA}$ catalyzes enol proton transfer, $\mathrm{P=O}$ can form a hydrogen bond complex with the hydroxyl group, and the bulky groups of the phosphate play a stereoselective role](1)</Checkpoint>

Stereoselectivity comes from enol protonation, statements 1,4 are incorrect.
一级胺与羟基酮类反应生成亚胺化合物，亚胺可以发生互变异构形成图3结构：

<MultiModal>![Fig. 3，图中分子以SMILES表示为：CC(C=C1)=CC=C1N/C(C2=CC=CC=C2)=C(CCC)/O](/media/uploads/70f54fa4-d512-47c2-9b3b-c62682ad98de.png)</MultiModal>

<Checkpoint>[形成亚胺，亚胺互变异构形成SMILES描述的以下结构：CC(C=C1)=CC=C1N/C(C2=CC=CC=C2)=C(CCC)/O](1)</Checkpoint>

该结构正好也是烯醇结构，能够发生质子转移异构化生成酮类化合物。该过程涉及到手性中心的形成，质子由立体特异性的$\mathrm{SPA}$提供。$\mathrm{SPA}$的$\mathrm{P=O}$可以与羟基形成氢键络合，同时$\mathrm{P-OH}$靠近烯醇碳原子，发生一步协同质子转移生成酮类。该过程中磷酸酯上的大空阻基团起到了立体选择性作用，使得质子化只在双键平面的某一侧发生。

涉及到亚胺和酮类的互变异构，说法2正确。
若磷酸酯催化剂无$\mathrm{P=O}$，则无法与羰基形成氢键催化协同过渡态，难以发挥立体选择性，说法3错误。

<Checkpoint>[$\mathrm{SPA}$催化烯醇质子转移，$\mathrm{P=O}$可以与羟基形成氢键络合，磷酸酯大空阻基团起到立体选择性作用](1)</Checkpoint>

立体选择性来自于烯醇质子化，说法1,4错误。