Infer the mechanism and intermediate(s) of the organic reaction in Figure 1.

<MultiModal>![Fig. 1, the reaction in the figure is represented by SMILES: CN1C2CC=CC(CC2)C1=O>CN1[C@@H]2CC[C@H]1CC[C@H]2C(C3=CC=CC=C3)=O, the reaction conditions are first phenyllithium, tetrahydrofuran, and then aqueous sodium hydroxide solution](/media/uploads/df893e2c-846b-4de6-b685-d0b189da8f66.png)</MultiModal>

There are the following statements:

1. Phenyllithium is strongly basic and functions as a base in this reaction to deprotonate the reactant.
2. A carbanion-mediated elimination reaction occurs in the reaction.
3. After the addition of $\mathrm{NaOH}$, a nucleophilic attack by the carbanion occurs in the reaction.
4. If the methyl group on the nitrogen atom of the reactant is replaced with a hydrogen atom, the reaction may no longer occur.

Select the option that contains the most correct statement numbers.
推断图1有机反应的机理和经过的中间体。

<MultiModal>![Fig. 1，图中反应以SMILES表示：CN1C2CC=CC(CC2)C1=O>CN1[C@@H]2CC[C@H]1CC[C@H]2C(C3=CC=CC=C3)=O，反应条件为先苯基锂、四氢呋喃，然后氢氧化钠水溶液](/media/uploads/df893e2c-846b-4de6-b685-d0b189da8f66.png)</MultiModal>

有以下几种说法：

1. 苯基锂碱性较强，在该反应中作为碱起到给反应物去质子化的功能
2. 反应中发生了碳负离子介导的消除反应
3. 加入$\mathrm{NaOH}$后反应中发生了碳负离子的亲核进攻反应
4. 若将反应物氮原子上的甲基换成氢原子，该反应可能不再能发生

选出包含有最多正确说法编号的选项。
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
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
1,2,3,4
Phenyllithium may participate in the reaction as a base or as a nucleophile. Only the hydrogen atom on the β-carbon of the carbonyl group in the reactant has a certain acidity, but this carbon atom is a bridgehead carbon atom, and deprotonation to form a double bond will cause excessive bridge ring strain, making the reaction difficult to occur. Therefore, phenyllithium is most likely to attack the carbonyl carbon as a nucleophile, producing an intermediate as shown in Figure 2. Statement 1 is incorrect. If the methyl group on the nitrogen atom of the reactant is replaced with a hydrogen atom, phenyllithium may undergo deprotonation reaction with the amino group, generating a negatively charged conjugated amide-enolate, which makes the next nucleophilic attack of phenyllithium on the carbonyl group more difficult. Statement 4 is correct.

<MultiModal>![Fig. 2，图中分子以SMILES表示为：CN1[C@@H]2CC[C@H]([C@@]1(C3=CC=CC=C3)O[Li])C=CC2](/media/uploads/dd62a9bc-3963-4d02-86c6-f601713722fd.png)</MultiModal>

<Checkpoint>[Phenyllithium attacks the amide carbonyl as a nucleophile, forming an intermediate with SMILES representation: CN1[C@@H]2CC[C@H]([C@@]1(C3=CC=CC=C3)O[Li])C=CC2](1)</Checkpoint>

The enolate anion undergoes an elimination reaction in the next step, possibly similar to a reverse Michael addition reaction to generate a carbanion, but this carbanion is unstable. It may also be eliminated to form an amino anion, which is equivalent to amide hydrolysis. Therefore, the second intermediate is shown in Figure 3.

<MultiModal>![Fig. 3，图中分子以SMILES描述为：CN[C@H]1CC=CC(C(C2=CC=CC=C2)=O)CC1](/media/uploads/72280c56-e2bb-40bd-800d-8d6248983e4e.png)</MultiModal>

<Checkpoint>[Elimination of the oxygen anion causes the amino group to leave, forming an intermediate with SMILES representation: CN[C@H]1CC=CC(C(C2=CC=CC=C2)=O)CC1](1)</Checkpoint>

Under alkaline conditions, deprotonation of the carbonyl β-carbon causes double bond migration, forming a stable β-unsaturated ketone. The intramolecular active secondary amino group can further undergo intramolecular Michael addition, generating the product in the question.
Based on the above reaction mechanism, no carbanion-mediated elimination reaction occurs in the reaction, and no nucleophilic attack reaction of carbanions occurs after adding $\mathrm{NaOH}$, so statements 2 and 3 are incorrect.

<Checkpoint>[Secondary amino group undergoes intramolecular Michael addition to generate the product](1)</Checkpoint>
苯基锂可能作为碱或者作为亲核试剂参与反应。反应物中只有羰基β位碳上的氢原子具有一定酸性，但是该碳原子为桥头碳原子，去质子化形成双键会导致桥环张力太大，反应不易发生，因此苯基锂最可能作为亲核试剂进攻羰基碳，产生中间体如图2。说法1错误。若将反应物氮原子上的甲基换成氢原子，则苯基锂可能与氨基发生去质子化反应，生成带负电的共轭酰胺-烯醇，导致下一步苯基锂对羰基亲核进攻更困难，说法4正确。

<MultiModal>![Fig. 2，图中分子以SMILES表示为：CN1[C@@H]2CC[C@H]([C@@]1(C3=CC=CC=C3)O[Li])C=CC2](/media/uploads/dd62a9bc-3963-4d02-86c6-f601713722fd.png)</MultiModal>

<Checkpoint>[苯基锂作为亲核试剂进攻酰胺羰基，形成中间体以SMILES表示为：CN1[C@@H]2CC[C@H]([C@@]1(C3=CC=CC=C3)O[Li])C=CC2](1)</Checkpoint>

烯醇负离子下一步发生消除反应，可能类似逆Michael加成反应生成碳负离子，但是该碳负离子不稳定。也可能消除形成氨基负离子，该过程相当于酰胺水解。因此第二个中间体如图3。

<MultiModal>![Fig. 3，图中分子以SMILES描述为：CN[C@H]1CC=CC(C(C2=CC=CC=C2)=O)CC1](/media/uploads/72280c56-e2bb-40bd-800d-8d6248983e4e.png)</MultiModal>

<Checkpoint>[氧负离子消除使氨基离去，形成中间体以SMILES表示为：CN[C@H]1CC=CC(C(C2=CC=CC=C2)=O)CC1](1)</Checkpoint>

在碱性下，羰基β碳去质子化发生双键移位，形成稳定的β不饱和酮。分子内活性二级氨基可以进一步发生分子内Michael加成，生成题目中的产物。
综合以上反应机理，反应中未发生碳负离子介导的消除反应，加入$\mathrm{NaOH}$后也并没有发生碳负离子的亲核进攻反应，说法2、3错误。

<Checkpoint>[二级氨基发生分子内Michael加成，生成产物](1)</Checkpoint>