The following one-pot reaction achieves a Ugi and aza-Wittig reaction cascade

<MultiModal>![Represented by molecular SMILES, the reactants are O=CC1=CC=CC=C1, C1(CNCC2=CC=CC=C2)=CC=CC=C1, O=C(O)C1=CC=CC=C1, [C-]#[N+]N=P(C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3, and the reaction conditions are room temperature in dichloromethane for two hours, generating the product](/media/uploads/9e56d771-0f64-48aa-828d-449535936e66.png)</MultiModal>

Regarding this reaction, the following statements are made. Determine which are correct:

1.  The reaction is an ionic reaction.

2.  The product contains 4 rings.

3.  The product contains 3 nitrogen atoms, and the farthest distance between nitrogen atoms is 4 atoms.

4.  The molecular weight of the main byproduct is greater than 250.
如下一锅反应实现了Ugi和aza-Wittig反应的串联

<MultiModal>![以分子SMILES表示，反应物为O=CC1=CC=CC=C1，C1(CNCC2=CC=CC=C2)=CC=CC=C1，O=C(O)C1=CC=CC=C1，[C-]#[N+]N=P(C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3，反应条件为室温在二氯甲烷中反应两小时，生成产物](/media/uploads/9e56d771-0f64-48aa-828d-449535936e66.png)</MultiModal>

关于此反应，有以下说法，判断哪些是正确的：

1.该反应是离子反应

2.产物含有4个环

3.产物含有3个氮原子，且氮原子最远相距4个原子

4.反应主要副产物分子量大于250
All other options are incorrect
1
2
3
4
1,2
1,3
1,4
2,3
2,4
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
According to the question, this problem is a tandem one-pot reaction of $Ugi$ reaction and $aza-Wittig$. Analyze the reactants:

The most reactive substances in the system are the secondary amine and the aldehyde carbonyl group. The partially ionized carboxylic acid provides hydrogen ions. Under this action, the two easily undergo nucleophilic addition-elimination, losing water to condense into an imine ion.

<MultiModal>![The reaction is expressed via the SMILES notation as：O=C([H])C1=CC=CC=C1.C2(CNCC3=CC=CC=C3)=CC=CC=C2>[H+].O>C4(/C=[N+](CC5=CC=CC=C5)/CC6=CC=CC=C6)=CC=CC=C4](/media/uploads/4290289b-83d4-4494-b68f-c625348d4400.png)</MultiModal>

<Checkpoint>[Generation of imine ion intermediate C4(/C=[N+](CC5=CC=CC=C5)/CC6=CC=CC=C6)=CC=CC=C4](1)</Checkpoint>

At this time, the imine ion amplifies the electrophilicity of the original carbonyl site. Meanwhile, the isocyanide in the system has a strong affinity because its carbon atom is at the terminal group and carries a negative charge. The two then undergo nucleophilic addition to generate a nitrilium ion:

<MultiModal>![The reaction is expressed via the SMILES notation as：[C-]#[N+]N=P(C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3.C4(/C=[N+](CC5=CC=CC=C5)\CC6=CC=CC=C6)=CC=CC=C4>>C7(C(C#[N+]N=P(C8=CC=CC=C8)(C9=CC=CC=C9)C%10=CC=CC=C%10)N(CC%11=CC=CC=C%11)CC%12=CC=CC=C%12)=CC=CC=C7](/media/uploads/f93d53af-d747-48bc-8669-d513ac3c376e.png)</MultiModal>

<Checkpoint>[Generation of nitrilium ion intermediate C7(C(C#[N+]N=P(C8=CC=CC=C8)(C9=CC=CC=C9)C%10=CC=CC=C%10)N(CC%11=CC=CC=C%11)CC%12=CC=CC=C%12)=CC=CC=C7](1)</Checkpoint>

At this time, because the ionized carboxylate anion has a certain nucleophilicity and the carboxyl oxygen atom produces a certain coordination effect with the phosphorus atom, it is relatively easy to perform nucleophilic addition to the nitrilium ion to generate the second imine intermediate:

<MultiModal>![[O-]C(C1=CC=CC=C1)=O.C2(C(C#[N+]N=P(C3=CC=CC=C3)(C4=CC=CC=C4)C5=CC=CC=C5)N(CC6=CC=CC=C6)CC7=CC=CC=C7)=CC=CC=C2>>O=C(O/C(C(N(CC8=CC=CC=C8)CC9=CC=CC=C9)C%10=CC=CC=C%10)=N\N=P(C%11=CC=CC=C%11)(C%12=CC=CC=C%12)C%13=CC=CC=C%13)C%14=CC=CC=C%14](/media/uploads/321935ec-8065-4a58-8c5e-575fd6e52e3a.png)</MultiModal>

<Checkpoint>[Generation of the second imine intermediate O=C(O/C(C(N(CC8=CC=CC=C8)CC9=CC=CC=C9)C%10=CC=CC=C%10)=N\N=P(C%11=CC=CC=C%11)(C%12=CC=CC=C%12)C%13=CC=CC=C%13)C%14=CC=CC=C%14](1)</Checkpoint>

Finally, according to the question, a heteroatom $aza-Wittig$ reaction will occur in one step. We note that the intermediate contains a nitrogen-phosphorus double bond and a carbon-oxygen double bond. Because the phosphorus atom has an affinity for oxygen, and the formation of a five-membered ring and triphenoxyphosphine after the reaction is thermodynamically stable, this step of the $Wittig$ reaction is kinetically and thermodynamically favorable, so it occurs:

<MultiModal>![The reaction is expressed via the SMILES notation as：O=C(C1=CC=CC=C1)O/C(C(C2=CC=CC=C2)N(CC3=CC=CC=C3)CC4=CC=CC=C4)=N\N=P(C5=CC=CC=C5)(C6=CC=CC=C6)C7=CC=CC=C7>>O=P(C8=CC=CC=C8)(C9=CC=CC=C9)C%10=CC=CC=C%10.C%11(C(C%12=NN=C(C%13=CC=CC=C%13)O%12)N(CC%14=CC=CC=C%14)CC%15=CC=CC=C%15)=CC=CC=C%11](/media/uploads/9ee16eba-0870-46c5-b7e2-107dd1b6a5f8.png)</MultiModal>

<Checkpoint>[Obtain product C%11(C(C%12=NN=C(C%13=CC=CC=C%13)O%12)N(CC%14=CC=CC=C%14)CC%15=CC=CC=C%15)=CC=CC=C%11](1)</Checkpoint>

Therefore, this reaction is an ionic mechanism reaction.

The product contains four benzene rings and a newly formed heterocycle, for a total of five rings, not four rings.

Observing the product structure, the product contains 3 nitrogen atoms, and the nitrogen atoms are at most 3 atoms apart, not 4.

The reaction will produce a byproduct of triphenoxyphosphine, with a molecular weight of 278, which is greater than 250.

In summary, 1 and 4 are correct.
由题意，本题是$Ugi$反应和$aza-Wittig$的串联一锅煮反应。对反应物进行分析：

体系中反应性最高的是二级胺和醛羰基，部分电离的羧酸提供了氢离子，在此作用下，二者容易通过亲核加成-消除，失水缩合为亚胺离子。

<MultiModal>![反应用SMILES表示为：O=C([H])C1=CC=CC=C1.C2(CNCC3=CC=CC=C3)=CC=CC=C2>[H+].O>C4(/C=[N+](CC5=CC=CC=C5)/CC6=CC=CC=C6)=CC=CC=C4](/media/uploads/4290289b-83d4-4494-b68f-c625348d4400.png)</MultiModal>

<Checkpoint>[生成亚胺离子中间体C4(/C=[N+](CC5=CC=CC=C5)/CC6=CC=CC=C6)=CC=CC=C4](1)</Checkpoint>

此时亚胺离子放大了原有羰基位点的缺电性，而此时体系中异腈由于其碳原子处于端基且带一个负电荷，亲和性较强，二者随即发生亲核加成生成腈鎓离子：

<MultiModal>![反应用SMILES表示为：[C-]#[N+]N=P(C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3.C4(/C=[N+](CC5=CC=CC=C5)\CC6=CC=CC=C6)=CC=CC=C4>>C7(C(C#[N+]N=P(C8=CC=CC=C8)(C9=CC=CC=C9)C%10=CC=CC=C%10)N(CC%11=CC=CC=C%11)CC%12=CC=CC=C%12)=CC=CC=C7](/media/uploads/f93d53af-d747-48bc-8669-d513ac3c376e.png)</MultiModal>

<Checkpoint>[生成腈鎓离子中间体C7(C(C#[N+]N=P(C8=CC=CC=C8)(C9=CC=CC=C9)C%10=CC=CC=C%10)N(CC%11=CC=CC=C%11)CC%12=CC=CC=C%12)=CC=CC=C7](1)</Checkpoint>

此时由于电离的羧基负离子具有一定亲核性，并由于羧基氧与磷原子产生一定的配位作用，所以较为容易地对腈鎓离子进行亲核加成生成第二个亚胺中间体：

<MultiModal>![[O-]C(C1=CC=CC=C1)=O.C2(C(C#[N+]N=P(C3=CC=CC=C3)(C4=CC=CC=C4)C5=CC=CC=C5)N(CC6=CC=CC=C6)CC7=CC=CC=C7)=CC=CC=C2>>O=C(O/C(C(N(CC8=CC=CC=C8)CC9=CC=CC=C9)C%10=CC=CC=C%10)=N\N=P(C%11=CC=CC=C%11)(C%12=CC=CC=C%12)C%13=CC=CC=C%13)C%14=CC=CC=C%14](/media/uploads/321935ec-8065-4a58-8c5e-575fd6e52e3a.png)</MultiModal>

<Checkpoint>[生成第二个亚胺中间体O=C(O/C(C(N(CC8=CC=CC=C8)CC9=CC=CC=C9)C%10=CC=CC=C%10)=N\N=P(C%11=CC=CC=C%11)(C%12=CC=CC=C%12)C%13=CC=CC=C%13)C%14=CC=CC=C%14](1)</Checkpoint>

最后，由题意会发生一步杂原子$aza-Wittig$反应，我们注意到该中间体含有氮磷双键和碳氧双键，由于磷原子亲氧的特性，且反应后生成五元环和三苯氧磷，热力学稳定，此步$Wittig$反应在动力学和热力学上均处于有利地位，故发生：

<MultiModal>![反应用SMILES表示为：O=C(OC(C(N(CC1=CC=CC=C1)CC2=CC=CC=C2)C3=CC=CC=C3)=NN=P(C4=CC=CC=C4)(C5=CC=CC=C5)C6=CC=CC=C6)C7=CC=CC=C7>>O=P(C8=CC=CC=C8)(C9=CC=CC=C9)C%10=CC=CC=C%10.C%11(C(C%12=NN=C(C%13=CC=CC=C%13)O%12)N(CC%14=CC=CC=C%14)CC%15=CC=CC=C%15)=CC=CC=C%11](/media/uploads/9ee16eba-0870-46c5-b7e2-107dd1b6a5f8.png)</MultiModal>

<Checkpoint>[得到产物C%11(C(C%12=NN=C(C%13=CC=CC=C%13)O%12)N(CC%14=CC=CC=C%14)CC%15=CC=CC=C%15)=CC=CC=C%11](1)</Checkpoint>

因此，此反应为离子机理反应。

产物含有四个苯环和一个新生成的杂环，共五个环，而非四个环。

观察产物结构，产物含有3个氮原子，且氮原子最远相距3个原子而非4个。

反应会产生副产物为三苯氧膦，分子量278大于250。

综上所述，1,4正确。