Pyridopyrimidinone is an important heterocyclic structure, and many drug molecules contain this ring system, so the construction of this ring system is very important in the pharmaceutical industry. In early 2023, researchers developed a one-pot synthesis method for this structure, which has high atom economy.

<MultiModal>![`COC1=CC=C(C=C1)C(C(=O)NC2=NC=CC=C2)C(=O)OC` first reacts with 1.2eq $Ph_3PO$, 1.15eq $(COCl)_2$ at 0 degrees Celsius, and then reacts with 3.0eq `CN1CCOCC1` at 0 degrees Celsius to obtain $\bf{A}$, $\bf{A}$ reacts with $PhSH$ to obtain $\bf{B}$, $\bf{B}$ loses one molecule of $MeOH$ to obtain $\bf{C}$, where the chemical formula of $\bf{A}$ is $C_{16}H_{14}N_2O_3$](/media/uploads/e2612996-f61a-4a76-b4f8-b236cc74eeec.png)</MultiModal>

The following statements are made:

1. In reaction (1), two gases are produced and they can be interconverted.

2. There are carbon atoms with $sp$ hybridization in $\bf{A}$.

3. $\bf{C}$ has 2 aromatic systems.

4. If the nucleophile in reaction (3) is replaced with aniline, the rate of formation of $\bf{C}$ from $\bf{B}$ will increase.

Select the option below that contains all the correct statements.
吡啶并嘧啶酮是一种重要的杂环结构，许多药物分子均含有这一环系，因此在制药行业该环系的构筑十分重要。2023年初，研究人员开发了一种一锅合成该结构的方法，具有很高的原子经济性。

<MultiModal>![`COC1=CC=C(C=C1)C(C(=O)NC2=NC=CC=C2)C(=O)OC`首先和1.2eq$Ph_3PO$、1.15eq$(COCl)_2$在0摄氏度下反应，随后同样在0摄氏度下与3.0eq `CN1CCOCC1`反应得到$\bf{A}$，$\bf{A}$与$PhSH$反应得到$\bf{B}$，$\bf{B}$失去一分子$MeOH$得到$\bf{C}$，其中$\bf{A}$的化学式为$C_{16}H_{14}N_2O_3$](/media/uploads/e2612996-f61a-4a76-b4f8-b236cc74eeec.png)</MultiModal>

有以下说法：

1.在反应（1）中，产生了两种气体且它们可相互转化

2.$\bf{A}$中存在杂化类型为$sp$的碳原子

3.$\bf{C}$中具有2个芳香体系

4.若将反应（3）的亲核试剂替换为苯胺，将会使$\bf{B}$生成$\bf{C}$的速率增加

选择以下选项中所含说法全部正确的一项
No correct statement.
1
2
3
4
1,2
1,3
1,4
2,4
3,4
1,2,4
1,2,3
1,3,4
2,3,4
1,2,3,4
无正确说法
1
2
3
4
1,2
1,3
1,4
2,3
3,4
1,2,4
1,2,3
1,3,4
2,3,4
1,2,3,4
First, triphenylphosphine is activated by oxalyl chloride to generate the active species `Cl[P+](C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3.[Cl-]`. This process also generates gases $CO$ and $CO_2$.

<Checkpoint>[Reaction (1) produces $CO$ and $CO_2$, and these two gases can be interconverted. Statement 1 is correct](1)</Checkpoint>

The phosphonium cation reacts with the amide oxygen to generate `COC1=CC=C(C(C(OC)=O)/C(O[P+](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4)=N/C5=NC=CC=C5)C=C1`, followed by elimination of triphenylphosphine under the action of base, to obtain $\bf{A}$: `COC1=CC=C(C(C(OC)=O)=C=NC2=NC=CC=C2)C=C1`.

<Checkpoint>[The structure of $\bf{A}$ is `COC1=CC=C(C(C(OC)=O)=C=NC2=NC=CC=C2)C=C1`](1)</Checkpoint>

<Checkpoint>[$\bf{A}$ has a cumulated diene structure with $sp$ hybridized carbon atoms. Statement 2 is correct](1)</Checkpoint>

The cumulated diene structure in $\bf{A}$ is susceptible to nucleophilic attack, reacting with thiophenol to give $\bf{B}$: `COC1=CC=C(C(C(OC)=O)/C(SC2=CC=CC=C2)=N/C3=NC=CC=C3)C=C1`.

<Checkpoint>[The structure of $\bf{B}$ is `COC1=CC=C(C(C(OC)=O)/C(SC2=CC=CC=C2)=N/C3=NC=CC=C3)C=C1`](1)</Checkpoint>

The pyridine nitrogen atom in $\bf{B}$ has nucleophilicity, nucleophilically attacks the methyl ester and leaves a methoxy anion. Subsequently, the α-hydrogen atom of the ester group is abstracted under the action of base, to obtain $\bf{C}$: `COC1=CC=C(C(C2=O)=C(SC3=CC=CC=C3)N=C4N2C=CC=C4)C=C1`. $\bf{C}$ has two phenyl groups. The tautomer of the pyridopyrimidinone ring system with nitrogen atoms conjugated to carbonyl groups gives the ring system 10 π electrons and also aromaticity, so it has three aromatic systems.

<Checkpoint>[The structure of $\bf{C}$ is `COC1=CC=C(C(C2=O)=C(SC3=CC=CC=C3)N=C4N2C=CC=C4)C=C1`, which has 3 aromatic systems. Statement 3 is incorrect](1)</Checkpoint>

If thiophenol is replaced with aniline, the structure of $\bf{B'}$ generated is `COC1=CC=C(C(C(OC)=O)C2=NC3=CC=CC=N3[H]N2C4=CC=CC=C4)C=C1`, where the hydrogen atom of aniline can form a six-membered ring hydrogen bond with the pyridine nitrogen, which reduces its nucleophilicity, thereby reducing the rate of generation of $\bf{C}$.

<MultiModal>![Shows the structure of the six-membered ring hydrogen bond in $\bf{B'}$, the structure of $\bf{B'}$ is COC1=CC=C(C(C(OC)=O)C2=NC3=CC=CC=N3[H]N2C4=CC=CC=C4)C=C1, the nucleophilic aniline hydrogen forms a six-membered ring hydrogen bond with the pyridine nitrogen](/media/uploads/a5e52c40-9f00-4849-98c4-b232de3c01cb.png)</MultiModal>

<Checkpoint>[The nucleophilic aniline hydrogen forms a six-membered ring hydrogen bond with the pyridine nitrogen, reducing its nucleophilicity and slowing down the rate of generation of $\bf{C}$. Statement 4 is incorrect](1)</Checkpoint>

Statements 1 and 2 are correct, choose F.

Finally, the structures of $\bf{A}$, $\bf{B}$, and $\bf{C}$ are shown:

<MultiModal>![$\bf{A}$: `COC1=CC=C(C(C(OC)=O)=C=NC2=NC=CC=C2)C=C1`; $\bf{B}$: `COC1=CC=C(C(C(OC)=O)/C(SC2=CC=CC=C2)=N/C3=NC=CC=C3)C=C1`; $\bf{C}$: `COC1=CC=C(C(C2=O)=C(SC3=CC=CC=C3)N=C4N2C=CC=C4)C=C1`](/media/uploads/b8b6afeb-d105-4906-940e-c68aac30a1d3.png)</MultiModal>
首先三苯氧磷被草酰氯活化，生成活性物种`Cl[P+](C1=CC=CC=C1)(C2=CC=CC=C2)C3=CC=CC=C3.[Cl-]`，这一过程还会产生气体$CO$和$CO_2$。

<Checkpoint>[反应（1）会产生$CO$和$CO_2$，这两种气体可相互转化，说法1正确](1)</Checkpoint>

鏻正离子与酰胺氧反应，生成`COC1=CC=C(C(C(OC)=O)/C(O[P+](C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4)=N/C5=NC=CC=C5)C=C1`随后在碱的作用下消去三苯氧磷，得到$\bf{A}$：`COC1=CC=C(C(C(OC)=O)=C=NC2=NC=CC=C2)C=C1`。

<Checkpoint>[$\bf{A}$的结构为`COC1=CC=C(C(C(OC)=O)=C=NC2=NC=CC=C2)C=C1`](1)</Checkpoint>

<Checkpoint>[$\bf{A}$存在累积双烯结构，有$sp$杂化的碳原子，说法2正确](1)</Checkpoint>

$\bf{A}$中的累积双烯结构易被亲核进攻，与苯硫酚反应得到$\bf{B}$：`COC1=CC=C(C(C(OC)=O)/C(SC2=CC=CC=C2)=N/C3=NC=CC=C3)C=C1`。

<Checkpoint>[$\bf{B}$的结构为`COC1=CC=C(C(C(OC)=O)/C(SC2=CC=CC=C2)=N/C3=NC=CC=C3)C=C1`](1)</Checkpoint>

$\bf{B}$中吡啶氮原子具有亲核性，亲核进攻甲酯后离去甲氧基负离子，随后在碱的作用下拔除该酯基α位氢原子，得到$\bf{C}$：`COC1=CC=C(C(C2=O)=C(SC3=CC=CC=C3)N=C4N2C=CC=C4)C=C1`。$\bf{C}$中具有两个苯基，吡啶并嘧啶酮环系氮原子与羰基共轭的互变异构体使得环系具有10个π电子，也具有芳香性，因此具有三个芳香体系。

<Checkpoint>[$\bf{C}$的结构为`COC1=CC=C(C(C2=O)=C(SC3=CC=CC=C3)N=C4N2C=CC=C4)C=C1`，具有3个芳香体系，说法3错误](1)</Checkpoint>

若将苯硫酚替换为苯胺，则生成$\bf{B'}$的结构为`COC1=CC=C(C(C(OC)=O)C2=NC3=CC=CC=N3[H]N2C4=CC=CC=C4)C=C1`，其中苯胺的氢原子可以与吡啶氮形成六元环氢键，使其亲核性降低，进而使产生$\bf{C}$的速率降低。

<MultiModal>![展示出$\bf{B'}$中六元环氢键的结构，$\bf{B'}$的结构为COC1=CC=C(C(C(OC)=O)C2=NC3=CC=CC=N3[H]N2C4=CC=CC=C4)C=C1，亲核的苯胺氢与吡啶氮形成六元环氢键](/media/uploads/a5e52c40-9f00-4849-98c4-b232de3c01cb.png)</MultiModal>

<Checkpoint>[亲核的苯胺氢与吡啶氮形成六元环氢键，降低其亲核性，使生成$\bf{C}$的速率降低，说法4错误](1)</Checkpoint>

说法1，2正确，选F

最后展示出$\bf{A}$、$\bf{B}$和$\bf{C}$的结构：

<MultiModal>![$\bf{A}$：`COC1=CC=C(C(C(OC)=O)=C=NC2=NC=CC=C2)C=C1`；$\bf{B}$：`COC1=CC=C(C(C(OC)=O)/C(SC2=CC=CC=C2)=N/C3=NC=CC=C3)C=C1`；$\bf{C}$：`COC1=CC=C(C(C2=O)=C(SC3=CC=CC=C3)N=C4N2C=CC=C4)C=C1`](/media/uploads/b8b6afeb-d105-4906-940e-c68aac30a1d3.png)</MultiModal>