The following three-component reaction forms a new ring system:

<MultiModal>![This image is a one-step chemical reaction equation. On the left side of the arrow are three reactants, numbered $\bf{A}$, $\bf{B}$, and $\bf{C}$, connected by + signs. Their SMILES structures are: $\bf{A}$: C1=CC=C2CNCCC2=C1; $\bf{B}$: O=C(N(CC)CC)C([N+]#[C-])CC1=CC=CC=C1; $\bf{C}$: C1=CC(=O)OC1=O. Above the arrow is the text "IBX (2.0 eq)", and below the arrow is the text "THF, reflux". On the right side of the arrow is the SMILES structure of a product: CCN(C1=CC2=C(N=C1CC3=CC=CC=C3)C4C5=C(CCN4C2=O)C=CC=C5)CC](/media/uploads/1d75fcac-10b8-435a-823f-1222c34a2214.png)</MultiModal>

Where IBX is 2-iodoxybenzoic acid, which can selectively oxidize alcohols, amines, oximes, allylic groups, etc. In this reaction, one reactant is first oxidized by IBX to generate a positive ion $\bf{X}$, which then reacts with another molecule to obtain $\bf{Y}$ containing three aromatic rings, and finally reacts with the third molecule to obtain the product through a series of processes.
Regarding this reaction, the following statements are correct:

①. Substance $\bf{A}$ undergoes a nucleophilic reaction with the positive ion $\bf{X}$.

②. $\bf{Y}$ contains an aromatic ring with two heteroatoms in the ring.

③. The last molecule involved in the reaction is $\bf{C}$.

④. The amide bond in $\bf{B}$ disappears due to nucleophilic addition.

⑤. The IBX in the reaction is in excess.

⑥. The pyridine ring of the product has only 2 carbon atoms from $\bf{C}$.
以下三组分反应中形成了新的环系：

<MultiModal>![本图是一个单步化学反应方程式，箭头左侧是编号为$\bf{A}$，$\bf{B}$，$\bf{C}$的三个反应物，之间用+号连接。它们的SMILES结构式分别为：$\bf{A}$：C1=CC=C2CNCCC2=C1；$\bf{B}$：O=C(N(CC)CC)C([N+]#[C-])CC1=CC=CC=C1；$\bf{C}$：C1=CC(=O)OC1=O。箭头上方是文字“IBX (2.0 eq)”，下方是文字“THF, reflux”。箭头右侧是SMILES结构式为CCN(C1=CC2=C(N=C1CC3=CC=CC=C3)C4C5=C(CCN4C2=O)C=CC=C5)CC的一种产物](/media/uploads/1d75fcac-10b8-435a-823f-1222c34a2214.png)</MultiModal>

其中IBX是2-碘酰苯甲酸，可以对醇、胺、肟、烯丙基等进行选择性氧化。在本反应中，某一反应物首先被IBX氧化生成正离子$\bf{X}$，随后再与另一个分子反应得到含有三个芳香环的$\bf{Y}$，最后与第三个分子反应，经一系列过程得到产物。
针对这一反应，下列说法正确的有：

①.物质$\bf{A}$对正离子$\bf{X}$进行了亲核反应

②.$\bf{Y}$中有一个环内有两个杂原子的芳环。

③.最后参与反应的分子是$\bf{C}$。

④.$\bf{B}$中的酰胺键因被亲核加成而消失。

⑤.反应中的IBX是过量的。

⑥.产物的吡啶环上有且仅有2个碳原子来自$\bf{C}$。
①③⑤
②④⑤
②③⑥
③④⑥
①②④⑤
①③④⑥
①③⑤⑥
②③⑤⑥
All other options are incorrect
①③⑤
②④⑤
②③⑥
③④⑥
①②④⑤
①③④⑥
①③⑤⑥
②③⑤⑥
其它选项均不正确
For the object of IBX oxidation, maleic anhydride $\bf{C}$ is difficult to oxidize. Both $\bf{A}$ and $\bf{B}$ have benzylic positions that can be oxidized, but the amino group in $\bf{A}$ makes the benzylic position easier to oxidize, and it will also produce an iminium ion as an oxidation product. $\bf{B}$, on the other hand, is difficult to obtain a positive ion after being oxidized. At the same time, according to the product structure analysis, the benzylic position of $\bf{B}$ should not have participated in the reaction. Therefore, the first step of the reaction should be the oxidation of $\bf{A}$ to an iminium ion by IBX, ① is incorrect.

<Checkpoint>[$\bf{X}$ is derived from the oxidation of $\bf{A}$, ① is incorrect](1)</Checkpoint>

The resulting structure of $\bf{X}$ is:

<MultiModal>![C12=CC=CC=C1C=[NH+]CC2](/media/uploads/b09daf9a-a51c-49ab-ac0e-f53dfb94159e.jpg)</MultiModal>

$\bf{X}$ is a good nucleophilic substrate. There is no good nucleophilic site in $\bf{C}$, but in the isonitrile structure of $\bf{B}$, the carbon atom has a negative formal charge, which can perform nucleophilic attack on the imine in $\bf{X}$. Subsequently, this carbon atom becomes a good electrophilic site again due to the positive formal charge of the adjacent nitrogen atom, which can be attacked by the oxygen atom in the amide bond to form a favorable five-membered ring structure, and then isomerizes to form an aromatic ring, obtaining $\bf{Y}$. Because the amide bond here participates in the reaction as a nucleophilic group rather than a site, ④ is incorrect.

<Checkpoint>[The amide bond participates in the reaction as a nucleophilic group rather than a site, ④ is incorrect](1)</Checkpoint>

The structure of $\bf{Y}$ is:

<MultiModal>![CCN(C1=C(N=C(O1)C2NCCC3=C2C=CC=C3)CC4=CC=CC=C4)CC](/media/uploads/1220841c-ef1e-4e34-9139-a5e0e0153ba3.jpg)</MultiModal>

There is an oxazole ring in it, ② is correct.

<Checkpoint>[The structure of $\bf{Y}$ is CCN(C1=C(N=C(O1)C2NCCC3=C2C=CC=C3)CC4=CC=CC=C4)CC, ② is correct](2)</Checkpoint>

At this time, both $\bf{A}$ and $\bf{B}$ have participated in the reaction, and maleic anhydride $\bf{C}$ participates in the reaction last, ③ is correct.

<Checkpoint>[$\bf{C}$ participates in the reaction last, ③ is correct](1)</Checkpoint>

The aromaticity of the oxazole ring is not very strong and it is very electron-rich. At the same time, $\bf{C}$ is an electron-deficient diene, and a Diels-Alder reaction can occur between the two.

<Checkpoint>[Diels-Alder reaction occurs between $\bf{Y}$ and $\bf{C}$](1)</Checkpoint>

At this time, there is a strongly electrophilic anhydride and a strongly nucleophilic secondary amine in the molecule, and nucleophilic substitution can occur between the two to obtain an amide structure, generating:

<MultiModal>![CCN(C1=C(N=C(O1)C2NCCC3=C2C=CC=C3)CC4=CC=CC=C4)CC](/media/uploads/a7e36759-ef6a-4577-b0b3-339de9aa8ad3.jpg)</MultiModal>

Then, the product is obtained through decarboxylation, ring-opening, and dehydration aromatization.

<Checkpoint>[The obtained molecule further undergoes amidation, decarboxylation, and dehydration processes to obtain the product.](1)</Checkpoint>

The pentavalent iodine in IBX will be reduced to trivalent iodine. The molar ratio of the oxidation of $\bf{A}$ and the consumption of IBX in the reaction is 1:1. In addition, IBX does not oxidize other molecules, so IBX is excessive, ⑤ is correct.

<Checkpoint>[IBX participates in the reaction in one equivalent, 2eq IBX is excessive, ⑤ is correct.](1)</Checkpoint>

The pyridine ring is derived from the Diels-Alder reaction. The nitrogen atom and 3 carbon atoms in it come from the oxazole ring in $\bf{Y}$, and the other two carbon atoms come from $\bf{C}$, ⑥ is correct.

<Checkpoint>[The two carbon atoms in the pyridine ring come from $\bf{C}$, ⑥ is correct.](1)</Checkpoint>

In summary, ②③⑤⑥ are correct, so option H should be selected.
对于IBX氧化的对象，马来酸酐$\bf{C}$很难被氧化，$\bf{A}$与$\bf{B}$中都存在可被氧化的苄位，但$\bf{A}$的氨基使得苄位更容易被氧化，也会产生亚胺正离子这一氧化产物。而$\bf{B}$被氧化后很难得到正离子，同时根据产物结构分析，$\bf{B}$的苄位应该并没有参与反应，因此反应的第一步应该是$\bf{A}$被IBX氧化为亚胺正离子，①错误。

<Checkpoint>[$\bf{X}$由$\bf{A}$氧化而来，①错误](1)</Checkpoint>

得到的$\bf{X}$结构为：

<MultiModal>![C12=CC=CC=C1C=[NH+]CC2](/media/uploads/b09daf9a-a51c-49ab-ac0e-f53dfb94159e.jpg)</MultiModal>

$\bf{X}$是很好的亲核底物，$\bf{C}$中并没有很好的亲核位点，而$\bf{B}$中的异腈结构中，碳原子上有负的形式电荷，可以对$\bf{X}$中的亚胺进行亲核。随后这个碳原子由于旁边氮原子的正形式电荷又变为了一个很好的亲电位点，可以被酰胺键中的氧原子亲核形成有利的五元环结构，再异构化后形成芳环，得到$\bf{Y}$。因为酰胺键在这里作为亲核基团而非位点参与反应，因此④错误。

<Checkpoint>[酰胺键作为亲核基团而非位点参与反应，④错误](1)</Checkpoint>

$\bf{Y}$的结构为：

<MultiModal>![CCN(C1=C(N=C(O1)C2NCCC3=C2C=CC=C3)CC4=CC=CC=C4)CC](/media/uploads/1220841c-ef1e-4e34-9139-a5e0e0153ba3.jpg)</MultiModal>

其中存在一个噁唑环，②正确。

<Checkpoint>[$\bf{Y}$的结构为CCN(C1=C(N=C(O1)C2NCCC3=C2C=CC=C3)CC4=CC=CC=C4)CC，②正确](2)</Checkpoint>

此时$\bf{A}$与$\bf{B}$均已参与反应，马来酸酐$\bf{C}$最后参与反应，③正确。

<Checkpoint>[$\bf{C}$最后参与反应，③正确](1)</Checkpoint>

噁唑环的芳香性不太强又很富电子，同时$\bf{C}$是一个缺电子双烯体，两者之间可以发生Diels-Alder反应。

<Checkpoint>[$\bf{Y}$与$\bf{C}$之间发生Diels-Alder反应](1)</Checkpoint>

此时分子内有一个强亲电性的酸酐，还有强亲核性的二级胺，两者可发生亲核取代得到酰胺结构，生成：

<MultiModal>![CCN(C1=C(N=C(O1)C2NCCC3=C2C=CC=C3)CC4=CC=CC=C4)CC](/media/uploads/a7e36759-ef6a-4577-b0b3-339de9aa8ad3.jpg)</MultiModal>

随后经脱羧、开环与脱水芳构化得到产物。

<Checkpoint>[D-A产物发生分子内酰胺化，脱羧，脱水过程得到产物。](1)</Checkpoint>


IBX中的五价碘会被还原为三价碘，反应中$\bf{A}$的氧化与IBX消耗的摩尔比为1:1，除此之外IBX不氧化其它分子，因此IBX是过量的，⑤正确。

<Checkpoint>[IBX以一当量参与反应， 2eq IBX是过量的，⑤正确。](1)</Checkpoint>

吡啶环由Diels-Alder反应得来，其中的氮原子与3个碳原子来源于$\bf{Y}$中的噁唑环，另外两个碳原子来源于$\bf{C}$，⑥正确。

<Checkpoint>[吡啶环中的两个碳原子来源于$\bf{C}$，⑥正确。](1)</Checkpoint>

综上，②③⑤⑥正确，应选择H选项。