The unnatural derivative of N-acetylmannosamine, $ManNAz$, can be metabolically converted into the unnatural sialic acid $SiaNAz$ on the cell surface, thereby introducing the azide group, which is very important in bioorthogonal reactions, to the cell surface (as shown in the figure below).

<MultiModal>![Adding `O[C@H]1[C@H](O)[C@H](NC(CN=[N+]=[N-])=O)C(O)O[C@@H]1CO` to the cell culture system, metabolic labeling can produce a structure of `O[C@@H]1[C@H]([C@@H](O[C@](O[*])(C1)C([O-])=O)[C@@H]([C@@H](CO)O)O)NC(CN=[N+]=[N-])=O` on the cell surface（[*] indicates bonding to the cell surface）](/media/uploads/872c542f-981f-4ce9-bb7e-bcd1c98fae67.png)</MultiModal>

Studies have shown that this metabolic process can be divided into the following steps:

1. Under the action of 1 molecule of $ATP$, the hydroxyl group at position 6 of $ManNAz$ undergoes a phosphorylation reaction to obtain intermediate $\bf{A}$, accompanied by the generation of 1 molecule of $ADP$.

2. Intermediate $\bf{A}$ isomerizes to its chain structure, and then reacts with PEP (phosphoenolpyruvate, `C=C(C([O-])=O)OP([O-])([O-])=O`) to obtain intermediate $\bf{B}$, accompanied by the generation of 1 molecule of phosphoric acid.

3. Intermediate $\bf{B}$ undergoes intramolecular nucleophilic addition, followed by hydrolysis of the phosphate group, to obtain intermediate $\bf{C}$.

4. The anomeric carbon hydroxyl group of intermediate $\bf{C}$ reacts with CTP and is then captured by the hydroxyl group on the cell surface, converting it into the unnatural sialic acid $SiaNAz$ on the cell surface.

The following statements are made:

1. $\bf{A}$ has 5 chiral carbon atoms.

2. Under physiological conditions, $\bf{B}$ carries two negative charges.

3. $\bf{B}$ has 6 chiral carbon atoms.

4. After introducing the azide group on the cell surface, $SiaNAz$ can undergo the following post-modification reaction (where the cell structure is replaced by the R group, and biotin refers to the biotin group, which can be considered not to participate in the reaction).

<MultiModal>![`O[C@@H]1[C@@H](NC(CN=[N+]=[N-])=O)[C@H]([C@H](O)[C@H](O)CO)O[C@](C(O)=O)(O[R])C1` reacts with `[biotin]C1=CC=C(C(OC)=O)C(P(C2=CC=CC=C2)C3=CC=CC=C3)=C1` in a biological system to obtain $\bf{D}$](/media/uploads/aaefd129-a766-49c9-bddb-470e719ee51d.png)</MultiModal>

Then $\bf{D}$ contains two nitrogen atoms and 11 oxygen atoms (not considering the biotin group and the R group).

Select the option that contains all the correct statements:
乙酰甘露糖胺的非天然衍生物$ManNAz$可以通过代谢转化为细胞表面的非天然唾液酸$SiaNAz$，从而向细胞表面引入生物正交反应中非常重要的叠氮基团(如下图所示)。

<MultiModal>![`O[C@H]1[C@H](O)[C@H](NC(CN=[N+]=[N-])=O)C(O)O[C@@H]1CO`加入细胞培养体系中，经代谢标记可在细胞表面产生结构为`O[C@@H]1[C@H]([C@@H](O[C@](O[*])(C1)C([O-])=O)[C@@H]([C@@H](CO)O)O)NC(CN=[N+]=[N-])=O`（[*]表示与细胞表面成键）的修饰](/media/uploads/872c542f-981f-4ce9-bb7e-bcd1c98fae67.png)</MultiModal>

研究表明，这一代谢过程可以分为以下几步：

1.在1分子$ATP$的作用下，$ManNAz$的6号位羟基发生磷酸化反应得到中间体$\bf{A}$，伴随1分子$ADP$的生成。

2.中间体$\bf{A}$异构为其链状结构，随后与PEP(磷酸烯醇式丙酮酸，`C=C(C([O-])=O)OP([O-])([O-])=O`)反应得到中间体$\bf{B}$，伴随1分子磷酸的生成。

3.中间体$\bf{B}$发生分子内亲核加成，随后磷酸基团水解，得到中间体$\bf{C}$。

4.中间体$\bf{C}$的异头碳羟基与CTP发生反应，随后被细胞表面的羟基捕获，转化为细胞表面的非天然唾液酸$SiaNAz$。

有以下说法：

1.$\bf{A}$中有5个手性碳原子

2.在生理条件下，$\bf{B}$带两个负电荷

3.$\bf{B}$中有6个手性碳原子

4.在细胞表面引入叠氮基团后，$SiaNAz$可以发生如下后修饰反应(其中细胞结构用R基团代替，biotin指生物素基团，可认为不参与反应)。

<MultiModal>![`O[C@@H]1[C@@H](NC(CN=[N+]=[N-])=O)[C@H]([C@H](O)[C@H](O)CO)O[C@](C(O)=O)(O[R])C1`在生物体系中与`[biotin]C1=CC=C(C(OC)=O)C(P(C2=CC=CC=C2)C3=CC=CC=C3)=C1`反应得到$\bf{D}$](/media/uploads/aaefd129-a766-49c9-bddb-470e719ee51d.png)</MultiModal>

则$\bf{D}$含有两个氮原子和11个氧原子（不考虑生物素基团和R基团）

选出下列选项中含有全部正确说法的一项：
All statements 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
所有说法均不正确
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
$ManNAz$ is phosphorylated at the six position hydroxyl group to obtain $\bf{A}$: `O[C@H]1[C@H](O)[C@H](NC(CN=[N+]=[N-])=O)C(O)O[C@@H]1COP([O-])([O-])=O`

<Checkpoint>[The structure of $\bf{A}$ is `O[C@H]1[C@H](O)[C@H](NC(CN=[N+]=[N-])=O)C(O)O[C@@H]1COP([O-])([O-])=O`, which has 5 chiral carbon atoms. Statement 1 is correct](1)</Checkpoint>

Subsequently, $\bf{A}$ isomerizes to an aldehyde, is added to by enol-PEP, and loses phosphate to yield $\bf{B}$: `OC(COP([O-])([O-])=O)C(O)C(O)C(NC(CN=[N+]=[N-])=O)C(CC(C([O-])=O)=O)O`

<Checkpoint>[The structure of $\bf{B}$ is `OC(COP([O-])([O-])=O)C(O)C(O)C(NC(CN=[N+]=[N-])=O)C(CC(C([O-])=O)=O)O`, which has 5 chiral carbon atoms. Statement 3 is incorrect](1)</Checkpoint>

Under physiological pH of approximately 7.4, both hydrogens of the phosphate group and the hydrogen of the carboxyl group will be ionized, and $\bf{B}$ carries 3 negative charges.

<Checkpoint>[Under physiological conditions, both hydrogens of the phosphate group and the hydrogen of the carboxyl group will be ionized, and $\bf{B}$ carries 3 negative charges. Statement 2 is incorrect](1)</Checkpoint>

Subsequently, an intramolecular nucleophilic addition occurs, where the free hydroxyl group attacks the carbonyl group to form a pyran six-membered ring, and hydrolysis of the phosphate group yields $\bf{C}$: `O[C@]1(C([O-])=O)C[C@H](O)[C@@H](NC(CN=[N+]=[N-])=O)[C@H]([C@H](O)[C@H](O)CO)O1`

<Checkpoint>[The structure of $\bf{C}$ is `O[C@]1(C([O-])=O)C[C@H](O)[C@@H](NC(CN=[N+]=[N-])=O)[C@H]([C@H](O)[C@H](O)CO)O1`](1)</Checkpoint>

In the reaction to generate $\bf{D}$, first, trivalent phosphorus undergoes a Staudinger reaction with the azide group, oxidizing itself to pentavalent while reducing the azide to an amine; subsequently, the amine attacks a spatially adjacent ester group, releasing methanol to produce an amide, yielding $\bf{D}$: `[biotin]C1=CC=C(C(NCC(N[C@H]2[C@H]([C@H](O)[C@H](O)CO)O[C@](C([O-])=O)(O[R])C[C@@H]2O)=O)=O)C(P(C3=CC=CC=C3)(C4=CC=CC=C4)=O)=C1`

<Checkpoint>[The structure of $\bf{D}$ is `[biotin]C1=CC=C(C(NCC(N[C@H]2[C@H]([C@H](O)[C@H](O)CO)O[C@](C([O-])=O)(O[R])C[C@@H]2O)=O)=O)C(P(C3=CC=CC=C3)(C4=CC=CC=C4)=O)=C1`, which has two nitrogen atoms and 11 oxygen atoms (not considering the biotin group and R group). Statement 4 is correct](1)</Checkpoint>

Statements 1 and 4 are correct, choose H

<MultiModal>![$\bf{A}$：`O[C@H]1[C@H](O)[C@H](NC(CN=[N+]=[N-])=O)C(O)O[C@@H]1COP([O-])([O-])=O`；$\bf{B}$：`OC(COP([O-])([O-])=O)C(O)C(O)C(NC(CN=[N+]=[N-])=O)C(CC(C([O-])=O)=O)O`；$\bf{C}$：`O[C@]1(C([O-])=O)C[C@H](O)[C@@H](NC(CN=[N+]=[N-])=O)[C@H]([C@H](O)[C@H](O)CO)O1`；$\bf{D}$：`[biotin]C1=CC=C(C(NCC(N[C@H]2[C@H]([C@H](O)[C@H](O)CO)O[C@](C([O-])=O)(O[R])C[C@@H]2O)=O)=O)C(P(C3=CC=CC=C3)(C4=CC=CC=C4)=O)=C1`](/media/uploads/57b83097-5cb7-4659-a090-d626f2614705.png)</MultiModal>
$ManNAz$的六号位羟基磷酸化得到$\bf{A}$：`O[C@H]1[C@H](O)[C@H](NC(CN=[N+]=[N-])=O)C(O)O[C@@H]1COP([O-])([O-])=O`

<Checkpoint>[$\bf{A}$的结构为`O[C@H]1[C@H](O)[C@H](NC(CN=[N+]=[N-])=O)C(O)O[C@@H]1COP([O-])([O-])=O`，其中有5个手性碳原子，说法1正确](1)</Checkpoint>

随后$\bf{A}$异构为醛，被烯醇式PEP加成，离去磷酸根得到$\bf{B}$：`OC(COP([O-])([O-])=O)C(O)C(O)C(NC(CN=[N+]=[N-])=O)C(CC(C([O-])=O)=O)O`

<Checkpoint>[$\bf{B}$的结构为`OC(COP([O-])([O-])=O)C(O)C(O)C(NC(CN=[N+]=[N-])=O)C(CC(C([O-])=O)=O)O`，其中有5个手性碳原子，说法3错误](1)</Checkpoint>

在生理pH约为7.4的条件下，磷酸根的两个氢和羧基氢均会电离，$\bf{B}$带3个负电荷。

<Checkpoint>[生理条件下磷酸根的两个氢和羧基氢均会电离，$\bf{B}$带3个负电荷，说法2错误](1)</Checkpoint>

随后发生分子内亲核加成，游离羟基进攻羰基形成吡喃六元环，磷酸根水解得到$\bf{C}$：`O[C@]1(C([O-])=O)C[C@H](O)[C@@H](NC(CN=[N+]=[N-])=O)[C@H]([C@H](O)[C@H](O)CO)O1`

<Checkpoint>[$\bf{C}$的结构为`O[C@]1(C([O-])=O)C[C@H](O)[C@@H](NC(CN=[N+]=[N-])=O)[C@H]([C@H](O)[C@H](O)CO)O1`](1)</Checkpoint>

生成$\bf{D}$的反应中，首先三价磷与叠氮基团发生Staudinger反应，自身被氧化到五价的同时还原其为氨基；随后氨基进攻空间上临近的酯基，离去甲醇产生酰胺，得到$\bf{D}$：`[biotin]C1=CC=C(C(NCC(N[C@H]2[C@H]([C@H](O)[C@H](O)CO)O[C@](C([O-])=O)(O[R])C[C@@H]2O)=O)=O)C(P(C3=CC=CC=C3)(C4=CC=CC=C4)=O)=C1`

<Checkpoint>[$\bf{D}$的结构为`[biotin]C1=CC=C(C(NCC(N[C@H]2[C@H]([C@H](O)[C@H](O)CO)O[C@](C([O-])=O)(O[R])C[C@@H]2O)=O)=O)C(P(C3=CC=CC=C3)(C4=CC=CC=C4)=O)=C1`，有两个氮原子和11个氧原子（不考虑生物素基团和R基团），说法4正确](1)</Checkpoint>

说法1，4正确，选H

<MultiModal>![$\bf{A}$：`O[C@H]1[C@H](O)[C@H](NC(CN=[N+]=[N-])=O)C(O)O[C@@H]1COP([O-])([O-])=O`；$\bf{B}$：`OC(COP([O-])([O-])=O)C(O)C(O)C(NC(CN=[N+]=[N-])=O)C(CC(C([O-])=O)=O)O`；$\bf{C}$：`O[C@]1(C([O-])=O)C[C@H](O)[C@@H](NC(CN=[N+]=[N-])=O)[C@H]([C@H](O)[C@H](O)CO)O1`；$\bf{D}$：`[biotin]C1=CC=C(C(NCC(N[C@H]2[C@H]([C@H](O)[C@H](O)CO)O[C@](C([O-])=O)(O[R])C[C@@H]2O)=O)=O)C(P(C3=CC=CC=C3)(C4=CC=CC=C4)=O)=C1`](/media/uploads/57b83097-5cb7-4659-a090-d626f2614705.png)</MultiModal>