<MultiModal>![This figure shows an organic reaction. The substrate is CC1=C(OCC2=CC=CC=C2)C=CC=C1[N+]([O-])=O, and the reaction conditions are CN(C)C(OC)OC.CO[Na].CN(C([H])=O)C.C3CCNC3; the products are $\bf{A,B,C}$.](/media/uploads/37eceb75-ef23-417b-b72c-627d78694800.png)</MultiModal>

The organic reaction in the figure above produced two products $\bf{A, B}$ and one byproduct $\bf{C}$.

Known:
1. The ratio of the two products is $\bf{A : B = \rm{15 : 1}}$.
2. The generation of the byproduct involves an intermolecular hydride transfer reaction.
3. The chemical formula of byproduct $\bf{C}$ is $\rm{C_{15}H_{13}NO_3}$.
4. The reaction producing the products is a condensation reaction.
5. The reaction temperature is $\rm{383 ~K}$

Regarding this reaction, the following statements are made:
1. The number of atoms in $\bf{A}$ is greater than that in $\bf{B}$.
2. The number of atoms in $\bf{A}$ is less than that in $\bf{B}$.
3. The number of atoms in $\bf{A}$ is equal to that in $\bf{B}$.
4. $\bf{B}$ contains a methyl group.
5. $\bf{C}$ contains a methyl group.
6. $\bf{A}$ contains a methyl group.
7. If the reaction temperature is lowered to $\rm{273 ~K}$, the ratio of reaction products will change to $\bf{B}$ being more than $\bf{A}$.
8. If the reaction temperature is lowered to $\rm{273 ~K}$, the ratio of reaction products will still be $\bf{B}$ being less than $\bf{A}$.
9. If all the hydrogen atoms on the methyl group of the substrate are labeled with $\rm{^{2}H}$, then the byproduct $\bf{C}$ contains two $\rm{^{2}H}$.

The following option contains the most correct statement numbers and no incorrect statement numbers:
<MultiModal>![该图为一有机反应，底物为CC1=C(OCC2=CC=CC=C2)C=CC=C1[N+]([O-])=O，反应条件为CN(C)C(OC)OC.CO[Na].CN(C([H])=O)C.C3CCNC3；产物为$\bf{A,B,C}$。](/media/uploads/37eceb75-ef23-417b-b72c-627d78694800.png)</MultiModal>

上图的有机反应产生了两种产物$\bf{A,B}$和一种副产物$\bf{C}$。

已知：
1. 两种产物的比例为$\bf{A : B = \rm{15 : 1}}$。
2. 副产物的产生经过了分子间负氢转移反应。
3. 副产物$\bf{C}$的化学式为$\rm{C_{15}H_{13}NO_3}$。
4. 产生产物的反应为缩合反应。
5. 反应温度为$\rm{383 ~K}$

关于该反应，有下列说法：
1. $\bf{A}$的原子数大于$\bf{B}$。
2. $\bf{A}$的原子数小于$\bf{B}$。
3. $\bf{A}$的原子数等于$\bf{B}$。
4. $\bf{B}$含有甲基。
5. $\bf{C}$含有甲基。
6. $\bf{A}$含有甲基。
7. 降低反应温度至$\rm{273 ~K}$，反应产物的比例会变化至$\bf{B}$多于$\bf{A}$。
8. 降低反应温度至$\rm{273 ~K}$，反应产物的比例仍旧是$\bf{B}$少于$\bf{A}$。
9. 若将底物的甲基上的氢均采用$\rm{^{2}H}$标记，则副产物$\bf{C}$含有两个$\rm{^{2}H}$。

下列选项中包含了最多正确的说法序号，且无错误的说法序号的选项是：
1，4，9
1，4，7
1，4，8
2，5，7
3，6，7
1，4，5，7
2，5，6，8
1，5，6，7
1，4，5，8
2，4，6，7
3，4，5，8，9
3，6，8，9
1，5，7，9
2，4，6，8
1，4，5，7，9
1，4，5，8，9
2，4，5，7，9
3，5，7，9
1，4，7，9
1，5，6，9
1，4，9
1，4，7
1，4，8
2，5，7
3，6，7
1，4，5，7
2，5，6，8
1，5，6，7
1，4，5，8
2，4，6，7
3，4，5，8，9
3，6，8，9
1，5，7，9
2，4，6，8
1，4，5，7，9
1，4，5，8，9
2，4，5，7，9
3，5，7，9
1，4，7，9
1，5，6，9
<MultiModal>![This figure shows the structure of the unknown species involved in this question. $\bf{A}$ is O=[N+](C1=CC=CC(OCC2=CC=CC=C2)=C1/C=C/N3CCCC3)[O-], $\bf{B}$ is O=[N+](C1=CC=CC(OCC2=CC=CC=C2)=C1/C=C/N(C)C)[O-], $\bf{C}$ is C=CC1=C(OCC2=CC=CC=C2)C=CC=C1[N+]([O-])=O, the imine intermediate structure is O=[N+](C1=CC=CC(OCC2=CC=CC=C2)=C1CC=[N+](C)C)[O-]](/media/uploads/883eecb2-304d-452a-adf3-b093e4a33256.png)</MultiModal>

This reaction is a simple condensation reaction; the benzylic position of the substrate is acidic under the influence of the strong electron-withdrawing nitro group in the ortho position and can be deprotonated by the strong base $\rm{MeONa}$ to form a carbanion;

<Checkpoint>[The benzylic position is acidic under the influence of the strong electron-withdrawing nitro group in the ortho position and can be deprotonated by the strong base $\rm{MeONa}$ to form a carbanion](1)</Checkpoint>

Another substrate molecule is an acetal structure, which is electrophilic. The generated carbanion undergoes nucleophilic attack on the acetal, leading to condensation, followed by the elimination of two molecules of methoxide ion to form the key imine intermediate, which has a structure of O=[N+](C1=CC=CC(OCC2=CC=CC=C2)=C1CC=[N+](C)C)[O-].

<Checkpoint>[Carbanion undergoes nucleophilic attack on the acetal, leading to condensation](1)</Checkpoint>

<Checkpoint>[Formation of imine intermediate, which has a structure of O=[N+](C1=CC=CC(OCC2=CC=CC=C2)=C1CC=[N+](C)C)[O-]](1)</Checkpoint>

Elimination of a molecule of hydrogen ion from this intermediate can form an enamine structure, which is one of the products. Its structure is O=[N+](C1=CC=CC(OCC2=CC=CC=C2)=C1/C=C/N(C)C)[O-].

<Checkpoint>[One of the products is O=[N+](C1=CC=CC(OCC2=CC=CC=C2)=C1/C=C/N(C)C)[O-]](1)</Checkpoint>

However, tetrahydro pyrrole exists in the reaction system; Since the above condensation reactions are all reversible, tetrahydro pyrrole can nucleophilically attack the imine intermediate, eliminating dimethylamine to form a similar imine intermediate, followed by elimination of a molecule of hydrogen ion to form an enamine structure. Therefore, the structure of the other product is O=[N+](C1=CC=CC(OCC2=CC=CC=C2)=C1/C=C/N3CCCC3)[O-]; this reaction is transamination.

<Checkpoint>[The other product is a transamination product, with a structure of O=[N+](C1=CC=CC(OCC2=CC=CC=C2)=C1/C=C/N3CCCC3)[O-]](1)</Checkpoint>

Considering the ratio of the two products, since the reaction temperature is $\rm{383 ~K}$, which is much higher than the boiling point of dimethylamine but has not reached the boiling point of tetrahydro pyrrole, dimethylamine will be distilled out of the system, thereby driving the equilibrium; therefore, the product with the larger proportion, $\bf{A}$, is the transamination product O=[N+](C1=CC=CC(OCC2=CC=CC=C2)=C1/C=C/N3CCCC3)[O-], and $\bf{B}$ is O=[N+](C1=CC=CC(OCC2=CC=CC=C2)=C1/C=C/N(C)C)[O-].

<Checkpoint>[The reaction temperature is $\rm{383 ~K}$, which is much higher than the boiling point of dimethylamine but has not reached the boiling point of tetrahydro pyrrole, thereby dimethylamine will be distilled out of the system, thereby driving the equilibrium](1)</Checkpoint>

<Checkpoint>[$\bf{A}$ is the transamination product O=[N+](C1=CC=CC(OCC2=CC=CC=C2)=C1/C=C/N3CCCC3)[O-]](1)</Checkpoint>

<Checkpoint>[$\bf{B}$ is O=[N+](C1=CC=CC(OCC2=CC=CC=C2)=C1/C=C/N(C)C)[O-]](1)</Checkpoint>

$\bf{A}$ has more atoms, statement 1 is correct, 2 and 3 are incorrect; $\bf{A}$ does not contain a methyl group while $\bf{B}$ contains a methyl group, statement 4 is correct, 6 is incorrect.

If the reaction temperature is lowered to $\rm{273 ~K}$, which is lower than the boiling point of dimethylamine, the transamination reaction is a side reaction at this time, and dimethylamine has a stronger nucleophilicity, so the product ratio $\bf{B}$ will be higher, statement 7 is correct, 8 is incorrect.

<Checkpoint>[If the reaction temperature is lowered to $\rm{273 ~K}$, which is lower than the boiling point of dimethylamine, dimethylamine has a stronger nucleophilicity, so the product ratio $\bf{B}$ will be higher](1)</Checkpoint>

Considering the byproduct $\bf{C}$; according to the chemical formula, $\bf{C}$ only has one more carbon atom than the substrate. Since only the benzylic position can react, the structure of the byproduct $\bf{C}$ can only be C=CC1=C(OCC2=CC=CC=C2)C=CC=C1[N+]([O-])=O. $\bf{C}$ does not have a methyl group, statement 5 is incorrect.

<Checkpoint>[The structure of $\bf{C}$ can only be C=CC1=C(OCC2=CC=CC=C2)C=CC=C1[N+]([O-])=O](1)</Checkpoint>

The formation mechanism of this product involves intermolecular hydride transfer, and can only be that the imine intermediate is reduced by the hydride of the substrate acetal. This hydrogen is not isotopically labeled; then β-elimination occurs to obtain an alkene, so the methyl group labeled with $\rm{^2H}$ will only have one $\rm{^2H}$ left after becoming an alkene, statement 9 is incorrect.

<Checkpoint>[The imine intermediate is reduced by the hydride of the substrate acetal, and this hydrogen is not isotopically labeled](1)</Checkpoint>

In summary, statements 1, 4, and 7 are correct, so option B is correct.

<Checkpoint>[Statements 1, 4, and 7 are correct](1)</Checkpoint>
<MultiModal>![该图展示了本题涉及到的未知物种结构。$\bf{A}$为O=[N+](C1=CC=CC(OCC2=CC=CC=C2)=C1/C=C/N3CCCC3)[O-]，$\bf{B}$为O=[N+](C1=CC=CC(OCC2=CC=CC=C2)=C1/C=C/N(C)C)[O-]，$\bf{C}$为C=CC1=C(OCC2=CC=CC=C2)C=CC=C1[N+]([O-])=O，亚胺中间体结构为O=[N+](C1=CC=CC(OCC2=CC=CC=C2)=C1CC=[N+](C)C)[O-]](/media/uploads/883eecb2-304d-452a-adf3-b093e4a33256.png)</MultiModal>

该反应为简单的缩合反应；底物的苄位在邻位的强吸电子基团硝基的作用下具有酸性，可以被强碱$\rm{MeONa}$拔出质子形成碳负离子；

<Checkpoint>[苄位在邻位的强吸电子基团硝基的作用下具有酸性，可以被强碱$\rm{MeONa}$拔出质子形成碳负离子](1)</Checkpoint>

另一分子底物为缩醛结构，具有亲电性，生成的碳负离子亲核进攻缩醛发生缩合，之后脱去两分子甲醇负离子形成关键的亚胺中间体，其结构为O=[N+](C1=CC=CC(OCC2=CC=CC=C2)=C1CC=[N+](C)C)[O-]。

<Checkpoint>[碳负离子亲核进攻缩醛发生缩合](1)</Checkpoint>

<Checkpoint>[生成亚胺中间体，其结构为O=[N+](C1=CC=CC(OCC2=CC=CC=C2)=C1CC=[N+](C)C)[O-]](1)</Checkpoint>

该中间体消除一分子氢离子即可形成烯胺结构，即某一种产物。其结构为O=[N+](C1=CC=CC(OCC2=CC=CC=C2)=C1/C=C/N(C)C)[O-]。

<Checkpoint>[某一种产物为O=[N+](C1=CC=CC(OCC2=CC=CC=C2)=C1/C=C/N(C)C)[O-]](1)</Checkpoint>

但反应体系中存在四氢吡咯；由于上述缩合反应均为可逆进行，四氢吡咯可以亲核亚胺中间体，脱去二甲胺形成类似的亚胺中间体，之后消除一分子氢离子形成烯胺结构。所以另一种产物结构为O=[N+](C1=CC=CC(OCC2=CC=CC=C2)=C1/C=C/N3CCCC3)[O-]；该反应即转氨基反应。

<Checkpoint>[另一种产物为转氨基产物，结构为O=[N+](C1=CC=CC(OCC2=CC=CC=C2)=C1/C=C/N3CCCC3)[O-]](1)</Checkpoint>

考虑两种产物的比例，由于反应温度为$\rm{383 ~K}$，远超过二甲胺的沸点但没有达到四氢吡咯的沸点，因此二甲胺会被蒸出体系，从而拉动平衡；因此比例大的产物$\bf{A}$为转氨基产物O=[N+](C1=CC=CC(OCC2=CC=CC=C2)=C1/C=C/N3CCCC3)[O-]，$\bf{B}$为O=[N+](C1=CC=CC(OCC2=CC=CC=C2)=C1/C=C/N(C)C)[O-]。

<Checkpoint>[反应温度为$\rm{383 ~K}$，远超过二甲胺的沸点但没有达到四氢吡咯的沸点，因此二甲胺会被蒸出体系，从而拉动平衡](1)</Checkpoint>

<Checkpoint>[$\bf{A}$为转氨基产物O=[N+](C1=CC=CC(OCC2=CC=CC=C2)=C1/C=C/N3CCCC3)[O-]](1)</Checkpoint>

<Checkpoint>[$\bf{B}$为O=[N+](C1=CC=CC(OCC2=CC=CC=C2)=C1/C=C/N(C)C)[O-]](1)</Checkpoint>

$\bf{A}$原子数更多，说法1正确，2，3均错误；$\bf{A}$不含甲基而$\bf{B}$含有甲基，说法4正确，6错误。

反应温度若降低至$\rm{273 ~K}$，小于二甲胺的沸点，此时转氨基反应为副反应，二甲胺亲核性更强，故产物比例$\bf{B}$会更多，说法7正确，8错误。

<Checkpoint>[反应温度若降低至$\rm{273 ~K}$，小于二甲胺的沸点，二甲胺亲核性更强，故产物比例$\bf{B}$会更多](1)</Checkpoint>

考虑副产物$\bf{C}$；根据化学式可知$\bf{C}$只比底物多出一个碳原子，由于只有苄位可以发生反应，副产物$\bf{C}$结构只能为C=CC1=C(OCC2=CC=CC=C2)C=CC=C1[N+]([O-])=O。$\bf{C}$没有甲基，说法5错误。

<Checkpoint>[$\bf{C}$结构只能为C=CC1=C(OCC2=CC=CC=C2)C=CC=C1[N+]([O-])=O](1)</Checkpoint>

该产物的生成机理经过分子间负氢转移，只能为亚胺中间体被底物缩醛的负氢还原，该氢没有被同位素标记；之后发生β消除得到烯烃，从而被$\rm{^2H}$标记的甲基变成烯烃后只会剩下一个$\rm{^2H}$，说法9错误。

<Checkpoint>[亚胺中间体被底物缩醛的负氢还原，该氢没有被同位素标记](1)</Checkpoint>

综上，说法1，4，7正确，故选项B正确。

<Checkpoint>[说法1，4，7正确](1)</Checkpoint>