A certain synthetic intermediate $\mathbf{A}$ can be reversibly converted into its more stable isomer $\mathbf{B}$ under KOH-EtOH sealed tube heating conditions. The structure of $\mathbf{A}$ is shown below:

<MultiModal>![[H][C@@]12NC3=CC=CC=C3[C@]1([C@@]45[C@@](NC6=CC=CC=C64)(NCC5)[H])CCN2](/media/uploads/6c76e53c-ceb3-44c2-b1fe-43b909dfe302.png)</MultiModal>

$\mathbf{B}$ reacts with excess iodomethane in the presence of ${K_2CO_3}$ as a base, and monomethylation occurs on the two most nucleophilic nitrogen atoms, respectively, to generate product $\mathbf{X}$, whose chemical formula is $C_{22}H_{26}N_4$. It is known that the number of rings of $\mathbf{B}$ and $\mathbf{X}$ are the same as that of $\mathbf{A}$, and no redox reaction occurs during the entire transformation process. $\mathbf{X}$ has two chiral carbon atoms with the same absolute configuration and directly connected to nitrogen atoms. Indicate the number of six-membered rings in $\mathbf{X}$, and the absolute configuration (R or S) of the chiral carbon atoms directly connected to the nitrogen atoms.
某合成中间体$\mathbf{A}$在KOH-EtOH封管加热条件下可以可逆地转化为其更稳定的同分异构体$\mathbf{B}$，$\mathbf{A}$的结构如下所示：

<MultiModal>![[H][C@@]12NC3=CC=CC=C3[C@]1([C@@]45[C@@](NC6=CC=CC=C64)(NCC5)[H])CCN2](/media/uploads/6c76e53c-ceb3-44c2-b1fe-43b909dfe302.png)</MultiModal>

$\mathbf{B}$在$K_2CO_3$作碱的条件下与过量的碘甲烷反应可以在亲核性最强的两个氮原子上分别发生一次甲基化，生成产物$\mathbf{X}$，其化学式为$C_{22}H_{26}N_4$。已知$\mathbf{B}$和$\mathbf{X}$的环数和$\mathbf{A}$相同，且整个转化过程中没有发生氧化还原反应，$\mathbf{X}$中有两个绝对构型相同的、与氮原子直接连接的手性碳原子。试指出$\mathbf{X}$中六元环的数量，以及与氮原子直接连接的手性碳原子的绝对构型（R或S）。
All other options are not completely correct.
0, S
1, S
2, S
3, S
4, S
5, S
6, S
7, S
8, S
0, R
1, R
2, R
3, R
4, R
5, R
6, R
7, R
8, R
其他选项均不完全正确。
0, S
1, S
2, S
3, S
4, S
5, S
6, S
7, S
8, S
0, R
1, R
2, R
3, R
4, R
5, R
6, R
7, R
8, R
Compound $\mathbf{A}$ can be regarded as the product of intramolecular condensation of a highly substituted butanedial with an amine on the substituent, and its properties should be similar to acetals (or hemiacetals).

<Checkpoint>[Intramolecular condensation product of butanedial](2)</Checkpoint>

<Checkpoint>[Properties should be similar to acetals (or hemiacetals)](2)</Checkpoint>

If we draw the hypothetical un-condensed form of $\mathbf{A}$, its structure is as follows (Fischer projection):

<MultiModal>![NC(C=CC=C1)=C1[C@@](CCN)([C@](CCN)(C2=C(N)C=CC=C2)C=O)C=O](/media/uploads/aa00f4f4-b1dd-487c-8d6d-f9b3c94254fd.png)</MultiModal>

According to the conditions of the question, since the transformation from $\mathbf{A}$ to $\mathbf{B}$ only involves an isomerization reaction, the whole process does not involve redox, and the reaction is reversible, it is likely that a retro-condensation (hydrolysis)-re-condensation reaction occurred, exchanging the amino groups of the acetal, so that all five-membered rings are converted into six-membered rings. This process is energetically favorable, so the equilibrium shifts toward the product $\mathbf{B}$.

<Checkpoint>[A retro-condensation (hydrolysis)-re-condensation reaction occurred, exchanging the amino groups of the acetal](3)</Checkpoint>

Due to conformational constraints, each alkylamine can only generate one type of bridged ring structure upon condensation.

<Checkpoint>[Due to conformational constraints, each alkylamine can only generate one type of bridged ring structure upon condensation](3)</Checkpoint>

Therefore, the structure of $\mathbf{B}$ can be deduced as follows:

<MultiModal>![C12=CC=CC=C1[C@]34[C@@]5(C(C=CC=C6)=C6N[C@H]3NCC5)[C@H](NCC4)N2](/media/uploads/78e79794-aed2-45fb-b5b7-c6514a209db8.png)</MultiModal>

Rather than the following structure with too much strain:

<MultiModal>![C12=CC=CC=C1[C@]34[C@@]5(C(C=CC=C6)=C6N[C@@H]3NCC5)[C@@H](NCC4)N2](/media/uploads/085702e8-d344-489c-ad22-9e64c3cf842c.png)</MultiModal>

<Checkpoint>[The structure of $\mathbf{B}$ is C12=CC=CC=C1[C@]34[C@@]5(C(C=CC=C6)=C6N[C@H]3NCC5)[C@H](NCC4)N2](3)</Checkpoint>

For the following methylation, since the basicity of potassium carbonate is low, the amino groups of $\mathbf{B}$ are all in the unprotonated state in the system.

<Checkpoint>[The basicity of potassium carbonate is low, so the amino groups of $\mathbf{B}$ are all in the unprotonated state in the system](2)</Checkpoint>

Obviously, at this time, the nucleophilicity of alkylamines is stronger than that of arylamines.

<Checkpoint>[At this time, the nucleophilicity of alkylamines is stronger than that of arylamines](1)</Checkpoint>

Therefore, methylation occurs on the nitrogens of the two alkylamines.

<Checkpoint>[Methylation occurs on the nitrogens of the two alkylamines](1)</Checkpoint>

So the structure of $\mathbf{X}$ is as follows:

<MultiModal>![CN(CC1)[C@@H]2NC3=C(C=CC=C3)[C@@]41[C@H]5NC6=CC=CC=C6[C@]42CCN5C](/media/uploads/d7a79835-a5ab-494a-8216-094be61211fa.png)</MultiModal>

<Checkpoint>[Deduce the structure of $\mathbf{X}$](2)</Checkpoint>

Since the alkylamine is methylated, this nitrogen atom is connected to two carbon-containing functional groups, while the arylamine has only one. Therefore, the alkylamine functional group in $\mathbf{X}$ has a higher priority than the arylamine in the chiral determination, so the symbol of this chiral center is S.

<Checkpoint>[The alkylamine functional group in $\mathbf{X}$ has a higher priority than the arylamine](2)</Checkpoint>

At the same time, since $\mathbf{X}$ has six six-membered rings, H is finally selected.

<Checkpoint>[$\mathbf{X}$ has six six-membered rings](2)</Checkpoint>
化合物$\mathbf{A}$可以看作一个高度取代的丁二醛与取代基上的胺分子内缩合的产物，其性质应当类似于缩醛（或半缩醛）。

<Checkpoint>[丁二醛分子内缩合产物](2)</Checkpoint>

<Checkpoint>[性质应当类似缩醛（或半缩醛）](2)</Checkpoint>

如果画出假想的$\mathbf{A}$的未缩合形体，则其结构如下（Fischer投影式）：

<MultiModal>![NC(C=CC=C1)=C1[C@@](CCN)([C@](CCN)(C2=C(N)C=CC=C2)C=O)C=O](/media/uploads/aa00f4f4-b1dd-487c-8d6d-f9b3c94254fd.png)</MultiModal>


由题干条件，由于$\mathbf{A}$到$\mathbf{B}$的转化仅仅是发生了同分异构化反应，整个过程不涉及氧化还原，且反应可逆，因此很可能是发生了逆缩合（水解）-重新缩合反应交换了缩醛的氨基，使所有五元环均转化为六元环。这一过程能量有利，故平衡向产物$\mathbf{B}$转化。

<Checkpoint>[发生了逆缩合（水解）-重新缩合反应交换了缩醛的氨基](3)</Checkpoint>

由于构象限制，每个烷基胺在缩合时都只能生成一种桥环结构。

<Checkpoint>[由于构象限制，每个烷基胺在缩合时都只能生成一种桥环结构](3)</Checkpoint>

因此，可以推出$\mathbf{B}$的结构如下：

<MultiModal>![C12=CC=CC=C1[C@]34[C@@]5(C(C=CC=C6)=C6N[C@H]3NCC5)[C@H](NCC4)N2](/media/uploads/78e79794-aed2-45fb-b5b7-c6514a209db8.png)</MultiModal>

而不是下面这个张力太大的结构：

<MultiModal>![C12=CC=CC=C1[C@]34[C@@]5(C(C=CC=C6)=C6N[C@@H]3NCC5)[C@@H](NCC4)N2](/media/uploads/085702e8-d344-489c-ad22-9e64c3cf842c.png)</MultiModal>

<Checkpoint>[$\mathbf{B}$的结构为C12=CC=CC=C1[C@]34[C@@]5(C(C=CC=C6)=C6N[C@H]3NCC5)[C@H](NCC4)N2](3)</Checkpoint>

对于接下来的甲基化，由于碳酸钾碱性较低，因此$\mathbf{B}$的氨基在体系中都是处于未脱质子状态。

<Checkpoint>[碳酸钾碱性较低，因此$\mathbf{B}$的氨基在体系中都是处于未脱质子状态](2)</Checkpoint>

显然，此时烷基胺亲核性强于芳胺。

<Checkpoint>[此时烷基胺亲核性强于芳胺](1)</Checkpoint>

因此甲基化发生在两个烷基胺的氮上。

<Checkpoint>[甲基化发生在两个烷基胺的氮上](1)</Checkpoint>

所以$\mathbf{X}$结构如下：

<MultiModal>![CN(CC1)[C@@H]2NC3=C(C=CC=C3)[C@@]41[C@H]5NC6=CC=CC=C6[C@]42CCN5C](/media/uploads/d7a79835-a5ab-494a-8216-094be61211fa.png)</MultiModal>

<Checkpoint>[推出$\mathbf{X}$的结构](2)</Checkpoint>

由于烷基胺被甲基化，该氮原子连有两个含碳官能团，而芳胺只有一个，故$\mathbf{X}$中的烷基胺官能团在手性判断的优先级上高于芳胺，因此该手性中心的记号为S。

<Checkpoint>[$\mathbf{X}$中的烷基胺官能团在手性判断的优先级上高于芳胺](2)</Checkpoint>

与此同时，由于$\mathbf{X}$有六个六元环，因此最终选择H。

<Checkpoint>[$\mathbf{X}$有六个六元环](2)</Checkpoint>