Vanadium complexes can be synthesized using the following process:

<MultiModal>![SMILES structure of the substrate: O=C(CCC1C(/C2=C/C3=N/C(C(C=C)=C3C)=C\C4=C(C(CC)=C5N4[Mg]N2/C1=C(C(C6=C/7C)=NC7=C/5)/C(C6=O)C(OC)=O)C)C)O[R], where R is $\rm{C_{20}H_{39}}$, undergoes seven steps of reaction 1)~7) (described in subsequent questions), generating intermediate products I/II/III/IV/V/VI and the final product X, with the chemical formula $\rm{C_{32}H_{32}N_4OV}](/media/uploads/eb383129-2355-47f0-9312-7d382fd90410.png)</MultiModal>

The following table lists the seven steps of the reaction and their reaction types:

| Step | Reaction Type |
| :--- | :--- |
| 1) | Metal Coordination Dissociation |
| 2) | Hydrolysis |
| 3) | Demethoxycarbonylation |
| 4) | Reduction (gaining 4 electrons) |
| 5) | Aromatization |
| 6) | Decarboxylation |
| 7) | Metal Chelation Coordination |

For the unknown intermediate products $\mathbf{I\sim VI}$, their chemical formulas can be written as:

$$\mathrm{C}_x\mathrm{H}_y\mathrm{N}_z\mathrm{O}_w$$

Calculate the product of the mass fractions of each element, denoted as:

$$ P_i = \prod\limits_{j \in \{\mathrm{C,H,O,N}\}}\omega_j,\quad i=\mathbf{I\sim VI}$$

Based on the chemical formulas of the reactants and final product, deduce the chemical formulas of the intermediate products, calculate their $P_i$ values, and finally, calculate the sum of all $P_i$:

$$ S = \sum_i{P_i},\quad i=\mathbf{I\sim VI}$$

Choose the correct result from the following options. (Keep three significant figures for the calculation result, and choose the option with a deviation of less than $0.5\%$ from the calculated result, otherwise choose **A**: None of the other options are correct)
钒配合物可以采用一下流程合成：

<MultiModal>![底物的SMILES结构：O=C(CCC1C(/C2=C/C3=N/C(C(C=C)=C3C)=C\C4=C(C(CC)=C5N4[Mg]N2/C1=C(C(C6=C/7C)=NC7=C/5)/C(C6=O)C(OC)=O)C)C)O[R]，其中，R为 $\rm{C_{20}H_{39}}$，经历了 1)~7) 七步反应（在之后的题目中会描述），分别生成中间产物 I/II/III/IV/V/VI 以及最终产物 X，其化学式为 $\rm{C_{32}H_{32}N_4OV}](/media/uploads/eb383129-2355-47f0-9312-7d382fd90410.png)</MultiModal>

下面的表格列出了反应的七个步骤及其反应种类：

| 步骤 | 反应种类 |
| :--- | :--- |
| 1) | 配位金属解离 |
| 2) | 水解 |
| 3) | 脱甲氧羰基 |
| 4) | 还原（得到4个电子） |
| 5) | 芳构化 |
| 6) | 脱羧 |
| 7) | 金属螯合配位 |

对于未知的中间产物 $\mathbf{I\sim VI}$，其化学式可以写作：

$$\mathrm{C}_x\mathrm{H}_y\mathrm{N}_z\mathrm{O}_w$$

计算出其中每种元素质量分数的乘积，记为：

$$ P_i = \prod\limits_{j \in \{\mathrm{C,H,O,N}\}}\omega_j,\quad i=\mathbf{I\sim VI}$$ 

请你请你根据反应物与最终产物的化学式，推断中间产物的化学式，计算其 $P_i$ 的值，最终，计算所有 $P_i$ 的和：

$$ S = \sum_i{P_i},\quad i=\mathbf{I\sim VI}$$

从以下选项中选择正确的结果。（计算结果保留三位有效数字，选择与计算结果偏差小于 $0.5\%$ 的选项，否则选 **A**: 其他选项均不正确）
All other options are incorrect
0.00288
0.00207
0.00113
0.000942
0.00235
0.00185
0.00208
0.00301
0.00262
0.00129
0.00101
其他选项均不正确
0.00288
0.00207
0.00113
0.000942
0.00235
0.00185
0.00208
0.00301
0.00262
0.00129
0.00101
To solve this problem, we need to analyze the entire reaction process, determine the chemical formula of each intermediate, and then calculate the sum of the products of the mass fractions as required by the question.

First, we need to deduce the exact chemical formulas of intermediates $\mathbf{I}$ to $\mathbf{VI}$ based on the given seven-step reaction. The starting material for the reaction is a chlorophyll derivative, and according to the image information, its chemical formula is $\mathrm{C_{55}H_{72}N_4O_5Mg}$.

<Checkpoint>[The chemical formula of the starting material is $\mathrm{C_{55}H_{72}N_4O_5Mg}$](1)</Checkpoint>

The first step is coordination metal dissociation, which means that the central magnesium ion is removed, usually replaced by two hydrogen atoms. Therefore, we generate intermediate $\mathbf{I}$ from the starting material, and its chemical formula is $\mathrm{C_{55}H_{74}N_4O_5}$.

<Checkpoint>[The chemical formula of intermediate $\mathbf{I}$ is $\mathrm{C_{55}H_{74}N_4O_5}$](1)</Checkpoint>

The second step is hydrolysis, which usually acts on the ester group in the molecule. There are two ester groups in the molecule that can be hydrolyzed: one $\mathrm{C_{20}H_{39}}$ ester group (commonly known as phytol group) and the other is a methyl ester group. If the methyl ester group is hydrolyzed in this step, the subsequent "demethoxycarbonyl" reaction cannot proceed.

<Checkpoint>[The methyl ester cannot be hydrolyzed in the second step](0.5)</Checkpoint>

Therefore, only the phytol group can be hydrolyzed in this step. Hydrolysis will form a carboxylic acid and release phytol, resulting in the formation of intermediate $\mathbf{II}$, and its chemical formula becomes $\mathrm{C_{35}H_{36}N_4O_5}$.

<Checkpoint>[The chemical formula of intermediate $\mathbf{II}$ is $\mathrm{C_{35}H_{36}N_4O_5}$](1)</Checkpoint>

The third step is demethoxycarbonylation, which removes another smaller methoxycarbonyl group ($\rm -COOCH_3$) from the molecule and replaces it with a hydrogen atom, resulting in a molecular formula of $\mathrm{C_{33}H_{34}N_4O_3}$, which is intermediate $\mathbf{III}$.

<Checkpoint>[The chemical formula of intermediate $\mathbf{III}$ is $\mathrm{C_{33}H_{34}N_4O_3}$](1)</Checkpoint>

The fourth step is a reduction reaction that gains 4 electrons. In porphyrin chemistry, this refers to the process of reducing a ketone group ($\rm C=O$) on the ring to a methylene group ($\rm CH_2$). This process consumes one oxygen atom and increases two hydrogen atoms, thus we obtain intermediate $\mathbf{IV}$, and its chemical formula is $\mathrm{C_{33}H_{36}N_4O_2}$.

<Checkpoint>[The chemical formula of intermediate $\mathbf{IV}$ is $\mathrm{C_{33}H_{36}N_4O_2}$](1)</Checkpoint>

The fifth step is aromatization. This process causes a partially saturated ring system to lose hydrogen atoms, forming a fully conjugated aromatic porphyrin macrocycle. Specifically, the molecule loses two hydrogen atoms, generating intermediate $\mathbf{V}$, and its chemical formula is $\mathrm{C_{33}H_{34}N_4O_2}$.

<Checkpoint>[The chemical formula of intermediate $\mathbf{V}$ is $\mathrm{C_{33}H_{34}N_4O_2}$](1)</Checkpoint>

The sixth step is decarboxylation, which removes the carboxyl group ($\rm -COOH$) generated by the hydrolysis of the phytol group in the second step in the form of carbon dioxide ($\rm CO_2$) and replaces it with a hydrogen atom, further changing the molecular formula to $\mathrm{C_{32}H_{34}N_4}$, which is intermediate $\mathbf{VI}$.

<Checkpoint>[The chemical formula of intermediate $\mathbf{VI}$ is $\mathrm{C_{32}H_{34}N_4}$](1)</Checkpoint>

Finally, intermediate $\mathbf{VI}$ coordinates with tetravalent $\rm VO^{2+}$ to obtain a vanadium complex with the chemical formula $\mathrm{C_{32}H_{32}N_4OV}$.

Next, we need to calculate the $P$ value of each intermediate, which is the product of the mass fractions of the four elements carbon, hydrogen, oxygen, and nitrogen it contains. We will use the following atomic weights for calculation: C: 12.011, H: 1.008, N: 14.007, O: 15.999.

For intermediate $\mathbf{I}$, $\mathrm{C_{55}H_{74}N_4O_5}$, its molar mass $M_{\mathbf{I}} = 871.220 \text{ g/mol}$. The corresponding product of mass fractions $P_{\mathbf{I}}$ is:

$$P_{\mathbf{I}} = \left(\frac{55 \times 12.011}{871.220}\right) \times \left(\frac{74 \times 1.008}{871.220}\right) \times \left(\frac{4 \times 14.007}{871.220}\right) \times \left(\frac{5 \times 15.999}{871.220}\right) \approx 0.0003834$$

<Checkpoint>[$P_{\mathbf{I}} = 0.0003834$](1)</Checkpoint>

For intermediate $\mathbf{II}$, $\mathrm{C_{35}H_{36}N_4O_5}$, its molar mass $M_{\mathbf{II}} = 592.696 \text{ g/mol}$. The corresponding product of mass fractions $P_{\mathbf{II}}$ is:

$$P_{\mathbf{II}} = \left(\frac{35 \times 12.011}{592.696}\right) \times \left(\frac{36 \times 1.008}{592.696}\right) \times \left(\frac{4 \times 14.007}{592.696}\right) \times \left(\frac{5 \times 15.999}{592.696}\right) \approx 0.0005541$$

<Checkpoint>[$P_{\mathbf{II}} = 0.0005541$](1)</Checkpoint>

For intermediate $\mathbf{III}$, $\mathrm{C_{33}H_{34}N_4O_3}$, its molar mass $M_{\mathbf{III}} = 534.660 \text{ g/mol}$. The corresponding product of mass fractions $P_{\mathbf{III}}$ is:

$$P_{\mathbf{III}} = \left(\frac{33 \times 12.011}{534.660}\right) \times \left(\frac{34 \times 1.008}{534.660}\right) \times \left(\frac{4 \times 14.007}{534.660}\right) \times \left(\frac{3 \times 15.999}{534.660}\right) \approx 0.0004470$$

<Checkpoint>[$P_{\mathbf{III}} = 0.0004470$](1)</Checkpoint>

For intermediate $\mathbf{IV}$, $\mathrm{C_{33}H_{36}N_4O_2}$, its molar mass $M_{\mathbf{IV}} = 520.677 \text{ g/mol}$. The corresponding product of mass fractions $P_{\mathbf{IV}}$ is:

$$P_{\mathbf{IV}} = \left(\frac{33 \times 12.011}{520.677}\right) \times \left(\frac{36 \times 1.008}{520.677}\right) \times \left(\frac{4 \times 14.007}{520.677}\right) \times \left(\frac{2 \times 15.999}{520.677}\right) \approx 0.0003508$$

<Checkpoint>[$P_{\mathbf{IV}} = 0.0003508$](1)</Checkpoint>

For intermediate $\mathbf{V}$, $\mathrm{C_{33}H_{34}N_4O_2}$, its molar mass $M_{\mathbf{V}} = 518.661 \text{ g/mol}$. The corresponding product of mass fractions $P_{\mathbf{V}}$ is:

$$P_{\mathbf{V}} = \left(\frac{33 \times 12.011}{518.661}\right) \times \left(\frac{34 \times 1.008}{518.661}\right) \times \left(\frac{4 \times 14.007}{518.661}\right) \times \left(\frac{2 \times 15.999}{518.661}\right) \approx 0.0003365$$

<Checkpoint>[$P_{\mathbf{V}} = 0.0003365$](1)</Checkpoint>

For intermediate $\mathbf{VI}$, $\mathrm{C_{32}H_{34}N_4}$, since its molecule does not contain oxygen atoms, the mass fraction of its oxygen element $\omega_O$ is zero. Therefore, the product of mass fractions $P_{\mathbf{VI}}$ must also be zero.

$$P_{\mathbf{VI}} = 0$$

<Checkpoint>[$P_{\mathbf{VI}} = 0$](1)</Checkpoint>

In the last step, we add up all the calculated $P$ values to get the final sum $S$.

$$S = P_{\mathbf{I}} + P_{\mathbf{II}} + P_{\mathbf{III}} + P_{\mathbf{IV}} + P_{\mathbf{V}} + P_{\mathbf{VI}}$$

$$S \approx 0.00207$$

Therefore, after a complete calculation, the sum $S$ of the $P$ values of all intermediates is $0.00207$.

<Checkpoint>[$S = 0.00207$](1)</Checkpoint>
要解决这个问题，我们需要分析整个反应流程，确定每个中间体的化学式，然后计算出题目所要求的质量分数乘积之和。

首先，我们需要根据给出的七步反应来推断中间体 $\mathbf{I}$ 到 $\mathbf{VI}$ 的确切化学式。反应的起始物是一个叶绿素衍生物，根据图片信息，其化学式为 $\mathrm{C_{55}H_{72}N_4O_5Mg}$。

<Checkpoint>[反应起始物的化学式为 $\mathrm{C_{55}H_{72}N_4O_5Mg}$](1)</Checkpoint>

第一步是配位金属解离，这意味着中心的镁离子被移除，通常由两个氢原子取代，因此，我们从起始物生成了中间体 $\mathbf{I}$，其化学式为 $\mathrm{C_{55}H_{74}N_4O_5}$。

<Checkpoint>[中间体 $\mathbf{I}$ 化学式为 $\mathrm{C_{55}H_{74}N_4O_5}$](1)</Checkpoint>

第二步是水解反应，这通常会作用于分子中的酯基，分子中有两个酯基可被水解：一个 $\mathrm{C_{20}H_{39}}$ 酯基（通常称为植醇基）、另一个是甲酯基。这一步如果水解甲酯基，则之后的“脱甲氧酰基”反应无法进行。

<Checkpoint>[第二步不能水解甲酯](0.5)</Checkpoint>

因此，这一步只能水解植醇基，水解会形成一个羧酸，并释放出植醇，从而生成了中间体 $\mathbf{II}$，其化学式变为 $\mathrm{C_{35}H_{36}N_4O_5}$。

<Checkpoint>[中间体 $\mathbf{II}$ 化学式为 $\mathrm{C_{35}H_{36}N_4O_5}$](1)</Checkpoint>

第三步是脱甲氧羰基，这会移除分子中另一个较小的甲氧基羰基（$\rm -COOCH_3$）并由一个氢原子取代，导致分子式变为 $\mathrm{C_{33}H_{34}N_4O_3}$，这便是中间体 $\mathbf{III}$。

<Checkpoint>[中间体 $\mathbf{III}$ 化学式为 $\mathrm{C_{33}H_{34}N_4O_3}$](1)</Checkpoint>

第四步是一个得到4个电子的还原反应，在卟啉化学中，这指环上的某个酮基（$\rm C=O$）被还原为亚甲基（$\rm CH_2$）的过程，这个过程会消耗一个氧原子并增加两个氢原子，由此我们得到中间体 $\mathbf{IV}$，其化学式为 $\mathrm{C_{33}H_{36}N_4O_2}$。

<Checkpoint>[中间体 $\mathbf{IV}$ 化学式为 $\mathrm{C_{33}H_{36}N_4O_2}$](1)</Checkpoint>

第五步是芳构化，这个过程会使部分饱和的环系失去氢原子，形成一个完全共轭的芳香性卟啉大环，具体表现为分子失去两个氢原子，生成了中间体 $\mathbf{V}$，其化学式为 $\mathrm{C_{33}H_{34}N_4O_2}$。

<Checkpoint>[中间体 $\mathbf{V}$ 化学式为 $\mathrm{C_{33}H_{34}N_4O_2}$](1)</Checkpoint>

第六步是脱羧反应，将第二步水解植醇基生成的羧基（$\rm -COOH$）以二氧化碳（$\rm CO_2$）的形式脱除，并由氢原子取代，分子式进一步变为 $\mathrm{C_{32}H_{34}N_4}$，这就是中间体 $\mathbf{VI}$。

<Checkpoint>[中间体 $\mathbf{VI}$ 化学式为 $\mathrm{C_{32}H_{34}N_4}$](1)</Checkpoint>

最终中间体 $\mathbf{VI}$ 与四价的 $\rm VO^{2+}$ 配位，即可得到化学式为 $\mathrm{C_{32}H_{32}N_4OV}$ 的钒络合物。

接下来，我们需要计算每个中间体的 $P$ 值，即其含有的碳、氢、氧、氮四种元素质量分数的乘积，我们将使用以下原子量进行计算：C: 12.011, H: 1.008, N: 14.007, O: 15.999。

对于中间体 $\mathbf{I}$, $\mathrm{C_{55}H_{74}N_4O_5}$，其摩尔质量 $M_{\mathbf{I}} = 871.220 \text{ g/mol}$。其对应的质量分数乘积 $P_{\mathbf{I}}$ 为：

$$P_{\mathbf{I}} = \left(\frac{55 \times 12.011}{871.220}\right) \times \left(\frac{74 \times 1.008}{871.220}\right) \times \left(\frac{4 \times 14.007}{871.220}\right) \times \left(\frac{5 \times 15.999}{871.220}\right) \approx 0.0003834$$

<Checkpoint>[$P_{\mathbf{I}} = 0.0003834$](1)</Checkpoint>

对于中间体 $\mathbf{II}$, $\mathrm{C_{35}H_{36}N_4O_5}$，其摩尔质量 $M_{\mathbf{II}} = 592.696 \text{ g/mol}$。其对应的质量分数乘积 $P_{\mathbf{II}}$ 为：

$$P_{\mathbf{II}} = \left(\frac{35 \times 12.011}{592.696}\right) \times \left(\frac{36 \times 1.008}{592.696}\right) \times \left(\frac{4 \times 14.007}{592.696}\right) \times \left(\frac{5 \times 15.999}{592.696}\right) \approx 0.0005541$$

<Checkpoint>[$P_{\mathbf{II}} = 0.0005541$](1)</Checkpoint>

对于中间体 $\mathbf{III}$, $\mathrm{C_{33}H_{34}N_4O_3}$，其摩尔质量 $M_{\mathbf{III}} = 534.660 \text{ g/mol}$。其对应的质量分数乘积 $P_{\mathbf{III}}$ 为：

$$P_{\mathbf{III}} = \left(\frac{33 \times 12.011}{534.660}\right) \times \left(\frac{34 \times 1.008}{534.660}\right) \times \left(\frac{4 \times 14.007}{534.660}\right) \times \left(\frac{3 \times 15.999}{534.660}\right) \approx 0.0004470$$

<Checkpoint>[$P_{\mathbf{III}} = 0.0004470$](1)</Checkpoint>

对于中间体 $\mathbf{IV}$, $\mathrm{C_{33}H_{36}N_4O_2}$，其摩尔质量 $M_{\mathbf{IV}} = 520.677 \text{ g/mol}$。其对应的质量分数乘积 $P_{\mathbf{IV}}$ 为：

$$P_{\mathbf{IV}} = \left(\frac{33 \times 12.011}{520.677}\right) \times \left(\frac{36 \times 1.008}{520.677}\right) \times \left(\frac{4 \times 14.007}{520.677}\right) \times \left(\frac{2 \times 15.999}{520.677}\right) \approx 0.0003508$$

<Checkpoint>[$P_{\mathbf{IV}} = 0.0003508$](1)</Checkpoint>

对于中间体 $\mathbf{V}$, $\mathrm{C_{33}H_{34}N_4O_2}$，其摩尔质量 $M_{\mathbf{V}} = 518.661 \text{ g/mol}$。其对应的质量分数乘积 $P_{\mathbf{V}}$ 为：

$$P_{\mathbf{V}} = \left(\frac{33 \times 12.011}{518.661}\right) \times \left(\frac{34 \times 1.008}{518.661}\right) \times \left(\frac{4 \times 14.007}{518.661}\right) \times \left(\frac{2 \times 15.999}{518.661}\right) \approx 0.0003365$$

<Checkpoint>[$P_{\mathbf{V}} = 0.0003365$](1)</Checkpoint>

对于中间体 $\mathbf{VI}$, $\mathrm{C_{32}H_{34}N_4}$，由于其分子中不含有氧原子，其氧元素的质量分数 $\omega_O$ 为零。因此，质量分数的乘积 $P_{\mathbf{VI}}$ 也必然为零。

$$P_{\mathbf{VI}} = 0$$

<Checkpoint>[$P_{\mathbf{VI}} = 0$](1)</Checkpoint>

最后一步，我们将所有计算出的 $P$ 值相加，得到最终的总和 $S$。

$$S = P_{\mathbf{I}} + P_{\mathbf{II}} + P_{\mathbf{III}} + P_{\mathbf{IV}} + P_{\mathbf{V}} + P_{\mathbf{VI}}$$

$$S \approx 0.00207$$

因此，经过完整的计算，所有中间体 $P$ 值的总和 $S$ 为 $0.00207$。

<Checkpoint>[$S = 0.00207$](1)</Checkpoint>