Students in a medicinal chemistry laboratory plan to use multicomponent reaction-based combinatorial chemistry to synthesize covalent compound libraries. The synthesis involves manual labeling and placement of reaction materials, followed by automated dispensing for batch addition. All reactants are grouped according to the functional groups involved in the multicomponent reaction and added to the reaction system in batches. Each parallel addition contains dozens of compounds with one type of functional group involved in the multicomponent reaction, and the covalent warhead is fixed on one of the compound components. Ultimately, a single experiment will yield a set of compounds with similar scaffolds but containing various different substituents and multiple different covalent warheads.

In one synthesis, the reactants in the reaction plan for one of the compounds are as follows:

1. <MultiModal>![[C-]#[N+]CC1=CC=CC=C1](/media/uploads/c024bc3d-3a47-4fb5-a7c7-cf6c5c405e8b.png)</MultiModal>
2. <MultiModal>![FC(C)(F)C1=CC=C(N)C=C1](/media/uploads/1f9cce00-6733-4a21-8cee-cdc920f2ea4d.png)</MultiModal>
3. <MultiModal>![O=S(C1=CC=C(C(O)=O)C=C1)(F)=O](/media/uploads/2c39f115-dbe8-492f-90d7-4d27ffef7399.png)</MultiModal>
4. <MultiModal>![O=C(N)C1=CC=C(C([H])=O)C=C1](/media/uploads/9c659185-afb5-43e6-8726-92584d9e16f6.png)</MultiModal>

The H-NMR characterization results of the product are:
`1H NMR (400 MHz, DMSO) δ 8.78 (t, J = 5.9 Hz, 1H), 7.87 (s, 1H), 7.63 (d, J = 8.1 Hz, 2H), 7.36 – 7.17 (m, 9H), 7.13 (d, J = 8.1 Hz, 2H), 6.23 (d, J = 17.9 Hz, 2H), 5.86 (dd, J = 16.8, 10.3 Hz, 1H), 5.60 (dd, J = 10.3, 2.4 Hz, 1H), 4.35 (d, J = 5.8 Hz, 2H), 1.77 (s, 1H), 1.72 (s, 2H), 1.67 (s, 1H)`.

After LCMS characterization, the molecular weight of the product was found to be **477**.

The product molecule contains only one type of covalent warhead.

Hint: Acrylamide is also a common covalent warhead, and errors in feeding materials sometimes occur in experiments.

Please select the correct statement from the following:
某药物化学实验室同学计划用基于多组分反应的组合化学进行共价化合物库的合成，合成由人工标注并放置反应原料，然后在自动化设备辅助下批量加样进行。所有的反应物是按照参与多组分反应的官能团分组并分批加入反应体系的，每次平行加入含有参与多组分反应的一种官能团的数十种化合物，并且共价弹头固定位于其中的某一种化合物组分上。最终一次实验中会得到骨架相似而含有多种不同取代基、多种不同共价头的一组化合物。

在一次合成中，其中一个化合物的合成反应计划中的反应物分别为：

1. <MultiModal>![[C-]#[N+]CC1=CC=CC=C1](/media/uploads/c024bc3d-3a47-4fb5-a7c7-cf6c5c405e8b.png)</MultiModal>
2. <MultiModal>![FC(C)(F)C1=CC=C(N)C=C1](/media/uploads/1f9cce00-6733-4a21-8cee-cdc920f2ea4d.png)</MultiModal>
3. <MultiModal>![O=S(C1=CC=C(C(O)=O)C=C1)(F)=O](/media/uploads/2c39f115-dbe8-492f-90d7-4d27ffef7399.png)</MultiModal>
4. <MultiModal>![O=C(N)C1=CC=C(C([H])=O)C=C1](/media/uploads/9c659185-afb5-43e6-8726-92584d9e16f6.png)</MultiModal>

产物的H-NMR表征结果为：  
`1H NMR (400 MHz, DMSO) δ 8.78 (t, J = 5.9 Hz, 1H), 7.87 (s, 1H), 7.63 (d, J = 8.1 Hz, 2H), 7.36 – 7.17 (m, 9H), 7.13 (d, J = 8.1 Hz, 2H), 6.23 (d, J = 17.9 Hz, 2H), 5.86 (dd, J = 16.8, 10.3 Hz, 1H), 5.60 (dd, J = 10.3, 2.4 Hz, 1H), 4.35 (d, J = 5.8 Hz, 2H), 1.77 (s, 1H), 1.72 (s, 2H), 1.67 (s, 1H)`。

产物经过LCMS表征后，发现分子量是**477**。

产物分子中仅包含了一种共价弹头。

提示：丙烯酰胺也是一种常见的共价弹头，实验中有时会出现加料错误的问题。

请选出下列说法中正确的选项：
None of the other options are correct
The product molecule is <MultiModal>![O=S(C1=CC=C(C(N(C(C2=CC=C(C(N)=O)C=C2)C(NCC3=CC=CC=C3)=O)C4=CC=C(C(F)(F)C)C=C4)=O)C=C1)(F)=O](/media/uploads/77baa057-1b7d-42ab-90a1-b092368ed92a.png)</MultiModal>
The protein residue covalently bound to the product molecule may be Cys, and the labeling process involves the formation of S-S bonds.
The protein residue covalently bound to the product molecule may be Lys
The product molecule has 28 carbon atoms.
The actual product molecule involves only three components participating in the reaction.
其他选项均不正确
产物分子为<MultiModal>![O=S(C1=CC=C(C(N(C(C2=CC=C(C(N)=O)C=C2)C(NCC3=CC=CC=C3)=O)C4=CC=C(C(F)(F)C)C=C4)=O)C=C1)(F)=O](/media/uploads/77baa057-1b7d-42ab-90a1-b092368ed92a.png)</MultiModal>
产物分子共价结合的蛋白残基可能是Cys，且标记过程涉及S-S键的形成
产物分子共价结合的蛋白残基可能是Lys
产物分子有28个碳原子
产物分子实际只有三个组分参与反应
If the Ugi reaction is carried out with the reactants specified in the problem, the product molecule corresponding to option B can be obtained:  

<MultiModal>![O=S(C1=CC=C(C(N(C(C2=CC=C(C(N)=O)C=C2)C(NCC3=CC=CC=C3)=O)C4=CC=C(C(F)(F)C)C=C4)=O)C=C1)(F)=O](/media/uploads/77baa057-1b7d-42ab-90a1-b092368ed92a.png)</MultiModal>  

<Checkpoint>[The theoretical product molecule is <MultiModal>![O=S(C1=CC=C(C(N(C(C2=CC=C(C(N)=O)C=C2)C(NCC3=CC=CC=C3)=O)C4=CC=C(C(F)(F)C)C=C4)=O)C=C1)(F)=O](/media/uploads/77baa057-1b7d-42ab-90a1-b092368ed92a.png)</MultiModal>](1)</Checkpoint>  

**Key H-NMR Signal Analysis**  
- The signals at δ 5.86 (dd, J = 16.8, 10.3 Hz, 1H) and δ 5.60 (dd, J = 10.3, 2.4 Hz, 1H) are characteristic of a terminal alkene ($-\mathrm{CH}{=}\mathrm{CH}_2$).  
- The signal at δ 6.23 (d, J = 17.9 Hz, 2H) likely corresponds to the amide $-\mathrm{NH}_2$ of the aldehyde component.  
- The signal at δ 4.35 (d, J = 5.8 Hz, 2H) matches the $\mathrm{CH}_2$ of the benzyl group, corresponding to the benzyl isocyanide.  
- The aromatic region shows a total of 13 protons, which is inconsistent with the theoretical product, possibly due to a missing phenyl ring.  
- The signal near δ 1.7 corresponds to methyl hydrogens from the amine component.  

<Checkpoint>[The product's H-NMR shows characteristic peaks of a terminal alkene ($-\mathrm{CH}{=}\mathrm{CH}_2$), which is inconsistent with the theoretical product](1)</Checkpoint>  

The theoretical product has a calculated molecular weight of 609, which does not match the mass spectrometry results.  

<Checkpoint>[The theoretical product has a molecular weight of 609, which does not match the mass spectrometry data](1)</Checkpoint>  

Neither the H-NMR nor the mass spectrometry results align with the proposed product. Additionally, none of the reactants inherently possess a structure that could reasonably form a terminal alkene. Therefore, we must consider whether an experimental error occurred. Given the combinatorial chemistry workflow, multiple compounds with the same functional group are typically pooled and added simultaneously (often prepared as a molecular library in solution, e.g., in microplates). A common mistake in this process is the accidental addition of the wrong reagent during library preparation, leading to an incorrect product.  

Based on the prominent terminal alkene signals in the H-NMR and the frequent use of covalent warheads such as acrylic acid/acrylamide, it is plausible that the fluorosulfonyl reagent was mistakenly replaced with acrylic acid.  

<Checkpoint>[Considering the H-NMR results and the combinatorial chemistry workflow, the fluorosulfonyl reagent may have been mistakenly replaced with acrylic acid or an alkenyl sulfone](1)</Checkpoint>  

If the carboxylic acid reactant is replaced with acrylic acid, the resulting product molecule is <MultiModal>![NC(C(C=C1)=CC=C1C(C(NCC2=CC=CC=C2)=O)N(C3=CC=C(C=C3)C(F)(C)F)C(C=C)=O)=O](/media/uploads/0e222098-eb73-4d28-ad3c-48755b20ad86.png)</MultiModal>, which matches the mass spectrometry result of 477, confirming the hypothesis. In contrast, an alkenyl sulfone would have an excessively high molecular weight.  

<Checkpoint>[When the carboxylic acid reactant is replaced with acrylic acid, the product’s molecular weight (477) matches the mass spectrometry data, supporting the hypothesis. An alkenyl sulfone would have an excessively high molecular weight and is unlikely](1)</Checkpoint>  
<Checkpoint>[The final product molecule contains no sulfur atoms, ruling out option C](1)</Checkpoint>  

Based on this, we can conclude that four compounds participated in the reaction, making option F incorrect.  

<Checkpoint>[Four compounds participated in the reaction](1)</Checkpoint>  

The product molecule contains an acrylamide group as the covalent warhead, which primarily targets Cys residues.  

<Checkpoint>[Acrylamide-based covalent warheads primarily target Cys residues and generally do not react with Lys, making option D incorrect](1)</Checkpoint>  

The product contains a total of 27 carbon atoms, so option E is incorrect.  

<Checkpoint>[The product contains 27 carbon atoms, ruling out option E](1)</Checkpoint>  

Thus, none of the options are correct.
如果按照题目中的反应物进行Ugi反应，可以得到选项B的产物分子<MultiModal>![O=S(C1=CC=C(C(N(C(C2=CC=C(C(N)=O)C=C2)C(NCC3=CC=CC=C3)=O)C4=CC=C(C(F)(F)C)C=C4)=O)C=C1)(F)=O](/media/uploads/77baa057-1b7d-42ab-90a1-b092368ed92a.png)</MultiModal>

<Checkpoint>[理论上的产物分子为<MultiModal>![O=S(C1=CC=C(C(N(C(C2=CC=C(C(N)=O)C=C2)C(NCC3=CC=CC=C3)=O)C4=CC=C(C(F)(F)C)C=C4)=O)C=C1)(F)=O](/media/uploads/77baa057-1b7d-42ab-90a1-b092368ed92a.png)</MultiModal>](1)</Checkpoint>


H-NMR关键核磁信号分析
- δ 5.86 (dd, J = 16.8, 10.3 Hz, 1H)和δ 5.60 (dd, J = 10.3, 2.4 Hz, 1H)的信号模式典型的是末端烯烃($-\mathrm{CH}{=}\mathrm{CH}_2$)。
- δ 6.23 (d, J = 17.9 Hz, 2H)可能是酰胺的$-\mathrm{NH}_2$，对应醛组分。
- δ 4.35 (d, J = 5.8 Hz, 2H)对应于苄基的$\mathrm{CH}_2$，对应苄异腈。
- 芳香区累计有13个H，与理论产物不符，可能少了苯环。
- δ 1.7附近为甲基氢，对应胺组分。

<Checkpoint>[产物H-NMR有典型的末端烯烃（$-\mathrm{CH}{=}\mathrm{CH}_2$）的特征峰，与理论上的产物不一致](1)</Checkpoint>

理论上的产物分子可以计算分子量为609，与质谱结果不一致。
<Checkpoint>[理论上的产物分子量为609，与质谱结果无法对应](1)</Checkpoint>


H-NMR与质谱结果都无法与上述产物对应，且注意到这几种反应物本身并不存在能产生末端烯烃的合理结构。因此我们需要推理是否在实验过程本身存在失误。考虑到组合化学的实验模式，含有同一官能团的多个不同化合物被一批加入，这一过程需要先将不同化合物准备成分子库的形式（常见的是在孔板中配成溶液）。此过程中一种常见的失误是做组合库时，操作失误导致加错原料，自然目标产物会被张冠李戴。

结合H-NMR中明显的末端烯烃，以及常用共价反应弹头丙烯酸/丙烯酰胺，猜测可能是氟磺酰原料加错成了丙烯酸。

<Checkpoint>[从H-NMR的结果和组合化学的实验流程推测，可能是氟磺酰原料加错成了丙烯酸或者烯基砜等](1)</Checkpoint>


如果将原料酸分子替换成丙烯酸，产物分子为<MultiModal>![NC(C(C=C1)=CC=C1C(C(NCC2=CC=CC=C2)=O)N(C3=CC=C(C=C3)C(F)(C)F)C(C=C)=O)=O](/media/uploads/0e222098-eb73-4d28-ad3c-48755b20ad86.png)</MultiModal>，与质谱结果477对应上，猜测正确；而烯基砜分子量过大，难以满足

<Checkpoint>[原料酸分子替换成丙烯酸时，产物分子量与质谱结果477对应，猜测正确；而烯基砜分子量过大，难以满足](1)</Checkpoint>
<Checkpoint>[最终产物分子中没有硫原子，C错误](1)</Checkpoint>

据此我们可以判断仍然有四个化合物参与了反应，F选项错误。

<Checkpoint>[有4个化合物参与了反应](1)</Checkpoint>

此时产物分子的共价反应官能团是丙烯酰胺，此类官能团主要靶向Cys残基。

<Checkpoint>[丙烯酰胺这类共价弹头主要靶向Cys残基，一般不与Lys反应，D错误](1)</Checkpoint>

产物一共有27个碳原子，E错误。

<Checkpoint>[产物一共有27个碳原子，E错误](1)</Checkpoint>

故所有选项均不正确。