In recent years, a research team developed a [3 + 2 + 1] strategy for synthesizing quinoline rings, with the model reaction as follows:  

<MultiModal>![The image shows an organic synthesis flowchart. The process is described as follows: First, two molecules of $\mathrm{Ph_3P=O}$ and one molecule of $\mathrm{Tf_2O}$ form $\mathbf{A}$. $\mathbf{A}$ reacts with an organic molecule (SMILES: C1=CC=C(C=C1)NC(=O)[R1]) to generate intermediate $\mathbf{B}$, during which $\mathbf{A}$ is converted to $\mathbf{C}$. Intermediate $\mathbf{B}$ then reacts with an organic molecule (SMILES: C(#C[R2])[R2]) to produce the final product shown on the far right, with the SMILES formula C1=CC=C2C(=C1)C(=C(C(=N2)[R1])[R2])[R2].](/media/uploads/15456655-d96d-4e37-bab9-89511939a47c.png)</MultiModal>  

The reaction below utilizes this model reaction to synthesize quinuclidine:  

<MultiModal>![The image shows an organic synthesis roadmap. The synthetic route is described as follows: The starting reactant on the left (SMILES: CC1=C(C=CC=C1C#N)C#N) forms $\mathbf{D}$ under the conditions $\mathrm{t-BuOK + (COOEt)_2}$. $\mathbf{D}$ reacts with $\mathrm{HCCCH_2OH}$ to form $\mathbf{E}$ under the conditions $\mathrm{PBu_3, DEAD}$. $\mathbf{E}$ is converted to $\mathbf{F}$ under the conditions of first adding $\mathrm{LiOH}$, then $\mathrm{(COCl)_2}$, and finally $\mathrm{PhNH_2}$. $\mathbf{F}$ is converted to $\mathbf{G}$ under the conditions $\mathrm{Ph_3P=O, Tf_2O}$.](/media/uploads/0db06ec5-ffda-4bfe-afd7-7377a6c37dd5.png)</MultiModal>  

Regarding the nine unknown species $\mathbf{A-G}$ in the two figures above, the known information is: Intermediate $\mathbf{B}$ is positively charged and contains no oxygen atoms.  

The correct options are:
近年来，某课题组开发了一种[3 + 2 + 1]策略合成喹啉环的方法，其模型反应如下：

<MultiModal>![该图片为一张有机合成流程图，流程描述如下：首先两分子$\mathrm{Ph_3P=O}$和一分子$\mathrm{Tf_2O}$生成$\mathbf{A}$，$\mathbf{A}$与SMILES为C1=CC=C(C=C1)NC(=O)[R1]的有机分子发生反应生成中间体$\mathbf{B}$，此过程中$\mathbf{A}$转化为$\mathbf{C}$；中间体$\mathbf{B}$与SMILES为C(#C[R2])[R2]的有机分子发生反应生成图最右边的最终产物，最终产物的SMILES为C1=CC=C2C(=C1)C(=C(C(=N2)[R1])[R2])[R2]。](/media/uploads/15456655-d96d-4e37-bab9-89511939a47c.png)</MultiModal>

下图的反应利用了该模型反应合成奎宁环：

<MultiModal>![该图片为一张有机合成路线图，合成路线描述如下：图左为最初的反应物，其SMILES为CC1=C(C=CC=C1C#N)C#N；该反应物生成$\mathbf{D}$，反应条件为$\mathrm{t-BuOK + (COOEt)_2}$；$\mathbf{D}$与$\mathrm{HCCCH_2OH}$生成$\mathbf{E}$，反应条件为$\mathrm{PBu_3, DEAD}$；$\mathbf{E}$生成$\mathbf{F}$，反应条件为首先加入$\mathrm{LiOH}$，再加入$\mathrm{(COCl)_2}$，最后加入$\mathrm{PhNH_2}$；$\mathbf{F}$生成$\mathbf{G}$，反应条件为$\mathrm{Ph_3P=O, Tf_2O}$。](/media/uploads/0db06ec5-ffda-4bfe-afd7-7377a6c37dd5.png)</MultiModal>

关于上面两张图中9个未知物种$\mathbf{A-G}$，已知信息：中间体$\mathbf{B}$带正电且不含氧原子。

下列选项正确的是：
The electrophilicity of intermediate $\mathbf{B}$ is very poor.
The predominant form of $\mathbf{D}$ in aqueous solution does not contain hydroxyl groups
The product $\mathbf{G}$ contains three rings
The molecular formula of $\mathbf{G}$ is $\mathrm{C_{20}H_{10}N_3O}$.
$\mathbf{G}$ undergoes complete hydrolysis under acidic conditions, and the products contain four carbonyl groups.
$\mathbf{G}$ contains a five-membered ring that is unstable under acidic conditions.
中间体$\mathbf{B}$的亲电性很差
$\mathbf{D}$在水溶液中的主要存在形式不含羟基
产物$\mathbf{G}$含有三个环
$\mathbf{G}$分子式为$\mathrm{C_{20}H_{10}N_3O}$
$\mathbf{G}$在酸性条件下完全水解，产物中含有四个羰基
$\mathbf{G}$含有在酸性条件下不稳定的五元环
First, analyze the model reaction. The reactants are 2 equivalents of triphenylphosphine oxide and 1 equivalent of trifluoromethanesulfonic anhydride. Tf₂O, as a strong electrophile and dehydrating agent, reacts with triphenylphosphine oxide to generate a phosphonium salt, specifically the bis(triphenylphosphine oxide) salt, commonly referred to as an analog of the Mukaiyama reagent. Its structure is $\mathrm{[(Ph₃P)_2O]^{2+}(TfO⁻)_2}$, denoted as $\mathbf{A}$. This is a highly potent oxygen-atom-activating reagent.

<Checkpoint>[The structure of $\mathbf{A}$ is $\mathrm{[(Ph₃P)_2O]^{2+}(TfO⁻)_2}$](1)</Checkpoint>

The substrate N-acylaniline reacts with reagent $\mathbf{A}$. The carbonyl oxygen of the amide acts as a nucleophile to attack one of the phosphorus atoms in $\mathbf{A}$. Subsequently, the lone pair on the nitrogen atom can delocalize toward the carbonyl carbon, expelling a triphenylphosphine oxide and yielding a positively charged nitrilium ion with the structure $\mathrm{R₁-C≡N⁺-Ph}$, which satisfies the property of being devoid of oxygen. Thus, this intermediate is $\mathbf{B}$. Due to its strong electrophilicity, option A is incorrect.

<Checkpoint>[The structure of $\mathbf{B}$ is $\mathrm{R₁-C≡N⁺-Ph}$](1)</Checkpoint>

Next, analyze the synthetic route. The starting material is 2,3-dicyanotoluene. Potassium tert-butoxide (t-BuOK), a strong base, abstracts a proton from the benzylic position, generating a stable benzylic anion.

<Checkpoint>[t-BuOK abstracts a proton from the benzylic position](1)</Checkpoint>

This anion acts as a nucleophile, undergoing nucleophilic addition to one of the carbonyl groups of diethyl oxalate to form a β-diketone structure, denoted as $\mathbf{D}$. Its SMILES is CCOC(=O)/C(=C/C1=C(C=CC=C1C#N)C#N)/O. Due to the presence of an intramolecular hydrogen bond in the six-membered ring, the enol form predominates, making option B incorrect.

<Checkpoint>[$\mathbf{D}$ is a β-diketone structure, predominantly in the enol form, with SMILES CCOC(=O)/C(=C/C1=C(C=CC=C1C#N)C#N)/O](1)</Checkpoint>

The formation of $\mathbf{E}$ involves the famous Mitsunobu reaction, where the enolic hydroxyl group of $\mathbf{D}$ condenses with propargyl alcohol, dehydrating to form an ether linkage. The net result is the replacement of the enolic hydroxyl hydrogen with a propargyl group. The SMILES structure is C#CCO/C(=C\C1=C(C=CC=C1C#N)C#N)/C(=O)OCC.

<Checkpoint>[$\mathbf{E}$ is formed by replacing the enolic hydroxyl hydrogen of $\mathbf{D}$ with a propargyl group, with SMILES C#CCO/C(=C\C1=C(C=CC=C1C#N)C#N)/C(=O)OCC](1)</Checkpoint>

The reaction to form $\mathbf{F}$ begins with the addition of lithium hydroxide, hydrolyzing the ethyl ester group derived from diethyl oxalate to a carboxylic acid. Subsequent treatment with oxalyl chloride converts the carboxylic acid to an acyl chloride, which is then reacted with aniline to form an amide. The final result is the conversion of the ethyl ester group to an N-phenylamide. The SMILES structure is C#CCO/C(=C\C1=C(C=CC=C1C#N)C#N)/C(=O)N[Ph].

<Checkpoint>[$\mathbf{F}$ is formed by converting the ethyl ester group of $\mathbf{E}$ to an N-phenylamide, with SMILES C#CCO/C(=C\C1=C(C=CC=C1C#N)C#N)/C(=O)N[Ph]](1)</Checkpoint>

The final step involves the application of the model reaction, where the N-phenylamide is converted to a nitrilium ion. Due to the presence of an alkyne group within the molecule, an intramolecular model reaction occurs, forming a quinoline ring followed by aromatization. The final product $\mathbf{G}$ has the SMILES structure C1=CC2=CC3=C(/C(=C\C4=C(C=CC=C4C#N)C#N)/OC3)N=C2C=C1.

<Checkpoint>[The SMILES structure of $\mathbf{G}$ is C1=CC2=CC3=C(/C(=C\C4=C(C=CC=C4C#N)C#N)/OC3)N=C2C=C1.](2)</Checkpoint>

Based on this structure: its molecular formula is $\mathrm{C_{20}H_{11}N_3O}$, containing four rings, among which the five-membered ring is a hemiketal structure. Under acidic conditions, it readily undergoes ring-opening to form an alcohol and a ketone, rendering it unstable.

<Checkpoint>[The five-membered ring in $\mathbf{G}$ is a hemiketal structure, which readily undergoes ring-opening under acidic conditions to form an alcohol and a ketone](1)</Checkpoint>

Under acidic hydrolysis, the five-membered ring opens to form a ketone carbonyl group, while the two cyano groups hydrolyze to form two carboxylic carbonyl groups, totaling three carbonyl groups. Thus, options C, D, and E are incorrect, and option F is correct.
首先分析模型反应，反应物是2当量的三苯基氧膦和1当量的三氟甲磺酸酐。Tf₂O作为一个强亲电试剂和脱水剂，与三苯基氧膦反应会生成一个鏻盐，即双(三苯基膦氧基)盐，通常称为Mukaiyama试剂的类似物。其结构为 $\mathrm{[(Ph₃P)_2O]^{2+}(TfO⁻)_2}$，即$\mathbf{A}$。这是一个非常强的氧原子活化试剂。

<Checkpoint>[$\mathbf{A}$结构为 $\mathrm{[(Ph₃P)_2O]^{2+}(TfO⁻)_2}$](1)</Checkpoint>

底物N-酰基苯胺与试剂$\mathbf{A}$反应。酰胺的羰基氧原子作为亲核试剂攻击$\mathbf{A}$中的一个磷原子，之后氮原子的孤对电子可以推向羰基碳，离去一个三苯基氧磷，得到带正电的腈鎓离子，其结构为$\mathrm{R₁-C≡N⁺-Ph}$，满足不带氧原子的性质，故该中间体即为$\mathbf{B}$。该结构具有强亲电性，故选项A错误。

<Checkpoint>[$\mathbf{B}$结构为$\mathrm{R₁-C≡N⁺-Ph}$](1)</Checkpoint>

之后解析合成路线。起始物是2,3-二氰基甲苯。叔丁醇钾是强碱，会夺取苯甲基上的质子，生成一个稳定的苯甲基负离子.

<Checkpoint>[t-BuOK夺取苯甲基上的质子](1)</Checkpoint>

该负离子作为亲核试剂，对草酸二乙酯的一个羰基进行亲核加成，得到一个β-二酮结构，即结构$\mathbf{D}$。其SMILES为CCOC(=O)/C(=C/C1=C(C=CC=C1C#N)C#N)/O，该结构由于六元环分子内氢键的存在，其烯醇式为主要存在形式，故选项B错误。

<Checkpoint>[$\mathbf{D}$为β-二酮结构，主烯醇式，SMILES为CCOC(=O)/C(=C/C1=C(C=CC=C1C#N)C#N)/O](1)</Checkpoint>

生成$\mathbf{E}$的反应为著名的Mitsunobu反应，将$\mathbf{D}$的烯醇羟基与反应物炔丙醇进行缩合，脱水生成醚键，净反应结果为烯醇羟基的H原子更换为炔丙基。SMILES结构式为C#CCO/C(=C\C1=C(C=CC=C1C#N)C#N)/C(=O)OCC

<Checkpoint>[$\mathbf{E}$结构为$\mathbf{D}$的烯醇羟基的H原子更换为炔丙基，SMILES结构式为C#CCO/C(=C\C1=C(C=CC=C1C#N)C#N)/C(=O)OCC](1)</Checkpoint>

生成$\mathbf{F}$的反应，首先加入氢氧化锂，水解掉来自于草酸二乙酯的乙酯基变为羧酸，之后加入草酰氯将羧酸变为酰氯，最后加入苯胺使酰氯变为酰胺，最终结果为乙酯基变为N-苯基酰胺。SMILES结构式为C#CCO/C(=C\C1=C(C=CC=C1C#N)C#N)/C(=O)N[Ph]

<Checkpoint>[$\mathbf{F}$为$\mathbf{E}$的乙酯基变为N-苯基酰胺，SMILES结构式为C#CCO/C(=C\C1=C(C=CC=C1C#N)C#N)/C(=O)N[Ph]](1)</Checkpoint>

最后一步反应即为模型反应的应用，N-苯基酰胺变为腈鎓离子，由于分子内存在炔基，会发生分子内的模型反应，形成喹啉环后芳构化。最终产物$\mathbf{G}$的SMILES结构式为C1=CC2=CC3=C(/C(=C\C4=C(C=CC=C4C#N)C#N)/OC3)N=C2C=C1。

<Checkpoint>[$\mathbf{G}$的SMILES结构式为C1=CC2=CC3=C(/C(=C\C4=C(C=CC=C4C#N)C#N)/OC3)N=C2C=C1。](2)</Checkpoint>

根据该结构：该结构分子式为$\mathrm{C_{20}H_{11}N_3O}$，含四个环，其中五元环为半缩酮结构，酸性条件下容易开环形成醇和酮，不稳定；

<Checkpoint>[$\mathbf{G}$的五元环为半缩酮结构，酸性条件下容易开环形成醇和酮](1)</Checkpoint>

在酸性条件下水解，五元环开环形成酮羰基，两个氰基水解形成两个羧羰基，共三个羰基。故选项C D E均错误，选项F正确。