<MultiModal>![This image describes an organic reaction. The substrate is O=C(CC1=CC=CC=C1COCC=C)N2CCCC2, which first reacts under $Tf_2O, 2,4,6-collidine, DCM$ conditions to produce an intermediate, and the intermediate then reacts with $NaHCO_3, H_2O$ to obtain product $\bf A$, with the structure O=C1OCC(C=CC=C2)=C2C1CC=C.](/media/uploads/3474f7fd-0675-4825-b781-0f4072884e7f.png)</MultiModal>

The process of the above reaction generating product $\bf A$ involves three key positively charged intermediates $\bf 1,2,3$, and undergoes a pericyclic reaction step.

The model reaction in the figure above has the following application:

<MultiModal>![This image describes an organic reaction, the substrate is O=C(CCCOCC=C)N1CCCC1, which generates product $\bf B$ under three-step conditions, the conditions being $1. Tf_2O, 2,4,6-collidine, 393K; 2. NaH, PhCH_2Br, DMF, 348K; 3. Na_2CO_3, H_2O$.](/media/uploads/17aa2dfb-6949-4097-b176-18637eae03ed.png)</MultiModal>

Regarding the structures of intermediates $\bf 1,2,3$ and product $\bf B$, the correct statement is:
<MultiModal>![该图片描述了一步有机反应。底物为O=C(CC1=CC=CC=C1COCC=C)N2CCCC2，首先在$Tf_2O, 2,4,6-collidine, DCM$条件下反应产生中间体，中间体再与$NaHCO_3, H_2O$反应得到产物$\bf A$，结构为O=C1OCC(C=CC=C2)=C2C1CC=C。](/media/uploads/3474f7fd-0675-4825-b781-0f4072884e7f.png)</MultiModal>

上图的反应生成产物$\bf A$的过程经历了三个关键的带正电中间体$\bf 1,2,3$，且经历一步周环反应。

上图的模型反应有如下应用：

<MultiModal>![该图描述了一步有机反应，底物为O=C(CCCOCC=C)N1CCCC1，在三步条件下生成产物$\bf B$，条件为$1. Tf_2O, 2,4,6-collidine, 393K; 2. NaH, PhCH_2Br, DMF, 348K; 3. Na_2CO_3, H_2O$。](/media/uploads/17aa2dfb-6949-4097-b176-18637eae03ed.png)</MultiModal>

关于中间体$\bf 1,2,3$和产物$\bf B$的结构，说法正确的是：
All other options are incorrect
$\bf 1,2,3$ each possesses two six-membered rings
The generation of $\bf A$ occurred via an electrocyclic reaction.
There exist two 6-membered rings in $\bf B$
$\bf B$ does not contain chiral carbon atoms.
$\bf B$ hydrolyzes under acidic conditions, and the hydrolysis product contain 18 hydrogen atoms.
其他选项均不正确
$\bf 1,2,3$均存在两个六元环
生成$\bf A$的过程发生了电环化反应
$\bf B$中存在两个六元环
$\bf B$不存在手性碳原子
$\bf B$在酸性条件下水解，水解产物含有18个氢原子
First, the added base abstracts the active hydrogen of the substrate to generate an enolate, and trifluoromethanesulfonic anhydride ($\rm Tf_2O$) captures the oxygen anion of the enolate to generate an $\rm OTf$ group. This group has extremely high leaving group ability, so the nitrogen atom of the amide promotes the departure of the $\rm OTf$ group to generate a positive ion allene structure $\bf 1$, with the structure C=CCOCC1=CC=CC=C1C=C=[N+]2CCCC2.

<Checkpoint>[$\rm Tf_2O$ captures the oxygen anion of the enolate to generate an $\rm OTf$ group](1)</Checkpoint>

<Checkpoint>[Therefore, the nitrogen atom of the amide promotes the departure of the $\rm OTf$ group to generate a positive ion allene](1)</Checkpoint>

<Checkpoint>[Intermediate $\bf 1$, with the structure C=CCOCC1=CC=CC=C1C=C=[N+]2CCCC2](1)</Checkpoint>

The $\rm sp$ carbon atom in the allene structure has strong electrophilicity. According to the oxa-six-membered ring in the product, the oxygen atom of the ether in the substrate nucleophilically attacks this carbon atom to generate a six-membered ring oxonium ion $\bf 2$, with the structure C=CC[O+](C(N1CCCC1)=C2)CC3=C2C=CC=C3.

<Checkpoint>[The oxygen atom of the ether nucleophilically attacks the allene to generate a six-membered ring oxonium ion](1)</Checkpoint>

<Checkpoint>[Intermediate $\bf 2$ with the structure C=CC[O+](C(N1CCCC1)=C2)CC3=C2C=CC=C3](1)</Checkpoint>

Observing the allyl position of product $\bf A$, it can be seen that the allyl group and the carbon atom of the six-membered ring form a new $\rm C-C$ bond. Observing the structure, it can be seen that a [3,3]-σ sigmatropic rearrangement occurs, causing the allyl group to migrate from the oxygen atom to the β-position of the oxygen atom. The structure after the migration is intermediate $\bf 3$, with the structure C=CCC1C(N2CCCC2)=[O+]CC3=C1C=CC=C3.

<Checkpoint>[A [3,3]-σ sigmatropic rearrangement occurs, causing the allyl group to migrate from the oxygen atom to the β-position of the oxygen atom](1)</Checkpoint>

<Checkpoint>[Intermediate $\bf 3$ with the structure C=CCC1C(N2CCCC2)=[O+]CC3=C1C=CC=C3](1)</Checkpoint>

Intermediate $\bf 3$ is hydrolyzed to enamine under basic conditions to obtain product $\bf A$.

According to the model reaction, the mechanism for the formation of $\bf B$ is similar. The substrate reacts with $\rm Tf_2O$ to generate a positive ion allene structure, which is nucleophilically attacked by the intramolecular ether to form a five-membered ring, followed by a [3,3]-σ sigmatropic rearrangement to migrate the allyl group. At this time, the obtained intermediate structure is C=CCC1CC[O+]=C1N2CCCC2.

<Checkpoint>[According to the model reaction, the structure of the intermediate for the formation of $\bf B$ is C=CCC1CC[O+]=C1N2CCCC2](1)</Checkpoint>

Before the enamine is hydrolyzed, sodium hydride and benzyl bromide are added. Sodium hydride abstracts the most acidic tertiary carbon hydrogen atom in the system, and then undergoes a substitution reaction with benzyl bromide; the final step is basic hydrolysis, and the structure of product $\bf B$ is O=C1OCCC1(CC2=CC=CC=C2)CC=C.

<Checkpoint>[Sodium hydride abstracts the most acidic tertiary carbon hydrogen atom in the system, and then undergoes a substitution reaction with benzyl bromide](1)</Checkpoint>

<Checkpoint>[The structure of product $\bf B$ is O=C1OCCC1(CC2=CC=CC=C2)CC=C](1)</Checkpoint>

According to the structure, options B-E are all incorrect.

$\bf B$ hydrolyzes the ester group under acidic conditions to generate an alcohol and an acid. This structure has a total of 18 hydrogen atoms, so option F is correct.

<MultiModal>![Intermediate $\bf 1$ with the structure C=CCOCC1=CC=CC=C1C=C=[N+]2CCCC2; intermediate $\bf 2$ with the structure C=CC[O+](C(N1CCCC1)=C2)CC3=C2C=CC=C3; intermediate $\bf 3$ with the structure C=CCC1C(N2CCCC2)=[O+]CC3=C1C=CC=C3; product $\bf B$ with the structure O=C1OCCC1(CC2=CC=CC=C2)CC=C; intermediate for the formation of $\bf B$ with the structure C=CCC1CC[O+]=C1N2CCCC2](/media/uploads/273da206-d7d8-4f20-8489-99191fd39a8c.png)</MultiModal>
第一步，加入的碱拔去底物的活泼氢生成烯醇负离子，三氟甲磺酸酐($\rm Tf_2O$)捕获该烯醇负离子的氧负生成$\rm OTf$基团。该基团具有极强的离去性，因此酰胺的氮原子促进$\rm OTf$基团离去生成正离子联烯结构$\bf 1$，结构为C=CCOCC1=CC=CC=C1C=C=[N+]2CCCC2。

<Checkpoint>[$\rm Tf_2O$捕获该烯醇负离子的氧负生成$\rm OTf$基团](1)</Checkpoint>

<Checkpoint>[因此酰胺的氮原子促进$\rm OTf$基团离去生成正离子联烯](1)</Checkpoint>

<Checkpoint>[中间体$\bf 1$，结构为C=CCOCC1=CC=CC=C1C=C=[N+]2CCCC2](1)</Checkpoint>

联烯结构中的$\rm sp$碳原子具有强亲电性，根据产物中的氧杂六元环可知底物中醚的氧原子亲核该碳原子生成六元环氧鎓离子$\bf 2$，结构为C=CC[O+](C(N1CCCC1)=C2)CC3=C2C=CC=C3.

<Checkpoint>[醚的氧原子亲核联烯生成六元环氧鎓离子](1)</Checkpoint>

<Checkpoint>[中间体$\bf 2$结构为C=CC[O+](C(N1CCCC1)=C2)CC3=C2C=CC=C3](1)</Checkpoint>

观察产物$\bf A$的烯丙基位置可知烯丙基与六元环的碳原子形成了新的$\rm C-C$键，观察结构可知发生了周环反应中的[3，3]-σ迁移反应使得烯丙基从氧原子上迁移到氧原子的β位，迁移后的结构即为中间体$\bf 3$，结构为C=CCC1C(N2CCCC2)=[O+]CC3=C1C=CC=C3。

<Checkpoint>[发生[3，3]-σ迁移反应，烯丙基从氧原子上迁移到氧原子的β位](1)</Checkpoint>

<Checkpoint>[中间体$\bf 3$结构为C=CCC1C(N2CCCC2)=[O+]CC3=C1C=CC=C3](1)</Checkpoint>

中间体$\bf 3$在碱性条件水解烯胺即得到产物$\bf A$。

根据模型反应，生成$\bf B$的机理类似，底物与$\rm Tf_2O$反应产生正离子联烯结构，被分子内的醚亲核生成五元环后发生[3，3]-σ迁移反应使烯丙基迁移，此时得到的中间体结构为C=CCC1CC[O+]=C1N2CCCC2。

<Checkpoint>[根据模型反应，生成$\bf B$中间体结构为C=CCC1CC[O+]=C1N2CCCC2](1)</Checkpoint>

该结构在水解烯胺前加入氢化钠和苄溴，氢化钠拔除体系酸性最强的三级碳氢原子，之后与苄溴发生取代反应；最后一步碱性水解，产物$\bf B$结构为O=C1OCCC1(CC2=CC=CC=C2)CC=C。

<Checkpoint>[氢化钠拔除体系酸性最强的三级碳氢原子，之后与苄溴发生取代反应](1)</Checkpoint>

<Checkpoint>[产物$\bf B$结构为O=C1OCCC1(CC2=CC=CC=C2)CC=C](1)</Checkpoint>

根据结构可知选项B-E均错误。

$\bf B$酸性条件水解酯基生成醇和酸，该结构共有18个氢原子，选项F正确。

<MultiModal>![中间体$\bf 1$结构为C=CCOCC1=CC=CC=C1C=C=[N+]2CCCC2；中间体$\bf 2$结构为C=CC[O+](C(N1CCCC1)=C2)CC3=C2C=CC=C3；中间体$\bf 3$结构为C=CCC1C(N2CCCC2)=[O+]CC3=C1C=CC=C3；产物$\bf B$结构为O=C1OCCC1(CC2=CC=CC=C2)CC=C；生成$\bf B$中间体结构为C=CCC1CC[O+]=C1N2CCCC2](/media/uploads/273da206-d7d8-4f20-8489-99191fd39a8c.png)</MultiModal>