<MultiModal>![CC(C)(C)C1=CC=NC=C1>$\mathrm{Tf_2O}$(1.2 equiv),$\mathrm{EtOAc}$, reaction at $-78\degree \mathrm{C}$ for 30 mins; then CC(CC(OCC)=O)=O(1.2 equiv),$\mathrm{NEt_3}$(3.0 equiv), at $-78\degree \mathrm{C}$; then rt, overnight>[$\mathbf{B}$].[$\mathbf{C}$], the ratio of products $\mathbf{B}$ and $\mathbf{C}$ is approximately 1:1](/media/uploads/9884fc45-e490-47d9-bafe-4f9798a7a3c6.png)</MultiModal>

Given that the reaction yields products $\mathbf{B}$ and $\mathbf{C}$ in an approximate ratio of 1:1 and their molecular formulas are different, and the molecular formula of product $\mathbf{C}$ is $\mathrm{C_{16}H_{22}F_3NO_5S}$, without considering enantiomers, provide the structural formula of product $\mathbf{C}$.
<MultiModal>![CC(C)(C)C1=CC=NC=C1>$\mathrm{Tf_2O}$(1.2 equiv),$\mathrm{EtOAc}$,在$-78\degree \mathrm{C}$反应30 mins；then CC(CC(OCC)=O)=O(1.2 equiv),$\mathrm{NEt_3}$(3.0 equiv),在$-78\degree \mathrm{C}$；then rt,overnight>[$\mathbf{B}$].[$\mathbf{C}$],产物$\mathbf{B}$和$\mathbf{C}$的比例约为1:1](/media/uploads/9884fc45-e490-47d9-bafe-4f9798a7a3c6.png)</MultiModal>

已知该反应得到产物$\mathbf{B}$和$\mathbf{C}$的比例大致为1:1且二者分子式不同，产物$\mathbf{C}$的分子式为$\mathrm{C_{16}H_{22}F_3NO_5S}$，不考虑对映异构的条件下，试给出产物$\mathbf{C}$的结构式
All other options are incorrect
<MultiModal>![CC(C)(C)C1=C[C@H](NS(=O)(C(F)(F)F)=O)[C@@](C(C)=O)(C(OCC)=O)C=C1](/media/uploads/bd114b6f-2f7c-40ee-80f0-28ee1d1fa966.png)</MultiModal>
<MultiModal>![CC(C)(C)C1=C[C@H](NS(=O)(C(F)(F)F)=O)[C@](C(C)=O)(C(OCC)=O)C=C1](/media/uploads/ef84a14e-9715-42d7-b71a-52a36389ab09.png)</MultiModal>
<MultiModal>![CC(C)(C)C1=CC(C(C(OCC)=O)C(C)=O)N(S(=O)(C(F)(F)F)=O)C=C1](/media/uploads/f9bf0d33-b22c-49b5-805c-9904e799fc40.png)</MultiModal>
<MultiModal>![CC(C)(C)C1[C@@]2([H])[C@@](C(C(OCC)=O)=C(C)O2)([H])N(S(=O)(C(F)(F)F)=O)C=C1](/media/uploads/e597add4-e096-4c0f-a317-ee58fe1281a7.png)</MultiModal>
<MultiModal>![CC(C)(C)C1[C@]2([H])[C@@](C(C(OCC)=O)=C(C)O2)([H])N(S(=O)(C(F)(F)F)=O)C=C1](/media/uploads/5867ff64-d758-4a39-b447-06a9b4912d19.png)</MultiModal>
<MultiModal>![CC(C)(C)C1=C[C@H](C(C)=O)[C@](C(OCC)=O)(NS(=O)(C(F)(F)F)=O)C=C1](/media/uploads/e450e821-84e7-42d1-bee2-6260849d31af.png)</MultiModal>
<MultiModal>![CC(C)(C)C1=C[C@H](C(C)=O)[C@@](C(OCC)=O)(NS(=O)(C(F)(F)F)=O)C=C1](/media/uploads/b8252507-278a-4a3d-8d14-5366e1f556a1.png)</MultiModal>
<MultiModal>![CC(C)(C)C1=C[C@H](C(OCC)=O)[C@](C(C)=O)(NS(=O)(C(F)(F)F)=O)C=C1](/media/uploads/2aa87de9-698b-4fa2-bfc0-bd100829e665.png)</MultiModal>
<MultiModal>![CC(C)(C)C1=C[C@H](C(OCC)=O)[C@@](C(C)=O)(NS(=O)(C(F)(F)F)=O)C=C1](/media/uploads/315bf702-0b12-4ae0-91bc-625e02d8ff40.png)</MultiModal>
其他选项均不正确
<MultiModal>![CC(C)(C)C1=C[C@H](NS(=O)(C(F)(F)F)=O)[C@@](C(C)=O)(C(OCC)=O)C=C1](/media/uploads/bd114b6f-2f7c-40ee-80f0-28ee1d1fa966.png)</MultiModal>
<MultiModal>![CC(C)(C)C1=C[C@H](NS(=O)(C(F)(F)F)=O)[C@](C(C)=O)(C(OCC)=O)C=C1](/media/uploads/ef84a14e-9715-42d7-b71a-52a36389ab09.png)</MultiModal>
<MultiModal>![CC(C)(C)C1=CC(C(C(OCC)=O)C(C)=O)N(S(=O)(C(F)(F)F)=O)C=C1](/media/uploads/f9bf0d33-b22c-49b5-805c-9904e799fc40.png)</MultiModal>
<MultiModal>![CC(C)(C)C1[C@@]2([H])[C@@](C(C(OCC)=O)=C(C)O2)([H])N(S(=O)(C(F)(F)F)=O)C=C1](/media/uploads/e597add4-e096-4c0f-a317-ee58fe1281a7.png)</MultiModal>
<MultiModal>![CC(C)(C)C1[C@]2([H])[C@@](C(C(OCC)=O)=C(C)O2)([H])N(S(=O)(C(F)(F)F)=O)C=C1](/media/uploads/5867ff64-d758-4a39-b447-06a9b4912d19.png)</MultiModal>
<MultiModal>![CC(C)(C)C1=C[C@H](C(C)=O)[C@](C(OCC)=O)(NS(=O)(C(F)(F)F)=O)C=C1](/media/uploads/e450e821-84e7-42d1-bee2-6260849d31af.png)</MultiModal>
<MultiModal>![CC(C)(C)C1=C[C@H](C(C)=O)[C@@](C(OCC)=O)(NS(=O)(C(F)(F)F)=O)C=C1](/media/uploads/b8252507-278a-4a3d-8d14-5366e1f556a1.png)</MultiModal>
<MultiModal>![CC(C)(C)C1=C[C@H](C(OCC)=O)[C@](C(C)=O)(NS(=O)(C(F)(F)F)=O)C=C1](/media/uploads/2aa87de9-698b-4fa2-bfc0-bd100829e665.png)</MultiModal>
<MultiModal>![CC(C)(C)C1=C[C@H](C(OCC)=O)[C@@](C(C)=O)(NS(=O)(C(F)(F)F)=O)C=C1](/media/uploads/315bf702-0b12-4ae0-91bc-625e02d8ff40.png)</MultiModal>
First, observe the first reaction. It is obvious that the substrate reacts with $\mathrm{Tf_2O}$ to obtain intermediate $\mathbf{1}$

<MultiModal>![Intermediate $\mathbf{1}$: CC(C)(C)C1=CC=[N+](S(=O)(C(F)(F)F)=O)C=C1](/media/uploads/b7797d09-fb57-4bfb-abd2-a8630a24fe1a.png)</MultiModal>

<Checkpoint>[Intermediate $\mathbf{1}$: CC(C)(C)C1=CC=[N+](S(=O)(C(F)(F)F)=O)C=C1](1)</Checkpoint>

Then, the second step is an addition reaction to obtain intermediate $\mathbf{2}$

<MultiModal>![Intermediate $\mathbf{2}$: CC(C)(C)C1=CC(C(C(OCC)=O)C(C)=O)N(S(=O)(C(F)(F)F)=O)C=C1](/media/uploads/4b2e0268-b115-4a70-a053-a37fad4a8350.png)</MultiModal>

<Checkpoint>[Intermediate $\mathbf{2}$: CC(C)(C)C1=CC(C(C(OCC)=O)C(C)=O)N(S(=O)(C(F)(F)F)=O)C=C1](1)</Checkpoint>

Under the catalysis of excess base, further elimination reaction occurs to obtain intermediate $\mathbf{3}$

<MultiModal>![Intermediate $\mathbf{3}$: CC(C)(C)C(/C=C\NS(=O)(C(F)(F)F)=O)=C/C=C(C(OCC)=O)\C(C)=O](/media/uploads/42e35fa9-2e5a-42b2-9419-16de38956941.png)</MultiModal>

<Checkpoint>[Intermediate $\mathbf{3}$: CC(C)(C)C(/C=C\NS(=O)(C(F)(F)F)=O)=C/C=C(C(OCC)=O)\C(C)=O (The cis/trans configuration of the double bond is not important)](1)</Checkpoint>

Then, a six-electron electrocyclic reaction occurs to generate a pair of diastereomers. At this time, the ratio of products is approximately 1:1, but the molecular formulas of the two are still the same.

<Checkpoint>[Then, a six-electron electrocyclic reaction occurs to generate a pair of diastereomers. At this time, the ratio of the obtained products is approximately 1:1, but the molecular formulas of the two are still the same.](1)</Checkpoint>

<MultiModal>![Intermediate $\mathbf{4}$: CC(C)(C)C1=C[C@H](NS(=O)(C(F)(F)F)=O)[C@@](C(C)=O)(C(OCC)=O)C=C1](/media/uploads/4fe463ba-f35d-4f3d-ae4d-49858502fff5.png)</MultiModal>

<Checkpoint>[Intermediate $\mathbf{4}$: CC(C)(C)C1=C[C@H](NS(=O)(C(F)(F)F)=O)[C@@](C(C)=O)(C(OCC)=O)C=C1](1)</Checkpoint>

<MultiModal>![Intermediate $\mathbf{5}$: CC(C)(C)C1=C[C@H](NS(=O)(C(F)(F)F)=O)[C@](C(C)=O)(C(OCC)=O)C=C1](/media/uploads/4b87adb1-c733-438f-9a37-2bc69f9a20a7.png)</MultiModal>

<Checkpoint>[Intermediate $\mathbf{5}$: CC(C)(C)C1=C[C@H](NS(=O)(C(F)(F)F)=O)[C@](C(C)=O)(C(OCC)=O)C=C1](1)</Checkpoint>

At this time, because the electrophilicity of the ketone is stronger than that of the ester, only intermediate $\mathbf{4}$ can further undergo a nucleophilic-elimination reaction to obtain product $\mathbf{B}$

<Checkpoint>[At this time, because the electrophilicity of the ketone is stronger than that of the ester, only intermediate $\mathbf{4}$ can further undergo a nucleophilic-elimination reaction to obtain product $\mathbf{B}$](1)</Checkpoint>

<MultiModal>![Product $\mathbf{B}$: CC(C)(C)C1=CC=C(C(OCC)=O)C=C1](/media/uploads/4947eaf9-c457-4e7a-b634-82eb9bd389cf.png)</MultiModal>

<Checkpoint>[Product $\mathbf{B}$: CC(C)(C)C1=CC=C(C(OCC)=O)C=C1](1)</Checkpoint>

Product $\mathbf{C}$ is intermediate $\mathbf{5}$
首先观察第一步反应，显然是底物与$\mathrm{Tf_2O}$ 发生反应得到中间体$\mathbf{1}$

<MultiModal>![中间体$\mathbf{1}$：CC(C)(C)C1=CC=[N+](S(=O)(C(F)(F)F)=O)C=C1](/media/uploads/b7797d09-fb57-4bfb-abd2-a8630a24fe1a.png)</MultiModal>

<Checkpoint>[中间体$\mathbf{1}$：CC(C)(C)C1=CC=[N+](S(=O)(C(F)(F)F)=O)C=C1](1)</Checkpoint>

接着第二步发生加成反应得到中间体$\mathbf{2}$

<MultiModal>![中间体$\mathbf{2}$：CC(C)(C)C1=CC(C(C(OCC)=O)C(C)=O)N(S(=O)(C(F)(F)F)=O)C=C1](/media/uploads/4b2e0268-b115-4a70-a053-a37fad4a8350.png)</MultiModal>

<Checkpoint>[中间体$\mathbf{2}$：CC(C)(C)C1=CC(C(C(OCC)=O)C(C)=O)N(S(=O)(C(F)(F)F)=O)C=C1](1)</Checkpoint>

在过量碱催化作用下，进一步发生消除反应得到中间体$\mathbf{3}$

<MultiModal>![中间体$\mathbf{3}$：CC(C)(C)C(/C=C\NS(=O)(C(F)(F)F)=O)=C/C=C(C(OCC)=O)\C(C)=O](/media/uploads/42e35fa9-2e5a-42b2-9419-16de38956941.png)</MultiModal>

<Checkpoint>[中间体$\mathbf{3}$：CC(C)(C)C(/C=C\NS(=O)(C(F)(F)F)=O)=C/C=C(C(OCC)=O)\C(C)=O（其中双键顺反不要紧）](1)</Checkpoint>

接着发生六电子电环化生成一对非对映异构体，此时产物比例恰好大致约为1:1，但是两者分子式依然相同

<Checkpoint>[接着发生六电子电环化生成一对非对映异构体，此时得到产物比例恰好大致约为1:1，但是两者分子式依然相同](1)</Checkpoint>

<MultiModal>![中间体$\mathbf{4}$：CC(C)(C)C1=C[C@H](NS(=O)(C(F)(F)F)=O)[C@@](C(C)=O)(C(OCC)=O)C=C1](/media/uploads/4fe463ba-f35d-4f3d-ae4d-49858502fff5.png)</MultiModal>

<Checkpoint>[中间体$\mathbf{4}$：CC(C)(C)C1=C[C@H](NS(=O)(C(F)(F)F)=O)[C@@](C(C)=O)(C(OCC)=O)C=C1](1)</Checkpoint>

<MultiModal>![中间体$\mathbf{5}$：CC(C)(C)C1=C[C@H](NS(=O)(C(F)(F)F)=O)[C@](C(C)=O)(C(OCC)=O)C=C1](/media/uploads/4b87adb1-c733-438f-9a37-2bc69f9a20a7.png)</MultiModal>

<Checkpoint>[中间体$\mathbf{5}$：CC(C)(C)C1=C[C@H](NS(=O)(C(F)(F)F)=O)[C@](C(C)=O)(C(OCC)=O)C=C1](1)</Checkpoint>

此时由于酮相对于酯的亲电性更强，因此只有中间体$\mathbf{4}$能进一步发生亲核-消除反应得到产物$\mathbf{B}$

<Checkpoint>[此时由于酮相对于酯的亲电性更强，因此只有中间体$\mathbf{4}$能进一步发生亲核-消除反应得到产物$\mathbf{B}$](1)</Checkpoint>

<MultiModal>![产物$\mathbf{B}$：CC(C)(C)C1=CC=C(C(OCC)=O)C=C1](/media/uploads/4947eaf9-c457-4e7a-b634-82eb9bd389cf.png)</MultiModal>

<Checkpoint>[产物$\mathbf{B}$：CC(C)(C)C1=CC=C(C(OCC)=O)C=C1](1)</Checkpoint>

产物$\mathbf{C}$即为中间体$\mathbf{5}$