<MultiModal>![OC(C1=CC=CO1)C2=CC=C(C(F)(F)F)C=C2.NC3=CC(Cl)=CC=C3O>$\mathrm{TsOH}$>[$\mathbf{A}$], the reaction proceeds at $80\degree \mathrm{C}$, $\mathbf{A}$ is the reaction product](/media/uploads/82d70779-3c61-499a-bfad-8598df748a74.png)</MultiModal>

Given that the product $\mathbf{A}$ contains a total of four rings, including two aromatic rings, and the molecular formula of $\mathbf{A}$ is $\mathrm{C_{18}H_{13}ClF_3NO_2}$, provide the structural formula of $\mathbf{A}$ without considering enantiomers.
<MultiModal>![OC(C1=CC=CO1)C2=CC=C(C(F)(F)F)C=C2.NC3=CC(Cl)=CC=C3O>$\mathrm{TsOH}$>[$\mathbf{A}$],反应在$80\degree \mathrm{C}$下进行，$\mathbf{A}为反应产物](/media/uploads/82d70779-3c61-499a-bfad-8598df748a74.png)</MultiModal>

已知产物$\mathbf{A}$中共含有四个环，其中包括两个芳香环，$\mathbf{A}$的分子式为$\mathrm{C_{18}H_{13}ClF_3NO_2}$，不考虑对映异构的条件下试给出$\mathbf{A}$的结构式。
All other options are incorrect
<MultiModal>![O=C1[C@H](C2=CC=C(C(F)(F)F)C=C2)[C@@H](NC3=C(O4)C=CC(Cl)=C3)[C@@H]4C1](/media/uploads/e336c844-4cb4-4a21-92c7-80095ef7044a.png)</MultiModal>
<MultiModal>![O=C1[C@@H](C2=CC=C(C(F)(F)F)C=C2)[C@H](NC3=C(O4)C=CC(Cl)=C3)[C@@H]4C1](/media/uploads/7c4094bf-439a-4684-b252-004f9a3ae425.png)</MultiModal>
<MultiModal>![O=C1[C@@H](C2=CC=C(C(F)(F)F)C=C2)[C@@H](NC3=C(O4)C=CC(Cl)=C3)[C@H]4C1](/media/uploads/732252a5-33dc-44c4-bacb-67bdac876d03.png)</MultiModal>
<MultiModal>![ClC1=CC(N2)=C(C=C1)OC2(C3=CC=CO3)C4=CC=C(C(F)(F)F)C=C4](/media/uploads/c96ab875-617c-494a-9627-04e40d0a18fb.png)</MultiModal>
<MultiModal>![ClC1=CC(NC(C2OC=CC2O3)C4=CC=C(C(F)(F)F)C=C4)=C3C=C1](/media/uploads/828f2d9c-de07-472b-bfe2-58fbd62acb78.png)</MultiModal>
<MultiModal>![OC(C=CC(Cl)=C1)=C1N2C(C3=CC=C(C(F)(F)F)C=C3)C(C=CC2)=O](/media/uploads/104ce32f-66ca-4e60-bead-4c6d31cf33c4.png)</MultiModal>
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<MultiModal>![O=C1[C@H](C2=CC=C(C(F)(F)F)C=C2)[C@@H](NC3=C(O4)C=CC(Cl)=C3)[C@@H]4C1](/media/uploads/e336c844-4cb4-4a21-92c7-80095ef7044a.png)</MultiModal>
<MultiModal>![O=C1[C@@H](C2=CC=C(C(F)(F)F)C=C2)[C@H](NC3=C(O4)C=CC(Cl)=C3)[C@@H]4C1](/media/uploads/7c4094bf-439a-4684-b252-004f9a3ae425.png)</MultiModal>
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<MultiModal>![ClC1=CC(N2)=C(C=C1)OC2(C3=CC=CO3)C4=CC=C(C(F)(F)F)C=C4](/media/uploads/c96ab875-617c-494a-9627-04e40d0a18fb.png)</MultiModal>
<MultiModal>![ClC1=CC(NC(C2OC=CC2O3)C4=CC=C(C(F)(F)F)C=C4)=C3C=C1](/media/uploads/828f2d9c-de07-472b-bfe2-58fbd62acb78.png)</MultiModal>
<MultiModal>![OC(C=CC(Cl)=C1)=C1N2C(C3=CC=C(C(F)(F)F)C=C3)C(C=CC2)=O](/media/uploads/104ce32f-66ca-4e60-bead-4c6d31cf33c4.png)</MultiModal>
First, according to the molecular formula $\mathrm{C_{18}H_{13}ClF_3NO_2}$ of $\mathbf{A}$, it can be deduced that the condensation reaction apparently only eliminates one molecule of $\mathrm{H_2O}$.

<Checkpoint>[the condensation reaction apparently only eliminates one molecule of $\mathrm{H_2O}$](1)</Checkpoint>

First, under the catalysis of acid, the substrate loses one molecule of water to obtain intermediate $\mathbf{1}$

<MultiModal>![FC(F)(F)C1=CC=C(/C=C2O[CH+]C=C/2)C=C1](/media/uploads/d3a29026-cf6d-4d99-b7c5-7de74936da3d.png)</MultiModal>

<Checkpoint>[intermediate $\mathbf{1}$：FC(C(C=C1)=CC=C1/C=C2[O+]=CC=C/2)(F)F](1)</Checkpoint>

Then, because the amino group has a stronger nucleophilicity than the hydroxyl group, the nitrogen atom attacks and opens the ring to obtain intermediate $\mathbf{2}$

<Checkpoint>[the amino group has a stronger nucleophilicity than the hydroxyl group](1)</Checkpoint>

<MultiModal>![OC(/C=C\C=N\C1=C(O)C=CC(Cl)=C1)=C/C2=CC=C(C(F)(F)F)C=C2](/media/uploads/548b9b84-dc13-4376-8142-41ef6148bbf7.png)</MultiModal>

<Checkpoint>[intermediate $\mathbf{2}$：OC(/C=C\C=N\C1=C(O)C=CC(Cl)=C1)=C/C2=CC=C(C(F)(F)F)C=C2](1)</Checkpoint>

It is known from the question that the reaction needs to form two more rings, so further cyclization reactions occur under acid catalysis to form a five-membered ring to obtain intermediate $\mathbf{3}$

<MultiModal>![O=C1[C@@H](C2=CC=C(C(F)(F)F)C=C2)[C@@H](NC3=C(O)C=CC(Cl)=C3)C=C1](/media/uploads/8112d116-f768-4557-a2ad-1e5b5a1a00fe.png)</MultiModal>

<Checkpoint>[intermediate $\mathbf{3}$：O=C1[C@@H](C2=CC=C(C(F)(F)F)C=C2)[C@@H](NC3=C(O)C=CC(Cl)=C3)C=C1](1)</Checkpoint>

The cis regioselectivity of the two groups on the five-membered ring comes from the $\pi$-$\pi$ interaction between the aromatic rings.

<Checkpoint>[The cis regioselectivity of the two groups on the five-membered ring comes from the $\pi$-$\pi$ interaction between the aromatic rings](1)</Checkpoint>

Finally, due to conformational reasons, the hydroxyl group tends to attack the conjugated double bond from the same side of the plane and undergo an addition reaction, forming a cis six-membered ring and a five-membered ring structure, to obtain product $\mathbf{A}$

<Checkpoint>[the hydroxyl group tends to attack the conjugated double bond from the same side of the plane and undergo an addition reaction, forming a cis six-membered ring and a five-membered ring structure](1)</Checkpoint>

<MultiModal>![O=C1[C@@H](C2=CC=C(C(F)(F)F)C=C2)[C@@H](NC3=C(O4)C=CC(Cl)=C3)[C@@H]4C1](/media/uploads/68c99055-417a-4eca-8af6-2bcf032ca2b4.png)</MultiModal>

<Checkpoint>[product $\mathbf{A}$：O=C1[C@@H](C2=CC=C(C(F)(F)F)C=C2)[C@@H](NC3=C(O4)C=CC(Cl)=C3)[C@@H]4C1](1)</Checkpoint>
首先根据$\mathbf{A}$的分子式$\mathrm{C_{18}H_{13}ClF_3NO_2}$可以推出，表观上该缩合反应仅脱去了一分子$\mathrm{H_2O}$。

<Checkpoint>[表观上该缩合反应仅脱去了一分子$\mathrm{H_2O}$](1)</Checkpoint>

首先在酸的催化下，底物脱去一分子水得到中间体$\mathbf{1}$

<MultiModal>![FC(F)(F)C1=CC=C(/C=C2O[CH+]C=C/2)C=C1](/media/uploads/d3a29026-cf6d-4d99-b7c5-7de74936da3d.png)</MultiModal>

<Checkpoint>[中间体$\mathbf{1}$：FC(C(C=C1)=CC=C1/C=C2[O+]=CC=C/2)(F)F](1)</Checkpoint>

接着由于氨基相对于羟基有着更强的亲核性，因此氮原子进攻开环得到中间体$\mathbf{2}$

<Checkpoint>[氨基相对于羟基有着更强的亲核性](1)</Checkpoint>

<MultiModal>![OC(/C=C\C=N\C1=C(O)C=CC(Cl)=C1)=C/C2=CC=C(C(F)(F)F)C=C2](/media/uploads/548b9b84-dc13-4376-8142-41ef6148bbf7.png)</MultiModal>

<Checkpoint>[中间体$\mathbf{2}$：OC(/C=C\C=N\C1=C(O)C=CC(Cl)=C1)=C/C2=CC=C(C(F)(F)F)C=C2](1)</Checkpoint>

由题干得知反应需要再形成两个环，因此在酸催化下进一步发生成环反应形成五元环得到中间体$\mathbf{3}$

<MultiModal>![O=C1[C@@H](C2=CC=C(C(F)(F)F)C=C2)[C@@H](NC3=C(O)C=CC(Cl)=C3)C=C1](/media/uploads/8112d116-f768-4557-a2ad-1e5b5a1a00fe.png)</MultiModal>

<Checkpoint>[中间体$\mathbf{3}$：O=C1[C@@H](C2=CC=C(C(F)(F)F)C=C2)[C@@H](NC3=C(O)C=CC(Cl)=C3)C=C1](1)</Checkpoint>

其中五元环上两个基团处于顺式的区域选择性来源于芳环之间的$\pi$-$\pi$相互作用

<Checkpoint>[其中五元环上两个基团处于顺式的区域选择性来源于芳环之间的$\pi$-$\pi$相互作用](1)</Checkpoint>

最后限于构象原因，羟基倾向于从平面同侧进攻共轭双键发生加成反应，形成顺式六元环并五元环结构，得到产物$\mathbf{A}$

<Checkpoint>[羟基倾向于从平面同侧进攻共轭双键发生加成反应，形成顺式六元环并五元环结构](1)</Checkpoint>

<MultiModal>![O=C1[C@@H](C2=CC=C(C(F)(F)F)C=C2)[C@@H](NC3=C(O4)C=CC(Cl)=C3)[C@@H]4C1](/media/uploads/68c99055-417a-4eca-8af6-2bcf032ca2b4.png)</MultiModal>

<Checkpoint>[产物$\mathbf{A}$：O=C1[C@@H](C2=CC=C(C(F)(F)F)C=C2)[C@@H](NC3=C(O4)C=CC(Cl)=C3)[C@@H]4C1](1)</Checkpoint>