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Given that the molecular formula of the reaction product $\mathbf{A}$ is $\mathrm{C_{18}H_{20}O_2}$ and $\mathbf{A}$ contains 3 rings. Without considering enantiomers, provide the structural formula of the thermodynamic product $\mathbf{A}$.
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已知反应产物$\mathbf{A}$的分子式为$\mathrm{C_{18}H_{20}O_2}$且$\mathbf{A}$中含有3个环。不考虑对映异构的情况下，试给出热力学产物$\mathbf{A}$的结构式
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First, according to the molecular formula $\mathrm{C_{18}H_{20}O_2}$ of the reaction product $\mathbf{A}$, it can be deduced that the product is formed by the condensation of 2 molecules of phenol and 2 molecules of acetone, with the removal of 2 molecules of water during the reaction.

<Checkpoint>[First, according to the molecular formula $\mathrm{C_{18}H_{20}O_2}$ of the reaction product $\mathbf{A}$, it can be deduced that the product is formed by the condensation of 2 molecules of phenol and 2 molecules of acetone, with the removal of 2 molecules of water during the reaction.](1)</Checkpoint>

The question states that $\mathbf{A}$ is the thermodynamic product, so the para position of phenol preferentially undergoes condensation reaction with acetone to obtain the intermediate

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<Checkpoint>[The para position of phenol preferentially undergoes condensation reaction with acetone](1)</Checkpoint>

<Checkpoint>[Intermediate 1: OC1=CC=C(C(C)(C)O)C=C1](1)</Checkpoint>

Eliminate one molecule of water under acid catalysis to obtain the intermediate

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<Checkpoint>[Intermediate 2: OC1=CC=C(C(C)=C)C=C1](1)</Checkpoint>

Intermediate 2 can be protonated to obtain intermediate 3

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<Checkpoint>[Intermediate 3: OC1=CC=C([C+](C)C)C=C1](1)</Checkpoint>

Intermediate 3 can be captured by intermediate 2 to obtain the intermediate

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<Checkpoint>[Intermediate 4: OC1=CC=C(C(C)(C)C[C+](C2=CC=C(O)C=C2)C)C=C1](1)</Checkpoint>

Based on the hint in the question that another ring needs to be formed, it is easy to think that the carbocation can be captured by the benzene ring to form a five-membered ring.

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<Checkpoint>[Reaction product $\mathbf{A}$: OC1=CC=C2C(C(C)(C3=CC=C(O)C=C3)CC2(C)C)=C1](1)</Checkpoint>
首先根据反应产物$\mathbf{A}$的分子式$\mathrm{C_{18}H_{20}O_2}$，可以推得产物由2分子苯酚与2分子丙酮缩合而成，反应过程中脱去2分子水。

<Checkpoint>[首先根据反应产物$\mathbf{A}$的分子式$\mathrm{C_{18}H_{20}O_2}$，可以推得产物由2分子苯酚与2分子丙酮缩合而成，反应过程中脱去2分子水。](1)</Checkpoint>

题目中说$\mathbf{A}$为热力学产物，因此苯酚对位优先与丙酮发生缩合反应，得到中间体

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<Checkpoint>[苯酚对位优先与丙酮发生缩合反应](1)</Checkpoint>

<Checkpoint>[中间体1：OC1=CC=C(C(C)(C)O)C=C1](1)</Checkpoint>

在酸催化下消去一分子水得到中间体

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<Checkpoint>[中间体2：OC1=CC=C(C(C)=C)C=C1](1)</Checkpoint>

中间体2可被质子化得到中间体3

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<Checkpoint>[中间体3：OC1=CC=C([C+](C)C)C=C1](1)</Checkpoint>

中间体3可被中间体2捕获得到中间体

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<Checkpoint>[中间体4：OC1=CC=C(C(C)(C)C[C+](C2=CC=C(O)C=C2)C)C=C1](1)</Checkpoint>

由题目提示需要再成一个环，容易想到该碳正离子可被苯环捕获形成五元环。

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<Checkpoint>[反应产物$\mathbf{A}$：OC1=CC=C2C(C(C)(C3=CC=C(O)C=C3)CC2(C)C)=C1](1)</Checkpoint>