<MultiModal>![O=C(C1(C(OC)=O)C(C2=CC=CC=C2)C1)OC.O=C=C(C3=CC=CC=C3)C>ClCCl.[$\mathrm{EtAlCl_2}$(15 mol%)].[$\mathrm{InBr_3}$(30 mol%)]>[$\mathbf{A}$], A is the thermodynamic product](/media/uploads/c23214ba-d3cd-4f0b-a08b-aa251ba82983.png)</MultiModal>

Given that the thermodynamic product $\mathbf{A}$ contains a five-membered ring, and the molecular formula of $\mathbf{A}$ is $\mathrm{C_{22}H_{22}O_5}$, without considering enantiomers, provide the structural formula of $\mathbf{A}$
<MultiModal>![O=C(C1(C(OC)=O)C(C2=CC=CC=C2)C1)OC.O=C=C(C3=CC=CC=C3)C>ClCCl.[$\mathrm{EtAlCl_2}$(15 mol%)].[$\mathrm{InBr_3}$(30 mol%)]>[$\mathbf{A}$],A是热力学产物](/media/uploads/c23214ba-d3cd-4f0b-a08b-aa251ba82983.png)</MultiModal>

已知该反应的热力学产物$\mathbf{A}$中含有一个五元环，$\mathbf{A}$的分子式为$\mathrm{C_{22}H_{22}O_5}$，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=C1C(C(OC)=O)(C(OC)=O)C[C@H](C2=CC=CC=C2)[C@]1(C3=CC=CC=C3)C](/media/uploads/761726fb-5719-44ff-9798-87accd1ef2f9.png)</MultiModal>
<MultiModal>![O=C1C(C(OC)=O)(C(OC)=O)C[C@@H](C2=CC=CC=C2)[C@]1(C3=CC=CC=C3)C](/media/uploads/d0f683b7-dfca-434b-8cef-10346fd4da13.png)</MultiModal>
<MultiModal>![C/C(C1=CC=CC=C1)=C2C(C(OC)=O)(C(OC)=O)C[C@H](C3=CC=CC=C3)O\2](/media/uploads/d050ecc7-8471-4ba5-878f-6dbd4e60a3f5.png)</MultiModal>
<MultiModal>![C/C(C1=CC=CC=C1)=C2C(C(OC)=O)(C(OC)=O)C[C@H](C3=CC=CC=C3)O/2](/media/uploads/deb6db8a-4ae6-4084-aa72-4838e05e5e14.png)</MultiModal>
其他选项均不正确
<MultiModal>![O=C1C(C(OC)=O)(C(OC)=O)C[C@H](C2=CC=CC=C2)[C@]1(C3=CC=CC=C3)C](/media/uploads/761726fb-5719-44ff-9798-87accd1ef2f9.png)</MultiModal>
<MultiModal>![O=C1C(C(OC)=O)(C(OC)=O)C[C@@H](C2=CC=CC=C2)[C@]1(C3=CC=CC=C3)C](/media/uploads/d0f683b7-dfca-434b-8cef-10346fd4da13.png)</MultiModal>
<MultiModal>![C/C(C1=CC=CC=C1)=C2C(C(OC)=O)(C(OC)=O)C[C@H](C3=CC=CC=C3)O\2](/media/uploads/d050ecc7-8471-4ba5-878f-6dbd4e60a3f5.png)</MultiModal>
<MultiModal>![C/C(C1=CC=CC=C1)=C2C(C(OC)=O)(C(OC)=O)C[C@H](C3=CC=CC=C3)O/2](/media/uploads/deb6db8a-4ae6-4084-aa72-4838e05e5e14.png)</MultiModal>
Based on the molecular formula of product $\mathbf{A}$, it is not difficult to see that the reaction should be, in appearance, a combination reaction between two substrates, and it is speculated that a one-step cycloaddition reaction should have occurred.

<Checkpoint>[A one-step cycloaddition reaction occurred](1)</Checkpoint>

Under the induction of Lewis acid, the 1,3-dipole $\mathbf{1}$ is first generated.

<MultiModal>![1,3-dipole $\mathbf{1}$: O=C([C-](C(OC)=O)C[CH+]C1=CC=CC=C1)OC](/media/uploads/0f770e35-c40b-4531-b555-1a9dec55b186.png)</MultiModal>

<Checkpoint>[1,3-dipole $\mathbf{1}$: O=C([C-](C(OC)=O)C[CH+]C1=CC=CC=C1)OC](1)</Checkpoint>

Subsequently, the 1,3-dipole undergoes a one-step [3+2] cycloaddition reaction with the ketene. According to the thermodynamic product hint given in the question, it is obvious that the ketone is more stable than the alkenyl ether. At the same time, in the reaction process, the Lewis acid $\mathrm{In^{3+}}$ may combine with the oxygen on the ketene, thereby reducing the nucleophilicity of the oxygen. In summary, options D and E are excluded.

<Checkpoint>[The ketone is more stable than the alkenyl ether](1)</Checkpoint>

<Checkpoint>[The Lewis acid $\mathrm{In^{3+}}$ may combine with the oxygen on the ketene, thereby reducing the nucleophilicity of the oxygen](1)</Checkpoint>

The repulsion is smaller when the aromatic rings are on opposite sides, so the thermodynamic product $\mathbf{A}$ of the reaction is obtained.

<MultiModal>![Thermodynamic product $\mathbf{A}$: O=C1C(C(OC)=O)(C(OC)=O)C[C@H](C2=CC=CC=C2)[C@]1(C3=CC=CC=C3)C](/media/uploads/761726fb-5719-44ff-9798-87accd1ef2f9.png)</MultiModal>

<Checkpoint>[The repulsion is smaller when the aromatic rings are on opposite sides](1)</Checkpoint>
根据产物$\mathbf{A}$的分子式不难看出，该反应表观上应为两个底物之间的化合反应，推测应发生了一步环加成反应

<Checkpoint>[发生了一步环加成反应](1)</Checkpoint>

在路易斯酸的诱导下，首先产生1,3-偶极子 $\mathbf{1}$

<MultiModal>![1,3-偶极子 $\mathbf{1}$：O=C([C-](C(OC)=O)C[CH+]C1=CC=CC=C1)OC](/media/uploads/0f770e35-c40b-4531-b555-1a9dec55b186.png)</MultiModal>

<Checkpoint>[1,3-偶极子 $\mathbf{1}$：O=C([C-](C(OC)=O)C[CH+]C1=CC=CC=C1)OC](1)</Checkpoint>

随后该1,3-偶极子与烯酮发生一步[3+2]环加成反应。根据题干中给出的热力学产物提示，显然酮比烯基醚更加稳定。同时在反应过程中路易斯酸$\mathrm{In^{3+}}$可能会与烯酮上的氧进行结合从而降低氧的亲核性。综上，排除选项D、E

<Checkpoint>[酮比烯基醚更加稳定](1)</Checkpoint>

<Checkpoint>[路易斯酸$\mathrm{In^{3+}}$可能会与烯酮上的氧进行结合从而降低氧的亲核性](1)</Checkpoint>

芳环处于异侧时排斥力更小，因此得到反应热力学产物$\mathbf{A}$

<MultiModal>![热力学产物$\mathbf{A}$：O=C1C(C(OC)=O)(C(OC)=O)C[C@H](C2=CC=CC=C2)[C@]1(C3=CC=CC=C3)C](/media/uploads/761726fb-5719-44ff-9798-87accd1ef2f9.png)</MultiModal>

<Checkpoint>[芳环处于异侧时排斥力更小](1)</Checkpoint>