<MultiModal>![O[C@@H](C)C(O)=O.C1(C2=CCCCO2)=CCCCO1>CCOCC.Cl>[M], M is the reaction product](/media/uploads/c31477f8-867d-40d4-ac8d-adbf116df066.png)</MultiModal>

When the two substrates shown in the figure are placed in an ether solution of $\mathrm{HCl}$, the reaction mainly yields the tricyclic product $\mathbf{M}$. Without considering enantiomers, give the structural formula of $\mathbf{M}$.
<MultiModal>![O[C@@H](C)C(O)=O.C1(C2=CCCCO2)=CCCCO1>CCOCC.Cl>[M],M是反应产物](/media/uploads/c31477f8-867d-40d4-ac8d-adbf116df066.png)</MultiModal>

将如图所示的两种底物置于$\mathrm{HCl}$的乙醚溶液中时，反应主要得到三环产物$\mathbf{M}$。不考虑对映异构的条件下，试给出$\mathbf{M}$的结构简式。
All other options are incorrect
<MultiModal>![O=C(O[C@]12OCCCC2)[C@H](C)O[C@]31CCCCO3](/media/uploads/bb254b16-58b8-49d8-9382-fe7d8870e047.png)</MultiModal>
<MultiModal>![O=C(O[C@]12OCCCC2)[C@H](C)O[C@]31OCCCC3](/media/uploads/985e570a-8b71-448f-919c-e445668b255c.png)</MultiModal>
<MultiModal>![O=C(O[C@]12CCCCO2)[C@H](C)O[C@]31CCCCO3](/media/uploads/9cb2ea4e-2d4d-4649-819c-05311e130a59.png)</MultiModal>
<MultiModal>![O=C(O[C@]12CCCCO2)[C@H](C)O[C@]31OCCCC3](/media/uploads/a64abdfd-c773-4bd3-b56a-bf0ba77bd01a.png)</MultiModal>
其他选项均不正确
<MultiModal>![O=C(O[C@]12OCCCC2)[C@H](C)O[C@]31CCCCO3](/media/uploads/bb254b16-58b8-49d8-9382-fe7d8870e047.png)</MultiModal>


<MultiModal>![O=C(O[C@]12OCCCC2)[C@H](C)O[C@]31OCCCC3](/media/uploads/985e570a-8b71-448f-919c-e445668b255c.png)</MultiModal>


<MultiModal>![O=C(O[C@]12CCCCO2)[C@H](C)O[C@]31CCCCO3](/media/uploads/9cb2ea4e-2d4d-4649-819c-05311e130a59.png)</MultiModal>
<MultiModal>![O=C(O[C@]12CCCCO2)[C@H](C)O[C@]31OCCCC3](/media/uploads/a64abdfd-c773-4bd3-b56a-bf0ba77bd01a.png)</MultiModal>
Acid-catalyzed, this condensation reaction is reversible, so it is speculated that $\mathbf{M}$ exists in its most stable form.

<Checkpoint>[Acid-catalyzed, this condensation reaction is reversible, so it is speculated that $\mathbf{M}$ exists in its most stable form.](1)</Checkpoint>

First, according to the hint of the tricyclic product, it is easy for us to think that this ketal structure can be formed through a simple two-step addition reaction.

<Checkpoint>[First, according to the hint of the tricyclic product, it is easy for us to think that this ketal structure can be formed through a simple two-step addition reaction.](1)</Checkpoint>

<MultiModal>![C[C@@H]1OC2(OCCCC2)C3(OC1=O)OCCCC3](/media/uploads/b4e44743-bbc0-4fd5-bcab-a51cb63d6f36.png)</MultiModal>

<Checkpoint>[C[C@@H]1OC2(OCCCC2)C3(OC1=O)OCCCC3](1)</Checkpoint>

Then analyze the stereoconformation. There are three chiral centers in this structure, so there should be 4 stereoisomers without considering enantiomers.

<MultiModal>![O=C(O[C@]12OCCCC2)[C@H](C)O[C@]31[X2]CCC[X1]3](/media/uploads/0e7d1123-6a84-4b88-9e8a-aa37d4b01bf8.png)</MultiModal>

When O is located at the $\mathrm{X_1}$ or $\mathrm{X_2}$ position, its endocyclic anomeric effect is equivalent.

<Checkpoint>[When O is located at the $\mathrm{X_1}$ or $\mathrm{X_2}$ position, its endocyclic anomeric effect is equivalent.](1)</Checkpoint>

But the difference is that when O is located at the $\mathrm{X_1}$ position, the lone pair electrons of the upright bond O can fill into the antibonding orbital of the $\mathrm{C-X_1}$ bond, reducing the system energy through the exocyclic anomeric effect.

<Checkpoint>[But the difference is that when O is located at the $\mathrm{X_1}$ position, the lone pair electrons of the upright bond O can fill into the antibonding orbital of the $\mathrm{C-X_1}$ bond, reducing the system energy through the exocyclic anomeric effect.](1)</Checkpoint>

<MultiModal>![O=C(O[C@]12[X4]CCC[X3]2)[C@H](C)O[C@]31CCCCO3](/media/uploads/de1621cb-a0a2-491a-9918-1ad1995585bd.png)</MultiModal>

Similarly, O at the $\mathrm{X_4}$ position has a set of exocyclic anomeric effects more than at the $\mathrm{X_3}$ position.

<Checkpoint>[Similarly, O at the $\mathrm{X_4}$ position has a set of exocyclic anomeric effects more than at the $\mathrm{X_3}$ position](1)</Checkpoint>

<MultiModal>![O=C(O[C@]12OCCCC2)[C@H](C)O[C@]31CCCCO3](/media/uploads/bb254b16-58b8-49d8-9382-fe7d8870e047.png)</MultiModal>

<Checkpoint>[Final product：O=C(O[C@]12OCCCC2)[C@H](C)O[C@]31CCCCO3](1)</Checkpoint>
酸催化下，该缩合反应是可逆的，因此推测$\mathbf{M}$以其最稳定的形式存在

<Checkpoint>[酸催化下，该缩合反应是可逆的，因此推测$\mathbf{M}$以其最稳定的形式存在](1)</Checkpoint>

首先根据三环产物的提示，我们很容易想到通过简单的两步加成反应可以形成该缩酮结构

<Checkpoint>[首先根据三环产物的提示，我们很容易想到通过简单的两步加成反应可以形成该缩酮结构](1)</Checkpoint>

<MultiModal>![C[C@@H]1OC2(OCCCC2)C3(OC1=O)OCCCC3](/media/uploads/b4e44743-bbc0-4fd5-bcab-a51cb63d6f36.png)</MultiModal>

<Checkpoint>[C[C@@H]1OC2(OCCCC2)C3(OC1=O)OCCCC3](1)</Checkpoint>

接着分析立体构象。该结构中有三个手性中心，因此在不考虑对映异构的情况下应当有4种立体异构体。

<MultiModal>![O=C(O[C@]12OCCCC2)[C@H](C)O[C@]31[X2]CCC[X1]3](/media/uploads/0e7d1123-6a84-4b88-9e8a-aa37d4b01bf8.png)</MultiModal>

O位于$\mathrm{X_1}$或$\mathrm{X_2}$位置时，其环内异头碳效应相当。

<Checkpoint>[O位于$\mathrm{X_1}$或$\mathrm{X_2}$位置时，其环内异头碳效应相当。](1)</Checkpoint>

但不同的是当O位于$\mathrm{X_1}$位置时，直立键O的孤对电子能填入$\mathrm{C-X_1}$键的反键轨道，通过环外异头碳效应降低了体系能量。

<Checkpoint>[但不同的是当O位于$\mathrm{X_1}$位置时，直立键O的孤对电子能填入$\mathrm{C-X_1}$键的反键轨道，通过环外异头碳效应降低了体系能量。](1)</Checkpoint>

<MultiModal>![O=C(O[C@]12[X4]CCC[X3]2)[C@H](C)O[C@]31CCCCO3](/media/uploads/de1621cb-a0a2-491a-9918-1ad1995585bd.png)</MultiModal>

同理，O位于$\mathrm{X_4}$位置比$\mathrm{X_3}$位置多一组环外异头碳效应

<Checkpoint>[同理，O位于$\mathrm{X_4}$位置比$\mathrm{X_3}$位置多一组环外异头碳效应](1)</Checkpoint>

<MultiModal>![O=C(O[C@]12OCCCC2)[C@H](C)O[C@]31CCCCO3](/media/uploads/bb254b16-58b8-49d8-9382-fe7d8870e047.png)</MultiModal>

<Checkpoint>[最终产物：O=C(O[C@]12OCCCC2)[C@H](C)O[C@]31CCCCO3](1)</Checkpoint>