Ozonolysis is often used for the determination of organic structures. Compound $\mathbf{M}\mathrm{（C_{15}H_{26}O）}$ is an alcohol that is difficult to hydrogenate catalytically. When $\mathbf{M}$ is treated with sulfur, an aromatic compound $\mathbf{R}\mathrm{（C_{15}H_{18}）}$ is ultimately obtained. Ozonolysis of $\mathbf{M}$ followed by additional treatments yields a naphthalene derivative, as shown in the following reaction:

<MultiModal>![The reaction is represented by SMILES: [M]>[O-][O+]=O.CSC>[S]>>[T]>Pd/C>CC(C)C1=CC=C(C)C2=CC=C(C)C(=C21)O, where [S] and [T] are compound designations rather than elemental symbols. The chemical formula of [S] is $\mathrm{C_{15}H_{26}O_3}$, and the conversion from [S] to [T] involves the elimination of two molecules of water](/media/uploads/a2e6391c-b186-4dcc-b382-4964cbe9287c.png)</MultiModal>

It is known that the hydroxyl group in $\mathbf{M}$ is located on a tertiary carbon outside the ring. Based on this, determine which of the following options are correct:

1. $\mathbf{M}$ has 3 chiral carbons.  
2. $\mathbf{M}$ has 4 chiral carbons.  
3. $\mathbf{M}$ contains two six-membered rings, and the double bond is in the cis configuration.  
4. $\mathbf{M}$ contains two six-membered rings, and the double bond is in the trans configuration.  
5. $\mathbf{M}$ contains one five-membered ring and one seven-membered ring, and the double bond is in the cis configuration.  
6. $\mathbf{M}$ contains one five-membered ring and one seven-membered ring, and the double bond is in the trans configuration.  
7. $\mathbf{M}$ contains one six-membered ring and two double bonds.  
8. In $\mathbf{R}$, a maximum of 12 carbons lie in the same plane.  
9. In $\mathbf{R}$, a maximum of 13 carbons lie in the same plane.  
10. The conversion from $\mathbf{T}$ to the product involves the elimination of 1 molecule of $\mathrm{H_2}$.  
11. The conversion from $\mathbf{T}$ to the product involves the elimination of 2 molecules of $\mathrm{H_2}$.  
12. The conversion from $\mathbf{T}$ to the product involves the elimination of 3 molecules of $\mathrm{H_2}$.
臭氧化反应常被用于有机物结构的判定。化合物$\mathbf{M}\mathrm{（C_{15}H_{26}O）}$是一种醇，其很难被催化加氢。用硫处理$\mathbf{M}$，最终可以得到芳香化合物$\mathbf{R}\mathrm{（C_{15}H_{18}）}$。而对$\mathbf{M}$进行臭氧化和一些其他处理后可得到萘的衍生物，反应如下：

<MultiModal>![反应用SMILES结构式表示为：[M]>[O-][O+]=O.CSC>[S]>>[T]>Pd/C>CC(C)C1=CC=C(C)C2=CC=C(C)C(=C21)O，其中[S],[T]均为化合物代号而非元素符号，[S]的化学式为$\mathrm{C_{15}H_{26}O_3}$，[S]到[T]脱去了两分子水](/media/uploads/a2e6391c-b186-4dcc-b382-4964cbe9287c.png)</MultiModal>

已知$\mathbf{M}$中的羟基位于环外的三级碳上，据此判断下列选项那些是正确的：

1. $\mathbf{M}$中有3个手性碳。
2. $\mathbf{M}$中有4个手性碳。
3. $\mathbf{M}$中有两个六元环，且双键为顺式。
4. $\mathbf{M}$中有两个六元环，且双键为反式。
5. $\mathbf{M}$中有1个五元环和1个七元环，且双键为顺式。
6. $\mathbf{M}$中有1个五元环和1个七元环，且双键为反式。
7. $\mathbf{M}$中有一个六元环和两个双键。
8. $\mathbf{R}$中最多有12个C位于同一平面上。
9. $\mathbf{R}$中最多有13个C位于同一平面上。
10. $\mathbf{T}$到产物脱去1分子$\mathrm{H_2}$。
11. $\mathbf{T}$到产物脱去2分子$\mathrm{H_2}$。
12. $\mathbf{T}$到产物脱去3分子$\mathrm{H_2}$。
1, 3, 8, 10
1, 4, 9, 11
1, 7, 10
1, 6, 11
1, 5, 9
1, 6, 9, 11
1, 4, 8, 11
1, 7, 12
2, 6, 9, 11
2, 6, 11
2, 5, 12
2, 3, 8, 10
2, 7, 12
3, 9, 10
6, 8, 11
7, 11
5, 12
None of the above options is correct
1, 3, 8, 10
1, 4, 9, 11
1, 7, 10
1, 6, 11
1, 5, 9
1, 6, 9, 11
1, 4, 8, 11
1, 7, 12
2, 6, 9, 11
2, 6, 11
2, 5, 12
2, 3, 8, 10
2, 7, 12
3, 9, 10
6, 8, 11
7, 11
5, 12
以上选项均不正确
$\mathbf{M}$ is converted to $\mathbf{S}$ via ozonolysis with unchanged carbon count, indicating the cleavage of an internal double bond in the ring.

<Checkpoint>[The cleavage from $\mathbf{M}$ to $\mathbf{S}$ involves an internal double bond in the ring](1)</Checkpoint>

Ozonolysis does not alter hydroxyl groups, so $\mathbf{S}$ still retains a hydroxyl group. It can be inferred that the two dehydration steps from $\mathbf{S}$ to $\mathbf{T}$ consist of one aldol condensation + elimination and one elimination reaction.

<Checkpoint>[The two dehydration steps from $\mathbf{S}$ to $\mathbf{T}$ consist of one aldol condensation + elimination and one elimination reaction](1)</Checkpoint>

From the chemical formula of $\mathbf{S}$, it is known that $\mathbf{T}$ has the formula $\mathrm{C_{15}H_{22}O}$, while the product formula is $\mathrm{C_{15}H_{18}O}$, indicating the loss of 2 molecules of $\mathrm{H_2}$. Thus, option 11 is correct, while options 10 and 12 are incorrect.

<Checkpoint>[The conversion from $\mathbf{T}$ to the product involves the loss of 2 molecules of $\mathrm{H_2}$](1)</Checkpoint>

It is evident that the ring skeleton of $\mathbf{T}$ remains unchanged in the product. According to the question, the hydroxyl group in $\mathbf{M}$ is located on a tertiary carbon outside the ring. Therefore, the hydroxyl group in $\mathbf{M}$ and $\mathbf{S}$ can only reside on the carbon corresponding to the sole tertiary carbon in the product.

<Checkpoint>[The hydroxyl group in $\mathbf{M}$ and $\mathbf{S}$ can only reside on the carbon corresponding to the sole tertiary carbon in the product](1)</Checkpoint>

The product is a fused ring system formed via aldol condensation. The shared C-C bond between the two rings can be cleaved, and the carbonyl oxygen is connected to the end farther from the -OH group. Based on the chemical formula and degree of unsaturation of $\mathbf{S}$, its structure is determined as:

<MultiModal>![CC1CCC(CC(=O)C(C)CCC1=O)C(C)(C)O](/media/uploads/54c8fb76-4ba9-47f4-83fe-d7d607d201d2.png)</MultiModal>

<Checkpoint>[$\mathbf{S}$ is CC1CCC(CC(=O)C(C)CCC1=O)C(C)(C)O](1)</Checkpoint>

Ozonolysis cleaves double bonds into two carbonyl groups, so the structure of $\mathbf{M}$ can be deduced from $\mathbf{S}$:

<MultiModal>![CC1CCC(CC2=C1CCC2C)C(C)(C)O](/media/uploads/c14b6461-280b-4dbc-8056-43579b5c6fdb.png)</MultiModal>

<Checkpoint>[$\mathbf{M}$ is CC1CCC(CC2=C1CCC2C)C(C)(C)O](1)</Checkpoint>

The difficulty in catalytic hydrogenation is due to the double bond in $\mathbf{M}$ being tetrasubstituted, with bulky groups hindering its interaction with the catalyst surface.

Evidently, after sulfur treatment, $\mathbf{M}$ transforms into the aromatic $\mathbf{R}$:

<MultiModal>![CC(C)C1=CC2=C(C)C=CC2=C(C)C=C1](/media/uploads/09fe7fb7-dd0b-49b1-8291-f8de4a8dcce8.png)</MultiModal>

<Checkpoint>[$\mathbf{R}$ is CC(C)C1=CC2=C(C)C=CC2=C(C)C=C1](1)</Checkpoint>

The correctness of the options can now be determined:

From $\mathbf{M}$ being CC1CCC(CC2=C1CCC2C)C(C)(C)O, it is clear that $\mathbf{M}$ has 3 chiral carbons, making option 1 correct and 2 incorrect. $\mathbf{M}$ contains one five-membered ring and one seven-membered ring, with the double bond in a trans configuration, so option 6 is correct, while options 3, 4, 5, and 7 are incorrect.

<Checkpoint>[$\mathbf{M}$ has 3 chiral carbons](1)</Checkpoint>

<Checkpoint>[$\mathbf{M}$ contains one five-membered ring and one seven-membered ring, with the double bond in a trans configuration](1)</Checkpoint>

From $\mathbf{R}$ being CC(C)C1=CC2=C(C)C=CC2=C(C)C=C1 and the tetrahedral geometry of $sp^3$ carbons, it follows that a maximum of 14 carbons in $\mathbf{R}$ can lie in the same plane. Thus, options 8 and 9 are incorrect.

<Checkpoint>[A maximum of 14 carbons in $\mathbf{R}$ can lie in the same plane](1)</Checkpoint>

In summary, options 1, 6, and 11 are correct, and the answer is D.
$\mathbf{M}$经由臭氧化反应到$\mathbf{S}$，碳数不变，说明断裂的是环内双键。

<Checkpoint>[$\mathbf{M}$到$\mathbf{S}$断裂的是环内双键](1)</Checkpoint>

臭氧化反应不改变羟基，因此$\mathbf{S}$中仍存在羟基，可以判断$\mathbf{S}$到$\mathbf{T}$的两分子脱水分别为一次羟醛缩合+消除和一次消除反应。

<Checkpoint>[$\mathbf{S}$到$\mathbf{T}$的两分子脱水分别为一次羟醛缩合+消除和一次消除反应](1)</Checkpoint>

通过$\mathbf{S}$的化学式可以知道$\mathbf{T}$的化学式为$\mathrm{C_{15}H_{22}O}$，而产物化学式为$\mathrm{C_{15}H_{18}O}$，可知脱去2分子$\mathrm{H_2}$，11正确，10、12错误。

<Checkpoint>[$\mathbf{T}$到产物脱去2分子$\mathrm{H_2}$](1)</Checkpoint>

容易知道$\mathbf{T}$到产物环骨架不发生改变，由题目可知，$\mathbf{M}$中的羟基位于环外的三级碳上，因此$\mathbf{M}$与$\mathbf{S}$中的羟基只能存在于产物中唯一的三级碳所对应的碳上。

<Checkpoint>[$\mathbf{M}$与$\mathbf{S}$中的羟基只能存在于产物中唯一的三级碳所对应的碳上](1)</Checkpoint>

产物为并环，并且连接是通过羟醛缩合得到，可以断开两环共用的碳碳键，并将羰基O连上远离-OH的一端，通过$\mathbf{S}$的化学式和不饱和度得到$\mathbf{S}$：

<MultiModal>![CC1CCC(CC(=O)C(C)CCC1=O)C(C)(C)O](/media/uploads/54c8fb76-4ba9-47f4-83fe-d7d607d201d2.png)</MultiModal>

<Checkpoint>[$\mathbf{S}$为CC1CCC(CC(=O)C(C)CCC1=O)C(C)(C)O](1)</Checkpoint>

臭氧化使得双键裂解为两个羰基，因此可从$\mathbf{S}$的结构倒推$\mathbf{M}$：

<MultiModal>![CC1CCC(CC2=C1CCC2C)C(C)(C)O](/media/uploads/c14b6461-280b-4dbc-8056-43579b5c6fdb.png)</MultiModal>

<Checkpoint>[$\mathbf{M}$为CC1CCC(CC2=C1CCC2C)C(C)(C)O](1)</Checkpoint>

至于难被催化加氢，则是因为$\mathbf{M}$中双键为四取代，且基团位阻较大，难以与催化剂表面结合加氢。

显然，经由硫处理后$\mathbf{M}$变为芳香性的$\mathbf{R}$：

<MultiModal>![CC(C)C1=CC2=C(C)C=CC2=C(C)C=C1](/media/uploads/09fe7fb7-dd0b-49b1-8291-f8de4a8dcce8.png)</MultiModal>

<Checkpoint>[$\mathbf{R}$为CC(C)C1=CC2=C(C)C=CC2=C(C)C=C1](1)</Checkpoint>

至此可以判断选项正误：

由$\mathbf{M}$为CC1CCC(CC2=C1CCC2C)C(C)(C)O可知$\mathbf{M}$中有3个手性碳，1正确，2错误；$\mathbf{M}$中有1个五元环和1个七元环，且双键为反式，故6正确，3、4、5、7错误。

<Checkpoint>[$\mathbf{M}$中有3个手性碳](1)</Checkpoint>

<Checkpoint>[$\mathbf{M}$中有1个五元环和1个七元环，且双键为反式](1)</Checkpoint>

由$\mathbf{R}$为CC(C)C1=CC2=C(C)C=CC2=C(C)C=C1以及$sp^3$碳原子的四面体构型可知，$\mathbf{R}$中最多有14个C位于同一平面上。因此8、9均不正确。

<Checkpoint>[$\mathbf{R}$中最多有14个C位于同一平面上](1)</Checkpoint>

综上，选项1, 6, 11正确，答案选D.