<MultiModal>![This image describes a one-step organic reaction where the substrate is O=C1C(CO)CCCC1, reacting with $\mathrm{NH_2OH}$ to produce two unknown species **M** and **N**](/media/uploads/d65e1e73-37bc-4f67-a49c-be1887fe620d.png)</MultiModal>

The **M** and **N** obtained from the reaction above are isomers and do not contain hydroxyl groups. To prepare pure **M**, the following reaction was designed:

<MultiModal>![This image describes a multi-step tandem organic reaction where the substrate is O=C1CCCCC1C(OCC)=O, first reacting with $\mathrm{(CH_2OH)_2}$ in the presence of hydrogen ions to produce **A**, **A** reacting with $\mathrm{LiAlH_4}$ to produce **B**, **B** reacting with $\mathrm{(CrO_3}$ in the presence of py to produce **C**, and **C** reacting with $\mathrm{NH_2OH}$ to produce **M**.](/media/uploads/5bce6f81-4e57-4a65-be12-a69c04bc89ef.png)</MultiModal>

Which of the following statements is correct?
<MultiModal>![该图描述了一步有机反应，底物为O=C1C(CO)CCCC1，与$\mathrm{NH_2OH}$反应生成两种未知物种**M**和**N**](/media/uploads/d65e1e73-37bc-4f67-a49c-be1887fe620d.png)</MultiModal>

上图的反应得到的**M**和**N**为同分异构体且不含羟基。为制备纯净的**M**，设计了以下反应：

<MultiModal>![该图片描述了一个有机多步串联反应，底物为O=C1CCCCC1C(OCC)=O，先与$\mathrm{(CH_2OH)_2}$在氢离子存在条件下反应生成**A**，**A**与$\mathrm{LiAlH_4}$反应生成**B**，**B**与$\mathrm{(CrO_3}$在py存在下反应生成**C**，**C**与$\mathrm{NH_2OH}$反应生成**M**。](/media/uploads/5bce6f81-4e57-4a65-be12-a69c04bc89ef.png)</MultiModal>

下列说法正确的是：
The chemical formula of **M** or **N** contains 11 hydrogens.
**M** contains a spirocyclic system
The product of complete hydrogenation of **N** has 13 hydrogen atoms.
**N** has hydrogen atoms with a chemical shift of 6-10.
None of the above options is correct
**M**或**N**的化学式中含有11个氢。
**M**中含有螺环体系
**N**彻底氢化的产物有13个氢原子
**N**拥有化学位移在8-10的氢原子
以上选项均不正确
Let's first analyze the reaction for the synthesis of **M**:

The substrate is added to ethylene glycol and acid, which are classic conditions for protecting ketone carbonyls, generating the five-membered ring ketal **A**, with the structural formula O=C(C1C2(OCCO2)CCCC1)OCC.

<Checkpoint>[**A** structural formula is O=C(C1C2(OCCO2)CCCC1)OCC](1)</Checkpoint>

**A** is reduced by lithium aluminum hydride, and the ester group is directly reduced to a primary alcohol, so the structural formula of **B** is OCC1C2(OCCO2)CCCC1.

<Checkpoint>[**B** structural formula is OCC1C2(OCCO2)CCCC1](1)</Checkpoint>

**B** is oxidized by chromium trioxide, and the primary alcohol is oxidized to a carbonyl group, generating an aldehyde, with the structural formula of **C** being O=CC1C2(OCCO2)CCCC1.

<Checkpoint>[**C** structural formula is O=CC1C2(OCCO2)CCCC1](1)</Checkpoint>

**C** reacts with hydroxylamine. First, the aldehyde group is attacked by the nitrogen atom of hydroxylamine to form an imine structure. Then, it can be found that the hydroxyl group of the imine can exchange with the ketal in the system, generating a new fused five- and six-membered ring system. At the same time, the newly formed five-membered ring has the driving force for aromatization and can lose ethylene glycol to form a five-membered aromatic heterocycle. Thus, the structural formula of **M** is C1(C=NO2)=C2CCCC1.

<Checkpoint>[The hydroxyl group of the imine can exchange with the ketal in the system, generating a new fused five- and six-membered ring system](1)</Checkpoint>

<Checkpoint>[The newly formed five-membered ring has the driving force for aromatization](1)</Checkpoint>

<Checkpoint>[**M** structural formula is C1(C=NO2)=C2CCCC1](1)</Checkpoint>

Analyzing the reaction for the synthesis of **M** and **N**, hydroxylamine reacts with the ketone carbonyl group in the substrate to form an imine. The hydroxyl group of the imine can continue to undergo nucleophilic substitution with the primary alcohol to form a five-membered ring, and aromatization occurs with the loss of water. At this point, the product is C12=CON=C1CCCC2, which is exactly an isomer of **M**. Therefore, the structural formula of **N** is C12=CON=C1CCCC2.

<Checkpoint>[**N** structural formula is C12=CON=C1CCCC2](1)</Checkpoint>

Analyzing the options: The chemical formula of **M** contains 9 hydrogens and is a fused five- and six-membered ring system, so options A and B are incorrect.

Complete hydrogenation of **N** will break the nitrogen-oxygen bond to form an amino group and a hydroxyl group, with the product being NC1C(CO)CCCC1, containing 15 hydrogen atoms, so option C is incorrect.

<Checkpoint>[Complete hydrogenation of **N** will break the nitrogen-oxygen bond, with the product being NC1C(CO)CCCC1](1)</Checkpoint>

**N** contains an aromatic five-membered heterocycle, so the hydrogen on the ring is located in the low-field region of the chemical shift, so option D is correct.

<Checkpoint>[**N** contains an aromatic five-membered heterocycle, so the hydrogen on the ring is located in the low-field region of the chemical shift](1)</Checkpoint>

In summary, option D is correct.

The following is the structure image of the answer to this question:

<MultiModal>![This image shows the structural formula of the unknown species in this question](/media/uploads/9d934517-c197-49f7-a039-56119702f1f8.png)</MultiModal>
先分析合成**M**的反应：

底物加入乙二醇和酸，是经典的保护酮羰基的条件，生成五元环缩酮**A**，结构式为O=C(C1C2(OCCO2)CCCC1)OCC。

<Checkpoint>[**A**结构式为O=C(C1C2(OCCO2)CCCC1)OCC](1)</Checkpoint>

**A**被氢化铝锂还原，酯基被直接还原为一级醇，故**B**结构式为OCC1C2(OCCO2)CCCC1。

<Checkpoint>[**B**结构式为OCC1C2(OCCO2)CCCC1](1)</Checkpoint>

**B**被三氧化铬氧化，一级醇被氧化为羰基，生成醛，**C**结构式为O=CC1C2(OCCO2)CCCC1。

<Checkpoint>[**C**结构式为O=CC1C2(OCCO2)CCCC1](1)</Checkpoint>

**C**与羟胺反应，首先醛基被羟胺的氮原子亲核变成亚胺结构；之后可以发现，亚胺的羟基可以与体系中的缩酮发生交换，生成新的五并六环系；同时新生成的五元环具有芳构化的动力，可以脱去乙二醇形成五元芳香杂环。从而**M**的结构式为C1(C=NO2)=C2CCCC1。

<Checkpoint>[亚胺的羟基可以与体系中的缩酮发生交换，生成新的五并六环系](1)</Checkpoint>

<Checkpoint>[新生成的五元环具有芳构化的动力](1)</Checkpoint>

<Checkpoint>[**M**的结构式为C1(C=NO2)=C2CCCC1](1)</Checkpoint>

分析合成**M**与**N**的反应，羟胺与底物中的酮羰基反应形成亚胺，亚胺的羟基可以继续与一级醇发生亲核取代形成五元环，并且芳构化脱去水。此时产物为C12=CON=C1CCCC2，刚好与**M**互为同分异构体；因此**N**结构式为C12=CON=C1CCCC2。

<Checkpoint>[**N**结构式为C12=CON=C1CCCC2](1)</Checkpoint>

分析选项：**M**化学式含有9个氢，且为五并六环系，选项A,B均错误。

**N**彻底氢化会打断氮氧键形成氨基和羟基，产物为NC1C(CO)CCCC1，含有15个氢原子，选项C错误。

<Checkpoint>[**N**彻底氢化会打断氮氧键，产物为NC1C(CO)CCCC1](1)</Checkpoint>

**N**含有芳香五元杂环，从而环上的氢位于化学位移低场，选项D正确。

<Checkpoint>[**N**含有芳香五元杂环，从而环上的氢位于化学位移低场](1)</Checkpoint>

综上，选项D正确。

下图为本题答案的结构图片：

<MultiModal>![此图为本题的未知物种结构式](/media/uploads/9d934517-c197-49f7-a039-56119702f1f8.png)</MultiModal>