The following transformation process is now presented:

<MultiModal>![The compound {SMILES code: C[C@]12CCC(=O)C=C2CCC1=O} is treated with $NaBH_4$ to yield compound A.

Compound A reacts with the compound {SMILES code: C=C(C)C} under $H_3PO_4$ catalysis to give compound B.

Compound B reacts with excess $CH_3I$ under $tBuOK$ catalysis to produce compound C.

Compound C reacts with $H_2$ under $Pt/C$ catalysis to yield compound D.

Compound D is treated with $HCl$ to give compound E.

Compound E reacts with PCC reagent in the presence of $NaOAc$ to obtain compound F.

Compound F reacts with the compound {SMILES code: OCCO} under $p-TsOH$ catalysis to give compound G.

Compound G reacts with $NH_2OH·HCl$ in $EtOH$ in the presence of $NaOAc$ to yield compound H.](/media/uploads/bc222f26-8194-4e24-afd8-145cb585ac01.png)</MultiModal>

Please analyze the structures of each compound and select the correct option.
现在有如下转化过程：

<MultiModal>![化合物{SMILES编码为C[C@]12CCC(=O)C=C2CCC1=O}通过$NaBH_4$处理后得到化合物A。

化合物A在$H_3PO_4$催化下和化合物{SMILES编码为C=C(C)C}反应，得到化合物B。

化合物B在$tBuOK$催化下和过量$CH_3I$反应得到化合物C。

化合物C在$Pt/C$催化下与$H_2$反应得到化合物D。

化合物D在$HCl$处理后得到化合物E。

化合物E在$NaOAc$存在条件下与PCC试剂反应得到化合物F。

化合物F在$p-TsOH$催化下和化合物{SMILES编码为OCCO}反应得到化合物G。

化合物G在$NaOAc$存在条件下和$NH_2OH·HCl$在$EtOH$中反应得到化合物H。](/media/uploads/bc222f26-8194-4e24-afd8-145cb585ac01.png)</MultiModal>

请分析各个化合物结构，并选出正确选项。
Compound **A** is  

<MultiModal>![C[C@@]12CCC(C=C1CCC2=O)O](/media/uploads/ba42be7c-c8b9-4ca6-bcb4-4c608759d1bc.png)</MultiModal>
The compound **B** is  

<MultiModal>![C[C@@]12CCC(C=C1CCC2OCC(C)C)=O](/media/uploads/ca2f76a5-e880-445f-a295-afab261cd0ea.png)</MultiModal>
Compound **C** is  

<MultiModal>![C[C@@]12CC=C(C=C1CCC2OC(C)(C)C)OC](/media/uploads/3bfa0cd4-9e3e-420a-a8fc-1575158f3232.png)</MultiModal>
The compound **D** is  

<MultiModal>![C[C@@]12CC=C(CC1CCC2OCC(C)C)OC](/media/uploads/b6676e91-414a-4279-9988-a82ec6d6ee0d.png)</MultiModal>
The compound **E** is  

<MultiModal>![C[C@@]12CCC(CC1CCC2O)=O](/media/uploads/5d48bf4d-8ac0-4aeb-9906-4c3a9f556ff0.png)</MultiModal>
The compound **F** is  

<MultiModal>![C[C@@]12CCC(C(C)(C)C1CCC2=O)=O](/media/uploads/72211071-a8e1-4d59-b409-d7d7647d9e14.png)</MultiModal>
The compound **G** is  

<MultiModal>![CC(C1CCC2=O)(C3(CC[C@]21C)OCCO3)C](/media/uploads/76175623-ae1e-46fa-ba54-de6c1da4e10e.png)</MultiModal>
Compound **H** is  

<MultiModal>![C[C@@]12CCC3(OCCO3)C(C)C1CC/C2=N\O](/media/uploads/8000c200-1634-48f1-8a06-a3d06c414234.png)</MultiModal>
化合物**A**为

<MultiModal>![C[C@@]12CCC(C=C1CCC2=O)O](/media/uploads/ba42be7c-c8b9-4ca6-bcb4-4c608759d1bc.png)</MultiModal>
化合物**B**为

<MultiModal>![C[C@@]12CCC(C=C1CCC2OCC(C)C)=O](/media/uploads/ca2f76a5-e880-445f-a295-afab261cd0ea.png)</MultiModal>
化合物**C**为

<MultiModal>![C[C@@]12CC=C(C=C1CCC2OC(C)(C)C)OC](/media/uploads/3bfa0cd4-9e3e-420a-a8fc-1575158f3232.png)</MultiModal>
化合物**D**为

<MultiModal>![C[C@@]12CC=C(CC1CCC2OCC(C)C)OC](/media/uploads/b6676e91-414a-4279-9988-a82ec6d6ee0d.png)</MultiModal>
化合物**E**为

<MultiModal>![C[C@@]12CCC(CC1CCC2O)=O](/media/uploads/5d48bf4d-8ac0-4aeb-9906-4c3a9f556ff0.png)</MultiModal>
化合物**F**为

<MultiModal>![C[C@@]12CCC(C(C)(C)C1CCC2=O)=O](/media/uploads/72211071-a8e1-4d59-b409-d7d7647d9e14.png)</MultiModal>
化合物**G**为

<MultiModal>![CC(C1CCC2=O)(C3(CC[C@]21C)OCCO3)C](/media/uploads/76175623-ae1e-46fa-ba54-de6c1da4e10e.png)</MultiModal>
化合物**H**为

<MultiModal>![C[C@@]12CCC3(OCCO3)C(C)C1CC/C2=N\O](/media/uploads/8000c200-1634-48f1-8a06-a3d06c414234.png)</MultiModal>
Sodium borohydride first reduces the non-conjugated carbonyl group, and since there are still acidic hydrogens at the α-position of the ketone in the system, only the non-conjugated carbonyl group should be reduced at this stage. Compound **A** is <MultiModal>![C[C@@]12CCC(C=C1CCC2O)=O](/media/uploads/c7b7e527-6acd-48cb-9741-41817fbd7b72.png)</MultiModal>  

<Checkpoint>[Compound **A** is C[C@@]12CCC(C=C1CCC2O)=O](1)</Checkpoint>  

Under acidic conditions, 2-methylpropene generates a tert-butyl cation, which then combines with the hydroxyl group. Thus, compound **B** is <MultiModal>![C[C@@]12CCC(C=C1CCC2OC(C)(C)C)=O](/media/uploads/88f2a099-078d-425d-bb88-9871ef1514d2.png)</MultiModal>  

<Checkpoint>[Compound **B** is C[C@@]12CCC(C=C1CCC2OC(C)(C)C)=O](1)</Checkpoint>  

Subsequently, under basic conditions, the most acidic hydrogen of the linear conjugated system is abstracted, followed by nucleophilic attack with excess methyl iodide twice. Thus, compound **C** is <MultiModal>![C[C@@]12CCC(C(C)(C)C1=CCC2OC(C)(C)C)=O](/media/uploads/f0858000-2e63-4069-851f-9c60d853ad0c.png)</MultiModal>  

<Checkpoint>[Compound **C** is C[C@@]12CCC(C(C)(C)C1=CCC2OC(C)(C)C)=O](1)</Checkpoint>  

Under palladium-on-carbon conditions, hydrogen gas can reduce the carbon-carbon double bond. Thus, compound **D** is <MultiModal>![C[C@@]12CCC(C(C)(C)C1CCC2OC(C)(C)C)=O](/media/uploads/31886fd9-0aa9-447d-a03a-c2f014bd0fe1.png)</MultiModal>  

<Checkpoint>[Compound **D** is C[C@@]12CCC(C(C)(C)C1CCC2OC(C)(C)C)=O](1)</Checkpoint>  

Under acid treatment, the tert-butyl protecting group is removed. Thus, compound **E** is <MultiModal>![C[C@@]12CCC(C(C)(C)C1CCC2O)=O](/media/uploads/109ad622-dc59-4231-82a1-95279cea0fbe.png)</MultiModal>  

<Checkpoint>[Compound **E** is C[C@@]12CCC(C(C)(C)C1CCC2O)=O](1)</Checkpoint>  

Pyridinium chlorochromate can oxidize the hydroxyl group to a carbonyl group. Thus, compound **F** is <MultiModal>![C[C@@]12CCC(C(C)(C)C1CCC2=O)=O](/media/uploads/4d1eace3-cef9-4325-b969-fd3d72a1d1f3.png)</MultiModal>  

<Checkpoint>[Compound **F** is C[C@@]12CCC(C(C)(C)C1CCC2=O)=O](1)</Checkpoint>  

Under acidic conditions, ethylene glycol can protect the carbonyl group, and further reactions involving the carbonyl group occur subsequently. At this stage, only one carbonyl group is protected, with the six-membered ring carbonyl being preferentially protected. Thus, compound **G** is <MultiModal>![C[C@@]12CCC3(OCCO3)C(C)(C)C1CCC2=O](/media/uploads/e35045a2-062e-4390-89c8-a6161f208d6b.png)</MultiModal>  

<Checkpoint>[Compound **G** is C[C@@]12CCC3(OCCO3)C(C)(C)C1CCC2=O](2)</Checkpoint>  

Finally, hydroxylamine can condense with the hydroxyl group, and considering steric hindrance, compound **H** is <MultiModal>![C[C@@]12CCC3(OCCO3)C(C)(C)C1CC/C2=N\O](/media/uploads/a6efa68e-4bae-4ed3-bb86-1f5d6fabece0.png)</MultiModal>  

<Checkpoint>[Compound **H** is C[C@@]12CCC3(OCCO3)C(C)(C)C1CC/C2=N\O](2)</Checkpoint>  

Thus, option F is correct.
硼氢化钠先还原不共轭羰基，且体系中后续仍有酮α位酸性氢，故而此时应只还原不共轭羰基，化合物**A**为<MultiModal>![C[C@@]12CCC(C=C1CCC2O)=O](/media/uploads/c7b7e527-6acd-48cb-9741-41817fbd7b72.png)</MultiModal>

<Checkpoint>[化合物**A**为C[C@@]12CCC(C=C1CCC2O)=O](1)</Checkpoint>

酸性条件下，2-甲基丙烯生成叔丁基正离子，进而与羟基结合，故化合物**B**为<MultiModal>![C[C@@]12CCC(C=C1CCC2OC(C)(C)C)=O](/media/uploads/88f2a099-078d-425d-bb88-9871ef1514d2.png)</MultiModal>

<Checkpoint>[化合物**B**为C[C@@]12CCC(C=C1CCC2OC(C)(C)C)=O](1)</Checkpoint>

接着碱性下拔出线性共轭的最强酸性的氢，对过量碘甲烷亲核进攻两次，进而化合物**C**为<MultiModal>![C[C@@]12CCC(C(C)(C)C1=CCC2OC(C)(C)C)=O](/media/uploads/f0858000-2e63-4069-851f-9c60d853ad0c.png)</MultiModal>

<Checkpoint>[化合物**C**为C[C@@]12CCC(C(C)(C)C1=CCC2OC(C)(C)C)=O](1)</Checkpoint>

而钯碳条件下氢气能还原碳碳双键，进而化合物**D**为<MultiModal>![C[C@@]12CCC(C(C)(C)C1CCC2OC(C)(C)C)=O](/media/uploads/31886fd9-0aa9-447d-a03a-c2f014bd0fe1.png)</MultiModal>

<Checkpoint>[化合物**D**为C[C@@]12CCC(C(C)(C)C1CCC2OC(C)(C)C)=O](1)</Checkpoint>

酸处理下，叔丁基保护基会去保护，进而化合物**E**为<MultiModal>![C[C@@]12CCC(C(C)(C)C1CCC2O)=O](/media/uploads/109ad622-dc59-4231-82a1-95279cea0fbe.png)</MultiModal>

<Checkpoint>[化合物**E**为C[C@@]12CCC(C(C)(C)C1CCC2O)=O](1)</Checkpoint>

吡啶氯铬酸盐可以将羟基氧化为羰基，故而化合物**F**为<MultiModal>![C[C@@]12CCC(C(C)(C)C1CCC2=O)=O](/media/uploads/4d1eace3-cef9-4325-b969-fd3d72a1d1f3.png)</MultiModal>

<Checkpoint>[化合物**F**为C[C@@]12CCC(C(C)(C)C1CCC2=O)=O](1)</Checkpoint>

而酸性下乙二醇能保护羰基，且后续还进一步发生了涉及羰基的反应。进而此时仅保护了一个羰基，而六元环上羰基会优先保护，进而化合物**G**为<MultiModal>![C[C@@]12CCC3(OCCO3)C(C)(C)C1CCC2=O](/media/uploads/e35045a2-062e-4390-89c8-a6161f208d6b.png)</MultiModal>

<Checkpoint>[化合物**G**为C[C@@]12CCC3(OCCO3)C(C)(C)C1CCC2=O](2)</Checkpoint>

最后羟胺能和羰基缩合，并考虑位阻，故而化合物**H**为<MultiModal>![C[C@@]12CCC3(OCCO3)C(C)(C)C1CC/C2=N\O](/media/uploads/a6efa68e-4bae-4ed3-bb86-1f5d6fabece0.png)</MultiModal>

<Checkpoint>[化合物**H**为C[C@@]12CCC3(OCCO3)C(C)(C)C1CC/C2=N\O](2)</Checkpoint>

从而选项F正确。