The following reaction pathway exists:

<MultiModal>![The diagram shows a multi-step synthetic pathway, $\mathbf{A}$ generates $\mathbf{B}$ under $\mathrm{(CH_3)_2SO_4,K_2CO_3,(CH_3)_2CO},reflux$ conditions, $\mathbf{B}$ successively undergoes: $(1)LDA(1eq),-78\degree C. (2)\mathrm{CH_3I}(1.05eq),-78\degree C~to~0\degree C. (3)LDA(1eq),-78\degree C. (4)\mathrm{CH_3I}(1.1eq),-78\degree C~to~10\degree C. $ to generate $\mathbf{C}$, $\mathbf{C}$ successively undergoes: $(1)\mathrm{CH_2CHMgBr},THF. (2)aq. \mathrm{H_2SO_4,(CH_3)_2CO},r.t.$ to generate $\mathbf{D}$, $\mathbf{D}$ generates $\mathbf{E}$ under $\mathrm{LiAlH_4,Et_2O}$ conditions, $\mathbf{E}$ generates $\mathbf{F}$ under $\mathrm{CH_3CCCOOH},DCC,DMAP,\mathrm{CH_2Cl_2}$ conditions, $\mathbf{F}$ generates the final product under $\mathrm{n-C_{10}H_{22}},reflux$ conditions, the final product is CC1=C2C(=O)O[C@H]3CC[C@](C)(C)C(=CC1)[C@]32C。](/media/uploads/0c155e4f-fa2b-4d87-af4e-0878bc393abb.png)</MultiModal>

Given that the chemical formula of $\mathbf{A}$ is $\mathrm{C_7H_{10}O_2}$, if the substance exists in tautomeric forms, the most stable one is taken. Then, which of the following statements is correct:
有如下的反应路径：

<MultiModal>![图中展示了一个多步合成路径，$\mathbf{A}$在$\mathrm{(CH_3)_2SO_4,K_2CO_3,(CH_3)_2CO},reflux$条件下生成$\mathbf{B}$，$\mathbf{B}$依次经过：$(1)LDA(1eq),-78\degree C. (2)\mathrm{CH_3I}(1.05eq),-78\degree C~to~0\degree C. (3)LDA(1eq),-78\degree C. (4)\mathrm{CH_3I}(1.1eq),-78\degree C~to~10\degree C. $生成$\mathbf{C}$，$\mathbf{C}$依次经过：$(1)\mathrm{CH_2CHMgBr},THF. (2)aq. \mathrm{H_2SO_4,(CH_3)_2CO},r.t.$生成$\mathbf{D}$，$\mathbf{D}$在$\mathrm{LiAlH_4,Et_2O}$条件下生成$\mathbf{E}$，$\mathbf{E}$在$\mathrm{CH_3CCCOOH},DCC,DMAP,\mathrm{CH_2Cl_2}$条件下生成$\mathbf{F}$，$\mathbf{F}$在$\mathrm{n-C_{10}H_{22}},reflux$条件下生成最终产物，最终产物为CC1=C2C(=O)O[C@H]3CC[C@](C)(C)C(=CC1)[C@]32C。](/media/uploads/0c155e4f-fa2b-4d87-af4e-0878bc393abb.png)</MultiModal>

已知$\mathbf{A}$的化学式为$\mathrm{C_7H_{10}O_2}$，若物质存在互变异构，则取其中最稳定的一个。则下列选项中说法正确的是：
All other options are incorrect
$\mathbf{A}$ contains two carbonyl groups.
The mass fraction of oxygen in $\mathbf{B}$ is 20.75%.
The configuration of the chiral carbon in $\mathbf{C}$ is R and S.
The mass fraction of carbon in $\mathbf{D}$ is 80.44%.
Two exocyclic double bonds exist in $\mathbf{E}$.
There exist two cycles in $\mathbf{F}$.
其他选项均不正确
$\mathbf{A}$中含有两个羰基。
$\mathbf{B}$中氧的质量分数为20.75%。
$\mathbf{C}$中手性碳的构型为R和S。
$\mathbf{D}$中碳的质量分数为80.44%。
$\mathbf{E}$中存在两个环外双键。
$\mathbf{F}$中存在两个环。
Solving this problem through retrosynthetic analysis, we first observe that $\mathbf{E}$ to $\mathbf{F}$ uses an alkyne, and the final product only has double bonds and contains a six-membered ring with two double bonds. Combining with the conditions of the last step, it can be inferred that a D-A reaction occurred in the last step. Therefore, $\mathbf{F}$ is:

<MultiModal>![CC#CC(O[C@H]1CCC(C)(C(C=C)=C1C)C)=O](/media/uploads/9ca03299-03f5-4ea9-a948-e9adda291853.png)</MultiModal>

<Checkpoint>[$\mathbf{F}$ is CC#CC(O[C@H]1CCC(C)(C(C=C)=C1C)C)=O](1)</Checkpoint>

The reaction conditions from $\mathbf{E}$ to $\mathbf{F}$ are typical substitution reaction conditions. Here, it is speculated that an esterification reaction occurs with the reactant. Therefore, $\mathbf{E}$ is:

<MultiModal>![O[C@H]1CCC(C)(C(C=C)=C1C)C](/media/uploads/3ddcb247-e3b0-4694-8b35-41205e8c6263.png)</MultiModal>

<Checkpoint>[$\mathbf{E}$ is O[C@H]1CCC(C)(C(C=C)=C1C)C](1)</Checkpoint>

$\mathbf{D}$ to $\mathbf{E}$ is under reduction reaction conditions, so it is speculated that this step is the reduction of the carbonyl group. Therefore, $\mathbf{D}$ is:

<MultiModal>![O=C1CCC(C)(C(C=C)=C1C)C](/media/uploads/c9a1f5ef-aa70-4af6-a9f4-6065d1a5bd02.png)</MultiModal>

<Checkpoint>[$\mathbf{D}$ is O=C1CCC(C)(C(C=C)=C1C)C](1)</Checkpoint>

At this point, it is more difficult to perform retrosynthesis. Therefore, we try to deduce the structure of $\mathbf{A}$ by combining the molecular formula. Observing that multiple methyl groups and one vinyl group are added during the process from $\mathbf{A}$ to $\mathbf{D}$, it is speculated that $\mathbf{A}$ is a 1,3-diketone structure. Combining with the position of the methyl group in the molecule, it can be speculated that $\mathbf{A}$ is:

<MultiModal>![CC1=C(O)CCCC1=O](/media/uploads/965629a9-32ec-43ee-8967-1219c39e298d.png)</MultiModal>

<Checkpoint>[$\mathbf{A}$ is CC1=C(O)CCCC1=O](1)</Checkpoint>

$\mathbf{A}$ to $\mathbf{B}$ is the methylation of the hydroxyl group, so $\mathbf{B}$ is:

<MultiModal>![CC1=C(OC)CCCC1=O](/media/uploads/84d1f677-7368-43e4-8f38-f06f8a5b8ed2.png)</MultiModal>

<Checkpoint>[$\mathbf{B}$ is CC1=C(OC)CCCC1=O](1)</Checkpoint>

Two reactions adding methyl groups are performed from $\mathbf{B}$ to $\mathbf{C}$. Combining with the structure of $\mathbf{D}$, it is easy to obtain that $\mathbf{C}$ is:

<MultiModal>![CC1=C(OC)CCC(C)(C)C1=O](/media/uploads/20b065df-2423-4eef-a280-1a9933e83292.png)</MultiModal>

<Checkpoint>[$\mathbf{C}$ is CC1=C(OC)CCC(C)(C)C1=O](1)</Checkpoint>

After obtaining the structure, a simple calculation reveals that the mass fraction of oxygen in $\mathbf{B}$ is 22.83%, and the mass fraction of carbon in $\mathbf{D}$ is 80.44%. The remaining options can be judged as incorrect by direct observation.

Therefore, choose option E.
通过逆合成分析来解决这个问题，首先观察到$\mathbf{E}$到$\mathbf{F}$中使用了一个炔烃，在最终产物中只有双键，并且有含有两个双键的六元环，结合最后一步的条件可以推断最后一步发生了D-A反应，因此得到$\mathbf{F}$为：

<MultiModal>![CC#CC(O[C@H]1CCC(C)(C(C=C)=C1C)C)=O](/media/uploads/9ca03299-03f5-4ea9-a948-e9adda291853.png)</MultiModal>

<Checkpoint>[$\mathbf{F}$为CC#CC(O[C@H]1CCC(C)(C(C=C)=C1C)C)=O](1)</Checkpoint>

$\mathbf{E}$到$\mathbf{F}$的反应条件为典型的取代反应条件，在这里推测是与反应物发生酯化反应，因此$\mathbf{E}$为：

<MultiModal>![O[C@H]1CCC(C)(C(C=C)=C1C)C](/media/uploads/3ddcb247-e3b0-4694-8b35-41205e8c6263.png)</MultiModal>

<Checkpoint>[$\mathbf{E}$为O[C@H]1CCC(C)(C(C=C)=C1C)C](1)</Checkpoint>

$\mathbf{D}$到$\mathbf{E}$为还原反应的条件，因此推测该步为还原羰基，因此$\mathbf{D}$为：

<MultiModal>![O=C1CCC(C)(C(C=C)=C1C)C](/media/uploads/c9a1f5ef-aa70-4af6-a9f4-6065d1a5bd02.png)</MultiModal>

<Checkpoint>[$\mathbf{D}$为O=C1CCC(C)(C(C=C)=C1C)C](1)</Checkpoint>

此时再进行逆推较为困难，因此尝试结合分子式推出$\mathbf{A}$的结构。观察到$\mathbf{A}$到$\mathbf{D}$的过程中添加了多个甲基和一个乙烯基，因此推测$\mathbf{A}$为1，3-二酮结构，结合分子中甲基的位置，可以推测$\mathbf{A}$为：

<MultiModal>![CC1=C(O)CCCC1=O](/media/uploads/965629a9-32ec-43ee-8967-1219c39e298d.png)</MultiModal>

<Checkpoint>[$\mathbf{A}$为CC1=C(O)CCCC1=O](1)</Checkpoint>

$\mathbf{A}$到$\mathbf{B}$为对羟基的甲基化，因此$\mathbf{B}$为：

<MultiModal>![CC1=C(OC)CCCC1=O](/media/uploads/84d1f677-7368-43e4-8f38-f06f8a5b8ed2.png)</MultiModal>

<Checkpoint>[$\mathbf{B}$为CC1=C(OC)CCCC1=O](1)</Checkpoint>

$\mathbf{B}$到$\mathbf{C}$进行了两步增加甲基的反应，结合$\mathbf{D}$的结构，很容易得到$\mathbf{C}$为：

<MultiModal>![CC1=C(OC)CCC(C)(C)C1=O](/media/uploads/20b065df-2423-4eef-a280-1a9933e83292.png)</MultiModal>

<Checkpoint>[$\mathbf{C}$为CC1=C(OC)CCC(C)(C)C1=O](1)</Checkpoint>

得到结构之后，进行简单的计算发现$\mathbf{B}$中氧的质量分数为22.83%，$\mathbf{D}$中碳的质量分数为80.44%。其余选项直接观察便可判断为错误。

因此选择E选项。